Compactness Conditions in the Theory of Nonlinear Differential … · 2019. 8. 7. · Yuxia Guo, China Qian Guo, China Chaitan P. Gupta, USA Uno Hamarik, Estonia¨ Ferenc Hartung,

Guest Editors Joacutezef Banas Mohammad Mursaleen Beata Rzepka and Kishin Sadarangani

Compactness Conditions in the Theory of Nonlinear Differential and Integral Equations

Abstract and Applied Analysis

Compactness Conditions inthe Theory of Nonlinear Differentialand Integral Equations



Guest Editors Jozef Banas Mohammad MursaleenBeata Rzepka and Kishin Sadarangani

Copyright q 2013 Hindawi Publishing Corporation All rights reserved

This is a special issue published in ldquoAbstract and Applied Analysisrdquo All articles are open access articles distributed underthe Creative Commons Attribution License which permits unrestricted use distribution and reproduction in anymediumprovided the original work is properly cited

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Massimo Furi ItalyGiovanni P Galdi USAIsaac Garcia SpainJose A Garcıa-Rodrıguez SpainLeszek Gasinski PolandGyorgy Gat HungaryVladimir Georgiev ItalyLorenzo Giacomelli ItalyJaume Gin SpainValery Y Glizer IsraelLaurent Gosse ItalyJean P Gossez BelgiumDimitris Goussis GreeceJose L Gracia SpainMaurizio Grasselli ItalyYuxia Guo ChinaQian Guo ChinaChaitan P Gupta USAUno Hamarik EstoniaFerenc Hartung HungaryBehnam Hashemi IranNorimichi Hirano JapanJiaxin Hu ChinaChengming Huang ChinaZhongyi Huang ChinaGennaro Infante ItalyIvan G Ivanov BulgariaHossein Jafari IranJaan Janno EstoniaAref Jeribi TunisiaUn C Ji KoreaZhongxiao Jia ChinaLucas Jodar SpainJong S Jung KoreaVarga K Kalantarov TurkeyHenrik Kalisch NorwaySatyanad Kichenassamy FranceTero Kilpelainen FinlandSung G Kim KoreaLjubisa Kocinac SerbiaAndrei Korobeinikov SpainPekka Koskela Finland

Victor Kovtunenko AustriaPavel Kurasov SwedenMiroslaw Lachowicz PolandKunquan Lan CanadaRuediger Landes USAIrena Lasiecka USAMatti Lassas FinlandChun-Kong Law TaiwanMing-Yi Lee TaiwanGongbao Li ChinaPedro M Lima PortugalElena Litsyn IsraelShengqiang Liu ChinaYansheng Liu ChinaCarlos Lizama ChileMilton C Lopes Filho BrazilJulian Lopez-Gomez SpainJinhu Lu ChinaGrzegorz Lukaszewicz PolandShiwang Ma ChinaWanbiao Ma ChinaEberhard Malkowsky TurkeySalvatore A Marano ItalyCristina Marcelli ItalyPaolo Marcellini ItalyJesus Marın-Solano SpainJose M Martell SpainMieczysław S Mastyło PolandMing Mei CanadaTaras Melrsquonyk UkraineAnna Mercaldo ItalyChangxing Miao ChinaStanislaw Migorski PolandMihai Mihailescu RomaniaFeliz Minhos PortugalDumitru Motreanu FranceRoberta Musina ItalyMaria Grazia Naso ItalyGaston M NrsquoGuerekata USASylvia Novo SpainMicah Osilike NigeriaMitsuharu Otani JapanTurgut Ozis TurkeyFilomena Pacella ItalyN S Papageorgiou Greece

Sehie Park KoreaAlberto Parmeggiani ItalyKailash C Patidar South AfricaKevin R Payne ItalyJosip E Pecaric CroatiaShuangjie Peng ChinaSergei V Pereverzyev AustriaMaria Eugenia Perez SpainDavid Perez-Garcia SpainAllan Peterson USAAndrew Pickering SpainCristina Pignotti ItalySomyot Plubtieng ThailandMilan Pokorny Czech RepublicSergio Polidoro ItalyZiemowit Popowicz PolandMaria M Porzio ItalyEnrico Priola ItalyVladimir S Rabinovich MexicoI Rachunkova Czech RepublicMaria A Ragusa ItalySimeon Reich IsraelWeiqing Ren USAAbdelaziz Rhandi ItalyHassan Riahi MalaysiaJuan P Rincon-Zapatero SpainLuigi Rodino ItalyYuriy Rogovchenko NorwayJulio D Rossi ArgentinaWolfgang Ruess GermanyBernhard Ruf ItalyMarco Sabatini ItalySatit Saejung ThailandStefan Samko PortugalMartin Schechter USAJavier Segura SpainSigmund Selberg NorwayValery Serov FinlandNaseer Shahzad Saudi ArabiaAndrey Shishkov UkraineStefan Siegmund GermanyA A Soliman EgyptPierpaolo Soravia ItalyMarco Squassina ItalyS Stanek Czech Republic

Stevo Stevic SerbiaAntonio Suarez SpainWenchang Sun ChinaRobert Szalai UKSanyi Tang ChinaChun-Lei Tang ChinaYoushan Tao ChinaGabriella Tarantello ItalyN Tatar Saudi ArabiaRoger Temam USASusanna Terracini ItalyGerd Teschke GermanyAlberto Tesei ItalyBevan Thompson AustraliaSergey Tikhonov SpainClaudia Timofte RomaniaThanh Tran AustraliaJuan J Trujillo SpainCiprian A Tudor FranceGabriel Turinici FranceMehmet Unal TurkeyS A van Gils The NetherlandsCsaba Varga RomaniaCarlos Vazquez SpainGianmaria Verzini ItalyJesus Vigo-Aguiar SpainYushun Wang ChinaXiaoming Wang USAJing Ping Wang UKShawn X Wang CanadaYouyu Wang ChinaPeixuan Weng ChinaNoemi Wolanski ArgentinaNgai-Ching Wong TaiwanPatricia J Y Wong SingaporeRoderick Wong Hong KongZili Wu ChinaYong Hong Wu AustraliaTiecheng Xia ChinaXu Xian ChinaYanni Xiao ChinaFuding Xie ChinaNaihua Xiu ChinaDaoyi Xu ChinaXiaodong Yan USA

Zhenya Yan ChinaNorio Yoshida JapanBeong I Yun KoreaVjacheslav Yurko RussiaA Zafer TurkeySergey V Zelik UK

Weinian Zhang ChinaChengjian Zhang ChinaMeirong Zhang ChinaZengqin Zhao ChinaSining Zheng ChinaTianshou Zhou China

Yong Zhou ChinaChun-Gang Zhu ChinaQiji J Zhu USAMalisa R Zizovic SerbiaWenming Zou China

Contents

Compactness Conditions in the Theory of Nonlinear Differential and Integral EquationsJozef Banas Mohammad Mursaleen Beata Rzepka and Kishin SadaranganiVolume 2013 Article ID 616481 2 pages

Compactness Conditions in the Study of Functional Differential and Integral EquationsJozef Banas and Kishin SadaranganiVolume 2013 Article ID 819315 14 pages

Existence of Mild Solutions for a Semilinear Integrodifferential Equation with Nonlocal InitialConditions Carlos Lizama and Juan C PozoVolume 2012 Article ID 647103 15 pages

On the Existence of Positive Periodic Solutions for Second-Order Functional DifferentialEquations with Multiple Delays Qiang Li and Yongxiang LiVolume 2012 Article ID 929870 13 pages

Solvability of Nonlinear Integral Equations of Volterra Type Zeqing Liu Sunhong Leeand Shin Min KangVolume 2012 Article ID 932019 17 pages

Existence of Solutions for Fractional Integro-Differential Equation with Multipoint BoundaryValue Problem in Banach Spaces Yulin Zhao Li Huang Xuebin Wang and Xianyang ZhuVolume 2012 Article ID 172963 19 pages

Application of Sumudu Decomposition Method to Solve Nonlinear System of PartialDifferential Equations Hassan Eltayeb and Adem KılıcmanVolume 2012 Article ID 412948 13 pages

Infinite System of Differential Equations in Some BK SpacesM Mursaleen and Abdullah AlotaibiVolume 2012 Article ID 863483 20 pages

Some Existence Results for Impulsive Nonlinear Fractional Differential Equations with ClosedBoundary Conditions Hilmi Ergoren and Adem KilicmanVolume 2012 Article ID 387629 15 pages

Some New Estimates for the Error of Simpson Integration RuleMohammad Masjed-Jamei Marwan A Kutbi and Nawab HussainVolume 2012 Article ID 239695 9 pages

Travelling Wave Solutions of the Schrodinger-Boussinesq SystemAdem Kılıcman and Reza AbazariVolume 2012 Article ID 198398 11 pages

Existence of Solutions for Nonlinear Mixed Type Integrodifferential Functional EvolutionEquations with Nonlocal Conditions Shengli XieVolume 2012 Article ID 913809 11 pages

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 616481 2 pageshttpdxdoiorg1011552013616481

EditorialCompactness Conditions in the Theory ofNonlinear Differential and Integral Equations

Joacutezef BanaV1 Mohammad Mursaleen2 Beata Rzepka1 and Kishin Sadarangani3

1 Department of Mathematics Rzeszow University of Technology Al Powstancow Warszawy 8 35-959 Rzeszow Poland2Department of Mathematics Aligarh Muslim University Aligarh 202 002 India3 Departamento de Matematicas Universidad de Las Palmas de Gran Canaria Campus de Tafira Baja35017 Las Palmas de Gran Canaria Spain

Correspondence should be addressed to Jozef Banas jbanasprzedupl

Received 10 January 2013 Accepted 10 January 2013

Copyright copy 2013 Jozef Banas et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The concept of the compactness appears very frequently inexplicit or implicit form in many branches of mathematicsParticulary it plays a fundamental role in mathematicalanalysis and topology and creates the basis of several inves-tigations conducted in nonlinear analysis and the theories offunctional differential and integral equations In view of thewide applicability of the mentioned branches the concept ofthe compactness is one of the most useful in several topicsof applied mathematics engineering mathematical physicsnumerical analysis and so on

Reasoning based on concepts and tools associated withthe concept of the compactness is very often used in fixedpoint theory and its applications to the theories of functionaldifferential and integral equations of various types More-over that concept is also applied in general operator theory Itis worthwhile mentioning several types of operators definedwith help of the compactness both in strong and in weaktopology

This special issue is devoted to discussing some aspectsof the above presented topics We focus here mainly on thepresentation of papers being the outcome of research andstudy associated with existence results concerning nonlineardifferential integral or functional differential and functionalintegral equations We describe roughly the results containedin the papers covering this special issue

One of those results contained in the paper ofQ Li andYLi is concerned with the existence of positive periodic solu-tions of a second-order functional differential equation with

multiple delay The investigations concerning that equationare located in the space of real functions defined continuousand periodic (with a fixed period) on the real line The maintool used in the proofs is the fixed point index theory in cones

The paper of C Lizama and J C Pozo presents alsoexistence results and is devoted to proving the existence ofmild solutions of a semilinear integrodifferential equation ina Banach space The arguments used in the study are basedon theory of semigroups and the technique of measures ofnoncompactness

The paper by Z Liu et al which we are going to describebriefly discusses existence results for an integral equation ofVolterra type The authors of that paper used some specialmeasure of noncompactness and the fixed point theorem ofDarbo type to obtain a result on the existence of asymptoticstable solutions of the equation in question

A review article concerning infinite systems of differentialequations in the so-called BK spaces presents a brief surveyof existence results concerning those systemsThe authorsMMursaleen and A Alotaibi present results obtained with helpof the technique of measures of noncompactness

In this special issue two research articles concerningthe existence of solutions of nonlinear fractional differentialequations with boundary value problems are included Thepaper authored by Y Zhao et al discusses a result onthe existence of positive solutions of a nonlinear fractionalintegrodifferential equation with multipoint boundary valueproblem The investigations are conducted in an ordered

2 Abstract and Applied Analysis

Banach space The main tools used in argumentations arerelated to a fixed point theorem for operators which arestrictly condensing with respect to the Kuratowski measureof noncompactness The research article of H Ergoren andA Kilicman deals with existence of solutions of impulsivefractional differential equations with closed boundary con-ditions The results of that paper are obtained with help of aBurton-Kirk fixed point theorem for the sum of a contractionand a completely continuous operator

The technique of measures of noncompactness also cre-ates the main tool utilized in the paper of S Xie devoted tothe existence of mild solutions of nonlinear mixed type inte-grodifferential functional evolution equations with nonlocalconditions Apart of the existence the controllability of thementioned solutions is also shown

The issue contains also three papers being only looselyrelated to compactness conditions The paper by H Eltayeband A Kilicman discusses approximate solutions of a non-linear system of partial differential equations with helpof Sumudu decomposition technique and with the use ofAdomian decomposition method The second paper of thementioned kind written by M Masjed-Jamei et al discussessome new estimates for the error of the Simpson integrationrule That study is conducted in the classical Lebesgue spaces1198711[119886 119887] and 119871infin[119886 119887] Finally the paper of A Kilicman and

R Abazari is dedicated to discussing exact solutions of theSchrodinger-Boussinesq system The authors of that paperconstruct solutions of the system in question in the form ofhyperbolic or trigonometric functions

This issue includes also a review paper by J Banas andK Sadarangani presenting an overview of several meth-ods involving compactness conditions Those methods aremainly related to some fixed point theorems (SchauderKrasnoselrsquoskii-Burton and Schaefer) and to the techniqueassociatedwithmeasures of noncompactness Examples illus-trating the applicability of those methods are also included

Acknowledgments

The guest editors of this special issue would like to expresstheir gratitude to the authors who have submitted papers forconsideration We believe that the results presented in thisissue will be a source of inspiration for researchers workingin nonlinear analysis and related areas of mathematics

Jozef BanasMohammad Mursaleen

Beata RzepkaKishin Sadarangani


Review ArticleCompactness Conditions in the Study of Functional Differentialand Integral Equations

Joacutezef BanaV1 and Kishin Sadarangani2

1 Department of Mathematics Rzeszow University of Technology Aleja Powstancow Warszawy 8 35-959 Rzeszow Poland2Departamento de Mathematicas Universidad de Las Palmas de Gran Canaria Campus de Tafira Baja35017 Las Palmas de Gran Canaria Spain


Received 13 December 2012 Accepted 2 January 2013

Academic Editor Beata Rzepka

Copyright copy 2013 J Banas and K Sadarangani This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We discuss some existence results for various types of functional differential and integral equations which can be obtained withthe help of argumentations based on compactness conditions We restrict ourselves to some classical compactness conditionsappearing in fixed point theorems due to Schauder Krasnoselrsquoskii-Burton and Schaefer We present also the technique associatedwithmeasures of noncompactness and we illustrate its applicability in proving the solvability of some functional integral equationsApart from this we discuss the application of the mentioned technique to the theory of ordinary differential equations in Banachspaces

1 Introduction

The concept of the compactness plays a fundamental role inseveral branches of mathematics such as topology mathe-matical analysis functional analysis optimization theory andnonlinear analysis [1ndash5] Numerous mathematical reasoningprocesses depend on the application of the concept of com-pactness or relative compactness Let us indicate only suchfundamental and classical theorems as the Weierstrass the-orem on attaining supremum by a continuous function on acompact set the Fredholm theory of linear integral equationsand its generalization involving compact operators as wellas a lot of fixed point theorems depending on compactnessargumentations [6 7] It is also worthwhile mentioning suchan important property saying that a continuous mappingtransforms a compact set onto compact one

Let us pay a special attention to the fact that severalreasoning processes and constructions applied in nonlinearanalysis depend on the use of the concept of the compactness[6] Since theorems and argumentations of nonlinear anal-ysis are used very frequently in the theories of functionaldifferential and integral equations we focus in this paperon the presentation of some results located in these theories

which can be obtained with the help of various compactnessconditions

We restrict ourselves to present and describe some resultsobtained in the last four decades which are related tosome problems considered in the theories of differentialintegral and functional integral equations Several resultsusing compactness conditions were obtained with the helpof the theory of measures of noncompactness Therefore wedevote one section of the paper to present briefly some basicbackground of that theory

Nevertheless there are also successfully used argumen-tations not depending of the concept of a measure onnoncompactness such as Schauder fixed point principleKrasnoselrsquoskii-Burton fixed point theorem and Schaefer fixedpoint theorem

Let us notice that our presentation is far to be completeThe reader is advised to follow the most expository mono-graphs in which numerous topics connected with compact-ness conditions are broadly discussed [6 8ndash10]

Finally let us mention that the presented paper has areview form It discusses some results described in details inthe papers which will be cited in due course


2 Selected Results of Nonlinear AnalysisInvolving Compactness Conditions

In order to solve an equation having the form

119909 = 119865119909 (1)

where 119865 is an operator being a self-mapping of a Banachspace 119864 we apply frequently an approach through fixedpoint theorems Such an approach is rather natural and ingeneral very efficient Obviously there exists a huge numberof miscellaneous fixed point theorems [6 7 11] dependingboth on order metric and topological argumentations

The most efficient and useful theorems seem to befixed point theorems involving topological argumentationsespecially those based on the concept of compactness Thereader can make an acquaintance with the large theory offixed point-theorems involving compactness conditions inthe above mentionedmonographs but it seems that themostimportant and expository fixed point theorem in this fashionis the famous Schauder fixed point principle [12] Obviouslythat theorem was generalized in several directions but tillnow it is very frequently used in application to the theoriesof differential integral and functional equations

At the beginning of our considerations we recall twowell-known versions of the mentioned Schauder fixed pointprinciple (cf [13]) To this end assume that (119864 || sdot ||) is a givenBanach space

Theorem 1 LetΩ be a nonempty bounded closed and convexsubset of 119864 and let 119865 Ω rarr Ω be a completely continuousoperator (ie 119865 is continuous and the image 119865Ω is relativelycompact) Then 119865 has at least one fixed point in the set Ω (thismeans that the equation 119909 = 119865119909 has at least one solution in theset Ω)

Theorem 2 IfΩ is a nonempty convex and compact subset of119864 and119865 Ω rarr Ω is continuous on the setΩ then the operator119865 has at least one fixed point in the set Ω

Observe thatTheorem 1 can be treated as a particular caseof Theorem 2 if we apply the well-known Mazur theoremasserting that the closed convex hull of a compact subset ofa Banach space 119864 is compact [14] The basic problem arisingin applying the Schauder theorem in the version presentedin Theorem 2 depends on finding a convex and compactsubset of 119864 which is transformed into itself by operator 119865

corresponding to an investigated operator equationIn numerous situations we are able to overcome the

above-indicated difficulty and to obtain an interesting resulton the existence of solutions of the investigated equations(cf [7 15ndash17]) Below we provide an example justifying ouropinion [18]

To do this let us denote by R the real line and put R+=

[0 +infin) Further let Δ = (119905 119904) isin R2 0 le 119904 le 119905Next fix a function 119901 = 119901(119905) defined and continuous on

R+with positive real values Denote by 119862

119901= 119862(R

+ 119901(119905)) the

space consisting of all real functions defined and continuouson R

+and such that

sup |119909 (119905)| 119901 (119905) 119905 ge 0 lt infin (2)

It can be shown that 119862119901forms the Banach space with respect

to the norm

119909 = sup |119909 (119905)| 119901 (119905) 119905 ge 0 (3)

For our further purposes we recall the following criterion forrelative compactness in the space 119862

119901[10 15]

Theorem 3 Let 119883 be a bounded set in the space 119862119901 If all

functions belonging to 119883 are locally equicontinuous on theinterval R

+and if lim

119879rarrinfinsup|119909(119905)|119901(119905) 119905 ge 119879 = 0

uniformly with respect to119883 then119883 is relatively compact in119862119901

In what follows if 119909 is an arbitrarily fixed function fromthe space119862

119901and if 119879 gt 0 is a fixed number we will denote by

120584119879(119909 120576) the modulus of continuity of 119909 on the interval [0 119879]

that is

120584119879(119909 120576) = sup |119909 (119905) minus 119909 (119904)| 119905 119904 isin [0 119879] |119905 minus 119904| le 120576

(4)

Further on we will investigate the solvability of thenonlinear Volterra integral equation with deviated argumenthaving the form

119909 (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 (5)

where 119905 isin R+ Equation (5) will be investigated under the

following formulated assumptions

(i) 119907 Δ times R rarr R is a continuous function and thereexist continuous functions 119899 Δ rarr R

+ 119886 R

+rarr

(0infin) 119887 R+rarr R+such that

|119907 (119905 119904 119909)| le 119899 (119905 119904) + 119886 (119905) 119887 (119904) |119909| (6)

for all (119905 119904) isin Δ and 119909 isin R

In order to formulate other assumptions let us put

119871 (119905) = int

119905

0

119886 (119904) 119887 (119904) 119889119904 119905 ge 0 (7)

Next take an arbitrary number 119872 gt 0 and consider thespace 119862

119901 where 119901(119905) = [119886(119905) exp(119872119871(119905) + 119905)]

minus1 Then we canpresent other assumptions

(ii) The function 119889 R+

rarr R is continuous and thereexists a nonnegative constant 119863 such that |119889(119905)| le

119863119886(119905) exp(119872119871(119905)) for 119905 ge 0

(iii) There exists a constant119873 ge 0 such that

int

119905

0

119899 (119905 119904) 119889119904 le 119873119886 (119905) exp (119872119871 (119905)) (8)

for 119905 ge 0

Abstract and Applied Analysis 3

(iv) 120593 R+rarr R+is a continuous function satisfying the

condition

119871 (120593 (119905)) minus 119871 (119905) le 119870 (9)

for 119905 isin R+ where 119870 ge 0 is a constant

(v) 119863+119873 lt 1 and 119886(120593(119905))119886(119905) le 119872(1minus119863minus119873) exp(minus119872119870)

for all 119905 ge 0

Now we can formulate the announced result

Theorem 4 Under assumptions (i)ndash(v) (5) has at least onesolution 119909 in the space 119862

119901such that |119909(119905)| le 119886(119905) exp(119872119871(119905))

for 119905 isin R+

Wegive the sketch of the proof (cf [18]) First let us definethe transformation 119865 on the space 119862

119901by putting

(119865119909) (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 119905 ge 0 (10)

In view of our assumptions the function (119865119909)(119905) is continu-ous on R

+

Further consider the subset 119866 of the space 119862119901consisting

of all functions 119909 such that |119909(119905)| le 119886(119905) exp(119872119871(119905)) for 119905 isin

R+ Obviously 119866 is nonempty bounded closed and convex

in the space 119862119901 Taking into account our assumptions for an

arbitrary fixed 119909 isin 119866 and 119905 isin R+ we get

|(119865119909) (119905)| le |119889 (119905)| + int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904)))1003816100381610038161003816 119889119904

le 119863119886 (119905) exp (119872119871 (119905))

+ int

119905

119900

[119899 (119905 119904) + 119886 (119905) 119887 (119904)1003816100381610038161003816119909 (120593 (119904))

1003816100381610038161003816] 119889119904

le 119863119886 (119905) exp (119872119871 (119905)) + 119873119886 (119905) exp (119872119871 (119905))

+ 119886 (119905) int

119905

0

119887 (119904) 119886 (120593 (119904)) exp (119872119871 (120593 (119904))) 119889119904

le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) int

119905

0

119872119886 (119904) 119887 (119904)

times exp (119872119871 (119904)) exp (minus119872119870) exp (119872119870) 119889119904

le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) exp (119872119871 (119905))

= 119886 (119905) exp (119872119871 (119905))

(11)

This shows that 119865 transforms the set 119866 into itselfNext we show that 119865 is continuous on the set 119866To this end fix 120576 gt 0 and take 119909 119910 isin 119866 such that ||119909 minus

119910|| le 120576 Next choose arbitrary 119879 gt 0 Using the fact that thefunction 119907(119905 119904 119909) is uniformly continuous on the set [0 119879]2 times

[minus120572(119879) 120572(119879)] where 120572(119879) = max119886(120593(119905)) exp(119872119871(120593(119905)))

119905 isin [0 119879] for 119905 isin [0 119879] we obtain1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)

1003816100381610038161003816

le int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904))) minus 119907 (119905 119904 119910 (120593 (119904)))1003816100381610038161003816 119889119904

le 120573 (120576)

(12)

where 120573(120576) is a continuous function with the propertylim120576rarr0

120573(120576) = 0Now take 119905 ge 119879 Then we get

1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)1003816100381610038161003816 [119886 (119905) exp (119872119871 (119905) + 119905)]

minus1

le |(119865119909) (119905)| +1003816100381610038161003816(119865119910) (119905)

1003816100381610038161003816

times [119886 (119905) exp (119872119871 (119905))]minus1

sdot 119890minus119905

le 2119890minus119905

(13)

Hence for 119879 sufficiently large we have1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)

1003816100381610038161003816 119901 (119905) le 120576 (14)

for 119905 ge 119879 Linking (12) and (14) we deduce that 119865 iscontinuous on the set 119866

The next essential step in our proof which enables us toapply the Schauder fixed point theorem (Theorem 1) is toshow that the set 119865119866 is relatively compact in the space 119862

119901

To this end let us first observe that the inclusion 119865119866 sub 119866 andthe description of 119866 imply the following estimate

|(119865119909) (119905)| 119901 (119905) le 119890minus119905 (15)

This yields that

lim119879rarrinfin

sup |(119865119909) (119905)| 119901 (119905) 119905 ge 119879 = 0 (16)

uniformly with respect to the set 119866On the other hand for fixed 120576 gt 0 119879 gt 0 and for 119905 119904 isin

[0 119879] such that |119905 minus 119904| le 120576 in view of our assumptions for119909 isin 119866 we derive the following estimate

|(119865119909) (119905) minus (119865119909) (119904)|

le |119889 (119905) minus 119889 (119904)|

+

10038161003816100381610038161003816100381610038161003816

int

119905

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119904 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

le 120584119879(119889 120576) + 120576max 119899 (119905 120591) + 119886 (120591) 119887 (120591)

times119901 (120593 (120591)) 0 le 120591 le 119905 le 119879

+ 119879120584119879(119907 (120576 119879 120572 (119879)))

(17)

where we denoted120584119879(119907 (120576 119879 120572 (119879)))

= sup |119907 (119905 119906 119907) minus 119907 (119904 119906 119907)| 119905 119904 isin [0 119879]

|119905 minus 119904| le 120576 119906 isin [0 119879] |119907| le 120572 (119879)

(18)


Taking into account the fact that lim120576rarr0

120584119879(119889 120576) =

lim120576rarr0

120584119879(119907(120576 119879 120572(119879))) = 0 we infer that functions belong-

ing to the set 119865119866 are equicontinuous on each interval [0 119879]Combining this fact with (16) in view of Theorem 3 weconclude that the set 119865119866 is relatively compact ApplyingTheorem 1 we complete the proof

Another very useful fixed point theorem using the com-pactness conditions is the well-known Krasnoselrsquoskii fixedpoint theorem [19]That theoremwas frequently modified byresearchers working in the fixed point theory (cf [6 7 20])but it seems that the version due to Burton [21] is the mostappropriate to be used in applications

Below we formulate that version

Theorem 5 Let 119878 be a nonempty closed convex and boundedsubset of the Banach space119883 and let119860 119883 rarr 119883 and 119861 119878 rarr

119883 be two operators such that

(a) 119860 is a contraction that is there exists a constant 119896 isin

[0 1) such that ||119860119909 minus 119860119910|| le 119896||119909 minus 119910|| for 119909 119910 isin 119883(b) 119861 is completely continuous(c) 119909 = 119860119909 + 119861119910 rArr 119909 isin 119878 for all 119910 isin 119878

Then the equation 119860119909 + 119861119909 = 119909 has a solution in 119878

In order to show the applicability of Theorem 5 we willconsider the following nonlinear functional integral equation[22]

119909 (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

+ int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(19)

where 119905 isin R+ Here we assume that 119891 and 119892 are given

functionsThe above equation will be studied in the space 119861119862(R

+)

consisting of all real functions defined continuous andbounded on the interval R

+and equipped with the usual

supremum norm

119909 = sup |119909 (119905)| 119905 isin R+ (20)

Observe that the space 119861119862(R+) is a special case of the

previously considered space 119862119901with 119901(119905) = 1 for 119905 isin

R+ This fact enables us to adapt the relative compactness

criterion contained inTheorem 3 for our further purposesIn what follows we will impose the following require-

ments concerning the components involved in (19)

(i) The function 119891 R+times R119899 rarr R is continuous and

there exist constants 119896119894isin [0 1) (119894 = 1 2 119899) such

that

1003816100381610038161003816119891 (119905 1199091 1199092 119909

119899) minus 119891 (119905 119910

1 1199102 119910

119899)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816

(21)

for all 119905 isin R+and for all (119909

1 1199092 119909

119899) (1199101 1199102

119910119899) isin R119899

(ii) The function 119905 rarr 119891(119905 0 0) is bounded on R+

with 1198650= sup|119891(119905 0 0)| 119905 isin R

+

(iii) The functions 120572119894 120574119895

R+

rarr R+are continuous

and 120572119894(119905) rarr infin as 119905 rarr infin (119894 = 1 2 119899 119895 =

1 2 119898)(iv) The function 120573 R

+rarr R+is continuous

(v) The function 119892 R2+times R119898 rarr R is continuous and

there exist continuous functions 119902 R2+

rarr R+and

119886 119887 R+rarr R+such that

1003816100381610038161003816119892 (119905 119904 1199091 119909

119898)1003816100381610038161003816 le 119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

10038161003816100381610038161199091198941003816100381610038161003816 (22)

for all 119905 119904 isin R+and (119909

1 1199092 119909

119898) isin R119898 Moreover

we assume that

lim119905rarrinfin

int

120573(119905)

0

119902 (119905 119904) 119889119904 = 0 lim119905rarrinfin

119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 = 0

(23)

Now let us observe that based on assumption (v) weconclude that the functions 119907

1 1199072 R+

rarr R+defined by

the formulas

1199071(119905) = int

120573(119905)

0

119902 (119905 119904) 119889119904 1199072(119905) = 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 (24)

are continuous and bounded on R+ Obviously this implies

that the constants11987211198722defined as

119872119894= sup 119907

119894(119905) 119905 isin R

+ (119894 = 1 2) (25)

are finiteIn order to formulate our last assumption let us denote

119896 = sum119899

119894=1119896119894 where the constants 119896

119894(119894 = 1 2 119899) appear in

assumption (i)

(vi) 119896 + 1198981198722lt 1

Then we have the following result [22] which was announcedabove

Theorem 6 Under assumptions (i)ndash(vi) (19) has at least onesolution in the space 119861119862(R

+) Moreover solutions of (19) are

globally attractive

Remark 7 In order to recall the concept of the global attrac-tivitymentioned in the above theorem (cf [22]) suppose thatΩ is a nonempty subset of the space 119861119862(R

+) and 119876 Ω rarr

119861119862(R+) is an operator Consider the operator equation

119909 (119905) = (119876119909) (119905) 119905 isin R+ (26)

We say that solutions of (26) are globally attractive if forarbitrary solutions 119909 119910 of this equation we have that

lim119905rarrinfin

(119909 (119905) minus 119910 (119905)) = 0 (27)

Let us mention that the above-defined concept was intro-duced in [23 24]


Proof of Theorem 6 We provide only the sketch of the proofConsider the ball 119861

119903in the space119883 = 119861119862(R

+) centered at the

zero function 120579 andwith radius 119903 = (1198650+1198721)[1minus(119896+119898119872

2)]

Next define two mappings 119860 119883 rarr 119883 and 119861 119861119903rarr 119883 by

putting

(119860119909) (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

(119861119909) (119905) = int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(28)

for 119905 isin R+ Then (19) can be written in the form

119909 (119905) = (119860119909) (119905) + (119861119909) (119905) 119905 isin R+ (29)

Notice that in view of assumptions (i)ndash(iii) the mapping119860 is well defined and for arbitrarily fixed function 119909 isin 119883the function 119860119909 is continuous and bounded on R

+ Thus

119860 transforms 119883 into itself Similarly applying assumptions(iii)ndash(v) we deduce that the function 119861119909 is continuous andbounded onR

+Thismeans that119861 transforms the ball119861

119903into

119883Now we check that operators119860 and119861 satisfy assumptions

imposed in Theorem 5 To this end take 119909 119910 isin 119883 Then inview of assumption (i) for a fixed 119905 isin R

+ we get

1003816100381610038161003816(119860119909) (119905) minus (119860119910) (119905)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816

le 1198961003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(30)

This implies that ||119860119909 minus 119860119910|| le 119896||119909 minus 119910|| and in view ofassumption (vi) we infer that 119860 is a contraction on119883

Next we prove that119861 is completely continuous on the ball119861119903 In order to show the indicated property of 119861 we fix 120576 gt 0

and we take 119909 119910 isin 119861119903with ||119909 minus 119910|| le 120576 Then taking into

account our assumptions we obtain

1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816

le int

120573(119905)

0

[1003816100381610038161003816119892 (119905 119904 119909 (120574

1(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816

+1003816100381610038161003816119892 (119905 119904 119910 (120574

1(119904)) 119910 (120574

119898(119904)))

1003816100381610038161003816] 119889119904

le 2 (1199071(119905) + 119903119898119907

2(119905))

(31)

Hence keeping inmind assumption (v) we deduce that thereexists 119879 gt 0 such that 119907

1(119905) + 119903119898119907

2(119905) le 1205762 for 119905 ge 119879

Combining this fact with (31) we get1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)

1003816100381610038161003816 le 120576 (32)

for 119905 ge 119879Further for an arbitrary 119905 isin [0 119879] in the similar way we

obtain

1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816 le int

120573(119905)

0

120596119879

119903(119892 120576) 119889119904 le 120573

119879120596119879

119903(119892 120576)

(33)

where we denoted 120573119879= sup120573(119905) 119905 isin [0 119879] and

120596119879

119903(119892 120576) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 1199092 119909

119898)

minus119892 (119905 119904 1199101 1199102 119910

119898)1003816100381610038161003816 119905 isin [0 119879]

119904 isin [0 120573119879] 119909119894 119910119894isin [minus119903 119903]

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816 le 120576 (119894 = 1 2 119898)

(34)

Keeping in mind the uniform continuity of the function 119892 onthe set [0 119879]times[0 120573

119879]times[minus119903 119903]

119898 we deduce that120596119879119903(119892 120576) rarr 0

as 120576 rarr 0 Hence in view of (32) and (33) we conclude thatthe operator 119861 is continuous on the ball 119861

119903

The boundedness of the operator 119861 is a consequence ofthe inequality

|(119861119909) (119905)| le 1199071(119905) + 119903119898119907

2(119905) (35)

for 119905 isin R+ To verify that the operator 119861 satisfies assumptions

of Theorem 3 adapted to the case of the space 119861119862(R+) fix

arbitrarily 120576 gt 0 In view of assumption (v) we can choose119879 gt 0 such that 119907

1(119905) + 119903119898119907

2(119905) le 120576 for 119905 ge 119879 Further take

an arbitrary function 119909 isin 119861119903 Then keeping in mind (35) for

119905 ge 119879 we infer that

|(119861119909) (119905)| le 120576 (36)

Next take arbitrary numbers 119905 120591 isin [0 119879] with |119905 minus 120591| le 120576Then we obtain the following estimate

|(119861119909) (119905) minus (119861119909) (120591)|

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

minus119892 (120591 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

le

1003816100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

[119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816] 119889119904

1003816100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

120596119879

1(119892 120576 119903) 119889119904

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119902 (119905 119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 119886 (119905)119898119903

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119887 (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 120573119879120596119879

1(119892 120576 119903)

(37)

where we denoted

120596119879

1(119892 120576 119903) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 119909

119898)

minus119892 (120591 119904 1199091 119909

119898)1003816100381610038161003816 119905 120591 isin [0 119879]

|119905 minus 119904| le 120576 119904 isin [0 120573119879]

100381610038161003816100381611990911003816100381610038161003816 le 119903

10038161003816100381610038161199091198981003816100381610038161003816 le 119903

(38)


From estimate (37) we get

|(119861119909) (119905) minus (119861119909) (120591)|

le 119902119879120584119879(120573 120576) + 119903119898119886

119879119887119879120584119879(120573 120576) + 120573

119879120596119879

1(119892 120576 119903)

(39)

where 119902119879

= max119902(119905 119904) 119905 isin [0 119879] 119904 isin [0 120573119879] 119886119879

=

max119886(119905) 119905 isin [0 119879] 119887119879

= max119887(119905) 119905 isin [0 120573119879]

and 120584119879(120573 120576) denotes the usual modulus of continuity of the

function 120573 on the interval [0 119879]Now let us observe that in view of the standard properties

of the functions 120573 = 120573(119905) 119892 = 119892(119905 119904 1199091 119909

119898) we infer

that 120584119879(120573 120576) rarr 0 and 120596119879

1(119892 120576 119903) rarr 0 as 120576 rarr 0 Hence

taking into account the boundedness of the image 119861119861119903and

estimates (36) and (39) in view of Theorem 3 we concludethat the set 119861119861

119903is relatively compact in the space 119861119862(R

+)

that is 119861 is completely continuous on the ball 119861119903

In what follows fix arbitrary 119909 isin 119861119862(R+) and assume

that the equality 119909 = 119860119909 + 119861119910 holds for some 119910 isin 119861119903 Then

utilizing our assumptions for a fixed 119905 isin R+ we get

|119909 (119905)| le |(119860119909) (119905)| +1003816100381610038161003816(119861119910) (119905)

1003816100381610038161003816

le1003816100381610038161003816119891 (119905 119909 (120572

1(119905)) 119909 (120572

119899(119905))) minus 119891 (119905 0 0)

1003816100381610038161003816

+1003816100381610038161003816119891 (119905 0 0)

1003816100381610038161003816

+ int

120573(119905)

0

119902 (119905 119904) 119889119904 + 1198981003817100381710038171003817119910

1003817100381710038171003817 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904

le 119896 119909 + 1198650+1198721+ 119898119903119872

2

(40)

Hence we obtain

119909 le1198650+1198721+ 119898119903119872

2

1 minus 119896 (41)

On the other hand we have that 119903 = (1198650+1198721+ 119898119903119872

2)(1 minus

119896) Thus ||119909|| le 119903 or equivalently 119909 isin 119861119903 This shows that

assumption (c) of Theorem 5 is satisfiedFinally combining all of the above-established facts and

applying Theorem 5 we infer that there exists at least onesolution 119909 = 119909(119905) of (19)

The proof of the global attractivity of solutions of (19) is aconsequence of the estimate

1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le

119899

sum

119894=1

119896119894max 1003816100381610038161003816119909 (120572

119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 2int

120573(119905)

0

119902 (119905 119904) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119910 (120574119894(119904))

1003816100381610038161003816) 119889119904

le 119896max 1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 21199071(119905) + 119898 (119909 +

10038171003817100381710038171199101003817100381710038171003817) 1199072 (119905)

(42)

which is satisfied for arbitrary solutions 119909 = 119909(119905) 119910 = 119910(119905) of(19)

Hence we get

lim sup119905rarrinfin

1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le 119896max1le119894le119899


1003816100381610038161003816119909 (120572119894(119905) minus 119910 (120572

119894(119905)))

1003816100381610038161003816

+ 2lim sup119905rarrinfin

1199071(119905) + 119898 (119909 +

10038171003817100381710038171199101003817100381710038171003817) lim sup119905rarrinfin

1199072(119905)

= 119896lim sup119905rarrinfin

1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

(43)

which implies that lim sup119905rarrinfin

|119909(119905) minus 119910(119905)| = lim119905rarrinfin

|119909(119905) minus

119910(119905)| = 0 This means that the solutions of (19) are globallyattractive (cf Remark 7)

It is worthwhile mentioning that in the literature one canencounter other formulations of the Krasnoselrsquoskii-Burtonfixed point theorem (cf [6 7 20 21]) In some of thoseformulations and generalizations there is used the conceptof a measure of noncompactness (both in strong and in weaksense) and simultaneously the requirement of continuityis replaced by the assumption of weak continuity or weaksequential continuity of operators involved (cf [6 20] forinstance)

In what follows we pay our attention to another fixedpoint theorem which uses the compactness argumentationNamely that theorem was obtained by Schaefer in [25]

Subsequently that theorem was formulated in other waysand we are going here to present two versions of that theorem(cf [11 26])

Theorem8 Let (119864 ||sdot||) be a normed space and let119879 119864 rarr 119864

be a continuous mapping which transforms bounded subsets of119864 onto relatively compact ones Then either

(i) the equation 119909 = 119879119909 has a solution

or

(ii) the set⋃0le120582le1

119909 isin 119864 119909 = 120582119879119909 is unbounded

The below presented version of Schaefer fixed pointtheorem seems to be more convenient in applications

Theorem 9 Let 119864 || sdot || be a Banach space and let 119879 119864 rarr 119864

be a continuous compact mapping (ie 119879 is continuous and119879 maps bounded subsets of 119864 onto relatively compact ones)Moreover one assumes that the set

⋃

0le120582le1

119909 isin 119864 119909 = 120582119879119909 (44)

is bounded Then 119879 has a fixed point


It is easily seen that Theorem 9 is a particular case ofTheorem 8

Observe additionally that Schaefer fixed point theoremseems to be less convenient in applications than Schauderfixed point theorem (cf Theorems 1 and 2) Indeed Schaefertheorem requires a priori bound on utterly unknown solu-tions of the operator equation 119909 = 120582119879119909 for 120582 isin [0 1] On theother hand the proof of Schaefer theorem requires the use ofthe Schauder fixed point principle (cf [27] for details)

It is worthwhile mentioning that an interesting result onthe existence of periodic solutions of an integral equationbased on a generalization of Schaefer fixed point theoremmay be found in [26]

3 Measures of Noncompactness andTheir Applications

Let us observe that in order to apply the fundamental fixedpoint theorem based on compactness conditions that is theSchauder fixed point theorem say the version of Theorem 2we are forced to find a convex and compact subset of a Banachspace 119864 which is transformed into itself by an operator 119865In general it is a hard task to encounter a set of such a type[16 28] On the other hand if we apply Schauder fixed pointtheorem formulated as Theorem 1 we have to prove that anoperator 119865 is completely continuous This causes in generalthat we have to impose rather strong assumptions on termsinvolved in a considered operator equation

In view of the above-mentioned difficulties starting fromseventies of the past century mathematicians working onfixed point theory created the concept of the so-calledmeasure of noncompactness which allowed to overcome theabove-indicated troubles Moreover it turned out that theuse of the technique of measures of noncompactness allowsus also to obtain a certain characterization of solutions ofthe investigated operator equations (functional differentialintegral etc) Such a characterization is possible provided weuse the concept of a measure of noncompactness defined inan appropriate axiomatic way

It is worthwhile noticing that up to now there haveappeared a lot of various axiomatic definitions of the conceptof a measure of noncompactness Some of those definitionsare very general and not convenient in practice Moreprecisely in order to apply such a definition we are oftenforced to impose some additional conditions on a measureof noncompactness involved (cf [8 29])

By these reasons it seems that the axiomatic definition ofthe concept of a measure of noncompactness should be notvery general and should require satisfying such conditionswhich enable the convenience of their use in concreteapplications

Below we present axiomatics which seems to satisfy theabove-indicated requirements That axiomatics was intro-duced by Banas and Goebel in 1980 [10]

In order to recall that axiomatics let us denote by M119864

the family of all nonempty and bounded subsets of a Banachspace 119864 and by N

119864its subfamily consisting of relatively

compact sets Moreover let 119861(119909 119903) stand for the ball with the

center at 119909 and with radius 119903 We write 119861119903to denote the ball

119861(120579 119903) where 120579 is the zero element in 119864 If 119883 is a subset of119864 we write 119883 Conv119883 to denote the closure and the convexclosure of 119883 respectively The standard algebraic operationson sets will be denoted by119883 + 119884 and 120582119883 for 120582 isin R

As we announced above we accept the following defini-tion of the concept of a measure of noncompactness [10]

Definition 10 A function 120583 M119864

rarr R+= [0infin) is said to

be ameasure of noncompactness in the space119864 if the followingconditions are satisfied

(1119900) The family ker 120583 = 119883 isin M

119864 120583(119883) = 0 is nonempty

and ker 120583 sub N

(2119900) 119883 sub 119884 rArr 120583(119883) le 120583(119884)

(3119900) 120583(119883) = 120583(Conv119883) = 120583(119883)

(4119900) 120583(120582119883+(1minus120582)119884) le 120582120583(119883)+(1minus120582)120583(119883) for 120582 isin [0 1]

(5119900) If (119883

119899) is a sequence of closed sets fromM

119864such that

119883119899+1

sub 119883119899for 119899 = 1 2 and if lim

119899rarrinfin120583(119883119899) = 0

then the set119883infin

= ⋂infin

119899=1119883119899is nonempty

Let us pay attention to the fact that from axiom (5119900)

we infer that 120583(119883infin) le 120583(119883

119899) for any 119899 = 1 2 This

implies that 120583(119883infin) = 0 Thus119883

infinbelongs to the family ker 120583

described in axiom (1119900) The family ker 120583 is called the kernel

of the measure of noncompactness 120583The property of the measure of noncompactness 120583 men-

tioned above plays a very important role in applicationsWith the help of the concept of a measure of noncom-

pactness we can formulate the following useful fixed pointtheorem [10] which is called the fixed point theoremofDarbotype

Theorem 11 Let Ω be a nonempty bounded closed andconvex subset of a Banach space 119864 and let 119865 Ω rarr Ω bea continuous operator which is a contraction with respect to ameasure of noncompactness 120583 that is there exists a constant 119896119896 isin [0 1) such that 120583(119865119883) le 119896120583(119883) for any nonempty subset119883 of the setΩ Then the operator 119865 has at least one fixed pointin the set Ω

In the sequel we show an example of the applicabilityof the technique of measures of noncompactness expressedby Theorem 11 in proving the existence of solutions of theoperator equations

Namely we will work on the Banach space 119862[119886 119887]

consisting of real functions defined and continuous on theinterval [119886 119887] and equipped with the standard maximumnorm For sake of simplicity we will assume that [119886 119887] =

[0 1] = 119868 so the space on question can be denoted by 119862(119868)One of the most important and handy measures of

noncompactness in the space 119862(119868) can be defined in the waypresented below [10]

In order to present this definition take an arbitrary set119883 isin M

119862(119868) For 119909 isin 119883 and for a given 120576 gt 0 let us put

120596 (119909 120576) = sup |119909 (119905) minus 119909 (119904)| 119905 119904 isin 119868 |119905 minus 119904| le 120576 (45)


Next let us define

120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883

1205960(119883) = lim

120576rarr0

120596 (119883 120576) (46)

It may be shown that the function 1205960(119883) is the measure of

noncompactness in the space 119862(119868) (cf [10]) This measurehas also some additional properties For example 120596

0(120582119883) =

|120582|1205960(119883) and 120596

0(119883 + 119884) le 120596

0(119883) + 120596

0(119884) provided 119883119884 isin

M119862(119868)

and 120582 isin R [10]In what follows we will consider the nonlinear Volterra

singular integral equation having the form

119909 (119905) = 1198911(119905 119909 (119905) 119909 (119886 (119905))) + (119866119909) (119905) int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(47)

where 119905 isin 119868 = [0 1] 119886 119868 rarr 119868 is a continuous function and119866 119876 are operators acting continuously from the space 119862(119868)

into itself Apart from this we assume that the function 1198912

has the form

1198912(119905 119904) = 119896 (119905 119904) 119892 (119905 119904) (48)

where 119896 Δ rarr R is continuous and 119892 is monotonic withrespect to the first variable and may be discontinuous on thetriangle Δ = (119905 119904) 0 le 119904 le 119905 le 1

Equation (47) will be considered in the space 119862(119868) underthe following assumptions (cf [30])

(i) 119886 119868 rarr 119868 is a continuous function(ii) The function 119891

1 119868 timesR timesR rarr R is continuous and

there exists a nonnegative constant 119901 such that

10038161003816100381610038161198911 (119905 1199091 1199101) minus 1198911(119905 1199092 1199102)1003816100381610038161003816 le 119901max 10038161003816100381610038161199091 minus 119909

2

1003816100381610038161003816 10038161003816100381610038161199101 minus 119910

2

1003816100381610038161003816

(49)

for any 119905 isin 119868 and for all 1199091 1199092 1199101 1199102isin R

(iii) The operator 119866 transforms continuously the space119862(119868) into itself and there exists a nonnegative constant119902 such that 120596

0(119866119883) le 119902120596

0(119883) for any set 119883 isin M

119862(119868)

where 1205960is the measure of noncompactness defined

by (46)(iv) There exists a nondecreasing function 120593 R

+rarr R+

such that ||119866119909|| le 120593(||119909||) for any 119909 isin 119862(119868)(v) The operator119876 acts continuously from the space119862(119868)

into itself and there exists a nondecreasing functionΨ R

+rarr R

+such that ||119876119909|| le Ψ(||119909||) for any

119909 isin 119862(119868)(vi) 119891

2 Δ rarr R has the form (48) where the function

119896 Δ rarr R is continuous(vii) The function 119892(119905 119904) = 119892 Δ rarr R

+occurring in

the decomposition (48) is monotonic with respect to119905 (on the interval [119904 1]) and for any fixed 119905 isin 119868 thefunction 119904 rarr 119892(119905 119904) is Lebesgue integrable over theinterval [0 119905] Moreover for every 120576 gt 0 there exists

120575 gt 0 such that for all 1199051 1199052isin 119868 with 119905

1lt 1199052and

1199052minus 1199051le 120575 the following inequalities are satisfied

10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

le 120576

int

1199052

1199051

119892 (1199052 119904) 119889119904 le 120576

(50)

The main result concerning (47) which we are going topresent now will be preceded by a few remarks and lemmas(cf [30]) In order to present these remarks and lemmas letus consider the function ℎ 119868 rarr R

+defined by the formula

ℎ (119905) = int

119905

0

119892 (119905 119904) 119889119904 (51)

In view of assumption (vii) this function is well defined

Lemma 12 Under assumption (vii) the function ℎ is continu-ous on the interval 119868

For proof we refer to [30]In order to present the last assumptions needed further

on let us define the constants 119896 1198911 ℎ by putting

119896 = sup |119896 (119905 119904)| (119905 119904) isin Δ

1198911= sup 10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816 119905 isin 119868

ℎ = sup ℎ (119905) 119905 isin 119868

(52)

The constants 119896 and 1198911are finite in view of assumptions (vi)

and (ii) while the fact that ℎ lt infin is a consequence ofassumption (vii) and Lemma 12

Now we formulate the announced assumption

(viii) There exists a positive solution 1199030of the inequality

119901119903 + 1198911+ 119896ℎ120593 (119903)Ψ (119903) le 119903 (53)

such that 119901 + 119896ℎ119902Ψ(1199030) lt 1

For further purposes we definite operators correspond-ing to (47) and defined on the space119862(119868) in the followingway

(1198651119909) (119905) = 119891

1(119905 119909 (119905) 119909 (120572 (119905)))

(1198652119909) (119905) = int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(119865119909) (119905) = (1198651119909) (119905) + (119866119909) (119905) (119865

2119909) (119905)

(54)


for 119905 isin 119868 Apart from this we introduce two functions119872 and119873 defined on R

+by the formulas

119872(120576) = sup10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119892)] 119889119904

10038161003816100381610038161003816100381610038161003816

1199051 1199052isin 119868

1199051lt 1199052 1199052minus 1199051le 120576

(55)

119873(120576) = supint1199052

1199051

119892 (1199052 119904) 119889119904 119905

1 1199052isin 119868 1199051lt 1199052 1199052minus 1199051le 120576

(56)

Notice that in view of assumption (vii) we have that119872(120576) rarr

0 and119873(120576) rarr 0 as 120576 rarr 0Now we can state the following result

Lemma 13 Under assumptions (i)ndash(vii) the operator 119865 trans-forms continuously the space 119862(119868) into itself

Proof Fix a function 119909 isin 119862(119868) Then 1198651119909 isin 119862(119868) which is a

consequence of the properties of the so-called superpositionoperator [9] Further for arbitrary functions 119909 119910 isin 119862(119868) invirtue of assumption (ii) for a fixed 119905 isin 119868 we obtain

1003816100381610038161003816(1198651119909) (119905) minus (1198651119910) (119905)

1003816100381610038161003816

le 119901max 1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816 1003816100381610038161003816119909 (119886 (119905)) minus 119910 (119886 (119905))

1003816100381610038161003816

(57)

This estimate in combination with assumption (i) yields

10038171003817100381710038171198651119909 minus 11986511199101003817100381710038171003817 le 119901

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (58)

Hence we conclude that 1198651acts continuously from the space

119862(119868) into itselfNext fix 119909 isin 119862(119868) and 120576 gt 0 Take 119905

1 1199052isin 119868 such that

|1199052minus 1199051| le 120576 Without loss of generality we may assume that

1199051le 1199052 Then based on the imposed assumptions we derive

the following estimate

1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816

int

1199052

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199051 119904) 119892 (119905

2 s) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

1 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le int

1199052

1199051

1003816100381610038161003816119896 (1199052 119904)1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199052 119904) minus 119896 (1199051 119904)

1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199051 119904)1003816100381610038161003816

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 |(119876119909) (119904)| 119889119904

le 119896Ψ (119909)119873 (120576) + 1205961(119896 120576) Ψ (119909) int

1199052

0

119892 (1199052 119904) 119889119904

+ 119896Ψ (119909) int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904

(59)

where we denoted

1205961(119896 120576) = sup 1003816100381610038161003816119896 (1199052 119904) minus 119896 (119905

1 119904)

1003816100381610038161003816

(1199051 119904) (119905

2 119904) isin Δ

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576

(60)

Since the function 119905 rarr 119892(119905 119904) is assumed to be monotonicby assumption (vii) we get

int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904 =

10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

(61)

This fact in conjunction with the above-obtained estimateyields

120596 (1198652119909 120576) le 119896Ψ (119909)119873 (120576) + ℎΨ (119909) 120596

1(119896 120576)

+ 119896Ψ (119909)119872 (120576)

(62)

where the symbol 120596(119910 120576) denotes the modulus of continuityof a function 119910 isin 119862(119868)

Further observe that 1205961(119896 120576) rarr 0 as 120576 rarr 0 which is

an immediate consequence of the uniform continuity of thefunction 119896 on the triangle Δ Combining this fact with theproperties of the functions 119872(120576) and 119873(120576) and taking intoaccount (62) we infer that119865

2119909 isin 119862(119868) Consequently keeping

in mind that 1198651

119862(119868) rarr 119862(119868) and assumption (iii) weconclude that the operator 119865 is a self-mapping of the space119862(119868)

In order to show that 119865 is continuous on 119862(119868) fixarbitrarily 119909

0isin 119862(119868) and 120576 gt 0 Next take 119909 isin 119862(119868) such

that ||119909 minus 1199090|| le 120576 Then for a fixed 119905 isin 119868 we get

1003816100381610038161003816(119865119909) (119905) minus (1198651199090) (119905)

1003816100381610038161003816

le1003816100381610038161003816(1198651119909) (119905) minus (119865

11199090) (119905)

1003816100381610038161003816

+1003816100381610038161003816(119866119909) (119905) (1198652119909) (119905) minus (119866119909

0) (119905) (119865

21199090) (119905)

1003816100381610038161003816

le10038171003817100381710038171198651119909 minus 119865

11199090

1003817100381710038171003817 + 1198661199091003816100381610038161003816(1198652119909) (119905) minus (119865

21199090) (119905)

1003816100381610038161003816

+100381710038171003817100381711986521199090

1003817100381710038171003817

1003817100381710038171003817119866119909 minus 1198661199090

1003817100381710038171003817

(63)


On the other hand we have the following estimate1003816100381610038161003816(1198652119909) (119905) minus (119865

21199090) (119905)

1003816100381610038161003816

le int

119905

0

|119896 (119905 119904)| 119892 (119905 119904)1003816100381610038161003816(119876119909) (119904) minus (119876119909

0) (119904)

1003816100381610038161003816 119889119904

le 119896 (int

119905

0

119892 (119905 119904) 119889119904)1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

le 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

(64)

Similarly we obtain

1003816100381610038161003816(11986521199090) (119905)1003816100381610038161003816 le 119896 (int

119905

0

119892 (119905 119904) 119889119904)10038171003817100381710038171198761199090

1003817100381710038171003817 le 119896ℎ10038171003817100381710038171198761199090

1003817100381710038171003817 (65)

and consequently

1003817100381710038171003817119865211990901003817100381710038171003817 le 119896ℎ

10038171003817100381710038171198761199090

1003817100381710038171003817 (66)

Next linking (63)ndash(66) and (58) we obtain the followingestimate

1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 1199011003817100381710038171003817119909 minus 119909

0

1003817100381710038171003817 + 119866119909 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎ1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

10038171003817100381710038171198761199090

1003817100381710038171003817

(67)

Further taking into account assumptions (iv) and (v) wederive the following inequality

1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 119901120576 + 120593 (10038171003817100381710038171199090

1003817100381710038171003817 + 120576) 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎΨ (10038171003817100381710038171199090

1003817100381710038171003817)1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

(68)

Finally in view of the continuity of the operators 119866 and 119876

(cf assumptions (iii) and (v)) we deduce that operator 119865 iscontinuous on the space 119862(119868) The proof is complete

Now we can formulate the last result concerning (47) (cf[30])

Theorem 14 Under assumptions (i)ndash(viii) (47) has at leastone solution in the space 119862(119868)

Proof Fix 119909 isin 119862(119868) and 119905 isin 119868 Then evaluating similarly as inthe proof of Lemma 13 we get

|(119865119909) (119905)| le10038161003816100381610038161198911 (119905 119909 (119905) 119909 (119886 (119905))) minus 119891

1(119905 0 0)

1003816100381610038161003816

+10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816

+ |119866119909 (119905)|

10038161003816100381610038161003816100381610038161003816

int

119905

0

119896 (119905 119904) 119892 (119905 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909)

(69)

Hence we obtain

119865119909 le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909) (70)

From the above inequality and assumption (viii) we infer thatthere exists a number 119903

0gt 0 such that the operator 119865 maps

the ball 1198611199030

into itself and 119901 + 119896ℎ119902Ψ(1199030) lt 1 Moreover by

Lemma 13 we have that 119865 is continuous on the ball 1198611199030

Further on take a nonempty subset119883 of the ball119861

1199030

and anumber 120576 gt 0Then for an arbitrary 119909 isin 119883 and 119905

1 1199052isin 119868with

|1199052minus1199051| le 120576 in view of (66) and the imposed assumptions we

obtain1003816100381610038161003816(119865119909) (1199052) minus (119865119909) (119905

1)1003816100381610038161003816

le10038161003816100381610038161198911 (1199052 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199052) 119909 (119886 (119905

2)))

1003816100381610038161003816

+10038161003816100381610038161198911 (1199051 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199051) 119909 (119886 (119905

1)))

1003816100381610038161003816

+1003816100381610038161003816(1198652119909) (1199052)

1003816100381610038161003816

1003816100381610038161003816(119866119909) (1199052) minus (119866119909) (1199051)1003816100381610038161003816

+1003816100381610038161003816(119866119909) (1199051)

1003816100381610038161003816

1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le 1205961199030

(1198911 120576) + 119901max 1003816100381610038161003816119909 (119905

2) minus 119909 (119905

1)1003816100381610038161003816

1003816100381610038161003816119909 (119886 (1199052)) minus 119909 (119886 (119905

1))1003816100381610038161003816

+ 119896ℎΨ (1199030) 120596 (119866119909 120576) + 120593 (119903

0) 120596 (119865

2119909 120576)

(71)

where we denoted

1205961199030

(1198911 120576) = sup 10038161003816100381610038161198911 (1199052 119909 119910) minus 119891

1(1199051 119909 119910)

1003816100381610038161003816 1199051 1199052 isin 119868

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576 119909 119910 isin [minus1199030 1199030]

(72)

Hence in virtue of (62) we deduce the estimate

120596 (119865119909 120576) le 1205961199030

(1198911 120576) + 119901max 120596 (119909 120576) 120596 (119909 120596 (119886 120576))

+ 119896ℎΨ (1199030) 120596 (119866119909 120576)

+ 120593 (1199030) Ψ (1199030) [119896119873 (120576) + ℎ120596

1(119896 120576) + 119896119872 (120576)]

(73)

Finally taking into account the uniform continuity of thefunction 119891

1on the set 119868 times [minus119903

0 1199030]2 and the properties of the

functions 119896(119905 119904) 119886(119905) 119872(120576) and 119873(120576) and keeping in mind(46) we obtain

1205960(119865119883) le 119901120596

0(119883) + 119896ℎΨ (119903

0) 1205960(119866119883) (74)

Linking this estimate with assumption (iii) we get

1205960(119865119883) le (119901 + 119896ℎ119902Ψ (119903

0)) 1205960(119883) (75)

The use of Theorem 11 completes the proof

4 Existence Results Concerning the Theory ofDifferential Equations in Banach Spaces

In this section we are going to present some classical resultsconcerning the theory of ordinary differential equations inBanach space We focus on this part of that theory in a which


the technique associated with measures of noncompactnessis used as the main tool in proving results on the existence ofsolutions of the initial value problems for ordinary differentialequationsOur presentation is basedmainly on the papers [3132] and the monograph [33]

The theory of ordinary differential equations in Banachspaces was initiated by the famous example of Dieudonne[34] who showed that in an infinite-dimensional Banachspace the classical Peano existence theorem is no longer trueMore precisely Dieudonne showed that if we consider theordinary differential equation

1199091015840= 119891 (119905 119909) (76)

with the initial condition

119909 (0) = 1199090 (77)

where 119891 [0 119879] times 119861(1199090 119903) rarr 119864 and 119864 is an infinite-dimen-

sional Banach space then the continuity of 119891 (and evenuniform continuity) does not guarantee the existence ofsolutions of problem (76)-(77)

In light of the example of Dieudonne it is clear that inorder to ensure the existence of solutions of (76)-(77) it isnecessary to add some extra conditions The first results inthis direction were obtained by Kisynski [35] Olech [36] andWa zewski [37] in the years 1959-1960 In order to formulatethose results we need to introduce the concept of the so-called Kamke comparison function (cf [38 39])

To this end assume that 119879 is a fixed number and denote119869 = [0 119879] 119869

0= (0 119879] Further assume that Ω is a nonempty

open subset of R119899 and 1199090is a fixed element of Ω Let 119891 119869 times

Ω rarr R119899 be a given function

Definition 15 A function 119908 119869 times R+

rarr R+(or 119908 119869

0times

R+

rarr R+) is called a Kamke comparison function provided

the inequality1003817100381710038171003817119891 (119905 119909) minus 119891 (119905 119910)

1003817100381710038171003817 le 119908 (1199051003817100381710038171003817119909 minus 119910

1003817100381710038171003817) (78)

for 119909 119910 isin Ω and 119905 isin 119869 (or 119905 isin 1198690) together with some

additional assumptions concerning the function 119908 guaran-tees that problem (76)-(77) has at most one local solution

In the literature one can encounter miscellaneous classesof Kamke comparison functions (cf [33 39]) We will notdescribe those classes but let us only mention that they aremostly associated with the differential equation 119906

1015840= 119908(119905 119906)

with initial condition 119906(0) = 0 or the integral inequality119906(119905) le int

119905

0119908(119904 119906(119904))119889119904 for 119905 isin 119869

0with initial condition

lim119905rarr0

119906(119905)119905 = lim119905rarr0

119906(119905) = 0 It is also worthwhilerecalling that the classical Lipschitz or Nagumo conditionsmay serve as Kamke comparison functions [39]

The above-mentioned results due to Kisynski et al [35ndash37] assert that if 119891 119869 times 119861(119909

0 119903) rarr 119864 is a continuous

function satisfying condition (78) with an appropriate Kamkecomparison function then problem (76)-(77) has exactly onelocal solution

Observe that the natural translation of inequality (78) interms of measures of noncompactness has the form

120583 (119891 (119905 119883)) le 119908 (119905 120583 (119883)) (79)

where 119883 denotes an arbitrary nonempty subset of the ball119861(1199090 119903) The first result with the use of condition (79) for

119908(119905 119906) = 119862119906 (119862 is a constant) was obtained by Ambrosetti[40] After the result of Ambrosetti there have appeared alot of papers containing existence results concerning problem(76)-(77) (cf [41ndash44]) with the use of condition (79) andinvolving various types of Kamke comparison functions Itturned out that generalizations of existence results concern-ing problem (76)-(77) with the use of more and more generalKamke comparison functions are in fact only apparent gen-eralizations [45] since the so-called Bompiani comparisonfunction is sufficient to give the most general result in thementioned direction

On the other hand we can generalize existence resultsinvolving a condition like (79) taking general measures ofnoncompactness [31 32] Below we present a result comingfrom [32] which seems to be the most general with respect totaking the most general measure of noncompactness

In the beginning let us assume that 120583 is a measure ofnoncompactness defined on a Banach space 119864 Denote by 119864

120583

the set defined by the equality

119864120583= 119909 isin 119864 119909 isin ker 120583 (80)

The set 119864120583will be called the kernel set of a measure 120583 Taking

into account Definition 10 and some properties of a measureof noncompactness (cf [10]) it is easily seen that119864

120583is a closed

and convex subset of the space 119864In the case when we consider the so-called sublinear

measure of noncompactness [10] that is a measure of non-compactness 120583 which additionally satisfies the following twoconditions

(6119900) 120583(119883 + 119884) le 120583(119883) + 120583(119884)

(7119900) 120583(120582119883) = |120582|120583(119883) for 120582 isin R

then the kernel set 119864120583forms a closed linear subspace of the

space 119864Further on assume that 120583 is an arbitrary measure of

noncompactness in the Banach space 119864 Let 119903 gt 0 be afixed number and let us fix 119909

0isin 119864120583 Next assume that

119891 119869 times 119861(1199090 119903) rarr 119864 (where 119869 = [0 119879]) is a given uniformly

continuous and bounded function say ||119891(119905 119909)|| le 119860Moreover assume that 119891 satisfies the following compari-

son condition of Kamke type

120583 (1199090+ 119891 (119905 119883)) le 119908 (119905 120583 (119883)) (81)

for any nonempty subset119883 of the ball 119861(1199090 119903) and for almost

all 119905 isin 119869Here we will assume that the function 119908(119905 119906) = 119908

1198690times R+

rarr R+(1198690= (0 119879]) is continuous with respect to

119906 for any 119905 and measurable with respect to 119905 for each 119906 Apartfrom this 119908(119905 0) = 0 and the unique solution of the integralinequality

119906 (119905) le int

119905

0

119908 (119904 119906 (119904)) 119889119904 (119905 isin 1198690) (82)

such that lim119905rarr0

119906(119905)119905 = lim119905rarr0

119906(119905) = 0 is 119906 equiv 0The following formulated result comes from [32]


Theorem 16 Under the above assumptions if additionallysup119905+119886(119905) 119905 isin 119869 le 1 where 119886(119905) = sup||119891(0 119909

0)minus119891(119904 119909)||

119904 le 119905 ||119909 minus 1199090|| le 119860119904 and 119860 is a positive constant such that

119860119879 le 119903 the initial value problem (76)-(77) has at least onelocal solution 119909 = 119909(119905) such that 119909(119905) isin 119864

120583for 119905 isin 119869

The proof of the above theorem is very involved and istherefore omitted (cf [32 33]) We restrict ourselves to give afew remarks

At first let us notice that in the case when 120583 is a sublinearmeasure of noncompactness condition (81) is reduced to theclassical one expressed by (79) provided we assume that 119909

0isin

119864120583 In such a case Theorem 16 was proved in [10]The most frequently used type of a comparison function

119908 is this having the form 119908(119905 119906) = 119901(119905)119906 where 119901(119905) isassumed to be Lebesgue integrable over the interval 119869 In sucha case comparison condition (81) has the form

120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (83)

An example illustratingTheorem 16 under condition (83) willbe given later

Further observe that in the case when 120583 is a sublinearmeasure of noncompactness such that 119909

0isin 119864120583 condition

(83) can be written in the form

120583 (119891 (119905 119883)) le 119901 (119905) 120583 (119883) (84)

Condition (84) under additional assumption ||119891(119905 119909)|| le 119875+

119876||119909|| with some nonnegative constants 119875 and 119876 is usedfrequently in considerations associated with infinite systemsof ordinary differential equations [46ndash48]

Now we present the above-announced example comingfrom [33]

Example 17 Consider the infinite system of differential equa-tions having the form

1199091015840

119899= 119886119899(119905) 119909119899+ 119891119899(119909119899 119909119899+1

) (85)

where 119899 = 1 2 and 119905 isin 119869 = [0 119879] System (85) will beconsidered together with the system of initial conditions

119909119899(0) = 119909

119899

0(86)

for 119899 = 1 2 We will assume that there exists the limitlim119899rarrinfin

119909119899

0= 119886(119886 isin R)

Problem (85)-(86) will be considered under the followingconditions

(i) 119886119899 119869 rarr R (119899 = 1 2 ) are continuous functions

such that the sequence (119886119899(119905)) converges uniformly

on the interval 119869 to the function which vanishesidentically on 119869

(ii) There exists a sequence of real nonnegative numbers(120572119899) such that lim

119899rarrinfin120572119899= 0 and |119891

119899(119909119899 119909119899+1

)| le

120572119899for 119899 = 1 2 and for all 119909 = (119909

1 1199092 1199093 ) isin 119897

infin(iii) The function 119891 = (119891

1 1198912 1198913 ) transforms the space

119897infin into itself and is uniformly continuous

Let usmention that the symbol 119897infin used above denotes theclassical Banach sequence space consisting of all real boundedsequences (119909

119899) with the supremum norm that is ||(119909

119899)|| =

sup|119909119899| 119899 = 1 2

Under the above hypotheses the initial value problem(85)-(86) has at least one solution 119909(119905) = (119909

1(119905) 1199092(119905) )

such that 119909(119905) isin 119897infin for 119905 isin 119869 and lim

119899rarrinfin119909119899(119905) = 119886 uniformly

with respect to 119905 isin 119869 provided 119879 le 1(119869 = [0 119879])As a proof let us take into account the measure of

noncompactness in the space 119897infin defined in the followingway

120583 (119883) = lim sup119899rarrinfin

sup119909isin119883

1003816100381610038161003816119909119899 minus 1198861003816100381610038161003816 (87)

for 119883 isin M119897infin (cf [10]) The kernel ker120583 of this measure is

the family of all bounded subsets of the space 119897infin consistingof sequences converging to the limit equal to 119886 with the samerate

Further take an arbitrary set119883 isin M119897infin Then we have

120583 (1199090+ 119891 (119905 119883))

= lim sup119899rarrinfin

sup119909isin119883

1003816100381610038161003816119909119899

0+ 119886119899(119905) 119909119899

+119891119899(119909119899 119909119899+1

) minus 1198861003816100381610038161003816

le lim sup119899rarrinfin

sup119909isin119883

[1003816100381610038161003816119886119899 (119905)

1003816100381610038161003816

10038161003816100381610038161199091198991003816100381610038161003816

+1003816100381610038161003816119891119899 (119909119899 119909119899+1 ) + 119909

119899

0minus 119886

1003816100381610038161003816]


sup119909isin119883

119901 (119905)1003816100381610038161003816119909119899 minus 119886

1003816100381610038161003816

+ lim sup119899rarrinfin

sup119909isin119883

1003816100381610038161003816119886119899 (119905)1003816100381610038161003816 |119886|


sup119909isin119883

1003816100381610038161003816119891119899 (119909119899 119909119899+1 )1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119909119899

0minus 119886

1003816100381610038161003816

(88)

where we denoted 119901(119905) = sup|119886119899(119905)| 119899 = 1 2 for 119905 isin 119869

Hence we get

120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (89)

which means that the condition (84) is satisfiedCombining this fact with assumption (iii) and taking into

account Theorem 16 we complete the proof

References

[1] Q H Ansari Ed Topics in Nonlinear Analysis and Optimiza-tion World Education Delhi India 2012


[2] N Bourbaki Elements of Mathematics General TopologySpringer Berlin Germany 1989

[3] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[4] W Rudin Real and Complex Analysis McGraw-Hill New YorkNY USA 1987

[5] W Rudin Functional Analysis McGraw-Hill New York NYUSA 1991

[6] R P Agarwal M Meehan and D OrsquoRegan Fixed PointTheory and Applications vol 141 Cambridge University PressCambridge UK 2001

[7] A Granas and J Dugundji Fixed Point Theory Springer NewYork NY USA 2003

[8] R R Akhmerov M I Kamenskiı A S Potapov A E Rod-kina and B N Sadovskiı Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhauser Basel Switzerland 1992

[9] J Appell and P P Zabrejko Nonlinear Superposition Operatorsvol 95 Cambridge University Press Cambridge UK 1990

[10] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces vol 60 ofLectureNotes in Pure andAppliedMathematicsMarcel Dekker New York NY USA 1980

[11] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[12] J Schauder ldquoZur Theorie stetiger Abbildungen in Funktion-alraumenrdquo Mathematische Zeitschrift vol 26 no 1 pp 47ndash651927

[13] J Banas ldquoMeasures of noncompactness in the study of solutionsof nonlinear differential and integral equationsrdquo Central Euro-pean Journal of Mathematics vol 10 no 6 pp 2003ndash2011 2012

[14] S Mazur ldquoUber die kleinste konvexe Menge die eine gegebenMenge enhaltrdquo Studia Mathematica vol 2 pp 7ndash9 1930

[15] C Corduneanu Integral Equations and Applications Cam-bridge University Press Cambridge UK 1991

[16] J Kisynski ldquoSur lrsquoexistence et lrsquounicite des solutions desproblemes classiques relatifs a lrsquoequation 119904 = 119865(119909 119910 119911 119901 119902)rdquovol 11 pp 73ndash112 1957

[17] P P Zabrejko A I KoshelevMA Krasnoselrsquoskii S GMikhlinL S Rakovschik and V J Stetsenko Integral Equations Nord-hoff Leyden Mass USA 1975

[18] J Banas ldquoAn existence theorem for nonlinear Volterra inte-gral equation with deviating argumentrdquo Rendiconti del CircoloMatematico di Palermo II vol 35 no 1 pp 82ndash89 1986

[19] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Pergamon New York NY USA1964

[20] J Garcia-Falset K Latrach E Moreno-Galvez and M-ATaoudi ldquoSchaefer-Krasnoselskii fixed point theorems using ausual measure of weak noncompactnessrdquo Journal of DifferentialEquations vol 252 no 5 pp 3436ndash3452 2012

[21] T A Burton ldquoA fixed-point theorem of Krasnoselskiirdquo AppliedMathematics Letters vol 11 no 1 pp 85ndash88 1998

[22] J Banas K Balachandran and D Julie ldquoExistence and globalattractivity of solutions of a nonlinear functional integralequationrdquo Applied Mathematics and Computation vol 216 no1 pp 261ndash268 2010

[23] J Banas and B Rzepka ldquoAn application of a measure ofnoncompactness in the study of asymptotic stabilityrdquo AppliedMathematics Letters vol 16 no 1 pp 1ndash6 2003

[24] X Hu and J Yan ldquoThe global attractivity and asymptoticstability of solution of a nonlinear integral equationrdquo Journal ofMathematical Analysis and Applications vol 321 no 1 pp 147ndash156 2006

[25] H Schaefer ldquoUber die Methode der a priori-SchrankenrdquoMathematische Annalen vol 129 pp 415ndash416 1955

[26] T A Burton and C Kirk ldquoA fixed point theorem ofKrasnoselskii-Schaefer typerdquo Mathematische Nachrichten vol189 pp 23ndash31 1998

[27] P Kumlin A Note on Fixed Point Theory MathematicsChalmers and GU 20032004

[28] J Banas and I J Cabrera ldquoOn existence and asymptoticbehaviour of solutions of a functional integral equationrdquo Non-linear Analysis Theory Methods amp Applications A vol 66 no10 pp 2246ndash2254 2007

[29] J Danes ldquoOn densifying and related mappings and their appli-cation in nonlinear functional analysisrdquo inTheory of NonlinearOperators pp 15ndash56 Akademie Berlin Germany 1974

[30] A Aghajani J Banas and Y Jalilian ldquoExistence of solutionsfor a class of nonlinear Volterra singular integral equationsrdquoComputers amp Mathematics with Applications vol 62 no 3 pp1215ndash1227 2011

[31] J Banas A Hajnosz and S Wedrychowicz ldquoOn the equation1199091015840

= 119891(119905 119909) in Banach spacesrdquo Commentationes MathematicaeUniversitatis Carolinae vol 23 no 2 pp 233ndash247 1982

[32] J Banas A Hajnosz and S Wedrychowicz ldquoOn existence andlocal characterization of solutions of ordinary differential equa-tions in Banach spacesrdquo in Proceedings of the 2nd Conferenceon Differential Equations and Applications pp 55ndash58 RousseBulgaria 1982

[33] J Banas ldquoApplications of measures of noncompactness tovarious problemsrdquo Zeszyty Naukowe Politechniki Rzeszowskiejno 5 p 115 1987

[34] J Dieudonne ldquoDeux examples singuliers drsquoequations differen-tiellesrdquo Acta Scientiarum Mathematicarum vol 12 pp 38ndash401950

[35] J Kisynski ldquoSur les equations differentielles dans les espaces deBanachrdquo Bulletin de lrsquoAcademie Polonaise des Sciences vol 7 pp381ndash385 1959

[36] C Olech ldquoOn the existence and uniqueness of solutions ofan ordinary differential equation in the case of Banach spacerdquoBulletin de lrsquoAcademie Polonaise des Sciences vol 8 pp 667ndash6731960

[37] T Wazewski ldquoSur lrsquoexistence et lrsquounicite des integrales desequations differentielles ordinaries au cas de lrsquoespace deBanachrdquoBulletin de lrsquoAcademie Polonaise des Sciences vol 8 pp 301ndash3051960

[38] K Deimling Ordinary Differential Equations in Banach Spacesvol 596 of Lecture Notes in Mathematics Springer BerlinGermany 1977

[39] WWalterDifferential and Integral Inequalities Springer BerlinGermany 1970

[40] A Ambrosetti ldquoUn teorema di esistenza per le equazionidifferenziali negli spazi di Banachrdquo Rendiconti del SeminarioMatematico dellaUniversita di Padova vol 39 pp 349ndash361 1967

[41] K Deimling ldquoOn existence and uniqueness for Cauchyrsquos prob-lem in infinite dimensional Banach spacesrdquo Proceedings of theColloquium on Mathematics vol 15 pp 131ndash142 1975

[42] K Goebel and W Rzymowski ldquoAn existence theorem for theequation 119909

1015840

= 119891(119905 119909) in Banach spacerdquo Bulletin de lrsquoAcademiePolonaise des Sciences vol 18 pp 367ndash370 1970


[43] BN Sadovskii ldquoDifferential equationswith uniformly continu-ous right hand siderdquo Trudy Nauchno-Issledovatelrsquonogo InstitutaMatematiki Voronezhskogo Gosudarstvennogo Universiteta vol1 pp 128ndash136 1970

[44] S Szufla ldquoMeasure of non-compactness and ordinary dif-ferential equations in Banach spacesrdquo Bulletin de lrsquoAcademiePolonaise des Sciences vol 19 pp 831ndash835 1971

[45] J Banas ldquoOn existence theorems for differential equations inBanach spacesrdquo Bulletin of the Australian Mathematical Societyvol 32 no 1 pp 73ndash82 1985

[46] J Banas and M Lecko ldquoSolvability of infinite systems ofdifferential equations in Banach sequence spacesrdquo Journal ofComputational andAppliedMathematics vol 137 no 2 pp 363ndash375 2001

[47] M Mursaleen and S A Mohiuddine ldquoApplications of mea-sures of noncompactness to the infinite system of differentialequations in ℓ

119901spacesrdquo Nonlinear Analysis Theory Methods amp

Applications A vol 75 no 4 pp 2111ndash2115 2012[48] M Mursaleen and A Alotaibi ldquoInfinite systems of differential

equations in some BK spacesrdquo Abstract and Applied Analysisvol 2012 Article ID 863483 20 pages 2012

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2012 Article ID 647103 15 pagesdoi1011552012647103

Research ArticleExistence of Mild Solutions fora Semilinear Integrodifferential Equation withNonlocal Initial Conditions

Carlos Lizama1 and Juan C Pozo2

1 Departamento de Matematica y Ciencia de la Computacion Universidad de Santiago de ChileCasilla 307 Correo 2 Santiago 9160000 Chile

2 Facultad de Ciencias Universidad de Chile Las Palmeras 3425 Santiago 7810000 Chile

Correspondence should be addressed to Carlos Lizama carloslizamausachcl

Received 26 October 2012 Accepted 14 December 2012

Academic Editor Jozef Banas

Copyright q 2012 C Lizama and J C Pozo This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

Using Hausdorff measure of noncompactness and a fixed-point argument we prove the existenceof mild solutions for the semilinear integrodifferential equation subject to nonlocal initial condi-tions uprime(t) = Au(t) +

int t0 B(tminus s)u(s)ds+ f(t u(t)) t isin [0 1] u(0) = g(u) whereA D(A) sube X rarr X

and for every t isin [0 1] the maps B(t) D(B(t)) sube X rarr X are linear closed operators defined ina Banach space X We assume further that D(A) sube D(B(t)) for every t isin [0 1] and the functionsf [0 1] times X rarr X and g C([0 1]X) rarr X are X-valued functions which satisfy appropriateconditions

1 Introduction

The concept of nonlocal initial condition has been introduced to extend the study of classicalinitial value problems This notion is more precise for describing nature phenomena than theclassical notion because additional information is taken into account For the importance ofnonlocal conditions in different fields the reader is referred to [1ndash3] and the references citedtherein

The earliest works related with problems submitted to nonlocal initial conditions weremade by Byszewski [4ndash7] In these works using methods of semigroup theory and theBanach fixed point theorem the author has proved the existence of mild and strong solutionsfor the first order Cauchy problem

uprime(t) = Au(t) + f(t u(t)) t isin [0 1]

u(0) = g(u)(11)


where A is an operator defined in a Banach space X which generates a semigroup T(t)t0and the maps f and g are suitable X-valued functions

Henceforth (11) has been extensively studied by many authors We just mention afew of these works Byszewski and Lakshmikantham [8] have studied the existence anduniqueness of mild solutions whenever f and g satisfy Lipschitz-type conditions Ntouyasand Tsamatos [9 10] have studied this problem under conditions of compactness for the semi-group generated by A and the function g Recently Zhu et al [11] have investigated thisproblemwithout conditions of compactness on the semigroup generated byA or the functionf

On the other hand the study of abstract integrodifferential equations has been anactive topic of research in recent years because it has many applications in different areasIn consequence there exists an extensive literature about integrodifferential equations withnonlocal initial conditions (cf eg [12ndash25]) Our work is a contribution to this theoryIndeed this paper is devoted to study the existence of mild solutions for the followingsemilinear integrodifferential evolution equation

uprime(t) = Au(t) +int t

0B(t minus s)u(s)ds + f(t u(t)) t isin [0 1]

u(0) = g(u)

(12)

where A D(A) sube X rarr X and for every t isin [0 1] the mappings B(t) D(B(t)) sube X rarr X arelinear closed operators defined in a Banach space X We assume further thatD(A) sube D(B(t))for every t isin [0 1] and the functions f [0 1] times X rarr X and g C([0 1]X) rarr X are X-valued functions that satisfy appropriate conditions which we will describe later In order toabbreviate the text of this paper henceforth wewill denote by I the interval [0 1] andC(IX)is the space of all continuous functions from I to X endowed with the uniform convergencenorm

The classical initial value version of (12) that is u(0) = u0 for some u0 isin X hasbeen extensively studied by many researchers because it has many important applications indifferent fields of natural sciences such as thermodynamics electrodynamics heat conduc-tion in materials with memory continuummechanics and population biology among othersFor more information see [26ndash28] For this reason the study of existence and other propertiesof the solutions for (12) is a very important problem However to the best of our knowledgethe existence of mild solutions for the nonlocal initial value problem (12) has not beenaddressed in the existing literature Most of the authors obtain the existence of solutions andwell-posedness for (12) by establishing the existence of a resolvent operator R(t)tisinI and avariation of parameters formula (see [29 30]) Using adaptation of the methods describedin [11] we are able to prove the existence of mild solutions of (12) under conditions ofcompactness of the function g and continuity of the function t rarr R(t) for t gt 0 Furthermorein the particular case B(t) = b(t)A for all t isin [0 1] where the operator A is the infinitesimalgenerator of a C0-semigroup defined in a Hilbert space H and the kernel b is a scalarmap which satisfies appropriate hypotheses we are able to give sufficient conditions forthe existence of mild solutions only in terms of spectral properties of the operator A andregularity properties of the kernel b We show that our abstract results can be applied to


concrete situations Indeed we consider an example with a particular choice of the functionb and the operator A is defined by

(Aw)(t ξ) = a1(ξ)part2

partξ2w(t ξ) + b1(ξ)

part

partξw(t ξ) + c(ξ)w(t ξ) (13)

where the given coefficients a1 b1 c satisfy the usual uniform ellipticity conditions

2 Preliminaries

Most of the notations used throughout this paper are standard So N Z R and C denote theset of natural integers and real and complex numbers respectively N0 = N cup 0 R

+ = (0infin)and R

+0 = [0infin)In this work X and Y always are complex Banach spaces with norms middot X and middot Y

the subscript will be dropped when there is no danger of confusion We denote the space ofall bounded linear operators from X to Y by L(XY ) In the case X = Y we will write brieflyL(X) Let A be an operator defined in X We will denote its domain by D(A) its domainendowed with the graph norm by [D(A)] its resolvent set by ρ(A) and its spectrum byσ(A) = C ρ(A)

As we have already mentioned C(IX) is the vector space of all continuous functionsf I rarr X This space is a Banach space endowed with the norm

∥∥f∥∥infin = sup

tisinI

∥∥f(t)∥∥X (21)

In the same manner for n isin N we write Cn(IX) for denoting the space of all functionsfrom I to X which are n-times differentiable Further Cinfin(IX) represents the space of allinfinitely differentiable functions from I to X

We denote by L1(IX) the space of all (equivalent classes of) Bochner-measurablefunctions f I rarr X such that f(t)X is integrable for t isin I It is well known that thisspace is a Banach space with the norm

∥∥f∥∥L1(IX) =

int

I

∥∥f(s)∥∥Xds (22)

We next include some preliminaries concerning the theory of resolvent operatorR(t)tisinI for (12)

Definition 21 Let X be a complex Banach space A family R(t)tisinI of bounded linearoperators defined in X is called a resolvent operator for (12) if the following conditions arefulfilled

(R1)For each x isin X R(0)x = x and R(middot)x isin C(IX)

(R2)The map R I rarr L([D(A)]) is strongly continuous


(R3) For each y isin D(A) the function t rarr R(t)y is continuously differentiable and

d

dtR(t)y = AR(t)y +

int t

0B(t minus s)R(s)yds

= R(t)Ay +int t

0R(t minus s)B(s)yds t isin I

(23)

In what follows we assume that there exists a resolvent operator R(t)tisinI for (12)satisfying the following property

(P) The function t rarr R(t) is continuous from (0 1] to L(X) endowed with theuniform operator norm middot L(X)

Note that property (P) is also named in different ways in the existing literature on thesubject mainly the theory of C0-semigroups namely norm continuity for t gt 0 eventuallynorm continuity or equicontinuity

The existence of solutions of the linear problem


0B(t minus s)u(s)ds + f(t) t 0

u(0) = u0 isin X(24)

has been studied by many authors Assuming that f [0+infin) rarr X is locally integrable itfollows from [29] that the function u given by

u(t) = R(t)u0 +int t

0R(t minus s)f(s)ds for t 0 (25)

is a mild solution of the problem (24) Motivated by this result we adopt the following con-cept of solution

Definition 22 A continuous function u isin C(IX) is called a mild solution of (12) if theequation

u(t) = R(t)g(u) +int t

0R(t minus s)f(s u(s))ds t isin I (26)

is verified

Themain results of this paper are based on the concept of measure of noncompactnessFor general information the reader can see [31] In this paper we use the notion of Hausdorffmeasure of noncompactness For this reason we recall a few properties related with thisconcept


Definition 23 Let S be a bounded subset of a normed space Y The Hausdorff measure ofnoncompactness of S is defined by

η(S) = infε gt 0 S has a finite cover by balls of radius ε

(27)

Remark 24 Let S1 S2 be bounded sets of a normed space Y The Hausdorff measure ofnoncompactness has the following properties

(i) If S1 sube S2 then η(S1) η(S2)

(ii) η(S1) = η(S1) where S1 denotes the closure of A

(iii) η(S1) = 0 if and only if S1 is totally bounded

(iv) η(λS1) = |λ|η(S1)with λ isin R

(v) η(S1 cup S2) = maxη(S1) η(S2)(vi) η(S1 + S2) η(S1) + η(S2) where S1 + S2 = s1 + s2 s1 isin S1 s2 isin S2(vii) η(S1) = η(co(S1)) where co(S1) is the closed convex hull of S1

We next collect some specific properties of the Hausdorff measure of noncompactnesswhich are needed to establish our results Henceforth when we need to compare the mea-sures of noncompactness in X and C(IX) we will use ζ to denote the Hausdorffmeasure ofnoncompactness defined in X and γ to denote the Hausdorffmeasure of noncompactness onC(IX) Moreover we will use η for the Hausdorff measure of noncompactness for generalBanach spaces Y

Lemma 25 LetW sube C(IX) be a subset of continuous functions IfW is bounded and equicontin-uous then the set co(W) is also bounded and equicontinuous

For the rest of the paper we will use the following notation LetW be a set of functionsfrom I to X and t isin I fixed and we denoteW(t) = w(t) w isin W The proof of Lemma 26can be found in [31]

Lemma 26 LetW sube C(IX) be a bounded set Then ζ(W(t)) γ(W) for all t isin I Furthermore ifW is equicontinuous on I then ζ(W(t)) is continuous on I and

γ(W) = supζ(W(t)) t isin I (28)

A set of functions W sube L1(IX) is said to be uniformly integrable if there exists apositive function κ isin L1(IR+) such that w(t) κ(t) ae for all w isinW

The next property has been studied by several authors the reader can see [32] for moredetails

Lemma 27 If unnisinNsube L1(IX) is uniformly integrable then for each n isin N the function t rarr

ζ(un(t)nisinN) is measurable and

ζ

(int t

0un(s)ds

infin

n=1

)

2int t

0ζ(un(s)infinn=1)ds (29)

The next result is crucial for our work the reader can see its proof in [33 Theorem 2]


Lemma 28 Let Y be a Banach space IfW sube Y is a bounded subset then for each ε gt 0 there existsa sequence unnisinN

subeW such that

η(W) 2η(uninfinn=1) + ε (210)

The following lemma is essential for the proof of Theorem 32 which is the main resultof this paper For more details of its proof see [34 Theorem 31]

Lemma 29 For all 0 m n denote Cnm = ( nm ) If 0 lt ε lt 1 and h gt 0 and let

Sn = εn + Cn1ε

nminus1h + Cn2ε

nminus2h2

2+ middot middot middot + hn

n n isin N (211)

then limnrarrinfinSn = 0

Clearly a manner for proving the existence of mild solutions for (12) is using fixed-point arguments The fixed-point theorem which we will apply has been established in [34Lemma 24]

Lemma 210 Let S be a closed and convex subset of a complex Banach space Y and let F S rarr Sbe a continuous operator such that F(S) is a bounded set Define

F1(S) = F(S) Fn(S) = F(co(Fnminus1(S)

)) n = 2 3 (212)

If there exist a constant 0 r lt 1 and n0 isin N such that

η(Fn0(S)) rη(S) (213)

then F has a fixed point in the set S

3 Main Results

In this section we will present our main results Henceforth we assume that the followingassertions hold

(H1)There exists a resolvent operator R(t)tisinI for (12) having the property (P)

(H2)The function g C(IX) rarr X is a compact map

(H3)The function f I times X rarr X satisfies the Caratheodory type conditions thatis f(middot x) is measurable for all x isin X and f(t middot) is continuous for almost all t isin I(H4)There exist a functionm isin L1(IR+) and a nondecreasing continuous functionΦ R

+ rarr R+ such that

∥∥f(t x)∥∥ m(t)Φ(x) (31)

for all x isin X and almost all t isin I


(H5)There exists a functionH isin L1(IR+) such that for any bounded S sube X

ζ(f(t S)

) H(t)ζ(S) (32)

for almost all t isin I

Remark 31 Assuming that the function g satisfies the hypothesis (H2) it is clear that g takesbounded set into bounded sets For this reason for each R 0 we will denote by gR thenumber gR = supg(u) uinfin R

The following theorem is the main result of this paper

Theorem 32 If the hypotheses (H1)ndash(H5) are satisfied and there exists a constant R 0 such that

KgR +KΦ(R)int1

0m(s)ds R (33)

where K = supR(t) t isin I then the problem (12) has at least one mild solution

Proof Define F C(IX) rarr C(IX) by

(Fu)(t) = R(t)g(u) +int t


for all u isin C(IX)We begin showing that F is a continuous map Let unnisinN

sube C(IX) such that un rarr uas n rarr infin (in the norm of C(IX)) Note that

F(un) minus F(u) K∥∥g(un) minus g(u)

∥∥ +Kint1

0

∥∥f(s un(s)) minus f(s u(s))∥∥ds (35)

by hypotheses (H2) and (H3) and by the dominated convergence theorem we have thatF(un) minus F(u) rarr 0 when n rarr infin

Let R 0 and denote BR = u isin C(IX) u(t) R forallt isin I and note that for anyu isin BR we have

(Fu)(t) ∥∥R(t)g(u)

∥∥ +

∥∥∥∥∥

int t

0R(t minus s)f(s u(s))ds

∥∥∥∥∥

KgR +KΦ(R)int1

0m(s)ds R

(36)

Therefore F BR rarr BR and F(BR) is a bounded set Moreover by continuity of thefunction t rarr R(t) on (0 1] we have that the set F(BR) is an equicontinuous set of functions

Define B = co(F(BR)) It follows from Lemma 25 that the set B is equicontinuous Inaddition the operator F B rarr B is continuous and F(B) is a bounded set of functions


Let ε gt 0 Since the function g is a compact map by Lemma 28 there exists a sequencevnnisinN

sub F(B) such that

ζ(F(B)(t)) 2ζ(vn(t)infinn=1) + ε 2ζ

(int t

0

R(t minus s)f(s un(s))

infinn=1ds

)

+ ε (37)

By the hypothesis (H4) for each t isin I we have R(t minus s)f(s un(s)) KΦ(R)m(s)Therefore by the condition (H5) we have

ζ(F(B)(t)) 4Kint t

0ζ(f(s un(s))

infinn=1ds + ε

4Kint t

0H(s)ζ(un(s)nisinN

) ds + ε

4Kγ(B)int t

0H(s)ds + ε

(38)

Since the functionH isin L1(IR+) for α lt 14K there exists ϕ isin C(IR+) satisfyingint10 |H(s) minus

ϕ(s)|ds lt α Hence

ζ(F(B)(t)) 4Kγ(B)[int t

0

∣∣H(s) minus ϕ(s)∣∣ds +int t

0ϕ(s)ds

]

+ ε

4Kγ(B)[α +Nt] + ε

(39)

whereN = ϕinfin Since ε gt 0 is arbitrary we have

ζ(F(B)(t)) (a + bt)γ(B) where a = 4αK b = 4KN (310)

Let ε gt 0 Since the function g is a compact map and applying the Lemma 28 there exists asequence wnnisinN

sube co(F(B)) such that

ζ(F2(B)(t)

) 2ζ

(int t

0

R(t minus s)f(swn(s))

infinn=1ds

)

+ ε

4Kint t

0ζf(swn(s))

infinn=1ds + ε

4Kint t

0H(s)ζ

(co(F1(B)(s)

))+ ε = 4K

int t

0H(s)ζ

(F1(B)(s)

)+ ε

(311)


Using the inequality (310)we have that

ζ(F2(B)(t)

) 4K

int t

0

[∣∣H(s) minus ϕ(s)∣∣ + ∣∣ϕ(s)

∣∣](a + bs)γ(B)ds + ε

4K(a + bt)γ(B)int t

0

∣∣H(s) minus ϕ(s)∣∣ds + 4KNγ(B)

(

at +bt2

2

)

+ ε

a(a + bt) + b

(

at +bt2

2

)

+ ε (

a2 + 2bt +(bt)2

2

)

γ(B) + ε

(312)

Since ε gt 0 is arbitrary we have

ζ(F2(B)(t)

)

(

a2 + 2bt +(bt)2

2

)

γ(B) (313)

By an inductive process for all n isin N it holds

ζ(Fn(B)(t)) (

an + Cn1a

nminus1bt + Cn2a

nminus2 (bt)2

2+ middot middot middot + (bt)n

n

)

γ(B) (314)

where for 0 m n the symbol Cnm denotes the binomial coefficient ( nm )

In addition for all n isin N the set Fn(B) is an equicontinuous set of functions Thereforeusing the Lemma 26 we conclude that

γ(Fn(B)) (

an + Cn1a

nminus1b + Cn2a

nminus2 b2

2+ middot middot middot + bn

n

)

γ(B) (315)

Since 0 a lt 1 and b gt 0 it follows from Lemma 27 that there exists n0 isin N such that

(

an0 + Cn01 a

n0minus1b + Cn02 a

n0minus2 b2

2+ middot middot middot + bn0

n0

)

= r lt 1 (316)

Consequently γ(Fn0(B)) rγ(B) It follows from Lemma 29 that F has a fixed pointin B and this fixed point is a mild solution of (12)

Our next result is relatedwith a particular case of (12) Consider the following Volterraequation of convolution type


0b(t minus s)Au(s)ds + f(t u(t)) t isin I

u(0) = g(u)

(317)


where A is a closed linear operator defined on a Hilbert space H the kernel b isin L1loc(R

+R)and the function f is an appropriateH-valued map

Since (317) is a convolution type equation it is natural to employ the Laplacetransform for its study

Let X be a Banach space and a isin L1loc(R

+R) We say that the function a is Laplacetransformable if there is ω isin R such that

intinfin0 eminusωt|a(t)|dt lt infin In addition we denote by

a(λ) =intinfin0 eminusλta(t)dt for Re λ gt ω the Laplace transform of the function aWe need the following definitions for proving the existence of a resolvent operator for

(317) These concepts have been introduced by Pruss in [28]

Definition 33 Let a isin L1loc(R

+R) be Laplace transformable and k isin N We say that thefunction a is k-regular if there exists a constant C gt 0 such that

∣∣∣λna(n)(λ)

∣∣∣ C|a(λ)| (318)

for all Re λ ω and 0 lt n k

Convolutions of k-regular functions are again k-regular Moreover integration and differen-tiation are operations which preserve k-regularity as well See [28 page 70]

Definition 34 Let f isin Cinfin(R+R) We say that f is a completely monotone function if andonly if (minus1)nf (n)(λ) 0 for all λ gt 0 and n isin N


+R) such that a is Laplace transformable We say that a iscompletely positive function if and only if

1λa(λ)

minusaprime(λ)(a(λ))2

(319)

are completely monotone functions

Finally we recall that a one-parameter family T(t)t0 of bounded and linear operat-ors is said to be exponentially bounded of type (Mω) if there are constantsM 1 andω isin R

such that

T(t) Meωt forallt 0 (320)

The next proposition guarantees the existence of a resolvent operator for (317)satisfying the property (P) With this purpose we will introduce the conditions (C1) and(C2)

(C1) The kernel a defined by a(t) = 1 +int t0 b(s)ds for all t 0 is 2-regular and

completely positive


(C2) The operatorA is the generator of a semigroup of type (Mω) and there existsμ0 gt ω such that

lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (321)

Proposition 36 Suppose thatA is the generator of aC0-semigroup of type (Mω) in a Hilbert spaceH If the conditions (C1)-(C2)are satisfied then there exists a resolvent operator R(t)tisinI for (317)having the property (P)

Proof Integrating in time (317)we get

u(t) =int t

0a(t minus s)Au(s)ds +

int t

0f(s u(s)) + g(u) (322)

Since the scalar kernel a is completely positive and A generates a C0-semigroup it followsfrom [28 Theorem 42] that there exists a family of operators R(t)tisinI strongly continuousexponentially bounded which commutes with A satisfying

R(t)x = x +int t

0a(t minus s)AR(s)xds forallx isin D(A) (323)

On the other hand using the condition (C2) and since the scalar kernel a is 2-regular itfollows from [35 Theorem 22] that the function t rarr R(t) is continuous for t gt 0 Furthersince a isin C1(R+R) it follows from (323) that for all x isin D(A) the map R(middot)x is differentiablefor all t 0 and satisfies

d

dtR(t)x = AR(t)x +

int t

0b(t minus s)AR(s)x ds t isin I (324)

From the quality (324) we conclude that R(t)tisinI is a resolvent operator for (317) havingthe property (P)

Corollary 37 Suppose that A generates a C0-semigroup of type (Mω) in a Hilbert space HAssume further that the conditions (C1)-(C2) are fulfilled If the hypotheses (H2)ndash(H5) are satisfiedand there exists R 0 such that

KgR +KΦ(R)int1

0m(s)ds R where K = supR(t) t isin I (325)

then (317) has at least one mild solution

Proof It follows from Proposition 36 that there exists a resolvent operator R(t)tisinI for theequation and this resolvent operator has the property (P) Since the hypotheses (H2)ndash(H5)are satisfied we apply Theorem 32 and conclude that (317) has at least one mild solution


4 Applications

In this section we apply the abstract results which we have obtained in the preceding sectionto study the existence of solutions for a partial differential equation submitted to nonlocalinitial conditions This type of equations arises in the study of heat conduction in materialswith memory (see [26 27]) Specifically we will study the following problem

partw(t ξ)partt

= Aw(t ξ) +int t

0βeminusα(tminuss)Aw(s ξ)ds + p1(t)p2(w(t ξ)) t isin I

w(t 0) = w(t 2π) for t isin I

w(0 ξ) =int1

0

int ξ

0qk(s ξ)w

(s y

)dsdy 0 ξ 2π

(41)

where k I times [0 2π] rarr R+ is a continuous function such that k(t 2π) = 0 for all t isin I the

constant q isin R+ and the constants α β satisfy the relation minusα β 0 α The operator A is

defined by



part


where the coefficients a1 b1 c satisfy the usual uniformly ellipticity conditions and D(A) =v isin L2([0 2π]R) vprimeprimeL2([0 2π]R) The functions p1 I rarr R

+ and p2 R rarr R satisfyappropriate conditions which will be specified later

Identifying u(t) = w(t middot) we model this problem in the space X = L2(TR) where thegroup T is defined as the quotient R2πZ We will use the identification between functionson T and 2π-periodic functions on R Specifically in what follows we denote by L2(TR) thespace of 2π-periodic and square integrable functions from R into R Consequently (41) isrewritten as



u(0) = g(u)

(43)

where the function g C(IX) rarr X is defined by g(w)(ξ) =int10

int ξ0 qk(s ξ)w(s y)dsdy and

f(t u(t)) = p1(t)p2(u(t)) where p1 is integrable on I and p2 is a bounded function satisfyinga Lipschitz type condition with Lipschitz constant L

We will prove that there exists q gt 0 sufficiently small such that (43) has a mildsolution on L2(TR)

With this purpose we begin noting that g q(2π)12(int2π0

int10 k(s ξ)

2dsdξ)12

Moreover it is a well-known fact that the g is a compact map

Further the function f satisfies f(t u(t)) p1(t)Φ(u(t)) with Φ(u(t)) equiv p2and f(t u1(t)) minus f(t u2(t)) Lp1(t)u1 minus u2 Thus the conditions (H2)ndash(H5) are fulfilled


Define a(t) = 1 +int t0 βe

minusαsds for all t isin R+0 Since the kernel b defined by b(t) = βeminusαt is

2-regular it follows that a is 2-regular Furthermore we claim that a is completely positiveIn fact we have

a(λ) =λ + α + βλ(λ + α)

(44)

Define the functions f1 and f2 by f1(λ) = 1(λa(λ)) and f2(λ) = minusaprime(λ)[a(λ)]2 respectivelyIn other words

f1(λ) =λ + α

λ + α + β f2(λ) =

λ2 + 2(α + β

)λ + αβ + α2

(λ + α + β

)2 (45)

A direct calculation shows that

f(n)1 (λ) =

(minus1)n+1β(n + 1)(λ + α + β

)n+1 f(n)2 (λ) =

(minus1)n+1β(α + β)(n + 1)

(λ + α + β

)n+2 for n isin N (46)

Since minusα β 0 α we have that f1 and f2 are completely monotone Thus the kernel a iscompletely positive

On the other hand it follows from [36] that A generates an analytic noncompactsemigroup T(t)t0 on L

2(TR) In addition there exists a constantM gt 0 such that

M = supT(t) t 0 lt +infin (47)

It follows from the preceding fact and the Hille-Yosida theorem that z isin ρ(A) for all z isin C

such that Re(z) gt 0 Let z = μ0 + iμ By direct computation we have

Re

(μ0 + iμ

b(μ0 + iμ

)+ 1

)

=μ30 + μ

20α + μ2

0

(α + β

)+ μ0α

(α + β

)+ μ0μ

2 minus μ2β(α + β

)2 + 2μ0(α + β

)+ μ2

0 + μ2

(48)

Hence Re((μ0 + iμ)(b(μ0 + iμ) + 1)) gt 0 for all z = μ0 + iμ such that μ0 gt 0 This implies that

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1

isin L(X) forallμ0 gt 0 (49)

Since the semigroup generated by A is an analytic semigroup we have

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥∥∥∥∥

∥∥∥∥

M

μ0 + iμ

∥∥∥∥ (410)


Therefore

lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (411)

It follows from Proposition 36 that (43) admits a resolvent operator R(t)tisinI satisfyingproperty (P)

Let K = supR(t) t isin I and c = (2π)12(int2π0

int10 k(s ξ)

2dsdξ)12

A direct computation shows that for each R 0 the number gR is equal to gR = qcRTherefore the expression (KgR +KΦ(R)

int10 m(s)ds) is equivalent to (qcKR+ p11LK)

Since there exists q gt 0 such that qcK lt 1 then there exists R 0 such that

qcKR +∥∥p1

∥∥1LK R (412)

From Corollary 37 we conclude that there exists a mild solution of (41)

Acknowledgments

C Lizama and J C Pozo are partially supported by FONDECYT Grant no 1110090 and RingProject ACT-1112 J C Pozo is also partially financed by MECESUP PUC 0711

References

[1] M Chandrasekaran ldquoNonlocal Cauchy problem for quasilinear integrodifferential equations inBanach spacesrdquo Electronic Journal of Differential Equations vol 2007 no 33 pp 1ndash6 2007

[2] Q Dong and G Li ldquoNonlocal Cauchy problem for functional integrodifferential equations in Banachspacesrdquo Nonlinear Funct Anal Appl vol 13 no 4 pp 705ndash717 2008

[3] X Xue ldquoNonlinear differential equations with nonlocal conditions in Banach spacesrdquo NonlinearAnalysis Theory Methods amp Applications vol 63 no 4 pp 575ndash586 2005

[4] L Byszewski ldquoStrong maximum principles for parabolic nonlinear problems with nonlocal inequal-ities together with arbitrary functionalsrdquo Journal of Mathematical Analysis and Applications vol 156no 2 pp 457ndash470 1991

[5] L Byszewski ldquoTheorem about existence and uniqueness of continuous solution of nonlocal problemfor nonlinear hyperbolic equationrdquo Applicable Analysis vol 40 no 2-3 pp 173ndash180 1991

[6] L Byszewski ldquoTheorems about the existence and uniqueness of solutions of a semilinear evolutionnonlocal Cauchy problemrdquo Journal of Mathematical Analysis and Applications vol 162 no 2 pp 494ndash505 1991

[7] L Byszewski ldquoExistence and uniqueness of solutions of semilinear evolution nonlocal Cauchy pro-blemrdquo Zeszyty Naukowe Politechniki Rzeszowskiej Matematyka i Fizyka vol 18 pp 109ndash112 1993

[8] L Byszewski and V Lakshmikantham ldquoTheorem about the existence and uniqueness of a solution ofa nonlocal abstract Cauchy problem in a Banach spacerdquo Applicable Analysis vol 40 no 1 pp 11ndash191991

[9] S Ntouyas and P Tsamatos ldquoGlobal existence for semilinear evolution equations with nonlocal con-ditionsrdquo Journal of Mathematical Analysis and Applications vol 210 no 2 pp 679ndash687 1997

[10] S Ntouyas and P Tsamatos ldquoGlobal existence for semilinear evolution integrodifferential equationswith delay and nonlocal conditionsrdquo Applicable Analysis vol 64 no 1-2 pp 99ndash105 1997

[11] T Zhu C Song and G Li ldquoExistence of mild solutions for abstract semilinear evolution equations inBanach spacesrdquo Nonlinear Analysis Theory Methods amp Applications vol 75 no 1 pp 177ndash181 2012

[12] K Balachandran and S Kiruthika ldquoExistence results for fractional integrodifferential equations withnonlocal condition via resolvent operatorsrdquo Computers amp Mathematics with Applications vol 62 no 3pp 1350ndash1358 2011


[13] K Balachandran S Kiruthika and J J Trujillo ldquoOn fractional impulsive equations of Sobolev typewith nonlocal condition in Banach spacesrdquo Computers amp Mathematics with Applications vol 62 no 3pp 1157ndash1165 2011

[14] M Benchohra and S K Ntouyas ldquoNonlocal Cauchy problems for neutral functional differential andintegrodifferential inclusions in Banach spacesrdquo Journal of Mathematical Analysis and Applications vol258 no 2 pp 573ndash590 2001

[15] X Fu and K Ezzinbi ldquoExistence of solutions for neutral functional differential evolution equationswith nonlocal conditionsrdquo Nonlinear Analysis Theory Methods amp Applications vol 54 no 2 pp 215ndash227 2003

[16] S Ji and G Li ldquoExistence results for impulsive differential inclusions with nonlocal conditionsrdquo Com-puters amp Mathematics with Applications vol 62 no 4 pp 1908ndash1915 2011

[17] A Lin and L Hu ldquoExistence results for impulsive neutral stochastic functional integro-differentialinclusions with nonlocal initial conditionsrdquo Computers amp Mathematics with Applications vol 59 no 1pp 64ndash73 2010

[18] Y P Lin and J H Liu ldquoSemilinear integrodifferential equations with nonlocal Cauchy problemrdquoNon-linear Analysis Theory Methods amp Applications vol 26 no 5 pp 1023ndash1033 1996

[19] R-N Wang and D-H Chen ldquoOn a class of retarded integro-differential equations with nonlocalinitial conditionsrdquo Computers amp Mathematics with Applications vol 59 no 12 pp 3700ndash3709 2010

[20] R-N Wang J Liu and D-H Chen ldquoAbstract fractional integro-differential equations involving non-local initial conditions in α-normrdquoAdvances in Difference Equations vol 2011 article 25 pp 1ndash16 2011

[21] J-R Wang X Yan X-H Zhang T-M Wang and X-Z Li ldquoA class of nonlocal integrodifferentialequations via fractional derivative and its mild solutionsrdquo Opuscula Mathematica vol 31 no 1 pp119ndash135 2011

[22] J-R Wang Y Zhou W Wei and H Xu ldquoNonlocal problems for fractional integrodifferential equa-tions via fractional operators and optimal controlsrdquo Computers amp Mathematics with Applications vol62 no 3 pp 1427ndash1441 2011

[23] Z Yan ldquoExistence of solutions for nonlocal impulsive partial functional integrodifferential equationsvia fractional operatorsrdquo Journal of Computational and Applied Mathematics vol 235 no 8 pp 2252ndash2262 2011

[24] Z Yan ldquoExistence for a nonlinear impulsive functional integrodifferential equation with nonlocalconditions in Banach spacesrdquo Journal of Applied Mathematics amp Informatics vol 29 no 3-4 pp 681ndash696 2011

[25] Y Yang and J-R Wang ldquoOn some existence results of mild solutions for nonlocal integrodifferentialCauchy problems in Banach spacesrdquo Opuscula Mathematica vol 31 no 3 pp 443ndash455 2011

[26] J Liang and T-J Xiao ldquoSemilinear integrodifferential equations with nonlocal initial conditionsrdquoComputers amp Mathematics with Applications vol 47 no 6-7 pp 863ndash875 2004

[27] R K Miller ldquoAn integro-differential equation for rigid heat conductors with memoryrdquo Journal ofMathematical Analysis and Applications vol 66 no 2 pp 313ndash332 1978

[28] J Pruss Evolutionary integral equations and applications vol 87 of Monographs in MathematicsBirkhauser Basel Switzerland 1993

[29] R Grimmer and J Pruss ldquoOn linear Volterra equations in Banach spacesrdquo Computers amp Mathematicswith Applications vol 11 no 1ndash3 pp 189ndash205 1985

[30] J Pruss ldquoPositivity and regularity of hyperbolic Volterra equations in Banach spacesrdquoMathematischeAnnalen vol 279 no 2 pp 317ndash344 1987

[31] J Banas and K GoebelMeasures of noncompactness in Banach spaces vol 60 of Lecture Notes in Pure andApplied Mathematics Marcel Dekker New York NY USA 1980

[32] H Monch ldquoBoundary value problems for nonlinear ordinary differential equations of second orderin Banach spacesrdquo Nonlinear Analysis Theory Methods amp Applications vol 4 no 5 pp 985ndash999 1980

[33] D Bothe ldquoMultivalued perturbations of m-accretive differential inclusionsrdquo Israel Journal of Math-ematics vol 108 pp 109ndash138 1998

[34] L Liu F Guo C Wu and Y Wu ldquoExistence theorems of global solutions for nonlinear Volterra typeintegral equations in Banach spacesrdquo Journal of Mathematical Analysis and Applications vol 309 no 2pp 638ndash649 2005

[35] C Lizama ldquoA characterization of uniform continuity for Volterra equations in Hilbert spacesrdquo Pro-ceedings of the American Mathematical Society vol 126 no 12 pp 3581ndash3587 1998

[36] K-J Engel and R Nagel One-Parameter Semigroups for Linear Evolution Equations Springer BerlinGermany 2000


Research ArticleOn the Existence of Positive Periodic Solutions forSecond-Order Functional Differential Equationswith Multiple Delays

Qiang Li and Yongxiang Li

Department of Mathematics Northwest Normal University Lanzhou 730070 China

Correspondence should be addressed to Yongxiang Li liyxnwnu163com

Received 30 August 2012 Accepted 1 November 2012

Academic Editor K Sadarangani

Copyright q 2012 Q Li and Y Li This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

The existence results of positive ω-periodic solutions are obtained for the second-order functionaldifferential equation with multiple delays uprimeprime(t) + a(t)u(t) = f(t u(t) u(t minus τ1(t)) u(t minus τn(t)))where a(t) isin C(R) is a positive ω-periodic function f R times [0+infin)n+1 rarr [0+infin) is a continuousfunction which is ω-periodic in t and τ1(t) τn(t) isin C(R [0+infin)) are ω-periodic functionsThe existence conditions concern the first eigenvalue of the associated linear periodic boundaryproblem Our discussion is based on the fixed-point index theory in cones

1 Introduction

In this paper we deal with the existence of positive periodic solution of the second-orderfunctional differential equation with multiple delays

uprimeprime(t) + a(t)u(t) = f(t u(t) u(t minus τ1(t)) u(t minus τn(t))) t isin R (11)

where a(t) isin C(R) is a positive ω-periodic function f R times [0+infin)n+1 rarr [0+infin) is acontinuous functionwhich isω-periodic in t and τ1(t) τn(t) isin C(R [0+infin)) areω-periodicfunctions ω gt 0 is a constant

In recent years the existence of periodic solutions for second-order functionaldifferential equations has been researched by many authors see [1ndash8] and references thereinIn some practice models only positive periodic solutions are significant In [4ndash8] theauthors obtained the existence of positive periodic solutions for some second-order functional


differential equations by using fixed-point theorems of cone mapping Especially in [5] Wuconsidered the second-order functional differential equation

uprimeprime(t) + a(t)u(t) = λf(t u(t minus τ1(t)) u(t minus τn(t))) t isin R (12)

and obtained the existence results of positive periodic solutions by using the Krasnoselskiifixed-point theorem of cone mapping when the coefficient a(t) satisfies the condition that0 lt a(t) lt π2ω2 for every t isin R And in [8] Li obtained the existence results of positiveω-periodic solutions for the second-order differential equation with constant delays

minusuprimeprime(t) + a(t)u(t) = f(t u(t minus τ1) u(t minus τn)) t isin R (13)

by employing the fixed-point index theory in cones For the second-order differentialequations without delay the existence of positive periodic solutions has been discussed bymore authors see [9ndash14]

Motivated by the paper mentioned above we research the existence of positiveperiodic solutions of (11) We aim to obtain the essential conditions on the existence ofpositive periodic solution of (11) by constructing a special cone and applying the fixed-pointindex theory in cones

In this paper we assume the following conditions

(H1) a isin C(R (0+infin)) is ω-periodic function and there exists a constant 1 le p le +infinsuch that

ap le K(2plowast) (14)

where ap is the p-norm of a in Lp[0 ω] plowast is the conjugate exponent of p definedby (1p) + (1plowast) = 1 and the function K(q) is defined by

K(q)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2πqω1+2q

(2

2 + q

)1minus2q( Γ(1q)

Γ(12 + 1q

)

)2

if 1 le q lt +infin

4ω if q = +infin

(15)

in which Γ is the Gamma function

(H2) f isin C(R times [0+infin)n+1 [0+infin)) and f(t x0 x1 xn) is ω-periodic in t

(H3) τ1(t) τn(t) isin C(R [0infin)) are ω-periodic functions

In Assumption (H1) if p = +infin since K(2) = π2ω2 then (14) implies that a satisfiesthe condition

0 lt a(t) le π2

ω2 t isin [0 ω] (16)

This condition includes the case discussed in [5]


The techniques used in this paper are completely different from those in [5] Ourresults are more general than those in [5] in two aspects Firstly we relax the conditionsof the coefficient a(t) appeared in an equation in [5] and expand the range of its valuesSecondly by constructing a special cone and applying the fixed-point index theory in coneswe obtain the essential conditions on the existence of positive periodic solutions of (11) Theconditions concern the first eigenvalue of the associated linear periodic boundary problemwhich improve and optimize the results in [5] To our knowledge there are very few workson the existence of positive periodic solutions for the above functional differential equationsunder the conditions concerning the first eigenvalue of the corresponding linear equation

Our main results are presented and proved in Section 3 Some preliminaries to discuss(11) are presented in Section 2

2 Preliminaries

In order to discuss (11) we consider the existence of ω-periodic solution of thecorresponding linear differential equation

uprimeprime+ a(t)u = h(t) t isin R (21)

where h isin C(R) is a ω-periodic function It is obvious that finding an ω-periodic solution of(21) is equivalent to finding a solution of the linear periodic boundary value problem

uprimeprime+ a(t)u = h(t) t isin [0 ω]

u(0) = u(ω) uprime(0) = uprime(ω)(22)

In [14] Torres show the following existence resulted

Lemma 21 Assume that (H1 ) holds then for every h isin C[0 ω] the linear periodic boundaryproblem (22) has a unique solution expressed by

u(t) =intω

0G(t s)h(s)ds t isin [0 ω] (23)

whereG(t s) isin C([0 ω]times[0 ω]) is the Green function of the linear periodic boundary problem (22)which satisfies the positivity G(t s) gt 0 for every (t s) isin [0 ω] times [0 ω]

For the details see [14 Theorem 21 and Corollary 23]Form isin N we useCm

ω (R) to denote themth-order continuous differentiableω-periodicfunctions space Let X = Cω(R) be the Banach space of all continuous ω-periodic functionsequipped the norm u = max0letleω|u(t)|

Let

K0 = u isin X | u(t) ge 0 t isin R (24)


be the cone of all nonnegative functions in X Then X is an ordered Banach space by the coneK0 K0 has a nonempty interior

int(K0) = u isin X | u(t) gt 0 t isin R (25)

Let

G = min0letsleω

G(t s) G = max0letsleω

G(t s) σ = GG (26)

Lemma 22 Assume that (H1) holds then for every h isin X (21) has a unique ω-periodic solutionu Let T h rarr u then T X rarr X is a completely continuous linear operator and when h isin K0 Thhas the positivity estimate

Th(t) ge σTh forallt isin R (27)

Proof Let h isin X By Lemma 21 the linear periodic boundary problem (22) has a uniquesolution u isin C2[0 ω] given by (23) We extend u to a ω-periodic function which is stilldenoted by u then u = Th isin C2

ω(R) is a unique ω-periodic solution of (21) By (23)

Th(t) =intω


From this we see that T maps every bounded set in X to a bounded equicontinuous set of XHence by the Ascoli-Arzela theorem T X rarr X is completely continuous

Let h isin K0 For every t isin [0 ω] from (28) it follows that

Th(t) =intω

0G(t s)h(s)ds le G

intω

0h(s)ds (29)

and therefore

Th le Gintω

0h(s)ds (210)

Using (28) and this inequality we have that

Th(t) =intω

0G(t s)h(s)ds ge G

intω

0h(s)ds

=(GG

)middotGintω

0h(s)ds

ge σTh

(211)

Hence by the periodicity of u (27) holds for every t isin R


From (27)we easily see that T(K0) sub int(K0) namely T X rarr X is a strongly positivelinear operator By the well-known Krein-Rutman theorem the spectral radius r(T) gt 0 is asimple eigenvalue of T and T has a corresponding positive eigenfunction φ isin K0 that is

Tφ = r(T)φ (212)

Since φ can be replaced by cφ where c gt 0 is a constant we can choose φ isin K0 such that

intω

0φ(t)dt = 1 (213)

Set λ1 = 1r(T) then φ = T(λ1φ) By Lemma 22 and the definition of T φ isin C2ω(R) satisfies

the differential equation

φprimeprime(t) + a(t)φ(t) = λ1φ(t) t isin R (214)

Thus λ1 is the minimum positive real eigenvalue of the linear equation (21) under the ω-periodic condition Summarizing these facts we have the following lemma

Lemma 23 Assume that (H1) holds then there exist φ isin K0 cap C2ω(R) such that (213) and (214)

hold

Let f R times [0infin)n+1 rarr [0infin) satisfy the assumption (H2) For every u isin X set

F(u)(t) = f(t u(t) u(t minus τ1) u(t minus τn)) t isin R (215)

Then F K0 rarr K0 is continuous Define a mapping A K0 rarr X by

A = T F (216)

By the definition of operator T theω-periodic solution of (11) is equivalent to the fixed pointof A Choose a subcone of K0 by

K = u isin K0 | u(t) ge σuC t isin R (217)

By the strong positivity (27) of T and the definition of A we easily obtain the following

Lemma 24 Assume that (H1) holds then A(K0) sub K and A K rarr K is completely continuous

Hence the positive ω-periodic solution of (11) is equivalent to the nontrivial fixedpoint ofA We will find the nonzero fixed point ofA by using the fixed-point index theory incones

We recall some concepts and conclusions on the fixed-point index in [15 16] Let Xbe a Banach space and K sub X a closed convex cone in X Assume that Ω is a bounded opensubset ofX with boundary partΩ andKcapΩ= empty LetA KcapΩ rarr K be a completely continuous


mapping IfAu=u for any u isin KcappartΩ then the fixed-point index i(AKcapΩ K) has definitionOne important fact is that if i(AKcapΩ K)= 0 thenA has a fixed point inKcapΩ The followingtwo lemmas in [16] are needed in our argument

Lemma 25 Let Ω be a bounded open subset of X with θ isin Ω and A K cap Ω rarr K a completelycontinuous mapping If μAu=u for every u isin K cap partΩ and 0 lt μ le 1 then i(AK capΩ K) = 1

Lemma 26 Let Ω be a bounded open subset of X and A K cap Ω rarr K a completely continuousmapping If there exists an e isin K θ such that u minusAu=μe for every u isin K cap partΩ and μ ge 0 theni(AK capΩ K) = 0

3 Main Results

We consider the existence of positive ω-periodic solutions of (11) Assume that f R times[0infin)n+1 rarr [0infin) satisfy (H2) To be convenient we introduce the notations

f0 = lim infxrarr 0+

mintisin[0ω]

f(t x0 x1 xn)x

f0 = lim supxrarr 0+

maxtisin[0ω]

f(t x0 x1 xn)x

finfin = lim infxrarr+infin

mintisin[0ω]

f(t x0 x1 xn)x

finfin = lim supxrarr+infin

maxtisin[0ω]

f(t x0 x1 xn)x

(31)

where x = maxx0 x1 xn and x = minx0 x1 xn Our main results are as follows

Theorem 31 Suppose that (H1)ndash(H3) hold If f satisfies the condition

(H4) f0 lt λ1 lt finfin

then (11) has at least one positive ω-periodic solution


(H5) finfin lt λ1 lt f0


In Theorem 31 the condition (H4) allows f(t x0 x1 xn) to be superlinear growthon x0 x1 xn For example

f(t x0 x1 xn) = b0(t)x20 + b1(t)x

21 + middot middot middot + bn(t)x2

n (32)

satisfies (H4) with f0 = 0 and finfin = +infin where b0 b1 bn isin Cω(R) are positive ω-periodicfunctions


In Theorem 32 the condition (H5) allows f(t x0 x1 xn) to be sublinear growth onx0 x1 xn For example

f(t x0 x1 xn) = c0(t)radic|x0| + c1(t)

radic|x1| + middot middot middot + cn(t)

radic|xn| (33)

satisfies (H5) with f0 = +infin and finfin = 0 where c0 c1 cn isin Cω(R) are positive ω-periodicsolution

Applying Theorems 31 and 32 to (12) we have the following

Corollary 33 Suppose that (H1)ndash(H3) hold If the parameter λ satisfies one of the followingconditions

(1) λ1finfin lt λ lt λ1f0

(2) λ1f0 lt λ lt λ1finfin


This result improves and extends [5 Theorem 13]

Proof of Theorem 31 Let K sub X be the cone defined by (217) and A K rarr K the operatordefined by (216) Then the positiveω-periodic solution of (11) is equivalent to the nontrivialfixed point of A Let 0 lt r lt R lt +infin and set

Ω1 = u isin X | u lt r Ω2 = u isin X | u lt R (34)

We show that the operator A has a fixed point in K cap (Ω2 Ω1) when r is small enough andR large enough

Since f0 lt λ1 by the definition of f0 there exist ε isin (0 λ1) and δ gt 0 such that

f(t x0 x1 xn) le (λ1 minus ε)x t isin [0 ω] x isin (0 δ) (35)

where x = maxx0 x1 xn and x = minx0 x1 xn Choosing r isin (0 δ) we prove thatA satisfies the condition of Lemma 25 in K cap partΩ1 namely μAu=u for every u isin K cap partΩ1

and 0 lt μ le 1 In fact if there exist u0 isin K cap partΩ1 and 0 lt μ0 le 1 such that μ0Au0 = u0 andsince u0 = T(μ0F(u0)) by definition of T and Lemma 22 u0 isin C2

ω(R) satisfies the differentialequation

uprimeprime0(t) + a(t)u0(t) = μ0F(u0)(t) t isin R (36)

where F(u) is defined by (215) Since u0 isin K cap partΩ1 by the definitions of K and Ω1 we have

0 lt σ u0 le u0(τ) le u0 = r lt δ forallτ isin R (37)

This implies that

0 lt maxu0(t) u0(t minus τ1(t)) u0(t minus τn(t)) lt δ t isin R (38)


From this and (35) it follows that

F(u0)(t) = f(t u0(t) u0(t minus τ1(t)) u0(t minus τn(t)))le (λ1 minus ε) middotminu0(t) u0(t minus τ1(t)) u0(t minus τn(t))le (λ1 minus ε)u0(t) t isin R

(39)

By this inequality and (36) we have

uprimeprime0(t) + a(t)u0(t) = μ0F(u0)(t) le (λ1 minus ε)u0(t) t isin R (310)

Let φ isin K cap C2ω(R) be the function given in Lemma 24 Multiplying the inequality (310) by

φ(t) and integrating on [0 ω] we have

intω

0

[uprimeprime0(t) + a(t)u0(t)

]φ(t)dt le (λ1 minus ε)

intω

0u0(t)φ(t)dt (311)

For the left side of the above inequality using integration by parts then using the periodicityof u0 and φ and (214) we have

intω

0


]φ(t)dt =

intω

0u0(t)

[φprimeprime(t) + a(t)φ(t)

]dt

= λ1

intω

0u0(x)φ(t)dt

(312)

Consequently we obtain that

λ1

intω

0u0(x)φ(t)dt le (λ1 minus ε)

intω

0u0(x)φ(t)dt (313)

Since u0 isin K cap partΩ1 by the definition of K and (213)

intω

0u0(x)φ(t)dt ge σu0

intω

0φ(t)dt = σu0 gt 0 (314)

From this and (313) we conclude that λ1 le λ1minusε which is a contradiction HenceA satisfiesthe condition of Lemma 25 in K cap partΩ1 By Lemma 25 we have

i(AK capΩ1 K) = 1 (315)

On the other hand since finfin gt λ1 by the definition of finfin there exist ε1 gt 0 andH gt 0such that

f(t x0 x1 xn) ge (λ1 + ε1) x t isin [0 ω] x gt H (316)


where x = maxx0 x1 xn and x = minx0 x1 xn Choose R gt maxHσ δ ande(t) equiv 1 Clearly e isin K θ We show thatA satisfies the condition of Lemma 26 inK cappartΩ2namely u minusAu=μφ for every u isin K cap partΩ2 and μ ge 0 In fact if there exist u1 isin K cap partΩ2 andμ1 ge 0 such that u1 minus Au1 = μ1e since u1 minus μ1e = Au1 = T(F(u1)) by the definition of T andLemma 22 u1 isin C 2

ω(R) satisfies the differential equation

uprimeprime1(t) + a(t)

(u1(t) minus μ1

)= F(u1)(t) t isin R (317)

Since u1 isin K cap partΩ2 by the definitions of K and Ω2 we have

u1(τ) ge σu1 = σR gt H forallτ isin R (318)

This means that

minu1(t) u1(t minus τ1(t)) u1(t minus τn(t)) gt H t isin R (319)

Combining this with (316) we have that

F(u1)(t) = f(t u1(t) u1(t minus τ1(t)) u1(t minus τn(t)))ge (λ1 + ε1) middotmaxu1(t) u1(t minus τ1(t)) u1(t minus τn(t))ge (λ1 + ε1)u1(t) t isin R

(320)

From this inequality and (317) it follows that

uprimeprime1(t) + a(t)u1(t) = μ1a(t) + F(u1)(t) ge (λ1 + ε1)u1(t) t isin R (321)

Multiplying this inequality by φ(t) and integrating on [0 ω] we have

intω

0

[u

primeprime1(t) + a(t)u1(t)

]φ(t)dt ge (λ1 + ε1)

intω

0u1(t)φ(t)dt (322)

For the left side of the above inequality using integration by parts and (214) we have

intω

0


]φ(t)dt =

intω

0u1(t)


]dt

= λ1

intω

0u1(x)φ(t)dt

(323)


λ1

intω

0u1(x)φ(t)dt ge (λ1 + ε1)

intω

0u1(x)φ(t)dt (324)


Since u1 isin K cap partΩ2 by the definition of K and (213) we have

intω


intω

0φ(t)dt = σu1 gt 0 (325)

Hence from (324) it follows that λ1 ge λ1 + ε1 which is a contradiction Therefore A satisfiesthe condition of Lemma 26 in K cap partΩ2 By Lemma 26 we have

i(AK capΩ2 K) = 0 (326)

Now by the additivity of the fixed-point index (315) and (326) we have

i(AK cap

(Ω2 Ω1

) K)= i(AK capΩ2 K) minus i(AK capΩ1 K) = minus1 (327)

HenceA has a fixed point inKcap(Ω2Ω1) which is a positiveω-periodic solution of (11)

Proof of Theorem 32 LetΩ1Ω2 sub X be defined by (34) We prove that the operatorA definedby (216) has a fixed point in K cap (Ω2 Ω1) if r is small enough and R is large enough

By f0 gt λ1 and the definition of f0 there exist ε gt 0 and δ gt 0 such that

f(t x0 x1 xn) ge (λ1 + ε)x t isin [0 ω] x isin (0 δ) (328)

where x = maxx0 x1 xn Let r isin (0 δ) and e(t) equiv 1 We prove that A satisfies thecondition of Lemma 26 in K cap partΩ1 namely u minusAu=μe for every u isin K cap partΩ1 and μ ge 0 Infact if there exist u0 isin KcappartΩ1 and μ0 ge 0 such that u0 minusAu0 = μ0e and since u0 minusμ0e = Au0 =T(F(u0)) by the definition of T and Lemma 22 u0 isin C2



(u0(t) minus μ0

)= F(u0)(t) t isin R (329)

Since u0 isin K cap partΩ1 by the definitions of K and Ω1 u0 satisfies (37) and hence (38) holdsFrom (38) and (328) it follows that

F(u0)(t) = f(t u0(t) u0(0 minus τ1(t)) u0(t minus τn(t)))ge (λ1 + ε) middotmaxu0(t) u0(t minus τ1(t)) u0(t minus τn(t))ge (λ1 + ε)u0(t) t isin R

(330)

By this and (329) we obtain that

uprimeprime0(t) + a(t)u0(t) = μ0a(t) + F(u0)(t) ge (λ1 + ε)u0(t) t isin R (331)


intω

0

[u


]φ(t)dt ge (λ1 + ε)

intω

0u0(t)φ(t)dt (332)


For the left side of this inequality using integration by parts and (214) we have

intω

0

[u


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(333)


λ1

intω


intω

0u1(x)φ(t)dt (334)

Since u0 isin K cap partΩ1 from the definition of K and (213) it follows that (314) holds By (314)and (334) we see that λ1 ge λ1 + ε which is a contradiction Hence A satisfies the conditionof Lemma 26 in K cap partΩ1 By Lemma 26 we have

i(AK capΩ1 K) = 0 (335)

Since finfin lt λ1 by the definition of finfin there exist ε1 isin (0 λ1) andH gt 0 such that

f(t x0 x1 xn) le (λ1 minus ε1)x t isin [0 ω] x gt H (336)

where x = minx0 x1 xn Choosing R gt maxHσ δ we show that A satisfies thecondition of Lemma 25 in K cap partΩ2 namely μAu=u for every u isin K cap partΩ2 and 0 lt μ le 1 Infact if there exist u1 isin K cap partΩ2 and 0 lt μ1 le 1 such that μ1Au1 = u1 since u1 = T(μ1F(u1))by the definition of T and Lemma 22 u1 isin C2

ω(Ω) satisfies the differential equation


Since u1 isin K cap partΩ2 by the definitions ofK andΩ2 u1 satisfies (318) and hence (319) holdsBy (319) and (336) we have

F(u1)(t) = f(t u1(t) u1(t minus τ1(t)) u1(t minus τn(t)))le (λ1 minus ε1) middotminu1(t) u1(t minus τ1(t)) u1(t minus τn(t))le (λ1 minus ε1)u1(t) t isin R

(338)


uprimeprime1(t) + a(t)u1(t) = μ1F(u1)(t) le (λ1 minus ε1)u1(t) t isin R (339)


intω

0

[u


]φ(t)dt le (λ1 minus ε1)

intω

0u1(t)φ(t)dt (340)



intω

0

[u


]φ(t)dt =

intω

0u1(t)

[φ

primeprime(t) + a(t)φ(t)

]dt

= λ1

intω

0u1(x)φ(t)dt

(341)


λ1

intω

0u1(x)φ(t)dt le (λ1 minus ε1)

intω

0u1(x)φ(t)dt (342)

Since u1 isin K cap partΩ2 by the definition of K and (213) we see that (325) holds From (325)and (342) we see that λ1 le λ1 minus ε1 which is a contradiction HenceA satisfies the conditionof Lemma 25 in K cap partΩ2 By Lemma 25 we have

i(AK capΩ2 K) = 1 (343)

Now from (335) and (343) it follows that

i(AK cap

(Ω2 Ω1

) K)= i(AK capΩ2 K) minus i(AK capΩ1 K) = 1 (344)


Acknowledgment

This research supported by NNSF of China (11261053 and 11061031)

References

[1] B Liu ldquoPeriodic solutions of a nonlinear second-order differential equation with deviatingargumentrdquo Journal of Mathematical Analysis and Applications vol 309 no 1 pp 313ndash321 2005

[2] J-W Li and S S Cheng ldquoPeriodic solutions of a second order forced sublinear differential equationwith delayrdquo Applied Mathematics Letters vol 18 no 12 pp 1373ndash1380 2005

[3] Y Wang H Lian and W Ge ldquoPeriodic solutions for a second order nonlinear functional differentialequationrdquo Applied Mathematics Letters vol 20 no 1 pp 110ndash115 2007

[4] J Wu and Z Wang ldquoTwo periodic solutions of second-order neutral functional differentialequationsrdquo Journal of Mathematical Analysis and Applications vol 329 no 1 pp 677ndash689 2007

[5] Y Wu ldquoExistence nonexistence and multiplicity of periodic solutions for a kind of functionaldifferential equation with parameterrdquo Nonlinear Analysis Theory Methods amp Applications vol 70 no1 pp 433ndash443 2009

[6] C Guo and Z Guo ldquoExistence of multiple periodic solutions for a class of second-order delaydifferential equationsrdquo Nonlinear Analysis Real World Applications vol 10 no 5 pp 3285ndash3297 2009

[7] B Yang R Ma and C Gao ldquoPositive periodic solutions of delayed differential equationsrdquo AppliedMathematics and Computation vol 218 no 8 pp 4538ndash4545 2011

[8] Y Li ldquoPositive periodic solutions of second-order differential equations with delaysrdquo Abstract andApplied Analysis vol 2012 Article ID 829783 13 pages 2012


[9] Y X Li ldquoPositive periodic solutions of nonlinear second order ordinary differential equationsrdquo ActaMathematica Sinica vol 45 no 3 pp 481ndash488 2002

[10] Y Li ldquoPositive periodic solutions of first and second order ordinary differential equationsrdquo ChineseAnnals of Mathematics B vol 25 no 3 pp 413ndash420 2004

[11] Y Li and H Fan ldquoExistence of positive periodic solutions for higher-order ordinary differentialequationsrdquo Computers amp Mathematics with Applications vol 62 no 4 pp 1715ndash1722 2011

[12] F Li and Z Liang ldquoExistence of positive periodic solutions to nonlinear second order differentialequationsrdquo Applied Mathematics Letters vol 18 no 11 pp 1256ndash1264 2005

[13] B Liu L Liu and Y Wu ldquoExistence of nontrivial periodic solutions for a nonlinear second orderperiodic boundary value problemrdquoNonlinear Analysis Theory Methods amp Applications vol 72 no 7-8pp 3337ndash3345 2010

[14] P J Torres ldquoExistence of one-signed periodic solutions of some second-order differential equationsvia a Krasnoselskii fixed point theoremrdquo Journal of Differential Equations vol 190 no 2 pp 643ndash6622003

[15] K Deimling Nonlinear Functional Analysis Springer-Verlag New York NY USA 1985[16] D J Guo and V Lakshmikantham Nonlinear Problems in Abstract Cones vol 5 Academic Press New

York NY USA 1988


Research ArticleSolvability of Nonlinear Integral Equations ofVolterra Type

Zeqing Liu1 Sunhong Lee2 and Shin Min Kang2

1 Department of Mathematics Liaoning Normal University Liaoning Dalian 116029 China2 Department of Mathematics and RINS Gyeongsang National University Jinju 660-701 Republic of Korea

Correspondence should be addressed to Sunhong Lee sunhonggnuackrand Shin Min Kang smkanggnuackr

Received 2 August 2012 Accepted 23 October 2012

Academic Editor Kishin B Sadarangani

Copyright q 2012 Zeqing Liu et al This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

This paper deals with the existence of continuous bounded solutions for a rather general nonlinearintegral equation of Volterra type and discusses also the existence and asymptotic stability ofcontinuous bounded solutions for another nonlinear integral equation of Volterra type Themain tools used in the proofs are some techniques in analysis and the Darbo fixed pointtheorem via measures of noncompactness The results obtained in this paper extend and improveessentially some known results in the recent literature Two nontrivial examples that explain thegeneralizations and applications of our results are also included

1 Introduction

It is well known that the theory of nonlinear integral equations and inclusions has becomeimportant in some mathematical models of real processes and phenomena studied in math-ematical physics elasticity engineering biology queuing theory economics and so on (see[1ndash3]) In the last decade the existence asymptotical stability and global asymptotical sta-bility of solutions for various Volterra integral equations have received much attention seefor instance [1 4ndash22] and the references therein

In this paper we are interested in the following nonlinear integral equations of Volterratype

x(t) = f

(

t x(a(t)) (Hx)(b(t))intα(t)

0u(t s x(c(s)))ds

)

forallt isin R+ (11)

x(t) = h

(

t x(t)intα(t)

0u(t s x(c(s)))ds

)



where the functions f h u a b c α and the operator H appearing in (11) are given whilex = x(t) is an unknown function

To the best of our knowledge the papers dealing with (11) and (12) are few Butsome special cases of (11) and (12) have been investigated by a lot of authors For exampleArias et al [4] studied the existence uniqueness and attractive behaviour of solutions for thenonlinear Volterra integral equation with nonconvolution kernels

x(t) =int t

0k(t s)g(x(s))ds forallt isin R+ (13)

Using the monotone iterative technique Constantin [13] got a sufficient condition whichensures the existence of positive solutions of the nonlinear integral equation

x(t) = L(t) +int t

0

[M(t s)x(s) +K(t s)g(x(s))

]ds forallt isin R+ (14)

Roberts [21] examined the below nonlinear Volterra integral equation

x(t) =int t

0k(t minus s)G((x(s) s))ds forallt isin R+ (15)

which arose from certain models of a diffusive medium that can experience explosivebehavior utilizing the Darbo fixed point theorem and the measure of noncompactness in [7]Banas and Dhage [6] Banas et al [8] Banas and Rzepka [9 10] Hu and Yan [16] and Liu andKang [19] investigated the existence andor asymptotic stability andor global asymptoticstability of solutions for the below class of integral equations of Volterra type

x(t) = (Tx)(t)int t

0u(t s x(s))ds forallt isin R+ (16)

x(t) = f(t x(t)) +int t


x(t) = f(t x(t))int t


x(t) = g(t x(t)) + x(t)int t


x(t) = f(t x(t)) + g(t x(t))int t


x(t) = f(t x(α(t))) +intβ(t)

0g(t s x

(γ(s)))ds forallt isin R+ (111)

respectively By means of the Schauder fixed point theorem and the measure of noncompact-ness in [7] Banas and Rzepka [11] studied the existence of solutions for the below nonlinearquadratic Volterra integral equation

x(t) = p(t) + f(t x(t))int t

0v(t s x(s))ds forallt isin R+ (112)


Banas and Chlebowicz [5] got the solvability of the following functional integral equation

x(t) = f1

(

t

int t

0k(t s)f2(s x(s))ds

)


in the space of Lebesgue integrable functions on R+ El-Sayed [15] studied a differential equa-tion of neutral type with deviated argument which is equivalent to the functional-integralequation

x(t) = f

(

t

intH(t)

0x(s)ds x(h(t))

)


by the technique linking measures of noncompactness with the classical Schauder fixed pointprinciple Using an improvement of the Krasnoselrsquoskii type fixed point theorem Taoudi [22]discussed the existence of integrable solutions of a generalized functional-integral equation

x(t) = g(t x(t)) + f1

(

t

int t

0k(t s)f2(s x(s))ds

)


Dhage [14] used the classical hybrid fixed point theorem to establish the uniform localasymptotic stability of solutions for the nonlinear quadratic functional integral equation ofmixed type

x(t) = f(t x(α(t)))

(

q(t) +intβ(t)

0u(t s x

(γ(s)))ds

)


The purpose of this paper is to prove the existence of continuous bounded solutions for(11) and to discuss the existence and asymptotic stability of continuous bounded solutionsfor (12) The main tool used in our considerations is the technique of measures of noncom-pactness [7] and the famous fixed point theorem of Darbo [23] The results presented in thispaper extend proper the corresponding results in [6 9 10 15 16 19] Two nontrivial exampleswhich show the importance and the applicability of our results are also included

This paper is organized as follows In the second section we recall some definitionsand preliminary results and prove a few lemmas which will be used in our investigations Inthe third section we state and prove our main results involving the existence and asymptoticstability of solutions for (11) and (12) In the final section we construct two nontrivial exam-ples for explaining our results from which one can see that the results obtained in this paperextend proper several ones obtained earlier in a lot of papers

2 Preliminaries

In this section we give a collection of auxiliary facts which will be needed further on LetR = (minusinfininfin) and R+ = [0infin) Assume that (E middot ) is an infinite dimensional Banach spacewith zero element θ and Br stands for the closed ball centered at θ and with radius r Let B(E)denote the family of all nonempty bounded subsets of E


Definition 21 LetD be a nonempty bounded closed convex subset of the space E A operatorf D rarr E is said to be a Darbo operator if it is continuous and satisfies that μ(fA) le kμ(A)for each nonempty subset A of D where k isin [0 1) is a constant and μ is a measure ofnoncompactness on B(E)

The Darbo fixed point theorem is as follows

Lemma 22 (see [23]) Let D be a nonempty bounded closed convex subset of the space E and letf D rarr D be a Darbo operator Then f has at least one fixed point in D

Let BC(R+) denote the Banach space of all bounded and continuous functions x R+ rarr R equipped with the standard norm

x = sup|x(t)| t isin R+ (21)

For any nonempty bounded subset X of BC(R+) x isin X t isin R+ T gt 0 and ε ge 0 define

ωT (x ε) = sup∣∣x(p) minus x(q)∣∣ p q isin [0 T] with

∣∣p minus q∣∣ le ε

ωT (X ε) = supωT (x ε) x isin X

ωT

0 (X) = limεrarr 0

ωT (X ε)

ω0(X) = limTrarr+infin

ωT0 (X) X(t) = x(t) x isin X

diam X(t) = sup∣∣x(t) minus y(t)∣∣ x y isin X

μ(X) = ω0(X) + lim suptrarr+infin

diam X(t)

(22)

It can be shown that the mapping μ is a measure of noncompactness in the space BC(R+) [4]

Definition 23 Solutions of an integral equation are said to be asymptotically stable if there existsa ball Br in the space BC(R+) such that for any ε gt 0 there exists T gt 0 with

∣∣x(t) minus y(t)∣∣ le ε (23)

for all solutions x(t) y(t) isin Br of the integral equation and any t ge T It is clear that the concept of asymptotic stability of solutions is equivalent to the con-

cept of uniform local attractivity [9]

Lemma 24 Let ϕ and ψ R+ rarr R+ be functions with

limtrarr+infin

ψ(t) = +infin lim suptrarr+infin

ϕ(t) lt +infin (24)

Then

lim suptrarr+infin

ϕ(ψ(t))= lim sup

trarr+infinϕ(t) (25)


Proof Let lim suptrarr+infin ϕ(t) = A It follows that for each ε gt 0 there exists T gt 0 such that

ϕ(t) lt A + ε forallt ge T (26)

Equation (24)means that there exists C gt 0 satisfying

ψ(t) ge T forallt ge C (27)

Using (26) and (27) we infer that

ϕ(ψ(t))lt A + ε forallt ge C (28)

that is (25) holds This completes the proof

Lemma 25 Let a R+ rarr R+ be a differential function If for each T gt 0 there exists a positive num-ber aT satisfying

0 le aprime(t) le aT forallt isin [0 T] (29)

then

ωT (x a ε) le ωa(T)(x aTε) forall(x ε) isin BC(R+) times (0+infin) (210)

Proof Let T gt 0 It is clear that (29) yields that the function a is nondecreasing in [0 T] andfor any t s isin [0 T] there exists ξ isin (0 T) satisfying

|a(t) minus a(s)| = aprime(ξ)|t minus s| le aT |t minus s| (211)

by the mean value theorem Notice that (29) means that a(t) isin [a(0) a(T)] sube [0 a(T)] foreach t isin [0 T] which together with (211) gives that

ωT (x a ε) = sup|x(a(t)) minus x(a(s))| t s isin [0 T] |t minus s| le εle sup

∣∣x(p) minus x(q)∣∣ p q isin [a(0) a(T)]

∣∣p minus q∣∣ le aTε

le ωa(T)(x aTε) forall(x ε) isin BC(R+) times (0+infin)

(212)

which yields that (210) holds This completes the proof

Lemma 26 Let ϕ R+ rarr R+ be a function with limtrarr+infin ϕ(t) = +infin and X be a nonemptybounded subset of BC(R+) Then


ωϕ(T)0 (X) (213)


Proof Since X is a nonempty bounded subset of BC(R+) it follows that ω0(X) =limTrarr+infinωT

0 (X) That is for given ε gt 0 there existsM gt 0 satisfying

∣∣∣ωT

0 (X) minusω0(X)∣∣∣ lt ε forallT gt M (214)

It follows from limtrarr+infin ϕ(t) = +infin that there exists L gt 0 satisfying

ϕ(T) gt M forallT gt L (215)

By means of (214) and (215) we get that

∣∣∣ω

ϕ(T)0 (X) minusω0(X)

∣∣∣ lt ε forallT gt L (216)

which yields (213) This completes the proof

3 Main Results

Now we formulate the assumptions under which (11)will be investigated

(H1) f R+ timesR3 rarr R is continuous with f(t 0 0 0) isin BC(R+) and f = sup|f(t 0 0 0)|

t isin R+(H2) a b c α R+ rarr R+ satisfy that a and b have nonnegative and bounded derivative

in the interval [0 T] for each T gt 0 c and α are continuous and α is nondecreasingand

limtrarr+infin

a(t) = limtrarr+infin

b(t) = +infin (31)

(H3) u R2+ times R rarr R is continuous

(H4) there exist five positive constants r M M0 M1 andM2 and four continuous fun-ctionsm1m2 m3g R+ rarr R+ such that g is nondecreasing and

∣∣f(t vw z) minus f(t p q y)∣∣ le m1(t)∣∣v minus p∣∣ +m2(t)

∣∣w minus q∣∣ +m3(t)∣∣z minus y∣∣

forallt isin R+ v p isin [minusr r] w q isin [minusg(r) g(r)] z y isin [minusMM](32)

limtrarr+infin

sup

m3(t)intα(t)

0|u(t s v(c(s))) minus u(t sw(c(s)))|ds vw isin Br

= 0 (33)

sup

m3(t)intα(t)

0|u(t s v(c(s)))|ds t isin R+ v isin Br

leM0 (34)

sup

intα(t)


leM (35)


supmi(t) t isin R+ leMi i isin 1 2 (36)

M1r +M2g(r) +M0 + f le r M1 +QM2 lt 1 (37)

(H5) H BC(R+) rarr BC(R+) satisfies that H Br rarr BC(R+) is a Darbo operator withrespect to the measure of noncompactness of μ with a constant Q and

|(Hx)(b(t))| le g(|x(b(t))|) forall(x t) isin Br times R+ (38)

Theorem 31 Under Assumptions (H1)ndash(H5) (11) has at least one solution x = x(t) isin Br

Proof Let x isin Br and define

(Fx)(t) = f

(


0u(t s x(c(s)))ds

)


It follows from (39) and Assumptions (H1)ndash(H5) that Fx is continuous on R+ and that

|(Fx)(t)| le∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minus f(t 0 0 0)∣∣∣∣∣+∣∣f(t 0 0 0)

∣∣

le m1(t)|x(a(t))| +m2(t)|(Hx)(b(t))| +m3(t)

∣∣∣∣∣

intα(t)

0u(t s x(c(s)))ds

∣∣∣∣∣+ f

leM1r +M2g(r) +M0 + f

le r forallt isin R+

(310)

which means that Fx is bounded on R+ and F(Br) sube Br We now prove that

μ(FX) le (M1 +M2Q)μ(X) forallX sube Br (311)

Let X be a nonempty subset of Br Using (32) (36) and (39) we conclude that

∣∣(Fx)(t) minus (Fy)(t)∣∣

=

∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minusf(

t y(a(t))(Hy)(b(t))

intα(t)

0u(t s y(c(s))

)ds

)∣∣∣∣∣


le m1(t)∣∣x(a(t)) minus y(a(t))∣∣ +m2(t)

∣∣(Hx)(b(t)) minus (Hy

)(b(t))

∣∣

+m3(t)intα(t)

0

∣∣u(t s x(c(s))) minus u(t s y(c(s)))∣∣ds

leM1∣∣x(a(t)) minus y(a(t))∣∣ +M2


)(b(t))

∣∣

+ sup

m3(t)intα(t)

0|u(t sw(c(s))) minus u(t s z(c(s)))|ds w z isin Br

forallx y isin X t isin R+

(312)

which yields that

diam(FX)(t) leM1 diamX(a(t)) +M2 diam(HX)(b(t))

+ sup

m3(t)intα(t)


forallt isin R+

(313)

which together with (33) Assumption (H2) and Lemma 24 ensures that

lim suptrarr+infin

diam(FX)(t)

leM1 lim suptrarr+infin

diamX(a(t)) +M2 lim suptrarr+infin

diam(HX)(b(t))

+ lim suptrarr+infin

sup

m3(t)intα(t)


=M1 lim suptrarr+infin

diamX(t) +M2 lim suptrarr+infin

diam(HX)(t)

(314)

that is

lim suptrarr+infin

diam(FX)(t) leM1 lim suptrarr+infin


diam(HX)(t) (315)

For each T gt 0 and ε gt 0 put

M3T = supm3(p) p isin [0 T]

uTr = sup∣∣u(p q v

)∣∣ p isin [0 T] q isin [0 α(T)] v isin [minusr r]ωTr (u ε) = sup

∣∣u(p τ v

) minus u(q τ v)∣∣ p q isin [0 T]∣∣p minus q∣∣ le ε τ isin [0 α(T)] v isin [minusr r]

ωT (u ε r)=sup∣∣u(p q v

) minus u(p qw)∣∣ p isin [0 T] q isin [0 α(T)] vw isin [minusr r] |vminusw| le ε

ωTr

(f ε g(r)

)= sup

∣∣f(p vw z

) minus f(q vw z)∣∣ p q isin [0 T]∣∣p minus q∣∣ le ε

v isin [minusr r] w isin [minusg(r) g(r)] z isin [minusMM]

(316)


Let T gt 0 ε gt 0 x isin X and t s isin [0 T] with |t minus s| le ε It follows from (H2) that there existaT and bT satisfying

0 le aprime(t) le aT 0 le bprime(t) le bT forallt isin [0 T] (317)

In light of (32) (36) (39) (316) (317) and Lemma 25 we get that

|(Fx)(t) minus (Fx)(s)|

le∣∣∣∣∣f

(


0u(t τ x(c(τ)))dτ

)

minusf(

t x(a(s)) (Hx)(b(s))intα(s)

0u(s τ x(c(τ)))dτ

)∣∣∣∣∣

+

∣∣∣∣∣f

(



)

minusf(

s x(a(s)) (Hx)(b(s))intα(s)


)∣∣∣∣∣

le m1(t)|x(a(t)) minus x(a(s))| +m2(t)|(Hx)(b(t)) minus (Hx)(b(s))|

+m3(t)

[∣∣∣∣∣

intα(t)

α(s)|u(t τ x(c(τ)))|dτ

∣∣∣∣∣+intα(s)

0|u(t τ x(c(τ))) minus u(s τ x(c(τ)))|dτ

]

+ sup∣∣f(p vw z



leM1ωT (x a ε) +M2ω

T ((Hx) b ε)+M3T |α(t) minus α(s)| sup

∣∣u(p τ v

)∣∣ p isin [0 T] τ isin [0 α(T)] v isin [minusr r]

+M3Tα(T) sup∣∣u(p τ v

)minusu(q τ v)∣∣ p q isin [0 T]∣∣pminusq∣∣leε τ isin [0 α(T)] visin[minusr r]

+ωTr

(f ε g(r)

)

leM1ωa(T)(x aTε) +M2ω

b(T)(Hx bTε

)+M3Tω

T (α ε)uTr

+M3Tα(T)ωTr (u ε) +ω

Tr

(f ε g(r)

)

(318)

which implies that

ωT (Fx ε) leM1ωa(T)(x aTε) +M2ω

b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)

) forallT gt 0 ε gt 0 x isin X

(319)


Notice that Assumptions (H1)ndash(H3) imply that the functions α = α(t) f = f(t p q v) andu = u(t y z) are uniformly continuous on the sets [0 T][0 T] times [minusr r] times [minusg(r) g(r)] times[minusMM] and [0 T] times [0 α(T)] times [minusr r] respectively It follows that

limεrarr 0

ωT (α ε) = limεrarr 0

ωTr (u ε) = lim

εrarr 0ωTr

(f ε g(r)

)= 0 (320)

In terms of (319) and (320) we have

ωT0 (FX) leM1ω

a(T)0 (X) +M2ω

b(T)0 (HX) (321)

letting T rarr +infin in the above inequality by Assumption (H2) and Lemma 26 we infer that

ω0(FX) leM1ω0(X) +M2ω0(HX) (322)

By means of (315) (322) and Assumption (H5) we conclude immediately that

μ(FX) = ω0(FX) + lim suptrarr+infin

diam(FX)(t)

leM1ω0(X) +M2ω0(HX) +M1 lim suptrarr+infin


diam(HX)(t)

=M1μ(X) +M2

(ω0(HX) + lim sup

trarr+infindiam(HX)(t)

)

le (M1 +M2Q)μ(X)(323)

that is (311) holdsNext we prove that F is continuous on the ball Br Let x isin Br and xnnge1 sub Br with

limnrarrinfin xn = x It follows from (33) that for given ε gt 0 there exists a positive constant Tsuch that

sup

m3(t)intα(t)


ltε

3 forallt gt T (324)

Since u = u(t s v) is uniformly continuous in [0 T] times [0 α(T)] times [minusr r] it follows from (316)that there exists δ0 gt 0 satisfying

ωT (u δ r) ltε

1 + 3M3Tα(T) forallδ isin (0 δ0) (325)

By Assumption (H5) and limnrarrinfin xn = x we know that there exists a positive integerN suchthat

(1 +M1)xn minus x +M2Hxn minusHx lt 12minε δ0 foralln gt N (326)


In view of (32) (36) (39) (324)ndash(326) and Assumption (H2) we gain that for any n gt Nand t isin R+

|(Fxn)(t) minus (Fx)(t)|

=

∣∣∣∣∣f

(

t xn(a(t)) (Hxn)(b(t))intα(t)

0u(t s xn(c(s)))ds

)

minusf(


0u(t s x(c(s)))ds

)∣∣∣∣∣

le m1(t)|xn(a(t)) minus x(a(t))| +m2(t)|(Hxn)(b(t)) minus (Hx)(b(t))|

+m3(t)intα(t)

0|u(t s xn(c(s))) minus u(t s x(c(s)))|ds

leM1xn minus x +M2Hxn minusHx

+max

supτgtT

sup

m3(τ)intα(τ)

0

∣∣u(τ s z(c(s))) minus u(τ s y(c(s)))∣∣ds z y isin Br

supτisin[0T]

sup

m3(τ)intα(τ)

0

∣∣u(τ s z(c(s))) minus u(τ s y(c(s)))∣∣ds z

y isin Br∥∥z minus y∥∥ le δ0

2

ltε

2+max

ε

3 supτisin[0T]

m3(τ)intα(τ)

0ωT

(uδ02 r

)ds

ltε

2+max

ε

3

M3Tα(T)ε1 + 3M3Tα(T)

=5ε6

(327)

which yields that

Fxn minus Fx lt ε foralln gt N (328)

that is F is continuous at each point x isin Br Thus Lemma 22 ensures that F has at least one fixed point x = x(t) isin Br Hence (11)

has at least one solution x = x(t) isin Br This completes the proof

Now we discuss (12) under below hypotheses

(H6) h R+ times R2 rarr R is continuous with h(t 0 0) isin BC(R+) and h = sup|h(t 0 0)| t isin

R+(H7) c α R+ rarr R+ are continuous and α is nondecreasing


(H8) there exist two continuous functionsm1 m3 R+ rarr R+ and four positive constantsrMM0 andM1 satisfying (33)ndash(35)

∣∣h(t vw) minus h(t p q)∣∣ le m1(t)

∣∣v minus p∣∣ +m3(t)

∣∣w minus q∣∣

forallt isin R+ v p isin [minusr r] w q isin [minusMM](329)

supm1(t) t isin R+ leM1 (330)

M0 + h le r(1 minusM1) (331)

Theorem 32 Under Assumptions (H3) and (H6)ndash(H8) (12) has at least one solution x = x(t) isinBr Moreover solutions of (12) are asymptotically stable

Proof As in the proof of Theorem 31 we conclude that (12) possesses at least one solutionin Br

Now we claim that solutions of (12) are asymptotically stable Note that rM0 andM1 are positive numbers and h ge 0 it follows from (331) thatM1 lt 1 In terms of (33) weinfer that for given ε gt 0 there exists T gt 0 such that

sup

m3(t)intα(t)


lt ε(1 minusM1) forallt ge T(332)

Let z = z(t) y = y(t) be two arbitrarily solutions of (12) in Br According to (329)ndash(332) wededuce that

∣∣z(t) minus y(t)∣∣

=

∣∣∣∣∣h

(

t z(t)intα(t)

0u(t τ z(c(τ)))dτ

)

minus h(

t y(t)intα(t)

0u(t τ y(c(τ))

)dτ

)∣∣∣∣∣

le m1(t)∣∣z(t) minus y(t)∣∣ +m3(t)

intα(t)

0

∣∣u(t τ z(c(τ))) minus u(t τ y(c(τ)))∣∣dτ

leM1∣∣z(t) minus y(t)∣∣ + sup

m3(t)intα(t)

0|u(t τ v(c(τ))) minus u(t τw(c(τ)))|dτ vw isin Br

lt M1∣∣z(t) minus y(t)∣∣ + ε(1 minusM1) forallt ge T

(333)

which means that

∣∣z(t) minus y(t)∣∣ lt ε (334)

whenever z y are solutions of (12) in Br and t ge T Hence solutions of (12) are asympto-tically stable This completes the proof


Remark 33 Theorems 31 and 32 generalize Theorem 31 in [6] Theorem 2 in [9] Theorem 3in [10] Theorem 1 in [15] Theorem 2 in [16] and Theorem 31 in [19] Examples 41 and 42in the fourth section show that Theorems 31 and 32 substantially extend the correspondingresults in [6 9 10 15 16 19]

4 Examples

In this section we construct two nontrivial examples to support our results

Example 41 Consider the following nonlinear integral equation of Volterra type

x(t) =1

10 + ln3(1 + t3)+tx2(1 + 3t2

)

200(1 + t)+

1

10radic3 + 20t + 3|x(t3)|

+1

9 +radic1 + t

⎛

⎜⎝

int t2

0

s2x(4s) sin(radic

1 + t3s5 minus (t minus s)4x3(4s))

1 + t12 +∣∣∣stx3(4s) minus 3s3 + 7(t minus 2s)2

∣∣∣ x2(4s)ds

⎞

⎟⎠

3

forallt isin R+

(41)

Put

f(t vw z) =1

10 + ln3(1 + t3)+

tv2

200(1 + t)+

1

10radic3 + 20t + |w|

+z3

9 +radic1 + t

u(t s p

)=

s2p sin(radic

1 + t3s5 minus (t minus s)4p3)

1 + t12 +∣∣∣stp3 minus 3s3 + 7(t minus 2s)2

∣∣∣p2

a(t) = 1 + 3t2 b(t) = t3 c(t) = 4t α(t) = t2 forallt s isin R+ vw z p isin R

(42)

Let

risin[9 minus radic

512

9 +

radic51

2

]

M=3 M0=9r10 M1=

r

100 M2=

1300

f =110

Q = 3 m1(t) =rt

100(1 + t) m2(t) =

1(10radic3 + 20t

)2 m3(t) =3M2

9 +radic1 + t

(Hy)(t) = 3y(t) g(t) = 3t forallt isin R+ y isin BC(R+)

(43)

It is easy to verify that (36) and Assumptions (H1)ndash(H3) and (H5) are satisfied Notice that

M1r +M2g(r) +M0 + f le r lArrrArr r isin[9 minus radic

512

9 +

radic51

2

]

∣∣f(t vw z) minus f(t p q y)∣∣


le t

200(1 + t)

∣∣∣v2 minus p2

∣∣∣ +

∣∣∣∣∣

1

10radic3 + 20t + |w|

minus 1

10radic3 + 20t +

∣∣q∣∣

∣∣∣∣∣

+1

9 +radic1 + t

∣∣∣z3 minus y3

∣∣∣ le m1(t)


∣∣w minus q∣∣ +m3(t)

∣∣z minus y∣∣

forallt isin R+ v p isin [minusr r] w q isin [minusg(r) g(r)] z y isin [minusMM]

sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic

1 + t3s5 minus (t minus s)4v3(4s))

1 + t12 +∣∣∣stv3(4s) minus 3s3 + 7(t minus 2s)2

∣∣∣v2(4s)

minuss2w(4s) sin

(radic1 + t3s5 minus (t minus s)4w3(4s)

)

1 + t12 +∣∣∣stw3(4s) minus 3s3 + 7(t minus 2s)2

∣∣∣w2(4s)

∣∣∣∣∣∣∣ds vw isin Br

⎫⎪⎬

⎪⎭

le 3M2

9 +radic1 + t

int t2

0

2s2(1 + t12

)2

[r(1 + t12

)+ r3(r3t3 + 3r6 + 7

(t + 2t2

)2)]ds

le 2M2t6(9 +

radic1 + t)(

1 + t12)2

r(1 + t12

)+ r3[r3t3 + 3r6 + 7

(t + 2t2

)2] minusrarr 0 as t minusrarr +infin

sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)

∣∣∣∣∣∣∣ds t isin R+ v isin Br

⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=rM2

20lt M0

sup

intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=r

6leM

(44)


that is (32)ndash(35) and (37) hold Hence all Assumptions of Theorem 31 are fulfilledConsequently Theorem 31 ensures that (41) has at least one solution x = x(t) isin Br HoweverTheorem 31 in [6] Theorem 2 in [9]Theorem 3 in [10] Theorem 1 in [15] Theorem 2 in [16]and Theorem 31 in [19] are unapplicable for (41)


x(t) =3 + sin4

(radic1 + t2

)

16+x2(t)8 + t2

+1

1 + t

(intradict

0

sx3(1 + s2)

1 + t2 + s cos2(1 + t3s7x2(1 + s2))ds

)2

forallt isin R+

(45)

Put

h(t vw) =3 + sin4

(radic1 + t2

)

16+

v2

8 + t2+

w2

1 + t

u(t s p

)=

sp3

1 + t2 + s cos2(1 + t3s7p2

)

α(t) =radict c(t) = 1 + t2 m1(t) =

18 + t2

m3(t) =4

1 + t forallt s isin R+ vw p isin R

r =12 M = 2 M0 =M1 =

18 h =

14

(46)

It is easy to verify that (330) (331) and Assumptions (H6) and (H7) are satisfied Notice that

∣∣h(t vw) minus h(t p q)∣∣

le 18 + t2


∣∣∣ +

11 + t

∣∣∣w2 minus q2

∣∣∣

le m1(t)∣∣v minus p∣∣ +m3(t)

∣∣z minus y∣∣ forallt isin R+ v p isin [minusr r] w q isin [minusMM]

sup

m3(t)intα(t)

0|u(t s v(c(s))) minus u(t sw(c(s)))|ds x y isin Br

= sup

4

1 + t

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))

minus sw3(1 + s2)

1 + t2 + s cos2(1 + t3s7w2(1 + s2))ds

∣∣∣∣∣ds v w isin Br

le4r3t(1 +

radict + t2

)

(1 + t)(1 + t2)2minusrarr 0 as t minusrarr +infin


sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣

sv3(1 + s2)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds

∣∣∣∣∣ t isin R+ v isin Br

le sup

4r3

1 + t

intradict

0

s

1 + t2ds t isin R

+

le sup

t

4(1 + t)(1 + t2) t isin R

+

le 18=M1

sup

intα(t)


= sup

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

r3intradic

t

0

s

1 + t2ds t isin R

+

=132

lt M

(47)

which yield (33)ndash(35) That is all Assumptions of Theorem 32 are fulfilled ThereforeTheorem 32 guarantees that (45) has at least one solution x = x(t) isin Br Moreover solutionsof (45) are asymptotically stable But Theorem 2 in [9] Theorem 3 in [10] Theorem 1 in [15]Theorem 2 in [16] and Theorem 31 in [19] are invalid for (45)

Acknowledgments

The authors would like to thank the referees for useful comments and suggestions Thisresearch was supported by the Science Research Foundation of Educational Department ofLiaoning Province (L2012380)

References

[1] R P Agarwal D OrsquoRegan and P J Y Wong Positive Solutions of Differential Difference and IntegralEquations Kluwer Academic Publishers Dordrecht The Netherlands 1999

[2] T A Burton Volterra Integral and Differential Equations vol 167 Academic Press Orlando Fla USA1983

[3] D OrsquoRegan andM Meehan Existence Theory for Nonlinear Integral and Integrodifferential Equations vol445 Kluwer Academic Publishers Dordrecht The Netherlands 1998

[4] M R Arias R Benıtez and V J Bolos ldquoNonconvolution nonlinear integral Volterra equations withmonotone operatorsrdquo Computers and Mathematics with Applications vol 50 no 8-9 pp 1405ndash14142005


[5] J Banas and A Chlebowicz ldquoOn existence of integrable solutions of a functional integral equationunder Caratheodory conditionsrdquo Nonlinear Analysis vol 70 no 9 pp 3172ndash3179 2009

[6] J Banas and B C Dhage ldquoGlobal asymptotic stability of solutions of a functional integral equationrdquoNonlinear Analysis vol 69 no 7 pp 1945ndash1952 2008

[7] J Banas and K Goebel ldquoMeasures of noncompactness in banach spacesrdquo in Lecture Notes in Pure andApplied Mathematics vol 60 Marcel Dekker New York NY USA 1980

[8] J Banas J Rocha and K B Sadarangani ldquoSolvability of a nonlinear integral equation of Volterratyperdquo Journal of Computational and Applied Mathematics vol 157 no 1 pp 31ndash48 2003

[9] J Banas and B Rzepka ldquoOn existence and asymptotic stability of solutions of a nonlinear integralequationrdquo Journal of Mathematical Analysis and Applications vol 284 no 1 pp 165ndash173 2003

[10] J Banas and B Rzepka ldquoAn application of a measure of noncompactness in the study of asymptoticstabilityrdquo Applied Mathematics Letters vol 16 no 1 pp 1ndash6 2003

[11] J Banas and B Rzepka ldquoOn local attractivity and asymptotic stability of solutions of a quadraticVolterra integral equationrdquo Applied Mathematics and Computation vol 213 no 1 pp 102ndash111 2009

[12] T A Burton and B Zhang ldquoFixed points and stability of an integral equation nonuniquenessrdquoApplied Mathematics Letters vol 17 no 7 pp 839ndash846 2004

[13] A Constantin ldquoMonotone iterative technique for a nonlinear integral equationrdquo Journal ofMathematical Analysis and Applications vol 205 no 1 pp 280ndash283 1997

[14] B C Dhage ldquoLocal asymptotic attractivity for nonlinear quadratic functional integral equationsrdquoNonlinear Analysis vol 70 no 5 pp 1912ndash1922 2009

[15] W G El-Sayed ldquoSolvability of a neutral differential equation with deviated argumentrdquo Journal ofMathematical Analysis and Applications vol 327 no 1 pp 342ndash350 2007

[16] X L Hu and J R Yan ldquoThe global attractivity and asymptotic stability of solution of a nonlinearintegral equationrdquo Journal of Mathematical Analysis and Applications vol 321 no 1 pp 147ndash156 2006

[17] J S Jung ldquoAsymptotic behavior of solutions of nonlinear Volterra equations and mean pointsrdquoJournal of Mathematical Analysis and Applications vol 260 no 1 pp 147ndash158 2001

[18] Z Liu and S M Kang ldquoExistence of monotone solutions for a nonlinear quadratic integral equationof Volterra typerdquo The Rocky Mountain Journal of Mathematics vol 37 no 6 pp 1971ndash1980 2007

[19] Z Liu and S M Kang ldquoExistence and asymptotic stability of solutions to a functional-integralequationrdquo Taiwanese Journal of Mathematics vol 11 no 1 pp 187ndash196 2007

[20] D OrsquoRegan ldquoExistence results for nonlinear integral equationsrdquo Journal of Mathematical Analysis andApplications vol 192 no 3 pp 705ndash726 1995

[21] C A Roberts ldquoAnalysis of explosion for nonlinear Volterra equationsrdquo Journal of Computational andApplied Mathematics vol 97 no 1-2 pp 153ndash166 1998

[22] M A Taoudi ldquoIntegrable solutions of a nonlinear functional integral equation on an unboundedintervalrdquo Nonlinear Analysis vol 71 no 9 pp 4131ndash4136 2009

[23] G Darbo ldquoPunti uniti in trasformazioni a codominio non compattordquo Rendiconti del SeminarioMatematico della Universita di Padova vol 24 pp 84ndash92 1955


Research ArticleExistence of Solutions for FractionalIntegro-Differential Equation with MultipointBoundary Value Problem in Banach Spaces

Yulin Zhao1 Li Huang1 Xuebin Wang1 and Xianyang Zhu2

1 School of Science Hunan University of Technology Zhuzhou 412007 China2 Department of Mathematics Jinggangshan Uiversity Jirsquoan 343009 China

Correspondence should be addressed to Yulin Zhao zhaoylchsinacom



Copyright q 2012 Yulin Zhao et al This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

By means of the fixed-point theorem in the cone of strict-set-contraction operators we considerthe existence of a nonlinear multi-point boundary value problem of fractional integro-differentialequation in a Banach space In addition an example to illustrate the main results is given

1 Introduction

The purpose of this paper is to establish the existence results of positive solution to nonlinearfractional boundary value problem

Dq

0+u(t) + f(t u uprime u(nminus2) Tu Su

)= θ 0 lt t lt 1 n minus 1 lt q le n

u(0) = uprime(0) = middot middot middot = u(nminus2)(0) = θ u(nminus2)(1) =mminus2sum

i=1

aiu(nminus2)(ηi

) (11)

in a Banach space E where θ is the zero element of E and n ge 2 0 lt η1 lt middot middot middot lt ηmminus2 lt 1 ai gt0 (i = 1 2 m minus 2) Dα

0+ is Riemann-Liouville fractional derivative and

Tu(t) =int t

0K(t s)u(s)ds Su(t) =

int1

0H(t s)u(s)ds (12)


where K isin C[BR+] B = (t s) isin I times I t ge s H isin C[I times I R+] I = [0 1] and R+ denotesthe set of all nonnegative numbers

Fractional differential equations have gained importance due to their numerousapplications in many fields of science and engineering including fluid flow rheologydiffusive transport akin to diffusion electrical networks and probability For details see [1ndash3]and the references therein In recent years there are some papers dealing with the existenceof the solutions of initial value problems or linear boundary value problems for fractionaldifferential equations by means of techniques of nonlinear analysis (fixed-point theoremsLeray-Schauder theory lower and upper solutions method and so forth) see for example[4ndash23]

In [8] by means of the fixed-point theorem for the mixed monotone operator theauthors considers unique existence of positive to singular boundary value problems forfractional differential equation

Dq

0+u(t) + a(t)f(t u uprime u(nminus2)

)= 0 0 lt t lt 1

u(0) = uprime(0) = middot middot middot = u(nminus2)(0) = u(nminus2)(1) = 0(13)

where Dq

0+ is Riemann-Liouville fractional derivative of order n minus 1 lt q le n n ge 2In [11] El-Shahed and Nieto study the existence of nontrivial solutions for a multi-

point boundary value problem for fractional differential equations

Dq

0+u(t) + f(t u(t)) = 0 0 lt t lt 1 n minus 1 lt q le n n isinN

u(0) = uprime(0) = middot middot middot = u(nminus2)(0) = 0 u(1) =mminus2sum

i=1

aiu(ηi)

(14)

where n ge 2 ηi isin (0 1) ai gt 0 (i = 1 2 m minus 2) and Dq

0+ is Riemann-Liouville fractionalderivative Under certain growth conditions on the nonlinearity several sufficient conditionsfor the existence of nontrivial solution are obtained by using Leray-Schauder nonlinearalternative And then Goodrich [24] was concerned with a partial extension of the problem(13)

Dq

0+u(t) = f(t u(t)) 0 lt t lt 1 n minus n lt q le n minus 2

u(i)(0) = 0 0 le i le n minus 2 Dp

0+u(1) = 0 1 le p le n minus 2(15)

and the authors derived the Green function for the problem (15) and showed that it satisfiescertain properties

By the contraction mapping principle and the Krasnoselskiirsquos fixed-point theoremZhou andChu [13] discussed the existence and uniqueness results for the following fractionaldifferential equation with multi-point boundary conditions

CDq

0+u(t) + f(t uKu Su) = 0 0 lt t lt 1 1 lt q lt 2

a1u(0) minus b1uprime(0) = d1u(η1) a2u(1) minus b2uprime(1) = d2u

(η2)

(16)


where CDq

0+ is the Caputorsquos fractional derivative a1 a2 b1 b2 d1 and d2 are real numbers0 lt η1 and η2 lt 1

In [20] Stanek has discussed the existence of positive solutions for the singular frac-tional boundary value problem

Dqu(t) + f(t u uprime Dpu

)= 0 2 lt q lt 3 0 lt p lt 1

u(0) = 0 uprime(0) = uprime(1) = 0(17)

However to the best of the authorrsquos knowledge a few papers can be found in theliterature dealing with the existence of solutions to boundary value problems of fractionaldifferential equations in Banach spaces In [25] Salem investigated the existence of Pseudosolutions for the following nonlinearm-point boundary value problem of fractional type

Dq

0+u(t) + a(t)f(t u(t)) = 0 0 lt t lt 1


i=1

ζiu(ηi) (18)

in a reflexive Banach space E whereDq

0+ is the Pseudo fractional differential operator of ordern minus 1 lt q le n n ge 2

In [26] by the monotone iterative technique and monch fixed-point theorem Lvet al investigated the existence of solution to the following Cauchy problems for differentialequation with fractional order in a real Banach space E

CDqu(t) = f(t u(t)) u(0) = u0 (19)

where CDqu(t) is the Caputorsquos derivative order 0 lt q lt 1

By means of Darborsquos fixed-point theorem Su [27] has established the existence resultof solutions to the following boundary value problem of fractional differential equation onunbounded domain [0+infin)

Dq

0+u(t) = f(t u(t)) t isin [0+infin) 1 lt q le 2

u(0) = θ Dqminus10+ u(infin) = uinfin

(110)

in a Banach space E Dq

0+ is the Riemann-Liouville fractional derivativeMotivated by the above mentioned papers [8 13 24 25 27 28] but taking a quite

different method from that in [26ndash29] By using fixed-point theorem for strict-set-contractionoperators and introducing a new cone Ω we obtain the existence of at least two positivesolutions for the BVP (11) under certain conditions on the nonlinear term in Banach spacesOur results are different from those of [8 13 24 25 28 30] Note that the nonlinear term fdepends on u and its derivatives uprime uprimeprime u(nminus2)


2 Preliminaries and Lemmas

Let the real Banach space Ewith norm middot be partially ordered by a cone P of E that is u le vif and only if v minus u isin P and P is said to be normal if there exists a positive constantN suchthat θ le u le v implies u le Nv where the smallestN is called the normal constant of P For details on cone theory see [31]

The basic space used in this paper is C[I E] For any u isin C[I E] evidently (C[I E] middotC) is a Banach space with norm uC = suptisinI |u(t)| and P = u isin C[I E] u(t) ge θ for t isin Iis a cone of the Banach space C[I E]

Definition 21 (see [31]) Let V be a bounded set in a real Banach space E and α(V ) = infδ gt0 V = cupmi=1Vi all the diameters of Vi le δ Clearly 0 le α(V ) lt infin α(V ) is called theKuratovski measure of noncompactness

We use α αC to denote the Kuratowski noncompactness measure of bounded sets inthe spaces E C(I E) respectively

Definition 22 (see [31]) Let E1 E2 be real Banach spaces S sub E1 T S rarr E2 is a continuousand bounded operator If there exists a constant k such that α(T(S)) le kα(S) then T is calleda k-set contraction operator When k lt 1 T is called a strict-set-contraction operator

Lemma 23 (see [31]) If D sub C[I E] is bounded and equicontinuous then α(D(t)) is continuouson I and

αC(D) = maxtisinI

α(D(t)) α

(int

I

u(t)dt u isin D)

leint

I

α(D(t))dt (21)

where D(t) = u(t) u isin D t isin I

Definition 24 (see [2 3]) The left-sided Riemann-Liouville fractional integral of order q gt 0of a function y R0

+ rarr R is given by

Iq

0+y(t) =1

Γ(q)int t

0(t minus s)qminus1y(s)ds (22)

Definition 25 (see [2 3]) The fractional derivative of order q gt 0 of a function y R0+ rarr R is

given by

Dq

0+y(t) =1

Γ(n minus q)

(d

dt

)n int t

0(t minus s)nminusqminus1y(s)ds (23)

where n = [q] + 1 [q] denotes the integer part of number q provided that the right side ispointwise defined on R0

+

Lemma 26 (see [2 3]) Let q gt 0 Then the fractional differential equation

Dq

0+y(t) = 0 (24)

has a unique solution y(t) = c1tqminus1 + c2tqminus2 + middot middot middot + cntqminusn ci isin R i = 1 2 n here n minus 1 lt q le n


Lemma 27 (see [2 3]) Let q gt 0 Then the following equality holds for y isin L(0 1) Dq

0+y isinL(0 1)

Iq

0+Dq

0+y(t) = y(t) + c1tqminus1 + c2tqminus2 + middot middot middot + cNtqminusN (25)

for some ci isin R i = 1 2 N hereN is the smallest integer greater than or equal to q

Lemma 28 (see [31]) Let K be a cone in a Banach space E Assume that Ω1Ω2 are open subsetsof E with 0 isin Ω1 Ω1 sub Ω2 If T K cap (Ω2 Ω1) rarr K is a strict-set-contraction operator such thateither

(i) Tx le x x isin K cap partΩ1 and Tx ge x x isin K cap partΩ2 or

(ii) Tx ge x x isin K cap partΩ1 and Tx le x x isin K cap partΩ2

then T has a fixed point in K cap (Ω2 Ω1)

3 Main Results

For convenience we list some following assumptions

(H1) There exist a isin C[I R+] and h isin C[Rn+1+ R+] such that

∥∥f(t u1 un+1)∥∥ le a(t)h(u1 un+1) forallt isin I uk isin P k = 1 n + 1 (31)

(H2) f I times Pn+1r rarr P for any r gt 0 f is uniformly continuous on I times Pn+1r and thereexist nonnegative constants Lk k = 1 n + 1 with

2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1

Lk(n minus 2 minus k) + Lnminus1 +

alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

lt 1 (32)

such that

α(f(tD1 D2 Dn+1)

) len+1sum

k=1

Lkα(Dk) forallt isin I bounded sets Dk isin Pr (33)

where alowast = maxK(t s) (t s) isin B blowast = maxH(t s) (t s) isin I times I Pr = u isin P u ler ηlowast = 1 minussummminus2

i=1 aiηqminusn+1i

Lemma 31 Given y isin C[I E] and 1 minussummminus2i=1 aiη

qminusn+1i = 0 hold Then the unique solution of

Dqminusn+20+ x(t) + y(t) = 0 0 lt t lt 1 n minus 1 lt q le n n ge 2

x(0) = 0 x(1) =mminus2sum

i=1

aix(ηi)

(34)


is

x(t) =int1

0G(t s)y(s)ds (35)

where

G(t s) = g(t s) +summminus2

i=1 aig(ηi s)

ηlowasttqminusn+1 (36)

g(t s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(t(1 minus s))qminusn+1 minus (t minus s)qminusn+1Γ(q minus n + 2

) s le t(t(1 minus s))qminusn+1Γ(q minus n + 2

) t le s(37)

Proof Deduced from Lemma 27 we have

x(t) = minusIqminusn+20+ y(t) + c1tqminusn+1 + c2tqminusn (38)

for some c1 c2 isin R Consequently the general solution of (34) is

x(t) = minusint t

0

(t minus s)qminusn+1Γ(q minus n + 2

)y(s)ds + c1tqminusn+1 + c2tqminusn (39)

By boundary value conditions x(0) = 0 x(1) =summminus2

i=1 aix(ηi) there is c2 = 0 and

c1 =1

1 minussummminus2i=1 aiη

qminusn+1i

int1

0

(1 minus s)qminusn+1Γ(q minus n + 2

)y(s)ds minussummminus2

i=1 ai


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

(310)

Therefore the solution of problem (34) is

x(t) = minusint t

0


)y(s)ds +tqminusn+1

1 minussummminus2i=1 aiηiqminusn+1

int1

0


)y(s)ds

minussummminus2

i=1 aitqminusn+1


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

=int1

0G(t s)y(s)ds

(311)

The proof is complete

Moreover there is one paper [8] in which the following statement has been shown


Lemma 32 The function g(t s) defined in (37) satisfying the following properties

(1) g(t s) ge 0 is continuous on [0 1] times [0 1] and g(t s) le tqminusn+1Γ(q minus n + 2) g(t s) leg(s s) for all 0 le t s le 1

(2) there exists a positive function ρ0 isin C(0 1) such that minγletleδ g(t s) ge ρ0(s)g(t s) s isin(0 1) where 0 lt γ lt δ lt 1 and

ρ0(s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(δ(1 minus s))qminusn+1 minus (δ minus s)qminusn+1(s(1 minus s))qminusn+1

s isin (0 ξ]

(γ

s

)qminusn+1 s isin [ξ 1)

(312)

where γ lt ξ lt δ is the solution of

(δ(1 minus ξ))qminusn+1 minus (δ minus ξ)qminusn+1 = (γ(1 minus ξ))qminusn+1 (313)

For our purpose one assumes that

(H3) ηlowast = 1 minussummminus2i=1 aiη

qminusn+1i gt 0 and 0 lt γ le min2δ minus 1 δ2 23 le δ lt 1 where γ δ are

the constants in (2) of Lemma 32

Remark 33 We note that if (H3) holds then the function G(t s) defined in (34) is satisfyingthe following properties

(i) G(t s) ge 0 is continuous on [0 1] times [0 1] and G(t s) le Δtqminusn+1 for all 0 le t s le 1where Δminus1 = ηlowastΓ(q minus n + 2)

(ii) G(t s) le G(s) for all 0 le t s le 1 where

G(s) = g(s s) +summminus2

i=1 aig(ηi s)

ηlowast (314)

Indeed it is obvious from (1) of Lemma 32 and (36) that

G(t s) le g(s s) + 1ηlowast

mminus2sum

i=1

aig(ηi s)= G(s)

le tqminusn+1

Γ(q minus n + 2

) +summminus2

i=1 aiηqminusn+1i

ηlowastΓ(q minus n + 2

) tqminusn+1 le Δtqminusn+1

(315)

Lemma 34 Let u(t) = Inminus20+ x(t) x isin C[I E] Then the problem (11) can be transformed into thefollowing modified problem

Dqminusn+20+ x(t) + f

(s Inminus20+ x(s) I10+x(s) x(s) T

(Inminus20+ x(s)

) S(Inminus20+ x(s)

))= θ

x(0) = θ x(1) =mminus2sum

i=1

aix(ηi)

(316)


where 0 lt t lt 1 n minus 1 lt q le n n ge 2 Moreover if x isin C[I E] is a solutions of problem (316) thenthe function u(t) = Inminus20+ x(t) is a solution of (11)

The proof follows by routine calculationsTo obtain a positive solution we construct a cone Ω by

Ω =x(t) isin P x(t) ge λ

3x(s) t isin Ilowast s isin I

(317)

where P = x isin C[I E] x(t) ge θ t isin I λ = minminγletleδρ(t) γqminusn+1 Ilowast = [γ δ]Let

(Ax)(t) =int1

0G(t s)f

(s Inminus20+ x(s) x(s) T

(Inminus20+ x(s)

) S(Inminus20+ x(s)

))ds 0 le t le 1 (318)

Lemma 35 Assume that (H1)ndash(H3) hold Then A Ω rarr Ω is a strict-set-contraction operator

Proof Let x isin Ω Then it follows from Remark 33 that

(Ax)(t) leint1

0G(s)f

(s Inminus20+ xj(s) xj(s) T

(Inminus20+ xj(s)

) S(Inminus20+ xj(s)

))ds

=

(int γ

0+intδ

γ

+int1

δ

)

G(s)f(s Inminus20+ xj(s) xj(s) T

(Inminus20+ xj(s)


))ds

le 3intδ

γ


(Inminus20+ xj(s)


))ds

(319)

here by (H3) we know that γ lt δ minus γ and δ minus γ gt 1 minus δFrom (36) and (318) we obtain

mintisin[γ δ]

(Ax)(t) = mintisin[γ δ]

int1

0G(t s)f


(Inminus20+ xj(s)


))ds

geintδ

γ

(

g(t s) +tqminusn+1

ηlowast

mminus2sum

i=1

aig(ηi s))

times f(s Inminus20+ xj(s) xj(s) T

(Inminus20+ xj(s)


))ds


geintδ

γ

(

ρ0(s)g(s s) +γqminusn+1

ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds

ge λ

intδ

γ


(Inminus20+ xj(s)


))ds

ge λ

3(Ax)

(tprime) tprime isin I

(320)

which implies that (Ax)(t) isin Ω that is A(Ω) sub ΩNext we prove that A is continuous on Ω Let xj x sub Ω and xj minus xΩ rarr 0 (j rarr

infin) Hence xj is a bounded subset ofΩ Thus there exists r gt 0 such that r = supjxjΩ ltinfinand xΩ le r It is clear that

∥∥(Axj)(t) minus (Ax)(t)

∥∥ =int1

0G(t s)

∥∥∥f(s Inminus20+ xj(s) xj(s) T

(Inminus20+ xj(s)


))

minusf(s Inminus20+ x(s) x(s) T

(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds

le Δtqminusn+1∥∥∥f(s Inminus20+ xj(s) xj(s) T

(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds

(321)

According to the properties of f for all ε gt 0 there exists J gt 0 such that


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ ltε

Δ

(322)

for j ge J for all t isin ITherefore for all ε gt 0 for any t isin I and j ge J we get


∥∥ lt tqminusn+1ε le ε (323)

This implies that A is continuous on ΩBy the properties of continuous of G(t s) it is easy to see that A is equicontinuous

on I


Finally we are going to show that A is a strict-set-contraction operator Let D sub Ω bebounded Then by condition (H1) Lemma 31 implies that αC(AD) = maxtisinI α((AD)(t)) Itfollows from (318) that

α((AD)(t)) le α(coG(t s)f


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))

s isin [0 t] t isin I x isin D)

le Δ middot α(f(s Inminus20+ x(s) x(s) T

(Inminus20+ x(s)

) S(Inminus20+ x(s)

))


le Δ middot α(f(I times(Inminus20+ D

)(I) times middot middot middot timesD(I) times T

(Inminus20+ D

)(I) times S

(Inminus20+ D

)(I)))

le Δ middotnminus2sum

k=1

Lkα((Inminus1minusk0+ D

)(I))+ Lnminus1α(D(I)) + alowastLnα

(T(Inminus20+ D

)(I))

+blowastLn+1α(S(Inminus20+ D

)(I))

(324)

which implies

αC(AD) le Δ middotnminus2sum

k=1


)(I))+ Lnminus1α(D(I))

+alowastLnα(T(Inminus20+ D

)(I))+ blowastLn+1α

(S(Inminus20+ D

)(I))

(325)

Obviously

α(Inminus1minusk0+ D

)(I) = α

(int s

0

(s minus τ)nminus2minusk(n minus 2 minus k)x(τ)dτ τ isin [0 s] s isin I k = 1 n minus 2

)

le 1(n minus 2 minus k)α(D(I))

(326)

α(T(Inminus20+ D

))(I) = α

(int t

0K(t s)

(ints

0

(s minus τ)nminus2(n minus 3)

u(τ)dτ

)

ds u isin D t isin I)

le alowast

(n minus 3)α(u(t) t isin I u isin D) le alowast

(n minus 3)α(D(I))

(327)

α(S(Inminus20+ D

))(I) = α

(int1

0H(t s)

(int s

0


u(τ)dτ

)


le blowast

(n minus 3)α(u(t) t isin I u isin D) le blowast

(n minus 3)α(D(I))

(328)


Using a similar method as in the proof of Theorem 211 in [31] we have

α(D(I)) le 2αC(D) (329)

Therefore it follows from (326)ndash(329) that

αC(AD) le 2Δ middot(

nminus2sum

k=1


alowastLn(n minus 3)

+blowastLn+1(n minus 3)

)

αC(D) (330)

Noticing that (33) we obtain that T is a strict-set-contraction operator The proof is complete

Theorem 36 Let cone P be normal and conditions (H1)sim(H3) hold In addition assume that thefollowing conditions are satisfied

(H4) There exist ulowast isin P θ c1 isin C[Ilowast R+] and h1 isin C[Pn+1 R+] such that

f(t u1 un+1) ge c1(t)h1(u1 unminus1)ulowast forallt isin Ilowast uk isin P

h1(u1 unminus1)sumnminus1

k=1ukminusrarr infin as

nminus1sum

k=1

uk minusrarr infin uk isin P(331)

(H5) There exist ulowast isin P θ c2 isin C[Ilowast R+] and h2 isin C[Pnminus1 R+] such that



k=1uiminusrarr infin as

nminus1sum

k=1

uk minusrarr 0 uk isin P(332)

(H6) There exists a β gt 0 such that

NMβ

int1

0G(s)a(s)ds lt β (333)

whereMβ = maxukisinPβh(u1 un+1) Then problem (11) has at least two positive solutions

Proof Consider condition (H4) there exists an r1 gt 0 such that

h1(u1 unminus1) ge3N2sumnminus1

k=1ukλ2intδγ G(s)c1(s)ds middot ulowast

foralluk isin Pnminus1sum

k=1

uk ge r1 (334)


Therefore

f(t u1 middot un+1) ge3N2sumnminus1


middot c1(t)ulowast foralluk isin Pnminus1sum

k=1

uk ge r1 (335)

Take

r0 gt max3Nλminus1r1 β

(336)

Then for t isin [γ δ] xΩ = r0 we have by (318)

x(t) ge λ

3NxΩ ge λ

3Nr0 gt r1 (337)

Hence

(Ax)(t) geintδ

γ

G(t s)f(s Inminus20+ x(s) x(s) T

(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2

λintδγ G(s)c1(s)ds middot ulowast

intδ

γ

G(s)

(nminus2sum

k=1

∥∥∥Inminus1minusk0+ x(s)∥∥∥ + x(s)

)

c1(s)ds middot ulowast

ge 3N2


intδ

γ

G(s)c1(s)x(s)ds middot ulowast

ge Nintδγ G(s)c1(s)ds middot ulowast

xΩ(intδ

γ

G(s)c1(s)ds

)

middot ulowast

=1

intδγ G(s)c1(s)ds middot ulowast

(intδ

γ

G(s)c1(s)dsulowast)

middot NxΩulowast ulowast

ge NxΩulowast middot ulowast

(338)

and consequently

AxΩ ge xΩ forallx isin Ω xΩ = r0 (339)

Similarly by condition (H5) there exists r2 gt 0 such that


k=1ukλintδγ G(ξ s)c2(s)ds middot ulowast

foralluk isin P 0 ltnminus1sum

k=1

uk le r2 (340)


where ξ is given in (2) of Lemma 32 Therefore

f(t u1 un+1) ge3N2sumnminus1


middot c2(t)ulowast foralluk isin P 0 ltnminus1sum

k=1

uk le r2 (341)

Choose

0 lt r lt min

⎧⎨

⎩

(nminus2sum

k=0

1k

)minus1r2 β

⎫⎬

⎭ (342)

Then for t isin [γ δ] x isin Ω xΩ = r we have

(Ax)(ξ) =int1

0G(ξ s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

geintδ

γ

G(ξ s)f(s Inminus20+ x(s) x(s) T

(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2

λintδγ G(ξ s)c2(s)ds middot ulowast

intδ

γ

G(ξ s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ

G(ξ s)x(s)c2(s)ds middot ulowast

ge Nintδγ G(ξ s)c2(s)ds middot ulowast

x(s)Ωintδ

γ

G(ξ s)c2(s)ds middot ulowast

=1

intδγ G(ξ s)c2(s)ds middot ulowast

x(s)Ω(intδ

γ

G(ξ s)c2(s)dsulowast)

middot Nx(s)Ωulowast ulowast

ge Nx(s)Ωulowast middot ulowast

(343)

which implies

(Ax)(ξ)Ω ge x(s)Ω forallx isin Ω xΩ = r (344)

that is

AxΩ ge x(s)Ω forallx isin Ω xΩ = r (345)


On the other hand according to (ii) of Remark 33 and (318) we get

(Ax)(t) leint1

0G(s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds (346)

By condition (H1) for t isin I x isin Ω xΩ = β we have

∥∥f(t u1 un+1)

∥∥ le a(t)h(u1 un+1) leMβa(t) (347)

Therefore

(Ax)(t)Ω leNMβ middotint1

0G(s)a(s)ds lt β = xΩ (348)

Applying Lemma 27 to (339) (345) and (348) yields that T has a fixed-point xlowast isinΩrβ r le xlowast le β and a fixed-point xlowastlowast isin Ωβr0 β le xlowastlowast le r0 Noticing (348) we getxlowast= β and xlowastlowast= β This and Lemma 34 complete the proof

Theorem 37 Let cone P be normal and conditions (H1)sim(H4) hold In addition assume that thefollowing condition is satisfied

(H7)

h(u1 un+1)sumn+1

k=1ukminusrarr 0 as uk isin P

n+1sum

k=1

uk minusrarr 0+ (349)

Then problem (11) has at least one positive solution

Proof By (H4) we can choose r0 gt 3Nλminus1r1 As in the proof of Theorem 36 it is easy to seethat (339) holds On the other hand considering (349) there exists r3 gt 0 such that

h(u1 un+1) le ε0n+1sum

k=1

uk for t isin I uk isin P 0 ltn+1sum

k=1

uk le r3 (350)

where ε0 gt 0 satisfies

ε0 =

(

N

nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

int1

0G(s)a(s)ds

)minus1 (351)


Choose 0 lt rlowast lt min(sumnminus1k=1(1k)+(a

lowast+blowast)(nminus3))minus1r3 r0 Then for t isin I x isin Ω xΩ = rlowastit follows from (346) that

(Ax)(t) leNint1

0G(s)∥∥∥f(s Inminus20+ x(s) x(s) T

(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))∥∥∥ds

leNint1

0G(s)a(s)h

(∥∥∥Inminus20+ x(s)

∥∥∥ x(s)

∥∥∥T(Inminus20+ x

)(s)∥∥∥∥∥∥S(Inminus20+ x

)(s)∥∥∥)ds

leNε0

int1

0G(s)a(s)

(nminus1sum

k=1

∥∥∥Inminus1minusk0+ x(s)

∥∥∥ +∥∥∥T(Inminus20+ x

)(s)∥∥∥ +∥∥∥S(Inminus20+ x

)(s)∥∥∥

)

ds

leNε0

(nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

)

rlowastint1

0G(s)a(s)ds = rlowast

(352)

and consequently

(Ax)(t)Ω le xΩ forallx isin Ω xΩ le rlowast (353)

Since 0 lt rlowast lt r0 applying Lemma 27 to (339) and (353) yield that T has a fixed-pointxlowast isin Ωrlowastr0 r

lowast le xlowast le r0 This and Lemma 34 complete the proof

4 An Example

Consider the following system of scalar differential equations of fractional order

minusD52uk(t) =(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎡

⎣u2k(t) + uprime3k(t) +infinsum

j=1

u2j(t) +infinsum

j=1

uprimej(t)

⎤

⎦

3

+

⎛

⎝3uk(t) + 3uprimek+1(t) +infinsum

j=1

uj(t) +infinsum

j=1

uprime2j(t)

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5

(int t

0eminus(1+t)suk(s)ds

)23

+1 + t2

24k4

(int1

0eminusssin2(t minus s)πu2k(s)ds

)

t isin I

uk(0) = uprimek(0) = 0 uprimek(1) =18uprimek

(14

)+12uprimek

(49

) k = 1 2 3

(41)

Conclusion The problem (41) has at least two positive solutions


Proof Let E = l1 = u = (u1 u2 uk ) suminfin

k=1 |uk| lt infin with the norm u =suminfin

k=1 |uk|and P = (u1 uk ) uk ge 0 k = 1 2 3 Then P is a normal cone in E with normalconstant N = 1 and system (41) can be regarded as a boundary value problem of the form(11) In this situation q = 52 n = 3 a1 = 18 a2 = 12 η1 = 14 η2 = 49 ηlowast =2948 K(t s) = eminus(1+t)s H(t s) = eminusssin2(tminus s)π u = (u1 uk ) f = (f1 f2 fk )in which

fk(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛

⎝u2k + v3k +infinsum

j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛

⎝3uk + 3vk+1 +infinsum

j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭+1 + t3

36k5x15k

+1 + t2

24k4y2k

(42)

Observing the inequalitysuminfin

k=1(1k3) lt 32 we get by (42)

∥∥f(t u v x y

)∥∥ =infinsum

k=1

∣∣fk(t u v x y

)∣∣

le (1 + t)3

2

(140

(u + v)3 + 1160

(u + v)12 + 112

x15 + 18∥∥y∥∥)

(43)

Hence (H1) is satisfied for a(t) = (1 + t)32 and

h(u v x y

)=

140

(u + v)3 +1160

(u + v)12 +112x15 +

18y (44)

Now we check condition (H2) Obviously f I times P 4r rarr P for any r gt 0 and f is

uniformly continuous on I times P 4r Let f = f (1) + f (2) where f (1) = (f (1)

1 f(1)k ) and f (2) =

(f (2)1 f

(2)k ) in which

f(1)k

(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5x15k (k = 1 2 3 )

f(2)k

(t u v x y

)=

1 + t2

24k4y2k (k = 1 2 3 )

(45)


For any t isin I and bounded subsets Di sub E i = 1 2 3 4 from (45) and by the diagonalmethod we have

α(f (1)(tD1 D2 D3 D4)

)= 0 forallt isin I bounded sets Di sub E i = 1 2 3 4

α(f (2)(ID1 D2 D3 D4)

)le 1

12α(D4) forallt isin IDi sub E i = 1 2 3 4

(46)

It follows from (46) that

α(f(ID1 D2 D3 D4)

) le 112α(D4) forallt isin IDi sub E i = 1 2 3 4

2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

asymp 01565 lt 1(47)

that is condition (H2) holds for L1 = L2 = L3 = 0 L4 = 112On the other hand take γ = 14 δ = 34 Then 14 = γ le minδ2 2δ minus 1 =

38 23 lt δ which implies that condition (H3) holds By (42) we have

fk(t u v x y

) ge (1 + t)3

960k3(u + v)3 forallt isin Ilowast u v x y isin P (k = 1 2 3 )

fk(t u v x y

) ge (1 + t)3

960k3

radicu + v forallt isin Ilowast u v x y isin P (k = 1 2 3 )

(48)

Hence condition (H4) is satisfied for

c1(t) =(1 + t)3

960 h1 k(u v) = (u + v)3 ulowast =

(1

1k3

) (49)

in this situation

h1 k = limu+vrarrinfin

(u + v)3u + v = infin (410)

And condition (H5) is also satisfied for

c2(t) =(1 + t)3

960 h2 k(u v) =

radicu + v ulowast =

(1

1k3

) (411)

in this situation

h2k = limu+vrarr 0

radicu + vu + v = infin (412)


Finally choose β = 1 it is easy to check that condition (H6) is satisfied In this case Mβ asymp04162 and so

NMβ

int1

0G(s)a(s)ds asymp 07287 lt β = 1 (413)

Hence our conclusion follows from Theorem 36

Acknowledgments

This research is supported by the National Natural Science Foundation of China (1116102411271372 and 11201138) Hunan Provincial Natural Science Foundation of China (12JJ2004)NSF of Jiangxi province (2010GZC0115) and Scientific Research Fund of Hunan ProvincialEducation Department (11C0412)

References

[1] K S Miller and B Ross An Introduction to the Fractional Calculus and Fractional Differential EquationsJohn Wiley amp Sons New York NY USA 1993

[2] I Podlubny Fractional Differential Equations vol 198 of Mathematics in Science and EngineeringAcademic Press London UK 1999

[3] A A Kilbas H M Srivastava and J J Trujillo Theory and Applications of Fractional DifferentialEquations vol 204 Elsevier Science Amsterdam The Netherlands 2006

[4] V Lakshmikantham and A S Vatsala ldquoBasic theory of fractional differential equationsrdquo NonlinearAnalysis vol 69 no 8 pp 2677ndash2682 2008

[5] V Lakshmikantham ldquoTheory of fractional functional differential equationsrdquo Nonlinear Analysis vol69 no 10 pp 3337ndash3343 2008

[6] V Lakshmikantham and A S Vatsala ldquoGeneral uniqueness and monotone iterative technique forfractional differential equationsrdquo Applied Mathematics Letters vol 21 no 8 pp 828ndash834 2008

[7] Z Bai and H Lu ldquoPositive solutions for boundary value problem of nonlinear fractional differentialequationrdquo Journal of Mathematical Analysis and Applications vol 311 no 2 pp 495ndash505 2005

[8] S Zhang ldquoPositive solutions to singular boundary value problem for nonlinear fractional differentialequationrdquo Computers amp Mathematics with Applications vol 59 no 3 pp 1300ndash1309 2010

[9] Z Bai ldquoOn positive solutions of a nonlocal fractional boundary value problemrdquo Nonlinear Analysisvol 72 no 2 pp 916ndash924 2010

[10] C F Li X N Luo and Y Zhou ldquoExistence of positive solutions of the boundary value problem fornonlinear fractional differential equationsrdquo Computers amp Mathematics with Applications vol 59 no 3pp 1363ndash1375 2010

[11] M El-Shahed and J J Nieto ldquoNontrivial solutions for a nonlinear multi-point boundary valueproblem of fractional orderrdquo Computers ampMathematics with Applications vol 59 no 11 pp 3438ndash34432010

[12] Y Zhao H Chen and L Huang ldquoExistence of positive solutions for nonlinear fractional functionaldifferential equationrdquo Computers amp Mathematics with Applications vol 64 no 10 pp 3456ndash3467 2012

[13] W-X Zhou and Y-D Chu ldquoExistence of solutions for fractional differential equations with multi-point boundary conditionsrdquo Communications in Nonlinear Science and Numerical Simulation vol 17no 3 pp 1142ndash1148 2012

[14] G Wang B Ahmad and L Zhang ldquoSome existence results for impulsive nonlinear fractionaldifferential equations with mixed boundary conditionsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1389ndash1397 2011

[15] C Yuan ldquoTwo positive solutions for (nminus1 1)-type semipositone integral boundary value problems forcoupled systems of nonlinear fractional differential equationsrdquo Communications in Nonlinear Scienceand Numerical Simulation vol 17 no 2 pp 930ndash942 2012


[16] B Ahmad and S Sivasundaram ldquoExistence of solutions for impulsive integral boundary valueproblems of fractional orderrdquo Nonlinear Analysis vol 4 no 1 pp 134ndash141 2010

[17] B Ahmad and S Sivasundaram ldquoOn four-point nonlocal boundary value problems of nonlinearintegro-differential equations of fractional orderrdquo Applied Mathematics and Computation vol 217 no2 pp 480ndash487 2010

[18] S Hamani M Benchohra and J R Graef ldquoExistence results for boundary-value problems withnonlinear fractional differential inclusions and integral conditionsrdquo Electronic Journal of DifferentialEquations vol 2010 pp 1ndash16 2010

[19] A Arikoglu and I Ozkol ldquoSolution of fractional integro-differential equations by using fractionaldifferential transform methodrdquo Chaos Solitons and Fractals vol 40 no 2 pp 521ndash529 2009

[20] S Stanek ldquoThe existence of positive solutions of singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 62 no 3 pp 1379ndash1388 2011

[21] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocal boundary value problems of higher-order nonlinear fractional differential equationsrdquo Abstract and Applied Analysis vol 2009 Article ID494720 9 pages 2009

[22] M A Darwish ldquoOn a perturbed functional integral equation of Urysohn typerdquo Applied Mathematicsand Computation vol 218 no 17 pp 8800ndash8805 2012

[23] M A Darwish and S K Ntouyas ldquoBoundary value problems for fractional functional differentialequations of mixed typerdquo Communications in Applied Analysis vol 13 no 1 pp 31ndash38 2009

[24] C S Goodrich ldquoExistence of a positive solution to a class of fractional differential equationsrdquo AppliedMathematics Letters vol 23 no 9 pp 1050ndash1055 2010

[25] A H Salem ldquoOn the fractional order m-point boundary value problem in reflexive banach spacesand weak topologiesrdquo Journal of Computational and Applied Mathematics vol 224 no 2 pp 565ndash5722009

[26] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy problem for differential equations in Banachspaces with fractional orderrdquo Computers amp Mathematics with Applications vol 62 no 3 pp 1303ndash13112011

[27] X Su ldquoSolutions to boundary value problem of fractional order on unbounded domains in a banachspacerdquo Nonlinear Analysis vol 74 no 8 pp 2844ndash2852 2011

[28] Y-L Zhao and H-B Chen ldquoExistence of multiple positive solutions for m-point boundary valueproblems in banach spacesrdquo Journal of Computational and Applied Mathematics vol 215 no 1 pp 79ndash90 2008

[29] Y Zhao H Chen and C Xu ldquoExistence of multiple solutions for three-point boundary-valueproblems on infinite intervals in banach spacesrdquo Electronic Journal of Differential Equations vol 44pp 1ndash11 2012

[30] X Zhang M Feng and W Ge ldquoExistence and nonexistence of positive solutions for a class of nth-order three-point boundary value problems in Banach spacesrdquo Nonlinear Analysis vol 70 no 2 pp584ndash597 2009

[31] D J Guo V Lakshmikantham and X Z Liu Nonlinear Integral Equations in Abstract Spaces vol 373Kluwer Academic Publishers Dordrecht The Netherlands 1996


Research ArticleApplication of Sumudu DecompositionMethod to Solve Nonlinear System of PartialDifferential Equations

Hassan Eltayeb1 and Adem Kılıcman2

1 Mathematics Department College of Science King Saud University PO Box 2455Riyadh 11451 Saudi Arabia

2 Department of Mathematics and Institute for Mathematical Research Universiti Putra MalaysiaSerdang 43400 Selangor Malaysia

Correspondence should be addressed to Adem Kılıcman akilicmanputraupmedumy


Academic Editor M Mursaleen

Copyright q 2012 H Eltayeb and A Kılıcman This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We develop a method to obtain approximate solutions of nonlinear system of partial differentialequations with the help of Sumudu decomposition method (SDM) The technique is based on theapplication of Sumudu transform to nonlinear coupled partial differential equations The nonlinearterm can easily be handled with the help of Adomian polynomials We illustrate this techniquewith the help of three examples and results of the present technique have close agreement withapproximate solutions obtained with the help of Adomian decomposition method (ADM)

1 Introduction

Most of phenomena in nature are described by nonlinear differential equations So scientistsin different branches of science try to solve them But because of nonlinear part of thesegroups of equations finding an exact solution is not easy Different analytical methods havebeen applied to find a solution to them For example Adomian has presented and developeda so-called decomposition method for solving algebraic differential integrodifferentialdifferential-delay and partial differential equations In the nonlinear case for ordinarydifferential equations and partial differential equations the method has the advantage ofdealing directly with the problem [1 2] These equations are solved without transformingthem to more simple ones The method avoids linearization perturbation discretization orany unrealistic assumptions [3 4] It was suggested in [5] that the noise terms appears alwaysfor inhomogeneous equations Most recently Wazwaz [6] established a necessary condition


that is essentially needed to ensure the appearance of ldquonoise termsrdquo in the inhomogeneousequations In the present paper the intimate connection between the Sumudu transformtheory and decomposition method arises in the solution of nonlinear partial differentialequations is demonstrated

The Sumudu transform is defined over the set of the functions

A =f(t) existMτ1 τ2 gt 0

∣∣f(t)

∣∣ lt Metτj if t isin (minus1)j times [0infin)

(11)

by the following formula

G(u) = S[f(t)u

]

=intinfin

0f(ut)eminustdt u isin (minusτ1 τ2)

(12)

The existence and the uniqueness were discussed in [7] for further details andproperties of the Sumudu transform and its derivatives we refer to [8] In [9] somefundamental properties of the Sumudu transform were established

In [10] this new transform was applied to the one-dimensional neutron transportequation In fact one can easily show that there is a strong relationship between doubleSumudu and double Laplace transforms see [7]

Further in [11] the Sumudu transform was extended to the distributions and some oftheir properties were also studied in [12] Recently Kılıcman et al applied this transform tosolve the system of differential equations see [13]

A very interesting fact about Sumudu transform is that the original function and itsSumudu transform have the same Taylor coefficients except a factor n Thus if

f(t) =infinsum

n=0

antn (13)

then

F(u) =infinsum

n=0

nantn (14)

see [14]Similarly the Sumudu transform sends combinations C(mn) into permutations

P(mn) and hence it will be useful in the discrete systems Further

S(H(t)) = pound(δ(t)) = 1

pound(H(t)) = S(δ(t)) =1u

(15)

Thus we further note that since many practical engineering problems involvemechanical or electrical systems where action is defined by discontinuous or impulsive


forcing terms Then the Sumudu transform can be effectively used to solve ordinarydifferential equations as well as partial differential equations and engineering problemsRecently the Sumudu transform was introduced as a new integral transform on a timescale T to solve a system of dynamic equations see [15] Then the results were applied onordinary differential equations when T = R difference equations when T = N0 but also forq-difference equations when T = qN0 where qN0 = qt t isin N0 for q gt 1 or T = qZ = qZ cup 0for q gt 1 which has important applications in quantum theory and on different types of timescales like T = hN0 T = N

20 and T = Tn the space of the harmonic numbers During this study

we use the following Sumudu transform of derivatives

Theorem 11 Let f(t) be in A and let Gn(u) denote the Sumudu transform of the nth derivativefn(t) of f(t) then for n ge 1

Gn(u) =G(u)un

minusnminus1sum

k=0

f (k)(0)unminusk

(16)

For more details see [16]

We consider the general inhomogeneous nonlinear equation with initial conditionsgiven below

LU + RU +NU = h(x t) (17)

where L is the highest order derivative which is assumed to be easily invertible R is a lineardifferential operator of order less than LNU represents the nonlinear terms and h(x t) is thesource term First we explain themain idea of SDM themethod consists of applying Sumudutransform

S[LU] + S[RU] + S[NU] = S[h(x t)] (18)

Using the differential property of Laplace transform and initial conditions we get

1unS[U(x t)] minus 1

unU(x 0) minus 1

unminus1Uprime(x 0) minus middot middot middot minus Unminus1(x 0)

u+ S[RU] + S[NU] = S[h(x t)]

(19)

By arrangement we have

S[U(x t)] = U(x 0) + uUprime(x 0) + middot middot middot + unminus1Unminus1(x 0) minus unS[RU] minus unS[NU] + unS[h(x t)](110)

The second step in Sumudu decomposition method is that we represent solution as aninfinite series

U(x t) =infinsum

i=0

Ui(x t) (111)


and the nonlinear term can be decomposed as

NU(x t) =infinsum

i=0

Ai (112)

where Ai are Adomian polynomials [6] of U0 U1 U2 Un and it can be calculated byformula

Ai =1idi

dλi

[

Ninfinsum

i=0

λiUi

]

λ=0

i = 0 1 2 (113)

Substitution of (111) and (112) into (110) yields

S

[ infinsum

i=0

Ui(x t)

]

= U(x 0) + uUprime(x 0) + middot middot middot + unminus1Unminus1(x 0) minus unS[RU(x t)]

minus unS[ infinsum

i=0

Ai

]

+ unS[h(x t)]

(114)

On comparing both sides of (114) and by using standard ADM we have

S[U0(x t)] = U(x 0) + uUprime(x 0) + middot middot middot + unminus1Unminus1(x 0) + unS[h(x t)] = Y (x u) (115)

then it follows that

S[U1(x t)] = minusunS[RU0(x t)] minus unS[A0]

S[U2(x t)] = minusunS[RU1(x t)] minus unS[A1](116)

In more general we have

S[Ui+1(x t)] = minusunS[RUi(x t)] minus unS[Ai] i ge 0 (117)

On applying the inverse Sumudu transform to (115) and (117) we get

U0(x t) = K(x t)

Ui+1(x t) = minusSminus1[unS[RUi(x t)] + unS[Ai]] i ge 0(118)

where K(x t) represents the term that is arising from source term and prescribed initialconditions On using the inverse Sumudu transform to h(x t) and using the given conditionwe get

Ψ = Φ + Sminus1[h(x t)] (119)


where the function Ψ obtained from a term by using the initial condition is given by

Ψ = Ψ0 + Ψ1 + Ψ2 + Ψ3 + middot middot middot + Ψn (120)

the termsΨ0Ψ1Ψ2Ψ3 Ψn appears while applying the inverse Sumudu transform on thesource term h(x t) and using the given conditions We define

U0 = Ψk + middot middot middot + Ψk+r (121)

where k = 0 1 n r = 0 1 n minus k Then we verify that U0 satisfies the original equation(17) We now consider the particular form of inhomogeneous nonlinear partial differentialequations

LU + RU +NU = h(x t) (122)


U(x 0) = f(x) Ut(x 0) = g(x) (123)

where L = part2partt2 is second-order differential operator NU represents a general non-lineardifferential operator where as h(x t) is source term The methodology consists of applyingSumudu transform first on both sides of (110) and (123)

S[U(x t)] = f(x) + ug(x) minus u2S[RU] minus u2S[NU] + u2S[h(x t)] (124)

Then by the second step in Sumudu decomposition method and inverse transform asin the previous we have

U(x t) = f(x) + tg(t) minus Sminus1[u2S[RU] minus u2S[NU]

]+ Sminus1

[u2S[h(x t)]

] (125)

2 Applications

Now in order to illustrate STDM we consider some examples Consider a nonlinear partialdifferential equation

Utt +U2 minusU 2x = 0 t gt 0 (21)

with initial conditions

U(x 0) = 0

Ut(x 0) = ex(22)


By taking Sumudu transform for (21) and (22)we obtain

S[U(x t)] = uex + u2S[U2x minusU2

] (23)

By applying the inverse Sumudu transform for (23) we get

[U(x t)] = tex + Sminus1[u2S

[U2x minusU2

]](24)

which assumes a series solution of the function U(x t) and is given by

U(x t) =infinsum

i=0

Ui(x t) (25)

Using (24) into (25)we get

infinsum

i=0

Ui(x t) = tex + Sminus1[

u2S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

(26)

In (26) Ai(u) and Bi(u) are Adomian polynomials that represents nonlinear terms SoAdomian polynomials are given as follows

infinsum

i=0

Ai(U) = U2x

infinsum

i=0

Ai(U) = U2

(27)

The few components of the Adomian polynomials are given as follows

A0(U) = U20x A1(U) = 2U0xU1x Ai(U) =

isum

r=0

UrxUiminusrx

B0(U) = U20 B1(U) = 2U0U1 Bi(U)

isum

r=0

UrUiminusr

(28)

From the above equations we obtain

U0(x t) = tex

Ui+1(x t) = Sminus1[

S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

n ge 0(29)


Then the first few terms ofUi(x t) follow immediately upon setting

U1(x t) = Sminus1[

u2S

[ infinsum

i=0

A0(U) minusinfinsum

i=0

B0(U)

]]

= Sminus1[u2S

[U2

0x minusU20

]]= Sminus1

[u2S

[t2e2x minus t2e2x

]]

= Sminus1[u2S[0]

]= 0

(210)

Therefore the solution obtained by LDM is given as follows

U(x t) =infinsum

i=0

Ui(x t) = tex (211)

Example 21 Consider the system of nonlinear coupled partial differential equation

Ut

(x y t

) minus VxWy = 1

Vt(x y t

) minusWxUy = 5

Wt

(x y t

) minusUxVy = 5

(212)


U(x y 0

)= x + 2y

V(x y 0

)= x minus 2y

W(x y 0

)= minusx + 2y

(213)

Applying the Sumudu transform (denoted by S)we have

U(x y u

)= x + 2y + u + uS

[VxWy

]

V(x y u

)= x minus 2y + 5u + uS

[WxUy

]

W(x y u

)= minusx + 2y + 5u + uS

[UxVy

]

(214)

On using inverse Sumudu transform in (214) our required recursive relation is givenby

U(x y t

)= x + 2y + t + Sminus1[uS

[VxWy

]]

V(x y t

)= x minus 2y + 5t + Sminus1[uS

[WxUy

]]

U(x y t

)= minusx + 2y + 5t + Sminus1[uS

[UxVy

]]

(215)


The recursive relations are

U0(x y t

)= t + x + 2y

Ui+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ci(VW)

]]

i ge 0

V0(x y t

)= 5t + x minus 2y

Vi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Di(UW)

]]

i ge 0

W0(x y t

)= 5t minus x + 2y

Wi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ei(UV )

]]

i ge 0

(216)

where Ci(VW) Di(UW) and Ei(UV ) are Adomian polynomials representing the nonlin-ear terms [1] in above equations The few components of Adomian polynomials are given asfollows

C0(VW) = V0xW0y

C1(VW) = V1xW0y + V0xW1y

Ci(VW) =isum

r=0

VrxWiminusry

D0(UW) = U0yW0x

D1(UW) = U1yW0x +W1xU0y

Di(UW) =isum

r=0

WrxUiminusry

E0(UV ) = U0xV0y

E1(UV ) = U1xV0y +U0xV1y

Ei(VW) =isum

r=0

UrxViminusry

(217)


By this recursive relation we can find other components of the solution

U1(x y t

)= Sminus1[uS[C0(VW)]] = Sminus1[uS

[V0xW0y

]]= Sminus1[uS[(1)(2)]] = 2t

V1(x y t

)= Sminus1[uS[D0(UW)]] = Sminus1[uS

[W0xU0y

]]= Sminus1[uS[(minus1)(2)]] = minus2t

W1(x y t

)= Sminus1[uS[E0(UV )]] = Sminus1[uS

[U0xV0y

]]= Sminus1[uS[(1)(minus2)]] = minus2t

U2(x y t


[V1xW0y + V0xW1y

]]= 0

V2(x y t


[U0yW1x +U1yW0x

]]= 0

W2(x y t

)= Sminus1[uS[D1(UV )]] = Sminus1[uS

[U1xV0y +U0xV1y

]]= 0

(218)

The solution of above system is given by

U(x y t

)=

infinsum

i=0

Ui

(x y t

)= x + 2y + 3t

V(x y t

)=

infinsum

i=0

Vi(x y t

)= x minus 2y + 3t

W(x y t

)=

infinsum

i=0

Wi

(x y t

)= minusx + 2y + 3t

(219)

Example 22 Consider the following homogeneous linear system of PDEs

Ut(x t) minus Vx(x t) minus (U minus V ) = 2

Vt(x t) +Ux(x t) minus (U minus V ) = 2(220)


U(x 0) = 1 + ex V (x 0) = minus1 + ex (221)

Taking the Sumudu transform on both sides of (220) then by using the differentiationproperty of Sumudu transform and initial conditions (221) gives

S[U(x t)] = 1 + ex minus 2u + uS[Vx] + uS[U minus V ]

S[V (x t)] = minus1 + ex minus 2u minus uS[Ux] + uS[U minus V ](222)

Ux(x t) =infinsum

i=0

Uxi(x t) Vx(x t) =infinsum

i=0

Vxi(x t) (223)


Using the decomposition series (223) for the linear terms U(x t) V (x t) and Ux Vxwe obtain

S

[ infinsum

i=0

Ui(x t)

]

= 1 + ex minus 2u + uS

[ infinsum

i=0

Vix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

S

[ infinsum

i=0

Vi(x t)

]

= minus1 + ex minus 2u minus uS[ infinsum

i=0

Uix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

(224)

The SADM presents the recursive relations

S[U0(x t)] = 1 + ex minus 2u

S[V0(x t)] = minus1 + ex minus 2u

S[Ui+1] = uS[Vix] + uS[Ui minus Vi] i ge 0

S[Vi+1] = minusuS[Uix] + uS[Ui minus Vi] i ge 0

(225)

Taking the inverse Sumudu transform of both sides of (225) we have

U0(x t) = 1 + ex minus 2t

V0(x t) = minus1 + ex minus 2t

U1 = Sminus1[uS[V0x] + uS[U0 minus V0]] = Sminus1[uex + 2u] = tex + 2t

V1 = Sminus1[minusuS[U0x] + uS[U0 minus V0]] = Sminus1[minusuex + 2u] = minustex + 2t

U2 = Sminus1[u2ex

]=t2

2ex

V2 = Sminus1[u2ex

]=t2

2ex

(226)

and so on for other components Using (111) the series solutions are given by

U(x t) = 1 + ex(

1 + t +t2

2+t3

3middot middot middot

)

V (x t) = minus1 + ex(

1 minus t + t2

2minus t3


)

(227)

Then the solutions follows

U(x t) = 1 + ex+t

V (x t) = minus1 + exminust(228)


Example 23 Consider the system of nonlinear partial differential equations

Ut + VUx +U = 1

Vt minusUVx minus V = 1(229)


U(x 0) = ex V (x 0) = eminusx (230)

On using Sumudu transform on both sides of (229) and by taking Sumudu transformfor the initial conditions of (230)we get

S[U(x t)] = ex + u minus uS[VUx] minus uS[U]

S[V (x t)] = ex + u + uS[UVx] + uS[V ](231)

Similar to the previous example we rewrite U(x t) and V (x t) by the infinite series(111) then inserting these series into both sides of (231) yields

S

[ infinsum

i=0

Ui(x t)

]

= ex + u minus uS[ infinsum

i=0

Ai

]

minus uS[ infinsum

i=0

Ui

]

S

[ infinsum

i=0

Vi(x t)

]

= eminusx + u + uS

[ infinsum

i=0

Bi

]

minus uS[ infinsum

i=0

Vi

]

(232)

where the terms Ai and Bi are handled with the help of Adomian polynomials by (112)that represent the nonlinear terms VUx and UVx respectively We have a few terms of theAdomian polynomials for VUx andUVx which are given by

A0 = U0xV0 A1 = U0xV1 +U1xV0

A2 = U0xV2 +U1xV1 +U2xV0

B0 = V0xU0 B1 = V0xU1 + V1xU0

B2 = V0xU2 + V1xU1 + V2xU0

(233)


By taking the inverse Sumudu transform we have

U0 = ex + t

V0 = eminusx + t(234)

Ui+1 = Sminus1[u] minus Sminus1[uS[Ai]] minus Sminus1[uS[Ui]]

Vi+1 = Sminus1[u] + Sminus1[uS[Bi]] + Sminus1[uS[Vi]](235)

Using the inverse Sumudu transform on (235)we have

U1 = minust minus t2

2minus tex minus t2

2ex

V1 = minust minus t2

2+ teminusx minus t2

2eminusx

U2 =t2

2+t2

2ex middot middot middot

V2 =t2

2+t2

2eminusx middot middot middot

(236)

The rest terms can be determined in the same way Therefore the series solutions aregiven by

U(x t) = ex(

1 minus t + t2

2minus t3


)

V (x t) = eminusx(

1 + t +t2

2+t3


)

(237)

Then the solution for the above system is as follows

U(x t) = exminust V (x t) = eminusx+t (238)

3 Conclusion

The Sumudu transform-Adomian decomposition method has been applied to linear andnonlinear systems of partial differential equations Three examples have been presented thismethod shows that it is very useful and reliable for any nonlinear partial differential equationsystems Therefore this method can be applied to many complicated linear and nonlinearPDEs


Acknowledgment

The authors would like to express their sincere thanks and gratitude to the reviewers fortheir valuable comments and suggestions for the improvement of this paper The first authoracknowledges the support by the Research Center College of Science King Saud UniversityThe second author also gratefully acknowledges the partial support by the University PutraMalaysia under the Research University Grant Scheme (RUGS) 05-01-09-0720RU and FRGS01-11-09-723FR

References

[1] E Alizadeh M Farhadi K Sedighi H R Ebrahimi-Kebria and A Ghafourian ldquoSolution of theFalkner-Skan equation for wedge by Adomian Decomposition methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 3 pp 724ndash733 2009

[2] A-M Wazwaz ldquoThe modified decomposition method and Pade approximants for a boundary layerequation in unbounded domainrdquo Applied Mathematics and Computation vol 177 no 2 pp 737ndash7442006

[3] N Bellomo and R A Monaco ldquoA comparison between Adomianrsquos decomposition methods andperturbation techniques for nonlinear random differential equationsrdquo Journal of Mathematical Analysisand Applications vol 110 no 2 pp 495ndash502 1985

[4] R Race ldquoOn the Adomian decomposition method and comparison with Picardrsquos methodrdquo Journal ofMathematical Analysis and Applications vol 128 pp 480ndash483 1987

[5] G Adomian and R Rach ldquoNoise terms in decomposition solution seriesrdquo Computers amp Mathematicswith Applications vol 24 no 11 pp 61ndash64 1992

[6] A M Wazwaz ldquoNecessary conditions for the appearance of noise terms in decomposition solutionseriesrdquo Applied Mathematics and Computation vol 81 pp 1718ndash1740 1987

[7] A Kılıcman andH Eltayeb ldquoA note on integral transforms and partial differential equationsrdquoAppliedMathematical Sciences vol 4 no 3 pp 109ndash118 2010

[8] M A Asiru ldquoSumudu transform and the solution of integral equations of convolution typerdquo Inter-national Journal of Mathematical Education in Science and Technology vol 32 no 6 pp 906ndash910 2001

[9] M A Asiru ldquoFurther properties of the Sumudu transform and its applicationsrdquo International Journalof Mathematical Education in Science and Technology vol 33 no 3 pp 441ndash449 2002

[10] A Kadem and A Kılıcman ldquoNote on transport equation and fractional Sumudu transformrdquoComputers amp Mathematics with Applications vol 62 no 8 pp 2995ndash3003 2011

[11] H Eltayeb A Kılıcman and B Fisher ldquoA new integral transform and associated distributionsrdquo Inte-gral Transforms and Special Functions vol 21 no 5 pp 367ndash379 2010

[12] A Kılıcman and H Eltayeb ldquoOn the applications of Laplace and Sumudu transformsrdquo Journal of theFranklin Institute vol 347 no 5 pp 848ndash862 2010

[13] A Kılıcman H Eltayeb and R P Agarwal ldquoOn Sumudu transform and system of differentialequationsrdquo Abstract and Applied Analysis vol 2010 Article ID 598702 11 pages 2010

[14] A Kılıcman H Eltayeb and K AM Atan ldquoA note on the comparison between Laplace and Sumudutransformsrdquo Bulletin of the Iranian Mathematical Society vol 37 no 1 pp 131ndash141 2011

[15] H Agwa F Ali and A Kılıcman ldquoA new integral transform on time scales and its applicationsrdquoAdvances in Difference Equations vol 2012 article 60 2012

[16] F BM Belgacem A A Karaballi and S L Kalla ldquoAnalytical investigations of the Sumudu transformand applications to integral production equationsrdquoMathematical Problems in Engineering vol 2003 no3 pp 103ndash118 2003


Review ArticleInfinite System of Differential Equations inSome BK Spaces

M Mursaleen1 and Abdullah Alotaibi2

1 Department of Mathematics Aligarh Muslim University Aligarh 202002 India2 Department of Mathematics King Abdulaziz University Jeddah 21589 Saudi Arabia

Correspondence should be addressed to Abdullah Alotaibi mathker11hotmailcom

Received 26 July 2012 Accepted 1 October 2012


Copyright q 2012 M Mursaleen and A Alotaibi This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

The first measure of noncompactness was defined by Kuratowski in 1930 and later the Hausdorffmeasure of noncompactness was introduced in 1957 by Goldenstein et al These measures ofnoncompactness have various applications in several areas of analysis for example in operatortheory fixed point theory and in differential and integral equations In particular the Hausdorffmeasure of noncompactness has been extensively used in the characterizations of compactoperators between the infinite-dimensional Banach spaces In this paper we present a brief surveyon the applications of measures of noncompactness to the theory of infinite system of differentialequations in some BK spaces c0 c p(1 le p le infin) and n(φ)

1 FK and BK Spaces

In this section we give some basic definitions and notations about FK and BK spaces forwhich we refer to [1ndash3]

We will write w for the set of all complex sequences x = (xk)infink=0 Let ϕ infin c and c0

denote the sets of all finite bounded convergent and null sequences respectively and cs bethe set of all convergent series We write p = x isin w

suminfink=0 |xk|p lt infin for 1 le p lt infin By e

and e(n) (n isin N) we denote the sequences such that ek = 1 for k = 0 1 and e(n)n = 1 and

e(n)k = 0 (k =n) For any sequence x = (xk)

infink=0 let x

[n] =sumn

k=0 xke(k) be its n-section

Note that infin c and c0 are Banach spaces with the norm xinfin = supkge0|xk| and p(1 lep ltinfin) are Banach spaces with the norm xp = (

suminfink=0 |xk|p)1p

A sequence (b(n))infinn=0 in a linear metric spaceX is called Schauder basis if for every x isin Xthere is a unique sequence (λn)

infinn=0 of scalars such that x =

suminfinn=0 λnb

(n) A sequence space Xwith a linear topology is called aKspace if each of the maps pi X rarr C defined by pi(x) = xi


is continuous for all i isin N A K space is called an FK space if X is complete linear metricspace a BK space is a normed FK space An FK space X sup ϕ is said to have AK if everysequence x = (xk)

infink=0 isin X has a unique representation x =

suminfink=0 xke

(k) that is x = limnrarrinfinx[n]A linear space X equipped with a translation invariant metric d is called a linear metric

space if the algebraic operations on X are continuous functions with respect to d A completelinear metric space is called a Frechet space If X and Y are linear metric spaces over the samefield then we write B(XY ) for the class of all continuous linear operators from X to Y Further ifX and Y are normed spaces thenB(XY ) consists of all bounded linear operators L X rarr Y which is a normed space with the operator norm given by L = supxisinSXL(x)Y forall L isin B(XY ) where SX denotes the unit sphere in X that is SX = x isin X x = 1 Alsowe write BX = x isin X x le 1 for the closed unit ball in a normed space X In particularif Y = C then we write Xlowast for the set of all continuous linear functionals on X with the normf = supxisinSX |f(x)|

The theory of FK spaces is the most powerful and widely used tool in thecharacterization of matrix mappings between sequence spaces and the most important resultwas that matrix mappings between FK spaces are continuous

A sequence space X is called an FK space if it is a locally convex Frechet space withcontinuous coordinates pn X rarr C(n isin N) where C denotes the complex field and pn(x) =xn for all x = (xk) isin X and every n isin N A normed FK space is called a BK space that is aBK space is a Banach sequence space with continuous coordinates

The famous example of an FK space which is not a BK space is the space (wdw)where

dw(x y

)=

infinsum

k=0

12k

( ∣∣xk minus yk∣∣

1 +∣∣xk minus yk

∣∣

)

(x y isin w)

(11)

On the other hand the classical sequence spaces are BK spaces with their naturalnorms More precisely the spaces infin c and c0 are BK spaces with the sup-norm given byxinfin = supk|xk| Also the space p (1 le p ltinfin) is a BK space with the usual p-norm definedby xp = (

sumk |xk|p)1p Further the spaces bs cs and cs0 are BK spaces with the same norm

given by xbs = supnsumn

k=0 |xk| and bv is a BK space with xbv =sum

k |xk minus xkminus1|An FK space X sup φ is said to have AK if every sequence x = (xk) isin X has a unique

representation x =suminfin

k=0 xke(k) that is limnrarrinfin(

sumnk=0 xke

(k)) = x This means that (e(k))infink=0 isa Schauder basis for any FK space with AK such that every sequence in an FK space withAK coincides with its sequence of coefficients with respect to this basis

Although the space infin has no Schauder basis the spaces w c0 c and p all haveSchauder bases Moreover the spaces w c0 and p have AK where 1 le p ltinfin

There are following BK spaces which are closely related to the spaces p (1 le p le infin)Let C denote the space whose elements are finite sets of distinct positive integers

Given any element σ of C we denote by c(σ) the sequence cn(σ) such that cn(σ) = 1 forn isin σ and cn(σ) = 0 otherwise Further

Cs =

σ isin C infinsum

n=1

cn(σ) le s

(12)


that is Cs is the set of those σ whose support has cardinality at most s and define

Φ =φ =

(φk

) isin w 0 lt φ1 le φn le φn+1 (n + 1)φn ge nφn+1 (13)

For φ isin Φ the following sequence spaces were introduced by Sargent [4] and furtherstudied in [5ndash8]

m(φ)=

x = (xk) isin w xm(φ) = supsge1

supσisinCs

(1φs

sum

kisinσ|xk|

)

ltinfin

n(φ)=

x = (xk) isin w xn(φ) = supuisinS(x)

( infinsum

k=1

|uk|Δφk)

ltinfin

(14)

where S(x) denotes the set of all sequences that are rearrangements of x

Remark 11 (i) The spacesm(φ) and n(φ) are BK spaces with their respective norms(ii) If φn = 1 for all n isin N then m(φ) = l1 n(φ) = linfin and if φn = n for all n isin N then

m(φ) = linfin n(φ) = l1(iii) l1 sube m(φ) sube linfin[linfin supe n(φ) supe l1] for all φ of Φ

2 Measures of Noncompactness

The first measure of noncompactness was defined and studied by Kuratowski [9] in 1930 TheHausdorffmeasure of noncompactness was introduced by Goldenstein et al [10] in 1957 andlater studied by Goldenstein and Markus [11] in 1965

Here we shall only consider the Hausdorffmeasure of noncompactness it is the mostsuitable one for our purposes The basic properties of measures of noncompactness can befound in [12ndash14]

Let S andM be subsets of a metric space (X d) and let ε gt 0 Then S is called an ε-netof M in X if for every x isin M there exists s isin S such that d(x s) lt ε Further if the set S isfinite then the ε-net S ofM is called a finite ε-net ofM and we say thatM has a finite ε-net inX A subsetM of a metric space X is said to be totally bounded if it has a finite ε-net for everyε gt 0 and is said to be relatively compact if its closure M is a compact set Moreover if themetric space X is complete thenM is totally bounded if and only ifM is relatively compact

Throughout we shall write MX for the collection of all bounded subsets of a metricspace (X d) IfQ isin MX then the Hausdorff measure of noncompactness of the setQ denoted byχ(Q) is defined to be the infimum of the set of all reals ε gt 0 such that Q can be covered bya finite number of balls of radii lt ε and centers in X This can equivalently be redefined asfollows

χ(Q) = infε gt 0 Q has a finite ε-net (21)

The function χ MX rarr [0infin) is called the Hausdorff measure of noncompactness


If Q Q1 and Q2 are bounded subsets of a metric space X then we have

χ(Q) = 0 if and only ifQ is totally bounded

Q1 sub Q2 impliesχ(Q1) le χ(Q2)(22)

Further if X is a normed space then the function χ has some additional propertiesconnected with the linear structure for example

χ(Q1 +Q2) le χ(Q1) + χ(Q2)

χ(αQ) = |α| χ(Q) forallα isin C(23)

Let X and Y be Banach spaces and χ1 and χ2 be the Hausdorff measures ofnoncompactness on X and Y respectively An operator L X rarr Y is said to be (χ1χ2)-bounded if L(Q) isin MY for all Q isin MX and there exist a constant C ge 0 such thatχ2(L(Q)) le Cχ1(Q) for all Q isin MX If an operator L is (χ1χ2)-bounded then the numberL(χ1χ2) = infC ge 0 χ2(L(Q)) le Cχ1(Q) for all Q isin MX is called the (χ1χ2)-measure ofnoncompactness of L If χ1 = χ2 = χ then we write L(χ1χ2) = Lχ

The most effective way in the characterization of compact operators between theBanach spaces is by applying the Hausdorff measure of noncompactness This can beachieved as follows let X and Y be Banach spaces and L isin B(XY ) Then the Hausdorffmeasure of noncompactness of L denoted by Lχ can be determined by

Lχ = χ(L(SX)) (24)

and we have that L is compact if and only if

Lχ = 0 (25)

Furthermore the function χ is more applicable whenX is a Banach space In fact thereare many formulae which are useful to evaluate the Hausdorff measures of noncompactnessof bounded sets in some particular Banach spaces For example we have the following resultof Goldenstein et al [10 Theorem 1] which gives an estimate for the Hausdorff measureof noncompactness in Banach spaces with Schauder bases Before that let us recall that if(bk)

infink=0 is a Schauder basis for a Banach space X then every element x isin X has a unique


k=0 φk(x)bk where φk (k isin N) are called the basis functionals Moreoverthe operator Pr X rarr X defined for each r isin N by Pr(x) =

sumrk=0 φk(x)bk(x isin X) is called

the projector onto the linear span of b0 b1 br Besides all operators Pr and I minus Pr areequibounded where I denotes the identity operator on X


Theorem 21 Let X be a Banach space with a Schauder basis (bk)infink=0 E isin MX and Pn X rarr

X(n isin N) be the projector onto the linear span of b0 b1 bn Then one has

1amiddot lim sup

nrarrinfin

(

supxisinQ

(I minus Pn)(x))

le χ(E) le lim supnrarrinfin

(

supxisinQ

(I minus Pn)(x))

(26)

where a = lim supnrarrinfinI minus Pn

In particular the following result shows how to compute the Hausdorff measure ofnoncompactness in the spaces c0 and p (1 le p ltinfin)which are BK spaces with AK

Theorem 22 Let E be a bounded subset of the normed space X where X is p for 1 le p lt infin orc0 If Pn X rarr X(n isin N) is the operator defined by Pn(x) = x[n] = (x0 x1 xn 0 0 ) for allx = (xk)

infink=0 isin X then one has

χ(E) = limnrarrinfin

(

supxisinQ

(I minus Pn)(x))

(27)

It is easy to see that for E isin Mp


(

supxisinQ

sum

kgen|xk|p

)

(28)

Also it is known that (e e(0) e(1) ) is a Schauder basis for the space c and everysequence z = (zn)

infinn=0 isin c has a unique representation z = ze +

suminfinn=0(zn minus z)e(n) where

z = limnrarrinfinzn Thus we define the projector Pr c rarr c (r isin N) onto the linear span ofe e(0) e(1) e(r) by

Pr(z) = ze +rsum

n=0(zn minus z)e(n) (r isin N) (29)

for all z = (zn) isin c with z = limnrarrinfinzn In this situation we have the following

Theorem 23 Let Q isin Mc and Pr c rarr c (r isin N) be the projector onto the linear span ofe e(0) e(1) e(r) Then one has

12middot limrrarrinfin

(

supxisinQ

(I minus Pr)(x)infin)

le χ(Q) le limrrarrinfin

(

supxisinQ


(210)

where I is the identity operator on c

Theorem 24 Let Q be a bounded subset of n(φ) Then

χ(Q) = limkrarrinfin

supxisinQ

(

supuisinS(x)

( infinsum

n=k

|un|Δφn))

(211)


The idea of compact operators between Banach spaces is closely related to the Hausdorffmeasure of noncompactness and it can be given as follows

Let X and Y be complex Banach spaces Then a linear operator L X rarr Y is said tobe compact if the domain of L is all of X that is D(L) = X and for every bounded sequence(xn) in X the sequence (L(xn)) has a convergent subsequence in Y Equivalently we say thatL is compact if its domain is all of X and L(Q) is relatively compact in Y for every Q isin MX

Further we write C(XY ) for the class of all compact operators from X to Y Let usremark that every compact operator in C(XY ) is bounded that is C(XY ) sub B(XY ) Moreprecisely the class C(XY ) is a closed subspace of the Banach space B(XY )with the operatornorm

The most effective way in the characterization of compact operators between theBanach spaces is by applying the Hausdorff measure of noncompactness This can beachieved as follows


Let X and Y be Banach spaces and L isin B(XY ) Then the Hausdorff measure ofnoncompactness of L denoted by Lχ can be given by

Lχ = χ(L(SX)) (212)

and we have

L is compact if and only if Lχ = 0 (213)

Since matrix mappings between BK spaces define bounded linear operators betweenthese spaces which are Banach spaces it is natural to use the Hausdorff measure ofnoncompactness to obtain necessary and sufficient conditions for matrix operators betweenBK spaces to be compact operators This technique has recently been used by several authorsin many research papers (see for instance [5 6 15ndash32])

3 Applications to Infinite Systems of Differential Equations

This section is mainly based on the work of Banas and Lecko [33] Mursaleen andMohiuddine [26] and Mursaleen [32] In this section we apply the technique of measuresof noncompactness to the theory of infinite systems of differential equations in some Banachsequence spaces c0 c p(1 le p ltinfin) and n(φ)

Infinite systems of ordinary differential equations describe numerous world realproblems which can be encountered in the theory of branching processes the theory ofneural nets the theory of dissociation of polymers and so on (cf [34ndash38] eg) Let usalso mention that several problems investigated in mechanics lead to infinite systems ofdifferential equations [39ndash41] Moreover infinite systems of differential equations can bealso used in solving some problems for parabolic differential equations investigated viasemidiscretization [42 43] The theory of infinite systems of ordinary differential equationseems not to be developed satisfactorily up to now Indeed the existence results concerningsystems of such a kind were formulated mostly by imposing the Lipschitz condition on


right-hand sides of those systems (cf [10 11 39 40 44ndash50]) Obviously the assumptionsformulated in terms of the Lipschitz condition are rather restrictive and not very useful inapplications On the other hand the infinite systems of ordinary differential equations can beconsidered as a particular case of ordinary differential equations in Banach spaces Until nowseveral existence results have been obtained concerning the Cauchy problem for ordinarydifferential equations in Banach spaces [33 35 51ndash53] A considerable number of those resultswere formulated in terms of measures of noncompactness The results of such a type havea concise form and give the possibility to formulate more general assumptions than thoserequiring the Lipschitz continuity But in general those results are not immediately applicablein concrete situations especially in the theory of infinite systems of ordinary differentialequations

In this section we adopt the technique of measures of noncompactness to the theory ofinfinite systems of differential equations Particularly we are going to present a few existenceresults for infinite systems of differential equations formulated with the help of convenientand handy conditions

Consider the ordinary differential equation

xprime = f(t x) (31)


x(0) = x0 (32)

Then the following result for the existence of the Cauchy problem (31)-(32)was givenin [34]which is a slight modification of the result proved in [33]

Assume that X is a real Banach space with the norm middot Let us take an intervalI = [0 T] T gt 0 and B(x0 r) the closed ball in X centered at x0 with radius r

Theorem A (see [34]) Assume that f(t x) is a function defined on I times X with values in X suchthat

∥∥f(t x)∥∥ le Q + Rx (33)

for any x isin X where Q and R are nonnegative constants Further let f be uniformly continuous onthe set I1 timesB(x0 r) where r = (QT1 +RT1x0)(1minusRT1) and I1 = [0 T1] sub I RT1 lt 1 Moreoverassume that for any nonempty set Y sub B(x0 s) and for almost all t isin I the following inequality holds

μ(f(t Y )

) le q(t)μ(Y ) (34)

with a sublinear measure of noncompactness μ such that x0 isin kerμ Then problem (31)-(32)has a solution x such that x(t) isin kerμ for t isin I1 where q(t) is an integrable function on I andkerμ = E isin MX μ(E) = 0 is the kernel of the measure μ

Remark 31 In the case when μ = χ (the Hausdorff measure of noncompactness) theassumption of the uniform continuity on f can be replaced by the weaker one requiring onlythe continuity of f


Results of Sections 31 and 32 are from [33] Section 33 from [26] and Section 34 from[32]

31 Infinite Systems of Differential Equations in the Space c0

From now on our discussion is exactly same as in Section 3 of [33]In this section we study the solvability of the infinite systems of differential equations

in the Banach sequence space c0 It is known that in the space c0 the Hausdorff measure ofnoncompactness can be expressed by the following formula [33]


(

supxisinE

supkgen

|xk|)

(35)

where E isin Mc0 We will be interested in the existence of solutions x(t) = (xi(t)) of the infinite systems

of differential equations

xprimei = fi(t x0 x1 x2 ) (36)


xi(0) = x0i (37)

(i = 0 1 2 ) which are defined on the interval I = [0 T] and such that x(t) isin c0 for eacht isin I

An existence theorem for problem (36)-(37) in the space c0 can be formulated bymaking the following assumptions

Assume that the functions fi (i = 1 2 ) are defined on the set I times Rinfin and take real

values Moreover we assume the following hypotheses

(i) x0 = (x0i ) isin c0

(ii) the map f = (f1 f2 ) acts from the set I times c0 into c0 and is continuous

(iii) there exists an increasing sequence (kn) of natural numbers (obviously kn rarr infinas n rarr infin) such that for any t isin I x = (xi) isin c0 and n = 1 2 the followinginequality holds

∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

(38)

where (pi(t)) and (qi(t)) are real functions defined and continuous on I such thatthe sequence (pi(t)) converges uniformly on I to the function vanishing identicallyand the sequence (qi(t)) is equibounded on I


Now let us denote

q(t) = supnge1

qn(t)

Q = suptisinI

q(t)

P = suppn(t) t isin I n = 1 2

(39)

Then we have the following result

Theorem 32 (see [33]) Under the assumptions (i)ndash(iii) initial value problem (36)-(37) has atleast one solution x = x(t) = (xi(t)) defined on the interval I1 = [0 T1] whenever T1 lt T andQT1 lt 1 Moreover x(t) isin c0 for any t isin I1

Proof Let x = (xi) be any arbitrary sequence in c0 Then by (i)ndash(iii) for any t isin I and for afixed n isin N we obtain

∣∣fn(t x)∣∣ =

∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

le P +Q supigekn

|xi| le P +Qxinfin(310)

Hence we get

∥∥f(t x)∥∥ le P +Qxinfin (311)

In what follows let us take the ball B(x0 r) where r is chosen according to TheoremA Then for a subset Y of B(x0 r) and for t isin I1 we obtain

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x)∣∣)

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x1 x2 )∣∣)

le limnrarrinfin

supxisinY

(

supigen

pi(t) + qi(t)supjgekn

∣∣xj∣∣)

le limnrarrinfin

supigen

pi(t) + q(t) limnrarrinfin

supxisinY

(

supigen

supjgekn

∣∣xj∣∣)

(312)

Hence by assumptions we get

χ(f(t Y )

) le q(t)χ(Y ) (313)


Now using our assumptions and inequalities (311) and (313) in view of Theorem A andRemark 31 we deduce that there exists a solution x = x(t) of the Cauchy problem (36)-(37)such that x(t) isin c0 for any t isin I1

This completes the proof of the theorem

We illustrate the above result by the following examples

Example 33 (see [33]) Let kn be an increasing sequence of natural numbers Consider theinfinite system of differential equations of the form

xprimei = fi(t x1 x2 ) +

infinsum

j=ki+1

aij(t)xj (314)


xi(0) = x0i (315)

(i = 1 2 t isin I = [0 T])We will investigate problem (314)-(315) under the following assumptions

(i) x0 = (x0) isin c0(ii) the functions fi I times R

ki rarr R (i = 1 2 ) are uniformly equicontinuous and thereexists a function sequence (pi(t)) such that pi(t) is continuous on I for any i isin N and(pi(t)) converges uniformly on I to the function vanishing identically Moreover thefollowing inequality holds

∣∣fi(t x1 x2 xki)∣∣ le pi(t) (316)

for t isin I(x1 x2 xki) isin Rki and i isin N

(iii) the functions aij(t) are defined and continuous on I and the function seriessuminfinj=ki+1 aij(t) converges absolutely and uniformly on I (to a function ai(t)) for any

i = 1 2

(iv) the sequence (ai(t)) is equibounded on I

(v) QT lt 1 where Q = supai(t) i = 1 2 t isin IIt can be easily seen that the assumptions of Theorem 32 are satisfied under

assumptions (i)ndash(v) This implies that problem (314)-(315) has a solution x(t) = (xi(t)) onthe interval I belonging to the space c0 for any fixed t isin I

As mentioned in [33] problem (314)-(315) considered above contains as a specialcase the infinite system of differential equations occurring in the theory of dissociationof polymers [35] That system was investigated in [37] in the sequence space infin undervery strong assumptions The existing result proved in [35] requires also rather restrictiveassumptions Thus the above result is more general than those quoted above

Moreover the choice of the space c0 for the study of the problem (314)-(315) enablesus to obtain partial characterization of solutions of this problem since we have that xn(t) rarr 0when n rarr infin for any fixed t isin [0 T]


On the other hand let us observe that in the study of the heat conduction problem viathe method of semidiscretization we can obtain the infinite systems of form (314) (see [42]for details)

Example 34 (see [33]) In this example we will consider some special cases of problem (314)-(315) Namely assume that ki = i for i = 1 2 and aij equiv 0 on I for all i j Then system (314)has the form

xprime1 = f1(t x1) xprime

2 = f2(t x1 x2)

xprimei = fi(t x1 x2 xi)

(317)

and is called a row-finite system [35]Suppose that there are satisfied assumptions from Example 33 that is x0 = (x0

i ) isin c0and the functions fi act from I times R

i into R (i = 1 2 ) and are uniformly equicontinuous ontheir domains Moreover there exist continuous functions pi(t) (t isin I) such that

∣∣fi(t x1 x2 xki)∣∣ le pi(t) (318)

for t isin I and x1 x2 xi isin R (i = 1 2 ) We assume also that the sequence pi(t) convergesuniformly on I to the function vanishing identically

Further let | middot |i denote the maximum norm in Ri(i = 1 2 ) Take fi = (f1 f2 fi)

Then we have

∣∣∣fi(t x)∣∣∣i= max

∣∣f1(t x1)∣∣∣∣f2(t x1 x2)

∣∣ ∣∣fi(t x1 x2 xi)

∣∣

le maxp1(t) p2(t) pi(t)

(319)

Taking Pi(t) = maxp1(t)p2(t) pi(t) then the above estimate can be written in the form

∣∣∣fi(t x)∣∣∣ile Pi(t) (320)

Observe that from our assumptions it follows that the initial value problem uprime =Pi(t) u(0) = x0

i has a unique solution on the interval I Hence applying a result from [33] weinfer that Cauchy problem (317)-(315) has a solution on the interval I Obviously from theresult contained in Theorem 32 and Example 33 we deduce additionally that the mentionedsolution belongs to the space c0

Finally it is noticed [33] that the result described above for row-finite systems of thetype (317) can be obtained under more general assumptions

In fact instead of inequality (318) we may assume that the following estimate holdsto be true

∣∣fi(t x1 x2 xi)∣∣ le pi(t) + qi(t)max|x1| |x2| |xi| (321)

where the functions pi(t) and qi(t) (i = 1 2 ) satisfy the hypotheses analogous to thoseassumed in Theorem 32


Remark 35 (see [33]) Note that in the birth process one can obtain a special case of the infinitesystem (317) which is lower diagonal linear infinite system [35 43] Thus the result provedabove generalizes that from [35 37]

32 Infinite Systems of Differential Equations in the Space c

Now we will study the solvability of the following perturbed diagonal system of differentialequations

xprimei = ai(t)xi + gi(t x1 x2 ) (322)


xi(0) = x0i (323)

(i = 1 2 ) where t isin I = [0 T]From now we are going exactly the same as in Section 4 of [33]We consider the following measures μ of noncompactness in c which is more

convenient regular and even equivalent to the Hausdorff measure of noncompactness [33]For E isin Mc

μ(E) = limprarrinfin

supxisinE

supnmgep

|xn minus xm|


(

supxisinQ

supkgen

|xk|)

(324)

Let us formulate the hypotheses under which the solvability of problem (322)-(323)will be investigated in the space c Assume that the following conditions are satisfied



(i) x0 = (x0i ) isin c

(ii) the map g = (g1 g2 ) acts from the set I times c into c and is uniformly continuouson I times c

(iii) there exists sequence (bi) isin c0 such that for any t isin I x = (xi) isin c and n = 1 2 the following inequality holds

∣∣gn(t x1 x2 )∣∣ le bi (325)

(iv) the functions ai(t) are continuous on I such that the sequence (ai(t)) convergesuniformly on I


Further let us denote

a(t) = supige1

ai(t)

Q = suptisinI

a(t)(326)

Observe that in view of our assumptions it follows that the function a(t) is continuouson I Hence Q ltinfin

Then we have the following result which is more general than Theorem 32

Theorem 36 (see [33]) Let assumptions (i)ndash(iv) be satisfied If QT lt 1 then the initial valueproblem (312)-(313) has a solution x(t) = (xi(t)) on the interval I such that x(t) isin c for each t isin I

Proof Let t isin I and x = (xi) isin c and

fi(t x) = ai(t)xi + gi(tx)

f(t x) =(f1(t x) f2(tx)

)=(fi(t x)

)

(327)

Then for arbitrarily fixed natural numbers nmwe get

∣∣fn(t x) minus fm(t x)∣∣ =

∣∣an(t)xn + gn(tx) minus am(t)xm minus gm(tx)∣∣

le ∣∣an(t)xn + gn(tx)∣∣ minus ∣∣am(t)xm + gm(tx)

∣∣

le |an(t)xn minus an(t)xm| + |an(t)xm minus am(t)xm| + bn + bmle |an(t)||xn minus xm| + |an(t) minus am(t)||xm| + bn + bmle |an(t)||xn minus xm| + xinfin|an(t) minus am(t)| bn + bm

(328)

By assumptions (iii) and (iv) from the above estimate we deduce that (fi(t x)) is a realCauchy sequence This implies that (fi(t x)) isin c

Also we obtain the following estimate

∣∣fi(t x)∣∣ le |ai(t)||xi| +

∣∣gi(t x)∣∣

le Q|xi| + bi le Qxinfin + B(329)

where B = supige1bi Hence

∥∥f(t x)∥∥ le Qxinfin + B (330)

In what follows let us consider the mapping f(t x) on the set I times B(x0 r) where r istaken according to the assumptions of Theorem A that is r = (BT + QTx0infin)(1 minus QT)


Further fix arbitrarily t s isin I and x y isin B(x0 r) Then by our assumptions for a fixed i weobtain

∣∣fi(t x) minus fi

(s y

)∣∣ =∣∣ai(t)xi + gi(t x) minus ai(s)yi minus gi

(s y

)∣∣

le ∣∣ai(t)xi minus ai(s)yi

∣∣ +

∣∣gi(t x) minus gi

(s y

)∣∣

le |ai(t) minus ai(s)||xi| + |ai(s)|∣∣xi minus yi

∣∣ +


(s y

)∣∣

(331)

Then

∥∥f(t x) minus f(s y)∥∥ = sup

ige1


(s y

)∣∣

le (r + x0infin) middot supige1

|ai(t) minus ai(s)|

+Q∥∥x minus y∥∥infin +

∥∥g(t x) minus g(s y)∥∥

(332)

Hence taking into account that the sequence (ai(t)) is equicontinuous on the interval Iand g is uniformly continuous on I times c we conclude that the operator f(t x) is uniformlycontinuous on the set I times B(x0 r)

In the sequel let us take a nonempty subset E of the ball B(x0 r) and fix t isin I x isin XThen for arbitrarily fixed natural numbers nm we have

∣∣fn(t x) minus fm(t x)∣∣ le |an(t)||xn minus xm| + |xm||an(t) minus am(t)| +

∣∣gn(t x)∣∣ +

∣∣gm(t x)∣∣

le a(t)|xn minus xm| + (r + x0infin)|an(t) minus am(t)| + bn + bm(333)

Hence we infer the following inequality

μ(f(t E)

) le a(t)μ(E) (334)

Finally combining (330) (334) and the fact (proved above) that f is uniformlycontinuous on I times B(x0 r) in view of Theorem A we infer that problem (322)-(323) issolvable in the space c


Remark 37 (see [33 Remark 4]) The infinite systems of differential equations (322)-(323)considered above contain as special cases the systems studied in the theory of neural sets (cf[35 pages 86-87] and [37] eg) It is easy to notice that the existence results proved in [35 37]are obtained under stronger and more restrictive assumptions than our one

33 Infinite Systems of Differential Equations in the Space p

In this section we study the solvability of the infinite systems of differential equations (36)-(37) in the Banach sequence space p (1 le p ltinfin) such that x(t) isin p for each t isin I


An existence theorem for problem (36)-(37) in the space p can be formulated bymaking the following assumptions

(i) x0 = (x0i ) isin p

(ii) fi I times Rinfin rarr R (i = 0 1 2 )maps continuously the set I times p into p

(iii) there exist nonnegative functions qi(t) and ri(t) defined on I such that

∣∣fi(t x)

∣∣p =

∣∣fi(t x0 x1 x2 )

∣∣p le qi(t) + ri(t)|xi|p (335)

for t isin Ix = (xi) isin p and i = 0 1 2

(iv) the functions qi(t) are continuous on I and the function seriessuminfin

i=0 qi(t) convergesuniformly on I

(v) the sequence (ri(t)) is equibounded on the interval I and the function r(t) =lim supirarrinfinri(t) is integrable on I

Now we prove the following result

Theorem 38 (see [26]) Under the assumptions (i)ndash(v) problem (36)-(37) has a solution x(t) =(xi(t)) defined on the interval I = [0 T] whenever RT lt 1 where R is defined as the number

R = supri(t) t isin I i = 0 1 2 (336)

Moreover x(t) isin p for any t isin I

Proof For any x(t) isin p and t isin I under the above assumptions we have

∥∥f(t x)∥∥pp =

infinsum

i=0

∣∣fi(t x)∣∣p

leinfinsum

i=0

[qi(t) + ri(t)|xi|p

]

leinfinsum

i=0

qi(t) +

(

supige0

ri(t)

)( infinsum

i=0|xi|p

)

le Q + Rxpp

(337)

where Q = suptisinI(suminfin

i=0 qi(t))Now choose the number s = (QT + RTx0pp)(1 minus RT) as defined in Theorem A

Consider the operator f = (fi) on the set I timesB(x0 s) Let us take a set Y isin Mp Then by using(28) we get

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(sum

igen

∣∣fi(t x1 x2 )∣∣p)

le limnrarrinfin

(sum

igenqi(t) +

(

supigen

ri(t)

)(sum

igen|xi|p

) )

(338)


Hence by assumptions (iv)-(v) we get

χ(f(t Y )

) le r(t)χ(Y ) (339)

that is the operator f satisfies condition (34) of Theorem A Hence by Theorem A andRemark 31 we conclude that there exists a solution x = x(t) of problem (36)-(37) such thatx(t) isin p for any t isin I


Remark 39 (I) For p = 1 we get Theorem 5 of [33]

(II) It is easy to notice that the existence results proved in [44] are obtained understronger and more restrictive assumptions than our one

(III) We observe that the above theorem can be applied to the perturbed diagonal infinitesystem of differential equations of the form

xprimei = ai(t)xi + gi(t x0 x1 x2 ) (340)


xi(0) = x0i (341)

(i = 0 1 2 ) where t isin IAn existence theorem for problem (36)-(37) in the space p can be formulated by

making the following assumptions


(ii) the sequence (|ai(t)|) is defined and equibounded on the interval I = [0 T]Moreover the function a(t) = lim supirarrinfin sup |ai(t)| is integrable on I

(iii) the mapping g = (gi) maps continuously the set I times p into p(iv) there exist nonnegative functions bi(t) such that

∣∣fi(t x0 x1 x2 )∣∣p le bi(t) (342)


(v) the functions bi(t) are continuous on I and the function seriessuminfin

i=0 bi(t) convergesuniformly on I

34 Infinite Systems of Differential Equations in the Space n(φ)

An existence theorem for problem (36)-(37) in the space n(φ) can be formulated by makingthe following assumptions

(i) x0 = (x0i ) isin n(φ)


(ii) fi I times Rinfin rarr R (i = 1 2 )maps continuously the set I times n(φ) into n(φ)

(iii) there exist nonnegative functions pi(t) and qi(t) defined on I such that

∣∣fi(t u)

∣∣ =

∣∣fi(t u1 u2 )

∣∣ le pi(t) + qi(t)|ui| (343)

for t isin Ix = (xi) isin n(φ) and i = 1 2 where u = (ui) is a sequence ofrearrangement of x = (xi)

(iv) the functions pi(t) are continuous on I and the function seriessuminfin

i=1 pi(t)Δφiconverges uniformly on I

(v) the sequence (qi(t)) is equibounded on the interval I and the function q(t) =lim supirarrinfinqi(t) is integrable on I


Theorem 310 (see [32]) Under the assumptions (i)ndash(v) problem (36)-(37) has a solution x(t) =(xi(t)) defined on the interval I = [0 T] whenever QT lt 1 where Q is defined as the number

Q = supqi(t) t isin I i = 1 2

(344)

Moreover x(t) isin n(φ) for any t isin I

Proof For any x(t) isin n(φ) and t isin I under the above assumptions we have

∥∥f(t x)∥∥n(φ) = sup

uisinS(x)

infinsum

i=1

∣∣fi(t u)∣∣Δφi

le supuisinS(x)

infinsum

i=1

[pi(t) + qi(t)|ui|

]Δφi

leinfinsum

i=1

pi(t)Δφi +

(

supi

qi(t)

)(

supuisinS(x)

infinsum

i=1

|ui|Δφi)

le P +Qxn(φ)

(345)

where P = suptisinIsuminfin

i=1 pi(t)ΔφiNow choose the number r defined according to Theorem A that is r = (PT +

QTx0n(φ))(1 minus QT) Consider the operator f = (fi) on the set I times B(x0 r) Let us take aset X isin M n(φ) Then by using Theorem 24 we get

χ(f(t X)

)= lim

krarrinfinsupxisinX

(

supuisinS(x)

( infinsum

n=k

∣∣fn(t u1 u2 )∣∣Δφn

))

le limkrarrinfin

( infinsum

n=k

pn(t)Δφn +

(

supngek

qn(t)

)(

supuisinS(x)

infinsum

n=k

|un|Δφn))

(346)



χ(f(t X)

) le q(t)χ(X) (347)

that is the operator f satisfies condition (34) of Theorem A Hence the problem (36)-(37)has a solution x(t) = (xi(t))


Remark 311 Similarly we can establish such type of result for the spacem(φ)

References

[1] F Basar Summability Theory and Its Applications Bentham Science Publishers e-books MonographsIstanbul Turkey 2011

[2] M Mursaleen Elements of Metric Spaces Anamaya Publishers New Delhi India 2005[3] A Wilansky Summability through Functional Analysis vol 85 of North-Holland Mathematics Studies

North-Holland Publishing Amsterdam The Netherlands 1984[4] W L C Sargent ldquoOn compact matrix transformations between sectionally bounded BK-spacesrdquo

Journal of the London Mathematical Society vol 41 pp 79ndash87 1966[5] E Malkowsky and Mursaleen ldquoMatrix transformations between FK-spaces and the sequence spaces

m(ϕ) and n(ϕ)rdquo Journal of Mathematical Analysis and Applications vol 196 no 2 pp 659ndash665 1995[6] E Malkowsky and M Mursaleen ldquoCompact matrix operators between the spaces m(φ) and n(φ)rdquo

Bulletin of the Korean Mathematical Society vol 48 no 5 pp 1093ndash1103 2011[7] M Mursaleen R Colak and M Et ldquoSome geometric inequalities in a new Banach sequence spacerdquo

Journal of Inequalities and Applications vol 2007 Article ID 86757 6 pages 2007[8] MMursaleen ldquoSome geometric properties of a sequence space related to lprdquo Bulletin of the Australian

Mathematical Society vol 67 no 2 pp 343ndash347 2003[9] K Kuratowski ldquoSur les espaces completsrdquo Fundamenta Mathematicae vol 15 pp 301ndash309 1930[10] L S Golden stein I T Gohberg and A S Markus ldquoInvestigations of some properties of bounded

linear operators with their q-normsrdquo Ucenie Zapiski Kishinevskii vol 29 pp 29ndash36 1957[11] L S Goldenstein and A S Markus ldquoOn the measure of non-compactness of bounded sets and of

linear operatorsrdquo in Studies in Algebra and Mathematical Analysis pp 45ndash54 Kishinev 1965[12] R R Akhmerov M I Kamenskiı A S Potapov A E Rodkina and B N Sadovskiı Measures

of Noncompactness and Condensing Operators vol 55 of Operator Theory Advances and ApplicationsBirkhauser Basel Switzerland 1992

[13] J Banas and K GoeblMeasures of Noncompactness in Banach Spaces vol 60 of Lecture Notes in Pure andApplied Mathematics Marcel Dekker New York NY USA 1980

[14] E Malkowsky and V Rakocevic ldquoAn introduction into the theory of sequence spaces and measuresof noncompactnessrdquo Zbornik Radova vol 9(17) pp 143ndash234 2000

[15] F Basar and E Malkowsky ldquoThe characterization of compact operators on spaces of stronglysummable and bounded sequencesrdquo Applied Mathematics and Computation vol 217 no 12 pp 5199ndash5207 2011

[16] M Basarir and E E Kara ldquoOn compact operators on the Riesz Bmmdashdifference sequence spacesrdquoIranian Journal of Science amp Technology A vol 35 no 4 pp 279ndash285 2011

[17] M Basarir and E E Kara ldquoOn some difference sequence spaces of weighted means and compactoperatorsrdquo Annals of Functional Analysis vol 2 no 2 pp 114ndash129 2011

[18] I Djolovic and E Malkowsky ldquoThe Hausdorff measure of noncompactness of operators on thematrix domains of triangles in the spaces of strongly C1 summable and bounded sequencesrdquo AppliedMathematics and Computation vol 216 no 4 pp 1122ndash1130 2010

[19] I Djolovic and E Malkowsky ldquoCharacterizations of compact operators on some Euler spaces ofdifference sequences of ordermrdquo Acta Mathematica Scientia B vol 31 no 4 pp 1465ndash1474 2011

[20] E E Kara and M Basarir ldquoOn compact operators and some Euler B(m)-difference sequence spacesrdquoJournal of Mathematical Analysis and Applications vol 379 no 2 pp 499ndash511 2011


[21] B de Malafosse E Malkowsky and V Rakocevic ldquoMeasure of noncompactness of operators andmatrices on the spaces c and c0rdquo International Journal of Mathematics and Mathematical Sciences vol2006 Article ID 46930 5 pages 2006

[22] B de Malafosse and V Rakocevic ldquoApplications of measure of noncompactness in operators on thespaces sα s0α s

(c)α l

pαrdquo Journal of Mathematical Analysis and Applications vol 323 no 1 pp 131ndash145

2006[23] E Malkowsky and I Djolovic ldquoCompact operators into the spaces of strongly C1 summable and

bounded sequencesrdquoNonlinear Analysis Theory Methods amp Applications vol 74 no 11 pp 3736ndash37502011

[24] M Mursaleen V Karakaya H Polat and N Simsek ldquoMeasure of noncompactness of matrixoperators on some difference sequence spaces of weighted meansrdquo Computers amp Mathematics withApplications vol 62 no 2 pp 814ndash820 2011

[25] MMursaleen and A Latif ldquoApplications of measure of noncompactness in matrix operators on somesequence spacesrdquo Abstract and Applied Analysis vol 2012 Article ID 378250 10 pages 2012

[26] M Mursaleen and S A Mohiuddine ldquoApplications of measures of noncompactness to the infinitesystem of differential equations in p spacesrdquo Nonlinear Analysis Theory Methods amp Applications vol75 no 4 pp 2111ndash2115 2012

[27] M Mursaleen and A K Noman ldquoCompactness by the Hausdorff measure of noncompactnessrdquoNonlinear Analysis Theory Methods amp Applications vol 73 no 8 pp 2541ndash2557 2010

[28] MMursaleen and A K Noman ldquoApplications of the Hausdorffmeasure of noncompactness in somesequence spaces of weighted meansrdquo Computers amp Mathematics with Applications vol 60 no 5 pp1245ndash1258 2010

[29] M Mursaleen and A K Noman ldquoThe Hausdorffmeasure of noncompactness of matrix operators onsome BK spacesrdquo Operators and Matrices vol 5 no 3 pp 473ndash486 2011

[30] M Mursaleen and A K Noman ldquoOn σ-conservative matrices and compact operators on the spaceVσ rdquo Applied Mathematics Letters vol 24 no 9 pp 1554ndash1560 2011

[31] MMursaleen and A K Noman ldquoCompactness of matrix operators on some new difference sequencespacesrdquo Linear Algebra and Its Applications vol 436 no 1 pp 41ndash52 2012

[32] M Mursaleen ldquoApplication of measure of noncompactness to infinite systems of differentialequa-tionsrdquo Canadian Mathematical Bulletin In press

[33] J Banas and M Lecko ldquoSolvability of infinite systems of differential equations in Banach sequencespacesrdquo Journal of Computational and Applied Mathematics vol 137 no 2 pp 363ndash375 2001

[34] R BellmanMethods of Nonliner Analysis II vol 12 Academic Press New York NY USA 1973[35] K Deimling Ordinary Differential Equations in Banach Spaces vol 596 of Lecture Notes in Mathematics

Springer Berlin Germany 1977[36] E Hille ldquoPathology of infinite systems of linear first order differential equations with constant

coefficientsrdquo Annali di Matematica Pura ed Applicata vol 55 pp 133ndash148 1961[37] M N Oguztoreli ldquoOn the neural equations of Cowan and Stein Irdquo Utilitas Mathematica vol 2 pp

305ndash317 1972[38] O A Zautykov ldquoDenumerable systems of differential equations and their applicationsrdquoDifferentsial-

cprime nye Uravneniya vol 1 pp 162ndash170 1965[39] K P Persidskii ldquoCountable system of differential equations and stability of their solutionsrdquo Izvestiya

Akademii Nauk Kazakhskoj SSR vol 7 pp 52ndash71 1959[40] K P Persidski ldquoCountable systems of differential equations and stability of their solutions III

Fundamental theorems on stability of solutions of countable many differential equationsrdquo IzvestiyaAkademii Nauk Kazakhskoj SSR vol 9 pp 11ndash34 1961

[41] K G Zautykov and K G Valeev Beskonechnye Sistemy Differentsialnykh Uravnenii Izdat ldquoNaukardquoKazah SSR Alma-Ata 1974

[42] A Voigt ldquoLine method approximations to the Cauchy problem for nonlinear parabolic differentialequationsrdquo Numerische Mathematik vol 23 pp 23ndash36 1974

[43] W Walter Differential and Integral Inequalities Springer New York NY USA 1970[44] M Frigon ldquoFixed point results for generalized contractions in gauge spaces and applicationsrdquo

Proceedings of the American Mathematical Society vol 128 no 10 pp 2957ndash2965 2000[45] G Herzog ldquoOn ordinary linear differential equations in CJ rdquo Demonstratio Mathematica vol 28 no 2

pp 383ndash398 1995[46] G Herzog ldquoOn Lipschitz conditions for ordinary differential equations in Frechet spacesrdquo

Czechoslovak Mathematical Journal vol 48(123) no 1 pp 95ndash103 1998


[47] R Lemmert and A Weckbach ldquoCharakterisierungen zeilenendlicher Matrizen mit abzahlbaremSpektrumrdquoMathematische Zeitschrift vol 188 no 1 pp 119ndash124 1984

[48] R Lemmert ldquoOn ordinary differential equations in locally convex spacesrdquoNonlinear Analysis TheoryMethods amp Applications vol 10 no 12 pp 1385ndash1390 1986

[49] W Mlak and C Olech ldquoIntegration of infinite systems of differential inequalitiesrdquo Annales PoloniciMathematici vol 13 pp 105ndash112 1963

[50] K Moszynski and A Pokrzywa ldquoSur les systemes infinis drsquoequations differentielles ordinaires danscertains espaces de Frechetrdquo Dissertationes Mathematicae vol 115 29 pages 1974

[51] H Monch and G-F von Harten ldquoOn the Cauchy problem for ordinary differential equations inBanach spacesrdquo Archives Mathematiques vol 39 no 2 pp 153ndash160 1982

[52] D OrsquoRegan andM Meehan Existence Theory for Nonlinear Integral and Integrodifferential Equations vol445 of Mathematics and Its Applications Kluwer Academic Publishers Dordrecht The Netherlands1998

[53] S Szufla ldquoOn the existence of solutions of differential equations in Banach spacesrdquo LrsquoAcademiePolonaise des Sciences vol 30 no 11-12 pp 507ndash515 1982


Research ArticleSome Existence Results for ImpulsiveNonlinear Fractional Differential Equations withClosed Boundary Conditions

Hilmi Ergoren1 and Adem Kilicman2

1 Department of Mathematics Faculty of Sciences Yuzuncu Yil University 65080 Van Turkey2 Department of Mathematics and Institute for Mathematical Research University Putra Malaysia43400 Serdang Malaysia

Correspondence should be addressed to Adem Kilicman akilicmanputraupmedumy

Received 30 September 2012 Accepted 22 October 2012


Copyright q 2012 H Ergoren and A Kilicman This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We investigate some existence results for the solutions to impulsive fractional differentialequations having closed boundary conditions Our results are based on contracting mappingprinciple and Burton-Kirk fixed point theorem

1 Introduction

This paper considers the existence and uniqueness of the solutions to the closed boundaryvalue problem (BVP) for the following impulsive fractional differential equation

CDαx(t) = f(t x(t)) t isin J = [0 T] t = tk 1 lt α le 2

Δx(tk) = Ik(x(tminusk)) Δxprime(tk) = Ilowastk

(x(tminusk)) k = 1 2 p

x(T) = ax(0) + bTxprime(0) Txprime(T) = cx(0) + dTxprime(0)

(11)

where CDα is Caputo fractional derivative f isin C(J times RR) Ik Ilowastk isin C(RR)

Δx(tk) = x(t+k) minus x(tminusk

) (12)


with

x(t+k)= lim

hrarr 0+x(tk + h) x

(tminusk)= lim

hrarr 0minusx(tk + h) (13)

and Δxprime(tk) has a similar meaning for xprime(t) where

0 = t0 lt t1 lt t2 lt middot middot middot lt tp lt tp+1 = T (14)

a b c and d are real constants with Δ = c(1 minus b) + (1 minus a)(1 minus d)= 0The boundary value problems for nonlinear fractional differential equations have been

addressed by several researchers during last decades That is why the fractional derivativesserve an excellent tool for the description of hereditary properties of various materials andprocesses Actually fractional differential equations arise in many engineering and scientificdisciplines such as physics chemistry biology electrochemistry electromagnetic controltheory economics signal and image processing aerodynamics and porous media (see [1ndash7]) For some recent development see for example [8ndash14]

On the other hand theory of impulsive differential equations for integer orderhas become important and found its extensive applications in mathematical modeling ofphenomena and practical situations in both physical and social sciences in recent years Onecan see a noticeable development in impulsive theory For instance for the general theoryand applications of impulsive differential equations we refer the readers to [15ndash17]

Moreover boundary value problems for impulsive fractional differential equationshave been studied by some authors (see [18ndash20] and references therein) However to thebest of our knowledge there is no study considering closed boundary value problems forimpulsive fractional differential equations

Here we notice that the closed boundary conditions in (11) include quasi-periodicboundary conditions (b = c = 0) and interpolate between periodic (a = d = 1 b = c = 0) andantiperiodic (a = d = minus1 b = c = 0) boundary conditions

Motivated by the mentioned recent work above in this study we investigate theexistence and uniqueness of solutions to the closed boundary value problem for impulsivefractional differential equation (11) Throughout this paper in Section 2 we present somenotations and preliminary results about fractional calculus and differential equations to beused in the following sections In Section 3 we discuss some existence and uniqueness resultsfor solutions of BVP (11) that is the first one is based on Banachrsquos fixed point theoremthe second one is based on the Burton-Kirk fixed point theorem At the end we give anillustrative example for our results

2 Preliminaries

Let us set J0 = [0 t1] J1 = (t1 t2] Jkminus1 = (tkminus1 tk] Jk = (tk tk+1] J prime = [0 T] t1 t2 tpand introduce the set of functions

PC(J R) = x J rarr R x isin C((tk tk+1] R) k = 0 1 2 p and there exist x(t+k) andx(tminusk) k = 1 2 p with x(tminusk) = x(tk) and

PC1(J R) = x isin PC(J R) xprime isin C((tk tk+1] R) k = 0 1 2 p and there exist xprime(t+k)

and xprime(tminusk) k = 1 2 p with xprime(tminusk) = xprime(tk) which is a Banach space with the norm x =suptisinJxPC xprimePCwhere xPC = sup|x(t)| t isin J

The following definitions and lemmas were given in [4]


Definition 21 The fractional (arbitrary) order integral of the function h isin L1(J R+) of orderα isin R+ is defined by

Iα0h(t) =1

Γ(α)

int t

0(t minus s)αminus1h(s)ds (21)

where Γ(middot) is the Euler gamma function

Definition 22 For a function h given on the interval J Caputo fractional derivative of orderα gt 0 is defined by

CDα0+h(t) =

1Γ(n minus α)

int t

0(t minus s)nminusαminus1h(n)(s)ds n = [α] + 1 (22)

where the function h(t) has absolutely continuous derivatives up to order (n minus 1)

Lemma 23 Let α gt 0 then the differential equation

CDαh(t) = 0 (23)

has solutions

h(t) = c0 + c1t + c2t2 + middot middot middot + cnminus1tnminus1 ci isin R i = 0 1 2 n minus 1 n = [α] + 1 (24)

The following lemma was given in [4 10]

Lemma 24 Let α gt 0 then

IαCDαh(t) = h(t) + c0 + c1t + c2t2 + middot middot middot + cnminus1tnminus1 (25)

for some ci isin R i = 0 1 2 n minus 1 n = [α] + 1

The following theorem is known as Burton-Kirk fixed point theorem and proved in[21]

Theorem 25 Let X be a Banach space and A D X rarr X two operators satisfying

(a) A is a contraction and

(b) D is completely continuous

Then either

(i) the operator equation x = A(x) +D(x) has a solution or(ii) the set ε = x isin X x = λA(xλ) + λD(x) is unbounded for λ isin (0 1)

Theorem 26 (see [22] Banachrsquos fixed point theorem) Let S be a nonempty closed subset of aBanach space X then any contraction mapping T of S into itself has a unique fixed point


Next we prove the following lemma

Lemma 27 Let 1 lt α le 2 and let h J rarr R be continuous A function x(t) is a solution of thefractional integral equation

x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

0

(t minus s)αminus1Γ(α)

h(s)ds + Ω1(t)Λ1 + Ω2(t)Λ2 t isin J0int t

tk


h(s)ds + [1 + Ω1(t)]ksum

i=1

int ti

timinus1

(ti minus s)αminus1Γ(α)

h(s)ds

+[(T minus tk)Ω1(t) + Ω2(t) + (t minus tk)]ksum

i=1

int ti

timinus1

(ti minus s)αminus2Γ(α minus 1)

h(s)ds

+[1 + Ω1(t)]kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds + Ω1(t)intT

tk

(T minus s)αminus1Γ(α)

h(s)ds

+Ω2(t)intT

tk

(T minus s)αminus2Γ(α minus 1)

h(s)ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))

+ Ilowastk(x(tminusk))]

+Ω1(t)kminus1sum

i=1

(T minus ti)Ilowasti(x(tminusi))

+ Ω2(t)kminus1sum

i=1

Ilowasti(x(tminusi))

+kminus1sum

i=1

(t minus ti)Ilowasti(x(tminusi)) t isin Jk k = 1 2 p

(26)

if and only if x(t) is a solution of the fractional BVP

CDαx(t) = h(t) t isin J primeΔx(tk) = Ik

(x(tminusk)) Δxprime(tk) = Ilowastk

(x(tminusk))


(27)

where

Λ1 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(T minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1


+ Ilowastk(x(tminusk))


Λ2 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1


Ω1(t) = minus (1 minus d)Δ

minus ct

TΔ

Ω2(t) =(1 minus b)T

Δminus (1 minus a)t

Δ

(28)

Proof Let x be the solution of (27) If t isin J0 then Lemma 24 implies that

x(t) = Iαh(t) minus c0 minus c1t =int t

0


h(s)ds minus c0 minus c1t

xprime(t) =int t

0

(t minus s)αminus2Γ(α minus 1)

h(s)ds minus c1(29)

for some c0 c1 isin RIf t isin J1 then Lemma 24 implies that

x(t) =int t

t1


h(s)ds minus d0 minus d1(t minus t1)

xprime(t) =int t

t1


h(s)ds minus d1(210)

for some d0 d1 isin R Thus we have

x(tminus1)=int t1

t0

(t1 minus s)αminus1Γ(α)

h(s)ds minus c0 minus c1t1 x(t+1)= minusd0

xprime(tminus1)=int t1

t0

(t1 minus s)αminus2Γ(α minus 1)

h(s)ds minus c1 xprime(t+1)= minusd1

(211)

Observing that

Δx(t1) = x(t+1) minus x(tminus1

)= I1(x(tminus1)) Δxprime(t1) = xprime(t+1

) minus xprime(tminus1)= Ilowast1(x(tminus1)) (212)


then we have

minusd0 =int t1

t0


h(s)ds minus c0 minus c1t1 + I1(x(tminus1))

minusd1 =int t1

t0


h(s)ds minus c1 + Ilowast1(x(tminus1))

(213)

hence for t isin (t1 t2]

x(t) =int t

t1


h(s)ds +int t1

t0


h(s)ds

+ (t minus t1)int t1

t0


h(s)ds + I1(x(tminus1))

+ (t minus t1)Ilowast1(x(tminus1)) minus c0 minus c1t

xprime(t) =int t

t1


h(s)ds +int t1

t0


h(s)ds + Ilowast1(x(tminus1)) minus c1

(214)

If t isin J2 then Lemma 24 implies that

x(t) =int t

t2


h(s)ds minus e0 minus e1(t minus t2)

xprime(t) =int t

t2


h(s)ds minus e1(215)

for some e0 e1 isin R Thus we have

x(tminus2)=int t2

t1


h(s)ds +int t1

t0


h(s)ds

+ (t2 minus t1)int t1

t0



+ (t2 minus t1)Ilowast1(x(tminus1)) minus c0 minus c1t2

x(t+2)= minus e0


t1


h(s)ds +int t1

t0



xprime(t+2)= minus e1

(216)

Similarly we observe that





thus we have

minuse0 = x(tminus2)+ I2(x(tminus2))

minuse1 = xprime(tminus2)+ Ilowast2(x(tminus2))

(218)

Hence for t isin (t2 t3]

x(t) =int t

t2


h(s)ds +int t1

t0


h(s)ds +int t2

t1


h(s)ds


t0


h(s)ds

+ (t minus t2)[int t1

t0


h(s)ds +int t2

t1


h(s)ds

]

+ I1(x(tminus1))

+ I2(x(tminus2))

+ (t minus t1)Ilowast1(x(tminus1))

+ Ilowast2(x(tminus2)) minus c0 minus c1t

(219)

By a similar process if t isin Jk then again from Lemma 24 we get

x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(t minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1

(t minus ti)Ilowasti(x(tminusi))

+ Ilowastk(x(tminusk)) minus c0 minus c1t

xprime(t) =

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1

Ilowasti(x(tminusi)) minus c1

(220)

Now if we apply the conditions

x(T) = ax(0) + bTxprime(0) Txprime(T) = cx(0) + dTxprime(0) (221)

we have

Λ1 = (1 minus a)c0 + T(1 minus b)c1

Λ2 = minus cTc0 + (1 minus d)c1

minusc0 = minus (1 minus d)Λ1

Δ+(1 minus b)TΛ2

Δ

minusc1 = minuscΛ1

TΔminus (1 minus a)Λ2

Δ

(222)


In view of the relations (28) when the values of minusc0 and minusc1 are replaced in (29) and (220)the integral equation (27) is obtained

Conversely assume that x satisfies the impulsive fractional integral equation (26)then by direct computation it can be seen that the solution given by (26) satisfies (27) Theproof is complete

3 Main Results

Definition 31 A function x isin PC1(J R) with its α-derivative existing on J prime is said to bea solution of (11) if x satisfies the equation CD

αx(t) = f(t x(t)) on J prime and satisfies the

conditions


(x(tminusk))

x(T) = ax(0) + bTxprime(0) Txprime(T) = cx(0) + dTxprime(0)(31)

For the sake of convenience we define

Ωlowast1 = sup

tisinJ|Ω1(t)| Ωlowast

2 = suptisinJ

|Ω2(t)| Ωlowastlowast1 = sup

tisinJ

∣∣Ωprime1(t)∣∣ Ωlowastlowast

2 = suptisinJ

∣∣Ωprime2(t)∣∣ (32)

The followings are main results of this paper

Theorem 32 Assume that

(A1) the function f J times R rarr R is continuous and there exists a constant L1 gt 0 such thatf(t u) minus f(t v) le L1u minus v for all t isin J and u v isin R

(A2) Ik Ilowastk R rarr R are continuous and there exist constants L2 gt 0 and L3 gt 0 suchthat Ik(u) minus Ik(v) le L2u minus v Ilowast

k(u) minus Ilowast

k(v) le L3u minus v for each u v isin R and

k = 1 2 p

Moreover consider the following

[L1T

α

Γ(α + 1)(1 + Ωlowast

1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

lt 1

(33)

Then BVP (11) has a unique solution on J

Proof Define an operator F PC1(J R) rarr PC1(J R) by

(Fx)(t) =int t

tk


f(s x(s))ds + [1 + Ω1(t)]ksum

i=1

int ti

timinus1


f(s x(s))ds

+ [(T minus tk)Ω1(t) + Ω2(t) + (t minus tk)]ksum

i=1

int ti

timinus1


f(s x(s))ds


+ [1 + Ω1(t)]kminus1sum

i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk


f(s x(s))ds

+ Ω2(t)intT

tk


f(s x(s))ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1

Ilowasti(x(tminusi))+

kminus1sum

i=1


(34)

Now for x y isin PC(J R) and for each t isin J we obtain

∣∣(Fx)(t) minus (Fy)(t)∣∣ leint t

tk


∣∣f(s x(s)) minus f(s y(s))∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



+ |(T minus tk)Ω1(t) + Ω2(t) + (t minus tk)|

timesksum

i=1

int ti

timinus1



+ |1 + Ω1(t)|kminus1sum

i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



+ |1 + Ω1(t)|[

ksum

i=1

∣∣Ii(x(tminusi)) minus Ii

(y(tminusi))∣∣ +

∣∣Ilowastk(x(tminusk)) minus Ilowastk

(y(tminusk))∣∣]

+ |Ω1(t)|kminus1sum

i=1

(T minus ti)∣∣Ilowasti(x(tminusi)) minus Ilowasti

(y(tminusi))∣∣


i=1

∣∣Ilowasti(x(tminusi)) minus Ilowasti

(y(tminusi))∣∣

+kminus1sum

i=1

|t minus ti|∣∣Ilowasti(x(tminusi)) minus Ilowasti

(y(tminusi))∣∣


∥∥(Fx)(t) minus (Fy)(t)∥∥ le

[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]∥∥x(s) minus y(s)∥∥

(35)

Therefore by (33) the operator F is a contraction mapping In a consequence of Banachrsquosfixed theorem the BVP (11) has a unique solution Now our second result relies on theBurton-Kirk fixed point theorem

Theorem 33 Assume that (A1)-(A2) hold and

(A3) there exist constantsM1 gt 0M2 gt 0M3 gt 0 such that f(t u) le M1 Ik(u) le M2Ilowast

k(u) leM3 for each u v isin R and k = 1 2 p

Then the BVP (11) has at least one solution on J

Proof We define the operators AD PC1(J R) rarr PC1(J R) by

(Ax)(t) = [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(Dx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds


i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk


f(s x(s))ds + Ω2(t)intT

tk


f(s x(s))ds

(36)

It is obvious that A is contraction mapping for

(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3 lt 1 (37)

Now in order to check that D is completely continuous let us follow the sequence of thefollowing steps


Step 1 (D is continuous) Let xn be a sequence such that xn rarr x in PC(J R) Then for t isin J we have

|(Dxn)(t) minus (Dx)(t)| leint t

tk


∣∣f(s xn(s)) minus f(s x(s))

∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



∣∣ds


timesksum

i=1

int ti

timinus1



∣∣ds


i=1

(tk minus ti)int ti

timinus1


∣∣f(s xn(s)) minus f(s x(s))∣∣ds

+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



(38)

Since f is continuous function we get

(Dxn)(t) minus (Dx)(t) minusrarr 0 as n minusrarr infin (39)

Step 2 (D maps bounded sets into bounded sets in PC(J R)) Indeed it is enough to showthat for any r gt 0 there exists a positive constant l such that for each x isin Br = x isin PC(J R) x le r we have D(x) le l By (A3) we have for each t isin J

|(Dx)(t)| leint t

tk


∣∣f(s x(s))∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



+ |(T minus tk)Ω1(t) + Ω2(t) + (t minus tk)|ksum

i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1




+ |Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + |Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

D(x) le M1Tα


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)= l

(310)

Step 3 (D maps bounded sets into equicontinuous sets in PC1(J R)) Let τ1 τ2 isin Jk 0 le k le pwith τ1 lt τ2 and let Br be a bounded set of PC1(J R) as in Step 2 and let x isin Br Then

∣∣(Dy

)(τ2) minus (D)(τ1)

∣∣ leint τ2

τ1

∣∣∣(Dy)prime(s)

∣∣∣ds le L(τ2 minus τ1) (311)

where

∣∣(Dx)prime(t)∣∣ leint t

tk


∣∣f(s x(s))∣∣ds +

∣∣Ωprime1(t)∣∣

ksum

i=1

int ti

timinus1



+∣∣(T minus tk)Ωprime

1(t) + Ωprime2(t) + 1

∣∣ksum

i=1

int ti

timinus1



+∣∣Ωprime

1(t)∣∣kminus1sum

i=1

(tk minus ti)int ti

timinus1



+∣∣Ωprime

1(t)∣∣intT

tk


∣∣f(s x(s))∣∣ds +

∣∣Ωprime2(t)∣∣intT

tk



le M1Tα

Γ(α + 1)Ωlowast

1

(1 + p + 2pα

)+M1T

αminus1

Γ(α)(1 + Ωlowast

2)(1 + p

)= L

(312)

This implies that A is equicontinuous on all the subintervals Jk k = 0 1 2 p Thereforeby the Arzela-Ascoli Theorem the operator D PC1(J R) rarr PC1(J R) is completelycontinuous

To conclude the existence of a fixed point of the operator A + D it remains to showthat the set

ε =x isin X x = λA

(xλ

)+ λD(x) for some λ isin (0 1)

(313)

is boundedLet x isin ε for each t isin J

x(t) = λD(x)(t) + λA(xλ

)(t) (314)


Hence from (A3) we have

|x(t)| le λ

int t

tk


∣∣f(s x(s))

∣∣ds + λ|1 + Ω1(t)|

ksum

i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|(T minus tk)Ω1(t) + Ω2(t) + (t minus tk)|ksum

i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|1 + Ω1(t)|kminus1sum

i=1

(tk minus ti)int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + λ|Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

+ λ|1 + Ω1(t)|[

ksum

i=1

∣∣∣∣∣Ii

(x(tminusi)

λ

)∣∣∣∣∣+

∣∣∣∣∣Ilowastk

(x(tminusk

)

λ

)∣∣∣∣∣

]

+ λ|Ω1(t)|kminus1sum

i=1

(T minus ti)∣∣∣∣∣Ilowasti

(x(tminusi)

λ

)∣∣∣∣∣


i=1

∣∣∣∣∣Ilowasti

(x(tminusi)

λ

)∣∣∣∣∣+ λ

kminus1sum

i=1

(t minus ti)∣∣∣∣∣Ilowasti

(x(tminusi)

λ

)∣∣∣∣∣

x(t) le[M1T

α


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pM2 +M3

)+(Ωlowast

1T + Ωlowast2 + T

)pM3

]

(315)

Consequently we conclude the result of our theorem based on the Burton-Kirk fixed pointtheorem

4 An Example

Consider the following impulsive fractional boundary value problem

CD32x(t) =

sin 2t|x(t)|(t + 5)2(1 + |x(t)|)

t isin [0 1] t =13

Δx(13

)=

|x(1minus3)|15 + |x(1minus3)| Δxprime

(13

)=

|xprime(1minus3)|10 + |xprime(1minus3)|

x(1) = 5x(0) minus 2xprime(0) xprime(1) = x(0) + 4xprime(0)

(41)


Here a = 5 b = minus2 c = 1 d = 4 α = 32 T = 1 p = 1 Obviously L1 = 125 L2 = 115L3 = 110 Ωlowast

1 = 415 Ωlowast2 = 715 Further

[L1T

α


1

)(1 + p(1 + L2) + 2pα + L3

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

=464

1125radicπ

+173450

lt 1

(42)

Since the assumptions of Theorem 32 are satisfied the closed boundary value problem (41)has a unique solution on [0 1] Moreover it is easy to check the conclusion of Theorem 33

Acknowledgments

The authors would like to express their sincere thanks and gratitude to the reviewer(s) fortheir valuable comments and suggestions for the improvement of this paper The secondauthor also gratefully acknowledges that this research was partially supported by theUniversity Putra Malaysia under the Research University Grant Scheme 05-01-09-0720RU

References

[1] K Diethelm The Analysis of Fractional Differential Equations Springer 2010[2] N Heymans and I Podlubny ldquoPhysical interpretation of initial conditions for fractional differential

equationswith Riemann Liouville fractional derivativesrdquo Rheologica Acta vol 45 no 5 pp 765ndash7722006

[3] R Hilfer Applications of Fractional Calculus in Physics World Scientific Singapore 2000[4] A A Kilbas H M Srivastava and J J Trujillo Theory and Applications of Fractional Differential

Equations vol 204 of North-Holland Mathematics Studies Elsevier Science BV Amsterdam TheNetherlands 2006

[5] V Lakshmikantham S Leela and J Vasundhara Devi Theory of Fractional Dynamic SystemsCambridge Academic Publishers Cambridge UK 2009

[6] J Sabatier O P Agrawal and J A T Machado Eds Advances in Fractional Calculus TheoreticalDevelopmentsand Applications in Physics and Engineering Springer Dordrecht The Netherlands 2007

[7] S G Samko A A Kilbas and O I Marichev Fractional Integrals and Derivatives Theory andApplications Gordon and Breach Science Yverdon Switzerland 1993

[8] R P Agarwal M Benchohra and S Hamani ldquoBoundary value problems for fractional differentialequationsrdquo Georgian Mathematical Journal vol 16 no 3 pp 401ndash411 2009

[9] R P Agarwal M Benchohra and S Hamani ldquoA survey on existence results for boundary valueproblems of nonlinear fractional differential equations and inclusionsrdquoActa ApplicandaeMathematicaevol 109 no 3 pp 973ndash1033 2010


[11] M Belmekki J J Nieto and R Rodrıguez-Lopez ldquoExistence of periodic solution for a nonlinearfractional differential equationrdquo Boundary Value Problems vol 2009 Article ID 324561 18 pages 2009

[12] M Benchohra S Hamani and S K Ntouyas ldquoBoundary value problems for differential equationswith fractional order and nonlocal conditionsrdquo Nonlinear Analysis vol 71 no 7-8 pp 2391ndash23962009

[13] H A H Salem ldquoOn the fractional orderm-point boundary value problem in reflexive Banach spacesand weak topologiesrdquo Journal of Computational and Applied Mathematics vol 224 no 2 pp 565ndash5722009

[14] W Zhong and W Lin ldquoNonlocal and multiple-point boundary value problem for fractionaldifferential equationsrdquo Computers amp Mathematics with Applications vol 59 no 3 pp 1345ndash1351 2010


[15] V Lakshmikantham D D Baınov and P S Simeonov Theory of Impulsive Differential Equations WorldScientific Singapore 1989

[16] Y V Rogovchenko ldquoImpulsive evolution systems main results and new trendsrdquo Dynamics ofContinuous Discrete and Impulsive Systems vol 3 no 1 pp 57ndash88 1997

[17] A M Samoılenko and N A Perestyuk Impulsive Differential Equations World Scientific Singapore1995


[19] M Benchohra and B A Slimani ldquoExistence and uniqueness of solutions to impulsive fractionaldifferential equationsrdquo Electronic Journal of Differential Equations vol 10 pp 1ndash11 2009


[21] T A Burton and C Kirk ldquoA fixed point theorem of Krasnoselskii-Schaefer typerdquo MathematischeNachrichten vol 189 pp 23ndash31 1998

[22] A Granas and J Dugundji Fixed Point Theory Springer Monographs in Mathematics Springer NewYork NY USA 2003


Research ArticleSome New Estimates for the Error ofSimpson Integration Rule

Mohammad Masjed-Jamei1 Marwan A Kutbi2and Nawab Hussain2

1 Department of Mathematics K N Toosi University of Technology Tehran 19697 Iran2 Department of Mathematics King AbdulAziz University Jeddah 21589 Saudi Arabia

Correspondence should be addressed to Nawab Hussain nhusainkauedusa


Academic Editor Mohammad Mursaleen

Copyright q 2012 Mohammad Masjed-Jamei et al This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We obtain some new estimates for the error of Simpson integration rule which develop availableresults in the literature Indeed we introduce three main estimates for the residue of Simpsonintegration rule in L1[a b] and Linfin[a b] spaces where the compactness of the interval [a b] playsa crucial role

1 Introduction

A general (n + 1)-point-weighted quadrature formula is denoted by

intb

a

w(x)f(x)dx =nsum

k=0

wkf(xk) + Rn+1[f] (11)

where w(x) is a positive weight function on [a b] xknk=0 and wknk=0 are respectivelynodes and weight coefficients and Rn+1[f] is the corresponding error [1]

LetΠd be the set of algebraic polynomials of degree at most d The quadrature formula(11) has degree of exactness d if for every p isin Πd we have Rn+1[p] = 0 In addition ifRn+1[p]= 0 for some Πd+1 formula (11) has precise degree of exactness d

The convergence order of quadrature rule (11) depends on the smoothness of thefunction f as well as on its degree of exactness It is well known that for given n + 1 mutuallydifferent nodes xknk=0 we can always achieve a degree of exactness d = n by interpolating


at these nodes and integrating the interpolated polynomial instead of f Namely taking thenode polynomia

Ψn+1(x) =nprod

k=0

(x minus xk) (12)

by integrating the Lagrange interpolation formula

f(x) =nsum

k=0

f(xk)L(xxk) + rn+1(f x) (13)

where

L(xxk) =Ψn+1(x)

Ψprimen+1(xk)(x minus xk)

(k = 0 1 n) (14)

we obtain (11) with

wk =1

Ψprimen+1(xk)

intb

a

Ψn+1(x)w(x)x minus xk dx (k = 0 1 n)

Rn+1[f]=intb

a

rn+1(f x)w(x)dx

(15)

Note that for each f isin Πn we have rn+1(f x) = 0 and therefore Rn+1[f] = 0Quadrature formulae obtained in this way are known as interpolatory Usually the

simplest interpolatory quadrature formula of type (11)with predetermined nodes xknk=0 isin[a b] is called a weighted Newton-Cotes formula For w(x) = 1 and the equidistant nodesxknk=0 = a + khnk=0 with h = (b minus a)n the classical Newton-Cotes formulas are derivedOne of the important cases of the classical Newton-Cotes formulas is the well-knownSimpsonrsquos rule

intb

a

f(t)dt =b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)+ E(f) (16)

In this direction Simpson inequality [2ndash7] gives an error bound for the above quadraturerule There are few known ways to estimate the residue value in (16) The main aim of thispaper is to give three new estimations for E(f) in L1[a b] and Linfin[a b] spaces

Let Lp[a b] (1 le p ltinfin) denote the space of p-power integrable functions on the inter-val [a b]with the standard norm

∥∥f∥∥p =

(intb

a

∣∣f(t)∣∣pdt

)1p

(17)

and Linfin[a b] the space of all essentially bounded functions on [a b]with the norm∥∥f∥∥infin = ess sup

xisin[ab]

∣∣f(x)∣∣ (18)


If f isin L1[a b] and g isin Linfin[a b] then the following inequality is well known

∣∣∣∣∣

intb

a

f(x)g(x)dx

∣∣∣∣∣le ∥∥f∥∥1

∥∥g∥∥infin (19)

Recently in [8] a main inequality has been introduced which can estimate the error ofSimpson quadrature rule too

Theorem A Let f I rarr R where I is an interval be a differentiable function in the interior I0of Iand let [a b] sub I0 If α0 β0 are two real constants such that α0 le f prime(t) le β0 for all t isin [a b] then forany λ isin [12 1] and all x isin [(a + (2λ minus 1)b)2λ (b + (2λ minus 1)a)2λ] sube [a b] we have

∣∣∣∣∣f(x) minus 1

λ(b minus a)intb

a

f(t)dt minus f(b) minus f(a)b minus a x +

(2λ minus 1)a + b2λ(b minus a) f(b) minus a + (2λ minus 1)b

2λ(b minus a) f(a)

∣∣∣∣∣

le β0 minus α04(b minus a)

λ2 + (1 minus λ)2λ

((x minus a)2 + (b minus x)2

)

(110)

As is observed replacing x = (a + b)2 and λ = 23 in (110) gives an error bound for theSimpson rule as

∣∣∣∣∣

intb

a

f(t)dt minus b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5

72(b minus a)2(β0 minus α0

) (111)

To introduce three new error bounds for the Simpson quadrature rule in L1[a b] and Linfin[a b]spaces we first consider the following kernel on [a b]

K(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

t minus 5a + b6

t isin[aa + b2

]

t minus a + 5b6

t isin(a + b2

b

]

(112)

After some calculations it can be directly concluded that

intb

a

f prime(t)K(t)dt =b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt (113)

maxtisin[ab]

|K(t)| = 13(b minus a) (114)


2 Main Results

Theorem 21 Let f I rarr R where I is an interval be a function differentiable in the interior I0of Iand let [a b] sub I0 If α(x) le f prime(x) le β(x) for any α β isin C[a b] and x isin [a b] then the followinginequality holds

m1 =int (5a+b)6

a

(t minus 5a + b

6

)β(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)α(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)β(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)α(t)dt

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt le

M1 =int (5a+b)6

a

(t minus 5a + b

6

)α(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)β(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)α(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)β(t)dt

(21)

Proof By referring to the kernel (112) and identity (113)we first have

intb

a

K(t)(f prime(t) minus α(t) + β(t)

2

)dt

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus 12

(intb

a

K(t)(α(t) + β(t)

)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)

(22)

On the other hand the given assumption α(t) le f prime(t) le β(t) results in

∣∣∣∣fprime(t) minus α(t) + β(t)

2

∣∣∣∣ leβ(t) minus α(t)

2 (23)


Therefore one can conclude from (22) and (23) that

∣∣∣∣∣b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a


2

)dt

∣∣∣∣∣leintb

a

|K(t)| β(t) minus α(t)2

dt

=12

(int (a+b)2

a

∣∣∣∣ t minus

5a + b6

∣∣∣∣(β(t) minus α(t))dt +

intb

(a+b)2


a + 5b6

∣∣∣∣(β(t) minus α(t))dt

)

(24)

After rearranging (24) we obtain

m1 =int (a+b)2

a

((t minus 5a + b

6minus∣∣∣∣ t minus

5a + b6

∣∣∣∣

)β(t)2

+(t minus 5a + b

6+∣∣∣∣ t minus

5a + b6

∣∣∣∣

)α(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)β(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)α(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)β(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)α(x)dx

M1 =int (a+b)2

a

((t minus 5a + b

6minus∣∣∣∣t minus

5a + b6

∣∣∣∣

)α(t)2

+(t minus 5a + b

6+∣∣∣∣t minus

5a + b6

∣∣∣∣

)β(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)α(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)β(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)α(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)β(x)dx

(25)

The advantage of Theorem 21 is that necessary computations in bounds m1 and M1

are just in terms of the preassigned functions α(t) β(t) (not f prime)


Special Case 1

Substituting α(x) = α1x + α0 = 0 and β(x) = β1x + β0 = 0 in (21) gives

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5(b minus a)2

144((β1 minus α1

)(a + b) + 2

(β0 minus α0

))

(26)

In particular replacing α1 = β1 = 0 in above inequality leads to one of the results of [9] as

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (27)

Remark 22 Although α(x) le f prime(x) le β(x) is a straightforward condition in Theorem 21however sometimes one might not be able to easily obtain both bounds of α(x) and β(x) forf prime In this case we can make use of two analogue theorems The first one would be helpfulwhen f prime is unbounded from above and the second one would be helpful when f prime is unbound-ed from below

Theorem 23 Let f I rarr R where I is an interval be a function differentiable in the interior I0 ofI and let [a b] sub I0 If α(x) le f prime(x) for any α isin C[a b] and x isin [a b] then

int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt minus b minus a

3

(

f(b) minus f(a) minusintb

a

α(t)dt

)

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t) dt

leint (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

+b minus a3

(


a

α(t)dt

)

(28)

Proof Since

intb

a

K(t)(f prime(t) minus α(t))dt = b minus a

6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)α(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)

(29)


so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a

K(t)(f prime(t) minus α(t))dt


a

|K(t)|(f prime(t) minus α(t))dt

le maxtisin[ab]

|K(t)|intb

a

(f prime(t) minus α(t))dt = b minus a

3

(


a

α(t)dt

)

(210)

After rearranging (210) the main inequality (28)will be derived

Special Case 2

If α(x) = α1x + α0 = 0 then (28) becomes

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le (b minus a)2

3

(f(b) minus f(a)

b minus a minus(α0 +

a + b2

α1

))

(211)

if and only if α1x + α0 le f prime(x) for all x isin [a b] In particular replacing α1 = 0 in aboveinequality leads to [10 Theorem 1 relation (4)] as follows

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)

b minus a minus α0) (212)

Theorem 24 Let f I rarr R where I is an interval be a function differentiable in the interior I0 ofI and let [a b] sub I0 If f prime(x) le β(x) for any β isin C[a b] and x isin [a b] then

int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt minus b minus a

3

(intb

a

β(t)dt minus f(b) + f(a))

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

leint (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt +

b minus a3

(intb

a


(213)


Proof Since

intb

a

K(t)(f prime(t) minus β(t))dt

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)β(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)

(214)

so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

int

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a

|K(t)|(β(t) minus f prime(t))dt

le maxtisin[ab]

|K(t)|intb

a

(β(t) minus f prime(t)

)dt =

b minus a3

(int b

a


(215)

After rearranging (215) the main inequality (213) will be derived

Special Case 3

If β(x) = β1x + β0 = 0 in (213) then

∣∣∣∣∣

intb

a

f(t) dt minus b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

) ∣∣∣∣∣le (b minus a)2

3

(β0 +

a + b2

β1 minusf(b) minus f(a)

b minus a)

(216)

if and only if f prime(x) le β1x + β0 for all x isin [a b] In particular replacing β1 = 0 in aboveinequality leads to [10 Theorem 1 relation (5)] as follows

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 minus

f(b) minus f(a)b minus a

) (217)


Acknowledgment

The second and third authors gratefully acknowledge the support provided by the Deanshipof Scientific Research (DSR) King Abdulaziz University during this research

References

[1] W Gautschi Numerical Analysis An Introduction Birkhauser Boston Mass USA 1997[2] P Cerone ldquoThree points rules in numerical integrationrdquo Nonlinear Analysis Theory Methods amp

Applications vol 47 no 4 pp 2341ndash2352 2001[3] D Cruz-Uribe and C J Neugebauer ldquoSharp error bounds for the trapezoidal rule and Simpsonrsquos

rulerdquo Journal of Inequalities in Pure and Applied Mathematics vol 3 no 4 article 49 pp 1ndash22 2002[4] S S Dragomir R P Agarwal and P Cerone ldquoOn Simpsonrsquos inequality and applicationsrdquo Journal of

Inequalities and Applications vol 5 no 6 pp 533ndash579 2000[5] S S Dragomir P Cerone and J Roumeliotis ldquoA new generalization of Ostrowskirsquos integral inequality

for mappings whose derivatives are bounded and applications in numerical integration and forspecial meansrdquo Applied Mathematics Letters vol 13 no 1 pp 19ndash25 2000

[6] S S Dragomir J Pecaric and S Wang ldquoThe unified treatment of trapezoid Simpson and Ostrowskitype inequality for monotonic mappings and applicationsrdquoMathematical and Computer Modelling vol31 no 6-7 pp 61ndash70 2000

[7] I Fedotov and S S Dragomir ldquoAn inequality of Ostrowski type and its applications for Simpsonrsquosrule and special meansrdquoMathematical Inequalities amp Applications vol 2 no 4 pp 491ndash499 1999

[8] M Masjed-Jamei ldquoA linear constructive approximation for integrable functions and a parametricquadrature model based on a generalization of Ostrowski-Gruss type inequalitiesrdquo ElectronicTransactions on Numerical Analysis vol 38 pp 218ndash232 2011

[9] M Matic ldquoImprovement of some inequalities of Euler-Gruss typerdquo Computers amp Mathematics withApplications vol 46 no 8-9 pp 1325ndash1336 2003

[10] N Ujevic ldquoNew error bounds for the Simpsonrsquos quadrature rule and applicationsrdquo Computers ampMathematics with Applications vol 53 no 1 pp 64ndash72 2007


Research ArticleTravelling Wave Solutions of theSchrodinger-Boussinesq System

Adem Kılıcman1 and Reza Abazari2

1 Department of Mathematics and Institute of Mathematical Research Universiti Putra Malaysia 43400UPM Serdang Selangor Malaysia

2 Young Researchers Club Ardabil Branch Islamic Azad University PO Box 5616954184 Ardabil Iran




Copyright q 2012 A Kılıcman and R Abazari This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We establish exact solutions for the Schrodinger-Boussinesq System iut + uxx minus auv = 0 vtt minus vxx +vxxxx minus b(|u|2)xx = 0 where a and b are real constants The (GprimeG)-expansion method is used toconstruct exact periodic and soliton solutions of this equation Our work is motivated by the factthat the (GprimeG)-expansion method provides not only more general forms of solutions but alsoperiodic and solitary waves As a result hyperbolic function solutions and trigonometric functionsolutions with parameters are obtained These solutions may be important and of significance forthe explanation of some practical physical problems

1 Introduction

It is well known that the nonlinear Schrodinger (NLS) equation models a wide rangeof physical phenomena including self-focusing of optical beams in nonlinear media themodulation of monochromatic waves propagation of Langmuir waves in plasmas and soforth The nonlinear Schrodinger equations play an important role in many areas of appliedphysics such as nonrelativistic quantum mechanics laser beam propagation Bose-Einsteincondensates and so on (see [1]) Some properties of solutions for the nonlinear Schrodingerequations on R

n have been extensively studied in the last two decades (eg see [2])The Boussinesq-type equations are essentially a class of models appearing in physics

and fluidmechanics The so-called Boussinesq equationwas originally derived by Boussinesq[3] to describe two-dimensional irrotational flows of an inviscid liquid in a uniformrectangular channel It also arises in a large range of physical phenomena including thepropagation of ion-sound waves in a plasma and nonlinear lattice waves The study on the


soliton solutions for various generalizations of the Boussinesq equation has recently attractedconsiderable attention frommanymathematicians and physicists (see [4]) We should remarkthat it was the first equation proposed in the literature to describe this kind of physicalphenomena This equation was also used by Zakharov [5] as a model of nonlinear stringand by Falk et al [6] in their study of shape-memory alloys

In the laser and plasma physics the Schrodinger-Boussinesq system (hereafter referredto as the SB-system) has been raised Consider the SB-system

iut + uxx minus auv = 0

vtt minus vxx + vxxxx minus b(|u|2)

xx= 0

(11)

where t gt 0 x isin [0 L] for some L gt 0 and a b are real constants Here u and v arerespectively a complex-valued and a real-valued function defined in space-time [0 L] times RThe SB-system is considered as a model of interactions between short and intermediatelong waves which is derived in describing the dynamics of Langmuir soliton formationand interaction in a plasma [7] and diatomic lattice system [8] The short wave termu(x t) [0 L] times R rarr C is described by a Schrodinger type equation with a potentialv(x t) [0 L] times R rarr R satisfying some sort of Boussinesq equation and representing theintermediate long wave The SB-system also appears in the study of interaction of solitonsin optics The solitary wave solutions and integrability of nonlinear SB-system has beenconsidered by several authors (see [7 8]) and the references therein

In the literature there is a wide variety of approaches to nonlinear problems forconstructing travelling wave solutions such as the inverse scattering method [9] Bcklundtransformation [10] Hirota bilinear method [11] Painleve expansion methods [12] and theWronskian determinant technique [13]

With the help of the computer software most of mentionedmethods are improved andmany other algebraic method proposed such as the tanhcothmethod [14] the Exp-functionmethod [15] and first integral method [16] But most of the methods may sometimes fail orcan only lead to a kind of special solution and the solution procedures become very complexas the degree of nonlinearity increases

Recently the (GprimeG)-expansion method firstly introduced by Wang et al [17] hasbecome widely used to search for various exact solutions of NLEEs [17ndash19]

The main idea of this method is that the traveling wave solutions of nonlinearequations can be expressed by a polynomial in (GprimeG) where G = G(ξ) satisfies the secondorder linear ordinary differential equation Gprimeprime(ξ) + λGprime(ξ) + μG(ξ) = 0 where ξ = kx +ωt andk ω are arbitrary constants The degree of this polynomial can be determined by consideringthe homogeneous balance between the highest order derivatives and the non-linear termsappearing in the given non-linear equations

Our first interest in the present work is in implementing the (GprimeG)-expansionmethodto stress its power in handling nonlinear equations so that one can apply it to models ofvarious types of nonlinearity The next interest is in the determination of exact traveling wavesolutions for the SB-system (11)


2 Description of the (GprimeG)-Expansion Method

The objective of this section is to outline the use of the (GprimeG)-expansion method for solvingcertain nonlinear partial differential equations (PDEs) Suppose that a nonlinear equationsay in two independent variables x and t is given by

P(u ux ut uxx uxt utt ) = 0 (21)

where u(x t) is an unknown function P is a polynomial in u(x t) and its various partialderivatives in which the highest order derivatives and nonlinear terms are involved Themain steps of the (GprimeG)-expansion method are the following

Step 1 Combining the independent variables x and t into one variable ξ = kx+ωt we supposethat

u(x t) = U(ξ) ξ = kx +ωt (22)

The travelling wave variable (22) permits us to reduce (21) to an ODE for u(x t) = U(ξ)namely

P(U kUprime ωUprime k2Uprimeprime kωUprimeprime ω2Uprimeprime

)= 0 (23)

where prime denotes derivative with respect to ξ

Step 2 We assume that the solution of (23) can be expressed by a polynomial in (GprimeG) asfollows

U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0 (24)

where m is called the balance number α0 and αi are constants to be determined later G(ξ)satisfies a second order linear ordinary differential equation (LODE)

d2G(ξ)dξ2

+ λdG(ξ)dξ

+ μG(ξ) = 0 (25)

where λ and μ are arbitrary constants The positive integer m can be determined byconsidering the homogeneous balance between the highest order derivatives and nonlinearterms appearing in ODE (23)

Step 3 By substituting (24) into (23) and using the second order linear ODE (25) collectingall terms with the same order of (GprimeG) together the left-hand side of (23) is converted intoanother polynomial in (GprimeG) Equating each coefficient of this polynomial to zero yields aset of algebraic equations for kω λ μ α0 α1 αm


Step 4 Assuming that the constants kω λ μ α0 α1 αm can be obtained by solving thealgebraic equations in Step 3 since the general solutions of the second order linear ODE (25)is well known for us then substituting kω λ μ α0 αm and the general solutions of (25)into (24)we have more travelling wave solutions of the nonlinear evolution (21)

In the subsequent section we will illustrate the validity and reliability of this methodin detail with complex model of Schrodinger-Boussinesq System (11)

3 Application

To look for the traveling wave solution of the Schrodinger-Boussinesq System (11) we usethe gauge transformation

u(x t) = U(ξ)eiη

v(x t) = V (ξ)(31)

where ξ = kx +ωt η = px + qt and p q ω are constants and i =radicminus1 We substitute (31) into

(11) to obtain nonlinear ordinary differential equation

k2Uprimeprime minus i(2kp +ω)Uprime minus aUV minus(p2 + q

)U = 0 (32)

(ω2 minus k2

)V primeprime + k4V (4) minus bk2

(U2)primeprime

= 0 (33)

In order to simplify integrating (33) twice and taking integration constant to zero the system(32)-(33) reduces to the following system


)U = 0

(ω2 minus k2

)V + k4V primeprime minus bk2U2 = 0

(34)

Suppose that the solution of the nonlinear ordinary differential system (34) can be expressedby a polynomial in (GprimeG) as follows

U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0

V (ξ) =nsum

i=1

βi

(Gprime

G

)i+ β0 βn = 0

(35)

where m n are called the balance number αi (i = 0 1 m) and βj (j = 0 1 n) areconstants to be determined later G(ξ) satisfies a second order linear ordinary differentialequation (25) The integersm n can be determined by considering the homogeneous balance


between the highest order derivatives and nonlinear terms appearing in nonlinear ordinarydifferential system (34) as follow

m + n = m + 2

2m = n + 2(36)

so thatm = n = 2 We then suppose that (34) has the following formal solutions

U(ξ) = α2(Gprime

G

)2

+ α1(Gprime

G

)+ α0 α2 = 0

V (ξ) = β2(Gprime

G

)2

+ β1(Gprime

G

)+ β0 β2 = 0

(37)

Substituting (37) along with (25) into (34) and collecting all the terms with the same powerof (GprimeG) together equating each coefficient to zero yields a set of simultaneous algebraicequations for k ω λ μ αj and βj (j = 0 1 2) as follows

2k2α2μ2 + iωα1μ + k2α1λμ minus aα0β0 minus p2α0 minus qα0 + 2ikpα1μ = 0

ω2β0 + 2k4β2μ2 minus k2(minusμλβ1k2 + bα02 + β0

)= 0

k2α1λ2 minus p2α1 +

(4ikα2μ + 2ikα1λ

)p +(2iα2μ + iα1λ

)ω minus qα1

minus aα0β1 + 2k2α1μ minus aα1β0 + 6k2α2λμ = 0

ω2β1 + 6k4β2λμ minus k2(minusk2λ2β1 minus 2k2μβ1 + β1 + 2bα1α0

)= 0

(4ikα2λ + 2ikα1)p + (2iα2λ + iα1)ω minus p2α2 +(4λ2k2 minus q minus aβ0 + 8μk2

)α2

+ 3k2α1λ minus aα1β1 minus aα0β2 = 0

ω2β2 + k2(4λ2k2 + 8μk2 minus 1

)β2 minus k2

(2bα2α0 + bα12 minus 3λβ1k2

)= 0

4ikpα2 + 2iωα2 +(10λk2 minus aβ1

)α2 minus α1

(minus2k2 + aβ2

)= 0

10k4β2λ minus 2k2(minusk2β1 + bα2α1

)= 0

(6k2 minus aβ2

)α2 = 0

6k4β2 minus bk2α22 = 0

(38)

Then explicit and exact wave solutions can be constructed through our ansatz (37) via theassociated solutions of (25)

In the process of constructing exact solutions to the Schrodinger-Boussinesq system(11) (25) is often viewed as a key auxiliary equation and the types of its solutions determine


the solutions for the original system (11) indirectly In order to seek more new solutions to(11) we here combine the solutions to (25) which were listed in [18 19] Our computationresults show that this combination is an efficient way to obtain more diverse families ofexplicit exact solutions

Solving (38) by use of Maple we get the following reliable results

λ = plusmn13

radic3ω2 minus 3k2

k2 μ =

16k2 minusω2

k4 p = minus1

2ω

k

q =124

24k4 + 30ω4 minus 54k2ω2

k2(k2 minusω2) α0 = 0

α1 = plusmn2radic

1ab

radic3ω2 minus 3k2 α2 = 6k2

radic1ab

β0 = 0 β1 = plusmn2radic3ω2 minus 3k2

a β2 =

6k2

a

(39)

where k and ω are free constant parameters and k = plusmnω Therefore substitute the above casein (37) we get

U(ξ) = 6k2radic

1ab

(Gprime

G

)2

plusmn 2

radic1ab

radic3ω2 minus 3k2

(Gprime

G

)

V (ξ) =6k2

a

(Gprime

G

)2

plusmn 2

radic3ω2 minus 3k2

a

(Gprime

G

)

(310)

Substituting the general solutions of ordinary differential equation (25) into (310) we obtaintwo types of traveling wave solutions of (11) in view of the positive and negative of λ2 minus 4μ

WhenD = λ2minus4μ = (ω2minusk2)k4 gt 0 using the general solutions of ordinary differentialequation (25) and relationships (310) we obtain hyperbolic function solutions uH(x t) andvH(x t) of the Schrodinger-Boussinesq system (11) as follows

uH(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2


plusmnradic3ω2 minus 3k2

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vH(x t) =6k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(311)

where D = (ω2 minus k2)k4 gt 0 ξ = kx + ωt η = minus(12)(ωxk) + (124)((24k4 + 30ω4 minus54k2ω2)k2(k2 minusω2))t and kωC1 and C2 are arbitrary constants and k = plusmnω

It is easy to see that the hyperbolic solutions (311) can be rewritten at C21 gt C2

2 asfollows

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312a)

vH(x t) =12ω2 minus k2ak2

⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312b)

while at C21 lt C

22 one can obtain

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312c)


⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312d)

where ξ = kx + ωt η = minus(12)(ωxk) + (124)((24k4 + 30ω4 minus 54k2ω2)k2(k2 minus ω2))t ρH =tanhminus1(C1C2) and kω are arbitrary constants


Now whenD = λ2 minus4μ = (ω2 minusk2)k4 lt 0 we obtain trigonometric function solutionsuT and vT of Schrodinger-Boussinesq system (11) as follows

uT(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2


⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vT(x t) =6k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(313)

where D = (ω2 minus k2)k4 lt 0 ξ = kx + ωt η = minus(12)(ωxk) + (124)((24k4 + 30ω4 minus54k2ω2)k2(k2 minus ω2))t and k ω C1 and C2 are arbitrary constants and k = plusmn ω Similaritythe trigonometric solutions (313) can be rewritten at C2

1 gt C22 and C

21 lt C

22 as follows

uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314a)

vT(x t) =12k2 minusω2

ak2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314b)


|u H|

tx

1

05

0

minus054

20

minus2minus4 minus4

minus20

24

Figure 1 Soliton solution |uH(x t)| of the Schrodinger-Boussinesq system (312a) for α = 1 β = 16 k =1 ω = minus2 and ρH = tanhminus1(14)

0

02

04

06

08

tx

42

0minus2

minus4 minus4minus2

02

4

|u H|2

Figure 2 Soliton solution |uH(x t)|2 of the Schrodinger-Boussinesq system (312a) for α = 1 β = 16 k =1 ω = minus2 and ρH = tanhminus1(14)

uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314c)


ak2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314d)

where ξ = kx + ωt η = minus(12)(ωxk) + (124)((24k4 + 30ω4 minus 54k2ω2)k2(k2 minus ω2))t ρT =tanminus1(C1C2) and k ω are arbitrary constants


0

200

400

600

800

tx

42

0minus2

minus4 minus4minus2

02

4

vH

Figure 3 Soliton solution |vH(x t)| of the Schrodinger-Boussinesq system (312b) for α = 1 β = 16 k =1 ω = minus2 and ρH = tanhminus1(14)

4 Conclusions

This study shows that the (GprimeG)-expansion method is quite efficient and practically wellsuited for use in finding exact solutions for the Schrodinger-Boussinesq system With theaid of Maple we have assured the correctness of the obtained solutions by putting them backinto the original equation To illustrate the obtained solutions the hyperbolic type of obtainedsolutions (312a) and (312b) are attached as Figures 1 2 and 3 We hope that they will beuseful for further studies in applied sciences

Acknowledgments

The authors would like to express their sincere thanks and gratitude to the reviewers for theirvaluable comments and suggestions for the improvement of this paper

References

[1] C Sulem and P-L Sulem The Nonlinear Schrodinger Equation Self-Focusing and Wave Collapse vol 139of Applied Mathematical Sciences Springer New York NY USA 1999

[2] T Kato ldquoOn nonlinear Schrodinger equationsrdquo Annales de lrsquoInstitut Henri Poincare Physique Theoriquevol 46 no 1 pp 113ndash129 1987

[3] J Boussinesq ldquoTheorie des ondes et des remous qui se propagent le long dun canal rectangulairehorizontal en communiquant au liquide continu dans 21 ce canal des vitesses sensiblement pareillesde la surface au fondrdquo Journal de Mathematiques Pures et Appliquees vol 17 no 2 pp 55ndash108 1872

[4] O V Kaptsov ldquoConstruction of exact solutions of the Boussinesq equationrdquo Journal of AppliedMechanics and Technical Physics vol 39 pp 389ndash392 1998

[5] V Zakharov ldquoOn stochastization of one-dimensional chains of nonlinear oscillatorsrdquo Journal ofExperimental and Theoretical Physics vol 38 pp 110ndash108

[6] F Falk E Laedke and K Spatschek ldquoStability of solitary-wave pulses in shape-memory alloysrdquoPhysical Review B vol 36 no 6 pp 3031ndash3041 1978

[7] V Makhankov ldquoOn stationary solutions of Schrodinger equation with a self-consistent potentialsatisfying Boussinesqs equationsrdquo Physics Letters A vol 50 pp 42ndash44 1974


[8] N Yajima and J Satsuma ldquoSoliton solutions in a diatomic lattice systemrdquo Progress of TheoreticalPhysics vol 62 no 2 pp 370ndash378 1979

[9] M J Ablowitz and H Segur Solitons and the Inverse Scattering Transform vol 4 SIAM PhiladelphiaPa USA 1981

[10] M R Miurs Backlund Transformation Springer Berlin Germany 1978[11] R Hirota ldquoExact solution of the Korteweg-de Vries equation for multiple collisions of solitonsrdquo

Physical Review Letters vol 27 pp 1192ndash1194 1971[12] JWeissM Tabor andG Carnevale ldquoThe Painleve property for partial differential equationsrdquo Journal

of Mathematical Physics vol 24 no 3 pp 522ndash526 1983[13] N C Freeman and J J C Nimmo ldquoSoliton solutions of the Korteweg-de Vries and the Kadomtsev-

Petviashvili equations the Wronskian techniquerdquo Proceedings of the Royal Society of London Series Avol 389 no 1797 pp 319ndash329 1983

[14] W Malfliet and W Hereman ldquoThe tanh methodmdashI Exact solutions of nonlinear evolution and waveequationsrdquo Physica Scripta vol 54 no 6 pp 563ndash568 1996

[15] F Xu ldquoApplication of Exp-function method to symmetric regularized long wave (SRLW) equationrdquoPhysics Letters A vol 372 no 3 pp 252ndash257 2008

[16] F Tascan A Bekir and M Koparan ldquoTravelling wave solutions of nonlinear evolution equations byusing the first integral methodrdquo Communications in Nonlinear Science and Numerical Simulation vol 14pp 1810ndash1815 2009

[17] M Wang X Li and J Zhang ldquoThe (GprimeG)-expansion method and travelling wave solutions ofnonlinear evolution equations in mathematical physicsrdquo Physics Letters A vol 372 no 4 pp 417ndash423 2008

[18] R Abazari ldquoThe (GprimeG)-expansion method for Tzitzeica type nonlinear evolution equationsrdquoMathematical and Computer Modelling vol 52 no 9-10 pp 1834ndash1845 2010

[19] R Abazari ldquoThe (GprimeG )-expansion method for the coupled Boussinesq equationsrdquo ProcediaEngineering vol 10 pp 2845ndash2850 2011


Research ArticleExistence of Solutions for Nonlinear Mixed TypeIntegrodifferential Functional Evolution Equationswith Nonlocal Conditions

Shengli Xie

Department of Mathematics and Physics Anhui University of Architecture Anhui Hefei 230022 China

Correspondence should be addressed to Shengli Xie slxieaiaieducn

Received 29 June 2012 Revised 14 August 2012 Accepted 1 September 2012


Copyright q 2012 Shengli Xie This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

Using Monch fixed point theorem this paper proves the existence and controllability of mild solu-tions for nonlinear mixed type integrodifferential functional evolution equations with nonlocalconditions in Banach spaces some restricted conditions on a priori estimation andmeasure of non-compactness estimation have been deleted our results extend and improve many known resultsAs an application we have given a controllability result of the system

1 Introduction

This paper related to the existence and controllability of mild solutions for the following non-linear mixed type integrodifferential functional evolution equations with nonlocal conditionsin Banach space X

xprime(t) = A(t)x(t) + f

(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J

x0 = φ + g(x) t isin [minusq 0](11)

where q gt 0 J = [0 b] A(t) is closed linear operator on X with a dense domain D(A) whichis independent of t x0 isin X and xt [minusq 0] rarr X defined by xt(θ) = x(t+θ) for θ isin [minusq 0] andx isin C(JX) f J timesC([minusq 0] X)timesX timesX rarr XK ΔtimesC([minusq 0] X) rarr X Δ = (t s) isin J times J s le t H J times J times C([minusq 0] X) rarr X g C([0 b] X) rarr C([minusq 0] X) φ [minusq 0] rarr X aregiven functions


For the existence and controllability of solutions of nonlinear integrodifferential func-tional evolution equations in abstract spaces there are many research results see [1ndash13] andtheir references However in order to obtain existence and controllability of mild solutions inthese study papers usually some restricted conditions on a priori estimation and compact-ness conditions of evolution operator are used Recently Xu [6] studied existence of mildsolutions of the following nonlinear integrodifferential evolution systemwith equicontinuoussemigroup

(Ex(t))prime +Ax(t) = f

(

t x(σ1(t))int t

0k(t s)h(s x(σ2(s))ds)

)

t isin [0 b]

x(0) + g(x) = x0

(12)

Some restricted conditions on a priori estimation andmeasure of noncompactness estimation

(1 minus αβMc

)N

αβM(d + x0) + αMθ1L1Ω1(N +Kθ2L1Ω2(N))gt 1

2αθ1L1M(1 + 2Kθ2L1) lt 1

(13)

are used and some similar restricted conditions are used in [14 15] But estimations (315)and (321) in [15] seem to be incorrect Since spectral radius σ(B) = 0 of linear Volterraintegral operator (Bx)(t) =

int t0 k(t s)x(s)ds in order to obtain the existence of solutions for

nonlinear Volterra integrodifferential equations in abstract spaces by using fixed point theoryusually some restricted conditions on a priori estimation and measure of noncompactnessestimation will not be used even if the infinitesimal generator A = 0

In this paper using Monch fixed point theorem we investigate the existence and con-trollability of mild solution of nonlinear Volterra-Fredholm integrodifferential system (11)some restricted conditions on a priori estimation and measure of noncompactness estima-tion have been deleted our results extend and improve the corresponding results in papers[2ndash20]

2 Preliminaries

Let (X middot ) be a real Banach space and let C([a b] X) be a Banach space of all continuousX-valued functions defined on [a b]with norm x[ab] = suptisin[ab]x(t) for x isin C([a b] X)B(X) denotes the Banach space of bounded linear operators from X into itself

Definition 21 The family of linear bounded operators R(t s) 0 le s le t lt +infin on X is saidan evolution system if the following properties are satisfied

(i) R(t t) = I where I is the identity operator in X

(ii) R(t s)R(s τ) = R(t τ) for 0 le s le t lt +infin

(iii) R(t s) isin B(X) the space of bounded linear operator on X where for every (t s) isin(t s) 0 le s le t lt +infin and for each x isin X the mapping (t s) rarr R(t s)x is contin-uous


The evolution system R(t s) is said to be equicontinuous if for all bounded set Q sub Xs rarr R(t s)x x isin Q is equicontinuous for t gt 0 x isin C([minusq b] X) is said to be a mild solu-tion of the nonlocal problem (11) if x(t) = φ(t) + g(x)(t) for t isin [minusq 0] and for t isin J it satis-fies the following integral equation

x(t) = R(t 0)[φ(0) + g(x)(0)

]+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

(21)

The following lemma is obvious

Lemma 22 Let the evolution system R(t s) be equicontinuous If there exists a ρ isin L1[JR+] suchthat x(t) le ρ(t) for ae t isin J then the set int t0 R(t s)x(s)ds is equicontinuous

Lemma 23 (see [21]) Let V = xn sub L1([a b] X) If there exists σ isin L1([a b]R+) such thatxn(t) le σ(t) for any xn isin V and ae t isin [a b] then α(V (t)) isin L1([a b]R+) and

α

(int t

0xn(s)ds n isin N

)

le 2int t

0α(V (s))ds t isin [a b] (22)

Lemma 24 (see [22]) Let V sub C([a b] X) be an equicontinuous bounded subset Then α(V (t)) isinC([a b]R+) (R+ = [0infin)) α(V ) = maxtisin[ab]α(V (t))

Lemma 25 (see [23]) LetX be a Banach spaceΩ a closed convex subset inX and y0 isin Ω Supposethat the operator F Ω rarr Ω is continuous and has the following property

V sub Ω countable V sub co(y0 cup F(V )

)=rArr V is relatively compact (23)

Then F has a fixed point in Ω

Let V (t) = x(t) x isin C([minusq b] X) sub X (t isin J) Vt = xt x isin C([minusq b] X) subC([minusq 0] X) α(middot) and αC(middot) denote the Kuratowski measure of noncompactness in X andC([minusq b] X) respectively For details on properties of noncompact measure see [22]

3 Existence Result

We make the following hypotheses for convenience

(H1) g C([0 b] X) rarr C([minusq 0] X) is continuous compact and there exists a constantN such that g(x)[minusq0] leN

(H2) (1)f J times C([minusq 0] X) times X times X rarr X satisfies the Carathodory conditions that isf(middot x y z) is measurable for each x isin C([minusq b] X) y z isin X f(t middot middot middot) is continuousfor ae t isin J (2) There is a bounded measure function p J rarr R

+ such that

∥∥f(t x y z

)∥∥ le p(t)(x[minusq0] +

∥∥y∥∥ + z

) ae t isin J x isin C([minusq 0] X) y z isin X (31)


(H3) (1) For each x isin C([minusq 0] X) K(middot middot x)H(middot middot x) J times J rarr X are measurable andK(t s middot)H(t s middot) C([minusq 0] X) rarr X is continuous for ae t s isin J (2) For each t isin (0 b] there are nonnegative measure functions k(t middot) h(t middot) on [0 b]such that

K(t s x) le k(t s)x[minusq0] (t s) isin Δ x isin C([minusq 0] X)

H(t s x) le h(t s)x[minusq0] t s isin J x isin C([minusq 0] X)(32)

andint t0 k(t s)ds

intb0 h(t s)ds are bounded on [0 b]

(H4) For any bounded set V1 sub C([minusq 0] X) V2 V3 sub X there is bounded measurefunction li isin C[JR+] (i = 1 2 3) such that

α(f(t V1 V2 V3) le l1(t) sup

minusqleθle0α(V1(θ)) + l2(t)α(V2) + l3(t)α(V3) ae t isin J

α

(int t

0K(t s V1)ds

)

le k(t s) supminusqleθle0

α(V1(θ)) t isin J

α

(intb

0H(t s V1)ds

)

le h(t s) supminusqleθle0

α(V1(θ)) t isin J

(33)

(H5) The resolvent operator R(t s) is equicontinuous and there are positive numbersM ge 1 and

w = maxMp0(1 + k0 + h0) + 1 2M

(l01 + 2l02k0 + 2l03h0

)+ 1 (34)

such that R(t s) le Meminusw(tminuss) 0 le s le t le b where k0 = sup(ts)isinΔint t0 k(t s)ds h0 =

suptsisinJintb0 h(t s)ds p0 = suptisinJp(t) l

0i = suptisinJ li(t) (i = 1 2 3)

Theorem 31 Let conditions (H1)ndash(H5) be satisfied Then the nonlocal problem (11) has at least onemild solution

Proof Define an operator F C([minusq b] X) rarr C([minusq b] X) by

(Fx)(t)

=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

φ(t) + g(x)(t) tisin[minusq 0]R(t 0)

[φ(0) + g(x)(0)

]

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(35)


We have by (H1)ndash(H3) and (H5)

(Fx)(t) le ∥∥φ∥∥[minusq0] +N leM(∥∥φ∥∥[minusq0] +N

)= L t isin [minusq 0]

(Fx)(t) le L +Mint t

0ew(sminust)

∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥ds

le L +Mint t

0ew(sminust)p(s)

(

xs[minusq0] +ints

0K(s r xr)dr +

intb

0H(s r xr)dr

)

ds

le L +Mp0

int t

0ew(sminust)

(

xs[minusq0] +int s

0k(s r)xr[minusq0]dr +

intb

0h(s r)xr[minusq0]dr

)

ds

le L +Mp0(1 + k0 + h0)wminus1x[minusqb] t isin [0 b](36)

Consequently

(Fx)(t) le L +Mp0(1 + k0 + h0)wminus1x[minusqb] = L + ηx[minusqb] t isin [minusq b] (37)

where 0 lt η =Mp0(1 + k0 + h0)wminus1 lt 1 Taking R gt L(1 minus η)minus1 let

BR =x isin C([minusq b] X) x[minusqb] le R

(38)

Then BR is a closed convex subset in C([minusq b] X) 0 isin BR and F BR rarr BR Similar to theproof in [14 24] it is easy to verify that F is a continuous operator from BR into BR Forx isin BR s isin [0 b] (H2) and (H3) imply

∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥le p(s)(1 + k0 + h0)R (39)

We can show that from (H5) (39) and Lemma 22 that F(BR) is an equicontinuous inC([minusq b] X)

Let V sub BR be a countable set and

V sub co(0 cup (FV )) (310)


From equicontinuity of F(BR) and (310) we know that V is an equicontinuous subset inC([minusq b] X) By (H1) it is easy to see that α((FV )(t)) = 0 t isin [minusq 0] By properties of non-compact measure (H4) and Lemma 23 we have

α((FV )(t)) le 2int t

0R(t minus s)α

(

f

(

s Vs

int s

0K(s r Vr)dr

intb

0H(s r Vr)dr

))

ds

le 2Mint t

0ew(sminust)

[

l1(s) supminusqleθle0

α(Vs(θ)) + l2(s)α(ints

0K(s r Vr)dr

)

+l3(s)α

(intb

0H(s r Vr)dr

)]

ds

le 2Mint t

0ew(sminust)

[

l01 supminusqleθle0

α(V (s + θ)) + 2l02

int s

0k(s r) sup

minusqleθle0α(V (r + θ))dr

+2l03

intb

0h(s r) sup


]

ds

le 2M(l01 + 2l02k0 + 2l03h0

)int t

0ew(sminust)ds sup

minusqleτlebα(V (τ))

le 2M(l01 + 2l02k0 + 2l03h0

)wminus1α C(V ) t isin [0 b]

(311)

Consequently

αC(FV ) = supminusqletleb

α((FV )(t)) le 2M(l01 + 2l02k0 + 2l03h0

)wminus1αC(V ) (312)

Equations (310) (312) and Lemma 24 imply

αC(V ) le αC(FV ) le δαC(V ) (313)

where δ = 2M(l01 + 2l02k0 + 2l03h0)wminus1 lt 1 Hence αC(V ) = 0 and V is relative compact in

C([minusq b] X) Lemma 25 implies that F has a fixed point in C([minusq b] X) then the system(11) (12) has at least one mild solution The proof is completed

4 An Example

In this section we give an example to illustrate Theorem 31


Let X = L2([0 π]R) Consider the following functional integrodifferential equationwith nonlocal condition

ut(t y)= a1

(t y)uyy(t y)

+ a2(t)

[

sinu(t + θ y

)ds +

int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

0 le t le b

u(t y)= φ(t y)+intπ

0

intb

0F(r y)log(1 + |u(r s)|12

)dr ds minusq le t le 0 0 le y le π

u(t 0) = u(t π) = 0 0 le t le b(41)

where functions a1(t y) is continuous on [0 b] times [0 π] and uniformly Holder continuous int a2(t) is bounded measure on [0 b] φ [minusq 0] times [0 π] a3 [minusq b] and F [0 b] times [0 π] arecontinuous respectively Taking u(t y) = u(t)(y) φ(t y) = φ(t)(y)

f

(

t ut

int t

0K(t s us)ds

intb

0H(t s us)ds

)(y)

= a2(t)

[

sinu(t + θ y

)+int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

K(t s us)(y)=ints

minusqa3(s + τ)

u(τ y)dτ

(1 + t) H(t s us)

(y)=

u(s + θ y

)

(1 + t)(1 + s)2

g(u)(y)=intπ

0

intb

0F(r y)log(1 + |u(r s)|12

)dr ds

(42)

The operator A defined by A(t)w = a1(t y)wprimeprime with the domain

D(A) =w isin X wwprime are absolutely continuous wprimeprime isin X w(0) = w(π) = 0

(43)

Then A(t) generates an evolution system and R(t s) can be deduced from the evolution sys-tems so that R(t s) is equicontinuous and R(t s) le Meβ(tminuss) for some constants M and β(see [24 25]) The system (41) can be regarded as a form of the system (11) (12) We haveby (42)

∥∥f(t u v z)∥∥ le |a2(t)|

(u[minusq0] + v + z

)

K(t s u) leints

minusq|a3(s + τ)|dτ

u[minusq0](1 + t)

H(t s u) leu[minusq0]

(1 + t)(1 + s)

(44)


for u isin C([minusq 0] X) v z isin X

∥∥g(u)

∥∥[minusq0] le bπ max

(ry)isin[0b]times[0π]

∣∣F(r y)∣∣(u[0b] +

radicπ) (45)

and g C([0 b] X) rarr C([minusq b] X) is continuous and compact (see the example in [7])w gt 0 andM ge 1 can be chosen such that R(t s) le Mew(tminuss) 0 le s le t le b In addition forany bounded set V1 sub C([minusq 0] X) V2 V3 sub X we can show that by the diagonal method

α(f(t V1 V2 V3)

) le |a2(t)|(

supminusqleθle0

α(V1(θ)) + α(V2) + α(V3)

)

t isin J

α(K(t s V1)) le 11 + t

supminusqleθle0

α(V1(θ)) t s isin Δ

α(H(t s V1)) le 1

(1 + t)(1 + s)2sup

minusqleθle0α(V1(θ)) t s isin [0 b]

(46)

It is easy to verify that all conditions of Theorem 31 are satisfied so the system (51) has atleast one mild solution

5 An Application

As an application of Theorem 31 we shall consider the following system with control para-meter


(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J


where C is a bounded linear operator from a Banach spaceU to X and v isin L2(JU) Then themild solution of systems (51) is given by

x(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

φ(t) + g(x)(t) t isin [minusq 0]R(t 0)

[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(52)

where the resolvent operator R(t s) isin B(X) fKH g and φ satisfy the conditions stated inSection 3

Definition 51 The system (52) is said to be controllable on J = [0 b] if for every initial func-tion φ isin C([minusq 0] X) and x1 isin X there is a control v isin L2(JU) such that the mild solutionx(t) of the system (51) satisfies x(b) = x1


To obtain the controllability result we need the following additional hypotheses

(Hprime5) The resolvent operator R(t s) is equicontinuous and R(t s) le Meminusw(tminuss) 0 le s let le b forM ge 1 and positive number

w = maxMp0(1 + k0 + h0)(1 +MM1b) + 1 2M

(l01 + 2l02k0 + 2l03h0

)(1 +MM1b) + 1

(53)

where k0 h0 p0 l0i (i = 1 2 3) are as before

(H6) The linear operatorW from L2(JU) into X defined by

Wv =intb

0R(b s)(Cv)(s)ds (54)

has an inverse operatorWminus1 which takes values in L2(JU)kerW and there exists a positiveconstantM1 such that CWminus1 leM1

Theorem 52 Let the conditions (H1)ndash(H4) (Hprime5) and (H6) be satisfied Then the nonlocal problem

(11) (12) is controllable

Proof Using hypothesis (H6) for an arbitrary x(middot) define the control

v(t) =Wminus1(

x1 minus R(b 0)[φ(0) + g(x)(0)

]

+intb

0R(b s)f

(

s xs

ints

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

)

(t) t isin [0 b]

(55)

Define the operator T C([minusq b] X) rarr C([minusq b] X) by

(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(56)

Now we show that when using this control T has a fixed point Then this fixed point is asolution of the system (51) Substituting v(t) in (56) we get

(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)CWminus1(x1 minus R(b 0)

[φ(0) + g(x)(0)

]

+intb

0R(b τ)f

(

τ xτ

int τ

0K(τ r xr)dr

intb

0H(τ r xr)dr

)

dτ

)

(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(57)


Clearly (Tx)(b) = x1 which means that the control v steers the system (51) from the giveninitial function φ to the origin in time b provided we can obtain a fixed point of nonlinearoperator T The remaining part of the proof is similar to Theorem 31 we omit it

Remark 53 Since the spectral radius of linear Fredholm type integral operator may be greaterthan 1 in order to obtain the existence of solutions for nonlinear Volterra-Fredholm typeintegrodifferential equations in abstract spaces by using fixed point theory some restrictedconditions on a priori estimation andmeasure of noncompactness estimation will not be usedeven if the generator A = 0 But these restrictive conditions are not being used in Theorems31 and 52

Acknowledgments

The work was supported by Natural Science Foundation of Anhui Province (11040606M01)and Education Department of Anhui (KJ2011A061 KJ2011Z057) China

References

[1] R Ravi Kumar ldquoNonlocal Cauchy problem for analytic resolvent integrodifferential equations inBanach spacesrdquo Applied Mathematics and Computation vol 204 no 1 pp 352ndash362 2008

[2] B Radhakrishnan and K Balachandran ldquoControllability results for semilinear impulsive integrodif-ferential evolution systems with nonlocal conditionsrdquo Journal of Control Theory and Applications vol10 no 1 pp 28ndash34 2012

[3] J Wang and W Wei ldquoA class of nonlocal impulsive problems for integrodifferential equations inBanach spacesrdquo Results in Mathematics vol 58 no 3-4 pp 379ndash397 2010

[4] M B Dhakne and K D Kucche ldquoExistence of a mild solution of mixed Volterra-Fredholm functionalintegrodifferential equation with nonlocal conditionrdquo Applied Mathematical Sciences vol 5 no 5ndash8pp 359ndash366 2011

[5] N Abada M Benchohra and H Hammouche ldquoExistence results for semilinear differential evolutionequations with impulses and delayrdquo CUBO A Mathematical Journal vol 12 no 2 pp 1ndash17 2010

[6] X Xu ldquoExistence for delay integrodifferential equations of Sobolev type with nonlocal conditionsrdquoInternational Journal of Nonlinear Science vol 12 no 3 pp 263ndash269 2011

[7] Y Yang and J Wang ldquoOn some existence results of mild solutions for nonlocal integrodifferentialCauchy problems in Banach spacesrdquo Opuscula Mathematica vol 31 no 3 pp 443ndash455 2011

[8] Q Liu and R Yuan ldquoExistence of mild solutions for semilinear evolution equations with non-localinitial conditionsrdquo Nonlinear Analysis vol 71 no 9 pp 4177ndash4184 2009

[9] A Boucherif and R Precup ldquoSemilinear evolution equations with nonlocal initial conditionsrdquoDynamic Systems and Applications vol 16 no 3 pp 507ndash516 2007

[10] D N Chalishajar ldquoControllability of mixed Volterra-Fredholm-type integro-differential systems inBanach spacerdquo Journal of the Franklin Institute vol 344 no 1 pp 12ndash21 2007

[11] J H Liu and K Ezzinbi ldquoNon-autonomous integrodifferential equations with non-local conditionsrdquoJournal of Integral Equations and Applications vol 15 no 1 pp 79ndash93 2003

[12] K Balachandran and R Sakthivel ldquoControllability of semilinear functional integrodifferential sys-tems in Banach spacesrdquo Kybernetika vol 36 no 4 pp 465ndash476 2000

[13] Y P Lin and J H Liu ldquoSemilinear integrodifferential equations with nonlocal Cauchy problemrdquoNon-linear Analysis vol 26 no 5 pp 1023ndash1033 1996

[14] K Malar ldquoExistence of mild solutions for nonlocal integro-di erential equations with measure of non-compactnessrdquo Internationnal Journal of Mathematics and Scientic Computing vol 1 pp 86ndash91 2011

[15] H-B Shi W-T Li and H-R Sun ldquoExistence of mild solutions for abstract mixed type semilinearevolution equationsrdquo Turkish Journal of Mathematics vol 35 no 3 pp 457ndash472 2011

[16] T Zhu C Song and G Li ldquoExistence of mild solutions for abstract semilinear evolution equations inBanach spacesrdquo Nonlinear Analysis vol 75 no 1 pp 177ndash181 2012


[17] Z Fan Q Dong and G Li ldquoSemilinear differential equations with nonlocal conditions in Banachspacesrdquo International Journal of Nonlinear Science vol 2 no 3 pp 131ndash139 2006

[18] X Xue ldquoExistence of solutions for semilinear nonlocal Cauchy problems in Banach spacesrdquo ElectronicJournal of Differential Equations vol 64 pp 1ndash7 2005

[19] S K Ntouyas and P Ch Tsamatos ldquoGlobal existence for semilinear evolution equations with nonlocalconditionsrdquo Journal of Mathematical Analysis and Applications vol 210 no 2 pp 679ndash687 1997

[20] L Byszewski and H Akca ldquoExistence of solutions of a semilinear functional-differential evolutionnonlocal problemrdquo Nonlinear Analysis vol 34 no 1 pp 65ndash72 1998

[21] H-P Heinz ldquoOn the behaviour of measures of noncompactness with respect to differentiation andintegration of vector-valued functionsrdquo Nonlinear Analysis vol 7 no 12 pp 1351ndash1371 1983

[22] D Guo V Lakshmikantham and X Liu Nonlinear Integral Equations in Abstract Spaces vol 373 ofMathematics and Its Applications Kluwer Academic Dordrecht The Netherlands 1996

[23] H Monch ldquoBoundary value problems for nonlinear ordinary differential equations of second orderin Banach spacesrdquo Nonlinear Analysis vol 4 no 5 pp 985ndash999 1980

[24] R Ye Q Dong and G Li ldquoExistence of solutions for double perturbed neutral functional evolutionequationrdquo International Journal of Nonlinear Science vol 8 no 3 pp 360ndash367 2009

[25] A Friedman Partial Differential Equations Holt Rinehart and Winston New York NY USA 1969
















Contents



































Acknowledgments













1 Introduction











119909 = 119865119909 (1)












119901= 119862(R

+ 119901(119905)) the


+and such that

sup |119909 (119905)| 119901 (119905) 119905 ge 0 lt infin (2)


to the norm

119909 = sup |119909 (119905)| 119901 (119905) 119905 ge 0 (3)


119901[10 15]



+and if lim

119879rarrinfinsup|119909(119905)|119901(119905) 119905 ge 119879 = 0





that is


(4)


119909 (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 (5)




+ 119886 R

+rarr


|119907 (119905 119904 119909)| le 119899 (119905 119904) + 119886 (119905) 119887 (119904) |119909| (6)



119871 (119905) = int

119905

0

119886 (119904) 119887 (119904) 119889119904 119905 ge 0 (7)


119901 where 119901(119905) = [119886(119905) exp(119872119871(119905) + 119905)]




119863119886(119905) exp(119872119871(119905)) for 119905 ge 0


int

119905

0

119899 (119905 119904) 119889119904 le 119873119886 (119905) exp (119872119871 (119905)) (8)

for 119905 ge 0



condition

119871 (120593 (119905)) minus 119871 (119905) le 119870 (9)



for all 119905 ge 0



119901such that |119909(119905)| le 119886(119905) exp(119872119871(119905))

for 119905 isin R+


119901by putting

(119865119909) (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 119905 ge 0 (10)


+






|(119865119909) (119905)| le |119889 (119905)| + int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904)))1003816100381610038161003816 119889119904

le 119863119886 (119905) exp (119872119871 (119905))

+ int

119905

119900

[119899 (119905 119904) + 119886 (119905) 119887 (119904)1003816100381610038161003816119909 (120593 (119904))

1003816100381610038161003816] 119889119904

le 119863119886 (119905) exp (119872119871 (119905)) + 119873119886 (119905) exp (119872119871 (119905))

+ 119886 (119905) int

119905

0

119887 (119904) 119886 (120593 (119904)) exp (119872119871 (120593 (119904))) 119889119904

le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) int

119905

0

119872119886 (119904) 119887 (119904)


le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) exp (119872119871 (119905))

= 119886 (119905) exp (119872119871 (119905))

(11)





1003816100381610038161003816

le int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904))) minus 119907 (119905 119904 119910 (120593 (119904)))1003816100381610038161003816 119889119904

le 120573 (120576)

(12)



1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)1003816100381610038161003816 [119886 (119905) exp (119872119871 (119905) + 119905)]

minus1

le |(119865119909) (119905)| +1003816100381610038161003816(119865119910) (119905)

1003816100381610038161003816

times [119886 (119905) exp (119872119871 (119905))]minus1


le 2119890minus119905

(13)


1003816100381610038161003816 119901 (119905) le 120576 (14)



119901


|(119865119909) (119905)| 119901 (119905) le 119890minus119905 (15)

This yields that

lim119879rarrinfin

sup |(119865119909) (119905)| 119901 (119905) 119905 ge 119879 = 0 (16)



|(119865119909) (119905) minus (119865119909) (119904)|

le |119889 (119905) minus 119889 (119904)|

+

10038161003816100381610038161003816100381610038161003816

int

119905

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119904 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

le 120584119879(119889 120576) + 120576max 119899 (119905 120591) + 119886 (120591) 119887 (120591)

times119901 (120593 (120591)) 0 le 120591 le 119905 le 119879

+ 119879120584119879(119907 (120576 119879 120572 (119879)))

(17)

where we denoted120584119879(119907 (120576 119879 120572 (119879)))

= sup |119907 (119905 119906 119907) minus 119907 (119904 119906 119907)| 119905 119904 isin [0 119879]

|119905 minus 119904| le 120576 119906 isin [0 119879] |119907| le 120572 (119879)

(18)



120584119879(119889 120576) =

lim120576rarr0











119909 (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

+ int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(19)



+)



supremum norm

119909 = sup |119909 (119905)| 119905 isin R+ (20)








that

1003816100381610038161003816119891 (119905 1199091 1199092 119909

119899) minus 119891 (119905 119910

1 1199102 119910

119899)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816

(21)


1 1199092 119909

119899) (1199101 1199102

119910119899) isin R119899


with 1198650= sup|119891(119905 0 0)| 119905 isin R

+


R+



1 2 119898)(iv) The function 120573 R




rarr R+and


1003816100381610038161003816119892 (119905 119904 1199091 119909

119898)1003816100381610038161003816 le 119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

10038161003816100381610038161199091198941003816100381610038161003816 (22)

for all 119905 119904 isin R+and (119909

1 1199092 119909


we assume that

lim119905rarrinfin

int

120573(119905)

0

119902 (119905 119904) 119889119904 = 0 lim119905rarrinfin

119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 = 0

(23)


1 1199072 R+

rarr R+defined by

the formulas

1199071(119905) = int

120573(119905)

0

119902 (119905 119904) 119889119904 1199072(119905) = 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 (24)



119872119894= sup 119907

119894(119905) 119905 isin R

+ (119894 = 1 2) (25)


119896 = sum119899


119894(119894 = 1 2 119899) appear in

assumption (i)

(vi) 119896 + 1198981198722lt 1




globally attractive




119909 (119905) = (119876119909) (119905) 119905 isin R+ (26)


lim119905rarrinfin

(119909 (119905) minus 119910 (119905)) = 0 (27)




119903in the space119883 = 119861119862(R

+) centered at the


2)]


putting

(119860119909) (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

(119861119909) (119905) = int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(28)


119909 (119905) = (119860119909) (119905) + (119861119909) (119905) 119905 isin R+ (29)


+ Thus



119903into



+ we get

1003816100381610038161003816(119860119909) (119905) minus (119860119910) (119905)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816

le 1198961003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(30)





1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816

le int

120573(119905)

0

[1003816100381610038161003816119892 (119905 119904 119909 (120574

1(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816

+1003816100381610038161003816119892 (119905 119904 119910 (120574

1(119904)) 119910 (120574

119898(119904)))

1003816100381610038161003816] 119889119904

le 2 (1199071(119905) + 119903119898119907

2(119905))

(31)


1(119905) + 119903119898119907

2(119905) le 1205762 for 119905 ge 119879


1003816100381610038161003816 le 120576 (32)


obtain

1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816 le int

120573(119905)

0

120596119879

119903(119892 120576) 119889119904 le 120573

119879120596119879

119903(119892 120576)

(33)


120596119879

119903(119892 120576) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 1199092 119909

119898)

minus119892 (119905 119904 1199101 1199102 119910

119898)1003816100381610038161003816 119905 isin [0 119879]

119904 isin [0 120573119879] 119909119894 119910119894isin [minus119903 119903]

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816 le 120576 (119894 = 1 2 119898)

(34)


119879]times[minus119903 119903]



119903


|(119861119909) (119905)| le 1199071(119905) + 119903119898119907

2(119905) (35)




1(119905) + 119903119898119907




|(119861119909) (119905)| le 120576 (36)


|(119861119909) (119905) minus (119861119909) (120591)|

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

minus119892 (120591 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

le

1003816100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

[119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816] 119889119904

1003816100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

120596119879

1(119892 120576 119903) 119889119904

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119902 (119905 119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 119886 (119905)119898119903

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119887 (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 120573119879120596119879

1(119892 120576 119903)

(37)

where we denoted

120596119879

1(119892 120576 119903) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 119909

119898)

minus119892 (120591 119904 1199091 119909

119898)1003816100381610038161003816 119905 120591 isin [0 119879]

|119905 minus 119904| le 120576 119904 isin [0 120573119879]

100381610038161003816100381611990911003816100381610038161003816 le 119903

10038161003816100381610038161199091198981003816100381610038161003816 le 119903

(38)



|(119861119909) (119905) minus (119861119909) (120591)|

le 119902119879120584119879(120573 120576) + 119903119898119886

119879119887119879120584119879(120573 120576) + 120573

119879120596119879

1(119892 120576 119903)

(39)

where 119902119879

= max119902(119905 119904) 119905 isin [0 119879] 119904 isin [0 120573119879] 119886119879

=

max119886(119905) 119905 isin [0 119879] 119887119879

= max119887(119905) 119905 isin [0 120573119879]



of the functions 120573 = 120573(119905) 119892 = 119892(119905 119904 1199091 119909

119898) we infer

that 120584119879(120573 120576) rarr 0 and 120596119879





+)





|119909 (119905)| le |(119860119909) (119905)| +1003816100381610038161003816(119861119910) (119905)

1003816100381610038161003816

le1003816100381610038161003816119891 (119905 119909 (120572

1(119905)) 119909 (120572

119899(119905))) minus 119891 (119905 0 0)

1003816100381610038161003816

+1003816100381610038161003816119891 (119905 0 0)

1003816100381610038161003816

+ int

120573(119905)

0

119902 (119905 119904) 119889119904 + 1198981003817100381710038171003817119910

1003817100381710038171003817 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904

le 119896 119909 + 1198650+1198721+ 119898119903119872

2

(40)

Hence we obtain

119909 le1198650+1198721+ 119898119903119872

2

1 minus 119896 (41)


2)(1 minus





1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le

119899

sum

119894=1

119896119894max 1003816100381610038161003816119909 (120572

119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 2int

120573(119905)

0

119902 (119905 119904) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119910 (120574119894(119904))

1003816100381610038161003816) 119889119904

le 119896max 1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 21199071(119905) + 119898 (119909 +

10038171003817100381710038171199101003817100381710038171003817) 1199072 (119905)

(42)


Hence we get


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le 119896max1le119894le119899


1003816100381610038161003816119909 (120572119894(119905) minus 119910 (120572

119894(119905)))

1003816100381610038161003816


1199071(119905) + 119898 (119909 +


1199072(119905)


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

(43)



|119909(119905) minus








or






⋃

0le120582le1

119909 isin 119864 119909 = 120582119879119909 (44)























119864 120583(119883) = 0 is nonempty


(2119900) 119883 sub 119884 rArr 120583(119883) le 120583(119884)

(3119900) 120583(119883) = 120583(Conv119883) = 120583(119883)


(5119900) If (119883


119864such that

119883119899+1

sub 119883119899for 119899 = 1 2 and if lim

119899rarrinfin120583(119883119899) = 0


= ⋂infin

119899=1119883119899is nonempty



119899) for any 119899 = 1 2 This

















Next let us define

120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883

1205960(119883) = lim

120576rarr0

120596 (119883 120576) (46)



0(120582119883) =

|120582|1205960(119883) and 120596

0(119883 + 119884) le 120596

0(119883) + 120596

0(119884) provided 119883119884 isin

M119862(119868)



119909 (119905) = 1198911(119905 119909 (119905) 119909 (119886 (119905))) + (119866119909) (119905) int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(47)



has the form

1198912(119905 119904) = 119896 (119905 119904) 119892 (119905 119904) (48)







2

1003816100381610038161003816 10038161003816100381610038161199101 minus 119910

2

1003816100381610038161003816

(49)



0(119866119883) le 119902120596


119862(119868)



+rarr R+



+rarr R


119909 isin 119862(119868)(vi) 119891



+occurring in



1lt 1199052and


10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

le 120576

int

1199052

1199051

119892 (1199052 119904) 119889119904 le 120576

(50)



ℎ (119905) = int

119905

0

119892 (119905 119904) 119889119904 (51)





119896 = sup |119896 (119905 119904)| (119905 119904) isin Δ

1198911= sup 10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816 119905 isin 119868

ℎ = sup ℎ (119905) 119905 isin 119868

(52)





119901119903 + 1198911+ 119896ℎ120593 (119903)Ψ (119903) le 119903 (53)

such that 119901 + 119896ℎ119902Ψ(1199030) lt 1


(1198651119909) (119905) = 119891

1(119905 119909 (119905) 119909 (120572 (119905)))

(1198652119909) (119905) = int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(119865119909) (119905) = (1198651119909) (119905) + (119866119909) (119905) (119865

2119909) (119905)

(54)



+by the formulas

119872(120576) = sup10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119892)] 119889119904

10038161003816100381610038161003816100381610038161003816

1199051 1199052isin 119868

1199051lt 1199052 1199052minus 1199051le 120576

(55)

119873(120576) = supint1199052

1199051

119892 (1199052 119904) 119889119904 119905

1 1199052isin 119868 1199051lt 1199052 1199052minus 1199051le 120576

(56)






1003816100381610038161003816(1198651119909) (119905) minus (1198651119910) (119905)

1003816100381610038161003816


1003816100381610038161003816

(57)


10038171003817100381710038171198651119909 minus 11986511199101003817100381710038171003817 le 119901

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (58)







1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816

int

1199052

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199051 119904) 119892 (119905

2 s) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

1 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le int

1199052

1199051

1003816100381610038161003816119896 (1199052 119904)1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199052 119904) minus 119896 (1199051 119904)

1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199051 119904)1003816100381610038161003816

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 |(119876119909) (119904)| 119889119904

le 119896Ψ (119909)119873 (120576) + 1205961(119896 120576) Ψ (119909) int

1199052

0

119892 (1199052 119904) 119889119904

+ 119896Ψ (119909) int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904

(59)

where we denoted

1205961(119896 120576) = sup 1003816100381610038161003816119896 (1199052 119904) minus 119896 (119905

1 119904)

1003816100381610038161003816

(1199051 119904) (119905

2 119904) isin Δ

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576

(60)


int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904 =

10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

(61)


120596 (1198652119909 120576) le 119896Ψ (119909)119873 (120576) + ℎΨ (119909) 120596

1(119896 120576)

+ 119896Ψ (119909)119872 (120576)

(62)










1003816100381610038161003816(119865119909) (119905) minus (1198651199090) (119905)

1003816100381610038161003816

le1003816100381610038161003816(1198651119909) (119905) minus (119865

11199090) (119905)

1003816100381610038161003816

+1003816100381610038161003816(119866119909) (119905) (1198652119909) (119905) minus (119866119909

0) (119905) (119865

21199090) (119905)

1003816100381610038161003816

le10038171003817100381710038171198651119909 minus 119865

11199090

1003817100381710038171003817 + 1198661199091003816100381610038161003816(1198652119909) (119905) minus (119865

21199090) (119905)

1003816100381610038161003816

+100381710038171003817100381711986521199090

1003817100381710038171003817

1003817100381710038171003817119866119909 minus 1198661199090

1003817100381710038171003817

(63)



21199090) (119905)

1003816100381610038161003816

le int

119905

0

|119896 (119905 119904)| 119892 (119905 119904)1003816100381610038161003816(119876119909) (119904) minus (119876119909

0) (119904)

1003816100381610038161003816 119889119904

le 119896 (int

119905

0

119892 (119905 119904) 119889119904)1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

le 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

(64)

Similarly we obtain

1003816100381610038161003816(11986521199090) (119905)1003816100381610038161003816 le 119896 (int

119905

0

119892 (119905 119904) 119889119904)10038171003817100381710038171198761199090

1003817100381710038171003817 le 119896ℎ10038171003817100381710038171198761199090

1003817100381710038171003817 (65)

and consequently

1003817100381710038171003817119865211990901003817100381710038171003817 le 119896ℎ

10038171003817100381710038171198761199090

1003817100381710038171003817 (66)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 1199011003817100381710038171003817119909 minus 119909

0

1003817100381710038171003817 + 119866119909 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎ1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

10038171003817100381710038171198761199090

1003817100381710038171003817

(67)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 119901120576 + 120593 (10038171003817100381710038171199090

1003817100381710038171003817 + 120576) 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎΨ (10038171003817100381710038171199090

1003817100381710038171003817)1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

(68)






|(119865119909) (119905)| le10038161003816100381610038161198911 (119905 119909 (119905) 119909 (119886 (119905))) minus 119891

1(119905 0 0)

1003816100381610038161003816

+10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816

+ |119866119909 (119905)|

10038161003816100381610038161003816100381610038161003816

int

119905

0

119896 (119905 119904) 119892 (119905 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909)

(69)

Hence we obtain

119865119909 le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909) (70)



the ball 1198611199030




1199030


1 1199052isin 119868with


obtain1003816100381610038161003816(119865119909) (1199052) minus (119865119909) (119905

1)1003816100381610038161003816

le10038161003816100381610038161198911 (1199052 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199052) 119909 (119886 (119905

2)))

1003816100381610038161003816

+10038161003816100381610038161198911 (1199051 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199051) 119909 (119886 (119905

1)))

1003816100381610038161003816

+1003816100381610038161003816(1198652119909) (1199052)

1003816100381610038161003816

1003816100381610038161003816(119866119909) (1199052) minus (119866119909) (1199051)1003816100381610038161003816

+1003816100381610038161003816(119866119909) (1199051)

1003816100381610038161003816

1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le 1205961199030

(1198911 120576) + 119901max 1003816100381610038161003816119909 (119905

2) minus 119909 (119905

1)1003816100381610038161003816

1003816100381610038161003816119909 (119886 (1199052)) minus 119909 (119886 (119905

1))1003816100381610038161003816

+ 119896ℎΨ (1199030) 120596 (119866119909 120576) + 120593 (119903

0) 120596 (119865

2119909 120576)

(71)

where we denoted

1205961199030

(1198911 120576) = sup 10038161003816100381610038161198911 (1199052 119909 119910) minus 119891

1(1199051 119909 119910)

1003816100381610038161003816 1199051 1199052 isin 119868

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576 119909 119910 isin [minus1199030 1199030]

(72)


120596 (119865119909 120576) le 1205961199030

(1198911 120576) + 119901max 120596 (119909 120576) 120596 (119909 120596 (119886 120576))

+ 119896ℎΨ (1199030) 120596 (119866119909 120576)

+ 120593 (1199030) Ψ (1199030) [119896119873 (120576) + ℎ120596

1(119896 120576) + 119896119872 (120576)]

(73)





1205960(119865119883) le 119901120596

0(119883) + 119896ℎΨ (119903

0) 1205960(119866119883) (74)


1205960(119865119883) le (119901 + 119896ℎ119902Ψ (119903

0)) 1205960(119883) (75)







1199091015840= 119891 (119905 119909) (76)


119909 (0) = 1199090 (77)









rarr R+(or 119908 119869

0times

R+



1003817100381710038171003817 le 119908 (1199051003817100381710038171003817119909 minus 119910

1003817100381710038171003817) (78)




1015840= 119908(119905 119906)


119905

0119908(119904 119906(119904))119889119904 for 119905 isin 119869


lim119905rarr0

119906(119905)119905 = lim119905rarr0






120583 (119891 (119905 119883)) le 119908 (119905 120583 (119883)) (79)





120583


119864120583= 119909 isin 119864 119909 isin ker 120583 (80)



120583is a closed



(6119900) 120583(119883 + 119884) le 120583(119883) + 120583(119884)

(7119900) 120583(120582119883) = |120582|120583(119883) for 120582 isin R








120583 (1199090+ 119891 (119905 119883)) le 119908 (119905 120583 (119883)) (81)



1198690times R+



119906 (119905) le int

119905

0

119908 (119904 119906 (119904)) 119889119904 (119905 isin 1198690) (82)


119906(119905)119905 = lim119905rarr0




0)minus119891(119904 119909)||



120583for 119905 isin 119869



0isin



120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (83)





120583 (119891 (119905 119883)) le 119901 (119905) 120583 (119883) (84)





1199091015840

119899= 119886119899(119905) 119909119899+ 119891119899(119909119899 119909119899+1

) (85)


119909119899(0) = 119909

119899

0(86)


119909119899

0= 119886(119886 isin R)






119899rarrinfin120572119899= 0 and |119891

119899(119909119899 119909119899+1

)| le

120572119899for 119899 = 1 2 and for all 119909 = (119909

1 1199092 1199093 ) isin 119897






119899)|| =

sup|119909119899| 119899 = 1 2


1(119905) 1199092(119905) )






sup119909isin119883

1003816100381610038161003816119909119899 minus 1198861003816100381610038161003816 (87)




120583 (1199090+ 119891 (119905 119883))


sup119909isin119883

1003816100381610038161003816119909119899

0+ 119886119899(119905) 119909119899

+119891119899(119909119899 119909119899+1

) minus 1198861003816100381610038161003816


sup119909isin119883

[1003816100381610038161003816119886119899 (119905)

1003816100381610038161003816

10038161003816100381610038161199091198991003816100381610038161003816

+1003816100381610038161003816119891119899 (119909119899 119909119899+1 ) + 119909

119899

0minus 119886

1003816100381610038161003816]


sup119909isin119883

119901 (119905)1003816100381610038161003816119909119899 minus 119886

1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119886119899 (119905)1003816100381610038161003816 |119886|


sup119909isin119883

1003816100381610038161003816119891119899 (119909119899 119909119899+1 )1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119909119899

0minus 119886

1003816100381610038161003816

(88)


Hence we get

120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (89)



References













































1015840























1 Introduction




u(0) = g(u)(11)







u(0) = g(u)

(12)







part



2 Preliminaries


+ = (0infin)and R





tisinI

∥∥f(t)∥∥X (21)



∥∥f∥∥L1(IX) =

int

I

∥∥f(s)∥∥Xds (22)







d

dtR(t)y = AR(t)y +

int t


= R(t)Ay +int t


(23)







u(0) = u0 isin X(24)








is verified





(27)
















ζ

(int t

0un(s)ds

infin

n=1

)

2int t





subeW such that




Sn = εn + Cn1ε

nminus1h + Cn2ε

nminus2h2


n n isin N (211)





)) n = 2 3 (212)


η(Fn0(S)) rη(S) (213)


3 Main Results





+ rarr R+ such that

∥∥f(t x)∥∥ m(t)Φ(x) (31)




ζ(f(t S)

) H(t)ζ(S) (32)





KgR +KΦ(R)int1

0m(s)ds R (33)








∥∥ +Kint1

0




(Fu)(t) ∥∥R(t)g(u)

∥∥ +

∥∥∥∥∥

int t


∥∥∥∥∥

KgR +KΦ(R)int1

0m(s)ds R

(36)





sub F(B) such that


(int t

0


infinn=1ds

)

+ ε (37)


ζ(F(B)(t)) 4Kint t

0ζ(f(s un(s))

infinn=1ds + ε

4Kint t

0H(s)ζ(un(s)nisinN

) ds + ε

4Kγ(B)int t

0H(s)ds + ε

(38)




0


0ϕ(s)ds

]

+ ε


(39)





ζ(F2(B)(t)

) 2ζ

(int t

0


infinn=1ds

)

+ ε

4Kint t

0ζf(swn(s))

infinn=1ds + ε

4Kint t

0H(s)ζ

(co(F1(B)(s)

))+ ε = 4K

int t

0H(s)ζ

(F1(B)(s)

)+ ε

(311)



ζ(F2(B)(t)

) 4K

int t

0

[∣∣H(s) minus ϕ(s)∣∣ + ∣∣ϕ(s)

∣∣](a + bs)γ(B)ds + ε


0


(

at +bt2

2

)

+ ε

a(a + bt) + b

(

at +bt2

2

)

+ ε (

a2 + 2bt +(bt)2

2

)

γ(B) + ε

(312)


ζ(F2(B)(t)

)

(

a2 + 2bt +(bt)2

2

)

γ(B) (313)


ζ(Fn(B)(t)) (

an + Cn1a

nminus1bt + Cn2a

nminus2 (bt)2


n

)

γ(B) (314)



γ(Fn(B)) (

an + Cn1a

nminus1b + Cn2a

nminus2 b2


n

)

γ(B) (315)


(

an0 + Cn01 a

n0minus1b + Cn02 a

n0minus2 b2


n0

)

= r lt 1 (316)





u(0) = g(u)

(317)













∣∣∣ C|a(λ)| (318)






1λa(λ)


(319)



such that




completely positive



lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (321)



u(t) =int t


int t

0f(s u(s)) + g(u) (322)


R(t)x = x +int t



d

dtR(t)x = AR(t)x +

int t




KgR +KΦ(R)int1





4 Applications


partw(t ξ)partt

= Aw(t ξ) +int t



w(0 ξ) =int1

0

int ξ

0qk(s ξ)w

(s y

)dsdy 0 ξ 2π

(41)



defined by



part







u(0) = g(u)

(43)






int10 k(s ξ)

2dsdξ)12







a(λ) =λ + α + βλ(λ + α)

(44)


f1(λ) =λ + α

λ + α + β f2(λ) =

λ2 + 2(α + β

)λ + αβ + α2

(λ + α + β

)2 (45)


f(n)1 (λ) =

(minus1)n+1β(n + 1)(λ + α + β

)n+1 f(n)2 (λ) =

(minus1)n+1β(α + β)(n + 1)

(λ + α + β








Re

(μ0 + iμ

b(μ0 + iμ

)+ 1

)

=μ30 + μ

20α + μ2

0

(α + β

)+ μ0α

(α + β

)+ μ0μ


)2 + 2μ0(α + β

)+ μ2

0 + μ2

(48)


(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1



∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1


∥∥∥∥

M

μ0 + iμ

∥∥∥∥ (410)


Therefore

lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (411)



int10 k(s ξ)

2dsdξ)12




qcKR +∥∥p1

∥∥1LK R (412)


Acknowledgments


References















































1 Introduction
















K(q)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2πqω1+2q

(2

2 + q

)1minus2q( Γ(1q)

Γ(12 + 1q

)

)2

if 1 le q lt +infin

4ω if q = +infin

(15)





0 lt a(t) le π2

ω2 t isin [0 ω] (16)





2 Preliminaries








u(t) =intω





Let





Let

G = min0letsleω


G(t s) σ = GG (26)





Th(t) =intω




Th(t) =intω

0G(t s)h(s)ds le G

intω

0h(s)ds (29)

and therefore

Th le Gintω

0h(s)ds (210)


Th(t) =intω

0G(t s)h(s)ds ge G

intω

0h(s)ds

=(GG

)middotGintω

0h(s)ds

ge σTh

(211)




Tφ = r(T)φ (212)


intω

0φ(t)dt = 1 (213)






hold




A = T F (216)











3 Main Results



mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x


mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x

(31)









f(t x0 x1 xn) = b0(t)x20 + b1(t)x


n (32)






radic|xn| (33)



















This implies that





(39)





intω

0



intω

0u0(t)φ(t)dt (311)


intω

0


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(312)


λ1

intω


intω

0u0(x)φ(t)dt (313)


intω


intω

0φ(t)dt = σu0 gt 0 (314)


i(AK capΩ1 K) = 1 (315)







(u1(t) minus μ1

)= F(u1)(t) t isin R (317)



This means that




(320)




intω

0

[u



intω

0u1(t)φ(t)dt (322)


intω

0


]φ(t)dt =

intω

0u1(t)


]dt

= λ1

intω

0u1(x)φ(t)dt

(323)


λ1

intω


intω

0u1(x)φ(t)dt (324)



intω


intω

0φ(t)dt = σu1 gt 0 (325)


i(AK capΩ2 K) = 0 (326)


i(AK cap

(Ω2 Ω1









(u0(t) minus μ0

)= F(u0)(t) t isin R (329)



(330)




intω

0

[u



intω

0u0(t)φ(t)dt (332)



intω

0

[u


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(333)


λ1

intω


intω

0u1(x)φ(t)dt (334)


i(AK capΩ1 K) = 0 (335)








(338)




intω

0

[u



intω

0u1(t)φ(t)dt (340)



intω

0

[u


]φ(t)dt =

intω

0u1(t)

[φ


]dt

= λ1

intω

0u1(x)φ(t)dt

(341)


λ1

intω


intω

0u1(x)φ(t)dt (342)


i(AK capΩ2 K) = 1 (343)


i(AK cap

(Ω2 Ω1



Acknowledgment


References

















York NY USA 1988










1 Introduction



x(t) = f

(


0u(t s x(c(s)))ds

)


x(t) = h

(

t x(t)intα(t)

0u(t s x(c(s)))ds

)





x(t) =int t



x(t) = L(t) +int t

0




x(t) =int t



x(t) = (Tx)(t)int t











0g(t s x







x(t) = f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f

(

t

intH(t)

0x(s)ds x(h(t))

)



x(t) = g(t x(t)) + f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f(t x(α(t)))

(

q(t) +intβ(t)

0u(t s x

(γ(s)))ds

)




2 Preliminaries












ωT

0 (X) = limεrarr 0

ωT (X ε)





diam X(t)

(22)







limtrarr+infin



Then

lim suptrarr+infin

ϕ(ψ(t))= lim sup












then









(212)




ωϕ(T)0 (X) (213)




∣∣∣ωT





∣∣∣ω




3 Main Results





limtrarr+infin


b(t) = +infin (31)






limtrarr+infin

sup

m3(t)intα(t)


= 0 (33)

sup

m3(t)intα(t)


leM0 (34)

sup

intα(t)


leM (35)








(Fx)(t) = f

(


0u(t s x(c(s)))ds

)



|(Fx)(t)| le∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minus f(t 0 0 0)∣∣∣∣∣+∣∣f(t 0 0 0)

∣∣


∣∣∣∣∣

intα(t)

0u(t s x(c(s)))ds

∣∣∣∣∣+ f



(310)





=

∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minusf(

t y(a(t))(Hy)(b(t))

intα(t)

0u(t s y(c(s))

)ds

)∣∣∣∣∣




)(b(t))

∣∣

+m3(t)intα(t)

0




)(b(t))

∣∣

+ sup

m3(t)intα(t)



(312)

which yields that


+ sup

m3(t)intα(t)


forallt isin R+

(313)


lim suptrarr+infin

diam(FX)(t)



diam(HX)(b(t))


sup

m3(t)intα(t)




diam(HX)(t)

(314)

that is

lim suptrarr+infin



diam(HX)(t) (315)





∣∣u(p τ v




ωTr

(f ε g(r)

)= sup

∣∣f(p vw z



(316)






le∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣

+

∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣


+m3(t)

[∣∣∣∣∣

intα(t)


∣∣∣∣∣+intα(s)


]

+ sup∣∣f(p vw z





∣∣u(p τ v




+ωTr

(f ε g(r)

)


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)

)

(318)

which implies that


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)


(319)



limεrarr 0


ωTr (u ε) = lim

εrarr 0ωTr

(f ε g(r)

)= 0 (320)


ωT0 (FX) leM1ω

a(T)0 (X) +M2ω

b(T)0 (HX) (321)


ω0(FX) leM1ω0(X) +M2ω0(HX) (322)



diam(FX)(t)



diam(HX)(t)

=M1μ(X) +M2

(ω0(HX) + lim sup


)

le (M1 +M2Q)μ(X)(323)



sup

m3(t)intα(t)


ltε



ωT (u δ r) ltε







=

∣∣∣∣∣f

(


0u(t s xn(c(s)))ds

)

minusf(


0u(t s x(c(s)))ds

)∣∣∣∣∣


+m3(t)intα(t)



+max

supτgtT

sup

m3(τ)intα(τ)

0


supτisin[0T]

sup

m3(τ)intα(τ)

0



2

ltε

2+max

ε

3 supτisin[0T]

m3(τ)intα(τ)

0ωT

(uδ02 r

)ds

ltε

2+max

ε

3


=5ε6

(327)

which yields that














M0 + h le r(1 minusM1) (331)




sup

m3(t)intα(t)





=

∣∣∣∣∣h

(

t z(t)intα(t)


)

minus h(

t y(t)intα(t)

0u(t τ y(c(τ))

)dτ

)∣∣∣∣∣


intα(t)

0



m3(t)intα(t)



(333)

which means that





4 Examples



x(t) =1

10 + ln3(1 + t3)+tx2(1 + 3t2

)

200(1 + t)+

1

10radic3 + 20t + 3|x(t3)|

+1

9 +radic1 + t

⎛

⎜⎝

int t2

0

s2x(4s) sin(radic



∣∣∣ x2(4s)ds

⎞

⎟⎠

3

forallt isin R+

(41)

Put

f(t vw z) =1

10 + ln3(1 + t3)+

tv2

200(1 + t)+

1

10radic3 + 20t + |w|

+z3

9 +radic1 + t

u(t s p

)=

s2p sin(radic



∣∣∣p2


(42)

Let

risin[9 minus radic

512

9 +

radic51

2

]

M=3 M0=9r10 M1=

r

100 M2=

1300

f =110

Q = 3 m1(t) =rt

100(1 + t) m2(t) =

1(10radic3 + 20t

)2 m3(t) =3M2

9 +radic1 + t


(43)



512

9 +

radic51

2

]



le t

200(1 + t)


∣∣∣ +

∣∣∣∣∣

1

10radic3 + 20t + |w|

minus 1

10radic3 + 20t +

∣∣q∣∣

∣∣∣∣∣

+1

9 +radic1 + t


∣∣∣ le m1(t)





sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)

minuss2w(4s) sin


)


∣∣∣w2(4s)


⎫⎪⎬

⎪⎭

le 3M2

9 +radic1 + t

int t2

0

2s2(1 + t12

)2

[r(1 + t12

)+ r3(r3t3 + 3r6 + 7

(t + 2t2

)2)]ds

le 2M2t6(9 +

radic1 + t)(

1 + t12)2

r(1 + t12

)+ r3[r3t3 + 3r6 + 7

(t + 2t2


sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=rM2

20lt M0

sup

intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=r

6leM

(44)




x(t) =3 + sin4

(radic1 + t2

)

16+x2(t)8 + t2

+1

1 + t

(intradict

0

sx3(1 + s2)

1 + t2 + s cos2(1 + t3s7x2(1 + s2))ds

)2

forallt isin R+

(45)

Put

h(t vw) =3 + sin4

(radic1 + t2

)

16+

v2

8 + t2+

w2

1 + t

u(t s p

)=

sp3

1 + t2 + s cos2(1 + t3s7p2

)

α(t) =radict c(t) = 1 + t2 m1(t) =

18 + t2

m3(t) =4


r =12 M = 2 M0 =M1 =

18 h =

14

(46)



le 18 + t2


∣∣∣ +

11 + t


∣∣∣



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))

minus sw3(1 + s2)

1 + t2 + s cos2(1 + t3s7w2(1 + s2))ds


le4r3t(1 +

radict + t2

)



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣

sv3(1 + s2)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

4r3

1 + t

intradict

0

s

1 + t2ds t isin R

+

le sup

t

4(1 + t)(1 + t2) t isin R

+

le 18=M1

sup

intα(t)


= sup

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

r3intradic

t

0

s

1 + t2ds t isin R

+

=132

lt M

(47)


Acknowledgments


References


































1 Introduction


Dq




i=1

aiu(nminus2)(ηi

) (11)



Tu(t) =int t


int1

0H(t s)u(s)ds (12)





Dq


)= 0 0 lt t lt 1


where Dq



Dq



i=1

aiu(ηi)

(14)



Dq



0+u(1) = 0 1 le p le n minus 2(15)



CDq



(η2)

(16)


where CDq




)= 0 2 lt q lt 3 0 lt p lt 1

u(0) = 0 uprime(0) = uprime(1) = 0(17)


Dq

0+u(t) + a(t)f(t u(t)) = 0 0 lt t lt 1


i=1

ζiu(ηi) (18)




CDqu(t) = f(t u(t)) u(0) = u0 (19)



Dq



(110)












αC(D) = maxtisinI

α(D(t)) α

(int

I

u(t)dt u isin D)

leint

I

α(D(t))dt (21)




Iq

0+y(t) =1

Γ(q)int t



given by

Dq

0+y(t) =1

Γ(n minus q)

(d

dt

)n int t



+


Dq

0+y(t) = 0 (24)




0+y isinL(0 1)

Iq

0+Dq







3 Main Results





2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

lt 1 (32)

such that

α(f(tD1 D2 Dn+1)

) len+1sum

k=1



i=1 aiηqminusn+1i





i=1

aix(ηi)

(34)


is

x(t) =int1

0G(t s)y(s)ds (35)

where


i=1 aig(ηi s)


g(t s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩



) t le s(37)




x(t) = minusint t

0





c1 =1


qminusn+1i

int1

0



i=1 ai


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

(310)


x(t) = minusint t

0


)y(s)ds +tqminusn+1


int1

0


)y(s)ds

minussummminus2

i=1 aitqminusn+1


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

=int1

0G(t s)y(s)ds

(311)







ρ0(s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


s isin (0 ξ]

(γ

s


(312)











i=1 aig(ηi s)

ηlowast (314)



mminus2sum

i=1

aig(ηi s)= G(s)

le tqminusn+1

Γ(q minus n + 2

) +summminus2

i=1 aiηqminusn+1i



(315)




(Inminus20+ x(s)

) S(Inminus20+ x(s)

))= θ


i=1

aix(ηi)

(316)






(317)


(Ax)(t) =int1

0G(t s)f


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))ds 0 le t le 1 (318)



(Ax)(t) leint1

0G(s)f


(Inminus20+ xj(s)


))ds

=

(int γ

0+intδ

γ

+int1

δ

)


(Inminus20+ xj(s)


))ds

le 3intδ

γ


(Inminus20+ xj(s)


))ds

(319)


mintisin[γ δ]


int1

0G(t s)f


(Inminus20+ xj(s)


))ds

geintδ

γ

(

g(t s) +tqminusn+1

ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds


geintδ

γ

(


ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds

ge λ

intδ

γ


(Inminus20+ xj(s)


))ds

ge λ

3(Ax)


(320)




∥∥ =int1

0G(t s)


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds

(321)



(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ ltε

Δ

(322)





on I





(Inminus20+ x(s)

) S(Inminus20+ x(s)

))



(Inminus20+ x(s)

) S(Inminus20+ x(s)

))




(Inminus20+ D

)(I) times S

(Inminus20+ D

)(I)))


k=1



(T(Inminus20+ D

)(I))


)(I))

(324)

which implies


k=1





(S(Inminus20+ D

)(I))

(325)

Obviously


)(I) = α

(int s

0


)


(326)

α(T(Inminus20+ D

))(I) = α

(int t

0K(t s)

(ints

0


u(τ)dτ

)


le alowast


(n minus 3)α(D(I))

(327)

α(S(Inminus20+ D

))(I) = α

(int1

0H(t s)

(int s

0


u(τ)dτ

)


le blowast


(n minus 3)α(D(I))

(328)



α(D(I)) le 2αC(D) (329)



nminus2sum

k=1




)

αC(D) (330)







nminus1sum

k=1






nminus1sum

k=1



NMβ

int1







k=1

uk ge r1 (334)


Therefore




k=1

uk ge r1 (335)

Take


(336)


x(t) ge λ

3NxΩ ge λ

3Nr0 gt r1 (337)

Hence

(Ax)(t) geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



xΩ(intδ

γ

G(s)c1(s)ds

)

middot ulowast

=1


(intδ

γ

G(s)c1(s)dsulowast)



(338)

and consequently






k=1

uk le r2 (340)






k=1

uk le r2 (341)

Choose

0 lt r lt min

⎧⎨

⎩

(nminus2sum

k=0

1k

)minus1r2 β

⎫⎬

⎭ (342)


(Ax)(ξ) =int1

0G(ξ s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(ξ s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



x(s)Ωintδ

γ


=1


x(s)Ω(intδ

γ




(343)

which implies


that is




(Ax)(t) leint1

0G(s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds (346)


∥∥f(t u1 un+1)


Therefore





(H7)

h(u1 un+1)sumn+1


n+1sum

k=1





k=1


k=1

uk le r3 (350)


ε0 =

(

N

nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

int1

0G(s)a(s)ds

)minus1 (351)




(Ax)(t) leNint1


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))∥∥∥ds

leNint1

0G(s)a(s)h


∥∥∥ x(s)



)(s)∥∥∥)ds

leNε0

int1

0G(s)a(s)

(nminus1sum

k=1



)(s)∥∥∥ +∥∥∥S(Inminus20+ x

)(s)∥∥∥

)

ds

leNε0

(nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

)

rlowastint1


(352)

and consequently




4 An Example



960k3

⎧⎪⎨

⎪⎩

⎡


j=1

u2j(t) +infinsum

j=1

uprimej(t)

⎤

⎦

3

+

⎛


j=1

uj(t) +infinsum

j=1

uprime2j(t)

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5

(int t


)23

+1 + t2

24k4

(int1


)

t isin I


(14

)+12uprimek

(49

) k = 1 2 3

(41)






fk(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭+1 + t3

36k5x15k

+1 + t2

24k4y2k

(42)


k=1(1k3) lt 32 we get by (42)

∥∥f(t u v x y

)∥∥ =infinsum

k=1

∣∣fk(t u v x y

)∣∣

le (1 + t)3

2

(140

(u + v)3 + 1160

(u + v)12 + 112

x15 + 18∥∥y∥∥)

(43)


h(u v x y

)=

140

(u + v)3 +1160

(u + v)12 +112x15 +

18y (44)



1 f(1)k ) and f (2) =

(f (2)1 f

(2)k ) in which

f(1)k

(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5x15k (k = 1 2 3 )

f(2)k

(t u v x y

)=

1 + t2

24k4y2k (k = 1 2 3 )

(45)



α(f (1)(tD1 D2 D3 D4)


α(f (2)(ID1 D2 D3 D4)

)le 1


(46)


α(f(ID1 D2 D3 D4)


2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

asymp 01565 lt 1(47)



fk(t u v x y

) ge (1 + t)3


fk(t u v x y

) ge (1 + t)3

960k3


(48)


c1(t) =(1 + t)3

960 h1 k(u v) = (u + v)3 ulowast =

(1

1k3

) (49)

in this situation


(u + v)3u + v = infin (410)


c2(t) =(1 + t)3

960 h2 k(u v) =


(1

1k3

) (411)

in this situation

h2k = limu+vrarr 0




NMβ

int1

0G(s)a(s)ds asymp 07287 lt β = 1 (413)


Acknowledgments


References











































1 Introduction






∣∣f(t)


(11)


G(u) = S[f(t)u

]

=intinfin


(12)





f(t) =infinsum

n=0

antn (13)

then

F(u) =infinsum

n=0

nantn (14)





(15)







Gn(u) =G(u)un

minusnminus1sum

k=0

f (k)(0)unminusk

(16)



LU + RU +NU = h(x t) (17)


S[LU] + S[RU] + S[NU] = S[h(x t)] (18)



unU(x 0) minus 1


u+ S[RU] + S[NU] = S[h(x t)]

(19)




U(x t) =infinsum

i=0

Ui(x t) (111)



NU(x t) =infinsum

i=0

Ai (112)


Ai =1idi

dλi

[

Ninfinsum

i=0

λiUi

]

λ=0

i = 0 1 2 (113)


S

[ infinsum

i=0

Ui(x t)

]


minus unS[ infinsum

i=0

Ai

]

+ unS[h(x t)]

(114)









U0(x t) = K(x t)



Ψ = Φ + Sminus1[h(x t)] (119)







LU + RU +NU = h(x t) (122)


U(x 0) = f(x) Ut(x 0) = g(x) (123)





]+ Sminus1

[u2S[h(x t)]

] (125)

2 Applications




U(x 0) = 0

Ut(x 0) = ex(22)




] (23)



[U2x minusU2

]](24)


U(x t) =infinsum

i=0

Ui(x t) (25)


infinsum

i=0


u2S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

(26)


infinsum

i=0

Ai(U) = U2x

infinsum

i=0

Ai(U) = U2

(27)


A0(U) = U20x A1(U) = 2U0xU1x Ai(U) =

isum

r=0

UrxUiminusrx

B0(U) = U20 B1(U) = 2U0U1 Bi(U)

isum

r=0

UrUiminusr

(28)


U0(x t) = tex


S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

n ge 0(29)



U1(x t) = Sminus1[

u2S

[ infinsum

i=0

A0(U) minusinfinsum

i=0

B0(U)

]]

= Sminus1[u2S

[U2

0x minusU20

]]= Sminus1

[u2S

[t2e2x minus t2e2x

]]

= Sminus1[u2S[0]

]= 0

(210)


U(x t) =infinsum

i=0

Ui(x t) = tex (211)


Ut

(x y t

) minus VxWy = 1

Vt(x y t

) minusWxUy = 5

Wt

(x y t

) minusUxVy = 5

(212)


U(x y 0

)= x + 2y

V(x y 0

)= x minus 2y

W(x y 0

)= minusx + 2y

(213)


U(x y u

)= x + 2y + u + uS

[VxWy

]

V(x y u


[WxUy

]

W(x y u


[UxVy

]

(214)


U(x y t


[VxWy

]]

V(x y t


[WxUy

]]

U(x y t


[UxVy

]]

(215)



U0(x y t

)= t + x + 2y

Ui+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ci(VW)

]]

i ge 0

V0(x y t

)= 5t + x minus 2y

Vi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Di(UW)

]]

i ge 0

W0(x y t

)= 5t minus x + 2y

Wi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ei(UV )

]]

i ge 0

(216)


C0(VW) = V0xW0y


Ci(VW) =isum

r=0

VrxWiminusry

D0(UW) = U0yW0x


Di(UW) =isum

r=0

WrxUiminusry

E0(UV ) = U0xV0y


Ei(VW) =isum

r=0

UrxViminusry

(217)



U1(x y t


[V0xW0y

]]= Sminus1[uS[(1)(2)]] = 2t

V1(x y t


[W0xU0y


W1(x y t


[U0xV0y


U2(x y t


[V1xW0y + V0xW1y

]]= 0

V2(x y t


[U0yW1x +U1yW0x

]]= 0

W2(x y t


[U1xV0y +U0xV1y

]]= 0

(218)


U(x y t

)=

infinsum

i=0

Ui

(x y t

)= x + 2y + 3t

V(x y t

)=

infinsum

i=0

Vi(x y t

)= x minus 2y + 3t

W(x y t

)=

infinsum

i=0

Wi

(x y t

)= minusx + 2y + 3t

(219)





U(x 0) = 1 + ex V (x 0) = minus1 + ex (221)




Ux(x t) =infinsum

i=0


i=0

Vxi(x t) (223)



S

[ infinsum

i=0

Ui(x t)

]


[ infinsum

i=0

Vix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

S

[ infinsum

i=0

Vi(x t)

]


i=0

Uix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

(224)






(225)






U2 = Sminus1[u2ex

]=t2

2ex

V2 = Sminus1[u2ex

]=t2

2ex

(226)


U(x t) = 1 + ex(

1 + t +t2

2+t3


)


1 minus t + t2

2minus t3


)

(227)


U(x t) = 1 + ex+t




Ut + VUx +U = 1



U(x 0) = ex V (x 0) = eminusx (230)





S

[ infinsum

i=0

Ui(x t)

]


i=0

Ai

]

minus uS[ infinsum

i=0

Ui

]

S

[ infinsum

i=0

Vi(x t)

]

= eminusx + u + uS

[ infinsum

i=0

Bi

]

minus uS[ infinsum

i=0

Vi

]

(232)


A0 = U0xV0 A1 = U0xV1 +U1xV0


B0 = V0xU0 B1 = V0xU1 + V1xU0

B2 = V0xU2 + V1xU1 + V2xU0

(233)



U0 = ex + t






2minus tex minus t2

2ex



2eminusx

U2 =t2

2+t2


V2 =t2

2+t2


(236)


U(x t) = ex(

1 minus t + t2

2minus t3


)

V (x t) = eminusx(

1 + t +t2

2+t3


)

(237)



3 Conclusion



Acknowledgment


References


























1 FK and BK Spaces







infink=0 let x

[n] =sumn






suminfinn=0 λnb





suminfink=0 xke






dw(x y

)=

infinsum

k=0

12k



∣∣

)

(x y isin w)

(11)








sumnk=0 xke





Cs =

σ isin C infinsum

n=1

cn(σ) le s

(12)



Φ =φ =

(φk



m(φ)=


supσisinCs

(1φs

sum

kisinσ|xk|

)

ltinfin

n(φ)=


( infinsum

k=1

|uk|Δφk)

ltinfin

(14)
















χ(Q1 +Q2) le χ(Q1) + χ(Q2)




Lχ = χ(L(SX)) (24)


Lχ = 0 (25)










1amiddot lim sup

nrarrinfin

(

supxisinQ

(I minus Pn)(x))


(

supxisinQ

(I minus Pn)(x))

(26)






(

supxisinQ

(I minus Pn)(x))

(27)



(

supxisinQ

sum

kgen|xk|p

)

(28)





Pr(z) = ze +rsum





(

supxisinQ



(

supxisinQ


(210)




supxisinQ

(

supuisinS(x)

( infinsum

n=k

|un|Δφn))

(211)








Lχ = χ(L(SX)) (212)

and we have












x(0) = x0 (32)




∥∥f(t x)∥∥ le Q + Rx (33)


μ(f(t Y )

) le q(t)μ(Y ) (34)









(

supxisinE

supkgen

|xk|)

(35)





xi(0) = x0i (37)





(i) x0 = (x0i ) isin c0



∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

(38)



Now let us denote

q(t) = supnge1

qn(t)

Q = suptisinI

q(t)


(39)




∣∣fn(t x)∣∣ =

∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

le P +Q supigekn


Hence we get



χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x)∣∣)

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x1 x2 )∣∣)

le limnrarrinfin

supxisinY

(

supigen


∣∣xj∣∣)

le limnrarrinfin

supigen


supxisinY

(

supigen

supjgekn

∣∣xj∣∣)

(312)


χ(f(t Y )

) le q(t)χ(Y ) (313)







infinsum

j=ki+1

aij(t)xj (314)


xi(0) = x0i (315)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (316)



i = 1 2










2 = f2(t x1 x2)


(317)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (318)



Then we have


∣∣f1(t x1)∣∣∣∣f2(t x1 x2)

∣∣ ∣∣fi(t x1 x2 xi)

∣∣


(319)


∣∣∣fi(t x)∣∣∣ile Pi(t) (320)













xi(0) = x0i (323)




supxisinE

supnmgep

|xn minus xm|


(

supxisinQ

supkgen

|xk|)

(324)







∣∣gn(t x1 x2 )∣∣ le bi (325)




a(t) = supige1

ai(t)

Q = suptisinI

a(t)(326)






f(t x) =(f1(t x) f2(tx)

)=(fi(t x)

)

(327)





∣∣


(328)



∣∣fi(t x)∣∣ le |ai(t)||xi| +

∣∣gi(t x)∣∣








(s y


(s y

)∣∣


∣∣ +


(s y

)∣∣


∣∣ +


(s y

)∣∣

(331)

Then


ige1


(s y

)∣∣


|ai(t) minus ai(s)|



(332)




∣∣gn(t x)∣∣ +

∣∣gm(t x)∣∣



μ(f(t E)

) le a(t)μ(E) (334)











∣∣fi(t x)

∣∣p =

∣∣fi(t x0 x1 x2 )

∣∣p le qi(t) + ri(t)|xi|p (335)







R = supri(t) t isin I i = 0 1 2 (336)



∥∥f(t x)∥∥pp =

infinsum

i=0


leinfinsum

i=0

[qi(t) + ri(t)|xi|p

]

leinfinsum

i=0

qi(t) +

(

supige0

ri(t)

)( infinsum

i=0|xi|p

)

le Q + Rxpp

(337)




χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(sum

igen

∣∣fi(t x1 x2 )∣∣p)

le limnrarrinfin

(sum

igenqi(t) +

(

supigen

ri(t)

)(sum

igen|xi|p

) )

(338)



χ(f(t Y )

) le r(t)χ(Y ) (339)








xi(0) = x0i (341)






∣∣fi(t x0 x1 x2 )∣∣p le bi(t) (342)










∣∣fi(t u)

∣∣ =

∣∣fi(t u1 u2 )

∣∣ le pi(t) + qi(t)|ui| (343)








(344)




uisinS(x)

infinsum

i=1


le supuisinS(x)

infinsum

i=1

[pi(t) + qi(t)|ui|

]Δφi

leinfinsum

i=1

pi(t)Δφi +

(

supi

qi(t)

)(

supuisinS(x)

infinsum

i=1

|ui|Δφi)

le P +Qxn(φ)

(345)




χ(f(t X)

)= lim

krarrinfinsupxisinX

(

supuisinS(x)

( infinsum

n=k


))

le limkrarrinfin

( infinsum

n=k

pn(t)Δφn +

(

supngek

qn(t)

)(

supuisinS(x)

infinsum

n=k

|un|Δφn))

(346)



χ(f(t X)

) le q(t)χ(X) (347)




References























(c)α l












































1 Introduction






(11)



) (12)


with

x(t+k)= lim

hrarr 0+x(tk + h) x

(tminusk)= lim










2 Preliminaries








Iα0h(t) =1

Γ(α)

int t




CDα0+h(t) =

1Γ(n minus α)

int t




CDαh(t) = 0 (23)

has solutions










Then either






x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

0



tk



i=1

int ti

timinus1


h(s)ds


i=1

int ti

timinus1


h(s)ds


i=1

(tk minus ti)int ti

timinus1


h(s)ds + Ω1(t)intT

tk


h(s)ds

+Ω2(t)intT

tk


h(s)ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


+kminus1sum

i=1


(26)




(x(tminusk))


(27)

where

Λ1 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(T minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1




Λ2 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1



minus ct

TΔ



Δ

(28)



0



xprime(t) =int t

0


h(s)ds minus c1(29)


x(t) =int t

t1



xprime(t) =int t

t1




x(tminus1)=int t1

t0




t0



(211)

Observing that





then we have

minusd0 =int t1

t0



minusd1 =int t1

t0



(213)


x(t) =int t

t1


h(s)ds +int t1

t0


h(s)ds


t0




xprime(t) =int t

t1


h(s)ds +int t1

t0



(214)


x(t) =int t

t2



xprime(t) =int t

t2




x(tminus2)=int t2

t1


h(s)ds +int t1

t0


h(s)ds


t0




x(t+2)= minus e0


t1


h(s)ds +int t1

t0




(216)






thus we have



(218)


x(t) =int t

t2


h(s)ds +int t1

t0


h(s)ds +int t2

t1


h(s)ds


t0


h(s)ds


t0


h(s)ds +int t2

t1


h(s)ds

]

+ I1(x(tminus1))

+ I2(x(tminus2))



(219)


x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(t minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1



xprime(t) =

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1


(220)



we have




Δ+(1 minus b)TΛ2

Δ

minusc1 = minuscΛ1


Δ

(222)




3 Main Results



conditions


(x(tminusk))



Ωlowast1 = sup


2 = suptisinJ


tisinJ


2 = suptisinJ

∣∣Ωprime2(t)∣∣ (32)





k(u) minus Ilowast


k = 1 2 p


[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

lt 1

(33)



(Fx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds



i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk


f(s x(s))ds

+ Ω2(t)intT

tk


f(s x(s))ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(34)



tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




timesksum

i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



+ |1 + Ω1(t)|[

ksum

i=1




(y(tminusk))∣∣]


i=1


(y(tminusi))∣∣


i=1


(y(tminusi))∣∣

+kminus1sum

i=1


(y(tminusi))∣∣



[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3


(35)







(Ax)(t) = [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(Dx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds


i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk



tk


f(s x(s))ds

(36)


(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3 lt 1 (37)





tk



∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



∣∣ds


timesksum

i=1

int ti

timinus1



∣∣ds


i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



(38)




|(Dx)(t)| leint t

tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1




+ |Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + |Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

D(x) le M1Tα


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)= l

(310)


∣∣(Dy


∣∣ leint τ2

τ1



where


tk


∣∣f(s x(s))∣∣ds +


ksum

i=1

int ti

timinus1




1(t) + Ωprime2(t) + 1

∣∣ksum

i=1

int ti

timinus1



+∣∣Ωprime


i=1

(tk minus ti)int ti

timinus1



+∣∣Ωprime

1(t)∣∣intT

tk


∣∣f(s x(s))∣∣ds +


tk



le M1Tα

Γ(α + 1)Ωlowast

1

(1 + p + 2pα

)+M1T

αminus1

Γ(α)(1 + Ωlowast

2)(1 + p

)= L

(312)




(xλ


(313)



)(t) (314)



|x(t)| le λ

int t

tk


∣∣f(s x(s))

∣∣ds + λ|1 + Ω1(t)|

ksum

i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

(tk minus ti)int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + λ|Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

+ λ|1 + Ω1(t)|[

ksum

i=1

∣∣∣∣∣Ii

(x(tminusi)

λ

)∣∣∣∣∣+


(x(tminusk

)

λ

)∣∣∣∣∣

]


i=1


(x(tminusi)

λ

)∣∣∣∣∣


i=1


(x(tminusi)

λ

)∣∣∣∣∣+ λ

kminus1sum

i=1


(x(tminusi)

λ

)∣∣∣∣∣

x(t) le[M1T

α


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pM2 +M3

)+(Ωlowast

1T + Ωlowast2 + T

)pM3

]

(315)


4 An Example


CD32x(t) =

sin 2t|x(t)|(t + 5)2(1 + |x(t)|)

t isin [0 1] t =13

Δx(13

)=


(13

)=



(41)




[L1T

α


1

)(1 + p(1 + L2) + 2pα + L3

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

=464

1125radicπ

+173450

lt 1

(42)


Acknowledgments


References

































1 Introduction


intb

a

w(x)f(x)dx =nsum

k=0

wkf(xk) + Rn+1[f] (11)






Ψn+1(x) =nprod

k=0

(x minus xk) (12)


f(x) =nsum

k=0

f(xk)L(xxk) + rn+1(f x) (13)

where

L(xxk) =Ψn+1(x)


(k = 0 1 n) (14)

we obtain (11) with

wk =1

Ψprimen+1(xk)

intb

a


Rn+1[f]=intb

a

rn+1(f x)w(x)dx

(15)



intb

a

f(t)dt =b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)+ E(f) (16)



∥∥f∥∥p =

(intb

a

∣∣f(t)∣∣pdt

)1p

(17)


xisin[ab]

∣∣f(x)∣∣ (18)



∣∣∣∣∣

intb

a

f(x)g(x)dx

∣∣∣∣∣le ∥∥f∥∥1





λ(b minus a)intb

a



2λ(b minus a) f(a)

∣∣∣∣∣




)

(110)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (111)


K(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

t minus 5a + b6

t isin[aa + b2

]

t minus a + 5b6

t isin(a + b2

b

]

(112)


intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt (113)

maxtisin[ab]

|K(t)| = 13(b minus a) (114)


2 Main Results


m1 =int (5a+b)6

a

(t minus 5a + b

6

)β(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)α(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)β(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)α(t)dt

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt le

M1 =int (5a+b)6

a

(t minus 5a + b

6

)α(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)β(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)α(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)β(t)dt

(21)


intb

a


2

)dt

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus 12

(intb

a

K(t)(α(t) + β(t)

)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)

(22)



2


2 (23)




(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a


2

)dt


a


dt

=12

(int (a+b)2

a


5a + b6


intb

(a+b)2


a + 5b6


)

(24)


m1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)β(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)α(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)β(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)α(x)dx

M1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)α(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)β(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)α(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)β(x)dx

(25)




Special Case 1


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


144((β1 minus α1

)(a + b) + 2

(β0 minus α0

))

(26)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (27)



int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(


a

α(t)dt

)

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t) dt

leint (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

+b minus a3

(


a

α(t)dt

)

(28)

Proof Since

intb

a


6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)α(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)

(29)


so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


3

(


a

α(t)dt

)

(210)


Special Case 2


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)


a + b2

α1

))

(211)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)



int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(intb

a


le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

leint (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt +

b minus a3

(intb

a


(213)


Proof Since

intb

a


=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)β(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)

(214)

so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

int

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


)dt =

b minus a3

(int b

a


(215)


Special Case 3

If β(x) = β1x + β0 = 0 in (213) then

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 +

a + b2


b minus a)

(216)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 minus


) (217)


Acknowledgment


References





















1 Introduction









xx= 0

(11)













u(x t) = U(ξ) ξ = kx +ωt (22)



)= 0 (23)



U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0 (24)


d2G(ξ)dξ2

+ λdG(ξ)dξ

+ μG(ξ) = 0 (25)






3 Application


u(x t) = U(ξ)eiη

v(x t) = V (ξ)(31)




)U = 0 (32)

(ω2 minus k2


(U2)primeprime

= 0 (33)



)U = 0

(ω2 minus k2


(34)


U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0

V (ξ) =nsum

i=1

βi

(Gprime

G

)i+ β0 βn = 0

(35)




m + n = m + 2

2m = n + 2(36)


U(ξ) = α2(Gprime

G

)2

+ α1(Gprime

G

)+ α0 α2 = 0

V (ξ) = β2(Gprime

G

)2

+ β1(Gprime

G

)+ β0 β2 = 0

(37)




)= 0




)ω minus qα1



)= 0


)α2



)β2 minus k2


)= 0


)α2 minus α1

(minus2k2 + aβ2

)= 0


)= 0

(6k2 minus aβ2

)α2 = 0


(38)






λ = plusmn13

radic3ω2 minus 3k2

k2 μ =

16k2 minusω2

k4 p = minus1

2ω

k

q =124

24k4 + 30ω4 minus 54k2ω2


α1 = plusmn2radic

1ab


radic1ab


a β2 =

6k2

a

(39)


U(ξ) = 6k2radic

1ab

(Gprime

G

)2

plusmn 2

radic1ab

radic3ω2 minus 3k2

(Gprime

G

)

V (ξ) =6k2

a

(Gprime

G

)2

plusmn 2

radic3ω2 minus 3k2

a

(Gprime

G

)

(310)



uH(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2



⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vH(x t) =6k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(311)



2 asfollows

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312a)


⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312b)

while at C21 lt C

22 one can obtain

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312c)


⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312d)




uT(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2


⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vT(x t) =6k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(313)


1 gt C22 and C

21 lt C

22 as follows

uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314a)


ak2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314b)


|u H|

tx

1

05

0

minus054

20

minus2minus4 minus4

minus20

24


0

02

04

06

08

tx

42

0minus2

minus4 minus4minus2

02

4

|u H|2


uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314c)


ak2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314d)



0

200

400

600

800

tx

42

0minus2

minus4 minus4minus2

02

4

vH


4 Conclusions


Acknowledgments


References























Shengli Xie







1 Introduction



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J






(

t x(σ1(t))int t


)

t isin [0 b]

x(0) + g(x) = x0

(12)


(1 minus αβMc

)N


2αθ1L1M(1 + 2Kθ2L1) lt 1

(13)





2 Preliminaries








x(t) = R(t 0)[φ(0) + g(x)(0)

]+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

(21)




α

(int t

0xn(s)ds n isin N

)

le 2int t








3 Existence Result




+ such that

∥∥f(t x y z


∥∥y∥∥ + z






andint t0 k(t s)ds





α

(int t

0K(t s V1)ds

)


α(V1(θ)) t isin J

α

(intb

0H(t s V1)ds

)


α(V1(θ)) t isin J

(33)


w = maxMp0(1 + k0 + h0) + 1 2M

(l01 + 2l02k0 + 2l03h0

)+ 1 (34)






(Fx)(t)

=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(35)






0ew(sminust)

∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥ds

le L +Mint t

0ew(sminust)p(s)

(

xs[minusq0] +ints

0K(s r xr)dr +

intb

0H(s r xr)dr

)

ds

le L +Mp0

int t

0ew(sminust)

(

xs[minusq0] +int s


intb


)

ds


Consequently




(38)


∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥le p(s)(1 + k0 + h0)R (39)







0R(t minus s)α

(

f

(

s Vs

int s

0K(s r Vr)dr

intb

0H(s r Vr)dr

))

ds

le 2Mint t

0ew(sminust)

[



0K(s r Vr)dr

)

+l3(s)α

(intb

0H(s r Vr)dr

)]

ds

le 2Mint t

0ew(sminust)

[


α(V (s + θ)) + 2l02

int s

0k(s r) sup


+2l03

intb

0h(s r) sup


]

ds

le 2M(l01 + 2l02k0 + 2l03h0

)int t

0ew(sminust)ds sup


le 2M(l01 + 2l02k0 + 2l03h0


(311)

Consequently


α((FV )(t)) le 2M(l01 + 2l02k0 + 2l03h0






4 An Example




ut(t y)= a1

(t y)uyy(t y)

+ a2(t)

[

sinu(t + θ y

)ds +

int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

0 le t le b


0

intb

0F(r y)log(1 + |u(r s)|12


u(t 0) = u(t π) = 0 0 le t le b(41)


f

(

t ut

int t

0K(t s us)ds

intb

0H(t s us)ds

)(y)

= a2(t)

[

sinu(t + θ y

)+int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

K(t s us)(y)=ints

minusqa3(s + τ)

u(τ y)dτ

(1 + t) H(t s us)

(y)=

u(s + θ y

)

(1 + t)(1 + s)2

g(u)(y)=intπ

0

intb

0F(r y)log(1 + |u(r s)|12

)dr ds

(42)



(43)


∥∥f(t u v z)∥∥ le |a2(t)|

(u[minusq0] + v + z

)

K(t s u) leints


u[minusq0](1 + t)


(1 + t)(1 + s)

(44)



∥∥g(u)



∣∣F(r y)∣∣(u[0b] +

radicπ) (45)


α(f(t V1 V2 V3)

) le |a2(t)|(

supminusqleθle0

α(V1(θ)) + α(V2) + α(V3)

)

t isin J

α(K(t s V1)) le 11 + t

supminusqleθle0


α(H(t s V1)) le 1

(1 + t)(1 + s)2sup


(46)


5 An Application



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J



x(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(52)






w = maxMp0(1 + k0 + h0)(1 +MM1b) + 1 2M

(l01 + 2l02k0 + 2l03h0

)(1 +MM1b) + 1

(53)



Wv =intb

0R(b s)(Cv)(s)ds (54)





v(t) =Wminus1(

x1 minus R(b 0)[φ(0) + g(x)(0)

]

+intb

0R(b s)f

(

s xs

ints

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

)

(t) t isin [0 b]

(55)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(56)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t


[φ(0) + g(x)(0)

]

+intb

0R(b τ)f

(

τ xτ

int τ

0K(τ r xr)dr

intb

0H(τ r xr)dr

)

dτ

)

(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(57)




Acknowledgments


References









































Contents



































Acknowledgments













1 Introduction











119909 = 119865119909 (1)












119901= 119862(R

+ 119901(119905)) the


+and such that

sup |119909 (119905)| 119901 (119905) 119905 ge 0 lt infin (2)


to the norm

119909 = sup |119909 (119905)| 119901 (119905) 119905 ge 0 (3)


119901[10 15]



+and if lim

119879rarrinfinsup|119909(119905)|119901(119905) 119905 ge 119879 = 0





that is


(4)


119909 (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 (5)




+ 119886 R

+rarr


|119907 (119905 119904 119909)| le 119899 (119905 119904) + 119886 (119905) 119887 (119904) |119909| (6)



119871 (119905) = int

119905

0

119886 (119904) 119887 (119904) 119889119904 119905 ge 0 (7)


119901 where 119901(119905) = [119886(119905) exp(119872119871(119905) + 119905)]




119863119886(119905) exp(119872119871(119905)) for 119905 ge 0


int

119905

0

119899 (119905 119904) 119889119904 le 119873119886 (119905) exp (119872119871 (119905)) (8)

for 119905 ge 0



condition

119871 (120593 (119905)) minus 119871 (119905) le 119870 (9)



for all 119905 ge 0



119901such that |119909(119905)| le 119886(119905) exp(119872119871(119905))

for 119905 isin R+


119901by putting

(119865119909) (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 119905 ge 0 (10)


+






|(119865119909) (119905)| le |119889 (119905)| + int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904)))1003816100381610038161003816 119889119904

le 119863119886 (119905) exp (119872119871 (119905))

+ int

119905

119900

[119899 (119905 119904) + 119886 (119905) 119887 (119904)1003816100381610038161003816119909 (120593 (119904))

1003816100381610038161003816] 119889119904

le 119863119886 (119905) exp (119872119871 (119905)) + 119873119886 (119905) exp (119872119871 (119905))

+ 119886 (119905) int

119905

0

119887 (119904) 119886 (120593 (119904)) exp (119872119871 (120593 (119904))) 119889119904

le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) int

119905

0

119872119886 (119904) 119887 (119904)


le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) exp (119872119871 (119905))

= 119886 (119905) exp (119872119871 (119905))

(11)





1003816100381610038161003816

le int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904))) minus 119907 (119905 119904 119910 (120593 (119904)))1003816100381610038161003816 119889119904

le 120573 (120576)

(12)



1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)1003816100381610038161003816 [119886 (119905) exp (119872119871 (119905) + 119905)]

minus1

le |(119865119909) (119905)| +1003816100381610038161003816(119865119910) (119905)

1003816100381610038161003816

times [119886 (119905) exp (119872119871 (119905))]minus1


le 2119890minus119905

(13)


1003816100381610038161003816 119901 (119905) le 120576 (14)



119901


|(119865119909) (119905)| 119901 (119905) le 119890minus119905 (15)

This yields that

lim119879rarrinfin

sup |(119865119909) (119905)| 119901 (119905) 119905 ge 119879 = 0 (16)



|(119865119909) (119905) minus (119865119909) (119904)|

le |119889 (119905) minus 119889 (119904)|

+

10038161003816100381610038161003816100381610038161003816

int

119905

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119904 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

le 120584119879(119889 120576) + 120576max 119899 (119905 120591) + 119886 (120591) 119887 (120591)

times119901 (120593 (120591)) 0 le 120591 le 119905 le 119879

+ 119879120584119879(119907 (120576 119879 120572 (119879)))

(17)

where we denoted120584119879(119907 (120576 119879 120572 (119879)))

= sup |119907 (119905 119906 119907) minus 119907 (119904 119906 119907)| 119905 119904 isin [0 119879]

|119905 minus 119904| le 120576 119906 isin [0 119879] |119907| le 120572 (119879)

(18)



120584119879(119889 120576) =

lim120576rarr0











119909 (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

+ int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(19)



+)



supremum norm

119909 = sup |119909 (119905)| 119905 isin R+ (20)








that

1003816100381610038161003816119891 (119905 1199091 1199092 119909

119899) minus 119891 (119905 119910

1 1199102 119910

119899)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816

(21)


1 1199092 119909

119899) (1199101 1199102

119910119899) isin R119899


with 1198650= sup|119891(119905 0 0)| 119905 isin R

+


R+



1 2 119898)(iv) The function 120573 R




rarr R+and


1003816100381610038161003816119892 (119905 119904 1199091 119909

119898)1003816100381610038161003816 le 119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

10038161003816100381610038161199091198941003816100381610038161003816 (22)

for all 119905 119904 isin R+and (119909

1 1199092 119909


we assume that

lim119905rarrinfin

int

120573(119905)

0

119902 (119905 119904) 119889119904 = 0 lim119905rarrinfin

119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 = 0

(23)


1 1199072 R+

rarr R+defined by

the formulas

1199071(119905) = int

120573(119905)

0

119902 (119905 119904) 119889119904 1199072(119905) = 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 (24)



119872119894= sup 119907

119894(119905) 119905 isin R

+ (119894 = 1 2) (25)


119896 = sum119899


119894(119894 = 1 2 119899) appear in

assumption (i)

(vi) 119896 + 1198981198722lt 1




globally attractive




119909 (119905) = (119876119909) (119905) 119905 isin R+ (26)


lim119905rarrinfin

(119909 (119905) minus 119910 (119905)) = 0 (27)




119903in the space119883 = 119861119862(R

+) centered at the


2)]


putting

(119860119909) (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

(119861119909) (119905) = int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(28)


119909 (119905) = (119860119909) (119905) + (119861119909) (119905) 119905 isin R+ (29)


+ Thus



119903into



+ we get

1003816100381610038161003816(119860119909) (119905) minus (119860119910) (119905)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816

le 1198961003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(30)





1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816

le int

120573(119905)

0

[1003816100381610038161003816119892 (119905 119904 119909 (120574

1(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816

+1003816100381610038161003816119892 (119905 119904 119910 (120574

1(119904)) 119910 (120574

119898(119904)))

1003816100381610038161003816] 119889119904

le 2 (1199071(119905) + 119903119898119907

2(119905))

(31)


1(119905) + 119903119898119907

2(119905) le 1205762 for 119905 ge 119879


1003816100381610038161003816 le 120576 (32)


obtain

1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816 le int

120573(119905)

0

120596119879

119903(119892 120576) 119889119904 le 120573

119879120596119879

119903(119892 120576)

(33)


120596119879

119903(119892 120576) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 1199092 119909

119898)

minus119892 (119905 119904 1199101 1199102 119910

119898)1003816100381610038161003816 119905 isin [0 119879]

119904 isin [0 120573119879] 119909119894 119910119894isin [minus119903 119903]

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816 le 120576 (119894 = 1 2 119898)

(34)


119879]times[minus119903 119903]



119903


|(119861119909) (119905)| le 1199071(119905) + 119903119898119907

2(119905) (35)




1(119905) + 119903119898119907




|(119861119909) (119905)| le 120576 (36)


|(119861119909) (119905) minus (119861119909) (120591)|

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

minus119892 (120591 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

le

1003816100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

[119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816] 119889119904

1003816100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

120596119879

1(119892 120576 119903) 119889119904

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119902 (119905 119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 119886 (119905)119898119903

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119887 (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 120573119879120596119879

1(119892 120576 119903)

(37)

where we denoted

120596119879

1(119892 120576 119903) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 119909

119898)

minus119892 (120591 119904 1199091 119909

119898)1003816100381610038161003816 119905 120591 isin [0 119879]

|119905 minus 119904| le 120576 119904 isin [0 120573119879]

100381610038161003816100381611990911003816100381610038161003816 le 119903

10038161003816100381610038161199091198981003816100381610038161003816 le 119903

(38)



|(119861119909) (119905) minus (119861119909) (120591)|

le 119902119879120584119879(120573 120576) + 119903119898119886

119879119887119879120584119879(120573 120576) + 120573

119879120596119879

1(119892 120576 119903)

(39)

where 119902119879

= max119902(119905 119904) 119905 isin [0 119879] 119904 isin [0 120573119879] 119886119879

=

max119886(119905) 119905 isin [0 119879] 119887119879

= max119887(119905) 119905 isin [0 120573119879]



of the functions 120573 = 120573(119905) 119892 = 119892(119905 119904 1199091 119909

119898) we infer

that 120584119879(120573 120576) rarr 0 and 120596119879





+)





|119909 (119905)| le |(119860119909) (119905)| +1003816100381610038161003816(119861119910) (119905)

1003816100381610038161003816

le1003816100381610038161003816119891 (119905 119909 (120572

1(119905)) 119909 (120572

119899(119905))) minus 119891 (119905 0 0)

1003816100381610038161003816

+1003816100381610038161003816119891 (119905 0 0)

1003816100381610038161003816

+ int

120573(119905)

0

119902 (119905 119904) 119889119904 + 1198981003817100381710038171003817119910

1003817100381710038171003817 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904

le 119896 119909 + 1198650+1198721+ 119898119903119872

2

(40)

Hence we obtain

119909 le1198650+1198721+ 119898119903119872

2

1 minus 119896 (41)


2)(1 minus





1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le

119899

sum

119894=1

119896119894max 1003816100381610038161003816119909 (120572

119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 2int

120573(119905)

0

119902 (119905 119904) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119910 (120574119894(119904))

1003816100381610038161003816) 119889119904

le 119896max 1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 21199071(119905) + 119898 (119909 +

10038171003817100381710038171199101003817100381710038171003817) 1199072 (119905)

(42)


Hence we get


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le 119896max1le119894le119899


1003816100381610038161003816119909 (120572119894(119905) minus 119910 (120572

119894(119905)))

1003816100381610038161003816


1199071(119905) + 119898 (119909 +


1199072(119905)


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

(43)



|119909(119905) minus








or






⋃

0le120582le1

119909 isin 119864 119909 = 120582119879119909 (44)























119864 120583(119883) = 0 is nonempty


(2119900) 119883 sub 119884 rArr 120583(119883) le 120583(119884)

(3119900) 120583(119883) = 120583(Conv119883) = 120583(119883)


(5119900) If (119883


119864such that

119883119899+1

sub 119883119899for 119899 = 1 2 and if lim

119899rarrinfin120583(119883119899) = 0


= ⋂infin

119899=1119883119899is nonempty



119899) for any 119899 = 1 2 This

















Next let us define

120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883

1205960(119883) = lim

120576rarr0

120596 (119883 120576) (46)



0(120582119883) =

|120582|1205960(119883) and 120596

0(119883 + 119884) le 120596

0(119883) + 120596

0(119884) provided 119883119884 isin

M119862(119868)



119909 (119905) = 1198911(119905 119909 (119905) 119909 (119886 (119905))) + (119866119909) (119905) int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(47)



has the form

1198912(119905 119904) = 119896 (119905 119904) 119892 (119905 119904) (48)







2

1003816100381610038161003816 10038161003816100381610038161199101 minus 119910

2

1003816100381610038161003816

(49)



0(119866119883) le 119902120596


119862(119868)



+rarr R+



+rarr R


119909 isin 119862(119868)(vi) 119891



+occurring in



1lt 1199052and


10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

le 120576

int

1199052

1199051

119892 (1199052 119904) 119889119904 le 120576

(50)



ℎ (119905) = int

119905

0

119892 (119905 119904) 119889119904 (51)





119896 = sup |119896 (119905 119904)| (119905 119904) isin Δ

1198911= sup 10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816 119905 isin 119868

ℎ = sup ℎ (119905) 119905 isin 119868

(52)





119901119903 + 1198911+ 119896ℎ120593 (119903)Ψ (119903) le 119903 (53)

such that 119901 + 119896ℎ119902Ψ(1199030) lt 1


(1198651119909) (119905) = 119891

1(119905 119909 (119905) 119909 (120572 (119905)))

(1198652119909) (119905) = int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(119865119909) (119905) = (1198651119909) (119905) + (119866119909) (119905) (119865

2119909) (119905)

(54)



+by the formulas

119872(120576) = sup10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119892)] 119889119904

10038161003816100381610038161003816100381610038161003816

1199051 1199052isin 119868

1199051lt 1199052 1199052minus 1199051le 120576

(55)

119873(120576) = supint1199052

1199051

119892 (1199052 119904) 119889119904 119905

1 1199052isin 119868 1199051lt 1199052 1199052minus 1199051le 120576

(56)






1003816100381610038161003816(1198651119909) (119905) minus (1198651119910) (119905)

1003816100381610038161003816


1003816100381610038161003816

(57)


10038171003817100381710038171198651119909 minus 11986511199101003817100381710038171003817 le 119901

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (58)







1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816

int

1199052

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199051 119904) 119892 (119905

2 s) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

1 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le int

1199052

1199051

1003816100381610038161003816119896 (1199052 119904)1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199052 119904) minus 119896 (1199051 119904)

1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199051 119904)1003816100381610038161003816

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 |(119876119909) (119904)| 119889119904

le 119896Ψ (119909)119873 (120576) + 1205961(119896 120576) Ψ (119909) int

1199052

0

119892 (1199052 119904) 119889119904

+ 119896Ψ (119909) int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904

(59)

where we denoted

1205961(119896 120576) = sup 1003816100381610038161003816119896 (1199052 119904) minus 119896 (119905

1 119904)

1003816100381610038161003816

(1199051 119904) (119905

2 119904) isin Δ

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576

(60)


int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904 =

10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

(61)


120596 (1198652119909 120576) le 119896Ψ (119909)119873 (120576) + ℎΨ (119909) 120596

1(119896 120576)

+ 119896Ψ (119909)119872 (120576)

(62)










1003816100381610038161003816(119865119909) (119905) minus (1198651199090) (119905)

1003816100381610038161003816

le1003816100381610038161003816(1198651119909) (119905) minus (119865

11199090) (119905)

1003816100381610038161003816

+1003816100381610038161003816(119866119909) (119905) (1198652119909) (119905) minus (119866119909

0) (119905) (119865

21199090) (119905)

1003816100381610038161003816

le10038171003817100381710038171198651119909 minus 119865

11199090

1003817100381710038171003817 + 1198661199091003816100381610038161003816(1198652119909) (119905) minus (119865

21199090) (119905)

1003816100381610038161003816

+100381710038171003817100381711986521199090

1003817100381710038171003817

1003817100381710038171003817119866119909 minus 1198661199090

1003817100381710038171003817

(63)



21199090) (119905)

1003816100381610038161003816

le int

119905

0

|119896 (119905 119904)| 119892 (119905 119904)1003816100381610038161003816(119876119909) (119904) minus (119876119909

0) (119904)

1003816100381610038161003816 119889119904

le 119896 (int

119905

0

119892 (119905 119904) 119889119904)1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

le 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

(64)

Similarly we obtain

1003816100381610038161003816(11986521199090) (119905)1003816100381610038161003816 le 119896 (int

119905

0

119892 (119905 119904) 119889119904)10038171003817100381710038171198761199090

1003817100381710038171003817 le 119896ℎ10038171003817100381710038171198761199090

1003817100381710038171003817 (65)

and consequently

1003817100381710038171003817119865211990901003817100381710038171003817 le 119896ℎ

10038171003817100381710038171198761199090

1003817100381710038171003817 (66)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 1199011003817100381710038171003817119909 minus 119909

0

1003817100381710038171003817 + 119866119909 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎ1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

10038171003817100381710038171198761199090

1003817100381710038171003817

(67)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 119901120576 + 120593 (10038171003817100381710038171199090

1003817100381710038171003817 + 120576) 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎΨ (10038171003817100381710038171199090

1003817100381710038171003817)1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

(68)






|(119865119909) (119905)| le10038161003816100381610038161198911 (119905 119909 (119905) 119909 (119886 (119905))) minus 119891

1(119905 0 0)

1003816100381610038161003816

+10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816

+ |119866119909 (119905)|

10038161003816100381610038161003816100381610038161003816

int

119905

0

119896 (119905 119904) 119892 (119905 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909)

(69)

Hence we obtain

119865119909 le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909) (70)



the ball 1198611199030




1199030


1 1199052isin 119868with


obtain1003816100381610038161003816(119865119909) (1199052) minus (119865119909) (119905

1)1003816100381610038161003816

le10038161003816100381610038161198911 (1199052 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199052) 119909 (119886 (119905

2)))

1003816100381610038161003816

+10038161003816100381610038161198911 (1199051 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199051) 119909 (119886 (119905

1)))

1003816100381610038161003816

+1003816100381610038161003816(1198652119909) (1199052)

1003816100381610038161003816

1003816100381610038161003816(119866119909) (1199052) minus (119866119909) (1199051)1003816100381610038161003816

+1003816100381610038161003816(119866119909) (1199051)

1003816100381610038161003816

1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le 1205961199030

(1198911 120576) + 119901max 1003816100381610038161003816119909 (119905

2) minus 119909 (119905

1)1003816100381610038161003816

1003816100381610038161003816119909 (119886 (1199052)) minus 119909 (119886 (119905

1))1003816100381610038161003816

+ 119896ℎΨ (1199030) 120596 (119866119909 120576) + 120593 (119903

0) 120596 (119865

2119909 120576)

(71)

where we denoted

1205961199030

(1198911 120576) = sup 10038161003816100381610038161198911 (1199052 119909 119910) minus 119891

1(1199051 119909 119910)

1003816100381610038161003816 1199051 1199052 isin 119868

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576 119909 119910 isin [minus1199030 1199030]

(72)


120596 (119865119909 120576) le 1205961199030

(1198911 120576) + 119901max 120596 (119909 120576) 120596 (119909 120596 (119886 120576))

+ 119896ℎΨ (1199030) 120596 (119866119909 120576)

+ 120593 (1199030) Ψ (1199030) [119896119873 (120576) + ℎ120596

1(119896 120576) + 119896119872 (120576)]

(73)





1205960(119865119883) le 119901120596

0(119883) + 119896ℎΨ (119903

0) 1205960(119866119883) (74)


1205960(119865119883) le (119901 + 119896ℎ119902Ψ (119903

0)) 1205960(119883) (75)







1199091015840= 119891 (119905 119909) (76)


119909 (0) = 1199090 (77)









rarr R+(or 119908 119869

0times

R+



1003817100381710038171003817 le 119908 (1199051003817100381710038171003817119909 minus 119910

1003817100381710038171003817) (78)




1015840= 119908(119905 119906)


119905

0119908(119904 119906(119904))119889119904 for 119905 isin 119869


lim119905rarr0

119906(119905)119905 = lim119905rarr0






120583 (119891 (119905 119883)) le 119908 (119905 120583 (119883)) (79)





120583


119864120583= 119909 isin 119864 119909 isin ker 120583 (80)



120583is a closed



(6119900) 120583(119883 + 119884) le 120583(119883) + 120583(119884)

(7119900) 120583(120582119883) = |120582|120583(119883) for 120582 isin R








120583 (1199090+ 119891 (119905 119883)) le 119908 (119905 120583 (119883)) (81)



1198690times R+



119906 (119905) le int

119905

0

119908 (119904 119906 (119904)) 119889119904 (119905 isin 1198690) (82)


119906(119905)119905 = lim119905rarr0




0)minus119891(119904 119909)||



120583for 119905 isin 119869



0isin



120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (83)





120583 (119891 (119905 119883)) le 119901 (119905) 120583 (119883) (84)





1199091015840

119899= 119886119899(119905) 119909119899+ 119891119899(119909119899 119909119899+1

) (85)


119909119899(0) = 119909

119899

0(86)


119909119899

0= 119886(119886 isin R)






119899rarrinfin120572119899= 0 and |119891

119899(119909119899 119909119899+1

)| le

120572119899for 119899 = 1 2 and for all 119909 = (119909

1 1199092 1199093 ) isin 119897






119899)|| =

sup|119909119899| 119899 = 1 2


1(119905) 1199092(119905) )






sup119909isin119883

1003816100381610038161003816119909119899 minus 1198861003816100381610038161003816 (87)




120583 (1199090+ 119891 (119905 119883))


sup119909isin119883

1003816100381610038161003816119909119899

0+ 119886119899(119905) 119909119899

+119891119899(119909119899 119909119899+1

) minus 1198861003816100381610038161003816


sup119909isin119883

[1003816100381610038161003816119886119899 (119905)

1003816100381610038161003816

10038161003816100381610038161199091198991003816100381610038161003816

+1003816100381610038161003816119891119899 (119909119899 119909119899+1 ) + 119909

119899

0minus 119886

1003816100381610038161003816]


sup119909isin119883

119901 (119905)1003816100381610038161003816119909119899 minus 119886

1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119886119899 (119905)1003816100381610038161003816 |119886|


sup119909isin119883

1003816100381610038161003816119891119899 (119909119899 119909119899+1 )1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119909119899

0minus 119886

1003816100381610038161003816

(88)


Hence we get

120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (89)



References













































1015840























1 Introduction




u(0) = g(u)(11)







u(0) = g(u)

(12)







part



2 Preliminaries


+ = (0infin)and R





tisinI

∥∥f(t)∥∥X (21)



∥∥f∥∥L1(IX) =

int

I

∥∥f(s)∥∥Xds (22)







d

dtR(t)y = AR(t)y +

int t


= R(t)Ay +int t


(23)







u(0) = u0 isin X(24)








is verified





(27)
















ζ

(int t

0un(s)ds

infin

n=1

)

2int t





subeW such that




Sn = εn + Cn1ε

nminus1h + Cn2ε

nminus2h2


n n isin N (211)





)) n = 2 3 (212)


η(Fn0(S)) rη(S) (213)


3 Main Results





+ rarr R+ such that

∥∥f(t x)∥∥ m(t)Φ(x) (31)




ζ(f(t S)

) H(t)ζ(S) (32)





KgR +KΦ(R)int1

0m(s)ds R (33)








∥∥ +Kint1

0




(Fu)(t) ∥∥R(t)g(u)

∥∥ +

∥∥∥∥∥

int t


∥∥∥∥∥

KgR +KΦ(R)int1

0m(s)ds R

(36)





sub F(B) such that


(int t

0


infinn=1ds

)

+ ε (37)


ζ(F(B)(t)) 4Kint t

0ζ(f(s un(s))

infinn=1ds + ε

4Kint t

0H(s)ζ(un(s)nisinN

) ds + ε

4Kγ(B)int t

0H(s)ds + ε

(38)




0


0ϕ(s)ds

]

+ ε


(39)





ζ(F2(B)(t)

) 2ζ

(int t

0


infinn=1ds

)

+ ε

4Kint t

0ζf(swn(s))

infinn=1ds + ε

4Kint t

0H(s)ζ

(co(F1(B)(s)

))+ ε = 4K

int t

0H(s)ζ

(F1(B)(s)

)+ ε

(311)



ζ(F2(B)(t)

) 4K

int t

0

[∣∣H(s) minus ϕ(s)∣∣ + ∣∣ϕ(s)

∣∣](a + bs)γ(B)ds + ε


0


(

at +bt2

2

)

+ ε

a(a + bt) + b

(

at +bt2

2

)

+ ε (

a2 + 2bt +(bt)2

2

)

γ(B) + ε

(312)


ζ(F2(B)(t)

)

(

a2 + 2bt +(bt)2

2

)

γ(B) (313)


ζ(Fn(B)(t)) (

an + Cn1a

nminus1bt + Cn2a

nminus2 (bt)2


n

)

γ(B) (314)



γ(Fn(B)) (

an + Cn1a

nminus1b + Cn2a

nminus2 b2


n

)

γ(B) (315)


(

an0 + Cn01 a

n0minus1b + Cn02 a

n0minus2 b2


n0

)

= r lt 1 (316)





u(0) = g(u)

(317)













∣∣∣ C|a(λ)| (318)






1λa(λ)


(319)



such that




completely positive



lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (321)



u(t) =int t


int t

0f(s u(s)) + g(u) (322)


R(t)x = x +int t



d

dtR(t)x = AR(t)x +

int t




KgR +KΦ(R)int1





4 Applications


partw(t ξ)partt

= Aw(t ξ) +int t



w(0 ξ) =int1

0

int ξ

0qk(s ξ)w

(s y

)dsdy 0 ξ 2π

(41)



defined by



part







u(0) = g(u)

(43)






int10 k(s ξ)

2dsdξ)12







a(λ) =λ + α + βλ(λ + α)

(44)


f1(λ) =λ + α

λ + α + β f2(λ) =

λ2 + 2(α + β

)λ + αβ + α2

(λ + α + β

)2 (45)


f(n)1 (λ) =

(minus1)n+1β(n + 1)(λ + α + β

)n+1 f(n)2 (λ) =

(minus1)n+1β(α + β)(n + 1)

(λ + α + β








Re

(μ0 + iμ

b(μ0 + iμ

)+ 1

)

=μ30 + μ

20α + μ2

0

(α + β

)+ μ0α

(α + β

)+ μ0μ


)2 + 2μ0(α + β

)+ μ2

0 + μ2

(48)


(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1



∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1


∥∥∥∥

M

μ0 + iμ

∥∥∥∥ (410)


Therefore

lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (411)



int10 k(s ξ)

2dsdξ)12




qcKR +∥∥p1

∥∥1LK R (412)


Acknowledgments


References















































1 Introduction
















K(q)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2πqω1+2q

(2

2 + q

)1minus2q( Γ(1q)

Γ(12 + 1q

)

)2

if 1 le q lt +infin

4ω if q = +infin

(15)





0 lt a(t) le π2

ω2 t isin [0 ω] (16)





2 Preliminaries








u(t) =intω





Let





Let

G = min0letsleω


G(t s) σ = GG (26)





Th(t) =intω




Th(t) =intω

0G(t s)h(s)ds le G

intω

0h(s)ds (29)

and therefore

Th le Gintω

0h(s)ds (210)


Th(t) =intω

0G(t s)h(s)ds ge G

intω

0h(s)ds

=(GG

)middotGintω

0h(s)ds

ge σTh

(211)




Tφ = r(T)φ (212)


intω

0φ(t)dt = 1 (213)






hold




A = T F (216)











3 Main Results



mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x


mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x

(31)









f(t x0 x1 xn) = b0(t)x20 + b1(t)x


n (32)






radic|xn| (33)



















This implies that





(39)





intω

0



intω

0u0(t)φ(t)dt (311)


intω

0


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(312)


λ1

intω


intω

0u0(x)φ(t)dt (313)


intω


intω

0φ(t)dt = σu0 gt 0 (314)


i(AK capΩ1 K) = 1 (315)







(u1(t) minus μ1

)= F(u1)(t) t isin R (317)



This means that




(320)




intω

0

[u



intω

0u1(t)φ(t)dt (322)


intω

0


]φ(t)dt =

intω

0u1(t)


]dt

= λ1

intω

0u1(x)φ(t)dt

(323)


λ1

intω


intω

0u1(x)φ(t)dt (324)



intω


intω

0φ(t)dt = σu1 gt 0 (325)


i(AK capΩ2 K) = 0 (326)


i(AK cap

(Ω2 Ω1









(u0(t) minus μ0

)= F(u0)(t) t isin R (329)



(330)




intω

0

[u



intω

0u0(t)φ(t)dt (332)



intω

0

[u


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(333)


λ1

intω


intω

0u1(x)φ(t)dt (334)


i(AK capΩ1 K) = 0 (335)








(338)




intω

0

[u



intω

0u1(t)φ(t)dt (340)



intω

0

[u


]φ(t)dt =

intω

0u1(t)

[φ


]dt

= λ1

intω

0u1(x)φ(t)dt

(341)


λ1

intω


intω

0u1(x)φ(t)dt (342)


i(AK capΩ2 K) = 1 (343)


i(AK cap

(Ω2 Ω1



Acknowledgment


References

















York NY USA 1988










1 Introduction



x(t) = f

(


0u(t s x(c(s)))ds

)


x(t) = h

(

t x(t)intα(t)

0u(t s x(c(s)))ds

)





x(t) =int t



x(t) = L(t) +int t

0




x(t) =int t



x(t) = (Tx)(t)int t











0g(t s x







x(t) = f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f

(

t

intH(t)

0x(s)ds x(h(t))

)



x(t) = g(t x(t)) + f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f(t x(α(t)))

(

q(t) +intβ(t)

0u(t s x

(γ(s)))ds

)




2 Preliminaries












ωT

0 (X) = limεrarr 0

ωT (X ε)





diam X(t)

(22)







limtrarr+infin



Then

lim suptrarr+infin

ϕ(ψ(t))= lim sup












then









(212)




ωϕ(T)0 (X) (213)




∣∣∣ωT





∣∣∣ω




3 Main Results





limtrarr+infin


b(t) = +infin (31)






limtrarr+infin

sup

m3(t)intα(t)


= 0 (33)

sup

m3(t)intα(t)


leM0 (34)

sup

intα(t)


leM (35)








(Fx)(t) = f

(


0u(t s x(c(s)))ds

)



|(Fx)(t)| le∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minus f(t 0 0 0)∣∣∣∣∣+∣∣f(t 0 0 0)

∣∣


∣∣∣∣∣

intα(t)

0u(t s x(c(s)))ds

∣∣∣∣∣+ f



(310)





=

∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minusf(

t y(a(t))(Hy)(b(t))

intα(t)

0u(t s y(c(s))

)ds

)∣∣∣∣∣




)(b(t))

∣∣

+m3(t)intα(t)

0




)(b(t))

∣∣

+ sup

m3(t)intα(t)



(312)

which yields that


+ sup

m3(t)intα(t)


forallt isin R+

(313)


lim suptrarr+infin

diam(FX)(t)



diam(HX)(b(t))


sup

m3(t)intα(t)




diam(HX)(t)

(314)

that is

lim suptrarr+infin



diam(HX)(t) (315)





∣∣u(p τ v




ωTr

(f ε g(r)

)= sup

∣∣f(p vw z



(316)






le∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣

+

∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣


+m3(t)

[∣∣∣∣∣

intα(t)


∣∣∣∣∣+intα(s)


]

+ sup∣∣f(p vw z





∣∣u(p τ v




+ωTr

(f ε g(r)

)


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)

)

(318)

which implies that


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)


(319)



limεrarr 0


ωTr (u ε) = lim

εrarr 0ωTr

(f ε g(r)

)= 0 (320)


ωT0 (FX) leM1ω

a(T)0 (X) +M2ω

b(T)0 (HX) (321)


ω0(FX) leM1ω0(X) +M2ω0(HX) (322)



diam(FX)(t)



diam(HX)(t)

=M1μ(X) +M2

(ω0(HX) + lim sup


)

le (M1 +M2Q)μ(X)(323)



sup

m3(t)intα(t)


ltε



ωT (u δ r) ltε







=

∣∣∣∣∣f

(


0u(t s xn(c(s)))ds

)

minusf(


0u(t s x(c(s)))ds

)∣∣∣∣∣


+m3(t)intα(t)



+max

supτgtT

sup

m3(τ)intα(τ)

0


supτisin[0T]

sup

m3(τ)intα(τ)

0



2

ltε

2+max

ε

3 supτisin[0T]

m3(τ)intα(τ)

0ωT

(uδ02 r

)ds

ltε

2+max

ε

3


=5ε6

(327)

which yields that














M0 + h le r(1 minusM1) (331)




sup

m3(t)intα(t)





=

∣∣∣∣∣h

(

t z(t)intα(t)


)

minus h(

t y(t)intα(t)

0u(t τ y(c(τ))

)dτ

)∣∣∣∣∣


intα(t)

0



m3(t)intα(t)



(333)

which means that





4 Examples



x(t) =1

10 + ln3(1 + t3)+tx2(1 + 3t2

)

200(1 + t)+

1

10radic3 + 20t + 3|x(t3)|

+1

9 +radic1 + t

⎛

⎜⎝

int t2

0

s2x(4s) sin(radic



∣∣∣ x2(4s)ds

⎞

⎟⎠

3

forallt isin R+

(41)

Put

f(t vw z) =1

10 + ln3(1 + t3)+

tv2

200(1 + t)+

1

10radic3 + 20t + |w|

+z3

9 +radic1 + t

u(t s p

)=

s2p sin(radic



∣∣∣p2


(42)

Let

risin[9 minus radic

512

9 +

radic51

2

]

M=3 M0=9r10 M1=

r

100 M2=

1300

f =110

Q = 3 m1(t) =rt

100(1 + t) m2(t) =

1(10radic3 + 20t

)2 m3(t) =3M2

9 +radic1 + t


(43)



512

9 +

radic51

2

]



le t

200(1 + t)


∣∣∣ +

∣∣∣∣∣

1

10radic3 + 20t + |w|

minus 1

10radic3 + 20t +

∣∣q∣∣

∣∣∣∣∣

+1

9 +radic1 + t


∣∣∣ le m1(t)





sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)

minuss2w(4s) sin


)


∣∣∣w2(4s)


⎫⎪⎬

⎪⎭

le 3M2

9 +radic1 + t

int t2

0

2s2(1 + t12

)2

[r(1 + t12

)+ r3(r3t3 + 3r6 + 7

(t + 2t2

)2)]ds

le 2M2t6(9 +

radic1 + t)(

1 + t12)2

r(1 + t12

)+ r3[r3t3 + 3r6 + 7

(t + 2t2


sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=rM2

20lt M0

sup

intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=r

6leM

(44)




x(t) =3 + sin4

(radic1 + t2

)

16+x2(t)8 + t2

+1

1 + t

(intradict

0

sx3(1 + s2)

1 + t2 + s cos2(1 + t3s7x2(1 + s2))ds

)2

forallt isin R+

(45)

Put

h(t vw) =3 + sin4

(radic1 + t2

)

16+

v2

8 + t2+

w2

1 + t

u(t s p

)=

sp3

1 + t2 + s cos2(1 + t3s7p2

)

α(t) =radict c(t) = 1 + t2 m1(t) =

18 + t2

m3(t) =4


r =12 M = 2 M0 =M1 =

18 h =

14

(46)



le 18 + t2


∣∣∣ +

11 + t


∣∣∣



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))

minus sw3(1 + s2)

1 + t2 + s cos2(1 + t3s7w2(1 + s2))ds


le4r3t(1 +

radict + t2

)



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣

sv3(1 + s2)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

4r3

1 + t

intradict

0

s

1 + t2ds t isin R

+

le sup

t

4(1 + t)(1 + t2) t isin R

+

le 18=M1

sup

intα(t)


= sup

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

r3intradic

t

0

s

1 + t2ds t isin R

+

=132

lt M

(47)


Acknowledgments


References


































1 Introduction


Dq




i=1

aiu(nminus2)(ηi

) (11)



Tu(t) =int t


int1

0H(t s)u(s)ds (12)





Dq


)= 0 0 lt t lt 1


where Dq



Dq



i=1

aiu(ηi)

(14)



Dq



0+u(1) = 0 1 le p le n minus 2(15)



CDq



(η2)

(16)


where CDq




)= 0 2 lt q lt 3 0 lt p lt 1

u(0) = 0 uprime(0) = uprime(1) = 0(17)


Dq

0+u(t) + a(t)f(t u(t)) = 0 0 lt t lt 1


i=1

ζiu(ηi) (18)




CDqu(t) = f(t u(t)) u(0) = u0 (19)



Dq



(110)












αC(D) = maxtisinI

α(D(t)) α

(int

I

u(t)dt u isin D)

leint

I

α(D(t))dt (21)




Iq

0+y(t) =1

Γ(q)int t



given by

Dq

0+y(t) =1

Γ(n minus q)

(d

dt

)n int t



+


Dq

0+y(t) = 0 (24)




0+y isinL(0 1)

Iq

0+Dq







3 Main Results





2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

lt 1 (32)

such that

α(f(tD1 D2 Dn+1)

) len+1sum

k=1



i=1 aiηqminusn+1i





i=1

aix(ηi)

(34)


is

x(t) =int1

0G(t s)y(s)ds (35)

where


i=1 aig(ηi s)


g(t s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩



) t le s(37)




x(t) = minusint t

0





c1 =1


qminusn+1i

int1

0



i=1 ai


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

(310)


x(t) = minusint t

0


)y(s)ds +tqminusn+1


int1

0


)y(s)ds

minussummminus2

i=1 aitqminusn+1


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

=int1

0G(t s)y(s)ds

(311)







ρ0(s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


s isin (0 ξ]

(γ

s


(312)











i=1 aig(ηi s)

ηlowast (314)



mminus2sum

i=1

aig(ηi s)= G(s)

le tqminusn+1

Γ(q minus n + 2

) +summminus2

i=1 aiηqminusn+1i



(315)




(Inminus20+ x(s)

) S(Inminus20+ x(s)

))= θ


i=1

aix(ηi)

(316)






(317)


(Ax)(t) =int1

0G(t s)f


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))ds 0 le t le 1 (318)



(Ax)(t) leint1

0G(s)f


(Inminus20+ xj(s)


))ds

=

(int γ

0+intδ

γ

+int1

δ

)


(Inminus20+ xj(s)


))ds

le 3intδ

γ


(Inminus20+ xj(s)


))ds

(319)


mintisin[γ δ]


int1

0G(t s)f


(Inminus20+ xj(s)


))ds

geintδ

γ

(

g(t s) +tqminusn+1

ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds


geintδ

γ

(


ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds

ge λ

intδ

γ


(Inminus20+ xj(s)


))ds

ge λ

3(Ax)


(320)




∥∥ =int1

0G(t s)


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds

(321)



(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ ltε

Δ

(322)





on I





(Inminus20+ x(s)

) S(Inminus20+ x(s)

))



(Inminus20+ x(s)

) S(Inminus20+ x(s)

))




(Inminus20+ D

)(I) times S

(Inminus20+ D

)(I)))


k=1



(T(Inminus20+ D

)(I))


)(I))

(324)

which implies


k=1





(S(Inminus20+ D

)(I))

(325)

Obviously


)(I) = α

(int s

0


)


(326)

α(T(Inminus20+ D

))(I) = α

(int t

0K(t s)

(ints

0


u(τ)dτ

)


le alowast


(n minus 3)α(D(I))

(327)

α(S(Inminus20+ D

))(I) = α

(int1

0H(t s)

(int s

0


u(τ)dτ

)


le blowast


(n minus 3)α(D(I))

(328)



α(D(I)) le 2αC(D) (329)



nminus2sum

k=1




)

αC(D) (330)







nminus1sum

k=1






nminus1sum

k=1



NMβ

int1







k=1

uk ge r1 (334)


Therefore




k=1

uk ge r1 (335)

Take


(336)


x(t) ge λ

3NxΩ ge λ

3Nr0 gt r1 (337)

Hence

(Ax)(t) geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



xΩ(intδ

γ

G(s)c1(s)ds

)

middot ulowast

=1


(intδ

γ

G(s)c1(s)dsulowast)



(338)

and consequently






k=1

uk le r2 (340)






k=1

uk le r2 (341)

Choose

0 lt r lt min

⎧⎨

⎩

(nminus2sum

k=0

1k

)minus1r2 β

⎫⎬

⎭ (342)


(Ax)(ξ) =int1

0G(ξ s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(ξ s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



x(s)Ωintδ

γ


=1


x(s)Ω(intδ

γ




(343)

which implies


that is




(Ax)(t) leint1

0G(s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds (346)


∥∥f(t u1 un+1)


Therefore





(H7)

h(u1 un+1)sumn+1


n+1sum

k=1





k=1


k=1

uk le r3 (350)


ε0 =

(

N

nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

int1

0G(s)a(s)ds

)minus1 (351)




(Ax)(t) leNint1


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))∥∥∥ds

leNint1

0G(s)a(s)h


∥∥∥ x(s)



)(s)∥∥∥)ds

leNε0

int1

0G(s)a(s)

(nminus1sum

k=1



)(s)∥∥∥ +∥∥∥S(Inminus20+ x

)(s)∥∥∥

)

ds

leNε0

(nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

)

rlowastint1


(352)

and consequently




4 An Example



960k3

⎧⎪⎨

⎪⎩

⎡


j=1

u2j(t) +infinsum

j=1

uprimej(t)

⎤

⎦

3

+

⎛


j=1

uj(t) +infinsum

j=1

uprime2j(t)

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5

(int t


)23

+1 + t2

24k4

(int1


)

t isin I


(14

)+12uprimek

(49

) k = 1 2 3

(41)






fk(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭+1 + t3

36k5x15k

+1 + t2

24k4y2k

(42)


k=1(1k3) lt 32 we get by (42)

∥∥f(t u v x y

)∥∥ =infinsum

k=1

∣∣fk(t u v x y

)∣∣

le (1 + t)3

2

(140

(u + v)3 + 1160

(u + v)12 + 112

x15 + 18∥∥y∥∥)

(43)


h(u v x y

)=

140

(u + v)3 +1160

(u + v)12 +112x15 +

18y (44)



1 f(1)k ) and f (2) =

(f (2)1 f

(2)k ) in which

f(1)k

(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5x15k (k = 1 2 3 )

f(2)k

(t u v x y

)=

1 + t2

24k4y2k (k = 1 2 3 )

(45)



α(f (1)(tD1 D2 D3 D4)


α(f (2)(ID1 D2 D3 D4)

)le 1


(46)


α(f(ID1 D2 D3 D4)


2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

asymp 01565 lt 1(47)



fk(t u v x y

) ge (1 + t)3


fk(t u v x y

) ge (1 + t)3

960k3


(48)


c1(t) =(1 + t)3

960 h1 k(u v) = (u + v)3 ulowast =

(1

1k3

) (49)

in this situation


(u + v)3u + v = infin (410)


c2(t) =(1 + t)3

960 h2 k(u v) =


(1

1k3

) (411)

in this situation

h2k = limu+vrarr 0




NMβ

int1

0G(s)a(s)ds asymp 07287 lt β = 1 (413)


Acknowledgments


References











































1 Introduction






∣∣f(t)


(11)


G(u) = S[f(t)u

]

=intinfin


(12)





f(t) =infinsum

n=0

antn (13)

then

F(u) =infinsum

n=0

nantn (14)





(15)







Gn(u) =G(u)un

minusnminus1sum

k=0

f (k)(0)unminusk

(16)



LU + RU +NU = h(x t) (17)


S[LU] + S[RU] + S[NU] = S[h(x t)] (18)



unU(x 0) minus 1


u+ S[RU] + S[NU] = S[h(x t)]

(19)




U(x t) =infinsum

i=0

Ui(x t) (111)



NU(x t) =infinsum

i=0

Ai (112)


Ai =1idi

dλi

[

Ninfinsum

i=0

λiUi

]

λ=0

i = 0 1 2 (113)


S

[ infinsum

i=0

Ui(x t)

]


minus unS[ infinsum

i=0

Ai

]

+ unS[h(x t)]

(114)









U0(x t) = K(x t)



Ψ = Φ + Sminus1[h(x t)] (119)







LU + RU +NU = h(x t) (122)


U(x 0) = f(x) Ut(x 0) = g(x) (123)





]+ Sminus1

[u2S[h(x t)]

] (125)

2 Applications




U(x 0) = 0

Ut(x 0) = ex(22)




] (23)



[U2x minusU2

]](24)


U(x t) =infinsum

i=0

Ui(x t) (25)


infinsum

i=0


u2S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

(26)


infinsum

i=0

Ai(U) = U2x

infinsum

i=0

Ai(U) = U2

(27)


A0(U) = U20x A1(U) = 2U0xU1x Ai(U) =

isum

r=0

UrxUiminusrx

B0(U) = U20 B1(U) = 2U0U1 Bi(U)

isum

r=0

UrUiminusr

(28)


U0(x t) = tex


S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

n ge 0(29)



U1(x t) = Sminus1[

u2S

[ infinsum

i=0

A0(U) minusinfinsum

i=0

B0(U)

]]

= Sminus1[u2S

[U2

0x minusU20

]]= Sminus1

[u2S

[t2e2x minus t2e2x

]]

= Sminus1[u2S[0]

]= 0

(210)


U(x t) =infinsum

i=0

Ui(x t) = tex (211)


Ut

(x y t

) minus VxWy = 1

Vt(x y t

) minusWxUy = 5

Wt

(x y t

) minusUxVy = 5

(212)


U(x y 0

)= x + 2y

V(x y 0

)= x minus 2y

W(x y 0

)= minusx + 2y

(213)


U(x y u

)= x + 2y + u + uS

[VxWy

]

V(x y u


[WxUy

]

W(x y u


[UxVy

]

(214)


U(x y t


[VxWy

]]

V(x y t


[WxUy

]]

U(x y t


[UxVy

]]

(215)



U0(x y t

)= t + x + 2y

Ui+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ci(VW)

]]

i ge 0

V0(x y t

)= 5t + x minus 2y

Vi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Di(UW)

]]

i ge 0

W0(x y t

)= 5t minus x + 2y

Wi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ei(UV )

]]

i ge 0

(216)


C0(VW) = V0xW0y


Ci(VW) =isum

r=0

VrxWiminusry

D0(UW) = U0yW0x


Di(UW) =isum

r=0

WrxUiminusry

E0(UV ) = U0xV0y


Ei(VW) =isum

r=0

UrxViminusry

(217)



U1(x y t


[V0xW0y

]]= Sminus1[uS[(1)(2)]] = 2t

V1(x y t


[W0xU0y


W1(x y t


[U0xV0y


U2(x y t


[V1xW0y + V0xW1y

]]= 0

V2(x y t


[U0yW1x +U1yW0x

]]= 0

W2(x y t


[U1xV0y +U0xV1y

]]= 0

(218)


U(x y t

)=

infinsum

i=0

Ui

(x y t

)= x + 2y + 3t

V(x y t

)=

infinsum

i=0

Vi(x y t

)= x minus 2y + 3t

W(x y t

)=

infinsum

i=0

Wi

(x y t

)= minusx + 2y + 3t

(219)





U(x 0) = 1 + ex V (x 0) = minus1 + ex (221)




Ux(x t) =infinsum

i=0


i=0

Vxi(x t) (223)



S

[ infinsum

i=0

Ui(x t)

]


[ infinsum

i=0

Vix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

S

[ infinsum

i=0

Vi(x t)

]


i=0

Uix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

(224)






(225)






U2 = Sminus1[u2ex

]=t2

2ex

V2 = Sminus1[u2ex

]=t2

2ex

(226)


U(x t) = 1 + ex(

1 + t +t2

2+t3


)


1 minus t + t2

2minus t3


)

(227)


U(x t) = 1 + ex+t




Ut + VUx +U = 1



U(x 0) = ex V (x 0) = eminusx (230)





S

[ infinsum

i=0

Ui(x t)

]


i=0

Ai

]

minus uS[ infinsum

i=0

Ui

]

S

[ infinsum

i=0

Vi(x t)

]

= eminusx + u + uS

[ infinsum

i=0

Bi

]

minus uS[ infinsum

i=0

Vi

]

(232)


A0 = U0xV0 A1 = U0xV1 +U1xV0


B0 = V0xU0 B1 = V0xU1 + V1xU0

B2 = V0xU2 + V1xU1 + V2xU0

(233)



U0 = ex + t






2minus tex minus t2

2ex



2eminusx

U2 =t2

2+t2


V2 =t2

2+t2


(236)


U(x t) = ex(

1 minus t + t2

2minus t3


)

V (x t) = eminusx(

1 + t +t2

2+t3


)

(237)



3 Conclusion



Acknowledgment


References


























1 FK and BK Spaces







infink=0 let x

[n] =sumn






suminfinn=0 λnb





suminfink=0 xke






dw(x y

)=

infinsum

k=0

12k



∣∣

)

(x y isin w)

(11)








sumnk=0 xke





Cs =

σ isin C infinsum

n=1

cn(σ) le s

(12)



Φ =φ =

(φk



m(φ)=


supσisinCs

(1φs

sum

kisinσ|xk|

)

ltinfin

n(φ)=


( infinsum

k=1

|uk|Δφk)

ltinfin

(14)
















χ(Q1 +Q2) le χ(Q1) + χ(Q2)




Lχ = χ(L(SX)) (24)


Lχ = 0 (25)










1amiddot lim sup

nrarrinfin

(

supxisinQ

(I minus Pn)(x))


(

supxisinQ

(I minus Pn)(x))

(26)






(

supxisinQ

(I minus Pn)(x))

(27)



(

supxisinQ

sum

kgen|xk|p

)

(28)





Pr(z) = ze +rsum





(

supxisinQ



(

supxisinQ


(210)




supxisinQ

(

supuisinS(x)

( infinsum

n=k

|un|Δφn))

(211)








Lχ = χ(L(SX)) (212)

and we have












x(0) = x0 (32)




∥∥f(t x)∥∥ le Q + Rx (33)


μ(f(t Y )

) le q(t)μ(Y ) (34)









(

supxisinE

supkgen

|xk|)

(35)





xi(0) = x0i (37)





(i) x0 = (x0i ) isin c0



∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

(38)



Now let us denote

q(t) = supnge1

qn(t)

Q = suptisinI

q(t)


(39)




∣∣fn(t x)∣∣ =

∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

le P +Q supigekn


Hence we get



χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x)∣∣)

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x1 x2 )∣∣)

le limnrarrinfin

supxisinY

(

supigen


∣∣xj∣∣)

le limnrarrinfin

supigen


supxisinY

(

supigen

supjgekn

∣∣xj∣∣)

(312)


χ(f(t Y )

) le q(t)χ(Y ) (313)







infinsum

j=ki+1

aij(t)xj (314)


xi(0) = x0i (315)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (316)



i = 1 2










2 = f2(t x1 x2)


(317)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (318)



Then we have


∣∣f1(t x1)∣∣∣∣f2(t x1 x2)

∣∣ ∣∣fi(t x1 x2 xi)

∣∣


(319)


∣∣∣fi(t x)∣∣∣ile Pi(t) (320)













xi(0) = x0i (323)




supxisinE

supnmgep

|xn minus xm|


(

supxisinQ

supkgen

|xk|)

(324)







∣∣gn(t x1 x2 )∣∣ le bi (325)




a(t) = supige1

ai(t)

Q = suptisinI

a(t)(326)






f(t x) =(f1(t x) f2(tx)

)=(fi(t x)

)

(327)





∣∣


(328)



∣∣fi(t x)∣∣ le |ai(t)||xi| +

∣∣gi(t x)∣∣








(s y


(s y

)∣∣


∣∣ +


(s y

)∣∣


∣∣ +


(s y

)∣∣

(331)

Then


ige1


(s y

)∣∣


|ai(t) minus ai(s)|



(332)




∣∣gn(t x)∣∣ +

∣∣gm(t x)∣∣



μ(f(t E)

) le a(t)μ(E) (334)











∣∣fi(t x)

∣∣p =

∣∣fi(t x0 x1 x2 )

∣∣p le qi(t) + ri(t)|xi|p (335)







R = supri(t) t isin I i = 0 1 2 (336)



∥∥f(t x)∥∥pp =

infinsum

i=0


leinfinsum

i=0

[qi(t) + ri(t)|xi|p

]

leinfinsum

i=0

qi(t) +

(

supige0

ri(t)

)( infinsum

i=0|xi|p

)

le Q + Rxpp

(337)




χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(sum

igen

∣∣fi(t x1 x2 )∣∣p)

le limnrarrinfin

(sum

igenqi(t) +

(

supigen

ri(t)

)(sum

igen|xi|p

) )

(338)



χ(f(t Y )

) le r(t)χ(Y ) (339)








xi(0) = x0i (341)






∣∣fi(t x0 x1 x2 )∣∣p le bi(t) (342)










∣∣fi(t u)

∣∣ =

∣∣fi(t u1 u2 )

∣∣ le pi(t) + qi(t)|ui| (343)








(344)




uisinS(x)

infinsum

i=1


le supuisinS(x)

infinsum

i=1

[pi(t) + qi(t)|ui|

]Δφi

leinfinsum

i=1

pi(t)Δφi +

(

supi

qi(t)

)(

supuisinS(x)

infinsum

i=1

|ui|Δφi)

le P +Qxn(φ)

(345)




χ(f(t X)

)= lim

krarrinfinsupxisinX

(

supuisinS(x)

( infinsum

n=k


))

le limkrarrinfin

( infinsum

n=k

pn(t)Δφn +

(

supngek

qn(t)

)(

supuisinS(x)

infinsum

n=k

|un|Δφn))

(346)



χ(f(t X)

) le q(t)χ(X) (347)




References























(c)α l












































1 Introduction






(11)



) (12)


with

x(t+k)= lim

hrarr 0+x(tk + h) x

(tminusk)= lim










2 Preliminaries








Iα0h(t) =1

Γ(α)

int t




CDα0+h(t) =

1Γ(n minus α)

int t




CDαh(t) = 0 (23)

has solutions










Then either






x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

0



tk



i=1

int ti

timinus1


h(s)ds


i=1

int ti

timinus1


h(s)ds


i=1

(tk minus ti)int ti

timinus1


h(s)ds + Ω1(t)intT

tk


h(s)ds

+Ω2(t)intT

tk


h(s)ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


+kminus1sum

i=1


(26)




(x(tminusk))


(27)

where

Λ1 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(T minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1




Λ2 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1



minus ct

TΔ



Δ

(28)



0



xprime(t) =int t

0


h(s)ds minus c1(29)


x(t) =int t

t1



xprime(t) =int t

t1




x(tminus1)=int t1

t0




t0



(211)

Observing that





then we have

minusd0 =int t1

t0



minusd1 =int t1

t0



(213)


x(t) =int t

t1


h(s)ds +int t1

t0


h(s)ds


t0




xprime(t) =int t

t1


h(s)ds +int t1

t0



(214)


x(t) =int t

t2



xprime(t) =int t

t2




x(tminus2)=int t2

t1


h(s)ds +int t1

t0


h(s)ds


t0




x(t+2)= minus e0


t1


h(s)ds +int t1

t0




(216)






thus we have



(218)


x(t) =int t

t2


h(s)ds +int t1

t0


h(s)ds +int t2

t1


h(s)ds


t0


h(s)ds


t0


h(s)ds +int t2

t1


h(s)ds

]

+ I1(x(tminus1))

+ I2(x(tminus2))



(219)


x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(t minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1



xprime(t) =

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1


(220)



we have




Δ+(1 minus b)TΛ2

Δ

minusc1 = minuscΛ1


Δ

(222)




3 Main Results



conditions


(x(tminusk))



Ωlowast1 = sup


2 = suptisinJ


tisinJ


2 = suptisinJ

∣∣Ωprime2(t)∣∣ (32)





k(u) minus Ilowast


k = 1 2 p


[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

lt 1

(33)



(Fx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds



i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk


f(s x(s))ds

+ Ω2(t)intT

tk


f(s x(s))ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(34)



tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




timesksum

i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



+ |1 + Ω1(t)|[

ksum

i=1




(y(tminusk))∣∣]


i=1


(y(tminusi))∣∣


i=1


(y(tminusi))∣∣

+kminus1sum

i=1


(y(tminusi))∣∣



[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3


(35)







(Ax)(t) = [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(Dx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds


i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk



tk


f(s x(s))ds

(36)


(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3 lt 1 (37)





tk



∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



∣∣ds


timesksum

i=1

int ti

timinus1



∣∣ds


i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



(38)




|(Dx)(t)| leint t

tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1




+ |Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + |Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

D(x) le M1Tα


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)= l

(310)


∣∣(Dy


∣∣ leint τ2

τ1



where


tk


∣∣f(s x(s))∣∣ds +


ksum

i=1

int ti

timinus1




1(t) + Ωprime2(t) + 1

∣∣ksum

i=1

int ti

timinus1



+∣∣Ωprime


i=1

(tk minus ti)int ti

timinus1



+∣∣Ωprime

1(t)∣∣intT

tk


∣∣f(s x(s))∣∣ds +


tk



le M1Tα

Γ(α + 1)Ωlowast

1

(1 + p + 2pα

)+M1T

αminus1

Γ(α)(1 + Ωlowast

2)(1 + p

)= L

(312)




(xλ


(313)



)(t) (314)



|x(t)| le λ

int t

tk


∣∣f(s x(s))

∣∣ds + λ|1 + Ω1(t)|

ksum

i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

(tk minus ti)int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + λ|Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

+ λ|1 + Ω1(t)|[

ksum

i=1

∣∣∣∣∣Ii

(x(tminusi)

λ

)∣∣∣∣∣+


(x(tminusk

)

λ

)∣∣∣∣∣

]


i=1


(x(tminusi)

λ

)∣∣∣∣∣


i=1


(x(tminusi)

λ

)∣∣∣∣∣+ λ

kminus1sum

i=1


(x(tminusi)

λ

)∣∣∣∣∣

x(t) le[M1T

α


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pM2 +M3

)+(Ωlowast

1T + Ωlowast2 + T

)pM3

]

(315)


4 An Example


CD32x(t) =

sin 2t|x(t)|(t + 5)2(1 + |x(t)|)

t isin [0 1] t =13

Δx(13

)=


(13

)=



(41)




[L1T

α


1

)(1 + p(1 + L2) + 2pα + L3

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

=464

1125radicπ

+173450

lt 1

(42)


Acknowledgments


References

































1 Introduction


intb

a

w(x)f(x)dx =nsum

k=0

wkf(xk) + Rn+1[f] (11)






Ψn+1(x) =nprod

k=0

(x minus xk) (12)


f(x) =nsum

k=0

f(xk)L(xxk) + rn+1(f x) (13)

where

L(xxk) =Ψn+1(x)


(k = 0 1 n) (14)

we obtain (11) with

wk =1

Ψprimen+1(xk)

intb

a


Rn+1[f]=intb

a

rn+1(f x)w(x)dx

(15)



intb

a

f(t)dt =b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)+ E(f) (16)



∥∥f∥∥p =

(intb

a

∣∣f(t)∣∣pdt

)1p

(17)


xisin[ab]

∣∣f(x)∣∣ (18)



∣∣∣∣∣

intb

a

f(x)g(x)dx

∣∣∣∣∣le ∥∥f∥∥1





λ(b minus a)intb

a



2λ(b minus a) f(a)

∣∣∣∣∣




)

(110)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (111)


K(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

t minus 5a + b6

t isin[aa + b2

]

t minus a + 5b6

t isin(a + b2

b

]

(112)


intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt (113)

maxtisin[ab]

|K(t)| = 13(b minus a) (114)


2 Main Results


m1 =int (5a+b)6

a

(t minus 5a + b

6

)β(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)α(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)β(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)α(t)dt

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt le

M1 =int (5a+b)6

a

(t minus 5a + b

6

)α(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)β(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)α(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)β(t)dt

(21)


intb

a


2

)dt

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus 12

(intb

a

K(t)(α(t) + β(t)

)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)

(22)



2


2 (23)




(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a


2

)dt


a


dt

=12

(int (a+b)2

a


5a + b6


intb

(a+b)2


a + 5b6


)

(24)


m1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)β(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)α(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)β(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)α(x)dx

M1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)α(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)β(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)α(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)β(x)dx

(25)




Special Case 1


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


144((β1 minus α1

)(a + b) + 2

(β0 minus α0

))

(26)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (27)



int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(


a

α(t)dt

)

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t) dt

leint (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

+b minus a3

(


a

α(t)dt

)

(28)

Proof Since

intb

a


6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)α(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)

(29)


so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


3

(


a

α(t)dt

)

(210)


Special Case 2


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)


a + b2

α1

))

(211)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)



int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(intb

a


le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

leint (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt +

b minus a3

(intb

a


(213)


Proof Since

intb

a


=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)β(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)

(214)

so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

int

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


)dt =

b minus a3

(int b

a


(215)


Special Case 3

If β(x) = β1x + β0 = 0 in (213) then

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 +

a + b2


b minus a)

(216)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 minus


) (217)


Acknowledgment


References





















1 Introduction









xx= 0

(11)













u(x t) = U(ξ) ξ = kx +ωt (22)



)= 0 (23)



U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0 (24)


d2G(ξ)dξ2

+ λdG(ξ)dξ

+ μG(ξ) = 0 (25)






3 Application


u(x t) = U(ξ)eiη

v(x t) = V (ξ)(31)




)U = 0 (32)

(ω2 minus k2


(U2)primeprime

= 0 (33)



)U = 0

(ω2 minus k2


(34)


U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0

V (ξ) =nsum

i=1

βi

(Gprime

G

)i+ β0 βn = 0

(35)




m + n = m + 2

2m = n + 2(36)


U(ξ) = α2(Gprime

G

)2

+ α1(Gprime

G

)+ α0 α2 = 0

V (ξ) = β2(Gprime

G

)2

+ β1(Gprime

G

)+ β0 β2 = 0

(37)




)= 0




)ω minus qα1



)= 0


)α2



)β2 minus k2


)= 0


)α2 minus α1

(minus2k2 + aβ2

)= 0


)= 0

(6k2 minus aβ2

)α2 = 0


(38)






λ = plusmn13

radic3ω2 minus 3k2

k2 μ =

16k2 minusω2

k4 p = minus1

2ω

k

q =124

24k4 + 30ω4 minus 54k2ω2


α1 = plusmn2radic

1ab


radic1ab


a β2 =

6k2

a

(39)


U(ξ) = 6k2radic

1ab

(Gprime

G

)2

plusmn 2

radic1ab

radic3ω2 minus 3k2

(Gprime

G

)

V (ξ) =6k2

a

(Gprime

G

)2

plusmn 2

radic3ω2 minus 3k2

a

(Gprime

G

)

(310)



uH(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2



⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vH(x t) =6k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(311)



2 asfollows

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312a)


⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312b)

while at C21 lt C

22 one can obtain

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312c)


⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312d)




uT(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2


⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vT(x t) =6k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(313)


1 gt C22 and C

21 lt C

22 as follows

uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314a)


ak2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314b)


|u H|

tx

1

05

0

minus054

20

minus2minus4 minus4

minus20

24


0

02

04

06

08

tx

42

0minus2

minus4 minus4minus2

02

4

|u H|2


uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314c)


ak2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314d)



0

200

400

600

800

tx

42

0minus2

minus4 minus4minus2

02

4

vH


4 Conclusions


Acknowledgments


References























Shengli Xie







1 Introduction



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J






(

t x(σ1(t))int t


)

t isin [0 b]

x(0) + g(x) = x0

(12)


(1 minus αβMc

)N


2αθ1L1M(1 + 2Kθ2L1) lt 1

(13)





2 Preliminaries








x(t) = R(t 0)[φ(0) + g(x)(0)

]+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

(21)




α

(int t

0xn(s)ds n isin N

)

le 2int t








3 Existence Result




+ such that

∥∥f(t x y z


∥∥y∥∥ + z






andint t0 k(t s)ds





α

(int t

0K(t s V1)ds

)


α(V1(θ)) t isin J

α

(intb

0H(t s V1)ds

)


α(V1(θ)) t isin J

(33)


w = maxMp0(1 + k0 + h0) + 1 2M

(l01 + 2l02k0 + 2l03h0

)+ 1 (34)






(Fx)(t)

=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(35)






0ew(sminust)

∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥ds

le L +Mint t

0ew(sminust)p(s)

(

xs[minusq0] +ints

0K(s r xr)dr +

intb

0H(s r xr)dr

)

ds

le L +Mp0

int t

0ew(sminust)

(

xs[minusq0] +int s


intb


)

ds


Consequently




(38)


∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥le p(s)(1 + k0 + h0)R (39)







0R(t minus s)α

(

f

(

s Vs

int s

0K(s r Vr)dr

intb

0H(s r Vr)dr

))

ds

le 2Mint t

0ew(sminust)

[



0K(s r Vr)dr

)

+l3(s)α

(intb

0H(s r Vr)dr

)]

ds

le 2Mint t

0ew(sminust)

[


α(V (s + θ)) + 2l02

int s

0k(s r) sup


+2l03

intb

0h(s r) sup


]

ds

le 2M(l01 + 2l02k0 + 2l03h0

)int t

0ew(sminust)ds sup


le 2M(l01 + 2l02k0 + 2l03h0


(311)

Consequently


α((FV )(t)) le 2M(l01 + 2l02k0 + 2l03h0






4 An Example




ut(t y)= a1

(t y)uyy(t y)

+ a2(t)

[

sinu(t + θ y

)ds +

int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

0 le t le b


0

intb

0F(r y)log(1 + |u(r s)|12


u(t 0) = u(t π) = 0 0 le t le b(41)


f

(

t ut

int t

0K(t s us)ds

intb

0H(t s us)ds

)(y)

= a2(t)

[

sinu(t + θ y

)+int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

K(t s us)(y)=ints

minusqa3(s + τ)

u(τ y)dτ

(1 + t) H(t s us)

(y)=

u(s + θ y

)

(1 + t)(1 + s)2

g(u)(y)=intπ

0

intb

0F(r y)log(1 + |u(r s)|12

)dr ds

(42)



(43)


∥∥f(t u v z)∥∥ le |a2(t)|

(u[minusq0] + v + z

)

K(t s u) leints


u[minusq0](1 + t)


(1 + t)(1 + s)

(44)



∥∥g(u)



∣∣F(r y)∣∣(u[0b] +

radicπ) (45)


α(f(t V1 V2 V3)

) le |a2(t)|(

supminusqleθle0

α(V1(θ)) + α(V2) + α(V3)

)

t isin J

α(K(t s V1)) le 11 + t

supminusqleθle0


α(H(t s V1)) le 1

(1 + t)(1 + s)2sup


(46)


5 An Application



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J



x(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(52)






w = maxMp0(1 + k0 + h0)(1 +MM1b) + 1 2M

(l01 + 2l02k0 + 2l03h0

)(1 +MM1b) + 1

(53)



Wv =intb

0R(b s)(Cv)(s)ds (54)





v(t) =Wminus1(

x1 minus R(b 0)[φ(0) + g(x)(0)

]

+intb

0R(b s)f

(

s xs

ints

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

)

(t) t isin [0 b]

(55)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(56)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t


[φ(0) + g(x)(0)

]

+intb

0R(b τ)f

(

τ xτ

int τ

0K(τ r xr)dr

intb

0H(τ r xr)dr

)

dτ

)

(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(57)




Acknowledgments


References






































Contents



































Acknowledgments













1 Introduction











119909 = 119865119909 (1)












119901= 119862(R

+ 119901(119905)) the


+and such that

sup |119909 (119905)| 119901 (119905) 119905 ge 0 lt infin (2)


to the norm

119909 = sup |119909 (119905)| 119901 (119905) 119905 ge 0 (3)


119901[10 15]



+and if lim

119879rarrinfinsup|119909(119905)|119901(119905) 119905 ge 119879 = 0





that is


(4)


119909 (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 (5)




+ 119886 R

+rarr


|119907 (119905 119904 119909)| le 119899 (119905 119904) + 119886 (119905) 119887 (119904) |119909| (6)



119871 (119905) = int

119905

0

119886 (119904) 119887 (119904) 119889119904 119905 ge 0 (7)


119901 where 119901(119905) = [119886(119905) exp(119872119871(119905) + 119905)]




119863119886(119905) exp(119872119871(119905)) for 119905 ge 0


int

119905

0

119899 (119905 119904) 119889119904 le 119873119886 (119905) exp (119872119871 (119905)) (8)

for 119905 ge 0



condition

119871 (120593 (119905)) minus 119871 (119905) le 119870 (9)



for all 119905 ge 0



119901such that |119909(119905)| le 119886(119905) exp(119872119871(119905))

for 119905 isin R+


119901by putting

(119865119909) (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 119905 ge 0 (10)


+






|(119865119909) (119905)| le |119889 (119905)| + int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904)))1003816100381610038161003816 119889119904

le 119863119886 (119905) exp (119872119871 (119905))

+ int

119905

119900

[119899 (119905 119904) + 119886 (119905) 119887 (119904)1003816100381610038161003816119909 (120593 (119904))

1003816100381610038161003816] 119889119904

le 119863119886 (119905) exp (119872119871 (119905)) + 119873119886 (119905) exp (119872119871 (119905))

+ 119886 (119905) int

119905

0

119887 (119904) 119886 (120593 (119904)) exp (119872119871 (120593 (119904))) 119889119904

le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) int

119905

0

119872119886 (119904) 119887 (119904)


le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) exp (119872119871 (119905))

= 119886 (119905) exp (119872119871 (119905))

(11)





1003816100381610038161003816

le int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904))) minus 119907 (119905 119904 119910 (120593 (119904)))1003816100381610038161003816 119889119904

le 120573 (120576)

(12)



1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)1003816100381610038161003816 [119886 (119905) exp (119872119871 (119905) + 119905)]

minus1

le |(119865119909) (119905)| +1003816100381610038161003816(119865119910) (119905)

1003816100381610038161003816

times [119886 (119905) exp (119872119871 (119905))]minus1


le 2119890minus119905

(13)


1003816100381610038161003816 119901 (119905) le 120576 (14)



119901


|(119865119909) (119905)| 119901 (119905) le 119890minus119905 (15)

This yields that

lim119879rarrinfin

sup |(119865119909) (119905)| 119901 (119905) 119905 ge 119879 = 0 (16)



|(119865119909) (119905) minus (119865119909) (119904)|

le |119889 (119905) minus 119889 (119904)|

+

10038161003816100381610038161003816100381610038161003816

int

119905

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119904 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

le 120584119879(119889 120576) + 120576max 119899 (119905 120591) + 119886 (120591) 119887 (120591)

times119901 (120593 (120591)) 0 le 120591 le 119905 le 119879

+ 119879120584119879(119907 (120576 119879 120572 (119879)))

(17)

where we denoted120584119879(119907 (120576 119879 120572 (119879)))

= sup |119907 (119905 119906 119907) minus 119907 (119904 119906 119907)| 119905 119904 isin [0 119879]

|119905 minus 119904| le 120576 119906 isin [0 119879] |119907| le 120572 (119879)

(18)



120584119879(119889 120576) =

lim120576rarr0











119909 (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

+ int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(19)



+)



supremum norm

119909 = sup |119909 (119905)| 119905 isin R+ (20)








that

1003816100381610038161003816119891 (119905 1199091 1199092 119909

119899) minus 119891 (119905 119910

1 1199102 119910

119899)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816

(21)


1 1199092 119909

119899) (1199101 1199102

119910119899) isin R119899


with 1198650= sup|119891(119905 0 0)| 119905 isin R

+


R+



1 2 119898)(iv) The function 120573 R




rarr R+and


1003816100381610038161003816119892 (119905 119904 1199091 119909

119898)1003816100381610038161003816 le 119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

10038161003816100381610038161199091198941003816100381610038161003816 (22)

for all 119905 119904 isin R+and (119909

1 1199092 119909


we assume that

lim119905rarrinfin

int

120573(119905)

0

119902 (119905 119904) 119889119904 = 0 lim119905rarrinfin

119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 = 0

(23)


1 1199072 R+

rarr R+defined by

the formulas

1199071(119905) = int

120573(119905)

0

119902 (119905 119904) 119889119904 1199072(119905) = 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 (24)



119872119894= sup 119907

119894(119905) 119905 isin R

+ (119894 = 1 2) (25)


119896 = sum119899


119894(119894 = 1 2 119899) appear in

assumption (i)

(vi) 119896 + 1198981198722lt 1




globally attractive




119909 (119905) = (119876119909) (119905) 119905 isin R+ (26)


lim119905rarrinfin

(119909 (119905) minus 119910 (119905)) = 0 (27)




119903in the space119883 = 119861119862(R

+) centered at the


2)]


putting

(119860119909) (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

(119861119909) (119905) = int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(28)


119909 (119905) = (119860119909) (119905) + (119861119909) (119905) 119905 isin R+ (29)


+ Thus



119903into



+ we get

1003816100381610038161003816(119860119909) (119905) minus (119860119910) (119905)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816

le 1198961003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(30)





1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816

le int

120573(119905)

0

[1003816100381610038161003816119892 (119905 119904 119909 (120574

1(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816

+1003816100381610038161003816119892 (119905 119904 119910 (120574

1(119904)) 119910 (120574

119898(119904)))

1003816100381610038161003816] 119889119904

le 2 (1199071(119905) + 119903119898119907

2(119905))

(31)


1(119905) + 119903119898119907

2(119905) le 1205762 for 119905 ge 119879


1003816100381610038161003816 le 120576 (32)


obtain

1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816 le int

120573(119905)

0

120596119879

119903(119892 120576) 119889119904 le 120573

119879120596119879

119903(119892 120576)

(33)


120596119879

119903(119892 120576) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 1199092 119909

119898)

minus119892 (119905 119904 1199101 1199102 119910

119898)1003816100381610038161003816 119905 isin [0 119879]

119904 isin [0 120573119879] 119909119894 119910119894isin [minus119903 119903]

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816 le 120576 (119894 = 1 2 119898)

(34)


119879]times[minus119903 119903]



119903


|(119861119909) (119905)| le 1199071(119905) + 119903119898119907

2(119905) (35)




1(119905) + 119903119898119907




|(119861119909) (119905)| le 120576 (36)


|(119861119909) (119905) minus (119861119909) (120591)|

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

minus119892 (120591 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

le

1003816100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

[119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816] 119889119904

1003816100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

120596119879

1(119892 120576 119903) 119889119904

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119902 (119905 119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 119886 (119905)119898119903

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119887 (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 120573119879120596119879

1(119892 120576 119903)

(37)

where we denoted

120596119879

1(119892 120576 119903) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 119909

119898)

minus119892 (120591 119904 1199091 119909

119898)1003816100381610038161003816 119905 120591 isin [0 119879]

|119905 minus 119904| le 120576 119904 isin [0 120573119879]

100381610038161003816100381611990911003816100381610038161003816 le 119903

10038161003816100381610038161199091198981003816100381610038161003816 le 119903

(38)



|(119861119909) (119905) minus (119861119909) (120591)|

le 119902119879120584119879(120573 120576) + 119903119898119886

119879119887119879120584119879(120573 120576) + 120573

119879120596119879

1(119892 120576 119903)

(39)

where 119902119879

= max119902(119905 119904) 119905 isin [0 119879] 119904 isin [0 120573119879] 119886119879

=

max119886(119905) 119905 isin [0 119879] 119887119879

= max119887(119905) 119905 isin [0 120573119879]



of the functions 120573 = 120573(119905) 119892 = 119892(119905 119904 1199091 119909

119898) we infer

that 120584119879(120573 120576) rarr 0 and 120596119879





+)





|119909 (119905)| le |(119860119909) (119905)| +1003816100381610038161003816(119861119910) (119905)

1003816100381610038161003816

le1003816100381610038161003816119891 (119905 119909 (120572

1(119905)) 119909 (120572

119899(119905))) minus 119891 (119905 0 0)

1003816100381610038161003816

+1003816100381610038161003816119891 (119905 0 0)

1003816100381610038161003816

+ int

120573(119905)

0

119902 (119905 119904) 119889119904 + 1198981003817100381710038171003817119910

1003817100381710038171003817 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904

le 119896 119909 + 1198650+1198721+ 119898119903119872

2

(40)

Hence we obtain

119909 le1198650+1198721+ 119898119903119872

2

1 minus 119896 (41)


2)(1 minus





1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le

119899

sum

119894=1

119896119894max 1003816100381610038161003816119909 (120572

119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 2int

120573(119905)

0

119902 (119905 119904) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119910 (120574119894(119904))

1003816100381610038161003816) 119889119904

le 119896max 1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 21199071(119905) + 119898 (119909 +

10038171003817100381710038171199101003817100381710038171003817) 1199072 (119905)

(42)


Hence we get


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le 119896max1le119894le119899


1003816100381610038161003816119909 (120572119894(119905) minus 119910 (120572

119894(119905)))

1003816100381610038161003816


1199071(119905) + 119898 (119909 +


1199072(119905)


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

(43)



|119909(119905) minus








or






⋃

0le120582le1

119909 isin 119864 119909 = 120582119879119909 (44)























119864 120583(119883) = 0 is nonempty


(2119900) 119883 sub 119884 rArr 120583(119883) le 120583(119884)

(3119900) 120583(119883) = 120583(Conv119883) = 120583(119883)


(5119900) If (119883


119864such that

119883119899+1

sub 119883119899for 119899 = 1 2 and if lim

119899rarrinfin120583(119883119899) = 0


= ⋂infin

119899=1119883119899is nonempty



119899) for any 119899 = 1 2 This

















Next let us define

120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883

1205960(119883) = lim

120576rarr0

120596 (119883 120576) (46)



0(120582119883) =

|120582|1205960(119883) and 120596

0(119883 + 119884) le 120596

0(119883) + 120596

0(119884) provided 119883119884 isin

M119862(119868)



119909 (119905) = 1198911(119905 119909 (119905) 119909 (119886 (119905))) + (119866119909) (119905) int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(47)



has the form

1198912(119905 119904) = 119896 (119905 119904) 119892 (119905 119904) (48)







2

1003816100381610038161003816 10038161003816100381610038161199101 minus 119910

2

1003816100381610038161003816

(49)



0(119866119883) le 119902120596


119862(119868)



+rarr R+



+rarr R


119909 isin 119862(119868)(vi) 119891



+occurring in



1lt 1199052and


10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

le 120576

int

1199052

1199051

119892 (1199052 119904) 119889119904 le 120576

(50)



ℎ (119905) = int

119905

0

119892 (119905 119904) 119889119904 (51)





119896 = sup |119896 (119905 119904)| (119905 119904) isin Δ

1198911= sup 10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816 119905 isin 119868

ℎ = sup ℎ (119905) 119905 isin 119868

(52)





119901119903 + 1198911+ 119896ℎ120593 (119903)Ψ (119903) le 119903 (53)

such that 119901 + 119896ℎ119902Ψ(1199030) lt 1


(1198651119909) (119905) = 119891

1(119905 119909 (119905) 119909 (120572 (119905)))

(1198652119909) (119905) = int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(119865119909) (119905) = (1198651119909) (119905) + (119866119909) (119905) (119865

2119909) (119905)

(54)



+by the formulas

119872(120576) = sup10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119892)] 119889119904

10038161003816100381610038161003816100381610038161003816

1199051 1199052isin 119868

1199051lt 1199052 1199052minus 1199051le 120576

(55)

119873(120576) = supint1199052

1199051

119892 (1199052 119904) 119889119904 119905

1 1199052isin 119868 1199051lt 1199052 1199052minus 1199051le 120576

(56)






1003816100381610038161003816(1198651119909) (119905) minus (1198651119910) (119905)

1003816100381610038161003816


1003816100381610038161003816

(57)


10038171003817100381710038171198651119909 minus 11986511199101003817100381710038171003817 le 119901

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (58)







1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816

int

1199052

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199051 119904) 119892 (119905

2 s) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

1 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le int

1199052

1199051

1003816100381610038161003816119896 (1199052 119904)1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199052 119904) minus 119896 (1199051 119904)

1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199051 119904)1003816100381610038161003816

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 |(119876119909) (119904)| 119889119904

le 119896Ψ (119909)119873 (120576) + 1205961(119896 120576) Ψ (119909) int

1199052

0

119892 (1199052 119904) 119889119904

+ 119896Ψ (119909) int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904

(59)

where we denoted

1205961(119896 120576) = sup 1003816100381610038161003816119896 (1199052 119904) minus 119896 (119905

1 119904)

1003816100381610038161003816

(1199051 119904) (119905

2 119904) isin Δ

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576

(60)


int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904 =

10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

(61)


120596 (1198652119909 120576) le 119896Ψ (119909)119873 (120576) + ℎΨ (119909) 120596

1(119896 120576)

+ 119896Ψ (119909)119872 (120576)

(62)










1003816100381610038161003816(119865119909) (119905) minus (1198651199090) (119905)

1003816100381610038161003816

le1003816100381610038161003816(1198651119909) (119905) minus (119865

11199090) (119905)

1003816100381610038161003816

+1003816100381610038161003816(119866119909) (119905) (1198652119909) (119905) minus (119866119909

0) (119905) (119865

21199090) (119905)

1003816100381610038161003816

le10038171003817100381710038171198651119909 minus 119865

11199090

1003817100381710038171003817 + 1198661199091003816100381610038161003816(1198652119909) (119905) minus (119865

21199090) (119905)

1003816100381610038161003816

+100381710038171003817100381711986521199090

1003817100381710038171003817

1003817100381710038171003817119866119909 minus 1198661199090

1003817100381710038171003817

(63)



21199090) (119905)

1003816100381610038161003816

le int

119905

0

|119896 (119905 119904)| 119892 (119905 119904)1003816100381610038161003816(119876119909) (119904) minus (119876119909

0) (119904)

1003816100381610038161003816 119889119904

le 119896 (int

119905

0

119892 (119905 119904) 119889119904)1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

le 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

(64)

Similarly we obtain

1003816100381610038161003816(11986521199090) (119905)1003816100381610038161003816 le 119896 (int

119905

0

119892 (119905 119904) 119889119904)10038171003817100381710038171198761199090

1003817100381710038171003817 le 119896ℎ10038171003817100381710038171198761199090

1003817100381710038171003817 (65)

and consequently

1003817100381710038171003817119865211990901003817100381710038171003817 le 119896ℎ

10038171003817100381710038171198761199090

1003817100381710038171003817 (66)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 1199011003817100381710038171003817119909 minus 119909

0

1003817100381710038171003817 + 119866119909 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎ1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

10038171003817100381710038171198761199090

1003817100381710038171003817

(67)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 119901120576 + 120593 (10038171003817100381710038171199090

1003817100381710038171003817 + 120576) 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎΨ (10038171003817100381710038171199090

1003817100381710038171003817)1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

(68)






|(119865119909) (119905)| le10038161003816100381610038161198911 (119905 119909 (119905) 119909 (119886 (119905))) minus 119891

1(119905 0 0)

1003816100381610038161003816

+10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816

+ |119866119909 (119905)|

10038161003816100381610038161003816100381610038161003816

int

119905

0

119896 (119905 119904) 119892 (119905 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909)

(69)

Hence we obtain

119865119909 le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909) (70)



the ball 1198611199030




1199030


1 1199052isin 119868with


obtain1003816100381610038161003816(119865119909) (1199052) minus (119865119909) (119905

1)1003816100381610038161003816

le10038161003816100381610038161198911 (1199052 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199052) 119909 (119886 (119905

2)))

1003816100381610038161003816

+10038161003816100381610038161198911 (1199051 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199051) 119909 (119886 (119905

1)))

1003816100381610038161003816

+1003816100381610038161003816(1198652119909) (1199052)

1003816100381610038161003816

1003816100381610038161003816(119866119909) (1199052) minus (119866119909) (1199051)1003816100381610038161003816

+1003816100381610038161003816(119866119909) (1199051)

1003816100381610038161003816

1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le 1205961199030

(1198911 120576) + 119901max 1003816100381610038161003816119909 (119905

2) minus 119909 (119905

1)1003816100381610038161003816

1003816100381610038161003816119909 (119886 (1199052)) minus 119909 (119886 (119905

1))1003816100381610038161003816

+ 119896ℎΨ (1199030) 120596 (119866119909 120576) + 120593 (119903

0) 120596 (119865

2119909 120576)

(71)

where we denoted

1205961199030

(1198911 120576) = sup 10038161003816100381610038161198911 (1199052 119909 119910) minus 119891

1(1199051 119909 119910)

1003816100381610038161003816 1199051 1199052 isin 119868

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576 119909 119910 isin [minus1199030 1199030]

(72)


120596 (119865119909 120576) le 1205961199030

(1198911 120576) + 119901max 120596 (119909 120576) 120596 (119909 120596 (119886 120576))

+ 119896ℎΨ (1199030) 120596 (119866119909 120576)

+ 120593 (1199030) Ψ (1199030) [119896119873 (120576) + ℎ120596

1(119896 120576) + 119896119872 (120576)]

(73)





1205960(119865119883) le 119901120596

0(119883) + 119896ℎΨ (119903

0) 1205960(119866119883) (74)


1205960(119865119883) le (119901 + 119896ℎ119902Ψ (119903

0)) 1205960(119883) (75)







1199091015840= 119891 (119905 119909) (76)


119909 (0) = 1199090 (77)









rarr R+(or 119908 119869

0times

R+



1003817100381710038171003817 le 119908 (1199051003817100381710038171003817119909 minus 119910

1003817100381710038171003817) (78)




1015840= 119908(119905 119906)


119905

0119908(119904 119906(119904))119889119904 for 119905 isin 119869


lim119905rarr0

119906(119905)119905 = lim119905rarr0






120583 (119891 (119905 119883)) le 119908 (119905 120583 (119883)) (79)





120583


119864120583= 119909 isin 119864 119909 isin ker 120583 (80)



120583is a closed



(6119900) 120583(119883 + 119884) le 120583(119883) + 120583(119884)

(7119900) 120583(120582119883) = |120582|120583(119883) for 120582 isin R








120583 (1199090+ 119891 (119905 119883)) le 119908 (119905 120583 (119883)) (81)



1198690times R+



119906 (119905) le int

119905

0

119908 (119904 119906 (119904)) 119889119904 (119905 isin 1198690) (82)


119906(119905)119905 = lim119905rarr0




0)minus119891(119904 119909)||



120583for 119905 isin 119869



0isin



120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (83)





120583 (119891 (119905 119883)) le 119901 (119905) 120583 (119883) (84)





1199091015840

119899= 119886119899(119905) 119909119899+ 119891119899(119909119899 119909119899+1

) (85)


119909119899(0) = 119909

119899

0(86)


119909119899

0= 119886(119886 isin R)






119899rarrinfin120572119899= 0 and |119891

119899(119909119899 119909119899+1

)| le

120572119899for 119899 = 1 2 and for all 119909 = (119909

1 1199092 1199093 ) isin 119897






119899)|| =

sup|119909119899| 119899 = 1 2


1(119905) 1199092(119905) )






sup119909isin119883

1003816100381610038161003816119909119899 minus 1198861003816100381610038161003816 (87)




120583 (1199090+ 119891 (119905 119883))


sup119909isin119883

1003816100381610038161003816119909119899

0+ 119886119899(119905) 119909119899

+119891119899(119909119899 119909119899+1

) minus 1198861003816100381610038161003816


sup119909isin119883

[1003816100381610038161003816119886119899 (119905)

1003816100381610038161003816

10038161003816100381610038161199091198991003816100381610038161003816

+1003816100381610038161003816119891119899 (119909119899 119909119899+1 ) + 119909

119899

0minus 119886

1003816100381610038161003816]


sup119909isin119883

119901 (119905)1003816100381610038161003816119909119899 minus 119886

1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119886119899 (119905)1003816100381610038161003816 |119886|


sup119909isin119883

1003816100381610038161003816119891119899 (119909119899 119909119899+1 )1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119909119899

0minus 119886

1003816100381610038161003816

(88)


Hence we get

120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (89)



References













































1015840























1 Introduction




u(0) = g(u)(11)







u(0) = g(u)

(12)







part



2 Preliminaries


+ = (0infin)and R





tisinI

∥∥f(t)∥∥X (21)



∥∥f∥∥L1(IX) =

int

I

∥∥f(s)∥∥Xds (22)







d

dtR(t)y = AR(t)y +

int t


= R(t)Ay +int t


(23)







u(0) = u0 isin X(24)








is verified





(27)
















ζ

(int t

0un(s)ds

infin

n=1

)

2int t





subeW such that




Sn = εn + Cn1ε

nminus1h + Cn2ε

nminus2h2


n n isin N (211)





)) n = 2 3 (212)


η(Fn0(S)) rη(S) (213)


3 Main Results





+ rarr R+ such that

∥∥f(t x)∥∥ m(t)Φ(x) (31)




ζ(f(t S)

) H(t)ζ(S) (32)





KgR +KΦ(R)int1

0m(s)ds R (33)








∥∥ +Kint1

0




(Fu)(t) ∥∥R(t)g(u)

∥∥ +

∥∥∥∥∥

int t


∥∥∥∥∥

KgR +KΦ(R)int1

0m(s)ds R

(36)





sub F(B) such that


(int t

0


infinn=1ds

)

+ ε (37)


ζ(F(B)(t)) 4Kint t

0ζ(f(s un(s))

infinn=1ds + ε

4Kint t

0H(s)ζ(un(s)nisinN

) ds + ε

4Kγ(B)int t

0H(s)ds + ε

(38)




0


0ϕ(s)ds

]

+ ε


(39)





ζ(F2(B)(t)

) 2ζ

(int t

0


infinn=1ds

)

+ ε

4Kint t

0ζf(swn(s))

infinn=1ds + ε

4Kint t

0H(s)ζ

(co(F1(B)(s)

))+ ε = 4K

int t

0H(s)ζ

(F1(B)(s)

)+ ε

(311)



ζ(F2(B)(t)

) 4K

int t

0

[∣∣H(s) minus ϕ(s)∣∣ + ∣∣ϕ(s)

∣∣](a + bs)γ(B)ds + ε


0


(

at +bt2

2

)

+ ε

a(a + bt) + b

(

at +bt2

2

)

+ ε (

a2 + 2bt +(bt)2

2

)

γ(B) + ε

(312)


ζ(F2(B)(t)

)

(

a2 + 2bt +(bt)2

2

)

γ(B) (313)


ζ(Fn(B)(t)) (

an + Cn1a

nminus1bt + Cn2a

nminus2 (bt)2


n

)

γ(B) (314)



γ(Fn(B)) (

an + Cn1a

nminus1b + Cn2a

nminus2 b2


n

)

γ(B) (315)


(

an0 + Cn01 a

n0minus1b + Cn02 a

n0minus2 b2


n0

)

= r lt 1 (316)





u(0) = g(u)

(317)













∣∣∣ C|a(λ)| (318)






1λa(λ)


(319)



such that




completely positive



lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (321)



u(t) =int t


int t

0f(s u(s)) + g(u) (322)


R(t)x = x +int t



d

dtR(t)x = AR(t)x +

int t




KgR +KΦ(R)int1





4 Applications


partw(t ξ)partt

= Aw(t ξ) +int t



w(0 ξ) =int1

0

int ξ

0qk(s ξ)w

(s y

)dsdy 0 ξ 2π

(41)



defined by



part







u(0) = g(u)

(43)






int10 k(s ξ)

2dsdξ)12







a(λ) =λ + α + βλ(λ + α)

(44)


f1(λ) =λ + α

λ + α + β f2(λ) =

λ2 + 2(α + β

)λ + αβ + α2

(λ + α + β

)2 (45)


f(n)1 (λ) =

(minus1)n+1β(n + 1)(λ + α + β

)n+1 f(n)2 (λ) =

(minus1)n+1β(α + β)(n + 1)

(λ + α + β








Re

(μ0 + iμ

b(μ0 + iμ

)+ 1

)

=μ30 + μ

20α + μ2

0

(α + β

)+ μ0α

(α + β

)+ μ0μ


)2 + 2μ0(α + β

)+ μ2

0 + μ2

(48)


(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1



∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1


∥∥∥∥

M

μ0 + iμ

∥∥∥∥ (410)


Therefore

lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (411)



int10 k(s ξ)

2dsdξ)12




qcKR +∥∥p1

∥∥1LK R (412)


Acknowledgments


References















































1 Introduction
















K(q)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2πqω1+2q

(2

2 + q

)1minus2q( Γ(1q)

Γ(12 + 1q

)

)2

if 1 le q lt +infin

4ω if q = +infin

(15)





0 lt a(t) le π2

ω2 t isin [0 ω] (16)





2 Preliminaries








u(t) =intω





Let





Let

G = min0letsleω


G(t s) σ = GG (26)





Th(t) =intω




Th(t) =intω

0G(t s)h(s)ds le G

intω

0h(s)ds (29)

and therefore

Th le Gintω

0h(s)ds (210)


Th(t) =intω

0G(t s)h(s)ds ge G

intω

0h(s)ds

=(GG

)middotGintω

0h(s)ds

ge σTh

(211)




Tφ = r(T)φ (212)


intω

0φ(t)dt = 1 (213)






hold




A = T F (216)











3 Main Results



mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x


mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x

(31)









f(t x0 x1 xn) = b0(t)x20 + b1(t)x


n (32)






radic|xn| (33)



















This implies that





(39)





intω

0



intω

0u0(t)φ(t)dt (311)


intω

0


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(312)


λ1

intω


intω

0u0(x)φ(t)dt (313)


intω


intω

0φ(t)dt = σu0 gt 0 (314)


i(AK capΩ1 K) = 1 (315)







(u1(t) minus μ1

)= F(u1)(t) t isin R (317)



This means that




(320)




intω

0

[u



intω

0u1(t)φ(t)dt (322)


intω

0


]φ(t)dt =

intω

0u1(t)


]dt

= λ1

intω

0u1(x)φ(t)dt

(323)


λ1

intω


intω

0u1(x)φ(t)dt (324)



intω


intω

0φ(t)dt = σu1 gt 0 (325)


i(AK capΩ2 K) = 0 (326)


i(AK cap

(Ω2 Ω1









(u0(t) minus μ0

)= F(u0)(t) t isin R (329)



(330)




intω

0

[u



intω

0u0(t)φ(t)dt (332)



intω

0

[u


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(333)


λ1

intω


intω

0u1(x)φ(t)dt (334)


i(AK capΩ1 K) = 0 (335)








(338)




intω

0

[u



intω

0u1(t)φ(t)dt (340)



intω

0

[u


]φ(t)dt =

intω

0u1(t)

[φ


]dt

= λ1

intω

0u1(x)φ(t)dt

(341)


λ1

intω


intω

0u1(x)φ(t)dt (342)


i(AK capΩ2 K) = 1 (343)


i(AK cap

(Ω2 Ω1



Acknowledgment


References

















York NY USA 1988










1 Introduction



x(t) = f

(


0u(t s x(c(s)))ds

)


x(t) = h

(

t x(t)intα(t)

0u(t s x(c(s)))ds

)





x(t) =int t



x(t) = L(t) +int t

0




x(t) =int t



x(t) = (Tx)(t)int t











0g(t s x







x(t) = f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f

(

t

intH(t)

0x(s)ds x(h(t))

)



x(t) = g(t x(t)) + f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f(t x(α(t)))

(

q(t) +intβ(t)

0u(t s x

(γ(s)))ds

)




2 Preliminaries












ωT

0 (X) = limεrarr 0

ωT (X ε)





diam X(t)

(22)







limtrarr+infin



Then

lim suptrarr+infin

ϕ(ψ(t))= lim sup












then









(212)




ωϕ(T)0 (X) (213)




∣∣∣ωT





∣∣∣ω




3 Main Results





limtrarr+infin


b(t) = +infin (31)






limtrarr+infin

sup

m3(t)intα(t)


= 0 (33)

sup

m3(t)intα(t)


leM0 (34)

sup

intα(t)


leM (35)








(Fx)(t) = f

(


0u(t s x(c(s)))ds

)



|(Fx)(t)| le∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minus f(t 0 0 0)∣∣∣∣∣+∣∣f(t 0 0 0)

∣∣


∣∣∣∣∣

intα(t)

0u(t s x(c(s)))ds

∣∣∣∣∣+ f



(310)





=

∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minusf(

t y(a(t))(Hy)(b(t))

intα(t)

0u(t s y(c(s))

)ds

)∣∣∣∣∣




)(b(t))

∣∣

+m3(t)intα(t)

0




)(b(t))

∣∣

+ sup

m3(t)intα(t)



(312)

which yields that


+ sup

m3(t)intα(t)


forallt isin R+

(313)


lim suptrarr+infin

diam(FX)(t)



diam(HX)(b(t))


sup

m3(t)intα(t)




diam(HX)(t)

(314)

that is

lim suptrarr+infin



diam(HX)(t) (315)





∣∣u(p τ v




ωTr

(f ε g(r)

)= sup

∣∣f(p vw z



(316)






le∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣

+

∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣


+m3(t)

[∣∣∣∣∣

intα(t)


∣∣∣∣∣+intα(s)


]

+ sup∣∣f(p vw z





∣∣u(p τ v




+ωTr

(f ε g(r)

)


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)

)

(318)

which implies that


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)


(319)



limεrarr 0


ωTr (u ε) = lim

εrarr 0ωTr

(f ε g(r)

)= 0 (320)


ωT0 (FX) leM1ω

a(T)0 (X) +M2ω

b(T)0 (HX) (321)


ω0(FX) leM1ω0(X) +M2ω0(HX) (322)



diam(FX)(t)



diam(HX)(t)

=M1μ(X) +M2

(ω0(HX) + lim sup


)

le (M1 +M2Q)μ(X)(323)



sup

m3(t)intα(t)


ltε



ωT (u δ r) ltε







=

∣∣∣∣∣f

(


0u(t s xn(c(s)))ds

)

minusf(


0u(t s x(c(s)))ds

)∣∣∣∣∣


+m3(t)intα(t)



+max

supτgtT

sup

m3(τ)intα(τ)

0


supτisin[0T]

sup

m3(τ)intα(τ)

0



2

ltε

2+max

ε

3 supτisin[0T]

m3(τ)intα(τ)

0ωT

(uδ02 r

)ds

ltε

2+max

ε

3


=5ε6

(327)

which yields that














M0 + h le r(1 minusM1) (331)




sup

m3(t)intα(t)





=

∣∣∣∣∣h

(

t z(t)intα(t)


)

minus h(

t y(t)intα(t)

0u(t τ y(c(τ))

)dτ

)∣∣∣∣∣


intα(t)

0



m3(t)intα(t)



(333)

which means that





4 Examples



x(t) =1

10 + ln3(1 + t3)+tx2(1 + 3t2

)

200(1 + t)+

1

10radic3 + 20t + 3|x(t3)|

+1

9 +radic1 + t

⎛

⎜⎝

int t2

0

s2x(4s) sin(radic



∣∣∣ x2(4s)ds

⎞

⎟⎠

3

forallt isin R+

(41)

Put

f(t vw z) =1

10 + ln3(1 + t3)+

tv2

200(1 + t)+

1

10radic3 + 20t + |w|

+z3

9 +radic1 + t

u(t s p

)=

s2p sin(radic



∣∣∣p2


(42)

Let

risin[9 minus radic

512

9 +

radic51

2

]

M=3 M0=9r10 M1=

r

100 M2=

1300

f =110

Q = 3 m1(t) =rt

100(1 + t) m2(t) =

1(10radic3 + 20t

)2 m3(t) =3M2

9 +radic1 + t


(43)



512

9 +

radic51

2

]



le t

200(1 + t)


∣∣∣ +

∣∣∣∣∣

1

10radic3 + 20t + |w|

minus 1

10radic3 + 20t +

∣∣q∣∣

∣∣∣∣∣

+1

9 +radic1 + t


∣∣∣ le m1(t)





sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)

minuss2w(4s) sin


)


∣∣∣w2(4s)


⎫⎪⎬

⎪⎭

le 3M2

9 +radic1 + t

int t2

0

2s2(1 + t12

)2

[r(1 + t12

)+ r3(r3t3 + 3r6 + 7

(t + 2t2

)2)]ds

le 2M2t6(9 +

radic1 + t)(

1 + t12)2

r(1 + t12

)+ r3[r3t3 + 3r6 + 7

(t + 2t2


sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=rM2

20lt M0

sup

intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=r

6leM

(44)




x(t) =3 + sin4

(radic1 + t2

)

16+x2(t)8 + t2

+1

1 + t

(intradict

0

sx3(1 + s2)

1 + t2 + s cos2(1 + t3s7x2(1 + s2))ds

)2

forallt isin R+

(45)

Put

h(t vw) =3 + sin4

(radic1 + t2

)

16+

v2

8 + t2+

w2

1 + t

u(t s p

)=

sp3

1 + t2 + s cos2(1 + t3s7p2

)

α(t) =radict c(t) = 1 + t2 m1(t) =

18 + t2

m3(t) =4


r =12 M = 2 M0 =M1 =

18 h =

14

(46)



le 18 + t2


∣∣∣ +

11 + t


∣∣∣



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))

minus sw3(1 + s2)

1 + t2 + s cos2(1 + t3s7w2(1 + s2))ds


le4r3t(1 +

radict + t2

)



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣

sv3(1 + s2)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

4r3

1 + t

intradict

0

s

1 + t2ds t isin R

+

le sup

t

4(1 + t)(1 + t2) t isin R

+

le 18=M1

sup

intα(t)


= sup

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

r3intradic

t

0

s

1 + t2ds t isin R

+

=132

lt M

(47)


Acknowledgments


References


































1 Introduction


Dq




i=1

aiu(nminus2)(ηi

) (11)



Tu(t) =int t


int1

0H(t s)u(s)ds (12)





Dq


)= 0 0 lt t lt 1


where Dq



Dq



i=1

aiu(ηi)

(14)



Dq



0+u(1) = 0 1 le p le n minus 2(15)



CDq



(η2)

(16)


where CDq




)= 0 2 lt q lt 3 0 lt p lt 1

u(0) = 0 uprime(0) = uprime(1) = 0(17)


Dq

0+u(t) + a(t)f(t u(t)) = 0 0 lt t lt 1


i=1

ζiu(ηi) (18)




CDqu(t) = f(t u(t)) u(0) = u0 (19)



Dq



(110)












αC(D) = maxtisinI

α(D(t)) α

(int

I

u(t)dt u isin D)

leint

I

α(D(t))dt (21)




Iq

0+y(t) =1

Γ(q)int t



given by

Dq

0+y(t) =1

Γ(n minus q)

(d

dt

)n int t



+


Dq

0+y(t) = 0 (24)




0+y isinL(0 1)

Iq

0+Dq







3 Main Results





2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

lt 1 (32)

such that

α(f(tD1 D2 Dn+1)

) len+1sum

k=1



i=1 aiηqminusn+1i





i=1

aix(ηi)

(34)


is

x(t) =int1

0G(t s)y(s)ds (35)

where


i=1 aig(ηi s)


g(t s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩



) t le s(37)




x(t) = minusint t

0





c1 =1


qminusn+1i

int1

0



i=1 ai


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

(310)


x(t) = minusint t

0


)y(s)ds +tqminusn+1


int1

0


)y(s)ds

minussummminus2

i=1 aitqminusn+1


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

=int1

0G(t s)y(s)ds

(311)







ρ0(s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


s isin (0 ξ]

(γ

s


(312)











i=1 aig(ηi s)

ηlowast (314)



mminus2sum

i=1

aig(ηi s)= G(s)

le tqminusn+1

Γ(q minus n + 2

) +summminus2

i=1 aiηqminusn+1i



(315)




(Inminus20+ x(s)

) S(Inminus20+ x(s)

))= θ


i=1

aix(ηi)

(316)






(317)


(Ax)(t) =int1

0G(t s)f


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))ds 0 le t le 1 (318)



(Ax)(t) leint1

0G(s)f


(Inminus20+ xj(s)


))ds

=

(int γ

0+intδ

γ

+int1

δ

)


(Inminus20+ xj(s)


))ds

le 3intδ

γ


(Inminus20+ xj(s)


))ds

(319)


mintisin[γ δ]


int1

0G(t s)f


(Inminus20+ xj(s)


))ds

geintδ

γ

(

g(t s) +tqminusn+1

ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds


geintδ

γ

(


ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds

ge λ

intδ

γ


(Inminus20+ xj(s)


))ds

ge λ

3(Ax)


(320)




∥∥ =int1

0G(t s)


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds

(321)



(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ ltε

Δ

(322)





on I





(Inminus20+ x(s)

) S(Inminus20+ x(s)

))



(Inminus20+ x(s)

) S(Inminus20+ x(s)

))




(Inminus20+ D

)(I) times S

(Inminus20+ D

)(I)))


k=1



(T(Inminus20+ D

)(I))


)(I))

(324)

which implies


k=1





(S(Inminus20+ D

)(I))

(325)

Obviously


)(I) = α

(int s

0


)


(326)

α(T(Inminus20+ D

))(I) = α

(int t

0K(t s)

(ints

0


u(τ)dτ

)


le alowast


(n minus 3)α(D(I))

(327)

α(S(Inminus20+ D

))(I) = α

(int1

0H(t s)

(int s

0


u(τ)dτ

)


le blowast


(n minus 3)α(D(I))

(328)



α(D(I)) le 2αC(D) (329)



nminus2sum

k=1




)

αC(D) (330)







nminus1sum

k=1






nminus1sum

k=1



NMβ

int1







k=1

uk ge r1 (334)


Therefore




k=1

uk ge r1 (335)

Take


(336)


x(t) ge λ

3NxΩ ge λ

3Nr0 gt r1 (337)

Hence

(Ax)(t) geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



xΩ(intδ

γ

G(s)c1(s)ds

)

middot ulowast

=1


(intδ

γ

G(s)c1(s)dsulowast)



(338)

and consequently






k=1

uk le r2 (340)






k=1

uk le r2 (341)

Choose

0 lt r lt min

⎧⎨

⎩

(nminus2sum

k=0

1k

)minus1r2 β

⎫⎬

⎭ (342)


(Ax)(ξ) =int1

0G(ξ s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(ξ s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



x(s)Ωintδ

γ


=1


x(s)Ω(intδ

γ




(343)

which implies


that is




(Ax)(t) leint1

0G(s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds (346)


∥∥f(t u1 un+1)


Therefore





(H7)

h(u1 un+1)sumn+1


n+1sum

k=1





k=1


k=1

uk le r3 (350)


ε0 =

(

N

nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

int1

0G(s)a(s)ds

)minus1 (351)




(Ax)(t) leNint1


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))∥∥∥ds

leNint1

0G(s)a(s)h


∥∥∥ x(s)



)(s)∥∥∥)ds

leNε0

int1

0G(s)a(s)

(nminus1sum

k=1



)(s)∥∥∥ +∥∥∥S(Inminus20+ x

)(s)∥∥∥

)

ds

leNε0

(nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

)

rlowastint1


(352)

and consequently




4 An Example



960k3

⎧⎪⎨

⎪⎩

⎡


j=1

u2j(t) +infinsum

j=1

uprimej(t)

⎤

⎦

3

+

⎛


j=1

uj(t) +infinsum

j=1

uprime2j(t)

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5

(int t


)23

+1 + t2

24k4

(int1


)

t isin I


(14

)+12uprimek

(49

) k = 1 2 3

(41)






fk(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭+1 + t3

36k5x15k

+1 + t2

24k4y2k

(42)


k=1(1k3) lt 32 we get by (42)

∥∥f(t u v x y

)∥∥ =infinsum

k=1

∣∣fk(t u v x y

)∣∣

le (1 + t)3

2

(140

(u + v)3 + 1160

(u + v)12 + 112

x15 + 18∥∥y∥∥)

(43)


h(u v x y

)=

140

(u + v)3 +1160

(u + v)12 +112x15 +

18y (44)



1 f(1)k ) and f (2) =

(f (2)1 f

(2)k ) in which

f(1)k

(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5x15k (k = 1 2 3 )

f(2)k

(t u v x y

)=

1 + t2

24k4y2k (k = 1 2 3 )

(45)



α(f (1)(tD1 D2 D3 D4)


α(f (2)(ID1 D2 D3 D4)

)le 1


(46)


α(f(ID1 D2 D3 D4)


2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

asymp 01565 lt 1(47)



fk(t u v x y

) ge (1 + t)3


fk(t u v x y

) ge (1 + t)3

960k3


(48)


c1(t) =(1 + t)3

960 h1 k(u v) = (u + v)3 ulowast =

(1

1k3

) (49)

in this situation


(u + v)3u + v = infin (410)


c2(t) =(1 + t)3

960 h2 k(u v) =


(1

1k3

) (411)

in this situation

h2k = limu+vrarr 0




NMβ

int1

0G(s)a(s)ds asymp 07287 lt β = 1 (413)


Acknowledgments


References











































1 Introduction






∣∣f(t)


(11)


G(u) = S[f(t)u

]

=intinfin


(12)





f(t) =infinsum

n=0

antn (13)

then

F(u) =infinsum

n=0

nantn (14)





(15)







Gn(u) =G(u)un

minusnminus1sum

k=0

f (k)(0)unminusk

(16)



LU + RU +NU = h(x t) (17)


S[LU] + S[RU] + S[NU] = S[h(x t)] (18)



unU(x 0) minus 1


u+ S[RU] + S[NU] = S[h(x t)]

(19)




U(x t) =infinsum

i=0

Ui(x t) (111)



NU(x t) =infinsum

i=0

Ai (112)


Ai =1idi

dλi

[

Ninfinsum

i=0

λiUi

]

λ=0

i = 0 1 2 (113)


S

[ infinsum

i=0

Ui(x t)

]


minus unS[ infinsum

i=0

Ai

]

+ unS[h(x t)]

(114)









U0(x t) = K(x t)



Ψ = Φ + Sminus1[h(x t)] (119)







LU + RU +NU = h(x t) (122)


U(x 0) = f(x) Ut(x 0) = g(x) (123)





]+ Sminus1

[u2S[h(x t)]

] (125)

2 Applications




U(x 0) = 0

Ut(x 0) = ex(22)




] (23)



[U2x minusU2

]](24)


U(x t) =infinsum

i=0

Ui(x t) (25)


infinsum

i=0


u2S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

(26)


infinsum

i=0

Ai(U) = U2x

infinsum

i=0

Ai(U) = U2

(27)


A0(U) = U20x A1(U) = 2U0xU1x Ai(U) =

isum

r=0

UrxUiminusrx

B0(U) = U20 B1(U) = 2U0U1 Bi(U)

isum

r=0

UrUiminusr

(28)


U0(x t) = tex


S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

n ge 0(29)



U1(x t) = Sminus1[

u2S

[ infinsum

i=0

A0(U) minusinfinsum

i=0

B0(U)

]]

= Sminus1[u2S

[U2

0x minusU20

]]= Sminus1

[u2S

[t2e2x minus t2e2x

]]

= Sminus1[u2S[0]

]= 0

(210)


U(x t) =infinsum

i=0

Ui(x t) = tex (211)


Ut

(x y t

) minus VxWy = 1

Vt(x y t

) minusWxUy = 5

Wt

(x y t

) minusUxVy = 5

(212)


U(x y 0

)= x + 2y

V(x y 0

)= x minus 2y

W(x y 0

)= minusx + 2y

(213)


U(x y u

)= x + 2y + u + uS

[VxWy

]

V(x y u


[WxUy

]

W(x y u


[UxVy

]

(214)


U(x y t


[VxWy

]]

V(x y t


[WxUy

]]

U(x y t


[UxVy

]]

(215)



U0(x y t

)= t + x + 2y

Ui+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ci(VW)

]]

i ge 0

V0(x y t

)= 5t + x minus 2y

Vi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Di(UW)

]]

i ge 0

W0(x y t

)= 5t minus x + 2y

Wi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ei(UV )

]]

i ge 0

(216)


C0(VW) = V0xW0y


Ci(VW) =isum

r=0

VrxWiminusry

D0(UW) = U0yW0x


Di(UW) =isum

r=0

WrxUiminusry

E0(UV ) = U0xV0y


Ei(VW) =isum

r=0

UrxViminusry

(217)



U1(x y t


[V0xW0y

]]= Sminus1[uS[(1)(2)]] = 2t

V1(x y t


[W0xU0y


W1(x y t


[U0xV0y


U2(x y t


[V1xW0y + V0xW1y

]]= 0

V2(x y t


[U0yW1x +U1yW0x

]]= 0

W2(x y t


[U1xV0y +U0xV1y

]]= 0

(218)


U(x y t

)=

infinsum

i=0

Ui

(x y t

)= x + 2y + 3t

V(x y t

)=

infinsum

i=0

Vi(x y t

)= x minus 2y + 3t

W(x y t

)=

infinsum

i=0

Wi

(x y t

)= minusx + 2y + 3t

(219)





U(x 0) = 1 + ex V (x 0) = minus1 + ex (221)




Ux(x t) =infinsum

i=0


i=0

Vxi(x t) (223)



S

[ infinsum

i=0

Ui(x t)

]


[ infinsum

i=0

Vix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

S

[ infinsum

i=0

Vi(x t)

]


i=0

Uix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

(224)






(225)






U2 = Sminus1[u2ex

]=t2

2ex

V2 = Sminus1[u2ex

]=t2

2ex

(226)


U(x t) = 1 + ex(

1 + t +t2

2+t3


)


1 minus t + t2

2minus t3


)

(227)


U(x t) = 1 + ex+t




Ut + VUx +U = 1



U(x 0) = ex V (x 0) = eminusx (230)





S

[ infinsum

i=0

Ui(x t)

]


i=0

Ai

]

minus uS[ infinsum

i=0

Ui

]

S

[ infinsum

i=0

Vi(x t)

]

= eminusx + u + uS

[ infinsum

i=0

Bi

]

minus uS[ infinsum

i=0

Vi

]

(232)


A0 = U0xV0 A1 = U0xV1 +U1xV0


B0 = V0xU0 B1 = V0xU1 + V1xU0

B2 = V0xU2 + V1xU1 + V2xU0

(233)



U0 = ex + t






2minus tex minus t2

2ex



2eminusx

U2 =t2

2+t2


V2 =t2

2+t2


(236)


U(x t) = ex(

1 minus t + t2

2minus t3


)

V (x t) = eminusx(

1 + t +t2

2+t3


)

(237)



3 Conclusion



Acknowledgment


References


























1 FK and BK Spaces







infink=0 let x

[n] =sumn






suminfinn=0 λnb





suminfink=0 xke






dw(x y

)=

infinsum

k=0

12k



∣∣

)

(x y isin w)

(11)








sumnk=0 xke





Cs =

σ isin C infinsum

n=1

cn(σ) le s

(12)



Φ =φ =

(φk



m(φ)=


supσisinCs

(1φs

sum

kisinσ|xk|

)

ltinfin

n(φ)=


( infinsum

k=1

|uk|Δφk)

ltinfin

(14)
















χ(Q1 +Q2) le χ(Q1) + χ(Q2)




Lχ = χ(L(SX)) (24)


Lχ = 0 (25)










1amiddot lim sup

nrarrinfin

(

supxisinQ

(I minus Pn)(x))


(

supxisinQ

(I minus Pn)(x))

(26)






(

supxisinQ

(I minus Pn)(x))

(27)



(

supxisinQ

sum

kgen|xk|p

)

(28)





Pr(z) = ze +rsum





(

supxisinQ



(

supxisinQ


(210)




supxisinQ

(

supuisinS(x)

( infinsum

n=k

|un|Δφn))

(211)








Lχ = χ(L(SX)) (212)

and we have












x(0) = x0 (32)




∥∥f(t x)∥∥ le Q + Rx (33)


μ(f(t Y )

) le q(t)μ(Y ) (34)









(

supxisinE

supkgen

|xk|)

(35)





xi(0) = x0i (37)





(i) x0 = (x0i ) isin c0



∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

(38)



Now let us denote

q(t) = supnge1

qn(t)

Q = suptisinI

q(t)


(39)




∣∣fn(t x)∣∣ =

∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

le P +Q supigekn


Hence we get



χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x)∣∣)

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x1 x2 )∣∣)

le limnrarrinfin

supxisinY

(

supigen


∣∣xj∣∣)

le limnrarrinfin

supigen


supxisinY

(

supigen

supjgekn

∣∣xj∣∣)

(312)


χ(f(t Y )

) le q(t)χ(Y ) (313)







infinsum

j=ki+1

aij(t)xj (314)


xi(0) = x0i (315)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (316)



i = 1 2










2 = f2(t x1 x2)


(317)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (318)



Then we have


∣∣f1(t x1)∣∣∣∣f2(t x1 x2)

∣∣ ∣∣fi(t x1 x2 xi)

∣∣


(319)


∣∣∣fi(t x)∣∣∣ile Pi(t) (320)













xi(0) = x0i (323)




supxisinE

supnmgep

|xn minus xm|


(

supxisinQ

supkgen

|xk|)

(324)







∣∣gn(t x1 x2 )∣∣ le bi (325)




a(t) = supige1

ai(t)

Q = suptisinI

a(t)(326)






f(t x) =(f1(t x) f2(tx)

)=(fi(t x)

)

(327)





∣∣


(328)



∣∣fi(t x)∣∣ le |ai(t)||xi| +

∣∣gi(t x)∣∣








(s y


(s y

)∣∣


∣∣ +


(s y

)∣∣


∣∣ +


(s y

)∣∣

(331)

Then


ige1


(s y

)∣∣


|ai(t) minus ai(s)|



(332)




∣∣gn(t x)∣∣ +

∣∣gm(t x)∣∣



μ(f(t E)

) le a(t)μ(E) (334)











∣∣fi(t x)

∣∣p =

∣∣fi(t x0 x1 x2 )

∣∣p le qi(t) + ri(t)|xi|p (335)







R = supri(t) t isin I i = 0 1 2 (336)



∥∥f(t x)∥∥pp =

infinsum

i=0


leinfinsum

i=0

[qi(t) + ri(t)|xi|p

]

leinfinsum

i=0

qi(t) +

(

supige0

ri(t)

)( infinsum

i=0|xi|p

)

le Q + Rxpp

(337)




χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(sum

igen

∣∣fi(t x1 x2 )∣∣p)

le limnrarrinfin

(sum

igenqi(t) +

(

supigen

ri(t)

)(sum

igen|xi|p

) )

(338)



χ(f(t Y )

) le r(t)χ(Y ) (339)








xi(0) = x0i (341)






∣∣fi(t x0 x1 x2 )∣∣p le bi(t) (342)










∣∣fi(t u)

∣∣ =

∣∣fi(t u1 u2 )

∣∣ le pi(t) + qi(t)|ui| (343)








(344)




uisinS(x)

infinsum

i=1


le supuisinS(x)

infinsum

i=1

[pi(t) + qi(t)|ui|

]Δφi

leinfinsum

i=1

pi(t)Δφi +

(

supi

qi(t)

)(

supuisinS(x)

infinsum

i=1

|ui|Δφi)

le P +Qxn(φ)

(345)




χ(f(t X)

)= lim

krarrinfinsupxisinX

(

supuisinS(x)

( infinsum

n=k


))

le limkrarrinfin

( infinsum

n=k

pn(t)Δφn +

(

supngek

qn(t)

)(

supuisinS(x)

infinsum

n=k

|un|Δφn))

(346)



χ(f(t X)

) le q(t)χ(X) (347)




References























(c)α l












































1 Introduction






(11)



) (12)


with

x(t+k)= lim

hrarr 0+x(tk + h) x

(tminusk)= lim










2 Preliminaries








Iα0h(t) =1

Γ(α)

int t




CDα0+h(t) =

1Γ(n minus α)

int t




CDαh(t) = 0 (23)

has solutions










Then either






x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

0



tk



i=1

int ti

timinus1


h(s)ds


i=1

int ti

timinus1


h(s)ds


i=1

(tk minus ti)int ti

timinus1


h(s)ds + Ω1(t)intT

tk


h(s)ds

+Ω2(t)intT

tk


h(s)ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


+kminus1sum

i=1


(26)




(x(tminusk))


(27)

where

Λ1 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(T minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1




Λ2 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1



minus ct

TΔ



Δ

(28)



0



xprime(t) =int t

0


h(s)ds minus c1(29)


x(t) =int t

t1



xprime(t) =int t

t1




x(tminus1)=int t1

t0




t0



(211)

Observing that





then we have

minusd0 =int t1

t0



minusd1 =int t1

t0



(213)


x(t) =int t

t1


h(s)ds +int t1

t0


h(s)ds


t0




xprime(t) =int t

t1


h(s)ds +int t1

t0



(214)


x(t) =int t

t2



xprime(t) =int t

t2




x(tminus2)=int t2

t1


h(s)ds +int t1

t0


h(s)ds


t0




x(t+2)= minus e0


t1


h(s)ds +int t1

t0




(216)






thus we have



(218)


x(t) =int t

t2


h(s)ds +int t1

t0


h(s)ds +int t2

t1


h(s)ds


t0


h(s)ds


t0


h(s)ds +int t2

t1


h(s)ds

]

+ I1(x(tminus1))

+ I2(x(tminus2))



(219)


x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(t minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1



xprime(t) =

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1


(220)



we have




Δ+(1 minus b)TΛ2

Δ

minusc1 = minuscΛ1


Δ

(222)




3 Main Results



conditions


(x(tminusk))



Ωlowast1 = sup


2 = suptisinJ


tisinJ


2 = suptisinJ

∣∣Ωprime2(t)∣∣ (32)





k(u) minus Ilowast


k = 1 2 p


[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

lt 1

(33)



(Fx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds



i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk


f(s x(s))ds

+ Ω2(t)intT

tk


f(s x(s))ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(34)



tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




timesksum

i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



+ |1 + Ω1(t)|[

ksum

i=1




(y(tminusk))∣∣]


i=1


(y(tminusi))∣∣


i=1


(y(tminusi))∣∣

+kminus1sum

i=1


(y(tminusi))∣∣



[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3


(35)







(Ax)(t) = [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(Dx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds


i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk



tk


f(s x(s))ds

(36)


(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3 lt 1 (37)





tk



∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



∣∣ds


timesksum

i=1

int ti

timinus1



∣∣ds


i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



(38)




|(Dx)(t)| leint t

tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1




+ |Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + |Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

D(x) le M1Tα


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)= l

(310)


∣∣(Dy


∣∣ leint τ2

τ1



where


tk


∣∣f(s x(s))∣∣ds +


ksum

i=1

int ti

timinus1




1(t) + Ωprime2(t) + 1

∣∣ksum

i=1

int ti

timinus1



+∣∣Ωprime


i=1

(tk minus ti)int ti

timinus1



+∣∣Ωprime

1(t)∣∣intT

tk


∣∣f(s x(s))∣∣ds +


tk



le M1Tα

Γ(α + 1)Ωlowast

1

(1 + p + 2pα

)+M1T

αminus1

Γ(α)(1 + Ωlowast

2)(1 + p

)= L

(312)




(xλ


(313)



)(t) (314)



|x(t)| le λ

int t

tk


∣∣f(s x(s))

∣∣ds + λ|1 + Ω1(t)|

ksum

i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

(tk minus ti)int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + λ|Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

+ λ|1 + Ω1(t)|[

ksum

i=1

∣∣∣∣∣Ii

(x(tminusi)

λ

)∣∣∣∣∣+


(x(tminusk

)

λ

)∣∣∣∣∣

]


i=1


(x(tminusi)

λ

)∣∣∣∣∣


i=1


(x(tminusi)

λ

)∣∣∣∣∣+ λ

kminus1sum

i=1


(x(tminusi)

λ

)∣∣∣∣∣

x(t) le[M1T

α


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pM2 +M3

)+(Ωlowast

1T + Ωlowast2 + T

)pM3

]

(315)


4 An Example


CD32x(t) =

sin 2t|x(t)|(t + 5)2(1 + |x(t)|)

t isin [0 1] t =13

Δx(13

)=


(13

)=



(41)




[L1T

α


1

)(1 + p(1 + L2) + 2pα + L3

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

=464

1125radicπ

+173450

lt 1

(42)


Acknowledgments


References

































1 Introduction


intb

a

w(x)f(x)dx =nsum

k=0

wkf(xk) + Rn+1[f] (11)






Ψn+1(x) =nprod

k=0

(x minus xk) (12)


f(x) =nsum

k=0

f(xk)L(xxk) + rn+1(f x) (13)

where

L(xxk) =Ψn+1(x)


(k = 0 1 n) (14)

we obtain (11) with

wk =1

Ψprimen+1(xk)

intb

a


Rn+1[f]=intb

a

rn+1(f x)w(x)dx

(15)



intb

a

f(t)dt =b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)+ E(f) (16)



∥∥f∥∥p =

(intb

a

∣∣f(t)∣∣pdt

)1p

(17)


xisin[ab]

∣∣f(x)∣∣ (18)



∣∣∣∣∣

intb

a

f(x)g(x)dx

∣∣∣∣∣le ∥∥f∥∥1





λ(b minus a)intb

a



2λ(b minus a) f(a)

∣∣∣∣∣




)

(110)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (111)


K(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

t minus 5a + b6

t isin[aa + b2

]

t minus a + 5b6

t isin(a + b2

b

]

(112)


intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt (113)

maxtisin[ab]

|K(t)| = 13(b minus a) (114)


2 Main Results


m1 =int (5a+b)6

a

(t minus 5a + b

6

)β(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)α(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)β(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)α(t)dt

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt le

M1 =int (5a+b)6

a

(t minus 5a + b

6

)α(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)β(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)α(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)β(t)dt

(21)


intb

a


2

)dt

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus 12

(intb

a

K(t)(α(t) + β(t)

)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)

(22)



2


2 (23)




(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a


2

)dt


a


dt

=12

(int (a+b)2

a


5a + b6


intb

(a+b)2


a + 5b6


)

(24)


m1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)β(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)α(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)β(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)α(x)dx

M1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)α(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)β(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)α(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)β(x)dx

(25)




Special Case 1


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


144((β1 minus α1

)(a + b) + 2

(β0 minus α0

))

(26)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (27)



int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(


a

α(t)dt

)

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t) dt

leint (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

+b minus a3

(


a

α(t)dt

)

(28)

Proof Since

intb

a


6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)α(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)

(29)


so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


3

(


a

α(t)dt

)

(210)


Special Case 2


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)


a + b2

α1

))

(211)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)



int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(intb

a


le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

leint (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt +

b minus a3

(intb

a


(213)


Proof Since

intb

a


=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)β(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)

(214)

so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

int

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


)dt =

b minus a3

(int b

a


(215)


Special Case 3

If β(x) = β1x + β0 = 0 in (213) then

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 +

a + b2


b minus a)

(216)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 minus


) (217)


Acknowledgment


References





















1 Introduction









xx= 0

(11)













u(x t) = U(ξ) ξ = kx +ωt (22)



)= 0 (23)



U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0 (24)


d2G(ξ)dξ2

+ λdG(ξ)dξ

+ μG(ξ) = 0 (25)






3 Application


u(x t) = U(ξ)eiη

v(x t) = V (ξ)(31)




)U = 0 (32)

(ω2 minus k2


(U2)primeprime

= 0 (33)



)U = 0

(ω2 minus k2


(34)


U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0

V (ξ) =nsum

i=1

βi

(Gprime

G

)i+ β0 βn = 0

(35)




m + n = m + 2

2m = n + 2(36)


U(ξ) = α2(Gprime

G

)2

+ α1(Gprime

G

)+ α0 α2 = 0

V (ξ) = β2(Gprime

G

)2

+ β1(Gprime

G

)+ β0 β2 = 0

(37)




)= 0




)ω minus qα1



)= 0


)α2



)β2 minus k2


)= 0


)α2 minus α1

(minus2k2 + aβ2

)= 0


)= 0

(6k2 minus aβ2

)α2 = 0


(38)






λ = plusmn13

radic3ω2 minus 3k2

k2 μ =

16k2 minusω2

k4 p = minus1

2ω

k

q =124

24k4 + 30ω4 minus 54k2ω2


α1 = plusmn2radic

1ab


radic1ab


a β2 =

6k2

a

(39)


U(ξ) = 6k2radic

1ab

(Gprime

G

)2

plusmn 2

radic1ab

radic3ω2 minus 3k2

(Gprime

G

)

V (ξ) =6k2

a

(Gprime

G

)2

plusmn 2

radic3ω2 minus 3k2

a

(Gprime

G

)

(310)



uH(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2



⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vH(x t) =6k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(311)



2 asfollows

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312a)


⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312b)

while at C21 lt C

22 one can obtain

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312c)


⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312d)




uT(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2


⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vT(x t) =6k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(313)


1 gt C22 and C

21 lt C

22 as follows

uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314a)


ak2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314b)


|u H|

tx

1

05

0

minus054

20

minus2minus4 minus4

minus20

24


0

02

04

06

08

tx

42

0minus2

minus4 minus4minus2

02

4

|u H|2


uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314c)


ak2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314d)



0

200

400

600

800

tx

42

0minus2

minus4 minus4minus2

02

4

vH


4 Conclusions


Acknowledgments


References























Shengli Xie







1 Introduction



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J






(

t x(σ1(t))int t


)

t isin [0 b]

x(0) + g(x) = x0

(12)


(1 minus αβMc

)N


2αθ1L1M(1 + 2Kθ2L1) lt 1

(13)





2 Preliminaries








x(t) = R(t 0)[φ(0) + g(x)(0)

]+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

(21)




α

(int t

0xn(s)ds n isin N

)

le 2int t








3 Existence Result




+ such that

∥∥f(t x y z


∥∥y∥∥ + z






andint t0 k(t s)ds





α

(int t

0K(t s V1)ds

)


α(V1(θ)) t isin J

α

(intb

0H(t s V1)ds

)


α(V1(θ)) t isin J

(33)


w = maxMp0(1 + k0 + h0) + 1 2M

(l01 + 2l02k0 + 2l03h0

)+ 1 (34)






(Fx)(t)

=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(35)






0ew(sminust)

∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥ds

le L +Mint t

0ew(sminust)p(s)

(

xs[minusq0] +ints

0K(s r xr)dr +

intb

0H(s r xr)dr

)

ds

le L +Mp0

int t

0ew(sminust)

(

xs[minusq0] +int s


intb


)

ds


Consequently




(38)


∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥le p(s)(1 + k0 + h0)R (39)







0R(t minus s)α

(

f

(

s Vs

int s

0K(s r Vr)dr

intb

0H(s r Vr)dr

))

ds

le 2Mint t

0ew(sminust)

[



0K(s r Vr)dr

)

+l3(s)α

(intb

0H(s r Vr)dr

)]

ds

le 2Mint t

0ew(sminust)

[


α(V (s + θ)) + 2l02

int s

0k(s r) sup


+2l03

intb

0h(s r) sup


]

ds

le 2M(l01 + 2l02k0 + 2l03h0

)int t

0ew(sminust)ds sup


le 2M(l01 + 2l02k0 + 2l03h0


(311)

Consequently


α((FV )(t)) le 2M(l01 + 2l02k0 + 2l03h0






4 An Example




ut(t y)= a1

(t y)uyy(t y)

+ a2(t)

[

sinu(t + θ y

)ds +

int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

0 le t le b


0

intb

0F(r y)log(1 + |u(r s)|12


u(t 0) = u(t π) = 0 0 le t le b(41)


f

(

t ut

int t

0K(t s us)ds

intb

0H(t s us)ds

)(y)

= a2(t)

[

sinu(t + θ y

)+int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

K(t s us)(y)=ints

minusqa3(s + τ)

u(τ y)dτ

(1 + t) H(t s us)

(y)=

u(s + θ y

)

(1 + t)(1 + s)2

g(u)(y)=intπ

0

intb

0F(r y)log(1 + |u(r s)|12

)dr ds

(42)



(43)


∥∥f(t u v z)∥∥ le |a2(t)|

(u[minusq0] + v + z

)

K(t s u) leints


u[minusq0](1 + t)


(1 + t)(1 + s)

(44)



∥∥g(u)



∣∣F(r y)∣∣(u[0b] +

radicπ) (45)


α(f(t V1 V2 V3)

) le |a2(t)|(

supminusqleθle0

α(V1(θ)) + α(V2) + α(V3)

)

t isin J

α(K(t s V1)) le 11 + t

supminusqleθle0


α(H(t s V1)) le 1

(1 + t)(1 + s)2sup


(46)


5 An Application



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J



x(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(52)






w = maxMp0(1 + k0 + h0)(1 +MM1b) + 1 2M

(l01 + 2l02k0 + 2l03h0

)(1 +MM1b) + 1

(53)



Wv =intb

0R(b s)(Cv)(s)ds (54)





v(t) =Wminus1(

x1 minus R(b 0)[φ(0) + g(x)(0)

]

+intb

0R(b s)f

(

s xs

ints

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

)

(t) t isin [0 b]

(55)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(56)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t


[φ(0) + g(x)(0)

]

+intb

0R(b τ)f

(

τ xτ

int τ

0K(τ r xr)dr

intb

0H(τ r xr)dr

)

dτ

)

(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(57)




Acknowledgments


References




































Contents



































Acknowledgments













1 Introduction











119909 = 119865119909 (1)












119901= 119862(R

+ 119901(119905)) the


+and such that

sup |119909 (119905)| 119901 (119905) 119905 ge 0 lt infin (2)


to the norm

119909 = sup |119909 (119905)| 119901 (119905) 119905 ge 0 (3)


119901[10 15]



+and if lim

119879rarrinfinsup|119909(119905)|119901(119905) 119905 ge 119879 = 0





that is


(4)


119909 (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 (5)




+ 119886 R

+rarr


|119907 (119905 119904 119909)| le 119899 (119905 119904) + 119886 (119905) 119887 (119904) |119909| (6)



119871 (119905) = int

119905

0

119886 (119904) 119887 (119904) 119889119904 119905 ge 0 (7)


119901 where 119901(119905) = [119886(119905) exp(119872119871(119905) + 119905)]




119863119886(119905) exp(119872119871(119905)) for 119905 ge 0


int

119905

0

119899 (119905 119904) 119889119904 le 119873119886 (119905) exp (119872119871 (119905)) (8)

for 119905 ge 0



condition

119871 (120593 (119905)) minus 119871 (119905) le 119870 (9)



for all 119905 ge 0



119901such that |119909(119905)| le 119886(119905) exp(119872119871(119905))

for 119905 isin R+


119901by putting

(119865119909) (119905) = 119889 (119905) + int

119905

0

119907 (119905 119904 119909 (120593 (119904))) 119889119904 119905 ge 0 (10)


+






|(119865119909) (119905)| le |119889 (119905)| + int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904)))1003816100381610038161003816 119889119904

le 119863119886 (119905) exp (119872119871 (119905))

+ int

119905

119900

[119899 (119905 119904) + 119886 (119905) 119887 (119904)1003816100381610038161003816119909 (120593 (119904))

1003816100381610038161003816] 119889119904

le 119863119886 (119905) exp (119872119871 (119905)) + 119873119886 (119905) exp (119872119871 (119905))

+ 119886 (119905) int

119905

0

119887 (119904) 119886 (120593 (119904)) exp (119872119871 (120593 (119904))) 119889119904

le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) int

119905

0

119872119886 (119904) 119887 (119904)


le (119863 + 119873) 119886 (119905) exp (119872119871 (119905))

+ (1 minus 119863 minus 119873) 119886 (119905) exp (119872119871 (119905))

= 119886 (119905) exp (119872119871 (119905))

(11)





1003816100381610038161003816

le int

119905

0

1003816100381610038161003816119907 (119905 119904 119909 (120593 (119904))) minus 119907 (119905 119904 119910 (120593 (119904)))1003816100381610038161003816 119889119904

le 120573 (120576)

(12)



1003816100381610038161003816(119865119909) (119905) minus (119865119910) (119905)1003816100381610038161003816 [119886 (119905) exp (119872119871 (119905) + 119905)]

minus1

le |(119865119909) (119905)| +1003816100381610038161003816(119865119910) (119905)

1003816100381610038161003816

times [119886 (119905) exp (119872119871 (119905))]minus1


le 2119890minus119905

(13)


1003816100381610038161003816 119901 (119905) le 120576 (14)



119901


|(119865119909) (119905)| 119901 (119905) le 119890minus119905 (15)

This yields that

lim119879rarrinfin

sup |(119865119909) (119905)| 119901 (119905) 119905 ge 119879 = 0 (16)



|(119865119909) (119905) minus (119865119909) (119904)|

le |119889 (119905) minus 119889 (119904)|

+

10038161003816100381610038161003816100381610038161003816

int

119905

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

119904

0

119907 (119905 120591 119909 (120593 (120591))) 119889120591 minus int

119904

0

119907 (119904 120591 119909 (120593 (120591))) 119889120591

10038161003816100381610038161003816100381610038161003816

le 120584119879(119889 120576) + 120576max 119899 (119905 120591) + 119886 (120591) 119887 (120591)

times119901 (120593 (120591)) 0 le 120591 le 119905 le 119879

+ 119879120584119879(119907 (120576 119879 120572 (119879)))

(17)

where we denoted120584119879(119907 (120576 119879 120572 (119879)))

= sup |119907 (119905 119906 119907) minus 119907 (119904 119906 119907)| 119905 119904 isin [0 119879]

|119905 minus 119904| le 120576 119906 isin [0 119879] |119907| le 120572 (119879)

(18)



120584119879(119889 120576) =

lim120576rarr0











119909 (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

+ int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(19)



+)



supremum norm

119909 = sup |119909 (119905)| 119905 isin R+ (20)








that

1003816100381610038161003816119891 (119905 1199091 1199092 119909

119899) minus 119891 (119905 119910

1 1199102 119910

119899)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816

(21)


1 1199092 119909

119899) (1199101 1199102

119910119899) isin R119899


with 1198650= sup|119891(119905 0 0)| 119905 isin R

+


R+



1 2 119898)(iv) The function 120573 R




rarr R+and


1003816100381610038161003816119892 (119905 119904 1199091 119909

119898)1003816100381610038161003816 le 119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

10038161003816100381610038161199091198941003816100381610038161003816 (22)

for all 119905 119904 isin R+and (119909

1 1199092 119909


we assume that

lim119905rarrinfin

int

120573(119905)

0

119902 (119905 119904) 119889119904 = 0 lim119905rarrinfin

119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 = 0

(23)


1 1199072 R+

rarr R+defined by

the formulas

1199071(119905) = int

120573(119905)

0

119902 (119905 119904) 119889119904 1199072(119905) = 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904 (24)



119872119894= sup 119907

119894(119905) 119905 isin R

+ (119894 = 1 2) (25)


119896 = sum119899


119894(119894 = 1 2 119899) appear in

assumption (i)

(vi) 119896 + 1198981198722lt 1




globally attractive




119909 (119905) = (119876119909) (119905) 119905 isin R+ (26)


lim119905rarrinfin

(119909 (119905) minus 119910 (119905)) = 0 (27)




119903in the space119883 = 119861119862(R

+) centered at the


2)]


putting

(119860119909) (119905) = 119891 (119905 119909 (1205721(119905)) 119909 (120572

119899(119905)))

(119861119909) (119905) = int

120573(119905)

0

119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904))) 119889119904

(28)


119909 (119905) = (119860119909) (119905) + (119861119909) (119905) 119905 isin R+ (29)


+ Thus



119903into



+ we get

1003816100381610038161003816(119860119909) (119905) minus (119860119910) (119905)1003816100381610038161003816 le

119899

sum

119894=1

119896119894

1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816

le 1198961003817100381710038171003817119909 minus 119910

1003817100381710038171003817

(30)





1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816

le int

120573(119905)

0

[1003816100381610038161003816119892 (119905 119904 119909 (120574

1(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816

+1003816100381610038161003816119892 (119905 119904 119910 (120574

1(119904)) 119910 (120574

119898(119904)))

1003816100381610038161003816] 119889119904

le 2 (1199071(119905) + 119903119898119907

2(119905))

(31)


1(119905) + 119903119898119907

2(119905) le 1205762 for 119905 ge 119879


1003816100381610038161003816 le 120576 (32)


obtain

1003816100381610038161003816(119861119909) (119905) minus (119861119910) (119905)1003816100381610038161003816 le int

120573(119905)

0

120596119879

119903(119892 120576) 119889119904 le 120573

119879120596119879

119903(119892 120576)

(33)


120596119879

119903(119892 120576) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 1199092 119909

119898)

minus119892 (119905 119904 1199101 1199102 119910

119898)1003816100381610038161003816 119905 isin [0 119879]

119904 isin [0 120573119879] 119909119894 119910119894isin [minus119903 119903]

1003816100381610038161003816119909119894 minus 119910119894

1003816100381610038161003816 le 120576 (119894 = 1 2 119898)

(34)


119879]times[minus119903 119903]



119903


|(119861119909) (119905)| le 1199071(119905) + 119903119898119907

2(119905) (35)




1(119905) + 119903119898119907




|(119861119909) (119905)| le 120576 (36)


|(119861119909) (119905) minus (119861119909) (120591)|

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

1003816100381610038161003816119892 (119905 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

minus119892 (120591 119904 119909 (1205741(119904)) 119909 (120574

119898(119904)))

1003816100381610038161003816 119889119904

le

1003816100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

[119902 (119905 119904) + 119886 (119905) 119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816] 119889119904

1003816100381610038161003816100381610038161003816100381610038161003816

+ int

120573(120591)

0

120596119879

1(119892 120576 119903) 119889119904

le

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119902 (119905 119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 119886 (119905)119898119903

100381610038161003816100381610038161003816100381610038161003816

int

120573(119905)

120573(120591)

119887 (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

+ 120573119879120596119879

1(119892 120576 119903)

(37)

where we denoted

120596119879

1(119892 120576 119903) = sup 1003816100381610038161003816119892 (119905 119904 119909

1 119909

119898)

minus119892 (120591 119904 1199091 119909

119898)1003816100381610038161003816 119905 120591 isin [0 119879]

|119905 minus 119904| le 120576 119904 isin [0 120573119879]

100381610038161003816100381611990911003816100381610038161003816 le 119903

10038161003816100381610038161199091198981003816100381610038161003816 le 119903

(38)



|(119861119909) (119905) minus (119861119909) (120591)|

le 119902119879120584119879(120573 120576) + 119903119898119886

119879119887119879120584119879(120573 120576) + 120573

119879120596119879

1(119892 120576 119903)

(39)

where 119902119879

= max119902(119905 119904) 119905 isin [0 119879] 119904 isin [0 120573119879] 119886119879

=

max119886(119905) 119905 isin [0 119879] 119887119879

= max119887(119905) 119905 isin [0 120573119879]



of the functions 120573 = 120573(119905) 119892 = 119892(119905 119904 1199091 119909

119898) we infer

that 120584119879(120573 120576) rarr 0 and 120596119879





+)





|119909 (119905)| le |(119860119909) (119905)| +1003816100381610038161003816(119861119910) (119905)

1003816100381610038161003816

le1003816100381610038161003816119891 (119905 119909 (120572

1(119905)) 119909 (120572

119899(119905))) minus 119891 (119905 0 0)

1003816100381610038161003816

+1003816100381610038161003816119891 (119905 0 0)

1003816100381610038161003816

+ int

120573(119905)

0

119902 (119905 119904) 119889119904 + 1198981003817100381710038171003817119910

1003817100381710038171003817 119886 (119905) int

120573(119905)

0

119887 (119904) 119889119904

le 119896 119909 + 1198650+1198721+ 119898119903119872

2

(40)

Hence we obtain

119909 le1198650+1198721+ 119898119903119872

2

1 minus 119896 (41)


2)(1 minus





1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le

119899

sum

119894=1

119896119894max 1003816100381610038161003816119909 (120572

119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 2int

120573(119905)

0

119902 (119905 119904) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119909 (120574119894(119904))

1003816100381610038161003816) 119889119904

+ 119886 (119905) int

120573(119905)

0

(119887 (119904)

119898

sum

119894=1

1003816100381610038161003816119910 (120574119894(119904))

1003816100381610038161003816) 119889119904

le 119896max 1003816100381610038161003816119909 (120572119894(119905)) minus 119910 (120572

119894(119905))

1003816100381610038161003816 119894 = 1 2 119899

+ 21199071(119905) + 119898 (119909 +

10038171003817100381710038171199101003817100381710038171003817) 1199072 (119905)

(42)


Hence we get


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

le 119896max1le119894le119899


1003816100381610038161003816119909 (120572119894(119905) minus 119910 (120572

119894(119905)))

1003816100381610038161003816


1199071(119905) + 119898 (119909 +


1199072(119905)


1003816100381610038161003816119909 (119905) minus 119910 (119905)1003816100381610038161003816

(43)



|119909(119905) minus








or






⋃

0le120582le1

119909 isin 119864 119909 = 120582119879119909 (44)























119864 120583(119883) = 0 is nonempty


(2119900) 119883 sub 119884 rArr 120583(119883) le 120583(119884)

(3119900) 120583(119883) = 120583(Conv119883) = 120583(119883)


(5119900) If (119883


119864such that

119883119899+1

sub 119883119899for 119899 = 1 2 and if lim

119899rarrinfin120583(119883119899) = 0


= ⋂infin

119899=1119883119899is nonempty



119899) for any 119899 = 1 2 This

















Next let us define

120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883

1205960(119883) = lim

120576rarr0

120596 (119883 120576) (46)



0(120582119883) =

|120582|1205960(119883) and 120596

0(119883 + 119884) le 120596

0(119883) + 120596

0(119884) provided 119883119884 isin

M119862(119868)



119909 (119905) = 1198911(119905 119909 (119905) 119909 (119886 (119905))) + (119866119909) (119905) int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(47)



has the form

1198912(119905 119904) = 119896 (119905 119904) 119892 (119905 119904) (48)







2

1003816100381610038161003816 10038161003816100381610038161199101 minus 119910

2

1003816100381610038161003816

(49)



0(119866119883) le 119902120596


119862(119868)



+rarr R+



+rarr R


119909 isin 119862(119868)(vi) 119891



+occurring in



1lt 1199052and


10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

le 120576

int

1199052

1199051

119892 (1199052 119904) 119889119904 le 120576

(50)



ℎ (119905) = int

119905

0

119892 (119905 119904) 119889119904 (51)





119896 = sup |119896 (119905 119904)| (119905 119904) isin Δ

1198911= sup 10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816 119905 isin 119868

ℎ = sup ℎ (119905) 119905 isin 119868

(52)





119901119903 + 1198911+ 119896ℎ120593 (119903)Ψ (119903) le 119903 (53)

such that 119901 + 119896ℎ119902Ψ(1199030) lt 1


(1198651119909) (119905) = 119891

1(119905 119909 (119905) 119909 (120572 (119905)))

(1198652119909) (119905) = int

119905

0

1198912(119905 119904) (119876119909) (119904) 119889119904

(119865119909) (119905) = (1198651119909) (119905) + (119866119909) (119905) (119865

2119909) (119905)

(54)



+by the formulas

119872(120576) = sup10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119892)] 119889119904

10038161003816100381610038161003816100381610038161003816

1199051 1199052isin 119868

1199051lt 1199052 1199052minus 1199051le 120576

(55)

119873(120576) = supint1199052

1199051

119892 (1199052 119904) 119889119904 119905

1 1199052isin 119868 1199051lt 1199052 1199052minus 1199051le 120576

(56)






1003816100381610038161003816(1198651119909) (119905) minus (1198651119910) (119905)

1003816100381610038161003816


1003816100381610038161003816

(57)


10038171003817100381710038171198651119909 minus 11986511199101003817100381710038171003817 le 119901

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (58)







1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le

10038161003816100381610038161003816100381610038161003816

int

1199052

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199052 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

2 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

10038161003816100381610038161003816100381610038161003816

int

1199051

0

119896 (1199051 119904) 119892 (119905

2 s) (119876119909) (119904) 119889119904

minusint

1199051

0

119896 (1199051 119904) 119892 (119905

1 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le int

1199052

1199051

1003816100381610038161003816119896 (1199052 119904)1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199052 119904) minus 119896 (1199051 119904)

1003816100381610038161003816 119892 (1199052 119904) |(119876119909) (119904)| 119889119904

+ int

1199051

0

1003816100381610038161003816119896 (1199051 119904)1003816100381610038161003816

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 |(119876119909) (119904)| 119889119904

le 119896Ψ (119909)119873 (120576) + 1205961(119896 120576) Ψ (119909) int

1199052

0

119892 (1199052 119904) 119889119904

+ 119896Ψ (119909) int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904

(59)

where we denoted

1205961(119896 120576) = sup 1003816100381610038161003816119896 (1199052 119904) minus 119896 (119905

1 119904)

1003816100381610038161003816

(1199051 119904) (119905

2 119904) isin Δ

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576

(60)


int

1199051

0

1003816100381610038161003816119892 (1199052 119904) minus 119892 (119905

1 119904)

1003816100381610038161003816 119889119904 =

10038161003816100381610038161003816100381610038161003816

int

1199051

0

[119892 (1199052 119904) minus 119892 (119905

1 119904)] 119889119904

10038161003816100381610038161003816100381610038161003816

(61)


120596 (1198652119909 120576) le 119896Ψ (119909)119873 (120576) + ℎΨ (119909) 120596

1(119896 120576)

+ 119896Ψ (119909)119872 (120576)

(62)










1003816100381610038161003816(119865119909) (119905) minus (1198651199090) (119905)

1003816100381610038161003816

le1003816100381610038161003816(1198651119909) (119905) minus (119865

11199090) (119905)

1003816100381610038161003816

+1003816100381610038161003816(119866119909) (119905) (1198652119909) (119905) minus (119866119909

0) (119905) (119865

21199090) (119905)

1003816100381610038161003816

le10038171003817100381710038171198651119909 minus 119865

11199090

1003817100381710038171003817 + 1198661199091003816100381610038161003816(1198652119909) (119905) minus (119865

21199090) (119905)

1003816100381610038161003816

+100381710038171003817100381711986521199090

1003817100381710038171003817

1003817100381710038171003817119866119909 minus 1198661199090

1003817100381710038171003817

(63)



21199090) (119905)

1003816100381610038161003816

le int

119905

0

|119896 (119905 119904)| 119892 (119905 119904)1003816100381610038161003816(119876119909) (119904) minus (119876119909

0) (119904)

1003816100381610038161003816 119889119904

le 119896 (int

119905

0

119892 (119905 119904) 119889119904)1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

le 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

(64)

Similarly we obtain

1003816100381610038161003816(11986521199090) (119905)1003816100381610038161003816 le 119896 (int

119905

0

119892 (119905 119904) 119889119904)10038171003817100381710038171198761199090

1003817100381710038171003817 le 119896ℎ10038171003817100381710038171198761199090

1003817100381710038171003817 (65)

and consequently

1003817100381710038171003817119865211990901003817100381710038171003817 le 119896ℎ

10038171003817100381710038171198761199090

1003817100381710038171003817 (66)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 1199011003817100381710038171003817119909 minus 119909

0

1003817100381710038171003817 + 119866119909 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎ1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

10038171003817100381710038171198761199090

1003817100381710038171003817

(67)


1003817100381710038171003817119865119909 minus 1198651199090

1003817100381710038171003817 le 119901120576 + 120593 (10038171003817100381710038171199090

1003817100381710038171003817 + 120576) 119896ℎ1003817100381710038171003817119876119909 minus 119876119909

0

1003817100381710038171003817

+ 119896ℎΨ (10038171003817100381710038171199090

1003817100381710038171003817)1003817100381710038171003817119866119909 minus 119866119909

0

1003817100381710038171003817

(68)






|(119865119909) (119905)| le10038161003816100381610038161198911 (119905 119909 (119905) 119909 (119886 (119905))) minus 119891

1(119905 0 0)

1003816100381610038161003816

+10038161003816100381610038161198911 (119905 0 0)

1003816100381610038161003816

+ |119866119909 (119905)|

10038161003816100381610038161003816100381610038161003816

int

119905

0

119896 (119905 119904) 119892 (119905 119904) (119876119909) (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909)

(69)

Hence we obtain

119865119909 le 1198911+ 119901 119909 + 119896ℎ120593 (119909)Ψ (119909) (70)



the ball 1198611199030




1199030


1 1199052isin 119868with


obtain1003816100381610038161003816(119865119909) (1199052) minus (119865119909) (119905

1)1003816100381610038161003816

le10038161003816100381610038161198911 (1199052 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199052) 119909 (119886 (119905

2)))

1003816100381610038161003816

+10038161003816100381610038161198911 (1199051 119909 (119905

2) 119909 (119886 (119905

2)))

minus1198911(1199051 119909 (1199051) 119909 (119886 (119905

1)))

1003816100381610038161003816

+1003816100381610038161003816(1198652119909) (1199052)

1003816100381610038161003816

1003816100381610038161003816(119866119909) (1199052) minus (119866119909) (1199051)1003816100381610038161003816

+1003816100381610038161003816(119866119909) (1199051)

1003816100381610038161003816

1003816100381610038161003816(1198652119909) (1199052) minus (1198652119909) (1199051)1003816100381610038161003816

le 1205961199030

(1198911 120576) + 119901max 1003816100381610038161003816119909 (119905

2) minus 119909 (119905

1)1003816100381610038161003816

1003816100381610038161003816119909 (119886 (1199052)) minus 119909 (119886 (119905

1))1003816100381610038161003816

+ 119896ℎΨ (1199030) 120596 (119866119909 120576) + 120593 (119903

0) 120596 (119865

2119909 120576)

(71)

where we denoted

1205961199030

(1198911 120576) = sup 10038161003816100381610038161198911 (1199052 119909 119910) minus 119891

1(1199051 119909 119910)

1003816100381610038161003816 1199051 1199052 isin 119868

10038161003816100381610038161199052 minus 1199051

1003816100381610038161003816 le 120576 119909 119910 isin [minus1199030 1199030]

(72)


120596 (119865119909 120576) le 1205961199030

(1198911 120576) + 119901max 120596 (119909 120576) 120596 (119909 120596 (119886 120576))

+ 119896ℎΨ (1199030) 120596 (119866119909 120576)

+ 120593 (1199030) Ψ (1199030) [119896119873 (120576) + ℎ120596

1(119896 120576) + 119896119872 (120576)]

(73)





1205960(119865119883) le 119901120596

0(119883) + 119896ℎΨ (119903

0) 1205960(119866119883) (74)


1205960(119865119883) le (119901 + 119896ℎ119902Ψ (119903

0)) 1205960(119883) (75)







1199091015840= 119891 (119905 119909) (76)


119909 (0) = 1199090 (77)









rarr R+(or 119908 119869

0times

R+



1003817100381710038171003817 le 119908 (1199051003817100381710038171003817119909 minus 119910

1003817100381710038171003817) (78)




1015840= 119908(119905 119906)


119905

0119908(119904 119906(119904))119889119904 for 119905 isin 119869


lim119905rarr0

119906(119905)119905 = lim119905rarr0






120583 (119891 (119905 119883)) le 119908 (119905 120583 (119883)) (79)





120583


119864120583= 119909 isin 119864 119909 isin ker 120583 (80)



120583is a closed



(6119900) 120583(119883 + 119884) le 120583(119883) + 120583(119884)

(7119900) 120583(120582119883) = |120582|120583(119883) for 120582 isin R








120583 (1199090+ 119891 (119905 119883)) le 119908 (119905 120583 (119883)) (81)



1198690times R+



119906 (119905) le int

119905

0

119908 (119904 119906 (119904)) 119889119904 (119905 isin 1198690) (82)


119906(119905)119905 = lim119905rarr0




0)minus119891(119904 119909)||



120583for 119905 isin 119869



0isin



120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (83)





120583 (119891 (119905 119883)) le 119901 (119905) 120583 (119883) (84)





1199091015840

119899= 119886119899(119905) 119909119899+ 119891119899(119909119899 119909119899+1

) (85)


119909119899(0) = 119909

119899

0(86)


119909119899

0= 119886(119886 isin R)






119899rarrinfin120572119899= 0 and |119891

119899(119909119899 119909119899+1

)| le

120572119899for 119899 = 1 2 and for all 119909 = (119909

1 1199092 1199093 ) isin 119897






119899)|| =

sup|119909119899| 119899 = 1 2


1(119905) 1199092(119905) )






sup119909isin119883

1003816100381610038161003816119909119899 minus 1198861003816100381610038161003816 (87)




120583 (1199090+ 119891 (119905 119883))


sup119909isin119883

1003816100381610038161003816119909119899

0+ 119886119899(119905) 119909119899

+119891119899(119909119899 119909119899+1

) minus 1198861003816100381610038161003816


sup119909isin119883

[1003816100381610038161003816119886119899 (119905)

1003816100381610038161003816

10038161003816100381610038161199091198991003816100381610038161003816

+1003816100381610038161003816119891119899 (119909119899 119909119899+1 ) + 119909

119899

0minus 119886

1003816100381610038161003816]


sup119909isin119883

119901 (119905)1003816100381610038161003816119909119899 minus 119886

1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119886119899 (119905)1003816100381610038161003816 |119886|


sup119909isin119883

1003816100381610038161003816119891119899 (119909119899 119909119899+1 )1003816100381610038161003816


sup119909isin119883

1003816100381610038161003816119909119899

0minus 119886

1003816100381610038161003816

(88)


Hence we get

120583 (1199090+ 119891 (119905 119883)) le 119901 (119905) 120583 (119883) (89)



References













































1015840























1 Introduction




u(0) = g(u)(11)







u(0) = g(u)

(12)







part



2 Preliminaries


+ = (0infin)and R





tisinI

∥∥f(t)∥∥X (21)



∥∥f∥∥L1(IX) =

int

I

∥∥f(s)∥∥Xds (22)







d

dtR(t)y = AR(t)y +

int t


= R(t)Ay +int t


(23)







u(0) = u0 isin X(24)








is verified





(27)
















ζ

(int t

0un(s)ds

infin

n=1

)

2int t





subeW such that




Sn = εn + Cn1ε

nminus1h + Cn2ε

nminus2h2


n n isin N (211)





)) n = 2 3 (212)


η(Fn0(S)) rη(S) (213)


3 Main Results





+ rarr R+ such that

∥∥f(t x)∥∥ m(t)Φ(x) (31)




ζ(f(t S)

) H(t)ζ(S) (32)





KgR +KΦ(R)int1

0m(s)ds R (33)








∥∥ +Kint1

0




(Fu)(t) ∥∥R(t)g(u)

∥∥ +

∥∥∥∥∥

int t


∥∥∥∥∥

KgR +KΦ(R)int1

0m(s)ds R

(36)





sub F(B) such that


(int t

0


infinn=1ds

)

+ ε (37)


ζ(F(B)(t)) 4Kint t

0ζ(f(s un(s))

infinn=1ds + ε

4Kint t

0H(s)ζ(un(s)nisinN

) ds + ε

4Kγ(B)int t

0H(s)ds + ε

(38)




0


0ϕ(s)ds

]

+ ε


(39)





ζ(F2(B)(t)

) 2ζ

(int t

0


infinn=1ds

)

+ ε

4Kint t

0ζf(swn(s))

infinn=1ds + ε

4Kint t

0H(s)ζ

(co(F1(B)(s)

))+ ε = 4K

int t

0H(s)ζ

(F1(B)(s)

)+ ε

(311)



ζ(F2(B)(t)

) 4K

int t

0

[∣∣H(s) minus ϕ(s)∣∣ + ∣∣ϕ(s)

∣∣](a + bs)γ(B)ds + ε


0


(

at +bt2

2

)

+ ε

a(a + bt) + b

(

at +bt2

2

)

+ ε (

a2 + 2bt +(bt)2

2

)

γ(B) + ε

(312)


ζ(F2(B)(t)

)

(

a2 + 2bt +(bt)2

2

)

γ(B) (313)


ζ(Fn(B)(t)) (

an + Cn1a

nminus1bt + Cn2a

nminus2 (bt)2


n

)

γ(B) (314)



γ(Fn(B)) (

an + Cn1a

nminus1b + Cn2a

nminus2 b2


n

)

γ(B) (315)


(

an0 + Cn01 a

n0minus1b + Cn02 a

n0minus2 b2


n0

)

= r lt 1 (316)





u(0) = g(u)

(317)













∣∣∣ C|a(λ)| (318)






1λa(λ)


(319)



such that




completely positive



lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (321)



u(t) =int t


int t

0f(s u(s)) + g(u) (322)


R(t)x = x +int t



d

dtR(t)x = AR(t)x +

int t




KgR +KΦ(R)int1





4 Applications


partw(t ξ)partt

= Aw(t ξ) +int t



w(0 ξ) =int1

0

int ξ

0qk(s ξ)w

(s y

)dsdy 0 ξ 2π

(41)



defined by



part







u(0) = g(u)

(43)






int10 k(s ξ)

2dsdξ)12







a(λ) =λ + α + βλ(λ + α)

(44)


f1(λ) =λ + α

λ + α + β f2(λ) =

λ2 + 2(α + β

)λ + αβ + α2

(λ + α + β

)2 (45)


f(n)1 (λ) =

(minus1)n+1β(n + 1)(λ + α + β

)n+1 f(n)2 (λ) =

(minus1)n+1β(α + β)(n + 1)

(λ + α + β








Re

(μ0 + iμ

b(μ0 + iμ

)+ 1

)

=μ30 + μ

20α + μ2

0

(α + β

)+ μ0α

(α + β

)+ μ0μ


)2 + 2μ0(α + β

)+ μ2

0 + μ2

(48)


(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1



∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1


∥∥∥∥

M

μ0 + iμ

∥∥∥∥ (410)


Therefore

lim|μ|rarrinfin

∥∥∥∥∥∥

1

b(μ0 + iμ

)+ 1

(μ0 + iμ

b(μ0 + iμ

)+ 1

minusA)minus1∥∥

∥∥∥∥= 0 (411)



int10 k(s ξ)

2dsdξ)12




qcKR +∥∥p1

∥∥1LK R (412)


Acknowledgments


References















































1 Introduction
















K(q)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2πqω1+2q

(2

2 + q

)1minus2q( Γ(1q)

Γ(12 + 1q

)

)2

if 1 le q lt +infin

4ω if q = +infin

(15)





0 lt a(t) le π2

ω2 t isin [0 ω] (16)





2 Preliminaries








u(t) =intω





Let





Let

G = min0letsleω


G(t s) σ = GG (26)





Th(t) =intω




Th(t) =intω

0G(t s)h(s)ds le G

intω

0h(s)ds (29)

and therefore

Th le Gintω

0h(s)ds (210)


Th(t) =intω

0G(t s)h(s)ds ge G

intω

0h(s)ds

=(GG

)middotGintω

0h(s)ds

ge σTh

(211)




Tφ = r(T)φ (212)


intω

0φ(t)dt = 1 (213)






hold




A = T F (216)











3 Main Results



mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x


mintisin[0ω]

f(t x0 x1 xn)x


maxtisin[0ω]

f(t x0 x1 xn)x

(31)









f(t x0 x1 xn) = b0(t)x20 + b1(t)x


n (32)






radic|xn| (33)



















This implies that





(39)





intω

0



intω

0u0(t)φ(t)dt (311)


intω

0


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(312)


λ1

intω


intω

0u0(x)φ(t)dt (313)


intω


intω

0φ(t)dt = σu0 gt 0 (314)


i(AK capΩ1 K) = 1 (315)







(u1(t) minus μ1

)= F(u1)(t) t isin R (317)



This means that




(320)




intω

0

[u



intω

0u1(t)φ(t)dt (322)


intω

0


]φ(t)dt =

intω

0u1(t)


]dt

= λ1

intω

0u1(x)φ(t)dt

(323)


λ1

intω


intω

0u1(x)φ(t)dt (324)



intω


intω

0φ(t)dt = σu1 gt 0 (325)


i(AK capΩ2 K) = 0 (326)


i(AK cap

(Ω2 Ω1









(u0(t) minus μ0

)= F(u0)(t) t isin R (329)



(330)




intω

0

[u



intω

0u0(t)φ(t)dt (332)



intω

0

[u


]φ(t)dt =

intω

0u0(t)


]dt

= λ1

intω

0u0(x)φ(t)dt

(333)


λ1

intω


intω

0u1(x)φ(t)dt (334)


i(AK capΩ1 K) = 0 (335)








(338)




intω

0

[u



intω

0u1(t)φ(t)dt (340)



intω

0

[u


]φ(t)dt =

intω

0u1(t)

[φ


]dt

= λ1

intω

0u1(x)φ(t)dt

(341)


λ1

intω


intω

0u1(x)φ(t)dt (342)


i(AK capΩ2 K) = 1 (343)


i(AK cap

(Ω2 Ω1



Acknowledgment


References

















York NY USA 1988










1 Introduction



x(t) = f

(


0u(t s x(c(s)))ds

)


x(t) = h

(

t x(t)intα(t)

0u(t s x(c(s)))ds

)





x(t) =int t



x(t) = L(t) +int t

0




x(t) =int t



x(t) = (Tx)(t)int t











0g(t s x







x(t) = f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f

(

t

intH(t)

0x(s)ds x(h(t))

)



x(t) = g(t x(t)) + f1

(

t

int t

0k(t s)f2(s x(s))ds

)



x(t) = f(t x(α(t)))

(

q(t) +intβ(t)

0u(t s x

(γ(s)))ds

)




2 Preliminaries












ωT

0 (X) = limεrarr 0

ωT (X ε)





diam X(t)

(22)







limtrarr+infin



Then

lim suptrarr+infin

ϕ(ψ(t))= lim sup












then









(212)




ωϕ(T)0 (X) (213)




∣∣∣ωT





∣∣∣ω




3 Main Results





limtrarr+infin


b(t) = +infin (31)






limtrarr+infin

sup

m3(t)intα(t)


= 0 (33)

sup

m3(t)intα(t)


leM0 (34)

sup

intα(t)


leM (35)








(Fx)(t) = f

(


0u(t s x(c(s)))ds

)



|(Fx)(t)| le∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minus f(t 0 0 0)∣∣∣∣∣+∣∣f(t 0 0 0)

∣∣


∣∣∣∣∣

intα(t)

0u(t s x(c(s)))ds

∣∣∣∣∣+ f



(310)





=

∣∣∣∣∣f

(


0u(t s x(c(s)))ds

)

minusf(

t y(a(t))(Hy)(b(t))

intα(t)

0u(t s y(c(s))

)ds

)∣∣∣∣∣




)(b(t))

∣∣

+m3(t)intα(t)

0




)(b(t))

∣∣

+ sup

m3(t)intα(t)



(312)

which yields that


+ sup

m3(t)intα(t)


forallt isin R+

(313)


lim suptrarr+infin

diam(FX)(t)



diam(HX)(b(t))


sup

m3(t)intα(t)




diam(HX)(t)

(314)

that is

lim suptrarr+infin



diam(HX)(t) (315)





∣∣u(p τ v




ωTr

(f ε g(r)

)= sup

∣∣f(p vw z



(316)






le∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣

+

∣∣∣∣∣f

(



)

minusf(



)∣∣∣∣∣


+m3(t)

[∣∣∣∣∣

intα(t)


∣∣∣∣∣+intα(s)


]

+ sup∣∣f(p vw z





∣∣u(p τ v




+ωTr

(f ε g(r)

)


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)

)

(318)

which implies that


b(T)(Hx bTε

)+M3Tω

T (α ε)uTr


Tr

(f ε g(r)


(319)



limεrarr 0


ωTr (u ε) = lim

εrarr 0ωTr

(f ε g(r)

)= 0 (320)


ωT0 (FX) leM1ω

a(T)0 (X) +M2ω

b(T)0 (HX) (321)


ω0(FX) leM1ω0(X) +M2ω0(HX) (322)



diam(FX)(t)



diam(HX)(t)

=M1μ(X) +M2

(ω0(HX) + lim sup


)

le (M1 +M2Q)μ(X)(323)



sup

m3(t)intα(t)


ltε



ωT (u δ r) ltε







=

∣∣∣∣∣f

(


0u(t s xn(c(s)))ds

)

minusf(


0u(t s x(c(s)))ds

)∣∣∣∣∣


+m3(t)intα(t)



+max

supτgtT

sup

m3(τ)intα(τ)

0


supτisin[0T]

sup

m3(τ)intα(τ)

0



2

ltε

2+max

ε

3 supτisin[0T]

m3(τ)intα(τ)

0ωT

(uδ02 r

)ds

ltε

2+max

ε

3


=5ε6

(327)

which yields that














M0 + h le r(1 minusM1) (331)




sup

m3(t)intα(t)





=

∣∣∣∣∣h

(

t z(t)intα(t)


)

minus h(

t y(t)intα(t)

0u(t τ y(c(τ))

)dτ

)∣∣∣∣∣


intα(t)

0



m3(t)intα(t)



(333)

which means that





4 Examples



x(t) =1

10 + ln3(1 + t3)+tx2(1 + 3t2

)

200(1 + t)+

1

10radic3 + 20t + 3|x(t3)|

+1

9 +radic1 + t

⎛

⎜⎝

int t2

0

s2x(4s) sin(radic



∣∣∣ x2(4s)ds

⎞

⎟⎠

3

forallt isin R+

(41)

Put

f(t vw z) =1

10 + ln3(1 + t3)+

tv2

200(1 + t)+

1

10radic3 + 20t + |w|

+z3

9 +radic1 + t

u(t s p

)=

s2p sin(radic



∣∣∣p2


(42)

Let

risin[9 minus radic

512

9 +

radic51

2

]

M=3 M0=9r10 M1=

r

100 M2=

1300

f =110

Q = 3 m1(t) =rt

100(1 + t) m2(t) =

1(10radic3 + 20t

)2 m3(t) =3M2

9 +radic1 + t


(43)



512

9 +

radic51

2

]



le t

200(1 + t)


∣∣∣ +

∣∣∣∣∣

1

10radic3 + 20t + |w|

minus 1

10radic3 + 20t +

∣∣q∣∣

∣∣∣∣∣

+1

9 +radic1 + t


∣∣∣ le m1(t)





sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)

minuss2w(4s) sin


)


∣∣∣w2(4s)


⎫⎪⎬

⎪⎭

le 3M2

9 +radic1 + t

int t2

0

2s2(1 + t12

)2

[r(1 + t12

)+ r3(r3t3 + 3r6 + 7

(t + 2t2

)2)]ds

le 2M2t6(9 +

radic1 + t)(

1 + t12)2

r(1 + t12

)+ r3[r3t3 + 3r6 + 7

(t + 2t2


sup

m3(t)intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=rM2

20lt M0

sup

intα(t)


= sup

⎧⎪⎨

⎪⎩

3M2

9 +radic1 + t

int t2

0

∣∣∣∣∣∣∣

s2v(4s) sin(radic



∣∣∣v2(4s)


⎫⎪⎬

⎪⎭

le sup

3M2

10

int t

0

s2ds

1 + t12 t isin R

+

=r

6leM

(44)




x(t) =3 + sin4

(radic1 + t2

)

16+x2(t)8 + t2

+1

1 + t

(intradict

0

sx3(1 + s2)

1 + t2 + s cos2(1 + t3s7x2(1 + s2))ds

)2

forallt isin R+

(45)

Put

h(t vw) =3 + sin4

(radic1 + t2

)

16+

v2

8 + t2+

w2

1 + t

u(t s p

)=

sp3

1 + t2 + s cos2(1 + t3s7p2

)

α(t) =radict c(t) = 1 + t2 m1(t) =

18 + t2

m3(t) =4


r =12 M = 2 M0 =M1 =

18 h =

14

(46)



le 18 + t2


∣∣∣ +

11 + t


∣∣∣



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))

minus sw3(1 + s2)

1 + t2 + s cos2(1 + t3s7w2(1 + s2))ds


le4r3t(1 +

radict + t2

)



sup

m3(t)intα(t)


= sup

4

1 + t

intradict

0

∣∣∣∣∣

sv3(1 + s2)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

4r3

1 + t

intradict

0

s

1 + t2ds t isin R

+

le sup

t

4(1 + t)(1 + t2) t isin R

+

le 18=M1

sup

intα(t)


= sup

intradict

0

∣∣∣∣∣sv3(1 + s2

)

1 + t2 + s cos2(1 + t3s7v2(1 + s2))ds


le sup

r3intradic

t

0

s

1 + t2ds t isin R

+

=132

lt M

(47)


Acknowledgments


References


































1 Introduction


Dq




i=1

aiu(nminus2)(ηi

) (11)



Tu(t) =int t


int1

0H(t s)u(s)ds (12)





Dq


)= 0 0 lt t lt 1


where Dq



Dq



i=1

aiu(ηi)

(14)



Dq



0+u(1) = 0 1 le p le n minus 2(15)



CDq



(η2)

(16)


where CDq




)= 0 2 lt q lt 3 0 lt p lt 1

u(0) = 0 uprime(0) = uprime(1) = 0(17)


Dq

0+u(t) + a(t)f(t u(t)) = 0 0 lt t lt 1


i=1

ζiu(ηi) (18)




CDqu(t) = f(t u(t)) u(0) = u0 (19)



Dq



(110)












αC(D) = maxtisinI

α(D(t)) α

(int

I

u(t)dt u isin D)

leint

I

α(D(t))dt (21)




Iq

0+y(t) =1

Γ(q)int t



given by

Dq

0+y(t) =1

Γ(n minus q)

(d

dt

)n int t



+


Dq

0+y(t) = 0 (24)




0+y isinL(0 1)

Iq

0+Dq







3 Main Results





2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

lt 1 (32)

such that

α(f(tD1 D2 Dn+1)

) len+1sum

k=1



i=1 aiηqminusn+1i





i=1

aix(ηi)

(34)


is

x(t) =int1

0G(t s)y(s)ds (35)

where


i=1 aig(ηi s)


g(t s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩



) t le s(37)




x(t) = minusint t

0





c1 =1


qminusn+1i

int1

0



i=1 ai


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

(310)


x(t) = minusint t

0


)y(s)ds +tqminusn+1


int1

0


)y(s)ds

minussummminus2

i=1 aitqminusn+1


qminusn+1i

intηi

0

(ηi minus s

)qminusn+1

Γ(q minus n + 2

)y(s)ds

=int1

0G(t s)y(s)ds

(311)







ρ0(s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


s isin (0 ξ]

(γ

s


(312)











i=1 aig(ηi s)

ηlowast (314)



mminus2sum

i=1

aig(ηi s)= G(s)

le tqminusn+1

Γ(q minus n + 2

) +summminus2

i=1 aiηqminusn+1i



(315)




(Inminus20+ x(s)

) S(Inminus20+ x(s)

))= θ


i=1

aix(ηi)

(316)






(317)


(Ax)(t) =int1

0G(t s)f


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))ds 0 le t le 1 (318)



(Ax)(t) leint1

0G(s)f


(Inminus20+ xj(s)


))ds

=

(int γ

0+intδ

γ

+int1

δ

)


(Inminus20+ xj(s)


))ds

le 3intδ

γ


(Inminus20+ xj(s)


))ds

(319)


mintisin[γ δ]


int1

0G(t s)f


(Inminus20+ xj(s)


))ds

geintδ

γ

(

g(t s) +tqminusn+1

ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds


geintδ

γ

(


ηlowast

mminus2sum

i=1

aig(ηi s))


(Inminus20+ xj(s)


))ds

ge λ

intδ

γ


(Inminus20+ xj(s)


))ds

ge λ

3(Ax)


(320)




∥∥ =int1

0G(t s)


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds


(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ds

(321)



(Inminus20+ xj(s)


))


(Inminus20+ x(s)

) S(Inminus20+ x(s)

))∥∥∥ ltε

Δ

(322)





on I





(Inminus20+ x(s)

) S(Inminus20+ x(s)

))



(Inminus20+ x(s)

) S(Inminus20+ x(s)

))




(Inminus20+ D

)(I) times S

(Inminus20+ D

)(I)))


k=1



(T(Inminus20+ D

)(I))


)(I))

(324)

which implies


k=1





(S(Inminus20+ D

)(I))

(325)

Obviously


)(I) = α

(int s

0


)


(326)

α(T(Inminus20+ D

))(I) = α

(int t

0K(t s)

(ints

0


u(τ)dτ

)


le alowast


(n minus 3)α(D(I))

(327)

α(S(Inminus20+ D

))(I) = α

(int1

0H(t s)

(int s

0


u(τ)dτ

)


le blowast


(n minus 3)α(D(I))

(328)



α(D(I)) le 2αC(D) (329)



nminus2sum

k=1




)

αC(D) (330)







nminus1sum

k=1






nminus1sum

k=1



NMβ

int1







k=1

uk ge r1 (334)


Therefore




k=1

uk ge r1 (335)

Take


(336)


x(t) ge λ

3NxΩ ge λ

3Nr0 gt r1 (337)

Hence

(Ax)(t) geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



xΩ(intδ

γ

G(s)c1(s)ds

)

middot ulowast

=1


(intδ

γ

G(s)c1(s)dsulowast)



(338)

and consequently






k=1

uk le r2 (340)






k=1

uk le r2 (341)

Choose

0 lt r lt min

⎧⎨

⎩

(nminus2sum

k=0

1k

)minus1r2 β

⎫⎬

⎭ (342)


(Ax)(ξ) =int1

0G(ξ s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

geintδ

γ


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds

ge 3N2


intδ

γ

G(ξ s)

(nminus2sum

k=1


)


ge 3N2


intδ

γ



x(s)Ωintδ

γ


=1


x(s)Ω(intδ

γ




(343)

which implies


that is




(Ax)(t) leint1

0G(s)f


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))ds (346)


∥∥f(t u1 un+1)


Therefore





(H7)

h(u1 un+1)sumn+1


n+1sum

k=1





k=1


k=1

uk le r3 (350)


ε0 =

(

N

nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

int1

0G(s)a(s)ds

)minus1 (351)




(Ax)(t) leNint1


(Inminus20+ x

)(s) S

(Inminus20+ x

)(s))∥∥∥ds

leNint1

0G(s)a(s)h


∥∥∥ x(s)



)(s)∥∥∥)ds

leNε0

int1

0G(s)a(s)

(nminus1sum

k=1



)(s)∥∥∥ +∥∥∥S(Inminus20+ x

)(s)∥∥∥

)

ds

leNε0

(nminus1sum

k=1

1k

+alowast + blowast

(n minus 3)

)

rlowastint1


(352)

and consequently




4 An Example



960k3

⎧⎪⎨

⎪⎩

⎡


j=1

u2j(t) +infinsum

j=1

uprimej(t)

⎤

⎦

3

+

⎛


j=1

uj(t) +infinsum

j=1

uprime2j(t)

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5

(int t


)23

+1 + t2

24k4

(int1


)

t isin I


(14

)+12uprimek

(49

) k = 1 2 3

(41)






fk(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭+1 + t3

36k5x15k

+1 + t2

24k4y2k

(42)


k=1(1k3) lt 32 we get by (42)

∥∥f(t u v x y

)∥∥ =infinsum

k=1

∣∣fk(t u v x y

)∣∣

le (1 + t)3

2

(140

(u + v)3 + 1160

(u + v)12 + 112

x15 + 18∥∥y∥∥)

(43)


h(u v x y

)=

140

(u + v)3 +1160

(u + v)12 +112x15 +

18y (44)



1 f(1)k ) and f (2) =

(f (2)1 f

(2)k ) in which

f(1)k

(t u v x y

)=

(1 + t)3

960k3

⎧⎪⎨

⎪⎩

⎛


j=1

u2j +infinsum

j=1

vj

⎞

⎠

3

+

⎛


j=1

uj +infinsum

j=1

v2j

⎞

⎠

12⎫⎪⎬

⎪⎭

+1 + t3

36k5x15k (k = 1 2 3 )

f(2)k

(t u v x y

)=

1 + t2

24k4y2k (k = 1 2 3 )

(45)



α(f (1)(tD1 D2 D3 D4)


α(f (2)(ID1 D2 D3 D4)

)le 1


(46)


α(f(ID1 D2 D3 D4)


2Γ(q minus n + 2

)ηlowast

(nminus2sum

k=1


alowast

(n minus 3)Ln +

blowast

(n minus 3)Ln+1

)

asymp 01565 lt 1(47)



fk(t u v x y

) ge (1 + t)3


fk(t u v x y

) ge (1 + t)3

960k3


(48)


c1(t) =(1 + t)3

960 h1 k(u v) = (u + v)3 ulowast =

(1

1k3

) (49)

in this situation


(u + v)3u + v = infin (410)


c2(t) =(1 + t)3

960 h2 k(u v) =


(1

1k3

) (411)

in this situation

h2k = limu+vrarr 0




NMβ

int1

0G(s)a(s)ds asymp 07287 lt β = 1 (413)


Acknowledgments


References











































1 Introduction






∣∣f(t)


(11)


G(u) = S[f(t)u

]

=intinfin


(12)





f(t) =infinsum

n=0

antn (13)

then

F(u) =infinsum

n=0

nantn (14)





(15)







Gn(u) =G(u)un

minusnminus1sum

k=0

f (k)(0)unminusk

(16)



LU + RU +NU = h(x t) (17)


S[LU] + S[RU] + S[NU] = S[h(x t)] (18)



unU(x 0) minus 1


u+ S[RU] + S[NU] = S[h(x t)]

(19)




U(x t) =infinsum

i=0

Ui(x t) (111)



NU(x t) =infinsum

i=0

Ai (112)


Ai =1idi

dλi

[

Ninfinsum

i=0

λiUi

]

λ=0

i = 0 1 2 (113)


S

[ infinsum

i=0

Ui(x t)

]


minus unS[ infinsum

i=0

Ai

]

+ unS[h(x t)]

(114)









U0(x t) = K(x t)



Ψ = Φ + Sminus1[h(x t)] (119)







LU + RU +NU = h(x t) (122)


U(x 0) = f(x) Ut(x 0) = g(x) (123)





]+ Sminus1

[u2S[h(x t)]

] (125)

2 Applications




U(x 0) = 0

Ut(x 0) = ex(22)




] (23)



[U2x minusU2

]](24)


U(x t) =infinsum

i=0

Ui(x t) (25)


infinsum

i=0


u2S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

(26)


infinsum

i=0

Ai(U) = U2x

infinsum

i=0

Ai(U) = U2

(27)


A0(U) = U20x A1(U) = 2U0xU1x Ai(U) =

isum

r=0

UrxUiminusrx

B0(U) = U20 B1(U) = 2U0U1 Bi(U)

isum

r=0

UrUiminusr

(28)


U0(x t) = tex


S

[ infinsum

i=0

Ai(U) minusinfinsum

i=0

Bi(U)

]]

n ge 0(29)



U1(x t) = Sminus1[

u2S

[ infinsum

i=0

A0(U) minusinfinsum

i=0

B0(U)

]]

= Sminus1[u2S

[U2

0x minusU20

]]= Sminus1

[u2S

[t2e2x minus t2e2x

]]

= Sminus1[u2S[0]

]= 0

(210)


U(x t) =infinsum

i=0

Ui(x t) = tex (211)


Ut

(x y t

) minus VxWy = 1

Vt(x y t

) minusWxUy = 5

Wt

(x y t

) minusUxVy = 5

(212)


U(x y 0

)= x + 2y

V(x y 0

)= x minus 2y

W(x y 0

)= minusx + 2y

(213)


U(x y u

)= x + 2y + u + uS

[VxWy

]

V(x y u


[WxUy

]

W(x y u


[UxVy

]

(214)


U(x y t


[VxWy

]]

V(x y t


[WxUy

]]

U(x y t


[UxVy

]]

(215)



U0(x y t

)= t + x + 2y

Ui+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ci(VW)

]]

i ge 0

V0(x y t

)= 5t + x minus 2y

Vi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Di(UW)

]]

i ge 0

W0(x y t

)= 5t minus x + 2y

Wi+1(x y t

)= Sminus1

[

uS

[ infinsum

i=0

Ei(UV )

]]

i ge 0

(216)


C0(VW) = V0xW0y


Ci(VW) =isum

r=0

VrxWiminusry

D0(UW) = U0yW0x


Di(UW) =isum

r=0

WrxUiminusry

E0(UV ) = U0xV0y


Ei(VW) =isum

r=0

UrxViminusry

(217)



U1(x y t


[V0xW0y

]]= Sminus1[uS[(1)(2)]] = 2t

V1(x y t


[W0xU0y


W1(x y t


[U0xV0y


U2(x y t


[V1xW0y + V0xW1y

]]= 0

V2(x y t


[U0yW1x +U1yW0x

]]= 0

W2(x y t


[U1xV0y +U0xV1y

]]= 0

(218)


U(x y t

)=

infinsum

i=0

Ui

(x y t

)= x + 2y + 3t

V(x y t

)=

infinsum

i=0

Vi(x y t

)= x minus 2y + 3t

W(x y t

)=

infinsum

i=0

Wi

(x y t

)= minusx + 2y + 3t

(219)





U(x 0) = 1 + ex V (x 0) = minus1 + ex (221)




Ux(x t) =infinsum

i=0


i=0

Vxi(x t) (223)



S

[ infinsum

i=0

Ui(x t)

]


[ infinsum

i=0

Vix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

S

[ infinsum

i=0

Vi(x t)

]


i=0

Uix

]

+ uS

[ infinsum

i=0

Ui minusinfinsum

i=0

Vi

]

(224)






(225)






U2 = Sminus1[u2ex

]=t2

2ex

V2 = Sminus1[u2ex

]=t2

2ex

(226)


U(x t) = 1 + ex(

1 + t +t2

2+t3


)


1 minus t + t2

2minus t3


)

(227)


U(x t) = 1 + ex+t




Ut + VUx +U = 1



U(x 0) = ex V (x 0) = eminusx (230)





S

[ infinsum

i=0

Ui(x t)

]


i=0

Ai

]

minus uS[ infinsum

i=0

Ui

]

S

[ infinsum

i=0

Vi(x t)

]

= eminusx + u + uS

[ infinsum

i=0

Bi

]

minus uS[ infinsum

i=0

Vi

]

(232)


A0 = U0xV0 A1 = U0xV1 +U1xV0


B0 = V0xU0 B1 = V0xU1 + V1xU0

B2 = V0xU2 + V1xU1 + V2xU0

(233)



U0 = ex + t






2minus tex minus t2

2ex



2eminusx

U2 =t2

2+t2


V2 =t2

2+t2


(236)


U(x t) = ex(

1 minus t + t2

2minus t3


)

V (x t) = eminusx(

1 + t +t2

2+t3


)

(237)



3 Conclusion



Acknowledgment


References


























1 FK and BK Spaces







infink=0 let x

[n] =sumn






suminfinn=0 λnb





suminfink=0 xke






dw(x y

)=

infinsum

k=0

12k



∣∣

)

(x y isin w)

(11)








sumnk=0 xke





Cs =

σ isin C infinsum

n=1

cn(σ) le s

(12)



Φ =φ =

(φk



m(φ)=


supσisinCs

(1φs

sum

kisinσ|xk|

)

ltinfin

n(φ)=


( infinsum

k=1

|uk|Δφk)

ltinfin

(14)
















χ(Q1 +Q2) le χ(Q1) + χ(Q2)




Lχ = χ(L(SX)) (24)


Lχ = 0 (25)










1amiddot lim sup

nrarrinfin

(

supxisinQ

(I minus Pn)(x))


(

supxisinQ

(I minus Pn)(x))

(26)






(

supxisinQ

(I minus Pn)(x))

(27)



(

supxisinQ

sum

kgen|xk|p

)

(28)





Pr(z) = ze +rsum





(

supxisinQ



(

supxisinQ


(210)




supxisinQ

(

supuisinS(x)

( infinsum

n=k

|un|Δφn))

(211)








Lχ = χ(L(SX)) (212)

and we have












x(0) = x0 (32)




∥∥f(t x)∥∥ le Q + Rx (33)


μ(f(t Y )

) le q(t)μ(Y ) (34)









(

supxisinE

supkgen

|xk|)

(35)





xi(0) = x0i (37)





(i) x0 = (x0i ) isin c0



∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

(38)



Now let us denote

q(t) = supnge1

qn(t)

Q = suptisinI

q(t)


(39)




∣∣fn(t x)∣∣ =

∣∣fn(t x1 x2 )∣∣ le pn(t) + qn(t)

(

supigekn

|xi|)

le P +Q supigekn


Hence we get



χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x)∣∣)

χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(

supigen

∣∣fi(t x1 x2 )∣∣)

le limnrarrinfin

supxisinY

(

supigen


∣∣xj∣∣)

le limnrarrinfin

supigen


supxisinY

(

supigen

supjgekn

∣∣xj∣∣)

(312)


χ(f(t Y )

) le q(t)χ(Y ) (313)







infinsum

j=ki+1

aij(t)xj (314)


xi(0) = x0i (315)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (316)



i = 1 2










2 = f2(t x1 x2)


(317)




∣∣fi(t x1 x2 xki)∣∣ le pi(t) (318)



Then we have


∣∣f1(t x1)∣∣∣∣f2(t x1 x2)

∣∣ ∣∣fi(t x1 x2 xi)

∣∣


(319)


∣∣∣fi(t x)∣∣∣ile Pi(t) (320)













xi(0) = x0i (323)




supxisinE

supnmgep

|xn minus xm|


(

supxisinQ

supkgen

|xk|)

(324)







∣∣gn(t x1 x2 )∣∣ le bi (325)




a(t) = supige1

ai(t)

Q = suptisinI

a(t)(326)






f(t x) =(f1(t x) f2(tx)

)=(fi(t x)

)

(327)





∣∣


(328)



∣∣fi(t x)∣∣ le |ai(t)||xi| +

∣∣gi(t x)∣∣








(s y


(s y

)∣∣


∣∣ +


(s y

)∣∣


∣∣ +


(s y

)∣∣

(331)

Then


ige1


(s y

)∣∣


|ai(t) minus ai(s)|



(332)




∣∣gn(t x)∣∣ +

∣∣gm(t x)∣∣



μ(f(t E)

) le a(t)μ(E) (334)











∣∣fi(t x)

∣∣p =

∣∣fi(t x0 x1 x2 )

∣∣p le qi(t) + ri(t)|xi|p (335)







R = supri(t) t isin I i = 0 1 2 (336)



∥∥f(t x)∥∥pp =

infinsum

i=0


leinfinsum

i=0

[qi(t) + ri(t)|xi|p

]

leinfinsum

i=0

qi(t) +

(

supige0

ri(t)

)( infinsum

i=0|xi|p

)

le Q + Rxpp

(337)




χ(f(t Y )

)= lim

nrarrinfinsupxisinY

(sum

igen

∣∣fi(t x1 x2 )∣∣p)

le limnrarrinfin

(sum

igenqi(t) +

(

supigen

ri(t)

)(sum

igen|xi|p

) )

(338)



χ(f(t Y )

) le r(t)χ(Y ) (339)








xi(0) = x0i (341)






∣∣fi(t x0 x1 x2 )∣∣p le bi(t) (342)










∣∣fi(t u)

∣∣ =

∣∣fi(t u1 u2 )

∣∣ le pi(t) + qi(t)|ui| (343)








(344)




uisinS(x)

infinsum

i=1


le supuisinS(x)

infinsum

i=1

[pi(t) + qi(t)|ui|

]Δφi

leinfinsum

i=1

pi(t)Δφi +

(

supi

qi(t)

)(

supuisinS(x)

infinsum

i=1

|ui|Δφi)

le P +Qxn(φ)

(345)




χ(f(t X)

)= lim

krarrinfinsupxisinX

(

supuisinS(x)

( infinsum

n=k


))

le limkrarrinfin

( infinsum

n=k

pn(t)Δφn +

(

supngek

qn(t)

)(

supuisinS(x)

infinsum

n=k

|un|Δφn))

(346)



χ(f(t X)

) le q(t)χ(X) (347)




References























(c)α l












































1 Introduction






(11)



) (12)


with

x(t+k)= lim

hrarr 0+x(tk + h) x

(tminusk)= lim










2 Preliminaries








Iα0h(t) =1

Γ(α)

int t




CDα0+h(t) =

1Γ(n minus α)

int t




CDαh(t) = 0 (23)

has solutions










Then either






x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

0



tk



i=1

int ti

timinus1


h(s)ds


i=1

int ti

timinus1


h(s)ds


i=1

(tk minus ti)int ti

timinus1


h(s)ds + Ω1(t)intT

tk


h(s)ds

+Ω2(t)intT

tk


h(s)ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


+kminus1sum

i=1


(26)




(x(tminusk))


(27)

where

Λ1 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(T minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1




Λ2 =intT

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1



minus ct

TΔ



Δ

(28)



0



xprime(t) =int t

0


h(s)ds minus c1(29)


x(t) =int t

t1



xprime(t) =int t

t1




x(tminus1)=int t1

t0




t0



(211)

Observing that





then we have

minusd0 =int t1

t0



minusd1 =int t1

t0



(213)


x(t) =int t

t1


h(s)ds +int t1

t0


h(s)ds


t0




xprime(t) =int t

t1


h(s)ds +int t1

t0



(214)


x(t) =int t

t2



xprime(t) =int t

t2




x(tminus2)=int t2

t1


h(s)ds +int t1

t0


h(s)ds


t0




x(t+2)= minus e0


t1


h(s)ds +int t1

t0




(216)






thus we have



(218)


x(t) =int t

t2


h(s)ds +int t1

t0


h(s)ds +int t2

t1


h(s)ds


t0


h(s)ds


t0


h(s)ds +int t2

t1


h(s)ds

]

+ I1(x(tminus1))

+ I2(x(tminus2))



(219)


x(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds

+ksum

i=1

(t minus tk)int ti

timinus1


h(s)ds+kminus1sum

i=1

(tk minus ti)int ti

timinus1


h(s)ds

+ksum

i=1

Ii(x(tminusi))+

kminus1sum

i=1



xprime(t) =

int t

tk


h(s)ds+ksum

i=1

int ti

timinus1


h(s)ds+kminus1sum

i=1


(220)



we have




Δ+(1 minus b)TΛ2

Δ

minusc1 = minuscΛ1


Δ

(222)




3 Main Results



conditions


(x(tminusk))



Ωlowast1 = sup


2 = suptisinJ


tisinJ


2 = suptisinJ

∣∣Ωprime2(t)∣∣ (32)





k(u) minus Ilowast


k = 1 2 p


[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

lt 1

(33)



(Fx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds



i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk


f(s x(s))ds

+ Ω2(t)intT

tk


f(s x(s))ds + [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(34)



tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




timesksum

i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



+ |1 + Ω1(t)|[

ksum

i=1




(y(tminusk))∣∣]


i=1


(y(tminusi))∣∣


i=1


(y(tminusi))∣∣

+kminus1sum

i=1


(y(tminusi))∣∣



[L1T

α


1

)(1 + p + 2pα

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3


(35)







(Ax)(t) = [1 + Ω1(t)]

[ksum

i=1

Ii(x(tminusi))


+ Ω1(t)kminus1sum

i=1


+ Ω2(t)kminus1sum

i=1


kminus1sum

i=1


(Dx)(t) =int t

tk



i=1

int ti

timinus1


f(s x(s))ds


i=1

int ti

timinus1


f(s x(s))ds


i=1

(tk minus ti)int ti

timinus1


f(s x(s))ds

+ Ω1(t)intT

tk



tk


f(s x(s))ds

(36)


(1 + Ωlowast

1

)(pL2 + L3

)+(Ωlowast

1T + Ωlowast2 + T

)pL3 lt 1 (37)





tk



∣∣ds

+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1



∣∣ds


timesksum

i=1

int ti

timinus1



∣∣ds


i=1

(tk minus ti)int ti

timinus1



+ |Ω1(t)|intT

tk



+ |Ω2(t)|intT

tk



(38)




|(Dx)(t)| leint t

tk



+ |1 + Ω1(t)|ksum

i=1

int ti

timinus1




i=1

int ti

timinus1




i=1

(tk minus ti)int ti

timinus1




+ |Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + |Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

D(x) le M1Tα


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)= l

(310)


∣∣(Dy


∣∣ leint τ2

τ1



where


tk


∣∣f(s x(s))∣∣ds +


ksum

i=1

int ti

timinus1




1(t) + Ωprime2(t) + 1

∣∣ksum

i=1

int ti

timinus1



+∣∣Ωprime


i=1

(tk minus ti)int ti

timinus1



+∣∣Ωprime

1(t)∣∣intT

tk


∣∣f(s x(s))∣∣ds +


tk



le M1Tα

Γ(α + 1)Ωlowast

1

(1 + p + 2pα

)+M1T

αminus1

Γ(α)(1 + Ωlowast

2)(1 + p

)= L

(312)




(xλ


(313)



)(t) (314)



|x(t)| le λ

int t

tk


∣∣f(s x(s))

∣∣ds + λ|1 + Ω1(t)|

ksum

i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

int ti

timinus1


∣∣f(s x(s))

∣∣ds


i=1

(tk minus ti)int ti

timinus1


∣∣f(s x(s))

∣∣ds

+ λ|Ω1(t)|intT

tk


∣∣f(s x(s))

∣∣ds + λ|Ω2(t)|

intT

tk


∣∣f(s x(s))

∣∣ds

+ λ|1 + Ω1(t)|[

ksum

i=1

∣∣∣∣∣Ii

(x(tminusi)

λ

)∣∣∣∣∣+


(x(tminusk

)

λ

)∣∣∣∣∣

]


i=1


(x(tminusi)

λ

)∣∣∣∣∣


i=1


(x(tminusi)

λ

)∣∣∣∣∣+ λ

kminus1sum

i=1


(x(tminusi)

λ

)∣∣∣∣∣

x(t) le[M1T

α


1

)(1 + p + 2pα

)+M1T

αminus1

Γ(α)Ωlowast

2(1 + p

)

+(1 + Ωlowast

1

)(pM2 +M3

)+(Ωlowast

1T + Ωlowast2 + T

)pM3

]

(315)


4 An Example


CD32x(t) =

sin 2t|x(t)|(t + 5)2(1 + |x(t)|)

t isin [0 1] t =13

Δx(13

)=


(13

)=



(41)




[L1T

α


1

)(1 + p(1 + L2) + 2pα + L3

)+L1T

αminus1

Γ(α)Ωlowast

2(1 + p

)+(Ωlowast

1T + Ωlowast2 + T

)pL3

]

=464

1125radicπ

+173450

lt 1

(42)


Acknowledgments


References

































1 Introduction


intb

a

w(x)f(x)dx =nsum

k=0

wkf(xk) + Rn+1[f] (11)






Ψn+1(x) =nprod

k=0

(x minus xk) (12)


f(x) =nsum

k=0

f(xk)L(xxk) + rn+1(f x) (13)

where

L(xxk) =Ψn+1(x)


(k = 0 1 n) (14)

we obtain (11) with

wk =1

Ψprimen+1(xk)

intb

a


Rn+1[f]=intb

a

rn+1(f x)w(x)dx

(15)



intb

a

f(t)dt =b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)+ E(f) (16)



∥∥f∥∥p =

(intb

a

∣∣f(t)∣∣pdt

)1p

(17)


xisin[ab]

∣∣f(x)∣∣ (18)



∣∣∣∣∣

intb

a

f(x)g(x)dx

∣∣∣∣∣le ∥∥f∥∥1





λ(b minus a)intb

a



2λ(b minus a) f(a)

∣∣∣∣∣




)

(110)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (111)


K(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

t minus 5a + b6

t isin[aa + b2

]

t minus a + 5b6

t isin(a + b2

b

]

(112)


intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt (113)

maxtisin[ab]

|K(t)| = 13(b minus a) (114)


2 Main Results


m1 =int (5a+b)6

a

(t minus 5a + b

6

)β(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)α(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)β(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)α(t)dt

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt le

M1 =int (5a+b)6

a

(t minus 5a + b

6

)α(t)dt +

int (a+b)2

(5a+b)6

(t minus 5a + b

6

)β(t)dt

+int (a+5b)6

(a+b)2

(t minus a + 5b

6

)α(t)dt +

intb

(a+5b)6

(t minus a + 5b

6

)β(t)dt

(21)


intb

a


2

)dt

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus 12

(intb

a

K(t)(α(t) + β(t)

)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)

(22)



2


2 (23)




(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus 12

(int (a+b)2

a

(t minus 5a + b

6

)(α(t) + β(t)

)dt +

intb

(a+b)2

(t minus a + 5b

6

)(α(t) + β(t)

)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a


2

)dt


a


dt

=12

(int (a+b)2

a


5a + b6


intb

(a+b)2


a + 5b6


)

(24)


m1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)β(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)α(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)β(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)α(x)dx

M1 =int (a+b)2

a

((t minus 5a + b


5a + b6

∣∣∣∣

)α(t)2

+(t minus 5a + b


5a + b6

∣∣∣∣

)β(t)2

)dt

+intb

(a+b)2

((t minus a + 5b


a + 5b6

∣∣∣∣

)α(t)2

+(t minus a + 5b


a + 5b6

∣∣∣∣

)β(t)2

)dt

=int (5a+b)6

a

(x minus 5a + b

6

)α(x)dx +

int (a+b)2

(5a+b)6

(x minus 5a + b

6

)β(x)dx

+int (a+5b)6

(a+b)2

(x minus a + 5b

6

)α(x)dx +

intb

(a+5b)6

(x minus a + 5b

6

)β(x)dx

(25)




Special Case 1


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


144((β1 minus α1

)(a + b) + 2

(β0 minus α0

))

(26)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)

)∣∣∣∣∣le 5


) (27)



int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(


a

α(t)dt

)

le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t) dt

leint (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

+b minus a3

(


a

α(t)dt

)

(28)

Proof Since

intb

a


6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)α(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)

(29)


so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)α(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)α(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


3

(


a

α(t)dt

)

(210)


Special Case 2


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)


a + b2

α1

))

(211)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(f(b) minus f(a)



int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6


3

(intb

a


le b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

leint (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt +

b minus a3

(intb

a


(213)


Proof Since

intb

a


=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt minus(intb

a

K(t)β(t)dt

)

=b minus a6

(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)

(214)

so we have


(f(a) + 4f

(a + b2

)+ f(b)

)minusintb

a

f(t)dt

minus(int (a+b)2

a

int

a

(t minus 5a + b

6

)β(t)dt +

intb

(a+b)2

(t minus a + 5b

6

)β(t)dt

)∣∣∣∣∣

=

∣∣∣∣∣

intb

a



a


le maxtisin[ab]

|K(t)|intb

a


)dt =

b minus a3

(int b

a


(215)


Special Case 3

If β(x) = β1x + β0 = 0 in (213) then

∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 +

a + b2


b minus a)

(216)


∣∣∣∣∣

intb

a


(f(a) + 4f

(a + b2

)+ f(b)


3

(β0 minus


) (217)


Acknowledgment


References





















1 Introduction









xx= 0

(11)













u(x t) = U(ξ) ξ = kx +ωt (22)



)= 0 (23)



U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0 (24)


d2G(ξ)dξ2

+ λdG(ξ)dξ

+ μG(ξ) = 0 (25)






3 Application


u(x t) = U(ξ)eiη

v(x t) = V (ξ)(31)




)U = 0 (32)

(ω2 minus k2


(U2)primeprime

= 0 (33)



)U = 0

(ω2 minus k2


(34)


U(ξ) =msum

i=1

αi

(Gprime

G

)i+ α0 αm = 0

V (ξ) =nsum

i=1

βi

(Gprime

G

)i+ β0 βn = 0

(35)




m + n = m + 2

2m = n + 2(36)


U(ξ) = α2(Gprime

G

)2

+ α1(Gprime

G

)+ α0 α2 = 0

V (ξ) = β2(Gprime

G

)2

+ β1(Gprime

G

)+ β0 β2 = 0

(37)




)= 0




)ω minus qα1



)= 0


)α2



)β2 minus k2


)= 0


)α2 minus α1

(minus2k2 + aβ2

)= 0


)= 0

(6k2 minus aβ2

)α2 = 0


(38)






λ = plusmn13

radic3ω2 minus 3k2

k2 μ =

16k2 minusω2

k4 p = minus1

2ω

k

q =124

24k4 + 30ω4 minus 54k2ω2


α1 = plusmn2radic

1ab


radic1ab


a β2 =

6k2

a

(39)


U(ξ) = 6k2radic

1ab

(Gprime

G

)2

plusmn 2

radic1ab

radic3ω2 minus 3k2

(Gprime

G

)

V (ξ) =6k2

a

(Gprime

G

)2

plusmn 2

radic3ω2 minus 3k2

a

(Gprime

G

)

(310)



uH(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2



⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vH(x t) =6k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicD2

⎛

⎜⎝C1 sinh

((radicD2

)ξ)+ C2 cosh

((radicD2

)ξ)

C1 cosh((radic

D2)ξ)+ C2 sinh

((radicD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(311)



2 asfollows

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312a)


⎛

⎝3tanh2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312b)

while at C21 lt C

22 one can obtain

uH(x t) =12

radic1ab

ω2 minus k2k2

⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠eiη (312c)


⎛

⎝3coth2

⎛

⎝12

radicω2 minus k2k4

ξ + ρH

⎞

⎠ minus 1

⎞

⎠ (312d)




uT(x t) = 2

radic1ab

⎡

⎢⎣3k2

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2


⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

⎤

⎥⎦eiη

vT(x t) =6k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠ ∓ 1

6

radic3ω2 minus 3k2

k2

⎞

⎟⎠

2

plusmn 2

radic3ω2 minus 3k2

a

⎛

⎜⎝

radicminusD2

⎛

⎜⎝

minusC1 sin((radic

minusD2)ξ)+ C2 cos

((radicminusD2

)ξ)

C1 cos((radic

minusD2)ξ)+ C2 sin

((radicminusD2

)ξ)

⎞

⎟⎠

∓16

radic3ω2 minus 3k2

k2

⎞

⎟⎠

(313)


1 gt C22 and C

21 lt C

22 as follows

uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314a)


ak2

⎛

⎝3tan2

⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314b)


|u H|

tx

1

05

0

minus054

20

minus2minus4 minus4

minus20

24


0

02

04

06

08

tx

42

0minus2

minus4 minus4minus2

02

4

|u H|2


uT(x t) =12

radic1ab

k2 minusω2

k2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠eiη (314c)


ak2

⎛

⎝3cot2⎛

⎝12

radick2 minusω2

k4ξ + ρT

⎞

⎠ + 1

⎞

⎠ (314d)



0

200

400

600

800

tx

42

0minus2

minus4 minus4minus2

02

4

vH


4 Conclusions


Acknowledgments


References























Shengli Xie







1 Introduction



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J






(

t x(σ1(t))int t


)

t isin [0 b]

x(0) + g(x) = x0

(12)


(1 minus αβMc

)N


2αθ1L1M(1 + 2Kθ2L1) lt 1

(13)





2 Preliminaries








x(t) = R(t 0)[φ(0) + g(x)(0)

]+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

(21)




α

(int t

0xn(s)ds n isin N

)

le 2int t








3 Existence Result




+ such that

∥∥f(t x y z


∥∥y∥∥ + z






andint t0 k(t s)ds





α

(int t

0K(t s V1)ds

)


α(V1(θ)) t isin J

α

(intb

0H(t s V1)ds

)


α(V1(θ)) t isin J

(33)


w = maxMp0(1 + k0 + h0) + 1 2M

(l01 + 2l02k0 + 2l03h0

)+ 1 (34)






(Fx)(t)

=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(35)






0ew(sminust)

∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥ds

le L +Mint t

0ew(sminust)p(s)

(

xs[minusq0] +ints

0K(s r xr)dr +

intb

0H(s r xr)dr

)

ds

le L +Mp0

int t

0ew(sminust)

(

xs[minusq0] +int s


intb


)

ds


Consequently




(38)


∥∥∥∥∥f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)∥∥∥∥∥le p(s)(1 + k0 + h0)R (39)







0R(t minus s)α

(

f

(

s Vs

int s

0K(s r Vr)dr

intb

0H(s r Vr)dr

))

ds

le 2Mint t

0ew(sminust)

[



0K(s r Vr)dr

)

+l3(s)α

(intb

0H(s r Vr)dr

)]

ds

le 2Mint t

0ew(sminust)

[


α(V (s + θ)) + 2l02

int s

0k(s r) sup


+2l03

intb

0h(s r) sup


]

ds

le 2M(l01 + 2l02k0 + 2l03h0

)int t

0ew(sminust)ds sup


le 2M(l01 + 2l02k0 + 2l03h0


(311)

Consequently


α((FV )(t)) le 2M(l01 + 2l02k0 + 2l03h0






4 An Example




ut(t y)= a1

(t y)uyy(t y)

+ a2(t)

[

sinu(t + θ y

)ds +

int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

0 le t le b


0

intb

0F(r y)log(1 + |u(r s)|12


u(t 0) = u(t π) = 0 0 le t le b(41)


f

(

t ut

int t

0K(t s us)ds

intb

0H(t s us)ds

)(y)

= a2(t)

[

sinu(t + θ y

)+int t

0

ints

minusqa3(s + τ)

u(τ y)dτds

(1 + t)+intb

0

u(s + θ y

)ds

(1 + t)(1 + s)2

]

K(t s us)(y)=ints

minusqa3(s + τ)

u(τ y)dτ

(1 + t) H(t s us)

(y)=

u(s + θ y

)

(1 + t)(1 + s)2

g(u)(y)=intπ

0

intb

0F(r y)log(1 + |u(r s)|12

)dr ds

(42)



(43)


∥∥f(t u v z)∥∥ le |a2(t)|

(u[minusq0] + v + z

)

K(t s u) leints


u[minusq0](1 + t)


(1 + t)(1 + s)

(44)



∥∥g(u)



∣∣F(r y)∣∣(u[0b] +

radicπ) (45)


α(f(t V1 V2 V3)

) le |a2(t)|(

supminusqleθle0

α(V1(θ)) + α(V2) + α(V3)

)

t isin J

α(K(t s V1)) le 11 + t

supminusqleθle0


α(H(t s V1)) le 1

(1 + t)(1 + s)2sup


(46)


5 An Application



(

t xt

int t

0K(t s xs)ds

intb

0H(t s xs)ds

)

t isin J



x(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(52)






w = maxMp0(1 + k0 + h0)(1 +MM1b) + 1 2M

(l01 + 2l02k0 + 2l03h0

)(1 +MM1b) + 1

(53)



Wv =intb

0R(b s)(Cv)(s)ds (54)





v(t) =Wminus1(

x1 minus R(b 0)[φ(0) + g(x)(0)

]

+intb

0R(b s)f

(

s xs

ints

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds

)

(t) t isin [0 b]

(55)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t

0R(t s)(Cv)(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(56)


(Tx)(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩


[φ(0) + g(x)(0)

]+int t


[φ(0) + g(x)(0)

]

+intb

0R(b τ)f

(

τ xτ

int τ

0K(τ r xr)dr

intb

0H(τ r xr)dr

)

dτ

)

(s)ds

+int t

0R(t s)f

(

s xs

int s

0K(s r xr)dr

intb

0H(s r xr)dr

)

ds t isin [0 b]

(57)




Acknowledgments


References



























Documents

Compactness Conditions in the Theory of Nonlinear Differential … · 2019. 8. 7. · Yuxia Guo, China Qian Guo, China Chaitan P. Gupta, USA Uno Hamarik, Estonia¨ Ferenc Hartung,