Journal of Computational Analysis and Applications · Editorial Board Associate Editors. of. Journal of Computational Analysis and Applications. Francesco Altomare . Dipartimento

Volume 22, Number 7 June 15, 2017 ISSN:1521-1398 PRINT,1572-9206 ONLINE

Journal of Computational Analysis and Applications EUDOXUS PRESS,LLC

Journal of Computational Analysis and Applications ISSNno.’s:1521-1398 PRINT,1572-9206 ONLINE SCOPE OF THE JOURNAL An international publication of Eudoxus Press, LLC (fourteen times annually) Editor in Chief: George Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152-3240, U.S.A [email protected] http://www.msci.memphis.edu/~ganastss/jocaaa The main purpose of "J.Computational Analysis and Applications" is to publish high quality research articles from all subareas of Computational Mathematical Analysis and its many potential applications and connections to other areas of Mathematical Sciences. Any paper whose approach and proofs are computational,using methods from Mathematical Analysis in the broadest sense is suitable and welcome for consideration in our journal, except from Applied Numerical Analysis articles. Also plain word articles without formulas and proofs are excluded. The list of possibly connected mathematical areas with this publication includes, but is not restricted to: Applied Analysis, Applied Functional Analysis, Approximation Theory, Asymptotic Analysis, Difference Equations, Differential Equations, Partial Differential Equations, Fourier Analysis, Fractals, Fuzzy Sets, Harmonic Analysis, Inequalities, Integral Equations, Measure Theory, Moment Theory, Neural Networks, Numerical Functional Analysis, Potential Theory, Probability Theory, Real and Complex Analysis, Signal Analysis, Special Functions, Splines, Stochastic Analysis, Stochastic Processes, Summability, Tomography, Wavelets, any combination of the above, e.t.c. "J.Computational Analysis and Applications" is a peer-reviewed Journal. See the instructions for preparation and submission of articles to JoCAAA. Assistant to the Editor: Dr.Razvan Mezei,Lenoir-Rhyne University,Hickory,NC 28601, USA. Journal of Computational Analysis and Applications(JoCAAA) is published by EUDOXUS PRESS,LLC,1424 Beaver Trail Drive,Cordova,TN38016,USA,[email protected] http://www.eudoxuspress.com. Annual Subscription Prices:For USA and Canada,Institutional:Print $700, Electronic OPEN ACCESS. Individual:Print $350. For any other part of the world add $130 more(handling and postages) to the above prices for Print. No credit card payments. Copyright©2017 by Eudoxus Press,LLC,all rights reserved.JoCAAA is printed in USA. JoCAAA is reviewed and abstracted by AMS Mathematical Reviews,MATHSCI,and Zentralblaat MATH. It is strictly prohibited the reproduction and transmission of any part of JoCAAA and in any form and by any means without the written permission of the publisher.It is only allowed to educators to Xerox articles for educational purposes.The publisher assumes no responsibility for the content of published papers.

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Editorial Board

Associate Editors of Journal of Computational Analysis and Applications

Francesco Altomare Dipartimento di Matematica Universita' di Bari Via E.Orabona, 4 70125 Bari, ITALY Tel+39-080-5442690 office +39-080-3944046 home +39-080-5963612 Fax [email protected] Approximation Theory, Functional Analysis, Semigroups and Partial Differential Equations, Positive Operators. Ravi P. Agarwal Department of Mathematics Texas A&M University - Kingsville 700 University Blvd. Kingsville, TX 78363-8202 tel: 361-593-2600 [email protected] Differential Equations, Difference Equations, Inequalities George A. Anastassiou Department of Mathematical Sciences The University of Memphis Memphis, TN 38152,U.S.A Tel.901-678-3144 e-mail: [email protected] Approximation Theory, Real Analysis, Wavelets, Neural Networks, Probability, Inequalities. J. Marshall Ash Department of Mathematics De Paul University 2219 North Kenmore Ave. Chicago, IL 60614-3504 773-325-4216 e-mail: [email protected] Real and Harmonic Analysis Dumitru Baleanu Department of Mathematics and Computer Sciences, Cankaya University, Faculty of Art and Sciences, 06530 Balgat, Ankara, Turkey, [email protected]

Fractional Differential Equations Nonlinear Analysis, Fractional Dynamics Carlo Bardaro Dipartimento di Matematica e Informatica Universita di Perugia Via Vanvitelli 1 06123 Perugia, ITALY TEL+390755853822 +390755855034 FAX+390755855024 E-mail [email protected] Web site: http://www.unipg.it/~bardaro/ Functional Analysis and Approximation Theory, Signal Analysis, Measure Theory, Real Analysis.

Martin Bohner Department of Mathematics and Statistics, Missouri S&T Rolla, MO 65409-0020, USA [email protected] web.mst.edu/~bohner Difference equations, differential equations, dynamic equations on time scale, applications in economics, finance, biology. Jerry L. Bona Department of Mathematics The University of Illinois at Chicago 851 S. Morgan St. CS 249 Chicago, IL 60601 e-mail:[email protected] Partial Differential Equations, Fluid Dynamics Luis A. Caffarelli Department of Mathematics The University of Texas at Austin Austin, Texas 78712-1082 512-471-3160 e-mail: [email protected] Partial Differential Equations George Cybenko Thayer School of Engineering

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Dartmouth College 8000 Cummings Hall, Hanover, NH 03755-8000 603-646-3843 (X 3546 Secr.) e-mail:[email protected] Approximation Theory and Neural Networks Sever S. Dragomir School of Computer Science and Mathematics, Victoria University, PO Box 14428, Melbourne City, MC 8001, AUSTRALIA Tel. +61 3 9688 4437 Fax +61 3 9688 4050 [email protected] Inequalities, Functional Analysis, Numerical Analysis, Approximations, Information Theory, Stochastics. Oktay Duman TOBB University of Economics and Technology, Department of Mathematics, TR-06530, Ankara, Turkey, [email protected] Classical Approximation Theory, Summability Theory, Statistical Convergence and its Applications Saber N. Elaydi Department Of Mathematics Trinity University 715 Stadium Dr. San Antonio, TX 78212-7200 210-736-8246 e-mail: [email protected] Ordinary Differential Equations, Difference Equations Christodoulos A. Floudas Department of Chemical Engineering Princeton University Princeton,NJ 08544-5263 609-258-4595(x4619 assistant) e-mail: [email protected] Optimization Theory&Applications, Global Optimization J .A. Goldstein Department of Mathematical Sciences The University of Memphis Memphis, TN 38152 901-678-3130 [email protected]

Partial Differential Equations, Semigroups of Operators H. H. Gonska Department of Mathematics University of Duisburg Duisburg, D-47048 Germany 011-49-203-379-3542 e-mail: [email protected] Approximation Theory, Computer Aided Geometric Design John R. Graef Department of Mathematics University of Tennessee at Chattanooga Chattanooga, TN 37304 USA [email protected] Ordinary and functional differential equations, difference equations, impulsive systems, differential inclusions, dynamic equations on time scales, control theory and their applications Weimin Han Department of Mathematics University of Iowa Iowa City, IA 52242-1419 319-335-0770 e-mail: [email protected] Numerical analysis, Finite element method, Numerical PDE, Variational inequalities, Computational mechanics Tian-Xiao He Department of Mathematics and Computer Science P.O. Box 2900, Illinois Wesleyan University Bloomington, IL 61702-2900, USA Tel (309)556-3089 Fax (309)556-3864 [email protected] Approximations, Wavelet, Integration Theory, Numerical Analysis, Analytic Combinatorics Margareta Heilmann Faculty of Mathematics and Natural Sciences, University of Wuppertal Gaußstraße 20 D-42119 Wuppertal, Germany, [email protected]

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Approximation Theory (Positive Linear Operators) Xing-Biao Hu Institute of Computational Mathematics AMSS, Chinese Academy of Sciences Beijing, 100190, CHINA [email protected] Computational Mathematics Jong Kyu Kim Department of Mathematics Kyungnam University Masan Kyungnam,631-701,Korea Tel 82-(55)-249-2211 Fax 82-(55)-243-8609 [email protected] Nonlinear Functional Analysis, Variational Inequalities, Nonlinear Ergodic Theory, ODE, PDE, Functional Equations. Robert Kozma Department of Mathematical Sciences The University of Memphis Memphis, TN 38152, USA [email protected] Neural Networks, Reproducing Kernel Hilbert Spaces, Neural Percolation Theory Mustafa Kulenovic Department of Mathematics University of Rhode Island Kingston, RI 02881,USA [email protected] Differential and Difference Equations Irena Lasiecka Department of Mathematical Sciences University of Memphis Memphis, TN 38152 PDE, Control Theory, Functional Analysis, [email protected]

Burkhard Lenze Fachbereich Informatik Fachhochschule Dortmund University of Applied Sciences Postfach 105018 D-44047 Dortmund, Germany e-mail: [email protected] Real Networks, Fourier Analysis, Approximation Theory

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National Technical University of Athens Zografou campus, 157 80 Athens, Greece tel:: +30(210) 772 1722 Fax +30(210) 772 1775 [email protected] Partial Differential Equations, Probability Choonkil Park Department of Mathematics Hanyang University Seoul 133-791 S. Korea, [email protected] Functional Equations Svetlozar (Zari) Rachev, Professor of Finance, College of Business, and Director of Quantitative Finance Program, Department of Applied Mathematics & Statistics Stonybrook University 312 Harriman Hall, Stony Brook, NY 11794-3775 tel: +1-631-632-1998, [email protected] Alexander G. Ramm Mathematics Department Kansas State University Manhattan, KS 66506-2602 e-mail: [email protected] Inverse and Ill-posed Problems, Scattering Theory, Operator Theory, Theoretical Numerical Analysis, Wave Propagation, Signal Processing and Tomography Tomasz Rychlik Polish Academy of Sciences Instytut Matematyczny PAN 00-956 Warszawa, skr. poczt. 21 ul. Śniadeckich 8 Poland [email protected] Mathematical Statistics, Probabilistic Inequalities Boris Shekhtman Department of Mathematics University of South Florida Tampa, FL 33620, USA Tel 813-974-9710 [email protected]

Approximation Theory, Banach spaces, Classical Analysis T. E. Simos Department of Computer Science and Technology Faculty of Sciences and Technology University of Peloponnese GR-221 00 Tripolis, Greece Postal Address: 26 Menelaou St. Anfithea - Paleon Faliron GR-175 64 Athens, Greece [email protected] Numerical Analysis H. M. Srivastava Department of Mathematics and Statistics University of Victoria Victoria, British Columbia V8W 3R4 Canada tel.250-472-5313; office,250-477-6960 home, fax 250-721-8962 [email protected] Real and Complex Analysis, Fractional Calculus and Appl., Integral Equations and Transforms, Higher Transcendental Functions and Appl.,q-Series and q-Polynomials, Analytic Number Th. I. P. Stavroulakis Department of Mathematics University of Ioannina 451-10 Ioannina, Greece [email protected] Differential Equations Phone +3-065-109-8283 Manfred Tasche Department of Mathematics University of Rostock D-18051 Rostock, Germany [email protected] rostock.de Numerical Fourier Analysis, Fourier Analysis, Harmonic Analysis, Signal Analysis, Spectral Methods, Wavelets, Splines, Approximation Theory Roberto Triggiani Department of Mathematical Sciences University of Memphis Memphis, TN 38152 PDE, Control Theory, Functional

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Journal of Computational Analysis and Applications An international publication of Eudoxus Press, LLC, of TN.

Editor in Chief: George Anastassiou

Department of Mathematical Sciences University of Memphis

Memphis, TN 38152-3240, U.S.A.

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ON QUADRATIC ρ-FUNCTIONAL INEQUALITIES IN FUZZY NORMED

SPACES

CHOONKIL PARK, SUN YOUNG JANG, AND SUNGSIK YUN∗

Abstract. In this paper, we solve the following quadratic ρ-functional inequalities

N(f(x+ y) + f(x− y)− 2f(x)− 2f(y)− ρ

(2f(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)

), t)

≥ t

t+ ϕ(x, y), (0.1)

where ρ is a fixed real number with ρ 6= 2, and

N(

2f(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)− ρ (f(x+ y) + f(x− y)− 2f(x)− 2f(y)) , t

)≥ t

t+ ϕ(x, y), (0.2)

where ρ is a fixed real number with ρ 6= 12.

Using the direct method, we prove the Hyers-Ulam stability of the quadratic ρ-functionalinequalities (0.1) and (0.2) in fuzzy Banach spaces.

1. Introduction and preliminaries

Katsaras [14] defined a fuzzy norm on a vector space to construct a fuzzy vector topologicalstructure on the space. Some mathematicians have defined fuzzy norms on a vector space fromvarious points of view [9, 16, 37]. In particular, Bag and Samanta [2], following Cheng andMordeson [6], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metricis of Kramosil and Michalek type [15]. They established a decomposition theorem of a fuzzynorm into a family of crisp norms and investigated some properties of fuzzy normed spaces [3].

We use the definition of fuzzy normed spaces given in [2, 19, 20] to investigate the Hyers-Ulamstability of quadratic ρ-functional inequalities in fuzzy Banach spaces.

Definition 1.1. [2, 19, 20, 21] Let X be a real vector space. A function N : X × R→ [0, 1] iscalled a fuzzy norm on X if for all x, y ∈ X and all s, t ∈ R,

(N1) N(x, t) = 0 for t ≤ 0;(N2) x = 0 if and only if N(x, t) = 1 for all t > 0;(N3) N(cx, t) = N(x, t

|c|) if c 6= 0;

(N4) N(x+ y, s+ t) ≥ minN(x, s), N(y, t);(N5) N(x, ·) is a non-decreasing function of R and limt→∞N(x, t) = 1.(N6) for x 6= 0, N(x, ·) is continuous on R.

The pair (X,N) is called a fuzzy normed vector space.The properties of fuzzy normed vector spaces and examples of fuzzy norms are given in [18, 19].

Definition 1.2. [2, 19, 20, 21] Let (X,N) be a fuzzy normed vector space. A sequence xn in Xis said to be convergent or converge if there exists an x ∈ X such that limn→∞N(xn − x, t) = 1

2010 Mathematics Subject Classification. Primary 46S40, 39B52, 39B62, 26E50, 47S40.Key words and phrases. fuzzy Banach space; quadratic ρ-functional inequality; Hyers-Ulam stability.

∗Corresponding author.

J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 22, NO.7, 2017, COPYRIGHT 2017 EUDOXUS PRESS, LLC

1189 CHOONKIL PARK et al 1189-1197

C. PARK, S. Y. JANG, AND S. YUN

for all t > 0. In this case, x is called the limit of the sequence xn and we denote it byN -limn→∞ xn = x.

Definition 1.3. [2, 19, 20, 21] Let (X,N) be a fuzzy normed vector space. A sequence xnin X is called Cauchy if for each ε > 0 and each t > 0 there exists an n0 ∈ N such that for alln ≥ n0 and all p > 0, we have N(xn+p − xn, t) > 1− ε.

It is well-known that every convergent sequence in a fuzzy normed vector space is Cauchy. Ifeach Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzynormed vector space is called a fuzzy Banach space.

We say that a mapping f : X → Y between fuzzy normed vector spaces X and Y is continuousat a point x0 ∈ X if for each sequence xn converging to x0 in X, then the sequence f(xn)converges to f(x0). If f : X → Y is continuous at each x ∈ X, then f : X → Y is said to becontinuous on X (see [3]).

The stability problem of functional equations originated from a question of Ulam [36] con-cerning the stability of group homomorphisms.

The functional equation f(x+ y) = f(x) + f(y) is called the Cauchy equation. In particular,every solution of the Cauchy equation is said to be an additive mapping. Hyers [11] gave afirst affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem wasgeneralized by Aoki [1] for additive mappings and by Th.M. Rassias [27] for linear mappings byconsidering an unbounded Cauchy difference. A generalization of the Th.M. Rassias theoremwas obtained by Gavruta [10] by replacing the unbounded Cauchy difference by a general controlfunction in the spirit of Th.M. Rassias’ approach.

The functional equation f(x+y)+f(x−y) = 2f(x)+2f(y) is called the quadratic functionalequation. In particular, every solution of the quadratic functional equation is said to be aquadratic mapping. The stability of quadratic functional equation was proved by Skof [35] formappings f : E1 → E2, where E1 is a normed space and E2 is a Banach space. Cholewa [7]noticed that the theorem of Skof is still true if the relevant domain E1 is replaced by an Abeliangroup. Czerwik [8] proved the Hyers-Ulam stability of the quadratic functional equation. The

functional equation f(x+y2

)+f

(x−y2

)= 1

2f(x)+ 12f(y) is called a Jensen type quadratic equation.

The stability problems of several functional equations have been extensively investigated by anumber of authors and there are many interesting results concerning this problem (see [5, 12,13, 17, 24, 25, 26, 28, 29, ?, 30, 31, 32, 33, 34]).

Park [22, 23] defined additive ρ-functional inequalities and proved the Hyers-Ulam stabilityof the additive ρ-functional inequalities in Banach spaces and non-Archimedean Banach spaces.

In Section 2, we solve the quadratic ρ-functional inequality (0.1) and prove the Hyers-Ulamstability of the quadratic ρ-functional inequality (0.1) in fuzzy Banach spaces by using the directmethod.

In Section 3, we solve the quadratic ρ-functional inequality (0.2) and prove the Hyers-Ulamstability of the quadratic ρ-functional inequality (0.2) in fuzzy Banach spaces by using the directmethod.

Throughout this paper, assume that X is a real vector space and (Y,N) is a fuzzy Banachspace.

2. Quadratic ρ-functional inequality (0.1)

In this section, we prove the Hyers-Ulam stability of the quadratic ρ-functional inequality(0.1) in fuzzy Banach spaces. Let ρ be a real number with ρ 6= 2. We need the following lemmato prove the main results.



QUADRATIC ρ-FUNCTIONAL INEQUALITIES IN FUZZY NORMED SPACES

Lemma 2.1. Let f : X → Y be a mapping satisfying f(0) = 0 and

f(x+ y) + f(x− y)− 2f(x)− 2f(y) = ρ

(2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)

)(2.1)

for all x, y ∈ X. Then f : X → Y is quadratic.

Proof. Replacing y by x in (2.1), we get f(2x)− 4f(x) = 0 and so f(2x) = 4f(x) for all x ∈ X.Thus

f(x+ y) + f(x− y)− 2f(x)− 2f(y) = ρ

(2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)

)=

ρ

2(f(x+ y) + f(x− y)− 2f(x)− 2f(y))

and so f(x+ y) + f(x− y) = 2f(x) + 2f(y) for all x, y ∈ X, as desired.

Theorem 2.2. Let ϕ : X2 → [0,∞) be a function such that

Φ(x, y) :=∞∑j=1

4jϕ

(x

2j,y

2j

)<∞ (2.2)

for all x, y ∈ X. Let f : X → Y be a mapping satisfying f(0) = 0 and

N(f(x+ y) + f(x− y)− 2f(x)− 2f(y) (2.3)

−ρ(

2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)

), t) ≥ t

t+ ϕ(x, y)

for all x, y ∈ X and all t > 0. Then Q(x) := N -limn→∞ 4nf(x2n)

exists for each x ∈ X anddefines a quadratic mapping Q : X → Y such that

N (f(x)−Q(x), t) ≥ t

t+ 14Φ(x, x)

(2.4)

for all x ∈ X and all t > 0.

Proof. Letting y = x in (2.3), we get

N (f (2x)− 4f(x), t) ≥ t

t+ ϕ(x, x)(2.5)

and so N(f (x)− 4f

(x2

), t)≥ t

t+ϕ(x2,x2 )

for all x ∈ X. Hence

N

(4lf

(x

2l

)− 4mf

(x

2m

), t

)(2.6)

≥ min

N

(4lf

(x

2l

)− 4l+1f

(x

2l+1

), t

), · · · , N

(4m−1f

(x

2m−1

)− 4mf

(x

2m

), t

)= min

N

(f

(x

2l

)− 4f

(x

4l+1

),t

4l

), · · · , N

(f

(x

2m−1

)− 4f

(x

2m

),

t

4m−1

)

≥ min

t4l

t4l

+ ϕ(

x2l+1 ,

x2l+1

) , · · · , t4m−1

t4m−1 + ϕ

(x2m ,

x2m)

= min

t

t+ 4lϕ(

x2l+1 ,

x2l+1

) , · · · , t

t+ 4m−1ϕ(x2m ,

x2m)

≥ t

t+ 14

∑mj=l+1 4jϕ

(x2j, x2j

)




for all nonnegative integers m and l with m > l and all x ∈ X and all t > 0. It follows from (2.2)and (2.6) that the sequence 4nf( x

2n ) is a Cauchy sequence for all x ∈ X. Since Y is complete,the sequence 4nf( x

2n ) converges. So one can define the mapping Q : X → Y by

Q(x) := N - limn→∞

4nf(x

2n)

for all x ∈ X. Moreover, letting l = 0 and passing the limit m→∞ in (2.6), we get (2.4).By (2.3),

N

(4n(f

(x+ y

2n

)+ f

(x− y

2n

)− 2f

(x

2n

)− 2f

(y

2n

))− ρ

(4n(

2f

(x+ y

2n+1

)+ 2f

(x− y2n+1

)− f

(x

2n

)− f

(y

2n

))), 4nt

)≥ t

t+ ϕ(x2n ,

y2n)

for all x, y ∈ X, all t > 0 and all n ∈ N. So

N

(4n(f

(x+ y

2n

)+ f

(x− y

2n

)− 2f

(x

2n

)− 2f

(y

2n

))− ρ

(4n(

2f

(x+ y

2n+1

)+ 2f

(x− y2n+1

)− f

(x

2n

)− f

(y

2n

))), t

)≥

t4n

t4n + ϕ

(x2n ,

y2n) =

t

t+ 4nϕ(x2n ,

y2n)

for all x, y ∈ X, all t > 0 and all n ∈ N. Since limn→∞t

t+4nϕ( x2n, y2n )

= 1 for all x, y ∈ X and all

t > 0,

Q(x+ y) +Q(x− y)− 2Q(x)− 2Q(y) = ρ

(2Q

(x+ y

2

)+ 2Q

(x− y

2

)−Q(x)−Q(y)

)for all x, y ∈ X. By Lemma 2.1, the mapping Q : X → Y is quadratic, as desired.

Corollary 2.3. Let θ ≥ 0 and let p be a real number with p > 2. Let X be a normed vectorspace with norm ‖ · ‖. Let f : X → Y be a mapping satisfying f(0) = 0 and

N

(f(x+ y) + f(x− y)− 2f(x)− 2f(y)− ρ

(2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)

), t

)≥ t

t+ θ(‖x‖p + ‖y‖p)(2.7)

for all x, y ∈ X and all t > 0. Then Q(x) := N -limn→∞ 4nf( x2n ) exists for each x ∈ X and

defines a quadratic mapping Q : X → Y such that

N (f(x)−Q(x), t) ≥ (2p − 4)t

(2p − 4)t+ 2θ‖x‖p

for all x ∈ X.

Proof. The proof follows from Theorem 2.2 by taking ϕ(x, y) := θ(‖x‖p + ‖y‖p) for all x, y ∈ X,as desired.


Φ(x, y) :=∞∑j=0

1

4jϕ(2jx, 2jy

)<∞




for all x, y ∈ X. Let f : X → Y be a mapping satisfying f(0) = 0 and (2.3). Then Q(x) := N -limn→∞

14n f (2nx) exists for each x ∈ X and defines a quadratic mapping Q : X → Y such

that

N (f(x)−Q(x), t) ≥ 1

t+ 14Φ(x, x)


Proof. It follows from (2.5) that N(f(x)− 1

4f(2x), 14 t)≥ t

t+ϕ(x,x) and so

N

(f(x)− 1

4f(2x), t

)≥ 4t

4t+ ϕ(x, x)=

t

t+ 14ϕ(x, x)

for all x ∈ X and all t > 0.The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.5. Let θ ≥ 0 and let p be a real number with 0 < p < 2. Let X be a normedvector space with norm ‖ · ‖. Let f : X → Y be a mapping satisfying f(0) = 0 and (2.7). ThenQ(x) := N -limn→∞

14n f(2nx) exists for each x ∈ X and defines a quadratic mapping Q : X → Y

such that

N (f(x)−Q(x), t) ≥ (4− 2p)t

(4− 2p)t+ 2θ‖x‖p

for all x ∈ X.


3. Quadratic ρ-functional inequality (0.2)

In this section, we prove the Hyers-Ulam stability of the quadratic ρ-functional inequality(0.2) in fuzzy Banach spaces. Let ρ be a real number with ρ 6= 1

2 . We need the following lemmato prove the main results.

Lemma 3.1. Let f : X → Y be a mapping satisfying f(0) = 0 and

2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y) = ρ (f(x+ y) + f(x− y)− 2f(x)− 2f(y)) (3.1)

for all x, y ∈ X. Then f : X → Y is quadratic.

Proof. Letting y = 0 in (3.1), we get 4f(x2

)− f(x) = 0 and so f(2x) = 4f(x) for all x ∈ X.

Thus

1

2f(x+ y)− 1

2f(x− y)− f(x)− f(y) = 2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)

= ρ(f(x+ y) + f(x− y)− 2f(x)− 2f(y))

and so f(x+ y) + f(x− y) = 2f(x) + 2f(y) for all x, y ∈ X, as desired.


Φ(x, y) :=∞∑j=0

4jϕ

(x

2j,y

2j

)<∞ (3.2)





N

(2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y)− ρ (f(x+ y) + f(x− y)− 2f(x)− 2f(y)) , t

)≥ t

t+ ϕ(x, y)(3.3)

for all x, y ∈ X and all t > 0. Then Q(x) := N -limn→∞ 4nf(x2n)

exists for each x ∈ X anddefines a quadratic mapping Q : X → Y such that

N (f(x)−Q(x), t) ≥ t

t+ Φ(x, 0)(3.4)


Proof. Letting y = 0 in (3.3), we get

N

(f(x)− 4f

(x

2

), t

)= N

(4f

(x

2

)− f(x), t

)≥ t

t+ ϕ(x, 0)(3.5)

for all x ∈ X. Hence

N

(4lf

(x

2l

)− 4mf

(x

2m

), t

)(3.6)

≥ min

N

(4lf

(x

2l

)− 4l+1f

(x

2l+1

), t

), · · · , N

(4m−1f

(x

2m−1

)− 4mf

(x

2m

), t

)= min

N

(f

(x

2l

)− 4f

(x

2l+1

),t

4l

), · · · , N

(f

(x

2m−1

)− 4f

(x

2m

),

t

4m−1

)

≥ min

t4l

t4l

+ ϕ(x2l, 0) , · · · , t

4m−1

t4m−1 + ϕ

(x

2m−1 , 0)

= min

t

t+ 4lϕ(x2l, 0) , · · · , t

t+ 4m−1ϕ(

x2m−1 , 0

)

≥ t

t+∑m−1j=l 4jϕ

(x2j, 0)

for all nonnegative integers m and l with m > l and all x ∈ X and all t > 0. It follows from (3.2)and (3.6) that the sequence 4nf( x

2n ) is a Cauchy sequence for all x ∈ X. Since Y is complete,the sequence 4nf( x

2n ) converges. So one can define the mapping Q : X → Y by

Q(x) := N - limn→∞

4nf(x

2n)

for all x ∈ X. Moreover, letting l = 0 and passing the limit m→∞ in (3.6), we get (3.4).By (3.2),

N

(4n(

2f

(x+ y

2n+1

)+ 2f

(x− y2n+1

)− f

(x

2n

)− f

(y

2n

))−ρ

(4n(f

(x+ y

2n

)+ f

(x− y

2n

)− 2f

(x

2n

)− 2f

(y

2n

))), 4nt

)≥ t

t+ ϕ(x2n ,

y2n)





N

(4n(

2f

(x+ y

2n+1

)+ 2f

(x− y2n+1

)− f

(x

2n

)− f

(y

2n

))−ρ

(4n(f

(x+ y

2n

)+ f

(x− y

2n

)− 2f

(x

2n

)− 2f

(y

2n

))), t

)≥

t4n

t4n + ϕ

(x2n ,

y2n) =

t

t+ 4nϕ(x2n ,

y2n)

for all x, y ∈ X, all t > 0 and all n ∈ N. Since limn→∞t

t+4nϕ( x2n, y2n )


t > 0,

2Q

(x+ y

2

)+ 2

(x− y

2

)−Q(x)−Q(y) = ρ (Q (x+ y) +Q (x− y)− 2Q(x)− 2Q(y))

for all x, y ∈ X=. By Lemma 3.1, the mapping Q : X → Y is quadratic, as desired.

Corollary 3.3. Let θ ≥ 0 and let p be a real number with p > 2. Let X be a normed vectorspace with norm ‖ · ‖. Let f : X → Y be a mapping satisfying f(0) = 0 and

N(2f

(x+ y

2

)+ 2f

(x− y

2

)− f(x)− f(y) (3.7)

−ρ (f(x+ y) + f(x− y)− 2f(x)− 2f(y)) , t) ≥ t

t+ θ(‖x‖p + ‖y‖p)for all x, y ∈ X and all t > 0. Then Q(x) := N -limn→∞ 4nf( x

2n ) exists for each x ∈ X anddefines a quadratic mapping Q : X → Y such that

N (f(x)−Q(x), t) ≥ (2p − 4)t

(2p − 4)t+ 2pθ‖x‖p

for all x ∈ X.



Φ(x, y) :=∞∑j=1

1

4jϕ(2jx, 2jy

)<∞

for all x, y ∈ X. Let f : X → Y be a mapping satisfying f(0) = 0 and (3.3). Then Q(x) := N -limn→∞

14n f (2nx) exists for each x ∈ X and defines a quadratic mapping Q : X → Y such

that

N (f(x)−Q(x), t) ≥ t

t+ Φ(x, 0)


Proof. It follows from (3.5) that N(f(x)− 1

4f(2x), t4

)≥ t

t+ϕ(2x,0) and so

N

(f(x)− 1

4f(2x), t

)≥ 4t

4t+ ϕ(2x, 0)=

t

t+ 14ϕ(2x, 0)

for all x ∈ X and all t > 0.The rest of the proof is similar to the proof of Theorem 3.2.




Corollary 3.5. Let θ ≥ 0 and let p be a real number with 0 < p < 2. Let X be a normedvector space with norm ‖ · ‖. Let f : X → Y be a mapping satisfying f(0) = 0 and (3.7). ThenQ(x) := N -limn→∞

14n f(2nx) exists for each x ∈ X and defines a quadratic mapping Q : X → Y

such that

N (f(x)−Q(x), t) ≥ (4− 2p)t

(4− 2p)t+ 2pθ‖x‖p

for all x ∈ X.


Acknowledgments

This research was supported by Hanshin University Research Grant.

References

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397–407.[24] C. Park, K. Ghasemi, S. G. Ghaleh and S. Jang, Approximate n-Jordan ∗-homomorphisms in C∗-algebras,

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Choonkil ParkResearch Institute for Natural Sciences, Hanyang University, Seoul 04763, KoreaE-mail address: [email protected]

Sun Young JangDepartment of Mathematics, University of Ulsan, Ulsan 44610, KoreaE-mail address: [email protected]

Sungsik YunDepartment of Financial Mathematics, Hanshin University, Gyeonggi-do 18101, KoreaE-mail address: [email protected]



ON A DOUBLE INTEGRAL EQUATION INCLUDING A SET OFTWO VARIABLES POLYNOMIALS SUGGESTED BY LAGUERRE

POLYNOMIALS

M. AL·I ÖZARSLAN AND CEMAL·IYE KÜRT

Abstract. In this paper, we introduce general classes of bivariate and Mittag-Le er functions E(;;;;) 1; 2 (x; y) and Laguerre polynomials L(;; ;;)n;m (x; y).We investigate double fractional integrals and derivative properties of theabove mentioned classes. We further obtain linear generating function forL(;; ;;)n;m (x; y) in terms of E(;;;;) 1; 2 (x; y). Finally, we calculate doubleLaplace transforms of the above mentioned classes and then we consider a gen-eral singular integral equation with L(;; ;;)n;m (x; y) in the kernel and obtain

the solution in terms of E(;;;;) 1; 2 (x; y).

1. Introduction

The special function of the form [7]

E(z) =1Xk=0

zk

(k + 1)(1.1)

( 2 C;Re() > 0; z 2 C)

and more general function [12] of (1.1)

E;(z) =1Xk=0

zk

(k + )(1.2)

(; 2 C;Re();Re() > 0; z 2 C)

are known as Mittag-Le er functions the rst of which was introduced by Swedishmathematician G. Mittag-Le er and the second one by Wiman.Setting = = 1, the equation (1.2) becomes the exponential function ez.

When 0 < < 1, it bridges an interpolation between the pure exponential functionez and a geometric function

1

1 z =1Xn=0

zn:

(jzj < 1)

Key words and phrases. Double fractional integrals and derivatives, Bivariate Mittag-Le erfunction, Bivariate Laguerre polynomials, Double generating functions, Singular double integralequation, Double Laplace integral.

2010 Mathematics Subject Classication. 33E12, 33C45, 45E10.

1


1198 OZARSLAN et al 1198-1207

2 M. AL·I ÖZARSLAN AND CEMAL·IYE KÜRT

A further generalization of (1.2) was introduced by Prabhakar (see [9]) as

E ;(z) =1Xn=0

( )n(n+ )

zn

n!(1.3)

(; ; 2 C;Re();Re() > 0)

where the Pochammer symbol [11], ( )n, is dened as

( )n =( + n)

( )=

1 ;n = 0; 6= 0

( + 1) ( + n 1) ;n = 1; 2;

In the special case, we have the polynomials Zn (x; k) (see [6],[10]) which weredened by

Zn (x; k) =(kn+ + 1)

n!Enk;+1(x

k) .

(Re() > 0; k 2 Z0+)

Note that, in [6] and [10], generating functions, integrals and recurrence relationswere developed for the polynomials Zn (x; k) of degree n in x

k, which form oneset of the biorthogonal pair corresponding to the weight function exx over theinterval (0;1).For k = 1, we have Zn (x; 1) = L

n(x) where L

n(x) is the usual Laguerre polyno-

mial which were given as follows

Ln(x) =(1 + )nn!

1F1(n; 1 + ;x)

where

1F1(n; 1 + ;x) =nXk=0

(n)k(1 + )k

xk

k!:

Very recently, a class of polynomials Z()n1; ;njx1; ; xj ; 1; ; j

(see [8])

suggested by the multivariate Laguerre polynomials were dened by

Z()n1; ;nj

x1; ; xj ; 1; ; j

(1.4)

=1n1 + + jnj + + 1

n1! nj !

n1; ;njXk1; ;kj=0

(n1)k1 (nj)kj x1k11 xjkjj

1k1 + + jkj + + 1

k1! kj !

:

(; 1; ; j 2 C; Re(i) > 0 (i = 1; ; j))

Obviously Z()n1; ;njx1; ; xj ; 1; ; j

gives L()n1; ;nj (x1; ; xj) when 1 =

= j = 1, where L()n1; ;nj (x1; ; xj) is the multivariable Laguerre polynomial

[2] given by

L()n1; ;nj (x1; ; xj) =

(n1 + + nj + + 1)n1! nj !

n1; ;njXk1; ;kj=0

(n1)k1 (nj)kj xk11 x

kjj

(k1 + + kj + + 1) k1! kj !:



DOUBLE INTEGRAL EQUATION INCLUDING TWO VARIABLES LAGUERRE POLYNOMIALS3

It is known that the multivariate Mittag-Le er functions were dened by the mul-tiple series as [13]

E( 1; ; j)1; ;j ; (x1; ; xj) =

1Xk1; ;kj=0

( 1)k1 jkjxk11 x

kjj

1k1 + + jkj +

k1! kj !

:(1.5)

(; 1; ; j ; 1; ; j 2 C; Re(i) > 0 (i = 1; ; j))

Note that the function in (1.5) is a special case of the generalized Lauricella seriesin several variables introduced and investigated by Srivastava and Daoust [16] (seealso see [14],[15]). Also, when j = 1; 1 = ; = ; 1 = ; x1 = z, the function(1.5) reduces to (1.3).The polynomials Z()n1; ;nj

x1; ; xj ; 1; ; j

can be represented in terms of

the multivariate Mittag-Le er functions as follows (see [8]):

Z()n1; ;nj

x1; ; xj ; 1; ; j

(1.6)

=1n1 + + jnj + + 1

n1! nj !

E(n1; ;nj)1; ;j ;+1 (x

11 ; ; x

jj ):

Clearly, setting 1 = 2 = = j = 1 in (1.6) gives

L()n1; ;nj (x1; ; xj) =

(n1 + + nj + + 1)n1! nj !

E(n1; ;nj)1; ;1;+1 (x1; ; xj):

Very recently, a slight motivated form of the multivariate Mittag-Le er functionswere introduced and investigated in [3].On the other hand, a nontrivial two variables Mittag-Le er functions were de-

ned in [4] by

E1(x; y) = E1

1; 1; 2; 1

1; 2; 2; 2; 3; 3; 3

xy

=1Xm=0

1Xn=0

( 1)1m( 2)1n

(1 + 2m+ 2n)

xm

(2 + 3m)

yn

(3 + 3n):

( 1; 2; 1; 2; 3; x; y 2 C;minf1; 2; 3; 1; 2; 3g > 0g)Motivated essentially by the above denitions and investigations, in this paper,

we introduce a class of bivariate Mittag-Le er function

(1.7) E(;;;;) 1; 2(x; y) =

1Xk1=0

1Xk2=0

( 1)k1( 2)k2xk1yk2

(k1 + k2 + )(k2 + )k1!k2!

where 1; 2; ; ; ; ; 2 C;Re(+ ) > 0 and Re() > 0:According to the convergence conditions investigated by Srivastava and Daoust

([15], p. 155) for the generalized Lauricella series in two variables, the series in(1.7) converges absolutely for Re(+ ) > 0 and Re() > 0.We also introduce a general class of bivariate Laguerre polynomials

L(;; ;;)n;m (x; y)(1.8)

=(n+ m+ + 1)

( + m)

nXk1=0

mXk2=0

(n)k1(m)k2xk1yk2(k1 + k2 + + 1)(k2 + )k1!k2!

where ; ; ; ; 2 C;Re();Re();Re();Re() > 0;Re( ) > 1:




Comparing (1.7) and (1.8), we see that

(1.9) L(;; ;;)n;m (x; y) =(n+ m+ + 1)

( + m)E(;;;;)n;m (x; y):

This paper is organized as follows. In section 2, we calculate the double fractionalintegrals and derivatives of the above mentioned classes (1.7) and (1.8). Lineargenerating functions for L(;; ;;)n;m (x; y) are given in terms of E(;;;;) 1; 2 (x; y)in Section 3. In the last section, we rst investigate double Laplace transformsof the above mentioned classes and then we consider a general singular integralequation with L(;; ;;)n;m (x; y) in the kernel and obtain the solution by means ofE(;;;;) 1; 2 (x; y).

2. Fractional integrals and derivatives

This section aims to provide the fractional integral formulas of the functionsE(;;;;) 1; 2 (x; y) and L(;; ;;)n;m (x; y) : Throughout this section, we assume thatRe();Re() > 0;Re();Re() > 0;Re( ) > 1:

Denition 2.1. ([1],[8])Let = [a; b] be a nite interval of the real axis. TheRiemann-Liouville fractional integral of order 2 C (Re () > 0) is dened by

xIa+ [f ] =

1

()

Z x

a

f(t) dt

(x t)1 : (x > a;Re () > 0)

Similarly, the partial fractional integrals of a function f(x; t), where (x; t) 2 RRis dened as follows:

xIa+f(x; t) =

1

()

Z x

a

(x )1f(; t)d ; (x > a;Re () > 0)

tIa+f(x; t) =

1

()

Z t

b

(t )1f(x; )d ; (t > b;Re () > 0)

tIb+xI

a+f(x; t)

=1

()()

Z t

b

Z x

a

(t )1(x )1f(; )dd : (x > a; y > b;Re () > 0;Re () > 0)

Denition 2.2. ([1],[8])The Riemann-Liouville fractional derivative of order 2C (Re () 0) is dened by

xDa+ [f ] =

d

dx

n1

(n )

Z x

a

(x )n1f()d ; (n = [Re()] + 1; x > a)

where, as usual, [Re()] means the integral part of Re():




Similarly, the partial fractional derivatives of a function f(x; t), where (x; t) 2R R is dened as follows:

xDa+f(x; t) =

d

dx

n1

(n )

Z x

a

(x )n1f(; t)d ; (n = [Re()] + 1; x > a)

tDb+f(x; t) =

d

dt

m1

(m )

Z t

b

(t )m1f(x; )d ; (m = [Re()] + 1; t > b)

tDb+xD

a+f(x; t)

=

d

dt

md

dx

n1

(n )1

(m )

Z t

b

Z x

a

(t )m1(x )n1f(; )dd :

(n = [Re()] + 1;m = [Re()] + 1; t > b; x > a)

Theorem 2.1. We have for Re(+ ) > 0 and Re() > 0 and () > 0, that

yI0+xI

0+

hx1y1E(;;;;) 1; 2

x; xy

i= x+1y+1E(;;;+;+) 1; 2

x; xy

Proof. Because of the hypothesis of the Theorem, we have a right to interchangeof the order of series and fractional integral operators, which yields

yI0+xI

0+

hx1y1E(;;;;) 1; 2

x; xy

i=

Z y

0

Z x

0

(y )1(x t)1()()

t1 1E(;;;;) 1; 2

t; t

dtd

=1

()()

1Xk1=0

1Xk2=0

( 1)k1( 2)k2(k1 + k2 + )(k2 + )k1!k2!

Z y

0

(y )1k2+1dZ x

0

(x t)1tk1+k2+1dt

=1

()()

1Xk1=0

1Xk2=0

( 1)k1( 2)k2xk1+k2++1yk2++1

(k1 + k2 + + )(k2 + + )k1!k2!

= x+1y+1E(;;;+;+) 1; 2

x; xy

In a similar manner, we have the following corollary:

Corollary 2.2. For Re() > 0 and Re() > 0, that

yI0+xI

0+

hx y1L(;; ;;)n;m

x; xy

i=(n+ m+ + 1)

( + m)

(+ + m)

(n+ m+ + + 1)x+ y+1L(;; +;;+)n;m

x; xy

:

Theorem 2.3. For Re(+ ) > 0,Re() > 0 and () > 0, that

yD0+xD

0+

hx1y1E(;;;;) 1; 2

x; xy

i= x1y1E(;;;;) 1; 2

x; xy

:




Proof. Because of the hypothesis of the Theorem, we have a right to interchangeof the order of series and fractional derivate operators, which yields

yD0+xD

0+

hx1y1E(;;;;) 1; 2

x; xy

i= yD

0+xD

0+

"x1y1

1Xk1=0

1Xk2=0

( 1)k1( 2)k2xk1xk2y

k2

(k1 + k2 + )(k2 + )k1!k2!

#

=1Xk1=0

1Xk2=0

( 1)k1( 2)k2xD0+yD

0+ [x

k1+k2+1yk2+1

]

(k1 + k2 + )(k2 + )k1!k2!

= (d

dy)m(

d

dx)n

1

(n )1

(m )

1Xk1=0

1Xk2=0

( 1)k1( 2)k2(k1 + k2 + )(k2 + )k1!k2!

Z y

0

(y )m1k2+1

d

Z x

0

(x )n1k1+k2+1d

= x1y11Xk1=0

1Xk2=0

( 1)k1( 2)k2xk1+k2y

k2

(k1 + k2 + )(k2 + )k1!k2!

= x1y1E(;;;;) 1; 2

x; xy

:

In a similar manner, we have the following corollary:

Corollary 2.4. For Re(+ ) > 0 and Re() > 0, that

yD0+xD

0+ [x

y1L(;; ;;)n;m

x; xy

]

=(n+ m+ + 1)

( + m)

( + m)(n+ m+ + 1)x

y1L(;; ;;)n;m

x; xy

:

3. Linear generating function

In this section, we provide a linear generating function for the polynomialsL(;; ;;)n;m (x; y) by means of two variables analogue of Mittag-Le er functionsdened in (1.7):

Theorem 3.1. For jt1j < 1 and jt2j < 1, 1; 2 2 C and ; ; ; ; 2 C, we have

1Xn=0

1Xm=0

( 1)n( 2)mL(;; ;;)n;m (x; y) ( + m)

(n+ m+ + 1)n!m!tn1 t

m2

= (1 t1) 1(1 t2) 2E(;;;; +1) 1; 2

xt11 t1

;yt21 t2

:




Proof. Direct calculations yield that

1Xn=0

1Xm=0

( 1)n( 2)mL(;; ;;)n;m (x; y) ( + m)

(n+ m+ + 1)n!m!tn1 t

m2

=1X

n;m=0

n;mXk1;k2=0

( 1)n( 2)m(n)k1(m)k2xk1yk2(k1 + k2 + + 1)( + k2)k1!k2!n!m!

tn1 tm2

=1X

n;m=0

n;mXk1;k2=0

(1)k1+k2( 1)n( 2)mxk1yk2(k1 + k2 + + 1)( + k2)k1!k2!(n k1)!(m k2)!

tn1 tm2 :

Letting n! n+ k1 and m! m+ k2, we get

1Xn;m=0

1Xk1;k2=0

(1)k1+k2( 1)n+k1( 2)m+k2xk1yk2(k1 + k2 + + 1)( + k2)k1!k2!(n)!(m)!

tn+k11 tm+k22 :

Since ( 1)n+k1 = ( 1 + k1)n( 1)k1 and ( 2)m+k2 = ( 2 + k2)m( 2)k2, we have

1Xk1;k2=0

( 1)k1( 2)k2(xt1)k1(yt2)k2(k1 + k2 + + 1)( + k2)k1!k2!

1Xn;m=0

( 1 + k1)n( 2 + k2)m(n)!(m)!

tn1 tm2

= (1 t1) 1(1 t2) 2E(;;;; +1) 1; 2

xt11 t1

;yt21 t2

:

Note that, because of the uniform converge of the series under the conditions jt1j < 1and jt2j < 1; we have interchanged the order of summations.

4. Singular double integral equation

In this section, we rst obtain the double Laplace transform of the functionsE(;;;;) 1; 2 (x; y) and L(;; ;;)n;m (x; y) : Then, we compute the double integral in-

volving the product of two E(;;;;) 1; 2 (x; y) functions in the integrand. Finally, we

solve a double integral equation with L(;; ;;)n;m (x; y) in the kernel, in terms of thetwo E(;;;;) 1; 2 (x; y) functions:As usual [5],

L2[f(x; t)] =Z 1

0

epxZ 1

0

estf(x; t)dtdx(4.1)

(x; t > 0; p; s 2 C)

denotes the double Laplace transform of f:

Lemma 4.1. For Re(1);Re(2);Re( + ) > 0;Re() > 0;Re(s1);Re(s2) > 0

and1s1 ; 2s1 s

2

< 1, we haveL2[x1y1E(;;;;) 1; 2

((1x); (2x

y))](s1; s2) =1

s1

1

s2(1

1

s1) 1(1 2

s1 s2

) 2 :




Proof. Using denition (4.1) and taking into account that1s1 < 1 and 2s1 s

2

< 1,we get

L2[x1y1E(;;;;) 1; 2((1x)

; (2xy))](s1; s2)

=1Xk1=0

1Xk2=0

( 1)k1( 2)k2k11 k22

(k1 + k2 + )(k2 + )k1!k2!

Z 1

0

xk1+k2+1es1xdx

Z 1

0

yk2+1es2ydy

=1

s1

1

s2

1Xk1=0

( 1)k1k1!

(1s1)k1

1Xk2=0

( 2)k2k2!

(2

s1 s2

)k2 =1

s1

1

s2(1

1

s1) 1(1 2

s1 s2

) 2 :

We deduce the following result from Lemma 4:1 by setting 1 = and usingequation (1.9).

Corollary 4.2. For Re(1);Re(2);Re();Re();Re();Re(s1);Re(s2) > 0 and1s1 ; 2s1 s2

< 1; we haveL2[t 1L(;; ;;)n;m ((1t); (2t

))](s1; s2)

=1

s +11

1

s2

(n+ m+ + 1)

(m+ )(1

1

s1)n(1 2

s1 s2

)m:

Theorem 4.3. Let 1; 2 2 C;Re(+ ) > 0 and Re() > 0: Then

Z y

0

Z x

0

h(x t)1(y )1E(;;;;) 1; 2

(1 (x t); 2 (x t)(y ))

t 1 1E(;;;; ) 3; 4(1 t

; 2 t)dtd

i= x+ y+E(;;;;) 1; 2

(1x; 2x

y)E(;;;;) 3; 4(1x

; 2xy):

Proof. Using the convolution theorem for the Laplace transform we have,

L2Z y

0

Z x

0

(x t)1(y )1E(;;;;) 1; 2(1 (x t);

2 (x t)(y ))t 1 1

E(;;;; ) 3; 4(1 t

; 2 t)dtd

i(s1; s2)

= L2[x1y1E(;;;;) 1; 2(1 x

; 2xy)]L2[x 1y1E(;;;; ) 3; 4

(1 t; 2 t

)](s1; s2)

=1

s1

1

s2

1

1

s1

1 1 2

s1 s2

! 21

s 1

1

s2

1

1

s1

3 1 2

s1 s2

! 4




We have for Re(s1);Re(s2) > 0

L2Z y

0

Z x

0

(x t)1(y )1E(;;;;) 1; 2(1 (x t);

2 (x t)(y ))(4.2)

t 1 1E(;;;; ) 1; 2(1 t

; 2 t)dtd

i(s1; s2)

=1

s+ 1

1

s+2

1

1

s1

1 1 2

s1 s2

! 2 1

1

s1

3 1 2

s1 s2

! 4= L2

hx+ y+E(;;;;) 1; 2

(1x; 2x

y)E(;;;;) 3; 4(1x

; 2xy)

i(s1; s2):

Taking inverse Laplace transform on both sides of (4.2), the result follows.

The next assertation follows from Theorem 4:5 by letting 1 = and takinginto account (1.9).

Corollary 4.4. Let 1; 2 2 C;Re();Re();Re( );Re() > 0: ThenZ y

0

Z x

0

(x t) (y )1L(;; 1;;1)n1;m1(1(x t); 2(x t)(y )

))

t 1L(;; 2;;2)n2;m2(1t; 2t

)dtd

= x 1+ 2+1y1+21L(;; 1;;1)n1;m1(1x; 2xy

)L(;; 2;;2)n2;m2

(1x; 2xy ):

Now, we consider the following double convolution equation:

(4.3)Z y

0

Z x

0

(x t) 1(y )1L(;; ;;)n;m ((1x); (2x

y))(t; )dtd = (x; y)

where Re( ) > 1:For the solution of the integral equation (4.3), we have the following theorem:

Theorem 4.5. The singular double integral equation (4.3) admits a locally inte-grable solution

(t; ) =(m+ )

(n+ m+ + 1)

Z y

0

Z x

0

(x t)1 2(y )21E(;;;;) 1; 2((1x)

; (2xy))[I10+ I20+ (t; )]dtd :

Proof. Applying double Laplace transform on both sides of (4.3), then using doubleconvolution theorem, we get

1

s +11

1

s2

(n+ m+ + 1)

(m+ )(1

1

s1)n(1 2

s1 s2

)mL2[(t; )](s1; s2) = L2[(t; )](s1; s2)

Therefore, we have,

L2[(t; )](s1; s2) =(m+ )

(n+ m+ + 1)

(s1) 1+1(s2)2(11s1)n(1 2

s1 s2

)mfs11 s22 L2[(t; )](s1; s2)g




Finally taking inverse Laplace transform on both sides and using Lemma 3:2 of[1] and Lemma 4:1, we get

(t; ) =(m+ )

(n+ m+ + 1)

Z y

0

Z x

0

(x t)1 2(y )21E(;;;;) 1; 2((1x)

; (2xy))[I10+ I20+ (t; )]dtd

and the proof is completed.

References

[1] Anwar, AMO, Jarad, F, Baleanu, D, Ayaz, F: Fractional Caputo Heat equation within thedouble Laplace transform, Rom. Journ. Phys., 58, 15-22 (2013).

[2] Carlitz, L: Bilinear generating functions for Laguerre and Lauricella polinomials in severalvariables, Rend. Sem. Mat. Univ. Padova, 43, 269-276 (1970).

[3] Gaboury, S, Özarslan, MA: Singular integral equation involving a multivariable analog ofMittag-Le er function, Advances in Di¤erence Equations, 252 (2014)

[4] Garg, M, Manohar, P, Kalla, SL: A Mittag-Le er-type function of two variables, Int. Trans.Special Funct., 24 (11), 934-944 (2013).

[5] K¬l¬cman, A, Gadain, HE: On the applications of Laplace and Sumudu transforms, Journalof the Franklin Institude, 347 (5), 848-862 (2010).

[6] Konhauser, JDE: Biorthogonal polynomials suggested by the Laguerre polynomials, PasicJ. Math., 21, 303-314 (1967).

[7] Mittag-Le er, GM:Sur la nouvelle function E(x), C. R. Acad. Sci. Paris, 137, 554-558(1903).

[8] Ozarslan MA: On a singular integral equation including a set of multivariate polynomialssuggested by Laguerre polynomials, Appl. Math. Comput., 229, 350-358 (2014).

[9] Prabhakar, TR: A singular integral equation with a generalized Mittag-Le er function in thekernel, Yokohama Math. J., 19, 7-15 (1971).

[10] Prabhakar, TR: On a set of polynomials suggested by Laguerre polynomials, Pacic J. Math.35 (1), 213-219 (1970).

[11] Rainville, ED: Special Functions, Macmillan Company, New York, (1960).[12] Saxena, RK, Mathai, AM, Haubold, HJ: On fractional kinetic equations. Astrophys. Space

Sci. 282(1), 281-287 (2002).[13] Saxena, RK, Kalla, SL, Saxena, R: Multivariate analogue of generalized Mittag-Le er func-

tion, Int. Trans. Special Funct., 22 (7), 533-548 (2011).[14] Srivastava, HM, Karlsson, PW: Multiple Gaussian Hypergeometric Series, Ellis Horwood

Limited, New York, (1985).[15] Srivastava, HM, Daust, MC: A note on the convergence of Kampé de Feriéts double hyper-

geometric series, Math. Nachr., 53, 151-159 (1972) .[16] Srivastava, HM, Daust, MC: Certain generalized Neumann expansion associated with Kampé

de Feriet function, Nederl. Akad., Wetensch. Proc. Ser. A, 72 31 (1969) 449-457 (Indag.Math.).

Eastern Mediterranean University, Department of Mathematics, Gazimagusa, TRNC,Mersin 10, Turkey, Email: [email protected], [email protected]



Generalized Inequalities of the type of Hermite-Hadamard-Fejer withQuasi-Convex Functions by way of k-Fractional Derivatives

A. Ali(1), G. Gulshan(1), R. Hussain(1), A. Latif(1), M. Muddassar(2 ∗)(1) Department of Mathematics, Mirpur University of Science & Technology, Mirpur, Pakistan,

[email protected], [email protected], [email protected],

[email protected](2) Department of Mathematics,University of Engineering & Technology, Taxila, Pakistan,

[email protected] (∗ for correspondence)

Abstract: In this article, Hermite-Hadamard-Fejer type inequalities are discussed with

quasi-convex functions and obtained the generalized results of the type using k-fractional

derivatives. And proposed some new bounds in terms of some special means.

Keywords: Hermite-Hadamard inequality, Hermite-Hadamard-Fejer inequality, quasi con-

vex functions, k-Riemann-Liouville fractional derivatives, Holder’s integral inequality,

Power mean inequality.

1. INTRODUCTION

The function f : I ⊂ R −→ R is said to be convex on I if for every x, y ∈ I and t ∈ [0, 1],

we get

f(tx+ (1− t)y) ≤ tf(x) + (1− t)f(y).

Let f : I ⊂ R −→ R be a convex function on the interval I of real numbers and a, b ∈ Iwith a < b, f satisfies the following well-known Hermite-Hadamard type inequality

f

(a+ b

2

)≤ 1

b− a

∫ b

a

f(t)dt ≤ f(a) + f(b)

2.

Definition 1. The function f : I ⊂ R −→ R is said to be quasi-convex if

f(tx+ (1− t)y) ≤ max f(x), f(y) ,

for every x, y ∈ I and t ∈ [0, 1] (see [4]).

In [3] Mubeen and Habibullah introduced the following class of fractional derivatives.

1


1208 A. Ali et al 1208-1219

2 A. Ali, G. Gulshan, R. Hussain, A. Latif, M. Muddassar and J. Park

Definition 2. Let f ∈ L[a, b], then k-Riemann-Liouville fractional derivatives kJαa+f(x)

and kJαb−f(x) of order α > 0 are defined by

kJαa+f(x) =

1

kΓk(α)

∫ x

a

(x− t)αk−1f(t)dt, (0 ≤ a < x < b)

and

kJαb−f(x) =

1

kΓk(α)

∫ b

x

(t− x)αk−1f(t)dt, (0 ≤ a < x < b)

respectively, where k > 0 and Γk(α) is the k-gamma function given as Γk(α) =∫∞0tα−1e

−tkk dt.

Furthermore Γk(α+ k) = αΓk(α) and kJ0a+f(x) = kJ

0b−f(x) = f(x).

In [1] Fejer established the following inequality.

Lemma 1. Let f : [a, b] ⊂ R −→ R be a convex function, the inequality

f

(a+ b

2

)∫ b

a

g(x)dx ≤∫ b

a

f(x)g(x)dx ≤ f(a) + f(b)

2

∫ b

a

g(x)dx

holds, where g : [a, b] ⊂ R −→ R is non-negative integrable and symmetric to a+b2 . This

inequality is called Hermite-Hadamard-Fejer inequality.

Lemma 2. ([7]) For 0 < t ≤ 1 and 0 ≤ a < b, we get

|at − bt| ≤ (b− a)t.

E. Set et al. established the following Lemma in [6].

Lemma 3. Let f : [a, b] ⊂ R −→ R be a differentiable mapping on (a, b) and g : [a, b] ⊂R −→ R. If f

′, g ∈ L[a, b], the following identity for fractional derivatives holds

f

(a+ b

2

)[Jα( a+b2 )

−g(a) + Jα( a+b2 )

+g(b)

]−[Jα( a+b2 )

−(fg)(a) + Jα( a+b2 )

+(fg)(b)

]=

1

Γ(α)

∫ b

a

m(t)f′(t)dt (1.1)

where

m(t) =

∫ ta(s− a)α−1g(s)ds t ∈

[a, a+b2

)−∫ bt

(b− s)α−1g(s)ds t ∈[a+b2 , b

].

Iscan obtained the following lemma in [2].


1209 A. Ali et al 1208-1219

Generalized Inequalities of the type of Hermite-Hadamard-Fejer with Quasi-Convex Functions by way of k-Fractional Derivatives 3

Lemma 4. Let f : [a, b] ⊂ R −→ R be a differentiable mapping on (a, b) and f′ ∈ L[a, b].

If g : [a, b] −→ R is integrable and symmetric to a+b2 , the following identity for fractional

derivatives holds

f(a) + f(b)

2[Jαa+g(b) + Jαb−g(a)]− [Jαa+(fg)(b) + Jαb−(fg)(a)]

=1

Γ(α)

∫ b

a

(∫ t

a

(b− s)α−1g(s)ds−∫ b

t

(s− a)α−1g(s)ds

)f′(t)dt (1.2)

where α > 0.

In the present paper motivated by the recent results given in [5] we established some

Hermite-Hadamard-Fejer type inequalities for quasi-convex functions via k-fractional deriva-

tives.

2. MAIN FINDINGS

Throughout this paper, let I be an interval on R and let ||g||[a,b],∞ = supt∈[a,b]g(t) for

continuous function g : [a, b]R −→ R.

The following identity is the generalization of identity (1.1) in Lemma 3 for k-fractional

derivatives.

Lemma 5. Let f : [a, b] ⊂ R −→ R be a differentiable mapping on (a, b) and g : [a, b] ⊂R −→ R. If f

′, g ∈ L[a, b], the following identity for k-fractional derivatives holds

f

(a+ b

2

)[kJ

α

( a+b2 )−g(a) + kJ

α

( a+b2 )+g(b)

]−[kJ

α

( a+b2 )−(fg)(a) + kJ

α

( a+b2 )+(fg)(b)

]=

1

Γk(α)

∫ b

a

m(t)f′(t)dt

where

m(t) =

∫ ta(s− a)

αk−1g(s)ds t ∈

[a, a+b2

)−∫ bt

(b− s)αk−1g(s)ds t ∈[a+b2 , b

].

Here the identity (1.2) of Lemma 4 is also generalized for k-fractional derivatives.

Lemma 6. Let f : [a, b] ⊂ R −→ R be a differentiable mapping on (a, b) and f′ ∈ L[a, b].

If g : [a, b] ⊂ R −→ R is integrable and symmetric to a+b2 , the following for k-fractional


1210 A. Ali et al 1208-1219


derivatives holds

f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

=1

Γk(α)

∫ b

a

(∫ t

a

(b− s)αk−1g(s)ds−∫ b

t

(s− a)αk−1g(s)ds

)f′(t)dt

where αk > 0.

Theorem 1. Let f : I ⊂ R −→ R be a differentiable mapping on Io and f′ ∈ L[a, b] and

g : [a, b] ⊂ R −→ R is continuous. If |f ′ |q is quasi-convex function on [a, b], q ≥ 1, the

following inequality for k-fractional derivatives holds

∣∣∣∣f (a+ b

2

)[kJ

α

( a+b2 )−g(a) + kJ

α

( a+b2 )+g(b)

]−[kJ

α

( a+b2 )−(fg)(a) + kJ

α

( a+b2 )+(fg)(b)

]∣∣∣∣≤

(b− a)αk+1||g||[a,b],∞

2αk

(αk + 1

)Γk(α+ k)

(max

|f′(a)|q, |f

′(b)|q

) 1q

where αk > 0.

Proof. Since |f ′ |q is quasi-convex on [a, b], we know that for t ∈ [a, b]

|f′(t)|q =

∣∣∣∣f ′ ( b− tb− aa+

t− ab− a

b

)∣∣∣∣q ≤ max|f′(a)|q, |f

′(b)|q

.

Using lemma 5, power mean inequality and the fact that |f ′ |q is quasi-convex function on

[a, b], it follows that

∣∣∣∣f (a+ b

2

)[kJ

α

( a+b2 )−g(a) + kJ

α

( a+b2 )+g(b)

]−[kJ

α

( a+b2 )−(fg)(a) + kJ

α

( a+b2 )+(fg)(b)

]∣∣∣∣≤ 1

Γk(α)

(∫ a+b2

a

∣∣∣∣∫ t

a


∣∣∣∣ dt)1− 1

q(∫ a+b

2

a

∣∣∣∣∫ t

a


∣∣∣∣ |f ′(t)|qdt) 1q

+1

Γk(α)

(∫ b

a+b2

∣∣∣∣∣∫ b

t

(b− s)αk−1g(s)ds

∣∣∣∣∣ dt)1− 1

q(∫ b

a+b2

∣∣∣∣∣∫ b

t


∣∣∣∣∣ |f ′(t)|qdt) 1q


1211 A. Ali et al 1208-1219


≤||g||[a, a+b2 ],∞

Γk(α)

(∫ a+b2

a

∣∣∣∣∫ t

a

(s− a)αk−1ds

∣∣∣∣ dt)1− 1

q(∫ a+b

2

a

∣∣∣∣∫ t

a

(s− a)αk−1ds

∣∣∣∣ |f ′(t)|qdt) 1q

+||g||[ a+b2 ,b],∞

Γk(α)

(∫ b

a+b2

∣∣∣∣∣∫ b

t

(b− s)αk−1ds

∣∣∣∣∣ dt)1− 1

q(∫ b

a+b2

∣∣∣∣∣∫ b

t

(b− s)αk−1ds

∣∣∣∣∣ |f ′(t)|qdt) 1q

≤ 1

Γk(α+ k)

((b− a)

αk+1

2αk+1

(αk + 1

))1− 1q(

(b− a)αk+1

2αk+1

(αk + 1

)) 1q (||g||[a, a+b2 ],∞ + ||g||[ a+b2 ,b],∞

)(

max|f′(a)|q, |f

′(b)|q

) 1q

≤(b− a)

αk+1||g||[a,b],∞

2αk

(αk + 1

)Γk(α+ k)

(max

|f′(a)|q, |f

′(b)|q

) 1q

where

∫ a+b2

a

∣∣∣∣∫ t

a

(s− a)αk−1ds

∣∣∣∣ dt =

∫ b

a+b2

∣∣∣∣∣∫ b

t

(b− s)αk−1ds

∣∣∣∣∣ dt =(b− a)

αk+1

2αk+1

(αk + 1

)αk

Which completes the proof.

Corollary 1. If we choose g(x) = 1 and α = k in Theorem 1, we get

∣∣∣∣∣ 1

b− a

∫ b

a

f(x)dx− f(a+ b

2

)∣∣∣∣∣ ≤ b− a4

(max

|f′(a)|q, |f

′(b)|q

) 1q

.

Theorem 2. Let f : I ⊂ R −→ R be a differentiable mapping on Io and f′ ∈ L[a, b] and

g : [a, b] ⊂ R −→ R is continuous. If |f ′ |q is quasi-convex function on [a, b], q > 1, the

following inequality for k-fractional derivatives holds

∣∣∣∣f (a+ b

2

)[kJ

α

( a+b2 )−g(a) + kJ

α

( a+b2 )+g(b)

]−[kJ

α

( a+b2 )−(fg)(a) + kJ

α

( a+b2 )+(fg)(b)

]∣∣∣∣≤ (b− a)

αk+1||g||∞

2αk

(αk p+ 1

) 1p Γk(α+ k)

(max

|f′(a)|q, |f

′(b)|q

) 1q

where 1p + 1

q = 1.

Proof. Using Lemma 5, Holder’s inequality and the fact that |f ′ |q is quasi-convex function

on [a, b], it follows that


1212 A. Ali et al 1208-1219


∣∣∣∣f (a+ b

2

)[kJ

α

( a+b2 )−g(a) + kJ

α

( a+b2 )+g(b)

]−[kJ

α

( a+b2 )−(fg)(a) + kJ

α

( a+b2 )+(fg)(b)

]∣∣∣∣≤ 1

Γk(α)

(∫ a+b2

a

∣∣∣∣∫ t

a


∣∣∣∣ |f ′(t)|dt+

∫ b

a+b2

∣∣∣∣∣∫ b

t


∣∣∣∣∣ |f ′(t)|dt)

≤ 1

Γk(α)

(∫ a+b2

a

∣∣∣∣∫ t

a


∣∣∣∣p dt) 1p(∫ a+b

2

a

|f′(t)|qdt

) 1q

+1

Γk(α)

(∫ b

a+b2

∣∣∣∣∣∫ b

t


∣∣∣∣∣p

dt

) 1p (∫ b

a+b2

|f′(t)|qdt

) 1q

=1

Γk(α)

(∫ a+b2

a

∣∣∣∣∫ t

a


∣∣∣∣p dt) 1p

(∫ a+b2

a

|f′(t)|qdt

) 1q

+

(∫ b

a+b2

|f′(t)|qdt

) 1q

≤ ||g||∞

Γk(α)

((b− a)

αk p+1

2αk p+1

(αk p+ 1

) (αk

)p) 1p

(∫ a+b2

a

max|f′(a)|q, |f

′(b)|q

dt

) 1q

+

(∫ b

a+b2

max|f′(a)|q, |f

′(b)|q

dt

) 1q

=

(b− a)αk+1||g||∞

2αk

(αk p+ 1

) 1p Γk(α+ k)

(max

|f′(a)|q, |f

′(b)|q

) 1q

.

Where ∫ a+b2

a

∣∣∣∣∫ t

a

(s− a)αk−1ds

∣∣∣∣p dt =(b− a)

αk p+1

2αk p+1

(αk p+ 1

) (αk

)p .

Corollary 2. If we choose g(x) = 1 and α = k in Theorem 2, then we get∣∣∣∣∣ 1

b− a

∫ b

a

f(x)dx− f(a+ b

2

)∣∣∣∣∣ ≤ b− a2(p+ 1)

1p

(max

|f′(a)|q, |f

′(b)|q

) 1q

.

Theorem 3. Let f : I ⊂ R −→ R be a differentiable mapping on Io and f′ ∈ L[a, b]. If

|f ′ | is quasi-convex function on [a, b] and g : [a, b] ⊂ R −→ R is continuous and symmet-

ric to a+b2 , the following inequality for k-fractional derivatives holds


1213 A. Ali et al 1208-1219


∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 2(b− a)

αk+1||g||∞(

αk + 1

)Γk(α+ k)

(1− 1

2αk

)max

|f′(a)|, |f

′(b)|

where αk > 0.

Proof. From Lemma 6, we get

∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 1

Γk(α)

∫ b

a

∣∣∣∣∣∫ t

a


t


∣∣∣∣∣ |f ′(t)|dt.Since |f ′ | is quasi-convex on [a, b], we know that for t ∈ [a, b]

|f′(t)| =

∣∣∣∣f ′ ( b− tb− aa+

t− ab− a

b

)∣∣∣∣ ≤ max|f′(a)|, |f

′(b)|

and since g : [a, b] ⊂ R −→ R is continuous and symmetric to a+b2 we can write

∫ b

t

(s− a)αk−1g(s)ds =

∫ a+b−t

a

(b− s)αk−1g(a+ b− s)ds

=

∫ a+b−t

a


therefore we get

!

∣∣∣∣∣∫ t

a


t


∣∣∣∣∣=

∫ a+b−t

t


≤

∫ a+b−tt

∣∣(b− s)αk−1g(s)∣∣ ds, t ∈

[a, a+b2

]∫ ta+b−t

∣∣(b− s)αk−1g(s)∣∣ ds, t ∈

[a+b2 , b

].

(2.3)


1214 A. Ali et al 1208-1219


Therefore we get∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 1

Γk(α)

[∫ a+b2

a

∫ a+b−t

t

∣∣(b− s)αk−1g(s)∣∣ dsdt+

∫ b

a+b2

∫ t

a+b−t

∣∣(b− s)αk−1g(s)∣∣ dsdt](

max|f′(a)|, |f

′(b)|)

≤ ||g||∞Γk(α+ k)

(∫ a+b2

a

[(b− t)αk − (t− a)

αk

]dt+

∫ b

a+b2

[(t− a)

αk − (b− t)αk

]dt

)(

max|f′(a)|, |f

′(b)|)

=2(b− a)

αk+1||g||∞(

αk + 1

)Γk(α+ k)

(1− 1

2αk

)(max

|f′(a)|, |f

′(b)|)

since ∫ a+b2

a

(b− t)αk dt =

∫ b

a+b2

(t− a)αk dt =

(b− a)αk+1

(2αk+1 − 1

)2αk+1

(αk + 1

)and ∫ a+b

2

a

(t− a)αk dt =

∫ b

a+b2

(b− t)αk dt =(b− a)

αk+1

2αk+1

(αk + 1

) .

Corollary 3. In Theorem 3, if we take g(x) = 1, we get the inequality∣∣∣∣f(a) + f(b)

2− Γk(α+ k)

2(b− a)αk

[kJαa+f(b) + kJ

αb−f(a)]

∣∣∣∣≤ b− a(

αk + 1

) (1− 1

2αk

)max

|f′(a)|, |f

′(b)|.

.


|f ′ |q, q ≥ 1 is quasi-convex function on [a, b] and g : [a, b] ⊂ R −→ R is continuous and

symmetric to a+b2 , the following inequality for k-fractional derivatives holds∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 2(b− a)

αk+1||g||∞(

αk + 1

)Γk(α+ k)

(1− 1

2αk

)(max

|f′(a)|q, |f

′(b)|q

) 1q

where αk > 0.


1215 A. Ali et al 1208-1219


Proof. From Lemma 6, power mean inequality, inequality (2.3) and the quasi-convexity

of |f ′ |q , we get

∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 1

Γk(α)

(∫ b

a

∣∣∣∣∣∫ a+b−t

t


∣∣∣∣∣ dt)1− 1

q(∫ b

a

∣∣∣∣∣∫ a+b−t

t


∣∣∣∣∣ |f ′(t)|qdt) 1q

≤ 1

Γk(α)

(∫ a+b2

a

[∫ a+b−t

t

∣∣(b− s)αk−1g(s)∣∣ ds] dt+

∫ b

a+b2

[∫ t

a+b−t

∣∣(b− s)αk−1g(s)∣∣ ds] dt)1− 1

q

×

(∫ a+b2

a

[∫ a+b−t

t

|(b− s)αk−1g(s)|ds

]|f′(t)|qdt+

∫ b

a+b2

[∫ t

a+b−t|(b− s)αk−1g(s)|ds

]|f′(t)|qdt

) 1q

≤ 2(b− a)αk+1||g||∞(

αk + 1

)Γk(α+ k)

(1− 1

2αk

)(max

|f′(a)|q, |f

′(b)|q

) 1q

where∫ a+b2

a

[∫ a+b−t

t

|(b− s)αk−1g(s)|ds

]dt+

∫ b

a+b2

[∫ t

a+b−t|(b− s)αk−1g(s)|ds

]dt

=2(b− a)

αk+1

αk

(αk + 1

) (1− 1

2αk

).

Theorem 5. Let f : I ⊂ R −→ R be a differentiable mapping on Io and f′ ∈ L[a, b].

If |f ′ |q, q > 1 is quasi-convex function on [a, b], and g : [a, b] −→ R is continuous and

symmetric to a+b2 , the following inequality for k-fractional derivatives holds

∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 2

1p (b− a)

αk+1||g||∞(

αk p+ 1

) 1p Γk(α+ k)

(1− 1

2αk p

) 1p (

max|f′(a)|q, |f

′(b)|q

) 1q

,

where αk > 0.


1216 A. Ali et al 1208-1219


Proof. From Lemma 6, Holder’s inequality, inequality (2.3) and the quasi-convexity of

|f ′ |q , we get

∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ 1

Γk(α)

(∫ b

a

∣∣∣∣∣∫ a+b−t

t


∣∣∣∣∣p

dt

) 1p(∫ b

a

|f′(t)|qdt

) 1q

≤ ||g||∞Γk(α+ k)

(∫ a+b2

a

[(b− t)αk − (t− a)

αk

]pdt+

∫ b

a+b2

[(t− a)

αk − (b− t)αk

]pdt

) 1p

(∫ b

a

max|f′(a)|q, |f

′(b)|q

dt

) 1q

≤ ||g||∞(b− a)αk+1

Γk(α+ k)

(∫ 12

0

[(1− t)αk − tαk

]pdt+

∫ 1

12

[tαk − (1− t)αk

]pdt

) 1p

(max

|f′(a)|q, |f

′(b)|q

) 1q

≤ ||g||∞(b− a)αk+1

Γk(α+ k)

(∫ 12

0

[(1− t)αk p − tαk p

]dt+

∫ 1

12

[tαk p − (1− t)αk p

]dt

) 1p

(max

|f′(a)|q, |f

′(b)|q

) 1q

≤ 21p (b− a)

αk+1||g||∞(

αk p+ 1

) 1p Γk(α+ k)

(1− 1

2αk p

) 1p (

max|f′(a)|q, |f

′(b)|q

) 1q

.

Where[(1− t)αk − tαk

]p ≤ (1− t)αk p − tαk p, for t ∈[0,

1

2

]and [

tαk − (1− t)αk

]p ≤ tαk p − (1− t)αk p, for t ∈[

1

2, 1

]

which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. Hence the

proof is complete.


|f ′ |q, q > 1 is quasi-convex function on [a, b], and g : [a, b] ⊂ R −→ R is continuous and

symmetric to a+b2 , the following inequality for k-fractional derivatives holds


1217 A. Ali et al 1208-1219


∣∣∣∣f(a) + f(b)

2[kJ

αa+g(b) + kJ

αb−g(a)]− [kJ

αa+(fg)(b) + kJ

αb−(fg)(a)]

∣∣∣∣≤ (b− a)

αk+1||g||∞(

αk p+ 1

) 1p Γk(α+ k)

(max

|f′(a)|q, |f

′(b)|q

) 1q

where 0 < αk ≤ 1, 1p + 1

q = 1.

Proof. The inequality can also be proved by using Lemma 6, Holder’s inequality, inequal-

ity (2.3), the quasi-convexity of |f ′ |q and Lemma 2.

3. APPLICATIONS TO SPECIAL MEANS

We now consider the means of arbitrary real numbers ξ, η (ξ 6= η). We take

Arithmetic mean

A(ξ, η) =ξ + η

2, ξ, η ∈ R.

Logarithmic mean

L(ξ, η) =ξ − η

ln|ξ| − ln|η|, ξ, η ∈ R, ξ 6= η, |ξ| 6= |η|, ξη 6= 0.

Genralised log-mean

Ln(ξ, η) =

[ξn+1 − ηn+1

(n+ 1)(ξ − η)

] 1n

, n ∈ Z \ −1, 0, ξ, η ∈ R, ξ 6= η.

Proposition 1. Let a, b ∈ R \ 0, a < b, and n ∈ Z \ −1, 0, then we have

|An(a, b)A(a, b)− Lnn(a, b)| ≤ kb (b− a)

4

(max

|nan−1|q, |nbn−1|q

) 1q .

|An(a, b)A(a, b)− Lnn(a, b)| ≤ kb (b− a)

2(p+ 1)1p

(max


) 1q .

Proof. The assertion follows from Theorem 1 and Theorem 2, applied to f(x) = xn, x ∈R, g(x) = x and α = k.

Proposition 2. Let a, b ∈ R, a < b, and n ∈ Z \ −1, 0 is odd, then for every q ≥ 1,we

have∣∣∣∣A(an, bn)[(a−A(a, b))2+(b−A(a, b))2]− 2

n+ 1[(a−A(a, b))n+1 + (b−A(a, b))n+1]

∣∣∣∣≤ k|b− a|3

4max

|nan−1|, |nbn−1|


1218 A. Ali et al 1208-1219

12 A. Ali, G. Gulshan, R. Hussain, A. Latif, M. Muddassar and J. Park∣∣∣∣A(an, bn)[(a−A(a, b))2+(b−A(a, b))2]− 2

n+ 1[(a−A(a, b))n+1 + (b−A(a, b))n+1]

∣∣∣∣≤ k|b− a|3

4

(max


) 1q

∣∣∣∣A(an, bn)[(a−A(a, b))2+(b−A(a, b))2]− 2

n+ 1[(a−A(a, b))n+1 + (b−A(a, b))n+1]

∣∣∣∣≤ 2

1p−1k|b− a|3

(p+ 1)1p

(1− 1

2p

) 1p (

max|nan−1|q, |nbn−1|q

) 1q

∣∣∣∣A(an, bn)[(a−A(a, b))2+(b−A(a, b))2]− 2

n+ 1[(a−A(a, b))n+1 + (b−A(a, b))n+1]

∣∣∣∣≤ k|b− a|3

(p+ 1)1p

(max


) 1q

Proof. The assertion follows from Theorems 3, 4, 5 and 6 respectively, applied to

f(x) = xn, x ∈ R, g(x) = |x− a+b2 | and α = k.

Note: If n ∈ Z \ −1, 0 is even in Proposition 2, then the term (a − A(a, b))n+1 in

the left hand side of each of above inequalities will bear negative sign instead of positive.

REFERENCES

[1] L. Fejer, Uberdie Fourierreihen, II, Math. Naturwise. Anz Ungar. Akad., Wiss, 24 (1906), 369-390, (in

Hungarian).

[2] I. Iscan, Hermite-Hadamard-Fejer type inequalities for convex functions via fractional integrals,

arXiv:1404.7722v1.(http://rgmia.org/monographs.php)

[3] S. Mubeen and G. M. Habibullah, k-fractional integrals and applications, Int. J. Contemp. Math. Sciences,

7(2), (2012), 89-94.

[4] J. E. Pecaric, F. Proschan and Y. L. Tong, Convex functions, partial ordering and statistical applications,

Academic Press, Inc, Boston/London, 1992.

[5] E. Set, I. Iscan and S. Paca, Hermite-Hadamard-Fejer type inequalities for quasi-convex functions via frac-

tional integrals, Malaya J. Mat. 3(3), (2015), 241-249.

[6] E. Set, I. Iscan, M. E. Ozdemir and M. Z. Sarikaya, Some New Hermite-Hadamard-Fejer type inequalities

for convex functions via fractional integrals, Applied Mathematics and Computation 259,(2015), 875-881.

[7] J. Wang, C. Zhu and Y. Zhou, New generalized Hermite-Hadamard inequalities and applications to special

means, J. Inequal. Appl., 2013(325) (2013), 15 pages.


1219 A. Ali et al 1208-1219

NONLINEAR DIFFERENTIAL POLYNOMIALS OFMEROMORPHIC FUNCTIONS WITH REGARD TOMULTIPLICITY SHARING A SMALL FUNCTION

JIANREN LONG

Abstract. We study the uniqueness problem of nonlinear differentialpolynomials of meromorphic functions that share one small function. Auniqueness result which related to multiplicity of meromorphic functionis proved in this paper.

1. Introduction and main results

Let f be a nonconstant meromorphic function in the complex plane C.

We will assume that the reader is familiar with the standard notation of

the Nevanlinna’s theory of meromorphic functions, such as T (r, f),m(r, f),

N(r, f) and N(r, f), see [9, 14, 16] for more details. The notation S(r, f)

is defined to be any quantity satisfying S(r, f) = o(T (r, f)) as r → ∞, r ∈E, where E is a set of positive real number of finite linear measure, not

necessarily the same at each occurrence. The notations T (r) and S(r) are

defined respectively by

T (r) = maxT (r, f), T (r, g), S(r) = o (T (r)) as r →∞, r ∈ E,

for any two nonconstant meromorphic functions f and g. A meromorphic

function h is called a small function with respect to f , proved that T (r, a) =

S(r, f). Moreover, GCD(n1, n2, . . . , nk) denotes the greatest common divisor

of positive integers n1, n2, . . . , nk.

Let f and g be two nonconstant meromorphic functions, and let a ∈ C.

We say that f and g share the value a CM (counting multiplicities), provided

that f − a and g − a have the same zeros with the same multiplicities. If

f − a and g − a have the same zeros, then we say that f and g share a

IM (ignoring multiplicities). Similarly, we immediately get the definitions

of f and g share h IM (or CM), where h is a small function of f and g. In

addition, we also need the following notation, for any a ∈ C = C ∪ ∞,

Θ(a, f) = 1− lim supr→∞

N(r, a; f)

T (r, f).

2010 Mathematics Subject Classification. Primary 30D30; Secondary 30D35.Key words and phrases. Uniqueness, Meromorphic functions, Small function, Differ-

ential polynomials, Sharing value.1


1220 JIANREN LONG 1220-1230

2 J. R. LONG

Hayman [10] and Clunie [5] proved the following result.

Theorem 1.1. Let f be a transcendental entire function, n ≥ 1 be a positive

integer. Then fnf ′ = 1 has infinitely many zeros.

Remark 1. The similar result of Theorem 1.1 in which entire function is

replaced with meromorphic function is proved in [2] and [4].

Fang and Hua [8], Yang and Hua [15] obtained a uniqueness theorem

corresponding to Theorem 1.1.

Theorem 1.2. Let f and g be two nonconstant entire (meromorphic) func-

tions, and let n ≥ 6 (n ≥ 11) be a positive integer. If fnf ′ and gng′ share

1 CM, then either f(z) = c1ecz, g(z) = c2e

−cz, where c1, c2 and c are three

constants satisfying (c1c2)n+1c2 = −1, or f = tg for a constant t such that

tn+1 = 1.

Fang [7] considered the case of the kth derivative, and proved the fol-

lowing result.

Theorem 1.3. Let f and g be two nonconstant entire functions, and let n, k

be two positive integers with n ≥ 2k+ 8. If (fn(f − 1))(k) and (gn(g− 1))(k)

share 1 CM, then f = g.

Zhang et al. [18] considered some general differential polynomials. They

proved the following results.

Theorem 1.4. Let f and g be two nonconstant entire functions. Let n, k

and m be three positive integers with n ≥ 3m+2k+5 and let P (z) = amzm+

am−1zm−1 + · · · + a1z + a0, where a0 = 0, a1, . . . , am−1, am = 0 are complex

constants. If (fnP (f))(k) and (gnP (g))(k) share 1 CM, then either f = tg for

a constant t such that td = 1, where d =GCD(n+m, . . . , n+m− i, . . . , n),

am−i = 0 for some i = 0, 1, . . . ,m, or f and g satisfying the algebraic

function equation R(f, g) = 0, where R(w1, w2) = wn1 (amw

m1 + am−1w

m−11 +

· · ·+ a0)− wn2 (amw

m2 + am−1w

m−12 + · · ·+ a0).

Theorem 1.5. Let f and g be two nonconstant meromorphic functions,

and h(≡ 0,∞) be a small function with respect to f and g. Let n, k and

m be three positive integers with n > 3k + m + 8 and P (z) be defined as

in Theorem 1.4. If (fnP (f))(k) and (gnP (g))(k) share h(z) CM, then one of

the following three cases holds:

(i) f = tg for a constant t such that td = 1, where d =GCD(n +

m, . . . , n+m− i, . . . , n), am−i = 0 for some i = 0, 1, . . . ,m;



NONLINEAR DIFFERENTIAL POLYNOMIALS SHARING A SMALL FUNCTION 3

(ii) f and g satisfying the algebraic function equation R(f, g) = 0, where

R(w1, w2) = wn1 (amw

m1 +am−1w

m−11 +· · ·+a0)−wn

2 (amwm2 +am−1w

m−12 +

· · ·+ a0);

(iii) (fnP (f))(k) (gnP (g))(k) = h2.

In 2011 Dyavanal [6] considered the uniqueness problem of meromorphic

function related to the value sharing of two nonlinear differential polyno-

mials in which the multiplicities of zeros and poles of f and g are taken

into account. In 2013, Bhoosnurmath and Kabbur [3] proved the following

uniqueness theorem by using the idea from Dyavanal [6].

Theorem 1.6. Let f and g be two nonconstant meromorphic functions,

whose zeros and poles are of multiplicities at least s, where s is a posi-

tive integer. Let n and m be two positive integers with (n − m − 1)s ≥max10, 2m + 3, and let P (z) be defined as in Theorem 1.4. If fnP (f)f ′

and gnP (g)g′ share 1 CM, then either f = tg for a constant t such that

td = 1, where d =GCD(n+m+ 1, · · · , n+m+ 1− i, · · · , n+ 1), am−i = 0

for some i = 0, 1, · · · ,m or f and g satisfy the algebraic function equation

R(f, g) = 0, where R(x, y) = xn+1( amn+m+1

xm + am−1

n+mxm−1 + · · · + a0

n+1) −

yn+1( amn+m+1

ym + am−1

n+mym−1 + · · ·+ a0

n+1).

Similar Theorem 1.5 in which a small function and kth derivative are

considered, what can we say when the condition sharing 1 and the first

derivative in Theorem 1.6 are replaced with sharing a small function and

kth derivative respectively? In this paper, we will study the problem and

establish the following uniqueness theorem.

Theorem 1.7. Let f and g be two transcendental meromorphic functions,


tive integer. Let n and m be two positive integers with n −m > max2 +2ms, (n+2)(k+4)

ns, Θ(∞, f) + Θ(∞, g) > 4

nand let P (z) be defined as in Theo-

rem 1.4. If (fnP (f))(k) and (gnP (g))(k) share h(z) CM, where h(z)(≡ 0,∞)

is a small function of f and g, then one of the following three cases hold:

(i) (fnP (f))(k) (gnP (g))(k) = h2;

(ii) f = tg for a constant t such that td = 1, where d = GCD(n+m,n+

m− 1, . . . , n+m− i, . . . , n+ 1, n), am−i = 0 for i = 0, 1, . . . ,m;

(iii) f and g satisfy the algebraic equation R(f, g) = 0, where R(f, g) =

fnP (f)− gnP (g).

The possibility (fnP (f))(k) (gnP (g))(k) = h2 does not occur for k = 1.



4 J. R. LONG

2. Auxiliary results

For the proof of our result, we need the following lemmas and definitions.

Definition 2.1. [11] Let a ∈ C. We use N(r, a; f | = 1) to denote the

counting function of simple a−points of f . For a positive integer p we denote

by N(r, a; f | ≤ p) the counting function of those a−points of f (counted

with proper multiplicities) whose multiplicities are not greater than p. By

N(r, a; f | ≤ p) we denote the corresponding reduced counting function.

Similarly, we can define N(r, a; f | ≥ p) and N(r, a; f | ≥ p).

Definition 2.2. [1] Let a ∈ C, and let k be a nonnegative integer. We denote

by Nk(r, 1f−a

) the counting function of a−points of f , where an a−point of

multiplicity m is counted m times if m ≤ k and k times if m > k. Then

Nk(r,1

f − a) = N(r,

1

f − a) +N(r, a; f | ≥ 2) + · · ·+N(r, a; f | ≥ k).

(2.1)

Obviously N1(r,1

f−a) = N(r, 1

f−a).

Lemma 2.1. [15] Let f and g be two nonconstant meromorphic functions

that share 1 CM. Then one of the following cases hold:

(i) T (r) ≤ N2(r,1f) +N2(r,

1g) +N2(r, f) +N2(r, g) + S(r);

(ii) f = g;

(iii) fg = 1.

Lemma 2.2. [17] Let f be a nonconstant meromorphic function, and p, k

be positive integers. Then

Np(r,1

f (k)) ≤ T (r, f (k))− T (r, f) +Np+k(r,

1

f) + S(r, f),(2.2)

Np(r,1

f (k)) ≤ kN(r, f) +Np+k(r,

1

f) + S(r, f).(2.3)

Lemma 2.3. [13] Let f be a nonconstant meromorphic function, and let

Pn(f) =n∑

j=0

ajfj be a polynomial in f , where an = 0, an−1, · · · , a1, a0

satisfying T (r, aj) = S(r, f). Then

T (r, Pn) = nT (r, f) + S(r, f).(2.4)

Lemma 2.4. Let f and g be two nonconstant meromorphic functions such

that

Θ(∞, f) + Θ(∞, g) >4

n




for all integer n ≥ 3. Then

fn(af + b) = gn(ag + b)

implies f = g, where a and b are two finite nonzero complex constants.

Proof. By using similar way in [12], we can obtain the lemma.

Lemma 2.5. Let f and g be two nonconstant meromorphic functions,


tive integer and let n, k be two positive integers. Let F = (fnP (f))(k) and

G = (gnP (g))(k), where P (z) be defined as in Theorem 1.4. If there ex-

ist two nonzero constants b1 and b2 such that N(r, 1F

) = N(r, 1G−b1

) and

N(r, 1G

) = N(r, 1F−b2

), then n−m ≤ (k+1)(n+2)ns

.

Proof. By the second fundamental theorem of Nevanlinna’s theory,

T (r, F ) ≤ N(r,1

F) +N(r, F ) +N(r,

1

F − b2) + S(r, F )

≤ N(r,1

F) +N(r, F ) +N(r,

1

G) + S(r, F ).(2.5)

Combining (2.2), (2.3), (2.5) and Lemma 2.3, we get

(n+m)T (r, f) ≤ T (r, F )−N(r,1

F) +Nk+1(r,

1

fnP (f)) + S(r, f)

≤ N(r,1

G) +N(r, f) +Nk+1(r,

1

fnP (f)) + S(r, f)

≤ Nk+1(r,1

fnP (f)) +Nk+1(r,

1

gnP (g)) +N(r, f)

+ kN(r, g) + S(r, f) + S(r, g)

≤ (k + 1 + n

ns+m)T (r, f) + (

k + 1 + nk

ns+m)T (r, g)

+ S(r, f) + S(r, g)

≤ ((k + 1)(n+ 2)

ns+ 2m)T (r) + S(r).(2.6)

Similarly, for the case of g,

(n+m)T (r, g) ≤ ((k + 1)(n+ 2)

ns+ 2m)T (r) + S(r).(2.7)

It follows from (2.6) and (2.7) that

(n− (k + 1)(n+ 2)

ns−m)T (r) ≤ S(r),(2.8)

which gives n−m ≤ (k+1)(n+2)ns

. This completes the proof.

Lemma 2.6. Let f and g be two transcendental meromorphic functions,

whose zeros and poles are of multiplicities at least s, where s is a positive



6 J. R. LONG

integer. Let P (z) be defined as in Theorem 1.4, and n,m, k be three positive

integers, and h(≡ 0,∞) be small function of f and g. Then

(fnP (f))(k) (gnP (g))(k) ≡ h2(2.9)

holds for k = 1 and (n + m − 2)p > 2m(1 + 1s), where p is the number of

distinct roots of P (z) = 0.

Proof. If (2.9) is possible for k = 1, i.e.,

(fnP (f))′(gnP (g))′ = h2.

Then

fn−1Q(f)f ′gn−1Q(g)g′ = h2,(2.10)

where Q(z) =m∑j=0

bjzj, bj = (n+ j)aj, j = 0, 1, . . . ,m. Denote Q(z) as

Q(z) = bm(z − d1)l1(z − d2)l2 · · · (z − dp)lp ,

where

p∑i=1

li = m, 1 ≤ p ≤ m, di = dj, i = j, 1 ≤ i, j ≤ p, di are nonzero

constants and li are positive integers, i = 1, 2, . . . p.

Suppose that z1 ∈ S0 is a zero of f with multiplicity s1(≥ s), where S0

is a set defined as

S0 = z : h(z) = 0 ∪ z : h(z) =∞.

Then z1 is a pole of g with multiplicity q1(≥ s). We deduce from (2.10) that

ns1 − 1 = (n+m)q1 + 1

and so

mq1 + 2 = n(s1 − q1).(2.11)

From (2.11) we get q1 ≥ n−2m

, so

s1 ≥n+m− 2

m.

Hence,

N(r,1

f) ≤ m

n+m− 2N(r,

1

f) + S(r, f).(2.12)

Suppose that z2 ∈ S0 is a zero of Q(f) with multiplicity s2 and is a zero

of f − di of order qi, i = 1, 2, . . . p. Then s2 = liqi, i = 1, 2, · · · p. Then z2 is

a pole of g with multiplicity q(≥ s). It follows from (2.10) that

qili + qi − 1 = (n+m)q + 1 ≥ (n+m)s+ 1.




So

qi ≥(n+m)s+ 2

li + 1, i = 1, 2, . . . p.

Hence,

N(r,1

f − di) ≤ li + 1

(n+m)s+ 2N(r, di, f) + S(r, f), i = 1, 2, . . . p.

By this and the first fundamental theorem of Nevanlinna’s theory, we have

p∑i=1

N(r,1

f − di) ≤ m+ p

(n+m)s+ 2T (r, f) + S(r, f).(2.13)

Suppose that z3 ∈ S0 is a pole of f . Then we know that z3 is either a

zero of gn−1Q(g) or a zero of g′ by (2.10). Therefore,

N(r, f) ≤ N(r,1

g) +

p∑i=1

N(r,1

g − di) +N0(r,

1

g′) + S(r, f) + S(r, g)

≤ (m

n+m− 2+

m+ p

(n+m)s+ 2)T (r, g) +N0(r,

1

g′)

+ S(r, f) + S(r, g),(2.14)

where N0(r,1g′

) denote the reduce counting function of those zeros of g′

which are not the zeros of gQ(g).

By (2.12)-(2.14), and the second fundamental theorem of Nevanlinna’s

theory,

pT (r, f) ≤ N(r, f) +N(r,1

f) +

p∑i=1

N(r,1

f − di)−N0(r,

1

f ′ ) + S(r, f)

≤ (m

n+m− 2+

m+ p

(n+m)s+ 2)(T (r, f) + T (r, g)) +N0(r,

1

g′)

−N0(r,1

f ′ ) + S(r, f) + S(r, g).(2.15)


pT (r, g) ≤ (m

n+m− 2+

m+ p

(n+m)s+ 2)(T (r, f) + T (r, g)) +N0(r,

1

f ′ )

−N0(r,1

g′) + S(r, f) + S(r, g).(2.16)


(p− 2m

n+m− 2− 2(m+ p)

(n+m)s+ 2)(T (r, f) + T (r, g)) ≤ S(r, f) + S(r, g).

This is a contradiction with our assumption that (n+m−2)p > 2m(1 + 1s),

and hence the proof is complete.



8 J. R. LONG

3. Proof of Theorem 1.7

Let F =(fnP (f))(k)

h, G =

(gnP (g))(k)

h. Then F and G share 1 CM.

Applying Lemma 2.3,

T (r, h) = o(T (r, F )) = S(r, f), T (r, h) = o(T (r,G)) = S(r, g).(3.1)


N2(r,1

F) ≤ N2(r,

1

(fnP (f))(k)) +N2(r, h) + S(r, f)

≤ N2(r,1

(fnP (f))(k)) + S(r, f)

≤ T (r, (fnP (f))− (n+m)T (r, f) +Nk+2(r,1

fnP (f)) + S(r, f)

≤ T (r, F )− (n+m)T (r, f) +Nk+2(r,1

fnP (f)) + S(r, f).(3.2)

We deduce from (2.3) that

N2(r,1

F) ≤ N2(r,

1

(fnP (f))(k)) + S(r, f)

≤ kN(r, fnP (f))(k)) +Nk+2(r,1

fnP (f)) + S(r, f)

≤ kN(r, f) +Nk+2(r,1

fnP (f)) + S(r, f).(3.3)

It follows from (3.2) that

(n+m)T (r, f) ≤ T (r, F ) +Nk+2(r,1

fnP (f))−N2(r,

1

F) + S(r, f).(3.4)

Suppose that (i) of Lemma 2.1 holds, i.e.,

maxT (r, F ), T (r,G) ≤ N2(r,1

F) +N2(r,

1

G) +N2(r, F ) +N2(r,G)

+ S(r, f) + S(r, g).




Combining this, (3.3) and (3.4),

(n+m)T (r, f) ≤ T (r, F ) +Nk+2(r,1

fnP (f))−N2(r,

1

F) + S(r, f) + S(r, g)

≤ N2(r,1

G) +N2(r, F ) +N2(r,G) +Nk+2(r,

1

fnP (f))

+ S(r, f) + S(r, g)

≤ Nk+2(r,1

fnP (f)) +Nk+2(r,

1

gnP (g)) + kN(r, g)

+ 2N(r, f) + 2N(r, g) + S(r, f) + S(r, g)

≤ (k + 2 + 2n

ns+m)T (r, f) + (

(k + 2)(n+ 1)

ns+m)T (r, g)

+ S(r, f) + S(r, g)

≤ (k(n+ 2) + 4(n+ 1)

ns+ 2m)T (r) + S(r).(3.5)


(n+m)T (r, g) ≤ (k(n+ 2) + 4(n+ 1)

ns+ 2m)T (r) + S(r).(3.6)


(n+m)T (r) ≤ (k(n+ 2) + 4(n+ 1)

ns+ 2m)T (r) + S(r).

This implies that

(n−m− k(n+ 2) + 4(n+ 1)

ns)T (r) ≤ S(r).(3.7)

This contradicts with our assumption that (n−m) > max2+2ms, (n+2)(k+4)

ns.

So, we conclude that either FG = 1 or F = G by Lemma 2.1. Suppose that

FG = 1, then

(fnP (f))(k)(gnP (g))(k) = h2.

This is a contradiction when k = 1 by Lemma 2.6. So F = G, this implies

that

(fnP (f))(k) = (gnP (g))(k).(3.8)

Integrating for (3.8), we have

(fnP (f))(k−1) = (gnP (g))(k−1) + bk−1,(3.9)

where bk−1 is constant. If bk−1 = 0, we obtain n−m ≤ (k+1)(n+2)ns

< (k+4)(n+2)ns

by Lemma 2.5. This is a contradiction with our assumption that (n−m) >

max2 + 2ms, (n+2)(k+4)

ns. Thus bk−1 = 0. By repeating k−times,

fnP (f) = gnP (g).(3.10)



10 J. R. LONG

If m = 1 in (3.10), then f = g by Lemma 2.4. Suppose that m ≥ 2 and

b = fg. If b is a constant, putting f = bg in (3.10), we get

amgn+m(bn+m − 1) + am−1g

n+m−1(bn+m−1 − 1) + · · ·+ a0gn(bn − 1) = 0,

(3.11)

which implies bd = 1, where d = GCD(n+m,n+m−1, . . . , n+1, n). Hence

f = tg for a constant t such that td = 1, d = GCD(n+m,n+m−1, . . . , n+

m− i, . . . , n+ 1, n), i = 0, 1, . . . ,m.

If b is not a constant, then we can see that f and g satisfy the algebraic

function equation R(f, g) = 0 by (3.10), where R(f, g) = fnP (f)− gnP (g).

This completes the proof of theorem.

Acknowledgements. This research work is supported in part by the Foun-

dation of Science and Technology of Guizhou Province (Grant No. [2015]2112),

the National Natural Science Foundation of China (Grant No. 11501142).

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Jianren LongSchool of Mathematics and Computer Science, Guizhou Normal University,Guiyang, 550001, P.R. China.School of Mathematical Sciences, Xiamen University, Xiamen, 361005, P.R.China.

E-mail address: [email protected], [email protected]



Impulsive hybrid fractional quantum difference equations

Bashir Ahmada, Sotiris K. Ntouyasb,a, Jessada Tariboonc,Ahmed Alsaedia, Wafa Shammakha

aNonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics,Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabiae-mail: bashirahmad−[email protected], [email protected], [email protected] of Mathematics, University of Ioannina, 451 10 Ioannina, Greecee-mail: [email protected] Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science,King Mongkut’s University of Technology North Bangkok, Bangkok, 10800 Thailande-mail: [email protected]

Abstract

This paper is concerned with the existence of solutions for impulsive hybrid fractional q-differenceequations involving a q-shifting operator of the type aΦq(m) = qm + (1− q)a. A hybrid fixed pointtheorem for two operators in a Banach algebra due to Dhage [29] is applied to obtain the existenceresult. An example illustrating the main result is also presented.

Key words and phrases: Quantum calculus; impulsive fractional q-difference equations; hybriddifferential equations; existence; fixed point theoremAMS (MOS) Subject Classifications: 34A08; 34A12; 34A37

1 Introduction

Fractional differential equations have been extensively investigated by several researchers in the recentyears. The overwhelming interest in this branch of mathematics is due to the application of fractional-order operators in the mathematical modelling of several phenomena occurring in a variety of disciplinesof applied sciences and engineering such as biomathematics, signal and image processing, control theory,dynamical systems, etc.

Hybrid fractional differential equations dealing with the fractional derivative of an unknown functionhybrid with the nonlinearity depending on it is another interesting field of research. For some recentworks on this topic, we refer the reader to a series of papers ([1]-[6]).

The subject of q-difference calculus or quantum calculus dates back to the beginning of the 20thcentury, when Jackson [7] introduced the concept of q-difference operator. Afterwards, this field ofresearch flourished with the contributions of researchers from different parts of the world, for instance,see ([8]-[15]). The intensive development of fractional calculus motivated several investigators to considerfractional q-difference calculus. Now a great deal of work on initial and boundary value problemsinvolving nonlinear fractional q-difference equations is available, for example, see [16]-[24] and thereferences therein.

The quantum calculus, known as the calculus without limits, provides a descent approach to studynondifferentiable functions in terms of difference operators. Quantum difference operators appear indifferent areas of mathematics such as orthogonal polynomials, basic hyper-geometric functions, combi-natorics, the calculus of variations, mechanics and the theory of relativity. For the fundamental conceptsof quantum calculus, we refer the reader to a text by Kac and Cheung [25].

More recently, the topic of qk-calculus has also gained consideration attention. The notions of qk-derivative and qk-integral for a function f : Jk := [tk, tk+1] → R, together with their properties can


1231 Bashir Ahmad et al 1231-1240

B. Ahmad, S.K. Ntouyas, J. Tariboon, A. Alsaedi and W. Shammakh

be found in [26, 27]. In [28], new concepts of fractional quantum calculus were defined via a q-shiftingoperator of the form: aΦq(m) = qm+ (1− q)a.

The purpose of the present work is to study the following impulsive hybrid fractional quantumdifference equations:

ctkDαk

qk

[x(t)

f(t, x(t))

]= g(t, x(t)), t ∈ Jk ⊆ [0, T ], t 6= tk,

∆x(tk) = ϕk (x(tk)) , k = 1, 2, . . . ,m,

x(0) = µ,

(1)

where 0 = t0 < t1 < · · · < tm < tm+1 = T , ctkDαk

qkdenotes the Caputo fractional qk-derivative of

order αk on intervals Jk, J0 = [0, t1], Jk = (tk, tk+1], 0 < αk ≤ 1, 0 < qk < 1, k = 0, 1, . . . ,m,J = [0, T ], f ∈ C(J × R,R \ 0), ϕk ∈ C(R,R), k = 1, 2, . . . ,m, µ ∈ R and ∆x(tk) = x(t+k ) − x(tk),x(t+k ) = limθ→0+ x(tk + θ), k = 1, 2, . . . ,m. Here, we emphasize that the above initial value problemcontains the new q-shifting operator aΦq(m) = qm+ (1− q)a [28].

The papers is organized as follows. In Section 2, we recall some preliminary concepts and present anauxiliary lemma which is used to convert the impulsive problem (1) into an equivalent integral equation.An existence result for the problem (1) obtained by means of a hybrid fixed point theorem due to Dhage[29] is presented in Section 3, which is well illustrated with the aid of an example.

2 Preliminaries

For the convenience of the reader, we recall some preliminary concepts from [28].First of all, we define a q-shifting operator as

aΦq(m) = qm+ (1− q)a (2)

such thataΦk

q (m) = aΦk−1q (aΦq(m)) and aΦ0

q(m) = m,

for any positive integer k. The power law for q-shifting operator is

a(n−m)(0)q = 1, a(n−m)(k)q =

k−1∏i=0

(n− aΦi

q(m)), k ∈ N ∪ ∞.

In case γ ∈ R, the above power law takes the form

a(n−m)(γ)q = n(γ)

∞∏i=0

1− anΦi

q(m/n)

1− anΦγ+i

q (m/n).

The q-derivative of a function h on interval [a, b] is defined by

(aDqh)(t) =h(t)− h(aΦq(t))(1− q)(t− a)

, t 6= a, and (aDqh)(a) = limt→a

(aDqh)(t),

while the higher order q-derivative is given by the formula

(aD0qf)(t) = f(t) and (aD

kq f)(t) = aD

k−1q (aDqf)(t), k ∈ N.

The product and quotient formulas for q-derivative are

aDq(h1h2)(t) = h1(t)aDqh2(t) + h2(aΦq(t))aDqh1(t) = h2(t)aDqh1(t) + h1(aΦq(t))aDqh2(t),

aDq

(h1

h2

)(t) =

h2(t)aDqh1(t)− h1(t)aDqh2(t)h2(t)h2(aΦq(t))

,



Impulsive hybrid fractional q-difference equations

where h1 and h2 are well defined on [a, b] with h2(t)h2(aΦq(t)) 6= 0.The q-integral of a function h defined on the interval [a, b] is given by

(aIqh)(t) =∫ t

a

h(s)ads = (1− q)(t− a)∞∑

i=0

qih(aΦqi(t)), t ∈ [a, b], (3)

with(aI

0qh)(t) = h(t) and (aI

kq h)(t) = aI

k−1q (aIqh)(t), k ∈ N.

The fundamental theorem of calculus applies to the operator aDq and aIq, that is,

(aDqaIqh)(t) = h(t),

and if h is continuous at t = a, then

(aIqaDqh)(t) = h(t)− h(a).

The q-integration by parts formula on the interval [a, b] is∫ b

a

f(s)aDqg(s)adqs = (fg)(t)∣∣∣ba−∫ b

a

g(aΦq(s))aDqf(s)adqs.

Let us now define Riemann-Liouville fractional q-derivative and q-integral on interval [a, b] andoutline some of their properties [28].

Definition 2.1 The fractional q-derivative of Riemann-Liouville type of order ν ≥ 0 on interval [a, b]is defined by (aD

0qh)(t) = h(t) and

(aDνqh)(t) = (aD

lqaI

l−νq h)(t), ν > 0,

where l is the smallest integer greater than or equal to ν.

Definition 2.2 Let α ≥ 0 and h be a function defined on [a, b]. The fractional q-integral of Riemann-Liouville type is given by (aI

0qh)(t) = h(t) and

(aIαq h)(t) =

1Γq(α)

∫ t

aa(t− aΦq(s))(α−1)

q h(s)adqs, α > 0, t ∈ [a, b].

From [28], we have the following formulas

aDαq (t− a)β =

Γq(β + 1)Γq(β − α+ 1)

(t− a)β−α, (4)

aIαq (t− a)β =

Γq(β + 1)Γq(β + α+ 1)

(t− a)β+α. (5)

Lemma 2.3 Let α, β ∈ R+ and f be a continuous function on [a, b], a ≥ 0. The Riemann-Liouvillefractional q-integral has the following semi-group property

aIβq aI

αq h(t) = aI

αq aI

βq h(t) = aI

α+βq h(t).

Lemma 2.4 Let h be a q-integrable function on [a, b]. Then the following equality holds

aDαq aI

αq h(t) = h(t), for α > 0, t ∈ [a, b].

Lemma 2.5 Let α > 0 and p be a positive integer. Then for t ∈ [a, b] the following equality holds

aIαq aD

pqh(t) = aD

pq aI

αq h(t)−

p−1∑k=0

(t− a)α−p+k

Γq(α+ k − p+ 1)aDkqh(a).




We define Caputo fractional q-derivative as follows.

Definition 2.6 The fractional q-derivative of Caputo type of order α ≥ 0 on interval [a, b] is definedby (c

aD0qf)(t) = h(t) and

(caD

αq h)(t) = (aI

n−αq aD

nq h)(t), α > 0,

where n is the smallest integer greater than or equal to α.

Lemma 2.7 Let α > 0 and n be the smallest integer greater than or equal to α. Then for t ∈ [a, b] thefollowing equality holds

aIαq

caD

αq h(t) = h(t)−

n−1∑k=0

(t− a)k

Γq(k + 1)aDkqh(a).

Proof. From Lemma 2.5, for α = p = m, where m is a positive integer, we have

aImq aD

mq h(t) = aD

mq aI

mq h(t)−

m−1∑k=0

(t− a)k

Γq(k + 1)aDkqh(a) = h(t)−

m−1∑k=0

(t− a)k

Γq(k + 1)aDkqh(a).

Then, by Definition 2.6, we have

aIαq

caD

αq h(t) = aI

αq aI

n−αq aD

nq h(t) = aI

nq aD

nq h(t) = h(t)−

n−1∑k=0

(t− a)k

Γq(k + 1)aDkqh(a).

Now we present a lemma which plays a pivotal role in the forthcoming analysis.

Lemma 2.8 Assume that the map x 7→ x

f(t, x)is injection for each t ∈ J. x ∈ PC(J,R) is the solution

of (1) if and only if x is a solution of the impulsive integral equation

x(t) = f(t, x(t))

(µ

f(0, µ)

k∏i=1

f(ti, x(ti))f(ti, x(t+i ))

+k∑

i=1

∏i≤j≤k

ti−1Iαi−1qi−1

g(ti, x(ti))f(tj , x(tj))f(tj , x(t+j ))

+k∑

i=1

∏i<j≤k

ϕi(x(ti))f(ti, x(t+i ))

· f(tj , x(tj))f(tj , x(t+j ))

+ tkIαkqkg(t, x(t))

), (6)

where∑

b<a(·) = 0,∏

b<a(·) = 1 for b > a and for t ∈ Jk,

tkIαkqkg(t, x(t)) =

1Γqk

(αk)

∫ t

tk

tk(t− tk

Φqk(s))(αk−1)

qkg(s, x(s))tk

dqks. (7)

Proof. Applying Riemann-Liouville fractional q0-integral operator of order α0 to both sides of the firstequation of (1) for t ∈ J0 and using Lemma 2.7, we get

t0Iα0q0

ct0D

α0q0

[x(t)

f(t, x(t))

]=

x(t)f(t, x(t))

− x(0)f(0, x(0))

= t0Iα0q0g(t, x(t)),

which, in view of the initial condition, takes the form

x(t) = f(t, x(t))[

µ

f(0, µ)+ t0I

α0q0g(t, x(t))

].

At t = t1, we have

x(t1) = f(t1, x(t1))[

µ

f(0, µ)+ t0I

α0q0g(t1, x(t1))

]. (8)




For t ∈ J1, operating the Riemann-Liouville fractional q1-integral of order α1 on (1) and using theabove process together with impulsive condition, we obtain

x(t)f(t, x(t))

=x(t+1 )

f(t+1 , x(t+1 ))

+ t1Iα1q1g(t, x(t)) =

x(t1) + ϕ1(x(t1))f(t+1 , x(t

+1 ))

+ t1Iα1q1g(t, x(t)). (9)

By the continuity of f with respect to the variable t, the expression f(t+1 , x(t+1 )) can be written as

f(t1, x(t+1 )). Substituting (8) into (9) yields

x(t) = f(t, x(t))

(µ

f(0, µ)· f(t1, x(t1))f(t1, x(t+1 ))

+f(t1, x(t1))f(t1, x(t+1 )) t0I

α0q0g(t1, x(t1))

+ϕ1(x(t1))f(t1, x(t+1 ))

+ t1Iα1q1g(t, x(t))

).

Also, for t ∈ J2, we have

x(t) = f(t, x(t))

(µ

f(0, µ)· f(t1, x(t1))f(t1, x(t+1 ))

· f(t2, x(t2))f(t2, x(t+2 ))

+f(t1, x(t1))f(t1, x(t+1 ))

· f(t2, x(t2))f(t2, x(t+2 )) t0I

α0q0g(t1, x(t1)) +

f(t2, x(t2))f(t2, x(t+2 )) t1I

α1q1g(t2, x(t2))

+ϕ1(x(t1))f(t1, x(t+1 ))

· f(t2, x(t2))f(t2, x(t+2 ))

+ϕ2(x(t2))f(t2, x(t+2 ))

+ t2Iα2q2g(t, x(t))

).

Repeating the above process, for t ∈ J , we obtain (6).Conversely, we assume that x(t) is a solution of (6). Dividing by f(t, x(t)) and applying c

tkDαk

qkon

both sides of (6) for t ∈ Jk, t 6= tk k = 0, 1, . . . ,m, we get

ctkDαk

qk

[x(t)

f(t, x(t))

]= g(t, x(t)).

It is easy to see that ∆x(tk) = x(t+k )− x(tk) = ϕk(x(tk)). Since f(0, x(0)) 6= 0, and using the fact thatthe map x 7→ x

f(t, x)is injection for each t ∈ J, we have x(0) = µ. This completes the proof.

Now we state a hybrid fixed point theorem due to Dhage [29], which we need to prove our mainexistence result.

Lemma 2.9 Let S be a nonempty, closed convex and bounded subset of the Banach algebra E andlet A : E → E and B : S → E be two operators such that (a) A is Lipschitzian with Lipschitzconstant δ; (b) B is completely continuous; (c) x = AxBy ⇒ x ∈ S for all y ∈ S; (d) δM < 1, whereM = ‖B(S)‖ = sup‖B(x)‖ : x ∈ S. Then the operator equation x = AxBx has a solution in S.

3 Main Result

Let PC(J,R) = x : J → R : x(t) is continuous everywhere except for some tk at which x(t+k ) andx(t−k ) exist and x(t−k ) = x(tk), k = 1, 2, . . . ,m. Define a norm ‖ · ‖ and a multiplication in PC(J,R)by ‖x‖ = supt∈J |x(t)| and (xy)(t) = x(t)y(t), ∀t ∈ J.Clearly PC(J,R) is a Banach algebra with respect to above supremum norm and the multiplication init.

Now, we are in the position to present the main existence result.

Theorem 3.1 Assume that the map x 7→ x

f(t, x)is injection for each t ∈ J. In addition we suppose

that:




(H1) The function f : J × R → R \ 0 is bounded, continuous and there exists a positive function φwith bound ‖φ‖ such that

|f(t, x(t))− f(t, y(t))| ≤ φ(t)|x(t)− y(t)|, for t ∈ J andx, y ∈ R. (10)

(H2) There exist a function p ∈ C(J,R+) and a continuous nondecreasing function ψ : [0,∞) → (0,∞)such that

|g(t, x(t))| ≤ p(t)ψ(|x|), (t, x) ∈ J × R. (11)

(H3) The functions ϕi : R → R, i = 1, 2, . . . ,m, are bounded and continuous.

(H4) There exists a number r > 0 such that

r ≥ Ω1

(|µ|

|f(0, µ)|

(Ω1

Ω2

)m

+ ‖p‖ψ(r)m+1∑i=1

(ti − ti−1)αi−1

Γqi−1(αi−1 + 1)

(Ω1

Ω2

)m+1−i

+Ω3

Ω2

m∑i=1

(Ω1

Ω2

)m−i), (12)

and

‖φ‖

(|µ|

|f(0, µ)|

(Ω1

Ω2

)m

+ ‖p‖ψ(r)m+1∑i=1


Γqi−1(αi−1 + 1)

(Ω1

Ω2

)m+1−i

+Ω3

Ω2

m∑i=1

(Ω1

Ω2

)m−i)< 1,

where Ω1 = sup|f(t, x)| : (t, x) ∈ J × R, Ω2 = inf|f(t, x)| : (t, x) ∈ J × R and Ω3 =maxsup |ϕi(x)| : x ∈ R, i = 1, 2, . . . ,m.

Then the impulsive initial value problem (1) has at least one solution on J .

Proof. Let us introduce a subset S of PC(J,R) by

S = x ∈ PC(J,R) : ‖x‖ ≤ r,

where r satisfies inequality (12). Clearly S is closed, convex and bounded subset of the Banach spacePC(J,R). In view of Lemma 2.8, the problem (1) is equivalent to the integral equation (6). Let usdefine two operators A : PC(J,R) → PC(J,R) by

Ax(t) = f(t, x(t)), t ∈ J, (13)

and B : S → PC(J,R) by

Bx(t) =µ

f(0, µ)

k∏i=1


+k∑

i=1

∏i≤j≤k



+k∑

i=1

∏i<j≤k



+ tkIαkqkg(t, x(t)), t ∈ J. (14)

Then, the problem (1) is transformed into an operator equation as

x = AxBx. (15)

Under our assumptions, we will show that the operators A and B satisfy all the conditions of Lemma2.9. This will be achieved in a series of steps.

Step 1. The operator A is Lipschitzian on PC(J,R).

Let x, y ∈ PC(J,R). Then by (H1), for t ∈ J , we have

|Ax(t)−Ay(t)| = |f(t, x(t))− f(t, y(t))| ≤ φ(t)|x(t)− y(t)|.




Taking supremum over t, we obtain ‖Ax−Ay‖ ≤ ‖φ‖‖x− y‖ for all x, y ∈ PC(J,R). This show thatA is a Lipschitzian on PC(J,R) with Lipschitz constant ‖φ‖.

Step 2. The operator B is completely continuous on S.

In this step, we first show that the operator B is continuous on S. Let xn be a sequence in Sconverging to a point x ∈ S. Then, for all t ∈ J , we have

limn→∞

Bxn(t)

= limn→∞

µ

f(0, µ)

k∏i=1

f(ti, xn(ti))f(ti, xn(t+i ))

+ limn→∞

k∑i=1

∏i≤j≤k


g(ti, xn(ti))f(tj , xn(tj))f(tj , xn(t+j ))

+ limn→∞

k∑i=1

∏i<j≤k

ϕi(xn(ti))f(ti, xn(t+i ))

· f(tj , xn(tj))f(tj , xn(t+j ))

+ limn→∞ tk

Iαkqkg(t, xn(t)) = Bx(t),

which implies that Bxn → Bx point-wise on J . Further it can be shown that Bxn is an equicontinuoussequence of functions. So Bxn → Bx uniformly and the operator B is continuous on S.

Next we will prove that B is a compact operator on S. It is enough to show that the set B(S) isuniformly bounded and equicontinuous in PC(J,R). For any x ∈ S, on account of (5), we get

|Bx(t)| ≤ |µ||f(0, µ)|

k∏i=1

|f(ti, x(ti))||f(ti, x(t+i ))|

+k∑

i=1

∏i≤j≤k


|g(ti, x(ti))||f(tj , x(tj))||f(tj , x(t+j ))|

+k∑

i=1

∏i<j≤k

|ϕi(x(ti))||f(ti, x(t+i ))|

· |f(tj , x(tj))||f(tj , x(t+j ))|

+ tkIαkqk|g(t, x(t))|

≤ |µ||f(0, µ)|

m∏i=1

|f(ti, x(ti))||f(ti, x(t+i ))|

+m∑

i=1

∏i≤j≤m


|g(ti, x(ti))||f(tj , x(tj))||f(tj , x(t+j ))|

+m∑

i=1

∏i<j≤m

|ϕi(x(ti))||f(ti, x(t+i ))|

· |f(tj , x(tj))||f(tj , x(t+j ))|

+ tmIαmqm|g(T, x(T ))|

≤ |µ||f(0, µ)|

(Ω1

Ω2

)m

+ ‖p‖ψ(r)m+1∑i=1


Γqi−1(αi−1 + 1)

(Ω1

Ω2

)m+1−i

+Ω3

Ω2

m∑i=1

(Ω1

Ω2

)m−i

:= K,

for all t ∈ J . Taking supremum over t, we have ‖Bx‖ ≤ K for all x ∈ S. This shows that B is uniformlybounded on S.

Further, we will show that B(S) is an equicontinuous set in PC(J,R). Let τ1, τ2 ∈ J with τ1 < τ2and x ∈ S. Then we have

|Bx(τ2)− Bx(τ1)|

=

∣∣∣∣∣ µ

f(0, µ)

k∏i=1


+k∑

i=1

∏i≤j≤k



+k∑

i=1

∏i<j≤k



+ tkIαkqkg(τ2, x(τ2))

− µ

f(0, µ)

n∏i=1


−n∑

i=1

∏i≤j≤n



−n∑

i=1

∏i<j≤n



− tnIαnqng(τ1, x(τ1))

∣∣∣∣∣,




for some n ≤ k, n, k ∈ 0, 1, 2, . . . ,m. Further

|Bx(τ2)− Bx(τ1)| =∣∣tkIαkqkg(τ2, x(τ2))− tk

Iαkqkg(τ1, x(τ1))

∣∣≤ ‖p‖ψ(r)

∣∣∣∣ (τ2 − tk)αk

Γqk(αk + 1)

− (τ1 − tk)αk

Γqk(αk + 1)

∣∣∣∣→ 0,

independent of x ∈ S as τ1 → τ2. This shows that B(S) is an equicontinuous set in PC(J,R). Therefore,it follows by the Arzela-Ascoli theorem that B is a completely continuous operator on S.

Step 3. The hypothesis (c) of Lemma 2.9 is satisfied.

Let x ∈ PC(J,R) and y ∈ S be arbitrary elements such that x = AxBy. Then we have

|x(t)|≤ |Ax(t)||By(t)|

≤ |f(t, x(t))|

(|µ|

|f(0, µ)|

k∏i=1

|f(ti, y(ti))||f(ti, y(t+i ))|

+k∑

i=1

∏i≤j≤k


|g(ti, y(ti))||f(tj , y(tj))||f(tj , y(t+j ))|

+k∑

i=1

∏i<j≤k

|ϕi(y(ti))||f(ti, y(t+i ))|

· |f(tj , y(tj))||f(tj , y(t+j ))|

+ tkIαkqk|g(t, y(t))|

)

≤ Ω1

(|µ|

|f(0, µ|)

m∏i=1

|f(ti, y(ti))||f(ti, y(t+i ))|

+m∑

i=1

∏i≤j≤m


|g(ti, y(ti))||f(tj , y(tj))||f(tj , y(t+j ))|

+m∑

i=1

∏i<j≤m

|ϕi(y(ti))||f(ti, y(t+i ))|

· |f(tj , y(tj))||f(tj , y(t+j ))|

+ tmIαmqm|g(T, y(T ))|

)

≤ Ω1

(|µ|

|f(0, µ)|

(Ω1

Ω2

)m

+ ‖p‖ψ(r)m+1∑i=1


Γqi−1(αi−1 + 1)

(Ω1

Ω2

)m+1−i

+Ω3

Ω2

m∑i=1

(Ω1

Ω2

)m−i).

Taking supremum over t, we have

‖x‖ ≤ Ω1

(|µ|

|f(0, µ)|

(Ω1

Ω2

)m

+ ‖p‖ψ(r)m+1∑i=1


Γqi−1(αi−1 + 1)

(Ω1

Ω2

)m+1−i

+Ω3

Ω2

m∑i=1

(Ω1

Ω2

)m−i)≤ r.

Thus we deduce that x ∈ S.

Step 4. We show that the condition (d) of Lemma 2.9 holds.

As

M = ‖B(S)‖

≤

(|µ|

|f(0, µ)|

(Ω1

Ω2

)m

+ ‖p‖ψ(r)m+1∑i=1


Γqi−1(αi−1 + 1)

(Ω1

Ω2

)m+1−i

+Ω3

Ω2

m∑i=1

(Ω1

Ω2

)m−i),

therefore, by (H4), we have δM < 1 with δ = ‖φ‖.Thus all the conditions of Lemma 2.9 are satisfied and hence the operator equation x = AxBx has

a solution in S. In consequence, we infer that the problem (1) has a solution on J . This completes theproof.




Example 3.2 Consider the following impulsive hybrid fractional quantum difference equation with ini-tial condition

ctkD

k2+2k+1k2+2k+3k2+3k+12k2+3k+2

x(t)|x(t)|+30|x(t)|+35 + 1

25

(t− 1

2

)2 =

1 + sin2 t

2(t+ 5)

(x2(t)

4(1 + |x(t)|)+e−|x(t)|

2

),

t ∈ [0, 3/2] \ t1, . . . , t5,

∆x(tk) =|x(tk)|+ 1

(k + 1)(|x(tk)|+ 2), tk =

k

4, k = 1, . . . , 5,

x(0) =13.

(16)

Here αk = (k2 + 2k + 1)/(k2 + 2k + 3), qk = (k2 + 3k + 1)/(2k2 + 3k + 2), k = 0, 1, . . . , 5, tk = k/4,k = 1, 2, . . . , 5, m = 5, T = 3/2, µ = 1/3, f(t, x) = ((|x| + 30)/(|x| + 35)) + (1/25)(t − (1/2))2 andg(t, x) = ((1 + sin2 t)/(2(t + 5)))((x2/(4(1 + |x|))) + (e−|x|/2)). With the given values, we find thatΩ1 = 26/25, Ω2 = 6/7. Also, we have

|f(t, x)− f(t, y)| ≤ 1245

|x− y|, |g(t, x)| ≤ 1t+ 5

(|x|4

+12

), |ϕk(x)| ≤ 1

(k + 1), k = 1, 2, . . . , 5.

Clearly ‖φ‖ = 1/245, Ω3 = 1/2, ‖p‖ = 1/5 and ψ(|x|) = (|x|/4) + (1/2). Hence, there exists a constantr such that 6.611569689 < r < 1092.541483 satisfying (H4). Thus all the conditions of Theorem 3.1are satisfied. Therefore, the conclusion of Theorem 3.1 implies that the problem (16) has at least onesolution on [0, 3/2].Acknowledgment. This project was funded by the Deanship of Scientific Research (DSR) at KingAbdulaziz University, Jeddah, under grant no. RG-14-130-36. The authors, therefore, acknowledgewith thanks DSR for technical and financial support.

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A FIXED POINT ALTERNATIVE TO THE STABILITY OF A

QUADRATIC α-FUNCTIONAL EQUATION IN FUZZY BANACH SPACES

CHOONKIL PARK, JUNG RYE LEE∗, AND DONG YUN SHIN∗

Abstract. In this paper, we solve the following quadratic α-functional equation

N(2f(x) + 2f(y) − f(x+ y) − α−2f(α(x− y)), t

)≥ t

t+ φ(x, y)(0.1)

in fuzzy normed spaces, where ρ is a fixed real number with α−1 = ±√

3.Using the fixed point method, we prove the Hyers-Ulam stability of the quadratic α-

functional equation (0.1) in fuzzy Banach spaces.

1. Introduction and preliminaries

Katsaras [22] defined a fuzzy norm on a vector space to construct a fuzzy vector topological

structure on the space. Some mathematicians have defined fuzzy norms on a vector space from

various points of view [13, 26, 51]. In particular, Bag and Samanta [2], following Cheng and

Mordeson [9], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric

is of Kramosil and Michalek type [25]. They established a decomposition theorem of a fuzzy

norm into a family of crisp norms and investigated some properties of fuzzy normed spaces [3].

We use the definition of fuzzy normed spaces given in [2, 30, 31] to investigate the Hyers-Ulam

stability of additive ρ-functional inequalities in fuzzy Banach spaces.

Definition 1.1. [2, 30, 31, 32] Let X be a real vector space. A function N : X ×R→ [0, 1] is

called a fuzzy norm on X if for all x, y ∈ X and all s, t ∈ R,

(N1) N(x, t) = 0 for t ≤ 0;

(N2) x = 0 if and only if N(x, t) = 1 for all t > 0;

(N3) N(cx, t) = N(x, t|c|) if c = 0;

(N4) N(x+ y, s+ t) ≥ minN(x, s), N(y, t);(N5) N(x, ·) is a non-decreasing function of R and limt→∞N(x, t) = 1.

(N6) for x = 0, N(x, ·) is continuous on R.

The pair (X,N) is called a fuzzy normed vector space.

The properties of fuzzy normed vector spaces and examples of fuzzy norms are given in

[29, 30].

2010 Mathematics Subject Classification. Primary 46S40, 39B52, 47H10, 39B62, 26E50, 47S40.Key words and phrases. fuzzy Banach space; quadratic α-functional equation; fixed point method; Hyers-

Ulam stability.∗Corresponding authors.



C. PARK, J. R. LEE, AND D. Y. SHIN

Definition 1.2. [2, 30, 31, 32] Let (X,N) be a fuzzy normed vector space. A sequence xn in

X is said to be convergent or converge if there exists an x ∈ X such that limn→∞N(xn−x, t) = 1

for all t > 0. In this case, x is called the limit of the sequence xn and we denote it by N -

limn→∞ xn = x.

Definition 1.3. [2, 30, 31, 32] Let (X,N) be a fuzzy normed vector space. A sequence xnin X is called Cauchy if for each ε > 0 and each t > 0 there exists an n0 ∈ N such that for all

n ≥ n0 and all p > 0, we have N(xn+p − xn, t) > 1− ε.

It is well-known that every convergent sequence in a fuzzy normed vector space is Cauchy. If

each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy

normed vector space is called a fuzzy Banach space.

We say that a mapping f : X → Y between fuzzy normed vector spaces X and Y is

continuous at a point x0 ∈ X if for each sequence xn converging to x0 in X, then the

sequence f(xn) converges to f(x0). If f : X → Y is continuous at each x ∈ X, then

f : X → Y is said to be continuous on X (see [3]).

The stability problem of functional equations originated from a question of Ulam [50]

concerning the stability of group homomorphisms.

The functional equation f(x+ y) = f(x) + f(y) is called the Cauchy equation. In particular,

every solution of the Cauchy equation is said to be an additive mapping. Hyers [18] gave a

first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was

generalized by Aoki [1] for additive mappings and by Th.M. Rassias [42] for linear mappings by

considering an unbounded Cauchy difference. A generalization of the Th.M. Rassias theorem

was obtained by Gavruta [14] by replacing the unbounded Cauchy difference by a general

control function in the spirit of Th.M. Rassias’ approach. The stability problems of several

functional equations have been extensively investigated by a number of authors and there are

many interesting results concerning this problem (see [8, 12, 16, 17, 19, 21, 23, 24, 27, 35, 36,

37, 38, 39, 40, 43, 44, 45, 46, 47, 48, 49]).

We recall a fundamental result in fixed point theory.

Let X be a set. A function d : X × X → [0,∞] is called a generalized metric on X if d

satisfies

(1) d(x, y) = 0 if and only if x = y;

(2) d(x, y) = d(y, x) for all x, y ∈ X;

(3) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Theorem 1.4. [5, 10] Let (X, d) be a complete generalized metric space and let J : X → X

be a strictly contractive mapping with Lipschitz constant L < 1. Then for each given element

x ∈ X, either

d(Jnx, Jn+1x) =∞

for all nonnegative integers n or there exists a positive integer n0 such that




(1) d(Jnx, Jn+1x) <∞, ∀n ≥ n0;(2) the sequence Jnx converges to a fixed point y∗ of J ;

(3) y∗ is the unique fixed point of J in the set Y = y ∈ X | d(Jn0x, y) <∞;(4) d(y, y∗) ≤ 1

1−Ld(y, Jy) for all y ∈ Y .

In 1996, G. Isac and Th.M. Rassias [20] were the first to provide applications of stability

theory of functional equations for the proof of new fixed point theorems with applications. By

using fixed point methods, the stability problems of several functional equations have been

extensively investigated by a number of authors (see [4, 6, 7, 11, 15, 29, 33, 34, 40, 41]).

In this paper, we solve the quadratic α-functional equation (0.1) and prove the Hyers-Ulam

stability of the quadratic α-functional equation(0.1) in fuzzy Banach spaces by using the fixed

point method.

Throughout this paper, assume that X is a real vector space and (Y,N) is a fuzzy Banach

space. Assume that α is a real number with α−1 = ±√

3.

2. Quadratic α-functional equation (0.1)

In this section, we prove the Hyers-Ulam stability of the quadratic α-functional equation

(0.1) in fuzzy Banach spaces.

We need the following lemma to prove the main results.

We solve the quadratic α-functional equation (0.1) in vector spaces.

Lemma 2.1. Let X and Y be vector spaces. If a mapping f : X → Y satisfies

2f(x) + 2f(y) = f(x+ y) + α−2f(α(x− y)) (2.1)

for all x, y ∈ X, then f : X → Y is quadratic.

Proof. Assume that f : X → Y satisfies (2.1).

Letting x = y = 0 in (2.1), we get 3f(0) = α−2f(0). So f(0) = 0.

Letting y = 0 in (2.1), we get f(x) = α−2f(αx) and so f(αx) = α2f(x) for all x ∈ X. Thus

2f(x) + 2f(y) = f(x+ y) + α−2f(α(x− y)) = f(x+ y) + f(x− y)

for all x, y ∈ X, as desired.

Using the fixed point method, we prove the Hyers-Ulam stability of the quadratic α-functional

equation (2.1) in fuzzy Banach spaces.

Theorem 2.2. Let φ : X2 → [0,∞) be a function such that there exists an L < 1 with

φ(x, y) ≤ L

4φ (2x, 2y)


N(2f(x) + 2f(y)− f(x+ y)− α−2f(α(x− y)), t

)≥ t

t+ φ(x, y)(2.2)




for all x, y ∈ X and all t > 0. Then Q(x) := N -limn→∞ 4nf(

x2n)exists for each x ∈ X and


N (f(x)−Q(x), t) ≥ (4− 4L)t

(4− 4L)t+ Lφ(x, x)(2.3)


Proof. Letting y = x in (2.2), we get

N (f (2x)− 4f(x), t) ≥ t

t+ φ(x, x)(2.4)

for all x ∈ X.

Consider the set

S := g : X → Y

and introduce the generalized metric on S:

d(g, h) = inf

µ ∈ R+ : N(g(x)− h(x), µt) ≥ t

t+ φ(x, x), ∀x ∈ X, ∀t > 0

,

where, as usual, inf ϕ = +∞. It is easy to show that (S, d) is complete (see [28, Lemma 2.1]).

Now we consider the linear mapping J : S → S such that

Jg(x) := 4g

(x

2

)for all x ∈ X.

Let g, h ∈ S be given such that d(g, h) = ε. Then

N(g(x)− h(x), εt) ≥ t

t+ φ(x, x)

for all x ∈ X and all t > 0. Hence

N(Jg(x)− Jh(x), Lεt) = N

(4g

(x

2

)− 4h

(x

2

), Lεt

)= N

(g

(x

2

)− h

(x

2

),L

4εt

)≥

Lt4

Lt4 + φ

(x2 ,

x2

) ≥ Lt4

Lt4 + L

2φ(x, x)

=t

t+ φ(x, x)

for all x ∈ X and all t > 0. So d(g, h) = ε implies that d(Jg, Jh) ≤ Lε. This means that

d(Jg, Jh) ≤ Ld(g, h)

for all g, h ∈ S.


N

(f(x)− 4f

(x

2

),L

4t

)≥ t

t+ φ(x, x)

for all x ∈ X and all t > 0. So d(f, Jf) ≤ L4 .




By Theorem 1.4, there exists a mapping Q : X → Y satisfying the following:

(1) Q is a fixed point of J , i.e.,

Q

(x

2

)=

1

4Q(x) (2.5)

for all x ∈ X. The mapping Q is a unique fixed point of J in the set

M = g ∈ S : d(f, g) <∞.

This implies that Q is a unique mapping satisfying (2.5) such that there exists a µ ∈ (0,∞)

satisfying

N(f(x)−Q(x), µt) ≥ t

t+ φ(x, x)for all x ∈ X;

(2) d(Jnf,Q)→ 0 as n→∞. This implies the equality

N - limn→∞

4nf

(x

2n

)= Q(x)

for all x ∈ X;

(3) d(f,Q) ≤ 11−Ld(f, Jf), which implies the inequality

d(f,Q) ≤ L

4− 4L.

This implies that the inequality (2.3) holds.

By (2.2),

N

(4n(

2f

(x

2n

)+ 2f

(y

2n

)− f

(x+ y

2n

)− α−2f

(αx− y

2n

)), 4nt

)≥ t

t+ φ(

x2n ,

y2n)


N

(4n(

2f

(x

2n

)+ 2f

(y

2n

)− f

(x+ y

2n

)− α−2f

(αx− y

2n

)), t

)≥

t4n

t4n + Ln

4n φ (x, y)

for all x, y ∈ X, all t > 0 and all n ∈ N. Since limn→∞t4n

t4n

+Ln

4nφ(x,y)


t > 0,

2Q(x) + 2Q(y)−Q(x+ y)− α−2Q(α(x− y)) = 0

for all x, y ∈ X. By Lemma 2.1, the mapping Q : X → Y is quadratic, as desired.

Corollary 2.3. Let θ ≥ 0 and let p be a real number with p > 2. Let X be a normed vector

space with norm ∥ · ∥. Let f : X → Y be a mapping satisfying f(0) = 0 and

N(2f(x) + 2f(y)− f(x+ y)− α−2f(α(x− y)), t

)≥ t

t+ θ(∥x∥p + ∥y∥p)(2.6)

for all x, y ∈ X and all t > 0. Then Q(x) := N -limn→∞ 4nf( x2n ) exists for each x ∈ X and


N (f(x)−Q(x), t) ≥ (2p − 4)t

(2p − 4)t+ 2θ∥x∥p





Proof. The proof follows from Theorem 2.2 by taking φ(x, y) := θ(∥x∥p+∥y∥p) for all x, y ∈ X.

Then we can choose L = 22−p, and we get the desired result.

Theorem 2.4. Let φ : X2 → [0,∞) be a function such that there exists an L < 1 with

φ(x, y) ≤ 4Lφ

(x

2,y

2

)for all x, y ∈ X. Let f : X → Y be a mapping satisfying f(0) = 0 and (2.2). Then Q(x) := N -

limn→∞14n f (2nx) exists for each x ∈ X and defines a quadratic mapping Q : X → Y such

that

N (f(x)−Q(x), t) ≥ (4− 4L)t

(4− 4L)t+ φ(x, x)(2.7)


Proof. Let (S, d) be the generalized metric space defined in the proof of Theorem 2.2.

Now we consider the linear mapping J : S → S such that

Jg(x) :=1

4g (2x)

for all x ∈ X.


N

(f(x)− 1

4f(2x),

1

4t

)≥ t

t+ φ(x, x)

for all x ∈ X and all t > 0. So d(f, Jf) ≤ 14 . Hence

d(f,A) ≤ 1

4− 4L,

which implies that the inequality (2.7) holds.

The rest of the proof is similar to the proof of Theorem 2.2.

Corollary 2.5. Let θ ≥ 0 and let p be a real number with 0 < p < 2. Let X be a normed

vector space with norm ∥ · ∥. Let f : X → Y be a mapping satisfying f(0) = 0 and (2.6).

Then Q(x) := N -limn→∞14n f(2nx) exists for each x ∈ X and defines a quadratic mapping

Q : X → Y such that

N (f(x)−Q(x), t) ≥ (4− 2p)t

(4− 2p)t+ 2θ∥x∥p


Proof. The proof follows from Theorem 2.4 by taking φ(x, y) := θ(∥x∥p+∥y∥p) for all x, y ∈ X.

Then we can choose L = 2p−2, and we get the desired result.




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297–300.[43] J. Ratz, On inequalities associated with the Jordan-von Neumann functional equation, Aequationes Math.

66 (2003), 191–200.[44] L. Reich, J. Smıtal and M. Stefankova, Singular solutions of the generalized Dhombres functional equation,

Results Math. 65 (2014), 251–261.[45] S. Schin, D. Ki, J. Chang and M. Kim, Random stability of quadratic functional equations: a fixed point

approach, J. Nonlinear Sci. Appl. 4 (2011), 37–49.[46] S. Shagholi, M. Bavand Savadkouhi and M. Eshaghi Gordji, Nearly ternary cubic homomorphism in ternary

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133 (2003), 389–399.

Choonkil ParkResearch Institute for Natural Sciences, Hanyang University, Seoul 04763, KoreaE-mail address: [email protected]

Jung Rye LeeDepartment of Mathematics, Daejin University, Kyeonggi 11159, Korea

E-mail address: [email protected]

Dong Yun ShinDepartment of Mathematics, University of Seoul, Seoul 02504, Korea

E-mail address: [email protected]



Four-point impulsive multi-orders fractional boundary valueproblems

N. I. Mahmudov and H. MahmoudEastern Mediterranean University

Gazimagusa, TRNC, Mersiin 10, TurkeyEmail: [email protected]; [email protected]

Abstract

Four-point boundary value problem for impulsive multi-orders fractional di¤erential equation is stud-ied. The existence and uniqueness results are obtained for impulsive multi-orders fractional di¤erentialequation with four-point fractional boundary conditions by applying standard xed point theorems. Anexample for the illustration of the main result is presented.

Keywords: fractional di¤erential equations, xed point theorems, multi-orders, impulse.

1 Introduction

Impulsive di¤erential equations have extensively been studied in the past two decades. Impulsive di¤erentialequations are used to describe the dynamics of processes in which sudden, discontinuous jumps occur.Such processes are naturally seen in harvesting, earthquakes, diseases, and so forth. Recently, fractionalimpulsive di¤erential equations have attracted the attention of many researchers. For the general theory andapplications of such equationswe refer the interested reader to see [1]-[18] and the references therein.In [8], Kosmatov considered the following two impulsive problems:8<:

CDu (t) = f (t; u (t)) ; 1 < < 2; t 2 [0; 1] n ft1; t2; :::; tpg ;CD u

t+kC D u

tk= Ik

utk; tk 2 (0; 1) ; k = 1; :::; p;

u (0) = u0; u0 (0) = u0; 0 < < 1;

and 8<:LDu (t) = f (t; u (t)) ; 0 < < 1; t 2 [0; 1] n ft1; t2; :::; tpg ;LD u

t+k LD u

tk= Ik

utk; tk 2 (0; 1) ; k = 1; :::; p;

I1u (0) = u0; 0 < < < 1:

In [4], Feckan et al. studied the impulsive problem of the following form:8<:CDu (t) = f (t; u (t)) ; 0 < < 1; t 2 [0; 1] n ft1; t2; :::; tpg ;ut+k u

tk= Ik

utk; tk 2 (0; 1) ; k = 1; :::; p;

u (0) = u0; 0 < < < 1:

Wang et al. [17] obtained some existence and uniqueness results for the following impulsive multipoint frac-tional integral boundary value problem involving multi-orders fractional derivatives and deviating argument8>><>>:

CDktku (t) = f (t; u (t) ; u ( (t))) ; 1 < k 2; t 2 [0; T ] n ft1; t2; :::; tpg ;

u (tk) = Ikutk; u0 (tk) = Jk

utk; tk 2 (0; T ) ; k = 1; :::; p;

u (0) =Pp

k=0 kIktku (k) ; tk < k < tk+1;

u0 (0) = 0:


1249 N. I. Mahmudov et al 1249-1260

Yukunthorn et.al. [18] studied the similar problem for multi-order CaputoHadamard fractional di¤erentialequations equipped with nonlinear integral boundary conditions.

Motivated by the above works, in this paper, we study the existence of solutions for the four-pointnonlocal boundary value problems of nonlinear impulsive equations of fractional order8>><>>:

CDktku (t) = f (t; u (t) ; u0 (t)) ; 1 + k 2; t 2 [tk; tk+1) ;


u0tk; tk 2 (0; T ) ; k = 1; :::; p;

u (0) + 1CD

0+u (0) = 1u (1) ; 0 < 1 < t1 < T;

u (T ) + 2CD

tpu (T ) = 2u (2) ; 0 < tp < 2 < T; 0 < < 1;

(1)

where CDt is the Caputo derivative, f : [0; T ] R R ! R is a continuous function, Ik; Jk : R ! R;

u (tk) = ut+k u

tk; u0 (tk) = u0

t+k u0

tk; u

t+kand u

tkrepresent the right hand limit

and the left hand limit of the function u (t) at t = tk; and the sequence ftkg satises that 0 = t0 < t1 < ::: <tp < tp+1 = T .To the best of our knowledge, there is no paper that consider the four-point impulsive boundary value

problem involving nonlinear di¤erential equations of fractional order (1). The main di¢ culty of this problemis that the corresponding integral equation is very complex because of the impulse e¤ects. In this paper, westudy the existence and uniqueness of solutions for four-point impulsive boundary value problem (1). By useof Banachs xed point theorem and Schauders xed point theorem, some existence and uniqueness resultsare obtained.

2 Preliminaries

Let [0; T ] = [0; T ] n ft1; t2; :::; tpg and

PC ([0; T ] ;R) = fx : [0; T ]! R : x (t) is continuous everywhere except for sometk at which x

t+k; xtkexist and x

tk= x (tk) ; k = 1; :::; p

;

and

PC1 ([0; T ] ;R) = fx 2 PC ([0; T ] ;R) : x0 (t) is continuous everywhere except for sometk at which x0

t+k; x0tkexist and x0

tk= x0 (tk) ; k = 1; :::; p

:

PC ([0; T ] ;R) and PC1 ([0; T ] ;R) are Banach spaces with the norms kxkPC = sup fjx (t)j : t 2 [0; T ]g andkxkPC1 = max fkxkPC ; kx0kPCg ; respectively. Let X = PC1 ([0; T ] ;R)\C2

[0; T ]

;R: A function x 2 X

is called a solution of problem (1) if it satises (1).Throughout the paper we will use the following notations.

= 11 +

T + 2

T 1

(2 )

(1 1) ;

A0 =1

1 1 1

111 1

; B0 =1;

Ap =(1 1)

111 1

; Bp =1 1

:



Fk (y; u; u0) (t) =

1

(k)

Z t

tk

(t s)k1 y (s) ds

+kXj=1

1

(j1)

Z tj

tj1

(tj s)j11 y (s) ds+kXj=1

Ijutj

+kXj=1

(t tj)1

(j1 1)

Z tj

tj1


(t tj) Jju0tj;

Gk (y; u; u0) (t) =

1

(k )

Z t

tk

(t s)k1 y (s) ds

+t1

(2 )

kXj=1

1

(j1 1)

Z tj

tj1

(tj s)j12 y (s) ds

+t1

(2 )

kXj=1

Jju0tj:

F 0k (y; u; u0) (t) =

1

(k 1)

Z t

tk

(t s)k2 y (s) ds

+kXj=1

1

(j1 1)

Z tj

tj1


Jju0tj:

Lemma 1 Let y 2 C [0; T ]. A function u 2 PC1 [0; T ] is a solution of the boundary value problem8>><>>:CDk

tku (t) = y (t) ; 1 + < k 2; t 2 [0; T ] n ft1; t2; :::; tpg ;


u0tk; tk 2 (0; T ) ; k = 1; :::; p;

u (0) + 1CD

0+u (0) = 1u (1) ; 0 < 1 < t1 < T;

u (T ) + 2CD

tpu (T ) = 2u (2) = 2u (2) ; 0 < tp < 2 < T; 0 < < 1;

(2)

if and only if

u (t) = Fk (y; u) (t) +1

1 1F0 (y; u) (1)

1

111 1

+ t

F0 (y; u) (1)

+2 (1 1)

111 1

+ t

Fp (y; u) (2)

(1 1)

111 1

+ t

Fp (y; u) (T )

2 (1 1)

111 1

+ t

Gp (y; u) (T ) : (3)

Proof. Suppose that u is a solution of (2). For 0 t t1; we have

u (t) = I00+y (t) c1 c2t =1

(0)

Z t

0

(t s)01 y (s) ds c1 c2t; c1; c2 2 R: (4)

Then di¤erentiating (4), we get

D0+u (t) =

1

(0 )

Z t

0

(t s)01 y (s) ds c2t1

(2 ) ;

u0 (t) =1

(0 1)

Z t

0

(t s)02 y (s) ds c2:



If t1 < t t2; then for some d1; d2 2 R; we have

u (t) =1

(1)

Z t

t1

(t s)11 y (s) ds d1 d2 (t t1) ;

u0 (t) =1

(1 1)

Z t

t1

(t s)12 y (s) ds d2;

D

t+1u (t) =

1

(1 )

Z t

t1

(t s)11 y (s) ds d2t1

(2 ) :

Thus

ut1=

1

(0)

Z t1

0

(t1 s)01 y (s) ds c1 c2t1; ut+1= d1

u0t1=

1

(0 1)

Z t1

0

(t1 s)02 y (s) ds c2; u0t+1= d2:

In view ofut+1 u

t1= I1

ut1; u0

t+1 u0

t1= J1

u0t1;

we nd that

d1 =1

(0)

Z t1

0

(t1 s)01 y (s) ds+ I1ut1 c1 c2t1;

d2 =1

(0 1)

Z t1

0

(t1 s)02 y (s) ds+ J1u0t1 c2:

Hence we obtain for t1 < t t2

u (t) =1

(1)

Z t

t1

(t s)11 y (s) ds

+1

(0)

Z t1

0

(t1 s)01 y (s) ds+ I1ut1

+ (t t1)1

(0 1)

Z t1

0

(t1 s)02 y (s) ds+ (t t1) J1u0t1

c1 c2t; t1 < t t2:In a similar way, for k = 1; 2; :::; p we can obtain

u (t) =1

(k)

Z t

tk

(t s)k1 y (s) ds+kXj=1

1

(j1)

Z tj

tj1


Ijutj

+kXj=1

(t tj)1

(j1 1)

Z tj

tj1


(t tj) Jju0tj

c1 c2t; tk < t tk+1: (5)

Moreover,

CDtku (t) =

1

(k )

Z t

tk

(t s)k1 y (s) ds

+t1

(2 )

kXj=1

1

(j1 1)

Z tj

tj1

(tj s)j12 y (s) ds

+t1

(2 )

kXj=1

Jju0tj c2

t1

(2 ) :



Now applying the boundary conditions

u (0) + 1CD

0+u (0) = 1u (1) ; 0 < 1 < t1 < T;

u (T ) + 2CD

tpu (T ) = 2u (2) ; 0 < tp < 2 < T; 0 < < 1;

we get

c1 = 1F0 (y; u; u0) (1) 1c1 c211;

Fp (y; u; u0) (T ) c1 c2T + 2Gp (y; u) (T ) 2c2

T 1

(2 ) = 2Fp (y; u; u0) (2) :

Solving this system for c1, c2 and inserting these values into (5) we get

u (t) = Fk (y; u; u0) (t) +

11 1

F0 (y; u; u0) (1) c2

111 1

+ t

= Fk (y; u; u

0) (t) +1

1 1F0 (y; u; u

0) (1)1

111 1

+ t

F0 (y; u; u

0) (1)

+2 (1 1)

111 1

+ t

Fp (y; u; u

0) (2)

(1 1)

111 1

+ t

Fp (y; u; u

0) (T )

2 (1 1)

111 1

+ t

Gp (y; u; u

0) (T ) :

Conversely, assume that u is a solution of the impulsive fractional integral equation (3). Then by a directcomputation, it follows that the solution given by (3) satises (2). This completes the proof.

3 Existence and Uniqueness

In the sequel,we assume that

(A1) f : [0; T ] R R! R is continuous function and such that

jf (t; x; x1) f (t; y; y1)j lf (jx yj+ jx1 y1j) ; lf > 0; 0 t T; x; y; x1; y1 2 R:

(A2) Ik; Jk : R! R are continuous functions and satisfy

jIk (x) Ik (y)j l1 jx yj ;jJk (x) Jk (y)j l2 jx yj ; l1 > 0; l2 > 0; 0 t T; x; y 2 R:

For convenience, we will give some notations:

T = max fTk : 0 k pg ; = min f (k) : 0 k pg ;

1 =

pXj=1

(tj tj1)j1

(j1 + 1); 2 =

pXj=1

(T tj) (tj tj1)j11

(j1);

3 =T 1

(2 )

pXj=1

(tj tj1)j11

(j1); 4 =

pXj=1

(tj tj1)j11

(j1):



F := lfT

+ lf1 + lf2 + pl1 + l2pT;

G := lfT

+ lf1

T 1

(2 ) + l2pT 1

(2 ) ;

F 0 := lfT

+ lf4 + l2p:

Lemma 2 Fk (f; u; u0) and Gk (f; u; u0) are Lipschitzian operators.

jFk (f; u; u0) Fk (f; v; v0)j F ku vkPC1 ; LFk > 0;

jGk (f; u; u0)Gk (f; v; v0)j G ku vkPC1 ; LGk> 0; u; v 2 PC1 ([0; T ] ;R) :

Proof. For u; v 2 PC1 ([0; T ] ;R), we have

jFk (f; u; u0) (t) Fk (f; v; v0) (t)j

1

(k)

Z t

tk

(t s)k1 jf (s; u (s) ; u0 (s)) f (s; v (s) ; v0 (s))j ds

+kXj=1

1

(j1)

Z tj

tj1

(tj s)j11 jf (s; u (s) ; u0 (s)) f (s; v (s) ; v0 (s))j ds

+

kXj=1

Ij u tj Ij v tj +

kXj=1

(t tj) (j1 1)

Z tj

tj1


+kXj=1

(t tj)Jj u0 tj Jj v0 tj

lf1

(k)

Z t

tk

(t s)k1 (ju (s) v (s)j+ ju0 (s) v0 (s)j) ds

+ lf

kXj=1

1

(j1)

Z tj

tj1

(tj s)j11 (ju (s) v (s)j+ ju0 (s) v0 (s)j) ds

+ l1

kXj=1

u tj v tj + lf kXj=1

1

(j1 1)(t tj)

Z tj

tj1

(tj s)j12 (ju (s) v (s)j+ ju0 (s) v0 (s)j) ds

+ l2

kXj=1

(t tj)u0 tj v0 tj

F ku vkPC1 :



Similarly,

jGk (f; u; u0) (t)Gk (f; v; v0) (t)j

1

(k )

Z t

tk

(t s)k1 jf (s; u (s) ; u0 (s)) f (s; v (s) ; v0 (s))j ds

+T 1

(2 )

kXj=1

1

(j1 1)

Z tj

tj1


+t1

(2 )

kXj=1

Jj u0 tj Jj v0 tj

0@lf (T tk)k (k + 1)

+ lfT 1

(2 )

pXj=1

(tj tj1)j11

(j1)+

T 1

(2 ) l2

1A ku vkPC1

G ku vkPC1 :

Also, we havejF 0k (f; u; u0) (t) F 0k (f; v; v0) (t)j F 0 ku vkPC1 :

In view of Lemma 1 we dene an operator : X ! X by

(u) (t) = Fk (f; u) (t) (A0 B0t)F0 (f; u) (1)+ 2 (Ap +Bpt)Fp (f; u) (2) (Ap +Bpt)Fp (f; u) (T ) 2 (Ap +Bpt)Gp (f; u) (T ) ;

where

A0 =1

1 1 1

111 1

; B0 =1;

Ap =(1 1)

111 1

; Bp =1 1

:

Let := max fF ;G;F 0g :

Theorem 3 Suppose that the assumption (A1), (A2) are satised. If

:= max f(1 + jA0j+ jB0jT + (j2j+ j2j+ 1) (jApj+ jBpjT )); (1 + jB0j+ (j2j+ j2j+ 1) jBpj)g < 1;

then the boundary value problem (1) has a unique solution.

Proof. Let u; v 2 PC1 ([0; T ] ;R) : For u; v 2 (tk; tk+1] ; k = 0; :::; p; we have

j(u) (t) (v) (t)j jFk (f; u; u0) (t) Fk (f; v; v0) (t)j+ jA0 B0tj jF0 (f; u; u0) (1) F0 (f; v; v0) (1)j+ j2j jAp +Bptj jFp (f; u; u0) (2) Fp (f; v; v0) (2)j+ jAp +Bptj jFp (f; u; u0) (T ) Fp (f; v; v0) (T )j+ j2j jAp +Bptj jGp (f; u; u0) (T )Gp (f; v; v0) (T )j (1 + jA0j+ jB0jT + (j2j+ j2j+ 1) (jApj+ jBpjT )) ku vkPC1 :



Similarly, for u; v 2 (tk; tk+1] we have(u)0 (t) (v)0 (t) jF 0k (f; u; u0) (t) F 0k (f; v; v0) (t)j+ jB0j jF0 (f; u; u0) (1) F0 (f; v; v0) (1)j+ j2j jBpj jFp (f; u; u0) (2) Fp (f; v; v0) (2)j+ jBpj jFp (f; u; u0) (T ) Fp (f; v; v0) (T )j+ j2j jBpj jGp (f; u; u0) (T )Gp (f; v; v0) (T )j (1 + jB0j+ (j2j+ j2j+ 1) jBpj) ku vkPC1 :

It follows thatkuvkPC1 ku vkPC1 :

Since < 1; is a contraction. According to the Banach xed point theorem has a unique xed point,that is the problem (1) has a unique solution.

4 Existence

In this section, we assume that

(A3) f : [0; T ] R! R is continuous function and there exists h 2 C ([0; T ] ;R+) such thatjf (t; u; v)j h (t) + b1 juj + b2 jvj% ; (t; u; v) 2 [0; T ] R R; 0 < ; % < 1:

(A4) Ik; Jk : R! R are continuous functions and there L2 > 0; L3 > 0 such that

jIk (x)j L2; jJk (x)j L3; x 2 R:

For convenience, we will give some notations:

C1 := (1 + jA0j+ jB0jT + (j2j+ 1) (jApj+ jBpjT )) (pL2 + pTL3) khk

+ j2j (jApj+ jBpjT )T 1

(2 )L3 khk ;

C2 := (1 + jA0j+ jB0jT + (j2j+ 1) (jApj+ jBpjT ))T

+1 +2

+ j2j (jApj+ jBpjT )

T

+3

:

Lemma 4 IfR max

n3C1; (3b1C2)

11 ; (3b1C2)

11%o;

then maps B (0; R) :=u 2 PC1 ([0; T ] ;R) : kukPC1 R

into itself.

Proof. Assume thatR max

n3C1; (3b1C2)

11 ; (3b1C2)

11%o:

Then for t 2 (tk; tk+1] ; k = 0; :::; p; we havejFk (f; u; u0) (t)j

1

(k)

Z t

tk

(t s)k1 jf (s; u (s) ; u0 (s))j ds

+

kXj=1

1

(j1)

Z tj

tj1

(tj s)j11 jf (s; u (s) ; u0 (s))j ds+kXj=1

Ij u tj +

kXj=1

(t tj) (j1 1)

Z tj

tj1

(tj s)j12 jf (s; u (s) ; u0 (s))j ds+kXj=1

(t tj)Jj v0 tj ;



jFk (f; u; u0) (t)j

1

(k)

Z t

tk

(t s)k1h (s) + b1 ju (s)j + b2 ju0 (s)j%

ds

+kXj=1

1

(j1)

Z tj

tj1

(tj s)j11h (s) + b1 ju (s)j + b2 ju0 (s)j%

ds

+kXj=1

Ij u tj +

kXj=1

(t tj) (j1 1)

Z tj

tj1

(tj s)j12 (h (s) + b1 ju (s)j + b2 jv (s)j%) ds

+kXj=1

(t tj)Jj u tj

Tk

(k + 1)

khk+ b1 kuk + b2 ku0k%

+

pXj=1

(tj tj1)j1

(j1 + 1)


+ pL2

+

pXj=1

(t tj) (tj tj1)j11

(j1)


+ pTL3

T

+1 +2


+ pL2 + pTL3;

jGk (y; u; u0) (t)j Tk

(k + 1)khk+ b1 kuk + b2 ku0k%

+

t1

(2 )

pXj=1

(tj tj1)j11

(j1)


+

t1

(2 )L3

T

+3


+

t1

(2 )L3;

jF 0k (y; u; u0) (t)j

0@Tk1 (k)

+

kXj=1

(tj tj1)j11

(j1)

1Akhk+ b1 kuk + b2 ku0k%+ L3:

It follows that

j(u) (t)j (1 + jA0j+ jB0jT + (j2j+ 1) (jApj+ jBpjT ))

T

+1 +2


+ pL2 + pTL3

+ j2j (jApj+ jBpjT )

T

+3


+

T 1

(2 )L3

C1 + C2b1 kuk + C2b2 ku0k% ;



and

(u)0 (t) 0@Tk1 (k)

+kXj=1

(tj tj1)j11

(j1)

1Akhk+ b1 kuk + b2 ku0k%+ L3+ (jB0j+ j2j jBpj+ jBpj)

T

+1 +2


+ pL2 + pTL3

+ jj jBpjT

+3


+

T 1

(2 )L3

C1 + C2b1 kuk + C2b2 ku0k% :

Thus

k(u)kPC1 C1 + C2b1R + C2b2R% R

3+R

3+R

3= R:

Theorem 5 Assume that the conditions (A3) and (A4) are satised. Then the problem (1) has at least onesolution.

Proof. Firstly, we prove that : PC1([0; T ] ; R) ! PC1([0; T ] ; R) is completely continuous operator. It isclear that, the continuity of functions f; Ikand Jk implies the continuity of the operator .Let PC1([0; T ] ; R) be bounded. Then there exist positive constants such that

jf(t; u; u0)j L1; jIk (u)j L2; jJk (u)j L3;

for all u 2 . Thus, for any u 2 ,we haveFk(f; u; u0) L1T +1 +2

+ pL2 + L3pT;

Similarly, Gk f; u; u0 (t) L1T + L11

T 1

(2 ) +T 1

(2 )pL3:

It follows thatj(u) (t)j 1(constant):

In a like manner, F 0k f; u; u0 (t) L1T +4+ L3p:

It follows that (u)0 (t) L1T +4

+ L3p

+(j0j+ j2j jBpj+ jBpj) F + j2j jBpjG =: 2

ThuskukPC1 1 + 2 = constant:

On the other hand, for 1 ; 2 2 [tk; tk+1] with 1 2 and we have

j(u)(1) (u)2j Z 2

1

(u)0 (s) ds (2 1) :



Similarly(u)

0(2) (u)

0(1) (2 1);

where is a constant. This implies that u is equicontinuous on all (tk; tk+1], k = 0; 1; :::; p. Consequently,Arzela-Ascoli theorem ensures us that the operator is a completely continuous operator and by Lemma 4: B(0; R) ! B(0; R). Hence, we conclude that : B(0; R) ! B(0; R) is completely continuous. It followsfrom the Schauder xed point theorem that has at least one xed point. That is problem (1) has at leastone solution.Example 1. For p = 1; t1 = 1

4 ; T = 1; =12 ; 1 = 2; 1 =

12 ; 2 = 3; 1 =

110 ; 1 =

15 ; 2 =

23 ; 0 =

32 ; k =

32 ; we consider the following impulsive multi-orders fractional di¤erential equation:8>>>><>>>>:

CDktku (t) = 1

100 cosu (t) +ju0(t)j

ju0(t)j+100 + t; 0 < t < 1; t 6= 14 ;

u14

=

ju( 14 )jju( 14 )j+50

; u014

=

ju0( 14 )jju0( 14 )j+70

;

u (0) + 2 CD0+u (0) =12u15

;

u (1) + 2 CD0+u (1) =12u23

:

(6)

It is clear that

jf (t; x; x1) f (t; y; y1)j 0:02 (jx yj+ jx1 y1j) ; 0 t 1; x; y; x1; y1 2 R:

One can easily calculate that = 0:2178 < 1:

Therefore, all the assumptions of Theorem 3 hold. Thus, by Theorem 3, the impulsive multi-orders fractionalboundary value problem (6) has a unique solution on [0; 1].

References

[1] B. Ahmad and S. Sivasundaram, Existence of solutions for impulsive integral boundary value problemsof fractional order,Nonlinear Analysis: Hybrid Systems, vol. 4, no. 1, pp. 134141, 2010.

[2] C. Bai, Existence result for boundary value problem of nonlinear impulsive fractional di¤erential equa-tion at resonance, Journal of Applied Mathematics and Computing, vol. 39, no. 1-2, pp. 421443,2012.

[3] J. Cao and H. Chen, Some results on impulsive boundary value problem for fractional di¤erentialinclusions,Electronic Journal of Qualitative Theory of Di¤erential Equations, vol. 2011, no. 11, pp.124, 2011.

[4] Feµckan, Y. Zhou, and J. R. Wang, On the concept and existence of solution for impulsive fractionaldi¤erential equations,Communications in Nonlinear Science and Numerical Simulation, vol. 17, no. 7,pp. 30503060, 2012.

[5] T. L. Guo and J. Wei, Impulsive problems for fractional di¤erential equations with boundary valueconditions,Computers & Mathematics with Applications, vol. 64, no. 10, pp. 32813291, 2012.

[6] Z. He and J. Yu, Periodic boundary value problem for rstorder impulsive ordinary di¤erential equa-tions,Journal of Mathematical Analysis and Applications, vol. 272, no. 1, pp. 6778, 2002.

[7] R. W. Ibrahim, Stability of sequential fractional di¤erential equations, Appl. Comput. Math., V.14,N.2, 2015, pp.141-149.



[8] N. Kosmatov, Initial value problems of fractional order with fractional impulsive conditions,Resultsin Mathematics, vol. 63, no. 3-4, pp. 12891310, 2013.

[9] V. Lakshmikantham, D. D. Bainov, and P. S. Simeonov, Theory of Impulsive Di¤erential Equations,vol. 6 of Series in Modern Applied Mathematics,World Scientic, Singapore, 1989.

[10] Z. Luo and J. Shen, Stability of impulsive functional di¤erential equations via the Liapunov functional,Applied Mathematics Letters, vol. 22, no. 2, pp. 163169, 2009.

[11] S. Tiwari, M. Kumari, An initial value tecgnique to solve two-point nonlinear singularly perturbedboundary value problems, Appl. Comput. Math., V.14, N.2, 2015, pp.150-157.

[12] H.Wang, Existence results for fractional functional di¤erential equations with impulses, Journal ofApplied Mathematics and Computing, vol. 38, no. 1-2, pp. 85101, 2012.

[13] G. Wang, B. Ahmad, and L. Zhang, Impulsive anti-periodic boundary value problem for nonlineardi¤erential equations of fractional order,Nonlinear Analysis: Theory, Methods & Applications, vol.74, no. 3, pp. 792804, 2011.

[14] G. Wang, B. Ahmad, and L. Zhang, Some existence results for impulsive nonlinear fractional di¤erentialequations with mixed boundary conditions,Computers & Mathematics with Applications, vol. 62, no.3, pp. 13891397, 2011.

[15] X.Wang, Existence of solutions for nonlinear impulsive higher order fractional di¤erential equations,Electronic Journal of QualitativeTheory of Di¤erential Equations, vol. 2011,no. 80, pp. 112, 2011.

[16] J.R.Wang, Y. Zhou, andM. Feµckan, On recent developments in the theory of boundary value problemsfor impulsive fractional di¤erential equations,Computers &Mathematics with Applications, vol. 64,no. 10, pp. 30083020, 2012.

[17] G. Wang, S. Liu, D. Baleanu, and L. Zhang, A New Impulsive Multi-Orders Fractional Di¤erentialEquation Involving Multipoint Fractional Integral Boundary Conditions, Abstract and Applied AnalysisVolume 2014, Article ID 932747, 10 pages, http://dx.doi.org/10.1155/2014/932747.

[18] W. Yukunthorna, B. Ahmad, S. K. Ntouyas, J. Tariboon, On CaputoHadamard type fractional impul-sive hybrid systems with nonlinear fractional integral conditions, Nonlinear Analysis: Hybrid Systems19 (2016) 7792.



Convergence of modification of the

Kantorovich-type q-Bernstein-Schurer operators

Qing-Bo Caia,∗and Guorong Zhoub

aSchool of Mathematics and Computer Science, Quanzhou Normal University,

Quanzhou 362000, ChinabSchool of Applied Mathematics, Xiamen University of Technology,

Xiamen 361024, China

E-mail: [email protected], [email protected]

Abstract. In this paper, we introduce a new modification of Kantorovich-type

Bernstein-Schurer operators K∗n,p,q(f ;x) based on the concept of q-integers. We

investigate statistical approximation properties, establish a local approximation theo-

rem, give a convergence theorem for the Lipschitz continuous functions and obtain a

Voronovskaja-type theorem. Furthermore, we also give some illustrative graphics and

some numerical examples for comparisons for the convergence of operators to some

function.

2000 Mathematics Subject Classification: 41A10, 41A25, 41A36.

Key words and phrases: q-integers, Bernstein-Schurer operators, A-statistical

convergence, rate of convergence, Lipschitz continuous functions.

1 Introduction

In 2015, Agrawal, Finta and Kumar [1] introduced a new Kantorovich-type general-

ization of the q-Bernstein-Schurer operators, they gave the basic convergence theorem,

obtained the local direct results, estimated the rate of convergence and so on. The oper-

ators are defined as

Kn,p,q(f ;x) = [n+ 1]q

n+p∑k=0

bn+p,k(q;x)q−k∫ [k+1]q

[n+1]q

[k]q[n+1]q

f(t)dRq t, (1)

where bn+p,k(q;x) is defined by

bn+p,k(q;x) =

[n+ p

k

]q

xk(1− x)n+p−kq . (2)

They obtained the following lemma of the moments.

∗Corresponding author.

1


1261 Qing-Bo Cai et al 1261-1272

Q. -B. CAI AND G. ZHOU

Lemma 1.1. (See [1], Lemma 2.1) The following equalities hold

Kn,p,q(1;x) = 1; (3)

Kn,p,q(t;x) =2q[n+ p]q[2]q[n+ 1]q

x+1

[2]q[n+ 1]q; (4)

Kn,p,q

(t2;x

)=

q2(1 + q + 4q2)[n+ p]q[n+ p− 1]q[2]q[3]q[n+ 1]2q

x2 +q(3 + 5q + 4q2)[n+ p]q

[2]q[3]q[n+ 1]2qx

+1

[3]q[n+ 1]2q. (5)

Apparently, these operators reproduce only constant functions. In present paper,

we will introduce a new modification of Kantorovich-type q-Bernstein-Schurer operators

K∗n,p,q(f ;x) which will be defined in (7). The advantage of these new operators is that

they reproduce not only constant functions but also linear functions. We will investigate

statistical approximation properties, establish a local approximation theorem, give a con-

vergence theorem for the Lipschitz continuous functions and obtain a Voronovskaja-type

theorem. Furthermore, we will give some illustrative graphics and some numerical exam-

ples for comparisons for the convergence of operators to some function. We may observe

that the new operators K∗n,p,q(f ;x) give a better approximation to f(x) than Kn,p,q(f ;x).

Before introducing the operators, we mention certain definitions based on q-integers,

detail can be found in [4, 5]. For any fixed real number 0 < q ≤ 1 and each nonnegative

integer k, we denote q-integers by [k]q, where

[k]q =

1−qk1−q , q 6= 1;

k, q = 1.

Also q-factorial and q-binomial coefficients are defined as follows:

[k]q! =

[k]q[k − 1]q...[1]q, k = 1, 2, ...;

1, k = 0,,

[n

k

]q

=[n]q!

[k]q![n− k]q!, (n ≥ k ≥ 0).

For x ∈ [0, 1] and n ∈ N0, we recall that

(1− x)nq =

1, n = 0;∏n−1

j=0

(1− qjx

)= (1− x)(1− qx)...

(1− qn−1x

), n = 1, 2, ...

.

The Riemann-type q-integral is defined by∫ b

af(t)dRq t = (1− q)(b− a)

∞∑j=0

f(a+ (b− a)qj

)qj , (6)

where the real numbers a, b and q satisfy that 0 ≤ a < b and 0 < q < 1.

For f ∈ C(I), I = [0, 1+p], p ∈ N0, q ∈ (0, 1) and n ∈ N, we introduce the modification

of Kantorovich-type q-Bernstein-Schurer operators as follows:

K∗n,p,q(f ;x) = [n+ 1]q

n+p∑k=0

bn+p,k (q;u(x)) q−k∫ [k+1]q

[n+1]q

[k]q[n+1]q

f(t)dRq t, (7)



CONVERGENCE OF MODIFICATION OF THE KANTOROVICH-TYPEq-BERNSTEIN-SCHURER OPERATORS

where bn+p,k(q;x) is defined by (2), and

u(x) =[2]q[n+ 1]qx− 1

2q[n+ p]q,

(1

[2]q[n+ 1]q≤ x ≤ 1 + 2q[n+ p]q

[2]q[n+ 1]q

). (8)

2 Auxiliary Results

In order to obtain the approximation properties, We need the following lemmas:

Lemma 2.1. For the modification of Kantorovich-type q-Bernstein-Schurer operators (7),

using lemma 1.1, by some easily computations we have

K∗n,p,q(1;x) = 1, (9)

K∗n,p,q(t;x) = x, (10)

K∗n,p,q(t2;x

)=

[2]q(1 + q + 4q2

)[n+ p− 1]qx

2

4[3]q[n+ p]q+

(3 + 5q + 4q2

)x

2[3]q[n+ 1]q

−(1 + q + 4q2

)[n+ p− 1]qx

2[3]q[n+ 1]q[n+ p]q+

(1 + q + 4q2

)[n+ p− 1]q

4[2]q[3]q[n+ 1]2q [n+ p]q

− 3 + 5q + 4q2

2[2]q[3]q[n+ 1]2q+

1

[3]q[n+ 1]2q. (11)

Remark 2.2. Let qn denotes a sequence such that 0 < qn < 1. Then, by Bohman and

Korovkin Theorem, for any f ∈ C(I), operators K∗n,p,q(f ;x) converge uniformly to f(x),

if and only if limn→∞ qn = 1.

Lemma 2.3. For the modification of Kantorovich-type q-Bernstein-Schurer operators (7),

we have

K∗n,p,q(t− x;x) = 0, (12)

K∗n,p,q((t− x)2;x

)≤

(q2 + 4q3 − 2q − 3

)x2

4[3]q+

(1 + 2q)x

[3]q[n+ 1]q+

(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q(13)

≤ (1 + 2q)x

[3]q[n+ 1]q+

(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q, (14)


)≤ O

(1

[n]2q

). (15)

Proof. By (9) and (10), we get (12). Using (10), (11) and some computations, we have


)= K∗n,p,q(t

2;x)− 2xK∗n,p,q(t;x) + x2

≤(q2 + 4q3 − 2q − 3

)x2

4[3]q+

(1 + 2q)[n+ p− 1]qx

[3]q[n+ 1]q[n+ p]q+qn+p−1(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q

≤(q2 + 4q3 − 2q − 3

)x2

4[3]q+

(1 + 2q)x

[3]q[n+ 1]q+

(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q.




Indeed, using the similar method for estimate Kn,p,q

((t− x)4;x

)in [1, P. 229], we have


)≤ 64

n+p∑k=0

bn+p,k(q;u(x))

([k]q

[n+ 1]q− [k]q

[n+ p]q

)4

+ 64

n+p∑k=0

bn+p,k(q;u(x))

([k]q

[n+ p]q− x)4

+8

[5]q

n+p∑k=0

bn+p,k(q;u(x))

(qk

[n+ 1]q

)4

≤ 64

n+p∑k=0

bn+p,k(q;u(x))

([k]q

[n+ p]q

)4 [qn([p]q − 1)

[n+ 1]q

]4

+64

n+p∑k=0

bn+p,k(q;u(x))

([k]q

[n+ p]q− [2]q[n+ 1]qx− 1

2q[n+ p]q+

[2]q[n+ 1]qx− 1

2q[n+ p]q− x)4

+8

[5]q

n+p∑k=0

bn+p,k(q;u(x))

(qk

[n+ 1]q

)4

≤ C1([p]q − 1)4

[n]2q+ 512

n+p∑k=0

bn+p,k(q;u(x))((t− u(x))4;x

)+512

n+p∑k=0

bn+p,k(q;u(x))

([2]q[n+ 1]qx− 1

2q[n+ p]q− x)4

+C2

[n]2q,

where u(x) is defined in (8), C1 and C2 are some positive constants. Thus,


)≤ C1

([p]q − 1)4

[n]2q+ 512

C3

[n]2q+ 512

[[2]q ([n]q + qn)x− 1− 2q ([n]q + qn[p]q)x

2q[n+ p]q

]4+

C2

[n]2q

= C1([p]q − 1)4

[n]2q+ 512

C3

[n]2q+ 512

[(1 + qn+1 − 2qn+1[p]q

)x− 1

2q[n+ p]q

]4+

C2

[n]2q= O

(1

[n]2q

),

where C3 is a positive constant, lemma 2.3 is proved.

3 Statistical approximation properties

In this section, we present the statistical approximation properties of the operator

K∗n,p,q(f ;x).

Let K be a subset of N, the set of all natural numbers. The density of K is defined

by δ(K) := limn1n

∑nk=1 χK(k) provided the limit exists, where χK is the characteristic

function of K. A sequence x := xn is called statistically convergent to a number L if,

for every ε > 0, δn ∈ N : |xn − L| ≥ ε = 0. Let A := (ajn), j, n = 1, 2, ... be an infinite

summability matrix. For a given sequence x := xn, the A−transform of x, denoted by

Ax := ((Ax)j), is given by (Ax)j =∑∞

k=1 ajnxn provided the series converges for each




j. We say that A is regular if limn(Ax)j = L whenever limx = L. Assume that A is a

non-negative regular summability matrix. A sequence x = xn is called A-statistically

convergent to L provided that for every ε > 0, limj∑

n:|xn−L|≥ε ajn = 0. We denote this

limit by stA − limn xn = L. For A = C1, the Cesaro matrix of order one, A-statistical

convergence reduces to statistical convergence. It is easy to see that every convergent

sequence is statistically convergent but not conversely.

We consider a sequence q := qn for 0 < qn < 1 satisfying

stA − limnqn = 1. (16)

If ei = ti, t ∈ R+, i = 0, 1, 2, ... stands for the ith monomial, then we have

Theorem 3.1. Let A = (ank) be a non-negative regular summability matrix and q := qnbe a sequence satisfying (16), then for all f ∈ C(I), x ∈ [0, 1], we have

stA − limn||K∗n,p,qf − f ||C(I) = 0. (17)

Proof. Obviously

stA − limn||K∗n,p,qn(ei)− ei||C(I) = 0. (i = 0, 1) (18)

By (11) and (13), we have∣∣K∗n,p,qn(e2;x)− e2(x)∣∣ ≤ 1 + 2qn

[3]qn [n+ 1]qn+

3 + 5qn + 4q2n2[3]qn [n+ 1]qn [n+ p]qn

.

Now for a given ε > 0, let us define the following sets:

U :=k :∣∣∣∣K∗n,p,qk(e2)− e2

∣∣∣∣C(I)≥ ε, U1 :=

k :

1 + 2qk[3]qk [n+ 1]qk

≥ ε

2

,

U2 :=

k :

3 + 5qk + 4q2k2[3]qk [n+ 1]qk [n+ p]qk

≥ ε

2

.

Then one can see that U ⊆ U1 ∪ U2, so we have

δk ≤ n : ||K∗n,p,qk(e2)− e2||C(I)

≤ δ

k ≤ n :

1 + 2qk[3]qk [n+ 1]qk

≥ ε

2

+δ

k ≤ n :

3 + 5qk + 4q2k2[3]qk [n+ 1]qk [n+ p]qk

≥ ε

2

,

since stA − limnqn = 1, we have

stA − limn

1 + 2qn[3]qn [n+ 1]qn

= 0, stA − limn

3 + 5qn + 4q2n2[3]qn [n+ 1]qn [n+ p]qn

= 0,

which implies that the right-hand side of the above inequality is zero, thus we have

stA − limn||K∗n,p,qn(e2)− e2||C(I) = 0. (19)

Combining (18) and (19), theorem 3.1 follows from the Korovkin-type statistical approx-

imation theorem established in [3], the proof is completed.




4 Local approximation properties

Let f ∈ C(I), endowed with the norm ||f || = supx∈I |f(x)|. The Peetre’s K−functional

is defined by

K2(f ; δ) = infg∈C2

||f − g||+ δ||g′′||

,

where δ > 0 and C2 = g ∈ C(I) : g′, g′′ ∈ C(I) . By [2, p.177, Theorem 2.4], there exits

an absolute constant C > 0 such that

K2(f ; δ) ≤ Cω2(f ;√δ), (20)

where

ω2(f ; δ) = sup0<h≤δ

supx,x+h,x+2h∈I

|f(x+ 2h)− 2f(x+ h) + f(x)|

is the second order modulus of smoothness of f ∈ C(I).

Now we give a direct local approximation theorem for the operators K∗n,p,q(f, x).

Theorem 4.1. For q ∈ (0, 1), x ∈ [0, 1] and f ∈ C(I), we have

∣∣K∗n,p,q(f, x)− f(x)∣∣ ≤ Cω2

(f ;

√(q2 + 4q3 − 2q − 3)x2

8[3]q+

(1 + 2q)x

2[3]q[n+ 1]q+

(3 + 5q + 4q2)x

4[3]q[n+ 1]q[n+ p]q

),

(21)

where C is a positive constant.

Proof. Let g ∈ C2. By Taylor’s expansion

g(t) = g(x) + g′(x)(t− x) +

∫ t

x(t− u)g′′(u)du,

and (12), we get

K∗n,p,q(g;x) = g(x) +K∗n,p,q

(∫ t

x(t− u)g′′(u)du;x

).

Hence, by (13), we have∣∣K∗n,p,q(g;x)− g(x)∣∣

≤∣∣∣∣K∗n,p,q (∫ t

x(t− u)g′′(u)du;x

)∣∣∣∣≤ K∗n,p,q

(∣∣∣∣∫ t

x(t− u)|g′′(u)|du

∣∣∣∣ ;x)≤ K∗n,p,q

((t− x)2;x

)||g′′||

≤

[(q2 + 4q3 − 2q − 3

)x2

4[3]q+

(1 + 2q)x

[3]q[n+ 1]q+

(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q

]||g′′||. (22)




On the other hand, using lemma 2.1, we have

∣∣K∗n,p,q(f ;x)∣∣ ≤ [n+ 1]q

n+p∑k=0

bn+p,k (q;u(x)) q−k∫ [k+1]q

[n+1]q

[k]q[n+1]q

|f(t)|dRq t ≤ ||f ||. (23)

Now (22) and (23) imply∣∣K∗n,p,q(f ;x)− f(x)∣∣

≤∣∣K∗n,p,q(f − g;x)− (f − g)(x)

∣∣+∣∣K∗n,p,q(g;x)− g(x)

∣∣≤ 2||f − g||+

[(q2 + 4q3 − 2q − 3

)x2

4[3]q+

(1 + 2q)x

[3]q[n+ 1]q+

(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q

]||g′′||.

Hence taking infimum on the right hand side over all g ∈ C2, we get

∣∣K∗n,p,q(f ;x)− f(x)∣∣ ≤ 2K2

(f ;

(q2 + 4q3 − 2q − 3

)x2

8[3]q+

(1 + 2q)x

2[3]q[n+ 1]q+

(3 + 5q + 4q2)x

4[3]q[n+ 1]q[n+ p]q

).

By (20), for every q ∈ (0, 1), we have

|K∗n,p,q(f ;x)−f(x)| ≤ Cω2

(f ;

√(q2 + 4q3 − 2q − 3)x2

8[3]q+

(1 + 2q)x

2[3]q[n+ 1]q+

(3 + 5q + 4q2)x

4[3]q[n+ 1]q[n+ p]q

).

This completes the proof of theorem 4.1.

Remark 4.2. For any fixed x ∈ [0, 1], p ∈ N0 and n ∈ N, let q := qn be a sequence

satisfying 0 < qn < 1 and limn qn = 1, we have

limn→∞

[(q2n + 4q3n − 2qn − 3

)x2

8[3]qn+

(1 + 2qn)x

2[3]qn [n+ 1]qn+

(3 + 5qn + 4q2n)x

4[3]qn [n+ 1]qn [n+ p]qn

]= 0.

These gives us a rate of pointwise convergence of the operators K∗n,p,qn(f ;x) to f(x).

Next we study the rate of convergence of the operators K∗n,p,q(f ;x) with the help of

functions of Lipschitz class LipM (α), where M > 0 and 0 < α ≤ 1. A function f belongs

to LipM (α) if

|f(y)− f(x)| ≤M |y − x|α (y, x ∈ R). (24)

We have the following theorem.

Theorem 4.3. Let q := qn be a sequence satisfying 0 < qn < 1, limnqn = 1 and f ∈

LipM (α), 0 < α ≤ 1. Then we have

∣∣K∗n,p,q(f ;x)− f(x)∣∣ ≤M [(

q2 + 4q3 − 2q − 3)x2

4[3]q+

(1 + 2q)x

[3]q[n+ 1]q+

(3 + 5q + 4q2)x

2[3]q[n+ 1]q[n+ p]q

]α2

.

(25)




Proof. Since K∗n,p,q is a linear positive operator and f ∈ LipM (α) (0 < α ≤ 1), we have

|K∗n,p,q(f ;x)− f(x)|≤ K∗n,p,q (|f(t)− f(x)|;x)

= [n+ 1]q

n+p∑k=0

bn+p,k(q;u(x))q−k∫ [k+1]q

[n+1]q

[k]q[n+1]q

|f(t)− f(x)|dRq t

≤ M [n+ 1]q

n+p∑k=0

bn+p,k(q;u(x))q−k∫ [k+1]q

[n+1]q

[k]q[n+1]q

|t− x|αdRq t

≤ M [n+ 1]q

n+p∑k=0

bn+p,k(q;u(x))q−k

(∫ [k+1]q[n+1]q

[k]q[n+1]q

[|t− x|α]2α dRq t

)α2(∫ [k+1]q

[n+1]q

[k]q[n+1]q

dRq t

) 2−α2

= M [n+ 1]q

n+p∑k=0

bn+p,k(q;u(x))q−k

(∫ [k+1]q[n+1]q

[k]q[n+1]q

(t− x)2dRq t

)α2 (

qk

[n+ 1]q

) 2−α2

= M

n+p∑k=0

bn+p,k(q;u(x))

(∫ [k+1]q[n+1]q

[k]q[n+1]q

(t− x)2dRq t

)α2 (

[n+ 1]qqk

)α2

= M

n+p∑k=0

[bn+p,k(q;u(x))]2−α2

([n+ 1]qbn+p,k(q;u(x))q−k

∫ [k+1]q[n+1]q

[k]q[n+1]q

(t− x)2dRq t

)α2

.

Applying Holder’s inequality for sums, we obtain

|K∗n,p,q(f ;x)− f(x)|

≤ M

(n+p∑k=0

bn+p,k(q;u(x))

) 2−α2(n+p∑k=0

[n+ 1]qbn+p,k(q;u(x))q−k∫ [k+1]q

[n+1]q

[k]q[n+1]q

(t− x)2dRq t

)α2

= M[K∗n,p,q

((t− x)2;x

)]α2 .

Thus, theorem 4.3 is proved.

Now, we give a Voronovskaja-type asymptotic formula for K∗n,p,q(f ;x) by means of the

second and fourth central moments.

Theorem 4.4. Let q := qn be a sequence satisfying 0 < qn < 1, limn qn = 1. For

f ∈ C2(I), (f(x) is a twice differentiable function in I), the following equality holds

limn→∞

[n]q(K∗n,p,q(f ;x)− f(x)

)=f ′′(x)

2x. (26)

Proof. Let x ∈ [0, 1] be fixed. By the Taylor formula, we may write

f(t) = f(x) + f ′(x)(t− x) +1

2f ′′(x)(t− x)2 + r(t;x)(t− x)2, (27)




where r(t;x) is the Peano form of the remainder, r(t;x) ∈ C(I), using L’Hopital’s rule,

we have

limt→x

r(t;x) = limt→x

f(t)− f(x)− f ′(x)(t− x)− 12f′′(x)(t− x)2

(t− x)2

= limt→x

f ′(t)− f ′(x)− f ′′(x)(t− x)

2(t− x)= lim

t→x

f ′′(t)− f ′′(x)

2= 0.

Since (12), applying K∗n,p,q(f ;x) to (27), we obtain

[n]q(K∗n,p,q(f ;x)− f(x)

)=

1

2[n]qf

′′(x)K∗n,p,q((t− x)2;x

)+ [n]qK

∗n,p,q

(r(t;x)(t− x)2;x

).

By the Cauchy-Schwarz inequality, we have

K∗n,p,q(r(t;x)(t− x)2;x

)≤√K∗n,p,q (r2(t;x);x)

√K∗n,p,q ((t− x)4;x). (28)

Since r2(x;x) = 0, then it is obtained easily that limnKn,p,q

(r2(t;x);x

)= r2(x;x) = 0 by

remark 2.2. Now, from (15), (28) and (14), we get immediately

limn→∞

[n]qK∗n,p,q

(r(t;x)(t− x)2;x

)= 0, lim

n→∞

1

2[n]qf

′′(x)K∗n,p,q((t− x)2;x

)=f ′′(x)

2x.

Thus, theorem 4.4 is proved.

x (for n = 50, p = 1)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

f(x)For q = 0.6For q = 0.9For q = 0.99

Figure 1: Convergence of K∗n,p,q(f ;x) for n = 50, p = 1 and different values of q.




x (for p = 1, q = 0.9)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

f(x)For n = 10For n = 20For n = 50

Figure 2: Convergence of K∗n,p,q(f ;x) for p = 1, q = 0.9 and different values of n.

x (for n = 10, p = 1, q = 0.9)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

f(x)

K*

n,p,q(f;x)

Kn,p,q

(f;x)

Figure 3: The graphs of K∗n,p,q(f ;x) (red) and Kn,p,q(f ;x) (blue) for n = 10, p = 1 and q = 0.9.




x (for n = 50, p = 1, q = 0.99)0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

f(x)

K*

n,p,q(f;x)

Kn,p,q

(f;x)

Figure 4: The graphs of K∗n,p,q(f ;x) (red) and Kn,p,q(f ;x) (blue) for n = 50, p = 1 and q = 0.99.

5 Graphical and numerical examples analysis

In this section, we give several graphs and numerical examples to show the convergence

of K∗n,p,q(f ;x) to f(x) with different values of n and q, and also compare the operators

K∗n,p,q(f ;x) with Kn,p,q(f ;x).

Let f(x) = 1− cos(4ex), for n = 50 and p = 1, the graphs of K∗n,p,q(f ;x) with different

values of q are shown in figure 1. Moreover, for p = 1 and q = 0.9, the graphs of K∗n,p,q(f ;x)

with different values of n are shown in figure 2.

Figure 3 shows the graphs of K∗n,p,q(f ;x) (red) and Kn,p,q(f ;x) (blue) for n = 10, p = 1

and q = 0.9. In figure 4, the values of n and q are replaced by 50 and 0.99, respectively.

In Table 1, we give the errors of the approximation to f(x) ofK∗n,p,q(f ;x) andKn,p,q(f ;x)

with different values of q and n. We may observe that operators K∗n,p,q(f ;x) give a better

estimate than Kn,p,q(f ;x).

Acknowledgement

This work is supported by the National Natural Science Foundation of China (Grant No.

61572020), the Startup Project of Doctor Scientific Research and Young Doctor Pre-Research

Fund Project of Quanzhou Normal University (Grant No. 2015QBKJ01), Fujian Provincial Key

Laboratory of Data Intensive Computing and Key Laboratory of Intelligent Computing and Infor-

mation Processing, Fujian Province University.




Table 1: The errors of the approximation to f(x) of K∗n,p,q(f ;x) and Kn,p,q(f ;x).

q = 1− 1/m‖f −K∗n,p,q(f)‖∞ ‖f −Kn,p,q(f)‖∞n = 10 n = 50 n = 10 n = 50

m = 100 0.4890 0.1318 0.5628 0.1587

m = 200 0.4856 0.1201 0.5638 0.1471

m = 300 0.4844 0.1163 0.5642 0.1436

m = 400 0.4838 0.1145 0.5645 0.1419

m = 500 0.4835 0.1134 0.5646 0.1409

m = 600 0.4832 0.1126 0.5647 0.1402

m = 700 0.4831 0.1121 0.5648 0.1397

m = 800 0.4829 0.1117 0.5648 0.1394

m = 900 0.4829 0.1114 0.5649 0.1391

m = 1000 0.4828 0.1112 0.5649 0.1389

References

[1] P. N. Agrawal, Z. Finta, A. S. Kumar, Bernstein-Schurer-Kantorovich operators based onq-integers, Appl. Math. Comput., 256 (2015), 222-231.

[2] R. A. DeVore, G. G. Lorentz, Constructive Approximation. Springer. Berlin 1993.

[3] O. Duman, A-statistical convergence of sequences of convolution operators, Taiwanese J.Math., 12(2) (2008), 523-536.

[4] G. Gasper, M. Rahman, Basic Hypergeometric Series, Encyclopedia of Mathematics and ItsApplications. Cambridge University press, Cambridge, UK 1990.

[5] V. G. Kac, P. Cheung, Quantum Calculus. Springer. New York 2002.



BARNES-TYPE DEGENERATE BERNOULLI AND EULER

MIXED-TYPE POLYNOMIALS

TAEKYUN KIM, DAE SAN KIM, HYUCKIN KWON, AND TOUFIK MANSOUR

Abstract. In this paper, we consider the Barnes-type degenerate Bernoulli and Eulermixed-type polynomials. We present several explicit formulas and recurrence relationsfor these polynomials. Also, we establish a connection between our polynomials andseveral known families of polynomials.

1. Introduction

The goals of this paper are to use umbral calculus to obtain several new and interestingidentities of Barnes-type degenerate Bernoulli and Euler mixed-type polynomials. Theuse of umbral calculus technique has been very attractive in numerous problems of math-ematics (for example, see [1, 6, 8, 14, 18–21, 24]) and used in different areas of physics (forexample, see [4, 5, 19]).

Let r, s ∈ Z>0. Throughout the paper we assume that a = a1, . . . , ar and b = b1, b2, . . . , bs.The Barnes-type degenerate Bernoulli and Euler mixed-type polynomials βEn(λ, x|a;b)with a1, . . . , ar; b1, . . . , bs = 0 are defined by the generating function

r∏i=1

(t

(1 + λt)ai/λ − 1

) s∏i=1

(2

(1 + λt)bi/λ + 1

)(1 + λt)x/λ =

∑n≥0

βEn(λ, x|a;b)tn

n!.(1)

If x = 0, βEn(λ|a;b) = βEn(λ, 0|a;b) are called the Barnes-type degenerate Bernoulli andEuler mixed-type numbers. Here, we recall that the polynomial βn(λ, x|a) with a1, . . . , ar =0 are given by

r∏i=1

(t

(1 + λt)ai/λ − 1

)(1 + λt)x/λ =

∑n≥0

βn(λ, x|a)tn

n!(2)

are called the Barnes-type degenerate Bernoulli polynomials and studied in [7]. We notehere that

limλ→0

βn(λ, x|a) = Bn(x|a),

limλ→∞

λ−nβn(λ, λx|a) = (a1a2 · · · ar)−1b(r)n (x),

where Bn(x|a) are the Barnes-type Bernoulli polynomials given by∏r

i=1

(t

eait−1

)etx =∑

n≥0Bn(x|a) tn

n! and b(r)n (x) are the Bernoulli polynomials of the second kind of order

2000 Mathematics Subject Classification. 05A40, 11B83.Key words and phrases. Euler polynomials, Bernoulli polynomials, Umbral calculus.

1


1273 TAEKYUN KIM et al 1273-1287

2 TAEKYUN KIM, DAE SAN KIM, HYUCKIN KWON, AND TOUFIK MANSOUR

r given by(

tlog(1+t)

)r(1 + t)x =

∑n≥0 b

(r)n (x) t

n

n! (see [12, 22]). Also, we recall that the

polynomial En(λ, x|b) with b1, . . . , bs = 0 are given by

s∏i=1

(2

(1 + λt)bi/λ + 1

)(1 + λt)x/λ =

∑n≥0

En(λ, x|b)tn

n!(3)

are called the Barnes-type degenerate Euler polynomials and studied in [11, 17, 25]. Wedenote En(λ, 0|b) by En(λ|b). We note here that

limλ→0En(λ, x|b) = En(x|b),

limλ→∞

λ−nEn(λ, λx|a) = (x)n = x(x− 1) · · · (x− n+ 1),

where En(x|a) are the Barnes-type Euler polynomials given by (see [3])

s∏i=1

(2

ebit + 1

)etx =

∑n≥0

En(x|b)tn

n!.

In order to study the Barnes-type degenerate Bernoulli and Euler mixed-type polynomials,we use the umbral calculus technique. We denote the algebra of polynomials in a singlevariable x over C by Π. Let Π∗ be the vector space of all linear functionals on Π. Let⟨L|p(x)⟩ be the action of a linear functional L ∈ Π∗ on a polynomial p(x), where we extendit as ⟨cL+ c′L′|p(x)⟩ = c⟨L|p(x)⟩+ c′⟨L′|p(x)⟩, where c, c′ ∈ C (see [22,23]). Define

H =

f(t) =∑k≥0

aktk

k!| ak ∈ C

(4)

to be the algebra of formal power series in a single variable t. The formal power seriesin the variable t defines a linear functional on Π by setting ⟨f(t)|xn⟩ = an, for all n ≥ 0(see [22,23]). By (4), we have

⟨tk|xn⟩ = n!δn,k, for all n, k ≥ 0, (see [22,23]),(5)

where δn,k is the Kronecker’s symbol. For fL(t) =∑

n≥0⟨L|xn⟩tn

n! , by (5), we have that

⟨fL(t)|xn⟩ = ⟨L|xn⟩. Thus, the map L 7→ fL(t) is a vector space isomorphism from Π∗

onto H, namely H is thought of as set of both formal power series and linear functionals.We call H the umbral algebra. The umbral calculus is the study of umbral algebra.

The order O(f(t)) of the non-zero power series f(t) is the smallest integer ℓ for which thecoefficient of tℓ does not vanish (see [22, 23]). If O(f(t)) = 1 (O(f(t)) = 0) then f(t) iscalled a delta (an invertable) series. If O(f(t)) = 1 and O(g(t)) = 0, then there exists aunique sequence sn(x) of polynomials such that ⟨g(t)(f(t))k|sn(x)⟩ = n!δn,k, where n, k ≥0. The sequence sn(x) is called the Sheffer sequence for (g(t), f(t)), and we write sn(x) ∼(g(t), f(t)) (see [22, 23]). For f(t) ∈ H and p(x) ∈ Π, we have that ⟨eyt|p(x)⟩ = p(y),⟨f(t)g(t)|p(x)⟩ = ⟨g(t)|f(t)p(x)⟩, f(t) =

∑n≥0⟨f(t)|xn⟩ tnn! and p(x) =

∑n≥0⟨tn|p(x)⟩xn

n! .Thus,

⟨tk|p(x)⟩ = p(k)(0), ⟨1|p(k)(x)⟩ = p(k)(0),(6)



BARNES-TYPE DEGENERATE BERNOULLI AND EULER MIXED-TYPE POLYNOMIALS 3

where p(k)(0) denotes the k-th derivative of p(x) with respect to x at x = 0. So, by (6), we

get that tkp(x) = p(k)(x) = dk

dxk p(x), for all k ≥ 0, (see [22, 23]). Let sn(x) ∼ (g(t), f(t)).Then we have

1

g(f(t))eyf(t) =

∑n≥0

sn(y)tn

n!,(7)

for all y ∈ C, where f(t) is the compositional inverse of f(t) (see [22, 23]). For sn(x) ∼(g(t), f(t)) and rn(x) ∼ (h(t), ℓ(t)), let sn(x) =

∑nk=0 cn,krk(x), then we have

cn,k =1

k!

⟨h(f(t))

g(f(t))(ℓ(f(t)))k|xn

⟩,(8)

(see [22,23]).

By the theory of Sheffer sequences, it is immediate that the Barnes-type degenerateBernoulli and Euler mixed-type polynomial is the Sheffer sequence for the pair g(t) =(

λeλt−1

)r∏ri=1

(eait − 1

)∏si=1

(ebit+1

2

)and f(t) = 1

λ(eλt − 1). Thus

βEn(λ, x|a;b) ∼

((λ

eλt − 1

)r r∏i=1

(eait − 1

) s∏i=1

(ebit + 1

2

),

1

λ(eλt − 1)

).(9)

The aim of the present paper is to present several new identities for Barnes-type degenerateBernoulli and Euler mixed-type polynomials by the use of umbral calculus.

2. Explicit Expressions

In this section we suggest several explicit formulas for the Barnes-type degenerate Bernoulliand Euler mixed-type polynomials. To do so, we recall that the Stirling numbers S1(n,m)of the first kind are defined as (x)n =

∑nm=0 S1(n,m)xm ∼ (1, et − 1) or 1

j!(log(1 + t))j =∑ℓ≥j S1(ℓ, j)

tℓ

ℓ! . Also, we recall that the Stirling numbers S2(n,m) of the second kind

are defined by (et−1)k

k! =∑

ℓ≥k S2(ℓ, k) tℓ

ℓ! . Define (x|λ)n = λn(x/λ)n to be (x|λ)n =

x(x− λ)(x− 2λ) · · · (x− (n− 1)λ) with (x|λ)0 = 1. Also, we define

Pr,s(t) =

r∏i=1

(t

(1 + λt)ai/λ − 1

) s∏i=1

(2

(1 + λt)bi/λ + 1

)and

Qr,s(t) =

r∏i=1

(t

eait − 1

) s∏i=1

(2

ebit + 1

).

Theorem 2.1. For all n ≥ 0,

βEn(λ, x|a;b) =

n∑j=0

n∑ℓ=j

(n

ℓ

)S1(ℓ, j)λ

ℓ−jβEn−ℓ(λ|a;b)

xj .




Proof. By applying the conjugation representation for sn(x) ∼ (g(t), f(t)), that is,

sn(x) =n∑

j=0

1

j!⟨g(f(t))−1f(t)j |xn⟩xj ,

(see [22,23]) we obtain

⟨g(f(t))−1f(t)j |xn⟩ =

⟨Pr,s(t)

(log(1 + λt)

λ

)j

|xn⟩

= λ−j⟨Pr,s(t)| (log(1 + λt))j xn

⟩= λ−j

⟨Pr,s(t)|j!

∑ℓ≥j

S1(ℓ, j)λℓ t

ℓ

ℓ!xn

⟩.

Thus,

1

j!⟨g(f(t))−1f(t)j |xn⟩ =

n∑ℓ=j

(n

ℓ

)S1(ℓ, j)λ

ℓ−j⟨Pr,s(t)|xn−ℓ

⟩

=

n∑ℓ=j

(n

ℓ

)S1(ℓ, j)λ

ℓ−j

⟨∑m≥0

βEm(λ|a;b)tm

m!|xn−ℓ

⟩

=

n∑ℓ=j

(n

ℓ

)S1(ℓ, j)λ

ℓ−jβEn−ℓ(λ|a;b),

which completes the proof.


βEn(λ, x|a;b) = λnn∑

m=0

m∑k=0

(mk

)(k+rr

)S1(n,m)S2(k + r, r)λk−mBEm−k(x|a;b),

where BEn(x|a;b) are the Barnes-type Bernoulli and Euler mixed-type polynomials with

r∏i=1

(t

eait − 1

) s∏i=1

(2

ebit + 1

)ext =

∞∑n=0

BEn(x|a;b)tn

n!

(see [26]).

Proof. By (9), we have(λ

eλt − 1

)r r∏i=1

(eait − 1

) s∏i=1

(ebit + 1

2

)βEn(λ, x|a;b) ∼

(1,

1

λ(eλt − 1)

),(10)




which implies

βEn(λ, x|a;b) =

n∑m=0

S1(n,m)λn−mQr,s(t)

(eλt − 1

λt

)r

xm

=n∑

m=0

S1(n,m)λn−mQr,s(t)

r!∑k≥0

S2(k + r, r)λk

(k + r)!tk

xm

= λnn∑

m=0

m∑k=0

(mk

)(k+rr

)S1(n,m)S2(k + r, r)λk−mQr,s(t)xm−k

= λnn∑

m=0

m∑k=0

(mk

)(k+rr

)S1(n,m)S2(k + r, r)λk−mBEm−k(x|a;b),



βEn(λ, x|a;b) =

n−1∑ℓ=0

n−ℓ∑k=0

(n−1ℓ

)(n−ℓk

)(k+rr

) λk+ℓS2(k + r, r)B(n)ℓ BEn−ℓ−k(x|a;b).

Proof. We proceed the proof by invoking the following transfer formula (see (7) and (8)):

for pn(x) ∼ (1, f(t)) and qn(x) ∼ (1, g(t)), then qn(x) = x(f(t)g(t)

)nx−1pn(x), for all n ≥ 1.

In our case, by xn ∼ (1, t) and (10), we have(λ

eλt − 1

)r r∏i=1

(eait − 1

) s∏i=1

(ebit + 1

2

)βEn(λ, x|a;b)

= x

(λt

eλt − 1

)n

xn−1 = x∑ℓ≥0

B(n)ℓ

λℓ

ℓ!tℓxn−1 =

n−1∑ℓ=0

(n− 1

ℓ

)λℓB

(n)ℓ xn−ℓ.

Thus,

βEn(λ, x|a;b) =

n−1∑ℓ=0

(n− 1

ℓ

)λℓB

(n)ℓ

(Qr,s(t)

(eλt − 1

λt

)r

xn−ℓ

)

=n−1∑ℓ=0

(n− 1

ℓ

)λℓB

(n)ℓ

Qr,s(t)∑k≥0

S2(k + r, r)r!λktk

(k + r)!xn−ℓ

=

n−1∑ℓ=0

n−ℓ∑k=0

(n−1ℓ

)(n−ℓk

)(k+rr

) λk+ℓS2(k + r, r)B(n)ℓ Qr,s(t)x

n−ℓ−k

=

n−1∑ℓ=0

n−ℓ∑k=0

(n−1ℓ

)(n−ℓk

)(k+rr

) λk+ℓS2(k + r, r)B(n)ℓ BEn−ℓ−k(x|a;b),

as claimed.




Note that the Barnes-type Daehee polynomials with λ-parameterDn,λ(x|a) with a1, . . . , ar =0 was defined as

r∏i=1

log(1 + λt)

λ(1 + λt)ai/λ − 1(1 + λt)x/λ =

∑n≥0

Dn,λ(x|a)tn

n!,(11)

see [15,16]. When x = 0 we write Dn,λ(a) = Dn,λ(0|a); the Barnes-type Daehee numbers.


βEn(λ, x|a;b) =

n∑ℓ=0

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λℓb

(r)ℓ (x/λ)Dk,λ(a)En−ℓ−k(λ|b)

=

n∑ℓ=0

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λℓb

(r)ℓ Dk,λ(x|a)En−ℓ−k(λ|b)

=

n∑ℓ=0

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λℓb

(r)ℓ Dk,λ(a)En−ℓ−k(x, λ|b).

Proof. By (9) we have

βEn(λ, x|a;b) =⟨Pr,s(t)(1 + λt)x/λ|xn

⟩=

⟨r∏

i=1

(log(1 + λt)

λ((1 + λt)ai/λ − 1)

) s∏i=1

(2

(1 + λt)bi/λ + 1

)(λt

log(1 + λt)

)r

(1 + λt)x/λ|xn⟩

=

⟨r∏

i=1

(log(1 + λt)

λ((1 + λt)ai/λ − 1)

) s∏i=1

(2

(1 + λt)bi/λ + 1

)|∑ℓ≥0

b(r)ℓ (x/λ)

λℓ

ℓ!tℓxn

⟩

=n∑

ℓ=0

(n

ℓ

)b(r)ℓ (x/λ)λℓ

⟨r∏

i=1

(log(1 + λt)

λ((1 + λt)ai/λ − 1)

) s∏i=1

(2

(1 + λt)bi/λ + 1

)|xn−ℓ

⟩.

Thus, by (11), we obtain

βEn(λ, x|a;b) =

n∑ℓ=0

(n

ℓ

)b(r)ℓ (x/λ)λℓ

⟨s∏

i=1

(2

(1 + λt)bi/λ + 1

)|∑k≥0

Dk,λ(a)tk

k!xn−ℓ

⟩

=

n∑ℓ=0

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λℓb

(r)ℓ (x/λ)Dk,λ(a)

⟨s∏

i=1

(2

(1 + λt)bi/λ + 1

)|xn−ℓ−k

⟩

=

n∑ℓ=0

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λℓb

(r)ℓ (x/λ)Dk,λ(a)En−ℓ−k(λ|b),

which proves the first formula. Similar techniques show the second and the third formulas.

3. Recurrence relations

In this section, we present several recurrence relations for the Barnes-type degenerateBernoulli and Euler mixed-type polynomials.





βEn(λ, x+ y|a;b) =

n∑j=0

(n

j

)βEj(λ, x|a;b)(y|λ)n−j .

Proof. By (9) we have(

λteλt−1

)r1

Qr,s(t)βEn(λ, x|a;b) = (x|λ)n ∼

(1, e

λt−1λ

), which implies

the result.

Theorem 3.1 with x = 0, gives the following result.

Corollary 3.2. For all n ≥ 0,

βEn(λ, x|a;b) =

n∑j=0

(n

j

)βEn−j(λ|a;b)(x|λ)j .


βEn(λ, x+ λ|a;b) = βEn(λ, x|a;b) + nλβEn−1(λ, x|a;b).

Proof. By (7) we have that f(t)sn(x) = nsn−1(x) when sn(x) ∼ (g(t), f(t)). In our case,from (9), we have

eλt − 1

λβEn(λ, x|a;b) = nβEn−1(λ, x|a;b),

which implies that βEn(λ, x+ λ|a;b)− βEn(λ, x|a;b) = nλβEn−1(λ, x|a;b), as required.


d

dxβEn(λ, x|a;b) = n!

n−1∑ℓ=0

(−λ)n−1−ℓ

ℓ!(n− ℓ)βEℓ(λ, x|a;b)

= nλn−1n−1∑ℓ=0

S1(n− 1, ℓ)λ−ℓBEℓ(x|a;b).

Proof. By (7) we have ddxsn(x) =

∑n−1ℓ=0

(nℓ

)⟨f(t)|xn−ℓ⟩sℓ(x) when sn(x) ∼ (g(t), f(t)). In

our case, from (9), we have

⟨f(t)|xn−ℓ⟩ = ⟨ 1λ

log(1 + λt)|xn−ℓ⟩ = λ−1⟨∑m≥1

(−1)m−1λmtm

m|xn−ℓ⟩

= λ−1(−1)n−ℓ−1λn−ℓ(n− ℓ− 1)! = (−λ)n−ℓ−1(n− ℓ− 1)!.

Thus ddxβEn(λ, x|a;b) = n!

∑n−1ℓ=0

(−λ)n−1−ℓ

ℓ!(n−ℓ) βEℓ(λ, x|a;b), as required.

To show the second formula, we note that (x|λ)n =∑n

ℓ=0 S1(n, ℓ)λn−ℓxℓ ∼

(1, e

λt−1λ

),

which shows that eλt−1λ (x|λ)n = n(x|λ)n−1. Thus

(eλt−1

λ

)r(x|λ)n = (n)r(x|λ)n−r, for all




n ≥ r. Thus, by (10), we have

dr

dxrβEn(λ, x|a;b) = trβEn(λ, x|a;b) = Qr,s(t)

(eλt − 1

λ

)r

(x|λ)n

= (n)rQr,s(t)(x|λ)n−r = (n)rλn−r

n−r∑m=0

S1(n− r,m)λ−mBEm(x|a;b),



(1− r/n)βEn(λ, x|a;b) =

x− r∑i=1

ai −s∑

j=1

bj

βEn−1(λ, x− λ|a;b)

− 1

n

r∑i=1

aiβEn(λ, x− λ|ai,a;b) +1

2

s∑i=1

biβEn−1(λ, x− λ|a; bi,b).

Proof. Let n ≥ 1. By (9), we have

βEn(λ, y|a;b) =⟨Pr,s(t)(1 + λt)y/λ|xn

⟩=

⟨d

dt

(r∏

i=1

(t

(1 + λt)ai/λ − 1

) s∏i=1

(2

(1 + λt)bi/λ + 1

)(1 + λt)y/λ

)|xn−1

⟩

=

⟨d

dt

r∏i=1

(t

(1 + λt)ai/λ − 1

) s∏i=1

(2

(1 + λt)bi/λ + 1

)(1 + λt)y/λ|xn−1

⟩(12)

+

⟨r∏

i=1

(t

(1 + λt)ai/λ − 1

)d

dt

s∏i=1

(2

(1 + λt)bi/λ + 1

)(1 + λt)y/λ|xn−1

⟩(13)

+

⟨r∏

i=1

(t

(1 + λt)ai/λ − 1

) s∏i=1

(2

(1 + λt)bi/λ + 1

)d

dt(1 + λt)y/λ|xn−1

⟩.(14)

The term in (14) is given by

y⟨Pr,s(t)(1 + λt)y/λ−1|xn−1

⟩= yβEn−1(λ, y − λ|a;b).(15)

In order to find the first term, namely (12), we note that

d

dt

r∏i=1

(t

(1 + λt)ai/λ − 1

)

=

r∏i=1

(t

(1 + λt)ai/λ − 1

) r∑i=1

(− ai

1 + λt+

1

t

(1− ai

1 + λt

t

(1 + λt)ai/λ − 1

)),




where the order of 1− ai1+λt

t(1+λt)ai/λ−1

is at least 1. Thus the term in (12) is given by

−r∑

i=1

ai

⟨Pr,s(t)(1 + λt)y/λ−1|xn−1

⟩+

⟨Pr,s(t)(1 + λt)y/λ|1

t

r∑i=1

(1− ai

1 + λt

t

(1 + λt)ai/λ − 1

)xn−1

⟩

which equals

−r∑

i=1

aiβEn−1(λ, y − λ|a;b) +r

n

⟨Pr,s(t)(1 + λt)y/λ|xn

⟩− 1

n

r∑i=1

ai

⟨t

(1 + λt)ai/λ − 1Pr,s(t)(1 + λt)y/λ−1|xn

⟩

= −r∑

i=1

aiβEn−1(λ, y − λ|a;b) +r

nβEn(λ, y|a;b)− 1

n

r∑i=1

aiβEn(λ, y − λ|ai,a;b).(16)

In order to find the second term, namely (13), we note that

d

dt

s∏i=1

(2

(1 + λt)bi/λ + 1

)

=s∏

i=1

(2

(1 + λt)bi/λ + 1

) s∑i=1

(− bi

1 + λt+

bi2(1 + λt)

2

(1 + λt)bi/λ + 1

).

Thus the term in (13) is given by

s∑i=1

bi

⟨(−1 +

1

(1 + λt)bi/λ + 1

)Pr,s(t)(1 + λt)y/λ−1|xn−1

⟩

= −s∑

i=1

biβEn−1(λ, y − λ|a;b) +1

2

s∑i=1

biβEn−1(λ, y − λ|a; bi,b).(17)

Altogether, namely by (15), (16) and (17), we complete the proof.

Theorem 3.6. For n ≥ 0,

βEn+1(λ, x|a;b) = xβEn(λ, x− λ|a;b)

− λnn∑

m=0

m∑k=0

k∑ℓ=0

λ−kS1(n,m)S2(m− k + r, r)(mk

)(kℓ

)(m−k+r

r

) ·

·(Bk−ℓ+1(1)

k − ℓ+ 1

(r∑

i=1

ak−ℓ+1i − rλk−ℓ+1

)+Ek−ℓ(1)

2

r∑j=1

bk−ℓ+1j

)BEℓ(x− λ|a;b).




Proof. By (7), we have that sn+1(x) =(x− g′(t)

g(t)

)1

f ′(t)sn(x) when sn(x) ∼ (g(t), f(t)). In

our case, see (9), we have

βEn+1(λ, x|a;b) = xβEn(λ, x− λ|a;b)− e−λt g′(t)

g(t)βEn(λ, x|a;b),

where

g′(t)

g(t)= (log g(t))′

=

r log λ− r log(eλt − 1) +r∑

i=1

log(eait − 1) +s∑

j=1

log(ebjt + 1)− s log 2

′

= − rλeλt

eλt − 1+

r∑i=1

aieait

eait − 1+

s∑j=1

bjebjt

ebjt + 1

=1

t

(− rλte

λt

eλt − 1+

r∑i=1

aiteait

eait − 1

)+

1

2

s∑j=1

2bjebjt

ebjt + 1

=1

t

−r∑ℓ≥0

Bℓ(1)λℓtℓ

ℓ!+

r∑i=1

∑ℓ≥0

Bℓ(1)aℓit

ℓ

ℓ!

+1

2

s∑j=1

∑ℓ≥0

Eℓ(1)bℓ+1j tℓ

ℓ!

=∑ℓ≥0

Bℓ+1(1)

(ℓ+ 1)!

(r∑

i=1

aℓ+1i − rλℓ+1

)tℓ +

1

2

∑ℓ≥0

Eℓ(1)

ℓ!

s∑j=1

bℓ+1j

tℓ.

Therefore, by Theorem 2.2, we obtain

g′(t)

g(t)βEn(λ, x|a;b)

= λnn∑

m=0

m∑k=0

λk−mS1(n,m)S2(k + r, r)(mk

)(k+rr

) m−k∑ℓ=0

Bℓ+1(1)

(ℓ+ 1)!

(r∑

i=1

aℓ+1i − rλℓ+1

)tℓBEm−k(x|a;b)

+λn

2

n∑m=0

m∑k=0

λk−mS1(n,m)S2(k + r, r)(mk

)(k+rr

) m−k∑ℓ=0

Eℓ(1)

ℓ!

s∑j=1

bℓ+1j

tℓBEm−k(x|a;b)

= λnn∑

m=0

m∑k=0

k∑ℓ=0


)(kℓ

)(m−k+r

r

) ·

·(Bk−ℓ+1(1)

k − ℓ+ 1

(r∑

i=1


)+Ek−ℓ(1)

2

r∑j=1

bk−ℓ+1j

)BEℓ(x|a;b).




Thus,

βEn+1(λ, x|a;b)

= xβEn(λ, x− λ|a;b)

− λnn∑

m=0

m∑k=0

k∑ℓ=0


)(kℓ

)(m−k+r

r

) ·

·(Bk−ℓ+1(1)

k − ℓ+ 1

(r∑

i=1


)+Ek−ℓ(1)

2

r∑j=1

bk−ℓ+1j

)BEℓ(x− λ|a;b),

as claimed.

4. Connections with families of polynomials

The Bernoulli polynomials B(α)n (x) of order α are defined by the generating function(

t

et − 1

)α

ext =∑n≥0

B(α)n (x)

tn

n!,

equivalently, B(α)n (x) ∼

((et−1t

)α, t)

(see [3, 9, 10]). In the next result, we express our

polynomials βEn(λ, x|a;b) in terms of Bernoulli polynomials of order α.


βEn(λ, x|a;b) =

n∑m=0

λ−mdn,mB(α)m (x),

where

dn,m =

n∑ℓ=m

n−ℓ∑k=0

[(n

ℓ

)(n− ℓk

)λk+ℓS1(ℓ,m)b

(α)k ·

·n−ℓ−k∑q=0

q∑p=0

(n−ℓ−k

q

)(q+αα

) S1(q + α, q − p+ α)S2(q − p+ α, α)λpβEn−ℓ−k−q(λ|a;b)

].

Proof. Let βEn(λ, x|a;b) =∑n

m=0 cn,mB(α)m (x). By (8) and (9), we have

cn,m =1

m!λm

⟨Pr,s(t)

((1 + λt)1/λ − 1

t

)α(λt

log(1 + λt)

)α

(log(1 + λt))m|xn⟩

=1

m!λm

⟨Pr,s(t)

((1 + λt)1/λ − 1

t

)α(λt

log(1 + λt)

)α

|m!∑ℓ≥m

S1(ℓ,m)λℓtℓ

ℓ!xn

⟩

= λ−mn∑

ℓ=m

(n

ℓ

)λℓS1(ℓ,m)

⟨Pr,s(t)

((1 + λt)1/λ − 1

t

)α

|(

λt

log(1 + λt)

)α

xn−ℓ

⟩

= λ−mn∑

ℓ=m

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λk+ℓS1(ℓ,m)b

(α)k

⟨Pr,s(t)|

((1 + λt)1/λ − 1

t

)α

xn−ℓ−k

⟩.




Before proceeding further, we note that((1 + λt)1/λ − 1

t

)α

=

(e

1λlog(1+λt) − 1

t

)α

=∑q≥0

q∑p=0

(q + α

α

)−1

S1(q + α, q − p+ α)S2(q − p+ α, α)λptq

q!.

Thus,

cn,m = λ−mn∑

ℓ=m

n−ℓ∑k=0

[(n

ℓ

)(n− ℓk

)λk+ℓS1(ℓ,m)b

(α)k ·

·n−ℓ−k∑q=0

q∑p=0

(n−ℓ−k

q

)(q+αα

) S1(q + α, q − p+ α)S2(q − p+ α, α)λp⟨Pr,s(t)|xn−ℓ−k−q

⟩],

which gives

cn,m = λ−mn∑

ℓ=m

n−ℓ∑k=0

[(n

ℓ

)(n− ℓk

)λk+ℓS1(ℓ,m)b

(α)k ·

·n−ℓ−k∑q=0

q∑p=0

(n−ℓ−k

q

)(q+αα

) S1(q + α, q − p+ α)S2(q − p+ α, α)λpβEn−ℓ−k−q(λ|a;b)

],


The degenerate Bernoulli polynomials β(α)n (λ, x) of order α are defined by the generating

function (t

(1 + λt)1/λ − 1

)α

(1 + λt)x/λ =∑n≥0

β(α)n (λ, x)tn

n!,

equivalently, B(α)n (λ, x) ∼

((λ(et−1)eλt−1

)α, 1λ(eλt − 1)

). Then by using similar arguments as

in the proof of Theorem 4.1, we obtain the following result.


βEn(λ, x|a;b) =

n∑m=0

(n

m

)dn,mβ

(α)m (λ, x),

where

dn,m =

n−m∑q=0

q∑p=0

(n−mq

)(q+αα

) S1(q + α, q − p+ α)S2(q − p+ α, α)λpβEn−m−q(λ|a;b).

The Frobenius-Euler polynomials of order α are defined by the generating function(1− µet − µ

)α

ext =∑n≥0

H(α)n (x|µ)

tn

n!,

equivalently, H(α)n (x|µ) ∼

((et−µ1−µ

)α, t)

(see [2, 13]). In the next result, we express our

polynomials βEn(λ, x|a;b) in terms of Frobenius-Euler polynomials.





βEn(λ, x|a;b) = (1− µ)−αn∑

m=0

λ−mdn,mH(α)m (x|µ),

where

dn,m =n∑

ℓ=m

n−ℓ∑k=0

α∑p=0

(n

ℓ

)(n− ℓk

)(α

p

)S1(ℓ,m)λℓ(−µ)α−pβEk(λ|a;b)(p|λ)n−ℓ−k.

Proof. Let βEn(λ, x|a;b) =∑n

m=0 cn,mH(α)m (x|µ). Then

cn,m =1

m!(1− µ)αλm

⟨((1 + λt)1/λ − µ)αPr,s(t)|(log(1 + λt))mxn

⟩=

1

m!(1− µ)αλm

⟨((1 + λt)1/λ − µ)αPr,s(t)|m!

∑ℓ≥m

S1(ℓ,m)λℓtℓ

ℓ!xn

⟩

=1

(1− µ)αλm

n∑ℓ=m

(n

ℓ

)λℓS1(ℓ,m)

⟨((1 + λt)1/λ − µ)α|Pr,s(t)x

n−ℓ⟩

=1

(1− µ)αλm

n∑ℓ=m

n−ℓ∑k=0

(n

ℓ

)(n− ℓk

)λℓS1(ℓ,m)βEk(λ|a;b)

⟨((1 + λt)1/λ − µ)α|xn−ℓ−k

⟩.

Note that⟨((1 + λt)1/λ − µ)α|xn−ℓ−k

⟩=

⟨α∑

p=0

(α

p

)(−µ)α−p(1 + λt)p/λ|xn−ℓ−k

⟩

=

α∑p=0

(α

p

)(−µ)α−p

⟨∑q≥0

(p|λ)qtq

q!|xn−ℓ−k

⟩

=

α∑p=0

(α

p

)(−µ)α−p(p|λ)n−ℓ−k.

Thus,

cn,m =1

(1− µ)αλm

n∑ℓ=m

n−ℓ∑k=0

α∑p=0

(n

ℓ

)(n− ℓk

)(α

p

)λℓ(−µ)α−pS1(ℓ,m)βEk(λ|a;b)(p|λ)n−ℓ−k,


The degenerate Euler polynomials E(α)n (λ, x) of order α are defined by the generatingfunction (

2

(1 + λt)1/λ + 1

)α

(1 + λt)x/λ =∑n≥0

E(α)n (λ, x)tn

n!,

equivalently, E(α)n (λ, x) ∼

((et+12

)α, e

λt−1λ

). Then by using similar arguments as in the

proof of Theorems 4.1 and 4.3, we obtain the following result.





βEn(λ, x|a;b) = 2−αn∑

m=0

(n

m

)dn,mE(α)m (λ, x),

where

dn,m =n−m∑q=0

α∑p=0

(n−mq

)(α

p

)βEn−m−q(λ|a;b)(p|λ)q.

Acknowledgemts. The first author is appointed as a chair professor at Tianjin Poly-technic University by Tianjin City in China from August 2015 to August 2019.

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calculus viewpoint, J. Inequal. Appl. 2014 (2014) 324.[18] Kim, D.S., Kim, T., Lee, S.-H. and Rim, S.-H., Frobenius-Euler polynomials and umbral calculus in

the p-adic case, Adv. Difference Equ. 2012 (2012) 222.[19] Kim, T., Kim, D.S., Mansour, T., Rim, S.-H. and Schork, M., Umbral calculus and Sheffer sequences

of polynomials, J. Math. Phys. 54 (2013), no. 8, 083504.[20] Kwasniewski, A.K., q-quantum plane, ψ(q)-umbral calculus, and all that, Quantum groups and inte-

grable systems (Prague, 2001), Czechoslovak J. Phys. 51 (2001), no. 12, 1368–1373.




[21] Levi, D., Tempesta, P. and Winternitz, P., Umbral calculus, difference equations and the discreteSchrodinger equation, J. Math. Phys. 45 (2004), no. 11, 4077–4105.

[22] Roman, S., More on the umbral calculus, with emphasis on the q-umbral calculus, J. Math. Anal.Appl. 107 (1985) 222–254.

[23] Roman, S., The umbral calculus, Dover Publ. Inc. New York, 2005.[24] Wilson, B.G. and Rogers, F.G., Umbral calculus and the theory of multispecies nonideal gases, Phys.

A 139 (1986) 359–386.[25] Kim, D.S. and Kim, T., Some identities of degenerate Euler polynomials arising from p-adic fermionic

integrals on Zp, Integral Transforms Spec. Funct., to appear.[26] Kim, D.S., Kim, T., Kwon, H.I. and Seo, J.-J., Identities of some special mixed-type polynomials,

Adv. Stud. Theor. Phys. 8:17 (2014) 745–754.

Department of Mathematics, College of Science, Tianjin Polytechnic University, TianjinCity, 300387, China, Department of Mathematics, Kwangwoon University, Seoul, S. Korea

E-mail address: [email protected], [email protected]

Department of Mathematics, Sogang University, Seoul 121-742, S. Korea

E-mail address: [email protected] (corresponding)

Department of Mathematics, Kwangwoon University, Seoul 139-701, S. Korea

Department of Mathematics, University of Haifa, 3498838 Haifa, Israel



Ground state solutions for second order nonlinearp-Laplacian difference equations with periodic

coefficients

Ali Mai Guowei Sun ∗

Department of Applied Mathematics, Yuncheng University

Shanxi, Yuncheng 044000, China

Abstract

We study the existence of homoclinic solutions for nonlinear p-Laplacian differenceequations with periodic coefficients. The proof of the main result is based on the crit-ical point theory in combination with the Nehari manifold approach. Under ratherweaker conditions, we obtain the existence of ground state solutions and considerablyimprove some existing ones even for some special cases.

Key words: P-Laplacian Difference equations; Nehari manifold; Ground state solu-tions; Critical point theory.

1 Introduction

Difference equations represent the discrete counterpart of ordinary differential equations,have been widely used in many fields such as computer science, economics, neural network,ecology, cybernetics, etc. In the past decades, the existence of homoclinic solutions for differ-ence equations with p-Laplacian has been extensively studied, The classical method used isfixed point theory, to mention a few, see [1–3] and references therein for details. As it is wellknown, the critical point theory is used to deal with the existence of solutions of differenceequations [4–10]. Here we mention the works of Cabada, Iannizzotto and Tersian [4], Jiangand Zhou [5], Long and Shi [6]. In these papers, critical point theory is applied on bounddiscrete intervals, which leads to the study of critical points of an energy functional definedon a finite-dimensional Banach space. For unbounded discrete intervals such as the whole setof integers Z, Ma and Guo used critical point theory in combination with periodic approxi-mation to deal with such problems [7]. In the present paper, under convenient assumption,

∗Corresponding author. E-mail address: [email protected]


1288 Ali Mai et al 1288-1297

without periodic approximation and without verifying Palais-Smale condition, we not onlyprove the existence of homoclinic solution, but also obtain the ground state solution. weextend [11] to the case of the p-lapacian difference equation with periodic coefficients.

In this paper, our work focus on the existence of homoclinic solution for the followingsecond order nonlinear difference equations with p-Laplacian

−∆[a(k)φp(∆u(k − 1))] + b(k)φp(u(k)) = f(k, u(k)), k ∈ Z, (1.1)

where φp(t) = |t|p−2t for all t ∈ R, p > 1. a(k), b(k) are positive and T−periodic sequences, Tis a fixed positive integer. f(k, u) : Z×R→ R is a continuous function on u and T−periodicon k. The forward difference operator ∆ is defined by

∆u(k − 1) = u(k)− u(k − 1), for all k ∈ Z.

where Z and R denote the set of all integers and real numbers, respectively.

In addition, we are interested in the existence of nontrivial homoclinic solution for (1.1),that is, solutions that are not equal to 0 identically. We call that a solution u = u(k) of(1.1) is homoclinic (to 0) if

lim|k|→∞

u(k) = 0. (1.2)

Throughout this paper, we always suppose that the following conditions hold.

(A) a(k) > 0 and a(k + T ) = a(k) for all k ∈ Z.

(B) b(k) > 0 and b(k + T ) = b(k) for all k ∈ Z.

(f1) f ∈ C(Z× R,R), and there exist C > 0, q ∈ (p,∞) such that

|f(k, u)| ≤ C(1 + |u|q−1), for all k ∈ Z, u ∈ R.

(f2) lim|u|→0

f(k, u)/|u|p−1 = 0 uniformly for k ∈ Z.

(f3) lim|u|→∞

F (k, u)/|u|p = +∞ uniformly for k ∈ Z, where F (k, u) is the primitive function

of f(k, u), i.e.,

F (k, u) =

∫ u

0

f(k, s)ds.

(f4) u 7→ f(k, u)/|u|p−1 is strictly increasing on (−∞, 0) and (0,∞).

The main result in this paper is the following theorem:

Theorem 1.1. Suppose conditions (A), (B) and (f1)−(f4) are satisfied. Then equation (1.1)has at least a nontrivial ground state solution.

Remark 1.1. In [7], Ma and Guo considered the special case of (1.1) with p = 2 andobtained the following theorem:


1289 Ali Mai et al 1288-1297

Theorem A Suppose conditions (A), (B), (f2) and the following generalized Ambrosetti-Rabinowitz superlinear condition are satisfied:

(GAR) there exists a constant µ > p such that

0 < µF (k, u) ≤ f(k, u)u, for all k and u 6= 0, (1.3)

Then equation (1.1) has a nontrivial ground state solution.

It is easy to see that (1.3) implies (f3). There exists a p-superlinear function, such as

f(k, u) = |u|p−2u ln(1 + |u|),

does not satisfy (1.3). However, it satisfies the condition (f1) − (f4). So our conditionsare weaker than conditions in [7]. And we do not need periodic approximation techniqueto obtain homoclinic solutions. Furthermore, we obtain the existence of a ground statesolution. Therefore, our result not only extends the main result in [7] to difference equationswith p-Laplacian but also improves it.

Remark 1.2. In [12], the authors considered the following second order nonlinear differenceequations with p-Laplacian

−∆φp(∆u(k − 1)) + b(k)φp(u(k)) = f(k, u(k)), k ∈ Z, (1.4)

without any periodic assumption, they obtained the homoclinic solutions of the equation.However, PS condition need to be proved in [12], in this paper, we only prove the coercivecondition (below Lemma 3.2) is satisfied.

Example 1.1. Let

f(k, u) =

0, u = 0,

|u|p−2u ln(1 + |u|), u 6= 0,

for all k ∈ Z, If (A) and (B) are satisfied, then it is easy to check that all the conditions ofour Theorem 1.1 are satisfied. Therefore, the nontrivial homoclinic solution is obtained atonce.

The rest of the paper is organized as follows: In Section 2, we establish the variationalframework associated with (1.1), then present the main results of this paper. Section 3 isdevoted to prove the main result.

2 Preliminaries

In this section, we shall establish the corresponding variational framework associated with(1.1). We are going to define a suitable space E and an energy functional J ∈ E, such thatcritical points of J in E are exactly solutions of (1.1).


1290 Ali Mai et al 1288-1297

Consider the real sequence spaces

lp ≡ lp(Z) =

u = u(k)k∈Z : ∀ k ∈ Z, u(k) ∈ R, ‖u‖lp =

(∑k∈Z

|u(k)|p) 1

p

<∞

. (2.1)

Then the following embedding between lp spaces holds,

lq ⊂ lp, ‖u‖lp ≤ ‖u‖lq , 1 ≤ q ≤ p ≤ ∞. (2.2)

Define the space

E := u ∈ lp :∑k∈Z

[a(k)|∆u(k − 1)|p + b(k)|u(k)|p] <∞.

Then E is a Hilbert space equipped with the norm

‖u‖p =∑k∈Z

[a(k)|∆u(k − 1)|p + b(k)|u(k)|p] .

| · | is the usual absolute value in R.

Now we consider the variational functional J defined on E by

J(u) =1

p

∑k∈Z

[a(k)|∆u(k − 1)|p + b(k)|u(k)|p]−∑k∈Z

F (k, u(k))

=1

p‖u‖p −

∑k∈Z

F (k, u(k)).

Then J ∈ C1(E,R) with for all v ∈ E,

(J ′(u), v) = limt→0

J(u+ tv)− J(u)

t

=∑k∈Z

[a(k)φp(∆u(k − 1))∆v(k − 1) + b(k)φp(u(k))v(k)]

−∑k∈Z

f(k, u(k))v(k).

and

∂J(u)

∂u(k)= −a(k)∆φp(∆u(k − 1)) + b(k)φp(u(k))− f(k, u(k)), k ∈ Z.

Thus, u is a critical point of J on E only if u is a homoclinic solutions of equation (1.1).

Letcmin = infJ(u) : J ′(u) = 0, u ∈ E\0.

Then u0 6= 0 with J(u0) = cmin is said to be a ground state solution of (1.1).


1291 Ali Mai et al 1288-1297

3 Proofs of main result

We define the Nehari manifold

N = u ∈ E \ 0 : (J ′(u), u) = 0.

To prove the main results, we need some lemmas.

Lemma 3.1. Assume that (B), (f1)− (f4) are satisfied. Then for each w ∈ E \ 0, thereexists a unique sw > 0 such that sww ∈ N .

Proof. Let I(u) =∑k∈Z

F (k, u(k)). By (f2), we have

I ′(u) = o(‖u‖p−1) as u→ 0. (3.1)

From (f4), for all u 6= 0 and s > 0, we have

s 7→ I ′(su)u/sp−1 is strictly increasing. (3.2)

Let W ⊂ E \ 0 be a weakly compact subset and s > 0, we claim that

I(su)/sp →∞ uniformly for u on W, as s→∞. (3.3)

Indeed, let un ⊂ W . It suffices to show that

if sn →∞, I(snun)/(sn)p →∞.

as n → ∞. Passing to a subsequence if necessary, un u ∈ E \ 0 and un(k) → u(k) forevery k, as n→∞.

Note that from (f2) and (f4), it is easy to get that

F (k, u) > 0, for all u 6= 0. (3.4)

Since |snun(k)| → ∞ and un 6= 0, by (f3) and (3.4), we have

I(snun)

(sn)p=∑k∈Z

F (k, snun(k))

|snun(k)|p|un(k)|p →∞ as n→∞.

Therefore, (3.3) holds.

Let g(s) := J(sw), s > 0. Then

g′(s) = J ′(sw)w = sp−1(‖w‖p − s1−pI ′(sw)w),

from (3.1)-(3.3), then there exists a unique sw, such that g′(s) > 0 whenever 0 < s < sw,g′(s) < 0 whenever s > sw and g′(sw) = J ′(sww)w = 0. So sww ∈ N .


1292 Ali Mai et al 1288-1297

Lemma 3.2. J is coercive on N , i.e., J(u)→∞ as ‖u‖ → ∞, u ∈ N .

Proof. Suppose by contradiction, there exists a sequence un ⊂ N such that ‖un‖ → ∞and J(un) ≤ d. Let vn = un

‖un‖ , then there exists a subsequence, still denoted by the same

notation, such that vn v and vn(k)→ v(k) for every k, as n→∞.

First we know that there exist δ > 0 and kj ∈ Z such that

|vn(kj)| ≥ δ. (3.5)

Indeed, if not, then vn → 0 in l∞ as n→∞. For r > p,

‖vn‖rlr ≤ ‖vn‖r−pl∞ ‖vn‖

plp

we have vn → 0 in all lr, r > p.

Note that by (f1) and (f2), for any ε > 0, there exists cε > 0 such that

|f(k, u)| ≤ ε|u|p−1 + cε|u|q−1 and |F (k, u)| ≤ ε|u|p + cε|u|q. (3.6)

Then for each s > 0, we have∑k∈Z

F (k, svn(k)) ≤ εsp‖vn‖plp + cεsq‖vn‖qlq

which implies that∑k∈Z

F (k, svn(k))→ 0 as n→∞. So

d ≥ J(un) ≥ J(svn) =sp

p‖v(k)‖p −

∑k∈Z

F (k, svn(k))→ sp

p, (3.7)

as n→∞. This is a contradiction with s > p√pd.

Due to periodicity of coefficients, we know J andN are both invariant under T-translation.Making such shifts, we can assume that 1 ≤ kj ≤ T − 1 in (3.5). Moreover, passing to asubsequence, we can assume that kj = k0 is independent of j.

Next we may extract a subsequence, still denoted by vn, such that vn(k) → v(k) forall k ∈ Z. Specially, for k = k0, inequality (3.5) shows that |v(k0)| ≥ δ, so v 6= 0. Since|un(k)| → ∞ as n→∞, it follows again from (f3) that

0 ≤ J(un)

‖un‖p=

1

p−∑k∈Z

F (k, un(k))

(un(k))p(vn(k))p → −∞ as n→∞,

a contradiction again.

Proof of Theorem 1.1.

The proof consists of five steps. The proof of step 1-3 is similar to [12], for readers’convenience, we give the proof.


1293 Ali Mai et al 1288-1297

step 1. we claim that N is homeomorphic to the unit sphere S in E.

By (3.1) and (3.3), g(s) > 0 for s > 0 small and g(s) < 0 for s > 0 large. So sw is aunique maximum of g(s) and sww is the unique point on the ray s 7→ sw (s > 0) whichintersects N . That is, u ∈ N is the unique maximum of J on the ray. Therefore, by Lemma3.1, we may define the mapping m : E \ 0 → N by setting

m(w) := sww.

Next we show the mapping m is continuous. Indeed, suppose wn → w 6= 0. Since m(tu) =m(u) for each t > 0, we may assume wn ∈ S for all n. Write m(wn) = swnwn. Then swnis bounded. If not, swn →∞ as n→∞.

Note that by (f4), for all u 6= 0,

1

pf(k, u)u− F (k, u) =

1

pf(k, u)u−

∫ u

0

f(k, s)ds

>1

pf(k, u)u− f(k, u)

up−1

∫ u

0

sp−1ds

= 0.

Therefore, for all u ∈ N , we have

J(u) = J(u)− 1

pJ ′(u)u =

∑k∈Z

(1

pf(k, u(k))u(k)− F (k, u(k))

)> 0. (3.8)

Combining with (f3) and Lemma 3.1, we have

0 <J(swnw)

(swn)p=

1

p‖w‖p −

∑k∈Z

F (k, swnw(k))

|swnw(k)|p|w(k)|p → −∞, as n→∞,

this is a contradiction. Therefore, swn → s > 0 after passing to a subsequence if needed.Since N is closed and m(wn) = swnwn → sw, sw ∈ N . Hence sw = sww = m(w) by theuniqueness of sw of Lemma 3.1. Therefore, m is continuous.

Then we define a mapping m : S → N by setting m := m|S, then m is a homeomorphismbetween S and N , and the inverse of m is given by m−1(u) = u

‖u‖ .

step 2. now we define the functional Ψ : E \ 0 → R and Ψ : S → R by

Ψ(w) := J(m(w)) and Ψ(w) := Ψ|S.

Then we have

Ψ ∈ C1(E \ 0,R) and Ψ ∈ C1(S,R). Moreover,

Ψ′(w)z =‖m(w)‖‖w‖

J ′(m(w))z for all w, z ∈ E,w 6= 0. (3.9)


1294 Ali Mai et al 1288-1297

Ψ′(w)z = ‖m(w)‖J ′(m(w))z for all z ∈ Tw(S) = v ∈ E : (w, v) = 0. (3.10)

In fact, let w ∈ E \ 0 and z ∈ E. By Lemma 3.1 and the mean value theorem, we obtain

Ψ(w + tz)− Ψ(w) = J(sw+tz(w + tz))− J(sww)

≤ J(sw+tz(w + tz))− J(sw+tz(w))

= J ′(sw+tz(w + τttz))sw+tztz,

where |t| is small enough and τt ∈ (0, 1). Similarly,

Ψ(w + tz)− Ψ(w) = J(sw+tz(w + tz))− J(sww)

≥ J(sw(w + tz))− J(sw(w))

= J ′(sw(w + ηttz))swtz,

where ηt ∈ (0, 1). Combining these two inequalities and the continuity of function w 7→ sw,we have

limt→0

Ψ(w + tz)− Ψ(w)

t= swJ

′(sww)z =‖m(w)‖‖w‖

J ′(m(w))z.

Hence the Gateaux derivative of Ψ is bounded linear in z and continuous in w. It followsthat Ψ is a class of C1 and (3.9) holds. Note only that since w ∈ S, m(w) = m(w), so (3.10)is clear.

step 3. wn is a Palais-Smale sequence for Ψ if and only if m(wn) is a Palais-Smalesequence for J .

Let wn be a Palais-Smale sequence for Ψ, and let un = m(wn) ∈ N . Since for everywn ∈ S we have an orthogonal splitting E = TwnS ⊕ Rwn, we have

‖Ψ′(wn)‖ = supz∈TwnS‖z‖=1

Ψ′(wn)z = ‖m(wn)‖ supz∈TwnS‖z‖=1

J ′(m(wn))z = ‖un‖ supz∈TwnS‖z‖=1

J ′(un)z.

Then

‖Ψ′(wn)‖ ≤ ‖un‖‖J ′(un)‖ = ‖un‖ supz∈TwnS, t∈R

z+tw 6=0

J ′(un)(z + tw)

‖z + tw‖

≤ ‖un‖ supz∈TwnS\0

J ′(un)(z)

‖z‖= ‖Ψ′(wn)‖,

Therefore

‖Ψ′(wn)‖ = ‖un‖‖J ′(un)‖. (3.11)


1295 Ali Mai et al 1288-1297

By (3.8), for un ∈ N , J(un) > 0, so there exists a constant c0 > 0 such that J(un) > c0.And since c0 ≤ J(un) = 1

p‖un‖p − I(un) ≤ 1

p‖un‖p, ‖un‖ ≥ p

√pc0. Together with Lemma

3.2, p√pc0 ≤ ‖un‖ ≤ supn ‖un‖ < ∞. Hence wn is a Palais-Smale sequence for Ψ if and

only if un is a Palais-Smale sequence for J .

step 4. by (3.11), Ψ′(w) = 0 if and only if J ′(m(w)) = 0. So w is a critical point of Ψ ifand only if m(w) is a nontrivial critical point of J . Moreover, the corresponding values of Ψand J coincide and infS Ψ = infN J .

If u0 ∈ N satisfies J(u0) = c := infu∈N J(u), then m−1(u0) ∈ S is a minimizer of Ψ andtherefore a critical point of Ψ, so u0 is a critical point of J . It remains to show that thereexists a minimizer u ∈ N of J |N .

Let wn ⊂ S be a minimizing sequence for Ψ. By Ekeland’s variational principle wemay assume Ψ(wn) → c, Ψ′(wn) → 0 as n → ∞, hence J(un) → c, J ′(un) → 0 as n → ∞,where un := m(wn) ∈ N .

We know that un is bounded in N by Lemma 3.2, then there exists a subsequence,still denoted by the same notation, such that un weakly converges to some u ∈ E. We claimthat there exist δ > 0 and kj ∈ Z such that

|un(kj)| ≥ δ. (3.12)

Indeed, if not, then un → 0 in l∞ as n→∞. From the simple fact that, for r > p,

‖un‖rlr ≤ ‖un‖r−pl∞ ‖un‖

plp

we have un → 0 in all lr, r > p. By (3.6), we know∑k∈Z

f(k, un(k))un(k) ≤ ε∑k∈Z

|un(k)|p−1 · |un(k)|+ cε∑k∈Z

|un(k)|q−1 · |un(k)|

≤ ε‖un‖plp + cε‖un‖q−1lq

which implies that∑k∈Z

f(k, un(k))un(k) = o(‖un‖) as n→∞. Then

o(‖un‖) = (J ′(un), un) = ‖un‖p −∑k∈Z

f(k, un(k))un(k) = ‖un‖p − o(‖un‖).

So ‖un‖p → 0, as n→∞, which contradicts with un ∈ N .

Since J and J ′ are both invariant under T -translation. Making such shifts, we can assumethat 1 ≤ kj ≤ T−1 in (3.12). Moreover passing to a subsequence, we can assume that kj = k0is independent of j. Extract a subsequence, still denoted by un, we have un u andun(k) → u(k) for all k ∈ Z. Specially, for k = k0, inequality (3.12) shows that |u(k0)| ≥ δ,so u 6= 0. Hence u ∈ N .

step 5. we need to show that J(u) = c. By Fatou’s lemma, we have

c = limn→∞

(J(un)− 1

2J ′(un)un

)= lim

n→∞

∑k∈Z

(1

2f(k, un(k))un(k)− F (k, un(k))

)≥

∑k∈Z

(1

2f(k, u(k))u(k)− F (k, u(k))

)= J(u)− 1

2J ′(u)u = J(u) ≥ c.


1296 Ali Mai et al 1288-1297

Hence J(u) = c. The proof of Theorem 1.1 is completed.

Acknowledgments

This work is Supported by National Natural Science Foundation of China(11526183,11371313, 11401121), the Natural Science Foundation of Shanxi Province (2015021015) andFoundation of Yuncheng University(YQ-2014011,XK-2014035).

References

[1] R. Avery and J. Henderson, Existence of three positive pseudo-symmetric solutions fora one dimensional discrete p-Laplacian, J. Difference Equ. Appl. 10 (2004) 529C539.

[2] Z. He, On the existence of positive solutions of p-Laplacian difference equations, J.Comput. Appl. Math. 161 (2003) 193C201.

[3] Y. Li and L. Lu, Existence of positive solutions of p-Laplacian difference equations,Appl. Math. Lett. 19 (2006) 1019C1023.

[4] A. Cabada, A. Iannizzotto, S. Tersian, Multiple solutions for discrete boundary valueproblems, J. Math. Anal. Appl. 356 (2009) 418-428.

[5] L. Jiang, Z. Zhou, Three solutions to Dirichlet boundary value problems for p-Laplaciandifference equations, Adv. Differential Equations 2008 (2008) 1-11.

[6] Y. Long, H. Shi, Multiple solutions for the discrete p-Laplacian boundary value prob-lems, Discrete Dyn. Nat. Soc. 2014 (2014) 1-6.

[7] M. Ma, Z. Guo, Homoclinic orbits for nonliear second order difference equations, Non-linear Anal. 67 (2007) 1737-1745.

[8] A. Mai, Z. Zhou, Ground state solutions for the periodic discrete nonlinear Schrodingerequations with superlinear nonlinearities. Abstr. Appl. Anal. 2013(2013) 1-11.

[9] A. Mai, Z. Zhou, Discrete solitons for periodic discrete nonlinear Schrodinger equations,Appl. Math. Comput, 222 (2013): 34-41.

[10] A. Mai, Z. Zhou, Homoclinic solutions for a class of nonlinear difference equations, J.Appl. Math., 2014: 1-8.

[11] G. Sun, On Standing Wave solutions for discrete nonlinear Schroinger equations, Abstr.Appl. Anal. 2013(2013) 1-6.

[12] G. Sun, Ali Mai, Infinitely many homoclinic solutions for second order nonlinear differ-ence equations with p-Laplacian, Sci. World J. 2014:1-6.

[13] P. Rabinowitz, Minimax Methods in Critical Point Theory with Applications to Differ-ential Equations, CBMS Regional Conference Series in Mathematics vol 65, AmericanMathematical Society, Providence, RI, 1986.


1297 Ali Mai et al 1288-1297

On a solutions of fourth order rational systems of differenceequations

E. M. Elsayed1,2, Abdullah Alotaibi1, and Hajar A. Almaylabi11King AbdulAziz University, Faculty of Science,Mathematics Department, P. O. Box 80203,

Jeddah 21589, Saudi Arabia.2Mathematics Department, Faculty of Science,Mansoura University, Mansoura 35516, Egypt.

E-mail: [email protected], [email protected],[email protected].

ABSTRACT

In this paper, we get the form of the solutions of the following difference equation systems of order four

xn+1 =ynxn−2

yn + yn−3, yn+1 =

xnyn−2±xn ± xn−3

, n = 0, 1, 2, · · · ,

where the initial conditions x−3, x−2, x−1, x0, y−3, y−2, y−1, y0 are arbitrary non zero real numbers.

Keywords: difference equations, recursive sequences, system of difference equations, stability, periodicity,boundedness.

Mathematics Subject Classification: 39A10.

–––––––––––––––––––

1. INTRODUCTION

Difference equations enter as approximations of continuous problems and as models describing life situations inmany directions. Recently there has been a great interest in studying difference equations, see, for instance [4],[11], [30] and references cited therein, as well as in studying systems of difference equations (see, e.g. [1], [3], [6],[8]-[10]).

Some of the systems of difference equations that are of considerable interest nowadays are symmetric or thoseobtained from symmetric ones by modifications of their parameters or the sequence coefficients appearing inthem (for the case of nonautonomous systems of difference equations). Such systems are studied, for example,in the following papers: Clark et al. [2] has investigated the global stability properties and asymptotic behaviorof solutions of the system

xn+1 =xn

a+ cyn, yn+1 =

ynb+ dxn

.

Din and Elsayed [5] investigated the boundedness character, persistence, local and global behavior of positivesolutions of following two directional interactive and invasive species model

xn+1 = α+ βxn + γxn−1e−yn , yn+1 = δ + yn + ζyn−1e

−xn .

Halim et al. [13] deal with the form of the solutions of the two following systems of rational difference equations

xn+1 =yn(xn−2 + yn−3)

yn−3 + xn−2 − yn, yn+1 =

xn−1(xn−1 + yn−2)

2xn−1 + yn−2,

xn+1 =(yn−3 − xn−2)ynyn−3 − xn−2 + yn

, yn+1 =(yn−2 − xn−1)xn−1

yn−2.


1298 E. M. Elsayed et al 1298-1308

Kurbanli [21] investigated the behavior of the solution of the difference equation system

xn+1 =xn−1

xn−1yn−1 , yn+1 =yn−1

yn−1xn−1 , zn+1 =1

znyn.

The authors in [27] have got the form of the solutions of some systems of the following rational differenceequations

xn+1 =xn−1

α− xn−1yn, yn+1 =

yn−1β + γyn−1xn

.

In [29] Papaschinnopoulos and Schinas studied the oscillatory behavior, the boundedness of the solutions, andthe global asymptotic stability of the positive equilibrium of the system of nonlinear difference equations

xn+1 = A+yn

xn−p, yn+1 = A+

xnyn−q

.

In [39], Yalcinkaya et al. studied the periodic character of the following two systems of difference equations

x(1)n+1 =

x(2)n

x(2)n −1

, x(2)n+1 =

x(3)n

x(3)n −1

, . . . , x(k)n+1 =

x(1)n

x(1)n −1

,

andx(1)n+1 =

x(k)n

x(k)n −1

, x(2)n+1 =

x(1)n

x(1)n −1

, . . . , x(k)n+1 =

x(k−1)n

x(k−1)n −1

,

where the initial values are nonzero real numbers for x(1)0 , x(2)0 , . . . , x

(k)0 6= 1.

In [42]-[43] Zhang et al. studied the boundedness, the persistence and global asymptotic stability of thepositive solutions of the systems of difference equations

xn+1 = A+yn−mxn

, yn+1 = A+xn−myn

,

andxn = A+

1

yn−p, yn = A+

yn−1xn−ryn−s

.

Similar to difference equations and nonlinear systems of rational difference equations were investigated see [12]-[45].

In this paper, we obtain the expressions of the solutions of the following nonlinear systems of differenceequations

xn+1 =ynxn−2

yn + yn−3, yn+1 =

xnyn−2±xn ± xn−3

, n = 0, 1, 2, · · · ,

where the initial values x−3, x−2, x−1, x0, y−3, y−2, y−1, y0 are arbitrary non zero real numbers, moreover,we take some numerical examples for the equation to illustrate the results.

2. ON THE SYSTEM XN+1 =YNXN−2YN+YN−3

, YN+1 =XNYN−2XN+XN−3

In this section,we study the solutions of the following system of difference equations

xn+1 =ynxn−2

yn + yn−3, yn+1 =

xnyn−2xn + xn−3

, (1)

where the initial values x−3, x−2, x−1, x0, y−3, y−2, y−1, y0 are arbitrary nonzero real numbers.

Theorem 1. Suppose that xn, yn are solutions of the system (1). Then for n = 0, 1, 2, · · · , we have thefollowing formula

x6n−3 =adnhnQn−1

i=0 (e+ (6i+ 3)h)(a+ (6i)d), x6n−2 =

bdnhnQn−1i=0 (e+ (6i+ 1)h)(a+ (6i+ 4)d)

,

x6n−1 =cdnhnQn−1

i=0 (e+ (6i+ 5)h)(a+ (6i+ 2)d), x6n =

dn+1hnQn−1i=0 (e+ (6i+ 3)h)(a+ (6i+ 6)d)

,


1299 E. M. Elsayed et al 1298-1308

x6n+1 =bdnhn+1

(e+ h)Qn−1

i=0 (e+ (6i+ 7)h)(a+ (6i+ 4)d), x6n+2 =

cdn+1hn

(a+ 2d)Qn−1

i=0 (e+ (6i+ 5)h)(a+ (6i+ 8)d),

y6n−3 =ednhnQn−1

i=0 (e+ (6i)h)(a+ (6i+ 3)d), y6n−2 =

fdnhnQn−1i=0 (e+ (6i+ 4)h)(a+ (6i+ 1)d)

,

y6n−1 =gdnhnQn−1

i=0 (e+ (6i+ 2)h)(a+ (6i+ 5)d), y6n =

dnhn+1Qn−1i=0 (e+ (6i+ 6)h)(a+ (6i+ 3)d)

,

y6n+1 =fdn+1hn

(a+ d)Qn−1

i=0 (e+ (6i+ 4)h)(a+ (6i+ 7)d), y6n+2 =

gdnhn+1

(e+ 2h)Qn−1

i=0 (e+ (6i+ 8)h)(a+ (6i+ 5)d),

where x−3 = a, x−2 = b, x−1 = c, x0 = d, y−3 = e, y−2 = f, y−1 = g, y0 = h.

Proof. By using mathematical induction. The result holds for n = 0. Suppose that the result holds for n− 1

x6n−7 =cdn−1hn−1Qn−2

i=0 (e+ (6i+ 5)h)(a+ (6i+ 2)d), x6n−6 =

dnhn−1Qn−2i=0 (e+ (6i+ 3)h)(a+ (6i+ 6)d)

,

x6n−5 =bdn−1hn

(e+ h)Qn−2

i=0 (e+ (6i+ 7)h)(a+ (6i+ 4)d), x6n−4 =

cdnhn−1

(a+ 2d)Qn−2

i=0 (e+ (6i+ 5)h)(a+ (6i+ 8)d),

y6n−7 =gdn−1hn−1Qn−2

i=0 (e+ (6i+ 2)h)(a+ (6i+ 5)d), y6n−6 =

dn−1hnQn−2i=0 (e+ (6i+ 6)h)(a+ (6i+ 3)d)

,

y6n−5 =fdnhn−1

(a+ d)Qn−2

i=0 (e+ (6i+ 4)h)(a+ (6i+ 7)d), y6n−4 =

gdn−1hn

(e+ 2h)Qn−2

i=0 (e+ (6i+ 8)h)(a+ (6i+ 5)d).

From system (1) we can prove as follow

x6n−3 =y6n−4x6n−6

y6n−4 + y6n−7=

³gdn−1hn

(e+2h) n−2i=0 (e+(6i+8)h)(a+(6i+5)d)

´³dnhn−1

n−2i=0 (e+(6i+3)h)(a+(6i+6)d)

´³

gdn−1hn

(e+2h) n−2i=0 (e+(6i+8)h)(a+(6i+5)d)

´+³

gdn−1hn−1n−2i=0 (e+(6i+2)h)(a+(6i+5)d)

´=

dnhnQn−2i=0 (e+ (6i+ 3)h)(a+ (6i+ 6)d)

³h+

(e+2h) n−2i=0 (e+(6i+8)h)

n−2i=0 (e+(6i+2)h)

´=

dnhn

(e+ (6n− 3)h)Qn−2

i=0 (e+ (6i+ 3)h)(a+ (6i+ 6)d)=

adnhnQn−1i=0 (e+ (6i+ 3)h)(a+ (6i)d)

,

y6n−3 =x6n−4y6n−6

x6n−4 + x6n−7=

cdnhn−1

(a+2d) n−2i=0 (e+(6i+5)h)(a+(6i+8)d)

dn−1hnn−2i=0 (e+(6i+6)h)(a+(6i+3)d)³

cdnhn−1

(a+2d) n−2i=0 (e+(6i+5)h)(a+(6i+8)d)

+ cdn−1hn−1n−2i=0 (e+(6i+5)h)(a+(6i+2)d)

´= dnhn

(a+2d) n−2i=0 (a+(6i+8)d)(e+(6i+6)h)(a+(6i+3)d)

d

(a+2d)n−2i=0

(a+(6i+8)d)+ 1

n−2i=0

(a+(6i+2)d)

= dnhn

n−2i=0 (e+(6i+6)h)(a+(6i+3)d) d+

(a+2d)n−2i=0

(a+(6i+8)d)

n−2i=0

(a+(6i+2)d)

=dnhnQn−2

i=0 (e+ (6i+ 6)h)(a+ (6i+ 3)d)³d+

(a+2d) n−2i=0 (a+(6i+8)d)

n−2i=0 (a+(6i+2)d)

´=

dnhnQn−2i=0 (e+ (6i+ 6)h)(a+ (6i+ 3)d) (a+ (6n− 3)d)

=ednhnQn−1

i=0 (e+ (6i)h)(a+ (6i+ 3)d).


1300 E. M. Elsayed et al 1298-1308

The other relations can be proved similarly, this completes the proof.

Lemma 1. Every positive solution of system (1) is bounded and limn→∞

xn = limn→∞

yn = 0.

Proof: It follows from system (1), that

xn+1 =ynxn−2

yn + yn−3<

ynxn−2yn

= xn−2, yn+1 =xnyn−2

xn + xn−3<

xnyn−2xn

= yn−2.

Then the subsequences x3n−2∞n=0, x3n−1∞n=0, x3n∞n=0, y3n−2∞n=0, y3n−1∞n=0, y3n∞n=0 are decreasingand so are bounded from above byM, N respectively sinceM = maxx−3, x−2, x−1, x0 ,N = maxy−3, y−2, y−1, y0.


, YN+1 =XNYN−2XN−XN−3

We study, in this section, the solutions formulas of the system of rational difference equations

xn+1 =ynxn−2

yn + yn−3, yn+1 =

xnyn−2xn − xn−3

, (2)


Theorem 2. Assume that xn, yn are solutions of system (2) with x−3 6= x0, x−3 6= 2x0 and y−3 6= ±y0. Thenfor n = 0, 1, 2, ...,

x12n−3 =d2nh2n

an−1(h+ e)n(h− e)n(2d− a)n, y12n−3 =

h2nd2n

e2n−1(d− a)2n,

x12n−2 =bd2nh2n

an(h+ e)n(h− e)n(2d− a)n, y12n−2 =

fh2nd2n

e2n(d− a)2n,

x12n−1 =cd2nh2n

an(h+ e)n(h− e)n(2d− a)n, y12n−1 =

gh2nd2n

e2n(d− a)2n,

x12n =d2n+1h2n

an(h+ e)n(h− e)n(2d− a)n, y12n =

h2n+1d2n

e2n(d− a)2n,

x12n+1 =bd2nh2n+1

an(h+ e)n+1(h− e)n(2d− a)n, y12n+1 =

fh2nd2n+1

e2n(d− a)2n+1,

x12n+2 =cd2n+1h2n

an(h+ e)n(h− e)n(2d− a)n+1, y12n+2 =

−gh2n+1d2ne2n+1(d− a)2n

,

x12n+3 =d2n+1h2n+1

an(h+ e)n(h− e)n+1(2d− a)n, y12n+3 =

−h2n+1d2n+1e2n(d− a)2n+1

,

x12n+4 =bd2n+1h2n+1

an+1(h+ e)n+1(h− e)n(2d− a)n, y12n+4 =

fh2n+1d2n+1

e2n+1(d− a)2n+1,

x12n+5 =cd2n+1h2n+1

an(h+ e)n+1(h− e)n(2d− a)n+1, y12n+5 =

−gh2n+1d2n+1e2n+1(d− a)2n+1

,

x12n+6 =d2n+2h2n+1

an(h+ e)n(h− e)n+1(2d− a)n+1, y12n+6 =

h2n+2d2n+1

e2n+1(d− a)2n+1,

x12n+7 =bd2n+1h2n+2

an+1(h+ e)n+1(h− e)n+1(2d− a)n, y12n+7 =

−fh2n+1d2n+2e2n+1(d− a)2n+2

,

x12n+8 =cd2n+2h2n+1

an+1(h+ e)n+1(h− e)n(2d− a)n+1, y12n+8 =

−gh2n+2d2n+1e2n+2(d− a)2n+1

.


1301 E. M. Elsayed et al 1298-1308

Proof. By using mathematical induction. The result holds for n = 0. Suppose that the result holds for n− 1

x12n−7 =cd2n−1h2n−1

an−1(h+ e)n(h− e)n−1(2d− a)n, y12n−7 =

−gh2n−1d2n−1e2n−1(d− a)2n−1

,

x12n−6 =d2nh2n−1

an−1(h+ e)n−1(h− e)n(2d− a)n, y12n−6 =

h2nd2n−1

e2n−1(d− a)2n−1,

x12n−5 =bd2n−1h2n

an(h+ e)n(h− e)n(2d− a)n−1, y12n−5 =

−fh2n−1d2ne2n−1(d− a)2n

,

x12n−4 =cd2nh2n−1

an(h+ e)n(h− e)n−1(2d− a)n, y12n−4 =

−gh2nd2n−1e2n(d− a)2n−1

,

From system (2) we have

x12n−3 =y12n−4x12n−6

y12n−4 + y12n−7=


d2nh2n−1

an−1(h+ e)n−1(h− e)n(2d− a)n


+−gh2n−1d2n−1

e2n−1(d− a)2n−1

=h2nd2n

an−1(h+ e)n−1(h− e)n(2d− a)n(h+ e)=

d2nh2n

an−1(h+ e)n(h− e)n(2d− a)n,

y12n−3 =x12n−4y12n−6

x12n−4 − x12n−7=

cd2nh2n−1

an(h+e)n(h−e)n−1(2d−a)nh2nd2n−1

e2n−1(d−a)2n−1cd2nh2n−1

an(h+e)n(h−e)n−1(2d−a)n −cd2n−1h2n−1

an−1(h+e)n(h−e)n−1(2d−a)n=

h2nd2n

e2n−1(d− a)2n,

x12n−2 =y12n−3x12n−5

y12n−3 + y12n−6

= d2nh2nbd2n−1h2n

e2n−1(d−a)2nan(h+e)n(h−e)n(2d−a)n−1£

d2nh2n

e2n(d−a)2n+d2n−1h2n

e2n−1(d−a)2n−1

¤ = bd2nh2n

an(h+e)n(h−e)n(2d−a)n ,

y12n−2 =x12n−3y12n−5

x12n−3 − x12n−6

= −h2nd2nfd2nh2n−1

an−1(h+e)n(h−e)n(2d−a)ne2n−1(d−a)2n£

h2nd2n

an−1(h+e)n(h−e)n(2d−a)n−h2n−1d2n

an−1(h+e)n−1(h−e)n(2d−a)n

¤ = fd2nh2n

e2n(d−a)2n .

So, we can prove the other relations and the proof is completed.

Lemma 2. Every positive solution of the equation xn+1 =ynxn−2

yn + yn−3is bounded and lim

n→∞xn = 0.

The following cases can be proved similarly.


, YN+1 =XNYN−2

−XN+XN−3

In this section, we study the solutions of the system of the difference equations

xn+1 =ynxn−2

yn + yn−3, yn+1 =

xnyn−2−xn + xn−3

, (3)

where the initial values x−3, x−2, x−1, x0, y−3, y−2, y−1, y0 are arbitrary nonzero real numbers withx−3 6= x0, and y−3 6= −y0.Theorem 3. Let xn, yn+∞n=−3 be solutions of system (3). Then for n = 0, 1, 2, ...,

x6n−3 =hndn

an−1(h+ e)n, y6n−3 =

hndn

en−1(−d+ a)n,

x6n−2 =bhndn

an(h+ e)n, y6n−2 =

fhndn

en(−d+ a)n,

x6n−1 =chndn

an(h+ e)n, y6n−1 =

ghndn

en(−d+ a)n,


1302 E. M. Elsayed et al 1298-1308

x6n =hndn+1

an(h+ e)n, y6n =

dnhn+1

en(−d+ a)n,

x6n+1 =bdnhn+1

an(h+ e)n+1, y6n+1 =

fhndn+1

en(−d+ a)n+1,

x6n+2 =chndn+1

an+1(h+ e)n, y6n+2 =

gdnhn+1

en+1(−d+ a)n,

Lemma 3. The system (3) has a periodic solutions of period 6 iff hd = e(a− d) = a(h+ e).

Proof. First, if hd = e(a− d) = a(h+ e), then from the form of the solutions of system (3), we see that

x6n−3 = an(h+ e)nan−1(h+ e)n = a , x6n−2 = b, x6n−1 = c, x6n = d, x6n+1 = hh+ e, x6n+2 = cda,

y6n−3 = e, y6n−2 = f, y6n−1 = g, y6n = h, y6n+1 = fda− d, y6n+2 = hge.

Thus system (3) has a periodic solution with period 6. Second:if we have a period 6 then

x6n−3 =hndn

an−1(h+ e)n= x−3 = a, x6n−2 =

bhndn

an(h+ e)n= x−2 = b, x6n−1 =

chndn

an(h+ e)n= x−1 = c,

x6n =hndn+1

an(h+ e)n= x0 = d, x6n+1 =

bdnhn+1

an(h+ e)n+1= x1 =

bh

h+ e, x6n+2 =

chndn+1

an+1(h+ e)n= x2 =

cd

a,

y6n−3 =hndn

en−1(−d+ a)n= y−3 = e, y6n−2 =

fhndn

en(−d+ a)n= y−2 = f,

y6n−1 =ghndn

en(−d+ a)n= y−1 = g, y6n =

dnhn+1

en(−d+ a)n= y0 = h,

y6n+1 =fhndn+1

en(−d+ a)n+1= y1 =

fd

a− d, y6n+2 =

gdnhn+1

en+1(−d+ a)n= y2 =

gh

e,

Then we get hd = a(h+ e), hd = e(a− d), and the proof is completed.


, YN+1 =XNYN−2

−XN−XN−3

In this section,we study the solutions of the system of the difference equations

xn+1 =ynxn−2

yn + yn−3, yn+1 =

xnyn−2−xn − xn−3

, (4)


Theorem 4. If xn, yn are solutions of difference equation system (4). Then for n = 0, 1, 2, ..., we have

x12n−3 =d2nh2n

a2n−1(h+ e)2n, y12n−3 =

(−1)nd2nh2nen−1(d+ a)n(d− a)n(2h+ e)n

,

x12n−2 =bd2nh2n

a2n(h+ e)2n, y12n−2 =

(−1)nfd2nh2nen(d+ a)n(d− a)n(2h+ e)n

,

x12n−1 =cd2nh2n

a2n(h+ e)2n, y12n−1 =

(−1)ngd2nh2nen(d+ a)n(d− a)n(2h+ e)n

,

x12n =d2n+1h2n

a2n(h+ e)2n, y12n =

(−1)nd2nh2n+1en(d+ a)n(d− a)n(2h+ e)n

,


1303 E. M. Elsayed et al 1298-1308

x12n+1 =bd2nh2n+1

a2n(h+ e)2n+1, y12n−3 =

(−1)n+1fd2n+1h2nen(d+ a)n+1(d− a)n(2h+ e)n

,

x12n+2 =−cd2n+1h2n

a2n+1(h+ e)2n, y12n+2 =

(−1)n+1gd2nh2n+1en(d+ a)n(d− a)n(2h+ e)n+1

,

x12n+3 =−d2n+1h2n+1a2n(h+ e)2n+1

, y12n+3 =(−1)n+1d2n+1h2n+1

en(d+ a)n(d− a)n+1(2h+ e)n,

x12n+4 =bd2n+1h2n+1

a2n+1(h+ e)2n+1, y12n+4 =

(−1)n+1fd2n+1h2n+1en+1(d+ a)n+1(d− a)n(2h+ e)n

,

x12n+5 =−cd2n+1h2n+1

a2n+1(h+ e)2n+1, y12n+5 =

(−1)ngd2n+1h2n+1en(d+ a)n+1(d− a)n(2h+ e)n+1

,

x12n+6 =d2n+2h2n+1

a2n+1(h+ e)2n+1, y12n+6 =

(−1)nd2n+1h2n+2en(d+ a)n(d− a)n+1(2h+ e)n+1

,

x12n+7 =−bd2n+1h2n+2

a2n+1(h+ e)2n+2, y12n+7 =

(−1)nfd2n+2h2n+1en+1(d+ a)n+1(d− a)n+1(2h+ e)n

,

x12n+8 =−cd2n+2h2n+1

a2n+2(h+ e)2n+1, y12n+8 =

(−1)ngd2n+1h2n+2en+1(d+ a)n+1(d− a)n(2h+ e)n+1

.

6. NUMERICAL EXAMPLES

Here, we consider interesting numerical examples in order to illustrate the results of the previous sections andto support our theoretical discussions.

Example 1. We consider numerical example for the difference system (1) with the initial conditions x−3 =2, x−2 = 14, x−1 = 6, x0 = 7, y−3 = 5, y−2 = 9, y−1 = 7 and y0 = −8. (See Fig. 1).

0 5 10 15 20 25 30−10

−5

0

5

10

15

20

25

30

35

40

n

x(n)

,y(n

)

plot of x(n+1)=x(n−2)y(n)/y(n)+y(n−3),y(n+1)=x(n)y(n−2)/x(n)+x(n−3)

x(n)y(n)

Figure 1.

Example 2. Assume for the system (2) with the initial conditions x−3 = 4, x−2 = 5, x−1 = 6, x0 = 3, y−3 =1.8, y−2 = 9, y−1 = 2 and y0 = 1.9. See Figure (2).


1304 E. M. Elsayed et al 1298-1308

0 5 10 15 20 25 30 35 40−1

−0.5

0

0.5

1

1.5x 10

4

n

x(n)

,y(n

)

plot of x(n+1)=x(n−2)y(n)/y(n)+y(n−3),y(n+1)=x(n)y(n−2)/x(n)−x(n−3)

x(n)y(n)

Figure 2.

Example 3. Figure (3) shows the behavior of the solution of the difference system (3) with the initial conditionsx−3 = 4, x−2 = 5, x−1 = 6, x0 = 10, y−3 = 8, y−2 = 9, y−1 = 2 and y0 = 2.

0 10 20 30 40 50 60 70−15

−10

−5

0

5

10

15

n

x(n)

,y(n

)

plot of x(n+1)=x(n−2)y(n)/y(n)+y(n−3),y(n+1)=x(n)y(n−2)/x(n−3)−x(n)

x(n)y(n)

Figure 3.

Example 4. We take the initial conditions, for the system (4), as follows x−3 = 3, x−2 = 5, x−1 = −9,x0 = 6, y−3 = 2, y−2 = 1.7, y−1 = 2.8 and y0 = 4. See Figure (4).


1305 E. M. Elsayed et al 1298-1308

0 10 20 30 40 50 60−100

−50

0

50

100

150

200

250

n

x(n)

,y(n

)

plot of x(n+1)=x(n−2)y(n)/y(n)+y(n−3),y(n+1)=x(n)y(n−2)/−x(n−3)−x(n)

x(n)y(n)

Figure 4.

Acknowledgements

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Theauthors, therefore, acknowledge with thanks DSR technical and financial support.

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[17] V. Khuong, and T. Thai, Asymptotic Behavior of the Solutions of System of Difference Equationsof Exponential Form, Journal of Difference Equations, Volume 2014, Article ID 936302, 6 pages.

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difference equations xn+1 =xn−1

xn−1yn − 1, yn+1 =

yn−1yn−1xn − 1

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zn−1yn, Applied Mathemat-

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[24] A. S. Kurbanli, C. Cinar and I. Yalçınkaya, On the behavior of positive solutions of the systemof rational difference equations, Math. Comput. Mod., 53 (2011), 1261-1267.

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[44] Q. Zhang and W. Zhang, On a system of two high-order nonlinear difference equations, Advancesin Mathematical Physics, 2014 (2014), Article ID 729273, 8 pages.

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1308 E. M. Elsayed et al 1298-1308

On the dynamics of higher Order di¤erenceequations xn+1 = axn +

αxnxn¡lβxn+γxn¡k

M. M. El-Dessoky1,2

1King Abdulaziz University, Faculty of Science,Mathematics Department, P. O. Box 80203,

Jeddah 21589, Saudi Arabia.2Department of Mathematics, Faculty of Science,

Mansoura University, Mansoura 35516, Egypt.E-mail: [email protected]

Abstract

The main objective of this paper is to study the global stability of thepositive solutions and the periodic character of the di¤erence equation

xn+1 = axn +αxnxn¡l

βxn + γxn¡k, n = 0, 1, ...,

where the parameters α, β, γ and a are positive real numbers and the initialconditions x¡t, x¡t+1 ..., x¡1 and x0 are positive real numbers where t =maxfl, kg. Numerical examples to the di¤erence equation are given to explainour results.

Keywords: di¤erence equations, stability, global stability, boundedness, periodicsolutions.Mathematics Subject Classi…cation: 39A10—————————————————

1 Introduction and Preliminaries

Our object in this paper is to study some qualitative behavior of the positive solutionsof the di¤erence equation


βxn+γxn¡k, n = 0, 1, ..., (1)

1


1309 M. M. El-Dessoky 1309-1322

where the parameters α, β, γand δ are positive real numbers and the initial condi-tions x¡t, x¡t+1 ..., x¡1 and x0 are positive real numbers where t = maxfl, kg. Inaddition, we obtain the solutions of some special cases of this equation.

Many researchers have Many researchers have studied the behavior of the solutionof di¤erence equations for example: Kalabušic et al. [1] studied the global characterof the solution of the nonlinear rational di¤erence equation

xn+1 =βxn¡l+δxn¡k

Bxn¡l+Dxn¡k, n = 0, 1, ...,

with positive parameters and non-negative initial conditions.Cinar [2] studied the solutions of the following di¤erence equation

xn+1 =axn¡1

1+bxnxn¡1, n = 0, 1, ...,

where a, b, x¡1 and x0 are non-negative real numbers.Yang et al. [3] studied the invariant intervals, the asymptotic behavior of the

solutions, and the global attractivity of equilibrium points of the recursive sequence

xn+1 =axn¡1+bxn¡2c+dxn¡1xn¡2

, n = 0, 1, ...,

where a ¸ 0, b, c, d > 0.In [4] kenneth et al. got the global asymptotic stability for positive solutions to

the di¤erence equation

yn+1 =yn¡k + yn¡m

1+yn¡kyn¡m, n = 0, 1, ...,

with y¡m, y¡m+1, ..., y¡1 2 (0, 1) and 1 · k · m.Raafat [5] investigated the global asymptotic stability of all solutions of the dif-

ference equationxn+1 =

Axn¡2B+Cxnxn¡1xn¡2

, n = 0, 1, ...,

where A, B, C are positive real numbers and the initial conditions x¡2, x¡1, x0are real numbers.

Also, Raafat [6] introduced an explicit formula and discuss the global behavior ofsolutions of the di¤erence equation

xn+1 =axn¡3

b+cxn¡1xn¡3, n = 0, 1, ...,

where a, b, c are positive real numbers and the initial conditions x¡3, x¡2, x¡1, x0are real numbers.

In [7] Elsayed studied the behavior of the solutions of the di¤erence equation

xn+1 = axn¡1 +bxnxn¡1

cxn+dxn¡2, n = 0, 1, ...,

where a, b, c are positive constant and the initial conditions x¡2, , x¡1, x0 arearbitrary positive real numbers.

2


1310 M. M. El-Dessoky 1309-1322

Zayed et al. [8] investigated some qualitative behavior of the solutions of thedi¤erence equation,

xn+1 = γxn¡k +axn+bxn¡k

cxn¡dxn¡k, n = 0, 1, ...,

where the coe¢cients γ, a, b, c and d are positive constants and the initial con-ditions x¡k, ..., x¡1, x0 are arbitrary positive real numbers, while k is a positiveinteger number.

Other related results on rational di¤erence equations can be found in refs. [11] -[24].

Let I be some interval of real numbers and let

F : I t+1 ! I,

be a continuously di¤erentiable function. Then for every set of initial conditionsx¡t, x¡t+1, ..., x0 2 I, the di¤erence equation

xn+1 = F (xn, xn¡1, ..., xn¡t), n = 0, 1, ..., (2)

has a unique solution fxng1n=¡t.

De…nition 1 The linearized equation of the di¤erence equation (2) about the equi-librium x is the linear di¤erence equation

yn+1 =tX

i=0

∂F (x, x, ..., x)

∂xn¡i

yn¡i. (3)

Now, assume that the characteristic equation associated with (3) is

p(λ) = p0λt + p1λ

t¡1 + ...+ pt¡1λ+ pt = 0, (4)

where

pi =∂F (x, x, ..., x)

∂xn¡i.

Theorem 1 [9]: Assume that pi 2 R, i = 1, 2, ..., t and t is non-negative integer.Then

tXi=1

jpij < 1,

is a su¢cient condition for the asymptotic stability of the di¤erence equation

xn+t + p1xn+t¡1 + ...+ ptxn = 0, n = 0, 1, ... .

3


1311 M. M. El-Dessoky 1309-1322

Theorem 2 [10, 11]: Let g : [a, b]t+1 ! [a, b], be a continuous function, where tis a positive integer, and where [a, b] is an interval of real numbers. Consider thedi¤erence equation

xn+1 = g(xn, xn¡1, ..., xn¡t), n = 0, 1, ... . (5)

Suppose that g satis…es the following conditions.

(1) For each integer i with 1 · i · t+ 1; the function g(z1, z2, ..., zt+1) is weaklymonotonic in zi for …xed z1, z2, ..., zi¡1, zi+1, ..., zt+1.

(2) If m, M is a solution of the system

m = g(m1, m2, ..., mt+1), M = g(M1, M2, ..., Mt+1),

then m = M , where for each i = 1, 2, ..., t+ 1, we set

mi =

½m, if g is non-decreasing in zi,

M, if g is non-increasing in zi,

¾,

and

Mi =

½M, if g is non-decreasing in zi,

m, if g is non-increasing in zi.

¾.

Then there exists exactly one equilibrium point ¹x of Equation (5), and everysolution of Equation (5) converges to ¹x.

2 Stability of the Equilibrium Point of Eq. (1)

2.1 Local stability

In this subsection, we study the local stability character of the equilibrium point ofEq. (1).

Eq. (1) has equilibrium point and is given by

¹x = a¹x+ α¹x2

β¹x+γ¹x, or ((1¡ a) (β+γ)¡ α) ¹x2 = 0,

if (1¡ a) (β+γ) 6= α, then the unique equilibrium point is x = 0.

Theorem 3 Assume that a+ 2αβ+γ

< 1,then equilibrium x of Eq. (1) is locally asymp-totically stable.

Proof: Let f : (0, 1)3 ¡! (0, 1) be a continuous function de…ned by

f (v0, v1, v2) = av0 +αv0v1

βv0 +γv2. (6)

4


1312 M. M. El-Dessoky 1309-1322

Therefore, it follows that

∂f (v0, v1, v2)∂v0

= a+ αv1(βv0+γv2)¡αβv0v1(βv0+γv2)

2 = a+αβv21

(βv0+γv2)2 ,

∂f (v0, v1, v2)∂v1

= αv0βv0+γv2

,

∂f (v0, v1, v2)∂v2

= ¡αv0v1(βv0+γv2)

2 = ¡ αγv0v1(βv0+γv2)

2 .

Then, we see that

∂f (¹x, ¹x, ¹x)∂v0

= a+ αβ

(β+γ)2, ∂f (¹x, ¹x, ¹x)

∂v1= α

β+γ, ∂f(¹x, ¹x, ¹x)

∂v2= ¡ αγ

(β+γ)2.

and the linearized equation of Eq. (1) about ¹x, is

yn+1 =³a+ αβ

(β+γ)2

ýn +

³α

β+γ

ýn¡l +

³¡αγ

(β+γ)2

ýn¡k,

Under the conditions, we get¯a+ αβ

(β+γ)2

¯+

¯α

β+γ

¯+

¯¡αγ

(β+γ)2

¯< 1,

and soa+ 2α

β+γ< 1.

According to Theorem 1, the proof is complete.

Example 1. The solution of the di¤erence equation (1) is local stability if l =2, k = 3, α = 0.1, β = 0.2, γ = 1, a = 0.2 and the initial conditions x¡3 = 0.6,x¡2 = 0.3, x¡1 = 0.4 and x0 = 0.8 (See Fig. 1).

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

n

x(n)

plot of x(n+1)= a X(n)+(alpha X(n) X(n-l)/(beta X(n)+gamma X(n-k)))

Fig. 1. Plot the behavior of the solution of equation (1).

5


1313 M. M. El-Dessoky 1309-1322

Example 2. See Figure (2) when we take the di¤erence equation (1) with l =2, k = 3, α = 1, β = 0.2, γ = 0.4, a = 0.5 and the initial conditions x¡3 = 0.6,x¡2 = 0.3, x¡1 = 0.4 and x0 = 0.8.

0 5 10 15 20 25 30 35 40 45 50-2

0

2

4

6

8

10

12x 10 7

n

x(n)


Fig. 2. Draw the behavior of the solution of equation (1).

2.2 Global Stability

In this subsection we study the global stability of the positive solutions of Eq. (1).

Theorem 4 The equilibrium point ¹x is a global attractor of equation (1) if

(1¡ a) (β¡ γ) 6= α.

Proof. Let r, s be nonnegative real numbers and assume that h : [r, s]3 ! [r, s]be a function de…ned by

h(v0, v1, v2) = av0 +αv0v1

βv0+γv2.

Then

∂h(v0, v1, v2)∂v0

= a+ αβv21(βv0+γv2)

2 ,∂h(v0, v1, v2)

∂v1= αv0

βv0+γv2and ∂h(v0, v1, v2)

∂v2= ¡ αγv0v1

(βv0+γv2)2 .

We can see that the function h(v0, v1, v2) increasing in v0, v1 and decreasing in v2.Suppose that (m, M) is a solution of the system

M = h(M, M, m) and m = h(m, m, M).

Then from Equation (1), we see that

M = aM + αM2

βM+γm, m = am+ αm2

βm+γM,

then

β(1¡ a)M +γ(1¡ a)m = αM,

β(1¡ a)m+γ(1¡ a)M = αm,

6


1314 M. M. El-Dessoky 1309-1322

Subtracting this two equations, we obtain

((1¡ a) (β¡ γ)¡ α) (M ¡ m) = 0,

under the condition (1¡ a) (β¡ γ) 6= α, we see that M = m. It follows from Theorem2 that ¹x is a global attractor of Equation (1).

Example 3. The solution of the di¤erence equation (1) is global stability ifl = 2, k = 3, α= 0.01, β= 0.2, γ= 0.4, a = 0.1 and the initial conditions x¡3 = 0.6,x¡2 = 0.3, x¡1 = 0.4 and x0 = 0.8 (See Fig. 3).

0 10 20 30 40 50 600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

n

x(n)


Fig. 3. Sketch the behavior of the solution of Eq. (1).

3 Boundedness of Solutions of Equation (1)

In this section we investigate the boundedness nature of the solutions of Equation(1).

Theorem 5 Every solution of Equation (1) is bounded if a < 1.

Proof. Let fxng1n=¡m be a solution of Equation (1). It follows from Equation (1)that


βxn+γxn¡k· axn +

αxnxn¡l

βxn= axn +

³αβ

´xn¡l.

By using a comparison, we can right hand side as follows

tn+1 = atn +³

αβ

´tn¡l.

and this equation is locally asymptotically stable if a < 1, and converges to theequilibrium point t = 0. Therefore

limn!1

sup xn 6 0.

7


1315 M. M. El-Dessoky 1309-1322

Example 4. Figure (4) shows that l = 4, k = 3, α= 0.1, β= 0.2, γ= 0.4, a =1.3, the solution of the di¤erence equation (1) with initial conditions x¡3 = 0.6,x¡2 = 0.3, x¡1 = 0.4 and x0 = 0.8 is unbounded.

0 10 20 30 40 50 600

1

2

3

4

5

6

7x 10 8

n

x(n)


Fig. 4. Polt the behavior of the solution of equation (1) when a > 1.

4 Existence of Periodic Solutions

In this section we investigate the existence of periodic solutions of Eq. (5).

Theorem 6 Equation (1) has no prime period two solutions if l and k are evenwhen a+α 6= 0 and β+γ 6= 0.

Proof. Suppose that there exists a prime period two solution ...p, q, p, q, ..., ofEquation (1). We see from Equation (1) when l and k are even that

p = aq +αq2

βq +γq, q = ap+

αp2

βp +γp.

(β+γ) pq = a (β+γ) q2 +αq2, (7)

(β+γ) pq = a (β+γ) p2 +αp2 (8)

Subtracting (7) from (8) gives

(a+α) (β+γ) (p2 ¡ q2) = 0,

Since a+α 6= 0 and β+γ 6= 0, then p = q. This is a contradiction. Thus, the proofis completed.

Theorem 7 Equation (1) has no prime period two solutions if l and k are odd whenγ 6= aβ.

Theorem 8 Equation (1) has no prime period two solutions if l is an even and kis an odd when α+γ 6= aβ.

Theorem 9 Equation (1) has no prime period two solutions if l is an odd and k isan even when a (β+γ) 6= 0.

8


1316 M. M. El-Dessoky 1309-1322

5 Special Cases of Equation (1)

5.1 First Equation When l = k = 1, a = 0 and α= β= γ= 1.

In this subsection we study the following special case of Eq. (1)

xn+1 =xnxn¡1

xn+xn¡1, (9)

where the initial conditions are arbitrary non zero real numbers.

Theorem 10 Let fxng1n=¡1 be a solution of Eq. (9). Then for n = 0, 1, 2, ...

xn =cb

fnb+fn+1c,

where x¡1 = c, x0 = b, ffng1n=1 = f1, 1, 2, 3, 5, 8, 13, ...g f0 = 0 and f¡1 = 1.

Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumptionholds for n ¡ 1 and n. Now, it follows

xn¡2 = cbfn¡2b+fn¡1c

and xn¡1 = cbfn¡1b+fnc

.

Now, it follows from Eq. (9) that

xn+1 = xnxn¡1xn+xn¡1 =

(cb

fnb+fn+1c

)(cb

fn¡1b+fnc

)(

cbfnb+fn+1c

)+

(cb

fn¡1b+fnc

) =(

c2b2

(fnb+fn+1c)(fn¡1b+fnc)

)(

cb(fn¡1b+fnc)+cb(fnb+fn+1c)(fnb+fn+1c)(fn¡1b+fnc)

)= c2b2

cb(fn¡1b+fnc)+cb(fnb+fn+1c)= cb

(fn¡1+fn)b+(fn+fn+1)c= cb

fn+1b+fn+2c.

Thus, the proof is completed.

5.2 Second Equation When l = k = 1, a = 0, α= β= 1 andγ= ¡1.


xn+1 =xnxn¡1

xn¡xn¡1, (10)



xn =(¡1)n+1cbfnb¡fn+1c

,

where x¡1 = c, x0 = b, and ffng1n=¡1 = f1, 0, 1, 1, 2, 3, 5, 8, 13, ...g.

9


1317 M. M. El-Dessoky 1309-1322


xn¡2 =(¡1)n¡1cb

fn¡2b¡fn¡1c and xn¡1 =(¡1)ncb

fn¡1b¡fnc.


xn+1 = xnxn¡1xn¡xn¡1 =

((¡1)n+1cbfnb¡fn+1c

)((¡1)ncb

fn¡1b¡fnc

)((¡1)n+1cbfnb¡fn+1c

)¡

((¡1)ncb

fn¡1b¡fnc

) =(

(¡1)2n+1c2b2

(fnb¡fn+1c)(fn¡1b¡fnc)

)(¡cb(fn¡1b¡fnc)¡cb(fnb¡fn+1c)

(fnb¡fn+1c)(fn¡1b¡fnc)

)= (¡1)2n+2c2b2

cb(fn¡1b¡fnc)+cb(fnb¡fn+1c)= (¡1)n+2cb

(fn¡1+fn)b¡(fn+1+fn)c= (¡1)n+2cb

fn+1b¡fn+2c.


5.3 Third Equation When l = k = 1, a = 0, α = γ= 1 andβ= ¡1.


xn+1 =xnxn¡1

¡xn+xn¡1 , (11)



x3n¡1 = (¡1)n c, x3n = (¡1)n b, and x3n+1 =(¡1)n+1bc

b¡c,

where x¡1 = c, x0 = b.


x3n¡4 = (¡1)n¡1 c, x3n¡3 = (¡1)n¡1 b, and x3n¡2 =(¡1)nbc

b¡c.


x3n+2 = x3n+1x3n

¡x3n+1+x3n=

((¡1)n+1bc

b¡c

)((¡1)nb)

¡((¡1)n+1bc

b¡c

)+(¡1)nb

=(¡1)n+1

(b2cb¡c

)( bc

b¡c+b)

= (¡1)n+1b2cb2

= (¡1)n+1 c,

x3n = x3n¡1x3n¡2¡x3n¡1+x3n¡2

=((¡1)nc)

((¡1)nbc

b¡c

)¡(¡1)nc+( (¡1)

nbcb¡c )

=(¡1)n

(bc2

b¡c

)(¡c+ bc

b¡c)= (¡1)nbc2

c2= (¡1)n b,

andx3n+4 =

x3nx3n¡1¡x3n+x3n¡1

= ((¡1)nb)((¡1)nc)¡(¡1)nb+(¡1)nc

= (¡1)nbc¡(b¡c)

= (¡1)n+1bcb¡c

.


10


1318 M. M. El-Dessoky 1309-1322

Theorem 13 Let fxng1n=¡1 be a solution of Eq. (11). Then every solution of Eq.(11) is a periodic with period six. Moreover fxng1n=¡1 takes the form form

fxng = ©c, b, ¡ bc

b¡c, ¡ c, ¡ b, bc

b¡c, c, b, ¡ bc

b¡c, ¡ c, ¡ b, bc

b¡c, ...

ª.


Example 5. Figure (5) shows the solution of Eq. (11) when the initial conditionsx¡1 = 0.3 and x0 = 0.6.

0 10 20 30 40 50 60-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

n

x(n)

plot of x(n+1)= X(n) X(n-1)/( - X(n)+ X(n-1))

Fig. 5. Draw the solution of equation (11) has a periodic with period six.

5.4 Fourth Equation When l = k = 1, a = 0, β= γ= 1 andα= ¡1.


xn+1 = ¡ xnxn¡1xn+xn¡1

, (12)



x3n¡1 = c, x3n = b, and x3n+1 = ¡ bcb+c

,



x3n¡4 = c, x3n¡3 = b, and x3n¡2 = ¡ bcb+c

.

11


1319 M. M. El-Dessoky 1309-1322


x3n+2 = ¡ x3n+1x3n

x3n+1+x3n= ¡(¡

bcb+c)(b)

(¡ bcb+c)+b

= ¡(¡ b2c

b+c

)(¡bc+b2+bc

b+c

) = b2cb2= c,

x3n = ¡ x3n¡1x3n¡2x3n¡1+x3n¡2 = ¡ (c)(¡ bc

b¡c)c+(¡bc

b¡c)=

(bc2

b¡c

)(

bc+c2¡bcb¡c

) = bc2

c2= b,

andx3n+4 =

x3nx3n¡1¡x3n+x3n¡1

= ¡ bcb+c

.


Theorem 15 Let fxng1n=¡1 be a solution of Eq. (12). Then every solution of Eq.(12) is a periodic with period three. Moreover fxng1n=¡1 takes the form form

fxng =©c, b, ¡ bc

b+c, c, b, ¡ bc

b+c, c, b, ¡ bc

b+c, ...

ª,


Example 6. The solution of Eq. (12) when the initial conditions x¡1 = 0.3 and x0 =0.6 (See Fig. 6).

0 10 20 30 40 50 60-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

n

x(n)

plot of x(n+1)= - X(n) X(n-1) / ( X(n) + X(n-1))

Fig. 6. Polt the solution of equation (12) has a periodic with period three.

AcknowledgementsThis article was funded by the Deanship of Scienti…c Research (DSR), King Ab-

dulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSRtechnical and …nancial support.

12


1320 M. M. El-Dessoky 1309-1322

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xn¡1 + xn¡2xn¡3 + a, Appl. Math. Comp., 171(2) (2005), 857-861.

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[13] Bratislav D. Iriµcanin, On a Higher-Order Nonlinear Di¤erence Equation, Abstr.Appl. Anal., Vol. 2010, (2010), Article ID 418273, 8 pages.

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[15] Mehmet G½um½us, The Periodicity of Positive Solutions of the Nonlinear Di¤erence

Equation xn+1 = α+xp

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14


1322 M. M. El-Dessoky 1309-1322

Applications of soft sets in BF -algebras

Jeong Soon Han1 and Sun Shin Ahn2,∗

1Department of Applied Mathematics, Hanyang University, Ahnsan 15588, Korea2Department of Mathematics Education, Dongguk University, Seoul 04620, Korea

Abstract. The aim of this article is to lay a foundation for providing a soft algebraic tool in considering

many problems that contain uncertainties. In order to provide these soft algebraic structures, the notion of an

intersectional soft subalgebra and an intersectional soft normal subalgebra of a BF -algebra are introduced, and

related properties are investigated. A quotient structure of a BF -algebra using an intersectional soft normal

subalgebra is constructed. The fundamental homomorphism of a quotient BF -algebra is established.

1. Introduction

The real world is inherently uncertain, imprecise and vague. Various problems in system identification involve

characteristics which are essentially non-probabilistic in nature [14]. In response to this situation Zadeh [15]

introduced fuzzy set theory as an alternative to probability theory. Uncertainty is an attribute of information.

In order to suggest a more general framework, the approach to uncertainty is outlined by Zadeh [16]. To solve

complicated problem in economics, engineering, and environment, we can’t successfully use classical methods

because of various uncertainties typical for those problems. There are three theories: theory of probability,

theory of fuzzy sets, and the interval mathematics which we can consider as mathematical tools for dealing with

uncertainties. But all these theories have their own difficulties. Uncertainties can’t be handled using traditional

mathematical tools but may be dealt with using a wide range of existing theories such as probability theory,

theory of (intuitionistic) fuzzy sets, theory of vague sets, theory of interval mathematics, and theory of rough

sets. However, all of these theories have their own difficulties which are pointed out in [11]. Maji et al. [10] and

Molodtsov [11] suggested that one reason for these difficulties may be due to the inadequacy of the parametrization

tool of the theory. To overcome these difficulties, Molodtsov [11] introduced the concept of soft set as a new

mathematical tool for dealing with uncertainties that is free from the difficulties that have troubled the usual

theoretical approaches. Molodtsov pointed out several directions for the applications of soft sets. Worldwide,

there has been a rapid growth in interest in soft set theory and its applications in recent years. Evidence of

this can be found in the increasing number of high-quality articles on soft sets and related topics that have been

published in a variety of international journals, symposia, workshops, and international conferences in recent years.

02010 Mathematics Subject Classification: 06F35; 03G25; 06D72.0Keywords: γ-inclusive set, int-soft (normal) subalgebra, BF -algebra.

∗ The corresponding author. Tel: +82 2 2260 3410, Fax: +82 2 2266 34090E-mail: [email protected] (J. S. Han); [email protected] (S. S. Ahn)


1323 Jeong Soon Han et al 1323-1331

2 Jeong Soon Han and Sun Shin Ahn

Maji et al. [10] described the application of soft set theory to a decision making problem. Maji et al. [9] also

studied several operations on the theory of soft sets. Aktas and Cagman [2] studied the basic concepts of soft set

theory, and compared soft sets to fuzzy and rough sets, providing examples to clarify their differences. They also

discussed the notion of soft groups. Jun [7] discussed the union soft sets with applications in BCK/BCI-algebras.

We refer the reader to the papers [1, 3, 5, 6, 13] for further information regarding algebraic structures/properties

of soft set theory.

In this paper, we discuss applications of the an intersectional soft sets in a (normal) subalgebra of a BF -algebra.

We introduce the notion of an intersectional (normal) soft subalgebra of a BF -algebra, and investigated related

properties. We consider a new construction of a quotient BF -algebra induced by an int-soft normal subalgebra.

Also we establish the fundamental homomorphism of a quotient BF -algebra.

2. Preliminaries

We review some definitions and properties that will be useful in our results (see [12]).

By a BF -algebra we mean an algebra (X, ∗, 0) of type (2,0) satisfying the following conditions:

(B1) x ∗ x = 0,

(B2) x ∗ 0 = x,

(B3) 0 ∗ (x ∗ y) = y ∗ x

for all x, y ∈ X.

A BF -algebra (X, ∗, 0) is called a BF1-algebra if it satisfies the following identity:

(BG) x = (x ∗ y) ∗ (0 ∗ y) for all x, y ∈ X.

A BF -algebra (X, ∗, 0) is called a BF2-algebra if it satisfies the following identity:

(BH) x ∗ y = y ∗ x = 0 imply x = y for all x, y ∈ X.

For brevity, we also call X a BF -algebra. If we can define a binary operation “ ≤ ” by x ≤ y if and only if

x ∗ y = 0. A non-empty subset A of a BF -algebra X is called a subalgebra of X if x ∗ y ∈ A for any x, y ∈ A. A

non-empty subset A of a BF -algebra X is said to be normal (or normal subalgebra) ([8]) of X if (x∗a)∗ (y ∗b) ∈ Afor any x ∗ y, a ∗ b ∈ A. Note that any normal subalgebra A of a BF -algebra X is a subalgebra of X, but the

converse need not be true. A mapping f : X → Y of BF -algebras is called a homomorphism if f(x∗y) = f(x)∗f(y)

for all x, y ∈ X.

Lemma 2.1. If X is a BF -algebra, then

(i) 0 ∗ (0 ∗ x) = x, for all x ∈ X.

(ii) 0 ∗ x = 0 ∗ y implied x = y for any x, y ∈ X.

(iii) if x ∗ y = 0, then y ∗ x = 0 for any x, y ∈ X.

Lemma 2.2. Let X be a BF -algebra and let N be a subalgebra of X. If x ∗ y ∈ N for any x, y ∈ N , then

y ∗ x ∈ N .



Applications of soft sets in BF -algebras 3

A BG-algebra (X; ∗, 0) is an algebra of type (2, 0) satisfying (B1), (B2) and (BG).

Theorem 2.3 Let X be a BF1-algebra. Then

(i) X is a BG-algebra.

(ii) x ∗ y = 0 implies x = y for any x, y ∈ X.

(iii) The right cancellation law holds in X, i.e., if x ∗ y = z ∗ y, then x = z for any x, y, z ∈ X.

(iv) The left cancellation law holds in X, i.e., if y ∗ x = y ∗ z, then x = z for any x, y, z ∈ X.

Molodtsov [11] defined the soft set in the following way: Let U be an initial universe set and let E be a

set of parameters. We say that the pair (U,E) is a soft universe. Let P(U) denotes the power set of U and

A,B,C, · · · ⊆ E.A fair (f , A) is called a soft set over U , where f is a mapping given by f : X →P(U).

In other words, a soft set over U is parameterized family of subsets of the universe U . For ε ∈ A, f(ε) may be

considered as the set of ε-approximate elements of the set (f , A). A soft set over U can bd represented by the set

of ordered pairs:

(f , A) = (x, f(x))|x ∈ A, f(x) ∈P(U),

where f : X →P(U) such that f(x) = ∅ if x /∈ A. Clearly, a soft set is not a set.

For a soft set (f , A) of X and a subset γ of U , the γ-inclusive set of (f , A), defined to be the set

iA(f ; γ) := x ∈ A|γ ⊆ f(x).

3. Intersectional soft subalgebras

In what follows let X denote a BF -algebra X unless otherwise specified.

Definition 3.1. A soft set (f , X) over U is called an intersectional soft subalgebra (briefly, int-soft subalgebra of

X if it satisfies:

(3.1) f(x) ∩ f(y) ⊆ f(x ∗ y) for all x, y ∈ X.

Proposition 3.2. Every int-soft subalgebra (f , X) of a BF -algebra X satisfies the following inclusion:

(3.2) f(x) ⊆ f(0) for all x ∈ X.

Proof. Using (3.1) and (B1), we have f(x) = f(x) ∩ f(x) ⊆ f(x ∗ x) = f(0) for all x ∈ X.

Example 3.3. Let (U = Z, X) where X = 0, 1, 2, 3 is a BF -algebra ([12]) with the following Cayley table:

∗ 0 1 2 3

0 0 1 2 3

1 1 0 1 1

2 2 1 0 1

3 3 1 1 0

Let (f , X) be a soft set over U defined as follows:




f : X →P(U), x 7→

Z if x = 0

2Z if x ∈ 1, 23Z if x = 3.

It is easy to check that (f , X) is an int-soft subalgebra over U .

Theorem 3.4. A soft set (f , X) of a BF -algebra X over U is an int-soft subalgebra of X over U if and only if

the γ-inclusive set iX(f ; γ) is a subalgebra of X for all γ ∈P(U) with iX(f ; γ) = ∅.

The subalgebra iX(f ; γ) in Theorem 3.4 is called the inclusive subalgebra of X.

Proof. Assume that (f , X) is an int-soft subalgebra over U . Let x, y ∈ X and γ ∈ P(U) be such that x, y ∈iX(f ; γ). Then γ ⊆ f(x) and γ ⊆ f(y). It follows from (3.1) that γ ⊆ f(x)∩ f(y) ⊆ f(x∗y) Hence x∗y ∈ iX(f ; γ).

Thus iX(f , X) is a subalgebra of X.

Conversely, suppose that iX(f ; γ) is a subalgebra X for all γ ∈ P(U) with iX(; γ) = ∅. Let x, y ∈ X, be such

that f(x) = γx and f(y) = γy. Take γ = γx ∩ γy. Then x, y ∈ iX(f ; γ) and so x ∗ y ∈ iX(f ; γ) by assumption.

Hence f(x) ∩ f(y) = γx ∩ γy = γ ⊆ f(x ∗ y). Thus (f , X) is an int-soft subalgebra over U .

Theorem 3.5. Every subalgebra of a BF -algebra can be represented as a γ-inclusive set of an int-soft subalgebra.

Proof. Let A be a subalgebra of a BF -algebra X. For a subset γ of U , define a soft set (f , X) over U by

f : X →P(U), x 7→γ if x ∈ A∅ if x /∈ A

Obviously, A = iX(f ; γ). We now prove that (f ; γ) is an int-soft subalgebra over U . Let x, y ∈ X. If x, y ∈ A,

then x ∗ y ∈ A because A is a subalgebra of X. Hence f(x) = f(y) = f(x ∗ y) = γ, and so f(x) ∩ f(y) ⊆ f(x ∗ y).

If x ∈ A and y /∈ A, then f(x) = γ and f(y) = ∅ which imply that f(x) ∩ f(y) = γ ∩ ∅ = ∅ ⊆ f(x ∗ y). Similarly,

if x /∈ A and y ∈ A, then f(x) ∩ f(y) ⊆ f(x ∗ y). Obviously, if x /∈ A and y /∈ A, then f(x) ∩ f(y) ⊆ f(x ∗ y).

Therefore (f , X) is an int-soft subalgebra over U .

Any subalgebra of a BF -algebra X may not be represented as a γ-inclusive set of an int-soft subalgebra (f , X)

over U in general (see the following example).

Example 3.6. Let E = X be the set of parameters, and let U = X be the initial universe set where where

X = 0, 1, 2, 3 is a BF -algebra ([12]) with the following Cayley table:

∗ 0 1 2 3

0 0 1 2 3

1 1 0 3 0

2 2 3 0 2

3 3 0 2 0

Consider a soft set (f , X) which is given by




f : X →P(U), x 7→0, 3 if x = 0

3 if x ∈ 1, 2, 3

Then (f , X) is an int-soft subalgebra over U . The γ-inclusive set of (f , X) are described as follows:

iX(f ; γ) =

X if γ ∈ ∅, 30 if γ ∈ 0, 0, 3∅ otherwise.

The subalgebra 0, 2 cannot be a γ-inclusive set iX(f ; γ) since there is no γ ⊆ U such that iX(f ; γ) = 0, 2.

We make a new int-soft subalgebra from old one.

Theorem 3.7. Let (f , X) be a soft set of a BF -algebra X over U . Define a soft set (f∗, X) of X over U by

f∗ : X →P(U), x 7→f(x) if x ∈ iX(f ; γ)

∅ otherwise

where γ is a non-empty subset subset of U . If (f , X) is an int-soft subalgebra of X, then so is (f∗, X).

Proof. If (f , X) is an int-soft subalgebra over U , then iX(f ; γ) is a subalgebra of X for all γ ⊆ U by Theorem 3.6.

Let x, y ∈ X. If x, y ∈ iX(f ; γ), then x ∗ y ∈ iX(f ; γ). Hence we have

f∗(x) ∩ f∗(y) = f(x) ∩ f(y) ⊆ f(x ∗ y) = f∗(x ∗ y).

If x /∈ iX(f ; γ) or y /∈ iX(f ; γ), then f∗(x) = ∅ or f∗(y) = ∅. Thus

f∗(x) ∩ f∗(y) = ∅ ⊆ f∗(x) ∗ f∗(y).

Therefore (f∗, X) is an int-soft subalgebra over U .

Definition 3.8. A soft set (f , X) over U is called an intersectional soft normal subalgebra (briefly, int-soft normal

subalgebra of X if it satisfies:

(3.3) f(x ∗ y) ∩ f(a ∗ b) ⊆ f((x ∗ a) ∗ (y ∗ b)) for all x, y, a, b ∈ X.

Proposition 3.9. Every int-soft subalgebra (f , X) of a BF -algebra X satisfies the following inclusion:

(3.4) f(x ∗ y) ⊆ f(y ∗ x) for all x, y ∈ X.

Proof. Using (B3), (3.1) and (3.2), we have

f(y ∗ x) = f(0 ∗ (x ∗ y)) ⊇ f(0) ∩ f(x ∗ y) = f(x ∗ y), ∀x, y ∈ X.

Proposition 3.10. Every int-soft normal subalgebra (f , X) of a BF -algebra X is an int-soft subalgebra of X.

Proof. Put y := 0, b := 0 and a := y in (3.3). Then f(x ∗ 0)∩ f(y ∗ 0) ⊆ f((x ∗ y) ∗ (0 ∗ 0)) for any x, y ∈ X. Using

(B2) and (B1), we have f(x) ∩ f(y) ⊆ f(x ∗ y). Hence (f , X) is an int-soft subalgebra of X.

The converse of Proposition 3.10 may not be true in general (see Example 3.11).




Example 3.11 Let E = X be the set of parameters where where X = 0, 1, 2, 3 is a BF -algebra with the

following Cayley table:

∗ 0 1 2 3

0 0 2 1 3

1 1 0 1 1

2 2 2 0 2

3 3 2 1 0

Let (f , X) be a soft set over U defined as follows:

f : X →P(U), x 7→

γ3 if x = 0

γ2 if x = 3

γ1 if x ∈ 1, 2.

where γ1, γ2 and γ3 are subsets of U with γ1 ⊊ γ2 ⊊ γ3. It is easy to check that (f , X) is an int-soft normal

subalgebra over U .

Let (g, X) be a soft set over U defined as follows:

g : X →P(U), x 7→

α3 if x = 0

α2 if x ∈ 1, 2α1 if x = 3.

where α1, α2 and α3 are subsets of U with α1 ⊊ α2 ⊊ α3. It is easy to check that (f , X) is an int-soft subalgebra

over U . But it is not an int-soft normal subalgebra over U since g(2 ∗ 3) ∩ g(2 ∗ 0) = g(2) ∩ g(2) = α2 ⊈ α1 =

g(3) = g((2 ∗ 2) ∗ (3 ∗ 0)).

Theorem 3.12. A soft set (f , X) of X over U is an int-soft normal subalgebra of X over U if and only if the

γ-inclusive set iX(f ; γ) is a normal subalgebra of X for all γ ∈P(U) with iX(f ; γ) = ∅.

Proof. Similar to Theorem 3.4.

The normal subalgebra iX(f ; γ) in Theorem 3.12 is called the inclusive normal subalgebra of X.

4. Quotient BF -algebras induces by soft sets

Let (f , X) be an int-soft normal subalgebra of a BF -algebra X. For any x, y ∈ X, we define a binary operation

“ ∼f ” on X as follows:

x ∼f y ⇔ f(x ∗ y) = f(0).

Lemma 4.1. The operation ∼f is an equivalence relation on a BF -algebra X.

Proof. Obviously, it is reflexive. Let x ∼f y. Then f(x ∗ y) = f(0). It follows from (3.4) and (3.2) that

f(0) = f(x ∗ y) ⊆ f(y ∗ x) ⊆ f(0). Hence f(y ∗ x) = f(0). Hence ∼f is symmetric. Let x, y, z ∈ X be such that




x ∼f y and y ∼f z. Then f(x ∗ y) = f(0) and f(y ∗ z) = f(0). Using (3.4), (3.3), (B1), (B2) and (3.2), we have

f(0) = f(x ∗ y)∩f(y ∗ z) ⊆ f(x ∗ y) ∩ f(z ∗ y)

⊆f((x ∗ z) ∗ (y ∗ y))

=f((x ∗ z) ∗ 0) = f(x ∗ z) ⊆ f(0).

Hence f(x ∗ z) = f(0), i.e., ∼f is transitive. Therefore “ ∼f ” is an equivalence relation on X.

Lemma 4.2. For any x, y, p, q ∈ X, if x ∼f y and p ∼f q, then x ∗ p ∼f y ∗ q.

Proof. Let x, y, p, q ∈ X be such that x ∼f y and p ∼f q. Then f(x∗y) = f(y ∗x) = f(0) and f(p∗q) = f(q ∗p) =

f(0). Using (3.3) and (3.2), we have

f(0) =f(x ∗ y) ∩ f(p ∗ q)

⊆f((x ∗ p) ∗ (y ∗ q)) ⊆ f(0).

Hence f((x ∗ p) ∗ (y ∗ q)) = f(0). By similar way, we get f((y ∗ q) ∗ (x ∗ p)) = f(0). Therefore x ∗ p ∼f y ∗ q. Thus

“ ∼f ” is a congruence relation on X.

Denote fx and X/f the set of all equivalence classes containing x and the set of all equivalence classes of X,

respectively, i.e.,

fx := y ∈ X|y ∼f x and X/f := fx|x ∈ X.

Define a binary relation • on X/f as follows:

fx • fy = fx∗y

for all fx, fy ∈ X/f . Then this operation is well-defined by Lemma 4.2.

Theorem 4.3. If (f , X) is an int-soft normal subalgebra of a BF -algebraX, then the quotientX/f := (X/f, •, f0)

is a BF -algebra.

Proof. Let fx, fy, fz ∈ X/f. Then we have fx • fx = fx∗x = f0, fx • f0 = fx∗0 = fx, f0 • (fx • fy) = f0∗(x∗y) =

fy∗x = fy • fx. Therefore X/f = (X/f, •, f0) is a BF -algebra.

Corollary 4.4. If (f , X) is an int-soft normal subalgebra of a BF2-algebra X, then the quotient X/f :=

(X/f, •, f0) is a BF2-algebra.

Proof. It is enough to show that X/f satisfies (BH). If fx • fy = f0 and fy • fx = f0 for any fx, fy ∈ X/f , then

fx∗y = f0 = fy∗x. Hence f(x ∗ y) = f(0) = f(y ∗ x) and so x ∼f y. Hence fx = fy. Therefore X/f = (X/f, •, f0)

is a BF2-algebra.

Proposition 4.5. Let µ : X → Y be a homomorphism of BF -algebras. If (f , Y ) is an int-soft normal subalgebra

of Y , then (f µ,X) is an int-soft normal subalgebra of X.




Proof. For any x, y, a, b ∈ X, we have

(f µ)((x ∗ a) ∗ (y ∗ b)) =f(µ((x ∗ a) ∗ (y ∗ b))

=f((µ(x) ∗ µ(a)) ∗ (µ(y) ∗ µ(b)))

⊇f(µ(x) ∗ µ(y)) ∩ f(µ(a) ∗ µ(b))

=f(µ(x ∗ y)) ∩ f(µ(a ∗ b))

=(f µ)(x ∗ y) ∩ (f µ)(a ∗ b).

Hence f µ is an int-soft normal subalgebra. Therefore (f µ,X) is an int-soft normal subalgebra of X.

Theorem 4.6. Let X := (X; ∗X , 0X) be a BF -algebra and Y := (Y ; ∗Y , 0Y ) be a BF2-algebra and let µ : X → Y

be an epimorphism. If (f , Y ) is an int-soft normal subalgebra of Y , then the quotient algebra X/(f µ) :=

(X/(f µ), •X , (f µ)0X ) is isomorphic to the quotient algebra Y/f := (Y/f , •Y , f0Y ).

Proof. By Theorem 4.3, Corollary 4.4, and Proposition 4.5, X/f µ : (X/(f µ), •X , (f µ)0X ) is a BF -algebra

and Y/f := (Y/f , •Y , f0Y ) is a BF2-algebra. Define a map

η : X/(f µ)→ Y/f , (f µ)x 7→ fµ(x)

for all x ∈ X. Then the function η is well-defined. In fact, assume that (f µ)x = (f µ)y for all x, y ∈ X. Then

we have

f(µ(x) ∗Y µ(y)) =f(µ(x ∗X y)) = (f µ)(x ∗X y)

=(f µ)(0X) = f(µ(0X)) = f(0Y )

and

f(µ(y) ∗Y µ(x)) =f(µ(y ∗X x)) = (f µ)(y ∗X x)

=(f µ)(0X) = f(µ(0X)) = f(0Y ).

Hence fµ(x) = fµ(y).

For any (f µ)x, (f µ)X ∈ X/(f µ), we have

η((f µ)x •X (f µ)y) =η((f µ)x∗y) = fµ(x∗Xy)

=fµ(x)∗Y µ(y) = fµ(x) • fµ(y)

=η((f µ)x) •Y η((f µ)y)).

Therefore η is a homomorphism.

Let fa ∈ Y/f . Then there exists x ∈ X such that µ(x) = a since µ is surjective. Hence η((f µ)X) = fµ(x) = fa

and so η is surjective.

Let x, y ∈ X be such that fµ(x) = fµ(y). Then we have

(f µ)(x ∗X y) =f(µ(x ∗X y)) = f(µ(x) ∗Y µ(y))

=f(0Y ) = f(µ(0X)) = (f µ)(0X)




and

(f µ)(y ∗X x) =f(µ(y ∗X x)) = f(µ(y) ∗Y µ(x))

=f(0Y ) = f(µ(0X)) = (f µ)(0X).

It follows that (f µ)X = (f µ)y. Thus η is injective. This completes.

The homomorphism π : X → X/f , x→ fX , is called the natural homomorphism of X onto X/f . In Theorem

4.6, if we define natural homomorphisms πX : X → X/f µ and πY : Y → Y/f then it is easy to show that

η πX = πY µ, i.e., the following diagram commutes.

Xµ−−−−→ Y

πX

y πY

yX/(f µ)

η−−−−→ Y/f .

References

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592-601.

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Math. Appl. 44 (2002) 1077-1083.

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Symmetric solutions for hybrid fractional differential equations

Jessada Tariboona,∗, Sotiris K. Ntouyasb,c, and Suthep Suantaid

a Nonlinear Dynamic Analysis Research Center, Department of Mathematics,Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok,Bangkok, ThailandE-mail: [email protected]

b Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece

c Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group,Department of Mathematics, Faculty of Science, King Abdulaziz University,P.O. Box 80203, Jeddah 21589, Saudi ArabiaE-mail: [email protected]

d Department of Mathematics, Faculty of Science, Chiang Mai University,Chiang Mai, 50200 ThailandE-mail: [email protected]

Abstract

In this paper we introduce a new class of symmetric functions and study the existence of symmet-ric solutions for hybrid Caputo fractional differential equations. A fixed point theorem in Banachalgebra for two operators is used. An example is presented to illustrate our result.

Keywords: Caputo fractional derivative; hybrid fractional differential equation; symmetric solution;fixed point theorem2010 Mathematics Subject Classifications: 34A08; 34A12.

1 Introduction

The aim of this manuscript is to study the existence at least one symmetric solution for hybrid Caputofractional differential equation subject to initial and symmetric conditions Dα

[x(t)

f(t, x(t))

]+ g(t, x(t)) = 0, t ∈ J := [0, T ],

x(0) = β, x(t) = x(T − t),(1.1)

where Dα denotes the Caputo fractional derivative of order α, 1 < α ≤ 2, f ∈ C(J × R, R \ 0),g ∈ C(J × R, R), β ∈ R. A function x ∈ C([0, T ], R) satisfying the relation x(t) = x(T − t), t ∈ [0, T ],is called symmetric on [0, T ].

Fractional differential equations have been of great interest recently. It is caused both by the intensivedevelopment of the theory of fractional calculus itself and by the applications of such constructions invarious science such as physics, mechanics, chemistry, and engineering. There have appeared lots ofworks, in which fractional derivatives are used for a better description of considered material properties.For details, and some recent results on the subject we refer to [1]-[17] and references cited therein.

Recently, many authors have focused on the existence of symmetric solutions for ordinary differen-tial equation boundary value problems; for example, see [18]-[21] and the references therein. In [22]

∗Corresponding author


1332 Jessada Tariboon et al 1332-1342

J. TARIBOON , S. K. NTOUYAS AND S. SUANTAI

the existence and uniqueness of symmetric solutions for a boundary value problem for nonlinear frac-tional differential equations with multi-order fractional integral boundary conditions was studied, byusing a variety of fixed point theorems (such as Banach contraction principle, nonlinear contractions,Krasnoselskii fixed point theorem and Leray-Schauder nonlinear alternative).

Hybrid fractional differential equations have also been studied by several researchers. This classof equations involves the fractional derivative of an unknown function hybrid with the nonlinearitydepending on it. Some recent results on hybrid differential equations can be found in a series of papers([23]-[28]).

In this paper we prove the existence of symmetric solutions for the hybrid Caputo fractional boundaryvalue problem (1.1). One new result is proved by using a hybrid fixed point theorem for two operatorsin a Banach algebra due to Dhage [29].

The rest of this paper is organized as follows: In Section 2 we present some preliminary notations,definitions and lemmas that we need in the sequel. Also we introduce a new class of symmetric functionsand prove some interesting properties, which are used to establish the Green function. In Section 3we establish the existence of symmetric solutions for the boundary value problem (1.1). An exampleillustrating the obtained result is also presented.

2 Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [1, 2] and presentpreliminary results needed in our proofs later. In addition, a new definition of α-symmetric function ispresented and also some properties are proved.

Definition 2.1 The Riemann-Liouville fractional integral of order α > 0 of a function g : (0,∞) → Ris defined by

Iαg(t) =∫ t

0

(t − s)α−1

Γ(α)g(s)ds,

provided the right-hand side is point-wise defined on (0,∞), where Γ is the Gamma function.

Definition 2.2 The Caputo fractional derivative of order α > 0 for an at least n-times differentiablefunction g : (0,∞) → R is defined by

Dαg(t) =1

Γ(n − α)

∫ t

0

g(n)(s)(t − s)α−n+1

ds, n − 1 < α < n,

where n = [α] + 1, [α] denotes the integer part of real number α.

From the definition of the Caputo fractional derivative, we can obtain the following lemmas.

Lemma 2.3 (see [1]) Let α > 0, the general solution of the fractional differential equation Dαy(t) = 0is given by

y(t) = c0 + c1t + · · · + cn−1tn−1,

where ci ∈ R, i = 0, 1, 2, . . . , n − 1, n = [α] + 1.

Definition 2.4 A function y ∈ C2(J, R) is called symmetric, if it satisfies the relation y(t) = y(T − t).

From Definition 2.4 we have y′(t) = −y′(T − t), y′′(t) = y′′(T − t) and∫ T−t

0

y(s)ds =∫ T

0

y(s)ds −∫ t

0

y(s)ds. (2.1)

Lemma 2.5 Let f ∈ L2(J, R) be symmetric function. Then we have

I1f(T ) =2T

I2f(T ). (2.2)



SYMMETRIC SOLUTIONS FOR HYBRID FRACTIONAL DIFFERENTIAL EQUATIONS

Proof. Since f is symmetric on [0, T ], we have

I1f(T ) =∫ T

0

f(s)ds =1T

∫ T

0

(T − s + s)f(s)ds

=1T

∫ T

0

(T − s)f(s)ds +1T

∫ T

0

sf(s)ds

=2T

∫ T

0

(T − s)f(s)ds =2T

I2f(T ).

Therefore, (2.2) holds. 2

Now, we define a new class of symmetric functions as follows:

Definition 2.6 A function f ∈ C1(J, R) is called α-symmetric if D2−αf(t) is symmetric function on[0, T ], where 1 < α ≤ 2.

Example 2.7 Let f : [0, 1] → R be defined as

f(t) =4

3√

πt

32

(1 − 4

5t

).

It easy to verify that

D2− 32 f(t) = D

12 f(t)

=4

3√

πD

12 t

32 − 16

15√

πD

12 t

52

= t(1 − t).

Therefore, f is 32 -symmetric function.

Remark 2.8 If α = 2, then the class of α-symmetric functions is reduced to the class of usual sym-metric functions.

Lemma 2.9 Let z ∈ C1(J, R) be an α-symmetric function. Then the symmetric solution of linearfractional differential equation

Dαy(t) = z(t), 1 < α ≤ 2, t ∈ J, (2.3)y(t) = y(T − t), (2.4)

is given by

y(t) = Iαz(t) − t

TIαz(T ) + c0, (2.5)

where c0 ∈ R.

Proof. By Lemma 2.3, we havey(t) = Iαz(t) + c1t + c0, (2.6)

where c0, c1 ∈ R. We apply symmetric condition to obtain

Iαz(t) + c1t + c0 = Iαz(T − t) + c1(T − t) + c0. (2.7)

Evidently, (2.7) becomes

c1(2t − T ) = Iαz(T − t) − Iαz(t)

=∫ T−t

0

(T − t − s)α−1

Γ(α)z(s)ds −

∫ t

0

(t − s)α−1

Γ(α)z(s)ds. (2.8)




Taking the first-order usual derivative with respect to t in (2.8), we get

2c1 = −Iα−1z(T − t) − Iα−1z(t)= −I1

(D2−αz

)(T − t) − I1

(D2−αz

)(t).

Since D2−αz(t) is symmetric on J , and z is symmetric, by (2.1), we have

I1(D2−αz

)(T − t) = I1

(D2−αz

)(T ) − I1

(D2−αz

)(t),

which leads to

2c1 = −I1(D2−αz

)(T )

= − 2T

I2(D2−αz

)(T ),

by using Lemma 2.5.Therefor, we obtain the constant c1 as

c1 = − 1T

Iαz(T ) = − 1T

∫ T

0

(T − s)α−1

Γ(α)z(s)ds.

Substituting the constant c1 in (2.6), we get the result in (2.5) as desired. 2

In the following we present the Green function of the hybrid fractional boundary value problem(1.1).

Lemma 2.10 Let h ∈ C1(J, R) be the α-symmetric function and f ∈ C(J × R, R \ 0). Then theunique solution of

Dα

[x(t)

f(t, x(t))

]+ h(t) = 0, t ∈ J, (2.9)

x(0) = β, x(t) = x(T − t), (2.10)

is given by

x(t) = f(t, x(t))

(∫ T

0

G(t, s)h(s)ds +β

f(0, β)

), (2.11)

where

G(t, s) =

t (T − s)α−1 − T (t − s)α−1

TΓ(α), 0 ≤ s ≤ t ≤ T,

t (T − s)α−1

TΓ(α), 0 ≤ t ≤ s ≤ T.

(2.12)

Proof. Applying Lemma 2.9, the equation (2.9) can be written as

x(t)f(t, x(t))

= −Iαh(t) +t

TIαh(T ) + c0, (2.13)

where c0 ∈ R. The condition x(0) = 0 implies that

c0 =β

f(0, β).

Therefore, the unique solution of problem (2.9)-(2.10) is

x(t) = f(t, x(t))(− 1

Γ(α)

∫ t

0

(t − s)α−1h(s)ds




+t

TΓ(α)

∫ T

0

(T − s)α−1h(s)ds

)+

β

f(0, β)f(t, x(t))

= f(t, x(t))∫ T

0

G(t, s)h(s)ds +β

f(0, β)f(t, x(t)).

The proof is completed. 2

Remark 2.11 The Green’s function G(t, s) defined by (2.12), is not positive for all t, s ∈ J . Forexample, if T = 5, t = 2, s = 1 and α = 3/2, then we have G(2, 1) = −2/(5

√π).

Lemma 2.12 The Green’s function G(t, s) in (2.12) satisfies the following inequalities

G(t, s) ≤ G(s, s) ≤ ((α − 1)T )α−1

αα−1Γ(α + 1)for all s, t ∈ J. (2.14)

Proof. Let us define two functions by

g1(t, s) = t (T − s)α−1 − T (t − s)α−1, 0 ≤ s ≤ t ≤ T,

andg2(t, s) = t (T − s)α−1

, 0 ≤ t ≤ s ≤ T.

Obviously, for 0 ≤ t ≤ s ≤ T , the function g2(t, s) satisfies

g2(t, s) ≤ g2(s, s) = s(T − s)α−1.

Let s ∈ [0, T ) be fixed. Differentiating with respect to t the function g1(t, s), we have

∂

∂tg1(t, s) = (T − s)α−1 − (α − 1)T (t − s)α−2

, s < t.

We can find that ∂g1/∂t = 0 if and only if

t = t∗ = s +(T − s)

α−1α−2

((α − 1)T )1

α−2.

It follows from ∂g1/∂t > 0 on (0, t∗) and ∂g1/∂t < 0 on (t∗, T ) that

g1(t, s) ≤ g1(t∗, s).

Simplifying the above inequality, we get

g1(t, s) ≤ g1(t∗, s)

= s(T − s)α−1 − (2 − α)T · (T − s)(α−1)2

α−2

((α − 1)T )α−1α−2

≤ s(T − s)α−1 = g1(s, s),

which implies the first inequality.Next, we will prove the second inequality. Taking the first derivative for g2(s, s) with respect to s on[0, T ), we have

g′2(s, s) = (T − s)α−2(T − αs).

Thus g′2(s, s) has a unique zero at the point s = s∗ = T/α such that s∗ ∈ (0, T ). Observe thatg′2(s, s) > 0 on (0, s∗) and g′2(s, s) < 0 on (s∗, T ). Hence

g2(s, s) ≤ g2

(T

α,T

α

)=

(α − 1)α−1

ααTα.




Then the second inequality is proved. 2

Let E = C([0, T ], R) be the Banach space endowed with the supremum norm ‖ · ‖. Define amultiplication in E by

(xy)(t) = x(t)y(t), ∀t ∈ J.

Clearly E is a Banach algebra with respect to above supremum norm and the multiplication in it. Themain result is based on the following fixed point theorem for two operators in Banach algebra due toDhage [29].

Lemma 2.13 Let S be a non-empty, closed convex and bounded subset of the Banach algebra E, letA : E → E and B : S → E be two operators such that:

(a) A is Lipschitzian with a Lipschitz constant δ,

(b) B is completely continuous,

(c) x = AxBy ⇒ x ∈ S for all y ∈ S, and

(d) Mδ < 1, where M = ‖B(S)‖ = sup‖B(x)‖ : x ∈ S.

Then the operator equation x = AxBx has a solution in S.

3 Main Result

Now, we are in the position to prove the existence of symmetric solutions for hybrid fractional problem(1.1).

Theorem 3.1 Assume that the following conditions are satisfied:

(H1) The functions f ∈ C(J × R, R \ 0) and g ∈ C1(J × R, R) are symmetric and α-symmetric onJ , respectively.

(H2) There exists a bounded function φ(t), with bound ‖φ‖, such that

|f(t, x) − f(t, y)| ≤ ‖φ‖ · |x − y|

for t ∈ J and x, y ∈ R.

(H3) There exist a function p ∈ C(J, R+) and a continuous nondecreasing function Ψ : [0,∞) → (0,∞)such that

|g(t, x)| ≤ p(t)Ψ(|x|), (t, x) ∈ J × R.

(H4) There exist a number r > 0 such that

r ≥F0

[(α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

]

1 − ‖φ‖

[(α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

] (3.1)

where F0 = supt∈J |f(t, 0)| and

‖φ‖

[(α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

]< 1. (3.2)

Then the problem (1.1) has at least one symmetric solution on J .




Proof. To prove our main result, we first define a subset S of E by

S = x ∈ E : ‖x‖ ≤ r,

where r satisfies (3.1). Clearly S is closed, convex and bounded subset of the Banach space E. ByLemma 2.10, we define two operators A : E → E by

Ax(t) = f(t, x(t)), t ∈ J, (3.3)

and

Bx(t) =∫ T

0

G(t, s)g(s, x(s))ds +β

f(0, β), t ∈ J. (3.4)

Hence, the problem (1.1) is transformed into an operator equation as

x = AxBx. (3.5)

Next, we shall show that the operators A and B satisfy all the conditions of Lemma 2.13 under ourassumptions. This will be achieved in the series of following steps.

Step 1. We first show that A is Lipschitzian on E.

Let x, y ∈ E. Then by (H2), for t ∈ J we have

|Ax(t) −Ay(t)| = |f(t, x(t)) − f(t, y(t))|≤ φ(t)|x(t) − y(t)|≤ ‖φ‖‖x − y‖,

which implies that ‖Ax−Ay‖ ≤ ‖φ‖‖x− y‖ for all x, y ∈ E. Therefore, A is a Lipschitzian on E withLipschitz constant δ = ‖φ‖.

Step 2. The operator B is completely continuous on S.

We first show that the operator B is continuous on S. Let xn be a sequence in S converging to apoint x ∈ S. Then by Lebesgue dominated convergence theorem, for all t ∈ J , we have

limn→∞

Bxn(t) = limn→∞

∫ T

0

G(t, s)g(s, xn(s))ds +β

f(0, β)

=∫ T

0

G(t, s) limn→∞

g(s, xn(s))ds +β

f(0, β)

=∫ T

0


f(0, β)= Bx(t).

This shows that Bxn converges to Bx pointwise on J .Next, we will show that Bxn is an equicontinuous sequence of functions in S. Let τ1, τ2 ∈ J be

arbitrary with τ1 < τ2. Then

|Bxn(τ2) − Bxn(τ1)| =

∣∣∣∣∣∫ T

0

G(τ2, s)g(s, xn(s))ds −∫ T

0

G(τ1, s)g(s, xn(s))ds

∣∣∣∣∣≤ ‖p‖Ψ(r)

∣∣∣∣∣∫ T

0

G(τ2, s)ds −∫ T

0

G(τ1, s)ds

∣∣∣∣∣≤ ‖p‖Ψ(r)

∣∣∣∣∣∫ τ2

0

τ2(T − s)α−1 − T (τ2 − s)α−1

TΓ(α)ds




+∫ T

τ2

τ2(T − s)α−1

TΓ(α)ds −

∫ T

τ1

τ1(T − s)α−1

TΓ(α)ds

−∫ τ1

0

τ1(T − s)α−1 − T (τ1 − s)α−1

TΓ(α)ds

∣∣∣∣∣= ‖p‖Ψ(r)

∣∣∣∣∣∫ T

0

τ2(T − s)α−1

TΓ(α)ds −

∫ τ2

0

(τ2 − s)α−1

Γ(α)ds

−∫ T

0

τ1(T − s)α−1

TΓ(α)ds +

∫ τ1

0

(τ1 − s)α−1

Γ(α)ds

∣∣∣∣∣≤ ‖p‖Ψ(r)

∫ T

0

(τ2 − τ1)(T − s)α−1

TΓ(α)ds

+ ‖p‖Ψ(r)∫ τ1

0

(τ2 − s)α−1 − (τ1 − s)α−1

Γ(α)ds

+ ‖p‖Ψ(r)∫ τ2

τ1

(τ2 − s)α−1

Γ(α)ds.

Consequently|Bxn(τ2) − Bxn(τ1)| → 0 as τ2 → τ1

uniformly for all n ∈ N. This shows that the convergence Bxn → Bx is uniformly and hence B is acontinuous operator on S.

Now we will prove that the set B(S) is a uniformly bounded in S. For any x ∈ S and using Lemma2.12, we have

|Bx(t)| =

∣∣∣∣∣∫ T

0


f(0, β)

∣∣∣∣∣≤

∫ T

0

|G(t, s)|p(s)Ψ(r)ds +|β|

|f(0, β)|

≤ (α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

:= K1,

for all t ∈ J . Therefore, ‖Bx‖ ≤ K1 which shows that B is uniformly bounded on S.Next, we will show that B(S) is an equicontinuous set in E. Let τ1, τ2 ∈ J with τ1 < τ2 and x ∈ S.

Then, as above, we have

|Bx(τ2) − Bx(τ1)| =

∣∣∣∣∣∫ T

0

G(τ2, s)g(s, x(s))ds −∫ T

0

G(τ1, s)g(s, x(s))ds

∣∣∣∣∣≤ ‖p‖Ψ(r)

∣∣∣∣∣∫ T

0

G(τ2, s)ds −∫ T

0

G(τ1, s)ds

∣∣∣∣∣≤ ‖p‖Ψ(r)

∫ T

0

(τ2 − τ1)(T − s)α−1

TΓ(α)ds

+ ‖p‖Ψ(r)∫ τ1

0

(τ2 − s)α−1 − (τ1 − s)α−1

Γ(α)ds

+ ‖p‖Ψ(r)∫ τ2

τ1

(τ2 − s)α−1

Γ(α)ds,

which is independent of x ∈ S. As τ1 → τ2, the right-hand side of the above inequality tends to zero.Therefore, it follows from the Arzela-Ascoli theorem that B is a completely continuous operator on S.

Step 3. The hypothesis (c) of Lemma 2.13 is satisfied.




Let x ∈ E and y ∈ S be arbitrary elements such that x = AxBy. Then we have

|x(t)| ≤ |Ax(t)||By(t)|

= |f(t, x(t))|

∣∣∣∣∣∫ T

0

G(t, s)g(s, y(s))ds +β

f(0, β)

∣∣∣∣∣≤ (|f(t, x(t)) − f(t, 0)| + |f(t, 0)|)

((α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

)≤ (|x(t)| · ‖φ‖ + F0)

((α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

),

which leads to

‖x‖ ≤ (‖x‖ · ‖φ‖ + F0)(

(α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

)≤ r.

Therefore, x ∈ S.

Step 4. Finally we show that δM < 1, that is (d) of Lemma 2.13 holds.

Since

M = ‖B(S)‖

= supx∈S

supt∈J

|Bx(t)|

≤ (α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

, (3.6)

by (3.2) we have

δ‖M ≤ ‖φ‖(

(α − 1)α−1Tα

αα−1Γ(α + 1)‖p‖Ψ(r) +

|β||f(0, β)|

)< 1,

with δ = ‖φ‖.Thus all the conditions of Lemma 2.13 are satisfied and hence the operator equation x = AxBx has

a solution in S. In consequence, the problem (1.1) has a symmetric solution on J . This completes theproof. 2

Next, we present an example to illustrate our result.

Example 3.2 Consider the following hybrid fractional differential equation with initial and symmetricconditions

D32

x(t)x2(t) + 2|x(t)|

2(5 + (t − 1)2)(|x(t)| + 1)+

12

+124

(1 + sin2

(8t

32

3√

π

(1 − 2

5t

)))

×(

x2(t)1 + |x(t)|

+ 1)

= 0, t ∈ [0, 2],

x(0) =13, x(t) = x(2 − t).

(3.7)

Here α = 3/2, T = 2 and β = 1/3. Since (t − 1)2 is symmetric on [0, 2] and (8t32 )(1 − (2/5)t)/(3

√π) is

3/2-symmetric by

D12

(8t

32

3√

π

(1 − 2

5t

))= t(2 − t), t ∈ [0, 2],




then we get that f(t, ·) and g(t, ·) are symmetric and 3/2-symmetric functions on [0, 2], respectively.With the above information, we find that

|f(t, x) − f(t, y)| =∣∣∣∣ x2 + 2|x|2(5 + (t − 1)2)(|x| + 1)

− y2 + 2|y|2(5 + (t − 1)2)(|y| + 1)

∣∣∣∣≤ 1

5 + (t − 1)2|x − y|,

and

|g(t, x)| ≤ 112

(|x| + 1),

and F0 = supt∈[0,2] |f(t, 0)| = 1/2. Choosing φ(t) = 1/(5+(t−1)2), p(t) = 1/12, we have ‖φ‖ = 1/5 and‖p‖ = 1/12. Setting Ψ(|x|) = |x| + 1, we can find that there exists 0.06962115393 < r < 45.01973321which is satisfied (3.1)-(3.2). Thus all the conditions of Theorem 3.1 are satisfied. Therefore, by theconclusion of Theorem 3.1, the problem (3.7) has at least one symmetric solution on [0, 2].

Acknowledgements

This paper was supported by the Thailand Research Fund under the project RTA5780007.

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ON THE k-TH DEGENERATION OF THE GENOCCHI POLYNOMIALS

LEE-CHAE JANG, C.S. RYOO, JEONG GON LEE, HYUCK IN KWON

Graduate School of Education,Konkuk University, Seoul 143-701, Republic of Korea

E-mail : [email protected]

Department of Mathematics,Hannam University, Daejeon 306-791, Republic of Korea


Division of Mathematics and Informational Statistics,and Nanoscale Science and Technology Institute,

Wonkwang University, Iksan 570-749, Republic of KoreaE-mail : [email protected]

Department of Mathematics,Kwangwoon University, Seoul 139-50, Republic of Korea


Abstract. Jeong-Rim-Kim(2015) studied the degenerate Cauchy numbers and polyno-

mials and investigated some properties of these k-times degenerate Cauchy numbers andpolynomials. In this paper, we define the degenerate Genocchi polynomials and the k-th

degeneration of Genocchi polynomials , and investigate some properties of these polyno-

mials.

1. Introduction

Let p be a fixed odd prime number. Throughout this paper, Zp, Qp, and Cp will, re-spectively, denote the rings of p-adic integers, the field of p-adic rational numbers, and thecompletion of algebraic closure of Qp. The p-adic norm | · | is normalized by |p|p = 1

p . Let

UD(Zp) be the space of uniformly differentiable functions on Zp. For f ∈ UD(Zp), thefermionic p-adic integral on Zp is defined by Kim as

I−1(f) =

∫Zpf(x)dµ−1(x) = lim

N→∞

pN−1∑x=0

f(x)(−1)x (1)

(see [7,8,11,13,14,16,17,20,22]). Then, by (1), we get

I−1(f) = −I−1(f) + 2f(0), (2)

where f1(x) = f(x+ 1).

1991 Mathematics Subject Classification. 11B68, 11S40.Key words and phrases. Genocchi polynomials, degenerate Genocchi polynomials, fermionic p-adic integral,

Higher order Daehee polynomials.

1


1343 LEE-CHAE JANG et al 1343-1349

2 LEE-CHAE JANG, C.S. RYOO, JEONG GON LEE, HYUCK IN KWON

From (2), we can derive the following integral equation

I−1(fn) = (−1)nI−1(f) + 2n−1∑l=0

(−1)n−1−lf(l), (3)

where fn(x) = f(x+ n), (n ∈ N).As is well known, the Euler polynomials are also defined by the generating function to be(

2

et + 1

)ext =

∞∑n=0

En(x)tn

n!(see [1, 2, 4− 22]). (4)

When x = 0, En = En(0) are called the Euler numbers.The degenerate Euler polynomials are also defined by the degenerating function to be

2

(1 + λt)1λ + 1

(1 + λt)xλ =

∞∑n=0

En(λ, x)tn

n!(see [1, 4, 8, 11, 13, 14, 16, 17, 20, 22]). (5)

When x = 0, En(λ) = En(λ, 0) are called the degenerate Euler number.Note that limx→0En(λ, x) = En(x). We recall that the Genocchi polynomials are defined

by the generating function to be

2

et + 1=∞∑n=0

Gn(x)tn

n!(see [18, 20, 22]). (6)

In recent years, many researchers have studied various types of special polynomials, forexamples, Barnes-type degenerate Euler polynomials, the degenerate Frobenius-Euler polyno-mials, the degenerate Frobenius-Genocchi polynomials, and degenerate Bernoulli polynomials(see [2,3,6,9,10,12,13,15]).

In particular, recently, Jeong-Rim-Kim ([5]) studied finite times degenerate Cauchy poly-nomials and investigated some properties of them.

Thus, our motivation in this paper is to define the degenerate Genocchi polynomials, todefine the k-th degeneration of Genocchi polynomials, and to investigate some properties ofthese k-th degeneration of Genocchi polynomials.

2. The k-th degeneration of Genocchi polynomials

In this section, we define the degenerate Genocchi polynomials which are defined by thegenerating function to be

2t

(1 + λt)1λ + 1

(1 + λt)xλ =

∞∑n=1

g(0)n (x|λ)tn

n!. (7)

When x = 0, g(0)n (λ) = g

(0)n (0|λ) are called the degenerate Genocchi number.

From (2), we easily obtain

2t

(1 + λt)1λ + 1

(1 + λt)xλ = t

∫Zp

(1 + λt)x+yλ dµ−1(y). (8)



3

We note that the Stirling number of the first kind is defined as

(x)n =n∑l=0

S1(n, l)xl (n ≥ 0) (9)

where (x)n = x(x− 1) · · · (x− n+ 1) and (x)0 = 1, and

(log(1 + t))n = n!∞∑m=n

S1(m,n)tm

m!(10)

and the Stirling number of the second kind is defined as

(et − 1)n = n!∞∑l=n

S2(l, n)tl

l!. (11)

From (7), we get

∞∑n=1

g(0)n (x|λ)tn

n!=

∞∑n=0

g(0)n+1(x|λ)

tn+1

(n+ 1)!

= t∞∑n=0

g(0)n+1(x|λ)

n+ 1

tn

n!. (12)

From (8), we get

2t

(1 + λt)1λ + 1

(1 + λt)xλ = t

∫Zp

(1 + λt)x+yλ dµ−1(y)

=∞∑n=0

λ−n∫Zp

(x+ y

λ

)n

dµ−1tn

n!

=∞∑n=0

∫Zp

(x+ y|λ)ndµ−1tn

n!. (13)

Thus, by (7), (12), and (13), we get

g(0)n+1(x|λ)

(n+ 1)=

∫Zp

(x+ y|λ)ndµ−1. (14)

In the viewpoint of (7), we consider the first degeneration of Genocchi polynomials whichare defined by the generating function to be

2 log(1 + λt)1λ

(1 + log(1 + λt))1λ + 1

(1 + log(1 + λt)xλ =

∞∑m=1

g(1)m (x|λ)tm

m!. (15)

By replacing t by log(1 + λt)1λ in (8), we get

2 log(1 + λt)1λ

(1 + log(1 + λt))1λ + 1

(1 + log(1 + λt)xλ

=1

λlog(1 + λt)

∫Zp

(1 + log(1 + λt))x+yλ dµ−1(y)

=1

λlog(1 + λt)

∞∑n=0

1

n!

∫Zpλ−n(x+ y|λ)ndµ−1(log(1 + λt))n

=∞∑n=0

1

n!

∫Zp

(x+ y|λ)nλ−n−1dµ−1(log(1 + λt))n+1




=∞∑n=0

∫Zp

(x+ y|λ)nλ−n−1dµ−1

(n+ 1)!

n!

∞∑m=n+1

λmS1(m,n+ 1)tm

m!

=∞∑m=1

m−1∑n=0

(n+ 1)λm−n−1S1(m,n+ 1)

∫Zp

(x+ y|λ)ndµ−1(y)tm

m!. (16)

Thus, by (14), (15), and (16), we get

g(1)m (x|λ) =m−1∑n=0

(n+ 1)λm−n−1S1(m,n+ 1)

∫Zp

(x+ y|λ)ndµ−1(y)

=m−1∑n=0

λ−m−n−1S1(m,n+ 1)g(0)n+1(x|λ). (17)

By (17), we obtain the following theorem.

Theorem 2.1. Let m ∈ N. Then we have

g(1)m (x|λ) =m−1∑n=0

λm−n−1S1(m,n+ 1)g(0)n+1(x|λ). (18)

Now, we consider the second degeneration of Genocchi polynomials as follows:

2 log(1 + log(1 + λt))1λ

(1 + log(1 + log(1 + λt)))1λ + 1

(1 + log(1 + log(1 + λt)))1λ

=∞∑m=1

g(2)m (x|λ)tm

m!. (19)

From (19), we get

2λ log(1 + log(1 + λt))

(1 + log(1 + log(1 + λt)))1λ + 1

(1 + log(1 + log(1 + λt)))xλ

=1

λlog(1 + log(1 + λt))

∫Zp

(1 + log(1 + log(1 + λt)))x+yλ dµ−1(y). (20)

From (20), we get

1


∫Zp

(1 + log(1 + log(1 + λt)))x+yλ dµ−1(y)

=1


∞∑n=0

1

n!

∫Zpλ−n(x+ y|λ)ndµ−1(y)(log(1 + log(1 + λt)))n

=∞∑n=0

1

n!

∫Zp

(x+ y|λ)nλ−n−1dµ−1(log(1 + log(1 + λt)))n

=∞∑n=0

1

n!

∫Zp

(x+ y|λ)nλ−n−1dµ−1(y)(n+ 1)!

∞∑m=n+1

S1(m,n+ 1)(log(1 + λt))m

m!

=∞∑n=0

∫Zp

(x+ y|λ)nλ−n−1dµ−1(y)(n+ 1)

∞∑m=n+1

S1(m,n+ 1)∞∑l=m

S1(l,m)λltl

l!

=∞∑

n3=0

n3∑n2=0

n2−1∑n1=0

λn3−n1−1S1(n3, n2)S1(n2, n1 + 1)g(0)n1+1(x|λ)

tn3

n3!. (21)

From (20) and (21), we obtain the following theorem.



5

Theorem 2.2. Let n3 ∈ N. Then we have

g(2)n3(x|λ) =

n3∑n2=0

n2−1∑n1=0

λn3−n1−1S1(n3, n1)S1(n2, n1 + 1)g(0)n1+1(x|λ). (22)

Inductively, we have the k-th degeneration of Genocchi polynomials as follows:

Theorem 2.3. Let k, nk ∈ N. Then we have

g(k−1)nk(x|λ) =

∑nk−1=0

· · ·n2−1∑n1−0

λnk−n1−1S1(nk, nk−1) · · ·S1(n2, n1 + 1)g(0)n1+1(x|λ). (23)

By replacing t by 1λ (eλt − 1) in (19) and (20).

We have∞∑n=1

g(2)n (x|λ)1

λn(eλt − 1)n

n!=

∞∑n=1

g(2)n (x|λ)∞∑l=n

λl−nS2(l, n)tl

l!

=∞∑l=0

(l∑

n=0

g(2)n (x|λ)λl−nS2(l, n))tl

l!. (24)

By (14) and (23), we obtain the following theorem.

Theorem 2.4. Let l ∈ N. Then we have

g(p)l (x|λ) =

l∑n=0

g(2)n (x|λ)λl−nS2(l, n). (25)

We note the Daehee polynomials of order r is defined by the generating function to be(log(1 + t)

t

)r(1 + t)x =

∞∑n=0

D(r)n (x)

tn

n!(26)

When x = 0, D(r)n = D

(r)n (0) are called the Daehee numbers of order r.

By replacing t by log(1 + λt)1λ in (7), we get

∞∑n=1

g(0)n (x|λ)(log(1 + λt)

1λ )n

n!=

2(log(1 + λt)1λ )

(1 + log(1 + λt))1λ + 1

− (1 + log(1 + λt)xλ

=

∞∑n=1

g(1)n (λ|x)tn

n!. (27)

and∞∑n=1

g(0)n (x|λ)

(log(1 + λt)

t

)nλ−n

tn

n!=

∞∑n=1

g(0)n (x|λ)

( ∞∑l=0

D(n)l

tl

l!

)λ−n

tn

n!

=∞∑n=1

∞∑l=0

g(0)n (x|λ)D(n)l λ−n

tl+n

l!n!

=∞∑m=1

(m∑n=0

g(0)n (x|λ)D(n)m−nλ

−n(n

m

))tm

m!. (28)

Thus, by (27) and (28), we obtain the following theorem.

Theorem 2.5. Let m ∈ N. Then we have

g(1)m (x|λ) =m∑n=0

(m

n

)λ−ng(0)n (x|λ)D

(n)m−n. (29)




By replacing t by 1λ (eλt − 1) in (15), we get

∞∑m=1

g(1)m (x|λ)

(eλt−1λ

)mm!

=2t

(1 + λt)1λ + 1

(1 + λt)xλ

=∞∑l=1

g(0)l (x|λ)

tl

l!. (30)

and∞∑m=1

g(1)m (x|λ)λ−m(eλt − 1

)mm!

=∞∑m=1

g(1)m (x|λ)λ−m∞∑l=m

S2(l,m)(λt)l

l!

=∞∑m=1

( ∞∑l=m

λl−mg(1)m (x|λ)S2(l,m)

)tl

l!

=∞∑l=0

(l∑

m=1

λl−mg(1)m (x|λ)S2(l,m)

)tl

l!. (31)

Thus, by (30) and (31), we obtain the following theorem.

Theorem 2.6. Let l ∈ N. Then we have

g(0)l (x|λ) =

l∑m=1

λl−mg(1)m (x|λ)S2(l,m). (32)

References

[1] A. Bayad, T. Kim, Identities involving values of Bernstein, q-Bernoulli, and q-Euler polynomials, Russ.

J. Math. Phys. 18(2) (2011) 133-143.[2] L. Carlitz, Degenerate Stirling, Bernoulli and Eulerian numbers, Utilitas Math. 15 (1979) 51-88.

[3] D. V. Dolgy, D. S. Kim, T. Kim, T. Mansour, Degenerate poly-Cauchy polynomials, Appl. Math. Comput.

269 (2015) 637-646.[4] S. Gaboury, R. Tremblay, B. J. Fugere, Some explicit formulas for certain new classes of Bernoulli, Euler

and Genocchi polynomials, Proc. Jangjeon Math. Soc. 17(1) (2014) 115-123.

[5] J. Jeong, S.-H. Rim, B. M. Kim, On finite-times degenerate Cauchy numbers and polynomials, Adv.Differece Equ. 2015 (2015) 2015:321.

[6] D. S. Kim, T. Kim, Higher-order Cauchy of the first kind and poly-Cauchy of the first kind mixed type

polynomials, Adv. Stud. Contemp. Math. 24 (2013) 512-527.[7] D. S. Kim, Identities associatied with generalized twisted Euler polynomials twisted by ramified roots of

unity, Adv. Stud. Contemp. Math. 22(3) (2012) 363-377.[8] D. S. Kim, Symmetry identities for generlaized twisted Euler polynomials twisted by unramified roots of

unity, Proc. Jangjeon Math. Soc. 15(3) (2012) 303-316.

[9] T. Kim, On the degenerate higher-order Cauchy numbers and polynomials, Adv. Stud. Contemp. Math.25(3) (2015) 417-421.

[10] T. Kim, Degenerate Bernoulli polynomials associated with p-adic invariant integral on Zp, Adv. Stud.

Contemp. Math. 25(3) (2015) 273-279.[11] T. Kim, D. S. Kim, D. V. Dolgy, Degenerate q-Euler polynomials, Adv. Difference Equ. 2015 (2015)

2015:246

[12] T. Kim, On the degenerate Cauchy numbers and polynomials, Proc. Jangjeon Math. Soc. 18(3) (2015)307-312.

[13] T. Kim, J. Y. Choi, J. Y. Sug, Extended q-Euler numbers and polynomials associated with fermionic

p-adic q-integral on Zp, Russ. J. Math. Phys. 14(2) (2007) 160-163.[14] T. Kim, Barnes’ type multiple degenerate Bernoulli and Euler polynomials, Appl. Math. Comput. 258

(2015) 556-564.[15] T. Kim, A note on q-Bernstein polynimials, Russ. J. Math. Phys. 18(1) (2011) 73-82.



7

[16] T. Kim, Some identities on the q-Euler polynomials of higher order and q-Stirling numbers by thefermionic p-adic integral on Zp, Russ. J. Math. Phys. 16(4) (2009) 484-491.

[17] T. Kim, New approach to q-Euler polynomials of higher order, Russ. J. Math. Phys. 17(2) (2010) 218-

225.[18] T. Kim, On the multiple q-Genocchi and Euler numbers, Russ. J. Math. Phys. 15(4) (2008) 481-486.

[19] T. Kim, q-Volkenborn integration Russ. J. Math. Phys. 9(3) (2002) 288-299.

[20] T. Kim, On Euler-Barnes multiple zeta functions, Russ. J. Math. Phys. 10(3) (2003) 261-267.[21] C. S. Ryoo, Calculating zeros of the twisted Genocchi polynomials, Adv. Stud. Contemp. Math. 17(2)

(2008) 147-159.

[22] J. J. Seo, T. Kim, Some symmetric identities of the modified q-Euler polynomials under symmetric groupof degree four, Proc. Jangjeon Math. Soc. 18(2) (2015) 211-216.



Regularization solutions of ill-posed Helmholtz-typeequations with fuzzy mixed boundary value†

Hong Yanga,b, Zeng-Tai Gonga,∗

aCollege of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, ChinabCollege of Technology and Engineering, Lanzhou University of Technology, Lanzhou 730050, China

Abstract In this study, we discuss the solutions of fuzzy Helmholtz-type equations(FHTEs) and theirill-posedness. A regularization method is required to recover the numerical stability. Moreover, the errorestimates and convergence of the method is considered. To support our study, one numerical example isillustrated.Keywords: Fuzzy numbers; Ill-posed problem; Helmholtz equation; Regularization method; Conver-gence estimate.

1. Introduction

The study of fuzzy partial differential equations (FPDEs) provides a suitable setting for the mathe-matical modeling of real-world problems that include uncertainty or vagueness. As a new and powerfulmathematical tool, FPDEs have been studied using several approaches. The first definition of an FPDEwas presented by Buckley and Feuring in [1]. In [2], the authors considered the application of FPDEsobtained using fuzzy rule-based systems. Furthermore, Oberguggenberger described weak and fuzzy so-lutions for FPDEs [3] and Chen et al. presented a new inference method with applications to FPDEs [4].In [5], an interpretation was provided of the use of FPDEs for modeling hydrogeological systems. Studiesof heat, wave, and Poisson equations with uncertain parameters were provided in [6]. Fuzzy solutions forheat equations based on generalized Hukuhara differentiability were considered in [7]. Several numericalmethods have been proposed to solve FPDEs. Such as Allahviranloo ([8]) proposed a difference methodfor solving FPDEs. The Adomian decomposition method was studied for finding the approximate so-lution of the fuzzy heat equation in [9]. Solving FPDEs by the differential transformation method wasconsidered in [10].

In this paper, we proposed a numerical method to solve ill-posed problems for the fuzzy Helmholtz-type equation (FHTEs). The Helmholtz equation arises in many areas, especially in practical physicalapplications, such as acoustic, wave propagation and scattering, vibration of the structure, electromag-netic scattering and so on, see [11, 12, 13, 14]. The direct problems, i.e. Dirichlet, Neumann or mixedboundary value problems for the Helmholtz equation have been studied extensively in the past century.However, in some practical problems, the boundary data on the whole boundary cannot be obtained.We only know the noisy data on a part of the boundary or at some interior points of the concerneddomain, which will lead to some inverse problems and severely ill-posed problems. In 1923, Hadamard[15] introduced the concept of a well-posed problem from philosophy where the mathematical model of aphysical problem must have properties where the solution exhibits uniqueness, existence, and stability. Ifone of the properties fails to hold, the problem is known as ill-posed. Numerical computation is difficultdue to the ill-posedness of the problem. That means the solution does not depend continuously on thegiven Cauchy data and any small perturbation in the given data may cause large change to the solution[15, 16, 17]. The present study addresses two issues. First, we consider the ill-posedness of FHTEs usingthe decomposition theorem. Second, we use the regularization method to recover the numerical stability.

The remainder of this paper is organized as follows. In Section2, we briefly introduce the necessarynotions related to fuzzy numbers and differentiability properties for fuzzy set-valued mappings. In Sec-tion3, we define the solution and ill-posedness of FHTEs. The regularization method and convergence

†Supported by the Natural Scientific Fund of China (11461062, 61262022).∗Corresponding Author:Zeng-Tai Gong. Tel.: +869317971430. E-mail addresses: [email protected]


1350 Hong Yang et al 1350-1360

Hong Yang and Zeng-Tai Gong: Regularization solutions of ill-posed Helmholtz-type...

estimates for the initial-boundary value problems of FHTEs are considered in Section4. In Section5, wepresent some numerical results and our conclusions are given in Section6.

2. Definitions and preliminaries

Let Pk(Rn) denote the family of all nonempty compact convex subset of Rn and define the additionand scalar multiplication in Pk(Rn) as usual. Let A and B be two nonempty bounded subset of Rn. Thedistance between A and B is defined by the Hausdorff metric

dH(A,B) = max

supa∈A

infb∈B||a− b||, sup

b∈Binfa∈A||b− a||

, (2.1)

where || · || denotes the usual Euclidean norm in Rn [18]. Then (Pk(Rn); dH) is a metric space.

DenoteEn = u : Rn → [0, 1]|u satisfies (1)-(4) below

is a fuzzy number space, where(1) u is normal, i.e. there exists an x0 ∈ Rn such that u(x0) = 1,(2) u is fuzzy convex, i.e. u(λx+ (1− λ)y) ≥ minu(x), u(y) for any x, y ∈ Rn and 0 ≤ λ ≤ 1,(3) u is upper semi-continuous,(4) [u]0 = clx ∈ Rn|u(x) > 0 is compact.Here, cl(X) denotes the closure of set X. For 0 < α ≤ 1, the α-level set of u (or simply the α-cut) is

defined by [u]α = x ∈ Rn|u(x) ≥ α. The core of u is the set of elements of Rn having membership grade1, i.e., [u]1 = x|x ∈ Rn, u(x) = 1. Then from above (1)-(4), it follows that the α-level set [u]α ∈ Pk(Rn)for all 0 < α ≤ 1. According to Zadeh’s extension principle, we have addition and scalar multiplicationin fuzzy number space En as follows:

[u+ v]α = [u]α + [v]α = x+ y|x ∈ [u]α, y ∈ [v]α,

[ku]α = k[u]α = kx|x ∈ [u]α, [0]α = 0.

where u, v ∈ En and 0 < α ≤ 1. The distance between two fuzzy numbers u and v is defined by

D(u, v) = supα∈[0,1]

dH([u]α, [v]α). (2.2)

We recall some differentiability properties for fuzzy set-valued mappings.

Definition 2.1 (See[19]) Let KC denote the family of all bounded closed intervals in R, the generalizedHukuhara difference of two intervals A,B ∈ KC (gH-difference, for short), denoted by A⊖gHB, is definedby

A⊖gH B = C ⇐⇒

(i)A = B + C;

or(ii)B = A+ (−C).(2.3)

Definition 2.2 (See[20]) The generalized Hukuhara difference of two fuzzy numbers u, v ∈ E1 (gH-difference, for short) is the fuzzy number ω, if it exists, such that

u⊖gH v = ω ⇐⇒

(i)u = v + ω;

or(ii)v = u+ (−ω).(2.4)

It is easy to show that (i) and (ii) are both valid if and only if w is a crisp number.It may happen that the gH-difference of two fuzzy numbers does not exist (see, for example, [21]). In

order to overcome this shortcoming, in [20, 21], a new difference between fuzzy numbers was proposed,which always exists.




Henceforth, T=]a, b[denotes an open interval in R. A function F : T → FC is said to be a fuzzyfunction. For each α ∈ [0, 1], associated to F, we define the family of interval-valued functions Fα : T →KC given by Fα(x) = [F (x)]α, for x ∈ T . For any ∈ [0, 1], we denote

Fα(x) = [fα(x), fα(x)]. (2.5)

Here, for each α ∈ [0, 1], the endpoint functions fα, fα : T → R are called upper and lower functions of F ,respectively. Next, we present the concept of gH-differentiable fuzzy functions based on the gH-differenceof fuzzy intervals.

Definition 2.3 (See [21]) The gH-derivative of a fuzzy function F : T → FC at x0 ∈ T is defined as

F ′(x0) = limh→0

1

h[F (x0 + h)⊖gH F (x0)]. (2.6)

If F (x0) ∈ FC satisfying (2.5) exists, we say that F is generalized Hukuhara differentiable (gH-differentiable,for short) at x0.

Theorem 2.1(See [22]) If F : T → FC is gH-differentiable at x0 ∈ T , then Fα is gH-differentiable at x0uniformly in α ∈ [0, 1] and

F ′α(x0) = [F ′(x0)]

α, (2.7)

for all α ∈ [0, 1].

Theorem 2.2 (See [22]) Let F: T → FC be a fuzzy function and x ∈ T . Then F is gH-differentiable atx if and only if one of the following four cases holds:

(a) fα

and fα are differentiable at x, uniformly in α ∈ [0, 1], (fα)′(x) is monotonic increasing and

(fα)′(x) is monotonic decreasing as functions of α and (fα)′(x) ≤ (fα)′(x). In this case,

F ′α(x) = [(f

α)′(x), (fα)′(x)]. (2.8),

for all α ∈ [0, 1].

(b) fα

and fα are differentiable at x, uniformly in α ∈ [0, 1], (fα)′(x) is monotonic increasing and

(fα)′(x) is monotonic decreasing as functions of α and (fα)′(x) ≤ (fα)′(x). In this case,

F ′α(x) = [fα)′(x), (f

α)′(x)]. (2.9),

for all α ∈ [0, 1].

(c) (fα)′+/−(x) and (fα)′+/−(x) exist uniformly in α ∈ [0, 1], (f

α)′+(x) = (fα)′−(x) is monotonic

increasing and (fα)′+(x) = (fα)′−(x) is monotonic decreasing as functions of α and (f

α)′+(x) ≤ (fα)′+(x).

In this case,F ′α(x) = [(f

α)′+(x), fα)′+(x)] = [fα)′−(x), (f

α)′−(x)]. (2.10),

for all α ∈ [0, 1].

(d) (fα)′+/−(x) and (fα)′+/−(x) exist uniformly in α ∈ [0, 1], (f

α)′+(x) = (fα)′−(x) is monotonic

increasing and (fα)′+(x) = (fα)′−(x) is monotonic decreasing as functions of α and (fα)′+(x) ≤ (f

α)′+(x).

In this case,F ′α(x) = [fα)′+(x), (f

α)′+(x)] = [(f

α)′−(x), fα)′−(x)]. (2.11),

for all α ∈ [0, 1].

Theorem 2.3(Decomposition Theorem[23]) If u ∈ En, then




u =∪

α∈[0,1]

(α · [u]α). (2.3)

The following well-known characterization theorem makes the connection between a fuzzy intervaland its LU-representation.

Theorem 2.4 (See[24]) Let u be a fuzzy number. Then the functions u, u: [0, 1] → R, defining theendpoints of the α-level sets of u, satisfy the following conditions:

(i) u is a bounded, non-decreasing, left-continuous function in (0, 1] and it is right-continuous at 0.(ii) u is a bounded, non-increasing, left-continuous function in (0, 1] and it is right-continuous at 0.(iii) u(1) ≤ u(1).

Reciprocally, given two functions that satisfy the above conditions, they uniquely determine a fuzzynumber.

3. Solutions of FHTEs and Ill-posedness

Now, we consider a Cauchy problem for the Helmholtz-type equation with fuzzy initial-boundaryvalue in a rectangle domain as follows

∂2u

∂x2+∂2u

∂y2+ k2u = 0, 0 < x < π, 0 < y < 1,

u(x, 0) = φ(x), 0 ≤ x ≤ π,∂u

∂y(x, 0) = 0, 0 ≤ x ≤ π,

u(0, y) = u(π, y) = 0, 0 ≤ y ≤ 1,

(3.1)

where where constant k > 0 is the wave number. u, ∂2u

∂x2 ,∂2u∂y2

, ∂u∂y , φ(x), 0 are fuzzy-number-valued functions

and their α-cut sets are shown as follows:

[u(x, y)]α = [u(x, y, α), u(x, y, α)],

[∂2u

∂x2(x, y)]α = [

∂2u

∂x2(x, y, α),

∂2u

∂x2(x, y, α)],

[∂2u

∂y2(x, y)]α = [

∂2u

∂y2(x, y, α),

∂2u

∂y2(x, y, α)],

[∂u

∂y(x, y)]α = [

∂u

∂y(x, y, α),

∂u

∂y(x, y, α)],

[φ(x)]α = [φ(x, α), φ(x, α)], [0]α = [0(α), 0(α)].

From Theorem 2.1 and Theorem 2.2, in order to investigate the solution of (3.1), we consider thefollowing two systems of two partial differential equations

∂2u

∂x2(x, y, α) +

∂2u

∂y2(x, y, α) + k2u(x, y, α) = 0, 0 < x < π, 0 < y < 1,

u(x, 0) = φ(x, α), 0 ≤ x ≤ π,∂u

∂y(x, 0, α) = 0(α), 0 ≤ x ≤ π,

u(0, y, α) = u(π, y) = 0(α), 0 ≤ y ≤ 1,

(3.2)




∂2u

∂x2(x, y, α) +

∂2u

∂y2(x, y, α) + k2u(x, y, α) = 0, 0 < x < π, 0 < y < 1,

u(x, 0, α) = φ(x, α), 0 ≤ x ≤ π,∂u

∂y(x, 0, α) = 0(α), 0 ≤ x ≤ π,

u(0, y) = u(π, y, α) = 0(α), 0 ≤ y ≤ 1,

(3.3)

Definition 3.1 (see [1]) Let u(x, y, α) and u(x, y, α) be solutions of equations (3.2) and (3.3), respec-tively. If [u(x, y, α), u(x, y, α)] defines the α-cut of a fuzzy number, for all (x, y) ∈ [0, π] × [0, 1], thenu(x, y) is a solution for (3.1).

By the method of separation of variables, it is easy to derive a solution of the direct problem (3.2)and (3.3), respectively as follows:

u(x, y, α) =

[k]∑n=1

cn sin(nx) cos(√k2 − n2y) +

∞∑n=[k]+1

cn sin(nx) cosh(√n2 − k2y) (3.4)

where

cn =2

π

∫ π

0φ(x, α) sin(nx)dx (3.5)

u(x, y, α) =

[k]∑n=1


∞∑n=[k]+1

cn sin(nx) cosh(√n2 − k2y) (3.6)

where

cn =2

π

∫ π

0φ(x, α) sin(nx)dx (3.7)

Obviously, for the solutions u(x, y, α) of the equations (3.2) and the solutions u(x, y, α) of the equations(3.3), [u(x, y, α), u(x, y, α)] satisfies the conditions of Theorem 2.2, [u(x, y, α), u(x, y, α)] determines asolution of problem (3.1) as follows:

u =∪

α∈[0,1]

(α · [u(x, y, α), u(x, y, α)]). (3.8)

Remark 3.1 If 0 < k < 1, the first term in Equations (3.4) and (3.6) is vanished.

In the following, we discuss the ill-posedness of problem (3.1).

Definition 3.2 (Hadamard’s definition of well-posedness [15]) If a deterministic solution problem ofFPDE satisfies the following properties (3.9-3.11), then it is well-posed.

For all admissible date, a solution exists. (3.9)For all admissible date, the solution is unique. (3.10)The solution depends continuously on the date. (3.11)Conversely, if one of the properties (3.9-3.11) does not satisfy for a deterministic solution problem of

FPDE, then it is ill-posed.

Next, we are always suppose that (3.9) and (3.10) hold for the convenience of discussion, (3.11) doesnot hold.

Definition 3.3 Problem of FHTEs (3.1) is said to be ill-posed if both problems of PDE (3.2) and PDE(3.3) are ill-posed.




The the systems of PDE (3.2) and (3.3) are highly ill-posed, see[16]. Thus, the systems (3.1) isill-posed.

Ill-posed problem means the solution does not depend continuously on the given Cauchy data andany small perturbation in the given data may cause large change to the solution. Thus regularizationtechniques are required to stabilize numerical computations. In general terms, regularization is theapproximation of an ill-posed problem by a family neighbouring well-posed problems.

4. Regularization and Convergence estimates

In this section, we use the solution of perturbation problems to approach the solution of problems(3.2) and (3.3). Thus the regularization solution of problems (3.1) be derived by (3.4).

For 0 < k < 1, we consider the following problem∆v(x, y) + k2v(x, y) = 0, 0 < x < π, 0 < y < 1,

v(x, 0) + βv(x, 1) = φδ1(x, α), 0 ≤ x ≤ π,vy(x, 0) = 0, 0 ≤ x ≤ π,v(0, y) = v(π, y) = 0, 0 ≤ y ≤ 1,

(4.1)

∆v(x, y) + k2v(x, y) = 0, 0 < x < π, 0 < y < 1,

v(x, 0) + βv(x, 1) = φδ2(x, α), 0 ≤ x ≤ π,vy(x, 0) = 0, 0 ≤ x ≤ π,v(0, y) = v(π, y) = 0, 0 ≤ y ≤ 1,

(4.2)

where 0 < α ≤ 1 is α-level set parameter, and β > 0 is a regularization parameter. The measureddata of equations (3.1) is fuzzy-number-valued function φ(x), and its α-level set is defined as [φ(x)]α =[φ(x, α), φ(x, α)]. φδ1 ∈ L2(0, π), φδ2 ∈ L2(0, π) satisfies

∥φδ1 − φ∥ ≤ δ1, (4.3)

∥φδ2 − φ∥ ≤ δ2, (4.4)

in which the constant δ1 > 0 and δ2 > 0 is called an error level and ∥ · ∥ denotes the L2-norm. Furtherassume that there exists a constant E > 0 such that the following a priori bound exists

∥u(·, 1)∥ ≤ E. (4.5)

By the method of separation of variables, it is easy to derive a solution of direct problem (4.1) and(4.2) as follows, respectively

v(x, y, α) =

∞∑n=1

cδ1ncosh(

√n2 − k2y)

1 + β cosh(√n2 − k2)

sin(nx), (4.6)

where

cδ1n =2

π

∫ π

0φδ1(x, α) sin(nx)dx. (4.7)

v(x, y, α) =

∞∑n=1

cδ2ncosh(

√n2 − k2y)

1 + β cosh(√n2 − k2)

sin(nx), (4.8)

where

cδ2n =2

π

∫ π

0φδ2(x, α) sin(nx)dx. (4.9)




For k ≥ 1, we define a regularized solution v as follows:

v(x, y, α) =

[k]∑n=1

cδ1n cosh(√n2 − k2y) +

∞∑n=[k]+1

cδ1ncosh(

√n2 − k2y)

1 + β cosh(√n2 − k2)

sin(nx), (4.10)

where cδ1n is defined by Equation (4.7).

v(x, y, α) =

[k]∑n=1

cδ2n cosh(√n2 − k2y) +

∞∑n=[k]+1

cδ2ncosh(

√n2 − k2y)

1 + β cosh(√n2 − k2)

sin(nx), (4.11)

where cδ2n is defined by Equation (4.9).

Remark 4.1 (see[25]) For k ≥ 1, the regularized solution (4.10) and (4.11) be not an exact solutionof the problem (4.1) and (4.2), respectively, but a modified solution. This is done to avoid the case1 + β cos(

√n2 − k2) = 0 for k ≥ 1 and n < k and prove a convergence result.

In the following results shall show that the regularization solution v given by Equation (4.6)and(4.10),and v given by Equation (4.8) and (4.11) are a stable approximation to the exact solution u and u givenby Equation (3.4) and (3.6),respectively. The regularization solution v and v depends continuously onthe measured data φδ1 and φδ2 for a fixed parameter β > 0, respectively.

Theorem 4.1 (see[25]) Suppose that u and u is defined by Equation (3.4) and (3.6) with the exact dataφ and φ, respectively. Suppose that v is defined by Equation (4.6) for the case 0 < k < 1 or Equation

(4.10)for the case k ≥ 1 with the measured data φδ1 , v is defined by Equation (4.8) for the case 0 < k < 1

or Equation (4.11) for the case k ≥ 1 with the measured data φδ2 . Let the measured data φδ1 and φδ2

satisfy Equation (4.3) and (4.4), respectively. Let the exact solution u at y = 1 satisfy Equation (4.5). Ifthe regularization parameter β is chosen as, respectively

β =δ1E, (4.11)

β =δ2E, (4.12)

then for fixed 0 < y < 1, we have the following convergence estimate

∥v(·, y)− u(·, y)∥ ≤ δ1 + 2CyEyδ1−y

1 . (4.13)

∥v(·, y)− u(·, y)∥ ≤ δ2 + 2CyEyδ1−y

2 . (4.14)

where Cy = 1−y

( 2y(1−y)

)y .

However, the convergence estimate in Equation (4.13) and (4.14) is not useful for y = 1. In order toobtain a convergence rate at y = 1, we need a stronger a priori assumption

∥∂pu(·, 1)

∂yp∥ ≤ E, (4.15)

where p ≥ 1 is an integer. We have the following convergence estimate.

Theorem 4.2 (see[25]) Suppose that u and u is defined by Equation (3.4) and (3.6) with the exact dataφ and φ, respectively. Suppose that v is defined by Equation (4.6) for the case 0 < k < 1 or Equation

(4.10) for the case k ≥ 1 with the measured data φδ1 , v is defined by Equation (4.8) for the case 0 < k < 1

or Equation (4.11) for the case k ≥ 1 with the measured data φδ2 . Let the measured data φδ1 and φδ2




satisfy Equation (4.3) and (4.4), respectively. Let the exact solution u at y = 1 satisfy Equation (4.15).If the regularization parameter β is chosen as, respectively

β =δ1E, (4.16)

β =δ2E, (4.17)

then we have the following convergence estimate at y = 1,

∥v(·, 1)− u(·, 1)∥ ≤ δ1 +√δ1E +

2E

1− e−2kmaxK−p(

δ1E

)13 , 2(

1

6lnE

δ1)−p. (4.18)

∥v(·, 1)− u(·, 1)∥ ≤ δ2 +√δ2E +

2E

1− e−2kmaxK−p(

δ2E

)13 , 2(

1

6lnE

δ2)−p. (4.19)

where K =√

([k] + 1)2 − k2 and [·] denotes the integer part of a real number.

Theorem 4.3 Suppose that u defined by Equation (3.8) is a solution of problem (3.1) and v is itsregularization solution. If u is defined by Equation (3.4) and v is its regularization solution defined byEquation (4.6) for the case 0 < k < 1 or Equation (4.10) for the case k ≥ 1, while u is defined by Equation(3.6) and v is its regularization solution defined by Equation (4.8) for the case 0 < k < 1 or Equation(4.11) for the case k ≥ 1. then v is a stable approximation to u, where

v =∪

α∈[0,1]

(α · [v(x, y, α), v(x, y, α)]). (4.20)

Proof By Equation(2.2), sinceD(u, v) = sup

α∈[0,1]dH([u]α, [v]α)

= supα∈[0,1]

max|u(α)− v(α)|, |u(α)− v(α)|, (4.21)

from Theorem 4.1 and 4.2, v(α) is a stable approximation to u(α) and v(α) is a stable approximation tou(α). Hence, From (4.21) we have that v is a stable approximation to u. The proof is complete.

5. Numerical examples

Consider the following direct problem for the Helmholtz equation with fuzzy mixed boundary value

∂2u

∂x2+∂2u

∂y2+ k2u = 0, 0 < x < π, 0 < y < 1,

u(x, 1) = f(x), 0 ≤ x ≤ π,∂u

∂y(x, 0) = 0, 0 ≤ x ≤ π,

u(0, y) = u(π, y) = 0, 0 ≤ y ≤ 1,

(5.1)

in which f : [0, π]→ E1.

f = v · 2x(π − x), x ∈ [0, π]. (5.2)

where v ∈ E1 is given by a triangular fuzzy number

v(t) =

t+ 1, t ∈ (−1, 0),

− t+ 1, t ∈ (0, 1),

0, t ∈ (−∞,−1] ∪ [1,+∞).

(5.3)




The α-cut set of f(x) is given by

[f(x)]α = [2x(π − x)v(t, α), 2x(π − x)v(t, α)]

= [2x(π − x)(α− 1), 2x(π − x)(1− α)]. (5.4)

In order to investigate the numerical solution of (5.1), we consider the following two systems of twopartial differential equations

∂2u

∂x2+∂2u

∂y2+ k2u = 0, 0 < x < π, 0 < y < 1,

u(x, 1) = 2x(π − x)(α− 1), 0 ≤ x ≤ π,∂u

∂y(x, 0) = 0, 0 ≤ x ≤ π,

u(0, y) = u(π, y) = 0, 0 ≤ y ≤ 1,

(5.5)

∂2u

∂x2+∂2u

∂y2+ k2u = 0, 0 < x < π, 0 < y < 1,

u(x, 1) = 2x(π − x)(1− α), 0 ≤ x ≤ π,∂u

∂y(x, 0) = 0, 0 ≤ x ≤ π,

u(0, y) = u(π, y) = 0, 0 ≤ y ≤ 1,

(5.6)

By the method of separation of variables, the solution of the direct problem (5.5) and (5.6) can beobtained as follows, respectively.

u(x, y, α) =

[k]∑n=1


∞∑n=[k]+1

cn sin(nx) cosh(√n2 − k2y), (5.7)

u(x, y, α) =

[k]∑n=1


∞∑n=[k]+1

cn sin(nx) cosh(√n2 − k2y), (5.8)

where φn

= 2π cosh(n)dn, dn =

∫ π0 2x(π − x)(α − 1) sin(nx)dx, φn = 2

π cosh(n)dn, dn =∫ π0 2x(π − x)(1 −

α) sin(nx)dx, and they can be computed by the Simpson formulation, respectively.

Remark 5.1 If 0 < k < 1, the first term in Equations (5.7) and (5.8) is vanished.

Then we choose the initial data φ(x) for equation (3.2) and φ(x) for equation (3.3) as follows,

φ(x) = u(x, 0) ≈25∑n=1

φn

sin(nx). (5.9)

φ(x) = u(x, 0) ≈25∑n=1

φn sin(nx). (5.10)

The measured data φδ1

and φδ2 is given by φδ1(xi) = φ(xi) + ε · rand(i), and φδ2(xi) = φ(xi) + ε · rand(i),

respectively, where ε is an error level,

δ1 := ∥φδ1− φ∥l2 =

(1

N1

N1∑i=1

∣∣∣φδ1

(xi)− φ(xi)∣∣∣2)1/2

. (5.11)




δ2 := ∥φδ2 − φ∥l2 =

(1

N1

N1∑i=1

∣∣φδ2(xi)− φ(xi)∣∣2)1/2

. (5.12)

The function rand(·) denotes a random number uniformly distributed in the interval [0, 1]. In our numer-ical computations, we always take N1 = 31. The regularization parameter β is chosen by (4.10),(4.11)and (4.15),(4.16) respectively.

The numerical results for u(·, y) and uδβ(·, y) with k = 12 , ε = 0.0001, α = 1

2 are shown in Figure1.

01

23

4

0

0.5

1−40

−30

−20

−10

0

x y

exac

t sol

utio

n

(a) exact solution u.

01

23

4

0

0.5

1−40

−30

−20

−10

0

x y

regu

lariz

atio

n so

lutio

n

(b) regularization solution uδβ.

01

23

4

0

0.5

10

10

20

30

40

x y

exac

t sol

utio

n

(c) exact solution u.

01

23

4

0

0.5

10

10

20

30

40

x y

regu

lariz

atio

n so

lutio

n

(d) regularization solution uδβ.

Figure 1: ε = 1× 10−4, α = 12 , k = 1

2 .

6. Conclusion

In this paper, we investigate a new numerical method of solution for inverse problem of FHTEs. Wedefined the ill-posedness for deterministic solution problem of FHTEs and the regularization method isproposed to solve a Cauchy problem for the ill-posed FHTEs. The convergence and stability estimatesfor 0 < y < 1, y = 1 have been obtained under a-priori bound assumptions for the exact solution. Finally,one example shows that our proposed numerical method is effective.

References

[1] J.J. Buckley, T. Feuring, Introduction to fuzzy partial differential equations, Fuzzy Sets Syst. 105(1999) 241-248.

[2] G.Q. Zhang, Fuzzy continuous function and its properties, Fuzzy Sets Syst. 43 (2) (1991) 159-171.




[3] M. Oberguggenberger, Fuzzy and weak solutions to differential equations, in: Proceedings of the10th International IPMU Conference, 2004.

[4] Y.Y. Chen, Y.T. Chang, B.S. Chen, Fuzzy solutions to partial differential equations: adaptiveapproach, IEEE Trans. Fuzzy Syst. 17 (1) (2009) 116-127.

[5] B.A. Faybishenko, Introduction to modeling of hydrogeologic systems using fuzzy partial differentialequation, in: M. Nikravesh, L. Zadeh, V. Korotkikh (Eds.), Fuzzy Partial Differential Equations andRelational Equations: Reservoir Characterization and Modeling, Springer, 2004.

[6] A.M. Bertone, R.M. Jafelice, L.C. de Barros, R.C. Bassanezi, On fuzzy solutions for partial differ-ential equations, Fuzzy Sets Syst. 219 (2013) 68-80.

[7] T. Allahviranlooa, Z. Gouyandeha., A. Armanda, A. Hasanoglub, On fuzzy solutions for heat equa-tion based on generalized Hukuhara differentiability, Fuzzy Sets Syst. 265 (2015) 1-23.

[8] T. Allahviranloo, Difference methods for fuzzy partial differential equations, Comput. Methods Appl.Math. 2 (3) (2002) 233-242

[9] T. Allahviranloo, N. Taheri, An analytic approximation to the solution of fuzzy heat equation byadomian decomposition method, Int. J. Contemp. Math. Sci. 4 (3) (2009).

[10] M.B. Ahmadi, N.A. Kiani, Solving fuzzy partial differential equation by differential transformationmethod, Journal of Applied Mathematics, Islamic Azad University of Lahijan, 7 (2011) 1-16.

[11] J.T. Chen, F.C.Wong, Dual formulation of multiple reciprocity method for the acoustic mode of acavity with a thin partition, J. Sound Vib. 217 (1) (1998) 75-95.

[12] W.S. Hall, X.Q. Mao, Boundary element investigation of irregular frequencies in electromagneticscattering, Eng. Anal. Bound. Elem. 16 (3) (1995) 245-252.

[13] I. Harari, P.E. Barbone, M. Slavutin, R. Shalom, Boundary infinite elements for the Helmholtzequation in exterior domains, Int. J. Numer. Meth. Eng. 41 (6) (1998) 1105-1131.

[14] L. Marin, L. Elliott, P.J. Heggs, D.B. Ingham, D. Lesnic, X. Wen, An alternating iterative algorithmfor the Cauchy problem associated to the Helmholtz equation, Comput. Meth. Appl. Mech. Eng.192 (5-6) (2003) 709-722.

[15] J. Hadamard, Lectures on Cauchy Problems in Linear PDE, New Haven: Yale University Press,1923.

[16] V. Isakov, Inverse problems for partial differential equations, Applied Mathematical Sciences, vol.127, Springer-Verlag, New York, 1998.

[17] A.N. Tikhonov, V.Y. Arsenin, Solutions of Ill-posed Problems, V.H. Winston and Sons, John Wileyand Sons, Washington, DC, New York, 1977.

[18] P. Diamond, P. Kloeden, Characterization of compact subsets of fuzzy sets, Fuzzy Sets Syst. 29(1989) 341-348.

[19] L. Stefanini, B. Bede, Generalized Hukuhara differentiability of interval-valued functions and intervaldifferential equations, Nonlinear Anal. 71 (2009) 1311–1328.

[20] L. Stefanini, A generalization of Hukuhara difference and division for interval and fuzzy arithmetic,Fuzzy Sets Syst. 161 (2010) 1564–1584.

[21] B. Bede, L. Stefanini, Generalized differentiability of fuzzy-valued functions, Fuzzy Sets Syst. 230(2013) 119–141.

[22] Y.Chalco-Canoa,., R.Rodriguez-Lopezb, M.D.Jimenez-Gameroc, Characterizations of generalizeddifferentiable fuzzy functions. Fuzzy Sets Syst. Accepted 4 September 2015.

[23] O. Kaleva, Fuzzy differential equations, Fuzzy Sets Syst. 24 (1987) 301-317.

[24] R. Goestschel, W. Voxman, Elementary fuzzy calculus, Fuzzy Sets Syst. 18 (1986) 31-34.

[25] H.W. Zhang, T. Wei, A quasi-reversibility regularization method for the Cauchy problem of theHelmholtz equation, Int J Comput Math. 88 (4) (2011) 839-850.



Behavior of the Difference Equations xn+1 = xn xn−1 − 1

Keying Liu1, 2 , Peng Li2 , Fei Han3 , Weizhou Zhong1, 4

1. School of Economics and Finance, Xi’an Jiaotong University, Xi’an 710061, China

2. School of Mathematics and Information Sciences, North China University of Water

Resources and Electric Power, Zhengzhou 450045, China

3. Nutrition Data Laboratory, United States Department of Agriculture, Beltsville,

Maryland 20705, USA

4. College of Business Administration, Huaqiao University, Quanzhou 362021, China

Correspondence should be addressed to Zhong Weizhou, [email protected].

Abstract

In this paper, the behavior of solutions of a kind of nonlinear difference equations was

studied. According to the first initial value, the regions of the second initial values was

partitioned by zeroes of auxiliary functions such that the asymptotical behavior of the

equation was determined, which was convergent or unbounded.

Key words: Nonlinear difference equations; Convergent; Unbounded

AMS 2000 Subject Classification: 39A10, 39A11

1 Introduction

In 2011, Kosmala[1] proposed a kind of nonlinear difference equations

xn+1 = xn−k xn−l − 1, n = 1, 2, . . . (1)

with k, l ∈ N and the initial values being real numbers. It stems from investigating periodic

difference equations.

Stevic and Iricanin [2] presented the first general result on the behavior of solutions of (1),

by describing the long-term behavior of the solutions of (1) for all values of parameters k and

l, where the initial values satisfy a special condition.

Moreover, some particular cases of (1) were investigated in [3–7]. Paper [3] investigated the

case where k = 1, l = 2; paper [4] and [7] investigated the case where k = 0, l = 1; paper [5]

investigated the case where k = 0, l = 2; paper [6] investigated the case where k = 0, l = 3.

The relatively simple appearance of (1) is deceiving in that its behavior changes signifi-

cantly for different choices of k and l. These results of (1) were mainly about the periodicity,

unboundedness and stability for particular choices of k and l.

In this paper, we consider a special case of (1), which was investigated in [4] and [7],

xn+1 = xn xn−1 − 1, n = 0, 1, 2, . . . (2)

1


1361 Keying Liu et al 1361-1370

with the initial values x−1 and x0 being real numbers. Note that the equilibria x of (2) are

x1 =1−√

5

2, x2 =

1 +√

5

2.

Furthermore, x1 was locally asymptotically stable and x2 is unstable[4].

We first summarize the main results[4, 7] on the solutions of (2).

(1) (C) If −1 < x−1, x0 < 0, then every solution of (2) converges to x1.

(2) (UB) If one of the following holds, then the solution of (2) is unbounded.

(i) x−1 > x2, x0 > x2;

(ii) x−1 < −1, x0 < −1;

(iii) x−1 < 0, x0 > 0;

(iv) 0 < x−1 < 1, 0 < x0 < 1,

x20x2−1 − 2x0x−1 + 1− x−1 > 0.

(3) (UB or C)

– If 1 < x−1, x0 < x2, then one of the following occurs.

(i) The solution of (2) is unbounded.

(ii) There exists n0 ≥ 1 such that xn ∈ (−1, 0) for all n ≥ n0.

– If x−1 > 0, x0 < 0, then the solution of (2) in certain cases is bounded and in other

cases is unbounded.

– If 0 < x−1, x0 < x2, then the solutions of (2) exhibit somewhat chaotic behavior

relative to the initial values. A little change in the initial conditions can cause a

drastic difference in the long-term behavior of the solutions.

For simplicity, we show them in Figure 1. For the initial values (x−1, x0) in different colored

regions, the solution of (2) is of three kinds: being convergent(C) and being unbounded(UB),

being unbounded or convergent(UB or C).

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

x

y

UB/C

UB/C

C

UB

UB

UB

UB

Figure 1: Different regions of the initial values of (2)

From the above results, one can see that these regions were presented from the perspective

of the relation of two initial values of (2). For the initial values in the green regions, the

corresponding solution is bounded and convergent. For the initial values in the red regions, the

solution is unbounded. As far as the initial values in the blue regions is concerned, the solution

was either unbounded or convergent and such a conclusion was not concise.

2



Specially, for the initial values in the blank regions, the behavior of the solution is unknown.

It is interesting to investigate the evolution of the solution according to the initial values

in the plane. In the following, we try to use a new method to consider the behavior of (2).

Different from the method in [4], we construct auxiliary functions and then use the zeroes of

them to create new partitions of the second initial value. In this way, for the first initial value

which is arbitrarily chosen, the corresponding solution is convergent only for the second initial

value in some intervals which are determined by the zeroes of auxiliary functions. And the

lengths of these intervals are decreasing to zero.

2 Main Results

In this section, we present the main result by investigating the behavior of solutions of (2).

First of all, from the results in [4], we made a little generalization.

Theorem 2.1.

(I) If there is an N≥0 such that −1<xN−1, xN <0, then xn of (2) converges to x1.

(II) If there is an N≥0 such that one of the following five conditions holds, then the solution

of (2) is unbounded.

1) xN−1 > x2, xN > x2;

2) xN−1 < −1, xN < −1;

3) xN−1 < 0, xN > 0;

4) 0 < xN−1 < 1, 0 < xN < 1,

x2Nx2N−1 − 2xNxN−1 +1− xN−1 > 0;

5) xN−1 > 0, xN < −1.

It is worth pointing out that the last case 5) is a direct result of the case 2) and it is crucial

for our main result.

Thus, the behavior of solutions of (2) depends on the location of its two consecutive terms

of xN−1 and xN being less than −1, greater than x2 or in the interval (−1, 0). However, it is

still complicated in terms of the boundedness of solutions of (2) for other cases.

By Remark 2.6 in [4], if the solution of (2) is not periodic or eventually periodic with minimal

period three, then the solution is either bounded, while inside (−1, 0), or unbounded.

Now, we present a necessary and sufficient condition on the existence of eventually prime

period-three solutions of (2).

Lemma 2.1. The solution xn of (2) is an eventually prime period-three solution if and only

if there is an N≥1 such that xN =0.

Proof. By Theorem 2.1 in [4], if the solution xn is an eventually prime period-three solution,

then there is an N≥1 such that xN =0.

On the other hand, if there is an N ≥ 1 such that xN = 0, then we have xN+1 =−1 and

xN+2 =−1 from (2). Thus, it is an eventually prime period-three solution.

In the following, letting the first initial value x−1 being fixed, we consider the behavior of

the solution for the second initial value x0, mainly on the convergence and unboundedness of

the corresponding solution of (2).

For simplicity, we could assume that x−1 =a and x0 =b, where a and b are real numbers.

3



Now, we introduce auxiliary functions Fi(b) = xi for i ≥ 1, from (2), which are

F1(b) = ab− 1, (3)

F2(b) = b F1(b)− 1 = ab2 − b− 1, (4)

F3(b) = F2(b)F1(b)− 1 = b (F 21 (b)− a), (5)

F4(b) = F3(b)F2(b)− 1 = F1(b) (F 22 (b)− b), (6)

F5(b) = F4(b)F3(b)− 1 = F2(b) (F 23 (b)− F1(b)), (7)

and by induction, for i ≥ 5, we have

Fi+1(b) = Fi(b)Fi−1(b)− 1 = Fi−2(b)(F 2i−1(b)− Fi−3(b)), (8)

from which we know that Fi(b) is a higher-degree polynomial of b.

By listing the roots of Fi(b) = 0 for each i ≥ 1, we consider the behavior of Fi(b) with b

in the intervals between these adjacent roots, which describes the long term behavior of the

solution of (2) with the second initial value x0 in such intervals for the first one x−1 being fixed.

In the following, we investigate the roots of Fi(b) = 0 step by step.

It is obvious that r11 = 1/a is the root of F1(b) = 0 if a = 0.

If a ≥ −0.25 and a = 0, then F2(b)=0 has two roots which are

r21 =1−√

1 + 4a

2a, r22 =

1 +√

1 + 4a

2a

and they satisfy r21 < r11 < r22 for a > 0.

It is noted that 0 is always a root of F3(b) = 0 (for convenience, denoted by itself) and for

a > 0, the other two roots are

r31 =1−√a

a, r32 =

1 +√a

a

satisfying 0 < r31 < r11 < r22 < r32 for 0 < a < 1 and r31 < 0 < r11 < r22 < r32 for a > 1.

From (6), we know that F4(b)=0 is equivalent to F1(b)=0 or F 22 (b)=b. Thus r11 is always

a root of F4(b)=0. From F 22 (b) = b, in view of the strict monotonicity of F2(b) for b>r11, there

are only two roots of F4(b)=0, satisfying r41 < r22 < r42 for a>0 and b>0.

Similarly, the other two roots of F5(b) = 0 satisfy r51 < r32 < r52 for a > 0 and b > r11,

which are different from r21 and r22.

Here and after, we only focus on these ”new” roots of Fi(b) = 0, which have not been labeled

by other smaller indices.

Now, we conclude the existence of two roots of Fi+1(b)=0 for i≥5.

Lemma 2.2. Fi+1(b)=0 has only two roots for a>0 and b>r(i−3)2 for i≥5.

Proof. Letting rij be the roots of Fi(b)=0 for i>1 and j=1, 2, from (8), we have

F ′i+1(b)=F ′

i (b)Fi−1(b) + Fi(b)F′i−1(b) > 0 (9)

for b>ri2 and thus Fi+1(b) is strictly increasing for b>ri2.

From (8), we have F 2i−1(b)=Fi−3(b) for i≥5. Hence, in view of the monotonicity of Fi−1(b)

for b>r(i−2)2 and the positivity of Fi−3(b) for b>r(i−3)2, by induction, there are only two roots

of Fi+1(b)=0 satisfying r(i+1)1 < r(i−1)2 < r(i+1)2 for a>0 and b>r(i−3)2.

Furthermore, we could conclude that ri1+∞i=2 and ri2+∞

i=2 are convergent.

4



Lemma 2.3.

limi→+∞

ri1 = limi→+∞

ri2. (10)

Proof. First, from the strict monotonicity of Fi+1(b) for b>ri2, we have ri2 < r(i+1)2 and thus

r22<r32<r42 < . . .. The convergence of ri2+∞i=2 is guaranteed by Fi(b) being a higher-degree

polynomial of b. Similarly, ri1+∞i=2 is convergent.

Denote

limi→+∞

ri1 = b = b(a), limi→+∞

ri2 = b = b(a).

In order to prove (10), we suppose that b < b. Then there exists N such that rN2 > b.

From the above, there exists a root r(N+2)1 of FN+2(b) = 0 near rN2. In view of ri2+∞i=2

being increasing, it is enough to choose such a rN2 that the corresponding r(N+2)1 > b, which

contradicts the convergence of ri1+∞i=2 . The case of b < b is similar. In fact, we can find such

an M that b < rM1 < rM2 and they are roots of FM (b)=0, which contradicts the convergence

of ri2+∞i=2 . Hence, (10) is true.

From the above, for the first initial value x−1 = a being fixed, we have obtained that

Fi(b)=0 has only two ”new” roots for i>3 and the sequences ri1+∞i=2 and ri2+∞

i=2 converge

to a same number.

To investigate the behavior of the solution of (2) with initial values in these intervals which

are partitioned by the adjacent roots rij , we consider three cases.

Case 1 a = 0

If −1<b<0, it follows that both F2(b)=−b− 1 and F3(b)=b are in the interval (−1, 0).

Thus xn of (2) converges to x1 by Theorem 2.1.

Case 2 a < 0

From r11 = 1/a < 0 and F2(b) = a b2 − b − 1, we have that F2(r11) =F2(0) =−1. Hence,

only two cases in the following are needed.

(i) If a <−0.25 and b ∈ (r11, 0), then −1< F1(b), F2(b)< 0. Thus, by Theorem 2.1,

xn of (2) converges to x1.

(ii) If −0.25 ≤ a < 0, then F2(b) = 0 has two roots satisfying r11 < r21 ≤ r22 < 0. For

b ∈ (r11, r21) ∪ (r22, 0), we have −1<F1(b), F2(b)<0 and xn of (2) converges to

x1 by Theorem 2.1. For b∈(−∞, r11), we have F2(b)<−1 and F3(b)<−1 and thus

xn of (2) is unbounded by Theorem 2.1. For b∈ (r21, r22), we have F1(b)<0 and

F2(b)>0 and thus xn of (2) is unbounded by Theorem 2.1.

Generally speaking, for a< 0, the solution xn of (2) converges to x1 only for two cases:

one is a<−0.25 and b ∈ (r11, 0), the other is −0.25 ≤a< 0 and b ∈ (r11, r21) ∪ (r22, 0).

Hence , for a≤0 and b>0, xn of (2) is unbounded. Thus, the dynamics of (2) is clear.

Case 3 a > 0

For this case, it is complicated to arrange these roots rij . We divide it into three cases.

3.1 0 < a < 1

In this case, we prove that

r21 < 0 < r31 < r41 < r11 < r22 < r51 < r61 < r32 < r42 < r52 < r62. (11)

5



From the above, we only need to show r31 < r41 < r11 and r22 < r51 < r61 < r32.

First, from r41 < r22 < r42, in view of F2(r31) < 0 and F ′3(r31) < 0, it follows that

F4(r31) = −1, F4(r11) = 0, F ′4(r31) > 0, F ′

4(r11) = a−1 < 0 (12)

for 0 < a < 1. Thus, r31 < r41 < r11.

Second, in order to prove r22 < r51 < r61 < r32, we only need to show r22 < r51 and

r61 < r32. Thus, the key is to compare ri2 with rj1 for i ≥ 2 and j= i+ 3.

From r51 < r32 < r52, we could conclude that r22 < r51.

In fact, in view of F5(r22) = 0 and F5(r32) = −1, we have

F ′5(r32) = −F ′

3(r32) < 0, (13)

F ′5(r22) = F ′

2(r22) (1− F1(r22)) > 0 (14)

which is guaranteed by

1− F1(r22) =2(2− a)

3 +√

1 + 4a> 0 (15)

for 0 < a < 2. Thus, the conclusion is true.

In a similar way, from r61 < r42 < r62, we conclude r51 < r61 < r32.

In fact, in view of F6(r32) = 0 and F6(r51) = −1, we have

F ′6(r51) = F ′

5(r51)F4(r51) > 0,

F ′6(r32) = F ′

3(r22) (1− F2(r32)) < 0(16)

which is guaranteed by

1− F2(r32) =a− 1

a+√a

< 0 (17)

for 0 < a < 1. Thus, r51 < r61 < r32 holds.

Hence, from the above, it is proved that (11) holds for 0 < a < 1.

Third, we analyze the behavior of xn of (2) with x0 being in the intervals parti-

tioned by these adjacent roots.

(1) For b ∈ (−∞, r21), we have F1(b)< 0, F2(b)> 0. Thus, xn is unbounded by

Theorem 2.1. It is also true for b∈(0, r31)∪(r41, r11)∪(r22, r51)∪(r61, r32) which

are listed in Table 1.

(2) For b∈ (r21, 0), we have −1<F2(b), F3(b)< 0. Thus, xn converges to x1 by

Theorem 2.1. It is also true for b ∈ (r31, r41) ∪ (r11, r22) ∪ (r51, r61) which are

listed in Table 2.

Table 1: Intervals of x0 such that xn is unbounded for 0 < x−1 < 1Intervals of x0 Reasons xn is

(−∞, r21) F1(b) < 0, F2(b) > 0 unbounded

(0, r31) F2(b) < 0, F3(b) > 0 unbounded

(r41, r11) F3(b) < 0, F4(b) > 0 unbounded



6



Table 2: Intervals of x0 such that xn is convergent for 0 < x−1 < 1Intervals of x0 Reasons xn

(r21, 0) −1<F2(b), F3(b)<0 converges to x1

(r31, r41) −1<F3(b), F4(b)<0 converges to x1



3.2 1 ≤ a < 2


r21 < r31 ≤ 0 < r11 ≤ r41 < r22 < r51 < r32 ≤ r61 < r42 < r52 < r62. (18)

In fact, for a = 1, in view of their expressions and F4(b) = (b− 1)2 (b3− b3− 2b− 1),

we have r31 =0, r32 =2=r61, r11 =1=r41 .

For 1 < a < 2, it is apparent that r21 < r31 < 0 < r11 < r22 < r32. From (12), (16)

and (17), we have

F ′4(r11) = a−1 > 0,

F ′6(r32) = F ′

3(r22) a−1a+

√a> 0.

(19)

It follows that r11 < r41 and r32 < r61.

And r22 < r51 follows from (14) and (15). Thus, (18) holds for 1 ≤ x−1 < 2.

It is worth pointing out that xn of (2) converges to x1 for 1 ≤ x−1 < 2 and

x0∈(r21, r31) ∪ (0, r11) ∪ (r41, r22) ∪ (r51, r32) which are listed in Table 3.

Table 3: Intervals of x0 such that xn is convergent for 1≤ x−1<2Intervals of x0 Reasons xn


(0, r11) −1<F3(b), F4(b)<0 converges to x1



3.3 a ≥ 2


r21 < r31 < 0 < r11 < r41 < r51 ≤ r22 < r32 < r61 < r42 < r52 < r62. (20)

Compared with (18), we only need to prove r22 ≥ r51 for a ≥ 2. In fact, from (14)

and (15), for a ≥ 2, we have that r22 ≥ r51.

It is worth pointing out that xn of (2) converges to x1 for x−1 ≥ 2 and x0 ∈(r21, r31) ∪ (0, r11) ∪ (r41, r51) ∪ (r22, r32) which are listed in Table 4.

From the above, we derive such intervals of x0 for x−1 such that xn of (2) is convergent. It

is worth pointing out that we couldn’t continue such a procedure because there are no explicit

expressions of r42 and so on. From the above procedures, we know that the key is how to

compare ri2 with rj1 where j= i+ 3 for i ≥ 4.

In fact, for such an interval Ii = (ri2, rj1) (or (rj1, ri2)) where j= i + 3 for i ≥ 4, in view

of auxiliary functions Fj(b), we have Fj−1(b) < 0 and Fj(b) > 0. Thus, for x−1 being fixed and

x0 ∈∪Ii (the union of Ii for i ≥ 4), xn of (2) is unbounded by Theorem 2.1.

7



Table 4: Intervals of x0 such that xn is convergent for x−1 ≥2Intervals of x0 Reasons xn


(0, r11) −1<F3(b), F4(b)<0 converges to x1



In view of Lemma 2.3, we obtain that the lengths of these open intervals Ii for i ≥ 4 tend to

zero as i tends to +∞. For x0 > b, the increasing property of xn of (2) leads to its divergence.

Therefore, we generalize the above results into the following theorem.

Theorem 2.2. The solution xn of (2) is unbounded only for its second initial value x0 in

such open intervals depending on the first initial valuex0, which are listed in Table 5, where the

endpoints rij are the roots of auxiliary functions Fi(b) = xi = 0 with x0 = b and x−1 = a for

i ≥ 1. And xn of (2) is an eventually prime period-three solution just at x0 = rij or x0 = 0.

For x0 belongs to the complementary set of such intervals except those endpoints, xn of (2)

is convergent to the negative equilibrium x1.

Table 5: Intervals of x0 for x−1 such that xn is unboundedx−1 Intervals of x0

(−∞, −0.25) (−∞, r11) ∪ (0, +∞)

[−0.25, 0) (−∞, r11) ∪ (r21, r22) ∪ (0, +∞)

0 (−∞, −1) ∪ (0, +∞)

(0, 1) (−∞, r21) ∪ (0, r31) ∪ (r41, r11) ∪ (b, +∞) ∪ (∪Ii)

[1, +∞) (−∞, r21) ∪ (0, r31) ∪ (r41, r11) ∪ (b, +∞) ∪ (∪Ii)

From Theorem 2.2 and Table 5, for x−1 and x0 greater than zero, solutions of (2) would ex-

hibit somewhat chaotic behavior[4], that is, xn is either unbounded or convergent alternately

for x0 depending on x−1, which is more concise from Table 5.

Now, we give some examples for particular x−1 which are listed in Table 6. Here, we only

present the former six intervals of x0 such that the solution xn of (2) is convergent. It is

noted that the numerical values of these endpoints of these intervals are approximated to the

values of the solutions of the auxiliary equations Fi(b)=0.

From Table 6, for x−1 =1.5, it is shown the former six intervals of x0 such that the solution

xn of (2) is convergent, which are on both sides of zero. If x0 =1.6 in (1.4975, 1.6073), then

the solution of (2) enters and then remains in the interval (−1, 0), and hence is bounded and

convergent. Whereas if x0 =1.61, then the solution is unbounded. It is clear for the third case

that the solution is UB or C.

3 Conclusion

The existence of prime period-three solutions of (2) is proved in [4] and the convergence of

(2) in its invariant interval (−1, 0) is proved in [7]. In this paper, we present a new method

to partition the intervals of x0 depending on x−1 to describe the behavior of solutions of (2)

and explain in detail that the solution of (2) exhibits somewhat chaotic behavior relative to the

8



Table 6: Intervals of x0 for x−1 > 0 such that xn is convergentx−1 Intervals of x0

0.1 (−0.9161, 0), (6.8377, 7.6946), (10, 10.9161),

(12.4540, 12.8553), (13.1623, 13.4675), (13.6396, 13.7755)

0.618 (−0.6985, 0), (0.3461, 1.4048), (1.6181, 2.3166),

(2.5350, 2.8614), (2.8902, 3.0690), (3.0996, 3.1756)

1 (−0.618, 0), (0, 1), (1, 1.6180), (1.7121, 2),

(2, 2.1479), (2.1637, 2.2237)

1.5 (−0.5486, −0.1498), (0, 0.6667), (0.7717, 1.2153),

(1.2447, 1.4832), (1.4975, 1.6073), (1.6149, 1.6633)

2.5 (−0.4633, −0.2325), (0, 0.4), (0.5711, 0.8476),

(0.8633, 1.0325), (1.0558, 1.1302), (1.1316, 1.1680)

10 (−0.2702, −0.2162), (0, 0.1), (0.2740, 0.3327),

(0.3702, 0.4162), (0.4383, 0.4596), (0.4630, 0.4752)

initial values. Compared with the known results[4], our results are much more accurate and

easy to obtain by computers to describe the evolution of (2) for the initial values in the plane.

We conclude that the solution of (2) is bounded and convergent only for x0 in particular

intervals depending on x−1, which are partitioned by the zeroes of auxiliary functions presented

in this paper. Specially, it is unbounded only for x0 in such open intervals listed in Table 5

which depend on x−1.

It is of great interest to continue the investigation of the monotonicity, periodicity, and

boundedness nature of solutions of (1) for different choices of parameters k and l and other

equations presented in [4]. We believe that prime-period solutions and the negative equilibrium

are crucial for the dynamics of difference equations (1). The future work is to extend our study

to a more generalized equation (1).

Conflict of Interests

The authors declare that they have no competing interests.

Acknowledgement

This work is financially supported by the Natural Science Foundation of China (No. 71271086,

71172184) and the Education Department of Henan Province (No. 12A110014).

References

[1] W. Kosmala,A period 5 difference equation, International Journal of Nonlinear Analysis and Ap-

plications, 2(1)( 2011): 82-84.

[2] S. Stevic, B. Iricanin, Unbounded solutions of the difference equation xn+1 = xn−l xn−k − 1 ,

Abstract and Applied Analysis, Article ID 561682, 2011.

[3] C. M. Kent, W. Kosmala, S. Stevic, Long-term behavior of solutions of the difference equation

xn+1 =xn−1 xn−2 − 1 , Abstract and Applied Analysis, Article ID 152378, 2010.

9



[4] C.M. Kent, W. Kosmala, M. Radin, S. Stevic, Solutions of the difference equation xn+1 =xn xn−1−1 , Abstract and Applied Analysis, Article ID 469683, 2010.

[5] C.M. Kent, W. Kosmala, S. Stevic, On the difference equation xn+1 =xn xn−2−1 , Abstract and

Applied Analysis, Article ID 815285, 2011.

[6] C.M. Kent, W. Kosmala, On the nature of solutions of the difference equation xn+1 =xn xn−3−1,

International Journal of Nonlinear Analysis and Applications, 2(2) (2011): 24-43.

[7] Yitao Wang, Yong Luo, Zhengyi Lu, Convergence of solutions of xn+1 =xn xn−1 − 1, Appl. Math.

E-Notes, 12 (2012): 153-157.

10



TABLE OF CONTENTS, JOURNAL OF COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 22, NO. 7, 2017

On Quadratic 𝜌-Functional Inequalities in Fuzzy Normed Spaces, Choonkil Park, Sun Young Jang, and Sungsik Yun, ………………………………………………………………………1189

On a Double Integral Equation Including a Set of Two Variables Polynomials Suggested by Laguerre Polynomials, M. Ali Özarslan and Cemaliye Kürt,………………………………...1198

Generalized Inequalities of the type of Hermite-Hadamard-Fejer with Quasi-Convex Functions by way of k-Fractional Derivatives, A. Ali, G. Gulshan, R. Hussain, A. Latif, and M. Muddassar,…………………………………………………………………………………….1208

Nonlinear Differential Polynomials of Meromorphic Functions with Regard to Multiplicity Sharing a Small Function, Jianren Long,……………………………………………………...1220

Impulsive Hybrid Fractional Quantum Difference Equations, Bashir Ahmad, Sotiris K. Ntouyas, Jessada Tariboon, Ahmed Alsaedi, and Wafa Shammakh,…………………………………...1231

A Fixed Point Alternative to the Stability of a Quadratic 𝛼-Functional Equation in Fuzzy Banach Spaces, Choonkil Park, Jung Rye Lee, and Dong Yun Shin,…………………………………1241

Four-Point Impulsive Multi-Orders Fractional Boundary Value Problems, N. I. Mahmudov and H. Mahmoud,…………………………………………………………………………………1249

Convergence of Modification of the Kantorovich-Type q-Bernstein-Schurer Operators, Qing-Bo Cai and Guorong Zhou,………………………………………………………………………..1261

Barnes-Type Degenerate Bernoulli and Euler Mixed-Type Polynomials, Taekyun Kim, Dae San Kim, Hyuckin Kwon, and Toufik Mansour,…………………………………………………..1273

Ground State Solutions for Second Order Nonlinear p-Laplacian Difference Equations with Periodic Coefficients, Ali Mai and Guowei Sun,……………………………………………..1288

On a Solutions of Fourth Order Rational Systems of Difference Equations, E. M. Elsayed, Abdullah Alotaibi, and Hajar A. Almaylabi,………………………………………………….1298

On the Dynamics of Higher Order Difference Equations 𝑥𝑛+1 = 𝑎𝑥𝑛 + 𝛼𝑥𝑛𝑥𝑛−𝑙𝛽𝑥𝑛+𝛾𝑥𝑛−𝑘

, M. M. El-

Dessoky,………………………………………………………………………………………1309

Applications of Soft Sets in BF-Algebras, Jeong Soon Han and Sun Shin Ahn,…………….1323

TABLE OF CONTENTS, JOURNAL OF COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 22, NO. 7, 2017

(continued)

Symmetric Solutions for Hybrid Fractional Differential Equations, Jessada Tariboon, Sotiris K. Ntouyas, and Suthep Suantai,…………………………………………………………………1332

On the k-th Degeneration of the Genocchi Polynomials, Lee-Chae Jang, C.S. Ryoo, Jeong Gon Lee, and Hyuck In Kwon,……………………………………………………………………..1343

Regularization Solutions of Ill-Posed Helmholtz-Type Equations with Fuzzy Mixed Boundary Value, Hong Yang and Zeng-Tai Gong,………………………………………………………1350

Behavior of the Difference Equations 𝑥𝑛+1 = 𝑥𝑛𝑥𝑛−1 − 1, Keying Liu, Peng Li, Fei Han, and Weizhou Zhong,……………………………………………………………………………….1361

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