Research Article Existence and Uniqueness of Periodic

Research ArticleExistence and Uniqueness of Periodic Solutions forSecond Order Differential Equations

Yuanhong Wei

College of Mathematics Jilin University Changchun 130012 China

Correspondence should be addressed to Yuanhong Wei yhweiamssaccn

Received 12 June 2014 Accepted 16 August 2014 Published 27 August 2014

Academic Editor Mohamed Abdalla Darwish

Copyright copy 2014 Yuanhong Wei This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study some second order ordinary differential equations We establish the existence and uniqueness in some appropriatefunction space By using Schauderrsquos fixed-point theorem new results on the existence and uniqueness of periodic solutions areobtained

1 Introduction

In this paper we are concerned with the existence anduniqueness of periodic solution for the nonlinear equation

11990910158401015840= 119891 (119905 119909 119909

1015840) (1)

where 119891 R times R times R rarr R is continuous periodic in 119905 withperiod 119879 and 1198621 with respect to (119909 1199091015840)

Because of wide interests in physics and engineeringperiodic solutions of second order differential equations havebeen investigated by many authors We refer the reader to [1ndash7] and the references cited therein

The purpose of this paper is to study the existence anduniqueness of periodic solution in some appropriate functionspace 1198621

119879 To be precise we first derive a result of existence

and uniqueness when the nonlinearity is a 1198621 functionwith respect to (119909 119909

1015840) Then a similar result concerning

the existence of periodic solutions is obtained when thenonlinearity is not a 1198621 function

Throughout this paper we use the following assumptions

(A1) There exist two continuous functions 119886(119905) and 119887(119905)such that for all 119905 isin [0 119879]

0 ≨ 119886 (119905) ⩽ 119891119909(119905 119909 119909

1015840) ⩽ 119887 (119905) (2)

where 0 ≨ 119886(119905)means 0 ⩽ 119886(119905) and 0 equiv 119886(119905)

(A2) There exists a positive constant 119872 such that for all119905 isin [0 119879]

100381610038161003816100381610038161198911199091015840 (119905 119909 119909

1015840)

10038161003816100381610038161003816⩽ 119872 (3)

We now present our main results of this paper

Theorem 1 Let assumptions (A1) and (A2) hold Then (1) hasa unique 119879-periodic solution

Remark 2 We point out that the condition 0 ≨ 119886(119905) ⩽ 119891119909

in (A1) is necessary Let us take the following equation forexample

11990910158401015840= 119862 (4)

where119862 is a constant and119862 = 0The equation has no periodicsolution

Remark 3 Consider the following example

11990910158401015840= sin 119905 sdot sin119909 sdot sin1199091015840 + 119909 (5)

We can easily check that assumptions (A1) and (A2) hold ByTheorem 1 the equation has a unique 2120587-periodic solution

We also consider the case when the right-hand side ofthe equation is only continuous In this case we establish theexistence of periodic solutions for the differential equation

11990910158401015840= 119891 (119905 119909 119909

1015840) 119909 + 119892 (119905 119909 119909

1015840) (6)

Hindawi Publishing CorporationJournal of Function SpacesVolume 2014 Article ID 246258 5 pageshttpdxdoiorg1011552014246258

2 Journal of Function Spaces

where 119891 119892 RtimesRtimesR rarr R are continuous and periodic in119905 with period 119879 We assume the following

(A3) There exist continuous functions 119886(119905) and 119887(119905) and apositive constant119872 such that

0 ≨ 119886 (119905) ⩽ 119891 (119905 119909 1199091015840) ⩽ 119887 (119905)

10038161003816100381610038161003816119892 (119905 119909 119909

1015840)

10038161003816100381610038161003816⩽ 119872

(7)

for all 119905 isin [0 119879]

Then we have the following result

Theorem4 Let (A3) holdThen (6) has at least one119879-periodicsolution


11990910158401015840= 2119909 + 119909 sin1199091015840 + cos 119905 sdot cos119909 (8)

By Theorem 4 the equation has at least one 2120587-periodicsolution

To establish the main results we introduce some appro-priate function space By using Schauderrsquos fixed-point theo-rem the existence and uniqueness of periodic solutions areobtained We know from the anonymous referee that theproof of this paper can be simplified by using the results in[8] The proof of this paper can be seen as an application ofSchauderrsquos fixed-point theorem

2 Preliminary

In this sectionWefirst introduce the function space inwhichwe will obtain the periodic solution for the problem Thensome preliminary lemma is introduced which is valuable forthe proof of our main results

Let 1198621119879be the space of continuously differentiable 119879-

periodic functions with norm sdot given by

119909 = max119905isin[0119879]

|119909 (119905)| + max119905isin[0119879]

100381610038161003816100381610038161199091015840

(119905)

10038161003816100381610038161003816 (9)

It is well known that 1198621119879is a Banach space

We now introduce the following lemma

Lemma 6 119906(119905) and V(119905) are 119879-periodic functions If 0 ≨ V(119905)for all 119905 isin 119877 then the equation

11990910158401015840= 119906 (119905) 119909

1015840+ V (119905) 119909 (10)

has a unique 119879-periodic solution 119909(119905) equiv 0

Proof Assume there exists a 119879-periodic solution 119909(119905)119909(119905) equiv 0 Since 0 ≨ V(119905) 119909(119905) is not a constant Otherwise wewill get a contradiction by substituting 119909 = 119862 = 0 into (10)Then we claim that there exist 119905

0and 1199051 1199050lt 1199051 such that

119909 (119905) gt 0 for 119905 isin (1199050 1199051)

1199091015840(1199050) gt 0 119909

1015840(1199051) = 0

(11)

Now we prove it

Case 1 119909(119905) has zero points Assume that 119909(120579) = 0 then wehave 1199091015840(120579) = 0 because the initial value problem of (10) has aunique solution If 1199091015840(120579) gt 0 then we let 119905

0= 120579 if 1199091015840(120579) lt 0

by the periodic condition there exists 1199050 1199050gt 120579 such that

119909 (119905) lt 0 119905 isin (120579 1199050) 119909 (119905

0) = 0 (12)

So 1199091015840(1199050) gt 0 Also by the periodic condition there exists 120578

120578 gt 1199050 such that

119909 (119905) gt 0 119905 isin (1199050 120578) 119909 (120578) = 0 (13)

So we can find a 1199051isin (1199050 120578) such that 1199091015840(119905

1) = 0 Thus under

Case 1 we verify (11) holds

Case 2 119909(119905) has no zero point Then 119909(119905) has a constant signSuppose 119909(119905) gt 0 (if not we can consider the 119879-periodicsolution 119909(119905) = minus119909(119905) 119909(119905) gt 0) Since 119909(119905) is periodicand is not a constant there exists 119905

0such that 1199091015840(119905

0) gt 0

119909(1199050) = 119909(119905

0+ 119879) So we can find a 119905

1 1199051gt 1199050 1199091015840(1199051) = 0

Then we prove that (11) holdsMultiplying both sides of (10) by expminus int1199051

1199050

119906(119904)119889119904 andintegrating between 119905

0and 1199051we get

0 gt minus1199091015840(1199050) = int

1199051

1199050

V (119905) 119909 (119905) expminusint1199051

1199050

119906 (119904) 119889119904 119889119905 ⩾ 0

(14)

which leads to a contradiction This completes the proof ofLemma 6

Remark 7 We can also prove the following If 119906(0) = 119906(119879)V(0) = V(119879) 0 ≨ V(119905) ae 119905 isin [0 119879] then the followingperiodic boundary value problem

11990910158401015840= 119906 (119905) 119909

1015840+ V (119905) 119909 119905 isin [0 119879]

119909 (0) = 119909 (119879) 1199091015840

(0) = 1199091015840

(119879)

(15)

also has a unique solution 119909(119905) equiv 0

3 Proof of the Main Results

Proof of Theorem 1 First prove the uniqueness Assume that1199091(119905) and 119909

2(119905) are two 119879-periodic solutions of (1) Setting

119909 (119905) = 1199091(119905) minus 119909

2(119905) (16)

we get

119909 (119905 + 119879) = 119909 (119905)

11990910158401015840= 119891 (119905 119909

1 1199091015840

1) minus 119891 (119905 119909

2 1199091015840

2)

= 119891119909(119905 1199091+ 1205791(1199091minus 1199092) 1199091015840

1) 119909

+ 1198911199091015840 (119905 1199092 1199092+ 1205792(1199091minus 1199092)) 1199091015840

(17)

From assumption (A1) we know

0 ≨ 119886 (119905) ⩽ 119891119909(119905 1199091+ 1205791(1199091minus 1199092) 1199091015840

1) (18)

Hence by Lemma 6 119909(119905) equiv 0

Journal of Function Spaces 3

We next prove the existence Rewrite (1) in the followingform

11990910158401015840= 119891 (119905 119909 119909

1015840)

= (119891 (119905 119909 1199091015840) minus 119891 (119905 119909 0))

+ (119891 (119905 119909 0) minus 119891 (119905 0 0)) + 119891 (119905 0 0)

= int

1

0

1198911199091015840 (119905 119909 120579

11199091015840) 11988912057911199091015840

+ int

1

0

119891119909(119905 1205792119909 0) 119889120579

2119909 + 119891 (119905 0 0)

(19)

From Lemma 6 for each 119909 isin 1198621119879 the equation

11991010158401015840= int

1

0

1198911199091015840 (119905 119909 120579

11199091015840) 11988912057911199101015840

+ int

1

0

119891119909(119905 1205792119909 0) 119889120579

2119910 + 119891 (119905 0 0)

(20)

has a unique 119879-periodic solution because the correspondinghomogeneous equation only has trivial 119879-periodic solution119909(119905) equiv 0

We define operator

119875 1198621

119879997888rarr 119862

1

119879 (21)

for given 119909 isin 1198621

119879 119910(119905) = 119875[119909](119905) is the unique 119879-periodic

solution of (20)Then the existence of the119879-periodic solutionis equivalent to the existence of fixed point of 119875 in the space1198621

119879 We will prove that 119875 is continuous and compact and

119875(1198621

119879) is a bounded subset of 1198621

119879

Proof of Continuity For any convergent sequence 119909119896 sub 119862

1

119879

satisfying 119909119896rarr 1199090(119896 rarr infin) let 119910

119896= 119875119909119896 Then

11991010158401015840

119896= int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2119910119896+ 119891 (119905 0 0)

(22)

We claim that 119910119896 is a bounded sequence in 1198621

119879 If not we

can find a subsequence of 119910119896 (for convenience we also use

the same notations) such that 119910119896 rarr infin (119896 rarr infin) Let

120596119896= 119910119896119910119896 Then 120596

119896 sub 1198621

119879 120596119896 = 1

12059610158401015840

119896= int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 11988912057911205961015840

119896

+ int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2120596119896+

119891 (119905 0 0)

1003817100381710038171003817119910119896

1003817100381710038171003817

(23)

So 12059610158401015840119896 ⩽ 119872+max

119905isin[0119879]119887(119905)+1 lt infin (119896 rarr infin) Since 12059610158401015840

119896

is bounded and

1205961015840

119896(119905) = 120596

1015840

119896(0) + int

119905

0

12059610158401015840

119896(119904) 119889119904 (24)

1205961015840

119896 is bounded and equicontinuous sequence of functions

Furthermore by

120596119896(119905) = 120596

119896(0) + int

119905

0

1205961015840

119896(119904) 119889119904 (25)


By Ascoli-ArzelaTheorem 1205961015840119896 and 120596

119896 contain a uniformly

convergent subsequence respectively (for convenience wealso use the same notations) such that

120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0 (26)

where the notation ldquo 1997888rarrrdquo means uniform convergence Obvi-ously 120596

0 V0isin 1198621

119879 From (23) and (24) we obtain

1205961015840

119896(119905) = 120596

1015840

119896(0)

+ int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911205961015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2120596119896+

119891 (119904 0 0)

1003817100381710038171003817119910119896

1003817100381710038171003817

) 119889119904

(27)

Let 119896 rarr infin From (25) and (27) we get

1205960(119905) = 120596

0(0) + int

119905

0

V0(119904) 119889119904

V0(119905) = V

0(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 1198891205791V0(119904)

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

21205960(119904)) 119889119904

(28)

Hence

12059610158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911205961015840

0+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21205960

(29)

By Lemma 6 we have 1205960equiv 0 which is in contradiction

with 1205960 = 1 so 119910

119896 is a bounded sequence Then by

(22) we know 11991010158401015840119896 is bounded so 1199101015840

119896 and 119910

119896 are bounded

and equicontinuous sequences of functions By Ascoli-ArzelaTheorem 1199101015840

119896 and 119910

119896 contain a uniformly convergent

subsequence respectively (for convenience we also use thesame notations) such that

119910119896

1

997888rarr 1199100 119910

1015840

119896

1

997888rarr V0 (30)


We know

1199101015840

119896(119905) = 119910

1015840

119896(0) + int

119905

0

11991010158401015840

119896(119904) 119889119904

= 1199101015840

119896(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2119910119896

+119891 (119904 0 0) ) 119889119904

119910119896(119905) = 119910

119896(0) + int

119905

0

1199101015840

119896(119904) 119889119904

(31)

When 119896 rarr infin from (31) we obtain

11991010158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911199101015840

0

+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21199100+ 119891 (119905 0 0)

(32)

By the uniqueness we know 1199100= 119875119909

0 thus operator 119875 is

continuous

Proof of Compactness For each bounded set 119878 sub 1198621119879 we claim

that 119875(119878) is bounded in 1198621119879 If not by an analogous manner

as above we will reach a contradiction For every 119909 isin 119878119910 = 119875119909 is defined by (20) Since 1199101015840 119910 119891

119909 and 119891

1199091015840

are all bounded then 11991010158401015840 lt infin Proceeding as proof ofcontinuity we conclude that 1199101015840 and 119910 are bounded andequicontinuous sequences of functions By the Ascoli-ArzelaTheorem 119875 is a compact operator

We claim that 119875(1198621119879) is bounded in 1198621

119879 If not there exist

119909119896 119896 = 1 2 such that 119875119909

119896 rarr infin (119896 rarr infin)

Let 119910119896= 119875119909

119896 Then (22) holds Take 120596

119896= 119910119896119910119896 Then

120596119896 sub 119862

1

119879 120596119896 = 1 and (23) (24) (25) and (27) hold

By the above proof we know 1205961015840

119896 and 120596

119896 are bounded

and equicontinuous sequences of functions and contain auniformly convergent subsequence respectively (also use thesame notations) such that

120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0

10038171003817100381710038171205960

1003817100381710038171003817= 1 (33)

Sequences int

1

01198911199091015840(119905 119909119896 12057911199091015840

119896)1198891205791

infin

119896=1and int

1

0119891119909(119905 1205792119909119896

0)1198891205792infin

119896=1are bounded sequences in 1198712[0 119879] By the weakly

sequential compactness of 1198712 space both of them have aweakly convergent subsequence (also use the same notations)such that

int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 1198891205791

119908

997888rarr 1198911(119905)

int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2

119908

997888rarr 1198912(119905)

(34)

in 1198712[0 119879] where ldquo 119908997888rarrrdquo means weak convergence Obviously

10038161003816100381610038161198911(119905)1003816100381610038161003816⩽ 119872 0 ≨ 119886 (119905) ⩽ 119891

2(119905) ⩽ 119887 (119905)

ae 119905 isin [0 119879] (35)

When 119896 rarr infin from (25) and (27) for ae 119905 isin [0 119879] wehave

V10158400(119905) = 119891

1(119905) V0(119905) + 119891

2(119905) 1205960(119905) 120596

1015840

0(119905) = V

0(119905)

(36)

Then

12059610158401015840

0(119905) = 119891

1(119905) 1205961015840

0(119905) + 119891

2(119905) 1205960(119905)

1205960(0) = 120596

0(119879) 120596

1015840

0(0) = 120596

1015840

0(119879)

(37)

By Remark 3 we obtain 1205960equiv 0 which is in contradiction

with 1205960 = 1 Then there exists a constant 119870 gt 0 such that

119875119909 ⩽ 119870 119909 isin 1198621119879

Assume 119863 = 119909 isin 1198621

119879 119909 ⩽ 119870 + 1 By Schauderrsquos fixed-

point theorem 119875 119863 rarr 119863 has at least one fixed point Thiscompletes the proof of Theorem 1

Proof of Theorem 4 For each 119909 isin 1198621

119879 from Lemma 6 we

know the equation

11991010158401015840= 119891 (119905 119909 119909

1015840) 119910 + 119892 (119905 119909 119909

1015840) (38)


We define 119875 1198621119879rarr 1198621

119879 for each given 119909 isin 1198621

119879119910(119905) =

119875[119909](119905) is the unique 119879-periodic solution of (38) Hencethe existence of the periodic solutions is equivalent to theexistence of fixed points of 119875 in the Banach space 1198621

119879

Proceeding as the proof of Theorem 1 we can prove that119875 is a compact continuous operator and 119875(1198621

119879) is a bounded

subset of 1198621119879

Then there exists a constant 1198701gt 0 such that 119875119909 ⩽ 119870

1

for all 119909 isin 1198621119879 Let119863 = 119909 isin 119862

1

119879 119909 ⩽ 119870

1+ 1 By Schauderrsquos

fixed-point theorem 119875 119863 rarr 119863 has at least one fixed pointThis completes the proof of Theorem 4

4 Another Simple Proof

Actually by the anonymous referee we know that the proofof the theorem can be much simplified if we use the theoremin [8] In fact for equation

119906(119899)= 119891 (119905 119906 119906

(119899minus1)) (39)

where 119891 R119899+1 rarr R is continuous and 119879-periodic in 119905Theorem 22 of [8] implies the following propositions


Lemma 8 Let there exist continuous and 119879-periodic in thefirst argument functions 119891

119896 R119899+1 rarr R (119896 = 1 2 119899)

such that1003816100381610038161003816100381610038161003816100381610038161003816

119891 (119905 1199091 119909

119899) minus

119899

sum

119896=1

119891119896(119905 1199091 119909

119899) 119909119896

1003816100381610038161003816100381610038161003816100381610038161003816

⩽ 119903

1199011119896(119905) ⩽ 119891

119896(119905 1199091 119909

119899) ⩽ 1199012119896(119905) (119896 = 1 2 119899)

(40)

are satisfied on R119899+1 where 119903 = 119888119900119899119904119905 gt 0 and 119901119894119896

R rarr R (119894 = 1 2 119896 = 1 2 119899) are continuous119879-periodic functions Let moreover for any continuous 119879-periodic functions 119901

119896 R rarr R (119896 = 1 2 119899) satisfying

1199011119896(119905) ⩽ 119901

119896(119905) ⩽ 119901

2119896(119905) (119896 = 1 2 119899) (41)

the equation

119906(119899)=

119899

sum

119896=1

119901119896(119905) 119906(119896minus1) (42)

have no nontrivial 119879-periodic solution Then equation has atleast one 119879-periodic solution

Lemma 9 Let the function 119891 in the last 119899 arguments havecontinuous partial derivatives satisfying

1199011119896(119905) ⩽

120597119891119896(119905 1199091 119909

119899)

120597119909119896

⩽ 1199012119896(119905) (119896 = 1 2 119899)

(43)

where 119901119894119896 R rarr R (119894 = 1 2 119896 = 1 2 119899) are continuous

119879-periodic functions Let moreover for any 119879-periodic 119901119896

R rarr R (119896 = 1 2 119899) satisfying (41) (42) has nonontrivial119879-periodic solutionThen the equation has a unique119879-periodic solution

On the basis of these theorems we can prove the theoremby checking the conditions in the previous lemma ByLemma 6 the conditions above can be easily proved

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This paper is supported by National Natural Science Founda-tion of China (Grant no 11301209) The author would like tothank the anonymous referee for the valuable comments andsuggestions on the paper especially for the considerations inSection 4 The author would like to thank Professor Yong Lifor his helpful instruction and valuable suggestions

References

[1] F Cong ldquoPeriodic solutions for second order differential equa-tionsrdquo Applied Mathematics Letters vol 18 no 8 pp 957ndash9612005

[2] J Ward ldquoPeriodic solutions for a class of ordinary differentialequationsrdquo Proceedings of the American Mathematical Societyvol 78 no 3 pp 350ndash352 1980

[3] R Kannan ldquoExistence of periodic solutions of nonlinear dif-ferential equationsrdquo Transactions of the AmericanMathematicalSociety vol 217 pp 225ndash236 1976

[4] K Schmitt ldquoA note on periodic solutions of second orderordinary differential equationsrdquo SIAM Journal onAppliedMath-ematics vol 21 pp 491ndash494 1971

[5] Q Yao ldquoPositive solutions of nonlinear second-order periodicboundary value problemsrdquoAppliedMathematics Letters vol 20no 5 pp 583ndash590 2007

[6] I Kiguradze and S Stanck ldquoOn periodic boundary valueproblem for the equation11990610158401015840 = 119891(119905 119906 1199061015840)with one-sided growthrestrictions on frdquo Nonlinear Analysis vol 48 pp 1065ndash10752002

[7] P Yan andMZhang ldquoHigher order non-resonance for differen-tial equations with singularitiesrdquo Mathematical Methods in theApplied Sciences vol 26 no 12 pp 1067ndash1074 2003

[8] I Kiguradze ldquoBoundary value problems for systems of ordinarydifferential equationsrdquo Itogi Nauki i Tekhniki Seriya Sovremen-nye Problemy Matematiki Noveishie Dostizheniya vol 30 pp3ndash103 1987 (Russian) English translation Journal of SovietMathematics vol 43 no 2 pp 2259ndash2339 1988

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Differential EquationsInternational Journal of

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where 119891 119892 RtimesRtimesR rarr R are continuous and periodic in119905 with period 119879 We assume the following

(A3) There exist continuous functions 119886(119905) and 119887(119905) and apositive constant119872 such that

0 ≨ 119886 (119905) ⩽ 119891 (119905 119909 1199091015840) ⩽ 119887 (119905)

10038161003816100381610038161003816119892 (119905 119909 119909

1015840)

10038161003816100381610038161003816⩽ 119872

(7)

for all 119905 isin [0 119879]

Then we have the following result

Theorem4 Let (A3) holdThen (6) has at least one119879-periodicsolution


11990910158401015840= 2119909 + 119909 sin1199091015840 + cos 119905 sdot cos119909 (8)

By Theorem 4 the equation has at least one 2120587-periodicsolution

To establish the main results we introduce some appro-priate function space By using Schauderrsquos fixed-point theo-rem the existence and uniqueness of periodic solutions areobtained We know from the anonymous referee that theproof of this paper can be simplified by using the results in[8] The proof of this paper can be seen as an application ofSchauderrsquos fixed-point theorem

2 Preliminary

In this sectionWefirst introduce the function space inwhichwe will obtain the periodic solution for the problem Thensome preliminary lemma is introduced which is valuable forthe proof of our main results

Let 1198621119879be the space of continuously differentiable 119879-

periodic functions with norm sdot given by

119909 = max119905isin[0119879]

|119909 (119905)| + max119905isin[0119879]

100381610038161003816100381610038161199091015840

(119905)

10038161003816100381610038161003816 (9)

It is well known that 1198621119879is a Banach space

We now introduce the following lemma

Lemma 6 119906(119905) and V(119905) are 119879-periodic functions If 0 ≨ V(119905)for all 119905 isin 119877 then the equation

11990910158401015840= 119906 (119905) 119909

1015840+ V (119905) 119909 (10)

has a unique 119879-periodic solution 119909(119905) equiv 0

Proof Assume there exists a 119879-periodic solution 119909(119905)119909(119905) equiv 0 Since 0 ≨ V(119905) 119909(119905) is not a constant Otherwise wewill get a contradiction by substituting 119909 = 119862 = 0 into (10)Then we claim that there exist 119905

0and 1199051 1199050lt 1199051 such that

119909 (119905) gt 0 for 119905 isin (1199050 1199051)

1199091015840(1199050) gt 0 119909

1015840(1199051) = 0

(11)

Now we prove it

Case 1 119909(119905) has zero points Assume that 119909(120579) = 0 then wehave 1199091015840(120579) = 0 because the initial value problem of (10) has aunique solution If 1199091015840(120579) gt 0 then we let 119905

0= 120579 if 1199091015840(120579) lt 0

by the periodic condition there exists 1199050 1199050gt 120579 such that

119909 (119905) lt 0 119905 isin (120579 1199050) 119909 (119905

0) = 0 (12)

So 1199091015840(1199050) gt 0 Also by the periodic condition there exists 120578

120578 gt 1199050 such that

119909 (119905) gt 0 119905 isin (1199050 120578) 119909 (120578) = 0 (13)

So we can find a 1199051isin (1199050 120578) such that 1199091015840(119905

1) = 0 Thus under

Case 1 we verify (11) holds

Case 2 119909(119905) has no zero point Then 119909(119905) has a constant signSuppose 119909(119905) gt 0 (if not we can consider the 119879-periodicsolution 119909(119905) = minus119909(119905) 119909(119905) gt 0) Since 119909(119905) is periodicand is not a constant there exists 119905

0such that 1199091015840(119905

0) gt 0

119909(1199050) = 119909(119905

0+ 119879) So we can find a 119905

1 1199051gt 1199050 1199091015840(1199051) = 0

Then we prove that (11) holdsMultiplying both sides of (10) by expminus int1199051

1199050

119906(119904)119889119904 andintegrating between 119905

0and 1199051we get

0 gt minus1199091015840(1199050) = int

1199051

1199050

V (119905) 119909 (119905) expminusint1199051

1199050

119906 (119904) 119889119904 119889119905 ⩾ 0

(14)

which leads to a contradiction This completes the proof ofLemma 6

Remark 7 We can also prove the following If 119906(0) = 119906(119879)V(0) = V(119879) 0 ≨ V(119905) ae 119905 isin [0 119879] then the followingperiodic boundary value problem

11990910158401015840= 119906 (119905) 119909

1015840+ V (119905) 119909 119905 isin [0 119879]

119909 (0) = 119909 (119879) 1199091015840

(0) = 1199091015840

(119879)

(15)

also has a unique solution 119909(119905) equiv 0

3 Proof of the Main Results

Proof of Theorem 1 First prove the uniqueness Assume that1199091(119905) and 119909

2(119905) are two 119879-periodic solutions of (1) Setting

119909 (119905) = 1199091(119905) minus 119909

2(119905) (16)

we get

119909 (119905 + 119879) = 119909 (119905)

11990910158401015840= 119891 (119905 119909

1 1199091015840

1) minus 119891 (119905 119909

2 1199091015840

2)

= 119891119909(119905 1199091+ 1205791(1199091minus 1199092) 1199091015840

1) 119909

+ 1198911199091015840 (119905 1199092 1199092+ 1205792(1199091minus 1199092)) 1199091015840

(17)

From assumption (A1) we know

0 ≨ 119886 (119905) ⩽ 119891119909(119905 1199091+ 1205791(1199091minus 1199092) 1199091015840

1) (18)

Hence by Lemma 6 119909(119905) equiv 0



11990910158401015840= 119891 (119905 119909 119909

1015840)

= (119891 (119905 119909 1199091015840) minus 119891 (119905 119909 0))

+ (119891 (119905 119909 0) minus 119891 (119905 0 0)) + 119891 (119905 0 0)

= int

1

0

1198911199091015840 (119905 119909 120579

11199091015840) 11988912057911199091015840

+ int

1

0

119891119909(119905 1205792119909 0) 119889120579

2119909 + 119891 (119905 0 0)

(19)


11991010158401015840= int

1

0

1198911199091015840 (119905 119909 120579

11199091015840) 11988912057911199101015840

+ int

1

0

119891119909(119905 1205792119909 0) 119889120579

2119910 + 119891 (119905 0 0)

(20)


We define operator

119875 1198621

119879997888rarr 119862

1

119879 (21)





119875(1198621


119879


1

119879


119896= 119875119909119896 Then

11991010158401015840

119896= int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2119910119896+ 119891 (119905 0 0)

(22)


119879 If not we



120596119896= 119910119896119910119896 Then 120596

119896 sub 1198621

119879 120596119896 = 1

12059610158401015840

119896= int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 11988912057911205961015840

119896

+ int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2120596119896+

119891 (119905 0 0)

1003817100381710038171003817119910119896

1003817100381710038171003817

(23)

So 12059610158401015840119896 ⩽ 119872+max


119896

is bounded and

1205961015840

119896(119905) = 120596

1015840

119896(0) + int

119905

0

12059610158401015840

119896(119904) 119889119904 (24)

1205961015840


Furthermore by

120596119896(119905) = 120596

119896(0) + int

119905

0

1205961015840

119896(119904) 119889119904 (25)





120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0 (26)


0 V0isin 1198621


1205961015840

119896(119905) = 120596

1015840

119896(0)

+ int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911205961015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2120596119896+

119891 (119904 0 0)

1003817100381710038171003817119910119896

1003817100381710038171003817

) 119889119904

(27)


1205960(119905) = 120596

0(0) + int

119905

0

V0(119904) 119889119904

V0(119905) = V

0(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 1198891205791V0(119904)

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

21205960(119904)) 119889119904

(28)

Hence

12059610158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911205961015840

0+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21205960

(29)


with 1205960 = 1 so 119910



119896 and 119910

119896 are bounded


119896 and 119910



119910119896

1

997888rarr 1199100 119910

1015840

119896

1

997888rarr V0 (30)


We know

1199101015840

119896(119905) = 119910

1015840

119896(0) + int

119905

0

11991010158401015840

119896(119904) 119889119904

= 1199101015840

119896(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2119910119896

+119891 (119904 0 0) ) 119889119904

119910119896(119905) = 119910

119896(0) + int

119905

0

1199101015840

119896(119904) 119889119904

(31)


11991010158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911199101015840

0

+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21199100+ 119891 (119905 0 0)

(32)



continuous




119909 and 119891

1199091015840




119909119896 119896 = 1 2 such that 119875119909


Let 119910119896= 119875119909

119896 Then (22) holds Take 120596

119896= 119910119896119910119896 Then

120596119896 sub 119862

1

119879 120596119896 = 1 and (23) (24) (25) and (27) hold


119896 and 120596

119896 are bounded


120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0

10038171003817100381710038171205960

1003817100381710038171003817= 1 (33)

Sequences int

1

01198911199091015840(119905 119909119896 12057911199091015840

119896)1198891205791

infin

119896=1and int

1

0119891119909(119905 1205792119909119896

0)1198891205792infin



int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 1198891205791

119908

997888rarr 1198911(119905)

int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2

119908

997888rarr 1198912(119905)

(34)


10038161003816100381610038161198911(119905)1003816100381610038161003816⩽ 119872 0 ≨ 119886 (119905) ⩽ 119891

2(119905) ⩽ 119887 (119905)

ae 119905 isin [0 119879] (35)


V10158400(119905) = 119891

1(119905) V0(119905) + 119891

2(119905) 1205960(119905) 120596

1015840

0(119905) = V

0(119905)

(36)

Then

12059610158401015840

0(119905) = 119891

1(119905) 1205961015840

0(119905) + 119891

2(119905) 1205960(119905)

1205960(0) = 120596

0(119879) 120596

1015840

0(0) = 120596

1015840

0(119879)

(37)



119875119909 ⩽ 119870 119909 isin 1198621119879

Assume 119863 = 119909 isin 1198621





know the equation

11991010158401015840= 119891 (119905 119909 119909

1015840) 119910 + 119892 (119905 119909 119909

1015840) (38)


We define 119875 1198621119879rarr 1198621


119879119910(119905) =


119879



subset of 1198621119879


1


1

119879 119909 ⩽ 119870





119906(119899)= 119891 (119905 119906 119906

(119899minus1)) (39)




119896 R119899+1 rarr R (119896 = 1 2 119899)

such that1003816100381610038161003816100381610038161003816100381610038161003816

119891 (119905 1199091 119909

119899) minus

119899

sum

119896=1

119891119896(119905 1199091 119909

119899) 119909119896

1003816100381610038161003816100381610038161003816100381610038161003816

⩽ 119903

1199011119896(119905) ⩽ 119891

119896(119905 1199091 119909

119899) ⩽ 1199012119896(119905) (119896 = 1 2 119899)

(40)




1199011119896(119905) ⩽ 119901

119896(119905) ⩽ 119901

2119896(119905) (119896 = 1 2 119899) (41)

the equation

119906(119899)=

119899

sum

119896=1

119901119896(119905) 119906(119896minus1) (42)



1199011119896(119905) ⩽

120597119891119896(119905 1199091 119909

119899)

120597119909119896

⩽ 1199012119896(119905) (119896 = 1 2 119899)

(43)







Acknowledgments


References
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











11990910158401015840= 119891 (119905 119909 119909

1015840)

= (119891 (119905 119909 1199091015840) minus 119891 (119905 119909 0))

+ (119891 (119905 119909 0) minus 119891 (119905 0 0)) + 119891 (119905 0 0)

= int

1

0

1198911199091015840 (119905 119909 120579

11199091015840) 11988912057911199091015840

+ int

1

0

119891119909(119905 1205792119909 0) 119889120579

2119909 + 119891 (119905 0 0)

(19)


11991010158401015840= int

1

0

1198911199091015840 (119905 119909 120579

11199091015840) 11988912057911199101015840

+ int

1

0

119891119909(119905 1205792119909 0) 119889120579

2119910 + 119891 (119905 0 0)

(20)


We define operator

119875 1198621

119879997888rarr 119862

1

119879 (21)





119875(1198621


119879


1

119879


119896= 119875119909119896 Then

11991010158401015840

119896= int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2119910119896+ 119891 (119905 0 0)

(22)


119879 If not we



120596119896= 119910119896119910119896 Then 120596

119896 sub 1198621

119879 120596119896 = 1

12059610158401015840

119896= int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 11988912057911205961015840

119896

+ int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2120596119896+

119891 (119905 0 0)

1003817100381710038171003817119910119896

1003817100381710038171003817

(23)

So 12059610158401015840119896 ⩽ 119872+max


119896

is bounded and

1205961015840

119896(119905) = 120596

1015840

119896(0) + int

119905

0

12059610158401015840

119896(119904) 119889119904 (24)

1205961015840


Furthermore by

120596119896(119905) = 120596

119896(0) + int

119905

0

1205961015840

119896(119904) 119889119904 (25)





120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0 (26)


0 V0isin 1198621


1205961015840

119896(119905) = 120596

1015840

119896(0)

+ int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911205961015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2120596119896+

119891 (119904 0 0)

1003817100381710038171003817119910119896

1003817100381710038171003817

) 119889119904

(27)


1205960(119905) = 120596

0(0) + int

119905

0

V0(119904) 119889119904

V0(119905) = V

0(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 1198891205791V0(119904)

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

21205960(119904)) 119889119904

(28)

Hence

12059610158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911205961015840

0+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21205960

(29)


with 1205960 = 1 so 119910



119896 and 119910

119896 are bounded


119896 and 119910



119910119896

1

997888rarr 1199100 119910

1015840

119896

1

997888rarr V0 (30)


We know

1199101015840

119896(119905) = 119910

1015840

119896(0) + int

119905

0

11991010158401015840

119896(119904) 119889119904

= 1199101015840

119896(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2119910119896

+119891 (119904 0 0) ) 119889119904

119910119896(119905) = 119910

119896(0) + int

119905

0

1199101015840

119896(119904) 119889119904

(31)


11991010158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911199101015840

0

+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21199100+ 119891 (119905 0 0)

(32)



continuous




119909 and 119891

1199091015840




119909119896 119896 = 1 2 such that 119875119909


Let 119910119896= 119875119909

119896 Then (22) holds Take 120596

119896= 119910119896119910119896 Then

120596119896 sub 119862

1

119879 120596119896 = 1 and (23) (24) (25) and (27) hold


119896 and 120596

119896 are bounded


120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0

10038171003817100381710038171205960

1003817100381710038171003817= 1 (33)

Sequences int

1

01198911199091015840(119905 119909119896 12057911199091015840

119896)1198891205791

infin

119896=1and int

1

0119891119909(119905 1205792119909119896

0)1198891205792infin



int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 1198891205791

119908

997888rarr 1198911(119905)

int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2

119908

997888rarr 1198912(119905)

(34)


10038161003816100381610038161198911(119905)1003816100381610038161003816⩽ 119872 0 ≨ 119886 (119905) ⩽ 119891

2(119905) ⩽ 119887 (119905)

ae 119905 isin [0 119879] (35)


V10158400(119905) = 119891

1(119905) V0(119905) + 119891

2(119905) 1205960(119905) 120596

1015840

0(119905) = V

0(119905)

(36)

Then

12059610158401015840

0(119905) = 119891

1(119905) 1205961015840

0(119905) + 119891

2(119905) 1205960(119905)

1205960(0) = 120596

0(119879) 120596

1015840

0(0) = 120596

1015840

0(119879)

(37)



119875119909 ⩽ 119870 119909 isin 1198621119879

Assume 119863 = 119909 isin 1198621





know the equation

11991010158401015840= 119891 (119905 119909 119909

1015840) 119910 + 119892 (119905 119909 119909

1015840) (38)


We define 119875 1198621119879rarr 1198621


119879119910(119905) =


119879



subset of 1198621119879


1


1

119879 119909 ⩽ 119870





119906(119899)= 119891 (119905 119906 119906

(119899minus1)) (39)




119896 R119899+1 rarr R (119896 = 1 2 119899)

such that1003816100381610038161003816100381610038161003816100381610038161003816

119891 (119905 1199091 119909

119899) minus

119899

sum

119896=1

119891119896(119905 1199091 119909

119899) 119909119896

1003816100381610038161003816100381610038161003816100381610038161003816

⩽ 119903

1199011119896(119905) ⩽ 119891

119896(119905 1199091 119909

119899) ⩽ 1199012119896(119905) (119896 = 1 2 119899)

(40)




1199011119896(119905) ⩽ 119901

119896(119905) ⩽ 119901

2119896(119905) (119896 = 1 2 119899) (41)

the equation

119906(119899)=

119899

sum

119896=1

119901119896(119905) 119906(119896minus1) (42)



1199011119896(119905) ⩽

120597119891119896(119905 1199091 119909

119899)

120597119909119896

⩽ 1199012119896(119905) (119896 = 1 2 119899)

(43)







Acknowledgments


References
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra










We know

1199101015840

119896(119905) = 119910

1015840

119896(0) + int

119905

0

11991010158401015840

119896(119904) 119889119904

= 1199101015840

119896(0) + int

119905

0

(int

1

0

1198911199091015840 (119904 119909119896 12057911199091015840

119896) 11988912057911199101015840

119896

+ int

1

0

119891119909(119904 1205792119909119896 0) 119889120579

2119910119896

+119891 (119904 0 0) ) 119889119904

119910119896(119905) = 119910

119896(0) + int

119905

0

1199101015840

119896(119904) 119889119904

(31)


11991010158401015840

0= int

1

0

1198911199091015840 (119905 1199090 12057911199091015840

0) 11988912057911199101015840

0

+ int

1

0

119891119909(119905 12057921199090 0) 119889120579

21199100+ 119891 (119905 0 0)

(32)



continuous




119909 and 119891

1199091015840




119909119896 119896 = 1 2 such that 119875119909


Let 119910119896= 119875119909

119896 Then (22) holds Take 120596

119896= 119910119896119910119896 Then

120596119896 sub 119862

1

119879 120596119896 = 1 and (23) (24) (25) and (27) hold


119896 and 120596

119896 are bounded


120596119896

1

997888rarr 1205960 120596

1015840

119896

1

997888rarr V0

10038171003817100381710038171205960

1003817100381710038171003817= 1 (33)

Sequences int

1

01198911199091015840(119905 119909119896 12057911199091015840

119896)1198891205791

infin

119896=1and int

1

0119891119909(119905 1205792119909119896

0)1198891205792infin



int

1

0

1198911199091015840 (119905 119909119896 12057911199091015840

119896) 1198891205791

119908

997888rarr 1198911(119905)

int

1

0

119891119909(119905 1205792119909119896 0) 119889120579

2

119908

997888rarr 1198912(119905)

(34)


10038161003816100381610038161198911(119905)1003816100381610038161003816⩽ 119872 0 ≨ 119886 (119905) ⩽ 119891

2(119905) ⩽ 119887 (119905)

ae 119905 isin [0 119879] (35)


V10158400(119905) = 119891

1(119905) V0(119905) + 119891

2(119905) 1205960(119905) 120596

1015840

0(119905) = V

0(119905)

(36)

Then

12059610158401015840

0(119905) = 119891

1(119905) 1205961015840

0(119905) + 119891

2(119905) 1205960(119905)

1205960(0) = 120596

0(119879) 120596

1015840

0(0) = 120596

1015840

0(119879)

(37)



119875119909 ⩽ 119870 119909 isin 1198621119879

Assume 119863 = 119909 isin 1198621





know the equation

11991010158401015840= 119891 (119905 119909 119909

1015840) 119910 + 119892 (119905 119909 119909

1015840) (38)


We define 119875 1198621119879rarr 1198621


119879119910(119905) =


119879



subset of 1198621119879


1


1

119879 119909 ⩽ 119870





119906(119899)= 119891 (119905 119906 119906

(119899minus1)) (39)




119896 R119899+1 rarr R (119896 = 1 2 119899)

such that1003816100381610038161003816100381610038161003816100381610038161003816

119891 (119905 1199091 119909

119899) minus

119899

sum

119896=1

119891119896(119905 1199091 119909

119899) 119909119896

1003816100381610038161003816100381610038161003816100381610038161003816

⩽ 119903

1199011119896(119905) ⩽ 119891

119896(119905 1199091 119909

119899) ⩽ 1199012119896(119905) (119896 = 1 2 119899)

(40)




1199011119896(119905) ⩽ 119901

119896(119905) ⩽ 119901

2119896(119905) (119896 = 1 2 119899) (41)

the equation

119906(119899)=

119899

sum

119896=1

119901119896(119905) 119906(119896minus1) (42)



1199011119896(119905) ⩽

120597119891119896(119905 1199091 119909

119899)

120597119909119896

⩽ 1199012119896(119905) (119896 = 1 2 119899)

(43)







Acknowledgments


References
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119896 R119899+1 rarr R (119896 = 1 2 119899)

such that1003816100381610038161003816100381610038161003816100381610038161003816

119891 (119905 1199091 119909

119899) minus

119899

sum

119896=1

119891119896(119905 1199091 119909

119899) 119909119896

1003816100381610038161003816100381610038161003816100381610038161003816

⩽ 119903

1199011119896(119905) ⩽ 119891

119896(119905 1199091 119909

119899) ⩽ 1199012119896(119905) (119896 = 1 2 119899)

(40)




1199011119896(119905) ⩽ 119901

119896(119905) ⩽ 119901

2119896(119905) (119896 = 1 2 119899) (41)

the equation

119906(119899)=

119899

sum

119896=1

119901119896(119905) 119906(119896minus1) (42)



1199011119896(119905) ⩽

120597119891119896(119905 1199091 119909

119899)

120597119909119896

⩽ 1199012119896(119905) (119896 = 1 2 119899)

(43)







Acknowledgments


References
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article Existence and Uniqueness of Periodic