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MALAYSIANMATHEMATICAL

SCIENCESSOCIETY http://math.usm.my/bulletin

Anti-Periodic Boundary Value Problems of Second-Order Functional Differential Equations

1JINGZHAO,2JINBOLIU AND3LIJINGFANG

1,3College of Science, Guangxi University for Nationalities, Nanning, Guangxi 530006, P. R. China

2Department of Mathematics, Changsha University of Science and Technology, Changsha 410114, Hunan, P. R.China

1[email protected],2[email protected],3[email protected]

Abstract. We deal with anti-periodic problems for second-order functional differential equations. The main tools in our study will be the Schauder’s fixed point theorem and the property of the continuous function space with anti-periodic conditions. Some new results on the existence and uniqueness of anti-periodic solutions are obtained, which generalize and extend previously known theorems.

2010 Mathematics Subject Classification: 34K13,47G20,45J05

Keywords and phrases: Anti-periodic solutions, second-order differential equations, delays.

1. Introduction

Consider the nonlinear second-order differential equations with delays of the form (1.1) u00+f(t,x0(t),x(t),x(t−τ(t))) =0, t∈R

where f :R4→Ris continuous function,τ isT-periodic with respect tot andT >0 is a constant.

Anti-periodic boundary value problems have been discussed in the past 20 years. Okochi [16,17] initiated the study for anti-periodic solutions of evolution equation in Hilbert spaces.

Following Okochi’s work, Chenet al.[6,7], studied by fixed point theorem the anti-periodic solution for first order semilinear evolution equations in a real separable Hilbert space.

Recently Liu in [13] studied the following anti-periodic problem of nonlinear evolution equations with nonmonotone perturbations

(u0(t) +Au(t) +Gu(t) =f, a.e.t∈(0,T), u(0) =−u(T),

Communicated byShangjiang Guo.

Received:August 18, 2011;Revised:January 13, 2012.

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in a real reflexive Banach spaceV. Ais monotone andGis not. Existence of solutions for anti-periodic problem has been obtained by using the theory of pseudomonotone per- turbations of maximal monotone mappings. Sufficient conditions for the existence of anti- periodic solutions of the first order differential equations, we also refer to [2, 3, 8, 12].

Nevertheless second order differential equations with anti-periodic boundary value con- ditions are discussed in few papers [1, 14, 18, 19]. Aftabizaden, Aizicovici and Pavel [1]

studied the anti-periodic solutions of second order evolution equations in Hilbert and Ba- nach spaces by using monotone and accretive operator theory. In [14] the authors have dis- cussed anti-periodic boundary value problems for second order differential equations, and sufficient conditions for existence of coupled solutions and a unique solution are obtained by using the monotone iterative technique. In [18, 19], the existence of at least one solution is obtained by using the Schauder fixed point theorem and the Leray-Schauder topological degree respectively.

Although second order differential equations with anti-periodic boundary conditions have been discussed in [1, 4, 10, 14, 18–20], as we can see, the function f is independent ofx0andx(t−τ(t)). Since anti-periodic boundary conditions appear in physics in a variety of situations (see for example, in [5, 11] and the references therein), the development of the general theory of the problem is timely.

However, to the best of our knowledge, few authors have considered the existence and uniqueness of anti-periodic boundary value problems of second order functional differential equations (1.1). Thus it is worth continuing the investigation of the existence and uniqueness of anti-periodic solutions of equation (1.1). A primary purpose of this paper is to study the existence and uniqueness of anti-periodic solutions of equation (1.1). We will establish some sufficient conditions for the existence and uniqueness of anti-periodic solutions of equation (1.1). Our results are different from the references listed above. In particular, an example is also to be given to illustrate the effectiveness of our results.

2. Preliminaries

To prove our existence theorem, we need the following set of hypotheses:

(H1) f∈C(R4,R),τ∈C(R,R), and for allt,x,y,z∈R, f

t+T

2,−x,−y,−z

=−f(t,x,y,z), τ

t+T 2

=τ(t).

(H2) There exist three nonnegative constantsa,b,csuch that aT

2π+bT2

2+c T2 4√

3π <1, and∀t,x1,x2,y1,y2,z1,z2∈R,

|f(t,x1,y1,z1)−f(t,x2,y2,z2)| ≤a|x1−x2|+b|y1−y2|+c|z1−z2|.

(H¯2) There exist three nonnegative constants ¯a,b,¯ c¯such that b¯

2π+ c¯ 2√

3 < a¯ T, and

(f(t,x1,y,z)−f(t,x2,y,z))(x1−x2)≥a|x¯ 1−x2|2,

|f(t,x,y1,z1)−f(t,x,y2,z2)| ≤b|y¯ 1−y2|+c|z¯ 1−z2|,

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∀t,x,y,z,x1,x2,y1,y2,z1,z2∈R.

(H3) There exist two nonnegative continuous functionsp(t),q(t)and a nonnegative con- stantLsuch that

|f(t,u,0,0)| ≤p(t)|u|+q(t), ∀t∈[0,T],|u|>L.

Letu(t):R→Rbe continuous int.u(t)is said to be anti-periodic onRif, u(t+T) =u(t), u

t+T

2

=−u(t), ∀t∈R.

We shall adopt the following notations:

CkT:={x∈Ck(R,R),xisT−periodic}, k={0,1,· · · }

|x|p= ( Z T

0

|x(t)|pdt)1p,∀p≥1, |x|0= max

t∈[0,T]|x(t)|, Ck,

1 2

T :={x∈CTk,x(t+T

2) =−x(t), ∀t∈R}, which is a linear normed space endowed with the normk · k

CTk,1/2 defined by kxkCTk,1/2 =max{|x|0,|x0|0,· · ·,|x(k)|0},∀x∈CTk,1/2.

The following lemmas will be useful to prove our main results in Section 3.

Lemma 2.1. [15]If x∈CT1andR0Tx(t)dt=0, then (2.1)

Z T 0

|x(t)|2dt≤ T2

2 Z T

0

|x0(t)|2dt (Wirtinger inequality) and

(2.2) |x(t)|20

T 12

Z T 0

|x0(t)|2dt (Sobolev inequality).

Lemma 2.2. [18]Letλ>0,δ(t)∈C0,1/2T and x is an anti-periodic solution of

(2.3) −x00(t) +λ2x(t) =δ(t)

if and only if x satisfies

(2.4) x(t) =

Z T

2

0 Z T

2

0

G(t,s)G(s,u)(−δ(u))duds, ∀t∈[0,T/2]

where

G(t,s) =





eλ(T2−t+s) eλT2+1

, 0≤s<t≤T2,

−eλ(s−t) eλT2+1

, 0≤t≤s≤T2,

G(t,s) =





eλ(t−s) eλT2+1

, 0≤s<t≤T2,

−eλ(T2+t−s)

eλT2+1

, 0≤t≤s≤T2. Lemma 2.3. [9]Let A be a continuous and compact mapping of a Banach space B into itself, and suppose there exists a constant M such that

kxkB<M

for all x∈B andσ∈[0,1]satisfying x=σAx. Then A has a fixed point.

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3. Main results

In this section we study the existence and uniqueness of anti-periodic solutions to problem (1.1).

Theorem 3.1. Assume that the condition (H1)and one of the two conditions (H2),(H¯2) hold. Then equation (1.1) has at most one anti-periodic solution.

Proof. Suppose thatx1(t)andx2(t)are two anti-periodic solutions of equation (1.1). Then, we have

(3.1) (x1(t)−x2(t))00+f(t,x01(t),x1(t),x1(t−τ(t)))−f(t,x02(t),x2(t),x2(t−τ(t))) =0.

SinceX(t) =x1(t)−x2(t)is an anti-periodic function onR, then Z T

0

X(t)dt= Z T

2

0

X(t)dt+ Z T

T 2

X(t)dt= Z T

2

0

X(t)dt+ Z T

2

0

X(t+T

2)dt=0.

By using the Sobolev inequality, we can get

(3.2) |X|0

rT 12|X0|2.

Now suppose that(H2)(or(H¯2)) holds. We shall consider two cases as follows.

Case (i)If(H2)holds, multiplying both sides of (3.1) by−X(t)and then integrating it from 0 to T, we have from (2.1), (2.2) and Schwarz inequality

|X0|22=− Z T

0

X00(t)X(t)dt

= Z T

0

[f(t,x01(t),x1(t),x1(t−τ(t)))−f(t,x02(t),x2(t),x2(t−τ(t)))]X(t)dt

≤ Z T

0

{a|X0(t)||X(t)|+b|X(t)|2+c|X(t−τ(t))||X(t)|}dt

≤a|X0|2|X|2+b|X|22+c|X|0 Z T

0

|X(t)|dt≤

aT

2π +b T2

2+c T2 4√

|X0|22. (3.3)

It follows from(H2)thatX(t)≡0,∀t∈R.Thusx1(t)≡x2(t),∀t∈R.

Case (ii)If(H¯2)holds, multiplying both sides of (3.1) byX0(t)and then integrating it from 0 to T, we obtain from (2.1), (2.2) and Schwarz inequality

a|X¯ 0|22≤ Z T

0

[f(t,x01(t),x1(t),x1(t−τ(t))−f(t,x02(t),x1(t),x1(t−τ(t)))]X0(t)dt

= Z T

0

[f(t,x02(t),x2(t),x2(t−τ(t)))−f(t,x02(t),x1(t),x1(t−τ(t)))]X0(t)dt

≤ Z T

0

{b|X¯ 0(t)||X(t)|+c|X(t¯ −τ(t))||X0(t)|}dt

≤b|X¯ 0|2|X|2+c|X¯ |0 Z T

0

|X0(t)|dt≤

b¯ T 2π +c¯ T

2√ 3

|X0|22.

It follows from(H¯2)thatX(t)≡0,∀t∈R.Thusx1(t)≡x2(t),∀t∈R. Therefore, equation (1.1) has at most one anti-periodic solution. The proof of Theorem 3.1 is now complete.

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Now we show the existence of solutions for the anti-periodic problem (1.1).

Theorem 3.2. Let (H1)hold. Assume that either the condition (H2) or the conditions (H¯2),(H3)are satisfied. Then equation (1.1) has at least one anti-periodic solution.

Proof. Define a mappingA:CT1,1/2→CT1,1/2by (3.4) Ax(t) =

Z T

2

0 Z T

2

0

G(t,s)G(s,u)(−f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u))duds,

∀t∈[0,T/2], whileAx(t) =−Ax(t−T/2)∀t∈[T/2,T]. Lemma 2.2 implies that solving problem (1.1) is equivalent to finding anx∈CT1,1/2such thatx=Ax.

In the following we shall use the well-known fixed point theorem, Lemma 2.3 to com- plete our proof.

We firstly show thatAis completely continuous.

(i)A:C1,1/2T →C1,1/2T is continuous.

Let{xn}be a sequence such thatxn→xinCT1,1/2asn→∞. Since f∈C(R4,R), we easily obtain∀u∈[0,T/2]

n→∞lim f(u,x0n(u),xn(u),xn(u−τ(u))) +λ2xn(u)

=f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u)].

(3.5)

From the definitions of the functionsG,Gin Lemma 2.2, it is easy to get that

|G(t,s)| ≤ eλT2 eλT2 +1

, |G(t,s)| ≤ eλT2 eλT2 +1

. In virtue of Governed Convergence Theorem, we obtain

(3.6) lim

n→∞|A(xn)−A(x)|0=0.

By (3.4), we have∀t∈[0,T/2]

Ax(t) = Z t

0

eλ(T2−t+s) eλT2 +1

Z T

2

0

G(s,u)(−f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u))duds

− Z T

2

t

eλ(s−t) eλT2 +1

Z T

2

0

G(s,u)(−f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u))duds.

Differentiating both sides of the above identity,we have (3.7) (Ax(t))0=−λAx(t) +

Z T

2

0

G(t,u)(−f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u))du.

(Ax(t))00=−λ(Ax(t))0+λ Z T2

0

G(t,u)(−f(u,x0(u),x(u),x(u−τ(u)))

−λ2x(u))du−f(t,x0(t),x(t),x(t−τ(t))−λ2x(t))−λ(Ax(t))0 +λ((0Ax(t)) +λAx(t))−f(t,x0(t),x(t),x(t−τ(t)))−λ2x(t))

2Ax(t)−f(t,x0(t),x(t),x(t−τ(t)))−λ2x(t)).

(3.8)

In virtue of (3.5)–(3.7) and Governed Convergence Theorem again, we have

(3.9) lim

n→∞|(A(xn))0−(A(x))0|0=0

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which shows thatA:CT1,1/2→CT1,1/2is continuous.

(ii) LetDbe a bounded set inCT1,1/2, that is, there exists ad>0 for anyx∈Dsuch that

|x|0≤dand|x0|0≤d. Thus,

|Ax|0= max

t∈

0,T2| Z T

2

0 Z T

2

0

G(t,s)G(s,u)(−f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u))duds|

≤ T2eλT 4(eλT2 +1)2

max

|f(s,u1,u2,u3)|+λ2u2|:s∈

0,T 2

,|ui| ≤d,i=1,2,3.

=:M1.

|(Ax)0|0

= max

t∈

0,T2| −λAx(t) + Z T2

0

G(t,u)(−f(u,x0(u),x(u),x(u−τ(u)))−λ2x(u))du|

≤λ|Ax|0+ TeλT2 2(eλT2+1)

max

|f(s,u1,u2,u3)|+λ2u2|:s∈

0,T 2

,|ui| ≤d,i=1,2,3.

≤M1+ TeλT2 2(eλT2+1)

max

|f(s,u1,u2,u3)|+λ2u2|:s∈

0,T 2

,|ui| ≤d,i=1,2,3.

=:M2.

By (3.8) and the above two inequalities, we obtain

|(Ax(t))00|02|Ax|0+max

|f(s,u1,u2,u3)|+λ2u2|:s∈

0,T 2

,|ui| ≤d,i=1,2,3.

2|x|0≤Const.

which implies thatA:C1,1/2T →C2,1/2T (⊂CT1,1/2) is bounded. Therefore, by compact em- bedding theorem we get thatA:CT1,1/2→C1,1/2T is compact. Hence we have shown thatA is completely continuous.

In view of the fixed point theorem of Lemma 2.3, Assume thatxis a solution of the equation

(3.10) x=σAx, σ∈(0,1].

Then, like (3.8) we easily get

(3.11) x00(t) +σf(t,x0(t),x(t),x(t−τ(t))) =0, t∈[0,T].

We shall consider two cases as follows:

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Case (i)If(H2)holds, multiplying both sides of (3.11) by−x(t)and then integrating it from 0 to T, we have from (2.1), (2.2) and Schwarz inequality

|x0|22=− Z T

0

x00(t)x(t)dt=σ Z T

0

(f(t,x0(t),x(t),x(t−τ(t)))x(t)dt

≤ Z T

0

(a|x0(t)||x(t)|+b|x(t)|2+c|x(t−τ(t))||x(t)|dt+ Z T

0

|f(t,0,0,0)x(t)|dt

≤a|x0|2|x|2+b|x|22+c|x|0 Z T

0

|x(t)|dt+ ( Z T

0

|f(t,0,0,0)|2dt)12|x|2

a T

2π+b T2

2+c T2 4√

|x0|22+ T 2π

Z T 0

|f(t,0,0,0)|2dt 12

|x0|2. (3.12)

It follows from(H2)that|x0|2is bounded ifx(t)is a solution (3.10). Therefore, we may assume that

(3.13) |x0|2≤C1

ifxis a solution (3.10), whereC1is a positive constant. Using inequality (2.2), we obtain

(3.14) |x|0

√ T 2√

3C1

wherexis any solutions to (3.10).

Now we are going to show|x0|0is also bounded ifxis a solution (3.10). Multiplying both sides of (3.11) byx00(t)and then integrating it from 0 to T,

|x00|22= Z T

0

|x00(t)|2dt=σ Z T

0

(f(t,x0(t),x(t),x(t−τ(t)))x00(t)dt

≤ Z T

0

|f(t,x0(t),x(t),x(t−τ(t)))x00(t)|dt.

(3.15)

By use of(H2), we get

|x00|22≤ Z T

0

|f(t,x0(t),x(t),x(t−τ(t)))x00(t)|dt

≤ Z T

0

[a|x0(t)|+b|x(t)|+c|x(t−τ(t))|]|x00(t)|dt+ Z T

0

|f(t,0,0,0)||x00(t)|dt

a|x0|2+b|x|2+c√ T|x|0+

Z T

0

|f(t,0,0,0)|2dt 12

|x00|2 (3.16)

which implies that|x00|2is bounded from (2.1), (3.13) and (3.14). It follows from (2.2) that there exists a positive constantC2such that

(3.17) |x0|0

√T 2√

3|x00|2≤C2.

Therefore, under the assumption(H2), we have shown that for anyx∈CT1,1/2which is a solution of (3.10), then

(3.18) kxk

C1,

12 T

<max √

T 2√

3C1,C2

+1.

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Case (ii)If(H¯2)and(H3)hold, multiplying both sides of (3.11) byx0(t)and then integrating it from 0 to T, we obtain from (2.1), (2.2) and Schwarz inequality

a|x¯ 0|22≤ Z T

0

[f(t,x0(t),x(t),x(t−τ(t)))−f(t,0,x(t),x(t−τ(t)))]x0(t)dt

= Z T

0

|f(t,0,x(t),x(t−τ(t)))x0(t)|dt

≤ Z T

0

{b|x(t)||x¯ 0(t)|+c|x(t¯ −τ(t))||x0(t)|dt+ Z T

0

|f(t,0,0,0)||x0(t)|dt

≤b|x¯ 0|2|x|2+c|x|¯ 0 Z T

0

|x0(t)|dt+ Z T

0

|f(t,0,0,0)||x0(t)|dt

b¯ T 2π +c¯ T

2√ 3

|x0|22+ Z T

0

|f(t,0,0,0)|2dt 12

|x0|2

It follows from(H¯2)that hat|x0|2is bounded ifx(t)is a solution (3.10). Therefore, we may assume that

(3.19) |x0|2≤C¯1

ifxis a solution (3.10), where ¯C1is a positive constant. Using inequality (2.2), we obtain

(3.20) |x|0

√ T 2√

3 C¯1

wherexis any solutions to (3.10).

Now we are going to show|x0|0is also bounded ifxis a solution (3.10).

Like (3.15), multiplying both sides of (3.11) byx00(t)and then integrating it from 0 to T,

|x00|22= Z T

0

|x00(t)|2dt=σ Z T

0

(f(t,x0(t),x(t),x(t−τ(t)))x00(t)dt

≤ Z T

0

|f(t,x0(t),x(t),x(t−τ(t)))x00(t)|dt.

(3.21)

In virtue of(H¯2)and(H3), we get

|x00|22≤ Z T

0

|f(t,x0(t),x(t),x(t−τ(t)))x00(t)|dt

≤ Z T

0

[b|x(t)|¯ +c|x(t¯ −τ(t))|]|x00(t)|dt+ Z T

0

|f(t,x0(t),0,0)||x00(t)|dt

b|x|¯ 2+c¯√

T|x|0]|x00|2+ Z T

0

(max{|f(t,u,0,0)|:|u| ≤L})2dt 12

|x00|2

+ Z T

0

[p(t)|x0(t)|+q(t)]|x00|dt

≤ {b|x|¯ 2+c¯

√ T|x|0+

Z T 0

(max{|f(t,u,0,0)|:|u| ≤L})2dt 12

+|p|0|x0(t)|2+|q(t)|2}|x00|2 (3.22)

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which implies that|x00|2is bounded from (2.1), (3.19) and (3.20). It follows from (2.2) that there exists a positive constant ¯C2such that

(3.23) |x0|0≤C¯2

Therefore, under the assumption(H¯2)and(H3), we also have shown that for anyx∈ CT1,1/2which is a solution of (3.10),then

(3.24) kxk

C1,

1 T2

<max{

√ T 2√

3

1,C¯2}+1.

In view of (3.18) and (3.24), we can choose a positive constantMsuch that kxk

C1,

12 T

<M,

for allxin the Banach spaceC1,1/2T andσ∈[0,1]satisfyingx=σAx. Therefore, by Lemma 2.3,Ahas a fixed point. The proof is complete.

4. An example

We conclude with a simple example which can be treated by the methods developed above.

Example 4.1. Consider the following nonlinear second-order differential equations with delays of the form (the Rayleigh equation with delays)

(4.1) x00(t) +1

2(sin2t)x0(t) +1

7x(t) +1+sin4t 2√

3π sin(x(t−sin2t))−cost=0.

Then the above problem has a unique anti-periodic solution with periodic 2π.

Proof. By (24), we have f(t,x0(t),x(t),x(t−τ(t))) = 1

2(sin2t)x0(t) +1

7x(t) +1+sin4t 2√

3π sin(x(t−sin2t))−cost, anda=1/2,b=1/7,c=1/(√

3π). It is obvious that assumption(H1)and(H2)hold.

Hence, by Theorem 3.1 and 3.2, equation (4.1) has a unique anti-periodic solution with periodic 2π.

Acknowledgement.Project supported by NNSF of China Grant Nos. 11271087, 61263006, scientific research project of Guangxi Education Department (No. 2013YB074) and scien- tific research project of State Ethnic Affairs Commission of the Peoples Republic of China (No.12GXZ001).

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