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## Academic research paper on topic "Antiperiodic Boundary Value Problems for Finite Dimensional Differential Systems"

﻿Hindawi Publishing Corporation Boundary Value Problems Volume 2009, Article ID 541435,11 pages doi:10.1155/2009/541435

Research Article

Antiperiodic Boundary Value Problems for Finite Dimensional Differential Systems

Y. Q. Chen,1 D. O'Regan,2 F. L. Wang,1 and S. L. Zhou1

1 Faculty of Applied Mathematics, Guangdong University of Technology, Guangzhou, Guangdong 510006, China

2 Department of Mathematics, National University of Ireland, Galway, Ireland

Correspondence should be addressed to D. O'Regan, donal.oregan@nuigalway.ie Received 16 March 2009; Accepted 28 May 2009 Recommended by Juan J. Nieto

We study antiperiodic boundary value problems for semilinear differential and impulsive differential equations in finite dimensional spaces. Several new existence results are obtained.

Copyright © 2009 Y. Q. Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

The study of antiperiodic solutions for nonlinear evolution equations is closely related to the study of periodic solutions, and it was initiated by Okochi . During the past twenty years, antiperiodic problems have been extensively studied by many authors, see [1-31] and the references therein. For example, antiperiodic trigonometric polynomials are important in the study of interpolation problems [32, 33], and antiperiodic wavelets are discussed in . Moreover, antiperiodic boundary conditions appear in physics in a variety of situations, see [35-40]. In Section 2 we consider the antiperiodic problem

where A is an n x n matrix, f : R x Rn ^ Rn is continuous, and f (t + T,x) = -f (t,x) for all (t, x) e R x Rn. Under certain conditions on the nondiagonal elements of A and f we prove an existence result for (E 1.1). In Section 3 we consider the antiperiodic boundary value problem

u'(t) = Au(t) + f (t,u(t)), t e R, u(t) = -u(t + T), t e R,

(E 1.1)

u'(t) = Gu(t)+ f (t,u(t)), a.e. t e J = [0,T], t / tk, u(0) = -u(T),

(E 1.2)

Au(tk ) = Ik (u(tk )), k = 1,2,...,p,

where G : Rn ^ Rn is a function satisfying GO = 0, and f : J x Rn ^ Rn is a Caratheodory function, Au(tk) = u(t+) - u(t-), and Ik € C(Rn,Rn). Under certain conditions on G, f, and Ik (u) for k = 1,2,...,p, we prove an existence result for (E 1.2).

2. Antiperiodic Problem for Differential Equations in Rn

Let | ■ | be the norm in Rn. In this section we study

u'(t) = Au(t)+ f (t,u(t)), t € R,

(E 2.1)

u(t) = -u(t + T).

First, we have the following result.

Theorem 2.1. Let A = (aij) be an n x n matrix, where a ij is the element of A in the ith row and jth column, f : R ^ Rn is continuous and f (t + T) = -f (t) for t € R. Suppose (T/2)ll1<i<j<nlaij -a^l < 1. Then the equation

u'(t) = Au(t)+ f (t), t € R,

(E 2.2)

u(t) = -u(t + T), t € R

has a unique solution.

Proof. Put Wa = M-) € C(R; Rn) : v(t) = -v(t + T)}. Then Wa is a Banach space under the norm |v(-)|TO = maxie[0/T]|v(t)|. For each v(-) € Wa, consider the following equation:

u'(t) = Av(t)+ f (t), t € R,

(E 2.3)

u(t) = -u(t + T), t € R.

It is easy to see that u(t) = -(1/2)^ [Av(s) +f (s)]ds+J0 [Av(s) +f (s)]ds is the unique solution of (E 2.3).

We define a mapping K : Wa ^ Wa as follows:

for any v(-) € Wa, Kv(-) = u(-), u(-) is the solution of (E2.3). (2.1)

First we prove that K is a continuous compact mapping. Now assume vn(-) € Wa, n = 1,2,..., and vn() ^ v(-) € Wa. Then Avn(-) - Av(-)|TO ^ 0 as n ^ to. This immediately implies that f] KKvn(t))' - (Kv(t))'^dt ^ 0 as n ^ to.

We have Kvn(t) - Kv(t) = (1/2){f0[(Kvn(s))' - (Kv(s))']ds - fTt[(Kvn(s))' -(Kv(s))']ds]}, and so Kvn(-) ^ Kv(-) in Wa.

Now since (Kv(t))' = Av(t) + f (t), t € R, it is easy to see that

T \ 1 /2 / T

jj(Kv(t))'\2dt) <VT |Av(-)|to + ( \\f (t)|

Thus K maps a bounded subset of Wa to a bounded equicontinuous subset in Wa, therefore K is completely continuous.

Next take ro > (1 - (T/l)^^^; - an\)-1(VT/2)(jT\f (t)\2dt)1/2. We show that Kv(-) /Xv(-) for all X > 1, and VO^ = r0. If this is not true, there exist X0 > 1, w(-) e Wa with MOU = r0 such that Kw(-) = X0w(-), that is, w(t) = -w(t + T), t e R and

Xow'(t) = Aw(t) + f (t), t e R. (2.3)

Multiply (2.3) by w'(t) (i.e., take inner product) and integrate over [0,T], and notice that \Twi(t)wj(t)dt = -^)w'i(t)wj(t)dt to get

AoJ" \w'(t)\2dt < 2i<i<j<n\aij - aß\j \wi(t)w'j (t)\dt + ^J \f (t)\2dtj ^J \w'(t)\2l

' (2.4)

where w(t) = (wi(t)), i = 1,2,...,n. Notice that w(t) = (1/2)[f0w'(s)ds - ¡Tw'(s)]ds, so we have

\w(-)\œ< ^({V^d^ . (2.5)

From (2.4), (2.5), we have

■ ç T \ 1/2 /çT \ 1/2

Ao( J Jw'(t)\2dtj < ^Zi<i<j<n\aij - aji\\w(-)\œ + M Jf(t)\2dt) . (2.6)

This with (2.5) gives

T \ 1/2

Ao\w(-)\œ< 22j\atl - an\\w(-)\œ + ^ \f (t)\2dt ) . (2.7)

As a result

—1 /— / ef \ 1/2

\w(-)\œ< (1 - \21<l<j<n\alj - aji\) ^Mjf(f)\2df ) , (2.8)

Thus the Leray-Schauder degree deg(! - K, B(0,r0), 0) = 1, where B(0,r0) is the open ball centered at 0 with radius r0 in Ca. Consequently, K has a fixed point in B(0,r0), that is, (E 2.2) has a solution. For the uniqueness, if u(-), v(-) are two solutions of (E 2.2), set w(t) = u(t) - v(t), then w'(t) = Aw(t), and w(t) = -w(t + T), for t e R. Following the obvious

strategy above (see the clear adjustment of (2.8)) gives |w(-)|TO = 0. Thus the solution of (E 2.2) is unique. □

From Theorem 2.1 we have immediately the following result.

Corollary 2.2. Let A = (aij) be an n x n symmetric matrix, f : R ^ Rn is continuous and f (t + T)= -f (t) for t € R. Then

u'(t) = Au(t)+ f (t), t € R,

(E 2.4)

u(t) = -u(t + T), t € R,

has a unique solution.

Using a proof similar to Theorem 2.1, we have the following result.

Theorem 2.3. Let A = (aij) bean n x n matrix, G : Rn ^ Rn is an even continuously differentiable function, and f (t,u) : R x Rn ^ Rn is continuous and f (t + T,u) = -f (t,u) for (t,u) € R x Rn. Suppose the following conditions are satisfied:

(1) f (t,x) < MX + g(t),for a.e. (t,x) € R x Rn, where M > 0 is a constant, and g(■) € L2(0,T);

(2) (T/l)^^^^^ - aj^ + M] < 1. Then

u'(t) = Au(t) + dGu(t) + f (t,u(t)), t € R,

(E 2.5)

u(t) = -u(t + T), t € R

has a solution.

Remark 2.4. Equation (E 2.5) was studied by Haraux  and Chen et al.  in the case A = 0, and also by Chen  with different assumptions on f and A.

3. Antiperiodic Boundary Value Problem for Impulsive ODE

In this section, we prove an existence result for the equation

u'(t) = Gu(t)+ f (t,u(t)), a.e. t € J = [0,T], t/tk,

u(0)= -u(T), (E 3.1)

Au(tk ) = Ik (u(tk)), k = 1,2,...,p,

where G : Rn ^ Rn is a Lipschitz function. We first introduce some notations. Let J = [0,T], and 0 = t0 < t1 < ■■■ < tp < tp+1 = T. PC(J) = {u : J ^ Rn, u{tk,tk+l] € C((tk,tk+i ],Rn),k = 0,1,...,p, u(t-) exist for k = 1,2,...,p, and u(0+) = u(0)}, and PW12(J) = {u € PC (J) : u(tkA+1) € W 1'2((tk,tk+1),Rn ),k = 1.....p}. It is clear that PC(J)

and PW1,2(J) are Banach spaces with the respective norm ||u||PC(J) = sup{|u(t)|, t e J}, and \\u\\pwn(j) = Xk=o llUklwi,2(tkrtk+1), where uk : (tk,tk+i] ^ R is defined by uk(t) = u(t) for t e (tk,tk+i], k = 0,1,...,p.

We say a function u is a solution of (E 3.1) if u e PW1,2(J) and u satisfies (E 3.1). We first prove the following result.

Lemma 3.1. Let Ii : Rn ^ Rn be continuous functions for i = 1,2,...,p, and ^pk=1IIk (xk )| < a{max1<k<pIxk|} + 6 for all xk e Rn, k = 1,2,...,p, where a,6 > 0 are constants, and a < 2. Suppose u e PW1,2(J) with u(0) = -u(T), and Au(ti) = Ii(u(ti)), for i = 1,2,...,p. Then

\\u\\pc{n < (1 - 2«

\u'(s)\2ds

Proof. By assumption, we have u(t) = u(0) + J"i0u'(s)ds for t e [0, ti), and

u(t) = u(0) - lk=1 Ii(u(ti)) - f u!(s)ds

for t e [tk,tk+1), k = 1,2,...,p. Since u(0) = -u(T), it follows that u(t) = -(1/2)[l/p=1Ii(u(ti)) + j^}u'(s)ds] + J"t0u'(s)ds for t e [0,t1), and

u(t) = - 2

K-1 Ii(u(ti))- ( u'(s)ds Ii(u(ti))- ( u'(s)ds (3.3)

J 0 J 0

for t e [tk,tk+1), k = 1,2,...,p. Hence we have

\\u\\pc(j) < 2\aWu\\pc(j)

[a\\u\\pc(j) - * - — ( J \u'(s^) \

u\\pc(j) < ^ - 2«) ^* -^y(j:\u'(s)\2ds

(3.5) □

Theorem 3.2. Let G : Rn ^ Rn be a function satisfying G0 = 0, and f : [0,T] ^ Rn such that f (■) e L2([0,T]), and let Ik : Rn ^ Rn be continuous functions for k = 1,2,...,p. Suppose the following conditions are satisfied:

(1) Gu - Gv| < L\u - v| for all u,v e Rn, and L> 0 is a constant;

(2) 2^=^ (xk )| < j {mаx1<k<p|xk |} + 6 for all xk e Rn, k = 1,2,...,p, where y,6 > 0 are constants;

(3) y + TL< 2.

Then the problem

u'(t) = Gu(t) + f (t), a.e. t e J = [0,T], t / tk, u(0) = -u(T), Au(tk ) = Ik (u(tk)), k = 1,2,...,p

(E 3.2)

has a solution.

Proof. For each v e PC(J), consider the problem

u'(t) = Gv(t) + f (t) a.e. t e J = [0,T], t /tk, u(0) = -u(T), Au(tk ) = Ik (v(tk )), k = 1,2,...,p.

(E 3.3)

One can easily show that the solution u of (E 3.3) is given by the following:

u(t) = -

u(t) = -

P=1 Ii(v(ti)) + \ (Gv(s)+ f(s))ds J 0

I" (G(v(s)) + f (s))ds, for t e [0, t), J 0

p=1 Ii(v(ti ))+ f (Gv(s)+ f (s))d

f (Gv(s)+ f(s))ds,

+ 2= h(v(ti))

for t e [tk,tk+i), k = 1,...,p.

Obviously, the solution of (E 3.3) is unique. Now we define K : PC(J) ^ PW1,2(J) c PC(J) by u = Kv. We prove that K is continuous. Let vn e PC(J) and vn ^ v in PC(J). It is easy to see that

çT ( T c T

\(Kvn(t) - Kv(t))'\2dt = \Gvn (t) - Gv(t)\2dt<L2\ \vn(t) - v(t)\2dt. (3.7)

J 0 J 0 J 0

Therefore (jf\(Kvn(t) - Kv(t))'\2dt)1/2 < -VTL\\vn - v\\pc{j) ^ 0 as n ^ œ.

Note that A(Kvn - Kv)(tk) = Ik(vn(tk)) - Ik(v(tk)), and we have

Kvn(t) - Kv(t) = -2

(vn(ti)) - Ii(v(ti))) + f (Kvn - Kv)'(s)ds J 0

+ f (Kvn - Kv)'(s)ds, for t e [0, ti),

Kvn(t) - Kv(t) = -

2Pl_1(Il(Vn(ti)) - Ii(v(ti)))+ f (Kvn - Kv)'(s)ds

+ Zh(Ii(vn(U)) - Ii(v(ti)))+ i (Kvn - Kv)'(s)ds

for t e [tk,tk+\), k = 1,2,...,p. From the continuity of Ii, i = 1,2,...,p, and J0\(Kvn(t) -Kv(t))'\2dt ^ 0 as n ^ go, we deduce that K is continuous. For each v e PC(J), notice that 0 = G0, so we have

\Kv\2dt"j < vtL\\v\\pc{J) + (jo\f(s)\2ds

T \ 1/2 2

From (3.9) and Lemma 3.1, we know that K maps bounded subsets of PC(J) to relatively compact subsets of PC(J).

Finally, for VX e (0,1], we prove that the set of solutions of u = XKu is bounded. If u = XKu for some X e (0,1), then

u'(t) = XGu(t) + Xf (t) a.e. t e J = [0,T], t / tk, u(0) = -u(T), Au(tk ) = XIk (u(tk )), k = 1,2,...,p.

(3.10)

Therefore we have

u(t) = - 2 X

2=1 i,<

i(ui(ti)) + I" (Gu(s) + f (s))ds + ^ (G(u(s)) + f (s))ds (3.11)

for t e [0,t1 ), and

u(t) = - 2 X

1 Ii(ui(ti)) + f (Gu(s) + f(s))d,

Xi" (G(u(s)) + f(s))ds

+X ihmm

(3.12)

for t e (tk,tk+1], k = 1,...,p. This implies that

mpc(j) < 2

Y\\u\\pc(j)- * - I" (|Gu(s)| -\f (s)\)ds

(3.13)

Since 0 = G0, and Gu < L|u|, so we have

mpc(j) < 2

1 - 2 (y - :L

-1 / r: *-

f\f (s)\ds

(3.14)

The Leray-Schauder principle guarantees a fixed point of K, which is easily seen to be a solution of (E 3.2). □

By using a similar method to Theorem 3.2, one can deduce the following result.

Theorem 3.3. Let G : Rn ^ Rn be a function satisfying G0 = 0, and f (t,x) : [0,T] x Rn ^ Rn a Caratheodory function, that is, f is measurable in t for each x e Rn, and f is continuous in x for each t e [0,T], such that f (t,x) < g(t) for (t,x) e [0,T] x Rn, where gO e L2([0,T]), and let Ik : Rn ^ Rn be continuous functions for k = 1,2,...,p. Suppose the following conditions are satisfied:

(1) IGu - Gvl < Liu - vl for all u,v e Rn, and L> 0 is a constant;

(2) ^pk=1IIk(xk)| < Y{max1<k<plxk|} - ô for all Xk e Rn, k = 1,2, constants;

..,p, where y,* > 0 are

(3) y - :L< 2.

Then the equation

u'(t) = Gu(t) - f (t,u(t)), a.e. t e J = [0,:], t / tk, u(0) = -u(: ), Au(tk ) = Ik (u(tk)), k = 1,2,...,k

(E 3.4)

has a solution.

4. Examples

In this section, we give examples to show the application of our results to differential and impulsive differential equations.

Example 4.1. Consider the antiperiodic problem

u[(t) = A1u1(t) - 5u2(t) - sinnt, t e R,

u'2(t) = 2ui(t) - A2u2(t) - cosnt, t e R, ui(t) = -u1(t - 1), u2(t) = -u2(t - 1), t e R.

(E 4.1)

f (t) =

sin nt cos nt

Now (E 4.1) is equivalent to

u'(t) = Au(t) - f (t), t e R, u(t) = -u(t - 1), t e R.

(E 4.2)

Also f (t) = -f (t + l),for t e R and (1 /2)\a12 - a21\ = 3/4. By Theorem 2.1, (E 4.2) has a unique solution, so (E 4.1) has a unique solution.

Example 4.2. Consider the antiperiodic boundary value problem

u[(t) = u2 (t) =

--^-[3u\(t) - 2u2(t)]- sinnt, t e (0,1),t /

2 - u\(t) - u2(t)

--^-[2ut(t)- 3u2(t)] - cosnt, t e (0,1),t =

2 - u2(t) - u2(t)

aui T =

V 5(1 - |u2(1/4)|)'

^ud 1 ) =

V 8(1 - |u1(1/4)|)'

u1(0) = -ui(1), u2(0) = -u2(1).

(E 4.3)

f (t) =

sin nt -cos nt

3ui - 2u2

2 - u2 - u22 2u1 - 3u2

,2 - u2 - u2

I (u)\ 5(1 -1|u2|) ,8(1 - |u1|).

It is easy to check that \Gu - Gv\ < (713/2)\u - v\ for u,v e R2, \I(u)\ < 2/5 for u e R2, and v^3/2 < 2. Now (E 4.3) is equivalent to the equation

By Theorem 3.2, we know that (E 4.4) has a solution, so (E 4.3) has a solution.

Acknowledgment

The first author is supported by an NSFC Grant, Grant no. 10871052.

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