Hindawi Publishing Corporation Boundary Value Problems Volume 2008, Article ID 585378,14 pages doi:10.1155/2008/585378

Research Article

Antiperiodic Boundary Value Problems for Second-Order Impulsive Ordinary Differential Equations

Chuanzhi Bai

Department of Mathematics, Huaiyin Teachers College, Huaian, Jiangsu 223300, China Correspondence should be addressed to Chuanzhi Bai, czbai8@sohu.com Received 23 July 2008; Accepted 23 November 2008 Recommended by Raul F. Manasevich

We consider a second-order ordinary differential equation with antiperiodic boundary conditions and impulses. By using Schaefer's fixed-point theorem, some existence results are obtained.

Copyright © 2008 Chuanzhi Bai. 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

Impulsive differential equations, which arise in biology, physics, population dynamics, economics, and so forth, are a basic tool to study evolution processes that are subjected to abrupt in their states (see [1-4]). Many literatures have been published about existence of solutions for first-order and second-order impulsive ordinary differential equations with boundary conditions [5-19], which are important for complementing the theory of impulsive equations. In recent years, the solvability of the antiperiodic boundary value problems of first-order and second-order differential equations were studied by many authors, for example, we refer to [20-32] and the references therein. It should be noted that antiperiodic boundary value problems appear in physics in a variety of situations [33, 34]. Recently, the existence results were extended to antiperiodic boundary value problems for first-order impulsive differential equations [35, 36]. Very recently, Wang and Shen [37] investigated the antiperiodic boundary value problem for a class of second-order differential equations by using Schauder's fixed point theorem and the lower and upper solutions method.

Inspired by [35-37], in this paper, we investigate the antiperiodic boundary value problem for second-order impulsive nonlinear differential equations of the form

u"(t) + f(t,u(t)) = 0, t e J0 = J \{t1,...,tm}, Au(tk) = Ik(u(tk )), k = 1,...,m,

Au'(tk) = I*k (u(tk)), k = 1,...,m, u(0) + u(T) = 0, u'(0) + u'(T) = 0, X)

where J = [0,T], 0 < tx <t2 < ••• <tm<T, f : [0,T] x R ^ R is continuous on (t, x) e J0 x R, f (t+,x) := limt^t+f (t,x), f (tk,x) := limt^t-f (t,x) exist, f (tk,x) = f (tk,x); Au(tk) = u(t+) -u(t-k), Au'(tk) = u'(t+) - u'(t-); Ik,I*k e C(R, R).

To the best of the authors knowledge, no one has studied the existence of solutions for impulsive antiperiodic boundary value problem (1). The following Schaefer's fixed-point theorem is fundamental in the proof of our main results.

Lemma 1.1 (see [38] (Schaefer)). Let E be a normed linear space with H : E ^ E a compact operator. If the set

S := {x e E | x = XHx, for some X e (0,1)} (1.2)

is bounded, then H has at least one fixed point.

The paper is formulated as follows. In Section 2, some definitions and lemmas are given. In Section 3, we obtain two new existence theorems by using Schaefer's fixed point theorem. In Section 4, an illustrative example is given to demonstrate the effectiveness of the obtained results.

2. Preliminaries

In order to define the concept of solution for (1), we introduce the following spaces of functions:

PC(J) = {u : J ^ R : u is continuous for any t e Jo, u(t+), u(t-) exist, and u(t-) = u(tk), k = 1,.. .,m},

PC1(J) = {u : J ^ R : u is continuously differentiable for any t e J0, u'(t+), u'(t-) exist, and u'(t-) = u'(tk), k = 1,...,m}.

PC(J) and PC1 (J) are Banach space with the norms

\\u\\pc = sup|u(t)|,

teJ (2.1) \\u\\pc1 = max {\\u\\pc, \\u'\\pc}.

A solution to the impulsive BVP (1) is a function u e PC1 (J) n C2(J0) that satisfies (1) for each t e J.

Consider the following impulsive BVP with X > 0

-u"(t) + X2u(t) = a(t), t e J0, Au(tk) = Ik(u(tk)), k = 1,...,m, (22)

Au'(tk) = Ik (u(tk)), k = 1,...,m, u(0) + u(T) = 0, u'(0) + u'(T) = 0,

where a e PC(J).

For convenience, we set Ik = Ik(u(tk)), Ik = Ik(u(tk))•

Lemma 2.1. u e PC1(J) n C2(/o) is a solution of (2.2) f and only if u e PC1(J ) is a solution of the impulsive integral equation

u(t) = G(t,s)a(s)ds + £ [G(t,tk)( - Fk) + W(t,tk)Ik],

J0 k=1

G(t's) = 2!

W (t,s) =

' e-!(t-s) eX(t-s)

1 + e-XT 1 + eXT,

eX(T+t-s) e-X(T+t-s)

I 1 + eXT 1 + e-XT

e-X(t-s) eX(t-s)

1 + e-XT 1 + eXT,

eX(T+t-s) e-X(T+t-s)

.-XT '

0 < s <t < T, 0 < t < s < T, 0<s<t<T, 0 < t<s<T.

Proof. If u e PC1 (J) n C2(/o)is a solution of (2.2), setting

v(t) = u' (t) + Xu(t),

then, by the first equation of (2.2) we have

v'(t) - Xv(t) = -o(t), t / tk.

Multiplying (2.6) by e~xt and integrating on [0,t1) and (t1,t] (t1 <t < t2), respectively, we get

e~Xi1 v(t-) - v(0) = - a(s)e-!sds, 10

e-Xtv(t) - e-Xt1 v(t+) = - C a(s)e^Xsds, t1 <t < t2.

v(t) = eXt fv(0) - I" e-!sa(s)ds + e-"1 Av(t1) 0

In the same way, we can obtain that

v(t) = eXt v(0) - [ e-Xsa(s)ds + ^ e-Xtk (I*k + XIk)

J 0 0<t,.<t

t1 <t < t2.

t e J,

where v(0) = u'(0) + Xu(0). Integrating (2.5), we have

u(t) = e

(0) + [ v(s)eXsds + £ extkIk

' 0 0<tk<t

t e J.

By (2.9), we get

v(s)eXsds = — 0 21

v(0)(e2Xt - 1) -{ (e2Xi - e2Xs)a(s)e-Xsds

+ X (e2At - emk)e-xtk(Fk + XIk) 0<tk<t

Substituting (2.11) into (2.10), we obtain

U(t) = 2X

(Xu(0) - U(0))e-xt + (u'(0) + Xu(0))ext

+ f {e-X(t-s) - eX(t-s))a(s)ds + £ eX(t-tk) (Ik + XIk) 0 0<t <t

- X e-X(t-tk)(Ik - XIk) 0<tk<t

0<tk<t t e J,

u'(t) =

- (Xu(0) - u(0))e-Xt + (u'(0) + Xu(0))eXt

- f (e-X(i-s) + eX(i-s))a(s)ds + £ eX(t-tk) (Ik + XIk)

0 0<tk<t k

0<tk<t

e-X(t-tk^Ik - XIk)

t e J.

In view of u(0) + u(T) = 0 and u'(0) + u'(T) = 0, we have

u'(0) + Xu(0) = Xu(0) - u'(0) =

1 + eXT 1

\ eX(T-s)o(s)ds - £ eX(T-tk)(Ik + XIk)

0 0<tk<T k

- e-X(T-s)a(s)ds + X e-X(T-h) (Ik - XIk)

0 0<tk<T k

(2.10)

(2.11)

(2.12)

(2.13)

(2.14)

Substituting (2.14) into (2.12), by routine calculation, we can get (2.3).

Conversely, if u is a solution of (2.3), then direct differentiation of (2.3) gives -u"(t) = a(t) -X2u(t), t = tk. Moreover, we obtain Au\t=tk = Ik(u(tk)), Au'l^ = Ik(u(tk)), u(0) +u(T) = 0 and u'(0) + u(T) = 0. Hence, u e PC1 (J) n C2kJ0) is a solution of (2.2). □

Remark 2.2. We call G(t, s) above the Green function for the following homogeneous BVP:

-u!'(t) + X2u(t) = 0, t e J, u(0) + u(T) = 0, u'(0) + u'(T) = 0.

Define a mapping A : PC1 (J) ^ PC1 (J) by

Au(t) = G(t,s) f(s,u(s)) + X2u(s)] ds + Y [G(t,tk)( - I*k) + W(t,tk)Ik], t e [0,T].

J0 k=1

(2.16)

In view of Lemma 2.1, we easily see that u is a fixed point of operator A if and only if u is a solution to the impulsive boundary value problem (1). It is easy to check that

eXT - 1 1 |G(',s)l < 2l(T+3T), |W(',s)l < 1 (117)

Lemma 2.3. If u e PC1 (J) and u(0) + u(T) = 0, then

\\u\\pc < |T|u'(s)|ds + ||Au(tk. (2.18)

Proof. Since u e PC1 (J), we have

u(t) = u(0) + V Au(tk) + [ u'(s)ds. (2.19)

0<tk<t ' 0

Set t = T, we obtain from u(0) + u(T) = 0 that

1 / m f

--(EMk +

2 \ k=1 .M

Substituting (2.20) into (2.19), we get

|u(t)| = ^f u'(s)ds - I" u'(s)ds) + U V Au(tk) - ^Au(tk) )

2 ^ 0 Jt / 2\0<tk<t t<tk J

< K f|u'(s)|ds + flu' (s)|ds) + if V |Au(tk)|+ VlAu(tk)^ (2.21)

2 0 t 2 0<t <t t<t

1 / TT m = -( |u'(s)|ds + £|Au(tk) l

2 V 0 k=1

The proof is complete. □

u(0) = -"(£Au(tk) + | u(s)ds ). (2.20)

3. Main results

In this section, we study the existence of solutions for BVP (1). For this purpose we assume that there exist constants 0 < n < 1, functions a,b,h e C(J, [0, +œ)), and nonnegative constants ak,j5k,jk,6k (k = 1,2,...,m) such that

(H1) \f (t,u)\ < a(t)\u\ + b(t)\u\n + h(t), and

(H2) \Ik(u)\ < ak\u\ + fa, \I*k(u)\ < Yk\u\ + ôk, k = 1,...,m

Remark 3.1. (Hi) means that the nonlinearity growths at most linearly in u, (H2) implies that the impulses are (at most) linear. For convenience, let

3 / rT m S

P1 = 2 M a(t)dt + X2T + ^ Yi

P2 = 3 fb(t)dt, (3.1)

jTh(t)dt + Xôi

Theorem 3.2. Suppose that conditions (Hi) and (H2) are satisfied. Further assume that

t*< 1, (3.2)

¿=1 ¿=1

holds, where q1 = j^a(t)dt + "£^1(p1ai + Y¿) and p1 as in (3.1). Then, BVP (1) has at least one solution.

Proof. It is easy to check by Arzela-Ascoli theorem that the operator A is completely continuous. Assume that u is a solution of the equation

u = ¡Au, ¡1 e (0,1). (3.3)

u"(t) = ¡i(Au)"(t) = ¡i[ - f (t,u(t)) - X2u(t) + X2(Au)(t)]

2 (3.4)

= -¡f(t,u(t)) - X2(i - 1)u(t), -u(t)u"(t) = ¡u(t)f (t,u(t)) + X2(i - 1)u2(t) < ¡u(t)f (t,u(t)). (3.5)

Integrating (3.4) from 0 to T, we get that

pi m fT Aim

u'(T) - u'(0)= u"(t)dt + ^ I* = -J f(t,u(t)) dt -A2(n - 1) u(t)dt + ^ I*. (3.6)

J 0 ¿=1 J 0 J 0 ¿=1

In view of u'(0) + u'(T) = 0, we obtain by (3.6) that

1 fT,.......... I2

1 r \2 r 1 m

|u'(0)| < 1 |f(t,u(t))|dt + - |u(t)|dt + ^I|. (3.7)

2 0 2 0 2 i=1

Integrating (3.4) from 0 to t, we obtain that

u'(t) - u'(0) = [ u"(s)ds + V It = -I f f (s,u(s))ds - A2(i - 1) [ u(s)ds + V Fi

' 0 0<t<t ' 0 ' 0 0<ti<t

From (3.7) and (3.8), we have

/T fT m

|u'(t)1 < |u'(0)1 + f (s,u(s)) |ds + A2 |u(s)|ds + Y |I*1

J 0 J 0 ¿=1

o i*T o i*T o m

< J (fl(t)|u(t)| + fc>(t)|u(t)|^ + h(t)) dt + - A2 |u(t)|dt + ^ (ri\\u\\Pc +

2.>0 2 J0 2 ¿=1

< Mpc^a(t)dt + \\u\\Pcj"TKt)dt + JTh(t)dt) + 2A2T\u\pc + 3 V (ri\\u\\pc + 6i),

that is,

3 / /"T m \ 3 />i 3 />i 3 m

\\u\\pc < 3 M Qa(t)dt + A2T + £ rn \\u\\PC + 2J o^(t)dt\u\Pc + ¿J qh(t)dt + 2£

i 1 i 1 (3.10)

\\u'\pc < p1\\u\\pc + p2\\u\\PC + p3, (3.11) where p1, p2, p3 are as in (3.1). Integrating (3.5) from 0 to T, we get that

- f u(t)u"(t)dt < i[ u(t)f(t,u(t))dt. (3.12)

In view of u(0) + u(T) = 0 and u'(0) + u'(T) = 0, we have

f u(t)u"(t)dt= f u(t)d(u'(t)) J 0 J 0

= u(t)d(u'(t)) + u(t)d(u'(t)) + •••^ u(t)d(u'(t))

J 0 J t1 J t„

= (t)u'(t)l01 - f1 (u'(t))2dt + u(t)u'(t)lit2i - i"2(u'(t))2dt

0 0 t1 t1

■u(t)u'(t)ll - f (u'(t))2dt

= u{t\ - 0)u'(t1 - 0) - u(0)u!(0)+u(t2 - 0)u'(t2 - 0) - u(t1 + 0)u'(h + 0)

+ ••• + u(T )u'(T) - u(t„ + 0) u'( tn + 0) - (u'(t)) 2dt

= u(h - 0)u'(t1 - 0) - u(t1 + 0)u'(t1 + 0)

+ ••• + u(t„ - 0) u'( t„ - 0) - u(t„ + 0) u'( t„ + 0) - (u'(t)) 2dt

= u(h - 0)u'(t1 - 0) - u(t1 - 0)u'(t1 + 0) + u(t1 - 0)u'(t1 + 0) - u(t1 + 0)u'(t1 + 0) + ••• + u(t„ - 0)u'(t„ - 0) - u(t„ - 0)u'(t„ + 0)

+ u(t„- 0)u'(t„ + 0) -u(t„ + 0)u'(t„ + 0)- (u'(t))2dt

(t„ - 0)u'(t„ + 0) - u(t„ + 0)v!(t„ + 0) - f (u'(t))

= -u(t1 - 0)I*k - u'(t1 + 0)I1-----u(t„ - 0)I„ - v! (t„ + 0)I„ - (u'(t)fdt

(3.13)

Substituting (3.13) into (3.12), we obtain by (H2), (H3), and (3.11) that

f (u'(t))2dt < ¡[ u(t)f (t,u(t))dt - u(h - 0)I*1

- u'(t1 + 0)I1-----u(t„ - 0)I„ - u'(t„ + 0)I„

< J u(t)f(t,u(t))dt + \\u\\PcX\Ik\ + Wu'WpcX\M 0 ¿=1 ¿=1

(a(t)u2(t) + b(t)\u(t)\1+n + h(t)\u(t)\) dt

0 ¿=1 ¿=1

< ( (a(t)u2(t) + b(t)\u(t)r'1 + h(t)\u(

WuWpcXj(vWu\\pc + 60 + Wu'WpcXj(^Wu\\pc + fa)

< WuWPc

¿=1 ¿=1 ,T C T C T

a(t)dt + MpM b(t)dt + \\u\\PC\ h(t)dt J 0 J 0 J 0

YiY\\u\\pc + X 6i\\u\\pc + (P1\\u\\pc + P2\\u\\npc + V3)Yi (a\\u\\pc + fa).

¿=1 ¿=1 ¿=1

(3.14)

Chuanzhi Bai Thus,

(u'(t))2dt < q1 \\u\\PC + q2\\u\\l+ + q3\\u\\Pc + q4\\u\\nPC + q5, (3.15)

q1 = a(t)dt + Y (p1ai + r), ■>0 i=1

q2 = b(t)dt + p^ ai,

J0 i=1

q3 = h(t)dt + Y (p1^i + p3ai + (3.16)

•>0 i=1

q4 = p^ ^ i=1

q5 = p^ A

By Lemma 2.3 and (3.15), we have

1 / r- \21 fT m 1 / m ^ 2

\\Pc < 1(J Ju'(t)|dtj + 2J0 |u'(t)|dtZ\Ii\ + 4.

T f T PT / C- \ 1/2 m m m

< ijo(u'(t^2dt+(u'(t))2d^ 2i\+mE\Ii\2

< - [q1\\u\\2PC + q2\u\PCI + q3\\u\\PC + q4\u\Pc + q5]

+ ^ [q1 \u\Pc + q2\N\Pc + q3\u\Pc + q4\\u\\Pc + q5 ]1/2

m mm m

x X(ai\\u\\Pc + + TX(a2\u\Pc + \\u\\pc + ¿2) i=1 4 i=1

/Tq1 VTq1V m V ^ „2

(-4- + % ai + 4^ a2) Wu\\Pc

(3.17)

It follows from the above inequality and (3.2) that there exists M1 > 0 such that \\u\\PC < M1. Hence, we get by (3.11) that

\\u'\\pc < aM1 + a2Mn + a3 := M2. (3.18)

Thus, \\u\\pC1 < max(M1,M2}. It follows from Lemma 1.1 that BVP (1) has at least one solution. The proof is complete. □

Theorem 3.3. Assume that (H2) holds. Suppose that there exist a continuous and nondecreasing function f : [0, <) ^ [0, <) and a nonnegative function c e C(J) with

\f (t,u)+X2u\< c(t)f(lul), t e J, u e R. (3.19)

Moreover suppose that

lim sup ^u^ < L (3.20)

holds, where

L := 1 - ((eXT - 1)/2X(1 + eXT))"m=Y - (1/2)Zm=q > 0 (

: ((eXT - 1)/2X(1 + eXT))iT0c(s)ds . '

Then, BVP (1) has at least one solution.

Proof. From (3.20), there exist 0 <e < L and M > 0 such that

f(v) < (L - e)v, v > M. (3.22)

Thus, there exists K > 0 such that

f (v) < (L - e)v + K, v > 0. (3.23) Assume that u is a solution of the equation

u = ¡Au, ¡1 e (0,1). (3.24) Then, we have by (3.19), (2.17), and (3.23) that

i(t) \ =

G(t,s)f (s,u(s)) + X2u(s))ds + £ [G(t, tk) ( - ID + W(t, tk)Ik]

J 0 k=i

eXT - 1 CT eXT - 1 m 1 m

0c(s)f (lul)ds + ^X^r+exTJ Z toWuWpc + 60 + 2 § ^¿WuWpc + ft.

(3.25)

Thus, we have

eAT- 1 f T

\\u\\pc < 2A(1 + gXT Jc(s)ds((L - e)\\u\\pc + K)

2A 1 + eAT

(3.26)

eAT 1 m 1 m eAT 1 m 1 m §ri + 2 § av "u"PC + w+eAT) g6i + 2 §Pi,

that is,

eAT - 1 CT eAT - 1 CT eAT - 1

' tZ 6 + ^ Pi,

i=1 i=1

(3.27)

T eAT 1 T eAT 1 m 1 m -TK c(s)ds\\u\\PC < rrr-c(s)ds + -^E6 + oX P^

2A 1 + eAT 0 2A 1 + eAT 0 2A 1 + eAT i=1 2 i=1

which implies that there exists M3 > 0 such that \\u\\pc < M3. By (3.7), (3.8), and (3.23), we

o a- o a- o m

|u'(t)| < 3J0 If (s,u(s))|ds + -A2^\u(s)\ds + -g|I*|

o /"T /"T o m

< - |f(s,u(s))+ A2u(s)|ds + 3AM \u(s)\ds + ^^

2 0 0 2 i=1 i

3 fT 3 m

< 3 c(s)^(\u(s)\)ds + 3A2 \u(s)\ds + -Y(ji\\u\\Pc + 6^ 2 0 0 2 i=1

3 /T 3 m

< J c(s)ds{L\\u\\Pc + K) + 3A2T\\u\\pc + ^(ri\u\Pc + 6,), 2 0 2 i=1

which implies that

^T 3 m \ 3K 3 m

c(s)ds + 3A2 T + - V ri)\\u\\pc + c(s)ds + - V 6i 0 2 i=1 2 0 2 i=1

3 / f — m \ 3K f — 3 m

< 2 (Lj 0 c(s)ds + 2A2T + £ r^ M3 + "2" J q c(s)ds + 2£ 6 := M4.

(3.28)

(3.29)

Hence, \\u\\pC1 < max{M3,M4}. It follows from Lemma 1.1 that BVP (1) has at least one solution. The proof is complete. □

4. Example

In this section, we give an example to illustrate the effectiveness of our results.

Example 4.1. Consider the problem

11 u"(t) + — u(t)cos2t + -etu1/2(t) + 1 + tan t = 0, t e n 2

Au(h) = 3 sin (u(h)) + 1, Au'(h) = 1 u(h) + 1, ti = n, (41) u(0) + u(T) = 0, u'(0) + u'(T) = 0,

Let f (t,u) = (1/n)ucos2t + (1/2)etu1/2 +1 + tan t, Ii(u) = (1/3) sin u + (1/4), I*1 (u) = (1/2)u + (1/3), T = (n/2), J = [0,n/2]. It is easy to show that

\f (t, u) \ < a(t)\u\ + b(t)\u\1/2 + h(t), (4.2)

where a(t) = (1/n)cos2t, b(t) = (1/2)el, h(t) = 1 + tan t. And

1 1 1 1

\h(u)\ < 1 \u\ + ^, \I11(u)\ < ^\u\ + 3. (4.3)

Thus, (H1) and (H2) hold. Obviously, a\ = 1/3, fa = 1/4, J1 = 1/2, 61 = 1/3, and m = 1. Let I2 = 1/4n, we have

3(J><«++m r)=2(4+4+2)=3 -

j a(t)dt + m (pia + y) = 1 + 3 ■1 +1 = 4 •

Therefore,

T-f + ^ma. + ma2s = 0.7522 < 1, (4.5)

i=1 i=1

which implies that (3.2) holds. So, all the conditions of Theorem 3.2 are satisfied. By Theorem 3.2, antiperiod boundary value problem (4.1) has at least one solution.

Acknowledgments

The author would like to thank the referees for their valuable suggestions and comments. This project is supported by the National Natural Science Foundation of China (10771212) and the Natural Science Foundation of Jiangsu Education Office (06KJB110010).

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