Boundary value problem for second-order impulsive functional differential equations

Applied Mathematics and Computation 191 (2007) 582–591 www.elsevier.com/locate/amc

Boundary value problem for second-order impulsive functional differential equations Haihua Wang, Haibo Chen

*

Department of Mathematics, Central South University, Changsha 410075, PR China

Abstract This paper is concerned with the existence of minimal and maximal solutions of boundary value problem for secondorder impulsive functional differential equations. The method of upper and lower solutions and the monotone iterative technique are used.  2007 Published by Elsevier Inc. Keywords: Impulsive functional differential equation; Boundary value problem; Upper and lower solutions; Monotone iterative technique; Existence of solutions

1. Introduction Impulsive differential equations describe processes which experience a sudden change of their state at certain moments. The theory of impulsive differential equations has become important in recent years in mathematical models of real processes rising in phenomena studied in physics, chemical technology, population dynamics, biotechnology and economics; see [1,3,7–10]. There has been a significant development in impulsive theory especially in the area of impulsive differential equations with fixed moments [2]. In this paper, we consider the second-order impulsive functional differential equations with boundary conditions,  x00 ðtÞ ¼ f ðt; xðtÞ; xðhðtÞÞÞ; Dxðtk Þ ¼ I k ðxðtk ÞÞ; Dx0 ðtk Þ ¼ I k ðxðtk ÞÞ;

t 6¼ tk ; t 2 J ¼ ½0; T ;

k ¼ 1; . . . ; m;

xð0Þ ¼ xðT Þ þ k1 x0 ðT Þ þ k 1 ;

ð1:1Þ

x0 ð0Þ ¼ k2 x0 ðT Þ þ k 2 :

In a recent paper [5], Ding et al. studied the period boundary value problem for second-order impulsive functional differential equations

*

Corresponding author. E-mail addresses: [email protected] (H. Wang), [email protected] (H. Chen).

0096-3003/$ - see front matter  2007 Published by Elsevier Inc. doi:10.1016/j.amc.2007.02.140

H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

 x00 ðtÞ ¼ f ðt; xðtÞ; xðaðtÞÞÞ; Dxðtk Þ ¼ I k ðxðtk ÞÞ;

583

t 6¼ tk ; t 2 J ¼ ½0; T ; ð1:2Þ

Dx0 ðtk Þ ¼ I k ðxðtk ÞÞ; k ¼ 1; . . . ; m; xð0Þ ¼ xðT Þ; x0 ð0Þ ¼ x0 ðT Þ:

The results in this paper are inspired by Jankowski in [6], Ding et al. in [5], Chen and Sun in [4]. The structure of the rest of this paper is as follows: First, we transform BVP (1.1) to an equivalent form BVP (1.3), then introduce the new definition of upper and lower solutions. In Section 2, we establish two comparison principles and discuss the existence and uniqueness of the solutions for a linear impulsive differential equation. In Section 3, by using the monotone iterative technique and the method of upper and lower solutions, we obtain the existence theorem of extremal solutions for BVP (1.1). Finally, we give an example to illustrate Theorem 3.1. Let k01 ¼ k1 =ð1 þ k2 Þ and k 01 ¼ k 1  k1 k 2 =ð1 þ k2 Þ, where k1 P 0, k2 > 0; then the boundary condition of BVP (1.1) is xð0Þ ¼ xðT Þ þ k01 ½x0 ð0Þ þ x0 ðT Þ þ k 01 , x0 ð0Þ ¼ k2 x0 ðT Þ þ k 2 , where k01 P 0. On the other hand, if xð0Þ ¼ xðT Þ þ k1 ½x0 ð0Þ þ x0 ðT Þ þ k 1 , x0 ð0Þ ¼ k2 x0 ðT Þ þ k 2 , where k1 P 0, k2 > 0; then it is equivalent to xð0Þ ¼ xðT Þ þ k1 ð1 þ k2 Þx0 ðT Þ þ k1 k 2 þ k 1 , x0 ð0Þ ¼ k2 x0 ðT Þ þ k 2 , where k1 ð1 þ k2 Þ P 0. So, through this paper, we consider the second-order impulsive functional differential equations with boundary conditions,  x00 ðtÞ ¼ f ðt; xðtÞ; xðhðtÞÞÞ; Dxðtk Þ ¼ I k ðxðtk ÞÞ;

t 6¼ tk ; t 2 J ¼ ½0; T ;

Dx0 ðtk Þ ¼ I k ðxðtk ÞÞ; k ¼ 1; . . . ; m; xð0Þ ¼ xðT Þ þ k1 ½x0 ð0Þ þ x0 ðT Þ þ k 1 ;

ð1:3Þ x0 ð0Þ ¼ k2 x0 ðT Þ þ k 2 ;

  0 0 þ 0  where f 2 CðJ  R2 ; RÞ, 0 6 hðtÞ 6 t, t 2 J , Dxðtk Þ ¼ xðtþ k Þ  xðt k Þ, Dx ðt k Þ ¼ x ðt k Þ  x ðt k Þ, I k ; I k 2 CðR; RÞ, k ¼ 1; . . . ; m, 0 ¼ t0 < t1 < t2 <    < tm < tmþ1 ¼ T , k1 P 0, k2 > 0, k 1 ; k 2 2 R: The boundary condition of BVP (1.3) is beneficial for us to construct generalized upper and lower solutions, and to prove Lemma 2.2.

Definition 1.1. A function a is called a lower solution of BVP (1.3) if  a00 ðtÞ 6 f ðt; aðtÞ; aðhðtÞÞÞ þ aðtÞ; Daðtk Þ 6 I k ðaðtk ÞÞ þ mk ; Da0 ðtk Þ P I k ðaðtk ÞÞ þ lk ; 0

t 6¼ tk ; t 2 J ;

k ¼ 1; . . . ; m; 0

að0Þ ¼ aðT Þ þ k1 ½a ð0Þ þ a ðT Þ þ k 1 ; where M > 0, N P 0, Lk P 0, Lk P 0 are constants, and ( 0; if a0 ð0Þ P k2 a0 ðT Þ þ k 2 ; aðtÞ ¼ ðMþN ÞT ðMtþN hðtÞÞ 0 ½a ð0Þ  k2 a0 ðT Þ  k 2 ; if a0 ð0Þ < k2 a0 ðT Þ þ k 2 ; k2 ( 0; if a0 ð0Þ P k2 a0 ðT Þ þ k 2 ; mk ¼ Lk ½a0 ð0Þk2 a0 ðT Þk2  ; if a0 ð0Þ < k2 a0 ðT Þ þ k 2 ; k2 ( 0; if a0 ð0Þ P k2 a0 ðT Þ þ k 2 ; lk ¼ Lk minftk ;T tk g½a0 ð0Þk2 a0 ðT Þk 2   ; if a0 ð0Þ < k2 a0 ðT Þ þ k 2 : k2 Definition 1.2. A function b is called an upper solution of BVP (1.3) if  b00 ðtÞ P f ðt; bðtÞ; bðhðtÞÞÞ  bðtÞ; Dbðtk Þ P I k ðbðtk ÞÞ  Db0 ðtk Þ 6 I k ðbðtk ÞÞ 

mk ; lk ;

t 6¼ tk ; t 2 J ;

k ¼ 1; . . . ; m;

bð0Þ ¼ bðT Þ þ k1 ½b0 ð0Þ þ b0 ðT Þ þ k 1 ;

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H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

where M > 0, N P 0, Lk P 0, Lk P 0 are constants, and ( 0; if b0 ð0Þ 6 k2 b0 ðT Þ þ k 2 ; bðtÞ ¼ ðMþN ÞT ðMtþN hðtÞÞ ½k2 b0 ðT Þ þ k 2  b0 ð0Þ; if b0 ð0Þ > k2 b0 ðT Þ þ k 2 ; k2 ( 0; if b0 ð0Þ 6 k2 b0 ðT Þ þ k 2 ;  mk ¼ Lk ½k2 b0 ðT Þþk2 b0 ð0Þ ; if b0 ð0Þ > k2 b0 ðT Þ þ k 2 ; k2 ( 0; if b0 ð0Þ 6 k2 b0 ðT Þ þ k 2 ;  0 0  lk ¼ L minftk ;T tk g½k2 b ðT Þþk 2 b ð0Þ ; if b0 ð0Þ > k2 b0 ðT Þ þ k 2 :  k k2 2. Preliminaries Denote J 0 ¼ J n ftk ; t ¼ 1; 2; . . . ; mg: Let PCðJ ; RÞ ¼ fx : J ! R; xðtÞ is continuous everywhere except some 0 þ  0 tk at which xðt k Þ and xðt k Þ exist and xðt k Þ ¼ xðt k Þ; k ¼ 1; 2; . . . ; mg. PC ðJ ; RÞ ¼ fx 2 PCðJ ; RÞ : x ðtÞ is contin0  0 þ 0  0 uous everywhere except some tk at which x ðtk Þ and x ðtk Þ exist and x ðtk Þ ¼ x ðtk Þ; k ¼ 1; 2; . . . ; mg: Evidently, E0 ¼ PC 0 ðJ ; RÞ is a Banach space with norm kxkPC0 ¼ maxfkxðtÞkPC ; kx0 ðtÞkPC g, where kxkPC ¼ supt2J jxðtÞj. A function x 2 E0 \ C 2 ðJ 0 ; RÞ is called a solution of problem (1.3) if it satisfies (1.3). In this section, we first establish two comparison principles, then discuss the existence and uniqueness of the solutions for a linear impulsive differential equation. For convenience, we denote J 0 ¼ ½t0 ; t1 ; J 1 ¼ ðt1 ; t2 ; . . . ; J m ¼ ðtm ; tmþ1 ;

a ¼ maxftkþ1  tk ; k ¼ 0; 1; . . . ; mg: k

Lemma 2.1. Let M > 0, N P 0, Lk P 0, Lk P 0 ðk ¼ 1; . . . ; mÞ, k1 P 0, k2 > 0, and p 2 E0 \ C 2 ðJ 0 ; RÞ. Assume that (A1) (A2) (A3) (A4) (A5)

p00 ðtÞ 6 MpðtÞ þ NpðhðtÞÞ, t 2 J 0 , Dpðtk Þ P Lk p0 ðtk Þ, Dp0 ðtk Þ 6 Lk pðtk Þ, k ¼ 1; . . . ; m; pð0Þ ¼PpðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ, p0 ð0Þ 6 k2 p0 ðT Þ, P 1 P ð mk¼1 Lk þ aðm þ 1ÞðM þ N ÞÞðk1 þ ð1 þ 1=k2 Þðk1 þ mk¼1 Lk þ aðm þ 1ÞÞÞ.

Then pðtÞ P 0 on J. Proof. Assume that pðtÞ P 0 on J is not true. Then, there are two cases: (a) There exists t1 2 J such that pðt1 Þ < 0, and pðtÞ 6 0 for t 2 J ; (b) There exists t ; t 2 J such that pðt Þ < 0 and pðt Þ > 0. In case (a): (A1) implies p00 ðtÞ 6 0 for t 6¼ tk , t 2 J , and Dp0 ðtk Þ 6 Lk pðtk Þ 6 0 for k ¼ 1; . . . ; m: This means that p0 ðtÞ is nonincreasing in J. If p0 ðtÞ > 0 for t 2 J , then Dpðtk Þ P Lk p0 ðtk Þ > 0, k ¼ 1; . . . ; m: We get pðtÞ is strictly increasing on J, which contradicts pð0Þ ¼ pðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ. So, there exists a t 2 J , such that p0 ðtÞ 6 0. We first consider p0 ðtÞ ¼ 0. Case (a1): If t 2 ð0; T Þ, then p0 ðtÞ P 0 for t < t, and p0 ðtÞ 6 0 for t > t. From p0 ðtÞ is nonincreasing in J, we have 0 6 p0 ð0Þ=k2 6 p0 ðT Þ 6 p0 ðtÞ 6 p0 ð0Þ 6 k2 p0 ðT Þ 6 0; which means p0 ðtÞ ¼ 0 on each J i , 0 6 i 6 m. So, pðtÞ ¼ C i 6 0, t 2 J i , 0 6 i 6 m, where Ci are constants. From the fact that there exists t1 2 J such that pðt1 Þ < 0, we know that there exist a Jj, such that pðtÞ ¼ C j < 0, t 2 J j . Therefore, 0 ¼ p00 ðtÞ 6 MC j < 0, t 2 ðtj ; tjþ1 Þ, a contradiction.

H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

585

Case (a2): If t ¼ 0, then p0 ðtÞ 6 0, t 2 J . From p0 ð0Þ 6 k2 p0 ðT Þ, we know that p0 ðT Þ P 0. Therefore p0 ðT Þ ¼ 0 and 0 ¼ p0 ðT Þ 6 p0 ðtÞ 6 0, which means p0 ðtÞ ¼ 0 on each J i , 0 6 i 6 m. The rest proof is the same as Case (a1). Case (a3): If t ¼ T , then p0 ðtÞ P 0, t 2 J . From p0 ð0Þ 6 k2 p0 ðT Þ, we know that p0 ð0Þ 6 0. Therefore p0 ð0Þ ¼ 0 and 0 6 p0 ðtÞ 6 p0 ð0Þ ¼ 0, which means p0 ðtÞ ¼ 0 on each J i , 0 6 i 6 m. The rest proof is the same as Case (a1).Now, we consider p0 ðtÞ < 0. Case (a4): If t ¼ 0, then p0 ð0Þ < 0. From limt!0þ pðtÞpð0Þ < 0, pð0Þ ¼ pðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ, we know that t there exist n 2 J j , j 2 f0; 1; . . . ; mg, such that pðnÞ ¼ mint2ð0;T Þ pðtÞ < 0. If n 2 ðtj ; tjþ1 Þ, we obtain p0 ðtÞ P p0 ðnÞ ¼ 0 for t 2 ½0; n, which contradicts p0 ð0Þ < 0.If n ¼ tjþ1 , from p0 ðnÞ ¼ limt!n pðtÞpðnÞ 6 0, p0 ðnþ Þ ¼ limt!nþ pðtÞpðnÞ P 0 and Dp0 ðnÞ 6 Lk pðnÞ, we get pðnÞ P 0, a tn tn contradiction. Case (a5): If t ¼ T , then p0 ðT Þ < 0. From p0 ð0Þ 6 k2 p0 ðT Þ, we know that p0 ð0Þ < 0. It is the Case (a4). Case (a6): If t 2 ð0; T Þ, then p0 ðT Þ < 0. It is the Case (a5). In case (b): Let supt2J pðtÞ ¼ b. Then b > 0, and there exists ti < t 6 tiþ1 for some i such that pðt Þ ¼ b or þ  00 0 pðtþ i Þ ¼ b and p ðtÞ 6 ðM þ N Þb, Dp ðtk Þ 6 Lk b. We may assume that pðt  Þ ¼ b (in case of pðti Þ ¼ b, the proof 0 is similar). We assert that there exists a t 2 J , such that p ðtÞ P 0 (when ti < t < tiþ1 , we take t ¼ t . When t ¼ tiþ1 , if p0 ðtÞ < 0 for t 2 J , then p(t) is strictly decreasing on J i , which contradicts pðt Þ ¼ bÞ. Let t 2 J l , l 2 f0; 1; . . . ; mg, by mean value theorem, we obtain  0 0 þ 00 þ p0 ðtÞ  p0 ðt l Þ  bLl 6 p ðtÞ  p ðt l Þ ¼ p ðsl Þðt  t l Þ 6 ðM þ N Þab;

sl 2 ðtl ; tÞ;

 0 0 þ 00 þ p0 ðtl Þ  p0 ðt l1 Þ  bLl1 6 p ðt l Þ  p ðt l1 Þ ¼ p ðsl1 Þðt l  t l1 Þb

6 ðM þ N Þab;

sl1 2 ðtl1 ; tl Þ;

.. .  0 0 þ 00 p0 ðt2 Þ  p0 ðt 1 Þ  bL1 6 p ðt 2 Þ  p ðt 1 Þ ¼ p ðs1 Þðt 2  t 1 Þ 6 ðM þ N Þab;

p0 ðt1 Þ  p0 ð0Þ ¼ p00 ðs0 Þt1 6 ðM þ N Þab;

s0 2 ð0; t1 Þ:

Sum up these inequalities, we obtain 0

0

0 6 p ðtÞ 6 p ð0Þ þ b

m X

Lk

s1 2 ðt1 ; t2 Þ;

!

þ aðm þ 1ÞðM þ N Þ :

ð2:1Þ

k¼1

Using the same steps, we get 0

0

p ðtÞ P p ðT Þ  b

m X

Lk

! þ aðm þ 1ÞðM þ N Þ

0

P p ð0Þ=k2  b

k¼1 0

P p ðtÞ=k2  ð1 þ 1=k2 Þb

m X

! Lk

P ð1 þ 1=k2 Þb

! Lk

þ aðm þ 1ÞðM þ N Þ

k¼1

þ aðm þ 1ÞðM þ N Þ

k¼1 m X

m X

!

Lk þ aðm þ 1ÞðM þ N Þ :

ð2:2Þ

k¼1

Let t 2 J j for some j. If t < t , then j P i. By mean value theorem, we get 0 0  þ 0 pðt Þ  pðtj Þ P pðt Þ  pðtþ j Þ þ Lj p ðt j Þ ¼ p ðr j Þðt  t j Þ þ Lj p ðt j Þ

P ða þ Lj Þð1 þ 1=k2 Þb

m X

! Lk

þ aðm þ 1ÞðM þ N Þ ;

rj 2 ðtj ; t Þ;

k¼1

pðtj Þ  pðtj1 Þ P pðtj Þ 

pðtþ j1 Þ

0 þ Lj1 p0 ðtj1 Þ ¼ p0 ðrj1 Þðtj  tþ j1 Þ þ Lj1 p ðt j1 Þ

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H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

P ða þ Lj1 Þð1 þ 1=k2 Þb

m X

! Lk

þ aðm þ 1ÞðM þ N Þ ;

rj1 2 ðtj1 ; tj Þ;

k¼1

.. . m X

pðtiþ1 Þ  pðt Þ P að1 þ 1=k2 Þb

! Lk

þ aðm þ 1ÞðM þ N Þ :

k¼1

Sum up these inequalities, we obtain m X



pðt Þ P pðt Þ  ð1 þ 1=k2 Þb

! Lk

þ aðm þ 1ÞðM þ N Þ

k¼1

¼ b  ð1 þ 1=k2 Þb

m X

Lk

!

þ aðm þ 1ÞðM þ N Þ

m X

m X

Lk

!

þ aðm þ 1ÞðM þ N Þ

k¼1

pðt Þ P pð0Þ  ð1 þ 1=k2 Þb

Lk þ aðm þ 1Þ !

Lk þ aðm þ 1Þ

ð2:3Þ

P 0;

k¼1

which contradicts pðt Þ < 0. If t > t , then j 6 i. By mean value theorem, we get



!

k¼1

k¼1

pðT Þ P pðt Þ  ð1 þ 1=k2 Þb

m X

m X

! Lk þ aðm  i þ 1Þ ;

k¼iþ1

! Lk

m X

þ aðm þ 1ÞðM þ N Þ

k¼1

i X

! Lk þ ai :

k¼2

According to above Pm two inequalities, using the boundary  condition  (2.2), we have Pm and Eqs. (2.1), pðt Þ P pðt Þ  b k¼1 Lk þ aðm þ 1ÞðM þ N Þ k1 þ ð1 þ 1=k2 Þ k1 þ k¼1 Lk þ aðm þ 1Þ , which also contradicts pðt Þ < 0. The proof of Lemma 2.1 is complete. h Lemma 2.2. Let M > 0, N P 0, Lk P 0, Lk P 0 ðk ¼ 1; . . . ; mÞ, k1 P 0, k2 > 0, and p 2 E0 \ C 2 ðJ 0 ; RÞ. Assume that (B1) (B2) (B3) (B4)

hðtÞÞ 0 p00 ðtÞ 6 MpðtÞ þ NpðhðtÞÞ  ðMþN ÞT ðMtþN ½p ð0Þ  k2 p0 ðT Þ; t 2 J 0 , k2 0 0 0 Dpðtk Þ P Lk p ðtk Þ þ ðLk =k2 Þ½p ð0Þ  k2 p ðT Þ, Dp0 ðtk Þ 6 Lk pðtk Þ  ðLk =k2 Þ minftk ; T  tk g½p0 ð0Þ  k2 p0 ðT Þ; k ¼ 1; . . . ; m; pð0Þ ¼ pðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ, p0 ð0Þ > k2 p0 ðT Þ,

and they satisfy (A5). Then pðtÞ P 0 on J. Proof. Let uðtÞ ¼ pðtÞ  minft; T  tg k12 ½p0 ð0Þ  k2 p0 ðT Þ, then u 2 E0 , pðtÞ P uðtÞ. uð0Þ ¼ pð0Þ ¼ pðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ ¼ uðT Þ þ k1 ½u0 ð0Þ þ u0 ðT Þ; u0 ð0Þ ¼ p0 ð0Þ  p0 ð0Þ=k2 þ p0 ðT Þ < k2 p0 ðT Þ þ p0 ð0Þ  k2 p0 ðT Þ ¼ k2 u0 ðT Þ; and it is easy to verify u00 ðtÞ 6 MuðtÞ þ NuðhðtÞÞ;

t 2 J 0;

Duðtk Þ P Lk u0 ðtk Þ; Du0 ðtk Þ 6 Lk uðtk Þ;

k ¼ 1; . . . ; m:

By Lemma 2.1 and pðtÞ P uðtÞ, we get pðtÞ P 0: The proof of Lemma 2.2 is complete.

h

Consider the following boundary value problem for a linear impulsive functional differential equation:

H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

 x00 ðtÞ þ MxðtÞ ¼ rðtÞ  NxðhðtÞÞ; t 6¼ tk ; t 2 J ¼ ½0; T ; Dxðtk Þ ¼ Lk x0 ðtk Þ þ I k ðgðtk ÞÞ  Lk g0 ðtk Þ; Dx0 ðtk Þ ¼ Lk xðtk Þ þ I k ðgðtk ÞÞ  Lk gðtk Þ; k ¼ 1; . . . ; m; xð0Þ ¼ xðT Þ þ k1 ½x0 ð0Þ þ x0 ðT Þ þ k 1 ; x0 ð0Þ ¼ k2 x0 ðT Þ þ k 2 ;

587

ð2:4Þ

where M > 0, N P 0, Lk P 0, Lk P 0 are constants I k ; I k 2 CðJ ; RÞ ðk ¼ 1; . . . ; mÞ, rðtÞ 2 PCðJ ; RÞ, k1 P 0, k2 > 0. By direct computation, we have the following results. Lemma 2.3. x 2 E0 \ C 2 ðJ 0 ; RÞ is a solution of (2.4) if and only if x 2 PCðJ ; RÞ is a solution of the following impulsive integral equation: Z T m X G1 ðt; sÞ½rðsÞ  NxðhðsÞÞds þ ½G1 ðt; tk ÞðLk xðtk Þ þ I k ðgðtk ÞÞ  Lk gðtk ÞÞ xðtÞ ¼ 0

k¼1

þ G2 ðt; tk ÞðLk x0 ðtk Þ þ I k ðgðtk ÞÞ  Lk g0 ðtk ÞÞ þ wðtÞ; t 2 J ; pffiffiffiffiffi   ~ ~ ~ ¼ M ; C ¼ ð1 þ k2 Þ 1  12 ð1 þ mk ~ 1 ÞemT ~ 1 ÞemT where m  12 ð1  mk , and 8   ~  ~ ~ ~ > ~ 1 ÞemT ~ 1 ÞemT þ 12 ð1 þ k2 Þð1  mk  1 emðtsÞ 1  12 ð1 þ k2 Þð1 þ mk emðstÞ > > > ~ Þ > > ~ 1 k2 þ mk ~ 1 þ k2 ÞÞemðsþtT þ 12 ð1  ðmk > > > ~ tsÞ ~ 1 k2 þ mk ~ 1  k2 ÞÞemðT ; 0 6 s < t 6 T; 1 <  12 ð1 þ ðmk 1    G1 ðt; sÞ ¼  mT ~ mðstÞ ~ ~ > 2 ð1 þ k2 Þð1  mk ~ > 2mC ~ 1 Þe ~ 1 ÞemT  k2 e þ k2  12 ð1 þ k2 Þð1 þ mk > > > ~ ~ Þ > ~ 1 k2 þ mk ~ 1 þ k2 ÞÞemðsþtT þ 12 ð1  ðmk emðtsÞ > > > :  1 ð1 þ ðmk mðT ~ tsÞ ~ 1 k2 þ mk ~ 1  k2 ÞÞe ; 0 6 t 6 s 6 T; 2 8    mðtsÞ ~ ~ ~ > ~ 1 ÞemT ~ 1 ÞemT þ 1  12 ð1 þ k2 Þð1  mk 1  12 ð1 þ k2 Þð1 þ mk emðstÞ e~ > > > ~ 1 mðsþtT Þ > > ~ 1 k2 þ mk ~ 1 þ k2 ÞÞe þ 2 ð1  ðmk > > > < ~ tsÞ 1 ~ 1 k2 þ mk ~ 1  k2 ÞÞemðT þ ð1 þ ðmk ; 0 6 s < t 6 T; 1 1 2    ~ G2 ðt; sÞ ¼  mT ~ mðstÞ ~ ~ > 2 ð1 þ k2 Þð1  mk 2C > ~ 1 Þe ~ 1 ÞemT  k2 e þ 12 ð1 þ k2 Þð1 þ mk  k2 emðtsÞ > > > ~ Þ > ~ 1 k2 þ mk ~ 1 þ k2 ÞÞemðsþtT þ 12 ð1  ðmk > > > : þ 1 ð1 þ ðmk ~ tsÞ mðT ~ 1 k2 þ mk ~ 1  k2 ÞÞe ; 0 6 t 6 s 6 T; 2 1 ~ ~ ~ 1 þ mk ~ 2 k1  k 2 Þemt ~ 1 þ mk ~ 2 k1 þ k 2 Þemt ððmk þ ðmk ~ 2mC ~ tÞ ~ Þ ~ 2 k1  mk ~ 1 k2 ÞemðT ~ 2 k1  k 2  mk ~ 1 k2 ÞemðtT þ ðk 2 þ mk þ ðmk Þ:

wðtÞ ¼

By Lemma 2.3 and Banach fixed-point theorem, it is easy to prove the following Lemma. Lemma 2.4. Let M > 0, N P 0, Lk P 0, Lk P 0. If ! m m X X  Lk þ g2 Lk < 1; g1 NT þ k¼1

g2 NT þ

m X k¼1

! Lk

þ Mg1

m X

Lk < 1;

k¼1

where ~ ~ maxf1; k2 gðemT þ 1Þ 2k1 emT þ ; 2 ~  1Þ þ mk ~ þ 1ÞÞ ~  1ÞÞ ~  1Þ þ mk ~ þ k2 ÞððemT ~ 1 ðemT mð1 ~ 1 ðe2mT ððemT ~ maxf1; k2 gðemT  1Þ j1  k2 j : þ g2 ¼ ~ ~ mT mT ~ 1 ðe þ 1ÞÞ ð1 þ k2 Þððe  1Þ þ mk C Then Eq. (2.4) has a unique solution x in E0 :

g1 ¼

ð2:5Þ

k¼1

ð2:6Þ

588

H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

3. Main result Theorem 3.1. Let (A5), (2.5), (2.6) and the following conditions hold. • (C1) The function a0 ; b0 are lower and upper solutions of BVP (1.3), respectively. • (C2) The function f 2 CðJ  R2 ; RÞ satisfies f ðt; x; yÞ  f ðt; x; yÞ P Mðx  xÞ  N ðy  yÞ for a0 ðtÞ 6 x 6 x 6 b0 ðtÞ, a0 ðhðtÞÞ 6 y 6 y 6 b0 ðhðtÞÞ, t 2 J , where M > 0, N P 0, Lk P 0, Lk P 0. • (C3) The functions I k ; I k 2 CðJ ; RÞ satisfy I k ðxðtk ÞÞ  I k ðyðtk ÞÞ P Lk ðx0 ðtk Þ  y 0 ðtk ÞÞ; I k ðxðtk ÞÞ  I k ðyðtk ÞÞ 6 Lk ðxðtk Þ  yðtk ÞÞ for a0 ðtk Þ 6 yðtk Þ 6 xðtk Þ 6 b0 ðtk Þ, k ¼ 1; . . . ; m. Then, there exist monotone sequences fan ðtÞg; fbn ðtÞg  E0 \ C 2 ðJ 0 ; RÞ such that lim an ðtÞ ¼ qðtÞ;

lim bn ðtÞ ¼ rðtÞ

n!1

n!1

uniformly on J, and q, r are minimal and maximal solutions of problem (1.3), respectively. Proof. Let ½a0 ; b0  ¼ fx 2 E0 \ C 2 ðJ 0 ; RÞ : a0 ðtÞ 6 xðtÞ 6 b0 ðtÞ; t 2 J g. For any g 2 ½a0 ; b0 ; consider BVP (2.4) with rðtÞ ¼ f ðt; gðtÞ; gðhðtÞÞÞ þ MgðtÞ þ N gðhðtÞÞ: By Lemma 2.4, (2.4) possesses a unique solution x 2 E0 \ C 2 ðJ 0 ; RÞ. We define an operator A by xðtÞ ¼ AgðtÞ, then A has the following properties: (i) a0 6 Aa0 ; Ab0 6 b0 ; (ii) For any a0 6 g1 6 g2 6 b0 , we have Ag1 6 Ag2 . To prove (i), we set p ¼ a1  a0 , where a1 ¼ Aa0 . p00 ðtÞ ¼ a001 ðtÞ  a000 ðtÞ 6 Ma1 ðtÞ þ N a1 ðhðtÞÞ  f ðt; a0 ðtÞ; a0 ðhðtÞÞÞ  Ma0 ðtÞ  N a0 ðhðtÞÞ þ f ðt; a0 ðtÞ; a0 ðhðtÞÞÞ þ aðtÞ ¼ MpðtÞ þ NpðhðtÞÞ þ ap ðtÞ;

t 6¼ tk ; t 2 J ;

Dpðtk Þ ¼ Da1 ðtk Þ  Da0 ðtk Þ P Lk a01 ðtk Þ þ I k ða0 ðtk ÞÞ  Lk a00 ðtk Þ  I k ða0 ðtk ÞÞ  mk ¼ Lk p0 ðtk Þ þ mpk ; 0

Dp ðtk Þ ¼

Da01 ðtk Þ



k ¼ 1; . . . ; m;

Da00 ðtk Þ

6 Lk a1 ðtk Þ þ I k ða0 ðtk ÞÞ  Lk a0 ðtk Þ  I k ða0 ðtk ÞÞ  lk ¼ Lk pðtk Þ þ lpk ;

k ¼ 1; . . . ; m;

pð0Þ ¼ a1 ð0Þ  a0 ð0Þ ¼ pðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ; where ap ðtÞ; mpk ; lpk ðk ¼ 1; . . . ; mÞ are given by ( 0; if p0 ð0Þ 6 k2 p0 ðT Þ; ap ðtÞ ¼  k12 ½ðM þ N ÞT  ðMt þ N hðtÞÞ½p0 ð0Þ  k2 p0 ðT Þ; if p0 ð0Þ > k2 p0 ðT Þ;

H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

( mpk ¼ ( lpk ¼

0;

if p0 ð0Þ 6 k2 p0 ðT Þ;

Lk ½p0 ð0Þk2 p0 ðT Þ ; k2

if p0 ð0Þ > k2 p0 ðT Þ; if p0 ð0Þ 6 k2 p0 ðT Þ;

0; 

589

Lk

minftk ;T tk g½p0 ð0Þk2 p0 ðT Þ k2

;

if p0 ð0Þ > k2 p0 ðT Þ:

By Lemma 2.1 and 2.2, we get pðtÞ P 0 on J, i.e., a0 6 Aa0 . The proof of Ab0 6 b0 is similar. To prove (ii), set a0 6 g1 6 g2 6 b0 and p ¼ u2  u1 , where u1 ¼ Ag1 ; u2 ¼ Ag2 . By C2, C3, we obtain p00 ðtÞ ¼ u002 ðtÞ  u001 ðtÞ ¼ Mu2 ðtÞ þ Nu2 ðhðtÞÞ  f ðt; g2 ðtÞ; g2 ðhðtÞÞÞ  Mg2 ðtÞ  N g2 ðhðtÞÞ  fMu1 ðtÞ þ Nu1 ðhðtÞÞ  f ðt; g1 ðtÞ; g1 ðhðtÞÞÞ  Mg1 ðtÞ  N g1 ðhðtÞÞg 6 MpðtÞ þ NpðhðtÞÞ; t 6¼ tk ; t 2 J : It is easy to verify that Dpðtk Þ P Lk p0 ðtk Þ; Dp0 ðtk Þ 6 Lk pðtk Þ; k ¼ 1; . . . ; m; pð0Þ ¼ pðT Þ þ k1 ½p0 ð0Þ þ p0 ðT Þ; p0 ð0Þ ¼ k2 p0 ðT Þ: In view of Lemma 2.1, we have pðtÞ P 0 on J, i.e., Ag1 6 Ag2 . We define the sequences fan ðtÞg; fbn ðtÞg such that an ¼ Aan1 ; bn ¼ Abn1 ðn ¼ 1; 2; . . .Þ: From (i) and (ii), we have a 0 6 a 1 6 a 2 6    6 a n 6    6 b n 6    6 b 2 6 b1 6 b0

on J

and each an ; bn (n ¼ 1; 2; . . . ; n) satisfies Z T m X G1 ðt; sÞ½rn1 ðsÞ  N an ðhðsÞÞds þ ½G1 ðt; tk ÞðLk an ðtk Þ an ðtÞ ¼ 0

k¼1

þ I k ðan1 ðtk ÞÞ  Lk an1 ðtk ÞÞ þ G2 ðt; tk ÞðLk a0n ðtk Þ þ I k ðan1 ðtk ÞÞ  Lk a0n1 ðtk ÞÞ þ wðtÞ; Z T m X bn ðtÞ ¼ G1 ðt; sÞ½~ rn1 ðsÞ  N bn ðhðsÞÞds þ ½G1 ðt; tk ÞðLk bn ðtk Þ 0

þ

t 2 J;

k¼1

I k ðbn1 ðtk ÞÞ



Lk bn1 ðtk ÞÞ

þ G2 ðt; tk ÞðLk b0n ðtk Þ þ I k ðbn1 ðtk ÞÞ  Lk b0n1 ðtk ÞÞ þ wðtÞ;

t 2 J;

where rn1 ðtÞ ¼ f ðt; an1 ðtÞ; an1 ðhðtÞÞÞ þ Man1 ðtÞ þ N an1 ðhðtÞÞ; ~n1 ðtÞ ¼ f ðt; bn1 ðtÞ; bn1 ðhðtÞÞÞ þ Mbn1 ðtÞ þ N bn1 ðhðtÞÞ: r Therefore, there exist q; r such that lim an ðtÞ ¼ qðtÞ;

n!1

lim bn ðtÞ ¼ rðtÞ

n!1

uniformly on J :

Clearly q; r satisfy BVP (1.3). Finally, we prove that q and r are minimal and maximal solutions of BVP (1.3). Let xðtÞ be any solution of BVP (1.3) such that x 2 ½a0 ; b0 . Suppose that there exist a positive integer n such that x 2 ½an ; bn . Setting p ¼ x  anþ1 , by (C2) and (C3), we have p00 ðtÞ ¼ x00 ðtÞ  a00nþ1 ðtÞ ¼ f ðt; xðtÞ; xðhðtÞÞÞ  fManþ1 ðtÞ þ N anþ1 ðhðtÞÞ  f ðt; an ðtÞ; an ðhðtÞÞÞ  Man ðtÞ  N an ðhðtÞÞg 6 MpðtÞ þ NpðhðtÞÞ;

t 6¼ tk ;

t 2 J;

590

H. Wang, H. Chen / Applied Mathematics and Computation 191 (2007) 582–591

Dpðtk Þ ¼ Dxðtk Þ  Danþ1 ðtk Þ ¼ I k ðxðtk ÞÞ  Lk a0nþ1 ðtk Þ  I k ðan ðtk ÞÞ þ Lk a0n ðtk Þ P Lk p0 ðtk Þ; Dp0 ðtk Þ ¼ Dx0 ðtk Þ  Da0nþ1 ðtk Þ 6 Lk pðtk Þ; 0

0

pð0Þ ¼ pðT Þ þ k1 ½p ð0Þ þ p ðT Þ;

k ¼ 1; . . . ; m;

0

p ð0Þ ¼ k2 p0 ðT Þ:

By Lemma 2.1, it follows that pðtÞ P 0 on J, that is, anþ1 6 x on J. Similarly, we obtain x 6 bnþ1 on J. Thus, an ðtÞ 6 xðtÞ 6 bn ðtÞ on J for every n. Therefore, we obtain qðtÞ 6 xðtÞ 6 rðtÞ on J by taking limit as n ! 1. The proof of Theorem 3.1 is complete. h Example. Consider the problem of   1 1 1 1 00 xðtÞ  x t þ t; t 6¼ ; t 2 ½0; 1;  x ðtÞ ¼  100 1000 2 2     1 1 0 1 x Dx ¼ ; 2 300 2     1 1 1 x ¼ ; Dx0 2 4000 2 101 0 201 299 x ð1Þ  ; x0 ð0Þ ¼ 100x0 ð1Þ  : xð0Þ ¼ xð1Þ þ 100 15000 453 We change BVP (3.1) to its equivalent BVP (3.2)   1 1 1 1 xðtÞ  x t þ t; t 6¼ ; t 2 ½0; 1;  x00 ðtÞ ¼  100 1000 2 2     1 1 0 1 x Dx ¼ ; 2 300 2     1 1 1 x ¼ ; Dx0 2 4000 2 1 0 15401 299 ½x ð0Þ þ x0 ð1Þ  ; x0 ð0Þ ¼ 100x0 ð1Þ  : xð0Þ ¼ xð1Þ þ 100 2265000 453

ð3:1Þ

ð3:2Þ

1 1 1 0 with T ¼ 1, k ¼ 1, t1 ¼ 12, where hðtÞ ¼ 12 t, f ðt; x; xðhðtÞÞÞ ¼ t  100 xðtÞ  1000 xðhðtÞÞ, I 1 ðxðt1 ÞÞ ¼ 300 x ðt1 Þ,  1 I 1 ðxðt1 ÞÞ ¼ 4000 xðt1 Þ. Setting ( ( 1 1 t  4; t 2 ½0; 12; t þ 100; t 2 ½0; 12; 151 151 aðtÞ ¼ bðtÞ ¼ 1 1 t  4; t 2 ð12 ; 1; t þ 100; t 2 ð12 ; 1; 150 150 1 1 , N ¼ 1000 , It is easy to verify that aðtÞ; bðtÞ are lower and upper solutions of BVP (3.2), with a 6 b. Let M ¼ 100  1 1 Lk ¼ 300, Lk ¼ 4000, the conditions of Theorem 3.1 are all satisfied. So, BVP (3.2) has minimal and maximal solutions of (3.2) in the segment ½aðtÞ; bðtÞ.

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