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Periodic boundary value problems for second-order impulsive integro-differential equations with integral jump conditions

Abstract

This paper is concerned with the existence of extremal solutions of periodic boundary value problems for second-order impulsive integro-differential equations with integral jump conditions. We introduce a new definition of lower and upper solutions with integral jump conditions and prove some new maximum principles. The method of lower and upper solutions and the monotone iterative technique are used.

MSC: 34B37, 34K10, 34K45.

1 Introduction

Differential equations which have impulse effects describe many evolution processes that abruptly change their state at a certain moment. In recent years, impulsive differential equations have become more important tools in some mathematical models of real processes and phenomena studied in physics, biotechnology, chemical technology, population dynamics and economics; see [15]. Many papers have been published about existence analysis of periodic boundary value problems of first and second order for impulsive ordinary or functional or integro-differential equations. We refer the readers to the papers [629]. More recent works on existence results of impulsive problems with integral boundary conditions can be found in [3035] and the reference therein. This literature has lead to significant development of a general theory for impulsive differential equations.

The monotone iterative technique coupled with the method of upper and lower solutions has been used to study the existence of extremal solutions of periodic boundary value problems for second-order impulsive equations; see, for example, [3641]. This method has been also used to study abstract nonlinear problems; see [42]. However, in most of these papers concerned with applications of the monotone iterative technique to second-order periodic boundary value problems with impulses, the authors assume that the jump conditions at impulse point t k of solution values and the derivative of solution values depend on the left-hand limits of solutions or the slope of solutions themselves, such as Δx( t k )= I k (x( t k )), Δx( t k )= I k ( x ( t k )), Δ x ( t k )= I k (x( t k )), Δ x ( t k )= I k ( x ( t k )).

In this paper, we consider the periodic boundary value problem for second-order impulsive integro-differential equation (PBVP) with integral jump conditions:

{ x ( t ) = f ( t , x ( t ) , ( K x ) ( t ) , ( S x ) ( t ) ) , t J = [ 0 , T ] , t t k , Δ x ( t k ) = I k ( t k δ k t k ε k x ( s ) d s ) , k = 1 , 2 , , m , Δ x ( t k ) = I k ( t k τ k t k σ k x ( s ) d s ) , k = 1 , 2 , , m , x ( 0 ) = x ( T ) , x ( 0 ) = x ( T ) ,
(1.1)

where 0= t 0 < t 1 < t 2 << t k << t m < t m + 1 =T, f:J× R 3 R is continuous everywhere except at { t k }× R 3 , f( t k + ,x,y,z), f( t k ,x,y,z) exist, f( t k ,x,y,z)=f( t k ,x,y,z), I k C(R,R), I k C(R,R), Δx( t k )=x( t k + )x( t k ), Δ x ( t k )= x ( t k + ) x ( t k ), 0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m,

(Kx)(t)= 0 t k(t,s)x(s)ds,(Sx)(t)= 0 T h(t,s)x(s)ds,

k(t,s)C(D, R + ), h(t,s)C(J×J, R + ), D={(t,s) R 2 ,0stT}, R + =[0,+), k 0 =max{k(t,s):(t,s)D}, h 0 =max{h(t,s):(t,s)J×J}.

In [43, 44], the authors discussed some kinds of first-order impulsive problems with the integral jump condition

Δx( t k )= I k ( t k τ k t k x ( s ) d s t k 1 t k 1 + σ k 1 x ( s ) d s ) ,
(1.2)

where 0< σ k 1 ( t k t k 1 )/2, 0 τ k 1 ( t k t k 1 )/2, k=1,2,,m. We note that the jump condition (1.2) depends on functionals of path history before impulse points t k and after the past impulse points t k 1 . The aim of our research is to deal with the integral jump conditions

Δx( t k )= I k ( t k δ k t k ε k x ( s ) d s ) ,Δ x ( t k )= I k ( t k τ k t k σ k x ( s ) d s ) ,
(1.3)

where 0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m. The integral jump condition (1.3) means that a sudden change of solution values and the derivative of solution values at impulse point t k depend on the area under the curves of x (t) and x(t) between t= t k δ k to t= t k ε k and t= t k τ k to t= t k σ k , respectively. It should be noticed that the impulsive effects of PBVP (1.1) have memory of the past states.

This paper is organized as follows. Firstly, we introduce a new concept of lower and upper solutions. After that, we establish some new comparison principles and discuss the existence and uniqueness of the solutions for second-order impulsive integro-differential equations with integral jump conditions. By using the method of upper and lower solutions and the monotone iterative technique, we obtain the existence of an extreme solution of PBVP (1.1). Finally, we give an example to illustrate the obtained results.

2 Preliminaries

Let J =J{ t 1 , t 2 ,, t m }, J 0 =[ t 0 , t 1 ], J k =( t k , t k + 1 ] for k=1,2,,m. Let PC(J,R)={x:JR;x(t) is continuous everywhere except for some  t k  at which x( t k + ) and x( t k ) exist andx( t k )=x( t k ),k=1,2,,m}, and P C 1 (J,R)={xPC(J,R); x (t) is continuous everywhereexcept for some  t k  at which  x ( t k + ) and  x ( t k ) exist and  x ( t k )= x ( t k ),k=1,2,,m}. PC(J,R) and P C 1 (J,R) are Banach spaces with the norms x P C =sup{x(t):tJ} and x P C 1 =max{ x P C , x P C }. Let E=P C 1 (J,R) C 2 ( J ,R). A function xE is called a solution of PBVP (1.1) if it satisfies (1.1).

Definition 2.1 We say that the functions α 0 , β 0 E are lower and upper solutions of PBVP (1.1), respectively, if there exist M>0, N0, L0, L k 0, L k 0, 0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , such that

{ α 0 ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) + a ( t ) , t J , Δ α 0 ( t k ) = I k ( t k δ k t k ε k α 0 ( s ) d s ) + m k , k = 1 , 2 , , m , Δ α 0 ( t k ) I k ( t k τ k t k σ k α 0 ( s ) d s ) + l k , k = 1 , 2 , , m , α 0 ( 0 ) = α 0 ( T ) ,

where

a ( t ) = { 0 if  α 0 ( 0 ) α 0 ( T ) , [ α 0 ( T ) α 0 ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] if  α 0 ( 0 ) < α 0 ( T ) , m k = { 0 if  α 0 ( 0 ) α 0 ( T ) , L k [ α 0 ( T ) α 0 ( 0 ) ] T t k δ k t k ε k ( T 2 s ) d s if  α 0 ( 0 ) < α 0 ( T ) , l k = { 0 if  α 0 ( 0 ) α 0 ( T ) , L k [ α 0 ( T ) α 0 ( 0 ) ] T t k τ k t k σ k ( s T s 2 ) d s if  α 0 ( 0 ) < α 0 ( T ) ,

and

{ β 0 ( t ) f ( t , β 0 ( t ) , ( K β 0 ) ( t ) , ( S β 0 ) ( t ) ) b ( t ) , t J , Δ β 0 ( t k ) = I k ( t k δ k t k ε k β 0 ( s ) d s ) m k , k = 1 , 2 , , m , Δ β 0 ( t k ) I k ( t k τ k t k σ k β 0 ( s ) d s ) l k , k = 1 , 2 , , m , β 0 ( 0 ) = β 0 ( T ) ,

where

b ( t ) = { 0 if  β 0 ( 0 ) β 0 ( T ) , [ β 0 ( 0 ) β 0 ( T ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] if  β 0 ( 0 ) > β 0 ( T ) , m k = { 0 if  β 0 ( 0 ) β 0 ( T ) , L k [ β 0 ( 0 ) β 0 ( T ) ] T t k δ k t k ε k ( T 2 s ) d s if  β 0 ( 0 ) > β 0 ( T ) , l k = { 0 if  β 0 ( 0 ) β 0 ( T ) , L k [ β 0 ( 0 ) β 0 ( T ) ] T t k τ k t k σ k ( s T s 2 ) d s if  β 0 ( 0 ) > β 0 ( T ) .

Now we are in the position to establish some new comparison principles which play an important role in the monotone iterative technique.

Lemma 2.1 Assume that xE satisfies

{ x ( t ) M x ( t ) + N 0 t k ( t , s ) x ( s ) d s + L 0 T h ( t , s ) x ( s ) d s , t J , Δ x ( t k ) = L k t k δ k t k ε k x ( s ) d s , k = 1 , 2 , , m , Δ x ( t k ) L k t k τ k t k σ k x ( s ) d s , k = 1 , 2 , , m , x ( 0 ) = x ( T ) , x ( 0 ) x ( T ) ,
(2.1)

whereM>0, N0, L0, L k 0, L k 0are constants and0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m, and they satisfy

[ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] 1.
(2.2)

Thenx(t)0, tJ.

Proof Suppose, to the contrary, that x(t)>0 for some tJ. We divide the proof into two cases:

Case (i). There exists a t ˜ J such that x( t ˜ )>0 and x(t)0 for all tJ.

From (2.1), we have x (t)0 for t J . Since Δ x ( t k ) L k t k τ k t k σ k x(s)ds0, then x (t) is nondecreasing in tJ and so x (0) x (T). However, by (2.1) x (0) x (T), then x (0)= x (T), which implies x (t)= constant for all tJ. Thus, 0= x (t)Mx( t ˜ )>0, a contradiction.

Case (ii). There exists t , t J such that x( t )>0, x( t )<0.

Let inf t J x(t)=λ<0, then there exists t J i , for some i{0,1,,m}, such that x( t )=λ or x( t i + )=λ. Without loss of generality, we only consider x( t )=λ. For the case x( t i + )=λ the proof is similar. It follows that

If x (t)>0 for all tJ, then Δx( t k )= L k t k δ k t k ε k x (s)ds0, k=1,2,,m. Hence, x(t) is strictly increasing on J, which contradicts x(0)=x(T). Then there exists a t ¯ J such that x ( t ¯ )0.

Let t ¯ J j , j{0,1,,m}. By mean value theorem, we have

Summing up the above inequalities, we obtain

x ( 0 ) x ( t ¯ ) + λ [ k = 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( t ¯ t 0 ) ] λ [ k = 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) t ¯ ] .
(2.3)

Let t J h , h{0,1,,m}. If t t ¯ by using the method to get (2.3), then we have

x ( t ) x ( t ¯ ) + λ [ k = h + 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( t ¯ t ) ] λ [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] .

If t> t ¯ , then the above method together with (2.1), (2.3) implies that

x ( t ) x ( T ) + λ [ k = h + 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( T t ) ] x ( 0 ) + λ [ k = h + 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( T t ) ] λ [ k = 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) t ¯ ] + λ [ k = h + 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( T t ) ] λ [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] .

Thus,

x (t)λ [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] ,t J .

Let t J r for some r{0,1,,m}. We first assume that t < t , then i<r. By the mean value theorem, we have

Summing up, we get

0<x ( t ) λ+λ [ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] .

Hence,

[ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] >1,

which contradicts (2.2).

For the case t < t , the proof is similar, and thus we omit it. This completes the proof. □

Lemma 2.2 Assume that xE satisfies

whereM>0, N0, L0, L k 0, L k 0are constants and0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m, and they satisfy (2.2). Thenx(t)0for alltJ.

Proof Let u(t)=[ t T t 2 T ][ x (T) x (0)], tJ, and define

w(t)=x(t)+u(t)=x(t)+ [ t T t 2 T ] [ x ( T ) x ( 0 ) ] .

Note that u(0)=u(T), u(t)0 for tJ. If we prove that w0, then x(t)x(t)+u(t)0 and the proof is complete. Since u (t)=[ T 2 t T ][ x (T) x (0)], then we get

w ( 0 ) = x ( 0 ) + u ( 0 ) = x ( T ) + u ( T ) = w ( T ) , w ( 0 ) = x ( 0 ) + u ( 0 ) = x ( 0 ) + x ( T ) x ( 0 ) = x ( T ) , w ( T ) = x ( T ) + u ( T ) = x ( T ) x ( T ) + x ( 0 ) = x ( 0 ) .

Hence, w (0)> w (T). Indeed, for k=1,2,,m,

Δ w ( t k ) = Δ x ( t k ) + Δ u ( t k ) = L k t k δ k t k ε k x ( s ) d s + L k [ x ( T ) x ( 0 ) ] T t k δ k t k ε k T 2 s d s = L k t k δ k t k ε k x ( s ) + u ( s ) d s = L k t k δ k t k ε k w ( s ) d s ,

and

Δ w ( t k ) = Δ x ( t k ) + Δ u ( t k ) L k t k τ k t k σ k x ( s ) d s + L k [ x ( T ) x ( 0 ) ] T t k τ k t k σ k s T s 2 d s L k t k τ k t k σ k x ( s ) + u ( s ) d s L k t k τ k t k σ k w ( s ) d s .

Meanwhile, for t t k , tJ,

w ( t ) M w ( t ) N 0 t k ( t , s ) w ( s ) d s L 0 T h ( t , s ) w ( s ) d s = x ( t ) M x ( t ) M t T [ x ( T ) x ( 0 ) ] N 0 t k ( t , s ) x ( s ) d s L 0 T h ( t , s ) x ( s ) d s [ x ( T ) x ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] 0 .

Then by Lemma 2.1, we get w(t)0 for all tJ, which implies that x(t)0, tJ. □

Consider the linear PBVP

{ x ( t ) = M x ( t ) + N 0 t k ( t , s ) x ( s ) d s + L 0 T h ( t , s ) x ( s ) d s g ( t ) , t J , Δ x ( t k ) = L k t k δ k t k ε k x ( s ) d s + γ k , k = 1 , 2 , , m , Δ x ( t k ) = L k t k τ k t k σ k x ( s ) d s + λ k , k = 1 , 2 , , m , x ( 0 ) = x ( T ) , x ( 0 ) = x ( T ) ,
(2.4)

where constants M>0, N0, L0, L k 0, L k 0, γ k , λ k are constants and gPC(J,R), 0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m.

Lemma 2.3xEis a solution of (2.4) if and only ifxP C 1 (J,R)is a solution of the following impulsive integral equation:

x ( t ) = 0 T G 1 ( t , s ) [ g ( s ) N 0 s k ( s , r ) x ( r ) d r L 0 T h ( s , r ) x ( r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) ( L k t k τ k t k σ k x ( s ) d s + λ k ) + G 2 ( t , t k ) ( L k t k δ k t k ε k x ( s ) d s + γ k ) ] ,
(2.5)

where

G 1 ( t , s ) = [ 2 M ( e M T 1 ) ] 1 { e M ( T t + s ) + e M ( t s ) , 0 s < t T , e M ( T + t s ) + e M ( s t ) , 0 t s T , G 2 ( t , s ) = [ 2 ( e M T 1 ) ] 1 { e M ( T t + s ) e M ( t s ) , 0 s < t T , e M ( T + t s ) + e M ( s t ) , 0 t s T .

This proof is similar to the proof of Lemma 2.1 in [36], and we omit it.

Lemma 2.4 LetM>0, N0, L0, L k 0, L k 0are constants and0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m. If

(2.6)
(2.7)

then (2.4) has a unique solution x in E.

Proof For any xE, we define an operator F by

( F x ) ( t ) = 0 T G 1 ( t , s ) [ g ( s ) N 0 s k ( s , r ) x ( r ) d r L 0 T h ( s , r ) x ( r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) ( L k t k τ k t k σ k x ( s ) d s + λ k ) + G 2 ( t , t k ) ( L k t k δ k t k ε k x ( s ) d s + γ k ) ] ,

where G 1 , G 2 are given by Lemma 2.3. Then FxP C 1 (J,R) and

( F x ) ( t ) = 0 T G 2 ( t , s ) [ g ( s ) N 0 s k ( s , r ) x ( r ) d r L 0 T h ( s , r ) x ( r ) d r ] d s + k = 1 m [ G 2 ( t , t k ) ( L k t k τ k t k σ k x ( s ) d s + λ k ) M G 1 ( t , t k ) ( L k t k δ k t k ε k x ( s ) d s + γ k ) ] .

By computing directly, we have

max ( t , s ) J × J { G 1 ( t , s ) } = 1 + e M T 2 M ( e M T 1 )

and

max ( t , s ) J × J { G 2 ( t , s ) } = 1 2 .

On the other hand, for x,yP C 1 (J,R), we have

( F x ) ( F y ) P C = sup t J | ( F x ) ( t ) ( F y ) ( t ) | = sup t J | 0 T G 1 ( t , s ) [ N 0 s k ( s , r ) [ x ( r ) y ( r ) ] d r L 0 T h ( s , r ) [ x ( r ) y ( r ) ] d r ] d s + k = 1 m [ G 1 ( t , t k ) ( L k t k τ k t k σ k [ x ( s ) y ( s ) ] d s ) + G 2 ( t , t k ) ( L k t k δ k t k ε k [ x ( s ) y ( s ) ] d s ) ] | sup t J 0 T G 1 ( t , s ) [ N | x ( s ) y ( s ) | 0 s k ( s , r ) d r + L | x ( s ) y ( s ) | 0 T h ( s , r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) L k ( τ k σ k ) | x ( t ) y ( t ) | + G 2 ( t , t k ) L k ( δ k ε k ) | x ( t ) y ( t ) | ] x y P C 1 [ sup t J 0 T G 1 ( t , s ) [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) L k ( τ k σ k ) + G 2 ( t , t k ) L k ( δ k ε k ) ] ] ψ x y P C 1 .

Similarly,

( F x ) ( F y ) P C = sup t J | 0 T G 2 ( t , s ) [ N 0 s k ( s , r ) [ x ( r ) y ( r ) ] d r L 0 T h ( s , r ) [ x ( r ) y ( r ) ] d r ] d s + k = 1 m [ G 2 ( t , t k ) ( L k t k τ k t k σ k [ x ( s ) y ( s ) ] d s ) M G 1 ( t , t k ) ( L k t k δ k t k ε k [ x ( s ) y ( s ) ] d s ) ] | sup t J 0 T G 2 ( t , s ) [ N | x ( s ) y ( s ) | 0 s k ( s , r ) d r + L | x ( s ) y ( s ) | 0 T h ( s , r ) d r ] d s + k = 1 m [ G 2 ( t , t k ) L k ( τ k σ k ) | x ( t ) y ( t ) | + M G 1 ( t , t k ) L k ( δ k ε k ) | x ( s ) y ( s ) | ] x y P C 1 [ sup t J 0 T G 2 ( t , s ) [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m [ G 2 ( t , t k ) L k ( τ k σ k ) + M G 1 ( t , t k ) L k ( δ k ε k ) ] ] μ x y P C 1 .

Thus,

( F x ) ( F y ) P C 1 max{ψ,μ} x y P C 1 .

By the Banach fixed-point theorem, F has a unique fixed point x P C 1 (J,R), and by Lemma 2.3, x is also the unique solution of (2.4). This completes the proof. □

3 Main results

In this section, we establish existence criteria for solutions of PBVP (1.1) by the method of lower and upper solutions and the monotone iterative technique. For α 0 , β 0 E, we write α 0 β 0 if α 0 (t) β 0 (t) for all tJ. In such a case, we denote [ α 0 , β 0 ]={xE: α 0 (t)x(t) β 0 (t),tJ}.

Theorem 3.1 Suppose that the following conditions hold:

(H1) α 0 and β 0 are lower and upper solutions for PBVP (1.1), respectively, such that α 0 β 0 .

(H2) The function f satisfies

f(t, x 2 , y 2 , z 2 )f(t, x 1 , y 1 , z 1 )M( x 2 x 1 )+N( y 2 y 1 )+L( z 2 z 1 ),

for alltJ, α 0 (t) x 1 x 2 β 0 (t), (K α 0 )(t) y 1 y 2 (K β 0 )(t), (S α 0 )(t) z 1 z 2 (S β 0 )(t).

(H3) M>0, N0, L0, L k 0, L k 0are constants, and0 ε k δ k t k t k 1 , 0 σ k τ k t k t k 1 , k=1,2,,m, and they satisfy (2.2), (2.6) and (2.7).

(H4) The functions I k , I k satisfy

I k ( t k δ k t k ε k x ( s ) d s ) I k ( t k δ k t k ε k y ( s ) d s ) = L k t k δ k t k ε k x ( s ) y ( s ) d s , I k ( t k τ k t k σ k x ( s ) d s ) I k ( t k τ k t k σ k y ( s ) d s ) L k t k τ k t k σ k x ( s ) y ( s ) d s ,

where t k τ k t k σ k α 0 (s)ds t k τ k t k σ k y(s)ds t k τ k t k σ k x(s)ds t k τ k t k σ k β 0 (s)ds, 0 σ k τ k t k t k 1 , 0 ε k δ k t k t k 1 , k=1,2,,m.

Then there exist monotone sequences{ α n },{ β n }Ewhich converge in E to the extreme solutions of PBVP (1.1) in[ α 0 , β 0 ], respectively.

Proof For any η[ α 0 , β 0 ], we consider linear PBVP (2.4) with

g ( t ) = f ( t , η ( t ) , ( K η ) ( t ) , ( S η ) ( t ) ) M η ( t ) N 0 t k ( t , s ) η ( s ) d s L 0 T h ( t , s ) η ( s ) d s , γ k = I k ( t k δ k t k ε k η ( s ) d s ) L k t k δ k t k ε k η ( s ) d s , k = 1 , 2 , , m , λ k = I k ( t k τ k t k σ k η ( s ) d s ) L k t k τ k t k σ k η ( s ) d s , k = 1 , 2 , , m .

By Lemma 2.4, PBVP (2.4) has a unique solution xE. We define an operator A from [ α 0 , β 0 ] to E by x(t)=Aη(t). We complete the proof in four steps.

Step 1. We claim that α 0 A α 0 and A β 0 β 0 . We only prove α 0 A α 0 since the second inequality can be proved in a similar manner.

Let α 1 =A α 0 and p= α 0 α 1 . Then α 1 satisfies

We finish Step 1 in two cases.

Case 1. α 0 (0) α 0 (T), which implies that

a(t)=0, α 0 (t)f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) .

As α 0 is a lower solution of PBVP (1.1), then for t J ,

p ( t ) M p ( t ) N ( K p ) ( t ) L ( S p ) ( t ) = α 0 ( t ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) α 1 ( t ) + M α 1 ( t ) + N ( K α 1 ) ( t ) + L ( S α 1 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) + M α 0 ( t ) + N ( K α 0 ) ( t ) + L ( S α 0 ) ( t ) = 0 ,

and

Then by Lemma 2.1, p(t)0, which implies that α 0 (t)A α 0 (t), i.e., α 0 A α 0 .

Case 2. α 0 (0)< α 0 (T), which implies that

a ( t ) = [ α 0 ( T ) α 0 ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] .

Hence,

p ( t ) M p ( t ) N ( K p ) ( t ) L ( S p ) ( t ) [ p ( T ) p ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] = α 0 ( t ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) [ α 0 ( T ) α 0 ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] α 1 ( t ) + M α 1 ( t ) + N ( K α 1 ) ( t ) + L ( S α 1 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) + M α 0 ( t ) + N ( K α 0 ) ( t ) + L ( S α 0 ) ( t ) = 0 ,

and

Δ p ( t k ) = Δ α 0 ( t k ) Δ α 1 ( t k ) = I k ( t k δ k t k ε k α 0 ( s ) d s ) + L k [ α 0 ( T ) α 0 ( 0 ) ] T t k δ k t k ε k T 2 s d s L k t k δ k t k ε k α 1 ( s ) d s I k ( t k δ k t k ε k α 0 ( s ) d s ) + L k t k δ k t k ε k α 0 ( s ) d s = L k t k δ k t k ε k p ( s ) d s + L k [ p ( T ) p ( 0 ) ] T t k δ k t k ε k T 2 s d s , k = 1 , 2 , , m ,

and

Δ p ( t k ) = Δ α 0 ( t k ) α 1 ( t k ) I k ( t k τ k t k σ k α 0 ( s ) d s ) + L k [ α 0 ( T ) α 0 ( 0 ) ] T t k τ k t k σ k s T s 2 d s L k t k τ k t k σ k α 1 ( s ) d s I k ( t k τ k t k σ k α 0 ( s ) d s ) + L k t k τ k t k σ k α 0 ( s ) d s = L k t k τ k t k σ k p ( s ) d s + L k [ p ( T ) p ( 0 ) ] T t k τ k t k σ k s T s 2 d s ,

k=1,2,,m, and

p(0)=p(T), p (0)= α 0 (0) α 1 (0)< α 0 (T) α 1 (T)= p (T).

Then by Lemma 2.2, p(t)0, which implies α 0 (t)A α 0 (t), i.e., α 0 A α 0 .

Step 2. We prove that if α 0 η 1 η 2 β 0 , then A η 1 A η 2 .

Let η 1 =A η 1 , η 2 =A η 2 , and p= η 1 η 2 , then for t J , and by (H2), we obtain

p ( t ) M p ( t ) N ( K p ) ( t ) L ( S p ) ( t ) = f ( t , η 1 ( t ) , ( K η 1 ) ( t ) , ( S η 1 ) ( t ) ) M η 1 ( t ) N ( K η 1 ) ( t ) L ( S η 1 ) ( t ) f ( t , η 2 ( t ) , ( K η 2 ) ( t ) , ( S η 2 ) ( t ) ) + M η 2 ( t ) + N ( K η 2 ) ( t ) + L ( S η 2 ) ( t ) 0 ( by (H ) 2 ) .

From (H3), we obtain

Δ p ( t k ) = L k t k δ k t k ε k p ( s ) d s , k = 1 , 2 , , m , Δ p ( t k ) L k t k τ k t k σ k p ( s ) d s , k = 1 , 2 , , m , p ( 0 ) = p ( T ) , p ( 0 ) = p ( T ) .

Applying Lemma 2.1, we get p(t)0, which implies A η 1 A η 2 .

Step 3. We show that PBVP (1.1) has solutions.

Let α n =A α n 1 , β n =A β n 1 , n=1,2, . Following the first two steps, we have

α 0 α 1 α n β n β 1 β 0 ,nN.

Obviously, each α i , β i (i=1,2,) satisfies

and

Thus, there exist x and x such that

lim i α i (t)= x (t), lim i β i (t)= x (t),uniformly on tJ.

Clearly, x , x satisfy PBVP (1.1).

Step 4. We show that x , x are extreme solutions of PBVP (1.1).

Let x(t) be any solution of PBVP (1.1), which satisfies α 0 (t)x(t) β 0 (t), tJ. Suppose that there exists a positive integer n such that for tJ, α n (t)x(t) β n (t). Setting p(t)= α n + 1 (t)x(t), then for t J ,

p ( t ) = α n + 1 x ( t ) = M α n + 1 ( t ) + N ( K α n + 1 ) ( t ) + L ( S α n + 1 ) ( t ) + f ( t , α n ( t ) , ( K α n ) ( t ) , ( S α n ) ( t ) ) M α n ( t ) N ( K α n ) ( t ) L ( S α n ) ( t ) f ( t , x ( t ) , ( K x ) ( t ) , ( S x ) ( t ) ) = M α n + 1 ( t ) + N ( K α n + 1 ) ( t ) + L ( S α n + 1 ) ( t ) M x ( t ) N ( K x ) ( t ) L ( S x ) ( t ) + f ( t , α n ( t ) , ( K α n ) ( t ) , ( S α n ) ( t ) ) f ( t , x ( t ) , ( K x ) ( t ) , ( S x ) ( t ) ) M ( α n ( t ) x ( t ) ) N ( K ( α n ) ( t ) ( K x ) ( t ) ) L ( ( S α n ) ( t ) ( S x ) ( t ) ) M p ( t ) + N ( K p ) ( t ) + L ( S p ) ( t ) ,

and

Δp( t k )=Δ α n + 1 ( t k )Δx( t k )= L k t k δ k t k ε k p (s)ds,k=1,2,,m,

and

Δ p ( t k ) = Δ α n + 1 ( t k ) Δ x ( t k ) = L k t k τ k t k σ k α n + 1 ( s ) d s + I k ( t k τ k t k σ k α n ( s ) d s ) L k t k τ k t k σ k α n ( s ) d s I k ( t k τ k t k σ k x ( s ) d s ) L k t k τ k t k σ k α n + 1 ( s ) d s + L k t k τ k t k σ k α n ( s ) d s L k t k τ k t k σ k α n ( s ) d s L k t k τ k t k σ k x ( s ) d s = L k t k τ k t k σ k p ( s ) d s , k = 1 , 2 , , m ,

and

p(0)=p(T), p (0)= p (T).

Still by Lemma 2.1, we have for all tJ, p(t)0, i.e., α n + 1 (t)x(t). Similarly, we can prove that x(t) β n + 1 (t), tJ. Therefore, α n + 1 (t)x(t) β n + 1 (t), for all tJ, which implies x (t)x(t) x (t). The proof is complete. □

4 An example

In this section, in order to illustrate our results, we consider an example.

Example 4.1 Consider the following PBVP:

{ u ( t ) = 1 4 t 3 ( u ( t ) 2 ) + 5 18 [ 0 t t 2 s 4 u ( s ) d s ] 2 u ( t ) = + 1 8 [ 0 1 t 3 s 2 u ( s ) d s ] 2 , t J = [ 0 , 1 ] , t 1 2 , Δ u ( 1 2 ) = 1 2 1 6 3 10 u ( s ) d s , k = 1 , Δ u ( 1 2 ) = 1 3 1 10 3 10 u ( s ) d s , k = 1 , u ( 0 ) = u ( 1 ) , u ( 0 ) = u ( 1 ) .
(4.1)

Set k(t,s)= t 2 s 4 , h(t,s)= t 3 s 2 , m=1, t 1 = 1 2 , δ 1 = 1 3 , ε 1 = 1 5 , τ 1 = 2 5 , σ 1 = 1 5 , T=1. Obviously, α 0 =0, β 0 =3 are lower and upper solutions for (4.1), respectively, and α 0 β 0 .

Let

f(t, x 1 , y 1 , z 1 )= 1 4 t 3 ( x 1 2)+ 5 18 y 1 2 + 1 8 z 1 2 ,

we have

f(t, x 2 , y 2 , z 2 )f(t, x 1 , y 1 , z 1 ) 1 4 ( x 2 x 1 )+ 1 3 ( y 2 y 1 )+ 1 4 ( z 2 z 1 ),

where α(t) x 1 x 2 β(t), (Kα)(t) y 1 y 2 (Kβ)(t), (Sα)(t) z 1 z 2 (Sβ)(t), tJ. It is easy to see that

I 1 ( 1 6 3 10 x ( s ) d s ) I 1 ( 1 6 3 10 y ( s ) d s ) = 1 2 1 6 3 10 x (s) y (s)ds,

and

I 1 ( 1 10 3 10 x ( s ) d s ) I 1 ( 1 10 3 10 y ( s ) d s ) = 1 3 1 10 3 10 x(s)y(s)ds,

whenever t k τ k t k σ k α(s)ds t k τ k t k σ k y(s)ds t k τ k t k σ k x(s)ds t k τ k t k σ k β(s)ds, k=1.

Taking M= 1 4 , N= 1 3 , L= 1 4 , L 1 = 1 2 , L 1 = 1 3 , it follows that

[ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] = [ 1 2 ( 1 3 1 5 ) + 1 ] [ 1 3 ( 2 5 1 5 ) + ( 1 4 + ( 1 3 ) ( 1 ) ( 1 ) + ( 1 4 ) ( 1 ) ( 1 ) ) 1 ] = 24 25 1 ,

and

ψ = 1 + e M T 2 M ( e M T 1 ) × [ 0 T [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m L k ( τ k σ k ) ] + 1 2 k = 1 m L k ( δ k ε k ) = 1 + e 1 2 2 ( 1 2 ) ( e 1 2 1 ) [ 0 1 [ 1 3 0 s s 2 r 4 d r + 1 4 0 1 s 3 r 2 d r ] d s + 1 3 ( 2 5 1 5 ) ] + ( 1 2 ) ( 1 2 ) ( 1 3 1 5 ) 0.4246196990 < 1 , μ = 1 2 [ 0 T [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m L k ( τ k σ k ) ] + M ( 1 + e M T ) 2 ( e M T 1 ) k = 1 m L k ( δ k ε k ) = 1 2 [ 0 1 [ 1 3 0 s s 2 r 4 d r + 1 4 0 1 s 3 r 2 d r ] d s + 1 3 ( 2 5 1 5 ) ] + 1 2 ( 1 + e 1 2 ) 2 ( e 1 2 1 ) ( 1 2 ) ( 1 3 1 5 ) 0.1159664694 < 1 .

Therefore, (4.1) satisfies all the conditions of Theorem 3.1. So, PBVP (4.1) has minimal and maximal solutions in the segment [ α 0 , β 0 ].

Substituting α 0 , β 0 into monotone iterative scheme, we obtain

{ α 1 ( t ) 1 4 α 1 ( t ) 1 3 0 t t 2 s 4 α 1 ( s ) d s 1 4 0 1 t 3 s 2 α 1 ( s ) d s = 1 2 t 3 , J = [ 0 , 1 ] , t 1 2 , Δ α 1 ( 1 2 ) = 1 2 1 6 3 10 α 1 ( s ) d s , k = 1 , Δ α 1 ( 1 2 ) = 1 3 1 10 3 10 α 1 ( s ) d s , k = 1 , α 1 ( 0 ) = α 1 ( T ) , α 1 ( 0 ) = α 1 ( T ) ,
(4.2)

and

{ β 1 ( t ) 1 4 β 1 ( t ) 1 3 0 t t 2 s 4 β 1 ( s ) d s 1 4 0 1 t 3 s 2 β 1 ( s ) d s = 1 10 t 14 1 5 t 7 + 1 8 t 6 3 4 , J = [ 0 , 1 ] , t 1 2 , Δ β 1 ( 1 2 ) = 1 2 1 6 3 10 β 1 ( s ) d s , k = 1 , Δ β 1 ( 1 2 ) = 1 3 1 10 3 10 β 1 ( s ) d s , k = 1 , β 1 ( 0 ) = β 1 ( T ) , β 1 ( 0 ) = β 1 ( T ) .
(4.3)

After using the variational iteration method [45] for (4.2), (4.3), the approximate solutions for α 1 and β 1 can be illustrated as Figure 1 and Figure 2, respectively.

Figure 1
figure 1

Time history of α 1 .

Figure 2
figure 2

Time history of β 1 .

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Acknowledgements

This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.

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Thaiprayoon, C., Samana, D. & Tariboon, J. Periodic boundary value problems for second-order impulsive integro-differential equations with integral jump conditions. Bound Value Probl 2012, 122 (2012). https://doi.org/10.1186/1687-2770-2012-122

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Keywords

  • impulsive integro-differential equation
  • lower and upper solutions
  • periodic boundary value problem
  • monotone iterative technique