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 1 2 3 4 5 . 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 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 . More recent works on existence results of impulsive problems with integral boundary conditions can be found in 30 31 32 33 34 35 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, 36 37 38 39 40 41 . 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 Δ x ( 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 − ) ) , 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 ) ,

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 ) , Δ x ( 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 ,

( K x ) ( t ) = ∫ 0 t k ( t , s ) x ( s ) d s , ( S x ) ( t ) = ∫ 0 T h ( t , s ) x ( s ) d s ,

k ( t , s ) ∈ C ( D , R + ) , h ( t , s ) ∈ C ( J × J , R + ) , D = { ( t , s ) ∈ R 2 , 0 ≤ s ≤ t ≤ T } , 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

Δ x ( 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 ) ,

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

Δ x ( 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 ) ,

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.

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 P C ( J , R ) = { x : J → R ; x ( t )  is continuous everywhere except 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 } , and P C 1 ( J , R ) = { x ∈ P C ( J , R ) ; x ′ ( t )  is continuous everywhere   except 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 } . P C ( J , R ) and P C 1 ( J , R ) are Banach spaces with the norms ∥ x ∥ P C = sup { x ( t ) : t ∈ J } 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 x ∈ E 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 , N ≥ 0 , L ≥ 0 , 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 x ∈ E 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 ) ,

where M > 0 , N ≥ 0 , L ≥ 0 , L k ≥ 0 , L k ∗ ≥ 0 are constants and 0 ≤ 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 .

Then x ( t ) ≤ 0 , t ∈ J .

Proof Suppose, to the contrary, that x ( t ) > 0 for some t ∈ J . 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 t ∈ J .

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 ) d s ≥ 0 , then x ′ ( t ) is nondecreasing in t ∈ J 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 t ∈ J . Thus, 0 = x ″ ( t ) ≥ M x ( 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 t ∈ J , then Δ x ( t k ) = L k ∫ t k − δ k t k − ε k x ′ ( s ) d s ≥ 0 , 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 ¯ ] .

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 x ∈ E satisfies

where M > 0 , N ≥ 0 , L ≥ 0 , L k ≥ 0 , L k ∗ ≥ 0 are constants and 0 ≤ 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). Then x ( t ) ≤ 0 for all t ∈ J .

Proof Let u ( t ) = [ t T − t 2 T ] [ x ′ ( T ) − x ′ ( 0 ) ] , t ∈ J , 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 t ∈ J . If we prove that w ≤ 0 , 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 , t ∈ J ,

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 t ∈ J , which implies that x ( t ) ≤ 0 , t ∈ J . □

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 ) ,

where constants M > 0 , N ≥ 0 , L ≥ 0 , L k ≥ 0 , L k ∗ ≥ 0 , γ k , λ k are constants and g ∈ P C ( 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 2.3 x ∈ E is a solution of (2.4) if and only if x ∈ P 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 ) ] ,

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 Let M > 0 , N ≥ 0 , L ≥ 0 , L k ≥ 0 , L k ∗ ≥ 0 are constants and 0 ≤ ε k ≤ δ k ≤ t k − t k − 1 , 0 ≤ σ k ≤ τ k ≤ t k − t k − 1 , k = 1 , 2 , … , m . If

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

Proof For any x ∈ E , 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 F x ∈ P 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 , y ∈ P 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 .