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Solutions for nth-order boundary value problems of impulsive singular nonlinear integro-differential equations in Banach spaces

Yanlai Chen1*, Tingqiu Cao1 and Baoxia Qin2

Author Affiliations

1 School of Economics, Shandong University, Jinan, Shandong, 250100, People’s Republic of China

2 Department of Mathematics, Qilu Normal University, Jinan, Shandong, 250013, People’s Republic of China

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Boundary Value Problems 2014, 2014:128  doi:10.1186/1687-2770-2014-128

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/128


Received:2 January 2014
Accepted:6 May 2014
Published:20 May 2014

© 2014 Chen et al.; licensee Springer.

This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Not considering the Green’s function, the present study starts to construct a cone formed by a nonlinear term in Banach spaces, and through the cone creates a convex closed set. We obtain the existence of solutions for the boundary values problems of nth-order impulsive singular nonlinear integro-differential equations in Banach spaces by applying the Mönch fixed point theorem. An example is given to illustrate the main results.

MSC: 45J05, 34G20, 47H10.

Keywords:
impulsive singular integro-differential equation; Banach spaces; boundary value problem; Mönch fixed point theorem; measure of noncompactness

1 Introduction and preliminaries

By using the Schauder fixed point theorem, Guo [1] obtained the existence of solutions of initial value problems for nth-order nonlinear impulsive integro-differential equations of mixed type on an infinite interval with infinite number of impulsive times in a Banach space. In [2], by using the fixed point theorem in a cone, Chen and Qin investigated the existence of multiple solutions for a class of boundary value problems of singular nonlinear integro-differential equations of mixed type in Banach spaces. For singular differential equations in Banach spaces please see [3-9]. Generally based on Green’s function to construct a cone, but using the cone to study different nonlinear terms, we encountered difficulties, especially in infinite dimensional Banach spaces. In this paper, informed by the characteristics of the nonlinear term we construct a new cone, and through this cone create a convex closed set. On the new convex closed set, we apply the Mönch fixed point theorem to investigate the existence of solutions for the boundary value problems of nth-order impulsive singular nonlinear integro-differential equations in Banach spaces. Finally, an example of scalar second-order impulsive integro-differential equations for an infinite system is offered. Because of difficulties of compactness arising from impulsiveness and the use of nth-order integro-differential equations, a space P C n 1 [ J , E ] is introduced. Let E be a real Banach space and J : = [ 0 , 1 ] . Let P C [ J , E ] := { u | u : J E u ( t ) continuous at t t k , left continuous at t = t k , and u ( t k + ) exists, k = 1 , 2 , , m }. Obviously P C [ J , E ] is a Banach space with norm

u : = sup t J u ( t ) .

Let P C n 1 [ J , E ] := { u P C [ J , E ] | u ( n 1 ) ( t ) exists and let it be continuous at t t k , let u ( n 1 ) ( t k + ) and u ( n 1 ) ( t k ) exist, k = 1 , 2 , , m }, where u ( n 1 ) ( t k + ) and u ( n 1 ) ( t k ) represent the right and the left limits of u ( n 1 ) ( t ) at t = t k , respectively. For u P C n 1 [ J , E ] , we have

u ( n 2 ) ( t k ϵ ) = u ( n 2 ) ( t ) + t t k ϵ u ( n 1 ) ( s ) d s , t k 1 < t < t k ϵ < t k ( ϵ > 0 ) , k = 1 , 2 , , m .

So observing the existence of u ( n 1 ) ( t k ) and taking limits as ϵ 0 + in the above equality, we see that u ( n 2 ) ( t k ) exists and

u ( n 2 ) ( t k ) = u ( n 2 ) ( t ) + t t k u ( n 1 ) ( s ) d s , t k 1 < t < t , k = 1 , 2 , , m .

Similarly, we can show that u ( n 2 ) ( t k + ) exists. In the same way, we get the existence of u ( n 3 ) ( t k ) , u ( n 3 ) ( t k + ) , , u ( t k ) , u ( t k + ) . Define u ( i ) ( t k ) = u ( i ) ( t k ) ( i = 1 , 2 , , n 1 , k = 1 , 2 , , m ). Then u ( i ) P C [ J , E ] ( i = 1 , 2 , , n 1 ), and, as is natural, in the following, u ( i ) ( t k ) is understood as u ( i ) ( t k ) . It is easy to see that P C n 1 [ J , E ] is a Banach space with norm

u P C n 1 : = max i = 0 , 1 , , n 1 { sup t J u ( i ) ( t ) } .

Let P be a cone in E which defines a partial ordering in E by x y if and only if y x P . P is said to be normal if there exists a positive constant N such that θ x y implies x N y , where the smallest N is called the normal constant of P. For convenience, let N = 1 . Let P 1 = { u P : u u 0 u } , in which u 0 P and 0 < u 0 < 1 . For r > 0 , we write P 1 r = { u P 1 : u < r } . We consider the following singular boundary value problem (SBVP for short) for an nth-order impulsive nonlinear integro-differential equation in E:

{ u ( n ) ( t ) = f ( t , u ( t ) , u ( t ) , , u ( n 1 ) ( t ) , ( T u ) ( t ) , ( S u ) ( t ) ) , 0 < t < 1 , t t k ( k = 1 , 2 , , m ) , u ( i ) | t = t k = I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) ( i = 0 , 1 , , n 2 ; k = 1 , 2 , , m ) , u ( n 1 ) | t = t k = I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) ( k = 1 , 2 , , m ) , u ( i ) ( 0 ) = θ ( i = 0 , 1 , , n 2 ) , u ( n 1 ) ( 1 ) = θ , (1)

where 0 < t 1 < t 2 < < t m < 1 ,

f C [ ( 0 , 1 ) × P 1 { θ } × P 1 { θ } × × P 1 { θ } × P 1 × P 1 n + 2 , P 1 ] ,

I i k C [ P 1 × P 1 × × P 1 n , P 1 ] ( i = 0 , 1 , , n 1 ; k = 1 , 2 , , m ), and

( T u ) ( t ) = 0 t k ( t , s ) u ( s ) d s , ( S u ) ( t ) = 0 1 h ( t , s ) u ( s ) d s , t J , (2)

with k C [ D , R + ] ( D = { ( t , s ) J × J : t s } ), h C [ J × J , R + ] . u ( i ) | t = t k denotes the jump of u ( i ) ( t ) at t = t k , i.e.,

u ( i ) | t = t k = u ( i ) ( t k + ) u ( i ) ( t k ) ,

and θ denotes the zero element of E.

f ( t , v 0 , v 1 , , v n , v n + 1 ) is singular at v i = θ ( i = 0 , 1 , , n 1 ), t = 0 and/or t = 1 if

lim v i θ f ( t , v 0 , , v n + 1 ) = + ( i = 0 , 1 , , n 1 ) ,

t ( 0 , 1 ) , v k P 1 ( k = n , n + 1 ), v j P 1 { θ } ( i , j = 1 , , n 1 ), and

lim t 0 + f ( t , v 0 , , v n + 1 ) = + and/or lim t 1 f ( t , v 0 , , v n + 1 ) = + ,

v i P 1 { θ } ( i = 0 , 1 , , n 1 ), v j P 1 ( j = n , n + 1 ).

Remark Obviously, P 1 P , and P 1 is a normal cone of E if P is a normal cone of E. P 1 and P has the same normal constant N.

In the following, we assume that P is a normal cone. Let J = J { t 1 , t 2 , , t m } . A map u P C n 1 [ J , E ] C n [ J , E ] is called a solution of SBVP (1) if it satisfies (1).

2 Several lemmas

To continue, let us formulate some lemmas.

Lemma 2.1If H P C n 1 [ J , E ] is bounded and the elements of H ( n 1 ) are equicontinuous on each J k ( k = 0 , 1 , , m ), then

α n 1 ( H ) = max i = 0 , 1 , , n 1 { sup t J α ( H ( i ) ( t ) ) } ,

in whichαdenotes the Kuratowski measure of noncompactness, H ( i ) ( t ) = { x ( i ) ( t ) : x H } ( i = 0 , 1 , , n 1 ).

Proof For i = 0 , 1 , , n 1 , it is easy to prove that

sup t J α ( H ( i ) ( t ) ) α ( H ( i ) ( J ) ) α ( H ( i ) ) .

Since u ( i ) u P C n 1 ( i = 0 , 1 , , n 1 ), we know α ( H ( i ) ) α n 1 ( H ) ( i = 0 , 1 , , n 1 ). Hence

max i = 0 , 1 , , n 1 { sup t J α ( H ( i ) ( t ) ) } α n 1 ( H ) . (3)

Next, we check that

max i = 0 , 1 , , n 1 { sup t J α ( H ( i ) ( t ) ) } α n 1 ( H ) .

In fact, for any ϵ > 0 , there is a division H ( i ) = l = 1 p H l ( i ) ( i = 0 , 1 , , n 1 ) such that

diam ( H l ( i ) ) < α ( H ( i ) ) + ϵ , i = 0 , 1 , , n 1 . (4)

By hypothesis, the elements of H ( n 1 ) are equicontinuous on each J k and there is a division:

0 = t 0 < t 1 < < t j 1 = t 1 < t j 1 + 1 < < t j 2 = t 2 < < t j m = t m < t j m + 1 < < t j m + 1 = 1 ,

such that

u ( i ) ( t ) u ( i ) ( t 1 ) < ϵ , u H , t [ t 0 , t 1 ] ( i = 0 , 1 , , n 1 ) (5)

and

u ( i ) ( t ) u ( i ) ( t j ) < ϵ , u H , t ( t j 1 , t j ] ( j = 2 , , j m + 1 , i = 0 , 1 , , n 1 ) . (6)

Let J 1 = [ 0 , t 1 ] , J j = ( t j 1 , t j ] ( j = 2 , , j m + 1 ). By virtue of (5) and (6), we know that

u ( i ) ( t ) u ( i ) ( t j ) < ϵ , u H , t J j ( j = 1 , 2 , , j m + 1 , i = 0 , 1 , , n 1 ) . (7)

Let B : = i = 0 n 1 j = 1 j m + 1 H ( i ) ( t j ) . There is a division B = l = 1 p B l such that

diam B l < α ( B ) + ϵ ( l = 1 , , p ) . (8)

Let F be the finite set of all maps { 0 , 1 , , n 1 } × { 1 , 2 , , j m + 1 } into { 1 , 2 , , p } ( μ : ( i , j ) μ ( i , j ) ). For μ F , let H μ : = { u H : u ( i ) ( t j ) B μ ( i , j ) , ( i , j ) { 0 , 1 , , n 1 } × { 1 , 2 , , j m + 1 } } . It is clear that H = μ F H μ . For any u , v H μ , t J , we have t J j for some j { 1 , 2 , , j m + 1 } , and so

u ( i ) ( t ) v ( i ) ( t ) u ( i ) ( t ) u ( i ) ( t j ) + u ( i ) ( t j ) v ( i ) ( t j ) + v ( i ) ( t j ) v ( i ) ( t ) < α ( B ) + 3 ϵ ( i = 0 , 1 , , n 1 ) . (9)

Consequently,

diam H μ α ( B ) + 3 ϵ , μ F ,

which implies α n 1 ( H ) α ( B ) + 3 ϵ . Since ϵ > 0 is arbitrary, we get

α n 1 ( H ) α ( B ) = max { α ( H ( i ) ( t j ) ) : j = 1 , 2 , , j m + 1 , i = 0 , 1 , , n 1 } max i = 0 , 1 , , n 1 { sup t J α ( H ( i ) ( t ) ) } . (10)

Finally, the conclusion follows from (3) and (10). For details of the Kuratowski measure of noncompactness, please see [10]. □

Lemma 2.2 (see [11])

Let us take a countable set D = { u n } L [ J , E ] ( n N ). For all u n D , there is g L [ J , R + ] such that u n ( t ) g ( t ) , a.e, t J . Then α ( D ( t ) ) L [ J , R + ] , and

α ( { 0 t u n ( s ) d s : n N } ) 2 0 t α ( D ( s ) ) d s .

Lemma 2.3Suppose H P C [ J , E ] is bounded and equicontinuous on each J k ( k = 0 , 1 , , m ). Then α ( H ( t ) ) P C [ J , R + ] , and

α ( { J u ( t ) d t : u H } ) J α ( H ( t ) ) d t .

Proof By Theorem 1.2.2 of [10], the conclusion is obvious. □

Lemma 2.4Let B 1 , B 2 P C n 1 [ J , E ] be two countable sets. Suppose u 0 P C n 1 [ J , E ] and B 1 ¯ = co ¯ ( { u 0 } B 2 ) . Then

B 1 ( i ) ( t ) ¯ = co ¯ ( { u 0 ( i ) ( t ) } B 2 ( i ) ( t ) ) , t J ( i = 0 , 1 , , n 1 ) .

Proof The conclusion is obvious by Lemma 6 of [12]. □

Lemma 2.5 (see [13]) (The Mönch fixed point theorem)

LetEbe a Banach space. Assume that D E is close and convex. Assume also that A : D D is continuous with the further property that for some u 0 D , we have C D countable, C ¯ = co ¯ ( { u 0 } A ( C ) ) implies thatCis relatively compact. ThenAhas a fixed-point inD.

3 Main theorem and example

For convenience, we list the following conditions:

(H1) There exist b C [ J , R + ] , a i C [ J , R + ] ( i = 0 , 1 , , n + 1 ), g i C [ ( 0 , + ) , ( 0 , + ) ] ( i = 0 , 1 , , n 1 ) and h i C [ [ 0 , + ) , [ 0 , + ) ] ( i = 0 , 1 , , n + 1 ) such that

f ( t , v 0 , v 1 , , v n 1 , v n , v n + 1 ) b ( t ) + i = 0 n 1 a i ( t ) ( g i ( v i ) + h i ( v i ) ) + a n ( t ) h n ( v n ) + a n + 1 ( t ) h n + 1 ( v n + 1 ) , t ( 0 , 1 ) , v i P 1 r { θ } ( i = 0 , 1 , , n 1 ) , v n , v n + 1 P 1 r ,

where g i is nonincreasing, h i g i ( i = 0 , 1 , , n 1 ) and h n , h n + 1 are nondecreasing. And there exist d i k 0 , c i k j 0 ( i , j = 0 , 1 , , n 1 , k = 1 , 2 , , m ) such that

I i k ( v 0 , v 1 , , v n 1 ) d i k + j = 0 n 1 c i k j v j ( i = 0 , 1 , , n 1 , k = 1 , 2 , , m ) ,

v j P 1 r ( j = 0 , 1 , , n 1 ).

(H2) There exists a φ P 1 ( P 1 denotes the dual cone of P 1 ) such that φ = 1 . And for any r > 0 , there exists a h r ( t ) L [ ( 0 , 1 ) , ( 0 , + ) ] such that

φ ( f ( t , v 0 , v 1 , , v n 1 , v n , v n + 1 ) ) h r ( t ) , t ( 0 , 1 ) , v i P 1 r { θ } ( i = 0 , 1 , , n 1 ) , v n , v n + 1 P 1 r .

(H3) There exists a R 0 > 0 1 h R 0 ( s ) d s such that

0 1 ( b ( s ) + i = 0 n 2 a i ( s ) g i ( φ ( u 0 ) s n 1 i ( n 1 i ) ! 0 1 τ h R 0 ( τ ) d τ ) ( 1 + h i ( R 0 ) g i ( R 0 ) ) + a n 1 ( s ) × g n 1 ( φ ( u 0 ) s 1 h R 0 ( τ ) d τ ) × ( 1 + h n 1 ( R 0 ) g n 1 ( R 0 ) ) + a n ( s ) h n ( k R 0 ) + a n + 1 ( s ) h n + 1 ( h R 0 ) ) d s + k = 1 m i = 0 n 2 ( 1 t k ) i i ! ( d i k + j = 0 n 1 c i k j R 0 ) + k = 1 m ( d n 1 k + j = 0 n 1 c n 1 k j R 0 ) R 0 ,

where b, a i ( i = 0 , 1 , , n + 1 ), g i ( i = 0 , 1 , , n 1 ), φ, h i ( i = 0 , 1 , , n + 1 ), d i k ( i = 0 , 1 , , n 1 ), c i k j ( i , j = 0 , 1 , , n 1 , k = 0 , 1 , , m ) and h R 0 are defined as in conditions (H1) and (H2), and k : = max ( t , s ) D { k ( t , s ) } , h : = max ( t , s ) J × J { h ( t , s ) } .

(H4) There exist L i ( t ) L [ ( 0 , 1 ) , R + ] ( i = 0 , 1 , , n + 1 ), b > a > 0 such that

α ( f ( t , B 0 , B 1 , , B n + 1 ) ) i = 0 n + 1 L i ( t ) α ( B i ) , t ( 0 , 1 ) ,

B i P 1 b ¯ P 1 a ( i = 0 , 1 , , n 1 ), B n , B n + 1 P 1 b ¯ . There exist M i k j 0 ( i , j = 0 , 1 , , n 1 , k = 1 , 2 , , m ) such that

α ( I i k ( B 0 , B 1 , , B n 1 ) ) j = 0 n 1 M i k j α ( B j ) , B j P 1 b ¯ ( j = 0 , 1 , , n 1 ) ( i = 0 , 1 , , n 1 , k = 1 , 2 , , m ) .

Remark Obviously, condition (H4) is satisfied automatically when E is finite dimensional.

Lemma 3.1Suppose conditions (H1), (H2) and (H3) are satisfied. ThenQdefined by

Q = : { u P C n 1 [ J , P ] : u ( i ) ( t ) u 0 u ( i ) ( t ) ( i = 0 , 1 , , n 1 ) , u ( i ) | t = t k θ ( i = 0 , 1 , , n 2 ) , φ ( u ( n 1 ) ( t ) ) φ ( u 0 ) t 1 h R 0 ( s ) d s , u P C n 1 R 0 , t J }

is a nonempty, convex and closed subset of P C n 1 [ J , E ] .

Proof Let

u ˜ ( t ) = u 0 ( 1 ( n 1 ) ! 0 t ( t s ) n 1 h R 0 ( s ) d s + t n 1 ( n 1 ) ! 0 1 h R 0 ( s ) d s + 0 < t k < t i = 0 n 2 ( t t k ) i i ! + m t n 1 ( n 1 ) ! 0 < t k < t ( t t k ) n 1 ( n 1 ) ! ) , t J .

For j = 0 , 1 , , n 1 ,

u ˜ ( j ) ( t ) = u 0 ( 1 ( n 1 j ) ! 0 t ( t s ) n 1 j h R 0 ( s ) d s + t n 1 j ( n 1 j ) ! 0 1 h R 0 ( s ) d s + 0 < t k < t i = j n 2 ( t t k ) i j ( i j ) ! + m t n 1 j ( n 1 j ) ! 0 < t k < t ( t t k ) n 1 j ( n 1 j ) ! ) , t J . (11)

It is clear that u ˜ ( t ) P C n 1 [ J , P ] . Since 0 < u 0 < 1 , for j = 0 , 1 , , n 1 , by (11), one can see that

u ˜ ( j ) ( t ) = u 0 | 1 ( n 1 j ) ! 0 t ( t s ) n 1 j h R 0 ( s ) d s + t n 1 j ( n 1 j ) ! 0 1 h R 0 ( s ) d s + 0 < t k < t i = j n 2 ( t t k ) i j ( i j ) ! + m t n 1 j ( n 1 j ) ! 0 < t k < t ( t t k ) n 1 j ( n 1 j ) ! | ( 1 ( n 1 j ) ! 0 t ( t s ) n 1 j h R 0 ( s ) d s + t n 1 j ( n 1 j ) ! 0 1 h R 0 ( s ) d s + 0 < t k < t i = j n 2 ( t t k ) i j ( i j ) ! + m t n 1 j ( n 1 j ) ! 0 < t k < t ( t t k ) n 1 j ( n 1 j ) ! ) ,

which implies u ˜ ( i ) ( t ) u 0 u ˜ ( i ) ( t ) ( i = 0 , 1 , , n 1 ) for t J .

By conditions (H1), (H2), (H3), and (11), we have

u ˜ ( i ) | t = t k θ ( i = 0 , 1 , , n 2 ) , u ˜ P C n 1 R 0

and φ ( u ˜ n 1 ( t ) ) φ ( u 0 ) t 1 h R 0 ( s ) d s . Therefore, u ˜ Q and Q is a nonempty set.

Now, we check that Q is a convex subset of P C n 1 [ J , E ] . In fact, for any u , v Q , 0 λ 1 , we write v ˜ = λ u + ( 1 λ ) v , which means v ˜ P C n 1 [ J , P ] . It is clear that

v ˜ ( i ) | t = t k = λ u ( i ) | t = t k + ( 1 λ ) v ( i ) | t = t k λ θ + ( 1 λ ) θ = θ ( i = 0 , 1 , , n 2 ) . (12)

By virtue of the characters of elements of Q and the characters of φ, we have

v ˜ ( i ) ( t ) = λ u ( i ) ( t ) + ( 1 λ ) v ( i ) ( t ) λ u 0 u ( i ) ( t ) + ( 1 λ ) u 0 v ( i ) ( t ) u 0 ( λ u ( i ) ( t ) + ( 1 λ ) v ( i ) ( t ) ) = u 0 v ˜ ( i ) ( t ) ( i = 0 , 1 , , n 1 ) . (13)

In the same way,

φ ( v ˜ ( n 1 ) ) = φ ( λ u ( n 1 ) + ( 1 λ ) v ( n 1 ) ) λ φ ( u 0 ) t 1 h R 0 ( s ) d s + ( 1 λ ) φ ( u 0 ) t 1 h R 0 ( s ) d s = φ ( u 0 ) t 1 h R 0 ( s ) d s (14)

and

v ˜ P C n 1 = λ u + ( 1 λ ) v P C n 1 λ R 0 + ( 1 λ ) R 0 = R 0 .

Therefore, v ˜ Q . Thus, Q is a convex subset of P C n 1 [ J , E ] . It is clear that Q is a closed subset of P C n 1 [ J , E ] . So the conclusion holds. □

Lemma 3.2Assume that conditions (H1), (H2) and (H3) are satisfied. Then A : Q Q , where the operatorAis defined by

( A u ) ( t ) = 1 ( n 1 ) ! 0 t ( t s ) n 1 f ( s , u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + t n 1 ( n 1 ) ! 0 1 f ( s , u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t i = 0 n 2 ( t t k ) i i ! I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) + t n 1 ( n 1 ) ! k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t ( t t k ) n 1 ( n 1 ) ! I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) , t J . (15)

Proof For any u Q , i.e.,

u ( i ) ( t ) u 0 u ( i ) ( t ) ( i = 0 , 1 , , n 1 ) (16)

and

φ ( u ( n 1 ) ( t ) ) φ ( u 0 ) t 1 h R 0 ( s ) d s , t J , u ( i ) | t = t k θ ( i = 0 , 1 , , n 2 ) , u P C n 1 R 0 .

For any u Q and t ( fixed ) J ,

( T u ) ( t ) = 0 t k ( t , s ) u ( s ) d s 0 t k ( t , s ) u 0 u ( s ) d s u 0 0 t k ( t , s ) u ( s ) d s = u 0 ( T u ) ( t ) (17)

and

( S u ) ( t ) = 0 t h ( t , s ) u ( s ) d s 0 1 h ( t , s ) u 0 u ( s ) d s u 0 0 1 h ( t , s ) u ( s ) d s = u 0 ( S u ) ( t ) . (18)

Because of φ P , φ = 1 and φ ( u ( n 1 ) ( t ) ) φ ( u 0 ) t 1 h R 0 ( s ) d s , t ( 0 , 1 ) , we know

u ( n 1 ) ( t ) = u ( n 1 ) ( t ) φ φ ( u ( n 1 ) ( t ) ) φ ( u 0 ) t 1 h R 0 ( s ) d s . (19)

Analogously, for i = 0 , 1 , , n 2 , it is easy to see

φ ( u ( j ) ( t ) ) = φ ( 0 < t k < t i = j n 1 ( t t k ) i j ( i j ) ! u ( i ) | t = t k + 1 ( n 2 j ) ! 0 t ( t s ) n 2 j u ( n 1 ) ( s ) d s ) 1 ( n 2 j ) ! 0 t ( t s ) n 2 j φ ( u ( n 1 ) ( s ) ) d s 1 ( n 2 j ) ! 0 t ( t s ) n 2 j φ ( u 0 ) s 1 h R 0 ( τ ) d τ d s t n 1 j ( n 1 j ) ! φ ( u 0 ) 0 1 ( 1 ( 1 s ) n 1 j ) h R 0 ( s ) d s t n 1 j ( n 1 j ) ! φ ( u 0 ) 0 1 s h R 0 ( s ) d s , t ( 0 , 1 ) . (20)

Hence,

u ( j ) ( t ) t n 1 j ( n 1 j ) ! φ ( u 0 ) 0 1 s h R 0 ( s ) d s , j = 0 , 1 , , n 2 , t ( 0 , 1 ) . (21)

Differentiating (15) n 1 times, we get

( A u ) ( n 1 ) ( t ) = t 1 f ( s , u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) , t J . (22)

Obviously, ( A u ) ( n 1 ) ( t i + ) ( i = 1 , 2 , , m ) exist and

( A u ) ( n 1 ) ( t i + ) = t i 1 f ( s , u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = i + 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) , t J ( i = 1 , 2 , , m ) , (23)

where k = i + 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) is understood as θ for i = m . Similarly, ( A u ) ( n 1 ) ( t i ) ( i = 1 , 2 , , m ) exist. Hence,

A u P C n 1 [ J , P ] . (24)

Let

G l ( t , s ) = : { t n 1 l ( t s ) n 1 l ( n 1 l ) ! , 0 s t 1 ; t n 1 l ( n 1 l ) ! , 0 t < s 1 ( l = 0 , 1 , , n 1 ) . (25)

Since

f C [ ( 0 , 1 ) × P 1 { θ } × P 1 { θ } × × P 1 { θ } × P 1 × P 1 n + 2 , P 1 ]

and I i k C [ P 1 × P 1 × × P 1 n , P 1 ] ( i = 0 , 1 , , n 1 ; k = 1 , 2 , , m ), it follows from (15), (16), (17) and (18) that

( A u ) ( l ) ( t ) = 0 1 G l ( t , s ) f ( u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t i = l n 2 ( t t k ) i l ( i l ) ! I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) + t n 1 l ( n 1 l ) ! k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t ( t t k ) n 1 l ( n 1 l ) ! I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 1 G l ( t , s ) u 0 f ( u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t i = l n 2 ( t t k ) i l ( i l ) ! u 0 I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) + ( t n 1 l ( n 1 l ) ! k = 1 m 1 k 0 < t k < t ( t t k ) n 1 l ( n 1 l ) ! ) × u 0 I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) u 0 0 1 G l ( t , s ) f ( u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t i = l n 2 ( t t k ) i l ( i l ) ! I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) + t n 1 l ( n 1 l ) ! k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t ( t t k ) n 1 l ( n 1 l ) ! I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) = u 0 ( A u ) ( l ) ( t ) , l = 0 , 1 , , n 1 , t J . (26)

It is clear that

( A u ) ( l ) | t = t k θ , l = 0 , 1 , , n 2 . (27)

Since φ ( u 0 ) φ u 0 1 , by (22), (26) and condition (H2), we have

φ ( ( A u ) ( n 1 ) ( t ) ) = φ ( t 1 f ( s , u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) ) t 1 h R 0 ( s ) d s φ ( u 0 ) t 1 h R 0 ( s ) d s , t J . (28)

Now, we show that

A u P C n 1 R 0 , u Q . (29)

By (15), (19), (21), conditions (H1) and (H3) imply

( A u ) ( l ) ( t ) = 0 1 G l ( t , s ) f ( u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t i = l n 2 ( t t k ) i l ( i l ) ! I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) + t n 1 l ( n 1 l ) ! k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t ( t t k ) n 1 l ( n 1 l ) ! I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 1 G l ( t , s ) f ( u ( s ) , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s + 0 < t k < t i = l n 2 ( t t k ) i l ( i l ) ! I i k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) + t n 1 l ( n 1 l ) ! k = 1 m I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 < t k < t ( t t k ) n 1 l ( n 1 l ) ! I n 1 k ( u ( t k ) , u ( t k ) , , u ( n 1 ) ( t k ) ) 0 1 ( b ( s ) + i = 0 n 2 a i ( s ) g i ( φ ( u 0 ) s n 1 i ( n 1 i ) ! 0 1 τ h R 0 ( τ ) d τ ) ( 1 + h i ( R 0 ) g i ( R 0 ) ) + a n 1 ( s ) g n 1 ( φ ( u 0 ) s 1 h R 0 ( τ ) d τ ) ( 1 + h n 1 ( R 0 ) g n 1 ( R 0 ) ) + a n ( s ) h n ( k R 0 ) + a n + 1 ( s ) h n + 1 ( h R 0 ) ) d s + k = 1 m i = 0 n 2 ( 1 t k ) i i ! ( d i k + j = 0 n 1 c i k j R 0 ) + k = 1 m ( d n 1 k + j = 0 n 1 c n 1 k j R 0 ) R 0 , l = 0 , 1 , , n 1 , t J , (30)

which implies that (29) is true. By (24), (26) to (29), the conclusion holds. □

Lemma 3.3Suppose conditions (H1) to (H4) are satisfied. Let

D l ( t ) = { 0 1 G l ( t , s ) f ( s , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s : u B } ( l = 0 , 1 , , n 2 )

and

D n 1 ( t ) = { t 1 f ( s , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s : u B } ,

in which t J , B ( countable ) Q . Then

α ( D l ( t ) ) 0 1 2 s ( i = 0 n 1 L i ( s ) α ( B ( i ) ( s ) ) + L n ( s ) k α ( B ( s ) ) + L n + 1 ( s ) h α ( B ( s ) ) ) d s ( l = 0 , 1 , , n 2 ) (31)

and

α ( D n 1 ( t ) ) 0 1 2 ( i = 0 n 1 L i ( s ) α ( B ( i ) ( s ) ) + L n ( s ) k α ( B ( s ) ) + L n + 1 ( s ) h α ( B ( s ) ) ) d s , (32)

in which B ( i ) ( s ) = { u ( i ) ( s ) : u B } ( i = 0 , 1 , , n 1 ).

Proof In order to avoid the singularity, given 1 2 > δ > 0 , let

D l δ ( t ) = : { δ 1 δ G l ( t , s ) f ( s , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s : u B } ( l = 0 , 1 , , n 2 ) , 0 < δ < 1 2 , t J .

By conditions (H1), (H2) and (H3), for any t J , u B , we have

0 1 G l ( t , s ) f ( s , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s δ 1 δ G l ( t , s ) f ( s , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s 0 δ ( b ( s ) + i = 0 n 2 a i ( s ) g i ( φ ( u 0 ) s n 1 i ( n 1 i ) ! 0 1 τ h R 0 ( τ ) d τ ) ( 1 + h i ( R 0 ) g i ( R 0 ) ) + a n 1 ( s ) g n 1 ( φ ( u 0 ) s 1 h R 0 ( τ ) d τ ) × ( 1 + h n 1 ( R 0 ) g n 1 ( R 0 ) ) + a n ( s ) h n ( k R 0 ) + a n + 1 ( s ) h n + 1 ( h R 0 ) ) d s + 1 δ 1 ( b ( s ) + i = 0 n 2 a i ( s ) g i ( φ ( u 0 ) s n 1 i ( n 1 i ) ! 0 1 τ h R 0 ( τ ) d τ ) ( 1 + h i ( R 0 ) g i ( R 0 ) ) + a n 1 ( s ) g n 1 ( φ ( u 0 ) s 1 h R 0 ( τ ) d τ ) ( 1 + h n 1 ( R 0 ) g n 1 ( R 0 ) ) + a n ( s ) h n ( k R 0 ) + a n + 1 ( s ) h n + 1 ( h R 0 ) ) d s . (33)

By virtue of absolute continuity of the Lebesgue integrable function, we have

d H ( D l δ ( t ) , D l ( t ) ) 0 , as  δ 0 , t J , (34)

in which, d H ( D l δ ( t ) , D l ( t ) ) denotes the Hausdorff distance between D l δ ( t ) and D l ( t ) . Therefore,

α ( D l ( t ) ) = lim δ 0 α ( D l δ ( t ) ) , t J . (35)

Now, we show

α ( D l δ ( t ) ) 0 1 2 s ( i = 0 n 1 L i ( s ) α ( B ( i ) ( s ) ) + L n ( s ) k α ( B ( s ) ) + L n + 1 ( s ) h α ( B ( s ) ) ) d s ( l = 0 , 1 , , n 1 ) .

In fact, by Lemma 2.2, we have

α ( D l δ ( t ) ) = α ( { δ 1 δ G l ( t , s ) f ( s , u ( s ) , , u ( n 1 ) ( s ) , ( T u ) ( s ) , ( S u ) ( s ) ) d s : u B } ) δ 1 δ 2 G l ( t , s ) α ( f ( s , B ( s ) , , B ( n 1 ) ( s ) , ( T B ) ( s ) , ( S B ) ( s ) ) ) d s ( l = 0 , 1 , , n 2 ) , (36)

where ( T B ) = { ( T u ) ( t ) : u B } , ( S B ) = { ( S u ) ( t ) : u B } .

On the other hand, for u B Q , it follows from (19) and (21) that

u ( j ) ( s ) δ n 1 j ( n 1 j ) ! φ ( u 0 ) 0 1 s h R 0 ( s ) d s , j = 0 , 1 , , n 2 , s ( δ , 1 δ ) (37)

and

u ( n 1 ) ( s ) φ ( u 0 ) 1 δ 1 h R 0 ( s ) d s , s ( δ , 1 δ ) . (38)

Taking a = min { min j = 0 , 1 , , n 2 { δ n 1 j ( n 1 j ) ! φ ( u 0 ) 0 1 h R 0 ( s ) d s } , δ n 1 j ( n 1 j ) ! φ ( u 0 ) 0 1 h R 0 ( s ) d s } , b = max { k , h , 1 } R 0 , by (16), (17) and (18), one can see that

B ( i ) ( s ) P 1 b ¯ P 1 a ( i = 0 , 1 , , n 1 ) , ( T B ) ( s ) , ( S B ) ( s ) P 1 b ¯ . (39)

Therefore, by condition (H4) and (36), for l = 0 , 1 , , n 2 , it is easy to get

α ( D l δ ( t ) ) 0 1 2 G l ( t , s ) ( i = 0 n 1 L i ( s ) α ( B i ( s ) ) + L n ( s ) α ( ( T B ) ( s ) ) + L n + 1 ( s ) α ( ( S B ) ( s ) ) ) d s , t J . (40)

Since B is a bounded set of P C n 1 [ J , E ] and B ( t ) is a bounded set, B ( t ) is equicontinuous on each J k ( k = 1 , 2 , , m ). By Lemma 2.3, it is easy to get

α ( ( T B ) ( s ) ) k 0 s α ( B ( τ ) ) d τ , α ( ( S B ) ( s ) ) h 0 1 α ( B ( τ ) ) d τ . (41)

Substituting (41) into (40), we get (31).

Similarly, we obtain (32) and our conclusion holds. □

Lemma 3.4Let conditions (H1) to (H3) be satisfied. u P C n 1 [ J , E ] C n [ J , E ] is a solution of SBVP (1), if and only if u Q is a fixed point of the operatorAdefined by (15).

Proof First of all, by mathematical induction, for u P C n 1 [ J , E ] C n [ J , E ] , Taylor’s formula with the integral remainder term holds,

u ( t ) = i = 0 n 1 t i i ! u ( i ) ( 0 ) + 0 < t k < t i = 0 n 1 ( t t k ) i i ! [ u ( i ) ( t k + ) u ( i ) ( t k ) ] + 1 ( n 1 ) ! 0 t ( t s ) n 1 u ( n ) ( s ) d s . (42)

In fact, as n = 1 , for u P C [ J , E ] C 1 [ J , E ] , let t k < t t k + 1 , it is easy to see that

u ( t 1 ) u ( 0 ) = 0 t 1 u ( s ) d s , u ( t 2 ) u ( t 1 + ) = t 1 t 2 u ( s ) d s , , u ( t k ) u ( t k 1 + ) = t k 1 t k u ( s ) d s , u ( t ) u ( t k + ) = t k t u ( s ) d s .

Adding these together, we get

u ( t ) u ( 0 ) i = 1 k [ u ( t i + ) u ( t i ) ] = 0 t u ( s ) d s ,

that is,

u ( t ) = u ( 0 ) + 0 < t k < t [ u ( t k + ) u ( t k ) ] + 0 t u ( s ) d s , t J . (43)

This proves that (42) is true for n = 1 .

Suppose (42) is true for n 1 , i.e., for u P C n 2 [ J , E ] C n 1 [ J , E ] , the next formula holds:

u ( t ) = i = 0 n 2 t i i ! u ( i ) ( 0 ) + 0 < t k < t i = 0 n 2 ( t t k ) i i ! [ u ( i ) ( t k + ) u ( i ) ( t k ) ] + 1 ( n 2 ) ! 0 t ( t s ) n 2 u ( n 1 ) ( s ) d s . (44)

Now we check that (42) is also true for n. In fact, suppose u P C n 1 [ J , E ] C n [ J , E ] . Then u ( n 1 ) P C [ J , E ] C 1 [ J , E ] , by (43), one can see

u ( n 1 ) ( t ) = u ( n 1 ) ( 0 ) + 0 < t k < t [ u ( n 1 ) ( t k + ) u ( n 1 ) ( t k ) ] + 0 t u ( n ) ( s ) d s , t J . (45)

Substituting the above equation into (44), we get

u ( t ) = i = 0 n 2 t i i ! u ( i ) ( 0 ) + 0 < t k < t i = 0 n 2 ( t t k ) i i ! [ u ( i ) ( t k + ) u ( i ) ( t k ) ] + 1 ( n 2 ) ! 0 t ( t s ) n 2 × { u ( n 1 ) ( 0 ) + 0 < t k < s [ u ( n 1 ) ( t k + ) u ( n 1 ) ( t k ) ] + 0 s u ( n ) ( τ ) d τ } d s = i = 0 n 2 t i i ! u ( i ) ( 0 ) + 0 < t k < t i = 0 n 2 ( t t k ) i i ! [ u ( i ) ( t k + ) u ( i ) ( t k ) ] + ( t s ) n 1 ( n 1 ) ! | t 0 u ( n 1 ) ( 0 ) 0 < t k < t