AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 931217 10.1155/2014/931217 931217 Research Article Solvability of a Third-Order Multipoint Boundary Value Problem at Resonance Du Zengji 1 Zhao Bensheng 1 http://orcid.org/0000-0002-5131-9252 Bai Zhanbing 2 Chu Jifeng 1 School of Mathematics and Statistics Xuzhou Jiangsu Normal University Jiangsu 221116 China ujs.edu.cn 2 School of Mathematics and Systems Science Shandong University of Science and Technology Qingdao 266510 China sdust.edu.cn 2014 2322014 2014 10 12 2013 13 01 2014 23 2 2014 2014 Copyright © 2014 Zengji Du et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We discuss a third-order multipoint boundary value problem under some appropriate resonance conditions. By using the coincidence degree theory, we establish the existence result of solutions. The emphasis here is that the dimension of the linear operator is equal to two. Our results supplement other results.

1. Introduction

In this paper, we are concerned with the following third-order ordinary differential equation: (1) x ′′′ ( t ) = f ( t , x ( t ) , x ( t ) , x ′′ ( t ) ) + e ( t ) , t ( 0,1 ) , with the boundary conditions (2) x ( 0 ) = i = 1 m - 2 α i x ( ξ i ) ,    x ( 0 ) = 0 ,    x ( 1 ) = β x ( η ) , where f : [ 0,1 ] × R 3 R is a Carathéodory function, e L 1 [ 0,1 ] , 0 < ξ 1 < < ξ m - 2 < 1 , α i R , i = 1 , , m - 2 , β 0 , and η ( 0,1 ) .

In recent years, many authors have paid much attention to the existence of solutions for multipoint boundary value problems at resonance: we refer the readers to see . If the linear equation L x = x ′′′ ( t ) = 0 with boundary conditions (2) has nontrivial solutions, that is, dimKer L 1 , the BVP (1)-(2) is called a resonance problem. In , the authors all discussed the case that dimKer L = 1 . In [2, 3], the authors established the existence results for resonance boundary value problems with the case of dimKer L = 2 . However, we will show that some conditions such as Λ 0 assumed in [2, 3] are not necessary. We establish existence of some solutions for BVP (1)-(2) by using the coincidence degree theory of Mawhin  at resonance.

According to the constant β , the BVP (1)-(2) is divided into the following five resonance cases:

0 < β < 1 , ( 1 - β ) i = 1 m - 2 α i ξ i 2 = ( 1 - i = 1 m - 2 α i ) ( β η 2 - 1 ) ;

β = 1 , i = 1 m - 2 α i ξ i 2 = 0 , i = 1 m - 2 α i = 1 ;

1 < β < 1 / η 2 , ( 1 - β ) i = 1 m - 2 α i ξ i 2 = ( 1 - i = 1 m - 2 α i ) ( β η 2 - 1 ) ;

β = 1 / η 2 , i = 1 m - 2 α i ξ i 2 = 0 , i = 1 m - 2 α i = 1 ;

1 / η 2 < β , ( 1 - β ) i = 1 m - 2 α i ξ i 2 = ( 1 - i = 1 m - 2 α i ) ( β η 2 - 1 ) .

Du et al.  studied the existence results of BVP (1)-(2) under the resonance conditions ( C 2 ) and ( C 4 ), that is, dimKer L = 1 , but they did not discuss the other three cases. In this paper, under the resonance conditions, ( C 1 ), ( C 3 ), or ( C 5 ), we could imply dimKer L = 2 ; thus we supplement the results in .

The layout of this paper is as follows. In Section 2, we briefly present some notations and an abstract existence result due to Mawhin. In Section 3, we study BVP (1)-(2) under the condition ( C 1 ) and obtain some existence results. In Section 4, we give an example of the existence results in Section 3.

2. Preliminary

Now, we briefly recall some notations and an abstract existence result by Mawhin .

Let Y , Z be real Banach spaces and let L : dom L Y Z be a linear operator which is a Fredholm map of index zero and let P : Y Y , Q : Z Z be continuous projectors such that Im P = Ker L ,   Ker Q = Im L   and  Y = Ker L Ker P ,   Z = Im L Im Q . It follows that L | dom L Ker P : dom L Ker P Im L is invertible; we denote the inverse of that map by K P . Let Ω be an open bounded subset of Y such that dom L Ω ϕ ; the map N : Y Z is said to be L - compact on Ω - if the map Q N ( Ω - ) is bounded and K P ( I - Q ) N : Ω - Y is compact. For more details we refer the reader to the lecture of Mawhin .

The theorem we use in this paper is Theorem IV.13 of .

Theorem 1.

Let L be a Fredholm map of index zero and let N be L - compact on Ω - . Assume that the following conditions are satisfied.

L x N x for every ( x , λ ) [ ( dom L Ker    L ) Ω ] × ( 0,1 ) .

N x Im L for every x    Ker L Ω .

deg    ( Q N    Ker L    , Ω    Ker L , 0 ) 0 , where Q : Z Z is a continuous projector as above with Im L = Ker Q .

Then the abstract equation L x = N x has at least one solution in dom L Ω - .

In the following, we will use the classical spaces C [ 0,1 ] , C 1 [ 0,1 ] , C 2 [ 0,1 ] , and L 1 [ 0,1 ] . For x C 2 [ 0,1 ] , we use the norms x = max t [ 0,1 ] | x ( t ) | and x = max { x , x , x ′′ } , denote the norm in L 1 [ 0,1 ] by · 1 , and define the Sobolev space W 3,1 ( 0,1 ) as (3) W 3,1 ( 0,1 ) = { x : [ 0,1 ] R x , x , x ′′       are    absolutely    continuous    on    [ 0,1 ] with       x ′′′ L 1 [ 0,1 ] } .

Let Y = C 2 [ 0,1 ] ,   Z = L 1 [ 0,1 ] , and define the linear operator L : dom L Y Z as L x = x ′′′ , x dom L , where (4) dom L = { x W 3,1 ( 0,1 ) :    x satisfies    boundary    conditions    ( 2 ) ( 2 ) x W 3,1 ( 0,1 ) :    } . We define N : Y Z as (5) N x = f ( t , x ( t ) , x ( t ) , x ′′ ( t ) ) , t ( 0,1 ) . Then BVP (1)-(2) can be written as L x = N x .

3. Existence Results Lemma 2.

If the condition ( C 1 ) holds, then there exist p { 1,2 , , m - 2 } , q N + , q p + 1 , such that (6) Λ ( p , q ) = i = 1 m - 2 α i ξ i q + 2 ( 1 - β η p + 2 ) - i = 1 m - 2 α i ξ i p + 2 ( 1 - β η q + 2 ) 0 , where N + = { 1,2 , } .

Proof.

We prove that, for any l N , there exists k l { l ( m - 2 ) + 1 , , ( l + 1 ) ( m - 2 ) } , such that i = 1 m - 2 α i ξ i k l + 2 0 .

If else, one has i = 1 m - 2 α i ξ i k l + 2 = 0 , any k l { l ( m - 2 ) + 1 ,    l ( m - 2 ) + 2 , , ( l + 1 ) ( m - 2 ) } ; that is, (7) ( ξ 1 l ( m - 2 ) + 3 ξ m - 2 l ( m - 2 ) + 3 ξ 1 ( l + 1 ) ( m - 2 ) + 2 ξ m - 2 ( l + 2 ) ( m - 2 ) + 2 ) ( α 1 α m - 2 ) = ( 0 0 ) . Since (8)    | ξ 1 l ( m - 2 ) + 3 ξ m - 2 l ( m - 2 ) + 3 ξ 1 ( l + 1 ) ( m - 2 ) + 2 ξ m - 2 ( l + 1 ) ( m - 2 ) + 2 | = j = 1 m - 2 ξ j l ( m - 2 ) + 3 |    1 1 ξ 1 ξ m - 2 ξ 1 m - 3 ξ m - 2 m - 3 | = j = 1 m - 2 ξ j l ( m - 2 ) + 3 1 i < j m - 2 ( ξ j - ξ i ) 0 , thus α i = 0 ,   i = 1,2 , , m - 2 .

It is clear that (9) ( 1 - β ) i = 1 m - 2 α i ξ i 2 = 0 , ( 1 - i = 1 m - 2 α i ) ( β η 2 - 1 ) = β η 2 - 1 < 0 , which is a contradiction to the condition ( C 1 ) .

Set (10) S = { k l N + : 1 - β η p + 2 = i = 1 m - 2 α i ξ i p + 2 ( 1 - β η k l + 2 ) i = 1 m - 2 α i ξ i k l + 2 } . Then S is a finite set.

If else, there exists a monotone sequence { k l r } , r = 1,2 , ,   k l r < k l r + 1 , such that (11) 1 - β η p + 2 = i = 1 m - 2 α i ξ i p + 2 ( 1 - β η k l r + 2 ) i = 1 m - 2 α i ξ i k l r + 2 . From 0 < β η p + 2 < 1 , we get (12) i = 1 m - 2 α i ξ i p + 2 0 . Thus (13) 1 - β η p + 2 = lim k l r i = 1 m - 2 α i ξ i p + 2 ( 1 - β η k l r + 2 ) i = 1 m - 2 α i ξ i k l r + 2 = .

So it is a contradiction. Thus the Lemma is proved.

Lemma 3.

Let ( C 1 ) hold and Λ ( p , q ) 0 ; then L : dom L Y Z is a Fredholm map of index zero. Furthermore, the linear continuous projector operator Q : Z Z can be defined by (14) Q y ( t ) = ( T 1 y ) t p - 1 + ( T 2 y ) t q - 1 , where (15) T 1 y = p ( p + 1 ) ( p + 2 ) Λ ( p , q ) × [ ( β η q + 2 - 1 ) Q 1 y + i = 1 m - 2 α i ξ i q + 2 Q 2 y ] , T 2 y = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) × [ ( β η p + 2 - 1 ) Q 1 y + i = 1 m - 2 α i ξ i p + 2 Q 2 y ] , Q 1 y = i = 1 m - 2 α i 0 ξ i 0 s 0 τ y ( v ) d v d τ d s , Q 2 y = 0 1 0 s 0 τ y ( v ) d v d τ d s - β 0 η 0 s 0 τ y ( v ) d v d τ d s . And the linear operator K P : Im L dom L Ker P can be written by (16) K P y ( t ) = 0 t 0 s 0 τ y ( v ) d v d τ d s , y Im L . Furthermore (17) K P y y 1 , y Im L .

Proof.

It is clear that Ker L = { x dom L : x = a + c t 2 , a = ( ( β η 2 - 1 ) / ( 1 - β ) ) c , c R } .

Now we show that (18) Im L = { y Z : Q 1 y = Q 2 y = 0 } . The equation (19) x ′′′ = y , has a solution x ( t ) satisfying (2) if and only if (20) Q 1 y = Q 2 y = 0 .

In fact, if (19) has a solution x ( t ) such that (2), then from (19) we have (21) x ( t ) = β η 2 - 1 1 - β c + c t 2 + 0 t 0 s 0 τ y ( v ) d v d τ d s . According to the condition ( C 1 ) , we obtain (22) Q 1 y = Q 2 y = 0 . On the other hand, if (20) holds, let (23) x ( t ) = β η 2 - 1 1 - β c + c t 2 + 0 t 0 s 0 τ y ( v ) d v d τ d s , where c is an arbitrary constant; then x ( t ) is a solution of (19) and (2). Hence (18) holds.

Set (24) T 1 y = p ( p + 1 ) ( p + 2 ) Λ ( p , q ) × [ ( β η q + 2 - 1 ) Q 1 y + i = 1 m - 2 α i ξ i q + 2 Q 2 y ] , T 2 y = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) × [ ( β η p + 2 - 1 ) Q 1 y + i = 1 m - 2 α i ξ i p + 2 Q 2 y ] . Then we define (25) Q y ( t ) = ( T 1 y ) t p - 1 + ( T 2 y ) t q - 1 . It is clear that dim Im Q = 2 .

Again from (26) T 1 ( ( T 1 y ) t p - 1 ) = p ( p + 1 ) ( p + 2 ) Λ ( p , q ) × [ + i = 1 m - 2 α i ξ i q + 2 Q 2 ( ( T 1 y ) t p - 1 ) ( β η q + 2 - 1 ) Q 1 ( ( T 1 y ) t p - 1 ) + i = 1 m - 2 α i ξ i q + 2 Q 2 ( ( T 1 y ) t p - 1 ) ] = 1 Λ ( p , q ) [ i = 1 m - 2 α i ξ i q + 2 Q 2 ( t p - 1 ) ( β η q + 2 - 1 ) p ( p + 1 ) × ( p + 2 ) Q 1 ( t p - 1 ) + p ( p + 1 ) ( p + 2 ) × i = 1 m - 2 α i ξ i q + 2 Q 2 ( t p - 1 ) ] × ( T 1 y ) = T 1 y , T 1 ( ( T 2 y ) t q - 1 ) = p ( p + 1 ) ( p + 2 ) Λ ( p , q ) × [ i = 1 m - 2 α i ξ i q + 2 Q 2 ( ( T 2 y ) t q - 1 ) ( β η q + 2 - 1 ) Q 1 ( ( T 2 y ) t q - 1 ) + i = 1 m - 2 α i ξ i q + 2 Q 2 ( ( T 2 y ) t q - 1 ) ] = 1 Λ ( p , q ) [ i = 1 m - 2 α i ξ i q + 2 Q 2 ( t q - 1 ) ( β η q + 2 - 1 ) p ( p + 1 ) × ( p + 2 ) Q 1 ( t q - 1 ) + p ( p + 1 ) ( p + 2 ) h h h h h h h h h × i = 1 m - 2 α i ξ i q + 2 Q 2 ( t q - 1 ) ] × ( T 2 y ) = 0 , T 2 ( ( T 1 y ) t p - 1 ) = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) × [ + i = 1 m - 2 α i ξ i p + 2 Q 2 ( ( T 1 y ) t p - 1 ) ( β η p + 2 - 1 ) Q 1 ( ( T 1 y ) t p - 1 ) + i = 1 m - 2 α i ξ i p + 2 Q 2 ( ( T 1 y ) t p - 1 ) ] = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) × [ i = 1 m - 2 α i ξ i p + 2 Q 2 ( t p - 1 ) ( β η p + 2 - 1 ) Q 1 ( t p - 1 ) + i = 1 m - 2 α i ξ i p + 2 Q 2 ( t p - 1 ) ] × ( T 1 y ) = 0 , T 2 ( ( T 2 y ) t q - 1 ) = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) × [ i = 1 m - 2 α i ξ i p + 2 Q 2 ( ( T 2 y ) t q - 1 ) ( β η p + 2 - 1 ) Q 1 ( ( T 2 y ) t q - 1 ) + i = 1 m - 2 α i ξ i p + 2 Q 2 ( ( T 2 y ) t q - 1 ) ] = T 2 y . One has (27) Q 2 y = Q ( ( T 1 y ) t p - 1 + ( T 2 y ) t q - 1 ) = T 1 ( ( T 1 y ) t p - 1 + ( T 2 y ) t q - 1 ) t p - 1 + T 2 ( ( T 1 y ) t p - 1 + ( T 2 y ) t q - 1 ) t q - 1 = T 1 ( ( T 1 y ) t p - 1 ) t p - 1 + T 1 ( ( T 2 y ) t q - 1 ) t p - 1 + T 2 ( ( T 1 y ) t p - 1 ) t q - 1 + T 2 ( ( T 2 y ) t q - 1 ) t q - 1 = ( T 1 y ) t p - 1 + ( T 2 y ) t q - 1 = Q y . Thus the operator Q is a projector.

Now we show that Ker Q = Im L . If y Ker Q , from Q y = 0 , we have (28) i = 1 m - 2 α i ξ i q + 2 Q 2 y + ( β η q + 2 - 1 ) Q 1 y = 0 , i = 1 m - 2 α i ξ i p + 2 Q 2 y + ( β η p + 2 - 1 ) Q 1 y = 0 . Because of (29) | i = 1 m - 2 α i ξ i q + 2 β η q + 2 - 1 i = 1 m - 2 α i ξ i p + 2 β η p + 2 - 1 | = - Λ ( p , q ) 0 , Q 1 y = Q 2 y = 0 , which yields y Im L . On the other hand, if y Im L , from Q 1 y = Q 2 y = 0 and the definition of Q , so Q y = 0 ; thus y Ker Q . Hence, Ker Q = Im L .

For y Z , from y = ( y - Q y ) + Q y , ( y - Q y ) Ker Q = Im L , Q y Im Q , we have Z = Im L + Im Q . And if y Im L Im Q , from y Im Q , there exist constants a , b R , such that y ( t ) = a t p - 1 + b t q - 1 .

From y Im L , we obtain (30) a q ( q + 1 ) ( q + 2 ) × i = 1 m - 2 α i ξ i p + 2 + b p ( p + 1 ) ( p + 2 ) i = 1 m - 2 α i ξ i q + 2 = 0 , a q ( q + 1 ) ( q + 2 ) ( 1 - β η p + 2 ) + b p ( p + 1 ) ( p + 2 ) ( 1 - β η q + 2 ) = 0 . In view of (31) | q ( q + 1 ) ( q + 2 ) i = 1 m - 2 α i ξ i p + 2 p ( p + 1 ) ( p + 2 ) i = 1 m - 2 α i ξ i q + 2 q ( q + 1 ) ( q + 2 ) ( 1 - β η p + 2 ) p ( p + 1 ) ( p + 2 ) ( 1 - β η q + 2 ) | = - q ( q + 1 ) ( q + 2 ) p ( p + 1 ) ( p + 2 ) Λ ( p , q )    0 , therefore (30) has a unique solution a = b = 0 , which implies Im L Im Q = { 0 } . So we have Z = Im L Im Q . Since imKer L = dimIm Q = codimIm L = 2 , thus L is a Fredholm map of index zero.

Let P : Y Y be defined by (32) P x ( t ) = x ( 0 ) + 1 2 x ′′ ( 0 ) t 2 , t [ 0,1 ] . Then, the generalized inverse K P : Im L dom L Ker P can be written by (33) K P y ( t ) = 0 t 0 s 0 τ y ( v ) d v d τ d s , y Im L .

In fact, for y Im L , we have (34) ( L K P ) y ( t ) = ( K P y ) ′′′ = y ( t ) , and for x dom L Ker P , we know (35) ( K P L ) x ( t ) = ( K P ) x ′′′ ( t ) = 0 t 0 s 0 τ x ′′′ ( v ) d v d τ d s = x ( t ) - [ x ( 0 ) + 1 2 x ′′ ( 0 ) t 2 ] = x ( t ) - P x ( t ) .

Taking note that x dom L Ker P , P x ( t ) = 0 , thus ( K P L ) x ( t ) = x ( t ) .

It is clear that K P y y 1 .

Theorem 4.

Let the condition ( C 1 ) hold and Λ ( p , q ) 0 . Assume the following.

( H 1 ) There exist functions α , β , γ , θ L 1 [ 0,1 ] , such that (36) | f ( t , x 1 , x 2 , x 3 ) + e ( t ) | α ( t ) | x 1 | + β ( t ) | x 2 | + γ ( t ) | x 3 | + θ ( t ) , α 1 + β 1 + γ 1 < 1 ,   where ( x 1 , x 2 , x 3 ) R 3 , t [ 0,1 ] .

( H 2 ) There exists a constant A > 0 such that for x    dom L , if | x ( t ) | > A or | x ′′ ( t ) | > A for all t [ 0,1 ] , then (37) Q 1 N ( x ( t ) ) 0             o r             Q 2 N ( x ( t ) ) 0 .

( H 3 ) There exists a constant B > 0 such that for a , c R , if | a | > B or | c | > B , then either (38) Q 1 N ( a + c t 2 ) + Q 2 N ( a + c t 2 ) > 0   or (39) Q 1 N ( a + c t 2 ) + Q 2 N ( a + c t 2 ) < 0 .

Then BVP (1)-(2) has at least one solution in C 2 [ 0,1 ] .

Proof.

We divide the proof into the following steps.

Step 1. The set Ω 1 = { x    dom    L Ker    L : L x = λ N x    for    some       λ [ 0,1 ] } is bounded.

For x Ω 1 , since L x = λ N x , so λ 0 , N x Im  L ; hence (40) Q 1 N ( x ( t ) ) = 0 , Q 2 N ( x ( t ) ) = 0 .

From ( H 2 ), there exist t 0 , t 1 [ 0,1 ] such that | x ( t 0 ) | A , | x ′′ ( t 1 ) | A . x , x , and x ′′ are absolutely continuous for all t [ 0,1 ] , and (41) x ( t ) = x ( t 0 ) + t 0 t x ( s ) d s ,       x ( t ) = x ( 0 ) + 0 t x ′′ ( s ) d s x ′′ ( t ) = x ′′ ( t 1 ) + t 1 t x ′′′ ( s ) d s , which imply (42) x A + x , x A + x ′′′ 1 , x ′′ A + x ′′′ 1 .

From ( H 1 ), we obtain (43) x ′′′ 1 = L x 1 N x 1 α 1 x + β 1 x + γ 1 x ′′ + θ 1 ( α 1 + β 1 + γ 1 ) x ′′′ 1 + A ( 2 α 1 + β 1 + γ 1 ) + θ 1 , x ′′′ 1 1 1 - ( α 1 + β 1 + γ 1 ) × [ A ( 2 α 1 + β 1 + γ 1 ) + θ 1 ] . So there exists a constant M 1 > 0 such that x M 1 ; that is, the set Ω 1 is bounded.

Step 2. The set Ω 2 = { x    Ker L : N x    Im L } is bounded.

For x Ω 2 , x    Ker    L implies that x = a + c t 2 , a = ( ( β η 2 - 1 ) / ( 1 - β ) ) c , t [ 0,1 ] , c R . From Q N x = 0 , we get Q 1 N ( a + c t 2 ) = Q 2 N ( a + c t 2 ) = 0 . From ( H 3 ), then x | a | + | c | 2 B ; that is, the set Ω 2 is bounded.

Step 3. The set Ω 3 = { x    Ker L : λ J x + ( 1 - λ ) Q N x = 0 , λ [ 0,1 ] } is bounded.

For any a , c R , we define the linear isomorphism J :    Ker L    Im Q by (44) J ( a + c t 2 ) = a t p - 1 + b t q - 1 Λ ( p , q ) , where (45) a = p ( p + 1 ) ( p + 2 ) [ ( β η q + 2 - 1 ) | a | + i = 1 m - 2 α i ξ i q + 2 | c | ] , b = - q ( q + 1 ) ( q + 2 ) [ ( β η p + 2 - 1 ) | a | + i = 1 m - 2 α i ξ i p + 2 | c | ] .

Set (46) μ = p ( p + 1 ) ( p + 2 ) Λ ( p , q ) ( β η q + 2 - 1 ) , ν = p ( p + 1 ) ( p + 2 ) Λ ( p , q ) i = 1 m - 2 α i ξ i q + 2 , ρ = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) ( β η p + 2 - 1 ) , ω = - q ( q + 1 ) ( q + 2 ) Λ ( p , q ) i = 1 m - 2 α i ξ i p + 2 .

For any x ( t ) = a + c t 2 Ω 3 , we obtain (47) μ [ λ | a | + ( 1 - λ ) Q 1 N ( a + c t 2 ) ] + ν [ λ | c | + ( 1 - λ ) Q 2 N ( a + c t 2 ) ] = 0 , ρ [ λ | a | + ( 1 - λ ) Q 1 N ( a + c t 2 ) ] + ω [ λ | c | + ( 1 - λ ) Q 2 N ( a + c t 2 ) ] = 0 . On account of (48) | μ ν ρ ω | = - p ( p + 1 ) ( p + 2 ) q ( q + 1 ) ( q + 2 ) Λ ( p , q ) 0 , therefore, we have (49) λ | a | + ( 1 - λ ) Q 1 N ( a + c t 2 ) = 0 , λ | c | + ( 1 - λ ) Q 2 N ( a + c t 2 ) = 0 .

If λ = 1 , then a = c = 0 . If λ 1 and | a | > B or | c | > B , from the above equality and (38), one has (50) λ ( | a | + | c | ) = - ( 1 - λ ) × [ Q 1 N ( a + c t 2 ) + Q 2 N ( a + c t 2 ) ] < 0 , which contradicts λ ( | a | + | c | ) 0 ; thus x | a | + | c | 2 B . So the set Ω 3 is bounded.

Step 4. If (39) holds, similar to the above argument, we can prove that the set (51) Ω 3 = { x    Ker L : - λ J x + ( 1 - λ ) Q N x = 0 , λ [ 0,1 ] } is bounded too, where J is defined in (44).

Now, we will prove that all conditions of Theorem 1 are satisfied.

Let Ω be an open bounded subset of Y such that i = 1 3 Ω - i Ω . By the Arzelá - Ascoli theorem, we can prove that K P ( I - Q ) N : Ω - Y is compact, so N is L - compact on Ω - .

Then by the above argument, we have

L x N x for every ( x , λ ) [ ( dom L Ker L ) Ω ] × ( 0,1 ) ;

N x Im L for every x    Ker L Ω

let H ( x , λ ) = ± λ J x + ( 1 - λ ) Q N x .

According to the above argument in Steps  3 and  4, we know H ( x , λ ) 0 for every x Ω    Ker L . Thus, by using the homotopy property of degree, we have (52) deg ( Q N |    Ker L    , Ω Ker L , 0 ) =    deg ( H ( · , 0 ) , Ω Ker L , 0 ) =    deg ( H ( · , 1 ) , Ω Ker L , 0 ) =    deg ( ± J , Ω Ker L , 0 ) = ± 1 0 . Then by Theorem 1, L x = N x has at least one solution in dom L Ω - ; that is, BVP (1)-(2) has at least one solution in C 2 [ 0,1 ] .

4. Example Example 1.

We consider the following boundary value problem: (53) x ′′′ ( t ) = t 2 + 4 + 3 8 x ( 1 + 1 2 sin x ) + 1 4 t cos ( x ′′ ) 2 ,    t ( 0,1 ) , x ( 0 ) = 68 59 x ( 1 2 ) , x ( 0 ) = 0 ,    x ( 1 ) = 1 2 x ( 1 3 ) .

Let (54) f ( t , x ( t ) , x ( t ) , x ′′ ( t ) ) = 4 + 3 8 x ( 1 + 1 2 sin x ) + 1 4 t cos ( x ′′ ) 2 , e ( t ) = t 2 , m = 3 , α 1 = 68 59 , β = 1 2 , ξ 1 = 1 2 , η = 1 3 . Then the condition ( C 1 ) holds.

From Lemma 2, one has Λ ( 1,2 ) = - 697 / 9558 0 . By Lemma 3, we define (55) Q 1 y = 68 59 0 1 / 2 0 s 0 τ y ( v ) d v d τ d s , Q 2 y = 0 1 0 s 0 τ y ( v ) d v d τ d s - 1 2 0 1 / 3 0 s 0 τ y ( v ) d v d τ d s , T 1 y = - 57348 697 ( - 1 162 Q 1 y + 17 236 Q 2 y ) , T 2 y = 229392 697 ( - 53 54 Q 1 y + 17 118 Q 2 y ) , K P y ( t ) = 0 t 0 s 0 τ y ( v ) d v d τ d s ,       y    Im    L . Q y ( t ) = ( T 1 y ) t 1 - 1 + ( T 2 y ) t 2 - 1 = T 1 y + ( T 2 y ) t .

Since | f ( t , x ( t ) , x ( t ) , x ′′ ( t ) ) + e ( t ) | ( 9 / 16 ) | x ( t ) | + t 2 + ( 1 / 4 ) t + 4 , then α ( t ) = 9 / 16 , β ( t ) = γ ( t ) = 0 , θ ( t ) = t 2 + ( 1 / 4 ) t + 4 .

If x ( t ) > 1 = A , | x ′′ ( t ) | > 1 = A , and B = 1 , one has (56) Q 1 N x = 68 59 0 1 / 2 0 s 0 τ ( v 2 + 4 + 3 8 x ( 1 + 1 2 sin x ) + 1 4 v cos ( x ′′ ) 2 ) d v d τ d s 68 59 0 1 / 2 0 s 0 τ ( v 2 + 4 + 3 16 - 1 4 v ) d v d τ d s = 68 59 0 1 / 2 0 s 0 τ [ ( v - 1 8 ) 2 + 267 64 ] d v d τ d s > 0 , Q 2 N x = 0 1 0 s 0 τ ( v 2 + 4 + 3 8 x ( 1 + 1 2 sin x ) + 1 4 v cos ( x ′′ ) 2 ) d v d τ d s - 1 2 0 1 / 3 0 s 0 τ ( v 2 + 4 + 3 8 x ( 1 + 1 2 sin x ) + 1 4 v cos ( x ′′ ) 2 ) d v d τ d s 1 2 0 1 / 3 0 s 0 τ ( v 2 + 4 + 3 8 x ( 1 + 1 2 sin x ) + 1 4 v cos ( x ′′ ) 2 ) d v d τ d s 1 2 0 1 / 3 0 s 0 τ ( v 2 + 4 + 3 16 - 1 4 v ) d v d τ d s = 1 2 0 1 / 3 0 s 0 τ [ ( v - 1 8 ) 2 + 267 64 ] d v d τ d s > 0 , Q 1 N ( 2 + t 2 ) + Q 2 N ( 2 + t 2 ) > 0 . Then BVP (53) satisfies Theorem 4. So it has at least one solution in C 2 [ 0,1 ] .

Remark 2.

By using a similar method as employed in the above proof, we could obtain some similar results under the condition ( C 3 ) or ( C 5 ), then we omit them.

Conflict of Interests

The authors declare that there is no conflict of interests regarding to the publication of this paper.

Acknowledgments

This paper is sponsored by the Natural Science Foundation of China (11071205, 11101349, 61201431), the Natural Science Foundation of Jiangsu Province, and PAPD of Jiangsu Higher Education Institutions.

Du Z. Lin X. Ge W. On a third-order multi-point boundary value problem at resonance Journal of Mathematical Analysis and Applications 2005 302 1 217 229 10.1016/j.jmaa.2004.08.012 MR2107359 ZBL1072.34012 Kosmatov N. A multi-point boundary value problem with two critical conditions Nonlinear Analysis: Theory, Methods & Applications 2006 65 3 622 633 10.1016/j.na.2005.09.042 MR2231078 ZBL1121.34023 Liu B. Zhao Z. A note on multi-point boundary value problems Nonlinear Analysis: Theory, Methods & Applications 2007 67 9 2680 2689 10.1016/j.na.2006.09.032 MR2345756 ZBL1127.34006 Lin X. Du Z. Meng F. A note on a third-order multi-point boundary value problem at resonance Mathematische Nachrichten 2011 284 13 1690 1700 10.1002/mana.200910844 MR2832676 ZBL1229.34028 Liu B. Solvability of multi-point boundary value problem at resonance. IV Applied Mathematics and Computation 2003 143 2-3 275 299 10.1016/S0096-3003(02)00361-2 MR1981696 ZBL1071.34014 Ge W. The Boundary Value Problems of Nonlinear Ordinary Differential Equations 2007 Beijing, China Science Press Du Z. Solvability of functional differential equations with multi-point boundary value problems at resonance Computers & Mathematics with Applications 2008 55 11 2653 2661 10.1016/j.camwa.2007.10.015 MR2416033 ZBL1142.34357 Liang S. Mu L. Multiplicity of positive solutions for singular three-point boundary value problems at resonance Nonlinear Analysis:Theory, Methods & Applications 2009 71 7-8 2497 2505 10.1016/j.na.2009.01.085 MR2532777 ZBL1188.34032 Lian H. Pang H. Ge W. Solvability for second-order three-point boundary value problems at resonance on a half-line Journal of Mathematical Analysis and Applications 2008 337 2 1171 1181 10.1016/j.jmaa.2007.04.038 MR2386366 ZBL1136.34034 Yang L. Shen C. On the existence of positive solution for a kind of multi-point boundary value problem at resonance Nonlinear Analysis: Theory, Methods & Applications 2010 72 11 4211 4220 10.1016/j.na.2010.01.051 MR2606778 ZBL1200.34018 Ma R. Yang Y. Existence result for a singular nonlinear boundary value problem at resonance Nonlinear Analysis: Theory, Methods & Applications 2008 68 3 671 680 10.1016/j.na.2006.11.030 MR2372375 ZBL1134.34013 Mawhin J. Topological Degree Methods in Nonlinear Boundary Value Problems 1979 40 Providence, RI, USA American Mathematical Society NSFCBMS Regional Conference Series in Mathematics MR525202