Open Access Research

Solvability for p-Laplacian boundary value problem at resonance on the half-line

Weihua Jiang

Author Affiliations

College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei, 050018, P.R. China

Boundary Value Problems 2013, 2013:207  doi:10.1186/1687-2770-2013-207


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/207


Received:5 April 2013
Accepted:23 August 2013
Published:11 September 2013

© 2013 Jiang; 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

The existence of solutions for p-Laplacian boundary value problem at resonance on the half-line is investigated. Our analysis relies on constructing the suitable Banach space, defining appropriate operators and using the extension of Mawhin’s continuation theorem. An example is given to illustrate our main result.

MSC: 70K30, 34B10, 34B15.

Keywords:
p-Laplacian; resonance; half-line; multi-point boundary value problem; continuation theorem

1 Introduction

A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Resonance problems can be expressed as an abstract equation L x = N x , where L is a noninvertible operator. When L is linear, Mawhin’s continuation theorem [1] is an effective tool in finding solutions for these problems, see [2-10] and references cited therein. But it does not work when L is nonlinear, for instance, p-Laplacian operator. In order to solve this problem, Ge and Ren [11] proved a continuation theorem for the abstract equation L x = N x when L is a noninvertible nonlinear operator and used it to study the existence of solutions for the boundary value problems with a p-Laplacian:

{ ( φ p ( u ) ) + f ( t , u ) = 0 , 0 < t < 1 , u ( 0 ) = 0 = G ( u ( η ) , u ( 1 ) ) ,

where φ p ( s ) = | s | p 2 s , p > 1 , 0 < η < 1 . φ p ( s ) is nonlinear when p 2 .

As far as the boundary value problems on unbounded domain are concerned, there are many excellent results, see [12-15] and references cited therein.

To the best of our knowledge, there are few papers that study the p-Laplacian boundary value problem at resonance on the half-line. In this paper, we investigate the existence of solutions for the boundary value problem

{ ( φ p ( u ) ) + f ( t , u , u ) = 0 , 0 < t < + , u ( 0 ) = 0 , φ p ( u ( + ) ) = i = 1 n α i φ p ( u ( ξ i ) ) , (1.1)

where α i > 0 , i = 1 , 2 , , n , i = 1 n α i = 1 .

In order to obtain our main results, we always suppose that the following conditions hold.

(H1) 0 < ξ 1 < ξ 2 < < ξ n < + , α i > 0 , i = 1 n α i = 1 .

(H2) f : [ 0 , + ) × R 2 R is continuous, f ( t , 0 , 0 ) 0 , t ( 0 , ) and for any r > 0 , there exists a nonnegative function h r ( t ) L 1 [ 0 , + ) such that

| f ( t , x , y ) | h r ( t ) , a.e.  t [ 0 , + ) , x , y R , | x | 1 + t r , | y | r .

2 Preliminaries

For convenience, we introduce some notations and a theorem. For more details, see [11].

Definition 2.1[11]

Let X and Y be two Banach spaces with the norms X , Y , respectively. A continuous operator M : X dom M Y is said to be quasi-linear if

(i) Im M : = M ( X dom M ) is a closed subset of Y,

(ii) Ker M : = { x X dom M : M x = 0 } is linearly homeomorphic to R n , n < , where domM denote the domain of the operator M.

Let X 1 = Ker M and X 2 be the complement space of X 1 in X, then X = X 1 X 2 . On the other hand, suppose that Y 1 is a subspace of Y, and that Y 2 is the complement of Y 1 in Y, i.e., Y = Y 1 Y 2 . Let P : X X 1 and Q : Y Y 1 be two projectors and Ω X an open and bounded set with the origin θ Ω .

Definition 2.2[11]

Suppose that N λ : Ω ¯ Y , λ [ 0 , 1 ] is a continuous operator. Denote N 1 by N. Let Σ λ = { x Ω ¯ : M x = N λ x } . N λ is said to be M-compact in Ω ¯ if there exist a vector subspace Y 1 of Y satisfying dim Y 1 = dim X 1 and an operator R : Ω ¯ × [ 0 , 1 ] X 2 being continuous and compact such that for λ [ 0 , 1 ] ,

(a) ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y ,

(b) Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ ,

(c) R ( , 0 ) is the zero operator and R ( , λ ) | Σ λ = ( I P ) | Σ λ ,

(d) M [ P + R ( , λ ) ] = ( I Q ) N λ .

Theorem 2.1[11]

LetXandYbe two Banach spaces with the norms X , Y , respectively, and Ω X an open and bounded nonempty set. Suppose that

M : X dom M Y

is a quasi-linear operator and N λ : Ω ¯ Y , λ [ 0 , 1 ] M-compact. In addition, if the following conditions hold:

(C1) M x N λ x , x Ω dom M , λ ( 0 , 1 ) ,

(C2) deg { J Q N , Ω Ker M , 0 } 0 ,

then the abstract equation M x = N x has at least one solution in dom M Ω ¯ , where N = N 1 , J : Im Q Ker M is a homeomorphism with J ( θ ) = θ .

3 Main result

Let X = { u | u C 1 [ 0 , + ) , u ( 0 ) = 0 , sup t [ 0 , + ) | u ( t ) | 1 + t < + , lim t + u ( t )  exists } with norm u = max { u 1 + t , u } , where u = sup t [ 0 , + ) | u ( t ) | . Y = L 1 [ 0 , + ) with norm y 1 = 0 + | y ( t ) | d t . Then ( X , ) and ( Y , 1 ) are Banach spaces.

Define operators M : X dom M Y and N λ : X Y as follows:

M u = ( φ p ( u ) ) , N λ u = λ f ( t , u , u ) , λ [ 0 , 1 ] , t [ 0 , + ) ,

where

dom M = { u X | φ p ( u ) A C [ 0 , + ) , ( φ p ( u ) ) L 1 [ 0 , + ) , φ p ( u ( + ) ) = i = 1 n α i φ p ( u ( ξ i ) ) } .

Then the boundary value problem (1.1) is equivalent to M u = N u .

Obviously,

Ker M = { a t | a R } , Im M = { y | y Y , i = 1 n α i ξ i + y ( s ) d s = 0 } .

It is clear that KerM is linearly homeomorphic to ℝ, and Im M Y is closed. So, M is a quasi-linear operator.

Define P : X X 1 , Q : Y Y 1 as

( P u ) ( t ) = u ( + ) t , ( Q y ) ( t ) = i = 1 n α i ξ i + y ( s ) d s i = 1 n α i e ξ i e t ,

where X 1 = Ker M , Y 1 = Im Q = { b e t | b R } . We can easily obtain that P : X X 1 , Q : Y Y 1 are projectors. Set X = X 1 X 2 , Y = Y 1 Y 2 .

Define an operator R : X × [ 0 , 1 ] X 2 :

R ( u , λ ) ( t ) = 0 t φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ u ( + ) t ,

where 1 p + 1 q = 1 , φ q = φ p 1 . By (H1) and (H2), we get that R : X × [ 0 , 1 ] X 2 is continuous.

Lemma 3.1[15]

V X is compact if { u ( t ) 1 + t | u V } and { u ( t ) | u V } are both equicontinuous on any compact intervals of [ 0 , + ) and equiconvergent at infinity.

Lemma 3.2 R : X × [ 0 , 1 ] X 2 is compact.

Proof Let Ω X be nonempty and bounded. There exists a constant r > 0 such that u r , u Ω ¯ . It follows from (H2) that there exists a nonnegative function h r ( t ) L 1 [ 0 , + ) such that

| f ( t , u ( t ) , u ( t ) ) | h r ( t ) , a.e.  t [ 0 , + ) , u Ω ¯ .

For any T > 0 , t 1 , t 2 [ 0 , T ] , u Ω ¯ , λ [ 0 , 1 ] , we have

| R ( u , λ ) ( t 1 ) 1 + t 1 R ( u , λ ) ( t 2 ) 1 + t 2 | | 1 1 + t 1 0 t 1 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ 1 1 + t 2 0 t 2 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ | + | t 1 1 + t 1 t 2 1 + t 2 | | u ( + ) | | 1 1 + t 1 t 2 t 1 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ | + | 1 1 + t 1 1 1 + t 2 | × | 0 t 2 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ | + | t 1 1 + t 1 t 2 1 + t 2 | r φ q [ h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) ] [ | t 1 t 2 | + T | 1 1 + t 1 1 1 + t 2 | ] + | t 1 1 + t 1 t 2 1 + t 2 | r .

Since { t , 1 1 + t , t 1 + t } are equicontinuous on [ 0 , T ] , we get that { R ( u , λ ) ( t ) 1 + t , u Ω ¯ } are equicontinuous on [ 0 , T ] .

| R ( u , λ ) ( t 1 ) R ( u , λ ) ( t 2 ) | = | φ q [ t 1 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] φ q [ t 2 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] | .

Let

g ( t , u ) = t + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) .

Then

| g ( t , u ) | h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) : = k , t [ 0 , T ] , u Ω ¯ . (3.1)

For t 1 , t 2 [ 0 , T ] , t 1 < t 2 , u Ω ¯ , we have

| g ( t 1 , u ) g ( t 2 , u ) | = | t 1 t 2 λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s | t 1 t 2 h r ( s ) + h r 1 i = 1 n α i e ξ i e s d s .

It follows from the absolute continuity of integral that { g ( t , u ) , u Ω ¯ } are equicontinuous on [ 0 , T ] . Since φ q ( x ) is uniformly continuous on [ k , k ] , by (3.1), we can obtain that { R ( u , λ ) ( t ) , u Ω ¯ } are equicontinuous on [ 0 , T ] .

For u Ω ¯ , since

| τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s | τ + h r ( s ) + h r 1 i = 1 n α i e ξ i e s d s , lim τ + τ + h r ( s ) + h r 1 i = 1 n α i e ξ i e s d s = 0 ,

and φ q ( x ) is uniformly continuous on [ r r p 1 , r + r p 1 ] , for any ε > 0 , there exists a constant T 1 > 0 such that if τ T 1 , then

| φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) | < ε 4 , u Ω ¯ . (3.2)

Since

| 0 T 1 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ u ( + ) T 1 | { φ q [ h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) ] + r } T 1 , (3.3)

there exists a constant T > T 1 such that if t > T , then

1 1 + t { φ q [ h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) ] + r } T 1 < ε 4 . (3.4)

For t 2 > t 1 > T , by (3.2), (3.3) and (3.4), we have

| R ( u , λ ) ( t 1 ) 1 + t 1 R ( u , λ ) ( t 2 ) 1 + t 2 | = | 1 1 + t 1 0 t 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ 1 1 + t 2 0 t 2 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | | 1 1 + t 1 0 T 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | + | 1 1 + t 1 T 1 t 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | + | 1 1 + t 2 0 T 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | + | 1 1 + t 2 T 1 t 2 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | < ε ,

and

| R ( u , λ ) ( t 1 ) R ( u , λ ) ( t 2 ) | | φ q [ t 1 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) | + | φ q [ t 2 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) | < ε .

By Lemma 3.1, we get that { R ( u , λ ) | u Ω ¯ , λ [ 0 , 1 ] } is compact. The proof is completed. □

In the spaces X and Y, the origin θ = 0 . In the following sections, we denote the origin by 0.

Lemma 3.3Let Ω X be nonempty, open and bounded. Then N λ isM-compact in Ω ¯ .

Proof By (H2), we know that N λ : Ω ¯ Y is continuous. Obviously, dim X 1 = dim Y 1 . For u Ω ¯ , since Q ( I Q ) is a zero operator, we get ( I Q ) N λ ( u ) Im M . For y Im M , y = Q y + ( I Q ) y = ( I Q ) y ( I Q ) Y . So, we have ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y . It is clear that

Q N λ u = 0 , λ ( 0 , 1 ) Q N u = 0

and R ( u , 0 ) = 0 , u X . u Σ λ = { u Ω ¯ : M u = N λ u } means that N λ u Im M and ( φ p ( u ) ) + λ f ( t , u , u ) = 0 , thus,

R ( u , λ ) ( t ) = 0 t φ q [ τ + ( φ p ( u ) ) d s + φ p ( u ( + ) ) ] d τ u ( + ) t = u ( t ) u ( + ) t = ( I P ) u ( t ) .

For u X , we have

M [ P + R ( u , λ ) ] ( t ) = λ f ( t , u ( t ) , u ( t ) ) + i = 1 n α i ξ i + λ f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e t = ( I Q ) N λ u ( t ) .

These, together with Lemma 3.2, mean that N λ is M-compact in Ω ¯ . The proof is completed. □

In order to obtain our main results, we need the following additional conditions.

(H3) There exist nonnegative functions a ( t ) , b ( t ) , c ( t ) with ( 1 + t ) p 1 a ( t ) , b ( t ) , c ( t ) Y and ( 1 + t ) p 1 a ( t ) 1 + b ( t ) 1 < 1 such that

| f ( t , x , y ) | a ( t ) | φ p ( x ) | + b ( t ) | φ p ( y ) | + c ( t ) , a.e.  t [ 0 , + ) .

(H4) There exists a constant d 0 > 0 such that if | d | > d 0 , then one of the following inequalities holds:

d f ( t , x , d ) < 0 , ( t , x ) [ 0 , + ) × R ; d f ( t , x , d ) > 0 , ( t , x ) [ 0 , + ) × R .

Lemma 3.4Assume that (H3) and (H4) hold. The set

Ω 1 = { u | u dom M , M u = N λ u , λ [ 0 , 1 ] }

is bounded inX.

Proof If u Ω 1 , then Q N λ u = 0 , i.e., i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r = 0 . By (H4), there exists t 0 [ 0 , + ) such that | u ( t 0 ) | d 0 . It follows from M u = N λ u that

φ p ( u ( t ) ) = t 0 t λ f ( s , u ( s ) , u ( s ) ) d s + φ p ( u ( t 0 ) ) .

Considering (H3), we have

| φ p ( u ( t ) ) | 0 + [ a ( t ) | φ p ( u ( t ) ) | + b ( t ) | φ p ( u ( t ) ) | + c ( t ) ] d t + φ p ( d 0 ) a ( t ) ( 1 + t ) p 1 1 φ p ( u 1 + t ) + b 1 φ p ( u ) + c 1 + φ p ( d 0 ) . (3.5)

Since u ( t ) = 0 t u ( s ) d s , we get

| u ( t ) 1 + t | t 1 + t u u .

Thus,

u 1 + t u . (3.6)

By (3.5), (3.6) and (H3), we get

φ p ( u ) c 1 + φ p ( d 0 ) 1 a ( t ) ( 1 + t ) p 1 1 b 1 .

So,

u φ q ( c 1 + φ p ( d 0 ) 1 a ( t ) ( 1 + t ) p 1 1 b 1 ) .

This, together with (3.6), means that Ω 1 is bounded. The proof is completed. □

Lemma 3.5Assume that (H4) holds. The set

Ω 2 = { u | u Ker M , Q N u = 0 }

is bounded inX.

Proof u Ω 2 means that u = a t , a R and Q N u = 0 , i.e.,

i = 1 n α i ξ i + f ( s , a s , a ) d s = 0 .

By (H4), we get that | a | d 0 . So, Ω 2 is bounded. The proof is completed. □

Theorem 3.1Suppose that (H1)-(H4) hold. Then problem (1.1) has at least one solution.

Proof Let Ω = { u X | u < d 0 } , where d 0 = max { d 0 , sup u Ω 1 u , sup u Ω 2 u } + 1 . It follows from the definition of Ω 1 and Ω 2 that M u N λ u , λ ( 0 , 1 ) , u Ω and Q N u 0 , u Ω Ker M .

Define a homeomorphism J : Im Q Ker M as J ( k e t ) = k t . If d f ( t , x , d ) < 0 for | d | > d 0 , take the homotopy

H ( u , μ ) = μ u + ( 1 μ ) J Q N u , u Ω ¯ Ker M , μ [ 0 , 1 ] .

For u Ω ¯ Ker M , we have u = k t . Then

H ( u , μ ) = μ k t ( 1 μ ) i = 1 n α i ξ i + f ( s , k s , k ) d s i = 1 n α i e ξ i t .

Obviously, H ( u , 1 ) 0 , u Ω Ker M . For μ [ 0 , 1 ) , u = k t Ω Ker M , if H ( u , μ ) = 0 , we have

i = 1 n α i ξ i + k f ( s , k s , k ) d s i = 1 n α i e ξ i = μ 1 μ k 2 0 .

A contradiction with d f ( t , x , d ) < 0 , | d | > d 0 . If d f ( t , x , d ) > 0 , | d | > d 0 , take

H ( u , μ ) = μ u ( 1 μ ) J Q N u , u Ω ¯ Ker M , μ [ 0 , 1 ] ,

and the contradiction follows analogously. So, we obtain H ( u , μ ) 0 , μ [ 0 , 1 ] , u Ω Ker M .

By the homotopy of degree, we get that

deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( I , Ω Ker M , 0 ) = 1 .

By Theorem 2.1, we can get that M u = N u has at least one solution in Ω ¯ . The proof is completed. □

4 Example

Let us consider the following boundary value problem at resonance

{ ( | u | 1 2 u ) + e 4 t 1 + t sin | u | + e 4 t | u | 1 2 u + 1 4 e 4 t = 0 , 0 < t < + , u ( 0 ) = 0 , | u ( + ) | 1 2 u ( + ) = i = 1 n α i | u ( ξ i ) | 1 2 u ( ξ i ) , (4.1)

where 0 < ξ 1 < ξ 2 < < ξ n < + , α i > 0 , i = 1 n α i = 1 .

Corresponding to problem (1.1), we have p = 3 2 , f ( t , x , y ) = e 4 t 1 + t sin | x | + e 4 t | y | 1 2 y + 1 4 e 4 t .

Take a ( t ) = e 4 t 1 + t , b ( t ) = e 4 t , c ( t ) = 1 4 e 4 t , d 0 = 4 . By simple calculation, we can get that conditions (H1)-(H4) hold. By Theorem 3.1, we obtain that problem (4.1) has at least one solution.

Competing interests

The author declares that she has no competing interests.

Author’s contributions

All results belong to WJ.

Acknowledgements

This work is supported by the National Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108). The author is grateful to anonymous referees for their constructive comments and suggestions, which led to the improvement of the original manuscript.

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