Open Access

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

Boundary Value Problems20132013:207

DOI: 10.1186/1687-2770-2013-207

Received: 5 April 2013

Accepted: 23 August 2013

Published: 11 September 2013

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 [210] 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 [1215] 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
  1. (i)

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

     
  2. (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 ] ,
  1. (a)

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

     
  2. (b)

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

     
  3. (c)

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

     
  4. (d)

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

     

Theorem 2.1 [11]

Let X and Y be 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 , 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 3.3 Let Ω X be nonempty, open and bounded. Then N λ is M-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 3.4 Assume that (H3) and (H4) hold. The set
Ω 1 = { u | u dom M , M u = N λ u , λ [ 0 , 1 ] }

is bounded in X.

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 3.5 Assume that (H4) holds. The set
Ω 2 = { u | u Ker M , Q N u = 0 }

is bounded in X.

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 3.1 Suppose 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.

Declarations

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.

Authors’ Affiliations

(1)
College of Sciences, Hebei University of Science and Technology

References

  1. Mawhin J NSFCBMS Regional Conference Series in Mathematics. In Topological Degree Methods in Nonlinear Boundary Value Problems. Am. Math. Soc., Providence; 1979.View ArticleGoogle Scholar
  2. Mawhin J: Resonance problems for some non-autonomous ordinary differential equations. Lecture Notes in Mathematics 2065. In Stability and Bifurcation Theory for Non-Autonomous Differential Equatons. Edited by: Johnson R, Pera MP. Springer, Berlin; 2013:103-184.View ArticleGoogle Scholar
  3. Feng W, Webb JRL: Solvability of m -point boundary value problems with nonlinear growth. J. Math. Anal. Appl. 1997, 212: 467-480. 10.1006/jmaa.1997.5520MathSciNetView ArticleMATHGoogle Scholar
  4. Ma R: Existence results of a m -point boundary value problem at resonance. J. Math. Anal. Appl. 2004, 294: 147-157. 10.1016/j.jmaa.2004.02.005MathSciNetView ArticleMATHGoogle Scholar
  5. Zhang X, Feng M, Ge W: Existence result of second-order differential equations with integral boundary conditions at resonance. J. Math. Anal. Appl. 2009, 353: 311-319. 10.1016/j.jmaa.2008.11.082MathSciNetView ArticleMATHGoogle Scholar
  6. Du Z, Lin X, Ge W: Some higher-order multi-point boundary value problem at resonance. J. Comput. Appl. Math. 2005, 177: 55-65. 10.1016/j.cam.2004.08.003MathSciNetView ArticleMATHGoogle Scholar
  7. Kosmatov N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ. 2010, 135: 1-10.MathSciNetMATHGoogle Scholar
  8. Jiang W: The existence of solutions to boundary value problems of fractional differential equations at resonance. Nonlinear Anal. TMA 2011, 74: 1987-1994. 10.1016/j.na.2010.11.005View ArticleMathSciNetMATHGoogle Scholar
  9. Jiang W: Solvability for a coupled system of fractional differential equations at resonance. Nonlinear Anal., Real World Appl. 2012, 13: 2285-2292. 10.1016/j.nonrwa.2012.01.023MathSciNetView ArticleMATHGoogle Scholar
  10. Jiang W:Solvability of ( k , n k ) conjugate boundary-value problems at resonance. Electron. J. Differ. Equ. 2012, 114: 1-10.MathSciNetGoogle Scholar
  11. Ge W, Ren J: An extension of Mawhin’s continuation theorem and its application to boundary value problems with a p -Laplacian. Nonlinear Anal. 2004, 58: 477-488. 10.1016/j.na.2004.01.007MathSciNetView ArticleMATHGoogle Scholar
  12. Liu Y, Li D, Fang M: Solvability for second-order m -point boundary value problems at resonance on the half-line. Electron. J. Differ. Equ. 2009., 2009: Article ID 13Google Scholar
  13. Liu Y: Boundary value problem for second order differential equations on unbounded domain. Acta Anal. Funct. Appl. 2002, 4(3):211-216. (in Chinese)MathSciNetMATHGoogle Scholar
  14. Agarwal RP, O’Regan D: Infinite Interval Problems for Differential, Difference and Integral Equations. Kluwer Academic, Dordrecht; 2001.View ArticleMATHGoogle Scholar
  15. Kosmatov N: Multi-point boundary value problems on an unbounded domain at resonance. Nonlinear Anal. 2008, 68: 2158-2171. 10.1016/j.na.2007.01.038MathSciNetView ArticleMATHGoogle Scholar

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© Jiang; licensee Springer 2013

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