Skip to main content

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

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.

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 Lx=Nx, 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 Lx=Nx 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 p2.

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,yR, | 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:XdomMY is said to be quasi-linear if

  1. (i)

    ImM:=M(XdomM) is a closed subset of Y,

  2. (ii)

    KerM:={xXdomM:Mx=0} is linearly homeomorphic to R n , n<, where domM denote the domain of the operator M.

Let X 1 =KerM 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 ΩX an open and bounded set with the origin θΩ.

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 ¯ :Mx= 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)

    (IQ) N λ ( Ω ¯ )ImM(IQ)Y,

  2. (b)

    Q N λ x=θ,λ(0,1)QNx=θ,

  3. (c)

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

  4. (d)

    M[P+R(,λ)]=(IQ) N λ .

Theorem 2.1 [11]

Let X and Y be two Banach spaces with the norms X , Y , respectively, and ΩX an open and bounded nonempty set. Suppose that

M:XdomMY

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

(C1) Mx N λ x, xΩdomM, λ(0,1),

(C2) deg{JQN,ΩKerM,0}0,

then the abstract equation Mx=Nx has at least one solution in domM Ω ¯ , where N= N 1 , J:ImQKerM 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)|dt. Then (X,) and (Y, 1 ) are Banach spaces.

Define operators M:XdomMY and N λ :XY as follows:

Mu= ( φ p ( u ) ) , N λ u=λf ( 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 Mu=Nu.

Obviously,

KerM={at|aR},ImM= { 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 ImMY is closed. So, M is a quasi-linear operator.

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

(Pu)(t)= u (+)t,(Qy)(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 =KerM, Y 1 =ImQ={b e t |bR}. 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]

VX is compact if { u ( t ) 1 + t |uV} and { u (t)|uV} 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 ΩX be nonempty and bounded. There exists a constant r>0 such that ur, 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 ) ds+ 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 ΩX 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(IQ) is a zero operator, we get (IQ) N λ (u)ImM. For yImM, y=Qy+(IQ)y=(IQ)y(IQ)Y. So, we have (IQ) N λ ( Ω ¯ )ImM(IQ)Y. It is clear that

Q N λ u=0,λ(0,1)QNu=0

and R(u,0)=0, uX. u Σ λ ={u Ω ¯ :Mu= N λ u} means that N λ uImM and ( φ p ( u ) ) +λf(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 uX, 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))dr=0. By (H4), there exists t 0 [0,+) such that | u ( t 0 )| d 0 . It follows from Mu= N λ u that

φ p ( u ( t ) ) = t 0 t λf ( s , u ( s ) , u ( s ) ) ds+ 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)ds, 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=at, aR and QNu=0, i.e.,

i = 1 n α i ξ i + f(s,as,a)ds=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={uX|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 Mu N λ u, λ(0,1), uΩ and QNu0, uΩKerM.

Define a homeomorphism J:ImQKerM as J(k e t )=kt. If df(t,x,d)<0 for |d|> d 0 , take the homotopy

H(u,μ)=μu+(1μ)JQNu,u Ω ¯ KerM,μ[0,1].

For u Ω ¯ KerM, we have u=kt. Then

H(u,μ)=μkt(1μ) 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, uΩKerM. For μ[0,1), u=ktΩKerM, 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 df(t,x,d)<0, |d|> d 0 . If df(t,x,d)>0, |d|> d 0 , take

H(u,μ)=μu(1μ)JQNu,u Ω ¯ KerM,μ[0,1],

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

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 Mu=Nu 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.

References

  1. Mawhin J NSFCBMS Regional Conference Series in Mathematics. In Topological Degree Methods in Nonlinear Boundary Value Problems. Am. Math. Soc., Providence; 1979.

    Chapter  Google 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.

    Chapter  Google 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.1997.5520

    Article  MathSciNet  MATH  Google 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.2004.02.005

    Article  MathSciNet  MATH  Google 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.2008.11.082

    Article  MathSciNet  MATH  Google 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.2004.08.003

    Article  MathSciNet  MATH  Google Scholar 

  7. Kosmatov N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ. 2010, 135: 1-10.

    MathSciNet  MATH  Google 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.2010.11.005

    Article  MathSciNet  MATH  Google 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.2012.01.023

    Article  MathSciNet  MATH  Google Scholar 

  10. Jiang W:Solvability of (k,nk) conjugate boundary-value problems at resonance. Electron. J. Differ. Equ. 2012, 114: 1-10.

    MathSciNet  Google 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.2004.01.007

    Article  MathSciNet  MATH  Google 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 13

    Google 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)

    MathSciNet  MATH  Google Scholar 

  14. Agarwal RP, O’Regan D: Infinite Interval Problems for Differential, Difference and Integral Equations. Kluwer Academic, Dordrecht; 2001.

    Book  MATH  Google 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.2007.01.038

    Article  MathSciNet  MATH  Google Scholar 

Download references

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.

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Weihua Jiang.

Additional information

Competing interests

The author declares that she has no competing interests.

Author’s contributions

All results belong to WJ.

Rights and permissions

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

Reprints and permissions

About this article

Cite this article

Jiang, W. Solvability for p-Laplacian boundary value problem at resonance on the half-line. Bound Value Probl 2013, 207 (2013). https://doi.org/10.1186/1687-2770-2013-207

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1687-2770-2013-207

Keywords