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# Solvability for nonlocal boundary value problems on a half line with $dim(kerL)=2$

## Abstract

The existence of at least one solution to the second-order nonlocal boundary value problems on a half line is investigated by using Mawhin’s continuation theorem.

MSC: 34B10, 34B40, 34B15.

## 1 Introduction

Boundary value problems on an infinite interval arise quite naturally in the study of radially symmetric solutions of nonlinear elliptic equations and in various applications such as an unsteady flow of gas through a semi-infinite porous media, theory of drain flows and plasma physics. For an extensive collection of results as regards boundary value problems on unbounded domains, we refer the reader to a monograph by Agarwal and O’Regan . For more recent results on unbounded domains, we refer the reader to – and the references therein.

A boundary value problem is called 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  is an efficient tool in finding solutions for these problems. Recently, there have been many works concerning the existence of solutions for multi-point boundary value problems at resonance. For example, see , – in the case that $dim(kerL)=1$, and see , – in the case that $dim(kerL)=2$.

In this paper, we consider the existence of solutions to the following second-order nonlinear differential equation with nonlocal boundary conditions that contain integral and multi-point boundary conditions:

(1)

where $0≤ ξ 1 <⋯< ξ m <∞$, $α i ∈(−∞,∞)$, $c∈C[0,∞)$, $g∈ L 1 (0,∞)$, and $f:[0,∞)×(−∞,∞)×(−∞,∞)→(−∞,∞)$ is a Carathéodory function, i.e., $f=f(t,u,v)$ is Lebesgue measurable in t for all $(u,v)∈(−∞,∞)×(−∞,∞)$ and continuous in $(u,v)$ for almost all $t∈[0,∞)$. Throughout this paper, we assume that the following conditions hold:

(H1) $∑ i = 1 m α i =1$, $∫ 0 ∞ g(s)ds=1$, $c(t)>0$ for $t∈(0,∞)$, and $1 c ∈ L loc 1 [0,∞)$;

(H2) let $w(t):= ∫ 0 t 1 c ( s ) ds$, and there exist non-negative measurable functions α, β, and γ such that $(1+w)α,β/c,γ∈ L 1 (0,∞)$, and $|f(t,u,v)|≤α(t)|u|+β(t)|v|+γ(t)$, a.e. $(t,u,v)∈[0,∞)×(−∞,∞)×(−∞,∞)$;

(H3) let $k(t)$ be the function such that $(1+w(⋅)) e − k ( ⋅ ) ∈ L 1 (0,∞)$. Then

$△= | Q 2 ( w ( ⋅ ) e − k ( ⋅ ) ) − Q 1 ( w ( ⋅ ) e − k ( ⋅ ) ) − Q 2 ( e − k ( ⋅ ) ) Q 1 ( e − k ( ⋅ ) ) | =: | a 11 a 12 a 21 a 22 | ≠0,$

where the linear operators $Q 1 , Q 2 : L 1 (0,∞)→(−∞,∞)$ will be defined later in Section 3.

If $1 c ∈ L loc 1 [0,∞)$, then w is continuous in $[0,∞)$, and in (H3), there exists a function k satisfying $(1+w(⋅)) e − k ( ⋅ ) ∈ L 1 (0,∞)$ (see, e.g., Remark 3.1(1)). The boundary conditions in problem (1) are crucial since the differential operator $Lu= ( c u ′ ) ′$ under the boundary conditions in (1) satisfies $dim(kerL)=2$. The purpose of this paper is to establish the sufficient conditions for the existence of solutions to problem (1) on a half line at resonance with $dim(kerL)=2$ by using Mawhin’s continuation theorem .

The remainder of this paper is organized as follows: some preliminaries are provided in Section 2, the main result is presented in Section 3, and finally an example to illustrate the main result is given in Section 4.

## 2 Preliminaries

In this section, we recall some notations and two theorems which will be used later. Let X and Y be two Banach spaces with the norms $∥ ⋅ ∥ X$ and $∥ ⋅ ∥ Y$, respectively. Let $L:dom(L)⊂X→Y$ be a Fredholm operator with index zero, and let $P:X→X$, $Q:Y→Y$ be projectors such that $ImP=kerL$ and $kerQ=ImL$. Then $X=kerL⊕kerP$ and $Y=ImL⊕ImQ$. It follows that $L | dom L ∩ ker P :domL∩kerP→ImL$ is invertible. We denote the inverse of it by $K P$. If Ω is an open bounded subset of X with $domL∩ Ω ¯ ≠∅$, then the map $N:X→Y$ will be called L-compact on $Ω ¯$ if $QN( Ω ¯ )$ is bounded in Y and $K P (I−Q)N: Ω ¯ →X$ is compact.

### Theorem 2.1

()

Let$L:dom(L)⊂X→Y$be a Fredholm operator with index zero and$N:X→Y$be L-compact on$Ω ¯$. Assume that the following conditions are satisfied:

1. (1)

$Lu≠λNu$ for every $(u,λ)∈[(domL∖kerL)∩∂Ω]×(0,1)$;

2. (2)

$Nu∉ImL$ for every $u∈kerL∩∂Ω$;

3. (3)

$deg(QN | ker L ,Ω∩kerL,0)≠0$, where $Q:Y→Y$ is a projector such that $ImL=kerQ$.

Then the equation$Lu=Nu$has at least one solution in$domL∩ Ω ¯$.

Since the Arzelá-Ascoli theorem fails in the noncompact interval case, we use the following result in order to show that $K P (I−Q)N: Ω ¯ →X$ is compact.

### Theorem 2.2

()

Let Z be the space of all bounded continuous vector-valued functions on$[0,∞)$and$S⊂Z$. Then S is relatively compact in Z if the following conditions hold:

1. (i)

S is bounded in Z;

2. (ii)

S is equicontinuous on any compact interval of $[0,∞)$;

3. (iii)

S is equiconvergent at ∞, that is, given $ϵ>0$, there exists a $T=T(ϵ)>0$ such that $∥ ϕ ( t ) − ϕ ( ∞ ) ∥ ( − ∞ , ∞ ) n <ϵ$ for all $t>T$ and all $ϕ∈S$.

## 3 Main results

Let X be the set of the functions $u∈C[0,∞)∩ C 1 (0,∞)$ such that $u ( t ) 1 + w ( t )$ and $(c u ′ )(t)$ are uniformly bounded in $[0,∞)$. Here, w is the function in the assumption (H2). Then X is a Banach space equipped with a norm $∥ u ∥ X = ∥ u ∥ 1 + ∥ u ∥ 2$, where

$∥ u ∥ 1 = sup t ∈ [ 0 , ∞ ) | u ( t ) | 1 + w ( t ) and ∥ u ∥ 2 = sup t ∈ [ 0 , ∞ ) | ( c u ′ ) (t)|.$

Let Y denote the Banach space $L 1 (0,∞)$ equipped with the usual norm, $∥ h ∥ Y = ∫ 0 ∞ |h(s)|ds$.

### Remark 3.1

1. (1)

For any non-negative continuous function $w(t)$, we can choose a function $k(t)$ which satisfies $(1+w(⋅)) e − k ( ⋅ ) ∈Y$. For example, put

$k(t)= ∫ 0 t ( 1 + w ( s ) ) ds.$

Then $(1+w(⋅)) e − k ( ⋅ ) ∈Y$.

1. (2)

If $1 c ∈Y$, then $w∈ L ∞ (0,∞)$, and the norm $∥ ⋅ ∥ X$ is equivalent to the norm $∥u∥= ∥ u ∥ ∞ + ∥ u ∥ 2$. Here,

$∥ u ∥ ∞ = sup t ∈ [ 0 , ∞ ) |u(t)|.$

Define $L:domL→Y$ by $Lu:= ( c u ′ ) ′$, where

$dom L : = { u ∈ X : ( c u ′ ) ′ ∈ Y , ( c u ′ ) ( 0 ) = ∫ 0 ∞ g ( s ) ( c u ′ ) ( s ) d s , lim t → ∞ ( c u ′ ) ( t ) = ∑ i = 1 m α i ( c u ′ ) ( ξ i ) } .$

Clearly, $kerL:={a+bw:a,b∈(−∞,∞)}$. Now we define the linear operators $Q 1 , Q 2 :Y→(−∞,∞)$ under the hypothesis (H3) as follows:

$Q 1 (y):= ∑ i = 1 m α i ∫ ξ i ∞ y(s)dsand Q 2 (y):= ∫ 0 ∞ g(s) ∫ 0 s y(τ)dτds.$

### Lemma 3.2

Assume that (H1) holds. Then

$ImL= { y ∈ Y : Q 1 ( y ) = Q 2 ( y ) = 0 } .$

### Proof

Let $y∈ImL$. Then there exists $x∈domL$ such that $( c x ′ ) ′ =y$. For $t∈[0,∞)$,

$( c x ′ ) (t)= ( c x ′ ) (0)+ ∫ 0 t y(s)ds$

and

$( c x ′ ) ( ξ i )= ( c x ′ ) (0)+ ∫ 0 ξ i y(s)ds,$

which imply that

$lim t → ∞ ( c x ′ ) (t)= ( c x ′ ) (0)+ ∫ 0 ∞ y(s)ds= ( c x ′ ) ( ξ i )+ ∫ ξ i ∞ y(s)ds,$

and thus $Q 1 (y)=0$ by the fact that $∑ i = 1 m α i =1$. In a similar manner, $Q 2 (y)=0$. On the other hand, let $y∈Y$ satisfying $Q 1 (y)= Q 2 (y)=0$. Take

$x(t)= ∫ 0 t 1 c ( s ) ∫ 0 s y(τ)dτds.$

Then $x∈domL$, and $( c x ′ ) ′ =y∈ImL$. Thus the proof is complete. □

By Lemma 3.2, $codim(ImL)=2$. Since $dim(kerL)=2$, L is a Fredholm operator with index 0. Let $T 1 , T 2 :Y→Y$ be linear operators which are defined as follows:

$T 1 y : = 1 △ ( a 11 Q 1 ( y ) + a 12 Q 2 ( y ) ) e − k ( ⋅ ) , T 2 y : = 1 △ ( a 21 Q 1 ( y ) + a 22 Q 2 ( y ) ) e − k ( ⋅ ) .$

Then, by simple calculations, $T 1 ( T 1 y)= T 1 y$, $T 1 (( T 2 y)w)=0$, $T 2 ( T 1 y)=0$, and $T 2 (( T 2 y)w)= T 2 y$. Define a bounded linear operator $Q:Y→Y$ by

$(Qy)(t):=( T 1 y)(t)+( T 2 y)(t)w(t),t∈(0,∞).$
(2)

Then $Q 2 y=Qy$, i.e., $Q:Y→Y$ is a linear projector. Since $△≠0$, $kerQ=ImL$, and $Y=ImL⊕ImQ$.

Define a continuous projector $P:X→X$ by

$Px:=x(0)+ ( c x ′ ) (0)w(⋅).$

Clearly, $ImP=kerL$, and consequently $X=kerL⊕kerP$. Define an operator $K P :ImL→domL∩kerP$ by

$( K P y)(t):= ∫ 0 t 1 c ( s ) ∫ 0 s y(τ)dτds,t∈[0,∞).$

Then $K P$ is the inverse operator $L | dom L ∩ ker P$, and it satisfies

$∥ K P y ∥ X ≤2 ∥ y ∥ Y .$
(3)

Let a nonlinear operator $N:X→Y$ be defined by $(Nx)(t):=f(t,x(t), x ′ (t))$, $t∈[0,∞)$. Then problem (1) is equivalent to $Lx=Nx$, $x∈domL$.

From now on, we consider the case $1 c ∉Y$. The case $1 c ∈Y$ can be dealt in a similar manner.

### Lemma 3.3

Let$1 c ∉Y$, and assume that (H1)-(H3) hold. Assume that Ω is a bounded open subset of X such that$domL∩ Ω ¯ ≠∅$. Then N is L-compact on$Ω ¯$.

### Proof

Since Ω is a bounded open subset of X, there exists a constant $r>0$ such that $∥ x ∥ X ≤r$ for any $x∈ Ω ¯$. For any $x∈ Ω ¯$ and for almost all $t∈[0,∞)$, by (H2),

$|(Nx)(t)|=|f ( t , x ( t ) , x ′ ( t ) ) |≤ ( ( 1 + w ( t ) ) α ( t ) + β c ( t ) ) r+γ(t)=: N r (t).$

Then $N r ∈Y$ and $∥ N x ∥ Y ≤ ∥ N r ∥ Y$ for all $x∈ Ω ¯$. Thus $N( Ω ¯ )$ is bounded in Y.

For any $x∈ Ω ¯$,

$| Q 1 (Nx)|≤ ( ∑ i = 1 m | α i | ) ∥ N x ∥ Y ≤ ( ∑ i = 1 m | α i | ) ∥ N r ∥ Y ,| Q 2 (Nx)|≤ ∥ N r ∥ Y .$
(4)

It follows from (2) and (4) that, for almost all $t∈[0,∞)$,

$| ( Q N x ) ( t ) | ≤ | ( T 1 N x ) ( t ) | + | ( T 2 N x ) ( t ) | w ( t ) ≤ 1 | △ | ( ( | a 11 | ∑ i = 1 m | α i | + | a 12 | ) + ( | a 21 | ∑ i = 1 m | α i | + | a 22 | ) w ( t ) ) e − k ( t ) ∥ N r ∥ Y = : Q r ( t ) .$

Then $Q r ∈Y$ and $∥ Q N x ∥ Y ≤ ∥ Q r ∥ Y$ for all $x∈ Ω ¯$. Thus, $QN( Ω ¯ )$ is bounded in Y.

Next we will prove that $K P (I−Q)N( Ω ¯ )$ is a relatively compact set in X. For $x∈ Ω ¯$, by (3),

$∥ K P ( I − Q ) N x ∥ X ≤2 ( ∥ N x ∥ Y + ∥ Q N x ∥ Y ) ≤2 ( ∥ N r ∥ Y + ∥ Q r ∥ Y ) .$

Thus $K P (I−Q)N( Ω ¯ )$ is bounded in X.

Let $T>0$ be given. For any $x∈ Ω ¯$ and let $t 1 , t 2 ∈[0,T]$ with $t 1 < t 2$,

$| ( K P ( I − Q ) N x ) ( t 1 ) 1 + w ( t 1 ) − ( K P ( I − Q ) N x ) ( t 2 ) 1 + w ( t 2 ) | ≤ | w ( t 2 ) − w ( t 1 ) ( 1 + w ( t 1 ) ) ( 1 + w ( t 2 ) ) | | ∫ 0 t 1 1 c ( s ) ∫ 0 s ( I − Q ) N x ( τ ) d τ d s | + | ∫ t 1 t 2 1 c ( s ) ∫ 0 s ( I − Q ) N x ( τ ) d τ d s 1 + w ( t 2 ) | ≤ 2 | w ( t 2 ) − w ( t 1 ) | ( ∥ N r ∥ Y + ∥ Q r ∥ Y )$

and

$| ( c ( K P ( I − Q ) N x ) ′ ) ( t 1 )− ( c ( K P ( I − Q ) N x ) ′ ) ( t 2 )|≤ ∫ t 1 t 2 ( N r ( τ ) + Q r ( τ ) ) dτ,$

which imply that

${ K P ( I − Q ) N x 1 + w : x ∈ Ω ¯ } and { c ( K P ( I − Q ) N x ) ′ : x ∈ Ω ¯ }$

are equicontinuous on $[0,T]$.

For any $x∈ Ω ¯$, by L’Hôspital’s rule,

$lim t → ∞ ( K P ( I − Q ) N x ) ( t ) 1 + w ( t ) = lim t → ∞ ∫ 0 t ( ( I − Q ) N x ) ( s ) d s = ∫ 0 ∞ ( ( I − Q ) N x ) ( s ) d s − lim t → ∞ ∫ t ∞ ( ( I − Q ) N x ) ( s ) d s$

and

$| ( c ( K P ( I − Q ) N x ) ′ ) (t)− ∫ 0 ∞ ( ( I − Q ) N x ) (τ)dτ|≤ ∫ t ∞ | ( ( I − Q ) N x ) (τ)|dτ.$

Since $|(I−Q)Nx|≤ N r + Q r$ for all $x∈ Ω ¯$,

$( K P ( I − Q ) N x ) ( t ) 1 + w ( t ) → ∫ 0 ∞ ( ( I − Q ) N x ) (s)ds$

and

$( c ( K P ( I − Q ) N x ) ′ ) (t)→ ∫ 0 ∞ ( ( I − Q ) N x ) (τ)dτ$

uniformly on $Ω ¯$ as $t→∞$. In view of Theorem 2.2, $K P (I−Q)N( Ω ¯ )$ is a relatively compact set in X, and thus N is L-compact on $Ω ¯$. □

The following theorem is the main result in this paper.

### Theorem 3.4

Let$1 c ∉Y$, and assume that (H1)-(H3) hold. Assume also that the following hold:

(H4) there exist positive constants A and B such that if$|x(t)|>A$for every$t∈[0,B]$or$|(c x ′ )(t)|>A$for every$t∈[0,∞)$, then$QNx≠0$;

(H5) there exists a positive constant C such that if$|a|>C$or$|b|>C$, then$QN(a+bw)≠0$, and only one of the following inequalities is satisfied:

$( i ) a Q 1 ( N ( a + b w ) ) + b Q 2 ( N ( a + b w ) ) ≥ 0 , ( ii ) a Q 1 ( N ( a + b w ) ) + b Q 2 ( N ( a + b w ) ) ≤ 0 .$

If α and β satisfy

$∥ ( 1 + w ) α ∥ Y + ∥ β c ∥ Y < 1 4 + w ( B ) ,$
(5)

then problem (1) has at least one solution in X.

### Proof

We divide the proof into four steps.

Step 1. Let

Then $Ω 1$ is bounded. In fact, $x∈ Ω 1$ means $λ∈(0,1)$ and $Nx∈ImL$. Thus, by Lemma 3.2, $Q 1 (Nx)= Q 2 (Nx)=0$. By (H4), there exist $t 0 ∈[0,B]$ and $t 1 ∈[0,∞)$ such that $|x( t 0 )|≤A$, $|(c x ′ )( t 1 )|≤A$. Then

$| ( c x ′ ) (0)|=| ( c x ′ ) ( t 1 )− ∫ 0 t 1 ( c x ′ ) ′ (s)ds|≤A+ ∥ N x ∥ Y .$

On the other hand,

$| x ( 0 ) | = | x ( t 0 ) − ∫ 0 t 0 1 c ( s ) [ ( c x ′ ) ( t 1 ) + ∫ t 1 s ( c x ′ ) ′ ( τ ) d τ ] d s | ≤ | x ( t 0 ) | + ( | ( c x ′ ) ( t 1 ) | + ∥ N x ∥ Y ) w ( B ) ≤ A ( 1 + w ( B ) ) + w ( B ) ∥ N x ∥ Y .$

Thus,

$∥ P x ∥ X ≤|x(0)|+2| ( c x ′ ) (0)|≤A ( 3 + w ( B ) ) + ( 2 + w ( B ) ) ∥ N x ∥ Y .$

Since $kerL=ImP$, $LPx=0$ for all $x∈X$, it follows from (3) and (H2) that

$∥ x ∥ X = ∥ P x + ( I − P ) x ∥ X = ∥ P x + K P L ( I − P ) x ∥ X ≤ ∥ P x ∥ X + 2 ∥ L x ∥ Y ≤ ∥ P x ∥ X + 2 ∥ N x ∥ Y ≤ A ( 3 + w ( B ) ) + ( 4 + w ( B ) ) ∥ N x ∥ Y ≤ A ( 3 + w ( B ) ) + ( 4 + w ( B ) ) ( ( ∥ ( 1 + w ) α ∥ Y + ∥ β c ∥ Y ) ∥ x ∥ X + ∥ γ ∥ Y ) ,$

and, by (5),

$∥ x ∥ X ≤ A ( 3 + w ( B ) ) + ( 4 + w ( B ) ) ∥ γ ∥ Y 1 − ( 4 + w ( B ) ) ( ∥ ( 1 + w ) α ∥ Y + ∥ β c ∥ Y ) ,$

which implies that $Ω 1$ is bounded.

Step 2. Set

$Ω 2 ={x∈kerL:Nx∈ImL}.$

Then $Ω 2$ is bounded. In fact, $x∈ Ω 2$ implies $x=a+bw$ and $Q 1 (Nx)= Q 2 (Nx)=0$. By (H5), we obtain $|a|≤C$ and $|b|≤C$. Thus $Ω 2$ is bounded.

Step 3. Define an isomorphism $J:kerL→ImQ$ by

$J ( a + b w ( ⋅ ) ) = 1 △ ( a 11 a + a 12 b + ( a 21 a + a 22 b ) w ( ⋅ ) ) e − k ( ⋅ ) .$

Assume first that (i) in (H5) holds. Let

Then $Ω 3$ is bounded. Indeed, $x∈ Ω 3$ means that there exist constants $a,b∈(−∞,∞)$ and $λ∈[0,1]$ such that $x=a+bw$ and $λJx+(1−λ)QNx=0$. If $λ=0$, then $QNx=0$. It follows from $△≠0$ that $Q 1 (Nx)= Q 2 (Nx)=0$. By (H5), we obtain $|a|≤C$ and $|b|≤C$. If $λ=1$, clearly $a=b=0$. For $λ∈(0,1)$, by the facts that $−λJx=(1−λ)QNx$ and $△≠0$, it follows that $−λa=(1−λ) Q 1 (Nx)$ and $−λb=(1−λ) Q 2 (Nx)$. If $|a|>C$ or $|b|>C$, by (i) in (H5),

$−λ ( a 2 + b 2 ) =(1−λ) ( a Q 1 N ( a + b w ) + b Q 2 N ( a + b w ) ) ≥0,$

which is a contradiction. Thus $Ω 3$ is bounded in X. On the other hand, if (ii) in (H5) holds, taking

one can show that $Ω 3$ is bounded in a similar manner.

Step 4. Take an open bounded set $Ω⊃ ⋃ i = 1 3 Ω ¯ i ∪{0}$. By Step 1 and Step 2, in view of Theorem 2.1, we only need to prove that $deg(QN | ker L ,Ω∩kerL,0)≠0$ in order to show that $Lx=Nx$ has at least one solution in $domL∩ Ω ¯$. Let

$H(x,λ)=±λJx+(1−λ)QNx.$

By Step 3, $H(x,λ)≠0$ for all $(x,λ)∈(kerL∩∂Ω)×[0,1]$. Thus, by the homotopy invariance property of degree,

$deg ( Q N | ker L , Ω ∩ ker L , 0 ) = deg ( H ( ⋅ , 0 ) , Ω ∩ ker L , 0 ) = deg ( H ( ⋅ , 1 ) , Ω ∩ ker L , 0 ) = deg ( ± J , Ω ∩ ker L , 0 ) ≠ 0 .$

By Theorem 2.1, $Lx=Nx$ has at least one solution in $domL∩ Ω ¯$, i.e., problem (1) has at least one solution in X. □

In the case that $1 c ∈Y$, $w∈ L ∞ (0,∞)$, and using the norm $∥⋅∥$ on X in Remark 3.1(2), we have a similar result to Theorem 3.4. We omit the proof.

### Theorem 3.5

Let$1 c ∈Y$, and assume that (H1)-(H5) hold. If α and β satisfy

$∥ α ∥ Y + ∥ β c ∥ Y < 1 2 + 3 ∥ w ∥ ∞ ,$

then problem (1) has at least one solution in X.

## 4 Example

Consider the following second-order nonlinear differential equation:

(6)

where $f(t,u,v)= α 1 (t)u+ β 1 (t)v+ γ 1 (t)$ for $(t,u,v)∈[0,∞)×(−∞,∞)×(−∞,∞)$, $γ 1 ∈Y$ satisfies $∫ 0 1 (1− t 2 ) γ 1 (t)dt= ∫ 1 ∞ γ 1 (t)dt=0$,

$α 1 ( t ) = { K e − t , t ∈ [ 0 , 1 ] , 0 , t ∈ ( 1 , ∞ ) , β 1 ( t ) = { 0 , t ∈ [ 0 , 1 ] , K ( e − 4 ) e − t , t ∈ ( 1 , ∞ ) , c ( t ) = { t ρ , t ∈ [ 0 , 1 ] , 1 , t ∈ ( 1 , ∞ ) , g ( t ) = { 2 t , t ∈ [ 0 , 1 ] , 0 , t ∈ ( 1 , ∞ ) , w ( t ) = { 1 1 − ρ t 1 − ρ , t ∈ [ 0 , 1 ] , t + ρ 1 − ρ , t ∈ ( 1 , ∞ ) .$

We also assume that the constants K and ρ satisfy

$0

and $ρ∈[0,1)$. Then $1/c∉Y$, and (H1)-(H2) hold for $m= α 1 = ξ 1 =1$, $α= α 1$, $β=| β 1 |$, and $γ=| γ 1 |$. For $y∈Y$,

$Q 1 (y)= ∫ 1 ∞ y(s)dsand Q 2 (y)= ∫ 0 1 ( 1 − s 2 ) y(s)ds.$

Taking $k(t)=t$, then

$a 11 = 1 1 − ρ ∫ 0 1 ( 1 − t 2 ) t 1 − ρ e − t d t , a 12 = ρ − 2 1 − ρ e − 1 , a 21 = 1 − 4 e − 1 , a 22 = e − 1$

and

$△ = a 11 a 22 − a 12 a 21 = 1 ( 1 − ρ ) e ∫ 0 1 ( 1 − t 2 ) t 1 − ρ e − t d t + 2 − ρ 1 − ρ e − 1 ( 1 − 4 e − 1 ) = e − 4 e 2 + 1 ( 1 − ρ ) e ( e − 4 e + ∫ 0 1 ( 1 − t 2 ) t 1 − ρ e − t d t ) .$

Since, for any $ρ∈[0,1)$,

$e − 4 e + ∫ 0 1 ( 1 − t 2 ) t 1 − ρ e − t dt≤ e − 4 e + ∫ 0 1 ( 1 − t 2 ) e − t dt=0,$

and $△<0$. Thus (H3) holds for any $ρ∈[0,1)$.

Take $B=1$ and let $A>0$ be arbitrary. Then (H4) holds. In fact, if $|x(t)|>A$ for $t∈[0,1]$, then

$Q 2 (Nx)=K ∫ 0 1 ( 1 − t 2 ) e − t x(t)dt≠0.$

If $|(c x ′ )(t)|>A$ for $t∈[0,∞)$, then $| x ′ (t)|>A$ for $t∈[1,∞)$, and

$Q 1 (Nx)=K(e−4) ∫ 1 ∞ e − t x ′ (t)dt≠0.$

Since

$Q 1 ( N ( a + b w ) ) =K ( 1 − 4 e − 1 ) b$

and

$Q 2 ( N ( a + b w ) ) =K ( ( 4 e − 1 − 1 ) a + b 1 − ρ ∫ 0 1 ( 1 − t 2 ) e − t t 1 − ρ d t )$

for $(a,b)≠(0,0)$,

$| Q 1 ( N ( a + b w ) ) |+| Q 2 ( N ( a + b w ) ) |≠0,$

i.e., $QN(a+bw)≠0$ for $(a,b)≠(0,0)$, and

$a Q 1 ( N ( a + b w ) ) +b Q 2 ( N ( a + b w ) ) = b 2 K 1 − ρ ∫ 0 1 ( 1 − t 2 ) e − t t 1 − ρ dt≥0$

for all $(a,b)∈(−∞,∞)×(−∞,∞)$. Therefore (H5) holds for any $C>0$. Since $0, (5) is satisfied, and thus there exists at least one solution to problem (6) in view of Theorem 3.4.

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## Acknowledgements

This study was supported by the Research Fund Program of Research Institute for Basic Sciences, Pusan National University, Korea, 2014, Project No. RIBS-PNU-2014-101.

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Correspondence to Eun Kyoung Lee.

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All authors contributed equally to the manuscript and read and approved the final draft.

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