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# The existence of solutions for p-Laplacian boundary value problems at resonance on the half-line

Boundary Value Problems20152015:179

https://doi.org/10.1186/s13661-015-0439-9

• Accepted: 14 September 2015
• Published:

## Abstract

By using the extension of the continuation theorem of Ge and Ren and constructing suitable Banach spaces and operators, we investigate the existence of solutions for a p-Laplacian boundary value problem with integral boundary condition at resonance on the half-line.

## Keywords

• continuation theorem
• integral boundary condition
• resonance
• p-Laplacian
• boundary value problem

• 34B40

## 1 Introduction

A boundary value problem is said to be at resonance one if the corresponding homogeneous boundary value problem has non-trivial solutions. Mawhin’s continuation theorem  is an effective tool to investigate the boundary value problems at resonance with linear or semilinear differential operators (see  and the references cited therein). Boundary value problems with p-Laplacian have been widely studied owing to their importance in theory and application of mathematics and physics (see ). But the p-Laplacian boundary value problems at resonance cannot be solved by Mawhin’s continuation theorem. In order to solve these problems, Ge and Ren extended Mawhin’s continuation theorem and used it to study boundary value problems with p-Laplacian . In their new theorem, two projectors (Definition 2.2) P and Q must be constructed. But it is difficult to give the projector Q in many boundary value problems with p-Laplacian. In , the author extended the theorem in  and studied the problem
$$\left \{ \textstyle\begin{array}{@{}l} (\varphi_{p}(u'') )'(t)=f (t,u(t),u'(t),u''(t) ),\quad t\in [0,1],\\ u''(0)=0,\qquad u'(0)=\int_{0}^{1}g(t)u'(t)\,dt,\qquad u'(1)=\int_{0}^{1}h(t)u'(t)\,dt \end{array}\displaystyle \right .$$
in finite interval, where Q is not a projector but satisfies suitable conditions, $$\varphi_{p}(s)=|s|^{p-2}s$$, $$p>1$$, $$\int_{0}^{1}g(t)\,dt=1$$, $$\int_{0}^{1}h(t)\,dt=1$$.
Boundary value problems on the half-line arise in various applications such as in the study of the unsteady flow of gas through a semi-infinite porous medium, in analyzing the heat transfer in radial flow between circular disks, in the study of plasma physics, in an analysis of the mass transfer on a rotating disk in a non-Newtonian fluid, etc. . In , using the continuation theorem of Ge and Ren , the author investigated the existence of solutions for the problem
$$\left \{ \textstyle\begin{array}{@{}l} (\varphi_{p}(u'))'+f(t,u,u')=0,\quad 0< t< +\infty,\\ u(0)=0, \qquad\varphi_{p}(u'(+\infty))=\sum_{i=1}^{n}\alpha_{i}\varphi _{p}(u'(\xi_{i})) \end{array}\displaystyle \right .$$
on the half-line, where Q is a projector, $$\alpha_{i}>0$$, $$i=1,2,\ldots,n$$, $$\sum_{i=1}^{n}\alpha_{i}=1$$.
In this paper, we study the boundary value problem
$$\left \{ \textstyle\begin{array}{@{}l} (\varphi_{p}(u') )'(t)=\psi(t)f (t,u(t),u'(t) ),\quad t\in [0,+\infty),\\ u'(+\infty)=0,\qquad u(0)=\int_{0}^{+\infty}h(t)u(t)\,dt \end{array}\displaystyle \right .$$
(1.1)
in infinite interval, where Q is not a projector, $$\varphi_{p}(s)=|s|^{p-2}s$$, $$p>1$$. To the best of our knowledge, this is the first paper to study the boundary value problems at resonance on the half-line where the operator Q is not a projector.
In this paper, we will always suppose that the following conditions hold.
(H1):

$$\int_{0}^{+\infty}h(t)\,dt=1$$, $$th(t)\in L^{1}[0,+\infty)$$, $$\psi(t)\in L^{1}[0,+\infty)\cap C[0,+\infty)$$, $$h(t)\geq0$$, $$\psi(t)> 0$$, $$t\in[0,+\infty)$$.

(H2):

$$f(t,u,v)$$ is continuous in $$[0,\infty)\times\mathbb{R}^{2}$$. For any $$r>0$$, there exists a constant $$M_{r}>0$$ such that if $$\frac{|u|}{1+t}\leq r$$, $$|v|\leq r$$, $$t\in [0,+\infty)$$ then $$|f(t,u,v)|\leq M_{r}$$, and for any $$\varepsilon>0$$ there exists $$\delta>0$$ such that $$|f(t,u_{1},v_{1})-f(t,u_{2},v_{2})|<\varepsilon$$ for $$t\in [0,+\infty)$$, $$u_{i}, v_{i}\in\mathbb{R}$$, $$i=1,2$$, satisfying $$\frac{|u_{1}-u_{2}|}{1+t}<\delta$$, $$|v_{1}-v_{2}|<\delta$$ and $$\frac{|u_{i}|}{1+t}\leq r$$, $$|v_{i}|\leq r$$.

The paper is organized as follows. The first section provides a short overview of the problem. Section 2 recalls some preliminary facts. Section 3 contains the main result of the paper.

## 2 Preliminaries

### Definition 2.1

()

Let X and Z be two Banach spaces with norms $$\|\cdot\|_{X}$$, $$\|\cdot\|_{Z}$$, respectively. An operator $$M: X\cap \operatorname{dom}M\rightarrow Z$$ is said to be quasi-linear if
1. (i)

$$\operatorname{Im}M:=M(X \cap \operatorname{dom}M)$$ is a closed subset of Z,

2. (ii)

$$\operatorname{Ker}M:=\{x\in X \cap \operatorname{dom}M: Mx=0\}$$ is linearly homeomorphic to $$\mathbb{R}^{n}$$, $$n<\infty$$,

where domM denotes the domain of the operator M.

In this paper, an operator $$T:X\rightarrow Z$$ is said to be bounded if $$T(V)\subset Z$$ is bounded for any bounded subset $$V\subset X$$.

### Definition 2.2

P is a projector if $$P:Y\rightarrow Y$$ is linear and $$P^{2}x=Px$$, where Y is a vector space.

Let $$X_{1}=\operatorname{Ker}M$$, $$P:X\rightarrow X_{1}$$ be a projector and $$X_{2}$$ be the complement space of $$X_{1}$$ in X with $$X=X_{1}\oplus X_{2}$$. Let $$\Omega\subset X$$ be an open and bounded set with the origin $$\theta\in\Omega$$.

### Definition 2.3

()

Suppose that $$N_{\lambda}:\overline{\Omega}\rightarrow Z$$, $$\lambda\in[0,1]$$ is a continuous and bounded operator. Denote $$N_{1}$$ by N. Let $$\Sigma{_{\lambda}}=\{x\in\overline{\Omega}:Mx=N_{\lambda}x\}$$. $$N_{\lambda}$$ is said to be M-quasi-compact in Ω̅ if there exists a vector subspace $$Z_{1}$$ of Z satisfying $$\operatorname{dim} Z_{1}=\operatorname{dim} X_{1}$$ and two operators Q and R such that the following conditions hold:
1. (a)

$$\operatorname{Ker}Q=\operatorname{Im}M$$,

2. (b)

$$QN_{\lambda} x=\theta$$, $$\lambda\in(0,1)\Leftrightarrow QNx=\theta$$,

3. (c)

$$R(\cdot,0)$$ is the zero operator and $$R(\cdot,\lambda)|_{\Sigma{_{\lambda}}}=(I-P)|_{\Sigma{_{\lambda}}}$$,

4. (d)

$$M[P+R(\cdot,\lambda)]=(I-Q)N_{\lambda}$$,

where $$Q:Z\rightarrow Z_{1}$$ is continuous, bounded with $$Q(I-Q)=0$$, $$QZ=Z_{1}$$ and $$R:\overline{\Omega}\times[0,1]\rightarrow X_{2}$$ is continuous and compact with $$Pu+R(u,\lambda)\in \operatorname{dom} M$$, $$u\in\overline{\Omega}$$, $$\lambda\in[0,1]$$.

### Theorem 2.1

()

Let X and Z be two Banach spaces with the norms $$\|\cdot\|_{X}$$, $$\|\cdot\|_{Z}$$, respectively, and $$\Omega\subset X$$ be an open and bounded nonempty set. Suppose that
$$M:X\cap \operatorname{dom}M\rightarrow Z$$
is a quasi-linear operator and that $$N_{\lambda}:\overline{\Omega}\rightarrow Z$$, $$\lambda\in[0,1]$$ is M-quasi-compact. In addition, if the following conditions hold:
(C1):

$$Mx\neq N_{\lambda}x$$, $$\forall x\in\partial\Omega\cap \operatorname{dom}M$$, $$\lambda\in(0,1)$$,

(C2):

$$\operatorname{deg}\{JQN, \Omega\cap \operatorname{Ker}M, 0\}\neq0$$,

then the abstract equation $$Mx=Nx$$ has at least one solution in $$\operatorname{dom}M\cap\overline{\Omega}$$, where $$N=N_{1}$$, $$J:\operatorname{Im}Q\rightarrow \operatorname{Ker}M$$ is a homeomorphism with $$J(\theta)=\theta$$, $$deg$$ is the Brouwer degree.

### Remark

In the proof of Theorem 2.1 in , the continuity of the operator M is not needed. And domM may not be a linear space. But the operators P, R satisfy $$Pu+R(u,\lambda)\in \operatorname{dom}M$$, $$u\in\overline{\Omega}$$, $$\lambda\in[0,1]$$.

### Lemma 2.1

()

Let $$\varphi_{p} : \mathbb{R}\rightarrow \mathbb{R}$$ be a function given by the formula $$\varphi_{p}(s) = |s|^{p-2}s$$, where $$p > 1$$. Then, for any $$u, v \geq0$$, we have
1. (1)

$$\varphi_{p}(u+v)\leq\varphi_{p}(u)+\varphi_{p}(v)$$, $$1< p\leq2$$.

2. (2)

$$\varphi_{p}(u+v)\leq2^{p-2}(\varphi_{p}(u)+\varphi_{p}(v))$$, $$p\geq 2$$.

## 3 Main results

In the following, we will always assume that q satisfies $$1/p+1/q=1$$.

Let $$X= \{u\in C^{1}[0,+\infty):u'(+\infty)=0, \sup_{t\in [0,+\infty)}\frac{|u(t)|}{1+t}<+\infty \}$$ be endowed with the following norm $$\|u\|_{X}=\max \{\|u'\|_{\infty}, \Vert \frac {u(t)}{1+t}\Vert _{\infty}\}$$, $$Y=\{y\in C[0,+\infty):\sup_{t\in [0,+\infty)}|y(t)|<\infty\}$$ be endowed with the following norm $$\|y\|_{Y}=\sup_{t\in[0,+\infty)}|y(t)|:=\|y\|_{\infty}$$. Take $$Z=\{\psi y:y\in Y\}\times\mathbb{R}$$, with norm $$\|(\psi y,c)\|_{Z}=\max\{\|y\|_{\infty},|c|\}$$. We know that $$(X,\|\cdot\|_{X})$$ and $$(Z,\|\cdot\|_{Z})$$ are Banach spaces.

Define operator $$T:Y\rightarrow\mathbb{R}$$ by $$Ty=c$$, for $$y\in Y$$, where c satisfies
$$\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl(y(r)-c\bigr)\,dr \biggr)\,ds\,dt=0.$$
(3.1)

The following lemma shows that the operator T is well defined.

### Lemma 3.1

For $$y\in Y$$, there is only one constant $$c\in \mathbb{R}$$ such that $$Ty=c$$ with $$|c|\leq\|y\|_{\infty}$$, $$T:Y\rightarrow\mathbb{R}$$ is continuous and $$T(ky)=kT(y)$$, $$k\in \mathbb{R}$$.

### Proof

For $$y\in Y$$, let
$$F(c)=\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl(y(r)-c\bigr)\,dr \biggr)\,ds\,dt.$$
Obviously, F is continuous and strictly decreasing in $$\mathbb{R}$$. If y is a constant, the results hold, clearly. Assume y is not a constant. Take $$a=\inf_{t\in[0,+\infty)}y(t)$$, $$b=\sup_{t\in[0,+\infty)}y(t)$$. It is easy to see that $$F(a)> 0$$, $$F(b)< 0$$. So, there exists a unique constant $$c\in(a,b)$$ such that $$F(c)=0$$, i.e., there is only one constant $$c\in\mathbb{R}$$ such that $$Ty=c$$ with $$|c|\leq \|y\|_{\infty}$$.
For $$y_{1}, y_{2}\in Y$$, assume $$Ty_{1}=a$$, $$Ty_{2}=b$$. By $$h(t)\geq0$$, $$\psi(t)> 0$$, $$\int_{0}^{+\infty}h(t)\,dt=1$$ and $$\varphi_{q}$$ being strictly increasing, we obtain that if $$b-a> \sup_{t\in[0,+\infty)}(y_{2}(t)-y_{1}(t))$$, then
\begin{aligned} 0&=\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl(y_{2}(r)-b\bigr)\,dr \biggr)\,ds\,dt \\ &=\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl[\bigl(y_{1}(r)-a\bigr)+\bigl(y_{2}(r)-y_{1}(r)-(b-a) \bigr)\bigr]\,dr \biggr)\,ds\,dt \\ &< \int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl(y_{1}(r)-a\bigr)\,dr \biggr)\,ds\,dt=0, \end{aligned}
a contradiction. On the other hand, if $$b-a< \inf_{t\in[0,+\infty)}(y_{2}(t)-y_{1}(t))$$, then
\begin{aligned} 0&=\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl(y_{2}(r)-b\bigr)\,dr \biggr)\,ds\,dt \\ &=\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl[\bigl(y_{1}(r)-a\bigr)+\bigl(y_{2}(r)-y_{1}(r)-(b-a) \bigr)\bigr]\,dr \biggr)\,ds\,dt \\ &>\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r) \bigl(y_{1}(r)-a\bigr)\,dr \biggr)\,ds\,dt=0, \end{aligned}
a contradiction, too. So, we have $$\inf_{t\in[0,+\infty)}(y_{2}(t)-y_{1}(t))\leq b-a\leq \sup_{t\in[0,+\infty)}(y_{2}(t)-y_{1}(t))$$, i.e., $$| b-a|\leq \|y_{2}-y_{1}\|_{\infty}$$. So, $$T:Y\rightarrow\mathbb{R}$$ is continuous. Obviously, $$T(ky)=kT(y)$$, $$k\in\mathbb{R}$$. The proof is completed. □
Define operators $$M:X\cap \operatorname{dom}M\rightarrow Z$$, $$N_{\lambda}:X\rightarrow Z$$ as follows:
$$Mu(t)= \biggl[ \bigl(\varphi_{p}\bigl(u'\bigr) \bigr)'(t), T \biggl(\frac{ (\varphi_{p}(u') )'(t)}{\psi(t)} \biggr) \biggr],\qquad N_{\lambda}u(t)= \bigl[\lambda \psi(t)f \bigl(t,u(t),u'(t) \bigr), 0 \bigr],$$
where $$\operatorname{dom}M= \{u\in X |\frac{(\varphi_{p}(u'))'}{ \psi(t)}\in Y \}$$.

### Definition 3.1

u is a solution of (1.1) if $$u\in \operatorname{dom}M$$ satisfies (1.1).

It is clear that $$u\in \operatorname{dom}M$$ is a solution of (1.1) if and only if it satisfies $$Mu=Nu$$, where $$N=N_{1}$$.

### Lemma 3.2

M is a quasi-linear operator.

### Proof

It is easy to get that $$\operatorname{Ker}M=\{c | c\in \mathbb{R}\}:=X_{1}$$.

For $$u\in X\cap \operatorname{dom}M$$, if $$Mu=(\psi y,c)$$, then c satisfies (3.1) with y. On the other hand, if $$y\in Y$$, $$Ty=c$$, take
$$u(t)=-\int_{0}^{t}\varphi_{q} \biggl( \int_{s}^{+\infty}\psi(r)y(r)\,dr \biggr)\,ds.$$
By a simple calculation, we get $$u\in X\cap \operatorname{dom}M$$ and $$Mu=(\psi y,c)$$. Thus
$$\operatorname{Im}M=\bigl\{ (\psi y,Ty) | y\in Y\bigr\} = \bigl\{ (\psi y,c) | y\in Y, c \mbox{ satisfying } (3.1) \mbox{ with } y \bigr\} .$$
By the continuity of T, we get that $$\operatorname{Im}M\subset Z$$ is closed. So, M is quasi-linear. The proof is completed. □
Take a projector $$P: X\rightarrow X_{1}$$ and an operator $$Q: Z\rightarrow Z_{1}$$ as follows:
$$(Pu) (t)=u(0),\qquad Q(\psi y, c)=(0,c-Ty),$$
where $$Z_{1}=\{(0,c) | c\in\mathbb{R}\}$$. Obviously, $$QZ=Z_{1}$$ and $$\operatorname{dim} Z_{1}=\operatorname{dim} X_{1}$$.
Define an operator R as
$$R(u,\lambda) (t)= -\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty }\lambda \psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\,ds, \quad(u,\lambda) \in X\times[0,1].$$

### Lemma 3.3

()

$$V\subset X$$ is relatively compact if $$\{\frac{u(t)}{1+t} | u\in V \}$$ and $$\{u'(t)| u\in V\}$$ are both bounded, equicontinuous on any compact intervals of $$[0,+\infty)$$ and equiconvergent at infinity.

### Lemma 3.4

$$R:\overline{\Omega}\times[0,1]\rightarrow X_{2}$$ is continuous and compact, $$Pu+R(u,\lambda)\in \operatorname{dom} M$$, $$u\in\overline{\Omega}$$, $$\lambda\in[0,1]$$, where $$X_{2}=\{u\in X:u(0)=0\}$$, $$\Omega\subset X$$ is an open bounded set.

### Proof

Firstly, we prove that $$R:\overline{\Omega}\times [0,1]\rightarrow X_{2}$$ and $$Pu+R(u,\lambda)\in \operatorname{dom} M$$, $$u\in\overline{\Omega}$$, $$\lambda\in[0,1]$$.

Obviously, $$R(u,\lambda)(t)\in C^{1}[0,+\infty)$$, $$R(u,\lambda)'(+\infty)=-\lim_{t\rightarrow+\infty}\varphi _{q}(\int_{t}^{+\infty}\lambda\psi(s) f(s,u(s), u'(s))\,ds)=0$$. By (H2), we get $$\frac{|R(u,\lambda)(t)|}{1+t}\leq \varphi_{q}(\|\psi\|_{1}M_{\|u\|_{X}})<+\infty$$, $$u\in X$$. Therefore, $$R(u,\lambda)\in X$$. It is clear that $$R(u,\lambda)(0)=0$$. Thus $$R(u,\lambda)\in X_{2}$$. Clearly, $$R(u,\lambda)+Pu\in X$$. It follows from $$(\varphi_{p}(R(u,\lambda)(t)+Pu(t))')'=\lambda\psi(t) f(t,u(t),u'(t))$$ and (H2) that $$\frac{(\varphi_{p}(R(u,\lambda)(t)+Pu(t))')'}{\psi (t)}=\lambda f(t,u(t),u'(t))\in Y$$. So, $$R(u,\lambda)+Pu\in \operatorname{dom}M$$.

Secondly, we show that R is continuous.

Since Ω is bounded, there exists a constant $$r>0$$ such that $$\|u\|_{X}\leq r$$, $$u\in\overline{\Omega}$$. By (H2), there exists a constant $$M_{r}>0$$ such that $$|f(t,u(t),u'(t))|\leq M_{r}$$, $$u\in \overline{\Omega}$$, $$t\in[0,+\infty)$$. So, we get
$$\biggl\vert \int_{t}^{+\infty}\psi(s) f \bigl(s,u(s),u'(s)\bigr)\,ds\biggr\vert \leq \|\psi \|_{1}M_{r},\quad t\in[0,+\infty), u\in\overline{\Omega}.$$
By the uniform continuity of $$\varphi_{q}(x)$$ in $$[-\|\psi\|_{1}M_{r}, \max\{1,\|\psi\|_{1}M_{r}\}]$$, we obtain that for any $$\varepsilon>0$$, there exists a constant $$\delta_{\varepsilon}>0$$ such that
$$\bigl|\varphi_{q}(x_{1})-\varphi_{q}(x_{2})\bigr|< \varepsilon, \quad|x_{1}-x_{2}|\leq\delta _{\varepsilon}, x_{1}, x_{2}\in\bigl[-\|\psi\|_{1}M_{r}, \max\bigl\{ 1,\|\psi\|_{1}M_{r}\bigr\} \bigr].$$
For $$\alpha= \frac{\delta_{\varepsilon}}{\|\psi\|_{1}}$$, by (H2), there exists a constant $$\delta_{\alpha}>0$$ such that if $$u,v\in\overline{\Omega}$$, $$\|u-v\|_{X}<\delta_{\alpha}$$, then $$|f(t,u(t),u'(t))-f(t,v(t),v'(t))|<\alpha$$, $$t\in[0,\infty)$$. So, we have
\begin{aligned} &\biggl\vert \int_{t}^{+\infty}\psi(s) f \bigl(s,u(s),u'(s)\bigr)\,ds-\int_{t}^{+\infty} \psi(s) f\bigl(s,v(s),v'(s)\bigr)\,ds\biggr\vert \\ &\quad\leq \int_{t}^{+\infty}\psi(s)\bigl|f \bigl(s,u(s),u'(s)\bigr)-f\bigl(s,v(s),v'(s)\bigr)\bigr|\,ds \leq \delta_{\varepsilon},\quad \|u-v\|_{X}< \delta_{\alpha}. \end{aligned}
Take $$\delta=\min\{\delta_{\varepsilon}, \delta_{\alpha}\}$$. For $$u, v\in\overline{\Omega}$$, $$\lambda, \mu\in[0,1]$$, if $$\|u-v\| _{X}<\delta$$, $$|\lambda-\mu|<\delta$$, then
\begin{aligned} &\bigl|R(u,\lambda)'(t)-R(v,\mu)'(t)\bigr| \\ &\quad=\biggl\vert \varphi_{q} \biggl(\int_{t}^{+\infty} \lambda\psi(s) f\bigl(s,u(s),u'(s)\bigr)\,ds \biggr)- \varphi_{q} \biggl(\int_{t}^{+\infty}\mu\psi(s) f\bigl(s,v(s),v'(s)\bigr)\,ds \biggr)\biggr\vert \\ &\quad=\biggl\vert \varphi_{q}(\lambda)\varphi_{q} \biggl(\int_{t}^{+\infty}\psi(s) f\bigl(s,u(s),u'(s) \bigr)\,ds \biggr)-\varphi_{q}(\mu)\varphi_{q} \biggl(\int _{t}^{+\infty }\psi(s) f\bigl(s,v(s),v'(s) \bigr)\,ds \biggr)\biggr\vert \\ &\quad\leq \bigl[1+\varphi_{q}\bigl(\|\psi\|_{1}M_{r}\bigr) \bigr]\varepsilon,\quad t\in[0,+\infty). \end{aligned}
This, together with
\begin{aligned} &\frac{|R(u,\lambda)(t)-R(v,\mu)(t)|}{1+t} \\ &\quad= \frac{\vert \int_{0}^{t}\varphi_{q} (\int_{s}^{+\infty}\lambda \psi(r) f(r,u(r),u'(r))\,dr )\,ds-\int_{0}^{t}\varphi_{q} (\int_{s}^{+\infty}\mu \psi(r) f(r,v(r),v'(r))\,dr )\,ds\vert }{1+t} \\ &\quad\leq \frac{\int_{0}^{t}\vert \varphi_{q} (\int_{s}^{+\infty}\lambda \psi(r) f(r,u(r),u'(r))\,dr )-\varphi_{q} (\int_{s}^{+\infty}\mu\psi(r) f(r,v(r),v'(r))\,dr )\vert \,ds}{1+t} \\ &\quad= \frac{\int_{0}^{t}|R(u,\lambda)'(s)-R(v,\mu)'(s)|\,ds}{1+t} \leq\bigl\| R(u,\lambda)'-R(v, \mu)'\bigr\| _{\infty}, \end{aligned}
means that $$R:\overline{\Omega}\times[0,1]\rightarrow X_{2}\cap \operatorname{dom}M$$ is continuous.

We will prove that $$R:\overline{\Omega}\times[0,1]\rightarrow X_{2}\cap \operatorname{dom}M$$ is compact.

It is easy to get that $$\{\frac{R(u,\lambda)(t)}{1+t}:u\in\overline{\Omega },\lambda\in [0,1] \}$$ and $$\{R(u,\lambda)'(t):u\in\overline{\Omega},\lambda \in [0,1]\}$$ are bounded.

For any $$T>0$$, $$t_{1}, t_{2}\in[0,T]$$, $$t_{1}>t_{2}$$, $$u\in \overline{\Omega}$$, $$\lambda\in[0,1]$$, we have
\begin{aligned} &\biggl\vert \frac{R(u,\lambda)(t_{1})}{1+t_{1}}-\frac{R(u,\lambda )(t_{2})}{1+t_{2}}\biggr\vert \\ &\quad=\biggl\vert \frac{1}{1+t_{2}}\int_{0}^{t_{2}} \varphi_{q} \biggl(\int_{s}^{+\infty}\lambda \psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\,ds\\ &\qquad{}-\frac{1}{1+t_{1}} \int_{0}^{t_{1}}\varphi_{q} \biggl(\int _{s}^{+\infty}\lambda\psi(r) f\bigl(r,u(r),u'(r) \bigr)\,dr \biggr)\,ds\biggr\vert \\ &\quad\leq\biggl\vert \frac{1}{1+t_{2}}- \frac {1}{1+t_{1}}\biggr\vert \varphi_{q}\bigl(M_{r}\|\psi\|_{1}\bigr)T + \varphi_{q}\bigl(M_{r}\|\psi\|_{1}\bigr)|t_{2}-t_{1}|. \end{aligned}
Since t and $$\frac{1}{1+t}$$ are uniformly continuous on $$[0,T]$$, we get that $$\{\frac{R(u,\lambda)(t)}{1+t}, u\in \overline{\Omega}, \lambda\in[0,1] \}$$ is equicontinuous on $$[0,T]$$.
\begin{aligned} &\bigl|R(u,\lambda)'(t_{1})-R(u,\lambda)'(t_{2})\bigr| \\ &\quad=\biggl\vert \varphi_{q} \biggl(\int_{t_{2}}^{+\infty} \lambda\psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)- \varphi_{q} \biggl(\int_{t_{1}}^{+\infty}\lambda \psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\biggr\vert . \end{aligned}
Take $$G(t)=\int_{t}^{+\infty}\lambda\psi(r) f(r,u(r),u'(r))\,dr$$. We have
$$\bigl|G(t)\bigr|\leq M_{r}\|\psi\|_{1},\qquad \bigl|G(t_{2})-G(t_{1})\bigr| \leq M_{r}\int_{t_{2}}^{t_{1}}\psi(r)\,dr.$$
It follows from the absolute continuity of integral and the uniform continuity of $$\varphi_{q}(t)$$ in $$[-M_{r}\|\psi\|_{1},M_{r}\|\psi\|_{1}]$$ that $$\{R(u,\lambda)'(t), u\in \overline{\Omega}, \lambda\in[0,1]\}$$ is equicontinuous on $$[0,T]$$.
For $$u\in \overline{\Omega}$$, since
$$\int_{s}^{+\infty}\lambda\psi(r) \bigl|f \bigl(r,u(r),u'(r)\bigr)\bigr|\,dr\leq M_{r} \int _{s}^{+\infty}\psi(r)\,dr$$
and
$$\lim_{s\rightarrow+\infty}\int_{s}^{+\infty}\psi(r) \,dr=0,$$
for any $$\varepsilon>0$$, there exists a constant $$T_{1}>0$$ such that
$$\varphi_{q} \biggl(\int_{s}^{+\infty}\lambda \psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr| \biggr)\,dr< \frac{\varepsilon}{4}, \quad s>T_{1}, u\in \overline{\Omega}, \lambda\in[0,1].$$
Obviously, there exists a constant $$T>T_{1}$$ such that, for any $$t>T$$,
$$\frac{1}{1+t}\varphi_{q}\bigl(M_{r}\|\psi \|_{1}\bigr)T_{1}< \frac{\varepsilon}{4}.$$
Thus, for any $$t_{1}, t_{2}>T$$, we have
\begin{aligned} & \biggl\vert \frac{R(u,\lambda)(t_{1})}{1+t_{1}}-\frac{R(u,\lambda )(t_{2})}{1+t_{2}}\biggr\vert \\ &\quad=\biggl\vert \frac{1}{1+t_{2}}\int_{0}^{t_{2}} \varphi_{q} \biggl(\int_{s}^{+\infty}\lambda \psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\,ds\\ &\qquad{}-\frac{1}{1+t_{1}} \int_{0}^{t_{1}}\varphi_{q} \biggl(\int _{s}^{+\infty}\lambda\psi(r) f\bigl(r,u(r),u'(r) \bigr)\,dr \biggr)\,ds\biggr\vert \\ &\quad\leq \frac{1}{1+t_{2}}\int_{0}^{T_{1}} \varphi_{q} \biggl(\int_{s}^{+\infty }\lambda \psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr|\,dr \biggr)\,ds\\ &\qquad{}+ \frac{1}{1+t_{2}}\int_{T_{1}}^{t_{2}}\varphi _{q} \biggl(\int_{s}^{+\infty}\lambda\psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr|\,dr \biggr)\,ds \\ &\qquad{}+ \frac{1}{1+t_{1}}\int_{0}^{T_{1}} \varphi_{q} \biggl(\int_{s}^{+\infty }\lambda \psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr|\,dr \biggr)\,ds\\ &\qquad{}+ \frac{1}{1+t_{1}}\int_{T_{1}}^{t_{1}}\varphi _{q} \biggl(\int_{s}^{+\infty}\lambda\psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr|\,dr \biggr)\,ds \\ &\quad\leq \frac{1}{1+t_{2}}\varphi_{q}\bigl(M_{r}\|\psi \|_{1}\bigr)T_{1}+\frac{t_{2}-T_{1}}{1+t_{2}}\frac {\varepsilon}{4} + \frac{1}{1+t_{1}}\varphi_{q}\bigl(M_{r}\|\psi \|_{1}\bigr)T_{1}+\frac{t_{1}-T_{1}}{1+t_{1}}\frac {\varepsilon}{4} < \varepsilon \end{aligned}
and
\begin{aligned} &\bigl|R(u,\lambda)'(t_{1})-R(u,\lambda)'(t_{2})\bigr| \\ &\quad=\biggl\vert \varphi_{q} \biggl(\int_{t_{2}}^{+\infty} \lambda\psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)- \varphi_{q} \biggl(\int_{t_{1}}^{+\infty}\lambda \psi(r) f\bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\biggr\vert \\ &\quad\leq\varphi_{q} \biggl(\int_{t_{2}}^{+\infty} \lambda\psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr|\,dr \biggr)+ \varphi_{q} \biggl(\int_{t_{1}}^{+\infty}\lambda \psi(r) \bigl|f\bigl(r,u(r),u'(r)\bigr)\bigr|\,dr \biggr) \\ &\quad< \varepsilon. \end{aligned}
By Lemma 3.3, we get that $$\{R(u,\lambda)| u\in \overline{\Omega}, \lambda\in[0,1]\}$$ is relatively compact. The proof is completed. □

### Lemma 3.5

Assume that $$\Omega\subset X$$ is an open bounded set. Then $$N_{\lambda}$$ is M-quasi-compact in Ω̅.

### Proof

It is clear that $$\operatorname{Im}P=\operatorname{Ker}M$$, $$\operatorname{Ker} Q=\operatorname{Im} M$$ and $$QN_{\lambda} x=\theta$$, $$\lambda\in(0,1)\Leftrightarrow QNx=\theta$$, i.e., Definition 2.3(a) and (b) are satisfied.

For $$u\in\Sigma{_{\lambda}}=\{\omega\in\overline{\Omega}\cap \operatorname{dom}M:M\omega=N_{\lambda}\omega\}$$, by (H1) and (H2), we get
\begin{aligned} R(u,\lambda)&= -\int_{0}^{t}\varphi_{q} \biggl(\int_{s}^{+\infty }\lambda\psi(r) f \bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\,ds \\ &=-\int_{0}^{t}\varphi_{q} \biggl(\int _{s}^{+\infty}\bigl(\varphi_{p} \bigl(u'\bigr)\bigr)'(r)\,dr \biggr)\,ds \\ &=u(t)-u(0)=(I-P)u. \end{aligned}
Clearly, $$R(\cdot,0)=0$$. Thus, Definition 2.3(c) is satisfied. For $$u\in\overline{\Omega}$$, we have
$$M\bigl[Pu+R(u,\lambda)\bigr](t)= \bigl[ \lambda\psi(t) f\bigl(t,u(t),u'(t) \bigr), T\bigl(\lambda f\bigl(t,u(t),u'(t)\bigr)\bigr) \bigr]=(I-Q)N_{\lambda}u(t).$$
So, Definition 2.3(d) is satisfied.
Considering (H2),
$$\bigl\| N_{\lambda}(u)-N_{\lambda}(v)\bigr\| _{Z}\leq \sup _{t\in[0,\infty)}\bigl|f\bigl(t,u(t),u'(t)\bigr)- f \bigl(t,v(t),v'(t)\bigr)\bigr|,\quad u, v\in\overline{\Omega},$$
and
$$\bigl\| N_{\lambda}(u)\bigr\| _{Z}\leq \sup_{t\in[0,\infty)}\bigl|f \bigl(t,u(t),u'(t)\bigr)\bigr|,\quad u\in \overline{\Omega},$$
we can obtain that $$N_{\lambda}$$ is continuous and bounded in Ω̅.

It follows from the continuity and boundedness of T that Q is continuous and bounded in Z. By a simple calculation, we can obtain that $$Q(I-Q)(\psi y,c)=(0,0)$$, $$(\psi y,c)\in Z$$.

These, together with Lemma 3.4, mean that $$N_{\lambda}$$ is M-quasi-compact in Ω̅. The proof is completed. □

### Theorem 3.1

Suppose that (H1), (H2) and the following conditions hold:
(H3):
There exist constants $$c_{0}>0$$ and $$l>0$$ such that
$$\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r)f \bigl(r,u(r),u'(r)\bigr)\,dr \biggr)\,ds\,dt\neq 0,\quad \bigl|u(t)\bigr|>c_{0}, t\in[0,l], u\in X.$$
(H4):
There exist nonnegative functions $$a(t)$$, $$b(t)$$, $$c(t)$$ with $$(1+t)^{p-1}a(t)\psi(t), b(t)\psi(t), c(t)\psi(t)\in L^{1}[0,+\infty)$$ such that
$$\bigl|f(t,x,y)\bigr|\leq a(t)\bigl|\varphi_{p}(x)\bigr|+b(t)\bigl|\varphi_{p}(y)\bigr|+c(t), \quad\textit{a.e. } t\in[0,+\infty),$$
where $$\|(1+t)^{p-1}a(t)\psi(t)\|_{1}l_{0}^{p-1}+\|b(t)\psi(t)\|_{1}<1$$, if $$1< p\leq2$$; $$2^{p-2}\|(1+t)^{p-1}a(t) \psi(t)\|_{1}l_{0}^{p-1}+\|b(t)\psi(t)\|_{1}<1$$, if $$p\geq2$$, where $$l_{0}=\max\{1,l\}$$.
(H5):
There exists a constant $$d_{0}>0$$ such that if $$|d|>d_{0}$$, then one of the following inequalities holds:
1. (1)

$$d f(t,d,0)>0$$, $$t\in[0,l)$$;

2. (2)

$$d f(t,d,0)<0$$, $$t\in[0,l)$$.

Then the boundary value problem (1.1) has at least one solution.

In order to prove Theorem 3.1, we show two lemmas.

### Lemma 3.6

Assume (H1)-(H4) hold. Then the set
$$\Omega_{1}=\bigl\{ u\in \operatorname{dom}M | Mu=N_{\lambda}u, \lambda\in(0,1)\bigr\}$$
is bounded in X.

### Proof

For $$u\in\Omega_{1}$$, we have $$QN_{\lambda}u=0$$, i.e., $$T(f(t,u(t),u'(t)))=0$$. By (H3), there exists a constant $$t_{0}\in[0,l]$$ such that $$|u(t_{0})|\leq c_{0}$$. Since $$u(t)=u(t_{0})+\int_{t_{0}}^{t}u'(s)\,ds$$, then
$$\frac{|u(t)|}{1+t}\leq\frac{c_{0}+|t-t_{0}|\|u'\|_{\infty}}{1+t}\leq c_{0}+\max \{1,l\} \bigl\| u'\bigr\| _{\infty}= c_{0}+l_{0} \bigl\| u'\bigr\| _{\infty},\quad t\in[0,+\infty).$$
Thus
$$\biggl\Vert \frac{u}{1+t}\biggr\Vert _{\infty}\leq c_{0}+l_{0}\bigl\| u'\bigr\| _{\infty}.$$
(3.2)
It follows from $$Mu=N_{\lambda}u$$, (H4) and (3.2) that
\begin{aligned} &\bigl|\varphi_{p}\bigl(u'(t)\bigr)\bigr|\\ &\quad\leq \int _{0}^{+\infty}\psi(t) \bigl[a(t)\bigl|\varphi_{p} \bigl(u(t)\bigr)\bigr|+b(t)\bigl|\varphi_{p}\bigl(u'(t)\bigr)\bigr|+c(t) \bigr]\,dt \\ &\quad\leq \bigl\| \psi(t)a(t) (1+t)^{p-1}\bigr\| _{1}\varphi_{p} \biggl(\biggl\Vert \frac{u}{1+t}\biggr\Vert _{\infty} \biggr)+\| \psi b\|_{1}\varphi_{p}\bigl(\bigl\| u' \bigr\| _{\infty}\bigr)+\|\psi c\|_{1} \\ &\quad\leq \bigl\| \psi(t)a(t) (1+t)^{p-1}\bigr\| _{1}\varphi_{p} \bigl( c_{0}+l_{0}\bigl\| u'\bigr\| _{\infty}\bigr)+ \|\psi b\|_{1}\varphi_{p}\bigl(\bigl\| u' \bigr\| _{\infty}\bigr)+\|\psi c\|_{1}. \end{aligned}
Whenever $$1< p\leq2$$, by Lemma 2.1, we get
$$\bigl\| u'\bigr\| _{\infty}\leq \varphi_{q} \biggl( \frac{\|\psi c\|_{1}+\|\psi(t)a(t)(1+t)^{p-1}\|_{1}\varphi_{p}( c_{0})}{1-\|\psi(t)a(t)(1+t)^{p-1}\|_{1}l_{0}^{p-1}-\|\psi b\|_{1}} \biggr).$$
Whenever $$p>2$$, by Lemma 2.1, we get
$$\bigl\| u'\bigr\| _{\infty}\leq \varphi_{q} \biggl( \frac{\|\psi c\|_{1}+2^{p-2} \|\psi(t)a(t)(1+t)^{p-1}\|_{1}\varphi_{p}( c_{0})}{1-2^{p-2}\|\psi(t)a(t)(1+t)^{p-1}\|_{1}l_{0}^{p-1}-\|\psi b\|_{1}} \biggr).$$

These, together with (3.2), mean that $$\Omega_{1}$$ is bounded in X. □

### Lemma 3.7

Assume (H1)-(H3) and (H5) hold. Then
$$\Omega_{2}=\{u\in \operatorname{Ker}M | QN u=0\}$$
is bounded in X, where $$N=N_{1}$$.

### Proof

For $$u\in\Omega_{2}$$, we have $$u=a$$, $$a\in\mathbb{R}$$ and $$Q(N u)=0$$, i.e.,
$$\int_{0}^{+\infty}h(t)\int_{0}^{t} \varphi_{q} \biggl(\int_{s}^{+\infty}\psi (r)f(r,a,0)\,dr \biggr)\,ds\,dt=0.$$
By (H5), we get that $$|a|\leq d_{0}$$. So, $$\Omega_{2}$$ is bounded. The proof is completed. □

### Proof of Theorem 3.1

Let $$\Omega=\{u\in X | \|u\|< r\}$$, where $$r>d_{0}$$ is large enough such that $$\Omega\supset \overline{\Omega}_{1}\cup\overline{\Omega}_{2}$$.

By Lemmas 3.6 and 3.7, we know $$Mu\neq N_{\lambda}u$$, $$u\in \operatorname{dom}M\cap\partial\Omega$$ and $$QN u\neq0$$, $$u\in \operatorname{Ker}M\cap\partial\Omega$$.

Let $$H(u,\delta)=\rho\delta u+(1-\delta)JQNu$$, $$\delta\in[0,1]$$, $$u\in \operatorname{Ker}M\cap\overline{\Omega}$$, where $$J:\operatorname{Im}Q\rightarrow \operatorname{Ker}M$$ is a homeomorphism with $$J(0,a)=a$$,
$$\rho=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} -1 &\mbox{if } (\mathrm{H}_{5})(1) \mbox{ holds},\\ 1 &\mbox{if } (\mathrm{H}_{5})(2) \mbox{ holds}. \end{array}\displaystyle \right .$$
Define a function
$$\operatorname{sgn}(x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1 &\mbox{if } x\geq0,\\ -1& \mbox{if } x< 0. \end{array}\displaystyle \right .$$
For $$u\in \operatorname{Ker}M\cap\partial\Omega$$, we have $$u=a\neq0$$. Thus
$$H(u,\delta)=\rho\delta a+(1-\delta) (-T\bigl(f(t,a,0)\bigr).$$
If $$\delta=1$$, $$H(u,1)=\rho a\neq0$$. If $$\delta=0$$, by $$QNu\neq0$$, we get $$H(u,0)=JQN(a)\neq0$$. For $$0<\delta<1$$, we now prove that $$H(u,\delta )\neq0$$. Otherwise, if $$H(u,\delta)=0$$, then
$$T\bigl(f(t,a,0)\bigr)= \frac{\rho\delta}{1-\delta}a.$$
(3.3)
Since $$\|u\|_{X}=|a|=r>d_{0}$$, for $$t\in[0,l)$$, by (H5) and Lemma 3.1, we have
$$\operatorname{sgn}\bigl(af(t,a,0)\bigr)=\operatorname{sgn}\bigl(T\bigl(af(t,a,0) \bigr)\bigr)=\operatorname{sgn} \bigl[a\bigl(T\bigl(f(t,a,0)\bigr)\bigr)\bigr]= \operatorname{sgn} \biggl( \frac{\rho\delta}{1-\delta }a^{2} \biggr)= \operatorname{sgn}(\rho),$$
a contradiction with the definition of ρ. So, $$H(u,\delta)\neq0$$, $$u\in \operatorname{Ker}M\cap\partial\Omega$$, $$\delta\in[0,1]$$.
By the homotopy of degree, we get that
\begin{aligned} &\operatorname{deg}\Bigl(JQN, \Omega\cap \operatorname{Ker}M,0\Bigr)\\ &\quad=\operatorname{deg}\Bigl(H(\cdot,0),\Omega\cap \operatorname{Ker}M,0 \Bigr)=\operatorname{deg}\Bigl( H(\cdot,1),\Omega\cap \operatorname{Ker}M,0\Bigr) \\ &\quad=\operatorname{deg}\Bigl(\rho I, \Omega\cap \operatorname{Ker}M,0 \Bigr)=\operatorname{deg}\Bigl(\pm I, \Omega\cap \operatorname{Ker}M,0 \Bigr)=\pm1\neq0. \end{aligned}

By Theorem 2.1, we can get that $$Mu=Nu$$ has at least one solution in Ω̅. The proof is completed. □

## 4 Example

### Example 4.1

Let us consider the following boundary value problem at resonance:
$$\left \{ \textstyle\begin{array}{@{}l} (\varphi_{p}(u') )'(t)=\psi(t)f (t,u(t),u'(t) ),\quad t\in [0,+\infty),\\ u'(+\infty)=0,\qquad u(0)=\int_{0}^{+\infty}h(t)u(t)\,dt, \end{array}\displaystyle \right .$$
(4.1)
where $$p= \frac{4}{3}$$, $$h(t)=e^{-t}$$,
$$f(t,u,v)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0,& t>l,\\ (l-t) (u^{\frac{1}{3}}+\sin ^{\frac{1}{3}}v ),& t\leq l, u,v\in\mathbb{R}, \end{array}\displaystyle \right .$$
$$\psi(t)= \frac{1}{3}(1+t)^{-\frac{4}{3}}(1+l_{0})^{-\frac {4}{3}}e^{-t}$$, $$l_{0}=\max\{1,l\}$$. Take $$a(t)=b(t)=(l-t)$$, $$c(t)=0$$, $$d_{0}=c_{0}=27$$. By a simple calculation, we can obtain $$\|(1+t)^{p-1}a(t)\psi(t)\|_{1}l_{0}^{p-1}+\|b(t)\psi(t)\|_{1}<1$$. It is easy to get that conditions (H1)-(H5) are satisfied. It follows from Theorem 3.1 that problem (4.1) has at least one solution.

## Declarations

### Acknowledgements

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