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 [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 $Lx=Nx$ 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:

$$\{\begin{array}{c}{({\phi }_p(u^{\prime }))}^{\prime }+f(t,u)=0,0<t<1,\hfill \\ u(0)=0=G(u(\eta ),u(1)),\hfill \end{array}$$

where ${\phi }_p(s)={|s|}^{p-2}s$, $p>1$, $0<\eta <1$. ${\phi }_p(s)$ is nonlinear when $p\ne 2$.

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

$$\{\begin{array}{c}{({\phi }_p(u^{\prime }))}^{\prime }+f(t,u,u^{\prime })=0,0<t<+\mathrm{\infty },\hfill \\ u(0)=0,{\phi }_p(u^{\prime }(+\mathrm{\infty }))={\sum }_{i=1}^n{\alpha }_i{\phi }_p(u^{\prime }({\xi }_i)),\hfill \end{array}$$

(1.1)

where ${\alpha }_i>0$, $i=1,2,\dots ,n$, ${\sum }_{i=1}^n{\alpha }_i=1$.

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

(H₁) $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_n<+\mathrm{\infty }$, ${\alpha }_i>0$, ${\sum }_{i=1}^n{\alpha }_i=1$.

(H₂) $f:[0,+\mathrm{\infty })\times {\mathbb{R}}^2\to \mathbb{R}$ is continuous, $f(t,0,0)\ne 0$, $t\in (0,\mathrm{\infty })$ and for any $r>0$, there exists a nonnegative function $h_r(t)\in L^1[0,+\mathrm{\infty })$ such that

$$|f(t,x,y)|\le h_r(t),\text{a.e. }t\in [0,+\mathrm{\infty }),x,y\in \mathbb{R},\frac{|x|}{1+t}\le r,|y|\le r.$$

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 ${\parallel \cdot \parallel }_X$, ${\parallel \cdot \parallel }_Y$, respectively. A continuous operator $M:X\cap domM\to Y$ is said to be quasi-linear if

• (i) $ImM:=M(X\cap domM)$ is a closed subset of Y,

• (ii) $KerM:=\{x\in X\cap domM:Mx=0\}$ is linearly homeomorphic to ${\mathbb{R}}^n$, $n<\mathrm{\infty }$, 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 $X=X_1\oplus X_2$. 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., $Y=Y_1\oplus Y_2$. Let $P:X\to X_1$ and $Q:Y\to Y_1$ be two projectors and $\mathrm{\Omega }\subset X$ an open and bounded set with the origin $\theta \in \mathrm{\Omega }$.

Definition 2.2 [11]

Suppose that $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$, $\lambda \in [0,1]$ is a continuous operator. Denote $N_1$ by N. Let ${\mathrm{\Sigma }}_{\lambda }=\{x\in \overline{\mathrm{\Omega }}:Mx=N_{\lambda }x\}$. $N_{\lambda }$ is said to be M-compact in $\overline{\mathrm{\Omega }}$ if there exist a vector subspace $Y_1$ of Y satisfying $dim{Y}_1=dim{X}_1$ and an operator $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2$ being continuous and compact such that for $\lambda \in [0,1]$,

• (a) $(I-Q)N_{\lambda }(\overline{\mathrm{\Omega }})\subset ImM\subset (I-Q)Y$,

• (b) $Q{N}_{\lambda }x=\theta ,\lambda \in (0,1)\iff QNx=\theta$,

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

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

Theorem 2.1 [11]

Let X and Y be two Banach spaces with the norms ${\parallel \cdot \parallel }_X$, ${\parallel \cdot \parallel }_Y$, respectively, and $\mathrm{\Omega }\subset X$ an open and bounded nonempty set. Suppose that

$$M:X\cap domM\to Y$$

is a quasi-linear operator and $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$, $\lambda \in [0,1]$ M-compact. In addition, if the following conditions hold:

(C₁) $Mx\ne N_{\lambda }x$, $\mathrm{\forall }x\in \partial \mathrm{\Omega }\cap domM$, $\lambda \in (0,1)$,

(C₂) $deg\{JQN,\mathrm{\Omega }\cap KerM,0\}\ne 0$,

then the abstract equation $Mx=Nx$ has at least one solution in $domM\cap \overline{\mathrm{\Omega }}$, where $N=N_1$, $J:ImQ\to KerM$ is a homeomorphism with $J(\theta )=\theta$.

3 Main result

Let $X=\{u|u\in C^1[0,+\mathrm{\infty }),u(0)=0,{sup}_{t\in [0,+\mathrm{\infty })}\frac{|u(t)|}{1+t}<+\mathrm{\infty },{lim}_{t\to +\mathrm{\infty }}{u}^{\prime }(t)\text{ exists}\}$ with norm $\parallel u\parallel =max\{{\parallel \frac{u}{1+t}\parallel }_{\mathrm{\infty }},{\parallel u^{\prime }\parallel }_{\mathrm{\infty }}\}$, where ${\parallel u\parallel }_{\mathrm{\infty }}={sup}_{t\in [0,+\mathrm{\infty })}|u(t)|$. $Y=L^1[0,+\mathrm{\infty })$ with norm ${\parallel y\parallel }_1={\int }_0^{+\mathrm{\infty }}|y(t)|dt$. Then $(X,\parallel \cdot \parallel )$ and $(Y,{\parallel \cdot \parallel }_1)$ are Banach spaces.

Define operators $M:X\cap domM\to Y$ and $N_{\lambda }:X\to Y$ as follows:

$$Mu={\left({\phi }_p\left(u^{\prime }\right)\right)}^{\prime },N_{\lambda }u=-\lambda f(t,u,u^{\prime }),\lambda \in [0,1],t\in [0,+\mathrm{\infty }),$$

where

$$\begin{array}{rcl}domM& =& \{u\in X|{\phi }_p\left(u^{\prime }\right)\in AC[0,+\mathrm{\infty }),{\left({\phi }_p\left(u^{\prime }\right)\right)}^{\prime }\in L^1[0,+\mathrm{\infty }),\\ {\phi }_p(u^{\prime }(+\mathrm{\infty }))=\sum _{i=1}^n{\alpha }_i{\phi }_p(u^{\prime }({\xi }_i))\}.\end{array}$$

Then the boundary value problem (1.1) is equivalent to $Mu=Nu$.

Obviously,

$$KerM=\{at|a\in \mathbb{R}\},ImM=\left\{y\right|y\in Y,\sum _{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}y(s)ds=0\}.$$

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

Define $P:X\to X_1$, $Q:Y\to Y_1$ as

$$(Pu)(t)=u^{\prime }(+\mathrm{\infty })t,(Qy)(t)=\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}y(s)ds}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-t},$$

where $X_1=KerM$, $Y_1=ImQ=\{b{e}^{-t}|b\in \mathbb{R}\}$. We can easily obtain that $P:X\to X_1$, $Q:Y\to Y_1$ are projectors. Set $X=X_1\oplus X_2$, $Y=Y_1\oplus Y_2$.

Define an operator $R:X\times [0,1]\to X_2$:

$$\begin{array}{rl}R(u,\lambda )(t)=& {\int }_0^t{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]d\tau -u^{\prime }(+\mathrm{\infty })t,\end{array}$$

where $\frac{1}{p}+\frac{1}{q}=1$, ${\phi }_q={\phi }_p^{-1}$. By (H₁) and (H₂), we get that $R:X\times [0,1]\to X_2$ is continuous.

Lemma 3.1 [15]

$V\subset X$ is compact if $\{\frac{u(t)}{1+t}|u\in V\}$ and $\{u^{\prime }(t)|u\in V\}$ are both equicontinuous on any compact intervals of $[0,+\mathrm{\infty })$ and equiconvergent at infinity.

Lemma 3.2 $R:X\times [0,1]\to X_2$ is compact.

Proof Let $\mathrm{\Omega }\subset X$ be nonempty and bounded. There exists a constant $r>0$ such that $\parallel u\parallel \le r$, $u\in \overline{\mathrm{\Omega }}$. It follows from (H₂) that there exists a nonnegative function $h_r(t)\in L^1[0,+\mathrm{\infty })$ such that

$$\left|f(t,u(t),u^{\prime }(t))\right|\le h_r(t),\text{a.e. }t\in [0,+\mathrm{\infty }),u\in \overline{\mathrm{\Omega }}.$$

For any $T>0$, $t_1,t_2\in [0,T]$, $u\in \overline{\mathrm{\Omega }}$, $\lambda \in [0,1]$, we have

$$\begin{array}{r}|\frac{R(u,\lambda )(t_1)}{1+t_1}-\frac{R(u,\lambda )(t_2)}{1+t_2}|\\ \le \left|\frac{1}{1+t_1}{\int }_0^{t_1}{\phi }_q\right[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]d\tau \\ -\frac{1}{1+t_2}{\int }_0^{t_2}{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))\left]d\tau \right|+|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}||u^{\prime }(+\mathrm{\infty })|\\ \le \left|\frac{1}{1+t_1}{\int }_{t_2}^{t_1}{\phi }_q\right[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))\left]d\tau \right|+|\frac{1}{1+t_1}-\frac{1}{1+t_2}|\\ \times \left|{\int }_0^{t_2}{\phi }_q\right[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))\left]d\tau \right|+|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}|r\\ \le {\phi }_q[{\parallel h_r\parallel }_1(1+\frac{1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}})+{\phi }_p(r)][|t_1-t_2|+T|\frac{1}{1+t_1}-\frac{1}{1+t_2}|]\\ +|\frac{t_1}{1+t_1}-\frac{t_2}{1+t_2}|r.\end{array}$$

Since $\{t,\frac{1}{1+t},\frac{t}{1+t}\}$ are equicontinuous on $[0,T]$, we get that $\{\frac{R(u,\lambda )(t)}{1+t},u\in \overline{\mathrm{\Omega }}\}$ are equicontinuous on $[0,T]$.

$$\begin{array}{r}|R{(u,\lambda )}^{\prime }(t_1)-R{(u,\lambda )}^{\prime }(t_2)|\\ =\left|{\phi }_q\right[{\int }_{t_1}^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]\\ -{\phi }_q[{\int }_{t_2}^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))\left]\right|.\end{array}$$

Let

$$g(t,u)={\int }_t^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds+{\phi }_p(u^{\prime }(+\mathrm{\infty })).$$

Then

$$|g(t,u)|\le {\parallel h_r\parallel }_1(1+\frac{1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}})+{\phi }_p(r):=k,t\in [0,T],u\in \overline{\mathrm{\Omega }}.$$

(3.1)

For $t_1,t_2\in [0,T]$, $t_1<t_2$, $u\in \overline{\mathrm{\Omega }}$, we have

$$\begin{array}{rl}|g(t_1,u)-g(t_2,u)|& =\left|{\int }_{t_1}^{t_2}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\right|\\ \le {\int }_{t_1}^{t_2}{h}_r(s)+\frac{{\parallel h_r\parallel }_1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s}ds.\end{array}$$

It follows from the absolute continuity of integral that $\{g(t,u),u\in \overline{\mathrm{\Omega }}\}$ are equicontinuous on $[0,T]$. Since ${\phi }_q(x)$ is uniformly continuous on $[-k,k]$, by (3.1), we can obtain that $\{R{(u,\lambda )}^{\prime }(t),u\in \overline{\mathrm{\Omega }}\}$ are equicontinuous on $[0,T]$.

For $u\in \overline{\mathrm{\Omega }}$, since

$$\begin{array}{c}\begin{array}{r}\left|{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\right|\\ \le {\int }_{\tau }^{+\mathrm{\infty }}{h}_r(s)+\frac{{\parallel h_r\parallel }_1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s}ds,\end{array}\hfill \\ \underset{\tau \to +\mathrm{\infty }}{lim}{\int }_{\tau }^{+\mathrm{\infty }}{h}_r(s)+\frac{{\parallel h_r\parallel }_1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s}ds=0,\hfill \end{array}$$

and ${\phi }_q(x)$ is uniformly continuous on $[-r-r^{p-1},r+r^{p-1}]$, for any $\epsilon >0$, there exists a constant $T_1>0$ such that if $\tau \ge T_1$, then

$$\begin{array}{r}|{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds+{\phi }_p(u^{\prime }(+\mathrm{\infty }))]\\ -u^{\prime }(+\mathrm{\infty })|<\frac{\epsilon }{4},\mathrm{\forall }u\in \overline{\mathrm{\Omega }}.\end{array}$$

(3.2)

Since

$$\begin{array}{r}\left|{\int }_0^{T_1}{\phi }_q\right[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]d\tau -u^{\prime }(+\mathrm{\infty })T_1|\\ \le \{{\phi }_q[{\parallel h_r\parallel }_1(1+\frac{1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}})+{\phi }_p(r)]+r\}{T}_1,\end{array}$$

(3.3)

there exists a constant $T>T_1$ such that if $t>T$, then

$$\frac{1}{1+t}\{{\phi }_q[{\parallel h_r\parallel }_1(1+\frac{1}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}})+{\phi }_p(r)]+r\}{T}_1<\frac{\epsilon }{4}.$$

(3.4)

For $t_2>t_1>T$, by (3.2), (3.3) and (3.4), we have

$$\begin{array}{r}|\frac{R(u,\lambda )(t_1)}{1+t_1}-\frac{R(u,\lambda )(t_2)}{1+t_2}|\\ =\left|\frac{1}{1+t_1}{\int }_0^{t_1}\right\{{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })\}d\tau \\ -\frac{1}{1+t_2}{\int }_0^{t_2}\left\{{\phi }_q\right[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })\}d\tau |\\ \le \left|\frac{1}{1+t_1}{\int }_0^{T_1}\right\{{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })\}d\tau |\\ +\left|\frac{1}{1+t_1}{\int }_{T_1}^{t_1}\right\{{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })\}d\tau |\\ +\left|\frac{1}{1+t_2}{\int }_0^{T_1}\right\{{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })\}d\tau |\\ +\left|\frac{1}{1+t_2}{\int }_{T_1}^{t_2}\right\{{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })\}d\tau |<\epsilon ,\end{array}$$

and

$$\begin{array}{r}|R{(u,\lambda )}^{\prime }(t_1)-R{(u,\lambda )}^{\prime }(t_2)|\\ \le \left|{\phi }_q\right[{\int }_{t_1}^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })|\\ +\left|{\phi }_q\right[{\int }_{t_2}^{+\mathrm{\infty }}\lambda (f(s,u(s),u^{\prime }(s))-\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-s})ds\\ +{\phi }_p(u^{\prime }(+\mathrm{\infty }))]-u^{\prime }(+\mathrm{\infty })|\\ <\epsilon .\end{array}$$

By Lemma 3.1, we get that $\{R(u,\lambda )|u\in \overline{\mathrm{\Omega }},\lambda \in [0,1]\}$ is compact. The proof is completed. □

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

Lemma 3.3 Let $\mathrm{\Omega }\subset X$ be nonempty, open and bounded. Then $N_{\lambda }$ is M-compact in $\overline{\mathrm{\Omega }}$.

Proof By (H₂), we know that $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$ is continuous. Obviously, $dim{X}_1=dim{Y}_1$. For $u\in \overline{\mathrm{\Omega }}$, since $Q(I-Q)$ is a zero operator, we get $(I-Q)N_{\lambda }(u)\in ImM$. For $y\in ImM$, $y=Qy+(I-Q)y=(I-Q)y\in (I-Q)Y$. So, we have $(I-Q)N_{\lambda }(\overline{\mathrm{\Omega }})\subset ImM\subset (I-Q)Y$. It is clear that

$$Q{N}_{\lambda }u=0,\lambda \in (0,1)\iff QNu=0$$

and $R(u,0)=0$, $\mathrm{\forall }u\in X$. $u\in {\mathrm{\Sigma }}_{\lambda }=\{u\in \overline{\mathrm{\Omega }}:Mu=N_{\lambda }u\}$ means that $N_{\lambda }u\in ImM$ and ${({\phi }_p(u^{\prime }))}^{\prime }+\lambda f(t,u,u^{\prime })=0$, thus,

$$\begin{array}{rl}R(u,\lambda )(t)& ={\int }_0^t{\phi }_q[{\int }_{\tau }^{+\mathrm{\infty }}-{\left({\phi }_p\left(u^{\prime }\right)\right)}^{\prime }ds+{\phi }_p(u^{\prime }(+\mathrm{\infty }))]d\tau -u^{\prime }(+\mathrm{\infty })t\\ =u(t)-u^{\prime }(+\mathrm{\infty })t=(I-P)u(t).\end{array}$$

For $u\in X$, we have

$$\begin{array}{rl}M[P+R(u,\lambda )](t)& =-\lambda f(t,u(t),u^{\prime }(t))+\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}\lambda f(r,u(r),u^{\prime }(r))dr}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}{e}^{-t}\\ =(I-Q)N_{\lambda }u(t).\end{array}$$

These, together with Lemma 3.2, mean that $N_{\lambda }$ is M-compact in $\overline{\mathrm{\Omega }}$. 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 ${(1+t)}^{p-1}a(t),b(t),c(t)\in Y$ and ${\parallel {(1+t)}^{p-1}a(t)\parallel }_1+{\parallel b(t)\parallel }_1<1$ such that

$$|f(t,x,y)|\le a(t)|{\phi }_p(x)|+b(t)|{\phi }_p(y)|+c(t),\text{a.e. }t\in [0,+\mathrm{\infty }).$$

(H₄) There exists a constant $d_0>0$ such that if $|d|>d_0$, then one of the following inequalities holds:

$$\begin{array}{c}df(t,x,d)<0,(t,x)\in [0,+\mathrm{\infty })\times \mathbb{R};\hfill \\ df(t,x,d)>0,(t,x)\in [0,+\mathrm{\infty })\times \mathbb{R}.\hfill \end{array}$$

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

$${\mathrm{\Omega }}_1=\{u|u\in domM,Mu=N_{\lambda }u,\lambda \in [0,1]\}$$

is bounded in X.

Proof If $u\in {\mathrm{\Omega }}_1$, then $Q{N}_{\lambda }u=0$, i.e., ${\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(r,u(r),u^{\prime }(r))dr=0$. By (H₄), there exists $t_0\in [0,+\mathrm{\infty })$ such that $|u^{\prime }(t_0)|\le d_0$. It follows from $Mu=N_{\lambda }u$ that

$${\phi }_p(u^{\prime }(t))=-{\int }_{t_0}^t\lambda f(s,u(s),u^{\prime }(s))ds+{\phi }_p(u^{\prime }(t_0)).$$

Considering (H₃), we have

$$\begin{array}{rl}\left|{\phi }_p(u^{\prime }(t))\right|& \le {\int }_0^{+\mathrm{\infty }}[a(t)\left|{\phi }_p(u(t))\right|+b(t)\left|{\phi }_p(u^{\prime }(t))\right|+c(t)]dt+{\phi }_p(d_0)\\ \le {\parallel a(t){(1+t)}^{p-1}\parallel }_1{\phi }_p\left({\parallel \frac{u}{1+t}\parallel }_{\mathrm{\infty }}\right)+{\parallel b\parallel }_1{\phi }_p\left({\parallel u^{\prime }\parallel }_{\mathrm{\infty }}\right)+{\parallel c\parallel }_1+{\phi }_p(d_0).\end{array}$$

(3.5)

Since $u(t)={\int }_0^t{u}^{\prime }(s)ds$, we get

$$\left|\frac{u(t)}{1+t}\right|\le \frac{t}{1+t}{\parallel u^{\prime }\parallel }_{\mathrm{\infty }}\le {\parallel u^{\prime }\parallel }_{\mathrm{\infty }}.$$

Thus,

$${\parallel \frac{u}{1+t}\parallel }_{\mathrm{\infty }}\le {\parallel u^{\prime }\parallel }_{\mathrm{\infty }}.$$

(3.6)

By (3.5), (3.6) and (H₃), we get

$${\parallel {\phi }_p\left(u^{\prime }\right)\parallel }_{\mathrm{\infty }}\le \frac{{\parallel c\parallel }_1+{\phi }_p(d_0)}{1-{\parallel a(t){(1+t)}^{p-1}\parallel }_1-{\parallel b\parallel }_1}.$$

So,

$${\parallel u^{\prime }\parallel }_{\mathrm{\infty }}\le {\phi }_q\left(\frac{{\parallel c\parallel }_1+{\phi }_p(d_0)}{1-{\parallel a(t){(1+t)}^{p-1}\parallel }_1-{\parallel b\parallel }_1}\right).$$

This, together with (3.6), means that ${\mathrm{\Omega }}_1$ is bounded. The proof is completed. □

Lemma 3.5 Assume that (H₄) holds. The set

$${\mathrm{\Omega }}_2=\{u|u\in KerM,QNu=0\}$$

is bounded in X.

Proof $u\in {\mathrm{\Omega }}_2$ means that $u=at$, $a\in \mathbb{R}$ and $QNu=0$, i.e.,

$$\sum _{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(s,as,a)ds=0.$$

By (H₄), we get that $|a|\le d_0$. So, ${\mathrm{\Omega }}_2$ 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 $\mathrm{\Omega }=\{u\in X|\parallel u\parallel <d_0^{\prime }\}$, where $d_0^{\prime }=max\{d_0,{sup}_{u\in {\mathrm{\Omega }}_1}\parallel u\parallel ,{sup}_{u\in {\mathrm{\Omega }}_2}\parallel u\parallel \}+1$. It follows from the definition of ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$ that $Mu\ne N_{\lambda }u$, $\lambda \in (0,1)$, $u\in \partial \mathrm{\Omega }$ and $QNu\ne 0$, $u\in \partial \mathrm{\Omega }\cap KerM$.

Define a homeomorphism $J:ImQ\to KerM$ as $J(k{e}^{-t})=kt$. If $df(t,x,d)<0$ for $|d|>d_0$, take the homotopy

$$H(u,\mu )=\mu u+(1-\mu )JQNu,u\in \overline{\mathrm{\Omega }}\cap KerM,\mu \in [0,1].$$

For $u\in \overline{\mathrm{\Omega }}\cap KerM$, we have $u=kt$. Then

$$H(u,\mu )=\mu kt-(1-\mu )\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}f(s,ks,k)ds}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}t.$$

Obviously, $H(u,1)\ne 0$, $u\in \partial \mathrm{\Omega }\cap KerM$. For $\mu \in [0,1)$, $u=kt\in \partial \mathrm{\Omega }\cap KerM$, if $H(u,\mu )=0$, we have

$$\frac{{\sum }_{i=1}^n{\alpha }_i{\int }_{{\xi }_i}^{+\mathrm{\infty }}kf(s,ks,k)ds}{{\sum }_{i=1}^n{\alpha }_i{e}^{-{\xi }_i}}=\frac{\mu }{1-\mu }{k}^2\ge 0.$$

A contradiction with $df(t,x,d)<0$, $|d|>d_0$. If $df(t,x,d)>0$, $|d|>d_0$, take

$$H(u,\mu )=\mu u-(1-\mu )JQNu,u\in \overline{\mathrm{\Omega }}\cap KerM,\mu \in [0,1],$$

and the contradiction follows analogously. So, we obtain $H(u,\mu )\ne 0$, $\mu \in [0,1]$, $u\in \partial \mathrm{\Omega }\cap KerM$.

By the homotopy of degree, we get that

$$\begin{array}{rl}deg(JQN,\mathrm{\Omega }\cap KerM,0)& =deg(H(\cdot ,0),\mathrm{\Omega }\cap KerM,0)\\ =deg(H(\cdot ,1),\mathrm{\Omega }\cap KerM,0)=deg(I,\mathrm{\Omega }\cap KerM,0)=1.\end{array}$$

By Theorem 2.1, we can get that $Mu=Nu$ has at least one solution in $\overline{\mathrm{\Omega }}$. The proof is completed. □

4 Example

Let us consider the following boundary value problem at resonance

$$\{\begin{array}{c}{({|u^{\prime }|}^{-\frac{1}{2}}{u}^{\prime })}^{\prime }+\frac{e^{-4t}}{\sqrt{1+t}}sin\sqrt{|u|}+e^{-4t}{|u^{\prime }|}^{-\frac{1}{2}}{u}^{\prime }+\frac{1}{4}{e}^{-4t}=0,0<t<+\mathrm{\infty },\hfill \\ u(0)=0,{|u^{\prime }(+\mathrm{\infty })|}^{-\frac{1}{2}}{u}^{\prime }(+\mathrm{\infty })={\sum }_{i=1}^n{\alpha }_i{|u^{\prime }({\xi }_i)|}^{-\frac{1}{2}}{u}^{\prime }({\xi }_i),\hfill \end{array}$$

(4.1)

where $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_n<+\mathrm{\infty }$, ${\alpha }_i>0$, ${\sum }_{i=1}^n{\alpha }_i=1$.

Corresponding to problem (1.1), we have $p=\frac{3}{2}$, $f(t,x,y)=\frac{e^{-4t}}{\sqrt{1+t}}sin\sqrt{|x|}+e^{-4t}{|y|}^{-\frac{1}{2}}y+\frac{1}{4}{e}^{-4t}$.

Take $a(t)=\frac{e^{-4t}}{\sqrt{1+t}}$, $b(t)=e^{-4t}$, $c(t)=\frac{1}{4}{e}^{-4t}$, $d_0=4$. 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.