\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 47, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2015/47\hfil Existence of solutions]
{Existence of solutions for Kirchhoff type equations}
\author[Q.-L. Xie, X.-P. Wu, C.-L. Tang \hfil EJDE-2015/47\hfilneg]
{Qi-Lin Xie, Xing-Ping Wu, Chun-Lei Tang}
\address{Qi-Lin Xie \newline
School of Mathematics and Statistics, Southwest University \\
Chongqing 400715, China.\newline
School of Mathematics Sciences LPMC, Nankai University \\
Tianjing 300071, China}
\email{xieqilinsxdt@163.com}
\address{Xing-Ping Wu (corresponding author) \newline
School of Mathematics and Statistics, Southwest University \\
Chongqing 400715, China}
\email{wuxp@swu.edu.cn}
\address{Chun-Lei Tang \newline
School of Mathematics and Statistics, Southwest University \\
Chongqing 400715, China}
\email{tangcl@swu.edu.cn, tangcl8888@sina.com}
\thanks{Submitted August 13, 2014. Published February 17, 2015.}
\subjclass[2000]{35J60, 47J30, 35J20}
\keywords{ Kirchhoff type equation; Dirichlet problem;
critical point; \hfill\break\indent Mountain Pass Theorem}
\begin{abstract}
In this article, we prove the existence of solutions for Kirchhoff type equations
with Dirichlet boundary-value condition. We use the Mountain Pass Theorem
in critical point theory, without the (PS) condition.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks
\section{Introduction and statement of main results}
Consider the Kirchhoff type problem
\begin{equation}\label{Eq.(5.1)} %\tag{$\mathcal{K}$}
\begin{gathered}
-(a+b\int_\Omega |\nabla u|^2dx)\Delta u=f(x,u), \quad\text{in } \Omega, \\
u=0, \quad\text{on } \partial \Omega,
\end{gathered}
\end{equation}
where $\Omega $ is a smooth bounded domain in $ {\mathbb {R}}^N$, $a,b>0$,
and $f(x,t): \overline{\Omega}\times {\mathbb R}$ is a continuous real
function and satisfies the subcritical condition
\begin{equation}\label{8.1}
|f(x,t)|\leq C(|t|^{p-1}+1)\quad \text{for some }
2
\max \{1, N/2\}$ and $C>0$ such that
$|f (t)|^\sigma\leq CG(t)|t|^\sigma$ for $|t|$ large (see \cite{AM});
or some limitation forms,
\item[(S3)] $\lim_{|t|\to \infty} [f(t)t-4F(t)]=\infty$ (see \cite{YYJ});
\item[(S4)] $ \liminf_{|t|\to\infty}\frac{f(x,t)t-4F(x,t)}{|t|^\tau}
>-\alpha$ uniformly in $x\in \Omega$, where $\tau \in [0, 2]$ and
$0< \alpha4$ such that
$f(x, t)t\geq \theta F(x,t)$ for $|t|$ large, where $F (x, t)=\int_0^tf(x, s)ds$.
\end{itemize}
We consider the nonlinear eigenvalue problem
\begin{equation}
\begin{gathered}
-\Big(\int_{\Omega} |\nabla u|^2 dx\Big)\Delta u=\mu u^3, \quad\text{in }
\Omega, \\
u=0, \quad\text{on } \partial\Omega,
\end{gathered}
\end{equation}
whose the eigenvalues are the critical values of the functional
\begin{equation}
J(u)=\|u\|^4 ,\quad
u\in S:=\big\{u \in H_0^1(\Omega): \int_\Omega |u|^4dx=1\big\},
\end{equation}
where $\|u\|=\big(\int_\Omega |\nabla u|^2dx\big)^{1/2}$.
We already know the first eigenvalue $\mu_1 > 0$ and the first eigenfunction
$\psi_1 > 0$ (see \cite{ZT}).
Now, we can state our main results.
\begin{theorem} \label{thm1.1}
Assume that $f\in C(\Omega\times{\mathbb {R}}, {\mathbb {R}})$ satisfies
\eqref{8.1} and
\begin{itemize}
\item[(F1)] ${\lim_{|t|\to\infty}\big(\frac{a\lambda_1}{2}t^2
+\frac{b\mu_1}{4}t^4-F(x,t)\big)=+\infty}$ uniformly in $x\in \Omega$;
\item[(F2)] there exists $\lambda>\lambda_1$ such that
$F(x,t)\geq\frac{a\lambda}{2}t^2$ for $|t|$ small.
\end{itemize}
Then \eqref{Eq.(5.1)} has at least one nontrivial solution.
\end{theorem}
\begin{remark} \label{rmk1} \rm
Theorem \ref{thm1.1} is a new for the case
$\liminf_{|t|\to\infty}\frac{F(x, t)}{t^4}\leq\frac{b\mu_1}{4}$.
The condition (F1) is weaker than (F3) in \cite{YYJ}.
So our theorem is different from their theorems and obtains one nontrivial
solution by adding the condition (F2) near zero.
\end{remark}
\begin{theorem} \label{thm1.2}
Assume that $f\in C(\Omega\times{\mathbb {R}}, {\mathbb {R}})$ satisfies
\eqref{8.1} and
\begin{itemize}
\item[(F3)] ${\liminf_{|t|\to\infty}\frac{F(x, t)}{t^4}>\frac{b\mu_1}{4}}$
uniformly in $x\in \Omega$;
\item[(F4)] ${\lim_{|t|\to \infty}\big(\frac{1}{4}f(x,t)t-F(x,t)
+\frac{a\lambda_1}{4}t^2\big)=+\infty}$ uniformly in $x\in \Omega$;
\item[(F5)] there exists $\mu<\mu_1$ such that
$F(x,t)\leq\frac{a\lambda_1}{2}t^2+\frac{b\mu}{4}t^4$ for $|t|$ small.
\end{itemize}
Then \eqref{Eq.(5.1)} has at least one nontrivial solution.
\end{theorem}
\begin{remark} \label{rmk2} \rm
Condition (F4) is a new condition for a class of function $f(x,t)$ and is
weaker than (S1)--(S5).
For example, let
$$
f(x,t)=\frac{a\lambda_1}{8}\Big(8t^3\ln(1+t^2)+\frac{4t^5}{1+t^2}
+4t^3\cos t^4\Big).
$$
A simple computation shows that
$$
\frac{1}{4}f(x,t)t-F(x,t)+\frac{a\lambda_1}{4}t^2
= \frac{a\lambda_1}{8}\Big(\frac{t^6 (1+\cos t^4)}{1+t^2}
+ \frac{t^4 (2+\cos t^4)+2t^2}{1+t^2}- \sin t^4\Big)
$$
and
$$
\lim_{|t|\to \infty}\Big(\frac{1}{4}f(x,t)t-F(x,t)+\frac{a\lambda_1}{4}t^2\Big)
=+\infty.
$$
Hence, $f(x,t)$ satisfies all the assumptions of Theorem \ref{thm1.2}, but
it does not satisfy any conditions of (S1)--(S5).
\end{remark}
\section{Preliminaries}
We consider $H:=H_0^1(\Omega)$ endowed with the norm
$\|u\|=\big(\int_\Omega |\nabla u|^2dx\big)^{1/2}$. We denote the usual
$L^p(\Omega)$-norm by $|\cdot|_p$. Since $\Omega $ is a bounded domain,
it is well known that $H \hookrightarrow L^p(\Omega)$ continuously for
$p\in[1, 2^*]$, and compactly for $p\in[1, 2^*)$. Moreover there
exists $\gamma_p>0$ such that
\begin{equation}\label{eq6.1}
|u|_p\leq\gamma_p\|u\|, \quad u\in H.
\end{equation}
Seeking a weak solution of problem \eqref{8.1} is equivalent to finding a
critical point of the $C^1$ functional
\begin{equation}\label{eq6.2}
I(u):=\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4-\int_\Omega F(x,u)dx, \quad u\in H,
\end{equation}
which implies that
\begin{equation}\label{eq6.3}
\langle I'(u),v \rangle =(a+b\|u\|^2)\int_\Omega \nabla u \cdot\nabla v dx-\int_\Omega f(x,u)vdx, \ \ u,v\in H.
\end{equation}
Let
$$
{E_j:=\oplus_{i\leq j} \ker(-\Delta-\lambda_i)},
$$
where $0<\lambda_1\leq\lambda_2\leq\lambda_3 \leq\dots\leq\lambda_i\leq\dots$
are the eigenvalues of $(-\Delta,H)$.
We denote a subsequence of a sequence $\{u_n\}$ as $\{u_n\}$ to simplify
the notation unless specified. We need the following
concept, which was introduced by Cerami \cite{GC} and is a weak version of
the (PS) condition.
\begin{definition}[\cite{GC}] \label{def2.1} \rm
Let $J\in C^1(X,\mathbb {R})$, we say that $J$ satisfies the Cerami condition
at the level $c\in \mathbb {R}$ ($(Ce)_c$ for short), if any sequence
$\{u_n\}\subset X$ with
$$
J(u_n)\to c, \quad (1+\|u_n\|)J'(u_n)\to 0\quad\text{as } n\to \infty,
$$
possesses a convergence subsequence in $X$; $J$ satisfies the $(Ce)$
condition if $J$ satisfies the $(Ce)_c$ for all $c\in \mathbb {R}$.
\end{definition}
The following lemma, which can be found in \cite{DG}, is our main tool in this
article.
\begin{lemma}[Mountain Pass Theorem] \label{lem2.1}
Let $H$ be a real Banach space and $I\in C^1(H, \mathbb R)$ satisfying
the $(Ce)$ condition. Suppose $I(0)=0$,
\begin{itemize}
\item[(i)] there are constants $\rho, \beta> 0$ such that
$I|_{\partial B_\rho}\geq\beta$ where
$$
B_\rho=\{u\in H : \|u\|\leq \rho\};
$$
\item[(ii)] there is $u_1\in H$ and $\|u_1\|>\rho$ such that $I(u_1)<0$.
\end{itemize}
Then $I$ possesses a critical value $c\geq \beta$. Moreover $c$ can be
characterized as
$$
c =\inf_{g\in\Gamma}\max_{u\in g([0,1])} I(u), \quad
\Gamma= \{g\in C([0, 1],H) : g(0) = 0, g(1) = u_1\}.
$$
\end{lemma}
We give a lemma about the $(Ce)$ condition which will play an important role
in the proof of our theorems.
\begin{lemma} \label{lem2.2}
Assume that $f(x,t)$ satisfies \eqref{8.1} and (F4), then $I$ satisfies
the $(Ce)$ condition.
\end{lemma}
\begin{proof}
Suppose that $\{u_n\} $ is a $(Ce)_c$ sequence for $c\in \mathbb {R}$
\begin{equation}\label{eq6.4}
I(u_n)\to c, \quad (1+\|u_n\|)I'(u_n)\to 0 \quad\text{as } n\to \infty.
\end{equation}
Now firstly, we prove that $\{u_n\} $ is a bounded sequence.
From \eqref{eq6.2}, \eqref{eq6.3} and \eqref{eq6.4}, we obtain
\begin{equation}\label{eq6.5}
1+c \geq I(u_n)-\frac{1}{4} I'(u_n)u_n
=\frac{a}{4}\|u_n\|^2+\int_\Omega \Big(\frac{1}{4}f(x,u_n)u_n-F(x,u_n)\Big)dx.
\end{equation}
By (F4), there exists $M>0$ such that
\begin{equation}\label{eq6.6}
\frac{1}{4}f(x,t)t-F(x,t)+\frac{a\lambda_1}{4}|t|^2\geq -M
\end{equation}
for all $x\in\Omega$ and $t\in\mathbb {R}$.
And let $u_n=\phi_n+w_n$, where $\phi_n\in E_1$ and $w_n\in E^{\perp}_1$.
From \eqref{eq6.5} and \eqref{eq6.6}, one obtains
\begin{equation} \label{eq6.7}
\begin{aligned}
1+c
&\geq I(u_n)-\frac{1}{4} I'(u_n)u_n \\
&=\frac{a}{4}\|u_n\|^2-\frac{a\lambda_1}{4}|u_n|^2_2
+\int_\Omega \Big(\frac{1}{4}f(x,u_n)u_n-F(x,u_n)
+\frac{a\lambda_1}{4}|u_n|^2\Big)dx \\
&\geq \frac{a}{4}\big(1-\frac{\lambda_1}{\lambda_2}\big)\|w_n\|^2-M|\Omega|
\end{aligned}
\end{equation}
which implies that $\|w_n\|$ is bounded. We claim that $\{u_n\}$ is a bounded
sequence. Otherwise, there is a subsequence of $\{u_n\}$ satisfying
$\|u_n\| \to+\infty$ as $n\to+\infty$.
Then we obtain
$$
\frac{w_n}{\|u_n\|} \to 0 \in H.
$$
Since $\phi_n/\|u_n\|$ is bounded in $E_1$ ($E_1$ has finite dimension),
we have
$\phi_n/\|u_n\|\to v$ in $E_1$.
By
$$
v_n:=\frac{u_n}{\|u_n\|}=\frac{\phi_n+w_n}{\|u_n\|}
=\frac{\phi_n}{\|u_n\|}+\frac{w_n}{\|u_n\|}\to v\in E_1,
$$
one has
\begin{equation}\label{eq6.8}
\frac{u_n(x)}{\|u_n\|}\to v(x)\quad\text{a.e. in }\Omega.
\end{equation}
From $\|v_n\|=1$, we obtain that $\|v\|=1$. And by $v\in E_1$, one has
that $v(x)>0$ or $v(x)<0$, which implies that
\begin{equation}\label{eq6.9}
|u_n(x)|\to +\infty\quad\text{as }n\to+\infty
\end{equation}
for all $x\in \Omega$ by \eqref{eq6.8}.
It follows from \eqref{eq6.7}, \eqref{eq6.9} and Fatou's lemma that
\begin{align*}
1+c
&\geq I(u_n)-\frac{1}{4} I'(u_n)u_n \\
&=\frac{a}{4}\|u_n\|^2+\int_\Omega \Big(\frac{1}{4}f(x,u_n)u_n-F(x,u_n)\Big)dx \\
&\geq \int_\Omega \Big(\frac{1}{4}f(x,u_n)u_n-F(x,u_n)+\frac{a\lambda_1}{4}|u_n|^2
\Big)dx \\
&\to+\infty\quad\text{as }n \to +\infty,
\end{align*}
which is a contradiction. Then we get that $\{u_n\}$ is bounded in $H$.
Since $f(x,t)$ is subcritical growth, we can easily obtain that
$\{u_n\}$ has a convergence subsequence. Hence, $I$ satisfies the $(Ce)$
condition.
\end{proof}
\section{Proof of main results}
\begin{proof}[Proof of Theorem \ref{thm1.1}]
Let
$$
\overline{u}=\Big(\int_\Omega \nabla u \cdot \nabla \phi_1 dx \Big)\phi_1, \quad
\widetilde{u}=u-\overline{u},
$$
where the $\phi_1$ is the first eigenfunction corresponding to $\lambda_1$.
The following statements come from \cite{SQ}.
First, there exist a real function $g\in L^1(\Omega)$, and
$G\in C(\mathbb{R},\ \mathbb{R})$ which is subadditive; that is,
$$
G(s+t)\leq G(s)+G(t)
$$
for all $s,\ t\in \mathbb{R}$, and coercive; that is,
$G(t)\to+\infty$ as $|t|\to\infty$, and satisfies
$$
G(t)\leq|t|+4
$$
for all $t\in \mathbb{R}$, such that
$$
F(x, t)-\frac{a\lambda_1}{2}t^2-\frac{b\mu_1}{4}t^4\leq-G(t)+g(x)
$$
for all $t\in \mathbb{R}$ and $x\in \Omega$.
Second, the functional $\int_\Omega G(v)dx$ is coercive on $E_1$
(this result also can be seen in \cite{SQ}). We claim that $I(u)$ is coercive.
\begin{align*}
&\int_\Omega \Big( F(x, u)-\frac{a\lambda_1}{2}u^2-\frac{b\mu_1}{4}u^4 \Big)dx\\
&\leq -\int_\Omega G(u)dx+\int_\Omega g(x)dx \\
& \leq -\int_\Omega\left( G(\overline{u})-G(-\widetilde{u})\right) dx
+\int_\Omega g(x)dx \\
& \leq -\int_\Omega G(\overline{u})dx +|\widetilde{u}|_{1}+4 |\Omega|
+\int_\Omega g(x)dx \\
& \leq -\int_\Omega G(\overline{u})dx +C_1(\|\widetilde{u}\|+1)
\end{align*}
for all $u\in H$ and some
$$
C_1=C+4 |\Omega|+\int_\Omega g(x)dx,
$$
where $C$ is a positive constant in Sobolev's inequality,
$$
|u|_{1}\leq C\|u\|,\quad |u|_{2}\leq C\|u\|
$$
for all $u\in H$. Hence we have
\begin{align*}
I(u_n)
&=\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4-\int_\Omega F(x,u)dx \\
&=\frac{a}{2}\|u_n\|^2-\frac{a\lambda_1}{2}|u_n|^2_2
+\frac{b}{4}\|u_n\|^4-\frac{b\mu_1}{4}|u_n|^4_4 \\
&\quad+\int_\Omega\Big(\frac{a\lambda_1}{2}|u_n|^2+\frac{b\mu_1}{4}|u_n|^4
-F(x,u_n)\Big)dx \\
&\geq\frac{a}{2}\|u_n\|^2-\frac{a\lambda_1}{2}|u_n|^2_2
+\int_\Omega\Big(\frac{a\lambda_1}{2}|u_n|^2+\frac{b\mu_1}{4}|u_n|^4-F(x,u_n)\Big)dx \\
&\geq \frac{a}{2}\big(1-\frac{\lambda_1}{\lambda_2}\big)
\|\widetilde{u}\|^2+\int_\Omega G(\overline{u})dx-C_1(\|\widetilde{u}\|+1)
\end{align*}
for all $u\in H$. By the coercivity of the functional
$\int_\Omega G(v)dx$ on $E_1$ and that fact
$$
\|u\|^2=\|\overline{u}\|^2+\|\widetilde{u}\|^2,
$$
which implies that the functional $I(u)$ is coercive. $I$ satisfies the $(Ce)$
condition and is bounded from below. By (F2), we have
$$
F(x,t)\geq \frac{a\lambda}{2}t^2-C|t|^p
$$
for all $x\in \Omega$ and $t\in\mathbb{R}$, which implies that
\begin{align*}
I(u)
&\leq\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4-\frac{a\lambda}{2}|u|_2^2+C|u|_p^p \\
&=\frac{a}{2}\big(1-\frac{\lambda}{\lambda_1}\big)\|u\|^2
+\frac{b}{4}\|u\|^4+C\|u\|^p
<0
\end{align*}
for $u\in E_1\cap B_\delta$, $\lambda>\lambda_1$, where $\delta>0$
small enough and $E_1$ is the subspace of $H$ spanned by $\phi_1$ the
eigenfunctions of $\lambda_1$. Then $I(u)$ achieves the negative infimum.
This completes the proof
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm1.2}]
By Lemmas \ref{lem2.1} and \ref{lem2.2}, it is sufficient to show that $I$ satisfies
(i) and (ii).
\smallskip
\noindent\textbf{Step 1.} There are constants $\rho,\ \beta> 0$ such
that $I(u)\geq\beta$ for all $\|u\|=\rho$.
In fact, by (F5), it is easy to see that
$$
F(x,t)\leq\frac{a\lambda_1}{2}t^2+\frac{b(\mu_1-\varepsilon)}{4}t^4+C|t|^p
$$
for all $t\in \mathbb{R}$ and $x\in \Omega$,
\begin{align*}
I(u)&\geq\frac{a}{2}\|u\|^2+\frac{b}{4}\|u\|^4
-\frac{a\lambda_1}{2}|u|_2^2-\frac{b(\mu_1-\varepsilon)}{4}|u|_4^4
-C\int_\Omega |u|^pdx \\
&\geq\frac{b}{4}\big(1-\frac{\mu_1-\varepsilon}{\mu_1}\big)\|u\|^4-C\gamma_p\|u\|^p.
\end{align*}
Note that $4< p < 2^*$, then for $\varepsilon$ small enough.
So there exists $\beta>0$ such that $I(u)\geq\beta$ for all $\|u\|=\rho$,
where $\rho>0$ small enough.
\smallskip
\noindent\textbf{Step 2.} There exists $u_1\in H$ and $\|u_1\|>\rho$ such that
$I(u_1)<0$.
Indeed, for small $\varepsilon > 0$, by the definition of $\mu_1$, we can
choose $u\in S$ satisfying
\begin{equation}\label{eq7.6}
\mu_1+\frac{\varepsilon}{2}\geq\|u\|^4.
\end{equation}
It follows from (F3) that
\begin{equation}\label{eq7.7}
F(x,t)\geq\frac{b(\mu_1+\varepsilon)}{4}t^4-C.
\end{equation}
Hence, by \eqref{eq7.6} and \eqref{eq7.7}, we have
\begin{equation} \label{eq7.8}
\begin{aligned}
I(tu)
&\leq\frac{a}{2}t^2\|u\|^2+\frac{b}{4}t^4\|u\|^4
-\frac{b}{4}t^4(\mu_1+\varepsilon)+C|\Omega| \\
&\leq\frac{a}{2}t^2\|u\|^2+\frac{b}{4}t^4\mu_1
+\frac{b\varepsilon}{8}t^4-\frac{b}{4}t^4(\mu_1+\varepsilon)+C|\Omega| \\
&=-\frac{b\varepsilon}{8}t^4+\frac{a}{2}t^2\|u\|^2+C|\Omega|.
\end{aligned}
\end{equation}
Thus, $I(tu)\to-\infty$ as $t\to\infty$.
Therefore, there is $u_1 \in H$ with $\|u_1\|>\rho$ such that $I(u_1)< 0$.
This completes the proof.
\end{proof}
\subsection*{Acknowledgements} The authors would like to thank the
anonymous referees and the editors for their valuable suggestions.
This research was supported by National Natural Science
Foundation of China (No.11471267).
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