\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 145, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/145\hfil Asymptotically linear problems]
{Asymptotically linear fourth-order elliptic
problems whose nonlinearity crosses several eigenvalues}

\author[E. Monteiro\hfil EJDE-2011/145\hfilneg]
{Evandro Monteiro}

\address{Evandro Monteiro \newline
UNIFAL-MG, Rua Gabriel Monteiro da Silva, 700. Centro\\
CEP 37130-000 Alfenas-MG, Brazil}
\email{evandromonteiro@unifal-mg.edu.br}

\thanks{Submitted February 15, 2011. Published November 2, 2011.}
\subjclass[2000]{35J30, 35J35}
\keywords{Asymptotically linear; Morse theory; shifting theorem;
\hfill\break\indent multiplicity of solutions}

\begin{abstract}
 In this article we prove the existence of multiple solutions for
 the fourth-order elliptic problem
 \begin{gather*}
 \Delta^2u+c\Delta u = g(x,u) \quad\text{in }  \Omega\\
 u =\Delta u= 0  \quad\text{on }  \partial \Omega,
 \end{gather*}
 where $\Omega \subset \mathbb{R}^N$ is a bounded domain,
 $g:\Omega\times\mathbb{R}\to \mathbb{R}$ is a function
 of class $C^1$ such that $g(x,0)=0$ and it
 is asymptotically linear at infinity.
 We study the cases when  the parameter $c$ is less than the
 first eigenvalue,  and between two consecutive eigenvalues
 of the Laplacian. To obtain solutions we use the Saddle Point Theorem,
 the Linking Theorem, and Critical Groups Theory.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction} \label{intro}

Let us consider the problem
\begin{equation}\label{prob}
\begin{gathered}
\Delta^2u+c\Delta u = g(x,u) \quad\text{in }  \Omega\\
        u =\Delta u= 0   \quad\text{on }  \partial \Omega,
\end{gathered}
\end{equation}
where $\Omega\subset \mathbb{R}^N$ is a bounded domain with smooth
boundary $\partial\Omega$, and $g:\Omega\times\mathbb{R}\to \mathbb{R}$ is a function
of class $C^1$ such that $g(x,0)=0$. Assume that
\begin{gather}\label{g_0}
g_0:=\lim_{t\to 0}\frac{g(x,t)}{t}, \quad \text{uniformly in }\Omega,\\
\label{g_infinito}
g_\infty:=\lim_{|t|\to \infty}\frac{g(x,t)}{t},\quad\text{uniformly in}
\Omega,
\end{gather}
where $g_0$ and $g_\infty$ are constants.

Denote by $0 < \lambda_{1} < \lambda_{2} \leq \dots \leq
\lambda_{j} \leq \dots$ the eigenvalues of $(-\Delta,H_0^1)$ and
$\mu_k(c)=\lambda_k(\lambda_k-c)$ the eigenvalues of
$(\Delta^2+c\Delta,H_0^1\cap H^2)$.  We also denote by $\varphi_j$
the eigenfunction associated with $\lambda_j$ and consequently with
$\mu_j$.

This fourth-order problem  with $g$ asymptotically linear has been
studied by   Qian and Li \cite{qian1}, where the
authors considered the case $c<\lambda_1$ and
$g_0< \mu_1< \mu_k <g_{\infty}<\mu_{k+1}$ and they obtained
three nontrivial solutions.  Tarantelo \cite{tarantello}
found a negative solution of \eqref{prob} with the nonlinearity
of the form $g(x,u)=b[(u+1)^+-1]$, where $b$ is a constant.
With the same type of nonlinearity Micheletti and Pistoia
 \cite{Micheletti1} showed that there exist two solutions
when $b>\lambda_1(\lambda_1-c)$ and three solutions when $b$
is close to $\lambda_k(\lambda_k-c)$.
 Micheletti and Pistoia \cite{Micheletti} showed the existence
of two solutions for  problem \eqref{prob}  with linear growth
at infinity by the classical Mountain Pass Theorem and a variation
of the Linking Theorem. In \cite{qian} the authors considered
the superlinear case and showed the existence of two nontrivial
solutions. Zhang \cite{zhang1} and
Zhang and Li \cite{zhang} proved the existence of solutions when
$f(x,u)$ is sublinear at $\infty$.

In our work we suppose that $c<\lambda_1$ and
$\mu_{k-1}\leq g_0<\mu_{k}\leq\mu_m<g_{\infty}\leq\mu_{m+1}$,
and we prove  the existence of two nontrivial solutions of \eqref{prob}.
 We also obtain results for the case when
$\lambda_\nu <c<\lambda_{\nu+1}$. The case
$\mu_{k-1}< g_\infty<\mu_{k}\leq\mu_m<g_0<\mu_{m+1}$ is also
considered.


The classical solutions of problem \eqref{prob} correspond to
critical points of the functional $F$ defined on
$V=H_0^1(\Omega)\cap H^2(\Omega)$, by
\begin{equation}
\label{funcional} F(u)=\frac{1}{2}\int_{\Omega}(|\Delta
u|^2-c|\nabla u|^2) dx-\int_{\Omega}G(x,u)dx, \quad  u\in V,
\end{equation}
where $G(x,t)=\int_0^tg(x,s)ds$. Notice that $V$ is a Hilbert space
with the usual inner product  $\int_\Omega(|\Delta v|^2+|\nabla
v|^2)dx$. Let $\|\cdot\|$ be norm induced by this inner product.
Under the above assumptions $F$ is a functional of class $C^2$.

For the convenience of the reader, we recall some notation  of Morse
Theory. Let $H$ be a Hilbert space and $F:H\to \mathbb{R}$ be a functional
of class $C^1$. We assume that the set of critical points of $F$,
denoted by $K$, is finite. Let $y\in H$ be a critical point of $F$
with $c = F(y)$. The group
$$
C_p(F, y) = H_p(F^c, F^c \setminus \{y\}), p = 0, 1, 2, \dots,
$$
is called the $p^{th}$ critical group of $F$ at $y$, where
$F^c = \{x \in H : F(x) \leq c\}$ and $H_p(\cdot,\cdot)$ is
the singular relative homology group with integer coefficients.

\section{Case $c<\lambda_1$}

We denote $\frac{d}{dt}g(x,t)$ by $g' (x,t)$. We start
with following  result.

\begin{theorem}\label{teo3}
Assume that $g'(x,t)\geq g(x,t)/t$ for all $x\in\Omega$
and $t\in\mathbb{R}$. Suppose that there exists $k\geq 2$, $m\geq k+1$ such
that $\mu_{k-1}\leq g_0< \mu_{k}$, $\mu_{k-1}<g(x,t)/t$ and
$\mu_{m}< g_\infty< \mu_{m+1}$. Then  problem  \eqref{prob} has
at least two nontrivial solutions.
\end{theorem}

First we will prove that the associated functional satisfies
the Palais-Smale condition.  We remind that
$V$ is a Hilbert space with the inner product
$$
(u,v)_0=\int_{\Omega}(|\Delta
u|^2-c|\nabla u|^2) dx.
$$
Indeed, $\|u\|_0=\sqrt{(u,u)_0}$ is equivalent to norm
 $\|u\|$, provided $c<\lambda_1$.


\begin{lemma}\label{lema1}
If there exists $m\geq 1$ such that $\mu_m<g_\infty <\mu_{m+1}$ then
the functional $F$ defined in \eqref{funcional} satisfies the
Palais-Smale condition.
\end{lemma}

\begin{proof}
Let $(u_n)\subset V$ be a Palais-Smale sequence; that is, a
sequence such that $F(u_n)\to  C$ and
$F'(u_n)\to  0$. Since $g$ is a sublinear function, it is
sufficient to prove that $(\|u_n\|_0)_{n\in\mathbb{N}}$ is bounded. By
contradiction we suppose that
$\lim_{n\to \infty}\|u_n\|_0=\infty$. Up to a subsequence
we can assume that $v_n=u_n/\|u_n\|_0$ converge to $v$ weakly in
$V$, strongly in $L^2(\Omega)$ and pointwise in $\Omega$.

Let $\phi\in V$. Then
$$
F'(u_n)\phi=\int_\Omega (\Delta u_n\Delta \phi
- c\nabla u_n\nabla\phi)dx-\int_\Omega g(x,u_n)\phi dx,
$$
thus
$$
\frac{F'(u_n)}{\|u_n\|_0}\phi=\int_\Omega (\Delta v_n\Delta \phi
- c\nabla v_n\nabla\phi)dx-\int_\Omega \frac{g(x,u_n)}{u_n}v_n\phi
dx.
$$
Taking the limit in the last expression and using the above
convergence, we obtain
\begin{equation}\label{eq2}
    \Delta^2 v+c\Delta v=g_\infty v .
\end{equation}
in the weak sense.

In fact, define $A_+=\{x\in\Omega; v(x)>0 \text{ a. e. }\}$ and
$A_-=\{x\in\Omega; v(x)<0 \text{ a. e. }\}$ then
$u_n(x)\to \infty$  a.e. if $x\in A_+$ and
$u_n(x)\to -\infty$  a.e. if $x\in A_-$. Using
$(g_\infty)$ and the fact that over
$A_0=\{x\in\Omega; v(x)=0 \text{ a.e. }\}$, we obtain
$\frac{g(x,u_n)}{u_n}$ is bounded.

Now we will prove that $v\neq 0$. Note that
$$
\frac{F(u_n)}{\|u_n\|_0^2}=\frac{1}{2}
-\int_\Omega \frac{G(x,u_n)}{\|u_n\|^2_0}dx
=\frac{1}{2}-\int_\Omega\frac{G(x,u_n)}{u_n^2}v_n^2dx.
$$
Taking the limit in this expression and using the fact
$F(u_n)\to  C$ as $n\to \infty$, we obtain
$$
\int_\Omega g_\infty v^2dx=\frac{1}{2},
$$
which proves that $v\neq 0$. Thus, we conclude that $g_\infty$ is
an eigenvalue  of $(\Delta^2+c\Delta, V)$, contradiction.
Therefore, $(\|u_n\|_0)_{n\in\mathbb{N}}$ is bounded. The proof is complete.
\end{proof}

For the next lemma, we split the space $V$ in the following way:
$V=H\oplus H_3$, where
$H=\operatorname{span}\{\varphi_1,\dots,\varphi_m\}$ and
$H_3=H^{\perp}$.

\begin{lemma}\label{lema2}
Suppose  there exists  $m\geq 1$ such that $\mu_m<g_\infty
<\mu_{m+1}$. Then:
\begin{itemize}
\item[(1)] $F(u)\to  -\infty$ as $\|u\|_0\to \infty$ for
 $u\in H$.

 \item[(2)] There exists $C_1>0$ such that $F(u)\geq -C_1 $
for all $u\in H_3$.
\end{itemize}
\end{lemma}

\begin{proof}
Because  $\mu_m<g_\infty$, there exist $\epsilon, C>0$ such that
\begin{equation}\label{eq1}
    G(x,t)\geq \frac{t^2}{2}(\mu_m+\epsilon)-C.
\end{equation}
Thus
\begin{align*}
  F(u)
&=  \frac{1}{2}\int_\Omega
 (|\Delta u|^2-c|\nabla u|^2)dx-\int_\Omega G(x,u)dx \\
&\leq  \frac{1}{2}\|u\|_0^2-\int_{\Omega}\frac{u^2}{2}(\mu_m+\epsilon)dx+C\int_\Omega |u|dx \\
&\leq
   \frac{1}{2}\|u\|_0^2(1-\frac{\mu_m+\epsilon}{\mu_m})+C|\Omega|,
\end{align*}
which proves (1).

Using the fact $g_\infty<\mu_{m+1}$ and a similar argument as
in the proof of (1),  we obtain  (2).
\end{proof}

Now, we  split  the  space $H$  as follows
$$
H=H_1\oplus H_2,
$$
where $H_1=\operatorname{span}\{\varphi_1\dots,\varphi_{k-1}\}$
and $H_2=\operatorname{span}\{\varphi_k,\dots,\varphi_m\}$.
 Thus $V=H_1\oplus H_2\oplus H_3$.

\begin{lemma}\label{lema3}
Suppose that  there are $\alpha,\delta>0$ such that
$\mu_{k-1}\leq g(x,t)/t\leq\alpha<\mu_k$, for $|t|<\delta$,
 $k\geq 2$, and $g'(x,t)\geq \mu_{k-1}$.
Moreover, assume that there exists
$m\geq k+1$ such that $ \mu_{m}< g_\infty< \mu_{m+1}$. The following
statements  hold:
\begin{itemize}
\item[(1)] There are $r>0$ and $A>0$ such that
$F(u)\geq A$ for all $u\in H_2\oplus H_3$  with $\|u\|_0=r$.

\item[(2)] $F(u)\to  -\infty$, as
$\|u\|_0\to \infty$ for all $u\in H_1\oplus H_2$.

\item[(3)] $F(u)\leq 0$ for all $u\in H_1$.
\end{itemize}
\end{lemma}

\begin{proof}
Let $H^{m+1}=\ker(\Delta^2 +c\Delta-\mu_{m+1}I)$.
Then $H_2\oplus H_3=U\oplus W$, where $U=H_2\oplus H^{m+1}$.
 For $v\in V$ put $v=u+w$, $u\in U$ and
$w\in W$. Since $\dim U<+\infty$ then $U$ is generated by
eigenfunctions which are $L^\infty(\Omega)$, then there exists
$r>0$ such that
$$
\sup_{x\in\Omega}|u(x)|\leq\frac{\gamma-\mu_k}{\gamma-\alpha}\delta
\quad\mbox{if }  \|u\|_0\leq r,
$$
where  $\gamma>\mu_k$ and $\int_{\Omega}(|\Delta w|^2-c|\nabla
w|^2)dx\geq \gamma\int_\Omega |w|^2dx$, for all $w\in W$.

Suppose that $\|u\|_0\leq r$. If $|u(x)+w(x)|\leq \delta$, then
\begin{align*}
&\frac{1}{2}\mu_2|u|^2+\frac{1}{4}\gamma|w|^2-G(x,u+w)\\
&\geq   \frac{1}{2}\mu_2|u|^2+\frac{1}{4}\gamma|w|^2-\frac{1}{2}\alpha(u+w)^2 \\
&=  -\frac{1}{4}\alpha|w|^2+\frac{1}{4}(\gamma-\alpha)|w|^2
 + \frac{1}{2}(\mu_2-\alpha)u^2-\alpha uw \\
&\geq  -\frac{1}{4}\mu_2|w|^2+\frac{1}{2}(\mu_2-\alpha)u^2
-\alpha uw
\end{align*}
If $|u(x)+w(x)|>\delta$, then
$$
|G(x,u+w)|\leq\frac{1}{2}\mu_k(u+w)^2
-\frac{1}{2}(\mu_k-\alpha)\delta^2.
$$
Thus,
\begin{align*}
&\frac{1}{2}\mu_2|u|^2+\frac{1}{4}\gamma|w|^2-G(x,u+w)\\
&\geq  \frac{1}{2}\mu_2|u|^2+\frac{1}{4}\gamma |w|^2
 -\frac{1}{2}\mu_k(u+w)^2
 + \frac{1}{2}(\mu_k-\alpha)\delta^2 \\
&=  -\frac{1}{4}\mu_k|w|^2+\frac{1}{2}(\mu_2-\alpha)|u|^2-\alpha uw
  +  \frac{1}{4}(\gamma-\mu_k)|w|^2+(\alpha-\mu_k)uw\\
 &\quad + \frac{1}{2}(\alpha-\mu_k)|u|^2
  + \frac{1}{2}(\mu_k-\alpha)\delta^2 \\
&\geq -\frac{1}{4}\mu_k|w|^2+\frac{1}{2}(\mu_2-\alpha)|u|^2-\alpha uw,
\end{align*}
where the last inequality follows from the fact that the quadrat
form below is positive (see \cite[p. 235]{depa}).
$$
\frac{1}{4}(\gamma-\mu_k)|w|^2+(\alpha-\mu_k)uw
+\frac{1}{2}(\alpha-\mu_k)|u|^2+\frac{1}{2}(\mu_k-\alpha)\delta^2.
$$
Therefore,
\begin{align*}
  F(v)
&=  \frac{1}{2}\|u+w\|_0^2-\int_\Omega G(x,u+w)dx \\
&\geq  \frac{1}{4}\|w\|_0^2-\frac{1}{4}\mu_k\int_\Omega |w|^2dx+\frac{1}{2}(\mu_2-\alpha)\int_\Omega |u|^2dx \\
&\geq  \min\big\{ \frac{1}{4}\big(1-\frac{\mu_k}{\gamma}\big),
\frac{\mu_2-\alpha}{2\mu_k}\big\}\|v\|_0^2,
\end{align*}
which proves assertion (1).

The proof of (2) follows by the same argument as in the proof of (1)
 of Lemma \ref{lema1}. For (3), observe that $g'(x,s)\geq\mu_{k-1}$
and so $G(x,t)\geq\mu_{k-1} t^2/2$. Thus, if $u\in H_1$ then
$u=\sum_{i=1}^{k-1}m_i\varphi_i$ for some constant $m\in\mathbb{R}$. Hence
\begin{align*}
  F(u)
&\leq  \sum_{i=1}^{k-1}\frac{1}{2}\int_\Omega (|\Delta \varphi_i|^2
-c|\nabla \varphi_i|^2)dx-\sum_{i=1}^{k-1}\mu_{k-1}
 \int_\Omega \frac{\varphi_i^2}{2}dx \\
&\leq  \sum_{i=1}^{k-1}\frac{m_i^2}{2}\Big(\|\varphi_i\|_0^2- \mu_i\int_\Omega
   \varphi_i^2\Big)
 =  0.
\end{align*}
which proves $(3)$. The proof of lemma is complete.
\end{proof}


\subsection*{Conclusion of de proof Theorem \ref{teo3}}
By  Lemmas \ref{lema1} and \ref{lema2}, we have that
the functional $F$ satisfies the (PS) condition and has the
geometry of Saddle Point Theorem. Therefore there exists $u_1$, a
critical point of $F$, such that
\begin{equation}\label{g-u1-i}
C_{m}(F,u_1)\neq 0.
\end{equation}
Moreover, by conditions $\mu_{k-1}\leq g_0< \mu_{k}$ and
$g'(x,t)\geq g(x,t)/t$ for all $x\in\Omega$ and $t\in\mathbb{R}$, we
verifies the hypotheses of Lemma \ref{lema3}.  It follows that  the
functional $F$ satisfies the geometry of Linking Theorem. Thus,
there is a critical point  $u_2$ of $F$ satisfying
$$
C_k(F,u_2)\neq 0.
$$
Since $\mu_{k-1}\leq g_0<\mu_k$, then $m(0)+n(0)\leq k-1$, and by
a corollary of Shifting Theorem
\cite[Corollary 5.1, Chapter 1]{chang},
we have $C_p(F,0)=0$ for all $p>k-1$. Therefore $u_1$
and $u_2$ are nontrivial critical points of $F$.
The theorem follows from the next claim.


\noindent\textbf{Claim:}  $C_p(F,u_2)=\delta_{pk}G$.

From \eqref{g-u1-i}) and the Shifting Theorem we have that
$m(u_2)\leq k$. We will show that $m(u_2)=k$. Indeed, by
$g(x,t)/t>\mu_{k-1}$ we have that
$\beta_i(g(x,u_2)/u_2)<\beta_i(\mu_{k-1})\leq 1$ for all $i\leq
k-1$. Now, we have that
$$
\Delta^2u_2+c\Delta u_2=\frac{g(x,u_2)}{u_2}u_2.
$$
This implies that $\beta_k(g(x,u_2)/u_2)\leq 1$.  Then, it follows
from $g'(x,t)\geq g(x,t)/t$, that $\beta_k(g'(x,u_2))<1$. This
implies that $m(u_2)\geq k$, then $m(u_2)=k$.  Again, the Shifting
Theorem and \eqref{g-u1-i}) imply the Claim.


\begin{theorem}\label{teo2}
Assume that $\mu_{k-1}\leq g'(x,t)<\mu_{m+1}$ for all
$x\in\Omega$ and $t\in\mathbb{R}$. Suppose that there exists $k\geq 2$,
$m\geq k+1$ such that $\mu_{k-1}< g_0< \mu_{k}$ and $\mu_{m}<
g_\infty< \mu_{m+1}$.  Then problem {\rm \eqref{prob})} has at
least two nontrivial solutions.
\end{theorem}

\begin{proof}
By hypotheses $\mu_{k-1}< g_0< \mu_{k}$ and
$\mu_{k-1}\leq g'(x,t)<\mu_{m+1}$ for all
$x\in\Omega$ and $t\in\mathbb{R}$, we verifies the Lemma \ref{lema3}.
Thus, as in the proof of the previous theorem there exists
critical points $u_1$ and $u_2$ such that
$$
C_m(F,u_1)\not =0 \quad\mbox{and}\quad
C_k(F,u_2)\not =0,
$$
moreover, we can conclude that $u_1$ and $u_2$ are nontrivial
solutions, provided $\mu_{k-1}\leq g_0<\mu_k$.

We will show that $u_1\not =u_2$. Since $g'(x,t)<\mu_{m+1}$
we obtain
\begin{align*}
  F''(u_1)(v,v)
&=  \int_\Omega (|\Delta v|^2-c|\nabla v|^2)dx-\int_\Omega g' (x,u_1)v^2dx\\
&> \int_\Omega (|\Delta v|^2-c |\nabla
   v|^2)dx-\mu_{m+1}\int_\Omega |v|^2dx
\geq 0,
\end{align*}
for all $v\in \operatorname{span}\{\varphi_{m+1},\dots\}$. Hence
$m(u_1)+n(u_1)\leq m$. On the other hand, $C_m(F,u_1)\neq 0$.
Thus, by a corollary of Shifting Theorem
$C_p(F,u_1)=\delta_{pm}\mathbb{Z}$. Therefore $u_1\neq u_2$, which completes
the proof.
\end{proof}

\begin{theorem}\label{teo4}
Assume that $ \mu_1<g_\infty< \mu_{2}$ and there
exists $m\geq2$ such that $ \mu_{m}< g_0< \mu_{m+1}$. Then
\eqref{prob} has at least two nontrivial solutions.
\end{theorem}

\begin{proof}
By Lemmas \ref{lema1} and \ref{lema2}, we can apply the Saddle
Point Theorem to obtain a solution   $u_1\neq 0$  such
that $C_1(F,u_1)\neq 0$.


\noindent\textbf{Claim:}
$C_p(F,u_1)=\delta_{p1}\mathbb{Z}$.

Actually, we have that $m(u_1)\leq 1$.  If $m(u_1)=1$ the claim is
proved.  If $m(u_1)=0$, then we have that the first eigenvalue
$\beta_1$ of the problem
\begin{equation}
\label{ev2}
\begin{gathered}
\Delta^2v+c\Delta v  = \beta g'(x,u_1)v \quad\text{in }  \Omega  \\
  v=\Delta v  = 0   \quad\text{on }  \partial \Omega,
\end{gathered}
\end{equation}
satisfies $\beta_1=1$ and is simple. It follows that $n(u_1)=1$,
and so the claim follows by Shifting Theorem.

We also have that $C_p(F,0)=\delta_{pm}\mathbb{Z}$, provided
$\mu_m<g_0<\mu_{m+1}$. Now, suppose by contradiction that
$u_1$ and $0$ are the unique critical points of $F$. Thus the Morse
Inequality reads as
$$
(-1)=(-1)+(-1)^m.
$$
This is a contradiction.  So there is at least one more
nontrivial solution.
\end{proof}

\section{The case $\lambda_1<c<\lambda_2$}

Since $\lambda_1<c<\lambda_2$ the first eigenvalue of the problem
\begin{equation}\label{autovalor}
\begin{array}{ccl}
\Delta^2u+c\Delta u = \mu u \quad\text{in} & \Omega\\
        u =\Delta u= 0\ \   \quad\text{on} & \partial \Omega,
\end{array}
\end{equation}
is negative. Thus, $\int_\Omega(|\Delta
v|^2-c|\nabla v|^2)dx$ is not an inner product in $V$.
In this case, consider the following norm:
for all $\phi\in V$
\begin{align*}
  \|\phi\|^2_1
&=  \alpha_1^2\int_\Omega(|\Delta\varphi_1|^2+|\nabla
\varphi_1|^2)dx+\int_\Omega
(|\Delta\overline{\phi}|^2-c|\nabla\overline{\phi}|^2)dx \\
&=  \alpha_1^2(\lambda_1^2+\lambda_1)+\int_\Omega
(|\Delta\overline{\phi}|^2-c|\nabla\overline{\phi}|^2)dx \\
&=  \alpha_1^2(\lambda_1^2+\lambda_1)+\|\overline{\phi}\|^2_0
\end{align*}
where $\phi=\alpha_1\varphi_1+\overline{\phi}$ with
$\overline{\phi}\in \operatorname{span}\{\varphi_1\}^{\perp}$
 and $\|\cdot\|_0$ was defined in the previous section.
 Notice that $\|\cdot\|_0$ is a norm in
$\operatorname{span}\{\varphi_1\}^{\perp}$.

Clearly, the norm $\|\cdot\|_1$ is equivalent to usual
norm $\|\cdot\|$.

Next lemma will prove that the functional  \eqref{funcional}
with the above conditions satisfies the {\it Palais-Smale Condition},
(PS)-Condition.

\begin{lemma}\label{lema4}
Suppose that $g_\infty$ is not eigenvalue from
\eqref{autovalor}. Then the functional \eqref{funcional}
satisfies the (PS)-Condition.
\end{lemma}

\begin{proof}
Let $(u_n)\subset V$ be a Palais-Smale sequence, that is, a
sequence such that $F(u_n)\to  C$ and
$F'(u_n)\to  0$. This lemma is proved with the same
arguments used in Lemma  \ref{lema1}. By contradiction,
suppose that $\lim_{n\to \infty}\|u_n\|_1=\infty$. Up to a
subsequence we can assume that $v_n=u_n/\|u_n\|_1$ converge to $v$
weakly in $V$ strongly in $L^2(\Omega)$ and pointwise in $\Omega$.
Therefore
$$
\Delta^2 v+c\Delta v=g_\infty v .
$$
As in the proof of Lemma \ref{lema1} we have to show that
$v\neq 0$. In fact, let $u_n=t_1^n\varphi +\overline{\phi}_n$,
\begin{equation} \label{eq_PS}
\begin{split}
F(u_n)
&=  \frac{1}{2}\int_\Omega (|\Delta u_n|^2-c|\nabla u_n|^2)dx
 -\int_\Omega G(x,u_n)dx \\
&=  \frac{1}{2}\|u_n\|_1^2-\frac{1}{2}(t_1^n)^2(\lambda_1+c\lambda_1)
 -\int_\Omega  G(x,u_n)dx.
\end{split}
\end{equation}

Since $v_n\to  v$ in $L^2(\Omega)$ as $n\to \infty$
then  $\int v_n\varphi_1\to  \int v\varphi_1 =t_1$ as $n\to \infty$.
Taking limit in the expression
\begin{equation}\label{eq11}
    \frac{F(u_n)}{\|u_n\|_1^2}
=\frac{1}{2}-\frac{1}{2}\frac{(t_1^n)^2}{\|u_n\|^2_1}
(\lambda_1+c\lambda_1)
-\int_\Omega \frac{G(\,u_n)}{u_n^2}v_n^2dx,
\end{equation}
we obtain
\begin{equation}\label{eq12}
    0=\frac{1}{2}-\frac{1}{2}(t_1)^2(\lambda_1+c\lambda_1)
-\int_\Omega g_\infty v^2dx,
\end{equation}
 this implies $v\neq 0$. Thus, Lemma \ref{lema4} is proved.
\end{proof}

In the next result we obtain the functional geometry to establish
existence of two nontrivial solutions from \eqref{prob}).

\begin{lemma} \label{lema5}
Suppose that  $\mu_1<g_\infty <\mu_{2}$. Then
\begin{itemize}
\item[(i)] $F(t\varphi_1)\to  -\infty$, as
$t\to \infty$.

\item[(ii)] There exists $C_1>0$ such that $F(u)\geq -C_1 $
for all $u\in \operatorname{span}\{\varphi_1\}^{\perp}$.
\end{itemize}
\end{lemma}

\begin{proof}
(i).  Hence $\mu_1<g_\infty<\mu_2$ there exists $\epsilon>0$ and
$B>0$ such that
$$
G(x,s)\geq\frac{\mu_1+\epsilon}{2}s^2-B.
$$
So,
\[
  F(t\varphi_1)
\leq  \frac{1}{2}t^2(\lambda^2_1-c\lambda_1)
-\frac{\mu_1+\epsilon}{2}t^2\int_\Omega \varphi_1^2dx+B|\Omega|
=  -\frac{1}{2}t^2\epsilon+B|\Omega|.
\]
this implies $F(t\varphi_1)\to -\infty$ as $t\to
\infty$.

The proof of (ii) is analogous of (ii) of Lemma
\ref{lema2}.
\end{proof}


The next lemma is  analogous to Lemma  \ref{lema3}.

\begin{lemma}\label{lema6}
Suppose that  there are $\alpha,\delta>0$ such that
$\mu_{k-1}\leq g(x,t)/t\leq\alpha<\mu_k$, for $|t|<\delta$,
$k\geq 2$, and $g'(x,t)\geq \mu_{k-1}$. Moreover, assume
that there exists $m\geq k+1$ such that
$ \mu_{m}< g_\infty< \mu_{m+1}$. The following
statements  hold:
\begin{itemize}
\item[(i)] There exists  $r>0$  and $A>0$ such that
$F(u)\geq A$ for all $u\in H_2\oplus H_3$ with $\|u\|_1=r$.

\item[(ii)] $F(u)\to  -\infty$, as
$\|u\|_1\to \infty$ for all $u\in H_1\oplus H_2$.


\item[(iii)] $F(u)\leq 0$ for all $u\in H_1$.
\end{itemize}
\end{lemma}

\begin{proof}
The proof of  (i) is analogous to proof of  (i), Lemma
\ref{lema3}.

Proof of (ii). Let $u\in H_1\oplus H_2$. Then
$u=t\varphi_1+w$, where
$w\in \operatorname{span}\{\varphi_1\}^{\perp}$. By
$\mu_m<g_\infty$ there exists $\epsilon, C>0$ such that
$G(x,s)\geq ((\mu_m+\epsilon)/2) s^2-C$. Thus,
\begin{align*}
  F(u)
&=  \frac{1}{2}\int_\Omega (|\Delta u|^2-c|\nabla u|^2)dx
- \int_\Omega G(x,u)dx \\
&\leq  \frac{1}{2}\|w\|_0^2+\frac{1}{2}t^2\lambda_1(\lambda_1-c)
 -\frac{\mu_m+\epsilon}{2}\int_\Omega(t^2\varphi_1^2+w^2)dx+C|\Omega| \\
&\leq  \frac{1}{2}\|w\|_0^2\big(1-\frac{\mu_m+\epsilon}{\mu_m}\big)
+\frac{1}{2}t^2(\lambda_1^2-c\lambda_1)
-t^2\frac{\mu_m+\epsilon}{2}+C|\Omega|
\end{align*}
this implies $F(u)\to -\infty$ as
$\|u\|_1\to \infty$.

Proof of (iii). Since $g'(x,s)\geq\mu_1$ we obtain
$G(x,s)\geq \mu_1t^2/2$ and
\begin{align*}
  F(t\varphi_1)
&=  \frac{1}{2}t^2\int_\Omega (|\Delta \varphi_1|^2
 -c|\nabla \varphi_1|^2)dx-\int_\Omega G(x,t\varphi_1)dx \\
&\leq  \frac{t^2}{2}(\mu_1-\int_\Omega \mu_1\varphi_1^2dx)
 =  0.
\end{align*}
The proof is complete.
\end{proof}


From Lemmas \ref{lema5} and  \ref{lema6}, we find
analogous geometries as in Lemmas  \ref{lema2} and
\ref{lema3} for functional \eqref{funcional}. Furthermore, we
have the Palais-Smale  Condition  by Lemma \ref{lema4}. Thus, with
the same proofs of Theorems \ref{teo3},
\ref{teo2} and \ref{teo4}, we obtain the following results.

\begin{theorem} \label{teo11}
Assume that $g'(x,t)\geq g(x,t)/t$ for all
$x\in\Omega$ and $t\in\mathbb{R}$. Suppose that there exists $k\geq 2$,
$m\geq k+1$ such that $\mu_{k-1}\leq g_0< \mu_{k}$ and $\mu_{m}<
g_\infty< \mu_{m+1}$ and $\mu_{k-1}<g(x,t)/t$. Then
\eqref{prob} has at least two nontrivial solutions.
\end{theorem}


\begin{theorem} \label{teo7}
Assume that $\mu_{k-1}\leq g'(x,t)<\mu_{m+1}$ for all
$x\in\Omega$ and $t\in\mathbb{R}$. Suppose that there exists $k\geq 2$,
$m\geq k+1$ such that $\mu_{k-1}\leq g_0< \mu_{k}$ and
$\mu_{m}< g_\infty< \mu_{m+1}$.  Then \eqref{prob} has at
least two nontrivial solutions.
\end{theorem}

\begin{theorem}\label{teo5}
Assume that $\mu_1<g_\infty <\mu_{2}$. Suppose there exists $m\geq 2$
such that $\mu_m<g_0<\mu_{m+1}$. Then  \eqref{prob} has at
least two nontrivial solutions.
\end{theorem}


\section{The case $\lambda_\nu<c<\lambda_{\nu+1}$,  $\nu\geq 2$}

In this section we consider $\lambda_\nu<c<\lambda_{\nu+1}$.
Thus, the problem
\begin{equation}\label{autovalor2}
\begin{gathered}
\Delta^2u+c\Delta u = \mu u \quad\text{in }  \Omega\\
        u =\Delta u= 0   \quad\text{on } \partial \Omega,
\end{gathered}
\end{equation}
has $\nu$ first  negative eigenvalues. Therefore, we will define
the following norm in $V$:
\begin{align*}
  \|\phi\|^2_\nu
&=  \sum_{i=1}^{\nu}\alpha_i^2\int_\Omega(|\Delta\varphi_i|^2+|\nabla
\varphi_i|^2)dx+\int_\Omega
(|\Delta\overline{\phi}|^2-c|\nabla\overline{\phi}|^2)dx \\
   &=  \sum_{i=1}^{\nu} \alpha_i^2(\lambda_i^2+\lambda_i)+\int_\Omega
(|\Delta\overline{\phi}|^2-c|\nabla\overline{\phi}|^2)dx \\
&=  \sum_{i=1}^{\nu}
\alpha_i^2(\lambda_i^2+\lambda_i)+\|\overline{\phi}\|^2_0, \quad
\text{for all }  \phi\in V,
\end{align*}
where
$\phi=\alpha_1\varphi_1+\dots+\alpha_\nu\varphi_\nu+\overline{\phi}$
with $\overline{\phi}\in
\operatorname{span}\{\varphi_1,\dots,\varphi_\nu\}^{\perp}$.

In this section, results will be obtained with the same arguments used
in previous section. The Palais-Smale  Condition is proved as
Lemma \ref{lema4} with equation  \eqref{eq_PS} changed by
\begin{align*}
  F(u_n)
&=  \frac{1}{2}\int_\Omega (|\Delta u_n|^2-c|\nabla u_n|^2)dx
 -\int_\Omega G(x,u_n)dx \\
&=  \frac{1}{2}\|u_n\|_\nu^2-\sum_{i=1}^{\nu}\frac{1}{2}(t_i^n)^2
 (\lambda_i+c\lambda_i)-\int_\Omega G(x,u_n)dx.
\end{align*}
and the equation \eqref{eq12} changed by
$$
0=\frac{1}{2}-\frac{1}{2}\sum_{i=1}^{\nu}(t_i)^2(\lambda_i
+c\lambda_i)-\int_\Omega g_\infty v^2dx.
$$
Suppose $V$ as before and $\mu_m<g_\infty <\mu_{m+1}$. We can
split $V=H\oplus W$ where
$H=\operatorname{span}\{\varphi_1,\dots,\varphi_m\}$ and
 $W=H^{\perp}$.

Next lemma is  analogous to Lemma  \ref{lema5}.

\begin{lemma}\label{lema7}
Assume that $\mu_m<g_\infty <\mu_{m+1}$ and $\nu\leq m$. Then
\begin{itemize}
\item[(i)] $F(u)\to  -\infty$, as
$\|u\|_\nu\to \infty$, for $u\in H$.

\item[(ii)] There exists $C_1>0$ such that $F(w)\geq -C_1 $
for all $w\in W$.
\end{itemize}
\end{lemma}

\begin{proof}
The proof of (ii) is similar to the proof of Lemma
\ref{lema5}, (ii).

The proof of (i) follows from $g_\infty>\mu_m$. In fact, let
$u\in H$. Since $\nu\leq m$, we have
$u=\sum_{i=1}^{\nu}t_i\varphi_i+w $. Thus, we have two cases to
consider:

\textbf{Case 1:} $\nu<m$. Then there exists $\epsilon,
B>0$ such that
\begin{align*}
  F(u) &=  \frac{1}{2}\int_\Omega (|\Delta u|^2-c|\nabla u|^2)dx-\int_\Omega G(x,u)dx \\
&\leq  \frac{1}{2}\|w\|_0^2+\frac{1}{2}
 \sum_{i=1}^{\nu}t_i^2(\lambda^2_i-c\lambda_i)
 -\frac{\mu_m+\epsilon}{2}\Big(\sum_{i=1}^{\nu}t_i^2
 +\int_\Omega |w|^2 dx\Big)+B|\Omega| \\
&\leq \frac{1}{2}\|w\|_0^2(1-\frac{\mu_m+\epsilon}{\mu_m})
 +\frac{1}{2}\sum_{i=1}^{\nu}t_i^2(\lambda^2_i
 -c\lambda_i-(\mu_m+\epsilon))+B|\Omega|.
\end{align*}

\textbf{Case 2:} $\nu=m$. Then
\begin{align*}
  F(u)
&=  \frac{1}{2}\int_\Omega (|\Delta u|^2-c|\nabla u|^2)dx
 -\int_\Omega G(x,u)dx \\
&\leq \frac{1}{2}\sum_{i=1}^{\nu}t_i^2(\lambda^2_i-c\lambda_i
 -(\mu_\nu+\epsilon))+B|\Omega|.
\end{align*}

In both cases $F(u)\to -\infty$ as $\|u\|_\nu\to
\infty$, which completes the proof.
\end{proof}

From the  Palais-Smale Condition and Lemma  \ref{lema7},
we obtain the following result.

\begin{theorem}\label{teo8}
Assume that  $\mu_m<g_\infty <\mu_{m+1}$ and $\nu\leq m$. Suppose,
there exists $s\geq m+1$ such that $\mu_s<g_0<\mu_{s+1}$. Then
\eqref{prob} has at least one nontrivial solution.
\end{theorem}

To study multiplicity of solutions we have an analogous
lemma to Lemma \ref{lema6}.

\begin{lemma}\label{lema8}
Assume that $\nu\leq k$.  Suppose that  there are $\alpha,\delta>0$
such that $\mu_{k-1}\leq g(x,t)/t\leq\alpha<\mu_k$, for
$|t|<\delta$,  $k\geq 2$, and $g'(x,t)\geq \mu_{k-1}$.
Moreover, assume that there exists $m\geq k+1$ such that
$ \mu_{m}< g_\infty< \mu_{m+1}$. The following statements  hold:
\begin{itemize}

\item[(i)] There exists $r>0$  and $A>0$ such that $F(u)\geq
A$ for all $u\in H_2\oplus H_3$ with $\|u\|_\nu=r$.

\item[(ii)] $F(u)\to  -\infty$, as
$\|u\|_\nu\to \infty$ for $u\in H_1\oplus H_2$.

\item[(iii)] $F(u)\leq 0$ for all $u\in H_1$.
\end{itemize}
\end{lemma}

Thus we obtain the main theorem of this section.

\begin{theorem}\label{teo9}
Suppose there exist $k\in\mathbb{N}$, $m\geq k+1$ such that
$\mu_{k-1}< g_0<\mu_k$, $\mu_m<g_\infty<\mu_{m+1}$ and $\nu\leq
m$. Assume that $\mu_{k-1}\leq g'(x,t)\leq\mu_{m+1}$, for
all $x\in\Omega$ and $t\in\mathbb{R}$. If $\nu\leq k$  problem
\ref{prob} has at least two nontrivial solutions; If
$k+1\leq\nu$  problem \ref{prob} has at least one nontrivial
solution.
\end{theorem}

\begin{proof}
Since $\nu\leq k+1$ then, by Lemma \ref{lema7} and the
Palais-Smale  Condition,
we conclude that functional $F$  has the geometry of Saddle Point
Theorem. Then there exists $u_1$, a critical point of $F$, such
that
\begin{equation}\label{eq13}
C_m(F,u_1)\neq 0.
\end{equation}
On the other hand, from Lemma  \ref{lema8} there exists $u_2$
a critical point of $F$, such that
\begin{equation}\label{eq14}
C_k(F,u_2)\neq 0.
\end{equation}
The proof is completed with the same arguments as Theorem
\ref{teo2}.

If $k\leq\nu$ is immediate from  Lemma \ref{lema7} and
$\mu_{k-1}\leq g_0<\mu_k$ that there exists nontrivial solution
$u_1$.
\end{proof}

To finish, with the same arguments as in Theorem
\ref{teo3} we obtain the following result.

\begin{theorem}\label{teo12}
Suppose there exist $k\in\mathbb{N}$, $m\geq k+1$ such that
$\mu_{k-1}\leq g_0<\mu_k$, $\mu_m<g_\infty<\mu_{m+1}$ and
$\nu\leq m$. Assume that $g'(x,t)\geq g(x,t)/t$ for all $x\in\Omega$
and $t\in\mathbb{R}$; and $\mu_{k-1}\leq g'(x,t)$. Then: if $\nu\leq
k+1$  problem \ref{prob} has at least two nontrivial
solutions; if $k\leq\nu$  problem \ref{prob} has at least
one nontrivial solution.
\end{theorem}

\subsection*{Acknowledgements}
This work was done while the author was a Ph. D. student at the
Mathematics Department of the State University of Campinas.
The author would like to thank the anonymous referee for his/her
helpful comments and suggestions.


\begin{thebibliography}{00}

\bibitem{bcw}{ T. Bartsch, K. C. Chang}, Z.-Q. Wang,
{\it On the Morse indices
of sign changing solutions of nonlinear elliptic problems}. {Math.
Z.}, \textbf{233} (2000), 655-677.

\bibitem{chang}
{K. C. Chang}, {\it Infinite Dimensional Morse Theory and Multiple
Solutions Problems}, Birkh\"auser, Boston, 1993.

\bibitem{depa}
{F. O. V. De Paiva}, {\it Multiple Solutions for Asymptotically
Linear Ressonant Elliptics Problems}. {Topol. Methods Nonlinear
Anal.}, \textbf{21} (2003), 227-247.

\bibitem{Micheletti1}
{A. M. Micheletti},  A. Pistoia, {\it Multiplicity results for a
fourth-order semilinear elliptic problem}. {Nonlinear Analysis
Theory, Meth. \& Applications}, \textbf{31} 7 (1998), 895-908.

\bibitem{Micheletti}
{A. M. Micheletti},  A. Pistoia, {\it Nontrivial solutions for
some fourth-order semilinear elliptic problems}. {Nonlinear
Analysis}, \textbf{34} (1998), 509-523.

\bibitem{qian1}
{A. Qian},  S. Li {\it Multiple Solutions for a fourth-order
asymptotically linear elliptic problem}. { Acta Mathematica Sinica, English
Series}, vol. 22, no. 4 (2006), 1121-1126.

\bibitem{qian}
{A. Qian},  S. Li {\it On the existence of nontrivial solutions
for a fourth-order semilinear elliptic problems}.  Abstract and
App. Analysis, \textbf{6} (2005), 673-683.

\bibitem{tarantello}
{G. Tarantello}, {\it A note on a semilinear elliptic problem}.
{Diff. and Integral Equations}, vol. 5, no. 3 (1992),
561-565.

\bibitem{zhang1}
{J. Zhang}, {\it Existence results for some fourth-order nonlinear
elliptic problems}, {Nonlinear Analysis} \textbf{45} (2001), 29-36.

\bibitem{zhang}
{J. Zhang},  S. Li {\it Multiple nontrivial solutions for a some
foutrh-order semilinear elliptic problems}, { Nonlinear Analysis}
\textbf{60} (2005), 221-230.

\end{thebibliography}

\end{document}