\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 92, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/92\hfil Bifurcation for elliptic forth-order problems] {Bifurcation for elliptic forth-order problems with quasilinear source term} \author[S. S\^aanouni, N. Trabelsi \hfil EJDE-2016/92\hfilneg] {Soumaya S\^aanouni, Nihed Trabelsi} \address{Soumaya S\^aanouni \newline University of Tunis El Manar, Faculty of Sciences of Tunis, Department of Mathematics, Campus University 2092 Tunis, Tunisia} \email{saanouni.soumaya@yahoo.com} \address{Nihed Trabelsi \newline University of Tunis El Manar, Higher Institute of Medical Technologies of Tunis, 9 Street Dr. Zouhair Essafi 1006 Tunis, Tunisia} \email{nihed.trabelsi78@gmail.com} \thanks{Submitted December 3, 2015. Published April 6, 2016.} \subjclass[2010]{35B32, 35B65, 35B35, 35J62} \keywords{Bifurcation; regularity; stability; quasilinear} \begin{abstract} We study the bifurcations of the semilinear elliptic forth-order problem with Navier boundary conditions \begin{gather*} \Delta^2 u - \operatorname{div} ( c(x) \nabla u ) = \lambda f(u) \quad \text{in }\Omega, \\ \Delta u = u = 0 \quad\text{on } \partial \Omega. \end{gather*} Where $\Omega \subset \mathbb{R}^n$, $n \geq 2$ is a smooth bounded domain, $f$ is a positive, increasing and convex source term and $c(x)$ is a smooth positive function on $\overline{\Omega}$ such that the $L^\infty$-norm of its gradient is small enough. We prove the existence, uniqueness and stability of positive solutions. We also show the existence of critical value $\lambda^*$ and the uniqueness of its extremal solutions. \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} In the literature the term `bifurcation' is used in a general way to indicate stability changes, structural changes in a system etc.. The foundations of the theory has been laid by Poincar\'e who studied branching of solutions in many problem in celestial mechanics and bifurcation, i.e. splitting into two parts, of rotating fluid masses when the rotational velocity reached a certain value. The non-linearity of a phenomenon can have several origins. It often results from the geometry. Wish show very interesting characteristics, namely the existence of multiple solutions, the presence of bifurcations, the passage of a solution to another through loss of stability. The Bifurcations are one of the most interesting events and surprising of nonlinear systems. We say that a system has a bifurcation if an infinitesimal variation of its parameters causes a sudden change of regime. The main interest of non-linear physics lies in its ability to explain the evolution of the problems: a phenomenon usually depends on a number of parameters, called control parameters that control the evolution of the system. By variation of the parameters and the result of non-linearities, the system may undergo transitions. In math, they are bifurcations. Various authors have studied the existence of weak solutions for the bifurcation problem \begin{equation}\label{rad} \begin{gathered} - \Delta u = \lambda f(u) \quad \text{in } \Omega, \\ u > 0 \quad \text{in } \Omega, \\ u = 0 \quad \text{on } \partial \Omega, \end{gathered} \end{equation} where $\Omega$ is a bounded open subset of $\mathbb{R}^n$, $n\geq 2$. Mironescu and R\u{a}dulescu have proved in \cite{M.RAD} that there exists $0<\lambda^*<\infty$, a critical value of the parameter $\lambda$, such that \eqref{rad} has a minimal, positive, classical solution $u_\lambda$ for $0<\lambda<\lambda^*$ and does not have a weak solution for $\lambda>\lambda^*$. Abid et al generalized in \cite{I.M.N} the same result for the Bi-laplace operator. Now, let \[ a:={\lim_{t \to \infty}}\frac{f(t)}{t}. \] The value $a$ was be crucial in the study of $(E_{\lambda^*})$ and of the behavior of $u_\lambda$ when $\lambda$ approaches $\lambda^*$. Also in dimension 4, Wei in \cite{wei}, have studied the behavior of solutions to the following non-linear eigenvalue problem \eqref{rad}. More precisely, when $f(u) = e^u$, we can see that \eqref{rad} is issued from the geometry by prescribing the so-called $\mathcal{Q}$-curvature. For more details, see \cite{G.F.H.E} and \cite{T}. Our main interest here is the study of a bifurcation problem for $\lambda > 0$, \begin{equation} \label{ePl}\ \begin{gathered} \Delta^2 u - \operatorname{div} ( c(x) \nabla u ) = \lambda f(u) \quad \text{in } \Omega, \\ u > 0 \quad \text{in } \Omega, \\ \Delta u = u = 0 \quad \text{on } \partial \Omega, \\ \end{gathered} \end{equation} Where $\Omega$ be a smooth bounded domain in $\mathbb{R}^n$ $(n\geq 2)$, $c(x)$ is a smooth positive function on $\overline{\Omega}$ and $f$ is a positive, increasing and convex smooth function on $(0,+\infty)$, which verifies $$ {\lim_{t \to \infty}}\frac{f(t)}{t} = a \in (0,\infty). $$ \noindent In this paper, we show how the critical problem behaves when he is considered with the Navier boundary condition, we have to use the maximum principle which assured with smallest condition: There exists $\epsilon = \epsilon(n, \Omega)$ such that \[ \|\nabla c \|_\infty \ll \epsilon \] For more detail, see \cite{GGC}. Throughout this article, we denote by $\|\cdot\|_{2}$, the $L^{2}(\Omega)$-norm, whereas we denote by $\|\cdot\|$, the $H^2(\Omega) \cap H^1_0(\Omega)$-norm given by $$ \|u\|^2 = \int_\Omega |\Delta u|^2. $$ We say that $u \in H^2(\Omega) \cap H^1_0(\Omega)$ is a weak solution of \eqref{ePl}, if $f(u) \in L^1(\Omega)$ and $$ \int_\Omega \Delta u \cdot \Delta \varphi + \int_\Omega c(x) \nabla u \cdot \nabla \varphi = \lambda \int_\Omega f(u) \varphi , \quad \forall \varphi \in C^2(\overline{\Omega})\cap H^2(\Omega) \cap H^1_0(\Omega). $$ Such solutions are usually known as weak energy solutions. For short, we will refer to them simply as solutions wish is assuredly by the next lemma. \begin{lemma}\label{faible classique} Since $f(t)\leq at+f(0)$, if $u\in H^2(\Omega) \cap H^1_0(\Omega)$ is a weak solution of \eqref{ePl}, it is easily seen by a standard bootstrap argument that $u$ is always a classical solution. \end{lemma} For more details, see \cite[Proposition 7.15]{GGC}. In the rest of this article, we denote by a solution of \eqref{ePl} any weak or classical solution. \begin{definition}\label{sol min} \rm We say that a solution $u_\lambda$ of \eqref{ePl} is \emph{minimal} if $u_\lambda\leq u$ in $\Omega$ for any solution $u$ of \eqref{ePl}. \end{definition} \begin{definition}\label{def sub-sup sol}\rm We say that $u \in H^2(\Omega) \cap H^1_0(\Omega)$ is a supersolution (resp. subsolution) of \eqref{ePl} if $f(u) \in L^1(\Omega)$ and $$ \Delta^2 u - \operatorname{div} ( c(x) \nabla u ) \geq \lambda f(u) \quad (\text{resp. }\leq \lambda f(u)) \quad \text{in } \mathcal{D}'(\Omega). $$ \end{definition} \begin{definition}\label{sol stable}\rm A solution $u$ of \eqref{ePl} is stable if and only if the first eigenvalue of the linearized operator $$ v \mapsto L_\lambda(v):= \Delta^2 v - \operatorname{div} ( c(x) \nabla v ) - \lambda f'(u)v, $$ given by $$ \eta_1(\lambda,u):=\inf_{\varphi\in H^2(\Omega) \cap H^1_0(\Omega) - \{0\} }\frac{ \int_\Omega|\Delta\varphi|^2 + \int_\Omega c(x) |\nabla \varphi|^2 -\lambda \int_\Omega f'(u)\varphi^2}{\| \varphi\|^2_2}, $$ is nonnegative. If $\eta_1(\lambda, u)<0$, the solution $u$ is said to be \emph{unstable}. \end{definition} Let $v_1$ be a positive eigenfunction (see \cite[section 3.1.3]{GGC}) associated with the first eigenvalue $\lambda_1$ of the operator $\Delta^2 - \operatorname{div}(c(x) \nabla)$ with Navier boundary conditions, namely \begin{equation} \label{eq:v_1} \begin{gathered} \Delta^2 v_1 - \operatorname{div}(c(x) \nabla v_1 ) = \lambda_1 v_1 \quad \text{in } \Omega, \\ \Delta v_1 = v_1 = 0 \quad \text{on } \partial \Omega, \\ \| v_1 \|_2 = 1. \end{gathered} \end{equation} Next, we let $$ \Lambda:=\{\lambda>0: \eqref{ePl}\text{ admits a solution and } \lambda^*:=\sup\Lambda\leq +\infty. $$ We also let $$ r_0 := \inf_{t>0}\frac{f(t)}{t}. $$ The two values $a$ and $r_0$ that we have already defined will be important in the bifurcation phenomena. More precisely, in the frame of the critical value $\lambda^*$. \begin{theorem}\label{thm1} There exists a critical value $\lambda^*\in (0,\infty)$ such that the following properties hold: \begin{itemize} \item[(i)] For any $ \lambda\in (0,\lambda^*)$, problem \eqref{ePl} has a minimal solution $u_\lambda$, which is the unique stable solution of \eqref{ePl}. \item[(ii)] For any $\lambda\in (0,\lambda_1/a)$, $u_\lambda$ is the unique solution of problem \eqref{ePl}. \item[(iii)] The mapping $\lambda \longmapsto u_\lambda$ is increasing. \item[(iv)] $u^* := \lim_{\lambda \to \lambda^*} u_\lambda$ is a solution stable of the problem \eqref{ePl} with $\lambda$ in stead of $\lambda$. In particular, $\eta_1(\lambda^*,u^*)=0$. \end{itemize} \end{theorem} An important role in our arguments will be played by $$ l:= \lim_{t \to \infty} \big(f(t) - a t \big). $$ We distinguish two situations strongly depending on the sign of $l$. \begin{theorem}\label{thm2} Assume that $l\geq 0$. Then \begin{itemize} \item[(i)] $\lambda^*=\lambda_1/a$; \item[(ii)] problem \eqref{ePl} with $\lambda$ in stead of $\lambda$ has no solution; \item[(iii)] $ \lim_{\lambda\to \lambda^*} u_\lambda=\infty$ uniformly on compact subsets of $\Omega$. \end{itemize} \end{theorem} \begin{theorem}\label{thm3} Assume $l< 0$. Then the critical value $\lambda^*$ belongs to $(\lambda_1/a,\lambda_1/r_0)$ and \eqref{ePl} with $\lambda$ in stead of $\lambda$ has a unique solution $u^*$. In this case, \eqref{ePl} has an unstable solution $v_\lambda$ for any $\lambda\in(\lambda_1/a,\lambda^*)$ and the sequence $(v_\lambda)_\lambda$ has the following properties: \begin{itemize} \item[(i)] $ {\lim_{\lambda\to \lambda_1/a}v_\lambda=\infty}$ uniformly on compact subsets of $\Omega$; \item[(ii)] $ {\lim_{\lambda\to \lambda^*}v_\lambda=u^*}$ uniformly in $\Omega$. \end{itemize} \end{theorem} \section{Proof of Theorem \ref{thm1}} The basic idea is to apply the barrier method, when the existence of the critical value $\lambda^*$ is a consequence of the following auxiliary result. \begin{lemma}\label{lem1} Problem \eqref{ePl} has no solution for any $\lambda>\lambda_1/r_0$, but has at least one solution provided $\lambda$ is positive and small enough. \end{lemma} \begin{proof} To show that \eqref{ePl} has a solution, we use the barrier method. To this aim, let $\overline{w} \in H^4(\Omega) $ that satisfies \begin{gather*} \Delta^2 \overline{w} - \operatorname{div} (c(x) \nabla \overline{w} ) = 1 \quad \text{in } \Omega \\ \Delta \overline{w} = \overline{w} = 0 \quad \text{on } \partial\Omega. \end{gather*} The choice of $\overline{w}$ implies that $\overline{w}$ is a super-solution of \eqref{ePl} for $\lambda\leq 1/f(\|\overline{w}\|_\infty)$. Notice that for any $\lambda >0$, the function $\underline{w} \equiv 0$ is a sub-solution of \eqref{ePl} since $f(0)>0$. Next, we define a sequence $w_n\in H^4(\Omega)$ by \begin{equation}\label{eq:w_n} \begin{gathered} \Delta^2 w_{n+1} - \operatorname{div} (c(x) \nabla w_{n+1} ) = \lambda f(w_{n}) \quad \text{in } \Omega \\ \Delta w_{n+1} = w_{n+1} = 0 \quad \text{on } \partial\Omega. \end{gathered} \end{equation} The maximum principle (see \cite{GGC}] implies that $$ \underline{w} \leq w_n \leq w_{n+1} \leq \overline{w} \quad \text{for all } n \in \mathbb{N} , $$ so that the sequence $(w_n)_{n\geq0}$ is increasing and bounded, then it converges. It follows that problem \eqref{ePl} has a solution. Assume now that $u$ is a solution of \eqref{ePl} for some $\lambda>0$. Using $v_1$ given in \eqref{eq:v_1} as a test function and integrating by parts, we obtain \begin{align*} \lambda_1 \int_{\Omega} v_1 u &= \int_{\Omega} ( \Delta^2 v_1 - \operatorname{div}(c(x) \nabla v_1) ) u\\ &= \int_{\Omega} \Delta^2 u v_1 + \int_\Omega c(x) \nabla u \cdot \nabla v_1\\ &= \int_{\Omega} \Delta^2 u v_1 - \int_\Omega \operatorname{div} ( c(x) \nabla u ) v_1 \\ &= \lambda \int_{\Omega}f(u) v_1 \\ &\geq \lambda r_0 \int_{\Omega} u v_1 . \end{align*} This yields $$ (\lambda_1 - \lambda r_0)\int_{\Omega} v_1 u \geq 0. $$ Since $v_1>0$ and $u>0$, we conclude that the parameter $\lambda$ should belong to $(0,\lambda_1/ r_0)$. This completes our proof. \end{proof} Another useful result is stated in what follows. \begin{lemma}\label{lem2} Assume that \eqref{ePl} has a solution for some $\lambda\in(0,\lambda^*)$. Then there exists a minimal solution denoted by $u_\lambda$. Moreover, for any $\lambda'\in(0,\lambda)$, problem \eqref{ePl} with $\lambda'$ instead of $\lambda$ has a solution. \end{lemma} \begin{proof} Fix $\lambda\in(0,\lambda^*)$ and let $u$ be a solution of \eqref{ePl}. As above, we use the barrier method to obtain a minimal solution of \eqref{ePl}. The basic idea is to prove by induction that the sequence $(w_n)_{n\geq0}$ defined in \eqref{eq:w_n} is increasing and bounded by $u$, so it converges to some solution $u_\lambda$. Since $u_\lambda$ is independent of the choice of $u$, then it is a minimal solution. Now, if $u$ is a solution of \eqref{ePl}, then $u$ is a super-solution for the problem \eqref{ePl} with $\lambda'$ instead of $\lambda$ for any $\lambda'$ in $(0,\lambda)$ and $0$ can be used always as a sub-solution. These complete the proof. \end{proof} \begin{remark} \rm Thanks to lemmas \ref{lem1} and \ref{lem2}, the set $\Lambda$ is an interval bounded and not empty. \end{remark} \begin{proof}[Proof of (i) of Theorem \ref{thm1}] First, we claim that $u_\lambda$ is stable. Indeed, arguing by contradiction, i.e. the first eigenvalue $\eta_1(\lambda,u_\lambda)$ is negative. Then, there exists an eigenfunction $ \psi \in H^4(\Omega)$ such that \begin{gather*} \Delta^2\psi - \operatorname{div} ( c(x) \nabla \psi ) - \lambda f'(u_\lambda)\psi = \eta_1 \psi \quad \text{in } \Omega \\ \psi > 0 \quad \text{in } \Omega \\ \Delta \psi = \psi = 0 \quad \text{on } \partial\Omega. \end{gather*} Consider $u^\varepsilon : = u_\lambda - \varepsilon \psi$. Hence, by linearity, we have \begin{align*} &\Delta^2 u^\varepsilon - \operatorname{div} (c(x) \nabla u^\varepsilon ) - \lambda f(u^\varepsilon) \\ &= \lambda f(u_\lambda) - \varepsilon(\Delta^2 \psi - \operatorname{div} (c(x) \nabla \psi )) - \lambda f(u_\lambda - \varepsilon \psi) \\ &= \lambda f(u_\lambda) - \varepsilon(\lambda f'(u_\lambda) \psi + \eta_1 \psi) - \lambda f(u_\lambda - \varepsilon \psi) \\ &= \lambda \Big(- f(u_\lambda - \varepsilon \psi) + f(u_\lambda) - \varepsilon f'(u_\lambda) \psi \Big) - \varepsilon \eta_1 \psi \\ &= \lambda o_{\varepsilon}(\varepsilon \psi) - \varepsilon \eta_1 \psi \\ &= \varepsilon \psi (\lambda o_{\varepsilon}(1) - \eta_1). \end{align*} Since $\eta_1(\lambda,u_\lambda)<0$, for $\varepsilon>0$ small enough, we have $$ \Delta^2 u^\varepsilon - \operatorname{div} (c(x) \nabla u^\varepsilon )-\lambda f(u^\varepsilon)\geq 0 \quad \text{in } \Omega. $$ Then, for $\varepsilon>0$ small enough, we use the strong maximum principle to deduce that $u^\varepsilon\geq0$ is a super-solution of \eqref{ePl}. As before, we obtain a solution $u$ such that $u\leq u^\varepsilon$ and since $u^\varepsilon 0$ in $\Omega$. Thanks to the convexity of $f$, the term in the brackets is nonpositive, hence $$ f(v) - f(u_\lambda) - f'(v)(v - u_\lambda) = 0 \quad \text{in } \Omega, $$ which implies that $f$ is affine over $[u_\lambda,v]$ in $\Omega$. So, there exists two real numbers $\bar{a}$ and $b$ such that $$ f(x) = \bar{a} x + b \quad \text{in } [0,\max_\Omega v]. $$ Finally, since $u_\lambda$ and $v$ are two solutions to $\Delta^2w - \operatorname{div}(c(x) \nabla w) = \lambda \bar{a} w + \lambda b$, we obtain $$ 0 = \int_\Omega \Big(u_\lambda \Delta^2 v - v \Delta^2 u_\lambda \Big) - \int_\Omega \Big( u_\lambda \operatorname{div}(c(x) \nabla v) - v \operatorname{div}(c(x) \nabla u_\lambda) \Big) = \lambda b\int_\Omega (u_\lambda - v). $$ This is impossible since $b=f(0)>0$ and $w = v - u_\lambda$ is positive in $\Omega$. \end{proof} \begin{proof}[Proof of (ii) of Theorem \ref{thm1}] Recall that $\lambda_1$ is defined in \eqref{eq:v_1}. By the convexity of $f$, we deduce that $a=\sup_{\mathbb{R}_+}f'(t)$. Let $u$ be a solution to \eqref{ePl} for $\lambda \in (0,\lambda_1/a)$, we suppose that $u$ is unstable. Then, we can take $\varphi = v_1 \in H^2(\Omega) \cap H^1_0(\Omega)$ which satisfy $$ \lambda a \int_\Omega \varphi^2 \geq \lambda \int_\Omega f'(u) \varphi^2 > \int_\Omega |\Delta \varphi|^2 + \int_\Omega c(x) |\nabla \varphi|^2 = \lambda_1 \int_\Omega \varphi^2, $$ which shows that $$ ( \lambda a - \lambda_1) \int_\Omega \varphi^2 > 0. $$ That is impossible for $\lambda \in (0,\lambda_1/a)$. So, $\eta_1(\lambda, u) \geq 0$ and by (i), we obtain the uniqueness of $u$. For the existence, we consider the minimization problem $$ \min_{u \in H^2(\Omega) \cap H^1_0(\Omega)} \mathfrak{J}(u), $$ where \[ \mathfrak{J}(u):=\frac{1}{2}\int_{\Omega}|\Delta u|^2 + \frac{1}{2} \int_\Omega c(x) |\nabla u|^2 -\lambda\int_{\Omega} \mathfrak{F}(u), \] for all $u\in H^2(\Omega) \cap H^1_0(\Omega)$ with $$ u^+ := \max{(u,0)} \quad\text{and}\quad \mathfrak{F}(u):=\int^{u^+}_0f(s)ds. $$ If $\lambda\in (0,\lambda_1/a)$, there exist $\varepsilon>0$ and $A>0$ depending on $\lambda$ such that $$ 2\lambda \mathfrak{F}(t)\leq (\lambda_1-\varepsilon)t^2+A,\qquad\forall\,t\in\mathbb{R}. $$ Standard arguments imply that $\mathfrak{J}(u)$ is coercive, bounded from below and weakly lower semi-continuous in $ H^2(\Omega) \cap H^1_0(\Omega)$. Hence, the minimum of $\mathfrak{J}$ is attained by some function $u \in H^2(\Omega) \cap H^1_0(\Omega)$ and also by $u^+$ since $\mathfrak{J}(u^+) \leq \mathfrak{J}(u)$. So, the critical point $u$ of $\mathfrak{J}$ gives a solution of \eqref{ePl}. \end{proof} \begin{proof}[Proof of (iii) and (iv) in Theorem \ref{thm1}] By sub- and super-solution method, see \\ Lemma \ref{lem2}, we obtain that the mapping $\lambda\longmapsto u_\lambda$ is increasing and this proves (iii). Now we consider the nonlinear operator $ G:(0,+\infty)\times C^{4,\alpha}(\overline{\Omega})\cap E \to C^{0,\alpha}(\overline{\Omega})$, \[ (\lambda,u) \longmapsto \Delta^{2}u - \operatorname{div}(c(x) \nabla u) - \lambda f(u), \] where $\alpha\in (0,1)$ and $E$ is the function space \begin{equation}\label{E} E := \{u \in W^{4,2}(\Omega) : \Delta u = u = 0 \text{ on } \partial\Omega \}. \end{equation} Assume that \eqref{ePl} with $\lambda$ in stead of $\lambda$ has a solution $u$. Then for any $\lambda\in(0,\lambda^*)$, $u_\lambda\leq u$ in $\Omega$. Using the monotonicity of $u_\lambda$, we deduce that the function $$ u^* = \lim_{\lambda\to\lambda^*} u_\lambda $$ is well defined in $ \Omega$ and is a stable solution of problem \eqref{ePl} with $\lambda$ in stead of $\lambda$. Assuming that the first eigenvalue $\eta_1(\lambda^*,u^*)$ is positive, we can apply the implicit function theorem to the operator $G$. It follows that problem \eqref{ePl} has a solution for $\lambda$ in a neighborhood of $\lambda^{*}$. But this contradicts the definition of $\lambda^{*}$. So, $\eta_1(\lambda^*,u^*)=0$ and this completes the proof of Theorem \ref{thm1}. \end{proof} \begin{remark}\label{rmk2}\rm Thanks to Lemma \ref{lem1} and (ii) of Theorem \ref{thm1}, the critical value $\lambda^*$ satisfies $$ \lambda_1/a \leq \lambda^* \leq \lambda_1/r_0. $$ \end{remark} \section{Proof of Theorem \ref{thm2}} To prove this theorem, we show that the three assertions are equivalent. And finally, we prove that one hoolds. We first recall the following result which is due to H\"{o}rmander \cite{HOR}. \begin{lemma}\label{lem3} Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$, $n\geq 2 $ with smooth boundary. Let $(u_n)$ be a sequence of super-harmonic nonnegative functions defined on $\Omega$. Then the following alternative holds: \begin{itemize} \item[(i)] either $ \lim_{n\to \infty} u_n = \infty$ uniformly on compact subsets of $\Omega$, \item[(ii)] or $(u_n)$ contains a subsequence which converges in $L^1_{\rm loc}(\Omega)$ to some function $u$. \end{itemize} \end{lemma} \begin{remark}\label{rmk3}\rm The result by H\"{o}rmander is also true if $(u_n)$ is a sequence of a super-biharmonic nonnegative functions. \end{remark} First, we assume that $\lambda^* = \lambda_1/a$. If \eqref{ePl} with $\lambda$ in stead of $\lambda$ has a solution $u^*$, then, as we have already observed in (iv) of Theorem \ref{thm1}, $\eta_1(\lambda^*,u^*)=0$. Thus, there exists $\psi\in H^4(\Omega)$ satisfying: \begin{gather*} \Delta^2\psi - \operatorname{div} (c(x) \nabla \psi) - \lambda^* f'(u^*)\psi = 0 \quad \text{in } \Omega \\ \psi > 0 \quad \text{in } \Omega \\ \Delta \psi = \psi = 0 \quad \text{on } \partial\Omega. \end{gather*} Using $v_1$, given in \eqref{eq:v_1}, as a test function and integrating by parts, we obtain $$ \int_\Omega \Big(\Delta^2 v_1 - \operatorname{div}(c(x) \nabla v_1) \Big) \psi - \lambda^* \int_\Omega f'(u^*) \psi v_1 = 0; $$ therefore $$ \int_\Omega \big( \lambda_1 - \lambda^*f'(u^*) \big) \psi v_1 = 0. $$ Since $\lambda_1 - \lambda^*f'(u^*) \geq 0$, the above equation forces $\lambda_1 - \lambda^*f'(u^*)=0$. Hence $$ f'(u^*)\equiv a \quad \text{ in } \quad \Omega. $$ This implies that $f(t) = a t + b$ in $[0, \max_\Omega u^*]$ for some scalar $b>0$. But there is no positive function in $\Omega$ such that $u = \Delta u = 0$ on $\partial\Omega$ and $$ \Delta^2 u - \operatorname{div}(c(x) \nabla u) = \lambda^* a u + \lambda^* b \quad \text{in } \Omega. $$ If not, Using $v_1$ and integrating by parts, we have $$ \int_\Omega \Delta^2 u v_1 - \int_\Omega \operatorname{div}(c(x) \nabla u) v_1 = \lambda^* a \int_\Omega u v_1 + \lambda^* b \int_\Omega v_1 $$ then $$ \int_\Omega \Big( \Delta^2 v_1 - \operatorname{div}( c(x) \nabla v_1 ) \Big) u = \lambda_1 \int_\Omega u v_1 + \lambda^* b \int_\Omega v_1 $$ i.e. $$ 0 = \lambda^* b \int_\Omega v_1 \text{ which is impossible}. $$ Hence, problem \eqref{ePl} with $\lambda$ in stead of $\lambda$ has no solution and (i) implies (ii). Next, we assume that (ii) occurs and we claim that $ \lim_{\lambda\to \lambda^*} u_\lambda = \infty$ uniformly on compact subsets of $\Omega$. If not, by Lemma \ref{lem3} and up to a subsequence, $(u_\lambda)$ converges locally in $L^1(\Omega)$ to $u^*$ as $\lambda \to \lambda^*$. If $u_\lambda$ is not bounded in $L^2(\Omega)$, we define $$ u_\lambda := l_\lambda w_\lambda, $$ with $$ \|w_\lambda\|_2 = 1 \quad \text{and} \quad l_\lambda \to +\infty \quad \text{as } \lambda \to \lambda^*. $$ Since $f(t) \leq a t + f(0)$, we have \begin{align*} \int_\Omega|\Delta w_\lambda|^2 &\leq \int_\Omega |\Delta w_\lambda|^2 + \int_\Omega c(x) |\nabla w_\lambda|^2 \\ &= \int_\Omega \Delta^2 w_\lambda w_\lambda - \int_\Omega \operatorname{div}(c(x) \nabla w_\lambda) w_\lambda = \int_\Omega\frac{\lambda f(u_\lambda)}{l_\lambda}w_\lambda \\ &\leq \lambda^* \int_\Omega\Big(a\,w^2_\lambda +\frac{f(0)}{l_\lambda}w_\lambda\Big) \leq \lambda^* a + c_\lambda \int_\Omega w_\lambda \\ &\leq \lambda^* a + c_\lambda \sqrt{|\Omega|}, \end{align*} where $c_\lambda$ is a positive constant independent on $\lambda$. Recall that $w_\lambda$ satisfies $\Delta^2 w_\lambda - \operatorname{div}(c(x) \nabla w_\lambda) = \frac{\lambda f(l_\lambda w_\lambda)}{l_\lambda}$ and $f$ is quasilinear. These facts imply that $(w_\lambda)$ is bounded in $H^4(\Omega) $. Hence, up to a subsequence, we have $$ \text{$w_\lambda \rightharpoonup w$ weakly in $H^4(\Omega)$ and $w_\lambda \to w $ strongly in $H^3(\Omega)$ as $\lambda\to\lambda^*$}. $$ Moreover, by the trace theorem, \begin{equation}\label{tr} w = \Delta w = 0 \quad \text{on } \partial\Omega. \end{equation} We deduce that $$ \Delta^2 w_\lambda - \operatorname{div}(c(x) \nabla w_\lambda) = \frac{\lambda f(u_\lambda)}{l_\lambda} \to0 \quad \text{in } L^1_{\rm loc}(\Omega) \quad \text{as } \lambda \to \lambda^*. $$ This implies $\Delta^2 w - \operatorname{div}(c(x) \nabla w) = 0$ in $\mathcal{D}'(\Omega)$. So, by \eqref{tr}, we deduce that $w \equiv 0$ in $\Omega$. This contradicts the fact that $\|w\|_2 = \lim_{\lambda\to\lambda^*} \|w_\lambda\|_2 = 1$. Hence, $(u_\lambda)$ is bounded in $L^2(\Omega)$ and by the same arguments as above, it is bounded in $H^4(\Omega) $. This shows that (ii) implies (iii). Moreover, this simply shows that (ii) and (iii) are equivalent. Now, if \eqref{ePl} with $\lambda$ in stead of $\lambda$ has a solution $u^*$, then the sequence $(u_\lambda)$ converges to $u^*$ as $\lambda$ tends to $\lambda^*$, which cannot happen in the case where $ \lim_{\lambda\to \lambda^*}u_\lambda = \infty$. Hence, (iii) implies (i). Indeed, clearly if (ii) and (iii) occur, we have $\lim_{\lambda\to\lambda^*} \|u_{\lambda}\|_2 = \infty$. Set $$ u_\lambda = l_\lambda w_\lambda \quad \text{with } \|w_\lambda\|_2 = 1. $$ Then, up to a subsequence, we obtain $$ \text{$w_\lambda \rightharpoonup w$ weakly in $H^4(\Omega)$ and $w_\lambda \to w$ strongly in $H^3(\Omega)$ as $\lambda\to\lambda^*$}. $$ Moreover, $$ \Delta^2 w_\lambda - \operatorname{div}(c(x) \nabla w_\lambda) \to\Delta^2 w - \operatorname{div}(c(x) \nabla w) \quad \text{in } {\mathcal D}'(\Omega) \quad \text{as } \lambda\to\lambda^* $$ and $$ \frac{\lambda}{l_\lambda} f(l_\lambda w_\lambda) \to \lambda^* a w \quad \text{ in } L^2(\Omega) \quad \text{as } \lambda \to \lambda^*. $$ Then \begin{gather*} \Delta^2 w - \operatorname{div}(c(x) \nabla w) = \lambda^* a w \quad \text{in } \Omega, \\ \Delta w = w = 0 \quad \text{on } \partial \Omega. \end{gather*} Multiplying by $v_1$, which is defined in \eqref{eq:v_1}, we obtain \begin{align*} \int_\Omega \lambda^* a w v_1 &= \int_\Omega \Delta^2 w v_1 - \operatorname{div}(c(x) \nabla w) v_1 \\ &= \int_\Omega \Delta^2 v_1 w - \operatorname{div}(c(x) \nabla v_1) w = \int_\Omega \lambda_1 v_1 w. \end{align*} This proves (i). To finish the proof of Theorem \ref{thm2}, we need only to show that \eqref{ePl} with $\lambda_1/a$ in stead of $\lambda$ has no solution. Indeed, assume that $u$ is a solution of \eqref{ePl} with $\lambda_1/a$ in stead of $\lambda$. Since $f(t)-at \geq 0$, we have $$ \Delta^2 u - \operatorname{div}(c(x) \nabla u) = \frac{\lambda_1}{a} f(u) \geq \lambda_1 u \quad \text{in } \Omega. $$ Multiplying the previous equation by $v_1$ and integrating by parts, we obtain $f(u) = a u$ in $\Omega$, which contradicts $f(0)>0$. This concludes the proof of Theorem \ref{thm2}. \begin{figure}[htb] \begin{center} \setlength{\unitlength}{1cm} \begin{picture}(4.2,4.2)(0,0) \put(0,0){\vector(1,0){4}} \put(4.1,0){$\lambda$} \multiput(2.6,0)(0,0.4){11}{\line(0,1){0.1}} \put(2.5,-0.3) {$\lambda^*$} \put(0,0){\vector(0,1){4.2}} \put(0,4.3){$u_\lambda$} \qbezier(0.0,0.0)(2.2384,0.15) (2.5,4.0) \end{picture} \end{center} \caption{Behavior of the minimal solution.} \label{fig1} \end{figure} Finally, we see that the branch containing the minimal solution has the behavior shown in Figure \ref{fig1}. \begin{remark}\label{rmk4}\rm Observe that the equivalence of the assertions of Theorem \ref{thm2} does not depend on the sign of $l$. \end{remark} \section{Proof of Theorem \ref{thm3}} For the first part of Theorem \ref{thm3}, we have already seen in Remark \ref{rmk2} that $ \lambda_1/a \leq \lambda^* \leq \lambda_1/r_0$. Hence it suffices to prove that $\lambda^*\neq\lambda_1/a$ and $\lambda^*\neq\lambda_1/r_0$. First, assume that $\lambda^*=\lambda_1/a$. Let $u_\lambda$ be the minimal solution to \eqref{ePl}. Then, multiplying \eqref{ePl} by $v_1$ given in \eqref{eq:v_1} and integrating, we obtain \begin{align*} 0 = \int_\Omega \Big (\lambda_1 u_\lambda - \lambda f(u_\lambda)\Big) v_1 &= \int_\Omega \Big((\lambda_1 - a \lambda) u_\lambda - \lambda (f(u_\lambda)- a u_\lambda)\Big) v_1\\ &\geq - \lambda\int_\Omega \big(f(u_\lambda) - a u_\lambda \big) v_1. \end{align*} Passing to the limit in the last inequality as $\lambda$ tends to $\lambda^*$, we find $$ 0 \geq -l \lambda^* \int_\Omega v_1 > 0, $$ which is impossible. Now, assume that $\lambda^* = \lambda_1/r_0$ and let $u$ be a solution of problem \eqref{ePl} with $\lambda$ in stead of $\lambda$. Multiplying \eqref{ePl} with $\lambda$ in stead of $\lambda$ by $v_1$ and integrating by parts, we have $$ \lambda_1\int_\Omega u v_1 = \frac{\lambda_1}{r_0}\int_\Omega f(u) v_1 \geq \lambda_1\int_\Omega u v_1, $$ which forces $f(u) = r_0 u$ in $\Omega$, so that $f(t) = r_0 t$ in $[0, \max_{\Omega}u]$. As above, this contradicts the fact that $f(0)>0$. Since $\lambda^* > \lambda_1/a$, the existence of a solution to \eqref{ePl} with $\lambda$ in stead of $\lambda$ is assured by Remark \ref{rmk4}. Then, it remains to prove the uniqueness. Assume that $u$ is another solution to \eqref{ePl} with $\lambda$ in stead of $\lambda$ and let $w := u - u^*$. Since $u_{\lambda} 0$. Then by convexity of $f$ we have \[ \Delta^2 w - \operatorname{div}(c(x) \nabla w) = \lambda^* (f(u) - f(u^*)) \geq \lambda^* f'(u^*) w \; \text{ in } \Omega. \] Recall that $\eta_1(\lambda^*,u^*) = 0$, so let $\psi$ be the corresponding eigenfunction. Multiplying the last inequality by $\psi$ and integrating by parts, we find $$ 0 = \int_\Omega \lambda^* \Big( f(u) - f(u^*) - f'(u^*) w \Big) \psi \geq 0. $$ Therefore, we must have equality $f(u) - f(u^*) = f'(u^*) w$ \, in $\Omega$, which implies that $f$ is linear in $[0, \max_\Omega u]$ and this leads a contradiction as in the proof of Theorem \ref{thm2}. The second part of Theorem \ref{thm3} concerning the existence of a non stable solution $v_\lambda$ of \eqref{ePl} will be proved by using the mountain pass theorem of Ambrosetti and Rabinowitz \cite{AM.RA} in the following form. \begin{theorem} \label{thm4} Let $E$ be a real Banach space and $J\in C^1(E,\mathbb{R})$. Assume that $J$ satisfies the Palais-Smale condition and the following geometric assumptions: \begin{itemize} \item[(*)] there exist positive constants $R$ and $\rho$ such that $$ J(u) \geq J(u_0) + \rho, \text{ for all } u \in E \text{ with } \|u-u_0\| = R. $$ \item[(**)] there exists $v_0 \in E$ such that $\|v_0 - u_0\| > R$ and $J(v_0) \leq J(u_0)$. \end{itemize} Then the functional $J$ possesses at least a critical point. The critical value is characterized by $$ c := \inf_{g \in \Gamma} \max_{u \in g([0,1])} J(u), $$ where $$ \Gamma := \big\{ g \in C([0,1],E): g(0) = u_0,\, g(1) = v_0 \big\} $$ and satisfies $c \geq J(u_0) + \rho$. \end{theorem} In our case, $ J : E \to \mathbb{R}$ \[ u \longmapsto \frac{1}{2}\int_\Omega|\Delta u|^2 + \frac{1}{2}\int_\Omega c(x) |\nabla u|^2 - \int_\Omega F(u), \] where $E$ is the function space defined in \eqref{E} and $$ F(t)= \lambda\int_{0}^{t}f(s)ds, \quad \text{for all } t\geq 0. $$ We take $u_0$ as the stable solution $u_\lambda$ for each $\lambda\in(\lambda_1/a,\lambda^*)$. \begin{remark}\label{rem-ajt5}\rm The energy functional $J$ belongs to $C^1(E,\mathbb{R})$ and $$ \langle J'(u) , v \rangle = \int_\Omega \Delta u \cdot \Delta v + \int_\Omega c(x) \nabla u \cdot \nabla v - \lambda\int_\Omega f(u) v, \quad \text{for all } u,v \in E. $$ \end{remark} Since $\eta_1(\lambda,u_\lambda) > 0$, the function $u_\lambda$ is a strict local minimum for $J$, we apply the mountain pass theorem for $J$. Using the same arguments of Mironescu and R\u{a}dulescu in \cite[Lemma 9]{M.RAD}, we show in the next lemma that $J$ satisfies the Palais-Smale compactness condition. \begin{lemma}\label{lem4} Let $(u_n)\subset E$ be a Palais-Smale sequence; that is, \begin{gather} \sup_{n \in \mathbb{N}} |J(u_n)|<+\infty ,\label{eq:PS1} \\ \|J'(u_n)\|_{E^*} \to 0 \quad \text{as } n \to \infty . \label{eq:PS2} \end{gather} Then $(u_n)$ is relatively compact in $E$. \end{lemma} \begin{proof} Since any subsequence of $(u_n)$ verifies \eqref{eq:PS1} and \eqref{eq:PS2} it is enough to prove that $(u_n)$ contains a convergent subsequence. It suffices to prove that $(u_n)$ contains a bounded subsequence in $E$. Indeed, suppose we have proved this. Then, up to a subsequence, $u_n \to u$ weakly in $E$, strongly in $L^2(\Omega)$. Now \eqref{eq:PS2} gives $$ \Delta^2 u_n - \operatorname{div}(c(x) \nabla u_n) - \lambda f(u_n) \to0 \quad \text{in } \mathcal{D}'(\Omega) $$ Note that $f(u_n) \to f(u)$ in $L^2(\Omega)$ because $|f(u_n) - f(u)| \leq a |u_n - u|$. This shows that $$ \Delta^2 u_n - \operatorname{div}(c(x) \nabla u_n) \to\lambda f(u) \quad \text{in } \mathcal{D}'(\Omega). $$ That is $$ \Delta^2 u - \operatorname{div}(c(x) \nabla u) - \lambda f(u) = 0 . $$ The above equality multiplied by $u$ gives \begin{equation}\label{eq:PS3} \int_\Omega |\Delta u|^2 + \int_\Omega c(x) |\nabla u|^2 - \lambda \int_\Omega f(u) u = 0 . \end{equation} Now \eqref{eq:PS2} multiplied by $(u_n)$ gives \begin{equation}\label{eq:PS4} \int_\Omega |\Delta u_n|^2 + \int_\Omega c(x) |\nabla u_n|^2 - \lambda \int_\Omega f(u_n) u_n \to0 \end{equation} in view of the boundedness of $(u_n)$ and the $L^2(\Omega)$-convergence of $u_n$ and $f(u_n)$, we have $$ \lambda \int_\Omega f(u_n) u_n \to\lambda \int_\Omega f(u) u $$ Hence, \eqref{eq:PS3} and \eqref{eq:PS4} give $$ \int_\Omega |\Delta u_n|^2 \to\int_\Omega |\Delta u|^2 \quad\text{and}\quad \int_\Omega c(x) |\nabla u_n|^2 \to\int_\Omega c(x) |\nabla u|^2 $$ which insures us that $u_n \to u$ in $E$. Actually, it is enough to prove that $(u_n)$ is (up to a subsequence) bounded in $L^2(\Omega)$. Indeed, the $L^2(\Omega)$-boundedness of $(u_n)$ implies that $E$-boundedness of $(u_n)$ as it can be seen by examining \eqref{eq:PS1}. We shall conclude the proof obtaining a contradiction from the supposition that $\| u_n \|_2 \to \infty$. Let $u_n = k_n w_n$ with $k_n > 0$, $k_n \to\infty$ and $\|w_n\|_2 = 1$. Then $$ 0 = \lim_{n \to \infty} \frac{J(u_n)}{k_n^2} = \lim_{n \to \infty} \Big[ \frac{1}{2} \int_\Omega |\Delta w_n|^2 + \frac{1}{2} \int_\Omega c(x) |\nabla w_n|^2 - \frac{1}{k_n^2} \int_\Omega F(u_n) \Big] $$ However, since $|f(t)| \leq a |t| + b$, we have $$ |F(u_n)| = |F(k_n w_n)| \leq \frac{a \lambda}{2} k_n^2 w_n^2 + b \lambda |k_n w_n|. $$ This shows that $$ \frac{1}{k_n^2} \int_\Omega F(u_n) \leq \frac{a \lambda}{2} \int_\Omega w_n^2 + \frac{b \lambda}{k_n} \int_\Omega w_n < \infty. $$ We claim that \begin{equation}\label{eq:PS5} \Delta^2 w - \operatorname{div}( c(x) \nabla w) = a \lambda w^+ \quad \text{where } w^+ := \max \{0, w\}. \end{equation} Indeed, \eqref{eq:PS2} divided by $k_n$ gives \begin{equation}\label{eq:PS6} \int_\Omega \Delta w_n \cdot \Delta v + \int_\Omega c(x) \nabla w_n \cdot \nabla v - \lambda \int_\Omega \frac{f(u_n)}{k_n} v \to0 \end{equation} for each $v \in E$. Now $$ \int_\Omega \Delta w_n \cdot \Delta v + \int_\Omega c(x) \nabla w_n \cdot \nabla v \to \int_\Omega \Delta w \cdot \Delta v + \int_\Omega c(x) \nabla w \cdot \nabla v $$ Hence \eqref{eq:PS5} can be concluded from \eqref{eq:PS6} if we show that $1 / k_n f(u_n)$ converges (up to a subsequence) to $a w^+$ in $L^2(\Omega)$. Now $1 / k_n f(u_n) = 1 / k_n f(k_n w_n)$ and it is easy to see that the required limit is equal to $a w$ in the set $\{x \in \Omega : w_n(x) \to w(x) \not= 0\}$. If $w(x) = 0$ and $w_n(x) \to w(x)$, let $\varepsilon > 0$ and $n_0$ be such that $|w_n(x)| < \varepsilon$ for $n \geq n_0$. Then $$ \frac{f(k_n w_n)}{k_n} \leq a \varepsilon + \frac{b}{k_n} \quad \text{for such} n, $$ that is the required limit is $0$. Thus, $f(u_n) / k_n \to a w^+$ a.e. Here $b = f(0)$. Now $w_n \to w$ in $L^2(\Omega)$ and, thus, up to a subsequence, $w_n$ is dominated in $L^2(\Omega)$ (see \cite[Theorem IV.9]{BR}). Since $1 / k_n f(u_n) \leq a |w_n| + 1/ k_n b$, it follows that $1 / k_n f(u_n)$ is also dominated. Hence \eqref{eq:PS5} is now obtained. Now \eqref{eq:PS5} and the maximum principle imply that $w \geq 0$ and \eqref{eq:PS5} becomes \begin{equation} \label{eq:PS7} \begin{gathered} \Delta^2 w - \operatorname{div} (c(x) \nabla w) = \lambda a w \quad \text{in } \Omega, \\ w\geq 0 \quad \text{in } \Omega, \\ \|w\|_2 = 1 \quad \text{in } \Omega. \end{gathered} \end{equation} Thus from \eqref{eq:v_1}, we have $\lambda a = \lambda_1$ and $w = v_1$, which contradicts the fact that $\lambda \not= \lambda_1 / a$. This contradiction finishes the proof of the lemma \ref{lem4}. \end{proof} Now, we need only to check that the two geometric assumptions of theorem \ref{thm4} are fulfilled. First, since $u_\lambda$ is a local minimum of $J$, there exists $R>0$ such that for all $u\in E$ satisfying $\|u - u_\lambda\| = R $, we have $J(u)\geq J(u_\lambda)$ . Then $$ J(u) - J(u_\lambda) = J"(u_\lambda)(u - u_\lambda, u - u_\lambda) + \rho \quad \text{where} \rho > 0. $$ This makes $u_\lambda$ becomes a strict local minimal for $J$, which proves $(*)$. \smallskip Recall that $ \lim_{t\to +\infty} (f(t) - a \,t)$ is finite, then there exists $\beta\in\mathbb{R}$ such that $$ f(t) \geq a \, t + \beta, \quad \forall t > 0. $$ Hence \[ F(t)\geq \frac{a\,\lambda}{2}t^2+\beta\lambda t,\quad \forall t>0. \] This yields, using the definition of $v_1$ mentioned in \eqref{eq:v_1}, $$ J (t v_1) = \frac{\lambda_1 - a \lambda}{2} t^2 \int_\Omega v_1 ^2 - \beta \lambda t \int_\Omega v_1, $$ since $\|v_1\|_2 = 1$, then we have \begin{equation} \frac{J_\varepsilon(t v_1)}{t^2} = \frac{\lambda_1 - a \lambda }{2} - \frac{\beta \lambda}{t} \int_\Omega v_1 \label{eq:13} \end{equation} which implies $$ \limsup_{t\to +\infty}\;\frac{1}{t^2}\; J (t v_1) \leq \frac { \lambda_1 - a \lambda}{2} < 0,\quad \forall \lambda > \lambda_1/a. $$ Therefore $$ \lim_{t \to + \infty} \; J (t v_1) = -\infty. $$ So, there exists $v_0 \in E$ such that $ J (v_0) \leq J (u_\lambda)$ and $(**)$ is proved. Finally, let $\tilde{v}$ (respectively $\tilde{c}$) be the critical point (respectively critical value) of $J$, we recall that the function $\tilde{v}$ belongs to $ E$ and satisfies \[ \Delta^2 \tilde{v} - \operatorname{div} (c(x) \nabla \tilde{v}) = \lambda f(\tilde{v}) \quad \text{in } \Omega \quad\text{and}\quad J (\tilde{v}) = \tilde{c}. \] The next lemma states that the limit of a sequence of unstable solutions is also unstable (the proof is similar to that of \cite[Lemma 11]{M.RAD}). \begin{lemma}\label{lem5} Let $u_n\rightharpoonup u$ in $ H^2(\Omega) \cap H^1_0(\Omega)$ and $\mu_n\to \mu$ be such that $\eta_1(\mu_n,u_n)< 0$. Then, $\eta_1(\mu, u) < 0$. \end{lemma} \begin{proof} The fact that $\eta_1(\mu_n,u_n)< 0$ is equivalent to the existence of a $\varphi_n \in H^2(\Omega) \cap H^1_0(\Omega)$ such that \begin{equation}\label{r25} \int_\Omega |\Delta\varphi_n|^2 + \int_\Omega c(x) |\nabla \varphi_n|^2 \leq \mu_n \int_\Omega f'(u_n) \varphi_n^2 \quad \text{with } \int_\Omega \varphi_n^2 = 1 \end{equation} Since $f' \leq a$, \eqref{r25} shows that $(\varphi_n)$ is bounded in $ H^2(\Omega) \cap H^1_0(\Omega)$. Let $\varphi \in E$ be such that, up to a subsequence, $\varphi_n \rightharpoonup \varphi$ in $ H^2(\Omega) \cap H^1_0(\Omega)$. Then $$ \mu_n \int_\Omega f'(u_n) \varphi_n^2 \to\mu \int_\Omega f'(u) \varphi^2 $$ This can be seen by extracting from $(\varphi_n)$ a subsequence dominated in $L^2(\Omega)$) as in \cite[Theorem IV.9]{BR}. Now we have \begin{gather*} \int_\Omega |\Delta\varphi|^2 \leq \liminf \int_\Omega |\Delta\varphi_n|^2,\\ \int_\Omega c(x) |\nabla \varphi|^2 \leq \liminf \int_\Omega c(x) |\nabla \varphi_n|^2 \end{gather*} finally, since $\|\varphi\|_2 = 1$, we obtain $$ \int_\Omega |\Delta\varphi|^2 + \int_\Omega c(x) |\nabla \varphi|^2 \leq \mu \int_\Omega f'(u) \varphi^2. $$ \end{proof} Obviously, the fact that the function $v$ belongs to $C^4(\bar{\Omega}) \cap E$ follows from a bootstrap argument. \begin{proof}[Proof of (i) of Theorem \ref{thm3}] Thanks to Lemma \ref{lem3}, if (i) does not occur, then there is a sequence of positives scalars $(\mu_n)$ and a sequence $(v_n)$ of unstable solutions to $(P_{\mu_n})$ such that $v_n\to v$ in $L^1_{\rm loc}(\Omega)$ as $\mu_n \to \lambda_1/a$ for some function $v$. We first claim that $(v_n)$ cannot be bounded in $E$. Otherwise, let $w\in E$ be such that, up to a subsequence, $$ v_n \rightharpoonup w \text{ weakly in } E \quad \text{ and } \quad v_n \to w \text{ strongly in } L^2(\Omega). $$ Therefore, \begin{gather*} \Delta^2 v_n - \operatorname{div}( c(x) \nabla v_n) \to\Delta^2 w - \operatorname{div}( c(x) \nabla w) \text{ in } \mathcal{D'}(\Omega), \\ f(v_n) \to f(w) \text{ in } L^2(\Omega), \end{gather*} which implies that $\Delta^2 w - \operatorname{div}(c(x) \nabla w) = \frac{\lambda_1}{a} \,f(w)$ in $\Omega$. It follows that $w \in E$ and solves \eqref{ePl} with $\lambda_1/a$ in stead of $\lambda$. From Lemma \ref{lem5}, we deduce that \begin{equation} \eta_1 \Big(\frac{\lambda_1}{a},w\Big)\leq 0.\label{eq:400} \end{equation} Relation \eqref{eq:400} shows that $w \neq u_{\lambda_1/a}$ which contradicts the fact that \eqref{ePl} with $\lambda_1/a$ in stead of $\lambda$ has a unique solution. Now, since $\Delta^2 v_n - \operatorname{div}(c(x) \nabla v_n) = \mu_n f(v_n)$, the unboundedness of $(v_n)$ in $E$ implies that this sequence is unbounded in $L^2(\Omega)$, too. To see this, let $$ v_n = k_n w_n, \quad \text{where } k_n > 0, \quad \|w_n\|_2 = 1 \quad \text{and} \quad k_n \to \infty. $$ Then $$ \Delta^2 w_n - \operatorname{div}(c(x) \nabla w_n) = \frac{\mu_n}{k_n} f(v_n) \to0 \quad \text{in } L^1_{\rm loc}(\Omega). $$ So, we have convergence also in the sense of distributions and $(w_n)$ is seen to be bounded in $E$ with standard arguments. We obtain $$ \Delta^2 w - \operatorname{div}(c(x) \nabla w) = 0 \quad \text{and} \quad \|w\|_2 = 1. $$ The desired contradiction is obtained since $w \in E$. \end{proof} \begin{proof}[Proof of (ii) of Theorem \ref{thm3}] As before, it is sufficient to prove the $L^2(\Omega)$ boundedness of $v_\lambda$ near $\lambda^*$ and to use the uniqueness property of $u^*$. Assume that $\|v_n\|_2 \to \infty$ as $\mu_n \to \lambda^*$, where $v_n$ is a solution to $(P_{\mu_n})$. We write again $v_n = l_n w_n$. Then, \begin{equation} \Delta^2 w_n - \operatorname{div}(c(x) \nabla w_n) = \frac{\mu_n}{l_n} f(v_n).\label{eq:500} \end{equation} The fact that the right-hand side of \eqref{eq:500} is bounded in $L^2(\Omega)$ implies that $(w_n)$ is bounded in $E$. Let $(w_n)$ be such that (up to a subsequence) $$ w_n \rightharpoonup w \text{ weakly in } E \quad \text{and} \quad w_n \to w \text{ strongly in } L^2(\Omega). $$ A computation already done shows that $$ \Delta^2 w - \operatorname{div}(c(x) \nabla w) = \lambda^* a w, \quad w \geq 0 \text{ and } \|w\|_2 = 1, $$ which forces $\lambda^*$ to be $\lambda_1/a$. This contradiction concludes the proof. \end{proof} In the end, Figure \ref{fig2} gives the behavior of the solutions when $l$ is negative. \begin{figure}[htb] \begin{center} \setlength{\unitlength}{1.2cm} \begin{picture}(5.2,4.2)(0,0) \put(0,0){\vector(1,0){5}} \put(5.1,-0.2){$\lambda$} \put(-0.2,-0.2) {0} \put(2.9,-0.3) {$\lambda^*$} \put(1.1,-0.3) {$\frac{\lambda_1}{a}$} \put(4,-0.35) {$\frac{\lambda_1}{r_0}$} \multiput(3,-0.01)(0,0.4){4}{\line(0,1){0.1}} \multiput(1.2,-0.01)(0,0.4){11}{\line(0,1){0.1}} \multiput(4.1,-0.05)(0,0.4){1}{\line(0,1){0.1}} \put(0,0){\vector(0,1){4.2}} \put(0,4.2){$u_\lambda$} \put(-0.3,1.2) {$u^*$} \multiput(0,1.2)(0.4,0){8}{\line(1,0){0.1}} \qbezier(0.0,0.0)(2.5,0.5) (3,1.2) \qbezier (3,1.3)(2.8,1.7)(2,2.1) \qbezier (2,2.1)(1.4,2.6)(1.3,4) \end{picture} \end{center} \caption{Bifurcation branches in the case $l < 0$.} \label{fig2} \end{figure} \subsection*{Acknowledgments} The authors wish to express their gratitude to Professors Sami Baraket and Dong Ye for helpful discussions and suggestions. \begin{thebibliography}{99} \bibitem{I.M.N} I. 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