\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 $$\label{prob} \begin{gathered} \Delta^2u+c\Delta u = g(x,u) \quad\text{in } \Omega\\ u =\Delta u= 0 \quad\text{on } \partial \Omega, \end{gathered}$$ 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 $g_0< \mu_1< \mu_k \lambda_1(\lambda_1-c)$ 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 $\mu_{k-1}\leq g_0<\mu_{k}\leq\mu_m0 \text{ a. e. }\}$ 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 $\mu_m0$ such that $F(u)\geq -C_1$ for all $u\in H_3$. \end{itemize} \end{lemma} \begin{proof} Because $\mu_m0$ such that $$\label{eq1} G(x,t)\geq \frac{t^2}{2}(\mu_m+\epsilon)-C.$$ 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}\|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 $$\label{g-u1-i} C_{m}(F,u_1)\neq 0.$$ 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 $\mu_10$ such that $F(u)\geq -C_1$ for all $u\in \operatorname{span}\{\varphi_1\}^{\perp}$. \end{itemize} \end{lemma} \begin{proof} (i). Hence $\mu_10$ 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 $\mu_m0$ 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 $\mu_{k-1}0$ 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:} $\nu0$ 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 $\mu_m0$ 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