\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2009(2009), No. 93, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2009/93\hfil A biharmonic elliptic problem] {A biharmonic elliptic problem with dependence on the gradient and the Laplacian} \author[P. C. Carri\~ao, L. F. O. Faria, O. H. Miyagaki\hfil EJDE-2009/93\hfilneg] {Paulo C. Carri\~ao, Luiz F. O. Faria, Ol\'impio H. Miyagaki} % in alphabetical order \address{Paulo C. Carri\~ao \newline Departamento de Matem\'atica, Universidade Federal de Minas Gerais\\ 31270-010 Belo Horizonte (MG), Brazil} \email{carrion@mat.ufmg.br} \address{Luiz F. O. Faria\newline Departamento de Matem\'atica, Universidade Federal de Juiz de Fora\\ 36036-330 Juiz de Fora (MG), Brazil} \email{luiz.faria@ufjf.edu.br} \address{Ol\'impio H. Miyagaki \newline Departamento de Matem\'atica, Universidade Federal de Vi\c cosa \\ 36571-000 Vi\c cosa (MG), Brazil} \email{olimpio@ufv.br} \thanks{Submitted March 22, 2009. Published August 6, 2009.} \subjclass[2000]{31B30, 35G30, 35J40, 47H15} \keywords{Biharmonic; Navier boundary condition; truncation techniques; \hfill\break\indent iteration method} \begin{abstract} We study the existence of solutions for nonlinear biharmonic equations that depend on the gradient and the Laplacian, under Navier boundary condition. Our main tools are an iterative scheme of the mountain pass ``aproximated'' solutions, and the truncation method developed by de Figueiredo, Girardi and Matzeu. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} We prove the existence of nontrivial solutions for the equation \begin{equation}\label{nb} \begin{gathered} \Delta^{2}u+q\Delta u+\alpha(x)u=f(x,u,\nabla u,\Delta u)\quad \text{in } \Omega\\ u(x)=0, \quad \Delta u(x)=0\quad \text{on } \partial \Omega, \end{gathered} \end{equation} where $\Delta^2$ is the biharmonic operator and $\Omega\subset\mathbb{R}^{N}$, $N\geq 1$, is a bounded domain with smooth boundary $\partial \Omega$. The above fourth-order semilinear elliptic problem, when $f$ does not depend on derivatives of $u$, has been studied by many authors; see \cite{Arioli, Bernis, Noussair, pevorst} and references therein. In this case variational techniques are widely applied to obtain existence of solutions. When $\Omega=\mathbb{R}$ and $q >0$ the problem \eqref{nb} is is called the Swift-Hohenberg equation, and for $q>0$ it is called the extended Fisher-Kolmogorov equation. For this class of problems the existence of homoclinic, heteroclinic and periodic solutions have been obtained by several researchers mainly when $f$ does not depend on derivatives; see e. g. \cite{Terracini, Chap, GST, vivi, Berg, Tersian}. The reader is refereed to \cite{Aftabizadeh, Cosner-Fabry, Del pino-Manasevich, Lazer-Mckena2, Ma-Wang} for the case $\Omega = (0,1)$ and $f$ depending on the second order derivative but not on the first derivative of $u$. Recently, the authors in \cite{Luiz} studied a situation where $f$ depends on the first and second order derivatives. For studies with nonlinearities of the form $f(x,u,\Delta u)$ the reader is referred to \cite{ed-fort-serrin, Micheletti-Pistoia, qian-li, Tarantello} and references there in. In our case, due to the presence of the gradient and the Laplacian of $u$ in $f$, the problem is not variational whihc creates additional difficulties. For instance the critical point theory can not be applied directly. We recall that, to overcome this difficult, Xavier \cite{Xavier} and Yan \cite{Yan} handled some semilinear elliptic problems of the second order involving the gradient by using monotone iterative methods. We apply a technique developed by De Figueiredo, Girard and Matzeu \cite{FGM} (see also Girard and Matzeu \cite{gir}) which ``freezes'' the gradient variable and use truncation on the nonlinearity $f$. Thus the new problem becomes variational. The idea of this approach is to consider a class of problems through an iterative scheme where the approximated problem has a nontrivial solution via mountain pass Theorem. Then one obtains estimates in $H^2(\Omega) \cap H^{1}_{0}(\Omega) $-norm and $C^2$-norm. Passing to the limit in a sequence of the approximated solutions we gets a solution of the original problem. In general, a semilinear Navier fourth-order problem is equivalent to the semilinear Dirichlet problem for a system of two coupled second order equations but it is not clear that the truncation method works for the system. To state our results, let us assume the following conditions: \begin{itemize} \item[(A1)] $\alpha$ is a H\"older-continuous function. \item[(A2)] There are positive constants a, b verifying $00$, $p\in (1,\frac{N+4}{N-4})$, ($N\geq 5$), $r_{1}$ and $r_{2}$ , such that $r:= r_{1} + r_{2}<1$ and \[ |f(x,t,\xi_{1},\xi_{2})|\leq a_{1}(1+|t|^{p})(1+|\xi_{1}|^{r_{1}}) (1+|\xi_{2}|^{r_{2}}), \] for all $(x,t,\xi_{1},\xi_{2})\in \Omega\times\mathbb{R}^{N+2}$. \item[(F3)] There exist $\theta>2$ and $t_{0}>0$ such that \[ 0<\theta F(x,t,\xi_{1},\xi_{2})\leq tf(x,t,\xi_{1},\xi_{2}),\quad \forall x \in \Omega, |t|\geq t_{0}, (\xi_{1},\xi_{2}) \in \mathbb{R}^{N+1}, \] where $F(x,t,\xi_{1},\xi_{2})=\int_{0}^{t}f(x,s,\xi_{1},\xi_{2})ds$. \item[(F4)] There exists $a_{2},a_{3}>0$ such that \[ F(x,t,\xi_{1},\xi_{2})\geq a_{2}|t|^{\theta}-a_{3}, \quad \forall x\in \Omega, (t,\xi_{1},\xi_{2})\in \mathbb{R}^{N+2}.\] \end{itemize} Denote by $y^{i,j}$, $(i=1,2,3$) the vectors $$ y^{1,j}=(y^{j}_1,y_2,y_3),\quad y^{2,j}=(y_1,y^{j}_2,y_3),\quad y^{3,j}=(y_1,y_2, y^{j}_3) . $$ For $i,k=1,2,3$ and $j=1,2$, we define the numbers: $$ L_{\rho_{i}}=\sup\big\{ \frac{|f(x,y^{i,1})-f(x,y^{i,2})|}{|y^{1}_{i}-y^{2}_{i}|}: (x,y^{i,j}) \in A_i \big\}, $$ where $$ A_i=\{(x,y^{i,j})\in \Omega\times\mathbb{R}^{N+2}, \; |y^{j}_i|\leq\rho_{i},\; |y_k|\leq\rho_{k} (i \neq k )\}, $$ for some constants $\rho_i> 0$. \begin{itemize} \item[(F5)] There exist positive numbers $\rho_{i}$ ($i=1,2,3$) depending on $q,\theta,a_{1},a_{2}$ and $a_{3}$, in an explicit way, such that the above positive numbers $L_{\rho_{i}}$ $(i=1,2,3)$ satisfy the relation \[ (\tau_{1}L_{\rho_{1}}+\tau_{2}L_{\rho_{2}}+\tau_{3} L_{\rho_{3}})\tau_{1}< \gamma, \] where $\gamma$ is as in Lemma \ref{le0} and $\tau_{i}$ $(i=1,2,3)$ are the optimal constants (that is, the smaller constants) of the inequalities \[ \Big(\int_{\Omega}|u|^{2}dx\Big)^{1/2} \leq\tau_{1}\|u\|,\quad \Big(\int_{\Omega}|\nabla u|^{2}dx\Big)^{1/2}\leq\tau_{2}\|u\|,\quad \Big(\int_{\Omega}|\Delta u|^{2}dx\Big)^{1/2}\leq\tau_{3}\|u\|, \] where $\|u\|^{2}=(u,u)$ and $(u,v)=\int_{\Omega}(\Delta u\Delta v+\nabla u\nabla v+uv)dx$. \end{itemize} Under such hypotheses, we prove the following result. \begin{theorem}\label{teo1} If {\rm (A1)--(A3), (F0)--(F5)} hold, then there exists at least one classical solution of \eqref{nb}. \end{theorem} \noindent \textbf{Example:} Suppose $\beta$ and $\delta$ positive and continuous functions. If $f(x,t,\xi_1,\xi_2 )= \beta (x)|t|t(1+|\xi_1|)^{1/4}(1+|\xi_2|)^{1/4}+\delta(x)t^3$, then it satisfies all the conditions (F0)--(F5). \begin{remark} \label{rmk1} \rm If $\Omega= \delta \Omega'$, with $\delta > 0$ and $\Omega'$ is a bounded domain containing the origin, all functions verifying the growth conditions (F1)--(F4) satisfy the condition (F5) for $\delta$ small sufficient. It occurs because the constants $\tau_i$ ($i=2,3$) and $L_{\rho_i}$ ($i=1,2,3$) do not increase as $\delta$ approaches zero, and, by Poincar\'{e} inequality, we can choose $\tau_1 = (\frac{\delta (\mbox{diameter of }\Omega')}{w_N})^{1/N}$ with ${w_N}$ the measure of the unity ball in $\mathbb{R}^N$. \end{remark} \section{Notation and a technical result} Let $X\equiv H^2(\Omega) \cap H^{1}_{0}(\Omega)$, which is a Hilbert space with inner product and norm given in the previous section. Since (\ref{nb}), in general, is not variational we use a ``freezing'' technique whose formulation appears initially in \cite{FGM}. This technique consists of associating to the problem \eqref{nb} a family of problems without dependence of $f$ in the gradient and Laplacian of the solution. That is, for each $w\in X $ fixed we consider the ``freezed'' problem given by \begin{equation} \begin{gathered} \Delta^{2}u_{w}+q\Delta u_{w}+\alpha(x)u_{w}=f(x,u_{w},\nabla w,\Delta w)\quad\text{in } \Omega \\ u_{w}(x)=0, \quad \Delta u_{w}(x)=0 \quad \text{on } \partial \Omega. \end{gathered} \end{equation} The nonexistence of a priori estimates, with respect to the norms of the gradient and Laplacian of the solution, is the main difficulty for using variational techniques. Thus, we consider, for each $R>0$ fixed, the truncated ``functions'' \[ f_{R}(x,t,\xi_{1},\xi_{2}) =f(x,t,\xi_{1}\varphi_{R}(\xi_{1}),\xi_{2}\varphi_{R}(\xi_{2})), \] and \[ F_{R}(x,t,\xi_{1},\xi_{2}) = \int_{0}^{t}f_{R}(x,s,\xi_{1},\xi_{2} )ds, \] where $\varphi_{R} \in C^1(\mathbb{R})$, $| \varphi_{R} | \leq 1 $ and $$ \varphi_{R}(\xi) = \begin{cases} 1 &\text{if } |\xi|\leq R, \\ 0 & \text{if } | \xi|\geq R+1. \end{cases} $$ This argument appears initially in \cite{Jin}. See also \cite{gir}. \begin{remark} \label{obs-trunc}\rm Note that $|\xi\varphi_{R}(\xi)|\leq R+1$, for all $\xi\in\mathbb{R}$. \end{remark} Thus, for each $w\in X$ and $R>0$ fixed, we consider ``truncated'' and ``freezed'' problem, given by \begin{equation}\label{biarmonic-rt} \begin{gathered} \Delta^{2}u^{R}_{w}+q\Delta u^{R}_{w}+\alpha(x)u^{R}_{w}=f_{R}(x,u^{R}_{w},\nabla w,\Delta w)\quad\text{in } \Omega \\ u^{R}_{w}(x)=0, \quad \Delta u^{R}_{w}(x)=0 \quad \text{on } \partial \Omega. \end{gathered} \end{equation} The associated functional $I_{w}^{R}:X\to \mathbb{R}$ is \begin{equation}\label{func} I_{w}^{R}(v)=\frac{1}{2}\int_{\Omega}[(\Delta v)^{2} -q(\nabla v)^{2}+\alpha(x)v^{2} ]dx - \int_{\Omega}F_{R}(x,v,\nabla w,\Delta w)dx. \end{equation} The following technical Lemma gives us a new and equivalent norm in $X$. \begin{lemma}\label{le0} Suppose $\alpha$ and $q$ satisfy {\rm (A1)--(A3)}. Then there exist positive constants $\eta$ and $\gamma$ such that \[ \gamma\|u\|^{2}\leq \int_{\Omega}(\Delta u^{2}-q\nabla u^{2}+\alpha(x)u^{2} )dx\leq\eta\|u\|^{2}, \quad \forall u \in H^{2}(\Omega). \] \end{lemma} \begin{proof} The constant $\eta$ is obtained taking $\eta = \max\{-q, b, 1 \} $. To obtain $\gamma$, notice that if $q <0$, it is sufficient to take $\gamma = \min\{-q, a, 1 \}$. In the case where $q\geq 0 $, and $\Omega\subset \mathbb{R}$, $\gamma $ will be taken as in \cite[Lemma 8]{Tersian}. For $\Omega\subset \mathbb{R}^{N},N>1 $, the proof can be adapted from \cite[Lemma 8]{Tersian}. \end{proof} \section{Proof of main theorem} We assume $N\geq 5$; the case $N \in [1,4]$ is easier. The proof of Theorem \ref{teo1} is achieved with several lemmas. The following result establishes the mountain pass geometry for the functional $I_{w}^{R}$. \begin{lemma}\label{le1} Let $w\in X $ and $R>0$ be fixed. Then \begin{itemize} \item[(i)] there exist positive constants $\rho=\rho_{R}$ and $\alpha=\alpha_{R}$ such that $I_{w}^{R}(v)\geq\alpha$, for all $v\in X$ with $ \|v\|=\rho$. \item[ii)] fix $v_{0}$ with $\|v_{0}\|=1 $; there is a $T>0$ such that $I^{R}_{w}(tv_{0}) \leq 0$, for all $t>T$. \end{itemize} \end{lemma} \begin{proof} By (F1), given any $\varepsilon>0 $ there exists some $\delta>0 $ such that $|v|<\delta$, implies \begin{equation} \label{eq1} F_{R}(x,v,\nabla w,\Delta w)\leq \varepsilon\frac{v^{2}} {2}. \end{equation} Now, if $|v|\geq\delta $, by (F2) and by Remark \ref{obs-trunc}, there exists some constant $k=k(\delta)$ such that \begin{equation}\label{eq2} F_{R}(x,v,\nabla w,\Delta w)\leq k |v|^{p+1}(R+2)^{r}. \end{equation} Thus, by inequalities (\ref{eq1}) and \eqref{eq2} and by Lemma \ref{le0} we have \[ I_{w}^{R}(v) \geq \frac{\gamma}{2} \|v\|^{2}-\frac{\varepsilon}{2} \int_{\Omega}|v|^{2}dx-k(R+2)^{r}\int_{\Omega}|v|^{p+1}dx. \] So, by the Sobolev embedding Theorem we have \[ I_{w}^{R}(v) \geq \frac{1}{2} (\gamma -C\varepsilon) \|v\|^{2}-kC(R+2)^{r}\|v\|^{p+1}, \] for some positive constant $C$. Then for a $\varepsilon$ small sufficient, we can choose $\rho=\rho_{R}$ and $\alpha=\alpha_{R}$, both independent of $w$, such that the first part of the result holds. Now, take an arbitrary $v_{0}\in X$ with $\|v_{0}\|=1$. By (F4) and Lemma \ref{le0} \[ I_{w}^{R}(tv_{0}) \leq \frac{\eta |t|^{2}} {2}\|v_{0}\|^{2} - a_{2}|t|^{\theta}\int_{\Omega}|v_{0}|^{\theta}dx+a_{3}|\Omega |. \] Since $\theta >2$, it is possible to choose $T>0$ such that $I^{R}_{w}(tv_{0}) \leq 0$, for all $t>T $. \end{proof} \begin{lemma}\label{le2} For any $w\in X $, $R>0 $, problem \eqref{biarmonic-rt} has a nontrivial weak solution. \end{lemma} \begin{proof} First of all, from Lemma \ref{le0}, (F0), (F1) and (F2), the functional $I_{w}^{R}$ is in $C^1(X,\mathbb{R})$; see e.g. \cite{ambrab}. \subsection*{Claim} %\label{ob1} $I_{w}^{R}$ satisfies the Palais-Smale condition; that is, every sequence $(u_n)\subset X$ such that $$ I_{w}^{R}(u_n)\to c\quad\text{and}\quad I_{w}^{'R}(u_n)\to 0, \quad\text{as } n\to \infty, $$ for some constant $c$, contains a convergent subsequence. \subsection*{Verification of the Claim} Note that $$ I_{w}^{R}(u_n)-\frac{1}{\theta}\langle I_{w} ^ {' R}(u_n),u_n\rangle \leq c+\|u_n \|, \quad \forall n>n_{0}. $$ since $\theta>2 $, from Lemma \ref{le0}, it is standard to prove that, $$ \|u_{n}\|0. $$ By the Rellich-Kondrachov Theorem, up to a subsequence, there exists $u\in X$ such that \[ u_{n}\to u \quad\text{in }L^{p+1}(\Omega)\quad\text{as } n\to \infty. \] So, as $ n\to \infty$, we have $$ f_{R}(x,u_{n},\nabla w,\Delta w)\to f_{R}(x,u,\nabla w,\Delta w) \quad \text{in } L^{\frac{p+1}{p}} (\Omega). $$ Therefore, \begin{equation} \label{2b} \int_{\Omega} [f_R(x,u_n,\nabla w,\Delta w)-f_R(x,u,\nabla w, \Delta w)] (u_n -u)dx \to 0, \quad \text{as } n\to \infty. \end{equation} Since $(I_{w} ^ {' R}(u_n) -I_{w} ^ {' R}(u)) \to - I_{w}^ {' R}(u)$ and $u_{n}\rightharpoonup u$ weakly in $X$, we have \begin{equation} \label{2a} \langle I_{w} ^ {' R}(u_n) -I_{w} ^ {' R}(u), u_n-u\rangle \to 0, \quad \text{as } n\to \infty. \end{equation} Notice that by Lemma \ref{le0} \begin{align*} &\langle I_{w} ^ {' R}(u_n) -I_{w} ^ {' R}(u), u_n-u\rangle+ \int_{\Omega} [f_R(x,u_n,\nabla w,\Delta w)-f_R(x,u,\nabla w,\Delta w)] (u_n - u)dx \\ &\geq \gamma \|u_n -u\|^2. \end{align*} Using \eqref{2b} and \eqref{2a} in the above inequality, we obtain that $u_n \to u $(strong) in $X$ as $n\to \infty$. Thus, we conclude that the statement is true. Applying the mountain pass Theorem, due to Ambrosetti-Rabinowitz \cite{ambrab}, there exists $u_w^R \neq 0 $ weak solution to problem \eqref{biarmonic-rt} \end{proof} \begin{lemma}\label{le3} Let $R>0$ be fixed. Then there exist positive constants $d_{1}:=d_{1}(R) $, $d_{2}:=d_{2}(R) $, independent of $w$, such that \[ d_{2}\leq\|u_{w}^{R}\|\leq d_{1}, \] for all solution $u^{R}_{w}$ obtained in Lemma \ref{le2}. \end{lemma} \begin{proof} Notice that $$ I_{w}^{R}(u_w^R) \leq \max_{t\geq 0}I_{w}^{R}(tv_0), $$ with $v_{0}$ given as in Lemma \ref{le1}. From (F4) and Lemma \ref{le0} we obtain $$ I_{w}^{R}(tv_0)\leq \frac{t^2}{2}\eta -a_{2}|t|^{\theta} \int_{\Omega}|v_{0}|^{\theta}dx+a_{3}|\Omega |. $$ Since $\theta>2$ and $|v_{0}|_{\theta}\neq 0 $, the map $$ t\in\mathbb{R}\mapsto \eta\frac{t^{2}} {2}-a_{2}|t| ^{\theta}\int_{\Omega}|v_{0}|^{\theta}dx+a_{3}|\Omega| $$ attains a positive maximum, independent of $w$ and $R$. So we get a constant $C$ such that \begin{equation}\label{fuc} I^{R}_{w}(u_{w}^{R}) \leq C. \end{equation} Now, define $$ |||u|||^{2} = \int_{\Omega}(\Delta u^{2}-q\nabla u^{2} +\alpha(x)u^{2} )dx, $$ which by Lemma \ref{le0} is an equivalent norm in $X$. By (\ref{fuc}), we have \begin{equation} \label{3.55} \frac{1}{2}|||u_{w}^R|||^{2}\leq C +\int_{\Omega}F_{R}(x,u_{w}^R,\nabla w,\Delta w),\quad C>0. \end{equation} Let $t_{0}$ be as in condition (F3), and define $D:=\{x\in\Omega;|u_{w}^R(x)|>t_{0} \}$. Keeping in mind that $u_w^R$ is a solution from (F2) and (F3) and by Remark \ref{obs-trunc}, we have \begin{align*} \int_{\Omega}F_{R}(x,u_{w}^R,\nabla w,\Delta w)&=\int_{\Omega\backslash D}F_{R}(x,u_{w}^R,\nabla w,\Delta w)+\int_{D}F_{R}(x,u_{w}^R,\nabla w,\Delta w) \\ & \leq a_{1}(R+2)^{r}\Big(t_{0}+\frac{|t_{0}|^{p+1}} {p+1}\Big)|\Omega\backslash D|+\frac{1}{\theta}|||u_{w}^R|||^{2}. \end{align*} Returning to equation (\ref{3.55}) we have \[ \frac{1}{2} || | u_{w}^R|||^{2}\leq C+ a_{1}(R+2)^{r}\Big(t_{0}+\frac{|t_{0}|^{p+1}} {p+1}\Big)|\Omega\backslash D|+\frac{1}{\theta}|||u_{w}^R|||^{2}, \] where $|\Omega \backslash D|$ denotes the Lebesgue measure in $\mathbb{R}^N$ of the set $\Omega \backslash D$. Again by Lemma \ref{le0}, we have \[ \gamma\big(\frac{1}{2}-\frac{1}{\theta}\big) \| u_{w}^R\|^{2} \leq\big(\frac{1}{2}-\frac{1}{\theta}\big) ||| u_{w}^R|||^{2} \leq C+ a_{1}(R+2)^{r}\big(t_{0}+\frac{|t_{0}|^{p+1}} {p+1}\big)|\Omega\backslash D |. \] Thus, we can conclude that exists $c_1> 0$ such that $$ \gamma\big(\frac{1}{2}-\frac{1}{\theta}\big)\|u_w^R\|^2 0$. Now, we shall prove that there exists $d_2>0$ such that $\|u_w^R\|>d_2$. In fact, notice that \begin{equation} \label{2c} I_{w} ^ {'R}(u_w^R)(u_w^R)=0, \end{equation} and from (F1) and (F2), given $\varepsilon>0$, there exists $C_\varepsilon>0$ such that \begin{equation} \label{d} |f_R(x,u_w^R,\nabla w,\Delta w)|\leq \varepsilon |u_w^R|+C_{\varepsilon}|u_w^R|^{p}(R+2)^{r}. \end{equation} Inserting $(\ref{d})$ in $(\ref{2c})$ and using Lemma \ref{le0}, we have $$ \gamma\|u_{w}^R\|^{2}\leq C_{1}\varepsilon \|u_{w}^R\|^2+C_{2}C_{\varepsilon}\|u_{w}^R\|^{p+1}(R+2)^{R}, $$ for some constants $C_1, C_2 \geq 0$. Therefore, there exists $d_{2}>0$ such that $\|u_{w}^{R}\|\geq d_{2}$. This completes the proof. \end{proof} \begin{lemma}\label{le4} Choose $w\in C^{4,\alpha}(\overline{\Omega}) $, for some $\alpha\in(0,1)$, and let $R>0$ be fixed. If $u_{w}^{R}\in X$ is a weak solution of problem \eqref{biarmonic-rt}, then $u_{w}^{R}\in C^{4,\beta}(\overline{\Omega}) $, for some $\beta \in(0,1) $, and $\Delta(u_{w}^{R})(x)=0$ if $x\in \partial\Omega$. \end{lemma} \begin{proof} Let $u_w^R\in X$ be a weak solution of \eqref{biarmonic-rt}. Define $v=\Delta u_w^R $ and $$ g(x)=f_R(x,u_w^R,\nabla w, \Delta w)-q\Delta u_w^R-\alpha(x)u_w^R. $$ By hypotheses (F2), and the Sobolev embedding, notice that $g(x)\in L^2(\Omega)$. So, $v$ is a weak solution of $$ \Delta v=g(x),\quad\text{in } \Omega, $$ in the following sense: For $\phi\in C^\infty_c(\Omega) $, we have $$ \int_{\Omega}v\Delta \phi dx=\int_\Omega g\phi dx. $$ From Agmon \cite[Theorem 7.1']{Ag}, we have that $v\in H^2_{\rm loc}(\Omega)$. Therefore, $u_w^R\in H^4_{\rm loc}(\Omega)$. Fix $\phi\in C^\infty_c(\Omega)$, since $u_w^R\in X$ is a weak solution of problem \eqref{biarmonic-rt}, we have \begin{equation}\label{vb111} \int_\Omega\Delta u_w^R\Delta\phi dx -q\int_\Omega\nabla u_w^R\nabla\phi dx+\int_\Omega \alpha(x) u_w^R\phi dx=\int_\Omega f_R(x,u_w^R,\nabla w, \Delta w)\phi dx. \end{equation} But $\mathop{\rm supp}\phi\subset\subset\Omega $, so $$ \int_\Omega(\Delta^2 u_w^R + q\Delta u_w^R+ \alpha(x) u_w^R)\phi dx=\int_\Omega f_R(x,u_w^R,\nabla w, \Delta w)\phi dx. $$ From the denseness of $C^\infty_c(\Omega)$ in $X=H^2(\Omega)\cap H^1_0(\Omega)$ we conclude that \begin{equation}\label{vb11} \int_\Omega(\Delta^2 u_w^R +q\Delta u_w^R+ \alpha(x) u_w^R)\phi dx=\int_\Omega f_R(x,u_w^R,\nabla w, \Delta w)\phi dx\quad \forall\phi\in X. \end{equation} The Green identities guarantees \begin{gather}\label{vb22} \int_\Omega(\Delta u_w^R\Delta\phi - \phi\Delta^2u_w^R) dx=\int_{\partial\Omega}\Big(\Delta u_w^R\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial(\Delta u_w^R)} {\partial\nu}\Big) ds =\int_{\partial\Omega}\Delta u_w^R\frac{\partial\phi}{\partial\nu}ds,\\ \label{vb3} q\int_\Omega\Delta u_w^R\phi dx=q\int_{\partial\Omega}\phi\frac{\partial u_w^R}{\partial\nu}ds -q\int_\Omega\nabla u_w^R\nabla\phi dx. \end{gather} So, combining (\ref{vb111}), (\ref{vb11}), (\ref{vb22}) and (\ref{vb3}), we have \begin{equation}\label{vb333} \int_{\partial\Omega}\Delta u_w^R\frac{\partial\phi}{\partial\nu}ds=0. \end{equation} From (\ref{vb11}), we obtain that \begin{equation}\label{vb4} \Delta^2 u_w^R + q\Delta u_w^R+ \alpha(x) u_w^R = f_R(x,u_w^R,\nabla w, \Delta w) \quad\text{a.e. in }\Omega. \end{equation} By Green's identity, $$ \int_\Omega\Delta^2 u_w^R\Delta u_w^Rdx= -\int_\Omega (\nabla\Delta u_w^R)^2dx+\int_{\partial\Omega}\frac{\partial \Delta u_w^R} {\partial\nu}\Delta u_w^Rds. $$ By (\ref{vb333}), we obtain \begin{equation}\label{p} \int_\Omega\Delta^2 u_w^R\Delta u_w^Rdx=-\int_\Omega (\nabla\Delta u_w^R)^2dx. \end{equation} By Green's identity, $$ \int_\Omega\Delta u_w^R u_w^Rdx=-\int_\Omega (\nabla u_w^R)^2dx+ \int_{\partial\Omega}\frac{\partial u_w^R}{\partial\nu} u_w^Rds\,. $$ Since $u_w^R\in H_0^1(\Omega)$, we get \begin{equation}\label{o} \int_\Omega\Delta u_w^R u_w^Rdx=-\int_\Omega (\nabla u_w^R)^2dx. \end{equation} Multiplying $\Delta u_w^R$ equation (\ref{vb4}), and integrating by parts and using (\ref{p}) and (\ref{o}), we obtain \begin{equation}\label{vb2} \begin{aligned} &-\int_\Omega (\nabla\Delta u_w^R)^2 dx +q\int_\Omega(\Delta u_w^R)^2 dx-\int_\Omega \alpha(x)(\nabla u_w^R)^2 dx\\ &=\int_\Omega f_R(x,u_w^R,\nabla w, \Delta w)\Delta u_w^R dx \end{aligned} \end{equation} By (F2) and Sobolev embbedding we can assume that $$ \Big(\int_\Omega (f_R(x,u_w^R,\nabla w, \Delta w))^2 dx\Big)^{1/2} <\infty\,. $$ Thus, from (\ref{vb2}) we have $\|u_w^R\|_{W^{3,2}(\Omega)}<\infty$. Let be $\varphi\in C_c^1(\mathbb{R}^N)$. Integrating by parts, we have $$ \int_\Omega\Delta u_w^R \frac{\partial \varphi}{\partial x_i}dx=-\int_\Omega \varphi \frac{\partial \Delta u_w^R}{\partial x_i}+\int_{\partial\Omega}\Delta u_w^R \varphi \nu^ids. $$ Then, $$ \big|\int_\Omega\Delta u_w^R \frac{\partial \varphi}{\partial x_i}dx\big| \leq\big|\int_\Omega \varphi \frac{\partial \Delta u_w^R}{\partial x_i}dx\big| \leq \|u_w^R\|_{W^{3,2}(\Omega)}|\varphi|_{L^2(\Omega)}. $$ Now, by \cite[ Prop IX.18]{BrezisAF} we obtain \begin{equation}\label{equiv} \Delta u_w^R\in H_0^1(\Omega). \end{equation} Now, let us consider the following notation \begin{gather*} v=\Delta u_w^R+qu_w^R, \\ g(x)=f_R(x,u_w^R,\nabla w, \Delta w)-\alpha(x)u_w^R. \end{gather*} Notice that $v$ and $u_w^R $ are solutions in the weak sense of the respective differential equations with Dirichlet boundary condition, namely, \begin{equation} \label{nov1} \begin{gathered} \Delta v=g(x),\quad\text{in } \Omega \\ v(x)=0, \quad \text{on } \partial\Omega, \end{gathered} \end{equation} and \begin{equation} \label{nov} \begin{gathered} \Delta u_{w}^R+qu_w^R=v(x),\quad\text{in } \Omega \\ u_{w}^R(x)=0, \quad \text{on } \partial\Omega. \end{gathered} \end{equation} By the Sobolev embedding, we have $u_w^R\in L^{q}(\Omega)$, with $q=2N/(N-4)$. By (F2), we have that $g\in L^s(\Omega)$ with $s=\frac{q}{p}$, where $p$ is given in (F2). We want to show that $u_w^R\in W^{4,r}(\Omega) $, for some $r$ such that $4r>N $. If $4s>N $, it is sufficient to take $r=s$. In fact, applying Agmon \cite[Theorem 8.2]{Ag}, we have $u_w^R\in W^{4,r}(\Omega)$. Now, suppose that $4s 0 $ such that \[ s=(1+\epsilon)\frac{2N}{N+4}. \] Thus, \[ \frac{s_{1}} {s}=\frac{q_{1}} {q}=\frac{sN}{N-4s}\frac{N-4}{2N} =(1+\epsilon)\frac{N(N-4)}{(N+4)(N-4s)}. \] But notice that (it is sufficient we substitute $s=(1+\epsilon)2N/(N+4))$, $$ \frac{N(N-4)} {(N+4)(N-4s)} >1. $$ Therefore, $s_1/s>1+\epsilon$. This argument is known as a \emph{bootstrap}. If $4s_1 (1+\epsilon)\frac{N-4s}{N-4s_1}>(1+\epsilon). $$ We can repeat this last argument a finite times to obtain that $u_w^R\in W^{4,r}(\Omega) $, for some $r$ such that $4r\geq N $. For the case $4r=N $, since $g\in L^{r}(\Omega) $, we have that $g\in L^{k}(\Omega)$ for some $k r$. Applying again the bootstrap argument, we conclude that $u_w^R\in W^{4,r}(\Omega)$, with $4r>N $. Therefore, we can apply the Sobolev-Morrey Theorem to show that $u_w^R\in C^{\alpha}(\overline{\Omega}) $, for some $\alpha\in (0,1) $. By (F0) and $(A1) $, we have that $$ g(x)=f_R(x,u_w^R,\nabla w, \Delta w) -\alpha(x)u_w^R\in C^\beta(\overline{\Omega}), \quad \text{for some } \beta\in (0,1). $$ By applying the Schauder estimates in (\ref{nov1}), we obtain that $v \in C^{2,\beta}(\overline{\Omega})$. By applying the Schauder estimates again, in (\ref{nov}), we obtain \begin{equation}\label{equiv2} u \in C^{4,\beta}(\overline{\Omega}). \end{equation} To conclude, notice that by (\ref{equiv}) and (\ref{equiv2}), we have $\Delta u_w^R(x)=0$, if $x\in\partial\Omega$. \end{proof} \begin{lemma}\label{le5} There exist positive constants $\mu_{0} $, $\mu_{1}$ and $\mu_{2}$, independent of $R>0 $ and of $w\in X$, such that \begin{gather*} \|u_{w}^{R}\|_{C^{0}} \leq\mu_{0}(R+2)^{r}, \\ \|\nabla(u_{w}^{R}) \|_{C^{0}} \leq\mu_{1}(R+2)^{r}, \\ \|\Delta(u_{w}^{R}) \|_{C^{0}} \leq\mu_{2}(R+2)^{r}\,. \end{gather*} Also, there exists $\overline{R}>0$ such that $\mu_{i}(\overline{R}+2)^{r}\leq \overline{R}$, for $i=0,1,2$. \end{lemma} \begin{proof} This result follows combining Lemma \ref{le3} and the results of the Sobolev embedding by arguing as in the proof Lemma \ref{le4}. To obtain $\overline{R}>0 $ such that $\mu_{i}(\overline{R}+2)^{r}\leq \overline{R}$, it is sufficient to observe that $r <1 $, and therefore $$ \frac{\mu_{i}} {\overline{R}^{1-r}} \Big(\frac{\overline{R}+2}{\overline{R}} \Big)^{r} \leq 1 $$ for $\overline{R}$ sufficiently large. \end{proof} Now, let us ``construct'' a nontrivial solution for problem \eqref{nb}. Consider the following problem: Let $u_0\in X\cap C^{4,\lambda}(\overline{\Omega})$, $\lambda\in(0,1)$, and $u_{n}$ ($n=1,2,\ldots $) be a weak solution of the problem $(P_{n}) $, that is, problem \eqref{biarmonic-rt}, with $w=u_{n-1}$, which was found by the mountain pass Theorem in Lemma \ref{le2} and $R=\overline{R}$ obtained in Lemma \ref{le5}. Note that from Lemma \ref{le4} we have $u_{n} \in C^{4}(\overline{\Omega})$ and from Lemmas \ref{le3} and \ref{le5}, we infer that $\|u_{n}\|\geq d_{2}$ and \[ \|u_{n}\|_{C^{0}}, \| \nabla u_{n}\|_{C^{0}}, \| \Delta u_{n}\|_{C^{0}} \leq \overline{R}, \] respectively. Thus, \begin{align*} f_{\overline{R}} (x,u_n,\nabla u_{n-1},\Delta u_{n-1}) & = f(x,u_{n},\nabla u_{n-1}\varphi_{\overline{R}}(\nabla u_{n-1}), \Delta u_{n-1}\varphi_{\overline{R}}(\Delta u_{n-1})) \\ & = f(x,u_{n},\nabla u_{n-1},\Delta u_{n-1}). \end{align*} So, $u_{n}$ is a weak solution of problem $(P_{n}) $. \begin{remark} \label{rmk3} \rm When the diameter of $\Omega$ approaches zero, by an easy calculation in the proof of Lemma \ref{le5}, it is possible to choose $\mu_{i}$ ($i=0, 1, 2$) sufficiently small. \end{remark} \begin{lemma}\label{le7} In hypothesis {\rm (F5)}, let us take \begin{gather*} \rho_{1}=\inf\{k_{1}: \|u_{n}\|_{C^{0}} \leq k_{1},\;\forall\; n\in \mathbb{N} \} >0, \\ \rho_{2}=\inf\{k_{2}: \|\nabla u_{n}\|_{C^{0}} \leq k_{2},\;\forall\; n\in \mathbb{N} \} >0, \\ \rho_{3}=\inf\{k_{3}: \|\Delta u_{n}\|_{C^{0}} \leq k_{3},\;\forall\; n\in \mathbb{N} \}>0. \end{gather*} Then $\{u_{n} \} $ converges strongly in $X$. \end{lemma} \begin{remark} \label{rmk4} \rm We recall that the constant $d_1$ (Lemma\ \ref{le3}) is obtained using only the conditions (F1)--(F4), and the constants $\rho_1$, $\rho_2$, $\rho_3$ are exhibited combining the constant $d_1$ with the Sobolev embedding constants. Thus, as is pointed out in \cite{gir}, the condition (F5) can be read as a constraint on the growth coefficients of $f$ with respect to dimension $N$. \end{remark} \begin{proof}[Proof of Lemma \ref{le7}] In this proof we will use a similar argument that used in \cite{FGM} and \cite{gir}. Let $u_{n}$ and $u_{n+1}$ be a weak solutions of problems $(P_{n})$ and $(P_{n+1})$, respectively. Then, multiplying $(P_{n+1})$ resp. $(P_{n})$ by $(u_{n+1}-u_{n})$ and integrating by parts, and applying Lemma \ref{le0} we obtain \begin{align*} &\gamma\|u_{n+1}-u_{n}\|^{2}\\ & \leq \int_{\Omega}[f(x,u_{n+1},\nabla u_{n},\Delta u_{n}) -f(x,u_{n},\nabla u_{n},\Delta u_n)](u_{n+1}-u_{n} )dx \\ &\quad +\int_{\Omega}[f(x,u_{n},\nabla u_{n},\Delta u_{n}) -f(x,u_{n},\nabla u_{n-1},\Delta u_{n})](u_{n+1}-u_{n} )dx \\ &\quad +\int_{\Omega}[f(x,u_{n},\nabla u_{n-1},\Delta u_{n}) -f(x,u_{n},\nabla u_{n-1},\Delta u_{n-1})](u_{n+1}-u_{n} )dx. \end{align*} Thus, by (F5) and the H\"older inequality we obtain \begin{align*} \gamma\|u_{n+1}-u_{n}\|^{2} & \leq \tau_{1}^{2} L_{\rho_{1}} \|u_{n+1}-u_{n}\|^{2}+\tau_{1}\tau_{2}L_{\rho_{2}} \| u_{n} - u_{n-1} \| \| u_{n+1} - u_{n}\| \\ &\quad + \tau_{1}\tau_{3}L_{\rho_{3}} \|u_{n}-u_{n-1} \| \| u_{n+1} - u_{n} \|. \end{align*} Therefore, \begin{equation} \label{tt} \|u_{n+1}-u_{n}\|\leq \frac{(\tau_{1}\tau_{2}L_{\rho_{2} }+\tau_{1}\tau_{3}L_{\rho_{3} })} {\gamma - \tau_{1}^{2}L_{\rho_{1}}} \|u_{n}-u_{n-1} \|. \end{equation} Hence it follows that the sequence ${u_{n}}$ converges strongly to function $u$, in $X$. \end{proof} \subsection*{Proof of Theorem \ref{teo1}} First of all, as before, we obtain that $\|u_{n}\|\geq d_{2}>0 $. Also, we see that, \[ \|u_{n}\|_{C^{0}},\quad \|\nabla u_{n}\|_{C^{0}}, \quad \|\Delta u_{n}\|_{C^{0}} \] are uniformly bounded. Now, from $(P_{n}) $, notice that $v_{n}=\Delta u_{n}$ verifies the equation \[ \Delta v_{n}=h(x), \quad x \in \overline{\Omega}, \] where $$ h(x)=f(x,u_{n},\nabla u_{n-1},\Delta u_{n-1}) -q\Delta u_{n}-\alpha(x)u_{n}. $$ Since $\|h\|_{C^{\beta}} \leq C $, for some positive constant $C$, by the Schauder Theorem follows that there exists a constant $C> 0$ such that $\|v_{n}\|_{C^{2,\beta}} \leq C$; therefore, \[ \|u_{n}\|_{C^{4,\beta}} \leq C. \] From Arzela-Ascoli Theorem, passing to a subsequence, if necessary, we conclude that $$ \frac{\partial^{j}} {\partial x_{i}^{j} }u_{n}\to \frac{\partial^{j}} {\partial x_{i}^{j} }u, \quad\text{as } n\to \infty, $$ uniformly in $\overline{\Omega}$ for $j=0,1,\ldots,4$ and $i=1,\ldots,N $. 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