\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 261, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/261\hfil Semilinear elliptic PDE's]
{Solutions to semilinear elliptic PDE's with biharmonic operator
 and singular potential}

\author[M. Bhakta \hfil EJDE-2016/261\hfilneg]
{Mousomi Bhakta}

\address{Mousomi Bhakta \newline
Department of Mathematics,
Indian Institute of Science Education and Research,
Dr. Homi Bhaba road,  Pune-411008, India}
\email{mousomi@iiserpune.ac.in}

\thanks{Submitted February 5, 2016. Published September 28, 2016.}
\subjclass[2010]{35B09, 35B25, 35B35, 35G30, 35J91}
\keywords{Semilinear biharmonic equation;  singular potential;
\hfill\break\indent  Navier boundary condition; existence; nonexistence;
 blow-up phenomenon; stability;
\hfill\break\indent uniqueness of extremal solution}

\begin{abstract}
 We study the existence and nonexistence of positive solution to the problem
 \begin{gather*}
 \Delta^2u-\mu a(x)u=f(u)+\lambda b(x)\quad\text{in }\Omega,\\
 u>0 \quad\text{in }\Omega,\\
 u=0=\Delta u \quad\text{on }\partial\Omega,
 \end{gather*}
 where $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$.
 We show the existence of a value  $\lambda^*>0$ such that when
 $0<\lambda<\lambda^*$,  there is a solution
 and when  $\lambda>\lambda^*$  there is no solution
 in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
 Moreover as $\lambda\uparrow\lambda^*$, the minimal positive solution
 converges to a solution.
 We also prove that there exists $\tilde{\lambda}^*<\infty$ with
 $\lambda^*\leq\tilde{\lambda}^*$, and  for $\lambda>\tilde{\lambda}^*$,
 such that the above  problem does not have solution even in the distributional
 sense/very weak sense,  and there is a complete {\it blow-up}.
 Under an additional integrability condition  on $b$, we establish the
 uniqueness of positive solution. 
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction}

In this article  we  study  the semilinear fourth-order elliptic problem with
singular potential,
\begin{equation} \label{ePl}
\begin{gathered}
\Delta^2u-\mu a(x)u=f(u)+\lambda b(x)\quad\text{in }\Omega,\\
u>0 \quad\text{in } \Omega,\\
u=0=\Delta u \quad\text{on } \partial\Omega,
\end{gathered}
\end{equation}
where
$\Delta^2 u=\Delta(\Delta u)$, $\Omega$ is a smooth bounded domain in
$\mathbb{R}^N$, $N\geq 5$. $a, b, f$ are nonnegative functions.
$a\in L^1_{\rm loc}(\Omega)$, $ b\in L^2(\Omega)$, $b\not\equiv 0$.
 $\mu, \lambda$ are (small) positive constants.
We assume that
\begin{equation}\label{assum-f}
f:\mathbb{R}^{+}\to\mathbb{R}^{+} \text{ is a convex $C^1$ function with } f(0)=0=f'(0)
\end{equation}
and satisfying the growth conditions:
\begin{gather}\label{assum-f1}
\lim_{t\to\infty} \frac{f(t)}{t}=\infty, \\
\label{assum-f2}
\int_{1}^{\infty}g(s)ds<\infty \quad\text{and } sg(s)<1 \quad\text{for } s>1,
\end{gather}
where, for $s\geq 1$, we define
\begin{equation}\label{def-g}
g(s)=\sup_{t>0}\frac{f(t)}{f(ts)}.
\end{equation}
It is easy to see that $g$ is nonincreasing, nonnegative function.
Since by convexity $t\to\frac{f(t)}{t}$ is increasing and $f(0)=0$,
it follows that $s\to sg(s)$ is nonincreasing.

As in the literature, $W^{k,p}(\Omega)$ has the usual norm
 $\big(\int_{\Omega}\sum_{0\leq|\alpha|\leq k}|D^{\alpha}u|^p dx\big)^{1/p}$.
Thanks to interpolation theory, one can neglect intermediate derivatives
and see that
\begin{equation}\label{norm-1}
\|u\|_{W^{k,p}(\Omega)}=\Big(\int_{\Omega}|u|^p dx+\int_{\Omega}|D^k u|^p dx\Big)^{1/p},
\end{equation}
 defines a norm which is equivalent to the usual norm in $W^{k,p}(\Omega)$
(see \cite{A}). As $\Omega$ is a smooth bounded domain and $W^{k,p}_0(\Omega)$
is the closure of $C_0^{\infty}(\Omega)$ with respect to the norm in
$W^{k,p}(\Omega)$, invoking \cite[Theorem 2.2]{GGS} we find that
\begin{equation}\label{norm-2}
\|u\|_{W^{k,p}_0(\Omega)}=\Big(\int_{\Omega}| D^k u|^p dx\Big)^{1/p},
\end{equation}
defines an equivalent norm to \eqref{norm-1}. Now onwards we will consider
$W^{k,p}_0(\Omega)$ endowed with the norm defined in \eqref{norm-2}.
The inner product in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ is defined by
$$
(u, v)_{W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)}
=\int_{\Omega}\Delta u\Delta v dx,
$$
which induces the norm
\begin{equation}\label{norm-3}
\|u\|_{W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)}=|\Delta u|_{L^2(\Omega)},
\end{equation}
is equivalent to \eqref{norm-2} with $k=p=2$ (for details see \cite{GGS,PP}).

We assume $a\in L^1_{\rm loc}(\Omega)$ and
there exists a positive constant $\gamma>0$ such that
\begin{equation}\label{assum-a'}
\int_{\Omega} (|\Delta u|^2-a(x)^2u^2)dx\geq \gamma\int_{\Omega} u^2 \quad\forall
 u\in C^{\infty}_0(\Omega).
\end{equation}
Using Fatou's lemma and the standard density argument, it is easy to check that
\eqref{assum-a'} holds for every $u\in W^{2,2}\cap W^{1,2}_0(\Omega)$.
 Therefore we write
\begin{equation}\label{assum-a}
\int_{\Omega} (|\Delta u|^2-a(x)^2u^2)dx\geq \gamma\int_{\Omega} u^2 \quad\forall
 u\in W^{2,2}\cap W^{1,2}_0(\Omega).
\end{equation}
We  note that if $a(x)=\alpha/|x|^2$ where $\alpha<\bar\alpha:= \frac{N(N-4)}{4}$,
applying the following Rellich inequality \cite{Rel54,Rel69}:
\begin{equation}\label{Rellich}
\int_{\mathbb{R}^N}|\Delta u|^2dx
\geq \bar\alpha^2\int_{\mathbb{R}^N}|x|^{-4}|u|^2 dx\quad\forall
  u\in \mathcal C^{\infty}_{0}(\mathbb{R}^N),
\end{equation}
and the Poincare inequality along with the norm equivalence established above,
it is not difficult to check that  \eqref{assum-a} holds.
When $a(x)=\frac{\bar\alpha}{|x|^2}$, \eqref{assum-a} is the improved
 Hardy-Rellich inequality (see \cite{GGM}, \cite{TZ}).

We also assume
\begin{equation}\label{assum-mu}
0<\mu<\sqrt{\gamma}.
\end{equation}
Using \eqref{assum-a'} and \eqref{assum-mu} it follows that
\begin{equation}\label{mu-ga}
\mu\int_{\Omega} a(x)u^2 dx
\leq\mu\Big(\int_{\Omega} a(x)^2u^2 dx\Big)^{1/2}\Big(\int_{\Omega} u^2 dx\Big)^{1/2}
\leq\frac{\mu}{\sqrt{\gamma}}|\Delta u|^2_{L^2(\Omega)}
\end{equation}
for all $ u\in C^{\infty}_0(\Omega)$.
Therefore,
$$
\|u\|_{H}^2 :=\int_{\Omega}[|\Delta u|^2-\mu a(x)u^2]dx,
$$
is a norm in $C^{\infty}_0(\Omega)$ and completion of $C^{\infty}_0(\Omega)$
with respect to this norm yields the Hilbert space $H$.
By \eqref{mu-ga}, \eqref{assum-mu} and \eqref{norm-2}, it follows that
 $\|u\|_H$ is equivalent to $\|u\|_{W^{2,2}_0(\Omega)}$.
Thanks to \eqref{mu-ga},  the norm equivalence established above and the
Poincare inequality, there exists $\tilde{\gamma}>0$ such that
\begin{equation*}
\int_{\Omega}(|\Delta u|^2-\mu a(x)u^2)dx\geq \tilde{\gamma}\int_{\Omega} u^2 dx \quad\forall \   \  u\in \mathcal C^{\infty}_{0}(\Omega).
\end{equation*}
Using Fatou's lemma and the standard density argument it is easy to check that
\begin{equation}\label{tilde-ga}
\int_{\Omega}(|\Delta u|^2-\mu a(x)u^2)dx\geq \tilde{\gamma}\int_{\Omega} u^2 dx \quad\forall
  u\in W^{2,2}\cap W^{1,2}_0(\Omega).
\end{equation}

This inequality implies that the first eigenvalue of $\Delta^2-\mu a(x)$
is strictly positive.

\begin{definition}\label{d: sol} \rm
We say that $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ is a solution of
\eqref{ePl} if $u>0$ a.e.,
$f(u)\in L^2(\Omega)$ and $u$ satisfies
$$
\int_{\Omega} ( \Delta u \Delta\phi-\mu a(x)u\phi)dx=\int_{\Omega}(f(u)+\lambda b(x))\phi\, dx\quad
\forall  \phi\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega).
$$
Similarly $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ is called a
supersolution (subsolution) if $f(u)\in L^2(\Omega)$ and for all positive
$\phi\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$,
$$
\int_{\Omega} ( \Delta u \Delta\phi-\mu a(x)u\phi)dx
\geq (\leq)\int_{\Omega}(f(u)+\lambda b(x))\phi\ dx.
$$
\end{definition}

\begin{definition}\label{d:weak sol} \rm
We say that $u\in L^1(\Omega)$ is a distributional solution or very weak
solution of \eqref{ePl} if $u>0$ a.e.,
$\mu a(x)u+f(u)\in L^1_{\rm loc}(\Omega)$ and $u$ satisfies
\eqref{ePl} in the distributional sense, i.e.,
$$
\int_{\Omega}  u(\Delta^2 \phi-\mu a(x)\phi) dx
=\int_{\Omega}(f(u)+\lambda b(x))\phi\ dx \quad\forall  \phi\in C^{\infty}_{0}(\Omega).
$$
\end{definition}

Similar type of problem  with the Laplace operator in much more generalized
sense was extensively studied by Dupaigne and Nedev in \cite{DN}.
In \cite{DN}, the authors  proved a necessary and sufficient condition for
 the existence of $L^1$ solution and they have also established an estimate
from above and below for the solution. We also refer
\cite{BDT,BV,D} (and the references therein) for the related problems
in the second order case.

Higher order problems are quite different compared to the second order case.
In this case a possible failure of the maximum principle causes
several technical difficulties. Possibly because of this reason the knowledge
on higher order nonlinear problems is far from being reasonably complete,
as it is in the second-order case. In the case of fourth-order problem Navier
boundary conditions play an important role to prove existence
results as under this boundary condition, equation with biLaplacian operator
can be rewritten as a second order system with Dirichlet boundary value problems.
Then using classical elliptic theory,  one can easily prove a Maximum Principle.
 As a consequence,  one can deduce a Comparison Principle which plays as one of
the key factor in proving existence results. In a recent work \cite{PP}, an
equation similar to (\eqref{ePl}) with $a(x)=1/|x|^4$ and $f(u)=u^p$
has been studied. More precisely, in \cite{PP} the authors have studied the
optimal power $p$ for existence/nonexistence of distributional solutions.
In recent years there are many papers dealing with
$W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ solution of semilinear elliptic and
parabolic problem with biLaplacian operator and some specific nonlinearities.
We quote a few among them \cite{AGGM,BG,DDGM,EGP} (also see the references therein).
Semilinear elliptic equations with biharmonic operator arise in continuum mechanics,
bio- physics, differential geometry. In particular in the modeling of thin elastic
plates, clamped plates and in the study of the Paneitz-Branson equation and the
Willmore equation (see \cite{GGS} and the references therein for more details).

This article is organized as follows:
In Section 2 we recall some useful lemmas from \cite{PP} and prove some
important lemmas regarding existence. In Section 3 we prove our main
 existence result.  More precisely, under some hypothesis on $f$, we prove
there exists $\lambda^{*}>0$ such that if $0<\lambda<\lambda^{*}$, problem
\eqref{ePl} has a minimal solution $u_{\lambda}$ in
$W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$. Moreover, if
$\lambda>\lambda^{*}$, then \eqref{ePl} does not have any solution which
belongs to $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$. Under an additional mild
growth condition on $f$ at infinity, we also prove when $\lambda\uparrow \lambda^*$,
there exists $u^*\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ such that minimal
solution $u_{\lambda}$ of \eqref{ePl} converges to $u^*$ in
$W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and $u^*$ happens to be a solution of
\eqref{ePl} with $\lambda=\lambda^*$.
Section 4  deals with the case for which \eqref{ePl} does not have any
solution even in the very weak sense. In this case we establish
 {\it complete blow-up} phenomenon (see Definition \ref{d:blow-up}).
 Section 5 is devoted to the stability result where the minimal positive
solution in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ already exists.
 In this section, under some better integrability condition on $b$,
 we also prove \eqref{ePl} with $\lambda=\lambda^*$
has a unique solution in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.


\section{Preliminary lemmas}

\begin{definition} \rm
We say that $u\in L^1(\Omega)$ is a weak supersolution (subsolution) to
$$\Delta^2 u=g(x,u) \quad\text{in }\Omega,$$
in the sense of distribution if $g(x,u)\in L^1(\Omega)$ and for all positive
 $\phi\in C^{\infty}_{0}(\Omega)$, we have
$$
\int_{\Omega} u\Delta^2 \phi dx\geq(\leq)\int_{\Omega} g(x, u)\phi dx.
$$
If $u$ is a weak supersolution and as well a weak subsolution in the sense
of distribution, then we say that $u$ is a distributional solution.
\end{definition}

Next we recall three important lemmas from \cite{PP} which we will use
frequently in this paper.

\begin{lemma}[Strong Maximum Principle] \label{SMP}
Let $u$ be a nontrivial supersolution of
\begin{equation}\label{eq:smp}
\begin{gathered}
\Delta^2 u=0 \quad\text{in }\Omega,\\
u=0=\Delta u \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
Then $-\Delta u>0$ and $u>0$ in $\Omega$.
\end{lemma}

For a proof of the above lemma see \cite[Lemma 3.2]{PP}.

\begin{lemma}[Comparison Principle] \label{comp prin}
Let $u$ and $v$ satisfy the following:
\begin{equation}\label{eq:cp}
\begin{gathered}
\Delta^2 u\geq \Delta^2 v \quad\text{in }\Omega,\\
u\geq v \quad\text{on }\partial\Omega,\\
-\Delta u\geq -\Delta v \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
Then, $-\Delta u\geq -\Delta v$ and $u\geq v$ in $\Omega$.
\end{lemma}

For a proof of the above lemma see \cite[Lemma 3.3]{PP}.

\begin{lemma}[{Weak Harnack Principle \cite[Lemma 3.4]{PP}}] \label{WHP}
Let $u$ be a positive distributional supersolution to  \eqref{eq:smp}.
Then for any $B_R(x_0)\Subset\Omega$, there exists a positive constant
$C=C(\theta, \rho, q, R)$, $0<q<\frac{N}{N-2}$, $0<\theta<\rho<1$, such that
$$
\|u\|_{L^q(B_{\rho R}(x_0))}\leq C \operatorname{ess\,inf}_{B_{\theta R}(x_0)}u.
$$
\end{lemma}

\begin{lemma}\label{l:ex}
Let $a\in L^1_{\rm loc}(\Omega)$, $b\in L^2(\Omega)$ , $a, b\geq 0$ a.e.,
$b\not\equiv 0$, $\mu$ be a positive constant satisfying \eqref{assum-mu}
and $a$ satisfy \eqref{assum-a}.  Then the equation
\begin{equation}\label{eq-1}
\begin{gathered}
\Delta^2 u-\mu a(x)u=b \quad\text{in }\Omega,\\
u=0=\Delta u \quad\text{on }\partial\Omega,
\end{gathered}
\end{equation}
has a positive solution $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\end{lemma}

\begin{proof}
Given $b\in L^2(\Omega)$, we know there exists unique
$u_1\in W^{2,2}\cap W^{1,2}_0(\Omega)$ satisfying the following:
\begin{gather*}
\Delta^2 u_1=b \quad\text{in }\Omega,\\
u_1=0=\Delta u_1 \quad\text{on }\partial\Omega.
\end{gather*}
Applying strong maximum principle (Lemma \ref{SMP}) we obtain $u_1>0$.
Now define $u_n$ ($n\geq 2$) as follows:
\begin{equation}\label{u-n}
\begin{gathered}
\Delta^2 u_n=\mu a(x)u_{n-1}+b \quad\text{in }\Omega,\\
u_n=0=\Delta u_n \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
By \eqref{assum-a}, we have $\mu a(x)u_{n-1}\in L^2(\Omega)$.
This in turn implies the existence of unique
$u_n\in W^{2,2}\cap W^{1,2}_0(\Omega)$ which satisfies \eqref{u-n}.
Also by comparison principle we have
 $0<u_1\leq\dots\leq u_{n-1}\leq u_n\leq\dots$.
\smallskip

\noindent\textbf{Claim:}
$\{u_n\}$ is a Cauchy sequence in $W^{2,2}\cap W^{1,2}_0(\Omega)$.

To see this, we note that $\Delta^2(u_{n+1}-u_n)=\mu a(x)(u_{n}-u_{n-1})$.
By taking $(u_{n+1}-u_n)$ as a test function and using \eqref{assum-a}, we obtain
\begin{align*}
|\Delta(u_{n+1}-u_n)|_{L^2(\Omega)}^2
&=\mu\int_{\Omega} a(x)(u_{n}-u_{n-1})(u_{n+1}-u_n) dx\\
&\leq  \mu (\int_{\Omega} a(x)^2(u_{n}-u_{n-1})^2dx)^{1/2}(\int_{\Omega}(u_{n+1}-u_{n})^2dx)^{1/2}\\
&\leq \frac{\mu}{\sqrt{\gamma}}|\Delta(u_{n}-u_{n-1})|_{L^2(\Omega)}
 |\Delta(u_{n+1}-u_n)|_{L^2(\Omega)}.
\end{align*}
Therefore
\[
|\Delta(u_{n+1}-u_n)|_{L^2(\Omega)}\leq\frac{\mu}{\sqrt{\gamma}}
|\Delta(u_{n}-u_{n-1})|_{L^2(\Omega)}
\leq\dots\leq (\frac{\mu}{\sqrt{\gamma}})^{n-1}|\Delta(u_2-u_1)|_{L^2(\Omega)}.
\]
As $\mu <\sqrt{\gamma}$, from the above estimate we can conclude that $\{u_n\}$
is a Cauchy sequence in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
Hence, there exists $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ such that
$u_n\to u$ in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$. Moreover, $u>0$ since
$u_n>0$ for all  $n\geq 1$. As $u_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
solves \eqref{u-n}, we have
$$
\int_{\Omega} \Delta u_n\Delta \phi dx=\mu\int_{\Omega} a(x)u_{n-1}\phi dx+\int_{\Omega} b\phi dx\quad\forall
\phi \in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega).
$$
Taking the limit as $n\to\infty$, we obtain $u$ is a solution to \eqref{eq-1}.
\end{proof}

\begin{lemma}\label{l:supersol}
Let $a\in L^1_{\rm loc}(\Omega)$, $b\in L^2(\Omega)$,  $f:\mathbb{R}^{+}\to\mathbb{R}^{+}$
($f$ convex) be nonnegative functions. Let $\mu,\lambda>0$, $\mu<\sqrt{\gamma}$.
Suppose there exists a nonnegative supersolution
$\tilde{u}\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ of \eqref{ePl}
 (respectively for \eqref{eq-1}).
Then there exists a unique  solution $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
to \eqref{ePl} which satisfies $0\leq u\leq \tilde{w}$ for any
 supersolution $\tilde{w}\geq 0$ of  \eqref{ePl} (respectively for \eqref{eq-1}).
 $u$ is called the minimal nonnegative solution of \eqref{ePl}
 (respectively for \eqref{eq-1}). By strong maximum principle it also follows
that $u>0$ in $\Omega$.
\end{lemma}

\begin{remark}\label{r:1} \rm
We denote the minimal positive  solution of \eqref{eq-1} by $\zeta_1$ and
denote $G(b)=\zeta_1$.
The function $0<u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ solving
\eqref{ePl} (respectively \eqref{eq-1}) also solves \eqref{ePl} \eqref{eq-1}
 in the distributional sense (see definition \eqref{d:weak sol}).
\end{remark}

\begin{proof}
The proof is the same for both the equations \eqref{ePl} and \eqref{eq-1},
therefore we present only the proof for \eqref{ePl}.
First we will show that if minimal solution exists then it is unique.
To see this, let $u_1$ and $u_2$ are two solutions which satisfy
$0\leq u_i\leq\tilde{w}, ( i=1,2)$ for every nonnegative supersolution $\tilde{w}$.
Thus $u_1\leq u_2$ and $u_2\leq u_1$. Hence $u_1=u_2$.

Next, let $\tilde{u}\geq 0$ be a  supersolution to \eqref{ePl} and
$u_0\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ be a positive solution of
\begin{gather*}
\Delta^2 u_0=\lambda b \quad\text{in }\Omega,\\
u_0=0=\Delta u_0 \quad\text{on }\partial\Omega.
\end{gather*}
By comparison principle we obtain $0<u_0\leq\tilde{u}$ in $\Omega$.
Next, using iteration we will show that there exists
$u_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ for $n=1,2, \dots$ such that
$u_n$ solves the  problem
\begin{equation}\label{eq:un-1}
\begin{gathered}
\Delta^2 u_n=\mu a(x)u_{n-1}+f(u_{n-1})+\lambda b(x) \quad\text{in }\Omega,\\
u_n=0=\Delta u_n \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
Since $\tilde{u}$ is a weak supersolution to \eqref{ePl}, we have
 $f(\tilde{u})\in L^2(\Omega)$. Thanks to the fact that $0<u_0\leq\tilde{u}$
and $f$ is convex (thus $f$ is nondecreasing), we obtain $f(u_0)\leq f(\tilde{u})$.
Thus $f(u_0)+\lambda b(x)\in L^2(\Omega)$. Also, by \eqref{assum-a} it follows
that $\mu a(x)u_{0}\in L^2(\Omega)$. Therefore $u_1$ is well defined and by
comparison principle $0<u_0\leq u_1\leq \tilde{u}$. Using the induction method,
similarly we can show that $u_n$ is well defined and
$0<u_0\leq u_1\leq\dots\leq u_n\leq\dots\leq \tilde{u}$.
\smallskip

\noindent\textbf{Claim:}
$\{u_n\}$ is uniformly bounded in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.

To see this, let us note that from \eqref{eq:un-1} we can write
\begin{align*}
|\Delta u_n|^2_{L^2(\Omega)}
&= \int_{\Omega}(\mu a(x)u_{n-1}+f(u_{n-1})+\lambda b(x))u_n dx \\
&\leq \int_{\Omega} (\mu a(x)\tilde{u}^2+f(\tilde{u})\tilde{u}+\lambda b\tilde{u})dx\\
&\leq  \big[\mu |a(x)\tilde{u}|_{L^2(\Omega)}+|f(\tilde{u})|_{L^2(\Omega)}
+\lambda |b|_{L^2(\Omega)}\big]|\tilde{u}|_{L^2(\Omega)}
\leq  C.
\end{align*}
As a consequence there exists  $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
such that up to a subsequence $u_n\rightharpoonup u$ in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
and $u_n\to u$ in $L^2(\Omega)$. From \eqref{eq:un-1} we have,
$$
\int_{\Omega}\Delta u_n\Delta\phi dx=\int_{\Omega}[\mu a(x)u_{n-1}+f(u_{n-1})+\lambda b]\phi dx
\quad\forall \phi\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega).
$$
Using Vitaly's convergence theorem we can pass to the limit
$n\to\infty$ on the right-hand side and obtain $u$ is a solution to \eqref{ePl}.
Also $u>0$ since $u_n>0$ for all $n\geq 1$.

Let $\tilde{w}$ be another supersolution, then by comparison principle it
follows that $u_0\leq \tilde{w}$ and $u_{n}\leq\tilde{w}$ for every $n\geq 1$.
Taking the limit $n\to\infty$, it gives us that $u\leq\tilde{w}$. Hence the
lemma follows.
\end{proof}


\section{Existence and nonexistence results}

\begin{theorem}\label{main ex}
Assume $a\in L^1_{\rm loc}(\Omega)$, $0\not\equiv b\in L^2(\Omega)$, $a,b,f$
are nonnegative functions,  \eqref{assum-a}, \eqref{assum-mu}, \eqref{assum-f},
\eqref{assum-f1}, \eqref{assum-f2} and \eqref{def-g} are satisfied.
Let $G=(\Delta^2-\mu a(x))^{-1}$ and $\zeta_1=G(b)$, as proved in
Lemma \ref{l:ex} (also see Remark \ref{r:1}).
Suppose there exists constants $\epsilon>0$ and $C>0$ such that
\begin{equation}\label{ex-cond}
f(\epsilon\zeta_1)\in L^2(\Omega) \quad\text{and}\quad
G(f(\epsilon\zeta_1))\leq C\zeta_1 \quad a.e.
\end{equation}
Then there exists $0<\lambda^{*}=\lambda^{*}(N, a(x), b(x), f, \mu)$  such that
if $\lambda<\lambda^{*}$, then \eqref{ePl} has a minimal positive solution
 $u_{\lambda}\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and
$u_{\lambda}\geq\lambda\zeta_1$.

  If $\lambda>\lambda^{*}$ then \eqref{ePl} has no positive solution in
$W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.

Moreover, if $\lambda>0$ is  small then
$$
\lambda\zeta_1\leq u_{\lambda}\leq 2\lambda\zeta_1.
$$
\end{theorem}

The assumption \eqref{ex-cond} is motivated from the work of Dupaigne
and Nedev (see \cite[Theorem 1]{DN}).
To prove this theorem, first we need to prove a lemma and a  proposition.

\begin{lemma}\label{l:ex1}
Let the functions $a, b$ and the constant $\mu$ satisfy the assumptions in
Theorem \ref{main ex}. $\zeta_1=G(b)$ as in theorem \ref{main ex}
and assume that \eqref{assum-f} is satisfied.   If
$$
f(2\zeta_1)\in L^2(\Omega)\quad\text{and}\quad G(f(2\zeta_1))\leq \zeta_1,
$$
then \eqref{ePl} with $\lambda=1$  admits a solution
$u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\end{lemma}

\begin{proof}
Let $f(2\zeta_1)\in L^2(\Omega)$ and $G(f(2\zeta_1))\leq \zeta_1$.
 We define, $v :=G(f(2\zeta_1))+\zeta_1$. Clearly $v>0$ and
$v\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ since $\zeta_1$ and
$G(f(2\zeta_1))$ are in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega) $ by
Lemma \ref{l:ex}. Also,
$$
v-\zeta_1=G(f(2\zeta_1)), \quad  v\leq 2\zeta_1, \quad f(v)\in L^2(\Omega).
$$
Thus we have
$$
\Delta^2(v-\zeta_1)-\mu a(x)(v-\zeta_1)=f(2\zeta_1)\quad \text{in }\Omega,
$$
i.e.,
$$
\Delta^2 v-\mu a(x)v=f(2\zeta_1)+b\geq f(v)+b \quad\text{in } \Omega
$$
and $v=0=\Delta v$  on $\partial\Omega$. As a result, $v$ is a
positive supersolution of \eqref{ePl} with $\lambda=1$.
Finally, by applying Lemma \ref{l:supersol} we obtain the existence
of minimal positive solution  $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
of \eqref{ePl} with $\lambda=1$.
\end{proof}

\begin{proposition}\label{p:ex}
Suppose there exists $\tilde{\lambda}>0$ such that $(P_{\tilde{\lambda}})$
has a positive solution
$u_{\tilde{\lambda}}\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
Then for every $0<\lambda<\tilde{\lambda}$, \eqref{ePl} has a solution
in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\end{proposition}

\begin{proof}
Let $u_{\tilde{\lambda}}\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
denote a positive solution corresponding to
\eqref{ePl} with $\tilde{\lambda}$ instead of $\lambda$.
Therefore by definition (see Definition \ref{d: sol})
$f(u_{\tilde{\lambda}})\in L^2(\Omega)$. Define,
$v=\tilde{\lambda}\zeta_1$. Note that,
$$
\Delta^2(\frac{u_{\tilde{\lambda}}}{\tilde{\lambda}})
-\mu a(x)(\frac{u_{\tilde{\lambda}}}{\tilde{\lambda}})
= \frac{1}{\tilde{\lambda}}(f(u_{\tilde{\lambda}})+\tilde{\lambda}b )
=\frac{f(u_{\tilde{\lambda}})}{\tilde{\lambda}}+b\geq b \quad\text{in }\Omega.
$$
This implies, $\frac{u_{\tilde{\lambda}}}{\tilde{\lambda}}$ is a positive
supersolution to \eqref{eq-1}. Therefore by minimality of $\zeta_1$ it
follows, $\zeta_1\leq \frac{u_{\tilde{\lambda}}}{\tilde{\lambda}}$,
 which in turn implies  $v\leq u_{\tilde{\lambda}}$.
Let $0<\lambda<\tilde{\lambda}$ and define,
$w=u_{\tilde{\lambda}}-v+\lambda\zeta_1$. Clearly $w>0$.
Using the definition of $v$ and $\lambda$ we also get
$w\leq u_{\tilde{\lambda}}$. By convexity of $f$, it follows
$\frac{f(t)}{t}$ is increasing and thus $f$ is nondecreasing.
As a consequence, $f(w)\leq f(u_{\tilde{\lambda}})$ and hence
$f(w)\in L^2(\Omega)$. Also,
$$
\Delta^2 w-\mu a(x)w=f(u_{\tilde{\lambda}})
+\tilde{\lambda}b-(\tilde{\lambda}-\lambda)b
=f(u_{\tilde{\lambda}})+\lambda b\geq f(w)+\lambda b.
$$
As a result, $w\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
is a positive supersolution to \eqref{ePl}. Hence by Lemma \ref{l:supersol},
there exists minimal positive solution of \eqref{ePl}.
\end{proof}



\begin{proof}[Proof of Theorem \ref{main ex}]
 We assume \eqref{ex-cond} holds.
\smallskip

\noindent\textbf{Step 1:}
We show that if $\lambda>0$ is small then \eqref{ePl} has a positive a solution
$u_{\lambda}\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
We will prove this step in the spirit of \cite{DN}.
By Lemma \ref{l:ex1}, it follows that \eqref{ePl} has a solution as long
as it holds
\begin{equation}\label{ex-cond-1}
f(2\lambda\zeta_1)\in L^2(\Omega)  \quad\text{and}\quad G(f(2\lambda\zeta_1))\leq\lambda\zeta_1.
\end{equation}
From the definition of $g$ (see definition \eqref{def-g}), it follows that
$g(\frac{\epsilon}{2\lambda})\geq \frac{f(t)}{f(t\frac{\epsilon}{2\lambda})}$
for all $t>0$. Choosing $t=2\lambda\zeta_1$, we obtain
$f(2\lambda\zeta_1)\leq f(\epsilon\zeta_1)g(\frac{\epsilon}{2\lambda})$.
Applying \eqref{ex-cond}, we have $f(2\lambda\zeta_1)\in L^2(\Omega)$ and
$G(f(2\lambda\zeta_1))$ is well defined. Also by minimality of $G(f(2\lambda\zeta_1))$
and by assumption \eqref{ex-cond}, we obtain
$$
G(f(2\lambda\zeta_1))\leq g(\frac{\epsilon}{2\lambda})
G(f(\epsilon\zeta_1))\leq Cg(\frac{\epsilon}{2\lambda})\zeta_1.
$$
To show \eqref{ex-cond-1} holds for $\lambda>0$ small, it is enough to prove that
$$
\lim_{\lambda\to 0}\frac{1}{\lambda}g(\frac{\epsilon}{2\lambda})=0 \quad
\text{or equivalently}\quad \lim_{K\to\infty}Kg(K)=0.
$$
Since $s\to sg(s)$ is nonincreasing, the above limit is well defined,
i.e. there exists $C'\geq0$ such that $\lim_{K\to\infty}Kg(K)=C'$.
If $C'>0$, then $g(K)\sim\frac{C}{K}$ near $\infty$ and this contradicts
\eqref{assum-f2}. Hence $C'=0$ and \eqref{ex-cond-1} holds for $\lambda>0$ small.
\smallskip

\noindent\textbf{Step 2:}
 Define,
$$
\Lambda=\{\lambda>0:  (P_{\lambda})  \text{ has a minimal positive solution }
 u_{\lambda} \},
$$
By Step 1 and Proposition \ref{p:ex}, it follows that $\Lambda$ is a non-empty
interval. We define,
$$
\lambda^{*}=\sup \Lambda .
$$
Then it is easy to see that, if $\lambda<\lambda^{*}$, \eqref{ePl}
has a minimal positive solution and for $\lambda>\lambda^{*}$, \eqref{ePl}
does not have any positive solution in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\smallskip

\noindent\textbf{Step 3:}
 From $G(b)=\zeta_1$, it is easy to see that $G(\lambda b)=\lambda\zeta_1$.
If $\lambda<\lambda^{*}$ and $u_{\lambda}$ denotes the corresponding minimal
 positive solution of \eqref{ePl}, then it is not difficult to check that
 $u_{\lambda}$ is a supersolution to the equation satisfied by $\lambda\zeta_1$.
Therefore by minimality of $\lambda\zeta_1$, we obtain
\begin{equation}\label{u-low}
u_{\lambda}\geq\lambda\zeta_1.
\end{equation}
\smallskip

\noindent\textbf{Step 4:} We show that if $\lambda>0$ is small, then
$$
\lambda\zeta_1\leq u_{\lambda}\leq 2\lambda\zeta_1.
$$
By Step 1, \eqref{ex-cond-1} holds since $\lambda>0$ is small. Define,
$w=G(f(2\lambda\zeta_1))+\lambda\zeta_1$. Therefore
$$
w\leq 2\lambda\zeta_1 \quad\text{and}\quad w-\lambda\zeta_1=G(f(2\lambda\zeta_1)).
$$
As in the proof of Lemma \ref{l:ex1}, we can establish that
$w\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ is a positive
supersolution of \eqref{ePl}. Thus $u_{\lambda}\leq w\leq 2\lambda\zeta_1$.
Combining this with \eqref{u-low}, we have
$\lambda\zeta_1\leq u_{\lambda}\leq 2\lambda\zeta_1$.
\end{proof}


Define
\begin{equation}\label{u-star}
u^*(x)=\lim_{\lambda\uparrow\lambda^*}u_{\lambda}(x), \quad x\in\Omega.
\end{equation}


\begin{theorem} \label{thm3.5}
Assume  the assumptions in Theorem \ref{main ex} are satisfied,
$u_{\lambda}$ denotes the minimal positive solution of \eqref{ePl} for
$0<\lambda<\lambda^*$ and $u^*$ is as defined in \eqref{u-star}.
In addition suppose $f$ satisfies the  condition
\begin{equation}\label{assum-f3}
\lim_{s\to\infty}\frac{sf'(s)}{f(s)}>1.
\end{equation}
Then $u^{*}\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and $u^*$ is a
solution to \eqref{ePl} with $\lambda^*$ instead of $\lambda$.
Moreover, $u_{\lambda}\to u^*$ in  $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\end{theorem}

\begin{remark} \rm
Since $f$ is convex and $C^1$, \eqref{assum-f3} is a mild assumption.
It is easy to see that  if $f\in C^2$ and strictly convex,
then \eqref{assum-f3} is obvious.
\end{remark}

\begin{proof}[Proof of Theorem \ref{thm3.5}]
$u_{\lambda}$ begin a solution of \eqref{ePl} implies
\begin{equation}\label{eq:u-la}
\int_{\Omega}\Delta u_{\lambda}\Delta v=\mu\int_{\Omega} a(x)u_{\lambda}v
+\int_{\Omega} f(u_{\lambda})v+\lambda\int_{\Omega} b(x)v \quad \forall
v\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega).
\end{equation}
By Theorem \ref{t:stability}, it follows that $u_{\lambda}$ is a stable
solution of \eqref{ePl} (see Definition \ref{d:stable}).
Therefore $\int_{\Omega}(|\Delta u_{\lambda}|^2-\mu a(x)u_{\lambda}^2
-f'(u_{\lambda})u_{\lambda}^2)dx\geq 0$. Hence by taking
$v=u_{\lambda}$ in \eqref{eq:u-la} we have
\begin{equation}\label{eq:u-la1}
\int_{\Omega} f'(u_{\lambda})u_{\lambda}^2 dx
\leq\int_{\Omega}(|\Delta u_{\lambda}|^2-\mu a(x)u_{\lambda}^2)dx
=\int_{\Omega}(f(u_{\lambda})u_{\lambda}+\lambda b(x)u_{\lambda})dx.
\end{equation}
Moreover, using \eqref{assum-f3} we can write, for every $\epsilon>0$
there exists $C>0$ such that
\begin{equation}\label{eq:f}
(1+\epsilon)f(s)s\leq f'(s)s^2+C \quad\forall  s\geq 0.
\end{equation}
Hence combining \eqref{eq:u-la1} and \eqref{eq:f} we obtain
$$
(1+\epsilon)\int_{\Omega} (f'(u_{\lambda})u_{\lambda}^2-\lambda b(x)u_{\lambda})dx
\leq (1+\epsilon)\int_{\Omega} f(u_{\lambda})u_{\lambda} dx
\leq\int_{\Omega} (f'(u_{\lambda})u_{\lambda}^2+C)dx.
$$
As a result,
$$
\epsilon\int_{\Omega} f'(u_{\lambda})u_{\lambda}^2 dx
\leq C|\Omega|+(1+\epsilon)\lambda\int_{\Omega} b u_{\lambda}dx.
$$
Consequently,
\begin{equation}\label{eq:u-la2}
\int_{\Omega} f(u_{\lambda})u_{\lambda} dx\leq C_1+C_2\lambda\int_{\Omega} b u_{\lambda}dx,
\end{equation}
for some constants $C_1, C_2>0$.
Since $\lambda<\lambda^*$,  by taking $v=u_{\lambda}$ in \eqref{eq:u-la}
 and applying Holder inequality and \eqref{eq:u-la2} we have
\begin{align*}
\int_{\Omega} |\Delta u_{\lambda}|^2 dx
&= \mu\int_{\Omega} a(x)u^2_{\lambda}+\int_{\Omega} f(u_{\lambda})u_{\lambda}+\lambda\int_{\Omega} bu_{\lambda}\\
&\leq  \mu |a(x)u_{\lambda}|_{L^2(\Omega)}|u_{\lambda}|_{L^2(\Omega)}
+\lambda^{*}(1+C_2)\int_{\Omega} bu_{\lambda}dx+C_1.
\end{align*}
Applying \eqref{assum-a} and Cauchy-Schwartz inequality with $\delta>0$
on the above estimate, we obtain
\begin{align*}
\int_{\Omega} |\Delta u_{\lambda}|^2 dx
&\leq  \frac{\mu}{\sqrt{\gamma}}|\Delta u_{\lambda}|^2_{L^2(\Omega)}
 +C_3|b|_{L^2(\Omega)}|u_{\lambda}|_{L^2(\Omega)}+C_1\\
&\leq  \frac{\mu}{\sqrt{\gamma}}|\Delta u_{\lambda}|^2_{L^2(\Omega)}
 +\frac{C_3}{\sqrt{\gamma}}|b|_{L^2(\Omega)}|\Delta u_{\lambda}|_{L^2(\Omega)}+C_1\\
&\leq  \frac{\mu}{\sqrt{\gamma}}|\Delta u_{\lambda}|^2_{L^2(\Omega)}
 +\delta |\Delta u_{\lambda}|^2_{L^2(\Omega)}+c(\delta)|b|^2_{L^2(\Omega)}+C_1.
\end{align*}
Since $\mu<\sqrt{\gamma}$ (by  \eqref{assum-mu}), we can choose $\delta>0$ such that
$\frac{\mu}{\sqrt{\gamma}}+\delta<1$. Hence from the above estimate we have
$$
\int_{\Omega} |\Delta u_{\lambda}|^2 dx\leq C_4|b|^2_{L^2(\Omega)}+C_1\leq C',
$$
for some constant $C'>0$. This implies $\{u_{\lambda}\}$ is uniformly bounded
in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ for $\lambda<\lambda^*$.
Consequently, by \eqref{u-star} we conclude that $u_{\lambda}\rightharpoonup u^*$
in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
Passing to the limit $\lambda\to\lambda^*$ in \eqref{eq:u-la}, via
 Lebesgue monotone convergence theorem, it is easy to check that $u^*$ is a
solution to \eqref{ePl} with $\lambda^*$ instead of $\lambda$.
When $\lambda\to\lambda^*$, using monotone convergence theorem we also have
\begin{align*}
\|u_{\lambda}\|^2_{W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)}
&=\int_{\Omega}|\Delta u_{\lambda}|^2dx \\
&= \mu\int_{\Omega} a(x)u_{\lambda}^2+\int_{\Omega} f(u_{\lambda})u_{\lambda}+\lambda\int_{\Omega} bu_{\lambda}\\
&\to \mu\int_{\Omega} a(x){u^*}^2+\int_{\Omega} f(u^*)u^*+\lambda^*\int_{\Omega} bu^*\\
&=  \int_{\Omega}|\Delta u^*|^2dx=\|u^*\|^2_{W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)}
\end{align*}
Hence $\|u_{\lambda}\|_{W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)}
\to \|u^*\|_{W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)}$.
 Combining this along with the weak convergence, we conclude
$u_{\lambda}\to u^*$ in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\end{proof}

 We denote by $u_{\lambda^{*}}$, the minimal positive solution of
 \eqref{ePl} with $\lambda^*$ instead of $\lambda$.


\section{Nonexistence of very weak solution and complete blow-up}

Define
$$
\tilde{\lambda}^{*}=\sup\{\lambda>0:  \text{ \eqref{ePl} has a very weak
solution/distributional solution }\}.
$$
It is not difficult to check that if $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
is a solution to \eqref{ePl} in the sense of Definition \ref{d: sol},
then $u$ is a very weak solution of \eqref{ePl} as well.
Therefore $\tilde{\lambda}^{*}\geq\lambda^{*}$.

\begin{lemma}\label{l:blow-up}
$\tilde{\lambda}^{*}<\infty$.
\end{lemma}

\begin{proof}
Assume \eqref{ePl} has a very weak solution $u\in L^1(\Omega)$. Therefore
\begin{equation}\label{eq:dist}
\int_{\Omega}  u(\Delta^2 \phi-\mu a(x)\phi) dx
=\int_{\Omega}(f(u)+\lambda b(x))\phi\ dx \quad\forall  \phi\in C^{\infty}_{0}(\Omega).
\end{equation}
Let $\tilde{\Omega}\Subset\Omega$ and $\psi\in C^{\infty}_0(\Omega)$
be a  nonnegative function such that $\operatorname{supp}(\psi)\subset\tilde{\Omega}$.
We choose $\phi$ as follows:
\begin{gather*}
\Delta^2 \phi=\psi \quad\text{in }\Omega,\\
\phi=0=\Delta\phi \quad\text{on }\partial\Omega.
\end{gather*}
Clearly $\phi\in C^{\infty}(\Omega)$ and by strong maximum principle
$\phi> 0$ in $\Omega$. Thus there exists $c>0$ such that $\phi\geq c>0$
in $\tilde{\Omega}$. Substituting this $\phi$ in \eqref{eq:dist}, we have
\begin{equation}\label{eq:phi-psi}
\mu\int_{\Omega} a(x)u\phi\ dx+\int_{\Omega} f(u)\phi\ dx+\lambda\int_{\Omega} b(x)\phi\ dx
=\int_{\Omega} u\psi\ dx=\int_{\tilde{\Omega}}u\psi\ dx.
\end{equation}
Since $f$ satisfies \eqref{assum-f1}, it is easy to check that, for
$\epsilon>0$ there exists a constant $C_{\epsilon}>0$ such that
$$
u\leq C_{\epsilon}+\epsilon f(u).
$$
Therefore from the right-hand side of \eqref{eq:phi-psi} we obtain
\begin{equation*}
\int_{\tilde{\Omega}}u\psi\ dx
\leq C_{\epsilon}\int_{\Omega}\psi dx+\epsilon\int_{\tilde{\Omega}}f(u)\psi dx
\leq  C_{\epsilon}\int_{\Omega}\psi dx
 +\epsilon|\frac{\psi}{\phi}|_{L^{\infty}(\tilde{\Omega)}}\int_{\Omega}f(u)\phi dx
\end{equation*}
Now choose $\epsilon>0$ such that
$\epsilon|\frac{\psi}{\phi}|_{L^{\infty}(\tilde{\Omega)}}<1/2$.
Thus from \eqref{eq:phi-psi} we have
$$
\mu\int_{\Omega} a(x)u\phi\ dx+\frac{1}{2}\int_{\Omega} f(u)\phi\ dx+\lambda\int_{\Omega} b(x)\phi\ dx
\leq C\int_{\Omega}\psi dx\leq C'.
$$
This implies $\tilde{\lambda}^{*}<\infty$. In particular there are no
solutions of \eqref{ePl} for $\lambda>\tilde{\lambda}^{*}$,
even in the very weak sense.
\end{proof}

\begin{definition}\label{d:blow-up} \rm
Let $\{a_n(x)\}$, $\{b_n(x)\}$ and $\{f_n\}$ be increasing sequence of
bounded functions converging pointwise respectively to $a(x)$, $b(x)$ and $f$.
(Since $f\in C^1(\mathbb{R}^+)$, without loss of generality we can also assume
$f_n\in C(\mathbb{R}^+)$). Let $u_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$
be the minimal nonnegative solution of
\begin{equation}\label{a-n}
\begin{gathered}
\Delta^2 u_n-\mu a_n(x)u_n=f_n(u_n)+\lambda b_n(x) \quad\text{in }\Omega,\\
u_n=0=\Delta u_n \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
We say that there is a complete blow-up in \eqref{ePl}, if given any such
$\{a_n(x)\}$, $\{b_n(x)\}$, $\{f_n\}$ and $u_n$,
$$
u_n(x)\to\infty \quad\forall\ x\in\Omega.
$$
\end{definition}

We remark that the existence of $u_n$ follows from Theorem \ref{thmA.3}.
The next theorem is proved in the spirit of \cite{PP}.

\begin{theorem}\label{t:blow-up}
Fix $\lambda>0$. Suppose \eqref{ePl} does not have any solution,
even in the very weak sense. Then there is complete blow up.
\end{theorem}

\begin{proof}
Let $u_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ be the minimal
nonnegative solution of \eqref{a-n}. Using the monotonicity property of
$a_n, b_n$ and $f_n$, we obtain $u_{n+1}$ is a supersolution of the
equation satisfied by $u_n$. Thus $u_n\leq u_{n+1}$.
Therefore to establish the blow-up result, it is sufficient to show the
complete blow-up for the family of minimal solutions $u_n$.

We prove this by the method of contradiction.
Assume there exists $x_0\in\Omega$ and a positive constant $C$ such that
$u_n(x_0)\leq C$. Thus applying weak Harnack inequality (Lemma \ref{WHP})
we have
$$
|u_n|_{L^1(B_{\rho R}(x_0))}\leq C \operatorname{ess\,inf}_{B_{\theta R}(x_0)}u_n
\leq Cu_n(x_0)\leq C',
$$
where $0<\theta<\rho<1$.
Then following the same argument as in \cite{PP}, we can show that there
exists $r>0$ and a positive constant $C=C(r)$ such that
$$
\int_{B_r(0)}u_n dx\leq C, \quad\text{uniformly for } n\in\mathbb{N}.
$$
Therefore, applying the monotone convergence theorem we see that,
there exists $u\geq 0$ such that $u_n\to u$ in $L^1(B_r(0))$.

Let $\phi$ be the solution to the problem
\begin{gather*}
\Delta^2 \phi=\chi_{B_r(0)} \quad\text{in }\Omega,\\
\phi=0=\Delta\phi \quad\text{on }\partial\Omega.
\end{gather*}
Clearly $\phi\in W^{4,p}(\Omega)$ since $\chi_{B_r(0)}\in L^p(\Omega)$
for all $p\geq 1$. Taking $\phi$ as a test function in \eqref{a-n}, we have
$$
\int_{\Omega}(a_n(x)u_n\phi+f_n(u_n)\phi+\lambda b_n\phi )dx=\int_{B_r(0)}u_n dx\leq C.
$$
By monotone convergence theorem and Fatou's lemma, it follows that
\begin{gather*}
a_n(x)u_n\uparrow a(x)u \quad\text{in } L^1_{\rm loc}(B_r(0)), \\
f_n(u_n)\to f(u)\quad\text{in } L^1_{\rm loc}(B_r(0)) \quad\text{and}\quad
 b_n(x)\uparrow b(x) \quad\text{in } L^1_{\rm loc}(B_r(0)).
\end{gather*}
Hence as in \cite[Theorem 5.1]{PP}, we can conclude that $u$ is a very weak
solution to \eqref{ePl} in $B_{r_1}(0)\Subset B_r(0)$ and this contradicts
the assumption of this theorem.
\end{proof}

Combining Lemma \ref{l:blow-up} and Theorem \ref{t:blow-up}, we obtain the
following corollary.

\begin{corollary}
If $\lambda>\tilde{\lambda}^{*}$, then there is complete blow-up.
\end{corollary}

\section{Stability results}
\begin{definition}\label{d:stable} \rm
We say that $u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ is a stable solution,
if the first eigenvalue of the linearized operator of the equation \eqref{ePl}
is nonnegative, i.e., if
$$
\inf_{\phi\in C^{\infty}_0(\Omega)\setminus\{0\}}
\frac{\int_{\Omega} (|\Delta\phi|^2-\mu a(x)\phi^2-f'(u)\phi^2)dx}{\int_{\Omega}\phi^2 dx}\geq 0.
$$
\end{definition}

\begin{theorem}\label{t:stability}
Suppose all the assumptions in Theorem \ref{main ex} are satisfied and
for $0<\lambda<\lambda^{*}$,   let $u_{\lambda}$ denote the minimal positive
solution of \eqref{ePl}.  Then $u_{\lambda}$ is stable.
\end{theorem}

\begin{proof}
Following the idea of Dupaigne and Nedev \cite{DN}, we prove this theorem.
Let $a_n(x)=\text{min}(a(x), n)$, $b_n=\text{min}(b(x), n)$ and
$u_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ denote the minimal positive
solution of the problem
\begin{equation}\label{a-n1}
\begin{gathered}
\Delta^2 u_n-\mu a_n(x)u_n=f(u_n)+\lambda b_n(x) \quad\text{in }\Omega,\\
u_n=0=\Delta u_n \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
By Lemma \ref{l:supersol}, $u_n$ is well defined since $u_{\lambda}$
is a supersolution of \eqref{a-n1}. Let $\lambda_1^n(\Delta^2-\mu a_n(x)-f'(u_n))$
denote the 1st eigenvalue of the linearized operator $\Delta^2-\mu a_n(x)-f'(u_n)$.
\smallskip


\noindent\textbf{Claim:} $\lambda_1^n(\Delta^2-\mu a_n(x)-f'(u_n))\geq 0$.

To prove this claim, we choose $p>N$. Define,
$I: \mathbb{R}\times W^{4,p}(\Omega)\to L^p(\Omega)$ as follows
$$
I(\lambda, u)=\Delta^2 u-\mu a_n(x)u-f(u)-\lambda b_n.
$$
An easy computation using \eqref{tilde-ga} and implicit function theorem,
(see \cite{DN}) it follows that there exists a unique maximal curve
$\lambda\in [0, \lambda^{\#})\to u(\lambda)$ such that
$$
I(\lambda, u(\lambda))=0 \quad\text{and}\quad I_u(\lambda, u(\lambda))\in Iso(W^{4,p}. L^p).
$$
If $0<\lambda<\lambda^{\#}$, then $u_n\leq u(\lambda)$, since $u_n$ is the minimal
positive solution of \eqref{a-n1}. Thus $f(u_n)\leq f(u(\lambda))$.
Moreover, $I(\lambda, u(\lambda))=0$ implies
$f(u(\lambda))=\Delta^2 u(\lambda)-\mu a_n(x)u(\lambda)-\lambda b_n(x)\in L^p(\Omega)$,
which in turn implies $f(u_n)\in L^p(\Omega)$. Therefore by elliptic regularity
theory, $u_n$ is in the domain of $I$ and hence $u_n=u(\lambda)$.

Following the same method as in \cite{DN}, we can show that if
$0<\lambda<\lambda^{*}$, $u_n$ is in the domain of $I$. Thus
$\lambda^{\#}=\lambda^{*}$ (otherwise we could extend the curve $u(\lambda)$
beyond $\lambda^{\#}$ contradicting its maximality). We also claim that the
first eigenvalue of $I_u(\lambda, u_n)$ does not vanish for any $\lambda<\lambda^{*}$.
To see this, assume $\phi$ is an the eigenfunction corresponding to this
first eigenvalue. If the first eigenvalue vanishes for some $\lambda_0<\lambda^{*}$,
then we have $\Delta^2\phi-\mu a(x)\phi-f'(u_n)\phi=0$, i.e., $I_u(\lambda_0, u_n)=0$
but we know that $I_u(\lambda, u)$ can not vanish for any $\lambda<\lambda^{\#}$
(otherwise $u(\lambda)$ will not be the maximal curve). Consequently,
since $\lambda^{\#}=\lambda^{*}$, we can say that  the first eigenvalue of
$I_u(\lambda, u_n)$ does not vanish for any $\lambda<\lambda^{*}$.
Moreover, by \eqref{tilde-ga} we know first eigenvalue of $I_u(0,0)$
is strictly positive. Therefore we conclude that
$\lambda_1^n(\Delta^2-\mu a_n(x)-f'(u_n))\geq 0$ for every $\lambda\in[0,\lambda^{*})$.

Also, $\{u_n\}$ is a nondecreasing sequence and converges to a solution
of \eqref{ePl} in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
Since $u_n\leq u_{\lambda}$, $\lim_{n\to\infty}u_n$ has to be the minimal
solution $u_{\lambda}$. Therefore by monotone convergence theorem we conclude
the first eigenvalue $\lambda_1(\Delta^2-\mu a(x)-f'(u_{\lambda}))\geq 0$ which
completes the proof.
\end{proof}

\begin{theorem}
Suppose  the assumptions in Theorem \ref{main ex} hold and $u_{\lambda}$
is the minimal positive solution of \eqref{ePl}. Also assume \eqref{assum-f3}
is satisfied. If $\lambda=\lambda^{*}$ and $b\in L^p(\Omega)$ for some
$p>\frac{N}{3}$, then $u_{\lambda^*}$ is the only positive solution of
\eqref{ePl}, with $\lambda^{*}$ instead of $\lambda$, which belongs to
$\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
\end{theorem}

\begin{proof}
Suppose the theorem does not hold and $u$ and $v$ are two distinct positive
solutions of \eqref{ePl}, with $\lambda^{*}$ instead of $\lambda$,
 where $u, v\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega) $.
Let $u$ be the minimal positive solution. Therefore $u\leq v$.
 Applying strong maximal principle we can easily check that $u<v$ in $\Omega$.
Since $u$ and $v$ are solution, by Definition \eqref{d: sol} we have
$f(u), f(v)\in L^2(\Omega)$. Thus applying \eqref{assum-a}, we obtain
$\mu a(x)u+f(u)+\lambda^{*}b\in L^2(\Omega)$.
This together with the elliptic regularity theory gives
$u\in W^{4,2}(\Omega)\cap W^{1,2}_0(\Omega)$. Similarly same result holds
for $v$ as well. Define $w=\frac{u+v}{2}$. Then
$w\in W^{4,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and by convexity of $f$, we have
$$
f(w)= f\big(\frac{u+v}{2}\big) \leq\frac{f(u)+f(v)}{2}\in L^2(\Omega).
$$
Thus,
$$
\Delta^2 w-\mu a(x)w=\frac{f(u)+f(v)}{2}+\lambda^{*}b\geq f(w)+\lambda^{*}b.
$$
Thus $w$ is a supersolution of \eqref{ePl} with $\lambda^{*}$ instead of $\lambda$.
By Lemma \ref{lemA.1}, it follows that $w$ is a solution to $(P_{\lambda^{*}})$.
As a consequence, inequality on the above expression becomes equality
and by convexity of $f$ we conclude that $f$ is linear on $[u(x), v(x)]$
for almost every $x\in\Omega$. For $\epsilon\in(0,1)$, define
$\theta=\epsilon u+(1-\epsilon)v$. Therefore $f''(\theta(x))$ exists
for a.e $x\in\Omega$ and $f''(\theta(x))=0$ a.e. $x\in\Omega$.
This implies $\nabla(f'(\theta))=0$ a.e. in $\Omega$, which in turn implies
$f'(\theta)=C$ a.e. in $\Omega$ and $f(\theta)=C\theta+D$ a.e. in
$\Omega$ for some constant $C$ and $D$. Moreover, using convexity of $f$,
this implies $f(t)=Ct+D$ for $t\in[\text{ess inf}\ \theta, \operatorname{ess\,sup}
 \theta]$. Applying Lemma \ref{lemA.2}, we have $\operatorname{ess\,inf}\theta=0$.
Since $f(0)=0=f'(0)$, we obtain $f\equiv 0$ on $[0, \operatorname{ess\,sup} \theta]$.
As $\epsilon>0$  arbitrary, we can conclude $f\equiv 0$ on
$[0, \operatorname{ess\, sup} v]$. Therefore $u$ and $v$ both satisfy
\begin{gather*}
\Delta^2 u-\mu a(x)u=\lambda^{*} b(x) \quad\text{in }\Omega,\\
u=0=\Delta u \quad\text{on }\partial\Omega.
\end{gather*}
This in turn implies, $v-u$ satisfies
\begin{gather*}
\Delta^2(v-u)-\mu a(x)(v-u)=0 \quad\text{in }\Omega,\\
v-u=0=-\Delta(v- u) \quad\text{on }\partial\Omega.
\end{gather*}
This contradicts \eqref{tilde-ga} since
$v-u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$. Hence $u=v$.
\end{proof}

\section{Appendix}

\begin{lemma} \label{lemA.1}
If $b\in L^p(\Omega)$ for some $p>\max \{2,\frac{N}{3}\}$ and
$w\in W^{4,2}(\Omega)\cap W^{1,2}_0(\Omega)$ is a supersolution
of \eqref{ePl} with $\lambda^{*}$ instead of $\lambda$, then $w$
is a solution of \eqref{ePl} with $\lambda^{*}$ instead of $\lambda$.
\end{lemma}

\begin{proof}
Let $w$ be a supersolution of \eqref{ePl} with $\lambda^{*}$ instead of $\lambda$
and not a solution.  Define, $\nu\in\mathcal{D}'(\Omega)$ by
$$
\nu(\phi)=\int_{\Omega} w(\Delta^2\phi)-(\mu a(x)w+f(w)+\lambda^{*}b)\phi \quad\forall
 \phi\in C^{\infty}_0(\Omega).
$$
Since $w$ is a supersolution, by Definition \ref{d: sol} we have
$f(w)\in L^2(\Omega)$. Therefore thanks to \eqref{assum-a}, we obtain
$\nu\in L^2(\Omega)$. Moreover, $w$ is a supersolution implies $\nu\geq 0$.
$w$ is not a solution implies $\nu\not\equiv0$. Consider the problem
\begin{gather*}
\Delta^2\psi=\nu \quad\text{in }\Omega,\\
\psi=0=\Delta\psi \quad\text{on }\partial\Omega.
\end{gather*}
We can break this problem into system of second-order Dirichlet problem
by defining
\begin{gather*}
-\Delta\psi=\tilde{\psi} \quad\text{in }\Omega, \quad
\psi=0 \quad\text{on }\partial\Omega, \\
-\Delta\tilde{\psi}=\nu \quad\text{in }\Omega, \quad
\tilde{\psi}=0 \quad\text{on }\partial\Omega.
\end{gather*}
Then by the weak maximum principle it is easy to check that
$\psi>\epsilon\delta(x)$ for some $\epsilon>0$, where
$\delta(x)=\text{dist}(x,\partial\Omega)$.
 Next we consider the problem
\begin{gather*}
\Delta^2\eta=b \quad\text{in }\Omega,\\
\eta=0=\Delta\eta \quad\text{on }\partial\Omega.
\end{gather*}
As before we break this problem into system of equations as follows:
\begin{gather*}
-\Delta\eta=\tilde{\eta} \quad\text{in }\Omega, \quad
 \eta=0 \quad\text{on }\partial\Omega, \\
-\Delta\tilde{\eta}=b \quad\text{in }\Omega, \quad \tilde{\eta}=0
\quad\text{on }\partial\Omega.
\end{gather*}
Since $b\in L^p(\Omega)$ for some $p>\frac{N}{3}$, using theory of elliptic
regularity and Soblev embedding theorem, we obtain
$\tilde{\eta}\in L^{p^*}(\Omega)$ where $p^*=\frac{Np}{N-2p}>N$.
Therefore $\eta\in C^{1,\alpha}(\Omega)$ for some $\alpha\in(0,1)$.
Hence $\eta<C\delta(x)$ in $\Omega$ for some $C\in(0,\infty)$.
Define, $v=w+\epsilon C^{-1}\eta-\psi$. Clearly $v<w$ in $\Omega$
and $v\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$. Also,
\begin{align*}
\Delta^2v
&=\Delta^2w+\epsilon C^{-1}b-\nu \\
&=\mu a(x)w+f(w)+\lambda^{*}b+\nu+\epsilon C^{-1}b-\nu \\
&\geq \mu a(x)v+f(v)+(\lambda^{*}+\epsilon C^{-1})b.
\end{align*}
As a result, $v$ is a supersolution to
\eqref{ePl} with $\lambda^{*}+\epsilon C^{-1}$ instead of $\lambda$.
Hence \eqref{ePl}, with $\lambda^{*}+\epsilon C^{-1}$ instead of $\lambda$,
 has a solution contradicting the extremality of $\lambda^{*}$.
\end{proof}

The next lemma is in the spirit of \cite[Lemma 3.2]{DN}.

\begin{lemma} \label{lemA.2}
If $u\in L^1(\Omega)$ is an nonnegative distributional solution of
$\Delta^2 u=h$ in $\Omega$, where $h\in L^1(\Omega)$, then
$\operatorname{ess\,inf}u=0$.
\end{lemma}

\begin{proof}
Assume the lemma does not hold, that is, there exists $\epsilon>0$ such that
$u\geq \epsilon>0$ a.e. in $\Omega$. We extend $u$ and $h$ by $0$ in
$\mathbb{R}^N\setminus \Omega$. Let $\rho_n$ denote the standard molifier.
Define $u_n=u\star\rho_n$ and $h_n=h\star\rho_n$. Following the same
argument as in \cite[Lemma 3.2]{DN}, we can show that, there exists
$\alpha>0$ such that for $n$ large enough $u_n\geq \alpha\epsilon$
everywhere in $\Omega$ and given $\omega\Subset\Omega$ and $n$ large enough,
$\Delta^2 u_n=h_n$ everywhere in $\omega$. Let $\phi$ solve the following problem
\begin{equation}\label{app-phi}
\begin{gathered}
\Delta^2\phi=1 \quad\text{in }\omega,\\
\phi=0=\Delta\phi \quad\text{on }\partial\omega.
\end{gathered}
\end{equation}
Integrating by parts we obtain
\begin{align*}
\int_{\omega}u_n dx
&=\int_{\omega}u_n\Delta^2\phi dx\\
&=\int_{\omega}\Delta u_n\Delta\phi dx
 +\int_{\partial\omega}\frac{\partial}{\partial n}(\Delta\phi)u_n ds \\
&=\int_{\omega}h_n\phi dx+\int_{\partial\omega}\frac{\partial}{\partial n}(\Delta\phi)u_n ds.
\end{align*}
Thus,
$$
\int_{\omega}h_n\phi dx-\int_{\omega}u_n dx=-\int_{\partial\omega}
\frac{\partial}{\partial n}(\Delta\phi)u_n ds\leq -\alpha\epsilon|\omega|,
$$
since $\int_{\partial\omega}\frac{\partial}{\partial n}(\Delta\phi)ds=|\omega|$
(follows from \eqref{app-phi} after integrating by parts).
Since $u_n\to u$ in $L^1(\Omega)$, $h_n\to h$ in $L^1(\Omega)$ we obtain
$$
\int_{\omega}h\phi dx-\int_{\omega}u dx\leq -\alpha\epsilon|\omega|.
$$
Next we choose $\omega=\omega_n:=\{x\in\Omega:
\operatorname{dist}(x,\partial\Omega)>\frac{1}{n}\}$, $n\to\infty$.
Let $\phi_n$ denote the corresponding solution to \eqref{app-phi} in $\omega_n$.
Then $\phi_n\uparrow\phi$ where $\phi$ solves
\begin{gather*}
\Delta^2\phi=1 \quad\text{in }\Omega,\\
\phi=0=\Delta\phi \quad\text{on }\partial\Omega.
\end{gather*}
Taking the limit $n\to\infty$ in
$\int_{\omega_n}h\phi_n dx-\int_{\omega_n}u dx\leq -\alpha\epsilon|\omega_n|$
and using $\Delta^2 u=h \quad\text{in }\Omega$, we have
$0\leq-\alpha\epsilon|\Omega|$. This gives a contradiction.
\end{proof}

\begin{theorem} \label{thmA.3}
Assume \eqref{assum-mu} is satisfied.
Then problem \eqref{a-n} has a nonnegative minimal solution for every $\lambda>0$.
\end{theorem}

\begin{proof}
\textbf{Step 1:} Assume $a\in L^1_{\rm loc}(\Omega)$ which satisfies
\eqref{assum-a}.  Let $b\in L^{\infty}(\Omega)$ and
$f\in L^{\infty}(\mathbb{R}^+)\cap C(\mathbb{R}^+) $ be nonnegative functions,
$b\not\equiv 0$ and $\lambda>0$. Then there exists
$u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ such that $u$ solves \eqref{ePl}
for all $\lambda>0$.

To prove step 1, let $u_0\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ be a
positive solution to
\begin{gather*}
\Delta^2 u_0=\lambda b \quad\text{in }\Omega,\\
u_0=0=\Delta u_0 \quad\text{on }\partial\Omega.
\end{gather*}
Since $\lambda b\in L^{\infty}(\Omega)\subset L^2(\Omega)$ we obtain
$u_0\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.
Next, using iteration we will show that there exists
$u_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ for $n=1,2, \dots$ such that
$u_n$ solves the  problem
\begin{equation}\label{app:un-1}
\begin{gathered}
\Delta^2 u_n=\mu a(x)u_{n-1}+f(u_{n-1})+\lambda b(x) \quad\text{in }\Omega,\\
u_n=0=\Delta u_n \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
Thanks to \eqref{assum-a} and the assumptions that $f, b\in L^{\infty}(\Omega)$,
it follows that $\mu a(x)u_0+f(u_0)+\lambda b(x)\in L^2(\Omega)$.
Therefore $u_1$ is well defined. Moreover, by comparison principle
$0<u_0\leq u_1$. Using the induction method, similarly we can show $u_n$
is well defined and $0<u_0\leq u_1\leq\dots\leq u_n\leq\dots$.
\smallskip

\noindent\textbf{ Claim:}
$\{u_n\}$ is uniformly bounded in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$.

To see this,  note that from \eqref{app:un-1} we can write
\begin{equation}\label{app-un2}
|\Delta u_n|^2_{L^2(\Omega)} =\int_{\Omega}(\mu a(x)u_{n-1}+f(u_{n-1})+\lambda b(x))u_n dx .
\end{equation}
Using Holder inequality, \eqref{assum-a} and  Young's inequality,
the terms on the right-hand side can be simplified as follows
\begin{gather*}
\begin{aligned}
\lambda\int_{\Omega} bu_n dx
&\leq\lambda|b|_{L^{\infty}(\Omega)}|\Omega|^{1/2}|u_n|_{L^2(\Omega)}\\
&\leq\frac{C}{\sqrt{\gamma}}|b|_{L^{\infty}(\Omega)}|\Delta u_n|_{L^2(\Omega)}\\
&\leq \epsilon|\Delta u_n|^2_{L^2(\Omega)}+c(\epsilon)|b|^2_{L^{\infty}(\Omega)}\,,
\end{aligned} \\
\begin{aligned}
\int_{\Omega} f(u_{n-1})u_n dx
&\leq |f|_{L^{\infty}(\Omega)}|\Omega|^{1/2}|u_n|_{L^2(\Omega)} \\
&\leq  \frac{C}{\sqrt{\gamma}}|f|_{L^{\infty}(\Omega)}|\Delta u_n|_{L^2(\Omega)}\\
&\leq  \epsilon|\Delta u_n|^2_{L^2(\Omega)}+c(\epsilon)|f|^2_{L^{\infty}(\Omega)}\,,
\end{aligned} \\
\begin{aligned}
\mu\int_{\Omega} a(x)u_{n-1}u_n dx
&\leq\mu\int_{\Omega} a(x)u_n^2dx \\
& \leq\mu |a(x)u_n|_{L^2(\Omega)}|u_n|_{L^2(\Omega)} \\
& \leq \frac{\mu}{\sqrt{\gamma}}|\Delta u_n|^2_{L^2(\Omega)}\,.
\end{aligned}
\end{gather*}
Since $\mu/\sqrt{\gamma}<1$, we can choose $\epsilon>0$ such that
$2\epsilon+\frac{\mu}{\sqrt{\gamma}}<1$. Substituting this $\epsilon$
in above three inequalities and combining them with \eqref{app-un2}, we have
$$
|\Delta u_n|^2_{L^2(\Omega)}
\leq C(|b|_{L^{\infty}(\Omega)}+|f|_{L^{\infty}(\Omega)}).
$$
This proves the claim. As a consequence there exists
$u\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ such that up to a subsequence
$u_n\rightharpoonup u$ in $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and $u_n\to u$
in $L^2(\Omega)$. Therefore we can conclude the theorem as we did in
Lemma \ref{l:supersol}.
\smallskip

\noindent\textbf{Step 2:}
Let  $\{b_n(x)\}$ and $\{f_n\}$ be increasing sequence of bounded functions
converging pointwise respectively to  $b(x)$ and $f$ ($f_n$ is continuous
for $n=1,2, \dots$ ). Then by Step 1, there exists a nonnegative minimal
solution $v_n\in W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ of the  problem
\begin{equation}\label{app:a-n3}
\begin{gathered}
\Delta^2 v_n-\mu a(x)v_n=f_n(v_n)+\lambda b_n(x) \quad\text{in }\Omega,\\
v_n=0=\Delta v_n \quad\text{on }\partial\Omega.
\end{gathered}
\end{equation}
Clearly $v_n$ is a nonnegative supersolution to \eqref{a-n}.
Therefore the theorem follows from Lemma \ref{l:supersol}.
\end{proof}

\subsection*{Final Remark}
The results of this article can be easily extended to the equations of
the form
$$
\Delta^2 u-\mu a(x)u=c(x)f(u)+\lambda b(x) \quad\text{in }\Omega,
$$
where $c\in L^1_{\rm loc}(\Omega)$ is a nonnegative function.
In particular, \eqref{ex-cond} will be changed to
$$
c f(\epsilon\zeta_1)\in L^2(\Omega) \quad\text{and}\quad
 G(c(x)f(\epsilon\zeta_1))\leq C\zeta_1.
$$


\subsection*{Acknowledgements}
 The author would like to thank Dr. Anup Biswas for many
 fruitful discussions that led to various results in this work.
This work is supported by INSPIRE research grant DST/INSPIRE 04/2013/000152.

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\end{document}