\documentclass[reqno]{amsart}
 \usepackage{hyperref,graphicx,amssymb}

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 146, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.} 
\vspace{9mm}}

\begin{document} 

\title[\hfilneg EJDE-2005/146\hfil Multiplicity and symmetry breaking]
{Multiplicity and symmetry breaking for positive radial solutions
of semilinear elliptic equations modelling MEMS on annular
domains} 

\author[P. Feng, Z. Zhou\hfil EJDE-2005/146\hfilneg]
{Peng Feng, Zhengfang Zhou}  % in alphabetical order

\address{Peng Feng \hfill\break
Department of Mathematics, Michigan State University, 
East Lansing, MI 48824, USA} 
\email{fengpeng@math.msu.edu}

\address{Zhengfang Zhou \hfill\break
Department of Mathematics, Michigan State University,
East Lansing, MI 48824, USA }
\email{zfzhou@math.msu.edu}


\date{}
\thanks{Submitted August 10, 2005. Published December 12, 2005.}
\subjclass[2000]{35J65, 35J60, 35B32}
\keywords{Radial solution; symmetry breaking; multiplicity; MEMS}


\begin{abstract}
 The use of electrostatic forces to provide actuation is a method
 of central importance in microelectromechanical system (MEMS)
 and in nanoelectromechanical  systems (NEMS).  
 Here, we study the electrostatic deflection  of an annular 
 elastic membrane.  We investigate the exact  number of positive 
 radial solutions and non-radially symmetric  bifurcation for the model
 $$
 -\Delta u=\frac{\lambda}{(1-u)^2}\quad\hbox{in }\Omega,
 \quad  u=0 \quad\hbox{on }\partial \Omega,
 $$
 where
 $\Omega=\{x\in \mathbb{R}^2: \epsilon<|x|<1\}$.
 The exact number of positive radial solutions maybe 0, 1, or 2
 depending on $\lambda$.
 It will be shown that the upper branch of radial solutions
 has non-radially symmetric bifurcation at infinitely many
 $\lambda_N\in (0,\lambda^*)$. The proof of the multiplicity
 result relies on the characterization of the shape of the time-map.
 The proof of the bifurcation result relies on a well-known theorem
 due to Kielh\"ofer.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}

\section{Introduction} \label{s1}

In this paper, we shall study the multiplicity and symmetry breaking
of positive radial solutions to the equation
\begin{equation}
-\Delta u = \frac{\lambda}{(1-u)^2} \quad \text{in }  \Omega,
\end{equation}
\begin{equation}
u = 0 \quad \text{on } \partial\Omega,
\end{equation}
where $\Omega=\{x\in \mathbb{R}^2: \epsilon<|x|<1\}$ is an annulus
in $\mathbb{R}^2$, $\lambda$ is a positive parameter and its meaning
 will become clear later in the paper.

This paper is motivated by the recent work of Pelesko, Bernstein
and McCuan \cite{Pele}. In \cite{Pele}, they showed that
asymmetric  solutions exist through numerical investigation. A
bifurcation diagram was obtained. They conjectured that there are
an infinite number of branches of asymmetric solutions
intersecting the upper radially symmetric solution branch.

In this paper, we use shooting method and time map to show the
exact  multiplicity of radial solutions, see for example
\cite{Hai99, Lin1, Lin2, Lin3, Smoller, Yadava}. A well-known
bifurcation theorem essentially due to H.Kielh\"ofer \cite{Kielh}
is used to show the radial symmetry breaking result.

We can establish the following theorems:

\begin{theorem} \label{thm1.1}
There exists a $\lambda^*$ such that the problem has no positive
radial solution for $\lambda>\lambda^*$, one and only one radial
solution for $\lambda=\lambda^*$ and exactly two radial solutions
for $0<\lambda<\lambda^*$.
\end{theorem}

\begin{theorem} \label{thm1.2}
There exists infinitely many $\lambda_k\in (0,\lambda^*)$ such
that  the upper branch of radially symmetric solutions has a
non-radially symmetric bifurcation at each $\lambda_k$,
$k=1,2,\dots$.
\end{theorem}

The paper is organized as follows. In section 2, we briefly
describe the model proposed in \cite{Pele}. For more information
on this topic, refer to \cite{Bryzek, Dario}. In section 3, we
show the existence results for small $\lambda$. In section 4, we
obtain the multiplicity results. In section 5, we study the radial
symmetry breaking problem and the conjecture in \cite{Pele} is
proved.


\section{Formulation of the model}\label{s2}

\begin{figure}
\setlength{\unitlength}{.02in}
\begin{picture}(150, 80)(72,-100)
\put(70,-90){\line(1,0){100}} \put(100,-100){\vector(-1,0){30}}
\put(120,-100){$L$} \put(140,-100){\vector(1,0){30}}
\put(70,-90){\line(1,1){20}} \put(90,-70){\line(1,0){100}}
\put(190,-70){\line(-1,-1){20}}

\qbezier(120,-40)(100,-45)(110,-50)
\qbezier(110,-50)(162,-55)(170,-45)
\qbezier(170,-45)(175,-34)(120,-40)
\put(140,-45){\oval(5,3)}
\put(100,-58){\vector(0,1){10}}
\put(100,-65){\vector(0,-1){15}}
\put(100,-62){$d$}

\put(95,-35){\vector(1,-1){10}}
\put(80,-33){Supported Boundary}

\put(120,-45){$\Omega$}

\put(150,-30){\vector(-1,-1){10}}
\put(140,-28){Elastic Membrane at Potential $V$}

\put(80,-62){\vector(1,0){10} $y'$}
\put(80,-62){\vector(0,1){10}}
\put(80,-52){$z'$}
\put(80,-62){\vector(-1,-1){7}}
\put(72,-72){$x'$}

\put(150,-62){Thickness of Membrane =$h$}
\put(150,-58){\vector(0,1){7}}

\put(120, -80){Fixed Ground Plate}
\end{picture}
\caption{A basic electrostatically actuated elastic membrane.
The prime coordinates indicate they have not yet been scaled.}
\label{mems}
\end{figure}

We model the device shown in Figure \ref{mems} which consists of
an  annular elastic membrane suspended above a rigid plate. The
membrane is supported along the inner and outer boundaries. A
voltage difference is applied across the device in order to cause
deflection of the membrane. In particular, the upper surface of
the membrane is held at potential $V$, while the ground plate is
held at zero potential.

We shall notice the fact that most MEMS devices are of small
aspect ratio, $d/L<<1$, and use thin components, $h/d<<1$. Here
$d$ is the distance between the membrane and the plate, $L$ is the
size of the plate and $h$ is the thickness of membrane. We derive
an approximate solution. For the completeness of the paper, we
have reproduced the model following \cite{Pele}.

We assume the electrostatic potential satisfies Laplace's equation
everywhere away from the membrane and the plate.
\begin{equation}
\Delta \phi=0.
\end{equation}
It also satisfies appropriate boundary conditions on the membrane.
\begin{gather*}
\phi=V \quad {\rm on\quad elastic \quad plate},\\
\phi=0 \quad {\rm on \quad ground\quad plate}.
\end{gather*}
We model the elastic membrane using the plate equation.
In particular, the deflection $u'$ of the membrane satisfies
\begin{equation*}
\rho h\frac{\partial^2 u'}{\partial^2 t'}
+a\frac{\partial u'}{\partial t'}-\mu \nabla_{\bot}^2 u'
+D\nabla_{\bot}^4 u' =-\frac{\epsilon_0}{2}|\nabla\phi|^2.
\end{equation*}
Here $\rho$ is the density of the membrane, $h$ is the thickness,
$\mu$ is the tension in the membrane, $D$ is the flexural
rigidity, and $\epsilon_0$ is the permittivity of free space.
$\nabla_{\bot}$ represents the differentiation with respect to
$x'$ and $y'$. The standard plate equation has been modified in
two ways. First, a damping term has been added. The parameter $a$
is the damping constant. Second, we have assumed $a$ is
proportional to velocity. We shall rescale the system and rewrite
in dimensionless form. We rescale the electrostatic potential with
the applied voltage, time with a damping timescale of the system,
the $x'$ and $y'$ with a characteristic length of the device, and
$z'$ and $u'$ with the size of the gap between the ground plate
and the elastic membrane. We define
\begin{equation}
u=\frac{u'}{d},\quad \phi=\frac{\phi}{V},\quad x=\frac{x'}{L},\quad
y=\frac{y'}{L},\quad z=\frac{z'}{d},\quad t=\frac{\mu t'}{aL^2}.
\end{equation}
In dimensionless form, we have
\begin{gather}
\epsilon^2\left({{\partial^2\phi}\over{\partial^2 x^2}}
+ {{\partial^2\phi}\over{\partial^2 y^2}}\right)
+{{\partial^2\phi}\over{\partial^2 z^2}}=0, \label{8}\\
\phi = 0 \quad \text{on the ground  plate},\\
\phi = 1 \quad \text{on the membrane},\\
\frac{1}{\alpha^2}\frac{\partial^2 u}{\partial^2 t}
+\frac{\partial u}{\partial t}- \nabla_{\bot}^2 u+\delta\nabla_{\bot}^4 u
=-\lambda\big[\epsilon^2|\nabla_{\bot}\phi|^2
+(\frac{\partial \phi}{\partial z})^2 \big]. \label{11}
\end{gather}
Here $\phi$ is a dimensionless potential scaled with respect to voltage $V$, $x$ and $y$ are scaled with respect to the length of the ground pate $L$, $z$ is scaled with respect to the gap size $d$. $\alpha=\frac{aL}{\sqrt{\rho h\mu}}$ is the inverse of the quality factor for the system. $\delta=\frac{D}{L^2\mu}$ measures the relative importance of tension and rigidity. $\epsilon=\frac{d}{L}$ is the aspect ratio of the system.
$\lambda=\epsilon_0V^2L^2/2Td^3$, where $T$ is the tension in the
membrane and $\epsilon_0$ is the permittivity of free space. Note
that $\lambda$ is a dimensionless number which characterizes the
relative strengths of electrostatic and mechanical forces in the
system. As $\lambda$ is proportional to the applied voltage, it
serves as a convenient bifurcation parameter.We assume the
displacement of the membrane $u$ satisfies
\begin{gather*}
\Delta u=\lambda\big[\delta^2 ({{\partial^2\phi}\over{\partial^2 x^2}}
+ {{\partial^2\phi}\over{\partial^2 y^2}})
+{{\partial^2\phi}\over{\partial^2 z^2}}\big],\\
u=0 \quad \text{on the boundary}.
\end{gather*}

Assuming $d\ll L$, that is $\epsilon \ll1$. Physically, this means
that the lateral dimension of the device are large compared to the
gap between the membrane and the ground plate. For many MEMS
systems this is an excellent approximation. We exploit the
small-aspect ratio by setting $\epsilon$ goes to zero in equation
(\ref{8}). This reduces the electrostatic problem to
\begin{equation}
\frac{\partial^2 \phi}{\partial z^2}=0,
\end{equation}
which we may solve to find the approximate potential,
$$ \phi\approx Az+B.
$$
We are primarily concerned with the field between the plates  and
hence apply the boundary condition on $\phi$ which is
\begin{gather*}
\phi(x,y,u,t)=1,\\
\phi(x,y,0,t)=0.
\end{gather*}
Hence
$$\phi\approx \frac{z}{u}.$$
Therefore, by sending $\epsilon$ goes to zero and use this
approximate potential in equation (\ref{11}), we find
\begin{equation}
\frac{1}{\alpha^2}\frac{\partial^2 u}{\partial^2 t} +\frac{\partial u}{\partial t}- \nabla_{\bot}^2 u+\delta\nabla_{\bot}^4 u =-\frac{\lambda}{u^2}.
\end{equation}
We shall focus on the equilibrium state deflection.
 For convenience, we change variable $u\mapsto 1-u$. The
result is the following semi-linear elliptic equation for the
displacement $u$:
\begin{gather*}
-\Delta u = \frac{\lambda}{(1-u)^2} \quad \text{in }  \Omega,
\\
u = 0 \quad \text{on } \partial\Omega.
\end{gather*}


\section{Existence}\label{s3}

In this section, we shall study the following semilinear elliptic equation with Dirichlet boundary condition.
\begin{gather}
 -\Delta u=\frac{\lambda}{(u-1)^2} \quad \text{in } \Omega, \label{eq:MEMS1}\\
 u=0 \quad \text{on } \partial \Omega. \label{eq:MEMS2}
\end{gather}

\begin{theorem} \label{thm3.1}
There exists a $\lambda^*$ such that when $\lambda>\lambda^*$ there
is no solution to \eqref{eq:MEMS1} and \eqref{eq:MEMS2}.
\end{theorem}

\begin{proof} Let $\lambda_1$ be the lowest eigenvalue of
\begin{gather}
-\Delta u=\lambda u \quad \text{in } \Omega,
\\
u=0 \quad \text{on } \partial \Omega,
\end{gather}
with $u_1$ the corresponding eigenfunction which can be chosen strictly
positive on $\Omega$.
Multiplying (\ref{eq:MEMS1}) by $u_1$ and integrating yields
$$
\int_{\Omega} -u\Delta u_1=\lambda \int_{\Omega} \frac{u_1}{(1-u)^2}.
$$
Or equivalently,
$$
\lambda_1\int_{\Omega} u u_1=\lambda\int_{\Omega} \frac{u_1}{(1-u)^2}.
$$
Since $\frac{1}{(1-u)^2}\geq \frac{27}{4}u$ for $0\leq u<1$, we have
$$
\lambda_1\int_{\Omega} u u_1=\lambda\int_{\Omega} \frac{u_1}{(1-u)^2}
\geq \frac{27\lambda}{4}\int_{\Omega} u_1u.
$$
Hence, $\lambda\leq \frac{\lambda_1}{27}$. This completes the proof.
\end{proof}

Next we shall obtain the existence result for small $\lambda$.
We have the following theorem.

\begin{theorem} \label{thm3.2}
There exists a solution to \eqref{eq:MEMS1} and \eqref{eq:MEMS2}
 for some small $\lambda$.
\end{theorem}

To prove this theorem, we should apply the method of upper and lower
 solution. We have the following definition.

\begin{definition} \rm
A function $\bar{u} \in C^2(\Omega)$ is called an uppersolution of
\eqref{eq:MEMS1} and \eqref{eq:MEMS2} if it satisfies the inequalities
\begin{gather*}
-\Delta u \geq \frac{\lambda}{(u-1)^2} \quad \text{on } \Omega
\\
 u \geq 0 \quad \text{on } \partial \Omega.
\end{gather*}
Similarly, $\underline{u}$ is called a lower solution if it satisfies
 all the reversed inequalities.
\end{definition}

The following two lemmas provide us with a proper choice of lower
and upper solutions.

\begin{lemma} \label{lem3.4}
Any constant $c<0$ is a lower solution.
\end{lemma}

\begin{lemma} \label{lem3.5}
$\bar{u}=\frac{1}{3} v_1$ is an upper solution when
$\lambda\leq \frac{4}{27} \alpha_1m $.
Here $\alpha_1$ and $v_1$ is the first eigenvalue and eigenfunction
for the  problem
\begin{gather}
-\Delta v=\alpha v \quad \text{in } \Omega',
\\
v=0 \quad \text{on } \partial \Omega',
\end{gather}
where $\Omega'$ is a proper domain with smooth boundary which
contains $\Omega$  and  has been chosen such that $m\leq v_1\leq 1$
on $\Omega$ and $m>1/2$.
\end{lemma}

\begin{proof}
It is sufficient to show that
$$
-\Delta \bar{u}\geq \frac{\lambda}{(1-\bar{u})^2} \quad \text{in } \Omega.
$$
In fact,
$$
-\Delta \bar{u}=-\frac{1}{3} \Delta v_1 =\frac{1}{3} \alpha_1v_1\geq
 \frac{1}{3}\alpha_1m \geq \frac{\lambda}{3}\cdot \frac{27}{4}
 \geq \frac{\lambda}{(1-1/3v_1)^2}=\frac{\lambda}{(1-\bar{u})^2}.
 $$
This completes the proof.
\end{proof}
The existence result follows from the above two Lemmas.


\section{Multiplicity} \label{s4}

In this section we are concerned with the multiplicity of positive
radial solutions.
A radial solution $u=u(r)$ of \eqref{eq:MEMS1} and \eqref{eq:MEMS2}
satisfies the following equations
\begin{gather*}
u''(r)+\frac{1}{r}u'(r)+\frac{\lambda}{(1-u)^2}=0,\quad r\in(\epsilon,1),
u(\epsilon)=u(1)=0.
\end{gather*}
Let $s=-\ln r, w(s)=u(r)$, then $w(s)$ satisfies
\begin{gather*}
w''+\lambda e^{-2s} \frac{1}{(1-w)^2}=0 \quad \text{in } (0,-\ln \epsilon),
\\
w(0)=w(-\ln\epsilon)=0.
\end{gather*}
Henceforth, we shall consider the following initial value problem
\begin{gather*}
u''(r)+\lambda e^{-2r} \frac{1}{(1-u(r))^2}=0 \quad \text{in }
(0,-\ln \epsilon),
\\
u(0)=0 \quad {\rm and}\quad u'(0)=p.
\end{gather*}
Let $u(\cdot)=u(\cdot,p,\lambda)$ be the solution and define
$$
R(p,\lambda)=min\{R>0: u(R,p,\lambda)=0\}.
$$
We shall prove in the next lemma that $R(p,\lambda)$ is well
defined for all $p$.

By the boundary condition, $u$ has exactly one critical point,
at which it takes the maximum value. We shall denote this critical
point by $\tau(p,\lambda)$. Hence
$$
u'(r)>0 \quad {\rm for} \quad r\in (0,\tau(p,\lambda))
\quad \text{and} \quad u'(r)<0 \quad \text{for }
 r\in(\tau(p,\lambda),R(p,\lambda)).
 $$
Also note that
$$
u(r)=pr+\lambda\int_0^{r}(s-r)e^{-2s}\frac{1}{(1-u(s))^2}ds.
$$
To prove our multiplicity result, we need to establish several useful lemmas.

\begin{lemma} \label{lem4.1}
$R(p,\lambda)$ is well defined.
\end{lemma}

\begin{proof}
First we claim that it is indeed well defined for sufficiently
large $p$ and sufficiently small $p$.
Suppose otherwise that $\lim_{r\to  +\infty} u'(r)=0.$
Multiplying equation (4.5) by $u'$ and integrating yields
$$
\int_{0}^r u''(s)u'(s)ds=-\lambda \int_0^r \frac{e^{-2s}}{(1-u(s))^2}u'(s)ds.
$$
Hence
$$
\frac{1}{2}u'(r)^2-\frac{1}{2}p^2=\lambda-\lambda\frac{e^{-2r}}{1-u(r)}
-2\lambda\int_0^r\frac{e^{-2s}}{1-u(s)}ds.
$$
Let $h(r)=\int_0^r\frac{e^{-2s}}{1-u(s)}ds$, then we have
$$
\lambda h'(r)+2\lambda h(r)-\frac{1}{2}p^2+\frac{1}{2}u'(r)^2-\lambda=0.
$$
Notice that when $r$ is sufficiently large,
\begin{align*}
h(r)&=\int_0^r\frac{e^{-2s}}{1-u(s)}ds \\
    &=-\frac{1}{\lambda}\int_0^r u''(s)(1-u(s)) ds \\
    &=\frac{1}{\lambda}\Big[-(1-u)u'+p-\int_0^ru'^2(s)ds\Big] \\
    &\leq\frac{p}{\lambda}.
\end{align*}
Hence for sufficiently large $r$,
\begin{align*}
\lambda h'(r)&=-2\lambda h(r)+\frac{1}{2}p^2-\frac{1}{2}u'(r)^2+\lambda\nonumber\\
             &\geq-2p+\frac{1}{2}p^2-\frac{1}{2}u'(r)^2+\lambda\nonumber\\
             &\geq c>0
\end{align*}
for some constant $c$ and $p$ sufficiently large or small.
Therefore,
$$
\frac{e^{-2r}}{1-u(r)}\geq \frac{c}{\lambda}>0
$$
for sufficiently large $r$.
It follows that,
$\lim_{r\to  +\infty} u(r)=1$ for $p$ sufficiently large or small.
Applying L'Hopital's rule we have
\begin{align*}
\lim_{r\to  +\infty} h'(r)
&=\lim_{r\to  +\infty}\frac{e^{-2r}}{1-u(r)}\\
&=\lim_{r\to  +\infty} \frac{2e^{-2r}}{u'(r)}\\
&=\lim_{r\to  +\infty}\frac{-4e^{-2r}}{u''(r)}\\
&=\lim_{r\to  +\infty}\frac{4e^{-2r}}{\frac{e^{-2r}}{\lambda(1-u)^2}}\\
&=\lim_{r\to  +\infty} 4(1-u(r))=0.
\end{align*}
This is a contradiction to the previous conclusion that
$h'(r)\geq \frac{c}{\lambda}>0$.
Hence $R(p,\lambda)$ is well defined for $p$ sufficiently large and small.
 By continuous dependence on parameters, $R(p,\lambda)$ is well
defined for all $p$. This completes the proof.
\end{proof}

\begin{lemma} \label{lem4.2}
$$ \lim_{p\to  0+} R(p,\lambda)
=\lim_{p\to  0+} \tau(p,\lambda)=0.
$$
\end{lemma}

\begin{proof}
Suppose otherwise, there exists a $\lambda>0$, $\epsilon>0$
and a sequence $p_k\to  0+$ such that
$$
R_k \equiv R(p_k,\lambda)\geq \epsilon.$$
Since
\begin{align*}
u(r,p_k)&=p_kr+\lambda\int_0^r (s-r)e^{-2s}\frac{1}{\big(1-u(s)\big)^2}ds \\
&\leq p_kr+\lambda\int_0^r(s-r)e^{-2s}ds \\
&=p_kr+\lambda\Big[-\frac{1}{2}e^{-2s}(s-r)|_{0}^r
+\int_0^r \frac{1}{2}e^{-2s}ds\Big]\\
&=p_kr+\lambda\Big[-\frac{r}{2}-\frac{e^{-2r}}{4}+\frac{1}{4}\Big]\\
&<p_kr-\frac{\lambda r^2}{4},
\end{align*}
thus
$R_k<\frac{4p_k}{\lambda}$.
Hence,
\begin{align*}
p_kR_k&= -\lambda\int_0^{R_k}(s-R_k)e^{-2s}\frac{1}{(1-u(s))^2}ds \\
&\geq \lambda\int_{0}^{\epsilon}(\epsilon-s)e^{-2s}ds >0.
\end{align*}
This is a contradiction. Hence $\lim_{p\to  0+} R(p,\lambda)=0$.
It follows that $\lim_{p\to  0+} \tau(p,\lambda)=0$.
This completes the proof.
\end{proof}


\begin{lemma} \label{lem4.3}
$$\lim_{p\to  +\infty} R(p,\lambda)
=\lim_{p\to  +\infty} \tau(p,\lambda)=0.
$$
\end{lemma}

\begin{proof}
 Suppose $\lim_{p\to  \infty}
\tau(p,\lambda)\neq 0$, then there exists a $\tau_0>0$ and a
sequence
$p_k\to  +\infty$ with $u_k(r)\equiv u(r,p_k,\lambda)>0$ and $u_k'(r)> 0$ in $(0,\tau_0)$.\\
Let $\bar{\tau}=\tau_0/2$, we claim
$$
\limsup_{k\to +\infty} u_k(\bar{\tau})=1.
$$
Otherwise, there exists $\epsilon>0$ such that
$0<u_k(\bar{\tau})\leq 1-\epsilon$. It follows that
\begin{align*}
u_k(\bar{\tau})
&=p_k\bar{\tau}+\lambda\int_0^{\bar{\tau}}(r-\bar{\tau})e^{-2r}
\frac{1}{(1-u_k(r))^2}dr  \\
&\geq p_k\bar{\tau}+\frac{\lambda}{\epsilon^2}\int_0^{\bar{\tau}}
(r-\bar{\tau})e^{-2r}dr
\end{align*}
which is impossible since $p_k\to  +\infty$. Hence choosing
a subsequence if necessary, we may assume
$$ \lim_{k\to  +\infty} u_k(\bar{\tau})=1.
$$
Note that $u_k$ satisfies
$$
u''(r)+\frac{\lambda e^{-2r}}{u_k(1-u_k)^2}u(r)=0 \quad text{in }
 (\bar{\tau},\tau_0).
$$
Let
$$ M_k=\inf\{\frac{1}{u_k(1-u_k)^2}: r\in (\bar{\tau},\tau_0)\},
$$
then
$$ \lim_{k\to  +\infty} M_k=\infty.
$$
Note that $\lambda e^{-2r}\geq \lambda e^{-2\tau_0}$ in
$(\bar{\tau},\tau_0)$. Let $v_k$ solves
$$
v''(r)+\lambda e^{-2\tau_0} M_kv(r)=0 \quad text{in }
 (\bar{\tau},\tau_0).
$$
It follows that $v_k$ has at least two zeros in $(\bar{\tau},\tau_0)$
when $k$ is sufficiently large. By Sturm Comparison Principle, $u_k$
has at least one zero in $(\bar{\tau},\tau_0)$. But this is impossible.
Hence
$$
\lim_{p\to  +\infty} \tau(p,\lambda)=0.
$$
Finally, we prove $\lim_{p\to  +\infty} R(p,\lambda)=0$.
Otherwise, there exists a point $r_0>0$ and a sequence
$p_k\to  +\infty$ with
$$
u_k(r)>0\quad \text{and}\quad u_k'(r)\leq 0 \quad text{in }
 (\tau_k,r_0)
 $$
where $u_k\equiv u(r,p_k,\lambda)$ and $\tau_k \equiv \tau(p_k,\lambda)$.
Let $\bar{r}=\frac{r_0}{2}$, in view of previous lemma that
$\lim_{p\to  +\infty} \tau(p,\lambda)=0$, we may assume
$\bar{r}>\tau_k$ for any $k$. We claim that
$$
\limsup_{k\to  +\infty} u_k(\bar{r})<1.
$$
Otherwise, by Sturm Comparison Principle again, $u_k$ has zeros
in $(\tau_k,\bar{r})$ when $k$ is sufficiently large which is
impossible since $\tau_k\to  0$ as $k\to  +\infty$.\\
Note that
$$
u_k'(r)=-\int_{\tau_k}^r \frac{\lambda e^{-2s}}{(1-u_k(s))^2} ds,
$$
and
\begin{equation}
\left({1\over 2}u'^2+\frac{\lambda e^{-2r}}{1-u(r)}\right)'
=-\frac{2\lambda e^{-2r}}{1-u(r)}. \label{1.41}
\end{equation}
Integrate equation (\ref{1.41}) on $(\tau_k, \bar{r})$, we have
$$
{1\over 2} u_k'(\bar{r})^2=-\frac{\lambda e^{-2\bar{r}}}{1-u_k(\bar{r})}+\lambda \frac{e^{-2\tau_k}}{1-u_k(\tau_k)}-\int_{\tau_k}^{\bar{r}} \frac{2\lambda e^{-2s}}{1-u_k(s)}ds.$$
On the other hand, we have
\begin{align*}
{1\over 2} u_k'(r)^2+\int_{\tau_k}^{\bar{r}}
\frac{2\lambda e^{-2s}}{1-u_k(s)}ds
&\leq {1\over 2} u_k'(r)^2+ \int_{\tau_k}^{\bar{r}}
\frac{2\lambda e^{-2s}}{(1-u_k(s))^2}ds  \\
&\leq  {1\over 2} u_k'(\bar{r})^2+2|u_k'(\bar{r})|.
\end{align*}
Hence
\begin{equation}
-\frac{\lambda e^{-2\bar{r}}}{1-u_k(\bar{r})}
+\lambda \frac{e^{-2\tau_k}}{1-u_k(\tau_k)}
\leq {1\over 2} u_k'(\bar{r})^2+2|u_k'(\bar{r})|. \label{1.42}
\end{equation}
Integrating equation (\ref{1.41}) on $(0, \tau_k)$, we have
$$
\frac{\lambda e^{-2\tau_k}}{1-u_k(\tau_k)}+\int_0^{\tau_k}
\frac{2\lambda e^{-2s}}{1-u_k(s)}ds ={1\over 2} p_k^2+\lambda.
$$
Therefore,
\begin{equation}
\frac{\lambda e^{-2\tau_k}}{1-u_k(\tau_k)}
\geq \frac{1}{2} ({1\over 2} p_k^2+\lambda).\label{1.43}
\end{equation}
Combining inequalities (\ref{1.42}) and (\ref{1.43}), we have
$u_k'(\bar{r})\to  -\infty$.
Thus for $r>\bar{r}$, we have
$$
u_k(r_0)<u_k(\bar{r})+u_k'(\bar{r})(r_0-\bar{r})\to  -\infty,
$$
a contradiction to $u_k(r_0)>0$. This completes the proof.
\end{proof}


\begin{lemma} \label{lem4.4}
Define $\tilde{R}=\tilde{R}(\lambda)=sup\{R(p,\lambda),p>0\}$.
Then $\tilde{R}(\lambda)$ is strictly decreasing.
\end{lemma}

\begin{proof}
Let $0<\lambda_1<\lambda_2$ and $u_2$ is a solution at $\lambda_2$
on $(0,\tilde{R}(\lambda_2))$.
Let $v(s)=cu_2(r)$ with $r=s/c$ where $c$ is some constant greater
 but close to 1.
  It's easy to see $v(0)=0$ and $v(\tilde{R}(\lambda_2)+\epsilon_1)=0$
 for $\epsilon_1=(c-1)\tilde{R}(\lambda_2)$.
We note that
\begin{align*}
v''+\lambda_1\frac{e^{-2s}}{(1-v(s))^2}
& = \frac{1}{c}u_2''(r)
+\lambda_1\frac{e^{-2s}}{(1-cu_2(r))^2} \nonumber\\
&=-\frac{1}{c}\left(\lambda_2\frac{e^{-2r}}{(1-u_2(r))^2}
-\lambda_1\frac{e^{-2s}}{(1-cu_2(r))^2} \right)
\leq 0
\end{align*}
when $c$ is sufficient close to 1. Hence $v$ is a lower solution for
\begin{gather*}
v''(r)+\lambda_1\frac{e^{-2r}}{(1-v)^2}=0,\\
v(0)=0,\quad v(\tilde{R}(\lambda_2)+\epsilon_1)=0.
\end{gather*}
Hence $\tilde{R}(\lambda_1) \geq \tilde{R}(\lambda_2)+\epsilon_1$.
Hence $\tilde{R}(\lambda)$ is strictly decreasing. This completes the proof.
\end{proof}

\begin{lemma} \label{lem4.5}
$\lim_{\lambda\to  0+} \tilde{R}(\lambda)=+\infty$,
$\lim_{\lambda\to  +\infty} \tilde{R}(\lambda)=0$.
\end{lemma}

\begin{proof}
Suppose $\lim_{\lambda\to  0+} \tilde{R}(\lambda) \neq +\infty$,
then there exists a number $R^*>0$ and a sequence $\lambda_k\to  0+$ with
$\lim_{k\to  +\infty} \tilde{R}(\lambda_k)=\lim_{k\to
+\infty} \tilde{R}(\lambda_k,p_k)=R^*$.
Let us write $u_k(r)=u(r,\lambda_k,p_k)$, then
\begin{align*}
0&=u_k(R^*)\\
&=p_kR^* +\lambda_k\int_0^{R^*}(s-R^*)e^{-2s}\frac{1}{(1-u_k(s))^2}ds \\
&\geq p_kR^*+\frac{\lambda_k}{\epsilon^2} \int_0^{R^*}(s-R^*)e^{-2s}ds.
\end{align*}
Hence $p_k\to  0+$ or $R^*=0$. But this contradicts the fact that
$\lim_{p\to  0+}R(p,\lambda)=0$ and $R(\lambda)$ is strictly decreasing.
Similarly we may prove the second statement. This completes the proof.
\end{proof}

Finally for any given $\lambda$, we study the shape of $R(p)$.
Notice that $R(p)$ is determined by the implicit equation
\begin{equation}
u(R(p),p)=0. \label{eq:TimeMap}
\end{equation}
Differentiating equation~(\ref{eq:TimeMap}) with respect to $p$ we get the following equations for the derivatives of $R$:
\begin{equation}
u_r(R(p),p)R'(p)+u_p(R(p),p)=0, \label{eq:TimeMap1}
\end{equation}
\begin{equation}
\begin{gathered}
 u_{rr}(R(p),p)R'(p)^2+2u_{rp}(R(p),p)R'(p)\\
 +u_r(R(p),p)R''(p)
+u_{pp}(R(p),p)=0.
\end{gathered}\label{eq:TimeMap2}
\end{equation}
If we write $h(r,p)=u_p(r,p)$, $z(r,p)=u_{pp}(r,p)$
and $v(r,p)=u_r(r,p)$, then we can rewrite \eqref{eq:TimeMap1} as
\begin{equation}
v(R(p),p)R'(p)+h(R(p),p)=0.
\end{equation}
Also notice that when $R'(p)=0$, from equation (\ref{eq:TimeMap2}) we have
\begin{equation}
v(R(p),p)R''(p)+z(R(p),p)=0 \label{eq:TimeMap3}
\end{equation}
We have the following important Lemma.

\begin{lemma}
For a given $\lambda$, if $R'(p)=0$, then $R''(p)<0$.
\end{lemma}

\begin{proof} \label{lem4.6}
Note that $h(r,p)$ satisfies the following initial value problem
\begin{gather}
h''+\frac{2\lambda e^{-2r}}{(1-u)^3}h(r,p)=0, \label{eq:H}
\\
h(0,p)=0,\quad h'(0,p)=1.
\end{gather}
If $R'(p)=0$, then equation (\ref{eq:TimeMap1}) gives us $h(R(p),p)=0$.

We claim that $h(r,p)>0$ on $(0,R(p))$. Otherwise let $h(\xi(p),p)=0$
and $h>0$ on $(0,\xi(p))$. Note that $v$ satisfies the following
\begin{gather}
v''+\frac{2\lambda e^{-2r}}{(1-u)^3}v-\frac{2\lambda e^{-2r}}{(1-u)^2}=0,
\\
v(0,p)=p, \quad v'(0,p)=-\lambda.
\end{gather}
Recall that $v(\tau(p),p)=0$.
If $\xi(p)\geq \tau(p)$, then $v<0$ on $(\xi(p),R(p))$.
By Sturm Comparison Theorem, $v$ should have a zero on $(\xi(p),R(p))$
since $h(R(p),p)=0$. This is impossible.

If $\xi(p)<\tau(p)$, then $v<0$ on $(\tau(p),R(p))$.
Since $0=v(\tau(p),p)>h(\tau(p),p)$, by Sturm Second Comparison
Theorem, $v>h$ on $(\tau(p),R(p))$ which is impossible since $h$
has to cross over $v$ and reaches zero at $R(p)$.

Next we claim $z(R(p),p)<0$. Note that
\begin{equation}
\begin{gathered}
z''+\frac{2\lambda e^{-2r}}{(1-u)^3}z
+\frac{6\lambda e^{-2r}}{(1-u)^4}h^2=0, 
\\
z(0,p)=0,\quad z'(0,p)=0.
\end{gathered}\label{eq:Z}
\end{equation}
We claim $z$ is negative in some neighborhood of 0. Otherwise by
observing equation (\ref{eq:Z}), we have $z''<0$.
It follows that $z'<0$ in the neighborhood of 0 since $z'(0,p)=0$.
This contradicts the assumption.

Next we claim $z<0$ in $(0,R(p)]$. Otherwise, let $z(r_1,p)=0$
 with $z<0$ in $(0,r_1)$. Comparing
equation (\ref{eq:H}) and equation (\ref{eq:Z}), it follows that $h$
must have a zero in $(0,r_1)$ which contradicts our previous statement.
Hence $z(R(p),p)<0$ and it follows from (\ref{eq:TimeMap3})
that $R''(p)<0$.
\end{proof}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.43\textwidth]{fig2} % timemap
\quad
\includegraphics[width=0.45\textwidth]{fig3} % timemap
\end{center}
\caption{Time map diagram and bifurcation diagram \label{bifur}}
\end{figure}

We are now in position to prove  Theorem \ref{thm1.1}.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
In view of the above lemmas, we may obtain the timemap diagram
as shown in Figure \ref{bifur}. %timemap. 
From which we can easily conclude
the theorem. In fact, for any given $\epsilon>0$, $\exists$ $\lambda^*$
such that $\tilde{R}(\lambda^*)=-\ln \epsilon$ and there is a unique
$p$ such that $R(\lambda^*, p)=-\ln \epsilon$, thus there exists a
unique radial solution at $\lambda=\lambda^*$. For $\lambda<\lambda^*$,
 we can find $p_1, p_2$ such that
$R(\lambda, p_1)=R(\lambda, p_2)=-\ln \epsilon$. The problem has two
radial solutions in this case. For $\lambda>\lambda^*$, since
$\tilde{R}(\lambda)<-\ln \epsilon$, there is no radial solution.
This result is shown in Figure \ref{bifur}.
\end{proof}

\section{Symmetry breaking} \label{s5}

In previous section, we studied the multiplicity of radial solutions.
Our purpose in this section is to study how radial symmetry can be broken,
that is, to describe the bifurcation of these radial solutions into
non-radial solutions. The bifurcation problem has been studied by
many authors, see \cite{Lin1, Lin2, Lin3, Smoller}.

We shall consider two real Banach Spaces, $U\subset V$, as well as
a nonlinear abstract operator
$$ F:\quad R\times U\to  V
$$
of the form
$$ F(\lambda,u)=L(\lambda)u+R(\lambda,u)
$$
and the associated nonlinear abstract equation
$$ F(\lambda,u)=0
$$
where the following assumptions are assumed to be satisfied:
\begin{itemize}
\item There exists $\lambda_0\in R$ and $a,b\in R,\quad a<\lambda_0<b$, such that $L(\lambda)$ is a linear operator from $U$ to $V$ for all $\lambda\in (a,b)$. Moreover, $exists$ $r\geq 2$ such that the map $\lambda\to  L(\lambda)$ is of class $C^r$ and $L(\lambda_0)$ is a Fredholm operator of index zero.
\item R is an operator of class $C^r$ such that $R(\lambda,0)=0$ and $D_uR(\lambda,0)=0$ for each $\lambda\in (a,b)$.
\end{itemize}

\begin{definition} \rm
$(\lambda_0,0)$ is a bifurcation point from the curve of $(\lambda,0)$
if there exists a sequence $(\lambda_n,u_n)\in (a,b)\times(U\setminus\{0\})$
such that $\lim_{n\to  infty}(\lambda_n,u_n)=(\lambda_0,0)$ and
$F(\lambda_n,u_n)=0$.
\end{definition}

\begin{definition} \rm
$\lambda_0$ is a nonlinear eigenvalue of $L(\lambda)$ if $(\lambda_0,0)$
is a bifurcation point from the curve $(\lambda,0)$ and $R(\lambda,u)$
satisfies the second assumption.
\end{definition}

On other word, $\lambda_0$ is a nonlinear eigenvalue of $L(\lambda)$
if the fact that bifurcation occurs is exclusively based on the linear part.

\begin{definition} \rm
We call zero  a simple eigenvalue of $L(\lambda_0)$ if
$N[L(\lambda_0)]\oplus R[L(\lambda_0)]=V$.
\end{definition}

\begin{definition} \rm
Define $\lambda_0$ as an eigenvalue of the pair $(L_0,L_1)$ if zero
is an eigenvalue of $L_0-\lambda_0L_1$.
\end{definition}

Let $a(\lambda)$ denote the classical eigenvalue of the family $L(\lambda)$
 perturbed from the zero eigenvalue of $L(\lambda_0)$.
If zero is a simple eigenvalue of $L(\lambda_0)$, then
$a'(\lambda_0)\neq 0$ if and only if zero is a simple eigenvalue of the
pair $(L_0,L_1)$.
As we recall Crandall's theorem which states that if zero is a simple
eigenvalue of the pair $(L_0,L_1)$, then $(\lambda_0,0)$ is a bifurcation
 point. In other word,  $(\lambda_0,0)$ is a bifurcation point if
$a'(\lambda_0)\neq 0$. This condition is usually referred to
 as ``transversality condition" or ``nondegeneracy condition".
 We shall remove this condition by the following theorem essentially
 due to Kielh\"ofer.

\begin{theorem} \label{thm5.5}
Assume $U\subset V$ and zero is a simple eigenvalue of $L(\lambda_0)$.
Then $\lambda_0$ is a nonlinear eigenvalue of $L(\lambda)$ if and
only if $a(\lambda)$ changes sign as $\lambda$ crosses $\lambda_0$.
\end{theorem}

With the aid of this result, we now study the symmetry breaking problem.
We shall consider the linearized problem about a given radial solution $u$
$$
\Delta w+\frac{2\lambda }{(1-u)^3}w=0.
$$
We may write $w$ in the spherical harmonic decomposition form:
$$
w=\sum_{N=0}^{\infty} a_N(r)\Phi_N(\theta),
$$
and $a_N$ satisfies the equation:
$$
a_N''+\frac{1}{r}a_N'+\Big(\frac{2\lambda }{(1-u)^3}-\frac{N^2}{r^2}
\Big)a_N=0
$$
together with the boundary conditions
$a_N(1)=0=a_N(\epsilon)$.

If the above equation admits a nonzero solution $a_N\neq 0$ for
some $N\geq 1$, then radial symmetry breaks.
We consider the  eigenvalue problem
$$
a_N''+\frac{1}{r}a_N'+\Big(\frac{2\lambda }{(1-u)^3}-\frac{N^2}{r^2}\Big)a_N
=-\mu_{N,k}a_N.
$$
In the context of our previous setting, we shall let
$U=C_0^2(\epsilon,1)$ and $V=C(\epsilon,1)$.
We have the following lemma.

\begin{lemma} \label{lem5.6}
If $u$ is a radial solution on the upper branch, then for arbitrary
positive integer $N$, $\mu_{N,1}(\lambda)<0$ for $\lambda$ sufficiently
close to zero.
\end{lemma}

\begin{proof}
The eigenvalue $\mu_{N,1}(\lambda)$ can be characterized as
$$
\mu_{N,1}=\inf_{\phi\in C_0^2([\epsilon,1])}
\Big\{\frac{\int_{\epsilon}^{1} r(\phi'^2-\frac{2\lambda }{(1-u)^3}\phi^2
+N^2r^{-2}\phi^2)dr}{\int_{\epsilon}^1 r\phi^2 dr}\Big\}.
$$
If $u$ is a positive radial solution, then
$$
\int_{\Omega}|\bigtriangledown u|^2=\lambda\int_{\Omega}
 \frac{u}{(1-u)^2}.
$$
Since $u$ is a solution on the upper branch, $||u||_{\infty}\to  1$
as $\lambda\to  0+$. Notice that for arbitrary $p>0$, there exists
$\alpha>0$ such that
$$ \frac{2u}{1-u}\geq p \quad \text{for }  u\geq 1-\alpha.
$$
We write
$$
Q(u)=\int_{\epsilon}^1 r\Big(u'^2-\frac{2\lambda}{(1-u)^3}u^2
+\frac{N^2}{r^2}u^2\Big).
$$
Hence
\begin{align*}
2\pi Q(u)
&=\lambda\int_{\epsilon}^1 \Big(\frac{1}{(1-u)^2}-\frac{2u}{(1-u)^3} \Big)u
+N^2\int_{\epsilon}^1\frac{u^2}{r^2} \\
&=\int_{\epsilon}^1 |\bigtriangledown u|^2
-\lambda\int_{\epsilon}^1\frac{2u}{(1-u)}\cdot
\frac{1}{(1-u)^2}u +N^2\int_{\epsilon}^1 \frac{u^2}{r^2}   \\
&\leq  (1-p) \int_{\epsilon}^1 |\bigtriangledown u|^2
+\frac{N^2}{\epsilon^2}\int_{\epsilon}^1 u^2 -\lambda\int_{u\leq 1
-\alpha}\frac{2u}{(1-u)}\cdot \frac{1}{(1-u)^2}u \\
&\leq \Big(1-p+\frac{N^2}{\epsilon^2 \nu_1}\Big)
\int_{\epsilon}^1 |\bigtriangledown u|^2 -M
\end{align*}
for some constant $M>0$ which is independent of $\lambda$.
Hence for any given $N>0$, $\mu_{N,1}(\lambda)<0$ for $\lambda$
 sufficiently close to zero since $p>0$ can always be chosen to be
 sufficiently large. This completes the proof.
\end{proof}

\begin{remark} \label{rmk1}\rm
It's easy to see that if $u$ is an upper branch solution, then
$$
\Big(\int_{\Omega} |\bigtriangledown u|^2\Big)^{1/2}
\geq \sqrt{\frac{2\pi\epsilon}{1-\epsilon}}
$$
as $\lambda\to  0$. In fact,
\begin{align*}
u(r)&=\int_{\epsilon}^r u'(s)ds \leq(1-\epsilon)^{1/2}
\Big(\int_{\epsilon}^{1} (u'(s))^2 ds \Big)^{1/2} \\
&\leq (1-\epsilon)^{1/2}\frac{1}{\sqrt{2\pi\epsilon}}
\Big(\int_{\Omega} |\bigtriangledown u|^2\Big)^{1/2}
\end{align*}
\end{remark}
We now prove the symmetry breaking result.

\begin{proof}[Proof of Theorem \ref{thm1.2}]
Since $\mu_{0,1}(\lambda^*)=0$, it follows that $\mu_{N,1}(\lambda^*)>0$
for $N\geq 1$. By Lemma \ref{lem5.6}, for any $N\geq 1$ there exists
$\lambda_N\in (0,\lambda^*)$ such that $\mu_{N,1}(\lambda_N)=0$ and
$\mu(\lambda)$ changes sign as $\lambda$ crosses $\lambda_N$.
Hence by Theorem \ref{thm5.5}, there is a bifurcation at $\lambda_N$ where
the radial symmetry breaks. The proof is completed.
\end{proof}

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\end{document}