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\markboth{\hfil The first eigenvalue of the p-Laplacian\hfil 
EJDE--2001/35}
{EJDE--2001/35\hfil Tilak Bhattacharya \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 35, pp. 1--15. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or 
http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
\vspace{\bigskipamount} \\
%
 Some observations on the first eigenvalue of the p-Laplacian and
 its connections with asymmetry
%
\thanks{ {\em Mathematics Subject Classifications:} 35J60, 35P30.
\hfil\break\indent
{\em Key words:} Asymmetry, De Giorgi perimeter, p-Laplacian,
first eigenvalue, \hfil\break\indent Talenti's inequality.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted September 3, 2000. Published May 16, 2001.} }
\date{}
%
\author{Tilak Bhattacharya}
\maketitle

\begin{abstract}
In this work, we present a lower bound for the first eigenvalue
of the $p$-Laplacian on bounded domains in $\mathbb{R}^2$.
Let $\lambda_1$
be the first eigenvalue and $\lambda_1^*$ be the first eigenvalue
for the  ball of the same volume. Then we show that
$\lambda_1\ge\lambda_1^*(1+C\alpha(\Omega)^{3})$, for some constant $C$,
where $\alpha$ is the asymmetry of the domain $\Omega$. This provides a
lower bound sharper than the bound in Faber-Krahn inequality.
\end{abstract}


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\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]


\section{Introduction}

Let $\Omega \subset \mathbb{R}^{n}$, be a bounded domain. For
$1<p<\infty$, let
$$\lambda_1=\lambda_1(p, \Omega)=
\inf\frac{\int_{\Omega}|Du|^{p}}{\int_{\Omega}|u|^{p}},$$
where the infimum is taken over all $u\in W^{1,p}_{0}(\Omega)$,
$u\ne 0$. It is well known that $\lambda_1 = \lambda_1 (\Omega,p)>0$ and
a non-zero
minimizer, which we continue to call as $u=u(p, \Omega)$, exists and
satisfies
\begin{equation}
\mathop{\rm div}(|Du|^{p-2}Du) + \lambda_1 |u|^{p-2}u = 0,
\quad \mbox {in } \Omega ,
\end{equation}
where $u\in W^{1,p}_0(\Omega)$.
The operator $\mathop{\rm div}(|Du|^{p-2}Du)$ is the $p$-Laplacian and
this is the usual Laplacian when $p=2$. For $p\ne 2$, this is a
quasi-linear and a degenerate elliptic operator. The equation in
(1.1) is to be interpreted in the weak sense, i. e.,
$$\int_{\Omega}|Du|^{p-2}Du\cdot
D\psi=\lambda_1\int_{\Omega}|u|^{p-2}u\psi,\;\;
\forall\;\psi\in W^{1,p}_0(\Omega).$$
We refer to $\lambda_1$ as the first eigenvalue and $u$ as
the first eigenfunction of the $p$-Laplacian on $\Omega$. It is
well known that $\lambda_1$ is simple and $u$ has one sign
\cite{b2,l1, l2}. The first eigenvalue is also known to be
isolated [10]. Moreover, if $\Omega$ is a ball then $u$ is
radial, decreasing and has only one critical point. It will be
useful to note that
$u$ is $L^{\infty}(\Omega)$ \cite{l2}. It is also quite well
known that $u$ is $C^{1,\alpha}_{loc}$ in $\Omega $
\cite{d1,m1,t3}. See [2] for a more detailed discussion of
matters related to the regularity of $u$. It should also be
pointed out
that unlike the case of the Laplacian, i. e., $p=2$,
a complete characterization of the set of critical points of $u$,
when $p\ne 2$, is still unknown. This fact or lack thereof
becomes
particularly important when working with level sets of $u$. The
boundaries of such sets need not be smooth thus necessitating the
use of the DeGiorgi perimeter. We discuss this further in section
2. Also see \cite{b1,b2,t1,t2}.

Let $D^*$ denote the symmetrized
domain for an open set $D$, i .e., $D^{*}$ is the ball,
centered at the origin, with volume equal to that of $D$.
Let $\lambda ^*_1= \lambda_1 (\Omega^*)$; then the Faber-Krahn
inequality states that
$ \lambda_1 \ge \lambda^{*}_1$, where equality holds if and only if
$\Omega$ is a ball \cite{b2}.
Our attempt, in this work, will be to characterize this lower
bound for
$\lambda_1$, for $1<p<\infty$,
in terms of asymmetry.  The notion of asymmetry, which
was
introduced in \cite{h3}, is a measure of how close a set is to
being a ball.
More precisely, if $D$ is a compact set in $\mathbb{R}^n$, $n \ge 2$,
then the
asymmetry of $D$, denoted by $\alpha (D)$, is defined to be
\begin{equation}
\alpha (D) = \inf _{x} \frac {\mathop{\rm vol} (D \backslash B(x, R))}
{\mathop{\rm vol} (D)} .
\end{equation}
Here $\mathop{\rm vol}$ stands for volume, $B(x, R)$ is the ball 
centered at $x$,
radius $R$, such that the volume of $B(x, R)$ is the same as that of
$D$.

In \cite{b1,h1,h2,h3} lower bounds for capacities, for planar
domains, were
obtained in terms of asymmetry while in [3], an analogous upper
bound for the
Green's function was derived. The works
in \cite{h3,n1} address the issue of the first eigenvalue of the
Laplacian and present a sharper version of the Faber-krahn
inequality in terms of deficiencies. This work generalizes the
estimate in [9] to the case of the $p$-Laplacian on planar
domains. We thank the referee whose comments have helped improved
the exposition of this work. We also thank Juan Manfredi for his
encouragement and interest in this work. We are also highly appreciative
of Tom Salisbury of Department of Mathematics, York University, who 
kindly extended
texing facilities to us.
\section{Statement of the main result}

For $D\subset \mathbb{R}^{2}$, let $|D|$ denote
the area of a set $D$ and $\partial D$ denote its boundary. Let
$L(\partial D )$ denote the length i.e. L is the Hausdorff
1-dimensional measure
if $\partial D$ is
smooth and the De Giorgi perimeter otherwise.
From here on $\Omega $   will be a   bounded
domain in $\mathbb{R} ^2$ with $\partial \Omega $ a finite union of
rectifiable   curves.
Let $\alpha = \alpha (\Omega )$ denote the asymmetry of $\Omega,\;u=u(p,
\Omega)$ be the
first eigenfunction of the $p$-Laplacian, $1<p<\infty$, and
$\lambda_1$ be the first eigenvalue.  We
will
take the first eigenfunction $u>0$, we will also assume throughout
that
$$\int  _\Omega u^p = 1.$$
For $0\le t\le \sup u$, set
\begin{eqnarray*}
\Omega _t = \{ x \in \Omega : u (x) > t
\},\;\;\mbox{and}\;\;\mu(t)=|\Omega_t|.
\end{eqnarray*}
Note that $\mu(t)$ is decreasing and right continuous.
It is easy to show that $\mu(t)$ is continuous if and only if
$|\{u=t\}|=0$. Clearly, $\mu(t)$ has at most countably many
discontinuities. Since $u$ is only known to be
$C^{1,\alpha}_{loc}$, it is not clear that $\mu(t)$ is continuous
every where when $p\ne 2$ \cite{b1,b2,t1}.
Let $u^*$ be the non-increasing
rearrangement (Schwarz symmetrization) of $u$, defined as follows.
First
set $u^{\#}(a)=\inf \{t>0;\mu(t)<a\}$.
Let $(x,y)$ denote the coordinate of a point in $\Omega^{*}$. For
such a
point define $u^{*}(x,y)=u^{\#}(\pi(x^2+y^2))=u^{*}(r)$,
where $r=\sqrt{x^2+y^2}$.
By $\lambda ^*_1$ we will
mean $\lambda_1 (\Omega ^*)$, lastly set $M =  {\sup _{\Omega}} u$.

We now state the main theorem of the paper.

\begin{theorem} \label{main-thm}
 Let $\Omega \subset \mathbb{R}^2$ be a bounded domain
and   $\alpha=\alpha(\Omega)$ be  its asymmetry, then there exists a
constant $C>0$,   independent of $\Omega $,   such that
\begin{equation}
\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^*) (1 + C \alpha ^3) .
\end{equation}
\end{theorem}

We adapt the method developed in \cite{b1,b3,h1} to achieve our
goal, i. e., we characterize the propagation of asymmetry $\alpha$
via the level sets of $u$. This is expressed in terms of the
isoperimetric inequality. See Lemmas 3.3 and 3.5 for a more
precise statement. Our result relies on several lemmas proven in
Section 3 and the proof of the Theorem appears in Section 4.
We mention that we make considerable use of the co-area formula in
our work. In this context we refer to \cite{b4,f1}.
The reader may find some overlap between this work and \cite{h3}
however we believe some aspects of our work may be of independent
interest.
Lastly, we are unable to determine whether or not the third power
appearing in (2.2) is optimal. However, in the case of the
Laplacian it has been conjectured that the above Theorem holds
with the second power and if true, it would then be optimal
\cite{b3,n1}.


\section {Preliminaries}

In this section we present five lemmas
which will lead to the proof of Theorem \ref{main-thm}. For compactness 
of
our presentation, we take
$|\Omega | = 1$.

\begin{lemma} % Lemma 3.1
Let $u(x)$ be a solution of (1.1); set $h(t)=
\int_{\Omega_t}|Du|^p$.  Then $h(t)$ is convex in $t$ and
for $0 < t < M $,
\begin{equation} \lambda_1 \Big(1 - t \int_\Omega u^{p-1}\Big)
\le h(t) \le \lambda_1 \left(1 - t/M\right)\int_{\Omega_t}u^p.
\end{equation}
\end{lemma}
\paragraph{Proof.}  A proof of this lemma can be worked by using the
co-area formula. However, we will provide a proof which uses
appropriate
test functions (also see \cite{b2}).
Recall the weak formulation in Section 1, i. e.,
\begin{equation}
\int_{\Omega}|Du|^{p-2}Du\cdot
D\psi=\lambda_1\int_{\Omega}u^{p-1}\psi
\end{equation}
where $\psi\in W^{1,p}_0(\Omega)$.
Using the test function $(u-t)^+$ in (3.2), we find that
$$ h(t)=\lambda_1 \int_{\Omega_{t}}u^{p-1}(u-t).$$
We now make some observations which will prove useful later.

For $\delta>0$, $t\le u<t+\delta$ in
$\Omega_{t}\backslash\Omega_{t+{\delta}}$.
Then
\begin{eqnarray}
h(t+\delta)-h(t)&=&\lambda_1
\Big\{ \int_{\Omega_{t+\delta}} u^p -(t+\delta)
u^{p-1}-\int_{\Omega_t}(u^p-tu^{p-1}) \Big\}\nonumber\\
&=& \lambda_1\Big\{
-\int_{\Omega_t\backslash\Omega_{t+\delta}}u^p
+t\int_{\Omega_t\backslash\Omega_{t+\delta}}u^{p-1}
-\delta\int_{\Omega_{t+\delta}} u^{p-1}\Big\}.\nonumber\\
&=&\lambda_1\Big\{\int_{\Omega_t\backslash \Omega_{t+\delta}}
u^{p-1}(t-u)-\delta\int_{\Omega_{t+\delta}}u^{p-1}\Big\}.\nonumber
\end{eqnarray}
It follows that
\begin{equation}
\frac{h(t+\delta)-h(t)}{\delta}\le -\lambda_1\int_{\Omega_{t+\
delta}}
u^{p-1}.
\end{equation}
Now rearranging the above expression on the right hand side,
we also see that
\begin{eqnarray}
\lefteqn{h(t+\delta)-h(t)}\\
&=& \lambda_1\Big\{-\int_{\Omega_t\backslash\Omega_{t+\delta}}u^p +
t\int_{\Omega_t\backslash\Omega_{t+\delta}}u^{p-1}
- \delta\int_{\Omega_t}u^{p-1} +\delta
\int_{\Omega_t\backslash\Omega_{t+\delta}} u^{p-1}\Big\}\nonumber\\
&=&\lambda_1\Big\{
\int_{\Omega_t\backslash\Omega_{t+\delta}}u^{p-1}(\delta+t-u)
-\delta \int_{\Omega_t}u^{p-1}\Big\}\nonumber\\
&\ge& -\lambda_1\delta\int_{\Omega_t}u^{p-1}.\nonumber
\end{eqnarray}
Clearly
\begin{equation}
\frac{h(t+\delta)-h(t)}{\delta}\geq
-\lambda_1\int_{\Omega_{t}}u^{p-1}.
\end{equation}
A similar argument also yields
$$ -\lambda_1\int_{\Omega_{t-\delta}}u^{p-1}\le
\frac{h(t)-h(t-\delta)}{\delta}
\le-\lambda_1\int_{\Omega_{t}}u^{p-1}.$$
Clearly, (3.3), (3.4) and the foregoing imply that
$$h^{\prime}(t)=-\lambda_1\int_{\Omega_{t}}u^{p-1}\;\;\mbox{a.
e. t}.$$
Equality will hold at every value of $t$ iff $|\{u=t\}|=0$, i. e.,
iff $\mu(t)$ is continuous for all $t$. However, this is not
known for $p\ne 2$. In this context also see \cite{b2,t1,t2}.
But since $\mu(t)$ is decreasing, equality holds except on a
countable ( possibly finite) set. Note that the right continuity
of $\mu(t)$
does show that right hand derivative of $h$ exists at every $t$
and equals $-\lambda_1\int_{\Omega_{t}}u^{p-1}$.

Now the inequalities (3.3) and (3.4) clearly imply that
$h(t)$ is a convex
function. If $\theta_1< \theta<\theta_2$; then the following
two inequalities
hold, namely,
\begin{equation}
\frac{h(\theta)-h(\theta_1)}{\theta-\theta_1} \le
-\lambda_1\int_{\Omega_{\theta}}u^{p-1}\le
\frac{h(\theta_2)-h(\theta)}{\theta_2-\theta}.
\end{equation}
Using the convexity of $h(t)$, we may now find lower and upper
bounds for $h$.
Noting that $h(0)=\lambda_1$ and $\mu(t)$ is right continuous,
we see that (3.5)
yields
\begin{eqnarray*}
h(t)\leq h(0) -\lambda_1 t\int_{\Omega_t}u^{p-1}=
\lambda_1\left(1-t\int_{\Omega_t}u^{p-1}\right),
\end{eqnarray*}
and
\begin{equation}
h(t)\ge \lambda_1\left(1-t\int_{\Omega} u^{p-1}\right).
\end{equation}
Clearly, convexity of $h(t)$ on $[0,M]$, its continuity at $t=0$
and the facts $h(0)=\lambda_1$ and $h(M)=0$ imply the easy
inequality
\begin{equation}
h(t)\le \lambda_1(1-t/M)
\end{equation}
However, a simple argument provides us with a better upper bound
for
$h(t)$.
Notice that
\begin{eqnarray*}
\frac{t}{M}\int_{\Omega_t}u^{p} \le t\int_{\Omega_t}
u^{p-1}\frac{u}{M}\le t\int_{\Omega_t}u^{p-1}.
\end{eqnarray*}
Thus
\begin{eqnarray*}
\int_{\Omega_t}u^{p-1}(u-t)=
\int_{\Omega_t}u^p - t\int_{\Omega_t}u^{p-1}\le
(1-t/M)\int_{\Omega_t}u^p.
\end{eqnarray*}
Clearly,
\begin{equation}
h(t)\le \lambda_1(1-t/M)\int_{\Omega_t}u^p.
\end{equation}
Putting together (3.6) and (3.8), we get the statement of the lemma.
\hfill$\diamondsuit$\medskip

We now provide a simple upper bound for $u$.

\begin{lemma} \label{lm3.2}
Let $u$ solve (1.1) and $\mu(t)$ be as defined before, then
\begin{equation}
t \le \left(\frac{\lambda_1}{(4\pi)^{p/2}}\right)^{1/(p-1)}
\left (1 -  \mu (t)^{(p^2-2p+2)/(2p(p-1))}\right)
\left(\frac{p^2-2p+2}{2p(p-1)}\right).
\end{equation}
In particular, $M(p)=  {\sup _{\Omega}}\; u \le
\left(\frac{\lambda_1}{ (4\pi)^
{p/2}}\right)^{1/(p-1)}\left(\frac{2p(p-1)}{p^2-2p+2}\right)$.
\end{lemma}

\paragraph{Proof.}
We will use Talenti's inequality \cite{t1}. Recall
that $\int_{\Omega}u^p=1$.
Then for a. e. $t$,
\begin{eqnarray}
(4\pi \mu(t))^{p/2(p-1)} &\le&  -\mu(t)^{\prime }
\left(-\frac{d}{dt}\int_{\Omega_{t}} |Du|^p\right)^{1/(p-1)}\\
&=&-\mu(t)^{\prime }\left(-\lambda_1\frac{d}{dt}
\int_{\Omega_t}u^{p-1}(u-t)\right)^{1/(p-1)}\nonumber
\end{eqnarray}
Using Holder's inequality and the fact that $\int_{\Omega}u^p=1$,
(3.10) yields
\begin{eqnarray*}
\Big(\frac{(4\pi\mu(t))^{p/2}}{\lambda_1}\Big)^{1/(p-1)}
&\le&   -\mu(t)^\prime \Big(\int_{\Omega_{t}} u^{p-1}\Big)^{1/(p-1)}\\
&\leq&-\mu(t)^{\prime } \Big(\int_{\Omega_{t}}u^p\Big)^{1/p}
\mu(t)^{1/p(p-1)} \\
&\le& -\mu(t)^{\prime } \mu(t)^{1/p(p-1)}
\end{eqnarray*}
Thus
$$
\left((4\pi)^{p/2}/\lambda_1\right)^{1/(p-1)} \mu(t)^{(p^2-2)/(
2p(p-1))}\le -\mu(t)^{\prime }.
$$
Hence,
$$\left(\frac{(4\pi)^{p/2}}{\lambda_1}\right)^{1/(p-1)}
\le \left(-\mu(t)^{(p^2-2p+2)/(2p(p-1))}\right)^{\prime }
\left(\frac{2p(p-1)}{p^2-2p+2}\right).
$$
But for all $p>1$ it is clear that $p^2-2p+2>0$ and
$-\mu(t)^{(p^2-2p+2)/(2p(p-1))}$ is increasing and
right continuous.
Integrating from $0$ to $t$, we find that
\begin{eqnarray*}
t\le \left(\frac{2p(p-1)}{
p^2-2p+2}\right)\left(\frac{\lambda_1}{(4\pi)^{p/2}}\right)^{1/(p-1)}
\left(1-\mu(t)^{(p^2-2p+2)/(2p(p-1))}\right)
\end{eqnarray*}
Thus we get the estimate in the statement of the lemma.
\hfill$\diamondsuit$

\paragraph{Remark.}   Lemma 3.2 leads to an upper bound under the
assumption $\lambda_1 \le 2\lambda ^*_1$.\medskip

The next three lemmas are crucial to proving the Theorem, they
indicate
how asymmetry enters into the calculations. In Lemmas 3.3 and 3.5,
$k$ stands for a positive constant in $(0,1/100)$, whose exact value
will determined in Section 4. The basic approach to proving the
Theorem is along the lines of \cite{b1,b3,h1} and this motivates
the following lemma.

\begin{lemma} \label{lm3.3}
Let $u$ solve (2.1), $\alpha$ be the asymmetry   of
$\Omega$.   Assume that there exists a $T$ with $0 < T < M$ such
that
for a. e. $t\in [0,T]$, there exists a constant $k, 0 < k <
1/100$,
with the property that
$$L(\partial \Omega _t)^2 \ge 4\pi (1 + k\alpha ^2 )\mu (t) .$$
Then
\begin{equation}
\lambda_1 (\Omega )\ge \lambda_1 (\Omega ^*)\left(1 + \frac
{Tp}{8M}k\alpha^2\right).
\end{equation}
\end{lemma}

\paragraph{Proof.}
Our starting point will be Lemma 2 in \cite{t1} and the outline of the
proof is quite similar to the method used in Lemma 1 in \cite{t2}.
From \cite[Lemma 2]{t1} for a. e. $t$,
$$
L(\partial\Omega_t)^{p/(p-1)}\le -\mu(t)'
\left(-\frac{d}{dt}\int_{\Omega_t}
|Du|^p\right)^{1/(p-1)},
$$
where $L(\partial\Omega_t)$ is the De Giorgi perimeter.
Employing the
isoperimetric inequality,
$$(4\pi\mu(t))^{p/2(p-1)}\le -\mu(t)'
\left(-\frac{d}{dt}\int_{\Omega_t}
|Du|^p\right)^{1/(p-1)}\;\;\mbox{a. e. t.}$$
Employing the hypothesis of the lemma, we get for a. e. $t$,
\begin{eqnarray*}
\left(4\pi(1+k\alpha^2)\mu(t)\right)^{p/2(p-1)}\le
-\mu(t)' \left(-\frac{d}{dt}\int_{\Omega_t}|Du|^p\right)^{1/(p-1)}
\end{eqnarray*}
Therefore,
\begin{equation}
-\frac{d}{dt}\int_{\Omega_t}|Du|^p \ge
\frac{(4\pi(1+k\alpha^2)\mu(t))^{p/2}}
{(-\mu(t)')^{p-1}}.
\end{equation}
Recall that from Lemma 3.1,
$\int_{\Omega_{t}}|Du|^{p}$ is convex and hence continuous on
$[0,M]$.
Since $-\int_{\Omega_{t}}|Du|^{p}$ is non-decreasing
integrating (3.12) from $0$ to $T$, we obtain
\begin{eqnarray*}
\int_{\Omega}|Du|^p -\int_{\Omega_T}|Du|^p \ge (1+k\alpha^2)^{p/2}
\int^T_0  \frac{4\pi\mu(s))^{p/2}}{(-\mu(s)')^{p-1}}ds
\end{eqnarray*}
Rewriting and using that $p$-Dirichlet integrals diminish under
symmetrization,
we obtain
\begin{eqnarray}
\lambda_1(\Omega)&\ge& \int_{\Omega_T^*}|Du^{*}|^p +
(1+k\alpha^2)^{p/2} \int^T_0
\frac{(4\pi\mu(s))^{p/2}}{(-\mu(s)')^{p-1}}ds
\end{eqnarray}
where $u^*$ is the Schwartz non-increasing radial rearrangement of
$u$.
Recall that $u$ is $C^{1,\alpha}_{loc}(\Omega)$, hence $u^*$ is
locally  Lipschitz continuous.  Since $u^*(x)=u^*(|x|)$, we define
$r(t)=\sqrt{\mu(t)/\pi}$. Clearly, $u^*(r(t))=t$ where
$r=|x|$. Thus the co-area formula yields
\begin{equation}
\int_{\Omega_t^*}|Du^*|^p=\int_t^{\infty}
\left(\int_{\partial\Omega_s^{*}}|Du^*|^{p-1}
\right)ds=\int_{t}^{\infty}
|Du^{*}|^{p-1}(r(s))L(\partial\Omega_{s}^{*})ds,
\end{equation}
where $r(s)=\sqrt{\mu(s)/\pi}$.
Thus for a. e. $t$,
\begin{equation}
\frac{d}{dt}\int_{\Omega_t^*}|Du^*|^p=
-|Du^*|^{p-1}(r(t))L(\partial\Omega^*_t),
\end{equation}
where $r=\sqrt{|\Omega^*_t|/\pi}=\sqrt{\mu(t)\pi}$. Note that the
above also
shows that $\int_{\Omega_{t}^*}|Du^{*}|^{p}$ is Lipschtiz continuous
in $t$.
However, using polar coordinates we may express
\begin{eqnarray*}
\int_{\Omega^*_t}|Du^*|^p =2\pi \int_0^{\sqrt{\mu(t)/\pi}}|Du^*|^p
r\,dr.
\end{eqnarray*}
Differentiating the above and using (3.15)
\begin{eqnarray*}
\frac{d}{dt}\int_{\Omega^*_t}|Du^*|^p&=&\sqrt{\pi}|Du^{*}|^{p}(t)
r(t)\mu(t)^{\prime}/\sqrt{\mu(t)}\\
&=&  |Du^*|^p\mu(t)'=-|Du^*|^{p-1} L(\partial\Omega^*_t).
\end{eqnarray*}
Simplifying we obtain for a. e. $t$ that
\begin{equation}
|Du^*|=-\frac{L(\Omega^*_t)}{\mu(t)'}.
\end{equation}
Employing (3.16) in the co-area formula (3.14) results in
\begin{equation}
\int_{\Omega^*_t}|Du^*|^p=\int^{\infty}_t \frac{(4\pi\mu(s))^{p/2}}{
(-\mu(s)')^{p-1}}ds.
\end{equation}
Now since $k<1/100$ and $\alpha\le 1$, clearly
$(1+k\alpha^2)^{p/2} \ge 1+ kp\alpha^2/4,$
for all $1<p<\infty$. Using Lemmas 3.1, 3.2, (3.13), (3.17), we see
that
\begin{eqnarray}
\lambda_1(\Omega)&\ge& \int_{\Omega^*_T}|Du^*|^p
+(1+k\alpha^2)^{p/2}\int_{\Omega^*\backslash\Omega^*_T}
|Du^*|^p\nonumber\\
&\ge&\int_{\Omega^*_T}|Du^*|^p+(1+kp\alpha^2/4)
\int_{\Omega^*\backslash\Omega_T^*}
|Du^*|^p\nonumber\\
&=&(1+kp\alpha^2/4)\int_{\Omega^*}|Du^*|^p -(kp\alpha^2/4)\int_{\Omega^*
_T}
|Du^*|^p\nonumber\\
&\ge&\lambda_1(\Omega^*)(1+kp\alpha^2/4)-(kp\alpha^2/4)
\int_{\Omega_T}|Du|^p\nonumber\\
&\ge&\lambda_1(\Omega^*)(1+kp\alpha^2/4)-
\lambda_1(\Omega)kp\alpha^2(1-T/M)/4.
\end{eqnarray}
Simplifying (3.18), we get the statement of the lemma, namely,\\
$\lambda_1(\Omega)\geq\lambda_1(\Omega^*)\left(1+kp\alpha^2T/8M\right)$.
\hfill$\diamondsuit$\smallskip

Next we prove a relationship between $\lambda_1$ and the level set
$\Omega_t$.

\begin{lemma} \label{lm3.4}
 With $u$ as before, $ t \ge 0$ and   $|\Omega|=1$, we have
$$\lambda_1 (\Omega ) \ge \frac {\lambda_1 (\Omega ^*)(1-t)^{p-1}}
{|\Omega _t|^{(p-1)/2}} .$$
\end{lemma}

\paragraph{Proof.}
We start with the weak formulation for $u$ i.e.,
\begin{equation}
\int  _\Omega |Du|^{p-2}Du\cdot D\phi
=\lambda_1 (\Omega ) \int  u^{p-1}
\phi ,
\;\; \forall \;\; \phi \in W_0 ^{1,p} (\Omega ).
\end{equation}
Let $\Omega _t =  {\bigcup ^{\infty}_{i = 1}} C_i$, where $C_i$'s
are  pairwise disjoint components of $\Omega _t$. In Remark 3.2 we will
show that there can only be finitely many components of $\Omega_t$.
Thus $\Omega_t= {\bigcup^{n_t}_{i=1}}C_i$.
Setting $\phi_i = (u-t)_+$ in $C_i$ and zero elsewhere,
(3.19) yields
\begin{equation}
\int  _{C_{i}} |Du|^p = \lambda_1(\Omega ) \int  _{C_{i}}
u^{p-1}(u-t)_+ , \;\; \forall \;\; i = 1,2,
\end{equation}
Also, for each $i = 1,2,\dots , \phi _i$ is a trial function
for the minimum
problem (i.e., the variational formulation of the eigenvalue problem)
on
$C_i$.  Thus
\begin{equation}
\int  _{C_{i}} |Du|^p \ge \lambda_1 (C_i) \int  _{C_{i}}
(u-t)^p_+ .
\end{equation}
Employing H\"{o}lder's inequality, (3.20) together with (3.21) yields
\begin{eqnarray*}
\int  _{C_{i}} |Du|^p &\le& \lambda_1 (\Omega ) \left ( \int
_{C_{i}} u^p \right ) ^{(p-1)/p}
\left ( \int  _{C_{i}} (u-t)^p_+ \right)^{1/p} \\
& \le & \lambda_1 (\Omega ) \left ( \int  _{C_{i}} u^p \right )
^{(p-1)/p} \left ( \frac {1}{\lambda_1 (C_i)}
\; \int  _{C_{i}} |Du|^p
\right )^{1/p}.
\end{eqnarray*}
Thus
$$\int  _{C_{i}} |Du|^p \le
\frac {\lambda_1 (\Omega )^{p/(p-1)}}{\lambda_1
(C_i)^{1/(p-1)}} \; \int  _{C_{i}} u^p . $$
Summing over $i$,
\begin{equation}
\int  _{\Omega _{t}} |Du|^p \le \lambda_1^{p/(p-1)}(\Omega )
\sum_i \; \frac{1}{\lambda_1 (C_i)^{1/(p-1)}} \;
\int  _{C_{i}} u^p .
\end{equation}
Let $L = \inf \; \{ \lambda_1 (C_1 ), \lambda_1 (C_2),\dots  \}$.
Let $C$ be an appropriate set in
$\{ C_i \}$ such that $\lambda_1 (C) = L$.  Then, (3.22) yields
$$ \int  _{\Omega _{t}} |Du|^p
\le \frac{\lambda_1^{p/(p-1)}(\Omega )}
{\lambda_1(C)^{1/(p-1)}}\int _{\Omega _{t}} u^p.$$
Rearranging terms and employing Lemma 3.1,
\begin{eqnarray}
\lambda_1 (\Omega )^{p/(p-1)}  &\ge&  \lambda_1 (C)^{1/(p-1)}
\left(\int  _{\Omega _{t}} |Du |^p\right)
\left(\int_{\Omega_t}u^p\right)^{-1}\nonumber\\
&\ge&  \lambda_1 (C)^{1/(p-1)} \lambda_1(\Omega)
\left(1 -
t\int_{\Omega}u^{p-1}\right)\left(\int_{\Omega_t}u^p\right)^{-1}.
\end{eqnarray}
Now observe that
\begin{eqnarray*}
\int_{\Omega}u^{p-1}\le \left(\int_{\Omega}u^p\right)^{(p-1)/p}
|\Omega|^{1/p}\le \left(\int_{\Omega}u^p\right)^{(p-1)/p}=1.
\end{eqnarray*}
Clearly (3.23) yields,
\begin{equation}
\lambda_1(\Omega)\ge
\lambda_1(C)(1-t)^{p-1}\left(\int_{\Omega_t}u^p\right)^{-(p-1)}.
\end{equation}
Now $\lambda_1 (C) \ge \lambda_1
(C^*) = \lambda_1 (\Omega^* )/|C|^{p/2}$.  The latter follows
from a scaling
argument ( noting that $|\Omega^{*}|=1$ and $\Omega\subset
\mathbb{R}^{2}$); also observe that $|C| \le |\Omega _t |$.  Replacing
$\lambda_1(C)$, in (3.24), by this lower bound and $|C|$ by $|\Omega_t|$
one is lead to
\begin{eqnarray*}
\lambda_1 (\Omega) \ge \lambda_1(\Omega ^*)\frac{
\Big[(1-t)\left(\int_{\Omega_{t}}u^p\right)^{-1}\Big]^{p-1}}{|\Omega_{t}|^{p/2}}
\ge\lambda_1(\Omega ^*)\frac{(1-t)^{p-1}}{|\Omega _t|^{p/2}}.
\end{eqnarray*}
We will apply Lemma 3.4 in the case when $t<\!< 1$. See Section 4.
\hfill$\diamondsuit$

\paragraph{Remark 3.2} For every $0<t<M$,
let $\Omega_t=\bigcup_iC_i$ where
$C_i=C_i(t)$ is a component of $\Omega_t$.
Then $\{C_i\}$ is a finite family.

{\bf Proof.} Suppose that this family is infinite.
Let $x_{M_i}\in C_i$ be such
that $u(x_{M_i})=M$. Let $x_i\in \partial C_i$ be such that
$|x_i-x_{M_i}|=\mathop{\rm dist}(x_{M_i}, \partial C_i)$. Clearly, 
$u(x_i)=t$.
Since $\bar{\Omega_t}$ is compact, the local regularity results
in \cite{d1,m1,t3} imply that
$\mbox{sup}_{\Omega_t}|Du|\le K(t)<\infty$. Thus
$$|u(x_i)-u(x_{M_i})|=|M-t|\le K(t)|x_i-x_{M_i}|.$$
Since $C_i$'s are infinite and $\sum_i|C_i|=|\Omega_t|<\infty$,
$|C_i|\to 0$ as
$i\to\infty$. This means that the right side of the above
inequality goes to zero leading to a contradiction. 
\hfill$\diamondsuit$\medskip

We now study the situation when Lemma 3.3 fails to hold, namely, that
for some $t$, $L( \partial \Omega _t )^2 < 4\pi (1 + k\alpha ^2 )|\Omega 
_t|$.
Our effort in the next lemma is to estimate the size of such
an $\Omega_t$ in terms of $\alpha$ and $k$.

\begin{lemma} \label{lm.5}
 Let $0 < t < M$ be such that
$L( \partial \Omega _t )^2 < 4\pi (1 + k\alpha ^2 )|\Omega _t |, $
where $k$ is the constant in Lemma 3.3.  Then
\begin{equation}
|\Omega _t | \le(1-\alpha)/(1-5\alpha \sqrt {k}/2) .
\end{equation}
\end{lemma}

\paragraph{Proof.} Clearly, $\Omega_t$ is a set of finite perimeter.
We first recall the definition of perimeter $L(\partial\Omega_t)$;
we refer to \cite{b4}. We define
$$L(\partial\Omega_t)=\inf (\liminf_{i\to\infty}
L(\partial S_i)),$$
where the infimum is taken over all sequences of polyhedra $S_i$'s
( polygonal regions in $\mathbb{R}^2$) with boundary $\partial S_i$ and
satisfying
$$\lim_{i\to\infty}\int_Q|\chi_{S_i}-\chi_{\Omega_t}|=0,$$
for every compact $Q\subset\mathbb{R}^2$. Here $\chi_D$ is the
characteristic
of a set $D$.
Clearly then we may choose a sequence of polygonal regions $S_i$ such
that
$L(\partial S_i)\to L(\partial\Omega_t)$
with $|(S_i\backslash\Omega_t)\cup(\Omega_t\backslash S_i)|\to 0$
as $i\to\infty$. Moreover, the sequence may be so chosen that
$L(\partial S_i)<4\pi(1+k\alpha^2)|S_i|.$
We will continue to work with $\Omega_t$ but with the understanding
that
the estimates to follow hold for $S_i$ and the final statement for
$\Omega_t$ comes from taking the limit. The proof is carried out in
three steps.
\\
(A) Let $\Omega _t = {\bigcup _{i=1}^{\infty}} C_i$, where
$C_i$'s are disjoint components ( while, in our case, there can only
be
finitely many $C_i$'s, the proof we provide here applies to the
infinite case as well).  Let $H_j , j = 1,2,\dots $ denote the holes
in $\Omega _t$, i.e., the set $\Omega _t \cup ({\bigcup _{j=1}^{\infty}}
H_j)$ consists of simply connected components, say $F_i , i =
1,2,\dots $
Here $F_i$ denotes the simply connected component obtained by
plugging the
holes of $C_i$. We first prove an estimate for the total area of the
holes
$H_j$.
Clearly,
$L( \partial F_i ) \le L ( \partial C_i )$, and
via the isoperimetric inequality,
\begin{eqnarray*}
4\pi (\Sigma_i |F_i|) &\le & \Sigma_i L(\partial F_i)^2
\le \Sigma_i L(\partial C_i)^2\le \left(\Sigma_i L(\partial
C_i)\right)^2=L(\Omega_t)^2\\
&<&  4\pi (1 + k\alpha ^2 )|\Omega_t|= 4\pi (1 + k\alpha ^2 )(\Sigma_i
|C_i|).
\end{eqnarray*}
The latter follows via the assumption made in the statement of the
lemma.
While this leads to an estimate for the area of the holes, the
argument we present next will gives us a much better estimate.
Set $H ={\bigcup _{j}}H_j$, then the usual isoperimetric inequality
implies
\begin{eqnarray*}
L(\Omega_t)^2&=&(\Sigma_i L(\partial C_i))^2
= ( \Sigma L(\partial F_i)+ \Sigma   L(\partial H_j))^2 \\
&\ge& 4\pi \left (\Sigma_i|F_i|^{1/2} + \Sigma_j|H_j|^{1/2}\right)^2
\ge  4\pi \left(|F|^{1/2} + |H|^{1/2}\right)^2,
\end{eqnarray*}
where $F = {\bigcup _{i}} F_i.$  By our assumption on $\Omega_t$,
$$4\pi \left (|F|^{1/2} + |H|^{1/2}\right )^2
\le L(\partial\Omega_t)^2< 4\pi (1 +
k\alpha^2
)|\Omega _t |.$$
Recalling that $|F| = |\Omega _t | + |H|$, and expanding the left side,
we have
\begin{eqnarray*}
4\pi (1 + k\alpha ^2 )|\Omega _t |&>& 4\pi \Big\{|F| + |H| + 2
\{|F||H|\}^{1/2}\Big\}\\
&=& 4\pi \Big\{|\Omega _t| + 2|H| +2[(|\Omega _t|+|H|)|H|]^{1/2}\Big\}.
\end{eqnarray*}
Simplifying,
$$2\Big\{|H|(|\Omega _t| + |H|)\Big\}^{1/2} \le k\alpha^2 |\Omega _t|. 
$$
One then easily obtains
\begin{equation}
|H| \le \frac {k^2\alpha ^4}{4} |\Omega _t |.
\end{equation}
(B) Our second step is to show that of the $C_i$'s all
but one have small
areas.  In
order to simplify  our computations, we set
$R_i = \sqrt {|C_i|/\pi}, \quad i = 1,2,\dots   .$
Label $R_i$'s such that $R_1 = \sup \{ R_i, i = 1,\dots  \}$.  This
supremum
is attained since $\Sigma|C_i| = \pi \Sigma R^2_i = |\Omega _t|< 
\infty$.
Also
note that $L(\partial C_i) \ge 2\pi R_i, \; \forall \;\; i$.  Thus
\begin{eqnarray*}
4\pi ^2 (\Sigma R_i)^2 \le  (\Sigma L(\partial C_i))
^2=(L(\partial\Omega_t))^2
< 4\pi ^2 (1 + k\alpha ^2 )(\Sigma R^2_i ) .
\end{eqnarray*}
Set $\varepsilon _i = R_i/R_1$, $i = 1,2,\dots$, then
$$\Big(1 +  {\sum _{i > 1}} \varepsilon _i\Big)^2 \le (1 + k\alpha
^2)\Big(1 + {\sum_{i>1} \varepsilon ^2_i} \Big).$$
Thus,
$$1 + 2  {\sum _{i > 1}} \varepsilon _i +
{\sum _{i > 1}} \varepsilon ^2_i \le (1
+ k\alpha ^2) (1 +  {\sum _{i > 1}} \varepsilon ^2_i ), $$
hence,
$$2 {\sum _{i > 1}} \varepsilon _i \le k\alpha ^2 (1 +  {\sum _{i > 1}}
\varepsilon
_i ^2); $$
now, together with the fact $ {\sum _{i > 1}} \varepsilon _i^2 \le
{\sum _{i > 1}} \varepsilon _i $, we get
$$ {\sum _{i > 1}} \varepsilon _i \le \frac {k\alpha ^2}{2-k\alpha ^2}
\le k\alpha ^2 .$$
Thus,
$$ {\sum _{i > 1}} \varepsilon _i^2 \le ( {\sum _{i > 1}} \varepsilon 
_i)^2
\le
k^2 \alpha ^4, $$
implying that
\begin{eqnarray*}
{\sum _{i>1}} |C_i| \le k^2\alpha ^4|C_1| \le k^2\alpha ^4 |\Omega_t|.
\end{eqnarray*}
It n easy to see that
\begin{equation}
|C_1| \ge \frac {|\Omega _t|}{1 + k^2\alpha ^4} \ge |\Omega _t| (1 - 
k^2\alpha^4).
\end{equation}
\noindent (C) We now work with $F_1$; by hypothesis of the lemma and 
(3.27)
\begin{eqnarray}
L(\partial F_1)^2 \le L(\partial C_1)^2 \le  L(\partial \Omega _t)^2
&\le & 4\pi (1 + k\alpha ^2)(1 + k^2\alpha ^4)|C_1|\nonumber\\
& \le & 4\pi \left(1 + \frac {10}{9} k\alpha ^2\right) |F_1| .
\end{eqnarray}
The last inequality follows from noting that $k < 1/100$.  Since $F_1$
is simply connected, we may calculate the inradius $I$ (see \cite{o1})
using (3.28),
\begin{eqnarray*}
I & \ge & \frac{L (\partial F_1) - \sqrt {L( \partial F_1)^2 - 4\pi
|F_1|}}{2\pi} \\
&\ge& \frac{\sqrt {4\pi |F_1|} - \sqrt {4\pi (1 + \frac {10}{9} 
k\alpha^2)
|F_1|-4\pi
|F_1|}}{2\pi} \\
& \ge & \sqrt {\frac {|F_1|}{\pi}}
\left ( 1 - \frac {11}{10} \sqrt {k} \alpha
\right )
\ge \sqrt {\frac {|\Omega _t|}{\pi}} \frac {\left ( 1 - \frac {11}{10}
\sqrt {k} \; \alpha \right )}{\sqrt {1 + k^2\alpha ^4}} \\
& \ge & \sqrt {\frac {|\Omega _t|}{\pi}}
\left(1 - \frac {12}{10} \sqrt {k} \alpha \right)
= R.
\end{eqnarray*}
Clearly the ball $B_R$ with an appropriate center lies in $F_1$, and
so
$B_R\backslash H$ lies in $C_1$.  We now estimate $\Omega_t$ by using
the properties of the in-ball, the definition of asymmetry $\alpha $
(see (1.2)) and (3.26),
\begin{eqnarray*}
\alpha |\Omega | & \le  & |\Omega \backslash (B_R\backslash H)|
= |\Omega | - |B_R
\backslash H| \\
&\le&  1 - \Big[ \big(1 - \frac {12}{10} \sqrt {k} \alpha\big)^2 -
\frac {k^2\alpha   ^4}{4}\Big]|\Omega _t| \\
&\le& 1 - \Big(1 - \frac {5}{2} \sqrt {k}\alpha \Big)|\Omega _t |
\end{eqnarray*}
Thus,
\begin{equation}
|\Omega _t | \le \frac {1-\alpha}{1-5\sqrt {k}\alpha /2}.
\end{equation}
We now recall the discussion at the beginning of our proof. The
inequality in (3.29) is derived for $S_i$ with
$\alpha=\alpha_i=\alpha(S_i)$.
As pointed out, taking the limit $i\to\infty$ provides
justification for validity of (3.29) for $\Omega_t$.
\hfill$\diamondsuit$

\paragraph{Remark 3.3}  If $t$ satisfies the conditions of
Lemma 3.5, thenLemma 3.4 implies
\begin{equation}
\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^*) \left(\frac {1-5\sqrt 
{k}\alpha/2}
{1-\alpha }\right)^{p/2} (1-t)^{p-1} .
\end{equation}

\section {Proof of the main result}

We take $k =1/625$; we recapitulate the above results as follows:
\\
(a)  If asymmetry propagates over a ``t'' interval $[0,T]$, i.e., $L(
\partial \Omega _t)^2 \ge 4 \pi (1 + k\alpha ^2)|\Omega _t|$ a. e. 
$t\in[0,T]$,
then
$$\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^* )
\left(1 + \frac {kT}{8M} \alpha ^2 \right)\;\;.
\eqno {(*)}$$
This follows from Lemma 3.3, and Remark 3.1.  Note that it is enough
toassume that $\lambda_1 \le 2\lambda ^*_1$,
for otherwise the theorem follows.
\\
(b)  If not, i.e., for some $t$ in $[0, T]$, we have
$L(\partial \Omega _t)^2 < 4\pi (1 + k\alpha ^2 )|\Omega _t |, $
then via Lemma 3.4, and Remark 3.3 with $k=1/625$, we have
\begin{eqnarray}
\lambda_1 (\Omega ) &\ge& \lambda_1 (\Omega ^*) \Big(\frac {1-5\sqrt
{k}\;
\alpha/2}{1-\alpha}\Big)^{p/2} (1-t)^{p-1} \\
&\ge& \lambda_1(\Omega^*)
\left(1+\frac{9\alpha}{10}\right)^{p/2}(1-t)^{p-1} \nonumber
\end{eqnarray}
We make the following simple observations keeping in mind that $0\le
\alpha
\le 1$.
Firstly
\begin{equation}
\left( 1+9\alpha/10\right)^{p/2} \ge
1+9p\alpha/40\; \mbox{when $1<p$}.
\end{equation}
Also
\begin{eqnarray}
(1-t)^{p-1}\ge \left\{\begin{array} {ll}
1-(p-1)t &\mbox{if $p\ge 2$ and $0<t\le 1/p$}.   \\[3pt]
  1-p(p-1)t &\mbox{if $1<p<2$ and}\\
& \mbox{$0<t\le\frac{(p-1)}{p}<1-(1/p)^{1/(2-p)}$}.\end{array} \right.
\end{eqnarray}
To achieve the proof of the Theorem we will adapt a technique taken
from \cite{h3}. \\
\noindent\textbf{Case 1}  Let $p\ge 2.$
We start by observing that, using (4.32) and (4.33), the left side of
(4.31) may be written as
\begin{eqnarray}
\left(1+\frac{9\alpha}{10}\right)^{p/2}(1-t)^{p-1}&\ge&
\left(1+\frac{9p\alpha}{40}
\right)\left(1-(p-1)t\right)\nonumber\\
&=& 1+p\left( \frac{9\alpha}{40} -\frac{(p-1)}{p} t - \frac{9(p-1)\alpha
t}{40}\right).
\end{eqnarray}
We now reason as follows.
\\
(i) Either asymmetry propagates over the ''
t'' interval $[0, \alpha/10p]$ (in a. e. sense), in which case $(*)$
implies
$$\lambda_1 (\Omega ) \ge \lambda_1 (\Omega ^*)
\left ( 1 +k \frac {\alpha ^3}{80pM} \right ) ;$$
\\
or (ii) it does not, i.e., for some $t \in [0,\alpha/10p]$,
(b) holds. Employing
(4.34) and noting that $\alpha\le 1,\; 2\le p$ and $(p-1)/p\le 1$,
we find that
\begin{eqnarray}
\lambda_1 (\Omega ) & \ge & \lambda_1 (\Omega^*)\left[1+p
\left(\frac{9\alpha}{40} -
\frac{(p-1)\alpha}{10p^{2}}
- \frac{9(p-1)\alpha^2}{400p}\right)\right]\nonumber\\
&=&\lambda_1(\Omega^*) \left(1+
\frac{61p}{400}\alpha\right).\nonumber
\end{eqnarray}
\textbf{Case 2} Now take $1<p<2$. Thus (4.31), (4.32) and (4.33)
for $0<t<(p-1)/p$, give us
\begin{eqnarray}
\left(1+\frac{9\alpha}{10}\right)^{p/2} (1-t)^{p-1} &\ge&
\left(1+\frac{9p\alpha}{40}\right)(1-p(p-1)t)\nonumber\\
&=& 1+ p\left(\frac{9\alpha}{40} -(p-1)t-\frac{9p(p-1)\alpha 
t}{40}\right).
\end{eqnarray}
Again, if (i)
asymmetry propagates over the ``$t$'' interval $[0,(p-1)\alpha/10p]$,
then (*) implies
$$\lambda_1 (\Omega ) \ge
\lambda_1 (\Omega ^*)\left(1 +k\frac {(p-1)\alpha ^3}{80pM}\right). $$
(ii) If not, then (4.35) together with the fact that $1<p<2$ implies
that
\begin{eqnarray*}
\lambda_1 (\Omega ) & \ge & \lambda_1 (\Omega ^*)\left( 1 +
\frac{41p\alpha}{400}\right).
\end{eqnarray*}
The statement of the Theorem follows from the conclusions in Cases 1 and 
2.

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\noindent\textsc{Tilak Bhattacharya}\\
Indian Statistical Institute \\
7, S.J.S. Sansanwal Marg \\
New Delhi 110 016  India\\
e-mail: tlk@isid.isid.ac.in \smallskip

\noindent Current address:\\
Mathematics Department, Central Michigan University\\
Mount Pleasant, MI 48859 USA\\
e-mail: hadronT@netscape.net
\end{document}