\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
2004-Fez conference on Differential Equations and Mechanics \newline
{\em Electronic Journal of Differential Equations},
Conference 11, 2004, pp. 135--142.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{135}

\begin{document}

\title[\hfilneg EJDE/Conf/11 \hfil Thermal equilibrium of superconductors]
{On a nonlinear problem modelling states of thermal
equilibrium of superconductors}

\author[M. El khomssi\hfil EJDE/Conf/11 \hfilneg]
{Mohammed El Khomssi}

\address{Mohammed El Khomssi \hfill\break
UFR MDA Faculty of Sciences and Technology, Fez, Morocco}
\email{elkhomssi@fstf.ma}

\date{}
\thanks{Published October 15,2004.}
\subjclass[2000]{35J60, 34L30, 35Q99}
\keywords{Equilibrium states; nonlinear; thermal equilibrium; superconductors}


\begin{abstract}
  Thermal equilibrium states of superconductors are governed
  by the nonlinear problem
  $$
  \sum_{i=1}^{i=N}\frac{\partial }{\partial x_{i}}
  \big(k(u) \frac{\partial u}{\partial x_{i}}\big)=\lambda
  F(u) \quad \hbox{in } \Omega \,,
  $$
  with boundary condition $u=0$. Here the domain $\Omega $ is an open
  subset of $\mathbb{R}^{N}$ with  smooth boundary.
  The field $u$ represents the thermal state, which we assume is in
  $H_{0}^{1}( \Omega )$. The state $u=0$ models the superconductor's
  state which is the unique physically meaningful solution.
  In previous works, the superconductor domain is unidirectional
  while in this paper we consider a domain with arbitrary
  geometry. We obtain the following results:
  A set of criteria that leads to uniqueness of a superconductor state,
  a study of the existence of normal states and the number of them,
  and  optimal criteria when the geometric dimension is 1.
\end{abstract}

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

\section{Statement of the problem}

\subsection*{General model}

In the framework of superconductivity, the energy conservation in a 
physical
volume $\Omega _{S}$, having as boundary the closed surface
$\partial \Omega _{S}$ can be written as
\begin{equation}
\frac{\partial }{\partial \widetilde{t}}\iiint_{\Omega_{S}}E\,dv
=-\iiint_{\Omega _{S}}\mathop{\rm div}( \overrightarrow{q})
\,dv+\iiint_{\Omega _{S}}W\,dv+\iiint_{\Omega _{S}}P\,\,dv.  
\label{eq1}
\end{equation}
Where the left hand side of the equation is made of the inner quantity 
of
accumulated energy inside $\Omega _{S}$ during $d\widetilde{t}$.
The first member of the second hand
side is the heat flux going by conduction in the closed
$\partial \Omega _{S}$, and $P$ is a parasite volume supply of heat of 
nature,
responsible partly, of the thermal perturbation of the environment.
The Fourier hypothesis relates the flux density, $\widetilde{q}$ at
temperature $T$ by
\begin{equation}
\widetilde{q}=-K( X,T) \cdot \mathop{\rm grad} T \,, \label{eq2}
\end{equation}
where $K$ is the tensor of thermal conductivity.
Note at this stage that, it is well known that the
application of first Principle of Thermodynamics Theory to a continuous
environment is reduced, without matter transfer to the heat equation. 
This later is
simplified and established by assuming that the previous equation must 
be
valid for any volume $\Omega _{S}$ that one considers and is written as
\begin{equation}
C( X,T) \frac{\partial T}{\partial \widetilde{t}}=\mathop{\rm div}(
K( X,T) \cdot \mathop{\rm grad} T) +W( X,T) +P( X,\widetilde{t}),   
\label{eq3}
\end{equation}
where $C$ is the heat's capacity of the solid. Note that, the 
characteristics
 of problem \eqref{eq3}, is
\begin{equation}
W=G-Q \,. \label{eq4}
\end{equation}
Note that $W$ represents the competition between $G$ and $Q$, which 
depends,
\`{a} priori, of the thermal field, representing respectively the power
consumed by volume unit of the conductor and the power exchange between 
the
conductor and the external environment. One should note that equation 
\eqref{eq3}
does not make any physical meaning except for conditions well defined 
and applied
to a domain space-time well defined also. These conditions are the 
reason
that specifies the evolution of thermal field. Hence, it is necessary 
to
know the initial distribution in any point of the environment as well 
as the
volume of the field on the boundary of the domain; this, actually, 
states
initial conditions, at $\widetilde{t}=0$ and limit conditions as well. 
In
practice, $T$ is given at any point of $\partial \Omega _{S}$; hence 
limit
conditions will be conditions of Dirichlet's type:
\begin{equation}
T( X,\widetilde{t}) =T_{b}\quad \text{on }\partial \Omega _{S}\times
\mathbb{R}^{+}  \label{eq5}
\end{equation}
and the initial condition is
\begin{equation}
T( X,0) =T_{0}( X) \quad \text{on }\Omega _{S}. \label{eq6}
\end{equation}
The thermal field $T_{b}$ is the cryogenic temperature. A dimensional 
analysis
based on the use of floating parameters, numerical characteristics of 
the
environment see for example \cite{b1,e1}, as well as the isotropy and
homogeneity of the environment \cite{m2} allow us to rewrite problem
\eqref{eq3}-\eqref{eq4}-\eqref{eq5} under
a reduced form: Find $u$ a field modelling $T$ defined on
$\Omega \times [ 0,+\infty[ $ so that
\begin{gather}
c(u) \frac{\partial u}{\partial t}-\sum_{i=1}^{i=N}
\frac{\partial }{\partial x_{i}}( k( u) \frac{\partial u}{\partial
x_{i}})  =\lambda F( u) +a p( x,t)
\quad \text{in}\Omega \times \mathbb{R}^{+}  \label{eq7} \\
u( x,t)  = 0\quad \text{in }\partial \Omega \times \mathbb{R}^{+}
\label{eq8} \\
u( x,0) = u_{0}( x) \quad \text{in }\Omega \times \{0\}\,.   
\label{eq9}
\end{gather}
The function $F$, containing all information on energy assessment in 
the domain $\Omega $, which is an open and bounded subset of 
$\mathbb{R}^{N}$ containing zero. The term $p$ is null in the stationary case
(no perturbation at initial instant). The thermal equilibrium states of the superconductor are solutions in
$H_{0}^{1}( \Omega )$ of
\begin{equation}
-\sum_{i=1}^{i=N}\frac{\partial }{\partial x_{i}}( k( u)
\frac{\partial u}{\partial x_{i}}) =\lambda F( u) \quad \text{in
}\Omega \text{ with }u\in H_{0}^{1}( \Omega ) \label{eq10}
\end{equation}

\subsection*{Hypotheses}

The term $F$  depends explicitly on the cooling process of the 
environment.
This helps in defining conditions that are satisfied by the classes of 
admissible
functions. This is stated as the hypothesis
\begin{itemize}
\item[(HG)]
\begin{enumerate}
\item $F\in C^{2}( \mathbb{R}^{+}) $, $F( u) =0$ for $u\leq 0$ and
$( \frac{dF}{du}) ( 0^{+}) \leq 0$

\item  $u_{1}$ is so that $F( u_{1}) =0$ and $0<u_{1}<1$.
\end{enumerate}
\end{itemize}
Additional hypotheses are stated as follows:
\begin{itemize}
\item[(H1)] $F( u) \leq 1$ for all $u\geq u_{1}$

\item[(H2)] There is $u_{2}>1$ satisfying $F( u_{2}) =0$ and
$( \frac{dF}{du}) ( u) \leq 0$ for all $u\geq u_{2}$.

\item[(H3)] There is $\gamma _{0}>0$ so that
$\lim_{u\rightarrow+\infty }( \frac{F( u) }{u}) \leq \gamma _{0}$ and
$\lim_{u\rightarrow +\infty } F( u) =+\infty $.
\end{itemize}
Denote by $U_{ad}^1$ the set of functions $F$ satisfying (HG) and (H1);
$U_{ad}^2$ the set of functions satisfying (HG) and (H2); and
$U_{ad}^3$ the set of functions satisfying (HG) and (H3).
Also set
$U_{ad}^{G} =\bigcup_{m=1}^{m=3}U_{ad}^{m}$\,.


\subsection*{Canonical transform and consequences}


A large part of the thermal stability analysis is based on the nature 
and
the number of the possible stationary solutions. It seems interesting 
to
transform the differentiable operator of \eqref{eq10} so as $k(u)$ does 
not appear.
This is possible due to the following Kirchhoff's transform, related to 
a
function $k$, and  defined by
\begin{equation}
y=Y( u) =\int_{0}^{u}k( \omega ) d\omega\,. \label{eq11}
\end{equation}

The function $k$ is continuous and strictly positive, this implies $Y$ 
is
strictly increasing sequence of positive numbers; so it is invertible.
It is possible to clear out $k(u)$ with the introduction of the 
following
transformation


We remark that $k( u) \partial _{i}u=\partial _{i}y$ and the
equation of problem \eqref{eq10} has a new form:
\begin{equation}
-\Delta y=\lambda \widetilde{F}( y) =\lambda F\circ Y^{-1}( y)
\qquad \forall x\in \Omega \text{ and }y=0\text{ on }\Gamma\,.
\label{eq12}
\end{equation}
The stationary problem \eqref{eq10}
is transformed in a more simple one \eqref{eq12}. It not difficult to 
check that

\begin{proposition}[\cite{e1}] \label{prop1}
Let $F$ $\in U_{ad}^{G}$. Then $F\in U_{ad}^{m}$ if and only if 
$\widetilde{F}$
$\in U_{ad}^{m}$
\end{proposition}


\begin{remark} \label{rmk3} \rm
The case, $F( u) \leq 0$, is an optimal physical case
since it shows the domination of Joule's effect by the cryogenic system
$(G( u) \leq Q( u) )$. Hence, the conductor stays always in the
superconductor state. This situation is known as Stekly
criterion; see \cite{e2}.

Classes $U_{ad}^{m}$ model physical reality; noticing that $U_{ad}^{1}$
represents a cooling system based an Helium II.

Furthermore, Proposition \ref{prop1}
shows that $y$  and $u$  have the same property (due to the properties 
of $Y$).
\end{remark}

\noindent\textbf{Definition.}
A fundamental state or a superconductor state is a state in which 
$y\equiv 0$.


\noindent\textbf{Definition.}
We say that a state is normal if every state $y$ which is not null
is a solution in $H_{0}^{1}( \Omega ) $ of the problem  \eqref{eq12}.

\section{Analysis of the equilibrium problem}

\subsection*{Uniqueness criterion of an equilibrium state}

Recall that the only interesting physical state is when  $y\equiv 0$.
Hence, we will be looking for possibilities to avoid any equilibrium
solution that is not zero.

\begin{theorem}
Let $\widetilde{F}$ be an element of $ U_{ad}$ such that
\begin{equation}
\widetilde{F}( y) \leq \frac{\lambda _{1}}{\lambda }y\quad  \forall
y\geq 0.  \label{eq13}
\end{equation}
Then, problem \eqref{eq12} has $y\equiv 0$ as a unique solution in
$H_{0}^{1}( \Omega ) $.
\end{theorem}

\begin{proof}
Let the domain $\Omega $ be a bounded open subset in $\mathbb{R}^{N}$,
$F\in U_{ad}^{G}$ satisfy (HG) and only one of the hypothesis
(H1), (H2), (H3). Hence, by
Proposition \ref{prop1}, $F\in U_{ad}^{G}$, problem \eqref{eq10} and 
problem
\eqref{eq12} are equivalent. The operator $Ay=-\Delta y$ is an elliptic
self-adjoint operator.

Next, we recall  that the first eigenvalue of $Ay=-\Delta y$ is 
characterized by
\begin{equation}
\lambda _{1}=\inf \big\{ \int_{\Omega }\left| \nabla v\right|
^{2}dx: v\in H_{0}^{2}( \Omega ) \mbox{ and }\| v\|_{L( \Omega ) 
}^{2}=1\big\}
 \label{eq14}
\end{equation}

Next consider the energy function associated with equation 
\eqref{eq12},
\begin{equation}
\Phi ( y) =\int_{\Omega }\left| \nabla y\right| ^{2}dx-\lambda
\int_{\Omega }\int_{0}^{y}\widetilde{F}( s) ds\,dx\,.  \label{eq15}
\end{equation}%
It is known that solutions of problem \eqref{eq10} are the critical 
points of $\Phi$.
Hence, problem \eqref{eq10} has a critical solution obtained as a 
minimal point
of $\Phi $. So, by hypothesis \eqref{eq8}, we have
\begin{equation}
\Phi ( y) \geq \frac{1}{2}( \| \nabla y\|
_{L( \Omega ) }^{2}-\lambda _{1}\| y\|
_{L( \Omega ) }^{2}). \label{eq16}
\end{equation}
Since $\Omega $ is bounded and $y\in $ $H_{1}^{2}( \Omega ) $

Now, the Poincar\'{e} inequality applied to \eqref{eq16} allows us to 
derive
$\Phi ( y) \geq 0$ for all $y\in H_{0}^{1}( \Omega ) $.
Hence, $y\equiv 0$ is the only critical point of $\Phi $ and since
$y\equiv 0$ is a trivial solution, the proof is complete.
\end{proof}

\begin{corollary}
Let $F\in U_{ad}^{G}$ and assume one of the following conditions holds:
\begin{enumerate}
\item $\widetilde{F}\in U_{ad}^{2}$ and $\int_{0}^{y_{2}}
\widetilde{F}( \omega ) d\omega \leq 0$ with
$y_{2}=\int_{0}^{u_{2}}$ $k( \omega ) d\omega $

\item $\widetilde{F}\in U_{ad}^{1}$ and
$\lambda \leq \frac{\lambda _{1}}{\sup ( 1,\widetilde{F}_{m}) }$ with
$\widetilde{F}_{m}=\sup_{y\geq 0}\widetilde{F}( y) $

\item $\widetilde{F}\in U_{ad}^{3}$ and
$\lambda \leq \lambda _{1}\sup ( \frac{1}{\gamma _{0}},\frac{1}{\gamma 
_{3}}) $
with $\gamma_{3}=\sup_{y_{3}\geq y\geq y_{1}}\widetilde{F}( y) $
\end{enumerate}
Where $y_{3}$ is the value so that
$\widetilde{F}( y) \leq \gamma _{0}y$ for all $y\geq y_{3}$.
Then, problem \eqref{eq10} has only one solution which is the 
fundamental state.
\end{corollary}

\begin{proof}
If (1) is satisfied, it would be enough to look at two  possibilities:
 $\max_{x\in \Omega } y( x) \leq y_{2}$ or
 $\max_{x\in \Omega } y( x) \geq y_{2}$. For these two cases,
$$
\Phi _{\widetilde{F}}( y) =\lambda \int_{\Omega
}\int_{0}^{y}\widetilde{F}( s) ds\,dx
\leq \Phi _{\widetilde{F}}( y_{1}) +\lambda \int_{\Omega 
}\int_{y_{1}}^{y}
\widetilde{F}( s) ds\,dx\,.
$$
Hence, by the previous theorem the proof is complete in this case.

If condition (2) (or 3) holds, we note that this a restatement of the 
hypothesis
$\widetilde{F}( y) \leq \frac{\lambda _{1}}{\lambda }y$ for all  $y\geq 
0$
for $U_{ad}^{1}$  (respectively for $U_{ad}^{3}$). Hence, the
proof of the corollary is complete.
\end{proof}

\subsection*{The existence of a condition of normal states}
Let $r_{\rm max }$ be the maximal radius of a ball of center $0$
included in $\Omega $, let $\Psi ( r) $ the function defined on $%
D=] 0,\sqrt[N]{2}[ $ by
\begin{equation}
\Psi ( r) =\Big( \frac{1+r}{r}\Big) ^{2}\Big( \frac{( 1+r)^{N}-1}{2-( 
1+r)
^{N}}\Big) \label{eq17}
\end{equation}

\begin{lemma} \label{lem8}
There is a unique $r_{0}\in D$, satisfying
$\Psi ( r_{0}) = \min_{r\in D}\Psi ( r) $ and
$\lim_{r\rightarrow 0}\Psi ( r) =\lim_{r\rightarrow \sqrt[N]{2}-1}\Psi 
( r)=+\infty$.
\end{lemma}

To justify this lemma, it is enough to study the sign of $\frac{d\Psi 
}{dr}$
and of $\frac{d^{2}\Psi }{dr^{2}}$.

\begin{lemma} \label{lem9}
Let $\widetilde{F}\in $ $U_{ad}$ be such that there exists $\omega 
_{0}>0$
satisfying
\begin{equation}
\int_{0}^{\omega _{0}}\widetilde{F}( s) ds\geq \omega _{0}(
r_{\rm max }\sqrt{2}) ^{-1}\Psi ( r_{0})\,. \label{eq19}
\end{equation}
Then, there is $\widetilde{y} \in H_{0}^{1}( \Omega ) $ so that
$\Phi ( \widetilde{y}) <0$.
\end{lemma}

\begin{proof}
Set
\begin{equation}
y_{a}( x) =\sum_{l=1}^{l=3}\alpha _{l}\Sigma _{l}( x) \,,
 \label{eq20}
\end{equation}
so that
 $\alpha _{1}=\omega _{0}$,
$\alpha _{2}=\omega _{0}[ 1-\frac{1+a}{ar_{\rm max}}( \| x\|
-\frac{r_{\rm max}}{1+a})]$
($\| x\| $ is a norm $\mathbb{R}^{N}$)
and $\alpha _{3}=0$.

On the other hand $\Sigma _{l}$ are defined as follows:
\begin{gather*}
\Sigma _{1}( x)=\begin{cases} 1 &\mbox{if }\| x\| \leq \frac{r_{\rm 
max}}{1+a}\\
0  &\mbox{if } \| x\| >\frac{r_{\rm max }}{1+a},
\end{cases}\\
\Sigma _{2}( x) =\begin{cases} 1 &\mbox{if }
\frac{r_{\rm max }}{1+a}<\| x\| \leq r_{\rm max }\\
0 &\mbox{otherwise},
\end{cases}\\
\Sigma _{3}( x) =\begin{cases} 1 &\mbox{if }x\in \Omega -B(0,r_{\rm 
max}) \\
0 &\mbox{if } x\in B( 0,r_{\rm max }). \end{cases}
\end{gather*}
By construction, the function $y_{a}$ is in $H_{0}^{1}( \Omega ) $ and
satisfies $y_{a}( x) \leq \omega _{0}$.
Next, condition \eqref{eq19} implies
\begin{equation}
\Phi ( y_{a}) \leq \frac{1}{2}\| \nabla y_{a}\| _{L^{2}( \Omega ) }^{2}
-\big( \frac{\omega _{0}}{ar_{\rm max }}) ^{2}\Psi ( r_{0})\,. 
\label{eq21}
\end{equation}
Also, calculating the norm of $\nabla y_{a}$ in $\mathbb{R}^{N}$ allows 
us to
write
\begin{equation}
\| \nabla y_{a}\| =( \frac{\omega _{0}(1+a) }{a\,r_{\rm max}}) ^{2}\| 
x\|
^{2}\Sigma _{4}( x)  \label{eq22}
\end{equation}
with $\Sigma _{4}( x) =1$ if $r_{1}=\frac{r_{\rm max}}{1+a}\leq
x\leq r_{2}=r_{\rm max}$ and $\Sigma _{4}( x) =0$
To finish the proof, it suffices to compute the primitive of
$( \frac{\omega _{0}( 1+a) }{a\text{ }r_{\rm max}}) ^{2}\| x\| ^{2}
\Sigma _{4}( x) $ and to choose the
constant $a=a_{0}$ so that
\[
\| \nabla y_{a_{0}}\|_{L^{2}( \Omega ) }^{2}<( \frac{\omega _{0}}{a_{0}
r_{\rm max}}) ^{2}\Psi ( r_{0})\,.
\]\end{proof}

\begin{theorem}
Assume the following two hypothesis
\begin{enumerate}
\item There exist $\omega _{0}>0$ such that
\begin{equation}
\int_{0}^{\omega _{0}}\widetilde{F}( s) ds\geq \omega _{0}(
r_{\rm max}\sqrt{2}) ^{-1}\Psi ( r_{0}) \label{eq23}
\end{equation}

\item There exists $\omega _{1}>0$ such that
\begin{equation}
\widetilde{F}( s) \leq \gamma _{0}y\ \text{\ \ }\forall y\geq
\omega _{1}\,.  \label{eq24}
\end{equation}
Then the problem \eqref{eq12} has two equilibrium states $y$ and 
$\widetilde{y}$.
\end{enumerate}
\end{theorem}

\begin{proof}
The proof of this theorem uses col's theorem
(Mountain Pass theorem, A. Ambrossett) , which in turn uses Palais 
Smalle condition,
results given in \cite{r1,r2} and \cite[Corollary 2.16]{r3}.
As a consequence of these works $\Phi $ has the following properties
\begin{enumerate}
\item $\Phi $ has a unique minimum in a non null point of 
$H_{0}^{1}(\Omega )$.

\item $\Phi $ is convex and $\lim_{\| y\|_{H_{0}^{1}( \Omega ) 
\rightarrow +\infty }}
\Phi (y) =+\infty $.
\end{enumerate}
Note that this result implies that there exists $\widetilde{y}\in
H_{0}^{1}( \Omega ) $ such that
\[
\min_{y\in H_{0}^{1}( \Omega ) } \Phi (y) \leq \Phi ( \widetilde{y}) <0
\]
Then the continuity of $\Phi $  gives the existence of two critical
points, which completes the proof.
\end{proof}

\section{One-dimensional case}

Assume a superconductor as a piece of length for which we
can assume that the thermal control. The space variables are reduced to
curvilinear coordinates. This will allow us to obtain a one dimensional
problem by integrating on a line section that is constant and of of 
diameter
very small with respect to the length. Thus,  problem \eqref{eq10} and 
hence
\eqref{eq12} become the differential equation
\begin{equation}
y_{xx}+\lambda \widetilde{F}( y) =0\quad \text{in }] 0,1[
\text{ with }y( 0) =y( 1) =0\,. \label{eq25}
\end{equation}%
Starting from the integral problem, we show that the existence and
the number of solutions of this differential equation depend on  the 
minimum of
the function
\begin{equation}
E( \eta ) =\eta \sqrt{\frac{2}{\lambda }}\int_{0}^{1}
\Big( \sqrt{\int_{\eta t}^{\eta }\widetilde{F}( s) ds}\Big)
^{-1}dt , \quad \eta \in D_{\widetilde{F}}\,. \label{eq26}
\end{equation}
where $\eta $ $\in D_{\widetilde{F}}=] \eta _{0},\eta
_{\infty }[ $, $\eta _{0}$ is the unique solution of
$\int_{0}^{\eta }F( s) ds=0$, and
\[
\eta _{\infty }=\begin{cases} y_{2}&\mbox{if } \widetilde{F}\in 
U_{ad}^{2}\\
 +\infty &\mbox{if } \widetilde{F}\in U_{ad}^{1}\cup U_{ad}^{3} .
\end{cases}
\]

\subsection*{Critical value and number of possible normal states}

\begin{lemma} \label{lem11}
If $D_{\widetilde{F}}\neq \emptyset $, then the function $E( \eta ) $
has a unique minimum and $\eta _{\rm min}\in D_{\widetilde{F}}$.
\end{lemma}

\begin{theorem}
Set $\lambda _{c}^{\ast }=E( \eta _{\rm min}) $. Then for every
$\widetilde{F}\in U_{ad}^{1}\cup U_{ad}^{2}$ we have

\begin{enumerate}
\item A necessary and sufficient condition for \eqref{eq25} to have at 
least
one non-null solution is that
\begin{equation}
\lambda _{c}^{\ast }\leq 1  \label{eq27}
\end{equation}

\item If $\lambda _{c}^{\ast }=1$, then \eqref{eq24} has a non-null 
solution
$y_{\rm min}$ with $\max_{x\in ] 0,1[ } y_{\rm min}( x) =\eta _{\rm 
min}$

\item If $\lambda _{c}^{\ast }<1$, then \eqref{eq24} has 2 solutions 
$y_{a}$ and
$y_{b}$ so that
\[
\max_{x\in ] 0,1[ } y_{a}( x) =\eta _{a}\quad\mbox{and}\quad
\max_{x\in ] 0,1[ } y_{b}( x) =\eta _{b}
\]
with $\eta _{a}$ and $\eta _{b}$
solutions to $\lambda _{c}^{\ast }=E( \eta )$

\item For all $\widetilde{F}\in U_{ad}^{3}$, there is
$\eta _{\ast }>\eta_{0}$ with
$\widetilde{F}( \eta _{\ast }) >\sqrt{\widetilde{F}_{\rm max}}$
then we have the same results as in 1, 2 and 3 above; otherwise there
is at most a non-null solution.
\end{enumerate}

\end{theorem}

\subsection*{Optimal criterion of unconditional stability}

\begin{theorem}[\cite{e1,g1}]
 Let $F\in U_{ad}^{G}$. A necessary and sufficient condition for the 
equilibrium
problem \eqref{eq25} to have only the fundamental state $y\equiv 0$
as a solution is
\begin{equation}
\lambda _{c}^{\ast }>\sup \Big( 1,\pi \sqrt{\gamma _{0}^{-1}}\Big)
\label{eq28}
\end{equation}
\end{theorem}

The proof of this theorem and hence of the optimal criterion is very
technical. Indeed, one can start by looking at zeros of a differential 
function or
try to study its convexity. By noticing the complexity of
$\frac{d^{2}E}{d\eta ^{2}}$, one can use instead a technic developed by 
Smoller and
Wasserman \cite{s1}. To solve a non linear differential equation
with $\widetilde{F}$ having a polynomial function of degree three.
We justify the existence of at least one extremum $\eta _{e}$ of
$E( \eta ) $. After computing $\frac{dE}{d\eta }$ and
$\frac{d^{2}E}{d\eta ^{2}}$, we look at the  sum
\[
\Lambda( \eta ) =a( \eta) \frac{dE}{d\eta }+b( \eta ) 
\frac{d^{2}E}{d\eta ^{2}},
\]
where $a( \eta ) $ and $b( \eta ) $ are real functions \`{a}
priori. Now, we may choose to simplify the expression of
$\Lambda ( \eta ) $, in some extremum $\eta _{e}$ which is a zero of
$E( \eta ) $. Hence, we get a simplified expression of
$\frac{dE}{d\eta }$ and $\frac{d^{2}E}{d\eta ^{2}}$. By studying this 
sign
one may conclude the convexity of $E( \eta ) $.

\subsection*{Concluding Remark}

In the theory of partial differential equations there is a very strong
relationship between dynamic
solutions and equilibrium solutions. The study of the equilibrium 
problem
\eqref{eq10} is a fundamental step towards the analysis and the study 
of evolution
problem. As application of these results, we can mention the
application of the Invariance Principle of Lassalle \cite{j1}. Then 
show,
under some regular conditions, that the dynamic solution converges in
$H_{0}^{1}( \Omega ) $ after some time $t\geq t_{0}$, toward a
superconductor state. Hence, we have its stability.

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thermal stability of a unidimensionnal superconducteur}, J Phys.
III, France 4. 1994.

\bibitem{e1} M. El khomssi: \textit{Etude des \'{e}quations
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