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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 38, 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}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/38\hfil On the dimension of the kernel]
{On the dimension of the kernel of the linearized thermistor operator}

\author[G. Cimatti \hfil EJDE-2013/38\hfilneg]
{Giovanni Cimatti}  % in alphabetical order

\address{Giovanni Cimatti \newline
Department of Matematics, University of Pisa, 
Largo Bruno Pontecorvo 5, 56127 Pisa, Italy}
\email{cimatti@dm.unipi.it}

\thanks{Submitted September 4, 2012. Published February 1, 2013.}
\subjclass[2000]{35B15, 35J66}
\keywords{Elliptic system; thermistor problem; existence; \hfill\break\indent
 uniqueness of solutions}

\begin{abstract}
 The elliptic system of partial differential equations of the
 thermistor problem is linearized to obtain the system
 \begin{gather*}
 \nabla\cdot(\sigma(\bar u)\nabla\Phi+\sigma'(\bar u)U\nabla\bar\varphi)=0
 \quad\text{in }\Omega,\quad  \Phi=0\quad\text{on }\Gamma\\
 \Delta U+\sigma'(\bar u)|\nabla\bar\varphi|^2 U+2\sigma(\bar u)\nabla\bar
 \varphi \cdot\nabla\Phi=0\quad
 \text{in }\Omega, \quad U=0\quad\text{on } \Gamma.
 \end{gather*}
 We study the existence of nontrivial solutions for this linear 
 boundary-value problem, which is useful in the study of the
 thermistor problem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

The name ``thermistor'' refers to a three-dimensional body
made up of substances conducting both heat and electricity (typically a
mixture of semiconductors) for which the electrical conductivity
depends sharply on the temperature. We shall represent the body of
the thermistor by $\Omega$, an open and bounded subset of $\mathbb{R}^3$.
The regular boundary $\Gamma$ of $\Omega$ consists of two disjoint  surfaces
$\Gamma_1$ and $\Gamma_2$, the electrodes, to which a difference of potential
$2V$ is applied.


\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\end{center}
\caption{Thermistor and its circuit}
\label{fig1}
\end{figure}

Under stationary conditions
the electric potential $\varphi(\mathbf{x})$, $\mathbf{x}=(x_1,x_2,x_3)$ and the temperature
$u(\mathbf{x})$ inside $\Omega$ are determined by the following nonlinear elliptic
boundary-value problem
\begin{equation} \label{eP}
\begin{gathered}
\nabla\cdot(\sigma(u)\nabla\varphi)=0\quad\text{in }\Omega,\\
\varphi=-V\quad\text{on }\Gamma_1,\quad \varphi=V\quad\text{on }\Gamma_2 , \\
\Delta u+\sigma(u)|\nabla\varphi|^2=0\quad\text{in }\Omega,\\
u=u_b\quad\text{on }\Gamma,
\end{gathered}
\end{equation}
where $V$ is a given constant and $u_b$ a given function on $\Gamma$.
If $u_b$ is an arbitrary boundary data, many papers give results of existence
of both classical and weak solutions (see \cite{A,SH,CP} and references therein).
However, the nonlinear structure of \eqref{eP} seems to be, in full
generality, an open problem. We quote, in this respect, the following
result \cite{GC}.

\begin{theorem} \label{thm1.1}
Let $\sigma(u)\in C^0(\mathbb{R}^1)$, $\sigma(u)>0$ satisfy
\[
 \int_0^\infty \frac{dt}{\sigma(t)}=\infty
\]
and suppose in the problem \eqref{eP},
\[%\label{2p2}
u=0\quad\text{on }\Gamma ,
\]
then  problem \eqref{eP} has one and only one classical solution.
\end{theorem}


For more comprehensive results a first step is certainly to linearize
the equations and to study the corresponding linear boundary value problem.
Thus we consider the following linear problem  in the
unknowns $(\Phi(\mathbf{x}),U(\mathbf{x}))$
\begin{gather} \label{1p2_2}
\nabla\cdot(\sigma(\bar u)\nabla\Phi+\sigma'(\bar u)U\nabla\bar\varphi)=0\quad\text{in }\Omega,\\
\label{2p2_2}
\Delta U+\sigma'(\bar u)|\nabla\bar\varphi|^2U+2\sigma(\bar u)\nabla\bar\varphi\cdot\nabla\Phi=0\quad
 \text{in }\Omega, \\
\label{3p2_2}
\Phi=0\quad\text{on }\Gamma ,\quad U=0\quad\text{on }\Gamma ,
\end{gather}
where $(\bar\varphi,\bar u)$ is a solution of  \eqref{eP}.
 We have the following result.

\begin{lemma} \label{lem1.2}
Let $(\bar\varphi(\mathbf{x}),\bar u(\mathbf{x}))\in (C^1(\bar\Omega))^2$ and assume
\begin{equation}\label{1p3}
\sigma_M\geq\sigma(u)\geq\sigma_m>0.
\end{equation}
Define
\begin{gather*}
\alpha=\sup\{|2\sigma(\bar u(\mathbf{x}))-\sigma'(\bar u(\mathbf{x}))|\, 
|\nabla\bar\varphi|,\; \mathbf{x}\in\Omega\},\\
\beta=\sup\{\sigma'(\bar u(\mathbf{x}))|\nabla\bar\varphi|^2,\; \mathbf{x}\in\Omega\}
\end{gather*}
and suppose that
\begin{equation} \label{4p3}
\sigma_m-\frac{\alpha}{2}>0,\quad 1-\frac{\alpha}{2\lambda_0}-\frac{\beta}{\lambda_0}>0,
\end{equation}
where $\lambda_0$ is the first eigenvalue of the laplacian with zero boundary
conditions. Then the problem \eqref{1p2_2}--\eqref{3p2_2} has only the
trivial solution.
\end{lemma}

\begin{proof}
Let us multiply \eqref{1p2_2} by $\Phi$ and \eqref{2p2_2} by $U$. Integrating by
parts over $\Omega$ and adding we obtain
\[
\int_\Omega(\sigma(\bar u)|\nabla\Phi|^2+|\nabla U|^2)dx=\int_\Omega(2\sigma(\bar u)-\sigma'(\bar
u))U\nabla\bar\varphi\cdot\nabla\Phi dx+\int_\Omega\sigma'(\bar u)U^2|\nabla\bar\varphi|^2dx.
\]
Hence, by \eqref{1p3} we have
\[
\sigma_m\int_\Omega|\nabla\Phi|^2+|\nabla U|^2dx\leq\alpha\int_\Omega|U||\nabla\Phi| dx+\beta\int_\Omega U^2dx.
\]
Using the Cauchy-Schwartz and the Poincar\`e inequalities we obtain
\[
(\sigma_m-\frac{\alpha}{2})\int_\Omega|\nabla\Phi|^2dx+(1-\frac{\alpha}{2\lambda_0}-\frac{\beta}{\lambda_0})\int_\Omega|\nabla
U|^2dx\leq 0.
\]
This implies  $\Phi=0$ and $U=0$ by \eqref{4p3}.
\end{proof}
As an application of Lemma \ref{lem1.2} we have the following lemma.

\begin{lemma} \label{lem1.3}
Assume $\sigma(u)\in C^1(\mathbb{R}^1)$ and
\begin{equation} \label{1p9_1}
\sigma_M\geq\sigma(u)\geq\sigma_m>0.
\end{equation}
Let  $(\bar\varphi,\bar u)$ be the unique corresponding solution of
\eqref{eP} when $u_b=0$. Suppose that
\begin{equation} \label{2p9_1}
\Phi=0, \quad U=0 
\end{equation}
is the only solution of \eqref{1p2_2}-\eqref{3p2_2}.

Let $u_b\in C^{0,\alpha}(\Gamma)$. Then there exists $\mu_0>0$ such that,
if $\| u_b\|_{C^{0,\alpha}(\Gamma)}\leq\mu_0$, the problem
\begin{gather*}
\nabla\cdot(\sigma(u)\nabla\varphi)=0\quad\text{in }\Omega,\quad
\varphi=-V\quad\text{on } \Gamma_1 ,\quad
 \varphi=V\quad\text{on }\Gamma_2  \\
\Delta u+\sigma(u)|\nabla\varphi|^2=0\quad\text{in }\Omega, \quad
u=u_b\quad\text{on }\Gamma
\end{gather*}
has one and only one solution.
\end{lemma}

\begin{proof}
Let $F:X\to Y$, where
\begin{gather*}
X=\{(\varphi(\mathbf{x}), u(\mathbf{x}))\in (C^{2,\alpha}(\bar\Omega))^2,\;
 \varphi=-V \text{ on }\Gamma_1,\;
 \varphi=V \text{ on }\Gamma_2\},
\\
Y=(C^{0,\alpha}(\bar\Omega))^2\times C^{2,\alpha}(\Gamma),
F((\varphi,u))=(\nabla\cdot(\sigma(u)\nabla\varphi),\ \Delta u+\sigma(u)|\nabla\varphi|^2,\ u\vert_\Gamma).
\end{gather*}
We apply the local inversion theorem at $(\bar\varphi,\bar u)$ \cite{AP}. We have
\begin{align*}
&F'((\bar\varphi,\bar u))[\Phi,U]\\
&=(\nabla\cdot(\sigma(\bar u)\nabla\Phi+\sigma'(\bar u)U\nabla\bar\varphi),\Delta
U+\sigma'(\bar u)|\nabla\bar\varphi|^2U+2\sigma(\bar u)
\nabla\bar\varphi\cdot\nabla\Phi,\ u\vert_\Gamma).
\end{align*}
We claim that the problem
\begin{gather*}
\nabla\cdot(\sigma(\bar u)\nabla\Phi+\sigma'(\bar u)U\nabla\bar\varphi)=f \\
\Delta U+\sigma'(\bar u)|\nabla\bar\varphi|^2U+2\sigma(\bar u)\nabla\bar\varphi\cdot\nabla\Phi=g \\
\Phi=0\quad\text{on }\Gamma ,\quad U=U_b\quad\text{on }\Gamma
\end{gather*}
has one and only one solution if $((f,g),U_b)\in Y$. If $(\Phi_1,U_1)$ and
$(\Phi_2,U_2)$ are two solutions we set $(\Psi,W)=(\Phi_1-\Phi_2, U_1-U_2)$ and use
\eqref{2p9_1}. This gives $(\Phi_1,U_1)=(\Phi_2,U_2)$. To prove existence we use
a continuity method ( se e.g. \cite{CH} page 336).
We construct a one-parameter family of problems depending on the
 parameter $t\in [0,1]$. Let $\mathbf{U}=(\Phi,U)$ and define
\[
\mathbf{L}_t[\mathbf{U}]=\begin{bmatrix}
(1-t)\Delta\Phi+t(\nabla\cdot(\sigma(\bar u)\nabla\Phi+\sigma'(\bar u)U\nabla\bar\varphi))\\
(1-t) \Delta U+t(\sigma'(\bar u)|\nabla\bar\varphi|^2U+2\sigma(\bar u)\nabla\bar\varphi\cdot\nabla\Phi)
\end{bmatrix}.
\]
By the Schauder's estimates \cite{GT} any solution of the problem
\begin{equation} \label{a}
\mathbf{L}_t[\mathbf{U}]=\mathbf{f}\quad\text{in }\Omega, \quad
\mathbf{f}=\begin{bmatrix}
 f\\
 g
\end{bmatrix},\quad
 \mathbf{U}=\begin{bmatrix}
 0\\
 U_b
\end{bmatrix}\quad\text{on }\Gamma
\end{equation}
satisfies
\begin{equation}\label{b}
\|\mathbf{U}\|_{C^{2,a}}\leq K_1\Bigl(\|\mathbf{f}\|_{C^\alpha}+\| U_b\|_{C^\alpha}\Bigl).
\end{equation}
 We call $T$ the set of those value of $t$ in the
unit interval $[0,1]$ for which  problem \eqref{a} is uniquely solvable.
 $T$ is not empty since it contains $t=0$. We prove that $T$ is an open set; i.e.,
for every $t_0\in T$ there exists $\epsilon(t_0)>0$ such that every $t\in [0,1]$,
for which $|t-t_0|<\epsilon(t_0)$, belongs to $T$. This can be seen with a
contraction mapping argument as follows. We rewrite \eqref{a} in the form
\[
\mathbf{L}_{t_{0}}[\mathbf{U}]=\mathbf{L}_{t_{0}}[\mathbf{U}]-\mathbf{L}_t[\mathbf{U}]+\mathbf{f}\quad\text{in }\Omega,
\quad \mathbf{U}=\begin{bmatrix}
0\\
U_b
\end{bmatrix} \quad\text{on }\Gamma
\]
or
\begin{equation}\label{c}
\mathbf{L}_{t_{0}}[\mathbf{U}]=(1-t)(\Delta \mathbf{U}
-\mathbf{L}_1[\mathbf{U}])+\mathbf{f}\quad\text{in }\Omega,\quad
\mathbf{U}=\begin{bmatrix}
0\\
U_b
\end{bmatrix} \quad\text{on }\Omega.
\end{equation}
Substituting any function $\mathbf{U}\in C^{2,\alpha}$ on the right hand side
of \eqref{c} we obtain a function $F\in C^\alpha$. Since $t_0\in T$
there exists $\mathbf{W}\in C^\alpha$ such that
\begin{equation} \label{d}
\mathbf{L}_{t_{0}}[\mathbf{W}]=\mathbf{F} \quad\text{in }\Omega ,\quad
\mathbf{W}=\begin{bmatrix}
0\\
U_b
\end{bmatrix}\quad\text{on }\Gamma .
\end{equation}
The problem \eqref{d} defines a transformation
\begin{equation} \label{e}
\mathbf{W}=\mathbf{A}(\mathbf{U}).
\end{equation}
We claim that there exists a fixed point of \eqref{e}
 if $|t-t_0|$ is sufficiently small. From \eqref{c} we have
\[%\label{f}
\|\mathbf{F}\|_{C^\alpha}\leq\Bigl(|t-t_0|\|\mathbf{U}\|_{C^{2,\alpha}}+\|\mathbf{f}\|_{C^\alpha}\Bigl).
\]
Using again the Schauder's estimates, we obtain
\begin{equation} \label{g}
\|\mathbf{W}\|_{C^{2,\alpha}}\leq K_1K_2|t-t_0|\|\mathbf{U}\|_{C^{2,\alpha}}
+K_1\|\mathbf{f}\|_{C^\alpha}+K_1\| U_b\|_{C^\alpha}.
\end{equation}
Hence, if we assume $2K_1K_2|t-t_0|\leq 1$, an inequality of the form
\[
\|\mathbf{U}\|_{C^{2,\alpha}}\leq 2K_1\bigl(\|\mathbf{f}\|_{C^\alpha}+\| U_b\|_{C^\alpha}\bigl)
\]
would imply
\[
\|\mathbf{W}\|_{C^{2,\alpha}}=\|\mathbf{A}(\mathbf{U})\|_{C^{2,\alpha}}
\leq K_1\bigl(\|\mathbf{f}\|_{C^{\alpha}}+\| U_b\|_{C^\alpha}\bigl).
\]
Moreover, if $\mathbf{W}_1=\mathbf{A}(\mathbf{U}_1)$ and $\mathbf{W}_2=\mathbf{A}(\mathbf{U}_2)$,
 $\mathbf{W}_1$-$\mathbf{W}_2$ is a solution of
\[
\mathbf{L}_{t_0}[\mathbf{W}_1-\mathbf{W}_2]=(t-t_0)\bigl(\Delta(\mathbf{U}_1-\mathbf{U}_2)-\mathbf{L}_1[\mathbf{U}_1-\mathbf{U}_2]),\quad
\mathbf{W}_1-\mathbf{W}_2=\begin{bmatrix}
0\\
0
\end{bmatrix}.
\]
Recalling \eqref{g} we conclude that, if
\begin{equation}\label{m}
 2K_1K_2|t-t_0|\leq 1,
\end{equation}
then
\begin{equation} \label{n}
\|\mathbf{W}_1-\mathbf{W}_2\|_{C^{2,\alpha}}\leq\frac{1}{2}\|\mathbf{U}_1-\mathbf{U}_2\|_{C^{2,\alpha}}.
\end{equation}
Therefore, if \eqref{m} holds the transformation $\mathbf{A}(\mathbf{U})$ maps
the set of functions satisfying
\[
\|\mathbf{U}\|_{C^{2,\alpha}}\leq 2K_2\|\mathbf{f}\|_{C^\alpha}
\]
into itself and, by \eqref{n}, is a contraction.
Thus \eqref{e} has a fixed point $\mathbf{U}$ which gives the desired solution
of \eqref{a} if $|t-t_0|\leq (1/2)K_1K_2$. Hence $T$ is open.
Moreover, $T$ is a closed set. For, let $\tilde t$ be a cluster point
of a sequence $\{t_n\}$ in $T$. Consider any $\mathbf{f}$ in $C^\alpha$ and
let $\{\mathbf{U}_n\}$ be the corresponding sequence of solutions in $C^{2,\alpha}$
such that
\[
\mathbf{L}_{t_{n}}[\mathbf{U}_n]=\mathbf{f}\quad\text{in }\Omega,\quad
\mathbf{U}_n=\begin{bmatrix}
0\\
U_b
\end{bmatrix}\quad\text{on }\Gamma .
\]
By \eqref{b} we have
\[
\|\mathbf{U}_n\|_{C^{2,\alpha}}\leq K_1(\|\mathbf{f}\|_{C^\alpha}+\| U_b\|_{C^\alpha}).
\]
Thus the sequence $\{\mathbf{U}_n\}$ and their first and second derivatives
are equibounded and equicontinuous in $\bar\Omega$. Let $\{\mathbf{U}_{n_{j}}\}$
be a subsequence converging with first and second derivatives.
If $\tilde{\mathbf{U}}$ is the limit function it gives a solution to the problem
\[
\mathbf{L}_{\tilde t}[\tilde{\mathbf{U}}]=\mathbf{f}\quad\text{in }\Omega, \quad
\tilde{\mathbf{U}}=\begin{bmatrix}
0\\
U_b
\end{bmatrix}\quad\text{on }\Gamma .
\]
Hence $\tilde t\in T$, therefore $T=[0,1]$.
\end{proof}

We may also consider the  problem
\begin{equation} \label{ePm}
\begin{gathered}
\nabla\cdot(\sigma(u)\nabla\varphi)=0\quad\text{in }\Omega\\
\varphi=-V\quad\text{on }\Gamma_1 ,\quad \varphi=V\quad\text{on }\Gamma_2   \\
\Delta u+\sigma(u)|\nabla\varphi|^2+\mu R(u,\varphi)=0\quad\text{in }\Omega\\
 u=0\quad\text{on }\Gamma ,
\end{gathered}
\end{equation}
where $R(u,\varphi)\in C^0(\mathbb{R}^2)$ is a temperature depending source and $\mu$
a numerical parameter.

\begin{lemma} \label{lem1.4}
Assume $\sigma(u)\in C^1(\mathbb{R}^1)$ and
\[
\sigma_M\geq\sigma(u)\geq\sigma_m>0.
\]
Let $(\bar\varphi,\bar u)$ be the solution (unique by Theorem \ref{thm1.1}) of the problem
\eqref{eP} when $u_b=0$.
Suppose that the problem \eqref{1p2_2}-\eqref{3p2_2} has only the trivial
solution.
Then there exists $\mu_0>0$ such that the problem \eqref{ePm} has one and
only one solution if $|\mu|<\mu_0$.
\end{lemma}

\begin{proof}
We apply the implicit function theorem. Let $\mathcal{F}:\mathcal{X}\times\mathbb{R}^1\to\mathcal{Y}$,
where
\begin{gather*}
\mathcal{X}=\{(\varphi(\mathbf{x}),u(\mathbf{x}))\in (C^{2,\alpha}(\bar\Omega))^2,\;
 \varphi=-V\text{ on }\Gamma_1 ,\;
 \varphi=V \text{ on }\Gamma_2 ,\; u=0\text{ on }\Gamma \},\\
\mathcal{Y}=(C^{0,\alpha}(\bar\Omega))^2, \\
\mathcal{F}((\varphi,u),\mu)=(\nabla\cdot(\sigma(u)\nabla\varphi),\Delta u+\sigma(u)|\nabla\varphi|^2+\mu R(u,\varphi)),\quad
(\varphi,u)\in\mathcal{X},\; \mu\in\mathbb{R}^1.
\end{gather*}
We have $\mathcal{F}((\bar\varphi,\bar u),0)=((0,0),0)$. Moreover, the partial derivative
of $\mathcal{F}$ with respect to $(\varphi,u)$ at $((\bar\varphi,\bar u),0)$ is
\begin{align*}
&\mathcal{F}_{(\varphi,u)}((\bar\varphi,\bar u),0)[\Phi,U]\\
&=(\nabla\cdot(\sigma(\bar u)\nabla\Phi
+\sigma'(\bar u)U\nabla\bar\varphi),\Delta
U+\sigma'(\bar u)|\nabla\bar\varphi|^2U+2\sigma(\bar u)\nabla\bar\varphi\cdot\nabla\Phi).
\end{align*}
Proceeding, with minor changes, as in
Lemma \ref{lem1.2} we can prove that the problem
\begin{gather*}
\nabla\cdot(\sigma(\bar u)\nabla\Phi+\sigma'(\bar u)U\nabla\bar\varphi)=f,\quad
 \Phi=0\quad\text{on }\Gamma\\
\Delta U+\sigma'(\bar u)|\nabla\bar\varphi|^2U+2\sigma(\bar u)\nabla\bar\varphi\cdot\nabla\Phi=g,\quad
 U=0\quad\text{on }\Gamma
\end{gather*}
has one and only one solution for every $(f,g)\in\mathcal{Y}$. Thus
$\mathcal{F}_{(\varphi,u)}((\bar\varphi,\bar u),0)$ is invertible and therefore there exists
$\mu_0>0$ such that the thesis holds.
\end{proof}

\section{The one-dimensional case}

In this Section we study the one-dimensional version of the problem
\eqref{eP}; i.e.,
\begin{gather}\label{1p14} % problem(P1)
(\sigma(u)\varphi')'=0,\\
\label{2p14}
\varphi(-L)=-V,\quad \varphi(L)=V,\quad L>0,\\
\label{3p14}
u''+\sigma(u)\varphi'^2=0,\\
\label{4p14}
u(-L)=0,\quad u(L)=0.
\end{gather}
 In the next Lemma we collect certain elementary properties of
the solution of \eqref{1p14}--\eqref{4p14}.

\begin{lemma} \label{lem2.1}
Let $\sigma(u)\in C^1(\mathbb{R}^1)$ and $\sigma(u)>0\ \hbox{for all}\ u\in\mathbb{R}^1$. Suppose
\begin{equation} \label{2p15}
\int_0^\infty\frac{dt}{\sigma(t)}=\infty.
\end{equation}
Under these hypotheses there exists one and only one solution $(\varphi(x),u(x))$
of the problem \eqref{1p14}--\eqref{4p14}, and the solution satisfies
\begin{gather}\label{1p15_1}
\varphi'(x)>0, \\
\label{2p15_1}
\varphi(x)=-\varphi(-x),\ u(x)=u(-x).
\end{gather}
Moreover, if we define
\begin{gather}\label{1p16}
F(u)=\int_0^u\frac{dt}{\sigma(t)},\\
\label{2p16}
\xi=G(\varphi)=\int_0^\varphi\sigma(F^{-1}(\frac{V^2}{2}-\frac{t^2}{2}))dt
\end{gather}
we have
\begin{equation}\label{3p16}
\frac{d\varphi}{dx}(x)=\frac{G(V)}{L\sigma(u(x))}.
\end{equation}
\end{lemma}

\begin{proof}
Let us define the transformation
\begin{equation}\label{4p16}
\theta=\frac{1}{2}\varphi^2+\frac{V}{2}\varphi+F(u).
\end{equation}
Therefore, by \eqref{1p16},
\[
 \sigma(u)\theta'=\sigma(u)\varphi\varphi'+\frac{V}{2}\sigma(u)\varphi'+u'.
\]
Recalling \eqref{1p14} and \eqref{3p14} we have
\[
(\sigma(u)\theta')'=0.
\]
Hence in terms of $\varphi$ and $\theta$ the problem \eqref{1p14}--\eqref{4p14}
can be restated as
\begin{equation}\label{1p17}
\begin{gathered}
(\sigma(u)\varphi')'=0,\\
\varphi(-L)=-V,\quad  \varphi(L)=V,\\
(\sigma(u)\theta')'=0,\\
\theta(-L)=0,\quad  \theta(L)=V^2.
\end{gathered}
\end{equation}
This suggests the existence of a functional relation between $\theta$
and $\varphi$, of the form
\[
\theta=\frac{V}{2}\varphi+\frac{V^2}{2}.
\]
Hence, by \eqref{4p16}, we have
\[
F(u)=\frac{V^2}{2}-\frac{\varphi^2}{2}.
\]
By \eqref{2p15}, $F$ is globally invertible and
\begin{equation} \label{1p18}
u=F^{-1}(\frac{V^2}{2}-\frac{\varphi^2}{2}).
\end{equation}
Thus we can write \eqref{1p17} in the form
\[
(\sigma(F^{-1}(\frac{V^2}{2}-\frac{\varphi^2}{2}))\varphi')'=0.
\]
Using \eqref{2p16}, we have
\begin{gather*}
\xi''=0,\\
\xi(-L)=G(-V)=-G(V),\quad \xi(L)=G(V).
\end{gather*}
Thus we obtain
\[
\xi(x)=\frac{G(V)}{L}x.
\]
The potential $\varphi(x)$ can be computed from
\begin{equation}
G(\varphi(x))=\frac{G(V)}{L}x
\label{s}
\end{equation}
which gives
\[
\varphi(x)=G^{-1}(\frac{G(V)}{L}x).
\]
Finally the temperature $u(x)$ is obtained from \eqref{1p18}.
The solution $(\varphi(x),u(x))$ of problem \eqref{1p14}--\eqref{4p14}
obtained in this way is also unique \cite{GC}.
Now we prove \eqref{3p16}. From \eqref{s} we have
\[
\frac{G(V)}{L}=\frac{dG}{d\varphi}(\varphi(x))\varphi'(x)
\]
and by \eqref{1p18} and \eqref{2p16}
\[
\frac{dG}{d\varphi}=\sigma(u).
\]
Hence \eqref{3p16} follows. From \eqref{1p14} we have $\sigma(u)\varphi'=c$ with
$c>0$ by \eqref{2p14}, thus we obtain \eqref{1p15_1}.
To prove \eqref{2p15_1}  we define
\[
\tilde\varphi(x)=-\varphi(-x),\quad \tilde u(x)=u(-x).
\]
As it is easily verified $(\tilde\varphi(x),\tilde u(x))$ satisfy
\eqref{1p14}-\eqref{4p14}. Therefore, by the uniqueness of the solution of the
problem \eqref{1p14}--\eqref{4p14} we obtain \eqref{2p15_1}.
\end{proof}

The linearized problem  corresponding, in the present one-dimensional
case, to \eqref{1p2_2}-\eqref{3p2_2} reads
\begin{gather}  \label{1p22} % (LP1)
(\sigma(\bar u)\Phi'+\sigma'(\bar u)U\bar\varphi')'=0,\quad
 \bigl(\sigma'=\frac{d\sigma}{du},\; \varphi'=\frac{d\varphi}{dx}\bigl),\\
\label{2p22}
\Phi(-L)=0,\quad \Phi(L)=0,\\
\label{3p22}
U''+\sigma'(\bar u)\bar\varphi'^2U+2\sigma(\bar u)\bar\varphi'\Phi'=0,\\
\label{4p22}
U(-L)=0,\quad U(L)=0,
\end{gather}
where $(\bar \varphi(x),\bar u(x))$ is a solution of the problem
\eqref{1p14}--\eqref{4p14}. Here we proceed
by direct integration the linear problem \eqref{1p22}--\eqref{4p22}. From
\eqref{1p22} we have
\begin{equation} \label{2p23}
\sigma(\bar u)\Phi'=c_1-\sigma'(\bar u)U\bar\varphi'.
\end{equation}
Substituting \eqref{2p23} in \eqref{3p22} we obtain, as a problem equivalent to
\eqref{1p22}--\eqref{4p22},
\begin{gather} \label{3p23}
\sigma'(\bar u)\bar\varphi'U+\sigma(\bar u)\Phi'=c_1,\\
\label{4p23}
\Phi(-L)=0,\quad \Phi(L)=0,\\
\label{5p23}
U''-\sigma'(\bar u)\bar\varphi'^2U=-2c_1\bar\varphi',\\
\label{6p23}
U(-L)=0,\quad U(L)=0.
\end{gather}
If $\mathcal{V}(x)$ is a solution of the auxiliary problem
\begin{gather} \label{1p24}
\mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=-2\bar\varphi',\\
\label{2p24}
\mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0,
\end{gather}
then the function
\begin{equation} \label{3p24}
U(x)=c_1\mathcal{V}(x)
\end{equation}
solves \eqref{5p23} and \eqref{6p23} and vice versa.
Substituting \eqref{3p24} into \eqref{3p23} we obtain
\[
\Phi'(x)=\frac{c_1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\mathcal{V}).
\]
Integrating, we have
\begin{equation} \label{3p25}
\Phi(x)=c_1\int_{-L}^x\frac{1-\sigma'(\bar u(t))\bar\varphi(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt.
\end{equation}
The condition $\Phi(L)=0$ becomes
\begin{equation} \label{4p25}
c_1\int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt=0.
\end{equation}
Let us assume that
\begin{itemize}
\item[(H0)]  the number 1 is not an eigenvalue of the problem
\begin{equation} \label{ep}
\mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=0,\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0.
\end{equation}
\end{itemize}

When (H0) holds, the auxiliary problem \eqref{1p24}, \eqref{2p24} has one
and only one solution $\mathcal{V}(x)$ and two possibilities occur: a generic
case,
\begin{equation} \label{2p26}
\int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt\neq 0,
\end{equation}
and a special case
\begin{equation} \label{3p26}
\int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt=0.
\end{equation}
Assume (H0) and \eqref{2p26} hold. Then, from \eqref{4p25}, $c_1=0$ and
\eqref{3p25} imply
\begin{equation}\label{2p27}
\Phi(x)=0.
\end{equation}
Moreover, from \eqref{5p23} and \eqref{6p23} we have
\begin{equation}\label{3p27}
U''-\sigma'(\bar u)\bar\varphi'^2U=0,\quad U(-L)=0,\quad U(L)=0.
\end{equation}
On the other hand, by (H0), the value $1$ is not an eigenvalue of \eqref{3p27},
hence $U(x)=0$. Therefore the problem \eqref{1p22}-\eqref{4p22} has only
the trivial solution and the one-dimensional version of  Lemma \ref{lem1.4}
applies.

We consider next the special case in which the assumption (H0) holds, but
\begin{equation} \label{1p35}
\int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt=0,
\end{equation}
where in \eqref{1p35} $\mathcal{V}(x)$ is the unique solution of problem
\eqref{1p24}-\eqref{2p24}. We have
\[
\Phi(x)=c_1\int_{-L}^x\frac{1-\sigma'(\bar u(t))\bar\varphi(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt,
\]
where $c_1$ is an arbitrary constant and, by \eqref{3p24},
$U(x)=c_1\mathcal{V}(x)$. Thus in this case the linear problem
\eqref{3p23}-\eqref{6p23} has nontrivial solutions, more precisely the space
of its solutions has dimension $1$.

\begin{example} \label{examp2.2} \rm
The problem \eqref{ep}  can be solved only in
special cases and it is therefore difficult to check the condition
(H0). However, this can be done for the physical relevant conductivity law
\begin{equation} \label{1p28}
\sigma(u)=\frac{K}{au+b},\ K>0,\ a>0,\ b>0
\end{equation}
which is quite accurate for metals. If \eqref{1p28} holds, we have,
using the notation of Lemma \ref{lem2.1},
\[
\xi=F(u)=\frac{1}{K}(\frac{au^2}{2}+bu),\quad
u=F^{-1}(\xi)=\frac{-b+\sqrt{b^2+2a\xi K}}{a}.
\]
Moreover,
\[
\sigma(F^{-1}(\frac{V^2}{2}-\frac{t^2}{2}))=\frac{K}{\sqrt{b^2+aK(V^2-t^2)}}
\]
and
\begin{equation}
\label{2p29}
G(V)=\frac{\sqrt{K}}{\sqrt{a}}\arctan\frac{\sqrt{aK}V}{b}.
\end{equation}
Problem \eqref{ep} can be restated, in view of
\eqref{3p16}, in the  form
\begin{equation} \label{3p29}
\mathcal{V}''-\frac{\sigma'(\bar u)(G(V))^2}{L^2(\sigma(\bar u))^2}\mathcal{V}=0,\quad
 \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0.
\end{equation}
If \eqref{1p28} holds, we have
\[
\frac{\sigma'(\bar u)}{(\sigma(\bar u))^2}=-\frac{a}{K}.
\]
Hence, by \eqref{2p29}, the equation in \eqref{3p29} becomes
\[
\mathcal{V}''+\frac{1}{L^2}\bigl(\arctan\frac{\sqrt{aK}V}{b}\bigl)^2\mathcal{V}=0.
\]
Recalling that $\mu_0=\frac{\pi^2}{4L^2}$ is the first eigenvalue of the problem
\[
\mathcal{V}''+\mu\mathcal{V}=0,\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0
\]
 and taking into account that
\begin{equation}
\frac{1}{L^2}\bigl(\arctan\frac{\sqrt{aK}V}{b}\bigl)^2<\frac{\pi^2}{4L^2}
\label{j}
\end{equation}
we conclude that $1$ is not an eigenvalue of the problem \eqref{ep}
 if \eqref{1p28} holds. Hence the condition (H0) is certainly
verified. Moreover, in view of \eqref{j} the operator
\[
\frac{d^2}{dx^2}+\frac{1}{L^2}\bigl(\arctan\frac{\sqrt{aK}V}{b}\bigl)^2
\]
is a ``maximum principle operator''. Thus the unique solution of the
problem
\[
\mathcal{V}''+\frac{1}{L^2}(\arctan\frac{\sqrt{aK}V}{b})^2\mathcal{V}=-2\bar\varphi'(x),\
\mathcal{V}(-L)=0,\ \mathcal{V}(L)=0
\]
is positive in $(-L,L)$ since $\varphi'(x)>0$ by \eqref{1p15_1}. It follows
\[
\int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt>0.
\]
Therefore,  the condition \eqref{2p26} is satisfied. It follows that the
problem \eqref{1p22}-\eqref{4p22} has only the trivial solution and the
one-dimensional version of Lemma \ref{lem1.4} applies.
\end{example}

\begin{example} \label{examp2.3} \rm
If $\sigma'(u)\geq 0$ the problem \eqref{ep} has only the trivial solution
 $\mathcal{V}(x)=0$ \cite{AP},
therefore (H0) is verified. However, in this case, we have by the
maximum principle, from
\eqref{1p24}-\eqref{2p24} and, in view of \eqref{1p15_1}, $\mathcal{V}(x)>0$.
Thus the cases \eqref{2p26} and \eqref{3p26} are, in principle, both possible.
\end{example}

To treat the case in which $1$ is an eigenvalue of  \eqref{ep},
 we recall \cite{B} the following  result on the
 eigenvalues and eigenfunctions of the problem
\begin{equation} \label{1p37}
v''+\lambda p(x)v=0,\quad v(L)=0,\quad (L)=0.
\end{equation}

\begin{lemma} \label{lem2.4}
If $p(x)\in C^0([-L,L])$ and $p(x)>0$, then the eigenvalues $\lambda_n$, 
$n=0,1,2,\dots$
of problem \eqref{1p37} are all simple. When the eigenvalues are arranged
in increasing order, the eigenfunctions $v_n(x)$
(determined except for a constant multiplier) possess exactly $n$
zeros in $(-L,L)$. In particular, the first eigenvalue $v_0(x)$
 has constant sign.
\end{lemma}

\begin{lemma} \label{lem2.5}
Let $p(x)\in C^0([-L,L])$ be even and $p(x)>0$. Then the eigenfunctions
$v_n(x)$ of \eqref{1p37} with an even index are even, and the eigenfunctions
with an odd index are odd.
\end{lemma}

\begin{proof}
All eigenfunctions of \eqref{1p37} are either even or odd.  Let
$v(x)$ be an eigenfunction corresponding to the eigenvalue $\lambda$. Let $v(0)\neq
0$ and define
\begin{equation} \label{1p42}
W(x)=v(-x).
\end{equation}
It is easily seen that $W(x)$ is also an eigenfunction corresponding to
$\lambda$. Thus $W(x)=Cv(x)$ and $W(0)=v(0)$ by \eqref{1p42}. Hence $C=1$ and
therefore $v(-x)=v(x)$. Let $v(0)=0$. We have $v'(0)=\alpha\neq 0$ since $v'(0)=0$
would imply $v(x)=0$. Define $W(x)=-v(-x)$. $W(x)$ is an eigenfunction corresponding to $\lambda$. On the other hand,
$W(0)=-v(0)=0$. Therefore $W(x)=v(x)$ and $v(x)=-v(-x)$. To prove that
$v_0(x)$ is even we simply note that $v_0(x)\neq
0$. We prove that $v_1(x)$ is
odd. By Lemma \ref{lem2.4}, $v_1(x)$ has only one zero $x^*$ in $(-L,L)$  with
$v_1'(x^*)\neq 0$. We claim that $x^*=0$. Let $x^*\neq 0$, thus either
$v_1(x^*)=0$ and $v_1(-x^*)=0$ or  $v(x^*)=0$ and $-v_1(-x^*)=0$ and this
cannot be since $v_1(x)$ has only one zero in $(-L,L)$.
Suppose, by contradiction, $v_1(x)$ to be even. This implies
\begin{equation}\label{1p45}
v_1'(0)=0.
\end{equation}
But $v_1(0)=0$ and that, together with \eqref{1p45}, would imply
$v_1(x)=0$. Hence $v_1(x)$ is odd. In a similar vein we can prove the general
result: $v_n(x)$ is even if $n$ is even and $v_n(x)$ is odd if $n$ is odd.
\end{proof}

\begin{lemma} \label{lem2.6}
Let $p(x)\in C^0([-L,L]),\quad f(x)\in C^0([-L,L])$
 be even functions and $p(x)>0$. Consider the two-point problem
\begin{equation} \label{2p38}
v''+\lambda p(x)v=f(x),\quad v(L)=0,\quad v(L)=0.
\end{equation}
Let $\lambda$ be an eigenvalue of odd index of the problem \eqref{1p37}
and $\tilde v(x)$ the corresponding (odd) eigenfunction.
Then the solutions of \eqref{2p38} can be written as follows
\[
v(x)=C\tilde v(x)+w(x),
\]
where $C$ is an arbitrary constant and $w(x)$ is the only solution of
\eqref{2p38} which is even and satisfies the condition
\begin{equation} \label{1p40}
\int_{-L}^{L}w(x)\tilde v(x)dx=0.
\end{equation}
\end{lemma}

\begin{proof}
The condition of solvability of problem \eqref{2p38},
i. e., $\int_{-L}^Lf(x)\tilde v(x)dx=0$ is satisfied in view of the assumptions
on $f(x)$ and on the eigenvalue $\lambda$.
Let us normalize the eigenfunction $\tilde v(x)$ assuming $\int_{-L}^{L}\tilde
v^2dx=1$. The solutions of problem \eqref{2p38} are given by
\[%\label{qw}
v(x)=C\tilde v(x)+v^*(x),
\]
where $v^*(x)$ is an arbitrary function which satisfies
\[ %\label{qwe}
\frac{d^2v^*}{dx^2}+\lambda v^*=f(x),\ v^*(-L)=0,\ v^*(L)=0.
\]
Define
\[
w(x)=v^*(x)-\int_{-L}^{L}v^*(t)\tilde v(t)dt\ \tilde v(x).
\]
We have
\[
\int_{-L}^Lw(x)\tilde v(x)dx=0.
\]
On the other hand, if $w_1(x)$ and $w_2(x)$ both satisfy \eqref{2p38} and
\eqref{1p40} and if we define $h(x)=w_1(x)-w_2(x)$, we have $h(x)=C\tilde
v(x)$. If $C=0$ we have done. If $C\neq 0$ we have
\[
\int_{-L}^Lh(x)\tilde v(x)dx=0,\quad  C\int_{-L}^L\tilde v^2(x)dx=0
\]
which cannot be. Thus there exists only one solution of \eqref{2p38} which
satisfies \eqref{1p40}. We claim that $w(x)$ is even. Define
$z(x)=w(-x)$. Since $\tilde v(x)$ is odd we have
\[
\int_{-L}^Lz(x)\tilde v(x)dx=\int_{-L}^Lw(x)\tilde v(x)dx=0.
\]
Also we have
\[
\frac{d^2z}{dx^2}+\lambda p(x)z=f(x),\quad z(-L)=0,\quad z(L)=0
\]
since $p(x)$ and $f(x)$ are even functions. By uniqueness we conclude that
$w(x)$ is even.
\end{proof}
We assume now
\begin{itemize}
\item[(H1)] $\sigma'(\bar u(x))<0$ and the number 1 is the first eigenvalue
of the problem
\[
\mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=0,\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0.
\]
\end{itemize}
Denote by $\mathcal{V}_0(x)$ the corresponding eigenfunction normalized with
the condition $\int_{-L}^L\mathcal{V}^2(x)dx=1$.
Then we have
\[
\int^L_{-L}\bar\varphi'(x)\mathcal{V}_0(x)dx\neq 0
\]
since $\mathcal{V}_0(x)\neq 0$ by Lemma \ref{lem2.4} and $\bar\varphi'(x)>0$
by Lemma \ref{lem2.1}.
Hence the auxiliary problem
\[
\mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=-2\bar\varphi',\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0
\]
has no solution. Therefore the problem
\[
U''-\sigma'(\bar u)\bar\varphi'^2U=-2c_1\bar\varphi',\quad U(-L)=0,\quad U(L)=0
\]
has solutions only when
\begin{equation}\label{1p47_1}
c_1=0
\end{equation}
and these solutions are
\begin{equation}\label{2p47_1}
U(x)=\gamma\mathcal{V}_0(x),\quad \gamma\in \mathbb{R}^1.
\end{equation}
 From \eqref{2p23}, \eqref{1p47_1} and \eqref{2p47_1} we have
\begin{gather}\label{3p47_1}
\sigma(\bar u)\Phi'=-\gamma\sigma'(\bar u)\mathcal{V}_0(x)\bar\varphi'(x),\\
\label{4p47_1}
\Phi(-L)=0,\quad \Phi(L)=0.
\end{gather}
The condition of solvability of \eqref{3p47_1} and \eqref{4p47_1}
is thus given by
\[ %\label{1p47_2}
\gamma\int_{-L}^L\frac{\sigma'(\bar u(t))\mathcal{V}_0(t)\bar\varphi'(t)dt}{\sigma(\bar u(t))}=0.
\]
On the other hand, by (H1), Lemma \ref{lem2.1} and
Lemma \ref{lem2.4}, we have
\[%\label{2p47_2}
\int_{-L}^L\frac{\sigma'(\bar u(t))\mathcal{V}_0(t)\bar\varphi'(t)dt}{\sigma(\bar u(t))}\neq 0
\]
which implies $\gamma=0$ and $U(x)=0$ and, from \eqref{3p47_1} and \eqref{4p47_1},
$\Phi(x)=0$. Therefore, the problem
\eqref{3p23}-\eqref{6p23} has only the trivial solution and Lemma \ref{lem1.4}
applies.

Next we examine the case when
\begin{itemize}
\item[(H2)] $\sigma'(\bar u(x))<0$  and the number 1 is the second eigenvalue
of the problem
\[
\mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=0,\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0.
\]
\end{itemize}

Let $\mathcal{V}_1(x)$ be the corresponding eigenvalue which is normalized with 
the condition $\int_{-L}^L\mathcal{V}_1^2(x)dx=1$.
By Lemma \ref{lem2.5}, $\mathcal{V}_1(x)$ is an odd function. Thus we have, recalling
that $\bar\varphi'(x)$ is an even function,
\[
\int_{-L}^L\bar\varphi'(x)\mathcal{V}_1(x)dx=0.
\]
Thus, by Lemma \ref{lem2.6}, the solutions of
\begin{equation} \label{2p49}
\mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=-2\bar\varphi',\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0
\end{equation}
are given by
\[
\mathcal{V}(x)=C\mathcal{V}_1(x)+\tilde{\mathcal{V}}(x),
\]
where $C$ is an arbitrary constant and $\tilde{\mathcal{V}}(x)$ the only solution of
\eqref{2p49} which is even and satisfies
\[
\int_{-L}^L\tilde{\mathcal{V}}(x)\mathcal{V}_1(x)dx=0.
\]
Therefore, the solutions of
\[
U''-\sigma'(\bar u)\bar\varphi'^2U=-2c_1\bar\varphi',\quad U(-L)=0,\quad U(L)=0
\]
are given by
\begin{equation}\label{2p50}
U(x)=c_1C\mathcal{V}_1(x)+c_1\tilde{\mathcal{V}}(x)
\end{equation}
or, if we put $K=c_1C$,
\begin{equation}\label{3p50}
U(x)=K\mathcal{V}_1(x)+c_1\tilde{\mathcal{V}}(x).
\end{equation}
 From \eqref{2p23}, using \eqref{3p50}, we have
\[
\Phi'(x)=\frac{c_1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\tilde{\mathcal{V}})
-\frac{K}{\sigma(\bar u)}\sigma'(\bar u)\bar\varphi'\mathcal{V}_1.
\]
Hence
\begin{equation} \label{2p51}
\Phi(x)=c_1\int_{-L}^x\frac{1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\tilde{\mathcal{V}})dt
-K\int_{-L}^x\frac{1}{\sigma(\bar u)}\sigma'(\bar u)\bar\varphi'\mathcal{V}_1dt.
\end{equation}
The condition $\Phi(L)=0$ gives
\[
c_1\int_{-L}^L\frac{1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\tilde{\mathcal{V}})dt=0
\]
if we take into account that
$\frac{\sigma'(\bar u(x))\bar\varphi'\mathcal{V}_1(x)}{\sigma(\bar u(x))}$ is
an odd function of $x$ . Thus we need to distinguish a generic case when
\[
\int_{-L}^L\frac{1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\tilde{\mathcal{V}})dt\neq 0.
\]
This implies $c_1=0$. From \eqref{3p50} we have
\[
U(x)=K\mathcal{V}_1(x)
\]
and from \eqref{2p51},
\[
\Phi(x)=-K\int_{-L}^x\frac{1}{\sigma(\bar u)}\sigma'(\bar u)\bar\varphi'\mathcal{V}_1dt.
\]
Therefore, in the generic case the kernel of the linearized operator has
dimension $1$. On the other hand, if
\[
\int_{-L}^L\frac{1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\tilde{\mathcal{V}})dt=0
\]
the kernel has dimension $2$, since \eqref{2p50} and \eqref{2p51} hold.


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\end{document}