\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 $$\label{eP} \begin{gathered} \nabla\cdot(\sigma(u)\nabla\varphi)=0\quad\text{in }\Omega,\\ \varphi=-V\quad\text{on }\Gamma_1,\quad \varphi=V\quad\text{on }\Gamma_2 , \\ \Delta u+\sigma(u)|\nabla\varphi|^2=0\quad\text{in }\Omega,\\ u=u_b\quad\text{on }\Gamma, \end{gathered}$$ 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 $$\label{1p3} \sigma_M\geq\sigma(u)\geq\sigma_m>0.$$ 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 $$\label{4p3} \sigma_m-\frac{\alpha}{2}>0,\quad 1-\frac{\alpha}{2\lambda_0}-\frac{\beta}{\lambda_0}>0,$$ 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 $$\label{1p9_1} \sigma_M\geq\sigma(u)\geq\sigma_m>0.$$ Let $(\bar\varphi,\bar u)$ be the unique corresponding solution of \eqref{eP} when $u_b=0$. Suppose that $$\label{2p9_1} \Phi=0, \quad U=0$$ 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 $$\label{a} \mathbf{L}_t[\mathbf{U}]=\mathbf{f}\quad\text{in }\Omega, \quad \mathbf{f}=\begin{bmatrix} f\\ g \end{bmatrix},\quad \mathbf{U}=\begin{bmatrix} 0\\ U_b \end{bmatrix}\quad\text{on }\Gamma$$ satisfies $$\label{b} \|\mathbf{U}\|_{C^{2,a}}\leq K_1\Bigl(\|\mathbf{f}\|_{C^\alpha}+\| U_b\|_{C^\alpha}\Bigl).$$ 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 $$\label{c} \mathbf{L}_{t_{0}}[\mathbf{U}]=(1-t)(\Delta \mathbf{U} -\mathbf{L}_1[\mathbf{U}])+\mathbf{f}\quad\text{in }\Omega,\quad \mathbf{U}=\begin{bmatrix} 0\\ U_b \end{bmatrix} \quad\text{on }\Omega.$$ 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 $$\label{d} \mathbf{L}_{t_{0}}[\mathbf{W}]=\mathbf{F} \quad\text{in }\Omega ,\quad \mathbf{W}=\begin{bmatrix} 0\\ U_b \end{bmatrix}\quad\text{on }\Gamma .$$ The problem \eqref{d} defines a transformation $$\label{e} \mathbf{W}=\mathbf{A}(\mathbf{U}).$$ 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 $$\label{g} \|\mathbf{W}\|_{C^{2,\alpha}}\leq K_1K_2|t-t_0|\|\mathbf{U}\|_{C^{2,\alpha}} +K_1\|\mathbf{f}\|_{C^\alpha}+K_1\| U_b\|_{C^\alpha}.$$ 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 $$\label{m} 2K_1K_2|t-t_0|\leq 1,$$ then $$\label{n} \|\mathbf{W}_1-\mathbf{W}_2\|_{C^{2,\alpha}}\leq\frac{1}{2}\|\mathbf{U}_1-\mathbf{U}_2\|_{C^{2,\alpha}}.$$ 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 $$\label{ePm} \begin{gathered} \nabla\cdot(\sigma(u)\nabla\varphi)=0\quad\text{in }\Omega\\ \varphi=-V\quad\text{on }\Gamma_1 ,\quad \varphi=V\quad\text{on }\Gamma_2 \\ \Delta u+\sigma(u)|\nabla\varphi|^2+\mu R(u,\varphi)=0\quad\text{in }\Omega\\ u=0\quad\text{on }\Gamma , \end{gathered}$$ 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 $$\label{2p15} \int_0^\infty\frac{dt}{\sigma(t)}=\infty.$$ 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 $$\label{3p16} \frac{d\varphi}{dx}(x)=\frac{G(V)}{L\sigma(u(x))}.$$ \end{lemma} \begin{proof} Let us define the transformation $$\label{4p16} \theta=\frac{1}{2}\varphi^2+\frac{V}{2}\varphi+F(u).$$ 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 $$\label{1p17} \begin{gathered} (\sigma(u)\varphi')'=0,\\ \varphi(-L)=-V,\quad \varphi(L)=V,\\ (\sigma(u)\theta')'=0,\\ \theta(-L)=0,\quad \theta(L)=V^2. \end{gathered}$$ 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 $$\label{1p18} u=F^{-1}(\frac{V^2}{2}-\frac{\varphi^2}{2}).$$ 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 $$G(\varphi(x))=\frac{G(V)}{L}x \label{s}$$ 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 $$\label{2p23} \sigma(\bar u)\Phi'=c_1-\sigma'(\bar u)U\bar\varphi'.$$ 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 $$\label{3p24} U(x)=c_1\mathcal{V}(x)$$ 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 $$\label{3p25} \Phi(x)=c_1\int_{-L}^x\frac{1-\sigma'(\bar u(t))\bar\varphi(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt.$$ The condition $\Phi(L)=0$ becomes $$\label{4p25} c_1\int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt=0.$$ Let us assume that \begin{itemize} \item[(H0)] the number 1 is not an eigenvalue of the problem $$\label{ep} \mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=0,\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0.$$ \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, $$\label{2p26} \int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt\neq 0,$$ and a special case $$\label{3p26} \int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt=0.$$ Assume (H0) and \eqref{2p26} hold. Then, from \eqref{4p25}, $c_1=0$ and \eqref{3p25} imply $$\label{2p27} \Phi(x)=0.$$ Moreover, from \eqref{5p23} and \eqref{6p23} we have $$\label{3p27} U''-\sigma'(\bar u)\bar\varphi'^2U=0,\quad U(-L)=0,\quad U(L)=0.$$ 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 $$\label{1p35} \int_{-L}^L\frac{1-\sigma'(\bar u(t))\bar\varphi'(t)\mathcal{V}(t)}{\sigma(\bar u(t))}dt=0,$$ 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 $$\label{1p28} \sigma(u)=\frac{K}{au+b},\ K>0,\ a>0,\ b>0$$ 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 $$\label{2p29} G(V)=\frac{\sqrt{K}}{\sqrt{a}}\arctan\frac{\sqrt{aK}V}{b}.$$ Problem \eqref{ep} can be restated, in view of \eqref{3p16}, in the form $$\label{3p29} \mathcal{V}''-\frac{\sigma'(\bar u)(G(V))^2}{L^2(\sigma(\bar u))^2}\mathcal{V}=0,\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0.$$ 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 $$\frac{1}{L^2}\bigl(\arctan\frac{\sqrt{aK}V}{b}\bigl)^2<\frac{\pi^2}{4L^2} \label{j}$$ 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 $$\label{1p37} v''+\lambda p(x)v=0,\quad v(L)=0,\quad (L)=0.$$ \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 $$\label{1p42} W(x)=v(-x).$$ 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 $$\label{1p45} v_1'(0)=0.$$ 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 $$\label{2p38} v''+\lambda p(x)v=f(x),\quad v(L)=0,\quad v(L)=0.$$ 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 $$\label{1p40} \int_{-L}^{L}w(x)\tilde v(x)dx=0.$$ \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 $$\label{1p47_1} c_1=0$$ and these solutions are $$\label{2p47_1} U(x)=\gamma\mathcal{V}_0(x),\quad \gamma\in \mathbb{R}^1.$$ 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 $$\label{2p49} \mathcal{V}''-\sigma'(\bar u)\bar\varphi'^2\mathcal{V}=-2\bar\varphi',\quad \mathcal{V}(-L)=0,\quad \mathcal{V}(L)=0$$ 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 $$\label{2p50} U(x)=c_1C\mathcal{V}_1(x)+c_1\tilde{\mathcal{V}}(x)$$ or, if we put $K=c_1C$, $$\label{3p50} U(x)=K\mathcal{V}_1(x)+c_1\tilde{\mathcal{V}}(x).$$ 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 $$\label{2p51} \Phi(x)=c_1\int_{-L}^x\frac{1}{\sigma(\bar u)}(1-\sigma'(\bar u)\bar\varphi'\tilde{\mathcal{V}})dt -K\int_{-L}^x\frac{1}{\sigma(\bar u)}\sigma'(\bar u)\bar\varphi'\mathcal{V}_1dt.$$ 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 $x$ . 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