\documentclass[reqno]{amsart} \usepackage{graphicx} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 84, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2003 Southwest Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/84\hfil Asymptotic behavior for a heat conduction problem] {Asymptotic behavior for a heat conduction problem with perfect-contact boundary condition} \author[A. Barrea \& C. Turner \hfil EJDE--2003/84\hfilneg] {Andr\'es Barrea \& Cristina Turner } \address{Andr\'es Barrea \hfill\break Fa.M.A.F.- Universidad Nacional de C\'ordoba \\ CIEM-CONICET, C\'ordoba, Argentina} \email{abarrea@mate.uncor.edu} \address{Cristina Turner \hfill\break Fa.M.A.F.- Universidad Nacional de C\'ordoba \\ CIEM-CONICET, C\'ordoba, Argentina} \email{turner@mate.uncor.edu} \date{} \thanks{Submitted July 15, 2002. Published August 14, 2003.} \subjclass[2000]{35K60, 35R35} \keywords{Heat conduction, phase change, free boundary, perfect contact} \begin{abstract} In this paper we consider the heat conduction problem for a slab represented by the interval $[0,1]$. The initial temperature is a positive constant, the flux at the left end is also a positive constant, and at the right end there is a perfect contact condition: $u_{x}(1,t)+\gamma u_{t}(1,t)=0$. We analyze the asymptotic behavior of these problems as $\gamma$ approaches infinity, and present some numerical calculations. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{example}[theorem]{Example} \section{Introduction and preliminaries} When two bodies $A$ and $B$ are in perfect thermal contact at a boundary $S$, the boundary conditions are \begin{gather*} K_1 \frac{\partial V}{\partial \eta}=K_2\frac{\partial v}{\partial \eta},\quad \mbox{on } S,\\ V=v,\quad \mbox{on }S; \end{gather*} where $V$, $v$ are the temperatures of $A$ and $B$, $K_1$ and $K_2$ are their thermal conductivities, and $\eta$ is the outer unit normal. Assuming that $K_1\gg K_2$, we can consider that $V=V(t)=v|_S$. Then by means of an energy balance, we get a boundary condition for $v$: $$\label{pc} K_2\int \int_S \frac{\partial v}{\partial \eta}dS + Mc'\frac{\partial v}{\partial t}=0,$$ where $M$ and $c'$ denote the mass and the specific heat of the body $A$. This kind of boundary condition has been widely investigated; see for example \cite{citation3,citation4,citation5}. We consider a one-dimensional slab $[0,\ell]$ with its face $y=\ell$ in perfect thermal contact with mass $M_f$ per unit area of a well-stirred fluid (or a perfect conductor) of specific heat $c_f$. In this case the condition (\ref{pc}) is given by $$kv_{y}(\ell,\tau)+M_fc_fv_{\tau}(\ell,\tau)=0.$$ This condition appears in several interesting applications such as heat condensers \cite{citation3}. We consider the heat conduction problem $$\begin{gathered}\label{P} kv_{yy}=\rho c v_{\tau}, \quad \hbox{in } [0, \ell] \times (0, \mathcal{T}],\\ v(y,0)=V_0>0, \quad 0\leq y \leq \ell,\\ kv_{y}(0,\tau)=q_0>0, \quad 0<\tau \leq \mathcal{T},\\ kv_{y}(\ell,\tau)+M_fc_fv_{\tau}(\ell,\tau)=0, \quad 0<\tau \leq \mathcal{T}, \end{gathered}$$ where $k$ is the thermal conductivity, $\rho$ the density, $q_0$ the heat flux, and $c$ the specific heat of the material. All of these constants are positive. With the change of variables $$x=\frac{y}{\ell},\quad t=\frac{k \tau}{\rho c \ell^{2}}, \quad v(y,\tau)=cu(x,t).$$ problem \eqref{P} is transformed into the problem $$\label{cpu} \begin{gathered} u_{xx}=u_{t}, \quad \hbox{in } [0, 1] \times (0, T],\\ u(x,0)=M>0, \quad 0\leq x \leq 1,\\ u_{x}(0,t)=q>0, \quad 0< t \leq T,\\ \gamma u_{t}(1,t) + u_x(1,t)=0, \quad 0< t \leq T, \end{gathered}$$ where $$M=c V_0,\quad q=\frac{c \ell q_0}{k},\quad \gamma= \frac{M_fc_f}{\rho c}, \quad T=\frac{k \mathcal{T}}{\rho c \ell^2}\,.$$ Roughly speaking, we expect that as $\gamma \to +\infty$ the solution to \eqref{cpu} converge to the solution of the problem $$\label{Pinf} \begin{gathered} u_{xx}=u_{t}, \quad \hbox{in } [0, 1] \times (0, T],\\ u(x,0)=M>0, \quad 0\leq x \leq 1,\\ u_{x}(0,t)= q>0, \quad 0< t \leq T,\\ u_t(1,t)= 0. \quad 0< t \leq T. \end{gathered}$$ In the case of models of heat conduction in material media it is natural to attempt to determine the temporary range of validity (i.e. the solution remains positive). Here an important limitation of this range is imposed by the change of phase phenomena. An extensive bibliography on phase-change problem can be found in \cite{citation6}. In \cite{citation7} the authors studied this problem with temperature and convective boundary conditions at $x=1$. They obtained an explicit expression for the approximation of the time of phase change $t_{ch}$ for the problem \eqref{Pinf}, namely \begin{eqnarray}\label{tch} t_{ch}=\big(\frac{\sqrt{\pi}M}{2q}\big)^2. \end{eqnarray} Here we prove that the solution of the problem \eqref{cpu} converges to the solution of the problem \eqref{Pinf} using (\ref{tch}). This relation was obtained in \cite{citation8} using Laplace transforms. Next we prove that the solution to the problem \eqref{cpu} converges to the solution of the problem \eqref{Pinf} in the $L^{\infty}$ norm. In fact, by using a numerical scheme we visualize this convergence. The formulation and the results of the numerical scheme are provided. \section{Asymptotic behavior of problem \eqref{cpu}} \begin{lemma} \label{lm1} The solution to \eqref{cpu} satisfies \begin{enumerate} \item If $u_x(x,t)\geq 0$ then $u(0,t) \leq u(x,t)$, for $0 \leq x \leq 1$, $0< t \leq T$. \item For all $(x,t) \in [0,1] \times (0,T]$, $u_t(x,t) \leq 0$. \end{enumerate} \end{lemma} \begin{proof} Let $v=u_x$. Then $v$ satisfies $$\label{v1} \begin{gathered} v_{xx}=v_{t}, \quad \hbox{in } [0,1] \times (0, T],\\ v(x,0)=0, \quad 0\leq x \leq 1,\\ v(0,t)=q, \quad 0< t \leq T,\\ v(1,t)+\gamma v_x(1,t)=0, \quad 0< t \leq T. \end{gathered}$$ By the maximum principle \cite{citation10}, $\min v(x,t)=\min \{q,0,v(1,t)\}$ for $0 \leq x \leq 1$ and $t>0$. Assuming that $v(1,t)<0$ (we remark that $q>0$) it follows that $\min v(x,t)=v(1,t)$. By Hopf's lemma (see \cite{citation10}), $v_x(1,t)<0$, which contradicts the last equation in (\ref{v1}) ($\gamma >0$). Therefore, $u_x(x,t) \geq 0$. This proves part 1. Let $v^{\varepsilon}(x,t)=u(x,t+\varepsilon)-u(x,t)$. Hence $$\label{w1} \begin{gathered} v^{\varepsilon}_{xx}=v^{\varepsilon}_{t}, \quad \hbox{in } D=[0, 1] \times (0, T],\\ v^{\varepsilon}(x,0)=u(x,\varepsilon)-M, \quad 0\leq x \leq 1,\\ v^{\varepsilon}_x(0,t)=0, \quad ,0 < t \leq T\\ v^{\varepsilon}_x(1,t)+\gamma v^{\varepsilon}_t(1,t)=0, \quad t>0. \end{gathered}$$ Let us show that $u(x,\varepsilon)-M \leq 0$ for all $\varepsilon \geq 0$. By the maximum principle and Hopf's lemma we have $$\max v^{\varepsilon}(x,t)=\max \{u(x,\varepsilon)-M,v^{\varepsilon}(1,t)\},$$ for $0 \leq x \leq 1$ and $0 < t \leq T$. Assuming that $\max v^{\varepsilon}(x,t)=v^{\varepsilon}(1,t_0)>0$, then it follows that $v^{\varepsilon}_x(1,t_0)>0$. From (\ref{w1}) it follows that $v^{\varepsilon}_t(1,t_0)<0$, which implies that $v^{\varepsilon}(1,t)$ decreases in $(t_0-\varepsilon,t_0)$. This contradiction proves that that $v^{\varepsilon}(x,t) \leq 0$. Hence $$\lim_{\varepsilon \to 0}\frac{v^{\varepsilon}(x,t)}{\varepsilon}= u_t(x,t) \leq 0,$$ which completes the proof. \end{proof} Since $u(x,t)-M$ satisfies \eqref{cpu} with zero initial condition, by the maximum principle and Hopf's lemma, it follows that $u(x,\varepsilon)-M \leq 0$ for all $\varepsilon \geq 0$. \begin{lemma} \label{lm2} \begin{enumerate} \item Let $u_{\gamma_i}$ be solutions to Problem \eqref{cpu} with $\gamma_1$ and $\gamma_2$ respectively. If $\gamma_1 \leq \gamma_2$ then $u_{\gamma_1} \leq u_{\gamma_2}$. \item Let $u_{\infty}$ be the solution to Problem \eqref{Pinf}. Then $u_{\gamma} \leq u_{\infty}$ for all $\gamma > 0$. \end{enumerate} \end{lemma} \begin{proof} Let $z=u_{\gamma_2}-u_{\gamma_1}$. Then for $x =1$ the function $z$ satisfies \begin{equation*} \begin{aligned} z_x(1,t)&=u_{\gamma_{2_x}}(1,t)-u_{\gamma_{1_x}}(1,t)\\ &=-\gamma_2u_{\gamma_{2_t}}(1,t)+\gamma_1u_{\gamma_{1_t}}(1,t)\\ &=-\gamma_2(u_{\gamma_{2_t}}(1,t)- u_{\gamma_{1_t}}(1,t))+(\gamma_1-\gamma_2)u_{\gamma_{1_t}}(1,t)\\ &=-\gamma_2z_{t}(1,t)+(\gamma_1-\gamma_2)u_{\gamma_{1_t}}(1,t). \end{aligned} \end{equation*} By Lemma \ref{lm1}, $z(x,t)$ satisfies: $$\label{z1} \begin{gathered} z_{xx}=z_{t}, \quad \hbox{in } [0, 1] \times (0, \leq T],\\ z(x,0)=0, \quad 0\leq x \leq 1,\\ z_x(0,t) =0, \quad 0< t \leq T,\\ z_x(1,t)+\gamma_2 z_t(1,t)\geq 0, \quad 0< t \leq T. \end{gathered}$$ By the maximum principle and Hopf's lemma, for $0 \leq x \leq 1$ and $0< t \leq T$ we have $$\min z(x,t)=\min \{0,z(1,t)\}.$$ Assume that $\min z(x,t)=z(1,t_0)<0$, then using Hopf's lemma, $$z_x(1,t_0)<0.$$ From (\ref{z1}) ($\gamma_2 >0$) it follows that $z_t(1,t_0)>0$. This implies that $z(1,t)$ increases in $(t_0-\varepsilon,t_0)$, which contradicts that $z(1,t_0)$ is a minimum. Therefore we obtain that $z(x,t) \geq 0$. Setting $z=u_{\infty}-u_{\gamma}$, we obtain \begin{align*} z_x(1,t)+\gamma z_t(1,t) &= u_{\infty_x}(1,t)-u_{\gamma_x}(1,t)+\gamma(u_{\infty_t}(1,t)-u_{\gamma_t}(1,t)),\\ &=-(u_{\gamma_x}(1,t)+\gamma u_{\gamma_t}(1,t))+u_{\infty_x}(1,t)+\gamma u_{\infty_t}(1,t)\\ &=u_{\infty_x}(1,t). \end{align*} The function $\theta(x,t)=u_{\infty_x}(x,t)$ satisfies the heat conduction problem: \begin{gather*} \theta_{xx}=\theta_{t}, \quad D=\{(x,t): 0\leq x \leq 1, 0< t \leq T\},\\ \theta(x,0)=0, \quad 0\leq x \leq 1,\\ \theta(0,t)=q, \quad 0< t \leq T,\\ \theta_x(1,t)=0, \quad 0< t \leq T, \end{gather*} Using the maximum principle and Hopf's lemma we obtain $$0 \leq \theta(x,t)=u_{\infty_x}(x,t) \leq q.$$ From the above inequality, we conclude that $$0 \leq z_x(1,t)+\gamma z_t(1,t) \leq q.$$ Using maximum principle and Hopf's lemma, we deduce that $z(x,t) \geq 0$. \end{proof} \noindent {\bf Remark} From lemma \ref{lm2}, we can assure the existence of a function $u^*(x,t)$ such that $\lim_{\gamma \to \infty} u_{\gamma}(x,t)=u^*(x,t)$ a.e. $x \in [0,1]$, and $u^*(x,t) \leq u_{\infty}(x,t)$. \smallskip Let $\|\cdot \|_{\infty}$ be the $L^{\infty}([0,1]\times[0,T])$ norm: \begin{eqnarray*} \|u(x,t)\|_{\infty}=\sup_{[0,1]\times[0,T]}|u(x,t)|. \end{eqnarray*} In the next lemma, we prove that the convergence is uniform and that $u^*(x,t)=u_{\infty}(x,t)$ a.e. $x \in [0,1]$. For this proof we use Laplace Transforms \cite{citation1}. \begin{lemma} \label{lm3} With the above notation, $$\|u_{\infty}(x,t)-u_{\gamma}(x,t)\|_{\infty} \leq \frac{qT}{\gamma}.$$ \end{lemma} \begin{proof} Taking $w=u_{\infty}-u_{\gamma}$, we have \begin{gather} \label{w11} w_{xx}=w_{t}, \quad [0,1] \times (0, T], \\ \label{w22} w(x,0)=0, \quad 0\leq x \leq 1,\\ \label{w33} w_x(0,t)=0, \quad 0 < t \leq T,\\ \label{w44} 0 \leq w_x(1,t)+\gamma w_t(1,t)\leq q, \quad 0 < t \leq T. \end{gather} Note that since $u_{\infty}$ and $u_{\gamma}$ are increasing functions and $u_{\gamma} \leq u_{\infty}$, we have $$\label{eq1} \|u_{\infty}(x,t)-u_{\gamma}(x,t)\|_{\infty}=u_{\infty}(1,t)-u_{\gamma}(1,t),$$ Applying the Laplace Transform, \begin{gather} \label{W1} W_{xx}(s,x)-sW(s,x)=0,\\ \label{W2} W_x(s,0)=0,\\ \label{W3} 0 < W_x(s,1)+s\gamma W(1,t) < \frac{q}{s}, \end{gather} where $s$ is a positive parameter. The general solution to this problem is $$\label{W4} W(s,x)=C(s,q,\gamma)\cosh(\sqrt{s}x).$$ Replacing (\ref{W4}) in (\ref{W3}), $C(s,q,\gamma)\leq \frac{q}{s(\sqrt{s}\sinh(\sqrt{s})+\gamma s\cosh(\sqrt{s}))}\\ \leq \frac{q}{\gamma s^{2} \cosh(\sqrt{s})}.$ Therefore, $$\label{des1} W(s,x) \leq \frac{q}{\gamma s^2}.$$ To obtain a bound for $w(1,t)$, we apply the inverse Laplace Transform at $x=1$ to (\ref{des1}): $$\label{des2} w(1,t) \leq \frac{qt}{\gamma}.$$ From (\ref{eq1}) and (\ref{des2}) we obtain $$\|u_{\infty}(x,t)-u_{\gamma}(x,t)\|_{\infty} \leq \frac{qT}{\gamma},$$ which proves the proof. \end{proof} \section{Numerical scheme and results} We preset a short description of our numerical scheme, for problem \eqref{cpu}, and refer the reader to \cite{citation2} for more details. First, we consider the weak formulation for the problem \eqref{cpu}: \begin{align*} \int_0^1u_t(x,t) \phi(x) dx + \int_0^1 u_x(x,t) \phi_x(x) dx &=u_x(1,t)\phi(1)-u_x(0,t)\phi(0)\\ &=-\gamma u_t(1,t)\phi(1)-q\phi(0), \end{align*} where $\phi(x)$ belongs to $H^1(0,1)$. We will consider a finite element method for the discretization of the space variable. Let $x_i=i/N$ for $0\leq i\leq N$ be a partition of the interval $[0,1]$ into subintervals $I_i=[x_i,x_{i+1}]$, of length $h=1/N$. Let $V_h$ the set of continuous functions which are linear on each $I_i$. We consider the basis functions of $V_h$ taking as usual $\phi_i$, with $\phi_i(x_j)=\delta_{ij}$. We define a partition \$\{0=t_0