\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{graphicx} \AtBeginDocument{{\noindent\small Seventh Mississippi State - UAB Conference on Differential Equations and Computational Simulations, {\em Electronic Journal of Differential Equations}, Conf. 17 (2009), pp. 39--49.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2009 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \setcounter{page}{39} \title[\hfilneg EJDE-2009/Conf/17\hfil A boundary control problem] {A boundary control problem with a nonlinear reaction term} \author[J. R. Cannon, M. Salman \hfil EJDE/Conf/17 \hfilneg] {John R. Cannon, Mohamed Salman} % in alphabetical order \address{John R. Cannon \newline University of Central Florida, Department of Mathematics, Orlando, FL 32816, USA} \email{jcannon@pegasus.cc.ucf.edu} \address{Mohamed Salman \newline Tuskegee University, Department of Mathematics, Tuskegee, AL 36088, USA} \email{msalmanz@gmail.com} \thanks{Published April 15, 2009.} \subjclass[2000]{35K57, 35K55} \keywords{Reaction-diffusion, Parabolic, Nonlocal boundary conditions} \begin{abstract} The authors study the problem $u_t=u_{xx}-au$, $00$; $u(x,0)=0$, and $-u_x(0,t)=u_x(1,t)=\phi(t)$, where $a=a(x,t,u)$, and $\phi(t)=1$ for $t_{2k} < t0, \\ -u_x(0,t)=u_x(1,t)=\phi(t), \quad t\ge 0, \\ u(x,0)=0, \quad 0\le x\le 1, \end{gathered} where$a(x,t,u)$is a continuous function and $$0\leq \alpha\leq a(x,t,u)\leq\beta \label{H1}$$ for$(x ,t)\in[0,1]\times [0,T]$and$u\in \mathbb R$, and $$\phi(t)= \begin{cases} 1, &t_{2n} \leq t\leq t_{2n+1},\\ 0, &t_{2n+1} \leq t\leq t_{2n+2}, \end{cases} \label{3.2}$$ where$\{ t_n\}$depends on $$\mu (t)=\int_0^1 u(x,t)dx, \label{3.3}$$ where \begin{gather*} {2} \mu (t_{2n} )= m, \quad n=1,2, \dots, \\ \mu (t_{2n+1} ) =M, \quad n=0,1, \dots, \end{gather*} with$00, \quad 0\leq x\leq 1 \end{gather*} is positive for all $(x,t)\in \overline{D}$ where $\gamma$ is a positive constant. \end{lemma} \begin{proof} Let $w=e^{\gamma t}v$. Then \begin{gather*} w_t=w_{xx}, \quad \text{in } D,\\ w_x(0,t)=w_x(1,t)=0, \quad t\ge 0,\\ w(x,0)=v(x,0)>0, \quad 0\leq x\leq 1. \end{gather*} If $w\leq 0$, then $w$ has a minimum that's not positive either at $x=0$ or $x=1$ for some $t=t_0\in (0,T]$, which implies by the strong maximum principle \cite{prot}, $w_x(0,t_0)>0$ or $w_x(1,t_0)<0$, which is a contradiction. Hence, $w>0$, and therefore, $v>0$. \end{proof} \begin{lemma} \label{lem3.3} The solution $z(x,t,\epsilon )$ of \begin{gather*} z_{t}=z_{xx} \quad \text{in }D, \\ z_{x}(0,t)=\epsilon,\quad 0\le t\le T, \\ z_{x}(1,t)=-\epsilon, \quad 0\le t\le T, \\ z(x,0)=0,\quad 0\le x\le 1, \end{gather*} satisfies the inequality \begin{equation*} -C\epsilon 0$in$\overline{D}$. Consider$w=u-v-z$. Clearly,$w$satisfies \begin{gather*} w_t=w_{xx}-aw-(a-\beta )v \quad \text{in } D,\\ w_x(0,t)=-\epsilon , \quad 0\le t\le T,\\ w_x(1,t)=\epsilon , \quad 0\le t\le T,\\ w(x,0)=0, \quad 0\le x \le 1. \end{gather*} Suppose$w<0$somewhere in$\overline{D}$. Then the boundary conditions force a negative minimum in$D$, where $$w_t-w_{xx}+aw+(a-\beta )v<0,$$ which contradicts the equation $$w_t-w_{xx}+aw+(a-\beta )v=0 \quad\text{in}\;\;D.$$ Thus,$w\geq 0$which implies that $$u(x,t)\geq v(x,t,\beta )+z(x,t,\epsilon )$$ for all$\epsilon >0$sufficiently small. Hence, $$u(x,t)\geq v(x,t;\beta ) .$$ Likewise, by considering$w=v-z-u$, the inequality $$v(x,t;\alpha )\geq u(x,t),$$ follows by a similar argument. \end{proof} \begin{theorem} \label{thm3.5} The solution$u$of \eqref{3.1} is nonnegative. \end{theorem} \begin{proof} By applying, successively, Lemma \ref{lem3.1} and \ref{lem3.4} on each time stage where we keep the flux$u_x$either zero or one, and the conclusion follows. \end{proof} \section{Existence of the Time Switches} If we formally differentiate \eqref{3.3}, we obtain $$\mu '(t)=2\phi (t)- \int_0^1 a(x,t,u)udx . \label{3.7}$$ To prove the existence of$t_1$, let$\phi (t)=1$for$t>0$. In view of hypothesis \eqref{H1}, equation (\ref{3.7}) implies the estimate \begin{equation*} \mu '(t)\geq 2-\beta \int_0^1 u \, dx; \end{equation*} that is, \begin{equation*} \mu '(t)\geq 2-\beta\mu (t),\quad t\geq 0. \end{equation*} By applying Gronwal's inequality, we get \begin{equation*} \mu (t)\geq \frac{2}{\beta} [1-e^{-\beta t}]. \end{equation*} Since$\mu (t)$is continuous, then there exists a$t_1>0$such that \begin{equation*} \mu (t_1)=M, \end{equation*} for any$0t_1$. This implies \begin{equation*} \mu '(t)=-\int_0^1 a(x,t,u)u(x,t)dx, \quad t>t_1. \end{equation*} Using the estimate on$a(x,t,u)$, we obtain \begin{equation*} \mu '(t)\leq -\alpha \mu (t),\quad t\geq t_1. \end{equation*} Gronwal's inequality implies $\mu (t) \leq \mu (t_1)e^{-\alpha (t-t_1)} = Me^{-\alpha (t-t_1)} , \quad t\geq t_1.$ Since$\mu (t)$is continuous, then there exists a$t_2>t_1$such that \begin{equation*} \mu (t_2)=m, \end{equation*} where$0t_2$, we take$\phi (t)=1$. This gives the estimate \begin{equation*} \mu '(t)\geq 2-\beta\mu (t),\quad t\geq t_2. \end{equation*} Using the condition$\mu (t_2)=m$and Gronwal's inequality, we get \begin{equation*} \mu (t)\geq \frac{2}{\beta}-\big(\frac{2}{\beta} -m\big) e^{-\beta (t-t_2)}, \quad t\geq t_2. \end{equation*} Note that the coefficient$\frac{2}{\beta}-m$is positive, which implies the existence of$t_3>t_2$such that \begin{equation*} \mu(t_3)=M. \end{equation*} We inductively get for$t>t_{2n}$and$\phi (t)=1$, $$\mu (t)\geq \frac{2}{\beta} -\big(\frac{2}{\beta} -m\big) e^{-\beta (t-t_{2n})} , \quad t\geq t_{2n}, \label{3.8}$$ which implies the existence of$t_{2n+1}$such that$\mu(t_{2n+1})=M$. Also, for$t>t_{2n+1}$and$\phi(t)=0$, we have, $$\mu (t)\leq Me^{-\alpha (t-t_{2n+1})}, \quad t\geq t_{2n+1} , \label{3.9}$$ which ensures the existence of$t_{2n+2}$such that$\mu (t_{2n+2})=m\$. Estimate (\ref{3.8}) implies \begin{equation*} M=\mu (t_{2n+1})\geq \frac{2}{\beta}-\big(\frac{2}{\beta} -m\big) e^{-\beta (t_{2n+1}-t_{2n})}, \end{equation*} which gives rise to $$t_{2n+1}-t_{2n}\leq\frac{1}{\beta} \ln \frac{2-m\beta}{2-M\beta} . \label{3.10}$$ Similarly, if we employ (\ref{3.9}), we can get \begin{equation*} t_{2n+2}-t_{2n+1}\leq \frac{1}{\alpha} \ln \frac{M}{m}. \end{equation*} \section{Numerical Example} In this section, we consider a finite difference method to discretize the problem \begin{gather*} u_t=c u_{xx}- \sin u, \quad 0