\documentclass[twoside]{article} \usepackage{amsmath} \pagestyle{myheadings} \markboth{\hfil A three-point boundary-value problem \hfil EJDE--2002/62} {EJDE--2002/62\hfil S. Mesloub \& S. A. Messaoudi \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 62, pp. 1--13. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % A three-point boundary-value problem for a hyperbolic equation with a non-local condition % \thanks{\emph{Mathematics Subject Classifications:} 35L20, 35L67, 34B15. \hfil\break\indent \emph{Key words:} Wave equation, Bessel operator, nonlocal condition. \hfil\break \indent \copyright 2002 Southwest Texas State University. \hfil\break \indent Submitted April 16, 2002. Published July 3, 2002.} } \date{} \author{Said Mesloub \& Salim A. Messaoudi} \maketitle \begin{abstract} We use an energy method to solve a three-point boundary-value problem for a hyperbolic equation with a Bessel operator and an integral condition. The proof is based on an energy inequality and on the fact that the range of the operator generated is dense. \end{abstract} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \numberwithin{equation}{section} \allowdisplaybreaks \section{Introduction} In this paper, we investigate a boundary-value problem for a one-dimensional hyperbolic equation with a weighted nonlocal boundary integral condition of the form $\int_{l_{1}}^{l}\xi u(\xi ,t)d\xi =E(t),\quad 00,\mbox{ }t>0, \\ u(x,0)=\varphi (x),\quad x>0, \\ u(0,t)=g(t), \\ \int_{0}^{x(t)}u(\xi ,t)dx=f(t), \end{gathered} \label{e1} for x(t) and f(t) given functions. Introducing g\equiv u(0,t) as the unknown, it is proved in [2] that (1.1) is equivalent to a Volterra integral equation of the second kind for the function g. The author proved the existence and uniqueness of the solution with the aid of the integral equation. Shi [11] considered weak solutions of the problem \begin{gathered} u_{t}-u_{xx}=f+g_{x},\quad (x,t)\in (0,1)\times (0,T), \\ u(x,0)=\varphi (x),\quad 00 satisfies T-s_{0}=1/4\delta , (3.16) implies \begin{eqnarray} \lefteqn{\int_{0}^{l}\rho (x)v_{tt}^{2}(x,s)dx+\int_{0}^{l}\rho (x)\eta _{x}^{2}(x,s)dx} \nonumber \label{e65} \\ &\leq &4\delta \Big\{ \int_{s}^{T}\int_{0}^{l}\rho (x)v_{tt}^{2}\,dx\,dt+\int_{s}^{T}\int_{0}^{l}\rho (x)\eta _{x}^{2}(x,t)\,dx\,dt\Big\} \end{eqnarray} for all s\in [T-s_{0},T]. If, in (3.17) we put \[ g(s)=\int_{s}^{T}\int_{0}^{l}\rho (x)v_{tt}^{2}\,dx\,dt+\int_{s}^{T}\int_{0}^{l}\rho (x)\eta _{x}^{2}(x,t)\,dx\,dt,$ then we have $\frac{-dg}{ds}\leq 4\delta g(s)$, from which it follows that $%\label{e66} \frac{-d}{ds}\left( g(s)\exp (4\delta s)\right) \leq 0.$ Integrating this equation over $(s,T)$ and taking in account that $g(T)=0$, we obtain $%67 g(s)\exp (4\delta s)\leq 0.$ This inequality guarantees that $g(s)=0$ for all $s\in [T-s_{0},T]$, which implies that $v_{tt}=0$ on $Q_{s}$ where $s\in [T-s_{0},T]$. Hence it follows, from (3.6), that $\omega \equiv 0$ almost everywhere on $Q_{T-s_{0}}$. Proceeding this way step by step along the rectangle with side $s_{0}$, we prove that $\omega \equiv 0$ almost everywhere on $Q$. This completes the proof of the Proposition 3.3. \paragraph{Proof of Theorem 3.1} Suppose that for some $W=(\omega ,\omega _{1},\omega _{2})\in R(L)^{\perp }$% , $$(\mathcal{L}v,\omega )_{L_{\theta }^{2}(Q)}+(\ell _{1}v,\omega _{1})_{H_{\theta }^{1}((0,l))}+(\ell _{2}v,\omega _{2})_{L_{\theta }^{2}((0,l))}=0. \label{e68}$$ Then we must prove that $W\equiv 0$. Putting $v\in D_{0}(L)$ into (3.18), we have $(\mathcal{L}v,\omega )_{L_{\theta }^{2}(Q)}=0.$ Hence Proposition 3.3 implies that $\omega \equiv 0$. Thus (3.18) takes the form $$(\ell _{1}v,\omega _{1})_{H_{\theta }^{1}((0,l))}+(\ell _{2}v,\omega _{2})_{L_{\theta }^{2}((0,l))}=0,\quad \forall v\in D(L). \label{e69}$$ Since the quantities $\ell _{1}v$ and $\ell _{2}v$ can vanish independently and the ranges of the trace operators $\ell _{1}$ and $\ell _{2}$ are dense in the spaces $H_{\theta }^{1}((0,l))$ and $L_{\theta }^{2}((0,l))$ respectively, the equation (3.19) implies that $\omega _{1}\equiv 0$, $% \omega _{2}\equiv 0$. Hence $W\equiv 0$. The proof of Theorem 3.1 is established. \paragraph{Acknowledgment} This work was completed while the first author was visiting the Mathematical Sciences Department at KFUPM. Both authors would like to thank KFUPM for its support. \begin{thebibliography}{99} \bibitem{b1} A. Bouziani, \textit{Solution forte d'un probl\`{e}me mixte avec condition int\'{e}grale pour une classe d'\'{e}quations hyperboliques, Bull. CL. Sci., Acad. Roy. Belg.} \textbf{8} (1997), 53 - 70. \bibitem{c1} J. R. Cannon, \textit{The solution of heat equation subject to the specification of energy, Quart. Appl. Math}. \textbf{21} (1963), 155 - 160. \bibitem{d1} M. Denche and A. L. Marhoune, \textit{A Three-point boundary value problem with an integral condition for parabolic equations with the Bessel operator, Appl. Math. Letters} \textbf{13} (2000), 85 - 89. \bibitem{g1} L. Garding, \textit{Cauchy's problem for hyperbolic equations, }University of Chicago, 1957. \bibitem{i1} N. I. 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