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\markboth{\hfil A non-local problem with integral conditions \hfil EJDE--1999/45}
{EJDE--1999/45\hfil L. S. Pulkina \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 1999}(1999), No. 45, pp. 1--6. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
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  A non-local problem with integral conditions for hyperbolic equations
\thanks{ {\em 1991 Mathematics Subject Classifications:} 35L99, 35D05.
\hfil\break\indent
{\em Key words and phrases:} Non-local problem, generalized solution.
\hfil\break\indent
\copyright 1999 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted July 29, 1999. Published November 15, 1999.} }
\date{}
%
\author{L. S. Pulkina}
\maketitle

\begin{abstract} 
A linear second-order hyperbolic equation with forcing
and integral constraints on the solution is converted to a
non-local hyperbolic problem. Using the Riesz representation 
theorem and the Schauder fixed point theorem, we prove the 
existence and uniqueness of a generalized solution. 
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}


\section{Introduction}
Certain problems arising in: plasma physics [1], heat conduction [2, 3],
dynamics of ground waters [4, 5], thermo-elasticity [6], can be reduced to the
non-local problems with integral conditions. The above-mentioned papers 
consider problems with parabolic equations. However, some problems concerning
the dynamics of ground waters are described in terms of hyperbolic equations [4].
Motivated by this, we study the equation
\begin{equation}
\label{eq1}
Lu\equiv u_{xy} + A(x,y)u_x + B(x,y)u_y + C(x,y)u = f(x,y)
\end{equation}   
with smooth coefficients in the rectangular domain 
$$ D=\{ (x,y): 0<x<a, 0<y<b \}, $$
bounded by the characteristics of equation (\ref{eq1}), with the conditions
\begin{equation}
\label{cond2}
\int_0^\alpha u(x,y)\,dx=\psi (y), \quad \int_0^\beta u(x,y)\,dy=\phi(x).
\end{equation}
where $\phi (x)$, $\psi (y)$ are given functions and $ 0<\alpha <a, 0<\beta <b$.
The special case $\alpha =a$, $\beta =b$ is considered by author in [7].
The consistency condition assumes the form
$$ \int_0^\alpha \phi (x)\,dx = \int_0^\beta \psi(y)\,dy.$$

\section{A problem for a loaded equation}

Since the integral conditions (\ref{cond2}) are not homogeneous, we construct 
a function $ K(x,y)={1\over \alpha}\psi (y)+{1\over \beta}\phi(x) $
$ - {1\over \alpha \beta}\int_0^\alpha \phi (x)\,dx$, satisfying the conditions 
(\ref{cond2}), and
introduce a new unknown function $ \bar u(x,y)=u(x,y)-K(x,y)$.
Then  (\ref{eq1}) is converted into a similar equation $ L\bar u=\bar f$, 
where $\bar f=f-LK$, while the corresponding integral data are now 
 homogeneous. Now we construct another function
$$ 
M(x,y)={1\over a}\int_\alpha^a \bar u(x,y)\,dx+{1\over b}\int_\beta^b 
\bar u(x,y)\,dy-{1\over ab}\int_\beta^b\int_\alpha^a\bar u(x,y)\,dx\,dy\,,
$$
which satisfies the  conditions 
$$\int_0^a M(x,y)\,dx=\int_\alpha^a\bar u(x,y)\,dx,
\quad  \int_0^b M(x,y)\,dy=\int_\beta^b\bar u(x,y)\,dy\,.$$
Let $\bar u(x,y)=w(x,y)+M(x,y)$, where $ w(x,y)$ satisfies a differential 
equation to be determined.
To find the form of this equation, we consider the previous
equality as an integral equation with respect to $ \bar u $
\begin{equation}
\label{eq3}
\bar u(x,y)-{1\over a}\int_\alpha^a\bar u(x,y)\,dx-{1\over b}
\int_\beta^b\bar u(x,y)\,dy+{1\over ab}\int_\beta^b\int_\alpha^a 
\bar u(x,y)\,dx\,dy = w(x,y)\,.
\end{equation}
It is not difficult to show that
\begin{equation}
\label{eq4}
\bar u(x,y)=w(x,y)+{1\over \alpha} \int_\alpha^aw(x,y)\,dx
+{1\over \beta}\int_\beta^bw(x,y)\,dy+{1\over \alpha \beta}
\int_\beta^b\int_\alpha^aw(x,y)\,dx\,dy\,.
\end{equation}
If we substitute (\ref{eq4}) into the left-hand side of the equation 
$ L\bar u=\bar f$, then we obtain the so called loaded equation with respect 
to $ w(x,y) $,
\begin{eqnarray}
\label{eq5}
\bar Lw\equiv w_{xy}+A(w+{1\over \beta} \int_\beta^b w(x,y)\,dy)_x+
B(w+{1\over \alpha} \int_\alpha^a w(x,y)\,dx)_y \nonumber\\
+C(w+{1\over \alpha} \int_\alpha^a w(x,y)\,dx+{1\over \beta}\int_\beta^b w(x,y)
\,dy \\
+{1\over \alpha \beta}\int_\beta^b\int_\alpha^a w(x,y)\,dx\,dy)&=&\bar f(x,y)
\nonumber \end{eqnarray}
and integral conditions
\begin{equation}
\label{eq6}
\int_0^a w(x,y)\,dx=0, \int_0^b w(x,y)\,dy=0.
\end{equation}

\section{Generalized solution}

Define the function $S$ by
 \begin{eqnarray*}
Sw&=&A(w+{1\over \beta}\int_\beta^b w\,dy)_x + B(w+{1\over\alpha}
\int_\alpha^a w\,dx)_y \\
&&+C(w+{1\over \alpha}\int_\alpha^a w\,dx+{1\over \beta}\int_\beta^b w\,dy
+{1\over\alpha\beta}\int_\beta ^b\int_\alpha^a w\,dx\,dy) 
\end{eqnarray*}
and $ F(x,y,Sw)=\bar f(x,y)-Sw$. Then (\ref{eq5}) can be assumed to have the
form
$$ w_{xy}=F(x,y,Sw). $$
We introduce the function space
$$ 
V=\{ w: w\in C^1(\bar D),\exists w_{xy} \in C(\bar D), 
\int_0^a w\,dx=\int_0^b w\,dy=0\}\,. 
$$
The completion of this  space, with respect to the norm
$$
\|w\|^2_1=\int_0^b\int_0^a (w^2+w_x^2+w_y^2)\,dx\,dy 
$$
is denoted by $ \tilde {H}^1(D)$. Notice that $ \tilde{H}^1(D)$ is Hilbert 
space with 
$$ (w,v)_1 = \int_0^b\int_0^a (wv+w_xv_x+w_yv_y)\,dx\,dy\,. $$
For $ v\in \tilde{H}^1$ define the operator $ l$ by
$$ 
lv\equiv \int_0^y v_x(x,\tau)d\tau + \int_0^x v_y(t,y)dt - 
\int_0^y \int_0^x v(t,\tau)\,dt\,d\tau \,.
$$
Consider the scalar product $ (w_{xy}, lv)_{L_2}$. Employing integration by parts and taking account
of $ w\in V, v\in \tilde{H}^1$, we can see that $ (w_{xy}, v)_{L_2}=(w, v)_1$.

\paragraph{Definition.} A function $ w \in \tilde {H}^1(D)$ is called a
generalized solution of the problem (\ref{eq5})-(\ref{eq6}), if 
$ (w,v)_1=(F(x,y,Sw), lv)_{L_2} $
for every $ v \in \tilde{H}^1(D)$.

\section{Subsidiary problem}

Consider the problem with integral conditions (\ref{eq6}) for the equation
$$ w_{xy}=F(x,y).$$

\begin{theorem} Let $ F(x,y)\in L_2(D)$. Then there exists one and only one 
generalized solution $ w_0$ of the problem 
$$ \displaylines{
 w_{xy} = F(x,y) \cr
 \int_0^a w\,dx = 0, \quad \int_0^b w\,dy = 0,
} $$
where for some positive constant $c_1$,
\begin{equation}
\label{eq7}
c_1\|w_0\|_1 \leq \|F\|_{L_2}\,.
\end{equation}
\end{theorem}

\paragraph{Proof.} 
For $F(x,y)\in L_2(D)$, $\Psi(v)= (F, lv)_{L_2}$ is a bounded linear
functional on $ \tilde{H}^1(D)$. Indeed, 
$$ |(F, lv)|\leq \|F\|_{L_2}\|lv\|_{L_2}
\leq 3\max\{a^2,b^2,a^2b^2\}\|F\|_{L_2}\|v\|_1.$$
Thus by the Riesz-representation theorem there exists a unique
$ w_0 \in \tilde {H}^1(D) $
such that $ \Psi (v) = (F, lv)_{L_2} = (w_0, v)_1$. 
Hence $ (w, v)_1 = (w_0, v)_1 $
for every $ v \in \tilde{H}^1(D)$, i.e., $w_0$ is generalized solution.
Letting $ {1\over c_1} = 3 \max\{ a^2, b^2, a^2 b^2\}$, we obtain 
inequality (\ref{eq7}). \hfill$\diamondsuit$

\begin{lemma} Operator $S: \tilde H^1 \rightarrow L_2$ is bounded,
that is, there exists a positive constant $c_2$ such that 
$ \|Sw\|_{L_2}\leq c_2 \|w\|_1$.
\end{lemma}

\paragraph{Proof.} Let $ |A(x,y)|\leq A_0$, $|B(x,y)|\leq B_0$, and
$|C(x,y)|\leq C_0$.
Then $ Sw=A\bar u_x+B\bar u_y+C\bar u$, and
\begin{eqnarray*}
 \|Sw\|^2_{L_2}&=&\int_0^b\int_0^a (A\bar u_x+B\bar u_y+C\bar u)^2\,dx\,dy  \\
&\leq& 3(A_0^2\|\bar u_x\|^2_{L_2}+B_0^2\|\bar u_y\|^2_{L_2}
+C_0^2\|\bar u\|^2_{L_2})\,.
\end{eqnarray*}
Now by straightforward calculation, using the inequality
$ 2ab\leq a^2+b^2$, and H\"older's inequality, we find that
$$\displaylines{
 \|\bar u\|_{L_2}^2 \leq c_3\|w\|^2_{L_2}, \cr
\mbox{with } c_3=4\biggl(1+\frac{(a-\alpha )a}{\alpha^2}
+\frac{(b-\beta)b}{\beta^2}+
\frac{(b-\beta)(a-\alpha)ab}{\alpha^2\beta^2}\biggr); \cr
 \|\bar u_x\|^2_{L_2}\leq c_4\|w_x\|^2_{L_2}, 
\mbox{ with } c_4=2\biggl(1+\frac{(b-\beta)b}{\beta^2}\biggr);\cr
 \|\bar u_y\|^2_{L_2}\leq c_5\|w_y\|^2_{L_2}, 
\mbox{ with } c_5=2\biggl(1+\frac{(a-\alpha)a}{\alpha^2}\biggr)\,.
}$$
Hence $ \|Sw\|^2_{L_2}\leq c_2 \|w\|^2_1$, where $ c_2=3\max\{A_0^2c_4,\ 
B_0^2c_5, \ C_0^2c_3\}$. 
Indeed, 
\begin{eqnarray*}
\|Sw\|^2_{L_2}&\leq& 3(A_0^2c_4\|w_x\|^2_{L_2}+B_0^2c_5\|w_y\|^2_{L_2}
+C_0^2c_3\|w\|^2_{L_2}) \\
&\leq& c_2(\|w_x\|^2_{L_2}+\|w_y\|^2_{L_2}+\|w\|^2_{L_2})\\
&=&c_2\|w\|^2_1\,.
\end{eqnarray*}
\ \hfill$\diamondsuit$\smallskip
  
  
As $ S$ is linear $S(\sqrt{2} \lambda w)={\sqrt{2}\lambda} S(w)$
for arbitrary $\lambda$.  Let $ \lambda > {1\over c_1}$, and let
$$ S_\lambda (w)=S(\sqrt 2 \lambda w)\,.$$ 

\begin{theorem} If $ \bar f(x,y) \in L_2(D)$ and 
$|\bar f(x,y)|\leq {P\over \sqrt 2}$, then there exists at least one
generalized solution $ w_0 \in \tilde {H}^1(D)$ to 
problem~(\ref {eq5})-(\ref{eq6}), where 
$\|w_0\|^2_1 \leq {P^2 \over \eta ^2}$, with
$ \eta ^2 = c_1^2 - {1\over \lambda ^2}$. 
Furthermore, the solution is uniquely determined, if $c_2<c_1$.
\end{theorem}

\paragraph{Proof.} Consider the closed ball 
$$ W = \{ S_\lambda\omega :  S_\lambda\omega \in L_2(D),\  
\|S_\lambda \omega \|^2_{L_2}\leq {{P^2 ab}\over \eta ^2}\}\,. 
$$ 
Then 
$$ |F(x,y,S\omega)|\leq |\bar f(x,y)|+ \sqrt{{c_1^2-\eta^2}\over 2} 
|S_\lambda \omega|\,, 
$$ 
and for all $ S_\lambda \omega \in W$ we have
$$ \|F(x,y,S \omega)\|^2 \leq {{c_1^2 P^2ab}\over \eta^2}. $$ 

From Theorem 1 there exists a unique generalized solution of the problem
$$ w_{xy}=F(x,y,S\omega),\ \int_0^aw(x,y)\,dx = 0, \ \int_0^bw(x,y)\,dy = 0 
$$
so that $ (w,v)_1 = (F, lv)_{L_2}$ and 
$ \|w\|^2_1 \leq {1\over{c_1^2}}\|F\|^2 \leq {{P^2 ab}\over \eta^2}$.
Define an operator $ T: S\omega \in W \rightarrow w=TS\omega \in \tilde{H}^1(D)$,
$T(W)\subset W$. Notice that $T$ is a continuous operator. To see this, let
$ (S\omega)_n, \ (S\omega)_0 \in W$ and 
$ \|(S\omega)_n - (S\omega)_0\|\rightarrow 0 $
as $ n\rightarrow \infty$. Then for $ w_n=T(S\omega)_n, \ w_0=T(S\omega)_0$ 
we have
$$ (w_n-w_0, v)=(F(x,y,(S\omega)_n)-F(x,y,(S\omega)_0), 
\ lv)_{L_2}=((S\omega)_n-(S\omega)_0, lv)_{L_2}\,.$$
Now from Theorem~1
$$ \|w_n-w_0\|_1 \leq {1\over c_1}\|(S\omega)_n - (S\omega)_0\|_{L_2} 
\rightarrow 0,\ \ n\rightarrow \infty\,. $$
Furthermore, $T$ is a compact operator. In order to show this, we take a 
sequence $ \{(S\omega)_n\} \subset W$, that is 
$ \|(S\omega)_n\|^2_{L_2}\leq {P^2ab\over \eta^2}$.
For $ w_n=T(S\omega)_n$ we have $\|w_n\|^2 \leq {P^2ab\over \eta^2}$, so a 
sequence $ \{ w_n\}$ is bounded in $ \tilde {H}^1(D)$, therefore there exists 
a subsequence weakly convergent in $ \tilde {H}^1(D)$. Since  any bounded set in
$ \tilde {H}^1$ is compact in $ L_2$, then there exists a subsequence, which 
we again denote by $ \{ w_n\}$, strongly convergent in $ L_2(D)$ to $w_0$, as
$ n\rightarrow \infty $.  Now $w_0$ satisfies 
the inequality $\|w_0\|^2_{L_2}\leq P^2ab/ \eta^2$.
As $ S$ is a bounded operator, $T$ is completely continuous and so $TS$ is
completely continuous. Thus from Schauder's fixed-point theorem there exists 
at least one $ w_0 \in W$ such that $ w_0=TSw_0$ and
$$ (w_0, v)_1 = (F(x,y,Sw_0), lv)_{L_2} 
$$ 
for all$ v \in \tilde {H}^1(D)$. 

Assume that $ w_1, \ w_2$ are distinct generalized solutions, then
$$ (w_1-w_2, v)_1=(F(x,y,Sw_1)-F(x,y,Sw_2), lv)_{L_2}. $$
>From (\ref{eq7}) and Lemma 1 we have
$$ \|w_1-w_2\|_1 \leq {1\over c_1}\|Sw_1-Sw_2\|_{L_2}
\leq {c_2\over c_1}\|w_1-w_2\|_1.
$$
Thus, if $ c_2<c_1$ then it gives a contradiction; therefore, $ w_1=w_2$.


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\end{thebibliography}


\noindent{\sc Ludmila S. Pulkina}  \\
Department of Mathematics \\ 
Samara State University \\
443011, 1, Ac.Pavlov st.\\
Samara, Russia. \\
e-mail: pulkina@ssu.samara.ru \& louise@valhalla.hippo.ru

\end{document}