\documentclass[reqno]{amsart} 

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2001(2001), No. 76, 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 2001 Southwest Texas State University.} 
\vspace{1cm}}

\begin{document} 

\title[\hfilneg EJDE--2001/76\hfil 
Wave equation with a nonlocal condition]
{Existence of solutions for one-dimensional wave equations with
nonlocal conditions} 

\author[Sergei A. Beilin\hfil EJDE--2001/76\hfilneg]
{ Sergei A. Beilin }

\address{Sergei A. Beilin \hfill\break
Department of Mathematics\\
Samara State University\\
1, Ac.Pavlov st.\\
443011 Samara Russia}
\email{sbeilin@narod.ru, awr@ssu.samara.ru}

\date{}
\thanks{Submitted August 21, 2001. Published December 10, 2001.}
\subjclass[2000]{35L99, 35L05, 35L20}
\keywords{Mixed problem, non-local conditions, wave equation}


\begin{abstract}
  In this article we study an initial and boundary-value problem 
  with a nonlocal integral condition for a one-dimensional wave equation.
  We prove existence and uniqueness of classical solution and find
  its Fourier representation. The basis used consists of a 
  system of eigenfunctions and adjoint functions.
\end{abstract}

\maketitle

\newtheorem{theorem}{Theorem}[section] % theorems numbered with section #
\newtheorem{lemma}[theorem]{Lemma}
\numberwithin{equation}{section}
\renewcommand{\phi}{\varphi}

\section{Introduction}

Certain problems of modern physics and technology can be
effectively described in terms of nonlocal problems for
partial differential equations. These nonlocal conditions arise
mainly when the data on the boundary cannot be measured directly.

The first paper, devoted to second-order partial differential 
equations with non-local integral conditions goes back to Cannon \cite{Cannon}.
Later, the problems with non-local integral conditions for
parabolic equations were investigated by Kamynin \cite{Kamynin}, Ionkin \cite{Ionkin}, 
Yurchuk \cite{Yurchuk}, Bouziani \cite{Bouziani1}; 
problems for elliptic equations with operator nonlocal conditions were 
considered by Mikhailov and Guschin \cite{MikhailovGuschin}, Scubachevski \cite{Scubachevski}, Paneiah \cite{Paneiah}.

Then, Gordeziani and Avalishvili \cite{GordezianiAvalishvili}, Bouziani \cite{Bouziani2} devoted a few
papers to nonlocal problems for hyperbolic equations. Pulkina \cite{PulkinaEJDE,PulkinaDiffUr} studied the nonlocal 
analogue to classical Goursat problem. 

In this paper we investigate the nonlocal analogue to classical mixed problem, which
involves initial, boundary and nonlocal integral conditions.
In the rectangular domain $ D=\{(x,t): 0<x<l, \ 0<t<T\}$, we
consider the equation
\begin{equation}
\label{eq1}
\mathcal{L}U\equiv U_{tt}-U_{xx}=F(x,t)
\end{equation}
with initial data
\begin{equation}
\label{eq2}
U(x,0)=\Phi(x), \quad U_t(x,0)=\Psi(x),
\end{equation}
Dirichlet boundary condition
\begin{equation}
\label{eq3}
U(0,t)=0
\end{equation}
and the  nonlocal condition
\begin{equation}
\label{eq4}
\int_0^lU(x,t)\,dx=0,
\end{equation}
where $ \Phi(x)$, $\Psi(x) $ are given, $\Phi(x)\in C[0,l]\cap C^2(0,l)$,
$\Psi(x)\in C[0,l]\cap C^1(0,l) $ and satisfy the compatibility conditions
$$ 
\Phi(0)=0, \quad \Psi(0)=0, \quad \int_0^l\Phi(x)\,dx=\int_0^l\Psi(x)\,dx=0.
$$
Note that we do not lose generality by assuming that (\ref{eq3}) and 
(\ref{eq4}) are homogeneous. Indeed, if $ U(0,t)=m(t)$ and  
$ \int_0^lU(x,t)\,dx=n(t)$, we 
introduce a new unknown function $ v(x,t)=U(x,t)-W(x,t)$, where
$$ 
W(x,t)=(1-\frac{2x}{l})m(t)+\frac{2x}{l^2}n(t).
$$
Then (\ref{eq1}) is converted into the similar equation
$$ 
v_{tt}-v_{xx}=g(x,t), \quad g(x,t)=F(x,t)-\mathcal{L}W,
$$
while the Dirichlet and integral conditions are now homogeneous.

The presence of integral conditions complicates the application of standard 
techniques.
Therefore, we first reduce (\ref{eq1})-(\ref{eq4}) to an equivalent
problem. 

\begin{lemma} \label{lm1} %{\bf Lemma 1.} 
Problem  (\ref{eq1})-(\ref{eq4}) is equivalent to (\ref{eq1})-(\ref{eq3}) and
\begin{equation}
\label{eq5}
U_x(0,t)-U_x(l,t)=\int_0^lF(x,t)\,dx.
\end{equation}
\end{lemma}

\noindent{\bf Proof.} Let $ U(x,t)  $ is a solution of (\ref{eq1})-(\ref{eq4}).
Integrating (\ref{eq1}) with respect to $ x $ over $ (0,l)$, and taking in 
account (\ref{eq4}), we obtain
$$ U_x(0,t)-U_x(l,t)=\int_0^lF(x,t)\,dx.
$$
Let now $ U(x,t) $ be a solution of (\ref{eq1})-(\ref{eq3}), (\ref{eq5}).
We need only to show that $ \int_0^lU(x,t)\,dx=0$.
For this end we integrate again (\ref{eq1}) and obtain
$$ \frac{d^2}{dt^2}\int_0^lU(x,t)\,dx=0.
$$
By virtue of the compatibility conditions,
$$ \int_0^lU(x,0)\,dx=0, \quad \int_0^lU_t(x,0)\,dx=0\,.
$$
Then $ \int_0^lU(x,t)\,dx=0 $ is a unique solution to homogeneous Cauchy 
problem. \hfill$\Box$ \medskip

Introduce a new unknown function $ u(x,t)=U(x,t)-w(x,t)$, where
$ w(x,t)=-\frac{x^2}{2l}\int_0^lF(x,t)\,dx.$
Then  (\ref{eq1})-(\ref{eq3}), (\ref{eq5}) is transformed  now into 
\begin{gather}
\label{eq6}
u_{tt}-u_{xx}=g(x,t), \\
\label{eq7}
u(x,0)=\varphi(x), \quad  u_t(x,0)=\psi(x), \\
\label{eq8}
u(0,t)=0, \\
\label{eq9}
u_x(0,t)=u_x(l,t),
\end{gather}
where
\begin{gather*}
 g(x,t)=F(x,t)+\frac{x^2}{2l}\int_0^lF_{tt}(x,t)\,dx
 -\frac{1}{l}\int_0^lF(x,t)\,dx, \\
\varphi(x)=\Phi(x)+\frac{x^2}{2l}\int_0^lF(x,0)\,dx, \\
\psi(x)=\Psi(x)+\frac{x^2}{2l}\int_0^lF_t(x,0)\,dx.
\end{gather*}

\section{Uniqueness}

\begin{theorem} \label{thm1}
There exists at most one solution to  (\ref{eq6})-(\ref{eq9}).
\end{theorem}

\noindent{\bf Proof.} 
Let $ u_1(x,t)$,  $u_2(x,t)$ be two different solutions of 
(\ref{eq6})-(\ref{eq9}). Then $ u(x,t)=u_1(x,t)-u_2(x,t) $ is a nontrivial 
solution to the homogeneous problem
\begin{gather*}
 u_{tt}-u_{xx}=0, \\
u(x,0)=0, \quad  u_t(x,0)=0, \\
u(0,t)=0, \quad u_x(0,t)=u_x(l,t).
\end{gather*}
As $ u\in C^1(\bar D)\cap C^2(D)$, then $ u(x,t)$ takes on certain value 
for $ x=l$.
Let $ u(l,t)=\mu(t)$. Consider mixed problem for the equation $ u_{tt}-u_{xx}=0 $
with homogeneous initial data and the boundary conditions
$$ u(0,t)=0, \quad u(l,t)=\mu(t).
$$
Note, that $ \mu(t)$ is required to satisfy the compatibility 
conditions $ \mu(0)=0$ and $\mu '(0)=0$.

For all conditions to be homogeneous, we let $ \tilde u=u-\frac{x}{l}\mu(t)$.
Then, taking in account the compatibility conditions for $ \mu(t)$, we obtain
\begin{gather*}
 \tilde u_{tt}-\tilde u_{xx}=\frac{x}{l}\mu ''(t), \\
 \tilde u(x,0)=0, \ \ \tilde u_t(x,0)=0, \\
 \tilde u(0,t)=0, \ \ \tilde u(l,t)=0.
\end{gather*}
It is well known that there exists unique solution $ \tilde u(x,t) $ to this
 problem \cite{Bizadze}, hence $ u(x,t) $ assumes the form
$$
 u(x,t)=\frac{2l}{\pi^2}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}\left( \int_0^t\mu ''(\tau)
\sin {\frac{k\pi (t-\tau)}{l}}\,d\tau \right) \sin{\frac{k\pi x}{l}}
+\frac{x}{l}\mu(t).
$$
Now we find that
\begin{gather*} 
u_x(0,t)=\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\int_0^t
\mu ''(\tau)\sin{\frac{k\pi (t-\tau)}{l}}\,d\tau +\frac{1}{l}\mu(t),
\\
 u_x(l,t)=\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{1}{k}\int_0^t
\mu ''(\tau)\sin{\frac{k\pi (t-\tau)}{l}}\,d\tau +\frac{1}{l}\mu(t)
\end{gather*}
and consider
$$ u_x(l,t)-u_x(0,t)=\frac{4}{\pi}\int_0^t\mu ''(\tau)\sum_{m=1}^{\infty}
\frac{1}{2m-1}\sin{\frac{(2m-1)\pi(t-\tau)}{l}}\,d\tau.$$
As in \cite{GradshteinRyzhik}
$$ \sum_{m=1}^{\infty}\frac{\sin{(2m-1)x}}{2m-1}=\left\{
\begin{array}{ll}
\pi/4, & \mbox{if } 0 < x <\pi \\[2pt]
-\pi/4, & \mbox{if } \pi < x< 2\pi\,.
\end{array} \right. 
$$
Then by  (\ref{eq9}) we can  write
$$ 0=|u_x(l,t)-u_x(0,t)|=\int_0^t\mu ''(\tau)\,\,d\tau.
$$
Taking into account the compatibility conditions $ \mu (0)=\mu '(0)=0$, we easily 
obtain $ \mu(t)\equiv 0$.  Now from the uniqueness theorem \cite{Bizadze}, 
we obtain $ u(x,t)\equiv 0$. \hfill$\Box$

\section{Existence}

Obviously, the solution to the problem (\ref{eq6})-(\ref{eq9}), if it exists, 
is a sum of solutions to the following two problems:\\
{\bf Problem $\mathcal{H}$ }
\begin{gather*} 
 u_{tt}-u_{xx}=0,\\
 u(x,0)=\varphi(x), \quad u_t(x,0)=\psi(x), \\
 u(0,t)=0, \quad u_x(0,t)=u_x(l,t) 
\end{gather*}
{\bf Problem $\mathcal{N}\mathcal{H}$}
\begin{gather*}
 u_{tt}-u_{xx}=g(x,t), \\
 u(x,0)=u_t(x,0)=0, \quad u(0,t)=0, \quad u_x(0,t)=u_x(l,t).
\end{gather*}

First consider problem $\mathcal{H}$ and use separation of variables. 
Let $ u(x,t)=X(x)T(t)$. Substituting in the equation $ u_{tt}-u_{xx}=0 $ 
and taking into account (\ref{eq8}), (\ref{eq9}), we obtain 
\begin{equation}
\label{eq10}
X''(x)+\lambda X(x)=0,\quad  X(0)=0,\quad  X'(0)=X'(l).
\end{equation}
Note that problem (\ref{eq10}) is not self-adjoint: The adjoint problem is 
\begin{equation}
\label{eq11}
Y''(x)+\bar \lambda Y(x)=0,\quad  Y'(l)=0,\quad  Y(l)=Y(0).
\end{equation}
The eigenvalues and eigenfunctions of problem (\ref{eq10}) are
\begin{gather}
\label{eq12}
\lambda_k=({2\pi k \over l})^2,\quad  k=1,2,\ldots \\
\label{eq13}
X_0=x,\quad X_k=\sin{2\pi kx \over l}
\end{gather}
respectively.
Note, that for $ k>0 $ the functions (\ref{eq13}) are not orthonormal with 
$ X_0$. To construct a basis in $ L_2 $, we complete (\ref{eq13}) by
using adjoint functions.

Following  M. Keldysh \cite{Keldysh}, we define an adjoint function
 $ \tilde X_k$, corresponding
eigenvalue $ \lambda_k $ from (\ref{eq12}), as a solution to the boundary-valued problem
\begin{equation}
\label{eq14}
\tilde X''_k(x)+\lambda_k\tilde X_k(x)=-2\sqrt{\lambda_k}X_k(x), 
\quad \tilde X_k(0)=0, \quad \tilde X'_k(0)=\tilde X'_k(l).
\end{equation}
We obtain 
$$ \tilde X_k(x)=x\cos{\frac{2\pi kx}{l}}, \quad k=1,2,\ldots
$$
Rewrite now a system of eigenvalue and adjoint functions of (\ref{eq10}) as
\begin{equation}
\label{eq15}
X_0=x, \quad X_{2k-1}(x)= x\cos{\frac{2\pi kx}{l}}, 
\quad X_{2k}(x)=\sin{\frac{2\pi kx}{l}}.
\end{equation}
In a similar way we find the system of eigenvalue and adjoint functions 
(\ref{eq11}):
\begin{equation}
\label{eq16}
 Y_0(x)=\frac{2}{l^2},  \quad Y_{2k-1}(x)=\frac{4}{l^2}\cos{\frac{2\pi kx}{l}},
 \quad  Y_{2k}(x)=\frac{4(l-x)}{l^2}\sin{\frac{2\pi kx}{l}}\,,
\end{equation}
where for every $ \lambda_k$ with  $k>0$,  $X_{2k}(x)$,  $Y_{2k}(x) $ are 
eigenvalue functions, $ X_{2k-1}(x)$, $Y_{2k-1}(x)$ are adjoint functions 
of the problems (\ref{eq10}) and (\ref{eq11}) respectively.
Direct calculations show that (\ref{eq15}) and (\ref{eq16}) form a biorthogonal 
system for $ x\in (0,l)$:
$$ (X_i,Y_j)=\int_0^lX_i(x)Y_j(x)\,dx=\delta_{ij}.
$$

As it was shown in \cite{Ilyin} the system (\ref{eq15}) is complete and 
forms a basis in $ L_2(0,l)$.
Hence, an arbitrary function $ f(x)\in L_2(0,l) $ may be expanded as 
$$ 
f(x)=A_0X_0(x)+\sum_{k=1}^{\infty}(A_{2k}X_{2k}(x)+A_{2k-1}X_{2k-1}(x)),
$$
where
\begin{equation}
\label{eq17}
 A_i=\int_0^lf(x)Y_i(x)\,dx. 
\end{equation}

Returning to the separation variables technique,  for $T(t) $ we obtain
$$ 
T_k(t)=a_k\sin{\frac{2\pi kt}{l}}+b_k\cos{\frac{2\pi kt}{l}}.
$$
We assume now that a solution to $ \mathcal{H} $ is of the form
\begin{equation}\label{solutionH}
 u(x,t)=A_0X_0+\sum_{k=1}^{\infty}\left( (A_{2k}X_{2k}+A_{2k-1}X_{2k-1})T_k-
\frac{lt}{2\pi k}A_{2k-1}X_{2k}T'_k\right).
\end{equation}
Substitute $ T_k(t) $ and rewrite the coefficients. Then 
\begin{equation} \begin{aligned}
 u(x,t)=&C_0X_0+\sum_{k=1}^{\infty}( X_{2k}(C_{2k}\sin{\frac{2\pi kt}{l}}+
D_{2k}\cos{\frac{2\pi kt}{l}}) \\
&+X_{2k-1}(C_{2k-1}\sin{\frac{2\pi kt}{l}}+D_{2k-1}\cos{\frac{2\pi kt}{l}}) 
\\
&-tX_{2k}(C_{2k-1}\cos{\frac{2\pi kt}{l}}-D_{2k-1}\sin{\frac{2\pi kt}{l}})).
\end{aligned}
\end{equation}
The initial data (\ref{eq7}) give us the following two equalities
\begin{gather*}
 \varphi(x)=C_0X_0+\sum_{k=1}^{\infty}(D_{2k}X_{2k}+D_{2k-1}X_{2k-1}),\\
 \psi(x)=\sum_{k=1}^{\infty}\left( (\frac{2\pi k}{l}C_{2k}-C_{2k-1})X_{2k}
+\frac{2\pi k}{l}C_{2k-1}X_{2k-1}\right), 
\end{gather*}
and the coefficients can be found via formula (\ref{eq17}).

Assume a solution to the problem $ \mathcal{N}\mathcal{H} $ is of the form
\begin{equation}\label{solutionNH}
u(x,t)=V_0(t)X_0(x)+\sum_{k=1}^{\infty} \left( V_{2k}(t)X_{2k}(x)+V_{2k-1}(t)X_{2k-1}(x) \right),
\end{equation}
where $ V_i(t) $ are unknown coefficients satisfying the initial conditions
$ V_i(0)=V'_i(0)=0$.
Substitute (\ref{solutionNH}) into the equation $ u_{tt}-u_{xx}=g(x,t)$, 
where $ g(x,t) $ has been expanded as a biorthogonal series:
$$ 
g(x,t)=g_0(t)X_0(x)+\sum_{k=1}^{\infty}(g_{2k}(t)X_{2k}(x)
+g_{2k-1}(t)X_{2k-1}(x)),
$$
with coefficients
$$ g_i(t)=\int_0^lg(x,t)Y_i(x)\,dx, \quad i=0,1,\dots 
$$
We obtain
\begin{eqnarray*}
\lefteqn{ V''_0(t)x+\sum_{k=1}^{\infty}\left(V''_{2k}(t)
+\frac{4\pi^2k^2}{l^2}V_{2k}(t)\right)\sin{\frac{2\pi kx}{l}} 
}\\
\lefteqn{+\sum_{k=1}^{\infty}\left(V''_{2k-1}(t)+\frac{4\pi^2k^2}{l^2}V_{2k-1}(t)\right)
x\cos{\frac{2\pi kx}{l}}  }\\
\lefteqn{ +\sum_{k=1}^{\infty}V_{2k-1}(t)\frac{4\pi k}{l}\sin{
\frac{2\pi kx}{l}} }\\
&=& g_0(t)X_0(x)+\sum_{k=1}^{\infty}(g_{2k}(t)X_{2k}(x)+g_{2k-1}(t)
X_{2k-1}(x)).
\end{eqnarray*}
Thus we have a Cauchy problem for the system of ordinary differential equations
\begin{gather*}
 V''_0(t)=g_0(t) \\
 V''_{2k}+\frac{4\pi k}{l}(\frac{\pi k}{l}V_{2k}(t)+V_{2k-1}(t))=g_{2k}(t) \\
 V''_{2k-1}(t)+\frac{4\pi^2k^2}{l^2}V_{2k-1}(t)=g_{2k-1}(t)
\end{gather*} 
with initial data
$$ V_0(0)=V'_0(0)=0, \quad V_{2k}(0)=V'_{2k}(0)=0, \quad 
V_{2k-1}(0)=V'_{2k-1}(0)=0,
$$
which has a unique solution 
\begin{gather*}
 V_0(t) = \int_0^t (t-\tau) g_0(\tau) \,d\tau, \\
 V_{2k-1}(t) = {1\over k\pi} \int_0^t g_{2k-1}(\tau) 
 \sin {k\pi (t-\tau) \over l} \,d\tau,  \\
 V_{2k}(t) = {1\over k\pi} \int_0^t \left( g_{2k}(\tau) 
 - 4\pi k V_{2k-1}(\tau) \right) \sin {k\pi (t-\tau) \over l}\,d\tau. 
\end{gather*}


\begin{theorem} \label{thm2}
 Let:
\begin{enumerate}
\item $ g(x,t) \in C^2(D)$, $g_x(x,t) \in C[0,l]$ for all $t\in (0,T)$, 
$|g(x,t)| \leq P, (x,t) \in D $

\item $ \varphi \in C[0,l] \cap C^2(0,l)$,  $\psi \in C[0,l]$, $\phi(0)=0$, 
$\phi'(0)=\phi'(l)$,  $\psi(0) = 0$.
\end{enumerate}
Then there exists the solution to  (\ref{eq6})--(\ref{eq9}),
$$
u(x,t) \in C(\bar D) \cap C^1(\bar D \setminus \{t=T\}) \cap C^2(D) 
$$
which has the form of a sum of (\ref{solutionH}) and (\ref{solutionNH}).
\end{theorem}

\noindent{\bf Series Proof.} 
It is sufficient to prove uniform convergence of the series (\ref{solutionH}) 
and (\ref{solutionNH})
and the series, obtained with formal differentiation.
Let $| \phi'(x)|\leq M_1$,  $|\phi''(x)|\leq M_2$, $|\psi(x)|\leq N$, 
$|\psi'(x)|\leq N_1$, $|g_x|\leq P_1$,  $|g_{xx}| \leq P_2$.

Integrating $C_i, \ D_i, \ V_i$ by parts and taking in account the 
abovementioned assumptions, we obtain:
\begin{gather*}
|D_{2k}|\leq {1\over k^2} {l(lM_2+2M_1)\over \pi^2}, \quad 
|D_{2k-1}|\leq {1\over k^2} {M_2 l \over \pi^2},  \\
|C_{2k}|\leq {1\over k^2} {l(N_1+2N) \over 2\pi^2}, \quad 
|C_{2k-1}|\leq {1\over k^2} {N_1 l \over \pi^2}, \\
|V_{2k}|\leq {1\over k^2} {4T^2(2p_1+P_2l) \over \pi^2}, \quad 
|V_{2k-1}|\leq {1\over k^2} {2TP_1 \over \pi^2}, 
\end{gather*}
and hence the series (\ref{solutionH}) and (\ref{solutionNH}) and the series, 
obtained with formal differentiation, converge uniformly. 
\hfill$\Box$

 
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\end{document}