We study initial boundary value problems for linear scalar partial differential equations with constant coefficients, with spatial derivatives of {\em arbitrary order}, posed on the domain $\{t>0,\;0<x<L\}$. We first show that by analysing the so-called {\em global relation}, which is an algebraic relation defined in the complex $k$-plane coupling all boundary values of the solution, it is possible to identify how many boundary conditions must be prescribed at each end of the space interval in order for the problem to be well posed. We then show that the solution can be expressed as an integral in the complex $k$-plane. This integral is defined in terms of an $x$-transform of the initial condition and a $t$-transform of the boundary conditions. For particular cases, such as the case of periodic boundary conditions, or the case of boundary value problems for {\em second} order PDEs, the integral can be rewritten as an infinite series. However, there exist initial boundary value problems for which the only representation is an integral which {\em cannot} be written as an infinite series. An example of such a problem is provided by the linearised version of the KdV equation. Thus, contrary to common belief, the solution of many linear initial boundary value problems on a finite interval {\em cannot} be expressed in terms of an infinite series.
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\begin{document}
\bibliographystyle{plain}
\begin{flushright}
{\small submitted to {\em IMA J. Appl. Math.}}
\end{flushright}

\vspace{3mm}
\begin{center}
\Large
{\bf A transform method for evolution PDEs on a finite interval}

\vspace{3mm}

\large
A.S. Fokas$^{*}$ and B. Pelloni$^{**}$

\vspace{5mm}

*Department of Applied Mathematics and Theoretical Physics

Cambridge University

Cambridge CB3 0WA, UK.

{\em t.fokas@damtp.cam.ac.uk}

\vspace{2mm}

**Department of Mathematics

University of Reading

Reading RG6 6AX, UK

{\em b.pelloni@rdg.ac.uk}

\vspace{3mm}
\today
\end{center}
\normalsize

\begin{abstract}
We study initial boundary value problems for linear scalar partial
differential equations with constant coefficients, with spatial
derivatives of {\em arbitrary order}, posed on the domain
$\{t>0,\;0<x<L\}$.
We first show that by analysing the so-called {\em global relation},
which is an algebraic relation defined in the complex $k$-plane
coupling all boundary values of the solution, it is possible to
identify how many boundary conditions must be prescribed at each end
of the space interval in order for the problem to be well posed. We
then show that the solution can be expressed
as an  integral in the
complex $k$-plane. This
integral is
defined in terms of
an $x$-transform of the initial condition and a  $t$-transform
of the boundary conditions.

For particular cases, such as the case of periodic boundary
conditions, or the case of boundary value problems for {\em
second}
order PDEs, the integral can be rewritten as an infinite series. 
However,
there exist initial boundary value problems for which the only
representation is an integral which
{\em cannot} be written as an infinite series. An
example of such a problem is provided by the linearised version of the
KdV equation. Thus, contrary to common belief,  the solution of many
linear initial boundary value problems on a finite interval {\em
cannot} be expressed in terms of an infinite series.
\end{abstract}

\section{Introduction}
An evolution partial differential equation in one space dimension is 
characterised by its
{\em symbol}, which we denote by $\omega(k)$. This means that a
particular solution of the equation is given by
$$
E(x,t)={\rm e}^{ikx-\omega(k)t}.
%\label{fundsol}
$$
Physically significant examples of scalar evolution equations are:

{\bf (a)} the Schr\"odinger equation with zero potential
\be
iq_{t}+q_{xx}=0,\quad \omega(k)=ik^{2};
\label{eq1}
\ee
{\bf (b)} the diffusion-convection equation
\be
q_{t}=q_{xx}+q_{x},\quad \omega(k)=k^{2}-ik;
\label{eq2}
\ee
{\bf (c)} the Stokes equation
\be
q_{t}+q_{xxx}+q_{x}=0,\quad \omega(k)=i(k-k^{3}).
\label{eq3}
\ee

We note that equations (\ref{eq1}) and (\ref{eq3}) are the linearised
versions of the nonlinear Schr\"odinger and of the Korteweg-deVries
equations respectively.

\subsection*{ Notations}
\begin{itemize}
\item[(i)]
 $q_{0}(x)$ denotes the given initial condition, and
$\hat{q}_{0}(k)$ is the Fourier transform of $q_{0}(x)$:
\be
q_{0}(x)=q(x,0),\;\; 0<x<L;
\;\;\;\hat{q}_{0}(k)=\int_{0}^{L}q_{0}(x){\rm
e}^{-ikx}dx,\;\;k\in{\mathbb C}.
\label{q0}
\ee
\item[(ii)]
$\{f_{j}(t)\}_{0}^{n-1}$ and $\{g_{j}(t)\}_{0}^{n-1}$ denote the
{\em boundary values} of the solution at $x=0$ and $x=L$ 
respectively, while
$\{\tilde{f}_{j}(t,k)\}_{0}^{n-1}$ and
$\{\tilde{g}_{j}(t,k)\}_{0}^{n-1}$ denote certain $t$-transforms of
$f_{j}$ and $g_{j}$:
\be
f_{j}(t)=\partial_{x}^{j}q(0,t),\;\;t>0; \;\;\;
\tilde{f}_{j}(t,k)=\int_{0}^{t}{\rm
e}^{\omega(k)s}f_{j}(s)ds,\;\;k\in{\mathbb C},\;t>0,
\label{fj}
\ee
\be
g_{j}(t)=\partial_{x}^{j}q(L,t),\;\;t>0; \;\;\;
\tilde{g}_{j}(t,k)=\int_{0}^{t}{\rm
e}^{\omega(k)s}g_{j}(s)ds,\;\;k\in{\mathbb C},\;t>0,
\label{gj}
\ee
where $j=0,\ldots,n-1$.
\item[(iii)]
The domain $D$ in the complex $k$-plane is defined by
\be
D=\{k\in{\mathbb C}:\;{\rm Re}\omega(k)\leq 0\}.
\label{Ddef}
\ee
$D^{+}$ and $D^{-}$ denote the part of $D$ in the upper and lower half
of the complex $k$-plane respectively:
\be
D^{+}=\{k\in{\mathbb C}:k\in\,D, \;{\rm Im}(k)\geq 0\},\quad
D^{-}=\{k\in{\mathbb C}:k\in\,D, \;{\rm Im}(k)\leq 0\}.
\label{D+-}
\ee
The oriented boundary of $D^{+}$ and $D^{-}$ is denoted by $\partial
D^{+}$ and $\partial D^{-}$, where the orientation is such that the
interior of the domain $D$ is always on the left of the positive
direction.
\end{itemize}

\subsection*{Statement of the problem and assumptions}
{\em Let $q(x,t)$ satisfy a linear evolution equation with symbol
$\omega(k)$ in the domain
$$\left\{t>0,\quad 0<x<L\right\},$$
where $L$ is a finite positive constant. We assume that $\omega(k)$ 
is a
polynomial of degree $n$
such that the equation $\omega(k)=0$ has $n$ distinct roots, and
that Re$\omega(k)\geq 0$ for $k
\in\,{\mathbb R}$.
Assume that the initial
condition $q_{0}(x)$ is a given, sufficiently smooth, function.

{\bf (i)} Determine the number of boundary conditions that must be
prescribed at $x=0$ and $x=L$ in order to define a well posed problem.

{\bf (ii)} Given appropriate boundary conditions at the two ends of
the space interval, assuming that these given functions have 
sufficient
smoothness and  are compatible with $q_{0}(x)$ at $x=0$ and
$x=L$, construct the solution $q(x,t)$.
}

\subsection*{The classical approaches}
The form of the particular solution $E(x,t)$ suggests that the most 
convenient representation is the one obtained by an $x$-transform.

\vspace{2mm}
{\bf 1. An ${\bf x}$-transform}

\vspace{2mm}
Separation of variables gives rise to an $n$-th order differential 
operator, acting on the interval $0<x<L$.

$\bullet \;{\bf n=2}$

For second order differential 
operators there exists a well known spectral theory which states that 
if one boundary condition is prescribed at each end of the interval, 
then it is always possible to expand a function in terms of the 
complete set of the associated eigenfunctions \cite{tit}. Thus for PDEs of 
second order, using separation of variables it is always possible to 
solve boundary value problems on the finite interval using 
generalised 
Fourier series. For example, it is elementary to solve equation 
(\ref{eq1})  with Dirichlet or Neumann boundary conditions in terms 
of a sine or a cosine series respectively. Similarly, the Dirichlet 
problem for equation (\ref{eq2}) can be solved in terms of  
a generalised Fourier series over the complex eigenvalues 
$\frac{m\pi}{L}+\frac{i}{2}$, $m\in\,{\mathbb Z}$.


$\bullet \;{\bf n=3}$

For third order differential operators, there exists a complete basis 
of eigenfunctions only for {\em particular} types of boundary 
conditions \cite{akhie}. 
These do {\em not} include the uncoupled boundary conditions arising 
in the typical  boundary value problems for third order evolution 
PDEs. Thus it is {\em not} possible to solve equation (\ref{eq3}) 
with uncoupled boundary conditions by an $x$-transform. Furthermore, 
it is {\em not} possible to use the spectral theory of the associated 
differential operators $Lf=f'''+f'$ to decide how many boundary 
conditions must be prescribed at each end of the interval in order to 
guarantee that the resulting \bvp\, is well posed.

\vspace{2mm}
{\bf 2. The Laplace transform}

\vspace{2mm}
The particular solution $E(x,t)$ can be rewritten in the form
${\rm e}^{st+ik(s)x}$, where $k$ satisfies the $n$-th order equation
$$
s+\omega(k)=0.
$$

$\bullet \;{\bf n=2}$

Evolution PDEs with spatial derivatives of up to second order can  
be solved by a {\em Laplace transform}. For example, for 
equation (\ref{eq2}) one can construct a representation involving the 
exponential 
$$
{\rm e}^{st+i[1+(1+4s)^{1/2}]\frac{x}{2}}.
$$
However, it is easier to construct the solution using the associated 
$x$-transform. 

$\bullet \;{\bf n=3}$

The application of the Laplace transform in this case {\em fails}. 
Consider for example equation (\ref{eq3}). First, it is not 
immediately clear how many boundary conditions must be imposed at 
each 
end of the interval to guarantee well posedness. Furthermore, even if 
one prescribes one boundary condition at $x=0$  and two at $x=L$ 
(which is the correct number of conditions), one obtains a
representation for the Laplace transform $\tilde{q}(x,s)$ which has 
singularities along the real line, see Appendix (when
$n=2$, the singularities in the complex $s$-plane are along the half 
line arg$(s)=\varphi$, with $\pi/2 <\varphi<3\pi/2$). 


% Regarding equations (\ref{eq1})-(\ref{eq3}), we make the following
% remarks.
% 
% {\bf (1)} It is elementary to solve equation (\ref{eq1}) with 
%Dirichlet
% boundary  conditions, and the
% solution is given by a sine series. Similarly, the solution of
% the Neumann problem is given by a cosine series.
% 
% {\bf (2)} The Dirichlet problem for equation (\ref{eq2}) can also be
% solved in terms of an infinite series.
% Indeed, separation of variables gives rise to the second order
% operator $Lf=f''+f'$, which for {\em uncoupled} boundary conditions
% {\em cannot} be made self-adjoint. However, an eigenfunction 
%expansion
% corresponding to an
% infinite set of {\em complex} eigenvalues can be found. The
% relevant derivation is classical but tedious, see for example
% \cite{dym}.
% 
% {\bf (3)} For equation (\ref{eq3}), one must first identify the
% number of boundary conditions that must be prescribed at each end in
% order to have a well posed problem. It turns out that in this case
% one boundary condition at $x=0$ and two boundary conditions at $x=L$
% must be prescribed. Separation of variables gives rise to the
% operator $Lf=f'''+f'$, which {\em cannot} be symmetrised. Thus
% equation (\ref{eq3}) cannot be solved by an $x$-transform. One might
% attempt to use the Laplace transform in $t$. However, since $t$ is
% finite and the boundary conditions can have arbitrary growth as
% $t\to\infty$, this is {\em not} an appropriate transform. Even when
% boundary conditions decay as $t\to\infty$, and one can in principle
% use the Laplace transform, the relevant formulation becomes
% intricate (it is necessary to invert a Laplace transform 
%characterised through
% the solution of the cubic equation $k^{3}-k+is=0$).


\vspace{3mm}
The new method used here to analyse boundary value problems for
equations with spatial derivatives of {\em arbitrary} order
is mathematically straightforward, still it yields
results which are either very difficult or cannot be obtained by the
standard approaches. This method, which is the implementation to this
class of problems of the general approach introduced by
one of the authors \cite{fok}, involves the  steps outlined below.


\subsubsection{Reformulation of the PDE}
A given evolution PDE with symbol  $\omega(k)$ can be written in
the form \be \left({\rm e}^{-ikx+\omega(k)t}q\right)_{t}-
\left({\rm e}^{-ikx+\omega(k)t} X\right)_{x}=0, \quad k\in{\mathbb
C}, \label{consform0} \ee where $q(x,t)$ is a solution of the PDE,
and the function $X(x,t,k)$ is given by the formula \be
X(x,t,k)=\sum_{j=0}^{n-1}c_{j}(k)\partial_{x}^{j}q(x,t),
\label{bigx0} \ee where the coefficients $c_{j}(k)$ are known
polynomials in $k$. For example, for equation (\ref{eq1}), $n=2$,
$c_{1}(k)=-k$ and $c_{2}(k)=i$. The explicit form of $c_{j}(k)$
for an arbitrary $\omega(k)$ is given in section \ref{sec2}.

\subsubsection{The global relation and well posedness}
Equations (\ref{consform0})-(\ref{bigx0}), and Green's theorem applied
to the domain $\{[0,L]\times [0,t]\}$, imply
\be
\sum_{0}^{n-1} c_{j}(k)\left(\tilde{f}_{j}(t,k)-{\rm
e}^{-ikL}\tilde{g}_{j}(t,k)\right)=\hat{q}_{0}(k)-{\rm
e}^{\omega(k)t}\hat{q}(t,k),\quad k\in{\mathbb C},
\label{gr1}
\ee
where $\hat{q}_{0}$, $\tilde{f}_{j}$ and $\tilde{g}_{j}$ are defined
in the notations (equations (\ref{q0})-(\ref{gj}))and $\hat{q}(t,k)$
denotes the Fourier transform of $q(x,t)$.

Equation (\ref{gr1}) is the so-called {\em global relation}. The
crucial observation is that the functions $\tilde{f}_{j}$ and
$\tilde{g}_{j}$ depend on $k$ only through $\omega(k)$ \cite{best}. 
It follows
that these functions are {\em invariant} under any transformation of
the complex $k$-plane
that leaves $\omega(k)$ invariant. These transformations are 
determined
by the roots of the equation
\be
\omega(k)=\omega(\lambda)\;\;{\rm given}\;{\rm by}\;\quad 
\lambda_{0}(k)=k,
\lambda_{1},\ldots,\lambda_{n-1}.
\label{roots}
\ee
Replacing $k$ by $\lambda(k)$ in equation (\ref{gr1}) we obtain a
system of $n$ equations
\be
\sum_{j=0}^{n-1} c_{j}(\lambda_{l}(k))\left(\tilde{f}_{j}(t,k)-{\rm
e}^{-i\lambda_{l}(k)L}\tilde{g}_{j}(t,k)\right)=
\hat{q}_{0}(\lambda_{l}(k))-{\rm
e}^{\omega(k)t}\hat{q}(t,\lambda_{l}(k)),\quad
%k\in{\mathbb C},\quad
l=0,..,n-1.
\label{gr1n}
\ee
Ignoring for the moment the  {\em unknown} function
$\hat{q}(t,k)$, equations (\ref{gr1n}) can be considered as $n$
equations coupling the $2n$ unknown functions $\{\tilde{f}_{j},
\tilde{g}_{j}\}_{0}^{n-1}$. Therefore we expect that  if $n$ boundary
conditions are prescribed, hence that $n$ of the
functions $\{\tilde{f}_{j}(t,k),\,\tilde{g}_{j}(t,k)\}$ are known,
the remaining $n$ unknown functions can be obtained by solving
this system of $n$ equations. However,
not {\em all} choices of $n$ boundary conditions yield a well posed
problem.
Indeed, it turns out that the contribution of the additional unknown
function $\hat{q}(t,k)$ vanishes
if and only if a set of $N$ and $n-N$ boundary conditions are
prescribed at $x=0$ and $x=L$, respectively, where
$N$ depends in general only on the symbol $\omega(k)$: If $n$ is
even, then $N=n/2$, while if $n$ is odd then $N$ is either $(n+1)/2$
or $(n-1)/2$ depending on the sign of the highest derivative.
%(the requirement Re$(\omega(k)\geq 0$ makes this sign unambiguous).

We will show in section \ref{sec3} that $N$ can be computed as
follows: the set of prescribed boundary conditions must be such that
when solving for the unknown functions in the set
$\{\tilde{f}_{j}(t,k)\}$, the expression involving $\hat{q}(t,k)$ must
be bounded as $k\to\infty$ when $k\in\,D^{+}$, and
when solving for the unknown functions in the set
$\{\tilde{g}_{j}(t,k)\}$, the expression involving $\hat{q}(t,k)$ must
be bounded as $k\to\infty$ when $k\in\,D^{-}$.

\subsection*{The representation of the solution}
{\bf (i)} For problems where an $x$-transforms exists, equation 
(\ref{gr1}) 
porvides an {\em alternative} way of constructing the solution.
Consider for example equation (\ref{eq1}) and assume for
simplicity that homogeneous Dirichlet boundary conditions are
prescribed. Then $\tilde{f}_{0}=\tilde{g}_{0}=0$ and
$\lambda_{1}=-k$, hence equations (\ref{gr1n}) become
\bea
i\tilde{f}_{1}-i{\rm e}^{-ikL} \tilde{g}_{1}&=&\hat{q}_{0}(k)-{\rm
e}^{ik^{2}t}\hat{q}(t,k),\nonumber
\\
i\tilde{f}_{1}-i{\rm e}^{ikL} \tilde{g}_{1}&=&\hat{q}_{0}(-k)-{\rm
e}^{ik^{2}t}\hat{q}(t,-k).
\nonumber
\eea
Subtracting these equations we obtain
$$
i({\rm e}^{ikL}-{\rm 
e}^{-ikL})\tilde{g}_{1}=\hat{q}_{0}(k)-\hat{q}_{0}(-k)
-{\rm
e}^{ik^{2}t}\left(\hat{q}(t,k)-\hat{q}(t,-k)\right).
$$
Evaluating this equation at the values of $k$ for which the
coefficient of $\tilde{g}_{1}$ vanishes, i.e. 
$k=k_{m}:=\frac{m\pi}{L}$,
$m\in{\mathbb Z}$, and using the definition of $\hat{q}_{0}(k)$ and
$\hat{q}(t,k)$ we find
\be
\hat{q}^{(sin)}(t,k_{m})={\rm
e}^{-ik_{m}^{2}t}\hat{q}_{0}^{(sin)}(k_{m}),\quad
\hat{q}^{(sin)}(k_{m})=\int_{0}^{L}\sin(k_{m}x)q(x)dx.
\label{sintr}
\ee
Equation (\ref{sintr}a) can be inverted by the well known formula and
then $q(x,t)$ is expressed in the form of a sine series.

Similarly, consider  equation (\ref{eq2}) and
assume for simplicity that homogeneous Dirichlet boundary conditions
are prescribed. Then proceeding as in {\bf (i)} above, in analogy
with (\ref{sintr}), we find
\be
\hat{q}(t,k_{m})-\hat{q}(t,i-k_{m})= {\rm
e}^{-(k_{m}^{2}-ik_{m})t}\left(\hat{q}_{0}(k_{m})-\hat{q}_{0}(i-k_{m})\right),
\quad k_{m}=\frac{m\pi}{L}+\frac{i}{2},\;\;m\in{\mathbb Z}.
\label{2tr}
\ee
The left hand side of equation (\ref{2tr}a) involves the difference
$\left[{\rm
e}^{-ik_{m}x}-{\rm
e}^{-i(i-k_{m})x}\right]$. Using the orthogonality of this expression 
with
respect to ${\rm
e}^{ik_{n}x}-{\rm
e}^{i(i-k_{n})x}$, $n\neq m$, equation (\ref{2tr}a) can be inverted,
yielding
\be
q(x,t)=\frac{1}{2L}\sum_{m\in{\mathbb Z}}\left({\rm
e}^{ik_{m}x}-{\rm
e}^{i(i-k_{m})x}\right){\rm e}^{-(k_{m}^{2}-ik_{m})t}
\left(\hat{q}_{0}(k_{m})-\hat{q}_{0}(i-k_{m})\right).
\label{2invtr}
\ee

{\bf (ii)}
For problems where there does {\em not} exist an $x$-transform, the 
alternative approach also {\em fails}.  Consider for example 
equation (\ref{eq3}) with $q(0,t)=q(L,t)=q_x(L,t)=0$. In this case, 
substituting in the global relation 
$\tilde{f}_{0}=\tilde{g}_{0}=\tilde{g}_{1}=0$, and evaluating the 
resulting equation at $k$, $\lambda_{1}(k)$ and $\lambda_{2}(k)$, 
we obtain a system of three algebraic equations for 
$\tilde{f}_{1}$, $\tilde{f}_{2}$ and $\tilde{g}_{2}$. The determinant
of this system vanishes at $k=k_{m}$, $m\in\,{\mathbb Z}$, where for 
large $k$, arg$(k_{m})$ is either $\pi/6$, $5\pi/6$ or $3\pi/2$ (see 
section \ref{sec4}) The functions
$\tilde{f}_{1}$, $\tilde{f}_{2}$ and $\tilde{g}_{2}$ involve ${\rm 
e}^{i(k-k^{3})\tau}$, and for large $k$ these exponentials become 
{\em 
unbounded} at $k=k_{m}$. Hence, we {\em cannot} evaluate this system 
at the zeros of its determinant. 

It will be shown in section \ref{sec4} that the solution of an
{\em arbitrary \bvp\, for an evolution equation with symbol 
$\omega(k)$} can
be expressed in the form
\bea
q(x,t)&=&\frac{1}{2\pi}\left\{\int_{-\infty}^{\infty}{\rm
e}^{ikx-\omega(k)t}\hat{q}_{0}(k)dk-
\int_{\partial D^{+}}{\rm e}^{ikx-\omega(k)t}\sum_{j=0}^{n-1}
c_{j}(k)\tilde{f}_{j}(t,k)dk\right.
\nonumber \\
&-&\left.
\int_{\partial D^{-}}{\rm e}^{ik(x-L)-\omega(k)t}\sum_{j=0}^{n-1}
c_{j}(k)\tilde{g}(t,k)dk\right\},
\label{intrep1}
\eea
where $\hat{q}_{0}$, $\partial D^{+}$,$\partial D^{-}$ are defined in
the notations and $\{\tilde{f}_{j},\tilde{g}_{j}\}_{0}^{n-1}$ can be
computed as follows.

Let $f_{p}(t)$ and $g_{r}(t)$
denote the prescribed boundary conditions, where $p$ and $r$ take $N$
and $n-N$ integer values respectively. Then
$\tilde{f}_{p}$ and $\tilde{g}_{r}$ can be computed immediately (see
equations (\ref{fj}) and (\ref{gj})). Let $f_{P}(t)$ and $g_{R}(t)$
denote the remaining unknown boundary values. Solving the $n$ 
algebraic
equation (\ref{gr1n})  for $\tilde{f}_{P}$ and $\tilde{g}_{R}$ it
follows that these unknown functions can be expressed in terms of
$\hat{q}_{0}$, $\{\tilde{f}_{p},\tilde{g}_{r}\}$, and of a term
involving $\hat{q}(t,\lambda(k))$ and $1/\Delta(k)$, where
$\Delta(k)$ denotes the determinant of the relevant system. Because of
our characterisation of well posedness, the
expressions of $\hat{q}(t,\lambda(k))$ appearing in the representation
of $\tilde{f}_{P}$ are bounded as $k\to\infty$, $k\in\,D^{+}$, 
therefore,
using Cauchy theorem in the domain $D^{+}$, it follows that these 
expressions
give a zero contribution if $\Delta(k)\neq 0$, $k\in\,D^{+}$.
%, and
%give a residue contribution if
If $\Delta(k)$ has zeros in $D^{+}$, then the contour $\partial D^+$ 
must
be deformed to pass above all these zeros.
The contribution of the zeros can alternatively be written in terms of
the associated residues.
%The associated residues can be computed explicitly by evaluating
%equation (\ref{gr1n}) at the corresponding zeros.
The analysis of
the terms $\tilde{g}_{R}$ is similar.

Regarding equation (\ref{eq3}), we note that the associated
determinant $\Delta(k)$  has {\em no} zeros in $D$.
%, thus the
%expression (\ref{intrep1})  does {\em not}
%involve any residues.
Actually it will be shown in section \ref{sec4} that the solution of
the \bvp\, for this equation with the boundary conditions
\be
q(0,t)=f_{0}(t), \quad q(L,t)=g_{0}(t),\quad
q_{x}(L,t)=g_{1}(t),\quad t>0,
\label{bcc}
\ee
 is given by
$$
q(x,t)=\frac{1}{2\pi}\int_{\infty}^{\infty}{\rm
e}^{ikx-i(k-k^{3})t}\hat{q}_{0}(k)dk+\frac{1}{2\pi}\int_{\partial 
D^{+}}{\rm
e}^{ikx-i(k-k^{3})t}(1-k^2)\tilde{f}_{0}(t,k)dk
$$
\be
+\frac{1}{2\pi}\int_{\partial D^{-}}{\rm
e}^{ik(x-L)-i(k-k^{3})t}
%\frac{
(1-k^2)\tilde{g}_0(t,k)+ik\tilde{g}_{1}(t,k)
%}{\Delta(k)}
%dk
\label{eq3ex}
\ee
$$
+\frac{1}{2\pi}\int_{\partial D^{+}}{\rm
e}^{ikx-i(k-k^{3})t}
%\frac{
ik\tilde{f}_{1}(t,k)+\tilde{f}_{2}(t,k)
%}{\Delta(k)}
dk
+\frac{1}{2\pi}\int_{\partial D^{-}}{\rm
e}^{ik(x-L)-i(k-k^{3})t}
%\frac{
\tilde{g}_{2}(t,k)
%}{\Delta(k)
dk,
$$
where the contours $\partial D^{+}$ and $\partial D^{-}$ are shown in
figure \ref{fig1}, 
and the functions $\tilde{f}_{1}$, $\tilde{f}_2$ and
$\tilde{g}_{2}$ are given in terms of the given  initial and boundary
conditions by
\bea
i\tilde{f}_{1}&=&\frac{1}{\Delta(k)} \left[{\rm
e}^{-ikL}(N(\lambda_{2},t)-N(\lambda_{1},t))+{\rm
e}^{-i\lambda_{1}L}(N(k,t)-N(\lambda_{2},t))+{\rm
e}^{-i\lambda_{2}L}(N(\lambda_{1},t)-N(k,t))\right],
\nonumber \\
\tilde{f}_{2}&=&\frac{1}{\Delta(k)} \left[{\rm
e}^{-ikL}(\lambda_{2}N(\lambda_{1},t)-\lambda_{1}N(\lambda_{2},t))+{\rm
e}^{-i\lambda_{1}L}(kN(\lambda_{2},t)-\lambda_{2}N(k,t))
\right.\nonumber \\
&&\left.\hspace{30pt}+{\rm
e}^{-i\lambda_{2}L}(\lambda_{1}N(k,t)-kN(\lambda_{1},t))\right],
\nonumber \\
\tilde{g}_{2}&=&\frac{1}{\Delta(k)} \left[(N(k,t)(\lambda_2-\lambda_1)
+N(\lambda_{1},t)(\lambda_2-k)+N(\lambda_{2},t))(k-\lambda_1)\right],
\nonumber
\eea
where $\Delta(k)$ and $N(k,t)$ are defined by
\be
\Delta(k)={\rm e}^{-ikL}(\lambda_{1}-\lambda_{2})+{\rm
e}^{-i\lambda_{2}L}(k-\lambda_{1})+{\rm
e}^{-i\lambda_{1}L}(\lambda_{2}-k),
\label{detc1}
\ee
\be
N(k,t)=(1-k^{2})(\tilde{f}_{0}(t,k)-{\rm e}^{-ikL}\tilde{g}_{0}(t,k))-
ik{\rm e}^{-ikL}\tilde{g}_{1}(t,k)+\hat{q}_{0}(k),
\label{N3}\ee
and $\lambda_{1}$, $\lambda_{2}$ are the two roots of equation
(\ref{roots}).

\begin{figure}
\begin{center}
\includegraphics[angle=360]{IMAfig1.eps}
% %\centerline[angle=270]{\epsfbox{s2003fig2.eps}}
\caption{The domain $D$ and the contours $\partial D^{+}$ (in red) 
and $\partial D^{-}$ (in green)  for
equation (\ref{eq3}). Note that $D^{-}$ has two connected components.}
\label{fig1}
\end{center}
\end{figure}


\subsubsection{Advantages of the method}
The method
introduced by one of the authors \cite{fok} does {\em not} rely on
separation of variables. Thus this method can be used to obtain
appropriate integral representations of the solution of PDEs, for
which the spectral theory of the associated linear differential
operator fails.

From the above discussion, it appears that the advantage of the new
method presented here manifests itself mainly for equations of order 
{\em
greater} than two. However this method has also certain advantages
for the classical second order equations, particularly in the case
of more complicated boundary conditions, such as Robin conditions.
This is illustrated in section \ref{sec5.2} and further discussed in
section \ref{sec6}.



\section{A reformulation of evolution equations}\label{sec2}
\setcounter{equation}{0}
It can be verified (see \cite{ima1}) that a PDE with
%dispersion relation
symbol $\omega(k)$ can be written in the form
\be
\left({\rm e}^{-ikx+\omega(k)t}q\right)_{t}-
\left({\rm e}^{-ikx+\omega(k)t}
X\right)_{x}=0, \quad k\in{\mathbb C},
\label{consform}
\ee
where the function
$X(x,t,k)$ is given by the formula
\be
X(x,t,k)=i\frac{\omega(k)-
\omega(-i\partial_{x})}{k+i\partial_{x}}q(x,t).
\label{bigx}
\ee
\subsection{Examples}
\subsubsection{(a) The equation (\ref{eq1})}
The symbol is
$$\omega(k)=ik^{2}.$$
Indeed, it can immediately be verified that ${\rm
e}^{ikx-ik^{2}t}$ satisfies this equation. Hence
\be
X(x,t,k)=i\frac{ik^{2}+i\partial_{x}^{2}}
{k+i\partial_{x}}q=-\frac{(k+i\partial_{x})(k-i\partial_{x})}
{k+i\partial_{x}}q=iq_{x}-kq.
\label{bigX1}
\ee
This PDE is thus equivalent to the expression
\be
\left({\rm e}^{-ikx+ik^{2}t}q\right)_{t}-
\left({\rm e}^{-ikx+ik^{2}t}
(iq_{x}-kq)\right)_{x}=0, \quad k\in{\mathbb C}.
\label{eq1cs}
\ee

\subsubsection{(b) The equation (\ref{eq2})}
The symbol is
$$\omega(k)=k^{2}-ik.$$
Therefore
\be
X(x,t,k)=i\frac{-ik+k^{2}+\partial_{x}^{2}+\partial_{x}}
{k+i\partial_{x}}q=i\frac{(k+i\partial_{x})(k-i\partial_{x}-i)}
{k+i\partial_{x}}q=(1+ik)q+q_{x}.
\label{bigX2}
\ee
This PDE is thus equivalent to the expression
\be
\left({\rm e}^{-ikx+(k^{2}-ik)t}q\right)_{t}-
\left({\rm e}^{-ikx+(k^{2}-ik)t}
((1+ik)q+q_{x})\right)_{x}=0, \quad k\in{\mathbb C}.
\label{eq2cs}
\ee
\subsubsection{(c) The equation (\ref{eq3})}

The symbol is
$$\omega(k)=i(k-k^{3}).$$
Hence
\bea
X(x,t,k)&=&i\frac{i(k-k^{3})-(\partial_{x}+\partial_{x}^{3})}
{k+i\partial_{x}}q=-\frac{(k+i\partial_{x}-k^{3}-i\partial_{x}^{3})}{k+i\partial_{x}}
q
\nonumber \\
&=&\left\{-1+\frac{(k+i\partial_{x})(k^{2}-\partial_{x}^{2}-ik\partial_{x})}
{k+i\partial_{x}}\right\}q=(k^{2}-1)q-ikq_{x}-q_{xx}.
\label{bigX3}
\eea
This PDE is thus equivalent to the expression
\be
\left({\rm e}^{-ikx+i(k-k^{3})t}q\right)_{t}-
\left({\rm e}^{-ikx+i(k-k^{3})t}
((k^{2}-1)q-ikq_{x}-q_{xx})\right)_{x}=0, \quad k\in{\mathbb C}.
\label{eq3cs}
\ee



\section{Well posed problems}\label{sec3}
\label{wp}
\setcounter{equation}{0}
\subsection{The global relation}
We first derive the global relation for the case of
equations (\ref{eq1})-(\ref{eq3}).

\subsubsection{Example (a)}
Since $X(x,t,k)=iq_{x}(x,t)-kq(x,t)$, it follows that
the global relation is
\be
i\tilde{f}_{1}(t,k)-k\tilde{f}_{0}(t,k)-{\rm
e}^{-ikL}\left(i\tilde{g}_{1}(t,k)-k\tilde{g}_{0}(t,k)\right)
=\hat{q}_{0}(k)-{\rm
e}^{ik^{2}t}\hat{q}(t,k).
\label{agr1}
\ee


\subsubsection{Example (b)}
Since $X(x,t,k)=(1+ik)q+q_{x}$,
the global relation is
\be
(1+ik)\tilde{f}_{0}(t,k)+\tilde{f}_{1}(t,k)-{\rm
e}^{-ikL}\left((1+ik)\tilde{g}_{0}(t,k)+\tilde{g}_{1}(t,k)\right)
=\hat{q}_{0}(k)-{\rm
e}^{(k^{2}-ik)t}\hat{q}(t,k).
\label{agr2}
\ee

\subsubsection{Example (c)}
Since $X(x,t,k)=(k^{2}-1)q-ikq_{x}-q_{xx}$,
the global relation is
\be
\left[(k^{2}-1)\tilde{f}_{0}-ik\tilde{f}_{1}-\tilde{f}_{2}\right]-{\rm
e}^{-ikL}\left[(k^{2}-1)\tilde{g}_{0}-ik\tilde{g}_{1}-\tilde{g}_{2}\right]
=\hat{q}_{0}(k)-{\rm
e}^{i(k-k^{3})t}\hat{q}(t,k).
\label{agr3}
\ee

\subsection{The analysis of the global relation}
The roots of the equation $\omega(k)=\omega(\lambda)$, which are
defined in terms of $k$ (see equation (\ref{roots})),
can
be distinguished by fixing their asymptotic behaviour:
$$
\lambda_{m}(k)\sim {\rm e}^{2\pi i m/n},\;\;k\to \infty,
\quad 1\leq m\leq n-1.
$$
For the examples (a) - (c) the roots different from $k$ are given
below:

\begin{tabular}{llll}
\\
{\bf (a)}&
$i\lambda^{2}-ik^{2}=(\lambda-k)(\lambda+k)=0,$&%$\rightarrow$ &
$\lambda_{1}=-k$.
\\ \\
{\bf (b)}&
$\lambda^{2}-i\lambda-k^{2}+ik=(\lambda 
-k)(\lambda+k-i)=0,$&%$\rightarrow$ &
 $\lambda_{1}=i-k$.
%\label{rootsb}
\\ \\
{\bf (c)}&$
(\lambda-\lambda^{3})-(k-k^{3})=(k-\lambda)(\lambda^{2}+\lambda 
k+k^{2}-1)=0,$&%$\rightarrow$&
% $\left\{ \right.$
\begin{tabular}{l}$\lambda_{1}=\frac{\sqrt{4-3k^{2}}}{2}-\frac{k}{2}$,
\\
 $\lambda_{2}=-\frac{\sqrt{4-3k^{2}}}{2}-\frac{k}{2}$.
 \end{tabular}
% \label{rootsc}
\end{tabular}


\vspace{5mm}
Before turning to the examples (a)-(c), we make the
following observations, valid for an arbitrary $\omega(k)$:

{\bf (i)} The functions $\hat{q}_{0}(k)$ and $\hat{q}(t,k)$, which
are entire functions of $k$, are bounded as $k\to\infty$ in
${\mathbb C}^{-}$, the lower half of the complex $k$-plane.

{\bf (ii)} The functions ${\rm e}^{ikL}\hat{q}_{0}(k)$ and ${\rm
e}^{ikL}\hat{q}(t,k)$ are bounded as $k\to\infty$ in
${\mathbb C}^{+}$, the upper half of the complex $k$-plane.


{\bf (iii)}
The functions $\tilde{f}_{j}$ and $\tilde{g}_{j}$, which are entire
functions of $k$, are bounded as $k\to\infty$ if Re\,$\omega(k)\leq 
0$.


\subsubsection{Example (a)}
In this case, since
$$
{\rm Re}\,\omega(k)={\rm Re}(ik^{2})=-2{\rm Re}(k){\rm Im}(k),
$$
it follows that
\be
D=\{k\in{\mathbb C}^{+}:{\rm Re}(k){\rm Im}(k)\geq 0\}.
\label{D1}
\ee
Hence $D$ is the union of the first and third quadrant in the complex
plane.
Since $\lambda_{1}=-k$, equations (\ref{gr1n}) become
\bea
i\tilde{f}_{1}(t,k)-k\tilde{f}_{0}(t,k)-{\rm
e}^{-ikL}\left(i\tilde{g}_{1}(t,k)-k\tilde{g}_{0}(t,k)\right)&=&
\hat{q}_{0}(k)-{\rm
e}^{ik^{2}t}\hat{q}(t,k),
\nonumber \\
i\tilde{f}_{1}(t,k)+k\tilde{f}_{0}(t,k)-{\rm
e}^{ikL}\left(i\tilde{g}_{1}(t,k)+k\tilde{g}_{0}(t,k)\right)&=&
\hat{q}_{0}(-k)-{\rm
e}^{ik^{2}t}\hat{q}(t,-k).
\label{sys1}
\eea
We consider equations (\ref{sys1}) as two equations relating the four
terms $\tilde{f}_{0}$, $\tilde{f}_{1}$, $\tilde{g}_{0}$,
$\tilde{g}_1$, treating for the moment the function $\hat{q}(t,k)$
as a known function. The dependence of the solution of the system
(\ref{sys1}) on the terms involving $\hat{q}(t,k)$ is the crucial
factor that determines its well posedness.
%effective solvability.

To explain how to compute $N$, we distinguish two illustrative
cases.

\vspace{2mm}
{\bf (a.1) ${\bf q(0,t)=f_{0}(t)}$, ${\bf q(L,t)=g_{0}(t)}$}
%given  ($ N=1$)}

In this case, the two functions $\tilde{f}_{0}(t,k)$ and
$\tilde{g}_{0}(t,k)$ are known, and we must solve the system
(\ref{sys1}) for $\tilde{f}_{1}$ and $\tilde{g}_{1}$. We concentrate
only on the dependence of the solution on the term $\hat{q}(t,k)$.
This
yields
\bea
i\tilde{f}_{1}&=&(known)+{\rm e}^{ik^{2}t}\left[\frac{{\rm
e}^{2ikL}\hat{q}(t,k)-\hat{q}(t,-k)}{1-{\rm e}^{2ikL}}\right],
\nonumber \\
i\tilde{g}_{1}&=&(known)+{\rm
e}^{ik^{2}t}\left[\frac{\E\hat{q}(t,k)-\E\hat{q}(t,-k)}{1-{\rm
e}^{2ikL}}\right].
\label{asy1}
\eea
The terms  involving the function $\hat{q}(t,k)$ in equation
(\ref{asy1}a) are
{\em bounded in $k$} as $k\to\infty$
if $k\in\,D^{+}=D\cap {\mathbb C}^{+}$.

Indeed, for
$k$ in ${\mathbb C}^{+}$,
 $$
\frac{{\rm e}^{2ikL}\hat{q}(t,k)-\hat{q}(t,-k)}
{1-{\rm e}^{2ikL}}\sim {\rm e}^{2ikL}\hat{q}(t,k)-\hat{q}(t,-k),
$$
and both these terms are bounded in  ${\mathbb C}^{+}$. Since the term
${\rm e}^{ik^{2}t}$ is bounded in $D$, our claim follows.

Similarly, the terms involving the function $\hat{q}(t,k)$ in
equation (\ref{asy1}b) are bounded as $k
\to\infty$ if $k\in\,D^{-}=D\cap {\mathbb C}^{-}$. Indeed,
 $$
\frac{\E\hat{q}(t,k)-\E\hat{q}(t,-k)}{1-{\rm
e}^{2ikL}}\sim {\rm e}^{-ikL}\hat{q}(t,k)- {\rm 
e}^{-ikL}\hat{q}(t,-k),
$$
and both terms are bounded in ${\mathbb C}^{-}$.

\vspace{2mm}
{\bf (a.2) ${\bf q(0,t)=f_{0}(t)}$, ${\bf q_{x}(0,t)=f_{1}(t)}$}
%given ($ N=2$)}

In this case we obtain a system for the two functions
$\tilde{g}_{0}(t,k)$ and
$\tilde{g}_{1}(t,k)$. Considering explicitly only the terms involving
the function $\hat{q}(t,k)$, we
find
\bea
k\tilde{g}_{0}&=&known+{\rm e}^{ik^{2}t}\left[\frac{-{\rm
e}^{2ikL}\hat{q}(t,k)+\hat{q}(t,-k)}{2\E}\right],
\nonumber \\
i\tilde{g}_{1}&=&(known)+{\rm
e}^{ik^{2}t}\left[\frac{{\rm
e}^{2ikL}\hat{q}(t,k)+\hat{q}(t,-k)}{2\E}\right].
\nonumber
\eea
In this case  {\em not all} the terms  containing the function
$\hat{q}(t,k)$ are bounded as $k\to\infty$ for
$k\in\,D^{-}=D\cap {\mathbb C}^{-}$.
For example,
$$
\frac{-{\rm e}^{2ikL}\hat{q}(t,k)+\hat{q}(t,-k)}{2{\rm
e}^{ikL}}=-\frac{\E\hat{q}(t,k)}{2}+{\rm e}^{-ikL}\hat{q}(t,-k)$$
and the second term is bounded   in ${\mathbb C}^{-}$, but the
first term is bounded  in ${\mathbb C}^{+}$. Thus in the above
example, as expected, $N=1$, i.e. one boundary condition must be
prescribed at each end of the interval.
% For this reason,  it will not be possible to compute
% effectively the spectral functions
% $\tilde{g}_{i}$, and it follows then that the problem is not well 
%posed.

\subsubsection{Example (b)}



\begin{figure}
\begin{center}
 \includegraphics[angle=360]{IMAfig2.eps}
% %\centerline[angle=270]{\epsfbox{s2003fig2.eps}}
\caption{The domain $D$
and the contours $\partial D^{+}$ (in red)  and $\partial D^{-}$ (in 
green)
%the line where the zeros of the determinant $\Delta(k)$ lie
for
equation (\ref{eq2})}
\label{zerosb}
\end{center}
\end{figure}


The domain $D$ is in this case given by the interior of the
hyperbola
(see figure \ref{zerosb})
\be
D=\{k\in{\mathbb C}:\;
{\rm Re}(k)^{2}-{\rm
Im}(k)^{2}+{\rm Im}(k)\leq 0\}.
\label{hypb}
\ee
$D^{+}=D\cap {\mathbb C}^{+}$ is the interior of the upper branch
of this hyperbola, and $D^{-}=D\cap {\mathbb C}^{-}$ is the interior
%in ${\mathbb C}^{-}$
of its lower branch.


Since $\lambda_{1}=i-k$, equations (\ref{gr1n}) yield
\bea
(1+ik)\tilde{f}_{0}(t,k)+\tilde{f}_{1}(t,k)-
{\rm 
e}^{-ikL}\left[(1+ik)\tilde{g}_{0}(t,k)+\tilde{g}_{1}(t,k)\right]&=&
\hat{q}_{0}(k)-{\rm 
e}^{(k^{2}-ik)t}\hat{q}(t,k),
 \nonumber
 \\
 -ik\tilde{f}_{0}(t,k)+\tilde{f}_{1}(t,k)-{\rm 
e}^{-i(i-k)L}\left(-ik\tilde{g}_{0}(t,k)+\tilde{g}_{1}(t,k)\right)&=&
\hat{q}_{0}(i-k)-{\rm
e}^{(k^{2}-ik)t}\hat{q}(t,i-k).
 \nonumber \\
 \label{sys2}
 \eea
As for the previous example,  the analysis of the boundedness of the 
terms
containing the function $\hat{q}(t,k)$  implies that one boundary 
condition
must be
prescribed at  each end of the interval, hence $N=1$.
This analysis  is similar to
the analysis performed in the case of  Example (a).


\subsubsection{Example (c)}
The domain $D$ is
%the part in ${\mathbb C}^{+}$ of the interior of the following
%hyperbola
(see figure \ref{fig1})
\be
D=\{k\in\,{\mathbb C}: \,{\rm Im}\,(k)\left[3{\rm Re}\,(k)^{2}-
{\rm Im}\,(k)^{2}-1\right]\leq 0\}.
\label{hypc}
\ee


Evaluating equations (\ref{gr1n})  at $k$ , $\lambda_{1}$ and 
$\lambda_{2}$,
we find
\bea
\left[(k^{2}-1)\tilde{f}_{0}-ik\tilde{f}_{1}-\tilde{f}_{2}\right]-
{\rm 
e}^{-ikL}\left[(k^{2}-1)\tilde{g}_{0}-ik\tilde{g}_{1}-\tilde{g}_{2}\right]
&=&\hat{q}_{0}(k)-{\rm
e}^{i(k-k^{3})t}\hat{q}(t,k),
\nonumber \\
\left[(\lambda_{1}^{2}-1)\tilde{f}_{0}-i\lambda_{1}\tilde{f}_{1
}-\tilde{f}_{2}\right]-{\rm
e}^{-i\lambda_{1}L}\left[(\lambda_{1}^{2}-1)\tilde{g}_{0}-
i\lambda_{1}\tilde{g}_{1}-\tilde{g}_{2}\right]
&=&\hat{q}_{0}(\lambda_{1})-{\rm
e}^{i(k-k^{3})t}\hat{q}(\lambda_{1},t),
\nonumber \\
\left[(\lambda_{2}^{2}-1)\tilde{f}_{0}-i\lambda_{2}\tilde{f}_{1}
-\tilde{f}_{2}\right]-{\rm
e}^{-i\lambda_{2}L}\left[(\lambda_{2}^{2}-1)\tilde{g}_{0}
-i\lambda_{2}\tilde{g}_{1}-\tilde{g}_{2}\right]
&=&\hat{q}_{0}(\lambda_{2})-{\rm
e}^{i(k-k^{3})t}\hat{q}(\lambda_{2},t).
\nonumber \\
\label{sys3}
\eea
It turns out that, for a well posed problem,
one boundary condition must be
prescribed
at $x=0$ and two boundary conditions must be prescribed at $x=L$,
hence $N=1$.


%  \begin{figure}
% % \includegraphics[angle=360]{Dc.eps}
%  %\centerline[angle=270]{\epsfbox{s2003fig2.eps}}
%  \caption{The domain $D$ for equation (\ref{eq3})}
%  \label{domainc}
%  \end{figure}
%
%
We illustrate this by two particular examples. The first set of
boundary conditions, with $N=1$, yields a well posed problems.
The second, with $N=2$, does
not.

\vspace{2mm}
{\bf (c.1) ${\bf q(0,t)=f_{0}(t)}$, ${\bf q(L,t)=g_{0}(t)}$, ${\bf
q_{x}(L,t)=g_{1}(t)}$}
%given}

In this case the unknown functions
$\tilde{f}_{1}$, $\tilde{f}_{2}$ and $\tilde{g}_{2}$ are given by
\bea
&&i\tilde{f}_{1}=(known \;\;terms)
\nonumber \\
&&+{\rm e}^{i(k-k^{3})t}\left[\frac{
{\rm
e}^{-i\lambda_{1}L}(\hat{q}(k)-\hat{q}(\lambda_{2}))+{\rm
e}^{-i\lambda_{2}L}(\hat{q}(\lambda_{1})-\hat{q}(k))+
{\rm e}^{-ikL}(\hat{q}(\lambda_{1})-\hat{q}(\lambda_{2}))
}{\Delta(k)}\right]
\nonumber \\
&&\tilde{f}_{2}=(known \;\;terms)
\nonumber \\
&&+{\rm e}^{i(k-k^{3})t}\left[\frac{{\rm
e}^{-i\lambda_{1}L}(k\hat{q}(\lambda_{2})-\lambda_{2}\hat{q}(k))+{\rm
e}^{-i\lambda_{2}L}(\lambda_{1}\hat{q}(k)-k\hat{q}(\lambda_{1}))+
{\rm e}^{-ikL}
(\lambda_{1}\hat{q}(\lambda_{2})-\lambda_{2}\hat{q}(\lambda_{1}))}
{\Delta(k)}
\right]
\nonumber\\
&&\tilde{g}_{2}=(known \;\;terms)
\nonumber \\
&&+{\rm
e}^{i(k-k^{3})t}\left[\frac{\hat{q}(\lambda_{1})(k-\lambda_{1})+
\hat{q}(\lambda_{2})(\lambda_{2}-k)+
\hat{q}(k)(\lambda_{2}-\lambda_{1})}
{\Delta(k)}\right]
\label{asy3} \eea
where
$\Delta(k)$ is given by equation (\ref{detc1}) and
to simplify the
notation we have suppressed the $t$ dependence in $\hat{q}(t,k)$.

The terms  containing the function $\hat{q}(k)$ in equations
(\ref{asy3}a,b) are
{\em bounded in $k$} as $k\to\infty$
if $k\in\,D^{+}=D\cap {\mathbb C}^{+}$. This follows from the
observation that if
$k\in D^{+}$, then $\lambda_{1}$ and $\lambda_{2}$ are in
${\mathbb C}^{-}$.
For example, consider the function $\tilde{f}_{1}(k)$. Since ${\rm
e}^{ikL}$, ${\rm
e}^{-i\lambda_{1}L}$  and ${\rm
e}^{-i\lambda_{2}L}$  are all bounded for $k$ in ${\mathbb C}^{+}$, 
the
%square
bracket appearing on the right hand side of $i\tilde{f}_{1}$, as
$k\to\infty$, is asymptotically given by
$$
\frac{({\rm e}^{i(k-\lambda_{1})L}-
{\rm e}^{i(k-\lambda_{2})L})\hat{q}(k)
+(1-{\rm e}^{i(k-\lambda_{2})L})\hat{q}(\lambda_{1})
+({\rm
e}^{i(k-\lambda_{1})L}-1)\hat{q}(\lambda_{2})}{\lambda_{1}-\lambda_{2}}.
$$
All terms in this expression are bounded when $k\in\,{\mathbb C}^{+}$.
In addition, ${\rm
e}^{i(k-k^{3})t}$ is bounded for all $k
\in D$, and the claim follows.
Similarly, the terms containing $\hat{q}(k)$ in equation
(\ref{asy3}c) are
{\em bounded in $k$} as $k\to\infty$
if $k\in\,D^{-}=D\cap {\mathbb C}^{-}$.

\vspace{2mm}
{\bf (c.2) ${\bf q(0,t)=f_{0}(t)}$, ${\bf q(L,t)=g_{0}(t)}$,
${\bf q_{x}(0,t)=f_{1}(t)}$}
% given}

In this case, the unknown functions $\tilde{f}_{2}$, $\tilde{g}_{1}$
and $\tilde{g}_{2}$ are given by
\bea
&&\tilde{f}_{2}=(known \;\;terms)
\nonumber \\
&&+{\rm e}^{i(k-k^{3})t}\left[\frac{{\rm
e}^{ikL}(\lambda_{2}-\lambda_{1})\hat{q}(k)+{\rm
e}^{i\lambda_{2}L}(\lambda_{1}-k)\hat{q}(\lambda_{2})+
{\rm
e}^{i\lambda_{1}L}(k-\lambda_{2})\hat{q}(\lambda_{1})}{\Delta(k)}\right]
\label{asy3b} \\
&&i\tilde{g}_{1}=(known \;\;terms)
\nonumber \\
&&+{\rm e}^{i(k-k^{3})t}\left[\frac{{\rm
e}^{-i\lambda_{1}L}(\hat{q}(k)-\hat{q}(\lambda_{2}))+
{\rm
e}^{-i\lambda_{2}L}(\hat{q}(\lambda_{1})-\hat{q}(k))+
{\rm e}^{-ikL}(\hat{q}(\lambda_{2})-\hat{q}(\lambda_{1}))
}
{\Delta(k)}\right]
\nonumber \\
&&\tilde{g}_{2}=(known \;\;terms)
 \nonumber\\
&&+{\rm
e}^{i(k-k^{3})t}\left[\frac{{\rm
e}^{-i\lambda_{1}L}\lambda_{1}(\hat{q}(\lambda_{2})-\hat{q}(k))
+{\rm
e}^{-i\lambda_{2}L}\lambda_{2}(\hat{q}(k)-\hat{q}(\lambda_{1}))
+{\rm e}^{-ikL}k(\hat{q}(\lambda_{1})-\hat{q}(\lambda_{2})
}
{\Delta(k)}\right]
\nonumber
\eea
where $\Delta(k)$ is given by
$$
\Delta(k)={\rm e}^{ikL}(\lambda_{2}-\lambda_{1})
+{\rm e}^{i\lambda_{2}L}(\lambda_{1}-k)+{\rm
e}^{i\lambda_{1}L}(k-\lambda_{2}).
$$
As for example (a.2),  {\em not all} the terms containing the
unknown function $\hat{q}(t,k)$ in equations (\ref{asy3b})
are bounded for all $k\in\,D^{+}$ or $k\in\,D^{-}$. As an example,
consider the terms
in (\ref{asy3b}b), which  should be bounded as $k\to\infty$
for all  $k\in D^{-}$. Choose $k$ such that $\lambda(k)\in\,D^{-}$ and
$\lambda_{2}(k)\in\,D^{+}$. Then $\E$ and ${\rm e}^{i\lambda_{1}L}$ 
are
not bounded as $k\to\infty$, while ${\rm e}^{i\lambda_{2}L}$
is bounded. Since asymptotically $\lambda_1\sim {\rm e}^{2\pi i/3}k$, 
it is easy to verify
that, for $k\in \,D^-$ such that $\lambda_1(k)\in \,D^-$,
the dominant term in the denominator is ${\rm e}^{i\lambda_{1}L}$.
Hence (\ref{asy3b}b), as $k\to\infty$, is  given by
 $$
{\rm e}^{i(k-k^{3})t}
\left[{\rm
e}^{-2i\lambda_{1}L}(\hat{q}(k)-\hat{q}(\lambda_{2}))+
{\rm
e}^{ikL}(\hat{q}(\lambda_{1})-\hat{q}(k))+
{\rm e}^{i\lambda_2L}\hat{q}(\lambda_{2})-\hat{q}(\lambda_{1})\right]
 $$
% $$
% ={\rm e}^{i(k-k^{3})t}\left[{\rm
% e}^{i\lambda_{2}L}(\hat{q}(k)-\hat{q}(\lambda_{2}))+
% {\rm
% e}^{i\lambda_1 L}(\hat{q}(\lambda_{1})-\hat{q}(k))+
% {\rm
% e}^{2i(\lambda_{1}+\lambda_{2})L}(\hat{q}(\lambda_{2})-
% \hat{q}(\lambda_{1}))\right]
% $$
and
the term $\E\hat{q}(k)\sim {\rm e}^{ik(L-x)}$
is {\em not} bounded for  $k\in\,D^{-}$
(to simplify the
notation we have again suppressed the $t$ dependence in 
$\hat{q}(t,k)$).

\section{The integral representation of the solution}\label{sec4}
\setcounter{equation}{0}
We first derive equation (\ref{intrep1}). The global relation
(\ref{gr1}) is
$$
\hat{q}(t,k)={\rm e}^{-\omega(k)t}\hat{q}_{0}(k)-{\rm
e}^{-\omega(k)t}\left[\tilde{f}(t,k)-{\rm 
e}^{-ikL}\tilde{g}(t,k)\right],$$
where
$$\tilde{f}(t,k)=\sum_{0}^{n-1}c_{j}(k)\tilde{f}_{j}(t,k),\quad
\tilde{g}(t,k)=\sum_{0}^{n-1}c_{j}(k)\tilde{g}_{j}(t,k),
$$
$$
\hat{q}(t,k)=\int_{0}^{L}{\rm e}^{-ikx}q(x,t)dx,\quad
\hat{q}_{0}(k)=\hat{q}(0,k).$$
Taking the inverse Fourier transform of $\hat{q}(t,k)$, we obtain
$$
q(x,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\rm
e}^{ikx-\omega(k)t}\left\{\hat{q}_{0}(k)-\tilde{f}(t,k)+{\rm
e}^{-ikL}\tilde{g}(t,k)\right\}dk.
$$
It follows from the  definition of $D$  that for
$k\notin D$ and for any $t>0$, the functions
${\rm e}^{-\omega(k)t}\tilde{f}(t,k)$ and
${\rm e}^{-\omega(k)t}\tilde{g}(t,k)$ are  bounded
as $k\to\infty$.
Thus

\vspace{2mm}
\begin{tabular}{ll}
${\rm e}^{ikx-\omega(k)t}\tilde{f}(k)$&is
analytic and bounded for $k\in\,{\mathbb C}^{+}\setminus D$,
 \\
 \\
 ${\rm e}^{ik(x-L)-\omega(k)t}\tilde{g}(k)$&is
 analytic and bounded for $k\in\,{\mathbb C}^{-}\setminus D$.
 \\
 \end{tabular}

 \vspace{2mm}
 An application of Cauchy theorem thus yields
 \bea
 \int_{-\infty}^{\infty}{\rm
 e}^{ikx-\omega(k)t}\tilde{f}(t,k)dk&=&\int_{\partial D^{+}}{\rm
 e}^{ikx-\omega(k)t}\tilde{f}(t,k)dk,
 \nonumber \\
 \int_{-\infty}^{\infty}{\rm
 e}^{ik(x-L)-\omega(k)t}\tilde{g}(t,k)dk&=&-\int_{\partial D^{-}}{\rm
 e}^{ik(x-L)-\omega(k)t}\tilde{g}(t,k)dk,
 \nonumber
 \eea
 and (\ref{intrep1}) follows.
 
 We now derive the solution representation for our three examples. In 
 the case of 
 the two second order examples, we 
 will not need to use the general integral representation as a 
 series representation can in these case be deduced in a more direct way. 
 However, the integral representation can alternatively be derived, 
 and we will show in the next section how it can be reduced to the one   
 obtained here.
 
\subsection{Example (a)}
Consider equation (\ref{eq1}) with the boundary conditions
\be
q(0,t)=f_{0}(t),\quad q(L,t)=g_{0}(t),\quad t>0.
\label{bca1}
\ee
Equations
(\ref{sys1}) become
\bea
i\tilde{f}_{1}-{\rm e}^{-ikL}i\tilde{g}_{1}&=&N(k,t)-{\rm
e}^{ik^{2}t}\hat{q}(t,k),
\nonumber
\\
i\tilde{f}_{1}-{\rm e}^{ikL}i\tilde{g}_{1}&=&N(-k,t)-{\rm
e}^{ik^{2}t}\hat{q}(t,-k),
\label{A.2}
\eea
where $N(k,t)$ is the known function
\be
N(k,t)=k\left(\tilde{f}_{0}(t,k) -{\rm e}^{-ikL}\tilde{g}_{0}(t,k)
\right)+\hat{q}_{0}(k).
\label{Nk}
\ee
Subtracting equations (\ref{A.2}) we obtain
\be
({\rm e}^{ikL}-{\rm
e}^{-ikL})i\tilde{g}_{1}=N(k,t)-N(-k,t)
-{\rm
e}^{ik^{2}t}(\hat{q}(t,k)-\hat{q}(t,-k)),\quad k_{m}\in\,{\mathbb R}.
\label{lasteq}
\ee
We evaluate this equation at the values of $k$ for which ${\rm 
e}^{ikL}-{\rm
e}^{-ikL}=0$, i.e.
$$k=k_{m}=\frac{m\pi}{L}, \;m\in\,{\mathbb Z}.
$$
The definition of $\hat{q}(t,k)$ implies
$$
\int_{0}^{L}\sin(k_{m}x)q(x,t)dx=\frac{i}{2}{\rm
e}^{-ik_{m}^{2}t}\left[N(k_m,t)-N(-k_m,t)\right].
$$
Hence
\be
q(x,t)=\frac{i}{4L}\sum_{m}\sin(k_{m}x){\rm
e}^{-ik_{m}^{2}t}\left[N(k_m,t)-N(-k_m,t)\right],\quad
k_{m}=\frac{m\pi}{L}.
\label{sin1}
\ee

\subsection{Example (b)}
Consider equation (\ref{eq2}) with the boundary conditions
(\ref{bca1}).
Equations
(\ref{sys2}) become
\bea
\tilde{f}_{1}-
{\rm e}^{-ikL}\tilde{g}_{1}&=&N(k,t)-{\rm
 e}^{(k^{2}-ik)t}\hat{q}(t,k),
 \nonumber
 \\
 \tilde{f}_{1}-{\rm
 e}^{-i(i-k)L}\tilde{g}_{1}&=&N(i-k,t)-{\rm
e}^{(k^{2}-ik)t}\hat{q}(t,i-k),
 \nonumber \\
 \label{sys2b}
 \eea
 where $N(k,t)$ is the known function
 \be
 N(k,t)=-(1+ik)\left(\tilde{f}_{0}(k)-{\rm
 e}^{-ikL}\tilde{g}_{0}(k)\right)+\hat{q}_{0}(k).
 \label{N2}
 \ee
 Subtracting equations (\ref{sys2b}) we find
 \be
 \left({\rm
 e}^{-i(i-k)L}-{\rm
 e}^{-ikL}\right)\tilde{g}_{1}=N(k,t)-N(i-k,t)-
 {\rm
 e}^{(k^{2}-ik)t}\left(\hat{q}(t,k)-\hat{q}(t,i-k)\right).
 \label{sin2}
 \ee
 We evaluate this equation at the values of $k$ for which  ${\rm
 e}^{-i(i-k)L}-{\rm
 e}^{-ikL}=0$, i.e.
%  i.e. $k=k_{m}$,
\be
k=k_{m}=\frac{m\pi}{L}+\frac{i}{2}, \;m\in\,{\mathbb Z}.
\label{zeros2}
\ee
Equation (\ref{sin2}) and the definition of $\hat{q}(t,k)$ yield
\be
\int_{0}^{L}\left( {\rm
 e}^{-ik_{m}x}-{\rm
 e}^{-i(i-k_{m})x}\right)q(x,t)dx={\rm
 e}^{-(k_m^{2}-ik_m)t}\left(N(k_m,t)-N(i-k_m,t)\right).
 \label{sin3}
 \ee
 This equation can be solved for $q(x,t)$ using orthogonality, i.e. 
the fact that if
 $k_{m}$ is defined by (\ref{zeros2}), then
 \be
 \int_{0}^{L}\left( {\rm
 e}^{-ik_{m}x}-{\rm
 e}^{-i(i-k_{m})x}\right)\left( {\rm
 e}^{ik_{n}x}-{\rm
 e}^{i(i-k_{n})x}\right)dx=2L\delta_{mn}.
 \label{ortho2}
 \ee
Thus equation (\ref{sin3}) yields
\be
q(x,t)=\frac{1}{2L}\sum_{m}{\rm
 e}^{-(k_m^{2}-ik_m)t}\left( {\rm
 e}^{ik_{m}x}-{\rm
 e}^{i(i-k_{m})x}\right)\left(N(k_m,t)-N(i-k_m,t)\right).
 \label{sinrep2}
 \ee


\subsection{Example (c)}
Consider equation (\ref{eq3}) with the boundary conditions 
(\ref{bcc}).
Equations (\ref{sys3}) become
\bea
-\tilde{f}_{2}-ik\tilde{f}_{1}+{\rm
e}^{-ikL}\tilde{g}_{2}&=&N(k,t)-{\rm e}^{i(k-k^{3})t}\hat{q}(t,k),
\nonumber \\
-\tilde{f}_{2}-i\lambda_{1}\tilde{f}_{1}+{\rm
e}^{-i\lambda_{1}L}\tilde{g}_{2}&=&N(t,\lambda_{1})-{\rm
e}^{i(k-k^{3})t}\hat{q}(t,\lambda_{1}),
\nonumber \\
-\tilde{f}_{2}-i\lambda_{2}\tilde{f}_{1}+{\rm
e}^{-i\lambda_{2}L}\tilde{g}_{2}&=&N(t,\lambda_{2})-{\rm
e}^{i(k-k^{3})t}\hat{q}(t,\lambda_{2}),
\label{sys3b} \eea
where the known function $N(k,t)$ is given by
(\ref{N3}).
In  this case,  the representation (\ref{intrep1}) becomes
\bea
q(x,t)&=&\frac{1}{2\pi}\left\{\int_{-\infty}^{\infty}{\rm
e}^{ikx-i(k-k^{3})t}\hat{q}_{0}(k)dk-
 \int_{\partial D^{+}}{\rm e}^{ikx-i(k-k^{3})t}((k^{2}-1)\tilde{f}_{0}
 -ik\tilde{f}_{1}-\tilde{f}_{2})dk\right.
 \nonumber\\
 &-&\left.
\int_{\partial D^{-}}{\rm 
e}^{ik(x-L)-(k^{2}-ik)t}((k^{2}-1)\tilde{g}_{0}
 -ik\tilde{g}_{1}-\tilde{g}_{2})dk\right\},
\label{intrepc}
\eea
where $D^{+}$ is the interior of the upper branch of the hyperbola
(\ref{hypc}), and $D^{-}$ is the exterior of its lower branch
 (see figure \ref{fig1}).

 Solving equations (\ref{sys3b}) for $\tilde{f}_{1}$, $\tilde{f}_{2}$
 and $\tilde{g}_{2}$ and substituting the resulting expressions in
 equation (\ref{intrepc})  we find equation (\ref{eq3ex}).  We
 emphasise that the solution of the system (\ref{sys3})
 includes also the terms appearing on the right hand side of
 equation (\ref{asy3}). However these terms do {\em not} contribute to
 the solution.  Indeed, after multiplication by ${\rm
 e}^{-i(k-k^{3})t}$  the terms in (\ref{asy3}a,b) are analytic and 
bounded as
 $k\to\infty$ in $D^{+}$, while the terms in (\ref{asy3}c) are
 analytic and
 bounded for  $k\to\infty$ in $D^{-}$. Then an application of Jordan's
 lemma implies that these terms gives zero contribution provided
 $\Delta(k)\neq 0$ for k in $D$. This is indeed the case, as it can be
 proven rigorously,  using the fact that this determinant is an
 entire functions of finite order (the {\em order} of an entire
 function is a measure of its rate of growth as $k\to\infty$ - in
 this case the order is in fact equal to one). For such entire
 functions there exists a rich and classical theory, starting with the 
Hadamard
 factorization theorem, which implies that such a function has
 infinitely many zeros accumulating at infinity.
 The determination of these zeros in the complex plane is more
 difficult: in the present case,  the general theory
 implies that the zeros must  lie ``near'' the lines $\{{\rm
 arg}(k)=\pi/6,  \;{\rm
 arg}(k)=5\pi/6,   \;{\rm
 arg}(k)=3\pi/2\}$, at least asymptotically. By using the special 
symmetry properties of
 $\Delta(k)$ one can then prove that asymptotically they lie {\em 
exactly}
 on these lines, and that they are always outside $D$.
%  , except possibly
%  in a ball of finite radius around the origin, which can always be
%  removed using analiticity.
 For the proof, which is tedious but
 elementary,  see
 \cite{pel2}.


\section{Alternative representations for second order problems and
more complicated boundary conditions}\label{sec5}
\setcounter{equation}{0}
\subsection{Alternative representations}
The general representation (\ref{intrep1}) implies alternative
representations for the Dirichlet \bvps\, associated with the 
equations
(\ref{eq1}) and (\ref{eq2}). Regarding these alternative
representations, we note the following:

{\bf (i)} For example (b), equation (\ref{intrep1}) becomes
\bea
q(x,t)&=&\frac{1}{2\pi}\left\{\int_{-\infty}^{\infty}{\rm
e}^{ikx-(k^{2}-ik)t}\hat{q}_{0}(k)dk-
 \int_{\partial D^{+}}{\rm e}^{ikx-(k^{2}-ik)t}((1+ik)\tilde{f}_{0}
 +\tilde{f}_{1})dk\right.
 \nonumber \\
 &-&\left.
\int_{\partial D^{-}}{\rm e}^{ik(x-L)-
(k^{2}-ik)t}((1+ik)\tilde{g}_{0}
 +\tilde{g}_{1})dk\right\},
\label{intrepb}
\eea
where $D^{+}$ and $D^{-}$ are the interior of the upper and lower
branch of the hyperbola defined by equation (\ref{hypb}). Solving
equations (\ref{sys2}) for $\tilde{f}_{1}$ and $\tilde{g}_{1}$ and
substituting the resulting expressions in (\ref{intrepb}) we find
\bea
q(x,t)&=&\frac{1}{2\pi}\left\{\int_{-\infty}^{\infty}{\rm
e}^{ikx-(k^{2}-ik)t}\hat{q}_{0}(k)dk
 \right.
\label{eq2ex}\\
&-&
 \int_{\partial D^{+}}{\rm
 e}^{ikx-(k^{2}-ik)t}\left[(i+ik)\tilde{f}_{0}(t,k)+\frac{{\rm
 e}^{-ikL}N(k,t)-{\rm
 e}^{-i(i-k)L}N(i-k,t)}{\Delta(k)}\right]dk
\nonumber
\\
&-&\left.\int_{\partial D^{-}}{\rm
 e}^{ik(x-L)-(k^{2}-ik)t}\left[(i+ik)\tilde{g}_{0}(t,k)
 +\frac{N(i-k,t)-N(k,t)}{\Delta(k)}\right]dk,
  \right\}
\nonumber \eea where \be \Delta(k)={\rm e}^{-ikL}-{\rm
e}^{-i(i-k)L}, \label{detb} \ee and $N(k,t)$ is defined by
(\ref{N2}). We note that just as with example (c), the determinant
$\Delta(k)$ is {\em not} zero for $k$ in $D$.

Equation (\ref{eq2ex}) shows that there exists a representation of
the solution of the Dirichlet problem for equation (\ref{eq2})
involving {\em only} integrals. It is of course possible to rewrite 
the
representation (\ref{eq2ex}) in the series form (\ref{sinrep2}) using
Cauchy's theorem.  Indeed, it is easy to verify by studying the
boundedness of each exponential involved in (\ref{eq2ex})
that the
integrals along $\partial D^{+}$ and
$\partial D^{-}$ can be deformed to the real line.  This deformation 
passes through
the line Im$(k)=\frac{1}{2}$ where the zeros
$k_{m}=\frac{m\pi}{L}+\frac{i}{2}$ of $\Delta(k)$ lie.
Computing explicitly the  residues at these poles, and manipulating
the result, it is easy to verify  that all integral terms cancel out.
Thus we obtain
% \bea
\be
q(x,t)=
%&=&\frac{1}{2\pi}\left\{\int_{-\infty}^{\infty}{\rm
% e}^{ikx-(k^{2}-ik)t}\hat{q}_{0}(k)dk-
%  \int_{-\infty}^{\infty}{\rm e}^{ikx-(k^{2}-ik)t}\frac{{\rm 
%e}^{2ikL+L}\hat{q}_{0}(k)
%  -\hat{q}_{0}(i-k)}{{\rm e}^{2ikL+L}-1}dk\right.
%  \nonumber \\
%  &+&\left.
% \int_{-\infty}^{\infty}{\rm e}^{ikx-(k^{2}-ik)t}
% \frac{\hat{q}_{0}(k)-\hat{q}_{0}(i-k)}{{\rm
% e}^{2ikL+L}-1}dk\right\}
% \label{seriesb} \\
% &+&
\frac{1}{2L}\sum_{m\in{\mathbb Z}}{\rm e}^{-(k^{2}_{m}-ik_{m})t}
 \left({\rm e}^{ik_{m}x}-{\rm e}^{i(i-k_{m})x}\right)
 \left(\hat{q}_{0}(k_{m})-\hat{q}_{0}(i-k_{m})\right).
\label{seriesb}
\ee
% % \;\;k_{m}=\frac{m\pi}{L}+\frac{i}{2}.
% \nonumber
% \eea
% Finally :
% $$
% \int_{-\infty}^{\infty}{\rm e}^{ikx-(k^{2}-ik)t}\left\{
% \hat{q}_{0}(k)-\frac{{\rm e}^{2ikL+L}\hat{q}_{0}(k)
%  -\hat{q}_{0}(i-k)}{{\rm 
%e}^{2ikL+L}-1}+\frac{\hat{q}_{0}(k)-\hat{q}_{0}(i-k)}{{\rm
% e}^{2ikL+L}-1}\right\}dk=0.
% $$

{\bf (ii)} For the example (a), $\Delta(k)=\E-{\rm e}^{-ikL}$,
thus  $\Delta(k)$ {\em has} zeros in $D$. Hence the
part of the contour $\partial D^{+}$ along the positive real axis must
be deformed to a small circle above each of the points
$k_{m}\in\,{\mathbb R}$, $m\in\,{\mathbb
Z}^{+}$. Similarly the part
$\partial D^{-}$ along the negative real axis must
be deformed to a small circle below each of the points
$k_{m}\in\,{\mathbb R}$, $m\in\,{\mathbb
Z}^{-}$.


\subsection{The Robin problem for the heat equation}\label{sec5.2}
Let $q(x,t)$ satisfy the heat equation
\be
q_{t}=q_{xx},\quad 0<x<L, \;\;t>0,
\label{heat}
\ee
with initial condition $q_{0}(x)$ and with Robin  boundary
conditions at both ends
\be
q_{x}(0,t)-\alpha q(0,t)=h_{1}(t),\quad
q_{x}(L,t)-\beta q(L,t)=h_{2}(t),\quad t>0,
\label{robin}
\ee
where $\alpha$ and $\beta$ are given real constants, $\alpha\neq 
\beta$.

In this case, $X(x,t,k)=q_x+ikq$, thus the global relation is
\be
\tilde{f}_{1}+ik\tilde{f}_{0}-{\rm
e}^{-ikL}(\tilde{g}_{1}+ik\tilde{g}_{0})=\hat{q}_{0}(k)-{\rm
e}^{k^{2}t}\hat{q}(t,k),\quad k\in{\mathbb C}.
\label{grheat}
\ee
Using the boundary conditions (\ref{robin}), the global
relation becomes
\be
(\alpha+ik)\tilde{f}_{0}-{\rm
e}^{-ikL}(\beta+ik)\tilde{g}_{0}=N(k,t)-{\rm
e}^{k^{2}t}\hat{q}(t,k),\quad k\in{\mathbb C},
\label{grheat2}
\ee
where the known function
$N(k,t)$ is given by
\be
N(k,t)=\hat{q}_{0}(k)+\int_{0}^{t}{\rm e}^{k^{2}s}\left({\rm
e}^{-ikL}h_{2}(s)-h_{1}(s)\right)ds.
\label{Nheat}
\ee
Since $\lambda_{1}=-k$, replacing $k$
by $-k$ in equation (\ref{grheat2}) we find
\be
(\alpha-ik)\tilde{f}_{0}-{\rm
e}^{ikL}(\beta-ik)\tilde{g}_{0}=N(-k,t)-{\rm
e}^{k^{2}t}\hat{q}(t,-k),\quad k\in{\mathbb C}.
\label{grheat3}
\ee
Solving the system (\ref{grheat2})-(\ref{grheat3}) we obtain
\bea
\Delta(k)\tilde{f}_{0}&=&{\rm
e}^{ikL}(\beta-ik)N(k,t)-{\rm
e}^{-ikL}(\beta+ik)N(-k,t)
\nonumber \\
&&-{\rm
e}^{k^{2}t}\left[{\rm
e}^{ikL}(\beta-ik)\hat{q}(t,k)-{\rm
e}^{-ikL}(\beta+ik)\hat{q}(t,-k)\right],
\label{solveheat1}
\\
\Delta(k)\tilde{g}_{0}&=&(\alpha-ik)N(k,t)-(\alpha+ik)N(-k,t)
\nonumber \\
&&
-{\rm
e}^{k^{2}t}\left[(\alpha-ik)\hat{q}(t,k)-(\alpha+ik)\hat{q}(t,-k)\right],
\label{solveheat2}
\eea
where
\be
\Delta(k)=(\alpha+ik)(\beta-ik)\E-(\alpha-ik)(\beta+ik){\rm
e}^{-ikL}.
\label{heatdet}
\ee
Let $k_{m}$ be defined by
\be
k_{m}:\quad (\alpha+ik_{m})(\beta-ik_{m}){\rm
e}^{ik_{m}L}=(\alpha-ik_{m})(\beta+ik_{m}){\rm
e}^{-ik_{m}L}.
\label{zerosheat}
\ee
Evaluating equation (\ref{solveheat1})
%and (\ref{solveheat2})
at $k=k_{m}$, where
$\Delta(k_{m})=0$, and using
the definition of
$\hat{q}(t,k_{m})$, we find
$$
\int_{0}^{L}
% {\rm e}^{-ik_{m}x}q(x,t)dx={\rm e}^{-k_{m}^{2}t}N(k_{m},t).
\left[{\rm
e}^{ik_{m}L}(\beta-ik_{m}){\rm e}^{-ik_{m}x}-{\rm
e}^{-ik_{m}L}(\beta+ik_{m}){\rm
e}^{ik_{m}x}\right]q(x,t)dx=
$$
\be {\rm e}^{-k_{m}^{2}t}\left[
{\rm
e}^{ik_{m}L}(\beta-ik_{m})N(k_{m},t)-{\rm
e}^{-ik_{m}L}(\beta+ik_{m})N(-k_{m},t)\right].
\label{hfin} \ee Equation (\ref{zerosheat}) cannot be solved directly 
for the $k_{m}$'s. Thus although the general theory implies that 
equation (\ref{hfin}) can be
solved for $q(x,t)$, it does not appear to yield an effective
representation for the solution of this problem. On the other
hand, the general representation (\ref{intrep1}) {\em does}
provide an effective representation:
$$
q(x,t)=\frac{1}{2\pi}\left\{\int_{\partial D^{+}}{\rm
e}^{ikx-k^{2}t}\left[(ik+\alpha)\frac{{\rm
e}^{-ikL}(\beta+ik)N(-k,t)
-{\rm
e}^{ikL}(\beta-ik)N(k,t) 
}{\Delta(k)}
%+\frac{
-\alpha\tilde{h_{1}}(t,k)
% } {\Delta(k)}
\right]dk+
\right.
$$
\be
\left.
\int_{\partial D^{-}}{\rm
 e}^{ik(x-L)-k^{2}t}\left[(ik+\beta)\frac{
(\alpha+ik)N(-k,t)-(\alpha-ik)N(k,t)}{\Delta(k)}-
% +\frac{
\beta\tilde{h_{2}}(t,k)
% }{\Delta(k)}
\right]dk+\int_{-\infty}^{\infty}{\rm
e}^{ikx-k^{2}t}\hat{q}_{0}(k)dk
\right\},
\label{eq4ex} \ee
where  $\partial D^{+}$ and  $\partial D^{-}$ are the boundaries of
the domains
$$
D^{+}={\{\frac{\pi}{4}\leq \rm arg}(k)\leq
\frac{3\pi}{4}\},\quad
D^{-}={\{\frac{5\pi}{4}\leq \rm arg}(k)\leq
\frac{7\pi}{4}\}, \quad D=D^{+}\cup D^{-},
$$
shown in figure \ref{heatdom}, and 
$$
\tilde{h_{1}}(t,k)=\int_{0}^{t}{\rm e}^{k^{2}s}h_{1}(s)ds, \quad
\tilde{h_{2}}(t,k)=\int_{0}^{t}{\rm e}^{k^{2}s}
% {\rm e}^{-ikL}
h_{2}(s)ds.
$$

\begin{figure}
\begin{center}
 \includegraphics[angle=180]{IMAfig3.eps}
 %\centerline[angle=270]{\epsfbox{s2003fig2.eps}}
 \caption{The domain $D$ and the contours $\partial D^{+}$ (in red) 
and
 $\partial D^{-}$ (in green) for the heat equation }
 \label{heatdom}
\end{center}
 \end{figure}

To derive equation (\ref{eq4ex}) we use the solutions
(\ref{solveheat1}), (\ref{solveheat2}) of
the system (\ref{grheat2})-(\ref{grheat3}).
In the expressions (\ref{solveheat1}) there appears
also a term involving the function $\hat{q}(t,k)$. This term, when
multiplied by ${\rm e}^{ikx-k^{2}t}$, is analytic and
 bounded for  $k\to\infty$ in $D^{+}$. Similarly, the term in 
(\ref{solveheat2})
 involving the function $\hat{q}(t,k)$, when
multiplied by ${\rm e}^{ik(x-L)-k^{2}t}$, is analytic and
 bounded for  $k\to\infty$ in $D^{-}$.
An application of Jordan's
lemma therefore  implies that these terms give a zero contribution, 
provided
 $\Delta(k)\neq 0$ for k in $D$.
This is indeed the case, and can be seen as follows.  The function
$\Delta(k)$ defined by
(\ref{heatdet}) is entire and of finite order, hence it has infinitely
many zeros accumulating at infinity. Moreover, a simple computation 
shows that this function has infinitely many zeros on the real axis, and
that all zeros are asymptotically on ${\mathbb R}$ \cite{pel2}. It follows that 
the
only one of these zero in $D$ is $k=0$, where the numerator clearly
vanishes. Hence  the unknown
function $\hat{q}(t,k)$ does not  contribute to the solution
representation.




\section{Conclusions}\label{sec6}
We have illustrated the applicability of a {\em transform method} by
solving several concrete boundary value problems. It appears that the method is both
general and simple to implement. Indeed, the only mathematical tools
used in this paper are the Fourier transform and Cauchy's theorem.

An effort has been made to minimise technical considerations. In
particular, the given initial and boundary conditions are assumed to
be ``sufficiently smooth''. It is possible to work in a less
restrictive functional class; for problems on the half line $x>0$
this is done in \cite{fsung2}, where general theorems are proven in
appropriate Sobolev spaces.

The general result about \bvps on an interval is that $q(x,t)$ can be
expressed as an {\em integral in the complex $k$-plane}, see
equation (\ref{intrep1}). This integral
involves the Fourier transform $\hat{q}_{0}(k)$ of the initial
condition $q_{0}(x)$ and the $t$-transforms
$\{\tilde{f}_{j},\tilde{g}_{j}\}$ of {\em all} boundary values
$\{\partial_{x}^{j}q(0,t),\partial_{x}^{j}q(L,t)\}$. A subset of
these boundary values can be prescribed as boundary conditions. Thus
a subset $\tilde{f}_{p}$ and $\tilde{g}_{q}$, where $p$ takes $N$
values and $q$ takes $n-N$ values, can be computed immediately. The
remaining $\tilde{f}_{j}$'s and $\tilde{g}_{j}$'s can be expressed
through the solution of a system of $n$ algebraic equations obtained
from the global relation and from the equations derived from the
global relation by replacing $k$ with $\lambda_{j}(k),
\;j=1,\ldots,n-1$. The relevant expressions involve the unknown
function $\hat{q}(t,k)$ and the function $1/\Delta(k)$, where
$\Delta(k)$ is the determinant of the associated system of $n$
algebraic equations. However, using the integral representation
of $q(x,t)$ (equation (\ref{intrep1})) it can be shown that
{\bf (i)}:  If $\Delta(k)\neq 0$ in the domain $D$ defined by
equation (\ref{Ddef}), then the contribution of the terms involving
$\hat{q}(t,k)$  vanishes.
{\bf (ii)}: If $\Delta(k)$ has zeros in $D^{+}$ (hence also in 
$D^{-}$), then the
contours $\partial D^{+}$ and $\partial D^{-}$ must be deformed to
pass above or below these zeros. For some \bvps\, it is convenient
to rewrite the contribution of these zeros in terms of a sum of
residues.
In these
particular cases there exists an alternative representation
consisting only of an infinite series.

\vspace{2mm}
From the above discussion it follows that there exists three distinct
cases. The basic examples (\ref{eq1})-(\ref{eq3}) were chosen
in order to illustrate these three cases:

{\bf (a)} For equation (\ref{eq1}), the domain $D$ is the union of
the first and third quadrant of the complex $k$-plane. For
Dirichlet (or Neumann) boundary conditions, $\Delta(k)=0$ for
$k=\frac{m\pi}{L}$, $m\in{\mathbb Z}$, thus the contours along the
positive and negative real axes must be deformed. Furthermore, it
can be shown that the relevant integral can be computed entirely
in terms of a sum of residues. Thus the solution can be
rewritten as an infinite series. This is consistent with the well
known form of the solution as given by a sine series.

{\bf (b)} For equation ({\ref{eq2}), the domain $D$ is shown in
figure (\ref{zerosb}). For Dirichlet boundary conditions,
$\Delta(k)=0$ for $k=\frac{m\pi}{L}+\frac{i}{2}$, $m\in{\mathbb
Z}$, thus $\Delta(k)\neq 0$ for $k$ in $D$. However, just as in
(a) above, it is possible to rewrite these integrals in terms of
an infinite sum, see equation (\ref{eq2ex}). This is consistent
with the solution that can be obtained using the set of
eigenfunctions corresponding to the complex eigenvalues of the problem, 
see \cite{dym}.

{\bf (c)} For equation ({\ref{eq3}), the domain $D$ is shown in
figure (\ref{fig1}). For the case that the given  boundary conditions 
are
$q(0,t)$, $q(L,t)$ and $q_{x}(L,t)$, the determinant $\Delta(k)$ is
given by equation (\ref{detc1}). In can be shown that the zeros of
$\Delta(k)$ are asymptotically on the three lines $\{k:\;{\rm 
arg}(k)=\pi/6,
\;{\rm
arg}(k)=5\pi/6\;{\rm
arg}(k)=3\pi/2\}$, and that  $\Delta(k)$ has {\em no} zeros in $D$.


\vspace{2mm} We note that even in the cases when it is possible to
express the integral representation of the solution in terms of an
infinite sum, the integral form may have some advantages. For
example, the integral representation, in contrast with the series
one, is {\em uniformly convergent}  both at $x=0$ and at $x=L$. In
addition, integrals are more convenient than sums for studying the
long time asymptotic behaviour of the solution. Furthermore, for
many concrete examples it is possible to use the integral
representation and Cauchy's theorem to compute $q(x,t)$
explicitly. It is interesting that in these computations, one does
{\em not} compute the relevant integral by using the residues
associated with $\Delta(k)=0$, but one precisely {\em avoids} these
zeros.

We emphasise that the main advantage of the representation
(\ref{intrep1}) is that the general representation of the solution
can be constructed {\em without} the need to analyse in detail the
zeros of $\Delta(k)$. This is to be contrasted with the classical
approach: since the zeros of $\Delta(k)$ define the discrete
spectrum of the associated $x$-differential operator,
a {\em detailed} characterisation of this set is crucial for the 
derivation
of the associated basis of eigenfunctions. This important
advantage of the method presented here is illustrated in section
\ref{sec5.2}.

\vspace{2mm}
We conclude with some remarks:
% {\bf (1)} For the Dirichlet case for both equation (\ref{eq1}) and
% for the heat equation, it is possible to express the solution as an
% inifinite sine series. However, our computation indicates that there
% exists a significant difference between these two cases. The 
%solution
% of equation (\ref{eq1}) contains always the contribution of an 
%infinite sum, while
% the solution of the heat equation can be written {\em without}
% an infinite sum.  This seems to indicate that
% the discrete spectrum of the associated $x$ differential operator
% plays a crucial role for equation (\ref{eq1}), but that this is 
%{\em not}
% the case for the heat equation.

{\bf (1)}
Coupled boundary conditions can also be analysed using our
method.  For example, for the
\bvp\,
$$
q_{t}+q_{xxx}=0,\quad q(x,0)=q_{0}(x),\quad q(0,t)=q(L,t)=0,
q_{x}(L,t)+\alpha q_{x}(0,t)=0,
$$
the solution {\em can} be expressed in a series of eigenfunctions
\cite{zhang}. This is a consequence of the fact that the
associated differential operator with the above boundary
conditions can be made self-adjoint. In this case, it is possible
to deform the integrals appearing in our representation and to
rewrite them in terms of an infinite series.
% In this case $\Delta(k)$ has zeros in $D$, hence the
% basic integral representation (\ref{intrep1}) for this problem
% contains the residue sum. Work on such coupled boundary conditions 
%in
% is progress.

{\bf (2)} Earlier work on two-point boundary value problems for
linear evolution PDEs has appeared in \cite{fp4}, but the existence of
zeros of $\Delta(k)$ was not investigated. The
determination of conditions for well
posedness is presented in \cite{pel}. The role of the zeros of the 
determinant
$\Delta(k)$ and their determination in some particular cases are
studied in
\cite{pel2}.




\section*{Appendix}
\stepcounter{section}
Let $\tilde{q}(x,s)$ be the Laplace transform of $q(x,t)$, i.e. 
$$
\tilde{q}(x,s)=\int_0^{\infty} {\rm e}^{-st}q(x,t)dt,\quad {\rm 
Re}\,(s)>0.
\quad \eqno(A.1)
$$
Applying  the Laplace transform to equation (\ref{eq3}) we find
$$
\tilde{q}_{xxx}+\tilde{q}_x+s\tilde{q}=q_0(x), \quad {\rm Re}\,(s)>0.
\quad \eqno(A.2)
$$
The solutions of the homogeneous version of this equation are given by
$$
\tilde{q}(x,s)={\rm e}^{\lambda x}, \quad \lambda^3+\lambda+s=0.
\quad \eqno(A.3)
$$
 We distinguish the three roots of the cubic equation (A.3) by their 
large $s$ behaviour:
$$
\lambda_1\sim -s^{\frac{1}{3}},\quad \lambda_2\sim -\alpha 
s^{\frac{1}{3}},
\quad \lambda_3\sim -\alpha^2s^{\frac{1}{3}}, \quad \alpha={\rm 
e}^{2\pi i/3}.
\quad \eqno(A.4)
$$
If $-\pi/2<{\rm arg}(s)<\pi/3$ then, for large $s$,
$$
-\frac{\pi}{6}<{\rm arg}(s)<\frac{\pi}{6},\quad 
\frac{\pi}{2}<{\rm arg}(\alpha s)<\frac{5\pi}{6},\quad
\frac{7\pi}{6}<{\rm arg}(\alpha^2s)<\frac{3\pi}{2}.
$$
Thus, for large $s$,
$$
{\rm Re}\,(\lambda_1)<0,\quad {\rm Re}\,(\lambda_j)>0,\;\;j=2,3.
\quad \eqno(A.5)
$$
A solution of the inhomogeneous equation (A.2) is given by
$$
\tilde{q}(x,s)=\sum_{j-1}^3 \beta_j\int_0^x {\rm 
e}^{\lambda_j(x-\xi)}q_0(\xi)d\xi,
\quad \eqno(A.6)
$$
where the constants $\{\beta_j\}_1^3$ satisfy the algebraic conditions
$$
\sum_1^3 \beta_j=0,\quad
\sum_1^3 \lambda_j\beta_j=0,\quad\sum_1^3(1+\lambda_j^2) \beta_j=0.
\quad \eqno(A.7)
$$
In the integral 
$$\int_0^x {\rm e}^{\lambda_j(x-\xi)}q_0(\xi)d\xi
$$
we have  $x-\xi>0$, thus the integrand is bounded as $s\to\infty$, 
if Re$(\lambda_j)<0$. On the other hand, 
if we replace $\int_0^x$ with $-\int_x^{\infty}$, then $x-\xi<0$, and 
the integrand  is bounded if Re$(\lambda_j)>0$. Hence, equations 
(A.5) imply that we must choose the following solution of equation 
(A.2):
$$
\tilde{q}(x,s)=c_1{\rm e}^{\lambda_1(x-L)} +c_2{\rm e}^{\lambda_2x} 
+c_3{\rm e}^{\lambda_3x}
$$
$$
-\beta_1\int_x^L {\rm e}^{\lambda_1(x-\xi)}q_0(\xi)d\xi
+\beta_2\int_0^x{\rm e}^{\lambda_2(x-\xi)}q_0(\xi)d\xi+
\beta_3\int_0^x{\rm e}^{\lambda_3(x-\xi)}q_0(\xi)d\xi,
\quad \eqno(A.8)
$$
where the $c_j$'s are constants independent of $x$.

Let $\tilde{f}_0$, $\tilde{g}_0$ and $\tilde{g}_1$ denote the Laplace 
transforms of the given boundary conditions $f_0(t)$, $g_0(t)$ and 
$g_1(t)$ respectively,
$$
\tilde{f}_0(s)=\int_0^{\infty}{\rm e}^{-st}f_0(t)dt,\quad 
\tilde{g}_0(s)=\int_0^{\infty}{\rm e}^{-st}g_0(t)dt,\quad 
\tilde{g}_1(s)=\int_0^{\infty}{\rm e}^{-st}g_1(t)dt. 
\quad \eqno(A.9)
$$
The definitions (A.1) and (A.9) imply
$$
\tilde{q}(0,s)=\tilde{f}_0(s), \quad
\tilde{q}(L,s)=\tilde{g}_0(s), \quad
\tilde{q}_x(L,s)=\tilde{g}_1(s).
\quad \eqno(A.10)
$$
Using equation (A.8) to evaluate $\tilde{q}(0,s)$, 
$\tilde{q}(L,s)$ and $\tilde{q}_x(L,s)$ we find the following set of 
three algebraic equations for $\{c_j\}_1^3$:
$$
{\rm 
e}^{-\lambda_1L}c_1+c_2+c_3=\beta_1\hat{q}_0(-\lambda_1)+\tilde{f}_0(s),
$$
$$
c_1+{\rm e}^{\lambda_2L}c_2+{\rm e}^{\lambda_3L}c_3=-\beta_2{\rm 
e}^{\lambda_2L}\hat{q}_0(-\lambda_2)-\beta_3{\rm 
e}^{\lambda_3L}\hat{q}_0(-\lambda_3)
+\tilde{g}_0(s),
$$
$$
\lambda_1c_1+\lambda_2{\rm e}^{\lambda_2L}c_2+\lambda_3
{\rm e}^{\lambda_3L}c_3=
-\beta_2\lambda_2{\rm e}^{\lambda_2L}\hat{q}_0(-\lambda_2)-
\beta_3\lambda_3{\rm e}^{\lambda_3L}\hat{q}_0(-\lambda_3)
+\tilde{g}_1(s),
\quad \eqno(A.11)
$$
where 
$$
\hat{q}_0(\lambda_j)=\int_0^L{\rm e}^{\lambda_j\xi}q_0(\xi)d\xi.
\quad \eqno(A.12)
$$
The determinant of the system (A.11) is given by
$$
\Delta=(\lambda_3-\lambda_2){\rm e}^{(\lambda_2+\lambda_3-\lambda_1)L}
+(\lambda_1-\lambda_3){\rm e}^{\lambda_3L}+
(\lambda_2-\lambda_1){\rm e}^{\lambda_2L}.
\quad \eqno(A.13)
$$
It can be shown that, as $s\to\infty$, the zeros of $\Delta$ are on 
the real axis. Indeed, recall that the zeros of
the analogous determinant in the complex $k$ plane   lie,  for large 
$k$, 
on the half lines arg$(k)=\pi/6$, arg$(k)=5\pi/6$ and 
arg$(k)=3\pi/2$.
Letting
$
s=i(k^3-k)$, the determinant (\ref{detc1}) reduces to the determinant 
(A.13), and furthermore the above three rays are mapped to the real 
axis in the complex $s$ plane.

Since $\tilde{q}(x,s)$ has infinitely many  singularities for 
Re$(s)>0$, we {\em cannot} define an inverse Laplace transform to 
compute the solution $q(x,t)$ and the method {\em fails}.

\subsubsection{Remarks}
1. Solving equations (A.11) for the $c_j$'s and replacing the 
resulting expressions in equation (A.8) we find an expression for 
$\tilde{q}(x,s)$ which, as $s\to\infty$, can be shown to be bounded 
for Re$(s)>0$, Im$(s)\neq 0$. Actually, the relevant proof is 
identical to the proof of section \ref{sec4} which establishes that 
for this problem well-posedness demands one boundary condition at 
$x=0$ and two boundary conditions at $x=L$. 

2. Although the classical  
Laplace transform method fails, it is possible to use the following 
transform instead of the Laplace transform:
$$
\tilde{q}(x,s,t)=\int_0^t{\rm e}^{-s\tau}q(x,\tau)d\tau.
\quad \eqno(A.14)
$$
This gives rise to the additional unknown term ${\rm e}^{-st}q(x,t)$. 
One must use this term to control the singularities of $\Delta$ and 
then one must eliminate $q(x,t)$ using analyticity arguments. 

3. The application of the Laplace transform involves the solution of 
two sets of algebraic equations, namely equations (A.7) and (A.11), 
while the application of the transform method used in this paper uses 
only the solution of one set of algebraic equations. Furthermore, the 
complex representation is based on the three roots of the {\em 
cubic} equation (A.3), while the representation (\ref{intrep1}) is 
based on the two roots of the {\em quadratic} equation 
$$\lambda^2+\lambda k+k^2-1=0.$$

4. If one uses the transform (A.14)  
instead of the Laplace transform, and if one then uses the change of 
variables
$s=i(k^3-k)$, one can then obtain the representation (\ref{eq3ex}) 
after applying certain analyticity arguments to combine terms 
involving $\int_0^x$ and $-\int_x^{\infty}$ to terms of the form 
$\int_0^{\infty}$.
  
  \section*{Acknowlegdements}
The authors wish to thank A.R. Its and D. Powers for many useful
discussions.

BP gratefully acknowledges the support of the Nuffield foundation 
grant
NAL/00608/GR.

%\newpage
\begin{thebibliography}{1}

% \bibitem{afok}
% M.J Ablowitz and A.S. Fokas,
% \newblock {\em Introduction and applications of complex variables},
% \newblock Cambridge University Press, 1997.

\bibitem{akhie}
N.I.Akhiezer and I.M.Glazman,
\newblock
{\em Theory of linear operators in Hilbert spaces},
\newblock
Nauka, 1966.

\bibitem{cg}
T. Colin and J.M. Ghidaglia,
\newblock
An initial-boundary value problem for the Korteweg-deVries equation
posed on a finite interval,
\newblock {\em Adv. in Diff. Eq.}, {\bf 6(12)} 2001, 1463-1492.

% \bibitem{dp}
% V. A. Dougalis and  B. Pelloni,
% \newblock
% Numerical modelling of two-way propagation of non-linear dispersive 
waves,
% \newblock {\em Math. and Comp. Sim.}. 55: 595-606, 2001.

\bibitem{dym}
H. Dym and  H.P. McKean,
\newblock
{\em Fourier series and integrals},
\newblock Academic Press, 1972.

\bibitem{fok}
A.S. Fokas,
\newblock A unified transform method for solving linear and certain
nonlinear
  {PDE}'s,
\newblock {\em Proc. Royal Soc. Series A}, {\bf 453} 1997, 1411--1443.

\bibitem{best}
A.S. Fokas,
\newblock Two dimensional linear {P}{D}{E}'s in a convex polygon
\newblock {\em Proc. Royal Soc. Series A}, {\bf  457} 2001, 371--393.

\bibitem{ima1}
A.S. Fokas,
\newblock A new transform method for evolution PDEs,
\newblock {\em IMA J. Appl. Math}, {\bf 67} 2002, 559-590.

% \bibitem{newfok2}
% A.S. Fokas,
% \newblock Integrable nonlinear evolution equations on the half line,
% \newblock {\em Comm. Math. Phys.}, 230: 1-39, 2002.
%
%
% \bibitem{fp2}
% A.S. Fokas and B.~Pelloni,
% \newblock Integral transform, spectral representations and the d-bar
% problem,
% \newblock {\em Proc. Royal Soc. Series A}, 456:805--833, 2000.
%
% \bibitem{fp3}
% A.S. Fokas and B.~Pelloni,
% \newblock Moving boundary value problems\ldots
% \newblock {\em Phys. Rev. Lett.}, \ldots, 2001.

\bibitem{fp4}
A.S. Fokas and B.~Pelloni,
\newblock Two-point boundary value problems for linear evolution
equations
\newblock {\em Proc. Camb. Phil. Soc.} {\bf 17} 2001,
919-935.

% \bibitem{fp5}
% A.S. Fokas and B. Pelloni,
% \newblock
% Boundary value problems for Boussinesq type systems,
% \newblock submitted (2003).

\bibitem{fsung2}
A.S. Fokas and L.Y. Sung,
\newblock Initial boundary value problems for linear evolution
equations on the
  half line
\newblock (preprint) 2002

\bibitem{pel}
B.~Pelloni.
\newblock Well-posed boundary value problems for
linear evolution equations on a finite interval,
\newblock  {\em Proc. Camb. Phil. Soc.} (in press).

\bibitem{pel2}
B.~Pelloni.
\newblock
The spectral representation of two-point boundary value problems for
linear PDEs
\newblock submitted (2003)

\bibitem{tit}
C.E. Titchmarsh,
\newblock {\em The theory of functions},
\newblock Oxford University Press, 1962.

\bibitem{zhang}
B.Y. Zhang,
\newblock  Exact boundary controllability of the Korteweg-de
   Vries equation,
\newblock {\em SIAM J. Control Optim.} {\bf 37} 1999, 543-565 .

\end{thebibliography}

\end{document}


math0412027.extracted_bib
\begin{thebibliography}{1}

% \bibitem{afok}
% M.J Ablowitz and A.S. Fokas,
% \newblock {\em Introduction and applications of complex variables},
% \newblock Cambridge University Press, 1997.

\bibitem{akhie}
N.I.Akhiezer and I.M.Glazman,
\newblock
{\em Theory of linear operators in Hilbert spaces},
\newblock
Nauka, 1966.

\bibitem{cg}
T. Colin and J.M. Ghidaglia,
\newblock
An initial-boundary value problem for the Korteweg-deVries equation
posed on a finite interval,
\newblock {\em Adv. in Diff. Eq.}, {\bf 6(12)} 2001, 1463-1492.

% \bibitem{dp}
% V. A. Dougalis and  B. Pelloni,
% \newblock
% Numerical modelling of two-way propagation of non-linear dispersive 
waves,
% \newblock {\em Math. and Comp. Sim.}. 55: 595-606, 2001.

\bibitem{dym}
H. Dym and  H.P. McKean,
\newblock
{\em Fourier series and integrals},
\newblock Academic Press, 1972.

\bibitem{fok}
A.S. Fokas,
\newblock A unified transform method for solving linear and certain
nonlinear
  {PDE}'s,
\newblock {\em Proc. Royal Soc. Series A}, {\bf 453} 1997, 1411--1443.

\bibitem{best}
A.S. Fokas,
\newblock Two dimensional linear {P}{D}{E}'s in a convex polygon
\newblock {\em Proc. Royal Soc. Series A}, {\bf  457} 2001, 371--393.

\bibitem{ima1}
A.S. Fokas,
\newblock A new transform method for evolution PDEs,
\newblock {\em IMA J. Appl. Math}, {\bf 67} 2002, 559-590.

% \bibitem{newfok2}
% A.S. Fokas,
% \newblock Integrable nonlinear evolution equations on the half line,
% \newblock {\em Comm. Math. Phys.}, 230: 1-39, 2002.
%
%
% \bibitem{fp2}
% A.S. Fokas and B.~Pelloni,
% \newblock Integral transform, spectral representations and the d-bar
% problem,
% \newblock {\em Proc. Royal Soc. Series A}, 456:805--833, 2000.
%
% \bibitem{fp3}
% A.S. Fokas and B.~Pelloni,
% \newblock Moving boundary value problems\ldots
% \newblock {\em Phys. Rev. Lett.}, \ldots, 2001.

\bibitem{fp4}
A.S. Fokas and B.~Pelloni,
\newblock Two-point boundary value problems for linear evolution
equations
\newblock {\em Proc. Camb. Phil. Soc.} {\bf 17} 2001,
919-935.

% \bibitem{fp5}
% A.S. Fokas and B. Pelloni,
% \newblock
% Boundary value problems for Boussinesq type systems,
% \newblock submitted (2003).

\bibitem{fsung2}
A.S. Fokas and L.Y. Sung,
\newblock Initial boundary value problems for linear evolution
equations on the
  half line
\newblock (preprint) 2002

\bibitem{pel}
B.~Pelloni.
\newblock Well-posed boundary value problems for
linear evolution equations on a finite interval,
\newblock  {\em Proc. Camb. Phil. Soc.} (in press).

\bibitem{pel2}
B.~Pelloni.
\newblock
The spectral representation of two-point boundary value problems for
linear PDEs
\newblock submitted (2003)

\bibitem{tit}
C.E. Titchmarsh,
\newblock {\em The theory of functions},
\newblock Oxford University Press, 1962.

\bibitem{zhang}
B.Y. Zhang,
\newblock  Exact boundary controllability of the Korteweg-de
   Vries equation,
\newblock {\em SIAM J. Control Optim.} {\bf 37} 1999, 543-565 .

\end{thebibliography}