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\begin{document}

\title[Adjoint approach to the identification of an unknown diffusion
coefficient]{Analysis of an adjoint problem approach to the identification
of an unknown diffusion coefficient}

\author{P. DuChateau\dag\S, R. Thelwell\dag\ and G. Butters\ddag}

\address{\dag\ Department of Mathematics, Colorado State University, Fort
Collins, Colorado 80523.
The research of R. Thelwell is partially supported by the
National Science Foundation, DUE-0324175 and 0221595.
}

\address{\ddag\ Department of Soil and Crop Sciences, Colorado State University, Fort Collins, Colorado 80523}

\ead{\S pauld@math.colostate.edu}

\begin{abstract}
An inverse problem for the
identification of an unknown coefficient in a quasilinear parabolic partial
differential equation is considered. We present an approach based on
utilizing adjoint versions of the direct problem in order to derive
equations explicitly relating changes in inputs (coefficients) to changes in
outputs (measured data). Using these equations it is possible to show that
the coefficient to data mappings are continuous, strictly monotone and
injective. The equations are further exploited to construct an approximate
solution to the inverse problem and to analyze the error in the
approximation. Finally, results of some numerical experiments are displayed.

\noindent
{\footnotesize \textbf{Keywords.} Inverse problems, quasilinear parabolic
equations, parameter identification, adjoint method, ill-posed problems}
\end{abstract}

\ams{35R30}

%\maketitle

\section{Introduction}

\noindent Using partial differential equations to model physical systems is
one of the oldest activities in applied mathematics. A complete model
requires certain state inputs in the form of initial and/or boundary data
together with what might be called structure inputs such as coefficients or
source terms which are related to the physical properties of the system.
Obtaining a unique solution for the associated well posed problem
constitutes what we will call solving the direct problem. Solving the direct
problem permits the computation of various system outputs of physical
interest. On the other hand, when some of the required inputs are not
available we may instead be able to determine the missing inputs from
outputs that are measured rather than computed by formulating and solving an
appropriate inverse problem. In particular, when the missing inputs are one
or more unknown coefficients in the partial differential equation, the
problem is called a coefficient identification problem. The identification
of a diffusion coefficient in a quasilinear diffusion equation is chosen
here as a prototype coefficient identification problem that has been
approached by various methods.

The most common technique for identifying an unknown coefficient from
some measured output is the method of output least squares $\left[ 1,4,8,9,10%
\right] .$ Here the unknown coefficient, $C,$ is chosen from an appropriate
space $K$ and the output, $\Phi[C],$ is computed by solving
the direct problem. One defines an error functional, $J[C]
=\left| \left| \,\Phi[C]-f \right| \right|_{F}^{2}$ ,
comparing the computed output to the measured value, $f,$ in the norm of the
output space, $F,$ and seeks to minimize $J$ over $K.$ Output least squares
(OLS) methods are very general and can be efficiently programmed for
computer implementation. Typically there are problems with lack of
uniqueness, convergence to false minima, and instability under parameter
mesh refinement, although a skillful user may be able to incorporate a
priori information about the solution into the parametric description of the
unknown coefficient in order to lessen some of these difficulties $\left[ 1,9%
\right] $. Since the connection between the inputs and outputs is expressed
only indirectly through the solver, general information about an input to
output mapping is not readily available from OLS methods.

An alternative to coefficient identification by output
least squares is the so called equation error method $\left[ 3,6,7,11,12%
\right] .$ Here the measured overspecification is used as input to the
differential equation in the direct problem which is viewed then as an
equation for the unknown coefficient. This equation expresses a direct
relationship between the unknown coefficient values and the measured data
values. Since the relationship is frequently quite complicated, it is not
easy to discern from it properties of an input to output mapping. Equation
error methods are quite problem dependent and produce varying degrees of
success.

The method described in this paper is based on an integral equation
relating changes in the unknown coefficient to corresponding changes in the
measured output. The integral equation is derived by exploiting a problem
which is adjoint to the direct problem, an idea close to the techniques
often used to estimate sensitivity in the OLS approach $\left[ 8,9\right] .$
However, this integral equation provides information about the input/output
mapping itself rather than the error functional. It is possible then to
prove that the input to output map is continuous, monotone and injective.
Moreover, it is shown that when the input/output map is restricted to a
(finite dimensional) space of polygonal coefficients, it is explicitly
invertible. This observation provides the basis for a method for numerically
approximating the unknown coefficient. It is shown that a unique polygonal
approximation to the unknown coefficient is obtained by solving a triangular
system of linear algebraic equations. Error estimates show that the accuracy
of the approximation is limited by the precision of the data measurements so
that there is an optimal attainable accuracy but exact determination of the
coefficient is never possible.

Results of a few numerical experiments are provided here to
illustrate the working of the method. A more extensive presentation of
numerical experiments will be included in a later publication.

\section{Analysis of the Direct and Inverse Problems}

\noindent Consider the following IBVP for a quasilinear conduction/diffusion
equation on the domain $Q_{T}=\{ 0<x<1,\,0<t<T\} $,\qquad \qquad 
\begin{eqnarray}
\label{direct}
\partial _{t}u( x,t) &=&\partial _{x}( D( u)\partial _{x}u(x,t))=\partial _{xx}B(u(x,t)) \quad \mbox{ on } Q_{T}  \\
u(x,0) &=&f(0)  \qquad  0<x<1 \nonumber \\
u(0,t) &=&f(t); \quad \partial _{x} u(1,t) =0, \qquad 0<t<T.  \nonumber
\end{eqnarray}

Here \hspace{2in} $\disp{ B(u) = \int_{f(0)}^{u}D(s) \ ds }$

and we suppose

\begin{equation} 
\label{condf}
f\in C^{1}[0,T] \mbox{ and }f'(t)>0 \mbox{ for } t>0
\end{equation}

For f satisfying (\ref{condf}), we let $J=[ f(0),f(T)],$ and then suppose that for positive constants, $D_* \leq  D^*$ and $K,$

\begin{equation}
\label{condD}
\begin{array}{rllc}
D_* \leq  D(u) & \leq D^*, & \mbox{ for } u\in J & \mbox{(i)}\\
| D(\mu_2)- D(\mu_1)| & \leq  K| \mu_2-\mu_1| & \forall \mu_1,\mu_2
\in J & \mbox{(ii)} 
\end{array}
\end{equation}

Note that any polygonal function (i.e., a continuous and piecewise linear
function) satisfies (\ref{condD}i) and that any function satisfying 
both conditions of (\ref{condD}) is bounded away from zero and has at most finitely many zeroes on J.

Given $f$ satisfying (\ref{condf}) and $D(u)$ satisfying (\ref{condD}), 
the so called direct problem (\ref{direct}) has a unique weak 
solution $u=u(x,t)$ satisfying

\[
\begin{array}{l}
{\phantom{\partial_t}}u \in  L^2 [0,T:H^1(0,1) \cap C[0,T:L^2(0,1)] \\ 
\partial_t u \in  L^2[0,T:H^{-1}(0,1)].
\end{array} 
\]

Here, we consider the inverse problem in which the coefficient $ D=D(u)$
is to be identified from measured output data. There
are a variety of output measurements that are experimentally feasible in any
given physical setting; we are going to base our identification on one or
the other of the following observations at the boundary,
\begin{eqnarray*}
g(t) &=&-D(u(0,t)) \partial_x u(0,t) \quad \mbox{ or} \\
h(t) &=&u(1,t), \quad 0<t<T
\end{eqnarray*}

If we denote the class of uniformly positive, Lipschitz coefficients $D$
satisfying (\ref{condD}) by $W(J),$ then for a fixed $f$ satisfying 
(\ref{condf}), we can define mappings
\begin{eqnarray*}
\Phi \mbox{ and } \Psi &:&W(J) \longrightarrow L^{2}[0,T] \\
\Phi [f,D] &=&g \\
\Psi [f,D] &=&h,
\end{eqnarray*}

which assign to a coefficient $D$ from $W(J)$, the flux data, $
g$ or the function value data, $h,$ obtained by solving the direct problem
(\ref{direct}) with inputs $f$ and $D$. Then solving the inverse
problem will amount to inverting these mappings.

We begin with a result about the IBVP (\ref{direct}).

\begin{lemma} \rm
\label{L:lem1}
Suppose $f$ and $D$ satisfy (\ref{condf}) and (\ref{condD}) and let $u=u(x,t)$ denote the corresponding solution of (\ref{direct}). Then

\noindent a) for each $t \in ( 0,T),\quad f(0) \leq u(x,t) \leq f(t), \quad 0\leq x\leq 1$

\noindent b) $\partial_{x}u(x,t)<0 \quad$ a.e. on $Q_{T}$
\end{lemma}

\begin{proof}
It follows from (\ref{direct}) that

$ \begin{array}{rll}
\partial_t [f(t)-u(x,t)]-\partial_{xx}[B(f(t)) - B(u(x,t))] =&f'(t) &
\mbox{ on } Q_{T} \\
f(0)-u(x,0) =& 0 & 0<x<1 \\
f(t)-u(0,t) =&0, & 0<t<T, \\
\partial_x[f(t)-u(1,t) ] =&0, & 0<t<T.
\end{array}$

Then we multiply the equation by an arbitrary test function, $\psi(x,t),$
and integrate by parts,

\begin{eqnarray*}
&&-\int \int_{Q_T}[(f-u)\partial_t \psi +(B(f)-B(u)) \partial_{xx}\psi ] \
dxdt \\
&&+\int_{0}^{1}(f-u) \psi |_{t=0}^{t=T} \ dx \\
&&-\int_{0}^{T}[ \psi \partial_x (B(f)-B(u))-(B(f)-B(u)) \partial_x \psi]_{x=0}^{x=1} \ dt \\
&=&\int \int_{Q_T} f'(t) \psi \ dxdt.
\end{eqnarray*}

Note that $  \ \qquad \qquad B(f(t)) -B(u(x,t))=k(x,t)(f-u)$

where we define $\qquad \qquad   k(x,t)=D(\mu(x,t)) $
for $\mu(x,t)$ between $f(t)$ and $u(x,t).$

Next we require $\psi(x,t)$ to solve the adjoint problem, 
\[
\begin{array}{rll}
\partial_t \psi(x,t) +k(x,t) \partial_{xx}\psi(x,t) =&F(x,t) & \mbox{
in  } Q_T, \\
\psi(x,T) =&0, & 0<x<1, \\
\psi(0,t) =&0, \quad \partial _{x}\psi(1,t) =0, & 0<t<T,
\end{array}
\]

for a smooth function, $F(x,t).$ Then the integral expression
above reduces to

\begin{equation}
\label{rexp1}
-\int \int_{Q_T}(f-u)\,F(x,t) \ dxdt = \int \int_{Q_T} f'(t)\psi(x,t) \ dxdt
\end{equation}

The smoothness of $k(x,t)$ and $F(x,t)$ imply that
the strong maximum principle can be applied to the adjoint problem to
conclude that if the function $F( x,t) $ is positive in $Q_{T},$
then $\psi ( x,t) <0\;in\;Q_{T}.$ Since f satisfies (\ref{condf}), it
follows that for every function $F(x,t) $ which is positive in $
Q_{T}$ the right side of (\ref{rexp1}) is negative, which is to say, 
for every $F(x,t),$ smooth and positive in $Q_{T},$

\[
\int \int_{Q_T}(f-u) \ F(x,t) \ dxdt>0.
\]

But this is just the assertion that $f(t)-u(x,t)$ is positive in the sense of distributions on $Q_T.$ Given the
smoothness of the solution $u(x,t)$ this means $f(t)>u(x,t)$ almost everywhere on $Q_T.$
Applying the same reasoning to $u(x,t)-f(0),$ we arrive at the expression

\[
 -\int \int_{Q_T}(u(x,t) -f(0)) \ F(x,t) dxdt=\int_0^T B(f(t)) \partial_x
\psi (0,t) \ dt. 
\]

where we again use that $\psi(x,t)<0$ in $Q_T$ if the
function $F(x,t)$ is positive in $Q_T.$ Now this fact,
together with the adjoint boundary conditions imply that $\partial
_x \psi(0,t)<0,$ for $0<t<T.$ Then the conclusion follows as
before. This completes the proof of $(a).$

To prove $(b)$, multiply both sides of (\ref{direct}) by 
$\partial_x \phi(x,t)$ for an arbitrary test function $\phi(x,t)$ 
and use integration by parts to arrive at

\begin{eqnarray*}
0 &=&\int \int_{Q_T}\partial_{x} u[\partial_t \phi + D(u) \partial_{xx}
\phi] \ dxdt \\
&&+\int_{0}^{T} \phi \partial_t u|_{x=0}^{x=1} \ dt -\int_0^1 \phi
\partial_x u |_{t=0}^{t=T} \ dx \\
&&-\int_0^T \partial_x \phi \partial_x B(u)|_{x=0}^{x=1} \ dt.
\end{eqnarray*}

Now require that $\phi(x,t) $ satisfies the adjoint problem 
\begin{equation*}
\begin{array}{rcll}
\partial _{t}\phi(x,t) +D(u(x,t))
\partial_{xx}\phi(x,t) &=&F(x,t) & \mbox{ in } Q_{T}, \\
\phi(x,T) &=&0, & 0<x<1, \\
\partial_{x} \phi(0,t) &=&0, \quad \phi(1,t)=0, & 0<t<T.
\end{array}
\end{equation*}

Then the preceding integral expression reduces to

\[
\int \int_{Q_T}\partial_{x}u(x,t) F(x,t) dxdt= \int_0^T\phi(0,t)f'(t)
dt 
\]

The maximum principle can be applied to the adjoint problem to
conclude that $\phi(x,t) <0$ in $Q_T$ if the continuous
function $F(x,t)$ is positive in $Q_T.$ In particular, $\phi(0,t) <0$
for $0<t<T$ and since f satisfies (\ref{condf}), it
follows that for every function $F(x,t)$ which is positive in 
$Q_T$ the right side of the expression is negative. Then it follows as in
the proof of part a) that $\partial_x u(x,t) <0$ almost everywhere in
$Q_T.$
\end{proof}

\smallskip
The results of this lemma are crucial to the proof of,

\begin{lemma} \rm
\label{L:monotone}
Suppose $f$ satisfies (\ref{condf}) and 
$D_1, \ D_2$ both satisfy (\ref{condD}).
Then if $D_1( u) >D_2( u) $ for $u \in J= [f(0),f(T)]$ it follows that,

$\begin{array}{crll}
(a) & \Phi[f,D_1](t) >& \Phi[f,D_2](t),& 0<t<T, \\
(b) & \Psi[f,D_1](t) <& \Psi[f,D_2](t),& 0<t<T.
\end{array} $
\end{lemma}

\begin{proof}
For $w \in J,$ let $B'_j(w) = D_j(w), j=1,2.$, and let 
$u_{1},u_{2}$ denote the solutions for the direct problem with 
coefficients $D_{1},D_{2},$ respectively. Then 

\[
\partial _{t}( u_{1}-u_{2}) -\partial _{xx}( B_{1}(
u_{1}) -B_{1}( u_{2}) ) =\partial _{xx}(
B_{1}( u_{2}) -B_{2}(u_{2})) 
\]
and, for an arbitrary test function \ $\phi =\phi ( x,t) ,$ and
arbitrary $\tau, \  0<\tau \leq T,$

\begin{eqnarray*}
&&\int_0^{\tau }\int_{0}^{1}\left[ \partial _{t}( u_{1}-u_{2})
-\partial _{xx}( B_{1}( u_{1}) -B_{1}( u_{2})
) \right] \,\phi \,dxdt \\
&=&\int_{0}^{\tau }\int_{0}^{1}\phi \,\partial _{xx}( B_{1}(
u_{2}) -B_{2}(u_{2})) dxdt.
\end{eqnarray*}

Apply integration by parts on the left side of this equation,\qquad

\begin{eqnarray*}
&&\int_{0}^{\tau }\int_{0}^{1}\left[ \partial _{t}( u_{1}-u_{2})
-\partial _{xx}( B_{1}( u_{1}) -B_{1}( u_{2})
) \right] \,\phi \,dxdt \\
&=&\ -\int_{0}^{\tau }\int_{0}^{1}( u_{1}-u_{2}) \left\{ \partial
_{t}\phi +D_{1}( \mu ( x,t) ) \partial _{xx}\phi
\right\} dx\,dt \\
&&+\int_{0}^{1}( u_{1}-u_{2}) \phi \,|_{t=0}^{t=\tau }dx \\
&&-\int_{0}^{\tau }\left[ \phi \,\partial _{x}( B_{1}(
u_{1}) -B_{1}( u_{2}) ) -\partial _{x}\phi \,(
B_{1}( u_{1}) -B_{1}( u_{2}) ) \right]
_{x=0}^{x=1}dt,
\end{eqnarray*}

and on the right side as well,

\begin{eqnarray*}
&&\int_{0}^{\tau }\int_{0}^{1}\phi \left\{ \partial _{xx}( B_{1}(
u_{2}) -B_{2}(u_{2})) \right\} dx\,dt \\
&=&\int_{0}^{\tau }\left[ \phi \,\partial _{x}( B_{1}(
u_{2}) -B_{2}(u_{2})) \right] _{x=0}^{x=1}dt \\
&&-\int_{0}^{\tau }\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \partial _{x}\phi \partial _{x}u_{2}\,dx\,dt
\end{eqnarray*}

where for all $( x,t) \in Q_{\tau },\;\mu ( x,t) \;$%
lies\ between$\;u_{1}( x,t) \;and\;u_{2}( x,t) $ such
that for $( x,t) \in Q_{\tau }$

\[
B_1( u_1( x,t)) -B_1(u_2(x,t)) =D_1(\mu(x,t))[u_1( x,t)-u_2(x,t)].
\]

Then we obtain the following integral expression,

\begin{eqnarray*}
&&-\int_{0}^{\tau }\int_{0}^{1}( u_{1}-u_{2}) \left\{ \partial
_{t}\phi +D_{1}( \mu ( x,t) ) \partial _{xx}\phi
\right\} dx\,dt \\
&&+\int_{0}^{1}( u_{1}-u_{2}) \phi \,|_{t=0}^{t=\tau }dx \\
&&-\int_{0}^{\tau }\left[ \phi \,\partial _{x}( B_{1}(
u_{1}) -B_{2}(u_{2})) -\partial _{x}\phi \,( B_{1}(
u_{1}) -B_{1}( u_{2}) ) \right] _{x=0}^{x=1}dt \\
&=&-\int_{0}^{\tau }\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \partial _{x}\phi \,\partial _{x}u_{2}\,dx\,dt.
\end{eqnarray*}

The boundary and initial conditions of the direct problem cause this
expression to reduce to,
\begin{equation}
\label{weak}
\begin{array}{rcl}
&&\disp{-\int_{0}^{\tau }\int_{0}^{1}( u_{1}-u_{2}) \left\{ \partial
_{t}\phi +D_{1}( \mu ( x,t) ) \partial _{xx}\phi
\right\} dx\,dt}  \\
&&\disp{+\int_{0}^{1}( u_{1}-u_{2}) ( x,\tau ) \phi (
x,\tau ) \,dx}  \\
&&\disp{+\int_{0}^{\tau }\phi ( 0,t) \,\partial _{x}( B_{1}(
u_{1}) -B_{2}(u_{2})) dt } \\
&&\disp{ -\int_{0}^{\tau }\partial _{x}\phi ( 1,t) \,[B_{1}(
u_{1}) -B_{1}( u_{2}) ]\,dt } \\
&=&\disp{ -\int_{0}^{\tau }\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \partial _{x}\phi \,\partial _{x}u_{2}\,dx\,dt }
\end{array}
\end{equation}

Now require the arbitrary function $\phi(x,t)$ to solve the
so-called g-adjoint problem,

\begin{equation}
\label{gadj}
\begin{array}{rcll}
\partial _{t}\phi +D_{1}( \mu ( x,t) ) \partial_{xx}\phi &=&0 & \mbox{ in }\;Q_{\tau } \\
\phi ( x,\tau ) &=&0 & 0<x<1, \\
\phi ( 0,t) &=&\theta ( t), & 0<t<\tau ,  \\
\partial _{x}\phi ( 1,t) &=&0, & 0<t<\tau ,
\end{array}
\end{equation}

where $\theta(t) = F(\tau -t) $ and $F$ is any function satisfying (\ref{condf}). Then $(\ref{weak})$ reduces to 

\begin{equation}
\label{gint}
\int_{0}^{\tau }\,\theta ( t) [g_{1}( t) -g_{2}(
t) ]\,dt=\int_{0}^{\tau }\int_{0}^{1}( D_{1}( u_{2})
-D_{2}( u_{2}) ) \partial _{x}\phi \,\partial
_{x}u_{2}\,dx\,dt. 
\end{equation}

An argument similar to the one used in the proof of the previous lemma,
applied to (\ref{gadj}), shows that the assumption on the adjoint input, $\theta,$ implies 
$\partial_x \phi(x,t)<0$ on $Q_{\tau }.$ Since $\partial_x u_{2}
<0$ on $Q_{T}$ and $D_{1}( u_{2}) >D_{2}( u_{2})$ it follows that 
the right side of the last
expression is positive. Since $(\ref{gint})$ holds for all $\theta(t) =F(\tau -t)
, $ such that $F$ satisfies (\ref{condf}), it follows that
\begin{eqnarray*}
\qquad \  g_{1}( t) -g_{2}( t)>0 \mbox{ for } 0<t<T \\
i.e. \quad  g_{1}( t) =\Phi [f,D_1](t)> \Phi[f,D_2](t) =g_2(t).
\end{eqnarray*}

To see that this is true, note first that if $D_1(u)>D_2(u)$ for $u \in J,$ 
then existence of an interval 
$(0,t_{1}) $ with $g_{1}(t) < g_{2}(t)$ for 
$0<t<t_1$ is precluded by (\ref{gint}) simply by choosing $\tau =t_1.$ Suppose
then that there exists $t_{2}>t_{1}>0$ such that $g_1(t) \geq
g_{2}(t)$ for $0<t \leq t_1$ and $g_1(t) < g_2(t)$ for\ $t_1<t<t_2.$ 
Then choosing $\tau =t_{2}$ in (\ref{gint}) implies that for any admissible $\theta(t) ,$
\begin{eqnarray*}
\int_{t_{1}}^{t_{2}}\,\theta ( t) [g_{1}( t)
-g_{2}( t) ]\,dt &=&\int_{t_{1}}^{t_{2}}\int_{0}^{1}(
D_{1}( u_{2}) -D_{2}( u_{2}) ) \partial _{x}\phi
\,\partial _{x}u_{2}\,dx\,dt \\
&&+\int_{0}^{t_{1}}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \partial _{x}\phi \,\partial _{x}u_{2}\,dx\,dt \\
&&-\int_{0}^{t_{1}}\,\theta ( t) [g_{1}( t)
-g_{2}( t) ]\,dt.
\end{eqnarray*}

By applying equality (\ref{gint}) with $\tau = t_1$, the last two
terms of the previous equation vanish.  By assumption, the right side of the
resulting expression is strictly positive, while a suitable choice of
$\theta(t)$ makes the left side non-negative.  This contradicts (\ref{gint}).

%Now choose $\theta(t)$ so as to have

%\[
%\int_{0}^{t_{1}}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
%u_{2}) ) \partial _{x}\phi \,\partial _{x}u_{2}\,dx\,dt\geq
%\int_{0}^{t_{1}}\,\theta ( t) [g_{1}( t) -g_{2}(
%t) ]\,dt 
%\]
%
%in order to obtain a contradiction with (\ref{gint}).

Suppose now that we choose $\phi$ in (\ref{weak}) to solve a problem different from (\ref{gadj}). This problem will be called the h-adjoint
problem,

\begin{equation}
\label{hadj}
\begin{array}{rcll}
\partial _{t}\phi +D_{1}( \mu ( x,t) ) \partial
_{xx}\phi &=&0& \mbox{ in } Q_{\tau}, \\
\phi ( x,\tau ) &=&0, & 0<x<1, \\
\phi ( 0,t) &=&0, &  0<t<\tau, \\
D(\mu(1,t)) \partial_x \phi(1,t) &=&\beta(t), & 0<t<\tau.
\end{array}
\end{equation}

Here, choose $\beta(t)=F(\tau -t)$ where $F$ 
is any function satisfying (\ref{condf}). Then (\ref{weak}) reduces to

\begin{eqnarray*}
&&\int_{0}^{\tau }D( \mu ( 1,t) ) \partial _{x}\phi
(1,t\,)( u_{1}( 1,t) -u_{2}( 1,t) ) \,dt \\
&=&\int_{0}^{\tau }\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}\phi \,\partial _{x}u_{2}\,dx\,dt,
\end{eqnarray*}

or 
\begin{equation}
\label{hint}
\fl \qquad
\int_{0}^{\tau }\beta ( t) \,\left[ h_{1}( t)
-h_{2}( t) \right] \,dt=\int_{0}^{\tau }\int_{0}^{1}(
D_{1}( u_{2}) -D_{2}( u_{2}) ) \,\partial
_{x}\phi \,\partial _{x}u_{2}\,dx\,dt.
\end{equation}

In this case, the hypotheses on $\beta(t)$ imply that 
$\partial _{x}\phi ( x,t) >0$ on $Q_{\tau }$ and since 
$\partial _{x}u_{2}<0$ and $D_{1}( u_{2}) >D_{2}(
u_{2}) \quad  \forall u_{2}\in J,$ it follows that the right side of
(\ref{hint}) is negative. Since this holds with $\beta(t)
=F( \tau -t)$ for any F satisfying (\ref{condf}), it follows that

\[
\Psi \left[ f,D_{1}\right] ( t) =u_{1}( 1,t)
<u_{2}( 1,t) =\Psi \left[ f,D_{2}\right] ( t) \qquad
for\qquad 0<t<\tau.
\]

Finishing the argument as in the previous case, we see that this holds for 
$ \tau \leq T.$
\end{proof}

The conclusions of lemma \ref{L:monotone} assert that input to output mappings 
$\Phi$ and $\Psi $ are monotone mappings. More precisely, the mapping $\Phi$
is {\it isotone} while the mapping $\Psi $ is an {\it antitone}
mapping.

Now suppose $D_{1}(u_1)$ and $D_2( u_{1})$ are any two coefficients, 
both satisfying (\ref{condD}). Let $u_{1}(x,t), u_{2}(x,t)$ denote 
the solutions of (\ref{direct}) when
the coefficient is, respectively, $D_1(u)$ and $D_{2}(u),$ and 
for $i=1,2,$ let

\begin{eqnarray*}
g_{i}(t) &=& \Phi[f,D_i] \qquad \mbox{ and } \\
h_{i}(t) &=& \Psi[f,D_i],\qquad 0<t<T.
\end{eqnarray*}

Now choose the data in the adjoint problems (\ref{gadj}) and (\ref{hadj}) as,

\[
\qquad \qquad \qquad \qquad \;\;\phi ( 0,t) =\theta (
t) =\dfrac{g_{1}( t) -g_{2}( t) }{\left| \left|
\,g_{1}-g_{2}\right| \right| _{L^{2}\left[ 0,T\right] }}, 
\qquad \mbox{ in } (\ref{gadj})
\]

and 

\[
\qquad D_{1}( \mu ( 1,t) ) \partial _{x}\psi (
1,t) =\beta ( t) =\dfrac{h_{1}( t) -h_{2}(
t) }{\left| \left| \,h_{1}-h_{2}\right| \right| _{L^{2}\left[ 0,T%
\right] }} \qquad \mbox{ in } (\ref{hadj}).
\] 

It follows at once from (\ref{gint}) that

\begin{eqnarray*}
\left| \left| \,g_{1}-g_{2}\right| \right| _{L^{2}\left[ 0,T\right] } &\leq
&\left| \int_{0}^{T}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt\right|
\\
&\leq &C\,\left| \left| D_{1}-D_{2}\right| \right| _{\infty }
\end{eqnarray*}

and from (\ref{hint}) that

\qquad \qquad \qquad 
\begin{eqnarray*}
\left| \left| \,h_{1}-h_{2}\right| \right| _{L^{2}\left[ 0,T\right] } &\leq
&\left| \int_{0}^{T}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\psi \,dx\,dt\right|
\\
&\leq &C\,\left| \left| D_{1}-D_{2}\right| \right| _{\infty }
\end{eqnarray*}

Evidently, this is just the assertion that $\Phi$ and $\Psi$ are
continuous as a function of $D$ from $W(J)$ into $L^{2}[0,T]$; i.e.,

\smallskip

$\left| \left| \,g_{1}-g_{2}\right| \right| _{L^{2}\left[ 0,T\right]
}=\left| \left| \Phi ( f,D_{1}) -\Phi ( f,D_{2})
\right| \right| _{L^{2}\left[ 0,T\right] }\leq C\,\left| \left|
D_{1}-D_{2}\right| \right|_{\infty }$

$\left| \left| \,h_{1}-h_{2}\right| \right| _{L^{2}\left[ 0,T\right]
}=\left| \left| \Psi ( f,D_{1}) -\Psi ( f,D_{2})
\right| \right| _{L^{2}\left[ 0,T\right] }\leq C\,\left| \left|
D_{1}-D_{2}\right| \right|_{\infty }.$

\medskip

Having shown that $\Phi $ and $\Psi $ are continuous and
strictly monotone, one is encouraged to believe that this inverse problem is
not so badly ill posed and that $\Phi $ and $\Psi $ might be continuously
invertible. Such a strong result seems to be unlikely without a simple
ordering on the domain and range of these maps but it is at least true that
the input/output maps $\Phi $ and $\Psi $ are injective as the following
lemma shows.

\medskip
\begin{lemma} \rm
\label{L:injective}
For a fixed f satisfying (\ref{condf}) and coefficients $
D_{1},D_{2}\in W( J) $ let $g_k(t) =\Phi[f,D_k]$ and $h_k(t) =\Psi [f,D_kt],$ for $k=1,2.$

\smallskip
Then

$a) \ \Phi[f,D_{1}] = \Phi[f,D_{2}], 0<t<T, \ $ implies $ \ D_{1}(u) =D_{2}(u)$ for $u \in J.$

$b) \ \Psi[f,D_{1}] = \Psi[f,D_{2}], 0<t<T, \ $ implies $\ D_{1}( u) =D_{2}(u)$ for $u \in J.$
\end{lemma}

\begin{proof}
Suppose first that $D_{1}(f(0)) =D_{2}(f(0)).$ 
Now, since $D_1$ and $D_2$ both satisfy
(\ref{condD}), their difference satisfies (\ref{condD}) and if these 
functions are not
identical on J then there exists a positive time $t_{1}$, $0<t_{1}\leq T,$
where the difference,$\;$ $D_{1}( f( t) ) -D_{2}(
f( t) ) $ is of one sign on $\left[ 0,t_{1}\right] .$ Then
lemma \ref{L:lem1}(a) implies $D_{1}( u_{2}( x,t) )
-D_{2}( u_{2}( x,t) ) $ is of one sign on $(
0,1) \times ( 0,t_{1}) .$ Using the identity (\ref{gint}), we have

\begin{eqnarray*}
&&\int_{0}^{t_{1}}\int_{0}^{1}( D_{1}( u_{2}( x,t)
) -D_{2}( u_{2}( x,t) ) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&=&\int_{0}^{t_{1}}( g_{1}( t) -g_{2}( t) )
\,\theta ( t) \,dt,
\end{eqnarray*}

where $\phi$ solves (\ref{gadj}) with $\tau =$ $t_{1}$. Then the 
hypotheses imply the right side of this equation vanishes; i.e.,

\vspace{1pt}

\qquad \qquad $\int_{0}^{t_{1}}\int_{0}^{1}( D_{1}( u_{2})
-D_{2}( u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\phi
\,dx\,dt=0,$

\vspace{1pt}

and this holds independently of the data $\theta(t)$ chosen as
input to the adjoint problem. It is clearly possible to choose $\theta
( t) $ so that $\partial _{x}\phi <0$ on $(0,1)
\times ( 0,t_{1}) $ \ and in view of lemma \ref{L:lem1}(b) it is also the
case that, $\,\partial _{x}u_{2}<0$ on $(0,1) \times (
0,t_{1}).$ Then the vanishing integral above has an integrand which
is of one sign over the domain of integration and vanishes on no positive
measure subset of the domain. This contradiction is in opposition to the
assumption that $D_{1}$ and $D_{2}$ are not identical.

If we suppose $D_1(f(0)) \neq D_{2}(f(0))$ then it follows that 
either there is a smallest
time $t_{1}, 0<t_{1}<T,$ where the difference $D_1(f(t))-D_2(f(t))$ 
is zero, or else $t_1=T$ and the difference is of one sign on 
$[0,T]$. In either case, it is evident that $D_1(f(t))
-D_{2}(f(t))$ is of one sign on $[0,t_1], \ 0<t_1 \leq T,$ and the 
argument can be completed as before. A
similar argument, using the identity in (\ref{hint}), 
establishes conclusion(b).
\end{proof}

Formally, we can write,
\begin{eqnarray*}
( \Phi \left[ f,D_{1}\right] -\Phi \left[ f,D_{2}\right]
,\theta ) _{L^{2}}&\stackrel{def}{=}&( \delta \Phi \left[
D_{1},D_{2}\right] \Delta D,\theta ) _{L^{2}} \\
&=&\left\langle \Delta
D,^{t}\delta \Phi \left[ D_{1},D_{2}\right] \,\theta \right\rangle _{W(
J) \times W( J) ^{\ast }}.
\end{eqnarray*}
In view of (\ref{gint}),
\begin{eqnarray*}
( \Phi \left[ f,D_{1}\right] -\Phi \left[ f,D_{2}\right] ,\theta
) _{L^{2}} &=&\int_{0}^{T}( g_{1}( t) -g_{2}(
t) ) \theta ( t) \,dt \\
&=&\int_{0}^{T}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt, \\
&=&\left\langle \Delta D,^{t}\delta \Phi \left[ D_{1},D_{2}\right] \,\theta
\right\rangle _{W( J) \times W( J) ^{\ast }}
\end{eqnarray*}
Similarly,
\begin{eqnarray*}
( \Psi \left[ f,D_{1}\right] -\Psi \left[ f,D_{2}\right] ,\beta
) _{L^{2}}&\stackrel{def}{=}&( \delta \Psi \left[ D_{1},D_{2}\right]
\Delta D,\beta ) _{L^{2}} \\
&=&\left\langle \Delta
D,^{t}\delta \Psi \left[ D_{1},D_{2}\right] \,\beta \right\rangle _{W(
J) \times W( J) ^{\ast }},
\end{eqnarray*}
and, referring to (\ref{hint}),
\begin{eqnarray*}
( \Psi \left[ f,D_{1}\right] -\Psi \left[ f,D_{2}\right] ,\beta )
_{L^{2}} &=&\int_{0}^{T}( h_{1}( t) -h_{2}( t)
) \beta ( t) \,dt \\
&=&\int_{0}^{T}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\psi \,dx\,dt \\
&=&\left\langle \Delta D,^{t}\delta \Psi \left[ D_{1},D_{2}\right] \,\beta
\right\rangle _{W( J) \times W( J) ^{\ast }}
\end{eqnarray*}

\bigskip

Evidently, (\ref{gint}),(\ref{hint}) provide realizations for 
$^{t}\delta \Phi[D_1,D_2]$ and $^{t}\delta \Psi[D_1,D_2],$ the
Gateaux derivatives with respect to $D$ of the mappings $\Phi$ and $\Psi.$
It will be shown in the next section that 
$^{t}\delta \Phi[D_1,D_2]$ and $^{t}\delta \Psi[D_1,D_2]$ are invertible
in an approximate sense. More precisely we will devise a restriction of the
coefficient to data maps that induces a mapping from $\R$ into $\R.$ The
restriction inherits the strict monotonicity and continuity from the
coefficient to data map hence the restriction defines a homeomorphism from
its domain onto its range. Inversion of this mapping leads to an approximate
inverse for the coefficient to data map.

\section{The Approximate Solution of the Inverse Problem}

We consider the inverse problem in which the coefficient  
$D=D(u)$ is to be identified from data which is assumed to be recorded at
fixed nodes$ 0=t_0<t_1< \cdots <t_N=T$ in the interval $[0,T]$:

\begin{equation*}
data(f,g) \left\{
\begin{array}{rcll}
f( t_{k}) &=&\mu _{k} &   \\
g( t_{k}) &=&-D( \mu _{k}) \,\partial _{x}u_{1}(
0,t_{k}) =\gamma _{k}, & \quad k=0,1,...,N  
\end{array} \right.
\end{equation*}

We are also interested in the identification of $D=D( u_{1}) $
based on the alternative data,

\begin{equation*}
data(f,h) \left\{
\begin{array}{rcll}
f(t_k) &=&\mu _{k} &  \\
h( t_{k}) &=&u_{1}( 1,t_{k}) =\eta _{k}, & \qquad \qquad \qquad k=0,1,...,N
\end{array} \right.
\end{equation*}

More precisely, we are going use one or the other of these data sets
to construct a polygonal (i.e. piecewise linear and continuous)
approximation to the unknown coefficient $D( u) .$ The data set,\
\ $f_{k}=f( t_{k}) ,\;k=0,1,...,N,$ is assumed to be given at
fixed nodes which define a partition, $0=t_{0}<t_{1}<\cdots <t_{N}=T,$ of$%
\,\,$\ the interval $\ I=\left[ 0,T\right] $.$\;$This partition of $I$
\thinspace will be called the ''inner mesh''. We then define an associated
(but coarser) partition of $J=\left[ f( 0) ,f( T) %
\right] $, the domain of the coefficient $D$. This partition will be called
the ''outer mesh'' and is given by $f( 0) =\mu _{0}<\mu
_{1}<\cdots <\mu _{M}=f( T) ,$ ; i.e.,$\;\mu _{0}=f_{0},\;$and $%
\;\mu _{M}=f_{N}\;$\ and for each $j=1,...,M<N,\;$we have $\mu _{j}=f_{k}\;$%
for some $k\geq j.\;$

It is necessary for the outer mesh to be coarser than the inner mesh
since on each subinterval in the outer mesh, we will need to compute
interior values of the solution $u( x,t) ,$ for the direct
problem in order to be able to evaluate the integrals which appear in the
identities used in the identification. Between two outer mesh knots $\mu
_{j}=f( t_{k})$ and $\mu _{j+1},$ there must occur several inner
mesh knots and this fact prevents the outer mesh from being made arbitrarily
fine in order to improve the accuracy of the identification.

We can now consider a family of polygonal functions, $\hat{D},$
associated with the partition of $J$. Each member of the family is
characterized by its values at the nodes \ $\mu _{k};\;i.e.,\;for\;\;d_{k}=%
\hat{D}( \mu _{k}) ;$ More precisely, we define

\vspace{1pt} 
\begin{equation}
\label{Dparam1}
\hat{D}( u) =\sum\limits_{k=1}^{M}\left[ d_{k-1}\,\rho
_{k-1}( u) +d_{k}\,\lambda _{k}( u) \right] 
\end{equation}

\noindent where

$\begin{array}{ccc}
\rho_{k}(u) 
&  =\left\{  \begin{array}{cr} 
                     \dfrac{\mu _{k+1}-u}{\mu _{k+1}-\mu _{k}} & \mbox{
if } \mu _{k}\leq u\leq \mu _{k+1} \\ 
                     0 & \mbox{ otherwise }
                \end{array}
     \right\} & \quad 1 \leq k\leq M-1 \\
\ & \ & \ \\
\lambda_{k}(u) 
& =\left\{ \begin{array}{cr}
                    \dfrac{u-\mu _{k-1}}{\mu _{k}-\mu _{k-1}} & \mbox{ if } \mu _{k-1}\leq u\leq \mu _{k} \\ 
                    0 & \mbox{ otherwise }
\end{array}
    \right\}  & 1\leq k\leq M.
\end{array} $


Equivalently, we could write \\  for $1\leq k\leq M,$

\begin{equation}
\label{Dparam2}
\hat{D}(u) =d_{k-1}\,\rho _{k-1}( u) +d_{k}\,\lambda
_{k}( u) \mbox{ for } \mu_{k-1}\leq u\leq \mu _{k}
\end{equation}

\vspace{1pt}

We will introduce several notations:

\begin{itemize}
\item $\hat{D}(u) =P_{M}[d_0,d_1,\ldots,d_M]$ denotes the polygonal coefficient given by (\ref{Dparam1}) based on nodal
values $[d_0,d_1,\ldots ,d_M] $.

\item $u( x,t;D,f)$ \quad denotes the solution of the direct
problem (\ref{direct}) with coefficient $D$ and data, $f.$

\item  $\phi(x,t,D,\theta)$ \quad denotes the solution of the
adjoint problem (\ref{gadj}) with coefficient 
$D(x,t) \stackrel{def}{=} D(\mu(x,t))$ and data, $\theta(t).$

\item  $\psi(x,t,D,\beta)$ \quad  denotes the solution of the
adjoint problem (\ref{hadj}) with coefficient 
$D(x,t) \stackrel{def}{=} D(\mu(x,t))$ and data, $\beta(t)$
\end{itemize}

\vspace{1pt}

For a given $f(t)$ satisfying (\ref{condf}), an unknown
coefficient $D=D(u)$ satisfying (\ref{condD}) and measured flux data 
$g(t)=\Phi[f,D],$ we assume there is a fixed outer
partition, $\Pi =\{ 0=\mu _0<\mu_1<\cdots <\mu_M=f(T)\}$ of J. 
Then we will define a polygonal coefficient
approximation to $D$ by the following recursive algorithm based on 
$(f,g) -data, \quad \{f(t), g(t) \}$ :

\begin{itemize}
\item  $d_{0}$ is assumed to be given

\item  for $k=1,2,..., d_{k}$ is determined from 
$d_{0},d_{1},...,d_{k-1}$ by

%\item  
%\begin{eqnarray}
%\label{alg}
%&( d_{k}-d_{k-1}) \int_{T_{k-1}}^{T_{k}}\lambda _{k}(
%u_{2}) \partial _{x}u_{2}\partial _{x}\phi dxdt  \\
%&=&-\int_{T_{k-1}}^{T_{k}}( g( t) -g_{2}( t)
%) \theta ( t) dt,  \nonumber
%\end{eqnarray}

\item  
\begin{equation}
\label{alg}
\fl
(d_{k}-d_{k-1}) \int_{T_{k-1}}^{T_{k}}\lambda _{k}(u_{2}) \partial _{x}u_{2}\partial _{x}\phi dxdt 
=
-\int_{T_{k-1}}^{T_{k}}( g( t) -g_{2}( t)
) \theta(t) dt, 
\end{equation}

\end{itemize}

where
\begin{eqnarray*}
D_{1}(u)& =& P_{M}\left[ d_{0},d_{1},\ldots ,d_{k-1},d_{k}\right] \\
D_{2}(u) &=&P_{M}\left[ d_{0},d_{1},\ldots ,d_{k-1},d_{k-1}\right] \\
u_{2}(x,t) &=&u( x,t;D_{2},f), \\
g_{2}( t) &=&-D_{2}(f(t))\partial_x u_{2}(0,t) \ 0\leq x\leq 1, \ 0 \leq t\leq T_{k}, \\
\phi(x,t) &=&\phi(x,t;D_1,f(T-t)),\ for \ 0\leq x\leq 1,\ 0\leq t\leq T_{k}.\\
\end{eqnarray*}

The approximation of $D( u) $ based on $(f,h)
-data, \{f(t), h(t) \},$ is analogous. We can show then,

\vspace{1pt}

\begin{lemma} \rm
\label{L:bound1}
For $f(t)$ satisfying (\ref{condf}), for coefficient $D$
satisfying (\ref{condD}) and for a fixed partition, 
$\Pi ={ 0=\mu _{0}<\mu_1 < \cdots <\mu_M =f(T)}$ of J, let the nodal values 
$[ d_0,d_1, \ldots ,d_M] $ be determined by the algorithm
(\ref{alg}). Then for $k=1,2,...,M,$
\begin{equation}
\label{boundD}
\left| D( \mu _{k}) -d_{k}\right| \leq C\left| \mu
_{k}-\mu _{k-1}\right|
\end{equation}
\end{lemma}

\begin{proof}
We are going to assume that the initial nodal value, \ $%
D( \mu _{0}) =D( f( 0) ) =d_{0}$, \ is
known and that the remaining values $d_{1},\ldots ,d_{M}$ are determined by
the algorithm (\ref{alg}). Consider first, the value $d_{1}.$ If we apply the
identity (\ref{gint}) with $\tau =T_{1},\,$and

\begin{itemize}
\item  $on\;\;J_{1}=\left[ \mu _{0},\mu _{1}\right] ,\;D_{1}( u)
=P_{M}\left[ d_{0},d_{1}\right] ,\;$and $\;D_{2}( u) =P_{M}\left[
d_{0},d_{0}\right] ,\;$

\item  $on\;Q_{1}=( 0,1) \times ( 0,T_{1})
\;u_{1}( x,t) =u( x,t;D_{1},f) \;$and$\;\,u_{2}(
x,t) =u( x,t;D_{2},f) \;\;$
\end{itemize}

then we have 

\[
\int_{0}^{T_{1}}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\phi
\,dx\,dt=\int_{0}^{T_{1}}( g( t) -g_{2}( t)
) \theta ( t) \,dt. 
\]

Here $g(t)$ is the measured flux data and 
$g_2(t)$ is the output generated by solving (\ref{direct}) with the
coefficient $D(u)= D_2(u)$; i.e., $g_2= \Phi[f,D_2].$ The functions 
$\theta(t)$ and $\phi(x,t)$ denote the data and solution respectively for the
g-adjoint problem. Since the function $f(t)$ in the direct
problem satisfies (\ref{condf}), it follows from Lemma $\ref{L:lem1}(a)$ that 
$u_2$ satisfies

$f( 0) =\mu _{0}\leq u_{2}( x,t) \leq \mu
_{1}=f( T_{1}) \;\;for\;\;( x,t) \in ( 0,1)
\times ( 0,T_{1}) .$


Then according to (\ref{Dparam2}) for 
$\quad u \in J_1={\mu_0 \leq u\leq \mu _1},$

\vspace{1pt}

\qquad \qquad \qquad \qquad \qquad $D_{1}( u) =d_{0}\,\rho
_{0}( u) +d_{1}\,\lambda _{1}( u) \;\;\;\ \ \;\;\;\;\;$

\qquad \qquad \qquad \qquad \qquad $D_{2}( u) =d_{0}\,\rho
_{0}( u) +d_{0}\,\lambda _{1}( u) =d_{0}\;\;\ \ $.

\vspace{1pt}

and so\qquad \qquad \qquad $D_{1}( u_{2}) -D_{2}(
u_{2}) =( d_{1}-d_{0}) \lambda _{1}( u_{2}) .$

\vspace{1pt}

\vspace{1pt}Note that for each nodal value, $\mu _{k},\;0\leq k\leq M,\;$we
have \ $u_{2}( x_{k}( t) ,t) =\mu _{k}$ along some
curve \ $x=x_{k}( t) ,$ with $x_{k}( 0) =\mu _{k}$ and 
$x_{k}( \tau _{k}) =1$ for some $\tau _{k}>\tau _{k-1}>\cdots
>\tau _{1}>0.$  Examples of such curves are shown in Figure (\ref{F:isotherm}).

\begin{figure}
\caption[]{ \label{F:isotherm} }
\begin{center}
\epsfig{file=figure1.eps,width=4in}
\end{center}
\end{figure}

Then we have \ $u_{2}( x( t) ,t) =\mu _{0}$
along a curve \ $x=x_{0}( t) ,$ with $x_{0}( 0) =0$
and $x_{0}( \tau _{1}) =1$ for some $\tau _{1}>0.$ We suppose
further that $T_{1}\;$is sufficiently small that $0<x_{0}( T_{1})
<1.$Then

$\lambda _{1}( u_{2}( x,t) ) =\left\{ 
\begin{array}{c}
\dfrac{\mu _{0}-u_{2}( x,t) }{\mu _{0}-\mu _{1}}%
\;\;\;\;if\;\;0\leq x\leq x_{0}( t) ,\;0\leq t\leq T_{1} \\ 
0\;\;\;\;\;\;\;\;\ \ \ \ \ \;if\;\;x>x_{0}( t) ,\;0\leq t\leq
T_{1}
\end{array}
\right\} $

and the integral identity reduces to

\qquad

\[
( d_{1}-d_{0}) \int_{0}^{T_{1}}\int_{0}^{x_{0}( t)
}\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi
\,dx\,dt=\int_{0}^{T_{1}}( g( t) -g_{2}( t)
) \theta ( t) \,dt;
\]

i.e.,

\[
d_{1}=d_{0}+\dfrac{\int_{0}^{T_{1}}( g( t)
-g_{2}( t) ) \theta ( t) \,dt}{%
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt}.
\]

This equation defines the first unknown nodal value $d_{1}.$ Now we will
establish the relationship between $d_{1}$ and $D(\mu_{1}).$
It follows from (\ref{gint}) that

\begin{eqnarray*}
&&\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }( D(
u_{2}) -D_{2}( u_{2}) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&=&\int_{0}^{T_{1}}( g( t) -g_{2}( t) )
\theta ( t) \,dt \\
&=&( d_{1}-d_{0}) \int_{0}^{T_{1}}\int_{0}^{x_{0}( t)
}\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi
\,dx\,dt.
\end{eqnarray*}

\vspace{1pt}

Let $\hat{D}_{M}( u) $ denote the polygonal coefficient on the
partition $\Pi $ \ which satisfies $\hat{D}_{M}( \mu _{k})
=D( \mu _{k}) $ for all k. Note that this coefficient does not,
in general, generate the given measured data, $g( t) ,$ and is
not then the polygonal coefficient with nodal values $\left\{ d_{k}\right\}
\;$generated by the algorithm$.$ However, these coefficients are related as
follows,

\begin{eqnarray*}
&&\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }( D(
u_{2}) -D_{2}( u_{2}) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&=&\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }( D(
u_{2}) -\hat{D}_{M}( u_{2}) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&&+\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }( \hat{D}_{M}(
u_{2}) -D_{2}( u_{2}) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&=&\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }( D(
u_{2}) -\hat{D}_{M}( u_{2}) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&&+( D( \mu _{1}) -d_{0})
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt.
\end{eqnarray*}

and by combining these two expressions it follows that

\begin{eqnarray*}
&&( d_{1}-D( \mu _{1}) )
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&=&\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }( D(
u_{2}) -\hat{D}_{M}( u_{2}) ) \,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&\leq &\max\limits_{J_{1}}\,\left| D-\hat{D}_{M}\right| \left|
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt\right|
\end{eqnarray*}

Now

\vspace{1pt}\qquad $\max\limits_{J_{1}}\,\left| D-\hat{D}_{M}\right|
=\,\left| D( \mu _{\ast }) -\hat{D}_{M}( \mu _{\ast })
\right| $ $\;\;\;$for some $\mu _{\ast }\in J_{1}.$

\vspace{1pt}

But

\qquad \thinspace \thinspace \qquad

$\left| D( \mu _{\ast }) -\hat{D}_{M}( \mu _{\ast })
\right| \leq \left| D( \mu _{\ast }) -D( \mu _{0})
\right| +\left| D( \mu _{0}) -\hat{D}_{M}( \mu _{\ast
}) \right| $

\qquad \qquad \qquad \qquad $\;\;\leq K\left| \mu _{\ast }-\mu _{0}\right|
+\left| \hat{D}_{M}( \mu _{0}) -\hat{D}_{M}( \mu _{\ast
}) \right| .$

\vspace{1pt}

In addition,\qquad $\left| \hat{D}_{M}( \mu _{0}) -\hat{D}%
_{M}( \mu _{\ast }) \right| \leq K\left| \mu _{\ast }-\mu
_{0}\right| ,$

and

\qquad \qquad \qquad $\left| D( \mu _{\ast }) -\hat{D}_{M}(
\mu _{\ast }) \right| \leq 2K\left| \mu _{\ast }-\mu _{0}\right| .$

Then

\vspace{1pt}\qquad $\left| \,d_{1}-D( \mu _{1}) \right| \leq 2K\,%
\dfrac{\left| \int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt\right| }{\left|
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt\right| }%
\,\left| \mu _{\ast }-\mu _{0}\right| .$

\vspace{1pt}

Since it is clear that for some $\lambda _{1}^{\ast },\;$ $0<\lambda
_{1}^{\ast }<1,$

\qquad $\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt=\lambda
_{1}^{\ast }\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\,\partial
_{x}u_{2}\,\partial _{x}\phi \,dx\,dt$

\vspace{1pt}we find

\qquad $1\leq \dfrac{\left| \int_{0}^{T_{1}}\int_{0}^{x_{0}( t)
}\,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt\right| }{\left|
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt\right| }\leq 
\dfrac{1}{\lambda _{1}^{\ast }}<\infty .$

Then,

\qquad \qquad $\left| \,d_{1}-D( \mu _{1}) \right| \leq \dfrac{2K%
}{\lambda _{1}^{\ast }}\,\left| \mu _{\ast }-\mu _{0}\right| \leq
C_{1}\left| \mu _{1}-\mu _{0}\right| .$

\vspace{1pt}

This is the result (\ref{boundD}) for $k=1.$

\qquad In determining the succeeding values $d_{k},$ we assume 
$d_0,d_1,...,d_{k-1}$ are known and we let,


\begin{itemize}
\item  on \quad $[\mu_0,\mu_k], \ D_1(u)=P_M[d_0,d_1,\ldots,d_{k-1},d_k],$

and $D_2(u)=P_M[d_0,d_1,\ldots,d_{k-1},d_{k-1}], $

\item  on \quad $Q_k=(0,1) \times (0,T_k), \ u_1(x,t)=u(x,t;D_1,f),$

and \ $u_{2}(x,t)=u(x,t;D_2,f)$
\end{itemize}

\vspace{1pt}

Then $D_{1}(u)$ and $D_{2}(u)$ are identical on 
$[\mu_0,\mu_{k-1}]$ and only differ on 
$J_{k}=[\mu_{k-1},\mu_k]$ where we have

\vspace{1pt}

\qquad $D_{1}( u) =d_{k-1}\,\rho _{k-1}( u)
+d_{k}\,\lambda _{k}( u) \;\;\;\ \ \qquad \;\;\;\;\;\;for\;\ \
\;\;\mu _{k-1}\leq u\leq \mu _{k},$

\qquad $D_{2}( u) =d_{k-1}\,\rho _{k-1}( u)
+d_{k-1}\,\lambda _{k}( u) =d_{k-1}\;\;\;\ \ for\;\ \ \;\;\mu
_{k-1}\leq u\leq \mu _{k}.$

Then 
\begin{eqnarray*}
&&\int_{0}^{T_{k}}\int_{0}^{1}( D_{1}( u_{2}) -D_{2}(
u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\phi \,dx\,dt \\
&=&\int_{T_{k-1}}^{T_{k}}\int_{0}^{1}( D_{1}( u_{2})
-D_{2}( u_{2}) ) \,\partial _{x}u_{2}\,\partial _{x}\phi
\,dx\,dt \\
&=&( d_{k}-d_{k-1}) \int_{T_{k-1}}^{T_{k}}\int_{0}^{x_{k-1}(
t) }\lambda _{k}( u_{2}) \,\partial _{x}u_{2}\,\partial
_{x}\phi \,dx\,dt.
\end{eqnarray*}

\vspace{1pt}and we have 
\begin{eqnarray*}
&&( d_{k}-d_{k-1}) \int_{T_{k-1}}^{T_{k}}\int_{0}^{x_{k-1}(
t) }\lambda _{k}( u_{2}) \,\partial _{x}u_{2}\,\partial
_{x}\phi \,dx\,dt \\
&=&\int_{T_{k-1}}^{T_{k}}( g( t) -g_{2}( t)
) \theta ( t) \,dt,
\end{eqnarray*}

as prescribed by (\ref{alg}). Now we proceed as in the first part of the proof to
show that

\vspace{1pt}

\vspace{1pt}\qquad \qquad \qquad $\qquad \left| \,d_{k}-D(\mu _{k})\right|
\leq C\left| \mu _{k}-\mu _{k-1}\right| .$

$\vspace{1pt}$

The proof of the analogous result based on the data $\left\{ f(
t_{k}) ,h( t_{k}) \right\} $ proceeds similarly.
\end{proof}

\vspace{1pt}

\qquad For $d_{0}$ fixed and $d_{1}>0,$ let $\;P_{1}(d_{1})%
( u) =d_{0}\,\rho _{0}( u) +d_{1}\,\lambda _{1}(
u) \;$for $u\in J_{1}.$ Then $P_{1}$ is a mapping from $\left[
0,\infty \right] $ into a one dimensional subspace of $W( J_{1})
. $ It follows from (\ref{alg}) in the case $k=1$ that 
\begin{eqnarray*}
&&\left\langle \Delta D( u_{2}) ,^{t}\delta \Phi \left[
P_{1}( d_{1}) ,P_{1}( d_{0}) \right] (\theta
)\right\rangle \\
&=&\left\langle ( d_{1}-d_{0}) \lambda _{1}( u_{2})
,^{t}\delta \Phi \left[ P_{1}( d_{1}) ,P_{1}( d_{0}) %
\right] (\theta )\right\rangle \\
&=&( d_{1}-d_{0}) \int_{0}^{T_{1}}\int_{0}^{x_{0}( t)
}\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\phi
\,dx\,dt
\end{eqnarray*}

This means that the double integral in the expression above is a
representation for the derivative with respect to the parameter $d,$ of the
coefficient-to-data mapping, $\Phi,$ restricted to the one dimensional
subspace of $W( J_{1}).$ Since the double integral can be shown
to be nonzero, it follows that the restricted input/output mapping is
locally approximately invertible. Lemma 3.1 asserts that, if we are given
the data, $\left\{ f( t_{k}) ,g( t_{k}) \right\} $ or $%
\left\{ f( t_{k}) ,h( t_{k}) \right\} ,$ then we can
compute the nodal values $\left\{ d_{k}\right\} $ which reproduce the
measured data in the sense of (\ref{alg}) and that these nodal values approach the
nodal values of the ``true coefficient'' $D( u_{1}) ,$ as the
mesh size of the outer mesh decreases. However, this conclusion ignores
certain difficulties:

\vspace{1pt}

\begin{itemize}
\item  it is not possible to know the coefficient $D_{1}( \mu (
x,t) ) \;$in the adjoint problems since $D_{1}$ is the
coefficient we wish to identify and $\mu $ is an indeterminate point between 
$u_{1}\;and$ $u_{2}$. This means we can only approximate the solution to the
adjoint problem and this will have an influence on the conclusions of
lemma 3.1.
\end{itemize}

\vspace{1pt}

\begin{itemize}
\item  the integrals in the identity can only be approximated by numerical
integrations for which there is only a limited degree of refinement
possible. This may further interfere with the agreement between $\,d_{k}\;$%
and $\,D(\mu _{k}).$
\end{itemize}

\qquad \qquad\ \ \ \qquad

\qquad We will consider both of these effects, starting with the effect of
the approximate adjoint solution.

\qquad Note first, that the algorithm (\ref{alg}) asserts that in determining the
nodal value $\mu _{k},$ it is necessary to solve the adjoint problem only on
the strip $S_{k}=\{( 0,1) \times ( T_{k-1},T_{k}) \}.$
Let $\hat{\phi}( x,t) \;$denote the adjoint solution we compute
using a convenient approximation for the unknown coefficient $D_{1}(
\mu ( x,t) ) $ on this strip. For example, suppose the
coefficient in the g-adjoint problem is chosen to have the known constant
value, $d_{k-1};$ i.e.,

$\qquad \qquad \qquad \qquad \qquad $

$\qquad \qquad \qquad \qquad D_{1}( \mu ( x,t) )
=d_{k-1}\qquad $ $\mu ( x,t) \in J_{k}=\left[ \mu _{k-1},\mu _{k}%
\right] .$

\vspace{1pt}

Then if we replace $\phi $ in (\ref{alg}) by $\hat{\phi}( x,t) ,$ we
can denote the resulting computed nodal value by $\hat{d}_{k}.$ Note that
with this choice for the coefficient, there is now no difficulty in solving
the adjoint problem (\ref{gadj}) for $\hat{\phi}$ on the strip, $( 0,1)
\times \left[ T_{k-1},T_{k}\right] $ and proceeding to compute $\hat{d}_{k}$
using (\ref{alg}). It remains to be seen how the values $\hat{d}_{k}$ compare to
the values $d_{k}.$ We begin with a lemma.

\vspace{1pt}

\begin{lemma} \rm
\label{L:bound2}
Let $f(t)$ satisfy (\ref{condf}), let coefficient $D$ satisfy
(\ref{condD}) and let $\Pi $ denote a fixed partition, 
$\ \Pi =\{\mu_k=f(T_k): \ k=0,1,...,M\}$ of J. For $k$ between 1 and
$M$ consider the following adjoint problem, 
\begin{equation*}
\begin{array}{rcll}
\partial _{t}\phi ( x,t) +c\,\,\partial _{xx}\phi (
x,t) &=&0,& \in S_{k} \\
\phi ( x,T_{k}) &=&0,& x\in ( 0,1) \\
\phi ( 0,t) &=&f( T_{k}-t) & t\in (T_{k-1},T_{k})  \\
\partial_{x}\phi ( 1,t) &=&0,&t\in (T_{k-1},T_{k}) 
\end{array}
\end{equation*}

\vspace{1pt}

Suppose \ $\{ \phi_{i},c_{i}\} , \ i=1,2$ \ denote
two solutions to the adjoint problem corresponding to distinct choices of
the coefficient $c.$ In particular, suppose $\phi_{1}=\phi (
x,t,c_{1},\theta ) $ for the constant $\ c_{1}=d_{k-1}$, while $\phi
_{2}=\phi ( x,t,c_{2},\theta ) $ corresponding to the choice, 
$\;c_{2}(x,t) =D(\mu(x,t)),$ where $\mu(x,t)$ denotes a function 
that is continuous on the strip 
$S_{k}=( 0,1) \times ( T_{k-1},T_{k}) $ with values in 
$J_{k}=[\mu_{k-1},\mu_{k}]$. Then

\begin{equation*}
\,\,\,\left| \left| \,\partial _{x}(\phi _{1}-\phi _{2})\right| \right|
_{L^{2}( S_{k}) }\leq C\,\left| \mu _{k}-\mu _{k-1}\right| 
\end{equation*}
\end{lemma}

\begin{proof}
Begin by noting that $\;\Delta \phi =\phi _{1}-\phi _{2}\;$%
satisfies, 
\begin{equation*}
\begin{array}{rcll}
\partial _{t}(\Delta \phi )+c_{1}\,\,\partial _{xx}(\Delta \phi )
&=&(c_{2}-c_{1})\,\partial _{xx}\phi _{2},& ( x,t) \in
S_{k} \\
\Delta \phi ( x,T_{k}) &=&0, &  x \in ( 0,1) \\
\Delta \phi ( 0,t) &=&0, & \in ( T_{k-1},T_{k}) ,
\\
\partial _{x}(\Delta \phi )( 1,t) &=&0, &  t\in (T_{k-1},T_{k}) ,
\end{array}
\end{equation*}
\qquad \qquad \qquad

and if $\psi $ \ denotes an arbitrary test function, then

\qquad $\iint\limits_{S_{k}}\left\{ \partial _{t}(\Delta \phi
)+c_{1}\,\,\partial _{xx}(\Delta \phi )\right\} \partial _{x}\psi
\,dxdt=\iint\limits_{S_{k}}\left\{ -\Delta c\,\,\partial _{xx}\phi
_{2}\right\} \partial _{x}\psi \,dxdt.$

Integration by parts yields, 
\begin{eqnarray*}
\iint\limits_{S_{k}}\partial _{t}(\Delta \phi )\partial _{x}\psi \,dxdt
&=&\iint\limits_{S_{k}}\partial _{x}(\Delta \phi )\partial _{t}\psi \,dxdt \\
&&+\int_{0}^{1}\Delta \phi \,\partial _{x}\psi |_{t=0}^{t=T}dx \\
&&-\int_{T_{k-1}}^{T_{k}}\Delta \phi \,\partial _{x}\psi |_{x=0}^{x=1}dt,
\end{eqnarray*}

\vspace{1pt}and\qquad 
\begin{eqnarray*}
\iint\limits_{S_{k}}\partial _{xx}(\Delta \phi )\,\partial _{x}\psi \,dxdt
&=&-\iint\limits_{S_{k}}\partial _{x}(\Delta \phi )\partial _{xx}\psi \,dxdt
\\
&&+\int_{T_{k-1}}^{T_{k}}\partial _{x}(\Delta \phi )\,\partial _{x}\psi
|_{x=0}^{x=1}dt,
\end{eqnarray*}

so\qquad 
\begin{eqnarray*}
&&\iint\limits_{S_{k}}\partial _{x}(\Delta \phi )[\partial _{t}\psi
-c_{1}\partial _{xx}\psi ]\,dxdt+\int_{0}^{1}\Delta \phi \,\partial _{x}\psi
|_{t=T_{k-1}}^{t=T_{k}}dx \\
&&-\int_{T_{k-1}}^{T_{k}}\Delta \phi \,\partial _{x}\psi
|_{x=0}^{x=1}dt+c_{1}\int_{T_{k-1}}^{T_{k}}\partial _{x}(\Delta \phi
)\,\partial _{x}\psi |_{x=0}^{x=1}dt \\
&=&\iint\limits_{S_{k}}\left\{ -\Delta c\,\,\partial _{xx}\phi _{2}\right\}
\partial _{x}\psi \,dxdt
\end{eqnarray*}

Now choose the test function $\psi$ to satisfy 

\begin{equation*}
\begin{array}{rcll}
\partial _{t}\psi -c_{1}\partial _{xx}\psi &=&\partial _{x}(\Delta \phi
), & (x,t) \in S_{k} \\
\psi(x,T_{k-1}) &=&0, &  x\in ( 0,1) \\
\partial _{x}\psi(0,t) &=&0, &  t\in (T_{k-1},T_{k}) \\
\psi ( 1,t) &=&0, & t\in ( T_{k-1},T_{k})
\end{array}
\end{equation*}

Then the previous integral identity reduces to

\[
\iint\limits_{S_{k}}[\partial _{x}(\Delta \phi
)]^{2}dxdt=\iint\limits_{S_{k}}\,(c_{2}-c_{1})\,\partial _{xx}\phi
_{2}\,\partial _{x}\psi \,dxdt.
\]

Now, $\psi $ is the solution to a linear problem with constant coefficients
so it can be expressed in terms of a Green's function, $\Gamma (
x,t) ,$

\[
\psi ( x,t) =\int_{T_{k-1}}^{t}\int_{0}^{1}\Gamma (
x-y,t-\tau ) \,\partial _{x}(\Delta \phi )( y,\tau )
\,dyd\tau ,\;( x,t) \in S_{k},
\]

and

\[
\partial _{x}\psi ( x,t)
=\int_{T_{k-1}}^{t}\int_{0}^{1}\partial _{x}\Gamma ( x-y,t-\tau )
\,\partial _{x}(\Delta \phi )( y,\tau ) \,dyd\tau.
\]

Then for all $(x,t) \in S_{k},$ 

\begin{tabular}{c}
\ms
$\left| \,\partial _{x}\psi (x,t) \right| \leq
\int_{T_{k-1}}^{t}\int_{0}^{1}|\partial _{x}\Gamma ( x-y,t-\tau )
\,\partial _{x}(\Delta \phi )( y,\tau ) |\,dyd\tau ,$ \\
$ \leq \left( \int_{T_{k-1}}^{T_{k}}\int_{0}^{1}|\partial _{x}\Gamma
( x-y,t-\tau ) |\,^{2}dyd\tau \right) ^{1/2}\,\cdot \left(
\int_{T_{k-1}}^{T_{k}}\int_{0}^{1}|\partial _{x}(\Delta \phi )( y,\tau
) |^{2}\,dyd\tau \right) ^{1/2} $
\end{tabular}

and

\[
\max\limits_{( x,t) \in S_k}\left| \,\partial _{x}\psi
( x,t) \right| \leq C\,\,\left| \left| \,\partial _{x}(\Delta
\phi )\right| \right| _{L^{2}( S_{k}) }.
\]

\smallskip
Then it follows that

\begin{tabular}{rcl}
$\iint\limits_{S_{k}}[\partial _{x}(\Delta \phi
)]^{2}dxdt$  &$=$& $\left| \iint\limits_{S_{k}}\,(c_{2}-c_{1})\,\partial _{xx}\phi _{2}\,\partial _{x}\psi \,dxdt\right|$ \\
&$\leq$&
$\max\limits_{S_{k}}\,\left| \Delta c( x,t) \right|
\iint\limits_{S_{k}} \ |\partial _{xx}\phi _{2}\,\partial _{x}\psi| \ dxdt $\\ 
&$\leq$&
$ \max\limits_{S_{k}}\,\left|
\Delta c( x,t) \right| \,\left| \left| \,\partial _{xx}\phi
_{2}\,\right| \right| _{L^{1}}\,\left| \left| \,\partial _{x}\psi \right|
\right|_{\infty} $
\end{tabular}

and

\[
\left| \left| \,\partial _{x}(\Delta \phi
)\right| \right| _{L^{2}( S_{k}) }\leq
C\,\,\max\limits_{S_{k}}\,\left| \Delta c( x,t) \right|.
\] 

Also 

\begin{tabular}{rl}
$\max\limits_{S_{k}}\,\left| \Delta c(x,t) \right|$
&$=\,\max\limits_{S_{k}}\,|d_{k-1}-D( \mu ( x,t) ) |$ \\
&$\leq |d_{k-1}-D( \mu
_{k-1}) |+\max\limits_{S_{k}}\,|D(\mu _{k-1})-D( \mu(x,t))|$ \\
&$\leq 2K\left| \,\mu _{k}-\mu _{k-1}\right| .$ 
\end{tabular}

Then, it follows that,

\[
\left| \left| \,\partial _{x}(\Delta \phi )\right| \right|
_{L^{2}( S_{k}) }\leq C\,\left| \mu _{k}-\mu _{k-1}\right|.
\] 
\end{proof}

Now we will use this estimate in considering the effect of using the
approximate adjoint solution in the determination of the first nodal value, $%
d_{1}.$ It follows from (\ref{alg}) that the difference between the value, $d_{1},$
computed using the correct but unknown adjoint solution and the value, $\hat{%
d}_{1},$ computed using an incorrect but computable adjoint solution is
given by, 
\begin{eqnarray*}
\hat{d}_{1}-d_{1} &=&\dfrac{\int_{0}^{T_{1}}( g( t)
-g_{2}( t) ) \theta ( t) \,dt.}{%
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\hat{\phi}\,dx\,dt} \\
&&-\dfrac{\int_{0}^{T_{1}}( g( t) -g_{2}( t)
) \theta ( t) \,dt.}{\int_{0}^{T_{1}}\int_{0}^{x_{0}(
t) }\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,\partial
_{x}\phi \,dx\,dt} \\
&=&\dfrac{( g-g_{2},\theta ) }{II( \hat{\phi}) }-%
\dfrac{( g-g_{2},\theta ) }{II( \phi ) } \\
&=&( g-g_{2},\theta ) \left\{ \dfrac{1}{II( \hat{\phi}%
) }-\dfrac{1}{II( \phi ) }\right\}
\end{eqnarray*}

\vspace{1pt}

\qquad 
\begin{equation*}
\hat{d}_{1}-d_{1}=( d_{1}-d_{0}) \left\{ \dfrac{II( \phi
) -II( \hat{\phi}) }{II( \hat{\phi}) }\right\} 
\end{equation*}

Here

\vspace{1pt}\qquad \qquad \qquad $II( \hat{\phi})
=\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\hat{\phi}\,dx\,dt.$

\vspace{1pt}

We wish to show that as the outer mesh is refined, the discrepancy $II(
\phi ) -II( \hat{\phi}) $ that is due to solving the
adjoint problem with the wrong coefficient decreases to zero. On the other
hand, $II( \hat{\phi}) $ also decreases toward zero as the mesh
is refined. To see whether $II( \hat{\phi}) $ decreases more or
less rapidly than $II( \phi ) -II( \hat{\phi}) ,$ it
is necessary to examine the asymptotic behavior of $II( \hat{\phi}%
) $. We assume that $x_{0}( T_{1}) <1$ since if this is not
the case, we can always refine the outer partition to shrink the width of
the strip $S_{1}$ so as to make it true. Then the domain of integration for $%
II( \hat{\phi}) $ is the approximately triangular region $\left\{
0\leq x\leq x_{0}( t) ,0\leq t\leq T_{1}\right\} .$ An exact
analysis of the asymptotic rate of convergence of $II( \hat{\phi}%
) $ as $T_{1}$ tends to zero is difficult, but if we assume that $%
f( t) =At$ for a positive constant A, then it is possible to
solve explicitly for $u_{2}( x,t) $ and $\hat{\phi}(
x,t) .$ Using arguments like those in $\left[ 1\right] $, one finds
that $g( t) =-D( u_{1}( 0,t) ) \,\partial
_{x}u_{1}( 0,t) $ and\ $g_{2}( t) =-d_{k}\partial
_{x}u_{2}( 0,t) \;$behave asymptotically like $\sqrt{t}.$

This leads to 
\begin{equation}
\label{estI}
\begin{array}{rcl}
\int_{0}^{T_1}(g(t)-g_2( t) ) \theta(t) \ dt &=&
\int_0^{T_{1}}(g(t) -g_2(t)) \,A( T_{1}-t) dt  \\
&\approx &CT_{1}^{5/2} 
\end{array}
\end{equation}

A similar crude estimate for $\partial _{x}u_{2}\,\partial _{x}\hat{\phi}$
on $0\leq x\leq 1,\;0\leq t\leq T_{1},$ is the following 
\begin{equation*}
\partial _{x}u_{2}\,\partial _{x}\hat{\phi}( x,t) \thickapprox 
\sqrt{t}\,m( x) \,\sqrt{T_{1}-t}\,m( x) \;
\end{equation*}

where $m( x) $ denotes a decreasing function with $m(
0) =1$ and $m( 1) =0.$ In addition, for $T_{1}$ small, one
can suppose $x_{0}( t) \thickapprox at$ for a positive constant $%
a,$ and this leads to
\begin{eqnarray}
\label{estII}
\quad II(\hat{\phi}) &=& \int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda_1(
u_2) \,\partial _x u_2\,\partial _x\hat{\phi} \ dx \ dt \nonumber \\
&\approx& \int_{0}^{T_1}\int_{0}^{at}
\frac{u_{2}( x,t) }{AT_{1}}\sqrt{t}\,m( x) \,\sqrt{T_1-t}\,m( x) \,dx\,dt \nonumber \\
i.e. \quad  II(\hat{\phi}) &\approx& CT_{1}^{5/2}. 
\end{eqnarray}

Since this estimate (\ref{estII}) is rather rough, the quantity $II( \hat{\phi}%
) $ was computed numerically for a sequence of values for $T_{1}$
decreasing to zero. The result of this numerical asymptotic estimate
supported the estimate (\ref{estII}) which asserts that $II( \hat{\phi}) $
decreases like the $\frac{5}{2}$ power of $T_{1}\;$as $T_{1}$ tends to zero.

\qquad Now

\qquad \qquad \qquad $\hat{d}_{1}-d_{1}=( d_{1}-d_{0}) \left\{ 
\dfrac{II( \phi ) -II( \hat{\phi}) }{II( \hat{\phi%
}) }\right\} $

and,


$ \;\;\;\left| II( \phi ) -II( \hat{\phi}%
) \right| =\left| \int_{0}^{T_{1}}\int_{0}^{x_{0}( t)
}\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,[\partial _{x}\phi
-\partial _{x}\hat{\phi}]\,dx\,dt\right| $

\qquad \qquad \qquad \qquad \qquad $\leq C( T_{1}) \,\left|
\left| \,\partial _{x}(\Delta \phi )\right| \right| _{L^{2}(
S_{1}) }\leq C( T_{1}) \,\,\left| \mu _{1}-\mu _{0}\right|
. $

Also,

\[
\left| \,d_{1}-d_{0}\right| =\left| \,D(\mu
_{1})-D( \mu _{0}) \right| \leq K\left| \mu _{1}-\mu _{0}\right|
, \]

and hence 

\[
\left| \hat{d}_{1}-d_{1}\right| \leq \left|
d_{1}-d_{0}\right| \left| \dfrac{II( \phi ) -II( \hat{\phi}%
) }{II( \hat{\phi}) }\right| 
\leq \dfrac{K\,C(
T_{1}) }{II( \hat{\phi}) }\,\left| \mu _{1}-\mu _{0}\right|
^{2}.
\]
Then for $T_{1}$ sufficiently small,
\begin{eqnarray*}
\left| \hat{d}_{1}-d_{1}\right| &\leq& \dfrac{
K\,C( T_{1}) }{CT_{1}^{5/2}}\,\left| \mu _{1}-\mu _{0}\right|
^{2}  \\
&\leq& \dfrac{K\,f'(\tau)^2}{C}T_{1}^{-1/2} \mbox{ for some } \tau >0.
\end{eqnarray*}
In general, we have

\begin{lemma} \rm
\label{L:meshest}
For $f(t)=At,\;A>0$, for coefficient $D$ satisfying
(2.3) and for a fixed partition,$\;\Pi =\left\{ \mu
_{k}=AT_{k}:k=0,1,...,M\right\} $ of J, fix $k$ between 1 and M. Let $\hat{%
\phi}=\phi ( x,t,d_{k-1},A( T_{k}-t) ) $ and $\phi
=\phi ( x,t,c,A( T_{k}-t) ) $ corresponding to the
coefficients, $d_{k-1}\;$and $c( x,t) =D( \mu (
x,t) ) ,$ respectively. Finally, let $\hat{d}_{k}$ and $d_{k}$
denote the nodal values determined from (\ref{alg}) using the values $\left[
d_{0},d_{1},...,d_{k-1}\right] $ and the adjoint solutions $\hat{\phi}$ and $%
\phi ,$ respectively. Then
\[
\left| \hat{d}_{k}-d_{k}\right| \leq \dfrac{K}{II( \hat{\phi}%
) }\,\left| \mu _{k}-\mu _{k-1}\right| ^{2}\leq \dfrac{K\,f\,^{\prime
}( \tau ) ^{2}}{C}|T_{k}-T_{k-1}|^{-1/2}.
\]
\end{lemma}

This lemma implies that the error introduced into the identification
by solving the adjoint problem with an approximate coefficient has an
increasing effect as the outer mesh is refined. As the mesh is refined, the
discrepancy $II( \phi ) -II( \hat{\phi}) $ does tend
to zero like the square of the mesh size. However, as the mesh size tends to
zero, we find also that $II( \hat{\phi}) ,$ which can be viewed
as an approximation to the Gateaux derivative of the mapping $\Phi $
restricted to a one dimensional subspace of $W( J_{k}) ,$ tends
to zero even faster, (like the $\frac{5}{2}$ power of the mesh size). It is
likely that the means of approximating the adjoint solution could be
improved so that $II( \phi ) -II( \hat{\phi}) $ would
approach zero sufficiently rapidly that $\left| \hat{d}_{k}-d_{k}\right| $
would tend to zero as the mesh size goes to zero. However, the next result
will show that such an improvement does not improve the convergence of the
approximate solution.

We wish finally to consider the effect of numerical integration
errors on the calculation of $\hat{d}_{k}.$ We begin by considering $k=1.$
We have,

\[
\hat{d}_{1}=d_{0}+\dfrac{\int_{0}^{T_{1}}( g( t)
-g_{2}( t) ) \theta ( t) \,dt.}{%
\int_{0}^{T_{1}}\int_{0}^{x_{0}( t) }\,\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\hat{\phi}\,dx\,dt}=d_{0}+%
\dfrac{I( g-g_{2}) }{II( \hat{\phi}) }
\]
and
\[
\hat{d}_{1}^{\ast }=d_{0}+\dfrac{I^{\ast }(
g-g_{2}) }{II^{\ast }( \hat{\phi}) }
\]

where $I^*(g-g_{2})$ and $II^*(\hat{\phi})$ denote, respectively, the 
computed results using the inner mesh
to numerically approximate the corresponding exact single and double
integrals. Then
\begin{eqnarray*}
\hat{d}_{1}^{\ast }&=&d_{0}+\dfrac{I^{\ast }(
g-g_{2}) -I( g-g_{2}) +I( g-g_{2}) }{II^{\ast
}( \hat{\phi}) -II( \hat{\phi}) +II( \hat{\phi}%
) }\\
&=& d_{0}+\dfrac{I( g-g_{2}) }{II( 
\hat{\phi}) }\dfrac{1+\varepsilon _{1}}{1+\varepsilon _{2}}
\end{eqnarray*}
where
\[
\varepsilon _{1}=|\dfrac{I-I^{\ast }}{I}|
\mbox{ and }
\varepsilon _{2}=|\dfrac{II-II^{\ast }}{II}|.
\]
Now \qquad \qquad $\dfrac{1+\varepsilon _{1}}{1+\varepsilon _{2}}%
\approx 1+\varepsilon _{1}+\varepsilon _{2}$

so

\qquad \qquad $\hat{d}_{1}^{\ast }=d_{0}+\dfrac{I( g-g_{2}) }{%
II( \hat{\phi}) }\dfrac{1+\varepsilon _{1}}{1+\varepsilon _{2}}%
\approx d_{0}+\dfrac{I( g-g_{2}) }{II( \hat{\phi}%
) }( 1+\varepsilon _{1}+\varepsilon _{2}) ,$

and

\qquad \qquad $\left| \hat{d}_{1}-\hat{d}_{1}^{\ast }\right| \leq \left| 
\dfrac{I( g-g_{2}) }{II( \hat{\phi}) }\right| \,(
\varepsilon _{1}+\varepsilon _{2}) =\left| \hat{d}_{1}-d_{0}\right|
( \varepsilon _{1}+\varepsilon _{2}) .$

The numerical integration errors are estimated by terms of the form,

$\qquad \qquad \qquad \;\left| I-I^{\ast }\right| \leq C(\Delta t)^{2}\;$\ \
\ \ \ \ for\ $\Delta t=$inner\ mesh size$\;$

and$\;\qquad \qquad \left| II-II^{\ast }\right| \leq C( \Delta
x\,\Delta t) =C(\Delta t)^{2}\;\;$

\vspace{1pt}

\vspace{1pt}Use of higher order integration schemes is limited by the fact
that reducing the mesh size of the outer or $J-mesh$ in order to achieve
identification accuracy absorbs $I-mesh$ node points into the $J-mesh$
leaving only enough points in the inner mesh to perform low order numerical
integrations.

\qquad It follows from (\ref{estI}) and (\ref{estII}) that

\vspace{1pt}

\qquad \qquad \qquad \qquad $I=\int_{0}^{T_{1}}( g( t)
-g_{2}( t) ) \,A( T_{1}-t) dt\thickapprox
T_{1}^{5/2},$

\qquad \qquad \qquad\ \ \ \ $II=\int_{0}^{T_{1}}\int_{0}^{x_{0}(
t) }\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,\partial
_{x}\phi \,dx\,dt\thickapprox T_{1}^{5/2}$

\vspace{1pt}

Then, since $T_{1}=k\Delta t,$ we find

\qquad \qquad \qquad $\left| \hat{d}_{1}-\hat{d}_{1}^{\ast }\right| \leq
\left| \dfrac{I( g-p_{M}) }{II( \hat{\phi}) }\right|
( \varepsilon _{1}+\varepsilon _{2}) \,$

$\qquad \qquad \qquad \qquad \qquad \;\;\leq \left| \hat{d}_{1}-d_{0}\right| 
\dfrac{C_{1}(\Delta t)^{2}}{C_{2}\,(k\Delta t)^{5/2}}\leq C(\Delta
t)^{-1/2}. $

\vspace{1pt}

\vspace{1pt}More generally, we have

\vspace{1pt}

\begin{lemma} \rm
\label{L:ratioerr} \rm
Under the conditions of (\ref{L:meshest}), let $\hat{d}_{k}^{\ast
}$ reflect the error induced in $\hat{d}_{k}$ by numerically approximating
the integrals needed for (\ref{alg}). Then, as the (inner and outer) mesh size
tends to zero, 
\begin{equation*}
\left| \hat{d}_{k}-\hat{d}_{k}^{\ast }\right| \leq C(\Delta t)^{-1/2} 
\end{equation*} 
\end{lemma}

This estimate suggests that as the outer mesh is refined in order to
improve the accuracy of the identification of the nodal values of $D(
u_{1}) ,$ more and more node points of the inner mesh are absorbed
into the outer mesh, resulting in numerical integration errors, $\left|
I-I^{\ast }\right| $ and $\left| II-II^{\ast }\right| ,$ that are of order $%
\Delta t^{2}.$ At the same time, the approximate Gateaux derivative $%
II( \hat{\phi}) $ tends to zero like $\Delta t^{5/2}$ so the
effect of approximating the integrals becomes magnified at $\Delta t$ tends
to zero. Evidently, at some point the values of the integrals used to
compute $d_{k}$ become of the same order of magnitude as the numerical
integration errors and the computation then no longer contains information.
Further decreasing the mesh size then only increases the error.

Finally, we can combine lemmas \ref{L:bound1}, \ref{L:meshest} and \ref{L:ratioerr} to write

\qquad $\qquad \qquad \left| D( \mu _{k}) -\hat{d}_{k}^{\ast
}\right| =\left| D( \mu _{k}) -d_{k}+d_{k}-\hat{d}_{k}+\hat{d}%
_{k}-\hat{d}_{k}^{\ast }\right| $

$\qquad \qquad \qquad \qquad \qquad \qquad \,\leq \left| D( \mu
_{k}) -d_{k}\right| +\left| d_{k}-\hat{d}_{k}\right| +\left| \hat{d}%
_{k}-\hat{d}_{k}^{\ast }\right| $

and, 
\begin{equation}
\label{finalest}
\left| D( \mu _{k}) -\hat{d}_{k}^{\ast }\right| \leq C_{1}\Delta
t+C_{2}(\Delta t)^{-1/2}.
\end{equation}

$\vspace{1pt}$

Evidently the error in identifying $d_{k}$ does not tend to zero as
\thinspace $\Delta t$ tends to zero but is minimized by an optimal $\Delta t$
different from zero.

\vspace{1pt}

\section{Numerical Experiments}

In the numerical experiments we describe here, we chose $f( t)
=At $ for some positive constant $A$ and defined the node points $\mu _{k}$
for the outer mesh by $\mu _{k}=AT_{k},\;k=0,1,...,M.$ Here, for each k, $%
T_{k}=t_{j}$ for some $j>k$ where $0=t_{0}<t_{1}<\cdots <t_{N}=T$ denotes
the (inner) partition of $\left[ 0,T\right] .$ The unknown nodal values for
the coefficient $D(u)$ are given by $d_{k}=D( \mu _{k}) ,$ and we
assume that $d_{0}$ is known. Since the initial state for the direct
problem, $u(x,0),$ is constant and $f(t)$ is monotone increasing, the domain 
$Q_{T}$ consists of a sequence of non-overlapping strips, $S_{k},$ with only
one nodal value active on each strip. The algorithm to identify $D$ from the 
$( f,g) -data$ then proceeds as follows.

\textbf{The Algorithm}

To begin, we apply the g-integral identity (\ref{gint}) on $Q_{1}.$ Since
the solution of the direct problem satisfies lemma \ref{L:lem1}(a), we have $\mu
_{0}\leq u_{1}( x,t) \leq \mu _{1}$ for $( x,t) \in
Q_{1}.$ Then only the known nodal value $d_{0}$ and the unknown nodal value $%
d_{1}$ are active on this strip. We are going to compute the unknown nodal
values iteratively and we denote the $i-th$ iteration for $d_{k}$ by $%
d_{k}^{( i) }.$ We set $d_{1}^{( 0) }=d_{0}.$

We apply the integral identity (\ref{gint}) on $Q_{1}$ with,

$\qquad \qquad D_{1}=P_{1}\left[ d_{0},d_{1}^{( 1) }\right]
\qquad \mbox{ and }\qquad D_{2}=P_{1}\left[ d_{0},d_{1}^{( 0) }\right] $

$\qquad \qquad u_{2}( x,t) =u( x,t;D_{2},At) \qquad
and\qquad g_{2}( t) =\Phi \left[ f,D_{2}\right] $

$\qquad \qquad \hat{\phi}( x,t) =\phi ( x,t;D_{2},A(
T_{1}-t) ) $

$\qquad \qquad $

\qquad We compute\qquad \qquad $A_{11}=\int_{0}^{T_{1}}\int_{0}^{x_{0}(
t) }\,\lambda _{1}( u_{2}) \,\partial _{x}u_{2}\,\partial
_{x}\hat{\phi}\,dx\,dt,\;$\qquad

\qquad \qquad \qquad \qquad

\qquad \qquad \qquad \qquad \ \ \ \thinspace \qquad\ \ $b_{1}=%
\int_{0}^{T_{1}}( g( t) -g_{2}( t) )
A( T_{1}-t) \,dt$

\qquad and solve

\qquad \qquad \qquad \qquad \qquad $A_{11}( d_{1}^{( 1)
}-d_{0}) =b_{1}.$

Note that $A_{11}$ and $b_{1}$ are computed from $u_{2},\;\hat{\phi},\;g_{2}$
which are all based on the known coefficient $D_{2}.$

\vspace{1pt}\qquad To continue, we apply the g-integral identity (\ref{gint}) first
on $Q_{1},$ where only $d_{0},d_{1}$ are active, and then we apply the
g-integral identity (\ref{gint}) again, but now on $Q_{2}$ where $d_{0},d_{1},d_{2}$
are active. That is,

\vspace{1pt}

\qquad \qquad on $Q_{1}\qquad \qquad D_{1}=P_{1}\left[ d_{0},d_{1}^{(
2) }\right] \;\;d_{1}^{( 2) }$ is unknown,

$\qquad \;\;\;\;\;$\ \ \thinspace and$\qquad \qquad \;\;\,D_{2}=P_{1}\left[
d_{0},d_{1}^{( 1) }\right] \;\;d_{1}^{( 1) }$ is known,

$\vspace{1pt}$

$\qquad $\ \ \ \ and\ we compute \qquad $A_{11}$ and $\;b_{1}\;\;$as before$%
. $

Note that $u_{2},\;\hat{\phi},\;g_{2}$ are based on the updated coefficient $%
D_{2}$ so that, in general, $d_{1}^{( 2) }$ will not be the same
as $d_{1}^{( 1) }.$

\vspace{1pt}

\qquad $\;\;\;\,\,$On $Q_{2}\qquad \qquad D_{1}=P_{2}\left[
d_{0},d_{1}^{( 2) },d_{2}^{( 1) }\right] \;\;$

$\qquad $\ $\;\;$\ and$\qquad \qquad \;\;\,\,\,$\thinspace $D_{2}=P_{2}\left[
d_{0},d_{1}^{( 1) },d_{2}^{( 0) }\right]
\;\;\;\;\;\;note:\;d_{2}^{( 0) }=d_{1}^{( 1) }$

\vspace{1pt}

\qquad \ \ \ we compute\qquad $A_{2,1}=\int \int_{Q_{21}}\lambda _{1}(
u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\hat{\phi}\,dx\,dt,\;\;\;$

$\qquad \qquad \qquad \qquad \qquad $\ $\;\;\;\;\;\;\;\;\;\;\;Q_{21}=\left\{
\mu _{0}\leq u_{2}( x,t) \leq \mu _{1},\,0\leq t\leq
T_{2}\right\} $

$\qquad \qquad \qquad \qquad \qquad
\;A_{2,1}=\int_{T_{1}}^{T_{2}}\int_{0}^{x_{1}( t) }\,\lambda
_{2}( u_{2}) \,\partial _{x}u_{2}\,\partial _{x}\hat{\phi}%
\,dx\,dt\;\;$

$\;\;\;\;\,\;\qquad \qquad \qquad \qquad \;\;b_{2}=\int_{0}^{T_{2}}(
g( t) -g_{2}( t) ) A( T_{2}-t) \,dt$

\vspace{1pt}

\qquad \ \ and we solve\qquad $\left[ 
\begin{array}{ll}
A_{11} & 0 \\ 
A_{2,1} & A_{2,2}
\end{array}
\right] \left[ 
\begin{array}{l}
d_{1}^{( 2) }-d_{1}^{( 1) } \\ 
d_{2}^{( 1) }-d_{2}^{( 0) }
\end{array}
\right] =\left[ 
\begin{array}{l}
b_{1} \\ 
b_{2}
\end{array}
\right] .$

\vspace{1pt}

We proceed in this way, where at the k-th stage we apply the integral
identity k times, once on each of the strips $Q_{1}$ to $Q_{k}.$ Of course
this produces k equations, one for each strip. On each strip, $Q_{j}$ there
are only j unknown active node values $d_{1}^{( p)
},...,d_{j}^{( q) },$ at various stages of iteration, hence the
j-th equation contains only the first $j$ unknowns. This leads to a k by k
lower triangular system for the differences $d_{j}^{( p)
}-d_{j}^{( p-1) }.$ At the k-th stage of the algorithm we are
solving for the first iterate for $d_{k},$ for the second iterate of $%
d_{k-1},$ etc. This algorithm, which we will call the iterative
algorithm, differs from the non-iterative algorithm described in the
preceding section. The non-iterative algorithm amounts to suppressing 
the iterative
feature so that for each $k,$ the nodal value $d_{k}$ is obtained by solving
just a single equation, $A_{kk}( d_{k}-d_{k-1}) =b_{k}.$
Suppressing the iteration leads to cascading errors in the sequentially
computed nodal values $d_{k}$ as shown in Figure \ref{F:arctan}. 
The coefficient shown in this figure,

\begin{equation}
\label{E:arctan}
D( u) =2-\arctan\left[ 6(u - 1/2) \right] ,\qquad 0<u<1,
\end{equation}

\begin{figure}
\caption[]{ \label{F:arctan} }
\begin{center}
\epsfig{file=iter.eps,width=9cm}
\end{center}
\end{figure}

\noindent
was recovered in two ways.  In the first, the non-iterative algorithm applied to the data $\{f,g\}$ to produce the dashed line plot,  while the iterative algorithm was applied to produce the solid line plot. 
The data was generated by solving the direct problem (\ref{direct})
using a functional form of the coefficient (\ref{E:arctan}) on a
mesh of 70 nodes with the Matlab solver {\tt ode15s}. The flux, 
$g(t),$ was then computed using a difference formula.  
This flux data was submitted to the recovery algorithms, both
which used a 40 node mesh and {\tt ode15s} to compute solutions to 
the direct and adjoint problem.  It is clear from the figure that the errors in
non-iterated nodal values for $D(u)$ accumulate as the values are
sequentially determined.  We point out that determining $d_{k}$ we are
obliged to integrate over the approximately triangular region 
$\{ 0<x<x_{k}( t) ,\,T_{k-1}<t<T_{k}\}.$ However, the
algorithm must numerically approximate $x_{0}( t_{j}) $ on the
inner mesh, and this leads to a systematic overestimation of the value of $%
A_{kk}$ which, in turn leads to a correction term that is too small.
The fact that $D$ is a decreasing function of $u$, as given in
equation (\ref{E:arctan}), leads to a negative $\Delta g(t)$
and a negative correction, $b_{k}/A_{kk}.$ This is evident in the
dashed-line plot of Figure \ref{F:arctan}. The fact that $A_{kk}$ is too
large causes the negative correction to be too small so that the graph of
the computed polygonal function lies above the graph of the true
coefficient. Since the integrals for $A_{kk}\;and\;b_{k}$ involve only the
interval $[T_{k-1},T_{k}],$ each identified value, $d_{k},$ can
do nothing to diminish errors in previously identified values, hence the
identification error accumulates. 

This suggests that iteration might prove useful.
The solid line plot in Figure \ref{F:arctan} shows the result of identifying the coefficient {\ref{E:arctan}) but
now iterating as follows. We use the identity (\ref{gint}) on $Q_{1}$ together with the known value, $d_{0},$ to identify $d_{1}^{( 1) }.$ Here the
known value, $d_{0},$ is used to compute $u_{2}( x,t) ,$ $%
g_{2}( t) $ and $\hat{\phi}( x,t) .$ Next we use the
identity (\ref{gint}) on $Q_{1}$ and $Q_{2}$ together with known values, $d_{0},$ $%
d_{1}^{( 1) }$ to identify $d_{1}^{( 2) }$ and $%
d_{2}^{( 1) }.$ In the next step, we use the identity (\ref{gint}) on 
$Q_1,Q_2$ and $Q_3$ together with known values, $d_{0}, d_{1}^{(
2) }$ and $d_{2}^{( 1) }$ to identify $d_{1}^{(3)},d_{2}^{(2)}$ 
and $d_{3}^{(1)}.$ At each stage, the known nodal values are used to compute 
$u_{2}(x,t),g_{2}(t)$ and $\hat{\phi}( x,t) .$ Continuing in
this way, we eventually obtain $d_{M}^{( 1) },d_{M-1}^{(
2) },...,d_{1}^{( M) }.$ It is evident from the solid line plot in
Figure \ref{F:arctan} that
as a result of the iteration, the errors no longer exhibit the cumulative
character seen in the dashed line plot, where iteration was not applied.

Here the coefficient 
\begin{equation}
\label{E:sine}
D(u) =1 + \frac{1}{2}\sin(2 \pi u) ,\qquad 0<u<1, 
\end{equation}
was used to generate flux data as in the previous example, although
here the Matlab solver {\tt ode23s} was used.  
This data was passed to the iterative recovery algorithm, 
the results of which are plotted in Figure \ref{F:sine}
\begin{figure}
\caption[]{ \label{F:sine} }
\begin{center}
\epsfig{file=sine.eps,width=9cm}
\end{center}
\end{figure}

The qualitative agreement between the computed and true coefficient
appears reasonable in this figure.  Notice that the 
approximation initially lies above the
plot of the true coefficient (\ref{E:sine}) in regions where $D$ is increasing, 
which is in agreement with the analysis of the previous experiment.
The value at the last nodal is not iterated in this scheme, and is
visibly less accurate than the computed values on other nodes.

\begin{figure}
\caption[]{\label{F:freedom}}
\begin{center}
\epsfig{file = D0.eps , width=5in}
\end{center}
\end{figure}

Figure \ref{F:freedom} displays the effect of refining the outer mesh by
increasing $M,$ the number of nodes in order to identify the coefficient

\[
D(u) =1+u,\qquad 0<u<1, 
\]

The results for $M=2,5$ and $9$ are shown in addition to a plot of the 
$L^2-error$ versus M. This last display shows the error decreasing with
increasing M up to about $M=5,$ at which point the error begins again to
increase. This result is in qualitative agreement with (\ref{finalest}).

\begin{minipage}[b]{7 cm}
\begin{figure}[H]
\caption[]{\label{F:noiseD}}
\epsfig{file=noiseD.eps,width=6.5cm}
\end{figure}
\end{minipage}
\begin{minipage}[b]{7cm}
\begin{figure}[H]
\caption{\label{F:noiseG}}
\epsfig{file=noiseG.eps,width=6.5cm}
\end{figure}
\end{minipage}

Figure \ref{F:noiseD} represents coefficient recovery in which the
data contained induced error. 
A relative uniform random error of $10\%$ was induced in the flux data,
and the iterative algorithm was applied.
The flux data used for recovery is plotted in Figure \ref{F:noiseG}.  
The recovered coefficient, plotted in Figure
\ref{F:noiseD} appears to capture the general structure of the true 
coefficient.  No preprocessing was applied to this data, which was
possible since the error had mean zero. The integration of the of
$g$ data in (\ref{alg}) allows much of this error to cancel.

\section{Conclusions}

The integral identities (\ref{gint}) and (\ref{hint}) are equations providing
explicit representations for the input to output mappings associated with
the inverse problem of identifying an unknown diffusion coefficient from
overspecified data measured on the boundary. These equations provide a means
for proving that the input output maps are continuous, injective and
strictly monotone. Such information is not so readily obtained from an
output least squares approach nor from equation error techniques. The
equations may also be the basis for a numerical approximation procedure
although here it might be more difficult to compete with sophisticated OLS
implementations.

\medskip
{\bf Acknowledgment.}  The authors want to thank the referees
for their valuable insights and suggestions.
\section{References}

{\footnotesize \noindent }

\begin{itemize}
\item[\lbrack 1]] H. B Ameur, G. Chavent, J.\ Jaffre, Refinement and
coarsening of parameterization for the estimation of hydraulic
transmissivity, Proceedings of Inverse Problems in Engineering, 3rd Int'l
Conf on Inverse Problems in Engineering, 1999, Port Ludlow Wa.

\item[\lbrack 2]] J. Cannon and P. DuChateau, Some asymptotic boundary
behavior of solutions of nonlinear parabolic initial boundary value
problems, JMAA vol 68, no. 2, (1997) pp536-547
\end{itemize}

\begin{enumerate}
\item[\lbrack 3]] J. Cannon and P. DuChateau, Design of an experiment for
the determination of a coefficient in a nonlinear diffusion equation, Int.
J. Eng. Sci., vol 25, No. 8, pp1067-78, 1987
\end{enumerate}

\begin{itemize}
\item[\lbrack 4]] G. Chavent, and P. Lemonnier, Identification de la
non-linearite d'une equation parabolique quasilineaire, Appl. Math. Optim., 1%
$( 1974) $ pp121-162

\item[\lbrack 5]] P. DuChateau, An inverse problem for the hydraulic
properties of porous media, SIAM J Math Anal 28(1997) pp611-632.

\item[\lbrack 6]] M. Hanke and O. Scherzer, Error analysis of an equation
error method for the identification of the diffusion coefficient in a
quasilinear parabolic differential equation, SIAM J. Appl. Math. 59 (1999),
1012-1027

\item[\lbrack 7]] D. Hinestroza and D. Murio, Identification of
transmissivity coefficients by mollification techniques. Part I:
1-dimensional elliptic and parabolic problems, Comp Math Applic, 25(1993) pp
59-79

\item[\lbrack 8]] Y. Jarny, M. N. Ozisik and J P Bardon, A general
optimization method using adjoint equation for solving multidimensional
inverse heat conduction. Int. J. Heat Mass Transfer, Vol 34, No 11, 1991
pp2911-2918

\item[\lbrack 9]] P. Knabner and S. Bitterlich, An Efficient Method for
Solving an Inverse Problem for the Richards Equation. Journal of
Computational and Applied Mathematics, Vol. 147, Issue 1, 2002, pp153-173

\item[\lbrack 10]] R. Nabakov, An inverse problem for porous medium
equation, in Parameter Identification and Inverse Problems in Hydrology,
Geology and Ecology, J. Gottlieb and P DuChateau, eds, Kluwer, Dordrecht,
Boston, London 1996 pp155-163.

\item[\lbrack 11]] Reeve, D E; Spivack, M; Recovery of a variable
coefficient in a coastal evolution equation, Journal of Computational
Physics, Vol. 151, pp 585-596, 1999.
\end{itemize}

\begin{enumerate}
\item[\lbrack 12]] G. R. Richter, Numerical identification of a spatially
varying diffusion coefficient, Math Comp 36(1981) pp375-386.
\end{enumerate}

{\footnotesize \noindent }

\end{document}
