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\markboth{\hfil Uniqueness for a Boundary Identification Problem \hfil}%
{\hfil Kurt Bryan \& Lester F. Caudill, Jr. \hfil}
\begin{document}
\setcounter{page}{23}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Differential Equations and Computational Simulations III}\newline
J. Graef, R. Shivaji, B. Soni J. \& Zhu (Editors)\newline
Electronic Journal of Differential Equations, Conference~01, 1997, pp. 23--39.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp 147.26.103.110 or 129.120.3.113 (login: ftp)}
\vspace{\bigskipamount} \\
Uniqueness for a Boundary Identification Problem in Thermal Imaging
\thanks{ {\em 1991 Mathematics Subject Classifications:} 35A40, 35J25, 35R30.
\hfil\break\indent
{\em Key words and phrases:} Inverse problems, non-destructive testing, thermal imaging.
\hfil\break\indent
\copyright 1998 Southwest Texas State University and University of
North Texas. \hfil\break\indent
Published November 12, 1998. \hfil\break\indent
Partially supported by NSF Grant DMS-9623279. } }
\date{}
\author{Kurt Bryan \& Lester F. Caudill, Jr.}
\maketitle
\begin{abstract}
An inverse problem for an initial-boundary value problem is considered.
The goal is to determine an unknown portion of the boundary of a region
in ${\mathbb R}^n$ from measurements of Cauchy data on a known portion of
the boundary. The dynamics in the interior of the region are governed
by a differential operator of parabolic type. Utilizing a unique
continuation result for evolution operators, along with the method of
eigenfunction expansions, it is shown that uniqueness holds for a large
and physically reasonable class of Cauchy data pairs.
\end{abstract}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\def\rf#1{(\ref{#1})}
\def\pderiv#1#2{\frac{\partial #1}{\partial #2}}
\section{Introduction}
\label{sec.intro}
The goal of non-destructive evaluation is to gather information about the
interior or other inaccessible portion of some material object from exterior
measurements. Thermal imaging is one approach to this problem; a prescribed
heat flux is applied to a portion of the surface of the object and
the resulting surface temperature response is measured.
From this information one attempts to determine the internal thermal
properties of the object, or the shape of some unknown, inaccessible portion
of the boundary. Thermal imaging holds promise as a tool for
corrosion detection in aircraft, and has found utility in industrial
applications. The interested
reader is referred to \cite{BC}, and the references therein, for a
discussion of some of these applications. Thermal imaging methods have also
found application in the medical field. For example, infrared thermography has
been used to investigate the distribution and structure of skin blood vessels;
this has implications regarding potential recovery from burn injuries (\cite{CSCJ}),
and also bears on the selection of donor sites for skin grafts (\cite{IA}).
We are interested in the use of thermal imaging for the detection of so-called
``back surface'' corrosion and damage. The most elementary model of such a process
is simple material loss which leads to a change in the surface profile of
the object's boundary. This is the model we
have chosen for this paper. Our long-term goal is to develop a reliable
method for determining the presence and extent of corrosion. Of course,
any such method requires a sound theoretical foundation. To this end,
our focus in this work is on the
issue of uniqueness---under what conditions do the proposed data measurements
provide sufficient information from which to determine the shape of an unknown
portion of the object's boundary?
This problem may be formulated mathematically
as an inverse problem for the heat equation.
More precisely, let $\Omega\subseteq{\mathbb R}^n$ represent the object to be
imaged. We assume that the surface $\partial\Omega$ of $\Omega$ is piecewise $C^2$. We use
$\Gamma$ to denote the ``known,'' accessible portion of $\partial\Omega$, and
we assume that both $\Gamma$ and $\partial\Omega\setminus\Gamma$ have nonzero surface measure
as subsets of $\partial\Omega$. Let
$S_0$ denote some open portion of $\Gamma$ with positive measure and let
the applied heat flux $g(t,x)$ be defined for
each $(t,x)\in{\mathbb R}^+; \partial\Omega$ with support in $S_0$.
With some rescaling we model the propagation of heat through $\Omega$ with
an initial-boundary value problem for the heat equation,
\begin{eqnarray}
\label{eqn.bvp1a}
u_t(t,x)-\Delta_x u(t,x) &=& 0\,,\quad {\rm for} \,\,t\in{\mathbb R}^+,\,x\in\Omega\,,\\
\label{eqn.bvp1b}
\pderiv{u}{\eta}(t,x) &=& g(t,x)\,,\quad {\rm for}\,\,t\in{\mathbb R}^+,\,x\in\partial\Omega\,, \\
\label{eqn.bvp1c}
u(0,x) &=& u_0(x)\,,\quad {\rm for}\,\,x\in\Omega.
\end{eqnarray}
Here $u(t,x)$ denotes the temperature in the domain $\Omega$ at the point $x$ at time
$t$, $u_t$ the derivative of $u$ with respect to $t$, and $\eta$ an outward
unit normal vector field on $\partial\Omega$. Throughout this paper,
we will refer to \rf{eqn.bvp1a}-\rf{eqn.bvp1c} collectively as (IBVP).
Let $S_1\subset\Gamma$ denote the portion of the boundary on which we
take temperature measurements.
We consider the following inverse problem: Does knowledge of
$u(t,x)$ on $S_1$ for some time period $t_0 < t < t_1$ uniquely
determine $\partial\Omega\setminus\Gamma$? Specifically, suppose $\Omega_1\subseteq{\mathbb R}^n$
and $\Omega_2\subseteq{\mathbb R}^n$ with $\Gamma$ contained in $\partial\Omega_1
\cap\partial\Omega_2$. For $j=1,\,2$, let $u_j(t,x)$ be the solution
of \rf{eqn.bvp1a}-\rf{eqn.bvp1c} with $\Omega$ replaced by $\Omega_j$.
If, $u_1=u_2$ on $(t_0,t_1)\times S_1$, must it be true
that $\partial\Omega_1\setminus\Gamma=\partial\Omega_2\setminus\Gamma$?
\paragraph{Remark.} Implicit in the formulation of (IBVP) is the assumption that
on the unknown part of the boundary the condition $\pderiv{u}{\eta}=0$ holds,
so that the back surface acts as a perfect insulator. This is only a first
approximation in most situations. In Section \ref{sec.newtonbcs} we discuss
other boundary conditions in which the back surface loses heat to the ambient
environment.
\medskip
The answer to the uniqueness problem posed in the present paper will be
seen to depend on certain properties of the domain $\Omega$, the
initial condition
$u_0$, and the flux $g$. We will show that uniqueness holds for
constant $u_0$ and any non-zero flux $g$. For non-constant $u_0$ one can impose
reasonable conditions on the flux $g$ to ensure uniqueness, provided $\Omega$
is bounded. The case in which the flux $g$ is time-periodic
was analyzed in \cite{BC}.
This paper is organized as follows. The case of constant initial condition
and non-constant flux is analyzed in \S \ref{sec.constantIC}, where uniqueness is
proved. In \S \ref{sec.efn.expansion} we derive a useful
eigenfunction representation and associated estimates for solutions of (IBVP),
which are used in \S \ref{sec.second.unique} to prove a uniqueness
result for bounded domains. In
\S \ref{sec.newtonbcs}, we extend our results to include
other possibilities for the boundary
conditions on $\Gamma$.
The fact that uniqueness for the inverse problem fails without additional
hypotheses on the ingredients in \rf{eqn.bvp1a}-\rf{eqn.bvp1c} may be
illustrated by a simple example in ${\mathbb R}^2$. Let $\Omega_1$ be the rectangle
defined by $00$, then $u_1=u_2$ on $(0,T)\times \Omega'$.
\end{lemma}
In proving this lemma,
we will make use of the following unique continuation result for parabolic
equations. Its proof is based on the derivation of inequalities of Carleman
type, and is omitted here. The interested reader is referred to the work
of Saut and Scheurer \cite{SS}.
\begin{lemma}
\label{lemma.unique.continuation}
Let $\Omega$ be a connected open set in ${\mathbb R}^n$
and $Q=(-T,T)\times\Omega$.
Let ${u\in L^2\left((-T,T);\,H^2_{loc}\left(\Omega\right)\right)}$ be a
solution of $u_t-\Delta u=0$ which vanishes in some open subset ${\cal O}$
of $Q$. Then $u$ vanishes in the horizontal component of ${\cal O}$.
\end{lemma}
\paragraph{Note:} Following Nirenberg \cite{Ni}, we define
the {\it horizontal component} ${\cal O}_h$ of ${\cal O}$ to be
the union of all open hyperplanes of the form $t=constant$ in $Q$ which
have nonempty intersection with ${\cal O}$.
\paragraph{Proof of Lemma \ref{lemma.agree.both.defined}.}
The function $w\equiv u_1-u_2$ obeys
\begin{eqnarray*}
w_t-\Delta w &=& 0\,,\quad {\rm on}\,\,(0,T)\times\Omega', \\
w=\pderiv{w}{\eta} &=& 0\,,\quad {\rm on}\,\,(0,T)\times S_1, \\
w(x,0) &=& 0\,,\quad {\rm on}\,\,\Omega'.
\end{eqnarray*}
We can choose some open connected subset $I$ with $\bar{I}\subset S_1$ and
open ball $B\subset{\mathbb R}^n$ such that
$B\cap\partial\Omega'=I$. Let $\Omega_B=B\setminus \Omega'$ set
${\tilde\Omega}\equiv\Omega'\cup\Omega_B$. Define the function
$${\tilde w}\equiv\cases{w,&$(0,T)\times\Omega'$;\cr
0,&$(0,T)\times\Omega_B$;\cr
0,&$(-T,0]\times {\tilde\Omega}$.\cr}$$
For a smooth test function $\phi$,
$$\int_0^T\int_{{\tilde\Omega}}{\tilde w}\left[\phi_t+\Delta\phi\right]dxdt=0\,,$$
so that ${\tilde w}$ satisfies \rf{eqn.bvp1a} on ${\tilde\Omega}$.
Using standard parabolic regularity arguments (see, e.g., \cite{LSU}), one can
show that
${{\tilde w}\in H^1\left((-T,T);\,H^2\left({\tilde\Omega}\right)\right)}$.
Make the identifications $Q\equiv(-T,T)\times{\tilde\Omega}$
and ${\cal O}\equiv(-T,T)\times{\rm int}(\Omega_B)$ (in this
case, the horizontal component of ${\cal O}$ is $Q$) to see that ${\tilde w}$
satisfies the hypotheses of Lemma \ref{lemma.unique.continuation}.
We conclude that ${\tilde w}$ vanishes on $Q$ and so
$u_1=u_2$ on $(0,T)\times\Omega'$. \hfill$\Box$\medskip
We now present the main result of this section.
\begin{thm}
\label{thm.first.unique}
Let $(u_1,\Omega_1)$ and $(u_2,\Omega_2)$ be solutions of
\rf{eqn.bvp1a}-\rf{eqn.bvp1c},
with $(S_0\cup S_1)\subseteq (\partial\Omega_1\cap\partial\Omega_2)$. Suppose $u_0(x)=u_0$, a
constant, and suppose that there is some time $T>0$ for which the applied flux
$g(t,x)$ is not identically zero on $(0,T)\times S_0$. If $u_1=u_2$ on
$(0,T)\times S_1$ then $\Omega_1=\Omega_2$ and $u_1=u_2$
on $\Omega_1=\Omega_2$.
\end{thm}
\paragraph{proof}
By replacing $u_j$ with $u_j-u_0$, for $j=1,2$, if necessary, it suffices
to consider the case $u_0=0$. Suppose that $\Omega_1\ne \Omega_2$, and let
$\Omega'$ be as above. Then
there exists some nonempty connected component $D$, sharing a portion of
its boundary with $\partial\Omega'$, of either
$\Omega_1\setminus\Omega_2$ or $\Omega_2\setminus\Omega_1$. Let us
suppose the latter, so that $u_2$ is defined and satisfies \rf{eqn.bvp1a}
on $D$. The boundary $\partial D$ of $D$ is comprised of a
portion $\Gamma_1$ of $\partial\Omega_1$ and $\Gamma_2$ of $\partial\Omega_2$.
On $\Gamma_2$ we know that the normal derivative of $u_2$ is identically
zero; on $\Gamma_1$, we know that the normal derivative (from inside
$\Omega_1$) of $u_1$ is zero, and
since $u_2\equiv u_1$ on $\Omega'$ (by
Lemma \ref{lemma.agree.both.defined}) and $u_2$ is smooth across $\Gamma_1$,
we conclude that the normal derivative of $u_2$ vanishes on the boundary
of $D$. Since $u_2$ satisfies equation \rf{eqn.bvp1a} with zero initial data
on $D$, this forces $u_2\equiv 0$ on $(0,T)\times D$. Finally,
by extending $u_2$ to be zero on $(-T,0]\times(\Omega'\cup D)$, we may
appeal to Lemma \ref{lemma.unique.continuation} to
conclude that $u_2\equiv 0$ on
$(0,T)\times(\Omega'\cup D)$. This in turn implies that the flux $g$ is
identically zero on $(0,T)\times S_0$, a contradiction, and we must conclude that
$\Omega_1=\Omega_2$, as asserted. \hfill$\Box$
\section{Eigenfunction Expansion}
\label{sec.efn.expansion}
In this section we record a useful eigenfunction expansion for the function
$u(t,x)$ which satisfies the parabolic initial-boundary value problem (IBVP)
(\ref{eqn.bvp1a})-(\ref{eqn.bvp1c}). The technique, as well as the derivation
of the accompanying estimates, are standard, and we have spared
the reader the details. Instead, we defer to \cite{St}, or
virtually any text on classical PDE.
We assume that the initial condition $u_0$ belongs to $L^2(\Omega)$ and that
for all $t>0$ the applied flux $g(t,x)$ belongs to $C^1((0,T); L^2(\partial\Omega))$,
the space of continuously differentiable functions from $(0,T)$ to
$L^2(\partial\Omega)$. We seek a solution $u(t,x)$ to (IBVP) in the
space $C((0,T); L^2(\Omega))$; for such a solution the derivatives of $u$
with respect to $t$ and $x$ are not well-defined, and so we cast (IBVP)
into a weak form. Multiply equation (\ref{eqn.bvp1a})
by a smooth test function $\phi(t,x)$ with $\phi(T,x)\equiv 0$ and
$\frac{\partial\phi}{\partial\eta}=0$ on
$\partial\Omega$, and then integrate over $(0,T)\times\Omega$. Integrate the term
involving $\phi u_t$ by parts in $t$ use Green's second
identity on the term involving $\phi\bigtriangleup u$ to obtain
\begin{equation}
\label{weak}
\int_{\Omega} u_0(x)\phi(0,x)\,dx + \int_0^T\int_{\Omega} u\left (\phi_t
+ \Delta \phi\right )\,dx\,dt + \int_0^T\int_{\partial\Omega} \phi g\,dS_x\,dt = 0.
\end{equation}
The restriction of the $L^2(\Omega)$ function $u$ to $\partial\Omega$ is not
well-defined, but since $\pderiv{\phi}{\eta}=0$ on $\partial\Omega$
the boundary integral involving $u \pderiv{\phi}{\eta}$ vanishes. This is our
weak form of (\ref{eqn.bvp1a}) - (\ref{eqn.bvp1c}).
In preparation for the eigenfunction expansion, let
$\{\lambda_k,\,\psi_k(x)\}$, $k=0,1,\ldots$ be an eigensystem for $-\Delta$ on
$\Omega$ with homogeneous Neumann boundary conditions, so that
\begin{eqnarray*}
\Delta\psi_k+\lambda_k\psi_k & = & 0\;\;{\rm in}\;\;\Omega,\\
\pderiv{\psi_k}{\eta} & = & 0 \;\;{\rm on}\;\;\partial\Omega ..
\end{eqnarray*}
The eigenvalues $\lambda_k$ are non-negative; order them by magnitude, so
$\lambda_k \leq \lambda_{k+1}$. With the boundary condition
$\pderiv{\psi_k}{\eta}=0$, the first eigenvalue $\lambda_0=0$, is simple,
and has a constant eigenfunction. We normalize
the eigenfunctions so that $\|\psi_k\|_{L^2(\Omega)}=1$ for all $k$, and so
obtain an orthonormal basis for $L^2(\Omega)$. The function $\psi_0(x)$ is
constant and $\psi_0(x)=1/\sqrt{|\Omega |}$. Orthogonality of the eigenfunctions
then implies that
$$\int_{\Omega}\psi_k(x)\,dx=0\,,\quad k\ge 1\,.$$
In later sections we will make use of the following
standard estimate for solutions to Poisson's equation with Neumann boundary
conditions.
\begin{lemma}
\label{lemma.v.bound} Let $f_1\in L^2(\Omega)$, $f_2\in L^2(\partial\Omega)$,
and let $\psi(x)\in H^1(\Omega)$ satisfy
\begin{eqnarray*}
\bigtriangleup \psi & = & f_1\;\;\mbox{in}\;\;\Omega,\\
\pderiv{\psi}{\eta} & = & f_2\;\;\mbox{on}\;\;\partial\Omega, \\
\int_{\Omega}\psi(x)\,dx & = & 0,\nonumber
\end{eqnarray*}
Then
$$
\|\psi\|_{H^1(\Omega)} \leq C (\|f_2\|_{L^2(\partial\Omega)} + \|f_1\|_{L^2(\Omega)})
$$
where $C$ depends on the domain $\Omega$.
\end{lemma}
The main result of this section is
\begin{lemma}
\label{replemma}
The solution $u(t,x)$ to (\ref{weak}) is unique in
$C({\mathbb R}^+; L^2(\Omega))$, and can be expanded as
\begin{equation}
\label{ueqn}
u(t,x) = v(t,x) + \frac{d_0}{\sqrt{|\Omega |}} + \frac{1}{|\Omega |}
\int_0^t G(s)\,ds + \sum_{k=1}^{\infty} T_k(t)\psi_k(x)
\end{equation}
where $v(t,x)$ defined on
${\mathbb R}^+\times\Omega$ denotes the unique function which satisfies the
family of elliptic problems (indexed by $t$)
\begin{eqnarray}
\bigtriangleup_x v & = & \frac{1}{|\Omega|} G(t)\;\;\mbox{in}\;\;\Omega,
\label{veqn}\\
\pderiv{v}{\eta} & = & g(t,x)\;\;\mbox{on}\;\;\partial\Omega,
\nonumber\\
\int_{\Omega}v(t,x)\,dx & = & 0,\nonumber
\end{eqnarray}
and
\begin{eqnarray}
G(t) & = & \int_{\partial\Omega} g(t,x)\,dS_x, \label{Geqn}\\
d_k & = & \int_{\Omega}(u_0(x)-v(0,x))\psi_k(x)\,dx,\, \label{eqn.dk}\\
c_k(t)& = & -\int_{\Omega} v_t\psi_k(x)\,dx\,,\quad k>0\,,\label{ceqn}\\
c_0(t) & = & \frac{G(t)}{\sqrt{|\Omega |}},\nonumber\\
T_k(t) & = & d_ke^{-\lambda_k t} + \int_0^t c_k(s)e^{-\lambda_k (t-s)}\,ds,
\;\;k>0, \label{Teqn}
\end{eqnarray}
where $|\Omega |$ denotes the measure of $\Omega$, $dS_x$ denotes surface
measure on $\partial\Omega$, and $v_t$ is the derivative of
$v(t,x)$ with respect to $t$ (which exists with the given hypotheses).
We also have the estimate
\begin{equation}
\label{growthest}
\sum_{k=1}^{\infty} T_k^2(t) \leq C\left (e^{-2\lambda_1 t}\|u_0\|_{L^2(\Omega)}
+ \frac{t\|g_t(t,\cdot)\|^2_{L^2(\partial\Omega)}}{2\lambda_1}\right )
\end{equation}
where $C$ is a constant which depends on $\Omega$.
\end{lemma}
\section{Uniqueness for Bounded Regions}
\label{sec.second.unique}
We now consider the more general case in which the initial condition $u_0$
need not be constant. Here we will assume that $\Omega$ is a bounded
region. The essential idea in this section is simple. We note from the proof of
Theorem \ref{thm.first.unique} that if
uniqueness fails then there must be some ``insulated'' region $D$ inside $\Omega$. Within
such a region, heat neither enters nor leaves, so that the average temperature
of $D$ cannot increase with time. This is the basis of the argument
that follows: intuitively, if the applied flux $g$ pumps enough heat into
$\Omega$ over a long enough period then no region $D$ can remain at the same
average temperature, and so a uniqueness result must hold. We make this
physical argument precise below.
\begin{thm}
\label{thm.second.unique}
Let $g(t,x)$ denote a flux in the class $C^1({\mathbb R};L^2(\partial\Omega))$
supported for $x\in S_0$ with $\|g(t,\cdot)\|_{L^2(\partial\Omega)} \leq M_0$ for
all $t>0$ and $\|g_t(t,\cdot)\|_{L^2(\partial\Omega)}\leq M_1$ for all $t>0$.
Suppose also that $G(t)$ defined by equation (\ref{Geqn}) satisfies
$G(t)\geq G_0>0$ for all $t$.
Let $u(t,x)$ be the solution to (\ref{eqn.bvp1a})-(\ref{eqn.bvp1c}) or its
weak form (\ref{weak}) (the initial condition $u_0$ is not considered known).
Then knowledge of $u(t,x)$ for $00$ such that measurements of $u_1$ and $u_2$
on $(0,T)\times S_1$ must differ.
We proceed by contradiction.
Assume that $u_1\equiv u_2$ on $(0,\infty)\times S_1$, and as before, let
$\Omega'$ denote the connected component of $\Omega_1\cap\Omega_2$ for
which $\Gamma\subseteq\partial\Omega'$.
Let $w=u_1-u_2$. We will show that
$w(t,x)\equiv 0$ on $(0,\infty)\times\Omega'$. To see this note that
the function $w$ satisfies
\begin{equation}
\label{weqn}
\frac{\partial w}{\partial t} - \bigtriangleup w = 0
\,,\;\;\mbox{in}\;\;\Omega'\times (0,\infty)\,,
\end{equation}
with $w=\frac{\partial w}{\partial\eta} =0$ on $S_1 \times (0,\infty)$.
Let $p$ be a point in $S_1$ and $B$ a ball centered at $p$ such that $B\cap\partial
\Omega'\subset S_1$. Let $B_0$ denote that portion of $B$ which lies outside $\Omega'$.
Define $\tilde{w}(t,x)$ on $\Omega'\cup B_0$ as
$$
\tilde{w}(t,x) = \left \{
\begin{array}{ll}
w(t,x), & x\in\Omega'\\
0 & x\in B_0
\end{array}
\right .
$$
Standard regularity results (see \cite{LSU}) show that $w\in L^2((0,T);
H^2(\Omega'))$ for any $T>0$.
Since $w=\frac{\partial w}{\partial\eta}\equiv 0$ on $S_1$,
it is easy to check that $\tilde{w}\in L^2((0,T); H^2(\Omega'\cup B_0))$.
The function
$\tilde{w}$ vanishes on $B_0\times (0,\infty)$ and we conclude from Lemma
\ref{lemma.unique.continuation} (with the
minor alteration $-T\rightarrow 0$) that $\tilde{w}$ vanishes
on $\Omega'\times (0,\infty)$. This shows that
$u_1\equiv u_2$ on $\Omega'\times (0,\infty)$. Also,
since (\ref{weqn}) has a unique solution for given initial and boundary
conditions, we conclude that $u_0=\tilde{u}_0$ on $\Omega'$.
If $\Omega_1\neq \Omega_2$ then either $\Omega_1\setminus\Omega_2$ or
$\Omega_2\setminus\Omega_1$ contains a nonempty connected component $D$ for
which $\partial D\cap\partial\Omega'$ has positive surface measure.
For specificity, we assume
that $D\subset (\Omega_2\setminus\Omega_1)$. The boundary of $D$
consists of portions of $\partial\Omega_1\setminus (S_0\cup S_1)$ and $\partial
\Omega_2\setminus (S_0\cup S_1)$. On these portions of the boundary the applied
flux $g$ is identically zero. Standard regularity results then show that $u_2$
is a classical solution to the heat equation and smooth on $\bar{D}$, and we therefore have
$\frac{\partial u_2}{\partial\eta}\equiv 0$ on $\partial D$. Since
$D$ is bounded and $u_2$ is smooth,
$$
\frac{d}{dt}\int_{D}u_2(t,x)\,dx = \int_D \frac{\partial u_2}{\partial t}\,dx
= \int_D \bigtriangleup u_2\,dx = \int_{\partial D}\frac{\partial u_2}{\partial\eta}
\,dS_x = 0.
$$
The integral $\int_{D}u_2(t,x)\,dx$ on the left is just
the total thermal energy inside $D$. We have thus shown that the assumption
that $u_1=u_2$ on $(0,\infty)
\times S_1$ and $\Omega_1\neq\Omega_2$ force the existence of an insulated
region $D$ for which $\int_D u_2(t,x)\,dx$ is constant.
We will show that this is impossible for an applied flux
$g(t,x)$ of the form specified in the statement of the theorem.
Let $u_2(t,x)$ be expressed via an eigenfunction expansion as in equation (\ref{ueqn}).
Integrating over $D$ shows that
\begin{equation}
\label{eqnu2}
\int_D u_2(t,x)\,dx = \int_D v(t,x)\,dx + \frac{d_0|D|}{\sqrt{|\Omega_2 |}}
+ \frac{|D|}{|\Omega_2 |} \int_0^t G(s)\,ds + \int_D \sum_{k=1}^{\infty} T_k(t)\psi_k(x)\,dx
\end{equation}
where $T_k(t)$ is defined by equation (\ref{Teqn}), $d_0$ by equation (\ref{eqn.dk}),
and $v$ satisfies (\ref{veqn}) with $\Omega$ replaced by $\Omega_2$.
Since $G(t)\geq G_0$ for all $t$, the integral $\int_0^t G(s)\,ds$
grows at least as fast as $G_0t$;
however, the other terms in the equation can be shown to be $o(t)$ as
$t\to\infty$,
and this will show that $\int_D u_2(t,x)\,dx$ cannot be constant.
The first integral on the right side of equation \rf{eqnu2}
is bounded in $t$, which can be proved by noting that
\begin{eqnarray*}
\left|\int_D v(t,x)\,dx \right|& \leq & \sqrt{|D|}\|v(t,\cdot)\|_{L^2(D)}\\
& \leq & \sqrt{|D|}\|v(t,\cdot)\|_{L^2(\Omega)\,,}
\end{eqnarray*}
and applying Lemma \ref{lemma.v.bound} with the fact that
$\|g(t,\cdot)\|_{L^2(\partial\Omega)} \leq M_0$.
The second term in equation (\ref{eqnu2}) is constant and, therefore, bounded
in $t$. The last term can be estimated by noting that
\begin{eqnarray}
\left|\int_D \sum_{k=1}^{\infty} T_k(t)\psi_k(x)\,dx \right| & \leq & \sqrt{|D|}
\left\|\sum_{k=1}^{\infty} T_k(t)\psi_k(x)\,dx\right\|_{L^2(D)},\nonumber\\
& \leq & \sqrt{|D|} \left\|\sum_{k=1}^{\infty} T_k(t)\psi_k(x)\,dx\right\|_{L^2(\Omega)},
\nonumber\\
& = & \sqrt{|D|}\sqrt{\sum_{k=1}^{\infty} T^2_k(t)}\label{Tbound}.
\end{eqnarray}
Combining (\ref{Tbound}) with the estimate (\ref{growthest}) in Lemma \ref{replemma}
shows that
\begin{equation}
\label{T2bound}
\int_D \sum_{k=1}^{\infty} T_k(t)\psi_k(x)\,dx \leq C\sqrt{|D|}\left (e^{-2\lambda_1 t}
\|u_0\|^2_{L^2(\Omega)} + \frac{M_1t}{2\lambda_1}\right)^{1/2}.
\end{equation}
The quantity on the right side of (\ref{T2bound}) is clearly $o(t)$, and so grows more
slowly than $\int_0^t G(s)\,ds$. Equation (\ref{eqnu2}) then
shows that for sufficiently large $t$ the integral
$\int_D u_2(t,x)\,dx$ must increase, a contradiction that proves Theorem
\ref{thm.second.unique}. \hfill$\Box$\medskip
If in addition to the conditions above $g$ is analytic in $t$ (for example, if
$g$ is independent of $t$, so $g=g(x)$) then we can do better. Suppose that
$g(t,x) \in C^{\omega}((0,T); L^2(\partial\Omega))$, i.e. for each $t_0 > 0$
there is some $\delta >0$ such that $g(t,x)$ can be written as
$$
g(t,x) = \sum_{k=0}^{\infty} \frac{(t-t_0)^k}{k!}g_k(x)
$$
for all $t$ with $|t-t_0|<\delta$, where $g_k\in L^2(\partial\Omega)$.
In this case the solution to
(\ref{eqn.bvp1a})-(\ref{eqn.bvp1c}) is analytic in $t$,
$$
u(t,x) = \sum_{k=0}^{\infty} \frac{(t-t_0)^k}{k!}u_k(x)
$$
where $u_k\in L^2(\Omega)$.
Suppose that two domains $\Omega_1$ and $\Omega_2$ give rise to the same
temperature measurements on $(t_1,t_2)\times S_1$ with $t_10$ and $\|g_t(t,\cdot)\|_{L^2(\partial\Omega)} \leq M_1$ for $t>0$.
Suppose also that $G(t)$ defined by equation (\ref{Geqn}) satisfies
$G(t)\geq G_0>0$ for all $t$.
Let $u(t,x)$ be the solution to the IBVP (\ref{eqn.bvp1a})-(\ref{eqn.bvp1c}) or its
weak form (\ref{weak}) (the initial condition $u_0$ is not considered known).
Then knowledge of $u(t,x)$ for any open time interval $0 0$ and $g$ supported for $x\in S_0\subset\partial\Omega$. The Robin
boundary condition
$\frac{\partial u}{\partial n} + \alpha u = 0$ corresponds to a Newton-cooling type
of heat loss on the boundary with ambient temperature scaled to zero; note that we have
assumed that the loss term $-\alpha u$
applies even on $S_0$, where the flux $g$ is applied.
The solution $u$ to the initial-boundary value
problem \rf{newtoncool}-\rf{newtoncool.3}
can be represented with an eigenfunction expansion, as
\begin{equation}
\label{newtonrep}
u(t,x) = v(t,x) + \sum_{k=0}^{\infty} T_k(t)\psi_k(x)
\end{equation}
where $v(t,x)$ satisfies the family of elliptic problems (indexed by $t$)
\begin{eqnarray}
\bigtriangleup_x v & = & 0 \;\;\mbox{in}\;\;\Omega,
\label{newveqn}\\
\frac{\partial v}{\partial n} + \alpha v
& = & g\;\;{\rm on}\;\;\partial\Omega, \label{newveqn.2}
\end{eqnarray}
$T_k(t)$ is defined by
\begin{equation}
\label{newTeqn}
T_k(t) = d_ke^{-\lambda_k t} + e^{-\lambda_k t}\int_0^t c_k(s)e^{\lambda_k s}
\,ds,
\end{equation}
and
\begin{eqnarray}
c_k(t) & = & -\int_{\Omega} v_t \psi_k(x)\,dx,
\label{newceqn}\\
d_k & = & \int_{\Omega}(u_0(x)-v(0,x))\psi_k(x)\,dx\nonumber
\end{eqnarray}
and finally, $\{\lambda_k,\psi_k(x)\}$ is an orthonormal eigensystem for
$-\bigtriangleup$ with the Robin boundary conditions, so for each $k$,
\begin{eqnarray*}
\bigtriangleup\psi_k+\lambda_k\psi_k & = & 0\;\;{\rm in}\;\;\Omega,\\
\frac{\partial \psi_k}{\partial n} + \alpha \psi_k & = & 0,\;\;{\rm on}\;\;\partial\Omega\,.
\end{eqnarray*}
We order the eigenvalues by magnitude. It is easy to check that all eigenvalues are strictly
positive.
The next result gives sufficient conditions on the induced flux $g$ which
guarantee uniqueness. As before, we assume that we have measurements
of $u(t,x)$ for $x\in S_1\subset\partial\Omega$.
Loosely speaking, we require the flux to be
nonnegative and decaying in time, but not too quickly. More precisely, we
have
\begin{thm}
\label{newtonunique}
Let $(u_1,\Omega_1)$ and $(u_2,\Omega_2)$ be solutions
of \rf{newtoncool}-\rf{newtoncool.3} with
$(S_0\cup S_1)\subseteq
(\partial\Omega_1\cap\partial\Omega_2)$. Suppose that the applied flux $g(t,x) \in
C^1({\mathbb R};L^2(\partial\Omega))$ and is supported in $S_0$ for each $t$. Also,
assume that
\begin{enumerate}
\item $g(t,x)$ is not identically zero.
\item $g(t,x), \frac{\partial g}{\partial t}(t,x) \geq 0$ for all $x$ and $t$.
\item $\left\|g_t(t,\cdot)\right\|_{L^2(\partial\Omega)}
\rightarrow 0$ as $t\rightarrow\infty$ in such a way that \newline
${\sup_{s>t}\|g_t(s,\cdot)\|^2_{L^2(\partial\Omega)} = o\left(\frac{1}{\ln t}\right)}$ as $t\to\infty$.
\end{enumerate}
Then $u_1\equiv u_2$ on ${\mathbb R}^+\times S_1$ implies that $\Omega_1=\Omega_2$.
\end{thm}
In the case where $\Omega$ belongs to ${\mathbb R}^2$ or ${\mathbb R}^3$, one may
relax the decay condition 3. in this result. More precisely,
\begin{thm}
\label{newtonuniqueprime}
In space dimension 2 or 3, Theorem \ref{newtonunique} holds, with
hypothesis 3. relaxed to
\begin{itemize}
\item[$3'.$] $\left\|g_t(t,\cdot)\right\|_{L^2(\partial\Omega)}
\rightarrow 0$ as $t\rightarrow\infty$.
\end{itemize}
\end{thm}
We shall prove Theorem \ref{newtonunique} first. Afterward, we will indicate
the changes necessary to establish Theorem \ref{newtonuniqueprime}.
\paragraph{Proof of Theorem \ref{newtonunique}.}
Our proof proceeds by contradiction. Suppose $\Omega_1\neq \Omega_2$.
The same reasoning as in the proof of Theorem \ref{thm.first.unique} shows that
we must have some nonempty region
$D\subset\Omega$ (where $\Omega$ is $\Omega_1$ or $\Omega_2$), with
$\partial D\cap\partial\Omega'$ having positive surface measure, on which
\begin{eqnarray*}
\frac{\partial u}{\partial t} - \bigtriangleup u & = & 0\;\;{\rm on}\;\;
{\mathbb R}^+ \times D\\
\frac{\partial u}{\partial n} + \alpha u & = & 0,\;\;{\rm on}\;\;
{\mathbb R}^+\times \partial D\\
u(0,x) & = & u_0(x)\;\;{\rm on}\;\; D
\end{eqnarray*}
(where $u$ is either $u_1$ or $u_2$). (This is the same $\Omega'$ as in
the proof of Theorem \ref{thm.second.unique}.)
We first observe that the integral $\int_D u(t,x)\,dx$
must tend exponentially rapidly to zero as $t\to\infty$. To
see this, note that $u(t,x)$ can be expanded on $D$ in terms of eigenfunctions
\begin{equation}
\label{newueqn}
u(t,x) = \sum_{k=0}^{\infty} d_k e^{-\tilde{\lambda}_k t}\tilde{\psi}_k(x)
\end{equation}
with
$$
d_k = \int_{D}u_0(x)\tilde{\psi}_k(x)\,dx.
$$
and $\{\tilde{\lambda}_k,\tilde{\psi}_k(x)\}$ is an eigensystem
for $-\bigtriangleup$ on $D$ with boundary
conditions \mbox{$\frac{\partial \tilde{\psi}_k}{\partial n} + \alpha \tilde{\psi}_k = 0$}
on $\partial D$.
Again, the eigenvalues are strictly positive. From the representation (\ref{newueqn})
\begin{eqnarray}
\left | \int_D u(t,x)\,dx \right | & \leq & \sqrt{|D|}\|u\|_{L^2(D)}\nonumber\\
& = & O(e^{-\tilde{\lambda}_0 t}) \label{newthing}
\end{eqnarray}
where ${\tilde{\lambda}_0>0}$ is the smallest eigenvalue for the above
eigensystem.
We shall complete the proof of Theorem \ref{newtonunique} by contradicting
relation \rf{newthing} in the following way: We shall show that, under the
hypotheses on $g$,
\begin{eqnarray}
& & \lim_{t\rightarrow\infty} \int_D u(t,x)\,dx\longrightarrow \int_D v(t,x)\,dx
\label{assert1}.\\
& & \int_D v(t,x)\,dx\;\;{\rm is\; bounded\; away\; from\; zero,\;\; uniformly\; in}\;\; t
\label{assert2}.
\end{eqnarray}
From these two facts, it is clear that ${\int_D u(t,x)\,dx}$ must be bounded away
from zero, uniformly in $t$, the desired contradiction to \rf{newthing}.
To establish (\ref{assert1}), we first show that ${\|u-v\|_{L^2(\Omega)}\to 0}$ as
$t\to\infty$. Note that \rf{newtonrep} and \rf{newTeqn} imply
\begin{eqnarray}
\|u-v\|^2_{L^2(\Omega)} & = & \sum_{k=0}^{\infty}T^2_k(t) \nonumber\\
& \le & 2\sum_{k=0}^{\infty}\left(\int_0^tc_k(s)e^{-\lambda_k (t-s)}ds\right)^2 +o(1)
\label{uvdiff}
\end{eqnarray}
where the last equality follows from the fact that $\lambda_k>0$ for each $k$.
The integral appearing inside the sum on the right can be bounded as
\begin{eqnarray}
\left(\int_0^t c_k(s)e^{-\lambda_k (t-s)}ds\right )^2 & = &
\left(\int_0^{\beta} c_k(s)e^{-\lambda_k (t-s)}ds
+ \int_{\beta}^t c_k(s)e^{-\lambda_k (t-s)}ds \right )^2\nonumber\\
& \le & \left (\frac{e^{-2\lambda_k (t-\beta)} - e^{-2\lambda_k t}}{\lambda_k}\right )
\int_0^{\beta} c^2_k(s)\,ds\nonumber\\
& & + \left (\frac{1 - e^{-2\lambda_k (t-\beta)}}{\lambda_k}\right ) \int_{\beta}^t
c^2_k(s)\,ds\label{messy}
\end{eqnarray}
where $\beta=\beta(t)\in (0,t)$ is to be specified in a moment.
From the bounds (\ref{uvdiff}) and (\ref{messy}) we conclude that
\begin{eqnarray}
\|u-v\|^2_{L^2(\Omega)}
& \leq & C \left( \left(\frac{e^{-2\lambda_0 (t-\beta)} - e^{-2\lambda_0 t}}{\lambda_0}\right)
\int_0^{\beta}\sum_{k=0}^{\infty}
c_k^2(s)\,ds\right . \nonumber\\
& & +\left . \left(\frac{1 - e^{-2\lambda_0 (t-\beta)}}{\lambda_0}\right)\int_{\beta}^t \sum_{k=0}^{\infty} c_k^2(s)\,ds\right)
\label{uvbound2}
\end{eqnarray}
for some constant $C$, where we have interchanged the summation and integral for the
convergent series and used that fact that
${\frac{e^{-\lambda_k t} - e^{-2\lambda_k t}}{\lambda_k}}$
and ${\frac{1-e^{-\lambda_k t}}{\lambda_k}}$ are decreasing functions
of $\lambda_k$ for $\lambda_k,t>0$.
Next, note that from equation (\ref{newceqn}) we have
$\sum_{k=0}^{\infty} c_k(t)^2 = \|v_t(t,\cdot)\|^2_{L^2(\Omega)}$, where $v_t$ satisfies
\begin{eqnarray}
\bigtriangleup v_t & = & 0\;\;{\rm in}\;\;\Omega,\label{newvteqn}\\
\frac{\partial v_t}{\partial\eta} + \alpha v_t & = & g_t\;\;{\rm on}\;\;\partial\Omega
\label{newvteqn.2}
\end{eqnarray}
Standard estimates similar to Lemma \ref{lemma.v.bound} show that we can
bound
\begin{equation}
\label{othervbound}
\|v_t\|_{L^2(\Omega)} \leq C \|g_t\|_{L^2(\partial\Omega)}.
\end{equation}
It then follows from (\ref{uvbound2}) and (\ref{othervbound}) that
\begin{eqnarray}
\|u-v\|^2_{L^2(\Omega)} & \leq & C \left ( \right . (e^{-2\lambda_0 (t-\beta)} -
e^{-2\lambda_0 t}) \int_0^{\beta} \|g_t(s,\cdot)\|^2_{L^2(\partial\Omega)}\,ds
\nonumber\\
& & + (1 - e^{-2\lambda_0 (t-\beta)})
\int_{\beta}^t \|g_t(s,\cdot)\|^2_{L^2(\partial\Omega)}\,ds \left . \right )
\nonumber\\
&\leq & C \left ( \right . (e^{-2\lambda_0 (t-\beta)}-
e^{-2\lambda_0 t}) \beta\cdot\sup_{0~~r} \|g_t(s,\cdot)\|^2_{L^2(\partial\Omega)}$,
and observe that $\epsilon(r)$ has the following properties:
\begin{itemize}
\item For each $r$, $\epsilon(r)\ge 0$ and $\epsilon'(r)\le 0$.
\item $\epsilon(0)<\infty$ and $\epsilon(r)\to 0$ as $r\to\infty$.
\end{itemize}
From equation \rf{funstuff}, we have, in terms of $\epsilon(r)$,
\begin{eqnarray}
\|u-v\|^2_{L^2(\Omega)} & \leq & C \left ( \right . (e^{-2\lambda_0 (t-\beta)}-
e^{-2\lambda_0 t}) \beta\cdot\epsilon(0)
\nonumber\\
& & + (1 - e^{-2\lambda_0 (t-\beta)})(t-\beta)\epsilon(\beta)\left . \right )
\label{morefunstuff}
\end{eqnarray}
We now specify, for each fixed $t>0$, a corresponding value of $\beta$
implicitly by the relation
$$\beta + K\ln\beta = t\,,$$
where $K\in{\mathbb R}^+$ satisfies $K>\frac{1}{2\lambda_0}$. It is simple to
check that this relation defines $\beta$ uniquely as a function of $t$,
and that $\beta(t)\to\infty$ as $t\to\infty$. Furthermore,
$$t-\beta = K\ln\beta\to\infty$$
as $t\to\infty$. With $\beta(t)$ so defined, we note that
\begin{equation}\label{eqn.goaway1}
(t-\beta)\epsilon(\beta) = K\ln(\beta)\epsilon(\beta)\to 0
\end{equation}
as $t$ (and hence $\beta$) $\to\infty$, by virtue of the decay condition
on $\epsilon$. Also,
\begin{equation}\label{eqn.goaway2}
\beta e^{-2\lambda_0(t-\beta)} = \beta e^{-2\lambda_0K\ln\beta} = \beta^{1-2\lambda_0K}\to 0
\end{equation}
as $t\to\infty$, since $2\lambda_0 K>1$. In light of \rf{eqn.goaway1} and
\rf{eqn.goaway2}, we see that the right-hand side of \rf{morefunstuff}
decays to $0$ as $t\to\infty$. From this, we conclude that
${\|u-v\|_{L^2(\Omega)}\to 0}$ as $t\to\infty$. In fact,
${\|u-v\|_{L^2(D)}\to 0}$ for any $D\subset\Omega$, so that
$$
\left|\int_D (u-v)\,dx\right| \leq \|u-v\|_{L^1(D)}
\leq \sqrt{|D|}\|u-v\|_{L^2(D)} \rightarrow 0\,,
$$
from which we conclude
\begin{equation}\label{eqn.utov}
\int_D u(t,x)\,dx \rightarrow \int_D v(t,x)\,dx
\end{equation}
as $t\rightarrow \infty$, which is (\ref{assert1}).
It remains only to establish (\ref{assert2}). To this end, let us first
consider the case in which $g(t,x)\in C^1({\mathbb R}; C^2(\partial\Omega))$. Then
the function $v(t,\cdot)\in C^2(\bar{\Omega})$ for all $t$.
We have by the maximum principle that the
minimum value of $v(t,x)$ on $\bar{\Omega}$ occurs at a point on $\partial\Omega$ at
which $\frac{\partial v}{\partial\eta} \leq 0$. At such a point,
$\alpha v = g - \frac{\partial v}{\partial\eta} \geq 0,$ from which we conclude
that $v(t,x)\geq 0$ for $x\in\Omega$. In particular, for
any $D\subset\Omega$ we have $\int_D v(t,x)\,dx \geq 0$, with equality if and only
if $v(t,x)\equiv 0$. Since $v\equiv 0$ if and only if $g\equiv 0$ (from
\rf{newvteqn}-\rf{newvteqn.2}), the hypotheses on $g$ imply that
${\int_D v(t,x)\,dx > 0}$ for each $t$. (In particular,
${\int_D v(0,x)\,dx > 0}$.) Furthermore, since
$g_t\geq 0$, the same reasoning shows that
$\int_D v_t(t,x)\,dx \geq 0$ for any $D\subset\Omega$ and all $t>0$.
Consequently, for each $t>0$,
\begin{eqnarray*}
\int_D v(t,x)\,dx &=& \int_D v(0,x)\,dx + \int_0^t \pderiv{}{s}\int_D v(s,x)\,dxds \\
&=& \int_D v(0,x)\,dx + \int_0^t \int_D v_t(s,x)\,dxds \\
&\geq& \int_D v(0,x)\,dx > 0\,,
\end{eqnarray*}
where we have used the fact that $D$ is bounded and smooth enough to interchange
the order of integration and differentiation. This
establishes (\ref{assert2}) for $g(t,x)\in C^1({\mathbb R}; C^2(\partial\Omega))$.
Finally, we note that the same conclusion holds
if $g(t,\cdot)$ is merely $L^2(\partial\Omega)$, rather
than $C^2(\partial\Omega)$, for
we can approximate any non-negative $g\in L^2(\partial\Omega)$ arbitrarily closely
(in $L^2(\partial\Omega)$) with a non-negative function $\tilde{g}(t,\cdot) \in C^2
(\partial\Omega)$. From the standard estimate
$$
\|v - \tilde{v}\|_{L^2(\Omega)} \leq C \|g - \tilde{g}\|_{L^2(\partial\Omega)}
$$
where $\tilde{v}$ satisfies the boundary value problem \rf{newveqn}-\rf{newveqn.2}
with $g$ replaced by $\tilde{g}$, we conclude that $\|v - \tilde{v}\|_{L^2(D)}$ can
be made arbitrarily small. Since
$$
\left|\int_D v(t,x)\,dx - \int_D \tilde{v}(t,x)\,dx \right| \leq \sqrt{|D|}
\|v - \tilde{v}\|_{L^2(D)} \leq C \|g - \tilde{g}\|_{L^2(\partial\Omega)}\,,
$$
and since $\int_D \tilde{v}(t,x)\,dx>0$ uniformly in $t$, we conclude that
$\int_D v(t,x)\,dx>0$ uniformly in $t$ also. This establishes (\ref{assert2}), and completes
the proof. \hfill$\Box$
\paragraph{Proof of Theorem \ref{newtonuniqueprime}.}
Noting that the decay condition ${o\left(\frac{1}{\ln t}\right)}$ was
used only to establish \rf{eqn.utov}, the preceding proof also works for
Theorem \ref{newtonuniqueprime}, provided we show that this relation still
holds. Let $\Omega\subseteq{\mathbb R}^n$ with $n=2$ or $n=3$, and set
$$\phi(t)\equiv\sum_{k=0}^{\infty}\left(\int_0^tc_k(s)e^{-\lambda_k(t-s)}ds\right)^2\,.$$
In light
of \rf{uvdiff}, it suffices to show that
$$\lim_{t\to\infty}\phi(t) = 0\,.$$
To this end, set
$$M(t)\equiv \sup_{t~~~~t$ and all $k$. We can estimate
\begin{eqnarray}
\nonumber
\phi(t) & = & \sum_{k=0}^{\infty} \left (\int_0^{t/2} c_k(s)e^{-\lambda_k (t-s)}\,ds +
\int_{t/2}^{t} c_k(s)e^{-\lambda_k (t-s)}\,ds\right )^2 \\
& \leq & 2M^2(0)e^{-\lambda_0 t/2}\sum_{k=0}^{\infty}\frac{1}{\lambda^2_k}
+ 2M^2(t/2)\sum_{k=0}^{\infty}\frac{1}{\lambda^2_k}
\label{eqn.new}
\end{eqnarray}
In ${\mathbb R}^2$ or ${\mathbb R}^3$, we
have ${\sum_{k=0}^{\infty}\frac{1}{\lambda^2_k}<\infty\,,}$ so that
\rf{eqn.new} yields
$$\phi(t) \le C_1e^{-\lambda_0 t/2} + C_2M^2(t/2)\,.$$
By hypothesis $3'.$, $M\left(\frac{t}{2}\right)\to 0$ as $t\to\infty$, so
that \rf{eqn.utov} holds. \hfill$\Box$
\section{Concluding Remarks}
We have examined a variety of settings in which the Cauchy data for the heat
equation uniquely determines the shape of the region on which the heat equation
is defined. Specifically, if the initial condition is constant over the region
of interest, then the
Cauchy data---temperature and heat flux---on any open portion of the boundary of
the region over any time interval determines the shape of the region. In the
more general case in which the initial
condition is not necessarily constant, the Cauchy data for the time interval $(0,
\infty)$ uniquely determines the shape of a bounded region $\Omega$, provided that
the data satisfy certain reasonable conditions. For insulate boundary
conditions $\frac{\partial u}{\partial \eta} = 0$ on the unknown portion $B$ of the
boundary, the prescribed
flux $g(t,x)$ on $\partial\Omega\setminus B$ must provide a net positive flux at all times
and be bounded away from zero, and $g_t$ must be bounded. For the Robin boundary
condition $\frac{\partial u}{\partial\eta} + \alpha u = 0$, we require the flux to be
positive at all points and times, and obey a certain decay property in time.
The techniques employed in this analysis can be used to investigate other types
of boundary conditions. The choices presented here reflect sensible conditions
within the context of
the particular physical situation---remote corrosion detection---which motivates
our study of this model. We also note that these techniques may be extended in
a straightforward fashion to include more general parabolic equations. For
example, one could incorporate a spatially-varying thermal conductivity, and
conduct the preceding analysis within the framework of appropriately-weighted
Hilbert spaces, with similar results.
While a uniqueness result holds, this inverse problem is most certainly ill-posed;
the shape of the region will not be a continuous function of the measured data in
any reasonable norm. A next logical step would be to examine and quantify the nature
of the ill-posedness and identify the features of the boundary which can be
stably estimated from the Cauchy data. This should give insight into useful and
practical reconstruction algorithms. Such an algorithm
might be based on the ideas in \cite{BC}---linearize
the forward problem and examine the linearized map from the ``boundary shape''
space to the measured temperature data. The forward map will be given as an integral
operator with smooth kernel, and will have an unbounded inverse.
We are currently investigating such an approach to gain
an understanding of stability and reconstruction possibilities.
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\newpage
{\sc Kurt Bryan} \\
Department of Mathematics\\
Rose-Hulman Institute of Technology\\
Terre Haute, IN 47803, USA.\\
E-mail: kurt.bryan@rose-hulman.edu \\
\medskip
{\sc Lester F. Caudill, Jr.} \\
Department of Mathematics and Computer Science\\
University of Richmond\\
Richmond, VA 23173, USA\\
E-mail: lcaudill@richmond.edu
\end{document}
~~