\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 37, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/37\hfil An inverse boundary-value problem]
{An inverse boundary-value problem for semilinear elliptic
equations}

\author[Z. Sun\hfil EJDE-2010/37\hfilneg]
{Ziqi Sun} 

\address{Ziqi Sun \newline
Department of Mathematics and Statistics,
Wichita State University, Wichita, KS 67260-0033, USA}
\email{ziqi.sun@wichita.edu}

\thanks{Submitted January 31, 2010. Published March 14, 2010.}
\subjclass[2000]{35R30}
\keywords{Inverse Problem; Dirichlet to Neumann map}

\begin{abstract}
 We show that in dimension two or greater, a certain equivalence
 class of the scalar coefficient $a(x,u)$ of the semilinear elliptic
 equation $\Delta u\,+a(x,u)=0$ is uniquely determined by
 the Dirichlet to Neumann map of the equation on a
 bounded domain with smooth boundary. We also show that
 the coefficient $a(x,u)$  can be determined by the Dirichlet to Neumann
 map under some additional hypotheses.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]

\section{Introduction}

In this article, we study the inverse boundary-value problem (IBVP)
for the semilinear equation
\begin{equation} \label{e1.1}
\begin{gathered}
L_a(u):=\Delta u+a(x,u)=0\quad\text{in } \Omega\subset \mathbb{R}^n,\\
u|_{\partial\Omega}=f,\quad f\in C^{2,\alpha}(\partial\Omega),
\end{gathered}
\end{equation}
where $0<\alpha<1$ and $\Omega\subset \mathbb{R}^n$ is a bounded
domain with smooth boundary. We assume that the coefficient
of the equation satisfies
\begin{gather}
a(x,u), a_u(x,u)\in C^\alpha(\bar\Omega\times R), \label{e1.2}\\
a_u(x,u)\leq 0. \label{e1.3}
\end{gather}
Then the Dirichlet problem \eqref{e1.1} has an unique solution
$u\in C^{2,\alpha}(\bar\Omega)$ \cite{GT,LaUr}.
We define the nonlinear Dirichlet to Neumann map $\Lambda_a$:
\[
\Lambda_a(f)=\frac{\partial u}{\partial\nu}\Big|_{\partial\Omega},
\]
where $\nu$ is the unit outer normal on the boundary $\partial\Omega$.
The inverse problem is to recover $a(x,u)$ from knowledge of
$\Lambda_a$.

It was shown in \cite{IS} that if $a(x,u)$ satisfies the condition
\begin{equation} \label{e1.4}
a(x,0)=0,
\end{equation}
then the uniqueness holds for the above inverse problem.

In this paper we shall study the above inverse problem without
the assumption \eqref{e1.4}. We first observe that in the
general case, the Dirichlet to Neumann map $\Lambda_a$ does not
determine the coefficient $a$ uniquely.

To see the nonuniqueness, let $a$ be a coefficient satisfying
\eqref{e1.2} and \eqref{e1.3} and let $\phi$ be a function satisfying
\begin{equation} \label{e1.5}
\phi(x)\in C^{2,\alpha}(\bar\Omega), \quad
\phi|_{\partial \Omega}=\nabla \phi|_{\partial\Omega}=0.
\end{equation}
Define the transformation $T_\phi$ by
\begin{equation} \label{e1.6}
(T_\phi a)(x,u)=a(x,u+\phi(x))+\Delta\phi(x)
\end{equation}
Then the new coefficient $T_\phi a$ satisfies the same
assumptions \eqref{e1.2} and \eqref{e1.3}. It is easy to check that
$L_{T_\phi a}(u-\phi)=0$, and the assumption
$\phi|_{\partial \Omega}=\nabla\phi|_{\partial\Omega}=0$ implies
\[
(u-\phi)|_{\partial\Omega}=u|_{\partial\Omega},\quad
\frac{\partial(u-\phi)}{\partial\nu}\Big|_{\partial\Omega}
=\frac{\partial u}{\partial\nu}\Big|_{\partial\Omega}.
\]
Therefore,
\begin{equation} \label{e1.7}
\Lambda_{T_\phi a}=\Lambda_a.
\end{equation}

We define in the set of coefficients satisfying \eqref{e1.2}
and \eqref{e1.3} an equivalence relation induced by $T_\phi$ as follows:
\begin{equation} \label{e1.8}
a\sim\tilde{a} \quad\text{if}\quad \tilde{a}=T_\phi a.
\end{equation}
Then we see from the above discussion that $\Lambda_a$ remains
the same for any coefficient in the equivalence class $[a]$.
Therefore, the correct uniqueness question for
 \eqref{e1.1} in the general setting is to ask whether
$\Lambda_a$ determines $[a]$ uniquely.

The main purpose of this article is to give an affirmative answer
to this question.
To state the result, let us define for each coefficient
 $a$, a set $E_a\in \mathbb{R}^n\times R$ by
\begin{equation} \label{e1.9}
E_a=((x,u)\subset\Omega\times R; \,\exists\text{ solution $u$
of \eqref{e1.1} with }u=u(x)),
\end{equation}
and the transformation of $E_a$ by $T_\phi$ by
\begin{equation} \label{e1.10}
T_\phi E_a =((x,u+\phi(x))\subset\Omega\times R; \,\exists\text{
solution $u$ of \eqref{e1.1} with }u=u(x)).
\end{equation}

\begin{theorem} \label{thm1}
Given $a(x,u)$ and $\tilde{a}(x,u)$ satisfying the conditions
\eqref{e1.2} and \eqref{e1.3}. If $\Lambda_a=\Lambda_{\tilde{a}}$,
then there is a function $\phi$ satisfying \eqref{e1.5} such that
\begin{gather} \label{e1.11}
E_{\tilde{a}}=T_{-\phi} E_a, \\
 \label{e1.12}
\tilde{a}(x,u)=T_\phi a(x,u)\quad \text{on }E_{\tilde{a}}.
\end{gather}
\end{theorem}

As the example illustrates in \cite{IS},
 in general the set $E_a$ in \eqref{e1.9} may be a proper subset,
 and thus \eqref{e1.12} is the best one can hope for.

Another purpose of this article is to generalize the uniqueness
result proven in \cite{IS}. The condition \eqref{e1.4} implies that
zero is a constant solution of the equation \eqref{e1.1}.
Thus, the equation \eqref{e1.1} with the coefficient $a$
satisfying \eqref{e1.4} must carry a common solution $u\equiv0$.
 We shall show that the uniqueness holds in the general case as
long as a common solution, not necessarily $u\equiv0$, exists.

\begin{theorem} \label{thm2}
Given $a(x,u)$ and $\tilde{a}(x,u)$ satisfying the conditions
\eqref{e1.2} and \eqref{e1.3}. Assume that the equation \eqref{e1.1}
carries a common solution for both coefficients $a$ and $\tilde{a}$.
If $\Lambda_a=\Lambda_{\tilde{a}}$, then
\begin{gather} \label{e1.13}
E_a=E_{\tilde{a}}, \\
 \label{e1.14}
a(x,u)=\tilde{a}(x,u)\quad \text{on }  E_a.
\end{gather}
\end{theorem}


Similar problems have been studied  for various semilinear
and quasilinear elliptic equations and systems
\cite{I1, I2, IN, Su, SuU, HSu, M}. We refer to the survey papers
\cite{Su2, U}  for other recent developments in the field of inverse boundary
value problems for semilinear and quasilinear elliptic equations.

The proof of both theorems are based on a linearization argument and the
uniqueness result for the linear elliptic equations. In
the next section, we give a proof of Theorems \ref{thm1} and \ref{thm2}.

\section{Proofs of Theorems}

Let $u_f$ be the unique solution to \eqref{e1.1}.
Using the argument in \cite{Su2} that is based on Schauder's estimate,
 we can show that the map $f\to  u_f$ is differentiable
in the space $C^{2,\delta}(\bar\Omega)$ for any $\delta$
with $0<\delta<\alpha$.

Let $g\in C^{2,\alpha}(\partial\Omega)$. Denote by
$u^\ast$ the unique solution to the linear problem
\begin{equation} \label{e2.1}
\Delta u^\ast+a_u(x,u_f)u^\ast=0,\quad  u^\ast|_{\partial\Omega}=g.
\end{equation}
Then for any $\delta$, $0<\delta<\alpha$,
\begin{equation} \label{e2.2}
\lim_{t\to 0} \|\frac{u_{f+tg}-u_f}{t}
-u^\ast\|_{C^{2,\delta}(\bar\Omega)}=0.
\end{equation}

We denote by $\dot{u}_{f,g}$ the solution $u^\ast$ in
\eqref{e2.1} as the derivative of $u$ at $f$ in the direction $g$.
Similarly, we have that $u_{f+tg}$ is differentiable in $t$ at any
value of $t$ under the
$C^{2,\delta}(\bar\Omega)$ norm, $0<\delta<\alpha$, and
the derivative, denoted by $\dot{u}_{f+tg,g}$, satisfies
\begin{equation} \label{e2.3}
\Delta\dot{u}_{f+tg}+a_u(x,\nabla
u_{f+tg})\cdot\nabla\dot{u}_{f+tg,g}=0,\quad
\dot{u}_{f+tg,g}|_{\partial\Omega}=g.
\end{equation}

\begin{proof}[Proof of Theorem \ref{thm1}]
Given $a(x,u)$ and
$\tilde{a}(x,u)$ satisfying the conditions
\eqref{e1.2} and \eqref{e1.3}.
We denote by $u_f$ the unique
solution of \eqref{e1.1} and by $\tilde{u}_f$ the
unique solution of \eqref{e1.1} with $a$ replaced by
$\tilde{a}$, where $a$ and $\tilde{a}$
are two semilinear coefficients assumed in Theorem \ref{thm1}. Under the
assumption that $\Lambda_a=\Lambda_{\tilde{a}}$, we have that
\begin{equation} \label{e2.4}
\frac{\partial u_f}{\partial
\nu}\Big|_{\partial\Omega}=\frac{\partial\,\tilde{u}_f}{\partial \nu}\Big|_{\partial\Omega}
\end{equation}
for each $f\in C^{2,\alpha}(\partial\Omega)$. Then for any
$g\in C^{2,\alpha}(\partial\Omega)$,
\begin{equation} \label{e2.5}
\frac{\partial u_{f+tg}}{\partial
\nu}\Big|_{\partial\Omega}=\frac{\partial\tilde{u}_{f+tg}}{\partial
\nu}\Big|_{\partial\Omega},\,\forall t\in\mathbb{R}.
\end{equation}
Differentiating \eqref{e2.5} in $t$ at $t=0$, we get
\begin{equation} \label{e2.6}
\frac{\partial\dot{u}_{f,g}}{\partial
\nu}\Big|_{\partial\Omega}=\frac{\partial\,\dot{\tilde{u}}_{f,g}}{\partial \nu}\Big|_{\partial\Omega},
\end{equation}
where $\dot{u}_{f,g}$ and $\dot{\tilde{u}}_{f,g}$ satisfy
\begin{gather} \label{e2.7}
\Delta\dot{u}_{f,g}+a_u(x,u_f)\dot{u}_{f,g}=0,\quad
\dot{u}_{f,g}|_{\partial\Omega}=g, \\
 \label{e2.8}
\Delta
\dot{\tilde{u}}_{f,g}+\tilde{a}_u(x,\tilde{u}_f)\dot{\tilde{u}}_{f,g}=0,\quad
\dot{\tilde{u}}_{f,g}|_{\partial\Omega}=g.
\end{gather}
Since for a fixed $f\in C^{2,\alpha}(\partial\Omega)$, \eqref{e2.6}
holds for all $g\in C^{2,\alpha}(\partial\Omega)$, we have that the
Dirichlet to Neumann maps of \eqref{e2.7} and \eqref{e2.8} must be
equal; i.e.
\begin{equation} \label{e2.9}
\Lambda_{a_u(x,u_f)}^\ast=\Lambda_{\tilde{a}_u(x,\tilde{u}_f)}^\ast.
\end{equation}
Then the uniqueness results established in \cite{NSU}
can be applied to obtain
\begin{equation} \label{e2.10}
a_u(x,u_f)=\tilde{a}_u(x,\tilde{u}_f) \quad \text{on } \Omega,
\end{equation}
and consequently,
\begin{equation} \label{e2.11}
\dot{u}_{f,g}(x)=
\dot{\tilde{u}}_{f,g}(x)\,\, \text{on}\, \Omega.
\end{equation}
Replacing $f$ by $tf$ and $g$ by $f$ in \eqref{e2.11} we get
\begin{equation} \label{e2.12}
\dot{u}_{tf,f}(x)=
\dot{\tilde{u}}_{tf,f}(x)\quad \text{on } \Omega,\; \forall t\in\mathbb{R}.
\end{equation}
In other words,
\[
d/dt(u_{tf}(x))=d/dt(\tilde{u}_{tf}(x))\quad \text{on }\Omega,\;
 \forall t\in\mathbb{R}.
\]
Thus, there is a function $\phi\in C^{2,\alpha}(\bar\Omega)$,
independent of $t$, such that
\begin{equation} \label{e2.13}
u_f(x)=\tilde u_f(x)+\phi(x),\quad x\in\Omega.
\end{equation}
Clearly, the function $\phi$ is independent of $f$,
since by \eqref{e2.12}, each $f$ carries the same $\phi$ as $f=0$ does.

Since \eqref{e2.13} holds for all $f$, we have that \eqref{e2.13}
implies \eqref{e1.11}. Also, combining \eqref{e2.4} with
\eqref{e2.13}, we see that $\phi$ satisfies the boundary condition
in \eqref{e1.5}.

Substituting the right hand side of \eqref{e2.13} in  \eqref{e1.1},
we obtain
\begin{equation} \label{e2.14}
\Delta (\tilde{u}_f+\phi)+a(x,\tilde{u}_f+\phi)=0.
\end{equation}
Since
\begin{equation} \label{e2.15}
\Delta \tilde{u}_f+\tilde{a}(x,\tilde{u}_f)=0,
\end{equation}
combining \eqref{e2.14} with \eqref{e2.15} yields
\[
\tilde{a}(x,\tilde{u}_f)=a(x,\tilde{u}_f+\phi)+\Delta \phi,
\]
which implies \eqref{e1.12}. This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
Repeating the  argument used in the proof of Theorem \ref{thm1},
yields that \eqref{e2.13} holds for all $f$.
Since there is a common solution, we have that the function $\phi$
 must be the zero function. Thus, for all $f$,
\begin{equation} \label{e2.16}
u_f(x)=\tilde u_f(x),\quad x\in\Omega.
\end{equation}
This shows that
$E_a=E_{\tilde{a}}$,
which is \eqref{e1.13}.
Substituting \eqref{e2.16} in  \eqref{e1.1}, we obtain  that for all $f$,
\[
a(x,u_f)=\tilde{a}(x,\tilde{u}_f), \quad x\in\Omega.
\]
Therefore,
\[
a(x,u)=\tilde{a}(x,u), \quad (x,u)\in E_a.
\]
This completest the proof.
\end{proof}

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\end{document}