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\begin{document}

\title[Boundary determination of a Riemannian metric]{Boundary determination of a Riemannian metric by the localized boundary distance function}

\author{Gunther Uhlmann}
\address{\hskip-\parindent
        Gunther Uhlmann\\
        Department of Mathematics\\
        University of Washington\\
        Seattle, WA 98195-4350\\
        USA}
\email{gunther@math.washington.edu}

\author{Jenn-Nan Wang}
\address{\hskip-\parindent
        Jenn-Nan Wang\\
        Department of Mathematics\\
        National Taiwan University\\
        Taipei 106, Taiwan}
\email{jnwang@math.ntu.edu.tw}

\thanks{This work was done when the authors were participating in the
Inverse Problems Program at MSRI in the fall 2001. The authors
would like to thank MSRI for providing a very stimulating research
environment and partial financial support, through the NSF grant
DMS-9810361. The second author was supported by NSF and a John
Simon Guggenheim fellowship.}

\begin{abstract}
In this paper we give a reconstruction procedure for the Taylor
series of a Riemannian metric on the boundary in boundary normal
coordinates from the localized boundary distance function.
\end{abstract}

\maketitle

\section{Introduction}\label{sec1}
\setcounter{equation}{0}

Let $(M,g)$ be a Riemannian manifold with smooth boundary
$\Gamma:=\partial M$. Define the boundary distance function
$$d_g(x,y)=\mbox{dist}(x,y)\quad\mbox{for}\quad x,y\in\Gamma$$
which is the geodesic distance between boundary points. An
interesting inverse problem is whether one can determine $g$ from
its associated boundary distance function $d_g$. It is easy to see
that uniqueness is not possible. Indeed, let $\psi:M\to M$ be a
diffeomorphism that leaves $\Gamma$ invariant, i.e.
$\psi|_{\Gamma}=Id$, then $d_{\psi^{\ast}g}=d_g$. Therefore, the
inverse problem one would like to address is whether this is the
only obstruction to uniqueness. This problem arose in geophysics
in attempting to determine the inner structure of the Earth. The
boundary distance function measures the travel times of seismic
waves going through the Earth and the metric is the index of
refraction \cite{cre}, \cite{h}. This problem is also called the
boundary rigidity problem in Riemannian geometry \cite{cr},
\cite{gr}, \cite{mi1}. The boundary rigidity problem has been
extensively studied in the last three decades. The reader is
referred to the article \cite{u} for a recent survey.

\smallskip
In this paper we would like to look at this rigidity problem from
the boundary perspective. That is, we want to investigate what
information of $g$ on $\Gamma$ can be determined from $d_g$. As
indicated above, unique determination of $g|_{\Gamma}$ is not
possible. Nevertheless, in this article we are able to show that
one can reconstruct the Taylor series of the metric $g$ on
$\Gamma$ in suitable coordinates from the \emph{localized}
boundary distance function. Precisely, we prove that for any
$x\in\Gamma$ all derivatives of $g$ at $x$ in boundary normal
coordinates can be determined from the knowledge of $d_g(x,y)$ for
all $y\in\Gamma$ sufficiently close to $x$. More importantly, we
will give explicit formulas for all the normal derivatives. To
state the main theorem, we first introduce the notion of
convexity. We say that the boundary $\Gamma$ is locally convex at
$p\in\Gamma$ if there exists an open neighborhood ${\mathcal O}$
of $p$ such that for any two points $x, y\in{\mathcal
O}\cap\Gamma, x\neq y,$ there is a unique geodesic joining $x$ and
$y$ and all inner points of this geodesic lie entirely in
${\mathcal O}$. Also, $M$ is said to be extendible near $p$ if we
can extend $M$ to another smooth manifold $\tilde{M}\supset M$
across ${\mathcal O}\cap\Gamma$. \claim{Main Theorem} Let $(M,g)$
be a Riemannian manifold of dimension $n$ with smooth boundary
$\Gamma$. Let $p\in\Gamma$ and $\Gamma$ be locally convex there.
Assume that $g$ is smooth up to $\Gamma$ near $p$ and also $M$ is
extendible near $p$. Let $\{x^1,\cdots,x^n\}$ be the boundary
normal coordinates near $p$. Then we can reconstruct
$\partial_x^{\gamma}g(p)$ for all $|\gamma|\geq 0$ from the
knowledge of $d_g(q,r)$ for all $q,r\in\Gamma$ sufficiently near
$p$.
\endclaim

Previous results on the boundary determination of the Taylor
series of $g$ from $d_g$ have been given in \cite{mi1} (up to
order 2), \cite{mi2} (infinite order but $n=2$) and recently in
\cite{lsu} (general case). But, none of these papers gave
reconstruction formulas. The main tool of this paper is an
identity which relates the metric $g$ to the lengths of geodesics.
This identity is similar to the one derived by Stefanov and
Uhlmann in \cite{stuh}. It should be noted that the identity
derived in \cite{stuh} relies on the hypotheses that two unknown
metrics coincide with the Euclidean one up to certain order near
the boundary and have the same boundary distance function. Since
we are dealing with the reconstruction problem here, we have only
one unknown metric with no boundary information available. To
utilize Stefanov and Uhlmann's arguments, we therefore need to
choose an appropriate reference metric to go with the unknown
metric (see Section~\ref{sec2}). With the key identity at hand, we
are able to get the boundary information of the metric by
differentiating the identity and letting the boundary distance go
to zero (see Section~\ref{sec3}).

\smallskip
For the readers' convenience, we now outline the reconstruction
procedure. It should be noted that we are working in boundary
normal coordinates near any boundary point, say $p\ (\in\Gamma)$,
throughout the whole reconstruction.
\begin{description}
\item[{\rm Step 1.}] Using Michel's arguments \cite{mi1}, we can determine $g(p)$
in the tangential directions.
\item[{\rm Step 2.}] Differentiating the key identity \eqref{mwin4},
where the reference metric $g_0$ is given in \eqref{goinv}, leads
to a new formula whose one side is solely determined by $d_g$. We
construct the first normal derivatives $g^{-1}$ at $p$ by taking
$d_g\to 0$ in the new formula. Therefore, we can recover the first
normal derivative of $g$ at $p$.
\item[{\rm Step 3.}] Inductively, by further differentiating the identity
\eqref{mwin4} and repeating the arguments in Step 2, we can
determine all normal derivatives of $g$ at $p$.
\end{description}

\smallskip
We remark that the result here is the pointwise boundary
reconstruction of the metric by the knowledge of the boundary
distance function. This problem is somehow related to the inverse
problem of determining the metric by boundary measurements from
the Dirichlet-to-Neumann map associated with the Laplace-Beltrami
operator $\Delta_g$. It is well-known that the
Dirichlet-to-Neumann map is a pseudo-differential operator of
order one. By computing its full symbol (and this is requires
knowledge only of the local Dirichlet-to-Neumann map), one can
determine the boundary value of the metric up to any order in
boundary normal coordinates \cite{leuh}. For another approach see
\cite{kayu}.

\section{Key identity}\label{sec2}
\setcounter{equation}{0}

In this section we will derive an identity which plays a key role
in the proof of the main theorem. Let $H_g$ denote a Hamiltonian
related to $g$ and
$H_g(x,\xi)=(1/2)(\sum_{i,j=1}^ng^{ij}\xi_i\xi_j-1)$ in local
coordinates, where $(g^{ij})=(g_{ij})^{-1}$. Here we are
interested in integral curves associated with the Hamiltonian
vector field induced by $H_g$. The projection of these integral
curves are geodesics. Let $\xz\in\Gamma$ and $\{{\mathcal O},x\}$
be a local chart near $\xz$. Assume that $\Gamma$ is locally
convex at $\xz$ and the Riemannian manifold $(M,g)$ satisfies the
assumptions near $\xz$ as in the main theorem. Let $\xiz\in
T_{\xz}^{\ast}M$ satisfy
\begin{equation}\label{income}
\nu(x^{(0)})\cdot g^{-1}(x^{(0)})\xiz<0
\end{equation}
and
\begin{equation}\label{unit}
\xiz\cdot g^{-1}(x^{(0)})\xiz=1,
\end{equation}
where $\nu(\xz)$ is the unit outer normal (co)vector (relative to
$g^{-1}$) to $\Gamma$ at $\xz$. Considered in this local
coordinates $\{x^1,\cdots,x^n\}$ near $\xz$, let $x_g(s,\xz,\xiz)$
and $\xi_g(s,\xz,\xiz)$ solve the Hamiltonian system
\begin{equation}\label{ham}
\begin{cases}
\frac{d x^m}{ds}=\sum_{j=1}^ng^{mj}\xi_j,&\frac{d\xi_m}{ds}=-\frac{1}{2}\sum_{i,j=1}^n\frac{\partial g^{ij}}{\partial x^m}\xi_i\xi_j\\
x|_{s=0}=x^{(0)},&\xi|_{s=0}=\xi^{(0)}\qquad m=1,\cdots,n.
\end{cases}
\end{equation}
Note that $x_g(s)$ is the geodesic with initial condition
$(\xz,g^{-1}(\xz)\xiz)$ and $s$ in \eqref{ham} is the arc length
parameter. Denote $\Xz=(\xz,\xiz)$ and $t:=t(\Xz)$ the length of
the geodesic issued from $\Xz$ with endpoint on ${\mathcal
O}\cap\Gamma$. Here we choose $\xiz$ appropriately such that all
inner points of $x_g(s)$ lie entirely in ${\mathcal O}$. This
requirement is obviously true if $\xiz$ is taken so that $t$ is
sufficiently small.

\smallskip
Now let us fix an $\Xz$ satisfying \eqref{income} and
\eqref{unit}. Assume that $g_0$ is a Riemannian metric in $M$ such
that \eqref{income} holds at $\Xz$ with $g$ being replaced by
$g_0$. Notice that we do not require that $\Xz$ satisfy
\eqref{unit} with respect to $g_0$. In the coordinates chart
$\{{\mathcal O},x\}$, we let
$(x_{g_0}(s,\xz,\xiz),\xi_{g_0}(s,\xz,\xiz))$ be the solution to
the Hamiltonian system \eqref{ham} with respect to $g_0$ having
the initial $\Xz$. It is readily seen that
$x_{g_0}(s,\xz,\xiz)\subset{\mathcal O}$ for $0<s\leq t$ provided
that $t$ is small. Denote $X=(x,\xi)$. The solutions to
\eqref{ham} related to $g$ and $g_0$ can be written as
$X_g(s,\Xz)=X_g(s,\xz,\xiz)$ and
$X_{g_0}(x,\Xz)=X_{g_0}(s,\xz,\xiz)$, respectively. Let
$F(s)=X_{g}(t-s,X_{g_0}(s,\Xz))$, then
\begin{equation}\label{mwin3}
\begin{array}{ll}
\int_0^tF'(s)ds&=F(t)-F(0)\\
{}&=X_{g}(0,X_{g_0}(t,\Xz))-X_{g}(t,X_{g_0}(0,\Xz))\\
{}&=X_{g_0}(t,\Xz)-X_g(t,\Xz).
\end{array}
\end{equation}
It should be noted that $x$ component of $F$ may not be in
$\Omega$. In order to make sense of $F$, we can extend $g$
smoothly to $\tilde{M}$. Thus, $\tilde{M}$ becomes a Riemannian
manifold carrying a metric which is a smooth extension of $g$.
Nevertheless, the integral in \eqref{mwin3} is independent of the
extension of $g$. It is shown in \cite{stuh} that
$$F'(s)=\frac{\partial X_g}{\partial\Xz}(t-s,X_{g_0}(s,\Xz))(V_{g_0}-V_g)(X_{g_0}(s,\Xz)),$$
where $V_{g_0}=(\partial H_{g_0}/\partial\xi,-\partial
H_{g_0}/\partial x)$ and $V_g$ is defined similarly. In
conclusion, we obtain that
\begin{equation}\label{mwin4}
\int_0^t\frac{\partial
X_g}{\partial\Xz}(t-s,X_{g_0}(s,\Xz))(V_{g_0}-V_g)(X_{g_0}(s,\Xz))ds=X_{g_0}(t,\Xz)-X_g(t,\Xz).
\end{equation}
Before leaving this section, we want to remark that the right-hand
side of \eqref{mwin4} is solely determined by $d_g$ near $\xz$.
This property will be verified in the following section.

\section{Proof of Main Theorem}\label{sec3}
\setcounter{equation}{0}

Assume that $(M,g)$ satisfies the assumptions of Main Theorem near
$\xz=0\in\Gamma$. Let $\tilde{\mathcal O}$ be an open neighborhood
of $0$ in $\tilde{M}$ and the metric $g$ has been extended
smoothly in $\tilde{\mathcal O}$, still denoted by $g$. We now
introduce the boundary exponential map
$\exp_{\Gamma}(s,p)=\exp_p(s\mu(p))$ near $0$, where
$p\in\tilde{\mathcal O}\cap\Gamma$ and $\mu(p)\in T_p\tilde{M}$ is
the unit inner normal to $\Gamma$ with respect to $g$. It is clear
that $\exp_{\Gamma}$ is a diffeomorphism if $\tilde{\mathcal O}$
is small. By virtue of this map, we can introduce coordinates,
still denoted by $\{x^1,\cdots,x^{n-1},x^n\}$, in $\tilde{\mathcal
O}$ such that $\tilde{\mathcal O}$ is mapped onto an open
neighborhood $U$ of $0$ and the boundary $\Gamma$ is determined by
$x^n=0$ and $x^n>0$ in $\tilde{\mathcal O}\cap M$. Moreover, in
this coordinates system, the metric $g$ is given by
$$g=\begin{pmatrix}{}&{}&{}&0\\{}&g_{\alpha\beta}&{}&\vdots\\{}&{}&{}&0\\0&\cdots&0&1\end{pmatrix}\quad\mbox{and}\quad g^{-1}=\begin{pmatrix}{}&{}&{}&0\\{}&g^{\alpha\beta}&{}&\vdots\\{}&{}&{}&0\\0&\cdots&0&1\end{pmatrix}.$$ Hereafter the indices $\alpha$ and $\beta$ run from $1$
and $n-1$.

\smallskip
First, we want to determine g(0). Clearly, in boundary normal
coordinates, it is enough to determine the information of
$g(v,v)|_{x=0}$ for any tangent vector $v$. We use here the
following argument of Michel's paper \cite{mi1}. Denote
$e_{\alpha}$ the $\alpha$-th standard basis of $\R^n$. Let us
define a set of vectors $V=\{v_{\alpha\beta},\ \alpha\leq\beta\},$
where $v_{\alpha\alpha}=e_{\alpha}$ and
$v_{\alpha\beta}=e_{\alpha}+e_{\beta}$ for $\alpha<\beta$. Let
$c:[0,\eps)\to U$ be a curve on $\{x^n=0\}$ with $c(0)=0$ and
$c'(0)=v$, where $v$ is an element of $V$.  Here we choose $\eps$
small enough so that all geodesics joining $0$ and $c(\tau)$ for
$0<\tau\leq\eps$ lie entirely in $U^+=\{x\in U: x_n>0\}$ except
two endpoints. It is easy to see that
$$\lim_{\tau\to 0}\frac{d_g(0,c(\tau))}{\tau}=\|v\|_{g}=g(v,v)^{1/2}.$$
Now by repeating the arguments for all $v\in V$, we can determine
$g_{\alpha\beta}(0)$ for all $\alpha,\beta$ and hence $g(0)$.
Clearly, using the same method, we can find $g(x',0)$ for
$|x'|<\delta$ with $\delta$ sufficiently small, where
$x'=(x^1,\cdots,x^{n-1})$.

\smallskip
Having found $g$ on $\Gamma_{\delta}:=\{(x',0)\ |\ |x'|<\delta\}$
for small $\delta$, we can determine the right-hand side of
\eqref{mwin4} by knowing $d_g(p,g)$ for all
$p,g\in\Gamma_{\delta}$. This can be seen, using the notation of
the previous section, by observing that $d_g$ is the generating
function of the canonical relation obtained by projecting the set
$\{ (X^{(0)}, X_g(t, X^{(0)}) \}$ onto $T^*\Gamma\times
T^*\Gamma.$ This set is called the scattering relation in
\cite{gu}. A more differential geometric way to see this is via
the formula derived in \cite{mi1}
\begin{equation}\label{mby3}
\gamma'(t(p,q))=i_{\ast}(\nabla'_q t(p,q))-\sqrt{1-\|\nabla'_q
t(p,q)\|^2_{i^{\ast}g}}\quad\mu(q),
\end{equation}
where $\gamma$ is the geodesic issued from $p$ and parametrized by
the arc length, $i:\Gamma_{\delta}\to U$ is the inclusion map and
$\nabla'$ is the gradient operator on the boundary $x^n=0$. Let
$\xiz(p,q)$ and $\xi_g(t(p,q),\xiz(p,q))=:\xi_g(p,q)$ be the
initial and final covectors related to the geodesic connecting
$p,q\in\Gamma_{\delta}$. Reinterpreting \eqref{mby3} in the
covector setting, we can see that for the geodesic joining $p$ and
$q$, $p\ne q$, $\xiz(p,q)$ and $\xi_g(p,q)$ satisfy
\begin{equation}\label{mby4}
\xiz(p,q)=g(p)i_{\ast}(\nabla'_pt(p,q))-\sqrt{1-\|\nabla'_pt(p,q)\|^2_{i^{\ast}g}}\quad
g(p)\mu(p)
\end{equation}
and
\begin{equation}\label{mby5}
\xi_g(p,q)=g(q)i_{\ast}(\nabla'_qt(p,q))-\sqrt{1-\|\nabla'_qt(p,q)\|^2_{i^{\ast}g}}\quad
g(q)\mu(q).
\end{equation}

\smallskip
Next we would like to determine all derivatives of $g$ at $0$.
Since $g^{-1}g=I$, it suffices to determine all derivatives of
$g^{-1}$ at $0$. Let $v\in V$ and define $\tilde{c}(\tau),
0\leq\tau\leq\tilde{\eps}$, be a curve on $\{x^n=0\}$ with
$\tilde{c}(0)=0$ and $\tilde{c}'(0)=g^{-1}(0)\eta$, where
$\eta=v(v\cdot g^{-1}(0)v)^{-1/2}$. As before, we choose
$\tilde{\eps}$ sufficiently small so that all geodesics joining
$0$ and $\tilde{c}(\tau)$ for $0<\tau\leq\tilde{\eps}$ lie
entirely in $U^+$ except two endpoints. Now we are at the position
to choose our reference Riemannian metric $g_0$ in $M$. The goal
here is to choose a $g_0$ such that $\nu(0)\cdot
g^{-1}_0(0)\eta<0$, i.e
\begin{equation}\label{nonzero}
\sum_{j=1}^ng_0^{nj}(0)v_j>0
\end{equation}
for all $v=(v_1,\cdots,v_n)\in V$. One possible choice is that
$g_0^{-1}$ is of the form
\begin{equation}\label{goinv}
g_0^{-1}=\begin{pmatrix}{}&{}&{}&\lambda_1\\{}&I_{n-1}&{}&\vdots\\{}&{}&{}&\lambda_{n-1}\\\lambda_1&\cdots&\lambda_{n-1}&1\end{pmatrix},
\end{equation}
where $I_{n-1}$ is the identity matrix of size $n-1$ and
$\lambda_{\alpha}>0$. Since
$\mbox{det}(g_0^{-1})=1-\sum_{\alpha=1}^{n-1}\lambda_{\alpha}^2$,
we can choose $\lambda_{\alpha}$ sufficiently small for all
$\alpha$ to guarantee the positive-definiteness of $g_0^{-1}$.
Here we want to point out that $\eta$ does not satisfy the
incoming direction \eqref{income} with respect to $g$, but it
satisfies \eqref{income} in terms of $g_0$. In fact, we can see
that $\nu(0)\cdot g^{-1}(0)\eta=0$. With this choice of $g_0$, we
obtain that the solution to the Hamiltonian system \eqref{ham}
with respect to $g_0$ can be written explicitly as
$$(x_{g_0}(s,0,\xiz),\xi_{g_0}(s,0,\xiz))=(sg_0^{-1}\xiz,\xiz),$$ where the initial $(0,\xiz)$ satisfies \eqref{income} in terms of $g_0$. Note that the curve $x_{g_0}(s,0,\xiz)$ lies entirely in $U^{+}$ for all small $s$.

\smallskip
Now consider the geodesic (relative to $g$) connecting $0$ and
$\tilde{c}(\tau)$ for $0<\tau<\tilde{\eps}$. In view of the
formulas \eqref{mby4} and \eqref{mby5}, it is readily seen that
given $\tilde{c}(\tau)$ we can determine $\xiz=\xiz(\tau)$ and
$X_g(t(\tau),\Xz(\tau))=X_g(\tau)$ from the boundary distance
function $d_g(0,\tilde{c}(\tau))=t(\tau)$. Notice that if
$t(\tau)$ is sufficiently small (i.e. $\tau$ is small), then
$\xiz(\tau)$ is close to $\eta$ and $(0,\xiz(\tau))$ satisfies the
incoming condition \eqref{income} related to $g_0$. Furthermore,
we can see that $\xiz(\tau)\to\eta$ as $\tau\to 0$. Expressing
every variable in the identity \eqref{mwin4} in terms of $\tau$,
we have that
\begin{equation}\label{mwin5}
\begin{array}{l}
\int_0^{t(\tau)}\frac{\partial X_g}{\partial\Xz}(t(\tau)-s,X_{g_0}(s,\Xz(\tau)))(V_{g_0}-V_g)(X_{g_0}(s,\Xz(\tau)))ds\\
=X_{g_0}(\tau)-X_g(\tau)=(t(\tau)g_0^{-1}\xiz(\tau)-x_g(\tau),\xiz(\tau)-\xi_g(\tau)).
\end{array}
\end{equation}
Differentiating both sides of \eqref{mwin5} in $\tau$ yields
\begin{equation}\label{mwin6}
\begin{array}{l}
\frac{\partial X_g}{\partial\Xz}(0,X_{g_0}(t(\tau),\Xz(\tau)))(V_{g_0}-V_g)(X_{g_0}(t(\tau),\Xz(\tau)))t'(\tau)\\
+\int_0^{t(\tau)}\frac{d}{d\tau}\{\frac{\partial X_g}{\partial\Xz}(t(\tau)-s,X_{g_0}(s,\Xz(\tau)))(V_{g_0}-V_g)(X_{g_0}(s,\Xz(\tau)))\}ds\\
=(t'(\tau)g_0^{-1}\xiz(\tau)+t(\tau)g_0^{-1}\xiz{}'(\tau)-x_g'(\tau),\xiz{}'(\tau)-\xi_g'(\tau)).
\end{array}
\end{equation}
Taking $\tau\to 0$ in \eqref{mwin6}, we obtain that
\begin{equation}\label{mwin7}
\begin{array}{l}
\frac{\partial X_g}{\partial\Xz}(0,X_{g_0}(0,\Xz(0)))(V_{g_0}-V_g)(X_{g_0}(0,\Xz(0)))t'(0)\\
=I_{2n\times 2n}\cdot(V_{g_0}-V_g)(0,\eta)t'(0)\\
=(t'(0)g_0^{-1}\eta-x_g'(0),\xiz{}'(0)-\xi_g'(0)),
\end{array}
\end{equation}
where $I_{2n\times 2n}$ is the identity matrix of size $2n$. It
has been shown previously that $t'(0)=\|\tilde{c}'(0)\|_g=1$.
Writing out the formula \eqref{mwin7}, we can see that
\begin{equation}\label{mwin8}
g^{-1}(0)\eta=g_0^{-1}(0)\eta-t'(0)g_0^{-1}\eta-x_g'(0)
\end{equation}
and
\begin{equation}\label{mwin9}
\frac{1}{2}\eta\cdot \partial_xg^{-1}(0)\eta=\xiz{}'(0)-\xi_g'(0).
\end{equation}
It should be noted that we do not use \eqref{mwin8} as the
reconstruction formula for $g(0)$ since we need to choose the
curve $\tilde{c}$ before proceeding the arguments. The curve
$\tilde{c}$ has already used the information $g(0)$. It follows
from \eqref{mwin9} that
\begin{equation}\label{mwin10}
v\cdot \partial_xg^{-1}(0)v=2(v\cdot
g^{-1}(0)v)(\xiz{}'(0)-\xi_g'(0)).
\end{equation}
By repeating the above arguments for each one element of $V$, we
can derive \eqref{mwin10} for all $v\in V$. In turn we are able to
determine $\partial_xg^{-1}(0)$. Using the same method, we can
find $\partial_xg^{-1}(x',0)$ for $(x',0)$ near $0$. Thus,
$\partial^k_{x^{\alpha}}\partial_{x^n}g^{-1}(0)$ is also
determined for all positive integer $k$.

\smallskip
To continue the proof, we differentiate \eqref{mwin6} in $\tau$
and set $\tau=0$. To end, we get that
\begin{equation}\label{mwin11}
\begin{array}{l}
I_{2n\times 2n}\ \sum_{j=1}^n\partial_{x^j}(V_{g_0}-V_g)(g_0^{-1}\eta)_j (t'(0))^2\\
=(t''(0)g_0^{-1}\eta+2t'(0)g_0^{-1}\xiz{}'(0)-x_{g}''(0),\xiz{}''(0)-\xi_g''(0))\\
\quad +\Psi,
\end{array}
\end{equation}
where $(g_0^{-1}\eta)_j$ is the $j$-th component of $g_0^{-1}\eta$
and $\Psi$ is a $2n$ vector which consists of terms containing
only $g^{-1}(0)$ and $\partial_x g^{-1}(0)$. Therefore, $\Psi$ is
a known vector-valued function. It is easily observed that only
the last $n$ components of
$\sum_{j=1}^{n}\partial_{x^j}(V_{g_0}-V_g)(g_0^{-1}\eta)_j$
contain the second derivatives of $g^{-1}$ at $0$. Moreover, since
we have found $\partial_{x^{\alpha}}\partial_{x^n}g^{-1}(0)$, the
only term yet to be determined is $\partial^2_{x^n}g^{-1}(0)$.
Singling out the last component of \eqref{mwin11}, we have
$$
\frac{1}{2}\eta\cdot
[\sum_{j=1}^n\partial_{x^j}\partial_{x^n}g^{-1}(0)(g_0^{-1}\eta)_j]\eta=(\xiz{}''(0)-\xi_g''(0))_n+(\Psi)_{2n}
$$
from which we get that
\begin{equation}\label{mby6}
\begin{array}{l}
\{v\cdot\partial^2_{x^n}g^{-1}(0)v\}(g_0^{-1}v)_n\\
=2(v\cdot g^{-1}(0)v)^{3/2}\{\eta\cdot
[\sum_{\alpha=1}^{n-1}\partial_{x^{\alpha}}\partial_{x^n}g^{-1}(0)(g_0^{-1}\eta)_{\alpha}]\eta+(\xiz{}''(0)-\xi_g''(0))_n+(\Psi)_{2n}\}.
\end{array}
\end{equation}
Since $(g_0^{-1}v)_n$ is not zero (see \eqref{nonzero}), we can
determine $v\cdot\partial^2_{x^n}g^{-1}(0)v$ from \eqref{mby6}.
Once again, repeating the arguments for all $v\in V$ and noting
that $(g_0^{-1}v)_n$ is never zero for any $v\in V$, we can
determine $v\cdot\partial^2_{x^n}g^{-1}(0)v$ for all $v\in V$ and
hence $\partial^2_{x_n}g^{-1}(0)$. Using the same procedure, we
can determine $\partial_{x^n}^2g^{-1}(x',0)$ for $|x'|<\delta$
with $\delta$ sufficiently small. In turn we can find
$\partial^{\alpha'}_{x'}\partial^2_{x^n}g^{-1}(0)$ for any
multi-index $\ \alpha'$.

\smallskip
Inductively, assume that we have determined
$\partial^{\alpha'}_{x'}\partial^l_{x^n}g^{-1}(0)$ with $0<l<\ell$
and arbitrary $\alpha'$. Now by differentiating \eqref{mwin5}
$\ell$ times in $\tau$ and setting $\tau=0$, we single out the
term containing $\partial^{\ell}_{x^n}g^{-1}(0)$ and find that
\begin{equation}\label{mby7}
\{v\cdot\partial^{\ell}_{x^n}g^{-1}(0)v\}(g_0^{-1}v)^{\ell-1}_n={\mathcal
R},
\end{equation}
where ${\mathcal R}$ is a known value which is determined by the
induction assumption. Deriving \eqref{mby7} for each $v\in V$ and
noting that $(g_0^{-1}v)_n$ is never zero, we can determine
$v\cdot\partial^{\ell}_{x^n}g^{-1}(0)v$ for all $v\in V$ and
therefore $\partial^{\ell}_{x^n}g^{-1}(0)$.

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\end{document}
