# Introduction to inverse problems for hyperbolic PDEs

Medet Nursultanov<EMAIL_ADDRESS>and  Lauri Oksanen
<EMAIL_ADDRESS>
###### Contents

  1. 1 Introduction
    1. 1.1 Boundary Control method
    2. 1.2 Geometric optics based approach
  2. 2 Boundary Control method in $1+1$ dimensions
    1. 2.1 Finite speed of propagation in the case $q=0$
    2. 2.2 Finite speed of propagation for general $q$
    3. 2.3 Approximate controllability
    4. 2.4 An integration by parts trick for the inverse problem
    5. 2.5 Solution to the inverse problem
  3. 3 Boundary Control method on a Riemannian manifold
    1. 3.1 Finite speed of propagation with nonconstant speed of sound
    2. 3.2 Main tools
    3. 3.3 Solution to the inverse problem
  4. 4 Geometric optics
    1. 4.1 Single term ansatz
    2. 4.2 Multi-term ansatz
    3. 4.3 Solving for the remainder
    4. 4.4 Proof of Lemma 4
  5. 5 Solving an inverse problem using geometric optics
    1. 5.1 Reduction to the light ray transform
    2. 5.2 Inversion of the light ray transform

These lecture notes were written for CIRM SMF School Spectral Theory, Control
and Inverse Problems, November 2022.

##  1\. Introduction

There are two main approaches to solve inverse coefficient determination
problems for wave equations: the Boundary Control method and an approach based
on geometric optics. These notes focus on the Boundary Control method, but we
will have a brief look at the geometric optics as well.

###  1.1. Boundary Control method

The Boundary Control method can be used to solve several coefficient
determination problems for wave equations. The method originates from [2], and
it allows for building a rather comprehensive theory of inverse problems of
coefficient determination type. Indeed, many such problems can be reduced to
inverse boundary spectral problems, which again can be reduced to coefficient
determination problems for wave equations. This is the case for inverse
problems for heat and non-stationary Schroödinger equations [10], as well as
for several time-fractional [12] and space-fractional [7] equations. Inverse
problems for linear elliptic equations on a wave guide [5], and some (very
special) non-linear elliptic equations [4] are also covered by this theory.

The method has been generalized for symmetric systems of wave equations with
scalar leading order [13] and for wave equations on Lorentzian manifolds
satisfying a curvature bound [1].

In these notes we give an introductory exposition of the method. For an in-
depth presentation, we refer to the monograph [11].

###  1.2. Geometric optics based approach

Geometric optics constructions are well-suited for solving inverse problems
with time-dependent coefficients. In the case of time-independent
coefficients, this approach has not reached the same geometric generality as
the Boundary Control method.

The geometric optics based approach goes back to [19]. It has been generalized
for wave equations on real analytic Lorentzian manifolds satisfying a convex
foliation condition [18, 17], as well as for certain static and stationary
Lorentzian manifolds [8, 9].

At the moment it is not clear which of the two methods works better in the
Lorentzian context. This context is natural since it respects the idea that
coordinates do not exist a priori in nature, and hence should play no role in
the mathematical formulation of physical models.

##  2\. Boundary Control method in $1+1$ dimensions

Consider the following initial-boundary value problem

(1) $\begin{cases}(\partial_{t}^{2}-\partial_{x}^{2}+q)u=0;&\text{for
}(t,x)\in\mathbb{R}^{+}\times(0,1);\\\ \left.u\right|_{x=0}=f;\\\
\left.u\right|_{x=1}=0;\\\
\left.u\right|_{t=0}=\left.\partial_{t}u\right|_{t=0}=0,\end{cases}$

where $f$ is a boundary source, $q$ is a potential, and $\mathbb{R}^{+}$
stands for the interval $(0,\infty)$. For simplicity, we will assume that
$q\in C^{\infty}(0,1)$, a smooth function up to the end points of the interval
$[0,1]$, and that $f\in C_{0}^{\infty}(\mathbb{R}^{+})$, a smooth compactly
supported function. Then (1) has a unique solution that satisfies $u\in
C^{\infty}(\mathbb{R}^{+}\times(0,1))$, see e.g. the theory in Section 7.2 of
[6].

We write $u^{f}=u$ for the solution to (1) to emphasize its dependence on the
source $f$. Define the Dirichlet-to-Neumann map

(2) $\Lambda f=\left.\partial_{x}u^{f}\right|_{x=0}.$

We will study the following inverse problem: determine the potential $q$ given
the operator $\Lambda$.

###  2.1. Finite speed of propagation in the case $q=0$

The finite speed of propagation property for the acoustic wave equation plays
a central role in both direct and inverse theory. To illustrate the property
in the simplest possible case, let us consider (1) with $q=0$. We set

$\mathcal{E}(t,x)=|\partial_{t}u(t,x)|^{2}+|\partial_{x}u(t,x)|^{2}.$

######  Theorem 1 (Conservation of energy).

Suppose that $u\in C^{2}(\overline{{\mathbb{R}}^{+}\times(0,1)})$ satisfies

$\begin{cases}(\partial_{t}^{2}-\partial_{x}^{2})u=0\quad\text{on
}\mathbb{R}^{+}\times(0,1)\\\
\left.u\right|_{x=0}=\left.u\right|_{x=1}=0.\end{cases}$

Then the global energy

$\displaystyle E(t)=\frac{1}{2}\int_{0}^{1}\mathcal{E}(t,x)dx$

satisfies $\partial_{t}E(t)=0$ for all $t>0$.

###### Proof.

We write,

$\partial_{t}E(t)=\frac{1}{2}\int_{0}^{1}\partial_{t}\mathcal{E}(t,x)dx\\\
=\int_{0}^{1}\left(\partial_{t}u(t,x)\partial_{t}^{2}u(t,x)+\partial_{x}u(t,x)\partial_{t}\partial_{x}u(t,x)\right)dx.$

By integration by parts, we obtain

$\partial_{t}E(t)=\left[\partial_{x}u(t,x)\partial_{t}u(t,x)\right]^{1}_{0}+\int_{0}^{1}\left(\partial_{t}^{2}u(t,x)-\partial_{x}^{2}u(t,x)\right)\partial_{t}u(t,x)dx.$

The first term is zero since $u(t,0)=u(t,1)=0$, and the second term is zero
since $u$ satisfies the wave equation. ∎

######  Theorem 2 (Finite speed of propagation).

Suppose that $u\in C^{2}(\overline{{\mathbb{R}}^{+}\times(0,1)})$ satisfies

(3) $\begin{cases}(\partial_{t}^{2}-\partial_{x}^{2})u=0\quad\text{on
}\mathbb{R}^{+}\times(0,1)\\\ \left.u\right|_{x=1}=0.\end{cases}$

Then the local energy

$\displaystyle E(t)=\frac{1}{2}\int_{t}^{1}\mathcal{E}(t,x)dx$

satisfies $\partial_{t}E(t)\leq 0$ for all $t>0$.

###### Proof.

We write, by using Leibniz integral rule,

$\displaystyle\partial_{t}E(t)$
$\displaystyle=-\frac{1}{2}\mathcal{E}(t,t)+\frac{1}{2}\int_{t}^{1}\partial_{t}\mathcal{E}(t,x)dx$
$\displaystyle=-\frac{1}{2}\mathcal{E}(t,t)+\int_{t}^{1}\left(\partial_{t}u(t,x)\partial_{t}^{2}u(t,x)+\partial_{x}u(t,x)\partial_{t}\partial_{x}u(t,x)\right)dx.$

An integration by parts gives

$\partial_{t}E(t)=-\frac{1}{2}\mathcal{E}(t,t)+\left[\partial_{x}u(t,x)\partial_{t}u(t,x)\right]^{1}_{t}\\\
+\int_{t}^{1}\left(\partial_{t}^{2}u(t,x)-\partial_{x}^{2}u(t,x)\right)\partial_{t}u(t,x)dx.$

Since $u$ is the solution of the wave equation the last integral is $0$.
Moreover, since $u(t,1)=0$, the second term is $0$ at $1$, so that

$\partial_{t}E(t)=-\frac{1}{2}\mathcal{E}(t,t)-\partial_{t}u(t,t)\partial_{x}u(t,t)=-\frac{1}{2}\left(\partial_{t}u(t,t)+\partial_{x}u(t,t)\right)^{2}\leq
0.$

∎

In particular, if $E(0)=0$ then $E(t)=0$ for all $t>0$. So if $u$ and
$\partial_{t}u$ vanish initially on $(0,1)$ then at time $t$ they vanish for
$x\in(t,1)$. To see that this statement is optimal, observe that

$\displaystyle u(t,x)=f(t-x)-f(t+x-2)$

is a solution to (3) for any $f\in C^{2}(\mathbb{R})$. Moreover, if $f(r)$
vanishes for $r<0$, then $E(0)=0$. But $u$ does not vanish near the line $t=x$
if $f(r)$ does not vanish for $r>0$ near $r=0$.

###  2.2. Finite speed of propagation for general $q$

Let us now consider a wave equation with a potential as in (1), and prove a
finite speed of propagation result in the very natural setting of a diamond,
see the set $K$ in Theorem 3 below. The set $K$ is visualized in Figure 1.

It is also possible to study the case where boundary conditions are posed, say
at $x=0$ and $x=1$ as before, and where the diamond intersects the boundary of
$\mathbb{R}^{+}\times(0,1)$. However, we leave this case to the reader.

######  Theorem 3 (Finite speed of propagation).

Let $X>0$ and define

$K=\\{(t,x)\in\mathbb{R}\times\mathbb{R}:X-|x|\geq|t|\\}.$

Let $q\in L^{\infty}(K)$ and let $u\in C^{2}(K)$ be a solution of the equation

$\begin{cases}\left(\partial_{t}^{2}-\partial_{x}^{2}+q(t,x)\right)u=0,&\text{on
}K;\\\ \left.u\right|_{t=0}=\left.\partial_{t}u\right|_{t=0}=0,&\text{on
}(-X,X).\end{cases}$

Then $u=0$ in $K$.

###### Proof.

We will consider the case $t>0$. The case $t<0$ is analogous and we omit its
proof. Let us set $I(t)=[-X+t,X-t]$ and define the energy

$E(t)=\frac{1}{2}\int_{I(t)}\mathcal{E}(t,x)dx.$

Using Leibniz integral rule, we obtain

$\displaystyle\partial_{t}E(t)=$
$\displaystyle-\frac{1}{2}\left(|\partial_{t}u(t,X-t)|^{2}+|\partial_{x}u(t,X-t)|^{2}\right)$
$\displaystyle-\frac{1}{2}\left(|\partial_{t}u(t,-(X-t))|^{2}+|\partial_{x}u(t,-(X-t))|^{2}\right)$
$\displaystyle+\int_{I(t)}\left(\partial_{t}u(t,x)\partial_{t}^{2}u(t,x)+\partial_{x}u(t,x)\partial_{t}\partial_{x}u(t,x)\right)dx.$

An integration by parts gives

$\displaystyle\partial_{t}E(t)=$
$\displaystyle-\frac{1}{2}\left(|\partial_{t}u(t,X-t)|^{2}+|\partial_{x}u(t,X-t)|^{2}\right)$
$\displaystyle-\frac{1}{2}\left(|\partial_{t}u(t,-(X-t))|^{2}+|\partial_{x}u(t,-(X-t))|^{2}\right)$
$\displaystyle+\partial_{t}u(t,X-t)\partial_{x}u(t,X-t)-\partial_{t}u(t,-(X-t))\partial_{x}u(t,-(X-t))$
$\displaystyle+\int_{I(t)}\partial_{t}u(t,x)\left(\partial_{t}^{2}u(t,x)-\partial_{x}^{2}u(t,x)\right)dx.$

Therefore,

$\displaystyle\partial_{t}E(t)=$
$\displaystyle-\frac{1}{2}\bigg{(}|\partial_{t}u(t,X-t)|-|\partial_{x}u(t,X-t)|\bigg{)}^{2}$
$\displaystyle-\frac{1}{2}\bigg{(}|\partial_{t}u(t,-(X-t))|+|\partial_{x}u(t,-(X-t))|\bigg{)}^{2}$
$\displaystyle+\int_{I(t)}\partial_{t}u(t,x)\left(\partial_{t}^{2}u(t,x)-\partial_{x}^{2}u(t,x)\right)dx.$

Since the first two terms are non-positive, it follows

$\partial_{t}E(t)\leq-\int_{I(t)}\partial_{t}u(t,x)q(t,x)u(t,x)dx\\\
+\int_{I(t)}\partial_{t}u(t,x)\left(\partial_{t}^{2}u(t,x)-\partial_{x}^{2}u(t,x)+q(t,x)u(t,x)\right)dx.$

Let us use the notation

$P=\partial_{t}^{2}-\partial_{x}^{2}+q.$

The Cauchy-Schwarz inequality imply

$E(t)=E(0)+\int_{0}^{t}\partial_{t}E(s)ds\\\ \leq
E(0)+2\int_{0}^{t}\int_{I(s)}\left(|\partial_{t}u(s,x)|^{2}+|q(s,x)u(s,x)|^{2}+|Pu(s,x)|^{2}\right)dxds.$

Hence,

(4) $E(t)\leq
E(0)+C\left(\int_{0}^{t}E(s)ds+\int_{0}^{t}\int_{I(s)}\left(|u(s,x)|^{2}+|Pu(s,x)|^{2}\right)dxds\right),$

where $C$ is a constant depending on $q$ and $X$, which will change from line
to line.

Let us now take care of the $|u|^{2}$ term caused by the potential $q$. We
will use arguments close to the proof in [14, Section 4.2]. For
$x\in\mathbb{R}$ fixed, we know

$u(t,x)=u(0,x)+\int_{0}^{t}\partial_{t}u(s,x)ds.$

Let us square it, and then, use Cauchy’s inequality and the Cauchy-Schwarz
inequality to obtain

(5) $\displaystyle|u(t,x)|^{2}\leq
2|u(0,x)|^{2}+2t\int_{0}^{t}|\partial_{t}u(s,x)|^{2}ds.$

Next, we define

$z(t)=E(t)+\int_{I(t)}|u(t,x)|^{2}dx.$

Using (4) and (5), we estimate

$z(t)\leq
E(0)+C\left(\int_{0}^{t}E(s)ds+\int_{0}^{t}\int_{I(s)}\left(|u(s,x)|^{2}+|Pu(s,x)|^{2}\right)dxds\right)\\\
+2\int_{I(t)}|u(0,x)|^{2}dx+2t\int_{I(t)}\int_{0}^{t}|\partial_{t}u(s,x)|^{2}dsdx.$

Since we consider $t\in[0,X]$, and since $I(t)\subset I(s)$ for $s\in[0,t]$,
the last term can be estimated from above by $2X\int_{0}^{t}E(s)ds$.
Therefore, we have

$z(t)\leq
Cz(0)+C\int_{0}^{t}z(s)ds+C\int_{0}^{t}\int_{I(s)}|Pu(s,x)|^{2}dxds.$

Using the Gronwall’s inequality in the integral form, see e.g. [6, Appendix
B.2.k], we obtain

(6) $\displaystyle z(t)\leq
C\left(z(0)+\int_{0}^{t}\int_{I(s)}|Pu(s,x)|^{2}dxds\right).$

Since $z(0)=0$ and $Pu=0$, we conclude that $z(t)=0$ for $t\in[0,X]$.
Recalling the definition of the function $z$, we see that $u(t,x)=0$ for
$t\in[0,X]$ and $x\in I(t)$. ∎

######  Theorem 4 (Unique continuation).

Let $T>0$ and define

$K=\\{(t,x)\in\mathbb{R}\times\mathbb{R}:|x|\leq T-|t|\\}.$

Let $q\in L^{\infty}(K)$ and let $u\in C^{2}(K)$ be a solution of the equation

$\begin{cases}\left(\partial_{t}^{2}-\partial_{x}^{2}+q(t,x)\right)u=0,&\text{on
}K;\\\ \left.u\right|_{x=0}=\left.\partial_{x}u\right|_{x=0}=0,&\text{on
}(-T,T).\end{cases}$

Then $u=0$ on $K$.

(a) Finite speed of the wave propagation (b) Unique continuation

Figure 1. $1+1$ dimensional case

###### Proof.

By interchanging the roles of $t$ and $x$, we see that the theorem coincides
with Theorem 3; see Figure 1. ∎

###  2.3. Approximate controllability

Let us recall that $u^{f}$ is the solution of (1), when the boundary source is
given by $f$. The following result is obtained by transposing unique
continuation.

######  Lemma 1 (Approximate controllability).

Let $0<s<T\leq 1$ and suppose that $s\leq 1$. Then the set

$\mathcal{B}(s,T)=\left\\{u^{f}(T,\cdot):f\in C_{0}^{\infty}(T-s,T)\right\\}$

is a dense subset of $L^{2}(0,s)$.

Here we use the identification

$\displaystyle L^{2}(0,s)=\\{\phi\in
L^{2}(0,1):\operatorname{supp}(\phi)=(0,1)\\}.$

Note that due to finite speed of propagation the function $u^{f}(T,\cdot)$ is
supported on $[0,s]$. Hence we may view $\mathcal{B}(s,T)$ as a subspace of
$L^{2}(0,s)$.

###### Proof.

To show the density, it is enough to prove that
$\mathcal{B}(s,T)^{\perp}=\\{0\\}$, where

$\mathcal{B}(s,T)^{\perp}=\left\\{v\in L^{2}(0,s):(v,u)_{L^{2}(0,s)}=0\text{
for all }u\in\mathcal{B}(s,T)\right\\}.$

Let $h\in\mathcal{B}(s,T)^{\perp}$. Let $\omega$ be the solution of the
equation

$\begin{cases}\left(\partial_{t}^{2}-\partial_{x}^{2}+q(t,x)\right)\omega(t,x)=0&\text{on
}(0,T)\times(0,1);\\\ \left.\omega\right|_{x=0,1}=0;\\\
\left.\omega\right|_{t=T}=0;\\\
\left.\partial_{t}\omega\right|_{t=T}=h.\end{cases}$

Let $f\in C_{0}^{\infty}(T-s,T)$. An integration by parts, together with the
boundary and initial conditions for $u^{f}$ and $\omega$, give

$0=\left(\left(\partial_{t}^{2}-\partial_{x}^{2}+q\right)u^{f},\omega\right)_{L^{2}((0,T)\times(0,1))}-\left(u^{f},\left(\partial_{t}^{2}-\partial_{x}^{2}+q\right)\omega\right)_{L^{2}((0,T)\times(0,1))}\\\
=-\int_{0}^{1}u^{f}(T,x)\partial_{t}\omega(T,x)dx-\int_{0}^{T}u^{f}(t,0)\partial_{x}\omega(t,0)dt.$

Since $\left.\partial_{t}\omega\right|_{t=T}=h\in\mathcal{B}(s,T)^{\perp}$,
the first term of the right-hand side is zero, so that

$0=\int_{0}^{T}f(t)\partial_{x}\omega(t,0)dt,$

which is true for arbitrary $f\in C_{0}^{\infty}(T-s,T)$. Hence,
$\partial_{x}\omega(t,0)=0$ on $(T-s,T)$.

Let $\tilde{\omega}$ be the odd extension of $\omega$ to $(0,2T)\times(0,1)$,
more precisely,

$\tilde{\omega}(t,x)=\begin{cases}\omega(t,x)&\text{if }t\leq T\\\
-\omega(2T-t,x)&\text{otherwise}.\end{cases}$

Then $\tilde{\omega}$ satisfies

$\begin{cases}\left(\partial_{t}^{2}-\partial_{x}^{2}+q(t,x)\right)\tilde{\omega}(t,x)=0&\text{on
}(0,2T)\times(0,1);\\\ \left.\tilde{\omega}\right|_{x=0}=0&\text{on
}(0,2T);\\\ \left.\partial_{x}\tilde{\omega}\right|_{x=0}=0&\text{on
}(T-s,T+s).\end{cases}$

By unique continuation, see Theorem 4, we obtain that $\tilde{\omega}=0$ on

$\\{(t,x)\in(T-s,T+s)\times(0,1):|x|\leq s-|T-t|\\}.$

In particular, $\partial_{t}\omega(T,x)=h(x)=0$ on $(0,s)$, so that
$\mathcal{B}(s,T)^{\perp}=\\{0\\}$. ∎

###  2.4. An integration by parts trick for the inverse problem

Let us define the function

$W_{f,h}(t,s)=(u^{f}(t,\cdot),u^{h}(s,\cdot))_{L^{2}(0,1)}.$

Then the following holds

######  Lemma 2.

Let $f$, $h\in C_{0}^{\infty}(\mathbb{R}^{+})$. The operator $\Lambda$
determines $W_{f,h}(t,t)$ for all $t>0$.

###### Proof.

Since $u^{f}$ and $u^{h}$ are solutions of (1), there holds

$(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=\left(\partial_{x}^{2}u^{f}(t,\cdot)-qu^{f}(t,\cdot),u^{h}(s,\cdot)\right)_{L^{2}(0,1)}\\\
-\left(u^{f}(t,\cdot),\partial_{x}^{2}u^{h}(s,\cdot)-qu^{h}(s,\cdot)\right)_{L^{2}(0,1)}.$

The two terms with $q$ cancel out. Further, integration by parts gives

$(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=\partial_{x}u^{f}(t,1)u^{h}(s,1)-\partial_{x}u^{f}(t,0)u^{h}(s,0)\\\
-u^{f}(t,1)\partial_{x}u^{h}(s,1)+u^{f}(t,0)\partial_{x}u^{h}(s,0).$

Boundary conditions for $u^{f}$ and $u^{h}$ give

(7) $(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=f(t)\Lambda h(s)-\Lambda
f(t)h(s).$

Let us denote the right-hand side of (7) by $F(t,s)$, then $W_{f,h}$ solves

$\begin{cases}(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=F(t,s)&\text{on
}\mathbb{R}^{+}\times\mathbb{R}^{+},\\\
W_{f,h}(0,s)=\partial_{t}W_{f,h}(0,s)=0.\end{cases}$

Hence, $W_{f,h}$ is determined by $F$, and consequently, it is determined by
$\Lambda$. (When evaluating $W_{f,h}(t,t)$, we do not need to impose any
boundary condition at $s=0$ due to finite speed of propagation. Nonetheless,
we could impose, for example, $W_{f,h}(t,0)=0$ as this holds in view of the
initial conditions satisfied by $u^{h}(s,x)$.) ∎

Consider two potentials $q_{1}$ and $q_{2}$. Let us write $u_{1}^{f}$ and
$u_{2}^{f}$ for the solutions of (1) with $q$ replaced by $q_{1}$ and $q_{2}$,
respectively. The corresponding Dirichlet-to-Neumann operators, see (2), are
denoted by $\Lambda_{1}$ and $\Lambda_{2}$. It follows from the above lemma
that

$\displaystyle(u_{1}^{f}(t,\cdot),u_{1}^{h}(t,\cdot))_{L^{2}(0,1)}=(u_{2}^{f}(t,\cdot),u_{2}^{h}(t,\cdot))_{L^{2}(0,1)}$

whenever $\Lambda_{1}=\Lambda_{2}$.

###  2.5. Solution to the inverse problem

We will show that $\Lambda_{1}=\Lambda_{2}$ implies $q_{1}=q_{2}$. For a set
$S\subset\mathbb{R}$, we define the indicator function

$\displaystyle 1_{S}(x)=\begin{cases}1&x\in S,\\\
0&\text{otherwise}.\end{cases}$

######  Lemma 3.

Assume that $\Lambda_{1}=\Lambda_{2}$. Let $0<s<T$, let $f\in
C_{0}^{\infty}(\mathbb{R}^{+})$ and let

$\displaystyle f_{j}\in C_{0}^{\infty}(T-s,T),\quad j=1,2,\dots,$

be a sequence such that

(8) $u_{1}^{f_{j}}(T,\cdot)\rightarrow
1_{(0,s)}u_{1}^{f}(T,\cdot)\qquad\text{in }L^{2}(0,1).$

Then

$u_{2}^{f_{j}}(T,\cdot)\rightarrow 1_{(0,s)}u_{2}^{f}(T,\cdot)\qquad\text{in
}L^{2}(0,1).$

###### Proof.

Let $\tilde{f}\in C_{0}^{\infty}(T-s,T)$ and let $k=1,2$. We compute

$\displaystyle\left\|u_{k}^{\tilde{f}}(T,\cdot)-u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}=\left\|u_{k}^{\tilde{f}}(T,\cdot)-1_{(0,s)}u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}$
(9)
$\displaystyle\qquad+\left\|\left(1_{(0,s)}-1\right)u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}$
$\displaystyle\qquad+2\left(u_{k}^{\tilde{f}}(T,\cdot)-1_{(0,s)}u_{k}^{f}(T,\cdot),\left(1_{(0,s)}-1\right)u_{k}^{f}(T,\cdot)\right)_{L^{2}(0,1)}.$

Due to finite speed of propagation, we know that $u_{k}^{\tilde{f}}(T,\cdot)$
is supported in $(0,s)$. Therefore, the functions

$u_{k}^{\tilde{f}}(T,\cdot)-1_{(0,s)}u_{k}^{f}(T,\cdot)\qquad\left(1_{(0,s)}-1\right)u_{k}^{f}(T,\cdot)$

have disjoint supports, so that (2.5) becomes

(10)
$\left\|u_{k}^{\tilde{f}}(T,\cdot)-u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}=\left\|u_{k}^{\tilde{f}}(T,\cdot)-1_{(0,s)}u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}\\\
+\left\|\left(1_{(0,s)}-1\right)u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}.$

By approximate controllability,

(11) $\displaystyle\inf_{\tilde{f}\in
C_{0}^{\infty}(T-s,T)}\left\|u_{k}^{\tilde{f}}(T,\cdot)-u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}=\left\|\left(1_{(0,s)}-1\right)u_{k}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}.$

Let $f_{j}\in C_{0}^{\infty}(T-s,T)$ satisfy (8). Then

$\lim_{j\rightarrow\infty}\left\|u_{1}^{f_{j}}(T,\cdot)-u_{1}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}=\inf_{\tilde{f}\in
C_{0}^{\infty}(T-s,T)}\left\|u_{1}^{\tilde{f}}(T,\cdot)-u_{1}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}.$

Due to Lemma 2,

$\lim_{j\rightarrow\infty}\left\|u_{2}^{f_{j}}(T,\cdot)-u_{2}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}=\inf_{\tilde{f}\in
C_{0}^{\infty}(T-s,T)}\left\|u_{2}^{\tilde{f}}(T,\cdot)-u_{2}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}.$

Using (10) and (11) we see that

$\displaystyle\left\|u_{2}^{f_{j}}(T,\cdot)-1_{(0,s)}u_{2}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}$
$\displaystyle\quad=\left\|u_{2}^{f_{j}}(T,\cdot)-u_{2}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}-\left\|\left(1_{(0,s)}-1\right)u_{2}^{f}(T,\cdot)\right\|_{L^{2}(0,1)}^{2}\to
0.$

∎

######  Corollary 1.

Assume that $\Lambda_{1}=\Lambda_{2}$. Let $f,h\in
C_{0}^{\infty}(\mathbb{R}^{+})$ and let $0<s<T$. Then

(12)
$\displaystyle\left(1_{(0,s)}u_{1}^{f}(T,\cdot),u_{1}^{h}(T,\cdot)\right)_{L^{2}(0,1)}=\left(1_{(0,s)}u_{2}^{f}(T,\cdot),u_{2}^{h}(T,\cdot)\right)_{L^{2}(0,1)}.$

###### Proof.

By Lemma 1, there is a sequence $\\{f_{j}\\}\subset C_{0}^{\infty}(T-s,T)$
such that

$u_{k}^{f_{j}}(T,\cdot)\rightarrow 1_{(0,s)}u_{k}^{f}(T,\cdot)$

in $L^{2}(0,1)$ for $k=1$. By Lemma 3 this holds also for $k=2$. In view of
Lemma 2, we have

$\left(1_{(0,s)}u_{1}^{f}(T,\cdot),u_{1}^{h}(T,\cdot)\right)_{L^{2}(0,1)}=\lim_{j\rightarrow\infty}\left(u_{1}^{f_{j}}(T,\cdot),u_{1}^{h}(T,\cdot)\right)_{L^{2}(0,1)}\\\
=\lim_{j\rightarrow\infty}\left(u_{2}^{f_{j}}(T,\cdot),u_{2}^{h}(T,\cdot)\right)_{L^{2}(0,1)}=\left(1_{(0,s)}u_{2}^{f}(T,\cdot),u_{2}^{h}(T,\cdot)\right)_{L^{2}(0,1)}.$

∎

We will need the following lemma. Its proof is given in Section 4.

######  Lemma 4.

Let $T\geq 1$, then for any $x_{0}\in(0,1)$ there is $f\in
C_{0}^{\infty}(0,T)$ such that $u^{f}(T,x_{0})\neq 0$.

Now we are ready to prove the main result of this section

######  Theorem 5.

If $\Lambda_{1}=\Lambda_{2}$, then $q_{1}=q_{2}$.

###### Proof.

We consider $T>1$ and let $x\in(0,1)$. Let us differentiate the left-hand side
of (12):

$\partial_{x}\left(1_{(0,x)}(\cdot)u_{1}^{f}(T,\cdot),u_{1}^{h}(T,\cdot)\right)_{L^{2}(0,1)}\\\
=\partial_{x}\int_{0}^{x}u_{1}^{f}(T,y)u_{1}^{h}(T,y)dy=u_{1}^{f}(T,x)u_{1}^{h}(T,x).$

Therefore, by (12), we obtain

(13) $u_{1}^{f}(T,x)u_{1}^{h}(T,x)=u_{2}^{f}(T,x)u_{2}^{h}(T,x).$

Due to Lemma 4, for each $x\in(0,1)$ there is $f\in
C_{0}^{\infty}(\mathbb{R}^{+})$ such that $u_{1}^{f}(T,x)\neq 0$. Choosing
such $f$ we may define

$\displaystyle w(x)=\frac{u_{2}^{f}(T,x)}{u_{1}^{f}(T,x)}.$

We emphasize that the choice of $f$ depends on $x$, and it may appear that $w$
could be non-smooth. However, it is smooth. Indeed,

(14) $\displaystyle u_{1}^{h}(T,x)=w(x)u_{2}^{h}(T,x),$

and for each $x_{0}\in(0,1)$ there is $h\in C_{0}^{\infty}(\mathbb{R}^{+})$
such that $u_{2}^{h}(T,x_{0})\neq 0$. Thus, for $x$ near $x_{0}$,

$\displaystyle w(x)=\frac{u_{1}^{h}(T,x)}{u_{2}^{h}(T,x)}.$

As the right-hand side is smooth for $x$ near $x_{0}$, and as $x_{0}\in(0,1)$
is arbitrary, we see that $w$ is smooth.

Let us now take $f=h$ in (13) and use (14),

$\displaystyle(u_{2}^{h}(T,x))^{2}=(u_{1}^{h}(T,x))^{2}=w^{2}(x)(u_{2}^{h}(T,x))^{2}.$

Choosing again $x\in(0,1)$ and $h\in C_{0}^{\infty}(\mathbb{R}^{+})$ such that
$u_{2}^{h}(T,x)\neq 0$, we see that $w^{2}(x)=1$. The smoothness of $w$
implies that it is a constant function taking the value $1$ or $-1$.

To summarize

$\displaystyle u_{1}^{h}(T,x)=\pm u_{2}^{h}(T,x),$

for all $T>1$, $x\in(0,1)$ and $h\in C_{0}^{\infty}(\mathbb{R}^{+})$.

There holds

$\displaystyle
0=(\partial_{t}^{2}-\partial_{x}^{2}+q_{1})u_{1}^{h}=\pm(\partial_{t}^{2}-\partial_{x}^{2}+q_{1})u_{2}^{h}=\pm(q_{1}-q_{2})u_{2}^{h}.$

Choosing such $h\in C^{\infty}_{0}(\mathbb{R}^{+})$ that $u_{2}^{h}(T,x)\neq
0$, we get $q_{1}(x)=q_{2}(x)$. ∎

##  3\. Boundary Control method on a Riemannian manifold

The main advantage of the Boundary Control method is that it works in general
geometric settings. We refer to [16] for an introduction to Riemannian
geometry. A reader, who is not interesting in geometry, may simply replace
$(M,g)$ by a domain $M\subset\mathbb{R}^{n}$ with a smooth boundary $\partial
M$.

Let $(M,g)$ be a Riemannian manifold with boundary and $q\in C^{\infty}(M)$.
Consider the wave equation

(15) $\displaystyle\begin{cases}\partial_{t}^{2}u-\Delta_{g}u+qu=0&\text{on
}\mathbb{R}^{+}\times M,\\\ \left.u\right|_{x\in\partial M}=f,\\\
\left.u\right|_{t=0}=\left.\partial_{t}u\right|_{t=0}=0.\end{cases}$

Here $\Delta_{g}$ is the Laplacian. In local coordinates, it is given as
follows

$\Delta_{g}u=\frac{1}{\sqrt{\textrm{det}(g)}}\sum_{i,j=1}^{n}\partial_{i}\left(\sqrt{\textrm{det}(g)}g^{ij}\partial_{j}u\right).$

In particular, in the case that $M\subset\mathbb{R}^{n}$, the above expression
gives the usual Laplacian, since $g$ is the identity matrix.

We define the Dirichlet-to-Neumann map

$\displaystyle\Lambda:C_{0}^{\infty}(\mathbb{R}^{+}\times\partial
M)\rightarrow C^{\infty}(\mathbb{R}^{+}\times\partial M)$

as follows

$\Lambda f=\left.\partial_{\nu}u^{f}\right|_{\mathbb{R}^{+}\times\partial M},$

where $u^{f}$ is the solution in of (15). Here $\nu$ is the exterior unit
normal on $\partial M$. We will study the inverse problem: determine the
potential $q$ given $\Lambda$ and the Riemannian manifold $(M,g)$.

The finite speed of propagation will be formulated in terms of the distance
function $d_{g}$ on $(M,g)$. That is, it takes the time $d_{g}(x,y)$ for a
wave to propagate from a point $x\in M$ to a point $y\in M$. We recall the
definition of the distance on Riemannian manifold.

If $(M,g)$ is is a connected Riemannian manifold and $x,y\in M$, the distance
between $x$ and $y$ is defined by

$d_{g}(x,y)=\inf\\{l(\gamma):\gamma\in C_{x,y}\\},$

where

$\displaystyle C_{x,y}=\\{\gamma:[0,\ell]\rightarrow M:\gamma$ is a piecewise
smooth, continuous curve and $\displaystyle\gamma(0)=x,\;\gamma(\ell)=y\\}$

and

$l(\gamma)=\int_{0}^{\ell}\sqrt{g(\dot{\gamma}(t),\dot{\gamma}(t))}dt.$

Here $g$ is viewed as a bilinear form. When $M\subset\mathbb{R}^{n}$ is
convex, $d_{g}$ is the usual Euclidean distance.

Before entering into the multidimensional geometric case, let us see how non-
constant speed of sound gives the speed of propagation in $1+1$ dimensions.

###  3.1. Finite speed of propagation with nonconstant speed of sound

Let $c\in C^{\infty}([0,1])$ and suppose that $c(x)>0$ for all $x\in[0,1]$.
Consider a solution $u$ to

(16) $\displaystyle(\partial_{t}^{2}-c^{2}\partial_{x}^{2})u=0\quad\text{on
}\mathbb{R}^{+}\times(0,1),$

which satisfies

$\left.u\right|_{x=1}=0.$

We set

$\mathcal{E}(t,x)=c^{-2}(x)|\partial_{t}u(t,x)|^{2}+|\partial_{x}u(t,x)|^{2}.$

We want to find an increasing function $r\in C^{\infty}(\mathbb{R})$ with
$r(0)=0$ such that the energy

$E(t)=\frac{1}{2}\int_{r(t)}^{1}\mathcal{E}(t,x)dx$

satisfies

(17) $\partial_{t}E(t)\leq 0.$

The slower $r$ increases, the better finite speed of propagation result we
will get.

We write, by using Leibniz integral rule,

$\displaystyle\partial_{t}E(t)$
$\displaystyle=-\frac{1}{2}r^{\prime}(t)\mathcal{E}(t,r(t))+\frac{1}{2}\int_{r(t)}^{1}\partial_{t}\mathcal{E}(t,x)dx$
$\displaystyle=-\frac{1}{2}r^{\prime}(t)\mathcal{E}(t,r(t))+\int_{r(t)}^{1}\left(c^{-2}(x)\partial_{t}u(t,x)\partial_{t}^{2}u(t,x)+\partial_{x}u(t,x)\partial_{tx}u(t,x)\right)dx.$

Integration by parts gives

$\partial_{t}E(t)=-\frac{1}{2}r^{\prime}(t)\mathcal{E}(t,r(t))+\left[\partial_{x}u(t,x)\partial_{t}u(t,x)\right]^{1}_{r(t)}\\\
+\int_{r(t)}^{1}\left(c^{-2}\partial_{t}^{2}u(t,x)+\partial_{x}^{2}u(t,x)\right)\partial_{t}u(t,x)dx.$

Since $u$ is the solution of the wave equation the last integral is $0$.
Moreover, due to the boundary condition, the second term is $0$ at $x=1$, so
that

$\partial_{t}E(t)=-\frac{1}{2}r^{\prime}(t)\mathcal{E}(t,r(t))-\left.\left(\partial_{t}u(t,x)\partial_{x}u(t,x)\right)\right|_{x=r(t)}.$

Note that $r^{\prime}>0$ since $r$ is increasing. Therefore, using a simple
inequality

$2xy\leq\alpha x^{2}+\frac{1}{\alpha}y^{2},\qquad\text{for }\alpha,x,y>0,$

we know that (17) holds if

$\frac{1}{r^{\prime}(t)}=r^{\prime}(t)c^{-2}(r(t)),$

or equivalently,

(18) $r^{\prime}(t)=c(r(t)).$

This equation is solvable. Indeed, consider the following function

$\rho(x)=\int_{0}^{x}\frac{1}{c(y)}dy.$

Since $\rho$ is a strictly increasing function, its inverse function exists,
so we can set

$r(t)=\rho^{-1}(t).$

It is easy to check that this function indeed satisfies (18). Observe that if
$c=1$ identically, we obtain $r(t)=t$.

The function $\rho$ is the travel time between $0$ and $x$: the time necessary
for perturbation at $0$ to reach $x$. It also can be interpreted as the
distance from $0$ to $x$. Indeed, let us consider the metric

$g=c^{-2}dx,$

on $[0,1]$. Then, for $x\in[0,1]$, we get

$d_{g}(0,x)=\inf_{\gamma\in
C_{0,x}}\int_{0}^{b}\sqrt{g_{\gamma(t)}(\dot{\gamma}(t),\dot{\gamma}(t))}dt.$

Let $\gamma$ be a minimizer curve. Then $\gamma:[0,b]\rightarrow[0,x]$ needs
to be a bijection. By changing the coordinates $\tau=\gamma(t)$, we obtain

$d_{g}(0,x)=\int_{0}^{b}\dot{\gamma}(t)\frac{1}{|c(\gamma(t))|}dt=\int_{0}^{x}\frac{1}{|c(\tau)|}d\tau=\rho(x).$

###  3.2. Main tools

As in the one dimensional case, the main ingredients of solving the inverse
problem we are considering here are the finite speed of propagation and unique
continuation.

######  Theorem 6 (Finite speed of propagation).

Let $\Omega\subset M$ be open, and define

$\mathcal{C}=\\{(t,x)\in\mathbb{R}\times M:d_{g}(x,M\setminus\Omega)>|t|\\}.$

Suppose that $u\in C^{2}(\mathbb{R}\times M)$ is a solution of

$\begin{cases}\left(\partial_{t}^{2}-\Delta_{g}+q(x)\right)u=0,&\text{on
}\mathbb{R}\times M;\\\
\left.u\right|_{t=0}=\left.\partial_{t}u\right|_{t=0}=0,&\text{on
}\Omega.\end{cases}$

Suppose, furthermore, that
$\left.u\right|_{\mathcal{C}\cap(\mathbb{R}\times\partial M)}=0$. Then
$\left.u\right|_{\mathcal{C}}=0$.

We omit the proof and refer to [11, Theorem 2.47].

######  Theorem 7 (Unique continuation).

Let $\Gamma\subset\partial M$ be open, and define

$K=\\{(t,x)\in\mathbb{R}\times M:d_{g}(x,\Gamma)\leq T-|t|\\}.$

Suppose that $u\in C^{2}(\mathbb{R}\times M)$ is a solution of

$\begin{cases}\left(\partial_{t}^{2}-\Delta_{g}+q(x)\right)u=0,&\text{on
}\mathbb{R}\times M;\\\
\left.u\right|_{[-T,T]\times\Gamma}=\left.\partial_{\nu}u\right|_{[-T,T]\times\Gamma}=0.\end{cases}$

Then $\left.u\right|_{K}=0$.

We omit the proof and refer to [11, Theorem 3.16]. An introduction to the
concept of unique continuation for the wave operator and its applications can
be found in [15]. Applying Theorem 7 on $M\setminus\Omega$, where
$\Omega\subset M$ is a small open set with smooth boundary satisfying
$\partial\Omega\cap\partial M=\emptyset$, allows us to compare unique
continuation to finite speed of propagation, see Figure 2. Clearly these two
results are genuinely different in the higher dimensional case, in contrast to
the one dimensional case. (Recall that in the dimension one, finite speed of
propagation and unique continuation differ only by interchanging space and
time.)

(a) Finite speed of the wave propagation (b) Unique continuation

Figure 2. $n+1$ dimensional case

Let $\Gamma\subset\partial M$ be an open set. We define the domain of
influence

$M(\Gamma,T)=\\{x\in M:d_{g}(x,\Gamma)\leq T\\}$

and use the identification

$L^{2}(M(\Gamma,T))=\\{\phi\in L^{2}(M):\operatorname{supp}(\phi)\subset
M(\Gamma,T)\\}.$

Similarly, as we derived Lemma 1 from Theorem 4, Theorem 7 gives us the
following approximate controlability:

######  Lemma 5.

Let $T>0$, then the set

$\mathcal{B}(T,\Gamma)=\\{u^{f}(T,\cdot):f\in
C_{0}^{\infty}((0,T)\times\Gamma)\\}$

is a dense subset of $L^{2}(M(\Gamma,T))$.

Lemma 4 can be generalized for the multidimensional case:

######  Lemma 6.

Let $T>0$, then for any point $x_{0}$ in the interior of $M(\Gamma,T)$ there
is $f\in C_{0}^{\infty}((0,T)\times\Gamma)$ such that $u^{f}(T,x_{0})\neq 0$.

###  3.3. Solution to the inverse problem

Let $T>0$. For functions $f$, $h\in
C_{0}^{\infty}(\mathbb{R}^{+}\times\partial M)$, we define

$W_{f,h}(t,s)=(u^{f}(t,\cdot),u^{h}(s,\cdot))_{L^{2}(M)}.$

The following lemma is the higher dimensional analogue of Lemma 2.

######  Lemma 7.

Let $f$, $h\in C_{0}^{\infty}(\mathbb{R}^{+}\times\partial M)$. The operator
$\Lambda$ determines $W_{f,h}(t,t)$ for all $t>0$.

###### Proof.

We write

$(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=\left(\Delta_{g}u^{f}(t,\cdot)-qu^{f}(t,\cdot),u^{h}(s,\cdot)\right)_{L^{2}(M)}\\\
-\left(u^{f}(t,\cdot),\Delta_{g}u^{h}(s,\cdot)-qu^{h}(s,\cdot)\right)_{L^{2}(M)}.$

Further, by Green’s identity, we obtain

$(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=\int_{\partial
M}\partial_{\nu}u^{f}(t,x)u^{h}(s,x)dS(x)-\int_{\partial
M}u^{f}(t,x)\partial_{\nu}u^{h}(s,x)dS(x)\\\ =\left(\Lambda
f(t,\cdot),h(s,\cdot)\right)_{L^{2}(\partial M)}-\left(f(t,\cdot),\Lambda
h(s,\cdot)\right)_{L^{2}(\partial M)}.$

Let us denote the right-hand side by $F(t,s)$, then $W_{f,h}$ is the solution
of the equation

$\begin{cases}(\partial_{t}^{2}-\partial_{s}^{2})W_{f,h}(t,s)=F(t,s)&\text{on
}(0,T)\times(0,T),\\\ W_{f,h}(0,s)=\partial_{t}W_{f,h}(0,s)=0.\end{cases}$

Hence, $W_{f,h}$ is determined by $F$, and consequently, it is determined by
$\Lambda$. ∎

We have the analogue of Lemma 3.

######  Lemma 8.

Assume that $\Lambda_{1}=\Lambda_{2}$. Let $0<s<T$, let $\Gamma\subset\partial
M$ be open and let $\\{f_{j}\\}\subset C_{0}^{\infty}((T-s,T)\times\Gamma)$ be
a sequence such that

$u_{1}^{f_{j}}(T,\cdot)\rightarrow
1_{M(\Gamma,s)}u_{1}^{f}(T,\cdot)\qquad\text{in }L^{2}(M).$

Then

$u_{2}^{f_{j}}(T,\cdot)\rightarrow
1_{M(\Gamma,s)}u_{2}^{f}(T,\cdot)\qquad\text{in }L^{2}(M).$

We omit the proof as it coincides with that in the one dimensional case. As
before, we have also the following corollary

######  Corollary 2.

Assume that $\Lambda_{1}=\Lambda_{2}$. Let $0<s<T$, $t>0$ and let
$\Gamma\subset\partial M$ be open, then for any $f$, $h\in
C_{0}^{\infty}(\mathbb{R}^{+}\times\partial M)$, it follows

$\left(1_{M(\Gamma,s)}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}=\left(1_{M(\Gamma,s)}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}.$

Now we deviate from the one dimensional proof.

######  Corollary 3.

Assume that $\Lambda_{1}=\Lambda_{2}$. Let $s,\tilde{s}\in(0,T)$, let $t>0$
and consider two open $\Gamma$, $\tilde{\Gamma}\subset\partial M$. Then we
have for any $f$, $h\in C_{0}^{\infty}(\mathbb{R}^{+}\times\partial M)$

$\left(1_{M(\Gamma,s)}1_{M(\tilde{\Gamma},\tilde{s})}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}\\\
=\left(1_{M(\Gamma,s)}1_{M(\tilde{\Gamma},\tilde{s})}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}.$

###### Proof.

By Lemmas 5 and 8, there is a sequence $\\{f_{j}\\}\subset
C_{0}^{\infty}((T-s,T)\times\Gamma)$ such that

$u_{1}^{f_{j}}(T,\cdot)\rightarrow
1_{M(\Gamma,s)}u_{1}^{f}(T,\cdot)\quad\text{and}\quad
u_{2}^{f_{j}}(T,\cdot)\rightarrow 1_{M(\Gamma,s)}u_{2}^{f}(T,\cdot).$

Therefore, using Corollary 2, we obtain

$\displaystyle\left(1_{M(\Gamma,s)}1_{M(\tilde{\Gamma},\tilde{s})}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}$
$\displaystyle\quad=\lim_{j\rightarrow\infty}\left(1_{M(\tilde{\Gamma},\tilde{s})}u_{1}^{f_{j}}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}$
$\displaystyle\quad=\lim_{j\rightarrow\infty}\left(1_{M(\tilde{\Gamma},\tilde{s})}u_{2}^{f_{j}}(T,\cdot),u_{2}^{h}(t,\cdot)\right)_{L^{2}(M)}$
$\displaystyle\quad=\left(1_{M(\Gamma,s)}1_{M(\tilde{\Gamma},\tilde{s})}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}.$

∎

######  Theorem 8.

If $\Lambda_{1}=\Lambda_{2}$, then $q_{1}=q_{2}$.

###### Proof.

Let $x_{0}$ be an interior point of $M$ and write $s=d_{g}(x_{0},\partial M)$.
We choose $T>s$. Let $y\in\partial M$ be such that

$d_{g}(x,y)=s.$

Let $\Gamma\subset\partial M$ be a neighbourhood of $y$ and take
$\tilde{\Gamma}=\partial M$. Let also $0<\tilde{s}<s$ and set

$Z=M(\Gamma,s)\setminus M(\tilde{\Gamma},\tilde{s}).$

Then,

$\frac{1}{|Z|}\left(\left(1_{M(\Gamma,s)}u_{1}^{f}(T,\cdot),u_{1}^{h}(t,\cdot)\right)_{L^{2}(M)}-\left(1_{M(\tilde{\Gamma},\tilde{s})}1_{M(\Gamma,s)}u_{1}^{f}(T,\cdot),u_{1}^{h}(T,\cdot)\right)_{L^{2}(M)}\right)\\\
\rightarrow u_{1}^{f}(T,x_{0})u_{1}^{h}(T,x_{0})$

as $\tilde{s}\rightarrow s$ and $\Gamma\to\\{y\\}$. The same holds for
$u_{2}^{f}$ and $u_{2}^{h}$, so that by Corollaries 2 and 3, we know that

$u_{1}^{f}(T,x_{0})u_{1}^{h}(t,x_{0})=u_{2}^{f}(T,x_{0})u_{2}^{h}(t,x_{0}).$

This is the analogue of (13), and we conclude as in the proof of Theorem 5. ∎

##  4\. Geometric optics

In this section, we will construct solutions to

(19) $\displaystyle\partial_{t}^{2}u-\Delta u+qu=0,\quad\text{in
$(0,T)\times\mathbb{R}^{n}$},$

that concentrate on light rays, that is, lines of the form

$\beta(s)=(s,y+sv),\quad s\in\mathbb{R},$

where $y$ is a point in $\mathbb{R}^{n}$ and $v$ is a unit vector in
$\mathbb{R}^{n}$. We write

$S^{n-1}=\\{v\in\mathbb{R}^{n};\ |v|=1\\}$

for the set of unit vectors, that is, for the unit sphere. The name light ray
comes from the fact that the tangent vector $\dot{\beta}=(1,v)$ is light like
with respect to the Minkowski metric

(20) $g=\begin{pmatrix}-1\\\ &1\\\ &&\ddots\\\ &&&1\end{pmatrix},$

that is, $g(\dot{\beta},\dot{\beta})=0$.

The idea is to find first an approximate solution of the form

$e^{i\sigma\phi(t,x)}(a_{0}(t,x)+\sigma^{-1}a_{1}(t,x)+\sigma^{-2}a_{2}(t,x)+\dots),$

where $\sigma>0$ is a large parameter, and then an actual solution

$u=e^{i\sigma\phi}(a_{0}+\dots)+r_{\sigma},$

where the remainder $r_{\sigma}$ converges to zero as $\sigma\to\infty$. We
will begin with the single term approximation $e^{i\sigma\phi}a_{0}$ and write
$a_{0}=a$.

###  4.1. Single term ansatz

To simplify the notation, we write $\Box=\partial_{t}^{2}-\Delta$. The
equation $(\Box+q)u=0$ is equivalent with

(21) $(\Box+q)r_{\sigma}=-(\Box+q)(e^{i\sigma\phi}a),$

and we want to choose $\phi$ and $a$ so that

(“C”) $\displaystyle\Box(e^{i\sigma\phi}a)=e^{i\sigma\phi}\Box a.$

The rationale is that in this case the absolute value of the right-hand side
of (21) is independent of $\sigma$, and therefore $r_{\sigma}$ is at least not
blowing up as $\sigma\to\infty$.

It is a simple matter to expand the left-hand side of (“C”) but a useful
computational technique is to consider the conjugated wave operator

$e^{-i\sigma\phi}\Box
e^{i\sigma\phi}=e^{-i\sigma\phi}\partial_{t}^{2}e^{i\sigma\phi}+\dots=e^{-i\sigma\phi}\partial_{t}e^{i\sigma\phi}e^{-i\sigma\phi}\partial_{t}e^{i\sigma\phi}+\dots.$

Now
$e^{-i\sigma\phi}\partial_{t}e^{i\sigma\phi}=\partial_{t}+i\sigma(\partial_{t}\phi)$
and

$(e^{-i\sigma\phi}\partial_{t}e^{i\sigma\phi})^{2}=\partial_{t}^{2}+2i\sigma(\partial_{t}\phi)\partial_{t}-\sigma^{2}|\partial_{t}\phi|^{2}+i\sigma(\partial_{t}^{2}\phi).$

Treating the spacial derivatives in the same way we get

(22) $\displaystyle e^{-i\sigma\phi}\Box
e^{i\sigma\phi}=\Box+i\sigma(2(\partial_{t}\phi)\partial_{t}-2(\nabla\phi)\cdot\nabla+\Box\phi)-\sigma^{2}(|\partial_{t}\phi|^{2}-|\nabla\phi|^{2}).$

Therefore for $a\neq 0$, (“C”) is equivalent with the following two equations

(E) $\displaystyle|\partial_{t}\phi|^{2}-|\nabla\phi|^{2}=0,$ (T)
$\displaystyle 2(\partial_{t}\phi)\partial_{t}a-2(\nabla\phi)\cdot\nabla
a+(\Box\phi)a=0.$

It is natural to normalize $\phi$ so that (E) becomes
$|\partial_{t}\phi|^{2}=|\nabla\phi|^{2}=1$. There is some freedom when
choosing a solution to (E), but for our purposes it suffices to use the linear
solution $\phi(t,x)=t+v\cdot x$ where $v\in S^{n-1}$.

To simplify the notation, we may assume after a rotation that $v\cdot
x=-x^{1}$. The functions satisfying $|\nabla\phi|=1$ are often called distance
functions, and the particular choice $x^{1}$ is of course the signed distance
to the plane $x^{1}=0$.

The transport equation (T) simplifies now to

$\partial_{t}a+\partial_{x^{1}}a=0.$

The solutions to this are of the form

$a(t,x)=\chi(t-x^{1})\eta(x^{\prime})$

where $x^{\prime}=(x^{2},x^{3},\dots,x^{n})$. Taking $\chi\approx\delta$ and
$\eta\approx\delta$ we obtain $a$ that concentrates on the light ray
$\beta(s)=(s,s,0)$.

Neither $\phi$ nor $a$ depend on $q$ in the above construction. In order to
obtain information on $q$, in the context of inverse problems, there are two
typical approaches: use the difference of two solutions, corresponding to
different potentials, or use a multi-term approximation. We will take the
latter approach.

###  4.2. Multi-term ansatz

Let us consider the ansatz,

$e^{i\sigma\phi}A,\quad A=a_{0}+\sigma^{-1}a_{1}+\dots+\sigma^{-N}a_{N},$

and choose $\phi$ and $a_{0}=a$ as above. As we are using a more complicated
amplitude, we can ask for more than (“C”), namely

$\displaystyle(\Box+q)(e^{i\sigma\phi}A)=\mathcal{O}(\sigma^{-N}),\quad\sigma\gg
1.$

We use the conjugation formula (22), to obtain

$e^{-i\sigma\phi}(\Box+q)e^{i\sigma\phi}A=(\Box+q)A+2i(\partial_{t}+\partial_{x^{1}})(a_{1}+\dots+\sigma^{-N+1}a_{N}).$

This is of order $\sigma^{-N}$ whenever $a_{1},\dots,a_{N}$ solve the
transport equations

$\partial_{t}a_{j}+\partial_{x^{1}}a_{j}-\frac{i}{2}(\Box+q)a_{j-1}=0,\quad
j=1,\dots,N,$

or after the change of variables,

$s=\frac{t+x^{1}}{2},\quad r=\frac{t-x^{1}}{2},$

equivalently $\partial_{s}a_{j}=\frac{i}{2}(\Box+q)a_{j-1}$. Therefore we may
choose

(23) $\displaystyle
a_{j}(s,r,x^{\prime})=\frac{i}{2}\int_{-r}^{s}(\Box+q)a_{j-1}(s^{\prime},r,x^{\prime})ds^{\prime}.$

Note that $t=0$ is equivalent with $s=-r$. The choice of the lower limit $-r$
in the integration implies that $a_{j}=0$, $j=1,\dots,N$, when $t=0$.

###  4.3. Solving for the remainder

When $\chi\in C_{0}^{\infty}(\mathbb{R})$ and $\eta\in
C_{0}^{\infty}(\mathbb{R}^{n-1})$, the restrictions of all the amplitudes
$a_{j}$, $j=0,1,\dots,N$, are compactly supported in
$[0,T]\times\mathbb{R}^{n}$. We recall that the wave equation

$\displaystyle\begin{cases}\Box u+qu=F&\text{in
$(0,T)\times\mathbb{R}^{n}$},\\\ u|_{t=0}=\partial_{t}u|_{t=0}=0\end{cases}$

has a unique solution $u$ satisfying

$\lVert u\rVert_{C(0,T;H^{1}(\mathbb{R}^{n}))}+\lVert
u\rVert_{C^{1}(0,T;L^{2}(\mathbb{R}^{n}))}\leq C\lVert
F\rVert_{L^{2}((0,T)\times\mathbb{R}^{n})},$

see e.g. [6, Theorem 7.6]. We solve

$\displaystyle\begin{cases}\Box
r_{\sigma}+qr_{\sigma}=-(\Box+q)(e^{i\sigma\phi}A)&\text{in
$(0,T)\times\mathbb{R}^{n}$},\\\
r_{\sigma}|_{t=0}=\partial_{t}r_{\sigma}|_{t=0}=0.\end{cases}$

As the right-hand side is pointwise of order $\sigma^{-N}$ and compactly
supported, we see that $r_{\sigma}|_{t=T}=\mathcal{O}(\sigma^{-N})$ in
$H^{1}(\mathbb{R}^{n})$.

###  4.4. Proof of Lemma 4

If $\chi(0)\neq 0$ and $\eta(0)\neq 0$ then $a$ does not vanish anywhere on
the line given by $t=x^{1}$. For large enough $\sigma>0$, the same is true for
$e^{i\sigma\phi}A$ when restricted on $[0,T]\times\mathbb{R}^{n}$.

Let us now consider the case $n=1$. The Sobolev embedding
$H^{1}(\mathbb{R})\subset C(\mathbb{R})$, see e.g. [3, Theorem 8.8], implies
that $r_{\sigma}|_{t=T}=\mathcal{O}(\sigma^{-N})$ in $C(\mathbb{R})$.
Therefore $u(T,T)\neq 0$ for large $\sigma>0$. Lemma 4 follows after a
suitable translation in time.

Lemma 6 can be shown in a similar manner, but higher regularity energy
estimates, see e.g. [6, Theorem 6, p. 412], are needed to bound
$r_{\sigma}|_{t=T}$ in $H^{k}(\mathbb{R}^{n})$ with $k$ large enough so that
$H^{k}(\mathbb{R}^{n})\subset C(\mathbb{R}^{n})$.

##  5\. Solving an inverse problem using geometric optics

We will consider the inverse problem to determine a compactly supported
potential in a slab in the Minkowski space. This avoids some technicalities
appearing in more usual inverse boundary value problems, but it still allows
us to introduce the main techniques.

Let $n\geq 2$, $T>0$, $q\in C_{0}^{\infty}((0,T)\times\mathbb{R}^{n})$, and
consider the wave equation

$\displaystyle\begin{cases}\Box u+qu=0&\text{in
$(0,T)\times\mathbb{R}^{n}$},\\\ u|_{t=0}=u_{0},\
\partial_{t}u|_{t=0}=u_{1}.\end{cases}$

Define also the map

$L:(C_{0}^{\infty}(\mathbb{R}^{n}))^{2}\to
C_{0}^{\infty}(\mathbb{R}^{n}),\quad L(u_{0},u_{1})=u|_{t=T}.$

We will study how to solve the inverse problem to determine $q$ given $L$.

###  5.1. Reduction to the light ray transform

The function $u=e^{i\sigma\phi}A+r_{\sigma}$ constructed in Section 4 solves
(19) with

$u_{0}=(e^{i\sigma\phi}A)|_{t=0},\quad
u_{1}=\partial_{t}(e^{i\sigma\phi}A)|_{t=0}.$

As $q$ vanishes near $t=0$, we see that $u_{0}$ and $u_{1}$ are independent
from $q$. This again implies that $L$ determines $u|_{t=T}$. Moreover,

$\sigma(e^{-i\sigma\phi}u-a_{0})|_{t=T}\to a_{1}|_{t=T}\quad\text{as
$\sigma\to\infty$}.$

This determines $a_{1}|_{t=T}$. Recalling (23) and using the fact that $\Box
a_{0}$ does not depend on $q$, we find the integral

$\int_{-r}^{T-r}qa_{0}(s^{\prime},r,x^{\prime})ds^{\prime}.$

Here we used the fact that $t=T$ is equivalent with $s=T-r$.

In $(s,r,x^{\prime})$ coordinates $a_{0}=\chi(2r)\eta(x^{\prime})$ and the
above integral reduces at $(r,x^{\prime})=0$ to

$\int_{0}^{T}q(s^{\prime},0,0)ds^{\prime}\chi(0)\eta(0).$

As $q$ vanishes for $t<0$ and $t>T$, we can recover in $(s,r,x^{\prime})$
coordinates $\int_{\mathbb{R}}q(s,0,0)ds$, or equivalently, in
$(t,x^{1},x^{\prime})$ coordinates

$\int_{\mathbb{R}}q(\beta(s))ds,\quad\beta(s)=(s,s,0).$

Repeating the above argument, after using rotations and translations, we
obtain the light ray transform of $q$,

$\mathcal{L}q(y,v)=\int_{\mathbb{R}}q(\beta_{y,v}(s))ds,\quad\beta_{y,v}(s)=(s,y+sv),\
y\in\mathbb{R}^{n},\ v\in S^{n-1}.$

###  5.2. Inversion of the light ray transform

The above reduction works also when $n=1$, but inversion of $\mathcal{L}$
requires $n\geq 2$. Indeed, if $n=1$, then $\mathcal{L}q=0$ for
$q(s,r)=q_{0}(s)q_{1}(r)$ with $q_{j}$, $j=0,1$, integrating to zero.

For a fixed $v\in S^{n-1}$, consider the change of coordinates in
$\mathbb{R}^{1+n}$,

$(t,x)=(s,y+sv).$

Then $y=x-tv$. Using this, we obtain the Fourier slicing

$\int_{\mathbb{R}^{n}}e^{-i\eta\cdot
y}\mathcal{L}q(y,v)\,dy=\int_{\mathbb{R}^{1+n}}e^{-i\eta\cdot(x-tv)}q(t,x)\,dtdx=\widehat{q}(-\eta\cdot
v,\eta).$

Here $|\eta\cdot v|\leq|\eta|$. Moreover, as $n\geq 2$, we can choose a unit
vector $w$ that is orthogonal to $\eta$. Then for $a\in[-1,1]$ and $\eta\neq
0$, we may choose

$v=-\frac{a}{|\eta|}\eta+\sqrt{1-a^{2}}\,w\in S^{n-1}.$

This gives $-\eta\cdot v=a|\eta|$, and we see that the Fourier slicing allows
us to recover $\widehat{q}(a|\eta|,\eta)$ for any $a\in[-1,1]$.

As $q$ is compactly supported, $\hat{q}$ is analytic. We know $\hat{q}$ in a
non-empty open cone (in fact, in the cone of spacelike directions), and
therefore everywhere by analytic continuation. This shows that $\mathcal{L}q$
determines $q$. By the above reduction also $L$ determines $q$.

noline,color=white,size=]Unify the style for refs (maybe using bibtex), and
sort them alphabetically

## References

  * [1] Spyros Alexakis, Ali Feizmohammadi, and Lauri Oksanen.  Lorentzian Calderón problem under curvature bounds.  Invent. Math., 229(1):87–138, 2022.
  * [2] Mikhail Belishev.  An approach to multidimensional inverse problems for the wave equation.  Dokl. Akad. Nauk SSSR, 297(3):524–527, 1987.
  * [3] Haim Brezis.  Functional analysis, Sobolev spaces and partial differential equations.  Universitext. Springer, New York, 2011.
  * [4] Cătălin I. Cârstea, Ali Feizmohammadi, and Lauri Oksanen.  Remarks on the anisotropic calderón problem, 2022.
  * [5] David Dos Santos Ferreira, Yaroslav Kurylev, Matti Lassas, and Mikko Salo.  The Calderón problem in transversally anisotropic geometries.  J. Eur. Math. Soc. (JEMS), 18(11):2579–2626, 2016.
  * [6] Lawrence C. Evans.  Partial differential equations, volume 19 of Graduate Studies in Mathematics.  American Mathematical Society, Providence, RI, 1998.
  * [7] Ali Feizmohammadi, Tuhin Ghosh, Katya Krupchyk, and Gunther Uhlmann.  Fractional anisotropic calderón problem on closed riemannian manifolds, 2021.
  * [8] Ali Feizmohammadi, Joonas Ilmavirta, Yavar Kian, and Lauri Oksanen.  Recovery of time-dependent coefficients from boundary data for hyperbolic equations.  J. Spectr. Theory, 11(3):1107–1143, 2021.
  * [9] Ali Feizmohammadi, Joonas Ilmavirta, and Lauri Oksanen.  The light ray transform in stationary and static Lorentzian geometries.  J. Geom. Anal., 31(4):3656–3682, 2021.
  * [10] Aleksandr Kachalov, Yaroslav Kurylev, Matti. Lassas, and Niculae Mandache.  Equivalence of time-domain inverse problems and boundary spectral problems.  Inverse Problems, 20(2):419–436, 2004.
  * [11] Alexander Katchalov, Yaroslav Kurylev, and Matti Lassas.  Inverse boundary spectral problems, volume 123 of Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics.  Chapman & Hall/CRC, Boca Raton, FL, 2001.
  * [12] Yavar Kian, Lauri Oksanen, and Masahiro Soccorsi, Ericand Yamamoto.  Global uniqueness in an inverse problem for time fractional diffusion equations.  J. Differential Equations, 264(2):1146–1170, 2018.
  * [13] Yaroslav Kurylev, Lauri Oksanen, and Gabriel P. Paternain.  Inverse problems for the connection Laplacian.  J. Differential Geom., 110(3):457–494, 2018.
  * [14] Olga A. Ladyzhenskaya.  The boundary value problems of mathematical physics, volume 49 of Applied Mathematical Sciences.  Springer-Verlag, New York, 1985.  Translated from the Russian by Jack Lohwater [Arthur J. Lohwater].
  * [15] Camille Laurent and Matthieu Léautaud.  Lectures on unique continuation for waves.  Lecturenotes. This volume. 2023.
  * [16] John M. Lee.  Riemannian manifolds, volume 176 of Graduate Texts in Mathematics.  Springer-Verlag, New York, 1997.  An introduction to curvature.
  * [17] Plamen Stefanov.  Support theorems for the light ray transform on analytic Lorentzian manifolds.  Proc. Amer. Math. Soc., 145(3):1259–1274, 2017.
  * [18] Plamen Stefanov and Yang Yang.  The inverse problem for the Dirichlet-to-Neumann map on Lorentzian manifolds.  Anal. PDE, 11(6):1381–1414, 2018.
  * [19] Plamen D. Stefanov.  Uniqueness of the multi-dimensional inverse scattering problem for time dependent potentials.  Math. Z., 201(4):541–559, 1989.