INVERSE PROBLEMS IN BIOLUMINESCENCE TOMOGRAPHY AND WAVE IMAGING:
THEORY, ALGORITHM AND IMPLEMENTATION

By

Tianyu Yang

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

Computational Mathematics, Science and Engineering—Doctor of Philosophy

2024

ABSTRACT

Ultrasound modulated bioluminescence tomography (UMBLT) is a technique for imaging the 3D

distribution of biological objects such as tumors by using a bioluminescent source as a biomedical

indicator.

It uses bioluminescence tomography (BLT) with a series of perturbations caused by

acoustic vibrations. UMBLT outperforms BLT in terms of spatial resolution. The current UMBLT

algorithm in the transport regime requires measurement at every boundary point in all directions,

and reconstruction is computationally expensive. In Chapter 2, we will first introduce the UMBLT

model in both the diffusive and transport regimes, and then formulate the image reconstruction

problem as an inverse source problem using internal data. Second, we present an improved

UMBLT algorithm for isotropic sources in the transport regime. Third, we generalize an existing

UMBLT algorithm in the diffusive regime to the partial data case and quantify the error caused by

uncertainties in the prescribed optical coefficients.

The inverse boundary value problem (IBVP) of wave equation aims to recover medium distri-

bution via boundary measurement of wave propagation. Using an important identity that connects

boundary data and internal wave states, one can recover the medium’s interior structure by selecting

suitable boundary sources. In Chapter 3, we will first introduce the IBVP and the key identity.

Second, we present a direct wave speed reconstruction algorithm with vanished wave potential.

Third, we apply linearization on IBVPs to derive algorithms with nonvanishing parameters for both

wave speed and wave potential reconstruction.

Copyright by
TIANYU YANG
2024

ACKNOWLEDGEMENTS

First of all, I would like to express my deepest appreciation to my advisor and committee chair,

Professor Yang Yang, for his guidance, support, and encouragement though my Ph.D. study. He led

me into the area of inverse problems and helped me to establish a deep understanding of theoretical

and numerical PDE theory.

I would like to express my thanks to my dissertation committee

members, Professor Jianliang Qian, Professor Zhen Qiu, and Professor Adam Alessio, for their

guidance as well.

I would like to thank Dr. Albert Chua, Liantao Li and Henry Fessler for sharing their ideas and

research during group meetings. I would like to thank Remy Liu, Jiaxin Yang, Shuyang Qin, Dr.

He Lyu, Dr. Hao Wang, Dr. Qi Lyu, Kang Yu, Liantao Li, Dr. Danqi Qu, Dr. Ziyi Xi, Dr. Runze

Su, Dr. Wenjie Qi, Dr. Binbin Huang for the hotpot, barbecue, movies, mahjong, poker, and other

activities we enjoyed together.

Finally, I would like to thank my parents, Mr. Fashen Yang and Mrs. Lihua Chen, for their

support in pursuing a Ph.D. far away from home.

iv

TABLE OF CONTENTS

CHAPTER 1

INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

CHAPTER 2

ULTRASOUND MODULATED BIOLUMINESCENCE
TOMOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

1

4

CHAPTER 3

INVERSE BOUNDARY VALUE PROBLEMS FOR WAVE
EQUATIONS .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.

. 43

CHAPTER 4

CONCLUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. 96

BIBLIOGRAPHY .

.

.

.

.

. .

. .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

APPENDIX A

APPENDIX FOR CHAPTER 2 . . . . . . . . . . . . . . . . . . . . . . 104

APPENDIX B

APPENDIX FOR CHAPTER 3 . . . . . . . . . . . . . . . . . . . . . . 110

v

CHAPTER 1

INTRODUCTION

Inverse Problems are defined as the inverse of Forward Problems. Forward problems typically

involve determining the status of a physical system or predicting an outcome based on model

parameters, such as calculating a rocket’s trajectory or simulating the spread of water pollution.

The forward map of the physical system (model) is formulated as

𝔉 : 𝑋 → 𝑌 ,

𝔉( 𝑝) = 𝑑,

(1.1)

where the operator 𝔉 describes the relation between model parameter space 𝑋 with the observed

data space 𝑌 , 𝑝 denote the parameters and 𝑑 denote the data.

Inverse problems focus on determining model parameters 𝑝 from measurement data 𝑑, i.e.

inverting the operator 𝔉. Inverse problems are considered when the model parameters 𝑝 of interest

cannot be measured directly or practically, such as generating 3D images of internal organs or

reconstructing the seismic velocity structure of the Earth.

In mathematics, a well-posed problem, which is introduced by Jacques Hadamard (1865-1963),

satisfies the following three properties:

(1) Existence: The problem has a solution,

(2) Uniqueness: The solution is unique,

(3) Stability: The solution’s behavior changes continuously with the initial conditions.

Unlike well-posed problems, ill-posed problems violate at least one of the three properties. Forward

problems are usually well-posed, while inverse problems are usually ill-posed, which means that

either 𝔉 is not invertible or 𝔉−1 is not continuous. Assuming that 𝔉−1 exists, the model parameter

𝑝 is theoretically given by

𝑝 = 𝔉−1𝑚.

If the observation carry measurement error 𝜀 ∈ 𝑌 , the reconstructed parameter becomes

˜𝑝 = 𝔉−1 ˜𝑚 = 𝔉−1 [𝑚 + 𝜀] = 𝑝 + 𝔉−1𝜀.

1

The difference between ˜𝑝 and 𝑝 could be relatively large depending on the behavior of 𝔉−1, even if

the measurement error 𝜀 is relatively small. In these cases, the theory of regularization of ill-posed

problems have been considered by A. N. Tikhonov and his followers [48, 63, 77, 93].

When the physical model describes the propagation of lights, the 3D structure of an object

is reconstructed using light that has been transmitted and scattered. During light propagation,

two main effects occur: absorption and scattering. For example, in photoelectric absorption, the

photon is completely absorbed by the atom electron; in Rayleigh scattering, the photon is scattered,

with the effect proportional to the fourth power of its frequency.

In this case, it is common to

assume that the model is time and frequency independent. For Optical Tomography, a series of

near-infrared sources are attached on the surface of the object. The scattered field are measured

on the same surface in order to generate images of soft tissues [34, 50, 100]. For Bioluminescence

Tomography, biologists use transfection technology to generate an internal light source. The

distribution of different optical coefficients can be recovered by measuring the output flux through

the boundary [29,69,101]. However, both Optical Tomography and Bioluminescence Tomography

have poor resolution due to the ill-posedness [35]. The diffusive nature of the photons is cause for

the ill-posedness, which means the obtained images have low resolution and are very sensitive to

noise.

In order to improve the resolution, a high resolution modality is combined; these combina-

tions are known as Hybrid inverse problems, also called coupled-physics inverse problems, which

combine a high resolution modality with a high contrast modality for better reconstruction. For

example, in Ultrasound Modulated Optical Tomography [64, 80, 99] and Ultrasound Modulated

Bioluminescence Tomography [6, 8, 28], a series of ultrasound perturbations are performed to the

medium in order to generate more measurements, thereby overcoming the ill-posedness of the

original inverse problems.

In Chapter 2, we introduce one of the hybrid inverse problems, Ultrasound Modulated Biolu-

minescence Tomography(UMBL). Depending on the scattering effects of medium, we introduce

UMBLT under transport regime in Section 2.2 and diffusive regime in Section 2.3. The algorithms

2

in both full and partial data cases, as well as the uncertainty quantifications are given.

When the physical model describes the propagation of waves, the corresponding inverse problem

are concerned with extracting information about structural features from scattered wave measure-

ments. For example, in Ultrasound Computerized Tomography, a series of transducers are placed

around the object. Each transducer will send acoustic waves into the object and other transducers

will collect the waves. It can image soft tissue with high resolution [53,73,79]. If the wave are gen-

erated from external sources, the forward problems are formulated as boundary value problems of

wave equation, and the inverse problems are called inverse boundary value problems of wave equa-

tion. Under this framework, the Neumann boundary condition represent the external sources, and

the measurement is the Dirichlet value of the solution, or equivalently, the Neumann-to-Dirichlet

map Λ. Boundary Control Method is an effective approach to solve the inverse boundary value

problems [13, 47, 51, 88]. This method allow us generate a special input (Neumann boundary

condition) such that the solution will reach a desired state at a fixed time, and the existance of such

input is guaranteed by Tataru’s theorem in [90]. With Blagove˘s˘censki˘ı’s identity [21], one can

connect the boundary data with internal waves, which help recover the structure features, such as

wave speed.

In Chapter 3, we introduce nonlinear Inverse Boundary Value Problems of wave equation with

potential. In Section 3.2, we introduce a direct wave speed reconstruction algorithm with vanished

wave potential. In Section 3.3, we consider wave equation with non-vanished potential. We apply

linearization method to the inverse boundary value problems to derive algorithms for recovering

wave speed and wave potential, respectively.

3

CHAPTER 2

ULTRASOUND MODULATED BIOLUMINESCENCE TOMOGRAPHY

2.1

Introduction

Bioluminescence is the production and emission of light by a living organism; It is widely

occured in vertebrates and invertebrates, such as in firefly, anglerfish. This phenomenon can be

utilized in a medical imaging method called Bioluminescence tomography (BLT). BLT aims to

reconstruct images of biological objects, like tumors, by using the bioluminescent source as a

biomedical indicator. Specifically, scientists tag biological entities or process components (e.g.

bacteria, tumor cells, immune cells, or genes) with reporter genes that encode one of a number

of light-generating enzymes (luciferases) [29]. By measuring the light generated by the luciferin-

luciferase reaction, one can image the 3D distribution of the internal source, which can be used for

diagnosing diseases.

However, BLT have one main weakness, the spatial resolution of BLT is poor. This is because

changes of light source will only cause relatively small changes on the measured data, the mea-

surement error might have a large effect on the reconstruction. Even worse, sometimes different

sources can lead to same measurement [35]. In order to prevent the worst case, which means we

need to uniquely determine the source from the boundary measurement, BLT need additional in-

formation about the source, like its geometric aspect. One effective approach to enhance the spatial

resolution of BLT is to use the ultrasound modulation, and the new method is called Ultrasound

Modulated Bioluminescence tomography (UMBLT). This method performs BLT under series of

acoustic modulations to perturb the medium’s optical properties. With different type of ultrasound,

the measured data are different, which means the measured data are increased and different sources

can not give same measurement with all types of ultrasound modulations, which can help overcome

the weakness of BLT.

The first BLT scanner was developed by G. Wang et al. in 2003 [94] and the first small animal

study using BLT was conducted by Chaudhari et al. in 2005 [27]. In 2004, G. Wang et al. gave

uniqueness theorems of BLT under diffusion approximation [95], which is a simplified model for

4

strongly scattering medium based on diffusion equation. They proved the source can be uniquely

determined from the boundary measurement if the source is a linear combination of impulses or ball

sources. They also gave a summary of uniqueness theorems under different assumptions proposed

by other people. In 2007, the uniqueness theorem of BLT under transport regime was given [9],

they proved the source can be uniquely determined when the scattering kernel 𝑘 is invariant of

rotation and is relatively small under certain norm. One year later, Stefanov and Uhlmann gave a

generic result of BLT that when attenuation coefficient 𝜎 and scattering kernel 𝑘 are continuous,

the source can be uniquely determined from boundary measurement when (𝜎, 𝑘) is in a open dense

subset which contains a neighbor of (0, 0) [85].

The inverse problem of UMBLT was first considered under diffusion approximation by Bal

and Schotland [8]. They showed the well-posedness of the problem and developed an inversion

formula by assuming the diffusion and absorption coefficients are given. Two years later, the inverse

problem of UMBLT under transport regime was considered by Bal, Chung and Schotland [6]. They

assumed the scattering kernel is invariant under rotation, all optical coefficients are continuous and

𝜎, 𝑘 are known. They showed the well-posedness of the problem and give an algorithm to

reconstruct the source. They first derive an internal functional from the boundary measurements.

Theoretically the directional derivative of photon intensity 𝜃 · ∇𝑢 can be extract from the internal

functional, and then the interior value of the photon intensity 𝑢 can be calculated using 𝜃 · ∇𝑢 and

the corresponding boundary value, which can be used to calculate the source term. However, in this

proposed algorithm, the reconstruction of photon intensity needs measurement on every boundary

points with all directions and needs lots of computation. To this end we designed a more efficient

algorithm for UMBLT under transport regime when the source is isotropic [28], where the source

can be directly reconstruct by inverting a linear operator.

2.2 UMBLT under Transport Regime

The UMBLT method aims to image the distribution of light source.

In order to model the

UMBLT, we need to model the light propagation first. The propagation of light through a medium

𝑋 is affected by absorption, emission, and scattering processes, which can be modeled using

5

standard Radiative Transfer Equation (RTE). Such problem is called the UMBLT under transport

regime [6, 28]. In this section, we consider following RTE model:

𝜃 · ∇𝑢 + 𝜎(𝑥)𝑢 −

∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜃′)𝑢(𝑥, 𝜃′) d𝜃′ = 𝑆(𝑥, 𝜃),

𝑢|Γ− = 0,

(2.1)

(2.2)

where 𝑢 denote the intensity of light at spatial location 𝑥 traveling in direction 𝜃, 𝜎 is attenuation

coefficient, 𝑘 is the scattering kernel, 𝑆 is an isotropic source, 𝑋 is a bounded subset of R𝑛 with

smooth boundary, Γ± are the outgoing/incoming boundary, which are defined as

Γ± = (cid:8)(𝑥, 𝜃) ∈ 𝜕 𝑋 × S𝑛−1 | ±𝜃 · 𝜈 ≥ 0(cid:9) .

The vibration of the medium under acoustic modulation can be modeled using the time-harmonic

plane wave with frequency 𝜔 as

𝑝 = 𝐴 cos(𝜔𝑡) cos(𝑞 · 𝑥 + 𝜑),

(2.3)

where 𝑝 is the pressure, 𝑞 is the wave vector and 𝜑 is the phase. Since the pressure is related to

the changes of local density of the medium, the effect of the acoustic modulation on the optical

properties [7] can be modeled as

𝜎𝜀 (𝑥) (cid:66) (1 + 𝜀 cos(𝑞 · 𝑥 + 𝜑))𝜎(𝑥),

𝑘𝜀 (𝑥, 𝜃, 𝜃′) (cid:66) (1 + 𝜀 cos(𝑞 · 𝑥 + 𝜑))𝑘 (𝑥, 𝜃, 𝜃′),

𝑆𝜀 (𝑥, 𝜃) (cid:66) (1 + 𝜀 cos(𝑞 · 𝑥 + 𝜑))𝑆(𝑥, 𝜃),

(2.4)

(2.5)

(2.6)

where 0 ≤ 𝜀 ≪ 1 is a small parameter related to the amplitude, frequency, time, density and

acoustic wave speed.

With the acoustic modulation, the model of UMBLT is the following modulated RTE with

modulated optical properties and the same boundary condition

𝜃 · ∇𝑢𝜀 + 𝜎𝜀 (𝑥)𝑢𝜀 −

∫

S𝑛−1

𝑘𝜀 (𝑥, 𝜃, 𝜃′)𝑢𝜀 (𝑥, 𝜃′) d𝜃′ = 𝑆𝜀 (𝑥),

𝑢𝜀 |Γ− = 0.

6

(2.7)

(2.8)

The solution of the modulated RTE is denote as 𝑢𝜀, notice that when 𝜀 = 0, it is exactly the solution

of the standard RTE. Under different modulations, our measurement is the operator

𝑆 : R𝑛 × {0,
Λ𝜀

𝜋

2

} → 𝐶 (Γ+),

(𝑞, 𝜑) ↦→ 𝑢𝜀 |Γ+

,

𝜀 ≥ 0,

(2.9)

which is the light flows out through the boundary under different modulations. The goal is to

reconstruct source 𝑆 from the measurement assuming 𝜎𝜀, 𝑘𝜀, 𝑋 are given.

Throughout this section, we make the following assumptions to ensure well-posedness of some

forward boundary value problems.

(A1): 𝜎, 𝑘 and 𝑆 are continuous on 𝑋; moreover, 𝜎 ≥ 𝑐 > 0 and 𝑘 ≥ 𝑐 > 0 for some constant 𝑐

everywhere in 𝑋.

(A2): Set 𝜌 (cid:66)

(cid:13)
∫
(cid:13)
S𝑛−1
(cid:13)

𝑘 (𝑥, 𝜃, 𝜃′) d𝜃′(cid:13)
(cid:13)
(cid:13)𝐿∞ (𝑋×S𝑛−1)

, one of the following inequalities holds:

where 𝛼 > 0 is a positive constant, or

(cid:19)

𝜎

(cid:18)

inf
𝑥∈𝑋

− 𝜌 ≥ 𝛼

diam(𝑋) 𝜌 < 1

(2.10)

(2.11)

where diam(𝑋) (cid:66) sup{|𝑥 − 𝑦| : 𝑥, 𝑦 ∈ 𝑋 } is the diameter of 𝑋.

To ensure the integral in (2.1) is self-adjoint over 𝑋 × 𝑆𝑛−1, we assume the scattering kernel 𝑘

is invariant under rotation, which means

𝑘 (𝑥, 𝜃, 𝜃′) = 𝑘 (𝑥, 𝜃 · 𝜃′),

then we can derive an internal functional for reconstruction.

Under these assumptions, the following well-posedness theorem holds in the function space

𝐿 𝑝 (S𝑛−1, 𝐶 (𝑋)), where the norm is defined as

∥𝑢∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋))

(cid:66)

(cid:18)∫

S𝑛−1

∥𝑢(𝑥, 𝜃) ∥

𝑝
𝐶 (𝑋)

d𝜃

(cid:19) 1

𝑝

.

7

Proposition 2.1 ( [6, Theorem 2.1]). Suppose the assumptions (A1)(A2) hold. Then for any

𝑓− ∈ 𝐶 (Γ−), the RTE (2.1) has a unique solution 𝑢 ∈ 𝐿 𝑝 (S𝑛−1, 𝐶 (𝑋)) (1 ≤ 𝑝 ≤ ∞) with the

boundary condition 𝑢|Γ− = 𝑓−. Moreover, if (2.10) holds, we have the estimate

∥𝑢∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋)) ≤

(cid:16)

1
𝛼

(𝜌 + 𝛼) ∥ 𝑓−∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝜕 𝑋)) + ∥𝑆∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋))

(cid:17)

.

If instead (2.11) holds, we have the estimate

∥𝑢∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋)) ≤

1
1 − diam(𝑋) 𝜌

(cid:16)

∥ 𝑓−∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝜕 𝑋)) + diam(𝑋) ∥𝑆∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋))

(cid:17)

.

This well-posedness theorem can ensure the operators we defined later is well-defined, and can

be used in the estimation of the operator norm.

We consider the adjoint RTE for derivation. Let 𝑣 = 𝑣(𝑥, 𝜃) be the solution of the adjoint RTE

with prescribed outgoing boundary condition 𝑔:

−𝜃 · ∇𝑣 + 𝜎𝑣 −

∫

S𝑛−1

𝑣|Γ+ = 𝑔.

𝑘 (𝑥, 𝜃, 𝜃′)𝑣(𝑥, 𝜃′) d𝜃′ = 0

(2.12)

(2.13)

Theoretically, we know 𝑣 in the entire space 𝑋 × S𝑛−1. Multiply 𝑣 on both sides of (2.7), multiply

𝑢𝜀 on both sides of (2.12), subtract two equations and integrate over 𝑋 × S𝑛−1, integration by parts

gives

∫

∫

S𝑛−1

𝜕 𝑋

𝑢𝜀𝑣𝑛 · 𝜃 d𝑥 d𝜃 =

∫

∫

∫

(𝑘𝜀 − 𝑘)𝑣(𝑥, 𝜃)𝑢𝜀 (𝑥, 𝜃′) d𝜃 d𝜃′ d𝑥

𝑋
∫

S𝑛−1
∫

S𝑛−1

+

𝑋

S𝑛−1

𝑣𝑆𝜀 d𝜃 d𝑥 −

∫

∫

𝑋

S𝑛−1

(𝜎𝜀 − 𝜎)𝑢𝜀𝑣 d𝜃 d𝑥.

(2.14)

By the asymptotic expansion on 𝜀, write 𝑢𝜀 = 𝑢0 + 𝜀𝛿𝑢, the first order term gives

∫

∫

S𝑛−1

𝜕 𝑋

𝛿𝑢𝑣𝑛 · 𝜃 d𝑥 d𝜃

∫

∫

S𝑛−1

𝑋

∫

∫

∫

= −

+

𝑋

S𝑛−1

S𝑛−1

cos(𝑞 · 𝑥 + 𝜑)𝜎𝑢𝑣 d𝜃 d𝑥 +

∫

∫

𝑋

S𝑛−1

cos(𝑞 · 𝑥 + 𝜑)𝑣𝑆 d𝜃 d𝑥

(2.15)

cos(𝑞 · 𝑥 + 𝜑)𝑘 (𝑥, 𝜃, 𝜃′)𝑣(𝑥, 𝜃)𝑢(𝑥, 𝜃′) d𝜃 d𝜃′ d𝑥.

Notice that the LHS is known from the adjoint solution 𝑣 and the boundary measurement 𝑢𝜀, RHS

is an inner product of a cosine function and another function over 𝑋. By varying 𝑞 and 𝜑, the RHS

8

is exactly the Fourier coefficient of a function’s Fourier transform, and we denote the function as

𝐻𝑣:

𝐻𝑣 (𝑥) (cid:66) −

𝜎𝑢𝑣 d𝜃 +

∫

S𝑛−1

𝑣𝑆 d𝜃

∫

S𝑛−1

∫

∫

S𝑛−1

S𝑛−1

+

∫

=

S𝑛−1

𝑣(𝑥, 𝜃)𝜃 · ∇𝑢(𝑥, 𝜃) d𝜃,

𝑘 (𝑥, 𝜃, 𝜃′)𝑣(𝑥, 𝜃)𝑢(𝑥, 𝜃′) d𝜃′ d𝜃

(2.16)

Then we construct the internal functional 𝐻𝑣 from the boundary measurement Λ𝜀
𝑆.

2.2.1 Anisotropic Source

The optical coefficients in biological objects could depend on both location and direction. For

example, since each muscle is made up of muscle fiber groups, the ability of light to propagate

along fiber direction and perpendicular to fiber direction differs significantly. In this section, we

consider the reconstruction of anisotropic source 𝑆(𝑥, 𝜃), which depend both on spatial location

and direction [6].

Let

Since

𝜏+(𝑥, 𝜃) = min{𝑡 > 0|𝑥 + 𝑡𝜃 ∈ 𝜕 𝑋 }.

𝑢(𝑥, 𝜃) = 𝑢(𝑥 + 𝜏+𝜃, 𝜃) −

∫ 𝜏+ (𝑥,𝜃)

𝜃 · ∇𝑢(𝑥 + 𝑡𝜃, 𝜃) d𝑡,

(2.17)

(2.18)

0
Once 𝜃 · ∇𝑢(𝑥, 𝜃) can be reconstructed from the internal functional 𝐻𝑣, the forward RTE solution

𝑢(𝑥, 𝜃) can be calculated by (2.18), then the source 𝑆(𝑥, 𝜃) can be calculated by substituting 𝑢(𝑥, 𝜃)

into (2.1).

In order to reconstruct 𝜃 · ∇𝑢(𝑥, 𝜃), the controllability of RTE is required:

Proposition 2.2 ( [6, Theorem 1.3]). Suppose 𝑋, 𝑘 and 𝜎 are given. Then for any point 𝑥0 ∈ 𝑋
and any continuous function ℎ on 𝐿 𝑝 (S𝑛−1), there is a function 𝑔 ∈ 𝐿 𝑝 (S𝑛−1, 𝐿∞(𝜕 𝑋)) such

that the boundary value problem (2.12)(2.13) has a unique solution 𝑣 ∈ 𝐿 𝑝 (S𝑛−1, 𝐿∞(𝑋)) which

is continuous in a neighbourhood of 𝑥0, and satisfies the property that 𝑣(𝑥0, 𝜃) = ℎ(𝜃), for all
𝜃 ∈ S𝑛−1. Moreover, for any 1 ≤ 𝑝 ≤ ∞,

∥𝑣∥ 𝐿 𝑝 (S𝑛−1,𝐿∞ (𝑋)) + ∥𝑔∥ 𝐿 𝑝 (S𝑛−1,𝐿∞ (𝜕 𝑋)) ≤ 𝐶 ∥ℎ∥ 𝐿 𝑝 (S𝑛−1),

(2.19)

9

where 𝐶 depends only on 𝑋, 𝑘 and 𝜎.

With the controllability theorem, for any 𝑥0 ∈ 𝑋, we can arrange 𝑣(𝑥0, 𝜃) to be any continuous

function in 𝜃, then knowing 𝐻𝑣 (𝑥0, 𝜃) for any 𝑣 is equivalent to knowing 𝜃 · ∇𝑢(𝑥0, 𝜃), thus we can

recover 𝜃 · ∇𝑢(𝑥, 𝜃) from the internal functional 𝐻𝑣.

2.2.2

Isotropic Source

The inversion formula in Section 2.2.1 requires measurement on each boundary points (𝑥, 𝜃) ∈

Γ+, which is hard in practice. Besides, the reconstruction needs a lot of computations, since the

inversion formula is point-to-point. To this end, we focus on the inverse problem of UMBLT in

full RTE model with an isotropic source [28], i.e. the source 𝑆 = 𝑆(𝑥) is independent of the angle

𝜃. We built a more efficient inversion formula to reduce the requirement of measurement and

computation.

Since 𝑆 is independent of 𝜃, the internal functional can be written as

𝐻𝑣 (𝑥) =

=

=

∫

S𝑛−1

∫

S𝑛−1

∫

S𝑛−1

𝑣(𝑥, 𝜃)𝜃 · ∇𝑢(𝑥, 𝜃) d𝜃

𝑣(𝑥, 𝜃) [A𝑢(𝑥, 𝜃) + 𝑆(𝑥)] d𝜃,

(2.20)

𝑣(𝑥, 𝜃)A𝑢(𝑥, 𝜃) d𝜃 + 𝑆(𝑥)

∫

S𝑛−1

𝑣(𝑥, 𝜃) d𝜃,

where

A𝑢(𝑥, 𝜃) (cid:66) −𝜎(𝑥)𝑢(𝑥, 𝜃) +

∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜃′)𝑢(𝑥, 𝜃′) d𝜃′.

(2.21)

and the norm of 𝑆 in Proposition 2.1 becomes

∥𝑆∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋)) = Vol(S𝑛−1)

1

𝑝 ∥𝑆∥𝐶 (𝑋)

,

where Vol(S𝑛−1) denote the volume of (𝑛 − 1)-dimensional unit ball.

With (A1)(A2), we can choose a suitable adjoint solution satisfies 𝑣0 ≥ 𝑐 > 0 [28, Lemma 2].

Dividing (2.20) by ∫

S𝑛−1

𝑣0(𝑥, 𝜃) d𝜃 on both sides gives

𝐻𝑣0 (𝑥)
𝑣0(𝑥, 𝜃) d𝜃

∫
S𝑛−1

(cid:66) 𝑆(𝑥) +

∫
S𝑛−1 A𝑢(𝑥, 𝜃)𝑣0(𝑥, 𝜃) d𝜃

∫
S𝑛−1

𝑣0(𝑥, 𝜃) d𝜃

,

(2.22)

10

notice that the LHS is known, the first term on RHS is exactly the source 𝑆 we want to reconstruct

and the second term is linear in 𝑢, which is linearly depend on 𝑆, the RHS can be written as an

identity operator plus a linear operator act on 𝑆. To represent the linear operator, we define the

following three operators.

The first one is the source-to-solution operator S:

S : 𝐶 (𝑋) → 𝐿 𝑝 (S𝑛−1, 𝐶 (𝑋)),

𝑆 ↦→ 𝑢

(2.23)

with the norm estimation from the well-posedness Theorem 2.1

∥S∥𝐶 (𝑋)→𝐿 𝑝 (S𝑛−1,𝐶 (𝑋)) ≤

1
𝑝

Vol(S𝑛−1)
𝛼

diam(𝑋)Vol(S𝑛−1)
1−diam(𝑋) 𝜌

1
𝑝





(cid:19)

𝜎

(cid:18)

inf
𝑥∈𝑋

− 𝜌 ≥ 𝛼

diam(𝑋) 𝜌 < 1

(2.24)

The second operator is

K𝑣0 : 𝐿 𝑝 (S𝑛−1, 𝐶 (𝑋)) → 𝐶 (𝑋),

𝑢(𝑥, 𝜃) ↦→

∫

S𝑛−1

A𝑢(𝑥, 𝜃)𝑣0(𝑥, 𝜃) d𝜃

(2.25)

where the operator A is introduced in (2.21). K𝑣0 is a bounded operator and

∥K𝑣0 ∥ 𝐿 𝑝 (S𝑛−1,𝐶 (𝑋))→𝐶 (𝑋) ≤ ∥𝑣0∥𝐶 (𝑋) (∥𝜎∥𝐶 (𝑋) + 𝜌)Vol(S𝑛−1)1− 1

𝑝

(2.26)

The third operator is the multiplication operator

M𝑣0 : 𝐶 (𝑋) → 𝐶 (𝑋),

𝑓 (𝑥) ↦→

1

∫
S𝑛−1

𝑣0(𝑥, 𝜃) d𝜃

𝑓 (𝑥).

(2.27)

It is bounded since 𝑣0 is chosen in such a way that ∫

S𝑛−1

𝑣0(𝑥, 𝜃) d𝜃 is bounded away from zero. We

have

∥M𝑣0 ∥𝐶 (𝑋)→𝐶 (𝑋) ≤

inf𝑥∈𝑋

Then (2.22) can be represented as

1
𝑣0(𝑥, 𝜃) d𝜃

(cid:16)∫

S𝑛−1

.

(cid:17)

(2.28)

M𝑣0 [𝐻𝑣0] = (𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S) [𝑆].

To invert linear operator 𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S, there are two approaches. The first one is to prove

M𝑣0 ◦ K𝑣0 ◦ S is a contraction under certain norm, then we can use Neumann series to invert this

11

operator. The second approach is to prove M𝑣0 ◦ K𝑣0 ◦ S is compact over certain function space,
then 𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S is a Fredholm operator, we can use Fredolm inversion to solve 𝑆.

For the first approach, the inversion formula is given in the following theorem

Theorem 2.3 ( [28, Theorem 3]). Suppose the assumptions (A1)(A2) hold. If the following inequality

holds

∥𝑣0∥𝐶 (𝑋) (∥𝜎∥𝐶 (𝑋) + 𝜌)diam(𝑋)Vol(S𝑛−1)
(cid:17)
𝑣0(𝑥, 𝜃) d𝜃
(1 − diam(𝑋) 𝜌) inf𝑥∈𝑋
then the operator M𝑣0 ◦ K𝑣0 ◦ S is a contraction, and the source 𝑆 can be computed from the

< 1 when diam(𝑋) 𝜌 < 1,

(2.29)

S𝑛−1

(cid:16)∫

following Neumann series:

𝑆 =

∞
∑︁

𝑗=0

(−M𝑣0 ◦ K𝑣0 ◦ S) 𝑗 (M𝑣0 [𝐻𝑣0]).

For the second approach, we need additional assumptions on optical coefficients. For 𝑠 ∈ R, let

𝑊 𝑠,2 be the usual 𝐿2−based Sobolev space, for 1 ≤ 𝑝 ≤ ∞, let H 1

𝑝 be the following function space

H 1

𝑝 (cid:66) (cid:8)𝑢 ∈ 𝐿 𝑝 (𝑋 × S𝑛−1) | 𝜃 · ∇𝑢 ∈ 𝐿 𝑝 (𝑋 × S𝑛−1)(cid:9)

with norm

∥𝑢∥H 1

𝑝 =

(cid:18)∫

𝑋×S𝑛−1

|𝑢| 𝑝 + |𝜃 · ∇𝑢| 𝑝 d𝑥 d𝜃

(cid:19) 1

𝑝

.

we make following assumptions

(A3): 𝜎(𝑥) ≥ 𝜎0 > 0 everywhere in 𝑋 for some constant 𝜎0.

(A4): ∥ 1
𝜎(𝑥)

∫
S𝑛−1

𝑘 (𝑥, 𝜃, 𝜃′) d𝜃′∥ 𝐿∞ (𝑋×𝑆𝑛−1) ≤ 𝑘0 < 1 for some constant 𝑘0.

(A5): 𝜎(𝑥) ∈ 𝑊 1,2(𝑋), 𝑘 (𝑥, 𝜃, 𝜃′) ∈ 𝑊 1,2(𝑋) for any 𝜃, 𝜃′ ∈ S𝑛−1.

Here (A3) and (A4) are to ensure the well-posedness of forward RTE in H 1

2 , see Proposition 2.4.

(A5) is used in the averaging lemma [32].

Proposition 2.4 ( [1, Theorem 3.2]). For any 𝑆(𝑥) ∈ 𝐿2(𝑋), the boundary value problem (2.1) (2.2)

admits a unique solution 𝑢 ∈ H 1

2 . Moreover, the following estimate holds for some constants

𝐶,

˜𝐶 > 0 independent of 𝑆 and 𝑢 :

𝐶 ∥𝑆∥ 𝐿2 (𝑋) ≤ ∥𝑢∥H 1

2

≤ ˜𝐶 ∥𝑆∥ 𝐿2 (𝑋).

12

Proposition 2.5. The operator M𝑣0 ◦ K𝑣0 ◦ S : 𝐿2(𝑋) → 𝐿2(𝑋) is compact.

Proof. Since 𝑋 is bounded and 𝑆(𝑥) ∈ 𝐶 (𝑋), we have 𝑆(𝑥) ∈ 𝐿2(𝑋), hence 𝑢 ∈ H 1
tion 2.4. Similarly, we have 𝑣0, 𝜎𝑣0 ∈ 𝐿2(𝑋 × S𝑛−1). Moreover,

2 by Proposi-

(cid:12)
(cid:12)
(cid:12)
(cid:12)

1
2

(cid:32)∫

∫

𝑋

S𝑛−1

∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜃′)𝑣0(𝑥, 𝜃′) d𝜃′

(cid:33) 1

2

d𝜃 d𝑥

2

(cid:12)
(cid:12)
(cid:12)
(cid:12)

≤Vol(S𝑛−1)

(cid:18)∫

∫

∫

S𝑛−1
= sup |𝑘 |Vol(S𝑛−1)∥𝑣0∥ 𝐿2 (𝑋×S𝑛−1) < ∞,

S𝑛−1

𝑋

(sup |𝑘 |)2 |𝑣0(𝑥, 𝜃′)|2 d𝜃′ d𝜃 d𝑥

(cid:19) 1

2

then from (2.12), we have 𝜃 · ∇𝑣0(𝑥, 𝜃) ∈ 𝐿2(𝑋 × S𝑛−1), thus 𝑣0 ∈ H 1
2

.

The assumption (A5) ensures 𝜎𝑢𝑣0 ∈ H 1

2 and ∫

S𝑛−1

the Averaging Lemma (see [32, Theorem 1.1]) implies K𝑣0 ◦ S [𝑆] ∈ 𝑊 1

𝑘 (𝑥, 𝜃, 𝜃′)𝑢(𝑥, 𝜃)𝑣0(𝑥, 𝜃) d𝜃′ ∈ H 1
,2(𝑋).

2

2 , then

As the embedding 𝑊 1

,2(𝑋) ↩−→ 𝐿2(𝑋) is compact, the operator K𝑣0 ◦ S is a compact operator
from (𝐶 (𝑋), ∥ · ∥2) to 𝐿2(𝑋), which can be extend to be a compact operator defined on the entire

2

space 𝐿2(𝑋). We slightly abuse the notation and denote such extension again by K𝑣0 ◦ S.

On the other hand, the multiplication operator M𝑣0 can be extended to be a bounded operator
on 𝐿2(𝑋). Thus, the operator M𝑣0 ◦ K𝑣0 ◦ S : 𝐿2(𝑋) → 𝐿2(𝑋), as the composition of a bounded
□

operator with a compact operator, is compact as well.

We therefore have the following result due to the Fredholm alternative.

Theorem 2.6 ( [28, Theorem 5]). Suppose the assumptions (A1)~(A5) hold. If 0 is not an eigenvalue

of the Fredholm operator 𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S, then (𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S)−1 is a bounded linear
operator on 𝐿2(𝑋), and the source 𝑆 can be computed as

𝑆 = (𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S)−1(M𝑣0 [𝐻𝑣0]).

Then the stability estimation is immediately obtained from the inversion formulae

Corollary 2.7. Suppose the assumptions (A1)~(A5) hold. Let 𝑆 and ˜𝑆 be two different sources

with corresponding internal functional 𝐻𝑣0 and ˜𝐻𝑣0, respectively. If 0 is not an eigenvalue of the

13

operator 𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S, then the following stability estimate holds

∥𝑆 − ˜𝑆∥ 𝐿2 (𝑋) ≤ 𝐶 ∥𝐻𝑣0 − ˜𝐻𝑣0 ∥ 𝐿2 (𝑋)

for some constant 𝐶 > 0 depending on 𝜎, 𝑘, 𝑣0, 𝑋 yet independent of 𝑆 and ˜𝑆.

2.2.3 Numerical Experiment

In this section, we test our algorithm in Section 2.2.2. For the numerical experiment, we

consider the two dimension cases, 𝑋 is a square in R2, the coordinate is denoted as (𝑥1, 𝑥2), and we

choose the scattering kernel to be the Henyey-Greenstein function

𝑘 (𝑥, 𝜃, 𝜃′) =

1
2𝜋

1 − 𝑔2
1 + 𝑔2 − 2𝑔 cos 𝜙

,

where 𝜙 is the angle between 𝜃 and 𝜃′, and −1 ≤ 𝑔 ≤ 1 is the anisotropy parameter of the medium.

In this section, we present a number of numerical experiments. We discretize the spatial domain

into a 121 × 121 uniform grid and the angular space into 𝑀 = 8 directions for the forward issue.

In order to prevent the inverse crime, we interpolate the measurement using a spatial 61 × 61

uniform grid for the reconstruction. Employing a 121 × 121 spatial grid with a coarser angular

mesh 𝑀 = 8 and a finer mesh 𝑀 = 16, we compared the forward solutions. Next, a projection of

the 𝑀 = 16 solution onto the coarser mesh is made, and the results are compared with the previous

solution. Relative 𝐿2-error as a result is 0.0447%. All the numerical experiments are performed

on a Windows 10 laptop with Intel Core i7-9750H 2.6GHz CPU and 16GB RAM.

2.2.3.1 RTE Solver

In order to develop a RTE solver, since 𝑢 depend on the spatial domain 𝑋 and angular domain

S1, we uniformly discrete the angular space and use upwind scheme for spatial discretization.

The angular space [0, 2𝜋) is uniformly discretized into 𝑀 angles, denote as 𝜔𝑖 = (𝑖 − 1)Δ𝜔

with Δ𝜔 = 2𝜋

𝑀 . Then the integral over S1 can be discretized using the trapezoidal rule

∫

S1

𝑘 (𝑥, 𝜃′, 𝜃)𝑢(𝑥, 𝜃) d𝜃 ≈

𝑀
∑︁

𝑖=1

𝑘 (𝑥, 𝜃′, 𝜃𝑖)𝑢(𝑥, 𝜃𝑖)Δ𝜔.

14

The spatial discretization uses the upwind scheme, which is

𝜕𝑢
𝜕𝑥1
𝜕𝑢
𝜕𝑥2

(𝑥1, 𝑥2, 𝜃𝑖) ≈ sgn(cos 𝜔𝑖)

(𝑥1, 𝑥2, 𝜃𝑖) ≈ sgn(sin 𝜔𝑖)

𝑢(𝑥1 + sgn(cos 𝜔𝑖)Δ𝑥1, 𝑥2, 𝜃𝑖) − 𝑢(𝑥1, 𝑥2, 𝜃𝑖)
Δ𝑥1
𝑢(𝑥1, 𝑥2 + sgn(sin 𝜔𝑖)Δ𝑥2, 𝜃𝑖) − 𝑢(𝑥1, 𝑥2, 𝜃𝑖)
Δ𝑥2

.

,

where Δ𝑥1 and Δ𝑥2 are the spacings along the 𝑥1-direction and 𝑥2-direction, respectively. Although

the fraction in the scheme above is not an approximation of the derivative when sin 𝜔𝑖 or cos 𝜔𝑖 is

0, but in these cases, the product with the sign function is still 0 as expected.

With the discretization in spatial domain and angular spaces, we use the Jacobi iteration method

to solve the RTE and the adjoint RTE.

Given a known source 𝑆, we generate the measurement 𝐻𝑣 (𝑥) in the following steps. First, we

find the solution 𝑢(𝑥, 𝜃) by solving the forward problem (2.1) (2.2) using the RTE solver. This,

together with the known attenuation coefficient and scattering kernel, is employed to compute

A𝑢(𝑥, 𝜃) in (2.21). Finally, we solve the adjoint RTE (2.12) (2.13) to get 𝑣, and compute 𝐻𝑣 (𝑥)

in (2.20) with the trapezoidal rule. Since 𝜃 ∈ S1 is periodic, the discrete integration is simply

∫

S1

𝑓 (𝜃) d𝜃 ≈

𝑀
∑︁

𝑖=1

𝑓 (𝜃𝑖)Δ𝜃.

We test our algorithms in continuous and discontinuous cases based on the assumptions of

inversion formulae.

2.2.3.2 Neumann Series Inversion

The algorithm for Theorem 2.3 is simple. The operator S can be implemented using the

forward RTE solver, the operator K𝑣0 and M𝑣0 can be discretized using the trapezoidal rule, then

the reconstruction can be done by an iteration, see Algorithm 2.1

2.2.3.3 Fredholm Inversion

The Fredholm inversion in Theorem 2.6 boils down to solving the linear system (2.31). For

this purpose, we discretize the source 𝑆 with respect to some basis functions. Two types of basis

functions are used, one is polynomial functions of the form {𝑥𝑖
1

𝑥 𝑗
2

}𝑖, 𝑗 ≥0, 𝑖+ 𝑗 ≤10, which is used to

15

Data: adjoint RTE solution 𝑣0, measurement 𝐻𝑣0, scattering kernel 𝑘 (𝑥, 𝜃, 𝜃′), attenuation coeffi-

cient 𝜎(𝑥), domain 𝑋.

𝑆 ← 0;
Δ𝑆 ← M𝑣0 [𝐻𝑣0];
𝜀 ← 10−6;
while ∥Δ𝑆∥ 𝐿2 > 𝜀 do
𝑆 ← 𝑆 + Δ𝑆;
Δ𝑆 ← M𝑣0 ◦ K𝑣0 ◦ S [Δ𝑆];

end
return 𝑆;

Algorithm 2.1 Neumann Series Reconstruction.

represent the smooth feature of the source; the other is the following functions

𝑓𝑖 𝑗 = max

(cid:26)

1 − max

(cid:26)

20

(cid:12)
(cid:12)
(cid:12)
(cid:12)

𝑥1 −

𝑖

20

(cid:12)
(cid:12)
(cid:12)
(cid:12)

, 20

(cid:12)
(cid:12)
𝑥2 −
(cid:12)
(cid:12)

𝑗

20

(cid:27)

(cid:12)
(cid:12)
(cid:12)
(cid:12)

(cid:27)

,

, 0

𝑖, 𝑗 ∈ {0, 1, . . . , 20}.

which is inspired from the finite element basis functions. Note 𝑓𝑖 𝑗 a pyramid-shaped function with
the tip at ( 𝑖
20

𝑗
20 ), it captures some information of singularities. We write the expansion of a source

,

𝑆 with respect to these basis functions as

𝑆(𝑥1, 𝑥2) ≈

∑︁

𝑐𝑖 𝑗 𝑥𝑖
1

𝑥 𝑗
2

+

∑︁

𝑖 𝑗 𝑓𝑖 𝑗 (cid:67) ∑︁
𝑐′

˜𝑐𝑖𝑏𝑖,

(2.30)

where 𝑐𝑖 𝑗 , 𝑐′

0≤𝑖, 𝑗 ≤20
𝑖 𝑗 are the coefficients of the expansion. We use {𝑏𝑖 (𝑥1, 𝑥2)} to denote these basis

𝑖, 𝑗 ≥0,𝑖+ 𝑗 ≤10

𝑖

functions and { ˜𝑐𝑖} the correponding coefficients.

Denote T := 𝐼𝑑 + M𝑣0 ◦ K𝑣0 ◦ S, then the internal measurement (2.22) can be represented as

M𝑣0 [𝐻𝑣0] = T [𝑆] ≈

˜𝑐𝑖T [𝑏𝑖].

∑︁

𝑖

We can compute the inner product with T [𝑏 𝑗 ] as follows:

⟨M𝑣0 [𝐻𝑣0], T [𝑏 𝑗 ]⟩ ≈

∑︁

𝑖

˜𝑐𝑖 ⟨T [𝑏𝑖], T [𝑏 𝑗 ]⟩.

(2.31)

Solving the linear equation (2.31) gives the coefficient ˜𝑐𝑖, and then we can numerically reconstruct

the projection of the source 𝑆 on the chosen basis.

2.2.3.4 Experiment 1: Continuous Source

In this experiment, we choose the spatial domain 𝑋 = [0, 0.2]2 with attenuation coefficient

𝜎1(𝑥1, 𝑥2) = 0.1 + 0.1𝑥1 and anisotropy parameter 𝑔 = 0.5 in the scattering kernel. 𝑣0 is chosen as

16

the adjoint solution with boundary condition 𝑣0|Γ+ = 1. The source is chosen as

𝑆1(𝑥1, 𝑥2) = 𝑒−100[(𝑥1−0.08)2+(𝑥2−0.12)2],

see Figure 2.1.

Figure 2.1 Left: source 𝑆1. Right: attenuation coefficient 𝜎1.

Such choice of optical coefficients gives the following numerical values:

∥𝑣0∥𝐶 (𝑋) ≈ 1.2603,

(cid:18)∫

S1

inf
𝑥∈𝑋

(cid:19)

𝑣0(𝑥, 𝜃) d𝜃

≈ 6.4870.

On the other hand, 𝜌 ≡ 1 for any anisotropy parameter between −1 and 1, thus

∥𝑣0∥𝐶 (𝑋) (∥𝜎1∥𝐶 (𝑋) + 𝜌)diam(𝑋)Vol(S1)
(1 − diam(𝑋) 𝜌) inf𝑥∈𝑋

𝑣0(𝑥, 𝜃) d𝜃

(cid:16)∫

S1

(cid:17) ≈ 0.5392 < 1,

thus the optical coefficients satisfy the condition of Theorem 2.3. Applying Neumann series

inversion with different levels of Gaussian noise added to 𝐻𝑣0 gives the following result, see

Figure 2.2.

2.2.3.5 Experiment 2: Discontinuous Source

We choose the spatial domain 𝑋 = [0, 1]2 with attenuation coefficient 𝜎1(𝑥1, 𝑥2) = 0.1 + 0.1𝑥1.

The anisotropy parameter is still 𝑔 = 0.5 and 𝑣0 is chosen as the adjoint solution with boundary

condition 𝑣0|Γ+ = 1. The source 𝑆2 is chosen as the Shepp-Logan Phantom, see left and right panels

of Figure 2.3.

Apply Neumann series inversion with different levels of Gaussian noise added to 𝐻𝑣0 gives the

following result, see Figure 2.4.

17

00.050.10.150.200.020.040.060.080.10.120.140.160.180.20.10.20.30.40.50.60.70.80.9100.050.10.150.200.020.040.060.080.10.120.140.160.180.20.10.1020.1040.1060.1080.110.1120.1140.1160.1180.12Figure 2.2 Reconstructed 𝑆1 using Neumann series. For the first row, 0%, 1%, 2%, 5% random
noises are added to 𝐻𝑣0. The relative 𝐿2 errors of the reconstructions are 0.0268%, 1.0682%,
2.1759%, 5.4680%, respectively. The second row displays the corresponding differences between
the ground truth and the reconstructions.

Figure 2.3 Left: Shepp-Logan Phantom 𝑆2. Center: Smoothed Shepp-Logan Phantom 𝑆3. Right:
attenuation coefficient 𝜎1.

The errors of reconstruction is very large since we choose finitely many continuous basis to

approximate a discontinuous source. If we change the source to the Gaussian filtered Shepp-Logan

Phantom, see the middle panel of Figure 2.3, the reconstruction becomes better, see Figure 2.5.

2.2.3.6 More experiments:

We also test the performance of the inversion formulae in different cases, such as Neumann

inversion beyond the assumption of Theorem 2.3, and the Fredholm inversion in continuous case,

see [28, section 4.2] for more details.

2.3 UMBLT under Diffusive Regime

Biological objects are usually strongly scattering medium. The light propagation in strongly

scattering medium is diffusive. It can be simplified as diffusive regime, which is called the diffusion

approximation [5], see also Appendix A. The diffusive regime models the light propagation using

18

00.050.10.150.200.020.040.060.080.10.120.140.160.180.20.10.20.30.40.50.60.70.80.9100.050.10.150.200.020.040.060.080.10.120.140.160.180.20.10.20.30.40.50.60.70.80.9100.050.10.150.200.020.040.060.080.10.120.140.160.180.20.10.20.30.40.50.60.70.80.9100.050.10.150.200.020.040.060.080.10.120.140.160.180.20.10.20.30.40.50.60.70.80.911.100.050.10.150.200.020.040.060.080.10.120.140.160.180.212345610-400.050.10.150.200.020.040.060.080.10.120.140.160.180.20.0050.010.0150.020.0250.0300.050.10.150.200.020.040.060.080.10.120.140.160.180.20.010.020.030.040.050.060.0700.050.10.150.200.020.040.060.080.10.120.140.160.180.20.020.040.060.080.10.120.140.1600.20.40.60.8100.10.20.30.40.50.60.70.80.9100.10.20.30.40.50.60.70.80.9100.20.40.60.8100.10.20.30.40.50.60.70.80.9100.050.10.150.20.250.30.3500.20.40.60.8100.10.20.30.40.50.60.70.80.910.10.110.120.130.140.150.160.170.180.190.2Figure 2.4 Reconstructed 𝑆2 using Fredholm inversion. For the first row, 0%, 1%, 2%, 5% random
noises are added to 𝐻𝑣0. The relative 𝐿2 errors of the reconstructions are 57.5806%, 57.5818%,
57.5880%, 57.6199%, respectively. The second row displays the corresponding differences
between the ground truth and the reconstructions.

Figure 2.5 Reconstructed 𝑆3 using Fredholm inversion. For the first row, 0%, 1%, 2%, 5% random
noises are added to 𝐻𝑣0. The relative 𝐿2 errors of the reconstructions are 4.3211%, 4.3405%,
4.4136%, 5.0152%, respectively. The second row displays the corresponding differences between
the ground truth and the reconstructions.

the following diffusion equation [8]

−∇ · 𝐷 (𝑥)∇𝜙(𝑥) + 𝜎𝑎 (𝑥)𝜙(𝑥) = 𝑆(𝑥)

in 𝑋.

𝜙 + ℓ𝜈 · 𝐷∇𝜙 = 0

on 𝜕 𝑋.

(2.32)

(2.33)

where the positive definite matrix function 𝐷 is the diffusion coefficient, 𝜎𝑎 is the absorption

coefficient, ℓ is the extrapolation length, 𝜈 is the outer normal vector and 𝜙 is the angularly averaged

intensity of light.

19

00.20.40.60.8100.10.20.30.40.50.60.70.80.91-0.200.20.40.60.800.20.40.60.8100.10.20.30.40.50.60.70.80.91-0.200.20.40.60.800.20.40.60.8100.10.20.30.40.50.60.70.80.91-0.200.20.40.60.800.20.40.60.8100.10.20.30.40.50.60.70.80.91-0.200.20.40.60.800.20.40.60.8100.10.20.30.40.50.60.70.80.910.10.20.30.40.50.60.700.20.40.60.8100.10.20.30.40.50.60.70.80.910.10.20.30.40.50.60.700.20.40.60.8100.10.20.30.40.50.60.70.80.910.10.20.30.40.50.60.700.20.40.60.8100.10.20.30.40.50.60.70.80.910.10.20.30.40.50.60.700.20.40.60.8100.10.20.30.40.50.60.70.80.9100.050.10.150.20.250.30.3500.20.40.60.8100.10.20.30.40.50.60.70.80.9100.050.10.150.20.250.30.3500.20.40.60.8100.10.20.30.40.50.60.70.80.9100.050.10.150.20.250.30.3500.20.40.60.8100.10.20.30.40.50.60.70.80.9100.050.10.150.20.250.30.3500.20.40.60.8100.10.20.30.40.50.60.70.80.910.0050.010.0150.020.0250.0300.20.40.60.8100.10.20.30.40.50.60.70.80.910.0050.010.0150.020.02500.20.40.60.8100.10.20.30.40.50.60.70.80.910.0050.010.0150.020.02500.20.40.60.8100.10.20.30.40.50.60.70.80.910.0050.010.0150.020.0250.03The ultrasound modulation is modeled as [8]

𝐷𝜀 (𝑥) (cid:66) (1 + 𝜀(2𝛾 − 1) cos(𝑞 · 𝑥 + 𝜑))𝐷 (𝑥),

𝜎𝑎,𝜀 (𝑥) (cid:66) (1 + 𝜀(2𝛾 + 1) cos(𝑞 · 𝑥 + 𝜑))𝜎𝑎 (𝑥),

𝑆𝜀 (𝑥) (cid:66) (1 + 𝜀 cos(𝑞 · 𝑥 + 𝜑))𝑆(𝑥),

(2.34)

(2.35)

(2.36)

where 𝛾 is the elasto-optical constant, 𝑞 is the wave vector, 𝜑 is the phase and 0 ≤ 𝜀 ≪ 1 is a small

parameter related to the amplitude, frequency, time, density and acoustic wave speed.

Then the UMBLT under diffusion approximation is modeled as

−∇ · 𝐷𝜀 (𝑥)∇𝜙𝜀 (𝑥) + 𝜎𝑎,𝜀 (𝑥)𝜙𝜀 (𝑥) = 𝑆𝜀 (𝑥)

in 𝑋.

𝜙𝜀 + ℓ𝜈 · 𝐷𝜀∇𝜙𝜀 = 0

on 𝜕 𝑋.

(2.37)

(2.38)

Under different modulations, our measurement is the Neumann boundary value

−𝜈 · 𝐷𝜀∇𝜙𝜀 |G,

𝜀 ≥ 0,

which is considered as the output flux on boundary. Here G ⊂ 𝜕 𝑋 is a relatively open subset which

denotes the region of measurement. The goal is to reconstruct the distribution of 𝑆 assuming 𝐷,

𝜎𝑎, Γ and 𝑋 are given.

2.3.1 Full Data Case

In this section, we start with a simple case G = 𝜕 𝑋, which means we can make measurement on

the entire boundary. Such case is called UMBLT under diffusive regime with full data [8]. Similar

to the UMBLT under transport regime, we consider the adjoint diffusion equation with prescribed

Robin boundary condition 𝑔

−∇ · 𝐷 (𝑥)∇𝜓(𝑥) + 𝜎𝑎 (𝑥)𝜓(𝑥) = 0

in 𝑋.

𝜓 + 𝑙𝜈 · 𝐷∇𝜓 = 𝑔

on 𝜕 𝑋.

(2.39)

(2.40)

Since 𝐷 and 𝜎𝑎 are known, we know 𝜓 in the entire space 𝑋. Multiply 𝜓 on both sides of (2.37),

multiply 𝜙𝜀 on both sides of (2.39), subtract two equations and integrate over 𝑋. Integration by

20

parts gives

∫

𝜕 𝑋

𝑔𝜈 · 𝐷𝜀∇𝜙𝜀 d𝑥 =

∫

𝑋

(𝐷𝜀 − 𝐷0)∇𝜙𝜀 · ∇𝜓 + (𝜎𝑎,𝜀 − 𝜎𝑎,0)𝜙𝜀𝜓 − 𝜓𝑆𝜀 d𝑥,

(2.41)

By the asymptotic expansion on 𝜀, write 𝜈 · 𝐷𝜀∇𝜙𝜀 = 𝜈 · 𝐷∇𝜙0 + 𝜀𝛿Φ, the first order term gives

∫

𝜕 𝑋

𝑔𝛿Φ d𝑥 =

∫

𝑋

((2𝛾 − 1)𝐷∇𝜙0 · ∇𝜓 + (2𝛾 + 1)𝜎𝑎𝜙0𝜓 − 𝜓𝑆) cos(𝑞 · 𝑥 + 𝜑) d𝑥.

(2.42)

Notice that the LHS is theoretically known from the Neumann boundary measurement and the

adjoint boundary condition, RHS is an inner product of a cosine function and another function over

𝑋. By varying 𝑞 and 𝜑, the RHS is exactly the Fourier coefficient of a function’s Fourier transform,

and we denote the function as 𝐻𝜓:

𝐻𝜓 (cid:66) (2𝛾 − 1)𝐷∇𝜙0 · ∇𝜓 + (2𝛾 + 1)𝜎𝑎𝜙0𝜓 − 𝜓𝑆.

(2.43)

For a specific 𝜓0 > 0, dividing 𝜓0 on both sides and substitute (2.32) to replace the 𝑆 term, we have

the following PDE

𝐹𝜓0

(cid:66)

𝐻𝜓0
𝜓0

= (2𝛾 − 1)𝐷∇𝜙0 · ∇ log 𝜓0 + 2𝛾𝜎𝑎𝜙0 + ∇ · 𝐷 (𝑥)∇𝜙0

(2.44)

with boundary condition (2.33). Once 0 is not an eigenvalue of the PDE, the solution 𝜙0 can be

uniquely determined, substitute 𝜙0 to (2.32) will give us 𝑆.

2.3.2 Partial Data Case

In the proposed full data algorithms, see Section 2.3.1, the measurement is required on the

entire boundary 𝜕 𝑋. But in practice, it is hard to obtain all data on the boundary. For example,

for the brain imaging, the sensors could only be placed on the top half of head, the data in the

neck-direction will lost. A natural question is: Can we determine the source from measurements

only on a subset on the boundary? Such question is considered as inverse problem with partial data.

If the sources can be determined from partial data, then the full data measurement is redundant, the

partial data algorithm would be an improved algorithm for those sources.

Suppose we can only measure on a relatively open subset G ⊂ 𝜕 𝑋, i.e. 𝛿Φ in (2.42) is known

only on a subset G. A natural idea is to choose a special prescribed boundary condition 𝑔 such that

21

𝑔 is supported on G, then the left hand side of (2.42) is still known from the boundary measurement

and boundary condition 𝑔. In this case, we need to show that there exist such boundary condition

𝑔 support on G and gives positive adjoint solution 𝜓.

Instead of consider adjoint equation (2.39) with Robin boundary condition (2.40), we consider

the following diffusion equation with mixed boundary condition

−∇ · 𝐷 (𝑥)∇𝜓(𝑥) + 𝜎𝑎 (𝑥)𝜓(𝑥) = 0

in 𝑋.

𝜓 + ℓ𝜈 · 𝐷∇𝜓 = 0

on 𝜕 𝑋 \ G.

𝜓 = 𝑓

on G.

(2.45)

(2.46)

(2.47)

Once we find a positive solution 𝜓 to this mixed boundary value problem, we can simply take

𝑔 = (𝜓 + ℓ𝜈 · 𝐷∇𝜓)|𝜕 𝑋 to be the prescribed boundary condition in the adjoint problem (2.39)-

(2.40).

Throughout this section, the following hypotheses are made upon the anisotropic diffusion

coefficient 𝐷 (𝑥) and the scattering coefficient 𝜎𝑎 (𝑥):

H1 𝐷 (𝑥) = 𝐼 near 𝜕 𝑋, where 𝐼 is the identity matrix.

H2 𝜎𝑎 ∈ 𝐶𝛼 (𝑋), 𝐷𝑖 𝑗 ∈ 𝐶1,𝛼 (𝑋)

H3 𝐷 (𝑥) is uniformly elliptic for all 𝑥 ∈ 𝑋, that is, there exists a constant 𝜆 > 0 such that

1
𝜆

|𝜉 |2 ≥ 𝜉⊤𝐷 (𝑥)𝜉 ≥ 𝜆|𝜉 |2

a.e. on 𝑋

holds for any 𝜉 ∈ R𝑛.

H4 𝜎𝑎 ≥ 0 a.e. on 𝑋.

Theorem 2.8 ( [65, Theorem 1]). Assume that

𝜎𝑎 ∈ 𝐶𝛼 (𝑋),

𝐷𝑖 𝑗 ∈ 𝐶1,𝛼 (𝑋),

𝑓 ∈ 𝐶 (G) ∩ 𝐿∞(G),

then (2.45)-(2.47) has a unique solution 𝜓 ∈ 𝐶2(𝑋 \ G) ∩ 𝐶0(𝑋)

22

Theorem 2.9 ( [97]). Under the hypotheses H1-H4.

If the Dirichlet boundary condition 𝑓 ∈

𝐶 (G) ∩ 𝐿∞(G) is positive, then the mixed boundary value problem (2.45)-(2.47) admits a unique

solution 𝜓 ∈ 𝐶2(𝑋 \ G) ∩ 𝐶0(𝑋) which is positive on 𝑋.

Proof. Theorem 2.8 ensure that (2.45)-(2.47) have a unique solution 𝜓 ∈ 𝐶2(𝑋 \ G) ∩ 𝐶0(𝑋).

Assuming that 𝜓 has a negative value on 𝑋, the minimum is reached on the boundary 𝜕 𝑋,

according to the weak maximum principle [41, Section 6.4 Theorem 2]. The minimum is attained

on 𝜕 𝑋 \ G since 𝜓|G > 0. Assume that 𝜓(𝑥0) = inf𝑥∈𝑋 𝜓 < 0 and that 𝑥0 ∈ 𝜕 𝑋 \ G. The Robin
boundary condition (2.46) states that 𝜈 · 𝐷∇𝜓 > 0, meaning that 𝜕𝜈𝜓(𝑥0) > 0 because 𝐷 (𝑥) = 𝐼 is
close to 𝜕 𝑋. This contradicts the assumption 𝜓(𝑥0) = inf𝑥∈𝑋 𝜓, implying that 𝜓 ≥ 0.

The strong maximum principle [41, Section 6.4 Theorem 4] indicate that 𝜓| 𝑋 > 0, otherwise

𝜓 ≡ 0, which contradict to 𝜓|G = 𝑓 > 0. It remains to prove 𝜓|𝜕 𝑋 > 0.

Assume on 𝑥0 ∈ 𝜕 𝑋 \ G, 𝜓(𝑥0) = inf𝑥∈𝑋 𝜓 = 0. Apply the Hopf Lemma [41, Section 6.4
Lemma] on −𝜓 shows that 𝜕𝜈𝜓(𝑥0) < 0, which contradict to the boundary condition 𝜓(𝑥0) + ℓ𝜈 ·

𝐷∇𝜓 = 0, thus 𝜓|𝜕 𝑋 > 0.

Thus 𝜓 is a positive solution in 𝐶2(𝑋 \ G) ∩ 𝐶0(𝑋). Since 𝑋 is compact, 𝜓 have positive lower

bound.

□

The theorem above ensures that one can construct a suitable positive adjoint solution 𝜓, then

the source 𝑆 can be recovered under the same process as the full data case.

2.3.3 Uncertainty Quantification

The reconstruction methods described in Section 2.3.1 and Section 2.3.2 depend primarily

on precise prior knowledge of the optical coefficients (𝐷, 𝜎) to solve the elliptic equation (2.44)

(along with boundary conditions) for 𝜙0. The rationale is that these optical coefficients can be

measured beforehand using other imaging modalities, such as optical tomography [5]. In practice,

the imaging process in these additional modalities inevitably introduces inaccuracies in the optical

coefficients, which have an impact on UMBLT reconstruction. In the following two sections, we

will use the continuous and discretized models to quantify the impact on the reconstruction of the

23

bio-luminescence source 𝑆 caused by optical coefficient inaccuracies.

Let (𝐷, 𝜎𝑎) be the underlying true optical coefficients, and ( ˜𝐷, ˜𝜎𝑎) be the optical coefficients

that are reconstructed through additional imaging modalities before performing UMBLT. Observe

that ( ˜𝐷, ˜𝜎𝑎) do not play a role in the derivation of the internal data: This is because the boundary

integral on the left hand side of (2.42) remains the same, thus we can derive 𝐻𝜓 as before. Hereafter,

we will assume that the internal data 𝐻𝜓 has been accurately extracted, and focus on quantifying

the uncertainty of the reconstructed source 𝑆. The full data and partial data cases will be handled

in one shot, since the reconstruction process are identical once a suitable positive adjoint solution

𝜓0 > 0 is chosen.

2.3.3.1 Uncertainty Quantification with Continuous Diffusive Model

We record a regularity result for the diffusion equation with Robin boundary conditions.

Proposition 2.10 ( [33, Theorem 2.4]). Suppose 𝐷 is uniformly elliptic, 𝐷𝑖 𝑗 ∈ 𝐿∞(𝑋), 𝜎𝑎 ≥ 0 a.e.

For 𝑆 ∈ 𝐿2(𝑋) and 𝑔 ∈ 𝐻 1

2 (𝜕 𝑋), the following boundary value problem

−∇ · 𝐷 (𝑥)∇𝜙(𝑥) + 𝜎𝑎 (𝑥)𝜙(𝑥) = 𝑆(𝑥)

in 𝑋.

𝜙 + ℓ𝜈 · 𝐷∇𝜙 = 𝑔

on 𝜕 𝑋.

admits a unique solution 𝜙 ∈ 𝐻2(𝑋) with the estimation

∥𝜙∥𝐻2 (𝑋) ≤ 𝐶 (∥𝑆∥ 𝐿2 (𝑋) + ∥𝑔∥

)

𝐻

1
2 (𝜕 𝑋)

where 𝐶 is a constant independent of 𝜙.

(2.48)

(2.49)

(2.50)

Then we have the following global uncertainty quantification (UQ) estimate for the aforemen-

tioned UMBLT reconstruction in the diffusive regime.

Theorem 2.11 ( [97]). Suppose all optical coefficients and solutions satisfy

∥𝐷𝑖 𝑗 ∥𝑊 1,∞ (𝑋), ∥ ˜𝐷𝑖 𝑗 ∥𝑊 1,∞ (𝑋) ≤ 𝐶𝐷,

∥𝜙∥𝑊 2,∞ (𝑋), ∥ ˜𝜙∥𝑊 2,∞ (𝑋) ≤ 𝐶𝜙,

∥𝜓∥𝑊 2,∞ (𝑋), ∥ ˜𝜓∥𝑊 2,∞ (𝑋) ≤ 𝐶𝜓,

∥𝜎𝑎 ∥ 𝐿∞ (𝑋) ≤ 𝐶𝜎,

𝜓, ˜𝜓 ≥ 𝑐𝜓 > 0,

24

where 𝐶𝐷, 𝐶𝜙, 𝐶𝜓, 𝐶𝜎, 𝑐𝜓 are constants, and 0 is not eigenvalue of the following operators equipped

with the zero Robin boundary condition:

∇ · 𝐷∇ + (2𝛾 − 1)𝐷∇ log 𝜓0 · ∇ + 2𝛾𝜎𝑎, ∇ · ˜𝐷∇ + (2𝛾 − 1) ˜𝐷∇ log ˜𝜓0 · ∇ + 2𝛾 ˜𝜎𝑎,

then we can find constants 𝐶1𝑖 𝑗 , 𝐶2 > 0 such that

∥𝑆 − ˜𝑆∥ 𝐿2 (𝑋) ≤

∑︁

𝑖≤ 𝑗

𝐶1𝑖 𝑗 ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥𝐻1 (𝑋) + 𝐶2∥𝜎𝑎 − ˜𝜎𝑎 ∥ 𝐿2 (𝑋)

(2.51)

Proof. Let 𝜙 and ˜𝜙 solve the diffusion equations

𝑆 = −∇ · [𝐷∇𝜙] + 𝜎𝑎𝜙,

˜𝑆 = −∇ · [ ˜𝐷∇ ˜𝜙] + ˜𝜎𝑎 ˜𝜙,

respectively. Subtract these equations to get

𝑆 − ˜𝑆 = −∇ · [(𝐷 − ˜𝐷)∇𝜙] − ∇ · [ ˜𝐷∇(𝜙 − ˜𝜙)] + (𝜎𝑎 − ˜𝜎𝑎)𝜙 + ˜𝜎𝑎 (𝜙 − ˜𝜙).

Taking the 𝐿2-norms on both sides, we have

∥𝑆 − ˜𝑆∥ 𝐿2 (𝑋)

≤ ∥∇ · [(𝐷 − ˜𝐷)∇𝜙] ∥ 𝐿2 (𝑋) + ∥∇ · [ ˜𝐷∇(𝜙 − ˜𝜙)] ∥ 𝐿2 (𝑋) + ∥(𝜎𝑎 − ˜𝜎𝑎)𝜙∥ 𝐿2 (𝑋) + ∥ ˜𝜎𝑎 (𝜙 − ˜𝜙) ∥ 𝐿2 (𝑋)

∑︁

≤

𝑖 𝑗

∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋) ∥𝜕𝑖 (𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋) +

∑︁

𝑖 𝑗

∥𝜕𝑖 𝑗 𝜙∥ 𝐿∞ (𝑋) ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋)

∑︁

+

𝑖 𝑗

∥𝜕𝑖 ˜𝐷𝑖 𝑗 ∥ 𝐿∞ (𝑋) ∥𝜕𝑗 (𝜙 − ˜𝜙)∥ 𝐿2 (𝑋) +

∑︁

𝑖 𝑗

∥ ˜𝐷𝑖 𝑗 ∥ 𝐿∞ (𝑋) ∥𝜕𝑖 𝑗 (𝜙 − ˜𝜙) ∥ 𝐿2 (𝑋)

+ ∥𝜙∥ 𝐿∞ (𝑋) ∥𝜎𝑎 − ˜𝜎𝑎 ∥ 𝐿2 (𝑋) + ∥ ˜𝜎𝑎 ∥ 𝐿∞ (𝑋) ∥𝜙 − ˜𝜙∥ 𝐿2 (𝑋)

≤𝑐1∥𝜙 − ˜𝜙∥𝐻2 (𝑋) +

∑︁

𝑖≤ 𝑗

𝑐2𝑖 𝑗 ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥𝐻1 (𝑋) + 𝑐3∥𝜎𝑎 − ˜𝜎𝑎 ∥ 𝐿2 (𝑋)

where the constants 𝑐1, 𝑐2𝑖 𝑗 , 𝑐3 > 0 can be made explicit as follows:

(cid:118)(cid:117)(cid:117)(cid:116)

𝑐1 =

∥ ˜𝜎𝑎 ∥2

𝐿∞ (𝑋) +

(cid:34)

∑︁

∑︁

𝑗

𝑖

(cid:35) 2

∥𝜕𝑖 ˜𝐷𝑖 𝑗 ∥ 𝐿∞ (𝑋)

+ 4

∑︁

𝑖< 𝑗

∥ ˜𝐷𝑖 𝑗 ∥2

𝐿∞ (𝑋) +

∑︁

𝑖

∥ ˜𝐷𝑖𝑖 ∥2

𝐿∞ (𝑋)

(2.52)

𝑐2𝑖 𝑗 =

√︃

4∥𝜕𝑖 𝑗 𝜙∥2

𝐿∞ (𝑋) + (cid:0)∥𝜕𝑖 𝜙∥ 𝐿∞ (𝑋) + ∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋)

(cid:1) 2

(𝑖 < 𝑗)

√︃

𝑐2𝑖𝑖 =

∥𝜕𝑖𝑖 𝜙∥2

𝐿∞ (𝑋) + ∥𝜕𝑖 𝜙∥2

𝐿∞ (𝑋)

𝑐3 = ∥𝜙∥ 𝐿∞ (𝑋)

25

In order to estimate the term ∥𝜙 − ˜𝜙∥𝐻2 (𝑋), we turn to the second order elliptic equations

generated from the internal data 𝐻𝜓 = 𝐻 ˜𝜓:

𝐹𝜓 =

𝐹 ˜𝜓 =

𝐻𝜓
𝜓
𝐻𝜓
˜𝜓

= (2𝛾 − 1)𝐷∇𝜙 · ∇ log 𝜓 + 2𝛾𝜎𝑎𝜙 + ∇ · 𝐷∇𝜙

= (2𝛾 − 1) ˜𝐷∇ ˜𝜙 · ∇ log ˜𝜓 + 2𝛾 ˜𝜎𝑎 ˜𝜙 + ∇ · ˜𝐷∇ ˜𝜙.

Subtracting these equations gives

− ∇ · ˜𝐷∇[𝜙 − ˜𝜙] − 2𝛾 ˜𝜎𝑎 (𝜙 − ˜𝜙) − (2𝛾 − 1) ˜𝐷∇(𝜙 − ˜𝜙) · ∇ log 𝜓

=

𝐻𝜓
𝜓 ˜𝜓

(𝜓 − ˜𝜓) + (2𝛾 − 1)(𝐷 − ˜𝐷)∇𝜙 · ∇ log 𝜓

+ (2𝛾 − 1) ˜𝐷∇ ˜𝜙 · (∇ log 𝜓 − ∇ log ˜𝜓) + 2𝛾(𝜎𝑎 − ˜𝜎𝑎)𝜙 + ∇ · [𝐷 − ˜𝐷]∇𝜙,

This is a second order elliptic equation for 𝜙 − ˜𝜙 with zero Robin boundary condition, we have the

following regularity estimate by Proposition 2.10:

∥𝜙 − ˜𝜙∥ 𝐻2 (𝑋)
𝐻𝜓
𝜓 ˜𝜓

(cid:18)(cid:13)
(cid:13)
(cid:13)
(cid:13)

(𝜓 − ˜𝜓)

≤𝐶

(cid:13)
(cid:13)
(cid:13)
(cid:13)𝐿2 (𝑋)

+ |2𝛾 − 1|∥(𝐷 − ˜𝐷)∇𝜙 · ∇ log 𝜓∥ 𝐿2 (𝑋)

+ |2𝛾 − 1|∥ ˜𝐷∇ ˜𝜙 · (∇ log 𝜓 − ∇ log ˜𝜓)∥ 𝐿2 (𝑋) + ∥∇ · [𝐷 − ˜𝐷]∇𝜙∥ 𝐿2 (𝑋) + |2𝛾|∥ (𝜎𝑎 − ˜𝜎𝑎)𝜙∥ 𝐿2 (𝑋)
(cid:18) ∥𝐻𝜓 ∥ 𝐿∞ (𝑋)
𝑐2
𝜓

∥𝜕𝑖 log 𝜓∥ 𝐿∞ (𝑋) ∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋) ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋)

∥𝜓 − ˜𝜓∥ 𝐿2 (𝑋) + |2𝛾 − 1|

∑︁

𝑖 𝑗

≤𝐶

(cid:19)

+ |2𝛾 − 1|

∑︁

𝑖 𝑗

∥ ˜𝐷𝑖 𝑗 ∥ 𝐿∞ (𝑋) ∥𝜕𝑗 ˜𝜙∥ 𝐿∞ (𝑋) ∥𝜕𝑖 (log 𝜓 − log ˜𝜓)∥ 𝐿2 (𝑋) +

∑︁

𝑖 𝑗

∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋) ∥𝜕𝑖 (𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋)

∥𝜕𝑖 𝑗 𝜙∥ 𝐿∞ (𝑋) ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋) + |2𝛾|∥𝜙∥ 𝐿∞ (𝑋) ∥𝜎𝑎 − ˜𝜎𝑎 ∥ 𝐿2 (𝑋)

(cid:19)

∑︁

+

𝑖 𝑗

≤𝑐4∥𝜓 − ˜𝜓∥ 𝐻1 (𝑋) +

∑︁

𝑖 ≤ 𝑗

𝑐5𝑖 𝑗 ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐻1 (𝑋) + 𝑐6∥𝜎𝑎 − ˜𝜎𝑎 ∥ 𝐿2 (𝑋)

(2.53)

where in the last inequality, we used the upper bound ∥𝜕𝑖 log 𝜓∥ 𝐿∞ (𝑋) ≤ 1
𝑐 𝜓

∥𝜕𝑖𝜓∥ 𝐿∞ (𝑋) and

∥𝜕𝑖 (log 𝜓 − log ˜𝜓)∥ 𝐿2 (𝑋) ≤

≤

1
𝑐2
𝜓
1
𝑐2
𝜓

∥𝜓𝜕𝑖 ˜𝜓 − ˜𝜓𝜕𝑖𝜓∥ 𝐿2 (𝑋) =

1
𝑐2
𝜓

∥(𝜓 − ˜𝜓)𝜕𝑖 ˜𝜓 − ˜𝜓𝜕𝑖 (𝜓 − ˜𝜓) ∥ 𝐿2 (𝑋)

∥𝜕𝑖 ˜𝜓∥ 𝐿∞ (𝑋) ∥𝜓 − ˜𝜓∥ 𝐿2 (𝑋) +

1
𝑐2
𝜓

∥ ˜𝜓∥ 𝐿∞ (𝑋) ∥𝜕𝑖 (𝜓 − ˜𝜓) ∥ 𝐿2 (𝑋)

26

The constants 𝑐4, 𝑐5𝑖 𝑗 , 𝑐6 > 0 are defeind as

𝑐4 =

𝐶 |2𝛾 − 1|
𝑐2
𝜓

(cid:32)

∑︁

𝑗


∑︁






𝑖

∥ ˜𝐷𝑖 𝑗 ∥ 𝐿2 (𝑋) ∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋) ∥ ˜𝜓∥ 𝐿∞ (𝑋)

(cid:33) 2

(cid:32) ∥𝐻𝜓 ∥ 𝐿∞ (𝑋)
|2𝛾 − 1|

+

∑︁

+

𝑖 𝑗

∥ ˜𝐷𝑖 𝑗 ∥ 𝐿2 (𝑋) ∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋) ∥𝜕𝑖 ˜𝜓∥ 𝐿∞ (𝑋)

(cid:32)(cid:18)

𝑐5𝑖 𝑗 =𝐶 ·

2∥𝜕𝑖 𝑗 𝜙∥ 𝐿∞ (𝑋) +

|2𝛾 − 1|
𝑐 𝜓

∥𝜕𝑖𝜓∥ 𝐿∞ (𝑋) ∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋) +

1
2

(cid:33) 2





|2𝛾 − 1|
𝑐 𝜓

(cid:16)

+

∥𝜕𝑖 𝜙∥ 𝐿∞ (𝑋) + ∥𝜕𝑗 𝜙∥ 𝐿∞ (𝑋)

(cid:33) 1
2

(cid:17) 2

(𝑖 < 𝑗)

∥𝜕𝑗𝜓∥ 𝐿∞ (𝑋) ∥𝜕𝑖 𝜙∥ 𝐿∞ (𝑋)

(cid:19) 2

√︄(cid:18)

𝑐5𝑖𝑖 =𝐶 ·

∥𝜕𝑖𝑖𝜙∥ 𝐿∞ (𝑋) +

|2𝛾 − 1|
𝑐 𝜓

∥𝜕𝑖𝜓∥ 𝐿∞ (𝑋) ∥𝜕𝑖 𝜙∥ 𝐿∞ (𝑋)

(cid:19) 2

+ ∥𝜕𝑖 𝜙∥2

𝐿∞ (𝑋)

𝑐6 =|2𝛾|𝐶 · ∥𝜙∥ 𝐿∞ (𝑋)

It remains to estimate the term ∥𝜓 − ˜𝜓∥𝐻1 (𝑋). Let us consider the adjoint equations

−∇ · 𝐷∇𝜓 + 𝜎𝑎𝜓 = 0,

−∇ · ˜𝐷∇ ˜𝜓 + ˜𝜎𝑎 ˜𝜓 = 0.

(2.54)

Subtract these two equations to get

−∇ · ˜𝐷∇(𝜓 − ˜𝜓) + ˜𝜎𝑎 (𝜓 − ˜𝜓) = ∇ · (𝐷 − ˜𝐷)∇𝜓 − (𝜎𝑎 − ˜𝜎𝑎)𝜓

(2.55)

This is a second order elliptic equation for 𝜓 − ˜𝜓 with the zero Robin boundary condition. Again,

by the elliptic regularity result, we have

∥𝜓 − ˜𝜓∥𝐻1 (𝑋)

≤𝐶 (∥∇ · [(𝐷 − ˜𝐷)∇𝜓] ∥ 𝐿2 (𝑋) + ∥(𝜎𝑎 − ˜𝜎𝑎)𝜓∥ 𝐿2 (𝑋))

≤𝐶

(cid:18)∑︁

𝑖 𝑗

∑︁

+

𝑖 𝑗

∥𝜕𝑗 𝜓∥ 𝐿∞ (𝑋) ∥𝜕𝑖 (𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋)

∥𝜕𝑖 𝑗 𝜓∥ 𝐿∞ (𝑋) ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥ 𝐿2 (𝑋) + ∥𝜓∥ 𝐿∞ (𝑋) ∥(𝜎𝑎 − ˜𝜎𝑎) ∥ 𝐿2 (𝑋)

(2.56)

(cid:19)

≤

∑︁

𝑖≤ 𝑗

𝑐7𝑖 𝑗 ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥𝐻1 (𝑋) + 𝑐8∥𝜎𝑎 − ˜𝜎𝑎 ∥ 𝐿2 (𝑋)

27

with constants 𝑐7𝑖 𝑗 , 𝑐8 > 0, where

𝑐7𝑖 𝑗 = 𝐶 ·

𝑐7𝑖𝑖 = 𝐶 ·

√︃(cid:0)∥𝜕𝑖𝜓∥ 𝐿∞ (𝑋) + ∥𝜕𝑗 𝜓∥ 𝐿∞ (𝑋)
√︃

∥𝜕𝑖𝜓∥2

𝐿∞ (𝑋) + ∥𝜕𝑖𝑖𝜓∥2

𝐿∞ (𝑋)

(cid:1) 2 + 4∥𝜕𝑖 𝑗 𝜓∥2

𝐿∞ (𝑋)

(𝑖 < 𝑗)

𝑐8 = 𝐶 · ∥𝜓∥ 𝐿∞ (𝑋)

Combining (2.52) (2.53) (2.56), we conclude that

∥𝑆 − ˜𝑆∥ 𝐿2 (𝑋) ≤

∑︁

𝑖≤ 𝑗

𝐶1𝑖 𝑗 ∥(𝐷 − ˜𝐷)𝑖 𝑗 ∥𝐻1 (𝑋) + 𝐶2∥𝜎 − ˜𝜎∥ 𝐿2 (𝑋),

(2.57)

with 𝐶1𝑖 𝑗 = 𝑐1𝑐4𝑐7𝑖 𝑗 + 𝑐1𝑐5𝑖 𝑗 + 𝑐2𝑖 𝑗 and 𝐶2 = 𝑐1𝑐4𝑐8 + 𝑐1𝑐6 + 𝑐3. Note that all the constants in this

proof are explicit, except for the constant 𝐶 that comes from the estimate of elliptic regularity. □

Remark 2.12. Theorem 2.11 can be interpreted as follows. Squaring the estimate (2.51) gives

∥𝑆 − ˜𝑆∥2

𝐿2 (𝑋) ≤ ℭ

(cid:16)

∥𝐷 − ˜𝐷 ∥2

𝐻1 (𝑋) + ∥𝜎𝑎 − ˜𝜎𝑎 ∥2

𝐿2 (𝑋)

(cid:17)

where the constant ℭ is in terms of 𝐶1𝑖 𝑗 and 𝐶2. If we take 𝑆, 𝐷, 𝜎𝑎 to be the underlying ground-truth

parameters and ˜𝑆, ˜𝐷, ˜𝜎𝑎 the corresponding parameters with random uncertainty of mean zero, then

E[ ˜𝑆] = 𝑆, E[ ˜𝐷] = 𝐷, E[ ˜𝜎𝑎] = 𝜎𝑎. Therefore, the estimate provides a quantitative error bound on

the variance of the bioluminescent source.

2.3.3.2 Uncertainty Quantification with Discretized Diffusive Model

In the previous section, we considered the impact of inaccurate (𝐷, 𝜎𝑎) using continuous PDE

models. However, for the subsequent numerical simulation, the PDEs have to be discretized into

finite dimensional discrete models. This motivates us to study a similar UQ problem based on

the finite difference discretization of the PDE model. The analysis in this section provides a finite

dimensional counterpart of the infinite dimensional UQ estimate (2.51), bridging the gap between

the infinite dimensional analysis and the finite dimensional numerical experiments.

We will consider the discretization of three diffusion-type equations: the forward problem (2.37)

(2.38), the adjoint problem (2.39) (2.40), and the internal data problem (2.43) equipped with the

28

zero Robin boundary condition. These problems need to be discretized in order to implement the

reconstruction procedure outlined in Section 2.3.1. The discretization procedure requires numerical

evaluation of the terms ∇ · 𝐷∇𝜙0, 𝐷∇𝜙0 · ∇ log 𝜓0, and 𝜎𝑎𝜙0. The last term can be readily evaluated

on a grid. In the following, we explain how to discretize the first two differential operators using

the staggered grid scheme.

We take 𝑋 to be a 2D domain to agree with the setup of the subsequent numerical experiments.

The 2D coordinates are written as (𝑥, 𝑦). The problem in 3D can be considered likewise with

an additional spatial variable. Let Δ𝑥, Δ𝑦 denote the grid size on the 𝑥-direction and 𝑦-direction,

respectively. We will discretize the divergence-form diffusion operator using the staggered grid

scheme, see Figure 2.6. The black dots are indexed by (𝑖, 𝑗), where 𝑖 = 1, 2, . . . , 𝑁𝑥,

𝑗 =

1, 2, . . . , 𝑁𝑦, white dots are indexed by (𝑖 + 1
2

, 𝑗), where 𝑖 = 1, 2, . . . , 𝑁𝑥 − 1, 𝑗 = 1, 2, . . . , 𝑁𝑦 and

(𝑖, 𝑗 + 1

2), where 𝑖 = 1, 2, . . . , 𝑁𝑥, 𝑗 = 1, 2, . . . , 𝑁𝑦 − 1. For a function 𝑢, we use 𝑢𝑖, 𝑗 to represent
an approximate value of 𝑢(𝑥𝑖, 𝑦 𝑗 ), where 𝑥𝑖 = 𝑥1 + (𝑖 − 1)Δ𝑥 and 𝑦 𝑗 = 𝑦1 + ( 𝑗 − 1)Δ𝑦 are the

coordinates of the grid points.

Figure 2.6 The illustration of staggered grid scheme. The zero and second order terms are defined
on the grid points (black dots), the first order terms and 𝐷 are defined on the edges (while dots).

Discretization with Isotropic 𝐷. We begin the discretization with an isotropic diffusion coeffi-

cient, that is, 𝐷 = 𝐷 (𝑥) is a scalar function.

29

Discretization of the Forward Problem. First, we consider discretization of the forward

problem (2.37) (2.38). Using the staggered grid scheme, the operator ∇ · 𝐷∇ is discretized as

[∇ · 𝐷∇𝑢]𝑖, 𝑗 =[𝜕𝑥 𝐷𝜕𝑥𝑢 + 𝜕𝑦 𝐷𝜕𝑦𝑢]𝑖, 𝑗
[𝐷𝜕𝑥𝑢]𝑖+ 1

2

, 𝑗 − [𝐷𝜕𝑥𝑢]𝑖− 1
Δ𝑥

2

[𝐷𝜕𝑦𝑢]𝑖, 𝑗+ 1

2

, 𝑗

+

− [𝐷𝜕𝑦𝑢]𝑖, 𝑗− 1
Δ𝑦

2

, 𝑗 [𝑢𝑖+1, 𝑗 − 𝑢𝑖, 𝑗 ] − 𝐷𝑖− 1

2

, 𝑗 [𝑢𝑖, 𝑗 − 𝑢𝑖−1, 𝑗 ]

≈

𝐷𝑖+ 1

2

≈

Δ𝑥2
[𝑢𝑖, 𝑗+1 − 𝑢𝑖, 𝑗 ] − 𝐷𝑖, 𝑗− 1

2

[𝑢𝑖, 𝑗 − 𝑢𝑖, 𝑗−1]

(2.58)

=

𝑢𝑖+1, 𝑗 +

𝐷𝑖, 𝑗+ 1

2

+

2

(cid:35)

(cid:34) 𝐷𝑖+ 1
, 𝑗
Δ𝑥2
(cid:34) 𝐷𝑖+ 1
, 𝑗
Δ𝑥2

−

2

Δ𝑦2

(cid:35)

(cid:34) 𝐷𝑖− 1
, 𝑗
Δ𝑥2

2

𝑢𝑖−1, 𝑗 +

𝐷𝑖− 1
, 𝑗
Δ𝑥2

2

𝐷𝑖, 𝑗+ 1
Δ𝑦2

2

+

+

+

𝐷𝑖, 𝑗− 1
Δ𝑦2

2

2

(cid:34) 𝐷𝑖, 𝑗+ 1
Δ𝑦2
(cid:35)

𝑢𝑖, 𝑗 ,

(cid:35)

𝑢𝑖, 𝑗+1 +

(cid:35)

(cid:34) 𝐷𝑖, 𝑗− 1
Δ𝑦2

2

𝑢𝑖, 𝑗−1

where ≈ denotes the staggered grid scheme approximation.

For the Robin boundary condition on the four boundaries (excluding the four corners), it is

simply 𝑢 ± 2𝐷𝜕𝑥𝑢 on the right/left boundary, 𝑢 ± 2𝐷𝜕𝑦𝑢 on the top/bottom boundary. For the

four corner points, e.g.

the bottom left corner (Figure 2.7), the outgoing vector 𝜈 is chosen as

√
2
2

, −

√
2
2 ). For example,

(−

[𝑢 + ℓ𝜈 · 𝐷∇𝑢]1,1 =𝑢1,1 −

√

2ℓ
2
√
2ℓ
2
2ℓ𝐷
2Δ𝑥

[𝐷𝜕𝑥𝑢]1+ 1
𝐷

2

,1

1+ 1
2
Δ𝑥

1+ 1
2

,1

+

√

2ℓ
2

,1 −

[𝑢1,1 − 𝑢1,2] +
√

1,1+ 1
2

2ℓ𝐷
2Δ𝑦








2

√

[𝐷𝜕𝑦𝑢]1,1+ 1
2ℓ
2

𝐷

𝑢1,1 −

=𝑢1,1 +
√

=

1 +








1,1+ 1
2
Δ𝑦
√
2ℓ𝐷
2Δ𝑥

[𝑢1,1 − 𝑢2,1]
√

1+ 1
2

,1

𝑢1,2 −

1,1+ 1
2

2ℓ𝐷
2Δ𝑦

𝑢2,1.

(2.59)

This discretization gives rise to a linear system with the unknowns 𝑢𝑖, 𝑗 . In order to make this

linear system explicit, we introduce the index function I (𝑖, 𝑗) (cid:66) (𝑖−1)𝑁𝑦 + 𝑗 and use (𝑖, 𝑗) ∼ (𝑖′, 𝑗 ′)

to mean that the (𝑖′, 𝑗 ′)-point is a neighbor of (𝑖, 𝑗)-point. Denote by 𝐼 the set of interior points, by

𝐵 the set of non-corner boundary points, and by 𝐵𝑐 the set of four corner points. According to the

scheme (2.58), (2.59), the forward problem (2.37) (2.38) is discretized to yield the linear system

L𝝓0 = s

30

Figure 2.7 The outgoing vector at the corner.

where 𝝓0 consists of the vectorized values of the forward solution 𝜙0 at black dots such that

𝝓0I (𝑖, 𝑗) = 𝜙0(𝑥𝑖, 𝑦 𝑗 ).

LI (𝑖, 𝑗),I (𝑖′, 𝑗 ′) =






(cid:205)

(˜𝑖, ˜𝑗)∼(𝑖, 𝑗)
𝐷 𝑖+𝑖′
2

−

,
|𝑖−𝑖′ |Δ𝑥2+| 𝑗− 𝑗 ′ |Δ𝑦2

𝑗+ 𝑗′
2

1 + ℓ (cid:205)𝐼∋(˜𝑖, ˜𝑗)∼(𝑖, 𝑗)
𝐷 𝑖+𝑖′
2

|𝑖−𝑖′ |Δ𝑥+| 𝑗− 𝑗 ′ |Δ𝑦 ,
√

𝑗+ 𝑗′
2

−ℓ

,

(cid:205)

2ℓ
2

(˜𝑖, ˜𝑗)∼(𝑖, 𝑗)

𝐷 𝑖+𝑖′
2

,

𝑗+ 𝑗′
2

|𝑖−𝑖′ |Δ𝑥+| 𝑗− 𝑗 ′ |Δ𝑦 ,

2ℓ
2

1 +
√

−

0

𝐷

𝑖+˜𝑖
2

,

𝑗+ ˜𝑗
2

|𝑖−˜𝑖|Δ𝑥2+| 𝑗− ˜𝑗 |Δ𝑦2 + 𝜎𝑖, 𝑗 ,

(𝑖′, 𝑗 ′) = (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐼

,

(𝑖′, 𝑗 ′) ∼ (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐼

𝐷

𝑖+˜𝑖
,
2
|𝑖−˜𝑖|Δ𝑥+| 𝑗− ˜𝑗 |Δ𝑦

𝑗+ ˜𝑗
2

,

(𝑖′, 𝑗 ′) = (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐵

𝐼 ∋ (𝑖′, 𝑗 ′) ∼ (𝑖, 𝑗) ∈ 𝐵,

(2.60)

𝐷

𝑖+˜𝑖
,
2
|𝑖−˜𝑖|Δ𝑥+| 𝑗− ˜𝑗 |Δ𝑦

𝑗+ ˜𝑗
2

,

(𝑖′, 𝑗 ′) = (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐵𝑐,

(𝑖′, 𝑗 ′) ∼ (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐵𝑐,

others

sI (𝑖, 𝑗) =

𝑆𝑖, 𝑗 ,

(𝑖, 𝑗) ∈ 𝐼,

0,

(𝑖, 𝑗) ∈ 𝐵 ∪ 𝐵𝑐.

(2.61)





Before discussing further properties of the matrix L, we recall the definition of some special

matrices. Given a square matrix 𝐴 = ( 𝐴𝑘𝑙), its 𝑘-th row is said to be weakly diagonally dominant
(WDD) if | 𝐴𝑘 𝑘 | ≥ (cid:205)𝑙≠𝑘 | 𝐴𝑘𝑙 |, and the matrix 𝐴 is said to be WDD if all the rows are WDD.
Likewise, its 𝑘-th row is said to be strictly diagonally dominant (SDD) if ≥ is replaced by a strict

inequality >, and the matrix 𝐴 is said to be SDD if all the rows are WDD.

31

νDefinition 2.13. A square matrix 𝐴 = ( 𝐴𝑘𝑙) is said to be weakly chained diagonally dominant

(WCDD) if

• 𝐴 is WDD.

• For each row 𝑘 that is not SDD, there exists 𝑘1, 𝑘2, . . . , 𝑘 𝑝 such that 𝐴𝑘 𝑘1, 𝐴𝑘1𝑘2, . . . , 𝐴𝑘 𝑝−1𝑘 𝑝 ,

𝐴𝑘 𝑝𝑙 are nonzero and the row 𝐴𝑙,: is SDD.

Proposition 2.14. L is a WCDD matrix.

Proof. First, we show L is WDD. As 𝐷 > 0, 𝜎𝑎 ≥ 0 everywhere, all the off-diagonal terms (see

Row 2, 4, 6, 7 in (2.60)) are non-positive and all the diagonal terms (see Row 1, 3, 5 in (2.60)) are

non-negative. It suffices to show that

LI (𝑖, 𝑗),I (𝑖, 𝑗) ≥

∑︁

(𝑖′, 𝑗 ′)≠(𝑖, 𝑗)

−LI (𝑖, 𝑗),I (𝑖′, 𝑗 ′).

Move all the terms in this inequality to the left side. It suffices to show that any row sum of L is

non-negative. This is obvious from the definition of L in (2.60), where the row sum of the I (𝑖, 𝑗)-

th row is 𝜎𝑖, 𝑗 when (𝑖, 𝑗) ∈ 𝐼, and the row sum of the I (𝑖, 𝑗)-th row is 1 when (𝑖, 𝑗) ∈ 𝐵 ∪ 𝐵𝑐.

This proves that L is WDD. Moreover, the analysis shows that the I (𝑖, 𝑗)-th row is SDD when

(𝑖, 𝑗) ∈ 𝐵 ∪ 𝐵𝑐.

Next, we show the chain condition.

If the I (𝑖, 𝑗)-th row is not SDD, then (𝑖, 𝑗) ∈ 𝐼. As

the finite difference grid is connected, there exist (𝑖1, 𝑗1), . . . , (𝑖 𝑝, 𝑗 𝑝) such that (𝑖 𝑝, 𝑗 𝑝) ∈ 𝐵 ∪ 𝐵𝑐

and (𝑖, 𝑗) ∼ (𝑖1, 𝑗1) ∼ · · · ∼ (𝑖 𝑝, 𝑗 𝑝). Notice that the definition of L has the property that

LI (𝑖, 𝑗),I (𝑖′, 𝑗 ′) < 0 for (𝑖, 𝑗) ∼ (𝑖′, 𝑗 ′) (see Row 2,4,6 in (2.60)), we conclude the entries LI (𝑖, 𝑗),I (𝑖1, 𝑗1),
. . . , LI (𝑖 𝑝−1, 𝑗 𝑝−1),I (𝑖 𝑝, 𝑗 𝑝) are all negative, and the row LI (𝑖 𝑝, 𝑗 𝑝),: is SDD since (𝑖 𝑝, 𝑗 𝑝) ∈ 𝐵 ∪ 𝐵𝑐. □

Proposition 2.15 ( [82]). WCDD matrices are invertible.

As a result, the discretized forward problem admits a unique solution 𝝓0 = L−1s.

32

Discretization of the Adjoint Problem. The adjoint problem(2.39), (2.40) takes a similar

form as the forward problem, except that the source 𝑔 is imposed on the boundary. Therefore, the

adjoint problem can be discretized likewise to yield a linear system

L𝝍 = g

where L is the same finite difference matrix defined in (2.60), 𝝍 consists of the vectorized values

of the adjoint solution 𝜓 at black dots such that 𝝍I (𝑖, 𝑗) = 𝜓(𝑥𝑖, 𝑦 𝑗 ), and

gI (𝑖, 𝑗) =





0,

(𝑖, 𝑗) ∈ 𝐼,

𝑔(𝑥𝑖, 𝑦 𝑗 ),

(𝑖, 𝑗) ∈ 𝐵 ∪ 𝐵𝑐.

(2.62)

Discretization of the Internal Data Problem.

It remains to discretize the internal data

problem (2.43) along with the zero Robin boundary condition. This requires discretizing an

operator of the form 𝐷∇𝑢 · ∇𝑣 = 𝐷∇𝑣 · ∇𝑢. The staggered grid scheme gives

≈

≈

[𝐷∇𝑣 · ∇𝑢]𝑖, 𝑗
[𝐷𝜕𝑥𝑢𝜕𝑥𝑣]𝑖+ 1

2

, 𝑗 + [𝐷𝜕𝑥𝑢𝜕𝑥𝑣]𝑖− 1

2

2

[𝐷𝜕𝑥𝑣]𝑖+ 1

2

, 𝑗 [𝑢𝑖+1, 𝑗 − 𝑢𝑖, 𝑗] + [𝐷𝜕𝑥𝑣]𝑖− 1

2

, 𝑗

+

[𝐷𝜕𝑦𝑢𝜕𝑦𝑣]𝑖, 𝑗+ 1

+ [𝐷𝜕𝑦𝑢𝜕𝑦𝑣]𝑖, 𝑗 − 1
2
, 𝑗 [𝑢𝑖, 𝑗 − 𝑢𝑖−1, 𝑗]

2

2

[𝐷𝜕𝑦𝑣]𝑖, 𝑗+ 1

2

+

2Δ𝑥
[𝑢𝑖, 𝑗+1 − 𝑢𝑖, 𝑗] + [𝐷𝜕𝑦𝑣]𝑖, 𝑗 − 1
2Δ𝑦

2

[𝑢𝑖, 𝑗 − 𝑢𝑖, 𝑗 −1]

(cid:34) 𝐷𝑖+ 1

2

=

, 𝑗 [𝑣𝑖+1, 𝑗 − 𝑣𝑖, 𝑗]

(cid:35)

𝑢𝑖+1, 𝑗 +

(cid:34) 𝐷𝑖, 𝑗+ 1

2Δ𝑥2
[𝑣𝑖, 𝑗+1 − 𝑣𝑖, 𝑗]
2Δ𝑦2

2

(cid:35)

𝑢𝑖, 𝑗+1 +

(cid:34) 𝐷𝑖− 1

2

, 𝑗 [𝑣𝑖−1, 𝑗 − 𝑣𝑖, 𝑗]

(cid:35)

(cid:34) 𝐷𝑖, 𝑗 − 1

2Δ𝑥2
[𝑣𝑖, 𝑗 −1 − 𝑣𝑖, 𝑗]
2Δ𝑦2
, 𝑗 [𝑣𝑖−1, 𝑗 − 𝑣𝑖, 𝑗]

2

𝑢𝑖−1, 𝑗

(cid:35)

𝑢𝑖, 𝑗 −1

+

−

(cid:34) 𝐷𝑖+ 1

2

, 𝑗 [𝑣𝑖+1, 𝑗 − 𝑣𝑖, 𝑗]

2Δ𝑥2

𝐷𝑖− 1

2

+

2Δ𝑥2

𝐷𝑖, 𝑗+ 1

2

+

[𝑣𝑖, 𝑗+1 − 𝑣𝑖, 𝑗]
2Δ𝑦2

+

𝐷𝑖, 𝑗 − 1

2

(cid:35)

[𝑣𝑖, 𝑗 −1 − 𝑣𝑖, 𝑗]
2Δ𝑦2

𝑢𝑖, 𝑗,

The discretization of (2.43) becomes

A𝝍0 𝝓0 = h𝝍0

33

where 𝝓0 consists of the vectorized values of the forward solution 𝜙0 at black dots such that

𝝓0I (𝑖, 𝑗) = 𝜙(𝑥𝑖, 𝑦 𝑗 ), and

𝐷

𝑖+ ˜𝑖
2

,

[ 𝜓𝑖, 𝑗 + 2𝛾−1

𝑗+ ˜𝑗
2
|𝑖− ˜𝑖 |Δ𝑥2+| 𝑗 − ˜𝑗 |Δ𝑦2

2

[ 𝜓˜𝑖, ˜𝑗 − 𝜓𝑖, 𝑗 ] ]

+ 2𝛾𝜎𝑖, 𝑗𝜓𝑖, 𝑗,

(𝑖′, 𝑗 ′) = (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐼

( ˜𝑖, ˜𝑗 )∼(𝑖, 𝑗 )

,

(𝑖′, 𝑗 ′) ∼ (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐼

(A𝜓)I (𝑖, 𝑗),I (𝑖′, 𝑗 ′) =






− (cid:205)
𝐷 𝑖+𝑖′
2

,

[ 𝜓𝑖′ , 𝑗′ − 𝜓𝑖, 𝑗 ] ]

2

[ 𝜓𝑖, 𝑗 + 2𝛾−1

𝑗+ 𝑗′
2
|𝑖−𝑖′ |Δ𝑥2+| 𝑗 − 𝑗′ |Δ𝑦2
𝐷

𝑖+ ˜𝑖
,
2
|𝑖− ˜𝑖 |Δ𝑥+| 𝑗 − ˜𝑗 |Δ𝑦

𝑗+ ˜𝑗
2

1 + ℓ (cid:205)𝐼 ∋ ( ˜𝑖, ˜𝑗 )∼(𝑖, 𝑗 )
𝐷 𝑖+𝑖′
2

|𝑖−𝑖′ |Δ𝑥+| 𝑗 − 𝑗′ |Δ𝑦 ,

𝑗+ 𝑗′
2

−ℓ

,

1 +

√

2ℓ
2

(cid:205)

( ˜𝑖, ˜𝑗 )∼(𝑖, 𝑗 )
𝐷 𝑖+𝑖′
2

𝑗+ 𝑗′
2

,

|𝑖−𝑖′ |Δ𝑥+| 𝑗 − 𝑗′ |Δ𝑦 ,

√

2ℓ
2

−

0

𝐷

𝑖+ ˜𝑖
,
2
|𝑖− ˜𝑖 |Δ𝑥+| 𝑗 − ˜𝑗 |Δ𝑦

𝑗+ ˜𝑗
2

,

(𝑖′, 𝑗 ′) = (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐵

𝐼 ∋ (𝑖′, 𝑗 ′) ∼ (𝑖, 𝑗) ∈ 𝐵,

,

(𝑖′, 𝑗 ′) = (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐵𝑐,

(𝑖′, 𝑗 ′) ∼ (𝑖, 𝑗), (𝑖, 𝑗) ∈ 𝐵𝑐,

others

(h𝜓)I (𝑖, 𝑗) =

(𝐻𝜓)𝑖, 𝑗 ,

(𝑖, 𝑗) ∈ 𝐼,

0,

(𝑖, 𝑗) ∈ 𝐵 ∪ 𝐵𝑐.





Discrete Uncertainty Quantification Estimate.

In parallel to Theorem 2.11, we can derive

the following UQ estimate for the discretized model. Note that the uncertainties of the optical

parameters (𝐷, 𝜎𝑎) are implicitly encoded in the difference ˜L − L and ˜A ˜𝝓0

− A𝝓0.

Theorem 2.16 ( [97]). Suppose 0 is not an eigenvalue of A𝝍0 for some 𝝍0 > 0, then

∥˜s − s∥2 ≤∥h𝝓0 ∥2(∥A−1
𝜓0

∥2∥ ˜L − L∥2 + ∥ ˜L∥2∥ ˜A−1
˜𝝓0

∥2∥A−1
𝝓0

∥2∥ ˜A ˜𝝓0

− A𝝓0 ∥2).

(2.63)

Proof. Under the assumption, the matrix A𝝍0 is invertible for some 𝝍0 > 0. We can represent
𝝓0 = A−1
𝝍0

h𝜓0, then s = L𝝓0 = LA−1
𝝍0

h𝝍0. Therefore,

)h𝝓0 ∥2

− LA−1
𝝓0

∥˜s − s∥2 =∥( ˜L ˜A−1
˜𝝓0
≤∥ ˜L ˜A−1
˜𝝓0
≤(∥ ( ˜L − L)A−1
𝝓0

− LA−1
𝝓0

∥2∥h𝝓0 ∥2

(2.64)

∥2 + ∥ ˜L( ˜A−1
˜𝝓0

− A−1
𝝓0

) ∥2) ∥h𝝓0 ∥2

≤(∥ ˜L − L∥2∥A−1
𝝓0

∥2 + ∥ ˜L∥2∥ ˜A−1
˜𝝓0

− A−1
𝝓0

∥2) ∥h𝝓0 ∥2

where ∥ · ∥2 denotes the vector/matrix 2-norm. Using the relation 𝐴−1 − 𝐵−1 = 𝐴−1(𝐵 − 𝐴)𝐵−1,

we obtain the desired estimate.

□

34

Discretization with Anisotropic 𝐷. When 𝐷 is anisotropic, i.e, a symmetric positive definition

matrix-valued function, the operators ∇ · 𝐷∇ and 𝐷∇𝑣 · ∇ can be discretized as follows

[∇ · 𝐷∇𝑢]𝑖, 𝑗 =

[𝐷∇𝑣 · ∇𝑢]𝑖, 𝑗 =

[(𝐷∇𝑢)1]𝑖+ 1

, 𝑗 − [(𝐷∇𝑢)1]𝑖− 1
Δ𝑥
[(𝐷∇𝑣)1𝜕𝑥𝑢]𝑖+ 1

2

2

, 𝑗 + [(𝐷∇𝑣)2𝜕𝑥𝑢]𝑖− 1

, 𝑗

+

[(𝐷∇𝑢)2]𝑖, 𝑗+ 1

2

− [(𝐷∇𝑢)2]𝑖, 𝑗− 1
Δ𝑦
[(𝐷∇𝑣)2𝜕𝑦𝑢]𝑖, 𝑗+ 1

2

2

, 𝑗

+

2

2

2

+ [(𝐷∇𝑣)2𝜕𝑦𝑢]𝑖, 𝑗− 1
2

2

where (𝐷∇𝑢)1 (resp. (𝐷∇𝑢)2) denotes the first (resp. second) component of the vector 𝐷∇𝑢. The

discretization now differs from the isotropic case. This is because for an isotropic 𝐷

(𝐷∇𝑢)1 = 𝐷𝜕𝑥𝑢,

(𝐷∇𝑢)2 = 𝐷𝜕𝑦𝑢

which only requires to compute [𝜕𝑥𝑢]𝑖+ 1
anisotropic 𝐷:

2

, 𝑗 and [𝜕𝑦𝑢]𝑖, 𝑗+ 1

2

in the staggered grid. However, for an

(𝐷∇𝑢)1 = 𝐷11𝜕𝑥𝑢 + 𝐷12𝜕𝑦𝑢,

(𝐷∇𝑢)2 = 𝐷21𝜕𝑥𝑢 + 𝐷22𝜕𝑦𝑢

which requires to compute two additional terms [𝜕𝑥𝑢]𝑖, 𝑗+ 1

2

and [𝜕𝑦𝑢]𝑖+ 1

2

, 𝑗 . These additional terms

can be discretized as follows:

[𝜕𝑦𝑢]𝑖+ 1

2

, 𝑗 =

[𝜕𝑥𝑢]𝑖, 𝑗+ 1

2

=

[𝜕𝑦𝑢]𝑖, 𝑗 + [𝜕𝑦𝑢]𝑖+1, 𝑗
2
[𝜕𝑥𝑢]𝑖, 𝑗 + [𝜕𝑥𝑢]𝑖, 𝑗+1
2

=

=

𝑢𝑖+1, 𝑗+1 + 𝑢𝑖, 𝑗+1 − 𝑢𝑖, 𝑗−1 − 𝑢𝑖+1, 𝑗−1
4Δ𝑦
𝑢𝑖+1, 𝑗+1 + 𝑢𝑖+1, 𝑗 − 𝑢𝑖−1, 𝑗 − 𝑢𝑖−1, 𝑗+1
4Δ𝑥

,

,

see [45] for the detail. Once discretized, the rest of the steps are similar and one can derive the

estimate in Theorem

2.3.3.3 Numerical Experiment

In this section, we demonstrate numerical experiments to validate the reconstruction procedure

and quanfity the impact of inaccurate optical coefficients (𝐷, 𝜎𝑎) to the source recovery. We will

restrict the discussion in this section to isotropic 𝐷 for the ease of notations.

Uncertainty Generation We will utilize the generalized Polynomial Chaos Expansion (PCE)

to facilitate generation of uncertainty. PCE approximates a well-behaved random variable using

35

a series of polynomials under certain probability distribution. Specifically, let (𝑋, F , P) be a

probability space, and let 𝜉 (𝜔) be a random variable (where 𝜔 is a sample) with probability

density function 𝑝(𝑡). Suppose a deterministic ground-truth 𝔲 = 𝔲(x) is given, then the uncertainty

generated by PCE takes the form

𝔲(x, 𝜉 (𝜔)) =

∞
∑︁

𝑘=0

𝔲𝑘 (x)Φ𝑘 (𝜉 (𝜔)),

(𝑥, 𝜔) ∈ Ω × 𝑋

(2.65)

where 𝔲𝑘 (x)’s are the coefficients, 𝔲0 is the ground truth, Φ0 = 1, Φ𝑘 ’s are orthogonal polynomials,

that is,

∫

R

Φ𝑖 (𝑡)Φ 𝑗 (𝑡) 𝑝(𝑡) d𝑡 = 𝛿𝑖 𝑗 .

For the numerical experiments, 𝜉 is chosen to be uniformly distributed on the sample space

𝑋 = [−1, 1]; Φ𝑘 ’s are the Legendre polynomials on [−1, 1]; the PCE is truncated at 𝑘 = 𝐾𝑐. Then

E[𝔲] = 𝔲0,

Var[𝔲] =

𝐾𝑐∑︁

𝑘=1

𝑘 .
𝔲2

In the subsequent numerical experiments, we inject uncertainties into the optical coefficients

(𝐷, 𝜎𝑎) based on the following process:

(1) Generate the coefficients 𝔲𝐷 𝑘 , 𝔲𝜎𝑎 𝑘 using the truncated Fourier series in x:

𝔲𝐷 𝑘 =

𝔲𝜎𝑎 𝑘 =

∑︁

∥n∥∞=𝑘
∑︁

∥n∥∞=𝑘

𝑐1n sin(𝜋n · x) + 𝑐2n cos(𝜋n · x),

𝑐3n sin(𝜋n · x) + 𝑐4n cos(𝜋n · x).

Here n ∈ Z𝑛, the Fourier coefficients 𝑐1n, 𝑐2n, 𝑐3n, 𝑐4n are independently chosen from the
uniform distributions on [−1, 1]. Once generated, they are fixed so that the coefficients

𝔲𝐷 𝑘 , 𝔲𝜎𝑎 𝑘 are deterministic.

(2) Randomly generate 𝜉 from the uniform distribution on [−1, 1], then construct the uncertain-

ties 𝔲𝐷, 𝔲𝜎𝑎 according to (2.65) with 𝑘 = 1, 2, . . . , 10:

𝔲𝐷 (cid:66)

10
∑︁

𝑘=1

𝔲𝐷 𝑘 Φ𝑘 (𝜉 (𝜔)),

𝔲𝜎𝑎 (cid:66)

10
∑︁

𝑘=1

𝔲𝜎𝑎 𝑘 Φ𝑘 (𝜉 (𝜔))

Note that E[𝔲𝐷] = E[𝔲𝜎𝑎] = 0.

36

(3) Once the uncertainties are generated, we rescale the uncertainties based on prescribed relative

uncertainty levels 𝑒𝐷, 𝑒𝜎𝑎 to construct the optical coefficients with uncertainty ( ˜𝐷, ˜𝜎𝑎) as

follows:

˜𝐷 := 𝐷 +

˜𝜎𝑎 := 𝜎𝑎 +

∥𝐷 ∥𝐻1,

𝔲𝐷 𝑒𝐷
∥𝔲𝐷 ∥𝐻1
𝔲𝜎𝑎 𝑒𝜎𝑎
∥𝔲𝜎𝑎 ∥ 𝐿2

∥𝜎𝑎 ∥ 𝐿2.

(2.66)

The impact of the inaccuracy in the optical coefficients will be quantitatively measured by the

relative standard deviation defined as follows:

E𝑆 (cid:66)

√︃

E[∥ ˜𝑆 − 𝑆∥2

𝐿2]

∥𝑆∥ 𝐿2

, E𝐷 (cid:66)

√︃E[∥ ˜𝐷 − 𝐷 ∥2
∥𝐷 ∥𝐻1

𝐻1]

, E𝜎𝑎 (cid:66)

√︃E[∥ ˜𝜎𝑎 − 𝜎𝑎 ∥2
𝐿2]
∥𝜎𝑎 ∥ 𝐿2

.

(2.67)

Note that E𝐷 = 𝑒𝐷 and E𝜎𝑎 = 𝑒𝜎𝑎 are precisely the relative uncertainty levels that are used to

define ( ˜𝐷, ˜𝜎𝑎) in (2.66). This justifies that the relative standard deviation is a reasonable quantity

to measure the uncertainty. In the following, we will specify various uncertainty levels 𝑒𝐷, 𝑒𝜎𝑎 and

plot E𝑆 versus them, see Figure 2.12 and Figure 2.17.

Numerical Implementation. We choose the 2D computational domain Ω = [−1, 1]2, ℓ = 1. The

diffusion equation is solved using the staggered grid scheme as is outlined in Section 2.3.3.2. To

avoid the inverse crime, the forward problem is solved on a fine mesh with step size ℎ = 1

200, while

the inverse problem is solved on a coarse mesh with step size ℎ = 1

100 using re-sampled data. We
numerically calculate the noise-free 𝜙0 and 𝜓0 using ground truth 𝑆 and (𝐷, 𝜎𝑎), here we choose

𝜓0 > 0 by solving (2.39) with a positive Dirichlet boundary condition. Once we have 𝜙0 and 𝜓0,

we can calculate the internal data 𝐻𝜓0 through (2.43). Note that the internal data is derived from

the boundary measurement, hence is independent of the uncertainty on the optical coefficients.

Experiment 1.

In this experiment, we consider the case that the optical coefficients can be

represented using low-frequency Fourier basis. We choose

𝐷 = cos2(𝑥 + 2𝑦) − 3 sin2(3𝑥 − 4𝑦) + 5,

𝜎𝑎 = cos2(5𝑥) + sin2(5𝑦) + 1,

and the source 𝑆 to be the Shepp-Logan phantom, see Figure 2.8.

37

Figure 2.8 Left: Diffusion coefficient 𝐷. Middle: Absorption coefficient 𝜎𝑎. Right: Shepp-Logan
Source 𝑆.

Using the ground-truth (𝐷, 𝜎𝑎), we generate the uncertainties according to (2.66) to obtain

1000 samples of the inaccurate optical coefficients ( ˜𝐷, ˜𝜎𝑎). Set Δ𝐷 := ˜𝐷 − 𝐷 and Δ𝜎𝑎 =

˜𝜎𝑎 − 𝜎𝑎. We implemented the reconstruction procedure 1000 times to plot the distribution of

∥Δ𝑆∥ 𝐿2 versus ∥Δ𝐷 ∥𝐻1 and ∥Δ𝜎𝑎 ∥ 𝐿2, see Figure 2.9. It is clear that for fixed ∥Δ𝐷 ∥𝐻1, ∥Δ𝑆∥ 𝐿2

is more concentrated compared to fixed ∥Δ𝜎𝑎 ∥ 𝐿2, suggesting that the uncertainty in ˜𝐷 has larger

impact to the reconstruction than the uncertainty in ˜𝜎𝑎. Moreover, the distribution of the scatter

plot suggests that ∥Δ𝑆∥ 𝐿2 is locally Lipschitz stable with respect to ∥Δ𝐷 ∥𝐻1 for small Δ𝐷, agreeing

with the estimates in Theorem 2.11 and Theorem 2.16 One of the reconstructions is illustrated in

Figure 2.10, and the average of the 1000 reconstructed sources is illustrated in Figure 2.11. We

can see that the averaged ˜𝑆 is close to the ground truth 𝑆, which means the relation between ˜𝑆 and

( ˜𝐷, ˜𝜎𝑎) near (𝐷, 𝜎𝑎) have no sharp points, see Remark 2.17. It implies that the uncertainty in ˜𝑆

have certain regularity with respect to the uncertainties in ( ˜𝐷, ˜𝜎𝑎).

To better understand the relations between E𝑆 versus E𝐷 (resp. E𝑆 versus E𝜎𝑎), we take

Δ𝜎𝑎 = 0 (resp. Δ𝐷 = 0) and add 𝑒𝐷 = 2%, 4%, 6%, 8%, 10% of random noise to 𝐷 (resp.

𝑒𝜎𝑎 = 2%, 4%, 6%, 8%, 10% of random noise to 𝜎𝑎). The plots are shown in Figure 2.12. We

observe that E𝑆 depends linearly or superlinearly on E𝐷 and E𝜎𝑎, and the same level of relative

uncertainty on 𝐷 has larger impact than on 𝜎𝑎. We remark that the plotted curves are nonlinear

because the constant factors 𝐶1𝑖 𝑗 , 𝐶2 in Theorem 2.11 also depend on ( ˜𝐷, ˜𝜎𝑎).

Remark 2.17. Choose random variable 𝑋 ∼ 𝑁

(cid:16)

, 𝑓𝛼 (𝑥) = |𝑥|𝛼 (0 < 𝛼 < 1), we have

0 = 𝑓𝛼 (E[𝑋]) < E[ 𝑓𝛼 (𝑋)] =

Γ

(cid:19)

(cid:18) 𝛼 + 1
2

< 1,

(cid:17)

0, 1
100
2− 𝛼
2 5−𝛼
√
𝜋

38

-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.812.533.544.555.56-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.811.21.41.61.822.22.42.62.8-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.10.20.30.40.50.60.70.80.91Figure 2.9 The distribution of the error with respect to the inaccuracies in optical coefficients.

Figure 2.10 Reconstructed source ˜𝑆 and its error under 10% Gaussian random noise.

Figure 2.11 Averaged reconstructed source ˜𝑆 and its error under 10% Gaussian random noise.

and E[ 𝑓𝛼 (𝑋)] is monotonically converged to 1 as 𝛼 → 0+.

39

00.20.40.60.811.21.4|| D||0.040.050.060.070.080.090.10.110.120.130.14|| S||00.10.20.30.40.50.60.7||||0.040.050.060.070.080.090.10.110.120.130.14|| S||-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.20.40.60.811.2-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.020.040.060.080.10.120.140.16-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.20.40.60.81-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.010.020.030.040.050.060.070.080.090.1Figure 2.12 Left: E𝑆 versus E𝜎𝑎. Right: E𝑆 versus E𝐷.

Experiment 2:

In this experiment, we consider the case that the optical coefficients can not

be represented using the low frequency Fourier basis. We choose

𝐷 = 3 − max{|𝑥|, |𝑦|},

𝜎𝑎 =

(cid:18)

3
2

−

1
2

sgn

𝑥2 + 𝑦2 −

(cid:19)

,

4
5

and we choose the source 𝑆 to be the Shepp-Logan phantom, see Figure 2.13.

Figure 2.13 Left: Diffusion coefficient 𝐷. Middle: Absorption coefficient 𝜎𝑎. Right:
Shepp-Logan Source.

We choose the relative uncertainty level at 10% and run 1000 reconstructions to plot the

distribution of ∥ ˜𝑆 − 𝑆∥ 𝐿2 versus ∥ ˜𝐷 − 𝐷 ∥𝐻1 and ∥ ˜𝜎𝑎 − 𝜎𝑎 ∥ 𝐿2, see Figure 2.14. One of the

reconstructions is illustrated in Figure 2.15, and the average of 1000 reconstructed sources is

illustrated in Figure 2.16.

40

0.020.030.040.050.060.070.080.090.1E0.102250.10230.102350.10240.102450.10250.102550.10260.102650.10270.10275ES0.020.030.040.050.060.070.080.090.1ED0.10.150.20.250.30.350.40.450.5ES-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8122.12.22.32.42.52.62.72.82.93-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8111.11.21.31.41.51.61.71.81.92-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.10.20.30.40.50.60.70.80.91Figure 2.14 The distribution of the error with respect to the inaccuracies in optical coefficients.

Figure 2.15 Reconstructed source ˜𝑆 and its error under 10% Gaussian random noise.

Figure 2.16 Averaged reconstructed source ˜𝑆 and its error under 10% Gaussian random noise.

For the relation between the relative standard deviations, we fix 𝐷 and 𝜎𝑎 respectively and

add 2%, 4%, 6%, 8%, 10% Gaussian random noise to another optical coefficient. The relations are

shown in Figure 2.17. We can see that the average of ˜𝑆 is approximately 𝑆 and the same relative

41

00.10.20.30.40.50.60.70.80.9|| D||0.060.070.080.090.10.110.12|| S||00.10.20.30.40.50.6||||0.060.070.080.090.10.110.12|| S||-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.20.40.60.81-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.020.040.060.080.10.120.14-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.20.40.60.81-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8100.020.040.060.080.10.12uncertainty level on 𝐷 have larger impact than 𝜎𝑎 on ˜𝑆. From the scatter plot, we can see that the

uncertainty in ˜𝑆 is at least locally Lipschitz boundeded by uncertainties in ( ˜𝐷, ˜𝜎𝑎). The impact E𝑆

is also linearly or superlinearly depend on E𝐷 and E𝜎𝑎.

Figure 2.17 Left: E𝑆 versus E𝜎𝑎. Right: E𝑆 versus E𝐷.

Remark 2.18. Since we plot the distribution of norms, it is natural to have branches in the result.

A simple example is plotting |𝑦| versus |𝑥| with relation 𝑦 = (𝑥 + 1)2.

42

0.020.030.040.050.060.070.080.090.1E0.144850.14490.144950.1450.145050.14510.145150.14520.14525ES0.020.030.040.050.060.070.080.090.1ED0.1450.150.1550.160.1650.170.1750.18ESCHAPTER 3

INVERSE BOUNDARY VALUE PROBLEMS FOR WAVE EQUATIONS

3.1

Introduction

We begin with the formulation of the Inverse Boundary Value Problem (IBVP) for the wave

equation. Let 𝑇 > 0 be a constant and Ω ⊂ R𝑛 be a bounded open subset with smooth boundary

𝜕Ω. Consider the following boundary value problem for the acoustic wave equation with potential:




□𝜌,𝑞𝑢(𝑡, 𝑥) = 0,

in (0, 2𝑇) × Ω

𝜕𝜈𝑢 = 𝑓 ,

on (0, 2𝑇) × 𝜕Ω

(3.1)

𝑢(0, 𝑥) = 𝜕𝑡𝑢(0, 𝑥) = 0

𝑥 ∈ Ω.


Here □𝜌,𝑞 is a linear wave operator defined as

□𝜌,𝑞𝑢(𝑡, 𝑥) := 𝜌(𝑥)𝜕2

𝑡 𝑢(𝑡, 𝑥) − Δ𝑢(𝑡, 𝑥) + 𝑞(𝑥)𝑢(𝑡, 𝑥);

𝜌(𝑥) (cid:66) 𝑐−2(𝑥) ∈ 𝐶∞(Ω), where 𝑐(𝑥) is a smooth wave speed bounded away from 0 and ∞,

𝑞(𝑥) ∈ 𝐿∞(Ω) is a real-valued function referred to as the potential. We write the wave solution as

𝑢 = 𝑢 𝑓 (𝑡, 𝑥) whenever it is necessary to specify the Neumann data.

The wave equation with vanished potential are often used to describe mechanical wave, such as

acoustic wave, water wave, and seismic wave, whereas the wave equation with potential arise for

example in quantum mechanics in the context of the Klein-Gordon equation. In mathematics, we

formulate these equations together as (3.1).

Given 𝑓 ∈ 𝐶∞

𝑐 ((0, 2𝑇) × 𝜕Ω), the well-posedness of this problem is ensured by the standard

theory for second order hyperbolic partial differential equations [40]. As a result, the following

Neumann-to-Dirichlet map (ND map) is well defined:

Λ𝜌,𝑞 𝑓 := 𝑢 𝑓 | (0,2𝑇)×𝜕Ω.

(3.2)

The IBVP for the acoustic wave equation aims to recover the wave speed 𝑐(𝑥) or wave potential

𝑞(𝑥) from the knowledge of the ND map Λ𝜌,𝑞.

Many imaging technologies are based on this inverse problem with vanished potential 𝑞.

The Ultra-Sound Computed Tomography (USCT) is one significant example. During USCT, an

43

acoustic pulse is emitted from a known location outside the tissue by a point-like ultrasound source.

A group of nearby ultrasonic transducers records the wave field created when the acoustic wave

passes through the tissue. The aim of USCT is to reconstruct the acoustic wave speed throughout the

tissue by repeating this process numerous times for a large number of emitter locations. Figure 3.1

shows an example using 𝑀 transducers. Seismic tomography uses a similar data collection strategy

to find oil reservoirs by attempting to recover the underground wave speed.

In the continuous

formulation of USCT and seismic tomography, the measurement is the boundary values of the

Green’s function. However, it is well known [71] that such data is equivalent to knowledge of the

ND map under mild assumptions.

Figure 3.1 Data acquisition scheme in USCT [68].

In the literature, the IBVP for the acoustic wave equation has been thoroughly examined. For

variable 𝑐 and 𝑞 ≡ 0, Belishev [11] demonstrated that 𝑐 is uniquely determined by combining

Tataru’s unique continuation result [91] with the boundary control (BC) method. Since then, a

great deal of work has been done to extend the result to wave equations on Riemannian manifolds

with boundary [16, 36, 37, 38, 39, 43, 46, 49, 54, 55, 59, 62, 78, 81, 87]. Several studies have provided

stability estimates: [2, 10, 17, 18, 24, 66, 70, 83, 84, 86].

The wave speed has been numerically reconstructed using the BC method in [12], and later

in [15,31,76,96]. The implementations [12,15,31] involve solving unstable control problems, while

the implementations [76, 96] are based on solving stable control problems with target functions

exhibiting exponential growth or decay. The exponential behaviour leads to instability as well. On

the other hand, the linearized approach introduced in [75] is stable. It should be noted that the

one-dimensional case can be implemented steadily using the BC method [57]. See [22] on detection

44

of blockage in networks for an intriguing use of a variant of the method in the one-dimensional

case.

Under suitable geometric assumptions, it can be proven that the problem to recover the speed of

sound is Hölder stable [83, 84], even when the speed is given by an anisotropic Riemannian metric.

Moreover, a low-pass version of 𝑐 can be recovered in a Lipschitz stable manner [66].

The problem to recover 𝑞 is Hölder stable assuming, again, that the geometry is nice enough

[18, 70, 88]. To our knowledge, the method, based on using high frequency solutions to the wave

equation and yielding the latter three results, has not been implemented computationally. Stability

results applicable to general geometries have been proven using the BC method in [4], with an

abstract modulus continuity, and very recently in [25, 26], with a doubly logarithmic modulus of

continuity.

3.2 Nonlinear Inverse Boundary Value Problem

In this section, we consider the IBVP with vanished potential, i.e. 𝑞(𝑥) ≡ 0.

(3.1) can be

written as

𝑡 𝑢(𝑡, 𝑥) − 𝑐2(𝑥)Δ𝑢(𝑡, 𝑥) = 0,
𝜕2

in (0, 2𝑇) × Ω

𝜕𝜈𝑢 = 𝑓 ,

on (0, 2𝑇) × 𝜕Ω

(3.3)

𝑢(0, 𝑥) = 𝜕𝑡𝑢(0, 𝑥) = 0

𝑥 ∈ Ω.






This model is frequently used to describe the propagation of mechanical waves, and the inverse

problems focus on recovering the wave speed distribution in order to learn about the medium’s

internal structure.

The ND map is represented as Λ𝑐, and we aim at recover wave speed 𝑐(𝑥) from the ND map [96].

Given function 𝑢(𝑡, 𝑥), we write 𝑢(𝑡) to represent the spatial part 𝑢(𝑡, ·).

3.2.1 Derivation

Introduce the time reversal operator 𝑅 : 𝐿2( [0, 𝑇] × 𝜕Ω) → 𝐿2( [0, 𝑇] × 𝜕Ω),

𝑅𝑢(𝑡, ·) := 𝑢(𝑇 − 𝑡, ·),

0 < 𝑡 < 𝑇;

(3.4)

45

the low-pass filter 𝐽 : 𝐿2([0, 2𝑇] × 𝜕Ω) → 𝐿2( [0, 𝑇] × 𝜕Ω)

𝐽 𝑓 (𝑡, ·) :=

∫ 2𝑇−𝑡

𝑡

1
2

𝑓 (𝜏, ·) 𝑑𝜏,

0 < 𝑡 < 𝑇 .

the orthogonal projection operator 𝑃𝑇 : 𝐿2((0, 2𝑇) × 𝜕Ω) → 𝐿2((0, 𝑇) × 𝜕Ω)

𝑃𝑇 : 𝑓 ↦→ 𝑓 | (0,𝑇)×𝜕Ω

(3.5)

(3.6)

and its adjoint operator 𝑃∗

𝑇 : 𝐿2((0, 𝑇) × 𝜕Ω) → 𝐿2((0, 2𝑇) × 𝜕Ω), which is the extension by zero
from (0, 𝑇) to (0, 2𝑇). Let T𝐷 and T𝑁 be the Dirichlet and Neumann trace operators respectively,

that is,

T𝐷𝑢(𝑡, ·) = 𝑢(𝑡, ·)|𝜕Ω,

T𝑁𝑢(𝑡, ·) = 𝜕𝜈𝑢(𝑡, ·)|𝜕Ω.

Lemma 3.1. Let 𝑢 𝑓 be the solution of (3.3) with 𝑓 ∈ 𝐶∞

𝑐 ((0, 2𝑇) × 𝜕Ω). Suppose 𝑣(𝑡, 𝑥) ∈

𝐶∞((0, 2𝑇) × Ω) satisfies the wave equation

(𝜕2

𝑡 − 𝑐2(𝑥)Δ)𝑣(𝑡, 𝑥) = 0,

in (0, 2𝑇) × Ω

Then

(𝑢 𝑓 (𝑇), 𝑣(𝑇))𝐿2 (Ω,𝑐−2𝑑𝑥) = (𝑃𝑇 𝑓 , 𝐽T𝐷𝑣)𝐿2 ((0,𝑇)×𝜕Ω) − (𝑃𝑇 (Λ𝑐 𝑓 ), 𝐽T𝑁 𝑣)𝐿2 ((0,𝑇)×𝜕Ω).

where 𝜈 is the unit outer normal vector field on 𝜕Ω. Here, the weighted space 𝐿2(Ω, 𝑐−2
0

𝑑𝑥) is

defined as

Proof. Define

We compute

𝐿2(Ω, 𝑐−2
0

𝑑𝑥) :=

∫

(cid:26)

𝑢 :

Ω

|𝑢(𝑥)|2𝑐−2

0 (𝑥) d𝑥 < ∞

(cid:27)

.

𝐼 (𝑡, 𝑠) := (𝑢 𝑓 (𝑡), 𝑣(𝑠))𝐿2 (Ω,𝑐−2𝑑𝑥).

(𝜕2

𝑡 − 𝜕2

𝑠 )𝐼 (𝑡, 𝑠)

=(Δ𝑢 𝑓 (𝑡), 𝑣(𝑠))𝐿2 (Ω) − (𝑢 𝑓 (𝑡), Δ𝑣(𝑠))𝐿2 (Ω)

(3.7)

=( 𝑓 (𝑡), T𝐷𝑣(𝑠))𝐿2 (𝜕Ω) − (Λ𝑐 𝑓 (𝑡), T𝑁 𝑣(𝑠))𝐿2 (𝜕Ω),

46

where the last equality follows from integration by parts. Since 𝑢 𝑓 (0, 𝑥) = 𝜕𝑡𝑢 𝑓 (0, 𝑥) = 0, we

have 𝐼 (0, 𝑠) = 𝜕𝑡 𝐼 (0, 𝑠) = 0, thus (3.7) can be considered as a inhomogeneous 1D wave equation

together with initial conditions 𝐼 (0, 𝑠) = 𝜕𝑡 𝐼 (0, 𝑠) = 0. Solving this PDE gives

𝐼 (𝑇, 𝑇)
∫ 𝑇

1
2
0
∫ 𝑇

∫ 2𝑇−𝑡

𝑡

[( 𝑓 (𝑡), 1
2

0

=

=

(cid:2)( 𝑓 (𝑡), T𝐷𝑣(𝜎))𝐿2 (𝜕Ω) − (Λ𝑐 𝑓 (𝑡), T𝑁 𝑣(𝜎))𝐿2 (𝜕Ω)
∫ 2𝑇−𝑡

∫ 2𝑇−𝑡

T𝐷𝑣(𝜎) 𝑑𝜎)𝐿2 (𝜕Ω) − (Λ𝑐 𝑓 (𝑡), 1
2

𝑡

𝑡

(cid:3) 𝑑𝜎𝑑𝑡

T𝑁 𝑣(𝜎) 𝑑𝜎)𝐿2 (𝜕Ω)] 𝑑𝑡

=(𝑃𝑇 𝑓 , 𝐽T𝐷𝑣)𝐿2 ((0,𝑇)×𝜕Ω) − (𝑃𝑇 (Λ𝑐 𝑓 ), 𝐽T𝑁 𝑣)𝐿2 ((0,𝑇)×𝜕Ω).

□

Lemma 3.1 is used to derive two important results. The first is the Blagove˘s˘censki˘ı’s identity.

To this end, denote by Λ𝑐,𝑇 the truncated ND map defined as in (3.2), (3.3) with 2𝑇 replaced by 𝑇.

It can be easily verified from integration by parts that its adjoint operator (with respect to the inner

product in 𝐿2((0, 𝑇) × 𝜕Ω)) is Λ∗

𝑐,𝑇 = 𝑅Λ𝑐,𝑇 𝑅 where 𝑅 is the time reversal operator (3.4), see also

Appendix B.1.

Introduce the connecting operator

𝐾 := 𝐽Λ𝑐𝑃∗

𝑇 − 𝑅Λ𝑐,𝑇 𝑅𝐽𝑃∗
𝑇 ,

(3.8)

which is the principal object of the boundary control method [14]. The operator 𝐾 connects

inner-products between waves in the interior to measurements on the boundary, see Proposition 3.2.

Moreover, 𝐾 is a compact operator since Λ𝑐,𝑇 : 𝐿2((0, 𝑇) ×𝜕Ω) → 𝐻2/3((0, 𝑇) ×𝜕Ω) is smoothing,

see [92].

The Blagove˘s˘censki˘ı’s identity we will establish is slightly different from its original form [21].

Instead, it is a reformulation that has been previously used in [20, 30, 74].

Proposition 3.2. Let 𝑢 𝑓 , 𝑢ℎ be the solutions of (3.3) with Neumann traces 𝑓 , ℎ ∈ 𝐿2((0, 𝑇) × 𝜕Ω),

respectively. Then

(𝑢 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2𝑑𝑥) = ( 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

(3.9)

47

In particular if ℎ = 𝑓 , one has

∥𝑢 𝑓 (𝑇)∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) = ( 𝑓 , 𝐾 𝑓 )𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾 𝑓 , 𝑓 )𝐿2 ((0,𝑇)×𝜕Ω).

(3.10)

Proof. We first prove this for 𝑓 , ℎ ∈ 𝐶∞
𝑇 ℎ and T𝑁𝑢ℎ = 𝑃∗

notice that T𝐷𝑢ℎ = Λ𝑐𝑃∗

𝑇 ℎ. One has

𝑐 ((0, 𝑇) × 𝜕Ω). Apply Lemma 3.1 to 𝑢 𝑓 and 𝑣 = 𝑢ℎ and

(𝑢 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2𝑑𝑥)

=(𝑃𝑇 𝑓 , 𝐽Λ𝑐𝑃∗

𝑇 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) − (𝑃𝑇 (Λ𝑐 𝑓 ), 𝐽𝑃∗

𝑇 ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

=( 𝑓 , 𝐽Λ𝑐𝑃∗

𝑇 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) − (Λ𝑐,𝑇 𝑓 , 𝐽𝑃∗

𝑇 ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

=( 𝑓 , 𝐽Λ𝑐𝑃∗

𝑇 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) − ( 𝑓 , 𝑅Λ𝑐,𝑇 𝑅𝐽𝑃∗

𝑇 ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

=( 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

where we have used that 𝑃𝑇 (Λ𝑐 𝑓 ) = Λ𝑐,𝑇 𝑓 and that Λ∗

𝑐,𝑇 = 𝑅Λ𝑐,𝑇 𝑅 in 𝐿2((0, 𝑇) × 𝜕Ω). This

establishes the first equality in (3.9). Interchanging 𝑓 and ℎ yields the second equality in (3.9).

For general 𝑓 , ℎ ∈ 𝐿2((0, 𝑇) × 𝜕Ω), simply notice that 𝐾 is a continuous operator and that

compactly supported smooth functions are dense in 𝐿2. The proof is completed.

□

Notice that harmonic functions can be considered as time independent wave solutions, we can

establish an inner product between waves and harmonic functions from boundary data. Introduce

an operator 𝐵 as

𝐵 := 𝐽T𝐷 − 𝑅Λ𝑐,𝑇 𝑅𝐽T𝑁 .

(3.11)

Proposition 3.3. Let 𝑢 𝑓 be the solutions of (3.3) with Neumann traces 𝑓 ∈ 𝐿2((0, 𝑇) × 𝜕Ω). For

any harmonic function 𝜙 ∈ 𝐶∞(Ω), one has

(𝑢 𝑓 (𝑇), 𝜙)𝐿2 (Ω,𝑐−2𝑑𝑥) = ( 𝑓 , 𝐵𝜙)𝐿2 ((0,𝑇)×𝜕Ω).

Proof. Similar to the proof of Proposition 3.2, let 𝑓 ∈ 𝐶∞

𝑐 ((0, 𝑇) × Ω), apply Lemma 3.1 to 𝑢 𝑓 and

48

𝑣 = 𝜙. One has

(𝑢 𝑓 (𝑇), 𝜙)𝐿2 (Ω,𝑐−2𝑑𝑥) =( 𝑓 , 𝐽T𝐷 𝜙)𝐿2 ((0,𝑇)×𝜕Ω) − (𝑃𝑇 (Λ𝑐 𝑓 ), 𝐽T𝑁 𝜙)𝐿2 ((0,𝑇)×𝜕Ω)

=( 𝑓 , 𝐽T𝐷 𝜙)𝐿2 ((0,𝑇)×𝜕Ω) − (Λ𝑐,𝑇 𝑓 , 𝐽T𝑁 𝜙)𝐿2 ((0,𝑇)×𝜕Ω)

=( 𝑓 , 𝐽T𝐷 𝜙)𝐿2 ((0,𝑇)×𝜕Ω) − ( 𝑓 , 𝑅Λ𝑐,𝑇 𝑅𝐽T𝑁 𝜙)𝐿2 ((0,𝑇)×𝜕Ω).

by the continuity of 𝐵 and density of compactly supported functions in 𝐿2, we complete the

proof.

□

Suppose for any harmonic function 𝜓, one can find an explicit sequence 𝑓𝛼 such that 𝑢 𝑓𝛼 (𝑇) → 𝜓

as 𝛼 → 0 in 𝐿2(Ω, 𝑐−2𝑑𝑥), then according to Proposition 3.3:

(𝜓, 𝜙)𝐿2 (Ω,𝑐−2𝑑𝑥) = lim
𝛼→0

(𝑢 𝑓𝛼 (𝑇), 𝜙)𝐿2 (Ω,𝑐−2𝑑𝑥) = lim
𝛼→0

( 𝑓𝛼, 𝐵𝜙)𝐿2 ((0,𝑇)×𝜕Ω).

(3.12)

The right hand side can be computed from Λ𝑐, see (3.11). Thus the integral

(𝜓, 𝜙)𝐿2 (Ω,𝑐−2𝑑𝑥) =

∫

Ω

𝜓𝜙 𝑐−2(𝑥) 𝑑𝑥

(3.13)

is known for all harmonic functions 𝜓 and 𝜙. For any fixed vectors 𝜉, 𝜂 ∈ R𝑛 with |𝜉 | = |𝜂| and

𝜉 ⊥ 𝜂, choose the complex harmonic functions as

𝜓(𝑥) := 𝑒 𝑖

2 (−𝜉+𝑖𝜂)·𝑥,

𝜙(𝑥) := 𝑒 𝑖

2 (−𝜉−𝑖𝜂)·𝑥.

(3.14)

Then 𝜓𝜙 = 𝑒𝑖𝜉·𝑥 and one recovers F (𝑐−2), the Fourier transform of 𝑐−2, by varying 𝜉, the wave

speed 𝑐 is recovered.

In order to construct sequence 𝑓𝛼, we introduce the control operator

𝑊 𝑓 := 𝑢 𝑓 (𝑇).

where 𝑢 𝑓 is the solution of (3.3). According to [61], 𝑊 : 𝐿2((0, 𝑇) × 𝜕Ω) → 𝐿2(Ω) is a bounded

linear operator. Moreover, Tataru’s theorem in [90] implies that the range of 𝑊 is dense in 𝐿2(Ω).

It follows from Proposition 3.2 that 𝐾 = 𝑊 ∗𝑊. It is also easy to verify that 𝑊 ∗𝜓 = 𝐵𝜓 for any

harmonic function 𝜓. The control sequence 𝑓𝛼 is constructed using Tikhonov regularization, and

the following lemma is used to prove the convergence.

49

Lemma 3.4 ( [74, Lemma 1]). Let 𝐴 : 𝑋 → 𝑌 be a bounded linear operator between two Hilbert

spaces 𝑋 and 𝑌 . For any 𝑦 ∈ 𝑌 , let 𝛼 > 0 be a constant and 𝑥𝛼 := ( 𝐴∗ 𝐴 + 𝛼)−1 𝐴∗𝑦. Then

𝐴𝑥𝛼 → 𝑃

𝑦

𝑅𝑎𝑛( 𝐴)

as 𝛼 → 0

where 𝑃

𝑅𝑎𝑛( 𝐴)

𝑦 denotes the orthogonal projection of 𝑦 onto the closure of the range of 𝐴.

Proposition 3.5. For any harmonic function 𝜓, the following minimization problem with parameter

𝛼 > 0:

𝑓𝛼 := arg min

𝑓

∥𝑊 𝑓 − 𝜓∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) + 𝛼∥ 𝑓 ∥2

𝐿2 (0,𝑇)×𝜕Ω

.

has a unique solution 𝑓𝛼 ∈ 𝐿2((0, 𝑇) × 𝜕Ω). This solution satisfies the linear equation

(𝐾 + 𝛼) 𝑓𝛼 = 𝐵𝜓.

(3.15)

Moreover, 𝑢 𝑓𝛼 (𝑇) → 𝜓 as 𝛼 → 0 in 𝐿2(Ω, 𝑐−2𝑑𝑥).

Proof. The functional to be minimized is

𝐹𝛼 ( 𝑓 ) := ∥𝑊 𝑓 − 𝜓∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) + 𝛼∥ 𝑓 ∥2

𝐿2 ((0,𝑇)×𝜕Ω)

.

As 𝑊 : 𝐿2((0, 𝑇) × 𝜕Ω) → 𝐿2(Ω) is bounded and linear, [56, Theorem 2.11] claims that 𝐹𝛼 has a

unique minimizer, named 𝑓𝛼, in 𝐿2((0, 𝑇) × 𝜕Ω).

The functional to be minimized is

𝐹𝛼 ( 𝑓 ) :=∥𝑢 𝑓 (𝑇) − 𝜓∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) + 𝛼∥ 𝑓 ∥2

𝐿2 ((0,𝑇)×𝜕Ω)

=∥𝑢 𝑓 (𝑇)∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) − 2(𝑢 𝑓 (𝑇), 𝜓)𝐿2 (Ω,𝑐−2𝑑𝑥) + ∥𝜓∥2
=( 𝑓 , 𝐾 𝑓 )𝐿2 ((0,𝑇)×𝜕Ω) − 2( 𝑓 , 𝐵𝜓)𝐿2 ((0,𝑇)×𝜕Ω) + ∥𝜓∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) + 𝛼∥ 𝑓 ∥2
𝐿2 (Ω,𝑐−2𝑑𝑥) + 𝛼∥ 𝑓 ∥2

𝐿2 (0,𝑇)×𝜕Ω

.

𝐿2 ((0,𝑇)×𝜕Ω)

=( 𝑓 , (𝐾 + 𝛼) 𝑓 )𝐿2 ((0,𝑇)×𝜕Ω) − 2( 𝑓 , 𝐵𝜓)𝐿2 ((0,𝑇)×𝜕Ω) + ∥𝜓∥2

𝐿2 (Ω,𝑐−2𝑑𝑥)

The terms ∥𝑢 𝑓 (𝑇)∥2

𝐿2 (Ω,𝑐−2𝑑𝑥) and (𝑢 𝑓 (𝑇), 𝜙)𝐿2 ((Ω,𝑐−2𝑑𝑥)) are computed using Proposition 3.2 and

Proposition 3.3, respectively. This is a bilinear form of 𝑓 whose Frechét derivative is

𝐹′( 𝑓 ) = 2(𝐾 + 𝛼) 𝑓 − 2𝐵𝜓.

50

The minimizer satisfies 𝐹′( 𝑓𝛼) = 0, hence (3.15).

Finally, since 𝐾 = 𝑊 ∗𝑊 and 𝐵𝜓 = 𝑊 ∗𝜓 (see the content before Proposition 3.5), we conclude

from Lemma 3.4 that 𝑊 𝑓𝛼 → 𝑃

𝑅𝑎𝑛(𝑊)

𝜓 in 𝐿2(Ω, 𝑐−2𝑑𝑥) as 𝛼 → 0. Tataru’s theorem [90] claims

that the range of 𝑊 is dense in 𝐿2(Ω), hence 𝑃

𝜓 = 𝜓.

𝑅𝑎𝑛(𝑊)

□

Summarizing the discussion in this section, we have proved global convergence of the following

reconstruction algorithm, see Algorithm 3.1.

Input: low-pass filter 𝐽, time-reversal operator 𝑅, projection operator 𝑃𝑇 , ND map Λ𝑐
Output: wave speed 𝑐
1. Assemble the connecting operator 𝐾 = 𝐽Λ𝑐𝑃∗
𝑇 − 𝑅Λ𝑐,𝑇 𝑅𝐽𝑃∗
2. Assemble the operator 𝐵 = 𝐽T𝐷 − 𝑅Λ𝑐,𝑇 𝑅𝐽T𝑁 (see (3.11)).
3. Construct the harmonic function 𝜓(𝑥) = 𝑒 𝑖
(𝐾 + 𝛼) 𝑓𝛼 = 𝐵𝜓, (see (3.15)).
4. Construct the harmonic function 𝜙(𝑥) := 𝑒 𝑖
jection

𝑇 (see (3.8)).

2 (−𝜉−𝑖𝜂)·𝑥 (see (3.14)) and compute the Fourier pro-

2 (−𝜉+𝑖𝜂)·𝑥 (see (3.14)) and solve the linear system

∫

Ω

𝑒−𝑖𝜉·𝑥𝑐−2(𝑥) 𝑑𝑥 = lim
𝛼→0

( 𝑓𝛼, 𝐵𝜙)𝐿2 ((0,𝑇)×𝜕Ω)

through the limiting process, (see (3.12)).
5. Repeat the above steps with various 𝜉 to recover the Fourier transform F (𝑐−2).
6. Invert the Fourier transform to recover 𝑐−2, and eventually 𝑐.

Algorithm 3.1 Non-Iterative Reconstruction Algorithm for Acoustic IBVP.

3.2.2 Algorithm Implementation

In this section, we numerically implement the algorithm using finite difference scheme. We

choose the spatial domain to be Ω = [−1, 1]2. For the forward problem, we uniformly discretized

Ω into 101 × 101 grids, i.e. Δ𝑥 = Δ𝑦 = 1
𝑥𝑖 = −1 + 𝑖

100 , 0 ≤ 𝑖, 𝑗 ≤ 100. The time step size Δ𝑡 is chosen as Δ𝑡 = 2𝑇

50 . The grid points are represented as (𝑥𝑖, 𝑦 𝑗 ), where
Δ𝑥

100, 𝑦 𝑗 = −1 +

𝐿 ≤

√

𝑗

2
2𝑐max

to fulfill the Courant–Friedrichs–Lewy (CFL) condition, where 𝑐max denote the maximum of 𝑐(𝑥)

over Ω, 𝐿 is the number of time steps. Then the temporal grid points are labeled using 𝑡𝑙 = 𝑙Δ𝑡,

𝑙 = 0, 1, . . . , 𝐿. For simplicity, the values of 𝑢 on the grid points are denoted by

𝑢𝑙
𝑖 𝑗 := 𝑢(𝑡𝑙, 𝑥𝑖, 𝑦 𝑗 ),

𝑙 = 0, 1, . . . , 𝐿,

𝑖, 𝑗 = 0, 1, . . . , 100.

51

3.2.2.1 Discrete Wave Equation Solver

The finite difference solver for (3.3) requires the approximation of operators Δ, 𝜕2

𝑡 and 𝜕𝜈. We

use the second order approximation for interior grid points:

𝑡 𝑢(𝑡𝑙, 𝑥𝑖, 𝑦 𝑗 ) ≈
𝜕2

𝑖, 𝑗 + 𝑢𝑙+1
𝑖, 𝑗 − 2𝑢𝑙
𝑢𝑙−1
𝑖, 𝑗
Δ𝑡2

;

Δ𝑢(𝑡𝑙, 𝑥𝑖, 𝑦 𝑗 ) ≈

𝑖−1, 𝑗 + 𝑢𝑙
𝑢𝑙

𝑖+1, 𝑗 + 𝑢𝑙

𝑖, 𝑗−1
Δ𝑥2

+ 𝑢𝑙

𝑖, 𝑗+1

− 4𝑢𝑙
𝑖, 𝑗

,

which gives us the update formula for interior points:

𝑖, 𝑗 = 2𝑢𝑙
𝑢𝑙+1

𝑖, 𝑗 − 𝑢𝑙−1

𝑖, 𝑗 + 𝑐2(𝑥𝑖, 𝑦 𝑗 )

Δ𝑡2
Δ𝑥2

[𝑢𝑙

𝑖−1, 𝑗 + 𝑢𝑙

𝑖+1, 𝑗 + 𝑢𝑙

𝑖, 𝑗−1 + 𝑢𝑙

𝑖, 𝑗+1 − 4𝑢𝑙

𝑖, 𝑗 ].

The boundary points are updated from the discretization of Neumann derivatives. For instance

𝑖 = 0, we have

𝜕𝜈𝑢(𝑡𝑙, 𝑥0, 𝑦 𝑗 ) ≈ −

3𝑢𝑙

0, 𝑗 − 4𝑢𝑙
2Δ𝑥

1, 𝑗 + 𝑢𝑙
2, 𝑗

.

The initial condition 𝑢(0, 𝑥) = 𝑢𝑡 (0, 𝑥) = 0 is implemented by setting

𝑢0
𝑖, 𝑗 = 0,

𝑖, 𝑗 = 𝑢−1
𝑢1
𝑖, 𝑗 .

Notice that all the finite difference approximation above have second order accuracy, we have

an discrete wave solver.

3.2.2.2

Implementation of ND Map Λ𝑐

The spatial boundary 𝜕Ω consists of 400 boundary grid points, thus the temporal boundary

[0, 𝑇] × 𝜕Ω contains 400(𝐿 + 1) boundary grid points in total. These boundary grid points

are ordered in the lexicographical order to form a column vector, that is, a boundary grid point

(𝑡𝑙, 𝑥𝑖, 𝑦 𝑗 ) is ahead of another (𝑡𝑙′, 𝑥𝑖′, 𝑦 𝑗 ′) if and only if (1) 𝑙 < 𝑙′; or (2) 𝑙 = 𝑙′ and 𝑖 < 𝑖′; or (3)

𝑙 = 𝑙′, 𝑖 = 𝑖′, 𝑗 < 𝑗 ′. In this way, the discretized ND map is a 400(𝐿 + 1) × 400(𝐿 + 1) square

matrix, denoted by [Λ𝑐] ∈ R400(𝐿+1)×400(𝐿+1).

In order to find the matrix representation [Λ𝑐], we place a unit source 𝑓 𝑙

𝑖 𝑗 on each boundary grid

point as the Neumann data and utilize the forward solver to obtain the resulting Dirichlet data on all

52

the boundary grid points. Here 𝑓 𝑙

𝑖 𝑗 takes the value 1 on (𝑡𝑙, 𝑥𝑖, 𝑦 𝑗 ) and 0 on all the other boundary

grid points.

Once the matrices are generated, we re-sample the ND map on a coarser grid of size (𝐿 +

1) × 51 × 51 and implement Algorithm 3.1 to avoid the inverse crime, i.e.

changed to Δ𝑥 = Δ𝑦 = 1

25. Thus the ND map matrices are [Λ𝑐,𝑇 ] ∈ R200⌈ 𝐿+1

the spatial grid size
2 ⌉×200⌈ 𝐿+1

2 ⌉ and

[Λ𝑐] ∈ R200(𝐿+1)×200(𝐿+1).

3.2.2.3

Implementation of Connecting Operator 𝐾

In order to discretize 𝐾, we need to discretize 𝐽 first. The integration is calculated using

trapezoidal rule, which is

∫ 2𝑇−𝑡𝑙

𝑡𝑙

𝑓 (𝜏, ·) 𝑑𝜏 ≈

𝐿−𝑙−1
∑︁

𝑘=𝑙

𝑓 (𝑡𝑘 , ·) + 𝑓 (𝑡𝑘+1, ·)
2

Δ𝑡.

Since we rearrange the grid points into lexicographical order, the discretized filtering operator can
be written as a blocking matrix [𝐽] ∈ R200⌈ 𝐿+1
matrix. Here ⌈ 𝐿+1

2 ⌉ denote the smallest integer which is greater or equal to 𝐿+1
2 .

2 ⌉×200(𝐿+1), whose blocks are all 200 × 200 diagonal

Specifically, if 𝐿 is odd,

[𝐽] =

Δ𝑡
2

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

If 𝐿 is even,

[𝐼] 2[𝐼] 2[𝐼]

[𝐼]

2[𝐼]
. . .

. . .

. . .
. . .

. . .

. . .

. . .

. . .

. . . 2[𝐼] 2[𝐼]

[𝐼]

[𝐼]

. . . 2[𝐼]
. . .
. . .

[𝐼] 2[𝐼] 2[𝐼]

[𝐼]

[𝐼]

[𝐼]

,

(cid:170)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:172)

[𝐽] =

Δ𝑡
2

[𝐼] 2[𝐼] 2[𝐼]

[𝐼]

2[𝐼]
. . .

. . .

. . .

. . .

. . .
. . .

. . . 2[𝐼] 2[𝐼]

[𝐼]

[𝐼]

. . . 2[𝐼]
. . .
. . .

[𝐼] 2[𝐼]

[𝐼]

[𝑂]

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)

,

(cid:170)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:172)

where [𝐼] denote the identity matrix and [𝑂] denote the zero matrix.

53

Similarly the discretized time reversal operator [𝑅] ∈ R200⌈ 𝐿+1

2 ⌉×200⌈ 𝐿+1

2 ⌉ and discretized restric-

tion operator [𝑃𝑇 ] ∈ R200⌈ 𝐿+1

2 ⌉×200(𝐿+1) can be represented as

(cid:16)

[𝑃𝑇 ] =

[𝐼]200⌈ 𝐿+1

[𝐼]

,

(cid:17)

2 ⌉×200⌈ 𝐿+1
2 ⌉

[𝑂]

[𝑅] =

(cid:169)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:173)
(cid:171)
𝑇 ] is taken to be [𝑃𝑇 ]𝑡, the transpose of [𝑃𝑇 ]. It is obvious that

(cid:170)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:174)
(cid:172)

. . .

[𝐼]

.

The discrete extension operator [𝑃∗

[Λ𝑐,𝑇 ] = [𝑃𝑡] [Λ𝑐] [𝑃∗

𝑇 ], which implies [Λ𝑐,𝑇 ] is simply the top left submatrix of [Λ𝑐]. Finally, the

discretized 𝐾 is the following matrix product, according to (3.8):

[𝐾] = [𝐽] [Λ𝑐] [𝑃𝑇 ]𝑡 − [𝑅] [Λ𝑐,𝑇 ] [𝑅] [𝐽] [𝑃𝑇 ]𝑡

∈ R200⌈ 𝐿+1

2 ⌉×200⌈ 𝐿+1

2 ⌉ .

3.2.2.4

Implementation of the Operator 𝐵

The implementation of operator 𝐵 requires to calculate T𝐷, T𝑁 . Since the harmonic function 𝜓

is handcrafted and time independent, we can analytically calculate Dirichlet and Neumann value of

the known harmonic function and make 𝐿 + 1 copies to form vector [T𝐷𝜓], [T𝑁 𝜓] ∈ R200(𝐿+1)×1.

3.2.2.5 Solve Boundary Control Sequence 𝑓𝛼

Following The next step is to solve for [ 𝑓𝛼] from the discretized version of (3.15):

([𝐾] + 𝛼) [ 𝑓𝛼] = [𝐵] [𝜓|𝜕Ω].

(3.16)

Here [ 𝑓𝛼] is the discretized version of 𝑓𝛼 in (3.15); 𝜓 is an arbitrary harmonic function and

[𝜓|𝜕Ω] ∈ R200(𝐿+1)×1 denotes the vectorized boundary restriction 𝜓|𝜕Ω. Both [ 𝑓𝛼] and 𝜓|𝜕Ω are in

the lexicographical order as before. Since [𝐾] is calculated using matrix multiplication, it is not

guaranteed to be positive semidefinite. Instead of solving (3.15) with Tikhonov regularization, the

equation that we solve is

([𝐾]𝑡 [𝐾] + 𝛼) [ 𝑓𝛼] = [𝐾]𝑡 [𝐵] [𝜓|𝜕Ω]

(3.17)

where [𝐾]𝑡 is the transpose of [𝐾].

54

3.2.2.6 Solve Wave Speed 𝑐

In Algorithm 3.1, 𝑐 is recovered by creating appropriate complex exponential harmonic func-

tions (3.14) and inverting the Fourier transform. However, due to their propensity for explosive

growth in some directions, such harmonic functions are not appropriate for numerical implemen-

tation.

We build harmonic functions using the fundamental solutions method (FSM) as an alternative

to selecting complex exponential harmonic functions. Kupradze [58] was the first to propose this

method, which has the advantage of being easily implemented numerically. There has been research

on its suitability for elliptic boundary value problems in general [23]; additionally, see also the

review paper [42]. In FSM, the harmonic functions are of the form

𝑁
∑︁

𝑗=1

𝑎 𝑗 Φ(|𝑥 − 𝑥 ( 𝑗) |)

(3.18)

Φ(𝑟) = 1

Here Φ is the fundamental solution of the Laplace operator, i.e. Φ(𝑟) = log 𝑟 for 𝑛 = 2 and
𝑟 for 𝑛 ≥ 3, and 𝑎 𝑗 are real scalar coefficients. These functions are harmonic in R𝑛 except
at the singularities 𝑥 ( 𝑗), which are chosen to be outside the computational domain. It has been

shown [67] that an arbitrary 2D function which is harmonic inside the unit disk and continuous up

to the boundary can be approximated to any prescribed accuracy using functions of the form (3.18).

Under such construction, for a fixed sufficiently small 𝛼 > 0, the right-hand side of (3.12) can

be approximate using the trapezoidal rule:

( 𝑓𝛼, 𝐵𝜙)𝐿2 ((0,𝑇)×𝜕Ω) ≈

200⌈ 𝐿+1
2 ⌉
∑︁

𝑗=1

𝑤 𝑗 [ 𝑓𝛼] 𝑗 [𝐵𝜙] 𝑗 Δ𝑥Δ𝑡

(3.19)

where [ 𝑓𝛼] has been obtained from (3.17), [𝐵𝜙] is computed from the matrix multiplication, and
𝑤 ∈ R200⌈ 𝐿+1

2 ⌉ is the weight coefficient vector from the trapezoidal rule, whose first and last 200

elements are 1

2 and others are 1.

On the left-hand side of (3.12), we approximate the interior integral over Ω by successively

applying the trapezoidal rule first to 𝑦 and then to 𝑥. If we write ˜𝑤 = ( 1
2

, 1, . . . , 1, 1

2) ∈ R51 for the

55

coefficient vector of the trapezoidal rule, then

(𝜓, 𝜙)𝐿2 (Ω,𝑐−2𝑑𝑥) =

≈

∫ 1

∫ 1

𝜓(𝑥, 𝑦)𝜙(𝑥, 𝑦)𝑐−2(𝑥, 𝑦) 𝑑𝑥𝑑𝑦

−1

˜𝑤 𝑗 ˜𝑤 𝑘 𝜓(𝑥 𝑗 , 𝑦𝑘 )𝜙(𝑥 𝑗 , 𝑦𝑘 )𝑐−2(𝑥 𝑗 , 𝑦𝑘 ) (Δ𝑥)2.

(3.20)

−1
𝐼
∑︁

𝑗,𝑘=0

Equating (3.19) and (3.20) and inserting various harmonic functions of the form (3.18) gives

rise to a system of linear equations on the unknowns 𝑐−2(𝑥 𝑗 , 𝑦𝑘 ), 𝑗, 𝑘 = 0, 1, . . . , 50. Depending on

the number of harmonic functions, the linear system can be over determined or under determined,

we can apply different regression methods, such as least square regression, Tikhonov regularization,

to solve for the regularized unknowns.

3.2.3 Numerical Experiment

In this section, following (3.18), we choose harmonic basis to be

𝜙(1) = ln((𝑥 − 2.3)2 + (𝑦 − 2.2)2),

𝜙(2) = ln((𝑥 + 2.5)2 + (𝑦 − 2.1)2),

𝜙(3) = ln((𝑥 − 2.7)2 + (𝑦 + 1.9)2),

𝜙(4) = ln((𝑥 + 1.5)2 + (𝑦 + 2.5)2),

𝜙(5) = ln((𝑥 + 1.2)2 + (𝑦 + 2.5)2),

𝜙(6) = 1.

We denote the vector space generated by the products of these harmonic functions by 𝑆6, that is,

𝑆6 := span{𝜙(𝑖) 𝜙( 𝑗) : 𝑖, 𝑗 = 1, . . . , 6}.

If 𝑐−2 is in the vector space 𝑆6, we have following representation

𝑐−2 =

∑︁

𝑐𝑖 𝑗 𝜙(𝑖) 𝜙( 𝑗).

1≤𝑖≤ 𝑗 ≤6

Take the inner product (·, ·)𝐿2 (Ω) with 𝜙(𝑘) 𝜙(𝑙) respectively, 1 ≤ 𝑘 ≤ 𝑙 ≤ 6, to obtain the following

linear system

A ·

=

𝑐11

𝑐12
...

𝑐66





























(𝑐−2, 𝜙(1) 2)𝐿2 (Ω)
(𝑐−2, 𝜙(1) 𝜙(2))𝐿2 (Ω)
...
(𝑐−2, 𝜙(7) 2)𝐿2 (Ω)















.















56

(3.21)

where the 21 × 21 coefficient matrix A is

(𝜙(1) 2, 𝜙(1) 2)𝐿2 (Ω)
(𝜙(1) 𝜙(2), 𝜙(1) 2)𝐿2 (Ω)
...
(𝜙(6) 2, 𝜙(1) 2)𝐿2 (Ω)















(𝜙(1) 2, 𝜙(1) 𝜙(2))𝐿2 (Ω)
(𝜙(1) 𝜙(2), 𝜙(1) 𝜙(2))𝐿2 (Ω)
...
(𝜙(6) 2, 𝜙(1) 𝜙(2))𝐿2 (Ω)

. . .

. . .
. . .

. . .

(𝜙(1) 2, 𝜙(6) 2)𝐿2 (Ω)
(𝜙(1) 𝜙(2), 𝜙(6) 2)𝐿2 (Ω)
...
(𝜙(6) 2, 𝜙(6) 2)𝐿2 (Ω)















The coefficient matrix on the left-hand side can be analytically computed, and the components of

the vector on the right-hand side are exactly the inner products on the left hand side of (3.20). We

then solve the discretized version of this linear system to obtain the coefficients 𝑐𝑖 𝑗 .

The orthogonal projection of 𝑐−2 onto 𝑆6 is implied by the inner products on the right side

of (3.21) if 𝑐−2 is not in the vector space 𝑆6. The linear system can be solved to obtain an

orthogonal projection. Since harmonic function products are dense in 𝐿2(Ω), the orthogonal

projection should better approximate 𝑐−2 as the number of harmonic functions 𝜙( 𝑗) increases.

3.2.3.1 Experiment 1: 𝑐 ≡ 1, 𝑐−2 ∈ 𝑆6.

We start our experiment by evaluating the reconstruction for the simplest case, the wave speed

𝑐1 ≡ 1. Notice that 𝑐−2

1 ≡ 1 ∈ 𝑆6 since 𝜙(6) = 1. Figure 3.2 shows the reconstructed speed and
errors in the presence of 0%, 5%, and 50% of Gaussian random noises with zero mean and unit

variance, respectively. It is observed that the reconstructed wave speed are little affected by the

presence of random noise. This is because, in the definition (3.8) of 𝐾, the low-pass filter 𝐽 and the

ND map Λ𝑐 tend to smooth out the Gaussian random noise.

3.2.3.2 Experiment 2: 𝑐 is variable, 𝑐−2 ∈ 𝑆6.

The algorithm is then tested with a variable speed 𝑐2, where 𝑐−2

2 ∈ 𝑆6. In this experiment, the

random Gaussian noise is fixed at 5%. The ground-truth speed is generated as

𝑐−2
2 =

6
∑︁

𝑖

10

𝑖=1

𝜙(𝑖),

see the leftmost of Figure 3.3.

We execute the reconstructions using the first 2, 4, and 6 harmonic functions 𝜙(𝑖), correspond-

ingly, to demonstrate how the quality of the images improves with an increase in the number of basis

57

Figure 3.2 Experiment 1. Top row: reconstructed 𝑐. Bottom row: error between the
reconstruction and the ground truth. First column: 0% noise; the relative 𝐿2-error is 0.4769%.
Second column: 5% noise; the relative 𝐿2-error is 0.4873%. Third column: 50% noise; the
relative 𝐿2-error is 0.5454%. Grid: 283 × 51 × 51.

Figure 3.3 The variable speed 𝑐2.

functions. As the number of harmonic functions rises, we find that the reconstruction becomes

more accurate; Figure 3.4 provides numerical evidence for this observation.

3.2.3.3 Experiment 3: 𝑐 is variable, 𝑐−2 ∉ 𝑆6.

Next, we test the ability of the algorithm in recovering a variable speed

𝑐3(𝑥, 𝑦) = 1 + 0.08 sin 𝜋𝑥 + 0.06 cos 𝜋𝑦,

with 𝑐−2

3 ∉ 𝑆6. The wave speed 𝑐3 and its orthogonal projection on 𝑆6 are illustrated in Figure 3.5.
The reconstructions with the first 2, 4, and 6 harmonic functions 𝜙(𝑖) are shown in Figure 3.6.

It is not possible to recover the exact discrete version of 𝑐−2 in this scenario. The 𝐿2-orthogonal

projection of 𝑐−2 onto the subspace 𝑆6 is what the algorithm produces instead. This is because, in

58

Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.9940.9960.99811.0021.0041.0061.008Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.9940.9960.99811.0021.0041.0061.008Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.9920.9940.9960.99811.0021.0041.0061.008Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.811234567810-3Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.811234567810-3Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8112345678910-3Ground Truth c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.470.480.490.50.510.520.530.540.550.560.57Figure 3.4 Experiment 2. Top row: reconstructed 𝑐. Bottom row: error between the reconstruction
and the ground truth. First column: First 2 harmonic functions; the relative 𝐿2-error is 15.6987%.
Second column: First 4 harmonic functions; the relative 𝐿2-error is 0.7939%. Third column: All
6 harmonic functions; the relative 𝐿2-error is 0.7907%. Grid: 323 × 51 × 51.

Figure 3.5 Left: the variable speed 𝑐3. Right: orthogonal projection of 𝑐3 on 𝑆6.

order to solve for [𝑐−2], Tikhonov regularization was used. You can view the numerical validation

in Figure 3.6.

3.2.3.4 More Experiments

We also test the performance of the inversion formulae in different cases, such as discontinuous

wave speed 𝑐(𝑥) or partial data case, i.e. the NP map is set to zero at the region can not be measured.

See [96] for details.

3.3 Linearized Inverse Boundary Value Problem

Linearization, such as Born approximation, Kirchhoff approximation, is a method to give

linear approximation of a nonlinear model. It is widely used to solve nonlinear inverse problems

[3, 19, 60, 98]. The linearized inverse problems consider the model as a perturbation of a (known)

background model. One can construct a perturbed model which is linearly depend on the unknown

59

Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.450.50.550.60.65Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.470.480.490.50.510.520.530.540.550.560.57Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.470.480.490.50.510.520.530.540.550.560.57Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.020.040.060.080.10.120.140.160.180.2Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.811234567810-3Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.811234567810-3Ground Truth c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.90.9511.051.1Projection-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.90.920.940.960.9811.021.041.061.08Figure 3.6 Experiment 3. Top row: reconstructed 𝑐. Bottom row: error between the
reconstruction and the orthogonal projection of the ground truth. First column: First 2 harmonic
functions; the relative 𝐿2-error is 12.3535%. Second column: First 4 harmonic functions; the
relative 𝐿2-error is 0.4139%. Third column: All 6 harmonic functions; the relative 𝐿2-error is
0.3104%. Grid: 323 × 51 × 51.

parameters. In this section, we introduce linearized IBVP algorithms for both wave speed and wave

potential reconstruction.

3.3.1 Reconstruct Wave Speed through Linearization

In Section 3.2, we introduce an algorithm to reconstruct wave speed with full nonlinear treat-

ment. In this section, we apply linearization to the nonlinear IBVP to reconstruct the wave speed.

We assume the wave potential 𝑞 = 𝑞0(𝑥) ∈ 𝐶∞(Ω) is known, and we want to recover 𝑐(𝑥) from the

ND map Λ𝜌,𝑞0. For simplicity, we use Λ𝜌 to represent Λ𝜌,𝑞0 in this section.

We use the linearization to solve the IBVP. For the formal derivation, we write

𝜌(𝑥) = 𝜌0(𝑥) + 𝜀 (cid:164)𝜌(𝑥),

𝑢(𝑡, 𝑥) = 𝑢0(𝑡, 𝑥) + 𝜀 (cid:164)𝑢(𝑡, 𝑥)

where 𝜌0 = 𝑐−2

0 from a known background wave speed and 𝑢0 is the background solution. Substitute

these into (3.1). Equating the 𝑂 (1)-terms gives

□𝜌0,𝑞0

𝑢0(𝑡, 𝑥) = 0,

in (0, 2𝑇) × Ω

𝜕𝜈𝑢0 = 𝑓 ,

on (0, 2𝑇) × 𝜕Ω

(3.22)

𝑢0(0, 𝑥) = 𝜕𝑡𝑢0(0, 𝑥) = 0,

𝑥 ∈ Ω.






60

Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.911.11.21.31.4Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.920.940.960.9811.021.041.061.08Reconstructed c-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.90.920.940.960.9811.021.041.061.08Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.050.10.150.20.250.30.350.4Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.8124681012141610-3Error-1-0.500.51-1-0.8-0.6-0.4-0.200.20.40.60.810.511.522.533.5410-3Equating the 𝑂 (𝜀)-terms gives

□𝜌0,𝑞0 (cid:164)𝑢(𝑡, 𝑥) = − (cid:164)𝜌(𝑥)𝜕2
𝜕𝜈 (cid:164)𝑢 = 0,

𝑡 𝑢0(𝑡, 𝑥),

in (0, 2𝑇) × Ω

on (0, 2𝑇) × 𝜕Ω

(3.23)

(cid:164)𝑢(0, 𝑥) = 𝜕𝑡 (cid:164)𝑢(0, 𝑥) = 0

𝑥 ∈ Ω.






Write the ND map Λ𝜌 = Λ𝜌0 + 𝜀 (cid:164)Λ (cid:164)𝜌, where Λ𝜌0 denote the ND map for the unperturbed boundary
value problem (3.22), and (cid:164)Λ (cid:164)𝜌 is defined as

(cid:164)Λ (cid:164)𝜌 : 𝑓 ↦→ (cid:164)𝑢| (0,2𝑇)×𝜕Ω.

(3.24)

Since 𝜌0 and 𝑞0 are known, the unperturbed problem (3.22) can be explicitly solved to obtain 𝑢0
and Λ𝜌0. As in the previous section, we will write (cid:164)𝑢 = (cid:164)𝑢 𝑓 if it is necessary to specify the Neumann
data 𝑓 . Then the linearized IBVP concerns recovery of the speed (cid:164)𝜌 from (cid:164)Λ (cid:164)𝜌.

3.3.1.1 Derivation

Similar to Section 3.2, introduce the time reversal operator 𝑅, see (3.4), the low-pass filter 𝐽, see

(3.5). the orthogonal projection operator 𝑃𝑇 , see (3.6) and its adjoint operator 𝑃∗

𝑇 as the extension

by zero from (0, 𝑇) to (0, 2𝑇). Let T𝐷 and T𝑁 be the Dirichlet and Neumann trace operators

respectively, that is,

T𝐷𝑢(𝑡, ·) = 𝑢(𝑡, ·)|𝜕Ω,

T𝑁𝑢(𝑡, ·) = 𝜕𝜈𝑢(𝑡, ·)|𝜕Ω.

Introduce the connecting operator

𝐾 := 𝐽Λ𝑞𝑃∗

𝑇 − 𝑅Λ𝑞,𝑇 𝑅𝐽𝑃∗
𝑇

Similarly, the following Blagove˘s˘censki˘ı’s identity holds.

Proposition 3.6. Let 𝑢 𝑓 , 𝑢ℎ be the solutions of (3.1) with Neumann traces 𝑓 , ℎ ∈ 𝐿2((0, 𝑇) × 𝜕Ω),

respectively. Then

(𝑢 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = ( 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

(3.25)

61

Proof. The proof is similar to the proof of Proposition 3.2. We first prove this for 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇)×

𝜕Ω). Define

We compute

𝐼 (𝑡, 𝑠) := (𝑢 𝑓 (𝑡), 𝑢ℎ (𝑠))𝐿2 (Ω,𝑐−2

0

.

𝑑𝑥)

(𝜕2

𝑡 − 𝜕2

𝑠 )𝐼 (𝑡, 𝑠)

=((Δ + 𝑞)𝑢 𝑓 (𝑡), 𝑢ℎ (𝑠))𝐿2 (Ω) − (𝑢 𝑓 (𝑡), (Δ + 𝑞)𝑢ℎ (𝑠))𝐿2 (Ω)

=( 𝑓 (𝑡), Λ𝜌𝑃∗

𝑇 ℎ(𝑠))𝐿2 (𝜕Ω) − (Λ𝜌𝑃∗

𝑇 𝑓 (𝑡), ℎ(𝑠))𝐿2 (𝜕Ω),

(3.26)

where the last equality follows from integration by parts.

Notice that the expression in (3.26) is exactly the same as in (3.7), the following proof are

exactly the same as the proof of Proposition 3.2.

Corollary 3.7. Suppose 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω). Then

(Δ𝑢 𝑓 (𝑇) − 𝑞𝑢 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = (𝜕2

𝑡 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

Proof. Replacing 𝑓 by 𝜕2

𝑡 𝑓 in (3.25), we get

□

(3.27)

(𝑢𝜕2

𝑡 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = (𝜕2

𝑡 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

As both 𝑢𝜕2

𝑡 𝑓 and 𝜕2

𝑡 𝑢 𝑓 satisfy (3.1) with 𝑓 replaced by 𝜕2

𝑡 𝑓 , the well-posedness of the boundary

value problem ensures that

𝜌𝑢𝜕2

𝑡 𝑓 = 𝜌𝜕2

𝑡 𝑢 𝑓 = Δ𝑢 𝑓 − 𝑞𝑢 𝑓 .

Remember that we write Λ𝜌 = Λ𝜌0 + 𝜀 (cid:164)Λ (cid:164)𝜌 in the linearization setting, we have the following
linearization 𝐾 = 𝐾0 + 𝜀 (cid:164)𝐾. Here 𝐾0 is the connecting operator for the background wave equation

(3.22):

𝐾0 := 𝐽Λ𝜌0

𝑇 − 𝑅Λ𝜌0,𝑇 𝑅𝐽𝑃∗
𝑃∗
𝑇 .

(3.28)

□

62

𝐾0 can be explicitly computed since Λ𝜌0,𝑞0 is known.

(cid:164)𝐾 is the resulting perturbation in the

connecting operator:

(cid:164)𝐾 := 𝐽 (cid:164)Λ (cid:164)𝜌𝑃∗

𝑇 − 𝑅 (cid:164)Λ (cid:164)𝜌,𝑇 𝑅𝐽𝑃∗
𝑇 .

(3.29)

(cid:164)𝐾 can be explicitly computed once (cid:164)Λ (cid:164)𝜌 is given.

We write (cid:164)Λ for (cid:164)Λ (cid:164)𝜌 when there is no risk of confusion. Linearizing (3.25) and (3.27) gives the

following integral identity, which is essential to the development of the reconstruction procedure:

Proposition 3.8. Let 0 ≠ 𝜆 ∈ R be a nonzero real number. If 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) satisfy

[Δ − 𝑞0 + 𝜆𝜌0]𝑢 𝑓

0

(𝑇) = [Δ − 𝑞0 + 𝜆𝜌0]𝑢ℎ

0 (𝑇) = 0

in Ω,

(3.30)

then the following identity holds:

−( (cid:164)𝜌𝑢 𝑓
0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω) =

1
𝜆

[(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)].

(3.31)

Proof. For 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇) × 𝜕Ω), we will make use of (3.25) (3.27) to obtain some identities.

First, we substitute all linearizations into (3.25). Equating 𝑂 (1)-terms gives

(𝜌0𝑢 𝑓

0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω) = ( 𝑓 , 𝐾0ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾0 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

Equating 𝑂 (𝜀)-terms gives

( 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = ( (cid:164)𝐾 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω)
=( (cid:164)𝜌𝑢 𝑓
0

0 (𝑇))𝐿2 (Ω) + (𝜌0 (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω) + (𝜌0𝑢 𝑓

0

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω).

(3.32)

Similarly, we substitute all linearizations into (3.27). Equating 𝑂 (1)-terms gives

(Δ𝑢 𝑓
0

(𝑇) − 𝑞𝑢 𝑓
0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω) = (𝜕2

𝑡 𝑓 , 𝐾0ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾0𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

Equating 𝑂 (𝜀)-terms gives

(𝜕2

𝑡 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = ( (cid:164)𝐾𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

=((Δ − 𝑞0) (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

=( (cid:164)𝑢 𝑓 (𝑇), (Δ − 𝑞0)𝑢ℎ

0 (𝑇))𝐿2 (Ω) + ((Δ − 𝑞0)𝑢 𝑓
0 (𝑇))𝐿2 (Ω) − ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω) + ((Δ − 𝑞0)𝑢 𝑓

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω)

0

0

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω)

63

where the last inequality use the integration by parts and the fact that (cid:164)𝑢 𝑓 | (0,2𝑇)×𝜕Ω = (cid:164)Λ 𝑓 and
𝜕𝜈 (cid:164)𝑢 = 0. Add (3.32) multiplied by 𝜆 ∈ R to get

(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

=(𝜕2

𝑡 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω) + (𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

=( (cid:164)𝑢 𝑓 (𝑇), [Δ − 𝑞0 + 𝜆𝜌0]𝑢ℎ

0 (𝑇))𝐿2 (Ω) + ( [Δ − 𝑞0 + 𝜆𝜌0]𝑢 𝑓

0

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω)

+ 𝜆( (cid:164)𝜌𝑢 𝑓
0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω).

If [Δ − 𝑞0 + 𝜆𝜌0]𝑢 𝑓

0

(𝑇) = [Δ − 𝑞0 + 𝜆𝜌0]𝑢ℎ
0

(𝑇) = 0 in Ω, the first term and second term on the

right-hand side vanish, resulting in (3.31).

□

Notice that all parameters in (3.30) are known. For each 𝜆 ∈ R, we can construct functions

𝜙, 𝜓 satisfy (3.30). According to Proposition 3.8, once we find control sequence 𝑓𝛼, ℎ𝛼 such that
𝑢 𝑓𝛼 (𝑇) → 𝜙, 𝑢ℎ 𝛼 (𝑇) → 𝜓, we have the weighted inner product (𝜙, 𝜓)𝐿2 (Ω, (cid:164)𝜌 d𝑥) from (3.31). The

following proposition ensures the existance of such boundary controls:

Proposition 3.9. Let 𝑐0 ∈ 𝐶∞(Ω) be strictly positive and 𝑞0 ∈ 𝐶∞(Ω). Suppose that all maximal1
0 d𝑥2) have length strictly less than some fixed 𝑇 > 0. Then for any 𝜙 ∈ 𝐶∞(Ω),

geodesics on (Ω, 𝑐−2

there is 𝑓 ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) such that

𝑢 𝑓
0

(𝑇) = 𝜙 in Ω,

where 𝑢0 is the solution of (3.22). Moreover, there is 𝐶 > 0, independent of 𝜙, such that

∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω) ≤ 𝐶 ∥𝜙∥𝐻4 (Ω).

Proof. See [75, Proposition 4].

3.3.1.2 Stability and Reconstruction

(3.33)

(3.34)

□

The following reconstruction is mostly based on (3.30) (3.31).

1For a maximal geodesic 𝛾 : [𝑎, 𝑏] → Ω there may exists 𝑡 ∈ (𝑎, 𝑏) such that 𝛾(𝑡) ∈ 𝜕Ω. The geodesics are

maximal on the closed set Ω.

64

Case 1: 𝜌0 is constant Without loss of generality, we assume 𝜌0 = 1. When 𝑞0 is constant, we

choose 𝜆 ≥ 𝑞0. The equation (3.30) becomes the Helmholtz equation

[Δ + (𝜆 − 𝑞0)]𝑢 𝑓

0

(𝑇) = [Δ + (𝜆 − 𝑞0)]𝑢 𝑓

0

(𝑇) = 0

in Ω.

Without loss of generality, we take 𝑞0 = 0. Then the Helmholtz solutions are 𝑒𝑖
𝜃 ∈ S𝑛−1 is an arbitrary unit vector. Furthermore, Proposition 3.9 guarantees the existence of

√
𝜆𝜃·𝑥, where

𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) such that

𝑢 𝑓
0

(𝑇) = 𝑢ℎ

0 (𝑇) = 𝑒𝑖

√
𝜆𝜃·𝑥.

(3.35)

Theorem 3.10. Suppose 𝜆 > 0, 𝑐0 = 1 and 𝑞0 = 0. Then the Fourier transform ˆ(cid:164)𝜌 of (cid:164)𝜌 can be

reconstructed as follows:

√

𝜆𝜃) = −

ˆ(cid:164)𝜌(2

1
𝜆

[(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)]

(3.36)

where 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) are solutions to (3.35).

Proof. The formula is obtained by substitute (3.35) into (3.31). Since 𝜃 ∈ S𝑛−1 and 𝜆 ≥ 0 are

arbitrary, it gives the Fourier transform of (cid:164)𝜌 everywhere. See also Algorithm 3.2.

□

Remark 3.11. An explicit procedure to solve for 𝑓 and ℎ from (3.15) is explained in Section 3.3.1.3.

Input: low-pass filter 𝐽, time-reversal operator 𝑅, projection operator 𝑃𝑇 , linearized ND map (cid:164)Λ (cid:164)𝜌
Output: sound speed perturbation (cid:164)𝜌
1. Choose 𝜆 > 0 and 𝜃 ∈ S𝑛−1.
2. Solve the boundary control equations 𝑢 𝑓
0
3. Construct the linearized connecting operator (cid:164)𝐾 by (cid:164)𝐾 := 𝐽 (cid:164)Λ𝑃∗
4. Compute F [ (cid:164)𝜌] (2

√
𝜆𝜃·𝑥 for 𝑓 and ℎ.
𝑇 − 𝑅 (cid:164)Λ𝑇 𝑅𝐽𝑃∗
𝑇 .

(𝑇) = 𝑢ℎ
0

(𝑇) = 𝑒𝑖

𝜆𝜃) by

√

F [ (cid:164)𝜌] (2

√

𝜆𝜃) = −

1
𝜆

(cid:2)(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(cid:3) .

(3.37)

5. Repeat the above steps with various 𝜆 > 0 and 𝜃 ∈ S𝑛−1 to recover the Fourier transform F [ (cid:164)𝜌].
6. Invert the Fourier transform to recover (cid:164)𝜌.

Algorithm 3.2 Linearized Boundary Control Reconstruction of (cid:164)𝜌 when 𝑞 ≡ 0.

65

Theorem 3.12. Suppose 𝜆 > 0, 𝑐0 = 1 and 𝑞0 = 0. There exists a constant 𝐶 > 0, independent of

𝜆, such that

√

(cid:12)
(cid:12)
(cid:12)

ˆ(cid:164)𝜌(

2𝜆𝜃)

(cid:12)
(cid:12)
(cid:12)

≤ 𝐶 (1 +

√
2𝑇 (1 + 𝜆))𝜆3∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω)

Proof. For a bounded linear operator 𝑇 : X → Y between two Hilbert spaces X and Y, we write

∥𝑇 ∥X→Y for the operator norm of 𝑇. Let 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) be solutions of (3.35) obtained

from Proposition 3.9. We employ (3.36) to estimate:

𝜆

(cid:12)
(cid:12)
(cid:12)

F [ (cid:164)𝜌] (

√

2𝜆𝜃)

(cid:12)
(cid:12)
(cid:12)

≤∥𝜕2

𝑡 𝑓 + 𝜆 𝑓 ∥ 𝐿2 ((0,𝑇)×𝜕Ω) ∥ℎ∥ 𝐿2 ((0,𝑇)×𝜕Ω) ∥ (cid:164)𝐾 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω)

+∥ (cid:164)Λ 𝑓 (𝑇)∥ 𝐿2 (𝜕Ω) ∥ℎ(𝑇) ∥ 𝐿2 (𝜕Ω)

≤(1 + 𝜆)∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω) ∥ℎ∥ 𝐿2 ((0,𝑇)×𝜕Ω) ∥ (cid:164)𝐾 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω)

+∥ (cid:164)Λ 𝑓 ∥𝐻1 ((0,𝑇)×𝜕Ω) ∥ℎ∥𝐻1 ((0,𝑇)×𝜕Ω)

by the continuity of the trace operator.

In order to estimate ∥ (cid:164)Λ 𝑓 ∥𝐻1 ((0,𝑇)×𝜕Ω), we extend 𝑓 ∈ 𝐻2((0, 𝑇) × 𝜕Ω) to a function 𝐹 ∈
2 ((0, 𝑇) × Ω) so that 𝜕𝜈𝐹 | (0,𝑇)×𝜕Ω = 𝑓 and 𝐹 (𝑡, 𝑥) = 0 for 𝑥 ∈ Ω and 𝑡 close to 0 (recall that

𝐻2+ 3

𝑓 (𝑡, 𝑥) = 0 for 𝑡 near 0). Such 𝐹 can be chosen to fulfill

∥𝐹 ∥

𝐻3+ 1

2 ((0,𝑇)×Ω)

≤ 𝐶 ∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω)

Set 𝑣 := 𝐹 − 𝑢0 where 𝑢0 is the solution of (3.63), then 𝑣 satisfies

□𝜌0,𝑞0

𝑣 = □𝜌0,𝑞0

𝐹,

in (0, 2𝑇) × Ω

𝜕𝜈𝑣 = 0,

on (0, 2𝑇) × 𝜕Ω

𝑣|𝑡=0 = 𝜕𝑡𝑣|𝑡=0 = 0,

𝑥 ∈ Ω.






The regularity estimate for the wave equation [40] implies

∥𝑣∥

𝐻2+ 1

2 ((0,𝑇)×Ω)

≤ 𝐶 ∥□𝜌0,𝑞0

𝐹 ∥

𝐻1+ 1

2 ((0,𝑇)×Ω)

≤ 𝐶 ∥𝐹 ∥

𝐻3+ 1

2 ((0,𝑇)×Ω)

We conclude 𝑢0 ∈ 𝐻2+ 1

2 ((0, 𝑇) × Ω) and (cid:164)𝜌𝜕2

𝑡 𝑢0 ∈ 𝐻 1

2 ((0, 𝑇) × Ω). The same regularity estimate

for the wave equation applied to (3.23) implies

∥ (cid:164)𝑢∥

𝐻1+ 1

2 ((0,𝑇)×Ω)

≤ 𝐶 ∥ (cid:164)𝜌𝜕2

𝑡 𝑢0∥

𝐻

1
2 ((0,𝑇)×Ω)

≤ 𝐶 ∥𝑢0∥

𝐻2+ 1

2 ((0,𝑇)×Ω)

66

These inequalities together with the trace estimate yield

∥ (cid:164)Λ 𝑓 ∥𝐻1 ((0,𝑇)×𝜕Ω) ≤ 𝐶 ∥ (cid:164)𝑢∥

𝐻1+ 1

2 ((0,𝑇)×Ω)

≤ 𝐶 ∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω)

where the constant 𝐶 > 0 is independent of 𝑓 . Hence (cid:164)Λ : 𝐻2((0, 𝑇) × 𝜕Ω) → 𝐻1((0, 𝑇) × 𝜕Ω) is

a bounded linear operator.

It remains to estimate ∥ (cid:164)𝐾 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω). Consider the norms of each operator in

(3.29). It is clear that

∥𝑃∗

𝑇 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,2𝑇)×𝜕Ω) = 1,

∥𝑃𝑇 ∥ 𝐿2 ((0,2𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω) = 1,

∥𝑅∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω) = 1.

Since (cid:164)Λ𝑇 = 𝑃𝑇 (cid:164)Λ𝑃∗
𝑇 ,

∥ (cid:164)Λ𝑇 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω) ≤ ∥ (cid:164)Λ∥ 𝐿2 ((0,2𝑇)×𝜕Ω)→𝐿2 ((0,2𝑇)×𝜕Ω).

To estimate operator 𝐽, it is important to observe that

∥𝐽 𝑓 ∥2

𝐿2 ((0,𝑇)×𝜕Ω) =

∫ 𝑇

∫

0
∫ 𝑇

𝜕Ω
∫

(cid:12)
(cid:12)
(cid:12)
(cid:12)

∫ 2𝑇−𝑡

1
2
(cid:18)∫ 2𝑇−𝑡

𝑡

𝑓 (𝜏, 𝑥) d𝜏

2

(cid:12)
(cid:12)
(cid:12)
(cid:12)

d𝑥 d𝑡

| 𝑓 (𝜏, 𝑥)| d𝜏

(cid:19) 2

d𝑥 d𝑡

0
∫ 𝑇

𝜕Ω

∫

𝑡
(cid:18)∫ 2𝑇

| 𝑓 (𝜏, 𝑥)| d𝜏

(cid:19) 2

d𝑥 d𝑡

0
∫ 𝑇

𝜕Ω

∫

0

𝜕Ω

0
∫ 2𝑇

0

2𝑇

| 𝑓 (𝜏, 𝑥)|2 d𝜏 d𝑥 d𝑡

∥ 𝑓 ∥2

𝐿2 ((0,2𝑇)×𝜕Ω)

,

≤

1
4

≤

≤

=

1
4
1
4
𝑇 2

2

we have

Thus, we conclude

∥𝐽 ∥ 𝐿2 ((0,2𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω) ≤

𝑇
√

2

,

∥ (cid:164)𝐾 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω) ≤

√
2𝑇 ∥ (cid:164)Λ∥ 𝐿2 ((0,2𝑇)×𝜕Ω)→𝐿2 ((0,2𝑇)×𝜕Ω).

67

Finally, we can complete the stability estimate:

√

(cid:12)
(cid:12)
(cid:12)

2𝜆𝜃)

(cid:12)
F [ (cid:164)𝜌] (
(cid:12)
(cid:12)
1 + 𝜆
𝜆
1
∥ (cid:164)Λ 𝑓 ∥𝐻3 ((0,𝑇)×𝜕Ω) ∥ℎ∥𝐻1 ((0,𝑇)×𝜕Ω)
𝜆
√
2𝑇 (1 + 𝜆)
𝜆

≤

+

≤

∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω) ∥ℎ∥𝐻1 ((0,𝑇)×𝜕Ω) ∥ (cid:164)𝐾 ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω)

∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω) ∥ℎ∥𝐻1 ((0,𝑇)×𝜕Ω) ∥ (cid:164)Λ∥ 𝐿2 ((0,𝑇)×𝜕Ω)→𝐿2 ((0,𝑇)×𝜕Ω)

+

1
𝜆

∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω) ∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω) ∥ℎ∥𝐻1 ((0,𝑇)×𝜕Ω)

≤𝐶𝜆 ∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω)

where the constant 𝐶𝜆 satisfies (see (3.34))

√

𝐶𝜆 (cid:66) 1 +

≤𝐶 1 +

2𝑇 (1 + 𝜆)
𝜆
√
2𝑇 (1 + 𝜆)
𝜆

∥ 𝑓 ∥𝐻2 ((0,𝑇)×𝜕Ω) ∥ℎ∥𝐻1 ((0,𝑇)×𝜕Ω)

∥𝑒𝑖

√
𝜆𝜃·𝑥 ∥2

𝐻4 (Ω) ≤ 𝐶 (1 +

√

2𝑇 (1 + 𝜆))𝜆3

for some constant 𝐶 > 0 independent of 𝜆.

□

When 𝑞0 is variable, choose 𝜆 ≥ 0, then (3.30) becomes the perturbed Helmholtz equation

[Δ + 𝜆 − 𝑞0]𝑢 𝑓

0

(𝑇) = [Δ + 𝜆 − 𝑞0]𝑢ℎ

0 (𝑇) = 0

in Ω.

A class of solutions are total waves of the form

𝜙(𝑥) = 𝑒𝑖

√
𝜆𝜃·𝑥 + 𝑟 (𝑥; 𝜆)

with 𝜃 ∈ S𝑛−1 and the scattered wave 𝑟 (𝑥; 𝜆) satisfying

According to [75, Lemma 13],

for any 𝑠 ≥ 0.

(Δ + 𝜆 − 𝑞0)𝑟 = 𝑞0𝑒𝑖

√

𝜆𝜃·𝑥

in Ω.

∥𝑟 ∥𝐻𝑠 (R𝑛) = 𝑂 (𝜆 𝑠−1
2 )

as 𝜆 → ∞

68

(3.38)

(3.39)

(3.40)

One dimension:

In one dimension (1D), 𝜃 = ±1. Let us take 𝜃 = 1 and choose (3.38) to be

(𝑇). Substituting into (3.31) gives

𝜆) + 2( (cid:164)𝜌𝑒𝑖

√

𝜆𝜃·𝑥, 𝑟)𝐿2 (Ω) + ( (cid:164)𝜌𝑟, 𝑟)𝐿2 (Ω)

the value of 𝑢 𝑓
0

Since

(𝑇) and 𝑢ℎ
0
√

ˆ(cid:164)𝜌(2
1
𝜆

= −

[(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)]

(3.41)

|( (cid:164)𝜌𝑒𝑖

√

𝜆𝜃·𝑥, 𝑟)𝐿2 (Ω) | ≤ 𝐶 ∥ (cid:164)𝜌∥ 𝐿∞ (Ω) ∥𝑟 ∥ 𝐿2 (Ω),

|( (cid:164)𝜌𝑟, 𝑟)𝐿2 (Ω) | ≤ 𝐶 ∥ (cid:164)𝜌∥ 𝐿∞ (Ω) ∥𝑟 ∥2

𝐿2 (Ω)

.

The left hand side of (3.41) decay like 𝑂 (𝜆− 1

2 ) as 𝜆 → ∞ according to (3.40). The approximate

reconstruction follows in the high-frequency regime. Moreover, an approximate stability estimate

can be provided in the same way as in the proof of Theorem 3.12 with an extra 𝑂 (𝜆− 1

2 ) term:

(cid:12)
(cid:12)
(cid:12)

ˆ(cid:164)𝜌(

√

2𝜆𝜃)

(cid:12)
(cid:12)
(cid:12)

√

≤ 𝐶 (1 +

2𝑇 (1 + 𝜆))𝜆3∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω) + 𝑂 (𝜆− 1

2 ).

(3.42)

High Dimension:

In dimension 𝑛 ≥ 2, more freedom is available in choosing the wave

vectors. Let 𝜃, 𝜔 ∈ R𝑛 be two vectors such that 𝜃 ⊥ 𝜔. We take the following solutions:

𝜙(𝑥) :=𝜙0(𝑥) + 𝑟1(𝑥; 𝜆),

𝜙0(𝑥) := 𝑒𝑖(𝑘𝜃+𝑙𝜔)·𝑥

𝜓(𝑥) :=𝜓0(𝑥) + 𝑟2(𝑥; 𝜆),

𝜓0(𝑥) := 𝑒𝑖(𝑘𝜃−𝑙𝜔)·𝑥

where 𝑟1, 𝑟2 satisfy (3.40). Choose 𝑘 2 + 𝑙2 = 𝜆 such that (Δ + 𝜆)𝜙0 = (Δ + 𝜆)𝜓0 = 0. Proposition 3.9

asserts that there are 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) such that

𝑢 𝑓
0

(𝑇) = 𝜙 = 𝜙0 + 𝑟1,

𝑢ℎ
0 (𝑇) = 𝜓 = 𝜓0 + 𝑟2.

(3.43)

Inserting (3.43) into (3.31) gives

− ˆ(cid:164)𝜌(2𝑘𝜃) − ( (cid:164)𝜌𝑒𝑖(𝑘𝜃+𝑙𝜔)·𝑥, 𝑟2)𝐿2 (Ω) − ( (cid:164)𝜌𝑒𝑖(𝑘𝜃−𝑙𝜔)·𝑥, 𝑟1)𝐿2 (Ω) − ( (cid:164)𝜌𝑟1, 𝑟2)𝐿2 (Ω)
1
𝜆

𝑡 𝑓 + (𝑘 2 + 𝑙2) 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)]

[(𝜕2

=

(3.44)

If we fix 𝑘 and let 𝑙 → ∞, then 𝜆 → ∞. Due to the decay property (3.40), ∥𝑟1∥ 𝐿2 (Ω), ∥𝑟2∥ 𝐿2 (Ω) → 0.
We obtain the reconstruction formula for any 𝑘 ≥ 0 and any 𝜃 ∈ S𝑛−1:

ˆ(cid:164)𝜌(2𝑘𝜃) = − lim
𝑙→∞

1
𝑘 2 + 𝑙2

(cid:2)(𝜕2

𝑡 𝑓 + (𝑘 2 + 𝑙2) 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(cid:3) .

69

Moreover, we can obtain a Hölder-type stability estimate for ∥ (cid:164)𝜌∥𝐻 −𝑠 (R𝑛), where 𝑠 > 0 is an

arbitrary real number and 𝐻−𝑠 (R𝑛) is the 𝐿2-based Sobolev space of order −𝑠 over R𝑛.

Theorem 3.13. Suppose 𝑐0 = 1, 𝑞0 ∈ 𝐶∞(Ω) and 𝑞0 is not identically zero. For any 𝑠 > 0, there

exists a constant 𝐶 > 0 independent of 𝜆 such that

∥ (cid:164)𝜌∥𝐻 −𝑠 (R𝑛) ≤ 𝐶 ∥ (cid:164)Λ∥

2𝑠
9(𝑛+2𝑠)
𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω)

.

Proof. Write 𝜉 := 2𝑘𝜃 and

𝛿 := ∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω).

Let 𝜉0 > 0 be a sufficiently large number that is to be determined. We decompose

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (R𝑛) =

∫

|𝜉 |≤𝜉0

| ˆ(cid:164)𝜌(𝜉)|2
(1 + |𝜉 |2)𝑠

∫

𝑑𝜉 +

|𝜉 |>𝜉0

| ˆ(cid:164)𝜌(𝜉)|2
(1 + |𝜉 |2)𝑠

𝑑𝜉.

For the integral over high frequencies, we have

∫

|𝜉 |>𝜉0

| ˆ(cid:164)𝜌(𝜉)|2
(1 + |𝜉 |2)𝑠

𝑑𝜉 ≤

1
(1 + 𝜉2
0

)𝑠

∫

|𝜉 |>𝜉0

| ˆ(cid:164)𝜌(𝜉)|2 𝑑𝜉 ≤

For the integral over low frequencies, it is easy to see that:

∥ (cid:164)𝜌∥2

𝐿2 (R𝑛)

(1 + 𝜉2
0

)𝑠 ≤ 𝐶 1

𝜉2𝑠
0

.

∫

|𝜉 |≤𝜉0

| ˆ(cid:164)𝜌(𝜉)|2
(1 + |𝜉 |2)𝑠

∫

𝑑𝜉 ≤

|𝜉 |≤𝜉0

| ˆ(cid:164)𝜌(𝜉)|2 𝑑𝜉 ≤ 𝐶𝜉𝑛

0 ∥ ˆ(cid:164)𝜌∥2

𝐿∞ (𝐵(0,𝜉0))

.

The norm ∥ ˆ(cid:164)𝜌∥ 𝐿∞ (𝐵(0,𝜉0)) can be estimated using (3.82). Indeed, for |𝜉 | ≤ 𝜉0, we have

| ˆ(cid:164)𝜌(𝜉)| ≤

𝑡 𝑓 + (𝑘 2 + 𝑙2) 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω) | +
√

𝐶
√

𝜆

∥𝜙∥𝐻4 (Ω) ∥𝜓∥𝐻4 (Ω)𝛿 +

𝐶
√

𝜆

(cid:16)

(cid:16)

∥𝜙0∥𝐻4 (Ω) + ∥𝑟1∥𝐻4 (Ω)

(cid:17) (cid:16)

∥𝜓0∥𝐻4 (Ω) + ∥𝑟2∥𝐻4 (Ω)

(cid:17)

𝛿 +

𝐶
√

𝜆

𝜆2 + 𝜆 3

2

(cid:17) 2

𝛿 +

𝐶
√

𝜆

|(𝜕2

1
𝜆
≤ 𝐶 1 +

≤ 𝐶 1 +

≤ 𝐶 1 +

√

√

2𝑇 (1 + 𝜆)
𝜆
2𝑇 (1 + 𝜆)
𝜆
2𝑇 (1 + 𝜆)
𝜆

where the first and the last inequality is a consequence of (3.40), the second inequality follows from

the proof of Proposition 3.12. Utilizing the relation 𝜆 = 𝑘 2 + 𝑙2, we conclude

∥ ˆ(cid:164)𝜌∥2

𝐿∞ (𝐵(0,𝜉0)) ≤ 𝐶

√

(cid:20)

(1 +

2𝑇 (1 + 𝜆))2𝜆4(1 +

√

𝜆)4𝛿2 +

(cid:20)

≤ 𝐶

(cid:21)

1
𝜆

(𝜉2

0 + 𝑙2)8𝛿 +

(cid:21)

1
𝑙2

70

provided 𝜉0 > 0 is sufficiently large. Combining these estimates, we see that

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (R𝑛) ≤ 𝐶

(cid:34)

𝜉𝑛
0 (𝜉2

0 + 𝑙2)8𝛿2 +

𝜉𝑛
0
𝑙2

+

(cid:35)

.

1
𝜉2𝑠
0

Choosing 𝑙2 = 𝜉𝑛+2𝑠

0

and 𝜉0 = 𝛿−

2

9(𝑛+2𝑠) yields

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (R𝑛) ≤ 𝐶𝛿 4𝑠

9(𝑛+2𝑠) ,

where 𝐶 is a constant independent of 𝜆 and 𝛿 is sufficiently small.

□

Case2: 𝜌0 is variable When 𝜌0 = 𝜌0(𝑥) > 0 is non-constant, the equations (3.30) are no longer

perturbed Helmholtz equations, but Schrödinger’s equations with the potential −𝑞 + 𝜆𝜌0 ∈ 𝐿∞(Ω).

The idea is to employ Schrödinger solutions to probe based on the identity (3.31).

The class of solutions we will resort to are the complex geometric optics (CGO) solutions that

were first proposed in [89] for dimension 𝑛 ≥ 3. A CGO solution 𝜙 is a function of the form

𝜙(𝑥) := 𝑒𝑖𝜁 ·𝑥 (1 + 𝑟 (𝑥)).

where 𝜁 ∈ C𝑛 is a complex vector with 𝜁 · 𝜁 = 0, and the remainder term 𝑟 (𝑥) satisfies

Δ𝑟 + 2𝜁 · ∇𝑟 − (𝑞 − 𝜆𝜌0)𝑟 = 𝑞 − 𝜆𝜌0.

Moreover, 𝑟 → 0 in a certain function space as |𝜁 | → ∞.

(3.45)

(3.46)

The following proposition is a direct application of [89, Theorem 2.3 and Corollary 2.4] to the

Schrödinger’s equation (Δ − 𝑞 + 𝜆𝜌0)𝜙 = 0.

Lemma 3.14 ( [89, Theorem 2.3 and Corollary 2.4]). Let 𝑛 ≥ 3 and 𝑠 ∈ R a real number such that
2 . Let 𝜁 ∈ C𝑛 be a complex vector with 𝜁 · 𝜁 = 0 and |𝜁 | ≥ 𝜀0 > 0 for some positive constant

𝑠 > 𝑛

𝜀0. There exist positive constants 𝐶0, 𝐶1, depending on 𝑠, 𝑛, 𝜀0 and Ω, such that if

𝐶0∥𝑞 − 𝜆𝜌0∥𝐻𝑠 (Ω) < |𝜁 |,

then 𝜙 = 𝜙(𝑥) defined in (3.45) satisfies (Δ − 𝑞 + 𝜆𝜌0)𝜙 = 0; moreover

∥𝑟 ∥𝐻𝑠 (Ω) ≤

𝐶1
|𝜁 |

∥𝑞 − 𝜆𝜌0∥𝐻𝑠 (Ω)

(3.47)

71

We now construct specific CGO solutions that are useful for our purpose. Let 𝜉 ∈ R𝑛 (𝑛 ≥ 3) be

an arbitrary non-zero vector, and let 𝑒(1), 𝑒(2) ∈ S𝑛−1 be two real unit vectors such that {𝜉, 𝑒(1), 𝑒(2) }
forms an orthogonal set. Choose a positive number 𝑅 with 𝑅 ≥ |𝜉 |
√
2

. Define

𝜁 (1) := −

𝜉 + 𝑖

1
2

𝑅
√
2

𝑒(1) +

√︂ 𝑅2
2

−

|𝜉 |2
4

𝑒(2),

𝜁 (2) := −

𝜉 − 𝑖

1
2

𝑅
√

2

𝑒(1) −

√︂ 𝑅2
2

−

|𝜉 |2
4

𝑒(2).

It is easy to verify that

𝜁 (1) + 𝜁 (2) = −𝜉,

𝜁 ( 𝑗) · 𝜁 ( 𝑗) = 0,

|𝜁 ( 𝑗) | = 𝑅,

for 𝑗 = 1, 2.

If 𝑅 is sufficiently large, by Lemma 3.14, we can construct CGO solutions

𝜙 𝑗 (𝑥) = 𝑒𝑖𝜁 ( 𝑗 ) ·𝑥 (1 + 𝑟 𝑗 (𝑥))

where the remainder term 𝑟 𝑗 satisfies

∥𝑟 𝑗 ∥𝐻𝑠 (Ω) ≤

𝐶1
|𝜁 ( 𝑗) |

∥𝑞 − 𝜆𝜌0∥𝐻𝑠 (Ω) ≤

𝐶1
𝐶0

.

(Here, 𝐶0 is the constant in Lemma 3.14.)

Thus for 𝑠 > 𝑛
2 ,

(3.48)

(3.49)

∥𝜙 𝑗 ∥𝐻𝑠 (Ω) ≤ ∥𝑒𝑖𝜁 ( 𝑗 ) ·𝑥 ∥𝐻𝑠 (Ω) ∥1 + 𝑟 𝑗 ∥𝐻𝑠 (Ω) ≤

(cid:18)

|Ω|

1
2 +

(cid:19)

𝐶1
𝐶0

∥𝑒𝑖𝜁 ( 𝑗 ) ·𝑥 ∥𝐻𝑠 (Ω).

By choosing 𝜆0 such that for any 𝜆 > 𝜆0, we have

𝐶0∥𝑞 − 𝜆𝜌0∥𝐻𝑠 (Ω) ≥

1
√
𝑛

,

then |𝜁 ( 𝑗) | ≥ 1√

𝑛 . Noticing that for integer 𝑘 ≥ 0,

∥𝑒𝑖𝜁 ( 𝑗 ) ·𝑥 ∥2

𝐻 𝑘 (Ω) =

𝑘
∑︁

∑︁

𝑖=0

|𝛼|=𝑖

∥𝐷𝛼𝑒𝑖𝜁 ( 𝑗 ) ·𝑥 ∥2

𝐿2 (Ω) =

𝑘
∑︁

∑︁

(cid:32) 𝑛
(cid:214)

(cid:33)

|𝜁 ( 𝑗)

𝑚 |𝛼𝑚

𝑖=0

(cid:205)𝑛

𝑚=1

𝛼𝑚=𝑖

𝑚=1

∥𝑒Im𝜁 ( 𝑗 ) ·𝑥 ∥2

𝐿2 (Ω)

≤𝐶

𝑘
∑︁

𝑖=0

∥𝜁 ( 𝑗) ∥𝑖
1

𝑒

√
2𝑅 ≤ 𝐶

𝑘
∑︁

√

(

𝑛∥𝜁 ( 𝑗) ∥2)𝑖𝑒

√

2𝑅 = 𝐶

𝑖=0

𝑘
∑︁

𝑖=0

√

(

𝑛𝑅)𝑖𝑒

√

2𝑅 ≤ 𝐶 𝑅𝑘 𝑒

√

2𝑅

where 𝐶 only depend on 𝑘, 𝑛, Ω. Thus from interpolation formula, we have

∥𝜙 𝑗 ∥𝐻𝑠 (Ω) ≤

(cid:18)

|Ω|

1
2 +

(cid:19)

𝐶1
𝐶0

∥𝑒𝑖𝜁 ( 𝑗 ) ·𝑥 ∥𝐻𝑠 (Ω) ≤ 𝐶 𝑅 𝑠
2 𝑒

𝑅
√
2 .

(3.50)

72

We begin with a pointwise estimate for (cid:164)𝜌 in the Fourier domain. For simplicity, we denote

𝛿 := ∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻1 ((0,𝑇)×𝜕Ω).

Lemma 3.15. Let 𝑠 > 𝑛

2 with 𝑛 ≥ 3. Suppose there exists a constant 𝑀 > 0 such that

∥𝑞∥𝐻max(𝑠,4) (Ω) ≤ 𝑀,

∥ 𝜌0∥𝐻max(𝑠,4) (Ω) ≤ 𝑀.

Then there exists a constant 𝐶, independent of 𝜆 and 𝛿, such that

𝐶

(cid:104) 𝜆+1

𝜆 (𝜆 + 1)max(𝑠,4)𝑒

√
2𝑎0 (𝜆+1)𝛿 + 1
𝑎0

∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω)

(cid:105)

√

|𝜉 | ≤

2𝑎0(𝜆 + 1)

|𝜉 | ∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω)
for any 𝜆 > 0 and sufficiently small 𝛿. Here, 𝑎0 is a constant satisfying 𝑎0 ≥ 𝐶0𝑀, where 𝐶0 is the

𝜆 |𝜉 |max(𝑠,4)𝑒|𝜉 |𝛿 + 𝜆+1

2𝑎0(𝜆 + 1)

|𝜉 | ≥

𝐶

(cid:105)

√

(3.51)

| ˆ(cid:164)𝜌(𝜉)| ≤

(cid:104) 𝜆+1





constant in Lemma 3.14.

Proof. From Proposition 3.9, there exist boundary controls 𝑓 𝑗 such that 𝑢 𝑓 𝑗

0

(𝑇) = 𝜙 𝑗 for the CGO

solutions 𝜙 𝑗 defined in (3.45). With similar estimation as in the proof of Theorem 3.12,

(cid:12)
(cid:12)
(cid:12)
(cid:12)

∫

Ω

(cid:164)𝜌𝜙1𝜙2 d𝑥

(cid:12)
(cid:12)
(cid:12)
(cid:12)

≤𝐶𝜆𝛿

where the constant 𝐶𝜆 is

1 +

𝐶𝜆 :=

√

2𝑇 (1 + 𝜆)
𝜆

∥ 𝑓1∥𝐻2 ((0,𝑇)×𝜕Ω) ∥ 𝑓2∥𝐻2 ((0,𝑇)×𝜕Ω)

≤𝐶 1 + 𝜆
𝜆
≤𝐶 1 + 𝜆
𝜆
≤𝐶 1 + 𝜆
𝜆

∥𝜙1∥𝐻4 (Ω) ∥𝜙2∥𝐻4 (Ω)

∥𝜙1∥𝐻max(𝑠,4) (Ω) ∥𝜙2∥𝐻max(𝑠,4) (Ω)

𝑅max(𝑠,4)𝑒

√
2𝑅.

(3.52)

where the last inequality comes from (3.50). We obtain the estimate
(cid:12)
(cid:12)
(cid:12)
(cid:12)

(cid:164)𝜌𝑒−𝑖𝜉·𝑥 (𝑟1 + 𝑟2 + 𝑟1𝑟2) d𝑥

(cid:164)𝜌𝜙1𝜙2 d𝑥

| ˆ(cid:164)𝜌(𝜉)| ≤

(cid:12)
(cid:12)
(cid:12)
(cid:12)

(cid:12)
(cid:12)
(cid:12)
(cid:12)

(cid:12)
(cid:12)
(cid:12)
(cid:12)

∫

∫

+

Ω

Ω

≤𝐶𝜆𝛿 + ∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω) ∥𝑟1 + 𝑟2 + 𝑟1𝑟2∥𝐻𝑠 (Ω)

≤𝐶𝜆𝛿 + ∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω) (∥𝑟1∥𝐻𝑠 (Ω) + ∥𝑟2∥𝐻𝑠 (Ω) + ∥𝑟1∥𝐻𝑠 (Ω) ∥𝑟2∥𝐻𝑠 (Ω))

(3.53)

≤𝐶𝜆𝛿 + 𝐶 ∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω)

(cid:18) 2
𝑅

∥𝑞 − 𝜆𝜌0∥𝐻𝑠 (Ω) +

1
𝑅2

∥𝑞 − 𝜆𝜌0∥2

𝐻𝑠 (Ω)

(cid:19)

≤𝐶𝜆𝛿 + 𝐶 (𝜆 + 1)𝑅−1∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω),

73

where in the last inequality we used the estimate (3.49). This derivation holds for any 𝑅 ≥ |𝜉 |
√
2
particular, we choose 𝑅 = |𝜉 |
2𝑎0(𝜆+1)
√
2

2𝑎0(𝜆+1) when |𝜉 | ≤

2𝑎0(𝜆+1), and 𝑅 =

when |𝜉 | >

. In

√

√

√

to obtain (3.51).

The condition 𝑎0 ≥ 𝐶0𝑀 arises since

𝐶0∥𝑞 − 𝜆𝜌0∥𝐻𝑠 (Ω) ≤ 𝐶0(∥𝑞∥𝐻𝑠 (Ω) + 𝜆∥ 𝜌0∥𝐻𝑠 (Ω)) ≤ 𝐶0𝑀 (𝜆 + 1)

is a natural upper bound, thus we require |𝜁 ( 𝑗) | = 𝑅 > 𝐶0𝑀 (𝜆 + 1) to fulfill the assumption of
Lemma 3.14. For either choice of 𝑅 above, it holds that 𝑅 > 𝑎0(𝜆 + 1). It remains to require

𝑎0 ≥ 𝐶0𝑀.

□

With the help of Lemma 3.15, the following stability estimate can be established for (cid:164)𝜌.

Theorem 3.16. Let 𝑠 > 𝑛

2 with 𝑛 ≥ 3. Suppose there exists a constant 𝑀 > 0 such that

∥𝑞∥𝐻max(𝑠,4) (Ω) ≤ 𝑀,

∥ 𝜌0∥𝐻max(𝑠,4) (Ω) ≤ 𝑀,

∥ (cid:164)𝜌∥𝐻𝑠 (Ω) ≤ 𝑀.

and (cid:164)𝜌 is compact supported in Ω, then there exist a constant 𝐶 (independent of 𝜆 and 𝛿) and a

positive constant 𝜆0 > 0 such that

(cid:34)

∥ (cid:164)𝜌∥ 𝐿∞ (Ω) ≤ 𝐶

(𝜆 + 1)max(𝑠,4)𝑒𝐶 (𝜆+1)𝛿 +

(cid:35) 2𝑠−𝑛
8𝑠

(cid:19) 𝑛−2𝑠

2

(cid:18)

𝜆 + ln

1
𝛿

for any 𝜆 > 𝜆0 > 0 and 0 < 𝛿 ≤ 𝑒−1 (𝑒 = 2.71828... is the Euler’s number).

Remark 3.17. For any fixed 𝛿 > 0, it is clear that

(cid:16)

𝜆 + ln 1
𝛿

(cid:17) 𝑛−2𝑠

2 → 0 as 𝜆 → ∞ since 𝑛 − 2𝑠 < 0.

Therefore, for a large 𝜆 > 0, the estimate in Proposition 3.16 becomes a nearly Hölder-type

stability.

Proof. We follow the idea in the proof of the increasing stability result [72] and name all the

constants that are independet of 𝜆 and 𝛿 as 𝐶.

74

Let 𝜉0 be a constant such that 𝜉0 ≥

√

2𝑎0(𝜆 + 1), then

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω) =

=

∫

R𝑛

∫

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉 +

∫
√

2𝑎0 (𝜆+1)≤|𝜉 |≤𝜉0

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉

(3.54)

|𝜉 |>𝜉0
∫

+

|𝜉 |≤

√

2𝑎0 (𝜆+1)

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉

(cid:67)𝐼1 + 𝐼2 + 𝐼3.

We estimate 𝐼1, 𝐼2, 𝐼3 as follows. For 𝐼1, as (cid:164)𝜌 is compact supported in Ω, Hölder’s inequality gives
| ˆ(cid:164)𝜌(𝜉)| ≤ ∫
Ω

(cid:12) d𝑥 ≤ 𝐶 ∥ (cid:164)𝜌∥ 𝐿2 (Ω). Thus,

(cid:12) (cid:164)𝜌(𝑥)𝑒𝑖𝜉·𝑥(cid:12)
(cid:12)

𝐼1 :=

∫

|𝜉 |>𝜉0

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉

∫

≤ 𝐶 ∥ (cid:164)𝜌∥2

𝐿2 (Ω)

≤ 𝐶 ∥ (cid:164)𝜌∥2

𝐻𝑠 (Ω)

|𝜉 |>𝜉0
𝜉𝑛−2𝑠
0

(1 + |𝜉 |2)−𝑠 d𝜉

≤ 𝐶𝜉𝑛−2𝑠
0
(cid:32)
(cid:32)
(cid:125)
(cid:123)(cid:122)
(cid:124)
:=Φ1 (𝜉0)

where the last inequality follows from ∥ (cid:164)𝜌∥𝐻𝑠 (Ω) ≤ 𝑀. The function Φ1(𝜉0) denotes an upper

bound of 𝐼1.

For 𝐼3, we use | ˆ(cid:164)𝜌(𝜉)| ≤ ∥ ˆ(cid:164)𝜌∥ 𝐿∞ (𝐵(0,

√

2𝑎0 (𝜆+1))) (here 𝐵(0, 𝑡) means the unit ball of center 0 and

radius 𝑡) to get

𝐼3 :=

∫

|𝜉 |≤

√

2𝑎0 (𝜆+1)

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉

∫

R𝑛

(1 + |𝜉 |2)−𝑠 d𝜉

(3.55)

≤ ∥ ˆ(cid:164)𝜌∥2

𝐿∞ (𝐵(0,

√

2𝑎0 (𝜆+1)))

√

2𝑎0 (𝜆+1)))

≤ 𝐶 ∥ ˆ(cid:164)𝜌∥2
(cid:34)

≤ 𝐶

𝐿∞ (𝐵(0,
(𝜆 + 1)2
𝜆2

√
(𝜆 + 1)2 max(𝑠,4)𝑒2
2𝑎0 (𝜆+1)𝛿2 +

(cid:35)

.

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

1
𝑎2
0

where the last inequality is a consequence of (3.51) combined with the estimate (𝑎+𝑏)2 ≤ 2𝑎2+2𝑏2.

75

For 𝐼2, we apply the estimate (3.51) to get

𝐼2 :=

∫
√

(1 + |𝜉 |2)−𝑠 | ˆ(cid:164)𝜌(𝜉)|2 d𝜉

2𝑎0 (𝜆+1)≤|𝜉 |≤𝜉0
∫
√

2𝑎0 (𝜆+1)≤|𝜉 |≤𝜉0

∫
√

2𝑎0 (𝜆+1)≤|𝜉 |≤𝜉0

≤ 𝐶

≤ 𝐶

(1 + |𝜉 |2)−𝑠

(1 + |𝜉 |2)−𝑠

(cid:12)
𝜆 + 1
(cid:12)
(cid:12)
𝜆
(cid:12)
(cid:20) (𝜆 + 1)2
𝜆2

|𝜉 |max(𝑠,4)𝑒|𝜉 |𝛿 +

𝜆 + 1
|𝜉 |

∥ (cid:164)𝜌∥𝐻 −𝑠 (Ω)

2

(cid:12)
(cid:12)
(cid:12)
(cid:12)

d𝜉

|𝜉 |2 max(𝑠,4)𝑒2|𝜉 |𝛿2 +

(𝜆 + 1)2
|𝜉 |2

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

(cid:21)

d𝜉

= 𝐼21 + 𝐼22

Let 𝑡 := |𝜉 | be the radial variable, then

𝐼21 := 𝐶

∫
√

(1 + |𝜉 |2)−𝑠 (𝜆 + 1)2

𝜆2

|𝜉 |2 max(𝑠,4)𝑒2|𝜉 |𝛿2 d𝜉

(1 + 𝑡2)−𝑠𝑡2 max(𝑠,4)+𝑛−1𝑒2𝑡 𝑑𝑡

2𝑎0 (𝜆+1)≤|𝜉 |≤𝜉0
∫ 𝜉0
√

𝛿2

2𝑎0 (𝜆+1)
∫ 𝜉0

≤ 𝐶

≤ 𝐶

= 𝐶

(𝜆 + 1)2
𝜆2
(𝜆 + 1)2
𝜆2
(𝜆 + 1)2
𝜆2

𝑒2𝜉0𝛿2

𝑡2 max(𝑠,4)+𝑛−1−2𝑠 𝑑𝑡

0

𝜉2 max(𝑠,4)+𝑛−2𝑠
0

𝑒2𝜉0𝛿2;

and

𝐼22 := 𝐶

∫
√

2𝑎0 (𝜆+1)≤|𝜉 |≤𝜉0

= 𝐶 (𝜆 + 1)2∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

(1 + 𝑡2)−𝑠𝑡𝑛−3 𝑑𝑡

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω) d𝜉

(1 + |𝜉 |2)−𝑠 (𝜆 + 1)2
|𝜉 |2
∫ 𝜉0
√

2𝑎0 (𝜆+1)

∫ ∞
√

𝑡𝑛−3−2𝑠 𝑑𝑡

√

2𝑎0 (𝜆+1)
2𝑎0(𝜆 + 1)]𝑛−2−2𝑠

≤ 𝐶 (𝜆 + 1)2∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

≤ 𝐶 (𝜆 + 1)2∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω) [
1
𝑎2
0

0

≤ 𝐶 (𝜆0 + 1)𝑛−2𝑠𝑎𝑛−2𝑠

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

𝐶
𝑎2
0
Put together, we have the following upper bound for 𝐼2:

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

=

.

𝐼2 ≤ 𝐼21 + 𝐼22 ≤ 𝐶
(cid:124)

(𝜆 + 1)2
𝜆2

(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)

𝜉2 max(𝑠,4)+𝑛−2𝑠
0

𝑒2𝜉0𝛿2
(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)

+

𝐶
𝑎2
0

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

.

(3.56)

(cid:125)

(cid:123)(cid:122)
:=Φ2 (𝜉0)

76

Combining the estimate for 𝐼1, 𝐼2, 𝐼3, we conclude

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω) =𝐼1 + 𝐼2 + 𝐼3
(cid:34)

≤Φ1(𝜉0) +

Φ2(𝜉0) +

(cid:35)

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

𝐶
𝑎2
0

(cid:34)

+

𝐶

(𝜆 + 1)2
𝜆2

(𝜆 + 1)2 max(𝑠,4)𝑒2

√
2𝑎0 (𝜆+1)𝛿2 +

(3.57)

(cid:35)

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω)

𝐶
𝑎2
0

where the right hand side has been combined into three groups which are the upper bounds of
𝐼1, 𝐼2, 𝐼3, respectively. By choosing 𝑎0 sufficiently large, the 𝐻−𝑠 norm can be absorbed by the left

hand side to yield

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω) ≤Φ1(𝜉0) + Φ2(𝜉0) + 𝐶

(𝜆 + 1)2
𝜆2

√
(𝜆 + 1)2 max(𝑠,4)𝑒2

2𝑎0 (𝜆+1)𝛿2

(3.58)

The estimate will henceforth be split into two cases:
√

√

2𝑎0(𝜆 + 1). When 1

2 ln 1

𝛿 ≥

2𝑎0(𝜆 + 1), we choose 𝜉0 = 1

𝛿 ≥

1

2 ln 1
2 ln 1

𝛿 to get

√

2𝑎0(𝜆 + 1) and 1

2 ln 1

𝛿 <

Φ1(𝜉0) + Φ2(𝜉0) =𝐶𝜉𝑛−2𝑠

0
(cid:20)

=𝐶

1 +

𝜉2 max(𝑠,4)+𝑛−2𝑠
0

𝑒2𝜉0𝛿2

+ 𝐶

(𝜆 + 1)2
𝜆2
(𝜆 + 1)2
𝜆2

𝜉2 max(𝑠,4)
0

(cid:21)

𝑒2𝜉0𝛿2

(cid:34)

≤𝐶

1 +

(𝜆 + 1)2
𝜆2

(cid:18)

ln

1
𝛿

(cid:19) 2 max(𝑠,4)

𝛿

𝜉𝑛−2𝑠
0
(cid:35) (cid:18)

ln

1
𝛿

(cid:19) 𝑛−2𝑠

.

As lim𝛿→0+

(cid:17) 2 max(𝑠,4)

(cid:16)

𝛿

ln 1
𝛿

= 0 and lim𝜆→∞

(𝜆+1)2

𝜆2 = 1, the square parenthesis is bounded whenever

𝛿 ∈ (0, 𝑒−1] and 𝜆 ≥ 𝜆0 for some 𝜆0 > 0. Hence,

Φ1(𝜉0) + Φ2(𝜉0) ≤ 𝐶

(cid:18)

(cid:32)

ln

1
𝛿
√

(cid:19) 𝑛−2𝑠

= 𝐶

(cid:32)

ln 1
𝛿
𝜆 + ln 1
𝛿

(cid:33) 𝑛−2𝑠

(cid:18)

(cid:19) 𝑛−2𝑠

𝜆 + ln

1
𝛿

(cid:33) 𝑛−2𝑠

(cid:18)

(cid:19) 𝑛−2𝑠

𝜆 + ln

1
𝛿

≤ 𝐶

(cid:18)

≤ 𝐶

2

2𝑎0
2
√
2𝑎0 + 1
1
𝛿

𝜆 + ln

(cid:19) 𝑛−2𝑠

where the second but last inequality holds since the function ( 𝑡

𝜆+𝑡 )𝑛−2𝑠 is decreasing in 𝑡 > 0. When
2𝑎0(𝜆 + 1), then 𝐼2 = 0. As a result, we can simply choose

1

2 ln 1

𝛿 <

√

2𝑎0(𝜆 + 1), we choose 𝜉0 =

√

77

Φ2(𝜉0) = 0 as an upper bound of 𝐼2, hence

Φ1(𝜉0) + Φ2(𝜉0) = Φ1(𝜉0) = 𝐶𝜉𝑛−2𝑠
0
(cid:33) 𝑛−2𝑠

= 𝐶

In either case, we have

(cid:32) 𝜆 + 1
𝜆 + ln 1
𝛿

= 𝐶 (𝜆 + 1)𝑛−2𝑠

(cid:19) 𝑛−2𝑠

(cid:18)

𝜆 + ln

1
𝛿

≤ 𝐶

(cid:18)

1
√
1 + 2

2𝑎0

(cid:19) 𝑛−2𝑠 (cid:18)

𝜆 + ln

(cid:19) 𝑛−2𝑠

.

1
𝛿

Φ1(𝜉0) + Φ2(𝜉0) ≤ 𝐶

(cid:19) 𝑛−2𝑠

(cid:18)

𝜆 + ln

1
𝛿

for some constant 𝐶 > 0 that is independent of 𝜆 ∈ [𝜆0, ∞) and 𝛿 ∈ (0, 𝑒−1]. In view of (3.58), we

conclude

∥ (cid:164)𝜌∥2

𝐻 −𝑠 (Ω) ≤𝐶

(𝜆 + 1)2
𝜆2

√
2𝑎0 (𝜆+1)𝛿2 + 𝐶
(𝜆 + 1)2 max(𝑠,4)𝑒2

(cid:19) 𝑛−2𝑠

(cid:18)

𝜆 + ln

1
𝛿

≤𝐶 (𝜆 + 1)2 max(𝑠,4)𝑒𝐶 (𝜆+1)𝛿2 + 𝐶

(cid:18)

𝜆 + ln

1
𝛿

(cid:19) 𝑛−2𝑠

.

(3.59)

Finally, we interpolate to obtain an estimate for the infinity norm. Let 𝜂 > 0 such that 𝑠 = 𝑛

2 +2𝜂,

choose 𝑘0 = −𝑠, 𝑘1 = 𝑠, 𝑘 = 𝑛

2 + 𝜂 = 𝑠 − 𝜂. Then

𝑘 = (1 − 𝑝)𝑘0 + 𝑝𝑘1, where 𝑝 =

2𝑠 − 𝜂
2𝑠

.

Using the interpolation theorem and the Sobolev embedding, we have

∥ (cid:164)𝜌∥ 𝐿∞ (Ω) ≤𝐶 ∥ (cid:164)𝜌∥𝐻 𝑘 (Ω) ≤ 𝐶 ∥ (cid:164)𝜌∥1−𝑝

𝐻 −𝑠 (Ω) ∥ (cid:164)𝜌∥

𝑝

𝐻𝑠 (Ω) ≤ 𝐶 ∥ (cid:164)𝜌∥

2𝑠−𝑛
8𝑠
𝐻 −𝑠 (Ω)
(cid:19) 𝑛−2𝑠(cid:35) 2𝑠−𝑛

16𝑠

(cid:34)

(cid:34)

≤𝐶

≤𝐶

(𝜆 + 1)2 max(𝑠,4)𝑒𝐶 (𝜆+1)𝛿2 + 𝐶

(cid:18)

𝜆 + ln

1
𝛿

(𝜆 + 1)max(𝑠,4)𝑒𝐶 (𝜆+1)𝛿 + 𝐶

(cid:35) 2𝑠−𝑛
8𝑠

(cid:19) 𝑛−2𝑠

2

(cid:18)

𝜆 + ln

1
𝛿

(3.60)

□

3.3.1.3 Numerical Experiment

This section demonstrates the numerical implementation and validation of the reconstruction

formula (3.36) in a one-dimensional (1D) context, where 𝜌0 = 1 and 𝑞0 = 0.

78

Computing Boundary Controls with Time Reversal Notice that there is a second order deriva-

tive of 𝑓 in (3.36), the boundary controls should have at least second order differentiable. The 𝜕2
𝑡

operator will magnify the error caused by solving 𝑓 using similar methods as in Section 3.2.2.5.

In this section, we introduce an analytic method to construct boundary control with high-order

smoothness. In order to find boundary control 𝑓 such that 𝑢 𝑓
0

(𝑇) = 𝜙, we consider the following

backward initial value problem

□1,0𝑣(𝑡, 𝑥) = 0,

in (0, 𝑇) × R𝑛

𝑣(𝑇) = ˜𝜙,

𝜕𝑡𝑣(𝑇) = 0,

in R𝑛

in R𝑛.






where ˜𝜙 is the extension with compact support from Ω to R𝑛. If 𝑇 > 0 is sufficiently large and

dimension 𝑛 is odd, we would have 𝑣(0) = 𝜕𝑡𝑣(0) = 0 by the Huygen’s principle. This implies

𝑢 𝑓 −𝜕𝜈𝑣
0

(𝑇) = 𝑢 𝑓
0

(𝑇) − 𝑣(𝑇) = 0 in Ω.

As a result, we can take 𝑓 = 𝜕𝜈𝑣| [0,𝑇]×𝜕Ω. Note that 𝑣 can be explictly expressed using the

Kirchhoff’s formula [40], thus 𝜕2

𝑡 𝑓 = 𝜕𝜈𝜕2

𝑡 𝑣| [0,𝑇]×𝜕Ω can be analytically computed.

Recall that 𝑛 = 1, we take Ω = (𝑎, 𝑏). D’Alembert’s formula gives

𝑣(𝑡, 𝑥) =

1
2

[ ˜𝜙(𝑥 + 𝑡 − 𝑇) + ˜𝜙(𝑥 + 𝑇 − 𝑡)].

(3.61)

Thus

𝑡 𝑓 = 𝜕𝜈𝜕2
𝜕2

𝑡 𝑣| [0,𝑇]×𝜕Ω = ±

1
2

[ ˜𝜙′′′(𝑥 + 𝑡 − 𝑇) + ˜𝜙′′′(𝑥 + 𝑇 − 𝑡)] |𝑥=𝑎,𝑏,

where we take + when 𝑥 = 𝑏 and − when 𝑥 = 𝑎. We choose the following extension:

˜𝜙 :=






𝜙

𝑥 ∈ [𝑎, 𝑏],

𝜙 · exp{1 −

1
1−(𝑥−𝑎)2 𝑝 }

𝑥 ∈ (𝑎 − 1, 𝑎),

𝜙 · exp{1 −

1
1−(𝑥−𝑏)2 𝑝 }

𝑥 ∈ (𝑏, 𝑏 + 1),

0

𝑥 ∉ (𝑎 − 1, 𝑏 + 1),

79

where 𝑝 ≥ 1 is a positive integer. It is easy to verify that ˜𝜙 is 𝐶∞ almost everywhere, except for

two boundary points 𝑥 = 𝑎, 𝑏, where it only have 𝐶2𝑝−1 continuity. To guarantee the existence of

the second derivative of 𝑓 , we take 𝑝 ≥ 2.

Numerical Experiment We choose the spatial domain to be Ω = [−1, 1] with 𝑇 = 5. Since we

can analytically give the boundary control for each Helmholtz solution 𝑒𝑖

√
𝜆𝜃·𝑥, we do not have to

write all operators into matrices as in Section 3.2.2.

The solution of (3.22) is given by (3.61), and the forward problem (3.23) is solved using the

second order central difference scheme on a temporal-spatial grid of size 24999 × 501, i.e.

the

linearized ND map (3.24) is implemented by solving (3.23) with background solution (3.61).

Then (3.35) are inserted into (3.36) to recover the Fourier transform of (cid:164)𝑞 at 2

√

𝜆𝜃, where 𝑓 and

ℎ are computed using the time revesal method as in Section 3.3.1.3. The basis functions for the

prescribed Helmholtz solution 𝜙 in our experiments are

1, sin

(cid:17)

𝑥

(cid:16) 𝜋
2

, cos

(cid:17)

𝑥

(cid:16) 𝜋
2

, . . . , sin

(cid:19)

𝑥

, cos

(cid:18) 𝑁 𝜋

2

(cid:18) 𝑁 𝜋

2

(cid:19)

𝑥

(3.62)

with 𝑁 = 10. They correspond to Helmholtz solutions with

Noticing that the right hand side of (3.31) involves 1

Instead, we can take an arbitrary positive eigenvalue 𝜆 𝑗 =

√

𝜆 = 0, 𝜋
2

, . . . , 𝑁 𝜋
2 .
𝜆 , we can not directly compute F [ (cid:164)𝜌] (0).
for some 𝑗, compute the inner

𝑗 2𝜋2
4

products ( (cid:164)𝜌, cos2(

𝑗 𝜋
2

𝑥))𝐿2 (−1,1) and ( (cid:164)𝜌, sin2(

𝑗 𝜋
2

𝑥))𝐿2 (−1,1) using (3.31), then add them to get

F [ (cid:164)𝜌] (0) = ( (cid:164)𝜌, 1)𝐿2 (−1,1) = ( (cid:164)𝜌, cos2(

𝑗 𝜋

2

𝑥))𝐿2 (−1,1) + ( (cid:164)𝜌, sin2(

𝑗 𝜋

2

𝑥))𝐿2 (−1,1).

Experiment 1. We start with a continuous perturbation

(cid:164)𝜌 = sin(𝜋𝑥) + sin(2𝜋𝑥) − cos(5𝜋𝑥) + cos(7𝜋𝑥) − 1,

which is in the span of the Fourier basis functions (3.62). The graph of (cid:164)𝜌 is shown in Figure 3.7.

The Gaussian random noise are added to the measurement (cid:164)Λ by adding to the numerical solutions

on the boundary nodes. The reconstructions and corresponding errors with noise level 0%, 1%, 5%

are illustrated in Figure 3.8.

80

Figure 3.7 Ground truth (cid:164)𝜌.

Figure 3.8 Left: Reconstructed (cid:164)𝜌 with 0%, 1%, 5% Gaussian noise and the ground truth. Right:
The corresponding error between the reconstruction result and the ground truth. The relative
𝐿2-errors are 0.14%, 3.66% and 19.37%, respectively.

Experiment 2.

In this experiment, we consider a discontinuous perturbation

(cid:164)𝜌 = 𝜒

[−1,− 1

6 ] − 𝜒

[− 1
6

, 1
4 ]

,

where 𝜒 is the characteristic function. The Fourier series of (cid:164)𝜌 is given by

(cid:164)𝜌 =

5
24

+

(cid:34)

∞
∑︁

−

sin (cid:0) 𝑛𝜋
4

𝑛=1

(cid:1)

(cid:1) + 2 sin (cid:0) 𝑛𝜋
6
𝑛𝜋

cos(𝑛𝜋𝑥) +

cos(𝑛𝜋) + cos (cid:0) 𝑛𝜋
4
𝑛𝜋

(cid:1) + 2 cos (cid:0) 𝑛𝜋
6

(cid:1)

(cid:35)

sin(𝑛𝜋𝑥)

.

With the choice of the basis functions (3.62), we can only expect to reconstruct the orthogonal

projection:

(cid:164)𝜌𝑁 (cid:66) 5
24

+

(cid:34)

𝑁
∑︁

−

sin (cid:0) 𝑛𝜋
4

𝑛=1

(cid:1)

(cid:1) + 2 sin (cid:0) 𝑛𝜋
6
𝑛𝜋

cos(𝑛𝜋𝑥) +

cos(𝑛𝜋) + cos (cid:0) 𝑛𝜋
4
𝑛𝜋

(cid:1) + 2 cos (cid:0) 𝑛𝜋
6

(cid:1)

(cid:35)

sin(𝑛𝜋𝑥)

,

see Figure 3.9. We plot the reconstruction result and the corresponding error with respect to the

orthogonal projection (cid:164)𝜌𝑁 with different noise level in Figure 3.10.

81

-1-0.500.51-5-4-3-2-10123Ground Truth-1-0.500.51-5-4-3-2-10123ReconstructionGround TruthNo noise1% noise5% noise-1-0.500.5100.10.20.30.40.50.60.7ErrorNo noise1% noise5% noiseFigure 3.9 Ground truth (cid:164)𝜌.

Figure 3.10 Left: Reconstructed (cid:164)𝜌 with 0%, 1%, 5% Gaussian noise and the ground truth. Right:
The corresponding error between the reconstruction result and the ground truth. The relative
𝐿2-errors are 0.39%, 1.57% and 8.15%, respectively.

Experiment 3.

In this experiment, we apply the algorithm to the non-linear IBVP where

𝜌 = 𝜌0 + 𝜀 (cid:164)𝜌 + 𝜀2 (cid:165)𝜌,

with 𝜀 = 0.001 and

(cid:164)𝜌 = sin(𝜋𝑥) + sin(2𝜋𝑥) − cos(5𝜋𝑥) + cos(7𝜋𝑥) − 1,

(cid:165)𝜌 = 200 sin(25𝜋𝑥).

See Figure 3.11 for the graph of 𝜌. Since

Λ𝜌 − Λ𝜌0 ≈ 𝜀 (cid:164)Λ (cid:164)𝜌 = (cid:164)Λ𝜀 (cid:164)𝜌

when 𝜀 is small, we can use 𝜀−1(Λ𝜌 − Λ𝜌0) as an approximation of (cid:164)Λ (cid:164)𝜌 in (3.31). In this case, Λ𝜌 𝑓
𝑓 are computed by numerically solving the forward problem (3.22) with 𝜌 and 𝜌0 ≡ 1. We

and Λ𝜌0

then apply Algorithm 3.2 to find (cid:164)𝜌, and view 1 + (cid:164)𝜌 as an approximation of the ground truth 𝜌. In

82

-1-0.500.51-1.5-1-0.500.511.5Ground Truth-1-0.500.51-1.5-1-0.500.511.5ReconstructionProjectionNo noise1% noise5% noise-1-0.500.5100.020.040.060.080.10.120.140.160.18ErrorNo noise1% noise5% noisethe experiment, we added the Gaussian noise to the difference Λ𝜌 − Λ𝜌0 rather than to Λ𝜌 𝑓 and
𝑓 individually, see [75] for discussion of the difference. The reconstruction and the respective

Λ𝜌0

errors with 0%, 1%, 5% Gaussian noise are illustrated in Figure 3.12.

Figure 3.11 Ground truth 𝜌.

Figure 3.12 Left: Reconstructed 𝜌 with 0%, 1%, 5% Gaussian noise and the ground truth. Right:
The corresponding error between the reconstruction result and the ground truth. The relative
𝐿2-errors are 18.09%, 20.95% and 26.46%, respectively.

3.3.2 Reconstruct Wave Potential through Linearization

The wave potential 𝑞, similar to the potential energy function of quantum mechanics, describes

the reflection and transmission characteristics of the system [44]. However, the full nonlinear

treatment introduced in Section 3.2 is not sufficient for the potential reconstruction. In the algorithm

for reconstructing wave speed 𝑐(𝑥), the most important step is to construct a series of boundary

controls 𝑓𝛼 such that 𝑢 𝑓𝛼 (𝑇) converge to a handcrafted time independent wave solution 𝜓 as 𝛼 → 0.

83

-1-0.500.510.9750.980.9850.990.99511.0051.011.015Ground Truth-1-0.500.510.970.9750.980.9850.990.99511.0051.011.015ReconstructionGround TruthNo noise1% noise5% noise-1-0.500.5101234567810-3ErrorNo noise1% noise5% noiseHowever, for unknown 𝑞, the time independent wave solution 𝜓 of (3.1) satisfies equation

[−Δ + 𝑞(𝑥)]𝜓(𝑥) = 0,

which can not be construct explicitly. With linearization, the background potential 𝑞0 would give

us a series of time independent background wave solutions, which can help reconstructing the

perturbed wave potential [75]. In this section, we assume the wave speed 𝑐 = 𝑐0(𝑥) ∈ 𝐶∞(Ω) is

known, and we want to recover 𝑞(𝑥) from the ND map Λ𝜌0,𝑞. For simplicity, we use Λ𝑞 to represent

Λ𝜌0,𝑞 in this section.

We use the linearization to solve the IBVP. For the formal derivation, we write

𝑞(𝑥) = 𝑞0(𝑥) + 𝜀 (cid:164)𝑞(𝑥),

𝑢(𝑡, 𝑥) = 𝑢0(𝑡, 𝑥) + 𝜀 (cid:164)𝑢(𝑡, 𝑥)

where 𝑞0 is a known background potential and 𝑢0 is the background solution. Substitute these

into (3.1). Equating the 𝑂 (1)-terms gives

□𝜌0,𝑞0

𝑢0(𝑡, 𝑥) = 0,

in (0, 2𝑇) × Ω

𝜕𝜈𝑢0 = 𝑓 ,

on (0, 2𝑇) × 𝜕Ω

(3.63)

𝑢0(0, 𝑥) = 𝜕𝑡𝑢0(0, 𝑥) = 0,

𝑥 ∈ Ω.






Equating the 𝑂 (𝜀)-terms gives

□𝜌0,𝑞0 (cid:164)𝑢(𝑡, 𝑥) = −𝑢0(𝑡, 𝑥) (cid:164)𝑞(𝑥),

in (0, 2𝑇) × Ω

𝜕𝜈 (cid:164)𝑢 = 0,

on (0, 2𝑇) × 𝜕Ω

(3.64)

(cid:164)𝑢(0, 𝑥) = 𝜕𝑡 (cid:164)𝑢(0, 𝑥) = 0

𝑥 ∈ Ω.






Write the ND map Λ𝑞 = Λ𝑞0 + 𝜀 (cid:164)Λ (cid:164)𝑞, where Λ𝑞0 denote the ND map for the unperturbed boundary
value problem (3.63), and (cid:164)Λ (cid:164)𝑞 is defined as

(cid:164)Λ (cid:164)𝑞 : 𝑓 ↦→ (cid:164)𝑢| (0,2𝑇)×𝜕Ω.

(3.65)

Since 𝑐0 and 𝑞0 are known, the unperturbed problem (3.63) can be explicitly solved to obtain 𝑢0
and Λ𝑞0. As in the previous section, we will write (cid:164)𝑢 = (cid:164)𝑢 𝑓 if it is necessary to specify the Neumann
data 𝑓 . Then the linearized IBVP concerns recovery of the potential (cid:164)𝑞 from (cid:164)Λ (cid:164)𝑞.

84

3.3.2.1 Derivation

The derivation of the Blagove˘s˘censki˘ı’s identity is exactly the same. Introduce the connecting

operator

𝐾 := 𝐽Λ𝑞𝑃∗

𝑇 − 𝑅Λ𝑞,𝑇 𝑅𝐽𝑃∗
𝑇 .

Similar to the proof of Proposition 3.6 and Corollary 3.7, we have

(𝑢 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = ( 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

(Δ𝑢 𝑓 (𝑇) − 𝑞𝑢 𝑓 (𝑇), 𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = (𝜕2

𝑡 𝑓 , 𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

(3.66)

(3.67)

Remember that we write Λ𝑞 = Λ𝑞0 + 𝜀 (cid:164)Λ (cid:164)𝑞 in the linearization setting, we have the following
linearization 𝐾 = 𝐾0 + 𝜀 (cid:164)𝐾. Here 𝐾0 is the connecting operator for the background wave equation

(3.63):

𝐾0 := 𝐽Λ𝑞0

𝑇 − 𝑅Λ𝑞0,𝑇 𝑅𝐽𝑃∗
𝑃∗
𝑇 .

(3.68)

𝐾0 can be explicitly computed since Λ𝜌0,𝑞0 is known.

(cid:164)𝐾 is the resulting perturbation in the

connecting operator:

(cid:164)𝐾 := 𝐽 (cid:164)Λ (cid:164)𝑞𝑃∗

𝑇 − 𝑅 (cid:164)Λ (cid:164)𝑞,𝑇 𝑅𝐽𝑃∗
𝑇 .

(3.69)

(cid:164)𝐾 can be explicitly computed once (cid:164)Λ (cid:164)𝑞 is given.

We write (cid:164)Λ for (cid:164)Λ (cid:164)𝑞 when there is no risk of confusion. Linearizing (3.66) and (3.67) gives the

following integral identity, which is essential to the development of the reconstruction procedure:

Proposition 3.18. Let 𝜆 ∈ R be a real number. If 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) satisfy

[Δ − 𝑞0 + 𝜆]𝑢 𝑓

0

(𝑇) = [Δ − 𝑞0 + 𝜆]𝑢ℎ

0 (𝑇) = 0

in Ω,

then the following identity holds:

−( (cid:164)𝑞𝑢 𝑓
0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω) = (𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(3.70)

(3.71)

Proof. Similar to the proof of Proposition 3.8, substitute all linearizations into (3.66) and (3.67).

Equating 𝑂 (1)-terms gives

(𝑢 𝑓
0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = ( 𝑓 , 𝐾0ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾0 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

85

(Δ𝑢 𝑓
0

(𝑇) − 𝑞0𝑢 𝑓

0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω,𝑐−2

0

𝑑𝑥) = (𝜕2

𝑡 𝑓 , 𝐾0ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = (𝐾0𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω).

Equating 𝑂 (𝜀)-terms gives

( 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = ( (cid:164)𝐾 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω)
=( (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω,𝑐−2

𝑑𝑥) + (𝑢 𝑓

0

0

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2

.

𝑑𝑥)

(3.72)

0

(𝜕2

𝑡 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) = ( (cid:164)𝐾𝜕2

𝑡 𝑓 , ℎ)𝐿2 ((0,𝑇)×𝜕Ω)

=(Δ (cid:164)𝑢 𝑓 (𝑇) − (cid:164)𝑞𝑢 𝑓
0

(𝑇) − 𝑞0 (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω,𝑐−2

𝑑𝑥)

0

+ (Δ𝑢 𝑓
0
= (Δ (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)

(cid:124)

0 (𝑇))𝐿2 (Ω)
(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)
(cid:125)

𝑑𝑥)

0

0

(𝑇) − 𝑞0𝑢 𝑓

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω,𝑐−2
0 (𝑇))𝐿2 (Ω) − (𝑞0 (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ
(cid:123)(cid:122)
:=𝐼1
0 (𝑇))𝐿2 (Ω)
(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω) − (𝑞0𝑢 𝑓

(𝑇), 𝑢ℎ

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω)
(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)
(cid:125)

0

(cid:123)(cid:122)
:=𝐼2

(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)(cid:32)

− ( (cid:164)𝑞𝑢 𝑓
0
+ (Δ𝑢 𝑓
0
(cid:124)

.

(3.73)

Notice that

𝐼1 =(Δ (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

=( (cid:164)𝑢 𝑓 (𝑇), Δ𝑢ℎ

0 (𝑇))𝐿2 (Ω) − (𝑞0 (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ
0 (𝑇))𝐿2 (Ω) − ( (cid:164)Λ 𝑓 (𝑇), 𝜕𝜈𝑢ℎ

0 (𝑇))𝐿2 (Ω)
0 (𝑇))𝐿2 (𝜕Ω) − (𝑞0 (cid:164)𝑢 𝑓 (𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω)

=( (cid:164)𝑢 𝑓 (𝑇), [Δ − 𝑞0]𝑢ℎ

0 (𝑇))𝐿2 (Ω) − ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω).

Here we use the integration by parts and use the fact that (cid:164)𝑢 𝑓 | (0,2𝑇)×𝜕Ω = (cid:164)Λ 𝑓 and 𝜕𝜈 (cid:164)𝑢 = 0.

On the other hand, combing the terms in 𝐼2 gives

𝐼2 = ([Δ − 𝑞0]𝑢 𝑓

0

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω).

Insert these expressions for 𝐼1 and 𝐼2 into (3.73), then add (3.72) multiplied by 𝜆 ∈ R to get

(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

=( (cid:164)𝑢 𝑓 (𝑇), [Δ − 𝑞0 + 𝜆]𝑢ℎ
+ ([Δ − 𝑞0 + 𝜆]𝑢 𝑓

(𝑇), (cid:164)𝑢ℎ (𝑇))𝐿2 (Ω).

0

0 (𝑇))𝐿2 (Ω) − ( (cid:164)𝑞𝑢 𝑓

0

(𝑇), 𝑢ℎ

0 (𝑇))𝐿2 (Ω)

(3.74)

86

If [Δ − 𝑞0 + 𝜆]𝑢 𝑓

0

(𝑇) = [Δ − 𝑞0 + 𝜆]𝑢ℎ
0

(𝑇) = 0 in Ω, the first term and last term on the right-hand

side vanish, resulting in (3.71).

□

Notice that all parameters in (3.70) are known. For each 𝜆 ∈ R, we can construct functions 𝜙, 𝜓

satisfy (3.70). Proposition 3.9 ensures that we can find control sequence 𝑓 , ℎ such that 𝑢 𝑓 (𝑇) = 𝜙,
𝑢ℎ (𝑇) = 𝜓, therefore, we have the weighted inner product (𝜙, 𝜓)𝐿2 (Ω, (cid:164)𝑞 d𝑥) from (3.71).

3.3.2.2 Stability and Reconstruction

Observing that the right hand side of (3.31) and (3.71) differ only by a constant 𝜆, the discussion

is almost the same as in Section 3.3.1.2, see also [75].

Case 1: 𝑞0 is constant Without loss of generality, we take 𝑞0 = 0. By choosing 𝜆 ≥ 0, the

equation (3.70) becomes the Helmholtz equation

[Δ + 𝜆]𝑢 𝑓
0

(𝑇) = [Δ + 𝜆]𝑢 𝑓
0

(𝑇) = 0

in Ω.

Let 𝜃 ∈ S𝑛−1 be an arbitrary unit vector, Proposition 3.9 guarantees the existence of 𝑓 , ℎ ∈

𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) such that

𝑢 𝑓
0

(𝑇) = 𝑢ℎ

0 (𝑇) = 𝑒𝑖

√
𝜆𝜃·𝑥,

(3.75)

which is the Helmholtz solution. Similarly, we have following results:

Theorem 3.19. Suppose 𝑐0 = 1 and 𝑞0 = 0. Then the Fourier transform ˆ(cid:164)𝑞 of (cid:164)𝑞 can be reconstructed

as follows:

ˆ(cid:164)𝑞(2

√

𝜆𝜃) = −(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) − ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(3.76)

where 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) are solutions to (3.75).

Proof. The formula is obtained by substitute (3.75) into (3.71). Since 𝜃 ∈ S𝑛−1 and 𝜆 ≥ 0 are

arbitrary, it gives the Fourier transform of (cid:164)𝑞 everywhere.

□

Theorem 3.20. Suppose 𝑐0 = 1 and 𝑞0 = 0. There exists a constant 𝐶 > 0, independent of 𝜆, such

that

√

(cid:12)
(cid:12)
(cid:12)

ˆ(cid:164)𝑞(

2𝜆𝜃)

(cid:12)
(cid:12)
(cid:12)

≤ 𝐶 (1 +

√
2𝑇 (1 + 𝜆))𝜆4∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻3 ((0,𝑇)×𝜕Ω)

87

Proof. The proof is nearly a word-by-word repetition of Theorem 3.12, see also [75].

□

Case 2: 𝑞0 is variable Let 𝜆 ≥ 0, then (3.70) becomes the perturbed Helmholtz equation

[Δ + 𝜆 − 𝑞0]𝑢 𝑓

0

(𝑇) = [Δ + 𝜆 − 𝑞0]𝑢ℎ

0 (𝑇) = 0

in Ω.

For any 𝜃 ∈ S𝑛−1, we choose the solution to be

𝜙(𝑥) = 𝑒𝑖

√
𝜆𝜃·𝑥 + 𝑟 (𝑥; 𝜆)

the residual 𝑟 (𝑥; 𝜆) satisfies

According to [75, Lemma 13],

for any 𝑠 ≥ 0.

(Δ + 𝜆 − 𝑞0)𝑟 = 𝑞0𝑒𝑖

√

𝜆𝜃·𝑥

in Ω.

∥𝑟 ∥𝐻𝑠 (R𝑛) = 𝑂 (𝜆 𝑠−1
2 )

as 𝜆 → ∞

(3.77)

(3.78)

(3.79)

One dimension:

In one dimension (1D), 𝜃 = ±1. Let us take 𝜃 = 1 and choose (3.77) to be

the value of 𝑢 𝑓
0

(𝑇) and 𝑢ℎ
0

(𝑇). Substituting into (3.71) gives

− ˆ(cid:164)𝑞(2

√

𝜆) − 2( (cid:164)𝑞𝑒𝑖

√

𝜆𝜃·𝑥, 𝑟)𝐿2 (Ω) − ( (cid:164)𝑞𝑟, 𝑟)𝐿2 (Ω)

=(𝜕2

𝑡 𝑓 + 𝜆 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(3.80)

With similar analysis as (3.42), we have

(cid:12)
(cid:12)
(cid:12)

ˆ(cid:164)𝑞(

√

2𝜆𝜃)

(cid:12)
(cid:12)
(cid:12)

√

≤ 𝐶 (1 +

2𝑇 (1 + 𝜆))𝜆4∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻3 ((0,𝑇)×𝜕Ω) + 𝑂 (𝜆− 1

2 ).

High Dimension:

In dimension 𝑛 ≥ 2, let 𝜆 ≥ 0, 𝜃, 𝜔 ∈ R𝑛 be two vectors such that 𝜃 ⊥ 𝜔.

We take the following solutions:

𝜙(𝑥) :=𝜙0(𝑥) + 𝑟1(𝑥; 𝜆),

𝜙0(𝑥) := 𝑒𝑖(𝑘𝜃+𝑙𝜔)·𝑥

𝜓(𝑥) :=𝜓0(𝑥) + 𝑟2(𝑥; 𝜆),

𝜓0(𝑥) := 𝑒𝑖(𝑘𝜃−𝑙𝜔)·𝑥

88

where 𝑟1, 𝑟2 satisfy (3.79), 𝑘 2 + 𝑙2 = 𝜆 such that (Δ + 𝜆)𝜙0 = (Δ + 𝜆)𝜓0 = 0. Proposition 3.9 asserts

that there are 𝑓 , ℎ ∈ 𝐶∞

𝑐 ((0, 𝑇] × 𝜕Ω) such that

𝑢 𝑓
0

(𝑇) = 𝜙 = 𝜙0 + 𝑟1,

𝑢ℎ
0 (𝑇) = 𝜓 = 𝜓0 + 𝑟2.

(3.81)

Inserting (3.81) into (3.71) gives

− ˆ(cid:164)𝑞(2𝑘𝜃) − ( (cid:164)𝑞𝑒𝑖(𝑘𝜃+𝑙𝜔)·𝑥, 𝑟2)𝐿2 (Ω) − ( (cid:164)𝑞𝑒𝑖(𝑘𝜃−𝑙𝜔)·𝑥, 𝑟1)𝐿2 (Ω) − ( (cid:164)𝑞𝑟1, 𝑟2)𝐿2 (Ω)

=(𝜕2

𝑡 𝑓 + (𝑘 2 + 𝑙2) 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(3.82)

If we fix 𝑘 and let 𝑙 → ∞, we obtain the reconstruction formula for any 𝑘 ≥ 0 and any 𝜃 ∈ S𝑛−1:

ˆ(cid:164)𝑞(2𝑘𝜃) = − lim
𝑙→∞

(cid:2)(𝜕2

𝑡 𝑓 + (𝑘 2 + 𝑙2) 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω)

(cid:3) .

Similarly, we can obtain a Hölder-type stability estimate for ∥ (cid:164)𝑞∥𝐻 −𝑠 (R𝑛).

Theorem 3.21. Suppose 𝑐0 = 1, 𝑞0 ∈ 𝐶∞(Ω) and 𝑞0 is not identically zero. For any 𝑠 > 0, there

exists a constant 𝐶 > 0 independent of 𝜆 such that

∥ (cid:164)𝑞∥𝐻 −𝑠 (R𝑛) ≤ 𝐶 ∥ (cid:164)Λ∥

2𝑠
11(𝑛+2𝑠)
𝐻2 ((0,𝑇)×𝜕Ω)→𝐻3 ((0,𝑇)×𝜕Ω)

.

Proof. Write 𝜉 := 2𝑘𝜃 and

𝛿 := ∥ (cid:164)Λ∥𝐻2 ((0,𝑇)×𝜕Ω)→𝐻3 ((0,𝑇)×𝜕Ω).

Let 𝜌 > 0 be a sufficiently large number that is to be determined. We decompose

∥ (cid:164)𝑞∥2

𝐻 −𝑠 (R𝑛) =

∫

|𝜉 |≤𝜌

| ˆ(cid:164)𝑞(𝜉)|2
(1 + |𝜉 |2)𝑠

∫

𝑑𝜉 +

|𝜉 |>𝜌

| ˆ(cid:164)𝑞(𝜉)|2
(1 + |𝜉 |2)𝑠

𝑑𝜉.

For the integral over high frequencies, we have

∫

|𝜉 |>𝜌

| ˆ(cid:164)𝑞(𝜉)|2
(1 + |𝜉 |2)𝑠

𝑑𝜉 ≤

1
(1 + 𝜌2)𝑠

∫

|𝜉 |>𝜌

| ˆ(cid:164)𝑞(𝜉)|2 𝑑𝜉 ≤

For the integral over low frequencies, it is easy to see that:

∥ (cid:164)𝑞∥2

𝐿2 (R𝑛)

(1 + 𝜌2)𝑠 ≤ 𝐶 1

𝜌2𝑠

.

∫

|𝜉 |≤𝜌

| ˆ(cid:164)𝑞(𝜉)|2
(1 + |𝜉 |2)𝑠

∫

𝑑𝜉 ≤

|𝜉 |≤𝜌

| ˆ(cid:164)𝑞(𝜉)|2 𝑑𝜉 ≤ 𝐶 𝜌𝑛 ∥ ˆ(cid:164)𝑞∥2

𝐿∞ (𝐵(0,𝜌))

.

89

The norm ∥ ˆ(cid:164)𝑞∥ 𝐿∞ (𝐵(0,𝜌)) can be estimated using (3.82). Indeed, for |𝜉 | ≤ 𝜌, we have

| ˆ(cid:164)𝑞(𝜉)| ≤ |(𝜕2

𝑡 𝑓 + (𝑘 2 + 𝑙2) 𝑓 , (cid:164)𝐾 ℎ)𝐿2 ((0,𝑇)×𝜕Ω) + ( (cid:164)Λ 𝑓 (𝑇), ℎ(𝑇))𝐿2 (𝜕Ω) | +

𝐶
√

𝜆

≤ 𝐶 (1 +

√

2𝑇 (1 + 𝜆))∥𝜙∥𝐻4 (Ω) ∥𝜓∥𝐻4 (Ω)𝛿 +

𝐶
√

≤ 𝐶 (1 +

√

2𝑇 (1 + 𝜆))

≤ 𝐶 (1 +

√

2𝑇 (1 + 𝜆))

(cid:16)

(cid:16)

∥𝜙0∥𝐻4 (Ω) + ∥𝑟1∥𝐻4 (Ω)

𝜆2 + 𝜆 3

2

(cid:17) 2

𝛿 +

𝐶
√

𝜆

𝜆
(cid:17) (cid:16)

∥𝜓0∥𝐻4 (Ω) + ∥𝑟2∥𝐻4 (Ω)

(cid:17)

𝛿 +

𝐶
√

𝜆

where the first and the last inequality is a consequence of (3.79), the second inequality follows from

the proof of Proposition 3.20. Utilizing the relation 𝜆 = 𝑘 2 + 𝑙2, we conclude

∥ ˆ(cid:164)𝑞∥2

𝐿∞ (𝐵(0,𝜌)) ≤ 𝐶

(cid:20)

(1 +

√

2𝑇 (1 + 𝜆))2𝜆6(1 +

√

𝜆)4𝛿2 +

(cid:20)

≤ 𝐶

(cid:21)

1
𝜆

(𝜌2 + 𝑙2)10𝛿 +

(cid:21)

1
𝑙2

provided 𝜌 > 0 is sufficiently large. Combining these estimates, we see that

∥ (cid:164)𝑞∥2

𝐻 −𝑠 (R𝑛) ≤ 𝐶

(cid:20)

𝜌𝑛 (𝜌2 + 𝑙2)10𝛿2 +

𝜌𝑛
𝑙2

+

1
𝜌2𝑠

(cid:21)

.

Choosing 𝑙2 = 𝜌𝑛+2𝑠 and 𝜌 = 𝛿−

2

11(𝑛+2𝑠) yields

∥ (cid:164)𝑞∥2

𝐻 −𝑠 (R𝑛) ≤ 𝐶𝛿

4𝑠
11(𝑛+2𝑠) ,

where 𝐶 is a constant independent of 𝜆 and 𝛿 is sufficiently small.

□

3.3.2.3 Numerical Experiment

This section demonstrates the numerical implementation and validation of the reconstruction

formula (3.76) in a one-dimensional (1D) context, where 𝑐0 = 1 and 𝑞0 = 0.

The setting is the same as in Section 3.3.1.3. We choose the spatial domain to be Ω = [−1, 1]

with 𝑇 = 5. The forward problem (3.64) is solved using the second order central difference scheme

on a temporal-spatial grid of size 24999 × 501. The basis functions for the prescribed Helmholtz

solution 𝜙 in our experiments are

1, sin

(cid:17)

𝑥

(cid:16) 𝜋
2

, cos

(cid:17)

𝑥

(cid:16) 𝜋
2

, . . . , sin

(cid:19)

𝑥

, cos

(cid:18) 𝑁 𝜋

2

(cid:18) 𝑁 𝜋

2

(cid:19)

𝑥

90

with 𝑁 = 10. They correspond to Helmholtz solutions with

√

𝜆 = 0, 𝜋
2

, . . . , 𝑁 𝜋

2 . Boundary controls

are computed using the time revesal method as in Section 3.3.1.3.

Experiment 1. The first experiment aims to reconstruct the following smooth (cid:164)𝑞 using the for-

mula (3.76):

(cid:164)𝑞 = sin(𝜋𝑥) + 2 cos(2𝜋𝑥) + 4 sin(4𝜋𝑥) − 3.

The graph of (cid:164)𝑞 is shown in Figure 3.13. The measurement (cid:164)Λ (cid:164)𝑞 is added with 0%, 1%, and 5%

of Gaussian noise, respectively. The reconstructions and the corresponding errors are illustrated

in Figure 3.14. Notice that the reconstruction error with 5% noise is relatively larger, as can be

expected. When multiple measurements are available, we can repeat the reconstruction several

times and then take the average. This strategy effectively reduces the error, since the inverse

problem is linear and the Gaussian noise has zero mean, see Figure 3.15.

Figure 3.13 Ground truth (cid:164)𝑞 = sin(𝜋𝑥) + 2 cos(2𝜋𝑥) + 4 sin(4𝜋𝑥) − 3.

Experiment 2. The second experiment tests reconstruction of a discontinuous (cid:164)𝑞. we choose (cid:164)𝑞 to

be the Heaviside function

𝐻 (𝑥) =

The Fourier series of 𝐻 (𝑥) on Ω = [−1, 1] is

1

0





𝑥 ≥ 0,

𝑥 < 0.

𝐻 (𝑥) =

1
2

+

∞
∑︁

𝑛=1

2

(2𝑛 − 1)𝜋 sin((2𝑛 − 1)𝜋𝑥).

91

-1-0.500.51-10-8-6-4-2024Ground TruthFigure 3.14 Left: Reconstructed (cid:164)𝑞 with 0%, 1%, 5% Gaussian noise and the ground truth. Right:
The corresponding error between the reconstruction and the ground truth. The relative 𝐿2-errors
are 0.1%, 2.5%, and 23.9% respectively.

Figure 3.15 Left: Reconstructed (cid:164)𝑞 under 1, 7, 14, 21 times repetition with 5% Gaussian noise and
the ground truth. Right: The corresponding error between the reconstruction and the ground
truth. The relative 𝐿2-errors are 23.9%, 9.4%, 5.5%, and 4.0% respectively.

With the choice of the finite computational basis, we can only expect to reconstruct the following

orthogonal projection:

𝐻𝑁 (𝑥) :=

1
2

+

⌈ 𝑁
2 ⌉
∑︁

𝑛=1

2

(2𝑛 − 1)𝜋 sin((2𝑛 − 1)𝜋𝑥),

see Figure 3.16 for the graph of 𝐻 (𝑥) and 𝐻𝑁 (𝑥). The reconstruction formula (3.76) is implemented

with 0%, 1%, and 5% of Gaussian noise added to (cid:164)Λ (cid:164)𝑞, respectively. The reconstructions and

corresponding errors with a single measurement are illustrated in Figure 3.17. The averaged

reconstruction with 5% of Gaussian noise and multiple repeated measurements are illustrated in

Figure 3.18.

92

-1-0.500.51-10-505ReconstructionGround TruthNo noise1% noise5% noise-1-0.500.5100.511.522.53ErrorNo noise1% noise5% noise-1-0.500.51-10-8-6-4-20246Reconstruction1 time7 times14 times21 timesGround Truth-1-0.500.5100.511.522.53Error1 time7 times14 times21 timesFigure 3.16 Ground truth (cid:164)𝑞 = 𝐻 (𝑥) and its projection 𝐻𝑁 (𝑥).

Figure 3.17 Left: Reconstructed (cid:164)𝑞 with 0%, 1%, 5% Gaussian noise and the projection of the
ground truth. Right: The corresponding error between the reconstruction and the projection of the
ground truth. The relative 𝐿2-errors are 0.6%, 6.2%, and 33.8%, respectively.

Figure 3.18 Left: Reconstructed (cid:164)𝑞 under 1, 7, 14, 21 times repetition with 5% Gaussian noise and
the projection of the ground truth. Right: The corresponding error between the reconstruction and
the projection of the ground truth. The relative 𝐿2-errors are 33.8%, 7.2%, 6.9%, and 6.1%
respectively.

Experiment 3. This experiment aims to test the reconstruction in the case 𝑐0 = 1 and a small

𝑞0 ≠ 0. We choose

𝑞0 =

1
10

sin(𝜋𝑥),

93

-1-0.500.51-0.200.20.40.60.811.2Ground TruthH(x)HN(x)-1-0.500.51-0.500.511.5ReconstructionProjectionNo noise1% noise5% noise-1-0.500.5100.10.20.30.40.50.6ErrorNo noise1% noise5% noise-1-0.500.51-0.500.511.5Reconstruction1 time7 times14 times21 timesProjection-1-0.500.5100.10.20.30.40.50.6Error1 time7 times14 times21 timesand (cid:164)𝑞 to be the same Heaviside function as in Experiment 2, see Figure 3.19. We attempt to

reconstruct an approximate (cid:164)𝑞 based on (3.80) by neglecting the terms involving 𝑟. A computational

challenge is that we cannot find explicit form of 𝜕2

𝑡 𝑓 when 𝑞0 ≠ 0. Instead, we make use of the
𝑡 𝑓 as if 𝑞0 = 0. In the meanwhile, the operator (cid:164)𝐾 and
smallness of 𝑞0 to approximately construct 𝜕2
(cid:164)Λ (cid:164)𝑞 are still implemented using the exact 𝑞0 and (cid:164)𝑞. The reconstructions and corresponding errors

with a single measurement under 0%, 1%, and 5% of Gaussian noise are illustrated in Figure 3.20.

The averaged reconstruction with 5% of Gaussian noise and multiple repeated measurements are

illustrated in Figure 3.21. This experiment confirms that approximate reconstruction using (3.80)

remains possible for 𝑞0 ≠ 0 as long as it is small.

Figure 3.19 Left: 𝑞0 = 1

10 sin(𝜋𝑥). Right: Ground truth (cid:164)𝑞 = 𝐻 (𝑥) and its projection 𝐻𝑁 (𝑥).

Figure 3.20 Left: Reconstructed (cid:164)𝑞 with 0%, 1%, 5% Gaussian noise and the projection of the
ground truth. Right: The corresponding error between the reconstruction and the projection of the
ground truth. The relative 𝐿2-errors are 5.8%, 8.9%, and 24.2%, respectively.

94

-1-0.500.51-0.1-0.08-0.06-0.04-0.0200.020.040.060.080.1q0-1-0.500.51-0.200.20.40.60.811.2Ground TruthH(x)HN(x)-1-0.500.51-0.4-0.200.20.40.60.811.21.4ReconstructionProjectionNo noise1% noise5% noise-1-0.500.5100.050.10.150.20.25ErrorNo noise1% noise5% noiseFigure 3.21 Left: Reconstructed (cid:164)𝑞 under 1, 7, 14, 21 times repetition with 5% Gaussian noise and
the projection of the ground truth. Right: The corresponding error between the reconstruction and
the projection of the ground truth. The relative 𝐿2-errors are 24.2%, 9.38%, 9.87% and 9.93%,
respectively.

More Experiments We also apply the reconstruction formula (3.76) to measurement from the

non-linear IBVP, see [75] for more details.

95

-1-0.500.51-0.4-0.200.20.40.60.811.21.4Reconstruction1 time7 times14 times21 timesProjection-1-0.500.5100.050.10.150.20.250.30.350.4Error1 time7 times14 times21 timesCHAPTER 4

CONCLUSION

In Chapter 2, we introduce several algorithms for UMBLT. For UMBLT under transport regime,

see Section 2.2, we proposed an algorithm to reconstruct isotropic sources, which reduce the

measurement requirement and computational demand. However, our proposed algorithm still

based on the measurement over entire outgoing boundary, and the source are limited to isotropic

source. In future, we can generalize the algorithm to the partial data case similar to the partial data

case in diffusive regime. We can choose the adjoint outgoing boundary condition to be supported

on the measurement area and we can prove that we can choose such boundary condition to make the

adjoint RTE solution positive. For UMBLT under diffusive regime, see Section 2.3, we generalize

the algorithm from full data case to partial data case, and give uncertainty quantification based

on PDE theory. However, the numerical experiment start from the internal functional and lack of

real world data. In future, we could try to numerically recover internal functional from boundary

measurement to test the algorithm.

In Chapter 3, we introduce nonlinear IBVP for wave speed reconstruction, see Section 3.2, and

linearized IBVP for wave speed and wave potential reconstruction, see Section 3.3. We show that

the wave speed can be uniquely determined with vanished wave potential using stable, non-iterative

method. We also show that both wave speed and potential can be reconstructed using linearization

with at least Hölder stability. However, the nonlinear IBVP are limited to vanished wave potential,

the numerical experiment of linearized IBVP are limited to constant background speed and vanish

background potential due to the accuracy requirement of second order temporal derivative, and

we do not have algorithm to reconstruct wave speed and wave potential together from boundary

measurement. In future, we will consider the nonlinear IBVP of wave speed reconstruction with

nonvanished wave potential. We will also consider wave model with more parameters, such as

wave equation with absorption effect.

96

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103

APPENDIX A

APPENDIX FOR CHAPTER 2

A.1 Diffusion Approximation

In strong scattering medium, such as biological objects, the propagation of light is diffusive

and can be approximately described by diffusion equation. The standard way to accomplish

the approximation from RTE to diffusion equation is to expand functions in terms of spherical

harmonics and truncate the series.

In order to derive the diffusion approximation, we need following assumptions on the optical

coefficients:

1. 𝑘 (𝑥, 𝜃, 𝜗) = 𝑘 (𝑥, −𝜗, −𝜃) ≥ 0 for any 𝑥 ∈ 𝑋, 𝜃, 𝜗 ∈ S𝑛−1,

2. ∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜗) d𝜃 = ∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜗) d𝜗 = 𝜎𝑠 (𝑥) ≥ 0 for any 𝑥 ∈ 𝑋, 𝜃, 𝜗 ∈ S𝑛−1,

3. 𝜎(𝑥) ≥ 𝜎𝑠 (𝑥) for any 𝑥 ∈ 𝑋,

4. 𝑆(𝑥, 𝜃) is either independent of direction 𝜃, or compact supported on 𝑋 for any 𝜃 ∈ S𝑛−1.

We introduce the diffusion approximation under spherical harmonics up to the first order. The

space spanned by the spherical harmonics up to the first order in S𝑛−1 is

H1 = span{1, 𝜃1, 𝜃2, . . . , 𝜃𝑛} ⊂ 𝐿2(S𝑛−1),

(A.1)

where 𝜃𝑖 denote the 𝑖-th entry of 𝜃 ∈ S𝑛−1.

Lemma A.1 ( [52, Lemma 6.10]). The orthogonal projection

P : 𝐿2(S𝑛−1) → H1

is given as

⨏

P 𝑓 (𝜃) =

⨏

S𝑛−1

𝑓 (𝜗) d𝜗 + 𝑛

⨏

S𝑛−1

𝜃 · 𝜗 𝑓 (𝜗) d𝜗,

where

denote the average integral.

104

Let B denote the integro-differential operator on the right hand side of (2.1), i.e. the RTE can

be written as

B𝑢(𝑥, 𝜃) = 𝑆(𝑥, 𝜃).

The diffusion approximation is given by

PBP𝑢(𝑥, 𝜃) = P𝑆(𝑥, 𝜃).

Denote

P𝑢(𝑥, 𝜃) =

P𝑆(𝑥, 𝜃) =

⨏

S𝑛−1

⨏

S𝑛−1

𝑢(𝑥, 𝜗) d𝜗 + 𝑛

𝑆(𝑥, 𝜗) d𝜗 + 𝑛

⨏

S𝑛−1

⨏

S𝑛−1

𝜃 · 𝜗𝑢(𝑥, 𝜗) d𝜗 (cid:67) 𝜙(𝑥) + 𝑛𝜃 · 𝐽 (𝑥),

𝜃 · 𝜗𝑆(𝑥, 𝜗) d𝜗 (cid:67) 𝑆0(𝑥) + 𝑛𝜃 · 𝑆1(𝑥),

(A.2)

(A.3)

Lemma A.2. The explicit form is given by

PBP𝑢(𝑥, 𝜃) = (𝜎(𝑥) − 𝜎𝑠 (𝑥))𝜙(𝑥) + ∇ · 𝐽 (𝑥) + 𝑛𝜃 ·

∇𝜙(𝑥) + (𝜎(𝑥)𝐼 − 𝐵(𝑥))𝐽 (𝑥)

(cid:19)

,

(A.4)

(cid:18) 1
𝑛

where 𝐼 is 𝑛 × 𝑛 identity matrix, 𝐵(𝑥) is a 𝑛 × 𝑛 matrix with entries

𝐵𝑖 𝑗 (𝑥) =

𝑛
Vol(S𝑛−1)

∫

∫

S𝑛−1

S𝑛−1

𝜃𝑖𝜗 𝑗 𝑘 (𝑥, 𝜃, 𝜗) d𝜃 d𝜗.

Proof. We first introduce following identities from the symmetricity

⨏

S𝑛−1

⨏

S𝑛−1

⨏

S𝑛−1

𝜃𝑖 d𝜃 = 0,

𝜃𝑖𝜃 𝑗 d𝜃 =

𝛿𝑖 𝑗
𝑛

,

𝜃𝑖𝜃 𝑗 𝜃 𝑘 d𝜃 = 0,

where 1 ≤ 𝑖, 𝑗, 𝑘 ≤ 𝑛, 𝛿𝑖 𝑗 is the Kronecker delta.

For any fixed 𝑥, it is clear that 𝜃 · ∇𝜙(𝑥), 𝜎(𝑥) (𝜙(𝑥) + 𝑛𝜃 · 𝐽 (𝑥)) ∈ H1, we have

P [𝜃 · ∇𝜙(𝑥)] = 𝜃 · ∇𝜙(𝑥),

P [𝜎(𝑥) (𝜙(𝑥) + 𝑛𝜃 · 𝐽 (𝑥))] = 𝜎(𝑥) (𝜙(𝑥) + 𝑛𝜃 · 𝐽 (𝑥)).

(A.5)

Since

𝜃 · ∇(𝜃 · 𝐽 (𝑥)) = 𝜃⊤ 𝐴(𝑥)𝜃

105

where 𝐴(𝑥) denote the Jacobian of 𝐽 (𝑥), we have

𝜗 · ∇(𝜗 · 𝐽 (𝑥)) d𝜗 =

⨏

𝑛
∑︁

S𝑛−1

𝑖, 𝑗=1

⨏

S𝑛−1

⨏

(𝜃 · 𝜗)(𝜗 · ∇(𝜗 · 𝐽 (𝑥))) d𝜗 =

S𝑛−1

𝐴𝑖 𝑗 (𝑥)𝜗𝑖𝜗 𝑗 d𝜗 =

1
𝑛 tr𝐴(𝑥) =

1
𝑛

∇ · 𝐽 (𝑥),

⨏

𝑛
∑︁

S𝑛−1

𝑖, 𝑗,𝑘=1

𝐴𝑖 𝑗 (𝑥)𝜗𝑖𝜗 𝑗 𝜗𝑘 𝜃 𝑘 d𝜗 = 0,

thus

P [𝜃 · ∇(𝜙(𝑥) + 𝑛𝜃 · 𝐽 (𝑥))] = 𝜃 · ∇𝜙(𝑥) + ∇ · 𝐽 (𝑥).

(A.6)

Consider the integral operator in B. Since

∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜗) d𝜃 =

∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜗) d𝜗 = 𝜎𝑠 (𝑥),

we conclude

⨏

∫

⨏

S𝑛−1
∫

S𝑛−1

⨏

S𝑛−1
∫

S𝑛−1

S𝑛−1

⨏

S𝑛−1
∫

S𝑛−1

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜗)𝜙(𝑥) d𝜗 d𝜃 = 𝜎𝑠 (𝑥)𝜙(𝑥)

𝑘 (𝑥, 𝜃, 𝜗)𝜗 · 𝐽 (𝑥) d𝜗 d𝜃 = 𝜎𝑠 (𝑥)

𝜃 · 𝜗𝑘 (𝑥, 𝜗, 𝜗′)𝜙(𝑥) d𝜗′ d𝜗 = 𝜎𝑠 (𝑥)𝜙(𝑥)

⨏

𝜗 · 𝐽 (𝑥) d𝜗 = 0
S𝑛−1
⨏

𝜃 · 𝜗 d𝜗 = 0

S𝑛−1

𝜃 · 𝜗𝑘 (𝑥, 𝜗, 𝜗′)𝜗′ · 𝐽 (𝑥) d𝜗′ d𝜗 = 𝜃⊤

⨏

(cid:20)

∫

S𝑛−1

S𝑛−1

𝑘 (𝑥, 𝜗, 𝜗′)𝜗𝜗′⊤ d𝜗′ d𝜗

(cid:21)

𝐽 (𝑥)

=

1
𝑛

𝜃⊤𝐵(𝑥)𝐽 (𝑥)

thus

P

(cid:20)∫

S𝑛−1

𝑘 (𝑥, 𝜃, 𝜗)(𝜙(𝑥) + 𝑛𝜗 · 𝐽 (𝑥)) d𝜗

(cid:21)

= 𝜎𝑠 (𝑥)𝜙(𝑥) + 𝑛𝜃 · 𝐵(𝑥)𝐽 (𝑥)

Combining (A.5) (A.6) (A.7) gives (A.4).

(A.7)

□

Lemma A.3. 𝐵(𝑥) is positive definite with eigenvalues in [0, 𝜎𝑠 (𝑥)] for each 𝑥 ∈ 𝑋.

Proof. Since 𝑘 (𝑥, 𝜃, 𝜗) = 𝑘 (𝑥, −𝜗, −𝜃) ≥ 0, we have 𝐵𝑖 𝑗 (𝑥) = 𝐵 𝑗𝑖 (𝑥), i.e. 𝐵(𝑥) is symmetric. For

106

arbitrary vector 𝜔 ∈ S𝑛−1,

𝜔⊤𝐵(𝑥)𝜔 =

=

=

=

𝑛
Vol(S𝑛−1)
𝑛
Vol(S𝑛−1)

+

𝑛
Vol(S𝑛−1)
∬

𝑛
Vol(S𝑛−1)

+

𝑛
Vol(S𝑛−1)
∬

𝑛
Vol(S𝑛−1)

≥0,

(𝜔·𝜃)(𝜔·𝜗)≥0
∬

(𝜔·𝜃)(𝜔·𝜗)<0

(𝜔·𝜃)(𝜔·𝜗)≥0
∬

(𝜔·𝜃)(𝜔·𝜗)<0

(𝜔·𝜃)(𝜔·𝜗)≥0

∫

∫

S𝑛−1

S𝑛−1

∬

(𝜔 · 𝜃)𝑘 (𝑥, 𝜃, 𝜗) (𝜔 · 𝜗) d𝜃 d𝜗

(𝜔 · 𝜃)𝑘 (𝑥, 𝜃, 𝜗) (𝜔 · 𝜗) d𝜃 d𝜗

(𝜔 · 𝜃)𝑘 (𝑥, 𝜃, 𝜗) (𝜔 · 𝜗) d𝜃 d𝜗

(𝜔 · 𝜃)𝑘 (𝑥, 𝜃, 𝜗) (𝜔 · 𝜗) d𝜃 d𝜗

(𝜔 · 𝜃)𝑘 (𝑥, 𝜃, 𝜗) (𝜔 · −𝜗) d𝜃 d(−𝜗)

(𝜔 · 𝜃) [𝑘 (𝑥, 𝜃, 𝜗) + 𝑘 (𝑥, 𝜃, −𝜗)] (𝜔 · 𝜗) d𝜃 d𝜗

𝜔⊤𝐵(𝑥)𝜔
𝑛
Vol(S𝑛−1)
𝑛
Vol(S𝑛−1)

𝑛𝜎𝑠 (𝑥)
Vol(S𝑛−1)
𝑛𝜎𝑠 (𝑥)
Vol(S𝑛−1)

=

≤

≤

=

=𝜎𝑠 (𝑥).

∬

S𝑛−1×S𝑛−1

√︄∬

(cid:104)

(𝜔 · 𝜃)√︁𝑘 (𝑥, 𝜃, 𝜗)

(cid:105) (cid:104)

(𝜔 · 𝜗)√︁𝑘 (𝑥, 𝜃, 𝜗)

(cid:105)

d𝜃 d𝜗

(𝜔 · 𝜃)2𝑘 (𝑥, 𝜃, 𝜗) d𝜃 d𝜗

∬

S𝑛−1×S𝑛−1

(𝜔 · 𝜗)2𝑘 (𝑥, 𝜃, 𝜗) d𝜃 d𝜗

S𝑛−1×S𝑛−1

√︄∫

S𝑛−1

(𝜔 · 𝜃)2 d𝜃

∫

S𝑛−1

(𝜔 · 𝜗)2 d𝜗

∫

S𝑛−1

1 d𝜃
𝜃2

we conclude 𝐵(𝑥) is positive definite with eigenvalues in [0, 𝜎𝑠 (𝑥)].

□

Proposition A.4. The diffusion approximation of (2.1) is given by

−∇ · 𝐷 (𝑥)∇𝜙(𝑥) + 𝜎𝑎 (𝑥)𝜙(𝑥) = 𝑞(𝑥),

where

𝐷 (𝑥) =

1
𝑛

(𝜎(𝑥)𝐼 − 𝐵(𝑥))−1,

𝜎𝑎 (𝑥) = 𝜎(𝑥) − 𝜎𝑠 (𝑥),

𝑞(𝑥) = 𝑆0(𝑥) − 𝑛∇ · 𝐷 (𝑥)𝑆1(𝑥).

107

Proof. According to (A.4)

gives

PBP𝑢(𝑥, 𝜃) = P𝑆(𝑥, 𝜃),

(𝜎(𝑥) − 𝜎𝑠 (𝑥))𝜙(𝑥) + ∇ · 𝐽 (𝑥) + 𝑛𝜃 ·

∇𝜙(𝑥) + (𝜎(𝑥)𝐼 − 𝐵(𝑥))𝐽 (𝑥)

(cid:19)

= 𝑆0(𝑥) + 𝑛𝜃 · 𝑆1(𝑥),

(cid:18) 1
𝑛

thus

which gives

𝑆0(𝑥) = (𝜎(𝑥) − 𝜎𝑠 (𝑥))𝜙(𝑥) + ∇ · 𝐽 (𝑥) = 𝜎𝑎 (𝑥)𝜙(𝑥) + ∇ · 𝐽 (𝑥),

𝑆1(𝑥) =

1
𝑛

∇𝜙(𝑥) + (𝜎(𝑥)𝐼 − 𝐵(𝑥))𝐽 (𝑥) =

1
𝑛

(∇𝜙(𝑥) + 𝐷−1(𝑥)𝐽 (𝑥)),

𝑆0(𝑥) = 𝜎𝑎 (𝑥)𝜙(𝑥) + ∇ · 𝐽 (𝑥) = 𝜎𝑎 (𝑥)𝜙(𝑥) + ∇ · 𝐷 (𝑥) (𝑛𝑆1(𝑥) − ∇𝜙(𝑥)),

or equivalently

−∇ · 𝐷 (𝑥)∇𝜙(𝑥) + 𝜎𝑎 (𝑥)𝜙(𝑥) = 𝑞(𝑥).

Remark A.5. When 𝜎(𝑥) > 𝜎𝑠 (𝑥) or the eigenvalues of 𝐵(𝑥) are strictly smaller than 𝜎𝑠 (𝑥) for

any 𝑥 ∈ 𝑋, 𝐷 (𝑥) is well defined.

□

Proposition A.6. 𝐷 (𝑥) is isotropic if 𝑘 (𝑥, 𝜃, 𝜗) is invariant under rotation

Proof. When 𝑘 (𝑥, 𝜃, 𝜗) is invariant under rotation,

𝐵𝑖 𝑗 =

𝑛
Vol(S𝑛−1)

∫

∫

S𝑛−1

S𝑛−1

𝜃𝑖𝜗 𝑗 𝑘 (𝑥, 𝜃 · 𝜗) d𝜃 d𝜗.

From the symmetricity, 𝐵𝑖 𝑗 = 0 if 𝑖 ≠ 𝑗, and the diagonal terms are identical:

𝐵𝑖𝑖 (𝑥) =

=

1
𝑛 tr𝐵(𝑥) =
∫

S𝑛−1

1
Vol(S𝑛−1)

∫

∫

S𝑛−1

S𝑛−1

𝜃 · 𝜗𝑘 (𝑥, 𝜃 · 𝜗) d𝜃 d𝜗

𝜃 · 𝜗𝑘 (𝑥, 𝜃 · 𝜗) d𝜃 =

∫ 1

−1

𝑡𝑘 (𝑥, 𝑡)Vol(S𝑛−2) (1 − 𝑡2)

𝑛−3
2 d𝑡 (cid:67) 𝑏(𝑥).

Thus 𝐵(𝑥) = 𝑏(𝑥)𝐼 if 𝑘 (𝑥, 𝜃, 𝜗) is invariant under rotation, which implies 𝐷 (𝑥) is isotropic.

□

108

Proposition A.7. The diffusion approximation of (2.2) is given by

where

𝜙(𝑥) + 𝛾𝜈 · 𝐷 (𝑥)∇𝜙(𝑥) = 0,

𝛾 =

√

𝜋(𝑛 − 1)Γ
(cid:1)
2Γ (cid:0) 𝑛
2

(cid:17)

(cid:16) 𝑛−1
2

.

Proof. The photon flux intensity at 𝑥 ∈ 𝜕 𝑋 into the body is

Φ−(𝑥) = −

∫

𝜃·𝜈<0

𝑢(𝑥, 𝜃)𝜃 · 𝜈 d𝜃 = 0.

With the diffusion approximation, it is

∫

−

𝜃·𝜈<0

(𝜙(𝑥) + 𝑛𝜃 · 𝐽 (𝑥))𝜃 · 𝜈 d𝜃 = 0.

Since

∫

𝜃·𝜈<0

∫

−

𝜃·𝜈<0

𝜃 · 𝜈 d𝜃 =

∫

S𝑛−1

1
2

|𝜃1| d𝜃 =

1
2

∫ 1

−1

|𝑡|Vol(S𝑛−2) (1 − 𝑡2)

𝑛−3
2 d𝑡 =

Vol(S𝑛−2)
𝑛 − 1

,

𝜃 · 𝐽 (𝑥)𝜃 · 𝜈 d𝜃 = 𝑣⊤

(cid:20)∫

𝜃·𝜈<0

(cid:21)

𝜃𝜃⊤ d𝜃

𝐽 (𝑥) =

𝑣⊤

1
2

(cid:20)∫

S𝑛−1

(cid:21)

𝜃𝜃⊤ d𝜃

𝐽 (𝑥) =

Vol(S𝑛−1)
2𝑛

𝜈 · 𝐽 (𝑥),

we conclude

𝜙(𝑥) =

(𝑛 − 1)Vol(S𝑛−1)
2Vol(S𝑛−2)

𝜈 · 𝐽 (𝑥) =

(𝑛 − 1)Vol(S𝑛−1)
2Vol(S𝑛−2)

𝜈 · 𝐷 (𝑥) (𝑛𝑆1(𝑥) − ∇𝜙(𝑥)).

Since 𝑆(𝑥, 𝜃) is either independent of direction 𝜃, or compact supported on 𝑋 for any 𝜃 ∈ S𝑛−1,

𝑆1(𝑥) = 0, thus

𝜙(𝑥) +

(𝑛 − 1)Vol(S𝑛−1)
2Vol(S𝑛−2)

𝜈 · 𝐷 (𝑥)∇𝜙(𝑥) = 𝜙(𝑥) + 𝛾𝜈 · 𝐷 (𝑥)∇𝜙(𝑥) = 0.

□

Thus the diffusion approximation of RTE is given by the diffusion equation with Robin boundary

condition.

109

APPENDIX B

APPENDIX FOR CHAPTER 3

B.1 Adjoint of ND Map

Lemma B.1. Suppose Λ𝜌,𝑞 is the ND map of following wave equation

□𝜌,𝑞𝑢(𝑡, 𝑥) = 0

in [0, 2𝑇] × Ω,

𝑢(0, 𝑥) = 𝜕𝑡𝑢(0, 𝑥) = 0

on Ω,

(B.1)

𝜕𝜈𝑢(𝑡, 𝑥) = 𝑓

on [0, 2𝑇] × 𝜕Ω,






we have Λ∗

𝜌,𝑞 = 𝑅Λ𝜌,𝑞 𝑅.

Proof. Let 𝑣 denote the solution of the following adjoint wave equation

□𝜌,𝑞𝑣(𝑡, 𝑥) = 0

in [0, 2𝑇] × Ω,

𝑣(2𝑇, 𝑥) = 𝜕𝑡𝑣(2𝑇, 𝑥) = 0

on Ω,

(B.2)

𝜕𝜈𝑣(𝑡, 𝑥) = 𝑔

on [0, 2𝑇] × 𝜕Ω,






with 𝑔 in 𝐿2((0, 2𝑇) × 𝜕Ω), where 𝐿∗ = −∇ · 𝐴∇ − 𝑏 · ∇ + 𝑐 is the adjoint operator of 𝐿. Let Λ∗

denote the ND map of the adjoint equation.

The weak formulation gives

∫

∫ 2𝑇

Ω

0

Define

[𝜌(𝑥)𝜕2

𝑡 𝑢 𝑓 (𝑡, 𝑥) − Δ𝑢 𝑓 (𝑡, 𝑥) + 𝑞(𝑥)𝑢 𝑓 (𝑡, 𝑥)]𝑣(𝑡, 𝑥) d𝑡 d𝑥 = 0.

𝐼1 (cid:66)

∫

∫ 2𝑇

Ω

0

𝜌(𝑥)𝑢 𝑓

𝑡𝑡 (𝑡, 𝑥)𝑣(𝑡, 𝑥) d𝑡 d𝑥 =

∫

∫ 2𝑇

Ω

0

𝜌(𝑥)𝑢 𝑓 (𝑡, 𝑥)𝑣𝑡𝑡 (𝑡, 𝑥) d𝑡 d𝑥,

∫

∫ 2𝑇

Ω
∫

0
∫ 2𝑇

𝐼2 (cid:66)

=

[−Δ𝑢 𝑓 (𝑡, 𝑥) + 𝑞(𝑥)𝑢 𝑓 (𝑡, 𝑥)]𝑣(𝑡, 𝑥) d𝑡 d𝑥

[−Δ𝑣(𝑡, 𝑥) + 𝑞(𝑥)𝑣(𝑡, 𝑥)]𝑢 𝑓 (𝑡, 𝑥) d𝑡 d𝑥

0

Ω
+ (Λ𝜌,𝑞 𝑓 , 𝑔)𝐿2 ((0,2𝑇)×𝜕Ω) − ( 𝑓 , Λ∗

𝜌,𝑞𝑔)𝐿2 ((0,2𝑇)×𝜕Ω)

110

thus

𝐼1 + 𝐼2 = (Λ𝜌,𝑞 𝑓 , 𝑔)𝐿2 ((0,2𝑇)×𝜕Ω) − ( 𝑓 , Λ∗

𝜌,𝑞𝑔)𝐿2 ((0,2𝑇)×𝜕Ω) = 0,

which implies Λ∗

𝜌,𝑞 is the adjoint operator of Λ𝜌,𝑞 in 𝐿2((0, 2𝑇) × 𝜕Ω). Notice that the solution of

equation

□𝜌,𝑞𝑢(𝑡, 𝑥) = 0

in [0, 2𝑇] × Ω,

𝑢(0, 𝑥) = 𝜕𝑡𝑢(0, 𝑥) = 0

on Ω,

(B.3)

𝜕𝜈𝑢(𝑡, 𝑥) = 𝑅𝑔

on [0, 2𝑇] × 𝜕Ω,






is the time reversed adjoint solution of (B.2), which means the adjoint operator can be represent as

Λ∗

𝜌,𝑞 = 𝑅Λ𝜌,𝑞 𝑅

□

B.2 Frechét Differentiability of ND map Λ𝑞

In this section, we collect a few results that are used in the main text. First, we provide the

rigorous justification for the formal linearization process in the introduction to derive (3.63) (3.64)

(3.65). Recall that 𝑐0 ∈ 𝐶∞(Ω).

For 𝑓 ∈ 𝐿2((0, 2𝑇) × 𝜕Ω), the solution 𝑢 = 𝑢 𝑓 of the boundary value problem (3.1) satisfies

𝑢 ∈ 𝐶 ([0, 2𝑇]; 𝐻5/6−𝜀 (Ω)) for any 𝜀 > 0 with the norm estimate [61]

∥𝑢∥𝐶 ([0,2𝑇];𝐻5/6− 𝜀 (Ω)) ≤ 𝐶 ∥ 𝑓 ∥ 𝐿2 ((0,2𝑇)×𝜕Ω)

(B.4)

where ∥𝑢 𝑓 ∥𝐶 ([0,2𝑇];𝐻5/6− 𝜀 (Ω))
𝐿2((0, 2𝑇) × 𝜕Ω) → 𝐿2((0, 2𝑇) × 𝜕Ω) is a bounded linear operator.

:= ess sup0≤𝑡≤2𝑇 ∥𝑢(𝑡) ∥𝐻5/6− 𝜀 (Ω). As a result, the ND map Λ𝑞 :

Denote by L (𝐿2((0, 2𝑇)×𝜕Ω), 𝐿2((0, 2𝑇)×𝜕Ω)) the Banach space of bounded linear operators

over 𝐿2((0, 2𝑇) × 𝜕Ω). The IBVP aims to invert the following nonlinear map

F :

𝑞 ∈ 𝐿∞(Ω) ↦→ Λ𝑞 ∈ L (𝐿2((0, 2𝑇) × 𝜕Ω), 𝐿2((0, 2𝑇) × 𝜕Ω))

Suppose 𝑞 = 𝑞0 + (cid:164)𝑞 with 𝑞0 ∈ 𝐶∞(Ω). Define a linear operator (which will turn out to be the

Frechét differentiation of F ):

𝑑F :

(cid:164)𝑞 ∈ 𝐿∞(Ω) ↦→ (cid:164)Λ (cid:164)𝑞 ∈ L (𝐿2((0, 2𝑇) × 𝜕Ω), 𝐿2((0, 2𝑇) × 𝜕Ω)).

111

where (cid:164)Λ (cid:164)𝑞 is the operator defined in (3.65).

Proposition B.2. The nonlinear map F is Frechét differentiable at a fixed 𝑞0 ∈ 𝐶∞(Ω), and the
Frechét derivative along (cid:164)𝑞 ∈ 𝐿∞(Ω) is (cid:164)Λ (cid:164)𝑞.

Proof. It suffices to show that as ∥ (cid:164)𝑞∥ 𝐿∞ (Ω) → 0, we have

∥F (𝑞) − F (𝑞0) − 𝑑F ( (cid:164)𝑞) ∥L (𝐿2 ((0,2𝑇)×𝜕Ω),𝐿2 ((0,2𝑇)×𝜕Ω)) = 𝑂 (∥ (cid:164)𝑞∥2

𝐿∞ (Ω))

(or equivalently, ∥Λ𝑞 − Λ𝑞0 − (cid:164)Λ (cid:164)𝑞 ∥L (𝐿2 ((0,2𝑇)×𝜕Ω),𝐿2 ((0,2𝑇)×𝜕Ω)) = 𝑂 (∥ (cid:164)𝑞∥2
𝐿∞ (Ω))) to justify that 𝑑F
is indeed the Frechét differentiation of F . To this end, we will prove for any 𝑓 ∈ 𝐿2((0, 2𝑇) × 𝜕Ω)

that

∥Λ𝑞 𝑓 − Λ𝑞0

𝑓 − (cid:164)Λ (cid:164)𝑞 𝑓 ∥ 𝐿2 ((0,2𝑇)×𝜕Ω) ≤ 𝐶 ∥ (cid:164)𝑞∥2

𝐿∞ ∥ 𝑓 ∥ 𝐿2 ((0,2𝑇)×𝜕Ω)

(B.5)

for some constant 𝐶 > 0 that is independent of 𝑓 . For ease of notation, we will denote all the

constants independent of 𝑓 by 𝐶.

We continue to denote the solutions of (3.1) and (3.63) by 𝑢 and 𝑢0, respectively. Write

𝑢 = 𝑢0 + 𝛿𝑢. Then 𝛿𝑢 satisfies 𝛿𝑢| [0,2𝑇]×𝜕Ω = Λ𝑞 𝑓 − Λ𝑞0

𝑓 and

□𝑐0,𝑞0

𝛿𝑢(𝑡, 𝑥) = −𝑢 (cid:164)𝑞,

in (0, 2𝑇) × Ω

𝜕𝜈𝛿𝑢 = 0,

on (0, 2𝑇) × 𝜕Ω

(B.6)

𝛿𝑢(0, 𝑥) = 𝜕𝑡𝛿𝑢(0, 𝑥) = 0

𝑥 ∈ Ω.






Using the regularity estimate for the wave equation [40] and the trace theorem, we obtain

∥𝛿𝑢∥𝐻1 ((0,2𝑇)×Ω) ≤ 𝐶 ∥𝑢 (cid:164)𝑞∥ 𝐿2 ((0,2𝑇)×Ω) ≤ 𝐶 ∥𝑢∥ 𝐿2 ((0,2𝑇)×Ω) ∥ (cid:164)𝑞∥ 𝐿∞ (Ω).

(B.7)

Next, set 𝑤 := 𝛿𝑢 − (cid:164)𝑢, then 𝑤| [0,2𝑇]×𝜕Ω = Λ𝑞 𝑓 − Λ𝑞0

𝑓 − (cid:164)Λ (cid:164)𝑞 𝑓 , and 𝑤 satisfies

□𝑐0,𝑞0

𝑤(𝑡, 𝑥) = − (cid:164)𝑞𝛿𝑢,

in (0, 2𝑇) × Ω

𝜕𝜈𝑤 = 0,

on (0, 2𝑇) × 𝜕Ω

(B.8)

𝑤(0, 𝑥) = 𝜕𝑡𝑤(0, 𝑥) = 0

𝑥 ∈ Ω.






112

Applying the regularity estimate for the wave equation again yields

∥Λ𝑞 𝑓 − Λ𝑞0

𝑓 − Λ (cid:164)𝑞 𝑓 ∥ 𝐿2 ((0,2𝑇)×𝜕Ω) ≤ ∥𝑤∥𝐻1 ((0,2𝑇)×Ω) ≤ 𝐶 ∥ (cid:164)𝑞𝛿𝑢∥ 𝐿2 ((0,2𝑇)×Ω)

≤ ∥ (cid:164)𝑞∥ 𝐿∞ (Ω) ∥𝛿𝑢∥ 𝐿2 ((0,2𝑇)×Ω).

Combining the estimates (B.4) (B.7) (B.9) yields the desired estimate (B.5).

B.3 Frechét Differentiability of ND map Λ𝜌

According to [61, Theorem A], if 𝑓 (0) = 𝜕𝑡 𝑓 (0) = · · · = 𝜕 𝑘−1

𝑡

𝑓 (0) = 0, we have

∥𝑢∥

𝐶 ([0,2𝑇],𝐻 𝑘+ 3

5

− 𝜀

(Ω))

≤ 𝐶 ∥ 𝑓 ∥𝐻 𝑘 ((0,2𝑇)×𝜕Ω),

where 𝑘 ≥ 0, 𝜀 is an arbitrary positive real number. Thus we have

(B.9)

□

∥𝑢∥𝐻3 ((0,2𝑇)×Ω) ≤ 𝐶 ∥ 𝑓 ∥

,

5
2

𝐻

(B.10)

then the linearized ND map should be in Banach space

L (𝐻 5

2 ((0, 2𝑇) × 𝜕Ω), 𝐻 1

2 ((0, 2𝑇) × 𝜕Ω))

The IBVP aims to invert the following nonlinear map

F : 𝜌 ∈ 𝐶∞(Ω) ↦→ Λ𝜌 ∈ L (𝐻 3

2 ((0, 2𝑇) × 𝜕Ω), 𝐻 1

2 ((0, 2𝑇) × 𝜕Ω))

Assuming that 𝜌 = 𝜌0 + (cid:164)𝜌 with 𝜌0 ∈ 𝐶∞(Ω), define the following linear operator

dF : (cid:164)𝜌 ∈ 𝐶∞(Ω) ↦→ (cid:164)Λ (cid:164)𝜌 ∈ L (𝐻 3

2 ((0, 2𝑇) × 𝜕Ω), 𝐻 1

2 ((0, 2𝑇) × 𝜕Ω))

where (cid:164)Λ (cid:164)𝜌 is the linearized ND map defined in (3.65).

Proposition B.3. The nonlinear map F is Frechét differentiable at 𝜌0 ∈ 𝐶∞(Ω), and the Frechét
derivative along the direction (cid:164)𝜌 ∈ 𝐶∞(Ω) is (cid:164)Λ (cid:164)𝜌.

Proof. In order to show that F is Frechét differentiable, we need to show

∥F (𝜌) − F (𝜌0) − dF ( (cid:164)𝜌)∥

L (𝐻

5
2 ((0,2𝑇)×𝜕Ω),𝐻

1
2 ((0,2𝑇)×𝜕Ω))

(cid:16)

= 𝑂

(cid:17)

∥ (cid:164)𝜌∥2

𝑊 1,∞ (Ω)

113

as ∥ (cid:164)𝜌∥𝑊 1,∞ (Ω) → 0, which is equivalent to

∥Λ𝜌 𝑓 − Λ𝜌0

𝑓 − (cid:164)Λ (cid:164)𝜌 𝑓 ∥

𝐻

1
2 ((0,2𝑇)×𝜕Ω)

(cid:16)

= 𝑂

∥ (cid:164)𝜌∥2

𝑊 1,∞ (Ω)

∥ 𝑓 ∥

𝐻

5
2 ((0,2𝑇)×𝜕Ω)

(cid:17)

(B.11)

for any 𝑓 ∈ 𝐻 5

2 ((0, 2𝑇) × 𝜕Ω) as ∥ (cid:164)𝜌∥𝑊 1,∞ (Ω) → 0.

Write 𝑢 = 𝑢0 + 𝛿𝑢, where 𝑢 and 𝑢0 are the solutions of (3.1) and (3.63), respectively. Then 𝛿𝑢

satisfy equation

𝜌0𝛿𝑢𝑡𝑡 − Δ𝛿𝑢 + 𝑞𝛿𝑢 = − (cid:164)𝜌𝑢𝑡𝑡

in (0, 2𝑇) × Ω

𝜕𝜈𝛿𝑢 = 0

on (0, 2𝑇) × 𝜕Ω

(B.12)

𝛿𝑢(0, 𝑥) = 𝛿𝑢𝑡 (0, 𝑥) = 0

𝑥 ∈ Ω






Using the regularity estimate for wave euation and the trace theorem, we have

∥𝛿𝑢∥𝐻2 ((0,2𝑇)×Ω) ≤ 𝐶 ∥ (cid:164)𝜌𝑢𝑡𝑡 ∥𝐻1 ((0,2𝑇)×Ω) ≤ 𝐶 ∥𝑢∥𝐻3 ((0,2𝑇)×Ω) ∥ (cid:164)𝜌∥𝑊 1,∞ (Ω)

(B.13)

Denote 𝑤 (cid:66) 𝛿𝑢 − (cid:164)𝑢, then 𝑤| [0,2𝑇]×𝜕Ω = Λ𝜌 𝑓 − Λ𝜌0

𝑓 − (cid:164)Λ (cid:164)𝜌 𝑓 and satisfies

𝜌0𝑤𝑡𝑡 − Δ𝑤 + 𝑞𝑤 = − (cid:164)𝜌𝛿𝑢𝑡𝑡

in (0, 2𝑇) × Ω

𝜕𝜈𝑤 = 0

on (0, 2𝑇) × 𝜕Ω

(B.14)

𝑤(0, 𝑥) = 𝑤𝑡 (0, 𝑥) = 0

𝑥 ∈ Ω






Applying similar estimate yields

∥𝑤∥

𝐻

1
2 ((0,2𝑇)×𝜕Ω)

≤ ∥𝑤∥𝐻1 ((0,2𝑇)×Ω) ≤ 𝐶 ∥ (cid:164)𝜌𝛿𝑢𝑡𝑡 ∥ 𝐿2 ((0,2𝑇)×Ω) ≤ 𝐶 ∥𝛿𝑢∥𝐻2 ((0,2𝑇)×Ω) ∥ (cid:164)𝜌∥ 𝐿∞ (Ω)

Combining (B.10) (B.13) (B.15) gives (B.11).

(B.15)

□

114