Some Inverse Problems in Hyperbolic Partial Di erential ... · a lot during my PhD under his...

Some Inverse Problems in Hyperbolic PartialDifferential Equations

A Thesis

Submitted to the

Tata Institute of Fundamental Research, Mumbai

for the Degree of Doctor of Philosophyin Mathematics

by

MANMOHAN VASHISTH

Centre for applicable Mathematics

Tata Institute of Fundamental ResearchBangalore

July, 2018

http:// math.tifrbng.res.in/http:// math.tifrbng.res.in/http:// math.tifrbng.res.in/

Declaration of Authorship

This thesis is a presentation of my original research work. Whenever contribu-

tions of others are involved, every effort is made to indicate this clearly, with due

reference to the literature and acknowledgement of collaborative research and dis-

cussions.

The work was done under the supervision of Dr. Venkateswaran P. Krishnan at

TIFR Centre for Applicable Mathematics, Bangalore.

MANMOHAN VASHISTH

In my capacity as supervisor of the candidates thesis, I certify that the above

statements are true to the best of my knowledge.

VENKATESWARAN P. KRISHNAN

Date:

iii

Dedicated to

my beloved parents and teachers for their Belief,

Support, Encouragement and Love.

v

Abstract

In this thesis, we study three inverse problems related to linear and non-linear hy-

perbolic partial differential equations (PDEs). Our particular interest is in unique

determination of the coefficients appearing in these PDEs from the knowledge of

functionals that depend on the coefficients and solutions of the PDEs.

In the first problem, we address the inverse problem of determining the density co-

efficient of a medium by probing it with an external point source and by measuring

the responses at a single point for a certain period of time.

In the second problem, we consider the inverse problem of determining time-

dependent vector and scalar potentials appearing in the wave equation in space

dimension n ≥ 3 from information about the solution on a suitable subset of theboundary cylinder.

In the third problem, we consider an inverse boundary value problem for a non-

linear wave equation of divergence form with space dimension n ≥ 3. We study theunique determination of quadratic non-linearity appearing in the wave equation

from measurements of the solution at the boundary of spacial domain over finite

time interval.

Acknowledgements

First and foremost, I would like to express my sincerest gratitude to my thesis

supervisor Dr. Venkateswaran P. Krishnan for his continuous support, unending

patience, care and encouragement that I received from him during my PhD. I

must say that I am very lucky to have had such a wonderful supervisor. I learned

a lot during my PhD under his guidance, and due to him I got opportunity to

visit many places and learn from many well-known mathematicians. Beside his

deep knowledge in various fields of mathematics, I appreciate his writing skills and

grateful to him for helping me polish my manuscripts. Apart from research work,

he provided me many valuable suggestions while preparing for talks. One simply

could not have wished for better or friendlier supervisor.

I would like to thank Professor Gen Nakamura (Emeritus Professor at Hokkaido

University, Japan) for giving me the opportunity to work with him and providing

me the financial support through JSPS-grant for visiting his place. I am very

grateful to him for several disscussions that helped me widen my knowledge in the

field of inverse problems. Finally, I am very thankful to him for allowing me to

include our joint work in my thesis.

I would like to thank Professor Rakesh (Delaware University) for several useful

discussions during his visits to TIFR-CAM. I would also like to thank Dr. Lauri

Oksanen (University College London) and Professor Paul Sacks (Iowa State Uni-

versity) for stimulating discussions on hyperbolic inverse problems when we met

in conferences.

I would also like to thank the members of my thesis committee: Professor K.T.

Joseph and Professor A. S. Vasudeva Murthy for their insightful comments and

discussions during my informal presentations.

It has been a great privilege to pursue the integrated PhD programme at TIFR-

CAM. I would like to thank all my teachers at CAM for teaching courses on basic

and advanced level mathematics with utmost excellence. I learned a lot from

these courses which helped me immensely through my research. I am thankful

to Professor Adimurthi, Dr. Shyam Goshal and Dr. Venky Krishnan for giving

me teaching assistantship in their courses. I am especially thankful to Dr. Imran

and Prof. Mythily for boosting my confidence during the starting days in TIFR-

CAM. I would like to thank TIFR-CAM, Airbus (EADS) and Japan Society for

ix

the Promotion of Sciences (JSPS) for providing me the funding and opportunity

to attend various conferences and to go for various research visits both abroad and

in India, which further enhanced my knowledge of mathematics.

I am very grateful to all my teachers Dr. Mahabir Barak, Ranbir Kadiyan and

Dr. Amit Sehgal at Pt. Neki Ram Sharma Govt. College, Rohtak, Haryana. A

very special thanks to Dr. Amit Sehgal who was responsible for developing my

interest for mathematics. Without their excellent and systematic teaching, and

encouragement, I would not have been able to be part of TIFR-CAM.

Next I would like to thank my seniors, colleagues, post-docs and project-assistants

at TIFR-CAM without whom life would have been much more difficult at CAM.

I thank my seniors: Dr. Aekta Aggarwal, Dr. Sudarshan Kumar, Dr. Manas

Ranjan Sahoo, Dr. Arnab Jyoti Das Gupta, Dr. Souvik Roy, Dr. Tuhin Ghosh,

Dr. Debdip Ganguly, Dr. Abhishek Sarkar, Dr. Debanjana Mitra, Dr. Debayan

Maity, Dr. Debabrata Karmakar, Dr. Indranil Chowdhury, Dr. Deep Ray, Dr.

Sombuddha Bhattacharya and Dr. Rohit Kumar Mishra for providing a good

atmosphere in TIFR-CAM from the starting days of my life at CAM. I specially

thank Rohit, Sombuddha and Aekta for providing me help, making life more

enjoyable and support me both emotionally and academically during my journey

in TIFR. I also would like to thank my colleagues: Manish, Arka, Arnab, Amrita,

Sourav (Baba), Raman, Sakshi, Amit, Jayesh, Nilasis, Suman, Saibal, Saikatul,

Neeraj, Abhishek, Ganesh, Ankur, Sahil, Akash and Ritesh for making life at

TIFR enjoyable. A special thanks to Neeraj for being a special part of my life

at TIFR. Finally, I would thank all the post-docs: Dr. Gyula, Dr. Tanmay, Dr.

Gururaja, Dr. Daisy, Dr. Kamana, Dr. Divyang, Dr. Bidhan, Dr. Ameya, Dr.

Ramesh, Dr. Rakesh, Dr. Asha, Dr. Sanjib, Dr. Peeyush, Dr. Ruchi and project

assistants: Ashish, Deepak and Sumanth for making my life at TIFR memorable.

I express my sincere thanks to all the administrative and scientific staff of TIFR

for their help and service. I would like to thank all the canteen, security and

cleaning staff who always extended their help to me all these years.

I have been very fortunate to have some good friends. I take this opportunity to

thank all of them specially Surjeet Kaushik, Sunderam, Mannu Kadiyan, Satish

Dagar, Tinku Jangra, Pradeep, Deepak Deswal, Jitender, Jyoti, Ritu and Shyam

Sunder for being the shoulders I could always depend on.

Finally, I would like to thank my family and relatives without whom it would not

have been possible to be what I am today. I also express my sincere gratitude

to my late grandmother (Dadi) and late grandfather (Dada) for their love and

blessings.

Contents

Declaration of Authorship iii

Abstract vii

Acknowledgements ix

1 Introduction 1

2 An Inverse Problem for the Wave Equation with Source and Re-ceiver at Distinct Points 5

2.1 Introduction and statement of the main result . . . . . . . . . . . . 5

2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 Proof of Theorems 2.1.1 and 2.1.2 . . . . . . . . . . . . . . . . . . . 8

2.3.1 Proof of Theorem 2.1.1 . . . . . . . . . . . . . . . . . . . . . 13

2.3.2 Proof of Theorem 2.1.2 . . . . . . . . . . . . . . . . . . . . . 15

3 An Inverse Problem for the Relativistic Schrödinger Equationwith Partial Boundary Data 17

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2 Statement of the main result . . . . . . . . . . . . . . . . . . . . . . 19

3.2.1 Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2.2 Statement of the main result . . . . . . . . . . . . . . . . . . 20

3.3 Carleman Estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.4 Construction of geometric optics solutions . . . . . . . . . . . . . . 32

3.5 Integral Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.6 Proof of Theorem 3.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.6.1 Recovery of vector potential A . . . . . . . . . . . . . . . . . 383.6.2 Recovery of potential q . . . . . . . . . . . . . . . . . . . . . 45

4 Inverse Boundary Value Problem for a Non-linear HyperbolicPartial Differential Equations 47

4.1 Introduction and statement of the main result . . . . . . . . . . . . 47

4.2 �-expansion of the solution to the IBVP . . . . . . . . . . . . . . . 51

xiii

Contents xiv

4.3 Analysis of DN map in �-expansion . . . . . . . . . . . . . . . . . . 57

4.4 Proof of Theorem 4.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.4.1 Proof for uniqueness of γ . . . . . . . . . . . . . . . . . . . . 59

4.4.2 Proof for uniqueness of cjkl(x) . . . . . . . . . . . . . . . . . 59

4.5 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5 Conclusion 73

Bibliography 75

Chapter 1

Introduction

We consider inverse problems related to linear and non-linear hyperbolic PDEs.

These problems arise in the study of seismic imaging, thermo acoustic and photo

acoustic tomography (TAT), ultrasound imaging, vibrating string, piezoelectricity,

love waves etc. (see [22, 29, 37, 46, 62]). We briefly describe the three problems

related to these PDEs as follows:

• An inverse problem for the wave equation with source and receiver at distinctpoints [74] : In this work, we study the inverse problem of determining the

coefficient q appearing in the wave equation

(�− q(x))u(x, t) = δ(x, t), (x, t) ∈ R3 × R

u(x, t)|t

Chapter 1. Introduction 2

• An inverse problem for the relativistic Schrödinger equation with partialboundary data [36] : In this joint work with Venkateswaran P. Krishnan,

we establish the unique determination of the time-dependent vector and

scalar potentials A := (A0, A1, · · · , An) and q respectively, appearing in thefollowing relativistic Schrödinger

LA,qu(t, x) :=(

(∂t + A0(t, x))2 −

n∑j=1

(∂j + Aj(t, x))2 + q(t, x)

)u(t, x)

from the measurements of the solution on a part of the boundary. Let Ω ⊂ Rn

for n ≥ 3 be a bounded open set with C2 boundary. Let Q = (0, T )×Ω andΣ = (0, T )× ∂Ω. For a fixed ω ∈ Sn−1, define

∂Ω±,ω := {x ∈ ∂Ω : ±ω · ν(x) ≥ 0}

and denote by Σ±,ω := (0, T )× ∂Ω±,ω. Consider the following initial bound-ary value problem

LA,qu(t, x) = 0; (t, x) ∈ Q

u(0, x) = φ(x), ∂tu(0, x) = ψ(x); x ∈ Ω

u(t, x) = f(t, x); (t, x) ∈ Σ

(1.0.2)

where Aj ∈ C∞c (Q), for j = 0, 1, · · · , n and q ∈ L∞(Q). For φ ∈ H1(Ω), ψ ∈L2(Ω) and f ∈ H1(Σ) such that the compatibility condition f(0, x) = φ(x)∀ x ∈ ∂Ω is satisfied, it is shown in [32, 38] that Equation (1.0.2) admits aunique solution u with

u ∈ C([0, T ];H1(Ω)) ∩ C1([0, T ];L2(Ω)) and ∂νu ∈ L2(Σ).

Therefore we can define our input-output operator

ΛA,q : H1(Ω)× L2(Ω)×H1(Σ) → H1(Ω)× L2(G)

by

ΛA,q(φ, ψ, f) = (∂νu|G, u|t=T ) (1.0.3)


where u is the solution to (3.2.3) and G = (0, T ) × G′ with G′ a neigh-bourhood of ∂Ω−,ω in ∂Ω. We are interested in the unique-determination

of the coefficients A and q appearing in (1.0.2) from the knowledge of theinput-output operator ΛA,q. As follows from the work of [64], one cannot

determine the coefficients uniquely given ΛA,q because of gauge invariance

(see 3.2.2). In our work, we prove unique determination of scalar potential

q, and the vector potential term A modulo a gauge invariant term given theinput-output operator ΛA,q. For more details, we refer to Theorem 3.2.2 in

Chapter 3.

• Inverse boundary value problem for a non-linear hyperbolic partial differentialequations [44] : In this joint work with Gen Nakamura (Emeritus Professor

at Hokkaido University, Sapporo, Japan), we consider the inverse problem

of determining the quadratic non-linearity appearing in the wave equation∂2t u(t, x)−∇x · ~C(x,∇xu(t, x)) = 0, (t, x) ∈ QT := (0, T )× Ω,

u(0, x) = ∂tu(0, x) = 0, x ∈ Ω,

u = �f(t, x), (t, x) ∈ ∂QT := (0, T )× ∂Ω,

(1.0.4)

where ~C(x,∇xu) is given by

~C(x,∇xu) := γ(x)∇xu

+

(n∑

k,l=1

c1kl∂ku∂lu,n∑

k,l=1

c2kl∂ku∂lu, · · · ,n∑

k,l=1

cnkl∂ku∂lu

)︸︷︷︸

~P

+ ~R(x,∇xu), (1.0.5)

here c ≤ γ(x) ∈ C∞(Ω) for some constant c > 0 and each cikl ∈ C∞(Ω).

Now for f ∈ C∞c (∂QT ), we have for any T > 0, there exists �0 > 0 such thatfor any 0 < � < �0, the Equation (1.0.4) has a unique solution u (see [44, 46])

satisfying

u ∈m⋂j=0

Cj([0, T ];Hm−j(Ω)

).


We define the Dirichlet to Neumann (DN) map ΛT~C by

ΛT~C(�f) = ν(x) · ~C(x,∇xuf )|∂QT , f ∈ C∞c (∂QT ), 0 < � < �0 (1.0.6)

where uf (t, x) is the solution to (1.0.4) and ν(x) is the outward unit normal

to ∂Ω at x ∈ ∂Ω. The inverse problem we consider is the uniqueness ofidentifying ~C from the DN map ΛT~C . More precisely, we show that if the

DN maps ΛT~Ci, i = 1, 2 given by (1.0.6) for ~C = ~Ci, i = 1, 2 are same, then

(γi, ~Pi), i = 1, 2 are the same, where (γi, ~Pi), i = 1, 2 are (γ, ~P ) associated

to ~Ci, i = 1, 2. For more details, we refer to Theorem 4.1.2 in Chapter 4.

This thesis is organized as follows. In Chapter 2, we give details of our work on

inverse problems for the wave equation with source and receiver at distinct points.

Chapter 3 provides the details of our work related to unique-determination of the

time-dependent vector and scalar potentials in (1.0.2) from the measurements of

the solutions on a part of the boundary. In Chapter 4, we give details of our

work on determining the non-linearity appearing in the non-linear wave equation

(1.0.4).

Chapter 2

An Inverse Problem for the Wave

Equation with Source and

Receiver at Distinct Points

2.1 Introduction and statement of the main re-

sult

In this chapter, we address the inverse problem of determining the density coeffi-

cient of a medium by probing it with an external point source and by measuring

the responses at a single point for a certain period of time.

More precisely, consider the following initial value problem (IVP):

(�− q(x))u(x, t) = δ(x, t), (x, t) ∈ R3 × R

u(x, t)|t 1. Motivation for studying such problems arises in geophysics see [73]

and references therein. Geophysicists determine properties of the earth structure

5

Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 6

by sending waves from the surface of the earth and measuring the corresponding

scattered responses. Note that in the problem we consider here, the point source is

located at the origin, whereas the responses are measured at a different point. Since

the given data depends on one variable whereas the coefficient to be determined

depends on three variables, some additional restrictions on the coefficient q are

required to make the inverse problems tractable. There are several results related

to inverse problems with under-determined data, we refer to [52, 53, 59, 61].

There has been extensive work in the literature in the context of formally deter-

mined inverse problems involving the wave equation. For a partial list of works in

this direction, we refer to [7, 8, 10, 35, 41, 49–51, 54, 55, 60, 63, 67] .

We now state the main results of this chapter.

Theorem 2.1.1. Suppose qi ∈ C1(R3), i = 1, 2 with q1(x) ≥ q2(x) ∀ x ∈ R3. Letui(x, t) be the solution to the IVP

(�− qi(x))ui(x, t) = δ(x, t), (x, t) ∈ R3 × R

ui(x, t)|t 1 and where e = (1, 0, 0), thenq1(x) = q2(x) for all x with |e− x|+ |x| ≤ T .

Theorem 2.1.2. Suppose qi ∈ C1(R3), i = 1, 2 with qi(x) = ai(|x|+ |x− e|) withe = (1, 0, 0), for some C1 functions ai on (1− �,∞) for some 0 < � < 1. Let ui bethe solution to the IVP

(�− qi(x))ui(x, t) = δ(x, t), (x, t) ∈ R3 × R

ui(x, t)|t 1 and where e = (1, 0, 0), thenq1(x) = q2(x) for all x with |e− x|+ |x| ≤ T .

To the best of our knowledge, our results, Theorem 2.1.1 and 2.1.2, which treat

separated source and receiver, have not been studied earlier. Our result gener-

alize the work [52] who considered the aforementioned inverse problem but with

coincident source and receiver; see also [67].


The proofs of the above theorems are based on an integral identity derived using

the solution to an adjoint problem as used in [65] and [67]. Recently this idea was

used in [56] as well.

The chapter is organized as follows. In §2.2 we state the existence and uniquenessresults for the solution of Equation (2.1.1), the proof of which is given in [20, 39,

60]. §2.3 contains the proofs of Theorems 2.1.1 and 2.1.2.

2.2 Preliminaries

Proposition 2.2.1. [20, pp.139,140] Suppose q(x) is a C1 function on R3 andu(x, t) satisfies the following IVP

(�− q(x))u(x, t) = δ(x, t), (x, t) ∈ R3 × R

u(x, t)|t |x|, v(x, t) is a C2 solution ofthe characteristic boundary value problem (Goursat Problem)

(�− q(x))v(x, t) = 0, t > |x|

v(x, |x|) = 18π

1∫0

q(sx)ds.(2.2.3)

We will use the following version of this proposition. Consider the following IVP

(�− q(x))U(x, t) = δ(x− e, t), (x, t) ∈ R3 × R

U(x, t)|t


where V (x, t) = 0 for t < |x− e| and for t > |x− e|, V (x, t) is a C2 solution to thefollowing Goursat Problem

(�− q(x))V (x, t) = 0, t > |x− e|

V (x, |x− e|) = 18π

1∫0

q(sx+ (1− s)e)ds.(2.2.6)

We can see this by translating source by −e in Equation (2.2.4) and using theabove proposition.

2.3 Proof of Theorems 2.1.1 and 2.1.2

In this section, we prove Theorems 2.1.1 and 2.1.2. We will first show the following

three lemmas which will be used in the proof of the main results.

Lemma 2.3.1. Suppose q′is i = 1, 2 be C1 real-valued functions on R3. Let ui be

the solution to Equation (2.1.1) with q = qi and denote u(x, t) := u1(x, t)−u2(x, t)and q(x) := q1(x)− q2(x). Then we have the following integral identity

u(e, τ) =

∫R

∫R3

q(x)u2(x, t)U(x, τ − t)dxdt for all τ ∈ R (2.3.1)

where U(x, t) is the solution to the following IVP

(�− q1(x))U(x, t) = δ(x− e, t), (x, t) ∈ R3 × R

U(x, t)|t


Now since

u(e, τ) =

∫R3

∫R

u(x, t)δ(x− e, τ − t)dtdx,

using (2.3.2), we have

u(e, τ) =

∫R3

∫R

u(x, t)(�− q1(x))U(x, τ − t)dtdx.

Now by using integration by parts and Equations (2.3.2) and (2.3.3), also taking

into account that u(x, t) = 0 for t < |x| and that U(x, t) = 0 for |x − e| > t, weget

u(e, τ) =

∫R3

∫R

U(x, τ − t)(�− q1(x))u(x, t)dtdx

=

∫R3

∫R

q(x)u2(x, t)U(x, τ − t)dtdx.

This completes the proof of the lemma.

Lemma 2.3.2. Suppose q′is are as in Lemma 2.3.1 and ui is the solution to Equa-

tion (2.1.1) with q = qi and if u(e, t) := (u1 − u2)(e, t) = 0 for all t ∈ [0, T ], thenthere exists a constant K > 0 depending on the bounds on v2, V and T such that

the following inequality holds∣∣∣∣∣∫

|x−e|+|x|=2τ

q(x)

|2τx− |x|e|dSx

∣∣∣∣∣ ≤ K∫

|x−e|+|x|≤2τ

|q(x)||x||x− e|

dx, ∀τ ∈ (1/2, T/2].

(2.3.4)

Here dSx is the surface measure on the ellipsoid |x− e| + |x| = 2τ and v2, V aresolutions to the Goursat problem (see Equations (2.2.3) and (2.2.6)) corresponding

to q = qi.

Proof. From Lemma 2.3.1, we have

u(e, 2τ) =

∫R

∫R3

q(x)u2(x, t)U(x, 2τ − t)dxdt, for all τ ∈ R.


Now since u(e, 2τ) = 0 for all τ ∈ [0, T/2], and using Equations (2.2.2) and (2.2.5),we get

0 =

∫R3

∫R

q(x)δ(t− |x|)δ(2τ − t− |x− e|)

16π2|x||x− e|dtdx

+

∫R3

∫R

q(x)δ(t− |x|)V (x, 2τ − t)

4π|x|dtdx

+

∫R3

∫R

q(x)δ(2τ − t− |x− e|)

4π|x− e|v2(x, t)dtdx

+

∫R3

∫R

q(x)V (x, 2τ − t)v2(x, t)dtdx.

Now using the fact that v2(x, t) = 0 for t < |x|, V (x, t) = 0 for t < |x− e| and∫Rn

φ(x)δ(P )dx =

∫P (x)=0

φ(x)

|∇xP (x)|dSx

where dSx is the surface measure on the surface P = 0, we have that

0 =1

16π2

∫|x−e|+|x|=2τ

q(x)

|x||x− e||∇x(2τ − |x| − |x− e|)|dSx

+1

4π

∫|x|+|x−e|≤2τ

q(x)V (x, 2τ − |x|)|x|

dx

+1

4π

∫|x|+|x−e|≤2τ

q(x)v2(x, 2τ − |x− e|)|x− e|

dx

+

∫|x|+|x−e|≤2τ

2τ−|x−e|∫|x|

q(x)V (x, 2τ − t)v2(x, t)dtdx.

For simplicity, denote

F (τ, x) :=1

4π

(|x− e|V (x, 2τ − |x|) + |x|v2(x, 2τ − |x− e|)

+ 4π|x||x− e|2τ−|x−e|∫|x|

V (x, 2τ − t)v2(x, t)dt)


and using

|∇x(2τ − |x| − |x− e|)| =∣∣∣ x|x| + x− e|x− e| ∣∣∣ = ∣∣∣ |x− e|x+ (x− e)|x||x||x− e| ∣∣∣.

We have

1

16π2

∫|x−e|+|x|=2τ

q(x)

|2τx− |x|e|dSx = −

∫|x|+|x−e|≤2τ

q(x)

|x||x− e|F (τ, x)dx.

Note that τ ∈ [0, T/2] with T < ∞. Now using the boundedness of v2 and V oncompact subsets, we have |F (τ, x)| ≤ K on |x|+ |x− e| ≤ T .

Therefore, finally we have∣∣∣∣∣∫

|x−e|+|x|=2τ

q(x)

|2τx− |x|e|dSx

∣∣∣∣∣ ≤ K∫

|x−e|+|x|≤2τ

|q(x)||x||x− e|

dx, ∀τ ∈ (1/2, T/2].

The lemma is proved.

Lemma 2.3.3. Consider the solid ellipsoid |e−x|+|x| ≤ r, where e = (1, 0, 0) andx = (x1, x2, x3), then we have its parametrization in prolate-spheriodal co-ordinates

(ρ, θ, φ) given by

x1 =1

2+

1

2cosh ρ cosφ

x2 =1

2sinh ρ sin θ sinφ

x3 =1

2sinh ρ cos θ sinφ

(2.3.5)

with cosh ρ ≤ r, θ ∈ (0, 2π), φ ∈ (0, π) and the surface measure dSx on |e − x| +|x| = r and volume element dx on |e− x|+ |x| ≤ r, are given by

dSx =1

4sinh ρ sinφ

√cosh2 ρ− cos2 φdθdφ,

with cosh ρ = r, θ ∈ [0, 2π] and φ ∈ [0, π]

dx =1

8sinh ρ sinφ(cosh2 ρ− cos2 φ)dρdθdφ,

with cosh ρ ≤ r, θ ∈ [0, 2π] and φ ∈ [0, π].

(2.3.6)


Proof. The above result is well known, but for completeness, we will give the proof.

The solid ellipsoid |e− x|+ |x| ≤ r in explicit form can be written as

(x1 − 1/2)2

r2/4+

x22(r2 − 1)/4

+x23

(r2 − 1)/4≤ 1.

From this, we see that

x1 =1

2+

1

2cosh ρ cosφ

x2 =1


x3 =1


with cosh ρ ≤ r, θ ∈ [0, 2π] and φ ∈ [0, π]. This proves the first part of the lemma.

Now the parametrization of ellipsiod |e− x|+ |x| = r, is given by

F (θ, φ) =(1

2+

1

2cosh ρ cosφ,

1

2sinh ρ sin θ sinφ,

1


)with θ ∈ (0, 2π), φ ∈ (0, π), and cosh ρ = r.

Next, we have

∂F

∂θ=(

0,1

2sinh ρ cos θ sinφ,−1


)∂F

∂φ=(− 1

2cosh ρ sinφ,

1

2sinh ρ sin θ cosφ,

1

2sinh ρ cos θ cosφ

).

We have dSx = |∂F∂θ ×∂F∂φ|dθdφ, simple computation will gives us

dSx =1

4sinh ρ sinφ

√cosh2 ρ− cos2 φdθdφ,

with cosh ρ = r, θ ∈ (0, 2π) and φ ∈ (0, π).

Last part of the lemma follows from change of variable formula, which is given by

dx =∣∣∣∂(x1, x2, x3)∂(ρ, θ, φ)

∣∣∣dθdφdρ; with cosh ρ ≤ r, θ ∈ [0, 2π] and φ ∈ [0, π]


where ∂(x1,x2,x3)∂(ρ,θ,φ)

is given by

∂(x1, x2, x3)

∂(ρ, θ, φ)= det

∣∣∣∣∣∣∣∣∣∣

∂x1∂ρ

∂x1∂θ

∂x1∂φ

∂x2∂ρ

∂x2∂θ

∂x3∂φ

∂x3∂ρ

∂x3∂θ

∂x3∂φ

∣∣∣∣∣∣∣∣∣∣.

This gives

dx =1

8sinh ρ sinφ(cosh2 ρ− cos2 φ)dρdθdφ,

with cosh ρ ≤ r, θ ∈ [0, 2π] and φ ∈ [0, π].

2.3.1 Proof of Theorem 2.1.1

We first consider the surface integral in Equation (2.3.4) and denote it by Q(2τ):

Q(2τ) :=

∫|x−e|+|x|=2τ

q(x)

|2τx− |x|e|dSx. (2.3.7)

We have

|2τx− |x|e| = |(2τx1 − |x|, 2τx2, 2τx3)| =√

(2τx1 − |x|)2 + 4τ 2x22 + 4τ 2x23

=√

4τ 2|x|2 + |x|2 − 4τx1|x|.

From Equation (2.3.5) and using the fact that cosh ρ = 2τ , we have

|2τx− |x|e| = 12

√(2τ + cosφ){(4τ 2 + 1)(2τ + cosφ)− 4τ(1 + 2τ cosφ)}

=1

2

√(2τ + cosφ)(8τ 3 + 4τ 2 cosφ+ 2τ + cosφ− 4τ − 8τ 2 cosφ)

=1

2

√(2τ + cosφ)(8τ 3 − 4τ 2 cosφ− 2τ + cosφ)

=1

2

√(4τ 2 − cos2 φ)(4τ 2 − 1).


Using the above expression for |2τx− |x|e| and Equation (2.3.6), we get

Q(2τ) =1

8

π∫0

2π∫0

q(ρ, θ, φ) sinh ρ sinφ√

cosh2 ρ− cos2 φ√(4τ 2 − cos2 φ)(4τ 2 − 1)

dθdφ,

where we have denoted

q(ρ, θ, φ) = q

(1

2+

1

2cosh ρ cosφ,

1

2sinh ρ sin θ sinφ,

1


).

On using cosh ρ = 2τ , sinh ρ =√

4τ 2 − 1 and ρ = ln(2τ +√

4τ 2 − 1), we get

Q(2τ) =

π∫0

2π∫0

q(ln(2τ +√

4τ 2 − 1), θ, φ) sinφdθdφ. (2.3.8)

Now consider the integral ∫|x−e|+|x|≤2τ

q(x)

|x||x− e|dx.

Again using Equations (2.3.5) and (2.3.6) in the above integral, we have

∫|x−e|+|x|≤2τ

q(x)

|x||x− e|dx =

1

2

∫cosh ρ≤2τ

π∫0

2π∫0

q(ρ, θ, φ) sinh ρ sinφdθdφdρ.

After substituting cosh ρ = r and ρ = ln(r +√r2 − 1), we get

∫|x−e|+|x|≤2τ

q(x)

|x||x− e|dx =

1

2

2τ∫1

π∫0

2π∫0

q(ln(r +√r2 − 1), θ, φ) sinφdθdφdr.

Now using (2.3.8), we get

∫|x−e|+|x|≤2τ

q(x)

|x||x− e|dx ≤ C

2τ∫1

Q(r)dr. (2.3.9)


Now applying this inequality in Equation (2.3.4) and noting that q(x) = q1(x) −q2(x) ≥ 0, we get

Q(2τ) ≤ CK2τ∫

1

Q(r)dr. (2.3.10)

Now Equation (2.3.10) holds for all τ ∈ [1/2, T/2] and since q(x) ≥ 0 for allx ∈ R3, by Gronwall’s inequality, we have

Q(2τ) = 0, τ ∈ [1/2, T/2].

Now from Equation (2.3.7), again using q(x) ≥ 0, we have q(x) = 0, for all x ∈ R3

such that |x|+ |x− e| ≤ T . The proof is complete.

2.3.2 Proof of Theorem 2.1.2

Again first, we consider the surface integral in (2.3.4) and denote it by Q(2τ):

Q(2τ) :=

∫|x|+|x−e|=2τ

q(x)

|2τx− |x|e|dSx.

and q(x) := a(|x| + |x− e|). Now from Equations (2.3.5), (2.3.6) and (2.3.8) andhypothesis qi(x) = ai(|x|+ |x− e|) of the theorem, we get

Q(2τ) =1

8

π∫0

2π∫0

a(2τ) sinφdθdφ =π

2a(2τ). (2.3.11)

Now consider the integral ∫|x|+|x−e|≤2τ

q(x)

|x||x− e|dx.

Again using (2.3.5) and (2.3.6) in the above integral, we have

∫|x−e|+|x|≤2τ

q(x)

|x||x− e|dx =

1

2

∫cosh ρ≤2τ

π∫0

2π∫0

q(ρ, θ, φ) sinh ρ sinφdθdφdρ.


After substituting cosh ρ = r and ρ = ln(r +√r2 − 1), we get∣∣∣∣∣∣∣

∫|x−e|+|x|≤2τ

q(x)

|x||x− e|dx

∣∣∣∣∣∣∣ =∣∣∣∣∣∣12

2τ∫1

π∫0

2π∫0

a(r) sinφdθdφdr

∣∣∣∣∣∣≤ C

2τ∫1

|a(r)|dr.

Now using this inequality and Equation (2.3.11) in (2.3.4), we see

|a(2τ)| ≤ C2τ∫

1

|a(r)|dr. (2.3.12)

Now Equation (2.3.12) holds for all τ ∈ [1/2, T/2], so using Gronwall’s inequality,we have

a(2τ) = 0, τ ∈ [1/2, T/2].

Thus, we have q(x) = 0, for all x ∈ R3 such that |x|+ |x− e| ≤ T .

Chapter 3

An Inverse Problem for the

Relativistic Schrödinger Equation

with Partial Boundary Data

3.1 Introduction

Let Ω ⊂ Rn, n ≥ 3 be a bounded domain with C2 boundary. For T > diam(Ω), letQ := (0, T )×Ω and denote its lateral boundary by Σ := (0, T )×∂Ω. Consider thelinear hyperbolic partial differential operator of second order with time-dependent

coefficients:

LA,qu :={

(∂t + A0(t, x))2 −

n∑j=1

(∂j + Aj(t, x))2 + q(t, x)

}u = 0, (t, x) ∈ Q.

(3.1.1)

We denote A = (A0, · · · , An) and A = (A1, · · · , An). Then A = (A0, A). Weassume that A is R1+n valued with coefficients in C∞c (Q) and q ∈ L∞(Q). Theoperator (3.1.1) is known as the relativistic Schrödinger equation and appears in

quantum mechanics and general relativity [40, Chap. XII]. In this paper, we study

an inverse problem related to this operator. More precisely, we are interested in

determining the coefficients in (3.1.1) from certain measurements made on suitable

subsets of the topological boundary of Q.

17

Chapter 3. Relativistic Schrödinger Equation 18

Starting with the work of Bukhgèım and Klibanov [8], there has been extensive

work in the literature related to inverse boundary value problems for second order

linear hyperbolic PDE. For the case when A is 0 and q is time-independent, theunique determination of q from full lateral boundary Dirichlet to Neumann data

was addressed by Rakesh and Symes in [55]. Isakov in [27] considered the same

problem with an additional time-independent time derivative perturbation, that

is, with A = (A0(x), 0) and q(x) and proved uniqueness results. The results in[55] and [27] were proved using geometric optics solutions inspired by the work of

Sylvester and Uhlmann [72]. For the case of time-independent coefficients, another

powerful tool to prove uniqueness results is the boundary control (BC) method

pioneered by Belishev, see [2–4]. Later it was developed by Belishev, Kurylev,

Katchalov, Lassas, Eskin and others; see [32] and references therein. Eskin in [16,

18] developed a new approach based on the BC method for determining the time-

independent vector and scalar potentials assuming A0 = 0 in (3.1.1). Hyperbolic

inverse problems for time-independent coefficients have been extensively studied

by Yamamoto and his collaborators as well; see [1, 6, 12, 24, 25]

Inverse problems involving time-dependent first and zeroth order perturbations

focusing on the cases A = 0 or when A is of the form (A0, 0) have been wellstudied in prior works. We refer to [57, 58, 66] for some works in this direction.

Eskin in [17] considered full first and zeroth order time-dependent perturbations of

the wave equation in a Riemannian manifold set-up and proved uniqueness results

(for the first order term, uniqueness modulo a natural gauge invariance) from

boundary Dirichlet-to-Neumann data, under the assumption that the coefficients

are analytic in time. Salazar removed the analyticity assumption of Eskin in

[64], and proved that the unique determination of vector and scalar potential

modulo a natural gauge invariance is possible from Dirichlet-to-Neumann data on

the boundary. In a recent work of Stefanov and Yang in [47], proved stability

estimates for the recovery of light ray transforms of time-dependent first- and

zeroth-order perturbations for the wave equation in a Riemannian manifold setting

from certain local Dirichlet to Neumann map. Their results, in particular, would

give uniqueness results in suitable subsets of the domain recovering the vector field

term up to a natural gauge invariance and the zeroth-order potential term from

this data.


For the case of time-dependent perturbations, if one is interested in global unique-

ness results in a finite time domain, extra information in addition to Dirichlet-

to-Neumann data is required to prove uniqueness results. Isakov in [26] proved

unique-determination of time-dependent potentials (assuming A = 0 in (3.1.1))from the data set given by the Dirichlet-to-Neumann data as well as the solution

and the time derivative of the solution on the domain at the final time. Recently

Kian in [33] proved unique determination of time-dependent damping coefficient

A0(t, x) (with A of the form, A = (A0, 0)) and the potential q(t, x) from partialDirichlet to Neumann data together with information of the solution at the final

time.

In this article, we prove unique determination of time-dependent vector and scalar

potentials A(t, x) and q(t, x) appearing in (3.1.1) (modulo a gauge invariance forthe vector potential) from partial boundary data. Our work extends the results

of Salazar [64] in the sense that in [64], the uniqueness results are shown with full

Dirichlet to Neumann boundary data and assuming such boundary measurements

are available for all time. Additionally, it extends the recent work of Kian [33],

since we consider the full time-dependent vector field perturbation, whereas Kian

assumes only a time derivative perturbation.

The chapter is organized as follows. In §3.2, we state the main result of thearticle. In §3.3, we prove the Carleman estimates required to prove the existenceof geometric optics (GO) solutions, and in §3.4, we construct the required GOsolutions. In §3.5, we derive the integral identity using which, we prove the maintheorem in §3.6.

3.2 Statement of the main result

In this section, we state the main result of this article. We begin by stating

precisely what we mean by gauge invariance.

3.2.1 Gauge Invariance

Definition 3.2.1. The vector potentials A(1),A(2) ∈ C∞c (Q) are said to be gaugeequivalent if there exists a g(t, x) ∈ C∞(Q) such that g(t, x) = eΦ(t,x) with Φ ∈


C∞c (Q) and

(A(2) −A(1))(t, x) = −∇t,xg(t, x)g(t, x)

= −∇t,xΦ(t, x).

Now we state the following proposition proof of which is given in [64].

Proposition 3.2.2. [64] Suppose u(t, x) is a solution to the following IBVP

[(∂t + A

(1)0 (t, x))

2 −n∑j=1

(∂j + A(1)j (t, x))

2 + q1(t, x)]u(t, x) = 0 in Q (3.2.1)

u(0, x) = ∂tu(0, x) = 0 in Ω

u|Σ = f

and g(t, x) is as defined above, then v(t, x) = g(t, x)u(t, x) satisfies the following

IBVP

[(∂t + A

(2)0 (t, x))

2 −n∑j=1

(∂j + A(2)j (t, x))

2 + q1(t, x)]v(t, x) = 0 in Q (3.2.2)

v(0, x) = ∂tv(0, x) = 0 in Ω

v|Σ = f

with A(1) and A(2) gauge equivalent. In addition if Λ1 and Λ2 are Dirichlet toNeumann operator associated with (3.2.1) and (3.2.2) respectively, then

Λ1 = Λ2.

3.2.2 Statement of the main result

We introduce some notation. Following [9], fix an ω0 ∈ Sn−1, and define theω0-shadowed and ω0-illuminated faces by

∂Ω+,ω0 := {x ∈ ∂Ω : ν(x) · ω0 ≥ 0} , ∂Ω−,ω0 := {x ∈ ∂Ω : ν(x) · ω0 ≤ 0}

of ∂Ω, where ν(x) is outward unit normal to ∂Ω at x ∈ ∂Ω. Corresponding to∂Ω±,ω0 , we denote the lateral boundary parts by Σ±,ω0 := (0, T ) × ∂Ω±,ω0 . Wedenote by F = (0, T )×F ′ and G = (0, T )×G′ where F ′ and G′ are small enoughopen neighbourhoods of ∂Ω+,ω0 and ∂Ω−,ω0 respectively in ∂Ω.


Consider the IBVP

LA,qu(t, x) = 0; (t, x) ∈ Q

u(0, x) = φ(x), ∂tu(0, x) = ψ(x); x ∈ Ω

u(t, x) = f(t, x), (t, x) ∈ Σ.

(3.2.3)

For φ ∈ H1(Ω), ψ ∈ L2(Ω) and f ∈ H1(Σ), (3.2.3) has a unique solution u ∈C1([0, T ];L2(Ω)) ∩ C([0, T ];H1(Ω)) and furthermore ∂νu ∈ L2(Σ); see [32, 38].Thus we have u ∈ H1(Q). Therefore we can define our input-output operator ΛA,qby

ΛA,q(φ, ψ, f) = (∂νu|G, u|t=T ) (3.2.4)

where u is the solution to (3.2.3). The operator

ΛA,q : H1(Ω)× L2(Ω)×H1(Σ) → H1(Ω)× L2(G)

is a continuous linear map which follows from the well-posedness of the IBVP given

by Equation (3.2.3) (see [32, 38]). A natural question is whether this input-output

operator uniquely determines the time-dependent perturbations A and q. We nowstate our main result.

Theorem 3.2.3. Let(A(1), q1

)and

(A(2), q2

)be two sets of vector and scalar

potentials such that each A(i)j ∈ C∞c (Q) and qi ∈ L∞(Q) for i = 1, 2 and 0 ≤ j ≤ n.

Let ui be solutions to (3.2.3) when (A, q) = (A(i), qi) and ΛA(i),qi for i = 1, 2 be theinput-output operators defined by (3.2.4) corresponding to ui. If

ΛA(1),q1(φ, ψ, f) = ΛA(2),q2(φ, ψ, f), for all (φ, ψ, f) ∈ H1(Ω)× L2(Ω)×H1(Σ),

then there exists a function Φ ∈ C∞c (Q) such that

(A(1) −A(2))(t, x) = ∇t,xΦ(t, x); and q1(t, x) = q2(t, x), (t, x) ∈ Q.

Remark 3.2.4. We have stated the above result for vector potentials in C∞c (Q)

for simplicity. We can, in fact, prove for the case in which the vector potentials

A ∈ W 1,∞(Q) provided they are identical on the boundary ∂Q, see [33].


3.3 Carleman Estimate

We denote by H1scl(Q), the semiclassical Sobolev space of order 1 on Q with the

following norm

||u||H1scl(Q) = ||u||L2(Q) + ||h∇t,xu||L2(Q) ,

and for Q = R1+n we denote by Hsscl(R1+n), the Sobolev space of order s with thenorm given by

||u||2Hsscl(R1+n) = ||〈hD〉s u||2L2(R1+n) =

∫R1+n

(1 + h2τ 2 + h2|ξ|2)s |û(τ, ξ)|2 dτdξ.

In this section, we derive a Carleman estimate involving boundary terms for (3.1.1)

conjugated with a linear weight. We use this estimate to control boundary terms

over subsets of the boundary where measurements are not available. Our proof

follows from modifications of the Carleman estimate given in [33]. Since we work

in a semiclassical setting, we prefer to give the proof for the sake of completeness.

Theorem 3.3.1. Let ϕ(t, x) := t + x · ω, where ω ∈ Sn−1 is fixed. Assume thatAj ∈ C∞c (Q) for 0 ≤ j ≤ n and q ∈ L∞(Q). Then the Carleman estimate

h(e−ϕ/h∂νϕ∂νu, e

−ϕ/h∂νu)L2(Σ+,ω)

+ h(e−ϕ(T,·)/h∂tu(T, ·), e−ϕ(T,·)/h∂tu(T, ·)

)L2(Ω)

+ ‖e−ϕ/hu‖2L2(Q) + ‖he−ϕ/h∂tu‖2L2(Q) + ‖he−ϕ/h∇xu‖2L2(Q)≤ C

(‖he−ϕ/hLA,qu‖2L2(Q) +

(e−ϕ(T,·)/hu(T, ·), e−ϕ(T,·)/hu(T, ·)

)L2(Ω)

(3.3.1)

+ h(e−ϕ(T,·)/h∇xu(T, ·), e−ϕ(T,·)/h∇xu(T, ·)

)L2(Ω)

+ h(e−ϕ/h (−∂νϕ) ∂νu, e−ϕ/h∂νu

)L2(Σ−,ω)

)holds for all u ∈ C2(Q) with

u|Σ = 0, u|t=0 = ∂tu|t=0 = 0,

and h small enough.

Proof. To prove the estimate (3.3.1), we will use a convexification argument used

in [33]. Consider the following perturbed weight function

ϕ̃(t, x) = ϕ(t, x)− ht2

2ε. (3.3.2)


We first consider the conjugated operator

�ϕ,ε := h2e−ϕ̃/h�eϕ̃/h. (3.3.3)

For v ∈ C2(Q) satisfying v|Σ = v|t=0 = ∂tv|t=0 = 0, consider the L2 norm of �ϕ,ε:∫Q

|�ϕ,εv(t, x)|2 dxdt.

Expanding (3.3.3), we get,

�ϕ,εv(t, x) =(h2� + h�ϕ̃+

(|∂tϕ̃|2 − |∇xϕ̃|2

)+ 2h (∂tϕ̃∂t −∇xϕ̃ · ∇x)

)v(t, x).

We write this as

�ϕ,εv(t, x) = P1v(t, x) + P2v(t, x),

where

P1v(t, x) =(h2� + h�ϕ̃+

(|∂tϕ̃|2 − |∇xϕ̃|2

) )v(t, x)

=

(h2� +

h2t2

ε2− 2ht

ε− h

2

ε

)v(t, x),

and

P2v(t, x) = 2h (∂tϕ̃∂t −∇xϕ̃ · ∇x) v(t, x)

= 2h

((1− ht

ε

)∂t − ω · ∇x

)v(t, x).

Now∫Q

|�ϕ,εv(t, x)|2 dxdt ≥ 2∫Q

Re(P1v(t, x)P2v(t, x)

)dxdt

= 4h3∫Q

Re

(�v(t, x)

(1− ht

ε

)∂tv(t, x)

)dxdt

− 4h3∫Q

Re(�v(t, x)ω · ∇xv(t, x)

)dxdt

+ 4h

∫Q

Re

((h2t2

ε2− 2ht

ε− h

2

ε

)v(t, x)

(1− ht

ε

)∂tv(t, x)

)dxdt


− 4h∫Q

Re

((h2t2

ε2− 2ht

ε− h

2

ε

)v(t, x)ω · ∇xv(t, x)

)dxdt

:= I1 + I2 + I3 + I4.

We first simplify I1. We have

I1 = 4h3

∫Q

Re

(�v(t, x)

(1− ht

ε

)∂tv(t, x)

)dxdt

= 2h3∫Q

∂

∂t|∂tv(t, x)|2

(1− ht

ε

)dxdt− 4h3

∫Q

Re

(∆v(t, x)

(1− ht

ε

)∂tv(t, x)

)dxdt

= 2h3(

1− hTε

)∫Ω

(|∂tv(T, x)|2 + |∇xv(T, x)|2

)dx+

2h4

ε

∫Q

(|∂tv(t, x)|2 + |∇xv(t, x)|2

)dxdt.

In the above steps, we used integration by parts combined with the hypotheses

that v|Σ = v|t=0 = ∂tv|t=0 = 0. Note that v|Σ = 0 would imply that ∂tv = 0 on Σ.

Now we consider I2. We have

I2 = −4h3∫Q

Re(�v(t, x)ω · ∇xv(t, x)

)dxdt.

We have

I2 = −4h3Re∫Q

∂2t v(t, x)ω · ∇xv(t, x)dxdt+ 4h3Re∫Q

∆v(t, x)ω · ∇xv(t, x)dxdt

= −4h3Re∫Q

∂t

(∂tv(t, x)ω · ∇xv(t, x)

)dxdt+ 4h3Re

∫Q

∂tv(t, x)ω · ∇x∂tv(t, x)dxdt

+ 4h3Re

∫Q

∇x ·(∇xv(t, x)ω · ∇xv(t, x)

)dxdt− 4h3Re

∫Q

∇xv(t, x) · ∇x(ω · ∇xv(t, x)

)dxdt

= −4h3Re∫Ω

∂tv(T, x)ω · ∇xv(T, x)dx+ 2h3∫Q

∇x ·(ω|∂tv(t, x)|2

)dxdt

+ 2h3Re

∫Σ

∂νv(t, x)ω · ∇xv(t, x)dSxdt− 2h3∫Q

∇x ·(ω|∇xv|2

)dxdt

= −4h3Re∫Ω

∂tv(T, x)ω · ∇xv(T, x)dx+ 2h3∫Σ

ω · ν|∂tv(t, x)|2dSxdt+ 2h3∫Σ

ω · ν|∂νv|2dSxdt.


In deriving the above equation, we used the fact that

2h3Re

∫Σ

∂νv(t, x)ω · ∇xv(t, x)dSxdt = 2h3∫Σ

ω · ν|∂νv|2dSxdt,

since v = 0 on Σ. Also note that |∇xv| = |∂νv|.

Next we consider I3. We have

I3 = 4h

∫Q

Re

((h2t2

ε2− 2ht

ε− h

2

ε

)v(t, x)

(1− ht

ε

)∂tv(t, x)

)dxdt

= 2h

∫Q

(h2t2

ε2− 2ht

ε− h

2

ε

)(1− ht

ε

)∂t|v(t, x)|2dxdt

= 2

∫Ω

(h3T 2

ε2− 2h

2T

ε− h

3

ε

)(1− hT

ε

)|v(T, x)|2dx

− 2∫Q

[(2h3t

ε2− 2h

2

ε

)(1− ht

ε

)− h

2

ε

(h2t2

ε2− 2ht

ε− h

2

ε

)]|v(t, x)|2dxdt.

Finally, we consider I4. This is

I4 := −4h∫Q

Re

((h2t2

ε2− 2ht

ε− h

2

ε

)v(t, x)ω · ∇xv(t, x)

)dxdt

= −4h∫Q

(h2t2

ε2− 2ht

ε− h

2

ε

)∇ ·(ω|v|2

)dxdt

= 0 since v = 0 on Σ.

Therefore∫Q

|�ϕ,εv(t, x)|2dxdt ≥ 2h3(

1− hTε

)∫Ω

(|∂tv(T, x)|2 + |∇xv(T, x)|2

)dx

+2h4

ε

∫Q

(|∂tv(t, x)|2 + |∇xv(t, x)|2

)dxdt+ 2h3

∫Σ

ω · ν|∂tv(t, x)|2dSxdt

− 4h3Re∫Ω

∂tv(T, x)ω · ∇xv(T, x)dx+ 2h3∫Σ

ω · ν|∂νv|2dSxdt

+ 2

∫Ω

(h3T 2

ε2− 2h

2T

ε− h

3

ε

)(1− hT

ε

)|v(T, x)|2dx


− 2∫Q

[(2h3t

ε2− 2h

2

ε

)(1− ht

ε

)− h

2

ε

(h2t2

ε2− 2ht

ε− h

2

ε

)]|v(t, x)|2dxdt.

Choosing ε and h small enough, we have

∫Q

|�ϕ,εv(t, x)|2dxdt ≥2h4

ε

∫Q

|∂tv(t, x)|2 + |∇xv(t, x)|2dxdt

+ ch3 ∫Ω

|∂tv(T, x)|2dx

+ 2h3∫Σ

ω · ν(x)|∂νv(t, x)|2dSxdt− ch3∫Ω

|∇xv(T, x)|2dx

− ch2∫Ω

|v(T, x)|2dx+ ch2

ε

∫Q

|v(t, x)|2dxdt. (3.3.4)

Now we consider the conjugated operator Lϕ,ε := h2e−ϕ̃hLA,qe

ϕ̃h . We have

Lϕ,εv(t, x) = h2(e−ϕ̃/h (� + 2A0∂t − 2A · ∇x + q̃) eϕ̃/hv(t, x)

),

where

q̃ = q + |A0|2 − |A|2 + ∂tA0 −∇x · A.

We write

Lϕ,εv(t, x) = �ϕ,εv(t, x) + P̃ v(t, x),

where

P̃ v(t, x) = h2(e−ϕ̃/h (2A0∂t − 2A · ∇x + q̃) eϕ̃/hv(t, x)

). (3.3.5)

By triangle inequality,∫Q

|Lϕ,εv(t, x)|2 dxdt ≥1

2

∫Q

|�ϕ,εv(t, x)|2dxdt−∫Q

|P̃ v(t, x)|2dxdt. (3.3.6)

We have∫Q

∣∣∣P̃ v(t, x)∣∣∣2 dxdt = ∫Q

∣∣∣∣[h2 (2A0∂t − 2A · ∇x + q̃) v + h{2A0(1− htε)− 2ω · A

}v

]∣∣∣∣2 dxdt≤ Ch4

‖A0‖2L∞(Q) ∫Q

|∂tv(t, x)|2dxdt+ ‖A‖2L∞(Q)∫Q

|∇xv(t, x)|2dxdt


+ Ch2

((h2‖q̃‖2L∞(Q) + ‖A‖2L∞(Q)

) ∫Q

|v(t, x)|2dxdt

+ ‖A0‖2L∞(Q)∫Q

(1− ht

ε

)2|v(t, x)|2dxdt

).

Choosing h small enough, we have,

∫Q

∣∣∣P̃ v(t, x)∣∣∣2 dxdt ≤ Ch4‖A0‖2L∞(Q) ∫

Q

|∂tv(t, x)|2dxdt+ ‖A‖2L∞(Q)∫Q

|∇xv(t, x)|2dxdt

+ Ch2

(‖A0‖2L∞(Q) + ‖A‖2L∞(Q)

) ∫Q

|v(t, x)|2dxdt

+ Ch4‖q̃‖2L∞(Q)∫Q

|v(t, x)|2dxdt.

(3.3.7)

Using (3.3.4) and (3.3.7) in (3.3.6) and taking ε small enough, we have that there

exists a C > 0 depending only on ε, T , Ω, A and q such that

∫Q

|Lϕ,εv(t, x)|2dxdt ≥ Ch2{h2

∫Q

|∂tv(t, x)|2 + |∇xv(t, x)|2dxdt

+ ∫Q

|v(t, x)|2dxdt

−∫Ω

|v(T, x)|2dx

}+ Ch3

∫Ω

|∂tv(T, x)|2dx

+ h3∫Σ

ω · ν(x)|∂νv(t, x)|2dSxdt− Ch3∫Ω

|∇xv(T, x)|2dx,

and this inequality holds for h small enough. After dividing by h2, we get

C

∫Q

(|h∂tv(t, x)|2 + |h∇xv(t, x)|2

)dxdt+

∫Q

|v(t, x)|2dxdt

+ Ch

∫Ω

|∂tv(T, x)|2dx+ h∫Σ

ω · ν(x)|∂νv(t, x)|2dSxdt

≤ 1h2

∫Q

|Lϕ,εv(t, x)|2 dxdt+ Ch∫Ω

|∇xv(T, x)|2dx+ C∫Ω

|v(T, x)|2dx. (3.3.8)


Let us now substitute v(t, x) = e−ϕ̃hu(t, x). We have

he−ϕ/h∂tu(t, x) = he−t2/2ε∂tv + e

−ϕ/h(

1− htε

)u,

he−ϕ/h∇xu = he−t2/2ε∇xv + e−ϕ/hωu,

∂νv(t, x)|Σ = e−ϕ/h∂νu|Σ, since u = 0 on Σ.

Using the above equalities and the triangle inequality, we then have for h small

enough,

h

∫Ω

e−2ϕ/h|∂tu(T, x)|2dx+ h∫

Σ+,ω

e−2ϕ/h|∂νu(t, x)|2|ω · ν(x)|dSxdt

+ h2∫Q

e−2ϕ/h(|∂tu(t, x)|2 + |∇xu(t, x)|2

)dxdt+

∫Q

e−2ϕ/h|u(t, x)|2dxdt

≤ C

(h2∫Q

e−2ϕ/h|LA,qu(t, x)|2dxdt+ h∫Ω

e−2ϕ/h|∇xu(T, x)|2dx

+

∫Ω

e−2ϕ/h|u(T, x)|2dx+ h∫

Σ−,ω

e−2ϕ/h|∂νu(t, x)|2|ω · ν(x)|dSxdt

).

Finally,

h(e−ϕ/h∂νϕ∂νu, e

−φ/h∂νu)L2(Σ+,ω)

+ h(e−ϕ(T,·)/h∂tu(T, ·), e−ϕ(T,·)/h∂tu(T, ·)

)L2(Ω)

+ ‖e−ϕ/hu‖2L2(Q) + ‖he−ϕ/h∂tu‖2L2(Q) + ‖he−ϕ/h∇xu‖2L2(Q)

≤ C

(‖he−ϕ/hLA,qu‖2L2(Q) +

(e−ϕ(T,·)/hu(T, ·), e−ϕ(T,·)/hu(T, ·)

)L2(Ω)

+ h(e−ϕ(T,·)/h∇xu(T, ·), e−ϕ(T,·)/h∇xu(T, ·)

)L2(Ω)

+ h(e−ϕ/h (−∂νϕ) ∂νu, e−ϕ/h∂νu

)L2(Σ−,ω)

).

This completes the proof.

In particular, it follows from the previous calculations that for u ∈ C∞c (Q),

‖u‖H1scl(Q) ≤C

h‖Lϕu‖L2(Q). (3.3.9)

To show the existence of suitable solutions to (3.1.1), we need to shift the Sobolev

index by −1 in (3.3.9). This we do in the next lemma.


Lemma 3.3.2. Let ϕ(t, x) = t + x · ω and Lϕ := h2e−ϕ/hLA,qeϕ/h. There existsan h0 > 0 such that

‖v‖L2(R1+n) ≤C

h‖Lϕv‖H−1scl (R1+n), (3.3.10)

and

‖v‖L2(R1+n) ≤C

h‖L∗ϕv‖H−1scl (R1+n) (3.3.11)

for all v ∈ C∞c (Q), 0 < h ≤ h0.

Proof. We give the proof of the estimate in (3.3.10) and that of (3.3.11) follows

similarly. We follow arguments used in [15]. We again consider the convexified

weight

ϕ̃(t, x) = t+ x · ω − ht2

2ε,

and as before consider the convexified operator:

�ϕ,ε := h2e−ϕ̃/h�eϕ̃/h.

From the properties of pseudodifferential operators, we have

〈hD〉−1 (�ϕ,ε) 〈hD〉 = �ϕ,ε + hR1

where R1 is a semi-classical pseudo-differential operator of order 1. Now

‖�ϕ,ε〈hD〉v‖H−1scl (R1+n) = ‖〈hD〉−1�ϕ,ε〈hD〉v‖L2(R1+n).

and by the commutator property above, we get

‖�ϕ,ε〈hD〉v‖2H−1scl (R1+n) = ‖(�ϕ,ε + hR1) v‖2L2(R1+n) ≥

1

2‖�ϕ,εv‖2L2(R1+n)−‖hR1v‖2L2(R1+n).

Let Q ⊂⊂ Q̃, and for v ∈ C∞c (Q̃), using the estimate in (3.3.4) for C∞c functionscombined with estimates for pseudodifferential operators, we have,

‖�ϕ,ε〈hD〉v‖2H−1scl (R1+n) ≥Ch2

ε‖v‖2H1scl(R1+n) − h

2‖v‖2H1scl(R1+n). (3.3.12)

Using the expression for P̃ (see (3.3.5)), we get, for v ∈ C∞c (Q̃) and for h smallenough,

‖P̃ v‖2H−1scl (R1+n)

≤ Ch2‖v‖2L2(R1+n),


and therefore

‖P̃ 〈hD〉v‖2H−1scl (R1+n)

≤ Ch2‖〈hD〉v‖2L2(R1+n) = Ch2‖v‖2H1scl(R1+n).

Combining this with the estimate in (3.3.12) together with triangle inequality, we

get,

‖Lϕ,ε〈hD〉v‖2H−1scl (R1+n) ≥Ch2

ε‖v‖2H1scl(R1+n) (3.3.13)

for all v ∈ C∞c (Q̃).

Now to complete the proof, for any u ∈ C∞c (Q), consider v = χ〈hD〉−1u, whereχ ∈ C∞c (Q̃) with χ ≡ 1 on Q. Then from (3.3.13), we have

Ch2

ε‖χ〈hD〉−1u‖2H1scl(R1+n) ≤ ‖Lϕ,ε〈hD〉χ〈hD〉

−1u‖2H−1scl (R1+n)

.

The operator 〈hD〉χ〈hD〉−1 is a semiclassical pseudodifferential operator of order0, and therefore we have

Lϕ,ε〈hD〉χ〈hD〉−1u = 〈hD〉χ〈hD〉−1Lϕ,ε + hR1,

where R1 is a semiclassical pseudodifferential operator of order 1.

Ch2

ε‖χ〈hD〉−1u‖2H1scl(R1+n) ≤ ‖Lϕ,εu‖

2H−1scl (R1+n)

+ h2‖u‖2L2(R1+n).

Finally, write

〈hD〉−1u = χ〈hD〉−1u+ (1− χ)〈hD〉−1u,

where χ is as above. Then

‖〈hD〉−1u‖2H1scl(R1+n) ≥1

2‖χ〈hD〉−1u‖2H1scl(R1+n) − ‖(1− χ) 〈hD〉

−1u‖2H1scl(R1+n).

Since (1−χ)〈hD〉−1 is a smoothing semiclassical pseudodifferential operator, tak-ing h small enough, and arguing as in the proof of the Carleman estimate, we

get,

‖Lϕu‖2H−1scl (R1+n) ≥ Ch2‖u‖2L2(R1+n).


Cancelling out the h2 term, we finally have,

‖u‖L2(R1+n) ≤C

h‖Lϕu‖H−1scl (R1+n).

This completes the proof.

Proposition 3.3.3. Let ϕ, A and q be as in Theorem 3.3.1. For h > 0 smallenough and v ∈ L2(Q), there exists a solution u ∈ H1(Q) of

Lϕu = v,

satisfying the estimate

‖u‖H1scl(Q) ≤C

h‖v‖L2(R1+n),

where C > 0 is a constant independent of h.

Proof. The proof uses standard functional analysis arguments. Consider the space

S :={L∗ϕu : u ∈ C∞c (Q)

}as a subspace of H−1(R1+n) and define a linear form L

on S by

L(L∗ϕz) =∫Q

z(t, x)v(t, x)dxdt, for z ∈ C∞c (Q).

This is a well-defined continuous linear functional by the Carleman estimate (3.3.11)

. We have

∣∣L(L∗ϕz)∣∣ ≤ ‖z‖L2(Q)‖v‖L2(Q) ≤ Ch ‖v‖L2(Q)‖L∗ϕz‖H−1scl (R1+n), z ∈ C∞c (Q).By Hahn-Banach theorem, we can can extend L to H−1(R1+n) (still denoted as L)and it satisfies ‖L‖ ≤ C

h||v||L2(Q). By Riesz representation theorem, there exists a

unique u ∈ H1(R1+n) such that

L(f) = 〈f, u〉L2(R1+n) for all f ∈ H−1(R1+n) with ‖u‖H1scl(R1+n) ≤C

h‖v‖L2(Q).

Taking f = L∗ϕz, for z ∈ C∞c (Q), we get

L(L∗ϕz) = 〈L∗ϕz, u〉L2(R1+n) = 〈z,Lϕu〉L2(R1+n).


Therefore for all z ∈ C∞c (Q),

〈z,Lϕu〉 = 〈z, v〉.

Hence

Lϕu = v in Q with ‖u‖H1scl(Q) ≤C

h‖v‖L2(Q).

This completes the proof of the proposition.

3.4 Construction of geometric optics solutions

In this section we construct geometric optics solutions for LA,qu = 0 and its adjointoperator L∗A,qu = L−A,qu = 0.

Proposition 3.4.1. Let LA,q be as in (3.1.1).

1. (Exponentially decaying solutions) There exists an h0 > 0 such that for all

0 < h ≤ h0, we can find v ∈ H1(Q) satisfying L−A,qv = 0 of the form

vd(t, x) = e−ϕh (Bd(t, x) + hRd(t, x;h)) , (3.4.1)

where ϕ(t, x) = t+ x · ω,

Bd(t, x) = exp

− ∞∫0

(1,−ω) · A(t+ s, x− sω)ds

(3.4.2)with ζ ∈ (1,−ω)⊥ and Rd ∈ H1(Q) satisfies

‖Rd‖H1scl(Q) ≤ C. (3.4.3)

2. (Exponentially growing solutions) There exists an h0 > 0 such that for all

0 < h ≤ h0, we can find v ∈ H1(Q) satisfying LA,qv = 0 of the form

vg(t, x) = eϕh (Bg(t, x) + hRg(t, x;h)) , (3.4.4)


where ϕ(t, x) = t+ x · ω,

Bg(t, x) = e−iζ·(t,x) exp

∞∫0

(1,−ω) · A(t+ s, x− sω)ds

(3.4.5)with ζ ∈ (1,−ω)⊥ and Rg ∈ H1(Q) satisfies

‖Rg‖H1scl(Q) ≤ C. (3.4.6)

Proof. We have

LA,qv = �v + 2A0∂tv − 2A · ∇xv +(∂tA0 −∇x · A+ |A0|2 − |A|2 + q

)v.

Letting v of the form

v(t, x) = eϕh (Bg + hRg) ,

and setting the term involving h−1 to be 0, we get,

(1,−ω) · (∇t,xBg + (A0, A)Bg) = 0.

One solution of this equation is

Bg(t, x) = exp

∞∫0

(1,−ω) · A(t+ s, x− sω)ds

.Alternately, another solution is

Bg(t, x) = e−iζ·(t,x) exp

∞∫0

(1,−ω) · A(t+ s, x− sω)ds

,provided ζ ∈ (1,−ω)⊥.

Now we have

L∗A,q = L−A,q.

For this adjoint operator, the equation satisfied by Bd is

(1,−ω) · (∇t,xBg − (A0, A)Bg) = 0.


We let

Bd(t, x) = exp

− ∞∫0

(1,−ω) · A(t+ s, x− sω)ds

.Now Rg satisfies

LϕRg = −hLA,qBg.

Then using the estimate in Proposition 3.3.3, we get that

‖Rg‖H1scl(Q) ≤ C‖LA,qBg‖L2(Q).

Similarly,

‖Rd‖H1scl(Q) ≤ C‖L−A,qBd‖L2(Q).

The proof is complete.

3.5 Integral Identity

In this section, we derive an integral identity involving the coefficients A and qusing the geometric optics solutions described in the previous section.

Let ui be the solutions to the following initial boundary value problems with vector

field coefficient A(i) and scalar potential qi for i = 1, 2.LA(i),qiui(t, x) = 0; (t, x) ∈ Q

ui(0, x) = φ(x), ∂tui(0, x) = ψ(x); x ∈ Ω

ui(t, x) = f(t, x), (t, x) ∈ Σ.

(3.5.1)

Let us denote

u(t, x) := (u1 − u2)(t, x)

A(t, x) := A(2) −A(1)(t, x) := (A0(t, x), A1(t, x), · · · , An(t, x))

q̃i := ∂tA(i)0 −∇x · A(i) + |A

(i)0 |2 − |A(i)|2 + qi (3.5.2)

q̃ := q̃2 − q̃1.


Then u is the solution to the following initial boundary value problem:LA(1),q1u(t, x) = −2A · ∇xu2 + 2A0∂tu2 + q̃u2

u(0, x) = ∂tu(0, x) = 0, x ∈ Ω

u|Σ = 0.

(3.5.3)

Let v(t, x) of the form given by (3.4.1) be the solution to following equation

L−A(1),q1v(t, x) = 0 in Q. (3.5.4)

Also let u2 of the form given by (3.4.4) be solution to the following equation

LA(2),q2u2(t, x) = 0, in Q. (3.5.5)

By the well-posedness result from [32, 38], we have u ∈ H1(Q) and ∂νu ∈ L2(Σ).

Now we multiply (3.5.3) by v(t, x) ∈ H1(Q) and integrate over Q. We get, afterintegrating by parts, taking into account the following: u|Σ = 0, u(T, x) = 0,∂νu|G = 0, u|t=0 = ∂tu|t=0 = 0 and A(1) is compactly supported in Q:∫Q

LA(1),q1u(t, x)v(t, x)dxdt−∫Q

u(t, x)L−A(1),q1v(t, x)dxdt =∫Ω

∂tu(T, x)v(T, x)dx

−∫

Σ\G

∂νu(t, x)v(t, x)dSxdt.

Now using the fact that L−A(1),q1v(t, x) = 0 in Q and

LA(1),q1u(t, x) = −2A · ∇xu2 + 2A0∂tu2 + q̃u2,

we get,∫Q

(−2A · ∇xu2 + 2A0∂tu2 + q̃u2) v(t, x)dxdt =∫Ω


−∫

Σ\G

∂νu(t, x)v(t, x)dSxdt.


Lemma 3.5.1. Let ui for i = 1, 2 solutions to (3.5.1) with u2 of the form (3.4.4).

Let u = u1 − u2, and v be of the form (3.4.1). Then

h

∫Ω

∂tu(T, x)v(T, x)dx→ 0 as h→ 0+. (3.5.6)

h

∫Σ\G

∂νu(t, x)v(t, x)dSxdt→ 0 as h→ 0+. (3.5.7)

Proof. Using (3.4.1), (3.4.3) and Cauchy-Schwartz inequality, we get∣∣∣∣∣∣h∫Ω


∣∣∣∣∣∣ ≤∫Ω

h∣∣∣∂tu(T, x)e−ϕ(T,x)h (Bd(T, x) + hRd(T, x))∣∣∣ dx

≤ C

∫Ω

h2∣∣∣∂tu(T, x)e−ϕ(T,x)h ∣∣∣2 dx

12 ∫Ω

∣∣∣e−iξ·(T,x) + hRd(T, x)∣∣∣2 dx 12

≤ C

∫Ω

h2∣∣∣∂tu(T, x)e−ϕ(T,x)h ∣∣∣2 dx

12 (1 + ‖hRd(T, ·)‖2L2(Ω)) 12

≤ C

∫Ω

h2∣∣∣∂tu(T, x)e−ϕ(T,x)h ∣∣∣2 dx

12 .Now using the boundary Carleman estimate (3.3.1), we get,

h

∫Ω

∣∣∣∂tu(T, x)e−ϕ(T,x)h ∣∣∣2 dx ≤ C‖he−ϕ/hLA(1),q1u‖2L2(Q)= C‖he−ϕ/h (2A0∂tu2 − 2A · ∇xu2 + q̃u2) ‖2L2(Q).

Now substituting (3.4.4) for u2, we get,

he−ϕ/h (2A0∂tu2 − 2A · ∇xu2 + q̃u2) = (2A0∂tϕ− 2A · ∇xϕ+ q̃) (Bg + hRg)

+ h (2A0∂t − 2A · ∇x + q̃) (Bg + hRg)

Therefore

‖he−ϕ/h (2A0∂tu2 − 2A · ∇xu2 + q̃u2) ‖2L2(Q)≤ C,


uniformly in h.

Thus, we have h2 ∫Ω

∣∣∣∂tu(T, x)e−ϕ(T,x)h ∣∣∣2 dx 12 ≤ C√h.

Therefore

h

∫Ω

∂tu(T, x)v(T, x)dx→ 0 as h→ 0+.

For ε > 0, define

∂Ω+,ε,ω = {x ∈ ∂Ω : ν(x) · ω > ε}.

and

Σ+,ε,ω = (0, T )× ∂Ω+,ε,ω.

Next we prove (3.5.7). Since Σ \ G ⊆ Σ+,ε,ω for all ω such that |ω − ω0| ≤ ε,substituting v = vd from (3.4.1), in (3.5.7) we have∣∣∣∣∣∣∣

∫Σ\G

∂νu(t, x)v(t, x)dSxdt

∣∣∣∣∣∣∣ ≤∫

Σ+,ε,ω

∣∣∣∂νu(t, x)e−ϕh (Bd + hRd) (t, x)∣∣∣ dSxdt≤ C

(1 + ‖hRd‖2L2(Σ)

) 12

∫Σ+,ε,ω

∣∣∣∂νu(t, x)e−ϕh ∣∣∣2 dSxdt

with C > 0 is independent of h and this inequality holds for all ω such that

|ω − ω0| ≤ ε. Next by trace theorem, we have that ‖Rd‖L2(Σ) ≤ C‖Rd‖H1scl(Q)

.

Using this, we get∣∣∣∣∣∣∣∫

Σ\G

∂νu(t, x)v(t, x)dSxdt

∣∣∣∣∣∣∣ ≤ C ∫

Σ+,ε,ω

∣∣∣∂νu(t, x)e−ϕh ∣∣∣2 dSxdt 12 .

Now ∫Σ+,ε,ω

∣∣∣∂νu(t, x)e−ϕh ∣∣∣2 dSxdt = 1ε

∫Σ+,ε,ω

ε∣∣∣∂νu(t, x)e−ϕh ∣∣∣2 dSxdt


≤ 1ε

∫Σ+,ε,ω

∂νϕ∣∣∣∂νu(t, x)e−ϕh ∣∣∣2 dSxdt.

Using the boundary Carleman estimate (3.3.1), we have

h

ε

∫Σ+,ε,ω

∂νϕ∣∣∣∂νu(t, x)e−ϕh ∣∣∣2 dSxdt ≤ C‖he−ϕ/hLA(1),q1u‖2L2(Q).

We now proceed as before to conclude that

h

∫Σ\G

∂νu(t, x)v(t, x)dSxdt→ 0 as h→ 0+.

3.6 Proof of Theorem 3.2.3

In this section, we prove the uniqueness results.

3.6.1 Recovery of vector potential A

We consider the integral,∫Q

(−2A · ∇xu2 + 2A0∂xu2 + q̃u2) (t, x)v(t, x)dxdt.

Substituting (3.4.4) for u2 and (3.4.1) for v into the above equation, and letting

h→ 0+, we arrive at∫Q

(−ω · A+ A0)Bd(t, x)Bg(t, x)dxdt = 0 for all ω ∈ Sn−1 such that |ω − ω0| ≤ ε.

Denote ω̃ := (1,−ω), A = (A0, A), and using the expressions for Bd and Bg, see(3.4.1) and (3.4.4), we get

J :=

∫R1+n

ω̃ · A(t, x)e−iξ·(t,x) exp

∞∫0

ω̃ · A(t+ s, x− sω)

dxdt = 0,


where ξ ·(1,−ω) = 0 for all ω with |ω−ω0| < ε. We decompose R1+n = R(1,−ω)⊕(1,−ω)⊥. We then get

J =

∫(1,−ω)⊥

e−iξ·k

∫R

ω̃ · A(k + τ(1,−ω)) exp

∞∫τ

ω̃ · A(k + s(1,−ω))ds

√2dτ dk.

Here dk is the Lebesgue measure on the hyperplane (1,−ω)⊥.

= −√

2

∫(1,−ω)⊥

e−iξ·k

∫R

∂τ exp

∞∫τ

ω̃ · A(k + s(1,−ω))ds

dτ dk

= −√

2

∫(1,−ω)⊥

e−iξ·k

1− exp∫

R

ω̃ · A(k + s(1,−ω))ds

dk= CF(1,−ω)⊥

1− exp∫

R

ω̃ · A(k + s(1,−ω))ds

(ξ) for some constant C.Since the integral J = 0, we have that

F(1,−ω)⊥

1− exp∫

R

ω̃ · A(k + s(1,−ω))ds

(ξ) = 0, k ∈ (1,−ω)⊥.This gives us

exp

∫R

ω̃ · A(k + s(1,−ω))ds

= 1, for all k ∈ (1,−ω)⊥ and all ω with |ω − ω0| < ε.Thus we deduce that∫R

ω̃ · A(t+ s, x− sω)ds = 0, (t, x) ∈ (1,−ω)⊥ and for all ω with |ω − ω0| < ε.

(3.6.1)

Now we show that the orthogonality condition (t, x) ∈ (1,−ω)⊥, can be removedusing a change of variables as used in [64].


Consider any (t, x) ∈ R1+n. Then(t+ x · ω

2, x+

(t− x · ω)ω2

)is a point on (1,−ω)⊥, and we have∫

R

ω̃ · A(t+ x · ω

2+ s̃, x+

(t− x · ω)ω2

− s̃ω)

ds̃ = 0.

Consider the following change of variable in the above integral:

s =x · ω − t

2+ s̃.

Then we have ∫R

ω̃ · A (t+ s, x− sω) ds = 0.

Therefore, we have that∫R

(1,−ω)·A(t+s, x−sω)ds = 0 for all (t, x) ∈ R1+n and for all ω with |ω−ω0| < ε.

To conclude the uniqueness result for A, we prove the following lemma.

Lemma 3.6.1. Let n ≥ 3 and F = (F0, F1, · · · , Fn) be a real-valued vector fieldwhose components are C∞c (R

1+n) functions. Suppose

LF (t, x, ω) :=

∫R

(1, ω) · F (t+ s, x+ sω)ds = 0

for all ω ∈ Sn−1 near a fixed ω0 ∈ Sn−1 and for all (t, x) ∈ R1+n. Then there exista Φ ∈ C∞c (R1+n) such that F (t, x) = ∇t,xΦ(t, x).

Proof. The proof follows the analysis similar to the ones used in [48, 68], where

support theorems involving light ray transforms have been proved.

Denote ω = (ω1, · · · , ωn) ∈ Sn−1. We write

LF (t, x, ω) =

∫R

n∑i=0

ωiFi(t+ s, x+ sω)ds, where ω0 = 1.


Let η = (η0, η1, · · · , ηn) ∈ Rn+1 be arbitrary. We have

(η · ∇t,x)LF (t, x, ω) =∫R

n∑i,j=0

ωiηj∂jFi(t+ s, x+ sω)ds. (3.6.2)

By fundamental theorem of calculus, we have∫R

d

ds(η · F )(t+ s, x+ sω)ds = 0.

Butd

ds(η · F )(t+ s, x+ sω) =

n∑i,j=0

ωiηj∂iFj(t+ s, x+ sω).

with ∂0 = ∂t and ∂j = ∂xj for j = 1, 2, · · · , n. Therefore∫R

n∑i,j=0

ωiηj∂iFj(t+ s, x+ sω)ds = 0. (3.6.3)

Subtracting (3.6.3) from (3.6.2), we get,

(η · ∇t,x)LF (t, x, ω) =∫R

n∑i,j=0

ωiηj (∂jFi − ∂iFj) (t+ s, x+ sω)ds.

Since LF (t, x, ω) = 0 for all ω near ω0, and for all (t, x) ∈ R1+n, we have,

Ih(t, x, ω, η) :=

∫R

n∑i,j=0

ωiηjhij(t+ s, x+ sω)ds = 0. (3.6.4)

Next we will show that the (n + 1) dimensional Fourier transform ĥij(ζ) = 0 for

all space-like vectors ζ near the set {ζ : ζ · (1, ω) = 0, ω near ω0}.

We have

ωiηjĥij(ζ) =

∫R1+n

e−i(t,x)·ζωiηjhij(t, x)dtdx,

where ω, η are fixed and ζ ∈ (1, ω)⊥. Decomposing

R1+n = R(1, ω) + k, where k ∈ (1, ω)⊥,


we get,

ωiηjĥij(ζ) =√

2

∫(1,ω)⊥

e−ik·ζ∫R

ωiηjhij(k + s(1, ω)) dsdk.

Using (3.6.4), we get that,

n∑i,j=0

ωiηjĥij(ζ) = 0, for all ζ ∈ (1, ω)⊥, for all η ∈ R1+n, and for all ω near ω0 with ω0 = 1.

(3.6.5)

Let us take for η the standard basis vectors in R1+n.

Now {ej, 1 ≤ j ≤ n} be the standard basis of Rn. Let ω0 = e1, and assumethat ζ0 = (0, e2). Then this is a space-like vector that satisfies the condition

ζ0 ∈ (1, ω0)⊥. We will show that ĥij(ζ0) = 0 for all 0 ≤ i, j ≤ n. Consider thecollection of the following unit vectors for 3 ≤ k ≤ n:

ωk(α) = cos(α)e1 + sin(α)ek.

Note that for each 3 ≤ k ≤ n, ωk(α) is near ω0 for α near 0. Also ζ0 ∈ (1, ωk(α))⊥

for all such α and for all 3 ≤ k ≤ n. Let η be the collection of standard basisvectors in R1+n.

Substituting the above vectors ωk(α) and η, we get the following equations:

ĥ0j(ζ0)+cos(α)ĥ1j(ζ

0)+sin(α)ĥkj(ζ0) = 0 for all 3 ≤ k ≤ n, 0 ≤ j ≤ n and α near 0.

From this we have that

ĥ0j(ζ0) = ĥ1j(ζ

0) = ĥkj(ζ0) = 0 for all 3 ≤ k ≤ n and 0 ≤ j ≤ n.

Using the fact that

ĥij(ζ0) = −ĥji(ζ0) for all 0 ≤ i, j ≤ n,

we get

ĥij(ζ0) = 0, for all 0 ≤ i, j ≤ n with ζ0 = (0, e2). (3.6.6)


Now our goal is to show that the Fourier transform ĥij(ζ) vanishes for all light-

like vectors ζ in a small enough neighborhood of (0, e2). To show this, assume

ζ = (τ, ξ) be a space-like vector close to (0, e2). Without loss of generality, assume

that ζ is of the form

ζ = (τ, ξ) where |ξ| = 1 and |τ | < 1.

For instance, we can take ζ as

ζ = (− sinϕ, ξ),

where ϕ is close to 0 and ξ = (ξ1, · · · , ξn) is written in spherical coordinates as

ξ1 = sinϕ1 cosϕ2

ξ2 = cosϕ1

ξ3 = sinϕ1 sinϕ2 cosϕ3

...

ξn−1 = sinϕ1 · · · sinϕn−2 cosϕn−1ξn−1 = sinϕ1 · · · sinϕn−2 sinϕn−1.

Note that if ϕ1, · · · , ϕn−1 are close to 0, then ξ is close to e2.

Let

ωϕ = cos(ϕ)e1 + sin(ϕ)e2.

Also let

ωkϕ(α) = cos(α) cos(ϕ)e1 + sin(ϕ)e2 + sin(α) cos(ϕ)ek for k ≥ 3.

Since ϕ is close to 0, and letting α close enough to 0, we have that ωϕ and ωkϕ are

close to ω0.

Now, let A be an orthogonal transformation such that Ae2 = ξ, where ξ is as

above. With this A, consider the vectors

ωζ = Aωϕ and ωkζ = Aω

kϕ for k ≥ 3.


We have that

ζ ∈ (1, ωζ)⊥.

To see this, note that ζ = (− sinϕ, ξ) and we have

− sinϕ+ 〈Ae2, Aωϕ〉 = − sinϕ+ 〈e2, ωϕ〉 = 0.

Also similarly, we have that,

ζ ∈ (1, ωkζ )⊥ for all k ≥ 3.

Using the standard basis vectors for ηj and the vectors (1, ωkζ ) in (3.6.5), we get,

for all k ≥ 3,

ĥ0j(ζ)+sinϕ

(n∑i=1

ai2ĥij(ζ)

)+cosα cosϕ

(n∑i=1

ai1ĥij(ζ)

)+sinα cosϕ

(n∑i=1

aikĥij(ζ)

)= 0

This then implies that

ĥ0j(ζ) + sinϕ

(n∑i=1

ai2ĥij(ζ)

)= 0

cosϕ

(n∑i=1

ai1ĥij(ζ)

)= 0

cosϕ

(n∑i=1

aikĥij(ζ)

)= 0

Letting j = 0, since ĥ00(ζ) = 0, we have,

n∑i=1

aijĥi0(ζ) = 0 for all 1 ≤ j ≤ n.

Since A is an invertible matrix, we have that ĥi0(ζ) = 0 for all 1 ≤ i ≤ n.Now proceeding similarly and using the fact that hij is alternating, we have that

ĥij(ζ) = 0 for all 0 ≤ i, j ≤ n. The same argument as above also works for rζ,where ζ is as above and r > 0.

Since the support of all hij is a compact subset of R1+n, by Paley-Wiener theorem,we have ĥij(ζ) = 0 ∀ i, j = 0, 1, 2, ..., n. Hence, by Fourier inversion formula, wesee that hij(t, x) = 0 ∀ (t, x) ∈ R1+n, this gives us ∇t,xF (t, x) = 0 ∀ (t, x) ∈ R1+n.


Using Poincaré lemma, we have that there exists a Φ(t, x) ∈ C∞c (R1+n) such thatF = ∇t,xΦ.

This then gives that

A(t, x) :=(A(2) −A(1)

)(t, x) := (A0, A1, · · · , An) := ∇t,xΦ(t, x).

3.6.2 Recovery of potential q

In Section 3.6.1, we showed that there exist a Φ such that (A2 − A1)(t, x) =∇t,xΦ(t, x). After replacing the pair (A(1), q1) by (A(3), q3) where A(3) = A(1) +∇t,xΦ and q3 = q1, we conclude that A(3) = A(2). Therefore substituting (3.4.4)for u2 and (3.4.1) for v, and letting h→ 0+ in∫Q

q(t, x)u2(t, x)v(t, x)dxdt =

∫Ω

∂tu(T, x)v(T, x)dx−∫

Σ\G

∂νu(t, x)v(t, x)dSxdt,

we get, ∫R1+n

q(t, x)e−iξ·(t,x)dxdt = 0 for all ξ ∈ (1,−ω)⊥ and ω near ω0.

The set of all ξ such that ξ ∈ (1,−ω)⊥ for ω near ω0 forms an open cone and sinceq ∈ L∞(Q) has compact support, using Paley-Wiener theorem we conclude thatq1(t, x) = q2(t, x) for all (t, x) ∈ Q. This completes the proof of Theorem 3.2.3.

Chapter 4

Inverse Boundary Value Problem

for a Non-linear Hyperbolic

Partial Differential Equations

4.1 Introduction and statement of the main re-

sult

In this chapter, we consider the inverse boundary value problem for a non-linear

wave equation of divergence form with space dimension n ≥ 3. We will prove theunique determination of non-linearity from the boundary measurements so-called

the hyperbolic Dirichlet to Neumann map.

Let Ω ⊂ Rn be a bounded domain with smooth boundary ∂Ω. For T > 0, letQT := (0, T ) × Ω and denote its lateral boundary by ∂QT := (0, T ) × ∂Ω, andalso denote [∂QT ] := [0, T ] × ∂Ω. We will simply write Q = QT , ∂Q = ∂QT forT =∞.

Consider the following initial boundary value problem (IBVP):∂2t u(t, x)−∇x · ~C(x,∇xu(t, x)) = 0, (t, x) ∈ QT ,

u(0, x) = ∂tu(0, x) = 0, x ∈ Ω,

u(t, x) = �f(t, x), (t, x) ∈ ∂QT ,

(4.1.1)

47

Chapter 4. Inverse Problem for Non-Linear Hyperbolic PDE 48

where ∇x := (∂1, ∂2, · · · , ∂n) , ∂j = ∂xj for x = (x1, x2, · · · , xn). Here ~C(x, q) isgiven by

~C(x, q) := γ(x)q + ~P (x, q) + ~R(x, q) (4.1.2)

for vector q := (q1, q2, · · · , qn) ∈ Rn, C∞(Ω) 3 γ(x) ≥ C > 0 for some constant C,

~P (x, q) :=

(n∑

k,l=1

c1klqkql,

n∑k,l=1

c2klqkql,

n∑k,l=1

c3klqkql, · · · ,n∑

k,l=1

cnklqkql

), (4.1.3)

each cikl ∈ C∞(Ω) and ~R(x, q) ∈ C∞(Ω × H) with H := {q ∈ Rn : |q| ≤ h}for some constant h > 0, satisfies the following estimate: there exists a constant

C > 0 such that∣∣∣∂αq ∂βx ~R(x, q)∣∣∣ ≤ C|q|3−|α| for all multi-indices α, β with |α| ≤ 3. (4.1.4)Denote by B∞(∂QT ) the completion of C

∞0 (∂QT ) with respect to the topology

of the Fréchet space C∞([∂QT ]) with metric d(·, ·). Let m ≥ [n/2] + 3 with thelargest integer [n/2] not exceeding n/2 and BM := {f ∈ B∞(∂QT ) : d(0, f) ≤M}with a fixed constant M > 0, then there exists �0 = �0(h,m, T,M) > 0 such that

(4.1.1) has a unique solution u ∈ Xm :=⋂mj=0 C

j ([0, T ];Hm−j(Ω)) for any f ∈ BMand 0 < � < �0. We refer this by the unique solvability of (4.1.1). We will provide

some arguments about this together with the �-expansion of the solution u to the

IBVP (4.1.1) in Section 4.2.

Based on this, we define the Dirichlet to Neumann (DN) map ΛT~C by

ΛT~C(�f) = ν(x) · ~C(x,∇xuf )|∂QT , f ∈ BM , 0 < � < �0 (4.1.5)

where uf (t, x) is the solution to (4.1.1) and ν(x) is the outer unit normal vector

of ∂Ω at x ∈ ∂Ω.

The inverse problem we are going to consider is the uniqueness of identifying

γ = γ(x) and ~P = ~P (x, q) from the DN map ΛT~C . More precisely it is to show that

if the DN maps ΛT~Ci, i = 1, 2 given by (4.1.5) for ~C = ~Ci, i = 1, 2 are the same,

then (γi, ~Pi), i = 1, 2 are the same, where (γi, ~Pi), i = 1, 2 are (γ, ~P ) associated to

~Ci, i = 1, 2.


The non-linear wave equation of the form (4.1.1) arises as a model equation of a vi-

brating string with elasticity coefficient depending on strain and a model equation

describing the anti-plane deformation of a uniformly thin piezoelectric material

for the one spatial dimension ([46]), and as a model equation for non-linear Love

waves for the two spatial dimension ([62]).

There are several works on inverse problems for non-linear wave equations. For ex-

ample, Denisov [14], Grasselli [21] and Lorenzi-Paparoni[42] considered the inverse

problems related to non-linear wave equations, but non-linearity in their works is

in lower order terms. Under the same set up as our inverse problem except the

space dimension, Nakamura-Watanabe in [45] identified (γ, ~P ) by giving a recon-

struction formula in one space dimension which also gives uniqueness. We are

going to prove the uniqueness for our inverse problem when the space dimension

n ≥ 3.

We will also mention about some related works for elliptic and parabolic equa-

tions. For elliptic equations, Kang-Nakamura in [31] studied the uniqueness for

determining the non-linearity in conductivity equation. Our result can be viewed

as a generalization of [31] for non-linear wave equation. There are other works re-

lated to non-linear elliptic PDE, we refer to [23, 28, 30, 43, 69–71]. For parabolic

equations, we refer to [11, 28, 34].

In order to state our main result, we need to define the filling time T ∗.

Definition 4.1.1. Let ET be the maximal subdomain of Ω such that any solution

v = v(t, x) of ∂2t v − ∇x · (γ∇xv) = 0 in QT will become zero in this subdomainat t = 2T if the Cauchy data of v on ∂Q2T are zero. We call E

T the influence

domain. Then define the filling time T ∗ by T ∗ = inf{T > 0 : ET = Ω}.

By the Holmgren-John-Tataru unique continuation property of solutions of the

above equation for v in Definition 4.1.1, there exists a finite filling time T ∗ (see

[32] and the references there in for further details). Based on this we have the

following main theorem.

Theorem 4.1.2. For i = 1, 2, let

~P (i)(x, q) :=

(n∑

k,l=1

c1(i)kl qkql,

n∑k,l=1

c2(i)kl qkql,

n∑k,l=1

c3(i)kl qkql, · · · ,

n∑k,l=1

cn(i)kl qkql

)


and ~C(i)(x, q) = γi(x)p + ~P(i)(x, q) + ~R(i)(x, q) with γi, ~P

(i) and ~R(i), i = 1, 2

satisfying the same conditions as for γ, ~P and ~R. Further let u(i), i = 1, 2 be the

solutions to the following IBVP:∂2t u

(i)(t, x)−∇x · ~C(i)(x,∇xu(i)(t, x)) = 0, (t, x) ∈ QT ,

u(i)(0, x) = ∂tu(i)(0, x) = 0, x ∈ Ω,

u(i) = �f(t, x), (t, x) ∈ ∂QT

(4.1.6)

with any 0 < � < �0. Assume T > 2T∗ and let ΛT~C(1) and Λ

T~C(2)

be the DN maps as

defined in (4.1.5) corresponding to u(1) and u(2) respectively. Assume that

ΛT~C(1)(�f) = ΛT~C(2)

(�f), f ∈ BM , 0 < � < �0. (4.1.7)

Then we have

γ1(x) = γ2(x), cj(1)kl (x) = c

j(2)kl (x), x ∈ Ω, 1 ≤ j, k, l ≤ n.

The most difficult part of proving Theorem 4.1.2 is showing the uniqueness of iden-

tifying the quadratic nonlinear part ~P (x, q). The key ingredients for showing this

are to use the control with delay in time (see (4.4.6)) and the special polarization

for the difference of the quadratic nonlinear part with integration with respect to

the delay time (see (4.4.7), (4.4.11)) coming from two ~C(i)(x, q), i = 1, 2 so that

via the Laplace tranform with respect to t, we can relate the problem of identifying

the quadratic part to that for a nonlinear elliptic equation. The reduced problem

is almost the same as the one considered in [31].

The chapter is organized as follows. In §4.2, we will introduce the �-expansion ofthe IBVP to analyze the hyperbolic DN map. As a consequence, we will show

that the hyperbolic DN map determines the hyperbolic DN map associated with

the equation ∂2t v − ∇x · (γ∇xv) = 0 in (0, T ) × Ω. This immediately implies theuniqueness of identifying γ. §4.4 is devoted to proving the uniqueness of identifying~P (x, q).


4.2 �-expansion of the solution to the IBVP

Theorem 4.2.1. Let m ≥ [n/2] + 3 with the largest integer [n/2] not exceedingn/2 and f ∈ BM with a fixed constant M > 0, then for given T > 0, there exists�0 = �0(h,m,M) > 0 such that for any 0 < � < �0, (4.1.1) has a unique solution

u ∈ Xm where h and BM , Xm were defined in Section 4.1 right after (4.1.3) andthe paragraph after (4.1.4), respectively. Moreover, it admits an expansion which

we call �-expansion:

u = �u1 + �2u2 +O(�

3); (�→ 0) (4.2.1)

where u1 is a solution to∂2t u1(t, x)−∇x · (γ(x)∇xu1(t, x)) = 0, (t, x) ∈ QT

u1(0, x) = ∂tu1(0, x) = 0, x ∈ Ω

u1 = f(t, x), (t, x) ∈ ∂QT

(4.2.2)

and u2 is a solution to∂2t u2(t, x)−∇x · (γ(x)∇xu2(t, x)) = ∇x · ~P (x,∇xu1), (t, x) ∈ QT

u2(0, x) = ∂tu2(0, x) = 0, x ∈ Ω

u2 = 0, (t, x) ∈ ∂QT .

(4.2.3)

Remark 4.2.2. For the well-posedness of initial boundary value problems (4.2.2)

and (4.2.3), we refer to Theorem 2.45 of [32].

Remark 4.2.3. w� = O(�3) means that sup0≤t≤T

∑mk=0 ||w�(t)||2m−k = O(�3), where

||.||k is the norm of the usual Sobolev space W k,2(Ω).

Proof. We look for a solution u(t, x) to (4.1.1) of the form

u(t, x) = � {u1(t, x) + � (u2(t, x) + w(t, x))} , w = O(�) (�→ 0). (4.2.4)


Then, w(t, x) has to satisfy

∂2tw(t, x)−∇x (γ(x)∇xw(t, x)) = �−2∇x ·R(x, �∇xu1 + �2∇xu2 + �2∇xw)

+�∑n

j=1 ∂j

(∑nk,l=1 c

jkl(x) (∂ku1∂lu2 + ∂lu1∂ku2) (t, x)

)+�∑n

j=1 ∂j

(∑nk,l=1 c

jkl(x) (∂ku1∂lw + ∂lu1∂kw) (t, x)

)+�2

∑nj=1 ∂j

(∑nk,l=1 c

jkl (∂ku2∂lu2 + ∂ku2∂lw + ∂lu2∂kw + ∂kw∂lw)

)in QT ,

w(0, x) = ∂tw(0, x) = 0, x ∈ Ω and w|∂QT = 0.

By the mean value theorem, we have

R(x, �∇xu1 + �2∇xu2 + �2∇xw) = R(x, �∇xu1 + �2∇xu2)

+

1∫0

d

dθR(x, �∇xu1 + �2∇xu2 + θ�2∇xw)dθ

= R(x, �∇xu1 + �2∇xu2) + �3K(x, �∇xw; �)∇xw

where

�K(x, �∇xw; �) :=1∫

0

DqR(x, �∇xu1 + �2∇xu2 + θ�2∇xw)dθ

with DqR(x, q) = ((∂qiRj))1≤i,j≤n and Kij = ∂qiRj. After using this in previous

equation, we get

∂2tw −∇x (γ(x)∇xw) = �∑n

j=1 ∂j

(∑nk,l=1 c

jkl(x) (∂ku1∂lu2 + ∂lu1∂ku2)

)+�∑n

j=1

∑nk,l=1 ∂j

(cjkl(x)∂ku1

)∂lw + �

∑nj=1

∑nk,l=1

(cjkl(x)∂ku1

)∂2jlw

+�∑n

j=1

∑nk,l=1 ∂j

(cjkl(x)∂lu1

)∂kw + �

∑nj=1

∑nk,l=1

(cjkl(x)∂ku1

)∂2jkw

+�2∑n

j=1 ∂j

(∑nk,l=1 c

jkl(x)∂ku2∂lu2

)+ �2

∑nj=1

∑nk,l=1 ∂j

(cjkl(x)∂ku2

)∂lw

+�2∑n

j=1

∑nk,l=1

(cjkl(x)∂ku2

)∂2jlw + �

2∑n

j=1

∑nk,l=1 ∂j

(cjkl(x)∂lu2

)∂kw

+�2∑n

j=1

∑nk,l=1

(cjkl∂ku2

)∂2jkw + �

2∑n

j=1

∑nk,l=1 c

jkl

(∂2jkw∂lw + ∂kw∂

2jlw)

+�2∑n

j=1

∑nk,l=1 ∂jc

jkl∂kw∂lw + �

∑ni=1

∑nj=1 ∂iKij(x, �∇xw; �)∂jw

+�2∑n

i=1

∑nj,l=1 ∂qlKij(x, �∇xw; �)∂jw∂2ilw

+�∑n

i=1

∑nj=1Kij(x, �∇xw; �)∂2ijw + �−2∇x ·R(x, �∇xu1 + �∇xu2), in QT ,

w(0, x) = ∂tw(0, x) = 0, x ∈ Ω and w|∂QT = 0.


Thus, finally we have derived the following initial boundary value problem for w:

∂2tw −∇x · (γ(x)∇xw) = �∑n

j=1 ∂j

(∑nk,l=1 c

jkl(x) (∂ku1∂lu2 + ∂lu1∂ku2)

)+�∑n

j=1

∑nk,l=1 ∂j

(cjkl(x)∂ku1

)∂lw + �

∑nj=1

∑nk,l=1

(cjkl(x)∂ku1

)∂2jlw

+�∑n

j=1

∑nk,l=1 ∂j

(cjkl(x)∂lu1

)∂kw + �

∑nj=1

∑nk,l=1

(cjkl(x)∂lu1

)∂2jkw

+�2∑n

j=1 ∂j

(∑nk,l=1 c

jkl(x)∂ku2∂lu2

)+ �2

∑nj=1

∑nk,l=1 ∂j

(cjkl(x)∂ku2

)∂lw

+�2∑n

j=1

∑nk,l=1

(cjkl(x)∂ku2

)∂2jlw + �

2∑n

j=1

∑nk,l=1 ∂j

(cjkl(x)∂lu2

)∂kw

+�2∑n

j=1

∑nk,l=1

(cjkl∂ku2

)∂2jkw + �

2∑n

j=1

∑nk,l=1 c

jkl

(∂2jkw∂lw + ∂kw∂

2jlw)

+�2∑n

j=1

∑nk,l=1 ∂jc

jkl∂kw∂lw + �

−2∇x ·R(x, �∇xu1 + �∇xu2)

+�∑n

i=1

∑nj=1 ∂iKij(x, �∇xw; �)∂jw + �

∑ni=1

∑nj=1Kij(x, �∇xw; �)∂2ijw

+�2∑n

i=1

∑nj,l=1 ∂qlKij(x, �∇xw; �)∂jw∂2ilw, in QT ,

w(0, x) = ∂tw(0, x) = 0, x ∈ Ω and w|∂QT = 0.

(4.2.5)

In order to simplify the description of the above equation for w, let us introduce

the following notations:

A(w(t))w = ∇x · (γ(x)∇xw(t, x)) + �Γ(x,∇xw; �) · ∂2xw,

�Γ(x,∇xw; �) := �(∑n

k=1 cjkl(x)∂ku1

)1≤j,l≤n + �

(∑nl=1 c

jkl(x)∂lu1

)1≤j,k≤n

+�2(∑n

k=1 cjkl(x)∂ku2

)

Some Inverse Problems in Hyperbolic Partial Di erential ... · a lot during my PhD under his...

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