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)
Chapter 1. Introduction 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)
).
Chapter 1. Introduction 4
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].
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 7
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
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 8
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
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 9
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.
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 10
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)
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 11
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)
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 12
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
2sinh ρ sin θ sinφ
x3 =1
2sinh ρ cos θ sinφ
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
2sinh ρ cos θ sinφ
)with θ ∈ (0, 2π), φ ∈ (0, π), and cosh ρ = r.
Next, we have
∂F
∂θ=(
0,1
2sinh ρ cos θ sinφ,−1
2sinh ρ sin θ sinφ
)∂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]
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 13
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).
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 14
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
2sinh ρ cos θ sinφ
).
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)
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 15
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.
Chapter 2. An Inverse Problem with Source and Receiver at Distinct Points 16
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.
Chapter 3. Relativistic Schrödinger Equation 19
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 ∈
Chapter 3. Relativistic Schrödinger Equation 20
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.
Chapter 3. Relativistic Schrödinger Equation 21
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].
Chapter 3. Relativistic Schrödinger Equation 22
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)
Chapter 3. Relativistic Schrödinger Equation 23
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
Chapter 3. Relativistic Schrödinger Equation 24
− 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.
Chapter 3. Relativistic Schrödinger Equation 25
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
Chapter 3. Relativistic Schrödinger Equation 26
− 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
Chapter 3. Relativistic Schrödinger Equation 27
+ 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)
Chapter 3. Relativistic Schrödinger Equation 28
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.
Chapter 3. Relativistic Schrödinger Equation 29
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),
Chapter 3. Relativistic Schrödinger Equation 30
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).
Chapter 3. Relativistic Schrödinger Equation 31
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).
Chapter 3. Relativistic Schrödinger Equation 32
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)
Chapter 3. Relativistic Schrödinger Equation 33
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.
Chapter 3. Relativistic Schrödinger Equation 34
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.
Chapter 3. Relativistic Schrödinger Equation 35
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
∂tu(T, x)v(T, x)dx
−∫
Σ\G
∂νu(t, x)v(t, x)dSxdt.
Chapter 3. Relativistic Schrödinger Equation 36
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
∂tu(T, x)v(T, x)dx
∣∣∣∣∣∣ ≤∫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,
Chapter 3. Relativistic Schrödinger Equation 37
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
Chapter 3. Relativistic Schrödinger Equation 38
≤ 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,
Chapter 3. Relativistic Schrödinger Equation 39
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].
Chapter 3. Relativistic Schrödinger Equation 40
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.
Chapter 3. Relativistic Schrödinger Equation 41
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)⊥,
Chapter 3. Relativistic Schrödinger Equation 42
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)
Chapter 3. Relativistic Schrödinger Equation 43
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.
Chapter 3. Relativistic Schrödinger Equation 44
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.
Chapter 3. Relativistic Schrödinger Equation 45
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.
Chapter 4. Inverse Problem for Non-Linear Hyperbolic PDE 49
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
)
Chapter 4. Inverse Problem for Non-Linear Hyperbolic PDE 50
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).
Chapter 4. Inverse Problem for Non-Linear Hyperbolic PDE 51
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)
Chapter 4. Inverse Problem for Non-Linear Hyperbolic PDE 52
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.
Chapter 4. Inverse Problem for Non-Linear Hyperbolic PDE 53
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
)
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