STABILITY ESTIMATES FOR ELECTROMAGNETIC SCATTERING FROM OPEN
CAVITY
By
Qiong Zheng

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Applied Mathematics – Doctor of Philosophy
2013

ABSTRACT
STABILITY ESTIMATES FOR ELECTROMAGNETIC SCATTERING
FROM OPEN CAVITY
By
Qiong Zheng
The electromagnetic scattering from an open large cavity embedded in an infinite plane is
of practical importance due to its significant industrial and military applications. Examples
of cavities include jet engine inlet ducts, exhaust nozzles and cavity-backed antennas [3, 4].
In many practical applications, one is interested in the cavity problem with either a large
wave number k or a large diameter a, in which case the solution has a highly oscillatory
nature [7]. While the original time-harmonic problem is modeled by Helmholtz equation in
the unbounded domain, the reformulated model through Fourier transform is essentially a
Helmholtz equation in the bounded domain with mixed nonlocal boundary condition. Deriving an explicit dependency between the wave energy and the wave number is mathematically
interesting and challenging. The stability estimate is also important as it defines relations
between the wave number and the discretization parameters in the error analysis [16]. For
the open cavity problem, while the stability analysis for the rectangular cavity was derived
recently [8] as described , the stability results for more general shapes of cavities are to be
explored. The objective of this thesis work is to partially answer this question by imposing
some geometric assumptions.
We first start from considering a class of cavity with a strong geometric constraint. The
energy stability is established by careful choices of the parameters, and test functions, which
take full advantage of geometric properties. The arguments are based on the appropriate
usage of the real and imaginary part of the weak formulation of the problem, the separation of

lower frequency and higher frequency part, and connections between frequency components
and spatial components. The energy in cavity is bounded by the energy of incoming field
with coefficient in terms of powers of wave number. Next, we investigate the case where a
weaker geometric constraint is imposed. A new auxiliary function with compact support near
the boundary of the cavity is carefully constructed to reformulate the problem. However, the
original homogeneous Helmholtz equation is changed to a non-homogeneous one, all previous
work in homogeneous equation must be suitably modified, and the estimate in terms of wave
number k is obtained from detailed analysis of this auxiliary function. The energy norm
7
is proved to be at most in the order of k 10 , which is the same in terms of the power
of wave number k as the case with strong geometric conditions but with other additional
terms. Furthermore, we studied the case where the cavity domain is of rectangular-like
shape, where new test function is introduced and new inequalities are established to derive
the energy estimate.

ACKNOWLEDGMENTS

I would like to express my deepest gratitude to my advisor Dr. Zhengfang Zhou, who
provides exceptional guidance and tremendous support during my graduate study. Without
his guidance, this thesis would not have been possible. I feel very lucky to have the opportunity to study under the supervision of such a knowledgable and smart mathematician with
great personality. He always generously offers time to meet for questions and discussions,
he is very inspiring and patient in enhancing my mathematical understanding, and his view
on mathematics helps me to shape my own understanding. I am also very thankful for his
encouragements and suggestions when I meet with difficulties. His elegant personality and
persistent passion in mathematics & life will have a long time influence on me.
I would like to thank Dr. Gang Bao, Dr. Andrew Christlieb, Dr. Keith Promislow and
Dr. Baisheng Yan, for serving as members of my doctoral committee and for their invaluable
suggestions. In particular I want to thank Dr. Baisheng Yan for his consistent support and
great help during my graduate study.
I am grateful to the entire faculty and staff in the Department of Mathematics who have
taught and assisted me in one way or another. My special thanks go to Dr. Guowei Wei for
offering opportunities and research assistantship in research projects, as well as providing
valuable discussions, suggestions and assistance. And I would like to thank Dr. Casim
Abbas, Dr. Gabor Francsics, Dr. Jeanne Wald for their help and great support. Many
thanks to Dr. Clifford Weil for his help in providing useful information for formatting the
thesis. And I am grateful to Ms. Barbara Miller and Ms. Leslie Aitcheson for their assistance
in providing information and help.
Also, I want take this opportunity to thank my graduate fellows and friends for providing
iv

kind help, sharing with me your expertise knowledge, and sharing the challenges and happiness together. Particular thanks go to Cheryl Balm, Liangzhang Bao, Liping Chen, Yuqi
Hong, Jun Lai, Junshan Lin, Shu-man Lin, Xun Wang, Daniel Smith, Li Yang, Xin Yang,
Hai Zhang, Liangming Zhou.
Finally I would like to thank my parents and my husband for their endless love and
always standing besides me over the years.

v

TABLE OF CONTENTS

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii
Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1 Electromagnetic Scattering from Open Cavity . . . . . . . . . . . . .
1.2 Mathematical Formulation . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 2 Stability Estimates under a Strong Geometric Assumption
2.1 Main Theorem and Outlines of the Approach . . . . . . . . . . . .
2.2 Preliminary Lemmas for the Energy Identity . . . . . . . . . . . .
2.3 Conversion through Fourier Transform . . . . . . . . . . . . . . . .
2.4 Detailed Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Further Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

.
.
.
.
.
.

11
11
14
18
22
40

Chapter 3 Stability Estimates under a Weak Geometric Assumption . . .
3.1 Main Theorem and Outlines of the Approach . . . . . . . . . . . . . .
3.2 Some Basic Properties of the Auxiliary Function . . . . . . . . . . . .
3.2.1 The Construction of an Auxiliary Function . . . . . . . . . . .
3.2.2 The Existence and the Order of the Coefficients . . . . . . . .
3.3 Relations between New Source Terms and Original Source Term g
3.4 The Estimate for the Nonhomogeneous System . . . . . . . . . . . . .
3.4.1 Wave Energy Formulation . . . . . . . . . . . . . . . . . . . . . .
3.4.2 Analysis through Fourier Transform . . . . . . . . . . . . . . .
3.4.3 Auxiliary Lemmas and Final Estimates . . . . . . . . . . . . . .

42
42
46
46
49
57
64
66
69
70

Chapter 4 Stability Estimates for Rectangular-like Domains
4.1 Main Theorem and Sketches of Approach . . . . . . . .
4.2 Preliminary Lemmas . . . . . . . . . . . . . . . . . . . . .
4.3 Relation between D |wy |2 and D |w|2 . . . . . . . . . . .
4.4 Final Estimates . . . . . . . . . . . . . . . . . . . . . . . . .

.
.
.
.
.

78
78
80
87
88

Conclusion and Some Open Questions . . . . . . . . . . . . . . .

91

Chapter 5

BIBLIOGRAPHY

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vi

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1
6

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94

LIST OF FIGURES

Figure 1.1

Geometry of the problem. (For interpretation of the references to
color in this and all other figures, the reader is referred to the electronic version of this dissertation.) . . . . . . . . . . . . . . . . . . .

7

Figure 2.1

A cavity structure satisfying geometric assumption in (2.1). . . . . .

12

Figure 3.1

Domains of different shapes . . . . . . . . . . . . . . . . . . . . . . .

43

Figure 3.2

Cavity with subdomains where the auxiliary functions are supported

47

Figure 4.1

Rectangular-like domain . . . . . . . . . . . . . . . . . . . . . . . .

79

vii

Chapter 1
Introduction

1.1

Electromagnetic Scattering from Open Cavity

One of the important subjects studied in electromagnetics is Radar Cross Section (RCS),
which measures how detectable an object is with a radar. Accurate predictions of RCS of
complex objects are of great interest to the designers [4], for instance, a stealth aircraft is
expected to have low detectability such that it is less observable (or even invisible) while a
passenger airliner is designed to have a high RCS such that it is more visible. Furthermore,
it has been studied that the RCS of a B-1A bomber is two orders of magnitude less than
that of a B-52 [30, 21]. This is mainly achieved by two approaches: shaping the body of the
airplane and using radar absorbing material; meanwhile, the jet intake plays a significant
role among the remaining contributors to the total RCS of the airplane. The jet intake
is modeled as a cavity structure which has one open end and the other end such as the
blade could be modeled as the perfect electric conductor. Other examples of cavities include
exhaust nozzles and cavity-backed antennas [3, 4].
In many practical applications, one is interested in the cavity problem with either a
large wavenumber k or a large diameter a of the computational domain, which leads to
the large ka numbers [7]. It is challenging to solve for a large wavenumber or a large
diameter cavity problem due to the high oscillation of the solution. Over the past decade,
many computational approaches have been developed in the literatures to solve the open
1

cavity problems. If the cavity has an opening with length less than one wavelength, i.e,
the low frequency end, the integral equation formulation could be used for the computation.
However, if the aperture of the cavity has a length on the order of several wavelengths,
then high-frequency computational techniques are expected [24]. One of the approaches
proposed is waveguide modal analysis, which has been applied in early literatures [26, 21], and
still provides reference solutions to compare with results obtained from more approximate
methods. In waveguide analysis, the field inside the cavity is expressed in terms of the
known waveguide modes. The unknown modal coefficients are obtained through using the
reciprocity relationship and Kirchhoff’s approximation. Another popular approach is the
shooting and bouncing ray (SBR) method as introduced in [23]. This method involves
tracing a dense grid of geometrical optics rays originating from the incident wave into the
cavity through the front aperture. After multiple bounces in the interior walls of the cavity,
the rays eventually return to the opening of the cavity. A physical optics approximation is
used to calculate the scattered field from each exit ray, then the total scattered field results
from summing the scattered field due to individual rays. In [24], the two aforementioned
approaches are investigated for cavities with rectangular and circular cross sections. For
an aperture opening on the order of ten wavelengths, the computational results agree with
each other fairly well. At lower frequencies, the modal analysis results is accurate than
the SBR results ; however, the modal approach is limited only to cavities with uniform
cross sections whereas the SBR approach can be applied with much greater flexibility in
geometrical modeling. However, these methods are restricted to relatively shallow cavities,
due to the computational cost and the ray and beam distortion problems associated with deep
cavities, where a large number of internal reflections need to be considered. Another raybased method is the Generalized Ray Expansion (GRE) method [29]. The major difference
2

between SBR and GRE methods, as mentioned in [11], lies in that the SBR method includes
only the incident geometrical optics field that enters the cavity, while the GRE method
intrinsically includes the fields diffracted into the cavity by the edges at the end of the
cavity aperture. Furthermore, for each incidence angle SBR method needs a new set of
rays to be tracked, while in the GRE method only one set of rays needs to be tracked
regardless of the incidence angles. Other computational approaches include the Iterative
Physical Optics (IPO)[28] method and the Progressive Physical Optics (PPO) method [27],
where in both methods the magnetic field integral equation (MFIE) is obtained for the
equivalent currents in the interior cavity walls, and solved by different algorithms. Moreover,
many hybrid methods have been proposed by combining one of the preceding aysmptotic
approaches and some other more accurate methods, such as the Method of Moments (MoM),
the Finite Element Method (FEM) or the Finite-Difference Time-Domain method (FDTD)
[10, 13, 20, 32, 33]. Anastassiu [5] presented a comprehensive review on the methods for the
related electromagnetic scattering problems. However, theoretical analysis is quite limited,
especially we hope to establish relations between the energy in cavity, high wave number
and incoming field.
While the original time-harmonic problem is modeled by Helmholtz equations in the
unbounded domain, the reformulated model through Fourier analysis in the upper half plane
is essentially a Helmholtz equation in the bounded domain with nonlocal boundary condition
on the aperture of the cavity. Deriving an explicit dependence between the wave energy
and the wave number is mathematically interesting and challenging. The stability estimate
is also important as it defines relations between the wave number and the discretization
parameters in the error analysis [16]. Also, using the stability of the continuous problem
with perturbation argument, the stability result of a numerical method can be obtained
3

[15]. It turns out for the stability analysis on Helmholtz equations, the bounds on k highly
depends on the geometry of the domain and the type of boundary conditions [15]. Ihlenburg
and Babuˇka [18] considered the one dimensional Helmholtz equation on D = (0, 1) with
s
Dirichlet and nonreflecting boundary condition:


 u + k 2 u = −f in D,





 u(0) = 0,





 u (1) − iku(1) = 0.
They proved the stability estimate |u|1 ≤ Ck|f |−1 for the solution u in H 1 norm. Moreover,
they generalized the stability result under higher regularity assumptions in [19]. For l > 1,
f (x) ∈ H l−1 (0, 1), then u ∈ H l+1 (0, 1) and the estimate |u|l+1 ≤ Ck l−1 f l−1 . Melenk
[25] studied the two dimensional problem with the Robin boundary condition:



 − ∆u − k 2 u = f

in D,



 ∂n u + iku = g

on ∂D.

The geometric assumption is that the domain D is a bounded star-shaped domain with
1
smooth boundary or a bounded convex domain; then for any f ∈ L2 (D), g ∈ H 2 (∂D),

| u| 2
+ |k||u| 2
≤ C(D) |f | 2
+ |g| 2
,
L (D)
L (D)
L (D)
L (∂D)

where the constant C depends only on the domain.

This was extended to the three-

dimensional case by Cummings and Feng [14]. When g = 0, a sharp estimate is established as below, u

H j (D)

≤ Cj (k) f

, where Cj (k) = O(k j−1 ) + 1 , j = 0, 1, 2
L2 (D)
k2

4

for N ≥ 2 under the assumption that D is star-shaped when j = 0, 1, and D is convex polygons or smooth domains when j = 2. Moreover, Melenk [15] showed that for any
1
f ∈ L2 (D),g ∈ H 2 (∂D), assume D is a bounded Lipschitz domain, there exists a constant
C > 0 (independent of k) such that
5
.
| u| 2
+ |k||u| 2
≤ C(D) k 2 |f | 2
+ k 2 |g| 2
L (D)
L (D)
L (D)
L (∂D)

Hetmaniuk [16] presented the stability analysis for two dimensional and three dimensional
mixed boundary problems:


 − ∆u − k 2 u = f







u = 0


in D,
on Γd ,



 ∂n u = 0
on Γn ,







 ∂ u = (iβ − α)u + g on Γ .
n
r
Based on the following geometric assumption: there exists a point x0 and a constant γ such
that


 (x − x ) · n(x) ≤ 0 for ∀x ∈ Γ ,


0
d



 (x − x0 ) · n(x) = 0 for ∀x ∈ Γn ,





 (x − x ) · n(x) ≥ γ for ∀x ∈ Γr ;
0
the stability estimate is given as

| u| 2
+ |k||u| 2
≤ C(D) |f | 2
+ |g| 2
.
L (D)
L (D)
L (D)
L (∂D)

5

A major difference between the open cavity scattering problem and the results mentioned above is the boundary condition, where the cavity problem in the bounded domain
formulation involves the nonlocal boundary condition as indicated in the following section.
More specifically, part of the boundary condition is associated with Fourier transform. The
first stability result for the cavity problem was derived recently [8] for rectangular cavity.
Because of this particular shape of the cavity, the solution u can be expressed as the Fourier
1
−1
expansion. Two special norms, which behave like H 2 and H 2 on Γ, are introduced for
the technical analysis. And the stability estimate under the rectangular cavity of depth y0
in the y direction, is given by
13
1
15 7
4
2
| u| 2
+ |k||u| 2
≤ C k 4 y0 (log k) 4 + k 4 y0 |g| 2
L (Γ)
L (D)
L (D)

when f = 0, where Γ represents the aperture part of the cavity. In this thesis, we study the
stability estimates for open cavity embedded in an infinite ground plane in a more general
framework, where the shape of the cavity is under certain geometric constraints.

1.2

Mathematical Formulation

In this section, the formulation of the problem is reviewed and notations are introduced. Let
D be the cavity region, Γ denote the aperture boundary part of the cavity and Γ := [a, b] × {y2 },
ΓC = (R\[a, b]) × {y2 } and S = ∂D\Γ. Assume X ∗ = (x0 , y0 ) represents a reference point
on the plane for the convenience of introducing geometric assumptions in the following chapters.
For the TM (transverse magnetic) polarization, the cavity scattering problem in the

6

Figure 1.1: Geometry of the problem. (For interpretation of the references to color in this
and all other figures, the reader is referred to the electronic version of this dissertation.)
unbounded domain could be reduced to the bounded domain (the cavity) problem through
a Fourier transform and radiation condition [3], which is briefly reviewed here. Let E and H
denote the total electric and magnetic fields, then the following time harmonic (E(X, t) =
e−iωt E(X)) Maxwell equations are satisfied:

× E − iωµH = 0,
× H + iω E = 0,
where ω is wave frequency, µ is the magnetic permeability and

7

is the electric permittivity.

Denote (Ei , Hi ), (Er , Hr ),(Es , Hs ) as the incident, reflective and scattering fields respectively. In the TM polarization, the electric filed is E = (0, 0, u(x, y)). The total field contains
three components, that is, u = ui + ur + us . When ui = eiαx−iβ(y−y2 ) is a plane wave,
ur = −eiαx+iβ(y−y2 ) , where α = k sin θ, β = k cos θ, θ is the incident angel with respect
to y axis, and the wave number k = ω

√

µ. Moreover, us satisfies




 ∆us + k 2 us = 0 in D ∪ {y > y2 },


 us = −(u + ur ) on
S ∪ ΓC ,
i
with the radiation condition lim
r→∞

√

r(∂r us − ikus ) = 0.

By taking the Fourier transform with respect to x, and defining

1
us (ξ, y) = √
u(x, y)e−ixξ dx,
2π R
we have

(∂yy + (k 2 − ξ 2 ))us = 0 for y > y2 .

(1.1)

By solving Eq. (1.1) with the radiation condition and taking the inverse Fourier transform,
we obtain

1
i(y−y2 )
us = √
e
2π R

k 2 −ξ 2

8

us (ξ, y2 )eiξx dξ.

Therefore, on R × {y2 },
i
∂n us = √
2π R

k 2 − ξ 2 us (ξ, y2 )eiξx dξ.

To simplify notations, define an operator T on Γ as

i
T (ϕΓ ) = √
2π R

k 2 − ξ 2 ϕΓ eiξx dξ

1
for any ϕ with ϕ ∈ H 2 (R), and

ϕΓ :=



 ϕ on Γ,



 0 on ΓC .

Note that ui + ur = 0 when y = y2 , we have

∂n u = T (u) + g,

where g =

∂
(u + ur )|Γ . Therefore, the original problem in the unbounded domain (the
∂n i

cavity and the upper half plane, that is, D ∪ {y > y2 }) can be reduced to the following
problem in the bounded domain:


 ∆u + k 2 u = 0 in D,





on S,
u = 0





 ∂n u = T (u) + g on Γ.

(1.2)

Note that the reformulated model as shown in Equation (1.2) is essentially a Helmholtz

9

equation with mixed nonlocal boundary condition; in particular, part of the boundary condition is associated with Fourier transform. There were many papers studying the equation
with boundary condition ∂n u = iku + g, which is a much simplified and local form, but
the problem is not physical. The problem with nonlocal boundary condition is much harder
mathematically, but it is more significant physically. There are very few results in this direction. New methods have to be developed. Based on certain geometric assumptions of the
cavity domain, a few stability estimates are established in the following chapters.
The rest of the dissertation is arranged as follows. In Chapter 2, the stability estimate
for a class of cavity under a strong geometric constraint is established. In Chapter 3, we
derived the stability estimate for a class of cavity under a weaker geometric assumption. In
Chapter 4, we assume the cavity structure is of rectangular-like shape, where the geometric
assumptions introduced in Chapter 2 and 3 are not satisfied; the corresponding stability
estimate is obtained. In Chapter 5, a brief summary of the main contributions of the thesis
work is presented, followed by the discussion of future work.

10

Chapter 2
Stability Estimates under a Strong
Geometric Assumption

2.1

Main Theorem and Outlines of the Approach

Throughout this chapter, an open cavity that satisfies the following geometric assumption
is considered; that is, there is a point X ∗ = (x0 , y0 ) on the plane and positive constant p1
such that

(X − X ∗ ) · n ≤ −p1

on S,

(2.1)

where n is outnormal. We should remark that if S is given by y = h(x), x ∈ (a, b) and h is
C 1 , then Equation (??) is always satisfied. Without loss of generality, we assume X ∗ = (0, 0)
in this chapter; also for simplicity, we assume that b − a = π. It should be remarked that
if X ∗ = (0, 0), then y2 > y1 > 0, where y1 = min{y|(x, y) ∈ D} from condition (2.1), one
example is shown Figure 2.1.
A similar geometric constraint was presented by Hetmaniuk [16] to study a different mixed
boundary problem. In their case, p1 can be zero. We need a slightly stronger geometric
constraint for our arguments. If p1 = 0, some additional techniques will be developed as
indicated in the next chapter. This condition guarantees the positivity of one term which
11

(a): Triangular shaped

(b): Curly shaped

Figure 2.1: A cavity structure satisfying geometric assumption in (2.1).
lies on the same side as the wave energy terms in an inequality which will be derived later.
Furthermore, we assume that the medium for the upper half plane {(x, y) | y ≥ y2 } and the
cavity is homogeneous. The main result is Theorem 2.1.1 as stated below, which provides
an explicit dependence for the wave energy on the wave number k.
Theorem 2.1.1. Under the geometric assumption given in Equation (2.1), there exists a
positive constant C such that

u

L2 (D)

+ |k| u

L2 (D)

7
≤ C k 10 g

1
2 (Γ) + √ gx L2 (Γ) .
L
k

(2.2)

Essentially, our proof of the main theorem involves three major steps.
First, we consider the weak formulation of the problem, and adopt the test functions v
used in some of the earlier work [16, 17], which are v = u and v = X · u, then the standard
computation yields the following identity,
ξ 2 − k 2 |u|2 dξ −

u 22
+ k2 u 2 2
+
L (D)
L (D)
|ξ|>k
Γ

S

(X · n)| u|2 dx + k 2

gudx −

=

Γ

(X · n)| u|2
(2.3)

(X · n)|u|2 dx + 2
Γ
12

Γ

∂n u(X ·

u)dx.

Second, since the boundary condition on Γ is given in a simpler form in terms of the
Fourier transform, it is easier to evaluate the last three terms in Equation (2.3) in the
frequency domain. So the main focus of our second step is to convert those terms into the
frequency domain. After the conversion, the following inequality can be derived:
k2

( ξ 2 − k2 +
u 22
+ k2 u 2 2
+
L (D)
L (D)
|ξ|>k
≤
Γ
+2

gudx + y2

|ξ|≤k

|ξ|≤k

2(k 2 − ξ 2 )|u|2 dξ + 2y2

(X · n)| u|2

)|u|2 dξ −
S

ξ 2 − k2

(2.4)

T (u)g
Γ

−ξu (ξ) i k 2 − ξ 2 u(ξ)dξ + 2
Γ

xux gdx + y2

|g|2 dx.
Γ

Third, terms on the right hand side of Equation (2.4) will be named by I1 through I6 ,
and we have to estimate each term from I1 to I5 . The goal is to show that each term
is controlled by norms of g and the left hand side of Equation (2.4) multiplied by a small
coefficient. To illustrate the ideas and simplify notations, let

AL =
BH =

k 2 − ξ 2 |u|2 dξ,

|ξ|≤k
|ξ|>k

(

|g|2 dx,

G=
Γ

ξ 2 − k2 +

AH =
k2

|ξ|>k

ξ 2 − k 2 |u|2 dξ,

)|u|2 dξ,

ξ 2 − k2
| u|2 dS.

J=
S

For I1 , Lemma 2.4.1 will show that AL ≤ δBH + C(δ)G, i.e., the lower frequency part of
u is dominated by L2 norm of g and higher frequency part of u. Consequently the bound
of u|| 2
is obtained, which in turn will yield the estimate for I5 after the integration
L (Γ)
by parts. Since I2 involves only lower frequency part, Lemma 2.4.1 can be used to estimate
this term. For I3 , a new test function will be constructed to derive that I3 ≤ δ Energy +

13

δJ + C(δ) g

H1

. This step uses the fact that X · n ≤ −p1 < 0, and the detailed arguments

are provided in Lemma 2.4.4. I4 is also difficult to estimate, it will use Lemma 2.4.1 and
cancelation between positive and negative part of an introduced function. This part requires
7
careful choice of parameters and gives the reason why k 10 appears in Theorem 2.1.1. It is
proved in Lemma 2.4.5.

2.2

Preliminary Lemmas for the Energy Identity

In this section, we start from the weak formulation of the problem, and utilize appropriate
test functions and separate the real and imaginary components. An equality involving the
energy terms is derived as shown in Equation (2.16).
Define

1/2
H00 (Γ) = {ϕ ∈ H 1/2 (Γ) : ∃ϕ ∈ H 1/2 (R) such that ϕ = 0 on R\Γ and ϕ = ϕ|Γ },
where ϕ is called an extension of ϕ to H 1/2 (R). The weak formulation of the scattering
problem shown in Equation (1.2) is as follows:
1/2
1
Find u ∈ HS (D) = {φ ∈ H 1 (D), φ = 0 on S, φ ∈ H00 (Γ)} such that
1
a(u, φ) = (g, φ) ∀φ ∈ HS (D),

where
u·

a(u, φ) =

k 2 uφ −

φ−
D

D

T (u)φ.
Γ

If in particular, by choosing φ = u, the following lemma is obtained.

14

(2.5)

Lemma 2.2.1.

k 2 − ξ 2 |u|2 dξ = −

|ξ|≤k

u 22
− k2 u 2 2
+
L (D)
L (D)
|ξ|>k

i
Proof. Since T (u) = √
2π R
val’s Theorem,

(2.6)

ξ 2 − k 2 |u|2 dξ =

(2.7)

k 2 − ξ 2 u, and by Parse-

i k 2 − ξ 2 |u|2 dξ.

(T (u)) udξ =
R

gudx.
Γ

k 2 − ξ 2 ueiξx dξ, then T (u) = i

T (u)u =
Γ

gudx,
Γ

(2.8)

R

Now choosing φ = u in Equation (2.5) and using the identity in (2.8) yield that

| u|2 −
D

k 2 |u|2 −
D

|ξ|≤k

i

k 2 − ξ 2 |u|2 dξ +

|ξ|>k

ξ 2 − k 2 |u|2 dξ =

gu.
Γ

The identities (2.6) and (2.7) are obtained by taking the imaginary part and real part of the
equation above respectively.
While the identity Equation (2.7) extracted from the real part will be combined with the
result in the next lemma to form the energy equality, the imaginary part Equation (2.6) will
be used later for the estimate on the L2 norm of u as shown in Lemma 2.4.1. Note that
the sign for D |u|2 is −k 2 , not k 2 for the energy of u. This is natural from the partial
differential equation. Next, we try to get a formula with a “right” sign for D |u|2 by using
the test function X · u, which was used in the literatures [16, 25] for the stability estimates,

15

and in the well-known Pohozaev identity for nonlinear elliptic equations.
3+
H 2 (D), > 0, we have

1
Lemma 2.2.2. For all u ∈ HS (D)

D

Γ

S

(X · n)|u|2 + 2

= k2

|u|2

(X · n)| u|2 + 2k 2

(X · n)| u|2 +

−

∂D

∂u
(X ·
∂D ∂n

u·

Proof. Note that for any test function v,

u).

(2.9)

k 2 uv −

v−

D
D
u and plug into the above identity, it follows that

v=X·

u·

(X ·

k 2 u(X ·

u) −

u) −

D

D

∂u
(X ·
∂D ∂n

∂u
v = 0. Choose
∂D ∂n

u) = 0.

(2.10)

Two technical identities are used here to convert Equation (2.10) to an equation associated
with the energy terms, that is, D | u|2 and k 2 D |u|2 .

u·

2

(X ·

D

D
and

k 2 u(X ·

2

| u|2 +

u) = −(d − 2)

u) = k 2

D

(X · n)| u|2 . (2.11)
∂D

(X · n)|u|2 − dk 2
∂D

|u|2 .

(2.12)

D

where d = div(X) is the dimension. These two identities are the direct consequences from
Lemma 3.1 and 3.2 in [16], which also could be derived through the following standard
computation. We see that div(| u|2 X) = div(X)| u|2 +

(| u|2 ) · X =

d
xi
i=1

∂
(| u|2 ) =
∂xi

d

d
xi 2 (

i=1

d
uxj (
(xi uxi )xj − uxj ) = 2
i=1
j=1

d

uxj uxj xi ) = 2
j=1

d

=2

(| u|2 ) · X, and

u·

16

(X ·

u) − 2| u|2 .

d
uxj (
xi u xj xi )
j=1
i=1

It follows that

div(| u|2 X) = div(X)| u|2 + 2
2

u·

u·

u) − 2| u|2 ,

(X ·

u) = −(d − 2)| u|2 + div(| u|2 X).

(X ·

(2.13)

Integrate Equation (2.13) over D and use the divergence theorem, then Equation (2.11) can
be derived. Equation (2.12) can be obtained by the following direct computation.

div(|u|2 X) = div(X)|u|2 +
2

|u|2 · X = div(X)|u|2 + 2 (u u) · X,
u) = div(|u|2 X) − div(X)|u|2 .

u(X ·

(2.14)

Similarly, integrating Equation (2.14) over D and using the divergence theorem, we have
Equation (2.12). Then through multiplying Equation (2.10) by 2 and taking the real part,
applying Equation (2.11) and Equation (2.12), it follows that

(X · n)| u|2 + dk 2
∂D

D
| u|2 + k 2

= −(d − 2)
D
Note that

2

u=

|u|2
(X · n)|u|2 + 2
∂D

∂u
(X ·
∂D ∂n

u).

(2.15)

∂u
n on S, it implies that
∂n

∂u
(X ·
S ∂n

u) = 2

∂u
∂u
∂u
(X · ( n)) = 2 (X · n)| |2 = 2 (X · n)| u|2 .
∂n
∂n
S ∂n
S
S

Therefore,

(X · n)| u|2 − 2
S

∂u
(X ·
S ∂n
17

(X · n)| u|2 .

u) = −
S

Accordingly, Equation (2.15) can be written as Equation (2.9) by using the identity above
and the fact that u = 0 on S and d = 2, the lemma is proved.
Add Equation (2.7) and Equation (2.9) together, the following expression for the energy
of u is obtained, which is named the energy identity.
u 22
+ k2 u 2 2
+
L (D)
L (D)
|ξ|>k
gu −

=
Γ

2.3

S

(X · n)|u|2 + 2

(X · n)| u|2 + k 2
Γ

Γ

(X · n)| u|2

ξ 2 − k 2 |u|2 dξ −

∂u
(X ·
Γ ∂n

(2.16)
u).

Conversion through Fourier Transform

Notice that in Equation (2.16), the last term on the Left Hand side, that is − s (X · n)| u|2
will help us in the energy estimate, since it is positive under the geometric assumption
X · n ≤ −p1 < 0 on S. Note that

u on ∂D can not be controlled by energy directly,

it needs some further analysis. Since the operator T is simply a multiplication operator in
terms of Fourier transform, in this step, the main issue is how to convert the last three terms
in the right hand side of Equation (2.16) into the frequency domain. While the second and
third terms can be easily handled, the last term is not so trivial. So we first see how to
express the last term by Fourier transform, which is shown in the following lemma.
Lemma 2.3.1. In terms of Fourier transform, we have

2

∂u
(X ·
Γ ∂n

u) = 2
−

|ξ|≤k

−ξu (ξ) i k 2 − ξ 2 u(ξ)dξ
k2

|ξ|>k

|u|2 dξ + 2
Γ

ξ 2 − k2

18

xux gdx + 2y2

Γ

|uy |2

(2.17)

Proof. Note that on Γ, X · n = (x, y2 ) · (0, 1) = y2 and
∂u
(X ·
Γ ∂n

2

Since uy =

u) = 2
Γ

uy (xux + y2 uy ) = 2

∂u
= uy , it follows that
∂n

Γ

xux uy + 2y2

Γ

|uy |2 .

(2.18)

∂u
= T (u) + g on Γ, then
∂n

2
Γ

xux uy = 2

Γ

xux (T (u) + g) = 2

xux T (u) + 2
R

Γ

xux g.

(2.19)

Note that
b
xux =

a
d
=
dξ
=

b d
(iux e−iξx )dx
dξ
a
b
d
iue−iξx |b −i
(−iξ)ue−iξx dx =
a
dξ
a

xux e−iξx dx =

b
−

ξue−iξx dx

a

d
[−ξu(ξ)] .
dξ

Hence Equation (2.19) becomes

2
Γ

xux uy = −2

d
ξu(ξ) i
R dξ

k 2 − ξ 2 u(ξ)dξ + 2
Γ

xux g.

(2.20)

We rewrite the first term in the right hand side of Equation (2.20) in terms of |ξ| ≤ k and
|ξ| > k respectively. For the low frequency part,

−2
=−2
=−2

d
ξu(ξ) i
|ξ|≤k dξ
|ξ|≤k
|ξ|≤k

i

k 2 − ξ 2 u(ξ) dξ

k 2 − ξ 2 |u(ξ)|2 dξ +

ξu (ξ)i k 2 − ξ 2 u(ξ).

19

|ξ|≤k

ξu (ξ)i

k 2 − ξ 2 u(ξ)dξ

Similarly, for the high frequency part,

−2
=2

d
ξu(ξ) i
|ξ|>k dξ
|ξ|>k

=2

ξ 2 − k 2 |u(ξ)|2 dξ +

|ξ|>k

=2

ξ 2 − k 2 |u(ξ)|2 dξ −

|ξ|>k

=−

ξ 2 − k 2 u(ξ)dξ

ξ 2 − k 2 |u(ξ)|2 dξ + 2

|ξ|>k

=2

u(ξ) + ξu (ξ)

k 2 − ξ 2 u(ξ)dξ

k2
|ξ|>k

|ξ|>k
|ξ|>k

ξ

ξ 2 − k 2 u (ξ)u(ξ)dξ

ξ 2 − k2

ξ

d
(ξ
|ξ|>k dξ

d
(|u(ξ)|2 )dξ
dξ

ξ 2 − k 2 )|u(ξ)|2 dξ

|u|2 dξ.

ξ 2 − k2

Therefore, by applying the two equalities above, Equation (2.20) can be written as

2
Γ
=−2
=−2

xux uy
d
(ξu(ξ))i k 2 − ξ 2 u(ξ)dξ + 2
R dξ
|ξ|≤k

ξu (ξ)i

k 2 − ξ 2 u(ξ) −

Γ

xux g

|ξ|>k

k2
ξ 2 − k2

(2.21)
|u|2 dξ + 2
Γ

xux g.

Combing results in Equation (2.18) and Equation (2.21), the lemma is proved.
∂u
(X · u) actually contains lots
Γ ∂n
of information, essentially the higher frequency components is helpful in the estimate. Now
From Lemma 2.3.1, we can observe that the term 2

by using the result in Lemma 2.3.1 and through the Fourier transform and some additional
calculation, the following lemma can be derived which is in terms of the last three terms in
the energy equality Equation (2.16).

20

Lemma 2.3.2. For the last three terms of Equation (2.16), we have

(X · n)| u|2 + k 2

−

(X · n)|u|2 + 2

Γ
≤ 2y2

∂D
|ξ|≤k

(k 2 − ξ 2 )|u|2 dξ + 2y2

xux gdx + y2

+2
Γ

|g|2 dx −
Γ

∂u
(X ·
Γ ∂n

g T (u) + 2

|ξ|≤k

Γ
k2
|ξ|>k

u)
−ξu (ξ) i

k 2 − ξ 2 u(ξ)dξ

|u|2 dξ.

(2.22)

ξ 2 − k2

Proof. Since |uy |2 = |T (u) + g|2 = |T (u)|2 + |g|2 + 2Re [g T (u)] and X · n = y2 , then
(X · n)| u|2 + k 2

−
Γ
= y2

Γ

= y2

Γ

(X · n)|u|2 + 2
∂D

(−|ux |2 − |uy |2 + k 2 |u|2 ) + 2
(−|ux |2 + |uy |2 + k 2 |u|2 ) + 2

Γ
Γ

∂u
(X ·
Γ ∂n

xux uy + 2y2

Γ

u)
|uy |2

xux uy .

Using the fact that u and ux are compactly supported on Γ and Γ |uy |2 ≤ R |uy |2 , then

y2
≤ y2

Γ

(−|ux |2 + |uy |2 + k 2 |u|2 ) + 2

xux uy

(−ξ 2 |u|2 + |i k 2 − ξ 2 u|2 + k 2 |u|2 )dξ
R

+ 2y2
= y2

Γ

Γ

|ξ|≤k

g T (u) + y2

|g|2 + 2
Γ

2(k 2 − ξ 2 )|u|2 dξ + 2y2

Γ
Γ

xux uy

g T (u) + y2

|g|2 + 2
Γ

Γ

xux uy .

Plugging Equation (2.21) into the inequality above, Lemma 2.3.2 is obtained.
Combining Lemma 2.3.2 and Equation (2.16) yields the following inequality for the en-

21

ergy.
k2

( ξ 2 − k2 +
u 22
+ k2 u 2 2
+
L (D)
L (D)
|ξ|>k
≤
Γ
+2

gu + y2

|ξ|≤k

|ξ|≤k

2(k 2 − ξ 2 )|u|2 dξ + 2y2

(X · n)| u|2

)|u|2 dξ −
S

ξ 2 − k2

(2.23)

T (u)g
Γ

−ξu (ξ) i k 2 − ξ 2 u(ξ)dξ + 2
Γ

xux gdx + y2

|g|2 dx
Γ

So far, an inequality involving the energy terms and other additional terms is obtained,
and the term

u on Γ has been canceled out by using the lemmas shown in this section. In

the next section, we will estimate each term on the right hand side of Equation (2.23).

2.4

Detailed Estimates

To estimate each term on the right hand side of Equation (2.23) associated with u or u,
ˆ
the inequalities for

u 22
and
L (Γ)

k 2 − ξ 2 |u|2 dξ are needed, which is presented in

Lemma 2.4.1, this is obtained by using the technique of Lemma 3.5 in [8] and Lemma
2.2.1 . Recall that we denote AL = |ξ|≤k k 2 − ξ 2 |u|2 dξ, AH = |ξ|>k ξ 2 − k 2 |u|2 dξ,
k2
BH = |ξ|>k ( ξ 2 − k 2 +
)|u|2 dξ, and G = Γ |g|2 . In essence, AL , the lower
2 −k 2
ξ
frequency part of u, is relatively small compared to BH , higher frequency part of u, the
term

k2
| in BH played an important role.
ξ 2 −k 2

Lemma 2.4.1. There exists a positive constant C, such that

δ
1
AL ≤ BH + G,
k
δ

22

(2.24)

1
for any δ > 0 such that Cδ ≤ k 2 . Furthermore,
C2
2
G.
u 22
≤ BH +
L (Γ) k
k

(2.25)

Proof. To proof Equation (2.24), the function uα constructed in Lemma 3.5 in [8] is
2π
introduced. Define uα = u(x) − u(x −
[2k]), then, using the fact that the support of u is
k
2π
of length at most π, supp u(x) ∩ supp u(x −
[2k]) = ∅,
k
2π [2k]ξi
2 u 22
|2 |u|2 dξ.
= uα 2 2
=
|1 − e k
L (Γ)
L (R)
R

For |ξ| ≤ k, there exists a positive constant C such that
2π [2k]ξi
|2 ≤ k · 4 ≤ 2( ξ 2 − k 2 +
For |ξ| > k, k|1 − e k

2π [2k]ξi
|1 − e k
|2 ≤








C
√
k

k2 − ξ 2

2
 ( ξ 2 − k2 +

k


Therefore,

2π [2k]ξi
|2 ≤ C k 2 − ξ 2 .
k|1 − e k
k2
). It follows that
2 − k2
ξ

√

k2

for |ξ| ≤ k,
) for |ξ| > k.

ξ 2 − k2

C
1
u 22
≤ √ AL + BH .
L (Γ)
k
k

From Equation (2.6) and the inequality above, we have
1
δ
+
AL ≤ g 2
u 2
≤ u 22
g 22
L (Γ)
L (Γ) 2
L (Γ) 2δ
L (Γ)
δ C
1
1
g 22
≤ ( √ AL + BH ) +
L (Γ)
2 k
k
2δ
23

(2.26)

If Cδ ≤

√
k, then

1
Cδ
δ
1
AL ≤ (1 − √ )AL ≤
BH +
g 22 .
L (Γ)
2
2k
2δ
2 k
We finished the proof of Equation (2.24).
Plug Equation (2.24) into Equation (2.26), we have

C δ
1
1
u 22
≤ √ ( BH + g 2 2 ) + BH .
L (Γ)
L (Γ)
δ
k
k k
Choose Cδ =

√
k, then Equation (2.25) is obtained.

This is an important lemma as it determines the power of k in the estimate, and provides means to relate two integrals

|u|2 and

|ξ 2 − k 2 ||u|2 to the high frequency

Γ
|ξ|≤k
part BH and L2 norm of g . Now we will estimate each term on the right hand side of
Equation (2.23). Note that some terms are easier to estimate, the most difficult ones are:
2y2

T (u)g and 2

−ξu (ξ) i

k 2 − ξ 2 u(ξ)dξ. For the term 2y2

T (u)g,
Γ
|ξ|≤k
Γ
the direct usage of Schwarz inequality does not work, as after converted to the frequency
domain, it would lead to either the integral of ξ 2 |u|2 which can not be controlled by the
energy or ξ 2 |g|2 which may not be integrable as g may not be compactly supported on Γ.
Other technique are needed for the estimate as shown in Lemma 2.4.4.
Lemma 2.4.2. The first term in Equation (2.23) can be estimated as follows.

I1 =

Γ

gu ≤ ε1 BH + d1 G,

ε C2
where ε1 is any positive number, and d1 is given by d1 = 12 + 1 .
2ε1 k
24

(2.27)

Proof. Using Schwarz inequality and Lemma 2.4.1, we have

I1 =

ε1 k
1
2
2
2 (Γ) u L2 (Γ) ≤ 2 u L2 (Γ) + 2ε k g L2 (Γ)
L
1
2
ε k 2
C
1
g 22 )+
g 22
≤ 1 ( BH +
L (Γ)
L (Γ)
2 k
k
2ε1 k

gu ≤ g
Γ

then
ε C2
1
I1 ≤ ε1 BH + ( 1
+
)G = ε1 BH + d1 G.
2
2ε1 k

Lemma 2.4.3. For the second term of Equation (2.23), we have

I2 = y2

|ξ|≤k

2(k 2 − ξ 2 )|u|2 dξ ≤ ε2 BH + d2 kG,

(2.28)

4y 2
where ε2 is any positive number, and d2 = ε 2 .
2
Proof. Since |ξ| ≤ k, we have

I2 = y2

|ξ|≤k

k 2 − ξ 2 ≤ k, then

2(k 2 − ξ 2 )|u|2 dξ ≤ y2 2k ·

|ξ|≤k

k 2 − ξ 2 |u|2 dξ

δ
1
≤ 2y2 k( BH + G)
k
δ
ε2
Choose δ = 2y , where ε2 is a small positive number, then Equation (2.28) is obtained.
2
Next, we deal with I3 , which is more involved to estimate this term.

25

Lemma 2.4.4. For I3 , we have the following estimate.

I3 = 2y2
≤ ε31

T (u)g
Γ
| u|2 + k 2 |u|2 + ε32

D

e
∂u 2
| + d3 kG + 3
|gx |2 ,
k Γ
S ∂n
|

(2.29)

c

1
c
c
c
1
and e3 =
+
,
+ +1 +
3 k
ε31 k
ε32
ε31 ε32 k
c is a positive constant depending on the domain D.
where ε31 , ε32 are any positive numbers, d3 =

T (u)g =
T (u)g Γ =
T (u)(ξ)g Γ (ξ) =
i k 2 − ξ 2 u(ξ)g Γ (ξ), so
Γ
R
R
R
the direct usage of the H¨lder inequality does not work since gΓ (ξ) decays like 1 when ξ → ∞
o
ξ

Proof. Note that

if gΓ does not vanish on the boundary of Γ. Here another test function η is constructed for
the estimate of I3 .
To define η, first we extend g(x) on Γ to a function g(x) on the line y = y2 such that
g

H 1 (R)

≤c g

H 1 (Γ)

and g

≤ c g 2 . Moreover, let
L2 (R)
L (Γ)



 1 + k(y − y ) for y − 1 ≤ y ≤ y ,

2
2
2 k
h(y) :=


0
otherwise.
Then set η(x, y) = g(x)h(y), (x, y) ∈ D. For this particular choice of η, we have η = g on Γ
and |η(x, y)| ≤ |g(x)|.
Adopt η as the test function, then

u·
D

k 2 uη −

η−
D

(T (u) + g)η =
Γ

26

∂u
η,
S ∂n

Note that η = g on Γ, this leads to

u·

T (u)g =

k 2 uη −

η−

D

Γ

D

∂u
η−
|g|2 .
∂n
S
Γ

(2.30)

To estimate I3 , it suffices to estimate each term on the right hand side of Equation (2.30).
u·

We start from the estimate of

η. Since

D
u·

η=

D

D

ux g (x)h(y) + uy g(x)h (y) ,

the two terms on the right hand side of the identity above are handled separately as follows.
Choosing a rectangular domain D1 = [A, B] × [y1 , y2 ] such that D ⊂ D1 , considering
y
y
h(y) in D1 , we have y 2 |h(y)|2 ∼ O( 1 ) and y 2 |h(y)|2 ∼ O(k). Let χD denote the
k
1
1
characteristic function of D, then

D

ux g (x)h(y) =

D1
B

ux g (x)h(y)χD dydx
|g (x)|

≤

y1

A
B
|g (x)|

≤
A

y2

|ux ||h(y)|χD dydx

y2
y1

|ux χD |2 dy

1
2

y2

|h(y)|2 dy

1
2
dx

y1

B
y2
1
1
≤√
|g (x)|(
|ux χD |2 dy) 2 dy
k A
y1
1
1
2
B y2
B
2
2
1
√ |g (x)| dx
|ux χD |2 dydx
≤
k
A y1
A
ε
1 1 B
≤ 31
| u|2 +
|g (x)|2 dx.
2 D
ε31 k A

27

Therefore,

ε
c 1
| u|2 +
ux g (x)h(y) ≤ 31
2 D
ε31 k
D

|g(x)|2 .

|g (x)|2 +
Γ

(2.31)

Γ

Similarly,

D

uy g(x)h (y) =

D1
B

uy g(x)h (y)χD dydx
|g(x)|

≤
A

y2
y1

|uy ||h (y)|χD dydx

1
y2
1
2
|uy χD |2 dy) 2 (
|g(x)|
≤
|h (y)|2 dy
dx
y1
A
y1
y2
B
1
√
|uy χD |2 dy) 2 dx
≤ k
|g(x)|(
y1
A
1
1
B y2
B √
2
2
2
≤
|uy χD |2 dydx
k|g(x)| dx
A y1
A
B
ε
1
| u|2 +
|g(x)|2 dx.
k
≤ 31
2 D
ε31 A
B

y2

It follows that

ε
ck
uy g(x)h (y) ≤ 31
| u|2 +
|g(x)|2 .
2 D
ε31 Γ
D

(2.32)

Now we treat the second term on the right hand side of Equation (2.30). By the definition
|η|2 ≤

of η,
D

1 2
c
|g| ≤
|g|2 , where c depends on the length in the y direction, i.e.,
k
k Γ
D

28

y2 − y1 . It yields that

−
D

1
2 |η|2 2
k

1
2 |u|2 2
k

k 2 uη ≤
D

D
1
k 2 |u|2 +
≤ ε31
k 2 |η|2
ε31 D
D
c
≤ ε31
k 2 |u|2 +
k|g|2 .
ε31 Γ
D

(2.33)

|η|2 , which is derived as follows.
To treat the term − S ∂u η, we use an estimate for
∂n
S
By the divergence theorem, we have

div(|η|2 X) =

|η|2 (X · n),

D

∂D

this is equivalent to

|η|2 div(X) =

η η·X +

2
D

D

Γ

y2 |η|2 +

|η|2 (X · n).
S

It follows that
|η|2 (X · n) = −2

−
S

D
≤ a4
≤ a5

|η|2 div(X) +

η η·X −
D

k|η|2 +
D

Γ

y2 |η 2 |

1
| η|2 +
y2 |g|2
k
Γ
D

1
(|g|2 + |gx |2 ).
k2
Γ

Then use the fact that −(X · n) ≥ p1 > 0, one obtains

29

|η|2 ≤ c
S

1
(|g|2 + |gx |2 ).
k2
Γ

Hence

∂u
η≤
−
S ∂n

1
∂u 2 2
| |
S ∂n

S
c

1
2 2 ≤ε
|η|
32

∂u 2
1
| +
|η|2
ε32 S
S ∂n
|

(2.34)

∂u
1
| |2 +
≤ ε32
(|g|2 + |gx |2 ).
ε32 Γ
k2
S ∂n
Combining all the results in Equation (2.31), (2.32), (2.33) and (2.34), the estimate in the
lemma is arrived.
Now we are going to handle I4 , which yields the highest order of k in Theorem 2.1.1.
Lemma 2.4.5. For I4 , we have the following estimate.

I4 = 2

|ξ|≤k

−ξu (ξ) i

7
k 2 − ξ 2 u(ξ)dξ ≤ ε4 BH + d4 k 5 G,

(2.35)

where ε4 is a small positive number, and d4 is chosen accordingly.
Remark: By direct usage of Schwarz inequality, and use the fact that u (ξ) = (−ix)u(ξ),
we could arrive at an estimate of I4 ≤ ε4 BH + Ck 3 G. But we may lose some information
since the relation between u (ξ) and u relies on the entire domain, while in fact only the
portion |ξ| ≤ k is needed. Here our idea is to start directly from the frequency domain function defined in the lower frequency region, and introduce the corresponding spatial function
by using inverse fourier transform, then through possible cancelation between positive and
negative parts of the introduced function, it results in a lower power estimate in terms of k.
Proof. First I4 is written as the sum of two integrals, i.e. I4 = I41 + I42 as shown below,
where I41 represents the integral for the region 0 ≤ ξ ≤ k and I42 represents the integral

30

for the region −k ≤ ξ ≤ 0. Note that

I41 = 2

=2

Define f1 (x) =

k 1
1
1
1
ξ 2 (k 2 − ξ 2 ) 4 u(ξ) −iξ 2 (k 2 − ξ 2 ) 4 u (ξ) dξ
0
k 1
1
1
1
ξ 2 (k 2 − ξ 2 ) 4 u(ξ) iξ 2 (k 2 − ξ 2 ) 4 u (ξ) dξ.
0

k 1
1
ξ 2 (k 2 − ξ 2 ) 4 u(ξ)eiξx dξ, then through integration by parts, it yields
0

that
k

1
1
iξ 2 (k 2 − ξ 2 ) 4 u (ξ)

eiξx dξ = −

0

k
0

1
1
iξ 2 (k 2 − ξ 2 ) 4 u (ξ)ixeiξx dξ = xf1 (x).

Thus by Parseval’s identity,

I41 = 2

R

f1 (x)xf1 (x)dx = 2

R

x|f1 (x)|2 dx.

Furthermore, observe that
0
I42 = −2
=2
=2

Define f2 (x) =

1
iξ(k 2 − ξ 2 ) 2 u(ξ)u (ξ)dξ

−k
k
1
iξ(k 2 − ξ 2 ) 2 u(−ξ)u (−ξ)dξ
0
k 1
1
1
1
ξ 2 (k 2 − ξ 2 ) 4 u(−ξ) iξ 2 (k 2 − ξ 2 ) 4 u (−ξ) dξ.
0

k 1
1
ξ 2 (k 2 − ξ 2 ) 4 u(−ξ)eiξx dξ, then similar computation yields that
0

I42 = 2

R

x|f2 (x)|2 dx.

31

Therefore I4 = 2

x|f1 (x)|2 dx + 2

x|f2 (x)|2 dx.

R
R
Moreover, by the definitions of f1 (x) and f2 (x), we have

R

|f1 (x)|2 =

k
0

1
ξ(k 2 − ξ 2 ) 2 |u(ξ)|2 dξ ≤ kAL ,

R

|f2 (x)|2 ≤ kAL .

(2.36)

Note that by using Equation (2.36), for any given positive constant M , it follows that

δ
1
x|f1 (x)|2 + x|f2 (x)|2 ≤ M kAL ≤ M k( BH + G).
k
δ
|x|≤M

Thus to estimate I4 , it remains to estimate

|x|>M

(2.37)

x|f1 (x)|2 + x|f2 (x)|2 and choose an

appropriate M .
By using the fact that u(x, y2 ) is supported on the interval [a, b], it is easy to write f1 (x)
as follows,

f1 (x) =

k 1
k b 1
1
1
ξ 2 (k 2 − ξ 2 ) 4 u(ξ)eiξx dξ =
ξ 2 (k 2 − ξ 2 ) 4 eiξx u(s)e−iξs dsdξ. (2.38)
0
0 a

To simplify notations, define H(z) =

k 1
1
ξ 2 (k 2 − ξ 2 ) 4 eiξz dξ, then
0
b

f1 (x) =

Consequently |f1 (x)|2 =

b
a

b
a

u(s)H(x − s)ds.
a

u(s)u(t)H(x − s)H(x − t)dsdt. And |f1 (−x)|2 can be rep-

32

resented by

|f1 (−x)|2 =

b

b
u(s)u(t)H(−x − s)H(−x − t)dsdt

a a
b b
u(s)u(t) H(x + s)H(x + t)dsdt.

=
a

a

Then |x|>M x|f1 (x)|2 can be rewritten as a triple integral involving H term where
cancelations take place.
x|f (x)|2 =

x |f1 (x)|2 − |f1 (−x)|2

|x|>M
x>M
b b
u(s)u(t)
xH(x − s)H(x − t) − xH(x + t)H(x + s) dxdsdt.
=
x>M
a a
xH(x − s)H(x − t) − xH(x + t)H(x + s) dx. By changing varix>M
able for possible cancelation, it follows that
Define J1 =

∞
J1 =

M −s

∞
(x + s)H(x)H(x + s − t)dx −

∞
=

(x − t)H(x)H(x − t + s)dx
M +t

∞
sH(x)H(x + s − t)dx +

tH(x)H(x + s − t)dx

M −s
M +t
M +t
xH(x)H(x + s − t)dx.
+
M −s

Similar computations yield a representation of |x|>M x|f2 (x)|2 , that is,

|x|>M

x|f2 (x)|2 =

b
a

33

b
a

u(s)u(t)J2 dsdt,

where
∞
J2 = −

∞
sH(x)H(x − s + t)dx −

M −t

M +s
M +s

−

M −t

tH(x)H(x − s + t)dx

xH(x)H(x − s + t)dx.

Therefore the contribution to I4 due to |x| > M , that is, |x|>M x|f2 (x)|2 + x|f2 (x)|2 ,
can be expressed using the following integral form.

x|f2 (x)|2 + x|f2 (x)|2 =

|x|>M
b b

a

b
a

u(s)u(t)(J1 + J2 )dsdt

u(s)u(t) − u(t)u(s) J1 dsdt = T1 + T2 + T3 ,

=
a

a

b

where T1 , T2 , T3 are denoted as
b
T1 =
T2 =
T3 =

∞

b
u(s)u(t) − u(t)u(s)

a a
b b
u(s)u(t) − u(t)u(s)
a a
b b
u(s)u(t) − u(t)u(s)
a

a

M −s
∞

sH(x)H(x + s − t)dxdsdt,

tH(x)H(x + s − t)dxdsdt,
M +t
M +t
xH(x)H(x + s − t)dxdsdt.
M −s

Observe that the common term in T1 , T2 and T3 is H(x)H(x + s − t), so next we see how
to express H(x)H(x + s − t) in more details using the expression of H(x).
k 1
1
1 k
ξ 2 (k 2 − ξ 2 ) 4 eiξx dξ = −
ix 0
0
1 k k 2 − 2ξ 2
k
=−
eiξx dξ = −
3
ix 0 1
ix
ξ 2 (k 2 − ξ 2 ) 4

H(x) =

34

1
1
ξ 2 (k 2 − ξ 2 ) 4
1
0

eiξx dξ

F (η)eikηx dη,

1 − 2η 2

with F (η) =

1
3
η 2 (1 − η 2 ) 4

.
1

Consider the integral ψ(w) =
0

1
2

F (η)eiwη dη, and write ψ(w) = ψ1 (w) + ψ2 (w), where

1
F (η)eiwη dη and ψ2 (w) =
F (η)eiwη dη. To estimate ψ1 (w), note that on
1
0
2
1 − 2η 2
1
the interval [0, 1 ],
= 1 + F1 (η 2 ), where F1 (η 2 ) is continuous on [0, 2 ], thus
2
3
(1 − η 2 ) 4
ψ1 (w) =

ψ1 (w) =

0

1
1
2 −2
η
(1 + F1 (η 2 ))eiwη dη =

1
eiζ
1
2w √
dζ + O
=
1 0
ζ
w2

1
w

=

1
2 − 1 iwη
η 2e
dη +
0

0

∞ eiζ
1
√ dζ + O
1 0
ζ
w2

1
2 −1
η 2 F1 (η 2 )eiwη dη

1
w

.

It shows that
ψ1 (w) =

C1
+O
1
w2

1
.
w

1
1
iwη dη = 2 F (1 − η)eiw(1−η) dη, and on the interval [0, 1 ],
F (η)e
Furthermore, ψ2 (w) =
2
1
0
2
−1 + 4η − 2η 2
−3
−3
write F (1 − η) =
= η 4 −2 4 + F2 (η) , where F2 (η) is continu3
1
3
η 4 (1 − η) 2 (2 − η) 4
1
ous on [0, 2 ], thus
1
3
2 − 4 −iwη
e
dη +
η



−3
ψ2 (w) = eiw −2 4
1
− 3 iw 1
2w
= − 2 4e
1 0
w4

Therefore ψ2 (w) =

0

0

1 −iζ
e
dζ + O
3
ζ4

C2 eiw
+O
1
w4

1
w


1
3
2 −4
F2 (η)e−iwη dη 
η

∞
1
−3
= −2 4 eiw
1 0
w4

1 −iζ
e
dζ + O
3
ζ4

C eiw
C
1
. It follows that ψ(w) = 2
+ 1 +O
1
1
w
w4
w2

35

1
w

1
, and
w

the common term in T1 , T2 and T3 can be written as
k2
H(x)H(x + s − t) =
ψ(kx)ψ[k(x + s − t)]
x(x + s − t)


ikx
2
C1
C1
k
1   C2 e−ik(x+s−t)
 C2 e
=
+
+O
+
+
1
1
1
1
x(x + s − t)
kx
(kx) 4
(kx) 2
[k(x + s − t)] 4
[k(x + s − t)] 2
1
O
k(x + s − t)





2
−ik(s−t)
k
e
1 
1 
= |C2 |2
+O
+O
.
1
1 3
3
x2
(kx) 2
k 2 x2
(kx) 4
Accordingly, for |x| > M , we have




5
3
3
k2
 k4 
 k2 
|C2 |2 i sin k(t − s) + O 
xH(x)H(x + s − t) =
 + O 7 .
3
5
M2
M2
M4
3
k 2 |C |2 i sin k(t − s), correspondly, the dominant term in
Note that, the dominant term is
3 2
M2
T3 , denoted by T3d , is given by

T3d =

=

=

=

3
b b
k2
i
|C2 |2
u(s)u(t) − u(t)u(s) (t + s) sin k(t − s)dsdt
3
a a
M2
3
b b
2k 2
s u(s)u(t) − u(t)u(s) eikt e−iks dsdt
3
a a
M2
3
b
b
b
b
2k 2
su(s)e−iks ds
u(t)e−ikt dt −
su(s)eiks ds
u(t)eikt dt
3
a
a
a
a
M2
3
2k 2
su(k)u(k) − su(−k)u(−k) .
3
M2

36

Note that in T3d , su(ξ) ≤ u 2 , so it remains to consider u(k). Since
L (Γ)
b
|u(ξ) − u(k)| =
a
b
≤

u(x)[e−iξx − e−ikx ]dx
u(x)e−iξx 1 − ei(ξ−k)x dx ≤ u

a

L2 (Γ)

|ξ − k|.

Thus 1 |u(k)|2 − |u(ξ)|2 ≤ |u(ξ) − u(k)|2 ≤ u 2 2 |ξ − k|2 , it follows that
2
L (Γ)
1
|u(ξ)|2 ≥ |u(k)|2 − u 2 2 |ξ − k|2 .
L (Γ)
2

Integrate the above inequality from k − δ to k, then we have
k
AL ≥

k 2 − ξ 2 |u(ξ)|2 dξ

k−δ
k
k
1
≥ |u(k)|2
k 2 − ξ 2 dξ − u 2 2
|ξ − k|2
L (Γ) k−δ
2
k−δ
√ 7
√ 3
1
≥ |u(k)|2 kδ 2 − u 2 2
kδ 2 .
L (Γ)
2

k 2 − ξ 2 dξ

It provides an estimate for |u(k)|2 , which is bounded by u 2 2
and AL ,
L (Γ)
1
|u(k)|2 ≤ δ 2 u 2 2
+
A .
√ 3 L
L (Γ)
kδ 2
Using this inequality, T3 can be bounded by


3
3
3
1
k2
k2
1
1
2 ≤ k2 δ u 2
δ u
u 2
+
A .
|T3 | ≤
2 (Γ) + 1 3 AL
3
3
3 1 5 L
L (Γ)
L
L2 (Γ)
M2
k4δ4
M2
M 2 k2δ2

37

1
−1
In particular, choose M = k 5 and δ = k 5 , then T3 can be estimated by
6
7
k5
AL ≤ BH + Ck 5 G.
|T3 | ≤ k u 2 2
+
5
L (Γ)
2
∞
Note that for T1 , for the integral

M −s

sH(x)H(x + s − t)dx, the dominant term is

3
k2
s
|C |2 i sin k(t − s);
3 2
M2
∞
and in T2 , for the integral

tH(x)H(x + s − t)dx, the dominant term is
M +t
3
k2
t
|C |2 i sin k(t − s).
3 2
M2

Therefore, the dominant term for T1 + T2 is
3
b b
k2
u(s)u(t) − u(t)u(s) (t + s) sin k(t − s)dsdt,
i
|C2 |2
3
a a
M2
7
which is the same as T3d . Therefore, T1 and T2 can also be controlled by BH + Ck 5 G.
These estimates on T1 , T2 , and T3 and the chosen M provides the final estimate for I4 ,

|I4 | =

|x|≤M

x|f1 (x)|2 + x|f2 (x)|2 + |T1 | + |T2 | + |T3 |

−1
1
k 5
≤ k 5 k(
BH +
k

7
G) + BH + Ck 5 G
−1
k 5
1

7
≤ 4 BH + d4 k 5 G.
38

Finally, we deal with I5 , which follows from the integration by parts and previous estimate
on u

.
L2 (Γ)

Lemma 2.4.6. For any positive constant ε5 , we have

I5 = 2

xux gdx ≤ ε5 BH + d5 G + e5 k −1 gx 2 2 ,
L (Γ)
Γ

(2.39)

ε C 2 2M1
2M1
where d5 = 5
+
, and e5 =
.
2
ε5 k
ε5
Proof. Using Lemma 2.4.1, we have, for any ε5 > 0,

I5 = 2

Γ

xux gdx = −2

Γ

u(xg)x dx

1
2 dx 2
|u|

1
2
2 dx 2 ≤ ε5 k
|u|2 dx + k −1
≤2
|(xg)x |
|(xg)x |2 dx
2
ε5
Γ
Γ
Γ
Γ
2
ε
2
C
4
4
G) + k −1
≤ 5 k( BH +
|g|2 dx + k −1
|xgx |2 dx
2 k
k
ε5
ε5
Γ
Γ
2ε
C 5
4M1 −1
4
G + k −1 g 2 2
k
gx 2 2
≤ ε5 BH +
+
L (Γ)
L (Γ)
2
ε5
ε5
≤ ε5 BH + d5 G + e5 k −1 gx 2 2 .
L (Γ)
The lemma is proved.
By using the estimates obtained in Lemma 2.4.2 through 2.4.6, we can prove Theorem
2.1.1 as described below.

Proof of Theorem 2.1.1: Equation (2.23) can be estimated by

39

(
u 22
+ k2 u 2 2
+
L (D)
L (D)
|ξ|>k

ξ 2 − k2 +

k2

)|u|2 dξ −

(X · n)| u|2
S

ξ 2 − k2

| u|2 ,
| u|2 + k 2 |u|2 + ε32
≤ εBH + m1 G + m2 gx 2 2
+ ε31
L (Γ)
S
D

(2.40)

7
here ε = ε1 +ε2 +ε4 +ε5 , m1 = d1 +d2 k +d3 k +d4 k 5 +d5 +y2 and m2 = (e3 +e5 )k −1 . By
using the geometric assumption X · n ≤ −p1 < 0, we could choose ε32 such that ε32 < p1 ,
then by dropping the term associated with S | u|2 , Equation (2.40) becomes:

(1 − ε31 )(

(
u 22
+ k2 u 2 2
)+
L (D)
L (D)
|ξ|>k

≤ εBH + m1 G + m2 gx 2 2 .
L (Γ)

ξ 2 − k2 +

k2

)|u|2 dξ

ξ 2 − k2
(2.41)

7
Note here ε, ε31 can be chosen sufficiently small, choose C such that m1 ≤ Ck 5 and
m2 ≤ Ck −1 , then Equation (2.2) is obtained.

2.5

Further Remarks

For the case where the cavity domain is of rectangular shape with D = [0, π]×[0, π], consider
the solution u = sin(mx)sin( k 2 − m2 y), m ∈ Z and m < k . If in particular, k 2 − m2 =
j 2 , j ∈ Z, then u = 0 on the aperture of the cavity Γ where y = π, therefore T (u) = 0.
In this case, we have an explicit formula for g, that is, g = uy = k 2 − m2 sin(mx). It
2
2
follows that
u 22
+ k2 u 2 2
= π (m2 + (k 2 − m2 ) + k 2 ) = π k 2 , meanwhile
4
2
L (D)
L (D)
g 22
= π (k 2 − m2 ). It shows that when j = 1, i.e., k 2 − m2 = 1, we would need
2
L (Γ)

40

a coefficient of k 2 such that

u 22
+ k2 u 2 2
≤ Ck 2 g 2 2 . This particular
L (D)
L (D)
L (Γ)
2 in the estimate. Our estimate yields the order
example shows that at least we would need k
7
of k 5 , which may not be the optimal because of techniques we used. It would be interesting
to see whether the optimal order dependency on k is k for any g under the given geometric
assumption.

41

Chapter 3
Stability Estimates under a Weak
Geometric Assumption

3.1

Main Theorem and Outlines of the Approach

In this chapter, the cavity structure satisfies the following geometric assumptions:
(1) the angel θ between Γ and D satisfies θ ≥ θ0 > 0;
(2) there is a point X ∗ = (x0 , y0 ) on the plane and positive constants p such that

(X − X ∗ ) · n ≤ 0

on S.

(3.1)

(3) the domain D admits cusps of power sharpness 1 < τ < 2. One example is shown
Figure 3.1. For domain with cusps satisfying the geometric assumption (3), the related trace
theorem is established in [1], which states that for u ∈ H 1 (D), u
where the inclusion of H 1 (D) ⊂ Lr (D) for 2 ≤ r ≤

L2 (∂D)

≤C u

H 1 (D)

,

2(α+1)
α−1 for domain with cusps [2] is

used for the proof. The main result is Theorem 3.1.1 as stated below.
Theorem 3.1.1. Under the geometric assumptions given in Equation (3.1), there exists a
constant C such that for k > k0 ,

u

L2 (D)

+ |k| u

L2 (D)

≤C

7
k 10 g

6
1
+ √ gx 2
+ k 5 [|g(0)| + |g(π)|] .
L2 (Γ)
L (Γ)
k
42

(a) : (X − X ∗ ) · n ≤ 0

(b): (X − X ∗ ) · n ≤ −p1

Figure 3.1: Domains of different shapes
See Figure (3.1)(b) for example, note that in Figure (3.1)(a), there is no such X ∗ which
satisfies the geometric condition in Equation (2.1). It should be pointed out that even
though Theorem 3.1.1 implies Theorem 2.1.1 in the case of g(0) = g(π) = 0. In general,
6
7
1
+ √ gx 2 . This can be
k 10 [|g(0)| + |g(π)|] can not be controlled by k 5 g 2
L (Γ)
L (Γ)
k
seen when g(x) is smooth and is supported in a small subinterval of [0, π].
Remark 1: Note that in Chapter 2, the stronger geometric assumption is fully utilized in
the final estimate. Specifically, the following energy inequality was obtained:
k2

u 22
+ k2 u 2 2
+
( ξ 2 − k2 +
L (D)
L (D)
|ξ|>k
≤
Γ

gudx + y2

+2

|ξ|≤k

|ξ|≤k

2(k 2 − ξ 2 )|u|2 dξ + 2y2

)|u|2 dξ −

(X · n)| u|2
S

ξ 2 − k2

(3.2)

T (u) g
Γ

−ξu (ξ) i k 2 − ξ 2 u(ξ)dξ + 2
Γ

xux gdx + y2

|g|2 dx.
Γ

The integral − S (X · n)| u|2 on the left hand side played an important role to control the
term 2y2

Γ T (u)g is on the right hand side. Recall that in Lemma 2.4.4 , I is shown to
43

be estimated as follows,

I ≤ ε31

D

| u|2 + k 2 |u|2 + ε32

e
∂u 2
| + d3 k 2 G + 3
|gx |2 .
∂n
k Γ
S
|

Under the strong geometric assumption, −X · n ≥ p1 > 0 on S, the term ε32 S | ∂u |2
∂n
can be absorbed as long as ε32 is chosen small enough. Now we have a weaker geometric
assumption on S, thus the previous proof can not be extended directly.
Remark 2: Another natural way to estimate I is to rewrite the integral using Fourier
transform. That is,
|I| ≤ C

T (u) g = C
Γ

T (u) g
R

by extending g to be zero outside Γ. Hence

T (u) g =
R

T (u) gdξ =
R

i

k2 − ξ 2u g .

R
1
2 2
|k 2 − ξ 2 |u|

1
2

|k 2 − ξ 2 ||g| . Recall that
R
we only have control on
|k 2 − ξ 2 |u|2 , we need the convergence of
|k 2 − ξ 2 ||g|.
R
R
However this is not true if |g(0)2 |+|g(π)2 | = 0, since g ∼ 1 as |ξ| → ∞. If g(0) = g(π) = 0,
|ξ|
we will have convergence of
|k 2 − ξ 2 ||g| when gx (x) ∈ L2 (Γ). The new idea is to
R
introduce a new auxiliary function u∗ , and consider w = u + u∗ , we have
If Schwarz inequality is used, |I| ≤ C




 ∆w + k 2 w = ∆u∗ + k 2 u∗ ≡ f in D,




∗
(3.3)
 w|S = 0 if u |S = 0,





 wy | = uy + u∗ = T (u) + g + u∗ = T (w) − T (u∗ ) + g + u∗ = T (w) + g on Γ,
y
y
y
1
Γ

44

where g1 = g − T u∗ + u∗ . Using this construction, we want g1 (0) = g1 (π) = 0.
y
To prove Theorem 3.1.1 under the weaker geometric assumption Equation (3.1), three
main steps are involved.
Step 1. Introduce an auxiliary function u∗ such that
1) T u∗ (0, y2 ) = g(0) and T u∗ (π, y2 ) = g(π),
2) u∗ (0, y2 ) = u∗ (π, y2 ) = 0, u∗ (0, y2 ) = u∗ (π, y2 ) = 0,
y
y
3) u∗ = 0 on S. The trick is to construct such u∗ so that the norms of f and u∗ are as
small as possible. But the construction is not trivial since T u∗ is not a local operator. Once
u∗ is chosen, then w = u + u∗ satisfies Equation (3.3).
Step 2. Note that this formulation has similar structure as that in Equation (1.2) with
two major differences: 1) the governing equation here is nonhomogeneous with a right hand
side source term f ; 2) in the nonlocal boundary condition, g1 vanishes on the boundary of
Γ while g itself may not. It will be shown that
2
7
C
+ 2 g1x 2 2
+ C3 k 5
|f |2 . (3.4)
w 22
+ k2 w 2 2
≤ C 1 k 5 g1 2 2
L (Γ)
L (Γ)
L (D)
L (D)
k
D
The proof for the above inequality is presented in the last section since this part can be
viewed as a modification of results in Chapter 2, where most of the ideas are similar; however,
additional terms containing f are involved from the beginning of the proof, so each result
need to be restated. Furthermore, since here g1 vanishes on the boundary of Γ, an easier
estimate could be used to treat the term Γ T (w)g1 .
Step 3. To provide a final estimate for the wave energy on u, it remains to show that
g1 2 , g1x 2
and f 2
terms are controlled by norms of g with suitable
L (Γ)
L (Γ)
L (D)

45

powers of k, which turned out to be

|f |2 ≤ C |g(0)|2 + |g(π)|2 ,
D
Γ
Γ

|g1 |2 ≤ C |g(0)|2 + |g(π)|2 +

|g|2 ,
Γ

|g1x |2 ≤ Ck 2 |g(0)|2 + |g(π)|2 +

Γ

|gx |2 .

Furthermore, it could be shown that

C
u∗ 2 2
+ k 2 u∗ 2 2
≤
|g(0)|2 + |g(π)|2 ,
L (D)
L (D) k 2
hence the final estimate could be established accordingly.

3.2

Some Basic Properties of the Auxiliary Function

In this section, we focus on the construction of the auxiliary function u∗ , which essentially
is a linear combination of two compact supported functions. The aim is to construct a
function u∗ such that it satisfies 1) T u∗ (0, y2 ) = g(0) and T u∗ (π, y2 ) = g(π), 2) u∗ (0, y2 ) =
u∗ (π, y2 ) = 0, u∗ (0, y2 ) = u∗ (π, y2 ) = 0, and 3) u∗ = 0 on S. In this section, we introduce
y
y
two subdomains where u∗ is supported, and analyze some basic properties of u∗ .

3.2.1

The Construction of an Auxiliary Function

Define Ω1 and Ω2 as follows.
π
Ω1 = {(x, y) : x ∈ [0, ], y ∈ [y2 − α sin(jx), y2 ]},
j

46

and
π
Ω2 = {(x, y) : x ∈ [π − , π], y ∈ [y2 − α sin(jx), y2 ]},
j
where α is a small constant that will be chosen later to optimize the order in k, which
turns out to be α =

β
, this is explained in Section 4.4, j = µk, µ is a small constant as
k

indicated in the discussion in Section 3.2, and µ is chosen such that j is an integer. See
Figure 3.2 for an illustration of Ω1 and Ω2 .

Once these two subdomains are defined, two

Figure 3.2: Cavity with subdomains where the auxiliary functions are supported
corresponding functions u∗ and u∗ are introduced accordingly, which are supported in Ω1
1
2
and Ω2 respectively. Thereafter u∗ is defined as a linear combination of the two functions

47

in a way that T u∗ cancels out the value of g on the boundary of Γ. Specifically, we write

u∗ = a1 u∗ + a2 u∗ ,
2
1

where


 [y − y + α sin(jx)]3 (x, y) ∈ Ω ,

2
1
∗ (x, y) =
u1


 0,
otherwise.
and



 [y − y2 + α sin(jx)]3 (x, y) ∈ Ω2 ,
∗ (x, y) =
u2


 0,
otherwise,
Furthermore, a1 , a2 are constants chosen such that


 a T u∗ (0, y ) + a T u∗ (0, y ) = g(0)
 1 1
2
2 2
2


 a T u∗ (π, y ) + a T u∗ (π, y ) = g(π)
1 1
2
2 2
2

(3.5)

Note that in particular, u∗ = α3 sin3 (jx) for x ∈ [0, π ] and u∗ = α3 sin3 (jx) for x ∈
2
1
j
[π − π , π] on Γ, this fact is used in the next subsection to analyze properties on a1 and a2 .
j

48

3.2.2

The Existence and the Order of the Coefficients

To guarantee the existence of a1 and a2 in the auxiliary function, we show that the coefficient
matrix




∗
∗
 T u1 (0, y2 ) T u2 (0, y2 ) 


M =

T u∗ (π, y2 ) T u∗ (π, y2 )
2
1

(3.6)

is nonsingular. This could be proved by checking the leading order terms (in terms of
order in k) for each component in the coefficient matrix in Equation (3.6). Notice that
T φ(x) = √1 R i k 2 − ξ 2 φ(ξ)eiξx dξ, thus the Fourier transform of u∗ and u∗ are needed
1
2
2π
for the further analysis on T u∗ and T u∗ . First, we derive the formula for u∗ and u∗ .
1
2
1
2
Lemma 3.2.1. The fourier transform of u∗ and u∗ can be represented by
1
2

π

3 j 3 e−iξ j + 1

 ∗
 u (ξ, y ) = √1 6α

 1
2

2π ξ 2 − 9j 2 ξ 2 − j 2


π

j+1 e−iξπ + e−iξ(π− j )

(−1)



1 6α3 j 3
 ∗
 u (ξ, y ) = √
 2
2
2π ξ 2 − 9j 2
ξ2 − j2

(3.7)

Proof. Since on Γ, y = y2 , therefore u∗ (x, y) = α3 sin3 (jx) for (x, y) ∈ [0, π ] × {y = y2 }.
1
j
√
We may assume α3 = 2π from linearity of Fourier transform. u∗ (ξ, y2 ) can be expressed
1

49

as follows.
π
π
j
∗ (ξ, y ) = j sin3 (jx)e−iξx dx = 1
3 sin2 (jx) cos(jx)je−iξx dx
u1
2
iξ 0
0
π
1
j
=
[6 sin(jx) cos2 (jx)j 2 − 3 sin3 (jx)j 2 ]e−iξx dx
2 0
(iξ)
π
1
j
=
[6 sin(jx)j 2 − 9 sin3 (jx)j 2 ]e−iξx dx
2 0
−ξ
π
1
j
=
6 sin(jx)j 2 e−iξx dx − 9j 2 u∗ (ξ, y2 )]
[
1
2 0
−ξ
π
3j 2
2 + 9j 2 )u∗ (ξ, y ) = j 2 j 6 sin(jx)e−iξx dx. Denote C =
It shows that (−ξ
,
2
j
1
−ξ 2 +9j 2
0
then u∗ (ξ, y2 ) can be written as
1
π ijx
1 ei(j−ξ)x e−i(j+ξ)x
− e−ijx −iξx
j e
∗ (ξ, y ) = C
u1
e
dx = Cj
+
2
j
i
i (j − ξ)i
(j + ξ)i
0
π
i(j−ξ) π
−iξ π
j − 1)(j + ξ) + (e−i(j+ξ) j − 1)(j − ξ)
j +1
1 (e
e
=Cj
= 2Cj
.
i
(j 2 − ξ 2 )i
ξ2 − j2

π
j
0

u∗ (ξ, y2 ) can be obtained by a suitable substitution.
2
π
π
3 (jx)e−iξx dx = j sin3 [j(π − x )]e−iξ(π−x1 ) dx
∗ (ξ, y ) =
u2
sin
2
1
1
π− π
0
j
π
j
=
(−1)j+1 sin3 (jx1 )eiξx1 dx1 e−iξπ = (−1)j+1 e−iξπ u∗ (−ξ, y2 )
1
0
π
j+1 e−iξπ + e−iξ(π− j )
(−1)
= 2Cj

.

ξ2 − j2

Lemma 3.2.1 is proved.
By using the explicit expressions for u∗ (ξ, y2 ) and u∗ (ξ, y2 ), the following facts can be
1
2

50

obtained.
Corollary 3.2.2. Using the connection between u∗ (ξ, y2 ) and u∗ (ξ, y2 ) as shown in Equation
2
1
(3.8), we have the following relations.

 ∗
 T u (0, y ) = (−1)j+1 T u∗ (π, y ),
 2
2
2
1
 ∗

 T u (π, y ) = (−1)j+1 T u∗ (0, y ).
2
2
1
2

(3.8)

Proof. Since the following relate holds for u∗ (ξ, y2 ) and u∗ (−ξ, y2 ),
2
1
u∗ (ξ, y2 ) = (−1)j+1 e−iξπ u∗ (−ξ, y2 ),
2
1

(3.9)

it follows that
1
√
i
2π R

1
k 2 − ξ 2 u∗ (ξ, y2 )dξ = (−1)j+1 √
i k 2 − ξ 2 e−iξπ u∗ (−ξ, y2 )dξ
2
1
2π R
1
= (−1)j+1 √
i k 2 − ξ 2 eiξπ u∗ (ξ, y2 )dξ.
1
2π R

Thus the first identity in Equation (3.8) is proved, the second identity could be easily derived
after multiplying both sides of Equation (3.9) by eiξπ .
From this corollary, in order to show that the coefficient matrix M in Equation (3.6) is
nonsingular, note its determinant

det(M ) = (−1)j+1

T u∗ (0, y2 ) 2 − T u∗ (π, y2 ) 2 ,
1
1

thus the leading order for T u∗ (0, y2 ) and T u∗ (π, y2 ) in terms of power of k needs to be
1
1
determined.
51

Lemma 3.2.3. T u∗ (0, y2 ) and T u∗ (π, y2 ) can be expressed by the integrals containing only
1
1
the higher frequency part.

1
T u∗ (0, y2 ) = −
1
π ξ>k
1
T u∗ (π, y2 ) = −
1
π ξ>k

π
3 j 3 eiξ j + 1
6α
dξ,
ξ 2 − k2
2 − 9j 2 ξ 2 − j 2
ξ
π
3 j 3 eiξ(π− j ) + eiξπ
6α
ξ 2 − k2
dξ.
2 − 9j 2
ξ
ξ2 − j2

(3.10)

Proof. We first rewrite T u∗ (0, y2 ) using residue theorem. It should be pointed out that
1
i

k 2 − ξ 2 is not a restriction of an analytic function to the real line. However, we can

express this as

i

k 2 − ξ 2 :=



 A(ξ)


for ξ < k,



 A(ξ) − 2 ξ 2 − k 2

for ξ > k,

1
1
1
1 i argz
where A(z) = (z − k) 2 (z + k) 2 , z ≥ 0, z 2 = |z| 2 e 2 . A(z), 0 < argz < π, is an analytic function. Then
1
1
T u∗ (0, y2 ) = √
i k 2 − ξ 2 u∗ (ξ, y2 )dξ = √
i k 2 − ξ 2 u∗ (−ξ, y2 )dξ
1
1
1
2π R
2π R
2
1
A(ξ)u∗ (−ξ, y2 )dξ − √
ξ 2 − k 2 u∗ (−ξ, y2 )dξ
=√
1
1
2π R
2π ξ>k
=−

2
π ξ>k

ξ 2 − k 2 u∗ (−ξ, y2 )dξ,
1

1
1
∗
since √
A(ξ)u∗ (−ξ, y2 )dξ = √
lim
1
+ A(ξ)u1 (−ξ, y2 )dξ = 0 by residue the2π R
2π R→∞ ∂B
R

52

orem. Similarly,
1
i k 2 − ξ 2 u∗ (ξ, y2 )eiξπ dξ
T u∗ (π, y2 ) = √
1
1
2π R
2
1
=√
A(ξ)u∗ (ξ, y2 )eiξπ dξ − √
1
2π R
2π ξ>k
=−

2
π ξ>k

ξ 2 − k 2 u∗ (ξ, y2 )eiξπ dξ
1

ξ 2 − k 2 u∗ (ξ, y2 )eiξπ dξ.
1

Plug the expression of u∗ (ξ, y2 ) into the two identities above, the two equations in (3.10)
1
are obtained.
Using these expressions, we can compute the leading orders of T u∗ (0, y2 ) and T u∗ (π, y2 )
1
1
and therefore derive the bound for a1 and a2 in terms of k.
Lemma 3.2.4. For large wave number k, the matrix M is nonsingular. Furthermore,

|a1 | ≤

C
(|g(0)| + |g(π)|),
α3 k

|a2 | ≤

C
(|g(0)| + |g(π)|).
α3 k

(3.11)

Proof. To estimate the leading order of T u∗ (0, y2 ), rewrite the integral as the sum of two
1
integrals, i.e.,
ξ 2 − k2
iξ π
6
6
T u∗ (0, y2 ) = − α3 j 3
(e j + 1)dξ = − α3 j 3 (I1 + I2 ),
1
π
π
ξ>k (ξ 2 − 9j 2 )(ξ 2 − j 2 )
where

I1 =

ξ 2 − k2
ξ 2 − k2
iξ π
e j dξ and I2 =
dξ,
ξ>k (ξ 2 − 9j 2 )(ξ 2 − j 2 )
ξ>k (ξ 2 − 9j 2 )(ξ 2 − j 2 )

Assume that j is an integer such that j = µk and j < k in the following context. Consider
3
53

ξ 2 −k 2
I1 first, denote f (ξ) = 2
, then
(ξ −9j 2 )(ξ 2 −j 2 )

f (ξ) =

ξ
(ξ 2 − 9j 2 )(ξ 2 − j 2 ) −
2 −k 2
ξ

ξ 2 − k 2 2ξ(ξ 2 − 9j 2 + ξ 2 − j 2 )

(ξ 2 − 9j 2 )2 (ξ 2 − j 2 )2
ξ −3ξ 4 + (10j 2 + 4k 2 )ξ 2 + j 2 (9j 2 − 20k 2 )

=

.
ξ 2 − k 2 (ξ 2 − 9j 2 )2 (ξ 2 − j 2 )2

Consider the term q(x) = −3x2 + (10j 2 + 4k 2 )x + j 2 (9j 2 − 20k 2 ), x = ξ 2 on the numerator.
Use the fact that j < k, then q(0) = j 2 (9j 2 − 20k 2 ) < 0, q(k 2 ) = (k 2 − j 2 )(k 2 − 9j 2 ) >
0, q(4k 2 ) = −32k 4 + 20k 2 j 2 + 9j 4 < 0. It shows that the two zeros x1 , x2 of q(x) lie in
the following intervals: x1 ∈ (0, k 2 ) and x2 ∈ (k 2 , 4k 2 ). Consequently, there is a ξ0 ∈
√
(k, 2k), ξ0 = x2 , such that f (ξ) > 0 for ξ ∈ (k, ξ0 ) and f (ξ) < 0 for ξ ∈ (ξ0 , ∞).
This indicates that f (ξ) is a monotonic function in both the intervals (k, ξ0 ) (increasing)
and ((ξ0 , ∞) (decreasing), where ξ0 = µ0 k, µ0 ∈ (1, 2). Furthermore, note that f (k) = 0,
µ2 −1
0
, and f (ξ) → 0 as ξ → ∞, therefore, by the second mean
f (ξ0 ) = 2
2 )(µ2 −µ2 )k 3
(µ0 −9µ
0
value theorem, there exists M1 ∈ (k, ξ0 ) and M2 ∈ (ξ0 , ∞), such that
µ0 k iξ π
M2 iξ π
e j dξ + f (ξ0 )
e j dξ
M1
µ0 k
µ2 − 1
µ2 − 1
2j
2µk
0
0
≤
≤
2 − 9µ2 )(µ2 − µ2 )k 3 π
2 )(1 − µ2 )k 3
π (1 − 9µ
(µ0
0
4µ
.
≤
2 )(1 − µ2 )k 2
π(1 − 9µ

|I1 | = f (ξ0 )

µ1
4µ
Denote µ1 =
, then |I1 | ≤ 2 and µ1 is small for small µ. For I2 , rewrite
π(1−9µ2 )(1−µ2 )
k

54

it as I2 = I21 + I22 , where
ξ 2 − 9j 2
I21 =
dξ,
ξ>k (ξ 2 − 9j 2 )(ξ 2 − j 2 )

I22 =

We have |I21 | ≥

−k 2 + 9j 2
ξ>k (

ξ 2 − k2 +

1
dξ.
2 − 9j 2 ) (ξ 2 − 9j 2 )(ξ 2 − j 2 )
ξ

1
1
1
=
=
. Furthermore, since ( ξ 2 − 9j 2 )2 ≥ (ξ−
1
3
ξ
2k 2
ξ>k 2 2 2
ξ>k
(ξ ) ξ

3j)2 for ξ > k > 3j,

ξ 2 − 9j 2 ≥ ξ − 3j, consequently,

|I22 | ≤ (k 2 − 9j 2 )

1−9µ2
Denote µ2 =
, then |I22 | ≤
4(1−3c)4

1 − 9µ2 1
1
dξ =
.
4(1 − 3c)4 k 2
ξ>k (ξ − 3j)5
µ2
1
and µ2 is close to 4 when µ is small enough. From
k2

the estimates for I11 , I12 , I21 , I22 , it follows that
1
4µ
1
≥
,
| (I1 + I2 )| ≥ |I12 | − |I21 | − |I22 | ≥ ( − 2µ1 − µ2 )
2
2
k
k2
2µ
and | (I1 )| ≤
, when µ is chosen small enough. It indicates that for small µ , the real
k2
part dominates the imaginary part in the leading term in terms of order in k for T u∗ (0, y2 ).
1
On the other hand, note that
2 ξ 2 − k2
2 ξ2
1
C
|I1 + I2 | ≤
dξ ≤
dξ =
≤
.
2 − 9j 2 )(ξ 2 − j 2 )
2 − 9j 2 )2
2 − 9j 2
ξ
k2
ξ>k (ξ
ξ>k (ξ

55

It shows that the real part of |I1 + I2 | ∼ O( 1 ), which implies that
k2
6
T u∗ (0, y2 ) = − α3 j 3 (I1 + I2 ) ∼ O(α3 k),
1
π
where the real part dominates.
Next, we treat T u∗ (π, y2 ), where
1
ξ 2 − k2
iξ(π− π )
6
j + eiξπ )dξ = − 6 α3 j 3 J,
T u∗ (π, y2 ) = − α3 j 3
(e
1
π
π
ξ>k (ξ 2 − 9j 2 )(ξ 2 − j 2 )
Again by the second mean value theorem for integrals,
µ0 k iξ(π− π )
M2 iξ(π− π )
j + eiξπ )dξ + f (ξ )
j + eiξπ )dξ
(e
|J| = f (ξ0 )
(e
0
M1
µ0 k
µ2 − 1
8
2
2
8
0
≤
,
(
≤
π + π) ≤
π(1 − 9µ2 )(1 − µ2 )k 3
k3
(µ2 − 9µ2 )(µ2 − µ2 )k 3 π − j
0
0
6
when µ is small. This implies that T u∗ (π, y2 ) = − α3 j 3 J ≤ Cα3 . The arguments above
1
π
show that a1 , a2 are the solutions to the following linear system


α3 k

o(α3 k)










  a1   g(0) 
=
.


j+1 o(α3 k) (−1)j+1 α3 k
(−1)
a2
g(π)

(3.12)

Consequently, the matrix M is nonsingular and the result in the lemma as shown in Equation
(3.11) holds.
From the estimates of a1 and a2 in the lemma above, we can easily get the bound for
E(u∗ ). Moreover, the estimate of the energy terms involving w needs to be determined.
Remark: Note that w satisfies the nonhomogeneous equation in (3.3). To derive the
56

dependency of the energy E(w) on k, most of the ideas are similar to the previous work;
however, additional terms containing f are involved from the beginning of the proof. Furthermore, since here g1 vanishes on the boundary of Γ, a different estimate could be used
for the estimate of the term Γ T (u)g1 as stated in Lemma, that’s exactly the reason why
we could use a weaker geometric constraint which is stated in Equation (3.1) compared to
the previous work, which requires (X − X ∗ ) · n ≤ −p1 < 0 on S. So each result need to be
restated and this part will be proved in the last section. It will be shown in the last section
that

w 22
+ k2 w 2 2
≤C
L (D)
L (D)

3.3

2
7
1
+ g1x 2 2
+ k5
|f |2
k 5 g1 2 2
L (Γ) k
L (Γ)
D

. (3.13)

Relations between New Source Terms and Original
Source Term g

To obtain the main stability result for u associated with g instead of the new source contribution terms f , g1 , the dependence of f , g1 and g1x on g are needed, which are derived
respectively in the following three lemmas.
Lemma 3.3.1. For the right hand side source term f in Equation (3.3), we have

|f |2 ≤ C max(αk,
D
If in particular choose α =

1
) |g(0)|2 + |g(π)|2 .
3k3
α

β
, then
|f |2 ≤ C |g(0)|2 + |g(π)|2 .
k
D

57

(3.14)

Proof. Note that f is defined as

f = ∆u∗ + k 2 u∗ = a1 (u∗ + u∗ + k 2 u∗ ) + a2 (u∗ + u∗ + k 2 u∗ ).
2
2yy
2xx
1
1yy
1xx
From the definition of u∗ , we can see that u∗ , Du∗ , D2 u∗ vanish outside outside Ω1 . We
1
1
1
1
will treat u∗ first, and u∗ can be handled in a similar way. In Ω1 ,
1
2









u∗ = [y − y2 + α sin(jx)]3 ,
1

∗
2
 u1x = 3αj[y − y2 + α sin(jx)] cos(jx),



 ∗

2 2
2
2
2
u
1xx = 6α j [y − y2 + α sin(jx)] cos (jx) − 3αj [y − y2 + α sin(jx)] sin(jx),
and

 ∗
 u = 3[y − y + α sin(jx)]2 ,
 1y
2
 ∗

u
1yy = 6[y − y2 + α sin(jx)].
Note that in Ω1 , f = a1 (u∗ + u∗ + k 2 u∗ ), thus it follows that
1xx
1yy
1

Ω1

|f |2 = |a1 |2

π
j
0

y2
y2 −α sin(jx)

2
u∗ + u∗ + k 2 u∗ dydx.
1xx
1yy
1

(3.15)

Next, the expressions of u∗ ,u∗ ,and u∗ will be used to derive the estimate for Ω |f |2 .
1xx 1yy
1
1

58

For the first term in the expression of u∗ , we have
1xx
|a1 |2
=|a1 |2

π
j
0
π
j

y2
y2 −α sin(jx)

2
6α2 j 2 [y − y2 + α sin(jx)] cos2 (jx) dydx

12α4 j 4 [α sin(jx)]3 cos2 (jx)dx

(3.16)

0
π
C 2
≤
(|g(0)|2 + |g(π)|2 ) α7 j 4 ≤ Cαk(|g(0)|2 + |g(π)|2 ).
j
α3 k
Similarly, for the second term in the expression of u∗ , it follows that
1xx
|a1 |2

π
j
0

y2
y2 −α sin(jx)

2
3αj 2 [y − y2 + α sin(jx)]2 sin(jx) dydx
(3.17)

≤Cαk(|g(0)|2 + |g(π)|2 ).
Moreover, for u∗ ,
1yy

|a1 |2

π
j

y2

2

6[y − y2 + α sin(jx)] dydx = |a1 |2

0
y2 −α sin(jx)
2
C
π
C
≤
(|g(0)|2 + |g(π)|2 ) α3 ≤
(|g(0)|2 + |g(π)|2 ).
3k
3k3
j
α
α

π
j

12 [α sin(jx)]3 dx
0
(3.18)

And for the term k 2 u∗ , the following estimate holds,
1
|a1 |2

π
j

y2

2
k 2 [y − y2 + α sin(jx)]3 dydx = |a1 |2

0
y2 −α sin(jx)
C 2
π
≤
(|g(0)|2 + |g(π)|2 ) k 4 α7 ≤ Cαk(|g(0)|2 + |g(π)|2 ).
3k
j
α

Next, the integral involving g1 is computed.

59

π
j 1
[α sin(jx)]7 dx
7
0
(3.19)

Lemma 3.3.2. The relation between L2 norm of g1 and g can be expressed by

Γ

|g1 |2 ≤ C |g(0)|2 + |g(π)|2 +

|g|2 .
Γ

Proof. Recall that
g1 = g − T u∗ + u∗ ,
y
where T u∗ = a1 T u∗ + a2 T u∗ . For a1 T u∗ (x, y2 ), suitable contour integral and residue
1
2
1
theorem are used like the computation of T u∗ (0, y2 ) and T u∗ (π, y2 ) in Section 2. The
1
1
advantage of this is to convert the integral on R to ξ > k, which makes the estimate much
easier. If π ≤ x < π,then
j

1
|a1 T u∗ (x, y2 )| = − a1
1
π
ξ>k

iξ(x− π )
j + eiξx
6α3 j 3 e
dξ
ξ 2 − k2
ξ 2 − 9j 2
ξ2 − j2

ξ 2 − k2
6α3 j 3 |a1 |
2
dξ
π
ξ>k ξ 2 − 9j 2 ξ 2 − j 2
6α3 j 3 |a1 |
ξ
≤
dξ
2 − 9j 2 )2
π
ξ>k (ξ
6α3 j 3 |a1 |
1
=
≤ C(|g(0)| + |g(π)|).
π
k 2 − 9j 2
≤

If 0 < x < π ,then
j
−iξ(x− π )
j + eiξx
6α3 j 3 e
k2 − ξ 2
dξ
ξ 2 − 9j 2
ξ2 − j2
−iξ(x− π )
j + eiξx
−2a1
6α3 j 3 e
=
ξ 2 − k2
dξ
2π ξ>k
ξ 2 − 9j 2
ξ2 − j2

a
|a1 T u∗ (x, y2 )| = 1
i
1
2π R

≤ C(|g(0)| + |g(π)|).

60

Similarly, |a2 T u∗ (x, y2 )| ≤ C(|g(0)| + |g(π)|). Therefore,
2
|T u∗ |2 =
Γ

π
0

|a1 T u∗ |2 +
1

π
0

|a2 T u∗ |2 ≤ C |g(0)|2 + |g(π)|2 .
2

Moreover, note that u∗ = a1 u∗ + a2 u∗ , then
y
2y
1y
C 2π
C
|u∗ |2 ≤ [
] (|g(0)|2 + |g(π)|2 )α5 ≤
(|g(0)|2 + |g(π)|2 ).
y
3k j
3
α
αk
Γ
Again if α =

β
, then the result in this lemma is obtained.
k

Furthermore, we have the following estimate for the integral in terms of g1x .
Lemma 3.3.3. The L2 norm of g1x can be estimated by

Γ

|g1x |2 ≤ Ck 2 |g(0)|2 + |g(π)|2 +

Γ

|gx |2 .

Proof. Note that g1x = gx − T u∗ + u∗ , where (T u∗ )x = a1 (T u∗ )x + a2 (T u∗ )x ,
x
yx
1
2

iξ(x− π )
j + eiξx 
6α3 j 3 e
1

a1  i k 2 − ξ 2
dξ 
a1 T u∗ (x, y2 )x =
1
2π
ξ 2 − 9j 2
ξ2 − j2
R
x
iξ(x− π )
j + eiξx
1
6α3 j 3 e
=
dξ.
a1
(iξ)i k 2 − ξ 2
2π
ξ 2 − 9j 2
ξ2 − j2
R


61

Then if π ≤ x < π, using the residue theorem as before, then
j
iξ(x− π )
j + eiξx
a1
6α3 j 3 e
|a1 T u∗ (x, y2 )x | = −
(iξ) ξ 2 − k 2
dξ
1
π ξ>k
ξ 2 − 9j 2
ξ2 − j2
24α3 j 3 |a1 |
ξ2
≤
dξ
π
ξ>k ξ 2 (ξ − 3j)2
24α3 j 3 |a1 |
≤ Ck(|g(0)| + |g(π)|).
=
π(k − 3j)
Follow the similar argument, if 0 < x < π , then |a1 T u∗ (x, y2 )x | ≤ Ck(|g(0)| + |g(π)|). Also,
1
j
|a2 T u∗ (x, y2 )x | ≤ Ck(|g(0)| + |g(π)|).
2
Thus T u∗ (x)x ≤ Ck(|g(0)| + |g(π)|). It shows that

Γ

|(T u∗ )x |2 =

π
0

|a1 (T u∗ )x |2 +
1

π
0

|a2 (T u∗ )x |2 ≤ Ck 2 |g(0)|2 + |g(π)|2 .
2

Furthermore, we have u∗ = a1 u∗ + a2 u∗ , where
yx
1yx
2yx
|a1 u∗ | = 6a1 α cos(jx) sin(jx)j ≤ Cαk(|g(0)| + |g(π)|),
1yx

|a2 u∗ | = 6a2 α cos(jx) sin(jx)j ≤ Cαk(|g(0)| + |g(π)|).
2yx
Based on the above expressions,

C
|u∗ |2 ≤ α2 k 2 (|g(0)|2 + |g(π)|2 ) ≤ Cα2 k(|g(0)|2 + |g(π)|2 ).
yx
j
Γ
Therefore, if α is chosen such that α =

β
, then the lemma is proved.
k
62

β
, then the lowest power
k
in k to control f is obtained. (2) the reason that u∗ is chosen to be (α sin(jx))3 on Γ instead
1
of lower powers such as α sin(jx) or (α sin(jx))2 is that this guarantees the integrability of
Remark: (1) the proof in Lemma (3.3.1) indicates that if α =

(T u∗ )x .
Summarizing all the results proved in this section,

Ω2

Ω1

D

Γ
Γ

|f |2 ≤ C |g(0)|2 + |g(π)|2 ,

|f |2 +

|f |2 =

|g1 |2 ≤ C |g(0)|2 + |g(π)|2 +

|g|2 ,
Γ

|g1x |2 ≤ Ck 2 |g(0)|2 + |g(π)|2 +

Γ

|gx |2 .

Combining these inequalities with Equation (3.13) yields that

w 22
+ k2 w 2 2
≤C
L (D)
L (D)

7
k5

12
1
|g|2 +
|gx |2 + k 5 |g(0)|2 + |g(π)|2
k Γ
Γ

.

Note that w = u + u∗ , and similar as the computations in Lemma 3.3.1, we see that

C
|g(0)|2 + |g(π)|2 .
u∗ 2 2
+ k 2 u∗ 2 2
≤
L (D)
L (D) k 2
Hence, the final estimate could be established accordingly, that is,

u 22
+ k2 u 2 2
≤C
L (D)
L (D)

7
k5

12
1
|g|2 +
|gx |2 + k 5 |g(0)|2 + |g(π)|2
k Γ
Γ

63

.

3.4

The Estimate for the Nonhomogeneous System

In this last section, we establish the stability estimate for the formulation in Equation (3.3),
or in general, we provide the stability estimate for the following problem



 ∆w + k 2 w = f




w = 0





 ∂n w = T (w) + g
1

in D,
(3.20)

on S,
on Γ.

where g1 (a) = g1 (b) = 0, Γ = [a, b] × {y = y2 }, and T (w) = √i
2π R

k 2 − ξ 2 weiξx dξ,

then the following theorem holds for the nonhomogeneous case.
Theorem 3.4.1. Under the geometric assumptions given in Equation (3.1), there exists a
constant C such that

w 22
+ k2 w 2 2
≤C
L (D)
L (D)

2
7
1
+ g1x 2 2
+ k5 f 2 2
(3.21)
.
k 5 g1 2 2
L (Γ) k
L (Γ)
L (D)

Remark: The estimate for the case where f = 0 and g may not necessarily vanish on the
boundary of Γ is established in the previous chapter where the stronger geometric assumption
is imposed on S. Since most of the ideas are similar to the previous chapter, the arguments
here emphasize on the portion when the nonhomogeneous terms are involved.
To simplify notation, we may again assume X ∗ = (0, 0). Since in this chapter we use the
fact that on Γ x coordinate starts from 0 to π in the construction of u∗ which is related to
g1 . So if we set X ∗ = (0, 0), a translation of the coordinate system is applied here. That is
why we assumed that Γ = [a, b] × {y = y2 }.
The proof of Theorem 3.4.1 involves three major parts.
64

Part 1. The weak forms of the governing equation with two different test functions yield
the following identity,
w 22
+ k2 w 2 2
+
L (D)
L (D)
|ξ|>k
=
Γ

g1 w +

D

(3.22)

w)

D
(X · n)|w|2 + 2

(X · n)| w|2 + k 2

−

S

f (X ·

fw + 2

(X · n)| w|2

ξ 2 − k 2 |w|2 dξ −

Γ

Γ

∂w
(X ·
Γ ∂n

w).

Part 2. Appropriate usage of some basic facts related to Fourier transform, the nonlocal
boundary condition in Equation (3.20) and the geometric assumption on S yield the following
inequality

w 22
+ k2 w 2 2
+
(
L (D)
L (D)
|ξ|>k
≤
Γ

g1 w +

+ 2y2

Γ

+2
Γ

f (X ·

fw + 2
D

T (w)g1 − 2

xwx g1 dx + y2

ξ 2 − k2 +

D
|ξ|≤k
Γ

ξ w (ξ) i

k2

)|w|2 dξ

ξ 2 − k2
w) + 2y2

|ξ|≤k

(k 2 − ξ 2 )|w|2 dξ
(3.23)

k 2 − ξ 2 w(ξ)dξ

|g1 |2 dx.

Part 3. To fully use the higher frequency term on the left hand side in Equation (3.23), the
relations between the higher frequency term on the left and the lower frequency component
is needed, also the L2 norm of w is expected to be controlled by the higher frequency
component, these relations are derived in Lemma 3.4.5. Then by fully using the results in
Lemma 3.4.5 and other tools, we estimate each term in the right hand side of Equation
(3.23), which are named by I1 through I8 .

65

3.4.1

Wave Energy Formulation

Next, we start from Part 1, where the goal is to formulate the wave energy identity.
Lemma 3.4.2. For the lower frequency part of w, the following identity holds

k 2 − ξ 2 |w|2 dξ = −

|ξ|≤k

Γ

g1 w +

fw .

(3.24)

D

And for the higher frequency part of w, it can be shown that

w 22
− k2 w 2 2
+
L (D)
L (D)
|ξ|>k

ξ 2 − k 2 |u|2 dξ =
Γ

g1 w +

f w . (3.25)
D

Proof. The weak form of the governing equation is given by the following equality

w·

k 2 wv −

v−

D

D

∂D

∂n wv =

fv

(3.26)

D

for any v ∈ H 1 (D). Now choose v = w, then Equation (3.26) becomes

| w|2 −
D

k 2 |w|2 −
D

Γ

(T (w) + g1 )w =

f w,

(3.27)

D

where the boundary conditions w = 0 on S and ∂n w = T (w) + g1 on Γ are used. Since w
is supported on Γ, we can extend w such that w = 0 on ΓC , then Γ T (w)w = R T (w)w =
R T (w) w = R i

k 2 − ξ 2 |w|2 dξ. Hence Equation (3.27) can be written as

| w|2 −
D

k 2 |w|2 −
D

i k 2 − ξ 2 |w|2 dξ =
Γ

R

66

g1 w +

f w.
D

(3.28)

Considering the fact that

−

k 2 − ξ 2 |w|2 dξ = −

i
R

|ξ|≤k

k 2 − ξ 2 |w|2 dξ +

i

ξ 2 − k 2 |w|2 dξ,

|ξ|>k

and taking the imaginary part and real part of Equation (3.28) respectively, the results in
the lemma could be obtained consequently.
Notice that from Equation (3.25) in Lemma 3.4.2, the terms associated with the components in the wave energy, i.e.

w 22
and −k 2 w 2 2
are already there, unforL (D)
L (D)

tunately, the second term −k 2 w 2 2
carries a negative sign, while in the wave energy,
L (D)
we need the positive one, so another identity is needed such that both terms are positive.
Therefore, another test function v = X ·

w is introduced here, where X = (x, y), it yields

the following lemma.
Lemma 3.4.3. For

> 0, we have

(X · n)| w|2 + 2k 2

−
S
= 2

|w|2 +
D

∂w
(X ·
Γ ∂n

Γ
f (X ·

w) + 2

(X · n)| w|2 − k 2

(X · n)|w|2
Γ

w).

(3.29)

D

Proof. Plug this test function into Equation (3.26), it becomes

w·
D

(X ·

k 2 w(X ·

w) −

u) −

D

∂w
(X ·
∂D ∂n

f (X ·

w) =

w).

(3.30)

D

For the above equality, using divergence theorem, it is easy to check that the first two

67

terms on the left hand side can be written as

w·

2

(X ·

(X · n)| w|2 ,

w) =

(3.31)

∂D

D
and

k 2 w(X ·

−2
D

|w|2 .

(X · n)|w|2 + 2k 2

w) = −k 2

(3.32)

D

∂D

Then multiply Equation (3.30) by 2 and take its real part, and apply Equations.(3.31)
and (3.32), we could arrive at the following equality

(X · n)| w|2 − k 2
∂D

∂D
f (X ·

= 2

(X · n)|w|2 + 2k 2

|w|2 − 2
D

∂w
(X ·
∂D ∂n

w).

w)
(3.33)

D
Since ∂D contains two parts Γ and S with different boundary conditions, the two parts
will be considered separately. For S part, note that w = 0 on S,

(X · n)| w|2 − k 2
S

(X · n)|w|2 − 2
S

∂w
(X ·
S ∂n

w = ∂w n , hence
∂n
(X · n)| w|2 .

w) = −
S

Therefore, Equation (3.33) becomes the equality shown in the lemma.
Obviously the term 2k 2 D |w|2 in Equation (3.29) is helpful as this term provides positive
contribution to the wave energy, so add Equation (3.25) and Equation (3.29) together, the
identity involving the wave energy could be formulated as shown in Equation (3.22).

68

3.4.2

Analysis through Fourier Transform

On the left hand side of Equation (3.22), apart from the wave energy terms, the term
ξ 2 − k 2 |w|2 dξ will be helpful to control the higher frequency part of w, the term

|ξ|>k

− s (X · n)| w|2 is nonnegative under the geometric assumption X · n ≤ 0 on S, so it
could be dropped from the estimate. For the right hand side, the term Γ ∂w (X ·
∂n
is more involved, since

w)

w on Γ can not be controlled by the wave energy terms directly,

some further analysis is required. Since the operator T is simply a multiplication operator
in terms of Fourier transform. Using the boundary condition on Γ and some basic identity
from Fourier transform, the last 3 terms on the right hand side of Equation (3.22) can be
easily converted into the frequency domain. For the details of the proof, see Chapter 2.
Lemma 3.4.4. For the last three terms on the right hand side of Equation (3.22), it can be
shown that
(X · n)| w|2 + k 2

−
Γ
≤−

−2

(X · n)|w|2 + 2
Γ

k2
|ξ|>k

ξ 2 − k2

|ξ|≤k

|w|2 dξ2y2

ξ w (ξ) i

|ξ|≤k

∂w
(X ·
Γ ∂n

w)

(k 2 − ξ 2 )|w|2 dξ + 2y2

k 2 − ξ 2 w(ξ)dξ + 2
Γ

Γ

xwx g1 dx + y2

T (w)g1

Γ

From the result in Lemma 3.4.4, we can see that the original term 2

(3.34)

|g1 |2 dx.

∂w
Γ ∂n (X ·

w)

actually contains lots of useful information. Especially, it produced a negative higher grequency component term, − |ξ|>k

k2
|w|2 dξ, this can be moved to the left hand side
ξ 2 −k 2

and helps with the final estimate. Therefore, by dropping the term − s (X · n)| w|2 in
Equation (3.22) using the geometric assumption and also use the result in Lemma 3.4.4, the
result shown in Equation (3.23) is obtained.
69

Now we can see that the term

w has been canceled out by using the result shown in

lemma 3.4.4; as a consequence, we obtain an energy inequality which contains lots of other
terms on the right hand side.

3.4.3

Auxiliary Lemmas and Final Estimates

Next the estimate for each term on the right hand side of Equation (3.23) is derived. This
is not straightforward as no direct connection is there for the higher frequency component on the left hand side and the terms on the right. Therefore, the connection between them is established in Lemma 3.4.5. To simplify the notations, we denote AL =
k2
2
2 2
2
2
)|w|2 dξ . As stated in the
|ξ|≤k k − ξ |w| dξ, and BH = |ξ|>k ( ξ − k +
ξ 2 −k 2
following lemma, AL , the lower frequency part of w, could be controlled by BH , which is
connected to the higher frequency part of w.
Lemma 3.4.5. There exists a positive constant C, such that

1
δ
AL ≤ BH + ||g1 ||2 2
+2|
f w |,
L (Γ)
k
δ
D

(3.35)

1
where δ can be chosen such that Cδ ≤ k 2 . Furthermore,
2
C2
2C
w 22
≤ BH +
g1 2 2
+√ |
fw | .
L (Γ) k
L (Γ)
k
k
D
.

70

(3.36)

Proof. As shown in Lemma 2.4.1, there exists a constant C such that

1
C
w 22
≤ √ AL + BH .
L (Γ)
k
k

(3.37)

In addition, by using the identity Equation (3.24) and the above inequality, we have

δ
1
f w |≤ w 2 2
fw |
g1 2
w 2 +|
+
g 2
+|
L (Γ)
L (Γ)
L (Γ) 2δ 1 L2 (Γ)
2
D
D
1
δ C
1
≤
( √ AL + BH ) +
g 2
+|
fw |
2 k
k
2δ 1 L2 (Γ)
D

AL ≤

If δ is chosen such that Cδ ≤

√

k, then the following inequality holds,

1
Cδ
1
δ
AL ≤ (1 − √ )AL ≤
BH +
g 2
+|
fw | .
2
2k
2δ 1 L2 (Γ)
2 k
D
Equation (3.35) is obtained accordingly. Plugging Equation (3.35) into Equation (3.37), the
estimate for w 2 2
could be arrived as shown below.
L (Γ)
C
w 22
≤√
L (Γ)
k
Choose Cδ =

δ
1
1
BH + g 2 2
+2|
f w | + BH .
L (Γ)
k
δ
k
D

√
k, then Equation (3.36) holds.

Lemma 3.4.5 is crucial in the sense that it provides a way to relate the integrals of |w|2
and

|ξ 2 − k 2 ||w|2 to the high frequency part BH and also norms related to g1 ; it is also

important in determining the power of k in the estimate. Now we will estimate each term
on the right hand side of Equation (3.23), which are named I1 through I8 . Here similar
techniques could be used as in the previous work for the terms I1 , I3 , I5 and I6 , for other
new terms, i.e., I2 and I8 , they could be controlled by norms related to f and the energy
71

terms on the left, and for I4 , considering the fact that g1 = 0 on the boundary of Γ, a different
way as shown in Lemma 3.4.9 can be used, which is relatively straightforward comparing to
the previous work and here the geometric assumption is not needed for the estimate. The
following lemmas, i.e., Lemma 3.4.6 through Lemma 3.4.12, provides the estimate for terms
on the right hand side of the energy inequality.
Lemma 3.4.6. The first term on the right hand side of Equation (3.23) can be estimated by

I1 =

ε
d
1
g1 w ≤ √1 BH + √1 g1 2 2
+ ε1
k 2 |w|2 + e1
|f |2 ,
L (Γ)
k
k
k2 D
Γ
D

(3.38)

ε1 C 2
ε1 C 2
1
+
.
where ε1 is any small positive number, and d1 =
and e1 =
2
2ε1
4
Proof. By applying the relation between w 2 2
and BH as shown in Lemma 3.4.5, it
L (Γ)
follows that
√
ε1 k
1
√
I1 =
g1 w ≤ g1 2
w 2
≤
w 22
+
L (Γ)
L (Γ)
L (Γ) 2ε1 k
2
Γ
√
ε1 k 2
C2
2C
1
√ g1
BH +
g1 2 2
≤
+√ |
fw | +
L (Γ)
2
k
k
k
2ε1 k
D
d
ε
= √1 BH + √1 g1
k
k
ε
d
≤ √1 BH + √1 g1
k
k

g1 2 2
L (Γ)
2
L2 (Γ)

(3.39)

2
+ ε1 C |
fw |
L2 (Γ)
D
1
2
k 2 |w|2 + e1
|f |2 ,
2 (Γ) + ε1
2 D
L
k
D

where the constants are independent of k.
It is easy to obtain the estimate for the second term on the right hand side of Equation
(3.23) since a positive term k 2 |w|2 lies on the left hand side of the equation.

72

Lemma 3.4.7. For I2 , we have

I2 =

D

f w ≤ ε2

1
k 2 |w|2 + e2
|f |2 ,
k2 D
D

(3.40)

1
.
where ε2 is any small positive number, and e2 =
4ε2
Similarly, a direct usage of Schwarz inequality would lead to the estimate of I3 as shown
below.
Lemma 3.4.8.

I3 = 2

f (X ·
D

w) ≤ ε3

D

| w|2 + e3

|f |2 ,

(3.41)

D

M
where ε3 is any small positive number, and e3 = 4ε3 , M3 = maxx∈Γ {x2 }.
8
The estimate on I4 can be derived by using Lemma 3.4.5.
Lemma 3.4.9. I4 which contains only the lower frequency part can be estimated as follows.

I4 = 2y2

|ξ|≤k

(k 2 − ξ 2 )|w|2 dξ
(3.42)

≤ ε4 BH + d4 k g1 2 2
+ ε4
k 2 |w|2 + e4
|f |2 ,
L (Γ)
D
D
2
4y2

2
16y2

where ε4 is a small positive number, d4 =
and e4 =
.
ε4
ε4

73

Proof. Using the fact that for |ξ| ≤ k,

I4 = 2y2
≤ 2y2 k

|ξ|≤k

k 2 − ξ 2 < k , then

(k 2 − ξ 2 )|w|2 dξ ≤ 2y2 kAL

2y
ε3
fw |
BH + 2 g1 2 2
+2|
L (Γ)
2y2 k
ε4
D

(3.43)

= ε4 BH + d4 k g1 2 2
+ 4y2 k
fw |
L (Γ)
D
|f |2 .
k 2 |w|2 + e4
≤ ε4 BH + d4 k g1 2 2
+ ε4
L (Γ)
D
D

Lemma 3.4.10. It can be shown that I5 satisfies the following inequality

I5 = 2y2

Γ

T (w)g1

e
1
≤ ε5 BH + d5 g1 2 2
+ d5 g1 2 2
+ ε5
|f |2 , (3.44)
k 2 |w|2 + 4
L (Γ)
L (Γ)
k
k D
D
2
2
2
2y2
ε5 C 2 2y2
C 2 + 2y2 .
where ε5 is a small positive number, d5 = 2 + ε ,d5 = ε and e5 = 4ε
ε5
5
5
5
Proof. Define g1Γ in the following way,


g
 1 on Γ,
g1Γ :=


 0 on ΓC ,

74

then

I5 = 2y2
= 2y2

R

T (w)g1Γ = 2y2
i

R

R

T (w)g1Γ

k 2 − ξ 2 wg1Γ ≤ 2y2

1
2 dξ 2
|w|

R
2
2y
ε k
≤ 5 w 22
+ 2
|k 2 − ξ 2 ||g1Γ |2 dξ
L (Γ) ε5 k R
2
ε k 2
C2
2C
fw |
≤ 5
BH +
g1 2 2
+√ |
L (Γ)
2
k
k
k
D

R

1
2 − ξ 2 ||g |2 dξ 2
|k
1Γ

2
2y2

k 2 |g1Γ |2 dξ + ξ 2 ||g1Γ |2 dξ
ε5 k R
√
2y 2
2y 2
ε C2
= ε5 BH + 5
f w | + 2 k g1 2 2
g1 2 2
+ ε5 C k |
+ 2 g1 2 2
L (Γ)
L (Γ) ε5 k
L (Γ)
2
ε5
D
2
2
2y2
ε5 C 2
2
2 |w|2 + e5
2 + 2y2 k g 2
2
g1 2
+ ε5
k
|f |
≤ ε5 BH +
1 L2 (Γ) + ε k g1 L2 (Γ)
L (Γ)
2
k D
ε5
D
5
+

e
1
|f |2 .
≤ ε5 BH + d5 k g1 2 2
+ ε5
+ d5 g1 2 2
k 2 |w|2 + 5
L (Γ)
L (Γ)
k
k D
D

Lemma 3.4.11. By using Lemma 3.4.5 and further analysis in the frequency domain, we
have

I6 = −2

|ξ|≤k

ξ w (ξ) i

k 2 − ξ 2 w(ξ)dξ

2
7
≤ ε6 BH + d6 k 5 g1 2 2
+ ε6
k 2 |w|2 + e6 k 5
|f |2 ,
L (Γ)
D
D
where ε6 is any small positive number.

75

(3.45)

Proof. Following the conclusion in Lemma 2.4.5, the following inequality holds.
6
I6 = εk w 2 2
+ Ck 5 AL
L (Γ)
2
C2
2C
fw |
≤ εk
BH +
g1 2 2
+√ |
L (Γ)
k
k
k
D
6 ε
k
BH + g1 2 2
+2|
fw |
+ Ck 5
L (Γ)
k
ε
D
7
2
≤ ε6 BH + d6 k 5 g1 2 2
k 2 |w|2 + e6 k 5
|f |2 .
+ ε6
L (Γ)
D
D

Lemma 3.4.12. It is shown that I7 can be estimated by

I7 = 2

Γ

xwx gdx

d
1
≤ ε7 BH + d7 g1 2 2
+ 7 g1x 2 2
+ ε7
k 2 |w|2 + e7
|f |2 ,(3.46)
L (Γ)
L (Γ)
k
k D
D

ε7 can be chosen as an arbitrary positive number, d7 =
C 2ε
e7 = 4 7

2M3
ε7 C 2
4
2 + ε7 , d7 = ε7 , and

Proof. Using integration by parts and the connection between w 2 2
and BH in Lemma
L (Γ)
3.4.5, it implies that

I7 = 2

Γ

xwx g1 dx = −2

Γ

w(xg1 )x dx

ε k
2
≤ 7
|w|2 dx +
|(xg1 )x |2 dx
2 Γ
ε7 k Γ
ε k 2
C2
2C
BH +
g1 2 2
+√ |
≤ 7
fw |
L (Γ)
2
k
k
k
D

+

2M3
4
2
g1 2 2
+
g
L (Γ)
ε7 k
ε7 k 1x L2 (Γ)

d
1
+ 7 g1x 2 2
≤ ε7 BH + d7 g1 2 2
+ ε7
k 2 |w|2 + e7
|f |2 .
L (Γ)
L (Γ)
k
k D
D
76

By using the estimates obtained in Lemma 3.4.6 through 3.4.12, we can see that Equation
(3.23) can be estimated by:

( ξ 2 − k2 +
w 22
+ k2 w 2 2
+
L (D)
L (D)
|ξ|>k
7
1
+ε
≤εBH + dk 5 g1 2 2
+ d g1x 2 2
L (Γ)
L (Γ)
k

k2

)|w|2 dξ

ξ 2 − k2

k 2 |w|2 +
D

(3.47)
2
| w|2 + ek 5

D

|f |2 ,
D

here ε, d, d and e are chosen such that they are ≥ the sums of the corresponding components
in the estimates. Since ε can be chosen as small positive number, so the BH term on the
right hand side of Equation (3.47) can be absorbed by the BH term on the left, similarly,
the energy term on the right hand side can be absorbed by the energy term on the left hand
side. Therefore, it shows that there exists a constant C such that Theorem 3.4.1 holds.

77

Chapter 4
Stability Estimates for
Rectangular-like Domains

4.1

Main Theorem and Sketches of Approach

There are other cavity domains which do not satisfies the aforementioned geometric assumptions in Chapter 2 and Chapter 3, such as rectangular-like domains (domains composed of
rectangles of different sizes), one example is shown in Fig. 4.1. The main result is stated in
the following theorem, where the outnormal vector n = (nx , ny ).
Theorem 4.1.1. For the cavity of rectangular-like shapes with yny ≤ 0 on S, there exists a
constant C such that

u

7
3
2
2 (D) + |k| u L2 (D) ≤ C[k 4 g L2 (Γ) + k 4 gx L2 (Γ) + k (|g(0)| + |g(π)|)]. (4.1)
L

To prove the theorem, the same u∗ as in Chapter 3 is introduced, which satisfies: 1)
T u∗ (0, y2 ) = g(0, y2 ) and T u∗ (π, y2 ) = g(π, y2 ), 2) u∗ (0, y2 ) = u∗ (π, y2 ) = 0, 3) u∗ (0, y2 ) =
y
u∗ (π, y2 ) = 0, and 4) u∗ = 0 on S. Then we set w = u + u∗ , then w satisfies the following
y

78

Figure 4.1: Rectangular-like domain
equations


 ∆w + k 2 w = f






in D,

on S,
w = 0





 ∂n w = T w + g
1 on Γ,

(4.2)

where


 f = ∆u∗ + k 2 u∗ ,





∗
∗
 g1 = g − T u + uy ,





 g (0) = g (π) = 0.
1
2

79

(4.3)

The major difference lies in that we need to derive an energy estimate for w in the context
of rectangular-like domains. Here two main procedures are involved to derive this energy
estimate. First, we start from the weak formulation of w and also using appropriate test
functions, the following inequality could be arrived, for any ε > 0,
|wy |2 +

ξ 2 − k 2 |w|2 dξ

|ξ|>k
D
3
C
C
|f |2 ,
≤Ck 2 g1 2 2
+ √ g1x 2 2
|w|2 + k 2
+ε
L (Γ)
L (Γ)
ε
k
D
D

(4.4)

Note that this inequality only contains the estimate for wy , therefore, additional relations
involving other terms in the energy needs to be explored, that is,

|w|2 ≤ C
D

D

|wy |2 .

(4.5)

Combining these two inequalities and also the result in Lemma 4.2.4, the following energy
estimate for w is arrived:

k2 w 2 2
+
L (D)

7
3
w 22
≤ C k 2 g1 2 2
+ k 2 g1x 2 2
+ k4
|f |2 . (4.6)
L (D)
L (Γ)
L (Γ)
D

By using this energy estimate along with detailed analysis on u∗ , the main result could be
obtained.

4.2

Preliminary Lemmas

Since we use the same u∗ as applied in Chapter 3, so here the details related to u∗ are
skipped for simplification and the resulting estimates related to u∗ will be restated briefly

80

when needed as for the clarity of our further proof. The main effort is to establish a stability
estimate for the new function w by assuming the cavity region is of rectangular-like shape and
furthermore we assume yny ≤ 0. Here we start from the weak formulation of the problem
for w, and apply two different test functions to obtain an important inequality. The weak
form of the governing equation is given by the following equality

w·

k 2 wv −

v−
D

D

∂D

∂n wv =

f v.

(4.7)

D

Similar as before, by choosing v = w in Equation (4.7), and taking the imaginary part
and real part respectively, one has
Lemma 4.2.1.

|ξ|≤k

k 2 − ξ 2 |w|2 dξ = −Im

w 22
− k2 w 2 2
+
L (D)
L (D)
|ξ|>k

Γ

g1 w +

fw ,

(4.8)

D

ξ 2 − k 2 |u|2 dξ =
Γ

g1 w +

f w . (4.9)
D

Notice that in Equation (4.9), the terms associated with the components in the wave
energy, i.e.

w 22
and −k 2 w 2 2
are already there, unfortunately, the second
L (D)
L (D)

term −k 2 w 2 2
carries a negative sign, while in the wave energy, we need the positive
L (D)
one, so another identity is needed to cancel out the negative term. Therefore, another test
function v = ywy is introduced. Note that this is different from the test function used in
Chapter 2 and Chapter 3 since a different geometric assumption is imposed here.

81

Pluging this test function into Equation (4.7) yields

ywy −

w·
D

D

k 2 wywy −

∂w
ywy =
f ywy .
∂D ∂n
D

Multiply both sides of the above equation by 2 and take the real part, it becomes

w·

2

ywy − 2

D

D

k 2 wywy − 2

Γ

y2 |wy |2 = 2

D

∂w
ywy(4.10)
.
S ∂n

f ywy + 2

By adding appropriate additional terms and using the divergence theorem, it follows that

w·

2
D
D
=
∂D
Γ

D

y(| w|2 )y + 2|wy |2 =

=

=

ywy = 2

yny (| w|2 ) −

y2 (| w|2 ) +

D
=k 2
Γ

D

[y(| w|2 )]y − | w|2 + 2|wy |2

| w|2 + 2

(4.11)
|wy |2

D
D
∂w
ywy −
| w|2 + 2
|wy |2 .
S ∂n
D
D

2
D k wywy can be written as

Furthermore, the other term 2

2

[wx ywxy + wy ywyy ] + 2|wy |2

k 2 wywy =

y2 |w|2 −

D

k 2 y(|w|2 )y =

D

k 2 (y|w|2 )y − k 2 |w|2
(4.12)

k 2 |w|2 .
D

Plug the results in Equations (4.11) and (4.12) into Equation (4.10), it leads to

2
D

|wy |2 −

=−
Γ

| w|2 +
D

y2 (| w|2 ) + k 2

k 2 |w|2
D

Γ

y2 |w|2 + 2

Γ

y2 |wy |2 + 2

82

D

f ywy +

∂w
ywy .
S ∂n

(4.13)

Note the geometric assumption yny ≤ 0 on S guarantees that S ∂w ywy ≤ 0. Therefore
∂n
the following lemma could be obtained.
3+
H 2 (D), > 0 ,we have

1
Lemma 4.2.2. For all w ∈ HS (D)

2
D
≤ −
Γ

|wy |2 −

|w|2

| w|2 + k 2
D

D

y2 | w|2 + k 2

Γ

y2 |w|2 + 2

Γ

y2 |wy |2 + 2

D

f ywy .

(4.14)

Consequently, the negative term on the left hand side of the above equation could be
canceled out as shown in the following lemma.
Lemma 4.2.3. By adding Equation (4.9) and Equation (4.2.2) together, it yields that

2
D

|wy |2 +

≤
Γ

|ξ|>k

fw −

g1 w +

+2
Γ

ξ 2 − k 2 |w|2 dξ

D

y2 |wy |2 + 2

Γ
D

y2 | w|2 + k 2

Γ

y2 |w|2

(4.15)

f ywy .

Note that for the third to fifth term on the right hand side of the above identity, a direct
computation yields the following result.
Lemma 4.2.4. As detailed in Lemma 2.3.2 in Chapter 2, one has

−
Γ
≤2y2

y2 | w|2 + k 2

|ξ|≤k

Γ

y2 |w|2 + 2

(k 2 − ξ 2 )|w|2 dξ + 2y

2

Γ

y2 |wy |2

Γ

(4.16)
T (w)g1 + y2

Γ

|g1 |2 dx.

Use the result in Lemma 4.2.4 and apply it in Lemma 4.2.3, the following lemma holds.

83

Lemma 4.2.5.

2
D

|wy |2 +

≤
Γ
+ 2y2

|ξ|>k

g1 w +

Γ

ξ 2 − k 2 |w|2 dξ

D

f w + 2y2

T (w)g1 + y2

Γ

|ξ|≤k

(k 2 − ξ 2 )|w|2 dξ

|g1 |2 dx + 2

D

(4.17)

f ywy .

Observe that a higher frequency component lie on the left hand side of the above inequality, this could be used for the estimation on the right hand side, thus we need to build the
connections between the higher frequency component and lower frequency component. By
using the fact that w is supported on Γ and the result in Lemma 4.2.1 here, the connection
could be eatablished. To simplify the notations, denote AL = |ξ|≤k
AH = |ξ|>k

k 2 − ξ 2 |w|2 dξ, and

k 2 − ξ 2 |w|2 dξ . As stated in the following lemma, AL , the lower frequency

part of w, could be controlled by AH , which is connected to the higher frequency part of w.
Lemma 4.2.6. There exists positive constants C and d, such that

M
1
AL ≤ √ AH + ||g1 ||2 2
+2|
f w |,
L (Γ)
M
k
D

(4.18)

1
k 2 . Furthermore,
where M can be chosen such that M ≤
d
C
C
C
w 22
≤ √ AH +
g1 2 2
+√ |
fw | .
L (Γ)
L (Γ)
k
k
k
D

(4.19)

Remark. Note that the difference between this lemma and Lemma 3.4.5 lies in the power
1
of k, here the power associated with AH is √ while in 3.4.5 the power associated with BH
k
is 1 , since in BH additional higher order term exists. Here BH is not involved due to the
k
84

fact that a different test function is used. This power dependency difference essentially leads
to the different power in the final estimate.
Proof. Here we could start from the idea of Lemma 3.5 in [8], that is, there exists a constant
d, such that

d
≤ √ (AL + AH ).
w 22
L (Γ)
k

(4.20)

In addition, by using the identity (4.8) and the above inequality, we have

M
1
g1 2
w 2 +|
f w |≤
w 22
g 2
+
+|
fw |
L (Γ)
L (Γ)
L (Γ) 2M 1 L2 (Γ)
2
D
D
M C
1
√ (AL + AH ) +
≤
g 2
+|
fw |
2 k
2M 1 L2 (Γ)
D

AL ≤

If M is chosen such that dM ≤

√

k, then the following inequality holds,

1
dM
M
1
AL ≤ (1 − √ )AL ≤ √ AH +
g 22 + |
fw |
L (Γ)
2
2M
2 k
2 k
D
M
1
g 22
+2|
AL ≤ √ AH +
fw |
L (Γ)
M
k
D
Considering the results obtained in Equation (4.20) and Equation (4.21) together, the estimate for w 2 2
could be arrived as shown below.
L (Γ)
d
w 22
≤√
L (Γ)
k

M
1
d
√ AH +
g 22
+2|
f w | + √ AH .
L (Γ)
M
k
k
D

85

(4.21)

√
Choose M =

k , then
d
C
C
C
g 22
+√ |
fw | .
w 22
≤ √ AH +
L (Γ)
L (Γ)
k
k
k
D

This is an important lemma as it relates the integrals of |w|2 and

(4.22)

|ξ 2 − k 2 ||w|2 to the

high frequency part AH and also norms related to g1 . Now we will estimate each term on
the right hand side of Equation (4.17), which are named I1 through I6 . As most of the
details are similar to the previous chapter, the only difference is the power of k as stated in
the Remark, thus the details are skipped and only the estimates are given as below.
Lemma 4.2.7. Similar to the computations as shown in Chapter 3, one can show that for
ε > 0,

ε
C
C
ε
g1 w ≤ √1 AH +
g1 2 2
|w|2 +
|f |2 ,
+
L (Γ) 4 D
ε1 k
εk D
k
Γ
ε
C
fw ≤
I2 =
|w|2 +
|f |2 ,
4 D
ε D
D
I3 = 2y2
(k 2 − ξ 2 )|w|2 dξ
|ξ|≤k
ε
C
C 3
+
|w|2 + k 2
|f |2 ,
≤ ε1 AH + k 2 g1 2 2
L (Γ) 4 D
ε1
ε
D

I1 =

I4 = 2y2

Γ

(4.24)

T (w)g1

C
ε
C
√ ( g1 2 2
+ g1x 2 2 ) +
|w|2 +
|f |2 , (4.25)
L (Γ)
L (Γ)
4 D
ε D
ε1 k
C
|wy |2 +
|f |2 .
(4.26)
f ywy ≤ ε1
ε1 D
D
D

≤ ε1 AH +
I6 = 2

(4.23)

86

Combine all the estimates above, we have

(1 − ε1 )

D

|wy |2 + (1 − 3ε1 )

|ξ|>k

ξ 2 − k 2 |w|2 dξ

3
C
C
1
|w|2 + k 2
|f |2 ,
≤
k 2 g1 2 2
+ √ g1x 2 2
+ε
L (Γ)
L (Γ)
ε
ε
k
D
D

(4.27)

As long as ε1 is chosen as a positive number such that 1 − 3ε1 ≤ 0 , then we have
|wy |2 +

ξ 2 − k 2 |w|2 dξ

|ξ|>k
D
3
C
C
+ √ g1x 2 2
+ε
|w|2 + k 2
≤Ck 2 g1 2 2
|f |2 ,
L (Γ)
L (Γ)
ε
k
D
D

(4.28)

So far, on the left hand side of the inequality, notice that only partial energy term is there,
that is, D |wy |2 , and this inequality shows how D |wy |2 is controlled by D |w|2 and
other terms. To establish an energy estimate, the inequality of the reverse order, that is,
how D |w|2 could be controlled by D |wy |2 needs to be derived, which is the main goal of
the next section.

4.3

Relation between

D

|wy |2 and

D

|w|2

Using the divergence theorem, we see that

0=
∂D

(y − y2 )|w|2 ny =

D

[(y − y2 )|w|2 ]y =

D

|w|2 + 2 (y − y2 )wwy .

Then we can estimate D |w|2 by Schwarz inequality,

|w|2 = 2
D

D

(y2 − y)wwy ≤ ε

87

C
|w|2 +
|wy |2 .
ε D
D

Therefore, choose any ε < 1, it shows that

|w|2 ≤ C
D

4.4

D

|wy |2 .

(4.29)

Final Estimates

Note that earlier we have the following result,

D

|wy |2 +

|ξ|>k

ξ 2 − k 2 |w|2 dξ
(4.30)

3
C
1
|w|2 +
≤C k 2 g1 2 2
+ √ g1x 2 2
+ε
k 2 |f |2 .
L (Γ)
L (Γ)
ε D
k
D
Combine this with Equation (4.29) together, then

k 2 |w|2 ≤ Ck 2
D

|wy |2

D
3
7
C
|f |2 + ε
k 2 |w|2 .
+ k 2 g1x 2 2
+ k4
≤C k 2 g1 2 2
L (Γ)
L (Γ)
ε
D
D
Note that again ε D k 2 |w|2 could be absorbed by the left hand side term, thus follows that
the term D k 2 |w|2 could be estimated by
7
3
k 2 |w|2 ≤C k 2 g1 2 2
+ k 2 g1x 2 2
+ k4
|f |2 .
L (Γ)
L (Γ)
D
D

(4.31)

Recall the second identity in Lemma 4.2.1,

w 22
− k2 w 2 2
+
L (D)
L (D)
|ξ|>k

ξ 2 − k 2 |u|2 dξ =
Γ

88

g1 w +

f w . (4.32)
D

Use the above identity and Equation (4.31), we have

w 22
≤ k2 w 2 2
−
L (D)
L (D)
|ξ|>k

ξ 2 − k 2 |u|2 dξ +
Γ

g1 w +

fw
D

≤k 2 w 2 2
− AH + ε w 2 2
+ C g1 2 2
+ε w 22
+C f 22
L (D)
L (Γ)
L (Γ)
L (D)
L (D)
C
C
C
f w |)
g 22
+√ |
≤k 2 w 2 2
− AH + ε( √ AH +
L (Γ)
L (D)
k
k
k
D

(4.33)

+ C g1 2 2
+ε w 22
+C f 22
L (Γ)
L (D)
L (D)
7
3
≤C k 2 g1 2 2
+ k 2 g1x 2 2
+ k4
|f |2 .
L (Γ)
L (Γ)
D
Therefore, an estimate regarding to the wave energy of w is derived:

k2 w 2 2
+
L (D)

7
3
w 22
≤ C k 2 g1 2 2
+ k 2 g1x 2 2
+ k4
|f |2 . (4.34)
L (D)
L (Γ)
L (Γ)
D

Next, we will use this relation together with the result in Section 3 for further analysis. As
indicated in the previous subsection, an inequality related to the wave energy of w is derived,
which could be controlled by norms of f and g1 . Recall that in the previous chapter, the
following three inequalities are obtained, for which we can use to show the final estimate.

|f |2 =
D

|f |2 +
Ω1

Γ
Γ

|f |2 ≤ C(|g(0)|2 + |g(π)|2 ),
Ω2

|g1 |2 ≤ C(|g(0)|2 + |g(π)|2 ) +

|g1x |2 ≤ Ck 2 (|g(0)|2 + |g(π)|2 ) +

89

|g|2 ,
Γ
Γ

|gx |2 .

By using these facts in Equation (4.34), it yields that
3
7
|g|2 + k 2
|gx |2 + k 4 (|g(0)|2 + |g(π)|2 ) .
w 22
+ k2 w 2 2
≤ C k2
L (D)
L (D)
Γ
Γ
Note that w = u + u∗ , and for u∗ ,

C
u∗ 2 2
+ k 2 u∗ 2 2
≤
(|g(0)|2 + |g(π)|2 ),
L (D)
L (D) k 2
thus, the final estimate could be established accordingly, that is,
3
7
|g|2 + k 2
|gx |2 + k 4 (|g(0)|2 + |g(π)|2 ) .
u 22
+ k2 u 2 2
≤ C k2
L (D)
L (D)
Γ
Γ

90

Chapter 5
Conclusion and Some Open Questions
This work is a preliminary endeavor in studying the stability estimates of electromagnetic
scattering from open cavity. We focus on the study of TM case and assume the media is
homogeneous. Our main contribution is that explicit estimates are derived, on how the wave
energy in the cavity region depends on the wave number k and incoming fields. We have
provided the explicit relations in the context of different geometric features of the cavity
domain using different techniques. The stability estimates can provide guidance for the
numerical computations, as well as provide insights on the shape design of cavities.
On the other hand, the numerical results may give evidence on the dependency relations
and help to find the optimal power dependency in terms of wave number k, which is an interesting topic to explore in the future. Also, the closely related problem would be the stability
estimate for TE(Transverse Electric) polarization case in the two dimensional setting, where
the bounded domain formulation is given by



 ∆u + k 2 u = 0 in D,




on S,
 ∂n u = 0





 u = T (∂n u) + g on Γ,
i

(5.1)

∂n u. The major difference lies in two parts: one is the boundary
k2 − ξ 2
condition in particular the multiplication operator; another is that while in TM case u is
with T (∂n u) =

91

supported on Γ, here uy is supported on Γ while u and ux may not be, a natural extension of
the proof for TM case would fail in TE case; thus additional techniques are expected to derive
the stability estimates. As long as the stability estimate for TE case could be obtained given
the geometric assumption as in TM case, then the two dimensional problem would be solved
under same geometric setting since the general case could always be decomposed into the
sum of TM and TE case. For three dimensional problem, the bounded domain formulation
is given by [4]













× E − iωµH = 0,
× H + iω E = 0,
(5.2)



n × E = 0
on S,







 − n × (n × H) = P (−n × (n × E)) + g on Γ.
where
∗
−g i = −n × (n × (E i − peik0 q·x )) + P (−n × (n × (E i − peik0 q ·x )))

and

P (−n × (n × E))
=

1
{[−
(2π)2 ωµ0 R2

1

(ξ g + ξ2 g2 )ξ2 −
2 − ξ2 − ξ2 1 1
k0
1
2

1
(ξ g + ξ2 g2 )ξ1 −
2
2
2 1 1
k0 − ξ1 − ξ2

2
2
2
k0 − ξ1 − ξ2 g2 ],

2
2
2
k0 − ξ1 − ξ2 g1 ], 0}eiξ·x dξ,

with gj (x1 , x2 ) = E j (x1 , x2 , 0), j = 1, 2, 3. Existence and uniqueness of the solutions for the

92

model problem are established by a variational approach and the Hodge decomposition, the
stability estimates are not derived yet. Furthermore the electromagnetic scattering in layered
media, which is significantly important in many areas such as optics, geophysical probing,
communication, remote sensing[12, 31], etc. Another interesting problem is concerned with
the optimal design problem in inverse scattering, in aims to design the cavity shape and
material to reduce or enhance the radar cross section [22]. For inverse scattering, there are
some local stability result obtained for periodic structures and biperiodic structures [6, 9],
and Li [22] proved the local stability for one particular case of the cavity problem, where
the upper halfspace is filled with a lossless homogeneous medium above the flat ground
surface; while the interior of the cavity is assumed to be filled with a lossy homogeneous
medium accounting for the energy absorption; more stability results could be explored based
on different consideration of the medium. In a word, there are still many open questions
in terms of stability estimates in cavity problem, which are theoretically interesting and
challenging.

93

BIBLIOGRAPHY

94

BIBLIOGRAPHY

[1] G. Acostaa, M. G. Armentanob, R. G. Dur´b, and A. L. Lombardib. Nonhomogeneous
a
neumann problem for the poisson equation in domains with an external cusp. Journal
of Mathematical Analysis and Applications, 310(2):397–411, 2005.
[2] R. A. Adams. Sobolev Spaces. Academic Press, New York, 1975.
[3] H. Ammari, G. Bao, and A. W. Wood. Analysis of the electromagnetic scattering from
a cavity. Japan J. Indust. Appl. Math., 19(2):301–310, 2002.
[4] H. Ammari, G. Bao, and A. W. Wood. A cavity problem for maxwell’s equations.
Methods and applications of analysis, 9(2):249–260, 2002.
[5] H. Anastassiu. A review of electromagnetic scattering analysis for inlets, cavities, and
open ducts. IEEE Antennas Propag. Mag., 45(6):27–40, 2003.
[6] G. Bao and A. Friedman. Inverse problems for scattering by periodic structures. Arch.
Ration. Mech. Anal., 132:49–72, 1995.
[7] G. Bao and W. Sun. A fast algorithm for the electromagnetic scattering from a large
cavity. SIAM J. Sci. Comput., 27:553–574, 2005.
[8] G. Bao, K. Yun, and Z. Zhou. Stability of the scattering from a large electromagnetic
cavity in two dimensions. SIAM J. Math. Anal., 44(1):383–404, 2012.
[9] G. Bao and Z. Zhou. An inverse problem for scattering by a doubly periodic structure.
Trans. Amer. Math. Soc., 350:4089–4103, 1998.
[10] R. J. Burkholder. A fast and rapidly convergent iterative physical optics algorithm
for computing the rcs of open-ended cavities. Applied Computational Electromagnetics
Society Journal, 16(1):53–60, 2001.
[11] R. J. Burkholder, R. C. Chou, and P. H. Pathak. Two ray shooting methods for computing the EM scattering by large open-ended cavities. Computer Physics Communications,
68(1-3):353–365, 1991.
[12] W. C. Chew. Waves and Fields in Inhomogeneous Media. IEEE Press, 1995.
95

[13] T. T. Chia, R. J. Burkholder, and R. Lee. The application of fdtd in hybrid methods for
cavity scattering analysis. IEEE Trans. Antennas Propagat., 43(10):1082–1090, 1995.
[14] F. Cummings and F. X. Sharp regularity coefficient estimates for complex-valued acoustic and elastic helmholtz equations. Math. Models Methods. Appl. Sci., 16(1):139–160,
2006.
[15] S. Esterhazy and J. Melenk. On stability of discretizations of the helmholtz equation
(extended version). ASC Report, No. 01:1 – 49, 2011.
[16] U. Hetmaniuk. Stability estimates for a class of helmholtz problems. Commun. Math.
Sci., 5(3):665 – 678, 2007.
[17] R. Hiptmair, A. Moiola, and I. Perugia. Stability results for the time-harmonic maxwell
equations with impedance boundary conditions. Mathematical Models and Methods in
Applied Sciences (M3AS), 21(11):2263–2287, 2011.
[18] F. Ihlenburg and I. Babuska. Finite element solution of the Helmholtz equation with
high wave number part I: The h-Version of the FEM. Computers Math. Applic., 30(9):9
– 37, 1995.
[19] F. Ihlenburg and I. Babuska. Finite element solution of the helmholtz equation with
high wave number part II: The h-p Version of the FEM. SIAM J. Numer. Anal.,
34(1):315–358, 1997.
[20] J. M. Jin and J. Volakis. A finite-element-boundary integral formulation for scattering by three-dimensional cavity-backed apertures. IEEE Trans. Antennas Propagat.,
39(1):97–104, 1991.
[21] C. S. Lee and S. W. Lee. RCS of a coated circular waveguide terminated by a perfect
conductor. IEEE Trans. Antennas Propagat., 35(4):391–398, 1987.
[22] P. J. Li, H. J. Wu, and W. Y. Zheng. An overfilled cavity problem for maxwell’s
equations. Mathematical Methods in the Applied Sciences, 35(16):1951–1979, 2012.
[23] H. Ling, R. C. Chou, and S. W. Lee. Shooting and bouncing rays: calculating the rcs
of an arbitrarily shaped cavity. IEEE Trans. Antennas Propagat., 37(2):194–205, 1989.
[24] H. Ling, S. W. Lee, and R. C. Chou. High-frequency RCS of open cavities with rectangular and circular cross sections. IEEE Trans. Antennas Propagat., 37(5):648–654,
1989.
96

[25] J. Melenk. On Generalized Finite Element Methods. PhD thesis, The University of
Maryland, 1995.
[26] J. W. Moll and R. G. Seecamp. Calculation of radar reflecting properties of jet engine
intakes using a waveguide model. IEEE Trans. Aerospace Electron. Syst., AES-6(5):675–
683, 1970.
[27] F. Obelleiro, J. L. Rodriguez, and A. G. Pino. A progressive physical optics (PPO)
method for computing the electromagnetic scattering of large open-ended cavities. Microwave and optical technology letters, 14(3):166–169, 1997.
[28] F. Obelleiro-Basteiro, J. L. Rodr´
iuez, and R. J. Burkholder. An iterative physical optics
approach for analyzing the electromagnetic scattering by large open-ended cavities.
IEEE Trans. Antennas Propagat., 43(4):356–361, 1995.
[29] P. H. Pathak and R. J. Burkholder. High-frequency electromagnetic scattering by openended waveguide cavities. Radio Science, 26(1):211–218, 1991.
[30] H. Peot. An electronic umbrella: the B1B defensive avionics systems. Defense Syst.
Rev., 2:15–18, 1984.
[31] F. Seydou, R. Duraiswami, N. A. Gumerov, and T. Sepp¨nen. TM Electromagnetic
a
Scattering from 2D Multilayered Dielectric Bodies - Numerical Solution. Applied Computational Electromagnetics Society Journal, 19(2):100–107, 2004.
[32] T. Van and A. W. Wood. Analysis of transient electromagnetic scattering from overfilled
cavities. SIAM J. Appl. Math., 64(2):688–708, 2003.
[33] Y. Wang, K. Du, and W. Sun. A second-order method for the electromagnetic scattering
from a large cavity. Numer. Math. Theor. Meth. Appl., 1(4):357–382, 2008.

97