ANALYSIS OF TRANSIENT - Nanyang Technological University

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ANALYSIS OF TRANSIENT ELECTROMAGENTIC SCATTERING
FROM A THREE-DIMENSIONAL OPEN CAVITY
PEIJUN LI∗ , LI-LIAN WANG† , AND AIHUA WOOD‡
Abstract. This paper is concerned with the mathematical analysis of the time-domain Maxwell
equations in a three-dimensional open cavity. An exact transparent boundary condition (TBC) is
developed to reformulate the open cavity scattering problem in an unbounded domain equivalently
into an initial-boundary value problem in a bounded domain. The well-posedness and stability are
studied for the reduced problem. Moreover, an a priori estimate is established for the electric field
with a minimum regularity requirement for the data.
Key words. Time-domain Maxwell equations, transparent boundary conditions, three-dimensional
open cavity scattering problem, well-posedness, stability, a priori estimates
AMS subject classifications. 35Q61, 78A25, 78M30
1. Introduction. This paper is concerned with the mathematical analysis of
an electromagnetic open cavity scattering problem where the wave propagation is
governed by the time-domain Maxwell equations. As seen in Fig. 2.1, an open cavity
refers to as a compactly supported domain with its opening aligned with the infinite
ground plane. The cavity scattering problems have significant industry and military
applications such as the design of cavity-backed conformal antennas and the deliberate
control in the form of enhancement or reduction of radar cross section.
The time-harmonic problems have been widely investigated by numerous researchers in the engineering and mathematical communities [2–4,7,9,10,18,25,27,33].
A large amount of information is available regarding their solutions for both the twodimensional Helmholtz and the three-dimensional Maxwell equations [7, 8, 30, 32, 44].
We refer to [26] for a good introduction to the cavity scattering problem. One may
consult [16,17,34,35] for recent accounts of finite element methods and integral equation methods for general electromagnetic scattering problems.
The time-domain problems have attracted much attention due to their capability
of capturing wide-band signals and modeling more general material and nonlinearity
[11, 24, 29, 42, 43, 45]. However, rigorous mathematical analysis is very rare, especially
for time-domain three-dimensional Maxwell equations. The transient cavity scattering
problems have been mathematically studied in [22, 23, 37, 39, 40], where the focus was
largely on (i) temporal discretization by the Newmark scheme; (ii) the reduction of
the resulted system via frequency-domain TBC, and (iii) the analysis of the finiteelement method for the reduced problem. To the best of our knowledge, the theoretical
analysis of the time-domain Maxwell cavity problem itself was left undone and is still
lacking.
In this work, we aim to analyze the transient electromagnetic wave scattering by
an open cavity which is embedded in a perfectly conducting infinite ground plane. The
∗ Department of Mathematics,
Purdue University, West Lafayette, IN 47907, USA
(lipeijun@math.purdue.edu). The research was supported in part by the NSF grant DMS-1151308.
† Division of Mathematical Sciences, School of Physical and Mathematical Sciences, Nanyang Technological University, 637371, Singapore (lilian@ntu.edu.sg). The research was partially supported
by Singapore MOE AcRF Tier 1 Grant (RG 15/12), MOE AcRF Tier 2 Grant (MOE 2013-T2-1-095,
ARC 44/13), and Singapore A∗ STAR-SERC-PSF Grant (122-PSF-007).
‡ Department of Mathematics and Statistics, Air Force Institute of Technology, WPAFB, OH
45433, USA (aihua.wood@afit.edu). The research was supported in part by AFOSR Grants
F1ATA02059J001 and F1ATA01103J001.
1
2
P. Li, L. Wang, and A. Wood
problem geometry is shown in Fig. 2.1. The cavity is filled with an inhomogeneous,
isotropic, and non-dispersive medium which is allowed to protrude from its opening
to above the ground plane, while the upper half space outside of the cavity is filled
with a homogeneous, isotropic, and non-dispersive medium. Therefore, the analysis
in this work includes both planar and overfilled cavities. This problem is quite challenging due to the unbounded nature of the domain. In the frequency domain, various
approaches have been investigated such as developing TBCs and designing perfectly
matched layer techniques [1, 5, 12, 14, 20, 21, 28, 31, 32]. In contrast, much less study
is devoted to address this issue in the time domain. In [13], the well-posedness and
stability were shown for the electromagnetic obstacle scattering problem by using a
time-domain TBC on the sphere. Here, we present a new time-domain TBC on a
hemisphere enclosing the cavity, and prove well-posedness and stability of the underlying scattering problem. The proofs are based on examining the well-posedness of
the time-harmonic Maxwell equations with complex wavenumbers and applying the
abstract inversion theorem of the Laplace transform. Moreover, an a priori estimate,
featured with an explicit dependence on time and a minimum regularity requirement
of the data, is established for the electric field by studying directly the time-domain
Maxwell equations.
The outline of this paper is as follows. In section 2, we introduce the model
problem and exploit the time-domain TBC to reduce the scattering problem to an
initial-boundary value problem in a bounded domain. In section 3, we analyze two
auxiliary problems pertinent to the reduced time-domain Maxwell equations to pave
the way for the analysis of the main results in section 4. We study in section 4 the
well-posedness and stability of the reduced Maxwell equations, and derive an a prior
estimate with a minimum requirement for the regularity of the data. The paper is
concluded with some remarks and directions for future work in section 5. To avoid
distraction from the main results, we collect in the appendices some necessary notation and useful results on the Laplace transform, spherical harmonics, and functional
spaces.
2. Formulation and reduction of the problem. In this section, we introduce the mathematical model of interest, and exploit the exact TBC to reduce the
unbounded domain to a bounded one.
2.1. A model problem. We describe the setting of the cavity problem and
define some necessary notations. As seen in Fig. 2.1, denote by D the cavity embedded
in the perfectly electrically conducting infinite ground plane Γg . Let S = ∂D∩R3− , the
part of cavity wall below the ground, be Lipschitz continuous and perfectly electrically
conducting. The medium in the cavity is characterized by the dielectric permittivity
ε and the magnetic permeability µ, which satisfy
0 < εmin ≤ ε ≤ εmax < ∞,
0 < µmin ≤ µ ≤ µmax < ∞.
+
Here εmin , εmax , µmin , and µmax are constants. Let BR
and Γ+
R be the half-ball and
hemisphere above the ground plane, where the radius R is large enough to completely
+
contain the possibly overfilled portion of the cavity. The unbounded region R3+ \ B̄R
is filled with a homogeneous, isotropic, and non-dispersive medium with a constant
permittivity ε0 and a constant permeability µ0 . Throughout this paper, we assume
+
for simplicity in exposition that ε0 = µ0 = 1. Finally, we denote by Ω = BR
∪D
the bounded domain in which our reduced initial-boundary value problem will be
formulated. It is easy to note that ∂Ω = Γ+
R ∪ S is Lipschitz continuous.
Transient Electromagnetic Scattering
Ωe
3
eρ
Γ+
R
BR+
Γg
Γg
n
n
D
S
n
Fig. 2.1. A schematic diagram of the open cavity problem geometry.
As is shown in Fig. 2.1, it is evident that the problem geometry is not only
applicable to the open cavity problem but also to a broader class of scattering problems
where the surface S or a part of it may be above the ground plane.
Consider the system of time-domain Maxwell equations in R3+ ∪ D for t > 0:
(2.1)
(
∇ × E(r, t) + µ∂t H(r, t) = 0,
∇ × H(r, t) − ε∂t E(r, t) = J (r, t),
where r = (x, y, z) ∈ R3 , E is the electric field, H is the magnetic field, and J is
the electric current density which is assumed to be compactly supported in D. The
system is constrained by the initial conditions:
(2.2)
E |t=0 = E 0 ,
H |t=0 = H 0 in R3+ ∪ D,
where E 0 and H 0 are also assumed to be compactly supported in D. We consider
the perfectly electrical conducting boundary condition on the ground plane and cavity
wall:
(2.3)
n × E = 0 on Γg ∪ S, t > 0,
where n is the unit outward normal vector on Γg ∪ S. In addition, we impose the
Silver-Müller radiation condition:
(2.4)
r̂ × (∂t E × r̂) + r̂ × ∂t H = o(|r|−1 ),
as |r| → ∞, t > 0,
where r̂ = r/|r|.
The purpose of this paper is to study the well-posedness and establish the stability
for the time-domain electromagnetic cavity scattering problem (2.1)–(2.4). Hereafter,
the expression “a . b” stands for a ≤ Cb, where C is a generic positive constant
independent of any function and important parameters, which are clear from the
context.
2.2. Transparent boundary condition. We introduce a TBC to reformulate
the electromagnetic wave propagation problem into an equivalent initial-boundary
value problem in a bounded domain. The essential idea is to design a boundary
operator which maps the tangential component of the electric field to the tangential
trace of the magnetic field.
4
P. Li, L. Wang, and A. Wood
More precisely, we consider the reduced initial-boundary value problem:


∇ × E + µ∂t H = 0, ∇ × H − ε∂t E = J
in Ω, t > 0,




E |t=0 = E 0 , H |t=0 = H 0
in Ω,
(2.5)

n×E =0
on S, t > 0,




T [E + ] = H × n
on Γ+
R , t > 0,
Γ
R
where E Γ+ is the tangential trace of E on Γ+
R , and T is the time-domain electric-toR
magnetic (EtM) Calderon operator, as the counterpart of the time-harmonic setting,
for instance, with a spherical boundary (cf. [17, 34]).
In what follows, we derive the formulation of the operator T and analyze its
important properties. Equivalently, we aim to prove the well-posedness and stability
of the reduced problem (2.5). In particular, an a priori estimate is established with
a minimum requirement for the regularity of the data. The proofs are based on the
abstract inversion theorem of the Laplace transform and the a priori estimates for
the time-harmonic Maxwell equations with a complex wavenumber. These will be the
main topics of the forthcoming sections.
Since J is supported in D and ε0 = µ0 = 1, the Maxwell equations (2.1) become
(2.6)
∇ × E + ∂t H = 0,
+
∇ × H − ∂t E = 0 in Ωe := R3+ \ B̄R
, t > 0.
Let Ĕ(r, s) = L (E) and H̆(r, s) = L (H) be the Laplace transforms of E(r, t) and
H(r, t) with respect to t, respectively, where the complex variable
√
(2.7)
s = s1 + is2 with s1 , s2 ∈ R, s1 > 0, i = −1 ,
as seen in Appendix A. Recall that
L (∂t E) = sĔ − E 0 ,
L (∂t H) = sH̆ − H 0 .
Taking the Laplace transform of (2.6), and noting that E 0 , H 0 are supported in D,
we obtain the time-harmonic Maxwell equations with complex parameters:
∇ × Ĕ + sH̆ = 0,
(2.8)
∇ × H̆ − sĔ = 0 in Ωe , s1 > 0.
(1)
Let hn (z) be the spherical Hankel function of the first kind of order n (cf. [41]).
We introduce the vector wave functions
( m
(1)
M n (ρ, θ, ϕ) = ∇ × rhn (isρ)Xnm (θ, ϕ) ,
(2.9)
−1
Nm
∇ × Mm
n (θ, ϕ),
n (ρ, θ, ϕ) = −s
where (ρ, θ, ϕ) is the spherical coordinates, and Xnm is the rescaled spherical harmonic
function defined in (B.2). We refer to Appendix B for the properties of spherical
harmonics and the relevant calculus to be used throughout this paper.
The vector wave functions in (2.9) are the radiation solutions of (2.8) in R3 \ {0}
(cf. [34, Thm. 9.16]):
m
(2.10) ∇ × M m
n (ρ, θ, ϕ) + sN n (ρ, θ, ϕ) = 0,
m
∇ × Nm
n (ρ, θ, ϕ) − sM n (ρ, θ, ϕ) = 0.
It can be verified from (2.9) that the vector wave functions satisfy
(2.11)
(1)
m
Mm
n (ρ, θ, ϕ) = hn (isρ)∇Γ Xn (θ, ϕ) × eρ
5
Transient Electromagnetic Scattering
and
Nm
n (ρ, θ, ϕ)
(2.12)
p
n(n + 1) (1)
m
0
=−
hn (isρ) + isρ(h(1)
n ) (isρ) X n (θ, ϕ)
sρ
n(n + 1) (1)
−
hn (isρ)Ynm (θ, ϕ)eρ .
sρ
Once again, we refer to Appendix B for the notation and definition. A simple calculation yields
(
eθ × M m
for |m| ≤ n, m + n = even, n ∈ N,
n (ρ, π/2, ϕ) = 0
eθ × N m
n (ρ, π/2, ϕ) = 0
for |m| ≤ n, m + n = odd, n ∈ N.
Therefore, in the domain Ωe , the solution of the electric field Ĕ(ρ, θ, ϕ, s), which
satisfies the perfectly electric conducting condition n × Ĕ = 0 on Γg , i.e., eθ ×
Ĕ(ρ, π/2, ϕ, s) = 0, can be written in the series:
(2.13)
Ĕ(ρ, θ, ϕ, s) =
odd
X
αnm (s) N m
n (ρ, θ, ϕ) +
|m|≤n
even
X
βnm (s) M m
n (ρ, θ, ϕ),
|m|≤n
which is uniformly convergent on compact subsets in Ωe . The corresponding magnetic
field H̆ is given by
(2.14)
H̆ = −s−1 ∇ × Ĕ = −
odd
X
αnm (s) M m
n (ρ, θ, ϕ) +
even
X
βnm (s) N m
n (ρ, θ, ϕ).
|m|≤n
|m|≤n
Note that in the above, we compressed the summation notation as in Appendix B.
To deduce the explicit representation of the EtM Calderon operator, we need to
m
express Ĕ Γ+ = −eρ × (eρ × Ĕ) and H̆ × eρ on Γ+
R in terms of the coefficients αn and
R
m
βn . From the definition (2.11), one verifies that
p
m
(1)
−eρ × (eρ × M m
n (ρ, θ, ϕ)) = − n(n + 1) hn (isρ)Y n (θ, ϕ),
p
n(n + 1) (1)
(1) 0
−eρ × (eρ × N m
(ρ,
θ,
ϕ))
=
−
h
(isρ)
+
isρ
h
(isρ)
Xm
n
n
n
n (θ, ϕ),
sρ
and
p
m
n(n + 1) h(1)
n (isρ)X n (θ, ϕ),
p
n(n + 1) (1)
0
m
eρ × N m
hn (isρ) + isρ h(1)
n (ρ, θ, ϕ) = −
n (isρ) Y n (θ, ϕ).
sρ
eρ × M m
n (ρ, θ, ϕ) =
Therefore, by (2.13), the tangential component of the electric field along Γ+
R is
Ĕ Γ+ = −
R
odd
X
|m|≤n
p
n(n + 1) (1)
0
m
m
hn (isR) + isR h(1)
n (isR) αn (s) X n (θ, ϕ)
sR
+
even
X
|m|≤n
p
m
m
n(n + 1) h(1)
n (isR)βn (s) Y n (θ, ϕ),
6
P. Li, L. Wang, and A. Wood
and similarly, by (2.14), the tangential trace of the magnetic field along Γ+
R is
odd p
X
H̆ × eρ =
|m|≤n
+
even
X
|m|≤n
m
m
n(n + 1) h(1)
n (isR)αn (s) X n (θ, ϕ)
p
n(n + 1) (1)
0
m
m
hn (isR) + isRh(1)
n (isR) βn (s) Y n (θ, ϕ).
sR
Consequently, we have the following explicit representation of the frequency domain EtM Calderon operator B: given any tangential component of the electric field
along Γ+
R with the expansion:
(2.15)
u=
odd
X
even
X
αnm X m
n +
|m|≤n
βnm Y m
n,
|m|≤n
the tangential trace of the magnetic field on Γ+
R is
B[u] = −
(2.16)
odd
X
sR
|m|≤n
1 + rn (isR)
(1)
even
X
αnm X m
n −
|m|≤n
(1)
(1 + rn (isR)) m m
βn Y n ,
sR
where
(1) 0
rn(1) (z)
(2.17)
=
z hn (z)
(1)
.
hn (z)
We now analyze some properties of the EtM Calderon operator, and refer to
Appendix D for the definitions of the function spaces to be used hereafter.
−1/2
Lemma 2.1. The Calderon operator B : H −1/2 (curl, Γ+
(div, Γ+
R) → H
R ) is
continuous.
Proof. For any u, v ∈ H −1/2 (curl, Γ+
R ), we can expand
u=
odd
X
even
X
m
um
1n X n +
|m|≤n
m
um
2n Y n ,
v=
|m|≤n
odd
X
m
v1n
Xm
n +
|m|≤n
even
X
m
v2n
Ym
n.
|m|≤n
Then we have from (2.16) that
B[u] = −
odd
X
sR um
1n
|m|≤n
1 + rn (isR)
(1)
even
X
Xm
n −
|m|≤n
(1)
(1 + rn (isR))um
2n
Ym
n.
sR
To prove the lemma, it is required to estimate
(2.18)
hB[u], vi = −
odd
X
sR
|m|≤n
1 + rn (isR)
(1)
um
1n
m
v̄1n
−
even
X
|m|≤n
(1)
(1 + rn (isR)) m m
u2n v̄2n .
sR
It follows from the Cauchy-Schwarz inequality that

1/2
odd p
even
X
X |1 + rn(1) (isR)|2 |um |2
1
+
n(n
+
1)
2n 
2
p
|hB[u], vi| ≤ 
|sR|2 |um
1n | +
(1)
2
|sR|2
1
+
n(n
+
1)
|1
+
r
(isR)|
n
|m|≤n
|m|≤n

1/2
odd
even
X
X
p
1
m 2
m 2
p
×
|v1n
| +
1 + n(n + 1)|v2n
|
.
1
+
n(n
+
1)
|m|≤n
|m|≤n
7
Transient Electromagnetic Scattering
By Lemma C.3, we have
p
1 + n(n + 1)
1 + n(n + 1)
1
2
2
|sR|2 |um
|sR|2 |um
1n | = p
1n |
(1)
2
1 + n(n + 1) |1 + rn(1) (isR)|2
|1 + rn (isR)|
1
2
.p
|um
1n | ,
1 + n(n + 1)
and
2
2
p
|1 + rn (isR)|2 |um
|1 + rn (isR)|2 |um
2n |
2n |
p
=
1
+
n(n
+
1)
1 + n(n + 1) |sR|2
1 + n(n + 1) |sR|2
p
2
. 1 + n(n + 1) |um
2n | .
(1)
(1)
Combining the above estimates and using the expressions of the norms in Appendix
D yields
|hB[u], vi| . kukH −1/2 (curl, Γ+ ) kvkH −1/2 (curl, Γ+ ) ,
R
R
which completes the proof.
Another important property of the EtM Calderon operator is stated as follows.
Lemma 2.2. It holds that
∀ u ∈ H −1/2 (curl, Γ+
R ).
Re hB[u], ui ≥ 0,
(2.19)
Proof. From (2.18), we obtain
(2.20)
odd
X
sR
|m|≤n
1 + rn (isR)
−hB[u], ui =
(1)
By Lemma C.1 and Lemma C.2,
s1 Re 1 + rn(1) (isR) ≤ 0,
2
|um
1n | +
even
X
|m|≤n
(1)
1 + rn (isR) m 2
|u2n | .
sR
s2 Im 1 + rn(1) (isR) ≤ 0.
Taking the real part of (2.20) gives
−Re hB[u], ui =
(1)
(1)
odd
X
R s1 Re(1 + rn (isR)) + s2 Im(1 + rn (isR))
(1)
|m|≤n
+
even
X
|m|≤n
|1 + rn (isR)|2
2
|um
1n |
(1)
(1)
s1 Re(1 + rn (isR)) + s2 Im(1 + rn (isR)) m 2
|u2n | ≤ 0,
|s|2 R
which completes the proof.
With the aid of the frequency domain EtM Calderon operator, we obtain the
following TBC imposed on the hemisphere Γ+
R in the s-domain:
B[Ĕ Γ+ ] = H̆ × eρ ,
(2.21)
R
which maps the tangential component of the electric field to the tangential trace of
the magnetic field. Taking the inverse Laplace transform of (2.21) yields the TBC in
the time-domain:
(2.22)
T [E Γ+ ] = H × eρ ,
R
where T := L −1 ◦ B ◦ L .
8
P. Li, L. Wang, and A. Wood
Equivalently, we may eliminate the magnetic field and obtain an alternative TBC in
the s-domain:
(2.23)
s−1 (∇ × Ĕ) × n + B[Ĕ Γ+ ] = 0 on Γ+
R.
R
Correspondingly, by taking the inverse Laplace transform of (2.23), we may derive an
alternative TBC in the time-domain:
(2.24)
(∇ × E) × n + C [E Γ+ ] = 0
R
where C = L −1 ◦ sB ◦ L .
on Γ+
R,
3. Analysis of two auxiliary problems. In this section, we make necessary
preparations for the proof of the main results by considering two auxiliary problems
pertinent to (2.5).
3.1. Time-harmonic Maxwell equations with a complex wavenumber.
This subsection is devoted to the mathematical study of a time-harmonic Maxwell
scattering problem with a complex wavenumber, which may be viewed as a frequency
version of the initial-boundary value problem of the Maxwell equations under the
Laplace transform.
Consider the auxiliary boundary value problem:

−1

in Ω,
∇ × (sµ) ∇ × u + sεu = j

(3.1)
n×u=0
on S,


 −1
s (∇ × u) × n + B[uΓ ] = 0
on Γ+
R,
where s = s1 + is2 with s1 , s2 ∈ R, s1 > 0, and the applied current density j is
assumed to be supported in D ⊂ Ω.
By multiplying a test function v ∈ H S (curl, Ω), which is defined in Appendix
D, and integrating by parts, we arrive at the variational formulation of (3.1): find
u ∈ H S (curl, Ω) such that
Z
(3.2)
aTH (u, v) =
j · v̄ dr, ∀ v ∈ H S (curl, Ω),
Ω
where the sesquilinear form
Z
Z
(3.3) aTH (u, v) = (sµ)−1 (∇ × u) · (∇ × v̄) dr +
sε u · v̄ dr + hB[uΓ+ ], v Γ+ i.
Ω
R
Ω
R
Theorem 3.1. The variational problem (3.2) has a unique solution u ∈ H S (curl, Ω)
which satisfies
k∇ × ukL2 (Ω) + ksukL2 (Ω) . s−1
1 ksjkL2 (Ω) ,
where s = s1 + is2 with s1 , s2 ∈ R and s1 > 0.
Proof. It suffices to show the coercivity of aTH , since its continuity follows directly from the Cauchy-Schwarz inequality, Lemma 2.1, and Lemma D.1. A simple
calculation yields
Z
Z
(3.4)
aTH (u, u) = (sµ)−1 |∇ × u|2 dr +
sε|u|2 dr + hB[uΓ+ ], uΓ+ i.
Ω
Ω
R
R
9
Transient Electromagnetic Scattering
Taking the real part of (3.4) and using Lemma 2.2, we get
(3.5)
s1
k∇ × uk2L2 (Ω) + ksuk2L2 (Ω) .
2
|s|
Re{aTH (u, u)} &
It follows from the Lax-Milgram lemma that the variational problem (3.2) has a unique
solution u ∈ H S (curl, Ω). Moreover, we have from (3.2) that
(3.6)
|aTH (u, u)| ≤ |s|−1 kjkL2 (Ω) ksukL2 (Ω) .
Combining (3.5)–(3.6) leads to
k∇ × uk2L2 (Ω) + ksuk2L2 (Ω) . s−1
1 ksjkL2 (Ω) ksukL2 (Ω) ,
which completes the proof after applying the Cauchy-Schwarz inequality.
It is noteworthy that Theorem 3.1 gives the stability estimate with an explicit
dependence on the complex wavenumber s corresponding to a lossy medium with
s1 > 0. However, it is challenging to obtain a similar estimate for a lossless medium
with s1 = 0.
3.2. A second auxiliary problem. Before considering the reduced problem
(2.5), we need to study an auxiliary initial-boundary value problem and establish its
well-posedness and stability. Consider the system of time-domain Maxwell equations:


∇ × U + µ∂t V = 0, ∇ × V − ε∂t U = 0
in Ω, t > 0,


(3.7)
n×U =0
on S ∪ Γ+
R , t > 0,



U |t=0 = E 0 , V |t=0 = H 0
in Ω,
where E 0 , H 0 are assumed to be compactly supported in D as before.
Let Ŭ = L (U ) and V̆ = L (V ). Taking the Laplace transform of (3.7), we
obtain the boundary value problem:
(
∇ × (sµ)−1 ∇ × Ŭ + sε Ŭ = j̆
in Ω,
(3.8)
n × Ŭ = 0
on S ∪ Γ+
R,
where the current density function
j̆ = εE 0 + s−1 ∇ × H 0 .
The variational formulation for (3.8) is to find Ŭ ∈ H 0 (curl, Ω) such that
Z
(3.9)
aAP (Ŭ , v) =
j̆ · v̄ dr, ∀ v ∈ H 0 (curl, Ω),
Ω
where the sesquilinear form
Z
Z
(3.10)
aAP (u, v) = (sµ)−1 (∇ × u) · (∇ × v̄) dr +
sεu · v̄ dr.
Ω
Ω
Following the same proof as in Theorem 3.1, we can show the well-posedness of
the variational problem (3.9) and its stability, as stated below.
10
P. Li, L. Wang, and A. Wood
Lemma 3.2. The variational problem (3.9) has a unique solution Ŭ ∈ H 0 (curl, Ω)
which satisfies
−1
k∇ × Ŭ kL2 (Ω) + ksŬ kL2 (Ω) . s−1
1 |s|kE 0 kL2 (Ω) + s1 k∇ × H 0 kL2 (Ω) .
Theorem 3.3. The auxiliary problem (3.7) has a unique solution (U , V ), which
satisfies the stability estimates:
kU kL2 (Ω) + kV kL2 (Ω) . kE 0 kL2 (Ω) + kH 0 kL2 (Ω) ,
k∂t U kL2 (Ω) + k∂t V kL2 (Ω) . k∇ × E 0 kL2 (Ω) + k∇ × H 0 kL2 (Ω) ,
k∂t2 U kL2 (Ω) + k∂t2 V kL2 (Ω) . k∇ × (∇ × E 0 )kL2 (Ω) + k∇ × (∇ × H 0 )kL2 (Ω) .
Proof. Let Ŭ = L (U ) and V̆ = L (V ) as before. Taking the Laplace transform
of (3.7) leads to
(
∇ × Ŭ + sµ V̆ = µ H 0 , ∇ × V̆ − sε Ŭ = −ε E 0
in Ω,
(3.11)
n × Ŭ = 0
on S ∩ Γ+
R.
It follows from Lemma 3.2 that
−1
k∇ × Ŭ kL2 (Ω) + ksŬ kL2 (Ω) . s−1
1 |s|kE 0 kL2 (Ω) + s1 k∇ × H 0 kL2 (Ω) .
By (3.11), we have
−1
k∇× V̆ kL2 (Ω) +ksV̆ kL2 (Ω) . 1+s−1
1 |s| kE 0 kL2 (Ω) +kH 0 kL2 (Ω) +s1 k∇×H 0 kL2 (Ω) .
It follows from Lemma A.2 that Ŭ and V̆ are holomorphic functions of s, and the
inverse Laplace transform of Ŭ and V̆ exist and are supported in [0, ∞).
We next prove the stability. Define the energy function
e1 (t) = kε1/2 U (·, t)k2L2 (Ω) + kµ1/2 V (·, t)k2L2 (Ω) .
It follows from (3.7) and integration by parts that
Z
Z tZ
e01 (τ ) dτ = 2Re
ε∂t U · Ū + µ∂t V · V̄ drdτ
0
0
Ω
Z tZ
= 2Re
(∇ × V ) · Ū − (∇ × U ) · V̄ drdτ
0
Ω
Z tZ
= 2Re
(∇ × Ū ) · V − (∇ × U ) · V̄ drdτ = 0.
t
e1 (t) − e1 (0) =
0
Ω
Hence we have
kε1/2 U (·, t)k2L2 (Ω) + kµ1/2 V (·, t)k2L2 (Ω) = kε1/2 E 0 k2L2 (Ω) + kµ1/2 H 0 k2L2 (Ω) ,
which implies
kU kL2 (Ω) + kV kL2 (Ω) . kE 0 kL2 (Ω) + kH 0 kL2 (Ω) .
11
Transient Electromagnetic Scattering
Taking the first and second partial derivatives of (3.7) with respect to t yields


∇ × ∂t U + µ∂t2 V = 0, ∇ × ∂t V − ε∂t2 U = 0
in Ω, t > 0,


n × ∂t U = 0
on S ∪ Γ+
R , t > 0,



−1
−1
∂t U |t=0 = ε (∇ × H 0 ), ∂t V |t=0 = −µ (∇ × E 0 )
in Ω,
and

2
3
2
3

∇ × ∂t U + µ∂t V = 0, ∇ × ∂t V − ε∂t U = 0



n × ∂ 2 U = 0
t

∂t2 U |t=0 = −(εµ)−1 (∇ × (∇ × E 0 ))




∂ 2 V |
−1
(∇ × (∇ × H 0 ))
t=0 = −(εµ)
t
in Ω, t > 0,
on S ∪ Γ+
R , t > 0,
in Ω,
in Ω.
Considering the energy functions
e2 (t) = kε1/2 ∂t U (·, t)k2L2 (Ω) + kµ1/2 ∂t V (·, t)k2L2 (Ω) ,
and
e3 (t) = kε1/2 ∂t2 U (·, t)k2L2 (Ω) + kµ1/2 ∂t2 V (·, t)k2L2 (Ω)
for the above two problems, respectively, we can follow the same steps for proving the
first inequality to derive the other two inequalities.
4. The reduced Maxwell equations. In this section, we derive the main results of this work, which include the well-posedness of the reduced problem (2.5) and
the related a priori estimates.
4.1. Well-posedness. Let e = E − U and h = H − V . It follows from (2.5)
and (3.7) that e and h satisfy the system:


∇ × e + µ∂t h = 0, ∇ × h − ε∂t e = J
in Ω, t > 0,




n × e = 0
on S, t > 0,
(4.1)

e|t=0 = 0, h|t=0 = 0
in Ω,




T e + = h × n + V × n
on Γ+
R , t > 0.
Γ
R
Let ĕ = L (e) and h̆ = L (h). Taking the Laplace transform of (4.1) and eliminating
h̆, we obtain

∇ × ((sµ)−1 )∇ × ĕ) + sεĕ = −J̆
in Ω,



n × ĕ = 0
on S,
(4.2)


s−1 (∇ × ĕ) × n + Bĕ + = V̆ × n
on Γ+ .
ΓR
R
Lemma 4.1. The problem (4.2) has a unique weak solution ĕ ∈ H S (curl, Ω)
which satisfies
ksJ̆ kL2 (Ω) + ksV̆ × nkH −1/2 (div, Γ+ )
k∇ × ĕk2L2 (Ω) + ksĕk2L2 (Ω) . s−1
1
R
2
+k|s| V̆ × nkH −1/2 (div, Γ+ ) .
R
12
P. Li, L. Wang, and A. Wood
Proof. The well-posedness of the solution ĕ ∈ H S (curl, Ω) follows directly from
Theorem 3.1. Moreover, we have
Z
J̆ · ¯ĕ dr.
aTH (ĕ, ĕ) = hV̆ × n, ĕΓ+ i −
R
Ω
By the coercivity of aTH in (3.5) and the trace theorem in Lemma D.1,
s1 2
2
−1
+
ksĕk
J̆ kL2 (Ω) ksĕkL2 (Ω)
k∇
×
ĕk
2
2
L (Ω)
L (Ω) . ks
|s|2
+ kV̆ × nkH −1/2 (div, Γ+ ) kĕkH(curl, Ω)
R
−1
. ks
J̆ kL2 (Ω) ksĕkL2 (Ω) + kV̆ × nkH −1/2 (div, Γ+ ) k∇ × ĕkL2 (Ω)
R
+ ks−1 V̆ × nkH −1/2 (div, Γ+ ) ksĕkL2 (Ω) ,
R
which completes the proof.
To show the well-posedness of the reduced problem (2.5), we make the following
assumptions for the initial and boundary data:
(4.3)
E 0 , H 0 ∈ H(curl, Ω),
J ∈ H 1 (0, T ; L2 (Ω)),
J |t=0 = 0.
Theorem 4.2. The problem has a unique solution (E, H) which satisfies
E ∈ L2 (0, T ; H S (curl, Ω)) ∩ H 1 (0, T ; L2 (Ω)),
H ∈ L2 (0, T ; H(curl, Ω)) ∩ H 1 (0, T ; L2 (Ω)),
and the stability estimate:
max k∂t EkL2 (Ω) + k∇ × EkL2 (Ω) + k∂t HkL2 (Ω) + k∇ × HkL2 (Ω)
t∈[0,T ]
. kE 0 kH(curl,Ω) + kH 0 kH(curl,Ω) + kJ kH 1 (0,T ;L2 (Ω)) .
Proof. Recall the decomposition E = U + e and H = V + h where (U , V ) satisfy
(3.7) and (e, h) satisfy (4.1). Since
Z
T
0
Z
T
≤
0
k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt
e−2s1 (t−T ) k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt
Z
T
2s1 T
=e
Z
.
∞
0
−2s1 t
e
0
e−2s1 t k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt
k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt,
it suffices to estimate the integral
Z ∞
e−2s1 t k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt.
0
13
Transient Electromagnetic Scattering
Taking the Laplace transform of (4.1) yields


∇ × ĕ + sµh̆ = 0, ∇ × h̆ − sεĕ = J̆

n × ĕ = 0
(4.4)


Bĕ + = h̆ × n + V̆ × n
ΓR
in Ω,
on S,
on Γ+
R.
By Lemma 4.1,
(4.5)
ksJ̆ kL2 (Ω) + ksV̆ × nkH −1/2 (div, Γ+ )
k∇ × ĕk2L2 (Ω) + ksĕk2L2 (Ω) . s−1
1
R
2
+k|s| V̆ × nkH −1/2 (div, Γ+ ) .
R
By (4.4),
k∇ × h̆k2L2 (Ω) + ksh̆k2L2 (Ω) . s−1
kJ̆ kL2 (Ω) + ksJ̆ kL2 (Ω) + ksV̆ × nkH −1/2 (div, Γ+ )
1
R
(4.6)
+k|s|2 V̆ × nkH −1/2 (div, Γ+ ) .
R
It follows from [36, Lemma 44.1] that (ĕ, h̆) are holomorphic functions of s on the
half plane s1 > γ > 0, where γ is any positive constant. Hence we have from Lemma
A.2 that the inverse Laplace transform of ĕ and h̆ exist and are supported in [0, ∞].
Denote by e = L −1 (ĕ) and h = L −1 (h̆). Since ĕ = L (e) = F (e−s1 t e), where
F is the Fourier transform in s2 , we have from the Parseval identity and (4.5) that
Z ∞
Z ∞
e−2s1 t k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt = 2π
k∇ × ĕk2L2 (Ω) + ksĕk2L2 (Ω) ds2
0
−∞
Z ∞
Z ∞
−2
−2
2
. s1
ksJ̆ kL2 (Ω) ds2 + s1
ksV̆ × nk2H −1/2 (div, Γ+ )
R
−∞
−∞
+ k|s|2 V̆ × nk2H −1/2 (div, Γ+ ) ds2 .
R
Since J |t=0 = 0 in Ω, V × n|t=0 = ∂t (V × n)|t=0 = 0 on
we have L (∂t J) = sJ̆
+
in Ω and L (∂t (V × n)) = sV̆ × n on ΓR . It is easy to note that
Γ+
R,
|s|2 V̆ × n = (2s1 − s)sV̆ × n = 2s1 L (∂t (V × n)) − L (∂t2 (V × n)) on Γ+
R.
Hence we have
Z ∞
e−2s1 t k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt
0
Z ∞
−2
. s1
kL (∂t J )k2L2 (Ω) + kL (∂t2 V × n)k2H −1/2 (div, Γ+ ) ds2
R
−∞
Z ∞
+ (1 + s−2
kL (∂t V × n)k2H −1/2 (div, Γ+ ) ds2 .
1 )
−∞
R
Using the Parseval identity again gives
Z ∞
e−2s1 t k∇ × ek2L2 (Ω) + k∂t ek2L2 (Ω) dt
0
Z ∞
2
2
2
−2s1 t
dt
k∂
J
k
+
k∂
V
×
nk
e
. s−2
2
+
t
−1/2
t
1
L (Ω)
H
(div, ΓR )
0
Z ∞
e−2s1 t k∂t V × nk2H −1/2 (div, Γ+ ) dt,
+ (1 + s−2
1 )
0
R
14
P. Li, L. Wang, and A. Wood
which shows that
e ∈ L2 (0, T ; H S (curl, Ω)) ∩ H 1 (0, T ; L2 (Ω)).
Similarly, we can show from (4.6) that
h ∈ L2 (0, T ; H(curl, Ω)) ∩ H 1 (0, T ; L2 (Ω)).
We next prove the stability. Let Ẽ be the extension of E with respect to t in R
such that Ẽ = 0 outside the interval [0, t]. By Lemma A.1 and Lemma 2.2, we get
Z
t
Re
−2s1 t
Z
Z
e
Γ+
R
0
T [E Γ+ ] · Ē Γ+ dγdt = Re
R
R
=
1
2π
Z
∞
Z
Γ+
R
−∞
∞
0
¯ + dtdγ
e−2s1 t T [Ẽ Γ+ ] · Ẽ
Γ
R
R
˘ + ](s), Ẽ
˘ + (s)i ds ≥ 0,
RehB[Ẽ
2
Γ
Γ
R
R
which yields after taking s1 → 0 that
Z tZ
(4.7)
Re
0
Γ+
R
T [E Γ+ ] · Ē Γ+ dγdt ≥ 0.
R
R
For any 0 < t < T , consider an energy function
e(t) = kε1/2 E(·, t)k2L2 (Ω) + kµ1/2 H(·, t)k2L2 (Ω) .
It is easy to note that
Z
t
0
(4.8)
u0 (t)dt = kε1/2 E(·, t)k2L2 (Ω) + kµ1/2 H(·, t)k2L2 (Ω)
− kε1/2 E 0 k2L2 (Ω) + kµ1/2 H 0 k2L2 (Ω) .
On the other hand, we have from (2.5) and (4.7) that
Z
t
e0 (t)dt = 2Re
0
Z tZ
0
Z tZ
ε∂t E · Ē + µ∂t H · H̄ drdt
Ω
Z tZ
(∇ × H) · Ē − (∇ × E) · H̄ drdt − 2Re
J · Ē drdt
0
Ω
0
Ω
Z tZ
Z tZ
+
+
= −2Re
T [E Γ ] · Ē Γ dγdt − 2Re
J · Ē drdt
R
R
0
Γ
0
Ω
Z tZ
≤ −2Re
J · Ē drdt ≤ 2 max kE(·, t)kL2 (Ω) kJ kL1 (0,T ;L2 (Ω)) .
= 2Re
(4.9)
0
Ω
t∈[0,T ]
Taking the derivative of (2.5) with respect to t, we know that (∂t E, ∂t H) satisfy
the same set of equations with the source and the initial condition replaced by ∂t J ,
∂t E|t=0 = ε−1 ∇ × E 0 , ∂t H|t=0 = −µ−1 ∇ × H 0 . Hence we may follow the same steps
as above to obtain (4.9) for (∂t E, ∂t H), which completes the proof after combining
the above estimates.
15
Transient Electromagnetic Scattering
4.2. An a priori estimate. In what follows, we shall take a different route and
study the Maxwell equations directly in the time domain. The goal is to derive an a
priori stability estimate for the electric field with a minimum regularity requirement
for the data and an explicit dependence on the time variable.
Eliminating the magnetic field in (2.1) and using the transparent boundary condition (2.24), we consider the initial-boundary value problem in a bounded domain:


ε∂t2 E = −∇ × (µ−1 ∇ × E) − F
in Ω, t > 0,




E|t=0 = E 0 , ∂t E|t=0 = E 1
in Ω,
(4.10)

n×E =0
on S, t > 0,




(∇ × E) × n + C [E + ] = 0
on Γ+
R , t > 0,
Γ
R
where
E 1 = ε−1 (∇ × H 0 − J 0 ).
F = ∂t J ,
The variational problem is to find E ∈ H S (curl, Ω) for all t > 0 such that
Z
Z
ε∂t2 E · w̄ dr = −
µ−1 (∇ × E) · (∇ × w̄) dr
Ω
Ω
Z
(4.11)
F · w̄ dr for all w ∈ H S (curl, Ω).
− hC [E Γ+ ], wΓ+ i −
R
R
Ω
To show the stability of its solution, we follow the argument in [36] but with a
careful study of the TBC. The following lemma is useful for the subsequent analysis.
Lemma 4.3. Given ξ ≥ 0 and E ∈ L2 (0, ξ; H −1/2 (curl, Γ)), it holds that
Z ξ Z Z t
C [E Γ+ ](τ )dτ · Ē Γ+ (t) dγdt ≥ 0.
Re
Γ+
R
0
R
R
0
Proof. Let Ẽ be the extension of E with respect to t in R such that Ẽ = 0 outside
the interval [0, ξ]. We obtain from (A.1), Lemma A.2, and Lemma 2.2 that
Z t
Z Z ξ
−2s1 t
e
C [E Γ+ ](τ )dτ · Ē Γ+ (t) dtdγ
Γ+
R
0
Z
Z
=
Z
Γ+
R
−2s1 t
Z
1
2π
1
2π
Z
Z
0
∞
−∞
∞
C [Ẽ Γ+ ](τ )dτ
R
−∞
t
−1
Γ+
R
¯ + (t) dtdγ
· Ẽ
Γ
R
◦ sB ◦ L Ẽ Γ+ (τ )dτ
R
0
Z
t
L
e
Γ+
R
=
Z
0
∞
=
=
e−2s1 t
0
Z
R
R
0
∞
¯ + (t) dtdγ
· Ẽ
Γ
R
¯ + )(s) dγds
B ◦ L Ẽ Γ+ (s) · L (Ẽ
2
Γ
R
R
˘ + ], Ẽ
˘ + ds ≥ 0.
B[Ẽ
2
Γ
Γ
R
R
The proof is completed by taking s1 → 0 in the above inequality.
Theorem 4.4. Let E ∈ H S (curl, Ω) be the solution of (4.11). Given E 0 , E 1 ∈
L2 (Ω) and F ∈ L1 (0, T ; L2 (Ω)), for any T > 0, it holds that
(4.12)
kEkL∞ (0,T ;L2 (Ω)) . kE 0 kL2 (Ω) + T kE 1 kL2 (Ω) + T kF kL1 (0,T ;L2 (Ω))
16
P. Li, L. Wang, and A. Wood
and
(4.13) kEkL2 (0,T ;L2 (Ω)) . T 1/2 kE 0 kL2 (Ω) + T 3/2 kE 1 kL2 (Ω) + T 3/2 kF kL1 (0,T ;L2 (Ω)) .
Proof. Let 0 < ξ < T and define an auxiliary function
Z ξ
(4.14)
ψ(r, t) =
E(r, τ ) dτ, r ∈ Ω, 0 ≤ t ≤ ξ.
t
It is clear that
(4.15)
∂t ψ(r, t) = −E(r, t).
ψ(r, ξ) = 0,
For any φ(r, t) ∈ L2 (0, ξ; L2 (Ω)), we have
Z ξ
Z ξ Z t
(4.16)
φ(r, t) · ψ̄(r, t)dt =
φ(r, τ )dτ · Ē(r, t)dt.
0
0
0
Indeed, using integration by parts and (4.15), we have
!
Z ξ
Z ξ
Z ξ
φ(r, t) · ψ̄(r, t)dt =
φ(r, t) ·
Ē(r, τ )dτ dt
0
Z
ξ
0
Z
t
Z
ξ
Ē(r, τ )dτ · d
=
0
Z tZ
t
Z
t
φ(r, τ )dτ ·
=
0
0
Z ξ Z
=
0
ξ
ξ Z
Ē(r, τ )dτ +
0
t
φ(r, τ )dτ
0
φ(r, τ )dτ
t
t
Z
ξ
t
φ(r, τ )dτ
0
· Ē(r, t) dt
0
· Ē(r, t) dt.
0
Next, we take the test function w = ψ in (4.11) and get
Z
Z
µ−1 (∇ × E) · (∇ × ψ̄) dr
ε∂t2 E · ψ̄ dr = −
Ω
Z
ZΩ
(4.17)
F · ψ̄ dr.
C [E Γ+ ] · ψ̄ Γ+ dγ −
−
R
R
Γ+
R
Ω
It follows from (4.15) that
Z ξZ
Z Z ξ
Re
∂t2 E · ψ̄ drdt = Re
∂t (∂t E · ψ̄) + ∂t E · Ē dtdr
0
Ω
Ω 0
Z ξ ξ 1
= Re
(∂t E · ψ̄) + |E|2 dr
2
0
0
Ω
Z
1
1
(4.18)
= kE(·, ξ)k2L2 (Ω) − kE 0 k2L2 (Ω) − Re
E 1 (r) · ψ̄(r, 0) dr.
2
2
Ω
Integrating (4.17) from t = 0 to t = ξ and taking the real parts yields
Z
Z ξ
2
ε
1
ε
2
2
−1 kE(·, ξ)kL2 (Ω) − kE 0 kL2 (Ω) +
µ ∇ × E(r, t)dt dr
2
2
2 Ω
0
Z
Z ξZ
= εRe
E 1 (r) · ψ̄(r, 0) dr − Re
F · ψ̄ dr
Ω
(4.19)
Z
− Re
0
0
ξ
Z
Ω
Γ+
R
C [E Γ+ ] · ψ̄ Γ+ dγdt.
R
R
17
Transient Electromagnetic Scattering
In what follows, we estimate the three terms on the right hand side of (4.19) separately.
We derive from the Cauchy-Schwarz inequality that
Z
Z
E 1 (r) · ψ̄(r, 0) dr = Re
Re
Ω
(4.20)
Z
Ω
Z
ξ
E 1 (r) ·
Z
ξ
Ē(r, t) dt dr
0
Z
ξ
E 1 (r) · Ē(r, t) drdt ≤ kE 1 kL2 (Ω)
= Re
0
Ω
kE(·, t)kL2 (Ω) dt.
0
For 0 ≤ t ≤ ξ ≤ T , we have from (4.16) that
Z
ξ
Z Z
Z
ξ
Z
Ω
0
Z
ξ
Z tZ
ξ
0
t
Z
≤
0
ξ
≤
· Ē(r, t) dtdr
0
Ω
kF (·, τ )kL2 (Ω) dτ kE(·, t)kL2 (Ω) dt
0
!
Z ξ
kF (·, t)kL2 (Ω) dt kE(·, t)kL2 (Ω) dt
0
0
Z
! Z
ξ
≤
(4.21)
0
F (r, τ ) · Ē(r, t) drdτ dt
0
Z
F (r, τ )dτ
Ω
= Re
Z
t
F · ψ̄ drdt = Re
Re
kF (·, t)kL2 (Ω) dt
0
!
ξ
kE(·, t)kL2 (Ω) dt .
0
Using Lemma 4.3 and (4.16), we obtain
Z
ξ
Z
C (E Γ ) · ψ̄ Γ dγdt
Re
0
(4.22)
Z
Γ
Z
ξ
Z
= Re
Γ+
R
0
t
0
C [E Γ+ ](r, τ )dτ
R
· Ē Γ+ (r, t)dtdγ ≥ 0.
R
Substituting (4.20)–(4.22) into (4.19), we have for any ξ ∈ [0, T ] that
ε
1
kE(·, ξ)k2L2 (Ω) +
2
2
ε
(4.23) ≤ kE 0 k2L2 (Ω) +
2
Z
Z
µ−1 Ω
Z
ξ
2
∇ × E(r, t)dt dr
0
ξ
!Z
ξ
kF (·, t)kL2 (Ω) dt + εkE 1 kL2 (Ω)
0
kE(·, t)kL2 (Ω) dt.
0
Taking the L∞ -norm with respect to ξ on both sides of (4.23) yields
kEk2L∞ (0,T ;L2 (Ω)) . kE 0 k2L2 (Ω) + T kF kL1 (0,T ;L2 (Ω)) + kE 1 kL2 (Ω) kEkL∞ (0,T ;L2 (Ω)) ,
which gives the estimate (4.12) after applying the Cauchy-Schwarz inequality.
Integrating (4.23) with respect to ξ from 0 to T and using the Cauchy-Schwarz
inequality, we obtain
kEk2L2 (0,T ;L2 (Ω)) . T kE 0 k2L2 (Ω) +T 3/2 kF kL1 (0,T ;L2 (Ω)) + kE 1 kL2 (Ω) kEkL2 (0,T ;L2 (Ω)) ,
which implies the estimate (4.13) by using the Cauchy-Schwarz inequality again.
18
P. Li, L. Wang, and A. Wood
In Theorem 4.4, it is required that E 0 , E 1 ∈ L2 (Ω), and F ∈ L1 (0, T ; L2 (Ω)),
which can be satisfied if the data satisfy
E 0 ∈ L2 (Ω),
(4.24)
H 0 ∈ H(curl, Ω),
J ∈ H 1 (0, T ; L2 (Ω)).
It is important to point out that the estimates in Theorem 4.2 were derived from a
usual energy method, while the results in Theorem 4.4 were obtained by using different
test functions (cf. (4.14)).
5. Concluding remarks. In this paper, we studied the time-domain Maxwell
equations in a three-dimensional open cavity. The scattering problem was reduced
to an initial-boundary value problem in a bounded domain by using the exact TBC.
The reduced problem was shown to have a unique solution and its stability was also
presented. The proofs were based on the examination of the time-harmonic Maxwell
equations with a complex wavenumber and the abstract inversion theorem of the
Laplace transform. Moreover, by directly considering the variational problem of the
time-domain Maxwell equations, an a priori estimate was derived with an explicit
dependence on time for the electric field. Computationally, the variational approach
leads naturally to a class of finite element methods. As a time-dependent problem,
a fast and accurate marching technique shall be developed to deal with the temporal convolution in the TBC. We will report the work on its numerical analysis and
computation in a forthcoming paper.
Appendix A. Laplace transform.
For any s = s1 + is2 with s1 , s2 ∈ R, s1 > 0, define by ŭ(s) the Laplace transform
of the vector field u(t), i.e.,
Z ∞
ŭ(s) = L (u)(s) =
e−st u(t) dt.
0
It can be verified from the integration by parts that
Z t
(A.1)
u(τ ) dτ = L −1 s−1 ŭ(s) ,
0
−1
where L
is the inverse Laplace transform. One verifies from the formula of the
inverse Laplace transform that
(A.2)
u(t) = F −1 es1 t L (u)(s1 + is2 ) ,
where F −1 denotes the inverse Fourier transform with respect to s2 .
Recall the Plancherel or Parseval identity for the Laplace transform (cf., [15,
(2.46)]).
Lemma A.1. If ŭ = L (u) and v̆ = L (v), then
Z ∞
Z ∞
1
e−2s1 t u(t) · v(t)dt,
ŭ(s) · v̆(s)ds2 =
2π −∞
0
for all s1 > λ, where λ is the abscissa of convergence for the Laplace transform of u
and v.
The following theorem (cf., [36, Thm 43.1]) is an analog of the Paley-WienerSchwarz theorem for the Fourier transform of the distributions with compact support
in the case of Laplace transform.
Lemma A.2. Let w̆(s) denote a holomorphic function in the half plane s1 > σ0 ,
valued in the Banach space E. The following statements are equivalent:
Transient Electromagnetic Scattering
19
0
1. there is a distribution w ∈ D+
(E) whose Laplace transform is equal to w̆(s);
2. there is a σ1 with σ0 ≤ σ1 < ∞ and an integer m ≥ 0 such that for all
complex numbers s with s1 > σ1 , it holds kw̆(s)kE . (1 + |s|)m ,
0
where D+
(E) is the space of distributions on the real line which vanish identically in
the open negative half line.
Appendix B. Spherical harmonics on hemisphere.
The spherical coordinates (ρ, θ, ϕ) are related to the Cartesian coordinates r =
(x1 , x2 , x3 ) by x1 = ρ sin θ cos ϕ, x2 = ρ sin θ sin ϕ, x3 = ρ cos θ, with the local orthonormal basis {eρ , eθ , eϕ }:
eρ = (sin θ cos ϕ, sin θ sin ϕ, cos θ),
eθ = (cos θ cos ϕ, cos θ sin ϕ, − sin θ),
eϕ = (− sin ϕ, cos ϕ, 0),
where θ and ϕ are the Euler angles. Let Γ = {r : ρ = 1}, Γ+ = {r : ρ = 1, x3 ≥
0}, Γ− = {r : ρ = 1, x3 ≤ 0} be the unit sphere, upper unit hemisphere, and
lower unit hemisphere, respectively. Denote by ΓR = {r : ρ = R}, Γ+
R = {r : ρ =
R, x3 ≥ 0}, Γ−
R = {r : ρ = R, x3 ≤ 0} the whole sphere, upper hemisphere, and lower
hemisphere with radius R, respectively.
Let {Ynm (θ, ϕ), |m| ≤ n, n = 0, 1, 2, . . . } be an orthonormal sequence of spherical
harmonics of order n on the unit sphere Γ that satisfies
∆Γ Ynm + n(n + 1)Ynm = 0,
(B.1)
where
1 ∂
∆Γ =
sin θ ∂θ
∂
1 ∂2
sin θ
+
∂θ
sin2 θ ∂ϕ2
is the Laplace-Beltrami operator on Γ. Explicitly, the spherical harmonics of order n
is written as
s
2n + 1 (n − |m|)! |m|
Ynm (θ, ϕ) =
P (cos θ)eimϕ ,
4π (n + |m|)! n
where the associated Legendre functions are
Pnm (t) :=
p
dm Pn (t)
(1 − t2 )m
,
dtm
m = 0, 1, . . . , n.
Here Pn is the Legendre polynomial of degree n, which is an even function if n is even
and an odd function if n is odd.
Define a sequence of rescaled spherical harmonics of order n:
√
(B.2)
Xnm (θ, ϕ)
=
2 m
Y (θ, ϕ),
R n
which forms a complete orthonormal system in L2 (Γ+
R ) for |m| ≤ n, m + n = odd, n ∈
N (cf., [32, Lemma 3.1]).
20
P. Li, L. Wang, and A. Wood
+
Denote by L2 (Γ+
R ) the complex square integrable functions on the hemisphere ΓR .
For convenience, we take the following notation for double summations:
X
wnm : =
|m|≤n
odd
X
wnm ,
n=1 m=−n
wnm : =
|m|≤n
even
X
∞ X
n
X
∞
X
n
X
wnm ,
n=1
wnm : =
|m|≤n
m=−n
m+n=odd
∞
n
X
X
n=1
wnm .
m=−n
m+n=even
To describe vector wave functions on the hemisphere, we introduce some boundary
differential operators. For a smooth scalar function w defined on Γ+
R , let
∇Γ w =
∂w
1 ∂w
eθ +
eϕ
∂θ
sin θ ∂ϕ
be the tangential gradient on Γ+
R . The surface vector curl is defined by
curlΓ w = ∇Γ w × eρ .
Denote by divΓ and curlΓ the surface divergence and the surface scalar curl, respectively. For a smooth vector function w tangential to Γ+
R , it can be represented by its
coordinates in the local orthonormal basis:
w = wθ eθ + wϕ eϕ ,
where
wθ = w · eθ
and wϕ = w · eϕ .
The surface divergence and the surface scalar curl can be defined as
1
∂
∂wϕ
divΓ w =
(wθ sin θ) +
,
sin θ ∂θ
∂ϕ
1
∂
∂wθ
curlΓ w =
(wϕ sin θ) −
.
sin θ ∂θ
∂ϕ
It is known (cf., [35]) that these boundary differential operators satisfy
(B.3)
∆Γ = divΓ ∇Γ = −curlΓ curlΓ
and curlΓ ∇Γ = divΓ curlΓ = 0.
It is also know (cf., [16, Thm 6.23]) that an orthonormal basis for L2t (ΓR ) = {w ∈
L (ΓR ) : eρ · w = 0}, the tangential fields on ΓR , consists of functions of the form
2
Um
n (θ, ϕ) =
1
p
∇Γ Ynm (θ, ϕ)
R n(n + 1)
and
1
m
Vm
curlΓ Ynm
n (θ, ϕ) = eρ × U n (θ, ϕ) = − p
R n(n + 1)
Transient Electromagnetic Scattering
21
for |m| ≤ n, n ∈ N. It follows from (B.1) and (B.3) that
p
p
n(n + 1) m
n(n + 1) m
m
m
divΓ U n = −
Yn , curlΓ V n = −
Yn ,
R
R
and
m
curlΓ U m
n = divΓ V n = 0.
Define two sequences of tangential fields
(B.4)
√
1
∇Γ Xnm (θ, ϕ) = 2 U m
Xm
n (θ, ϕ)
n (θ, ϕ) = p
n(n + 1)
and
m
Ym
n (θ, ϕ) = eρ × X n (θ, ϕ) =
√
2V m
n (θ, ϕ).
Using the definition of the tangential gradient, and noticing that eθ × eϕ = eρ , eϕ ×
eρ = eθ , eρ × eθ = eϕ , we get
π
eθ × X m
n ( , ϕ) = 0
2
for |m| ≤ n, m + n = odd, n ∈ N,
π
eθ × Y m
n ( , ϕ) = 0
2
for |m| ≤ n, m + n = even, n ∈ N.
and
Define a subspace of complex square integrable tangential fields functions on the
hemisphere Γ+
R:
2
+
L2t (Γ+
R ) = {w ∈ L (ΓR ) : eρ · w = 0}.
It is shown (cf., [32, Lemma 3.2]) that the vector spherical harmonics {X m
n : m+n =
odd} and {Y m
n : m + n = even} for |m| ≤ n, n ∈ N form a complete orthonormal
system in L2t (Γ+
R ).
Appendix C. Hankel functions.
(1)
(2)
For ν ∈ R, the two Hankel functions Hν (z) and Hν (z) where z ∈ C, are two
fundamental solutions of the Bessel equation of order ν:
z2
d2 u
du
+z
+ (z 2 − ν 2 )u = 0.
dz 2
dz
Recall the Bessel functions of imaginary argument Kν (z), also called the modified
Bessel functions, which is the solution of the differential equation
z2
du
d2 u
+z
− (z 2 + ν 2 )u = 0.
dz 2
dz
(1)
It is connected with Hν (z) through the relation
(C.1)
Kν (z) =
1
1
πie 2 νπi Hν(1) (iz).
2
22
P. Li, L. Wang, and A. Wood
It is known (cf., [41, p. 511]) that Kν (z) has no zeros if |argz| ≤ 21 π, which implies
(1)
from (C.1) that Hν (z) has no zeros when Imz ≤ 0.
(1)
The spherical Hankel function hn (z) can also be defined by the Hankel function
of half integer order:
r
π (1)
(1)
(C.2)
hn (z) =
H 1 (z).
2z n+ 2
Combining (C.1) and (C.2) yields
r
h(1)
n (z)
=−
2π − 1 (n+ 1 )πi
2
ie 2
Kn+ 21 (−iz),
z
(1)
which implies that hn (z) has no zeros when Imz ≥ 0.
The following two lemmas on the spherical Hankel functions for the complex
number are proved in [13].
Lemma C.1. Let R > 0, n ∈ Z, s = s1 + is2 with s1 > 0. It holds that
Re 1 + rn(1) (isR) < 0,
where rn (z) = z(hn )0 (z)/hn (z).
Lemma C.2. Let R > 0, n ∈ Z, s = s1 + is2 with s1 > 0. It holds that
Im s2 (1 + rn(1) (isR)) ≤ 0.
(1)
(1)
(1)
Combining the result in Lemma C.1 for small value of n and the proof in [16]
(cf. [28, Lemma 3.1]) for large value of n, we may obtain the following estimate.
Lemma C.3. Let R > 0, n ∈ Z, s = s1 + is2 with s1 > 0. There exist two positive
constants C1 and C2 such that
C1 n ≤ |1 + rn(1) (isR)| ≤ C2 n.
Appendix D. Functional spaces.
+
∞
Denote by H s (Γ+
R ) the Sobolev space, the completion of C0 (ΓR ) in the norm
k · kH s (Γ+ ) characterized by
R
kwk2H s (Γ+ ) =
R
odd
X
s
(1 + n(n + 1)) |wnm |2 ,
|m|≤n
where
w(θ, ϕ) =
odd
X
wnm Xnm (θ, ϕ).
|m|≤n
Introduce three tangential trace spaces:
+ 3
s
H st (Γ+
R ) = w ∈ (H (ΓR )) , eρ · w = 0, eθ × w(π/2, ϕ) = 0 ,
−1/2
−1/2
H −1/2 (curl, Γ+
(Γ+
(Γ+
R) = w ∈ H t
R ), curlΓ w ∈ H
R) ,
−1/2
−1/2
H −1/2 (div, Γ+
(Γ+
(Γ+
R) = w ∈ H t
R ), divΓ w ∈ H
R) .
23
Transient Electromagnetic Scattering
For any tangential field w ∈ H st (Γ+
R ), it can be represented in the series expansion
odd
X
w=
m
Xm
w1n
n (θ, ϕ) +
even
X
m
w2n
Ym
n (θ, ϕ).
|m|≤n
|m|≤n
Using the series coefficients, the norm of the space H st (Γ+
R ) can be characterized by
kwk2H s (Γ+ ) =
t
R
odd
X
s
m 2
(1 + n(n + 1)) |w1n
| +
|m|≤n
even
X
s
m 2
(1 + n(n + 1)) |w2n
| ;
|m|≤n
the norm of the space H −1/2 (curl, Γ+
R ) can be characterized by
odd
X
kwk2H −1/2 (curl, Γ+ ) =
R
p
|m|≤n
1
1 + n(n + 1)
m 2
|w1n
| +
even
X
p
m 2
1 + n(n + 1) |w2n
| ;
|m|≤n
and the norm of the space H −1/2 (div, Γ+
R ) can be characterized by
kwk2H −1/2 (div, Γ+ ) =
R
odd p
even
X
X
1
m 2
m 2
p
1 + n(n + 1) |w1n
|w2n
| +
| .
1
+
n(n
+
1)
|m|≤n
|m|≤n
Define a dual pairing by
Z
odd
even
X
X
m
m
u · v̄ dγ =
um
um
hu, vi =
1n v̄1n +
2n v̄2n ,
Γ+
R
|m|≤n
|m|≤n
where
u=
odd
X
|m|≤n
m
um
1n X n +
even
X
m
um
2n Y n
and v =
|m|≤n
odd
X
m
v1n
Xm
n +
|m|≤n
even
X
m
v2n
Ym
n.
|m|≤n
Introduce three functional spaces
H(curl, Ω) = u ∈ L2 (Ω), ∇ × E ∈ L2 (Ω) ,
H S (curl, Ω) = u ∈ H(curl, Ω), n × u = 0 on S ,
H 0 (curl, Ω) = u ∈ H(curl, Ω), n × u = 0 on S ∪ Γ+
R ,
which are Sobolev spaces with the norm:
1/2
kukH(curl, Ω) = kuk2L2 (Ω) + k∇ × uk2L2 (Ω)
.
Given a vector field u on Γ+
R , denote by
uΓ+ = −eρ × (eρ × u)
R
the tangential component of u on Γ+
R , which satisfies the following trace estimate
(cf. [32, Lemma 3.3]).
Lemma D.1. For any u ∈ H S (curl, Ω), it holds that
kuΓ+ kH −1/2 (curl, Γ+ ) . kukH(curl, Ω) .
R
R
24
P. Li, L. Wang, and A. Wood
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