ULTISCALE ODEL. SIMUL Vol. 16, No. 2, pp. 727{751 …Apr 26, 2018 · WANGTAO LU domain...

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MULTISCALE MODEL. SIMUL. c© 2018 Society for Industrial and Applied MathematicsVol. 16, No. 2, pp. 727–751

EXTENDING BABICH’S ANSATZ FOR POINT-SOURCEMAXWELL’S EQUATIONS USING HADAMARD’S METHOD∗

WANGTAO LU† , JIANLIANG QIAN‡ , AND ROBERT BURRIDGE§

Abstract. Starting from Hadamard’s method, we extend Babich’s ansatz to the frequency-domain point-source (FDPS) Maxwell’s equations in an inhomogeneous medium in the high-frequencyregime. First, we develop a novel asymptotic series, dubbed Hadamard’s ansatz, to form the funda-mental solution of the Cauchy problem for the time-domain point-source (TDPS) Maxwell’s equationsin the region close to the source. Governing equations for the unknowns in Hadamard’s ansatz arethen derived. In order to derive the initial data for the unknowns in the ansatz, we further proposea condition for matching Hadamard’s ansatz with the homogeneous-medium fundamental solutionat the source. Directly taking the Fourier transform of Hadamard’s ansatz in time, we obtain anew ansatz, dubbed the Hadamard–Babich ansatz, for the FDPS Maxwell’s equations. Next, weelucidate the relation between the Hadamard–Babich ansatz and a recently proposed Babich-likeansatz for solving the same FDPS Maxwell’s equations. Finally, incorporating the first two terms ofthe Hadamard–Babich ansatz into a planar-based Huygens sweeping algorithm, we solve the FDPSMaxwell’s equations at high frequencies in the region where caustics occur. Numerical experimentsdemonstrate the accuracy of our method.

Key words. Hadamard’s method, Babich’s ansatz, Maxwell’s equation, caustics, fast Huygenssweeping, high-frequency waves

AMS subject classifications. 65N30, 65M60

DOI. 10.1137/17M1130381

1. Introduction. We consider the following frequency-domain point-source(FDPS) Maxwell’s equations in R3:

∇×∇×G(r; r0)− k20n2(r)G(r; r0) = Iδ(r− r0),(1.1)

where G(r; r0) is the so-called dyadic Green’s function at the source r0, I is the 3× 3identity dyad, k0 = ω

c is the freespace wavenumber, ω is the frequency, c is the speedof light in vacuum, r = (x1, x2, x3)T , and n = n(r) is the variable refractive index ofthe medium. Throughout this paper, we assume that n is smooth over R3 and 1− nhas a compact support. Tacitly, the Silver–Muller radiation condition is assumed atinfinity.

The dyadic Green’s function G is fundamental in many applications, ranging fromoptics, microwaves, and antennas to radars. For a generic inhomogeneous medium,an analytic form of G is in general not available. Numerical methods such as finite-difference time-domain (FDTD) methods or finite-element methods (FEMs) have thus

∗Received by the editors May 15, 2017; accepted for publication (in revised form) February 23,2018; published electronically April 26, 2018.

http://www.siam.org/journals/mms/16-2/M113038.htmlFunding: The first author’s research was partially supported by the Hundred Talents Program

of Zhejiang University and the National Thousand Young Talents Program, China. The secondauthor’s research was partially supported by the National Science Foundation (grants 1522249 and1614566).†School of Mathematical Sciences, Zhejiang University, Hangzhou, Zhejiang 310027, China

([email protected]).‡Department of Mathematics, Michigan State University, East Lansing, MI 48824 (qian@math.

msu.edu).§Department of Mathematics and Statistics, University of New Mexico, Albuquerque, NM 87131

([email protected]).

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http://www.siam.org/journals/mms/16-2/M113038.html

mailto:[email protected]





728 WANGTAO LU, JIANLIANG QIAN, AND ROBERT BURRIDGE

been developed to numerically compute G. However, such direct approaches becomevery costly at high frequencies since they cannot bypass the self-inflicted pollutioneffect [3]: in order to attain the same accuracy level, the number of mesh points perunit length in each spatial direction has to be superlinearly dependent on frequency.Therefore, we seek alternative methods, such as asymptotic methods or methods ofgeometrical optics (GO), to carry out scale separation to solve (1.1) when ω is large.

An intuitive approach is to use the following Wentzel–Kramers–Brillouin (WKB)GO ansatz

(1.2) G(r; r0) =

∞∑l=0

Al(r; r0)

ωleiωτ(r;r0),

where the unknowns Al and τ are independent of ω. Using the leading term of (1.2) toapproximate G, we have developed an Eulerian GO-based Huygens sweeping methodto solve (1.1) in [22]. There we have shown that keeping only the leading term of (1.2)fails to capture correct singularities of G at r = r0.

Moreover, such drawbacks cannot be easily resolved by using two or more termsin (1.2) since as in the situation of solving the (scalar) FDPS Helmholtz equation, acritical challenge is how to initialize Al at the source r0. To resolve these issues, Babichin [2] proposed an asymptotic series based on Hankel functions, dubbed Babich’sansatz, to expand the highly oscillatory wavefield. This new ansatz yields a uniformasymptotic solution as ω →∞ in the region of space containing the point source butno other caustics. It is worth mentioning that his method of finding such an ansatz isclosely bound up with Hadamard’s method of forming the fundamental solution of theCauchy problem for the time-domain point-source (TDPS) Helmholtz wave equation;details were given in [9] and then were outlined by Courant and Hilbert [5]. Recently,Babich’s ansatz has been applied to numerically solve the FDPS Helmholtz equationin [16, 25]. Nevertheless, Babich’s ansatz cannot be trivially extended to the FDPSMaxwell’s equations (1.1).

To address the issues mentioned above, in [15] we proposed for (1.1) a novelansatz, dubbed the Babich-like ansatz, based on the spherical Hankel functions, i.e.,

G(r; r0) =

∞∑l=0

Al(r; r0)

τ2l(r; r0)f l(τ(r; r0); k0), f l(τ ; k0) =

−τ l−2h(1)l−3(k0τ)

kl−10

,(1.3)

where the unknowns Al and τ are independent of k0 (and ω). We have demonstratedthat the Babich-like ansatz (1.3) gives a uniform asymptotic expansion of the under-lying solution in the region of space containing a point source but no other caustics.

The motivations of the current article are the following two questions. First, con-sidering the close relation between the Helmholtz equation and Maxwell’s equations,can we extend Hadamard’s method to produce an asymptotic series similar to Babich’sansatz for the FDPS Maxwell’s equations (1.1)? Second, if such a series exists, is itclosely related to our Babich-like ansatz (1.3)? As we shall see, both questions areanswered affirmatively in this paper.

The paper is organized as follows. First, we apply Hadamard’s method to developan asymptotic series, dubbed Hadamard’s ansatz, to form the fundamental solutionof the Cauchy problem for the TDPS Maxwell’s wave equations in a region close tothe source. Governing equations for the unknowns in Hadamard’s ansatz are thenderived. By comparing Hadamard’s ansatz with the homogeneous-medium funda-mental solution, we propose a matching condition at the source which in turn gives

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BABICH’S ANSATZ FOR MAXWELL’S EQUATIONS 729

the initial data for the unknowns. By taking the Fourier transform of Hadamard’sansatz in time, we immediately obtain the extension of Babich’s ansatz, which wedub the Hadamard–Babich ansatz, for the FDPS Maxwell’s equations (1.1). Next, weelucidate the relation between the new Hadamard–Babich ansatz and the previouslyproposed Babich-like ansatz (1.3), demonstrating their equivalence. Incorporating thefirst two terms of the Hadamard–Babich ansatz into a planar-based Huygens sweepingalgorithm in [22], we solve (1.1) when ω is high and when caustics occur. Numericalexperiments demonstrate the accuracy of the new method.

2. Hadamard’s method. As indicated in [2], to apply Hadamard’s method, weneed consider the following Cauchy problem of the TDPS Maxwell’s equations:

∇×∇×G(r, t; r0) + s2(r)G(r, t; r0) = Iδ(r− r0)δ(t),(2.1)

G(r, t; r0)|t<0 ≡ 0,(2.2)

where s(r) = n(r)/c is the slowness function, and G denotes the second-order timederivative of G. We seek the solution to (2.1) and (2.2) in terms of the (generalized)

functions f(λ)+ defined to be

(2.3) f(λ)+ (x) ≡

xλ+Γ(λ+ 1)

, with x+ =

x if x > 0,0 if x ≤ 0

for λ > −1; for other values of λ, fλ+ is defined by analytic continuation. Also, since

f(λ)′

+ = f(λ−1)+ ,(2.4)

f(λ)+ can be defined for negative integer values of λ by successive differentiation in the

sense of distributions. Since f(0)+ (x) = H(x), the Heaviside unit function, f

(−l−1)+ (x) =

δ(l)(x), the lth derivative of the δ-function for l = 0, 1, 2, . . . . For further discussion

of f(λ)+ and related functions, see Chapter I and sections 3.4 and 3.5 of [8].Equation (2.1) indicates that the current density satisfies

(2.5) − µ0∂J

∂t= Iδ(r− r0)δ(t),

which yields

(2.6) J(r, t; r0) = − 1

µ0Iδ(r− r0)H(t),

where µ0 denotes the magnetic permeability, assumed constant. This indicates thatthe jth column of the dyadic Green’s function G is the electric field due to a currentdensity produced by a sudden localized current at r0 of magnitude − 1

µ0along the jth

coordinate direction starting at t = 0 for j = 1, 2, 3.Using the relation

∇×∇×G = ∇(∇ ·G)−∆G,

(2.1) becomes

(2.7) ∇(∇ ·G)−∆G + s2(r)G(r, t; r0) = Iδ(r− r0)δ(t)

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so that we may rewrite it in the form

Gkj,ki −Gij,kk + s2Gij = δijδ(r− r0)δ(t),(2.8)

Gij = 0 for t < 0.(2.9)

Here, the subscript and comma notations are employed. Specifically, Gij denotes theijth entry of G, and Gij,k denotes the xk-derivative of Gij , so that Gij,kl denotes∂2xkxlGij for 1 ≤ i, j, k, l ≤ 3. Furthermore, the Einstein summation convention isassumed so that Gij,kk = ∆Gij , and so on.

Using Duhamel’s principle (p. 202 and pp. 729–730 in [5]), we may rewrite (2.8)for t > 0 as

Gkj,ki −Gij,kk + s2Gij = 0,(2.10)

with initial conditions

Gij(r, 0; r0) = 0,(2.11)

Gij(r, 0; r0) =1

s20δijδ(r− r0),(2.12)

where s0 = s(r0) is the slowness at the source point r0.

2.1. Homogeneous medium. In the case when s(r) ≡ s0, we give a closed formfor G. Following a suggestion of Friedlander [7], let us seek a solution to (2.10)–(2.12)in the form

(2.13) Gij = φ,ij + δijψ,kk − ψ,ij = δijψ,kk + (φ− ψ),ij ,

where φ and ψ are to be determined. On substituting this into (2.10), we obtain

(2.14) ∂i∂j [s20φ] + [δij∆− ∂i∂j ][s20ψ −∆ψ] = 0.

Let us now decompose the initial condition (2.12) in a similar way, using the factthat

δ(r− r0) = ∆

(−1

4πr

),(2.15)

where r = |r− r0|. Thus, we may write

δijδ(r− r0) = ∂i∂j

(−1

4πr

)+ (δij∆− ∂i∂j)

(−1

4πr

).(2.16)

Consequently, we see that Gij given by (2.13) will satisfy (2.10)–(2.12) if

s20φ = 0, t > 0,(2.17)

φ = 0, φ =−1

4πs20r, t = 0,(2.18)

and

s20ψ −∆ψ = 0, t > 0,(2.19)

ψ = 0, ψ =−1

4πs20r, t = 0.(2.20)

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We easily verify that (2.17)–(2.20) are satisfied by

φ =−t

4πs20r,(2.21)

ψ =R(t− s0r)

4πs20r− t

4πs20r=

−14πs0

, s0r ≤ t,

−t4πs20r

, s0r > t,(2.22)

where

R(ξ) =

ξ, ξ > 0,0, ξ ≤ 0.

(2.23)

Carrying out the differentiations indicated in (2.13), we finally arrive at

Gij =tδijδ(r− r0)

s20+δij − γiγj

4πrδ(t− s0r)

+δij − 3γiγj

4πs20

[s0r2H(t− s0r) +

1

r3R(t− s0r)

]=t(δij − 3γiγj)

4πs20r3

H(t− s0r) +δij − γiγj

4πrδ(t− s0r) +

tδijδ(r− r0)

s20for t ≥ 0,(2.24)

where γi = ri/r and r− r0 = (r1, r2, r3)T .Now, we need figure out the relation between Gij and the generalized functions

f(λ)+ in (2.3). It turns out that the second-order time derivative of Gij has a simple

relation with f(−3)+ and f

(−2)+ .

Proposition 2.1.

Gij =2s0π

[(s0r)

2(δij − γiγj)f (−3)+ (t2 − (s0r)2)− δijf (−2)+ (t2 − (s0r)

2)].(2.25)

Proof. To prove (2.25), we require the following properties of the δ-function: Fort, (s0r) ≥ 0,

δ(1)(t2 − (s0r)2) =

1

4(s0r)3(δ(t− s0r) + (s0r)δ

(1)(t− s0r)),(2.26)

δ(2)(t2 − (s0r)2) =

1

8(s0r)3δ(2)(t− s0r) +

3

2(s0r)2δ(1)(t2 − (s0r)

2).(2.27)

They indicate that

∂2t (tH(t− s0r)) = 4(s0r)3δ(1)(t2 − (s0r)

2),(2.28)

∂2t (δ(t− s0r)) = δ(2)(t− s0r) = 8(s0r)3δ(2)(t2 − (s0r)

2)

− 12s0rδ(1)(t2 − (s0r)

2).(2.29)

Consequently, one obtains

Gij =(δij − 3γiγj)

4πs20r3

(δ(t− (s0r)) + (s0r)δ(1)(t− (s0r))) +

δij − γiγj4πr

δ(2)(t− (s0r))

=2s0π

[(s0r)

2(δij − γiγj)δ(2)(t2 − (s0r)2)− δijδ(1)(t2 − (s0r)

2)]

=2s0π

[(s0r)

2(δij − γiγj)f (−3)+ (t2 − (s0r)

2)− δijf (−2)+ (t2 − (s0r)

2)].(2.30)

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2.2. Inhomogeneous medium. When s(r) varies with r, Proposition 2.1 mo-tivates us to seek the solution of (2.10) in the following form of asymptotic series:

G(r, t; r0) =

∞∑l=0

Bl(r; r0)f(−3+l)+ (t2 − T(r; r0))(2.31)

or, in subscript notation,

(2.32) Gij =

∞∑l=0

Blijf(−3+l)+ (t2 − T),

where we enforce all dyadic coefficients Bl = O(1) as r → 0+. In the following, wewill refer to the above series as Hadamard’s ansatz. Notice that due to the initialconditions (2.11) and (2.12), Hadamard’s ansatz (2.31) uniquely determines G. In ahomogeneous medium, Hadamard’s ansatz terminates at l = 1 with

T = (s0r)2,(2.33)

B0 =2s30r

2

π(I− rrT ),(2.34)

B1 = −2s0π

I,(2.35)

where I is the 3 × 3 identity matrix and r = (r − r0)/r. To derive the governingequations for Bk and T, we need the following two simple properties of fk+.

Proposition 2.2. We have for all k ∈ Z, x ∈ R the following:(a)

f(k)+

′(x) = f

(k−1)+ (x).(2.36)

(b)

xf(k)+ (x) = (k + 1)f

(k+1)+ (x).(2.37)

Taking the double time derivative of (2.10) yields

Gkj,ki − Gij,kk + s2∂4tGij = 0.(2.38)

On the other hand, taking the divergence of (2.10) (or, equivalently, summing up theith derivative of (2.10) over i) gives

(s2),iGij + s2Gij,i = 0.(2.39)

To write (2.38) in terms of the generalized functions f(k)+ , we compute the follow-

ing spatial derivatives of G using Proposition 2.2:

Gkj,k = (Blkjf(−3+l)+ (t2 − T)),k = Blkj,kf

(−3+l)+ (t2 − T)−Blkjf

(−3+l)+

′(t2 − T)T,k

= Blkj,kf(−3+l)+ (t2 − T)−BlkjT,kf

(−4+l)+ (t2 − T),

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Gkj,ki = (Blkj,kf(−3+l)+ (t2 − T)−BlkjT,kf

(−4+l)+ (t2 − T)),i

= Blkj,kif(−3+l)+ (t2 − T)− [Blkj,kT,i + (BlkjT,k),i]f

(−4+l)+ (t2 − T)

+BlkjT,kT,if(−5+l)+ (t2 − T),

and

Gij,kk = Blij,kkf(−3+l)+ (t2 − T)− [Blij,kT,k + (BlijT,k),k]f

(−4+l)+ (t2 − T)

+BlijT,kT,kf(−5+l)+ (t2 − T),

and the following double time derivative of G:

∂4tGij = Blij∂2t f

(−3+l)+ (t2 − T)

= Blij∂t[2tf(−4+l)+ (t2 − T)]

= 2Blijf(−4+l)+ (t2 − T) + 4t2Blijf

(−5+l)+ (t2 − T)

= [2 + 4(l − 4)]Blijf(−4+l)+ (t2 − T) + 4TBlijf

(−5+l)+ (t2 − T).

Inserting the above derivatives into (2.38) and (2.39) yields

0 =

∞∑l=0

(Blkj,ki −Blij,kk)f(−3+l)+ (t2 − T)

+ [2 + 4(l − 4)]s2Blij − [Blkj,kT,i + (BlkjT,k),i]

+ [Blij,kT,k + (BlijT,k),k]f (−4+l)+ (t2 − T)

+ [BlkjT,kT,i −BlijT,kT,k + 4TBlijs2]f

(−5+l)+ (t2 − T)

=

∞∑l=−∞

(Blkj,ki −Blij,kk) + [2 + 4(l − 3)]s2Bl+1

ij

− [Bl+1kj,kT,i + (Bl+1

kj T,k),i] + [Bl+1ij,kT,k + (Bl+1

ij T,k),k]

+Bl+2kj T,kT,i −Bl+2

ij T,kT,k + 4TBl+2ij s2

f(−3+l)+ (t2 − T)

and

0 =

∞∑l=0

(s2),iBlijf

(−3+l)+ + s2Blij,if

(−3+l)+ (t2 − T)− s2BlijT,if

(−4+l)+ (t2 − T)

=

∞∑l=−∞

(s2Blij),i − s2Bl+1ij T,if (−3+l)+ (t2 − T),

where we have made the convention that Bl ≡ 0 for l = −1,−2, . . . . In either of the

above two equations, the coefficient of f(−3+l)+ should be equated to zero since, in an

asymptotic series whose sum is asymptotically zero, each term must separately bezero. Consequently, we obtain the following two equations:

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0 = (Blkj,ki −Blij,kk) + [2 + 4(l − 3)]s2Bl+1ij − [Bl+1

kj,kT,i + (Bl+1kj T,k),i]

+ [Bl+1ij,kT,k + (Bl+1

ij T,k),k]

+Bl+2kj T,kT,i −Bl+2

ij T,kT,k + 4TBl+2ij s2,(2.40)

0 = (s2Blij),i − s2Bl+1ij T,i(2.41)

for l ∈ Z. Equation (2.41) further simplifies (2.40) into

0 = (Blkj,ki −Blij,kk) + [2 + 4(l − 3)]s2Bl+1ij − [Bl+1

kj,kT,i + (s−2(s2Blkj),k),i]

+ [Bl+1ij,kT,k + (Bl+1

ij T,k),k]

+ s−2(s2Bl+1kj ),kT,i + (4Ts2 − T,kT,k)Bl+2

ij .(2.42)

Now we are ready to discuss the governing equations of Bl and T.

2.2.1. Governing equation for T. Setting l = −2 in (2.42), we obtain

T,kT,kB0ij = 4s2TB0

ij .(2.43)

It is clear that B0ij is nonzero since otherwise Hadamard’s ansatz (2.32) cannot recover

the Green’s function G in a homogeneous medium. Consequently, we obtain thegoverning equation for T,

T,kT,k = 4s2T,(2.44)

or, in vector notation,

|∇T|2 = 4s2T.(2.45)

This further indicates that T is a nonnegative function in the sense that T = τ2 forsome nonnegative function τ(r; r0). To make sure that τ = s0r for a homogeneousmedium with s(r) ≡ s0, τ is in general a nonzero function. From (2.45), we mayfurther obtain the governing equation for τ ,

|∇τ |2 = s2,(2.46)

a.k.a. the eikonal equation.

2.2.2. Governing equation for Bl. Now by (2.44), equation (2.42) reduces to

2Bl+1ij,kT,k + [T,kk + 2 + 4(l − 3)s2]Bl+1

ij

=− s−2s2,kBl+1kj T,i +Blij,kk + (s−2s2,kB

lkj),i(2.47)

or, in vector notation but with l replaced by l − 1,

(∇T · ∇)Bl +

[∆T

2+ (2l − 7)s2

]Bl = −∇T∇s

2 ·Bl

2s2+

∆Bl−1

2+∇

(∇s2 ·Bl−1

2s2

).

(2.48)

Meanwhile, (2.41) becomes

∇T ·Bl =∇ · (s2Bl−1)

s2.(2.49)

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2.2.3. Initial conditions for T and Bl. Let us discuss the initial condition ofT first. Clearly, G itself should be singular (or undefined) when r = r0 and t = 0.This indicates that T(r0; r0) = 0 so that τ(r0; r0) = 0. Assuming that both s2 and T

are analytic at the source r0, it has been shown in [18, 25] that(T

s20r2

)α= 1 +

α

2S0

(S1

2+S2

3− |∇S1|2

48S0r2)

+α

4

(α2− 1) S2

1

4S20

+ O(r3),(2.50)

where Sk is the degree-k term in the Taylor expansion of s2 about r0 such that

s2(r) =

∞∑k=0

Sk(r; r0).(2.51)

Next, to derive initial conditions for Bl, we need to study the relation of G andG0 at the source point r0 first, where G0 is the Green’s function, with ijth entry(2.24) for 1 ≤ i, j ≤ 3 in a homogeneous medium with s ≡ s0. For convenience, wewrite again the governing equations of the two as follows:

∇×∇×G + s2G = Iδ(r− r0)δ(t),(2.52)

∇×∇×G0 + s20G0 = Iδ(r− r0)δ(t).(2.53)

Considering the difference between (2.52) and (2.53), we obtain

∇×∇× (G−G0) + s2(G− G0) = (s20 − s2)G0.(2.54)

In subscript notation, it becomes

(Gkj −G0kj),ki − (Gij −G0

ij),kk + s2(Gij − G0ij)

= (s20 − s2)G0,ij

= (s20 − s2)(δij − 3γiγj)

4πs20r3

(δ(t− s0r) + s0rδ(1)(t− s0r))

+ (s20 − s2)δij − γiγj

4πrδ(2)(t− s0r).(2.55)

Taking the double time derivative yields

(Gkj − G0kj),ki − (Gij − G0

ij),kk + s2∂2t (Gij − G0ij)

= (s20 − s2)(δij − 3γiγj)

4πs20r3

(δ(2)(t− s0r) + s0rδ(3)(t− s0r))


4πrδ(4)(t− s0r).(2.56)

For any function g ∈ C∞0 (R) with compact support [0, T ] for some T > 0, integratingg times the right-hand side of (2.56) over [0, T ] yields∫ T

0

(s20 − s2)

(δij − 3γiγj)

4πs20r3

(δ(2)(t− s0r) + s0rδ(3)(t− s0r))(2.57)


4πrδ(4)(t− s0r)

g(t)dt

= (s20 − s2)(δij − 3γiγj)

4πs20r3

[g′′(s0r)− s0rg(3)(s0r)] + (s20 − s2)δij − γiγj

4πrg(4)(s0r)

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= O

(1

r2

)as r → 0+.

Therefore, we have to enforce that as r → 0+,

O

(1

r2

)=

∫ T

0

(Gkj − G0

kj),ki − (Gij − G0ij),kk + s2∂2t (Gij − G0

ij)g(t)dt

=

∫ T

0

[(Gkj − G0

kj)g(t)],ki − [(Gij − G0ij)g(t)],kk

dt

+ s2∫ T

0

(Gij − G0ij)g(t)dt

=

[∫ T

0

(Gkj − G0kj)g(t)dt

],ki

−

[∫ T

0

(Gij − G0ij)g(t)dt

],kk

+ s2∫ T

0

(Gij − G0ij)g(t)dt.(2.58)

This motivates us to propose the following matching condition:∫ T

0

(Gij(r, t; r0)− G0ij(r, t; r0))g(t)dt = O(1)(2.59)

as r → 0+. Otherwise, if (2.59) is not satisfied, then the first term on the right-handside of (2.58) will make the singularity at r = 0 worse than O( 1

r2 ).As any function g ∈ C∞0 [0, T ] can be uniformly approximated by a polynomial

function due to the Weierstrass approximation theorem, the matching condition (2.59)in fact can be equivalently restated as follows: for any nonnegative integer m and anyT > 0, ∫ T

0

(G(r, t; r0)− G0(r, t; r0))tmdt = O(1)(2.60)

as r → 0+.Based on (2.60), we are ready to derive initial conditions for Bl. Inserting

Hadamard’s ansatz (2.31) and the closed form (2.25) of G0 into (2.60) yields

B0

∫ T

0

δ(2)(t2 − T)tmdt+ B1

∫ T

0

δ(1)(t2 − T)tmdt+ B2

∫ T

0

δ(t2 − T)tmdt

− 2s30r2

π(I− rrT )

∫ T

0

δ(2)(t2 − s20r2)tmdt+2s0I

π

∫ T

0

δ(1)(t2 − s20r2)tmdt

+

∞∑l=0

Bl+3

∫ T

0

f(l)+ (t2 − T)tmdt = O(1).(2.61)

As f(l)+ are bounded on [0, T ] for l = 0, 1, . . . ,∫ T

0

f(l)+ (t2 − T)tmdt = O(1).

Getting rid of all O(1) terms in (2.61), we get

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B0

∫ T

0

δ(2)(t2 − T)tmdt+B1

∫ T

0

δ(1)(t2 − T)tmdt+B2

∫ T

0

δ(t2 − T)tmdt

− 2s30r2

π(I− rrT )

∫ T

0

δ(2)(t2 − s20r2)tmdt+2s0I

π

∫ T

0

δ(1)(t2 − s20r2)tmdt = O(1).(2.62)

To further simplify (2.62), we need the following formula:∫ T

0

δ(k)[t2 − T]tmdt =

∫ T 2−T

−Tδ(k)(t)

1

2(t+ T(x))

m2 −

12 dt

=(−1)k

2

[(t+ T)

m2 −

12

](k)∣∣∣∣t=0

=(−1)k

2

k−1∏j=0

(m

2− 1

2− j)Tm2 −

12−k.(2.63)

For k ≤ 2, the integral in (2.63) is O(1) when m ≥ 5 since m2 −

12 − k ≥ 0. Thus,

considering only m = 0, 1, 2, 3, 4 respectively in (2.62) and using (2.63) with k ≤ 2,we obtain as r → 0+ that

3

8

[B0

T52

− 2s30r2

π(s0r)5(I− rrT )

]+

1

4

[B1

T32

+2s0I

π(s0r)3

]+

B2

2T12

=3

8

B0

T52

− 1

4

B1

T32

+B2

2T12

− I− 3rrT

4πs20r3

= O(1),(2.64)

B2

2= O(1),(2.65)

− 1

8

[B0

T32

− 2s30r2

π(s0r)3(I− rrT )

]− 1

4

[B1

T12

+2s0I

π(s0r)

]+

B2T12

2

=− 1

8

B0

T32

− 1

4

B1

T12

+B2T

12

2− I + rrT

4πr= O(1),(2.66)

− 1

2

[B1 +

2s0π

I

]+

1

2B2T = O(1),(2.67)

3

8

[B0

T12

− 2s30r2

πs0r(I− rrT )

]+

3

4

[B1T

12 +

2s20rI

π

]+

1

2B2T

32

=3

8

B0

T12

− 3

4B1T

12 +

1

2B2T

32 − 3s20r

4π(3I− rrT ) = O(1).(2.68)

Hiding all O(1) terms gives rise to

3

8

B0

T52

+1

4

B1

T32

+B2

2T12

− I− 3rrT

4πs20r3

= O(1),(2.69)

−1

8

B0

T32

− 1

4

B1

T12

− I + rrT

4πr= O(1),(2.70)

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3

8

B0

T12

= O(1).(2.71)

Considering (2.69)×T+(2.70),

B0

4T32

− T

4πs20r3

(I− 3rrT )− 1

4πr(I + rrT ) = O(1),(2.72)

so that we obtain an initial condition for B0:

B0 =T

52

πs20r3

(I− 3rrT ) +T

32

πr(I + rrT ) + O(T

32 ).(2.73)

Clearly, (2.73) implies (2.71) due to (2.50).

Next, (2.69)×4T32 leads to an initial condition for B1:

B1 =− 3B0

2T+

T32

πs20r3

(I− 3rrT )− 2B2T + O(T32 )

=− 3B0

2T+

T32

πs20r3

(I− 3rrT ) + O(T).(2.74)

Finally, (2.69)×2T12 gives an initial condition for B2:

B2 = −3B0

4T2− B1

2T+

T12

2πs20r3

(I− 3rrT ) + O(T12 ).(2.75)

Summarizing all the previous results, we have obtained the following theorem.

Theorem 2.1. Suppose a dyadic function G(r, t; r0) of the form (2.31) solves thetime-domain Maxwell’s equations (2.1) and (2.2). The following statements hold:

(a) The nonzero function T is governed by

|∇T|2 = 4s2T,(2.76)

with the initial condition T(r0; r0) = 0.(b) The dyadic coefficients Bl∞l=0 are governed by the following recursive equa-

tions:

(∇T · ∇)Bl +

[∆T

2+ (2l − 7)s2

]Bl = −∇T∇s

2 ·Bl

2s2+

∆Bl−1

2

+∇(∇s2 ·Bl−1

2s2

),(2.77)

∇T ·Bl =∇ · (s2Bl−1)

s2.(2.78)

To fulfill (2.60), B0, B1, and B2 asymptotically behave near r0 as follows:

B0 =T

52

πs20r3

(I− 3rrT ) +T

32

πr(I + rrT ) + O(T

32 ),(2.79)

B1 =− 3B0

2T+

T32

πs20r3

(I− 3rrT ) + O(T),(2.80)

B2 =− 3B0

4T2− B1

2T+

T12

2πs20r3

(I− 3rrT ) + O(T12 )(2.81)

as r → 0+.

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2.3. Hadamard–Babich ansatz in the frequency domain. The Fouriertransform of (2.1) in time yields the FDPS Maxwell’s equations (1.1) with

G(r; r0) =

∫ ∞0

G(r, t; r0)eiωtdt,(2.82)

so that, by (2.31), we get the following frequency-domain asymptotic ansatz:

G(r; r0) =

∞∑l=0

Bl(r; r0)

−ω2

∫ ∞0

eiωtf(−3+l)+ (t2 − τ2(r; r0))dt.(2.83)

The integral in (2.83) has the following closed form.

Proposition 2.3.∫ ∞τ

eiωtf(ν− 1

2 )+ (t2 − τ2)dt =

1

2i√π

(2τ

ω

)νeiπνH(1)

ν (ωτ) := fν(ω, τ).(2.84)

Here fν(ω, τ) is exactly the basis function used in Babich’s ansatz [2].

Proof. Magnus and Oberhettinger [19, p. 28] give the following integral represen-tation of the Hankel functions:

H(1,2)ν (x) =

Γ( 12 − ν)

(x2

)ν±iπΓ( 1

2 )[1− e∓2iπ(ν− 1

2 )]

∫ ∞1

e±ixt(t2 − 1)ν−12 dt.(2.85)

Here − 12 < Reν < 1

2 ; the upper signs are for H(1)ν and the lower signs are for H

(2)ν .

Notice that

1− e∓2iπ(ν− 12 ) = 1 + e∓2iπν

= e∓iπν [e±iπν + e∓iπν ]

= 2e∓iπν cosπν.(2.86)

Abramowitz and Stegun [1, Formula 6.1.17] give the reflection formula for the Γfunction:

Γ(z)Γ(1− z) = π cscπz.(2.87)

In (2.87) set z = ν + 12 to get

Γ

(ν +

1

2

)Γ

(1

2− ν)

= π cscπ

(ν +

1

2

)=

π

cosπν.(2.88)

So

cosπνΓ( 12 − ν)

π=

1

Γ(ν + 12 )

=1

(ν − 12 )!

.(2.89)

Using (2.86) and (2.89) in (2.85), we get

H1,2ν (x) =

2(x2

)νe∓iπν

±iΓ( 12 )

∫ ∞1

e±ixt(t2 − 1)ν−

12

(ν − 12 )!

dt,(2.90)

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and we have ∫ ∞1

e±ixtfν− 1

2+ (t2 − 1)dt = ±1

2i√π

(2

x

)νe±iπνH(1,2)

ν (x)(2.91)

when − 12 < Reν < 1

2 , and for other values of ν by analytic continuation. Setting

t = t′

τ , dt = dt′/τ , we see that∫ ∞τ

e±ixt′τf(ν− 1

2 )+ (t′2 − τ2)

τ2νdt′ = ±1

2i√π

(2

x

)νe±iπνH(1,2)

ν (x).(2.92)

Setting x = ωτ , dropping the primes on t′, and taking the upper sign, we get(2.84).

By (2.82) and Proposition 2.3, we immediately find that the frequency-domainHadamard’s ansatz should be

G(r; r0) =

∞∑l=0

Bl(r; r0)

−ω2f−3+l+ 1

2(ω, τ(r; r0)),(2.93)

which we refer to as the Hadamard–Babich ansatz in the following.Since

f−3+l+ 12(ω, τ(r; r0)) =

i√π

2

(2τ

ω

)−3+l+ 12

eiπ(−3+l+12 )H

(1)

−3+l+ 12

(ωτ)

=

√π

2

(2 τ

k0

)−3+l+ 12

(−1)lH(1)

−3+l+ 12

(k0 τ)

=

k20(−2)l−3

c2l−5

−τ l−2h(1)l−3(k0 τ)

kl−10

,(2.94)

where τ = cτ , we have

G(r; r0) =

∞∑l=0

−(−2)l−3

c2l−3Bl(r; r0)

−τ l−2h(1)l−3(k0 τ)

kl−10

=:

∞∑l=0

Bl(r; r0)f l(τ ; k0)

=:

∞∑l=0

Al(r; r0)

τ2lf l(τ ; k0),(2.95)

where we have defined

f l(τ ; k0) =−τ l−2h(1)l−3(k0 τ)

kl−10

,(2.96)

Bl(r; r0) =−(−2)l−3

c2l−3Bl(r; r0),(2.97)

Al(r; r0) = Blτ2l.(2.98)

Since Bl = O(1) as r → 0, Bl = O(1) so that Al = O(τ2l). We introduce Al here,since Bl may not be smooth at the source r0 for an inhomogeneous medium; this

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can be seen from (2.50) and from the initial conditions (2.79)–(2.81). The right-handside of (2.95) is exactly our Babich-like ansatz (1.3), so that we have established theequivalence of the Babich-like ansatz (1.3) and the Hadamard–Babich ansatz (2.93).For this reason, we will not differentiate (2.93) and (2.95) (or (1.3)), and we will callboth the Hadamard–Babich ansatz from now on. We will also refer to the unknownsAl and τ as the Hadamard–Babich ingredients.

Noticing that T = τ2 = τ2

c2 , we now can rewrite Theorem 2.1 in the framework ofthe Hadamard–Babich ansatz (2.95).

Theorem 2.2. Suppose a dyadic function G(r; r0) of the form (2.95) solves theFDPS Maxwell’s equations (1.1). The following statements hold:

(a) The nonzero function τ is governed by

|∇τ |2 = n2,(2.99)

with the initial condition τ(r0; r0) = 0.(b) The dyadic coefficients Al∞l=0 are governed by the following recursive equa-

tions:

(∇τ2 · ∇)Al +

[∆τ2

2− (2l + 7)n2

]Al = −∇τ2∇n

2 ·Al

2n2+ Rl−1,(2.100)

∇τ2 ·Al = −2τ2∇ · (n2Al−1)

n2+ 2(l − 1)∇τ2 ·Al−1,(2.101)

where we define for any l ≥ 0,

Rl−1 =− τ2∇(∇n2 ·Al−1

n2

)+ (l − 1)

∇n2 ·Al−1

n2∇τ2

− (l − 1)ln2Al−1 − τ2∆Al−1

+ (l − 1)(2∇Al−1 · ∇τ2 + Al−1∆τ2),(2.102)

and in particular R−1 ≡ 0. To fulfill the matching condition (2.60), A0, A1,and A2 behave asymptotically near r0 as follows:

A0 =(I− 3rrT )n0τ

2 + (I + rrT )n30r2

8π

(τ

n0r

)3

+ O(τ3),(2.103)

A1 =3A0 − 1

4π

(τ

n0r

)3

(I− 3rrT )τ2n0 + O(τ4),(2.104)

A2 =3A0 −A1 − 1

4π

(τ

n0r

)3

(I− 3rrT )τ2n0 + O(τ5),(2.105)

where n0 = n(r0) is the refractive index at the source r0.

With no effort, one sees immediately that Theorem 2.2 is the same as Theorem2.6 in [15] only in different notations. However, they are now completely derived fromthe TDPS Maxwell’s equations (2.1) using Hadamard’s method.

To conclude this section, we remark that although Al are governed by (2.100) and(2.101), the former implies the latter. As was shown, (2.101) or (2.78) is originatedfrom (2.39), which is just the divergence of (2.10) that gives (2.100). Alternatively,one can prove (2.101) by taking the dot product of (2.100) and ∇τ2 and by inductionon l; since the proof is tedious, we omit it here.

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3. Huygens sweeping method. We may develop two different approaches touse the proposed Hadamard–Babich ansatz: one is an ordinary differential equation(ODE)–based Lagrangian approach, and the other is a partial differential equation(PDE)–based Eulerian approach. By a Lagrangian approach, we mean that we solvea set of ODEs along rays to compute the Hadamard–Babich ingredients so as to con-struct asymptotic wavefields; certainly, different strategies will be employed accordingto where caustics occur along a ray path, and we will develop such Lagrangian ap-proaches to implement the ansatz in a future work.

Meanwhile, we remark that in [15] we have developed an Eulerian approach toimplement the Hadamard–Babich ansatz (2.95) by assuming that no caustic occursin a region of space containing a point source. However, as proved in [31], causticsoccur with high probability in a generic inhomogeneous medium. Therefore, inspiredby the work [22], assuming that the contributing rays emitted from a point sourcehave a consistent orientation so that waves propagate close to a single direction, wedevelop a fast Huygens sweeping method to deal with caustics by incorporating theHadamard–Babich ansatz (2.95) into the Huygens secondary-source principle.

As explained and illustrated in [16, 17, 22, 23], given a primary source point, we apriori partition the computational domain into several layers as sketched in Figure 1,where r0 denotes the primary source, Sm is the mth secondary source plane z = zm sothat Ωm+1 = (x, y, z)T |zm < z < zm+1, the wavefield associated with the primarysource r0 has no caustics in the region Ω0 but may develop caustics beyond Ω0, and,for all m, the wavefields associated with the secondary sources situated on Sm haveno caustics in Ωm+1 but may develop caustics beyond Ωm+1.

In Ω0, the Hadamard–Babich ansatz (2.95) is applicable for computing G(r; r0)since no caustics occur. Numerically, it is impossible to compute all dyadic amplitudesAl for l = 0, 1, 2, . . . . Following [15], we approximate G(r; r0) by the first two termsof the Hadamard–Babich ansatz (2.95), i.e.,

G(r; r0) ≈ A0(r; r0)f0(τ(r; r0); k0) +A1(r; r0)

τ2(r; r0)f1(τ(r; r0); k0).(3.1)

To compute (3.1) numerically, we need to solve the eikonal equation (2.99) to obtainτ and the vectorial Hamilton–Jacobi equations (2.100) and (2.101) for l = 0 andl = 1 to obtain A0 and A1, respectively. To obtain a first-order accurate G, weneed a fifth-order accurate τ , a third-order accurate A0, and a first-order accurateA1 since A1 depends on ∆τ2 and ∆A0. For the scalar eikonal equation (2.99), weemploy the Lax–Friedrichs weighted essentially nonoscillatory (LxF-WENO)–basedschemes [13, 17, 18, 23, 25, 28, 33, 34] to compute τ . For the vectorial equations

z

r0+0

+1

+2

S0

S1

S2

+3

Fig. 1. A two-dimensional sketch of Huygens sweeping method. The computational domainis partitioned into four layers. The primary source r0. The secondary source plane Sm and itsassociated region Ωm+1 for m = 0, 1, 2.

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(2.100) and (2.101), since the components of Al are coupled together, we need to firstuse a decoupling approach in [15] to transform the vectorial equations into severalscalar Hamilton–Jacobi equations, and we then solve those scalar equations by theLxF-WENO schemes to get A0 and A1; see [15] and the references therein for details.In passing, we remark that the high-order schemes for the eikonal and transportequations that we are using here were developed in [18, 23], which in turn are basedon the Lax–Friedrichs sweeping [12, 13, 26, 27, 28, 30, 33, 34], WENO finite differenceapproximation [10, 11, 14, 20], and factorization of the upwind source singularities[6, 18, 21, 32]. To treat the upwind singularity at the point source, an adaptivemethod for the eikonal and transport equations has been proposed in [24] as well.

Beyond Ω0, we construct G(r; r0) by using the Huygens–Kirchhoff formula ineach Ωm+1 in a layer-by-layer manner. Suppose now G(r; r0) becomes available forr ∈ Ωm so that G(r; r0) on Sm is known. As shown in [4, 22], for any r′ ∈ Ωm+1, theHuygens–Kirchhoff formula gives

Gj(r′; r0) = −

∮Sm

ν(r) · [Gj(r; r0)×∇×G(r; r′) +∇×Gj(r; r0)×G(r; r′)] dS(r),

(3.2)

where ν(r) denotes the unit normal vector at r on Sm and Gj is the jth column ofG for j = 1, 2, 3. To make use of (3.2), we need to compute G(r; r′) and its curl∇×G(r; r′), where the associated source r′ is in Ωm+1. Directly using the two-termansatz (3.1) to compute these two terms is costly, as we need to solve the governingequations (2.99)–(2.101) for each source point r′ in the three-dimensional manifoldΩm+1 at r on only a two-dimensional manifold Sm. To reduce the cost, we use thereciprocal relation to interchange the roles of r and r′ so that only the secondary sourcepoints on the two-dimensional manifold Sm are involved in the above computationalprocess.

Specifically, since G(r′; r) is assumed to have no caustics in r′ ∈ Ωm+1, by thereciprocal relation and by (3.1) we may approximate

(3.3) G(r; r′) = GT (r′; r) ≈ A0,T (r′; r)f0(τ(r′; r); k0) +A1,T (r′; r)

τ2(r′; r)f1(τ(r′; r); k0),

where the superscript T indicates the transpose. Then, using the properties of f l

listed in Lemma 2.2 in [15], the curl of G(r; r′) becomes

∇×G(r; r′) ≈[A0,T (r′; r)×∇rτ(r′; r)

]τ(r′; r)f−1(τ(r′; r); k0)

+

[∇r ×A0,T (r′; r) +

A1,T (r′; r)

τ(r′; r)×∇rτ(r′; r)

]f0(τ(r′; r); k0)

+∇r ×[A1,T (r′; r)

τ2(r′; r)

]f1(τ(r′; r); k0),(3.4)

where the subscript r in ∇r emphasizes that the gradient operator is applied withrespect to r. Since f l(τ(r′; r); k0) = O(k−l0 ) as k0 → ∞, retaining only the leading-order terms, i.e., the O(1) terms in (3.3) and the O(k0) terms in (3.4), we obtain

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G(r; r′) ≈A0,T (r′; r)

τ3(r′; r)eik0τ(r

′;r),(3.5)

∇×G(r; r′) ≈ ik0τ3(r′; r)

[A0,T (r′; r)×∇rτ(r′; r)

]eik0τ(r

′;r).(3.6)

Such approximations are adequately accurate for computing Gj(r′; r0) in (3.2) for the

following reason. The integrand in (3.2) is in general O(k0) and becomes O(1) onlywhen ∇ × G(r; r′) = O(1) and G(r; r′) = O(1/k0) (since ∇ × Gj(r; r0) = O(k0)).Therefore, the approximations lose accuracy only when the right-hand sides of (3.5)and (3.6) are simultaneously equal to 0, and the points with such a property constituteonly a set of measure 0 and will not affect the value of Gj(r

′; r0) in (3.2).As can be seen, we need the following three unknown ingredients: A0(r′; r),

τ(r′; r), and ∇rτ(r′; r) at each secondary source r ∈ Sm. Among these, only thegoverning equation of ∇rτ(r′; r) is unknown. According to [15],

∇rτ(r′; r) = t(0)(r′; r)n(r),(3.7)

where t(0) = (t(0)1 , t

(0)2 , t

(0)3 )T denotes the take-off direction of the ray from r to r′ and

is governed by

(∇r′ τ2 · ∇r′)t

(0)(r′; r) = 0(3.8)

with the initial condition

limr′→r

[t(0)(r′; r)− r′ − r

|r′ − r|

]= 0.(3.9)

We can apply the LxF-WENO schemes to compute t(0). Consequently, using (3.5)and (3.6), equation (3.2) can be approximated by

Gj(r′; r0) ≈

∮Sm

[ik0n(r)(ν(r) · t(0)(r′; r))(Gj(r; r0) ·A0,T (r′; r))

− ik0n(r)(Gj(r; r0) · t(0)(r′; r))(ν(r) ·A0,T (r′; r))

− (ν(r)×∇×Gj(r; r0)) ·A0,T (r′; r)]eik0τ(r′;r)τ3(r′; r)

dS(r)(3.10)

for r′ ∈ Ωm+1. Using (3.10) to sweep through all layers Ωm, we can obtain G(r; r0)in the whole computational domain. For implementation details of the above method,we refer readers to [22].

4. Numerical examples. In this section, we will study several numerical ex-amples. Unless otherwise stated, all computations were executed on a 20-core 2.5-Ghz Intel Xeon E5-2670v2 processor with 64 GB of RAM at the High PerformanceComputing Center at Michigan State University. To obtain a reference solution, ifnecessary, we apply the FDTD method [29] directly to the associated time-domainMaxwell’s equations (2.10)–(2.12) to obtain a numerical solution in the time domainand then the Fourier transform in time to compute a numerical solution in the fre-quency domain.

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Example 1. Constant refractive index model. In this example, the Green’sfunction is constructed with the following setup:

• The refractive index function is n(x, y, z) ≡ 1.• The computational domain is Ω = [0, 2] × [0, 2] × [0, 2], and the two-term

Hadamard–Babich ingredients are computed on a uniformly spacing grid withsize 51× 51× 51.

• The primary source point is r0 = (1.0, 1.0, 0.2)T .• One secondary source plane is placed at S0 : z = 1.2, and we update G in

Ω1 = [0, 2]× [0, 2]× [1.4, 2.0] by using the Huygens–Kirchhoff formula (3.10).As n is constant, we can use the exact solution of G (see (2.4) in [15]) to validate

our numerical solutions. Figures 2 and 3 show our numerical solutions and the exactsolutions of G11 and G31 at y = 1 for k0 = 24π. Detailed comparisons are shown inFigure 4. We see that they match well in the region close to the source and awayfrom it.

(a) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

(b) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

Fig. 2. Example 1: real part of G11 at y = 1 computed by (a) the Huygens sweeping method and(b) the exact solution. Mesh 101 × 101 × 101. Primary source r0 = (1.0, 1.0, 0.2)T and k0 = 24π.

(a) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

(b) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

Fig. 3. Example 1: real part of G31 at y = 1 computed by (a) the Huygens sweeping method and(b) the exact solution. Mesh 101 × 101 × 101. Primary source r0 = (1.0, 1.0, 0.2)T and k0 = 24π.

Example 2: Gaussian model. In this example, the Green’s function is con-structed with the following setup:

• The refractive index function is

n(x, y, z) =3

3− 1.75e−(x−1)2+(y−1)2+(z−1)2

0.64

.(4.1)Dow

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(a)0 0.5 1 1.5 2

−0.5

0

0.5

z

Wav

efie

ld

(b)0 0.5 1 1.5 2

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

z

Wav

efie

ld

(c)0 0.5 1 1.5 2

−0.4

−0.2

0

0.2

0.4

z

Wav

efie

ld

(d)0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

1.5

2

2.5

x 10−4

z

Wav

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Fig. 4. Example 1: the real part of G11 at (a) line x = 1 and y = 1 and (b) line x = 712

and

y = 1; the real part of the G31 at (c) line x = 1 and y = 1 and (d) line x = 712

and y = 1. Solid line:

the exact solution; circle line: the Huygens sweeping solution. Primary source r0 = (1.0, 1.0, 0.2)T

and k0 = 24π.

(a) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

(b) x

z

0 0.5 1 1.5

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

Fig. 5. Example 2: real part of G11 at y = 1 computed by (a) the Huygens sweeping methodand (b) the FDTD solution. Primary source r0 = (1.0, 1.0, 0.2)T and k0 = 10π. Mesh in (a):97 × 97 × 97; mesh in (b): 401 × 401 × 401.

• The computational domain is Ω = [0, 2] × [0, 2] × [0, 2], and the two-termHadamard–Babich ingredients are computed on a uniformly spacing grid withsize 51× 51× 51.

• The primary source point is r0 = (1.0, 1.0, 0.2)T .• One secondary source plane is placed at S0 : z = 1.2, and we update G in

Ω1 = [0, 2]× [0, 2]× [1.4, 2.0] by using the Huygens–Kirchhoff formula (3.10).To validate our numerical solutions, we first consider computing G at a low

frequency k0 = 10π so that accurate reference solutions can be obtained by the FDTDmethod. Figures 5 and 6 show our numerical solutions and the FDTD-based solutionsof G11 and G31 at y = 1 for k0 = 10π. Detailed comparisons are shown in Figure 7.We see that they have good agreement close to the source and away from it. However,

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(a) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

(b) x

z

0 0.5 1 1.5

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

Fig. 6. Example 2: the real part of G31 at y = 1 computed by (a) the Huygens sweeping methodand (b) the FDTD solution. Primary source r0 = (1.0, 1.0, 0.2)T and k0 = 10π. Mesh in (a):97 × 97 × 97; mesh in (b): 401 × 401 × 401.

(a)0 0.5 1 1.5 2

−1

−0.5

0

0.5

z

Wav

efie

ld

(b)0 0.5 1 1.5 2

−0.05

0

0.05

z

Wav

efie

ld

(c)1.2 1.4 1.6 1.8 2

−0.2

−0.1

0

0.1

0.2

z

Wav

efie

ld

(d)0 0.5 1 1.5 2

−0.04

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

z

Wav

efie

ld

Fig. 7. Example 2: the real part of G11 at (a) line x = 1 and y = 1 and (b) line x = 0.5 andy=1; the real part of G31 at (c) line x = 1 and y = 1 and (d) line x = 0.5 and y = 1. Solid line:FDTD solution; circle line: the Huygens sweeping solution. Primary source r0 = (1.0, 1.0, 0.2)T

and k0 = 10π.

we point out that to attain the same level of resolution of G, our asymptotic methodrequires only 97× 97× 97 points, roughly 70 times less than 401× 401× 401 pointsused in the FDTD method.

Moreover, we also use our asymptotic method to compute G at a higher frequencyk0 = 20π. Figure 8 shows our numerical solutions of G11 and G31 at y = 1 fork0 = 20π, where our method requires 193 × 193 × 193 points. Nevertheless, due tothe limited CPU and memory resource, the FDTD method is not able to produceaccurate results for G.

Example 3: Waveguide model. In this example, the Green’s function isconstructed with the following setup:

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(a) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

(b) x

z

0 0.5 1 1.5 2

0

0.5

1

1.5

2 −0.2

−0.1

0

0.1

0.2

Fig. 8. Example 2: the real part of (a) G11 at y = 1 and (b) G31 at y = 1, computed by theHuygens sweeping method. Primary source r0 = (1.0, 1.0, 0.2)T and k0 = 20π. Mesh 193×193×193.

(a)x

z

0 0.5 1 1.5 2

0

0.5

1

1.5−0.2

−0.1

0

0.1

0.2

(b)x

z

0 0.5 1 1.5

0

0.5

1

1.5−0.2

−0.1

0

0.1

0.2

Fig. 9. Example 3: the real part of G11 at y = 1 computed by (a) the Huygens sweeping methodand (b) the FDTD solution. Primary source r0 = (1.0, 1.0, 0.2)T and wavenumber k0 = 12π. Meshin (a): 97 × 97 × 77; mesh in (b): 401 × 401 × 321.

• The refractive index function is

n(x, y, z) =1

1− 0.5e−6((x−1)2+(y−1)2) .(4.2)

• The computational domain is Ω = [0, 2] × [0, 2] × [0, 1.6], and the two-termHadamard–Babich ingredients are computed on a uniformly spacing grid withsize 51× 51× 51.

• The primary source point is r0 = (1.0, 1.0, 0.2)T .• Two secondary source planes are placed at S0 : z = 0.6 and S1 : z = 1.0, and

we update G in Ω1 = [0, 2]× [0, 2]× [0.8, 1.2] and Ω2 = [0, 2]× [0, 2]× [1.2, 1.6],respectively, by using the Huygens–Kirchhoff formula (3.10).

To validate our numerical solutions, we first consider computing G at a low frequencyso that reference solutions can be obtained by the FDTD method. Figures 9 and 10show our numerical solutions and the FDTD-based solutions of G11 and G31 at y = 1for k0 = 12π. Detailed comparisons are shown in Figure 11. We see that they havegood agreement in the region close to the source and away from it. However, we pointout that to attain the same level of resolution of G, our asymptotic method requiresonly 97 × 97 × 77 points, roughly 70 times less than 401 × 401 × 321 points used inthe FDTD method.

Moreover, we also use our asymptotic method to compute G at a higher frequencyk0 = 24π. Figure 12 shows our numerical solutions of G11 and G31 at y = 1 fork0 = 24π, where we use 193× 193× 154 points.

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(a)x

z

0 0.5 1 1.5 2

0

0.5

1

1.5−0.2

−0.1

0

0.1

0.2

(b)x

z

0 0.5 1 1.5 2

0

0.5

1

1.5−0.2

−0.1

0

0.1

0.2

Fig. 10. Example 3: the real part of G31 at y = 1 computed by (a) the Huygens sweeping methodand (b) the FDTD solution. Primary source r0 = (1.0, 1.0, 0.2)T and wavenumber k0 = 12π. Meshin (a): 97 × 97 × 77; mesh in (b): 401 × 401 × 321.

(a)0 0.5 1 1.5

−1

−0.5

0

0.5

1

z

Wav

efie

ld

(b)0 0.5 1 1.5

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

z

Wav

efie

ld

(c)0.5 1 1.5

−0.2

−0.1

0

0.1

0.2

z

Wav

efie

ld

(d)0 0.5 1 1.5

−0.1

−0.05

0

0.05

0.1

z

Wav

efie

ld

Fig. 11. Example 3: the real part of G11 at (a) line x = 1 and y = 1 and (b) line x = 0.75 andy = 1; the real part of the G31 at (c) line x = 1 and y = 1 and (d) line x = 0.75 and y = 1. Solidline: FDTD solution; circle line: the Huygens sweeping solution.

5. Conclusion. Based on Hadamard’s method, we extended Babich’s ansatz tothe vectorial point-source Maxwell’s equations (1.1). We developed a new Hadamard’sansatz to form the fundamental solution of the Cauchy problem for the time-domainMaxwell’s wave equations (2.1) and (2.2) in the region close to the source. Governingequations for the unknowns in Hadamard’s ansatz were derived. The initial data forthose unknowns were then determined based on a condition of matching Hadamard’sansatz and the homogeneous-medium Green’s function. Consequently, the Fouriertransform of Hadamard’s ansatz in time directly gives the Hadamard–Babich ansatzfor the FDPS Maxwell’s equations (1.1). Next, we elucidated the relation betweenthe Hadamard–Babich ansatz (2.95) and the Babich-like ansatz (1.3). Finally, in-corporating the first two terms of the Hadamard–Babich ansatz into a planar-basedHuygens sweeping algorithm, we numerically solved the FDPS Maxwell’s equations

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(a)x

z

0 0.5 1 1.5 2

0

0.5

1

1.5−0.2

−0.1

0

0.1

0.2

(b)x

z

0 0.5 1 1.5 2

0

0.5

1

1.5−0.2

−0.1

0

0.1

0.2

Fig. 12. Example 3: the real part of (a) G11 at y = 1 and (b) G31 at y = 1, computed bythe Huygens sweeping method. Mesh 193 × 193 × 154. Primary source r0 = (1.0, 1.0, 0.2)T andk0 = 24π.

(1.1) at high frequencies in the region where caustics occur. Numerical experimentsdemonstrated the accuracy of our method.

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ULTISCALE ODEL. SIMUL Vol. 16, No. 2, pp. 727{751 …Apr 26, 2018 · WANGTAO LU domain...

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