ON THE COMPLETENESS OF EIGENMODES IN A PARALLEL PLATE WAVEGUIDE WITH A PERFECTLY MATCHED LAYER TERMINATION Luc Knockaert1 ,
Senior Member, IEEE
and Daniel De Zutter,
Fellow, IEEE
Abstract — An explicit proof of the completeness of the eigenmodes of a grounded parallel plate waveguide with a perfectly matched layer termination is given. The proof is based on a general theorem governing the completeness of sets of complex exponentials. The Green’s function of the structure is then obtained by means of a biorthogonalization procedure. A comparison with the Green’s function of a grounded halfspace indicates that a perfectly matched layer termination simulates the open halfspace efficiently. Index Terms — Perfectly matched layer, parallel plate waveguide, Green’s function, eigenmodes, completeness.
1
INTRODUCTION
In [1] Berenger proposed the perfectly matched layer (PML) to truncate computational domains for use in the numerical solution of Maxwell’s equations, without introducing reflections. The original split-field approach of Berenger was reformulated by Chew and Weedon [2] in terms of complex coordinate stretching and by Sacks et al. [3] in terms of perfectly matched anisotropic absorbers. It was shown by the authors [4] that the complex coordinate stretching and diagonal anisotropy formulations are equivalent in a general orthogonal coordinate system setting. Recently a new method [5] was proposed to derive an eigenmode series expansion for the twodimensional Green’s function of a planar substrate, by using a PML to turn the originally open configuration into a closed parallel plate waveguiding system. However, the proposed technique 1
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2 presents conceptual and theoretical difficulties, mainly due to the fact that complex stretching results in losing the self-adjoint nature of the underlying Sturm-Liouville problem [6]. In this paper we give an explicit proof of the completeness of the eigenmodes of a grounded parallel plate waveguide with a PML termination. The proof is based on a general theorem governing the completeness of sets of complex exponentials [7]. The resulting Green’s function can then be obtained by utilizing a biorthogonal expansion for the Dirac distribution. Comparison with the Green’s function of a grounded halfspace demonstrates that the PML termination judiciously simulates the open halfspace.
2
PARALLEL PLATE WAVEGUIDE WITH PML TERMINATION
Consider the Helmholtz equation governing the determination of the scalar Green’s function in a homogeneous parallel plate waveguide with propagation in the z−direction. We have ∂2G ∂2G + + k 2 G = δ(x − x0 )δ(z − z 0 ) ∂x2 ∂z 2
0 ≤ x, x0 ≤ a
− ∞ ≤ z, z 0 ≤ ∞
(1)
The Green’s function G(x, x0 ; z, z 0 ) is of the form G(x, x0 ; z − z 0 ) and hence with the Fourier transform ˜ x0 ; γ) = G(x,
Z
∞ −∞
0
eiγ(z−z ) G(x, x0 ; z − z 0 )dz
(2)
the defining equation (1) becomes ˜ ∂2G ˜ = δ(x − x0 ) 0 ≤ x, x0 ≤ a + β2G ∂x2 with β =
p
(3)
k 2 − γ 2 . Given appropriate boundary conditions, the Green’s function in the Fourier
domain can be readily obtained. For a parallel plate waveguide with Dirichlet boundary conditions at x = 0 and x = a, the Green’s function in the Fourier domain is ˜ p (x, x0 ; γ) = sin βx< sin β(x> − a) G β sin βa
(4)
with x< = min(x, x0 ), x> = max(x, x0 ). Hence 1 Gp (x, x ; z − z ) = 2π 0
0
Z
∞ −∞
0
˜ p (x, x0 ; γ)dγ e−iγ(z−z ) G
(5)
3 Unfortunately, the Fourier integral (5) is not known in closed form. Nevertheless the Green’s function can be obtained by means of an eigenmode expansion. Since the eigenmode sequence {sin(nπx/a)}, n = 1, 2, . . . forms a complete orthogonal basis in L2 [0, a] we may write ∞ X sin(nπx/a) sin(nπx0 /a) ˜ p (x, x0 ; γ) = 2 G a n=1 β 2 − (nπ/a)2
(6)
yielding
Gp (x, x0 ; z − z 0 ) =
∞ X
0
q
−i k 2 − (nπ/a)2 |z − z 0 |
sin(nπx/a) sin(nπx /a)e i q a n=1 k 2 − (nπ/a)2
(7)
Now if, while keeping the Dirichlet condition at x = 0, we add a perfectly matched layer in the interval [a, d], d > a and a Dirichlet condition at x = d, the equations in the Fourier domain become ˜ ∂2G ˜ = δ(x − x0 ) 0 ≤ x, x0 ≤ a + β2G ∂X 2 ˜ ∂2G ˜ = 0 a≤x≤d + β2G ∂X 2
(8) (9)
where X(x) is a stretched PML coordinate [4] satisfying X(x) = x for 0 ≤ x ≤ a,
X 0 (a+) = 1,
X 00 (a+) = 0
(10)
and which is complex and twice continuously differentiable in [a, d]. It is easy to show that the eigenmode sequence is now {sin(nπx/b)}, n = 1, 2, . . . , with b = X(d). It would therefore seem that formulas (6) and (7) remain valid by replacing a with b, but nothing is less true. The point is that, due to the complex stretching, the equations (8) and (9) taken together do not form a self-adjoint Sturm-Liouville problem [6], which implies that completeness and orthogonality are far from guaranteed. However in our case we have the following Theorem : The sequences {sin(nπx/b)}, n = 1, 2, . . . and {cos(nπx/b)}, n = 0, 1, . . . with b complex are complete over Lp [0, a] provided a ≤ |b|.
Proof : By Theorem 2 of the Appendix the sequence {enπix/b } is complete over Lp [−a, a] if a ≤ |b|. Now the even and odd functions in Lp [−a, a] are clearly spanned by the sequences
{cos(nπx/b)}, n = 0, 1, . . . and {sin(nπx/b)}, n = 1, 2, . . . respectively. Also, any function in Lp [0, a] has a unique even and odd extension in Lp [−a, a], and this completes the proof.
4 Next even when completeness is guaranteed, it is not possible, in the Hilbert space case p = 2, to write δ(x − x0 ) =
∞ 2X sin(nπx/b) sin(nπx0 /b) b n=1
p
0 ≤ x, x0 ≤ a
(11)
since the sequence { 2/b sin(nπx/b)}, n = 1, 2, . . . does not form a complete orthonormal basis for the Hilbert space L2 [0, a] with scalar product hf |gi =
Ra 0
f (x)g(x)dx. However, given a com-
plete sequence {φn (x)} we can find the biorthogonal sequence {ψn (x)} by the rule hφn |ψm i = δn,m . The reproducing kernel K(x, x0 ) =
X
φn (x)ψn (x0 )
(12)
n
exhibits the sieve property
Ra 0
K(x, x0 )φn (x0 )dx0 = φn (x), which makes it a representation of
the Dirac distribution δ(x − x0 ) with respect to this complete set. Hence if we take φn (x) = {sin(nπx/b)}, n = 1, 2, . . . , we have Gpml (X(x), x0 ; z − z 0 ) =
i 2
∞ X
φn (X(x))ψn (x0 )e q
n=1
q
−i k 2 − (nπ/b)2 |z − z 0 |
k 2 − (nπ/b)2
(13)
It is easily seen that the PML Green’s function satisfies the Dirichlet boundary conditions at x = 0 and x = d. Note that besides the condition for completeness a ≤ |b| we must also require sin βx< G β
(19)
The Green’s function for the PML terminated waveguide in the Fourier domain can also be written as ˜ pml (x, x0 ; γ) = sin βx< sin β(x> − b) G β sin βb
(20)
6 ˜ h and G ˜ pml to be close by the usual Lagrange procedure of variation of parameters. Hence, for G to one another, we should require sin β(x − b) ≈ −e−iβx sin βb
0≤x≤a β∈Γ
(21)
where Γ is the branch cut line formed by the negative imaginary axis and the interval [0, k]. A minimal asymptotic requirement is to have lim
β→−i∞
sin β(x − b) =0 sin βb
0 −∞ r→∞
(A1)
where T (r) =
Z
r 1
(Λ(t) − 2Dt)
dt 1 + log r t q
(A2)
1/p + 1/q = 1 and Λ(t) is the number of λn satisfying |λn | ≤ t. Theorem 1 allows the following specialization : Theorem 2 : The sequence {enαx } with α 6= 0 complex, n ∈ Z is complete Lp (p ≥ 1) on an interval of length smaller than or equal to 2π/|α|.
7 Proof : Defining the step function Υ(t) as Υ(t) = 0, t < 0, Υ(t) = 1, t ≥ 0 we can write with ρ = |α| Λ(t) = Υ(t) + 2
∞ X
n=1
Defining ξ(r, a) as ξ(r, a) =
Z
Υ(t − nρ)
(A3)
r 1
Υ(t − a)/t d t = Υ(r − max(a, 1)) log(r/ max(a, 1))
r≥1
(A4)
we can write T (r) = (1 + 1/q) log r − 2D(r − 1) + 2 = (1 + 1/q) log r − 2D(r − 1) + 2
∞ X
ξ(r, nρ)
(A5)
n=1
X
log(r/ max(1, nρ))
1≤n max(1,nρ)≤r
For r > ρ we distinguish two cases : • ρ≥1 We have br/ρc
T (r) = (1 + 1/q) log r − 2D(r − 1) + 2
X
log(r/nρ)
(A6)
n=1
= (1 + 1/q) log(ρ(m + θ)) − 2D(ρ(m + θ) − 1) + 2m log(m + θ) − 2
m X
log n
n=1
where m = br/ρc and θ are respectively the integer and fractional parts of r/ρ. Note that 0 ≤ θ < 1. From the asymptotic Stirling formula m X
1 1 1 log n = m log m − m − log m + log(2π) + O log m! = 2 2 m n=1
¶
(A7)
1 m
¶
(A8)
µ
we obtain T (r) = 2m(1 − Dρ) + (1 + 1/q) log(m + θ) + log m + C + O
µ
For m → ∞ T (r) tends to ∞ for D ≤ 1/ρ and to −∞ for D > 1/ρ and this proves the theorem. • ρ
