ON MAXWELL EQUATIONS WITH THE TRANSPARENT ...

Maxwell Equations with the Transparent Boundary Condition 287
For any Φ ∈ H(curl, B R ), ˆx × Φ| ΓR is in the trace space H −1/2 (Div; Γ R ), whose norm, for
any λ = ∑ ∞
n=1
∑ n
m=−n a nmU m n + b nm V m n , is defined by
‖ λ ‖ 2 H −1/2 (Div;Γ R) = ∞ ∑
n∑
n=1 m=−n
√
n(n + 1)|anm | 2 +
1
√
n(n + 1)
|b nm | 2 . (2.6)
It is also known that for Φ ∈ H(curl; B R ), the tangential component (ˆx ×Φ) × ˆx| ΓR belongs to
H −1/2 (Curl; Γ R ) which is the dual space of H −1/2 (Div; Γ R ) with respect to the scalar product
in L 2 t (Γ R) [10, Theorem 5.4.2, Lemma 5.3.1]. In the following we will always denote by 〈·, ·〉 ΓR
the duality pairing between H −1/2 (Div; Γ R ) and H −1/2 (Curl; Γ R ).
Let h (1)
n (z) be the spherical Hankel function of the first kind of order n. We introduce the
vector wave functions
M m n (r, ˆx) = ∇ × {xh (1)
n (kr)Y m n (ˆx)}, N m n (r, ˆx) = 1
ik ∇ × Mm n (r, ˆx),
which are the radiation solutions of the Maxwell equation (2.1) in R 3 \{0}.
Given the tangential vector λ = ∑ ∞ ∑ n
n=1 m=−n a nmU m n + b nm Vn m on Γ R , the solution E of
(2.1)-(2.3) in the domain R 3 \ ¯B R can be written as
E(r, ˆx) =
∞∑
n∑
n=1 m=−n h (1)
a nm M m n (r, ˆx)
n (kR) √ n(n + 1) + ikRb nmN m n (r, ˆx)
z n (1) (kR) √ n(n + 1) . (2.7)
The series in (2.7) converges uniformly for r > R if λ ∈ L 2 t(Γ R ) = {u ∈ L 2 (Γ R ) 3 : u · ˆx =
0 on Γ R } (cf., e.g., [9, Theorem 9.17]).
The Calderon operator G e : H −1/2 (Div; Γ R ) → H −1/2 (Div; Γ R ) is the Dirichlet to Neumann
operator defined by
G e (λ) = 1 ˆx × (∇ × E),
ik
where E satisfies (2.1)-(2.3). Since
we have
1
∞
ik ∇ × E = ∑
1
ik ∇ × Mm n = N n m, − 1
ik ∇ × Nm n = M m n ,
n∑
n=1 m=−n
a mn N m n
h (1)
n (kR) √ n(n + 1) −
ikRb mn M m n
z n (1) (kR) √ n(n + 1) .
On the other hand, it is easy to check that the vector wave functions satisfy
Thus
M m n (r, ˆx) = h(1) n (kr)∇ ∂B 1
Yn m (ˆx) × ˆx,
√
n(n + 1)
N m n (r, ˆx) = z n (1) (kr)U m n(n + 1)
n (ˆx) + h (1)
n (kr)Yn m (ˆx)ˆx.
ikr
ikr
ˆx × M m n = √ n(n + 1)h (1)
n (kr)U m n (ˆx), (2.8)
√
n(n + 1)
ˆx × N m n = z n (1) (kr)Vn m (ˆx), (2.9)
ikr