ON MAXWELL EQUATIONS WITH THE TRANSPARENT ...

Maxwell Equations with the Transparent Boundary Condition 289
This implies r|h (1)
n (kr)| 2 is a strictly decreasing function for r ∈ (0, ∞). Consequently,
d
[
r|h (1)
n
dr
(kr)|2]∣ ∣
∣r=R = |h (1)
n (kR)|2 + Re
This completes the proof.
]
[kRh (1)′
n (kR)h(1) n (kR) < 0.
□
Lemma 2.4. Let R > 0, n ∈ Z, k = k 1 + ik 2 , k 1 , k 2 ∈ R such that k 2 > 0, we have
( )
z n (1) (kR)
Im k 1 ≥ 0.
h (1)
n (kR)
Proof. By the definition of the vector wave function ∇ × M m n = ikN m n and (2.8)-(2.9) we
have
Thus we only need to prove
〈∇ × M m n × ˆx, ˆx × Mm n × ˆx〉 Γ R
= ik〈N m n × ˆx, ˆx × M m n × ˆx〉 ΓR = n(n + 1)Rz (1)
n (kR)h (1)
n (kR). (2.12)
Im (k 1 〈∇ × M m n × ˆx, ˆx × Mm n × ˆx〉 Γ R
) ≥ 0.
Since M m n
satisfies the Maxwell equation
∇ × ∇ × M m n − k2 M m n = 0 in R3 \{0}.
By multiplying the above equation by M m n and integrating over Ω R,ρ = B ρ \ ¯B R , we obtain
‖ ∇ × M m n ‖2 L 2 (Ω − R,ρ) k2 ‖M m n ‖2 L 2 (Ω − 〈∇ × R,ρ) Mm n × n,n × Mm n × n〉 Γ R∪Γ ρ
= 0.
Notice that Im(−k 1 k 2 ) = −2k1k 2 2 ≤ 0, we have
−Im (k 1 〈∇ × M m n × n,n × M m n × n〉 ΓR )
≥ Im ( k 1 〈∇ × M m n × n,n × Mm n × n〉 )
Γ ρ ,
which implies, since n = −ˆx on Γ R and n = ˆx on Γ ρ ,
Im (k 1 〈∇ × M m n × ˆx, ˆx × Mm n × ˆx〉 Γ R
)
≥ Im ( k 1 〈∇ × M m n × ˆx, ˆx × M m )
n × ˆx〉 Γρ . (2.13)
By (2.12)
Im ( k 1 〈∇ × M m n × ˆx, ˆx × Mm n × ˆx〉 )
Γ ρ
)
= n(n + 1)Im
(k 1 kρ 2 h (1)′
n (kρ)h(1) n (kρ) .
We are now going to show that
Since h (1)′
n (z) = − n+1
z h(1) n (z) − h (1)
n−1 (z), we have
|kρ 2 h (1)′
n (kρ)h(1) n (kρ)| → 0, as ρ → ∞. (2.14)
|kρ 2 h (1)′
n (kρ)h (1)
n (kρ)| ≤ (n + 1)ρ|h (1)
n (kρ)| 2 + |kρ 2 ||h (1)
n (kρ)||h (1)
n−1 (kρ)|.