ON MAXWELL EQUATIONS WITH THE TRANSPARENT ...

Maxwell Equations with the Transparent Boundary Condition 293
This completes the proof.
□
Theorem 3.1. Let the assumptions (H1)-(H2) be satisfied. Then the problem (1.1)-(1.4), (1.8)
has a unique solution (E,H) such that
E ∈ L 2 (0, T; H D (curl; Ω R )) ∩ H 1 (0, T; L 2 (Ω R ) 3 ),
H ∈ L 2 (0, T; H(curl, Ω R )) ∩ H 1 (0, T; L 2 (Ω R ) 3 ),
E| t=0 = E 0 , H| t=0 = H 0 , and
∫ T
[ ( ∂E
)
√ ]
ε
ε
∂t ,Φ − (H, ∇ × Φ) −
µ 〈(L −1 ◦ G e ◦ L )(ˆx × E),Φ〉 ΓR dt
0
∫ T
0
=
∫ T
0
(J,Φ)dt, ∀Φ ∈ L 2 (0, T; H D (curl; Ω R )), (3.9)
[ ( ∂H
) ]
µ
∂t ,Ψ + (∇ × E,Ψ) dt = 0, ∀Ψ ∈ L 2 (0, T; L 2 (Ω R ) 3 ). (3.10)
Here (L −1 ◦G◦L )(ˆx×E) ∈ L 2 (0, T; H −1/2 (Div; Γ R )). Moreover, (E,H) satisfies the following
stability estimate
(
‖ ∂t E ‖ L2 (Ω R) + ‖ ∇ × E ‖ L2 (Ω R) + ‖ ∂ t H ‖ L2 (Ω R) + ‖ ∇ × H ‖ L2 (Ω R))
max
0≤t≤T
≤ C‖(E 0 ,H 0 )‖ ΩR + C‖∂ t J‖ L 1 (0,T;L 2 (Ω R)), (3.11)
where
‖(E 0 ,H 0 )‖ ΩR = ‖E 0 ‖ H(curl;ΩR) + ‖H 0 ‖ H(curl;ΩR).
Proof. Our starting point is the solution E ′ L ,H′ L
of the following scattering problem
εsE ′ L − ∇ × H′ L = J L
in R 3 \ ¯D, (3.12)
µsH ′ L + ∇ × E′ L = 0 in R3 \ ¯D, (3.13)
n × E ′ L = 0 on Γ D, (3.14)
√ µ
ε ˆx × H′ L = G e(ˆx × E ′ L ) − √ µ
ε ˆx × ĤL on Γ R . (3.15)
Since k = i √ εµs, by Lemma 3.3, there exists a constant C independent of s such that
‖ ∇ × E ′ ‖ L L 2 (Ω R) + ‖ sE ′ ‖ L L 2 (Ω R)
≤ C (‖ sJ
s
L
‖ L 2 (Ω R) + ‖ sˆx × ĤL ‖ H −1/2 (Div;Γ R) + ‖ |s| 2ˆx )
× ĤL ‖ H −1/2 (Div;Γ R) . (3.16)
1
By (3.12)-(3.13),
‖ ∇ × H ′ ‖ L L 2 (Ω R) + ‖ sH ′ ‖ L L 2 (Ω R)
≤ C ( )
‖JL ‖ L
s 2 (Ω R) + ‖ sJ L
‖ L 2 (Ω R)
1
+ C (‖ sˆx ×
s ĤL ‖ H −1/2 (Div;Γ R) + ‖ |s| 2ˆx )
× ĤL ‖ H −1/2 (Div;Γ R) . (3.17)
1
By [12, Lemma 44.1], E ′ ,H′ are holomorphic functions of s on the half plane Re (s) > γ > 0,
L L
where γ is any positive constant. By Lemma 3.2 the inverse Laplace transform of E ′ ,H′ are L L