Existence and Uniqueness of Eddy Current Problems in Bounded ...

10 MICHAEL KOLMBAUER
unique solution of the problem: For given u 1 find u 2 , such that
⎧
curl curl u 2 = 0, in Ω 2
⎪⎨ div u 2 = 0, in Ω 2
(9)
u 2 × n = u 1 × n, on Γ I
⎪⎩
u 2 × n = 0, on ∂Ω 2
Here we use the notation u i := u| Ωi for i = 1, 2.
Lemma 9. The mapping H is bounded, i.e.
‖H(u 1 )‖ H(curl,Ω2) ≤ c‖u 1 ‖ H(curl,Ω1), ∀u 1 ∈ H(curl, Ω 1 )
Proof. Since H(u 1 ) = u 2 is the unique solution of (9), it is classical to deduce that
the following estimate holds
‖u 2 ‖ H(curl,Ω2) ≤ c‖u 1 × n‖
H
− 1 2
‖ (div Γ,Γ I ) .
Using the trace theorem the desired result follows:
‖u 1 × n‖ − 1 ≤ c‖u 1‖
H 2 (div ‖ Γ,Γ I ) H(curl,Ω1).
The mapping H allows to define the space of curl curl-harmonic extended functions
⎧
⎫
u ∈ H(curl, Ω) :u 1 ∈ H(curl, Ω 1 ),
⎪⎨
u 2 = H(u 1 ),
⎪⎬
Ṽ 0 :=
.
(u, w) L2(Ω) = 0, ∀w ∈ W(Ω 1 ),
⎪⎩
⎪⎭
u × n = 0 on ∂Ω
Using Ṽ 0 , we can state the variational problem as follows: Find u ∈ L 2 ((0, T ), Ṽ 0 )
with weak derivative ˙u ∈ L 2 ((0, T ), Ṽ0), ∗ such that
[
] ∫
∫
∂u
σ 1
∫Ω 1
∂t v + ν 1(|curl u|)curl u · curl v dx+ curl u·curl v dx = f 1 ·v dx
Ω 2 Ω 1
for all v ∈ Ṽ 0 . Consequently by using u 2 = H(u 1 ) we can reduce the problem to
one with support only in the conducting domain Ω 1 . By recalling the definition of
¯V (see (5)), we can state the variational form: Find u 1 ∈ L 2 ((0, T ), ¯V) with weak
derivative ˙u 1 ∈ L 2 ((0, T ), ¯V ∗ ), such that
[
∂u
]
1
σ 1
∫Ω
(10)
1
∂t v 1 + ν 1 (|curl u 1 |)curl u 1 · curl v 1 dx
∫
∫
+ curl H(u 1 ) · curl H(v 1 ) dx = f 1 · v 1 dx
Ω 2 Ω 1
for all v 1 ∈ ¯V.
In order to apply Theorem 1 to the variational setting (10) the crucial points are
to show boundedness and coercivity of the bilinear form. Boundedness follows by
the boundedness of the nonlinear part and the boundedness of the pde-harmonic
extension as stated in Lemma 9. In order to show coercivity, we proceed as in the
unbounded case. Since we have the non-negativity property
∫
curl H(u 1 ) · curl H(u 1 ) dx = ‖ curl H(u 1 )‖ 2 L ≥ 0
2(Ω 2)
Ω 2
we again obtain the estimate
∫
∫
[ν 1 (|curl u 1 |)curl u 1 · curl u 1 ] dx+ curl H(u 1 )·curl H(u 1 ) dx ≥ c‖ curl u 1 ‖ 2 L . 2(Ω 1)
Ω 1 Ω 2
□