Partial Differential Equations - Modelling and ... - ResearchGate

132 S. Lapin et al.
Ω
R
γ
Ω
2
Γ ext
Fig. 1. Computational domain.
Here ∇u = ( ∂u
∂x 1
, ∂u
∂x 2
), n is the unit outward normal vector on Γ ext .We
suppose that µ i = µ| Ωi , ε i = ε| Ωi are positive constants for all i =1, 2and
f i = f| Ωi ∈ C( ¯Ω i × [0,T]).
Let
{
{
ε 1 if x ∈ Ω 1 ,
µ 1 if x ∈ Ω 1 ,
ε(x) =
and µ(x) =
,ε 2 if x ∈ Ω 2 ,
µ 2 if x ∈ Ω 2 .
We define a weak solution of problem (1) as a function u such that
u ∈ L ∞ (0,T; H 1 ∂u
(Ω)),
∂t ∈ L∞ (0,T; L 2 ∂u
(Ω)),
∂t ∈ L2 (0,T; L 2 (Γ ext )) (2)
for a.a. t ∈ (0,T)andforallw ∈ H 1 (Ω) satisfying the equation
∫
√ ∫
ε(x) ∂2 u
Ω ∂t
∫Ω
2 wdx + µ −1 (x)∇u ·∇wdx + ε 1 µ −1 ∂u
1
Γ ext
∂t
∫Ω
wdΓ = fwdx
(3)
with the initial conditions
u(x, 0) = ∂u (x, 0) = 0.
∂t
Note that the first term in (3) means the duality between (H 1 (Ω)) ∗ and
H 1 (Ω).
Now, using the Faedo–Galerkin method (as in [DL92]), one can prove the
following:
Theorem 1. Under the assumptions (2) there exists a unique weak solution
of problem (1).