Partial Differential Equations - Modelling and ... - ResearchGate

136 S. Lapin et al.
Ω
R
γ
Fig. 3. Space Λ is the space of the piecewise constant functions defined on every
union of half-edges with common vertex.
∫
e
φ(x)dx ≈ 1 3∑
3 meas(e) φ(a i ) ≡ S e (φ),
where the a i ’s are the vertices of e and φ(x) is a continuous function on e.
Similarly,
∫
φ(x)dx ≈ 1 2∑
2 meas(∂e) φ(a i ) ≡ S ∂e (φ),
∂e
where a i ’s are the endpoints of the segment ∂e and φ(x) is a continuous
function on this segment.
We use the notations:
S i (φ) = ∑
S e (φ), i =1, 2, and S Γext (φ) = ∑
S ∂e (φ).
e∈T ih ∂e⊂Γ ext
Now, the fully discrete problem reads as follows: Let u 0 ih = u1 ih = 0,
i =1, 2. For n =1, 2,...,N − 1, find (u n+1
1h
,un+1 2h
,λn+1 h
) ∈ V 1h × V 2h × Λ h
such that
ε 1
∆t 2 S 1((u n+1
1h
+ 1
+
γ
∆t 2 S 2(ε(x)(u n+1
2h
√
ε 1 µ −1
1
S Γext ((u n+1
1h
2∆t
i=1
i=1
− 2un 1h + u n−1
1h
)w 1h)+S 1 (µ 1 −1 ∇un 1h ·∇w 1h )+
− 2un 2h + u n−1
2h )w 2h)+S 2 (µ −1 (x)∇u n 2h ·∇w 2h )+
∫
− un−1 1h
)w 1h)+ λ n+1
h
(w 2h − w 1h )dγ =
γ
= S 1 (f1 n w 1h )+S 2 (f2 n w 2h )
∫
for all w 1h ∈ V 1h ,w 2h ∈ V 2h , (9)
ζ h (u n+1
2h
− un+1 1h )dγ = 0 for all ζ h ∈ Λ h . (10)
Note that in S 2 (ε(x)(u n+1
2h
− 2u n 2h + un−1 2h
)w 2h) wetakeε(x) =ε 2 if a
triangle e ∈T 2h lies in Ω 2 and ε(x) =ε 1 if it lies in R \ Ω 2 , and similarly for
S 2 (µ −1 (x)∇u n 2h ∇w 2h).