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A Lagrange Multiplier Based Domain Decomposition Method 137
Denote by u 1 , u 2 and λ the vectors of the nodal values of the corresponding
functions u 1h , u 2h and λ h . Then, in order to find u n+1
1 , u n+1
2 and λ n+1 for a
fixed time t n+1 , we have to solve a system of linear equations such as
Au + B T λ = F, (11)
Bu =0, (12)
where matrix A is diagonal, positive definite and defined by
√
(Au, w) = ε 1
∆t 2 S 1(u 1h w 1h )+ 1
∆t 2 S 2(ε(x)u 2h w 2h )+
and where the rectangular matrix B is defined by
∫
(Bu, λ) = λ h (u 2h − u 1h )dΓ,
γ
ε 1 µ −1
1
S Γext (u 1h w 1h ),
2∆t
and vector F depends on the nodal values of the known functions u n 1h , un 2h ,
u n−1
1h
and u n−1
2h .
Eliminating u from the equation (11), we obtain
BA −1 B T λ = BA −1 F, (13)
with a symmetric matrix C ≡ BA −1 B T . Let us prove that C is positive
definite. Obviously, ker C =kerB T . Suppose, that B T λ = 0, then a function
λ h ∈ Λ h corresponding to vector λ satisfies
∫
I ≡ λ h u h dγ =0
γ
for all u h ∈ V 1h . Choose u h equal to λ h in the nodes of T 1h located on γ.
Direct calculations give
I = 1 ∑N λ
[
hi + h i+1
λ 2 (λ i + λ i+1 ) 2 ]
i + h i+1 ,
2 2
2
i=1
where N λ is the number of edges of T 1h on γ, h i is the length of i-th edge and
h Nλ +1 ≡ h 1 , λ Nλ +1 ≡ λ 1 . Thus, the equality I = 0 implies that λ = 0, i.e.
ker B T = {0}.
As a consequence we have
Theorem 2. The problem (9), (10) has a unique solution (u h ,λ h ).
Remark 3. A closely related domain decomposition method applied to the
solution of linear parabolic equations is discussed in [Glo03].