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Coupling at the
Interfaces (7)
⎧
T
⎪MU
1 1,link
−C1
Λ= 0
⎪
T
⎨MU
2 2,link
−C2Λ=
0
⎪ ∆ t ∆ t
⎪− CU − CU = CU + CU
⎩ 2 2
1 1,link 2 2,link 1 1,free 2 2,free
• By condensing on the Lagrange multipliers one obtains formally:
1 T
U
−
1,link
= M1 C1
Λ
1 T
U
−
2,link
= M2 C2
Λ
∆t
∆t
− CM C Λ− CM
2 2
C Λ = CU
+ CU
• By posing:
−1 T
−1
T
1 1 1 2 2 2 1 1,free 2 2,free
∆t
−1 T
−1
T
H ( C1M1 C1 + C2M2 C2
)
2
d CU
+ C U
1 1,free 2 2,free
one obtains the linear system: HΛ=
d
(Steklov-Poincaré operator)
which upon solution yields the Lagrange multipliers
the link accelerations
i,link
Λ
, then
115
U 116
Coupling at the Interfaces (8)
Summarizing, the problem may be written in matrix form:
• First, solve the problem
without links (free).
Symbolically:
⎡M1 0 0⎤⎡U
⎤
1,free ⎡F1⎤
⎢
0 M
2
0
⎥⎢
U
⎥ ⎢
2,free
F
⎥
⎢ ⎥⎢
⎥ =
⎢
2
⎥
⎢⎣ 0 0 0⎥⎦ ⎢ 0 ⎥
⎣ ⎦
⎢⎣ 0 ⎥⎦
Standard
problem! Use
single-domain
operators.
• Then, solve the problem with links (link). Symbolically:
T
⎡M1 0 −C ⎤⎡
1
U
⎤
1,link
⎡ 0 ⎤
∆t
⎢ T ⎥⎢ ⎥ ⎢ ⎥
0 M2 − C2 U
2,link
0
2
⎢ ⎥⎢
⎥ =
⎢ ⎥
⎢−C1 −C2 0 ⎥⎢
Λ ⎥ ⎢CU
⎣ 1 1,free
+ CU
⎥
⎣ ⎦⎣ ⎦
2 2,free⎦
Thanks to diagonal
mass matrix, the
problem with links is
confined only to the
interfaces!
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