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K )τ ψ V −Dϕ K )ψ

− = = + ψ
Dϕ 2 K)ϕ −( 1 I 2 −V
K)ϕ = =( 1 I v j, dsv j + v j, dsτ j, = =
+ n dsn · =(− 1 I 2 ·
=( 1 I 2 v j, dsv j, u Γ τ Γ v j, dsv j, n dsn

II n dsn · τ Γ · u 3(n−1) j=1 n dsn 3(n−1) j=1 · u · u Γ
Equations · τ L + =1 3(n−1) Γ j=1 L + =1 · τ Γ L + =1 3(n−1) Γ j=1 L + =1

Modified L + τ V =1 Γ K )τ − ( 1 I 2 L + Du =1 K)u + ( 1 I 2 L + τ V =1 K )τ − ( 1 I 2 L + Du =1 K)u − ( 1 I 2

=0
v j, ds

· u =0
Problems =0 = Γ =0 v j, ds =0, v j, ds n ds · u

Stokes =0 =0 v j, ds · u τ j, ds V · u =0 · τ Γ · u Γ = Γ

3–D 1) − n ds n ds and Γ Γ n ds and and Du 0 ds

the 1, 3(n · τ · τ and · τ −Dϕ K )ψ · u Γ
for = and Γ and Γ K )ψ − ψ V = and Γ = + and
Equations j 1,L , K)ϕ + Dϕ = =( 1 I 2 ω j, v j, K)ϕ + ω j, τ j, −( 1 I = 2 ψ −V =

Integral = I , =( 1 I 2 =1 ω n ω j, v j, 3(n−1) L j=1 =1 =(− 1 I 2 3(n−1) L j=1 =1 ω j, v j, L =1 ω n

Modified Equations L + =1 ω n + K )τ − 3(n−1) L + j=1 =1 + K)u + L + =1 ω n + K )τ − 3(n−1) L + j=1 =1 + K)u −

2.3.4. Modified τ (1)V (2)( 1 I 2 (1)Du (2)( 1 I 2 τ (1)V (2)( 1 I 2 (1)Du (2)( 1 I 2

Table

IDP INP EDP ENP