Partial Differential Equations - Modelling and ... - ResearchGate

138 S. Lapin et al.
4 Energy Inequality
Theorem 3. Let h min denote the minimal diameter of the triangles from T 1h ∪
T 2h . There exists a positive number c such that the condition
∆t ≤ c min{ √ ε 1 µ 1 , √ ε 2 µ 2 } h min (14)
ensures the positive definiteness of the quadratic form
( (u
E n+1 = 1 n+1
2 ε 1h
1S 1
∆t
( ∣
+ 1 2 S 1
(
− ∆t2
8 S 1
µ −1
1
µ −1
1
( u
n+1
∣ ∇ 1h
2
( u
n+1
∣ ∇ 1h
∆t
− un 1h
+ un 1h
− un 1h
)2 ) (
+ 1 ( u
n+1
2 S 2h
2 ε
∆t
)∣ ∣∣∣
2 ) (
+ 1 ∣ ∣∣∣
2 S 2 µ −1 ∇
)∣ ∣∣∣
2 ) ( ∣
− ∆t2 ∣∣∣
8 S 2 µ −1 ∇
which we call the discrete energy.
The system (9), (10) satisfies the energy identity
√
ε 1 µ −1
E n+1 −E n 1
+ S Γext ((u n+1
1h
4∆t
− un−1 1h )2 )=
= 1 2 S 1(f n 1 (u n+1
1h
− un 2h
( u
n+1
2h
( u
n+1
2h
− un−1 1h )) + 1 2 S 2(f2 n (u n+1
2h
)2 ) +
+ un 2h
2
− un 2h
∆t
)∣ ∣∣∣
2 ) −
)∣ ∣∣∣
2 ) ,
(15)
− un−1 2h
)) (16)
and the numerical scheme is stable: There exists a positive number M = M(T )
such that
n−1
∑
E n ≤ M∆t (S 1 ((f1 k ) 2 )+S 2 ((f2 k ) 2 )), ∀n. (17)
γ
k=1
Proof. Let n ≥ 1. From the equation (10) written for t n+1 and t n−1 we obtain
∫
ζ h ((u n+1
2h
− un−1 2h
) − (un+1 1h
− un−1 1h ))dγ = 0 for all ζ h ∈ Λ h . (18)
Choosing
in (9) and
w 1h = un+1 1h
− un−1 1h
, w 2h = un+1 2h
2
ζ h = − λn+1 h
2
in (18), we add these equalities. Using the identities
− un−1 2h
2