Untitled - Cdm.unimo.it

Time-Dependent Problems 241
(10.5.10)
d
dt
⎡
pn(η
⎢
⎣
(n)
0 ,t)
pn(η (n)
1 ,t)
pn(η (n)
2 ,t)
qn(η (n)
0 ,t)
qn(η (n)
⎤ ⎡
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ = ζ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
1 ,t) ⎥ ⎢
⎥ ⎢
⎦ ⎣
qn(η (n)
2 ,t)
− ˜ d (1)
00 − γ − ˜ d (1)
01 − ˜ d (1)
02 γL/ζ 0 0
− ˜ d (1)
10
− ˜ d (1)
20
− ˜ d (1)
11 − ˜ d (1)
12 0 0 0
− ˜ d (1)
21 − ˜ d (1)
22 0 0 0
0 0 0 ˜ d (1)
00
0 0 0 ˜ d (1)
10
0 0 γR/ζ ˜ d (1)
20
˜d (1)
01
˜d (1)
11
˜d (1)
21
˜d (1)
22
˜d (1)
02
˜d (1)
12
− γ
⎤⎡
pn(η
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎥⎢
⎦⎣
(n)
0 ,t)
pn(η (n)
1 ,t)
pn(η (n)
2 ,t)
qn(η (n)
0 ,t)
qn(η (n)
⎤
⎥
1 ,t) ⎥
⎦
qn(η (n)
2 ,t)
Convergence estimates for the error |R|pn − V 2 w + |L|qn − W 2 w, are given in
the Legendre case (w ≡ 1) when |RL| < 1 and γ := 1
2 ζn(n + 1) |RL|.
For the proof of stability, we argue as follows. Let us assume for simplicity that L > 0
and R > 0 with LR < 1. By (3.5.6), we first note that ˜w (n)
0
Then, formula (3.5.1) with w ≡ 1 yields ∀t ∈]0,T]
(10.5.11)
⎛
d
⎝R
dt
n
j=0
1
∂pn
= −2ζR
−1
− 2Rγ ˜w (n)
0 pn(−1,t) − Lqn(−1,t)
+
=
p 2 n(η (n)
j ,t) ˜w (n)
j
∂x pn dx + 2ζL
+ L
1
−1
n
pn(−1,t) − 2Lγ ˜w (n)
n
q
j=0
2 n(η (n)
j ,t) ˜w (n)
j
= ˜w(n)
n = ζ γ √ RL.
⎞
⎠
∂qn
∂x qn dx
qn(1,t) − Rpn(1,t) qn(1,t)
− ζRp 2 n(1,t) + 2γRL ˜w (n)
n pn(1,t)qn(1,t) + L(ζ − 2γ ˜w (n)
n )q 2
n(1,t)
− ζLq 2 n(−1,t) + 2γRL ˜w (n)
0 pn(−1,t)qn(−1,t) + R(ζ − 2γ ˜w (n)
0 )p 2
n(−1,t)
=