Partial Differential Equations - Modelling and ... - ResearchGate

74 J.Ch. Gilbert and P. Joly
+∞∑
lim Q m(x) =Q ∞ (x) ≡ x +2 (−1) l x l+1
m→+∞ (2l +2)! =2(1− cos √ x). (14)
l=1
Remark 3. Setting P ∞ (x) =2 1−cos √ x
x
and taking (formally) the limit of (7)
when m → +∞, we obtain the scheme
u n+1
h
− 2u n h + un−1 h
∆t 2 + A h P ∞ (∆t 2 A h )=0. (15)
This scheme is, in fact, an exact scheme for the differential equation (2). It
suffices to remark that
∣
sin ( A 1 2 h
t n+1) − 2 sin ( A 1 2
h
t n) + sin ( A 1 2
h
t n−1)
= −
[2 − cos ( A 1 2 h
∆t )] sin ( A 1 2
h
t n)
cos ( A 1 2 h
t n+1) − 2 cos ( A 1 2
h
t n) + cos ( A 1 2
h
t n−1)
[
∣ = − 2 − cos ( A 1 2
h
∆t )] cos ( A 1 2
h
t n) ,
so that any solution of (2), of the form (for some a and b in V h )
satisfies
u h (t) =cos ( A 1 2
h
t ) a + sin ( A 1 2
h
t ) b
u h (t n+1 ) − 2u h (t n )+u h (t n−1 )
∆t 2 = − ( A h ∆t 2) [ −1
2 − cos ( A 1 2
h
∆t )] A h u h (t n ),
that is to say
u h (t n+1 ) − 2u h (t n )+u h (t n−1 )
∆t 2 = −A h P ∞ (∆t 2 A h ).
⊓⊔
Since 0 ≤ Q ∞ (x) ≤ 4, if we define α ∞ by (19) for m =+∞ we have
α ∞ =+∞. Unfortunately, this does not mean, as we are going to see, that
α m → +∞ when m → +∞. In fact, to describe the behaviour of α m ,wehave
to distinguish between the even and odd sequences α 2m and α 2m+1 . Our first
observation is that the convergence of the sequences Q 2m (x) andQ 2m+1 (x) is
monotone. Indeed, for m ≥ 1
Q 2m−1 (x) − Q 2m+1 (x) =2 x2m
4m!
[
1 −
]
x
(4m + 1)(4m +2)
which shows that Q 2m+1 (x) is a strictly decreasing sequence for large m:
Q 2m+1 (x) x.