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Optimal Higher Order Time Discretizations 71
u h (t n+1 ) − 2u h (t n )+u h (t n−1 )
∆t 2
= d2 u h
dt 2 (tn )+ ∆t2 d 4 u h
12 dt 4 (tn )+O(∆t 4 ).
Using the equation satisfied by u h , we get the identity
u h (t n+1 ) − 2u h (t n )+u h (t n−1 )
∆t 2
= −A h u h (t n )+ ∆t2
12 A2 hu h (t n )+O(∆t 4 ),
which leads to the following fourth order scheme:
u n+1
h
− 2u n h + un−1 h
∆t 2
+ A h u n h − ∆t2
12 A2 hu n h =0. (6)
This one can be implemented in such a way that each time step involves only
two applications of the operator A h , using Horner’s rule,
( )
u n+1
h
=2u n h − u n−1
h
− ∆t 2 A h I − ∆t2
12 A h u n h.
More generally, an explicit centered scheme of order 2m is given by
u n+1
h
− 2u n h + un−1 h
∆t 2
where the polynomial P m (x) is defined by
Indeed, a Taylor expansion gives
so that
+ A (m)
h (∆t)un h =0, A (m)
h (∆t) =A hP m (∆t 2 A h ), (7)
m−1
∑
P m (x) =1+2 (−1) l x l
(2l +2)! . (8)
l=1
2m+1
∑
u h (t n±1 )=u h (t n )+ (±1) k ∆tk
k!
k=1
u h (t n+1 ) − 2u h (t n )+u h (t n−1 )
∆t 2 =2
m∑
k=1
d k u h
dt k (tn )+O(∆t 2m+2 )
∆t 2k−2
2k!
Since d2k u h
(t n )=(−1) k A k dt 2k
h u h(t n ), we also have
u h (t n+1 ) − 2u h (t n )+u h (t n−1 )
∆t 2 =
or equivalently
= −A h u h (t n )+2
m∑
k=2
(−1) k ∆t2k−2
2k!
d 2k u h
dt 2k (tn )+O(∆t 2m ).
A k hu h (t n )+O(∆t 2m ),