Partial Differential Equations - Modelling and ... - ResearchGate

Optimal Higher Order Time Discretizations 75
In particular, since (4m + 1)(4m +2)>π 2 for m ≥ 1:
Q ∞ (π 2 ) =
lim Q 2m+1(π 2 )=4 =⇒ Q 2m+1 (π 2 ) > 4,
m→+∞
which shows, using the definition (10), that
α 2m+1 ≤ π 2 , for m ≥ 1.
Moreover, by the definition of α m , we know that Q m (α m )=0or4.Onthe
other hand, since the sequence Q 2m+1 (x) is decreasing, for any x ∈ [0,π 2 ], we
have
Q 2m+1 (x) >Q ∞ (x) =2(1− cos √ x) in [0,π 2 ].
This makes impossible Q 2m+1 (α 2m+1 ) = 0, which implies that
Q 2m+1 (α 2m+1 )=4.
Finally, the inequality
implies
Q 2m+1 (x) 4.
Let α odd ∈ [4, 4π 2 ] be any accumulation point of α 2m+1 , since the convergence
of Q m to Q ∞ is uniform in any compact set, we get:
In the same way
Q ∞ (α odd ) =⇒ (since α odd ∈ [4,π 2 ]) α odd = π 2 .
x2m+1
Q 2m+2 (x) − Q 2m (x) =2
(4m +2)!
[
1 −
]
x
(4m + 3)(4m +4)
shows that the sequence Q 2m (x) is strictly increasing for large m:
Q 2m+2 (x) >Q 2m (x)
assoonas(4m + 3)(4m +4)>x.
In particular, as soon as m ≥ 1,
Q ∞ (4π 2 ) =
lim Q 2m(4π 2 )=0 =⇒ Q 2m (4π 2 ) < 0,
m→+∞
which shows that
α 2m ≤ 4π 2 , m ≥ 1,
while the inequality Q 2m (x) < 2(1 − cos √ x) ≤ 4in[0,π 2 ]form ≥ 1 implies
that
Q 2m (α 2m )=0.