Lecture Notes in Differential Equations - Bruce E. Shapiro

105
Thus
lim φ n = φ 0 + lim
n→∞ n→∞
n∑
∞∑
s n (t) = φ 0 + s n (t) = φ 0 + S(t) (12.36)
k=1
The left hand side of the equation exists if and only if the right hand side
of the equation exists. Hence lim n→∞ φ n exists if and only if S(t) exists,
i.e, S(t) converges.
k=1
Lemma 12.6. With s n , K, and M as previously defined,
|s n (t)| = |φ n − φ n−1 | ≤ K n−1 M |t − t 0| n
Proof. For n = 1, equation (12.37) says that
n!
(12.37)
|φ 1 − φ 0 | ≤ K 0 M |t − t 0| 1
= M|t − t 0 | (12.38)
1!
We have already proven that this is true in Lemma (12.4).
For n > 1, prove inductively. Assume that (12.37) is true and then prove
that
|φ n+1 − φ n | ≤ K n M |t − t 0| n+1
(12.39)
(n + 1)!
Using the definition of φ n and φ n+1 ,
∫ t
∫ t
|φ n+1 − φ n | =
∣ y 0 + f(s, φ n+1 (s))ds − y 0 − f(s, φ n (s))ds
∣ (12.40)
t 0 t 0
∫ t
=
∣ [f(s, φ n+1 (s))ds − f(s, φ n (s))]ds
∣ (12.41)
t 0
≤
≤
∫ t
t 0
|f(s, φ n+1 (s))ds − f(s, φ n (s))|ds (12.42)
∫ t
t 0
K|φ n+1 (s) − φ n (s)|ds (12.43)
where the last step follows because f is Lipshitz in y. Substituting equation
(12.37) gives
∫ t
|φ n+1 − φ n | ≤ K K n−1 M |s − t 0| n
ds (12.44)
t 0
n!
= Kn M
n!
= Kn M
n!
∫ t
|s − t 0 | n ds
t 0
(12.45)
|t − t 0 | n+1
n + 1
(12.46)