Lecture Notes in Differential Equations - Bruce E. Shapiro

96 LESSON 11. PICARD ITERATION
Continuing as before, use φ 3 to calculate f(s, φ 3 (s)) and then φ 4 :
f(s, φ 3 (s)) = −2sφ 3 (s) = −2s
(1 − s 2 + 1 2 s4 − 1 )
6 s6
(11.38)
φ 4 (s) = y 0 +
= 1 − 2
∫ t
0
∫ t
0
f(s, φ 3 (s))ds (11.39)
(s − s 3 + 1 2 s5 − 1 6 s7 )
ds (11.40)
= 1 − t 2 + t2 2 − t6 6 + t8
24
That pattern that appears to be emerging is that
φ n (t) = t2 · 0
0!
− t2·1
1!
+ t2·2
2!
− t2·3
3!
+ · · · + (−1)n t 2n
n!
=
n∑ (−1) k t 2k
k=0
k!
(11.41)
(11.42)
The only way to know if (11.42) is the correct pattern is to plug it in and
see if it works. First of all, it works for all of the n we’ve already calculated,
namely, n = 1, 2, 3, 4. To prove that it works for all n we use the principal
of mathematical induction: A statement P (n) is true for all n if and
only if (a) P (1) is true; and (b) P (n) =⇒ P (n − 1). We’ve already proven
(a). To prove (b), we need to show that
φ n+1 (t) =
n+1
∑
k=0
(−1) k t 2k
k!
(11.43)
logically follows when we plug equation (11.42) into (11.15). The reason for
using (11.15) is because it gives the general definition of any Picard iterate
φ n in terms of the previous iterate. From (11.15), then
φ n+1 (t) = y 0 +
∫ t
t 0
f(s, φ n (s))ds (11.44)
To evaluate (11.44) we need to know f(s, φ n (s)), based on the expression
for φ n (s) in (11.42):
f(s, φ n (s)) = −2sφ n (s) (11.45)
n∑ (−1) k s 2k
= −2s
k!
k=0
(11.46)
n∑ (−1) k s 2k+1
= −2
k!
(11.47)
k=0