Lecture Notes in Differential Equations - Bruce E. Shapiro

97
We can then plug this expression for f(s, φ n (s)) into (11.44) to see if it
gives us the expression for φ n+1 that we are looking for:
φ n+1 (t) = 1 − 2
n∑ (−1) k
k=0
k!
∫ t
0
s 2k+1 ds (11.48)
n∑ (−1) k t 2k+2
= 1 − 2
k! 2k + 2
k=0
(11.49)
n∑ (−1) k+1 t 2k+2
= 1 +
k! k + 1
k=0
(11.50)
= −10 t ( 2˙0)
n∑ (−1) k+1
+
0! (k + 1)! t2(k+1) (11.51)
The trick now is to change the index on the sum. Let j = k + 1. Then
k = 0 =⇒ j = 1 and k = n =⇒ j = n + 1. Hence
k=0
φ n+1 (t) = −10 t ( 2˙0)
0!
n+1
∑
+
j=1
(−1) j n+1
∑
t 2j (−1) j t 2j
=
(j)!
j!
j=0
(11.52)
which is identical to (11.43).
equation (11.42), is correct:
This means that our hypothesis, given by
n∑ (−1) k t 2k
φ n (t) =
k!
k=0
(11.53)
Our next question is this: does the series
φ(t) = lim
n→∞ φ n(t) =
∞∑ (−1) k t 2k
k=0
k!
∞∑ (−t 2 ) k
=
k!
k=0
(11.54)
converge? If the answer is yes, then the integral equation, and hence the
IVP, has a solution given by φ(t). Fortunately equation (11.54) resembles
a Taylor series that we know from calculus:
e x =
∞∑
k=0
x k
k!
(11.55)
Comparing the last two equations we conclude that the series does, in fact,
converge, and that
φ(t) = e −t2 (11.56)