Lecture Notes in Differential Equations - Bruce E. Shapiro

309
We will verify (31.54) by induction. We have already demonstrated it for
k = 1 (equation 31.53). We take (31.54) as an inductive hypothesis and
compute
∫ t
Φ k+1 (t) = y 0 + AΦ k (s)ds (31.55)
t 0
⎡
⎤
∫ t
k∑ 1
= y 0 + ⎣I +
k! Ak (s − t 0 ) k ⎦ y 0 ds (31.56)
t 0
A
j=1
∫ t ∫ t
= y 0 + AIy 0 ds +
t 0
= y 0 + AIy 0 (t − t 0 ) +
= y 0 + AIy 0 (t − t 0 ) +
k∑
t 0 j=1
k∑
j=1
k∑
j=1
1
k! Ak+1 (s − t 0 ) k y 0 ds (31.57)
∫
1
t
k! Ak+1
t 0
(s − t 0 ) k y 0 ds (31.58)
1
(k + 1)! Ak+1 (t − t 0 ) k+1 y 0 (31.59)
k+1
∑ 1
= y 0 + AIy 0 (t − t 0 ) +
j! Aj (t − t 0 ) j y 0 (31.60)
j=2
k+1
∑ 1
= y 0 +
j! Aj (t − t 0 ) j y 0 (31.61)
j=1
⎡
k+1
∑
= ⎣I +
j=1
⎤
1
k! Ak (t − t 0 ) k ⎦ y 0 (31.62)
which completes the proof of equation (31.54). The general existence theorem
(Picard) then says that
⎡
⎤
∞∑ 1
y = ⎣I +
k! Ak (t − t 0 ) k ⎦ y 0 (31.63)
j=1
If we define a matrix M = A(t − t 0 ) and observe that M 0 = I, then
⎡
⎤
∞∑
y = ⎣M 0 1
+
k! Mk ⎦ y 0 = e M y 0 = e A(t−t0) y 0 (31.64)
as required.
j=1