Lecture Notes in Differential Equations - Bruce E. Shapiro

92 LESSON 11. PICARD ITERATION
Let us change the variable t to s,
dφ(s)
ds
= f(s, φ(s)) (11.4)
If we multiply by ds and integrate from s = t 0 to s = t,
∫ t
t 0
∫
dφ(s) t
ds
ds = f(s, φ(s))ds (11.5)
t 0
By the fundamental theorem of calculus, since φ(s) is an antiderivative of
dφ(s)/ds, the left hand side becomes
∫ t
t 0
dφ(s)
ds ds = φ(t) − φ(t 0) = φ(t) − y 0 (11.6)
where the second equality follows from the second line of equation (11.3).
Comparing the right-hand sides of equations (11.5) and (11.6) we find that
φ(t) − y 0 =
∫ t
t 0
f(s, φ(s))ds (11.7)
Bringing the y 0 to the right hand side of the equation gives us equation
(11.2) which was the equation we needed to derive. This completes the
proof of part (a).
To prove part (b) we assume that equation (11.2) is true and need to show
that equation (11.1) follows as a direct consequence. If we differentiate
both sides of (11.2),
d
dt φ(t) = d (
y 0 +
dt
= dy 0
dt + d dt
∫ t
0
∫ t
0
)
f(s, φ(s))ds
(11.8)
f(s, φ(s))ds (11.9)
= φ(t, φ(t)) (11.10)
where the last equation follows from the fundamental theorem of calculus.
Changing the name of the variable from φ to y in (11.10) gives us y ′ =
f(t, y), which is the first line (11.1).
To prove that the second line of (11.1) follows from (11.2), we substitute
t = t 0 in (11.2).
∫ t0
φ(t 0 ) = y 0 + f(s, φ(s))ds = y 0 (11.11)
0