Lecture Notes in Differential Equations - Bruce E. Shapiro

114 LESSON 13. UNIQUENESS OF SOLUTIONS*
By the product rule,
d
[
F (t)e −G(t)] = F ′ (t)e −G(t) − F (t)G ′ (t)e −G(t)
dt
(13.53)
= [F ′ (t) − F (t)G ′ (t)]e −G(t) (13.54)
Integrating the left hand side of (13.53)
∫ t
a
≤ 0 (13.55)
d
[
F (s)e −G(s)] ds = F (t)e −G(t) − F (a)e −G(a) (13.56)
ds
= F (t)e −G(t) − K (13.57)
Since the integral of a negative function must be negative,
F (t)e −G(t) − K ≤ 0 (13.58)
which is equation (13.43).
F (t)e −G(t) ≤ K (13.59)
F (t) ≤ Ke G(t) (13.60)
Proof of Theorem (13.1) Suppose that y and z are two different solutions
of the initial value problem. Then
y(t) = y 0 +
z(t) = y 0 +
∫ t
t 0
f(s, y(s))ds (13.61)
∫ t
t 0
f(s, z(s))ds (13.62)
Therefore
∫ t
∫ t
|y(t) − z(t)| =
∣ f(s, y(s))ds − f(s, z(s))ds
∣ (13.63)
t 0 t 0
∫ t
=
∣ [f(s, y(s)) − f(s, z(s))]ds
∣ (13.64)
t 0
≤
∫ t
t 0
|f(s, y(s)) − f(s, z(s))|ds (13.65)
Since |∂f/∂y| is continuous on a closed interval it is bounded by some
number M, and hence f is Lipshitz with Lipshitz constant M. Thus
|f(s, y(s)) − f(s, z(s)| ≤ M|y(s) − z(s)| (13.66)