Lecture Notes in Differential Equations - Bruce E. Shapiro

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But δ(t) is an absolute value, so it can never take on a negative value. The
integral is the area under the curve from 0 to t. The only way this area can
be zero is if the
δ(t) = 0 (13.40)
for all t. Hence
for all t. Thus the solution is unique.
z(t) = y(t) (13.41)
Theorem 13.2. Gronwall Inequality. Let f, g be continuous, real functions
on some interval [a, b] that satisfy
f(t) ≤ K +
∫ t
for some constant K ≥ 0. Then
(∫ t
f(t) ≤ Kexp
Proof. Define the following functions:
a
f(s)g(s)ds (13.42)
a
)
g(s)ds
(13.43)
Then
F (t) = K +
G(t) =
∫ t
a
∫ t
a
f(s)g(s) (13.44)
g(s)ds (13.45)
F (a) = K (13.46)
G(a) = 0 (13.47)
F ′ (t) = f(t)g(t) (13.48)
G ′ (t) = g(t) (13.49)
By equation (13.42) we are given that
f(t) ≤ F (t) (13.50)
hence from equation (13.48)
F ′ (t) = f(t)g(t) ≤ F (t)g(t) = F (t)G ′ (t) (13.51)
where the last step follows from (13.49). Hence
F ′ (t) − F (t)G ′ (t) ≤ 0 (13.52)