Lecture Notes in Differential Equations - Bruce E. Shapiro

115
Substituting (13.66) into (13.65)
Let
Then
|y(t) − z(t)| ≤ M
∫ t
t 0
|y(s) − z(s)|ds (13.67)
f(t) = |y(t) − z(t)| (13.68)
f(t) ≤ M
∫ t
t 0
f(s)ds (13.69)
Then f satisfies the condition for the Gronwall inequality with K = 0 and
g(t) = M, which means
f(t) ≤ K exp
∫ t
a
g(s)ds = 0 (13.70)
Since f(t) is an absolute value it can never be negative so it must be zero.
for all t. Hence
0 = f(t) = |y(t) − z(t)| (13.71)
y(t) = z(t) (13.72)
for all t. Thus any two solutions are identical, i.e, the solution is unique.