Lecture Notes in Differential Equations - Bruce E. Shapiro

112 LESSON 13. UNIQUENESS OF SOLUTIONS*
where the next-to-last step follows because inside the integral |s| < 1.
Next, we define a function F (t) such that
F (t) =
∫ t
0
δ(s)ds (13.26)
Since F is an integral of an absolute value,
F (t) ≥ 0 (13.27)
Then
F ′ (t) = d dt
∫ t
0
δ(s)ds = δ(t) (13.28)
Since by equation (13.25) δ(t) ≤ F (t), we arrive at
F ′ (t) = δ(t) ≤ F (t) (13.29)
Therefore
From the product rule,
F ′ (t) − F (t) ≤ 0 (13.30)
d [
e −t F (t) ] = e −t F ′ (t) − e −t F (t)
dt
(13.31)
= e −t [F ′ (t) − F (t)] (13.32)
Integrating both sides of the equation from 0 to t,
∫ t
0
≤ 0 (13.33)
d [
e −s F (s)ds ] ≤ 0 (13.34)
dt
e −t F (t) − e 0 F (0) ≤ 0 (Fund. Thm. of Calc.) (13.35)
e −t F (t) ≤ 0 (F (0)=0) (13.36)
F (t) ≤ 0 (divide by the exponential) (13.37)
Now compare equations (13.27) and (13.37); the only consistent conclusion
is that
F (t) = 0 (13.38)
for all t. Thus
∫ t
0
δ(s)ds = 0 (13.39)