Lecture Notes in Differential Equations - Bruce E. Shapiro

111
The proof of theorem (13.1) is similar to the following example.
Example 13.2. Show that the initial value problem
}
y ′ (t, y) = ty
y(0) = 1
has a unique solution on the interval [−1, 1].
The solution itself is easy to find; the equation is separable.
∫ ∫ dy
y = tdt =⇒ y = Ce t2 /2
(13.15)
(13.16)
The initial condition tells us that C = 1, hence y = e t2 /2 .
The equivalent integral equation to (13.15) is
y(t) = y 0 +
∫ t
Since f(t, y) = ty, t 0 = 0, and y 0 = 1,
y(t) = 1 +
t 0
f(s, y(s))ds (13.17)
∫ t
0
sy(s)ds (13.18)
Suppose that z(t) is another solution; then z(t) must also satisfy
z(t) = 1 +
∫ t
0
sz(s)ds (13.19)
To prove that the solution is unique, we need to show that y(t) = z(t)
for all t in the interval [−1, 1]. To do this we will consider their difference
δ(t) = |z(t) − y(t)| and show that is must be zero.
δ(t) = |z(t) − y(t)| (13.20)
∫ t
∫ t
=
∣ 1 + sz(s)ds − 1 − sy(s)ds
∣ (13.21)
0
0
∫ t
=
∣ [sz(s)ds − sy(s)]ds
∣ (13.22)
≤
≤
=
0
∫ t
0
∫ t
0
∫ t
0
|s||z(s) − y(s)|ds (13.23)
|z(s) − y(s)|ds (13.24)
δ(s)ds (13.25)