Lecture Notes in Differential Equations - Bruce E. Shapiro

107
because the right hand side does not depend on n. But since φ n → φ, the
left hand side becomes |φ(s) − φ(t)|, i.e,
|φ(s) − φ(t)| ≤ M|s − t| (12.57)
To show that φ(t) is continuous we need to show that for every ɛ > 0, there
exists a δ > 0 such that whenever |s − t| < δ, then |φ(s) − φ(t)| < ɛ.
Let ɛ > 0 be given and define δ = ɛ/M. Then
|s − t| < δ =⇒ |φ(s) − φ(t)| ≤ M|s − t| ≤ Mδ = ɛ (12.58)
as required. Hence φ(t) is continuous.
Proof of the Fundamental Existence Theorem (theorem (12.1)). We
have already shown that the sequence φ n → φ converges to a continuous
function on R. To prove the existence theorem we need only to show that
φ satisfies the initial value problem (12.3), or equivalently, the integral
equation
∫ t
φ(t) = y 0 + f(s, φ(s))ds (12.59)
t 0
Let us define the function
F (t) = y 0 +
∫ t
t 0
f(s, φ(s))ds (12.60)
Since F (t 0 ) = y 0 , F satisfies the initial condition, and since
F ′ (t) = f(t, φ(t)) = φ ′ (t) (12.61)
F also satisfies the differential equation. If we can show that F (t) = φ(t)
then we have shown that φ solves the IVP.
We consider the difference
∫ t
∫ t
|F (t) − φ n+1 (t)| =
∣ y 0 + f(s, φ(s))ds − y 0 − f(s, φ n (s))ds
∣ (12.62)
t 0 t 0
∫ t
=
∣ (f(s, φ(s)) − f(s, φ n (s))) ds
∣ (12.63)
t 0
≤
∫ t
≤ K
t 0
|f(s, φ(s)) − f(s, φ n (s))| ds (12.64)
∫ t
t 0
|φ(s) − φ n (s)| ds (12.65)