First are some analytical preliminaries which are easy consequences of some of the considerations
presented earlier. These things have to do with existence and uniqueness of the initial value
problem
′
x = f (t,x),x(c) = x0
Suppose that f :
[a,b]
× Fn→ Fn satisfies the following two conditions.
|f (t,x)− f (t,x1)| ≤ K |x − x1|, (4.1)
(4.1)
f is continuous. (4.2)
(4.2)
The first of these conditions is known as a Lipschitz condition.
Lemma D.1.1Suppose x :
[a,b]
→ Fnis a continuous function and c ∈
[a,b]
. Then x is asolution to the initial value problem,
x′ = f (t,x),x(c) = x0 (4.3)
(4.3)
if and only if x is a solution to the integral equation,
∫ t
x (t) = x0 + f (s,x(s))ds. (4.4)
c
(4.4)
Proof: If x solves 4.4, then since f is continuous, we may apply the fundamental theorem of
calculus to differentiate both sides and obtain x′
(t)
= f
(t,x (t))
. Also, letting t = c on both sides,
gives x
(c)
= x0. Conversely, if x is a solution of the initial value problem, we may integrate both
sides from c to t to see that x solves 4.4. ■
Theorem D.1.2Let f satisfy 4.1and 4.2. Then there exists a unique solution to the initialvalue problem, 4.3on the interval
[a,b]
.
Proof:Let
||x||
λ≡ sup
{eλt|x(t)| : t ∈ [a,b]}
. Then this norm is equivalent to the usual
norm on BC