Linear Algebra

D.1 Theory Of Ordinary Differential Equations

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 :

× 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.

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^′

= f. Also, letting t = c on both sides, gives x = 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. ■

Proof: Let

_λ ≡ sup. Then this norm is equivalent to the usual norm on BC described in Example 13.6.2. This means that for the norm given there, there exist constants δ and Δ such that

||x||λδ ≤ ||x|| ≤ Δ ||x||λ

for all x ∈BC

. In fact, you can take δ ≡ e^λa and Δ ≡ e^λb in case λ > 0 with the two reversed in case λ < 0. Thus BC is a Banach space with this norm, _λ. Then let F : BC → BC be defined by

∫ t Fx (t) ≡ x0 + c f (s,x (s))ds.

Let λ < 0. It follows

If t ≥ c, this is no larger than

∫ t λ(t−s) 1 ||x − y ||λ Ke ds ≤ ||x− y||λ |λ-| c

∫ t K ≤ ||x − y ||λ Ke λ(t−s)ds ≤ ||x− y||λ --- a |λ|

If t < c, this equals

∫ c λ(t−s) λs Ke |x(s)− y(s)|e ds t

which is no larger than

|λ|(b−a) K- ||x − y ||λe |λ |

Therefore, it is always the case that

||F x− F y|| ≤ ||x − y||e|λ|(b− a)-K-. λ λ |λ|

If

is chosen larger than Ke

|λ|

(b−a)

, this implies F is a contraction mapping on BC. Therefore, there exists a unique fixed point. With Lemma D.1.1 this proves the theorem. ■