Number Theory: An Introduction to Mathematics

4 Metric Spaces 37

Proof Ifx(t) is a solution of the differential equation (1) which satisfies the initial
condition (2), then by integration we get

x(t 0 )=ξ 0 +

∫t

t 0

φ[τ,x(τ)]dτ.

Conversely, if acontinuousfunctionx(t)satisfies this relation then, sinceφis contin-
uous,x(t)is actually differentiable and is a solution of (1) that satisfies (2). Hence we
need only show that the mapFdefined by

(Fx)(t)=ξ 0 +

∫t

t 0

φ[τ,x(τ)]dτ

has a unique fixed point in the space of continuous functions.
There exist positive constantsM,Lsuch that

|φ(t,ξ)|≤M, |φ′(t,ξ)|≤L

for all(t,ξ)in a neighbourhood of(t 0 ,ξ 0 ), which we may take to beU.If(t,ξ 1 )∈U
and(t,ξ 2 )∈U,then

|φ(t,ξ 2 )−φ(t,ξ 1 )|=

∣

∣

∣

∣

∫ 1

0

φ′(t,( 1 −u)ξ 1 +uξ 2 )(ξ 2 −ξ 1 )du

∣

∣

∣

∣

≤L|ξ 2 −ξ 1 |.

Chooseδ>0 so that the box|t−t 0 |≤δ,|ξ−ξ 0 |≤Mδis contained inUand
alsoLδ<1. TakeI=[t 0 −δ,t 0 +δ]andletC(I)be the complete metric space of
all continuous functionsx:I→Rnwith the distance function

d(x 1 ,x 2 )=sup

t∈I

|x 1 (t)−x 2 (t)|.

The constant functionx 0 (t)=ξ 0 is certainly inC(I).LetEbe the subset of all
x∈C(I)such thatx(t 0 )=ξ 0 and d(x,x 0 )≤Mδ. Evidently ifxn∈Eandxn→x
inC(I),thenx∈E. HenceEis also a complete metric space with the same metric.
MoreoverF(E)⊆E, since ifx∈Ethen(Fx)(t 0 )=ξ 0 and, for allt∈I,

|(Fx)(t)−ξ 0 |=

∣

∣

∣

∣

∫t

t 0

φ[τ,x(τ)]dτ

∣

∣

∣

∣≤Mδ.

Furthermore, ifx 1 ,x 2 ∈E,thend(Fx 1 ,Fx 2 )≤Lδd(x 1 ,x 2 ), since for allt∈I,

|(Fx 1 )(t)−(Fx 2 )(t)|=

∣

∣

∣∣

∫t

t 0

{φ[τ,x 1 (τ)]−φ[τ,x 2 (τ)]}dτ

∣

∣

∣∣

≤Lδd(x 1 ,x 2 ).

SinceLδ<1, the result now follows from Proposition 26.