22.14. ORDINARY DIFFERENTIAL EQUATIONS 747

where Γ is any piecewise smooth curve from 0 to t. By the Cauchy integral theorem, thisdefinition is well defined and it is clear that ψ (0) = 0, ψ ′ (t) = φ (t) , and ψ is continuouson D1. This shows L is onto.

It only remains to show L is one to one. Suppose Lφ = 0. Since φ (0) = 0,

φ (t) =∫ 1

0φ′ (ts) tds = 0

if t ̸= 0. But φ (0) is given to equal zero. Thus L is one to one as claimed.

Theorem 22.14.2 Let Λ and Z be complex Banach spaces and let W be an open subset ofC×Z×Λ containing (0,y0,λ ). Also let f : W → Z be analytic. Then there exists a uniquey = y(y0,λ ) solving

y′ = f (t,y,λ ) ,y(0) = y0 (22.14.41)

valid for t ∈ Dα ≡ B(0, |α|) where α = α (y0,λ ). Furthermore, the map

(t,y0,λ )→ y(y0,λ )(t)

is analytic.

Proof: Let αs = t and define φ (s) ≡ y(t)− y0. Then y is a solution to 22.14.41 fort ∈ Dα if and only if φ is a solution for s ∈ D1 ≡ B(0,1) to the equations

φ′ (s) = α f (αs,φ (s)+ y0,λ ) , φ (0) = 0.

Let X , Y, and L be given above and define

W̃ ≡ {(α, ŷ0,µ,φ) ∈ C×Z×Λ×X :

for s ∈ D1,(sα, ŷ0 +φ (s) ,µ) ∈W}.

For a given (α, ŷ0,µ,φ) ∈ W̃ ,

{(sα, ŷ0 +φ (s) ,µ) : s ∈ D1}

is a compact subset of W . This is because you have s→ (α, ŷ0 +φ (s) ,µ) is the continuousimage of a compact set which is assumed to be in W . Consequently, the distance from thisset to WC is positive and so if (β ,y0,λ ,ψ) is sufficiently close to (α, ŷ0,µ,φ) in C×Z×Λ×X it follows (β ,y0,λ ,ψ) is also in W̃ . This shows W̃ is an open subset ofC×Z×Λ×X .

Now define F : W̃ → Y (Recall that Y was a space of functions.) by

F (α, ŷ0,µ,φ)(s)≡ Lφ (s)−α f (αs,φ (s)+ ŷ0,µ).

ThenF (0,y0,λ ,0) = Lφ = 0,

and F is analytic in W̃ . Also

D4F (0,y0,λ ,0)ψ = Lψ = ψ′

22.14. ORDINARY DIFFERENTIAL EQUATIONS 747where I is any piecewise smooth curve from 0 to t. By the Cauchy integral theorem, thisdefinition is well defined and it is clear that y(0) = 0, w’ (t) = @(t), and y is continuouson D,. This shows L is onto.It only remains to show L is one to one. Suppose L@ = 0. Since @ (0) = 0,16(t) = | 6! (ts)tds =00if t 4 0. But @ (0) is given to equal zero. Thus L is one to one as claimed. §§Theorem 22.14.2 Let A and Z be complex Banach spaces and let W be an open subset ofCxZ x A containing (0,yo,4). Also let f : W — Z be analytic. Then there exists a uniquey=y(yo,A) solvingy =f (t.y,A),y(0) =yo (22.14.41)valid for t € Dg = B(0,|a|) where & = a(yo,A). Furthermore, the map(t,y0,4) > y(yo,4) (t)is analytic.Proof: Let as = t and define ¢ (s) = y(t) — yo. Then y is a solution to 22.14.41 fort € Dg if and only if ¢ is a solution for s € D; = B(0, 1) to the equations$' (s) = af (@s,(s)+y0,A), 9 (0) =0.Let X, Y, and L be given above and defineW = {(0,50,M, 0) CCXZXAXX:for s € Dj, (sa, yo + (s),H) EC WH.For a given (@,¥0,U,0) € W,{(s0,Yo+9(s),M) 9 € Di}is a compact subset of W. This is because you have s + (@, 99 + @ (5), L) is the continuousimage of a compact set which is assumed to be in W. Consequently, the distance from thisset to WC is positive and so if (B,yo,A, W) is sufficiently close to (a, 9, U,) in CxZ xA xX it follows (B,yo,4, y) is also in W. This shows W is an open subset of CxZ x Ax X.Now define F :W — Y (Recall that Y was a space of functions.) byF (0,90,H, 9) (s) = Lo (s) _ af (as, (s) +0, L).ThenF (0,y0,4,0) = Loe = 0,and F is analytic in W. AlsoDaF (0,y0,4,0) y=Ly = y'