Kenneth Kuttler

814 CHAPTER 24. CRITICAL POINTS

By induction, yn (s) ,yn−1 (s) are still in B. This is because

∥yn (t)− y0∥ ≤n

∑k=1∥yk (t)− yk−1 (t)∥

≤n

∑k=1∥ f (y0)∥

1(k−1)!

tkKk−1

≤ ∥ f (y0)∥ teKt < R (24.1.4)

showing that yn (t) stays in B. Then since all values of the iterates remain in B, inductiongives

∥yn+1 (t)− yn (t)∥ ≤∫ t

0K ∥yn (s)− yn−1 (s)∥ds

≤ K∫ t

0∥ f (y0)∥

1(n−1)!

snKn−1ds = Kn 1(n−1)!

∥ f (y0)∥∫ t

0snds

= Kn 1n!∥ f (y0)∥ tn+1

which proves the claim. Since the inequality of the claim shows that ∥yn− yn−1∥ is sum-mable, it follows that {yn} is a Cauchy sequence in C ([0,T ] ,X). It satsifies ∥yn− y0∥< Rand so yn converges uniformly to some y ∈C ([0,T ] ,X) . Hence one can pass to a limit in24.1.3 and obtain

y(t) = y0 +∫ t

0f (y(s))ds

for t ∈ [0,T ]. Also ∥y(t)− y0∥ ≤ R and on B(y0,R) , f is Lipschitz continuous so Gron-wall’s inequality gives uniqueness of solutions which remain in B.

Here is an alternate proof which other than the ugly lemma, seems more elegant to me.However, it is a useful lemma.

Lemma 24.1.9 Define

γ (x)≡{

x if ∥x− y0∥ ≤ Ry0 +

x−y0∥x−y0∥

R if ∥x− y0∥> R

Then ∥γ (x)− γ (y)∥ ≤ 3∥x− y∥ for all x,y ∈ X. Thus

∥γ (x)− y0∥ ≤ R.

Proof: In case both of x,y are in B = B(y0,R), there is nothing to show. Suppose thenthat ∥y− y0∥ ≤ R but ∥x− y0∥> R. Then, assuming y− y0 ̸= 0,

∥γ (x)− γ (y)∥=∥∥∥∥y0 +

x− y0

∥x− y0∥R − y

∥∥∥∥= ∥∥∥∥ x− y0

∥x− y0∥R− (y− y0)

∥∥∥∥

814 CHAPTER 24. CRITICAL POINTSBy induction, yn (s) ,yn—1 (s) are still in B. This is becausenIlyn (t) —yol] < py Ile (t) — ye-1 (2). 1 k yek-1< t'K< ||f(yo)||te“’ <R (24.1.4)Ashowing that y, (t) stays in B. Then since all values of the iterates remain in B, inductiongivesmeasx1 sl K!— 1 nK" "ds = K" ——nl n= K 7 lf Golle +which proves the claim. Since the inequality of the claim shows that ||y, —y,—1|| is sum-mable, it follows that {y,} is a Cauchy sequence in C ([0, 7] ,X). It satsifies ||y, —yo|| <Rand so y, converges uniformly to some y € C((0,7],X). Hence one can pass to a limit in24.1.3 and obtain ;=y0+ [ FO())asfor t € [0,T]. Also ||y(t) —yo|| < R and on B(yo,R), f is Lipschitz continuous so Gron-wall’s inequality gives uniqueness of solutions which remain in B. §jHere is an alternate proof which other than the ugly lemma, seems more elegant to me.However, it is a useful lemma.Lemma 24.1.9 Definela) = xif lx —yol] SR= 1 yot PER Ff lx —yoll > RThen \|y(x) — ¥(y)|| < 3 ||x—y]| for all x,y © X. Thuslly) —yoll < R.Proof: In case both of x,y are in B = B(yo,R), there is nothing to show. Suppose thenthat ||y — yo|| < R but ||x—yo|| > R. Then, assuming y— yo 4 0,x— YOI|x—yollyx) — YO)I] = |[yo + = —(y—yo)Sm