Kenneth Kuttler

816 CHAPTER 24. CRITICAL POINTS

Thus, iterating this inequality, it follows that a large enough power of F is a contractionmap. Therefore, there is a unique fixed point. Now letting y be this fixed point,

∥y(t)− y0∥ ≤∫ t

03K ∥y(s)− y0∥ds+∥ f (y0)∥T

It follows that∥y(t)− y0∥ ≤ ∥ f (y0)∥Te3KT

Choosing T small enough, it follows that ∥y(t)− y0∥< R on [0,T ] and so γ has no effect.Thus this yields a local solution to the initial value problem.

In the case that U = X , the above argument shows that there exists a solution on some[0,T ) where T is maximal.

y(t) = y0 +∫ t

0f (y(s))ds, t < T

Suppose T < ∞. Suppose∫ T

0 ∥ f (y(s))∥ds < ∞. Then you can consider y0 +∫ T

0 f (y(s))dsas an initial condition for the equation and obtain a unique solution z valid on [T,T +δ ] .Then one could consider ŷ(t) = y(t) for t < T and for t ≥ T, ŷ(t) = z(t) . Then for t ∈[T,T +δ ] ,

ŷ(t) = z(t) = y0 +∫ T

0f (y(s))d +

∫ t

Tf (ŷ(s))ds

= y0 +∫ T

0f (ŷ(s))d +

∫ t

Tf (ŷ(s))ds

and so in fact, for all t ∈ [0,T +δ ] ,

ŷ(t) = y0 +∫ t

0f (ŷ(s))ds

contrary to the maximality of T. Hence it cannot be the case that T < ∞. Thus it must bethe case that

∫ T0 ∥ f (y(s))∥ds = ∞ if the solution is not global.

From the above observation, here is a corollary.

Corollary 24.1.10 Let f : X → X be locally Lipschitz where X is a Banach space. Thenthere exists a unique local solution to the IVP

y′ = f (y) , y(0) = y0

If f is bounded, then in fact the solutions exists on [0,T ] for any T > 0.

Proof: Say ∥ f (x)∥ ≤M for all M. Then letting [0, T̂ ) be the maximal interval, it mustbe the case that

∫ T̂0 ∥ f (y(t))∥dt = ∞, but this does not happen if f is bounded.

Note that this conclusion holds just as well if f has linear growth, ∥ f (u)∥ ≤ a+b∥u∥for a,b ≥ 0. One just uses an application of Gronwall’s inequality to verify a similar con-clusion.

One can also give a simple modification of these theorems as follows.

816 CHAPTER 24. CRITICAL POINTSThus, iterating this inequality, it follows that a large enough power of F is a contractionmap. Therefore, there is a unique fixed point. Now letting y be this fixed point,Iy@-yoll < [3K|ly(6) —yollas+IF 00) 17It follows thatlly (2) —yoll < LF Go) || Te**”Choosing T small enough, it follows that ||y (t) — yo|| < R on [0,7] and so y has no effect.Thus this yields a local solution to the initial value problem. JjIn the case that U = X, the above argument shows that there exists a solution on some(0,7) where T is maximal.®=r+ f FO0) \)ds,t<TSuppose T < co. Suppose fo \| f (y (s))|| ds < cc. Then you can consider yo + fo fv (s))dsas an initial condition for the equation and obtain a unique solution z valid on [T,T + 6].Then one could consider $(t) = y(t) for t < T and for t > T, $(t) = z(t). Then for t €[T,T + 4],S(t)nome k roe a+ [ aswf FG(9)at [F(9())asand so in fact, for all t € [0,7 +6],= [ £6(s))ascontrary to the maximality of T. Hence it cannot be the case that T < oo, Thus it must bethe case that {7 || f (y(s))||ds =~ if the solution is not global.From the above observation, here is a corollary.Corollary 24.1.10 Let f : X — X be locally Lipschitz where X is a Banach space. Thenthere exists a unique local solution to the IVPy=fl(y), y0)=y0If f is bounded, then in fact the solutions exists on |0,T| for any T > 0.Proof: Say || (x)|| <M for all M. Then letting [0,7) be the maximal interval, it mustbe the case that fj ||,f (y(t))||dt =~, but this does not happen if f is bounded.Note that this conclusion holds just as well if f has linear growth, ||f (u)|| <a+b||ul|for a,b > 0. One just uses an application of Gronwall’s inequality to verify a similar con-clusion.One can also give a simple modification of these theorems as follows.