# Application of inverse function theorem to get short time existence

I am reading a book on curve shortening flow. Optionally, please see this image for the page that is confusing me (I am not allowed to include it in this post since I'm new): http://i.stack.imgur.com/L54lm.png [Thanks to user Leonid from SE for the image. Page 17 of The Curve Shortening Problem by Kai Seng Chou and Xi-Ping Zhu]

The authors construct a map $\mathcal{F}$ from $\tilde{C}^{k+2, \alpha}(S^1 \times (0,t))$ to $\tilde{C}^{k, \alpha}(S^1 \times (0,t))$, find its Frechet derivative and show it's an isomorphism, so we can use the inverse function theorem. They say there exists a $t_0$, $\epsilon$ and $\delta$ such that for any $f$ with $\lVert f - \mathcal{F}(v) \rVert < \epsilon$ there exists a unique $u$ such that $\lVert u - v \rVert < \delta$ and $\mathcal{F}(u) = f$ for all $t \leq t_0$.

I am confused about the part they say that "there exists a $t_0$ ... such that $\mathcal{F}(u) = f$ for all $t \leq t_0$". How does this time dependence come into this from the inverse function theorem?

The inverse function theorem I know doesn't state anything about this time dependence. The proof is confusingly written (for me anyway). If they fix the space to be $\tilde{C}^{k, \alpha}(S^1 \times (0,t))$ then how can they only say that the solution exists within a neighbourhood of the $(0,t)$? I thought you don't get control of that, only the space of functions.

Can anyone explain this? Are they using some other theorem? Thanks.

Perhaps something can be gleaned from the analogous proof of short-time existence for scalar ODEs:

Consider the initial value problem $x' = f(x)$ with $x(0) = x_0$ and define $F(x)(t) = x(t) - x_0 - \int_0^t f(x(s))ds$ so that zeros of F correspond to solutions of the IVP. Regard the function F as acting on some space of functions whose elements obey $x(0) = x_0$. The Derivative of F is $(F'(x)y)(t) = y(t) - \int_0^t f'(x(s))y(s)ds$. To show that F' is an isomorphism, one wants to show that the norm of $y \mapsto \int_0^t f'(x(s))y(s)ds$ is less than one. If you are working on a space of functions from $[-T,T] \to R$, then a cheap estimate is given by T times the maximum value that the absolute value of $f'$ takes. This can be made smaller than one by choosing T sufficiently small. Of course f' has to have a maximum value in the first place. This is dealt with through a short song-and-dance in which one works in an open subset of the function space for which the functions x take a restricted set of values so that on these values f' does take a maximum absolute value.

I suspect that something similar is going on here.

More generally, given an operator A that can be regarded as acting on either a Banach space X or some other Banach space Y, the spectrum of A in general will depend upon the Banach space. That the note produced by a vibrating harp string depends on the length of the string furnishes an example of this phenomenon. (A is the second derivative, X is the set of functions from $[0,L_x]$ to R with Dirichlet boundary conditions and Y is the set of functions from $[0,L_y]$ to R with Dirichlet boundary conditions.)

In your situation again the different function spaces contain points which themselves are functions defined on shorter or longer time intervals.

• Indeed, the argument for short time existence of a parabolic PDE is essentially the same as the proof for the short time existence of a system of first order ODE's. The only difference is that the curve is a map into a carefully chosen Banach space instead of $R^n$. – Deane Yang Jun 21 '12 at 16:44

This will not answer your question, however, if you really want to get short-time existence, you can argue in the following way:

(1) Select some good $v$, for example, time derivatives of $\mathcal F(v)$ at $t=0$ vanish up to some order.

(2) There exists some $f \in \widetilde C^{k,α}(S^1×(0,t))$, some $\delta, \varepsilon>0$ such that $$f \equiv 0 \quad \forall t \in [0, \delta]$$ and $$\|\mathcal F(v) - f \|_{\widetilde C^{k,α}(S^1×(0,t))} \leqslant \varepsilon.$$

(3) Let $\widetilde u = \mathcal F^{-1} (f)$ be the inverse of $f$ under the mapping $\mathcal F$. Then $\mathcal F(\widetilde u) \equiv 0$ for all $t \in [0, \delta]$. From this you obtain existence in the short-time interval $[0, \delta]$.

Thank you.