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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): [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.

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up vote 1 down vote accepted

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.

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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
Thanks for the reply. – user24394 Jun 23 '12 at 17:46

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.

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