most recent 30 from http://mathoverflow.net 2013-06-19T23:55:40Z http://mathoverflow.net/feeds/question/100248 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100248/application-of-inverse-function-theorem-to-get-short-time-existence quentinknight 2012-06-21T14:26:22Z 2012-06-21T16:18:17Z

<p>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): <a href="http://i.stack.imgur.com/L54lm.png" rel="nofollow">http://i.stack.imgur.com/L54lm.png</a> [Thanks to user Leonid from SE for the image. Page 17 of <em>The Curve Shortening Problem</em> by Kai Seng Chou and Xi-Ping Zhu]</p> <p>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 &lt; \epsilon $ there exists a unique $u$ such that $\lVert u - v \rVert &lt; \delta$ and $\mathcal{F}(u) = f$ for all $t \leq t_0$.</p> <p><strong>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?</strong></p> <p>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.</p> <p>Can anyone explain this? Are they using some other theorem? Thanks. </p>

http://mathoverflow.net/questions/100248/application-of-inverse-function-theorem-to-get-short-time-existence/100257#100257 Aaron Hoffman 2012-06-21T16:18:17Z 2012-06-21T16:18:17Z

<p>Perhaps something can be gleaned from the analogous proof of short-time existence for scalar ODEs:</p> <p>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.</p> <p>I suspect that something similar is going on here.</p> <p>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.) </p> <p>In your situation again the different function spaces contain points which themselves are functions defined on shorter or longer time intervals.</p>