# Inverse function theorem to show existence of parabolic PDE

(See the image below if you want the actual text). My function space where the solution lives is $C^{k+2, \alpha}(S^1 \times (0,t_0)),$ where $t_0$ is "to be chosen later". If I have some parabolic PDE $$u_t = a(x,t,u, u_x, u_{xx}),$$ then I can look at the map $$F(u) = u_t - a(x, t, u, u_x, u_{xx})$$ and differentiate it and show that the derivative is an isomorphism. Note that $F:C^{k+2, \alpha}(S^1 \times (0,t_0)) \to C^{k, \alpha}(S^1 \times (0,t_0))$.

Then the inverse function says (according to the paper below) that there exists a $T$, $\epsilon$ and $\delta$ such that for any $\lVert f - F(v) \rVert < \epsilon$ there exists a unique $u$ such that $\lVert u - v \rVert < \delta$ and $F(u) = f$ for all $t \leq T$?

I am unclear about the "for all $t \leq T$" part. How does the time $T$ come into this? I still feel confused about this and am not confident in explaining it. Can anyone give me a clear explanation? Thanks.

[Image added by LK. Page 17 of The Curve Shortening Problem by Kai Seng Chou and Xi-Ping Zhu]

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Maybe $T$ is just a small number, the size of (one-sided) neighborhood in the time variable? Probably not. Maybe they are proving the existence of solution by the continuation method, using IFT to show that you always advance a little further from the solution you have so far. It would be be helpful to know who they are and where they make such insidious statements. –  user31373 Jun 18 '12 at 16:01
@LeonidKovalev I got it from "The Curve Shortening Problem" By Xi-Ping Zhu. It's hard to find a copy of it online unless your library subscribes to it. Anyway the author basically says what I wrote above. Also, he starts with "Let $Q(t) = S^1 \times (0,t), t \leq T$ where $T$ is to be chosen later" and defins the spaces associated with the map $F$ to be of the form $C^{2,1}(Q(t))$. I was thinking that he means a neighbourhood in the time period but this is not what I think of as the IFT. –  Court Jun 18 '12 at 17:01
I've found the book. At a first sight, the proof is badly written. It seems that $t$ is fixed, so that the size of the neighborhoods for invertibility should depend on $t$ itself. –  Siminore Jun 18 '12 at 17:17
@Siminore Are my following thoughts correct? Think of (say) $C(I\times (0,t))$ for $t \leq T$ as including functions that are continuous on $(0, t/5)$, functions continuous on $(0, t/2)$, etc etc etc. Then the IFT says that on some neighbourhood of this space $C(I \times (0,t))$ we have existence/uniqueness. So basically for some $T_1 \leq t$, the solution lies in $C(I \times (0,T_1))$. I still can't see where the $t \leq T$ thing comes in. (Does $T = T_1$?) Or does the statement "$F(u) = f$ for all $t \leq T$\$ just mean pointwise equals? –  Court Jun 18 '12 at 17:50
@LeonidKovalev Thanks for adding the picture! –  Court Jun 18 '12 at 17:51