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In the classic paper by Hamilton and Gage (see http://intlpress.com/JDG/archive/1986/23-1-69.pdf), they give the PDE problem:

Find $k:S^1 \times [0,T) \to \mathbb{R} \text{ s.t }$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\text{(i) } k \in C^{2+\alpha, 1+\alpha}(S^1 \times [0,T-\epsilon]) \text{ for all } \epsilon > 0$$

$$\text{(ii) }\frac{\partial k}{\partial t} = k^2 \frac{\partial^2 k}{\partial \theta^2} + k^3$$

$$\;\;\;\;\;\;\;\;\;\text{(iii) }k(\theta, 0) = \psi(\theta) \text{ satisfies }$$

$$\qquad \;\;\;\;\;\;\text{(a) } \psi \in C^{1+\alpha(S^1)}$$

$$\qquad \text{(b) } \psi(\theta) > 0$$

$$\qquad \;\;\;\;\;\;\;\;\;\;\;\;\text{(c) } \int_0^{2\pi} \frac{\cos(\theta)}{\psi(\theta))} = 0$$

And they say that is a shown from standard results. Every paper that I've seen that references this paper says that H-G prove this (but they don't, they just state it), and the ones that talk about it give references to papers that are difficult (maximal regularity and Volterra conditions and I don't know which spaces to use as it's quite an abstract result (by Herbert Amann)) that may not be necessary to use anyway. Does anyone know what the standard results are? I have a book on quasilinear equations but it looks like this equation doesn't fit the conditions that are required for the existence result it gives. It's all very confusing because there are lots of different requisites in different sources...

I appreciate any help.

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iii) is just an initial condition. I recall I did this some 10 years ago (make it 15). If I recall correctly (I may not) then the idea is to linearize ii), solve the linearized equation by using linear parabolic theory, and then (check the prerequisites of and) apply the (infinite dimensional) implicit function theorem to get the nonlinear equation solved locally. –  user20266 Jun 10 '12 at 17:30
    
@Thomas thanks for the reply. By linearize do you mean turn it into a fixed point problem and apply Schauder's theorem? Sounds like a lot of work for such a simple looking PDE. –  soup Jun 10 '12 at 18:40
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No. The PDE is a nonlinear operator between function spaces which are Banach Spaces. To linearize this means to differentiate that operator. No fixed point theorem. And regarding 'simple' -- if it is that simple, then why the question? (as a side remark: Hamilton stuff is rarely simple but almost always correct.) –  user20266 Jun 10 '12 at 19:01
    
@Thomas thanks, do you have any recommendations to learn this linearisation process you speak of? –  soup Jun 22 '12 at 10:55
    
I'm afraid that no. I think we did this is in a seminar or workshop which was a follow up to a lecture which covered the nonlinear FA basics. Apart from that we were studying papers. It't too long ago, sorry. –  user20266 Jun 22 '12 at 12:28

1 Answer 1

Gage and Hamilton prove this fact in the paper "The Heat Equation Shrinking Convex Plane Curves" see page 80 of this paper.. I confess that I didn't understand the argument too.. I can't access the link you furnished..

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