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I was given the PDE $C_t = (L/2\pi - 1/k)N(p,t) $ and $C(p,0) = C_0(p)$ where $k$ = curvature of the evolving curve, and $C(p,t)$ is the family of closed planar curves.

I was asked to show that this PDE does indeed evolve a planar curve to a circle while preserving its length, but I'm not sure how to show that $L'(t) = 0$ (Length preservation) or to show how it evolves a curve to a curve, because I'm rather weak in calculus. Please help.

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Did you mean $C(p,t)$ for $N(p,t)$? – Ilya Mar 6 '12 at 9:53
@Ilya: I don't think so. The $N(p,t)$ probably refers to the normal vector at the point $C(p,t)$. – Willie Wong Mar 7 '12 at 17:00

Since this is homework, I'm going to just give some signposts. There are a lot of articles on curve flows, which have many similar computations in them. They may help. Likely the best reference (certainly the most seminal) is Gage and Hamilton, but there is a huge literature on curve flows these days. (Shameless plug: a new article of mine which is available on the arxiv might even help!)

I'm only going to answer the question 'show the length is preserved'. To prove the flow converges to a circle is a much more complicated story, and one I can not believe would be posed as homework.

Let's use the notation from your question, and take $C:I\times[0,T)\rightarrow\mathbb{R}^2$ to be the evolving family of curves. The length of each curve is defined as $$ L(C(\cdot,t)) = \int_I |\partial_uC|\,du\,. $$ To solve your problem, complete the following.

Step 1: Compute $\partial_t\partial_uC$.

Step 2: Compute $\partial_tL(C(\cdot,t))$.

Step 3: Use integration by parts (recall that $C$ is closed) to simplify the answer from step 2. What do you notice?

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For convergence to a circle, aside from the annoying bit of the regularity (you need to use some sort of parabolic regularity argument), one probably can just use some sort of decay estimate (I'm guessing that $\int |C_t|^2$ is monotonically decreasing...) – Willie Wong Mar 7 '12 at 16:59
@Willie Right, while I would say showing that length is preserved is elementary, showing the parabolic regularity and global existence is not. (The classification of the limit is likely again elementary, like you say.) – Glen Wheeler Mar 7 '12 at 23:09

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