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(I am reading a paper called Shortening Complete Plane Curves by Kai-Seng Chou & Xi-Ping Zhu. It is available at http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.jdg/1214424967. Page 476 is relevant)

Consider the partial differential equation on the domain $\mathbb{R}\times [0,T]$: $$u_t - A(x,t)u_{xx} + \text{l.o.t} = f$$ where $$A(x,t) = \frac{(1-k_0(x)^2t)^2}{[(1-k_0(x)^2t)^2 + (k_{0_x}(x)t)^2]^2}$$ where $k_0(x)$ is the curvature of the curve $\gamma_0:\mathbb{R} \to \mathbb{R}^2$.

Recall that a PDE is uniformly parabolic if there exist positive constants $a$ and $b$ such that the term in the front of the Laplacian sits between $a$ and $b$, i.e., $a \leq A(x,t) \leq b$.

Questions:

1) Why is it true that

$A$ is bounded in $C^{k, \alpha}(\mathbb{R} \times [0,T])$ if $\gamma_0 \in C^{k+4, \alpha}(\mathbb{R})$?

Something to do with the fact that $k_0$ depends on $(\gamma_0)_{xx}$?

2) Why is it true that

If we restrict $T$ so that, for example, $$T < \frac{1}{2}\inf_x \frac{1}{1+k_0^2(x)},$$ then the PDE is uniformly parabolic.

I don't see where that comes from at all.

Thanks for any help.

1 Answers 1

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1) Yes, if in $k_0$ there are derivatives of up to the second order, then $\frac{\partial} {\partial x}A(x,t)$ contains $k_0$ derivatives of order no more than three etc. Also it is infinitely differentiable with respect to $t$. The only thing to be checked here is that the denominator is bounded away from zero. But generally it does not hold. For example, if $k_0(x_0)\ne0\;$ and ${k_0}_x(x_0)=0\;$ then $$ A(x_0,t) = \frac{1}{(1-k_0(x_0)^2t)^2}\to\infty, \quad t\to k_0^{-2}(x_0). $$ However, for small enough values of $t$ (as restricted in the question below), $$ 1-k_0(x_0)^2t \ge 1-\frac12\frac{k_0^2(x)}{1+k_0^2(x)}\ge\frac12, $$ thus the denominator is bounded away from zero by $1/16$.

2) Put $C_1=\sup_x |k_{0}(x)|\;$, $C_2=\sup_x |k_{0_x}(x)|\;$. Then we have $$ A(x,t)\ge \frac{\frac14}{\left(\frac14+C_2^2T^2\right)^2}=a, $$ $$ A(x,t)\le \frac {(1+C_1^2T)^2}{\frac1{16}}=b. $$

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    Thanks Andrew. I'll just think about it a little before awarding it to you.2012-07-23
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    How did you get the $\frac{1}{4}$ in the denominator for inequality involving $a$? I get a term involving $C_1$ there.2012-07-25
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    Also, am I right that I should use the inequality for $T$ given in the paper to get rid of $T$ in $a$ and $b$ as it doesn't seem to make sense to keep it in that form.2012-07-25
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    @blackcat $1/4$ is from the estimate $1-k_0(x_0)^2t \ge\frac12$ above. The maximal value of $T$ depend on $C_1$ but it is already taken into account in the condition $T < \frac{1}{2}\inf_x \frac{1}{1+k_0^2(x)}$. As for $T$ from this estimate it is just an explicitly pointed out length of time period (depending on $\gamma$) then the equation is guaranteed to be uniformly parabolic. The relevant thing here is perhaps that such an interval do exist and depends on a certain norm of $\gamma$ but not its exact value.2012-07-25
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    Thanks @Andrew but for the specfic inequality $A(x,t) \geq a$, the $1-k_0^2t$ is in the denominator of $A(x,t)$ so we can't substitute in the 1/2 (as then we'd get $A(x,t) \leq ..$.) But I think this can be fixed by using the $C_1$ .2012-07-25
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    @blackcat my bad, thanks. It should be $$A(x,t)\ge \frac{\frac14}{\left((1+C_1^2T^2)^2+C_2^2T^2\right)^2}.$$2012-07-25