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I asked this question on math.stackexchange.com, however I didn't get any answers so I'll try it here.

Let $M$ be a compact manifold without boundary. Consider $Lu:=\partial_tu-\Delta u$. Let $f\in C^{0,0,\alpha}((0,T)\times M,\mathbb{R})$, $u_0\in C^{2,\alpha}(M,\mathbb{R})$ and let $u\in C^{1,2,\alpha}((0,T)\times M,\mathbb{R})$ be the unique solution of

$$ \begin{cases} Lu=f \\ u(t,.)\to u_0(.) & \text{as $t\to 0$} \end{cases} $$

Here $C^{k,l,\alpha}$, $C^{k,\alpha}$ are the appropriate (parabolic) Hölder spaces.

Then we have

$$||u||_{C^{1,2,\alpha}((0,T)\times M,\mathbb{R})}\le C(||f||_{C^{0,0,\alpha}((0,T)\times M,\mathbb{R})}+||u_0||_{C^{2,\alpha}(M,\mathbb{R})})$$

For some $C>0$ independent of $f,$ $u$ and $u_0$.

Question: Is it possible to choose $C$ independent of $T$? If C depends on $T$, i.e. $C=C(T)$, is it possible to choose $C(T)$ in a way that $C(T)$ is bounded for $T\to 0$?

If one replaces $M$ by a suitable domain in $\mathbb{R}^n$ the answer of my question seems to be "Yes". However, I need the result for compact manifolds and was wondering if there occur any problems.

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  • $\begingroup$ What does mean $\mathbb R$ in ${C^{2,\alpha}(M,\mathbb{R})}$? $\endgroup$
    – Andrew
    Nov 27, 2015 at 11:04
  • $\begingroup$ It just refers to funtions $M\to\mathbb{R}$, i.e. the functions have values in $\mathbb{R}$ (and not e.g. another manifold $N$) $\endgroup$
    – alerte
    Nov 27, 2015 at 11:08

1 Answer 1

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The constant $C(T)$ depends on $T$, consider $f=1$, $u_0=0$ and the solution $u=t$. But it is bounded for $T\to0$. The rhs $f\in C^{0,\alpha}([0,T]\times M)$ can be continued to $\tilde f\in C^{0,0,\alpha}([0,1]\times M)$, $||\tilde f||_{C^{0,0,\alpha}((0,1)\times M)}= |f||_{C^{0,0,\alpha}((0,T)\times M)}$. Denote $\tilde u$ the solution with rhs $\tilde f$. Then $$ |u||_{C^{2,1,\alpha}((0,T)\times M)}\le |\tilde u||_{C^{2,1,\alpha}((0,1)\times M)}\le $$ $$ \le C(1)(||\tilde f||_{C^{0,0,\alpha}((0,1)\times M)}+||u_0||_{C^{2,\alpha}(M,\mathbb{R})}))\le C(1) (|f||_{C^{0,0,\alpha}((0,T)\times M)}+||u_0||_{C^{2,\alpha}(M,\mathbb{R})})). $$

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  • $\begingroup$ Thank you for your answer. I don't understand the last inequality. Why is $||\tilde{f}||_{C^{0,0,\alpha}((0,1)\times M}\le ||\tilde{f}||_{C^{0,0,\alpha}((0,T)\times M)}$ for $T\le 1$ $\endgroup$
    – alerte
    Nov 27, 2015 at 11:43
  • $\begingroup$ Consider the following continuation of $f$ to $[0,2T]$, symmetry with respect to the plane $t=T$: $\tilde f(t,x)=f(2T-t,x)$, $T<t\le 2T$. Then $||\tilde f||_{C^{0,0,\alpha}((0,2T)\times M)}=|f||_{C^{0,0,\alpha}((0,T)\times M)}$. То be repeated until $2^nT\ge1$. $\endgroup$
    – Andrew
    Nov 27, 2015 at 14:04
  • $\begingroup$ I apologize for asking this seemingly unrelated question here in an old answer of you, I hope you wouldn't mind. Although the original question asked about estimates, I would like to know how one proves an existence result for this equation with the same assumptions on regularity of $u_0$ and $f$? I am not familiar with compact manifold without boundary. $\endgroup$
    – BigbearZzz
    Mar 29, 2018 at 8:55
  • $\begingroup$ @BigbearZzz Existence follows from the existence of the fundamental solution: the solution of the Cauchy problem can be written as a sum of two potentials, one for $f$ and another for $u_0$. $\endgroup$
    – Andrew
    Mar 29, 2018 at 13:45
  • $\begingroup$ So the fundamental solution for compact manifold without boundary always exists? Is it a well-know result that I can find in a geometric-related PDE book? $\endgroup$
    – BigbearZzz
    Mar 29, 2018 at 14:05

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