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let $\mathbb{H}$ be the hyperbolic plane and let $k(t,x,y)$ be the associated heat kernel.

I am wondering, if for any fixed $y\in M$ and $\epsilon >0$ the function $u_t(x):=k(t,x,y)$ is continuous in $S:=\lbrace x\in M : d(x,y)>\epsilon \rbrace$ uniformly in $t\in(0,\infty)$?

For instance in Euclidean space $\mathbb{R}^n$ the heat kernel is given by $k(t,x,y)=\frac{1}{(4\pi t)^{n/2}}e^{-\frac{\Vert x-y \Vert^2}{4t}}$. Hence for $\mathbb{R}^n$ the claim is true because the function $u_t(x)$ has bounded derivative on S, i.e. $\sup_{t>0} \vert \nabla u_t(x) \vert\leq C$.

There is a formula for hyperbolic plane as well, given by

$$k(t,x,y)=\frac{\sqrt{2}}{(4\pi t)^{3/2}}e^{-\frac{t}{4}}\int\limits_{d(x,y)}^{\infty}\frac{s e^{-\frac{s^2}{4t}}}{\sqrt{\cosh{s}-\cosh{d(x,y)}}}ds$$ but I do not know how to estimate the derivative for this function.

Q: How can I prove/disprove that for any $x^{*}\in\mathbb{H}^2, x^{*}\neq y, \forall \epsilon>0$ $\exists\delta>0,$ s.t. $\vert u_t(x)-u_t(x^{*}) \vert\leq \epsilon$ for all $x\in \mathbb{H^2}$ with $d(x,x^{*})<\delta$? If the above statement is wrong, What happens if we restrict the domain to $t\in (0,T)$ for $T>0$ arbitrary and fixed?

Any help will be very appreciated!

Best wishes

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    $\begingroup$ Interesting question! Some observations: it would be enough to bound $\sup_{t >0}|\partial_d k(t;d)|$ under the assumption $d>\epsilon$. But we cannot apply the Leibniz integration rule because the integrand has an integrable singularity at $s=d$. Is this what you thought? Thinking it in physical terms what you ask should be true, since $\mathbb{H}$ is homogeneous, if $x$ and $x^*$ are very close to each other and very far from the heat source $y$ they should be heated up "at the same rate". Now I should try to turn this into a mathematical argument. $\endgroup$
    – Giovanni De Gaetano
    Commented Oct 14, 2015 at 13:54
  • $\begingroup$ Yes, your idea is nice! But it seems not clear to me how to compute the derivative $\partial_d k(r,d)$ or to estimate it. $\endgroup$
    – supersnail
    Commented Oct 14, 2015 at 20:13
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    $\begingroup$ I can only think to compute $$ \lim_{\epsilon \rightarrow 0} \partial_d \left( \frac{\sqrt{2}}{(4\pi t)^{3/2}} \int_{d+\epsilon}^\infty \frac{se^{-s^2/4t}}{\sqrt{\cosh(s) - \cosh(d)}} ds \right). $$ Where the inner derivative is computed by Leibniz rule and the exchange of limit and integrals is justified by uniform convergence of the derivative (this shouldn't be a problem). But this involves quite a few computations, perhaps I find the time to do them later, otherwise you can try and let me know what happens! :) $\endgroup$
    – Giovanni De Gaetano
    Commented Oct 15, 2015 at 11:05
  • $\begingroup$ I think there is a problem with your approach, since the integral $\int_d^{d+1}\frac{1}{(\cosh(s)-\cosh(d))^{3/2}} ds$ do not converge, so that it is not justified to change integral and derivative. $\endgroup$
    – supersnail
    Commented Oct 15, 2015 at 13:59

1 Answer 1

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It follows from the interior Schauder estimates for parabolic equations. Namely, put for simplicity's sake $y=0$, $k(t,x)=k(t,x,0)$ and for negative values of $t$ continue $k$ by zero: $k(x,t)=0$, $t<0.$ For $r>0$ denote $B_r(x_0)=\{x\in \mathbb{H}^2|d(x,x_0)<r\}$ a ball on the plane and $B_r=B_r(0)$. Then $k(x,t)$ is bounded on a cylindrical domain $H_r=\mathbb R\times (\mathbb{H}^2\backslash B_r)$. Now fixing $\alpha\in(0,1)$ and applying the interior Schauder estimate in the cylinder $G_r(x_0,t_0)=(t_0-r^2,t_0)\times B_r(x_0)$, $d(x_0,0)\ge 2r$, we have $$ \|k\|_{C^{2,\alpha}(G_{r/2}(x_0,t_0))}\le C_r\sup_{G_{r}(x_0,t_0)}|k|\le C_r\sup_{H_r}|k|. $$ It follows that $$ \sup_{H_{2r}}|\nabla_x k|\le C_r\sup_{H_r}|k|. $$

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  • $\begingroup$ Nice! I didn't know about these Schauder estimates. I still have a few questions though: are you using the interior Schauder estimate as in Wikipedia (en.wikipedia.org/wiki/Schauder_estimates#Interior_estimates)? (I ask because you mention parabolic equations) When you write $t_0$ you actually mean $t$? Otherwise I don't understand what is its role. Thanks! $\endgroup$
    – Giovanni De Gaetano
    Commented Oct 15, 2015 at 15:49
  • $\begingroup$ @GiovanniDeGaetano Yes, for parabolic equations there is an analogous estimate. Zeros in $(x_0,t_0)$ are to differ a fixed point (used to define a cylinder in which the interior estimate to be applied) from variables (x,t)$. $\endgroup$
    – Andrew
    Commented Oct 15, 2015 at 19:09

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