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For $x_0 \in \mathbb{R}^n$ and $R> 0$, set $B(x_0, R) = \{x \in \mathbb{R}^n \colon |x_0 - x| < R\}$. Let $\chi_{B(x_0, R)}$ denote the characteristic function for $B(x_0, R)$. The solution to the heat equation

$$ \cases{ \Delta f - f_t & $(x,t) \in \mathbb{R}^n \times (0, \infty)$,\cr f(x, 0) = \chi_{B(x_0, R)} & $ x \in \mathbb{R}^n$, } $$

is given by the convolution of $\chi_{B(x_0, R)}$ with the fundamental solution solution to the heat equation. That is, for $t > 0:$

$$\displaystyle f(x,t) = (4\pi t)^{-\frac{n}{2}}\int_{\mathbb{R}^n}\chi_{B(x_0, R)}(y)e^{-\frac{|x-y|^2}{4t}}dy =(4\pi t)^{-\frac{n}{2}}\int_{B(x_0, R)}e^{-\frac{|x-y|^2}{4t}}dy. $$

Prove that there exists some some $0 < A < 1$, independent of both $x_0$ and $R$, such that:

$$f(x_0, AR^2) = (4\pi AR^2)^{-\frac{n}{2}}\int_{B(x_0, R)}e^{-\frac{|x_0-y|^2}{4AR^2}}dy \ge \frac{1}{2}.$$

This question appears on an old PDE qual I am working through. So far, I have tried exploiting the spherical symmetry of the situation by converting this integral over $B(x_0, R)$ into a double integral:

$$f(x_0, AR^2) = (4\pi AR^2)^{-\frac{n}{2}}\int_0^R\int_{S(x_0,r)}e^{-\frac{r^2}{4AR^2}}d\sigma(y)dr = (4\pi AR^2)^{-\frac{n}{2}} \sigma_{n-1}\int_0^Rr^{n-1}e^{-\frac{r^2}{4AR^2}}dr.$$

Here, $S(x_0, r) = \{x \in \mathbb{R}^n \colon |x_0 - x| = r\}$, and $\sigma_{n-1}$ denotes the $(n-1)$-dimensional volume of $S(0,1) \subseteq \mathbb{R}^n$. One other thing I have noticed is that, for each $0 < r < R$, we have $\displaystyle e^{-\frac{r^2}{4AR^2}} \ge e^{-\frac{1}{4A}}$. So in fact I can write:

$$\displaystyle f(x_0, AR^2) \ge (4\pi AR^2)^{-\frac{n}{2}} \sigma_{n-1}\int_0^Rr^{n-1}e^{-\frac{1}{4A}}dr = (4\pi AR^2)^{-\frac{n}{2}}\frac{R^n}{n} \sigma_{n-1} e^{-\frac{1}{4A}} = \frac{\sigma_{n-1}}{n}(4\pi A)^{-\frac{n}{2}}e^{-\frac{1}{4A}}.$$

Hints or solutions are greatly appreciated.

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That last expression that you wrote down is independent of both $x_0$ and $R$. – Kirill Jan 6 '13 at 4:00
up vote 1 down vote accepted

This problem is made for a scaling argument. If $u=u(x,t)$ solves the heat equation, then so does the scaled function $u^{(\lambda)}(x,t):=(\lambda x, \lambda^2 t)$ for any scaling factor $\lambda>0$. Indeed, $(\frac{\partial}{\partial t}-\Delta)u^{(\lambda)}=\lambda^2 (\frac{\partial}{\partial t}-\Delta)u=0$.

Translation in the $x$-direction preserves the heat equation as well. Thus, by scaling and translation the problem reduces to looking at the solution with $f(\cdot,0)=\chi_{B(0,1)}$ and observing that $f(0,t)\to 1$ as $t\to 0$; by definition of the limit this gives you $f(0,t)\ge 1/2$ for all small $t$.

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Hi @5PM, just wanted to send a quick message and let you know that I passed this dreaded PDE qual that I was studying for. I really couldn't have done it without all your hints and answers from the past several weeks. Thanks a million! – JZShapiro Jan 16 '13 at 2:44
@jzshapiro Congratulations! – user53153 Jan 16 '13 at 3:08

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