How to prove that $\iint\frac{|y|^2}{s^2}\,dy\,ds=4$?

Question

Let $U\subset\mathbb{R}^n$ be an open set, $\Phi$ the fundamental solution of heat equation, $T>0$, $r>0$, $x\in\mathbb{R}^n$ and $t\in\mathbb{R}$. Defines $U_T=U\times(0,T]$ and $$E(x,t;r)=\{(y,s)\in\mathbb{R}^{n+1};\:s\leq t,\;\Phi(x-y,t-s)\geq r^{-n}\}.$$

Evans PDE book presents the following

Theorem (mean-value property for the heat equation): Let $u\in C^2_1(U_T)$ solve the heat equation. Then $$u(x,t)=\frac{1}{4r^n}\iint_{E(x,t;r)}u(y,s)\frac{|x-y|^2}{(t-s)^2}\,dy\,ds.$$

In the proof is used the following equality:

$$\iint_{E(0,0;1)}\frac{1}{s^2}\sum_{i=1}^n {y_i}^2\,dy\,ds=4$$

Could someone help me with details of this calculation?

Thanks.

Answer 1

I'm posting a compilation of Tomás answer and Felix Marin's answer.

For $x\in\mathbb{R}^n$ and $t>0$, we have $$\Phi(x,t)=\frac{1}{(4\pi t)^{n/2}}\mathrm{e}^{\frac{-|x|^2}{4t}}$$ Therefore, for $y\in\mathbb{R}^n$ and $s<0$, $$\begin{align*} \Phi(-y,-s)\geq 1\quad\Longleftrightarrow& \quad\frac{1}{(-4\pi s)^{n/2}}\mathrm{e}^{\frac{|y|^2}{4s}}\geq1\\\\ \Longleftrightarrow& \quad(-4\pi s)^{n/2}\leq\mathrm{e}^{\frac{|y|^2}{4s}}\\\\ \Longrightarrow& \quad(-4\pi s)^{n/2}\leq 1\\\\ \Longleftrightarrow& \quad s\geq -\frac{1}{4\pi} \end{align*}$$ and $$\begin{align*} \Phi(-y,-s)\geq 1\quad\Longleftrightarrow& \quad\ln\left(\frac{1}{(-4\pi s)^{n/2}}\mathrm{e}^{\frac{|y|^2}{4s}}\right)\geq0\\\\ \Longleftrightarrow& \quad(-n/2)\ln\left(-4\pi s\right)+\frac{|y|^2}{4s}\geq0\\\\ \Longleftrightarrow& \quad|y|^2\leq 2ns\ln(-4\pi s) \end{align*}$$ This implies that $$\begin{align*}E(0,0;1)&=\{(y,s)\in\mathbb{R}^{n+1}\mid s\leq 0,\;\Phi(-y,-s)\geq 1\}\\\\ &=\left\{(y,s)\in\mathbb{R}^{n+1}\;\Big|\; -\frac{1}{4\pi}\leq s\leq 0,\;|y|^2\leq 2ns\ln(-4\pi s)\right\}\end{align*}$$ Thus $$\begin{align*} \iint_{E(0,0;1)}\frac{|y|^2}{s^2}\;dyds&=\int_{-\frac{1}{4\pi}}^0\int_{|y|^2\leq2ns\ln(-4\pi s)}\frac{|y|^2}{s^2}\;dyds\\\\ &=\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\int_{\partial B(0,r)}\frac{|y|^2}{s^2}\;dS(y)drds \quad\text{(polar coordinates)}\\\\ &=\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\int_{\partial B(0,1)}\frac{|rw|^2}{s^2}r^{n-1}\;dS(w)drds \quad\text{(change $w=y/r$)}\\\\ &=\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\int_{\partial B(0,1)}\frac{r^{n+1}}{s^2}\;dS(w)drds\\\\ &=\text{med}\big(\partial B(0,1)\big)\int_{-\frac{1}{4\pi}}^0\int_0^{\sqrt{2ns\ln(-4\pi s)}}\frac{r^{n+1}}{s^2}drds\\\\ &=\text{med}\big(\partial B(0,1)\big)\int_{-\frac{1}{4\pi}}^0\frac{(2ns\ln(-4\pi s))^{(n+2)/2}}{(n+2)s^2}ds\\\\ &=\frac{(2n)^{(n+2)/2}}{(n+2)}\text{med}\big(\partial B(0,1)\big)\int_{-\frac{1}{4\pi}}^0\frac{(s\ln(-4\pi s))^{(n+2)/2}}{s^2}ds \end{align*}$$ By calculating this last integral, we get $$\begin{align*} \int_{-\frac{1}{4\pi}}^0\frac{(s\ln(-4\pi s))^{(n+2)/2}}{s^2}ds&=\int_{1}^0\frac{\left((-\frac{x}{4\pi})\ln(x)\right)^{(n+2)/2}}{(-\frac{x}{4\pi})^2}\left(-\frac{1}{4\pi}\right)dx\quad\text{(change $x=-4\pi s$)}\\\\ &=\int_{0}^\infty\frac{\left(\frac{\mathrm{e}^{-z}}{4\pi}z\right)^{(n+2)/2}}{(-\frac{\mathrm{e}^{-z}}{4\pi})^2}\left(-\frac{1}{4\pi}\right)(-\mathrm{e}^{-z})\;dz\quad\text{(change $z=-\ln(x)$)}\\\\ &=\frac{1}{(4\pi)^{n/2}}\int_{0}^\infty\mathrm{e}^{-nz/2}z^{(n+2)/2}\;dz\\\\ &=\frac{1}{(4\pi)^{n/2}}\int_{0}^\infty\mathrm{e}^{-w}\left(\frac{2w}{n}\right)^{(n+2)/2}\frac{2}{n}\;dw\quad\text{(change $w=nz/2$)}\\\\ &=\frac{1}{(4\pi)^{n/2}}\left(\frac{2}{n}\right)^{n/2+2}\int_{0}^\infty\mathrm{e}^{-w}w^{n/2+1}\;dw\\\\ &=\frac{1}{2^{n}\pi^{n/2}2^{-n/2-2}n^{n/2+2}}\int_{0}^\infty\mathrm{e}^{-w}w^{n/2+2-1}\;dw\\\\ &=\frac{1}{\pi^{n/2}2^{n/2-2}n^{n/2+2}}\Gamma\left(2+\frac{n}{2}\right)\quad\text{(definition of $\Gamma$)} \end{align*}$$ So, $$\begin{align*} \iint_{E(0,0;1)}\frac{|y|^2}{s^2}\;dyds&=\frac{(2n)^{(n+2)/2}}{(n+2)}\text{med}\big(\partial B(0,1)\big)\frac{1}{\pi^{n/2}2^{n/2-2}n^{n/2+2}}\Gamma\left(2+\frac{n}{2}\right)\\\\ &=\frac{8\;\text{med}\big(\partial B(0,1)\big)}{n(n+2)\pi^{n/2}}\Gamma\left(2+\frac{n}{2}\right)\\\\ &=\frac{8\;\text{med}\big(\partial B(0,1)\big)}{n(n+2)\pi^{n/2}}\left(1+\frac{n}{2}\right)\Gamma\left(1+\frac{n}{2}\right)\quad\text{(because $\Gamma(t+1)=t\Gamma(t)$)}\\\\ &=\frac{4\;\text{med}\big(\partial B(0,1)\big)}{n\pi^{n/2}}\Gamma\left(1+\frac{n}{2}\right)\\\\ &=\frac{4\;\text{med}\big(\partial B(0,1)\big)}{n\pi^{n/2}}\frac{n}{2}\Gamma\left(\frac{n}{2}\right)\quad\text{(because $\Gamma(t+1)=t\Gamma(t)$)}\\\\ &=4\ \text{med}\big(\partial B(0,1)\big)\frac{\Gamma\left(\tfrac{n}{2}\right)}{2\pi^{n/2}}\\\\ &=4\quad \text{(because med$(\partial B(0,1))=\tfrac{2\pi^{n/2}}{\Gamma(n/2)}$)} \end{align*}$$

• For definition and such properties of $\Gamma$, see pages 7 and 8 of Folland's book Introduction to partial differential equations (2nd edition).

Answer 2

First of all, let's try to understand what is $E(0,0;1)$. Remember that $$\Phi(x,t)=\frac{1}{(4\pi t)^{n/2}}e^{\frac{-|x|^2}{4t}},\ \forall\ x\in\mathbb{R}^n,\ \forall\ t>0$$

The function $e^{\frac{-|x|^2}{4t}}$ has range in $(0,1]$, hence, if $\Phi(-y,-t)\geq 1$, we can conclude that $(4\pi (-t))^{n/2}\leq 1$, which implies that $t\in \left(\frac{-1}{4\pi},0\right)$.

Moreover, applying $\log$ in both sides of the equation $\Phi(-y,-t)\geq 1$, we conclude that $$|y|^2\leq 2ns\log{(4\pi (-s))},\ \forall\ s\in \left(\frac{-1}{4\pi},0\right)\tag{2} $$

Note that $(2)$ is the domain of integration. We have that

$$ \int_{-1/4\pi}^0\int_{|y|^2\leq 2ns\log{4\pi (-s)}} \frac{|y|^2}{s^2}dyds= \int_{-1/4\pi}^0\int_{0}^{\sqrt{2ns\log{(4\pi (-s))}}}\frac{r^2}{s^2} r^{n-1}drds\int_{S(0,1)}d\omega $$

where $y=r\omega$ with $\omega\in S(0,1)=\{x\in\mathbb{R}^n:\ |x|=1\}$ (note that $dy=r^{n-1}drd\omega$). The last integral I calculated in Mathematica and got the result $4$.

Remark: (see here) $$\int_{S(0,1)}d\omega=\frac{2\pi^{n/2}}{\Gamma(n/2)}$$

Remark 1: This post may be of some help.