### Author Topic: Problem 2  (Read 6351 times)

#### Fanxun Zeng

• Sr. Member
•    • Posts: 36
• Karma: 11 ##### Problem 2
« on: November 28, 2012, 09:28:03 PM »
« Last Edit: November 29, 2012, 05:09:47 AM by Victor Ivrii »

#### Fanxun Zeng

• Sr. Member
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• Karma: 11 ##### Re: Problem 2
« Reply #1 on: November 28, 2012, 09:30:38 PM »
Problem 2 part c solution attached
« Last Edit: November 28, 2012, 09:38:35 PM by Peter Zeng »

#### Ian Kivlichan

• Sr. Member
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• Karma: 17 ##### Re: Problem 2
« Reply #2 on: November 28, 2012, 09:32:32 PM »
Peter: You've posted the problem, not the solution. ;P

#### Fanxun Zeng

• Sr. Member
•    • Posts: 36
• Karma: 11 ##### Re: Problem 2
« Reply #3 on: November 28, 2012, 09:42:03 PM »
Dear Ian Kivlichan: can you please check what I posted above is solution for part c?
I think it only asks us to formulate similar statements, without proof, so I type tyem.

#### Calvin Arnott ##### Re: Problem 2
« Reply #4 on: November 28, 2012, 09:51:31 PM »
Problem 2
Let: $\Delta u \ge 0$ in $B\left(y,r\right)$, $\omega_n$ be the volume of the $n$-dimensional unit ball, and $\sigma_n = n \omega_n$. Prove that:

Part a. $u\left(y\right)$ does not exceed the mean value of $u$ over the sphere $S\left(y,r\right)$ bounding the ball $B\left(y,r\right)$:
\begin{equation}
u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \label{2.a.0}
\end{equation}

In $\Delta u \ge 0$, $u$ is twice differentiable, so the spherical mean is continuous and we have:
\begin{equation}
\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \rightarrow u\left(y\right) \label{2.a.1}
\end{equation}
Thus, to prove $u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$, it is sufficient to show that $\frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$ increases with $r > 0$. By the divergence theorem we have for a $C^1$ function $F$ on a compact set $\Omega$ with boundary $\partial\Omega = \Sigma$:
$$\int_\Omega \left(\nabla \cdot F\right) \,dV = \int_\Sigma \left(F \cdot n\right) \, dS \text{, namely: }$$
$$\int_{B\left(y,r\right)} \left(\nabla \cdot \nabla \right)\left(x\right) \,dV = \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = \int_{S\left(y,r\right)} \left(\nabla u \cdot n\right)\left(x\right) \, dS = \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS$$
As $\Delta u \ge 0$, we then have:
$$0 \le \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS$$
Parametrizing by the normal $\nu = \frac{x-y}{|x-y|}$ on the boundary where $|x-y| = r$, we have: $\frac{\partial u}{\partial n} \left(x\right) = \partial_r u\left(y+ r \nu\right)$ on $T : \{ \nu : | \nu | = 1 \}$, and $dS = r^{n-1} d \nu$
What? See note below V.I..

Then:
$$\int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS = \int_{T} \partial_r\left( u\left(y+ r \nu\right)\right) r^{n-1} \, d \nu$$
By differentiating under the sign with $\partial_r$ and pulling the $r^{n-1}$ term out of the integral, we have:
$$= r^{n-1} \partial_r \int_{T} u\left(y+ r \nu\right) \, d \nu = r^{n-1} \partial_r \left(r^{1-n} \int_{S\left(y,r\right)} u\left(x\right) \, d S\right)$$
$$= r^{n-1} \partial_r\left(\sigma_n \frac{r^{1-n}}{\sigma_n } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
As $r>0$ and $\sigma_n>0$ by \eqref{2.a.1} we have:
$$0 \le \int_{S\left(y,r\right)} \frac{\partial u}{\partial n}\left(x\right) \, dS = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right) \implies$$
\begin{equation}
0 \le \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)  \label{2.a.2}
\end{equation}
Then by \eqref{2.a.1} as $r \rightarrow 0$ we have equality in \eqref{2.a.0}, and as $r$ increases the spherical mean is non-decreasing by \eqref{2.a.2}, $u\left(y\right)$ constant. Thus we have \eqref{2.a.0} for $r \ge 0$:
$$u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \phantom{O} \blacksquare$$

Part b. $u\left(y\right)$ does not exceed the mean value of $u$ over the ball $B\left(y,r\right)$:
\begin{equation}
u\left(y\right) \le \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \label{2.b.0}
\end{equation}

$$u\left(y\right) = \frac{\sigma_n \left(n r^n\right)}{\left(n \sigma_n\right) r^n} u\left(y\right) = \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u\left(y\right)$$
By \eqref{2.a.0} we have that:
$$u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$$
So because $\{r, \omega_n, \sigma_n\} \ge 0$ we have that:
$$\implies \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u\left(y\right) \le \frac{\sigma_n \int_{0}^{r} t^{n-1}\, dt}{\omega_n r^n} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$$
$$= \frac{1}{\omega_n r^n} \sigma_n\int_{0}^{r}t^{n-1} \,d t \frac{1}{\sigma_n r^{n-1}} \int_T u\left(y +r \nu\right) r^{n-1} \, d \nu$$
Where we parameterized as in part a. with $\nu = \frac{x-y}{|x-y|}$ on the boundary $|x-y| = r$, and $\left(x\right) = u\left(y+ r \nu\right)$ on $T : \{ \nu : | \nu | = 1 \}$, $dS = r^{n-1} d \nu$. Canceling terms gives:
$$= \frac{1}{\omega_n r^n} \int_{0}^{r}t^{n-1} \,d t \int_T u\left(y +r \nu\right) \, d \nu$$
Which allows us to change the order of integration to yield our result:
$$= \frac{1}{\omega_n r^n} \int_{0}^{r} t^{n-1} \int_T u\left(y +r \nu\right) \, d \nu d t = \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V$$
$$\implies u\left(y\right) \le \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \phantom{O} \blacksquare$$

Part c. Formulate similar statements for functions satisfying $\Delta u \le 0$.

In part a. we used the fact that $\Delta u \ge 0$, and $\{ \sigma_n, r \} \ge 0$ to show that the spherical mean is non-decreasing in $r$:
$$0 \le \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
$$\implies 0 \le \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
Because we have equality at $r \rightarrow 0$:
$$\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \rightarrow u\left(y\right)$$
and $u\left(y\right)$ is constant with respect to $r$, we showed that for all $r > 0$
$$u\left(y\right) \le \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S$$
If instead $\Delta u \le 0$, we would have the inequalities:
$$0 \ge \int_{B\left(y,r\right)} \Delta u\left(x\right) \,dV = r^{n-1} \sigma_n \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
$$\implies 0 \ge \partial_r\left( \frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S \right)$$
And $\frac{1}{\sigma_n r^{n-1} } \int_{S\left(y,r\right)} u\left(x\right) \, d S$ is non-increasing with $r$, which gives us by the same argument that for all $r>0$
$$u\left(y\right) \ge \frac{1}{\sigma_n r^{n-1}} \int_{S\left(y,r\right)} u\left(x\right) \, d S \phantom{O} \square$$
In part b. we merely integrated this result radially to extend the inequality for the interior of the sphere. Doing the same in the case of $\Delta u \le 0$ yields us:
$$u\left(y\right) \ge \frac{1}{\omega_n r^{n}} \int_{B\left(y,r\right)} u \, d V \phantom{O} \blacksquare$$
« Last Edit: November 29, 2012, 05:18:48 AM by Victor Ivrii »

#### Ian Kivlichan

• Sr. Member
•    • Posts: 51
• Karma: 17 ##### Re: Problem 2
« Reply #5 on: November 28, 2012, 09:53:34 PM »
Peter: I think you should justify how you got to your result, especially since showing how changing $\Delta u$ from positive-definite to negative-definite flips the sign of the inequality in 2.b) is very quick. (As Calvin has now done.)

#### Fanxun Zeng

• Sr. Member
•    • Posts: 36
• Karma: 11 ##### Re: Problem 2
« Reply #6 on: November 28, 2012, 11:41:49 PM »
Dear Ian: I strongly agree with you we should justify for math education purpose. For TA marking purpose only, I think what I posted at 21:30 before Calvin is correct to get full marks, as question part c only requires "to formulate similar statements".

#### Victor Ivrii ##### Re: Problem 2
« Reply #7 on: November 29, 2012, 05:26:23 AM »
I think Calvin meant

$dS=r^{n-1}d\theta$ where $d\theta$ is an area element on the unit sphere, expressed solely in coordinates on it. Then average is $\sigma_n^{-1}\int u\,d\theta$ and we can differentiate (we could not with $dS$ as it invoked $r$). Following Calvin
$$\partial_r \int_S u\,d\theta = \int_S \partial_r u\,d\theta= r^{1-n}\int _V \Delta u dV\ge 0.$$

(c) makes no sense without (a),(b) and solution is simple: $u\mapsto -u$.