### Author Topic: Solution of Question 2  (Read 2902 times)

#### Chaojie Li

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##### Solution of Question 2
« on: March 19, 2015, 09:06:43 PM »
Using the proof of mean value theorem prove that if $\Delta u\ge 0$ in $B(y,r)$ then

1.  $u(y)$ does not exceed the mean value of $u$ over the sphere $S(y,r)$ bounding this ball:

u(y)\le \frac{1}{\sigma_n r^{n-1}}\int_{S(y,r)} u\,dS.
\label{equ-H8.2}

2.  $u(y)$ does not exceed the mean value of $u$ over this ball $B(y,r)$:

u(y)\le \frac{1}{\omega_n   r^n}\int_{B(y,r)} u\,dV.
\label{equ-H8.3}

3.  Formulate similar statements for functions satisfying $\Delta u\le 0$ (in the next problem we refer to them as (a)' and (b)').

Definition

a.  Functions having property (a) (or (b) does not matter) of the  previous problem are called  subharmoni

b.  Functions having property (a)' (or (b)' does not matter) are called   superharmonic.

Well, so 9 pm is good to post solution right?
« Last Edit: March 20, 2015, 07:44:26 AM by Victor Ivrii »

#### Chaojie Li

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##### Re: Solution of Question 2
« Reply #1 on: March 19, 2015, 09:07:08 PM »
part b

#### Chaojie Li

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##### Re: Solution of Question 2
« Reply #2 on: March 19, 2015, 09:07:29 PM »
part c

#### Biao Zhang

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##### Re: Solution of Question 2
« Reply #3 on: March 19, 2015, 09:36:50 PM »
(a)

u(y) \leq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S

In $\Delta u \geqq 0$, $u$ is twice differentiable, so the spherical mean is continuous and we have:

\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \rightarrow u(y) \label{2.a.1}

Thus, to prove $u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S$, it is sufficient to show that $\frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S$ increases with $r > 0$. By the divergence theorem we have for a $C^1$ function $F$ on a compact set $\Omega$ with boundary $\partial\Omega = \Sigma$:

$$\int_\Omega (\nabla \cdot F) \,dV = \int_\Sigma (F \cdot n) \, dS \text{, namely: }$$

$$\int_{B(y,r)} (\nabla \cdot \nabla )(x) \,dV = \int_{B(y,r)} \Delta u(x) \,dV = \int_{S(y,r)} (\nabla u \cdot n)(x) \, dS = \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS$$

As $\Delta u \geqq 0$, we then have:

$$0 \leqq \int_{B(y,r)} \Delta u(x) \,dV = \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS$$

Parametrizing by the normal $\nu = \frac{x-y}{|x-y|}$ on the boundary where $|x-y| = r$, we have: $\frac{\partial u}{\partial n} (x) = \partial_r u(y+ r \nu)$ on $T : \{ \nu : | \nu | = 1 \}$, and $dS = r^{n-1} d \nu$

Then:

$$\int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS = \int_{T} \partial_r( u(y+ r \nu)) r^{n-1} \, d \nu$$

By differentiating under the sign with $\partial_r$ and pulling the $r^{n-1}$ term out of the integral, we have:

$$= r^{n-1} \partial_r \int_{T} u(y+ r \nu) \, d \nu = r^{n-1} \partial_r (r^{1-n} \int_{S(y,r)} u(x) \, d S)$$

$$= r^{n-1} \partial_r(\sigma_n \frac{r^{1-n}}{\sigma_n } \int_{S(y,r)} u(x) \, d S ) = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

As $r>0$ and $\sigma_n>0$ by (2) we have:

$$0 \leqq \int_{S(y,r)} \frac{\partial u}{\partial n}(x) \, dS = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S ) \implies$$

0 \leqq \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )

Then by (2) as $r \rightarrow 0$ we have equality in (1), and as $r$ increases the spherical mean is non-decreasing by (3), $u(y)$ constant. Thus we have (1) for $r \geqq 0$:

$$u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \phantom{O}$$

(b)

$$u(y) = \frac{\sigma_n (n r^n)}{(n \sigma_n) r^n} u(y) = \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u(y)$$

By (1) we have that:

$$u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S$$

So because $\{r, \omega_n, \sigma_n\} \geqq 0$ we have that:

$$\implies \frac{\sigma_n \int_{0}^{r} t^{n-1} \, dt}{\omega_n r^n} u(y) \leqq \frac{\sigma_n \int_{0}^{r} t^{n-1}\, dt}{\omega_n r^n} \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S$$

$$= \frac{1}{\omega_n r^n} \sigma_n\int_{0}^{r}t^{n-1} \,d t \frac{1}{\sigma_n r^{n-1}} \int_T u(y +r \nu) r^{n-1} \, d \nu$$

Where we parameterized as in part a. with $\nu = \frac{x-y}{|x-y|}$ on the boundary $|x-y| = r$, and $(x) = u(y+ r \nu)$ on $T : \{ \nu : | \nu | = 1 \}$, $dS = r^{n-1} d \nu$. Canceling terms gives:

$$= \frac{1}{\omega_n r^n} \int_{0}^{r}t^{n-1} \,d t \int_T u(y +r \nu) \, d \nu$$

Which allows us to change the order of integration to yield our result:

$$= \frac{1}{\omega_n r^n} \int_{0}^{r} t^{n-1} \int_T u(y +r \nu) \, d \nu d t = \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u \, d V$$

$$\implies u(y) \leqq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u \, d V \phantom{O}$$

(c)

In part a. we used the fact that $\Delta u \geqq 0$, and $\{ \sigma_n, r \} \geqq 0$ to show that the spherical mean is non-decreasing in $r$:

$$0 \leqq \int_{B(y,r)} \Delta u(x) \,dV = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

$$\implies 0 \leqq \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

Because we have equality at $r \rightarrow 0$:

$$\lim_{r \rightarrow 0^+} \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \rightarrow u(y)$$
and $u(y)$ is constant with respect to $r$, we showed that for all $r > 0$

$$u(y) \leqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S$$

If instead $\Delta u \leqq 0$, we would have the inequalities:

$$0 \geqq \int_{B(y,r)} \Delta u(x) \,dV = r^{n-1} \sigma_n \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

$$\implies 0 \geqq \partial_r( \frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S )$$

And $\frac{1}{\sigma_n r^{n-1} } \int_{S(y,r)} u(x) \, d S$ is non-increasing with $r$, which gives us by the same argument that for all $r>0$

$$u(y) \geqq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u(x) \, d S \phantom{O} \square$$

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 $\Delta u \leqq 0$ yields us:

$$u(y) \geqq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u \, d V \phantom{O}$$