### Author Topic: HA9-P2  (Read 4834 times)

#### Victor Ivrii

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##### HA9-P2
« on: November 14, 2015, 11:38:08 AM »

#### Rong Wei

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##### Re: HA9-P2
« Reply #1 on: November 14, 2015, 04:42:44 PM »
solution to a)

#### Rong Wei

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##### Re: HA9-P2
« Reply #2 on: November 14, 2015, 05:08:57 PM »
solution to b)

#### Chendong Lou

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##### Re: HA9-P2
« Reply #3 on: November 14, 2015, 05:44:35 PM »
This is the answer to part c
« Last Edit: November 14, 2015, 05:49:00 PM by Chendong Lou »

#### Rong Wei

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##### Re: HA9-P2
« Reply #4 on: November 17, 2015, 11:13:51 PM »
I found a mistake in part bï¼Œ the Wn should equal to nï¼Ïƒn

#### Jeremy Li 2

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##### Re: HA9-P2
« Reply #5 on: November 23, 2015, 12:30:29 AM »
Hi everyone. I want to learn LaTeX, so I'm trying to solve (a) problem (using 7.2.7, in a different way that the guy above me). I haven't figured it out fully, and I'm not even sure if this is right. Can someone help?

We begin with equation 7.2.7.

u(y)=\int_{\Omega}G(x,y) \Delta u(x) dV + \int_{\Sigma}-u(x) \frac{\partial G}{\partial \nu_x}(x,y) dS + \int_{\Sigma}G(x,y) \frac{\partial u}{\partial \nu}(x)dS

Let's look at the second term. Let $y$ be at the centre of the sphere, $\Omega$ is a ball, and $\Sigma$ is it's bounding sphere. We can pull $\frac{\partial G}{\partial v_x}$ out of the integral, since when $y$ is at the centre of the sphere, $\frac{\partial G}{\partial \nu_x}(x,y)$ is constant over $\Sigma$. The second term becomes:

\frac{1}{\sigma_n r^{n-1}} \int_{\Sigma}u(x)dS

This is straight out of the textbook. This term represents the average of $u(x,y)$ over the sphere.

All that's left is to prove the remaining terms are negative (or zero) when evaluated at $y$. I'm not sure how to solve this problem for $n<3$, but it is easy for $n\geq3$, because when $n\geq3$, $G(x,y)$ is always negative (see 7.2).

Let's look at the first term. Since $G(x,y)<0$ and $\Delta u(x)\geq0$ on $\Omega$, the integrand of the first term is always non-positive, so the integral is non-positive.

\int_{\Omega}G(x,y) \Delta u(x) dV \leq 0

The third term is easy too. $G(x,y)$ depends on $|x-y|$ only. Thus, if $y$ is at the centre of the sphere, $G(x,y)$ is constant over the sphere $\Sigma$. Pulling it out of the integral, we get

G(x,y) \int_{\Sigma} \frac{\partial u}{\partial \nu} (x) dS

By the divergence theorem, this term is

G(x,y) \int_{\Omega} \Delta u(x) dS

Since $\Delta u(x)$ is non-negative on the domain $\Omega$, so is the integral. Since $G(x,y)$ is always negative at $y$ (at least for $n\geq3$), this term is negative (or zero).

\int_{\Sigma}G(x,y) \frac{\partial u}{\partial \nu}(x)dS \leq 0

Any idea how to approach this problem for lower $n$? If 7.2.7 is true for $n=2$, it seems possible to create a function $u$ such both the first and third term are positive.
« Last Edit: November 23, 2015, 07:35:40 AM by Jeremy Li 2 »

#### Jeremy Li 2

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##### Re: HA9-P2
« Reply #6 on: November 23, 2015, 12:52:27 AM »
For (b),

Knowing that $u(y) \leq \frac{1}{\sigma_n\rho^{n-1}}\int_{S(y,\rho)} u(x) dS$

\int_{B(y,r)} u(x)dV = \int_0^r(\int_{S(y,\rho)}u(x)dS)d\rho \geq \int_0^r(\sigma_n \rho^{n-1} u(y))d\rho = u(y) (\sigma_n \int_0^r \rho^{n-1} d\rho)

So

u(y)\leq\frac{1}{\sigma_n \int_0^r \rho^{n-1} d\rho} \int_{B(y,r)} u(x)dV

Where $(\sigma_n \int_0^r \rho^{n-1} d\rho)$ is the volume of the ball.

#### Jeremy Li 2

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##### Re: HA9-P2
« Reply #7 on: November 23, 2015, 12:57:32 AM »
(c):

u(y) is at least the mean value of u over the sphere S(y,r) bounding the ball $B(y,r)$ in which $\Delta u=0$:

u(y) \geq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u dS

u(y) is at least the mean value of u over the ball $B(y,r)$ in which $\Delta u=0$:

u(y) \geq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u dV