Toronto Math Forum
APM346-2015F => APM346--Home Assignments => HA9 => Topic started by: Victor Ivrii on November 14, 2015, 11:38:08 AM
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Problem 2
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter7/S7.P.html#problem-7.P.2 (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter7/S7.P.html#problem-7.P.2)
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solution to a)
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solution to b)
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This is the answer to part c
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I found a mistake in part b， the Wn should equal to nï¼Ïƒn
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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.
\begin{equation}
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
\end{equation}
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:
\begin{equation}
\frac{1}{\sigma_n r^{n-1}} \int_{\Sigma}u(x)dS
\end{equation}
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.
\begin{equation}
\int_{\Omega}G(x,y) \Delta u(x) dV \leq 0
\end{equation}
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
\begin{equation}
G(x,y) \int_{\Sigma} \frac{\partial u}{\partial \nu} (x) dS
\end{equation}
By the divergence theorem, this term is
\begin{equation}
G(x,y) \int_{\Omega} \Delta u(x) dS
\end{equation}
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).
\begin{equation}
\int_{\Sigma}G(x,y) \frac{\partial u}{\partial \nu}(x)dS \leq 0
\end{equation}
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.
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For (b),
Knowing that $u(y) \leq \frac{1}{\sigma_n\rho^{n-1}}\int_{S(y,\rho)} u(x) dS$
\begin{equation}
\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)
\end{equation}
So
\begin{equation}
u(y)\leq\frac{1}{\sigma_n \int_0^r \rho^{n-1} d\rho} \int_{B(y,r)} u(x)dV
\end{equation}
Where $(\sigma_n \int_0^r \rho^{n-1} d\rho)$ is the volume of the ball.
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(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 $\Delta u=0$:
\begin{equation}
u(y) \geq \frac{1}{\sigma_n r^{n-1}} \int_{S(y,r)} u dS
\end{equation}
u(y) is at least the mean value of u over the ball $B(y,r)$ in which $\Delta u=0$:
\begin{equation}
u(y) \geq \frac{1}{\omega_n r^{n}} \int_{B(y,r)} u dV
\end{equation}