### Author Topic: HA9-P4  (Read 2177 times)

#### Victor Ivrii

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##### HA9-P4
« on: November 14, 2015, 11:43:00 AM »

#### Catch Cheng

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##### Re: HA9-P4
« Reply #1 on: November 14, 2015, 05:42:43 PM »
Please correct me if something is wrong, thanks.

#### Jeremy Li 2

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##### Re: HA9-P4
« Reply #2 on: November 23, 2015, 01:33:30 AM »
Hi again, this is my solution to part b. (Oops, I didn't realize P4 had 3 parts, so I did this problem before proving the maximum principle for subharmonic functions).

Let's prove that $u \geq v$. $u$ is harmonic and $v$ is subharmonic, so

\Delta{u} = 0 \\
\Delta{v} \geq 0

Define $g=v-u$. Then:

\Delta g \geq 0\\
g|_\Sigma = 0

We only need to prove that $g\leq0$ to obtain our desired result.

Let's assume, on the other hand that $g>0$. Then since $g$ is zero on the boundary, $g$ must have a maximum that is on the interior of $\Omega$.

Let $y$ denote the point on $\Omega$ where $g$ is maximum. Then we draw a ball around it $B(y,r)$ such that it remains inside the domain $\Omega$.

Now we assume the result of 2b, that $g(y)$ does not exceed the average of $g$ on this ball. But $g(y)$ is the global maximum, and also the maximum on this ball. The only way these can hold simultaneously is if $g = g(y)$ on this entire ball. If on $B(r,y)$, $g\neq g(y)$, then $g \leq g(y)$, and $g(y)$ would certainly exceed the average of $g$ on $B(r,y)$. Thus we conclude $g = g(y)$ on $B(r,y) \in \Omega$.

Now within this ball we take a new point $y'$. Then $g(y')$ is also the global maximum on $\Omega$. We repeat this over and over again and discover that on a connected $\Omega$, $g$ is identically equal to $g(y)$ and so $g>0$ on the boundary. This contradicts $g|_\Sigma = 0$. Therefore our original assumption cannot be correct, and so $g\leq0$. This proves that $u\geq v$.

Doing the exact same thing with $f=u-v$ proves that $v\geq u$
« Last Edit: November 23, 2015, 01:54:07 AM by Jeremy Li 2 »

#### Jeremy Li 2

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##### Re: HA9-P4
« Reply #3 on: November 23, 2015, 01:52:09 AM »
This is my solution to 1a).

Proof of the maximum principle for subharmonic functions

Let $u$ be subharmonic.

\Delta u \geq 0

To prove that

\max_{\Omega}u=\max_{\Sigma}u

Assume, on the contrary, that the maximum is not on the boundary. Then the maximum must lie in the interior of $\Omega$. Call this point $y$. Then we draw a ball around it $B(y,r)$ such that it remains inside the domain $\Omega$. $u(y)$ is the global maximum of $\Omega$.

Now we assume the result of 2b, that $u(y)$ does not exceed the average of $u$ on this ball. But $u(y)$ is the global maximum, and thus the maximum on this ball. The only way these can hold simultaneously is if $u = u(y)$ on this entire ball. For if on $B(r,y)$, $u\neq u(y)$, then $u \leq u(y)$, and $u(y)$ would certainly exceed the average of $u$ on $B(r,y)$. Thus we conclude $u = u(y)$ on $B(r,y) \in \Omega$.

Now within this ball we take a new point $y'$. Then $g(y')$ is also the global maximum on $\Omega$. We repeat this over and over again and discover that on a connected $\Omega$, $g$ is identically equal to $g(y)$ including on the boundaries. Therefore, the $g$ is constant on all of $\Omega$, and so the maximum is again on the boundary, contradicting our original assumption.

Proof of the maximum principle for subharmonic functions

This proof is very similar.
« Last Edit: November 23, 2015, 01:58:25 AM by Jeremy Li 2 »