Author Topic: Problem 3  (Read 4375 times)

Fanxun Zeng

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Problem 3
« on: November 28, 2012, 09:28:25 PM »
Problem 3 Solutions posting after 21:30 Edit: following professor's advice, web link:
http://www.math.toronto.edu/courses/apm346h1/20129/HA8.html#problem-8.3
« Last Edit: November 29, 2012, 05:20:51 AM by Peter Zeng »

Fanxun Zeng

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Re: Problem 3
« Reply #1 on: November 28, 2012, 09:31:20 PM »
Problem 3 part a solution attached

Fanxun Zeng

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Re: Problem 3
« Reply #2 on: November 28, 2012, 09:35:47 PM »
Problem 3 part a maximum principle solution attached mimimum principle is similar

Calvin Arnott

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Re: Problem 3
« Reply #3 on: November 28, 2012, 09:55:02 PM »
Problem 3

Part a. Using the proof of the maximum principle, prove the maximum principle for subharmonic functions, and the minimum principle for superharmonic functions.

Answer:
Maximum principle for subharmonic functions:
Let: $u\left(\mathbf{x}\right)$ be subharmonic, $ \Delta u\left(\mathbf{x}\right) \ge 0$, on a bounded domain $\Omega$ with boundary $\partial\Omega = \Sigma$, $\mathbf{x} \in \Omega$. Let: $v\left(\mathbf{x}\right) = u\left(\mathbf{x}\right) + \epsilon |\mathbf{x}|^2$ where $|\mathbf{x}|^2 = \sum_{i=1}^{n} x_i^2 $, $\epsilon > 0$. Let $\delta$ be the diameter of the set $\bar\Omega$, namely the largest value of $| \mathbf{x}|^2 = \delta^2$.
At any interior maximum point of a function $f$ we require $f_{x_i x_i} \le 0$, $\Delta f = \sum_{i=1}^{n}f_{x_i x_i} \le 0$ by the second derivative test. Notice that for $\mathbf{x} \in \Omega$:
$$ \Delta v\left(\mathbf{x}\right) = \Delta u\left(\mathbf{x}\right) + \Delta \epsilon |\mathbf{x}|^2 \ge 0 + 2 n \epsilon > 0  $$
So $v$ has no interior maximum on $\Omega$. But $\Omega$ is bounded so $v$ has a maximum somewhere on $\bar\Omega$. As there is no interior maximum of $v$ it must have its maximum on $\partial\Omega = \Sigma$. Let: $\mathbf{x_o}$ be the maximum of $v$ on $\bar\Omega$. Then, for $\mathbf{x} \in \bar\Omega$:
$$ u\left(\mathbf{x}\right) \le v\left(\mathbf{x}\right) \le v\left(\mathbf{x_0}\right) = u\left(\mathbf{x_0}\right) + \epsilon |\mathbf{x_0}|^2 \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) + \epsilon \delta^2 $$
As $\delta$ is some constant and we have this inequality for all $\epsilon > 0$, we take the limit as $\epsilon \rightarrow 0$ to yield:
$$ \lim_{\epsilon \rightarrow 0} \{u\left(\mathbf{x}\right) \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) + \epsilon \delta^2 \} \rightarrow \{ u\left(\mathbf{x}\right) \le \max_{\mathbf{x} \in \Sigma} u\left(\mathbf{x}\right) \} $$
We have the maximum of $u\left(\mathbf{x}\right)$ is attained on $\mathbf{x} \in \Sigma = \partial\Omega$, as needed. $\square$

Minimum principle for subharmonic functions:
Let: $u\left(\mathbf{x}\right)$ be superharmonic, $ \Delta u\left(\mathbf{x}\right) \le 0$, on a bounded domain $\Omega$ with boundary $\partial\Omega = \Sigma$, $\mathbf{x} \in \Omega$. Let: $v\left(\mathbf{x}\right) = - u\left(\mathbf{x}\right)$.
Then $v\left(\mathbf{x}\right)$ is subharmonic and attains its maximum on $\Sigma$, by the maximum principle for subharmonic functions.  This is equivalent to a minimum of $u$ and so $u\left(\mathbf{x}\right)$ attains its minimum on $\Sigma = \partial\Omega$. We are done. $\blacksquare$


Part b. Show that the minimum principle for subharmonic functions, and maximum principle for superharmonic functions do not hold.

Answer:
Let: $u\left(\mathbf{x}\right) = |\mathbf{x}|^2 = \sum_{i=1}^{n} x_i^2$ defined on some ball in $\mathbb{R}^n$ about the origin with radius larger than $0$.
Now: $\Delta u = \Delta |\mathbf{x}|^2 = 2 n > 0$ so $u$ is subharmonic. But clearly at $ \mathbf{x} = \mathbf{0}$, $u\left(\mathbf{x}\right) = |\mathbf{x}|^2 = 0$ but at any $\mathbf{x} \ne \mathbf{0}$, $ |\mathbf{x}|^2 > 0$. Then $u$ has an interior minimum and the minimum principle for subharmonic functions does not hold. $\square$
Let: $u\left(\mathbf{x}\right)$ be subharmonic, and $v\left(\mathbf{x}\right) = - u\left(\mathbf{x}\right)$. Then $v$ is superharmonic.
Because the minimum principle does not hold for subharmonic functions as shown, $u$ may have an interior minimum. This is equivalent to an interior maximum for $v$, so the maximum principle does not hold for superharmonic functions. We are done. $\blacksquare$


Part c. Prove that if $u, v, w$ are respectively harmonic, subharmonic, and superharmonic functions in the bounded domain $ \Omega$, coinciding on its boundary $ \{u \bigr|_\Sigma= v\bigr|_\Sigma= w\bigr|_\Sigma\}$, then $w \ge u \ge v$ in $\Omega$.

Answer:
Let: $\Delta u = 0$, $\Delta v \ge 0$, $ \Delta w \le 0$ on $ \Omega$, coinciding on its boundary $\partial\Omega = \Sigma$,  $ \{u \bigr|_\Sigma= v\bigr|_\Sigma, w\bigr|_\Sigma\}$.
Define: $f = v - u$, $g = w - u$.
Then: $\Delta f = \Delta v - \Delta u \ge  0$, $\Delta g = \Delta w - \Delta u \le 0$, and: $ \{u \bigr|_\Sigma= v\bigr|_\Sigma= w\bigr|_\Sigma\}$ so:
$$ \{ f \bigr|_\Sigma= \left(v-u\right)\bigr|_\Sigma = 0, \, g \bigr|_\Sigma= \left(w-u\right)\bigr|_\Sigma = 0 \} $$
$\Delta f \ge 0$, so $f$ is subharmonic and by the maximum principle attains its maximum on $\Sigma$. Then in $\Omega$, $f \le f \bigr|_\Sigma = 0$ so $f = v - u \le 0 \implies v \le u$ in $\Omega$.
Similarly, $\Delta g \le 0$ so $g$ is superharmonic and by the minimum principle attains its minimum on $\Sigma$. So in $\Omega$, $g \ge g \bigr|_\Sigma = 0$ so $g = w - u \ge 0 \implies w \ge u$ in $\Omega$.
We have in $\Omega$, $v \le u$, and $ w \ge u$. So $w \ge u \ge v$ inside of $\Omega$, as needed. $\blacksquare$
« Last Edit: November 28, 2012, 10:00:13 PM by Calvin Arnott »

Fanxun Zeng

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Re: Problem 3
« Reply #4 on: November 28, 2012, 11:25:34 PM »
Problem 3 b my own solution attached for 2 counter examples, different from Calvin's solution, I just typed it in Paint thanks! 
« Last Edit: November 28, 2012, 11:29:54 PM by Peter Zeng »

Victor Ivrii

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Re: Problem 3
« Reply #5 on: November 29, 2012, 05:34:53 AM »
Calvin, you proved that if $\Delta u\ge 0$ then minimum principle holds (a standard proof used in textbooks). However I gave you a weaker definition: just property of spherical means which allows even discontinuous functions.

Clearly with the counter-examples one better follow a stronger definition--but with the proof one needs a weaker one. Also after we proved property of spherical means methinks the proof in the Lecture Note is much simple.


We were always skipping proofs of existence as usually it requires a good Real Analysis command. I just want to mention that one of the possible proofs of the existence of solution to the problem $\Delta u=0$ in $\Omega$, $u=g$ on $\partial \Omega$ (boundary) is the following.

Note first that if we have a finite set  of subharmonic functions then $V(x)= \max _v v(x)$ will be also subharmonic. It follows trivially from Definition through spherical means: As $V(x)=v(x)$ for given $x$ then $V(x)$ does not exceed spherical means of $v$ and thus it does not exceed spherical means of $V$ (as V\ge v$). For infinite sets similar property holds but wit $\sup$ instead of $\max$.

Similar property for superharmonic functions (with $\min$ or $\inf$).

 Consider all subharmonic functions $v$ s.t. $v=g$ on $\Omega$ and all superharmonic functions $w$ s.t. $w=g$ on $\Omega$. Then $w\ge v$ for all choices.

Define  $V(x)=\sup _v v(x)$, $W(x)=\inf _w w(x)$ (where supremum and infimum are taken over all such functions). Then $W\ge V$.

Further one proves by contradiction that $V=W$ and therefore is both super and subharmonic. Finally one proves that then it must be harmonic.
« Last Edit: November 29, 2012, 05:58:53 AM by Victor Ivrii »