Author Topic: problem 5  (Read 4589 times)

Djirar

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problem 5
« on: December 20, 2012, 01:30:49 PM »
Suppose $\Delta u = 0$ and satisfies $|u| < 1000$ everywhere on $\mathbb{R}^2$.

Prove that $u$ is a constant function. In other words, show that there exists a constant $C$ so that $u (x) = C$ everywhere on $\mathbb{R}^2$.

Hint: Mean Value Theorem
« Last Edit: December 20, 2012, 01:59:57 PM by Victor Ivrii »

Victor Ivrii

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Re: problem 5
« Reply #1 on: December 20, 2012, 02:03:03 PM »
This was a proof that maximum cannot be reached in internal points unless $u$ is constant. Not a solution of our problem.

Jinchao Lin

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Re: problem 5
« Reply #2 on: December 20, 2012, 02:50:21 PM »
Hopeful solution for problem 5

Rouhollah Ramezani

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Re: problem 5
« Reply #3 on: December 20, 2012, 02:51:13 PM »
By Mean Value property for harmonic functions,
$$u(x)=\frac{1}{\pi r^2}\int_{B_r(x)}u dA$$
where $x \in R^2$ is a point, $B_r(x)$ is the ball of radius $r$ centred at $x$ and the integration is over area.
When $r \to \infty$, integral above becomes the average of $u$ over whole plain. Since this average is a constant, we conclude $u(x)$ should be a constant function.
To state this argument more rigorously;  Let $x_1$ and $x_2$ be distinct in $R^2$ and $|x_1-x_2|=a$:
Then by Mean Value theorem
\begin{equation}u(x_2)=\frac{1}{\pi r^2}\int_{B_r(x_2)}u dA
\end{equation}
\begin{equation}u(x_1)=\frac{1}{\pi (r+a)^2}\int_{B_{r+a}(x_1)}u dA
\end{equation}
Where ${B_r(x_2)}$ is the $r$-ball around $x_2$ and ${B_{r+a}(x)}$ is the ball with radius $r+a$ around $x_1$.

Rewriting (2) we get:
$$u(x_1)=\frac{1}{\pi (r+a)^2}\int_{B_{r}(x_2)}u dA + \frac{1}{\pi (r+a)^2}\int_{S}u dA$$
Where $S=B_{r+a}(x_1)$\ $ B_{r}(x_2) \in R^2$ is the hashed area in the graph.
We shall prove $u(x_2)-u(x_1)=0$. Since (1) and (2) are true for all $r$, it suffices to show $$lim_{r \to \infty} \frac{1}{\pi r^2}\int_{B_r(x_2)}u dA- \frac{1}{\pi (r+a)^2}\int_{B_{r+a}(x_1)}u dA$$
$$=lim_{r \to \infty} \frac{1}{\pi r^2}\int_{B_r(x_2)}u dA-\frac{1}{\pi (r+a)^2}\int_{B_{r}(x_2)}u dA - \frac{1}{\pi (r+a)^2}\int_{S}u dA=0$$
But first two terms cancel out in limit and remains
$$u(x_2)-u(x_1)=lim_{r \to \infty} \frac{-1}{\pi (r+a)^2}\int_{S}u dA=0$$
Since $u$ is bounded we can write
\begin{equation}
\begin{aligned}
lim_{r \to \infty} \frac{1}{\pi (r+a)^2}\int_{S}u dA & \leq lim_{r \to \infty} \frac{1}{\pi (r+a)^2} u_{max} S_a \\
& =lim_{r \to \infty} \frac{u_{max}[\pi (r+a)^2-\pi (r)^2]}{\pi (r+a)^2} \\
& =lim_{r \to \infty} \frac{u_{max}(a^2-2ar)}{(r+a)^2}=0 \\
\end{aligned}
\end{equation}
Where $u_{max}$ is the upper bound of $u$ and $S_a=\pi (r+a)^2-\pi (r)^2$ is the area of $S$.
Hence $u(x_2)-u(x_1)=0 \qquad \forall x_1,x_2 \in R^2$ and $u$ is constant.

Victor Ivrii

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Re: problem 5
« Reply #4 on: December 20, 2012, 05:11:18 PM »
This is a nice solution. Definitely you cannot presume a priori that an average exists but the analysis with 2 points does not require it. Obviously this solution generalizes to any dimension.

Alternative solution: from formula for solution of the Dirichlet problem in the disk
$$
u(x,y)= \frac{1}{2\pi} \int_0^{2\pi} \frac{R^2- (x^2+y^2)}{R^2+x^2+y^2 - 2R(x\cos (\theta)+y\sin(\theta))}\,d\theta.
$$
Then
$$
u_x(x,y)= \frac{1}{2\pi} \int_0^{2\pi} \frac{-2x\bigl[R^2+x^2+y^2 - 2R(x\cos (\theta)+y\sin(\theta))\bigr] +\bigl[R^2- (x^2+y^2)] [2x - 2R\cos (\theta)\bigr]}{\bigl[R^2+x^2+y^2 - 2R(x\cos (\theta)+y\sin(\theta))\bigr]^2}\,d\theta
$$
and as $R\to+\infty$  $u_x\to 0$. Similarly $u_y=0$.

One can prove

Theorem. If $\Delta u=0$ and $|u|\le M(1+|x|)^m $ for all $x\in \mathbb{R}^n$ then $u$ is a polynomial of degree $\le m$.