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**Final Exam / Re: problem 5**

« **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.

$$u(x)=\frac{1}{\pi r^2}\int_{B_r(x)}u dA$$

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.