Prove a function is harmonic(use Green formula)

Question

A real valued function $u$, defined in the unit disk, $D_1$ is harmonic if it satisfies the partial differential equation $\partial_{xx} u +\partial_{yy} u = 0$. Prove that a such function $u$ defined in $D_1$ is harmonic if and only if for each $(x, y) ∈ D_1$

$$ u(x,y)=\frac{1}{2\pi}\int_{0}^{2\pi}u(x+r~\cos{(\theta)},y+r~\sin{(\theta)})d\theta $$

for sufficiently small positive $r$.Hint: Recall Green’sformula: $$ \int_{D} \Delta u dA=\int_{b D}\partial_v\nu ds $$ I have no ideas on this question. Can you help me? Thank you!

Answer 1

Starting with Green's formula $\iint_{D} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} dxdy=\oint_{bD}Pdx+Qdy$, take $Q=u_x$, $P=-u_y$, and $D$ a small circle of radious r centered at $(x_0,y_0)$. We get:

$$ \iint_{D} \Delta u \space dA = \oint_{\partial D} \nabla u \cdot (dy,-dx) .$$

If we parametrized the boundary of $D$ as: $$ \begin{equation} x(\theta) = x_0 + r\cos(\theta) \\ y(\theta) = y_0 + r\sin(\theta) \\ \end{equation} $$ then $$ (dy,-dx) = r (\cos(\theta), \sin(\theta)) d\theta = r \nu d\theta $$ where $\nu$ is the exterior normal to $\partial D$, so the Green's formula you mentioned implies the formula in the problem suggestion.

Since $u$ is harmonic, the left side is $0$ and we get: $$ 0 = \oint_{\partial D} \nabla u \cdot (dy,-dx)=\\ \int_{0}^{2\pi} \{u_x(x_0 +r\cos(\theta), y_0 + r\sin(\theta)) r\cos(\theta) + \\ \space u_y(x_0 +r\cos(\theta), y_0 + r\sin(\theta)) r\sin(\theta)\} \space d\theta $$

Now we compare what we have and what we want to get. Let me point some obvious facts (hind side is 20/20...) :
* the last formula involves derivatives of $u$, but the problem has only $u$, without derivatives, which suggest we will need to differentiate something
* the right side of the formula involves $r$, but the left side has no $r$, so its derivative with respect to $r$ is $0$

You can give it the final kick.

$$\dots\dots\dots$$
Now we observed that
$$ 0 = r \space \int_{0}^{2\pi} \{u_x(x_0 +r\cos(\theta), y_0 + r\sin(\theta)) \cos(\theta) + \\ \space u_y(x_0 +r\cos(\theta), y_0 + r\sin(\theta)) \sin(\theta)\} \space d\theta =\\ = r \space \frac{\partial}{\partial r} \int_{0}^{2\pi} u(x_0 +r\cos(\theta), y_0 + r\sin(\theta)) \space d\theta .$$ Therefor the integral over the boundary of the little circle is constant, and the value can be found by taking limit as $r \rightarrow 0^+$

Answer 2

Here is a simple proof of the mean-value theorem for harmonic functions of two variables. We start with Green's Third Identity

$$u(\vec x_0)=\oint_C \left(u(\vec x')\frac{\partial G(\vec x_0,\vec x')}{\partial n'}-G(\vec x_0,\vec x')\frac{\partial u(\vec x')}{\partial n'}\right)\,d \ell' \tag 1$$

where $u(\vec x)$ is harmonic in a region $S$ bounded by the smooth contour $C$ and $G(\vec x_0,\vec x')=\frac{1}{2\pi}\log|\vec x_0-\vec x'|$ is the Green's Function for the Laplacian.

If we choose $S$ to be the region $|\vec x_0-\vec x'|\le r$, and thus $C$ to be the circle of radius $r$ and center $\vec x_0$, then on $C$ we have

$$\begin{align} G(\vec x_0,\vec x')&=\frac{1}{2\pi}\log(r) \tag 2\\\\ \frac{\partial G(\vec x_0,\vec x')}{\partial n'}&=\frac{1}{2\pi r} \tag 3 \end{align}$$

Using $(2)$ and $(3)$ in $(1)$ yields

$$\begin{align} u(\vec x_0)&=\frac{1}{2\pi r}\oint_C u(\vec x')\,d\ell'-\frac{1}{2\pi}\log(r)\oint_C \frac{\partial u(\vec x')}{\partial n'}d \ell' \tag 4\\\\ &=\frac{1}{2\pi r}0\int_C u(\vec x')\,d\ell'-\frac{1}{2\pi}\log(r)\int_{S} \nabla '\cdot \nabla 'u(\vec x')\,dS' \tag 5\\\\ &=\frac{1}{2\pi r}\oint_C u(\vec x')\,d\ell'-\frac{1}{2\pi}\log(r)\int_{S} \nabla '^2 u(\vec x')\,dS' \\\\ &=\frac{1}{2\pi r}\oint_C u(\vec x')\,d\ell' \tag 6\\\\ &=\frac{1}{2\pi }\int_0^{2\pi} u(x_0+r\cos(\phi),y_0+r\sin(\phi))\,d\phi \end{align}$$

as was to be shown!

In going from $(4)$ to $(5)$, we invoked the Divergence Theorem, while in arriving at $(6)$ we used the fact that $u$ is harmonic.