We have the Laplace equation in polar coordinates:
$$u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta}=0, 0 \leq r <a, 0 \leq \theta \leq 2 \pi$$

With the separation of variables, the solution is in the form $u(r \theta)=R(r) \Theta(\theta)$

Then after calculations, we get:
$$u(r, \theta)=\sum_{n=0}^{\infty}{[A_n \cos{(n \theta)}+B_n \sin{(n \theta)}]r^n}, 0 \leq \theta \leq 2 \pi, 0 \leq r <a$$

The boundary condition is $h(\theta)=u(r=a, \theta)$

$h(\theta)$ is a periodic function with period $2 \pi$, so we can write it as a Fourier series.

After calculations we get the following formula (Poisson formula):

$$u(r, \theta)=\frac{a^2-r^2}{2 \pi} \int_0^{2 \pi}{ \frac{h( \phi) d \phi}{a^2+r^2-2 a r \cos{(\theta-\phi)}}}$$

$|r’|=a, |\overrightarrow{r}-\overrightarrow{r’}|^2=|\overrightarrow{r}|^2+|\overrightarrow{r’}|^2-2|\overrightarrow{r}||\overrightarrow{r’}| \cos{(\theta-\phi)}=r^2+a^2-2ar \cos{(\theta – \phi)}$

So we can write the equation also:
$$u(r, \theta)=\frac{a^2-r^2}{2 \pi a} \int_{|\overrightarrow{r’}|=a}{\frac{u(\overrightarrow{r’})ds}{|\overrightarrow{r}-\overrightarrow{r’}|^2}}(*)$$
Could you tell me how we get to the relation $(*)$??

$\frac{ds}{a}= d \phi$

For $r=0$

$$u(0)=\frac{1}{2 \pi a} \int_{| \overrightarrow{r’}|=a}{u(\overrightarrow{r’})}ds$$
This is the mean value of the field at the circumference of the circle.

Could you explain me the sentence above?

#### Solutions Collecting From Web of "Question about Poisson formula"

For instance the following curve traces a circle of radius $a$:
$$\vec r(s) = a\cos \frac s {a} \hat \imath + a\sin \frac s {a} \hat \jmath$$
where in this case $s$ is the distance along the curve.

Integration of a function $f(\vec r) = f(r,\theta)$ over a circle with radius $a$ can then be written in the following ways:
$$\int_{|\vec r|=a} f(\vec r) |d\vec r| = \int_{|\vec r|=a} f(\vec r) ds = \int_{r=a} f(r,\theta) ds = \int_0^{2\pi} f(a,\theta)\ a d\theta$$

Substitute the relevant formulas to find (*).

The mean value of a function $f(\vec r)$ on a curve is:
$$\text{Mean value} = \frac 1 {\text{Length of curve}} \int_{\text{curve}} f(\vec r) ds$$