Welcome to our community

Be a part of something great, join today!

Dirichlet Problem for Laplace's Equation Outside of a Disc


New member
Feb 26, 2012
The Poisson Integral Formula is a representation of the bounded solution of the Dirichlet problem for Laplace's equation in the interior of the disc. Derive the corresponding formula
for the Dirichlet problem in the exterior of the disc, again assuming that the solution is bounded.

So we derived the expression for the interior part of a disc in class: $u(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}g(\phi) \frac{b^2-r^2}{b^2-2brcos(\theta-\phi)+r^2} d\phi$, where b is the radius of the disc and $u(b,\theta)=g(\theta)$. It was fairly straightforward, except that we were able to just assume the form of the solution $u(r,\theta)=\frac{1}{2}\gamma_0 +\sum_{n=1}^{\infty} (\frac{r}{b})^n[\gamma_ncos(n\theta)+\delta_nsin(n\theta)]$, where $\gamma_n=\frac{1}{\pi}\int_{0}^{2\pi}g(\phi)cos(n\phi)d\phi$ and $\delta_n=\frac{1}{\pi}\int_{0}^{2\pi}g(\phi)sin(n\phi)d\phi$. We then played around with the sums of the series of sin and cos and took the real part of that to get our answer.

Now, for the exterior of the disc, I'm just not sure how to start off with the form of the solution. It seems like the $(\frac{r}{b})^n$ should really be the only difference, but I can't justify randomly picking a form. Any suggestions?


Well-known member
MHB Math Helper
Jan 26, 2012
My first thought is to try the transformation $r \to \dfrac{1}{r}$ and see where that takes you.