Find the bounded solution of Laplace's equation in the region $\Omega=\{(r,\theta):r>1,0<\theta<\pi\}$ subject to the boundary conditions $u(r,\pi)=u(r,0)=0$ for $r>1$ and $u(1,\theta)=1$ for $0<\theta<\pi.$

I am not sure if I did this correct?

Laplace’s equation in polar coordinates $(r,\theta)$ is $$\frac{\partial^2u}{\partial r^2}+\frac1r\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2u}{\partial\theta^2}=0.$$ Using serpatation of variables and letting $u(r,\theta)=h(r)\phi(\theta)$ we have $$\frac{r^2h''(r)+rh'(r)}{h(r)}=-\frac{\phi''(\theta)}{\phi\theta}=\lambda.$$ Therefore, $$r^2h''+rh'-\lambda h=0$$$$\phi''+\lambda\phi=0.$$ I obtained, that $\phi_n(\theta)=\sin\sqrt{\lambda_n}\theta$, where $\lambda_n=n^2$, from the boundary conditions $u(r,\pi)=u(r,0)=0$.

Now for $r^2h''+rh'-\lambda h=0$ we have that $h(r)=a_nr^{\sqrt{\lambda_n}}+b_nr^{-\sqrt{\lambda_n}}$ and since our equation is bounded we have that $a_n=0$ thus $h(r)=b_nr^{-\sqrt{\lambda_n}}$. Therefore, $$u(r,\theta)=\sum_n\left[\sin\sqrt{\lambda_n}\theta\right]b_nr^{-\sqrt{\lambda_n}}$$ $$u(1,\theta)=\sum_n\left[\sin\sqrt{\lambda_n}\theta\right]b_n=1$$ $$b_n=\frac{2}{\pi}\int^{\pi}_0f(x)\sin\sqrt{\lambda_n}\theta d\theta.$$

  • $\begingroup$ Looks fine to me. $\endgroup$ – Gyu Eun Lee Nov 22 '14 at 7:11
  • $\begingroup$ @thank you for checking my work! $\endgroup$ – Robben Nov 22 '14 at 18:49
  • 1
    $\begingroup$ A well-asked question! +1 $\endgroup$ – Neal Nov 22 '14 at 18:53

Yes, except $f(x)$ should be $1$ in the last line: $$b_n={2\over \pi}\int_0^\pi 1\cdot \sin(\sqrt{\lambda_n}\theta)\,d\theta={2\over \pi}\int_0^\pi \sin(n\theta)\,d\theta,\ n=1,2,\dots $$

Side note: good job eliminating the $r^\sqrt{\lambda_n}=r^n$ terms from your solution since they blow up as $r\to\infty$ in this "exterior (semi)disk" problem.


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