PDE23

$\Delta u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta}=0\,$

$u_r(1,\theta) = \begin{cases}-1&-\pi<\theta<0 \\ 1 & 0<\theta<\pi\end{cases}\,$

$0<r<1,\,\,\,\,\, -\pi<\theta<\pi\,$

Let $u(r,\theta) = R(r)H(\theta)\,$

Plug into the original DE and seperate variables, set equal to a constant $\lambda\,$ . The two ODEs are

$r^2R'' + rR' - \lambda R = 0\,$

$H''+\lambda H = 0\,$

The solutions are

$R(r) = \begin{cases} c_1 + c_2 \log r & n=0 \\ c_3 r^n + c_4 r^{-n} & n=1,2,... \end{cases}\,$

$H(\theta) = \begin{cases} \frac{1}{2}A_0 & n=0 \\ A_n \cos(n \theta) + B_n \sin(n \theta) & n=1,2,... \end{cases}\,$

To eliminate infinite discontinuities as $r\rightarrow 0^+\,$ , set $c_2 = c_4 = 0\,$ . For mathematical ease, set $c_1 = 1, c_3 = 1\,$

$u(r,\theta) = \frac{1}{2}A_0 + \sum_{n=1}^\infty r^n(A_n\cos n\theta + B_n \sin n \theta)\,$

Evaluate this function at the boundary conditions.

$u_r(1,\theta) = \sum_{n=1}^\infty (A_n\cos n\theta + B_n \sin n \theta) = \begin{cases}-1&-\pi<\theta<0 \\ 1 & 0<\theta<\pi\end{cases}\,$

Evaluating the Fourier coefficients,

$A_n = \frac{1}{n} \frac{1}{\pi} \left[ \int_{-\pi}^0(-1)\cos n\theta\,d\theta + \int_0^\pi 1\cdot\cos n\theta\,d\theta\right]\,$

$= \frac{1}{n\pi}\left[\frac{-\sin n\theta}{n}\Bigg|_{-\pi}^0 + \frac{\sin n\theta}{n}\Bigg|_0^\pi\right] = 0\,$

$B_n = \frac{1}{n} \frac{1}{\pi} \left[ \int_{-\pi}^0(-1)\sin n\theta\,d\theta + \int_0^\pi 1\cdot\sin n\theta\,d\theta\right]\,$

$= \frac{1}{n\pi}\left[\frac{\cos n\theta}{n}\Bigg|_{-\pi}^0 + \frac{-\cos n\theta}{n}\Bigg|_0^\pi\right]\,$

$= \frac{2}{n\pi}\left[\frac{1-(-1)^n}{n}\right]\,$

The final solution is

$u(r,\theta) = \frac{1}{2} A_0 + \sum_{n=1}^\infty \frac{2}{n\pi}\left[\frac{1-(-1)^n}{n}\right]\sin n\theta r^n\,$

$= \frac{1}{2}A_0 + \frac{4}{\pi}\sum_{n=1}^\infty \frac{\sin(2n-1)\theta}{(2n-1)^2\pi}r^{2n-1}\,$