PDE21

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

$u_r(2,\theta) = 0\,$
$u_r(1,\theta) = \sin^2\theta\,$

$1<r<2,\,\,\,\,\, -\pi<\theta<\pi\,$

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

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

$H(\theta) = \begin{cases} \frac{1}{2}a_0 & n=0 \\ a_n \cos(n \theta) + b_n \sin(n \theta) & n=1,2,... \end{cases}\,$

The solution is

$u(r,\theta) = \frac{1}{2}(A_0 + B_0\log r) + \sum_{n=1}^\infty\left[(A_n r^n + B_n r^{-n}) \cos n\theta + (C_n r^n + D_n r^{-n})\sin n\theta\right]\,$

$u(1,\theta) = \frac{1}{2}A_0 + \sum_{n=1}^\infty \left[(A_n+B_n)\cos n \theta + (C_n+D_n)\sin n\theta \right] = \sin^2\theta\,$

Using the fact that $\sin^2\theta = \frac{1}{2}(1-\cos 2\theta)\,$ ,

We can compare the coefficients and get the identities:

$A_0 = 1, \,\, A_2+B_2=\frac{-1}{2}\,$

$A_n+B_n = 0,\,\, n=1,3,4,...\,$

Imposing the boundary conditions,

$u_r(r,\theta) = \frac{1}{2}\left(\frac{B_0}{r}\right) + \sum_{n=1}^\infty(n A_n r^{n-1} - n B_n r^{-n-1})\cos n \theta + (n C_n r^{n-1} - n D_n r^{-n-1})\sin n\theta\,$