PDE19

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

$u_r(\rho,\theta) = f(\theta)\,$

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

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 $c_1 = 1, c_3 = \frac{1}{\rho^n}\,$

$u(r,\theta) = \frac{1}{2}a_0 + \sum_{n=1}^\infty \left(r/\rho\right)^n(a_n\cos n\theta + b_n \sin n \theta)\,$

Evaluate this function at the boundary conditions.

$\frac{\partial u(r,\theta)}{\partial r} = \frac{1}{\rho} \sum_{n=1}^\infty \left(r/\rho\right)^{n-1}n(a_n\cos n\theta + b_n \sin n \theta)\,$

$\frac{\partial}{\partial r} u(\rho,\theta) = \frac{1}{\rho} \sum_{n=1}^\infty n a_n \cos n\theta + n b_n \sin n \theta = f(\theta)\,$

The Fourier coefficients are

$a_n = \frac{\rho}{n\pi}\int_{-\pi}^\pi f(\theta)\cos n\theta \,d\theta\,$

$b_n = \frac{\rho}{n\pi}\int_{-\pi}^\pi f(\theta)\sin n\theta \,d\theta\,$

$a_0\,$ is undetermined.

Partial Differential Equations

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