PDE17

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

u(r,-\pi) = u(r,\pi)\,
u_\theta(r,-\pi) = u_\theta(r,\pi)\,
\lim_{r\rightarrow 0^+}u(r,\theta) < \infty\,

u(\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 both sides 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(\frac{r}{\rho}\right)^n(a_n\cos n\theta + b_n \sin n \theta)\,

Evaluate this function at the boundary conditions.

u(\rho,\theta) = \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n \cos n\theta + b_n \sin n \theta = f(\theta)\,

The Fourier coefficients are:

a_0 = \frac{1}{\pi}\int_{-\pi}^\pi f(\theta)\,d\theta\,

a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(\theta) \cos n\theta

b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(\theta) \sin n\theta

Substituting these formulas back into the starred equation gives the Poisson integral formula for this problem:

u(r,\theta) = \frac{\rho^2-r^2}{2\pi}\int_{-\pi}^\pi \frac{f(x)}{\rho^2-2\rho r \cos(x-\theta) + r^2}\,dx\,

Partial Differential Equations

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