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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^{2}R''+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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