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\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^{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}=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}}\,

Partial Differential Equations

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