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

u(1,\theta )={\begin{cases}0&-\pi <\theta <0\\T_{0}&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 ),n=1,2,...\,

Evaluate this function at the boundary conditions.

u(1,\theta )={\frac  {1}{2}}a_{0}+\sum _{{n=1}}^{\infty }(a_{n}\cos n\theta +b_{n}\sin n\theta )={\begin{cases}0&-\pi <\theta <0\\T_{0}&0<\theta <\pi \end{cases}}\,

The Fourier coefficients are

a_{0}={\frac  {1}{\pi }}\int _{{0}}^{\pi }T_{0}\,d\theta ={\frac  {T_{0}\theta }{\pi }}{\Bigg |}_{0}^{\pi }=T_{o}\,

a_{n}={\frac  {1}{\pi }}\int _{{0}}^{\pi }T_{0}\cos(n\theta )\,d\theta ={\frac  {T_{0}}{n\pi }}\sin n\theta {\bigg |}_{0}^{\pi }=0,n=1,2,...\,

b_{n}={\frac  {1}{\pi }}\int _{{0}}^{\pi }T_{0}\sin(n\theta )\,d\theta ={\frac  {-T_{0}}{n\pi }}\cos n\theta {\bigg |}_{0}^{\pi }={\frac  {T_{0}}{n\pi }}(1-(-1)^{n}),n=1,2,...\,

u(r,\theta )={\frac  {1}{2}}T_{0}+\sum _{{n=1}}^{\infty }({\frac  {T_{0}}{n\pi }}\left[1-(-1)^{n}\right]\sin n\theta )r^{n}\,

Summing only the odd indices gives the final solution:

u(r,\theta )={\frac  {1}{2}}T_{0}+{\frac  {2T_{0}}{\pi }}\sum _{{n=1}}^{\infty }{\frac  {r^{{2n-1}}}{2n-1}}\sin(2n-1)\theta \,

Partial Differential Equations

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