PDE18

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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^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 = 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{2 T_0}{\pi}\sum_{n=1}^\infty \frac{r^{2n-1}}{2n-1}\sin(2n-1)\theta\,

Partial Differential Equations

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