My question:
Obtain an expression for the steady-state temperature distribution $T(r,φ)$ in a long, solid cylinder$ 0 \leq r\leq b, 0\leq φ \leq 2π$ for the following boundary conditions: The boundary at $r = b$ is subjected to a prescribed temperature distribution $f (φ)$.
My try:
Heat diffusion equation: $$\frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r} + \frac{1}{r^2} \frac{\partial T}{\partial \phi} = 0 $$ $$BC1: T=f(\phi) \text{ at } r=b, BC2: T(r \to 0) \text{ is finite}, BC3: T \text{ has } 2\pi \text{ period about } \phi, BC4: \frac{\partial T}{\partial \phi} \text{ has } 2\pi \text{ period.} $$ Let $T(r,\phi) = R(r) \Phi(\phi)$ $$ \frac{1}{R} \left( r^2 \frac{\partial ^2 R}{\partial r ^2} +r \frac{\partial R}{\partial r} \right)= - \frac{1}{\Phi} \frac{ \partial \Phi} { \partial \phi} = \lambda^2$$ $$ \Phi(\phi) = c_1 \cos \lambda \phi + c_2 \sin \lambda \phi $$ From BC3 and BC4, $\lambda_n = n (n=0,1,...,)$. $$ r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} - n^2 R = 0 $$ Put $R(r) = r^\gamma $, then $\gamma^2 -n^2=0$. If $n>0$, $$R_n(r) = c_1 r^n + c_2 r^{-n} $$ If $n=0$, $$R_0(r) = c_1 + c_2 \ln r $$ From BC2, $c_2=0$ and $R_n(r) =c_n r^n$. $$T(r,\phi)=\sum_{n=0}^\infty R_n (r) \Phi_n (\phi) = a_0 + \sum_{n=1}^\infty (a_n r^n \cos \lambda_n \phi + b_n r^n \sin \lambda_n \phi)$$ From the IC, $$a_0 =\frac{1}{2\pi} \int_0^{2\pi} f(\phi) d\phi$$ $$b^n a_n = \frac{\int_{0}^{2\pi} f(\phi) \cos{\lambda_n \phi} d\phi}{\int_0^{2\pi} \cos^2 \lambda_n \phi d\phi} $$ $$b^n b_n = \frac{\int_{0}^{2\pi} f(\phi) \sin \lambda_n \phi d\phi}{\int_0^{2\pi} \sin^2 \lambda_n \phi d\phi}$$ $$\int_0^{2\pi} \cos^2\lambda_n \phi d\phi = \int_0^{2\pi} \sin^2 \lambda_n \phi d\phi =\pi$$ Is it right?
