I try to solve the below problem: A long, hollow cylinder, $a \leq r \leq b$, is initially at a temperature of $T = F(r)$. For times $t > 0$ the boundaries at $r = a$ and $r = b$ are kept insulated. Obtain an expression for temperature distribution $T(r, t)$ in the solid for times $t > 0.$
My try:
$$IC: T(r,0)=F(r), BC1: \frac{\partial T}{\partial r} = 0 \text{ at } r=a, BC2: \frac{\partial T}{\partial r} = 0 \text{ at } r=b $$ Heat diffusion equation: $$\frac{\partial^2 T}{\partial r} + \frac{1}{r} \frac{\partial T}{\partial r} = \frac{1}{\alpha} \frac{ \partial T}{\partial t}$$ Let $T(r,t) = R(r) \Gamma(t) $. $$ \frac{1}{R} \left[ \frac{\partial^2 R}{\partial r^2} + \frac{1}{r} \frac{\partial R}{\partial r} \right]= \frac{1}{\alpha \Gamma} \frac{ \partial \Gamma} { \partial t} = - \lambda^2$$ $$ \frac{\partial^2 R}{\partial r^2} + \frac{1}{r} \frac{\partial R}{\partial r}+\lambda^2 R =0 $$ $$R(r) = C_1 J_0 ( \lambda r) + C_2 Y_0 ( \lambda r) $$ $$\frac{dR}{dr} = -C_1 \lambda J_1 (\lambda r) - C_2 \lambda Y_1 (\lambda r) $$ From BC1, BC2 $$-C_1 \lambda J_1 (\lambda r) - C_2 \lambda Y_1 (\lambda r) = 0 \text{ for } r=a,b$$ $$-\frac{C_1}{C_2} = \frac{Y_1(\lambda a)}{J_1(\lambda a)} = \frac{Y_1 (\lambda b)}{J_1 (\lambda b)} $$ $$ \lambda_n > 0 (n=1,2,...) \text{ is an eigen value iff } J_1(\lambda_n b) Y_1 (\lambda_na) -J_1(\lambda_n a) Y_1 (\lambda_n b) = 0 $$ $$R_n(r) = C_n(Y_1(\lambda_n a) J_0 (\lambda_n r) - J_1 (\lambda_n a) Y_0 (\lambda_n r)$$ From $\frac{1}{\alpha \Gamma} \frac{\partial \Gamma}{\partial t} = -\lambda^2$, $$\Gamma_n(t) = \exp (-\alpha \lambda_n^2 t) $$ $$T(r,t) = \sum_{n=1}^\infty R_n(r)\Gamma_n(t)$$ From the IC $$F(r) = \sum_{n=1}^\infty R_n(r) $$ $$C_n =\frac{\int_a^b rF(r)(Y_1(\lambda_n a) J_0 (\lambda_n r)-J_1 (\lambda_n a) Y_0(\lambda_n r)) dr}{\int_a^b r(Y_1(\lambda_n a) J_0 (\lambda_n r)-J_1 (\lambda_n a) Y_0(\lambda_n r))^2 dr}$$
Is it right?