# Separation of variables for heat equation in cylindrical shell

I already know how to apply the separation of variables method to solve transient radial heat equation inside a cylinder. But, when it comes to cylindrical shells, both Bessel J and Y functions appear in the solution and I don't know how to find the coefficients by taking advantage of orthogonality.

Here's the PDE: $$u_t = \frac{k}{r}(ru_r)_r$$

with the following boundary/initial conditions:

\begin{align} &r=r_3 &u_r = 0 \\ &r=r_2 &-ku_r + pu = 0 \\ &t=0 &u = u_0 \end{align}

After applying the separation of variables method, here's what I have:

\begin{align} u(r,t) &= R(r)T(t) \\ T(t) &= A e^{-k\lambda^2t} \\ R(r) &= B J_0(\lambda r) + C Y_0(\lambda r) \end{align}

I appreciate it if you could shed some light on how to proceed.

There are unique constants $$A,B$$ so that $$R(r)=AJ_0(\lambda r)+BY_0(\lambda r)$$ satisfies the endpoint conditions $$R'(r_3) = 0,\;\; R(r_3)=1.$$ This is because $$R(r_3)=0=R'(r_3)$$ implies $$R\equiv 0$$, and this is true for any constants $$A,B$$. Such a solution is $$R(r) = \frac{J_0(\lambda r)Y_0'(\lambda r_3)-J_0'(\lambda r_3)Y_0(\lambda r)}{J_0(\lambda r_3)Y_0'(\lambda r_3)-J_0'(\lambda r_3)Y_0(\lambda r_3)}$$ Indeed, the numerator $$N$$ is a solution of the Bessel equation with eigenvalue $$\lambda$$ such that $$N'(r_3)=0$$ and $$N(r_3)\ne 0$$. So $$R$$ is a solution of the Bessel equation with $$R'(r_3)=0$$ and $$R(r_3)=1$$. The eigenvalue equation for $$\lambda$$ is then determined by the condition $$-kR'(r_2)+pR(r_2) = 0.$$