Heat Equation in Cylindrical Coordinates
We now solve the full heat equation in cylindrical coordinates without assuming \( \theta \)-independence.
$$ \dot{T} = \alpha \nabla^2 T $$ $$ T(R,\theta,z,t) = 0 $$ $$ T(r,\theta,0,t) = T(r,\theta,H,t) = 0 $$ $$ \partial_{\theta} \, T(r,0,z,t) = \partial_{\theta} \, T(r,\pi,z,t) = 0 $$ $$ T(r,\theta,z,0) = T_i $$Assume
$$ T(r,\theta,z,t) = f(r) g(\theta) h(z) p(t) $$ $$ \frac{p'(t)}{p(t)} = \alpha \frac{\nabla^2(f(r) g(\theta) h(z))}{f(r) g(\theta) h(z)} = \alpha\lambda $$Since the left-hand side depends only on time while the right-hand side depends only on spatial variables, both sides must equal a constant.
$$ p(t) = e^{\alpha\lambda t} $$In cylindrical coordinates,
$$ \nabla^2 = \frac{1}{r}\frac{\partial}{\partial r} \left( r\frac{\partial}{\partial r} \right) + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{\partial^2}{\partial z^2} $$ $$ \frac{1}{f(r)} \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \frac{g''(\theta)}{g(\theta)} + \frac{h''(z)}{h(z)} = \lambda $$ $$ g'(0) = g'(\pi) = 0 $$ $$ g''(\theta) = A g(\theta) $$ $$ g_n(\theta) = \cos(n\theta), \quad A_n = -n^2, \quad n = 0,1,2,\dots $$These are the angular modes compatible with the insulating boundary conditions.
$$ h(0) = h(H) = 0 $$ $$ h''(z) = B h(z) $$ $$ h_m(z) = \sin\left(\frac{m\pi z}{H}\right), \quad B_m = -\left(\frac{m\pi}{H}\right)^2, \quad m = 1,2,3,\dots $$These are the axial standing-wave modes satisfying the vanishing boundary conditions
$$ \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial f}{\partial r} \right) + \frac{A_n}{r^2}f(r) + B_m f(r) = \lambda f(r) $$ $$ \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial f}{\partial r} \right) - \frac{n^2}{r^2}f(r) = (\lambda - B_m)f(r) $$Let
$$ \lambda - B_m = \lambda + \left(\frac{m\pi}{H}\right)^2 = - k^2 $$The axial eigenvalue therefore shifts the effective radial eigenvalue.
$$ \frac{1}{r} \frac{\partial}{\partial r} \left( r\frac{\partial f}{\partial r} \right) - \frac{n^2}{r^2}f(r) = - k^2 f(r) $$Rescale \( r \) by \( k \):
$$ x = kr $$ $$ \frac{\partial^2 f}{\partial x^2} + \frac{1}{x}\frac{\partial f}{\partial x} + \left( 1 - \frac{n^2}{x^2} \right)f = 0 $$This is the Bessel equation. Solutions finite at \( r = 0 \) are
$$ f(r) = J_n(kr) $$ $$ f(R) = J_n(kR) = 0 $$We solve
$$ J_n(kR) = 0, \quad k > 0 $$Label the solutions by \( k_{n,\ell} \)
$$ f_{n,\ell}(r) = J_n(k_{n,\ell} r) $$ $$ \lambda_{n,m,\ell} = - k_{n,\ell}^2 - \left(\frac{m\pi}{H}\right)^2 $$The general solution is
$$ T(r,\theta,z,t) = \sum_{n=0}^{\infty} \sum_{m=1}^{\infty} \sum_{\ell=1}^{\infty} c_{n,m,\ell} f_{n,\ell}(r) g_n(\theta) h_m(z) e^{\alpha\lambda_{n,m,\ell}t} $$ $$ = \sum_{n=0}^{\infty} \sum_{m=1}^{\infty} \sum_{\ell=1}^{\infty} c_{n,m,\ell} J_n(k_{n,\ell} r) \cos(n\theta) \sin\left(\frac{m\pi z}{H}\right) e^{\alpha\lambda_{n,m,\ell}t} $$The full solution is therefore constructed from products of radial, angular, and axial eigenmodes, each decaying exponentially in time.
Orthogonality of the eigenfunctions gives
$$ \langle f_{n,\ell}, f_{n,\tilde{\ell}} \rangle = \delta_{\ell\tilde{\ell}}\|f_{n,\ell}\|^2, \quad \langle u,v\rangle = \int_0^R u(r)v(r)r\,dr $$ $$ \langle g_n, g_{\tilde{n}} \rangle = \delta_{n\tilde{n}}\|g_n\|^2, \quad \langle u,v\rangle = \int_0^\pi u(\theta)v(\theta)\,d\theta $$ $$ \langle h_m, h_{\tilde{m}} \rangle = \delta_{m\tilde{m}}\|h_m\|^2, \quad \langle u,v\rangle = \int_0^H u(z)v(z)\,dz $$The initial condition is
$$ T(r,\theta,z,0) = \sum_{n=0}^{\infty} \sum_{m=1}^{\infty} \sum_{\ell=1}^{\infty} c_{n,m,\ell} f_{n,\ell}(r) g_n(\theta) h_m(z) = T_i $$Therefore,
$$ c_{n,m,\ell} = T_i \frac{\langle f_{n,\ell},1\rangle}{\langle f_{n,\ell},f_{n,\ell}\rangle} \frac{\langle g_n,1\rangle}{\langle g_n,g_n\rangle} \frac{\langle h_m,1\rangle}{\langle h_m,h_m\rangle} $$ $$ \langle g_n,1\rangle = \int_0^\pi \cos(n\theta)\,d\theta = \begin{cases} \pi & n = 0 \\ 0 & n > 0 \end{cases} $$ $$ \langle g_n,g_n\rangle = \int_0^\pi \cos^2(n\theta)\,d\theta = \begin{cases} \, \pi & n = 0 \\[0.4em] \dfrac{\pi}{2} & n > 0 \end{cases} $$ $$ \langle h_m,1\rangle = \int_0^H \sin\left(\frac{m\pi z}{H}\right)\,dz $$ $$ = \left[ -\frac{H}{m\pi} \cos\left(\frac{m\pi z}{H}\right) \right]_0^H $$ $$ = \frac{H}{m\pi} \left(1-\cos(m\pi)\right) $$ $$ = \begin{cases} \dfrac{2H}{m\pi} & m \text{ odd} \\[0.4em] \; \; 0 & m \text{ even} \end{cases} $$ $$ \langle h_m,h_m\rangle = \int_0^H \sin^2\left(\frac{m\pi z}{H}\right)\,dz = \frac{H}{2} $$Therefore only \( n = 0 \) and \( m \) odd survive. Thus the constant initial condition excites only radially symmetric modes with odd axial structure.
$$ c_{0,m,\ell} = T_i \frac{\langle f_{0,\ell},1\rangle}{\langle f_{0,\ell},f_{0,\ell}\rangle} \frac{\pi}{\pi} \frac{\frac{2H}{m\pi}}{\frac{H}{2}} $$ $$ c_{0,m,\ell} = T_i \frac{4}{m\pi} \frac{\langle f_{0,\ell},1\rangle}{\langle f_{0,\ell},f_{0,\ell}\rangle} $$ $$ = T_i \frac{4}{m\pi} \frac{ \int_0^R J_0(k_{0,\ell} r)\,r\,dr }{ \int_0^R J_0^2(k_{0,\ell} r)\,r\,dr } $$Hence
$$ T(r,\theta,z,t) = \frac{4T_i}{\pi} \sum_{\substack{m=1 \\ m\text{ odd}}}^{\infty} \sum_{\ell=1}^{\infty} \frac{1}{m} \frac{ \int_0^R J_0(k_{0,\ell} r)\,r\,dr }{ \int_0^R J_0^2(k_{0,\ell} r)\,r\,dr } J_0(k_{0,\ell} r) \sin\left(\frac{m\pi z}{H}\right) e^{\alpha\lambda_{0,m,\ell}t} $$where
$$ \lambda_{0,m,\ell} = - k_{0,\ell}^2 - \left(\frac{m\pi}{H}\right)^2 $$and \( k_{0,\ell} \) is determined by
$$ J_0(k_{0,\ell} R) = 0 $$Higher radial and axial modes decay more rapidly, so the temperature distribution gradually approaches the equilibrium state \( T = 0 \).