Heat Conduction Equation--Disk

To solve the Heat Conduction Equation on a 2-D disk of radius $R=1$, try to separate the equation using

$$T(r, \theta, t) = R(r)\Theta (\theta)T(t).$$ (1)

Writing the $\theta$ and $r$ terms of the Laplacian in Spherical Coordinates gives

$$\nabla^2={d^2R\over dr^2}+ {2\over r} {dR\over dr} + {1\over r^2} {d^2\Theta \over d\theta^2},$$ (2)

so the Heat Conduction Equation becomes

$${R\Theta\over \kappa}{d^2T\over dt^2} = {d^2R\over dr^2} \T... ...ver dr} \Theta T + {1\over r^2} {d^2\Theta\over d\theta^2} RT.$$ (3)

Multiplying through by $r^2/R\Theta T$ gives

$${r^2\over \kappa T}{d^2T\over dt^2} = {r^2\over R} {d^2R\ov... ...over R} {dR\over dr}+{d^2\Theta\over d\theta^2}{1\over\Theta}.$$ (4)

The $\theta$ term can be separated.

$${d^2\Theta \over d\theta^2} {1\over \Theta } = -n(n+1),$$ (5)

which has a solution

$$\Theta(\theta ) = A\cos\left[{\sqrt{n(n+1)}\, \theta}\right]+B\sin\left[{\sqrt{n(n+1)}\,\theta}\right].$$ (6)

The remaining portion becomes

$${r^2\over \kappa T}{d^2T\over dt^2}= {r^2\over R} {d^2R\over dr^2} + {2r\over R} {dR\over dr} -n(n+1).$$ (7)

Dividing by $r^2$ gives

$${1\over \kappa T}{d^2T\over dt^2} = {1\over R} {d^2R\over d... ...ver rR} {dR\over dr} -{n(n+1)\over r^2} = -{1\over \lambda^2},$$ (8)

where a Negative separation constant has been chosen so that the $t$ portion remains finite

$$T(t)=C e^{-\kappa t/\lambda^2}.$$ (9)

The radial portion then becomes

$${1\over R} {d^2R\over dr^2} + {2\over rR} {dR\over dr} -{n(n+1)\over r^2} +{1\over \lambda^2}=0$$ (10)

$$r^2{d^2R\over dr^2} + 2r {dR\over dr} +\left[{{r^2\over \lambda^2}-n(n+1)}\right]R=0,$$ (11)

which is the Spherical Bessel Differential Equation. If the initial temperature is $T(r,0)=0$ and the boundary condition is $T(1,t)=1$, the solution is

$$T(r,t)=1-2\sum_{n=1}^\infty {J_0(\alpha_n r)\over \alpha_nJ_1(\alpha_n)} e^{{\alpha_n}^2t},$$ (12)

where $\alpha_n$ is the $n$th Positive zero of the Bessel Function of the First Kind $J_0(x)$.