10.1: Temperature on a Disk

Let us now turn to a different two-dimensional problem. A circular disk is prepared in such a way that its initial temperature is radially symmetric, $$u(\rho,\phi,t=0) = f(\rho). \nonumber$$ Then it is placed between two perfect insulators and its circumference is connected to a freezer that keeps it at $0^\circ~\rm C$, as sketched in Fig. $\PageIndex{2}$.

Figure $\PageIndex{1}$: A circular plate, insulated from above and below.

Since the initial conditions do not depend on $\phi$, we expect the solution to be radially symmetric as well, $u(\rho,t)$, which satisfies the equation $$\begin{aligned} \frac{\partial u}{\partial t} &= k\left[\frac{\partial^2 u}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial u}{\partial \rho} \right],\nonumber\\ u(c,t) &=0,\nonumber\\ u(\rho,0) &= f(\rho).\end{aligned} \nonumber$$

Figure $\PageIndex{2}$: The initial temperature in the disk.

Once again we separate variables, $u(\rho,t)=R(\rho)T(t)$, which leads to the equation

$$\frac{1}{k} \frac{T'}{T} = \frac{R''+\frac{1}{\rho}R'}{R} = -\lambda. \nonumber$$ This corresponds to the two equations

$$\begin{aligned} \rho^2 R''+\rho R' + \lambda \rho^2 R &= 0,\quad R(c)=0m\nonumber\\ T'+\lambda k T = 0.\end{aligned} \nonumber$$

The radial equation (which has a regular singular point at $\rho=0$) is closely related to one of the most important equation of mathematical physics, Bessel’s equation. This equation can be reached from the substitution $\rho = x /\sqrt{\lambda}$, so that with $R(r)=X(x)$ we get the equation

$$x^2 \frac{d^2}{dx^2} X(x) + x \frac{d}{dx} X(x) + x^2 X(x) = 0, \qquad X(\sqrt \lambda c) =0. \nonumber$$