I'm faced with the problem of solving the Heat Equation on a two-dimensional disc:
$$\frac{1}{\kappa} \frac{\partial T}{\partial t}=\Delta T$$
The boundary conditions in polar coordinates $T(r,\theta,t)$ are:
$$T(0,\theta,t)=T_0$$
$$T(a,\theta,t)=T_1$$
The disc at $t=0$ should be at a uniform temperature $T_0$.
I want to solve first for the equilibrium temperature function $T_{eq}(r,\theta)$ and then solve for $W$ in: $$W(r,\theta,t)=T(r,\theta,t)-T_{eq}(r,\theta)$$
I would do this by substituting $T = T_{eq} + W$ into the heat equation and using the separation of variables technique to solve for $W$.
I'm running into a hitch on the first step however. I can't solve for $T_{eq}$.
$\frac{\partial T_{eq}}{\partial t} = 0$ so:
$$\Delta T_{eq} = 0$$
$$\frac{1}{r} \frac{\partial T_{eq}}{\partial r}+\frac{\partial^2 T_{eq}}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2 T_{eq}}{\partial \theta^2}=0$$
I now figure by radial symmetry that $\frac{\partial^2 T_{eq}}{\partial \theta^2}=0$:
$$\frac{1}{r} \frac{\partial T_{eq}}{\partial r}+\frac{\partial^2 T_{eq}}{\partial r^2}=0$$
Now I solve for $T_{eq}$:
$$\frac{\partial^2 T_{eq}}{\partial r^2}=-\frac{1}{r} \frac{\partial T_{eq}}{\partial r}$$
Therefore $\frac{\partial T_{eq}}{\partial t}=F_1(\theta)\frac{1}{r}$. Therefore $T_{eq}=F_1(\theta)(\ln(r) + F_2(\theta))$.
Now I try to fully determine $T_{eq}$ by applying the boundary conditions. First:
$$T_{eq}(0,\theta)=T_0$$
$$F_1(\theta)(\ln(0)+F_2(\theta)) = T_0$$
Uh oh! $\ln(0)$ is undefined!
Am I doing something wrong by assuming radial symmetry? I'm really not sure.
Edit:
If $T_{eq}$ is not radially symmetric, then how would one decide how one should 'rotate' the solution? The initial and boundary conditions do not discriminate depending on direction, so how would the system 'decide' in which direction it should settle at equilibrium?
Best Answer
If you do not assume radial symmetry, and you let $T_{eq}=R(r)\Theta(\theta)$, then by using the separation of variables, you will arrive at a Cauchy-Euler equation for $R$.
I think because $T_{eq}$ satisfies Laplace's equation rather than the heat equation, it is wrong to assume radial symmetry.
So use separation of variables, to obtain:
$\frac{1}{r}R'\Theta+R''\Theta+\frac{1}{r^{2}}R\Theta'',\,\,\,$ now divide by $R\Theta/r^2$
$\frac{rR'+r^2R''}{R}+\frac{\Theta''}{\Theta}=0$