Heat equation with a disc

Question

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 $T_0$.

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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?

Answer 1

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 $R\Theta/r^2$

$\frac{rR'+r^2R''}{R}+\frac{\Theta''}{\Theta}=0$

Answer 2

The problem that the OP is trying to solve can be made non-dimensional by defining the following quantities:

$$\phi = \frac{T - T_0}{T_1 - T_0}, \quad \xi = \frac{r}{a}, \quad \tau = \frac{\kappa t}{a^2}, $$ hence obtaining (prove it):

$$ \frac{\partial \phi}{\partial \tau} = \frac{1}{\xi} \frac{\partial }{ \partial \xi} \left( \xi \frac{\partial \phi}{\partial \xi} \right), \quad 0 < \xi < 1, \quad \tau > 0,$$ where we have assumed that the temperature distribution is axisymmetric, since $\partial_\theta = 0$ everywhere (see comments). This equation is to be solved with the following boundary and initial conditions:

$$\begin{align} |\phi(0,\tau ) | & \leq \infty, \\ \phi(1,\tau) & = 1, \\ \phi(\xi,0) & = 0. \end{align} $$ Since the boundary conditions are not homogenous, we can make use of the superposition principle and set $\phi = u + v$, where $u$ satisfies homogenous boundary conditions and $v$ satisfies $|v(0,\tau)| < \infty$ and $v(1,\tau) = 1$. We can therefore set $v = 1$ as a possible solution for $v$. The problem for $u$ is then given by:

$$ \frac{\partial u}{\partial \tau} = \frac{1}{\xi} \frac{\partial }{ \partial \xi} \left( \xi \frac{\partial u}{\partial \xi} \right), \quad 0 < \xi < 1, \quad \tau > 0,$$

$$\begin{align} |u(0,\tau ) | & \leq \infty, \\ u(1,\tau) & = 0, \\ u(\xi,0) & = T(\xi,0) - v(\xi,0) = -1, \end{align} $$ where we have made use of $\partial_t v= \partial_\xi v= 0$. Since the PDE for $u$ is linear, we can use the method of separation of variables to write: $u(\xi, \tau) = P(\xi)Q(\tau)$, for $P, Q$ non-zero functions of their respective arguments. Introducing this decomposition into the PDE yields:

$$ P Q' = \frac{Q}{\xi} \frac{d}{d \xi} \left( \xi \frac{d P}{d\xi} \right),$$ which can be reduced to:

$$ \frac{Q'}{Q} = \frac{1}{\xi \, P} \frac{d}{d \xi} \left( \xi \frac{d P}{d\xi} \right) = \lambda \in \mathbb{R}.$$ This equations represents two problems: one for $P$, which is the interesting one, and one for $Q$ which we will not worry about. The problem for $P$ reads:

$$\frac{d}{d \xi} \left( \xi \frac{d P}{d\xi} \right) - \lambda \xi P = 0, \quad 0 < \xi < 1, $$ and $|P(0)|< \infty, \ P(1) = 0$ (see the definition of $u$ to derive this boundary conditions). This equations has non-trivial BCs-satisfying solution for $\lambda < 0$ if I remember well, so that $\lambda = - |\lambda|$, and then:

$$P(\xi) = C_1 \, J_0 \left(\sqrt{|\lambda |} \xi \right) + C_2 \, Y_0 \left(\sqrt{| \lambda | } \xi \right),$$ where $J_0$ and $Y_0$ are the Bessel functions of first and second kind of order zero, respectively. Since $Y_0$ is singular at $\xi = 0$ and we must have bounded solutions for both $P$ and $u$, the constant of integration $C_2$ must vanish. The condition $P(1) = 0$ leads to:

$$ 0 = C_1 \, J_0 \left(\sqrt{|\lambda |} \right), $$ which holds whether $C_1 = 0$ (which yields the trivial, and not desired, solution $P = 0$) or $\sqrt{|\lambda|} = \chi_n$ is a zero of $J_0$. Note that $J_0$ has infinitely many zeros for $\chi_n = 0$. This values of $\chi_n$ are called the eigenvalues of the problem and allows us to write the solution as:

$$P(\xi) = C_n J_0 (\chi_n \xi) = J_0 (\chi_n \xi), \quad n = \{1,2, \ldots \} $$ where I have made $C_n = 1$ for reasons you are about to see.

The Sturm-Liouville theory tells us that we can expand the solution as a linear combination of the eigenfunctions as follows:

$$u(\xi, \tau ) = \sum_{n=1}^{\infty} Q_n(\tau) P_n(\xi),$$ where $P_n(\xi ) = J_0 (\chi_n \xi)$ and $Q_n(\tau)$ are the Fourier coefficients of the expansion, which are to be determined introducing this into the original PDE for $u$:

$$ \sum_{n=1}^{\infty} Q'_n(\tau) P_n(\xi) = \sum_{n=1}^{\infty} Q_n(\tau) \underbrace{\frac{1}{\xi} \frac{d}{d \xi} \left( \frac{1}{\xi} \frac{d P_n}{d \xi} \right) }_{- |\lambda_n| }$$

Work in progress