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.


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?

  • $\begingroup$ I think your troubles come up when you try to set the temperature at $r = 0$, which has nothing to do with the axisymmetrical behaviour of the solution. Indeed, in my opinion, the most appropriate boundary condition in the origin is that corresponding to bounding the solution, i.e., $|T(0,\theta)| < \infty$, which usually leads to the singular term to vanish. If you want me to ellaborate some more, just tell me! $\endgroup$ – Dmoreno Jun 8 '14 at 10:26
  • $\begingroup$ But if $|T(0,\theta)|<\infty$ then there must be some finite $T_0=T(0,\theta)$, right? I haven't even defined $T_0$ yet in my working and have not even needed to equate $T_{eq}(0,\theta)$ to the boundary temperature at the origin before running into a problem. $\endgroup$ – Myridium Jun 8 '14 at 10:33
  • $\begingroup$ Precisely. That temperature, $T(0,\theta)$ becomes a function of the problem and therefore cannot be imposed and should remain unknown. $\endgroup$ – Dmoreno Jun 8 '14 at 10:39
  • $\begingroup$ Two questions then: 1. What is the physical interpretation of not being able to impose a boundary condition at the origin. 2. Where else might be a natural place to set a boundary condition? And thank you very much! $\endgroup$ – Myridium Jun 8 '14 at 10:41
  • $\begingroup$ The first one is a good question and I'm not sure if I know the best answer; but I think this is related with the fact that $r=0$ is not a proper boundary of the domain and, since the laplacian is (geometrically) singular at $r = 0$, no boundary conditions can be considered except that of 'bounding' the solution. For the second question, the other boundary condition is set on the border of the circle, i.e., at $r = a$, as you have done. Cheers! $\endgroup$ – Dmoreno Jun 8 '14 at 10:44

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 $R\Theta/r^2$


  • $\begingroup$ Please see my edit on the question. $\endgroup$ – Myridium Jun 8 '14 at 10:13
  • $\begingroup$ I believe the rotation would be anticlockwise, since this is the standard orientation of a disk, also @Dmoreno is correct, usually with the Cauchy-Euler equation there is a singular part to the solution, so you set the coefficient of this part to zero. $\endgroup$ – Ellya Jun 8 '14 at 10:32
  • $\begingroup$ See my edit, hopefully it helps. $\endgroup$ – Ellya Jun 8 '14 at 10:42
  • $\begingroup$ Regarding the edit: since your boundary condition at $r = a$ is constant, it is hence axisymmetric so you would expect the solution to be axisymmetric too. The boundary conditions in the azimuthal direction would be, $T(\theta = 0) = T(\theta = 2 k \pi)$ and $T_\theta(\theta = 0) = T_\theta(\theta = 2 k \pi)$, $k \in \mathbb{Z}$, which is obviously satisfied for the axisymmetric condition $\partial_\theta = 0$ everywhere. $\endgroup$ – Dmoreno Jun 8 '14 at 10:55
  • $\begingroup$ If I understand, what you're saying here is that after using separation of variables on this equation, I would have to make sure that the $\theta$ part of $T_{eq}$ is $2\pi$ periodic. An axisymmetric solution will of course satisfy this condition. However, why can't I eliminate the $\frac{\partial^2 T_{eq}}{\partial \theta^2}$ term immediately instead of following this route? Isn't this partial derivative well-defined everywhere and equal to zero in a axisymmetric solution? $\endgroup$ – Myridium Jun 8 '14 at 11:11

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

  • $\begingroup$ It might take me a while to get through this as other things demand my attention and also this may be a bit beyond me. I appreciate it. $\endgroup$ – Myridium Jun 8 '14 at 12:01
  • $\begingroup$ It's ok @Myridium. Whether you want or don't want me to keep going through this approach, let me know! $\endgroup$ – Dmoreno Jun 8 '14 at 12:13

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