# Solutions to the heat equation with a ring of coolant

I'm interested in solutions to the heat equation for the following problem

$$\dfrac{\partial u}{\partial t} = c^2 \nabla^2 u$$

on $$\left\{ (x,y) \vert x^2+y^2 \leq R^2 \right\}$$

such that

$$D = \left\{ (x,y) \vert x^2+y^2 = r^2 \right\}$$

$$u(t,x,y) = h$$ if $$(x,y) \in D$$

and $$u(0,x,y) = H$$ if $$(x,y) \not\in D$$

and $$\dfrac{\partial u}{ \partial \hat{n}} = 0$$

If I have written this correctly, this is a PDE on an insulated disk on radius $$R$$ with a ring of radius $$r$$ on the interior held at a constant temperature of $$h$$.

My main question is: Are there analytic methods of solving problems of this form, or are numerical solutions required. I've searched my undergrad PDE texts, but all the problems studied a PDE with a point source of heat/cold.

This is a problem of personal interest, not homework or research. References are appreciated.

• Is $H$ constant? Dec 16, 2018 at 5:49
• Since there is no flux across the boundary $x^2+y^2=r^2$, I imagine that you can solve two separate problems: one is for the annulus $r^2 \leq x^2+y^2 \leq R^2$ and the other for the disc $0<x^2+y^2 \leq r^2$. Separation of variables might be a good method to use since you have simple geometries. Since your domain is 2D, your solution should be logarithmic with respect to the distance from the origin.
– D.B.
Dec 16, 2018 at 6:05
• There is something off about the second boundary condition for $H$. Not being in $D$ implies that that temperature is held on $\mathbb{R}^2 - D$ I think he meant to write a different formulation if the domain of interest is only on the disk. Dec 16, 2018 at 18:39

First convert to polar coordinates $$(x,y,t) \to (\rho,\phi,t)$$. The equation becomes

$$\frac{1}{c^2}u_t = u_{\rho\rho} + \frac{1}{\rho}u_{\rho} + \frac{1}{\rho^2}u_{\phi\phi}$$

Observe that $$u = h$$ is a steady-state solution satisfying the two boundary conditions, so we can write $$u = h + v$$, where $$v$$ also satisfies the heat equation but is homogeneous on the boundary

\begin{align} v(r,\phi,t) &= 0 \\ v_\rho(R,\phi, t) &= 0 \\ v(\rho,\phi,0) &= H(\rho,\phi) - h \equiv f(\rho,\phi) \end{align}

Now use separation of variables. Let $$v(\rho,\phi,t) = P(\rho)\Phi(\phi)T(t)$$, then

$$\frac{P''}{P} + \frac{P'}{\rho P} + \frac{\Phi''}{\rho^2 \Phi} = \frac{T'}{c^2 T} = -\lambda^2$$

which gives $$T(t) = e^{-c^2\lambda^2t}$$. Next, separate again

$$\frac{\rho^2P''}{P} + \frac{\rho P'}{P} + \lambda^2 \rho^2 = -\frac{\Phi''}{\Phi} = n^2$$

Here, $$n$$ must be an integer to satisfy the periodicity condition $$\Phi(\phi + 2\pi) = \Phi(\phi)$$. Hence, $$\Phi(\phi) = D\cos(n\phi) + E\sin(n\phi)$$

$$\rho^2 P'' + \rho P' + (\lambda^2\rho^2 - n^2)P = 0$$

The substitution $$z=\lambda\rho$$ leads to the Bessel equation, thus we can write

$$P(\rho) = A J_n(\lambda \rho) + B Y_n(\lambda \rho)$$

As stated in the comments, you have two separate problems due to the boundary condition at $$\rho = r$$. The solution on the smaller disk $$\rho \in (0,r)$$ is finite at the origin, so $$B=0$$ and

$$P_{n,m}(\rho) = J_n(\lambda_{n,m} \rho)$$

where the eigenvalue $$\lambda_{n,m}$$ is such that $$P(r) = J_n(\lambda_{n,m} r) = 0$$. Here $$m$$ refer to the order of the zeroes of $$J_n$$, which can be found numerically.

The solution on the annulus $$\rho \in (r,R)$$ has different eigenvalues, so I'll call it $$\mu$$ to avoid confusion. The radial function needs to satisfy $$P(r) = P'(R) = 0$$ therefore

\begin{align} A J_n(\mu r) + B Y_n(\mu r) &= 0 \\ A J_n'(\mu R) + B Y_n'(\mu R) &= 0 \end{align}

Eliminating either $$A$$ or $$B$$ leads to

$$J_n(\mu r) Y_n'(\mu R) - Y_n(\mu r)J_n'(\mu R) = 0$$

This is equation for $$\mu$$ that can be solved numerically. Once again, there are infinitely many solutions, so $$\mu = \mu_{n,m}$$. The derivatives of the Bessel function are given by the recurrence relations:

$$\begin{cases} J_0'(x) = -J_1(x) \\ J_n'(x) = \dfrac{J_{n-1}(x) - J_{n+1}(x)}{2} \end{cases}$$

$$P_{n,m}(\rho) = Y_n(\mu_{n,m} r) J_n(\mu_{n,m} \rho) - J_n(\mu_{n,m} r) Y_n (\mu_{n,m} \rho)$$

We can assemble the general solution

$$v(\rho,\phi,t) = \sum_{n=0}^\infty\sum_{m=0}^\infty P_{n,m}(\rho) \big[D_{n,m} \cos(n\phi) + E_{n,m}\sin(n\phi)\big]e^{-c^2\lambda_{n,m}^2 t}$$

Note that the function is piece-wise with different eigenvalues.

Matching the initial condition gives

$$u(\rho,\phi,0) = f(\rho,\phi) = \sum_{n=0}^\infty\sum_{m=0}^\infty P_{n,m}(\rho) \big[D_{n,m} \cos(n\phi) + E_{n,m}\sin(n\phi)\big]$$

You can use orthogonality to find the constants. For example

$$D_{n,m} = \dfrac{\displaystyle \int_0^{2\pi}\int_0^r f(\rho,\phi) J_n(\lambda_{n,m} \rho) \cos(n\phi) \ \rho\ d\rho\ d\phi}{\displaystyle \int_0^r \rho J_n^2(\lambda_{n,m}\rho)\ d\rho \int_0^{2\pi} \cos^2 (n\phi)\ d\phi}$$

• Incredible answer, thank you. Dec 16, 2018 at 18:31