Csdrhrt

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 $(x,y) notin D$

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.

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

$$ frac{1}{c^2}u_t = u_{rhorho} + frac{1}{rho}u_{rho} + frac{1}{rho^2}u_{phiphi} $$

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 $T(t) = e^{-c^2lambda^2t}$. 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 $Phi(phi + 2pi) = Phi(phi)$. Hence, $Phi(phi) = Dcos(nphi) + Esin(nphi)$

For the radial function

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

The substitution $z=lambdarho$ 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}

Then the radial function is

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

We can assemble the general solution

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

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

Matching the initial condition gives

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

You can use orthogonality to find the constants. For example

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