Finding steady-state solution for two-dimensional heat equation

Question

Find the steady-state solution for the following heat equation with a source:

$$\frac{\partial u}{\partial t}= \nabla^2u -12(x^2-y^2) $$

on a ring (two dimension) with $a \le r \le b$ with dirichlet boundary conditions:

$$u_{r=a}= u_{r=b} = 0 $$

I've tried solving this by switching to polar coordinates (since there's no dependance on z) and with separation of variables, arriving at a radial solution with the Bessel function. (of course after finding eigenvalues)

Plugging in the radial and angular solutions into the equations doesn't give me anything that I personally can solve.

Now the other method is to "guess" a solution and plug it in, but I'm not sure what I should look for in a "guess" and would appreciate a direction.

Answer 1

Guessing is always connected to the fact that one has already solved many of these problems. Using polar coordinates ($x=r \cos \phi, y = r\sin\phi$), you can observe that the inhomogeneity is given by $$x^2-y^2 = r^2 \cos (2\phi).$$

This provides the intuition to try the form $$u(x,y) = R(r) \cos(2\phi)$$ with $R$ a function that is yet to determine. You observe that $R$ has to fulfil the differential equation $$ \frac1r \frac{d}{dr}( r R'(r))= 12 r^2 + \frac{4 R(r)}{r^2}$$ with the boundary condition $R(a)=R(b)=0$. This equation is solved by $$ R(r) = \frac{a^4 b^4}{a^2+b^2} r^{-2}- \frac{a^4+a^2 b^2 + b^4}{a^2+b^2} r^2+r^4.$$

The final solution thus reads $$u(x,y) = R\left(\sqrt{x^2+y^2}\right) \frac{x^2-y^2}{x^2+y^2}.$$