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Being my first question in Math StackExchange, a difficulty arises when I attempted to solve a poisson equation on a ring-shaped domain

$$ \begin{cases} \triangle u = 12(x^2 - y^2),\quad u \in \Omega\\ u(x,y) = 1, \quad x^2 + y^2 = a^2,\\ \dfrac{\partial u}{\partial n} =0, \quad x^2 + y^2 = b^2\end{cases} $$ in which $\Omega := \{(x,y)| a^2 \leq x^2 + y^2 \leq b^2\}$,with real number $0 < a < b$.

Naturally I tried to use Separation of Variables, which indicates the ansatz $$ u(x,y) = R(r)\Theta(\theta) $$ and in polar coordinates we have $$ \triangle u(r,\theta) = \dfrac{\partial^2 u}{\partial r^2} + \dfrac{1}{r} \dfrac{\partial u}{\partial r} + \dfrac{1}{r^2} \dfrac{\partial^2 u}{\partial \theta^2} $$ Thus we get the equation $$ R''(r)\Theta(\theta) + \dfrac{1}{r}R'(r)\Theta(\theta) + \dfrac{1} {r^2}R(r)\Theta''(\theta) = 12 r^2 \cos 2\theta $$ by taking $R(r) = Ar^4$ we have $$ A[\Theta''(\theta) + 16\Theta(\theta)] = 12 \cos 2\theta $$ and I find $\Theta(\theta) = 1/A \cos 2\theta$ a natural soluion.

However, by giving the solution $u(r,\theta) = r^4\cos 2\theta$ , I find it hard to imagine a satisfaction of boundary condition $$u\bigg|_{x^2 + y^2 = a^2} =1$$,which is irrelevant to angle function $\Theta(\theta)$ and on which symmetry holds with $\theta$.

It is possible that the ansatz should be improved, but I have no idea about that. Any comments or suggestions will be greatly appreciated.

Update:

I think I've found the particular solution $u_p = r^4\cos 2\theta$, and a solution for homogeneous equation should be added. let $$ u = u_p + \sum_{k=1}^{+\infty} R_k(r)\Theta_k(\theta) $$ ,where the series expansion solves the Laplace equation.

By separation of Variable I get $$ r^2 R_k'' + rR_k' - R_k\lambda_k^2 = 0,\quad \Theta_k'' + \lambda_k^2\Theta_k = 0 $$ By solving the Euler-type equation and the harmonic oscillation equation I get $$ R_k = A_k^1 r^{\lambda_k} + A_k^2 r^{-\lambda_k},\quad \Theta_k = C_k^1 \cos \lambda_k \theta + C_k^2 \sin \lambda_k \theta $$ Can this series expansion satisfy the two boundary conditions?

Amelius 258
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    I think assuming a separated solution is already going too far. The inhomogeneity of the equation as well as the weird geometry and BCs make me think that this equation might not even have a closed form. – K.defaoite Dec 26 '21 at 18:57
  • As @Chee Han points out, solutions to the homogeneous equation $\frac{\partial^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}u}{\partial \theta^{2}}=0$ can be added on to your particular solution. For example $u(r,\theta)=1$ or $u(r,\theta)=-a^{2}r^{2}\cos(2\theta)$. In fact with $u(r,\theta)=1+\left(r^{4}-a^{2}r^{2}\right)\cos(2\theta)$ you have a solution that satisfies your first boundary condition though not, I think, your second. – Ali Dec 27 '21 at 00:08
  • yeah@Ali ,this weied boundary condition really make me confused. – Amelius 258 Dec 27 '21 at 00:08
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    As you may have worked out by now, the answer is $u(r,\theta)=1 +\left(r^{4}+s r^{2} + t r^{-2}\right)\cos(2\theta)$ where $\begin{pmatrix}a^{2}&a^{-2}\b&-b^{-3}\end{pmatrix} \begin{pmatrix}s\t\end{pmatrix}= \begin{pmatrix}-a^{4}\-2b^{3}\end{pmatrix}$. @K.defaoite was too pessimistic! Of course you'd never meet such a problem in the wild. – Ali Dec 27 '21 at 09:39

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I myself make this mistake countless of time! The general solution to an inhomogeneous linear PDE (or ODE) has two parts: the homogeneous solution $u_h$ and the particular solution $u_p$. What you found is actually "the particular solution", you still need to find the constant $A$ which can be found but choosing $\Theta = \cos(2\theta)$ in your equation above.

Chee Han
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  • Can you be more explicit please? I don't understand the role you see for $A$. It seems to be eliminated almost as soon as it is introduced. – Ali Dec 26 '21 at 23:22
  • Thank you! I quite agree with @K.defaoite that the wired equation even cannot be solved in closed form, a series expansion may be work, but I haven't get any valuable work yet. – Amelius 258 Dec 27 '21 at 00:07
  • @Amelius258 I didn't actually try and solve it with the boundary condition, did it really not work? By the way, I think I got $A = 16/21$. – Chee Han Dec 27 '21 at 00:13
  • @Amelius258 Ok, I have some time to kill so I tried to solve it, I did get an answer so I'll let you work it out on your own (: Any resources on solving Laplace's equation on an annulus should guide you on how to solve it, here's one example: http://www.math.usm.edu/lambers/mat417/class0419.pdf – Chee Han Dec 27 '21 at 00:57
  • Thank you@CheeHan ，some of my trial has been updated in the question. Quite similar lol – Amelius 258 Dec 27 '21 at 01:03
  • @Amelius258 Usually I would write out the answer but I feel like you got this (: You are missing two more terms in your homogeneous solution $A + B\ln r$. Also, I found the particular solution to be $(16/21)r^4\cos(2\theta)$. Anyway, let me know if you need more hint though! – Chee Han Dec 27 '21 at 01:22
  • Turns out I was wrong and yall are correct, the particular solution is $r^4\cos(2\theta)$ with coefficient $1$...... – Chee Han Dec 27 '21 at 15:21