Multipole expansion in cylindrical coordinates

Question

I am seeking the general solution for the Laplace equation in cylindrical coordinates or

$$\nabla^2 \omega = 0. $$

In several texts, the general solution can be found via separation of variables and I get the general solution

$$\omega = (A_0+B_0\theta)(C_0+D_0 \ln r) + \sum_{n = 1}^\infty (A_n\cos(\lambda_n\theta)+B_n\sin(\lambda_n\theta))(C_nr^{\lambda_n}+D_nr^{-\lambda_n})$$

In this general solution, most of the terms are represented by the exterior and interior multipole expansion except for $B_0D_0\theta\ln r$. So my first question is why does this term show up and why is it not included in the multipole expansion? Since the multipole expansion is an orthogonal basis shouldn't it cover all possible solutions?

Another problem I have is that I have found that

$$\omega = -\dfrac{2}{r} [A_{1L} \cos(\theta) + B_{1L} \sin(\theta) + C_{1L}(\theta \cos(\theta)- \sin(\theta) \ln r) + D_{1L}( \cos(\theta) \ln r + \theta \sin(\theta))]$$

is a solution to the Laplace equation. This was obtained by taking the Laplacian of a solution of $\psi$ where $\nabla^4 \psi = 0$. Specifically I see terms with $\dfrac{\ln r}{r}$ appear. Has this solution been discussed anywhere and how does it fit into the exterior/interior multipole expansion?

EDIT: Modified equation to clearly group harmonic terms

Answer 1

1. The expression you give is indeed the general solution for a harmonic function (i.e. $\nabla^2 f=0$) in two dimensions. The solution $f=\theta\ln(r)$ is usually omitted because it cannot be sustained as a periodic function over a $2\pi$ range in $\theta$.

   Moreover, even if you have a limited range in $\theta$, this term is singular at the origin, which reflects the fact that if you set two plates at an angle at different electrostatic potentials (say) then the solution will be singular because your boundary conditions are discontinuous.

   However, for regions which are also bounded in $r$ (i.e. $0\leq \theta\leq\theta_0<2\pi$, $r>r_0>0$) this term is a crucial part of the solution and it is trivial to construct boundary conditions that cannot be matched without it.

   If you have any books that claim otherwise -- i.e. that omit this term in situations which include a wedge with a limited range in $\theta$, and without appropriate boundary conditions to rule it out -- then they are wrong. Most resources I know don't fall into this category, but if you have specific examples then we can comment on the details for those.

2. (Apologies for having missed some terms in a previous version. Take this as a learning opportunity: displaying a big sum of terms without explicitly indicating which terms are repeated can, and will, make people misread your work. Communication is a two-way process but you need to make your expressions as easy to read (or as hard to misread) as possible.)

   The functions \begin{align} \omega_1 = \frac{\cos(\theta) \ln r + \theta \sin(\theta)}{r} \quad\text{and}\quad \omega_2 = \frac{ - \sin(\theta) \ln r +\theta \cos(\theta)\sin(\theta)}{r} \end{align} are indeed harmonic. They are not included explicitly in the multipolar expansion because they are not separable - they cannot be expressed in the form $\omega=R(r) \Theta(\theta)$. (The first two functions, in $A_{1L}$ and $B_{1L}$, are explicitly included in it.)

   Since the multipolar expansion is a basis, the two functions above can always be expressed in terms of it - i.e. they can be cast as a multipolar series if so desired. Note, however, that for this function to be allowed, you need a limited range in $\theta$, of the form $\theta_0<\theta<\theta_1<2\pi+\theta_0$, or you will have a discontinuous function (or, at best, a discontinuous derivative).

   Depending on the exact situation, you will also need to work in a domain that's bounded from below in $r$, or the $\omega_i$ will have infinite energy. Both of these constraints obviously affect the details of the orthogonality of the multipolar components, so you'll need to work a bit harder to make the expansion work. I won't post that process here because it depends on exactly what domain you want, and it's on you to do the drudgework. If you get stuck you can ask it here, of course.

Answer 2

$B_0 \theta$ is not periodic in $\theta$ so this term is always zero. Otherwise, it can't match the boundary conditions $\omega(\theta+2\pi)=\omega(\theta)$. The same logic eliminates $C$ and $D$ in your second solution. What remains is just a particular case of the general first solution.