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

  • $\begingroup$ Your first expression is the general solution. Can you point to references that miss the terms you care about? It would give a better idea of the context. It's unclear what your second expression is, though. Are you claiming it's a harmonic function? (It's not.) What equation, specifically, does it satisfy? $\endgroup$ Oct 26, 2015 at 23:30
  • $\begingroup$ @EmilioPisanty I was under the impression that a function that satisfies the laplace equation is harmonic. In cylindrical coordinates the second expression does satisfy $\nabla^2 \omega = 0$. The second expression was obtained from Theory of Elasticity by Borodich on page 228 available at this link: google.com/… $\endgroup$
    – Ragnar
    Oct 27, 2015 at 20:42
  • $\begingroup$ @EmilioPisanty Borodich provides a general solution to the biharmonic equation $\nabla^4 \psi = 0$. Thus if I define $\omega = \nabla^2 \psi$, then $\nabla^2 \omega = 0$. The second expression in my question was obtained by taking the laplacian of part of the general solution of $\psi$ $\endgroup$
    – Ragnar
    Oct 27, 2015 at 20:46
  • $\begingroup$ Quick comment - you don't really need to call out specific edits, unless you think it would lead to confusion. The revision history of every post is available by clicking on the edited ... ago link to the left of the poster username. You can also flag not-that-useful admin &etc comments, like this one, as obsolete once you read them, using the flag to the left, and it will help keep the site cleaner. You should also be able to vote up and down on posts now. $\endgroup$ Oct 28, 2015 at 20:35

2 Answers 2

  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.

  • $\begingroup$ 1. Thanks for the explanation regarding $\theta \ln r$. I suppose that it is mostly not included in discussion as most problems consider the region 0 to $2\pi$. Regarding the multivaluedness at the origin, I do not think it is a problem as the laplacian breaks down at $r = 0$ given that in cylindrical coordinates the laplacian contains a $1/r$ term. $\endgroup$
    – Ragnar
    Oct 27, 2015 at 23:38
  • 1
    $\begingroup$ 2: While it is true that the individual term $\dfrac{\cos(\theta)\ln(r)}{r}$ is not harmonic, the whole second expression for $\omega$ in my question is. I think that it is similar to how an individual term of the multipole expansion is not harmonic, eg the dipole $\left(\nabla^2 \dfrac{A\cos(\theta)+B\sin(\theta)}{r} \neq 0\right)$, while the summation of all terms converges to the exact harmonic solution. $\endgroup$
    – Ragnar
    Oct 27, 2015 at 23:40
  • $\begingroup$ If there are constraints between the coefficients, you should mention it in the question. Otherwise, I am free to set all coefficients but that one too zero and obtain a contradiction to your claim. $\endgroup$ Oct 27, 2015 at 23:51
  • $\begingroup$ It's not a multivaluedness issue, it's the fact that you have two metal plates at different potential touching each other. This is not a a physical condition - it requires infinite energy and an infinitely thin, infinitely resistive dielectric between them. It's useful to think about as a limit, but you pay for the simplicity in singularity. $\endgroup$ Oct 27, 2015 at 23:55
  • $\begingroup$ As far as I know, terms proportional to $\theta$ are ALWAYS omitted so that $\theta$ is single-valued (Jackson 3.8). It's the term $A_0D_0\ln{r}$ that is needed to match boundary conditions when a radius is excluded. If you can indeed trivially construct boundary conditions that require the linear $\theta$ term, it would be educational for me and perhaps others to see them. Since the Fourier functions are complete in $\theta$, I'm skeptical. Also, the function at $r=0$ is not single-valued if there is any $\theta$ dependence there, linear or otherwise. $\endgroup$
    – user27118
    Oct 28, 2015 at 18:18

$B_0 \theta$ 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.

  • $\begingroup$ Thanks for the answer, but why does $\omega$ have to periodic? What if I am only considering a wedge where $0 \leq \theta \leq 2\pi$? $\endgroup$
    – Ragnar
    Oct 26, 2015 at 18:09
  • $\begingroup$ Your multipole expansion would not be a basis for such a space, which is a very strange space indeed. For cylindrical coordinates, $\omega$ must accept all real values of $r$, $\theta$, and $z$. In a physical problem involving some wedge, we might seek a function subject to the conditions $\omega = \omega(\theta)$ in the wedge and constant (probably 0) outside. The linear $\theta$ terms must still be zero in that case. $\endgroup$
    – user27118
    Oct 26, 2015 at 19:35
  • $\begingroup$ If you prescribe a constant outside the wedge, it will likely result in a discontinuity at the boundaries of the wedge, which I think will cause problems. The way I have been approaching a wedge problem is to match the boundary conditions for the wedge defined by $0 \leq \theta \leq \phi$. The solution inside this wedge is valid, but I disregard the solution for $\theta > \phi$ or $\theta < 0$. I am not sure that this is any better but it ignores the discontinuity at $\theta + 2\pi$ $\endgroup$
    – Ragnar
    Oct 27, 2015 at 21:14
  • $\begingroup$ Actually I think that your method and my method may be similar in that they ignore anything outside the wedge, only you do it by setting it to 0. However, could you explain why setting the space outside the wedge to 0 would force the $\theta$ terms to 0? $\endgroup$
    – Ragnar
    Oct 27, 2015 at 21:16
  • $\begingroup$ What happens when you go around the circle completely? The constraint $0<\theta<\theta_0$ is unphysical - the actual constraint would have to be $0<\theta<\theta_0$ OR $0<\theta \pm 2\pi <\theta_0$ OR $0<\theta \pm 4\pi <\theta_0$ OR... $\endgroup$
    – user27118
    Oct 28, 2015 at 18:27

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