I am reading this paper about "Numerical approximation of the logarithmic capacity of domains", and there (on the third page) I found simple formulas for logarithmic capacity of simple figures like squares and equilateral triangles.

For a better understanding, I tried to recalculate those by myself several times, but I could not succeed. Is there a way that I can find those calculations or something similar that I can use as a guide?


In two dimensions, you have a powerful tool, the Riemann mapping. If you have a compact set in the plane whose complement is connected, knowing explicitly the map of the complement onto the exterior of the unit disk gives you the log capacity. Now take a book on complex variables with many examples of conformal maps, and you can calculate capacities of many planar sets. I recommend the book by

L. Volkovyskii, A collection of problems in complex analysis, there is an English Dover edition, which is a real encyclopedia of explicit conformal maps. (If you can read some Russian, there is another book: M. A. Evgrafov, Problems in the theory of analytic functions).

All examples given in the paper that you cite can be obtained with this method. For the disc, ellipse and the segment the conformal map is the Joukowski function, for the square and triangle use a modified Schwarz-Christoffel formula, and it leads to Gamma function. (All this is covered in the book of Volkovyskii cited above. You can easily generalize the last two examples to regular $n$-gons with any $n$. The answer is in terms of Gamma function; it is written in Polya, Szego, Isoperimetric problems of mathematical Physics.

It is more difficult in dimension 3 and higher (Newtonian potential). A good collection of explicitly computed capacities is contained in the book N. S. Landkof, Foundations of modern potential theory (there is an English translation).

EDIT. Here are some detail for the regular $n$-gon. We want to map the exterior region onto the exterior of the unit disk so that $\infty\to\infty$ by a function $f$. To do this, break the exterior region into $n$ congruent triangles with one vertex at infinity, and other two vertices the adjacent vertices of the polygon. The interior angles of a triangle are $2\pi/n$ at $\infty$ and $\pi/2+\pi/n$ at finite vertices. Now map one of these triangles onto the sector $\{ z: |\arg z|<\pi/n, |z|>1\}$. So that $\infty\to\infty$ and other two vertices go to $e^{\pm i\pi/n}$. By the symmetry principle this function will have an analytic continuation to the whole exterior and map it onto the exterior of the unit disk, so it is our $f$. The required map of the triangle onto a sector is performed with the Schwarz-Christoffel formula, and the integral which is involved is reduced to the Gamma-function. Then $f(z)\sim az,\;z\to\infty$ and capacity is $1/|a|$.


Here's a larger class of examples coming from dynamics. Let $$f(x)=ax^d+bx^{d-1}+\cdots+c\in\mathbb C[x]$$ be a polynomial of degree $d\ge2$. The filled Julia set of $f$ is the set $$ K_f := \bigl\{ z\in\mathbb C : \text{$|f^n(z)|$ is bounded as $n\to\infty$} \bigr\}. $$ (Here $f^n$ denotes the $n$'th iterate of $f$.) Then the Green function $G_f(z)$ of $\mathbb C\setminus K_f$ is the function $$ G_f(z) := \lim_{n\to\infty} \frac{1}{d^n}\log\bigl|f^n(z)\bigr|. $$ If we start with a "large" value of $z$, it's not hard to see that $$f^n(z)\approx a^{1+d+d^2+\cdots+d^{n-1}}z^{d^n} = a^{\frac{d^n-1}{d-1}}z^{d^n}. $$ (I'll let you make this rigorous.) Hence (with a bit of work) $$ G_f(z) = \log|z|+\frac{\log|a|}{d-1} + o(1), $$ leading to $$ -\log C(K_f) = \frac{\log|a|}{d-1}, $$ and hence $$ C(K_f) = |a|^{-1/(d-1)}.$$ To recover Carlo's example, notice that the filled Julia set of the polynomial $f(x)=ax^d$ is the disk of radius $|a|^{-1/(d-1)}$ centered at $0$

  • $\begingroup$ Very nice. Then the question reduced to construct a polynomial with a given filled Julia set. Is there any any that we can find a polynomial with a square (or a triangle) as the filled Julia set? $\endgroup$ – Bumblebee Nov 5 '17 at 20:53
  • $\begingroup$ @Bumblebee, "Is there any any that we can find a polynomial with a square (or a triangle) as the filled Julia set?" seems to be worth its own separate question. $\endgroup$ – J. M. is not a mathematician Nov 5 '17 at 21:22
  • 3
    $\begingroup$ This method gives capacities of very exotic sets (namely the (filled in) Julia sets and the Mandelbrot set). The only non-exotic among them are a segment and the a disk. (All other Julia sets cannot be defined by finitely many polynomial inequalities. The boundaries of other filled in Julia sets are curves without tangents anywhere.) $\endgroup$ – Alexandre Eremenko Nov 6 '17 at 3:10

For a compact set $K$ in the complex plane, find the Green function $G(z)$ (harmonic function in $\mathbb{C}\backslash K$ which vanishes on the boundary of $K$), and then the logarithmic capacity $C(K)$ follows from $$G(z)=\log|z|-\log C+o(1)\;\;\text{for}\;\;z\rightarrow\infty.$$ See Logarithmic Potential Theory with Applications to Approximation Theory (2010).

Two simple examples:

  • For a disc of radius $r$, center $w$, one has $$G(z)=\log\frac{|z-w|}{r}\Rightarrow C=r.$$
  • The line segment $[a,b]$ on the real axis has Green function $$G(z)=\log\left|z-\tfrac{1}{2}(a+b)+\sqrt{(a-z)(b-z)}\right|-\log\tfrac{1}{2}(b-a)\Rightarrow C=\tfrac{1}{4}(b-a).$$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.