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Let $A_r = \{z\in\mathbb{C}: |z|\leq 1\}\cup\{z\in\mathbb{C}: |z-1|\leq r\}$ be the unit disk with a small "bump" (I'm interested in the regime $r\to 0$). What can be said about the logarithmic capacity of $A_r$? I can prove that there exists a constant $\gamma>0$ such that $\mathop{\mathrm{cap}}(A_r)=\gamma r^2 + $terms of smaller order, but have no idea about how to obtain the value of $\gamma$.

Related to the above question: is there any useful explicit form of the conformal mapping of $\{z\in\mathbb{C}: \Im(z)\leq 0\}$ to $\{z\in\mathbb{C}: \Im(z)\leq 0\}\cup \{z\in\mathbb{C}: |z|\leq 1\}$?

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    $\begingroup$ Not sure about "useful", but it is definitely explicit: move $-1$ to $\infty$ by a fractional linear map and then shift the vertex of the resulting angle of aperture $3\pi/2$ to $0$ and apply $z^{2/3}$ to get a half-plane. $\endgroup$
    – fedja
    Commented Nov 8, 2017 at 22:03
  • $\begingroup$ $\mathrm{cap}(A_r)$ must tend to $1$, the capacity of the unit disk, not to $0$ as $r\to 0$, as you wrote. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 9, 2017 at 5:15

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Both questions have exact explicit answer, which is explained in any good textbook on analytic functions. A region on the Riemann sphere is called a digon if its boundary consists of two arcs of circles. Let $a$ and $b$ be the common endpoints of these arcs. Then $$f(z)=\frac{z-a}{z-b}$$ sends $a,b$ to $0,\infty$, and the arcs to rays. So the image of our region is a sector. The interior angle at a corner of the digon be $\pi\alpha$. Then the interior angle at the vertex of the sector is also $\pi\alpha$. Next $f_1(z)=z^{1/\alpha}$ maps this sector onto a half-plane. And the final ingredient is conformal map of this half-plane onto the unit disk.

EDIT 2. Asymptotics is $\mathrm{cap}(A_r)=1+r^2/2+O(r^3),$ as $r\to 0$. The exact formula is a bit complicated: $$\frac{1}{\mathrm{cap}(A_r)}=\frac{1}{4b}(1+\phi)(1+b^2)\sin\left(\frac{4}{1+\phi}\arctan b\right),$$ where $$\phi=\frac{2}{\pi}\arcsin(r/2),$$ and $$b=\frac{r}{2\sqrt{1-r^2/4}}.$$

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  • $\begingroup$ Thanks a lot for your answer! I've obtained a bit simpler formula, $\mathop{\mathrm{cap}}(A_r)=\frac{2\sin(\pi\phi)}{(1+\phi)\sin\frac{2\pi\phi}{1+\phi}}$ (at least, the asymptotics is the same). See docdro.id/GFTxNxN $\endgroup$
    – Serguei Popov
    Commented Nov 16, 2017 at 18:11

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