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I am looking to derive a conformal map for the problem illustrated in this image. I've read a bit about how to map a square onto a circle, but I'm struggling to extend the concepts for the domain at hand. I don't have a rigorous mathematical background (mech. engineer in computational fluid dynamics), so I would appreciate if someone here could advise me on the route that I should take in order to derive such a conformal mapping.

The end application is to generate a smooth computational mesh that looks like this. I have generated a mesh like this using other means, but the smoothness of the mesh vertices is not sufficient for extremely fine meshes. This results in spurious oscillations in the numerical problem I am trying to solve.

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    $\begingroup$ This is a question more appropriate for Mathematics StackExchange. This site is intended for research-level mathematics. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Nov 22, 2019 at 11:06
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    $\begingroup$ @Carl-Fredrik Broda: you think this is not research level mathematics? Perhaps you know the answer? $\endgroup$
    – Alexandre Eremenko
    Commented Nov 22, 2019 at 13:25
  • $\begingroup$ @AlexandreEremenko I do not know the answer, no. The problem seemed to me as one which was more appropriate for SE, as it seemed as one that might be readily solved by someone with a more "rigorous mathematical background'' in conformal mappings, as OP puts it, and there might be a higher chance of an answer to this question on SE. But my judgement on this might well be wrong! $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Nov 22, 2019 at 13:47
  • $\begingroup$ @Carl-Fredrik Nyberg Brodda, thanks for the suggestion! I have posted it on there as well. $\endgroup$
    – niran90
    Commented Nov 23, 2019 at 22:45
  • $\begingroup$ @niran90 It is an interesting question no doubt, so I hope you find the answer you're looking for! $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Nov 23, 2019 at 23:06

1 Answer 1

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Edited. Using symmetry lines, break the original $D$ (5 squares) into 8 trapezoids with angles $\pi/4,\pi/2,\pi/2,3\pi/4$. This trapezoid must be mapped conformally onto a sector which makes $1/8$-th of the disk. Under this map all angles at the corners are preserved, except $3\pi/4$ which becomes $\pi$. Then Christoffel-Schwarz leads to the following integral $$C\int_0^z \zeta^{-3/4}(\zeta-a)^{-1/2}(\zeta-1)^{-1/2}d\zeta,$$ where $a$ is the accessory parameter, and it is determined from the equation $$\int_0^a=-3i\int_a^1.$$ This equation can be easily solved numerically since the absolute values of right and left sides are monotone in the opposite directions (one increasing another decreasiing as functions of $a$). The definite integral involved in the equation for the accessory parameter can be expressed in terms of hypergeometric function, Gradshtein, Ryzhik, p. 317, item 3.197, but I am not sure how much this helps. I would just find $a$ numerically.

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  • $\begingroup$ Thanks so much for this info! I am not sure if I'm completely clear on some of the ideas though - do you mean 8 trapezoids or 8 right-angled triangles with angles 45,45,90? And if so, does the procedure you've proposed above ensure angle-preservation at the boundaries of the sectors too? I require that all angles, except for the ones surrounding the 4 points at which three square regions meet, preserve their orthogonality. $\endgroup$
    – niran90
    Commented Nov 23, 2019 at 22:44
  • $\begingroup$ I think just grasped the former part of your post. Do you think by vitue of the fact that 3PI/4 angle is mapped onto PI, the map cannot be conformal? Also, I am not entirely familiar with Christoffel-Schwartz yet, but what would be the procedure to derive the function that takes in a coordinate from the original domain and spits out a coordinate in the mapped space? $\endgroup$
    – niran90
    Commented Nov 23, 2019 at 23:51
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    $\begingroup$ "Conformal" means of course "conformal inside the region, not on the boundary. Otherwise how could you map even one square on a disk?? $\endgroup$
    – Alexandre Eremenko
    Commented Nov 24, 2019 at 4:01
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    $\begingroup$ Schwarz-Christoffel formula is derived or explained in most complex analysis textbooks. Of course you need some background on complex analysis which can be found in the same books. Take any of those books which has "Schwarz Symmetry Principle" and "Schwarz-Christoffel formula". $\endgroup$
    – Alexandre Eremenko
    Commented Nov 24, 2019 at 13:53
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    $\begingroup$ It is the preimage of one of the vertices. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 19, 2019 at 3:17

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