Conformal Mapping Wedge to Plate

Click For Summary
SUMMARY

The discussion focuses on the conformal mapping of a wedge with an interior angle α into a half-plane using the Schwarz-Christoffel transformation. The wedge W is defined by curves in polar coordinates, specifically c1: θ=0, c2: θ=α, and c3: r=R. The mapping involves two functions: f, which maps the angle θ to the horizontal axis, and g, which maps the radius r to the vertical axis. The logit function is suggested for f, while g can be represented by h(r) = tan((1-r/R)π/2), adjusted to ensure the derivative at r=0 is zero.

PREREQUISITES
  • Understanding of conformal mapping principles
  • Familiarity with the Schwarz-Christoffel transformation
  • Knowledge of polar coordinates and their applications
  • Basic calculus, particularly derivatives and function behavior
NEXT STEPS
  • Study the Schwarz-Christoffel transformation in detail
  • Explore the properties and applications of the logit function
  • Investigate the behavior of functions approaching infinity and their derivatives
  • Learn about polar coordinate transformations in complex analysis
USEFUL FOR

Mathematicians, physicists, and engineers interested in complex analysis, particularly those working with conformal mappings and fluid dynamics.

member 428835
Hi PF!

Does anyone know the conformal map that takes a wedge of some interior angle ##\alpha## into a half plane? I'm not talking about the potential flow, just the mapping for the shape.

Thanks!
 
Physics news on Phys.org
Yes, it’s Schwarz-Christoffel
 
Say the wedge ##W## is bounded by the curves, in polar coordinates, ##c1:\theta=0, c2:\theta=\alpha## and ##c3:r=R## with the last one, excluding its end points, included in ##W## but the other two not.

Then we could try to map ##c3## to the horizontal axis of the half plane, via a function ##g## that maps ##r## to ##y## and another function ##f## that maps ##\theta## to ##x##.

We require ##f:(0,\alpha)\to(-\infty,\infty)## and ##g:(0,R]\to [0,\infty)##.

The logit function is a natural choice for ##f##. Any cdf for a random variable distribution with support on the whole real line should also work.

For ##g## we can take any function ##h## that is zero at ##r=R## and approaches ##\infty## as ##r\to 0## with no turning points or inflections between, (eg ##h(r)=\tan ((1-r/R)\pi/2))##. I expect we want ##\frac{dg}{dr}(0)=0##, to which end we could deduct ##((1-r/R)\pi/2)\frac{dh}{dr}(0)## from ##h##. So a function that might work is ##y = g(r) = \tan((1-r/R)\pi/2) - (1-r/R)\pi/2##.
 
  • Like
Likes   Reactions: member 428835
Thank you both!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
7
Views
3K