Joukowski Conformal Ellipse and Critical Slit

worked-example

Let us calculate the image of a point under the Joukowski map w = z + 1/z. Consider a point on the circle of radius R = 2 at an angle of 90 degrees.

R = 2
theta = 90 degrees
z = 2 * (cos(90) + i*sin(90)) = 2i

w = z + 1/z
  = 2i + 1/(2i)
  = 2i - 0.5i
  = 1.5i

The coordinates of w are (0, 1.5).
The mapped X is 0.0.
The mapped Y is 1.5.

ellipseSemimajor = R + 1/R = 2 + 0.5 = 2.5
ellipseSemiminor = R - 1/R = 2 - 0.5 = 1.5

derivative = 1 - 1/z^2
           = 1 - 1/(2i)^2
           = 1 - 1/(-4)
           = 1 + 0.25 = 1.25
jacobian = |1.25|^2 = 1.5625

The preimage is uniquely exterior because R > 1.

experiment

In the Joukowski Conformal Ellipse and Critical Slit module, explore the conformal mapping of circles to ellipses and the degenerate critical slit.

To observe the critical endpoint, follow these steps to see where conformality fails:

1. Set the Circle radius control to 1.
2. Set the Angle degrees control to 0.
3. Note that the mapped coordinates become (2, 0) and the derivative becomes 0, indicating the conformal mapping fails here.

limitations

The Joukowski transform is conformal only where its derivative is non-zero, which is everywhere except at z = +/-1. On the unit circle, the mapping is not globally injective since points exp(i*theta) and exp(-i*theta) map to the same point on the critical slit. We do not claim global conformality or injectivity on the whole punctured plane.

references

For more information on the Joukowski mapping and its properties, including how circles map to airfoils when the center is shifted, see the comprehensive resource linked below:

• Joukowski Transformation and Aerofoils, Cambridge University