In Section 26.1 we saw that the real and imaginary parts of an analytic function each satisfies Laplace’s equation. We shall show now that the curves
intersect each other at right angles (i.e. are orthogonal ). To see this we note that along the curve we have . Hence
Thus, on these curves the gradient at a general point is given by
Similarly along the curve constant, we have
The product of these gradients is
where we have made use of the Cauchy-Riemann equations. We deduce that the curves are orthogonal.
As an example of the practical application of this work consider two-dimensional electrostatics. If constant gives the equipotential curves then the curves constant are the electric lines of force . Figure 2 shows some curves from each set in the case of oppositely-charged particles near to each other; the dashed curves are the lines of force and the solid curves are the equipotentials.
In ideal fluid flow the curves constant are the streamlines of the flow.
In these situations the function is the complex potential of the field.
A function can be regarded as a mapping, which maps a point in the -plane to a point in the -plane. Curves in the -plane will be mapped into curves in the -plane.
Consider aerodynamics where we are interested in the fluid flow in a complicated geometry (say flow past an aerofoil). We first find the flow in a simple geometry that can be mapped to the aerofoil shape (the complex plane with a circular hole works here). Most of the calculations necessary to find physical characteristics such as lift and drag on the aerofoil can be performed in the simple geometry - the resulting integrals being much easier to evaluate than in the complicated geometry.
Consider the mapping
The point maps to . The point lies on the intersection of the two lines and . To what curves do these map? To answer this question we note that a point on the line can be written as . Then
As usual, let , then
Eliminating we obtain:
so is the curve to which maps.
Onto what curve does the line map?
A point on the line is . Then
Hence and so that, eliminating we obtain
In Figure 3(a) we sketch the lines and and in Figure 3(b) we sketch the curves into which they map. Note these curves intersect at the point .
The angle between the original lines in (a) is clearly ; what is the angle between the curves in (b) at the point of intersection?
The curve has a gradient . Differentiating the equation implicitly we obtain
At the point .
Find for the curve and evaluate it at the point .
. At we obtain Note that the product of the gradients at is and therefore the angle between the curves at their point of intersection is also . Since the angle between the lines and the angle between the curves is the same we say the angle is preserved .
In general, if two curves in the -plane intersect at a point , and their image curves under the mapping intersect at and the angle between the two original curves at equals the angle between the image curves at we say that the mapping is conformal at .
An analytic function is conformal everywhere except where .
At which points is not conformal?
. Since this is never zero the mapping is conformal everywhere.
The mapping is called an inversion . It maps the interior of the unit circle in the -plane to the exterior of the unit circle in the -plane, and vice-versa. Note that
A line through the origin in the -plane will be mapped into a line through the origin in the -plane. To see this, consider the line , for constant. Then
so that , which is a line through the origin in the -plane.
Consider the line where . This represents a line in the -plane which does not pass through the origin. To what type of curve does it map in the -plane?
The mapped curve is
Hence . Dividing by we obtain the equation:
which is the equation of a circle in the -plane which passes through the origin.
Similarly, it can be shown that a circle in the -plane passing through the origin maps to a line in the -plane which does not pass through the origin. Also a circle in the -plane which does not pass through the origin maps to a circle in the -plane which does pass through the origin. The inversion mapping is an example of the bilinear transformation :
(If the mapping reduces to constant.)
Find the set of bilinear transformations which map to .
Any values of satisfying this equation will do provided .
Find the bilinear transformations for which is mapped to .
Find the bilinear transformation which maps
- to , and
- to , and
We have the answers to 1. and 2. from the previous two Tasks:
If is mapped to then so that . Substituting this last relation into the first two obtained we obtain
Solving these two in terms of we find and . Hence the transformation is:
(note that the ’s cancel in the numerator and denominator).
Some other mappings are shown in Figure 4.
As an engineering application we consider the Joukowski transformation
where is a constant.
It is used to map circles which contain as an interior point and which pass through into shapes resembling aerofoils. Figure 5 shows an example:
This creates a cusp at which the associated fluid velocity can be infinite. This can be avoided by adjusting the fluid flow in the -plane. Eventually, this can be used to find the lift generated by such an aerofoil in terms of physical characteristics such as aerofoil shape and air density and speed.
Find a bilinear transformation which maps
- gives so that
- gives so so .
- gives so that (using 1. and 2.)
We conclude from 3. that . We also know that .