2 Conformal mapping
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
$\phantom{\rule{2em}{0ex}}u\left(x,y\right)=\text{constant}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}v\left(x,y\right)=\text{constant}$
intersect each other at right angles (i.e. are orthogonal ). To see this we note that along the curve $u\left(x,y\right)=\text{constant}$ we have $du=0$ . Hence
$\phantom{\rule{2em}{0ex}}du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy=0$ .
Thus, on these curves the gradient at a general point is given by
$\phantom{\rule{2em}{0ex}}\frac{dy}{dx}=-\frac{\frac{\partial u}{\partial x}}{\frac{\partial u}{\partial y}}$ .
Similarly along the curve $v\left(x,y\right)=$ constant, we have
$\phantom{\rule{2em}{0ex}}\frac{dy}{dx}=-\frac{\frac{\partial v}{\partial x}}{\frac{\partial v}{\partial y}}$ .
The product of these gradients is
$\phantom{\rule{2em}{0ex}}\frac{\left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial v}{\partial x}\right)}{\left(\frac{\partial u}{\partial y}\right)\left(\frac{\partial v}{\partial y}\right)}=-\frac{\left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial u}{\partial y}\right)}{\left(\frac{\partial u}{\partial y}\right)\left(\frac{\partial u}{\partial x}\right)}=-1$
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 $u=$ constant gives the equipotential curves then the curves $v=$ 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.
Figure 2
In ideal fluid flow the curves $v=$ constant are the streamlines of the flow.
In these situations the function $w=u+\text{i}v$ is the complex potential of the field.
2.1 Function as mapping
A function $w=f\left(z\right)$ can be regarded as a mapping, which maps a point in the $z$ -plane to a point in the $w$ -plane. Curves in the $z$ -plane will be mapped into curves in the $w$ -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
$\phantom{\rule{2em}{0ex}}w={z}^{2}$ .
The point $z=2+\text{i}$ maps to $w={\left(2+\text{i}\right)}^{2}=3+4\text{i}$ . The point $z=2+\text{i}$ lies on the intersection of the two lines $x=2$ and $y=1$ . To what curves do these map? To answer this question we note that a point on the line $y=1$ can be written as $z=x+\text{i}$ . Then
$\phantom{\rule{2em}{0ex}}w={\left(x+\text{i}\right)}^{2}={x}^{2}-1+2x\text{i}$
As usual, let $w=u+\text{i}v$ , then
$\phantom{\rule{2em}{0ex}}u={x}^{2}-1\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}v=2x$
Eliminating $x$ we obtain:
$\phantom{\rule{2em}{0ex}}4u=4{x}^{2}-4={v}^{2}-4$ so ${v}^{2}=4+4u$ is the curve to which $y=1$ maps.
Example 5
Onto what curve does the line $x=2$ map?
Solution
A point on the line is $z=2+y\text{i}$ . Then
$\phantom{\rule{2em}{0ex}}w={\left(2+y\text{i}\right)}^{2}=4-{y}^{2}+4y\text{i}$
Hence $u=4-{y}^{2}$ and $v=4y$ so that, eliminating $y$ we obtain
$\phantom{\rule{2em}{0ex}}16u=64-{v}^{2}$ or ${v}^{2}=64-16u$
In Figure 3(a) we sketch the lines $x=2$ and $y=1$ and in Figure 3(b) we sketch the curves into which they map. Note these curves intersect at the point $\left(3,4\right)$ .
Figure 3
The angle between the original lines in (a) is clearly $9{0}^{0}$ ; what is the angle between the curves in (b) at the point of intersection?
The curve ${v}^{2}=4+4u$ has a gradient $\frac{dv}{du}$ . Differentiating the equation implicitly we obtain
$\phantom{\rule{2em}{0ex}}2v\frac{dv}{du}=4\phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}\frac{dv}{du}=\frac{2}{v}$
At the point $\left(3,4\right)$ $\frac{dv}{du}=\frac{1}{2}$ .
Task!
Find $\frac{dv}{du}$ for the curve ${v}^{2}=64-16u$ and evaluate it at the point $\left(3,4\right)$ .
$2v\frac{dv}{du}=-16\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{1em}{0ex}}\frac{dv}{du}=-\frac{8}{v}$ . At $v=4$ we obtain $\frac{dv}{du}=-2.$ Note that the product of the gradients at $\left(3,4\right)$ is $-1$ and therefore the angle between the curves at their point of intersection is also $9{0}^{0}$ . 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 $z$ -plane intersect at a point ${z}_{0}$ , and their image curves under the mapping $w=f\left(z\right)$ intersect at ${w}_{0}=f\left({z}_{0}\right)$ and the angle between the two original curves at ${z}_{0}$ equals the angle between the image curves at ${w}_{0}$ we say that the mapping is conformal at ${z}_{0}$ .
An analytic function is conformal everywhere except where ${f}^{\prime}\left(z\right)=0$ .
Task!
At which points is $w={e}^{z}$ not conformal?
${f}^{\prime}\left(z\right)={e}^{z}$ . Since this is never zero the mapping is conformal everywhere.
2.2 Inversion
The mapping $\phantom{\rule{1em}{0ex}}w=f\left(z\right)=\frac{1}{z}$ is called an inversion . It maps the interior of the unit circle in the $z$ -plane to the exterior of the unit circle in the $w$ -plane, and vice-versa. Note that
$\phantom{\rule{2em}{0ex}}w=u+\text{i}v=\frac{x}{{x}^{2}+{y}^{2}}-\frac{y}{{x}^{2}+{y}^{2}}\text{i}\phantom{\rule{1em}{0ex}}$ and similarly $z=x+\text{i}y=\frac{u}{{u}^{2}+{v}^{2}}-\frac{v}{{u}^{2}+{v}^{2}}\text{i}$
so that
$\phantom{\rule{2em}{0ex}}u=\frac{x}{{x}^{2}+{y}^{2}}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}v=-\frac{y}{{x}^{2}+{y}^{2}}$ .
A line through the origin in the $z$ -plane will be mapped into a line through the origin in the $w$ -plane. To see this, consider the line $y=mx$ , for $m$ constant. Then
$\phantom{\rule{2em}{0ex}}u=\frac{x}{{x}^{2}+{m}^{2}{x}^{2}}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}v=-\frac{mx}{{x}^{2}+{m}^{2}{x}^{2}}$
so that $v=-mu$ , which is a line through the origin in the $w$ -plane.
Task!
Consider the line $ax+by+c=0$ where $c\ne 0$ . This represents a line in the $z$ -plane which does not pass through the origin. To what type of curve does it map in the $w$ -plane?
The mapped curve is
$\phantom{\rule{2em}{0ex}}\frac{au}{{u}^{2}+{v}^{2}}-\frac{bv}{{u}^{2}+{v}^{2}}+c=0$
Hence $au-bv+c\left({u}^{2}+{v}^{2}\right)=0$ . Dividing by $c$ we obtain the equation:
$\phantom{\rule{2em}{0ex}}{u}^{2}+{v}^{2}+\frac{a}{c}u-\frac{b}{c}v=0$
which is the equation of a circle in the $w$ -plane which passes through the origin.
Similarly, it can be shown that a circle in the $z$ -plane passing through the origin maps to a line in the $w$ -plane which does not pass through the origin. Also a circle in the $z$ -plane which does not pass through the origin maps to a circle in the $w$ -plane which does pass through the origin. The inversion mapping is an example of the bilinear transformation :
$\phantom{\rule{2em}{0ex}}w=f\left(z\right)=\frac{az+b}{cz+d}\phantom{\rule{2em}{0ex}}\text{wherewedemandthat}\phantom{\rule{1em}{0ex}}ad-bc\ne 0$
(If $ad-bc=0$ the mapping reduces to $f\left(z\right)=$ constant.)
Task!
Find the set of bilinear transformations $w=f\left(z\right)=\frac{az+b}{cz+d}$ which map $z=2$ to $w=1$ .
$1=\frac{2a+b}{2c+d}.$ Hence $2a+b=2c+d.$
Any values of $a,b,c,d$ satisfying this equation will do provided $ad-bc\ne 0$ .
Task!
Find the bilinear transformations for which $z=-1$ is mapped to $w=3$ .
$3=\frac{-a+b}{-c+d}$ . Hence $-a+b=-3c+3d.$
Example 6
Find the bilinear transformation which maps
- $z=2$ to $w=1$ , and
- $z=-1$ to $w=3$ , and
- $z=0$ to $w=-5$
Solution
We have the answers to 1. and 2. from the previous two Tasks:
$$\begin{array}{rcll}2a+b& =& 2c+d& \text{}\\ -a+b& =& -3c+3d\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$$If $z=0$ is mapped to $w=-5$ then $-5=\frac{b}{d}$ so that $b=-5d$ . Substituting this last relation into the first two obtained we obtain
$$\begin{array}{rcll}2a-2c-6d& =& 0& \text{}\\ -a+3c-8d& =& 0\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$$Solving these two in terms of $d$ we find $2c=11d$ and $2a=17d$ . Hence the transformation is:
$\phantom{\rule{2em}{0ex}}w=\frac{17z-10}{11z+2}$ (note that the $d$ ’s cancel in the numerator and denominator).
Some other mappings are shown in Figure 4.
Figure 4
As an engineering application we consider the Joukowski transformation
$\phantom{\rule{2em}{0ex}}w=z-\frac{{\ell}^{2}}{z}$ where $\ell $ is a constant.
It is used to map circles which contain $z=1$ as an interior point and which pass through $z=-1$ into shapes resembling aerofoils. Figure 5 shows an example:
Figure 5
This creates a cusp at which the associated fluid velocity can be infinite. This can be avoided by adjusting the fluid flow in the $z$ -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.
Exercise
Find a bilinear transformation $w=\frac{az+b}{cz+d}$ which maps
- $z=0$ into $w=\text{i}$
- $z=-1$ into $w=0$
- $z=-\text{i}$ into $w=1$
- $z=0,\phantom{\rule{1em}{0ex}}w=\text{i}\phantom{\rule{1em}{0ex}}$ gives $\text{i}=\frac{b}{d}\phantom{\rule{1em}{0ex}}$ so that $b=d\text{i}$
- $z=-1,\phantom{\rule{1em}{0ex}}w=0$ gives $0=\frac{-a+b}{-c+d}$ so $-a+b=0$ so $a=b$ .
- $z=-\text{i},\phantom{\rule{1em}{0ex}}w=1\phantom{\rule{1em}{0ex}}$ gives $1=\frac{-a\text{i}+b}{-c\text{i}+d}\phantom{\rule{1em}{0ex}}$ so that $-c\text{i}+d=-a\text{i}+b=d+d\text{i}$ (using 1. and 2.)
We conclude from 3. that $-c=d$ . We also know that $a=b=d\text{i}$ .
Hence $w=\frac{d\text{i}z+d\text{i}}{-dz+d}=\frac{\text{i}z+\text{i}}{-z+1}$