11.9  The Schwarz-Christoffel Transformation

    In Section 10.4 we mentioned the Riemann mapping theorem:  If D is any simply connected domain in the plane (other than the entire plane itself), then there exists a one-to-one conformal mapping    that maps D onto the unit disk  .  The Mobius transformation is a one-to-one conformal mapping of the unit disk    onto the upper half-plane  , and the inverse is .  Hence the Riemann mapping theorem could have been stated:  If D is any simply connected domain in the w-plane, then there exists a one-to-one conformal mapping    that maps the upper half-plane   onto D.  In this section we will introduce the Schwarz-Christoffel formula for constructing a conformal mapping from the upper half plane onto a region G bounded by a polygonal curve.  At the end of this section we will introduce the remarkable mappings of the unit disk onto an equilateral triangle and a square.

    To proceed further, we must review the rotational effect of a conformal mapping at a point .  If the contour C has the parameterization  ,  then a vector tangent to C at the point is  

            .  

The image of C is a contour K given by  ,  and a vector T tangent to K at the point    is

            .  

If the angle of inclination of    is  ,  then the angle of inclination of T is  

            .  

Hence the angle of inclination of the tangent    to C at is rotated through the angle    to obtain the angle of inclination of the tangent T to K at the point .

    Many applications involving conformal mappings require the construction of a one-to-one conformal mapping from the upper half plane    onto a domain G in the w-plane where the boundary consists of straight line segments.  Let's consider the case where G is the interior of a polygon P with vertices    specified in the positive sense (counterclockwise).  We want to find a function    with the property  

(11-38)        ,  and  

            where

            .  

Two German mathematicians Herman Amandus Schwarz (1843-1921) and Elwin Bruno Christoffel (1829-1900) independently discovered a method for finding f, and that is our next theorem.

Theorem 11.6 (Schwarz-Christoffel Formula).  Let P be a polygon in the w plane with vertices    and exterior angles  ,  where  .  There exists a one-to-one conformal mapping    from the upper half plane   onto G that satisfies the boundary conditions  

              

The derivative is   

(11-39)          

and the function can be expressed as an indefinite integral,  

(11-40)        

where A and B are suitably chosen constants.  Two of the points    may be chosen arbitrarily, and the constants A and B determine the size and position of P.

 

Figure 11.70  A Schwarz-Christoffel mapping with and .  

 

Figure 11.71  A Schwarz-Christoffel mapping with and

Proof.

Proof of Theorem 11.6 is in the book.
Complex Analysis for Mathematics and Engineering

    Equation (11-40) gives a representation for f in terms of an indefinite integral.  Note that these integrals do not represent elementary functions unless the image is an infinite region.  Also, the integral will involve a multivalued function, and we must select a specific branch to fit the boundary values specified in the problem.  Table 11.2 is useful for our purposes.

    The following table of integrals is useful for hand computations and are compatible with the versions used in the textbook.  Occasionally they have been adjusted by a constant of integration which could be changed for indefinite integrands.  Also, depending on which quadrant you use it might be necessary to use various combinations of the branches of the multivalued terms.

            Table 11.2  Indefinite integrals.

Exploration

Example 11.26.  Use the Schwarz-Christoffel formula to verify that    maps the upper half plane    onto the semi infinite strip    shown in Figure 11.72.

Figure 11.72  The region with and .

Solution.  If we choose  ,  and  ,  then the exterior angles are  ,  and Equation (11-39) for becomes  

            

Then, using Table 11.2, the indefinite integral becomes

              

Using the image values    and  , we obtain the system   

            

and simplifies to be

            

which we can solve to obtain  .  Hence the required function is

              

Explore Solution 11.26.

Example 11.27.  Use the Schwarz Christoffel formula to verify that    maps the upper half plane    onto the upper half plane    slit along the line segment from  .   (Use the principal square root throughout.)

Solution.  If we choose    and  ,  then the formula

            

will determine a mapping from the upper half-plane    onto the portion of the upper half-plane    that lies outside the triangle with vertices    as indicated in Figure 11.73(a).  If we let ,  then  , and  .  

Figure 11.73   The region with and .

    The limiting formula for the derivative becomes    

            

which will determine a mapping from the upper half-plane    onto the upper half plane    slit along the line segment from   as indicated in Figure 11.73(b).  An easy integration reveals that is given by  

              

and the boundary values    lead to system of equations

               and   

the solution is easily found to be and the desired function  is  

            .

Explore Solution 11.27.

Example 11.28.  Show that    maps the upper half plane    onto the right angle channel in the first quadrant, which is bounded by the coordinate axes and the rays  , as depicted in Figure 11.74(b).

Solution.  If we choose    and  ,  then the formula  

            

will determine a mapping of the upper half-plane onto the domain indicated in Figure 11.74(a).  With  ,  we let ,  then    and  .  

Figure 11.74  The region with and .

    The limiting formula for the derivative becomes   

            

where  , which will determine a mapping from the upper half plane onto the channel as indicated in Figure 11.74(b).  

    Using integrals in Table 11.2, we obtain

           

    If we use the principal branch of the inverse sine function, then the boundary values    lead to the system  

               and   ,  

which we can solve to obtain .  Hence the required solution is

              

Explore Solution 11.28.

Extra Example 1.  Show that    maps the upper half-plane onto the domain indicated in Figure 11.77.  
Hint: Set    and   .

Figure 11.77  The region with and .

Solution.  The exterior angles are  ,  and the formula for the derivative is  

            
integrate and get

            
            
use the conditions    and    and solve the resulting system for A and B.  The desired result is

            .

Explore Extra Solution 1.

Extra Example 2.  Show that    maps the upper half-plane onto the domain indicated in Figure 11.78.   
Hint: Set    and  

Figure 11.78  The region with and .

Solution.  The exterior angles are  ,  and the formula for the derivative is  

              
integrate and get

            
            
use the conditions    and    and solve the resulting system for A and B.  The desired result is

            .

Explore Extra Solution 2.

Extra Example 3.  Show that    maps the upper half-plane onto the domain indicated in Figure 11.84.  
Hint: Set   and .  Use the change of variable    in the resulting integral.

Figure 11.84  The region with and .

Solution.  The exterior angles are  ,  and the formula for the derivative is  

               
integrate

            
            
use the substitutions

              
and get

              

Use the identity    and write the integral as

            

Now use the substitution and get

            
            
use the conditions  ,    and solve the resulting system for A and B.  The desired result is

            .

Explore Extra Solution 3.

Extra Example 4.  Show that    maps the upper half-plane onto the domain indicated in Figure 11.85.   
Hint: Set    and  .

Figure 11.85  The region with and .

Solution.  The exterior angles are  ,  and the formula for the derivative is  

            
integrate and get

            
            
use the conditions  ,    and solve the resulting system for A and B.  The desired result is

            .

Explore Extra Solution 4.

Extra Example 5.  Use the Schwarz Christoffel formula to show that    will map the upper-half-plane onto an equilateral triangle.
Hint: For convenience we have chosen  .

Solution.  The exterior angles of an equilateral triangle are  ,  and the formula for the derivative   is  

            
integrate and get

            

Explore Extra Solution 5.

Extra Example 6.  Use the Schwarz Christoffel formula to show that    will map the upper-half-plane onto a square.
Hint: For convenience we have chosen  .

Solution.  The exterior angles of a square are  ,  and the formula for the derivative   is  

            
integrate and get

              

Explore Extra Solution 6.

Exercises for Section 11.9.  The Schwarz-Christoffel Transformation

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

(c) 2012 John H. Mathews, Russell W. Howell