Module

for

The Schwarz-Christoffel Transformation

 

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  [Graphics:Images/SchwarzChristoffelMod_gr_1.gif]  that maps D onto the unit disk  [Graphics:Images/SchwarzChristoffelMod_gr_2.gif].  The Mobius transformation [Graphics:Images/SchwarzChristoffelMod_gr_3.gif] is a one-to-one conformal mapping of the unit disk  [Graphics:Images/SchwarzChristoffelMod_gr_4.gif]  onto the upper half-plane  [Graphics:Images/SchwarzChristoffelMod_gr_5.gif], and the inverse is [Graphics:Images/SchwarzChristoffelMod_gr_6.gif].  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  [Graphics:Images/SchwarzChristoffelMod_gr_7.gif]  that maps the upper half-plane  [Graphics:Images/SchwarzChristoffelMod_gr_8.gif] 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.

[Graphics:Images/SchwarzChristoffelMod_gr_9.gif]

 

 

    To proceed further, we must review the rotational effect of a conformal mapping [Graphics:Images/SchwarzChristoffelMod_gr_10.gif] at a point [Graphics:Images/SchwarzChristoffelMod_gr_11.gif].  If the contour C has the parameterization  [Graphics:Images/SchwarzChristoffelMod_gr_12.gif],  then a vector [Graphics:Images/SchwarzChristoffelMod_gr_13.gif] tangent to C at the point [Graphics:Images/SchwarzChristoffelMod_gr_14.gif] is  

            [Graphics:Images/SchwarzChristoffelMod_gr_15.gif].  

The image of C is a contour K given by  [Graphics:Images/SchwarzChristoffelMod_gr_16.gif],  and a vector T tangent to K at the point  [Graphics:Images/SchwarzChristoffelMod_gr_17.gif]  is

            [Graphics:Images/SchwarzChristoffelMod_gr_18.gif].  

If the angle of inclination of  [Graphics:Images/SchwarzChristoffelMod_gr_19.gif]  is  [Graphics:Images/SchwarzChristoffelMod_gr_20.gif],  then the angle of inclination of T is  

            [Graphics:Images/SchwarzChristoffelMod_gr_21.gif].  

Hence the angle of inclination of the tangent  [Graphics:Images/SchwarzChristoffelMod_gr_22.gif]  to C at [Graphics:Images/SchwarzChristoffelMod_gr_23.gif] is rotated through the angle  [Graphics:Images/SchwarzChristoffelMod_gr_24.gif]  to obtain the angle of inclination of the tangent T to K at the point [Graphics:Images/SchwarzChristoffelMod_gr_25.gif].

    Many applications involving conformal mappings require the construction of a one-to-one conformal mapping from the upper half plane  [Graphics:Images/SchwarzChristoffelMod_gr_26.gif]  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  [Graphics:Images/SchwarzChristoffelMod_gr_27.gif]  specified in the positive sense (counterclockwise).  We want to find a function  [Graphics:Images/SchwarzChristoffelMod_gr_28.gif]  with the property  

(11-38)        [Graphics:Images/SchwarzChristoffelMod_gr_29.gif],  and  [Graphics:Images/SchwarzChristoffelMod_gr_30.gif]

            where

            [Graphics:Images/SchwarzChristoffelMod_gr_31.gif].  

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  [Graphics:Images/SchwarzChristoffelMod_gr_32.gif]  and exterior angles  [Graphics:Images/SchwarzChristoffelMod_gr_33.gif],  where  [Graphics:Images/SchwarzChristoffelMod_gr_34.gif].  There exists a one-to-one conformal mapping  [Graphics:Images/SchwarzChristoffelMod_gr_35.gif]  from the upper half plane [Graphics:Images/SchwarzChristoffelMod_gr_36.gif]  onto G that satisfies the boundary conditions  

            [Graphics:Images/SchwarzChristoffelMod_gr_37.gif]  

The derivative [Graphics:Images/SchwarzChristoffelMod_gr_38.gif] is   

(11-39)        [Graphics:Images/SchwarzChristoffelMod_gr_39.gif]  

and the function [Graphics:Images/SchwarzChristoffelMod_gr_40.gif] can be expressed as an indefinite integral,  

(11-40)        [Graphics:Images/SchwarzChristoffelMod_gr_41.gif]

where A and B are suitably chosen constants.  Two of the points  [Graphics:Images/SchwarzChristoffelMod_gr_42.gif]  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 [Graphics:Images/SchwarzChristoffelMod_gr_43.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_44.gif].  

 

Figure 11.71  A Schwarz-Christoffel mapping with [Graphics:Images/SchwarzChristoffelMod_gr_45.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_46.gif]

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.

            [Graphics:Images/SchwarzChristoffelMod_gr_47.gif]

            [Graphics:Images/SchwarzChristoffelMod_gr_48.gif]
            
            [Graphics:Images/SchwarzChristoffelMod_gr_49.gif]
            
            [Graphics:Images/SchwarzChristoffelMod_gr_50.gif]
            
            [Graphics:Images/SchwarzChristoffelMod_gr_51.gif]

            Table 11.2  Indefinite integrals.

Exploration

 

Example 11.26.  Use the Schwarz-Christoffel formula to verify that  [Graphics:Images/SchwarzChristoffelMod_gr_74.gif]  maps the upper half plane  [Graphics:Images/SchwarzChristoffelMod_gr_75.gif]  onto the semi infinite strip  [Graphics:Images/SchwarzChristoffelMod_gr_76.gif]  shown in Figure 11.72.

[Graphics:Images/SchwarzChristoffelMod_gr_77.gif]

Figure 11.72  The region with [Graphics:Images/SchwarzChristoffelMod_gr_78.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_79.gif].

Solution.  If we choose  [Graphics:Images/SchwarzChristoffelMod_gr_80.gif],  and  [Graphics:Images/SchwarzChristoffelMod_gr_81.gif],  then the exterior angles are  [Graphics:Images/SchwarzChristoffelMod_gr_82.gif],  and Equation (11-39) for [Graphics:Images/SchwarzChristoffelMod_gr_83.gif] becomes  

            [Graphics:Images/SchwarzChristoffelMod_gr_84.gif]

Then, using Table 11.2, the indefinite integral becomes

            [Graphics:Images/SchwarzChristoffelMod_gr_85.gif]  

Using the image values  [Graphics:Images/SchwarzChristoffelMod_gr_86.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_87.gif], we obtain the system   

            [Graphics:Images/SchwarzChristoffelMod_gr_88.gif]

and simplifies to be

            [Graphics:Images/SchwarzChristoffelMod_gr_89.gif]

which we can solve to obtain  [Graphics:Images/SchwarzChristoffelMod_gr_90.gif].  Hence the required function is

            [Graphics:Images/SchwarzChristoffelMod_gr_91.gif]  

Explore Solution 11.26.

 

Example 11.27.  Use the Schwarz Christoffel formula to verify that  [Graphics:Images/SchwarzChristoffelMod_gr_106.gif]  maps the upper half plane  [Graphics:Images/SchwarzChristoffelMod_gr_107.gif]  onto the upper half plane  [Graphics:Images/SchwarzChristoffelMod_gr_108.gif]  slit along the line segment from  [Graphics:Images/SchwarzChristoffelMod_gr_109.gif].   (Use the principal square root throughout.)

Solution.  If we choose  [Graphics:Images/SchwarzChristoffelMod_gr_110.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_111.gif],  then the formula

            [Graphics:Images/SchwarzChristoffelMod_gr_112.gif]

will determine a mapping [Graphics:Images/SchwarzChristoffelMod_gr_113.gif] from the upper half-plane  [Graphics:Images/SchwarzChristoffelMod_gr_114.gif]  onto the portion of the upper half-plane  [Graphics:Images/SchwarzChristoffelMod_gr_115.gif]  that lies outside the triangle with vertices  [Graphics:Images/SchwarzChristoffelMod_gr_116.gif]  as indicated in Figure 11.73(a).  If we let [Graphics:Images/SchwarzChristoffelMod_gr_117.gif],  then  [Graphics:Images/SchwarzChristoffelMod_gr_118.gif], and  [Graphics:Images/SchwarzChristoffelMod_gr_119.gif].  

[Graphics:Images/SchwarzChristoffelMod_gr_120.gif]

Figure 11.73   The region with [Graphics:Images/SchwarzChristoffelMod_gr_121.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_122.gif].

    The limiting formula for the derivative [Graphics:Images/SchwarzChristoffelMod_gr_123.gif] becomes    

            [Graphics:Images/SchwarzChristoffelMod_gr_124.gif]

which will determine a mapping [Graphics:Images/SchwarzChristoffelMod_gr_125.gif] from the upper half-plane  [Graphics:Images/SchwarzChristoffelMod_gr_126.gif]  onto the upper half plane  [Graphics:Images/SchwarzChristoffelMod_gr_127.gif]  slit along the line segment from  [Graphics:Images/SchwarzChristoffelMod_gr_128.gif] as indicated in Figure 11.73(b).  An easy integration reveals that [Graphics:Images/SchwarzChristoffelMod_gr_129.gif] is given by  

            [Graphics:Images/SchwarzChristoffelMod_gr_130.gif]  

and the boundary values  [Graphics:Images/SchwarzChristoffelMod_gr_131.gif]  lead to system of equations

            [Graphics:Images/SchwarzChristoffelMod_gr_132.gif]   and   [Graphics:Images/SchwarzChristoffelMod_gr_133.gif]

the solution is easily found to be [Graphics:Images/SchwarzChristoffelMod_gr_134.gif] and the desired function  is  

            [Graphics:Images/SchwarzChristoffelMod_gr_135.gif].

Explore Solution 11.27.

 

Example 11.28.  Show that  [Graphics:Images/SchwarzChristoffelMod_gr_150.gif]  maps the upper half plane  [Graphics:Images/SchwarzChristoffelMod_gr_151.gif]  onto the right angle channel in the first quadrant, which is bounded by the coordinate axes and the rays  [Graphics:Images/SchwarzChristoffelMod_gr_152.gif], as depicted in Figure 11.74(b).

Solution.  If we choose  [Graphics:Images/SchwarzChristoffelMod_gr_153.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_154.gif],  then the formula  

            [Graphics:Images/SchwarzChristoffelMod_gr_155.gif]

will determine a mapping [Graphics:Images/SchwarzChristoffelMod_gr_156.gif] of the upper half-plane onto the domain indicated in Figure 11.74(a).  With  [Graphics:Images/SchwarzChristoffelMod_gr_157.gif],  we let [Graphics:Images/SchwarzChristoffelMod_gr_158.gif],  then  [Graphics:Images/SchwarzChristoffelMod_gr_159.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_160.gif].  

[Graphics:Images/SchwarzChristoffelMod_gr_161.gif]

Figure 11.74  The region with [Graphics:Images/SchwarzChristoffelMod_gr_162.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_163.gif].

    The limiting formula for the derivative [Graphics:Images/SchwarzChristoffelMod_gr_164.gif] becomes   

            [Graphics:Images/SchwarzChristoffelMod_gr_165.gif]

where  [Graphics:Images/SchwarzChristoffelMod_gr_166.gif], which will determine a mapping [Graphics:Images/SchwarzChristoffelMod_gr_167.gif] from the upper half plane onto the channel as indicated in Figure 11.74(b).  

    Using integrals in Table 11.2, we obtain

        [Graphics:Images/SchwarzChristoffelMod_gr_168.gif]   

    If we use the principal branch of the inverse sine function, then the boundary values  [Graphics:Images/SchwarzChristoffelMod_gr_169.gif]  lead to the system  

            [Graphics:Images/SchwarzChristoffelMod_gr_170.gif]   and   [Graphics:Images/SchwarzChristoffelMod_gr_171.gif],  

which we can solve to obtain [Graphics:Images/SchwarzChristoffelMod_gr_172.gif].  Hence the required solution is

            [Graphics:Images/SchwarzChristoffelMod_gr_173.gif]  

Explore Solution 11.28.

 

Extra Example 1.  Show that  [Graphics:Images/SchwarzChristoffelMod_gr_187.gif]  maps the upper half-plane onto the domain indicated in Figure 11.77.  
Hint: Set  [Graphics:Images/SchwarzChristoffelMod_gr_188.gif]  and   [Graphics:Images/SchwarzChristoffelMod_gr_189.gif].

[Graphics:Images/SchwarzChristoffelMod_gr_190.gif]

Figure 11.77  The region with [Graphics:Images/SchwarzChristoffelMod_gr_191.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_192.gif].

Solution.  The exterior angles are  [Graphics:Images/SchwarzChristoffelMod_gr_193.gif],  and the formula for the derivative [Graphics:Images/SchwarzChristoffelMod_gr_194.gif] is  

            
integrate and get

            [Graphics:Images/SchwarzChristoffelMod_gr_196.gif]
            
use the conditions  [Graphics:Images/SchwarzChristoffelMod_gr_197.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_198.gif]  and solve the resulting system for A and B.  The desired result is

            [Graphics:Images/SchwarzChristoffelMod_gr_199.gif].

Explore Extra Solution 1.

 

Extra Example 2.  Show that  [Graphics:Images/SchwarzChristoffelMod_gr_210.gif]  maps the upper half-plane onto the domain indicated in Figure 11.78.   
Hint: Set  [Graphics:Images/SchwarzChristoffelMod_gr_211.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_212.gif]

[Graphics:Images/SchwarzChristoffelMod_gr_213.gif]

Figure 11.78  The region with [Graphics:Images/SchwarzChristoffelMod_gr_214.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_215.gif].

Solution.  The exterior angles are  [Graphics:Images/SchwarzChristoffelMod_gr_216.gif],  and the formula for the derivative [Graphics:Images/SchwarzChristoffelMod_gr_217.gif] is  

              
integrate and get

            [Graphics:Images/SchwarzChristoffelMod_gr_219.gif]
            
use the conditions  [Graphics:Images/SchwarzChristoffelMod_gr_220.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_221.gif]  and solve the resulting system for A and B.  The desired result is

            [Graphics:Images/SchwarzChristoffelMod_gr_222.gif].

Explore Extra Solution 2.

 

Extra Example 3.  Show that  [Graphics:Images/SchwarzChristoffelMod_gr_233.gif]  maps the upper half-plane onto the domain indicated in Figure 11.84.  
Hint: Set  [Graphics:Images/SchwarzChristoffelMod_gr_234.gif] and .  Use the change of variable  [Graphics:Images/SchwarzChristoffelMod_gr_236.gif]  in the resulting integral.

[Graphics:Images/SchwarzChristoffelMod_gr_237.gif]

Figure 11.84  The region with [Graphics:Images/SchwarzChristoffelMod_gr_238.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_239.gif].

Solution.  The exterior angles are  [Graphics:Images/SchwarzChristoffelMod_gr_240.gif],  and the formula for the derivative [Graphics:Images/SchwarzChristoffelMod_gr_241.gif] is  

            [Graphics:Images/SchwarzChristoffelMod_gr_242.gif]   
integrate

            [Graphics:Images/SchwarzChristoffelMod_gr_243.gif]
            
use the substitutions

            [Graphics:Images/SchwarzChristoffelMod_gr_244.gif]  
and get

            [Graphics:Images/SchwarzChristoffelMod_gr_245.gif]  

Use the identity  [Graphics:Images/SchwarzChristoffelMod_gr_246.gif]  and write the integral as

            [Graphics:Images/SchwarzChristoffelMod_gr_247.gif]

Now use the substitution [Graphics:Images/SchwarzChristoffelMod_gr_248.gif] and get

            [Graphics:Images/SchwarzChristoffelMod_gr_249.gif]
            
use the conditions  [Graphics:Images/SchwarzChristoffelMod_gr_250.gif],  [Graphics:Images/SchwarzChristoffelMod_gr_251.gif]  and solve the resulting system for A and B.  The desired result is

            [Graphics:Images/SchwarzChristoffelMod_gr_252.gif].

Explore Extra Solution 3.

 

Extra Example 4.  Show that  [Graphics:Images/SchwarzChristoffelMod_gr_266.gif]  maps the upper half-plane onto the domain indicated in Figure 11.85.   
Hint: Set  [Graphics:Images/SchwarzChristoffelMod_gr_267.gif]  and  [Graphics:Images/SchwarzChristoffelMod_gr_268.gif].

[Graphics:Images/SchwarzChristoffelMod_gr_269.gif]

Figure 11.85  The region with [Graphics:Images/SchwarzChristoffelMod_gr_270.gif] and [Graphics:Images/SchwarzChristoffelMod_gr_271.gif].

Solution.  The exterior angles are  [Graphics:Images/SchwarzChristoffelMod_gr_272.gif],  and the formula for the derivative [Graphics:Images/SchwarzChristoffelMod_gr_273.gif] is  

            [Graphics:Images/SchwarzChristoffelMod_gr_274.gif]
integrate and get

            [Graphics:Images/SchwarzChristoffelMod_gr_275.gif]
            
use the conditions  [Graphics:Images/SchwarzChristoffelMod_gr_276.gif],  [Graphics:Images/SchwarzChristoffelMod_gr_277.gif]  and solve the resulting system for A and B.  The desired result is

            [Graphics:Images/SchwarzChristoffelMod_gr_278.gif].

Explore Extra Solution 4.

 

Extra Example 5.  Use the Schwarz Christoffel formula to show that  [Graphics:Images/SchwarzChristoffelMod_gr_289.gif]  will map the upper-half-plane onto an equilateral triangle.
Hint: For convenience we have chosen  [Graphics:Images/SchwarzChristoffelMod_gr_290.gif].

[Graphics:Images/SchwarzChristoffelMod_gr_291.gif]

Solution.  The exterior angles of an equilateral triangle are  [Graphics:Images/SchwarzChristoffelMod_gr_292.gif],  and the formula for the derivative  [Graphics:Images/SchwarzChristoffelMod_gr_293.gif] is  

            [Graphics:Images/SchwarzChristoffelMod_gr_294.gif]
integrate and get

            [Graphics:Images/SchwarzChristoffelMod_gr_295.gif]

Explore Extra Solution 5.

 

Extra Example 6.  Use the Schwarz Christoffel formula to show that  [Graphics:Images/SchwarzChristoffelMod_gr_314.gif]  will map the upper-half-plane onto a square.
Hint: For convenience we have chosen  [Graphics:Images/SchwarzChristoffelMod_gr_315.gif].

[Graphics:Images/SchwarzChristoffelMod_gr_316.gif]

Solution.  The exterior angles of a square are  [Graphics:Images/SchwarzChristoffelMod_gr_317.gif],  and the formula for the derivative  [Graphics:Images/SchwarzChristoffelMod_gr_318.gif] is  

            [Graphics:Images/SchwarzChristoffelMod_gr_319.gif]
integrate and get

            [Graphics:Images/SchwarzChristoffelMod_gr_320.gif]  

Explore Extra Solution 6.

 

Exercises for Section 11.9.  The Schwarz-Christoffel Transformation

 

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Schwarz-Christoffel Transformation

 

 

  

 

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