Exercise 9.   Use a Schwarz-Christoffel transformation to find a conformal mapping  [Graphics:Images/SourceSinkModHome_gr_524.gif]  that will map the flow

in the upper half-plane onto the flow in a right-angled channel,

as indicated in Figure 11.107.                     Figure 11.107.

Solution 9.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/SourceSinkModHome_gr_525.gif],   for convenience set  [Graphics:../Images/SourceSinkModHome_gr_526.gif].  

Integrate and get   [Graphics:../Images/SourceSinkModHome_gr_527.gif],  

which will map the flow in the upper half-plane from a source at  [Graphics:../Images/SourceSinkModHome_gr_528.gif]  onto the flow in a right-angled channel with  [Graphics:../Images/SourceSinkModHome_gr_529.gif], as shown in Figure 11.107.

Solution.   The details are given in the solution of Example 11.28 in Section 11.9.  

Using the Schwarz-Christoffel transformation method, if we choose  [Graphics:../Images/SourceSinkModHome_gr_530.gif]  and  [Graphics:../Images/SourceSinkModHome_gr_531.gif],  then the formula  

            [Graphics:../Images/SourceSinkModHome_gr_532.gif]

will determine a mapping  [Graphics:../Images/SourceSinkModHome_gr_533.gif] of the upper half-plane onto the domain indicated in Figure 11.74 (a).  

With  [Graphics:../Images/SourceSinkModHome_gr_534.gif],  we let [Graphics:../Images/SourceSinkModHome_gr_535.gif],  then  [Graphics:../Images/SourceSinkModHome_gr_536.gif]  and  [Graphics:../Images/SourceSinkModHome_gr_537.gif].  

    The limiting formula for the derivative [Graphics:../Images/SourceSinkModHome_gr_538.gif] becomes   

            [Graphics:../Images/SourceSinkModHome_gr_539.gif]

where  [Graphics:../Images/SourceSinkModHome_gr_540.gif], which will determine a mapping  [Graphics:../Images/SourceSinkModHome_gr_541.gif] from the upper half plane onto the channel as indicated in Figure 11.74 (b) and Figure 11.107.

    Using integrals in Table 11.2 in Section 11.9, we obtain

        [Graphics:../Images/SourceSinkModHome_gr_542.gif]   

    If we use the principal branch of the inverse sine function, then the boundary values  [Graphics:../Images/SourceSinkModHome_gr_543.gif]  lead to the system  

            [Graphics:../Images/SourceSinkModHome_gr_544.gif]   and   [Graphics:../Images/SourceSinkModHome_gr_545.gif],  

which we can solve to obtain  [Graphics:../Images/SourceSinkModHome_gr_546.gif].  Hence the required solution is

            [Graphics:../Images/SourceSinkModHome_gr_547.gif].    

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

 

[Graphics:../Images/SourceSinkModHome_gr_548.gif]

[Graphics:../Images/SourceSinkModHome_gr_549.gif]

Recall.  In the calculus course we used trigonometric substitutions to integrate integrals like [Graphics:../Images/SourceSinkModHome_gr_550.gif].

In particular, in calculus we introduced the trigonometric identities like  

                    [Graphics:../Images/SourceSinkModHome_gr_551.gif].

Hence, we can use this substitution in  [Graphics:../Images/SourceSinkModHome_gr_552.gif]  to obtain

                    [Graphics:../Images/SourceSinkModHome_gr_553.gif].    

 

[Graphics:../Images/SourceSinkModHome_gr_554.gif]

[Graphics:../Images/SourceSinkModHome_gr_555.gif]


[Graphics:../Images/SourceSinkModHome_gr_556.gif]


[Graphics:../Images/SourceSinkModHome_gr_557.gif]

[Graphics:../Images/SourceSinkModHome_gr_558.gif]

[Graphics:../Images/SourceSinkModHome_gr_559.gif]


We are really done.   

 

        We can let Mathematica graph some of the streamlines.

 

                    [Graphics:../Images/SourceSinkModHome_gr_560.gif]

                      The flow in a right-angled channel.

 

We are really really done.   

 

          We can use Mathematica to graph the conformal mapping   [Graphics:../Images/SourceSinkModHome_gr_561.gif].

 

          [Graphics:../Images/SourceSinkModHome_gr_562.gif]          [Graphics:../Images/SourceSinkModHome_gr_563.gif]

                      The conformal mapping   [Graphics:../Images/SourceSinkModHome_gr_564.gif].   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell