Exercise 10.   Use a Schwarz-Christoffel transformation to find a conformal mapping  [Graphics:Images/SourceSinkModHome_gr_565.gif]  that will map the flow

in the upper half-plane onto the flow from a channel back into a quadrant,

as indicated in Figure 11.108,                     Figure 11.108.

where   [Graphics:Images/SourceSinkModHome_gr_566.gif]  is the endpoint of the ray  [Graphics:Images/SourceSinkModHome_gr_567.gif].

Hint.  Set   [Graphics:Images/SourceSinkModHome_gr_568.gif],  [Graphics:Images/SourceSinkModHome_gr_569.gif],  [Graphics:Images/SourceSinkModHome_gr_570.gif],   and  [Graphics:Images/SourceSinkModHome_gr_571.gif],  [Graphics:Images/SourceSinkModHome_gr_572.gif],  [Graphics:Images/SourceSinkModHome_gr_573.gif],   respectively, and let   [Graphics:Images/SourceSinkModHome_gr_574.gif].    

Solution 10.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/SourceSinkModHome_gr_575.gif],   for convenience set  [Graphics:../Images/SourceSinkModHome_gr_576.gif].  
Integrate and get   [Graphics:../Images/SourceSinkModHome_gr_577.gif],  

which will map the flow in the upper half-plane onto the flow from a channel back into a quadrant, as shown in Figure 11.108,

where    [Graphics:../Images/SourceSinkModHome_gr_578.gif],  [Graphics:../Images/SourceSinkModHome_gr_579.gif],  [Graphics:../Images/SourceSinkModHome_gr_580.gif],   and  [Graphics:../Images/SourceSinkModHome_gr_581.gif],  [Graphics:../Images/SourceSinkModHome_gr_582.gif],  [Graphics:../Images/SourceSinkModHome_gr_583.gif],   respectively, and let   [Graphics:../Images/SourceSinkModHome_gr_584.gif].      

The initial conditions can be used to obtain  [Graphics:../Images/SourceSinkModHome_gr_585.gif] and  [Graphics:../Images/SourceSinkModHome_gr_586.gif],  and the answer is

            [Graphics:../Images/SourceSinkModHome_gr_587.gif].  

The logarithm term could also be written in the form  

            [Graphics:../Images/SourceSinkModHome_gr_588.gif].  

And if the inverse hyperbolic functions are used then this can be written as

            [Graphics:../Images/SourceSinkModHome_gr_589.gif].  

Solution.   The details are given in the solution of Exercise 15 in Section 11.9.  

Along the x-axis use the points   [Graphics:../Images/SourceSinkModHome_gr_590.gif].   

The exterior angles are  [Graphics:../Images/SourceSinkModHome_gr_591.gif],  and the formula for the derivative [Graphics:../Images/SourceSinkModHome_gr_592.gif] is  

            [Graphics:../Images/SourceSinkModHome_gr_593.gif]   

Integrate and get

            [Graphics:../Images/SourceSinkModHome_gr_594.gif].  

The first integral is easy to get

            [Graphics:../Images/SourceSinkModHome_gr_595.gif].

The second integral can be found using the suggested change of variable  

            [Graphics:../Images/SourceSinkModHome_gr_596.gif]
            
            [Graphics:../Images/SourceSinkModHome_gr_597.gif]

Make substitutions in the integral

            [Graphics:../Images/SourceSinkModHome_gr_598.gif]  

Now use the substitution  [Graphics:../Images/SourceSinkModHome_gr_599.gif],  

            [Graphics:../Images/SourceSinkModHome_gr_600.gif].

Now combine this with the first integral and get  

            [Graphics:../Images/SourceSinkModHome_gr_601.gif]  

            [Graphics:../Images/SourceSinkModHome_gr_602.gif]
        
            [Graphics:../Images/SourceSinkModHome_gr_603.gif]  

The initial conditions can be used to obtain  [Graphics:../Images/SourceSinkModHome_gr_604.gif] and  [Graphics:../Images/SourceSinkModHome_gr_605.gif],  and the answer is

            [Graphics:../Images/SourceSinkModHome_gr_606.gif].  

The logarithm term could also be written in the form  

            [Graphics:../Images/SourceSinkModHome_gr_607.gif].  

And if the inverse hyperbolic functions are used then this can be written as

            [Graphics:../Images/SourceSinkModHome_gr_608.gif].  

Remark.   If the computer algebra Mathematica is used to perform the integration then the answer is

            [Graphics:../Images/SourceSinkModHome_gr_609.gif].  

If the computer algebra Maple is used to perform the integration then the answer is

            [Graphics:../Images/SourceSinkModHome_gr_610.gif].  
            

Or if the second integral is treated separately, then Maple's answer will be  

            [Graphics:../Images/SourceSinkModHome_gr_611.gif].  

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

 

[Graphics:../Images/SourceSinkModHome_gr_612.gif]


[Graphics:../Images/SourceSinkModHome_gr_613.gif]

[Graphics:../Images/SourceSinkModHome_gr_614.gif]
[Graphics:../Images/SourceSinkModHome_gr_615.gif]


[Graphics:../Images/SourceSinkModHome_gr_616.gif]

[Graphics:../Images/SourceSinkModHome_gr_617.gif]


[Graphics:../Images/SourceSinkModHome_gr_618.gif]
[Graphics:../Images/SourceSinkModHome_gr_619.gif]

[Graphics:../Images/SourceSinkModHome_gr_620.gif]


We are really done.   

 

        We can let Mathematica draw a graph of the streamlines.

 

                    [Graphics:../Images/SourceSinkModHome_gr_621.gif]

                      The flow from a channel  [Graphics:../Images/SourceSinkModHome_gr_622.gif]  into a quadrant.           

 

We are really really done.   

 

        We can use Mathematica to graph the conformal mapping   [Graphics:../Images/SourceSinkModHome_gr_623.gif].  

 

          [Graphics:../Images/SourceSinkModHome_gr_624.gif]          [Graphics:../Images/SourceSinkModHome_gr_625.gif]

                                The conformal mapping   [Graphics:../Images/SourceSinkModHome_gr_626.gif].   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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