Solution 5.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/FluidFlowImageMod_gr_189.gif].   

Solution.   Along the x-axis use the points  [Graphics:../Images/FluidFlowImageMod_gr_190.gif],  [Graphics:../Images/FluidFlowImageMod_gr_191.gif],   and in the w-plane use   [Graphics:../Images/FluidFlowImageMod_gr_192.gif],  [Graphics:../Images/FluidFlowImageMod_gr_193.gif],   respectively.  

The exterior angles are   [Graphics:../Images/FluidFlowImageMod_gr_194.gif],  

and the formula for the derivative [Graphics:../Images/FluidFlowImageMod_gr_195.gif] is  given by the Schwarz-Christoffel formula  

            [Graphics:../Images/FluidFlowImageMod_gr_196.gif]   

Now integrate
            [Graphics:../Images/FluidFlowImageMod_gr_197.gif]

The integral can be found using the suggested change of variable  

            [Graphics:../Images/FluidFlowImageMod_gr_198.gif]
            
            [Graphics:../Images/FluidFlowImageMod_gr_199.gif]

Make substitutions in the integral

            [Graphics:../Images/FluidFlowImageMod_gr_200.gif]  

Now use the substitution  [Graphics:../Images/FluidFlowImageMod_gr_201.gif],  

            [Graphics:../Images/FluidFlowImageMod_gr_202.gif]

            [Graphics:../Images/FluidFlowImageMod_gr_203.gif]

            [Graphics:../Images/FluidFlowImageMod_gr_204.gif]

Solve for  [Graphics:../Images/FluidFlowImageMod_gr_205.gif],  and  [Graphics:../Images/FluidFlowImageMod_gr_206.gif],  and get the following solution

            [Graphics:../Images/FluidFlowImageMod_gr_207.gif].  

The first logarithm term can be rewritten in the following forms  

            [Graphics:../Images/FluidFlowImageMod_gr_208.gif],
            
            [Graphics:../Images/FluidFlowImageMod_gr_209.gif].         

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/FluidFlowImageMod_gr_210.gif]

[Graphics:../Images/FluidFlowImageMod_gr_211.gif]


[Graphics:../Images/FluidFlowImageMod_gr_212.gif]

[Graphics:../Images/FluidFlowImageMod_gr_213.gif]


We are really done.   

 

        We can let Mathematica graph some of the streamlines.

        The flow up an inclined ramp, as shown in Figure 11.92(a).   

 

                      [Graphics:../Images/FluidFlowImageMod_gr_214.gif]          [Graphics:../Images/FluidFlowImageMod_gr_215.gif]

                    The streamlines in the w-plane given by the parametric equation

                    [Graphics:../Images/FluidFlowImageMod_gr_216.gif],    for    [Graphics:../Images/FluidFlowImageMod_gr_217.gif].  

 

We are really really done.   

 

Aside.  We can extend the flow into the third and fourth quadrants using symmetry.

        The flow around a pointed object, as shown in Figure 11.92(b).   

 

          [Graphics:../Images/FluidFlowImageMod_gr_218.gif]          [Graphics:../Images/FluidFlowImageMod_gr_219.gif]

                    The streamlines in the w-plane given by the parametric equation

                    [Graphics:../Images/FluidFlowImageMod_gr_220.gif],    for    [Graphics:../Images/FluidFlowImageMod_gr_221.gif].  

 

We are really really really done.   

 

Aside.  We can use Mathematica to graph   

                    [Graphics:../Images/FluidFlowImageMod_gr_222.gif].  

 

          [Graphics:../Images/FluidFlowImageMod_gr_223.gif]          [Graphics:../Images/FluidFlowImageMod_gr_224.gif]

                    The image of the upper half plane  [Graphics:../Images/FluidFlowImageMod_gr_225.gif]  under the mapping   

                    [Graphics:../Images/FluidFlowImageMod_gr_226.gif].  

 

We are really really really really done.   

 

Aside.  We can extend the mapping into the third and fourth quadrants using symmetry.

 

          [Graphics:../Images/FluidFlowImageMod_gr_227.gif]          [Graphics:../Images/FluidFlowImageMod_gr_228.gif]

                    The image of the complex plane under the mapping   

                    [Graphics:../Images/FluidFlowImageMod_gr_229.gif].  

 

Aside.  We can graph the other version involving logarithms

                    [Graphics:../Images/FluidFlowImageMod_gr_230.gif].  

 

          [Graphics:../Images/FluidFlowImageMod_gr_231.gif]          [Graphics:../Images/FluidFlowImageMod_gr_232.gif]

                    The image of the upper half plane  [Graphics:../Images/FluidFlowImageMod_gr_233.gif]  under the mapping   

                    [Graphics:../Images/FluidFlowImageMod_gr_234.gif].  

 

We are really really really really really done.   

 

Remark 1.  If the computer algebra Mathematica is used to perform the integration then the answer is

            [Graphics:../Images/FluidFlowImageMod_gr_235.gif],   

which gives the correct result but uses the a specialized hypergeometric function.

[Graphics:../Images/FluidFlowImageMod_gr_236.gif]

[Graphics:../Images/FluidFlowImageMod_gr_237.gif]

Aside.  Here are two extra graphs using two different mappings:

                    [Graphics:../Images/FluidFlowImageMod_gr_238.gif]   and  

                    [Graphics:../Images/FluidFlowImageMod_gr_239.gif].

 

          [Graphics:../Images/FluidFlowImageMod_gr_240.gif]          [Graphics:../Images/FluidFlowImageMod_gr_241.gif]

                    The image of the upper half plane  [Graphics:../Images/FluidFlowImageMod_gr_242.gif]  under the mapping   

                    [Graphics:../Images/FluidFlowImageMod_gr_243.gif].  

 

          [Graphics:../Images/FluidFlowImageMod_gr_244.gif]          [Graphics:../Images/FluidFlowImageMod_gr_245.gif]

                    The image of the upper half plane  [Graphics:../Images/FluidFlowImageMod_gr_246.gif]  under the mapping   

                    [Graphics:../Images/FluidFlowImageMod_gr_247.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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