Exercise 9.   Consider the complex potential   .  

9 (a).   Show that    determines the ideal fluid flow through the aperture from     to  ,   (as shown in Figure 11.59).  

9 (b).   Show that the streamline      for the flow is a portion of the hyperbola   

                    .  

Solution 9.

See text and/or instructor's solution manual.

      The transformation      maps the plane slit along the two segments  

                        and    

onto the vertical strip  .     

Hence   

                       

maps the plane slit along the two segments  

                        and    

onto the horizontal strip  .     

The stream function is  

                    .

(The flow      would be in the opposite direction).

We are done.   

Aside.  We can make a plot of the stream function.  For illustration purposes, we choose A = 1.

                    Flow through a slit.  

                    The stream function is   .

                    The contour graph   ,   for   .  

We are really done.   

      The inverse of      is   

                    .  

                      The conformal mapping   .

We are really done.   

Remark.  The mapping      might not be as familiar as the standard trigonometric functions.   

The above graph in the z-plane will look similar to the a graph obtained using    .

However, the streamlines for      would be the images of the vertical lines  

and not the horizontal lines   .   

The relationship of these two mapping is discovered by writing out the real and imaginary parts:

                      

                      

Aside.   We can let Mathematica double check our work.

Aside.   We can explore the graph of    .

                    The conformal mapping   .

                    Caveat.  This is not the usual construction using the stream function.

This solution is complements of the authors.

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