Exercise 6.  Find the temperature function in the domain  ,  

that satisfies the following boundary values (shown in Figure 11.26).  

                      

Hint.  Use   .

Solution 6.

See text and/or instructor's solution manual.

Answer.   .  

A Short Solution.   This is almost identical to the boundary value problem given in Exercise 10 in Section 11.2 where

                    ,     
                    
                    

For this exercise, the nonzero boundary value on the semi-circle is      instead of   .  

The solution is   .  

Therefore,

                    .    

We are done.   

Aside.  For illustration purposes we can graph the function   .   

                     A contour graph of the function   

                     where      for   .  

                      A contour graph of the function   

                     where      for   .  

                    A graph of the function   ,   

                    

                    A graph of the function   ,   

                    

                    A graph of the function   ,  

                    

A More Detailed Solution.   Recall the solution for Exercise 10 in Section 11.2.     

Use the result of Example 11.9,  and the function  

                    ,    which has the boundary values

                      
                    
                    

Then construct  

                      

Then

                        has the boundary values

                      
                    
                    

        Apply the mapping    and find the image the region   .    

        The image of the region    under the mapping    is the half-disk  .

Observe that the upper semi-circle    is mapped onto itself.  This can be seen by considering the curve

                       for    

which parameterizes the upper semi-circle starting at  ,  passing through   ,   and ending up at  .     

Furthermore, the ray    is mapped onto the segment    ,

and the ray    is mapped onto the segment    .  

        Substituting     in     produces the desired function

Therefore,    has the boundary values

                    

We can manipulate the quantity as follows:

                       

Therefore,    

                    .  

Use the trigonometric identity    and write

                      

Therefore,    

                    .  

We are really done.   

Aside.  We can let Mathematica double check our work.

We are really really done.   

An Alternative Solution.   Example 10.3 showed that      maps the disk      onto the upper half plane   ,  

and a careful inspection of Example 10.3 reveals that      maps the infinite region      onto the lower half plane   .  

This can also be determined by using the points   ,  ,  and     and the image points  

,   ,   and      which give the lower half plane a positive orientation.  

Then use the three points   ,    and     which give the upper half plane a positive orientation,

and the image points   ,   ,   and      which give the right half plane a positive orientation.   

Hence the image of the domain      under the mapping      is the fourth quadrant   .  

                      The mapping   .  

                    Observe.  The ends of the semi-circles in the z-plane are perpendicular to the x-axis, and
                    the ends of the image semi-circles in the w-plane are perpendicular to the negative v-axis.

    Hence the image of the domain     under the mapping    is the fourth quadrant where the boundary values are

                        

Applying Example 11.2 in Section 11.1 we know that the form of the solution is  

                    .  

Use the values   ,  and     and write the system of equations  

                    

                    .  

Simplify them and obtain  

                        and   .  

Solving we get      and

                       

Now make the substitution   

Then

                    .   

Therefore,  

                    .  

We are really really really done.   

 Aside.  For illustration purposes we can graph the function    .

                     A contour graph of the function      

                     where      for   .  

                    A graph of the function   ,   

                    

                    A graph of the function   ,     

                    

We are really really really really done.   

        Use the trigonometric identity    and write

                       

Which is the same intermediate step we obtained in the previous method.

Continuing as before use the trigonometric identity    and write

                      

Therefore,    

                    

which is precisely the solution we obtained in the previous method.

We are really really really really really done.   

Aside.   We can graph the intermediate function  

                    A contour graph of the intermediate function  ,

                    where     for   .

                    A graph of the intermediate function  ,  

                        

This solution is complements of the authors.

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