Exercise 12.  Find the function    that is harmonic in the unit disk    and has the boundary values

Solution 12.

See text and/or instructor's solution manual.

Answer.   .  

Solution.   Apply the mapping      which is a rotation of  the unit disk  .

Here we have   ,    and      and   .  

Then the arc      is mapped onto the arc  ,

and the arc      is mapped onto the arc  .  

        This makes a new boundary value problem in the image unit disk  .  

Use the result of Example 11.8 and the function ,  where

                      
    
                    

        Substituting     in   produces the solution     which has the boundary values

                      
    
                    

We can manipulate the quantity as follows:

                      

Therefore,    

                    .  

We are done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

Aside.  For illustration purposes we can graph the function   .   

                     A contour graph of the function   

                     where      for   .  

We are really really done.   

                     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      

                    

We are really really really done.   

Extra Details.   Regarding the boundary value problem in the unit disk  ,   

                      
    
                    

Example 10.3 in Section 10.2 showed that the transformation    

maps the unit disk    onto the upper half-plane .   Furthermore,

the upper semi-circle is mapped onto the positive u-axis, i.e.  ,

and the lower semi-circle    is mapped onto the negative u-axis, i.e.  .  

This makes a new boundary value problem in the w-plane   

                    ,   for  ,   and      
            
                    ,   for  .        

And the solution is simply  .

Therefore,   

                       has the boundary values

                      
    
                    

This solution is complements of the authors.

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