Exercise 8.  Find the function  [Graphics:Images/DirichletProblemModHome_gr_295.gif]  that is harmonic in the unit disk  [Graphics:Images/DirichletProblemModHome_gr_296.gif]  and has the boundary values  

                     [Graphics:Images/DirichletProblemModHome_gr_297.gif][Graphics:Images/DirichletProblemModHome_gr_298.gif]

Solution 8.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/DirichletProblemModHome_gr_299.gif].  

Alternative Answer.   [Graphics:../Images/DirichletProblemModHome_gr_300.gif].  

Solution.   Use the result of Example 11.8 and the function [Graphics:../Images/DirichletProblemModHome_gr_301.gif],  where

                    [Graphics:../Images/DirichletProblemModHome_gr_302.gif]
    
                    [Graphics:../Images/DirichletProblemModHome_gr_303.gif]
    
Then it is easy to see that  [Graphics:../Images/DirichletProblemModHome_gr_304.gif].  

Therefore,   

                    [Graphics:../Images/DirichletProblemModHome_gr_305.gif].  

A More Detailed Solution.   Example 10.3 in Section 10.2 showed that the transformation  [Graphics:../Images/DirichletProblemModHome_gr_306.gif]

maps the unit disk [Graphics:../Images/DirichletProblemModHome_gr_307.gif] onto the upper half-plane [Graphics:../Images/DirichletProblemModHome_gr_308.gif].   Furthermore,

the upper semi-circle [Graphics:../Images/DirichletProblemModHome_gr_309.gif] is mapped onto the positive u-axis, i.e.  [Graphics:../Images/DirichletProblemModHome_gr_310.gif],

and the lower semi-circle [Graphics:../Images/DirichletProblemModHome_gr_311.gif] is mapped onto the negative u-axis, i.e.  [Graphics:../Images/DirichletProblemModHome_gr_312.gif].  

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

                    [Graphics:../Images/DirichletProblemModHome_gr_313.gif],   for  [Graphics:../Images/DirichletProblemModHome_gr_314.gif],   and      
            
                    [Graphics:../Images/DirichletProblemModHome_gr_315.gif],   for  [Graphics:../Images/DirichletProblemModHome_gr_316.gif].        

Apply Theorem 11.2 to construct a Dirichlet solution in the upper half-plane

use formula (11-5)   [Graphics:../Images/DirichletProblemModHome_gr_317.gif]   with   [Graphics:../Images/DirichletProblemModHome_gr_318.gif],  

                    [Graphics:../Images/DirichletProblemModHome_gr_319.gif].  

Now we substitute   [Graphics:../Images/DirichletProblemModHome_gr_320.gif]   and   [Graphics:../Images/DirichletProblemModHome_gr_321.gif]   to obtain  

                    [Graphics:../Images/DirichletProblemModHome_gr_322.gif].  

        Since   [Graphics:../Images/DirichletProblemModHome_gr_323.gif],   the solution  [Graphics:../Images/DirichletProblemModHome_gr_324.gif]  in the z-plane will be

                    [Graphics:../Images/DirichletProblemModHome_gr_325.gif].  

Therefore,   

                    [Graphics:../Images/DirichletProblemModHome_gr_326.gif].  

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

Enter the boundary values and construct the Dirichlet form of the solution.

[Graphics:../Images/DirichletProblemModHome_gr_327.gif]


[Graphics:../Images/DirichletProblemModHome_gr_328.gif]

[Graphics:../Images/DirichletProblemModHome_gr_329.gif]

[Graphics:../Images/DirichletProblemModHome_gr_330.gif]

We are really done.   

 

Remark.   For some situations it might be useful to use the trigonometric identity  [Graphics:../Images/DirichletProblemModHome_gr_331.gif].

This gives

                    [Graphics:../Images/DirichletProblemModHome_gr_332.gif]

which is a version that plots the boundary values better.  

Therefore,   

                    [Graphics:../Images/DirichletProblemModHome_gr_333.gif].  

 

We are really really done.   

 

Aside.  For illustration purposes we can graph the function   [Graphics:../Images/DirichletProblemModHome_gr_334.gif].   

                     [Graphics:../Images/DirichletProblemModHome_gr_335.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_336.gif]

                     where   [Graphics:../Images/DirichletProblemModHome_gr_337.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_338.gif].  

 

We are really really really done.   

 

                     [Graphics:../Images/DirichletProblemModHome_gr_339.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_340.gif]

                     where   [Graphics:../Images/DirichletProblemModHome_gr_341.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_342.gif].  

                     [Graphics:../Images/DirichletProblemModHome_gr_343.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_344.gif]

                     where   [Graphics:../Images/DirichletProblemModHome_gr_345.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_346.gif].  

 

                     [Graphics:../Images/DirichletProblemModHome_gr_347.gif]

                      A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_348.gif],  

                     [Graphics:../Images/DirichletProblemModHome_gr_349.gif]  

                     [Graphics:../Images/DirichletProblemModHome_gr_350.gif]

                      A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_351.gif],  

                     [Graphics:../Images/DirichletProblemModHome_gr_352.gif]  

                     [Graphics:../Images/DirichletProblemModHome_gr_353.gif]

                      A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_354.gif],  

                     [Graphics:../Images/DirichletProblemModHome_gr_355.gif]  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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