Exercise 2.  Find the function  [Graphics:Images/DirichletProblemModHome_gr_38.gif]  that is harmonic in the sector   [Graphics:Images/DirichletProblemModHome_gr_39.gif]   and has the boundary values  

                    [Graphics:Images/DirichletProblemModHome_gr_40.gif]     [Graphics:Images/DirichletProblemModHome_gr_41.gif]

Solution 2.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/DirichletProblemModHome_gr_42.gif].   

Solution.   Applying Example 11.2 we know that the form of the solution is  

                    [Graphics:../Images/DirichletProblemModHome_gr_43.gif].  

Use the values   [Graphics:../Images/DirichletProblemModHome_gr_44.gif],   and   [Graphics:../Images/DirichletProblemModHome_gr_45.gif]   and write the system of equations  

                    [Graphics:../Images/DirichletProblemModHome_gr_46.gif]

                    [Graphics:../Images/DirichletProblemModHome_gr_47.gif].  

Simplify them and obtain  

                     [Graphics:../Images/DirichletProblemModHome_gr_48.gif],    and    [Graphics:../Images/DirichletProblemModHome_gr_49.gif].  

Solving we get   [Graphics:../Images/DirichletProblemModHome_gr_50.gif]   and the desired solution

                    [Graphics:../Images/DirichletProblemModHome_gr_51.gif]   

Therefore,   

                    [Graphics:../Images/DirichletProblemModHome_gr_52.gif].   

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/DirichletProblemModHome_gr_53.gif]

[Graphics:../Images/DirichletProblemModHome_gr_54.gif]


[Graphics:../Images/DirichletProblemModHome_gr_55.gif]

[Graphics:../Images/DirichletProblemModHome_gr_56.gif]

Aside.   If you prefer to use the function  Arg,  then the solution can be written in the form   

                    [Graphics:../Images/DirichletProblemModHome_gr_57.gif].   

Aside.   If polar coordinates   [Graphics:../Images/DirichletProblemModHome_gr_58.gif]   are used, then the polar form of the solution is     

                    [Graphics:../Images/DirichletProblemModHome_gr_59.gif].   

[Graphics:../Images/DirichletProblemModHome_gr_60.gif]

[Graphics:../Images/DirichletProblemModHome_gr_61.gif]

We are really done.   

 

Aside.  For illustration purposes we can graph the function   [Graphics:../Images/DirichletProblemModHome_gr_62.gif].   

                     [Graphics:../Images/DirichletProblemModHome_gr_63.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_64.gif]

                     where   [Graphics:../Images/DirichletProblemModHome_gr_65.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_66.gif].  

 

We are really really done.   

 

                     [Graphics:../Images/DirichletProblemModHome_gr_67.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_68.gif]

                     where  [Graphics:../Images/DirichletProblemModHome_gr_69.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_70.gif].  

 

                     [Graphics:../Images/DirichletProblemModHome_gr_71.gif]

                    A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_72.gif]  

                    [Graphics:../Images/DirichletProblemModHome_gr_73.gif]

                     [Graphics:../Images/DirichletProblemModHome_gr_74.gif]

                    A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_75.gif]  

                    [Graphics:../Images/DirichletProblemModHome_gr_76.gif]

                     [Graphics:../Images/DirichletProblemModHome_gr_77.gif]

                    A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_78.gif]   

                    [Graphics:../Images/DirichletProblemModHome_gr_79.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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