Exercise 10.  Find the function  [Graphics:Images/DirichletProblemModHome_gr_420.gif]  that is harmonic in the portion of the upper half-plane  [Graphics:Images/DirichletProblemModHome_gr_421.gif]  

that lies outside the circle   [Graphics:Images/DirichletProblemModHome_gr_422.gif]  and has the boundary values

                    [Graphics:Images/DirichletProblemModHome_gr_423.gif][Graphics:Images/DirichletProblemModHome_gr_424.gif]

Solution 10.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/DirichletProblemModHome_gr_425.gif].  

Solution.   Use the result of Example 11.9,  and the function  

                    [Graphics:../Images/DirichletProblemModHome_gr_426.gif],    which has the boundary values

                    [Graphics:../Images/DirichletProblemModHome_gr_427.gif]  
                    
                    [Graphics:../Images/DirichletProblemModHome_gr_428.gif]

Then it is easy to see that  

                    [Graphics:../Images/DirichletProblemModHome_gr_429.gif]    has the boundary values

                    [Graphics:../Images/DirichletProblemModHome_gr_430.gif]  
                    
                    [Graphics:../Images/DirichletProblemModHome_gr_431.gif]

        Apply the mapping  [Graphics:../Images/DirichletProblemModHome_gr_432.gif]  and find the image the region   [Graphics:../Images/DirichletProblemModHome_gr_433.gif].    

                     [Graphics:../Images/DirichletProblemModHome_gr_434.gif]          [Graphics:../Images/DirichletProblemModHome_gr_435.gif]

                      The image the region   [Graphics:../Images/DirichletProblemModHome_gr_436.gif]   under the mapping   [Graphics:../Images/DirichletProblemModHome_gr_437.gif]

                    is the upper half disk   [Graphics:../Images/DirichletProblemModHome_gr_438.gif].  

Observe that the upper semi-circle  [Graphics:../Images/DirichletProblemModHome_gr_439.gif]  is mapped onto itself.  This can be seen by considering the curve

                    [Graphics:../Images/DirichletProblemModHome_gr_440.gif]   for    [Graphics:../Images/DirichletProblemModHome_gr_441.gif]

which parameterizes the upper semi-circle starting at  [Graphics:../Images/DirichletProblemModHome_gr_442.gif],  passing through   [Graphics:../Images/DirichletProblemModHome_gr_443.gif],   and ending up at  [Graphics:../Images/DirichletProblemModHome_gr_444.gif].     

Furthermore, the ray  [Graphics:../Images/DirichletProblemModHome_gr_445.gif]  is mapped onto the segment    [Graphics:../Images/DirichletProblemModHome_gr_446.gif],

and the ray  [Graphics:../Images/DirichletProblemModHome_gr_447.gif]  is mapped onto the segment    [Graphics:../Images/DirichletProblemModHome_gr_448.gif].  

        Substituting  [Graphics:../Images/DirichletProblemModHome_gr_449.gif]   in   [Graphics:../Images/DirichletProblemModHome_gr_450.gif]  produces the desired function.  

Therefore,  [Graphics:../Images/DirichletProblemModHome_gr_451.gif]  has the boundary values

                    [Graphics:../Images/DirichletProblemModHome_gr_452.gif]

We can manipulate the quantity [Graphics:../Images/DirichletProblemModHome_gr_453.gif] as follows:

                     [Graphics:../Images/DirichletProblemModHome_gr_454.gif]   

Therefore,    

                    [Graphics:../Images/DirichletProblemModHome_gr_455.gif].  

Use the trigonometric identity  [Graphics:../Images/DirichletProblemModHome_gr_456.gif]  and write

                    [Graphics:../Images/DirichletProblemModHome_gr_457.gif]  

Therefore,    

                    [Graphics:../Images/DirichletProblemModHome_gr_458.gif].  

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/DirichletProblemModHome_gr_459.gif]

[Graphics:../Images/DirichletProblemModHome_gr_460.gif]


[Graphics:../Images/DirichletProblemModHome_gr_461.gif]

[Graphics:../Images/DirichletProblemModHome_gr_462.gif]

We are really done.   

 

Aside.  For illustration purposes we can graph the function   [Graphics:../Images/DirichletProblemModHome_gr_463.gif].   

                     [Graphics:../Images/DirichletProblemModHome_gr_464.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_465.gif]

                     where   [Graphics:../Images/DirichletProblemModHome_gr_466.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_467.gif]

 

We are really really done.   

 

                     [Graphics:../Images/DirichletProblemModHome_gr_468.gif]

                     A contour graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_469.gif]

                     where   [Graphics:../Images/DirichletProblemModHome_gr_470.gif]   for   [Graphics:../Images/DirichletProblemModHome_gr_471.gif].  

 

                     [Graphics:../Images/DirichletProblemModHome_gr_472.gif]

                    A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_473.gif].   

                    [Graphics:../Images/DirichletProblemModHome_gr_474.gif]

                     [Graphics:../Images/DirichletProblemModHome_gr_475.gif]

                    A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_476.gif].   

                    [Graphics:../Images/DirichletProblemModHome_gr_477.gif]

                     [Graphics:../Images/DirichletProblemModHome_gr_478.gif]

                    A graph of the function   [Graphics:../Images/DirichletProblemModHome_gr_479.gif].  

                    [Graphics:../Images/DirichletProblemModHome_gr_480.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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