Solution 15.

See text and/or instructor's solution manual.

Solution.  The equipotentials   [Graphics:../Images/HarmonicFunctionModHome_gr_909.gif]   can be written as  [Graphics:../Images/HarmonicFunctionModHome_gr_910.gif].  

Then we get  [Graphics:../Images/HarmonicFunctionModHome_gr_911.gif],  and  then  [Graphics:../Images/HarmonicFunctionModHome_gr_912.gif].

Taking anti-logs this we have  [Graphics:../Images/HarmonicFunctionModHome_gr_913.gif] ,  which are

concentric circles centered at the origin with radii  [Graphics:../Images/HarmonicFunctionModHome_gr_914.gif].  

        The streamlines  [Graphics:../Images/HarmonicFunctionModHome_gr_915.gif]  can be written as  [Graphics:../Images/HarmonicFunctionModHome_gr_916.gif].   Using the polar form  [Graphics:../Images/HarmonicFunctionModHome_gr_917.gif],  we see that

this is the equation for rays emanating from the origin making an angle of  [Graphics:../Images/HarmonicFunctionModHome_gr_918.gif]  radians for  [Graphics:../Images/HarmonicFunctionModHome_gr_919.gif].  

We can let Mathematica do our work.

 

                                                            [Graphics:../Images/HarmonicFunctionModHome_gr_920.gif]

  

                    The equipotentials  [Graphics:../Images/HarmonicFunctionModHome_gr_921.gif]  are shown in blue for  [Graphics:../Images/HarmonicFunctionModHome_gr_922.gif].

 

                                                            [Graphics:../Images/HarmonicFunctionModHome_gr_923.gif]

                    The streamlines  [Graphics:../Images/HarmonicFunctionModHome_gr_924.gif]  for  [Graphics:../Images/HarmonicFunctionModHome_gr_925.gif]  are shown in blue for  [Graphics:../Images/HarmonicFunctionModHome_gr_926.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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