Exercise 5.   Find the electrostatic potential  [Graphics:Images/ElectrostaticsModHome_gr_209.gif]  in the domain  D  in the half-plane   [Graphics:Images/ElectrostaticsModHome_gr_210.gif],  

that lies to the left of the hyperbola   [Graphics:Images/ElectrostaticsModHome_gr_211.gif],   

and satisfies the following boundary values,  (shown in Figure 11.43)  

                    [Graphics:Images/ElectrostaticsModHome_gr_212.gif]  

Solution 5.

See text and/or instructor's solution manual.

Answer.   Map the given region onto a vertical strip with the mapping  [Graphics:../Images/ElectrostaticsModHome_gr_213.gif],  

then construct  [Graphics:../Images/ElectrostaticsModHome_gr_214.gif].   

Solution.   The transformation  [Graphics:../Images/ElectrostaticsModHome_gr_215.gif]  maps the domain  D  in the half-plane   [Graphics:../Images/ElectrostaticsModHome_gr_216.gif],  

that lies to the left of the hyperbola   [Graphics:../Images/ElectrostaticsModHome_gr_217.gif]   onto the infinite strip   [Graphics:../Images/ElectrostaticsModHome_gr_218.gif].

                     [Graphics:../Images/ElectrostaticsModHome_gr_219.gif]          [Graphics:../Images/ElectrostaticsModHome_gr_220.gif]

                      The mapping   [Graphics:../Images/ElectrostaticsModHome_gr_221.gif].  

 

Now construct the intermediate solution  [Graphics:../Images/ElectrostaticsModHome_gr_222.gif]  in the infinite strip in the w-plane that has the boundary values

                    [Graphics:../Images/ElectrostaticsModHome_gr_223.gif]  

Applying the method in Example 11.1 in Section 11.1, the solution takes on constant values along the vertical lines and has the form  

                    [Graphics:../Images/ElectrostaticsModHome_gr_224.gif]  

Substitute  [Graphics:../Images/ElectrostaticsModHome_gr_225.gif]  and obtain the intermediate solution

                    [Graphics:../Images/ElectrostaticsModHome_gr_226.gif]   

Now use   [Graphics:../Images/ElectrostaticsModHome_gr_227.gif]   and   [Graphics:../Images/ElectrostaticsModHome_gr_228.gif]   and construct   [Graphics:../Images/ElectrostaticsModHome_gr_229.gif].  

Therefore,  

                    [Graphics:../Images/ElectrostaticsModHome_gr_230.gif].   

 

We are done.   

 

        For computational purposes we can use the formulas for the real and imaginary parts of  [Graphics:../Images/ElectrostaticsModHome_gr_231.gif],  that were derived in Section 10.4.

                    [Graphics:../Images/ElectrostaticsModHome_gr_232.gif].  
                    
In particular,

(10-26)          [Graphics:../Images/ElectrostaticsModHome_gr_233.gif].  

Therefore,  

                    [Graphics:../Images/ElectrostaticsModHome_gr_234.gif].   

 

We are really done.   

 

Aside.  For illustration purposes we can graph the function   [Graphics:../Images/ElectrostaticsModHome_gr_235.gif].   

                     [Graphics:../Images/ElectrostaticsModHome_gr_236.gif]

                     A contour graph of the function   [Graphics:../Images/ElectrostaticsModHome_gr_237.gif]

                     where   [Graphics:../Images/ElectrostaticsModHome_gr_238.gif]   for   [Graphics:../Images/ElectrostaticsModHome_gr_239.gif].  

 

We are really really done.   

 

                     [Graphics:../Images/ElectrostaticsModHome_gr_240.gif]

                     A graph of the function   [Graphics:../Images/ElectrostaticsModHome_gr_241.gif],   

                    [Graphics:../Images/ElectrostaticsModHome_gr_242.gif]  

                     [Graphics:../Images/ElectrostaticsModHome_gr_243.gif]

                     A graph of the function   [Graphics:../Images/ElectrostaticsModHome_gr_244.gif],   

                    [Graphics:../Images/ElectrostaticsModHome_gr_245.gif]  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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