Solution 5 (b).

See text and/or instructor's solution manual.

Answer.  First verify that    satisfies Laplace's Equation.  Then construct  .  

Solution.  First verify that    is a harmonic function.  

                    ,   and  ,  

                    ,   and  ,  

                    ,  

which holds for all z.  Hence    is harmonic for all values of  (x,y).

Second, we follow the construction process of Theorem 3.9.  The first partial derivatives of    are

                        and   .  

To verify that    is harmonic, we compute the second partial derivatives and note that  

                    

so    satisfies Laplace's Equation.  

To construct  ,  we first use the Cauchy-Riemann equation    and integration with respect to y to get:

                      

Differentiating the left and right sides of this equation with respect to x and using and the Cauchy-Riemann    on the left side yields  

                      

Then an easy integration yields  

                      

where c is a constant.   Now we substitute into the previous expression for    and obtain the required solution

                      

Therefore the harmonic conjugate of     is    .

We are done.   

Aside.  We can let Mathematica double check our work.

Remark.  The function    is analytic.

In Section 5.4 we study the hyperbolic functions and learn about the identity  .  

Then we will be able to express  f(z)  in the form  .  

We are really done.   

Aside.  We can use the Milne-Thomson method (see Exercise 17) to construct the harmonic function  .

The function    is analytic and can be constructed as follows:

                       

Notice that the constant   is involved in this construction.  It is o.k. to use the analytic function

                   .  

Therefore  ,   or  

                  .  

Remarks.  In calculus we studied the hyperbolic identity  .  

In   Section 5.4  we will discover that there are relationships between the trigonometric and hyperbolic functions.  

In Exercise 5 (e) in  Section 5.4 we will ask you to establish the trigonometric identity  ,  

and this identity in turn can be used to show that  .  

We are really really done.   

Aside.  We can let Mathematica double check our work.

          The level curves    and  .  

          The orthogonal grid formed with    and  .    

We are really really really done.   

                    In Section 11.4 we will prove that the image of an orthogonal grid under an analytic function is an orthogonal grid.  

          It is best to worry about these concepts when we get there because for this example involves the inverse transformation  .

          The orthogonal grid formed by the composite image of several rectangular grids like the one on the left,

           under the multivalued inverse functions  .  

          The spacing between curves is not the same as in the previous figures because lines in the domain grid are equally spaced.

          A portion of the above grid where the mapping is  ,  and the principal value of    is used.    

          Remark.  In Section 5.5 we will introduce the formula for principal value of  .   

This solution is complements of the authors.

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