Solution 5 (c).

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 extend the construction process of Theorem 3.9 to the case when   is the given function.   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

                      

The harmonic conjugate of      is    .

We are done.   

Aside.  We can let Mathematica double check our work.

Remark. In Section 5.1 we will learn that    is the complex exponential function.  

This is not the function at hand.  However, after carefully pondering the situation we can see that our solution involves the function

                    .  

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 identities like  .  

In   Section 5.4  we will discover that there are relationships between the trigonometric and exponential functions,  like  

                    ,     

which will imply 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 this example involves the inverse transformation  .

          The orthogonal grid formed by the image of a rectangular grid under the multivalued inverse function  .  

          Remark.  In Section 5.2 we will introduce the formulas for the complex logarithm function.   

This solution is complements of the authors.

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