Exercise 9.  Find the temperature function [Graphics:Images/TemperaturesModHome_gr_468.gif] in the first quadrant  [Graphics:Images/TemperaturesModHome_gr_469.gif],   

that satisfies the following boundary conditions (shown in Figure 11.29).  

                    [Graphics:Images/TemperaturesModHome_gr_470.gif]  

Solution 9.

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/TemperaturesModHome_gr_471.gif][Graphics:../Images/TemperaturesModHome_gr_472.gif].  

A Short Solution.   Map the quadrant onto the upper half-plane with the function   [Graphics:../Images/TemperaturesModHome_gr_473.gif],  then multiply the boundary values in Example 11.17 by  75

and consider  [Graphics:../Images/TemperaturesModHome_gr_474.gif],   then construct  [Graphics:../Images/TemperaturesModHome_gr_475.gif].  

A Longer Solution.   The transformation  [Graphics:../Images/TemperaturesModHome_gr_476.gif]  maps the first quadrant  [Graphics:../Images/TemperaturesModHome_gr_477.gif],   

onto the upper half plane  [Graphics:../Images/TemperaturesModHome_gr_478.gif]   where the boundary values are

                    [Graphics:../Images/TemperaturesModHome_gr_479.gif]  

Use the result in Example 11.17 where we found that the function   

                    [Graphics:../Images/TemperaturesModHome_gr_480.gif]    has the boundary values  

                    [Graphics:../Images/TemperaturesModHome_gr_481.gif]  

We can easily obtain [Graphics:../Images/TemperaturesModHome_gr_482.gif] by multiplying  [Graphics:../Images/TemperaturesModHome_gr_483.gif]  by   75  and adding  25,

                    [Graphics:../Images/TemperaturesModHome_gr_484.gif].     

Now construct  

                    [Graphics:../Images/TemperaturesModHome_gr_485.gif]   

Therefore,

                    [Graphics:../Images/TemperaturesModHome_gr_486.gif].     

 

We are done.   

 

For computational purposes we can use the formulas for the real and imaginary parts of  [Graphics:../Images/TemperaturesModHome_gr_487.gif],  that were derived in Section 10.4.

                    [Graphics:../Images/TemperaturesModHome_gr_488.gif].  
                    
In particular,

(10-26)          [Graphics:../Images/TemperaturesModHome_gr_489.gif].  

If an explicit solution is required, then we can use Formula (10-26) in Section 10.4 and write  

                    [Graphics:../Images/TemperaturesModHome_gr_490.gif],    

where the function  [Graphics:../Images/TemperaturesModHome_gr_491.gif]  has range values satisfying   [Graphics:../Images/TemperaturesModHome_gr_492.gif].  

Now make the substitution   [Graphics:../Images/TemperaturesModHome_gr_493.gif]   and  

                    [Graphics:../Images/TemperaturesModHome_gr_494.gif]    and    [Graphics:../Images/TemperaturesModHome_gr_495.gif].

Thus,

                    [Graphics:../Images/TemperaturesModHome_gr_496.gif]      
                    
Therefore,

                    [Graphics:../Images/TemperaturesModHome_gr_497.gif].     

 

We are really done.   

 

Aside.   We can graph the function   [Graphics:../Images/TemperaturesModHome_gr_498.gif].   

                     [Graphics:../Images/TemperaturesModHome_gr_499.gif]

                    A contour graph of the function  [Graphics:../Images/TemperaturesModHome_gr_500.gif],

                    where  [Graphics:../Images/TemperaturesModHome_gr_501.gif]   for   [Graphics:../Images/TemperaturesModHome_gr_502.gif].

 

                     [Graphics:../Images/TemperaturesModHome_gr_503.gif]

                    A graph of the function  [Graphics:../Images/TemperaturesModHome_gr_504.gif],    

                    [Graphics:../Images/TemperaturesModHome_gr_505.gif]  

 

















 

This solution is complements of the authors.

 



































 

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