1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

452 CHAPTER 11 • APPLICATIONS OF H ARMONIC F UNCTIONS

T= -0.2

T = -0.4

T = -0.6

T = -1. 0 -1

y

T =0.2

T=0.4

T= 1.0

Figure 11. 21 The temperature T(x,y) with Ty (x,O) = 0, for - 1 < x < 1, a nd

boundary values T (x, 0) = - 1, for x < -1, a nd T (x, 0) = 1, for x > 1.

temperature T* ( u , v) that has the boundary conditions

r(i,v)= l, forv>O, and r·(~7r,·u)=-1, for ·u > O;

8T* -71" 1r

an = r; (u, O) = 0, for Z < u < z·

Using the result of Example 11.1, we can easily obtain the solution:

r· (u, v) = ~u.
1r

Therefore, the solution in D is

2. 
   

   T(x, y) = -Re(Arcsmz).
   1r
   If an explicit solution is required, then we can use Formula (10-26) to obtain

   
   


3. 
   

   V(x+l)
   2
   +y^2 - J(x- 1)
   2
   +y2
   T (x, y) = ;Arcsm
   2

   
   

,

where the function Arcsin t has range values satisfying - 2 " < Arcsin t < ~; see

Figure 11.21.

-------~EXERCISES FOR SECTION 11.5

1

1. Show that H (x, y, z) = satisfies Laplace's equation Hu + Hn +
   y'x2 +y2 + z2
   H,. = 0 in three-dimensional Cartesian space but that h(x,y) = k does
   :z;2+y2
   not satisfy equation h.,., + h.v = 0 in two-dimensional Cartesian space.