1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

11.5 • STEADY STATE TEMPERATURES 449

Solu t io n Since T(x,y) is a harmonic function, this problem is an example of
a Dirichlet problem. From Example 11.2, it follows that the solution is
T2-T 1
T (x, y) = T 1 + Argz.
1r

The isotherms T(x, y) = a are rays emanating from the origin. The conju-

gate harmonic function is S (x, y) = ~ (T 1 -T2) In lzl, and the heat flow lines

S(x, y) = f3 are semicircles centered at the origin. If T 1 > T2, then the heat

flows counterclockwise along the semicircles, as shown in Figure 11.18.

• EXAMPLE 11.16 Find the temperature T (x, y) at each point in t he upper
  half-disk H : Im (z) > 0, lzl < 1 if the temperatures at points on the boundary
  satisfy

T (x, y) = 100,

T(x, 0) = 50 ,

for x + iy = z = ei^8 , 0 < (} < 1rj

for -l<x<l.

Solution As discussed in Example 11 .9, the function

. i (1 - z) 2y. l -x^2 - y^2
u+iv = = +i--~--=--
1 + z (x + 1)^2 + y2 (x + 1)^2 + y2

(11-24)

is a one-to-one conformal mapping of the half-disk H onto the first quadrant

Q : u > 0, v > 0. The conformal map given by Equation (11-24) gives rise to

a new problem of finding the temperature T * (u, v) that satisfies the boundary

conditions

T (u, 0) = 100, for u > 0, and T (0, v) = 50, for v > O.

If we use Example 11.2, the harmonic function T* (u, v) is given by

50 -100 100 v

T* (u, v) = 100 + " Argw = 100 - - Arctan-.

)'

T(x,b) = T 2 for x<O

2 1r u

_T(x.y)=a

isothennals

-s(x,y>=P

beat flow lines

T(x, a)= T 1 for X> 0

(11-25)

Figure 11. 18 The temperature T (x, y) in the upper half-plane where T1 > T2.