1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

450 CHAPTER 11 • APPLI CATIONS OF HARMONIC FUNCTIONS

y

• I T(x,0)=50 for - I <x< I

Figure 11.19 The temperature T(x,y} in a half-disk.

Substituting the expressions for u and v from Equation (11-24) into Equation

(11-25) yields the desired solution:

100 1-x^2 - y^2

T (x, y) = 100 - -Arctan

1r 2 y

The isothermals T (x, y) = constant are circles that pass through the points ±1,

as shown in Figure 11.19.

11.5. 1 An Insulated Segment on the Boundary

We now turn to the problem of finding the steady state temperature function
T ( x, y) inside the simply connected domain D whose boundary consists of three
adjacent curves C 1 , C2, and C 3 , where T(x,y) = T 1 along C 1 , T(x,y) = T 2
along C2, and the region is insulated along C 3. Zero heat flowing across C 3
implies that

V (x , y)-N (x, y) =-KN (x, y) grad T(x, y) = 0,

where N (x, y) is perpendicular to C3. Thus the direction of heat flow must

be parallel to this portion of the boundary. In other words, C3 must be part

of a heat flow line S (x, y) = constant and the isothermals T (x, y) = constant

intersect C3 orthogonally.

We can solve this problem by finding a conformal mapping

w = f (z) = u (x, y) +iv (x , y) (11-26)

from D onto the semi-infinite strip G : 0 < u < 1, v > 0 so that the image of

the curve C, is the ray u = O; the image of the curve C2 is the ray given by

u = 1, v > O; and the thermally insulated curve C3 is mapped onto the thermally

insulated segment 0 < u < 1 of the u-axis, as shown in F igure 11.20.

The new problem in G is to find the steady state temperature function

r· (u,v) so that along the rays, we have the boundary values

r• (0, v) = T,, for v > 0, and for v > 0. (11- 27)