1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

11.5 • STEADY STA'l'E TEMPERATURES 447

y y

V(x,y+&y) V(x+lix,y+&y} S(x, Heat y) = fJ

flow

Jines

~

V(x, y) V(x + lix, y)

T(x. y) =a

Jsochermals

(a) The direction of heal flow. (b) Heat flow lines and isothermals.

Figure 11 .16 Steady state temperatures.

Expression (11-19), we find that the amount of heat flowing out the right edge

of the rectangle in Figure 11.16( a) is aJ?proxi.mately

V · N 1 .6.s 1 = -K [T., (x + t::.x, y) + iT 11 (x + t::.x, y)J · (1 + Oi)t::.y

=-KT., (x + t::.x, y) t::.y,

and the amount of heat flowing out the left edge is

V · N2 !::.s2 = - K [T,. (x, y) + iT 11 (x, y)] · (- 1 + Oi) t::.y

= KTx (x, y) t::.y.

(11-20)

(11- 21 )

If we add the contributions in Equations (11-20) and (11-21), the result is

-K [Tx (x + t::.x,;;-T., (x , y)] t::.x t::.y ~ -KTxx (x, y) !::.x t::.y. (11-22)

Similarly, the contribution for the amount of heat flowing out of the top and
bottom edges is

• K[T 11 (x ,y+t::.y)-T 11 (x,y)] A A - -K'T' ( ) A A
  !::.y uy uy - J.yy X , Y <..>X '-"Y· (11-23)

Adding the quantities in Equations (11-22) and (11-23), we find that the net

heat flowing out of the rectangle is approximated by the equation

• K [Txx (x, y) + T 1111 (x, y)J t::.x !::.y = 0,

which implies that T(x,y) satisfies Laplace's equation and is a harmonic func-
tion.
If the domain in which T (x, y) is defined is simply connected, then a conju-
gate harmonic function S (x, y) exists, and

F (z) = T (x, y) + iS (x, y)