1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

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11.5 • STEADY STATE TEMPERATURES 451

w =f(zl

0

v

ar•_ 0

on -

Figure 11 .20 Steady state temperatures with one boundary portion insulated.

The condition that a segment of the boundary is insulated can be expressed
mathematically by saying that the normal derivative of T* ( u, v ) is zero. That
is)

&T·

{}n = T: (u, 0) = 0, (11-28)

where n is a coordinate measured perpendicularly to the segment. We can easily

verify that t he function

satisfies the conditions stated in Equations (11-27) and (11-28) for region G.
Therefore, using Equation (11-26), we find that t he solution in D is

T(x, y) =Ti+ (T2 -Ti)u(x,y).

The isot hermals T (x, y) = constant and their images under w = f (z) are also

illustrated in Figure 11.20.

• EXAMPLE 11.17 Find the steady state temperature T (x, y) for the domain
  D consisting of the upper half-plane Im (z) > 0, where T (x, y) has the boundary
  conditions

T (x, O) = 1, for x > 1, and T(x, 0) = - 1, for x < -1;

&T

fJn = Tv (x, 0) = 0, for - l<x<l.

Solution The mapping w = Arcsinz conformally maps D onto the semi-infinite

strip v > 0, - 2 " < u < ~, where t he new problem is to find the steady state