**for**

**11.5 Steady State
Temperatures**

In the theory of heat conduction, an
assumption is made that heat flows in the direction of decreasing
temperature. Another assumption is that the time rate at
which heat flows across a surface area is proportional to the
component of the temperature gradient in the direction perpendicular
to the surface area. If the temperature
does not depend on time, then the heat flow at the point
is given by the vector

,

where K is the thermal conductivity of the medium and is assumed to
be constant. If
denotes a straight-line segment of length ,
then the amount of heat flowing across the segment per unit of time
is

(11-19)

where
is a unit vector perpendicular to the segment.

If we assume that no thermal energy is
created or destroyed within the region, then the net amount of heat
flowing through any small rectangle with sides of length
and
is identically zero (see Figure 11.16(a)). This leads to
the conclusion that
is a harmonic function. The following heuristic argument
is often used to suggest that
satisfies Laplace's equation. Using Expression
(11-19) we find that the amount of heat
flowing out of the right edge of the rectangle in Figure 11.16(a) is
approximately

(11-20)

and the amount of heat flowing out of the left edge
is

(11-21)

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

(11-22) .

Steady state temperatures.Figure 11.16

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

(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

,

which implies that

.

Hence satisfies
Laplace's equation and is a harmonic function.

If the domain in
which is
defined is simply connected, then a conjugate harmonic
function exists,
and

is an analytic function. The curves are
called isothermals and are lines connecting points of the same
temperature. The curves are
called the heat flow lines, and one can visualize the heat flowing
along these curves from points of higher temperature to points of
lower temperature. The situation is illustrated in Figure
11.16(b).

Boundary value problems for steady state temperatures are realizations of the Dirichlet problem where the value of the harmonic function is interpreted as the temperature at the point .

**Example
11.14.** Suppose that two parallel planes are
perpendicular to the z plane and pass
through the horizontal lines
and
and that the temperature is held constant at the
values and , respectively,
on these planes. Then
is given by

.

Solution. The two-dimensional solution is constructed at points in the horizontal strip in the complex plane. A reasonable assumption is that the temperature at all points on the plane passing through the line is constant. Hence , where is a function of y alone. Laplace's equation implies that , and an argument similar to that in Example 11.1, (see Section 11.1), will show that the solution has the form given in the preceding equation.

The isothermals are
easily seen to be horizontal lines. The conjugate harmonic
function is

,

and the heat flow lines are
vertical segments between the horizontal
lines. If , then
the heat flows along these segments from the plane through
to the plane through ,
as illustrated in Figure 11.17.

The temperature between parallel planes where .Figure 11.17

**Example 11.15.** Find
the temperature
at each point in the upper half plane , if
the temperature at points on the x-axis on the boundary satisfy

Solution. Since
is a harmonic function, this problem is an example of a Dirichlet
problem. From Example 11.2 (see Section
11.1), it follows that the solution is

The isotherms are
rays emanating from the origin. The conjugate harmonic
function is

,

and the heat flow lines are
semicircles centered at the
origin. If , then
the heat flows counterclockwise along the semicircles, as shown in
Figure 11.18.

The temperature in the upper half-plane where .Figure 11.18

**Example 11.16.** Find
the temperature
at each point in the upper half-disk , if
the temperature at points on the boundary satisfies

Solution. As discussed in Example 11.9, (see Section
11.2),the transformation

is a one-to-one conformal mapping of the half-disk H
onto the first quadrant ,
and can be written as

(11-24)

The conformal map given by Equation
(11-24) gives rise to a new problem of
finding the temperature
that satisfies the boundary conditions

If we use Example 11.2, (see Section
11.1), the harmonic function
is given by

(11-25) .

Substituting the expressions for u and v from Equation
(11-24) into Equation
(11-25) yields the desired
solution:

.

The isothermals are
circles that pass through the points ±1, as shown in Figure
11.19.

The temperature in a half-disk.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
inside the simply connected domain D whose boundary consists of three
adjacent curves ,
,
and ,
where along ; along , and
the region is insulated along . Zero
heat flowing across
implies that

,

where
is perpendicular to .
Thus the direction of heat flow must be parallel to this portion of
the boundary. In other words,
must be part of a heat flow line and
the isothermals intersect
orthogonally.

We can solve this problem by finding
a conformal mapping

(11-26)

from D onto the semi-infinite strip so
that the image of the curve
is the ray ; the
image of the curve
is the ray given by ; and
the thermally insulated curve
is mapped onto the thermally insulated segment of
the u axis, as shown in Figure
11.20.

The new problem in G is to find the steady
state temperature function
so that along the rays, we have the boundary values

(11-27)

The condition that a segment of the
boundary is insulated can be expressed mathematically by saying that
the normal derivative of
is zero. That is,

(11-28)

where n is a coordinate measured
perpendicular to the segment. We can easily verify that
the function

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

.

The isothermals and their images under are also illustrated in Figure 11.20.

Steady state temperatures with one boundary portion insulated.Figure 11.20

**Example 11.17.** Find
the temperature
for the domain D consisting of the the upper
half-plane where
the temperature at points on the boundary satisfies

Solution. The mapping conformally
maps D onto the semi-infinite strip , where
the new problem is to find the steady state
temperature
that has the boundary conditions

Using the result of Example 11.1, (see Section
11.1), we can easily obtain the solution:

.

Therefore the solution in D is

.

If an explicit solution is required, then we can use Formula
(10-26) to obtain

where the function has
range values satisfying ; see
Figure 11.21.

The temperature with ,Figure 11.21and boundary values , and .

**Using Previous Techniques**

The techniques of the N-value Dirichlet
problem in Section 11.2 and
Poisson's integral formula in Section
11.3 can be used to find steady state temperatures. We
recast two previous examples in the context of steady state
temperatures.

**Revisited Example
11.6.** Find the
temperature in
the upper half-plane where
the temperature at points on the boundary satisfies Figure
11.5. That is

Solution. This is a four-value Dirichlet problem in the upper
half-plane defined by . For
the z plane, the solution in Equation
(11-5) becomes

Here we have and , which
we substitute into equation for to
obtain

The Dirichlet solution for the steady state temperatures.Figure 11.5

**Explore
Revisited Solution 11.6.**

**Revisited Example
11.11.** Find the temperature
in the upper half-plane ,
where the temperature at points on the boundary
satisfies

Solution. Using Equation (11-12), we
obtain

Using techniques from calculus we have the integration
formula . We
obtain the solution as follows

**Explore
Revisited Solution 11.11.**

**Extra Example
1.** Find the temperature
in the upper half-plane ,
where the temperature at points on the boundary
satisfies

**Exercises
for Section 11.5. Steady State
Temperatures**** **

**The Next Module
is**

**Two-Dimensional
Electrostatics **

**Return to the Complex
Analysis Modules **

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to the Complex Analysis Project__

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

(c) 2012 John H. Mathews, Russell W. Howell