Steady State Temperatures


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 [Graphics:Images/TemperaturesMod_gr_1.gif] does not depend on time, then the heat flow at the point [Graphics:Images/TemperaturesMod_gr_2.gif] is given by the vector  


where K is the thermal conductivity of the medium and is assumed to be constant.  If [Graphics:Images/TemperaturesMod_gr_4.gif] denotes a straight-line segment of length [Graphics:Images/TemperaturesMod_gr_5.gif], then the amount of heat flowing across the segment per unit of time is

(11-19)        [Graphics:Images/TemperaturesMod_gr_6.gif]  

where [Graphics:Images/TemperaturesMod_gr_7.gif] 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 [Graphics:Images/TemperaturesMod_gr_8.gif] and [Graphics:Images/TemperaturesMod_gr_9.gif] is identically zero (see Figure 11.16(a)).  This leads to the conclusion that [Graphics:Images/TemperaturesMod_gr_10.gif] is a harmonic function.  The following heuristic argument is often used to suggest that [Graphics:Images/TemperaturesMod_gr_11.gif] 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)        [Graphics:Images/TemperaturesMod_gr_12.gif]  

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

(11-21)        [Graphics:Images/TemperaturesMod_gr_13.gif]  

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

(11-22)        [Graphics:Images/TemperaturesMod_gr_14.gif].

Figure 11.16  Steady state temperatures.

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

(11-23)        [Graphics:Images/TemperaturesMod_gr_15.gif].

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  [Graphics:Images/TemperaturesMod_gr_18.gif]  satisfies Laplace's equation and is a harmonic function.

    If the domain in which  [Graphics:Images/TemperaturesMod_gr_19.gif]  is defined is simply connected, then a conjugate harmonic function  [Graphics:Images/TemperaturesMod_gr_20.gif]  exists, and    


is an analytic function.  The curves  [Graphics:Images/TemperaturesMod_gr_22.gif]  are called isothermals and are lines connecting points of the same temperature.  The curves  [Graphics:Images/TemperaturesMod_gr_23.gif]  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 [Graphics:Images/TemperaturesMod_gr_24.gif] is interpreted as the temperature at the point [Graphics:Images/TemperaturesMod_gr_25.gif].


Example 11.14.  Suppose that two parallel planes are perpendicular to the z plane and pass through the horizontal lines [Graphics:Images/TemperaturesMod_gr_26.gif] and [Graphics:Images/TemperaturesMod_gr_27.gif] and that the temperature is held constant at the values  [Graphics:Images/TemperaturesMod_gr_28.gif]  and  [Graphics:Images/TemperaturesMod_gr_29.gif],  respectively, on these planes.  Then [Graphics:Images/TemperaturesMod_gr_30.gif] is given by  



Solution.  The two-dimensional solution is constructed at points in the horizontal strip  [Graphics:Images/TemperaturesMod_gr_33.gif]  in the complex plane.  A reasonable assumption is that the temperature at all points on the plane passing through the line [Graphics:Images/TemperaturesMod_gr_34.gif] is constant.  Hence  [Graphics:Images/TemperaturesMod_gr_35.gif],  where [Graphics:Images/TemperaturesMod_gr_36.gif] is a function of y alone.  Laplace's equation implies that  [Graphics:Images/TemperaturesMod_gr_37.gif],  and an argument similar to that in Example 11.1, (see Section 11.1), will show that the solution [Graphics:Images/TemperaturesMod_gr_38.gif] has the form given in the preceding equation.

    The isothermals  [Graphics:Images/TemperaturesMod_gr_39.gif]  are easily seen to be horizontal lines.  The conjugate harmonic function is  


and the heat flow lines  [Graphics:Images/TemperaturesMod_gr_41.gif]  are vertical segments between the horizontal lines.  If  [Graphics:Images/TemperaturesMod_gr_42.gif],  then the heat flows along these segments from the plane through [Graphics:Images/TemperaturesMod_gr_43.gif] to the plane through [Graphics:Images/TemperaturesMod_gr_44.gif], as illustrated in Figure 11.17.  

Figure 11.17  The temperature between parallel planes where  [Graphics:Images/TemperaturesMod_gr_45.gif].

Explore Solution 11.14.


Example 11.15.  Find the temperature [Graphics:Images/TemperaturesMod_gr_63.gif] at each point in the upper half plane  [Graphics:Images/TemperaturesMod_gr_64.gif]  ,  if the temperature at points on the x-axis on the boundary satisfy



Solution.  Since [Graphics:Images/TemperaturesMod_gr_67.gif] 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  [Graphics:Images/TemperaturesMod_gr_69.gif]  are rays emanating from the origin.  The conjugate harmonic function is  


and the heat flow lines  [Graphics:Images/TemperaturesMod_gr_72.gif]  are semicircles centered at the origin.  If  [Graphics:Images/TemperaturesMod_gr_73.gif],  then the heat flows counterclockwise along the semicircles, as shown in Figure 11.18.

Figure 11.18  The temperature [Graphics:Images/TemperaturesMod_gr_74.gif] in the upper half-plane where [Graphics:Images/TemperaturesMod_gr_75.gif].

Explore Solution 11.15.


Example 11.16.  Find the temperature [Graphics:Images/TemperaturesMod_gr_96.gif] at each point in the upper half-disk  [Graphics:Images/TemperaturesMod_gr_97.gif],  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 [Graphics:Images/TemperaturesMod_gr_101.gif], and can be written as

(11-24)            [Graphics:Images/TemperaturesMod_gr_102.gif]    

The conformal map given by Equation (11-24) gives rise to a new problem of finding the temperature [Graphics:Images/TemperaturesMod_gr_103.gif] that satisfies the boundary conditions  


If we use Example 11.2, (see Section 11.1), the harmonic function [Graphics:Images/TemperaturesMod_gr_105.gif] is given by  

(11-25)         [Graphics:Images/TemperaturesMod_gr_106.gif].  

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

The isothermals  [Graphics:Images/TemperaturesMod_gr_108.gif]  are circles that pass through the points ±1, as shown in Figure 11.19.

Figure 11.19  The temperature [Graphics:Images/TemperaturesMod_gr_109.gif] in a half-disk.

Explore Solution 11.16.



11.5.1  An Insulated Segment on the Boundary

    We now turn to the problem of finding the steady state temperature function [Graphics:Images/TemperaturesMod_gr_119.gif] inside the simply connected domain D whose boundary consists of three adjacent curves [Graphics:Images/TemperaturesMod_gr_120.gif], [Graphics:Images/TemperaturesMod_gr_121.gif], and [Graphics:Images/TemperaturesMod_gr_122.gif], where  [Graphics:Images/TemperaturesMod_gr_123.gif]  along  [Graphics:Images/TemperaturesMod_gr_124.gif];  [Graphics:Images/TemperaturesMod_gr_125.gif]  along  [Graphics:Images/TemperaturesMod_gr_126.gif],  and the region is insulated along [Graphics:Images/TemperaturesMod_gr_127.gif].  Zero heat flowing across [Graphics:Images/TemperaturesMod_gr_128.gif] implies that  

where  [Graphics:Images/TemperaturesMod_gr_130.gif] is perpendicular to [Graphics:Images/TemperaturesMod_gr_131.gif]. Thus the direction of heat flow must be parallel to this portion of the boundary.  In other words, [Graphics:Images/TemperaturesMod_gr_132.gif] must be part of a heat flow line  [Graphics:Images/TemperaturesMod_gr_133.gif]  and the isothermals  [Graphics:Images/TemperaturesMod_gr_134.gif]  intersect [Graphics:Images/TemperaturesMod_gr_135.gif] orthogonally.

     We can solve this problem by finding a conformal mapping  

(11-26)        [Graphics:Images/TemperaturesMod_gr_136.gif]

from D onto the semi-infinite strip  [Graphics:Images/TemperaturesMod_gr_137.gif]  so that the image of the curve [Graphics:Images/TemperaturesMod_gr_138.gif] is the ray  [Graphics:Images/TemperaturesMod_gr_139.gif];  the image of the curve [Graphics:Images/TemperaturesMod_gr_140.gif] is the ray given by  [Graphics:Images/TemperaturesMod_gr_141.gif];  and the thermally insulated curve [Graphics:Images/TemperaturesMod_gr_142.gif] is mapped onto the thermally insulated segment  [Graphics:Images/TemperaturesMod_gr_143.gif]  of the u axis, as shown in Figure 11.20.

    The new problem in G is to find the steady state temperature function [Graphics:Images/TemperaturesMod_gr_144.gif] so that along the rays, we have the boundary values  

(11-27)        [Graphics:Images/TemperaturesMod_gr_145.gif]  

    The condition that a segment of the boundary is insulated can be expressed mathematically by saying that the normal derivative of  [Graphics:Images/TemperaturesMod_gr_146.gif] is zero. That is,  

(11-28)        [Graphics:Images/TemperaturesMod_gr_147.gif]

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  [Graphics:Images/TemperaturesMod_gr_150.gif]  and their images under  [Graphics:Images/TemperaturesMod_gr_151.gif]  are also illustrated in Figure 11.20.

Figure 11.20  Steady state temperatures with one boundary portion insulated.


Example 11.17.  Find the temperature [Graphics:Images/TemperaturesMod_gr_152.gif] for the domain D consisting of the the upper half-plane  [Graphics:Images/TemperaturesMod_gr_153.gif]   where the temperature at points on the boundary satisfies  



Solution.  The mapping  [Graphics:Images/TemperaturesMod_gr_156.gif]  conformally maps D onto the semi-infinite strip  [Graphics:Images/TemperaturesMod_gr_157.gif],  where the new problem is to find the steady state temperature  [Graphics:Images/TemperaturesMod_gr_158.gif] 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  [Graphics:Images/TemperaturesMod_gr_163.gif]  has range values satisfying  [Graphics:Images/TemperaturesMod_gr_164.gif];  see Figure 11.21.

Figure 11.21  The temperature [Graphics:Images/TemperaturesMod_gr_165.gif] with [Graphics:Images/TemperaturesMod_gr_166.gif],

            and boundary values [Graphics:Images/TemperaturesMod_gr_167.gif], and [Graphics:Images/TemperaturesMod_gr_168.gif].

Explore Solution 11.17.



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  [Graphics:Images/TemperaturesMod_gr_185.gif]  in the upper half-plane  [Graphics:Images/TemperaturesMod_gr_186.gif]  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  [Graphics:Images/TemperaturesMod_gr_189.gif].  For the z plane, the solution in Equation (11-5) becomes  


Here we have  [Graphics:Images/TemperaturesMod_gr_191.gif]  and  [Graphics:Images/TemperaturesMod_gr_192.gif],  which we substitute into equation for  [Graphics:Images/TemperaturesMod_gr_193.gif]  to obtain  


Figure 11.5  The  Dirichlet solution for the steady state temperatures.

Explore Revisited Solution 11.6.


Revisited Example 11.11.  Find the temperature [Graphics:Images/TemperaturesMod_gr_205.gif] in the upper half-plane [Graphics:Images/TemperaturesMod_gr_206.gif], 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  [Graphics:Images/TemperaturesMod_gr_210.gif].  We obtain the solution as follows


Explore Revisited Solution 11.11.


Extra Example 1.  Find the temperature [Graphics:Images/TemperaturesMod_gr_223.gif] in the upper half-plane [Graphics:Images/TemperaturesMod_gr_224.gif], where the temperature at points on the boundary satisfies  



Explore Extra Solution 1.


Exercises for Section 11.5.  Steady State Temperatures


Library Research Experience for Undergraduates

Dirichlet Problem


Steady State Temperature





The Next Module is

Two-Dimensional Electrostatics  



Return to the Complex Analysis Modules  



Return 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