11.4  Two-Dimensional Mathematical Models

    We now consider problems involving steady state heat flow, electrostatics, and ideal fluid flow that can be solved with conformal mapping techniques.  Conformal mapping transforms a region in which the problem is posed to one in which the solution is easy to obtain.  As our solutions involve only two independent variables, x and y, we first mention a basic assumption needed for the validity of the model.

    The physical problems we just mentioned are real-world applications and involve solutions in three-dimensional Cartesian space.  Such problems generally would involve the Laplacian in three variables and the divergence and curl of three-dimensional vector functions.  Since complex analysis involves only x and y, we consider the special case in which the solution does not vary with the coordinate along the axis perpendicular to the xy plane.  For steady state heat flow and electrostatics, this assumption means that the temperature, T, or the potential, V, varies only with x, and y.  Thus for the flow of ideal fluids, the fluid motion is the same in any plane that is parallel to the z plane.  Curves drawn in the z plane are to be interpreted as cross sections that correspond to infinite cylinders perpendicular to the z plane.  An infinite cylinder is the limiting case of a "long" physical cylinder, so the mathematical model that we present is valid provided the three-dimensional problem involves a physical cylinder long enough that the effects at the ends can be reasonably neglected.

    In Section 11.2 and Section 11.3, we showed how to obtain solutions for harmonic functions.  For applications, we need to consider the family of level curves  

(10-16)        

and the conjugate harmonic function    and its family of level curves  

(10-17)        

For convenience, we introduce the term complex potential for the analytic function  

            .

We use Theorem 11.4, regarding the orthogonality of the families of level curves (10-16) and (10-17), to develop ideas concerning the physical applications that we will consider.

Theorem 11.4 (Orthogonal Families of Level Curves).  Let be harmonic in a domain D.  Let be the harmonic conjugate and let     be the complex potential.  Then the two families of level curves given in Equations  (10-16) and (10-17), respectively, are orthogonal in the sense that if is a point in common to the two curves    and  ,  and  if    ,  then these two curves intersect orthogonally.

Proof.

Proof of Theorem 11.4 is in the book.
Complex Analysis for Mathematics and Engineering

Exploration for Theorem 11.4.

    The complex potential    has many physical interpretations.  Suppose, for example, that we have solved a problem in steady state temperatures. Then we can obtain the solution to a similar problem with the same boundary conditions in electrostatics by interpreting the isothermals as equipotential curves and the heat flow lines as flux lines. This implies that heat flow and electrostatics correspond directly.

    Or suppose that we have solved a fluid flow problem.  Then we can obtain a solution to an analogous problem in heat flow by interpreting the equipotentials as isothermals and streamlines as heat flow lines.  Various interpretations of the families of level curves given in Equations  (10-16) and (10-17) and correspondences between families are summarized in Table 11.1.

            

                Table 11.1  Interpretations for the level curves of and .

Preview of Applications

    The following is a sample of applications taken from Section 11.5 Steady State Temperatures; Section 11.6 Two-Dimensional Electrostatics; Section 11.7 Two-Dimensional Fluid Flow; Section 11.8 The Joukowski Airfoil ; Section 11.9 The Schwarz-Christoffel Transformation; Section 11.10 Image of a Fluid Flow; Section 11.11 Sources and Sinks.

Example 11.16.  (See Section 11.5 for more.) 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, 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, 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.

            Figure 11.19  The temperature in a half-disk.

Explore Solution 11.16.

Example 11.21.  (See Section 11.6 for more.) Find the electrical potential    in the disk    that satisfies the boundary values

              

Solution.  The mapping    is a one-to-one conformal mapping of D onto the upper half-plane    with the property that is mapped onto the negative u axis and is mapped onto the positive u axis.  The potential in the upper half-plane that satisfies the new boundary values

              

is given by  

(11-29)            .  

A straightforward calculation shows that  

            

We substitute the real and imaginary parts,    and    from this equation, into Equation (11-29) to obtain the desired solution:   

             .  

The level curve    in the upper half-plane is a ray emanating from the origin, and the preimage    in the unit disk is an arc of a circle that passes through the points  .  Several level curves are illustrated in Figure 11.38.

            Figure 11.38  The potentials    and  .  

Explore Solution 11.21.

Example 11.24.  (See Section 11.7 for more.) Find the complex potential for an ideal fluid flowing from left to right across the complex plane and around the unit circle  .  

Solution.  We use the fact that the conformal mapping    maps the domain    one-to-one and onto the w plane slit along the segment  .  The complex potential for a uniform horizontal flow parallel to this slit in the w plane is  

            ,  

where A is a positive real number.  The stream function for the flow in the w plane is    so that the slit lies along the streamline  .

    The composite function    determines the fluid flow in the domain D, where the complex potential is

            ,  

where .  We can use polar coordinates to express as

            

The streamline     consists of the rays

               and   

along the x axis and the curve  ,  which is the unit circle  .  Thus the unit circle can be considered as a boundary curve for the fluid flow.

    The approximation    is valid for large values of z, so we can approximate the flow with a uniform horizontal flow having speed    at points that are distant from the origin.  The streamlines    and their images    under the mapping are illustrated in Figure 11.51.

            Figure 11.51  Fluid flow around a circle. 

Explore Solution 11.24.

Example 11.27.  (See Section 11.9 for more.) Use the Schwarz Christoffel formula to verify that    maps the upper half plane    onto the upper half plane    slit along the line segment from  .   (Use the principal square root throughout.)

Solution.  If we choose    and  ,  then the formula

            

will determine a mapping from the upper half-plane    onto the portion of the upper half-plane    that lies outside the triangle with vertices    as indicated in Figure 11.73(a).  If we let ,  then  , and  .  

            Figure 11.73   The region with and .

    The limiting formula for the derivative becomes    

            

which will determine a mapping from the upper half-plane    onto the upper half plane    slit along the line segment from   as indicated in Figure 11.73(b).  An easy integration reveals that is given by  

              

and the boundary values    lead to system of equations

               and   

the solution is easily found to be and the desired function  is  

            .

Explore Solution 11.27.

Example 11.29.  (See Section 11.10 for more.) Show that the mapping    maps the upper half plane   onto the domain in the w-plane that lies above the boundary curve consisting of the rays   and the segment     (see Figure 11.87).

            Figure 11.87  (a) Flow over a step.                    (b) Flow around a blunt object.

Hint:  Set    and   .  In the w-plane , and the exterior angles are  ,  and the formula for the derivative is  

            

Integrate and get  

            

Solve for the coefficients A and B and obtain

            

It maps the upper half-plane   onto the domain in the w plane that lies above the boundary curve consisting of the rays   and the segment  .  Furthermore, the image of horizontal streamlines in the z plane are curves in the w plane given by the parametric equation

        

for  .  The new flow is that of a step in the bed of a deep stream and is illustrated in Figure 11.87(a).  The function is also defined for values of z in the lower half-plane, and the images of horizontal streamlines that lie above or below the x axis are mapped onto streamlines that flow past a long rectangular obstacle, which is illustrated in Figure 11.87(b).

Explore Solution 11.29.

Example 11.30.  (Source and sink of equal strength)  (See Section 11.11 for more.) Let a source and sink of unit strength be located at the points and , respectively.  The complex potential for a fluid flowing from the source at    to the sink at    is  

            .  

            Figure 11.95 (a)  Source and sink of unit strength.

The velocity potential and stream function are  

            ,   and  

            ,  

respectively.  Solving for the streamline  ,  we start with   

              

and obtain the equation

            .

A straightforward calculation shows that points on the streamline must satisfy the equation  

            ,  

which is the equation of a circle with center at    that passes through the points  .  Several streamlines are indicated in Figure 11.95(a).

Explore Solution 11.30.