INTRODUCTION

In Sections 20.2, 20.3, and 20.5 we demonstrated how Laplace’s partial differential equation can be solved with conformal mapping methods, and we interpreted a solution u = u(x, y) of the Dirichlet problem as either the steady-state temperature at the point (x, y) or the equilibrium displacement of a membrane at the point (x, y). Laplace’s equation is a fundamental partial differential equation that arises in a variety of contexts. In this section we will establish a general relationship between vector fields and analytic functions and use our conformal mapping techniques to solve problems involving electrostatic force fields and two-dimensional fluid flows.

Vector Fields

A vector field F(x, y) = P(x, y)i + Q(x, y)j in a domain D can also be expressed in the complex form

F(x, y) = P(x, y) + iQ(x, y)

and thought of as a complex function. Recall from Chapter 9 that div F = ∂P/∂x + ∂Q/∂y and curl F = (∂Q/∂x − ∂P/∂y)k. If we require that both div F = 0 and curl F = 0, then

(1)

This set of equations is reminiscent of the Cauchy–Riemann criterion for analyticity presented in Theorem 17.5.2 and suggests that we examine the complex function g(z) = P(x, y) − iQ(x, y).

PROOF:

If u(x, y) and v(x, y) denote the real and imaginary parts of g(z), then u = P and v = −Q. Therefore the equations in (1) are equivalent to the equations

that is, (2)

The equations in (2) are the Cauchy–Riemann equations for analyticity. ≡

EXAMPLE 1 Vector Field Gives an Analytic Function

The vector field defined by F(x, y) = (−kq/z − z₀²)(z − z₀) may be interpreted as the electric field produced by a wire that is perpendicular to the z-plane at z = z₀ and carries a charge of q coulombs per unit length. The corresponding complex function is

Since g(z) is analytic for z ≠ z₀, div F = 0 and curl F = 0. ≡

EXAMPLE 2 Analytic Function Gives a Vector Field

The physical interpretation of the conditions div F = 0 and curl F = 0 depends on the setting. If F(x, y) represents the force in an electric field that acts on a unit test charge placed at (x, y), then, by Theorem 9.9.2, curl F = 0 if and only if the field is conservative. The work done in transporting a test charge between two points in a domain D must be independent of the path.

If C is a simple closed contour that lies in D, Gauss’s law asserts that the line integral (F · n) ds is proportional to the total charge enclosed by the curve C. If D is simply connected and all the electric charge is distributed on the boundary of D, then (F · n) ds = 0 for any simple closed contour in D. By the divergence theorem in the form (1) of Section 9.16,

(3)

where R is the region enclosed by C, and we can conclude that div F = 0 in D. Conversely, if div F = 0 in D, the double integral is zero and therefore the domain D contains no charge.

Potential Functions

Suppose that F(x, y) is a vector field in a simply connected domain D with both div F = 0 and curl F = 0. By Theorem 18.3.3, the analytic function g(z) = P(x, y) − iQ(x, y) has an antiderivative

(4)

in D, which is called a complex potential for the vector field F. Note that

and so (5)

Therefore, F = ∇ϕ and, as in Section 9.9, the harmonic function ϕ is called a (real) potential function for F.^* When the potential ϕ is specified on the boundary of a region R, we can use conformal mapping techniques to solve the resulting Dirichlet problem. The equipotential lines ϕ(x, y) = c can be sketched and the vector field F can be determined using (5).

EXAMPLE 3 Complex Potential

The potential ϕ in the half-plane x ≥ 0 satisfies the boundary conditions ϕ(0, y) = 0 and ϕ(x, 0) = 1 for x ≥ 1. See FIGURE 20.6.1(a). Determine a complex potential, the equipotential lines, and the force field F.

SOLUTION

We saw in Example 2 of Section 20.2 that the analytic function z = sin w maps the strip 0 ≤ u ≤ π/2 in the w-plane to the region R in question. Therefore, f(z) = sin⁻¹z maps R onto the strip, and Figure 20.6.1(b) shows the transferred boundary conditions. The simplified Dirichlet problem has the solution U(u, v) = (2/π)u, and so ϕ(x, y) = U(sin⁻¹z) = Re ((2/π) sin⁻¹z) is the potential function on D, and G(z) = (2/π)u sin⁻¹z is a complex potential for the force field F.

Note that the equipotential lines ϕ = c are the images of the equipotential lines U = c in the w-plane under the inverse mapping z = sin w. In Example 2 of Section 20.2 we showed that the vertical line u = a is mapped onto a branch of the hyperbola

Since the equipotential line U = c, 0 < c < 1, is the vertical line u = π/2c, it follows that the equipotential line ϕ = c is the right branch of the hyperbola

Since F = and d/dz sin⁻¹z = 1/(1 − z²)^1/2, the force field is given by

≡

Steady-State Fluid Flow

The vector V(x, y) = P(x, y) + iQ(x, y) may also be interpreted as the velocity vector of a two-dimensional steady-state fluid flow at a point (x, y) in a domain D. The velocity at all points in the domain is therefore independent of time, and all movement takes place in planes that are parallel to a z-plane.

The physical interpretation of the conditions div V = 0 and curl V = 0 was discussed in Section 9.7. Recall that if curl V = 0 in D, the flow is called irrotational. If a small circular paddle wheel is placed in the fluid, the net angular velocity on the boundary of the wheel is zero, and so the wheel will not rotate. If div V = 0 in D, the flow is called incompressible. In a simply connected domain D, an incompressible flow has the special property that the amount of fluid in the interior of any simple closed contour C is independent of time. The rate at which fluid enters the interior of C matches the rate at which it leaves, and consequently there can be no fluid sources or sinks at points in D.

If div V = 0 and curl V = 0, V has a complex velocity potential

G(z) = ϕ(x, y) + iψ(x, y)

that satisfies = V. In this setting, special importance is placed on the level curves ψ(x, y) = c. If z(t) = x(t) + iy(t) is the path of a particle (such as a small cork) that has been placed in the fluid, then

(6)

Hence, dy/dx = Q(x, y)/P(x, y) or −Q(x, y) dx + P(x, y) dy = 0. This differential equation is exact, since div V = 0 implies ∂(−Q)/∂y = ∂P/∂x. By the Cauchy–Riemann equations, ∂ψ/∂x = −∂ϕ/∂y = −Q and ∂ψ/∂y = ∂ϕ/∂x = P, and therefore all solutions of (6) satisfy ψ(x, y) = c. The function ψ(x, y) is therefore called a stream function and the level curves ψ(x, y) = c are streamlines for the flow.

EXAMPLE 4 Uniform Flow

The uniform flow in the upper half-plane is defined by V(x, y) = A(1, 0), where A is a fixed positive constant. Note that V = A, and so a particle in the fluid moves at a constant speed. A complex potential for the vector field is G(z) = Az = Ax + iAy, and so the streamlines are the horizontal lines Ay = c. See FIGURE 20.6.2(a). Note that the boundary y = 0 of the region is itself a streamline. ≡

EXAMPLE 5 Flow Around a Corner

Constructing Special Flows

The process of constructing an irrotational and incompressible flow that remains inside a given region R is called streamlining. Since the streamlines are described by ψ(x, y) = c, two distinct streamlines do not intersect. Therefore, if the boundary is itself a streamline, a particle that starts inside R cannot leave R. This is the content of the following theorem:

EXAMPLE 6 Flow Around a Cylinder

The analytic function G(z) = z + 1/z maps the region R in the upper half-plane and outside the circle z = 1 onto the upper half-plane v ≥ 0. The boundary of R is mapped onto the u-axis, and so v = ψ(x, y) = y − y/(x² + y²) is zero on the boundary of R. FIGURE 20.6.3 shows the streamlines of the resulting flow. The velocity field is given by = 1 − 1/, and so

It follows that V ≈ (1, 0) for large values of r, and so the flow is approximately uniform at large distances from the circle z = 1. The resulting flow in the region R is called flow around a cylinder. The mirror image of the flow can be adjoined to give a flow around a complete cylinder. ≡

If R is a polygonal region, we can use the Schwarz–Christoffel formula to find a conformal mapping z = f(w) from the upper half-plane R′ onto R. The inverse function G(z) = f⁻¹(z) maps the boundary of R onto the u-axis. Therefore, if G(z) = ϕ(x, y) + iψ(x, y), then ψ(x, y) = 0 on the boundary of R. Note that the streamlines ψ(x, y) = c in the z-plane are the images of the horizontal lines v = c in the w-plane under z = f(w).

EXAMPLE 7 Streamlines Defined Parametrically

The analytic function f(w) = w + Ln w + 1 maps the upper half-plane v ≥ 0 to the upper half-plane y ≥ 0 with the horizontal line y = π, x ≤ 0, deleted. See Example 4 in Section 20.4. If G(z) = f⁻¹(z) = ϕ(x, y) + iψ (x, y), then G(z) maps R onto the upper half-plane and maps the boundary of R onto the u-axis. Therefore, ψ (x, y) = 0 on the boundary of R.

It is not possible to find an explicit formula for the stream function ψ (x, y). The streamlines, however, are the images of the horizontal lines v = c under z = f(w). If we write w = t + ic, c > 0, then the streamlines can be represented in the parametric form

z = f(t + ic) = t + ic + Ln(t + ic) + 1;

that is, x = t + 1 + log_e (t² + c²), y = c + Arg(t + ic).

Graphing software was used to generate the streamlines in FIGURE 20.6.4. ≡

A stream function ψ (x, y) is harmonic but, unlike a solution to a Dirichlet problem, we do not require ψ (x, y) to be bounded (see Examples 4–6) or to assume a fixed set of constants on the boundary. Therefore, there may be many different stream functions for a given region that satisfy Theorem 20.6.2. This will be illustrated in the final example.

EXAMPLE 8 Streamlines Defined Parametrically

The analytic function f(w) = w + e^w + 1 maps the horizontal strip 0 ≤ v ≤ π onto the region R shown in Figure 20.6.4. Therefore, G(z) = f⁻¹(z) = ϕ(x, y) + iψ(x, y) maps R back to the strip and, from M-1 in the conformal mappings in Appendix D, maps the boundary line y = 0 onto the u-axis and maps the boundary line y = π, x ≤ 0, onto the horizontal line v = π. Therefore, ψ(x, y) is constant on the boundary of R.

The streamlines are the images of the horizontal lines v = c, 0 < c < π, under z = f(w). As in Example 7, a parametric representation of the streamlines is

z = f(t + ic) = t + ic + e^t+ic + 1

or x = t + 1 + e^t cos c, y = c + e^t sin c.

The streamlines are shown in FIGURE 20.6.5. Unlike the flow in Example 7, the fluid appears to emerge from the strip 0 ≤ y ≤ π, x ≤ 0. ≡

20.6 Exercises Answers to selected odd-numbered problems begin on page ANS-50.

In Problems 1–4, verify that div F = 0 and curl F = 0 for the given vector field F(x, y) by examining the corresponding complex function g(z) = P(x, y) − iQ(x, y). Find a complex potential for the vector field and sketch the equipotential lines.

1. F(x, y) = (cos θ₀) i + (sin θ₀) j
2. F(x, y) = −y i − x j
3. F(x, y) = j
4. F(x, y) = j
5. The potential ϕ on the wedge 0 ≤ Arg z ≤ π/4 satisfies the boundary conditions ϕ(x, 0) = 0 and ϕ(x, x) = 1 for x > 0. Determine a complex potential, the equipotential lines, and the corresponding force field F.
6. Use the conformal mapping f(z) = 1/z to determine a complex potential, the equipotential lines, and the corresponding force field F for the potential ϕ that satisfies the boundary conditions shown in FIGURE 20.6.6.

   

7. The potential ϕ on the semicircle z ≤ 1, y ≥ 0, satisfies the boundary conditions ϕ(x, 0) = 0, −1 < x < 1, and ϕ(e^iθ) = 1, 0 < θ < π. Show that

   ,

   and use the mapping properties of linear fractional transformations to explain why the equipotential lines are arcs of circles.

8. Use the conformal mapping C-1 in Appendix D to find the potential ϕ in the region outside the two circles z = 1 and z − 3 = 1 if the potential is kept at zero on z = 1 and one on z − 3 = 1. Use the mapping properties of linear fractional transformations to explain why the equipotential lines are, with one exception, circles.

In Problems 9–14, a complex velocity potential G(z) is defined on a region R.

1. Find the stream function and verify that the boundary of R is a streamline.
2. Find the corresponding velocity vector field V(x, y).
3. Use a graphing utility to sketch the streamlines of the flow.

9. G(z) = z⁴

   

10. G(z) = z^2/3

    

11. G(z) = sin z

    

12. G(z) = i sin⁻¹z

    

13. G(z) = z² + 1/z²

    

14. G(z) = e^z

    

In Problems 15–18, a conformal mapping z = f(w) from the upper half-plane v ≥ 0 to a region R in the z-plane is given and the flow in R with complex potential G(z) = f⁻¹(z) is constructed.

1. Verify that the boundary of R is a streamline for the flow.
2. Find a parametric representation for the streamlines of the flow.
3. Use a graphing utility to sketch the streamlines of the flow.

15. M-9 in Appendix D
16. M-4 in Appendix D; use a = 1
17. M-2 in Appendix D; use a = 1
18. M-5 in Appendix D
19. A stagnation point in a flow is a point at which V = 0. Find all stagnation points for the flows in Examples 5 and 6.
20. For any two real numbers k and x₁, the function G(z) = k Ln(z − x₁) is analytic in the upper half-plane and therefore is a complex potential for a flow. The real number x₁ is called a sink when k < 0 and a source for the flow when k > 0.
    1. Show that the streamlines are rays emanating from x₁.
    2. Show that V = (k/z − x₁²)(z − x₁) and conclude that the flow is directed toward x₁ precisely when k < 0.
21. If f(z) is a conformal mapping from a domain D onto the upper half-plane, a flow with a source at a point ξ₀ on the boundary of D is defined by the complex potential G(z) = k Ln(f(z) − f(ξ₀)), where k > 0. Determine the streamlines for a flow in the first quadrant with a source at ξ₀ = 1 and k = 1.
22. 1. Construct a flow on the horizontal strip 0 < y < π with a sink at the boundary point ξ₀ = 0. [Hint: See Problem 21.]
    2. Use a graphing utility to sketch the streamlines of the flow.
23. The complex potential G(z) = k Ln(z − 1) − k Ln(z + 1) with k > 0 gives rise to a flow on the upper half-plane with a single source at z = 1 and a single sink at z = −1. Show that the streamlines are the family of circles x² + (y − c)² = 1 + c². See FIGURE 20.6.13.

    

24. The flow with velocity vector V = is called a vortex at z = 0, and the geometric nature of the streamlines depends on the choice of a and b.
    1. Show that if z = x(t) + iy(t) is the path of a particle, then

       

       .

    2. Change to polar coordinates to establish that dr/dt = a/r and dθ/dt = b/r², and conclude that r = ce^aθ/b for b ≠ 0. [Hint: See (2) of Section 11.1.]
    3. Conclude that the logarithmic spirals in part (b) spiral inward if and only if a < 0, and the curves are traversed clockwise if and only if b < 0. See FIGURE 20.6.14.

       

*If F is an electric field, the electric potential function Φ is defined to be −ϕ and F = −∇Φ.