EXAMPLE 1 Potential Between Parallel Plates

Find the potential Φ of the field between two parallel conducting plates extending to infinity (Fig. 396), which are kept at potentials Φ₁ and Φ₂, respectively.

Solution. From the shape of the plates it follows that Φ depends only on x, and Laplace's equation becomes Φ″ = 0. By integrating twice we obtain Φ = ax + b, where the constants a and b are determined by the given boundary values of Φ on the plates. For example, if the plates correspond to x = −1 and x = 1, the solution is

The equipotential surfaces are parallel planes.

EXAMPLE 2 Potential Between Coaxial Cylinders

Find the potential Φ between two coaxial conducting cylinders extending to infinity on both ends (Fig. 397) and kept at potentials Φ₁ and Φ₂, respectively.

Solution. Here Φ depends only on for reasons of symmetry, and Laplace's equation r²u_rr + ru_r + u_θθ = 0 [(5), Sec. 12.10] with u_θθ = 0 and u = Φ becomes rΦ″ + Φ′ = 0. By separating variables and integrating we obtain

and a and b are determined by the given values of Φ on the cylinders. Although no infinitely extended conductors exist, the field in our idealized conductor will approximate the field in a long finite conductor in that part which is far away from the two ends of the cylinders.

EXAMPLE 3 Potential in an Angular Region

Find the potential Φ between the conducting plates in Fig. 398, which are kept at potentials Φ₁ (the lower plate) and Φ₂, and make an angle α, where 0 < α π. (In the figure we have α = 120° = 2π/3

Solution. θ = Arg z (z = x + iy ≠ 0) is constant on rays θ = const. It is harmonic since it is the imaginary part of an analytic function, Ln z (Sec. 13.7). Hence the solution is

with a and b determined from the two boundary conditions (given values on the plates)

Thus a = (Φ₂ + Φ₁)/2, b = (Φ₂ − Φ₁)/α. The answer is

Fig. 398. Potential in Example 3

EXAMPLE 4 Complex Potential

In Example 1, a conjugate is Ψ = ay. It follows that the complex potential is

and the lines of force are horizontal straight lines y = const parallel to the x-axis.

EXAMPLE 5 Complex Potential

In Example 2 we have Φ = a ln + r + b = a ln |z| + b. A conjugate is Ψ = a Arg z. Hence the complex potential is

and the lines of force are straight lines through the origin. F(z) may also be interpreted as the complex potential of a source line (a wire perpendicular to the xy-plane) whose trace in the xy-plane is the origin.

EXAMPLE 6 Complex Potential

In Example 3 we get F(z) by noting that i Ln z = i ln |z| − Arg z, multiplying this by −b, and adding a:

We see from this that the lines of force are concentric circles |z| = const. Can you sketch them?

EXAMPLE 7 Potential of a Pair of Source Lines (a Pair of Charged Wires)

Determine the potential of a pair of oppositely charged source lines of the same strength at the points z = c and z = −c on the real axis.

Solution. From Examples 2 and 5 it follows that the potential of each of the source lines is

respectively. Here the real constant K measures the strength (amount of charge). These are the real parts of the complex potentials

Hence the complex potential of the combination of the two source lines is

The equipotential lines are the curves

These are circles, as you may show by direct calculation. The lines of force are

We write this briefly (Fig. 399)

Now θ₁ − θ₂ is the angle between the line segments from z to c and −c (Fig. 399). Hence the lines of force are the curves along each of which the line segment S: −c x c appears under a constant angle. These curves are the totality of circular arcs over S, as is (or should be) known from elementary geometry. Hence the lines of force are circles. Figure 400 shows some of them together with some equipotential lines.

In addition to the interpretation as the potential of two source lines, this potential could also be thought of as the potential between two circular cylinders whose axes are parallel but do not coincide, or as the potential between two equal cylinders that lie outside each other, or as the potential between a cylinder and a plane wall. Explain this using Fig. 400.

PROBLEM SET 18.1

1–4 COAXIAL CYLINDERS

Find and sketch the potential between two coaxial cylinders of radii r₁ and r₂ having potential U₁ and U₂, respectively.

1. r₁ = 2.5 mm, r₂ = 4.0 cm, U₁ = 0 V, U₂ = 220 V
2. r₁ = 1 cm, r₂ = 2 cm, U₁ = 400 V, U₂ = 0 V
3. r₁ = 10 cm, r₂ = 1 m, U₁ = 10 kV, U₂ = −10 kV
4. If r₁ = 2 cm, r₂ = 6 cm and U₁ = 300 V, U₂ = 100 V, respectively, is the potential at r = 4 cm equal to 200 V? Less? More? Answer without calculation. Then calculate and explain.

5–7 PARALLEL PLATES

Find and sketch the potential between the parallel plates having potentials U₁ and U₂. Find the complex potential.

• 5. Plates at x₁ = −5 cm, x₂ = 5 cm, potentials U₁ = 250 V, U₂ = 500 V, respectively.
• 6. Plates at y = x and y = x + k, potentials U₁ = 0V, U₂ = 220 V, respectively.
• 7. Plates at x₁ = 12 cm, x₂ = 24 cm, potentials U₁ = 20 kV, U₂ = 8 kV, respectively.
• 8. CAS EXPERIMENT. Complex Potentials. Graph the equipotential lines and lines of force in (a)–(d) (four graphs, Re F(z) and Im F(z) on the same axes). Then explore further complex potentials of your choice with the purpose of discovering configurations that might be of practical interest.
  1. F(z) = z²
  2. F(z) = iz²
  3. F(z) = 1/z
  4. F(z) = i/z
• 9. Argument. Show that Φ = θ/π = (1/π) arctan (y/x) is harmonic in the upper half-plane and satisfies the boundary condition Φ(x, 0) = 1 if x < 0 and 0 if x > 0, and the corresponding complex potential is F(z) = −(i/π)Ln z.
• 10. Conformal mapping. Map the upper z-half-plane onto |w| 1 so that 0, ∞, −1 are mapped onto 1, i, −i, respectively. What are the boundary conditions on |w| = 1 resulting from the potential in Prob. 9? What is the potential at w = 0?
• 11. Text Example 7. Verify, by calculation, that the equipotential lines are circles.

12–15 OTHER CONFIGURATIONS

• 12. Find and sketch the potential between the axes (potential 500 V) and the hyperbola xy = 4 (potential 100 V).
• 13. Arccos. Show that F(z) = arccos z (defined in Problem Set 13.7) gives the potential of a slit in Fig. 401.

  

  Fig. 401. Slit

• 14. Arccos. Show that F(z) in Prob. 13 gives the potentials in Fig. 402.

  

  Fig. 402. Other apertures

• 15. Sector. Find the real and complex potentials in the sector −π/6 θ π/6 between the boundary θ = ±π/6, kept at 0 V, and the curve x³ − 3xy² = 1, kept at 220 V.

THEOREM 1 Harmonic Functions Under Conformal Mapping

Let Φ* be harmonic in a domain D* in the w-plane. Suppose that w = u + iv = f(z) is analytic in a domain D in the z-plane and maps D conformally onto D*. Then the function

is harmonic in D.

PROOF

The composite of analytic functions is analytic, as follows from the chain rule. Hence, taking a harmonic conjugate Ψ*(u, v) of Φ*, as defined in Sec. 13.4, and forming the analytic function F*(w) = Φ*(u, v) + iΨ*(u, v) we conclude that F(z) = F*(f(z)) is analytic in D. Hence its real part Φ(x, y) = Re F(z) is harmonic in D. This completes the proof.

We mention without proof that if D* is simply connected (Sec. 14.2), then a harmonic conjugate of Φ* exists. Another proof of Theorem 1 without the use of a harmonic conjugate is given in App. 4.

EXAMPLE 1 Potential Between Noncoaxial Cylinders

Model the electrostatic potential between the cylinders C₁: |z| = 1 and in Fig. 403. Then give the solution for the case that C₁ is grounded, U₁ = 0 V, and C₂ has the potential U₂ = 110 V.

Solution. We map the unit disk |z| = 1 onto the unit disk |w| = 1 in such a way that C₂ is mapped onto some cylinder By (3), Sec. 17.3, a linear fractional transformation mapping the unit disk onto the unit disk is

Fig. 403. Example 1: z-plane

Fig. 404. Example 1: w-plane

where we have chosen b = z₀ real without restriction. z₀ is of no immediate help here because centers of circles do not map onto centers of the images, in general. However, we now have two free constants b and r₀ and shall succeed by imposing two reasonable conditions, namely, that 0 and (Fig. 403) should be mapped onto r₀ and −r₀ (Fig. 404), respectively. This gives by (2)

a quadratic equation in r₀ with solutions r₀ = 2 (no good because r₀ < 1) and Hence our mapping function (2) with becomes that in Example 5 of Sec. 17.3,

From Example 5 in Sec. 18.1, writing w for z we have as the complex potential in the w-plane the function F*(w) = a Ln w + k and from this the real potential

This is our model. We now determine a and k from the boundary conditions. If |w| = 1, then Φ* = a ln 1 + k = 0, hence k = 0. If then Φ* = a ln hence a = 110/ln Substitution of (3) now gives the desired solution in the given domain in the z-plane

The real potential is

Can we “see” this result? Well, Φ(x, y) = const if and only if |(2z − 1)/(z − 2)| = const, that is, |w| = const by (2) with . These circles are images of circles in the z-plane because the inverse of a linear fractional transformation is linear fractional (see (4), Sec. 17.2), and any such mapping maps circles onto circles (or straight lines), by Theorem 1 in Sec. 17.2. Similarly for the rays arg w = const. Hence the equipotential lines Φ(x, y) = const are circles, and the lines of force are circular arcs (dashed in Fig. 404). These two families of curves intersect orthogonally, that is, at right angles, as shown in Fig. 404.

EXAMPLE 2 Potential Between Two Semicircular Plates

Model the potential between two semicircular plates P₁ and P₂ in Fig. 405 having potentials −3000 V and 3000 V, respectively. Use Example 3 in Sec. 18.1 and conformal mapping.

Solution. Step 1. We map the unit disk in Fig. 405 onto the right half of the w-plane (Fig. 406) by using the linear fractional transformation in Example 3, Sec. 17.3:

Fig. 405. Example 2: z-plane

Fig. 406. Example 2: w-plane

The boundary |z| = 1 is mapped onto the boundary u = 0 (the v-axis), with z = −1, i, 1 going onto w = 0, i, ∞, respectively, and z = −i onto w = −i. Hence the upper semicircle of |z| = 1 is mapped onto the upper half, and the lower semicircle onto the lower half of the v-axis, so that the boundary conditions in the w-plane are as indicated in Fig. 406.

Step 2. We determine the potential Φ*(u, v) in the right half-plane of the w-plane. Example 3 in Sec. 18.1 with α = π, U₁ = −3000, and U₂ = 3000 [with Φ*(u, v) instead of Φ(x, y)] yields

On the positive half of the imaginary axis (φ = π/2), this equals 3000 and on the negative half −3000, as it should be. Φ* is the real part of the complex potential

Step 3. We substitute the mapping function into F* to get the complex potential F(z) in Fig. 405 in the form

The real part of this is the potential we wanted to determine:

As in Example 1 we conclude that the equipotential lines Φ(x, y) = const are circular arcs because they correspond to Arg [(1 + z)/(1 − z)] = const, hence to Arg w = const. Also, Arg w = const are rays from 0 to ∞, the images of z = −1 and z = 1, respectively. Hence the equipotential lines all have −1 and 1 (the points where the boundary potential jumps) as their endpoints (Fig. 405). The lines of force are circular arcs, too, and since they must be orthogonal to the equipotential lines, their centers can be obtained as intersections of tangents to the unit circle with the x-axis, (Explain!)

PROBLEM SET 18.2

1. Derivation of (3) from (2). Verify the steps.
2. Second proof. Give the details of the steps given on p. A93 of the book. What is the point of that proof?

3–5 APPLICATION OF THEOREM 1

• 3. Find the potential Φ in the region R in the first quadrant of the z-plane bounded by the axes (having potential U₁) and the hyperbola y = 1/x (having potential U₂) by mapping R onto a suitable infinite strip. Show that Φ is harmonic. What are its boundary values?
• 4. Let Φ* = 4uv, w = f(z) = e^z, and D: x < 0, 0 < y < π. Find Φ. What are its boundary values?
• 5. CAS PROJECT. Graphing Potential Fields. Graph equipotential lines (a) in Example 1 of the text, (b) if the complex potential is F(z) = z², iz², e^z. (c) Graph the equipotential surfaces for F(z) = Ln z as cylinders in space.
• 6. Apply Theorem 1 to Φ*(u, v) = u² − v², w = f(z) = e^z, and any domain D, showing that the resulting potential Φ is harmonic.
• 7. Rectangle, sin z. Let D: , 0 y 1; D* the image of D under w = sin z; and Φ* = u² − v². What is the corresponding potential Φ in D? What are its boundary values? Sketch D and D*.
• 8. Conjugate potential. What happens in Prob. 7 if you replace the potential by its conjugate harmonic?
• 9. Translation. What happens in Prob. 7 if we replace by sin z by ?
• 10. Noncoaxial Cylinders. Find the potential between the cylinders C₁: |z| = 1 (potential U₁ = 0) and C₂: |z − c| = c (potential U₂ = 220 V), where 0 < c < . Sketch or graph equipotential lines and their orthogonal trajectories for Can you guess how the graph changes if you increase ?
• 11. On Example 2. Verify the calculations.
• 12. Show that in Example 2 the y-axis is mapped onto the unit circle in the w-plane.
• 13. At z = ±1 in Fig. 405 the tangents to the equipotential lines as shown make equal angles. Why?
• 14. Figure 405 gives the impression that the potential on the y-axis changes more rapidly near 0 than near ±i. Can you verify this?
• 15. Angular region. By applying a suitable conformal mapping, obtain from Fig. 406 the potential Φ in the sector such that Φ = −3 kV if and Φ = 3 kV if .
• 16. Solve Prob. 15 if the sector is
• 17. Another extension of Example 2. Find the linear fractional transformation z = g(Z) that maps |Z| 1 onto |z| 1 with Z = i/2 being mapped onto z = 0. Show that Z₁ = −0.6 + 0.8i is mapped onto z = −1 and Z₂ = −0.6 + 0.8i onto z = 1, so that the equipotential lines of Example 2 look in |Z| 1 as shown in Fig. 407.

  

  Fig. 407. Problem 17

• 18. The equipotential lines in Prob. 17 are circles. Why?
• 19. Jump on the boundary. Find the complex and real potentials in the upper half-plane with boundary values 5 kV if x < 2 and 0 if x > 2 on the x-axis.
• 20. Jumps. Do the same task as in Prob. 19 if the boundary values on the x-axis are V₀ when −a < x < a and 0 elsewhere.

EXAMPLE 1 Temperature Between Parallel Plates

Find the temperature between two parallel plates x = 0 and x = d in Fig. 408 having temperatures 0 and 100°C, respectively.

Solution. As in Example 1 of Sec. 18.1 we conclude that T(x, y) = ax + b. From the boundary conditions, b = 0 and a = 100/d. The answer is

The corresponding complex potential is F(z) = (100/d)z. Heat flows horizontally in the negative x-direction along the lines y = const.

EXAMPLE 2 Temperature Distribution Between a Wire and a Cylinder

Find the temperature field around a long thin wire of radius r₁ = 1 mm that is electrically heated to T₁ = 500°F and is surrounded by a circular cylinder of radius r₂ = 100 mm, which is kept at temperature T₂ = 60°F by cooling it with air. See Fig. 409. (The wire is at the origin of the coordinate system.)

Solution.T depends only on r, for reasons of symmetry. Hence, as in Sec. 18.1 (Example 2),

The boundary conditions are

Hence b = 500 (since ln 1 = 0) and a = (60 − b)/ln 100 = −95.54. The answer is

The isotherms are concentric circles. Heat flows from the wire radially outward to the cylinder. Sketch T as a function of r. Does it look physically reasonable?

EXAMPLE 3 A Mixed Boundary Value Problem

Find the temperature distribution in the region in Fig. 410 (cross section of a solid quarter-cylinder), whose vertical portion of the boundary is at 20°C, the horizontal portion at 50°C, and the circular portion is insulated.

Solution. The insulated portion of the boundary must be a heat flow line, since, by the insulation, heat is prevented from crossing such a curve, hence heat must flow along the curve. Thus the isotherms must meet such a curve at right angles. Since T is constant along an isotherm, this means that

Here ∂T/∂n is the normal derivative of T, that is, the directional derivative (Sec. 9.7) in the direction normal (perpendicular) to the insulated boundary. Such a problem in which T is prescribed on one portion of the boundary and ∂T/∂n on the other portion is called a mixed boundary value problem.

In our case, the normal direction to the insulated circular boundary curve is the radial direction toward the origin. Hence (2) becomes ∂T/∂r = 0, meaning that along this curve the solution must not depend on r. Now Arg z = θ satisfies (1), as well as this condition, and is constant (0 and π/2) on the straight portions of the boundary. Hence the solution is of the form

The boundary conditions yield a · π/2 + b = 20 and a · 0 + b = 50. This gives

The isotherms are portions of rays θ = const. Heat flows from the x-axis along circles r = const (dashed in Fig. 410) to the y-axis.

EXAMPLE 4 Another Mixed Boundary Value Problem in Heat Conduction

Find the temperature field in the upper half-plane when the x-axis is kept at T = 0°C for x < −1, is insulated for −1 < x < 1, and is kept at T = 20°C for x > 1 (Fig. 411).

Solution. We map the half-plane in Fig. 411 onto the vertical strip in Fig. 412, find the temperature T*(u, v) there, and map it back to get the temperature T(x, y) in the half-plane.

The idea of using that strip is suggested by Fig. 391 in Sec. 17.4 with the roles of z = x + iy and w = u + iv interchanged. The figure shows that z = sin w maps our present strip onto our half-plane in Fig. 411. Hence the inverse function

maps that half-plane onto the strip in the w-plane. This is the mapping function that we need according to Theorem 1 in Sec. 18.2.

The insulated segment −1 < x < 1 on the x-axis maps onto the segment −π/2 < u < π/2 on the u-axis. The rest of the x-axis maps onto the two vertical boundary portions u = −π/2 and π/2, v > 0, of the strip. This gives the transformed boundary conditions in Fig. 412 for T*(u, v), where on the insulated horizontal boundary, ∂T*/∂n = ∂T*/∂v = 0 because v is a coordinate normal to that segment.

Similarly to Example 1 we obtain

which satisfies all the boundary conditions. This is the real part of the complex potential F*(w) = 10 + (20/π)w. Hence the complex potential in the z-plane is

and T(x, y) = Re F(z) is the solution. The isotherms are u = const in the strip and the hyperbolas in the z-plane, perpendicular to which heat flows along the dashed ellipses from the 20°-portion to the cooler 0°-portion of the boundary, a physically very reasonable result.

PROBLEM SET 18.3

1. Parallel plates. Find the temperature between the plates y = 0 and y = d kept at 20 and 100°C, respectively. (i) Proceed directly. (ii) Use Example 1 and a suitable mapping.
2. Infinite plate. Find the temperature and the complex potential in an infinite plate with edges y = x − 4 and y = x + 4 kept at −20 and 40°C, respectively (Fig. 413). In what case will this be an approximate model?

   

   Fig. 413. Problem 2: Infinite plate

3. CAS PROJECT. Isotherms. Graph isotherms and lines of heat flow in Examples 2–4. Can you see from the graphs where the heat flow is very rapid?

4–18 TEMPERATURE T(x, y) IN PLATES

Find the temperature distribution T(x, y) and the complex potential F(z) in the given thin metal plate whose faces are insulated and whose edges are kept at the indicated temperatures or are insulated as shown.

• 4.
• 5.
• 6.
• 7.
• 8.
• 9.
• 10.
• 11.
• 12. Hint. Apply w = cosh z to Prob. 11.
• 13.
• 14.
• 15.
• 16.
• 17. First quadrant of the z-plane with y-axis kept at 100°C, the segment 0 < x < 1 of the x-axis insulated and the x-axis for x > 1 kept at 200°C. Hint. Use Example 4.
• 18. Figure 410, T(0, y) = −30°C, T(x, 0) = 100°C
• 19. Interpretation. Formulate Prob. 11 in terms of electrostatics.
• 20. Interpretation. Interpret Prob. 17 in Sec. 18.2 as a heat problem, with boundary temperatures, say, 10°C on the upper part and 200°C on the lower.

EXAMPLE 1 Flow Around a Corner

The complex potential F(z) = z² = x² − y² + 2ixy models a flow with

From (3) we obtain the velocity vector

The speed (magnitude of the velocity) is

The flow may be interpreted as the flow in a channel bounded by the positive coordinates axes and a hyperbola, say, xy = 1 (Fig. 415). We note that the speed along a streamline S has a minimum at the point P where the cross section of the channel is large.

EXAMPLE 2 Flow Around a Cylinder

Consider the complex potential

Using the polar form z = re^iθ, we obtain

Hence the streamlines are

In particular, Ψ(x, y) = 0 gives r − 1/r = 0 or sin θ = 0. Hence this streamline consists of the unit circle (r = 1/r gives r = 1) and the x-axis (θ = 0 and θ = π). For large |z| the term 1/z in F(z) is small in absolute value, so that for these z the flow is nearly uniform and parallel to the x-axis. Hence we can interpret this as a flow around a long circular cylinder of unit radius that is perpendicular to the z-plane and intersects it in the unit circle |z| = 1 and whose axis corresponds to z = 0.

The flow has two stagnation points (that is, points at which the velocity V is zero), at z = ±1. This follows from (3) and

THEOREM 1 Complex Potential of a Flow

If the domain of flow is simply connected and the flow is irrotational and incompressible, then the statements involving (2)–(5) hold. In particular, then the flow has a complex potential F(z), which is an analytic function. (Explanation of terms below.)

PROOF

We prove this theorem, along with a discussion of basic concepts related to fluid flow.

(a) First Assumption: Irrotational. Let C be any smooth curve in the z-plane given by z(s) = x(s) + iy(s), where s is the arc length of C. Let the real variable V_t be the component of the velocity V tangent to C (Fig. 417). Then the value of the real line integral

Fig. 417. Tangential component of the velocity with respect to a curve C

taken along C in the sense of increasing s is called the circulation of the fluid along C, a name that will be motivated as we proceed in this proof. Dividing the circulation by the length of C, we obtain the mean velocity¹ of the flow along the curve C. Now

Hence V_t is the dot product (Sec. 9.2) of V and the tangent vector dz/ds of C (Sec. 17.1); thus in (6),

The circulation (6) along C now becomes

As the next idea, let C be a closed curve satisfying the assumption as in Green's theorem (Sec. 10.4), and let C be the boundary of a simply connected domain D. Suppose further that V has continuous partial derivatives in a domain containing D and C. Then we can use Green's theorem to represent the circulation around C by a double integral,

The integrand of this double integral is called the vorticity of the flow. The vorticity divided by 2 is called the rotation

We assume the flow to be irrotational, that is, ω(x, y) ≡ 0 throughout the flow; thus,

To understand the physical meaning of vorticity and rotation, take for C in (8) a circle. Let r be the radius of C. Then the circulation divided by the length 2πr of C is the mean

velocity of the fluid along C. Hence by dividing this by r we obtain the mean angular velocity ω₀ of the fluid about the center of the circle:

If we now let r → 0, the limit of ω₀ is the value of ω at the center of C. Hence ω(x, y) is the limiting angular velocity of a circular element of the fluid as the circle shrinks to the point (x, y). Roughly speaking, if a spherical element of the fluid were suddenly solidified and the surrounding fluid simultaneously annihilated, the element would rotate with the angular velocity ω.

(b) Second Assumption: Incompressible. Our second assumption is that the fluid is incompressible. (Fluids include liquids, which are incompressible, and gases, such as air, which are compressible.) Then

in every region that is free of sources or sinks, that is, points at which fluid is produced or disappears, respectively. The expression in (11) is called the divergence of V and is denoted by div V. (See also (7) in Sec. 9.8.)

(c) Complex Velocity Potential. If the domain D of the flow is simply connected (Sec. 14.2) and the flow is irrotational, then (10) implies that the line integral (7) is independent of path in D (by Theorem 3 in Sec. 10.2, where F₁ = V₁, F₂ = V₂, F₃ = 0, and z is the third coordinate in space and has nothing to do with our present z). Hence if we integrate from a fixed point (a, b) in D to a variable point (x, y) in D, the integral becomes a function of the point (x, y), say, Φ(x, y):

We claim that the flow has a velocity potential Φ, which is given by (12). To prove this, all we have to do is to show that (4) holds. Now since the integral (7) is independent of path, V₁ dx + V₂ dy is exact (Sec. 10.2), namely, the differential of Φ, that is,

From this we see that V₁ = ∂Φ/∂x and V₂ = ∂Φ/∂y, which gives (4).

That Φ is harmonic follows at once by substituting (4) into (11), which gives the first Laplace equation in (5).

We finally take a harmonic conjugate Ψ of Φ. Then the other equation in (5) holds. Also, since the second partial derivatives of Φ and Ψ are continuous, we see that the complex function

is analytic in D. Since the curves Ψ(x, y) = const are perpendicular to the equipotential curves Φ(x, y) = const (except where F′(z) = 0), we conclude that Ψ(x, y) = const are the streamlines. Hence Ψ is the stream function and F(z) is the complex potential of the flow. This completes the proof of Theorem 1 as well as our discussion of the important role of complex analysis in compressible fluid flow.

PROBLEM SET 18.4

1. Differentiability. Under what condition on the velocity vector V in (1) will F(z) in (2) be analytic?
2. Corner flow. Along what curves will the speed in Example 1 be constant? Is this obvious from Fig. 415?
3. Cylinder. Guess from physics and from Fig. 416 where on the y-axis the speed is maximum. Then calculate.
4. Cylinder. Calculate the speed along the cylinder wall in Fig. 416, also confirming the answer to Prob. 3.
5. Irrotational flow. Show that the flow in Example 2 is irrotational.
6. Extension of Example 1. Sketch or graph and interpret the flow in Example 1 on the whole upper half-plane.
7. Parallel flow. Sketch and interpret the flow with complex potential F(z) = z.
8. Parallel flow. What is the complex potential of an upward parallel flow of speed K > 0 in the direction of y = x? Sketch the flow.
9. Corner. What F(z) would be suitable in Example 1 if the angle of the corner were π/4 instead of π/2?
10. Corner. Show that F(z) = iz² also models a flow around a corner. Sketch streamlines and equipotential lines. Find V.
11. What flow do you obtain from F(z) = −iKz, K positive real?
12. Conformal mapping. Obtain the flow in Example 1 from that in Prob. 11 by a suitable conformal mapping.
13. 60°- Sector. What F(z) would be suitable in Example 1 if the angle at the corner were π/3?
14. Sketch or graph streamlines and equipotential lines of F(z) = iz³. Find V. Find all points at which V is horizontal.
15. Change F(z) in Example 2 slightly to obtain a flow around a cylinder of radius r⁰ that gives the flow in Example 2 if r₀ → 1.
16. Cylinder. What happens in Example 2 if you replace z by z²? Sketch and interpret the resulting flow in the first quadrant.
17. Elliptic cylinder. Show that F(z) = arccos z gives confocal ellipses as streamlines, with foci at z = ±1, and that the flow circulates around an elliptic cylinder or a plate (the segment from −1 to 1 in Fig. 418).

    

    Fig. 418. Flow around a plate in Prob. 17.

18. Aperture. Show that F(z) = arccosh z gives confocal hyperbolas as streamlines, with foci at z = ±1, and the flow may be interpreted as a flow through an aperture (Fig. 419).

    

    Fig. 419. Flow through an aperture in Prob. 18.

19. Potential F(z) = 1/z. Show that the streamlines of F(z) = 1/z and circles through the origin with centers on the y-axis.
20. TEAM PROJECT. Role of the Natural Logarithm in Modeling Flows. (a) Basic flows: Source and sink. Show that F(z) = (c/2π) ln z with constant positive real c gives a flow directed radially outward (Fig. 420), so that F models a point source at z = 0 (that is, a source line x = 0, y = 0 in space) at which fluid is produced. c is called the strength or discharge of the source. If c is negative real, show that the flow is directed radially inward, so that F models a sink at z = 0, a point at which fluid disappears. Note that z = 0 is the singular point of F(z).

    

    Fig. 420. Point source

    • (b) Basic flows: Vortex. Show that F(z) = −(Ki/2π) ln z with positive real K gives a flow circulating counterclockwise around z = 0 (Fig. 421). z = 0 is called a vortex. Note that each time we travel around the vortex, the potential increases by K.
    • (c) Addition of flows. Show that addition of the velocity vectors of two flows gives a flow whose complex potential is obtained by adding the complex potentials of those flows.

      

      Fig. 421. Vortex flow

    • (d) Source and sink combined. Find the complex potentials of a flow with a source of strength 1 at z = −a and of a flow with a sink of strength 1 at z = a. Add both and sketch or graph the streamlines. Show that for small |a| these lines look similar to those in Prob. 19.
    • (e) Flow with circulation around a cylinder. Add the potential in (b) to that in Example 2. Show that this gives a flow for which the cylinder wall |z| = 1 is a streamline. Find the speed and show that the stagnation points are

      

      if K = 0 they are at ±1; as K increases they move up on the unit circle until they unite at z = i(K = 4π, see Fig. 422), and if K > 4π they lie on the imaginary axis (one lies in the field of flow and the other one lies inside the cylinder and has no physical meaning).

      

      Fig. 422. Flow around a cylinder without circulation (K = 0) and with circulation

EXAMPLE 1 Dirichlet Problem for the Unit Disk

Find the electrostatic potential Φ(r, θ) in the unit disk r < 1 having the boundary values

Solution. Since Φ(1, α) is even, b_n = 0, and from (8) we obtain and

Hence, a_n = −4/(n²π²) if n is odd, a_n = 0 if n = 2, 4, …, and the potential is

Figure 424 shows the unit disk and some of the equipotential lines (curves Φ = const).

PROBLEM SET 18.5

1. Give the details of the derivation of the series (7) from the Poisson formula (5).
2. Verify (3).
3. Show that each term of (7) is a harmonic function in the disk r < R.
4. Why does the series in Example 1 reduce to a cosine series?

5–18 HARMONIC FUNCTIONS IN A DISK

Using (7), find the potential Φ(r, θ) in the unit disk r < 1 having the given boundary values Φ(1, θ). Using the sum of the first few terms of the series, compute some values of Φ and sketch a figure of the equipotential lines.

• 5.
• 6. Φ(1, θ) = 5 − cos 2θ
• 7. Φ(1, θ) = a cos² 4θ
• 8. Φ(1, θ) = 4 sin³ θ
• 9. Φ(1, θ) = 8 sin⁴ θ
• 10. Φ(1, θ) = 16 cos³2θ
• 11. Φ(1, θ) = θ/π if −π < θ < π
• 12. Φ(1, θ) = k if 0 < θ < π and 0 otherwise
• 13. Φ(1, θ) = θ if and 0 otherwise
• 14. Φ(1, θ) = |θ|/π if −π < θ < π
• 15. Φ(1, θ) = 1 if and 0 otherwise
• 16.
• 17. Φ(1, θ) = θ²/π² if −π < θ < π
• 18.
• 19. CAS EXPERIMENT. Series (7). Write a program for series developments (7). Experiment on accuracy by computing values from partial sums and comparing them with values that you obtain from your CAS graph. Do this (a) for Example 1 and Fig. 424, (b) for Φ in Prob. 11 (which is discontinuous on the boundary!), (c) for a Φ of your choice with continuous boundary values, and (d) for Φ with discontinuous boundary values.
• 20. TEAM PROJECT. Potential in a Disk. (a) Mean value property. Show that the value of a harmonic function Φ at the center of a circle C equals the mean of the value of Φ on C (see Sec. 18.4, footnote 1, for definitions of mean values).
  • (b) Separation of variables. Show that the terms of (7) appear as solutions in separating the Laplace equation in polar coordinates.
  • (c) Harmonic conjugate. Find a series for a harmonic conjugate Ψ of Φ from (7). Hint. Use the Cauchy–Riemann equations.
  • (d) Power series. Find a series for F(z) = Φ + iΨ.

THEOREM 1 Mean Value Property of Analytic Functions

Let F(z) be analytic in a simply connected domain D. Then the value of F(z) at a point in D is equal to the mean value of F(z) on any circle in D with center at z0.

PROOF

In Cauchy's integral formula (Sec. 14.3)

we choose for C the circle z = z₀ + re^iα in D. Then z − z₀ = re^iα, dz = ire^iα dα, and (1) becomes

The right side is the mean value of F on the circle (= value of the integral divided by the length 2π of the interval of integration). This proves the theorem.

THEOREM 2 Two Mean Value Properties of Harmonic Functions

Let Φ(x, y) be harmonic in a simply connected domain D. Then the value of Φ(x, y) at a point (x₀, y₀) in D is equal to the mean value of Φ(x, y) on any circle in D with center at (x₀, y₀). This value is also equal to the mean value of Φ(x, y) on any circular disk in D with center (x₀, y₀). [See footnote 1 in Sec. 18.4.]

PROOF

The first part of the theorem follows from (2) by taking the real parts on both sides,

The second part of the theorem follows by integrating this formula over r from 0 to r₀ (the radius of the disk) and dividing by

The right side is the indicated mean value (integral divided by the area of the region of integration).

THEOREM 3 Maximum Modulus Theorem for Analytic Functions

Let F(z) be analytic and nonconstant in a domain containing a bounded region R and its boundary. Then the absolute value |F(z)| cannot have a maximum at an interior point of R. Consequently, the maximum of |F(z)| is taken on the boundary of R. If F(z) ≠ 0 in R, the same is true with respect to the minimum of |F(z)|.

PROOF

We assume that |F(z)| has a maximum at an interior point z₀ of R and show that this leads to a contradiction. Let |F(z₀)| = M be this maximum. Since F(z) is not constant, |F(z)| is not constant, as follows from Example 3 in Sec. 13.4. Consequently, we can find a circle C of radius r with center at z₀ such that the interior of C is in R and |F(z)| smaller than M at some point P of C. Since |F(z)| is continuous, it will be smaller than M on an arc C₁ of C that contains P (see Fig. 425), say,

Let C₁ have the length L₁. Then the complementary arc C₂ of C has the length 2πr − L₁. We now apply the ML-inequality (Sec. 14.1) to (1) and note that |z − z₀| = r. We then obtain (using straightforward calculation in the second line of the formula)

that is, M < M, which is impossible. Hence our assumption is false and the first statement is proved.

Next we prove the second statement. If F(z) ≠ 0 in R, then 1/F(z) is analytic in R. From the statement already proved it follows that the maximum of 1/|F(z)| lies on the boundary of R. But this maximum corresponds to the minimum of |F(z)|. This completes the proof.

THEOREM 4 Harmonic Functions

Let Φ(x, y) be harmonic in a domain containing a simply connected bounded region R and its boundary curve C. Then:

(I) (Maximum principle) If Φ(x, y) is not constant, it has neither a maximum nor a minimum in R. Consequently, the maximum and the minimum are taken on the boundary of R.

(II) If Φ(x, y) is constant on C, then Φ(x, y) is a constant.

(III) If h(x, y) is harmonic in R and on C and if h(x, y) = Φ(x, y) on C, then h(x, y) = Φ(x, y) everywhere in R.

PROOF

(I) Let Ψ(x, y) be a conjugate harmonic function of Φ(x, y) in R. Then the complex function F(z) = Φ(x, y) + iΦ(x, y) is analytic in R, and so is G(z) = e^F(z). Its absolute value is

From Theorem 3 it follows that |G(z)| cannot have a maximum at an interior point of R. Since e^Φ is a monotone increasing function of the real variable Φ, the statement about the maximum of Φ follows. From this, the statement about the minimum follows by replacing Φ by −Φ.

(II) By (I) the function Φ(x, y) takes its maximum and its minimum on C. Thus, if Φ(x, y) is constant on C, its minimum must equal its maximum, so that Φ(x, y) must be a constant.

(III) If h and Φ are harmonic in R and on C, then h − Φ is also harmonic in R and on C, and by assumption, h − Φ = 0 everywhere on C. By (II) we thus have h − Φ = 0 everywhere in R, and (III) is proved.

THEOREM 5 Uniqueness Theorem for the Dirichlet Problem

If for a given region and given boundary values the Dirichlet problem for the Laplace equation in two variables has a solution, the solution is unique.

PROBLEM SET 18.6

PROBLEMS RELATED TO THEOREMS 1 AND 2

1–4 Verify Theorem 1 for the given F(z), z₀, and circle of radius 1.

1. (z + 1)³,
2. 2z⁴, z₀ = −2
3. (3z − 2)², z₀ = 4
4. (z − 1)⁻², z₀ = −1
5. Integrate |z| around the unit circle. Does the result contradict Theorem 1?
6. Derive the first statement in Theorem 2 from Poisson's integral formula.

7–9 Verify (3) in Theorem 2 for the given Φ(x, y), (x₀, y₀), and circle of radius 1.

• 7. (x − 1)(y − 1), (2, −2)
• 8. x² − y², (3, 8)
• 9. x + y + xy, (1, 1)
• 10. Verify the calculations involving the inequalities in the proof of Theorem 3.
• 11. CAS EXPERIMENT. Graphing Potentials. Graph the potentials in Probs. 7 and 9 and for two other functions of your choice as surfaces over a rectangle in the xy-plane. Find the locations of the maxima and minima by inspecting these graphs.
• 12. TEAM PROJECT. Maximum Modulus of Analytic Functions. (a) Verify Theorem 3 for (i) F(z) = z² and the rectangle 1 x 5, 2 y 4, (ii) F(z) = sin z and the unit disk, and (iii) F(z) = e_z and any bounded domain.
  • (b) F(z) = 1 + |z| is not zero in the disk |z| 2 and has a minimum at an interior point. Does this contradict Theorem 3?
  • (c) F(x) = sin x (x real) has a maximum 1 at π/2. Why can this not be a maximum of |F(z)| = |sin z| in a domain containing z = π/2?
  • (d) If F(z) is analytic and not constant in the closed unit disk D: |z| 1 and |F(z)| = c = const on the unit circle, show that F(z) must have a zero in D.

13–17 MAXIMUM MODULUS

Find the location and size of the maximum of |F(z)| in the unit disk |z| 1.

• 13. F(z) = cos z
• 14. F(z) = exp z²
• 15. F(z) = sinh 2z
• 16. F(z) = az + b (a, b complex, a ≠ 0)
• 17. F(z) = 2z² − 2
• 18. Verify the maximum principle for Φ(x, y) = e^x sin y and the rectangle a x b, 0 y 2π.
• 19. Harmonic conjugate. Do Φ and a harmonic conjugate Ψ in a region R have their maximum at the same point of R?
• 20. Conformal mapping. Find the location (u₁, v₁) of the maximum of Φ* = e^u cos v in R*: |w| 1, v 0, where w = u + iv. Find the region R that is mapped onto R* by w = f(z) = z². Find the potential in R resulting from Ψ and the location (x₁, y₁) of the maximum. Is (u₁, v₁) the image of (x₁, y₁)? If so, is this just by chance?