Case III. Complex conjugate roots are of minor practical importance, and we discuss the derivation of real solutions from complex ones just in terms of a typical example
Step 3. Solution of the Entire Problem
18.2 Use of Conformal Mapping. Modeling
We have just explored the close relation between potential theory and complex analysis.
This relationship is so close because complex potentials can be modeled in complex analysis. In this section we shall explore the close relation that results from the use of conformal mapping in modeling and solving boundary value problems for the Laplace equation. The process consists of finding a solution of the equation in some domain, assuming given values on the boundary (Dirichlet problem, see also Sec. 12.6). The key idea is then to use conformal mapping to map a given domain onto one for which the solution is known or can be found more easily. This solution thus obtained is then mapped back to the given domain. The reason this approach works is due to Theorem 1, which asserts that harmonic functions remain harmonic under conformal mapping:
T H E O R E M 1 Harmonic Functions Under Conformal Mapping
Let be harmonic in a domain in the w-plane. Suppose that
is analytic in a domain D in the z-plane and maps D conformally onto Then the function
(1)
is harmonic in D.
P R O O F The composite of analytic functions is analytic, as follows from the chain rule. Hence, taking a harmonic conjugate of as defined in Sec. 13.4, and forming the analytic
function we conclude that is analytic in D.
Hence its real part is harmonic in D. This completes the proof.
We mention without proof that if is simply connected (Sec. 14.2), then a harmonic conjugate of exists. Another proof of Theorem 1 without the use of a harmonic
conjugate is given in App. 4. 䊏
£*
D*
£(x, y)⫽Re F(z)
F(z)⫽F*( f(z)) F*(w)⫽ £*(u, v)⫹i°*(u, v)
£*,
°*(u, v)
£(x, y)⫽ £*(u(x, y), v(x, y))
D*.
w⫽u⫹iv⫽f(z) D*
£*
13. Arccos. Show that (defined in Problem Set 13.7) gives the potential of a slit in Fig. 401.
F(z)⫽arccos z
Fig. 401. Slit
y
1 x –1
Fig. 402. Other apertures
y
x x
y
1 –1 1
E X A M P L E 1 Potential Between Noncoaxial Cylinders
Model the electrostatic potential between the cylinders and in Fig. 403. Then give the solution for the case that is grounded, and has the potential
Solution. We map the unit disk onto the unit disk in such a way that is mapped onto some cylinder By (3), Sec. 17.3, a linear fractional transformation mapping the unit disk onto the unit disk is
(2) w⫽ z⫺b
bz⫺1 C2*: ƒwƒ⫽r0.
C2
ƒwƒ⫽1 ƒzƒ⫽1
U2⫽110 V.
C2 U1⫽0 V, C1
C2: ƒz⫺25ƒ⫽25
C1: ƒzƒ⫽1
C1 U1 = 0
U2 = 110 V y
C2 x
Fig. 403. Example 1: z-plane
v
u U1 = 0 U2 = 110 V r0
Fig. 404. Example 1: w-plane
where we have chosen real without restriction. 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 band and shall succeed by imposing two reasonable conditions, namely, that 0 and (Fig. 403) should be mapped onto and
(Fig. 404), respectively. This gives by (2) and with this,
a quadratic equation in with solutions (no good because and Hence our mapping function (2) with becomes that in Example 5 of Sec. 17.3,
(3)
From Example 5 in Sec. 18.1, writing wfor zwe have as the complex potential in the w-plane the function and from this the real potential
This is our model. We now determine aand kfrom the boundary conditions. If then
hence . If then hence 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, if and only if that is,
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 Hence the equipotential lines 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. 䊏
£(x, y)⫽const
arg w⫽const.
b⫽12.
ƒwƒ⫽const ƒ(2z⫺1)>(z⫺2)ƒ⫽const,
£(x, y)⫽const
a⫽ ⫺158.7.
£(x, y)⫽Re F(z)⫽a ln `2z⫺1 z⫺2 `, F(z)⫽F*( f(z))⫽a Ln 2z⫺1
z⫺2 .
a⫽110>ln (12)⫽ ⫺158.7.
£*⫽a ln (12)⫽110, ƒwƒ⫽r0⫽12,
k⫽0
£*⫽a ln 1⫹k⫽0, ƒwƒ⫽1,
£*(u, v)⫽Re F*(w)⫽a ln ƒwƒ⫹k. F*(w)⫽a Ln w⫹k
w⫽f(z)⫽2z⫺1 z⫺2 . b⫽12
r0⫽12. r0⬍1)
r0⫽2 r0
⫺r0⫽ 45⫺b
4b>5⫺1⫽ 45⫺r0
4r0>5⫺1 , r0⫽0⫺b
0⫺1⫽b,
⫺r0
r0 4
5
r0
z0
b⫽z0
E X A M P L E 2 Potential Between Two Semicircular Plates
Model the potential between two semicircular plates and in Fig. 405 having potentials 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:
w⫽f(z)⫽1⫹z 1⫺z.
⫺3000 V P2
P1
y
x P2: 3 kV
P1: –3 kV 2
1 0 –1 –2 –3
Fig. 405. Example 2: z-plane
v
u 0 3 kV
–3 kV
–2 kV
–1 kV 1 kV 2 kV
Fig. 406. Example 2: w-plane
The boundary is mapped onto the boundary (the v-axis), with going onto
respectively, and onto Hence the upper semicircle of 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 in the right half-plane of the w-plane. Example 3 in Sec. 18.1 with
and [with instead of yields
On the positive half of the imaginary axis this equals 3000 and on the negative half as it should be. is the real part of the complex potential
Step 3. We substitute the mapping function into to get the complex potential 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 are circular arcs because they
correspond to hence to Also, are rays from 0
to the images of and respectively. Hence the equipotential lines all have 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!)
Further examples can easily be constructed. Just take any mapping in Chap. 17, a domain Din the z-plane, its image in the w-plane, and a potential in Then (1) gives a potential in D. Make up some examples of your own, involving, for instance, linear fractional transformations.
D*.
£* D*
w⫽f(z)
䊏
⫺1 z⫽1,
z⫽ ⫺1
⬁,
Arg w⫽const Arg w⫽const.
Arg [(1⫹z)>(1⫺z)]⫽const,
£(x, y)⫽const
£(x, y)⫽Re F(z)⫽6000 p Im Ln
1⫹z 1⫺z⫽6000
p Arg 1⫹z 1⫺z . F(z)⫽F*( f(z))⫽ ⫺6000 i
p Ln 1⫹z 1⫺z .
F(z) F*
F*(w)⫽ ⫺6000 i p Ln w.
£*
⫺3000,
( ⫽p>2),
⫽arctan v u .
£*(u, v)⫽6000 p ,
£(x, y)]
£*(u, v) U2⫽3000
a⫽p, U1⫽ ⫺3000,
£*(u, v)
ƒzƒ⫽1 w⫽ ⫺i.
z⫽ ⫺i
w⫽0, i, ⬁, z⫽ ⫺1, i, 1
u⫽0 ƒzƒ⫽1
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 in the region Rin the first quadrant of the z-plane bounded by the axes (having potential
) and the hyperbola (having potential by mapping Ronto a suitable infinite strip. Show that
is harmonic. What are its boundary values?
4. Let and
Find 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
(c)Graph the equipotential surfaces for as cylinders in space.
6. Apply Theorem 1 to
and any domain D, showing that the resulting potential is harmonic.
7. Rectangle, Let D:
the image of Dunder and
What is the corresponding potential in D? What are its boundary values? Sketch Dand
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
10. Noncoaxial Cylinders. Find the potential between
the cylinders (potential and
(potential where Sketch or graph equipotential lines and their orthogonal trajectories for Can you guess how the graph changes if you increase c (⬍12)?
c⫽14. 0⬍c⬍12.
U2⫽220 V), C2: ƒz⫺cƒ ⫽c
U1⫽0) C1: ƒzƒ⫽1
cos z⫽sin (z⫹12p)?
sin z
D*.
£
£*⫽u2⫺v2. w⫽sin z;
0⬉x⬉12p, 0⬉y⬉1; D*
sin z.
£ f(z)⫽ez,
w⫽
£*(u, v)⫽u2⫺v2, F(z)⫽Ln z F(z)⫽z2, iz2, ez.
£. 0⬍y⬍p.
D: x⬍0, w⫽f(z)⫽ez,
£*⫽4uv,
£
U2) y⫽1>x
U1
£
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 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 Can you verify this?
15. Angular region. By applying a suitable conformal mapping, obtain from Fig. 406 the potential in the
sector such that if
and if 16. Solve Prob. 15 if the sector is
17. Another extension of Example 2. Find the linear fractional transformation that maps onto with being mapped onto
Show that is mapped onto
and onto so that the
equipotential lines of Example 2 look in as shown in Fig. 407.
ƒZƒ ⬉1 z⫽1,
Z2⫽ ⫺0.6⫹0.8i
z⫽ ⫺1 Z1⫽0.6⫹0.8i
z⫽0.
Z⫽i>2 ƒzƒ ⬉1
ƒZƒ ⬉1 z⫽g(Z)
⫺18p⬍Arg z⬍18p. Arg z⫽14p.
£⫽3 kV
Arg z⫽ ⫺14p
£⫽ ⫺3 kV
⫺14p⬍Arg z⬍14p
£
⫾i.
z⫽ ⫾1
P R O B L E M S E T 1 8 . 2
Basic Comment on Modeling
We formulated the examples in this section as models on the electrostatic potential. It is quite important to realize that this is accidental. We could equally well have phrased everything in terms of (time-independent) heat flow; then instead of voltages we would have had temperatures, the equipotential lines would have become isotherms ( lines of constant temperature), and the lines of the electrical force would have become lines along which heat flows from higher to lower temperatures (more on this in the next section).
Or we could have talked about fluid flow; then the electrostatic lines of force would have become streamlines (more on this in Sec. 18.4). What we again see here is the unifying power of mathematics: different phenomena and systems from different areas in physics having the same types of model can be treated by the same mathematical methods. What differs from area to area is just the kinds of problems that are of practical interest.
⫽
Y
X
Z2 Z1
–3 kV
3 kV 2 1 0
Fig. 407. Problem 17