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
18.5 Poisson’s Integral Formula for Potentials
So far in this chapter we have seen powerful methods based on conformal mappings and complex potentials. They were used for modeling and solving two-dimensional potential problems and demonstrated the importance of complex analysis.
Now we introduce a further method that results from complex integration. It will yield the very important Poisson integral formula (5) for potentials in a standard domain if they are at as Kincreases they move up on the unit circle until they unite at see Fig. 422), and if 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).
K⬎4p
z⫽i (K⫽4p,
⫾1;
K⫽0
z⫽ iK 4p ⫾ B
⫺K2 16p2⫹1;
(d) Source and sink combined. Find the complex potentials of a flow with a source of strength 1 at and of a flow with a sink of strength 1 at Add both and sketch or graph the streamlines. Show that for small 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 is a streamline. Find the speed and show that the stagnation points are
ƒzƒ ⫽1 ƒaƒ
z⫽a.
z⫽ ⫺a (b) Basic flows: Vortex. Show that
ln zwith positive real Kgives a flow circulating coun- terclockwise around (Fig. 421). 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.
z⫽0 z⫽0
F(z)⫽ ⫺(Ki>2p) Fig. 420. Point source
y
x
Fig. 422. Flow around a cylinder without circulation (Kⴝ0) and with circulation
K = 0
K = 2.8π K = 4π
K = 6π
Fig. 421. Vortex flow
y
x
(a circular disk). In addition, from (5), we will derive a useful series (7) for these potentials.
This allows us to solve problems for disks and then map solutions conformally onto other domains.
Derivation of Poisson’s Integral Formula
Poisson’s formula will follow from Cauchy’s integral formula (Sec. 14.3) (1)
Here Cis the circle (counterclockwise, and we assume that is analytic in a domain containing Cand its full interior. Since
we obtain from (1) (2)
Now comes a little trick. If instead of zinside Cwe take a Zoutside C, the integrals (1) and (2) are zero by Cauchy’s integral theorem (Sec. 14.2). We choose
which is outside Cbecause From (2) we thus have
and by straightforward simplification of the last expression on the right,
We subtract this from (2) and use the following formula that you can verify by direct calculation cancels):
(3)
We then have (4)
From the polar representations of zand we see that the quotient in the integrand is real and equal to
R2⫺r2
(Reia⫺reiu)(Reⴚia⫺reⴚiu) ⫽ R2⫺r2
R2⫺2Rr cos (u⫺a)⫹r2 . z*
F(z)⫽ 1
2p冮02pF(z*) (z*z*z*⫺z)(⫺z*zz⫺z) da.
z*
z*⫺z ⫺ z
z⫺z*⫽ z*z*⫺zz (z*⫺z)(z*⫺z). (zz*
0⫽ 1
2p 冮02pF(z*) z⫺zz* da.
0⫽ 1
2p冮02pF(z*) z*z*⫺Z da⫽ 1
2p冮02pF(z*) z*
z*⫺z*z*
z da ƒZƒ ⫽R2>ƒzƒ ⫽R2>r⬎R.
Z⫽z*z*>z⫽R2>z, (z*⫽Reia, z⫽reiu).
F(z)⫽ 1
2p冮02pF(z*) z*z*⫺z da
dz*⫽iReia da⫽iz* da, F(z*) 0⬉a⬉2p),
z*⫽Reia
F(z)⫽ 1
2pi 冯C z*F(z*)⫺z dz*.
We now write and take the real part on both sides of (4). Then we obtain Poisson’s integral formula2
(5)
This formula represents the harmonic function in the disk in terms of its values on the boundary (the circle)
Formula (5) is still valid if the boundary function is merely piecewise continuous (as is practically often the case; see Figs. 405 and 406 in Sec. 18.2 for an example). Then (5) gives a function harmonic in the open disk, and on the circle equal to the given boundary function, except at points where the latter is discontinuous.
A proof can be found in Ref. [D1] in App. 1.
Series for Potentials in Disks
From (5) we may obtain an important series development of in terms of simple harmonic functions. We remember that the quotient in the integrand of (5) was derived from (3).
We claim that the right side of (3) is the real part of
Indeed, the last denominator is real and so is in the numerator, whereas in the numerator is pure imaginary. This verifies our claim.
Now by the use of the geometric series we obtain (develop the denominator) (6)
Since and we have
On the right, Hence from (6) we obtain
⫽1⫹2a
ⴥ n⫽1
ar
Rbn (cos nu cos na⫹sin nu sin na).
Re z*⫹z
z*⫺z ⫽1⫹2a
ⴥ n⫽1
Re a z z*bn (6*)
cos (nu⫺na)⫽cos nu cos na⫹sin nu sin na.
Re c az
z*bnd ⫽Re crn
Rn einueⴚinad ⫽ar
Rbncos (nu⫺na).
z*⫽Reia, z⫽reiu
z*⫹z
z*⫺z⫽ 1⫹(z>z*)
1⫺(z>z*) ⫽a1⫹ z z*b a
ⴥ n⫽0
a z
z*bn⫽1⫹2 a
ⴥ n⫽1
a z z*bn.
⫺z*z⫹zz*⫽2i Im (zz*)
z*z*⫺zz z*⫹z
z*⫺z⫽ (z*⫹z)(z*⫺z)
(z*⫺z)(z*⫺z) ⫽z*z*⫺zz⫺z*z⫹zz*
ƒz*⫺zƒ2 .
£
ƒzƒ ⫽R
£(R, a) ƒzƒ ⫽R.
£(R, a)
ƒzƒ ⬉R
£
£(r, u)⫽ 1
2p冮02p£(R, a) R2⫺2Rr cos (uR2⫺r⫺2 a)⫹r2 da.
F(z)⫽£(r, u)⫹i°(r, u)
2SIMÉON DENIS POISSON (1781–1840), French mathematician and physicist, professor in Paris from 1809.
His work includes potential theory, partial differential equations (Poisson equation, Sec. 12.1), and probability (Sec. 24.7).
This expression is equal to the quotient in (5), as we have mentioned before, and by inserting it into (5) and integrating term by term with respect to from 0 to we obtain
(7)
where the coefficients are [the 2 in cancels the 2 in in (5)]
(8)
the Fourier coefficients of see Sec. 11.1. Now, for the series (7) becomes the Fourier series of Hence the representation (7) will be valid whenever the given on the boundary can be represented by a Fourier series.
E X A M P L E 1 Dirichlet Problem for the Unit Disk
Find the electrostatic potential in the unit disk having the boundary values
(Fig. 423).
Solution. Since is even, and from (8) we obtain and
Hence, if nis odd, if and the potential is
Figure 424 shows the unit disk and some of the equipotential lines (curves £⫽const). 䊏
£(r, u)⫽1 2⫺ 4
p2 cr cos u⫹ r3
32cos 3u⫹r5
52cos 5u⫹Ád . n⫽2, 4,Á,
an⫽0 an⫽ ⫺4>(n2p2)
an⫽ 1
p c⫺冮ⴚ0ppa cos na da⫹ 冮0ppa cos na dad⫽ 2
n2p2 (cos np⫺1).
a0⫽12
bn⫽0,
£(1, a)
£(1, a)⫽b⫺a>p if ⫺p⬍a⬍0 a>p if 0⬍a⬍p
r⬍1
£(r, u)
£(R, a)
£(R, a).
r⫽R,
£(R, a);
bn⫽ 1
p冮02p£(R, a) sin na da,
n⫽1, 2,Á, a0⫽ 1
2p 冮02p£(R, a) da, an⫽ 1
p 冮02p£(R, a) cos na da, 1>(2p)
(6*)
£(r, u)⫽a0⫹ a
ⴥ n⫽1
ar
Rbn(an cos nu⫹bn sin nu)
2p a
1
π
0 α
Φ(1, α)
–π
Fig. 423. Boundary values in Example 1
0.1 0.2 0.3 0.4 0.6
0.7 Φ 0.8
= 0.9
x y
Fig. 424. Potential in Example 1
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
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 in the unit disk having the given boundary values Using the sum of the first few terms of the series, compute some values of and sketch a figure of the equipotential lines.
5.
6.
7.
8.
9.
10.
11.
12. and 0 otherwise
13. and 0 otherwise
14.
15. £(1, u)⫽1 if ⫺12p⬍u⬍12pand 0 otherwise
£(1, u)⫽ ƒuƒ>p if ⫺p⬍u⬍p
£(1, u)⫽u if ⫺12p⬍u⬍12p
£(1, u)⫽k if 0⬍u⬍p
£(1, u)⫽u>p if ⫺p⬍u⬍p
£(1, u)⫽16 cos3 2u
£(1, u)⫽8 sin4 u
£(1, u)⫽4 sin3 u
£(1, u)⫽a cos2 4u
£(1, u)⫽5⫺cos 2u
£(1, u)⫽32 sin 3u
£
£(1, u).
r⬍1
£(r, u) r⬍R.
16.
17.
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 in Prob. 11 (which is discontinuous on the boundary!), (c)for a of your choice with continuous boundary values, and (d)for with discontinuous boundary values.
20. TEAM PROJECT. Potential in a Disk. (a) Mean value property. Show that the value of a harmonic function at the center of a circle Cequals the mean of the value of 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 of from (7). Hint. Use the Cauchy–
Riemann equations.
(d) Power series.Find a series for F(z)⫽£⫹i°.
£
°
£
£
£
£
£
£(1, u)⫽b0 if ⫺p⬍u⬍0 u if 0⬍u⬍p
£(1, u)⫽u2>p2 if ⫺p⬍u⬍p
£(1, u)⫽bu⫹p if ⫺p⬍u⬍0 u⫺p if 0⬍u⬍p
P R O B L E M S E T 1 8 . 5
18.6 General Properties of Harmonic Functions.