ANALYTIC FUNCTIONS
26. HARMONIC FUNCTIONS
6. Use results in Sec. 23 to verify that the function
g(z)=lnr+iθ (r >0,0< θ <2π )
is analytic in the indicated domain of definition, with derivative g(z)=1/z. Then show that the composite function G(z)=g(z2+1) is analytic in the quadrant x >0, y >0, with derivative
G(z)= 2z z2+1.
Suggestion: Observe that Im(z2+1) >0 whenx >0, y >0.
7. Let a functionf be analytic everywhere in a domainD.Prove that if f (z)is real- valued for allz inD,thenf (z)must be constant throughtoutD.
sec. 26 Harmonic Functions 79
Txx(x, y)+Tyy(x, y)=0, T (0, y)=0, T (π, y)=0, T (x,0)=sinx, lim
y→∞T (x, y)=0,
which describe steady temperatures T (x, y) in a thin homogeneous plate in the xy plane that has no heat sources or sinks and is insulated except for the stated conditions along the edges.
The use of the theory of functions of a complex variable in discovering solu- tions, such as the one in Example 1, of temperature and other problems is described in considerable detail later on in Chap. 10 and in parts of chapters following it.∗ That theory is based on the theorem below, which provides a source of harmonic functions.
Theorem 1. If a functionf (z)=u(x, y)+iv(x, y)is analytic in a domainD, then its component functionsuandvare harmonic inD.
To show this, we need a result that is to be proved in Chap. 4 (Sec. 52). Namely, if a function of a complex variable is analytic at a point, then its real and imaginary components have continuous partial derivatives of all orders at that point.
Assuming thatf is analytic inD, we start with the observation that the first- order partial derivatives of its component functions must satisfy the Cauchy–Riemann equations throughoutD:
ux =vy, uy= −vx. (2)
Differentiating both sides of these equations with respect tox, we have uxx=vyx, uyx= −vxx.
(3)
Likewise, differentiation with respect toy yields uxy =vyy, uyy = −vxy. (4)
Now, by a theorem in advanced calculus,† the continuity of the partial derivatives of u and v ensures that uyx=uxy and vyx=vxy. It then follows from equations (3) and (4) that
uxx+uyy =0 and vxx+vyy =0.
That is,u andv are harmonic inD.
∗Another important method is developed in the authors’ “Fourier Series and Boundary Value Prob- lems,” 7th ed., 2008.
†See, for instance, A. E. Taylor and W. R. Mann, “Advanced Calculus,” 3d ed., pp. 199–201, 1983.
EXAMPLE 2. The function f (z)=e−ysinx−ie−ycosx is entire, as is shown in Exercise 1(c), Sec. 25. Hence its real component, which is the temperature function T (x, y)=e−ysinx in Example 1, must be harmonic in every domain of thexy plane.
EXAMPLE 3. Since the function f (z)=i/z2 is analytic whenever z=0 and since
i z2 = i
z2· z2 z2 = iz2
(zz)2 = iz2
|z|4 = 2xy+i(x2−y2) (x2+y2)2 , the two functions
u(x, y)= 2xy
(x2+y2)2 and v(x, y)= x2−y2 (x2+y2)2
are harmonic throughout any domain in thexyplane that does not contain the origin.
If two given functionsuandvare harmonic in a domainDand their first-order partial derivatives satisfy the Cauchy–Riemann equations (2) throughoutD, thenv is said to be a harmonic conjugate ofu. The meaning of the word conjugate here is, of course, different from that in Sec. 5, wherezis defined.
Theorem 2. A functionf (z)=u(x, y)+iv(x, y)is analytic in a domainD if and only ifvis a harmonic conjugate ofu.
The proof is easy. Ifv is a harmonic conjugate ofu inD, the theorem in Sec.
22 tells us thatf is analytic inD. Conversely, iff is analytic inD, we know from Theorem 1 thatuandv are harmonic inD; furthermore, in view of the theorem in Sec. 21, the Cauchy–Riemann equations are satisfied.
The following example shows that if v is a harmonic conjugate of u in some domain, it is not, in general, true that u is a harmonic conjugate of v there. (See also Exercises 3 and 4.)
EXAMPLE 4. Suppose that
u(x, y)=x2−y2 and v(x, y)=2xy.
Since these are the real and imaginary components, respectively, of the entire func- tionf (z)=z2, we know thatv is a harmonic conjugate ofu throughout the plane.
Butu cannot be a harmonic conjugate ofv since, as verified in Exercise 2(b), Sec.
25, the function 2xy+i(x2−y2)is not analytic anywhere.
In Chap. 9 (Sec. 104) we shall show that a function u which is harmonic in a domain of a certain type always has a harmonic conjugate. Thus, in such domains, every harmonic function is the real part of an analytic function. It is also true (Exercise 2) that a harmonic conjugate, when it exists, is unique except for an additive constant.
sec. 26 Exercises 81
EXAMPLE 5. We now illustrate one method of obtaining a harmonic con- jugate of a given harmonic function. The function
u(x, y)=y3−3x2y (5)
is readily seen to be harmonic throughout the entire xy plane. Since a harmonic conjugatev(x, y)is related tou(x, y)by means of the Cauchy–Riemann equations
ux=vy, uy= −vx, (6)
the first of these equations tells us that
vy(x, y)= −6xy.
Holdingx fixed and integrating each side here with respect toy, we find that v(x, y)= −3xy2+φ(x)
(7)
whereφis, at present, an arbitrary function ofx. Using the second of equations (6), we have
3y2−3x2=3y2−φ(x),
orφ(x)=3x2. Thusφ(x)=x3+C, whereCis an arbitrary real number. Accord- ing to equation (7), then, the function
v(x, y)= −3xy2+x3+C (8)
is a harmonic conjugate ofu(x, y).
The corresponding analytic function is
f (z)=(y3−3x2y)+i(−3xy2+x3+C).
(9)
The form f (z)=i(z3+C) of this function is easily verified and is suggested by noting that wheny=0, expression (9) becomesf (x)=i(x3+C).
EXERCISES
1. Show thatu(x, y)is harmonic in some domain and find a harmonic conjugatev(x, y) when
(a)u(x, y)=2x(1−y); (b)u(x, y)=2x−x3+3xy2; (c)u(x, y)=sinhxsiny; (d)u(x, y)=y/(x2+y2).
Ans. (a)v(x, y)=x2−y2+2y; (b)v(x, y)=2y−3x2y+y3; (c)v(x, y)= −coshxcosy; (d)v(x, y)=x/(x2+y2).
2. Show that ifvandV are harmonic conjugates ofu(x, y)in a domainD, thenv(x, y) andV (x, y)can differ at most by an additive constant.
3. Suppose that v is a harmonic conjugate of u in a domain D and also that u is a harmonic conjugate ofv in D. Show how it follows that bothu(x, y) andv(x, y) must be constant throughoutD.
4. Use Theorem 2 in Sec. 26 to show thatvis a harmonic conjugate ofuin a domain D if and only if −u is a harmonic conjugate of v in D. (Compare with the result obtained in Exercise 3.)
Suggestion:Observe that the functionf (z)=u(x, y)+iv(x, y)is analytic inD if and only if−if (z)is analytic there.
5. Let the functionf (z)=u(r, θ )+iv(r, θ ) be analytic in a domain D that does not include the origin. Using the Cauchy–Riemann equations in polar coordinates (Sec.
23) and assuming continuity of partial derivatives, show that throughoutDthe function u(r, θ )satisfies the partial differential equation
r2urr(r, θ )+rur(r, θ )+uθ θ(r, θ )=0,
which is the polar form of Laplace’s equation. Show that the same is true of the functionv(r, θ ).
6. Verify that the functionu(r, θ )=lnr is harmonic in the domainr >0,0< θ <2π by showing that it satisfies the polar form of Laplace’s equation, obtained in Exercise 5. Then use the technique in Example 5, Sec. 26, but involving the Cauchy–Riemann equations in polar form (Sec. 23), to derive the harmonic conjugatev(r, θ )=θ. (Com- pare with Exercise 6, Sec. 25.)
7. Let the functionf (z)=u(x, y)+iv(x, y)be analytic in a domainD, and consider the families oflevel curves u(x, y)=c1andv(x, y)=c2,wherec1 andc2are arbitrary real constants. Prove that these families are orthogonal. More precisely, show that if z0=(x0, y0)is a point in Dwhich is common to two particular curvesu(x, y)=c1
andv(x, y)=c2 and if f(z0)=0, then the lines tangent to those curves at(x0, y0) are perpendicular.
Suggestion: Note how it follows from the pair of equations u(x, y)=c1 and v(x, y)=c2that
∂u
∂x +∂u
∂y dy
dx =0 and ∂v
∂x+∂v
∂y dy dx =0.
8. Show that when f (z)=z2, the level curves u(x, y)=c1 and v(x, y)=c2 of the component functions are the hyperbolas indicated in Fig. 32. Note the orthogonality of the two families, described in Exercise 7. Observe that the curvesu(x, y)=0 and v(x, y)=0 intersect at the origin but are not, however, orthogonal to each other. Why is this fact in agreement with the result in Exercise 7?
9. Sketch the families of level curves of the component functions u and v when f (z)=1/z, and note the orthogonality described in Exercise 7.
10. Do Exercise 9 using polar coordinates.
11. Sketch the families of level curves of the component functionsuandvwhen f (z)= z−1
z+1, and note how the result in Exercise 7 is illustrated here.
sec. 27 Uniquely Determined Analytic Functions 83
c1 = 0 c1 = < 0
c1 > 0
x y
c2 > 0
c2 < 0 c1 = 0
FIGURE 32
27. UNIQUELY DETERMINED ANALYTIC FUNCTIONS