Definition and Examples Harmonic Conjugates Existence of Conjugates

Harmonic Functions

Ryan C. Daileda

Trinity University

Complex Variables

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Definition
The Laplacian (in two variables) is the differential operator

∂2 ∂2
∆ = ∇2 = + .
∂x 2 ∂y 2

Definition
Let Ω ⊂ R2 be a domain and u ∈ C 2 (Ω). We say u is harmonic
provided
∆u = 0
throughout Ω.

Recall: If u(x, y ) ∈ C 2 (Ω), then

∂2u ∂2u
=
∂x∂y ∂y ∂x
throughout Ω (Clairaut’s theorem).
Daileda Harmonic Functions
Definition and Examples Harmonic Conjugates Existence of Conjugates

Example 1
The function u(x, y ) = x 3 − 3xy 2 is harmonic on C.

Indeed, u is clearly C 2 and


∂u ∂2u
= 3x 2 − 3y 2 ⇒ = 6x, 


∂x ∂x 2
⇒ ∆u = 0.
∂u ∂2u 

= − 6xy ⇒ = − 6x, 
∂y ∂y 2

Example 2
The function u(x, y ) = arctan(y /x) is harmonic for x > 0.

In this case
∂u −y /x 2 −y ∂2u 2xy
= 2
= 2 2
⇒ 2
= 2
∂x 1 + (y /x) x +y ∂x (x + y 2 )2

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

and
∂u 1/x x ∂2u −2xy
= 2
= 2 2
⇒ 2
= 2 .
∂y 1 + (y /x) x +y ∂y (x + y 2 )2

Thus ∆u = 0 and u is harmonic.

More generally we have the following result.
Theorem 1
Let Ω ⊂ C be a domain, f : Ω → C, u = Re f and v = Im f . If f is
analytic and u, v ∈ C 2 (Ω), then u and v are harmonic on Ω.

Remarks.
1 The C 2 hypothesis is actually unnecessary. As we will see, if f
is analytic, then Re f and Im f are in fact C ∞ .
2 We will see that every harmonic function is (locally) the real
part of an analytic function.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Proof. By the C-R equations and Clairaut’s theorem we have

∂2u ∂ ∂u ∂ ∂v ∂ ∂v ∂2u
= = = = − ,
∂x 2 ∂x ∂x ∂x ∂y ∂y ∂x ∂y 2
which implies that ∆u = 0. Similarly, ∆v = 0.

Definition
Let Ω ⊂ R2 be a domain and let u : Ω → R be harmonic. We say
that v ∈ C 2 (Ω) is a harmonic conjugate of u (on Ω) provided u
and v satisfy the C-R equations on Ω:
∂u ∂v ∂u ∂v
= , =− .
∂x ∂y ∂y ∂x
Equivalently, provided f = u + iv is analytic on Ω.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Remarks.
1 If u is harmonic and v is a conjugate of u, then v is also
harmonic.

2 Being “a harmonic conjugate of” is not symmetric. One

cannot simply say that u and v are “harmonic conjugates of
one another.”

3 If v is a harmonic conjugate of u, then −u is a harmonic

conjugate of v : if f = u + iv is analytic, then so is
−if = v − iu.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Example 3
Show that v = 3x 2 y − y 3 is a harmonic conjugate of
u = x 3 − 3xy 2 .

We simply notice that if f (z) = z 3 , then f is entire and by the

binomial theorem
f (x + iy ) = (x + iy )3 = x 3 + 3ix 2 y − 3xy 2 − iy 3
= (x 3 − 3xy 2 ) +i (3x 2 y − y 3 ) .
| {z } | {z }
u v

Example 4
Find a harmonic conjugate for u = arctan(y /x), x > 0.

For x > 0, arctan(y /x) = Arg(x + iy ). Let f (z) = Log z.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Then −if (z) is analytic for x > 0 and is given by

−i Log z = − i (ln |z| + i Arg z) = Arg z −i ln |z|.

| {z }
u

Therefore a harmonic conjugate of u = Arg z is

1

v = − ln |z| = − ln x 2 + y 2 .
2

Harmonic conjugates are almost unique. To prove this we require a

preliminary result.
Lemma 1
Let Ω ⊂ C be a domain and f : Ω → C analytic. If the image of f
is contained in a line, then f is constant.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Proof. Let u = Re f and v = Im f , and suppose f (Ω) is a subset of

the line
aX + bY = c, (a, b) 6= (0, 0).
Then au(x, y ) + bv (x, y ) = c for all (x, y ) ∈ Ω. Differentiating
and applying the C-R equations we obtain
     
aux + bvx = 0, ux vx a 0
⇒ = ,
auy + bvy = − avx + bux = 0, −vx ux b 0

throughout Ω. Because (a, b) 6= (0, 0),

 
ux vx
0 = det = ux2 + vx2 = |f ′ (z)|2 ⇒ f ′ (z) = 0,
−vx ux

everywhere in Ω. As we have seen, this implies that f is constant.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Theorem 2
Let Ω ⊂ R2 be a domain and suppose u is harmonic on Ω. If v1
and v2 are harmonic conjugates of u on Ω, then there is an a ∈ R
so that v1 = v2 + a.

Proof. Let fj = u + ivj for j = 1, 2. Then f1 , f2 are analytic on Ω.

Therefore f1 − f2 = i (v1 − v2 ) is analytic and purely imaginary on
Ω. By the lemma, f1 − f2 is constant. The result follows.

This addresses the uniqueness of harmonic conjugates. What

about existence?

Example 5


Does u = ln x 2 + y 2 have a harmonic conjugate on C× ?

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Assume f = u + iv is analytic on C× . Then so is f /2.

Notice that Re(f /2) = u/2 = ln |z| = Re (Log z) on C \ (−∞, 0].

Therefore both v /2 and Arg z are conjugates of u/2 on the slit

plane.

Thus Arg z = a + v /2 for some a ∈ R.

But a + v /2 extends continuously to C× , whereas Arg z does not.

So u cannot have a conjugate on C× .

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

It turns out the problem in the preceding example is fact that C×

is multiply connected.

Theorem 3
Let Ω ⊂ C be a domain and let u : Ω → R be harmonic. If Ω is
simply connected, then u has a harmonic conjugate on Ω.

∂u ∂u
Sketch of Proof. Fix P0 ∈ Ω, let ω = − ∂y dx + ∂x dy and for
P ∈ Ω define Z P
v (P) = ω,
P0

where the integral is taken along any piecewise smooth curve in Ω

from P0 to P.

We first claim that v is well-defined, i.e. is path-independent.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Let C1 , C2 be paths in Ω from P0 to P. Then C1 − C2 is a loop in

Ω. Let R be the region enclosed by C1 − C2 .

Because Ω is simply connected, R ⊂ Ω. Green’s theorem then

implies
Z ZZ ZZ    
∂ ∂u ∂ ∂u
ω= dω = − − dA
C1 −C2 R R ∂x ∂x ∂y ∂y
ZZ
= ∆u dA = 0,
R

since u is harmonic. Thus

Z Z
ω= ω,
C1 C2

and v is well-defined.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

To differentiate v at P = (x, y ) ∈ Ω, let P ′ = (x + ∆x, y ).

Let C be any path in Ω from P0 to P and let C ′ = C + L, where L
is the horizontal segment from P to P ′ .
Then
Z
v (x + ∆x, y ) − v (x, y ) 1
vx (x, y ) = lim = lim ω
∆x→0 ∆x ∆x→0 ∆x L
Z ∆x
1
= lim −uy (x + t, y ) dt
∆x→0 ∆x 0

= lim −uy (x + h, y )
∆x→0

for some h ∈ [0, ∆x], by the Mean Value Theorem.

Since h → 0 as ∆x → 0, and uy is continuous, we find that

vx (x, y ) = −uy (x, y ).

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

By instead using a vertical segment one can compute vy (x, y ) in a

similar manner. The result is

vy (x, y ) = ux (x, y ).

Thus, v is a harmonic conjugate of u on Ω.

Remarks.
1 Using a different base point P1 yields a Rconjugate v1 that
P
differs from v by the additive constant P01 ω.

2 Strictly speaking, Green’s theorem only applies to simple

closed curves, a hypothesis we cannot assume for C1 − C2 .

3 A rigorous proof applies Green’s theorem locally, to

subdivisions of the homotopy between C1 and C2 in Ω.

Daileda Harmonic Functions

Definition and Examples Harmonic Conjugates Existence of Conjugates

Corollary 1
Harmonic conjugates always exist locally.

Assuming the C ∞ nature of analytic functions, we have the

following result as well.
Corollary 2
Let Ω ⊂ R2 be a domain. If u ∈ C 2 (Ω) and ∆u = 0 on Ω, then
u ∈ C ∞ (Ω).

Proof. Let P ∈ Ω. Choose an open disk D ⊂ Ω containing P.

Since D is simply connected and u is harmonic on D, there is an

analytic function f : D → C so that u = Re f .

It follows that u ∈ C ∞ (D). Since P ∈ Ω was arbitrary, we

conclude that u ∈ C ∞ (Ω).

Daileda Harmonic Functions