Chapter 6
Harmonic Functions
The shortest route between two truths in the real domain passes through the complex domain.
J. Hadamard
6.1
Definition and Basic Properties
We will now spend a chapter on certain functions defined on subsets of the complex plane which
are
real
valued.
The main motivation for studying them is that the partial di
ff
erential equation
they satisfy is very common in the physical sciences.
Recall from Section
??
the definition of a harmonic function:
Definition 6.1.
Let
G
⊆
C
be a region. A function
u
:
G
→
R
is
harmonic
in
G
if it has continuous
second partials in
G
and satisfies the
Laplace
1
equation
u
xx
+
u
yy
= 0
in
G
.
There are (at least) two reasons why harmonic functions are part of the study of complex
analysis, and they can be found in the next two theorems.
Proposition 6.2.
Suppose
f
=
u
+
iv
is holomorphic in the region
G
. Then
u
and
v
are harmonic
in
G
.
Proof.
First, by Corollary 5.2,
f
is infinitely di
ff
erentiable, and hence so are
u
and
v
. In particular,
u
and
v
have continuous second partials. By Theorem 2.13,
u
and
v
satisfy the Cauchy–Riemann
equations
u
x
=
v
y
and
u
y
=
−
v
x
in
G
. Hence
u
xx
+
u
yy
= (
u
x
)
x
+ (
u
y
)
y
= (
v
y
)
x
+ (
−
v
x
)
y
=
v
yx
−
v
xy
= 0
in
G
. Note that in the last step we used the fact that
v
has continuous second partials. The proof
that
v
satisfies the Laplace equation is completely analogous.
1
For more information about PierreSimon Laplace (1749–1827), see
http://wwwgroups.dcs.stand.ac.uk/
∼
history/Biographies/Laplace.html
.
63
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CHAPTER 6.
HARMONIC FUNCTIONS
64
Proposition 6.2 shouts for a converse theorem. There are, however, functions which are harmonic
in a region
G
but not the real part (say) of an holomorphic function in
G
(Exercise 3).
We do
obtain a converse of Proposition 6.2 if we restrict ourselves to
simply connected
regions.
Theorem 6.3.
Suppose
u
is harmonic on the simply connected region
G
.
Then there exists a
harmonic function
v
such that
f
=
u
+
iv
is holomorphic in
G
.
Remark.
The function
v
is called a
harmonic conjugate
of
u
.
Proof.
We will explicitly construct the holomorphic function
f
(and thus
v
= Im
f
). First, let
g
=
u
x
−
iu
y
.
The plan is to prove that
g
is holomorphic, and then to construct an antiderivative of
g
, which will
be almost the function
f
that we’re after. To prove that
g
is holomorphic, we use Theorem 2.13:
first because
u
is harmonic, Re
g
=
u
x
and Im
g
=
−
u
y
have continuous partials. Moreover, again
because
u
is harmonic, they satisfy the Cauchy–Riemann equations:
(Re
g
)
x
=
u
xx
=
−
u
yy
= (Im
g
)
y
and
(Re
g
)
y
=
u
xy
=
u
yx
=
−
(Im
g
)
x
.
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 Fall '11
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 Derivative, Sets, Holomorphic function, harmonic function, G. Proof, g G

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