Math 411: Honours Complex Variables - University of Alberta

for (x,y) ∈ R 2 \{(0,0)}. The partial derivatives of u with respect to x are
and
∂u
∂x =
x
x 2 +y 2.
∂2u ∂x2 = (x2 +y2 )−2x 2
(x2 +y2 ) 2
Moreover, since u is symmetric in x and y, we find
∂ 2 u
∂y 2 = −y2 +x 2
(x 2 +y 2 ) 2.
= −x2 +y 2
(x 2 +y 2 ) 2
Consequently, u is harmonic.
Now suppose that there is a holomorphic function f : C \ {0} → C such that
Ref = u. On C−, we then have that Ref = log|z|= ReLog. The Cauchy–Riemann
Equations thus yield
so that
∂(Imf)
(z) = −
∂x
∂Ref
∂y
(z) = −∂(ReLog)
∂y
(z) = ∂(ImLog)
(z),
∂x
f ′ (z) = ∂Ref
∂x (z)+i∂(Imf) (z) =
∂x
∂ReLog
(z)+i
∂x
∂(ImLog)
(z) = Log
∂x
′ z = 1
z
for z ∈ C−. By continuity, it follows that f ′ (z) = 1
for all z ∈ C\{0}, so that f is
z
an antiderivative of z ↦→ 1 on C\{0}. This is impossible (cf. page 24).
z
Definition. Let D ⊂ C be open, and let u: D → R be harmonic. We call a harmonic
function v: D → R a harmonic conjugate of u if u+iv is holomorphic.
Theorem 14.1 (Harmonic Conjugates). Let D ⊂ C be open and suppose that there
exists (x0,y0) ∈ D with the following property: for each (x,y) ∈ D, we have
• (x,t) ∈ D for each t between y and y0 and
• (s,y0) ∈ D for each s between x and x0.
Then every harmonic function on D has a harmonic conjugate.
Proof. Let u: D → R be harmonic. We will find a harmonic v: D → R such that
For (x,y) ∈ D, define
∂u
∂x
= ∂v
∂y
and
∂v
∂x
� y
∂u
v(x,y) =
∂x (x,t)dt+φ(x),
y0
95
= −∂u.
(∗)
∂y