# HARMONIC FUNCTIONS Let Ω C C be an open set. A twice ```HARMONIC FUNCTIONS
Let Ω ⊂ C be an open set. A twice continuously differentiable function u : Ω → R
is harmonic if it satisfies the Laplace equation
∆u = uxx + uyy = 0
everywhere on the set Ω.
Theorem 1. If f = u + iv is holomorphic on Ω then u and v are harmonic.
Proof. Since a holomorphic function is infinitely complex differentiable, u and v have
continuous partial derivatives of all orders. Indeed, f 0 = 2 ∂u
= ux − iuy , and since f 0 is
∂z
holomorphic and hence continuous, ux and uy exist and are both continuous. Similarly,
f 0 = 2i ∂v
= vy + ivx , so vx and vy exist and are both continuous. By differentiating f 0
∂z
and applying the same argument, we see that uxx , uxy , uyx , and uyy exist and are all
continuous, and similarly for all second partial derivatives of v. Continuing in this way
proves that u and v have continuous partial derivatives of all orders. In particular, we
have uxy = uyx and similarly for v.
By the Cauchy-Riemann equations, ux = vy and uy = −vx , thus
uxx = (ux )x = (vy )x = vyx = vxy = (vx )y = (−uy )y = −uyy ,
so uxx + uyy = 0, and u is harmonic. A similar argument shows v is also harmonic. Example 1. The function u(x, y) = log(x2 +y 2 ) is harmonic on C∗ (check this by explicit
computation). However, u is not the real part of a holomorphic function on C∗ . Indeed,
using the Cauchy-Riemann equations we calculate that if f = u + iv were holomorphic
on C∗ , then v = 2 arctan( xy ) + c, that is, v = θ + c where z = x + iy = |z|eiθ , so that
f = 2λ + ci where λ is a logarithm on C∗ . But by our results on complex logarithms,
we know that there does not exist a logarithm on C∗ , hence such a function f with real
part u cannot exist on C∗ .
Theorem 2. Let u be harmonic on a simply connected region Ω. Then u is the real
part of a holomorphic function on Ω.
Proof. First note that the function f = ux − iuy is holomorphic on Ω as it satisfies the
Cauchy-Riemann equations everywhere on Ω. Indeed, (ux )x = (−uy )y as u is harmonic,
and (ux )y = −(−uy )x as u is twice continuously differentiable by definition.
Since Ω is simply connected, f has a primitive F = U + iV on Ω. Then
∂U
Ux − iUy = 2
= F 0 = f = ux − iuy .
∂z
Hence U and u differ by a constant, u = U + c for some c ∈ R. Then u = Re(F + c). Corollary 1. A harmonic function u on an open set Ω ⊂ C is infinitely differentiable.
Proof. Cover Ω by open discs and apply the previous theorem to u on each disc.
Date: November 21, 2014.
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Example 2. Let u(x, y) = x − ex sin(y) on C. Then u is harmonic (check!), and since
C is simply connected there exists an entire function F = u + iv such that Re(F ) = u.
By the Cauchy-Riemann equations we have
vx = −uy = ex cos(y) ,
so
vy = ux = 1 − ex sin(y) ,
so
v = ex cos(y) + function of y ,
v = y + ex cos(y) + function of x .
Combining these constraints we see that we may take v = y + ex cos(y) and then F (z) =
F (x, y) = (x + iy) + (−ex sin(y) + iex cos(y)) = z + iez .
Let u, v : Ω → R be harmonic, where Ω ⊂ C is open. If u + iv is holomorphic then
we say that v is a harmonic conjugate for u on Ω.
If v1 , v2 are both harmonic conjugates for u on Ω, then (u+iv1 )−(u+iv2 ) = i(v1 −v2 )
is the difference of two holomorphic functions and is hence holomorphic. But i(v1 − v2 )
only takes purely imaginary values, so the image of Ω lies within the imaginary axis and
cannot be an open set. If Ω is connected then by the open mapping theorem, i(v1 − v2 )
is therefore a (purely imaginary) constant. Thus any two harmonic conjugates on a
region differ by a real constant.
Theorem 3 (Mean value theorem for holomorphic functions). Let f : DR (z0 ) → C be
holomorphic. Then
Z 2π
1
f (z0 ) =
f (z0 + reiθ )dθ
2π 0
for all 0 &lt; r &lt; R.
Proof.
1
f (z0 ) =
2πi
Z
|z−z0 |=r
1
f (z)
dz =
z − z0
2πi
Z
0
2π
f (z0 + reiθ ) iθ
rie dθ .
reiθ
Theorem 4 (Mean value theorem for harmonic functions). If u : DR (z0 ) → R is
harmonic then
Z 2π
1
u(z0 + reiθ )dθ
u(z0 ) =
2π 0
for all 0 &lt; r &lt; R.
Proof. Find a holomorphic function f : DR (z0 ) → C such that u = Re(f ). Apply
the mean value theorem for holomorphic functions to f and take the real part of both
sides.
Theorem 5 (Identity theorem for harmonic functions). Let u be harmonic on a region
Ω ⊂ C. If u = 0 at every point of a nonempty open subset U ⊂ Ω then u = 0 at every
point of Ω.
Proof. Consider the holomorphic function f = ux − iuy on Ω. Since f = 0 on U , f = 0
on all of Ω by the identity theorem for holomorphic functions. Hence u is constant. Note that the hypothesis that the harmonic function u vanishes on an open subset
U ⊂ Ω is stronger than was required in the identity theorem for holomorphic functions.
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For an example showing the stronger hypothesis is required, consider the harmonic function u(x, y) = y, which vanishes on the real axis y = 0, but does not vanish identically
on all of C.
Theorem 6 (Maximum and minimum principle for harmonic functions). If u is a nonconstant harmonic function in a region Ω ⊂ C then u has no maxima or minima in
Ω.
Proof. Let c ∈ Ω and find &gt; 0 such that D (c) ⊂ Ω. There is a holomorphic function
f on D (c) such that Re(f ) = u. By the identity theorem for harmonic functions, f is
not constant on D (c) (otherwise u would also be constant on D (c) and hence constant
on Ω by the identity theorem for harmonic functions). Hence f (D (c)) is an open set by
the open mapping theorem, and then u takes both larger and smaller values than u(c)
in D (c).
For details of the Poisson integral formula and the solution of the Dirichlet problem