Lecture 16:
Harmonic Functions
                                Dan Sloughter
                              Furman University
                                Mathematics 39
                                March 31, 2004
16.1     Harmonic functions
Suppose f is analytic in a domain D,
                         f (x + iy) = u(x, y) + iv(x, y),
and u and v have continuous partial derivatives of all orders. From the
Cauchy-Riemann equations we know that
                 ux (x, y) = vy (x, y) and uy (x, y) = −vx (x, y)
for all x + iy ∈ D. Differentiating with respect to x, we have
               uxx (x, y) = vyx (x, y) and uyx (x, y) = −vxx (x, y);
differentiating with respect to y, we have
               uxy (x, y) = vyy (x, y) and uyy (x, y) = −vxy (x, y).
Hence
   uxx (x, y) + uyy (x, y) = vyx (x, y) − vxy (x, y) = vxy (x, y) − vxy (x, y) = 0
for all x + iy ∈ D and
 vxx (x, y) + vyy (x, y) = −uyx (x, y) + uxy (x, y) = −uxy (x, y) + uxy (x, y) = 0
for all x + iy ∈ D.
                                         1
Definition 16.1. Suppose H : R2 → R has continuous second partial deriva-
tives on a domain D. We say H is harmonic in D if for all (x, y) ∈ D,
                          Hxx (x, y) + Hyy (x, y) = 0.
    Harmonic functions arise frequently in applications, such as in the study
of heat distributions and electrostatic potentials.
Theorem 16.1. If f is analytic in a domain D and
                        f (x + iy) = u(x, y) + iv(x, y),
then u and v are harmonic in D.
Proof. The result follows from the discussion above combined with a result
we will prove later: if f is analytic at z0 = x0 + iy0 , then u and v have
continuous partial derivatives of all orders at (x0 , y0 ).
Example 16.1. We know that f (z) = ez is entire. Since
                      f (x + iy) = ex cos(y) + iex sin(y),
it follows that u(x, y) = ex cos(y) and v(x, y) = ex sin(y) are both harmonic
in C (which is also easily checked directly).
Example 16.2. We know that
                                               1
                                   f (z) =
                                               z2
is analytic in {z ∈ C : z 6= 0}. Now
                         1    1 z̄ 2     x2 − y 2 − 2xyi
                            =          =                 ,
                         z2   z 2 z̄ 2     (x2 + y 2 )2
so
                                           x2 − y 2
                              u(x, y) =
                                          (x2 + y 2 )2
and
                                                    2xy
                             v(x, y) = −
                                              (x2   + y 2 )2
are harmonic in {(x, y) ∈ R2 : (x, y) 6= (0, 0)}.
                                          2
Definition 16.2. If u and v are harmonic in a domain D and satisfy the
Cauchy-Riemann equations, then we say v is a harmonic conjugate of u.
Example 16.3. It is easy to check that the function
                              u(x, y) = x3 − 3xy 2
is harmonic. To find a harmonic conjugate v of u, we must have
                               ux (x, y) = vy (x, y)
and
                              uy (x, y) = −vx (x, y).
From the first we have
                              vy (x, y) = 3x2 − 3y 2 ,
from which it follows that
                         v(x, y) = 3x2 y − y 3 + ϕ(x)
for some function ϕ of x. It now follows from the second equation that
                   −6xy = −vx (x, y) = −(6xy + ϕ0 (x)),
and so ϕ0 (x) = 0. Hence for any real number c, the function
                             v(x, y) = 3x2 y − y 3 + c
is a harmonic conjugate of u.