Lecture 16: Harmonic Functions - Furman
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.
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Definition 16.1. Suppose H : R2 R has continuous second partial derivatives 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 = 0}. Now
1 1 z?2 x2 - y2 - 2xyi z2 = z2 z?2 = (x2 + y2)2 , so
x2 - y2 u(x, y) = (x2 + y2)2 and
2xy v(x, y) = - (x2 + y2)2 are harmonic in {(x, y) R2 : (x, y) = (0, 0)}.
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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 - 3xy2
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 - 3y2,
from which it follows that
v(x, y) = 3x2y - y3 + (x)
for some function of x. It now follows from the second equation that
-6xy = -vx(x, y) = -(6xy + (x)),
and so (x) = 0. Hence for any real number c, the function v(x, y) = 3x2y - y3 + c
is a harmonic conjugate of u.
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