lacunary - Mathnotes

Harmonic Functions

A real-valued function ϕ(x,y) is said to be harmonic in a domain D if its second partial derivatives are continuous in D and if at each point of D, ϕ satisfies Laplace's equation

2ϕx2+2ϕy2=0

For an analytic function f(x,y)=u(x,y)+v(x,y)i, the second partial derivatives of u and x satisfy Laplace's equation. Therefore, we have the following theorem:

The real and imaginary parts of a function analytic in a domain D are harmonic in D.

When u(x,y) is harmonic in a domain D, a function v(x,y) such that u+vi is analytic in D is called a harmonic conjugate of u(x,y). This means that given a harmonic function u(x,y), we can always find a harmonic conjugate v(x,y) (this is not true for all domains though.)

Given a harmonic function u(x,y) on a suitable domain, we can use the Cauchy-Riemann equations to find a harmonic conjugate v(x,y), as follows.

From the Cauchy-Riemann equations we have ux=vy (using the subscript notation for partial derivatives). Then,

v(x,y)=uxdy=vydy(a)

is the function we're looking for, except the constant of integration will be some unknown h(x).

We'll say then that

v(x,y)=g(x,y)+h(x),(b)

where g(x,y) will explicitly be the antiderivative of vy with respect to y, but h(x) will be an unknown function of x.

Now, we can take

x(g(x,y)+h(x))=vx=g(x,y)+h(x,y).(c)

From the Cauchy-Riemann equations, vx=uy. Comparing g(x,y)+h(x) to uy, we find h(x) as the missing terms from g(x,y) in uy, if any. Integrating these gives the value of h(x), which we can plug into (b) to get v(x,y).

Note that this is the same mechnical procedure we use to solve first order exact differential equations.

We can use this to find analytic functions from a harmonic function by using the harmonic conjugates and the harmonic function as the real and imaginary parts of a complex function.

One interesting property of harmonic functions and their conjugates is that they define orthogonal families of curves - all of their intersections are at right angles to each other.