3.2 Harmonic Functions

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The fact that a function f from R2 to R2 is complex di_erentiable puts some

very strong conditions on it. These conditions turn out to have connections

with Laplace's equation which must be the most important Partial Di_erential

Equation (PDE) there is.

Recall the various PDE's you came across last year, in particular the di_usion

or heat equation and the wave equation. In steady state cases you had

functions satisfying Laplace's Equation arising in many cases. For those in

doubt, go to

http://maths.uwa.edu.au/~mike/m252alder.html

for some notes on second year calculus and PDE material. You should download

the vector calculus notes which have a part on Stoke's Theorem, and a

smaller part on PDEs at the end. I haven't the time to explain PDEs to you

again, so you should read this stu_ if you are confused and muddled about

PDEs.

I remind you that a function f : R2 􀀀! R is said to be harmonic or to satisfy

Laplace's Equation, if

@2f

@x2 +

@2f

@y2 = 0

98 CHAPTER 3. C - DIFFERENTIABLE FUNCTIONS

Thus x2􀀀y2 is harmonic, while x2+y2 is not. Notice that harmonic functions

on R2 have graphs which have opposite curvature in orthogonal directions. So

hyperboloids are in there with a chance of being harmonic, while paraboloids

can't ever be. Harmonic functions have the remarkable property that if you

draw a circle around a point and _nd the average value of the function around

the circle, it is always equal to the value of the function at the centre of the

circle. This holds true for every circle which has f de_ned everywhere inside

it and on its boundary. This gives a neat way of solving Laplace's equation

for f on some region of the plane when we are given the values of f on the

boundary (a Dirichlet Problem for the region). All we do is to _x f on the

boundary, give it random values on the interior, and then go through a cycle

of replacing the value at points inside the region with an average of the values

of neighbouring points on some _nite grid. This only gives an approximation,

but that is all you ever get anyway.

One of the ways of trying to understand functions from R2 to R2 is to think

of them as a pair of functions from R2 to R, the _rst giving the function

u(x; y) and the second v(x; y). This means that we can draw the graphs of

each function. While not entirely useless, this is not always illuminating. It

does have its merits however, when considering harmonic functions.

The reason is simple: if @u=@x = @v=@y then @2u=@x2 = @2v=@x@y, which

is equal to @2v=@y@x providing the mixed partial derivatives are equal. This

will be the case if f is analytic.

And if @u=@y = 􀀀@v=@x then @2u=@y2 = 􀀀@2v=@y@x.

Hence provided f is analytic we have:

@2u=@x2 + @2u=@y2 = 0

which is to say, u is harmonic.

It is trivial to check that v is harmonic by the same argument applied to v.

Not only are both functions harmonic, they are said to be conjugate harmonic

functions because they are related by the CR equations. For conjugate

harmonic functions a number of special properties hold: for example, their

product, and the di_erence of their squares, are also harmonic.

The argument is rather neat: If f is analytic, so is f2. This follows because

the product of analytic functions is analytic. If f = u + iv then f2 =

(u2 􀀀 v2) + i(2uv). Hence 2uv is harmonic, and so is any multiple of it for

3.2. HARMONIC FUNCTIONS 99

obvious reasons. Similarly u2 􀀀 v2 is harmonic, and so is its negative. They

are, of course, conjugate.

It is not true generally that the product of harmonic functions is harmonic.

Exercise 3.2.1 Find two harmonic functions the product of which is not

harmonic.

Quite a lot of investigation has gone on into working out which functions are

harmonic and which aren't. The reason for this is that if you are looking for

a solution to Laplace's Equation, then it helps if you don't have to look too

far, and if you have a 'dictionary' of them, you can save yourself some time.

The fact that they come streaming out of complex analytic functions makes

compiling such a dictionary easy.

Given a harmonic function, we can easily construct a conjugate harmonic

function to get back to a complex analytic function. Up to an additive

constant, the conjugate is unique. An example will make the procedure

clear.

Example 3.2.1 It is easy to verify that

u(x; y) = cosh(x) sin(y)

is harmonic.

Di_erentiating with respect to x,

@u=@x = sinh(x) sin(y) = @v=@y

giving, by integration,

v = 􀀀sinh(x) cos(y) + _(x)

Repeating this but di_erentiating with respect to y this time, we get:

v = 􀀀sinh(x) cos(y) +  (y)

From which we deduce that

v = 􀀀sinh(x) cos(y) +C

is a conjugate (for any real number C), and u+iv is easily seen to be analytic.

100 CHAPTER 3. C - DIFFERENTIABLE FUNCTIONS

Exercise 3.2.2 If f : C 􀀀! R is harmonic and g : C 􀀀! C is analytic,

show that f _ g is harmonic. We say that analytic maps preserve solutions

to Laplace's Equation, or Laplace's Equation is invariant under analytic

transforms.

3.2.1 Applications

Let us think about uid ow. (The uid might be the 'ux' of an electric

_eld, so don't imagine this has nothing to do with your _eld of study!)

We write a vector _eld in the plane as

V _ x

y _ = _ u

w _

where u and w are the components of some vector attached to _ x

y _.

Now if the uid is irrotational, the 'curl' of u dx+w dy is zero:

(@w=@x 􀀀 @u=@y) = 0

that is:

@w=@x = @u=@y (3.1)

This tells us that there is a potential function _ : R2 􀀀! R with u = @_=@x

and w = @_=@y

If there are no sources or sinks, then we also have that the divergence is zero:

@u=@x + @w=@y = 0

or

@u=@x = 􀀀@w=@y (3.2)

If you have trouble with this, take it out of two dimensions into three by

going to R3 where this makes more sense and assuming the dz component of

the vector _eld is zero.

Now equations 3.1 and 3.2 look rather like the CR conditions, but a sign has

gone wrong. This explains why I used w. I _x things up by saying that V is

3.2. HARMONIC FUNCTIONS 101

the wrong function to be concerned with, I really need _V , the conjugate. I

can then let

f : C 􀀀! C

be de_ned as

f(x + iy) = u+iv

where v = 􀀀w. Now we get

@v=@x = 􀀀@u=@y (3.3)

and

@u=@x = @v=@y (3.4)

This tells us that if V is an irrotational vector _eld with no sources and

sinks, then f = _ V is a di_erentiable complex function, and indeed an analytic

complex function if V is di_erentiable and the partial derivatives are

continuous.

This in turn tells us that the components of f are harmonic.

The function _ is the real part of an antiderivative of f There is an imaginary

part as well,   for later reference.

Example 3.2.2 Suppose V (x + iy) is the vector _eld 2x 􀀀 i2y. Find the

potential function.

Solution

_ V = f(x + iy) = 2(x+iy), i.e.

f(z) = 2z

This is well known to be the derivative of

F(z) = z2

This has real part x2 􀀀 y2 (and imaginary part 2xy). So

_(x + iy) = x2

􀀀y2

is the required potential function.

102 CHAPTER 3. C - DIFFERENTIABLE FUNCTIONS

Of course, we could have got the same answer by standard methods, but this

is rather neat.

We shall discover later that the curves _(x + iy) = C, the equipotentials

decompose the plane into a family of curves for various values of C, which

are orthogonal to the curves  (x + iy) = D for various values of D. This

means that we can look upon the latter curves as the streamlines of the ow.

It should be obvious to you for physical reasons that the ow should always

be orthogonal to the curves of constant potential. If it isn't obvious, ask.

In other words, the solutions to the vector _eld regarded as a system of ODEs

can be obtained directly from integrating a complex function. Thinking

about this leads to the conclusion that this is not too surprising, but again,

it is rather neat.

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