Reconstructing F from its divergence and curl

The first question we have to answer in this investigation is: Under what circumstances can two vector fields F(x,y) and G(x,y) have the same curl and divergence?

It is a bit more convenient to introduce

H(x,y) = F(x,y) - G(x,y)

Then since

div(H)(x,y) = div(F)(x,y) - div(G)(x,y) = 0

and

curl(H)(x,y) = curl(F)(x,y) - curl(G)(x,y) = 0

it is really the same theing to ask: Under what circumstances can a vector field H(x,y) have both zero curl and zero divergence?

The answer is pretty easy: We know that H(x,y) has zero curl if and only if it is the gradient of some potential function h(x,y). Then the fact that div(H)(x,y) = 0 implies that

div(gradh(x,y)) = 0.

If we write this out explicitily in terms of the partial derivatives involved, this is

hxx(x,y) + hyy(x,y) = 0

The object we have just run into is called the Laplacean of h. And the equation just above, with the Laplacean of a function set equal to zero, is called Laplace's equation.

The study of this equation is enormously important in mathematics and physics. And although it is an example of a partial differential equation -- that is, an equation involving the partial derivatives of an unknown function -- many of its most important properties can be deduced using nothing more than the vector calculus that we now know. And indeed, this is true of the properties we need to understand the reconstruction of a vector field from its divergecne and curl!

Let's introduce one more piece of terminology: We say that a function h(x,y) is harmonic if and only if it satifies Laplaces equation.

So here is where we are: If H(x,y) has both zero curl and zero divergence, then H(x,y) iss the gradient of a harmonic potential function h(x,y).

But wait! There's more! We are more interested in the P(x,y) and Q(x,y) in H(x,y) = (P(x,y),Q(x,y) than in the potential function h(x,y). But

P(x,y) = hx(x,y)

and so the Laplcaean of P(x,y) is

(hx)xx(x,y) + (hx)yy(x,y) =

(hxx(x,y) + hyy(x,y))x = 0

since switching the order in which e take partial derivatives has no efffect on the result of the computation.

The conclusion of this is that P(x,y) itself is harmonic. And of course, the same reaoning applies to Q(x,y), so both components must be harmonic.

We have our first theroem in this investigation:

Theorem on Equal Curl and Divergence

For any vector two fields F(x,y) and G(x,y) that have the same divergence and curl, let H(x,y) denote their difference

H(x,y) = F(x,y) - G(x,y) .

Let P(x,y) and Q(x,y) denote the components of H(x,y), and suppose that they have continuous second order partial derivatives in x and y throughout the region being considered.

Then P(x,y) and Q(x,y) are both harmonic. That is, the differnce between F(x,y) and G(x,y) is a vector field with components that satisfy Laplace's equation.

Now there are infinitely many harmonic functions: excos(y), x2-y2 and ex2+y2sin(2*x*y) are all harmonic. So the result above may not seem to pin down the possibilities very much.

But it does -- notice that all of these examples are unbounded. The functions listed above take on arbitrarily big positive and negative values. And this is true in general:

Liouville's Theorem

If h(x,y) is both harmonic and bounded over the whole plane, then it is constant. In particular if h(x,y) is harmonic, and it tends to zero as (x,y) tends to infinity, then h(x,y) = 0 everywhere.

This has the following corrolary:

Theorem Curl, Divergence and Boundary Conditions

Consider any vector two fields F(x,y) and G(x,y) that have the same divergence and curl over the whole plane.

Suppose further that both of them satisfy the boundary condition that their components both tend to zero as (x,y) tends to infinity.

Then F(x,y) = G(x,y) everywhere.

The proof is obvious -- the difference must be harmonic, and it too much have the property that its components both tend to zero as (x,y) tends to infinity. This makes them bounded. Hence by Liouville's Theorem, they both are zero everywhere, and there is no difference between F(x,y) and G(x,y) anywhere.

So the electrostatic version of Maxwell's equations, together with the boundary condition, really do specify unique electric and magnetic fields. In a little while, we'll even see how to find them.

But first, there is the issue of Liouville's Theorem, which we pulled out of a hat, as it were.

Actully, it can be easily proved using an even more fundamental fact about harmonic functions -- the Mean Value Theorem for Harmonic Functions. This is deeper than the plain-vanilla mean value theorems of the calculus, but it can be proved using nothing but the divergence theorem!

Mean Value Theorem for Harmonic Functions

For all harmonic functions f(x,y) and any base point (x0,y0) and any radius R

That is, the value of f(x,y) at the center of any circle or disk of any radius, is equal to the average value of f(x,y) over that circle or disk.

Notice that Rdt is the arc length element along the circle of radius R and the lotal length is 2*pi*R, so that is why the first integral on the right gives the average value of f(x,y) over the circle. The disk is even more clear.

To prove this, fix (x0,y0), and introduce the function

g(x,y) = (1/2)*ln((x-x0)2 + (y-y0)2)

which is defined everywhere except at (x0,y0).

Let G(x,,y) denote it's gradient. This is of course also defined everywhere except at (x0,y0). By simple computation, we find

G(x,,y) = ((x-x0)2 + (y-y0)2)-1*(x-x0,y-y0)

If you compute the divergence of this, you get zero everywhere it is defined. So g(x,y) is harmonic throught the region where it is defined.

Next, let F(x,y) denote the gradient of our harmonic function f(x,y).

Now the stage is set: Form the vectorfield g(x,y)F(x,,y) and compute its divergence. You get

dot(G,F) + g(x,y)div(F)(x,,y) = dot(G,F)

since div(F)(x,,y) = div(gradf)(x,,y) = 0, since f is harmonic.

The result and the hypothesis are pretty symmetric in f and g, so let's switch their roles:

Form the vector field f(x,y)G(x,,y) and compute its divergence. You get

dot(F,G) + f(x,y)div(G)(x,y) = dot(F,G)

since div(G)(x,y) = div(gradg)(x,y) = 0 since g is harmonic.

Of course dot(G,F) = dot(F,G) so the two vector fields have the same divergence within their field of definition.

Now since dot(F,G) is the divergence of two different vector fields namely f(x,y)G(x,y) and g(x,y)F(x,y), we can compute the integral of dot(F,G) over any domain A on which it is defined two different ways using the divergence theorem.

Here is tthe one we shall consider: Let A be the anulus about (x0,y0) with inner radius r, and outer radius R. Let Cr denote the inner part of the boundary, and CR the outer

Then, applying the divergence theorem one way gives:

The last equality is true because dot(G,N) = 1/R on CR, and dot(G,N) = -1/r on Cr, as you can easily see by parameterizing the circles and computing, or, bettter yet, staring at the diagram.

Applying the divergence theorem the other way gives us:

Here, we have used the fact that g has the constant value ln(R) on CR, and the constant value ln(r) on Cr to eliminate explicit reference to g from the computation. Then we used the divergence theorem again to go back to an area integral of the divergergence of F. Here DR denotes the disk bounded by CR, and Dr denotes the disk bounded by Cr. But since the divergence of F is zero, since f is harmonic, these integrals are zero.

Hence the difference between the two integrals that we got in our first computation is zero, and those two integrals are the same. This is true for any R and r. Let's take the limit in which Cr shrinks down around (x0,y0); i.e., the limit in which r tends to zero. Explicitly parameterizing Cr in the usual way, we find:

The conclusion is that

which is the first part of the theorem.

To see the second part, replace put r in place of R, and multiply both sides by r, and finally, integrate from 0 to R. The result is the second part.

Of course, this isn't really neessary, since it is geometrically obvious that if the average over all circles is the value at the center, the same is true for disks, and that is what the second part says.

Either way, this completes the proof of the mean value theorem for harmonic functions.

Now lwt's prove Lioville's theorem. A picture is almost enough to extract it from the mean value theorem.

To do so, considered a bounded hrmonic function h(x,y), and fix two points (x0,y0) and (x1,y1).

Now, what we want to show is that

h(x0,y0) = h(x1,y1)<

To see this replace these values by the averages of h over disks of radius centered at these two points:

As you can see, most of the area in each of the disks is contained in the overlap of the two disks, and as R increases to infinity the percentage of the overlap increases to one hundred percent. Since h(x,y) is bounded, the contribution to the average coming from the small bit outside the overlap is also a negligible percentage, just like the area itself. (If h(x,y) were not bounded, the values in the small non-overlap region could make an overwhelming contribution to the average, getting bigger as fast, or faster than, this non-overlap region shrinks. That's exactly what would happen with the unbounded harmonic functions we gave in our examples.)

But in our case, with h(x,y) bounded, this doesn't happen, and for very large values of R, we are averaging the ssame function over essentially the same region, and we get essentially the same average value. Since the mean value theorem holds for all R, we can take the limit in which R tends to infinity, and then we get exactly the same value; i.e.,

h(x0,y0) = h(x1,y1)

by the mean value theorem.

This proves Liouville's Theorem, and hence we have also proved our theorem saying that together with a boundary condition, the divergence and the curl do determine the vector field.