* Theorem.* If is a vector field in the plane,
and P and Q have continuous partial derivatives on a region. the
following four statements are equivalent:

1. for some function .

2. .

3. If is a * closed
curve* lying in the region --- i.e. a path which starts and ends
at the same point --- then

4. (* Path independence*) If and are paths in
the region which start at the same point and end at the same point,
then

To say that these statements are equivalent means that if one of them
is true, then all of them are true (and if one of them is false, all
of them are false). A field that satisfies any one of these
conditions is a * conservative field* --- or
sometimes a * gradient field*, or sometimes * path independent*.

Before I show that these statements are equivalent, I'll give a couple of examples.

* Example.* Show that is a gradient field.

I want a function f such that , i.e.

Integrate the first equation with respect to x:

Since the integral is with respect to x, y is constant, and must be included in the arbitrary constant --- hence . Differentiate with respect to y and set the result equal to above:

I get , so . This time, D is a *numerical*
constant. Since the derivative of a number is 0, and since I just
want *some* potential function (see the previous example), I
might as well take . Then , so

Thus, if , then

* Definition.* A function f such that is called a * potential
function* for .

* Example.* Let and

Show that

The velocity vector is

The field is

Hence,

So

* Theorem.* Suppose is a gradient field, and let for be a path.
Then

In other words, to evaluate the integral of a gradient field, just plug the endpoints of the path ( and ) into the potential function (f).

* Proof.* By the Chain Rule,

So

The antiderivative of the derivative of is just , so

All together,

* Example.* Compute , where

To compute this directly, you would need to do the integral

Yuck.

Instead, notice that if , then , which is the field in the integral. Hence, I can compute the integral by plugging the endpoints of the path into f.

So

(The fact that it comes out to 0, as opposed to a nonzero number, is a coincidence.) The important thing to notice is how easy it was to do the computation!

Now I will go back and prove that the four statements which define a conservative field are equivalent. To prove that they're equivalent, I must show that any one of them follows from any other. I will do it this way:

* Proof.* First, assume statement 1 is true, so
for some f. Since , this means that

Hence,

But the two second derivatives are equal by equality of mixed partials, so . This is statement 2.

Next, assume statement 2 is true, so . Take a closed curve . I want to show that the integral around
is 0. This follows from *
Green's theorem*, which I'll discuss in more detail later. For
now, note that the closed curve encloses a
region R.

Green's theorem says

But the double integral on the right is 0, because . Therefore, the integral around a closed curve is 0, and that is statement 3.

Before I do the next step, here is some notation. If is a curve, will denote the same curve traversed in the opposite direction.

If I traverse a curve backward, the line integral flips its sign:

Now assume statement 3 is true: The line integral around a closed curve is 0. Take curves and , both of which start at P and both of which end at Q. I need to show that

Note that is a closed curve. So the integral around is 0:

Do each path separately:

The second integral is the negative of the integral along :

Finally, move the second integral to the other side:

This is what I wanted to prove, so statement 4 follows from statement 3.

Finally, suppose statement 4 is true: The field is path independent.
I want to show that is a
gradient field. I need to find a function f such that . To do this, take *any* path from to and *define*

By path independence, it does not matter what path I choose. Now I have to show that and .

Consider the path from to made up of segments as shown below:

The horizontal part is for . The velocity vector is , and . So

Therefore, the integral for the horizontal segment is

The vertical part is for . The velocity vector is and . So

Therefore, the integral for the vertical segment is

The line integral along the whole path --- which by definition is f --- is

Remember that I was trying to show that and . Differentiate the last equation with respect to y.

The first integral only involves x, so its derivative with respect to y is 0. For the second integral, apply the Fundamental Theorem of Calculus: The y in the top limit replaces the t in the integrand and I get . So

That is, f has the right y derivative.

If you take a path made up of segments going from to the "other way" --- along the y-axis, then horizontally --- you can show in similar fashion that . This proves that , which is statement 1.

And that finishes the proof that the four statements are equivalent.

The last part of the proof gives a way of constructing a potential function for a gradient field. It's perhaps not the best way for a human being to do this, but would be a reasonable approach if you were doing this on a computer.

* Example.* Find a potential function for using the line integrals in
the proof of the theorem.

I can find f using the formula

Here , so . Likewise, , so . So

By the way, notice that is also a potential function for this field, since and . There are infinitely many potential functions for a gradient field; they differ by a numerical constant.

So far, I've discussed vector fields in , 2 dimensions. There are few surprises when you move up to 3 dimensions.

* Theorem.* Let be a 3-dimensional vector field,
and assume its components have continuous partial derivatives. Then
the following four statements are equivalent:

1. for some function .

2. .

3. If is a * closed
curve* --- i.e. a path which starts and ends at the same point
--- then

4. (* Path independence*) If and are paths which
start at the same point and end at the same point, then

Once again, a field that satisfies any one of these conditions is a
* conservative field* --- or sometimes a * gradient field*, or sometimes *
path independent*.

The only change in moving from 2 dimensions to 3 dimensions is in statement 2. To see how this is related to the for a 2-dimensional field, take a field and regard it as a 3-dimensional field by making the third component 0, so

Then

(Notice that, e.g., , because P does not contain any z's.) The condition is, in this case, the same as .

In fact, the components of should have
continuous first partials, *except perhaps at finitely many
points*. Here's the reason I can allow finitely many "bad
points" in this case but none in the 2-dimensional case.

Think of a closed curve as a piece of string. In a rough sense, the reason why the integral of a conservative field around a closed curve is 0 is that the string can be "reeled in" to the basepoint without changing the integral.

When the curve has been reeled in to a single point, the integral over the curve is obviously 0.

In 2 dimensions, a curve can "get stuck" on a single bad point as you reel it in.

However, in 3 dimensions, you have enough room to move the curve around finitely many bad points as you reel it in.

* Example.* Show that is
conservative, and find a potential function for .

Therefore, is conservative. A potential function f must satisfy

Integrate the first equation with respect to x:

Since the integral is with respect to x, y and z are constant, and must be included in the arbitrary constant . Now differentiate with respect to y and set the result equal to above:

I find that , so integrating with respect to y,

This time, z is constant and must be included in the arbitrary constant . Plug this back into f; it is

Finally, differentiate with respect to z and set the result equal to above:

I find that , so , where E is a numerical constant. As in an earlier example, I may take . This gives , so

* Example.* Compute , where

You could compute the integral directly --- would you want to? There must be an easier way ...

Since , the field is conservative. If I can find a potential function, I can compute the integral by simply plugging the endpoints of the path into the potential function.

A potential function f must satisfy

Integrate the first equation with respect to x:

Since the integral is with respect to x, y and z are constant, and must be included in the arbitrary constant . Now differentiate with respect to y and set the result equal to above:

I find that , so integrating with respect to y,

This time, z is constant and must be included in the arbitrary constant . Plug this back into f; it is

Finally, differentiate with respect to z and set the result equal to above:

I find that , so , where E is a numerical constant. As in an earlier example, I may take . This gives , so

The endpoints of the path are

Therefore,

* Example.* is a path
with positive components from a point P on to a point Q on . Compute , where

The path isn't given --- in fact, its endpoints aren't given.
Therefore, *the path must not matter*, i.e. the integral is
probably path independent. In fact, , for .
Therefore,

Now Q is on , so for this point . Likewise, P is on , so for this point . Hence,

Copyright 2018 by Bruce Ikenaga