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Math 32B Discussion Session Week 4 Notes April 19 and 21, 2016
We’ve changed gears this week. So far this quarter we’ve focused on integrating functions over two- and three-dimensional regions. Next we’d like to do vector integral calculus. First we’ll need some basic operations on vector fields — operations which we will later undo with integration.
Divergence
The first characteristic of a vector field we’d like to measure is the degree to which it is expanding or contracting at a given point. For example, consider the following vector fields on R^2 : It’s pretty clear to see that the vector field in (1a) is expanding at the origin and that
(a) Outward flow. (b) Inward flow.
(c) Cross flow. (d) Confused flow.
the vector field in (1b) is contracting a the origin (in fact they’re expanding and contracting, respectively, everywhere). We make this idea of expanding or contracting ever-so-slightly more accessible with the orange circle. In (1a) we see the vector field pushing out of the circle, and we consider this vector field to have positive divergence. The vector field in
(1b), on the other hand, is pushing into our circle, and we consider this vector field to have negative divergence.
The vector fields in (1c) and (1d) are a little different. In (1c), the vector field is flowing smoothly, neither expanding nor contracting. In particular, the vector field is pushing matter into our region in the third quadrant and at the same time pushing matter out of our region in the first quadrant at the same rate. The net change is zero, so we consider this vector field to have zero divergence. The vector field in (1d) also has zero divergence, but it’s less obvious. In this case, our vector field is pushing matter into the circle along the y-axis and out of the circle along the x-axis, and the net change is zero.
We would like to define an operator on vector fields, called the divergence operator, which measures the extent to which a vector field is pushing matter into or out of a particular point. For now we’ll skip a rigorous explanation of why the following definition is what does this, but we’ll try to touch on this in discussion session, and I may add an explanation to these notes later.
Definition. Let F(x, y) be a vector field on R^2. We define the divergence of F to be the scalar quantity
div(F) :=
∂F 1
∂x
∂F 2
∂y
where F 1 and F 2 are the component functions of F. Similarly, if F(x, y, z) is a vector field in R^3 , we define the divergence by
div(F) :=
∂F 1
∂x
∂F 2
∂y
∂F 3
∂z
A succinct way to write our definition is as
div(F) = ∇ · F,
where ∇ is the “vector”
∇ =
∂x
∂y
∂z
Example. Compute the divergence of the vector field on R^3 given by
F(x, y, z) := 〈xy, yz, y^2 − x^3 〉.
(Solution) We have
div(F) =
∂F 1
∂x
∂F 2
∂y
∂F 3
∂z
=
∂x
(xy) +
∂y
(yz) +
∂z
(y^2 − x^3 )
= y + z + 0 = y + z.
obvious. In our notation, the divergence of the curl vector field is given by the triple scalar product div(curl(F)) = ∇ · (∇ × F).
We know that when a vector makes two appearances in a triple scalar product, the product is zero. In this case, the “vector” that appears twice, ∇, is actually an operator, so we need to carefully check that this product does indeed vanish.
Conservative Vector Fields
Finally, we consider ∇ as an operator not on vector fields, but on functions. Given a function f (x, y, z), the gradient vector field, which you already encountered in 32A, of f is the vector field given by
F = ∇f =
∂f ∂x
∂f ∂y
∂f ∂z
We label the relationship between f and F by calling F a conservative vector field and saying that f is a potential function for F when ∇f = F. Not all vector fields are conservative (in fact, most aren’t), but those which are tend to have some very nice properties. One particular property is given below.
Theorem 1. If the vector field F(x, y, z) on R^3 is conservative, then curl(F).
Note that the converse of this statement is not true: there are vector fields F such that curl(F) = 0, but which are not conservative.
Example. For the vector field
F =
yz^2 , xz^2 , 2 xyz
either find a potential function f (x, y, z) or show that no such function exists.
(Solution) If we want to show that no potential function exists for F, we need to show that F is not conservative. At present, the only way we can do this is to show that curl(F) is nonzero. But curl(F) = 〈 2 xz − 2 xz, 2 yz − 2 yz, z^2 − z^2 〉 = 〈 0 , 0 , 0 〉,
so this approach won’t work. (Note that just because curl(F) = 0 does not mean that F is conservative.) Perhaps F does admit a potential function. If so, f is a function so that
∂f ∂x
= yz^2 ,
∂f ∂y
= xz^2 ,
∂f ∂z
= 2xyz.
From the first equation we see that
f (x, y, z) = xyz^2 + g(y, z),
where g is some function of y and z, independent of x. Then from the second equality,
xz^2 =
∂f ∂y
= xz^2 +
∂g ∂y
so ∂g/∂y = 0, meaning that g does not depend on y. Finally,
2 xyz =
∂f ∂z
= 2xyz +
∂g ∂z
meaning that ∂g/∂z = 0, and thus that g does not depend on z either. So g is constant, and we may write f (x, y, z) = xyz^2 + K.
A quick check confirms that ∇f = F.
