Given a function or a vector field , we can easily compute or . Here we ask a sort of “inverse†question. Given a vector field , can we determine:
does there exist some function such that ? If so, can we find ?
does there exist some vector field such that ? If so, can we find ?
We will discuss gradients first.
Vector Fields that are Gradients
Let be an open subset of for , and let be a continuous vector field. Then the following are equivalent:
There exists a function of class such that .
for any closed piecewise smooth oriented curve in .
for any two piecewise smooth oriented curves in that both start at the same point and end at the same point .
A continuous vector field is said to be conservative if any one of these conditions is satisfied.
The starting point of the proof is the fact that if for some of class , then for any oriented curve that starts at a point and ends at a point , We saw this in Section 5.1, where we called it the Fundamental Theorem of Line Integrals.
1 implies 2
Suppose that for some function . Then since a closed curve is just a curve that starts and ends at the same point , it follows from that the integral of over a closed curve equals .
2 implies 3(sketch) Let be two piecewise smooth curves in that both start at the same point and end at the same point .
Given these, we may define a piecewise smooth closed curve to be the curve formed by
following from to , then
following backwards, from to ,
thereby starting and ending at . Then and it follows from this that if (2) holds, then for all as above, i.e.(3) holds. Filling in some of the details we have glossed over is an exercise.
3 implies 1
This is the hardest and most interesting part of the theorem. Suppose that is a vector field that satisfies condition (3). If this function exists, it must satisfy . Thus we can try to use formula to reconstruct (if it exists) from . That is, we can fix some , and we can look for such that and . According to , such a function must satisfy Thus, given satisfying condition (3), we use equation to define, and then verify that . Condition (3) implies that the definition of makes sense (that is, is independent of the choice of path connecting to ). Thus the proof of the theorem is completed by…
satisfies . (sketch)
Let be the function defined in . We first claim that for any , any vector , and any such that the line segment from to is contained in , To see that this is true, fix some curve that starts at and ends at . Let be the line segment that starts at and ends at , and let be the curve obtained by
first following from to ;
then following from to .
Thus is a piecewise smooth curve that starts at and ends at . It follows from that If we express the line integral over in terms of the parametrization , , this reduces to .
Next, for any , if (as usual) denotes the unit vector in the th coordinate direction, then it follows from that It is then an exercise to prove (using our assumption that is continuous) then the limit exists and equals . Since is arbitrary, this proves that .
One drawback of Theorem 1 is that, given a vector field , it might be hard to check whether it satisfies condition (2) or (3). In order to do this, one would need to evaluate line integrals of over every possible closed curve (for (2)) or pair of curves that start and end at the same point (for (3)). However, in dimensions, and if is , there is sometimes a much easier way to check whether it is conservative:
If is a conservative vector field of class on an open set , then .
If is convex, then the converse is true:
If is a vector field, is convex, and , then is conservative.
However, on some non-convex sets, there exist non-conservative vector fields that satisfy . This is a special case of a much more general theorem that we will neither state nor discuss.
Sketch of proof.
We already know that if , then .
For the converse, suppose that is a vector field on a convex set such that .
Fix , and for , let denote the line segment starting at and ending at . Since is convex, this line segment is entirely contained in .
We now define This is clearly well-defined, since we have specified exactly which path we follow to get from to , for every .
In order to mimic the proof of Theorem 1, all we need is to be able to check that holds. If we know this, then every other argument in the earlier proof can be repeated with no change.
To prove it, let’s suppose that and are not colinear (in that case one can give a different, easier argument). Then they form the vertices of a triangle. Let’s call the triangle . For any choice of orientation (that is, of the direction of the unit normal), since by assumption. Thus Stokes’ Theorem implies that for any choice of the orientation of . Note however that consists of the three line segments connecting to to , and back to . If it is oriented in that order ( to to to ), then We obtain by rearranging this.
An example of a non-conservative vector field on a convex set that satisfies can be found in the exercises.
For conservative , how to find such that ?
Now suppose is a continuous vector field on an open set , and that we somehow know that is conservative. How can we find such that ?
One method is simply to carry out a concrete version of the abstract proof, sketched above, that uses formula to demonstrate of the existence of . This works particularly well if is a rectangle. Then given and , we can always join to as follows:
Change the first component from to , holding the other two components fixed. This straight-line path is parametrized by
Next change the second component from to , holding the other two components fixed. Since the first component has already been changed to , this straight-line path is parametrized by
Finally, change the third component from to , holding the other two components fixed. This straight-line path is parametrized by
Let be the piecewise linear curve obtained in this way. Then So one way to implement formula is by: fix , and define (Note, we could also change the components in a different order, if we prefer…)
The above theoretical considerations guarantee that if is conservative, then defined in this way must satisfy .
Example 1.
Let be defined by One can check that . Thus is conservative. Let us try to find such that . We will simply use the above formula, with . (We could choose any point, but is convenient.)
Then and By adding the contributions from these three terms, we conclude that It is easy to check that this is indeed the case. (The constant appears because, in choosing , we implicitly arranged that , and is the constant that makes this be the case. If we had chosen a different point , then chances are that we would have gotten a different constant.
Remark. Note that in writing down a concrete implementation of , we could have chosen any curve from to , such as (for example) a straight line. If the domain of is a convex set containing the origin, then we can take , and the straight line to is parametrized by for . This leads to a the general formula For example, for as above, this becomes (after some computation) This is easily evaluated by considering the two halves separately, and making th substitutions in the first half and in the second half.
Vector Fields that are Curls
There is a whole theory about vector fields (for an open subset of ) with the property that for some other vector field of class . It is very much parallel to the theory of gradient (= conservative) vector fields. However, we considered it in less detail.
Some main facts are summarized in the following:
If is a vector field of class and for some vector field of class , then
Suppose is a vector field in an open set such that . a. If is convex, then there exists a vector field such that . b. However, if is not convex, it may be the case that no such vector field exists.
About the proof
1. We already know that for all .
2b. Consider the vector field This is an example showing that on a nonconvex set (note that the domain of is ), there can exist vector fields with zero divergence that are not curls. Imdeed, you can check that for this . You can also check that if , oriented with the unit normal pointing outward, then whereas if in , then we would have . (See Section 5.6.)
2a. The way the proof works is illustrates below for a concrete example. After that we will discus the abstract proof.
Example 2.
Let One can check that , and the domain of is all of , which is convex. So it must be possible to write as the curl of some vector field .
It turns out that in this situation, it is always possible to find such that one of its components is zero everywhere. In this example, it turns out to be easiest to lok for of the form . (This can be discovered by trial and error.) Then the equation that we are trying to solve, , can be written as three equations: First we consider the first and third equations. For every fixed , we may take the antiderivative with respect to the variable to obtain where and are constants of integration, which are written in that way because they may depend on and . Substituting these into the second equation and simplifying, we obtain After a lot of cancellation, this reduces to and we can see by inspection that ths is solved for example by . We conclude that for
Proof of part 2a of Theorem 2
We will just follow the same procedure as in Example 2. We will also cheat a little cheat by proving the theorem not for an arbitrary convex set, but instead under the assumption that is a rectangle, e.g..
We suppose that is in and that . We want to find solving the system of equations We will look for such that . Then the first two equations reduce to We fix some and integrate both sides with respect to the third variable (calling it instead of for purposes of integration) getting constants of integration that depend on and : We now substitute these into the third equation, differentiate under the integral sign, then simplify using the fact that : Thus the equation reduces to and this can be solves for example by fixing and setting
Example 3
This is optional but interesting.
Suppose that is a regular region with piecewise smooth boundary, contained in for some positive number , and let
Below, an image of a portion of the graph of . For a suitable choice of the region , this is what would look like.
Also define It is a fact that you can check, if you are unusually fond of differentiation, that , and hence that there exists some such that . It follows by Stokes Theorem that for any other surface with the same boundary as .
It is easy to see that on (in fact exactly equals on ). As a result for any surface as above, Also, it is pretty clear that everywhere, and hence that everywhere on . Therefore Putting these together, we conclude that This shows that the particular surface pictured above is an example of what is called a minimal surface: a surface that has the smallest possible area, among all surfaces with a given boundary.
Problems
Basic
Given a vector field , determine whether whether there exists any functions such that , and if so, find one. For example:
.
.
, for all positive.
is a vector field of the form , where are all continuous functions of a single variable.
Given a vector field , determine whether there exist any vector fields such that , and if so find one. For example:
.
.
.
.
is a vector field of the form , where are all continuous functions of two variables.
Advanced
For , define by .
Prove that .
Prove that is not conservative.
Hint Consider , where is a circle around the -axis.
If we define , then we know that is conservative on . (why?) Find such that on .
Prove that if is a continuous function defined on an open set containing a point , then for unit vector , (This point comes up in the proofs of Theorems 1 and 2).
Suppose that is a vector field that is both a gradient and a curl.
Show that if is a function such that , then is harmonic, that is, , where we recall that the Laplacian is \(\nabla^2 f = \dfrac{\partial^2 f}{\partial x_1^2}+\dfrac{\partial^2 f}{\partial x_2^2\dfrac{\partial^2 f}{\partial x_3^2}.\)