5.7 Vector Fields that are Gradients or Curls

5.7 Vector Fields that are Gradients or Curls

  1. Vector Fields that are Gradients
  2. Vector Fields that are Curls
  3. Problems

  

Given a function f or a vector field F, we can easily compute f or curlF. Here we ask a sort of “inverse” question. Given a vector field G, can we determine:

We will discuss gradients first.

Vector Fields that are Gradients

Let U be an open subset of Rn for n2, and let G:URn be a continuous vector field. Then the following are equivalent:

  1. There exists a function f:UR of class C1 such that G=f.
  2. CGdx=0 for any closed piecewise smooth oriented curve C in U.
  3. C1Gdx=C2Gdx for any two piecewise smooth oriented curves C1,C2 in U that both start at the same point pU and end at the same point qU.
A continuous vector field G:URn is said to be conservative if any one of these conditions is satisfied.

The starting point of the proof is the fact that if G=f for some f of class C1, then for any oriented curve C that starts at a point p and ends at a point q, (1)CGdx=Cfdx=f(q)f(p). We saw this in Section 5.1, where we called it the Fundamental Theorem of Line Integrals.

1 implies 2
Suppose that G=f for some C1 function f. Then since a closed curve is just a curve that starts and ends at the same point p, it follows from that the integral of G=f over a closed curve equals f(p)f(p)=0.

2 implies 3(sketch) Let C1,C2 be two piecewise smooth curves in U that both start at the same point pU and end at the same point qU.

Given these, we may define a piecewise smooth closed curve C to be the curve formed by

thereby starting and ending at p. Then CGdx=C1GdxC2Gdx and it follows from this that if (2) holds, then C1GdxC2Gdx=0 for all C1,C2 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 G is a vector field that satisfies condition (3). If this function f exists, it must satisfy . Thus we can try to use formula to reconstruct f (if it exists) from G. That is, we can fix some pU, and we can look for f such that G=f and f(p)=0. According to (1), such a function f must satisfy (2)f(q)=Cp,qGdx for any curve Cp,q starting at p and ending at q. Thus, given G satisfying condition (3), we use equation (2) to define f:UR, and then verify that f=G. Condition (3) implies that the definition of f makes sense (that is, is independent of the choice of path Cp,q connecting p to q). Thus the proof of the theorem is completed by…

f satisfies f=G. (sketch)

Let f be the function defined in (2). We first claim that for any qU, any vector v, and any hR such that the line segment from q to q+hv is contained in U, (3)f(q+hv)=f(q)+0hG(q+tv)vdt. To see that this is true, fix some curve Cp,q that starts at p and ends at q. Let q,q+hv be the line segment that starts at q and ends at q+hv, and let Cp,q+hv be the curve obtained by

  • first following Cp,q from p to q;
  • then following q,q+hv from q to q+hv.

Thus Cp,q+hv is a piecewise smooth curve that starts at p and ends at q+hv. It follows from that f(q+hv)=Cp,q+hvGdx=Cp,qGdx+q,q+hvGdx=f(q)+q,q+hvGdx If we express the line integral over q,q+hv in terms of the parametrization g(t)=q+tv, 0th, this reduces to .

Next, for any j{1,,n}, if (as usual) ej denotes the unit vector in the jth coordinate direction, then it follows from that fxj(q)=limh0f(q+hej)f(q)h=limh01h0hG(q+tej)ejdt=limh01h0hGj(q+tej)dt. It is then an exercise to prove (using our assumption that G is continuous) then the limit exists and equals Gj(q). Since j is arbitrary, this proves that f(q)=G(q).

One drawback of Theorem 1 is that, given a vector field G, 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 G over every possible closed curve (for (2)) or pair of curves that start and end at the same point (for (3)). However, in 3 dimensions, and if G is C1, there is sometimes a much easier way to check whether it is conservative:

If G is a conservative vector field of class C1 on an open set UR3, then curlG=0.

If U is convex, then the converse is true:

If G:UR3 is a C1 vector field, U is convex, and curlG=0, then G is conservative.

However, on some non-convex sets, there exist non-conservative vector fields G that satisfy curlG=0. 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 G=gradf, then curlG=curlgradf=0.

For the converse, suppose that G is a C1 vector field on a convex set UR3 such that curlG=0.

Fix pU, and for qU, let p,q denote the line segment starting at p and ending at q. Since U is convex, this line segment is entirely contained in U.

We now define f(q)=p,qGdx. This is clearly well-defined, since we have specified exactly which path we follow to get from p to q, for every q.

In order to mimic the proof of Theorem 1, all we need is to be able to check that (4)f(q+hv)=f(q)+0hG(q+tv)vdt. 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 p,q and q+hv 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 S. For any choice of orientation (that is, of the direction n of the unit normal), S(curlG)ndA=0 since curlG=0 by assumption. Thus Stokes’ Theorem implies that SGdx=0 for any choice of the orientation of S. Note however that S consists of the three line segments connecting p to q to q+hv, and back to p. If it is oriented in that order (p to q to q+hv to p), then 0=SGdx = p,qGdx+q,q+hvGdx+q+hv,pGdx=f(q)+0hG(q+tv)vdtf(q+hv). We obtain by rearranging this.

An example of a non-conservative vector field G on a convex set that satisfies curlG=0 can be found in the exercises.

For conservative G, how to find f such that f=G?

Now suppose G is a continuous vector field on an open set URn, and that we somehow know that G is conservative. How can we find f such that f=G?

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 f. This works particularly well if U is a rectangle. Then given p=(a,b,c) and q=(x,y,z), we can always join p to q as follows:

  1. Change the first component from a to x, holding the other two components fixed. This straight-line path is parametrized by (t,b,c),t moves from a to x.
  2. Next change the second component from b to y, holding the other two components fixed. Since the first component has already been changed to x, this straight-line path is parametrized by (x,t,c),t moves from b to y.
  3. Finally, change the third component from c to z, holding the other two components fixed. This straight-line path is parametrized by (x,y,t),t moves from c to z.

Let Cp,q be the piecewise linear curve obtained in this way. Then Cp,qGdx=axG1(t,b,c)dt+byG2(x,t,c)dt+czG3(x,y,t)dt. So one way to implement formula is by: fix (a,b,c), and define (5)f(x,y,z)=axG1(t,b,c)dt+byG2(x,t,c)dt+czG3(x,y,t)dt. (Note, we could also change the components in a different order, if we prefer…)

The above theoretical considerations guarantee that if G is conservative, then f defined in this way must satisfy f=G.

Example 1.

Let G:R3R3 be defined by G(x,y,z)=(ysinxy,xsinxy+zcosyz,ycosyz). One can check that curlG=0. Thus G is conservative. Let us try to find f such that f=G. We will simply use the above formula, with (a,b,c)=(0,0,0). (We could choose any point, but (0,0,0) is convenient.)

Then axG1(t,b,c)dt=0x0dt=0, byG2(x,t,c)dt=0y(xsinxt+0)dt=cosxt|t=0t=y=cosxy1, and czG3(x,y,t)dt=0zycosytdt=sinyt|t=0z=sinyz. By adding the contributions from these three terms, we conclude that f(x,y,z)=cosxy+sinyz1 satisfies f=G. It is easy to check that this is indeed the case. (The constant 1 appears because, in choosing p=(0,0,0), we implicitly arranged that f(0,0,0)=0, and 1 is the constant that makes this be the case. If we had chosen a different point p=(a,b,c), then chances are that we would have gotten a different constant.

Remark. Note that in writing down a concrete implementation of (2), we could have chosen any curve from (a,b,c) to (x,y,z), such as (for example) a straight line. If the domain of G is a convex set containing the origin, then we can take (a,b,c)=(0,0,0), and the straight line to (x,y,z)is parametrized by g(t)=(tx,ty,tz) for 0t1. This leads to a the general formula f(x,y,z)=01G(tx,ty,tz)(x,y,z)dt. For example, for G(x,y,z)=(ysinxy,xsinxy+zcosyz,ycosyz) as above, this becomes (after some computation) f(x,y,z)=012txysin(t2xy)+2tyzcos(t2yz)dt This is easily evaluated by considering the two halves separately, and making th substitutions u=t2xy in the first half and u=t2yz in the second half.

Vector Fields that are Curls

There is a whole theory about vector fields G:UR3 (for U an open subset of R3) with the property that G=curlF for some other vector field F of class C1. 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:

  1. If G:UR3 is a vector field of class C1 and G=curlF for some vector field F:UR3 of class C2, then divG=0.

  2. Suppose G is a C1 vector field in an open set UR3 such that divG=0.
    a. If U is convex, then there exists a vector field F such that curlF=G.
    b. However, if U is not convex, it may be the case that no such vector field exists.

About the proof

1. We already know that divcurlF=0 for all F.

2b. Consider the vector field G(x,y,z)=(xr3,yr3,zr3), where  r=x2+y2+z2. This is an example showing that on a nonconvex set U (note that the domain of G is U=R3{0}), there can exist vector fields with zero divergence that are not curls. Imdeed, you can check that divG=0 for this G. You can also check that if S={(x,y,z)R3:x2+y2+z2=1}, oriented with the unit normal pointing outward, then SGndA=4π. whereas if G=curlF in U, then we would have SGndA=ScurlFndA=0. (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 G(x,y,z)=(xex2z26x,5y+2z,zzex2z2). One can check that divG=0, and the domain of G is all of R3, which is convex. So it must be possible to write G as the curl of some vector field F.

It turns out that in this situation, it is always possible to find F such that one of its components is zero everywhere. In this example, it turns out to be easiest to lok for F of the form F=(F1,0,F3). (This can be discovered by trial and error.) Then the equation that we are trying to solve, curlF=G, can be written as three equations: 2F3=G1=xex2z26x3F11F3=G2=5y+2z2F1=G3=zzex2z2. First we consider the first and third equations. For every fixed x,z, we may take the antiderivative with respect to the y variable to obtain F3(x,y,z)=(xex2z26x)dy=y(xex2z26x)+ϕ3(x,z),F1(x,y,z)=(zzex2z2)dy=y(zex2z2z)+ϕ1(x,z), where ϕ1(x,z) and ϕ3(x,z) are constants of integration, which are written in that way because they may depend on x and z. Substituting these into the second equation and simplifying, we obtain 5y+2z=G2=3F11F3=3ϕ11ϕ3+y(ex2z21)2yx2z2ex2z2[y(ex2z26)2yx2z2ex2z2]. After a lot of cancellation, this reduces to 3ϕ11ϕ3=2z and we can see by inspection that ths is solved for example by ϕ1(x,z)=0,ϕ3(x,z)=2xz. We conclude that curlF=G for F(x,y,z)=(yzex2z2yz, 0, yxex2z26xy2xz).

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 U is a rectangle, e.g. (a1,b1)×(a2,b2)×(a3,b3).

We suppose that G is C1 in U and that divG=0. We want to find F solving the system of equations 2F33F2=G13F11F3=G21F22F1=G3. We will look for F such that F3=0. Then the first two equations reduce to 3F2=G13F1=G2. We fix some a(a3,b3) and integrate both sides with respect to the third variable (calling it t instead of z for purposes of integration) getting constants of integration that depend on x and y: F2(x,y,z)=azG1(x,y,t)dt+ϕ(x,y)F1(x,y,z)=azG2(x,y,t)dt+ψ(x,y). We now substitute these into the third equation, differentiate under the integral sign, then simplify using the fact that divG=0: G3(x,y,z)=1F22F1=1ϕ2ψaz(1G1+2G2)(x,y,t)dt=1ϕ2ψ+az3G3(x,y,t) dt=1ϕ2ψ+G3(x,y,z)G3(x,y,a). Thus the equation reduces to 1ϕ(x,y)2ψ(x,y)=G3(x,y,a). and this can be solves for example by fixing a(a1,b1) and setting ψ(x,y)=0,ϕ(x,y)=axG3(t,y,a)dt .

Example 3

This is optional but interesting.

Suppose that RR2 is a regular region with piecewise smooth boundary, contained in [a,a]×[a,a] for some positive number a<π, and let S={(x,y,ϕ(x,y):(x,y)R} for ϕ(x,y)=ln(cosxcosy).

Below, an image of a portion of the graph of ϕ. For a suitable choice of the region R, this is what S would look like.

scherk

Also define G(x,y,z)=(xϕ,yϕ,1)1+|ϕ|2=(tanx,tany,1)1+tan2x+tan2y It is a fact that you can check, if you are unusually fond of differentiation, that G=0, and hence that there exists some F such that G=×F. It follows by Stokes Theorem that SGndA=SGndA for any other surface S with the same boundary as S.

It is easy to see that Gn=1 on S (in fact G exactly equals n on S). As a result for any surface S as above, area(S) = S1dA = SGndA = SGndA. Also, it is pretty clear that |G|=1 everywhere, and hence that |Gn||G||n|=1 everywhere on S. Therefore
SGndA  S|Gn|dA  S1dA=area(S). Putting these together, we conclude that area(S)area(S) for any surface S such that S=S. This shows that the particular surface S 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

  1. Given a vector field G:R3R3, determine whether whether there exists any functions f such that f=G, and if so, find one. For example:

  2. Given a vector field G:R3R3, determine whether there exist any vector fields F such that ×F=G, and if so find one. For example:

Advanced

  1. For U={(x,y,z)R3:x2+y2>0}, define G:UR3 by G(x,y,z)=(yx2+y2,xx2+y2,0).

  2. Prove that if g is a continuous function defined on an open set containing a point q, then for unit vector e, limh01h0hg(q+te)dt=g(q) (This point comes up in the proofs of Theorems 1 and 2).

  3. Suppose that G is a C2 vector field that is both a gradient and a curl.

  

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