$\newcommand{\curl}{\operatorname{curl}}$ $\newcommand{\div}{\operatorname{div}}$ $\newcommand{\grad}{\operatorname{grad}}$ $\newcommand{\R}{\mathbb R }$ $\newcommand{\N}{\mathbb N }$ $\newcommand{\Z}{\mathbb Z }$ $\newcommand{\bfa}{\mathbf a}$ $\newcommand{\bfb}{\mathbf b}$ $\newcommand{\bfc}{\mathbf c}$ $\newcommand{\bfe}{\mathbf e}$ $\newcommand{\bft}{\mathbf t}$ $\newcommand{\bfn}{\mathbf n}$ $\newcommand{\bff}{\mathbf f}$ $\newcommand{\bfF}{\mathbf F}$ $\newcommand{\bfk}{\mathbf k}$ $\newcommand{\bfg}{\mathbf g}$ $\newcommand{\bfG}{\mathbf G}$ $\newcommand{\bfh}{\mathbf h}$ $\newcommand{\bfu}{\mathbf u}$ $\newcommand{\bfv}{\mathbf v}$ $\newcommand{\bfx}{\mathbf x}$ $\newcommand{\bfp}{\mathbf p}$ $\newcommand{\bfq}{\mathbf q}$ $\newcommand{\bfy}{\mathbf y}$ $\newcommand{\ep}{\varepsilon}$

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 $\bfF$, we can easily compute $\grad f$ or $\curl \bfF$. Here we ask a sort of inverse question. Given a vector field $\bfG$, can we determine:

• does there exist some function $f$ such that $\bfG = \grad f$? If so, can we find $f$?
• does there exist some vector field $\bfF$ such that $\bfG = \curl \bfF$? If so, can we find $\bfF$?

We will discuss gradients first.

vector fields that are gradients

Theorem 1. Let $U$ be an open subset of $\R^n$ for $n\ge 2$, and let $\bfG:U\to \R^n$ be a continuous vector field. Then the following are equivalent:

(i) There exists a function $f:U\to \R$ of class $C^1$ such that $\bfG = \nabla f$.

(ii) $\int_C \bfG\cdot d\bfx = 0$ for any closed piecewise smooth oriented curve $C$ in $U$.

(iii) $\int_{C_1} \bfG\cdot d\bfx = \int_{C_2} \bfG\cdot d\bfx$ for any two piecewise smooth oriented curves $C_1, C_2$ in $U$ that both start at the same point $\bfp\in U$ and end at the same point $\bfq\in U$.

A continuous vector field $\bfG:U\to \R^n$ is said to be conservative if any one of these conditions is satisfied (and hence all three of them are satisfied).

The starting point of the proof is the fact that if $\bfG = \nabla f$ for some $f$ of class $C^1$, then for any oriented curve $C$ that starts at a point $\bfp$ and ends at a point $\bfq$, $$\begin{equation}\label{FFF} \int_C \bfG\cdot d\bfx = \int_C \nabla f\cdot d\bfx = f(\bfq) - f(\bfp). \end{equation}$$ (We saw this in Section 5.1, where we called it the Fundamental Theorem of Line Integrals.

(i) $\Rightarrow$ (ii). Assume that $\bfG = \nabla f$ for some $C^1$ function $f$. Then since a closed curve is just a curve that starts and ends at the same point $\bfp$, it follows from $\eqref{FFF}$ that the integral of $\bfG = \nabla f$ over a closed curve equals $f(\bfp)-f(\bfp) = 0$.

(ii) $\Rightarrow$ (iii). (sketch) Let $C_1, C_2$ be two peicewise smooth curves in $U$ that both start at the same point $\bfp\in U$ and end at the same point $\bfq\in U$.

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

• following $C_1$ from $\bfp$ to $\bfq$, then
• following $C_2$ backwards, from $\bfq$ to $\bfp$,

thereby starting and ending at $\bfp$. Then $$\int_C \bfG\cdot d\bfx = \int_{C_1} \bfG\cdot d\bfx - \int_{C_2} \bfG\cdot d\bfx$$ and it follows from this that if (2) holds, then $\int_{C_1} \bfG\cdot d\bfx - \int_{C_2} \bfG\cdot d\bfx =0$ for all $C_1, C_2$ as above, i.e. (3) holds. Filling in some of the details we have glossed over is an exercise.

(iii) $\Rightarrow$ (i). This is the hardest and most interesting part of the theorem. Suppose that $\bfG$ is a vector field that satisfies condition (3). If this function $f$ exists, it must satisfy $\eqref{FFF}$. Thus we can try to use formula $\eqref{FFF}$ to reconstruct $f$ (if it exists) from $\bfG$. That is, we can fix some $\bfp\in U$, and we can look for $f$ such that $\bfG=\nabla f$ and $f(\bfp)=0$. According to $\eqref{FFF}$, such a function $f$ must satisfy $$\begin{equation} f(\bfq) = \int_{C_{\bfp,\bfq}} \bfG\cdot d\bfx\qquad\mbox{ for any curve }C_{\bfp,\bfq}\mbox{ starting at }\bfp\mbox{ and ending at }\bfq. \label{FFFbis}\end{equation}$$ Thus, given $\bfG$ satisfying condition (3), we use equation $\eqref{FFFbis}$ to define $f:U\to \R$, and then verify that $\nabla f = \bfG$. Condition (3) implies that the definition of $f$ makes sense (that is, is independent of the choice of path $C_{\bfp, \bfq}$ connecting $\bfp$ to $\bfq$). Thus the proof of the theorem is completed by...

One drawback of Theorem 1 is that, given a vector field $\bfG$, 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 $\bfG$ 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 $\bfG$ is $C^1$, there is sometimes a much easier way to check whether it is conservative:

Theorem 2. If $\bfG$ is a conservative vector field of class $C^1$ on an open set $U\subset \R^3$, then $\curl \bfG = \bf0$.
If $U$ is convex, then the converse is true: if $\bfG:U\to \R^3$ is a $C^1$ vector field and $\curl \bfG = \bf0$, then $\bfG$ is conservative.
However, on some non-convex sets, there exist non-conservative vector fields $\bfG$ that satisfy $\curl \bfG = \bf 0$.

(This is a special case of a much more general theorem that we will neither state nor discuss.)

For conservative $\bfG$, how to find $f$ such that $\nabla f = \bfG$?

Now suppose $\bfG$ is a continuous vector field on an open set $U\subset \R^n$, and that we somehow know that $\bfG$ is conservative. How can we find $f$ such that $\nabla f = \bfG$?

One method is simply to carry out a concrete version of the abstract proof, sketched above, that uses formula $\eqref{FFFbis}$ to demonstrate of the existence of $f$. This works particularly well if $U$ is a rectangle. Then given $\bfp = (a,b,c)$ and $\bfq = (x,y,z)$, we can always join $\bfp$ to $\bfq$ as follows:

1. Change the first component from $a$ to $x$, holding the other two components fixed. This straight-line path is parametrized by $$(t,b,c), \qquad t\mbox{ moves from }a\mbox{ to }x.$$
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 $$(x,t,c), \qquad t\mbox{ moves from }b\mbox{ to }y.$$
3. Finally, change the third component from $c$ to $z$, holding the other two components fixed. This straight-line path is parametrized by $$(x,y,t), \qquad t\mbox{ moves from }c\mbox{ to }z.$$

Let $C_{\bfp, \bfq}$ be the piecewise linear curve obtained in this way. Then $$\int_{C_{\bfp, \bfq} }\bfG\cdot d\bfx = \int_a^x G_1(t,b,c)\, dt + \int_b^y G_2(x,t,c)\, dt + \int_c^z G_3(x,y,t)\, dt.$$ So one way to implement formula $\eqref{FFFbis}$ is by: fix $(a,b,c)$, and define $$\begin{equation}\label{inp} \boxed{ f(x,y,z) := \int_a^x G_1(t,b,c)\, dt + \int_b^y G_2(x,t,c)\, dt + \int_c^z G_3(x,y,t)\, dt.} \end{equation}$$ (Note, we could also change the components in a different order, if we prefer...)

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

Example 1. Let $\bfG:\R^3\to \R^3$ be defined by $$\bfG(x,y,z) = (-y \sin xy , -x \sin xy + z\cos yz, y\cos yz).$$ One can check that $\curl \bfG = \bf0$. Thus $\bfG$ is conservative. Let us try to find $f$ such that $\nabla f = \bfG$. 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 $$\int_a^x G_1(t,b,c)\, dt = \int_0^x 0\, dt = 0,$$ $$\left. \int_b^y G_2(x,t,c)\, dt = \int_0^y(-x\sin xt\, + 0)dt = \cos xt \right|_{t=0}^{t=y} = \cos xy - 1,$$ and $$\left. \int_c^zG_3(x,y,t)\, dt = \int_0^z y\cos yt \, dt = \sin yt\right|_{t=0}^z = \sin yz.$$ By adding the contributions from these three terms, we conclude that $$f(x,y,z) = \cos xy + \sin yz -1 \mbox{ satisfies }\nabla f = \bfG.$$ It is easy to check that this is indeed the case. (The constant $-1$ appears because, in choosing $\bfp = (0,0,0)$, 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 $\bfp = (a,b,c)$, then chances are that we would have gotten a different constant.

Remark. Note that in writing down a concrete implementation of $\eqref{FFFbis}$, 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 $\bfG$ 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 $\bfg(t) = (tx, ty, tz)$ for $0\le t\le 1$. This leads to a the general formula $$f(x,y,z) = \int_0^1 \bfG(tx, ty, tz)\cdot (x,y,z)\, dt.$$ For example, for $\bfG(x,y,z) = (-y \sin xy , -x \sin xy + z\cos yz, y\cos yz)$ as above, this becomes (after some computation) $$f(x,y,z) = \int_0^1 - 2txy\sin(t^2 xy) + 2tyz\cos(t^2yz)\, dt$$ This is easily evaluated by considering the two halves separately, and making th substitutions $u = t^2xy$ in the first half and $u =t^2yz$ in the second half.

vector fields that are curls

There is a whole theory about vector fields $\bfG:U\to \R^3$ (for $U$ an open subset of $\R^3$) with the property that $\bfG = \curl \bfF$ for some other vector field $\bfF$ of class $C^1$. 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:

Theorem 2.

1. If $\bfG:U\to \R^3$ is a vector field of class $C^1$ and $\bfG = \curl \bfF$ for some vector field $\bfF:U\to \R^3$ of class $C^2$, then $\operatorname{div}\bfG = 0.$

2. Suppose $\bfG$ is a $C^1$ vector field in an open set $U\subset \R^3$ such that $\operatorname{div}\bfG = 0$.
   a.. If $U$ is convex, then there exists a vector field $\bfF$ such that $\curl \bfF = \bfG$.
   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 $\div\curl \bfF = 0$ for all $\bfF$.

2b. Consider the vector field $$\bfG(x,y,z) = (\frac {x}{r^3},\frac {y}{r^3}, \frac {z}{r^3}) ,\qquad\mbox{ where }\ \quad r = \sqrt {x^2+y^2+z^2}.$$ This is an example showing that on a nonconvex set $U$ (note that the domain of $\bfG$ is $U = \R^3\setminus\{ {\bf 0}\}$), there can exist vector fields with zero divergence that are not curls. Imdeed, you can check that $\div \bfG = 0$ for this $\bfG$. You can also check that if $S := \{(x,y,z)\in \R^3 : x^2+y^2+z^2=1\}$, oriented with the unit normal pointing outward, then $$\iint_S \bfG\cdot \bfn \, dA = 4\pi.$$ whereas if $\bfG=\curl \bfF$ in $U$, then we would have $\iint_S \bfG\cdot \bfn \, dA =\iint_S \curl \bfF\cdot \bfn \, dA = 0$. (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 $$\bfG(x,y,z) = (xe^{-x^2z^2} -6x, 5y+2z, z - ze^{-x^2z^2}) .$$ One can check that $\operatorname{div}\bfG=0$, and the domain of $\bfG$ is all of $\R^3$, which is convex. So it must be possible to write $\bfG$ as the curl of some vector field $\bfF$.

It turns out (see Folland for a discussion) that in this situation, it is always possible to find $\bfF$ such that one of its components is zero everywhere. In this example, it turns out to be easiest to lok for $\bfF$ of the form $\bfF = (F_1, 0, F_3)$. (This can be discovered by trial and error.) Then the equation that we are trying to solve, $\curl \bfF = \bfG$, can be written as three equations: $$\begin{aligned} \partial_2 F_3 &= G_1 =xe^{-x^2z^2} -6x \\ \partial_3 F_1-\partial_1F_3 &= G_2 = 5y+2z\\ -\partial_2 F_1 &=G_3 = z - ze^{-x^2z^2}. \end{aligned}$$ 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 $$\begin{aligned} F_3(x,y,z) &= \int(xe^{-x^2z^2} -6x)dy = y(xe^{-x^2z^2} -6x) + \phi_3(x,z), \\ F_1(x,y,z) &= -\int(z - ze^{-x^2z^2})dy = y(ze^{-x^2z^2}-z) + \phi_1(x,z), \end{aligned}$$ where $\phi_1(x,z)$ and $\phi_3(x,z)$ 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 $$\begin{multline} 5y+2z = G_2 = \partial_3 F_1-\partial_1F_3 = \partial_3 \phi_1-\partial_1\phi_3 \\ + y(e^{-x^2z^2}-1) -2y x^2z^2 e^{-x^2z^2} - \left[ y(e^{-x^2z^2}- 6) - 2yx^2z^2 e^{-x^2z^2} \right]. \nonumber \end{multline}$$ After a lot of cancellation, this reduces to $$\partial_3 \phi_1-\partial_1\phi_3 = 2z$$ and we can see by inspection that ths is solved for example by $\phi_1(x,z) = 0, \phi_3(x,z) = -2xz$. We conclude that $\curl \bfF = \bfG$ for $$\bfF (x,y,z) = (yze^{-x^2z^2}-yz , \ 0 , \ yxe^{-x^2z^2} -6xy -2xz ).$$

Example 3. This is optional but interesting.

Assume that $R\subset \R^2$ is a regular region with piecewise smooth boundary, contained in $[-a,a]\times [-a,a]$ for some positive number $a<\pi$, and let $$S := \{ (x,y, \phi(x,y) : (x,y)\in R\} \quad\mbox{ for }\phi(x,y) = \ln(\frac {\cos x}{\cos y}).$$

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

$\qquad\qquad$

Also define $$\begin{aligned} \bfG(x,y,z) = \frac{(-\partial_x \phi, -\partial_y \phi , 1)}{\sqrt{1+|\nabla \phi|^2}}& = \frac{ (\tan x ,- \tan y, 1)}{ \sqrt{1+\tan^2 x +\tan^2 y}} %\\ %& = (\frac{\sin x}{\sqrt{1+\tan^2 y \cos^2x}} , \frac{-\sin y}{ %\sqrt{1+\tan^2 x\cos^2y}}, %\frac 1{\sqrt{1+\tan^2 x+\tan^2 y}}) \end{aligned}$$ It is a fact that you can check, if you are unusually fond of differentiation, that $\nabla \cdot \bfG = 0$, and hence that there exists some $\bfF$ such that $\bfG = \nabla\times \bfF$. It follows by Stokes Theorem that $$\iint_{S}\bfG\cdot\bfn \, dA = \iint_{S'}\bfG\cdot\bfn \, dA$$ for any other surface $S'$ with the same boundary as $S$.

It is easy to see that $\bfG \cdot \bfn = 1$ on $S$ (in fact $\bfG$ exactly equals $\bfn$ on $S$). As a result for any surface $S'$ as above, $$\mbox{area}(S) \ = \ \iint_S 1\, dA \ = \ \iint_S \bfG\cdot \bfn\, dA \ = \ \iint_{S'} \bfG\cdot \bfn\, dA .$$ Also, it is pretty clear that $|\bfG|=1$ everywhere, and hence that $|\bfG \cdot \bfn| \le |\bfG|\, |\bfn| = 1$ everywhere on $S'$. Therefore
$$\iint_{S'} \bfG\cdot \bfn\, dA \ \le \ \iint_{S'} |\bfG\cdot \bfn|\, dA \ \le \ \iint_{S'}1 \, dA = \mbox{area}(S').$$ Putting these together, we conclude that $$\mbox{area}(S) \le \mbox{area}(S')\mbox{ for any surface }S' \mbox{ such that }\partial S = \partial S'.$$ 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 skills

1. Given a vector field $\bfG:\R^3\to \R^3$, determine whether whether there exists any functions $f$ such that $\nabla f = \bfG$, and if so, find one. For example:

   • $\bfG(x,y,z) = ( \frac {e^z} {1+y^2}, - \frac {2xy e^z}{(1+y^2)^2}, \frac {xe^z}{1+y^2} +\sin z)$.

   • $\bfG(x,y,z) = (yz, xz,xy)$.

   • $\bfG(x,y,z) = (x\sin y \cos z , \frac 12 x^2 \cos y \cos z , -\frac 12 x^2 \sin y \sin z).$

   • $\bfG(x,y,z) = (\frac 12, \frac{y^2-xz}{y^2z} , \frac {-y}{z^2}) .$

   • $\bfG(x,y,z) = (\frac{\cos z}x , \frac {\cos z}y , -\sin z \ln(xyz) + \frac {\cos z}{z} )$, for $x,y,z$ all positive.

   • $\bfG$ is a vector field of the form $\bfG(x,y,z) = (f(x), g(y), h(z))$, where $f,g,h$ are all continuous functions of a single variable.

2. Given a vector field $\bfG:\R^3\to \R^3$, determine whether there exist any vector fields $\bfF$ such that $\nabla \times \bfF = \bfG$, and if so find one. For example:

   • $\bfG(x,y,z) = ( x,y , z )$.

   • $\bfG(x,y,z) = ( -y,x, 0)$.

   • $\bfG(x,y,z) = ( y, \frac 1 z \cos yz, \frac 1 y \cos yx)$.

   • $\bfG(x,y,z) = ( yz, xz, xy)$.

   • $\bfG$ is a vector field of the form $\bfG(x,y,z) = (f(y,z), g(x,z), h(x,y))$, where $f,g,h$ are all continuous functions of two variables.

Other questions

1. For $U := \{(x,y,z)\in \R^3 : x^2+y^2>0\}$, define $\bfG:U\to \R^3$ by $$\bfG(x,y,z) = (\frac {-y}{x^2+y^2}, \frac x{x^2+y^2}, 0)$$ .

   • Prove that $\curl \bfG = \bf 0$.
   • Prove that $\bfG$ is not conservative.
     (Hint: Consider $\int_C \bfG\cdot d\bfx$, where $C$ is a circle around the $z$-axis.))
   • possibly challenging, and not really recommended : if we define $U' := \{(x,y,z)\in \R^3 : x>0\}$, then we know that $\bfG$ is conservative on $U'$. (why?) Find $f$ such that $\nabla f = \bfG$ on $U'$.

2. Prove that if $g$ is a continuous function defined on an open set containing a point $\bfq$, then for unit vector $\bfe$, $$\lim_{h\to 0} \frac 1 h \int_0^h g(\bfq+t {\bf e})\, dt = g(\bfq)$$ (This point comes up in the proofs of Theorems 1 and 2).

3. Assume that $\bfG$ is a $C^2$ vector field that is both a gradient and a curl.

   • Show that if $f$ is a function such that $\nabla f = \bfG$, then $f$ is harmonic (that is, $\nabla^2 f = 0$, where we recall that $\nabla^2 f = \sum_{i=1}^3 \frac{\partial^2 f}{\partial x_i^2}$.)

   • Conclude that each component $G_i$ of $\bfG$ is also harmonic.

   $\Leftarrow$  $\Uparrow$  $\Rightarrow$