5.7 Vector Fields that are Gradients or Curls

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

$\Leftarrow$  $\Uparrow$

Given a function $f$ or a vector field $\mathbf{F}$, we can easily compute $\mathrm{\nabla }f$ or $\mathrm{curl}\mathbf{F}$. Here we ask a sort of “inverse” question. Given a vector field $\mathbf{G}$, can we determine:

• does there exist some function $f$ such that $\mathbf{G}=\mathrm{\nabla }f$? If so, can we find $f$?
• does there exist some vector field $\mathbf{F}$ such that $\mathbf{G}=\mathrm{curl}\mathbf{F}$? If so, can we find $\mathbf{F}$?

We will discuss gradients first.

Vector Fields that are Gradients

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

If $U$ is convex, then the converse is true:

However, on some non-convex sets, there exist non-conservative vector fields $\mathbf{G}$ that satisfy $\mathrm{curl}\mathbf{G}=\mathbf{0}$. This is a special case of a much more general theorem that we will neither state nor discuss.

For conservative $\mathbf{G}$, how to find $f$ such that $\mathrm{\nabla }f=\mathbf{G}$?

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

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 $\mathbf{p}=(a,b,c)$ and $\mathbf{q}=(x,y,z)$, we can always join $\mathbf{p}$ to $\mathbf{q}$ as follows:

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

Let $C_{\mathbf{p},\mathbf{q}}$ be the piecewise linear curve obtained in this way. Then $${\int }_{C_{\mathbf{p},\mathbf{q}}}\mathbf{G}\cdot d\mathbf{x}={\int }_a^x{G}_1(t,b,c)dt+{\int }_b^y{G}_2(x,t,c)dt+{\int }_c^z{G}_3(x,y,t)dt.$$ So one way to implement formula is by: fix $(a,b,c)$, and define $$\begin{array}{}\text{(5)}& \boxed{{\displaystyle f(x,y,z)={\int }_a^x{G}_1(t,b,c)dt+{\int }_b^y{G}_2(x,t,c)dt+{\int }_c^z{G}_3(x,y,t)dt.}}\end{array}$$ (Note, we could also change the components in a different order, if we prefer…)

The above theoretical considerations guarantee that if $\mathbf{G}$ is conservative, then $f$ defined in this way must satisfy $\mathrm{\nabla }f=\mathbf{G}$.

Example 1.

Let $\mathbf{G}:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be defined by $$\mathbf{G}(x,y,z)=(-y\sin xy,-x\sin xy+z\cos yz,y\cos yz).$$ One can check that $\mathrm{curl}\mathbf{G}=\mathbf{0}$. Thus $\mathbf{G}$ is conservative. Let us try to find $f$ such that $\mathrm{\nabla }f=\mathbf{G}$. 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 $${\int }_a^x{G}_1(t,b,c)dt={\int }_0^{x}0dt=0,$$ $${{\int }_b^y{G}_2(x,t,c)dt={\int }_0^y(-x\sin xt+0)dt=\cos xt|}_{t=0}^{t=y}=\cos xy-1,$$ and $${{\int }_c^z{G}_3(x,y,t)dt={\int }_0^{z}y\cos ytdt=\sin yt|}_{t=0}^z=\sin yz.$$ By adding the contributions from these three terms, we conclude that $$f(x,y,z)=\cos xy+\sin yz-1\text{ satisfies }\mathrm{\nabla }f=\mathbf{G}.$$ It is easy to check that this is indeed the case. (The constant $-1$ appears because, in choosing $\mathbf{p}=(0,0,0)$, 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 $\mathbf{p}=(a,b,c)$, then chances are that we would have gotten a different constant.

Remark. Note that in writing down a concrete implementation of $\text{(2)}$, 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 $\mathbf{G}$ 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 $\mathbf{g}(t)=(tx,ty,tz)$ for $0\le t\le 1$. This leads to a the general formula $$f(x,y,z)={\int }_0^1\mathbf{G}(tx,ty,tz)\cdot (x,y,z)dt.$$ For example, for $\mathbf{G}(x,y,z)=(-y\sin xy,-x\sin xy+z\cos yz,y\cos yz)$ as above, this becomes (after some computation) $$f(x,y,z)={\int }_0^1-2txy\sin (t^{2}xy)+2tyz\cos (t^{2}yz)dt$$ This is easily evaluated by considering the two halves separately, and making th substitutions $u=t^{2}xy$ in the first half and $u=t^{2}yz$ in the second half.

Vector Fields that are Curls

There is a whole theory about vector fields $\mathbf{G}:U\to {\mathbb{R}}^3$ (for $U$ an open subset of ${\mathbb{R}}^3$) with the property that $\mathbf{G}=\mathrm{curl}\mathbf{F}$ for some other vector field $\mathbf{F}$ 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:

Example 2.

Let $$\mathbf{G}(x,y,z)=(x{e}^{-x^2{z}^2}-6x,5y+2z,z-z{e}^{-x^2{z}^2}).$$ One can check that $\mathrm{div}\mathbf{G}=0$, and the domain of $\mathbf{G}$ is all of ${\mathbb{R}}^3$, which is convex. So it must be possible to write $\mathbf{G}$ as the curl of some vector field $\mathbf{F}$.

It turns out that in this situation, it is always possible to find $\mathbf{F}$ such that one of its components is zero everywhere. In this example, it turns out to be easiest to lok for $\mathbf{F}$ of the form $\mathbf{F}=(F_1,0,F_3)$. (This can be discovered by trial and error.) Then the equation that we are trying to solve, $\mathrm{curl}\mathbf{F}=\mathbf{G}$, can be written as three equations: $$\begin{array}{rl}{\partial }_2{F}_3& =G_1=x{e}^{-x^2{z}^2}-6x\\ {\partial }_3{F}_1-{\partial }_1{F}_3& =G_2=5y+2z\\ -{\partial }_2{F}_1& =G_3=z-z{e}^{-x^2{z}^2}.\end{array}$$ 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 $$\begin{array}{rl}{F}_3(x,y,z)& =\int (x{e}^{-x^2{z}^2}-6x)dy=y(x{e}^{-x^2{z}^2}-6x)+{\varphi }_3(x,z),\\ F_1(x,y,z)& =-\int (z-z{e}^{-x^2{z}^2})dy=y(z{e}^{-x^2{z}^2}-z)+{\varphi }_1(x,z),\end{array}$$ where ${\varphi }_1(x,z)$ and ${\varphi }_3(x,z)$ 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 $$\begin{array}{c}5y+2z=G_2={\partial }_3{F}_1-{\partial }_1{F}_3={\partial }_3{\varphi }_1-{\partial }_1{\varphi }_3\hfill \\ \hfill +y(e^{-x^2{z}^2}-1)-2y{x}^2{z}^2{e}^{-x^2{z}^2}-[y(e^{-x^2{z}^2}-6)-2y{x}^2{z}^2{e}^{-x^2{z}^2}].\end{array}$$ After a lot of cancellation, this reduces to $${\partial }_3{\varphi }_1-{\partial }_1{\varphi }_3=2z$$ and we can see by inspection that ths is solved for example by ${\varphi }_1(x,z)=0,{\varphi }_3(x,z)=-2xz$. We conclude that $\mathrm{curl}\mathbf{F}=\mathbf{G}$ for $$\mathbf{F}(x,y,z)=(yz{e}^{-x^2{z}^2}-yz,0,yx{e}^{-x^2{z}^2}-6xy-2xz).$$

Example 3

This is optional but interesting.

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

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

$$

Also define $$\begin{array}{rl}\mathbf{G}(x,y,z)=\frac{(-{\partial }_x\varphi ,-{\partial }_y\varphi ,1)}{\sqrt{1+|\mathrm{\nabla }\varphi {|}^2}}& =\frac{(\tan x,-\tan y,1)}{\sqrt{1+{\tan }^{2}x+{\tan }^{2}y}}\end{array}$$ It is a fact that you can check, if you are unusually fond of differentiation, that $\mathrm{\nabla }\cdot \mathbf{G}=0$, and hence that there exists some $\mathbf{F}$ such that $\mathbf{G}=\mathrm{\nabla }\times \mathbf{F}$. It follows by Stokes Theorem that $${\iint }_S\mathbf{G}\cdot \mathbf{n}dA={\iint }_{S^{\prime }}\mathbf{G}\cdot \mathbf{n}dA$$ for any other surface $S^{\prime }$ with the same boundary as $S$.

It is easy to see that $\mathbf{G}\cdot \mathbf{n}=1$ on $S$ (in fact $\mathbf{G}$ exactly equals $\mathbf{n}$ on $S$). As a result for any surface $S^{\prime }$ as above, $$\text{area}(S)={\iint }_{S}1dA={\iint }_S\mathbf{G}\cdot \mathbf{n}dA={\iint }_{S^{\prime }}\mathbf{G}\cdot \mathbf{n}dA.$$ Also, it is pretty clear that $|\mathbf{G}|=1$ everywhere, and hence that $|\mathbf{G}\cdot \mathbf{n}|\le |\mathbf{G}||\mathbf{n}|=1$ everywhere on $S^{\prime }$. Therefore
$${\iint }_{S^{\prime }}\mathbf{G}\cdot \mathbf{n}dA\le {\iint }_{S^{\prime }}|\mathbf{G}\cdot \mathbf{n}|dA\le {\iint }_{S^{\prime }}1dA=\text{area}(S^{\prime }).$$ Putting these together, we conclude that $$\text{area}(S)\le \text{area}(S^{\prime })\text{ for any surface }{S}^{\prime }\text{ such that }\partial S=\partial S^{\prime }.$$ 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 $\mathbf{G}:{\mathbb{R}}^3\to {\mathbb{R}}^3$, determine whether whether there exists any functions $f$ such that $\mathrm{\nabla }f=\mathbf{G}$, and if so, find one. For example:

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

   • $\mathbf{G}(x,y,z)=(yz,xz,xy)$.

   • $\mathbf{G}(x,y,z)=(x\sin y\cos z,\frac{1}{2}{x}^2\cos y\cos z,-\frac{1}{2}{x}^2\sin y\sin z).$

   • $\mathbf{G}(x,y,z)=(\frac{1}{2},\frac{y^2-xz}{y^{2}z},\frac{-y}{z^2}).$

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

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

2. Given a vector field $\mathbf{G}:{\mathbb{R}}^3\to {\mathbb{R}}^3$, determine whether there exist any vector fields $\mathbf{F}$ such that $\mathrm{\nabla }\times \mathbf{F}=\mathbf{G}$, and if so find one. For example:

   • $\mathbf{G}(x,y,z)=(x,y,z)$.

   • $\mathbf{G}(x,y,z)=(-y,x,0)$.

   • $\mathbf{G}(x,y,z)=(y,\frac{1}{z}\cos yz,\frac{1}{y}\cos yx)$.

   • $\mathbf{G}(x,y,z)=(yz,xz,xy)$.

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

Advanced

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

   • Prove that $\mathrm{curl}\mathbf{G}=\mathbf{0}$.
   • Prove that $\mathbf{G}$ is not conservative.
     Hint

     Consider ${\int }_C\mathbf{G}\cdot d\mathbf{x}$, where $C$ is a circle around the $z$-axis.
   • If we define $U^{\prime }=\{(x,y,z)\in {\mathbb{R}}^3:x>0\}$, then we know that $\mathbf{G}$ is conservative on $U^{\prime }$. (why?) Find $f$ such that $\mathrm{\nabla }f=\mathbf{G}$ on $U^{\prime }$.

4. Prove that if $g$ is a continuous function defined on an open set containing a point $\mathbf{q}$, then for unit vector $\mathbf{e}$, $$\underset{h\to 0}{lim}\frac{1}{h}{\int }_0^{h}g(\mathbf{q}+t\mathbf{e})dt=g(\mathbf{q})$$ (This point comes up in the proofs of Theorems 1 and 2).

5. Suppose that $\mathbf{G}$ is a $C^2$ vector field that is both a gradient and a curl.

   • Show that if $f$ is a function such that $\mathrm{\nabla }f=\mathbf{G}$, then $f$ is harmonic, that is, ${\mathrm{\nabla }}^{2}f=0$, where we recall that the Laplacian is \(\nabla^2 f = \dfrac{\partial^2 f}{\partial x_1^2}+\dfrac{\partial^2 f}{\partial x_2^2\dfrac{\partial^2 f}{\partial x_3^2}.\)

   • Conclude that each component $G_i$ of $\mathbf{G}$ is also harmonic.

$\Leftarrow$  $\Uparrow$

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