0
$\begingroup$

At my calculus II class we are studying multivariable functions and yesterday er talked about the curl operator (we used the definition in "usage" section here.

The typical exercise we got as homework, was like: given a function $F:\mathbb{R}^3\to\mathbb{R}^3$, compute $\operatorname{curl}(F)$ and so far so good. Today I got this homework which is exactly the opposite, i.e. $$"\text{Let } \ G=\frac{x}{|x|^3}. \text{Find an example of } \ F:\mathbb{R}^3\to\mathbb{R}^3 \text{such that } \ G=\operatorname{curl}(F)."$$

The exercise was proposed for $G=\frac{x}{|x|^3}$ and $G=\frac{x}{|x|^2}$, so I guess there is a similar strategy.

Could someone please help me in find an example even in only one of the cases?

I am thinking to examples like $F(x)= x/|x|$, or to some powers, but it is not working.

Thank you.

$\endgroup$
14
  • $\begingroup$ Consider $F(\vec x)=\dfrac1{\|\vec x\|}$ $\endgroup$
    – user170231
    Commented Feb 15, 2023 at 20:25
  • $\begingroup$ Hmm, it looks like divergence of $x / \lVert x \rVert^3$ is 0 as you'd expect from the curl of a vector field; but divergence of $x / \lVert x \rVert^4$ is nonzero so it can't be the curl of any vector field. $\endgroup$ Commented Feb 15, 2023 at 20:30
  • 1
    $\begingroup$ Sorry, I should have put $\langle1,1,1\rangle$ in the numerator, not $1$. In any case I'm not suggesting that that $F$ is a solution, rather a starting point, since its curl has $\|\vec x\|^3$ in the denominator. As for notation, if $\vec x=\langle x,y,z\rangle$, then $\|\vec x\|=\sqrt{x^2+y^2+z^2}$. $\endgroup$
    – user170231
    Commented Feb 15, 2023 at 20:59
  • 1
    $\begingroup$ After struggling for almost 24 hours with the Helmholtz decompositon theorem to solve this very interesting question I found an answer here: $$ F(x,y,z)=\frac{z-r}{rs}\begin{pmatrix}y\\ -x\\0\end{pmatrix}\,,\quad r=\sqrt{x^2+y^2+z^2}\,,\quad s=x^2+y^2\,. $$ I have checked numerically that it is correct. $\endgroup$
    – Kurt G.
    Commented Feb 16, 2023 at 16:37
  • 1
    $\begingroup$ Correct and fairly trivial because div rot is always zero. What I find a lot more interesting are the properties of those $F$ that can be found in $n=3$, One example is the answer below. $\endgroup$
    – Kurt G.
    Commented Mar 1, 2023 at 4:43

1 Answer 1

2
$\begingroup$

The vector field $$ G(x,y,z)=\frac{1}{(x^2+y^2+z^2)^{3/2}}\begin{pmatrix}x\\y\\z\end{pmatrix}\,,\quad\text{ on }\mathbb R^3\setminus\{0\}, $$ describes Coulomb's law and Newton's law of universal gravitation and is mathematically interesting in many ways: It is rotation free and divergence free. A scalar function whose gradient is $G$ is $$ H(x,y,z)=-\frac{1}{(x^2+y^2+z^2)^{1/2}}\,. $$ The divergence freeness of $G$ suggests that there should be a vector field $F$ whose curl is $G\,.$ It turns out that

  • $F$ exists but cannot be defined on all of $\mathbb R^3\setminus\{0\}\,.$

This was briefly remarked in this answer. After some confusion on my side an finally learning it from Ted Shifrin I'd like to sum it up a bit: If $F$ is defined on all of $\mathbb R^3\setminus\{0\}$ one can apply Stokes' theorem to each hemisphere separately to conclude that the flux of $G$ through the whole unit sphere must be zero but this cannot happen because the flux of $G$ is \begin{align} \oint_{S^2}\hat{\mathbf{n}}\cdot G\,dS&=\oint_{S^2}\frac{\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)}{(x^2+y^2+z^2)^{1/2}}\cdot \frac{\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right)}{(x^2+y^2+z^2)^{3/2}}\,dS\\ &=\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\varphi=4\pi\,. \end{align} To see that on the contrary Stokes' theorem leads to a flux of zero (assuming that $F$ is defined on all of $\mathbb R^3\setminus\{0\}$) note that the line integral of $F$ along the boundary of each hemisphere equals the flux of $\nabla\times F=G$ through that hemisphere. But these line integrals cancel out because of their opposite orientation which leads to zero flux. A contradiction.

  • Consequently, when $F$ whose curl is $G$ exists there must be at least one point on one of the two hemispheres on which $F$ is not defined. An example for such an $F$ is $$ F(x,y,z)=\frac1{r^2+rz}\begin{pmatrix}-y\\x\\0\end{pmatrix}\,,\quad r:=(x^2+y^2+z^2)^{1/2}\,, $$ which is a simpler notation for the solution found here. This $F$ is defined for all $x,y,z\in\mathbb R^3$ except on the ray $z\le 0,x=y=0$ (where $r+z$ becomes zero) and describes a clockwise circular motion around the $z$-axis and parallel to the $xy$-plane. A way to visualise it is to look at the speed of this motion which is $$ \|F\|=\frac{\varrho}{\varrho^2+z^2+z\sqrt{\varrho^2+z^2}}\,,\quad\varrho:=\sqrt{x^2+y^2}\,. $$ The following picture shows this speed as a function of the distance $\varrho$ to the $z$-axis for various levels of $z\,.$ For $z\le 0$ where $F$ is not defined on the $z$-axis the speed blows up when the $z$-axis is approached.

enter image description here

Another interesting picture shows the $[0,3\pi/2]$-sectors of the surfaces of constant speed $\|F\|\,.$ The blue surfaces has the slowest speed and the brown surface the highest. These surfaces converge at the origin.

enter image description here

Further properties of $F\,:$

  • The one-form $$ \omega=\frac{-y\,dx+x\,dy}{r^2+rz} $$ is closed if and only if $z=0\,.$ This follows from $(\nabla\times F)_3=\partial_xF_2-\partial_yF_1=G_3=z/r^3\,.$ For $z=0$ this form has interesting properties that I summarized here.

  • On the unit sphere $S^2$ the speed of the integral curves can be written more simply as $$ \|F\|=\sqrt{\frac{1-z}{1+z}}\,. $$ This is zero at the north pole $z=1$ and infinite at the south pole $z=-1\,.$ The vorticity of this vector field is, as we know, $\nabla\times F=G$ which on $S^2$ is simply $\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right).$ Due to the circular motion in planes parallel to the $xy$--axis this seems odd at points away from the poles. But this rotationally symmetric vorticity $G$ arises from different speeds of integral curves next to each other. In particular, near the equator we have practically a parallel flow with shear that has a nice animation here.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .