Let $u,v : U \subseteq \mathbb{R}^{3} \to \mathbb{R}$ be a pair of functions of class $C^2$ defined on an open set $U$. Is it possible to always find a field $F$ such that $\nabla u \times \nabla v = \text{curl} (F)$? My original intuition was that it wasn't possible to always do this and that the function $G: \mathbb{R}^{3} \setminus \{\vec{0}\} \to \mathbb{R}^{3}$ defined by $$G(x,y,z) := \left(\frac{x}{\left(x^2+y^2+z^2\right)^{3/2}},\frac{y}{\left(x^2+y^2+z^2\right)^{3/2}},\frac{z}{\left(x^2+y^2+z^2 \right)^{3/2}},\right)$$ would somehow help prove this fact.
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1"I am trying to show that the converse of this result is not true for non-contractible regions." Did you mean to write "converse of" here? – Jesse Madnick Feb 18 '23 at 22:34
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1Yes, that's what I meant, I even wrote it! Perhaps I didn't explain myself properly, but the objective of the question I asked is to figure out if it is possible to find functions of class $C^2$, $u$ and $v$, defined in a non-contractible open subset of $\mathbb{R}^3$, in such a way that the cross product $ \nabla u \times \nabla v$ is not the curl of a field. I think this could be achieved with the function $G$, but this belief is only sustained by the fact that it is the only vector field in $\mathbb{R}^{3}$ I know of that is not the curl of a field. – Peluso Feb 18 '23 at 22:40
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1Right, so the objective is to show that the result (not its converse) is false for non-contractible regions. – Jesse Madnick Feb 18 '23 at 22:45
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1You are absolutely right, I wrote the question badly. I'll correct it right away. Thank you! – Peluso Feb 18 '23 at 22:53
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1The title is a mess. It seems to me, upon further reading, that you are asking if a cross-product of two $C^1$ gradient fields can fail to be a curl. Your "From Clairaut's Theorem" is placed at the wrong point of your second sentence. – Ted Shifrin Feb 18 '23 at 23:42
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Thanks for your feedback Ted! I changed the title and wording of the question to reflect your comments. Thank you so much. – Peluso Feb 19 '23 at 00:14
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You know that your vector field $G$ cannot be a curl (say, because its flux across the unit sphere is nonzero). But note that $$\nabla u\times\nabla v = \text{curl}(u\nabla v),$$ so your $G$ cannot be written as the cross-product of two gradients. To be explicit, in your title you now say that you want the cross-product of two gradients to fail to be a curl. That is impossible.
Ted Shifrin
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I am a bit confused. The question about $G$ was raised recently and an answer was found here. In a more friendly notation the $F$ whose curl is $G$ should be $$F(x,y,z)=\frac{z-r}{rs}\begin{pmatrix}y\ -x\0\end{pmatrix},,\quad r=\sqrt{x^2+y^2+z^2},,\quad s=x^2+y^2,$$ I checked this numerically and the OP of the recent question verified it doing calculus. – Kurt G. Feb 19 '23 at 08:55
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@KurtG.: While it may be true that your $F$ satisfies $\text{curl}(F) = G$, it is not the case that $F$ is defined on all of $\mathbb{R}^3$, or even all of $\mathbb{R}^3 - {0}$. In fact, as the OP pointed out, there is no vector field $F$ defined on all of $\mathbb{R}^3 - {0}$ satisfying $\text{curl}(F) = G$. – Jesse Madnick Feb 19 '23 at 09:56
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@JesseMadnick . Then we agree. $G$ and $F$ are only defined on the punctured $\mathbb R^3$. BTW I checked again with a bit of wolfram alpha that curl$(F)=G$. Credit should go to user Anders Beta who found this four years ago in the link I posted. – Kurt G. Feb 19 '23 at 10:53
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1@KurtG. We don’t agree. The domain of $F$ omits the entire $z$-axis; this is not the punctured $\Bbb R^3$. The question you linked is also beyond sloppy regarding domains. – Ted Shifrin Feb 19 '23 at 15:40
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@KurtG.: To reiterate, there is no vector field defined on all of $\mathbb{R}^3 - {0}$ whose curl is $G$. In particular, as Ted Shifrin correctly points out, your vector field $F$ is not defined on all of $\mathbb{R}^3 - {0}$, but rather the subset $\mathbb{R}^3 - {z\text{-axis}}$. – Jesse Madnick Feb 19 '23 at 21:42
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The $z$-axis is only a coordinate singularity. Since $(r-z)(r+z)=r^2-z^2=s$ we can write $F$ as $$ F(x,y,z)=\frac{1}{r^2+rz}\begin{pmatrix}-y\x\0\end{pmatrix}, $$ and this is defined on the punctured $\mathbb R^3,.$ This should be expected because this is where $G$ is defined and $G$ is rotationally symmetric. Needless to say that I checked again multiple times that this $F$ also satisfies $\nabla \times F=G$. – Kurt G. Feb 21 '23 at 04:09
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$F$ is defined whenever $r=\sqrt{x^2+y^2+z^2}>0,.$ This rules out the origin but not the entire $z$-axis. Sorry I did not mean to be a pain. Just want to reach an agreement. – Kurt G. Feb 21 '23 at 04:18
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@KurtG. Then I guess Stokes’s Theorem fails. Your nonstandard notation messed me up, since i expect $r$ to be the polar coordinate. No agreement. As I stated at the outset, the flux of a curl across a closed surface must be $0$. – Ted Shifrin Feb 21 '23 at 04:24
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I hope Stokes' theorem does not fail. I will post my triple check as an answer to that recent related question. Lets see what comes out. – Kurt G. Feb 21 '23 at 04:26
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Indeed, @KurtG. For future purposes, make sure that you understand the proof of the result I claimed from the outset. Then you will be more self-critical when you make assertions contradicting it! – Ted Shifrin Feb 21 '23 at 05:04
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You are (honestly) the last person here in MSE I would like to disagree with. Just trying to learn. Please allow me one more question. The $G$ in OP should have zero divergence. I checked this and others did this too. So I am still a bit confused regarding the flux of it through a sphere. – Kurt G. Feb 21 '23 at 05:10
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@KurtG. The divergence theorem does not apply because of the singularity. You have to apply Stokes’s Theorem to the surface integral of a curl; cut the surface into two pieces. – Ted Shifrin Feb 21 '23 at 05:26