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Suppose we are given a fixed vector field $\mathbf{a}$. I am interested in the problem of determining a vector field $\mathbf{b}$ such that $$\nabla \times \mathbf{b} = \mathbf{b} \times \mathbf{a}.$$ This has another interpretation. Suppose $\alpha$ and $\beta$ are the 1-forms dual to $\mathbf{a}$ and $\mathbf{b}$. The above equation can be written as $$d \beta = \beta \wedge \alpha,$$ and so we can interpret this problem as finding, for a fixed 1-form $\alpha$, a foliation $\mathcal{F} = \text{ker}\ \beta$ such that $\alpha$ determines the Godbillon-Vey class of $\mathcal{F}$.

It seems unlikely to me that a solution always exists, but I have been unable to prove anything beyond the obvious fact that we must have $\mathbf{b} \cdot \nabla \times \mathbf{b} = 0$, and that $\mathbf{b} \cdot \nabla \times \mathbf{a} = \mathbf{a} \cdot \nabla \times \mathbf{b}$ (take the divergence), which implies (Mark's comment) that $\mathbf{b}$ is orthogonal to $\nabla \times \mathbf{a}$.

Are there any known results about such equations, or techniques one could use to construct a solution other than crunching through the PDEs for each component?

Joe
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    I can't go into any detail on this, but I've understood that geometric algebra/clifford algebra makes solving equations like this easy – ziggurism Dec 30 '17 at 20:56
  • We have $$\vec a\cdot \nabla \times \vec b=\vec b\cdot \nabla \times \vec b=0$$ – Mark Viola Dec 30 '17 at 20:58
  • Thanks for your comment, I'll look into it. Perhaps someone will elaborate. – Joe Dec 30 '17 at 20:58
  • What is known about $\mathbf{a}$? – md2perpe Dec 30 '17 at 23:14
  • @md2perpe In my case I know that $\mathbf{a} \cdot \nabla \times \mathbf{a} \neq 0$, but nothing else. I'd like to know what's possible in the greatest generality but if additional assumptions allow an easier solution that's interesting too. – Joe Dec 31 '17 at 10:49
  • Actually, I would also like to impose the constraint $\mathbf{b} = \mathbf{a} \times (\nabla \times \mathbf{a})$. With this extra constraint I don't think it's true that solutions always exist over the reals - setting $\mathbf{a} = (1,0,-y)$ I get $\mathbf{b} = (0, \pm i, 0)$. – Joe Dec 31 '17 at 19:06
  • If $\mathbf{a} = -\nabla\phi$ then $\mathbf{b} = e^\phi \mathbf{c}$ is a solution where $\mathbf{c}$ is a constant vector. – md2perpe Dec 31 '17 at 21:53
  • Interesting. Then perhaps we can use the Helmholtz decomposition to write $\mathbf{a} = -\nabla \phi + \nabla \times \mathbf{d}$ to make further progress. – Joe Dec 31 '17 at 22:37
  • @JoePollard, are given field $\vec a$ constant? – Michael Galuza Jan 03 '18 at 04:48
  • @MichaelGaluza It should depend on position. – Joe Jan 04 '18 at 07:49

1 Answers1

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Here is a small set of solutions. Let $p$ and $q$ be any numbers and $c$ any constant vector. Then $$ b(x) = |x|^p x\times (x\times c), \qquad\qquad a(x) = (p+3)\frac{x}{|x|^2}+qb(x) $$ is a solution. Here $x=(x_1,x_2,x_3)$ and $|x|$ is Euclidean norm.

Edit: your comment about $a$ rules out gradients, but I'll mention anyway more solutions when $a = \nabla g$. Using $\nabla\times(h\nabla f) = \nabla h\times\nabla f$ you get $$ \nabla\times(e^{-g}\nabla f) = e^{-g}\nabla f \times \nabla g. $$

Bob Terrell
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  • Thanks for your answer. I see that these work, but is there some particular technique you used to arrive at this answer? – Joe Jan 19 '18 at 08:50
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    When working with cross products I've sometimes found it useful to use linear combinations of the vector fields $x$, $x\times c$ and $x\times(x\times c)$. They are orthogonal for most values of $x$, and work well under crossing with $x$, since $x\times(x\times(x\times c))) = -|x|^2 x\times c$. Since these fields have been helpful more than once, does that qualify it as a "technique?" I started with $c=(1,0,0)$ because that is enough by linearity and symmetry, and is easy to work with. – Bob Terrell Jan 19 '18 at 13:02
  • I see, thank you for explaining. – Joe Jan 19 '18 at 13:51