Consider a vector field $v$ on $\mathbb{R}^3$ which satisfies $∇\cdot v=0$, and $∇×v=αv$ for some $α:\mathbb{R}^3 \to \mathbb{R}$. My aim is to compute $v\cdot ∇α$.
I let v(x,y,z) = (U,V,W) Using this, I computed $∇\cdot v$, and found that $\frac {\partial U}{\partial x} + \frac {\partial V}{\partial y} + \frac {\partial W}{\partial z} = 0$
Then I computed $∇×v$, and found that $(\frac {\partial W}{\partial y}- \frac {\partial V}{\partial z}, \frac {\partial U}{\partial z}- \frac {\partial W}{\partial x}, \frac {\partial V}{\partial x}- \frac {\partial U}{\partial y}) = αv$, and hence came up with the following three equations:
$\frac {\partial W}{\partial y}- \frac {\partial V}{\partial z} = αU$
$\frac {\partial U}{\partial z}- \frac {\partial W}{\partial x} = αV$
$\frac {\partial V}{\partial x}- \frac {\partial U}{\partial y} = αW$
From this I can maybe deduce that:
$$\frac {\frac {\partial W}{\partial y}- \frac {\partial V}{\partial z}}{U} = \frac{\frac {\partial U}{\partial z}- \frac {\partial W}{\partial x}}{V} = \frac{\frac {\partial V}{\partial x}- \frac {\partial U}{\partial y}}{W} = α $$
though I'm unsure of the validity of the above statement.
I'm unsure of what to do after this point and if what anything I have shown will help me to compute $v\cdot ∇α$. I am aware that I should be getting an answer of $0$.
Any assistance would be much appreciated. Thank you.
Note that
$\nabla \cdot \nabla \times v = 0; \tag 1$
the divergence of a curl is always $0$; thus the given
$\nabla \times v = \alpha v \tag 2$
yields
$\nabla \cdot (\alpha v) = 0; \tag 3$
also, by hypothesis
$\nabla \cdot v = 0; \tag 4$
now we have the identity (see this wikipedia entry)
$\nabla \cdot (\alpha v) = \nabla \alpha \cdot v + \alpha \nabla \cdot v; \tag 5$
(3), (4) and (5) together yield
$v \cdot \nabla \alpha = 0. \tag 6$