# Non-Linear Beltrami Vector fields

Consider two concentric toruses, and let $\Sigma$ be the domain interior to the greater torus and exterior to the smaller torus. Is it possible to find a vector field $\mathbf{b}$ satisfying the following conditions:

1. $\mathrm{div} \, \mathbf{b}=0$ in $\Sigma$;

2. $\mathbf{b}$ is everywhere tangent to $\partial \Sigma$; and

3. $\mathrm{curl} \, \mathbf{b} = \lambda(\mathbf{x}) \mathbf{b}$ in $\Sigma$.

Thank you

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Is $\lambda$ prescribed to us? If yes, what do we know about it? If no, have you tried to find a field with zero curl and divergence? – user53153 Jan 2 '13 at 23:59
No $\lambda$ is an arbitrary continuous function. In Wikipedia (page on Hairy ball theorem) there is a picture of field everywhere tangent to the torus, but I don't know if we can extend it to be Beltrami. – user48900 Jan 3 '13 at 0:02
Would not the two-dimensional example $\dfrac{x_2\mathbf{i}-x_1\mathbf{j}}{x_1^2+x_2^2}$ work just as well in three dimensions? – user53153 Jan 3 '13 at 0:10
Maybe after you compose it with a map $\pi$ which sends $\mathbb{R}^2$ to the torus. – user48900 Jan 3 '13 at 0:20
I don't why not. The flow lines of $F$ are circles centered on the $x_3$-axis. The domain is rotationally symmetric. Hence, flow lines do not exit the domain. – user53153 Jan 3 '13 at 0:35