Irrotational and incompressible?

Calculus Level 3

True or False?

There exists a nonlinear vectorfield $\vec{F}$ , defined on all of $\mathbb{R}^3$ , such that $\nabla \times \vec{F}=\vec{0}$ and $\nabla \cdot \vec{F}=0$ .

By "nonlinear" we mean that at least one of the component functions of $\vec{F}$ fails to be of the form $ax+by+cz+d$ .

(from a recent test on vector calculus)

No Yes Impossible to determine

1 solution

Otto Bretscher
Nov 20, 2018

A simple example is $\vec{F}=(2xy,x^2-y^2,0)$ . Thus the answer is $\boxed{Yes}$ .

More generally, for any entire function $f(z)$ on $\mathbb{C}$ we can take $\vec{F}=(\Im(f),\Re(f),0)$ . We get the example above from $f(z)=z^2$ .

Irrotational and incompressible?

1 solution

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