when $\text{curl}(\vec{F})=(x,y,z)$?

Question

For what vector fields $\vec{F}$ we have that $\text{curl}(\vec{F})=(x,y,z)$ ? or more generaly $\text{curl}(\vec{F})=a(x,y,z)$ for some nonzero constant a? Thanks!

Answer 1

Recall that $\text{curl} F = \nabla \times F$. Thus, you need to find $F$, such that :

$$\text{curl} \vec{F} = (x,y,z) \Rightarrow \nabla \times \vec{F} = (x,y,z) $$

$$\Leftrightarrow$$

$$\begin{vmatrix} \hat i & \hat j & \hat k \\ \partial_x & \partial_y & \partial_z \\ \vec{F}_x & \vec{F}_y & \vec{F}_z\end{vmatrix} = (x,y,z) = x \hat i + y \hat i + z \hat k$$

You can also use the hint :

$$(\rm div \circ \rm curl )(\vec F)=0 \Rightarrow \nabla (\nabla \times \vec F) =0$$