Is there an incompressible vector field $\mathbf{F}$ with $\nabla \times \mathbf{F} = (y^2, z^2, x^2)$?

Question

I want to find an incompressible vector field $\mathbf{F}$ such that $\nabla \times \mathbf{F} = (y^2,z^2,x^2)$.

After some attempts to find such $\mathbf{F}$, I think the vector field $\mathbf{F}$ with given conditions may not exist. However even for this side, I do not have an idea to proceed. Since $\mathbf{F}$ is incompressible we have $\mathbf{F} = \nabla \times \mathbf{G}$ for some $\mathbf{G}$, which leads to $\nabla \times(\nabla \times \mathbf{G}) = (y^2,z^2,x^2)$. However there is no additional property that I know about double curl to prove nonexistence of such $\mathbf{F}$.

Thanks in advance for any form of help, hint, or solution.

Answer 1

Yes, such incompressible vector field $\mathbf{F}$ exists. Consider $$\mathbf{G}=(-y^4/12,-z^4/12,-x^4/12)$$ then $$\mathbf{F}=\nabla \times \mathbf{G}=(z^3/3,x^3/3,y^3/3)$$ and finally $$\nabla\times \mathbf{F}= (y^2,z^2,x^2).$$ More generally, from the definition of curl in cartesian coordinates, $$\begin{align} \nabla \times(\nabla \times (f(y),g(z),h(x)))&= \nabla \times (-g'(z),-h'(x),-f'(y))\\ &=-(f''(y),g''(z),h''(x)).\end{align}$$

Answer 2

Hint $$ \nabla \cdot \mathbf{F} = 0 $$ for incompressible field. Thus $$ \partial_x F_x + \partial_y F_y + \partial_z F_z= 0 $$ we have from the curl $$ \left(\partial_y F_z - \partial_zF_y\right)\mathbf{e_x} - \left(\partial_x F_z - \partial_zF_x\right)\mathbf{e_y} + \left(\partial_x F_y - \partial_yF_x\right)\mathbf{e_z} $$ or $$ \partial_x F_y - \partial_yF_x = x^2\\ \partial_y F_z - \partial_zF_y = y^2\\ \partial_z F_x - \partial_x F_z = z^2 $$

we know from the incompressible requirement that $F_x = kx$ or $F_x =g(y,z)$ e.g. independent of $x$ and by symmetry the same for all other components.

Answer 3

There's an identity .

$\nabla \times (\nabla \times \vec{G})=\nabla(\nabla \cdot \vec{G})-\nabla ^2\vec{G}=(y^2,z^2,x^2)$.

To simply things, let's assume $\nabla \cdot \vec{G}=0$.

$-\nabla^2 G_x=\frac{-\partial^2 G_x}{\partial y ^2}=y^2\implies G_x=\frac{-y^4}{12}$. By symmetry you can reproduce Robert Z's answer.

A bit more difficult, you can find solutions where $\nabla \cdot \vec{G} \ne 0 $.