I've been stuck on a problem and I was wondering if anyone would be able to help me. I am trying to solve the following divergence problem explicitly for $\vec{A}$
$$\nabla \cdot \vec{A} = \nabla \phi \cdot \nabla (\nabla^{2} \phi) - (\nabla^{2} \phi)^{2} \tag{1}$$
where $\nabla = (\partial_{x}, \partial_{y})$ and $\phi$ (and hence $\vec{A}$) is periodic in space. I was wondering if there was a 'nice' factorisation of the RHS such that it can be written in the form $\nabla \cdot \vec{\Phi}$? I doubt that there is, as
$$\nabla \phi \cdot \nabla (\nabla^{2} \phi) - (\nabla^{2} \phi)^{2} = -(\nabla^{2} \phi)^{2} \nabla \cdot \left( \frac{\nabla \phi}{\nabla^{2} \phi} \right) \tag{2}$$
where the quotient is to be interpreted as elementwise division and the RHS of $(2)$, as far as I can tell, has no further factorisation that would put it in the form I'm after. Of course, I could just define
$$\vec{A} = (A_{1},A_{2}) = \left(c \int \nabla \phi \cdot \nabla (\nabla^{2} \phi) - (\nabla^{2} \phi)^{2} dx, \ (1 - c) \int \nabla \phi \cdot \nabla (\nabla^{2} \phi) - (\nabla^{2} \phi)^{2} dy \right)$$
for some $c \in \mathbb{R}$, which would satisfy $(1)$. However, I was hoping to find a nice simplification without integrals. We can also note that the first terms on the RHS of $(1)$ can be interpreted as $\mathcal{L}_{\nabla \phi} \nabla^{2} \phi$, the directional derivative of $\nabla^{2} \phi$ in the direction of $\nabla \phi$. I'm not sure if this is of much use though.
Also, if anyone else has seen the RHS of $(1)$ in any other contexts, please let me know.
Thanks.