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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.

Matthew Cassell
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1 Answers1

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The problem as you wrote it has no solutions unless $\phi\equiv c$. Observe that we would have $$\text{div} A=\text{div}(\Delta\phi\nabla\phi)-2(\Delta\phi)^2.$$ Since your space has no boundary, we can average and find $$\int_{\mathbb T^d}(\Delta\phi)^2=0$$ and we conclude.

If you fix this issue by, say, subtracting off the average of the right-hand side you will have an infinite-dimensional space of solutions. At this point, inverting the divergence in various spaces is a well-studied problem, see eg. here.

Matthew Cassell
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Funktorality
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  • Thanks heaps for your response. I actually do have an average term of the form $\langle(\nabla^{2} \phi)^{2} \rangle$ in the expression for $\text{div } A$, I just neglected it to simplify my problem above. Clearly that was a mistake. I will take a look at the paper you linked when I get a chance. Thanks again for your help! – Matthew Cassell Sep 30 '20 at 01:01