Very simply, I am trying to solve the following function in a 3D space: $$ \frac{\partial}{\partial F}(\nabla\cdot F) $$ where $F$ is a vector field. I THINK this should be zero based on physical reasoning, but every attempt to solve it rigorously has failed.

Approach 1:

Use a vector identity and rotational invariance, and solve in one dimension. Given $$ \nabla(\mathbf{A}\cdot\mathbf{B})=(\mathbf{A}\cdot\nabla)\mathbf{B}+(\mathbf{B}\cdot\nabla)\mathbf{A}+\mathbf{A}\times(\nabla\times\mathbf{B})+\mathbf{B}\times(\nabla\times\mathbf{A}) $$ I came out to a result that made little sense (though I think I see where I made the error while invoking rotational invariance): $$ =2\left(\frac{\partial}{\partial x}\cdot F\right)\frac{\partial}{\partial F} $$ clearly recovering an operator as a result doesn't work physically.

Approach 2:

Again, work in one dimension and play some chain rule games $$ \frac{\partial}{\partial F}(\nabla\cdot F)=\frac{\partial\left(\frac{\partial F_x}{\partial x}\right)}{\partial x}\frac{\partial x}{\partial F}=\frac{\partial^2F_x}{\partial x^2}\frac{\partial x}{\partial F} $$ but again, here I get lost.

Approach 3: Invoke Gauss' law, which gets the closest ultimately: $$ \int_v \frac{\partial}{\partial F}(\nabla\cdot F)dv=\int_{dv}\mathbf{\hat{n}}\cdot dS $$ which is surface area. So we arrive that the integral of this function is equal to the surface area of the bounding volume. What can we say about a function if its integral is a constant?

Any help or even a general direction would be appreciated.

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    $\begingroup$ I don't really understand what you are trying to do, but formally, if you differentiate a linear functional with respect to the variable you will get the functional back? $\endgroup$ – copper.hat Mar 6 '13 at 17:51
  • $\begingroup$ I don't understand either, but note that : 1. it makes no sense a priori to talk about the gradient of a vector field and 2. if the derivative of a map vanishes everywhere, then it is constant (if the domain of definition is connected, which seems to be the case here) $\endgroup$ – Glougloubarbaki Mar 6 '13 at 17:54
  • $\begingroup$ My apologies, the title should read "derivative of the DIVERGENCE of a function with respect to the function" $\endgroup$ – Andy K. Mar 6 '13 at 17:58
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    $\begingroup$ What do you mean by "derivative with respect to a function"? $\endgroup$ – mrf Mar 6 '13 at 18:00
  • $\begingroup$ It was erroneous nomenclature on my part--5pm has restated it correctly. $\endgroup$ – Andy K. Mar 6 '13 at 18:39

It would be more precise to say: the derivative of the divergence of a vector field with respect to that field. (You say "function" but then $F$ turns out to be a vector field.) In symbols, it could be expressed as $D_F (\nabla \cdot F)$ although notation varies. It helps to write fields in derivative notation, $$F=\sum_{i=1}^3 f_i \frac{\partial}{\partial x_i} \tag1$$ where $f_1,f_2,f_3$ are scalar functions, the components of the field. Then the divergence of $F$ is expressed by $$\nabla \cdot F=\sum_{i=1}^3 \frac{\partial f_i }{\partial x_i} \tag2$$ The derivative of (2) with respect to the vector field (1) comes out as the double sum $$D_F(\nabla \cdot F)=\sum_{i=1}^3 f_i \frac{\partial}{\partial x_i} \sum_{j=1}^3 \frac{\partial f_j }{\partial x_j} = \sum_{i,j=1}^3 f_i \frac{\partial^2 f_j }{\partial x_i\partial x_j} \tag3$$ The expression (3) is not zero in general. For example, take $f_i=x_i^2$ for $i=1,2,3$: then the sum (3) evaluates to $2(x_1^2+x_2^2+x_3^2)$.

  • $\begingroup$ This certainly helps, but can you please elaborate on (1) a bit? I have never encountered this--I found an article on Leibniz notation for vector fields; is this the same thing? $\endgroup$ – Andy K. Mar 6 '13 at 18:47
  • $\begingroup$ @AndyK. Probably this article. Yes, this is the notation I use in (1). It looks odd at first, but helps in computations because it tells exactly what the field does to functions. In differential geometry, vector fields are often defined via their action on functions, as linear functional that satisfy the Leibniz rule. $\endgroup$ – user53153 Mar 6 '13 at 18:53
  • $\begingroup$ This is very helpful, thank you (and that was indeed the article). My reasoning that it should be zero was predicated on a false assumption made before the statement of the problem (but that is a physics issue, not a maths one). $\endgroup$ – Andy K. Mar 6 '13 at 18:59
  • $\begingroup$ The link for the PlanetMath is now dead, use this one instead. $\endgroup$ – Joey Dumont Mar 27 '15 at 14:13

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