# Derivative of the divergence of a vector field with respect to the vector field

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.

• 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? – copper.hat Mar 6 '13 at 17:51
• 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) – Glougloubarbaki Mar 6 '13 at 17:54
• My apologies, the title should read "derivative of the DIVERGENCE of a function with respect to the function" – Andy K. Mar 6 '13 at 17:58
• What do you mean by "derivative with respect to a function"? – mrf Mar 6 '13 at 18:00
• It was erroneous nomenclature on my part--5pm has restated it correctly. – 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 $2(x_1^2+x_2^2+x_3^2)$.