I have, all-well-behaved vector fields $\vec{A},\vec{J}:\Bbb R^3 \to \Bbb R^3$ and the real function $f:\Bbb R\to\Bbb R$ and a scalar valued function $\rho:\Bbb R^3\to\Bbb R$. $\vec{J}$ does functionally depend on $f$, which I write like $\vec{J}[f]$. For these I have the variational condition: $$\frac{\delta}{\delta f}\big(\int_\Omega \vec{J}[f]\cdot\vec{A}d\vec{V}\big)^2 =0, $$ with $d\vec{V}=dx\,dy\,dz$ and $\Omega=\Bbb R^3$.
Does this simplifiy to any simpler expression for $f$?
The question arises in some chemical research context of mine.
Since the integral itself should be $\ne 0$ in general, I suspect $$ \frac{\delta \vec{J}[f]}{\delta f}\cdot\vec{A}=0$$ could result. But I am note sure, if I don't make a mistake.
I give a concrete example: \begin{align} \vec{r}=&(x,y,z)^T \\ r=&|\vec{r}|\\ \vec{A}=&(y,-x,0)^T\\ \vec{J}=&f(\rho(\vec{r}))(-y,x,0)^T\\ \rho(\vec{r})=&\exp{(-r)}\\ f(\zeta)=&\sin(\zeta)\\ \end{align} Then $$ \vec{J}[f]\cdot\vec{A}=-\sin(\exp{(-r)})(x^2+y^2)$$ The question I am interested in is how to choose $f$ in dependence of $\vec{A},\vec{J},\rho$ such that the integral from the first expression becomes extremal.