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Consider the following problem:

A vector field $\boldsymbol{F}(x)$ is defined over a finite region $V$. A functional of the form \begin{equation} U = \int_V u(\boldsymbol{F})\ d^3x \end{equation} is to undergo the variation with respect to both $x$ and $\boldsymbol{F}$ but with the constraint that the divergence of the vector field be a particular function, e.g. \begin{equation} \text{div} \boldsymbol{F}(x) \equiv F_{i,i}(x) = f(x). \end{equation}

My attempt is to introduce a Lagrange multiplier, say $\psi(x)$, and have the variation be of the form \begin{equation} \delta \{U - \int_V \psi(x) f(x) d^3x\} =0 \ \ \ \Rightarrow \ \ \ \delta U - \int_V\psi\delta (fd^3x) = \delta U - \int_V \psi \delta f (1 + \delta x_{k,k})d^3x = 0 \end{equation} where $(1+\delta x_{k,k})$ is the jacobian of the infinitesimal deformation. Doing so, one has \begin{equation} \delta U - \int_V\psi \delta (F_{i,i})(1+ \delta x_{k,k})d^3x =0 \end{equation}

BUT my instructor suggests instead to write \begin{equation} \delta U - \int_{V'} \psi \ \text{div}(\delta \boldsymbol{F}) d^3x =0 \end{equation} where $V'$ is the virtually deformed region of $V$.

I can't see why this is the case. I don't see them as equal, because I don't reckon $\delta$ and div commute in this case, unless $\delta \boldsymbol{F}$ in the latter excludes the convective term $\delta x \cdot \text{grad} \boldsymbol{F}$. The same question could be relevant to a curl problem. I mean, where a vector field's curl is regarded as a constraint.

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