Theorem: General Variation of Integral Functional/Dependent on n Variables

Let $\mathbf x$ be an $n$-dimensional vector.
Let $u = \map u {\mathbf x}$ be a real-valued function.
Let $J$ be a functional such that:
:$\ds J \sqbrk u = \int_R \map F {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x} } \rd x_1 \dotsm \rd x_n$
Let $\mathbf x^*, u^*$ be defined by the following transformations $\paren \star$:
{{begin-eqn}}
{{eqn | l = \mathbf x^*
      | r = \map {\boldsymbol \Phi} {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x}; \epsilon} = \mathbf x + \epsilon \map {\boldsymbol \phi} {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x} } + \map \OO {\epsilon^2}
}}
{{eqn | l = u^*
      | r = \map \Psi {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x}; \epsilon} = \mathbf x + \epsilon \map \psi {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x } } + \map \OO {\epsilon^2}
}}
{{end-eqn}}
where:
:$\boldsymbol \Phi$, $\Psi$ are differentiable {{WRT|Differentiation}} $\epsilon$
and:
{{begin-eqn}}
{{eqn | l = \map {\boldsymbol \Phi} {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x}; 0}
      | r = \mathbf x
}}
{{eqn | l = \map \Psi {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x}; 0}
      | r = u
}}
{{eqn | l = \map {\boldsymbol \phi} {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x} }
      | r = \valueat {\dfrac {\partial \boldsymbol \Phi} {\partial \epsilon} } {\epsilon \mathop = 0}
}}
{{eqn | l = \map \psi {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x} }
      | r = \valueat {\dfrac {\partial \Psi} {\partial \epsilon} } {\epsilon \mathop = 0}
}}
{{end-eqn}}
Then the variation of the functional $J$ due to the original mapping being transformed by $\paren \star$ reads:
{{explain|not clear what "aforementioned transformation" refers to. Best to identify it with a label and refer to that label.}}
:$\ds \delta J = \epsilon \int_R \paren {F_u - \dfrac {\partial F_{u_{\mathbf x} } } {\partial \mathbf x} } \overline \psi \rd x_1 \dotsm \rd x_n + \epsilon \int_R \map {\dfrac {\partial} {\partial \mathbf x} } {F_{u_x} \overline {\boldsymbol \psi} + F \boldsymbol \phi} \rd x_1 \dotsm \rd x_n$
where:
:$\overline \psi = \psi - u_{\mathbf x} \boldsymbol \phi$

Proof:
By definition:
{{begin-eqn}}
{{eqn | l = \Delta J
      | r = J \sqbrk {\map {u^*} {x^*} } - J \sqbrk {\map u x}
}}
{{eqn | r = \int_{R^*} \map F {\mathbf x^*, u^*, \frac {\partial u^*} {\partial \mathbf x^*} } \rd x_1^* \dotsm \rd x_n^* - \int_R \map F {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x} } \rd x_1 \dotsm \rd x_n
}}
{{eqn | r = \int_R \paren {\map F {\mathbf x^*, u^*, \dfrac {\partial u^*} {\partial \mathbf x^*} } \map \det {J_{\mathbf x^*} } - \map F {\mathbf x, u, \frac {\partial u} {\partial \mathbf x} } } \rd x_1 \dotsm \rd x_n
      | c = Volume element transformation
}}
{{end-eqn}}
From the definition of $\mathbf x^*$ a Jacobian matrix can be constructed:
:$\ds \dfrac {\partial x_i^*} {\partial x_j} = \delta_i^j + \epsilon \dfrac {\partial \phi_i} {\partial x_j} + \map \OO {\epsilon^2}$
By Taylor's theorem, The corresponding Jacobian determinant can be expanded {{WRT}} $\epsilon$.
The $\epsilon^0$ term is obtained from the diagonal product by opening brackets and collecting $\epsilon$-free terms.
The $\epsilon^1$ term is obtained from the same diagonal product by choosing non-constant term once.
:$\ds \map \det {J_{\mathbf x^*} } = 1 + \epsilon \dfrac {\partial \boldsymbol \phi} {\partial \mathbf x} + \map \OO {\epsilon^2}$
{{explain|make explanation more precise}}
:$\ds \Delta J = \int_R \paren {\map F {\mathbf x^*, u^* \dfrac {\partial u^*} {\partial \mathbf x^*} } \paren {1 + \epsilon \dfrac {\partial \boldsymbol \phi} {\partial \mathbf x} } - \map F {\mathbf x, u, \dfrac {\partial u} {\partial \mathbf x} } } \rd x_1 \dotsm \rd x_n + \map \OO {\epsilon^2}$
By definition, the principal part is:
:$\ds \delta J = \int_R \paren {F_{\mathbf x} \delta \mathbf x + F_u \delta u + F_{u_{\mathbf x} } \delta u_{\mathbf x} + \epsilon F \dfrac {\partial \boldsymbol \phi} {\partial \mathbf x} } \rd x_1 \dotsm \rd x_n$
{{explain|If use of square brackets has a particular significance and meaning here, explain it. If it is just a variant on conventional parenthesis, then best if $\paren \cdot$ is used instead. It's ultimately more clear to use one style of parenthesis than mix and match various different types.}}
Introduce the following differences:
{{begin-eqn}}
{{eqn | l = \Delta u
      | r = \map {u^*} {x^*} - \map u x
}}
{{eqn | l = \overline {\Delta u}
      | r = \map {u^*} x - \map u x
}}
{{eqn | l = \Delta x
      | r = x^*-x
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \overline {\Delta u}
      | r = \epsilon \overline \psi + \map \OO {\epsilon^2}
}}
{{eqn | l = \overline {\delta u}
      | r = \epsilon \overline \psi
}}
{{eqn | l = \Delta u
      | r = \dfrac {\partial u} {\partial \mathbf x} \delta \mathbf x + \overline {\delta u} + \map \OO {\epsilon^2}
}}
{{eqn | l = \delta u
      | r = \dfrac {\partial u} {\partial \mathbf x} \delta \mathbf x + \overline {\delta u}
}}
{{eqn | l = \paren {\delta u}_{x_i}
      | r = \paren {\overline {\delta u} }_{x_i} + u_{\mathbf x x_i} {\delta \mathbf x}_{x_i}
}}
{{eqn | l = \Delta \mathbf x
      | r = \epsilon \boldsymbol \phi + \map \OO {\epsilon^2}
}}
{{eqn | l = \delta \mathbf x
      | r = \epsilon \boldsymbol \phi
}}
{{end-eqn}}
Here $\delta$ difference stands for principal part, which is constant or linear {{WRT}} $\epsilon$.
Thus, the variation of $J$ reads:
:$\ds \delta J = \int_R \paren {F_{\mathbf x} \delta \mathbf x + F_u \overline {\delta u} + F_u u_{\mathbf x} \delta \mathbf x + F_{u_{\mathbf x} } {\overline {\delta u} }_{\mathbf x} + \sum_{i, j \mathop = 1}^n F_{u_{x_i} } u_{x_i x_j} \delta x_j + \map F {\delta \mathbf x}_{\mathbf x} } \rd x_1 \dotsm \rd x_n$
A few terms can be rewritten using:
:$\ds \dfrac {\partial} {\partial \mathbf x} \paren {F \delta \mathbf x} = F_{\mathbf x} \delta \mathbf x + \map F {\overline {\delta \mathbf x} }_{\mathbf x} + F_u u_{\mathbf x} \delta \mathbf x + \sum_{i, j\mathop = 1}^n F_{u_{x_i} } u_{x_i x_j} \delta x_j$
and:
:$\ds \map {F_{u_{\mathbf x} } } {\overline{\delta u} }_{\mathbf x} = \dfrac {\partial} {\partial \mathbf x} {F_{u_{\mathbf x} } \overline {\delta u} } - \dfrac {\partial F_{u_{\mathbf x} } } {\partial \mathbf x} \delta u$
Substitution of previous results into variation of $J$ leads to:
:$\ds \delta J = \int_R \paren {F_u - \dfrac {\partial F_{u_{\mathbf x} } } {\partial \mathbf x} } \overline {\delta u} \rd x_1 \dotsm \rd x_n + \int_R \map {\dfrac \partial {\partial \mathbf x} } {F_{u_{\mathbf x} } \overline {\delta u} + F \delta \mathbf x} \rd x_1 \dotsm \rd x_n$
Substitute expressions for $\overline {\delta u}$ and $\delta\mathbf x$ to obtain the desired result.
{{qed}}