Consider the boundary-value problem of nonlinear elastostatics: \[\tag{P} \begin{aligned} &\left. \begin{aligned} &{\operatorname{div}}{\mathbf S}+{\mathbf b}=0\\ &{\mathbf S}=\psi'({\mathbf F}),\\ &{\mathbf F}=\nabla{\mathbf y},\\ \end{aligned}\qquad \right\}\text{ in }\mathcal B\\ &\left.\begin{aligned} &{\mathbf y}={\mathbf y}_0,\quad\text{ on }\partial_c\mathcal B\\ &{\mathbf S}{\mathbf n}=\mathbf s_0 \quad\text{ on }\partial_f\mathcal B. \end{aligned}\right. \end{aligned}\] We suppose that the domain \(\mathcal B\) undergoes a perturbation \(\mathcal B\to\mathcal B_{\varepsilon}\), where \({\varepsilon}\) is a small parameter, and we consider the perturbed problem \[\tag*{$(\rm P)_{\varepsilon}$} \begin{aligned} &\left. \begin{aligned} &{\operatorname{div}}{\mathbf S}{_{{\varepsilon}}}+{\mathbf b}{_{{\varepsilon}}}=0\\ &{\mathbf S}{_{{\varepsilon}}}=\psi'({\mathbf F}{_{{\varepsilon}}}),\\ &{\mathbf F}{_{{\varepsilon}}}=\nabla{\mathbf y}{_{{\varepsilon}}},\\ \end{aligned}\qquad \right\}\text{ in }\mathcal B{_{{\varepsilon}}}\\ &\left.\begin{aligned} &{\mathbf y}_{\varepsilon}={\mathbf y}_{0,{\varepsilon}},\quad\text{on }\partial_c\mathcal B{_{{\varepsilon}}}\\ &{\mathbf S}{_{{\varepsilon}}}{\mathbf n}=\mathbf s_{\varepsilon}\quad\text{on }\partial_f\mathcal B{_{{\varepsilon}}}. \end{aligned}\right. \end{aligned}\] Given any \({\varepsilon}\)-dependent field \(\phi_{\varepsilon}\) on \(\mathcal B_{\varepsilon}\), and a point \(x\in\mathcal B\), we define the increment of \(\phi_{{\varepsilon}}(x)\) as¹ \[\label{eq:3} \delta\phi(x)=\left.\frac{\partial}{\partial{\varepsilon}}\right|_{{\varepsilon}=0}\phi_{\varepsilon}(x).\] We are going to show that the increments are solution of the following BPV: \[\tag*{$(\rm P)_{\varepsilon}$} \begin{aligned} &\left. \begin{aligned} &{\operatorname{div}}\delta{\mathbf S}+\delta{\mathbf b}=0\\ &\delta{\mathbf S}=\psi''({\mathbf F})[\delta{\mathbf F}],\\ &\delta{\mathbf F}=\nabla\delta{\mathbf y},\\ \end{aligned}\qquad \right\}\text{ in }\mathcal B\\ &\left.\begin{aligned} &\delta{\mathbf y}+{\mathbf F}\delta{\boldsymbol{\varphi}}=\delta{\mathbf y}_0,\quad\text{on }\partial_c\mathcal B\\ &\delta{\mathbf S}{\mathbf n}+\nabla{\mathbf S}[\delta{\boldsymbol{\varphi}}]{\mathbf n}+{\mathbf S}\delta{\mathbf n}=\delta\mathbf s_0\quad\text{on }\partial_f\mathcal B. \end{aligned}\right. \end{aligned}\] For simplicity, we prove this result in the case when \(\mathbf s_0\) vanishes.

As a start, we choose an arbitrary test function \(\mathbf v\) defined on \(\mathcal B\) such that \(\mathbf v=0\) on \(\partial_c\mathcal B\). Then, we let \(\mathbf v_{\varepsilon}\) be the unique test function on \(\mathcal B_\varepsilon\) such that \(\mathbf v_\varepsilon(\varphi_{\varepsilon}(x))=\mathbf v(x)\). Then, \[\int_{\mathcal B_{\varepsilon}}{\mathbf S}{_{{\varepsilon}}}\cdot\nabla\mathbf v_{\varepsilon}=\int_{\mathcal B{_{{\varepsilon}}}}{\mathbf b}{_{{\varepsilon}}}\cdot\mathbf v_{\varepsilon}.\] As done in the one-dimensional case, we change the domain of integration by considering a diffeomorphism \({\boldsymbol{\varphi}}_{\varepsilon}:\mathcal B\to\mathcal B_{\varepsilon}\), and by defining the functions \[\widetilde{\mathbf S}_{\varepsilon}(x)={\mathbf S}_{\varepsilon}({\boldsymbol{\varphi}}{_{{\varepsilon}}}(x)),\qquad \widetilde{\mathbf b}_{\varepsilon}(x)={\mathbf b}_{\varepsilon}({\boldsymbol{\varphi}}{_{{\varepsilon}}}(x)),\] we obtain \[\int_{\mathcal B}\widetilde{\mathbf S}{_{{\varepsilon}}}(\operatorname{Cof}\nabla{\boldsymbol{\varphi}}{_{{\varepsilon}}})\cdot\nabla{\mathbf v}=\int_{\mathcal B}(\det\nabla{\boldsymbol{\varphi}}{_{{\varepsilon}}})\widetilde{\mathbf b}{_{{\varepsilon}}}\cdot{\mathbf v}\qquad\forall {\mathbf v}=0\text{ on }\partial_c\mathcal B.\] It is important to keep in mindi that \({\boldsymbol{\varphi}}_{\varepsilon}\) is not defined uniquely. What matters is that it maps \(\mathcal B\) into \(\mathcal B_{\varepsilon}\).

We differentiate with respect to \({\varepsilon}\) at \({\varepsilon}=0\), and use the relation \[\delta\widetilde{\mathbf S}=\delta{\mathbf S}+\nabla{\mathbf S}[\delta{\boldsymbol{\varphi}}],\] as well as the analogous relation of \(\delta{\mathbf b}\) to obtain \[\label{eq:4} \int_{\mathcal B}\delta{\mathbf S}\cdot\nabla{\boldsymbol{v}}+\nabla\mathbf S[\delta{\boldsymbol{\varphi}}]\cdot\nabla \mathbf v+\mathbf S\delta(\operatorname{Cof}\nabla{\boldsymbol{\varphi}})\cdot\nabla\mathbf v=\int_{\mathcal B}\delta{\mathbf b}\cdot\mathbf v+(\nabla{\mathbf b}\delta{\boldsymbol{\varphi}})\cdot\mathbf v+{\mathbf b}\cdot(\delta(\det\nabla{\boldsymbol{\varphi}})\mathbf v).\] Now we can essentially repeat what has been done in the the one-dimensional case. Using integration by parts and the identity \(\operatorname{div}\delta(\operatorname{Cof}\nabla{\boldsymbol{\varphi}})=\mathbf 0\), we obtain \[\begin{gathered} \label{eq:2} \int_{\mathcal B}\left[\delta\mathbf S+\nabla\mathbf S[\delta{\boldsymbol{\varphi}}]+\mathbf S\delta(\operatorname{Cof}\nabla{\boldsymbol{\varphi}})\right]\cdot\nabla\mathbf v\\ = -\int_{\mathcal B}(\operatorname{div}\delta\mathbf S+(\nabla\operatorname{div}\mathbf S) \delta{\boldsymbol{\varphi}})\cdot\mathbf v+\operatorname{div}\mathbf S\cdot(\delta(\det\nabla{\boldsymbol{\varphi}})\mathbf v)+\int_{\partial\mathcal B}\left[\delta\mathbf S+\nabla\mathbf S[\delta{\boldsymbol{\varphi}}]+\mathbf S\delta(\operatorname{Cof}\nabla{\boldsymbol{\varphi}})\right]\mathbf n\cdot\mathbf v.\end{gathered}\] We substitute into . Then we observe that thanks to the equilibrium equation \(\operatorname{div}\mathbf S+\mathbf b=\mathbf 0\), several terms cancel out, and the final result is \[\label{eq:1} \int_{\mathcal B}(\operatorname{div}\delta\mathbf S+\delta{\mathbf b})\cdot\mathbf v+\int_{\partial\mathcal B}\left[\delta\mathbf S+\nabla\mathbf S[\delta{\boldsymbol{\varphi}}]+\mathbf S\delta(\operatorname{Cof}\nabla{\boldsymbol{\varphi}})\right]\mathbf n\cdot\mathbf v=0.\] ...