Consider the boundary-value problem of nonlinear elastostatics:

(P)

We suppose that the domain $ℬ$ undergoes a perturbation $ℬ\to {ℬ}_{\epsilon }$, where $\epsilon$ is a small parameter, and we consider the perturbed problem

${(\mathrm{P})}_{\epsilon }$

Given any $\epsilon$-dependent field ${\varphi }_{\epsilon }$ on ${ℬ}_{\epsilon }$, and a point $x\in ℬ$, we define the increment of ${\varphi }_{\epsilon }(x)$ as

$$\delta \varphi (x)={\frac{\partial }{\partial \epsilon }|}_{\epsilon =0}{\varphi }_{\epsilon }(x).$$ (1)

We are going to show that the increments are solution of the following BPV:

${(\mathrm{P})}_{\epsilon }$

For simplicity, we prove this result in the case when ${𝐬}_0$ vanishes.

As a start, we choose an arbitrary test function $𝐯$ defined on $ℬ$ such that $𝐯=0$ on ${\partial }_cℬ$. Then, we let ${𝐯}_{\epsilon }$ be the unique test function on ${ℬ}_{\epsilon }$ such that ${𝐯}_{\epsilon }({\phi }_{\epsilon }(x))=𝐯(x)$. Then,

$${\int }_{{ℬ}_{\epsilon }}{𝐒}_{\epsilon }\cdot \nabla {𝐯}_{\epsilon }={\int }_{{ℬ}_{\epsilon }}{𝐛}_{\epsilon }\cdot {𝐯}_{\epsilon }.$$ (2)

As done in the one-dimensional case, we change the domain of integration by considering a diffeomorphism ${𝝋}_{\epsilon }:ℬ\to {ℬ}_{\epsilon }$, and by defining the functions

$${\widetilde{𝐒}}_{\epsilon }(x)={𝐒}_{\epsilon }({𝝋}_{\epsilon }(x)),{\widetilde{𝐛}}_{\epsilon }(x)={𝐛}_{\epsilon }({𝝋}_{\epsilon }(x)),$$ (3)

we obtain

$${\int }_{ℬ}{\widetilde{𝐒}}_{\epsilon }(\mathrm{Cof}\nabla {𝝋}_{\epsilon })\cdot \nabla 𝐯={\int }_{ℬ}(det\nabla {𝝋}_{\epsilon }){\widetilde{𝐛}}_{\epsilon }\cdot 𝐯\mathit{\hspace{1em}\hspace{1em}}\forall 𝐯=0\text{ on }{\partial }_cℬ.$$ (4)

It is important to keep in mindi that ${𝝋}_{\epsilon }$ is not defined uniquely. What matters is that it maps $ℬ$ into ${ℬ}_{\epsilon }$.

We differentiate with respect to $\epsilon$ at $\epsilon =0$, and use the relation

$$\delta \widetilde{𝐒}=\delta 𝐒+\nabla 𝐒[\delta 𝝋],$$ (5)

as well as the analogous relation of $\delta 𝐛$ to obtain

$${\int }_{ℬ}\delta 𝐒\cdot \nabla 𝒗+\nabla 𝐒[\delta 𝝋]\cdot \nabla 𝐯+𝐒\delta (\mathrm{Cof}\nabla 𝝋)\cdot \nabla 𝐯={\int }_{ℬ}\delta 𝐛\cdot 𝐯+(\nabla 𝐛\delta 𝝋)\cdot 𝐯+𝐛\cdot (\delta (det\nabla 𝝋)𝐯).$$ (6)

Now we can essentially repeat what has been done in the the one-dimensional case. Using integration by parts and the identity $\mathrm{div}\delta (\mathrm{Cof}\nabla 𝝋)=\mathrm{𝟎}$, we obtain

$$\begin{array}{c}{\int }_{ℬ}\left[\delta 𝐒+\nabla 𝐒[\delta 𝝋]+𝐒\delta (\mathrm{Cof}\nabla 𝝋)\right]\cdot \nabla 𝐯\hfill \\ \hfill =-{\int }_{ℬ}(\mathrm{div}\delta 𝐒+(\nabla \mathrm{div}𝐒)\delta 𝝋)\cdot 𝐯+\mathrm{div}𝐒\cdot (\delta (det\nabla 𝝋)𝐯)+{\int }_{\partial ℬ}\left[\delta 𝐒+\nabla 𝐒[\delta 𝝋]+𝐒\delta (\mathrm{Cof}\nabla 𝝋)\right]𝐧\cdot 𝐯.\end{array}$$ (7)

We substitute (7) into (6). Then we observe that thanks to the equilibrium equation $\mathrm{div}𝐒+𝐛=\mathrm{𝟎}$, several terms cancel out, and the final result is

$${\int }_{ℬ}(\mathrm{div}\delta 𝐒+\delta 𝐛)\cdot 𝐯+{\int }_{\partial ℬ}\left[\delta 𝐒+\nabla 𝐒[\delta 𝝋]+𝐒\delta (\mathrm{Cof}\nabla 𝝋)\right]𝐧\cdot 𝐯=0.$$ (8)

…