Linearization and localization of nonconvex functionals motivated by nonlinear peridynamic models

Abstract

We consider a class of nonconvex energy functionals that lies in the framework of the peridynamics model of continuum mechanics. The energy densities are functions of a nonlocal strain that describes deformation based on pairwise interaction of material points and as such are nonconvex with respect to nonlocal deformation. We apply variational analysis to investigate the consistency of the effective behavior of these nonlocal nonconvex functionals with established classical and peridynamic models in two different regimes. In the regime of small displacement, we show the model can be effectively described by its linearization. To be precise, we rigorously derive what is commonly called the linearized bond-based peridynamic functional as a $\Gamma$-limit of nonlinear functionals. In the regime of vanishing nonlocality, the effective behavior of the nonlocal nonconvex functionals is characterized by an integral representation, which is obtained via $\Gamma$-convergence with respect to the strong $L^p$ topology. We also prove various properties of the density of the localized quasiconvex functional such as frame-indifference and coercivity. We demonstrate that the density vanishes on matrices whose singular values are less than or equal to one. These results confirm that the localization, in the context of $\Gamma$-convergence, of peridynamic-type energy functionals exhibits behavior quite different from classical hyperelastic energy functionals.

Appendix A: Characterizations of distance-preserving maps

Theorem A.1

Let $\Omega \subset \mathbb {R}^d$ be a bounded domain. Suppose $\textbf{v}: \Omega \rightarrow \mathbb {R}^d$ is measurable, and suppose $\textbf{v}$ satisfies

$$\begin{aligned} |\textbf{v}(\textbf{x})-\textbf{v}(\textbf{y})| = |\textbf{x}-\textbf{y}| \end{aligned}$$

(A.1)

for $\mathcal {L}^{2d}$-almost every $(\textbf{x},\textbf{y}) \in \Omega \times \Omega$, where in general $\mathcal {L}^N$ denotes N-dimensional Lebesgue measure. Then, there exists a constant matrix $\mathbb {F}\in \mathcal {O}(d)$ and $\textbf{b}\in \mathbb {R}^d$ such that $\textbf{v}(\textbf{x}) = \mathbb {F}\textbf{x}+ \textbf{b}$ for almost every $\textbf{x}\in \Omega$.

Proof

We prove the result first under the additional assumption that $\textbf{v}: \Omega \rightarrow \mathbb {R}^d$ is continuous. Then, the functions $f_1(\textbf{x},\textbf{y}):= |\textbf{v}(\textbf{x})-\textbf{v}(\textbf{y})|$ and $f_2(\textbf{x},\textbf{y}):=|\textbf{x}-\textbf{y}|$ are continuous on $\Omega \times \Omega$. Let $X \subset \Omega \times \Omega$ be the set where (A.1) holds. Then, $(\Omega \times \Omega ) \backslash X$ is dense in $\Omega \times \Omega$. So by density and continuity we have

$$\begin{aligned} |\textbf{v}(\textbf{x})-\textbf{v}(\textbf{y})| = |\textbf{x}-\textbf{y}| \quad \forall \, \textbf{x}, \textbf{y}\in \Omega \times \Omega . \end{aligned}$$

Since the relation (A.1) is translation- and shift-invariant, we can assume without loss of generality that $B({\textbf {0}},R) \subset \Omega$ for some $R>0$ and $\textbf{v}({\textbf {0}}) = {\textbf {0}}$. We will show that $\textbf{v}(\textbf{x}) = \mathbb {F}\textbf{x}$ for some constant matrix $\mathbb {F}\in \mathcal {O}(d)$, obtaining the result for continuous functions.

First, by (A.1) $|\textbf{v}(\textbf{x})|^2 = |\textbf{v}(\textbf{x})-\textbf{v}({\textbf {0}})|^2 = |\textbf{x}-{\textbf {0}}|^2 = |\textbf{x}|^2$. The identity

$$\begin{aligned} \left\langle \textbf{v}(\textbf{x}),\textbf{v}(\textbf{y}) \right\rangle = \left\langle \textbf{x},\textbf{y} \right\rangle \end{aligned}$$

(A.2)

for every $(\textbf{x},\textbf{y}) \in \Omega \times \Omega$ follows, since

$$\begin{aligned} \begin{aligned} |\textbf{x}|^2 - 2 \left\langle \textbf{x},\textbf{y} \right\rangle + |\textbf{y}|^2 =|\textbf{x}-\textbf{y}|^2 = |\textbf{v}(\textbf{x})-\textbf{v}(\textbf{y})|^2&= |\textbf{v}(\textbf{x})|^2 - 2 \left\langle \textbf{v}(\textbf{x}),\textbf{v}(\textbf{y}) \right\rangle + |\textbf{v}(\textbf{y})|^2 \\&= |\textbf{x}|^2 - 2\left\langle \textbf{v}(\textbf{x}),\textbf{v}(\textbf{y}) \right\rangle + |\textbf{y}|^2\,. \end{aligned} \end{aligned}$$

Define the $d \times d$ matrix

$$\begin{aligned} \mathbb {F}= (a^{jk}) = \frac{1}{R} (u_j( R \textbf{e}_k))\,, \end{aligned}$$

where $\textbf{e}_k$ is the vector in $\mathbb {R}^d$ with kth coordinate 1 and all other coordinates 0. Using this definition and using (A.2) with $R \textbf{e}_j$ in place of $\textbf{y}$,

$$\begin{aligned} {[}\mathbb {F}^T \textbf{v}(\textbf{x})]_j = \sum _{k=1}^{d} a^{kj} u_k(\textbf{x}) = \sum _{k=1}^{d} \frac{1}{R} u_k(R \textbf{e}_j) u_k(\textbf{x}) = \frac{1}{R} \left\langle \textbf{v}(\textbf{x}),\textbf{v}( R \textbf{e}_j) \right\rangle = \frac{1}{R} \left\langle \textbf{x},R \textbf{e}_j \right\rangle = x_j\,. \end{aligned}$$

This is true for any $j \in \{ 1, \ldots , d\}$, and so

$$\begin{aligned} \mathbb {F}^T \textbf{v}(\textbf{x}) = \textbf{x}\,. \end{aligned}$$

Again using the definition of $\mathbb {F}$ and using (A.2) with $(R \textbf{e}_j, R \textbf{e}_k)$ in place of $(\textbf{x},\textbf{y})$,

$$\begin{aligned} {[}\mathbb {F}^T \mathbb {F}]_{jk} = \sum _{\ell = 1}^n a^{\ell j} a_{\ell k} = \frac{1}{R^2} \sum _{\ell = 1}^n u_{\ell }(R \textbf{e}_j) u_{\ell } (R \textbf{e}_k) = \frac{1}{R^2}\left\langle \textbf{v}(R \textbf{e}_j),\textbf{v}(R \textbf{e}_k) \right\rangle = \frac{1}{R^2}\left\langle R \textbf{e}_j, R\textbf{e}_k \right\rangle = \delta _{jk}\,. \end{aligned}$$

Therefore, $\mathbb {F}^T \mathbb {F}= \mathbb {I}$, and we conclude that $\textbf{v}(\textbf{x}) = \mathbb {F}\textbf{x}$ with $\mathbb {F}\in \mathcal {O}(d)$.

Now suppose $\textbf{v}$ is measurable. By Lusin’s theorem for every $n \in \mathbb {N}$ there exists a closed set $K_n \subset \Omega$ with $\mathcal {L}^d(\Omega \setminus K_n) < \frac{1}{n}$ such that $\textbf{v}$ is continuous on $K_n$. By (A.1) $\textbf{v}$ is also Lipschitz on $K_n$, with Lipschitz constant 1. By Kirszbraun’s theorem there exists a function $\textbf{v}_n: \mathbb {R}^d \rightarrow \mathbb {R}^d$ that is 1-Lipschitz and coincides with $\textbf{v}$ on $K_n$. Therefore, by Rellich’s theorem there exists a subsequence (not relabeled) $\{\textbf{v}_n\}$ that converges uniformly on ${\overline{\Omega }}$ to a continuous and 1-Lipschitz function $\widetilde{\textbf{v}}$. By definition of the $\textbf{v}_n$ it follows that $\textbf{v}= \widetilde{\textbf{v}}$ almost everywhere on $\Omega$. Therefore, $\textbf{v}$ has a Lipschitz (hence continuous) representative, and the first part of the proof applies. $\square$

This rigidity result can be strengthened in the spirit of [25], as we demonstrate in the next theorem.

Theorem A.2

Let $m \ge 1$, let $\rho \in L^1(\mathbb {R}^d)$ be a nonnegative radial kernel satisfying $B({\varvec{0}},r) \subset {{\,\textrm{supp}\,}}\rho \subset B({\varvec{0}},R)$ for given $0< r < R$, and let $\Phi$ be a nondecreasing convex function satisfying (2.3). Suppose a sequence of vector fields $\{ \textbf{v}_n \}_n \subset \mathfrak {W}^{\rho ,p}(\Omega ;\mathbb {R}^d)$ satisfies

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \int _{\Omega } \int _{\Omega } \rho (\textbf{x}-\textbf{y}) \Phi \left( |s_m[\textbf{v}_n](\textbf{y},\textbf{x})| \right) \, \textrm{d}\textbf{y} \, \textrm{d}\textbf{x} = 0\,. \end{aligned}$$

Suppose additionally that there exists a function $\textbf{v}\in L^{1}(\Omega ;\mathbb {R}^d)$ such that $\textbf{v}_n \rightarrow \textbf{v}$ in $L^{1}(\Omega ;\mathbb {R}^d)$. Then, there exists a constant matrix $\mathbb {F}\in \mathcal {O}(d)$ and a vector $\textbf{b}\in \mathbb {R}^d$ such that $\textbf{v}(\textbf{x}) = \mathbb {F}\textbf{x}+ \textbf{b}$ for almost every $\textbf{x}\in \Omega$.

Proof

Since $\textbf{v}_n \rightarrow \textbf{v}$ in $L^{1}(\Omega ;\mathbb {R}^d)$ there exists a subsequence (not relabeled) $\{\textbf{v}_n\}_n$ that converges to $\textbf{v}$ $\mathcal {L}^d$-almost everywhere in $\Omega$. Since $\Phi$ is continuous,

$$\begin{aligned} \Phi \left( |s_m[\textbf{v}_n](\textbf{y},\textbf{x})| \right) \rightarrow \Phi \left( |s_m[\textbf{v}](\textbf{y},\textbf{x})| \right) \; \mathcal {L}^{2d}\text {-a.e. in } \Omega \times \Omega \,, \end{aligned}$$

and so by Fatou’s lemma

$$\begin{aligned} \int _{\Omega } \int _{\Omega } \rho (\textbf{x}-\textbf{y}) \Phi \left( |s_m[\textbf{v}](\textbf{y},\textbf{x})| \right) \, \textrm{d}\textbf{y} \, \textrm{d}\textbf{x} \le \mathop {\mathrm{lim\,inf}}\limits _{n \rightarrow \infty } \int _{\Omega } \int _{\Omega } \rho (\textbf{x}-\textbf{y}) \Phi \left( |s_m[\textbf{v}_n](\textbf{y},\textbf{x})| \right) \, \textrm{d}\textbf{y} \, \textrm{d}\textbf{x} = 0\,. \end{aligned}$$

Therefore, it must be that $\rho (\textbf{y}-\textbf{x})\Phi \left( |s_m[\textbf{v}](\textbf{y},\textbf{x})| \right) = 0 \text { for a.e. } \textbf{y}\in {{\,\textrm{supp}\,}}\rho + \textbf{x}\,, \textbf{x}\in \Omega \,.$ For any $\textbf{x}\in \Omega$, define $r_{\textbf{x}} = \min \{r,{{\,\textrm{dist}\,}}(\textbf{x},\partial \Omega )\}$. Then by assumption on $\rho$,

$$\begin{aligned} \Phi \left( |s_m[\textbf{v}](\textbf{y},\textbf{x})| \right) = 0 \qquad \text { for a.e. } \textbf{y}\in B(\textbf{x},r_{\textbf{x}})\,, \textbf{x}\in \Omega \,. \end{aligned}$$

Now fix $\textbf{x}_0 \in \Omega$ and fix $r_0 = r_{\textbf{x}_0}$. Then, by definition of $\Phi$ and $s_m$

$$\begin{aligned} |\textbf{v}(\textbf{y})-\textbf{v}(\textbf{x})|=|\textbf{y}-\textbf{x}| \qquad \text { for a.e. } \textbf{y}\in B(\textbf{x}_0,r_0/2)\,, \textbf{x}\in B(\textbf{x}_0,r_0/2)\,. \end{aligned}$$

The relation then holds for $\mathcal {L}^{2d}$-a.e. $(\textbf{x},\textbf{y}) \in B(\textbf{x}_0,r_0/2) \times B(\textbf{x}_0,r_0/2)$, and so Lemma A.1 applies on $B(\textbf{x}_0,r_0/2)$.

By covering $\Omega$ with sets of the form $B(\textbf{x}_0,r_0/2)$, we see that $\textbf{v}$ is a possibly piecewise affine map on $\Omega$. To conclude, note that since $\Omega$ is a domain there exists a finite chain of sets of the form $B(\textbf{x}_0,r_0/2)$ between any $\textbf{x}_1$ and $\textbf{x}_2$ in $\Omega$, and thus $\textbf{v}$ must be the same affine map at both points. $\square$

Theorem A.3

Suppose that $\textbf{v}\in C^2({\overline{\Omega }};\mathbb {R}^d)$ satisfies (1.8). Then there exists a constant matrix $\mathbb {F}\in \mathcal {O}(d)$ and a vector $\textbf{b}\in \mathbb {R}^d$ such that $\textbf{v}(\textbf{x}) = \mathbb {F}\textbf{x}+ \textbf{b}$ for every $\textbf{x}\in \Omega$.

Proof

First, since $\textbf{v}\in C^2{\overline{\Omega }};\mathbb {R}^d)$, $\det \nabla \textbf{v}\in C^1({\overline{\Omega }};\mathbb {R}^d)$, with

$$\begin{aligned} \partial _\ell [\det \nabla \textbf{v}(\textbf{x})] = \det \nabla \textbf{v}(\textbf{x}) \sum _{j,k =1}^d \partial _j u_k \cdot \partial _\ell [\partial _j u_k ] = \frac{\det \nabla \textbf{v}(\textbf{x}) }{2} \partial _\ell [ |\nabla \textbf{v}(\textbf{x})|^2 ]\,. \end{aligned}$$

But $|\nabla \textbf{v}(\textbf{x})|^2 = \textrm{tr}( \nabla \textbf{v}^T \nabla \textbf{v}) = d$, so therefore $\det \nabla \textbf{v}$ is constant in all of ${\overline{\Omega }}$. Thus, either $\nabla \textbf{v}= \textrm{cof} \nabla \textbf{v}$ for all $\textbf{x}\in \Omega$ or $\nabla \textbf{v}= -\textrm{cof} \nabla \textbf{v}$ for all $\textbf{x}\in \Omega$. In both cases it follows from the Piola identity $\textrm{div}\,\textrm{cof} \nabla \textbf{v}= {\textbf {0}}$ that $\textbf{v}$ is a harmonic function on $\Omega$. Thus $\textbf{v}\in C^{\infty }(\Omega )$, and so we can compute

$$\begin{aligned} 0 = \frac{1}{2} \Delta [ |\nabla \textbf{v}|^2 - d] = \nabla \textbf{v}: \Delta [ \nabla \textbf{v}] + |\nabla ^2 \textbf{v}|^2 = |\nabla ^2 \textbf{v}|^2\,. \end{aligned}$$

Thus, $\nabla \textbf{v}(\textbf{x})$ is constant in $\Omega$, and necessarily belongs to $\mathcal {O}(d)$. $\square$