Variational convergence of discrete geometrically-incompatible elastic models

Abstract

We derive a continuum model for incompatible elasticity as a variational limit of a family of discrete nearest-neighbor elastic models. The discrete models are based on discretizations of a smooth Riemannian manifold $({\mathcal {M}},\mathfrak {g})$, endowed with a flat, symmetric connection $\nabla$. The metric $\mathfrak {g}$ determines local equilibrium distances between neighboring points; the connection $\nabla$ induces a lattice structure shared by all the discrete models. The limit model satisfies a fundamental rigidity property: there are no stress-free configurations, unless $\mathfrak {g}$ is flat, i.e., has zero Riemann curvature. Our analysis focuses on two-dimensional systems, however, all our results readily generalize to higher dimensions.

Technical lemmas

Lemma 6

Let V and W be two-dimensional inner-product spaces. Let $A\in {\text {Hom}}(V,W)$. If there exist two independent vectors $x,y\in V$ such that

$$\begin{aligned} |A(x)| = |x|, \qquad |A(y)| = |y| \qquad \text { and }\qquad |A(x+y)| = |x+y|, \end{aligned}$$

then A is an isometry.

Proof

It follows from the polarization identity that $(A(x),A(y)) = (x,y)$, and therefore A preserves the inner-product, hence it is an isometry. $\square$

Lemma 7

Let V and W be two-dimensional inner-product spaces. Let $x,y\in V$ be independent vectors of equal length. Then, for every $A\in {\text {Hom}}(V,W)$,

$$\begin{aligned} |A|^2 \le \frac{2}{1-\cos \theta }\left( \frac{|Ax|^2}{|x|^2} + \frac{|Ay|^2}{|y|^2} + \frac{|A(x+y)|^2}{|x+y|^2}\right) , \end{aligned}$$

where $\theta$ is the angle between x and y.

Proof

It suffices to prove the lemma for unit vectors. Denote $c = \cos \theta = (x,y)$. The vectors x and $(y - c\, x)/\sqrt{1-c^2}$ are orthonormal, hence

$$\begin{aligned} |A|^2 = |Ax|^2 + \frac{|A(y - c\, x)|^2}{1-c^2} = \frac{|Ax|^2 + |Ay|^2 - 2c(Ax,Ay)}{1-c^2}. \end{aligned}$$

Now,

$$\begin{aligned} \begin{aligned} |A|^2&= \frac{2}{1-c} \cdot \frac{1}{2(1+c)} \left( |Ax|^2 + |Ay|^2 - 2c(Ax,Ay)\right) \\&= \frac{2}{1-c} \left( |Ax|^2 + |Ay|^2 + \frac{|A(x+y)|^2 - 2(1+c)(|Ax|^2 + |Ay|^2 + (Ax,Ay))}{|x+y|^2}\right) \\&\le \frac{2}{1-c} \left( |Ax|^2 + |Ay|^2 + \frac{|A(x+y)|^2}{|x+y|^2}\right) , \end{aligned} \end{aligned}$$

where in the last step we used the fact that $|Ax|^2 + |Ay|^2 + (Ax,Ay) > 0$. $\square$

Lemma 8

Let V and W be two-dimensional inner-product spaces. Let $x,y\in V$ be two independent vectors. Then, there exists a constant C depending continuously on the angle $\theta$ between x and y and the ratio of their lengths $r = |y|/|x|$, such that for every $A\in {\text {Hom}}(V,W)$,

$$\begin{aligned} |A|^2 \le C\left( \frac{|Ax|^2}{|x|^2} + \frac{|Ay|^2}{|y|^2} + \frac{|A(x+y)|^2}{|x+y|^2}\right) . \end{aligned}$$

Proof

Without loss of generality we can assume that $|y| > |x|$ (otherwise Lemma 7 applies). Set

$$\begin{aligned} v=x+\alpha y \qquad \text { and }\qquad w=(1-\alpha ) y, \end{aligned}$$

where

$$\begin{aligned} \alpha = \frac{r^2-1}{2r(r +\cos \theta )} \end{aligned}$$

is chosen such that $|v| = |w|$. Also, $v+w = x+y$. Note that $\alpha \in \left( (r-1)/2r, (r+1)/2r\right) \subset (0,1)$, and in particular, v and w are independent. The angle between v and w depends only on $\alpha$ and $\theta$, and therefore on r and $\theta$. By the previous lemma, there exists a $C = C(r,\theta )$ such that

$$\begin{aligned} \begin{aligned} |A|^2&\le C\left( \frac{|Av|^2}{|v|^2} + \frac{|Aw|^2}{|w|^2} + \frac{|A(v+w)|^2}{|v+w|^2}\right) \\&= C\left( \frac{|Ax|^2 + \alpha ^2|Ay|^2 + 2\alpha (Ax,Ay)}{(1-\alpha )^2|y|^2} + \frac{|Ay|^2}{|y|^2} + \frac{|A(x+y)|^2}{|x+y|^2}\right) \\&\le C\left( \frac{|Ax|^2 + \alpha ^2|Ay|^2 + \alpha (|Ax|^2 + |Ay|^2)}{(1-\alpha )^2|y|^2} + \frac{|Ay|^2}{|y|^2} + \frac{|A(x+y)|^2}{|x+y|^2}\right) \\&= C\left( \frac{1+\alpha }{(1-\alpha )^2r^2}\frac{|Ax|^2}{|x|^2} + \left( 1+\frac{\alpha ^2 +\alpha }{(1-\alpha )^2}\right) \frac{|Ay|^2}{|y|^2} + \frac{|A(x+y)|^2}{|x+y|^2}\right) \\&\le C'\left( \frac{|Ax|^2}{|x|^2} + \frac{|Ay|^2}{|y|^2} + \frac{|A(x+y)|^2}{|x+y|^2}\right) , \end{aligned} \end{aligned}$$

where in the passage to the third line we used the inequality $2ab\le a^2 + b^2$. $\square$

Lemma 9

Let V and W be d-dimensional inner-product spaces. Then, for every $A\in {\text {Hom}}(V,W)$,

$$\begin{aligned} {\text {dist}}^2(A,\text {SO}(V,W)) \le {\text {dist}}^2(A,\text {O}(V,W)) + 4|\det A|^{1/d} \mathbb {1}_{\left\{ \det A<0\right\} }. \end{aligned}$$

(41)

Proof

Let $\sigma _1\ge \sigma _2\ge \ldots \ge \sigma _d\ge 0$ be the singular values of A. Then

$$\begin{aligned} {\text {dist}}^2(A,\text {O}(V,W)) = \sum _{i=1}^{d} (\sigma _i - 1)^2 \quad \text { and }\quad |\det A| = \prod _{i=1}^{d} \sigma _i \end{aligned}$$

If $\det A\ge 0$, then

$$\begin{aligned} {\text {dist}}^2(A,\text {SO}(V,W)) = \sum _{i=1}^{d} (\sigma _i - 1)^2 = {\text {dist}}^2(A,\text {O}(V,W)), \end{aligned}$$

which shows the equality in (41) in this case. If $\det A<0$, then

$$\begin{aligned} \begin{aligned} {\text {dist}}^2(A,\text {SO}(V,W))&= \sum _{i=1}^{d-1} (\sigma _i - 1)^2 + (\sigma _d + 1)^2 \\&= {\text {dist}}^2(A,\text {O}(V,W)) + 4\sigma _d \le {\text {dist}}^2(A,\text {O}(V,W)) + 4|\det A|^{1/d}. \end{aligned} \end{aligned}$$

$\square$