Exact Constructions in the (Non-linear) Planar Theory of Elasticity: From Elastic Crystals to Nematic Elastomers

Abstract

In this article we deduce necessary and sufficient conditions for the presence of “Conti-type”, highly symmetric, exactly stress-free constructions in the geometrically non-linear, planar n-well problem, generalising results of Conti et al. (Proc R Soc A 473(2203):20170235, 2017). Passing to the limit $n\rightarrow \infty$, this allows us to treat solid crystals and nematic elastomer differential inclusions simultaneously. In particular, we recover and generalise (non-linear) planar tripole star type deformations which were experimentally observed in Kitano and Kifune (Ultramicroscopy 39(1–4):279–286, 1991), Manolikas and Amelinckx (Physica Status Solidi (A) 60(2):607–617, 1980; Physica Status Solidi (A) 61(1):179–188, 1980). Furthermore, we discuss the corresponding geometrically linearised problem.

Appendices

Appendix A: Necessary Relation Between the Radius of the Outer Polygon and the Radius of the Inner Polygon: The Solutions to (31)

In this first part of the appendix, we provide the remainder of the argument from Proposition 1.

To this end, we solve

$$\begin{aligned}&\Big (1+x^2-2x\cos \Big ( \frac{2\pi }{n}\alpha \Big )\Big ) \Big (1+x^2-2x\cos \Big (\frac{2\pi }{n}(1-\alpha ) \Big )\Big ) \cos \Big (\frac{(n-2)\pi }{2n }\Bigr )\nonumber \\&\quad = \Big (1+ x^2\cos \Big (\frac{2\pi }{n}\Big ) -2x\cos \Big (\frac{\pi }{n}\Big )\cos \Big (\frac{\pi }{n}(1-2\alpha )\Big )\Big )^2, \end{aligned}$$

(99)

which is (31) squared. We get the following four solutions of the equation (31) for x:

• $x=\frac{1}{\cos \frac{\pi }{n}}\left( \cos \left( \frac{\rho _n}{2}\right) - \sqrt{\sin (\frac{2\pi }{n}\alpha )\sin \left( \frac{2\pi }{n}(1-\alpha )\right) } \right)$,

• $x=\frac{1}{\cos \frac{\pi }{n}}\left( \cos \left( \frac{\rho _n}{2}\right) + \sqrt{\sin (\frac{2\pi }{n}\alpha )\sin \left( \frac{2\pi }{n}(1-\alpha )\right) } \right)$,

• $x=\frac{1}{\cos \bigl (\frac{3\pi }{n}\bigr )} \left( \cos \left( \frac{2\pi }{n}\right) \cos \left( \frac{\rho _n}{2} \right) -\sqrt{\cos ^2\left( \frac{2\pi }{n}\right) \cos ^2 \left( \frac{\rho _n}{2}\right) -\cos \left( \frac{3\pi }{n}\right) \cos \left( \frac{\pi }{n}\right) } \right)$,

• $x=\frac{1}{\cos \bigl (\frac{3\pi }{n}\bigr )}\left( \cos \left( \frac{2\pi }{n}\right) \cos \left( \frac{\rho _n}{2}\right) +\sqrt{\cos ^2\left( \frac{2\pi }{n}\right) \cos ^2\left( \frac{\rho _n}{2}\right) -\cos \left( \frac{3\pi }{n}\right) \cos \left( \frac{\pi }{n}\right) } \right)$,

where as in (iv), ${\rho _n}:=\frac{2\pi }{n}(1-2\alpha )$. We now claim that just the first solution is admissible for us. Here and below we define a solution x of (99) admissible if $x\in (0,1)$ and it satisfies (31). In order to prove our claim, we can assume without loss of generality that

$$\begin{aligned} \sqrt{4\cos ^2\Bigl (\frac{2\pi }{n}\Bigr )\cos ^2\Bigl (\frac{\rho _n}{2}\Bigr ) -4\cos \Bigl (\frac{3\pi }{n}\Bigr )\cos \Bigl (\frac{\pi }{n}\Bigr )} \end{aligned}$$

is real, otherwise the third and fourth solutions are not admissible. The proof of the claim is as follows:

• Second solution: We estimate

  $$\begin{aligned} x \geqq \frac{\cos \left( \frac{{\rho _n}}{2} \right) }{\cos \left( \frac{\pi }{n} \right) }. \end{aligned}$$

  Since $\alpha \in \left( 0,1\right)$ it is clear that the second solution is such that $x\geqq 1$ for any $\alpha \in (0,1)$, any $n\geqq 3$.

• Third solution: $x\geqq 1$ if $n=3,4$ and $\alpha \in [0,1].$ We can hence restrict to the case $n>4$. We now claim that

  $$\begin{aligned} 1+ x^2\cos \left( \frac{2\pi }{n}\right) -2x\cos \left( \frac{\pi }{n}\right) \cos \left( \frac{\rho _n}{2}\right) <0 \end{aligned}$$

  (100)

  for any $\alpha \in (0,1)$, and any $n\geqq 4$. Since the left-hand side of (31) is always non-negative, the claim would imply that the third solution of (99) does not satisfy (31), and is hence not admissible. We plot $1+ x^2\cos \bigl (\frac{2\pi }{n}\bigr )-2x\cos \bigl (\frac{\pi }{n}\bigr )\cos \bigl (\frac{{\rho _n}}{2}\bigr )$ for $n\in \{5,\dots ,50\}$ in Figure 18. For large n, we have that

  $$\begin{aligned} x = 1-\frac{2\pi }{n}\sqrt{(\alpha -\alpha ^2)} +\frac{2\pi ^2}{n^2}(-\alpha ^2+\alpha +1) + O(n^{-3}), \end{aligned}$$

  and, therefore,

  $$\begin{aligned}&1+ x^2\cos \left( \frac{2\pi }{n}\right) -2x\cos \left( \frac{\pi }{n}\right) \\&\quad \cos \left( \frac{\rho _n}{2}\right) = -\frac{4\pi ^3}{n^3}\sqrt{\alpha (1-\alpha )} +O(n^{-4}) < 0 \end{aligned}$$

  for any $\alpha \in (0,1),$ and for any n large enough.

• Fourth solution: It is easy to see that it is negative for any $\alpha \in [0,1]$ when $n=3,4,5$. Indeed, $\cos \frac{3\pi }{n}<0$. If $n=6$ we get $x=\infty ,$ while for $n>6$ we have $x>1$. Indeed, in this case,

  $$\begin{aligned} x\geqq \frac{2\cos \left( \frac{2\pi }{n}\right) \cos \left( \frac{\rho _n}{2}\right) }{2\cos \left( \frac{3\pi }{n}\right) } \geqq \frac{\cos \left( \frac{2\pi }{n}\right) \cos \left( \frac{\pi }{n}\right) }{\cos \left( \frac{3\pi }{n}\right) }= \frac{1}{2}\left( 1+\frac{1}{2\cos \left( \frac{2\pi }{n}\right) -1} \right) >1. \end{aligned}$$

  Therefore, for any $\alpha \in [0,1]$ and any $n\geqq 3$ the fourth solution is not admissible.

Fig. 18

Numerical verification of the fact that, for $n\in \{1,\dots ,50\}$ and $\alpha \in (0,1)$ we have (100). The bigger n is, the closer to zero the convex curves in the pictures are (colour figure online)

Appendix B: Proof of Corollary 2.4

In this part of the appendix we show that equation (43)

$$\begin{aligned} P_0 {H} P_0 = Q_{\alpha } {H} Q_{1-\alpha }^T \end{aligned}$$

is satisfied. In order to simplify calculations, we express all matrices with respect to the basis $(e_{11}, e_{11}^\perp )$ and thus have to show that

$$\begin{aligned} \begin{pmatrix} a &{} - \frac{a^{-1}-a}{\tan (\phi )} \\ 0 &{} a^{-1} \end{pmatrix} Q_{1-\alpha } = Q_{\alpha } \begin{pmatrix} a &{} \frac{a^{-1}-a}{\tan (\phi )} \\ 0 &{} a^{-1} \end{pmatrix}. \end{aligned}$$

(101)

We further recall that

$$\begin{aligned} a^2&= \frac{\sin \left( \frac{2\pi }{n}(1-\alpha )\right) }{\sin \left( \frac{2\pi }{n}\alpha \right) }, \ \frac{1}{\tan (\phi )}= \tan \left( \frac{\pi }{n}\right) . \end{aligned}$$

In particular, since $\alpha \in (0,1)$, we may multiply the claimed equation with $a \sin (\frac{2\pi }{n}\alpha ) \ne 0$ and for simplicity of notation introduce $t:=\frac{2\pi }{n}\alpha$ and $s=\frac{2\pi }{n}(1-\alpha ) = \frac{2\pi }{n}-t$. With this notation, we have to show that

$$\begin{aligned}&\begin{pmatrix} \sin (s) &{}\quad -(\sin (t)-\sin (s))\tan \left( \frac{\pi }{n}\right) \\ 0 &{}\quad \sin (t) \end{pmatrix} \begin{pmatrix} \cos (s) &{}\quad -\sin (s) \\ \sin (s) &{}\quad \cos (s) \end{pmatrix} \\&\quad = \begin{pmatrix} \cos (t) &{}\quad -\sin (t) \\ \sin (t) &{}\quad \cos (t) \end{pmatrix} \begin{pmatrix} \sin (s) &{}\quad (\sin (t)-\sin (s))\tan \left( \frac{\pi }{n}\right) \\ 0 &{} \quad \sin (t) \end{pmatrix}. \end{aligned}$$

We consider each matrix entry separately. The claimed equality for the upper left entry is given by

$$\begin{aligned} \sin (s)\cos (s)- \sin (s)\tan \left( \frac{\pi }{n}\right) (\sin (t) -\sin (s))=\cos (t)\sin (s). \end{aligned}$$

(102)

In order to show this, we may factor out $\sin (s)$ and use the angle addition formulas:

$$\begin{aligned} \cos (s)&=\cos (t)\cos \left( \frac{2\pi }{n}\right) + \sin (t) \sin \left( \frac{2\pi }{n}\right) , \\ \sin (s)&=\cos (t)\sin \left( \frac{2\pi }{n}\right) - \sin (t) \cos \left( \frac{2\pi }{n}\right) . \end{aligned}$$

We then collect terms involving $\cos (t)$ and $\sin (t)$ as

$$\begin{aligned}&\cos (t)\cos \left( \frac{2\pi }{n}\right) +\sin (t)\sin \left( \frac{2\pi }{n}\right) \\&\qquad -\tan \left( \frac{\pi }{n}\right) \left( \sin (t)-\cos (t)\sin \left( \frac{2\pi }{n}\right) +\sin (t)\cos \left( \frac{2\pi }{n}\right) \right) \\&\quad = \cos (t)\left( \cos \left( \frac{2\pi }{n}\right) +\tan \left( \frac{\pi }{n}\right) \sin \left( \frac{2\pi }{n}\right) \right) \\&\qquad + \sin (t)\left( \sin \left( \frac{2\pi }{n}\right) -\tan \left( \frac{\pi }{n}\right) (1+\cos \left( \frac{2\pi }{n}\right) \right) \\&\quad = \cos (t) 1+ \sin (t) 0 = \cos (t), \end{aligned}$$

where we used the half angle identities for $\cos (2x)$ and $\sin (2x)$ in the last equality.

The calculation for the bottom right-entry is analogous with the role of s and t and the sign of $(\sin (t)-\sin (s))\tan (\frac{\pi }{n})$) interchanged. The bottom left equality $\sin (t)\sin (s)=\sin (t)\sin (s)$ is always satisfied. It thus only remains to verify equality of the upper right entry, which can be simplified to read

$$\begin{aligned}&-\sin ^2(s)-\cos (s)(\sin (t)-\sin (s))\tan \left( \frac{\pi }{n}\right) \\&\quad = -\sin ^2(t)+ \cos (t)(\sin (t)-\sin (s))\tan \left( \frac{\pi }{n}\right) \\&\quad \Leftrightarrow \sin ^2(t)-\sin ^2(s) - (\cos (t)+\cos (s))(\sin (t) -\sin (s))\tan \left( \frac{\pi }{n}\right) =0. \end{aligned}$$

Factoring out the factor $(\sin (t)-\sin (s))$, it suffices to prove

$$\begin{aligned} \sin (t)+\sin (s) - (\cos (t)+\cos (s))\tan \left( \frac{\pi }{n}\right) =0. \end{aligned}$$

As above, the claimed equality then again follows by using angle addition formulas.

Appendix C: Reduction to Cauchy–Green Tensors Used in the Proof of Proposition 2.6

Last but not least, we provide the argument (used in the proof of Proposition 2.6) that it is possible to reduce the differential inclusion (23) to an inclusion for the associated Cauchy–Green tensors.

Lemma C.1

Suppose that $\det (M)=\det ({H})>0$, then the inclusion

$$\begin{aligned} M \in \bigcup _{P \in {\mathcal {P}}_n}SO(2) P^T {H} P \end{aligned}$$

(103)

is satisfied, if and only if

$$\begin{aligned} M^TM \in \bigcup _{P \in {\mathcal {P}}_n} P^T {H}^T {H} P. \end{aligned}$$

(104)

This characterisation follows from basic properties of the singular value decomposition.

Proof

We observe that (103) implies (104). Thus, we only consider the converse and assume that

$$\begin{aligned} M^TM = (P^T{H}^TP) (P^T {H} P)=: M_1^T M_1. \end{aligned}$$

for some $P \in {\mathcal {P}}_n$. Since $M^TM$ is symmetric, there exists $Q \in SO(2)$ and a diagonal matrix ${{\,\mathrm{diag}\,}}(\lambda _1,\lambda _2)$, with $\lambda _1 \lambda _2=\det (M)^2\ne 0$, $\lambda _1,\lambda _2>0$, such that

$$\begin{aligned} M^T M= Q^T {{\,\mathrm{diag}\,}}(\lambda _1,\lambda _2)Q. \end{aligned}$$

It follows that

$$\begin{aligned}&\tilde{M}:= M Q^T {{\,\mathrm{diag}\,}}\left( \frac{1}{\sqrt{\lambda _1}}, \frac{1}{\sqrt{\lambda _2}}\right) , \\&\tilde{M_1}:= M_1 Q^T {{\,\mathrm{diag}\,}}\left( \frac{1}{\sqrt{\lambda _1}}, \frac{1}{\sqrt{\lambda _2}}\right) , \end{aligned}$$

satisfy

$$\begin{aligned} \tilde{M}^T\tilde{M}= I= \tilde{M_1}^T\tilde{M_1} \end{aligned}$$

and thus $\tilde{M}, \tilde{M_1} \in SO(2)$. Here we used that $\det (M)=\det ({H})>0$. In particular,

$$\begin{aligned} M_1&= \tilde{M_1} Q^T{{\,\mathrm{diag}\,}}(\sqrt{\lambda _1}, \sqrt{\lambda _2}), \\ M&= \tilde{M} Q^T{{\,\mathrm{diag}\,}}(\sqrt{\lambda _1}, \sqrt{\lambda _2}) \\&= \tilde{M} \tilde{M}^T_1 M_1, \end{aligned}$$

where $\tilde{M}\tilde{M}^T_1 \in SO(2)$, which implies the result. $\quad \square$