Finite element approximation of large bending isometries

Abstract

A finite element scheme for the approximation of large isometric deformations with minimal bending energy is devised and analyzed. The convergence to a stationary point and energy decreasing property of an iterative algorithm for the numerical solution of the scheme is proved. Numerical experiments illustrate the performance of the iteration and show that the discretization leads to accurate approximations for large vertical loads and compressive boundary conditions.

Appendix A: Auxiliary results

1.1 A.1 Elementary differential geometry

Given a parametrized surface $y:\Omega \rightarrow \mathbb{R }^3$ the first fundamental form is given by $g_{ij} = \partial _i y \cdot \partial _j y$ and the second fundamental form by $h_{ij} = \partial _i b \cdot \partial _j y = - b \cdot \partial _i \partial _j y$, where $b= \partial _1 y\times \partial _2 y$. The inverse of $g$ has the entries $g^{ij}$. The Gaussian curvature is the determinant of the Weingarten map $L= (\sum _k h_{ik}g^{kj})$ and given by $K = \det h_{ij}/ \det g_{ij}$. The mean curvature is half of the trace of $L$ and given by $H=(h_{11}g_{22}-2h_{12}g_{12}+h_{22}g_{11}) /(2\det g_{ij})$. If the parametrization is an isometry, i.e., if $g_{ij}= \delta _{ij}$, then Gauss’s theorema egregium implies $K=0$. Moreover, we have $\text{ tr}II = \text{ tr}L = 2 H$ and

$$\begin{aligned} |II|^2&= \sum _{i, j} h_{ij}^2 = h_{11}^2 + h_{22}^2 + 2 h_{12}^2 = (h_{11}+h_{22})^2 - 2 h_{11}h_{22}+ 2 h_{12}^2\\&= 4 H^2 - 2 K = 4H^2. \end{aligned}$$

For a $C^2$ isometry with $|\partial _j y|^2 =1$, $j=1,2$, and $\partial _1 y \cdot \partial _2 y = 0$ we deduce that $\partial _1^2 y \cdot \partial _1 y = 0$ and $\partial _1^2 y \cdot \partial _2 y = - \partial _1 y \cdot \partial _1\partial _2 y = 0$. Analogously, we verify that $\partial _2^2 y \cdot \partial _1 y=-\partial _2 y\cdot \partial _1\partial _2y =0$ and $\partial _2^2 y \cdot \partial _2 y = 0$ so that $-\Delta y = \beta b$. Since $-\Delta y \cdot b = \text{ tr}II = 2H$ we verify that $-\Delta y = 2 H b$. The vectors $(\partial _1y, \partial _2y, b)$ form an orthonormal basis of $\mathbb{R }^3$ for every $x\in \Omega$ so that $|\partial _i\partial _j y| = |\partial _i\partial _j y\cdot b|$ and hence $|D^2y|^2=\sum _{i, j} |\partial _i\partial _j y \cdot b|^2=|II|^2$.

1.2 A.2 Proof of (2.3)

Given a weakly acute triangulation $\mathcal T _h$ we have for the entries $k_{zy} = (\nabla \varphi _z, \nabla \varphi _y)$ of the stiffness matrix that $k_{zy}\le 0$ if $z\ne y$ for all $z, y\in \mathcal N _h$. The symmetry $k_{zy}=k_{yz}$ and the identity $\sum _{y\in \mathcal N _h} k_{zy}=0$ for all $z\in \mathcal N _h$ show

$$\begin{aligned} \Vert \nabla v_h\Vert ^2&= \frac{1}{2} \sum _{z, y\in \mathcal N _h} k_{zy} v_h(z)\cdot (v_h(y)\!-v_h(z)) \!+ \frac{1}{2} \sum _{z, y\in \mathcal N _h} k_{zy} v_h(y)\cdot (v_h(z)-v_h(y)) \\&= -\frac{1}{2} \sum _{z, y\in \mathcal N _h} k_{zy} |v_h(z)-v_h(y)|^2. \end{aligned}$$

The assertion follows from the fact that the mapping $x\mapsto x/|x|$ is Lipschitz continuous with constant $1$ in $\{x\in \mathbb{R }^3: |x|\ge 1\}$.

1.3 A.3 Proof of Lemma 2.1

A discrete Poincaré inequality shows that $\Vert y_h\Vert \le C$ and hence there exists $y\in L^2(\Omega ;\,\mathbb{R }^3)$ with (after extraction of a subsequence) $y_h\rightharpoonup y$ in $L^2$. Since $\Vert \nabla _h y_h\Vert \le C$ there exists $\xi \in L^2(\Omega ;\,\mathbb{R }^{3\times 2})$ such that (after extraction of another subsequence) $\nabla _h y_h \rightharpoonup \xi$ in $L^2$. We have, using that $\int _E [y_h]\text{ d}s=0$ for all $E\in \mathcal E _h$ and that the (row-wise applied) Fortin interpolant $I_F \Psi$ of $\Psi \in C^\infty _0(\Omega ;\,\mathbb{R }^{3\times 2})$ on the Raviart–Thomas finite element space, cf. [10], satisfies that $(I_F \Psi ) \nu |_E$ is constant on each $E\in \mathcal E _h$, that

$$\begin{aligned} \int \limits _\Omega \nabla _h y_h : \Psi \text{ d}x&= -\int \limits _\Omega y_h \cdot \text{ div}\Psi \text{ d}x + \sum _{E\in \mathcal E _h} \int \limits _E [y_h] \cdot ([\Psi -I_F\Psi ] \nu )\text{ d}s \\&= \!-\!\int \limits _\Omega y_h \cdot \text{ div}\Psi \text{ d}x \!+\!\int \limits _\Omega \nabla _h y_h : [\Psi -I_F\Psi ] \text{ d}x \!+\!\int \limits _\Omega y_h \cdot \text{ div}[\Psi -I_F\Psi ] \text{ d}x. \end{aligned}$$

Since the last two terms on the right-hand side converge to zero as $h\rightarrow 0$ we deduce that $\xi = \nabla y$. The fact that $y|_{{\Gamma _\mathrm{D}}}=y_\mathrm{D}$ follows from an elementwise integration by parts as above provided that $y_h \rightarrow y$ in $L^2({\Gamma _\mathrm{D}})$.