Rigidity of thin domains

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Posted online: 2018-11-19 04:28:44Z by Davit Harutyunyan52

Cite as: P-181119.2

  • Analysis of PDEs
  • Differential Geometry
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Problem's Description

The problem of determining the rigidity of a thin domain is generally open in nonlinear elasticity. Given a $C^2$ compact surface $S\subset\mathbb R^3,$ a this domain $\Omega_h\subset\mathbb R^3$ around $S$ is defined as follows: One takes two Lipschitz functions $g_1^h,g_2^h\colon S\to\mathbb R$ that satisfy the conditions $$h\leq g_i^h\leq Ch,\quad c_1h\leq |\nabla g_i^h|\leq c_2^h,\quad I=1,2,$$ and defines $\Omega=\{x+tn(x) \ : \ x\in S, \ t\in (-g_1^h,g_2^h)\},$ where $h$ is a small parameter and $n(x)$ is the (the direction does not matter) unit normal to $S$ at $x.$ The following statement holds: There exists two constants $M=M(S)$ and $\alpha=\alpha(S),$ such that for any vector field $u\in H^1(\Omega,\mathbb R^3),$ there exists a proper rotation $R\in SO(3)$ such that $$\|\nabla u-R\|_{L^2(\Omega)}^2\leq\frac{M}{h^\alpha} \int_\Omega \mathrm{dist}^2(\nabla u(x), SO(3))dx.$$ The geometric rigidity of $\Omega$ is defined as the optimal constant $\alpha$ in the above inequality (the so-called geometric rigidity estimate). When $S$ is flat, the problem has been solved by Friesecke, James and Mueller in 2002, [2]. If $S$ is not flat, the problem is open. If $S$ has positive or negative Gaussian curvature, then the corresponding linear elasticity problem has been solved in [1] giving $\alpha=1$ and $\alpha=4/3.$ The conjecture is that these results hold also in the nonlinear setting.

  1. Article Gaussian curvature as an identifier of shell rigidity

    Archive for Rational Mechanics and Analysisfulltext

  2. Article A theorem on geometric rigidity and the derivation of nonlinear plate theory from three-dimensional elasticity.

    Communications on Pure and Applied Mathematicsfulltext


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  • Created at: 2018-11-19 04:28:44Z