Korn’s second inequality and geometric rigidity with mixed growth conditions

Abstract

Geometric rigidity states that a gradient field which is $L^p$-close to the set of proper rotations is necessarily $L^p$-close to a fixed rotation, and is one key estimate in nonlinear elasticity. In several applications, as for example in the theory of plasticity, energy densities with mixed growth appear. We show here that geometric rigidity holds also in $L^p+L^q$ and in $L^{p,q}$ interpolation spaces. As a first step we prove the corresponding linear inequality, which generalizes Korn’s inequality to these spaces.

Appendix: Extension

The subsequent extension theorem for functions with mixed growth follows immediately from the $L^2$-version in [21]. We include a sketch of the proof for the convenience of the reader.

Theorem 5.1

Let $\varphi \in \mathrm{Lip}(\mathbb R ^{n-1};\mathbb R )$ be a Lipschitz function with $\varphi (0)=0$ and Lipschitz constant $L$, let $R>0$ and set $\Omega =B(0,R)\cap \{(x^{\prime },x_n)\in \mathbb R ^{n-1}\times \mathbb R :x_n< \varphi (x^{\prime })\}$. Suppose that $1<p<q<\infty$ and that $u\in W^{1,1}(\Omega ;\mathbb R ^n)$ with

$$\begin{aligned} D u + D u^T= f + g, \end{aligned}$$

(5.1)

where $f \in L^p(\Omega ;\mathbb R ^{n\times n})$ and $g \in L^q(\Omega ;\mathbb R ^{n\times n})$. Then there exists for $r=R/(2\sqrt{1+L^2})$ a function $w\in W^{1,1}(B(0,r);\mathbb R ^n)$, and matrix fields $\widetilde{f}$, $\widetilde{g}$ such that $w=u$, $\widetilde{f}= f$, $\widetilde{g}=g$ on $\Omega \cap B(0,r)$ and

$$\begin{aligned} D w + D w^T= \widetilde{f}+ \widetilde{g}\quad \text{ on } B(0,r) \end{aligned}$$

with

$$\begin{aligned} \Vert \widetilde{f}\Vert _{L^p(B(0,r))}\le c \Vert f \Vert _{L^p(B(0,R)\,\cap \,\Omega )},\quad \Vert \widetilde{g}\Vert _{L^q(B(0,r))}\le c \Vert g \Vert _{L^q(B(0,R)\,\cap \,\Omega )}. \end{aligned}$$

(5.2)

The constant $c$ depends only on $n$, $p$, $q$, $\Omega$ but not on $u$, $f$, $g$.

Proof

Let $\delta \in C^2(B(0,R)\setminus \Omega )$ be a function such that

$$\begin{aligned} 2\mathrm{dist}(x,\Omega )\le \delta (x)\le C \mathrm{dist}(x,\Omega ) \end{aligned}$$

and

$$\begin{aligned} |D^\alpha \delta (x)| \le C \delta ^{1-|\alpha |}(x),\quad \alpha \in \mathbb N ^n, \end{aligned}$$

(5.3)

see, e.g., [30]. Fix a function $\psi \in C^1(\mathbb R )$ with

$$\begin{aligned} \int _1^2 \psi (\lambda )\,\mathrm{d}\lambda =1,\quad \int _1^2 \lambda \psi (\lambda )\,\mathrm{d}\lambda =0. \end{aligned}$$

(5.4)

We set $w=u$ on $\Omega$ and for $x\in B(0,r)\setminus \Omega$ we define

$$\begin{aligned} w(x)=\int _1^2 \psi (\lambda )\left[ u(x-\lambda \delta (x) e_n) - \lambda D\delta (x)u_n(x-\lambda \delta (x) e_n)\right] \,\mathrm{d}\lambda . \end{aligned}$$

For ease of notation we omit the arguments in the following calculations and write $\delta =\delta (x)$ and $u=u(x-\lambda \delta (x) e_n)$ with the same convention for their derivatives. By the chain rule

$$\begin{aligned} Dw(x)&=\int _1^2 \psi (\lambda )\left[ Du (\mathrm{Id}- \lambda e_n\otimes D\delta ) - \lambda D\delta \otimes Du_n (\mathrm{Id}- \lambda e_n\otimes D\delta ) -\lambda u_n D^2\delta \right] \,\mathrm{d}\lambda \\&= \int _1^2 \psi (\lambda )\left[ Du - \lambda D_nu \otimes D\delta - \lambda D\delta \otimes Du_n + \lambda ^2 D_nu_n D\delta \otimes D\delta -\lambda u_nD^2\delta \right] \,\mathrm{d}\lambda . \end{aligned}$$

Then the symmetric part of the gradient is given by

$$\begin{aligned}&\!\!\! Ew(x)=\\&\int _1^2 \psi (\lambda )\left[ Eu - \lambda (Eu e_n)\otimes D\delta - \lambda D\delta \otimes (Eu e_n) + \lambda ^2(Eu)_{nn} D\delta \otimes D\delta -\lambda u_nD^2\delta \right] \,\mathrm{d}\lambda . \end{aligned}$$

In the last term we write

$$\begin{aligned} u_n(x-\lambda \delta (x) e_n) = u_n(x- \delta (x) e_n) +\int _1^\lambda D_n u_n (x-s \delta (x) e_n)\delta (x) \,\mathrm{d}s. \end{aligned}$$

In view of the second property in (5.4) the weighted integral of $u_n(x- \delta (x) e_n)$ is equal to zero, and the other term only depends on $(Eu)_{nn}$. We recall (5.1) and define for $x\in B(0,r)\setminus \Omega$

$$\begin{aligned} \widetilde{f}(x)&=\int _1^2 \psi (\lambda )\left[ f (x-\lambda \delta (x)e_n ) - \lambda ( f (x-\lambda \delta (x) e_n)e_n)\otimes D\delta (x) \right] \,\mathrm{d}\lambda \\&\quad - \int _1^2 \psi (\lambda )\left[ \lambda D\delta (x) \otimes ( f (x-\lambda \delta (x) e_n)e_n) \right] \,\mathrm{d}\lambda \\&\quad + \int _1^2 \psi (\lambda )\left[ \lambda ^2 f _{nn}(x-\lambda \delta (x) e_n) D\delta (x) \otimes D\delta (x) \right] \,\mathrm{d}\lambda \\&\quad - \int _1^2 \psi (\lambda ) \lambda \int _1^\lambda f _{nn}(x-s \delta (x)e_n) \delta (x) \,\mathrm{d}s\, D^2\delta (x) \,\mathrm{d}\lambda , \end{aligned}$$

and use the analogous definition for $\widetilde{g}$ in $x\in B(0,r)\setminus \Omega$. On $B(0,r)\cap \Omega$ we set $\widetilde{f}= f$ and $\widetilde{g}=g$. It remains to show that

$$\begin{aligned} \Vert \widetilde{f}\Vert _{L^p(B(0,r)\setminus \Omega )}\le c \Vert f \Vert _{L^p(\Omega )},\quad \Vert \widetilde{g}\Vert _{L^q(B(0,r)\setminus \Omega )}\le c \Vert g\Vert _{L^q(\Omega )} \end{aligned}$$

with a constant which only depends on $n$, $p$, $q$ and $\Omega$. The calculation is identical to the proof of the estimate for the extension in [21, Lemma 4]. $\square$