= Korn inequality
{c}
{title2=$\|\nabla v\|_{L^2}\leq C\|\varepsilon(v)\|_{L^2}$}
The zero-boundary <Korn inequality> controls the full <displacement gradient tensor> by the <infinitesimal strain tensor> for $v\in H_0^1(\Omega;\mathbb R^d)$. On a bounded <Lipschitz domain>, combining it with the <Poincaré inequality> makes a positive <linear elasticity> energy a <coercive bilinear form> in the first-order <Sobolev norm>. The zero <Dirichlet boundary condition> removes the otherwise uncontrolled rigid displacements.
For this zero-boundary version one can prove an exact identity. For a smooth compactly supported <vector field>, two applications of <integration by parts> give
$$
\sum_{i,j}\int_\Omega\partial_jv_i\,\partial_iv_j=\int_\Omega(\operatorname{div}v)^2.
$$
Expanding the symmetric gradient and using density in the <zero-boundary Sobolev space> therefore yields
$$
2\|\varepsilon(v)\|_{L^2}^2=\|\nabla v\|_{L^2}^2+\|\operatorname{div}v\|_{L^2}^2.
$$
In three dimensions this also gives the stronger trace-free version
$$
\left\|\varepsilon(v)-\frac13(\operatorname{div}v)I\right\|_{L^2}^2=\frac12\|\nabla v\|_{L^2}^2+\frac16\|\operatorname{div}v\|_{L^2}^2\geq\frac12\|\nabla v\|_{L^2}^2.
$$
Here the matrix-valued <L2 norm> uses the <Frobenius inner product>. This explains why the trace-free form printed in the examination suffices for the usual <Korn inequality> estimate.
Back to article page
