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Let $\Omega \subset \subset \mathbb{R}^N$ have smooth boundary, $N \geqslant 2$ and

$ \mathcal{E} ( v) := \int_{\Omega} \sum_{i, j} \varepsilon_{i j} ( v) \varepsilon_{i j} ( v) \mathrm{d} x := \int_{\Omega} \sum_{i, j} \left( \frac{v_{i, j} + v_{j, i}}{2} \right)^2 \mathrm{d} x $

be defined in $H^1 ( \Omega, \mathbb{R}^N)$. Having just read for the umpteenth time that the reason that Korn's inequality

$ \mathcal{E} ( v) + \| v \|_{L^2}^2 \geqslant c \| v \|^2_{H^1} $

is not trivial is that the left hand side "involves only certain combinations of derivatives", I wonder whether this is actually true and if yes, whether I understand things correctly, because to me it's a matter of some products (the $v_{i,j}v_{j,i}$ for $i \neq j$ ) possibly being negative. Did I get lost among the indices?


Edit: in the context of linear elasticity it is often stated that this inequality is not a triviality (in the sense that it's not tautological), because of the different combinations of partial derivatives appearing at each side. Some authors affirm that only some (six) different partial derivatives appear at the left hand side (see [1], [2], [3]). However I see all partial derivatives at both sides, but combined differently (see my answer). I understand the actual difficulties in proving it and the implications for coercivity proofs for instance. It's this assertion that I find confusing.

[1] G. Duvaut and J.-L. Lions, Inequalities in mechanics and physics, vol. 219. Springer-Verlag, 1976.

[2] P. G. Ciarlet, An introduction to differential geometry with applications to elasticity. Springer, 2005, .

[3] F. Demengel and G. Demengel, Functional spaces for the theory of elliptic partial differential equations, vol. 8. Springer, 2012.

2 Answers 2

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Yes, it's "only" a matter of possible cancellation in the sum $v_{i,j}+v_{j,i}$. Like children, analysts can get excited about tiny little things of this kind.

But seriously, this is an amazing result. For example, if you want to bound the $H^1$ norm of a scalar function $u$, you have to integrate the square of every single partial derivative: $\int \sum_{i=1}^n u_i^{2}$, or use another positive definite quadratic form of the derivatives. No semidefinite form will do. If $Q$ is positive semidefinite and $Q(\xi,\xi)=0$ for some vector $\xi\ne 0$, then there is a function $u$ such that $\nabla u$ is always parallel to $\xi$, and therefore $\int Q(\nabla u,\nabla u)=0$ while $\|u\|_{H^1}$ can be huge (and $\|u\|_{L^2}$ will not control $\|u\|_{H^1}$ either).

In Korn's inequality we integrate the quadratic form $K(\xi,\xi)=\sum (\xi_{ij}+\xi_{ji})^2$. It is semidefinite with a huge kernel: there is an $n(n-1)/2$ dimensional subspace along which $K$ is zero (skew-symmetric matrices, to be precise). Recalling that in the scalar case even one-dimensional kernel killed the estimate, how can we expect that $\int K(\nabla v,\nabla v)$ will control $\|v\|_{H^1}$? Yet it does.

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    The fact that it "i$n$volves only certain combinations of derivatives" implies that it is only semidefinite as stated in the answer. Thus, there is no contradiction with Duvaut and Lions.2013-08-27
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As I see it, we have:

\begin{eqnarray*} \mathcal{E} ( v) + \| v \|_{L^2} & = & \int_{\Omega} \frac{1}{4} \sum_{i, j} ( v_{i, j} + v_{j, i})^2 \mathrm{d} x + \| v \|_{L^2}\\ & = & \int_{\Omega} \frac{1}{2} \sum_{i, j} ( v_{i, j}^2 + v_{i, j} v_{j, i}) \mathrm{d} x + \| v \|_{L^2}\\ & = & \frac{1}{2} \left( \| \nabla v \|^2_{L^2} + \underset{\geqslant 0}{\underbrace{\sum_{i = j} \| v_{i, j} \|^2_{L^2}}} + \sum_{i \neq j} \int_{\Omega} v_{i, j} v_{j, i} \mathrm{d} x \right) + \| v \|_{L^2}\\ & \geqslant & \frac{1}{2} \| v \|_{H^1}^2 + \frac{1}{2} \sum_{i \neq j} \int_{\Omega} v_{i, j} v_{j, i} \mathrm{d} x. \end{eqnarray*}

And the problem is that that last term could be negative. Reasoning similarly one also concludes that the opposite inequality is easy using brute-force-Hölder:

\begin{eqnarray*} \mathcal{E} ( v) + \| v \|_{L^2}^2 & \leqslant & \| v \|^2_{H^1} + \frac{1}{2} \sum_{i , j} \int_{\Omega} v_{i, j} v_{j, i} \mathrm{d} x\\ & \leqslant & \| v \|^2_{H^1} + \frac{N^2}{2} \| v \|_{H^1}^2\\ & \leqslant & N^2 \| v \|_{H^1}^2 . \end{eqnarray*}

Any comments/corrections are more than welcome...

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    I see your point. I should've phrased the question differently to stress that it's not its importance that I question, but the reasons alleged for the difficulty in proving it. This remains my question...2012-12-21