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


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.

  • $\begingroup$ I didn't question the importance of the inequality, but whether the reason that it's non trivial truly is that the left hand side "involves only certain combinations of derivatives". You do concur that this is not the reason, but then why would Duvaut & Lions state it in their book "Inequalities in mechanics and physics"? They aren't exactly a couple of random guys who happened to write on some subject unknown them... $\endgroup$
    – Miguel
    Dec 21, 2012 at 11:24
  • $\begingroup$ I just found a quote by Ciarlet (not exactly some random guy either...): "This inequality is truly remarkable, since only six different combinations of first-order partial derivatives, viz., $\frac{1}{2} ( \partial_i v_j + \partial_j v_i)$, occur in its right-hand side, 2 while all nine partial derivatives $\partial_i v_j$ occur in its left-hand side!" (P. G. Ciarlet, An introduction to differential geometry with applications to elasticity. Springer, 2005, p.138) $\endgroup$
    – Miguel
    Dec 21, 2012 at 11:35
  • $\begingroup$ The fact that it "involves 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. $\endgroup$ Aug 27, 2013 at 14:05

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...

  • $\begingroup$ This could have been a part of question, perhaps separated with <hr> or such thing. $\endgroup$
    – user53153
    Dec 19, 2012 at 7:49
  • $\begingroup$ I thought SE encouraged answering your own questions, that's why I did it. But it might have been a bad idea if that was the reason why it didn't get any attention... $\endgroup$
    – Miguel
    Dec 21, 2012 at 10:53
  • 1
    $\begingroup$ You did nothing wrong. But if it was me asking "why is Korn's inequality nontrivial" I would not be satisfies with the answer "if we expand it and move all terms to one side, some of them might be negative". $\endgroup$
    – user53153
    Dec 21, 2012 at 19:28
  • $\begingroup$ 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... $\endgroup$
    – Miguel
    Dec 21, 2012 at 22:53

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