$\textbf{Theorem}.1$ (The first Korn inequality) Suppose that $ \Omega $ is a bounded domain in $ \mathbb{R}^d $ with Lipschitz boundary. Then\ \begin{eqnarray} \sqrt{2}\left\|\triangledown u\right\|_{L^2(\Omega)}\leq \left\|\triangledown u+(\triangledown u)^T\right\|_{L^2(\Omega)} \end{eqnarray} for any $ u\in H_{0}^{1}(\Omega;\mathbb{R}^d) $, where $ (\triangledown u)^T $ denotes the transpose of $ \triangledown u $.

$\textbf{Theorem}.2$ (The second Korn inequality) Suppose that $ \Omega $ is a bounded domain in $ \mathbb{R}^d $ with Lipschitz boundary. If $ u\in H^{1}(\Omega,\mathbb{R}^d) $ is a function with the property that $ u\perp R $ in $ H^{1}(\Omega;\mathbb{R}^d) $, then
\begin{eqnarray} \int_{\Omega}|\triangledown u|^2dx\leq C\int_{\Omega}|\triangledown u+(\triangledown u)^T|^2dx \end{eqnarray} where $ R=\left\{\phi=Bx+b:B\in\mathbb{R}^{d\times d} \text{ is skew-symmetric and }b\in\mathbb{R}^d\right\} $ and $ C $ is a constant.

I recently see the two theorems in a book about elliptic equations. I tried to get the estimate for the second inequality by direct computation which works in the proof of the first Korn inequality, but for this inequality, I cannot combine the condition $ u\perp R $ with the final results. Can you give me some hints or references?

  • $\begingroup$ not my field and on a quick glance I couldnt find a matching statement, but this claims to have proofs (and it has references too) : fuchsbraun.homepage.t-online.de/media/… (J. A. NITSCHE - On Korn’s second inequality) $\endgroup$ Sep 4, 2021 at 5:21
  • $\begingroup$ @Calvin Khor Sorry, I cannot open your web link. Can you give me another? $\endgroup$ Sep 4, 2021 at 8:24
  • $\begingroup$ Sure, searching on zbMATH: https://zbmath.org/?q=J.+A.+NITSCHE+On+Korn’s+second+inequality one finds two links that should be open-access: one is on the EUDML, and one is on some site called esaim-m2an (DOI: doi.org/10.1051/m2an/1981150302371). There is also this on numdam. Hopefully one of these work for you $\endgroup$ Sep 4, 2021 at 8:29

2 Answers 2


You can find a full proof (to my knowledge the simpler one currently known) in the paper [1] and in the book [2], chapter I, §2.1 pp. 14-21. The original proof of Arthur Korn is so long and involved that K.O. Friedrichs, who gave a much simpler yet sophisticated proof, had doubts on his validity: starting from the work of Friedrichs, several authors gave their (in general quite complex) proofs, until Olga Oleĭnik gave a much shorter and simpler one (despite being still not elementary).


[1] Vladimir Alexandrovitch Kondratiev, Olga Arsenievna Oleĭnik, "On Korn’s inequalities" (English), Comptes Rendus de l’Académie des Sciences, Série I, 308, No. 16, pp. 483-487 (1989), MR0995908, Zbl 0698.35067.

[2] Olga Arsenievna Oleĭnik, Alexei Stanislavovich Shamaev, Grigorii Andronikovich Yosifian, Mathematical problems in elasticity and homogenization. (English) Studies in Mathematics and its Applications. 26. Amsterdam-London-New York-Tokyo: North- Holland, pp. xiii+398 (1992), ISBN: 0-444-88441-6, MR1195131, Zbl 0768.73003.


Let us assume Korn's inequality in the usual form $$\int_{\Omega} |\nabla u|^2 \leq C\left (Q(u)+\int_\Omega |u|^2 \right ),$$ with $Q(u)=\int_\Omega |\nabla u +\nabla u^{T}|^2$, for every $u \in H^1(\Omega)$ (you find it in the paper indicated by @Calvin Khor, for example, but the proof is not so elementary as in the case of $H^1_0(\Omega)$). Then the claim follows by contradiction. If not true, we find a sequence or vector fields $(u_n) \subset R^\perp$ such that $$1=\|u_n\| \geq n \,Q(u_n).$$ Then $(u_n)$ is bounded in $H^1$ and we may assume that is converges to $u_0$ weakly in $H^1$ and strongly in $L^2$, with $u_0 \in R^\perp$. Then $Q(u_0)=0$ which means that $\nabla u_0=-(\nabla u_0)^T$ and then $u_0 \in R$. Finally, $u_0 \in R \cap R^\perp$ gives $u_0=0$ but $\|u_0\|=1$.


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