 Solution

Write $V=H_0^1(\Omega ;{\mathbb{R}}^3)$. For any $v\in V$ and any real $t$, $u+tv$ is admissible. Expanding the functional givesSince the left side is nonnegative for positive and negative $t$, its derivative at zero vanishes. Thus the first variation gives the weak formulation

To identify its differential equation, define the stress tensorThis is a symmetric matrix field. The Frobenius inner product of a symmetric matrix with a skew-symmetric matrix is zero, so $\sigma (u):\nabla v=\sigma (u):\epsilon (v)$. Since $\mathrm{tr}\epsilon (v)=\mathrm{div}v$, we obtainFor smooth compactly supported test functions, integration by parts in each row givesHence $-\mathrm{div}\sigma (u)=f$ as a distributional identity, that is,Membership in $V$ supplies the zero Sobolev trace. This identifies precisely the required weak solution of the linear elasticity problem, with row-wise divergence.

For uniqueness, use the exact form of the Korn inequality supplied in the paper. Put $E=\epsilon (v)$ and $d=\mathrm{tr}E=\mathrm{div}v$. The deviatoric strain is $D=E-\frac{1}{3}dI$. Because $\mathrm{tr}D=0$ and $I:I=3$, orthogonality for the Frobenius inner product givesIf $C_K$ is the supplied Korn inequality constant and $K$ the permitted Poincaré inequality constant, thenConsequently,If $u_1,u_2$ are two weak solutions, their difference $d_u\in V$ satisfies $a(d_u,v)=0$ for every $v\in V$. Choosing $v=d_u$ and using this coercive bilinear form estimate forces $d_u=0$. The minimizer therefore gives the unique weak solution.

For later use, $|\epsilon (v){|}_F\le |\nabla v{|}_F$ and $|\mathrm{div}v|\le \sqrt{3}|\nabla v{|}_F$ also give the bounded bilinear form and linear functional estimatesTogether with the coercive bilinear form estimate and the Lax-Milgram theorem, these would establish existence independently of the assumed minimizer.