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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 giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere 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.

The infinitesimal strain tensor $\epsilon (v)=\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$ and the divergence are linear in $v$. Consequently the given $a$ is a symmetric bilinear form, andFor a matrix-valued function, $\Vert \cdot {\Vert }_0$ here uses the Frobenius inner product pointwise and then the L2 norm. All expressions are finite for $v\in H^1(\Omega ;{\mathbb{R}}^3)$.

Expand the quadratic term at $\theta v+(1-\theta )w$. The linearity of $\ell$ cancels its contribution to the difference, givingFor $0\le \theta \le 1$, this is nonnegative. Thereforeso $J$ is a convex functional on the whole Sobolev space. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the zero-boundary Sobolev space will instead use the Korn inequality in part (b).

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.