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The deviatoric strain is the trace-free part of the infinitesimal strain tensor. It removes the isotropic volume change. The decomposition is orthogonal for the Frobenius inner product, because a trace-free matrix has zero Frobenius inner product with $I$. ConsequentlyFor $\epsilon =\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$, its trace is $\mathrm{div}v$.

Under uniform compressive pressure $p$, isotropic Hooke's law gives each normal strain as $-p(1-2\nu )/E$. The volume strain is three times this, so the bulk modulus iswhere $E$ is Young's modulus and $\nu$ is Poisson's ratio. The shear modulus is $G_s=E/[2(1+\nu )]$.

For stable three-dimensional isotropic linear elasticity, both $B$ and $G_s$ must be positive. With finite $E>0$, these two positivity requirements giveFiniteness of $B$ alone does not establish the upper inequality: its positive sign is essential. The bounds come from positive strain energy density for both volumetric strain and deviatoric strain, rather than a prohibition on negative Poisson's ratio.

For example, $E=100\mathrm{GPa}$ and $\nu =-1/5$ give $B=500/21\mathrm{GPa}\simeq 23.8\mathrm{GPa}$ and $G_s=62.5\mathrm{GPa}$, both positive. A negative Poisson's ratio is compatible with stability. The limit $\nu \to 1/2$ at fixed $E$ sends the bulk modulus to infinity and represents an incompressible limit, rather than a finite positive-modulus point.

For a small displacement field $\mathbf{u}$, define $G_{ij}={\partial }_i{u}_j$ and the infinitesimal strain tensor $\epsilon =(G+G^T)/2$. The isotropic linear elasticity energy density isHere $\lambda ,\mu$ are Lamé parameters and the colon is the Frobenius inner product. The energy depends on symmetric strain, so an infinitesimal rigid rotation costs no elastic energy. A term involving only $G:G^T$ does not have this property. ^[ref]

Differentiation gives the Cauchy stress tensorVarying $\int WdV$ with fixed boundary displacement and no body force yields $\nabla \cdot \sigma =0$, orThese are the static Navier-Cauchy equations.

In two dimensions, separating trace and deviatoric strain givesThus $\mu >0$ and $\lambda +\mu >0$ make every nonzero two-dimensional strain cost positive energy. Rigid translations and infinitesimal rigid rotations remain zero-strain motions.

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.