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Fix a positive shear modulus $\mu$ and let the other Lamé parameter tend to infinity in $S_{\lambda }=\lambda \mathrm{tr}(e_{\lambda })I+2\mu e_{\lambda }$. If strain and stress converge to finite limits and differentiation commutes with the limit, taking the trace shows $\mathrm{tr}{e}_{\lambda }=O({\lambda }^{-1})$. Thus $\mathrm{div}u=0$ and $S=-pI+2\mu e$ for $p=-\lim \lambda \mathrm{tr}{e}_{\lambda }$. Force balance becomes the displayed Stokes equation, with an elastic shear modulus in place of fluid dynamic viscosity. The vanishing volumetric strain is the linearization of elastic incompressibility $\det Dy=1$.

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.

Assume homogeneous Lamé parameters, with $\mu >0$ and positive bulk modulus $K=\lambda +2\mu /3$. For spherical symmetry, write $q={\tau }_{rr}$ and $t={\tau }_{\theta \theta }={\tau }_{\varphi \varphi }$. Radial force balance is $q^{\prime }+2(q-t)/r=0$. The given linear elasticity relations implySubstitution into radial force balance givesThusThe volumetric strain is $e=u^{\prime }+2u/r$, and the trace of the stress tensor isSince $\lambda +2\mu =K+4\mu /3>0$ is constant, $e^{\prime }=0$. The radial elastic compatibility condition follows:This condition is local to a region governed by homogeneous linear elasticity; it is not a constitutive condition for a plastically deforming region.

In a homogeneous isotropic solid without body force, a radial displacement field satisfies $(\lambda +2\mu )(u^{\prime }+2u/r{)}^{\prime }=0$. For nonzero $\lambda +2\mu$, its volumetric strain is constant. Since Hooke's law gives $\mathrm{tr}\tau =(3\lambda +2\mu )(u^{\prime }+2u/r)$, its stress tensor has constant trace. After plasticity in continuum mechanics has occurred, this conclusion applies to purely elastic increments during elastic unloading, and need not apply to the total residual stress.