= Solution
Assume homogeneous <Lamé parameters>, with $\mu>0$ and positive <bulk modulus> $K=\lambda+2\mu/3$. For <spherical symmetry>, write $q=\tau_{rr}$ and $t=\tau_{\theta\theta}=\tau_{\phi\phi}$. Radial <force balance> is $q'+2(q-t)/r=0$. The given <linear elasticity> relations imply
$$
q-t=2\mu\left(u'-\frac ur\right),\qquad
q'=(\lambda+2\mu)u''+2\lambda\left(\frac{u'}r-\frac{u}{r^2}\right).
$$
Substitution into radial <force balance> gives
$$
0=(\lambda+2\mu)\left(u''+\frac{2u'}r-\frac{2u}{r^2}\right)
=\frac{d}{dr}\left[(\lambda+2\mu)\left(u'+\frac{2u}r\right)\right].
$$
Thus
$$
\boxed{\frac{d}{dr}\left[(\lambda+2\mu)\left(u'+\frac{2u}r\right)\right]=0.}
$$
The <volumetric strain> is $e=u'+2u/r$, and the <trace> of the <stress tensor> is
$$
q+2t=(3\lambda+2\mu)e.
$$
Since $\lambda+2\mu=K+4\mu/3>0$ is constant, $e'=0$. The <radial elastic compatibility> condition follows:
$$
\boxed{\frac{d}{dr}(\tau_{rr}+\tau_{\theta\theta}+\tau_{\phi\phi})=0.}
$$
This condition is local to a region governed by homogeneous <linear elasticity>; it is not a constitutive condition for a plastically deforming region.
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