= Solution
Differentiating the <displacement> gives
$$
\mathcal E=\alpha\operatorname{diag}(1,-\nu,-\nu),\qquad\operatorname{tr}\mathcal E=\alpha(1-2\nu).
$$
The <isotropic> <stress> is diagonal, with
$$
\tau_{xx}=\alpha[\lambda(1-2\nu)+2\mu],\qquad
\tau_{yy}=\tau_{zz}=\alpha[\lambda(1-2\nu)-2\mu\nu].
$$
The two transverse <traction>-free conditions set the latter entries to zero. Hence $\lambda=2\nu(\lambda+\mu)$ and
$$
\boxed{\nu=\frac\lambda{2(\lambda+\mu)},\qquad\tau=\operatorname{diag}(E\alpha,0,0),\qquad E=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu}.}
$$
Substitution shows $E=2\mu(1+\nu)$; solving for the two <Lamé parameters> gives
$$
\boxed{\mu=\frac E{2(1+\nu)},\qquad\lambda=\frac{\nu E}{(1+\nu)(1-2\nu)}.}
$$
To determine the allowed interval without assuming signs not given, put $K=\lambda+2\mu/3>0$. Then $\nu=(3K-2\mu)/[2(3K+\mu)]$, so
$$
1+\nu=\frac{9K}{2(3K+\mu)}>0,\qquad\frac12-\nu=\frac{3\mu}{2(3K+\mu)}>0.
$$
Therefore $\boxed{-1<\nu<1/2}$. The endpoints would require a degenerate <elastic modulus>, so they are excluded by the strict material inequalities.
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