 Solution

Choose zero energy in the unstrained state. The required strain energy density isIts variation is $\delta W=(\lambda \mathrm{tr}(e)I+2\mu e):\delta e$, giving the prescribed stress as the tensor gradient. This variational notation handles the symmetric off-diagonal components without a spurious factor of two.

Decompose $e=e^{\mathrm{dev}}+(\mathrm{tr}e)I/3$ into orthogonal deviatoric and spherical parts. ThenThe two squares vanish together only for $e=0$. Thus $\mu >0$ and $3\lambda +2\mu >0$ are sufficient for positive definiteness. They are necessary by testing a nonzero trace-free strain and a pure dilation separately. Equivalently both shear modulus $\mu$ and bulk modulus $K=\lambda +2\mu /3$ must be positive.