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For fixed nominal boundary tractions and dead body forces, use the total potentialThe dead-load terms are linear in the deformation map. Let $x(\epsilon )$ be a constraint-preserving perturbation path through equilibrium, with $H=\delta F$ and $K={\delta }^{2}F$. Differentiating $\varphi (F)=0$ twice gives ${\varphi }_{,F}:H=0$ and ${\varphi }_{,F}:K=-{\varphi }_{,FF}[H,H]$. Equilibrium has nominal stress tensor $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$. Its virtual-work relation, applied to ${\delta }^{2}x$, gives the otherwise remaining first-derivative contribution ${\Pi }^{\prime }[{\delta }^{2}x]=-\int q{\varphi }_{,F}:KdX$. Therefore the constrained second variation isFor a time-dependent linearized perturbation $y$, the incremental reaction stress contains $\delta q{\varphi }_{,F}$, but its power vanishes because ${\varphi }_{,F}:\nabla \dot{y}=0$. Multiply the incremental momentum equation by $\dot{y}$, integrate by parts, and use zero incremental dead-load traction or the prescribed displacement condition. One obtainsPositive $\mathcal{Q}$ is therefore the standard constrained second-variation stability in elasticity criterion: perturbations cannot grow in this quadratic energy norm. A statement in a specified displacement norm also uses the usual coercivity and boundary conditions removing uncontrolled rigid-motion modes; translations with $\delta F=0$ are not tested by the displayed condition. This is a small-perturbation energy argument, not a claim of global nonlinear stability.

For the cube use incompressibility in the form $\varphi =\det F-1$. A neo-Hookean solid has $W_{,F}=\mu F$, and hence the spatial-index-first transpose of nominal stress is $\mu F+q{F}^{-T}$. On the diagonal family, equality to $TI$ givesThe first relation immediately gives $\boxed{q=T\lambda -\mu {\lambda }^2}$. Eliminating $q$ and factoring yieldsThus $\boxed{\lambda =1\text{or}T/\mu =\lambda +{\lambda }^{-1/2}}$.

Let $t=T/\mu$ and $h(\lambda )=\lambda +{\lambda }^{-1/2}$. Its derivative $h^{\prime }=1-\frac{1}{2}{\lambda }^{-3/2}$ vanishes only at ${\lambda }_c=2^{-2/3}$, where $h^{\prime \prime }>0$ and $h_c=32^{-2/3}$. Also $h\to \infty$ at both ends of $(0,\infty )$. Therefore $h(\lambda )=t$ has two distinct positive roots for $t>h_c$, one repeated root at $h_c$, and none below. There is an exception to the printed assertion that both roots differ from one: at $t=2$, one root is exactly $\lambda =1$. Indeed, with $x=\sqrt{\lambda }$, the equation becomes $(x-1)(x^2+x-1)=0$, so the only other positive stretch is $\lambda =(3-\sqrt{5})/2$.

Evaluate the stability form directly on the prescribed uniform perturbationIt satisfies $\mathrm{tr}(F^{-1}H)=0$. The determinant Hessian identity isMeanwhile $W_{,FF}[H,H]=\mu (1+\frac{1}{2}{\lambda }^{-3})(\delta \lambda {)}^2$. For the unit reference cube,On the trivial branch, $q=T-\mu$ andThus the undeformed state is stable in this perturbation family for $t<2$, unstable for $t>2$, and quadratically neutral at $t=2$.

On a nontrivial branch, substitute $t=h(\lambda )$ and put $z={\lambda }^{-3/2}$:The smaller root always has $\lambda <{\lambda }_c$, hence $z>2$, and is stable. For $h_c<t<2$, the larger root lies between ${\lambda }_c$ and one, so $1<z<2$ and it is unstable; the trivial branch is the second stable equilibrium. For $t>2$, the larger root lies above one and has $0<z<1$, so it is stable; the trivial branch is now unstable. Whenever there are three distinct equilibria in this family, precisely two are stable and one unstable against the specified uniform perturbations. This does not establish stability to every nonuniform or symmetry-breaking perturbation.

The homogeneous tensile bifurcation of a neo-Hookean cube has a nontrivial-branch fold at $t=32^{-2/3}$ and a meeting with the undeformed branch at $\boxed{t=2}$. The latter is the bifurcation from the undeformed state. Stating both values distinguishes this branch intersection from the earlier creation of the two nontrivial solutions.