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For a static base state with zero incremental body force, homogeneous incremental tractions or homogeneous Dirichlet boundary conditions eliminate the boundary term in integration by parts. The displayed Rayleigh quotient then gives real squared frequencies because of major symmetry. Positive moduli give strictly positive squared frequencies for modes with nonzero displacement gradient. Pure dead loading leaves constant translations at zero; fixed displacement conditions remove those modes.

Write $B_{i\alpha }={\dot{A}}_{i\alpha }$ and assume that the material constraint is regular, so $F_{,A}\ne 0$. Differentiating the constraint shows that admissible rates of the deformation gradient satisfy $F_{,A_{i\alpha }}{B}_{i\alpha }=0$. The given stress-power identity says that $W_{,A}-N^T$ annihilates this tangent hyperplane. Its orthogonal complement is the one-dimensional span of $F_{,A}$, so a Lagrange multiplier gives the constraint reaction in hyperelastic stress:The sign of $q$ is a convention; the virtual work identity determines no constitutive value for this reaction. Regularity matters: a singular equation describing the same constraint need not have a nonzero gradient.

For incompressibility, take $F=\det A-1$. Differentiating the expression using the Levi-Civita symbol givesThe middle expression is the corresponding entry of the cofactor matrix. Since $\det A=1$, the nominal stress is thereforeFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$, so in the material-first convention $N=\mu A^T+q{A}^{-1}$. The identity matrix satisfies the prescribed dead loading with $q=T-\mu$, proving that the undeformed cube is always an equilibrium.

To obtain the homogeneous tensile bifurcation of a neo-Hookean cube, first seek $A=\mathrm{diag}(\lambda ,\lambda ,{\lambda }^{-2})$, with $\lambda >0$. This has unit determinant. The two distinct diagonal stress equations are $\mu \lambda +q/\lambda =T$ and $\mu {\lambda }^{-2}+q{\lambda }^2=T$. Eliminating $q$ givesThus the nontrivial branch isThe function $h(\lambda )=\lambda +{\lambda }^{-2}$ decreases on $(0,2^{1/3})$ and increases on $(2^{1/3},\infty )$, because $h^{\prime }=1-2{\lambda }^{-3}$. Its minimum is $h(2^{1/3})=3/2^{2/3}$. Above this threshold there are two positive roots ${\lambda }_{-}<2^{1/3}<{\lambda }_{+}$. For $t\ne 2$, neither root is one, and the three choices of the exceptional coordinate for each root give six distinct diagonal deformation gradients, in addition to $I$.

At $t=2$, however, ${\lambda }_{-}=1$ and the three smaller-root matrices coincide with $I$: it would be incorrect to count those as three new deformations. The requested existence of seven distinct homogeneous deformations nevertheless holds even at this value. For any unit vector $n$, putIts positive eigenvalues are ${\lambda }_{+},{\lambda }_{+},{\lambda }_{+}^{-2}$, and its determinant is one. Since $A(n)$ is an orthogonal conjugate of the diagonal solution, the same $q=\mu /{\lambda }_{+}$ gives $\mu A(n{)}^T+qA(n{)}^{-1}=TI$. Choose the six directionsTheir outer products are distinct, and ${\lambda }_{+}>2^{1/3}>1$, so the six $A(n)$ are distinct from one another and from $I$. Constant nominal stress satisfies equilibrium and the prescribed surface tractions. Hence seven distinct homogeneous equilibria exist whenever $T/\mu >3/2^{2/3}$. This is an existence count, not an exhaustive classification: rotated principal axes actually give a continuous family. The conventional seven axis-aligned positive-stretch branches coalesce to four distinct matrices at $T/\mu =2$.

For a static base state under dead loading, put $v_i=u_i(\xi )e^{i\omega t}$ and take zero incremental body force and traction. The admissible squared frequencies are the eigenvalues of the weighted boundary value problemMultiply by $\overline{u_i}$ and integrate over the reference body. The boundary term in integration by parts vanishes, giving the elastic normal-mode energy identityFor a nonzero mode and positive mass density, the denominator is real and positive. The numerator is real by major symmetry of the incremental elastic moduli. Explicitly, if $u=p+iq$ with real $p,q$, the two imaginary cross terms cancel, leaving the sum of the quadratic energies of $\nabla p$ and $\nabla q$. Consequently every admissible ${\omega }^2$ is real. A negative value, when allowed by indefinite moduli, corresponds to exponential growth or decay rather than an oscillation with real frequency.

If the moduli quadratic form is strictly positive on every nonzero real matrix, that same decomposition shows that the numerator is positive whenever $\nabla u$ is nonzero. Thus ${\omega }^2>0$ for every nontranslation mode. Under the usual assumptions of a bounded connected body, regular coefficients and uniform positivity, the Rayleigh quotient defines a self-adjoint operator with discrete positive squared frequencies on the mass-weighted mean-zero displacement space. Its eigenfunctions give the nontrivial small harmonic oscillations.

There is an unavoidable zero-frequency qualification for the pure traction problem as printed: $u_i=b_i$, for any nonzero constant vector $b$, has zero gradient, zero incremental nominal stress and ${\omega }^2=0$. These three constant translations obey all the homogeneous incremental boundary conditions. Under the strict quadratic-form hypothesis they are the entire zero eigenspace on a connected body. Therefore the unrestricted conclusion isStrict positivity for every nonzero displacement requires excluding these modes, for example by imposing ${\int }_{{\Omega }_0}{\rho }_{0}udV=0$, or by adding a displacement constraint. If harmonic perturbations are defined to have nonzero frequency from the outset, the requested positivity holds for all those perturbations.