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Use the material-first nominal stress tensor convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}{\dot{A}}_{i\alpha }$. At a regular constraint point, $F_{,A}\ne 0$, all admissible deformation rates satisfy $F_{,A_{i\alpha }}{\dot{A}}_{i\alpha }=0$. Equality of mechanical power and the derivative of the strain energy density givesfor every vector in this eight-dimensional tangent hyperplane. Its orthogonal complement is the span of $F_{,A}$. Consequently the constraint reaction in hyperelastic stress isThe scalar Lagrange multiplier $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular incompressibility constraint by $(\det A-1{)}^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.

For incompressibility use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ givesThe first expression is the cofactor entry; contracting it with $A_{j\alpha }$ gives $J{\delta }_{ij}$, proving the inverse formula. On $J=1$ this yieldsWith this plus-sign convention $q$ is the negative of the usual pressure multiplier.

For an equibiaxially stretched sheet, incompressibility requires the principal stretches $(\lambda ,\lambda ,{\lambda }^{-2})$. The printed sheet expression's ${\lambda }^{-1/2}$ is inconsistent with this constraint and must be ${\lambda }^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to principal stretch ${\lambda }_j$, evaluated on this constrained path. With no applied thickness-direction traction,Isotropy gives $W_1=W_2$ at equal in-plane stretches. HenceThe equibiaxial nominal tension of an incompressible sheet is thereforeThe two equal in-plane nominal tensions do work $2N_{11}d\lambda$ per reference volume; the thickness-direction nominal stress does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a neo-Hookean solid gives $N=\mu (\lambda -{\lambda }^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu /4$ at that state.

For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2{h}_0$. Its two tangential stretches are $\lambda =a/a_0$, while its thickness stretch is ${\lambda }^{-2}$ and its current thickness is $h=h_0{\lambda }^{-2}$. Define $\hat{W}(\lambda )=W(\lambda ,\lambda ,{\lambda }^{-2})$. Its total elastic energy at leading thin-shell order isDuring a quasistatic increment $da$, the enclosed volume changes by $d\mathcal{V}=4\pi a^{2}da$, and $d\lambda =da/a_0$. Equating energy change to work of the internal excess pressure givesThis is the inflation pressure of a thin incompressible elastic balloon. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll 1$. It also agrees with membrane force balance: $p=2h{\sigma }_t/a$, ${\sigma }_t=\lambda N$, and $N={\hat{W}}^{\prime }/2$. Here pressure is force per current area, while $N$ is force per reference area; confusing them would lose a factor of $\lambda$.

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 $T/\mu =2$.

Work in the fixed reference domain ${\Omega }_0$, and write $v_i=\delta x_i$. For the unconstrained hyperelastic material considered here, the material-first nominal stress is $N_{\alpha i}=W_{,A_{i\alpha }}$. The increment of the deformation gradient and the incremental elastic moduli areAll coefficients are evaluated at the base deformation. The Hessian symmetry is $c_{\alpha i\beta j}=c_{\beta j\alpha i}$; minor interchange symmetry is unnecessary. Linearizing the momentum equation givesThe reference mass density and reference outward unit normal vector are fixed, so the incremental boundary conditions areIn particular, no variation of a current surface normal belongs in this material-coordinate traction condition. If a material constraint were imposed, one would additionally linearize its constraint reaction in hyperelastic stress and its admissibility equation; the present moduli formula is the unconstrained one.