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The Jacobian matrix of a deformation map carries an infinitesimal reference line element to its current image. Its determinant $J=\det F$ is the local volume ratio. An orientation-preserving deformation has $J>0$, and incompressibility imposes $J=1$.

The factor $A$ can encode an internal relaxed configuration, while $F_{\ast }=F{A}^{-1}$ determines the instantaneous elastic energy. Differentiating $F_{\ast }$ at fixed $F$ or fixed $A$ produces the corresponding nominal stress tensor and thermodynamic force conjugate to an internal variable. This is a kinematic decomposition, not an assertion that each factor separately derives from a global deformation map.

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.

The inextensible fibre constraint gives $|{\mathbf{x}}_{,1}|=1$. Write the fibre direction and its transverse unit direction asThen $x_{1,1}=\cos \theta$ and $x_{2,1}=\sin \theta$. In plane strain, incompressibility gives $\det F=1$. Decomposing the second column of the deformation gradient along $e,n$ therefore gives ${\mathbf{x}}_{,2}=\gamma e+n$, where $\gamma ={\mathbf{x}}_{,2}\cdot e$. HenceThis is a local rotation followed by simple shear.

The compatibility of the deformation map requires ${\mathbf{x}}_{,12}={\mathbf{x}}_{,21}$. Since $e_{,j}={\theta }_{,j}n$ and $n_{,j}=-{\theta }_{,j}e$, this isTaking components gives the compatibility relations for incompressible inextensible-fibre plane strain:Along a reference curve,Thus $\theta$ is constant when $d{X}_2/d{X}_1=-1/\gamma$. The equivalent condition $d{X}_1+\gamma d{X}_2=0$ also covers $\gamma =0$. Its current tangent is proportional to $F(-\gamma ,1{)}^T=n$, so these transverse curves are straight wherever the deformation is regular.

To construct the stress with zero body force, specify the local shear stress $h$ as the material's function or functional of $\gamma$ under simple shear. The normal stresses are constraint reactions: incompressibility supplies pressure, and the inextensible fibre constraint supplies fibre tension. WriteThe compatible geometry has $\nabla \theta =\kappa e$ with $\kappa ={\theta }_{,1}$, so $\nabla \cdot e=0$, $\nabla \cdot n=-\kappa$, $(e\cdot \nabla )e=\kappa n$ and $(n\cdot \nabla )e=0$. Resolving $\nabla \cdot \sigma =0$ along the two directions givesThe triangular equilibrium construction for inextensible-fibre plane strain now determines $s$ by integration along fibres and then $t$ along transverse curves. Thus any regular compatible deformation has a local zero-body-force stress construction once the shear law and suitable boundary integration data are supplied. There is no additional constitutive normal-stress law to specify: the two normal components are fixed by force balance and boundary tractions. This does not assert that arbitrary separately prescribed boundary tractions can accompany an arbitrary deformation.

Put $c=a^2-a_0^2$ and use reference cylindrical coordinates $(\rho ,\Theta ,Z)$ and current coordinates $(r,\theta ,z)$. The deformation map is $r=\sqrt{{\rho }^2+c}$, $\theta =\Theta +\alpha Z$, $z=Z$. In the associated orthonormal bases its deformation gradient isIts determinant is one, as required by incompressibility. The radial principal stretch is ${\lambda }_0=\rho /r$. The other two squared principal stretches are the eigenvalues of the tangential-axial block of $F{F}^T$:Writing $K=(r/\rho {)}^2+{\alpha }^2{r}^2+1$, they areThe positive square roots give ${\lambda }_{\pm }$, and ${\lambda }_0{\lambda }_{+}{\lambda }_{-}=1$.

The strain energy density is per reference volume, so the energy per height isThe work rates per height are $M\dot{\alpha }$ from the end torque and $2\pi ap\dot{a}$ from the inner pressure. The outer surface contributes no work because it is traction free, and the fixed height has zero axial work. Thus the virtual work identities for twist and inflation of an incompressible tube are

For a neo-Hookean solid, an irrelevant energy constant can be omitted:Differentiating under the finite integral givesAn antiderivative divided by $\mu$ is $\log \rho -\frac{1}{2}\log ({\rho }^2+c)-c/[2({\rho }^2+c)]+{\alpha }^2{\rho }^2/2$. With $b^2=b_0^2+c$, the neo-Hookean tube pressure under twist and inflation is thereforeIt vanishes for an untwisted undeformed tube, and the twist contributes the last term at fixed radii.

In plane strain, the deformation map has $x_1=x_1(X_1,X_2)$, $x_2=x_2(X_1,X_2)$ and $x_3=X_3$. The out-of-plane stretch is one and out-of-plane shear vanishes. This kinematic condition need not make the out-of-plane stress vanish.