Finite elasticity

A spherical radial deformation is $y(x)=r(R)x/R$, with $R=|x|$ and $\mathbf{\theta }=x/R$. Differentiation gives the displayed symmetric gradient. Its radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $r/R$; when these are positive, they are the principal stretches. The determinant is $r^{\prime }(r/R{)}^2$. Positivity distinguishes an orientation-preserving radial elastic deformation from a map that collapses an interior sphere.

For a radial deformation $y=r(R)x/R$ with positive radial derivative, put $q=r/R$ and $\delta =r^{\prime }{q}^2=\det Dy$. The principal nominal stresses are ${\Phi }_1=\mu +h^{\prime }(\delta )q^2$ and ${\Phi }_2=\mu +h^{\prime }(\delta )r^{\prime }q$. Radial equilibrium reads ${\Phi }_1^{\prime }=2({\Phi }_2-{\Phi }_1)/R$. Since $q^{\prime }=(r^{\prime }-q)/R$, substitution cancels the terms proportional to $h^{\prime }$, leaving $h^{\prime \prime }(\delta )q^2{\delta }^{\prime }=0$. Strict convexity of $h$ and positivity of $q$ force a constant positive determinant $D$.

Integration of $r^2{r}^{\prime }=D{R}^2$ gives $r^3=D{R}^3+C$. Two spherical boundary radii fix $D,C$ uniquely. A positive increasing cube-radius interpolation verifies positivity of $r$ and $r^{\prime }$, so the algebraic solution is an admissible smooth radial equilibrium. The result is restricted to this energy family and radial class; strict convexity of its determinant term alone does not imply uniqueness among arbitrary three-dimensional deformations.

Consider an orientation-preserving classical radial deformation of an elastic solid on a connected spherical annulus. If its nonnegative energy is zero everywhere and the material is isotropic with strict Baker-Ericksen inequalities, every gradient is a natural state, so $r^{\prime }=r/R$. Thus $(r/R{)}^{\prime }=0$ and $y=vx$.

A separate rigidity argument applies when the zero wells are $SO(3)\cup {\bigcup }_{i=1}^{N}SO(3)U_i$, with finitely many positive definite symmetric $U_i$. The radial gradient $F$ is positive symmetric; if $F=Q{U}_i$, then $F^2=U_i^2$, so uniqueness of the positive square root gives $F=U_i$. The rotation well similarly gives $F=I$. Hence the continuous ratio $r(R)/R$ takes values in a finite set of eigenvalues and must be constant on the connected radial interval. Differentiating $r=vR$ finishes the proof. This version needs neither material isotropy nor Baker-Ericksen inequalities and does not require an assumed continuity of $r^{\prime }$ beyond the radial-map regularity used for the ratio.

Let ${\mathbf{x}}_0(\mathbf{X};\tau )$ be a family of static elastic equilibrium solutions for a homogeneous material with incompressibility under a fixed spatial body force. Replacing $\tau$ by time gives a dynamic solution of the elastic field equations exactly when the resulting material acceleration has a global acceleration potential $\xi$. At the same deformation gradient, the Cauchy stress tensor can change only by the scalar pressure reaction. Subtracting static and dynamic balance of linear momentum gives $\nabla q=\rho \mathbf{a}$, and constant mass density then gives $q=-\rho \xi +C(t)$. Surface tractions must match the adjusted Cauchy stress tensor.

The opening of an internal cavity under tensile loading in a solid. For spherically symmetric deformation of an isotropic hyperelastic material with incompressibility, and an empty cavity without surface tension, radial static elastic equilibrium gives the displayed critical signed external pressure, where $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$ and positive $P$ denotes compression. Indeed, the principal stress formula for incompressible hyperelasticity gives $d{t}_r/d\lambda =h^{\prime }(\lambda )/(1-{\lambda }^3)$. Integration between the free inner surface and the outer surface, followed by the vanishing-initial-hole and vanishing-current-cavity limits, gives the formula. For a neo-Hookean solid with shear modulus $\mu >0$, the tensile magnitude is $-P_{\mathrm{crit}}=5\mu /2$.

For the unit ball and energy $W(F)=|F{|}^2+(\det F-1{)}^2$, boundary dilation $y=tx$ admits a singular incompressible radial competitor $r(R)=(R^3+t^3-1{)}^{1/3}$. Its radial stretch is $R^2/r^2$ and its double tangential stretch is $r/R$. Although the latter diverges at zero, the three-dimensional volume element makes its squared gradient integrable. Its energy is $4\pi {\int }_0^1[R^6/r^4+2r^2],dR\sim 8\pi t^2$.

The smooth homogeneous equilibrium $y=tx$ has constant stress and energy $(4\pi /3)[3t^2+(t^3-1{)}^2]$, of sixth-order growth. It is therefore beaten for large $t$ in the finite-energy Sobolev space class admitting the cavity. This conclusion depends on that class: the singular map is not continuous at the center and cannot simply be substituted as a smooth cavity-free competitor.

For strain energy density $W=\mu [({\lambda }_r^2+{\lambda }_{\theta }^2+{\lambda }_{\varphi }^2{)}^{\alpha }-3^{\alpha }]/2$ with $\mu >0$ and $\alpha >0$, the critical loading for elastic cavitation is finite exactly in the displayed range. To prove this, put $S={\lambda }^{-4}+2{\lambda }^2$ in the reduced strain energy density. The critical-pressure integrand is $2\mu \alpha S^{\alpha -1}({\lambda }^{-2}+{\lambda }^{-5})$. It is finite at $\lambda =1$ and asymptotic to $\mu \alpha 2^{\alpha }{\lambda }^{2\alpha -4}$ at infinity. The comparison test for the resulting improper integral gives $2\alpha -4<-1$.

For reference radii $a_0,b_0$, current inner radius $a$ and uniform axial stretch $z>0$, incompressibility integrates to $r^2=a^2+(R^2-a_0^2)/z$. The principal stretches are $r_{,R}=1/(z\lambda )$, $\lambda =r/R$ and $z$. Write $\hat{W}(\lambda ,z)=W((z\lambda {)}^{-1},\lambda ,z)$ and $\mathcal{E}=2\pi {\int }_{a_0}^{b_0}\hat{W}RdR$ per reference length. Virtual work gives $p={\mathcal{E}}_{,a}/(2\pi az)$ and the total axial wall load $N_{\mathrm{wall}}={\mathcal{E}}_{,z}$, both derivatives holding the other displayed parameter fixed. For pressurized closed ends the independently applied axial load is $N_{\mathrm{ext}}=N_{\mathrm{wall}}-\pi a^{2}p$. The distinction follows from the axial pressure power $\pi a^{2}p\dot{z}$.

For an equilibrium with $P_{Ii}=W_{,F_{iI}}+q{\varphi }_{,F_{iI}}$ and dead loads, the constrained second variation on ${\varphi }_{,F}:\delta F=0$ isIndeed, a constraint-preserving perturbation path has ${\varphi }_{,F}:{\delta }^{2}F=-{\varphi }_{,FF}[\delta F,\delta F]$, and stationarity of the total potential supplies the multiplier term. Linearized kinetic energy plus $\mathcal{Q}/2$ is conserved. Positivity gives the usual energy-norm linear stability; control in a specified displacement norm additionally requires the corresponding coercivity and treatment of rigid-motion modes.

For axial stretch $\lambda$ and twist per reference length $\alpha$, the deformation in local cylindrical orthonormal bases hasThe principal stretches follow from the eigenvalues of $F^{T}F$. For a cylinder of reference radius $A$, stored energy per reference length is $\mathcal{E}=2\pi {\int }_0^{A}W(F)RdR$. Reversible power $M\dot{\alpha }+N\dot{\lambda }=\dot{\mathcal{E}}$ gives torque $M={\mathcal{E}}_{,\alpha }$ and axial force $N={\mathcal{E}}_{,\lambda }$. The relevant derivative holds the reference twist fixed, not the twist per current length.

A deformation map twisting every cross-section of a cylinder through an angle proportional to its reference axial position, without radial or axial extension. In the local current and reference cylindrical coordinate systems, the deformation gradient is $F=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& \tau r\\ 0& 0& 1\end{array}\right)$. Its determinant is one. The twist per unit reference length is $\tau$; an end separation $L$ gives end rotation $\phi =\tau L$. The physical torque is $N=2\pi {\int }_0^A{T}_{\theta z}{r}^{2}dr$.

For axial stretch $\lambda$ and twist $\alpha$ per reference length, an incompressible neo-Hookean solid has current radius $a=R/\sqrt{\lambda }$. In local cylindrical orthonormal bases the deformation gradient is $A=\left(\begin{array}{ccc}{\lambda }^{-1/2}& 0& 0\\ 0& {\lambda }^{-1/2}& \alpha r\\ 0& 0& \lambda \end{array}\right)$. The Cauchy stress tensor is $\mu A{A}^T+qI$. Radial equilibrium gives $q^{\prime }=\mu {\alpha }^{2}r$, and a traction-free curved surface gives $q=-\mu /\lambda +\mu {\alpha }^2(r^2-a^2)/2$. The nominal end traction is the displayed vector, measured per reference area. Its moment about the axis is $\int (x_1{t}_2-x_2{t}_1)d{S}_0=\mu \alpha \int r^{2}d{S}_0=\mu \pi R^4\alpha /(2\lambda )$. This calculation uses current moment arms with reference-area tractions.

In cylindrical orthonormal bases the deformation gradient is $F=\left(\begin{array}{ccc}\rho /r& 0& 0\\ 0& r/\rho & \alpha r\\ 0& 0& 1\end{array}\right)$ and has unit determinant. One principal stretch is $\rho /r$; the squares of the other two are the roots of $z^2-[(r/\rho {)}^2+{\alpha }^2{r}^2+1]z+(r/\rho {)}^2=0$. The reference-volume energy per height is $E=2\pi \int \rho Wd\rho$. Virtual work gives end torque $M={\partial }_{\alpha }E$ and inner pressure $p=(2\pi a{)}^{-1}{\partial }_{a}E$ at fixed reference radii and height.

Here $b^2=b_0^2+a^2-a_0^2$. Substitute the neo-Hookean solid energy into the twist and inflation of an incompressible tube work identity. Differentiating with respect to $a$ gives the integrand $\mu \rho [{\rho }^{-2}-{\rho }^2/({\rho }^2+c{)}^2+{\alpha }^2]$, with $c=a^2-a_0^2$. Integrating between the two reference radii proves the displayed pressure when the outer surface is traction free.

A unit reference fibre direction $m_0$ retains its length under this constraint on the deformation gradient. Its reaction contributes an undetermined tension along the current fibre direction, independently of the pressure reaction enforcing incompressibility.

For two distinct reference fibre directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0)$, with $0<\theta <\pi /2$, consider the volume-preserving deformation with right stretch tensor $U=\mathrm{diag}({\lambda }^{-1/2}\alpha ,{\lambda }^{-1/2}{\alpha }^{-1},\lambda )$, with positive $\lambda ,\alpha$. The inextensible fibre constraint gives ${\alpha }^2{\cos }^2\theta +{\alpha }^{-2}{\sin }^2\theta =\lambda$. Setting $b={\alpha }^2$ gives a quadratic with the displayed roots. There are two distinct positive branches for $\lambda >\sin 2\theta$. The smallest possible axial stretch is $\lambda =\sin 2\theta$, at which ${\alpha }^2=\tan \theta$ and $(U{l}_{+})\cdot (U{l}_{-})=0$. Since both image fibres have unit length, their mutual area cannot exceed one; their area is $\sin 2\theta /\lambda$, which gives the same lower bound. Coincident or antiparallel fibres at the angle endpoints do not give the two-family orthogonality conclusion.

Let $e=(\cos \theta ,\sin \theta )$ be the current fibre direction and $n=(-\sin \theta ,\cos \theta )$. Inextensibility fixes the first column of the deformation gradient to $e$, and incompressibility fixes the second to $\gamma e+n$. Commuting mixed derivatives gives ${\gamma }_{,1}={\theta }_{,1}$ and ${\theta }_{,2}=\gamma {\theta }_{,1}$. Thus $\gamma =\theta +f(X_2)$, and curves tangent to $(-\gamma ,1)$ map to straight transverse curves with constant $\theta$.

Write the Cauchy stress tensor in the local fibre frame as $\sigma =se\otimes e+h(e\otimes n+n\otimes e)+tn\otimes n$. The compatible geometry has $\nabla \theta =\kappa e$, $\nabla \cdot e=0$ and $\nabla \cdot n=-\kappa$. With zero body force, force balance becomes $e\cdot \nabla s=2\kappa h-n\cdot \nabla h$ and $n\cdot \nabla t-\kappa t=-e\cdot \nabla h-\kappa s$. For specified shear stress $h$, the first equation determines $s$ along fibres, then the second determines $t$ along transverse curves. Pressure and fibre tension supply these two normal constraint reactions. This is a local stress construction; boundary tractions must be chosen consistently with its integration data.

A hyperelastic material derives its elastic stress from a strain energy density. In reference-volume units its first Piola-Kirchhoff stress tensor is the derivative of that density with respect to the deformation gradient.

A hyperelastic energy well is a set of deformation gradients minimizing the stored-energy function at a fixed temperature. Material frame indifference makes each well invariant under left multiplication by proper rotations. A phase-changing crystal can have a single austenitic rotational well at high temperature and several martensitic wells below a transition. Continuous deformations cannot freely join arbitrary wells: planar interfaces require a rank-one connection between the selected gradients.

On positive-determinant deformation gradients, the first Piola-Kirchhoff stress is $\mu (F-J^{-1}{F}^{-T})$. Its zero set is $SO(3)$: multiplication by $F^T$ gives $F{F}^T=J^{-1}I$, and determinants give $J^5=1$. These are also the global minimizers, since $|F{|}^2\ge 3J^{2/3}$ and $3J^{2/3}/2+J^{-1}$ has its unique minimum $5/2$ at $J=1$. The minimum energy is $5\mu /2$ rather than zero; an additive constant changes no stress.

An unconstrained natural state is a stress-free homogeneous deformation gradient. For $J>0$, its equivalent conditions are $DW(F)=0$ and $J^{-1}DW(F)F^T=0$. An interior differentiable minimum of the strain energy density is a natural state, although a stationary state need not minimize the energy. Constraints require allowing their reaction stress; the unconstrained equivalence must then be modified. Under strict Baker-Ericksen inequalities, an isotropic natural state is a dilation times a proper rotation.

For a material isotropy and material frame indifference strain energy density $W(F)=h(I_1,I_2,I_3)$, take $B=F{F}^T$, $I_1=\mathrm{tr}B$, $I_2=((\mathrm{tr}B{)}^2-\mathrm{tr}{B}^2)/2$, $I_3=\det B=J^2$, $J=\det F>0$. The chain rule and $d(\det F)=J{F}^{-T}:dF$ give $DW=2h_{1}F+2h_2(I_{1}I-B)F+2I_3{h}_3{F}^{-T}$. Multiplying by $F^T/J$ yields the displayed Cauchy stress tensor. Material isotropy allows an identity term and a quadratic term in $B$, as well as the term proportional to $B$.

A hyperelastic material is homogeneous when its strain energy density $W(F)$ has no explicit dependence on the reference configuration position. Its first Piola-Kirchhoff stress tensor is $T=DW(F)$. Homogeneity of the material does not require the deformation gradient to be spatially constant.

For an isotropic hyperelastic material with incompressibility, the principal stresses are $t_i={\lambda }_i\partial W/\partial {\lambda }_i-p$. This follows by evaluating the Cauchy stress tensor formula $T=F({\partial }_{F}W{)}^T-pI$ in principal directions. For the reduced strain energy density $h(\lambda )=W({\lambda }^{-2},\lambda ,\lambda )$, the chain rule and equality of the tangential derivatives give $t_{\theta }-t_r=\lambda h^{\prime }(\lambda )/2$.

For a plane material-coordinate elastic wave with $u=(0,w(X_1,t),0)$, set $q=w_{,X_1}$. The longitudinal equation is $0={\partial }_{X_1}{P}_{11}(1,q,0)$. Arbitrary nonuniform pure transverse profiles thus require the displayed constitutive compatibility, which does not follow from isotropy alone. If it holds and ${\tau }^{\prime }(q)>0$, the shear equation closes as ${\rho }_0{w}_{tt}={\partial }_{X_1}\tau (q)$ with speed $c(q{)}^2={\tau }^{\prime }(q)/{\rho }_0$.

An explicit objective isotropic example is $\mathcal{W}=\mu (S-1)/2+h(S-1{)}^2/4+(\lambda +\mu )(J-1{)}^2/2$, with $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and $\lambda ,\mu >0$, $h\ge 0$. On $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, $S-1=(a-1{)}^2+q^2$ and $J=a$, so $P_{11}=0$ at $a=1$ while the shear nominal stress is the displayed cubic. The natural-state quadratic energy has Lamé parameters $\lambda ,\mu$, and the shear speed is $\sqrt{(\mu +3h{q}^2)/{\rho }_0}$.

This objective isotropic constitutive model uses the Green-Lagrange strain tensor $E=(A^{\mathsf{T}}A-I)/2$. It reproduces linear elasticity at small strain. Under simple shear $A=[(1,q,0{)}^{\mathsf{T}},e_2,e_3]$, its longitudinal nominal stress is $(\lambda +2\mu )q^2/2$. Nonuniform finite shear therefore generates longitudinal acceleration, even though the corresponding infinitesimal transverse mode is uncoupled.

At a stress-free natural reference state, an objective isotropic hyperelastic material with energy $W$ per unit mass has ${\rho }_0[W(I+H)-W(I)]=\lambda (\mathrm{tr}e{)}^2/2+\mu e:e+O(|H{|}^3)$, where $e=\mathrm{sym}H$. Twice differentiating gives the displayed Lamé parameters and the usual elastic stiffness tensor. A prestressed reference requires initial-stress corrections.

At leading thin-shell order, a spherical balloon with reference radius $a_0$, thickness $h_0$ and current radius $a=\lambda a_0$ has reference elastic energy $4\pi a_0^2{h}_{0}W(\lambda ,\lambda ,{\lambda }^{-2})$. Equating its differential to pressure work $p4\pi a^{2}da$ gives the displayed pressure difference. Its thickness becomes $h_0{\lambda }^{-2}$ by incompressibility. The same result follows from the spherical membrane force balance $p=2h{\sigma }_t/a$, with current tangential Cauchy stress tensor component ${\sigma }_t=\lambda N$ and the equibiaxial nominal tension of an incompressible sheet. Thin-shell validity requires $h/a\ll 1$.

For unconstrained hyperelastic material, the Hessian of the strain energy density gives $\delta N_{\alpha i}=c_{\alpha i\beta j}\delta A_{j\beta }$. Commuting the second derivatives gives major symmetry $c_{\alpha i\beta j}=c_{\beta j\alpha i}$. These material-coordinate moduli linearize about a finite base deformation; they need not have the minor symmetries of a small-strain elastic stiffness tensor. Material constraints require their additional reaction increments and admissibility equations.

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.

Strong ellipticity requires the displayed inequality for every nonzero displacement vector $m$ and material direction $\nu$. Equivalently, every acoustic tensor is positive-definite. It is weaker than positivity of the moduli on all matrices. A quadratic minor null Lagrangian vanishes on rank-one matrices, so adding such a term does not change the acoustic tensor or this condition.

The acoustic tensor determines the plane waves of a locally uniform incremental elastic medium. For a unit material direction $\nu$, its eigenvalues are ${\rho }_0{c}^2$ and its eigenvectors are displacement polarizations. Major symmetry of the incremental elastic moduli makes $Q$ a real symmetric matrix. Positivity on every rank-one matrix $m\otimes \nu$ makes all its wave speeds real and nonzero.

The elastic stiffness satisfies $c_{ijkl}{a}_i{p}_j{a}_k{p}_l>0$ for every nonzero polarization $a$ and propagation vector $p$. It makes the elastic acoustic tensor positive definite, ensuring positive squared wave speeds. Positive strain energy is a sufficient stronger condition in ordinary linear elasticity.

For a regular material constraint $F(A)=0$, admissible rates of the deformation gradient lie in the kernel of $F_{,A}$. The difference between the nominal stress and the derivative of the strain energy density annihilates that tangent space by virtual work, so it is a Lagrange multiplier times $F_{,A}$. For incompressibility, the derivative of the determinant is its cofactor matrix, and the material-first reaction is $q{A}^{-1}$ when $\det A=1$.

A thin isotropic hyperelastic material subject to incompressibility, with equal in-plane principal stretches $\lambda$ has transverse stretch ${\lambda }^{-2}$. With unloaded transverse faces, $N_{33}=W_3+q{\lambda }^2=0$. Hence $N_{11}=N_{22}=W_1-{\lambda }^{-3}{W}_3$, using isotropy to set $W_1=W_2$. The derivative along the constrained path is $W_1+W_2-2{\lambda }^{-3}{W}_3$, proving the displayed formula. The two equal nominal tensions do work $2Nd\lambda$ per reference volume, exactly the change in strain energy density.

An incompressible Mooney-Rivlin solid has strain energy density $W=C_1(I_1-3)+C_2(I_2-3)$, with right Cauchy-Green deformation tensor $C=F^{T}F$, $I_1=\mathrm{tr}C$ and $I_2=\mathrm{tr}{C}^{-1}$ when $\det C=1$. It extends the neo-Hookean solid by a second invariant. A convention $W={\mu }_1(I_1-3)/2-{\mu }_2(I_2-3)/2$ means $C_1={\mu }_1/2$, $C_2=-{\mu }_2/2$; the minus sign must be retained when comparing material constants.

For incompressible simple shear $F=I+\gamma {\mathbf{e}}_1\otimes {\mathbf{e}}_2$ with Mooney-Rivlin energy $\alpha (I_1-3)+\beta (I_2-3)$, zero traction on the third-coordinate faces fixes the pressure reaction. The remaining Cauchy stress has $T_{11}=2\alpha {\gamma }^2$, $T_{22}=-2\beta {\gamma }^2$ and $T_{12}=2(\alpha +\beta )\gamma$. The material face $x_1=\text{constant}$ has current unit normal $\mathbf{n}=(1,-\gamma ,0)/\sqrt{1+{\gamma }^2}$ and tangent $\mathbf{m}=(\gamma ,1,0)/\sqrt{1+{\gamma }^2}$. Resolving $T\mathbf{n}$ gives $N=-2{\gamma }^2(\alpha +2\beta +\beta {\gamma }^2)/(1+{\gamma }^2)$ and $S=2\gamma (\alpha +\beta )/(1+{\gamma }^2)$. Since $N<0$, the traction-normal angle satisfies $\tan (\theta -\pi /2)=-N/|S|$, proving the displayed formula. Consequently its two-sided small-shear slope is $(\alpha +2\beta )/(\alpha +\beta )$. The reference normal cannot replace the transformed normal in this calculation.

For an incompressible neo-Hookean solid, a convenient strain energy density is $W=\mu (\mathrm{tr}C-3)/2$, where $C$ is the right Cauchy-Green deformation tensor. An additive constant in $W$ does not affect the stress. The shear modulus $\mu$ fixes the small-strain stiffness.

Consider an incompressible neo-Hookean prism with energy $\alpha (|F{|}^2-3)$, zero nominal end traction normal to $e_3$, and equal nominal lateral tractions $TN$ on side faces whose normals enforce both in-plane principal stresses. Within positive diagonal homogeneous deformations the first Piola-Kirchhoff stress tensor has entries $t_i=2\alpha v_i-p/v_i$. Equilibrium makes $p$ constant; end freedom gives $p=2\alpha v_3^2>0$.

The function $2\alpha v-p/v$ is strictly increasing on positive $v$, so equal lateral stresses require $v_1=v_2$. The determinant constraint then gives $v_3=v^{-2}$ and the displayed load relation. Its derivative is $2\alpha (1+5v^{-6})>0$, with limits minus infinity and infinity at the two ends of the positive axis. Thus there is exactly one positive diagonal equilibrium for every signed load. This uniqueness is confined to the homogeneous diagonal class and does not rule out other modes or nonhomogeneous equilibria under a general dead-loading problem.

For the volume-preserving family $F=\mathrm{diag}(\lambda ,{\lambda }^{-1/2},{\lambda }^{-1/2})$, equal dead nominal tension $T$ gives potentialIts derivative factors as ${\Pi }^{\prime }=(1-{\lambda }^{-3/2})[\mu (\lambda +{\lambda }^{-1/2})-T]$. The nontrivial branch has a fold at $\lambda =2^{-2/3}$, $T/\mu =32^{-2/3}$, and meets the trivial branch at $T/\mu =2$. Stability in this one-parameter family is determined by ${\Pi }^{\prime \prime }$, not by branch count alone. At $T/\mu =2$ there is only one nontrivial positive solution, $\lambda =(3-\sqrt{5})/2$; the other branch value is $\lambda =1$.

In three-dimensional isotropic linear elasticity, decompose the symmetric strain as $e=pI+d$ with $p=\mathrm{tr}(e)/3$ and $\mathrm{tr}d=0$. The stored energy is $E=\mu {\sum }_{ij}{d}_{ij}^2+\frac{3}{2}(3\lambda +2\mu )p^2$. Its nonnegativity for every strain is equivalent to the displayed inequalities: the traceless and purely volumetric test strains prove necessity, and the sum of squares proves sufficiency.

Material isotropy makes the elastic response independent of a rotation of reference material directions: $\sigma (FR)=\sigma (F)$ for proper rotations $R$. Combined with material frame indifference, it yields coaxiality of isotropic elastic stress.

For a smooth objective isotropic hyperelastic material, the strict Baker-Ericksen inequalities require larger distinct principal stretches to have larger principal stresses. For $W=\Phi (v_1,v_2,v_3)$ and $J=v_1{v}_2{v}_3$, ${\sigma }_i=J^{-1}{v}_i{\Phi }_i$. The strict convention implies every stress-free state has equal stretches: if $v_i\ne v_j$ but both stresses are zero, the displayed product would vanish. Thus a natural state of a hyperelastic material has $F=vQ$, $v>0$, $Q\in SO(3)$. A weak non-strict inequality alone does not imply this rigidity.

The symmetry group of a hyperelastic material consists of invertible reference transformations $H$ preserving its strain energy density: $W(FH)=W(F)$ for every admissible deformation gradient $F$. A convention restricting reference transformations to determinant one restricts $H$ to the special linear group. Material isotropy requires all proper rotations in this group. This right action differs from the left action in material frame indifference.

The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

For simple shear, $B=\left(\begin{array}{cc}1+{\gamma }^2& \gamma \\ \gamma & 1\end{array}\right)$. The off-diagonal entry of $\sigma B-B\sigma$ is $\gamma ({\sigma }_{11}-{\sigma }_{22})-{\gamma }^2{\sigma }_{12}$. Coaxiality of isotropic elastic stress therefore proves the identity for $\gamma \ne 0$. At $\gamma =0$, material isotropy makes the stress isotropic, proving the same identity.