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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 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 $-P_{\mathrm{crit}}=5\mu /2$.

An elastic deformation is recoverable when the deforming loads are removed. In a local elastic constitutive model the stress depends on the present deformation. When this response also derives from a stored elastic energy, the material is a hyperelastic material. In the small-deformation approximation, the relevant kinematic variable is the infinitesimal strain tensor.

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.

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 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.

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.

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.

Use material coordinates $X$ and write the placement as $y=X+u$. The strain energy density $W$ in the printed convention is per unit reference mass. It is convenient to use the conventional first Piola-Kirchhoff stress tensor $P=s^{\mathsf{T}}$, soAssume the undeformed reference state is stress-free. With $H=A-I$ and infinitesimal strain tensor $e=(H+H^{\mathsf{T}})/2$, material isotropy and material frame indifference give the quadratic expansionConsequently the Lamé parameters from hyperelastic energy Hessian areIndeed, differentiating the quadratic expression twice givesthe elastic stiffness tensor of linear elasticity. A prestressed reference state requires the corresponding initial-stress corrections; the displayed formula concerns the natural state.

For a plane elastic wave, $u=u(X_1,t)$, put $r=e_1+u_{,X_1}$ and $\Psi (r)=W([r,e_2,e_3])$. Balance of momentum without body force becomesThis is a material-coordinate equation: the reference density cancels between inertia and nominal stress. If $C{a}_{\alpha }=c_{\alpha }^2{a}_{\alpha }$, the first-order system has characteristic speeds $\pm c_{\alpha }$ and right eigenvectors $(a_{\alpha },\mp c_{\alpha }{a}_{\alpha })$. Along a characteristic curve with $d{X}_1/dt=s=\pm c_{\alpha }$, its differential compatibility relation isThis follows by multiplying the system by the left eigenvector $(-s{a}_{\alpha }^{\mathsf{T}},a_{\alpha }^{\mathsf{T}})$. It does not assert the existence of globally integrable Riemann invariants for an arbitrary three-component constitutive law.

There is a necessary qualification to the finite-amplitude transverse request. Set $u=(0,w,0)$ and $q=w_{,X_1}$, so $r=(1,q,0)$. The longitudinal momentum equation requiresThus arbitrary nonuniform pure transverse profiles are possible only if the compatibility condition for pure nonlinear transverse waves holds:The third component vanishes by the reflection symmetry implied by isotropy and objectivity. The shear equation then closes:

General compressible hyperelastic material assumptions alone do not imply the required compatibility. For a concrete counterexample, take the Saint Venant–Kirchhoff hyperelastic energyFor $A=[(a,q,0{)}^{\mathsf{T}},e_2,e_3]$, this becomesAt $a=1$, $P_{11}=(\lambda +2\mu )q^2/2$. A varying shear therefore produces a longitudinal force $(\lambda +2\mu )q{q}_{,X_1}$ and cannot preserve $u_1=0$. Infinitesimal transverse waves do exist, because this coupling is second order, but a general pure finite-amplitude transverse wave does not.

The compatible nonlinear shear energy class is nonempty and can be genuinely nonlinear. Define $J=\det A$, $S=\mathrm{tr}(A^{\mathsf{T}}A)-2J$, and, for $h\ge 0$, takeThis is objective and isotropic. Near $A=I$, $S-1=2\mathrm{tr}(e^2)-(\mathrm{tr}e{)}^2+O(|H{|}^3)$, so it has the specified Lamé parameters. On the plane deformation above, $S-1=(a-1{)}^2+q^2$ and $J=a$. Hence $P_{11}=0$ at $a=1$, whileFor positive $\lambda ,\mu$, the relevant incremental plane moduli are positive. This supplies an explicit stable pure transverse family and, when $h>0$, amplitude-dependent wave speed.

For any compatible family, letThe equations $q_t=v_{,X_1}$ and $v_t=c(q{)}^2{q}_{,X_1}$ give the Riemann invariantsConsider the initially undisturbed half-space $X_1>0$. A right-going simple wave has the incoming invariant $v+H=0$, so $v=-H(q)$. Denote the prescribed boundary nominal stress by $T(t)=P_{21}(0,t)=\tau (q_b(t))$. The physical boundary traction, whose outward normal is $-e_1$, is $-T(t)e_2$. Invert the monotone law $\tau$ to obtain $q_b$. The wave state emitted at boundary time $s$ is transported alongAhead of the leading front $X_1=c(0)t$ the state is zero. Before characteristic crossing, these equations determine the unique emission time $s$ for each disturbed point. The displacement itself is also determined, rather than just its derivatives:Differentiating the implicit emission relation verifies $w_{,X_1}=q_b(s)$ and $w_t=-H(q_b(s))$, so this is the complete pre-shock simple wave solution.

Write $c_b(s)=c(q_b(s))$. The boundary-generated characteristic curves lose invertibility whenThus the characteristic crossing in a boundary-generated elastic simple wave criterion isHere ${\dot{c}}_b=c^{\prime }(q_b)T^{\prime }/({\rho }_0{c}_b^2)$. A smoothly increasing boundary stress therefore steepens into a shock wave only when later states travel faster: $c^{\prime }(q_b)T^{\prime }>0$. Increasing stress alone is insufficient. For the compatible example with $h>0$ and $q_b>0$, $c^{\prime }(q_b)=3h{q}_b/({\rho }_0{c}_b)>0$, so a nonzero increasing ramp produces finite-time crossing. For $h=0$ all rays are parallel and no such shock forms. This criterion locates the first gradient singularity; subsequent discontinuous evolution additionally requires the jump conditions and an admissibility condition.

For a general vector simple wave, replace the pure-shear line by an integral curveSubstitution gives $a_t+c_{\alpha }(a)a_{,X_1}=0$, with the same emission construction and crossing criterion. Boundary tractions must lie on this one-family curve. Generic finite-amplitude quasi-transverse waves may contain longitudinal displacement; arbitrary vector boundary forcing need not generate a single simple wave.

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.

For a hyperelastic material $T_R(F)=DW(F)$. Material frame indifference of this first Piola-Kirchhoff law meansThe spatial force components rotate under a superposed rigid rotation, whereas the reference normal does not.

If $W(QF)=W(F)$, differentiate with respect to $F$ in direction $H$ to obtain $DW(QF):(QH)=DW(F):H$, hence the displayed stress transformation. Conversely, assume that stress transformation. For fixed $Q$, the derivative of $W(QF)-W(F)$ with respect to $F$ vanishes. The positive-determinant matrix domain is connected, as follows from polar decomposition and the connectedness of $SO(3)$ and of the positive-definite cone. Thus $W(QF)-W(F)=c(Q)$, independent of $F$. Iterating givesThe assumed lower bound on $W$ forces $c(Q)=0$: either nonzero sign would tend to minus infinity for one direction of integer $n$. This proves the required equivalence and explains the use of the lower-bound hypothesis.

In the frame-indifferent setting, isotropy means invariance under reference rotations as well: $W(FQ)=W(F)$ for $Q\in SO(3)$. The distinction matters: left rotations encode changes of observer, and right rotations encode material symmetry.

An orientation-preserving singular value decomposition is $F=R\mathrm{diag}(v_1,v_2,v_3)S$, with $R,S\in SO(3)$ and $v_i>0$. Material frame indifference and isotropy remove $R,S$, leaving a function $\Phi$ of the stretches. Permuting the stretches using signed permutation rotations does not change the value, so $\Phi$ is symmetric. Conversely a symmetric stretch function is unchanged by both left and right rotations, in particular is isotropic.

The eigenvalues of $B=F{F}^T$ are $v_i^2$. Its three principal invariants areThey determine the unordered positive eigenvalues as the roots of ${\lambda }^3-I{\lambda }^2+II\lambda -III=0$. Therefore a symmetric stretch function defines a well-defined $h(I,II,III)$ on the admissible invariant triples. Conversely a function of these invariants is a symmetric stretch function, proving the equivalence of all three descriptions.

The printed domain of $\Phi$ must be $(0,\infty {)}^3$, since it takes three arguments. Also isotropy alone as right invariance would not imply the spectral representation without material frame indifference: $\mathrm{tr}(AF{F}^T)$ for a fixed nonscalar symmetric positive-definite $A$ is right invariant but not generally objective. The equivalence here uses the frame-indifferent constitutive setting.

For the specific energy, Newton identities for the three eigenvalues gives ${\sum }_i{v}_i^6=I^3-3III+3III$, while $v_1{v}_2{v}_3=J=\sqrt{III}$. ThusTo differentiate it, useThese follow from $I=|F{|}^2$, the displayed trace formula for $II$, and $III=J^2$. The three partial derivatives of $h$ are $h_I=\alpha +3\beta I^2-3\beta II$, $h_{II}=-3\beta I$ and $h_{III}=3\beta -\gamma II{I}^{-3/2}/2$. The chain rule therefore givesFinally $-6\beta II=-3\beta (\mathrm{tr}B{)}^2+3\beta \mathrm{tr}(B^2)$, and $BF=F{F}^{T}F$. This is precisely the requested expression, with the coefficient of $F$ separated from the $BF$ and cofactor terms. The positive sign of the $6\beta J$ term comes from the $3\beta III$ contribution in the energy.

Write the first strain invariant in components of the deformation gradient:Here lower-case and upper-case indices refer to the current configuration and reference configuration, respectively. Differentiating gives $\partial I_1/\partial F_{iI}=2F_{iI}$. The chain rule therefore givesSubstitution into the given hyperelastic material law, with its indicated matrix transpose, yieldsThe left Cauchy-Green deformation tensor is $\mathbf{B}=\mathbf{F}{\mathbf{F}}^T$. The scalar $p$ is the Lagrange multiplier enforcing incompressibility, rather than a function fixed by the strain energy density alone. The identity matrix $1$ makes this an isotropic constraint reaction in hyperelastic stress.

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 $t_{\theta }-t_r=\lambda h^{\prime }(\lambda )/2$.