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For a surface with unit normal $\mathbf{n}$, traction is $t_i={\tau }_{ij}{n}_j$. Thus ${\tau }_{ij}$ is the $i$th force-per-area component on a plane whose positive normal is ${\mathbf{e}}_j$. Diagonal entries are normal stresses and off-diagonal entries are shear stresses, with tensile normal stress positive in this convention.

The symmetric strain has six independent components and so does the symmetric stress. A general linear map between these two six-dimensional spaces therefore hasEquivalently choose the minor symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$. Components antisymmetric in the strain indices have no observable effect and are omitted from this representative.

A further distinction is important: if a strain energy density exists, $C_{ijkl}={\partial }^{2}W/(\partial e_{ij}\partial e_{kl})$ has the major symmetry $C_{ijkl}=C_{klij}$. Then the six-by-six map is symmetric and has $6\cdot 7/2=21$ independent coefficients. Thus 36 follows from the two stated tensor symmetries alone; 21 additionally uses the energy/reciprocity assumption, as in the variational setting later in the paper. These are the minor and major symmetries of linear elasticity.

The perturbation preserves the prescribed displacement because $\mathbf{w}=0$ on the boundary. Strain varies as $e_{ij}(\mathbf{u}+\alpha \mathbf{w})=e_{ij}(\mathbf{u})+\alpha (w_{i,j}+w_{j,i})/2$. Taylor expansion of the strain energy density givesSymmetry of the stress makes its contraction with the two symmetric terms equal, henceA minimizer must have zero first variation for both signs of $\alpha$. Integrating by parts and using the zero boundary variation yields ${\int }_B(-{\partial }_j{\tau }_{ij}-\rho g_i)w_{i}dV=0$. The fundamental lemma of the calculus of variations, applied to arbitrary compactly supported variations, givesThis is the static momentum equation. Fixed-displacement data remove the surface variation; prescribed tractions would instead appear through boundary work in the potential energy.

Write $W_i=\partial W/\partial {\alpha }_i$, evaluated at $({\alpha }^{-1/2},{\alpha }^{-1/2},\alpha )$. Applying the chain rule to the reduced strain energy density givesMultiply by $\alpha$ and use ${\alpha }_1={\alpha }_2={\alpha }^{-1/2}$:The transverse symmetry specified in the hint has $W_1=W_2$. ThusThe lateral faces have zero traction, so their diagonal Cauchy stress tensor components satisfy $t_1=t_2=0$. The hydrostatic pressure enforcing elastic incompressibility is therefore $p={\alpha }_1{W}_1={\alpha }_2{W}_2$, and the axial component simplifies toThis elimination of the Lagrange multiplier $p$ is what makes the reduced strain energy density useful for the spring balance.

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.

Interpret derivatives on the space of symmetric strains, with inner product $A:B=A_{ij}{B}_{ij}$. Assume a differentiable strain energy density with local stress $\sigma =D_{\epsilon }W$, and a differentiable stable equilibrium branch under the prescribed affine boundary displacement. Thus $u_E(x)=Ex+v_E(x)$ with $v_E=0$ on $\partial \Omega$; part(a) ensures that the mean strain is exactly $E$.

Vary $E$ in any symmetric direction $H$. The induced displacement variation has the form $\dot{u}=Hx+\dot{v}$, where $\dot{v}$ vanishes on the boundary. Differentiating the effective elastic energy and using static elastic equilibrium givesThe last term is zero by integration by parts, since the body force is zero and $\dot{v}$ has zero boundary trace. As $H$ is arbitrary,This is a work-conjugacy statement on symmetric tensors; treating off-diagonal entries as independent scalar coordinates without their contraction weights would introduce erroneous factors of two. If the minimized effective elastic energy has a nondifferentiable point, the corresponding formulation uses a subgradient rather than a claimed classical derivative.

In the Voigt elastic approximation, every local strain is $E$. The corresponding local stress is ${\sigma }_V(x)=D_{\epsilon }W(E,x)$. Differentiating the averaged strain energy density under the integral givesThe uniform strain need not generate an equilibrated local stress; this calculation concerns the stated approximation, not an assertion that it solves the heterogeneous boundary problem.

For the Reuss elastic approximation, use the convex conjugateUnder differentiable strictly convex duality, the local strain produced by uniform stress $S$ is $e_R(x)=D_S{W}^{\ast }(S,x)$. Set $F(S)=⟨W^{\ast }(S,x)⟩$. Averaging the local strains gives $E=DF(S)$. At the maximizing stress in $W_R(E)=F^{\ast }(E)$, the stationarity condition is precisely $DF(S)=E$. ConsequentlyThus the derivative of each approximate energy is its own mean stress, includingFor positive linear elasticity, $W=e:C(x)e/2$ and $W^{\ast }=S:C(x{)}^{-1}S/2$, giving the useful check $C_V=⟨C⟩$ and $C_R=⟨C^{-1}{⟩}^{-1}$. More general convex duality gives the same conjugacy relation with subgradients when differentiability or strict convexity is absent.

Let $u$ be the actual displacement field, let $u^{\prime }$ be an admissible trial displacement, and set $v=u^{\prime }-u$. On the displacement-prescribed part of the boundary, $v=0$. With actual stress $\sigma =D_{\epsilon }W(\epsilon )$, the supporting-plane inequality for the convex strain energy density givesBoth $\sigma$ and ${\sigma }_0$ satisfy static elastic equilibrium with the same body force, so $\nabla \cdot (\sigma -{\sigma }_0)=0$. On the traction-prescribed boundary their tractions agree. Since ${\epsilon }^{\prime }-\epsilon =\mathrm{sym}\nabla v$, integration by parts givesSubtracting the prescribed external-work representative proves the minimum potential energy principle in elasticity:Strict convexity gives uniqueness of the minimizing strain, although rigid displacements can remain when the boundary conditions allow them.

For the comparison bound, put $F(e,x)=W(e,x)-W_0(e)$ and use the lower conjugate, with its infimum sign convention,Its defining inequality is $F_{\ast }(\tau ,x)\le \tau :e-F(e,x)$, hence $W(e,x)\le W_0(e)+\tau :e-F_{\ast }(\tau ,x)$ for every $e$. Integrate this pointwise majorant for each compatible strain, then take the infimum over the same admissible set. This provesNo convexity of $F$ is used for this inequality. If $F_{\ast }(\tau ,x)=-\infty$, it yields only the extended-real upper bound $+\infty$. For example, $W_0=0$ and $W(e)=|e{|}^2/2$ give $F_{\ast }(\tau )=-\infty$ for every finite $\tau$. A useful finite bound requires a suitable comparison energy; choosing a sufficiently stiff quadratic comparison makes the usual excess energies concave. The derivative used below additionally presupposes finite differentiable duals, or an appropriate supergradient formulation.

Now normalize $|\Omega |=1$, take positive constant $C^0$, and write $F_r(e)=W_r(e)-e:C^{0}e/2$. For phasewise constant elastic polarization $\tau ={\sum }_r{\chi }_r{\tau }^r$, its lower conjugate contribution is constant within each phase. The remaining minimization isWrite the minimizing strain as $\hat{e}=E+\eta$, where $\eta$ comes from a zero-boundary displacement field. The static elastic equilibrium equation is $\nabla \cdot (C^0\eta +\tau )=0$, so the supplied elastic strain Green operator gives $\eta =-\Gamma \tau$. Its zero mean removes the cross term with the constant $E$. Using its reciprocity and projection property $\Gamma C^0\Gamma =\Gamma$, we obtainHere the brackets in the last two terms denote the spatial integral and tensor contraction, not just a phase average.

Let $c_r={\int }_{\Omega }{\chi }_r$. The zero output mean and reciprocity of $\Gamma$ imply that it annihilates constant elastic polarizations on either side. Thus subtracting $c_r$ and $c_s$ from the phase indicators does not change their bilinear integral. Define the fourth-order block tensorsConsequently $A_{rs}=A_{sr}^T$, ${\sum }_s{A}_{rs}=0$, andThese identities follow from reciprocity and ${\sum }_s{\chi }_s={\sum }_s{c}_s=1$; no phase independence assumption is needed. The phase polarization comparison bound, valid for every polarization with finite dual terms, is thereforeIn particular, it can be optimized over the phase elastic polarizations without changing the upper-bound direction.

To derive the final displayed form, impose the stationarity equations in the question. For $c_r>0$, put $e^r=D{F}_{r\ast }({\tau }^r)$; these are dual strains, initially distinct from the comparison solution's phase averages. Reciprocity of $A$ givesThus stationarity is exactlyThe comparison solution has phase meanso the dual strains become its phase means at stationarity. Multiplying the stationarity equations by ${\tau }^r$ and summing gives ${\sum }_{r,s}{\tau }^r:A_{rs}{\tau }^s={\sum }_r{c}_r{\tau }^r:(E-e^r)$.

For clarity, the concave biconjugate here uses another infimum: $F_{r\ast \ast }(e)={\inf }_s\{s:e-F_{r\ast }(s)\}$. The lower conjugate $F_{r\ast }$ is concave, since it is an infimum of affine functions. At a differentiable finite point, its supporting-plane inequality shows that $s={\tau }^r$ minimizes this last expression when $e=e^r$. HenceSubstitute both identities into $\mathcal{B}$ to obtainThis is the requested final bound with the accompanying phase equations. It remains valid at any well-defined stationary choice; asserting existence of such a differentiable choice for arbitrary $W_0$ would require extra hypotheses. Without stationarity, the earlier $\mathcal{B}(E,\tau )$ is the correct general bound. In the concave closed excess-energy case $F_{r\ast \ast }=F_r$; for general excess energies replacing the concave biconjugate by $F_r$ without proof would reverse the needed majorant logic.

For an orthogonal change of Cartesian coordinates $x_i^{\prime }=Q_{ij}{x}_j$, an ordinary rank-$n$ tensor transforms byAn isotropic tensor is invariant under such rotations. Orthogonality gives $Q_{ip}{Q}_{jq}{\delta }_{pq}={\delta }_{ij}$, so transforming any of the three products of two Kronecker deltas leaves it unchanged. Their linear combination with arbitrary scalar coefficients is therefore isotropic.

For completeness, these three products exhaust the rank-four isotropic tensors in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha =c_{1122}$, $\beta =c_{1212}$ and $\gamma =c_{1221}$. Rotating the first two axes by $\pi /4$ gives $X=\frac{1}{2}X+\frac{1}{2}(\alpha +\beta +\gamma )$, so $X=\alpha +\beta +\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity tensor.

Contracting with the symmetric strain givesThese are the Lamé parameters. Taking the trace determines the scalar part of the strain:Since ${\sum }_{ij}{e}_{ij}^2=3p^2+{\sum }_{ij}{d}_{ij}^2$, the stored strain energy density isA nonzero traceless test strain proves necessity of $\mu \ge 0$, and a pure scalar strain proves necessity of $3\lambda +2\mu \ge 0$. Conversely these two conditions make both terms nonnegative for every strain. Thus nonnegative isotropic elastic strain energy is equivalent to

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 $\lambda$.

Use orthonormal cylindrical bases in the reference and deformed configurations. The radial coordinate remains $\rho$, the angular coordinate becomes $\theta =\Theta +\alpha {\xi }_3$, and the axial coordinate is unchanged. Removing the rotation of the cylindrical basis, the deformation gradient isIt is a local simple shear with unit determinant. The squared principal stretches are the eigenvalues ofOne principal stretch is one. The remaining squared principal stretches have product one and sum $2+s^2$. Consequently they can be written ${\lambda }^2$ and ${\lambda }^{-2}$, whereFor negative $s$ their ordering reverses, which is immaterial to the strain energy density of an isotropic material.

Let the reference length be $L$. By material isotropy, the total elastic energy and end rotation areFor a quasistatic change, incompressibility makes the constraint reaction do no work. The velocity on an end section is its rigid rotational velocity; its traction power is the resultant torque times the angular velocity. Thus conservation of energy, or equivalently the torque and axial force of a twisted incompressible cylinder work identity, gives $Md{\Theta }_{\mathrm{end}}=dE$ andThe radius $a$ is fixed in this derivative. This argument computes the required end torque along the prescribed deformation family; proving that a suitable end-only traction distribution sustains that family is unnecessary here.

For a neo-Hookean solid, the identity $\lambda -{\lambda }^{-1}=s$ gives ${\lambda }^2+{\lambda }^{-2}-2=s^2$. ThereforeThis is the signed torque in the twist direction; its magnitude is $\pi \mu a^4|\alpha |/2$. For the positive twist convention the signed value equals the magnitude.

Use $z={\lambda }_z>0$. In cylindrical orthonormal bases the principal stretches are $r_{,R}$, $r/R$ and $z$. Incompressibility givesIntegrate from the inner surface to obtain the inflation and extension of an incompressible tube kinematicsIn particular $b^2=a^2+(b_0^2-a_0^2)/z$. The radial principal stretch is $1/(z\lambda )$, so the reference-volume strain energy density is $\hat{W}(\lambda ,z)$ with the three arguments $(1/(z\lambda ),\lambda ,z)$.

For reference length $L_0$, the stored energy is $L_0\mathcal{E}$, whereLet $N_{\mathrm{wall}}$ denote the total axial resultant acting on the material annulus. The work of the inner pressure on the cylindrical wall is $2\pi az{L}_{0}p\dot{a}$, and axial work is $N_{\mathrm{wall}}{L}_0\dot{z}$. Reversible virtual work therefore saysThe independent derivatives of the circumferential stretch areEquating the coefficients of $\dot{a}$ and $\dot{z}$ provesand the corresponding axial resultant isHere ${\hat{W}}_{,z}$ holds $\lambda$ fixed. If $N$ means the independently applied load on pressurized closed ends, the pressure also does axial end-cap work $\pi a^{2}p{L}_0\dot{z}$. Equivalently its complete work is $pd(\pi a^{2}z{L}_0)/dt$. In that convention,If $N$ includes the end-pressure resultant, it is $N_{\mathrm{wall}}$ itself. Stating this distinction is necessary because the ends are not specified; the pressure formula is the same in both conventions.

For the specified Mooney-Rivlin solid convention, differentiating the constrained strain energy density givesThusPut $c=z{a}^2-a_0^2$. If $c\ne 0$, $z{\lambda }^2-1=c/R^2$ yields $dR/R=-z\lambda d\lambda /(z{\lambda }^2-1)$. The factor cancels becauseThe limits are ${\lambda }_a=a/a_0$ and ${\lambda }_b=b/b_0$, so direct integration givesFor $c=0$, all circumferential stretches equal $z^{-1/2}$ and the original integrand vanishes, so this formula still holds, with $p=0$. This also verifies its continuous limiting value without dividing by zero.