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Subtract the two density operators. The difference is Hermitian with zero diagonal and off-diagonal entry $\alpha \overline{\beta }(1-\overline{c})$. A matrix $\left(\begin{array}{cc}0& a\\ \overline{a}& 0\end{array}\right)$ has eigenvalues $\pm |a|$, because its square is $|a{|}^{2}I$. Its two singular values are therefore both $|a|$. By the definition of the trace norm,For the intended real-overlap phase-flip channel, $1-c=2p$, givingThis vanishes for a computational basis state, since a phase flip then changes only an unobservable global phase. Its largest value at fixed $p$ is $p$, attained when $|\alpha |=|\beta |=1/\sqrt{2}$. The general formula remains correct even when the probabilistic premise of part 1 fails.

There is a necessary distinction between a failure bound averaged over the unknown state, a worst-case bound, and a separate bound for each candidate state. The last interpretation is false: a forger can discard the inputs and always prepare $|{\psi }_0{⟩}^{\otimes N}$. If the actual state is $|{\psi }_0⟩$, every authenticity test passes with probability one. The meaningful uniform statement concerns worst-case failure, or average failure under any fixed prior giving both states positive probability. The PDF does not specify that prior; the argument below gives an explicit bound for either interpretation.

Put $c=|⟨{\psi }_0|{\psi }_1⟩|\in (0,1)$. Any unconditional strategy producing $N$ output qubits, including ancillas, adaptive measurement in quantum measurements and correlated outputs, is a quantum channel $\mathcal{E}$. Its inputs and ideal outputs are the density operatorsThe projective measurements on the different banknotes commute, and their all-pass effect is exactly $P_i$. Therefore the probability of at least one rejection, conditional on input $i$, isThis remains true for entangled outputs; no independence of their individual test outcomes is assumed.

Use trace distance $D(\rho ,\sigma )=\frac{1}{2}\Vert \rho -\sigma {\Vert }_1$. The pure-state trace distance and fidelity identity givesFor completeness, two pure-state projectors differ by a matrix with nonzero eigenvalues $\pm \sqrt{1-|⟨u|v⟩{|}^2}$, which proves this identity. Also the pure-target upper bound on trace distance gives $D({\rho }_i,P_i)\le \sqrt{e_i}$. To see the mixed-state step, write ${\rho }_i={\sum }_j{q}_j|u_j⟩⟨u_j|$ and use convexity of the trace norm followed by the Cauchy-Schwarz inequality:Finally, trace-distance contraction under quantum channels gives $D({\rho }_0,{\rho }_1)\le D({\sigma }_0,{\sigma }_1)$. This follows from $D(\rho ,\sigma )={\max }_{0\le E\le I}\mathrm{Tr}[E(\rho -\sigma )]$: the adjoint channel is positive and unital, so its image of any effect $E$ is still an effect.

The triangle inequality now yieldsThus the cloning failure bound from trace-distance contraction isand in particular $\max (e_0,e_1)\ge {\Delta }^2/4$. For positive priors ${\pi }_0+{\pi }_1=1$, another Cauchy-Schwarz inequality givesso the average rejection probability satisfiesFor equal priors, one may take $p_0={\Delta }^2/8$; the same choice is a strict lower bound on worst-case rejection. This proves a quantitative version of the no-cloning theorem for the bank's all-pass test. It concerns an unconditional attempt: heralded success cannot be postselected while omitting failed attempts from the accounting.

Coordinate reflection in $x_i$ requires the $i$th component to be odd in $x_i$, and reflections in the other coordinates require it to be even in those coordinates. Thus no constant or quadratic terms are permitted. The only cubic monomials in component $i$ are $x_i^3$ and $x_i{x}_j^2$ for $j\ne i$. Coordinate permutations equate their coefficients in all components. After using the transverse linear eigenvalue as parameter, the normal form isThe remainder starts at fifth spatial order, together with parameter-dependent corrections such as $\mu |x{|}^3$. This is the cubic equivariants of the full cube symmetry group form. On a branch with $m=1,2,3$ equal nonzero absolute coordinates $s$, respectively,Hence the three leading branch amplitudes are $s^2=-\mu /a$, $-\mu /(a+b)$ and $-\mu /(a+2b)$. Their generic nonzero denominators determine on which side the branches occur; existence alone does not make them stable.

Set $g(x)=x^2+x^3/3$. Equilibria satisfy ${\mu }^2=g(x)$, and their scalar linear eigenvalue isIt vanishes only at $x=0$ or $x=-2$. Substitution in the equilibrium relation gives the three nonhyperbolic parameter-state pointsThe branches with $x<-2$ and $x>0$ are stable, while $-2<x<0$ is unstable. For $0<|\mu |<2/\sqrt{3}$ there are three equilibria; beyond the two outer saddle-node bifurcations there is just the positive stable one. At $\mu =0$ there is also the distinct stable equilibrium $x=-3$.

Near the crossing, use the smooth coordinate $y=x\sqrt{1+x/3}$, whose derivative is positive near zero. The equation becomes $\dot{y}=p(y)({\mu }^2-y^2)$ with $p>0$. A positive time rescaling removes $p$, and the parameter-dependent state shift $u=y-\mu$ givesThus the crossing at $\mu =0$ is a transcritical bifurcation, with parameter $-2\mu$. The two smooth branches $y=\pm \mu$ exchange dynamical stability; labeling the stable branch by $|\mu |$ instead would obscure that exchange. The parameter-dependent coordinate shift in a transcritical bifurcation explains why the original coordinate need not display a branch identically equal to zero.

Put $w=\dot{v}$. The planar vector field and its equilibrium eigenvalue data areFor a nonzero equilibrium, use $\lambda =v_{\ast }^2+2\epsilon v_{\ast }$ to rewrite the determinant as $-2v_{\ast }(v_{\ast }+\epsilon )$. These formulas determine the local bifurcations of the imperfect soft Duffing-van der Pol oscillator directly.

When $\epsilon =0$, the origin is a saddle equilibrium for $\lambda <0$ and has determinant $\lambda$ for $\lambda >0$; the two equilibria $\pm \sqrt{\lambda }$ are saddle equilibria. They meet the origin in a symmetry-protected pitchfork bifurcation at $\lambda =0$ for $\kappa \ne 0$. At $\kappa =0,\lambda >0$ the origin undergoes a supercritical Hopf bifurcation. To determine criticality, take $v=r\cos (\sqrt{\lambda }t)+\cdots$: averaging $\dot{H}=(\kappa -v^2)w^2$ gives $\dot{r}=\kappa r/2-r^3/8+\cdots$, so the stable small limit cycle has $r^2=4\kappa +\cdots$.

The potential energy of the conservative part and the exact energy balance areAt zero imperfection the stable limit cycle grows to a heteroclinic cycle connecting the two outer saddle equilibria. Near the double-zero point its leading connection curve can be calculated, rather than inferred just from symmetry. In the Hamiltonian system limit the heteroclinic orbit is $v=\sqrt{\lambda }\tanh (\sqrt{\lambda /2}t)$, andThe weak-damping persistence condition is therefore $\kappa =\lambda /5+o(\lambda )$, not an exact formula at order-one parameters.

Now take $\epsilon >0$; negative imperfection reflects the phase portrait. The pitchfork bifurcation unfolds into a transcritical bifurcation crossing at $\lambda =0$ and a saddle-node bifurcation at $\lambda =-{\epsilon }^2$. The latter occurs at $v=-\epsilon$. Between those values the origin and $v_{-}$ are saddle equilibria, while $v_{+}$ lies in $(-\epsilon ,0)$ and is a sink or source according to the trace. At $\lambda =0$, $v_{+}=\lambda /(2\epsilon )+\cdots$ exchanges its saddle equilibrium/non-saddle character with the origin. The double-zero points are now at $(\lambda ,\kappa )=(0,0)$ and $(-{\epsilon }^2,{\epsilon }^2)$.

The original Hopf bifurcation line $\kappa =0,\lambda >0$ remains, still supercritical. A second Hopf bifurcation curve occurs on the intermediate nonzero equilibrium:It is important not to assign the same criticality to its entire length. Write $s=\sqrt{{\epsilon }^2+\lambda }$, so $q=-\epsilon +s$. Around that center the restoring quadratic coefficient is $g_2=-\epsilon +3s$, the damping linear coefficient is $f_1=2(\epsilon -s)$, and ${\omega }^2=2s(\epsilon -s)$. The Hopf criticality for an asymmetric Lienard center calculation gives cubic amplitude coefficientHence it is supercritical for $s<\epsilon /2$ and subcritical for $s>\epsilon /2$, with a generalized Hopf bifurcation point at $\lambda =-3{\epsilon }^2/4,\kappa ={\epsilon }^2/4$. Higher-order terms are needed exactly there. For small imperfection the leading quintic coefficient is positive: writing $q_1=v-q$ and using the coordinate that normalizes the potential energy, $y=q_1\sqrt{1-a{q}_1-b{q}_1^2}$, with $a=2g_2/(3{\omega }^2)$ and $b=1/(2{\omega }^2)$, the coefficient of $y^4$ in $(f_1{q}_1-q_1^2)d{q}_1/dy$ at vanishing cubic coefficient is $10b/3>0$. Weak-damping averaging therefore supplies the nearby saddle-node bifurcation of stable and unstable limit cycles in the generalized Hopf bifurcation unfolding. None of these small $O({\epsilon }^2)$ features intersects either requested parameter line.

Globally, the two saddle equilibrium barriers are no longer equal. Their leading energy difference is $H(v_{+},0)-H(v_{-},0)=-4\epsilon {\lambda }^{3/2}/3+\cdots$. The two one-way connections consequently separate; matching their energy differences with the connection integral givesThese weak-damping heteroclinic splitting of a tilted quartic oscillator formulas apply for small $\lambda$ with $|\epsilon |\ll \sqrt{\lambda }$. Between the separated connection curves, the central stable limit cycle terminates in a homoclinic orbit to the lower-barrier saddle equilibrium, $v_{+}$ for positive imperfection. The simultaneous symmetric heteroclinic-cycle destruction is therefore replaced by two basin-changing one-way connections and a distinct homoclinic orbit loss of the limit cycle. Its period diverges at that loss.

Along $\kappa +\lambda =1$, decrease $\lambda$ from above one. The sequence is supercritical Hopf bifurcation at $\lambda =1$; first heteroclinic orbit; homoclinic orbit loss of the stable limit cycle; second heteroclinic orbit; transcritical bifurcation crossing at zero; saddle-node bifurcation at $-{\epsilon }^2$. The connection locations are global, not obtained by treating the Melnikov approximation as exact. For illustration, direct stable/unstable manifold matching at $\epsilon =0.04$ places them at approximately $\lambda =0.882,0.858,0.783$, respectively. The diagram below gives the qualitative ordering without prescribing those example values for arbitrary imperfection.

Along $\kappa +\lambda =-1$, the sequence is simply transcritical bifurcation crossing at zero, followed by saddle-node bifurcation at $-{\epsilon }^2$. Before the latter, $\kappa <0$ in the relevant region, so $H$ decreases strictly on nonstationary trajectories and there is no periodic orbit or saddle equilibrium loop. After it only a saddle equilibrium remains, which cannot be enclosed by a periodic orbit by the Poincare-index obstruction to a periodic orbit. The origin does not undergo a Hopf bifurcation at $\lambda =-1$, since its determinant there is negative.

Write $s={\alpha }^2+{\pi }^2$ and use modes satisfying all boundary conditions,The quadratic temperature distortion $c$ is retained for the saturation calculation; it does not enter the linear onset. Substitution of the primary modes givesThe determinant vanishes whenThe trace is $-[s+\sigma (s+Q{\pi }^2/s)]<0$. Thus no conjugate eigenvalues can cross the imaginary axis away from zero, and the initial instability is stationary, not oscillatory. Indeed for $R\ge 0$ the discriminant is $[\sigma (s+Q{\pi }^2/s)-s{]}^2+4\sigma R{\alpha }^2/s>0$. This is quasistatic vertical-field magnetoconvection: there is no independent magnetic evolution mode that could produce a magnetic-thermal Hopf bifurcation instability.

For a fixed box the other allowed horizontal wavenumbers are $n\pi /L$, and the actual first threshold is the minimum of their corresponding $R_c$ values. Vertical index $m$ replaces ${\pi }^2$ by $m^2{\pi }^2$ and strictly increases the threshold at fixed horizontal wavenumber, so $m=1$ is selected. The displayed formula describes the prescribed fundamental roll; part (b) optimizes its width.

The primary vorticity Jacobian determinant vanishes because ${\nabla }^2\psi =-s\psi$. The quadratic temperature Jacobian determinant isThe mean temperature correction therefore obeys $\dot{c}=-4{\pi }^{2}c-\alpha \pi ab/2$. Its Jacobian determinant with the primary streamfunction isProjecting on the retained primary modes gives, with $D=s+Q{\pi }^2/s$,For a nonzero steady roll, $b=(s^2+Q{\pi }^2)a/(R\alpha )$ and $c=-\alpha ab/(8\pi )$. Substitution in the second equation gives, within this truncation,The positive denominator ensures branches only on the supercritical side for every $Q\ge 0$. To verify dynamical stability as well as existence, the critical right eigenvector is $(1,\alpha /s)$; normalizing the left eigenvector gives linear eigenvalue derivativeQuadratic slaving gives $c=-{\alpha }^2{a}^2/(8\pi s)$, and projection of its feedback gives the cubic amplitude equationIts nonzero branches are stable in the center direction, while all other eigenvalues remain negative at a simple onset. Reflection about the mid-width sends $a\to -a$, so the bifurcation is a supercritical pitchfork bifurcation for all $Q$, in the generic case of a single critical roll mode.

The cubic saturation of quasistatic magnetoconvection result is identical in Fourier truncation and perturbation theory at the orders needed here. At quadratic order the only forced mode is precisely the retained horizontally uniform $\sin 2\pi z$ temperature mode. At cubic order its feedback projects on the primary mode with the same coefficient. The generated third vertical harmonic is orthogonal to that primary projection and contributes only to higher-order corrections. Thus both methods give the same linear threshold and cubic solvability condition; the truncated finite-amplitude branch is not an exact solution of the full PDE at all amplitudes.

Put $r=k/h^2$. The update isUnder $k\le h^2/(2\beta )$, all three weights are nonnegative and sum to one. With the imposed zero Dirichlet boundary conditions, this is a convex combination including any boundary values. Thus the monotone half-grid diffusion update givesDifferences of two computed solutions obey the same estimate, proving numerical stability in the discrete maximum norm, uniformly in $h,k,n$.

There is also an L2 norm proof. The diffusion matrix $L_h$ is symmetric and satisfiesSince $(v_{m+1}-v_m{)}^2\le 2(v_{m+1}^2+v_m^2)$, every eigenvalue lies in $[-4\beta /h^2,0]$. Hence the eigenvalues of $I+k{L}_h$ lie in $[-1,1]$ under the same step restriction, and the spectral theorem for real symmetric matrices gives an L2 norm contraction. These stability estimates are valid for the printed recurrence, although they cannot repair its inconsistency with the printed advection equation.

A constant-step linear multistep method has the formwith ${\alpha }_k\ne 0$. The characteristic polynomials of a linear multistep method encode both accuracy and error propagation. The method has order of a numerical method $p$ when insertion of a sufficiently smooth exact solution leaves an unnormalized local truncation error $O(h^{p+1})$. The exponential-symbol order criterion for a multistep method expresses this as $\rho (e^z)-z\sigma (e^z)=O(z^{p+1})$. In particular, first-order consistency of a numerical method requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Accuracy alone is insufficient. For $f=0$, perturbations obey the recurrence $\rho (E)e_n=0$. The root condition for a multistep method requires every root of a polynomial $\rho$ to lie in the closed unit disk, with each unit-circle root simple. This is zero-stability. A root outside the unit disk causes geometric amplification as $n\sim T/h$ increases; a repeated unit-circle root permits polynomial growth in $n$. Either can destroy convergence even for starting perturbations tending to zero.

The Dahlquist equivalence theorem states that, for a fixed-coefficient method and a well-posed ordinary differential equation with a suitably Lipschitz continuous right-hand side, consistency of a numerical method and zero-stability are equivalent to convergence for all convergent starting data. If the starting errors are $O(h^p)$ and the exact solution is sufficiently smooth, the global error is $O(h^p)$ on a fixed time interval. A stability estimate bounds the propagated sum of $O(h^{p+1})$ local defects over $O(1/h)$ steps; the discrete Gronwall inequality controls the dependence of $f$ on the accumulated error. Implicit methods also require the locally consistent, uniquely solvable update branch for sufficiently small $h$.

For example, the explicit Euler method has order one and $\rho (\zeta )=\zeta -1$, satisfying the root condition for a multistep method. The two-step Adams-Bashforth method has order two and $\rho (\zeta )=\zeta (\zeta -1)$, again satisfying it. The trapezoidal rule has order two with one step. The two-step Simpson multistep method, $Y_{n+2}-Y_n=h(f_{n+2}+4f_{n+1}+f_n)/3$, has order four and the simple roots $1,-1$: it is zero-stable, although its parasitic amplification root can spoil long-time decay. These examples distinguish finite-time numerical convergence from absolute stability.

The first Dahlquist barrier gives the maximum possible order of a convergent ordinary real $k$-step method:These bounds assume only first-derivative evaluations, fixed coefficients and zero-stability; the sixth-order multiderivative multistep method in solution is outside that class.

To obtain arbitrarily high convergent order within the ordinary class, increase the number of steps. The Adams-Bashforth method integrates a polynomial interpolation of $k$ past derivative values and has order $k$. The Adams–Moulton method includes the new endpoint and has order $k+1$ for $k$ steps. In both families $\rho (\zeta )={\zeta }^{k-1}(\zeta -1)$, so the root condition for a multistep method holds for every fixed $k$. Starting values can be generated by a Runge-Kutta method of sufficient accuracy. The backward differentiation formula instead differentiates an interpolating polynomial; its order-$k$ members are zero-stable only for $1\le k\le 6$.

Finally, high order and favorable stiff stability are separate design goals. By the Second Dahlquist barrier, an ordinary A-stable linear multistep method has order at most two. One may accept a smaller linear stability domain, use suitable backward differentiation formula members for stiff problems, or switch to high-order implicit Runge-Kutta methods or multiderivative multistep methods. A convergent high-order formula still needs time steps appropriate to its stability properties and the problem's eigenvalues.

A spatial finite difference method and time discretization for a linear evolution problem give a recurrence $U^{n+1}=B_{h,k}{U}^n$. To obtain numerical stability on $0\le nk\le T$, one needs a bound $\Vert B_{h,k}^n\Vert \le C_T$ uniform in the permitted grid sizes and time steps. More generally a growth estimate $\Vert B_{h,k}^n\Vert \le C{e}^{cnk}$ is sufficient. The finite-time stability versus power boundedness distinction matters: growth per step of $1+O(k)$ can be compatible with the latter finite-time estimate, whereas unrestricted geometric growth at a fixed Courant number is not.

The eigenvalue method diagonalizes $B_{h,k}$ into modes. If $B_{h,k}$ is a normal matrix, the spectral theorem gives $\Vert B_{h,k}^n{\Vert }_2={\max }_j|{\mu }_j{|}^n$. Thus $|{\mu }_j|\le 1$ gives contraction, and $|{\mu }_j|\le e^{ck}$ gives the preceding growth estimate. A diagonalizable matrix $B=V\Lambda V^{-1}$ has $\Vert B^n{\Vert }_2\le \Vert V{\Vert }_2\Vert V^{-1}{\Vert }_2{\max }_j|{\mu }_j{|}^n$, so its condition number must remain bounded uniformly as the mesh changes. A stable-looking list of eigenvalues alone is insufficient.

Indeed,Both eigenvalues are one, but the Jordan block gives growth proportional to $n\sim T/k$, violating uniform finite-time numerical stability. A nonnormal Fourier amplification matrix can similarly exhibit large amplification even with its eigenvalues inside the unit disk. Eigenvector conditioning and the full powers of the amplification operator are essential.

For constant-coefficient schemes on a periodic or infinite grid, a Fourier transform diagonalizes scalar translation-invariant stencils. This yields von Neumann stability analysis using the Fourier amplification symbol $G(\theta )$. The discrete Parseval identity turns a bound on $G(\theta {)}^n$ into a discrete L2 norm estimate. For systems, the symbol is a matrix; a uniform power bound for matrix Fourier symbols is required, rather than a bound on its spectral radius alone.

For the heat equation, a central finite difference in space and the explicit Euler method in time giveThe scalar condition $|G|\le 1$ for every frequency is exactly $0\le r\le 1/2$. With zero Dirichlet boundary conditions, the spatial matrix is symmetric and diagonalized by sine modes, leading to the same uniform sufficient restriction. By contrast, the Backward Euler diffusion scheme has $G(\theta )=[1+4r{\sin }^2(\theta /2){]}^{-1}$ and is stable for every $r\ge 0$.

For the advection equation $u_t+a{u}_x=0$, $a>0$, centered space with the explicit Euler method gives $G=1-i\nu \sin \theta$, $\nu =ak/h$, and $|G{|}^2=1+{\nu }^2{\sin }^2\theta$. At fixed positive Courant number, repeated powers are unbounded as the mesh is refined. The upwind finite difference scheme instead haswhich gives contraction precisely for $0\le \nu \le 1$. This comparison shows why the spatial stencil and time method must be analyzed together.

Boundary treatment can destroy the normality or mode decomposition available on a periodic grid, so a finite-interval scheme needs its own uniform operator estimate. Multilevel time schemes likewise require the root condition for a multistep method and uniform bounds for the associated amplification matrices; simple scalar root tests do not establish these bounds for arbitrary systems. Once consistency of a numerical method, a well-posed linear evolution problem and the required numerical stability estimate hold, the Lax equivalence theorem gives numerical convergence. Eigenvalues provide a powerful route to that estimate when the accompanying eigenvector and boundary analysis is justified.

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 $T/\mu =2$.

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.

For a uniform material and an affine static base deformation, the deformation gradient, incremental elastic moduli and reference mass density are constant. With zero incremental body force, the equation reduces to $c_{\alpha i\beta j}{v}_{j,\beta \alpha }={\rho }_0{\ddot{v}}_i$. Let $\nu$ be a unit material propagation direction, $m$ a constant polarization and $s=t-{\nu }_{\alpha }{\xi }_{\alpha }/c$. Direct differentiation of the plane wave givesThus every twice differentiable profile $f$ satisfies the incremental equation ifThis is the acoustic tensor eigenvalue problem. The acoustic tensor is a real symmetric matrix: interchange the index pairs in the incremental elastic moduli and then interchange the dummy material indices to obtain $Q_{ji}=Q_{ij}$. Its eigenvectors give the displacement polarizations. A positive eigenvalue gives a real nonzero speed $c$; arbitrary-profile propagation imposes precisely this algebraic condition. Boundary conditions determine which combinations of these local plane waves are allowed in a finite body.

For a static base state under dead loading, put $v_i=u_i(\xi )e^{i\omega t}$ and take zero incremental body force and traction. The admissible squared frequencies are the eigenvalues of the weighted boundary value problemMultiply by $\overline{u_i}$ and integrate over the reference body. The boundary term in integration by parts vanishes, giving the elastic normal-mode energy identityFor a nonzero mode and positive mass density, the denominator is real and positive. The numerator is real by major symmetry of the incremental elastic moduli. Explicitly, if $u=p+iq$ with real $p,q$, the two imaginary cross terms cancel, leaving the sum of the quadratic energies of $\nabla p$ and $\nabla q$. Consequently every admissible ${\omega }^2$ is real. A negative value, when allowed by indefinite moduli, corresponds to exponential growth or decay rather than an oscillation with real frequency.

If the moduli quadratic form is strictly positive on every nonzero real matrix, that same decomposition shows that the numerator is positive whenever $\nabla u$ is nonzero. Thus ${\omega }^2>0$ for every nontranslation mode. Under the usual assumptions of a bounded connected body, regular coefficients and uniform positivity, the Rayleigh quotient defines a self-adjoint operator with discrete positive squared frequencies on the mass-weighted mean-zero displacement space. Its eigenfunctions give the nontrivial small harmonic oscillations.

There is an unavoidable zero-frequency qualification for the pure traction problem as printed: $u_i=b_i$, for any nonzero constant vector $b$, has zero gradient, zero incremental nominal stress and ${\omega }^2=0$. These three constant translations obey all the homogeneous incremental boundary conditions. Under the strict quadratic-form hypothesis they are the entire zero eigenspace on a connected body. Therefore the unrestricted conclusion isStrict positivity for every nonzero displacement requires excluding these modes, for example by imposing ${\int }_{{\Omega }_0}{\rho }_{0}udV=0$, or by adding a displacement constraint. If harmonic perturbations are defined to have nonzero frequency from the outset, the requested positivity holds for all those perturbations.

At a fixed material point, freeze the incremental elastic moduli at their local values and fix a unit propagation direction $\nu$. For an arbitrary nonzero polarization $m$, the rank-one matrix $a_{i\alpha }=m_i{\nu }_{\alpha }$ annihilates the added quadratic minor null Lagrangian, sinceThe assumed positivity therefore givesThus the acoustic tensor is positive-definite; this is strong ellipticity in elasticity. Major symmetry also makes it a real symmetric matrix. By the spectral theorem for real symmetric matrices, it has three real positive eigenvalues $q_1,q_2,q_3$ and an orthonormal basis of real polarizations $m^{(1)},m^{(2)},m^{(3)}$. ThereforeEach polarization gives a real plane wave profile through the equation of part (a), in either sign of the propagation speed. Repeated eigenvalues still admit three independent polarizations; the claim is not that the three speeds must be distinct. For an inhomogeneous base deformation these are local, frozen-coefficient wave modes, rather than exact plane waves throughout the entire body.