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The printed normalization has its factors reversed. For the displayed map $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, the condition ${\sum }_k{A}_k{A}_k^{†}=I$ says $\Phi (I)=I$, which is unitality. Trace preservation instead requiresThese are distinct trace-preserving and unital Kraus conditions. To disprove the printed equivalence, take $0<\gamma <1$ andThey obey ${\sum }_j{B}_j{B}_j^{†}=I$, butThus their completely positive map is not trace preserving. They are the adjoints of the operators of an amplitude damping channel. That channel is trace preserving but sends $I$ to $\mathrm{diag}(1+\gamma ,1-\gamma )$, so it cannot satisfy the printed normalization in any Kraus representation either. The corrected characterization is proved next.

Work in finite-dimensional input and output Hilbert spaces, with a linear map $\Phi$ on operators. Suppose first that $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ and ${\sum }_k{A}_k^{†}{A}_k=I$. For every auxiliary dimension $m$ and every positive operator $W$ on the extended space,Each summand is positive because conjugation preserves positivity. This is exactly the definition of a completely positive map. Cyclicity of the trace givesso the map is also trace preserving.

Conversely, suppose $\Phi$ is completely positive and trace preserving, and let $|i⟩$ be an orthonormal input basis. Use the unnormalized entangled vector $|\Omega ⟩={\sum }_i|i⟩\otimes |i⟩$, with the reference factor first. Complete positivity makes the corresponding unnormalized Choi matrix positive:Its spectral decomposition can be written $J_{\Phi }={\sum }_k|v_k⟩⟨v_k|$, absorbing the nonnegative eigenvalues into the vectors. Define linear operators $A_k$ uniquely byExpanding the outer products and comparing their $(i,j)$ reference blocks yieldsThe matrix units span all operators, so linearity proves the Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ for every $X$. Finally, trace preservation givesfor all $X$. Taking matrix units as test operators shows that the matrix in parentheses is zero. This proves both directions of the corrected CPT characterization, including the normalization rather than assuming it.

For an input quantum state ensemble $\{p_x,{\rho }_x\}$, write ${\sigma }_x=\Phi ({\rho }_x)$ and $\bar{\sigma }={\sum }_x{p}_x{\sigma }_x$. Its output Holevo quantity isThe Holevo capacity is the optimized one-use quantitywhere the supremum ranges over finite input ensembles and $S$ is the Von Neumann entropy in bits.

For $n$ uses of a memoryless quantum channel, the map is ${\Phi }^{\otimes n}$. An $n$-use code consists of a finite message set ${\mathcal{M}}_n$ of size $M_n$, an encoding state ${\rho }_m^{(n)}$ on the $n$ inputs for each message, and a decoding POVM $\{D_m^{(n)}:m\in {\mathcal{M}}_n\}$ on the output space, with $D_m^{(n)}\ge 0$ and ${\sum }_m{D}_m^{(n)}=I$. The encoding states may be entangled across uses. The measurement outcome is the receiver's decoded message, with conditional probabilityFor uniformly distributed messages the average error and rate areA rate $R$ is achievable if there is a sequence of such codes with $P_e^{(n)}\to 0$ and ${\mathrm{liminf}}_n{R}_n\ge R$. The unassisted classical capacity of a quantum channel is the supremum of achievable rates; the sender and receiver do not share prior entanglement.

The operational content of the Holevo-Schumacher-Westmoreland theorem is that ${\chi }^{\ast }(\Phi )$ is the capacity for product-state input encodings with collective output measurements. Allowing arbitrary entangled block inputs gives the regularized formulaThe limit equals the supremum because superadditivity of Holevo capacity follows by taking product ensembles for two blocks and using additivity of the Von Neumann entropy on product states. The coding theorem makes every rate below the one-block Holevo quantity achievable through many independently encoded blocks, with collective decoding.

The converse can be seen explicitly from the Holevo bound and Fano's inequality. For a uniform transmitted message $M$ and decoded message $\hat{M}$,The first inequality is obtained by subtracting the Fano bound on $H(M\mid \hat{M})$ from $H(M)={\log }_2{M}_n$. Dividing by $n$ and letting the average error vanish bounds every achievable rate by the regularized expression. The block-coding achievability just described supplies the reverse bound.

Under the assumed additivity of Holevo capacity across tensor powers, ${\chi }^{\ast }({\Phi }^{\otimes n})=n{\chi }^{\ast }(\Phi )$ for every $n$. ThusThe stated conditional assertion is true: under this additivity assumption, entangled input states cannot increase the asymptotic unassisted classical rate beyond the rate already achievable with product inputs. The decoder may still use collective measurements across channel outputs.

A reachable set contains all states attainable from a specified initial state by admissible control inputs in the allowed times, and controllability means that every target in the declared state space is reachable from every initial state. For a closed quantum control system, distinguish pure-state controllability on state rays, density operator controllability on fixed-spectrum unitary orbits, and unitary operator controllability for entire propagators. State engineering asks for a particular state transfer; process engineering asks for an entire quantum gate, which must act correctly on every input. A closed Hamiltonian evolution preserves density-operator eigenvalues, so arbitrary mixed states of different spectra cannot be connected even by a fully operator-controllable system.

With $\hslash =1$, the free propagator is $U_0=\mathrm{diag}(e^{i{\omega }_{1}t},1,e^{i{\omega }_{2}t})$. If $E_j$ are its Quantum Hamiltonian eigenvalues, conjugation multiplies the $jk$ matrix element of $H_1$ by $e^{i(E_j-E_k)t}$. Therefore the interaction picture Quantum Hamiltonian isOpposite off-diagonal phases are complex conjugates, as required by Hermiticity. The upper $2$–$3$ element has the opposite phase sign to the upper $1$–$2$ element because the middle level lies above both outer levels.

For each fixed phase ${\varphi }_k$, the single-transition Quantum Hamiltonian is ${\Omega }_k(t)H_k({\varphi }_k)$, so its generators at different times commute. Its exact propagator depends only on ${\theta }_k(t)={\hslash }^{-1}{\int }_0^t{\Omega }_k(\tau )d\tau$. On the active two-level subspace, $H_k^2=I$; on the spectator state, $H_k=0$. Thus it is an embedded two-level quantum rotation, not an involution on the whole three-dimensional space. Explicitly,The spectator entries in both matrices are exactly one; the converted TeX damages the second matrix, so these entries use the PDF.

The proposed combined exponential is false in general for two independent envelopes. Direct multiplication givessoThe constant pulse-axis condition for removing time ordering holds if both envelopes are proportional to the same function, but not for generic independent controls. For example, turn on only the first transition, then only the second. The result is $e^{-i{\theta }_2{H}_2}{e}^{-i{\theta }_1{H}_1}$, whose Baker--Campbell--Hausdorff formula includes a nonzero commutator correction, rather than simply $e^{-i({\theta }_1{H}_1+{\theta }_2{H}_2)}$.

There is also a literal source defect: the PDF's right-hand exponent omits the factor $-i$ on its second term. With ${\theta }_1=0$ it would give $e^{{\theta }_2{H}_2}$, which has eigenvalues $e^{\pm {\theta }_2}$ and is not unitary for real nonzero ${\theta }_2$, whereas the true propagator is unitary. Even after repairing that missing factor, time ordering is still necessary unless the commutators vanish.

For real nonnegative envelopes, put $\Omega =\sqrt{{\Omega }_1^2+{\Omega }_2^2}$ and choose a continuous $\theta =atan2({\Omega }_1,{\Omega }_2)$. The state $|D⟩=\cos \theta |1⟩-\sin \theta |3⟩$ is dark because its two amplitudes feeding $|2⟩$ cancel: ${\Omega }_1\cos \theta -{\Omega }_2\sin \theta =0$. The orthogonal bright combination $|B⟩=\sin \theta |1⟩+\cos \theta |3⟩$ couples to $|2⟩$ with strength $\Omega$. Thus the other eigenstates are $(|B⟩\pm |2⟩)/\sqrt{2}$ with energies $\pm \Omega$.

Start with ${\Omega }_1=0$ and ${\Omega }_2\ne 0$, so $\theta =0$ and $|D⟩=|1⟩$. Slowly increase ${\Omega }_1/{\Omega }_2$ to infinity while maintaining a nonzero gap; end with ${\Omega }_2=0$ and ${\Omega }_1\ne 0$, so $|D⟩=-|3⟩$. This target-to-intermediate pulse first is the counterintuitive ordering of stimulated Raman adiabatic passage. Its quantitative slowness condition isIt follows from $|⟨\pm |\dot{D}⟩|=|\dot{\theta }|/\sqrt{2}$ and the gap $\Omega$. Avoid switching both envelopes off while $\theta$ is changing; switch off together only after the transfer angle has become stationary. The dark eigenvalue is zero, so there is no dynamical phase; with this real eigenvector $⟨D|\dot{D}⟩=0$ as well. The ideal adiabatic quantum control therefore transfers $|1⟩$ to $-|3⟩$, physically the same target ray.

Write $u=u_x$, $w=u_y$ and $B=B_y$. The isothermal equation of state replaces the adiabatic thermal equation by $p=c_s^2\rho$. Since the fields depend only on $x,t$, the solenoidal constraint gives ${\partial }_x{B}_x=0$ and the $x$ component of the ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_x=0$. Hence $B_x$ is constant in both space and time.

The continuity, two momentum and transverse induction equations areHere $B$ is the transverse magnetic component and the constant $B_x$ is the longitudinal component. The longitudinal magnetic force is the gradient of transverse magnetic pressure; the transverse force is magnetic tension. In primitive variables the isothermal coplanar magnetohydrodynamic characteristic matrix isThe imposed $z$ components stay zero because their corresponding equations have zero right-hand sides under this reduction.

Let $c=v-u$ be a characteristic speed relative to the fluid and define the Alfvén velocity ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$. To derive the characteristic polynomial, an eigenvector with components $(\delta \rho ,\delta u,\delta w,\delta B)$ obeysEliminating $\delta u$ and $\delta w$ on a nondegenerate branch givesTheir consistency yields $(c^2-c_s^2)(c^2-v_{ax}^2)=v_{ay}^2{c}^2$. Equivalently, direct expansion of $\det (\mathbf{A}-vI)$ gives the same polynomial without dividing by $c$ or a possibly vanishing field component:ThusThe discriminant is $(c_s^2-v_a^2{)}^2+4c_s^2{v}_{ay}^2\ge 0$, and both squared speeds are nonnegative. These are the fast magnetosonic wave and slow magnetosonic wave families, transported with the mean flow. The equations are hyperbolic; on a generic state the four distinct real eigenvalues give four characteristic directions. Special field alignments can make speeds coincide or the slow speed vanish. There is no separate entropy characteristic because the thermal model is isothermal, and the independent out-of-plane Alfvén wave polarization is excluded by $u_z=B_z=0$.

To construct a simple wave in magnetohydrodynamics, take $\mathbf{U}=\mathbf{U}(a)$ along a state-space curve whose tangent is a right eigenvector of one selected branch:Substitution into the partial differential equations then gives the scalar transport equationSuch curves exist locally by integrating the eigenvector field away from degeneracies. For example, parameterizing by $a=\rho$ where its eigenvector density component is nonzero gives the explicit state-curve equationswith signed $c$ equal to one of the four relative wave speeds. These formulas are used only where their denominators are nonzero.

A particularly transparent nonlinear example is an isothermal perpendicular magnetosonic simple wave. Set the allowed constant $B_x=0$, choose $B=K\rho$ and keep $w$ constant. For the fast branch,These relations satisfy the eigenvector equations and therefore all four original equations wheneverThe amplitude-dependent characteristic speed has derivativeThis already proves the existence of genuinely nonlinear waves in the stated system.

More generally, the method of characteristics applied to initial data $a(x,0)=a_0(x)$ givesIf the initial speed decreases towards increasing $x$ somewhere, the denominator reaches zero in finite time. When the negative minimum is attained, the first steepening time isFaster trailing states catch slower leading states, giving shock formation by characteristic intersection. For the right-moving perpendicular fast example, a negative density gradient provides such a compression. Beyond $t_{\ast }$ a smooth simple wave is no longer sufficient and an appropriate shock solution is required. Rarefactive profiles instead spread. A linearly degenerate characteristic field has $v^{\prime }=0$ along its state curve and avoids this mechanism; the conclusion is that compressive nonlinear magnetosonic simple waves steepen, not that every possible MHD waveform must do so.

Put $y=\dot{x}$. The planar autonomous differential equation isIts equilibria have $y=0$, $x^2=p$. There are none for $p<0$, one at $p=0$, and two $x_{\pm }=\pm \sqrt{p}$ for $p>0$. At an equilibrium $x_{\ast }$ the stability matrix isWriting $s=\sqrt{p}>0$, $x_{+}=s$ is always a saddle equilibrium. At $x_{-}=-s$, the determinant is $2s>0$ and the trace is $s-q$: it is attracting for $q>s$ and repelling for $q<s$.

At $p=0$ and $q\ne 0$, the eigenvalues are $0,-q$; the nonzero quadratic term and parameter unfolding give a saddle-node bifurcation. At $p>0$, the trace crosses zero at $q=s$, with eigenvalues $\pm i\sqrt{2s}$, giving a Hopf bifurcation. It is nondegenerate: around $x=-s$ the equation is $\ddot{\xi }+(q-s)\dot{\xi }+2s\xi -{\xi }^2+\xi \dot{\xi }=0$. At $q=s$, a small oscillatory expansion of leading amplitude $r$ has $⟨\xi {\dot{\xi }}^2⟩=r^4/8+O(r^5)$, so the first nonlinear damping is positive and nonzero. For example, its second-order oscillatory expansion is $\xi =r\cos (\Omega t)+r^2/(2{\Omega }^2)-r^2\cos (2\Omega t)/(6{\Omega }^2)-r^2\sin (2\Omega t)/(6\Omega )$ with ${\Omega }^2=2s$. Thus the small attracting limit cycle is on the $q<s$ side. At $(p,q)=(0,0)$ the matrix has a double zero eigenvalue and a nontrivial Jordan block, the Bogdanov–Takens bifurcation organizing the two curves. The local bifurcation curves areA change between node and focus away from these curves is not itself a local bifurcation.

For the requested global exclusion, suppose $x(t)$ were periodic with period $T$. Integrate the original equation over a period. The integrals of $\ddot{x}$, $q\dot{x}$, and $x\dot{x}=\frac{1}{2}(d/dt)x^2$ vanish. ThereforeThe left side is nonnegative and the right side is negative when $p<0$. This contradiction proves there are no periodic orbits for $p<0$, independently of $q$.

With $y=\dot{x}$, the feedback system isFor $p=0$, $(x,y,z)=(0,0,0)$ is an equilibrium. Its stability matrix and characteristic polynomial areThus there is always a zero eigenvalue, with eigenvector $(1,0,0)$, so it is a nonhyperbolic equilibrium. If $q\ne -1$, the other two eigenvalues are off the imaginary axis: for $q>-1$ their real parts are negative, while for $q<-1$ their product is negative. At $q=-1$, all three eigenvalues are zero, explaining the exclusion in the next part.

Adjoin $\dot{p}=0$ and use an extended centre manifold with local coordinates $(x,p)$. Count $p$ as order $x^2$, as appropriate to a steady-state fold, and writeThe centre-manifold invariance equation differentiates these graphs along $\dot{x}=y$. The derivatives $y_{x}y$ and $z_{x}y$ first contribute at weighted order three. Comparing the coefficients of $p$ and $x^2$ in the right sides therefore givesSince $q+1\ne 0$, their solution isHence the reduced normal form isBoth the quadratic and unfolding coefficients are nonzero, so the steady-state bifurcation is a saddle-node, with equilibria on the $p>0$ side. The exact steady equations confirm this: $y=0$, $z=x^2$, and $p=z+z^3=x^2+x^6$, giving two nearby equilibria for $p>0$ and none for $p<0$.

For $q>-1$ the transverse modes are attracting; the negative-$x$ branch is locally attracting and the positive-$x$ branch has one unstable direction. For $q<-1$, one transverse eigenvalue is already positive, so neither full equilibrium is asymptotically stable, although the reduced fold is still present with reversed flow orientation. The parameter $q=-1$ requires a higher-dimensional degeneracy analysis and is outside the stated assumption.

Assume a generic saddle-focus homoclinic orbit at $\mu =0$, a smooth nondegenerate global reinjection along the selected unstable branch, and a sufficiently small neighborhood in which the linearized local flow supplies the leading passage map. Use linearizing coordinates if available; otherwise the calculation is a leading local approximation, not an exact consequence of eigenvalues alone. Choose the incoming section near $(\rho ,0,0)$ in the plane $y=0$, with coordinates $(\xi ,z)$ where $r=\rho +\xi$ and $\rho >0$. This section is transverse since $\dot{y}=\omega r\ne 0$. Choose an outgoing section $z=z_0>0$. We mark one return after a global excursion; intermediate crossings of $y=0$ during the spiral are not counted as separate global returns.

In cylindrical coordinates the local flow isFor $0<z\ll z_0$, the time to the outgoing section is $t_{\ell }={\lambda }_{+}^{-1}\log (z_0/z)$. Set $\delta =-{\lambda }_{-}/{\lambda }_{+}>0$ and $\Omega =\omega /{\lambda }_{+}$. The outgoing coordinates areThe global part of the Poincaré return map is smooth around $(X,Y)=(0,0)$ and takes the unstable-manifold point to $({\xi }^{\prime },z^{\prime })=(0,-\mu )$. Expand it aswhere the coefficients may depend smoothly on $\mu$ and the matrix is generically nonsingular. Absorbing the fixed $z_0^{-\delta }$ into them gives the leading two-dimensional return mapThe remainder refers to the smooth global-map Taylor approximation with bounded $\xi$; corrections to the local linear approximation are also neglected. The global map supplies the additive splitting parameter and the local map supplies the power and logarithmic rotation. For an iterated excursion on this same unstable branch, the next coordinate must again satisfy $z^{\prime }>0$. A nonpositive image leaves this local return domain; the formula must not be iterated through negative logarithms.

Seek an equilibrium in the invariant coordinate plane $z=0$ with $x,y\ne 0$. Put $U=x^2$ and $W=y^2$. The two nonzero-component equations areSubtracting gives $eU+cW=0$, so positive $U,W$ require $ce<0$. With $D=a(c-e)-ce$, the solutions areThey are actual additional equilibria when $D$ has the sign of $c$; the signs of $x,y$ are independently free, and cyclic permutation supplies the other two coordinate planes. For example, $a=\mu =c=1$, $e=-1$ gives $D=3$ and $x^2=y^2=1/3$. Thus additional two-coordinate equilibria do exist for suitable opposite-sign cross-couplings. The inequality $ce<0$ is necessary for this type, but without the positivity conditions on the displayed denominator it need not be sufficient for every coefficient choice.

The isotropy of a generic such point is just $⟨{\kappa }_z⟩$, whose fixed-point subspace is two-dimensional. The Equivariant branching lemma guarantees axial branches with one-dimensional fixed spaces; it does not rule out these lower-symmetry branches. Their existence therefore gives no contradiction.

At fixed $\mu >0$, letting $e$ cross zero with $c\ne 0$ sends $x^2\to \mu /a$, $y^2\to 0$: the new pair of opposite-$y$ branches meets an axis equilibrium. Its $y$ eigenvalue is $e\mu /a$, so this is a secondary symmetry-forced pitchfork bifurcation. Similarly, as $c$ crosses zero with $e\ne 0$, the branches meet the $y$ axis, whose $x$ eigenvalue is $-c\mu /a$. The cubic formula gives $y^2\sim -\mu e/(ac)$ near the first crossing, and $x^2\sim -\mu c/(ae)$ near the second. Reflections force the two signs of the newly nonzero coordinate, while cyclic symmetry gives the corresponding pitchforks on the other axes. The simultaneous $c=e=0$ case is degenerate and is not a generic single-parameter pitchfork.

The equivariant Hopf theorem states that, under the usual nondegeneracy and transverse-crossing assumptions at an equivariant Hopf point, each complex axial isotropy subgroup of the spatial symmetry group combined with temporal phase produces a branch of periodic solutions with that spatiotemporal symmetry. Its common fixed-point space must have real dimension two, or complex dimension one. Restricting the Hopf normal form of a dynamical system to that plane gives a scalar complex amplitude equation; a nonzero real cubic coefficient provides a small periodic branch on the corresponding parameter side. The theorem supplies existence and symmetry, not stability.

Put $\zeta =e^{2\pi i/3}$ and write $T_{\theta }z=e^{i\theta }z$, so $T_{\theta }={\tau }_{\theta /{\omega }_0}$. Three complex axial fixed lines areFor the first generator, its eigenvalues are $1,{\zeta }^2$, so the fixed line is exactly $z_2=0$. Reflection supplies the conjugate line $z_1=0$, representing the opposite rotation direction. For the other two generators, swapping coordinates, with or without the common minus sign, gives the displayed one-complex-dimensional fixed spaces.

These subgroups are inequivalent: the rotating-wave spatial projection has order three, while the standing-wave projections have order two; the two reflection types differ by the central phase $T_{\pi }$, which conjugation cannot remove. Rotating the two standing lines by $\rho$ gives three orientations of each type, not new symmetry types. Therefore there are three distinct generic periodic branch types: rotating waves and two inequivalent standing-wave families. Their members have a period near $2\pi /{\omega }_0$, with the combined spatial–temporal operations determining their relative phase relations.

Without the threefold rotation, a complex linear map need only commute with coordinate swapping and common phase. Commuting with $m_x$ makes its two diagonal entries equal and its two off-diagonal entries equal; common phase still forbids conjugate-linear terms. Thus the additional leading coupling iswhere $\eta$ is a small complex symmetry-breaking coefficient and a small common diagonal shift can be absorbed into $\alpha$. The old rotation forbade these off-diagonal entries because the two coordinates have different rotation weights. Further symmetry-breaking cubic and quintic terms are also allowed; retaining only the new linear terms describes the leading onset splitting.

In symmetric and antisymmetric coordinates $z_{+}=(z_1+z_2)/\sqrt{2}$, $z_{-}=(z_1-z_2)/\sqrt{2}$, the eigenvalues are $\alpha +\eta$ and $\alpha -\eta$. With $\alpha =\mu +i{\omega }_0$, the two primary Hopf thresholds and frequencies are thereforeThe reflection-preserving standing line $z_1=z_2$ remains invariant. The line $z_1=-z_2$ also remains invariant because it is fixed by reflection combined with phase $\pi$. Their generic periodic branches survive, now starting at separate thresholds when $\mathrm{Re}\eta \ne 0$. If that real part vanishes, threshold coincidence is a nongeneric additional condition; frequency splitting may still occur.

The pure rotating-wave axes are not invariant: at $z_2=0$ the second equation has ${\dot{z}}_2=\eta z_1$. Hence the old pure rotating-wave branches cease to be primary one-component branches from the origin. At fixed nonzero symmetry breaking, a simple Hopf onset has just the appropriate symmetric or antisymmetric mode. Away from the degenerate onset, normally hyperbolic rotating-wave periodic orbits may persist as mixed-mode periodic states with both components nonzero, paired by reflection. In common unfoldings such a pair is reached through a secondary reflection-breaking pitchfork bifurcation of periodic orbits from a standing branch. Its precise attachment, any folds, and stability depend on the nonlinear coefficients and cannot be fixed by symmetry alone.

The three rotationally conjugate orientations within each original standing family are no longer equivalent. Only the lines aligned with the surviving reflection symmetries are forced primary branches; other members can deform, pin to preferred orientations, or connect through secondary bifurcations. The generic bifurcation diagram thus replaces the coincident multi-branch Hopf onset by two separated simple Hopf onsets, removes the pure traveling-wave invariance, and unfolds the rotational degeneracies. This is the reflection-preserving splitting of a dihedral Hopf bifurcation; it does not by itself prescribe a unique global diagram.

Write $h=\Delta x$, impose $u_0=u_{M+1}=0$, and use the discrete inner product $⟨u,v{⟩}_h=h{\sum }_{m=1}^M{\overline{u}}_m{v}_m$. Discrete integration by parts for the second difference givesFor the centered drift, reindexing with the zero endpoints givessince the two sums are complex conjugates. This is the fact that centered discrete advection is skew-adjoint. The semidiscrete energy estimate is thereforeThe finite-dimensional linear system has a unique solution for all times, and $\Vert u(t){\Vert }_h\le \Vert u(0){\Vert }_h$, uniformly in the mesh and in real $\kappa$. Differences of two semidiscrete solutions obey the same bound, which is the required stability.

One may also use the smallest eigenvalue ${\lambda }_h=4h^{-2}{\sin }^2(\pi /[2(M+1)])$ of the negative second-difference operator to obtain $\Vert u(t){\Vert }_h\le e^{-{\lambda }_{h}t}\Vert u(0){\Vert }_h$. This conclusion concerns the method of lines; a later choice of time integrator has its own stability conditions.

Away from the first parametric resonance, one forcing interaction changes a harmonic frequency by one and produces no resonant coupling of the two conjugate oscillations. Two interactions can couple frequency $+1$ to frequency $-1$, so the next parametric resonance is near $\omega =1$ and its effect first appears at order ${ϵ}^2$. This gives $p=2$ and the appropriate slow time $T={ϵ}^{2}t$.

Write $\omega =1+\kappa {ϵ}^2$ and $y_0=a(T)\cos t+b(T)\sin t$. The first correction satisfies $({\partial }_t^2+1)y_1=-\cos t{y}_0$, whose bounded particular solution isAt second order,The resonant part of $-\cos t{y}_1$ is $(5a/12)\cos t-(b/12)\sin t$. Applying the solvability condition in the method of multiple scales givesThe squared slow eigenvalue isThe second instability tongue of a weak Mathieu oscillator therefore has the leading intervalHere the two little-$o$ terms describe the respective exact edge locations; the strict inequality in $\kappa$ is the leading interior criterion. In particular $\omega =1$ is unstable, with physical growth rate ${ϵ}^2\sqrt{5}/24+o({ϵ}^2)$. The general leading solution for bounded $T$ again has $y_0=a(T)\cos t+b(T)\sin t$, where the matrix exponential of the displayed two-by-two amplitude system evolves arbitrary $a(0),b(0)$. Adding $ϵy_1$ identifies the first nonresonant harmonic correction. The leading expression alone has error $O(ϵ)$ on $t=O({ϵ}^{-2})$.

For comparison, at fixed $\omega \ne 1/2,1$, write $y_0=A(T)e^{i\omega t}+\overline{A}(T)e^{-i\omega t}$. The first forced harmonics have coefficients $A/[2(2\omega +1)]$ and $A/[2(1-2\omega )]$. Their return to the original harmonic at the next order givesIt is a phase correction, not amplitude growth. Consequently, away from the first two resonance neighborhoods, a general leading solution valid for bounded ${ϵ}^{2}t$ isNarrower higher-order instability intervals are invisible on this time scale. The detuned resonant calculation above is needed near $\omega =1$ because both the phase shift and the conjugate-mode coupling then matter.