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For an irrotational flow, write $\mathbf{u}=\nabla \varphi$. The incompressible flow condition then gives the Laplace equationAt leading order in a small disturbance this equation holds on $-h<y<0$. Including gravity, the Unsteady Bernoulli equation isChoose the time-dependent additive freedom in the velocity potential so that $B(t)=p_0/\rho$. Neglecting surface tension, pressure continuity at the free surface gives the exact dynamic conditionThe free surface is a material boundary: applying the material derivative to $y-\eta (x,t)=0$ yields the kinematic boundary condition for a free-surface graphFor a small disturbance, the Taylor expansion from $y=\eta$ to $y=0$ changes the first-order velocity potential and its partial derivatives only at second order. Both $|\nabla \varphi {|}^2$ and ${\varphi }_x{\eta }_x$ are also second-order terms. The linearized free-surface boundary conditions are thereforeThe fixed bed has the no-penetration boundary conditionThe required approximation is small surface elevation, small wave steepness, and small velocity perturbations. The stated $p_1\ll p_0$ alone does not ensure this approximation near resonance.

The denominator of the travelling response vanishes atFor the positive speeds in the paper, a finite travelling response exists whenever $c\ne c^{\ast }$. At $c=c^{\ast }$, the dynamic condition would reduce to $p_1/\rho =0$, which is impossible. The finite-depth dispersion relation of a surface gravity wave with wavenumber $k$ is ${\omega }^2=gk\tanh (kh)$. At $k=1$, its phase speed $\omega /k$ equals $c^{\ast }$: the moving pressure pattern drives a natural surface gravity wave at its own speed, producing resonance.

The restriction concerns a bounded travelling response. To distinguish this from existence of a solution of an initial value problem, set $T=\tanh h$, $P=p_1/\rho$, and ${\omega }_{\ast }^2=gT$. Ifthe linearized free-surface boundary conditions reduce to the forced harmonic oscillatorFor example, if the forcing is switched on at $t=0$ from a flat surface with zero initial velocity, then, for $c\ne {\omega }_{\ast }$,This satisfies $H(0)=\dot{H}(0)=0$ and contains both free oscillations and the forced component. At $c={\omega }_{\ast }$ a solution of the initial value problem still exists, but its particular solution containswhich grows linearly with time. Thus the ideal inviscid flow model predicts resonant growth rather than a finite constant amplitude. Additional free oscillations can set the initial conditions but cannot remove that secular growth.

Write $m=k_{\varphi }$, $B_{\mathbf{k}}=\mathbf{k}\cdot {\partial }_{\mathbf{J}}{f}_0$, and let the detuning from the rotating perturbation beSubstitute $\delta {\Phi }_{\mathbf{k}}={\Psi }_{\mathbf{k}}{e}^{-im{\Omega }_{\mathrm{p}}t}$ into the linear response in action-angle variables, then set $\tau =t-t^{\prime }$:At exact resonance the last quotient is interpreted by its continuous limit $it$. Inserting this response in c givesPair $\mathbf{k}$ with $-\mathbf{k}$. Because $B_{-\mathbf{k}}=-B_{\mathbf{k}}$, ${\Delta }_{-\mathbf{k}}=-{\Delta }_{\mathbf{k}}$, and $|{\Psi }_{-\mathbf{k}}{|}^2=|{\Psi }_{\mathbf{k}}{|}^2$, the imaginary cosine terms cancel. The finite-time torque isThis is a leading, second-order response in the perturbation amplitude.

For the long-time limit, assume a continuum of occupied action variables, sufficiently smooth integrable weights, convergent harmonic sums, and regular resonance surfaces ${\nabla }_{\mathbf{J}}{\Delta }_{\mathbf{k}}\ne 0$ wherever a contributing surface intersects the distribution. Assume also that the perturbation is weak enough to neglect nonlinear trapping and secular changes in $f_0$, and that the prescribed stellar bar pattern and its coefficients stay fixed. Then phase mixing removes nonresonant transients in the integrated response. The relevant long-time resonance kernel is the distributional convergenceIt is not a pointwise limit: at $x=0$ the kernel equals $t$. To justify its use under the integral over action variables, resolve the integral into detuning surfaces; the coarea formula givesSmoothness and decay of this surface weight reduce the limit to the one-dimensional long-time resonance kernel. Equivalently, a causal convergence factor $e^{-\eta \tau }$ gives $\mathrm{Re}{\int }_0^{\infty }{e}^{-\eta \tau -i\Delta \tau }d\tau =\eta /({\eta }^2+{\Delta }^2)\to \pi \delta (\Delta )$ as $\eta \downarrow 0$.

Therefore the resonant torque from a rotating perturbation isOnly orbital resonances contribute to this steady integrated torque. The limit is a formal weak-perturbation limit, or an intermediate-time approximation after phase mixing but before substantial secular redistribution. It does not prove an unchanged halo can absorb angular momentum at this rate forever. A discrete set of orbital frequencies, singular resonance surfaces, or strong trapping can invalidate the stated steady limit.

Apply the chain rule to $f_0=f_0(E)$ with $E=H_0(\mathbf{J})$:On the resonance selected by the Dirac delta function, $\mathbf{k}\cdot \mathbf{\Omega }=k_{\varphi }{\Omega }_{\mathrm{p}}$. ThusEvery weight is nonnegative, and $-f_0^{\prime }(E)>0$. Strict positivity additionally requires a populated nonaxisymmetric resonance with nonzero coupling on a set of positive surface measure. The unconditional statement is nonnegativity, rather than positive definiteness. As a concrete counterexample to strict positivity of this formula, retain one nonzero conjugate harmonic pair with $k_{\varphi }\ne 0$ and choose its smooth coefficient ${\Psi }_{\mathbf{k}}$ supported in an open region $|{\Delta }_{\mathbf{k}}|>d>0$. The perturbation is nonzero but has zero resonant coupling, so $\mathcal{T}=0$ even when $f_0^{\prime }(E)<0$. The even simpler zero-perturbation case also gives zero. Whether a particular physical stellar bar has nonzero resonant overlap is additional information.

For a coupled stellar bar and halo, the halo gains positive angular momentum and exerts the opposite torque on the stellar bar. If the halo mass is $M_h$ and the rigid stellar bar has constant positive moment of inertia $I_b$, thenThis bar-halo angular-momentum transfer is a form of dynamical friction: a resonantly coupled stellar bar tends to slow down while the dark matter halo spins up. Actual long-term evolution changes both the stellar bar pattern and the halo distribution function, so the fixed-pattern response must then be updated. If the stellar bar changes its moment of inertia, the general relation is $d(I_b{\Omega }_{\mathrm{p}})/dt=-M_h\mathcal{T}$; positive moment of inertia alone would not fix the sign of $d{\Omega }_{\mathrm{p}}/dt$ without that rigidity assumption.

Set $\kappa =|k|$ and first work with a representative positive $k=\kappa$. For $\kappa {\lambda }_{\mathrm{De}}\ll 1$, the thermal core is strongly screened. However, the causal dielectric function has weakly damped Langmuir wave roots ${\omega }_s+i{\gamma }_s$, with $s=\pm 1$, ${\omega }_s\simeq s{\omega }_{pe}$ and ${\gamma }_s<0$. Their phase velocities $v_s={\omega }_s/\kappa$ are in the tail. Both roots must be retained for each spatial Fourier series harmonic.

Write $ϵ(-i\omega +0^{+},\kappa )={ϵ}_R(\omega )+i{ϵ}_I(\omega )$. The supplied approximation givesExpanding $ϵ({\omega }_s+i{\gamma }_s)=0$ to first order gives the Landau damping rateIn particular, stable damping means ${\omega }_s{f}_0^{\prime }(v_s)<0$. Near either root,The narrow Lorentzian resonance approximation then gives the leading resonant contributionThis is a formula under the integral over velocity; it is not a pointwise approximation away from the resonances. Insert it into the heating formula in c. For one spatial harmonic, the contribution from one root isbecauseThus the tiny tail slope cancels the tiny Landau damping rate: weak absorption is compensated by a large noise-driven wave intensity. The two temporal roots supply two equal halves of the answer. Negative $k$ gives the same result by the conjugate response of a real electric field, without altering the original convention that the spatial sum includes both signs. ThereforeAs in c, equality means the leading result in the stated asymptotic limit. This calculation assumes nonzero damping, observation times $t\gg {\max }_s|{\gamma }_s{|}^{-1}$, and forcing weak enough that $f_0$ changes negligibly during that relaxation. Taking zero damping before the long-time limit invalidates the stationary calculation.

The long-wavelength limit in d is resonant heating. Most electrons screen the slowly varying spatial field. The noise nevertheless excites collective Langmuir waves, whose sharp temporal spectrum lies near $\omega =\pm {\omega }_{pe}$. Electrons with $v\simeq \omega /k$ remain in phase with a wave and exchange energy with it. A stable tail absorbs net wave energy through Landau damping. The heating is initially deposited in narrow tail regions of velocity space, even though the volume-integrated rate equals that in c.

The short-wavelength limit in c is stochastic heating in the terminology intended here. With $ϵ\simeq 1$, the imposed electric field is essentially unscreened. Its temporally uncorrelated kicks cause a random walk of electron velocity, so $⟨(\Delta v{)}^2⟩=2D_0\Delta t$. Electrons throughout the thermal population contribute to the resulting kinetic energy increase; no narrow collective wave peak selects a small tail population.

Both mechanisms are driven by white noise, and the formal condition $\omega =kv$ enters the stationary quasilinear diffusion calculation in both cases. The physical distinction is the sharp collective resonance in d versus the broad direct random forcing in c, rather than the mere presence of randomness. In collisionless theory this heating is an increase of the averaged kinetic energy; subsequent collisions can establish thermal equilibrium.

The restriction $k_{\phi }=\pm m$ describes an $m$-fold rotating perturbation of a galactic disk, for example a transient spiral density wave or, for $m=2$, a galactic bar. Its pattern speed is ${\Omega }_{\mathrm{p}}$, its amplitude peaks at $t_{\mathrm{peak}}$, and its Gaussian function envelope has duration of order $\tau$. The coefficients $u_{\mathbf{k}}(\mathbf{J})$ describe its strength along different stellar orbits. A real perturbing potential requires $u_{-\mathbf{k}}=u_{\mathbf{k}}^{\ast }$.

Write $g(t)=\exp [-(t-t_{\mathrm{peak}}{)}^2/(2{\tau }^2)]$. The product entering c isThe phase therefore measures detuning from the rotating pattern:Substitution givesThe strongest long-lived response occurs near an orbital resonance, ${\omega }_{\mathbf{k}}=0$, where successive force contributions do not rapidly cancel in phase.

At resonance, the Hamiltonian operator becomes $\hslash V{\sigma }_x/2$. ThereforeIt gives the requested positive rotation about the x-axis. In the ordered basis $(|1⟩,|0⟩)$, $|1⟩$ becomes $(|1⟩-i|0⟩)/\sqrt{2}$, whose Bloch vector is $(0,-1,0)$.

For the stated Hamiltonian operator, put $\mathbf{\Omega }=(V,0,-\delta )$ and $\Omega =\sqrt{V^2+{\delta }^2}$. The Bloch vector obeysThus it rotates right-handedly about the fixed axis $\mathbf{\Omega }$, preserving its length and its projection on that axis. For example, on resonance $\delta =0$, a north-pole Bloch vector follows $(0,-\sin Vt,\cos Vt)$ and reaches the south pole at $t=\pi /V$.

Take $V>0$, with the field phase chosen accordingly. The Hadamard gate is $H=({\sigma }_x+{\sigma }_z)/\sqrt{2}$. Align the axis with $(1,0,1)$ and rotate through $\pi$:The factor $-i$ is an irrelevant common global phase. This realizes the whole quantum gate, rather than merely its action on one input quantum state.

There is a basis-ordering convention to make explicit: this $H$ is the Hadamard gate matrix in the ordered basis $(|1⟩,|0⟩)$ chosen in part (a). If instead its definition is the usual ket rules $H_{01}|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H_{01}|1⟩=(|0⟩-|1⟩)/\sqrt{2}$, its matrix in our order is $({\sigma }_x-{\sigma }_z)/\sqrt{2}$. For that definition choose $\boxed{\delta =+V,\tau =\pi /(\sqrt{2}V)}$. Both conventions use the same pulse duration; the sign of the Rabi detuning selects the desired axis.

Quantize a single electromagnetic normal mode of angular frequency ${\omega }_c$ and represent one isolated atomic transition as a two-level system with raising operator ${\sigma }_{+}=|e⟩⟨g|$. The bare Hamiltonian operator isThe first main approximation is the electric-dipole approximation: the field varies negligibly across an atom, $k{a}_{\mathrm{atom}}\ll 1$, and the leading coupling is $H_d=-\mathbf{d}\cdot \mathbf{E}({\mathbf{r}}_0)$. For states of definite opposite parity, the relevant dipole part is off-diagonal. After choosing operator phases, this gives $H_d=\hslash g(a+a^{†})({\sigma }_{+}+{\sigma }_{-})$, with $g$ fixed by the transition dipole and the one-photon field wave amplitude.

In the interaction picture, $a{\sigma }_{+}$ and $a^{†}{\sigma }_{-}$ oscillate at the difference ${\omega }_a-{\omega }_c$, whereas $a{\sigma }_{-}$ and $a^{†}{\sigma }_{+}$ oscillate at the sum ${\omega }_a+{\omega }_c$. The second main approximation is the rotating-wave approximation: near resonance and for $g\sqrt{n+1}\ll {\omega }_a+{\omega }_c$, the latter terms average away on the interaction time scale. The resulting Jaynes-Cummings model hasThe two-level system and single-normal mode truncations are additional assumptions: other atomic transitions and optical normal modes must be sufficiently far off resonance or weakly coupled. Dissipation is neglected over the interaction time. The retained terms exchange one atomic excitation for one photon, conserving $a^{†}a+{\sigma }_{+}{\sigma }_{-}$.

Use the resonant interaction picture and define $g>0$ by the one-atom Jaynes-Cummings model in part (a). Equal coupling to both atoms gives the Tavis-Cummings modelThe initial state has two excitations and is symmetric under atom exchange. Excitation conservation and this symmetry confine the evolution toThe antisymmetric atomic spin singlet state is uncoupled and unpopulated. Applying the annihilation operator and creation operator givesThe first off-diagonal entry combines two indistinguishable atomic emission paths; the second also includes the $\sqrt{2}$ enhancement for creating the second photon. Its eigenvalues are $0,\pm \sqrt{6}g$. Put $\Omega =\sqrt{6}g$ and $K=H_I/\hslash$, so $K^3={\Omega }^{2}K$. Its exponential isApplying this to $|A⟩$ yieldsAll three states have the same bare energy on resonance, so restoring the free Schrödinger-picture evolution adds only a common global phase. The expression starts at $|ee;0⟩$, has initial derivative $-i\sqrt{2}g|B⟩$, and its three squared probability amplitudes sum to one.

The sensor must survive the stress range and repeated loading without fracture, fatigue or excessive mechanical compliance. The tensor character of the piezoelectric coefficient makes crystal orientation important; a ceramic's poling quality and domain stability also matter. Compare linearity, hysteresis, reproducibility, aging and the usable frequency bandwidth, including resonance and dynamic response.

Electrical design is equally important: leakage resistance and readout input impedance determine low-frequency loss, and cable/input capacitance loads a voltage output. A charge amplifier may favour a different material ranking from a passive open-circuit voltage measurement. temperature changes can alter the coefficients, produce a pyroelectric signal in a polar material and depole a ferroelectric near its transition. Humidity, electrode bonding, electrical breakdown, manufacturability, cost and environmental constraints such as lead content also affect a practical choice. Calibration should separate applied stress from temperature drift and readout loading.

Divide the external force by $m$. Away from resonance a particular solution has the forcing frequency, so the general forced harmonic oscillator solution isThe two homogeneous terms cannot be discarded in an undamped system: they do not decay. If the same initial conditions are imposed, then $A=x_0-C\cos \varphi$, $B=-C(\omega /{\omega }_0)\sin \varphi$, where $C=F_0/[m({\omega }_0^2-{\omega }^2)]$.

For $\omega \gg {\omega }_0$, the forced displacement approaches $-F_0\cos (\omega t-\varphi )/(m{\omega }^2)$: its amplitude decreases quadratically and it is in antiphase with the force. This statement concerns the forced component; a freely excited component at ${\omega }_0$ can remain.

At resonance the preceding denominator is zero and the appropriate solution isIts amplitude grows linearly because the external force supplies energy coherently. For $x(0)=x_0$, $\dot{x}(0)=0$, choose $A=x_0$ and $B=F_0\sin \varphi /(2m{\omega }_0^2)$. Physical damping or nonlinear response ultimately limits the ideal resonant growth.

For ${\beta }^2=1$, the reduced equation is $u^{\prime \prime }+u=1+\cos x$. The constant forcing has particular solution one, while the resonant cosine requires $u_p=(x/2)\sin x$. ThusThe secular factor $x$ is the signature of resonance. It must replace the divergent nonresonant coefficient $1/({\beta }^2-1)$; substituting ${\beta }^2=1$ directly into that coefficient is invalid.

The exponential substitution removing a constant first derivative, $y=e^{x}u$, reduces the equation to $u^{\prime \prime }+{\beta }^{2}u=1+\cos x$. For $\beta \ne 0$ and ${\beta }^2\ne 1$, solve the homogeneous harmonic equation and use a constant plus a cosine for the particular solution:There is a further exceptional case inside the printed condition ${\beta }^2\ne 1$: $\beta =0$ must be treated separately. It gives $u^{\prime \prime }=1+\cos x$, henceThis retains the constant forcing's resonance with the zero-frequency homogeneous equation. Substitution verifies both branches.

The natural normal modes belong to the homogeneous system with both end displacements fixed at zero. Seek $\mathbf{y}=\mathbf{a}\cos \Omega t$, soThe characteristic polynomial is $(3{\omega }_0^2-{\Omega }^2{)}^2-{\omega }_0^4=0$. The eigenvectors $(1,1)$ and $(1,-1)$ therefore giveThe symmetric mode leaves the middle segment horizontal, while the antisymmetric mode bends it strongly and has the higher restoring stiffness. Defining $q_{+}=(y_1+y_2)/2$ and $q_{-}=(y_1-y_2)/2$ diagonalizes the driven equations too:This shows that the boundary drive excites both natural modes; neither natural resonance is removed by a zero drive projection.

Away from the natural resonances, seek a periodic particular solution $y_j=A_j\cos \omega t$. Let $s={\omega }^2/{\omega }_0^2$. The amplitude equations areTheir determinant is $(3-s{)}^2-1=(2-s)(4-s)$, giving the signed displacement amplitudesA negative sign denotes a phase reversal relative to the prescribed cosine; physical amplitude magnitudes are $|A_j|$. In the quasistatic limit,These are the values of a straight string interpolating between endpoint displacements $0$ and $h_0$ at bead positions $L$ and $3L$ along a total length $4L$.

Because damping is explicitly absent, this periodic response is not a unique attracting late-time state. Arbitrary homogeneous oscillations at $\sqrt{2}{\omega }_0$ and $2{\omega }_0$ persist unless the initial conditions eliminate them. At either natural resonance, the drive produces secular growth rather than a finite bounded periodic response. Thus “steady state” here refers to the selected particular solution away from the poles.

At $\omega =\sqrt{3}{\omega }_0$, $s=3$, so the periodic amplitudes becomeThe second bead is stationary in this particular motion, although the endpoint oscillates. The first bead moves at twice the endpoint amplitude and in opposite phase. This is an antiresonance of the second-bead response, not a natural resonance of the complete two-bead system.

The force balance makes the cancellation explicit. At $y_2=0$, the middle segment supplies transverse force $T{y}_1/(2L)=-Th/L$, while the end segment supplies $Th/L$; these cancel on the second bead. The first bead then oscillates between two effectively fixed neighbours with restoring coefficient $3T/(2L)$, whose single-coordinate natural angular frequency is $\sqrt{3}{\omega }_0$.

As elsewhere in the undamped problem, this exact nodal motion presumes initial conditions selecting the periodic particular response. Additional free normal modes would move the second bead as well; they do not decay automatically.

Write $A=aI+B$, where $B=\left(\begin{array}{cc}1& -2\\ 1& -1\end{array}\right)$ has $B^2=-I$. One eigenpair isBecause $A$ is real, conjugating $Ae=\ell e$ gives $A\bar{e}=\bar{\ell }\bar{e}$. The eigenvector matrix $(e,\bar{e})$ has determinant $2i\ne 0$, so the eigenvectors form a basis. Reality of $z$ then forces conjugate coefficients, giving $z=\alpha e+\bar{\alpha }\bar{e}$.

Set $c=(1-i)/2$. The forcing has the same basis decomposition,Equating the coefficient of $e$ in the linear system of differential equations yieldsFor $a>0$, an integrating factor gives $\alpha =(c/a)e^{-it}+C{e}^{-(a+i)t}$. The zero initial vector means $\alpha (0)=0$, soTaking the real vector combination givesExact secular resonance occurs at $a=0$: the forcing frequency matches the undamped homogeneous mode. For positive $a$, damping prevents secular growth, and the late-time periodic amplitude is proportional to $1/a$. Thus there is a resonant small-damping limit, but no unbounded resonant response at fixed $a>0$.

Since $(1-e^{-at})/a\to t$ for each fixed $t$, the requested limit givesEquivalently $\alpha =ct{e}^{-it}$ at zero damping; its linear prefactor displays the resonance explicitly.

The spatial reflection $x\mapsto -x$ acts on the two complex amplitudes as $(A,B)\mapsto (A,-B)$, because the first spatial eigenfunction is even and the second is odd. The first equation must therefore be even in $B,\overline{B}$, and the second odd. For a nonresonant double Hopf bifurcation, the temporal phases give independent actions $A\mapsto e^{i{\phi }_1}A$, $B\mapsto e^{i{\phi }_2}B$ in the cubic normal form of a dynamical system. A cubic monomial in the first equation must have net $A$ phase weight one and net $B$ phase weight zero. Only $A|A{|}^2,A|B{|}^2$ satisfy this; the second equation similarly has $B|B{|}^2,B|A{|}^2$. Parametrizing the independent cross-saturation coefficients givesAll coefficients can be complex: their real parts determine amplitude growth and their imaginary parts frequency shifts. These are cubic truncations of a family of reduced vector fields on the four-real-dimensional extended centre manifold, with the two unfolding parameters entering ${\nu }_1,{\nu }_2$.

The nonresonance assumption must be made explicit. Merely having ${\omega }_1\ne {\omega }_2$ does not imply this form: for ${\omega }_1=2{\omega }_2$, the quadratic terms $B^2$ in $\dot{A}$ and $A\overline{B}$ in $\dot{B}$ satisfy both temporal resonance and spatial parity. Thus the initial form is the generic nonresonant double-Hopf form, rather than a statement for every possible unequal frequency pair.

At the one-to-one reflection-symmetric one-to-one Hopf resonance, only the common temporal phase action remains. Cubic terms must then have total phase weight one. Spatial reflection retains, in the first equation, the three terms $A|A{|}^2,A|B{|}^2,B^2\overline{A}$; the second retains $B|B{|}^2,B|A{|}^2,A^2\overline{B}$. Common phase covariance excludes all quadratic terms. The additional resonant terms are consequently $-{\gamma }_1{B}^2\overline{A}$ and $-{\gamma }_2{A}^2\overline{B}$, with independent complex coefficients. This is the reflection-symmetric one-to-one Hopf resonance.

In the specified equal-coefficient specialization, put $r=R^2$. A nonzero pure-$A$ periodic branch exists for $\mathrm{Re}{\nu }_1>0$ and hasIts radial eigenvalue is $-2{\alpha }_{R}r<0$, and its phase is neutral. For a transverse $B$ perturbation the linear equation isIts time dependence can be removed exactly by $B=C{e}^{i\Omega t}$. Using ${\nu }_1-\alpha r=i\Omega$ and ${\nu }_2={\nu }_1-d$ givesWriting $C=u+iv$, $d=d_R+i{d}_I$, $\beta ={\beta }_R+i{\beta }_I$ yields the constant real stability matrixLet $Q=\mathrm{Re}(\beta \overline{d})={\beta }_R{d}_R+{\beta }_I{d}_I$. The transverse characteristic polynomial isBoth transverse eigenvalues have negative real part exactly whenThese hold at $r=0$ because $d_R>0$. Thus a sufficiently small pure branch is transversely stable; it is stable modulo its neutral temporal phase until one of the two inequalities fails. The secondary steady and Hopf thresholds of a resonant pure mode are derived in the two parts below. They refer to accessible points on the pure branch; in a multi-parameter problem $d$ may also vary, so a proposed parameter path must actually cross the corresponding surface.