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Put $M=m_1+m_2$ and let $\mathbf{R}=(m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2)/M$ be the center of mass. Applying Newton's second law separately to the two point masses givesThus the center of mass has constant velocity, $\mathbf{R}(t)=\mathbf{R}(0)+t\dot{\mathbf{R}}(0)$. For the relative position $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$, subtraction givesHere $\mu$ is the reduced mass; it converts the relative motion into a one-body equation of motion.

For the given linear force, the relative position obeys simple harmonic motion, $\ddot{\mathbf{r}}+{\Omega }^2\mathbf{r}=0$, where ${\Omega }^2=k/\mu$. The initial rest condition givesFor $d>0$, collision first occurs when the relative position vanishes, namely when $\Omega t=\pi /2$. ThereforeThe separation $d$ cancels because the frequency of this simple harmonic motion is independent of its amplitude.

A kinematic redshift is the relativistic Doppler effect arising from relative motion of emitter and receiver. In a local inertial frame, recession decreases the received frequency and approach increases it. A cosmological redshift instead compares photons emitted and received by comoving observers at different cosmic times: expansion of the scale factor stretches their wavelengths. It exists even when both observers have zero peculiar velocity. Both are frequency ratios measured by observers, but a large cosmological redshift should not be inserted into a flat-spacetime radial-speed formula to define the galaxy's unique recession speed.

Let ${\nu }_c$ be the photon frequency in the local comoving frame just after emission. The kinematic redshift gives ${\nu }_e/{\nu }_c=1+z_{\mathrm{kin}}$, and propagation in the FRW metric gives ${\nu }_c/{\nu }_0=1+z_{\mathrm{cosm}}$. Multiplication givesEquivalently, $z_{\mathrm{tot}}=z_{\mathrm{cosm}}+z_{\mathrm{kin}}+z_{\mathrm{cosm}}{z}_{\mathrm{kin}}$. Both redshifts being small is what permits their approximate addition. Here the observer's own peculiar velocity is assumed already corrected to the Hubble flow frame.

For Planck law expressed per unit frequency, put $x=h\nu /(k_{B}T)$. At fixed temperature, the frequency dependence is $B_{\nu }\propto T^3{x}^3/(e^x-1)$. Differentiation givesThe positive interior maximum therefore satisfies $3(1-e^{-x_{\ast }})=x_{\ast }$, with $x_{\ast }=2.821439\dots$ independent of temperature. The zero solution of the rearranged equation is a boundary, not the interior maximum. Hence the frequency form of Wien's displacement law isThis frequency peak is not $c$ divided by the peak wavelength of $B_{\lambda }$: the two spectral densities differ by the frequency-to-wavelength Jacobian.

Motion toward incoming cosmic microwave background radiation produces a blueshift. The photon occupation function of a blackbody depends on $h\nu /(k_{B}T)$, so the same Doppler factor multiplies both frequency and effective temperature. Looking forward along the Sun's peculiar velocity, one obtainsTaking the mean cosmic microwave background temperature as about $2.73\mathrm{K}$ givesThe numerical total inherits the rounded mean temperature; the robust estimate is the positive millikelvin dipole increment.

Floquet theory suggests a sequence of instability tongues nearAt these limiting frequencies the unperturbed Floquet multipliers $e^{\pm 2\pi i\omega }$ coincide. Each multiplication by $\cos t$ changes a Fourier frequency by one, so $m$ forcing interactions are needed to couple $+m/2$ and $-m/2$. This explains both the sequence of parametric resonances and the successive powers of $ϵ$ in their widths.

There is no implication that the exact half-integers themselves lie in every finite-$ϵ$ growing interval. The tongue centers generally shift by $O({ϵ}^2)$, which exceeds the width when $m>2$. It is the narrow intervals near the shifted centers that contain exponentially growing solutions. Frequencies staying a fixed distance from every half-integer are stable for sufficiently small $ϵ$.

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.

Set $T=ϵt$ and use the method of multiple scales:Since ${\omega }^2=1/4+kϵ+O({ϵ}^2)$, the first correction satisfiesThe resonant cosine and sine coefficients must vanish to avoid secular terms. Using the product-to-sum formulas givesThe remaining forcing has frequency $3/2$ and permits the bounded particular correction $y_1=(a/4)\cos (3t/2)+(b/4)\sin (3t/2)$. The two slow amplitudes carry the two arbitrary constants of the original equation.

For an explicit general leading solution, putThe matrix exponential givesThe error is uniform on bounded intervals of $T$, for bounded initial data. When $\lambda$ is imaginary this expression uses ordinary sine and cosine; at $\lambda =0$ its continuous limit is $I+TM$.

The primary instability tongue of a weak Mathieu oscillator has a positive slow eigenvalue precisely whenThese are the leading edges, rather than an exact finite-$ϵ$ interval. At the leading endpoints the amplitude matrix is a nilpotent matrix, so the leading solution can grow linearly in $T$ and does not decide the exact exponential instability. For completeness, set $\omega =1/2+\beta ϵ+\gamma {ϵ}^2$. The periodic or antiperiodic edge calculation uses a pure cosine when $\beta =-1/2$ and a pure sine when $\beta =1/2$, with the third-harmonic correction just found. The next resonant coefficient is $-(\gamma +3/8)y_0$, soThus the exact growing interval is between these shifted edges. In particular, for positive small $ϵ$ the fixed leading value $k=-1/2$ lies just inside it, whereas $k=1/2$ lies just outside. Interior detunings bounded away from the edges are already resolved by the leading method of multiple scales.

Place the four fixed charges at $(sa,ta)$ with $s,t\in \{-1,1\}$, and write the moving particle's position as $(x,y)$. Introduce its mass $m>0$, which is needed for an oscillation frequency but is not named in the question. Let $C=Qq/(4\pi {ϵ}_0)>0$. By Coulomb's law and superposition, the moving particle's potential energy isThe electrostatic potential itself is $V/q$. Put $r^2=x^2+y^2$ and ${\delta }_{st}=-(sx+ty)/a+r^2/(2a^2)$. Near the centre, a Taylor expansion givesOn summing over the four corners, linear terms cancel. To quadratic order, $\sum {\delta }_{st}=2r^2/a^2$ and $\sum {\delta }_{st}^2=4r^2/a^2+O(r^4/a^4)$, since $\sum st=0$. Reflection symmetry removes all odd-degree terms. HencewhereThus the centre has zero planar force and a positive-definite planar Hessian matrix $\kappa I_2$. It is a strict local minimum of the constrained potential energy, so it is a stable equilibrium. More explicitly, conservation of the total energy confines sufficiently small-energy motions within a small neighbourhood of this isolated local minimum.

Keeping only the quadratic potential, the two coordinates obey independent small oscillations:Both planar modes have the same angular frequency, and the ordinary frequency isThe planar constraint is essential to this constrained electrostatic equilibrium. Along the perpendicular axis, $V(0,0,z)=4C/(2a^2+z^2{)}^{1/2}=V(0,0)-C{z}^2/(\sqrt{2}{a}^3)+O(z^4)$, so the same stationary point is unstable to unconstrained normal displacements. That also checks the cancellation of the three-dimensional Hessian trace required by Laplace's equation away from charges; it does not contradict the positive-definite restriction to the given plane.

Apply the method of multiple scales with independent fast time $t$ and slow time $T=\epsilon t$, and write $x=x_0(t,T)+\epsilon x_1(t,T)+\cdots$. The chain rule gives $d/dt={\partial }_t+\epsilon {\partial }_T$. At leading order,At first order,A forcing component at the natural frequency produces a secular term in $x_1$. Thus the solvability condition in the method of multiple scales is orthogonality to both $\sin \theta$ and $\cos \theta$. Define the fast-phase average, holding $R,\varphi ,T$ fixed, bySince $⟨{\sin }^2\theta ⟩=⟨{\cos }^2\theta ⟩=1/2$ and $⟨\sin \theta \cos \theta ⟩=0$, the amplitude-phase equations areThey are leading averaged equations, not exact equations for a purely sinusoidal solution of a general nonlinear oscillator.

For the cubic conservative force, $f=-R^3{\cos }^3\theta$, odd symmetry gives $⟨{\cos }^3\theta \sin \theta ⟩=0$, while $⟨{\cos }^4\theta ⟩=3/8$. HenceIf $x(0)=x_{\ast }$ and $x^{\prime }(0)=v_{\ast }$, leading initial matching givesand thereforeon bounded slow time intervals for fixed bounded initial data. The positive frequency shift agrees with the increased restoring force of $x^{\prime \prime }+x+\epsilon x^3=0$. The zero initial state is the exact zero solution, with irrelevant phase. If a first waveform correction is desired, the remaining first-order forcing is $-R_0^3\cos 3\theta /4$, so $x_1=R_0^3\cos 3\theta /32$ plus a homogeneous correction chosen to match the initial data at the next order. This exhibits explicitly why the leading harmonic ansatz is approximate.

Substitute $\varphi =e^{i(kx-\omega t)}$ into the mixed-derivative equation. Its left side is $k\omega \varphi$, so for $k\ne 0$ the dispersion relation, phase velocity and group velocity areThe graph of a function for each velocity is symmetric about $k=0$, singular at zero, and approaches the horizontal axis as $|k|$ grows. Energy packets travel right while their crests travel left.

At a distant fixed observer $x>0$, stationary components obey $x/t=c_g(k)$, so $|k|=\sqrt{\alpha t/x}$. As time increases their wavelength $2\pi /|k|$ decreases and their frequency magnitude $|\omega |=\sqrt{\alpha x/t}$ decreases; the crests pass toward decreasing $x$.

For the given initial spectrum the solution isThe value of a regular spectrum at the single point $k=0$ is irrelevant to this integral; a nonzero spatially constant mode is incompatible with the PDE. For asymptotic calculation assume the usual smooth localized spectrum, with sufficient integrability for the following stationary-phase steps.

At $x=Vt$, $V>0$, the phase $\Phi (k)=Vk+\alpha /k$ has stationary points $k=\pm k_0$, $k_0=\sqrt{\alpha /V}$. Their values are $\pm 2\sqrt{\alpha V}$ and their second derivatives are $\pm 2\alpha /k_0^3$. Expanding quadratically and using the oscillatory Gaussian integral gives the stationary phase leading termFor real data this is twice the real part of the positive-frequency term. If a stationary coefficient vanishes, the formula instead asserts that its contribution has smaller order. At $x=-Vt$, the phase derivative is $-V-\alpha /k^2<0$, so there is no real stationary point and no corresponding $t^{-1/2}$ wave packet. For smooth compactly supported spectra away from zero, repeated integration by parts gives decay faster than any power of $t$. More general spectra require endpoint/tail control; absence of stationary points alone does not justify that stronger rate for arbitrary $A$.

The bounded classical solution is the heat-kernel convolutionThe two-dimensional heat kernel has integral one. Boundedness of $g$ permits differentiation under the integral for $t>0$, proving $u_t=\Delta u$ and $|u|\le \Vert g{\Vert }_{\infty }$. Its approximate identity property gives $u(x,t)\to g(x)$, locally uniformly since $g$ is continuous. This bounded solution satisfies the required Gaussian-growth bound.

For heat-equation uniqueness under Gaussian growth, let $w$ be the difference of two solutions, so $w(x,0)=0$ and $|w|\le M_0{e}^{a|x{|}^2}$ on a finite time interval. Choose $b>a$ and a slab $0\le t\le \tau <1/(4b)$. The positive comparison solutionsatisfies $({\Phi }_b{)}_t=\Delta {\Phi }_b$. For any $\epsilon >0$, its faster spatial growth makes $|w|\le \epsilon {\Phi }_b$ on a sufficiently large lateral cylinder boundary. At the initial boundary, $w=0$. The heat equation maximum principle applied to $\pm w-\epsilon {\Phi }_b$ proves the same inequality inside. First allow the cylinder radius to increase, then let $\epsilon \downarrow 0$. Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of $g$ is required.

At time $T$, this is Gaussian filtering with covariance $2TI$, hence $\sigma =\sqrt{2T}$ per coordinate. With the angular-frequency Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, the Gaussian filtering Fourier multiplier isFor merely bounded $g$, this identity is understood through tempered distributions. The multiplier leaves the zero frequency unchanged and attenuates large frequencies exponentially: it suppresses rapid noise fluctuations but blurs image edges. In cycles-per-length frequency $k$, the multiplier is $e^{-4{\pi }^{2}T|k{|}^2}$.

For the printed Perona-Malik equation, retain the supplied conductance $c(s)=s{e}^{-s^2/(2{\lambda }^2)}$, which includes an extra factor $s$. In one dimension write $p=u_x$ and $F(p)=c(|p|)p=p|p|e^{-p^2/(2{\lambda }^2)}$. Then $u_t=F^{\prime }(u_x)u_{xx}$, withThe forward-backward threshold for gradient-weighted exponential diffusion is thereforeAt $|u_x|=0$ and $\sqrt{2}\lambda$ the coefficient vanishes, giving degenerate diffusion. Increasing $\lambda$ increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and ill-posedness, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by $e^{-s^2/(2{\lambda }^2)}$ would give threshold $\lambda$, but that is a different equation.

The four-neighbour mean expansion follows by Taylor expansion: opposing first and third derivatives cancel, soIf $u(x_0)={\mathrm{mean}}_h(u)(x_0)$ for every sufficiently small $h$, division by $h^2$ and passage to the limit give $\Delta u(x_0)=0$. This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.

For the area median over the disk, the disk-median curvature expansion, at a point where $|\nabla u|\ne 0$, isConsequently the analogous median fixed-point condition yieldsThis is the vanishing of the level-line curvature at $x_0$: to second order the level line is straight there. It need not be a straight segment. For example $u(x,y)=y+x^3$ has zero disk median at the origin for every radius, by odd symmetry, but its zero level line is the cubic $y=-x^3$. Vanishing curvature on an entire connected regular level arc would force that arc to be straight. The factor $1/6$ is for the disk's uniform area measure; a circle-boundary median has a different scale factor, while producing the same zero-curvature equation.

Assume $0<ϵ\ll 1$ and $X,Y,\alpha =O(1)$. For $n=2$, use the method of multiple scales with $T=ϵt$ and complex amplitudesAt first order the forcing equations areThe product has frequencies $\pm 1,\pm 3$. Its $e^{it}$ coefficient is $\overline{A}B$, while it has no $e^{2it}$ coefficient. Removing the secular terms givesThe leading initial conditions give $A(0)=X/2$ and $B=Y/2$. Put $b=Y/2$, $A=p+iq$. The primary parametric resonance of a two-to-one oscillator pair is governed byFor $b^2>{\alpha }^2$, choose $\lambda >0$. Solving with $p(0)=X/2,q(0)=0$ yieldsIf $b^2<{\alpha }^2$, let $\nu =\sqrt{{\alpha }^2-b^2}/2$; replace $\cosh (\lambda T)$ by $\cos (\nu T)$ and $\sinh (\lambda T)/\lambda$ by $\sin (\nu T)/\nu$. On the boundary $b^2={\alpha }^2$, take the continuous limit:Thus exponential growth occurs when $|Y|>2|\alpha |$ and $X\ne 0$, on the scale $t=O({ϵ}^{-1})$. Equality is a marginal case that can produce linear slow growth: for $b=\alpha \ne 0$ the displayed initial phase excites it, whereas for $b=-\alpha$ it does not. If $X=0$, $x$ stays identically zero and the other oscillator is exactly uncoupled. The leading $y$ amplitude is constant; its backreaction is a higher-order effect on this time scale.

These are leading approximations for bounded slow time $T$, hence $t=O({ϵ}^{-1})$, while the amplitudes stay $O(1)$. First-order corrections can enforce the initial velocities exactly; the leading formula alone has an $O(ϵ)$ initial-velocity mismatch. In the unstable case the expansion must not be extrapolated through arbitrarily large amplification, where weak coupling and omitted terms cease to be uniform.

For $n=1$, both leading oscillations have frequencies $\pm 1$, so their quadratic product contains only frequencies $0,\pm 2$. There is no first-order resonant coupling: the solvability condition in the method of multiple scales gives $A_T=i\alpha A/2$ and $B_T=0$, a frequency shift without first-order amplitude growth. The constant and second-harmonic corrections can feed back into the fundamental at second order. Consequently the first possible growth scale from this second-order resonance from quadratic oscillator coupling isThis identifies the possible slow scale, not a guaranteed instability for every parameter or initial state. In particular, the first-order $\alpha$ frequency shift may detune the second-order resonance unless appropriately tuned; a second-order amplitude analysis would decide actual growth.

In the local short-wavelength approximation, derivatives of the slowly varying coefficients are smaller than derivatives of the phase. For the perturbation $v{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, replace ${\partial }_t$ by $-i\omega$ and ${\mathbf{B}}_p\cdot \nabla$ by $i\mathbf{k}\cdot {\mathbf{B}}_p$ in the torsional Alfvén wave equation. Cancel the common nonzero amplitude and $R^2$ to obtainThus the local dispersion relation is an Alfvén wave relation with Alfvén velocity ${\mathbf{v}}_{A,p}={\mathbf{B}}_p/\sqrt{{\mu }_0\rho }$. The approximation requires wavelength small compared with the background variation scales. A wave vector exactly perpendicular to ${\mathbf{B}}_p$ gives zero leading frequency, so cannot simultaneously obey the assumed large-frequency limit.

The linearized azimuthal Euler momentum equation isFor the time dependence $e^{-i\omega t}$ and $u_R^{\prime }=-i\omega {\xi }_R$, it givesThis expresses conservation of the displaced element's specific angular momentum. It applies directly to nonzero-frequency modes, with the zero-frequency limit taken in the displacement formulation.

The radial advective acceleration supplies $-2\Omega v_{\varphi }$, while the pressure force perturbation is $\delta \rho \nabla p/{\rho }^2-\nabla \delta p/\rho$. Eliminate $v_{\varphi }$ and retain the vertical equation. Under the Cowling approximation, $\delta \Phi =0$, soThe continuity equation gives $\delta \rho =-\nabla \cdot (\rho \mathbf{\xi })$. The Lagrangian adiabatic relation $\Delta p/p=\gamma \Delta \rho /\rho$, with $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, givesHere $\kappa$ is the radial epicyclic frequency. These formulas define the axisymmetric adiabatic displacement operator.

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.

The dimensional linear friction coefficient is $\zeta$. Put $y=x-v_{T}t$ and $u(y)=F(y)/\zeta$. In overdamped particle dynamics the laboratory velocity is $\dot{x}=u(y)$, whereas the trap-frame velocity is $\dot{y}=u(y)-v_T$. The particle enters the optical tweezers at $y=X_R$ and leaves at $y=-X_L$ when the trap overtakes it. Write $w=X_R+X_L$ for this support width. The printed subtraction in the later definition of the width is a sign error: the distance between these endpoints is their sum. We use $w=2X_0$ in the circular calculation.

For a bounded continuous force, the strict conditionensures finite passage through a translating optical trap: the trap-frame coordinate decreases throughout the interaction. If $u(y)=v_T$ is encountered from the right, the deterministic trajectory cannot pass this stationary trap-frame point for a locally Lipschitz force; it approaches or stays at a locked state. At a smooth maximum the critical passage time diverges. More generally the passage criterion is $v_T-u>0$ on the traversed interval together with a finite integral below. Separating the trap-frame equation givesThese formulas account for both portions of the optical tweezers, including motion against the trap direction.

The forward-displacement result uses the ordinary localized potential energy interpretation of an optical trap: $F=-U^{\prime }$ and $U$ has the same value outside both ends, hence $\int Fdy=0$. Compact support of the force alone does not ensure that condition. Under the equal-endpoint condition,for a nonzero force and a passing trajectory. This is forward displacement from a translating localized potential. A zero force gives zero displacement. If the force is negative everywhere on its support, it instead gives $\Delta x<0$; thus an arbitrary compactly supported force does not satisfy the printed assertion.

There are two complementary explanations of the forward displacement from a translating localized potential. The forward push slows the relative passage and therefore acts longer, while the backward pull speeds up the relative passage and therefore acts for less time. An exact potential energy balance makes the same point: along the trajectoryThe moving optical trap supplies the positive work lost to linear drag, although the initial and final potential energy are equal.

For $v_T\gg \Vert u{\Vert }_{\infty }$, a uniformly convergent expansion of the passage integrals givesFor a localized potential energy well the first integral vanishes, soThe displacement decreases quadratically with the trap speed, rather than linearly. Without the equal-endpoint assumption the general expansion above remains valid.

For repeated kicks from a circular optical trap, let $L=2\pi R$ and interpret the force profile locally along the arc. This description requires a tangential constraint, a nonoverlapping force support $w<L$, and, if the straight profile is used geometrically, a small support compared with $R$. The condition $R\gg a$ alone does not specify that latter width hierarchy. During an encounter the particle advances by $\Delta x$ and the trap advances by $v_T\Delta t=w+\Delta x$. Between encounters the deterministic particle is stationary and the trap travels the remaining relative distance $L-w$. Thus the time between corresponding points of successive encounters isHere $f_T$ and $f_p$ are revolution frequencies, not angular velocities; the printed relation $f_T=v_T/(2\pi R)$ fixes this convention. The encounter frequency is $f_T-f_p$, so the same result follows from $L{f}_p=\Delta x(f_T-f_p)$.