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Put $x=z$, $y=\dot{z}$. The system isIts equilibria are $\boxed{(0,0),(1,0),(-1,0)}$. At $(\pm 1,0)$, the eigenvalues are $(-A\pm \sqrt{A^2+8})/2$, of opposite signs, so these are saddles for every $A$.

At the origin, the eigenvalues are $(-A\pm \sqrt{A^2-4})/2$. It is a stable focus for $0<A<2$, a stable node for $A>2$, an unstable focus for $-2<A<0$, and an unstable node for $A<-2$. At $A=\pm 2$ it is the corresponding repeated-eigenvalue improper node. At $A=0$, linearization gives a center, but nonlinear damping matters: withsmall closed energy sublevels are invariant and $\dot{E}=-y^4$ forces convergence to the origin. Thus it is a weak attracting focus, not a nonlinear center.

Local bifurcation. Write $a=-A$. Averaging the radial equation near the origin, or eliminating nonresonant cubic terms, givesThe $x^3$ restoring term has zero average radial contribution, while $⟨y^4⟩=3R^4/8$. Thus $A=0$ is a supercritical Hopf bifurcation as $a$ increases: a stable cycle appears for small $a>0$, with leading amplitude $R\simeq \sqrt{4a/3}$.

Global changes. Two further bifurcations must be retained. Define $a_h$ as the parameter where the upper unstable branch from $(-1,0)$ reaches $(1,0)$. Central symmetry supplies the lower connection in the opposite direction, forming a heteroclinic cycle. Define $a_f>a_h$ as the subsequent collision of an inner attracting and an outer repelling periodic orbit. The complete phase portraits for $|A|<1$ are:

Here escape means leaving the central well and becoming unbounded; it is not an additional finite equilibrium. The softening Rayleigh-Duffing phase portrait in the figure includes the two critical cases.

The return-map calculation underlying this classification is useful because the two global critical parameters are very close. On the section $x=0,y>0$, let $H_a(p)$ be minus the velocity at the next crossing of $x=0$, so a symmetric cycle is a root of $H_a(p)-p$. Any periodic orbit must have its extrema inside $-1<x<1$, since at a positive maximum $y=0$ requires $x^3-x<0$. Its index is one, so it surrounds the origin and neither saddle. Every such cycle is centrally symmetric: its reflected image cannot be a distinct nonintersecting cycle of equal enclosed area. Thus this half-return test detects all central cycles. A fixed point with $H_a^{\prime }<1$ is attracting and one with $H_a^{\prime }>1$ repelling; the full-return multiplier is $(H_a^{\prime }{)}^2$.

For the saddle connection, let $p_L(a)$ be the upper velocity at $x=0$ from the left saddle's unstable branch and $p_R(a)$ the upper velocity at $x=0$ on the right saddle's stable branch. While $y>0$, their shooting equation isComparison shows $p_L$ increases with $a$ and $p_R$ decreases. At $a=0$, energy dissipation gives $p_L<1/\sqrt{2}<p_R$. At $a=1$, the lower comparison curve $y=(x+1)/2$ on $[-1,0]$ gives $p_L>1/2$, while the upper comparison curve $y=(1-x^2)/2$ on $[0,1]$ gives $p_R<1/2$. The latter follows fromand the stable-branch expansion $y=(1-x)-3(1-x{)}^2/4+\cdots$ near $x=1$. Hence there is a unique $a_h\in (0,1)$ with $p_L=p_R$.

At that connection the saddle eigenvalues are ${\lambda }_u=(a_h+\sqrt{a_h^2+8})/2$ and ${\lambda }_s=(a_h-\sqrt{a_h^2+8})/2$, so $|{\lambda }_s|/{\lambda }_u<1$. The loop is therefore repelling. In particular, the attracting Hopf cycle cannot simply disappear in that connection. Near the upper endpoint $p_R$, the half-return map hasFor $a$ just above $a_h$, this produces a second, repelling fixed point near the endpoint, in addition to the attracting inner one. Continuing the two roots of $H_a(p)-p$ makes their collision criterion explicit:This is a saddle-node bifurcation of periodic orbits, not a disappearance of the equilibrium points.

For a reproducible numerical check of the global sketches, event-based integration of this half-return map and its variational equation givesThese values are computed by shooting the stated equations; their precision is unnecessary for the requested sketches. They explain why a coarse parameter sweep can miss the two-cycle interval entirely. The three bifurcation types are therefore Hopf at $A=0$, a two-saddle heteroclinic connection at $A=-a_h$, and a fold of cycles at $A=-a_f$. There is no pitchfork of equilibria because their locations remain fixed for every $A$.

Define $A=2\cot \alpha /3+8\mathrm{Re}/15$. The linearization about $h=1$ givesThe Galilean transformation $\xi =x-2t$, $\tau =t$ has ${\partial }_t={\partial }_{\tau }-2{\partial }_{\xi }$ and ${\partial }_x={\partial }_{\xi }$. It removes the leading advection, leavingFor a normal mode proportional to $e^{\sigma \tau +ik\xi }$, the second and fourth derivatives contribute $-k^2$ and $k^4$. The dispersion relation is thereforeFor $A>0$ and $S>0$, the unstable wavenumbers satisfy $0<|k|<\sqrt{A/S}$. In the laboratory coordinates, the temporal exponent is $\sigma -2ik$, so this linear stability calculation predicts growth together with downslope transport at dimensionless speed $2$.

For a scalar ordinary differential equation $N^{\prime }=f(N)$ and an equilibrium point $N^{\ast }$ with $f^{\prime }(N^{\ast })<0$, the linearization gives $n^{\prime }=f^{\prime }(N^{\ast })n$. Thus small perturbations decay like $e^{-t/\tau }$ with $\tau =-1/f^{\prime }(N^{\ast })$. This is a local exponential recovery time. In several dimensions, the slowest decaying stable eigenvalue typically controls asymptotic recovery, subject to additional polynomial factors for nontrivial Jordan blocks.

With $k_{\parallel }/k_{\perp }\sim \delta B/B_0\ll 1$, define $b=\delta B_{\parallel }/B_0$ and $\delta {\mathbf{B}}_{\perp }/B_0=\hat{\mathbf{z}}\times {\nabla }_{\perp }\Psi /v_A$ to leading order. Let ${\nabla }_{\parallel }={\partial }_z+\{\Psi ,\cdot \}/v_A$, where $\{f,g\}=f_x{g}_y-f_y{g}_x$. The leading electron velocities are ${\mathbf{u}}_{e\perp }=v_A{d}_i\hat{\mathbf{z}}\times {\nabla }_{\perp }b$ and $u_{e\parallel }=-d_i{\nabla }_{\perp }^2\Psi$. Substitution into frozen-in induction yieldsLinearization gives ${\omega }^2=v_A^2{d}_i^2{k}_{\parallel }^2{k}_{\perp }^2$, agreeing with the full EMHD relation to leading anisotropic order. If $b$ varies along the guide field, an order-smaller perpendicular gradient correction restores exact solenoidality; the two-field representation is not an exact decomposition at arbitrary anisotropy.

Release the upper rod at ${\theta }_1(0)=\alpha =0.06$ radians while keeping the lower rod vertical, ${\theta }_2(0)=0$, with both angular velocities zero. The initial condition is the sum $\mathbf{\theta }(0)=(\alpha /2){\mathbf{a}}_{-}+(\alpha /2){\mathbf{a}}_{+}$. The solution from linearization is thereforeAt release the lower angle vanishes by cancellation between the two normal modes. Their different angular frequencies then destroy that cancellation. Differentiating twice gives ${\ddot{\theta }}_1(0)=-2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)=2g\alpha /\ell$, agreeing with the small-angle limit of the exact double pendulum equations of motion.

The quadratic mechanical energy at release is $E_2=mg\ell {\alpha }^2$. Each normal mode contributes half, because $(4+2\sqrt{2})(2-\sqrt{2})=(4-2\sqrt{2})(2+\sqrt{2})=4$. Equal coordinate amplitudes need not mean equal energies in general; here the modal inertias and squared angular frequencies happen to compensate. Releasing instead along $\mathbf{\theta }(0)=\alpha {\mathbf{a}}_{-}$ or $\alpha {\mathbf{a}}_{+}$, with zero angular velocities, excites just one normal mode.

For $K\dot{p}=M-pV(p(t-\tau ))$ and a positive steady state, linearization gives $\dot{z}=-Az-Bz(t-\tau )$, where $A=V(p^{\ast })/K$ and $B=p^{\ast }{V}^{\prime }(p^{\ast })/K$. Splitting the characteristic equation of a delay differential equation at an imaginary root yields ${\omega }^2=B^2-A^2$ and the displayed first critical delay. The root crosses into the unstable half-plane with real derivative ${\omega }^2/[(1+A{\tau }_c{)}^2+{\omega }^2{\tau }_c^2]>0$. If $A\ll B$, the period near onset is approximately $4{\tau }_c$. For a Hill equation response, $B/A=n{\theta }^n/({\theta }^n+(p^{\ast }{)}^n)$ identifies whether delayed gain can exceed instantaneous damping.

A longer circulation time increases the delay between gas exchange and its sensed feedback. The American Thoracic Society's documented case relates a change in breathing pattern during one night to a longer estimated circulation time; the simple scalar model isolates a possible feedback mechanism rather than representing every determinant of that observation.^[ref]

For a shallow glacier with basal traction proportional to thickness and sliding law $u_b\propto {\tau }_b^m$, negligible internal-deformation flux gives $u_b\propto h^m$. After scaling, the thickness flux is $h^{m+1}$. Linearization around a steady thickness $h_{\ast }(x)$ gives ${\zeta }_t+[c(x)\zeta {]}_x=0$, $c=(m+1)h_{\ast }^m$. Thickness disturbances therefore propagate faster than the surface material speed $u_s\simeq h_{\ast }^m$ by factor $m+1$.

A steady accumulation profile and upstream discharge $q_{\mathrm{in}}$ give $h_{\ast }^{m+1}=q_{\mathrm{in}}+{\int }_0^{x}a(\xi )d\xi$. Along a disturbance ray, $dx/dt=c$ and $d\zeta /dt=-c^{\prime }\zeta$, so a spatially varying background also changes amplitude. The surface-speed ratio is specific to sliding-dominated flow; internal deformation and significant surface-slope corrections change the flux derivative.

For a fast-slow system $ϵ\dot{x}=f(x,y)$, $\dot{y}=g(x,y)$, the formal critical manifold is $f=0$. Where the fast linearization has eigenvalues with strictly negative real parts, trajectories rapidly approach a nearby attracting slow manifold and then evolve on the slow time. Near folds or loss of this spectral condition, simply imposing $f=0$ may fail.

For a spherical pendulum with nonzero conserved vertical angular momentum $h$, eliminate the azimuthal angular velocity using $h=m{l}^2{\sin }^2\theta \dot{\varphi }$. The polar effective potential is $V_{\mathrm{eff}}=h^2/(2m{l}^2{\sin }^2\theta )-mgl\cos \theta$. A conical pendulum at $0<{\theta }_0<\pi /2$ is a minimum satisfying $h^2=m^{2}g{l}^3{\sin }^4{\theta }_0/\cos {\theta }_0$. Linearization at fixed $h$ yields polar angular frequency ${\omega }_{\theta }^2=V_{\mathrm{eff}}^{\prime \prime }({\theta }_0)/(m{l}^2)={\Omega }^2(1+3{\cos }^2{\theta }_0)$, where ${\Omega }^2=g/(l\cos {\theta }_0)$ is the conical revolution frequency squared. Holding $\dot{\varphi }$ fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar $h=0$ limit requires separate treatment.

For $a=a_0(1+ϵ)$, equilibrium outer radius $b_0$, and a bubble gas obeying $p_0\pi a^2=\text{constant}$, the linearization isThe bubble therefore undergoes simple harmonic motion with

Consider fluid at rest with mean depth $d$ and a bottom displacement $a\sin (k(x-ct))$. The linearized free-surface boundary conditions at $y=0$ and moving-bottom kinematic boundary condition at $y=-d$ give a periodic particular response $\eta =b\sin (k(x-ct))$ and velocity potentialThe bottom derivative fixes the signed response coefficient to the title formula. Here $k>0$, $c\ne 0$, and the forcing must avoid $k^2{c}^2=gk\tanh (kd)$. This equality is resonance with the natural finite-depth surface gravity wave; nonzero undamped resonant forcing has no bounded periodic solution. The response reverses phase below resonance. General initial data can add free surface gravity waves, which do not decay in this inviscid model. Near resonance the resulting wave steepness and particle excursions, rather than the bottom displacement alone, determine whether the linearization remains valid.

For total plume flux $P(h)$, fixed ceiling exhaust $F$, zero ambient buoyancy, and uniform upper-layer reduced gravity $b$, the ideal two-layer balances areAt $P(h_{\ast })=F$ and $b_{\ast }=B/F$, linearization yields the decay times ${\tau }_h=A/P^{\prime }(h_{\ast })$ and ${\tau }_b=A(H-h_{\ast })/F$. For $P(h)\propto h^{5/3}$, ${\tau }_h=3A{h}_{\ast }/(5F)$. Upper-layer buoyancy relaxation usually dominates when $h_{\ast }\ll H$. Exact equilibrium is approached asymptotically; a fixed small relative tolerance takes order $\max ({\tau }_h,{\tau }_b)\log (1/\epsilon )$ after the initial transient.

A weak focus is an isolated planar equilibrium point with nonzero imaginary eigenvalues and nearby trajectories that spiral toward or away from it because of nonlinear terms. Its linearization is a center equilibrium, so linear analysis alone does not settle stability. The first nonzero radial drift in its Poincaré return map decides the direction.

For example, if the radial return map is $P(r)=r+\kappa r^{2m+1}+O(r^{2m+2})$ with $\kappa \ne 0$, sufficiently small positive $r$ decreases under successive returns when $\kappa <0$ and increases when $\kappa >0$. The first case gives asymptotic stability; the second gives instability. Equivalently, if a positive local function $W=r^2+O(r^3)$ satisfies $\dot{W}=c{r}^{2m+2}+O(r^{2m+3})$ and the angular velocity stays positive, then the sign of $c$ settles the same question. A quadratic planar system can have its cubic radial drift vanish and still have a nonzero fifth-order radial drift.

This real planar dynamical system has a saddle point at $(-1,0)$ and a repelling weak focus at $(1,0)$. The latter has purely imaginary linearized eigenvalues $\pm 2i$, so its linearization has closed elliptical orbits. Nonlinear terms nevertheless produce outward drift.

Set $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The cubic expansion isThe linear rotation acts on scalar functions by $L=Y{\partial }_X-X{\partial }_Y$. For $H_2=r^2/2$, its cubic derivative is $X{Y}^2-3X^{2}Y/4-Y^3/2$. The homogeneous correction $H_3=7X^3/12+X{Y}^2/2+Y^3/3$ satisfies $L{H}_3$ equal to the negative of that derivative. The remaining quartic derivative of $H_2+H_3$ is $3X^2{Y}^2/2-X^{3}Y$. Withdirect differentiation givesfor all sufficiently small nonzero $r$. Meanwhile $H_2+H_3+H_4=r^2/2+O(r^3)>0$. In reversed time it is a strict Lyapunov function, proving asymptotic stability of the reversed system and hence repulsion forward in time. The angular speed stays close to $-1$, so the nearby nonzero trajectories spiral rather than approaching along a fixed direction. This explicit calculation illustrates why a centre in the linearization need not be a nonlinear centre.

A saturated convection roll is not stable merely because its single-amplitude cubic coefficient is positive. Perturbations include phase sidebands, transverse bending, differently oriented convection rolls, oscillatory modes and large-scale Eulerian mean flow. Their growth rates follow from the corresponding enlarged amplitude equations or the full linearization about the convection roll. Küppers–Lortz instability and small-angle instability of rotating convection rolls are distinct orientation-dependent mechanisms.

In this idealized greenhouse-weathering model, temperature adjusts rapidly when $ϵ\ll 1$, giving $\theta \simeq \Gamma (P-1)$. The remaining carbon balance has equilibrium $\mu \log P_{\ast }+\Gamma (P_{\ast }-1)=\log v$. For positive $\mu ,\Gamma$, the left side has derivative $\mu /P+\Gamma >0$ and ranges from minus infinity to infinity, so there is exactly one positive equilibrium for every sustained positive source $v$.

Linearization of the reduced carbon equation gives dimensionless decay rate $v(\Gamma +\mu /P_{\ast })$, hence dimensional relaxation time $t_C/[v(\Gamma +\mu /P_{\ast })]$. Its negative feedback is increased weathering when either temperature or carbon pressure rises. The two-variable equilibrium is also linearly stable: its Jacobian has negative trace and determinant $v(\mu /P_{\ast }+\Gamma )/ϵ>0$. A fast radiative response therefore does not imply a fast carbon-cycle response. Linearized radiative coefficients must be calibrated at a consistent reference state before numerical use.

The linearization at a travelling wave $w_c$ is $A={\partial }_z^2+c{\partial }_z+k-2w_c$. Conjugation by $e^{\gamma z}$ changes this to ${\partial }_z^2+(c-2\gamma ){\partial }_z+{\gamma }^2-c\gamma +k-2w_c$. The choice $\gamma =c/2$ removes the first derivative and makes the operator self-adjoint. Its two limiting potentials are $-k-c^2/4$ and $k-c^2/4$, so essential-spectrum invariance under relatively compact perturbations gives ${\sigma }_{\mathrm{ess}}=(-\infty ,k-c^2/4]$. This lies strictly below zero for $c>2\sqrt{k}$ but touches zero at the critical speed. Stability of this essential spectrum alone is not a general nonlinear stability theorem.