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A Hopf bifurcation creates or destroys a small periodic orbit near an equilibrium point as a pair of complex conjugate eigenvalues crosses the imaginary axis. It explains how a resting state can give way to sustained oscillations: the linearization of a dynamical system supplies rotation and growth, while a nonlinear term can saturate that growth at a nonzero amplitude.

The simplest supercritical Hopf bifurcation is the exact Hopf normal formHere $z$ is a complex number, $\mu$ is the bifurcation parameter, and $\omega$ is the angular frequency. In polar coordinates $z=r{e}^{i\theta }$, the equations become $\dot{r}=\mu r-r^3$ and $\dot{\theta }=\omega$ for $r>0$. At $\mu <0$, each turn shrinks towards the equilibrium point. At $\mu >0$, small radii grow and large radii shrink towards $r=\sqrt{\mu }$, producing a limit cycle. For example, $\mu =0.04$ and $\omega =2$ give radius $0.2$, period $\pi$, and frequency $1/\pi$ cycles per unit time. The limiting angular frequency stays nonzero while the amplitude tends to zero: a tiny newborn oscillation can still repeat at a finite rate.

For a precise generic statement, consider a smooth function $f$ defining an autonomous differential equation $\dot{x}=f(x,\alpha )$ on ${\mathbb{R}}^n$, with a smooth branch of equilibrium points $x_{\ast }(\alpha )$ and critical parameter $\alpha =0$. Require:Then a locally unique branch of small periodic orbits, up to phase, emerges on the side $\beta a<0$. On the two-dimensional centre manifold, the branch attracts for $a<0$ and repels for $a>0$. Attraction in the full phase space additionally requires all remaining eigenvalues to have negative real part; any transverse unstable direction makes the periodic orbit unstable there.^[ref]

An eigenvalue-crossing bifurcation test alone cannot establish these nonlinear conclusions. For example, the purely linear equation $\dot{z}=(\mu +i\omega )z$ has the same crossing but no isolated periodic orbit: at $\mu =0$ every nonzero radius is constant, and at $\mu \ne 0$ none is. If the cubic coefficient vanishes, higher-order terms must be examined; a generalized Hopf bifurcation is one possible degeneracy. The usefulness of the generic theorem is that a large system can have its small oscillatory onset understood through one Hopf normal form, subject to these hypotheses and local limits.

A planar node whose Jacobian matrix has a repeated nonzero real eigenvalue and only one independent eigenvector is an improper node. In suitable real coordinates its linearization of a dynamical system has the displayed form, with solutionsIt is a stable node for $\lambda <0$ and an unstable node for $\lambda >0$. The factor $t$ makes generic trajectories tangent to the unique eigendirection as they approach the equilibrium in the attracting time direction. For a nonlinear autonomous system with this hyperbolic equilibrium, the sign of $\lambda$ still determines local stability; the local phase portrait has the same topological type as its linearization.

Small perturbations of a class-B laser exchange excitation between the upper-state population and cavity photon number. Linearization about a four-level lasing state gives $\ddot{w}+(r/{\tau }_2)\dot{w}+(r-1)w/({\tau }_c{\tau }_2)=0$. Its damped oscillation frequency is $(2\pi {)}^{-1}\sqrt{(r-1)/({\tau }_c{\tau }_2)-r^2/(4{\tau }_2^2)}$. Long population lifetime compared with photon lifetime permits underdamped relaxation.

Both components vanish atThe Jacobian matrix of the vector field isIts eigenvalues classify these equilibria by linearization of a dynamical system:All real parts are nonzero, so the linear classifications determine the local stability of the nonlinear autonomous system.

At the origin, the stable direction is the $y$-axis and the unstable direction is the $x$-axis. At $(1,0)$ the eigendirections are $(1,0)$ for $-1$ and $(1,-2)$ for $-3$. At $(0,1)$,has only one eigendirection, the $y$-axis; its linearization of a dynamical system has solutions $\delta x=A{e}^{2t}$, $\delta y=(B-At)e^{2t}$, explaining the improper node. At $(4,3)$, stable and unstable tangent vectors are respectively $(4,5-\sqrt{13})$ and $(4,5+\sqrt{13})$.

For the phase portrait, the axes are invariant, and the remaining nullclines are $y=x-1$ and $y=x/2+1$. Off these lines, the complete directional information isIn the positive quadrant, arrows point right above $y=x-1$ and left below it, and point up above $y=x/2+1$ and down below it. On the positive $x$-axis, trajectories tend to $(1,0)$. On the $y$-axis, $0<y<1$ flows toward the origin and $y>1$ flows upward. These signs and the saddle directions give the following sketch; the plotted curves illustrate the phase portrait, with the marked saddle curves tangent to the calculated eigenvectors.

A critical point of a dynamical system is a point where both components of the vector field vanish. For a continuously differentiable vector field, strictly negative real parts of both eigenvalues of its Jacobian matrix imply asymptotic stability of the equilibrium. If an isolated critical point is a center of a planar dynamical system and the eigenvalues are distinct, they must be a nonzero purely imaginary conjugate pair. Purely imaginary eigenvalues alone do not in general prove nonlinear stability; the nonlinear ordinary differential equation system must still be examined.

Here the critical points are $(0,0)$ and $(\pm 1,0)$, with Jacobian $\left(\begin{array}{cc}0& 1\\ 1-3x^2& 0\end{array}\right)$. At zero its eigenvalues are $1,-1$, giving an unstable saddle equilibrium. The unstable direction is $(1,1)$ and the stable direction is $(1,-1)$. Thus trajectories leaving the origin are tangent to $y=x$, and those returning are tangent to $y=-x$, with both opposite rays included.

An exact first integral isThe two homoclinic loops belong to $H=0$:They confirm the incoming and outgoing tangent directions. For example the right loop is $x=\sqrt{2}\mathrm{sech}(t-t_0)$, $y=-\sqrt{2}\mathrm{sech}(t-t_0)\tanh (t-t_0)$; changing both signs gives the left loop.

At $(1,0)$ put $x=1+X$, $y=Y$. The exact equations are $X^{\prime }=Y$, $Y^{\prime }=-2X-3X^2-X^3$, so the linearization of a dynamical system gives $X^{\prime \prime }+2X=0$. Its nonconstant solutions haveThe same linear classification holds at $(-1,0)$. For the nonlinear ordinary differential equation system, $H$ has strict local minima $-1/4$ at both points. Levels $-1/4<H<0$ are two families of smooth closed ovals, proving that both are stable centers, rather than asymptotically stable equilibria. These give actual periodic orbits of the nonlinear ordinary differential equation system, with small-amplitude periods tending to $\sqrt{2}\pi$.

A laser relaxation oscillation is the damped exchange between stored upper-level excitation and intracavity photon number after a small perturbation of a lasing steady state. Extra photons increase stimulated depletion of the upper level; the resulting smaller gain then reduces the photon population. Pumping subsequently replenishes the inversion, so the two populations overshoot their equilibrium values before damping settles them.

Laser spiking is the corresponding large-signal sequence of intense output pulses, often during switch-on. Excitation first accumulates while the optical field is small. Once gain exceeds loss, rapid stimulated emission releases this stored energy in a spike, depleting the inversion below threshold. Recovery can produce further spikes. This is distinct from an externally imposed Q-switching cycle, although both draw on stored excitation.

Many solid-state gain media have upper-state lifetimes far longer than the cavity photon lifetime. The optical polarization relaxes rapidly, leaving the two slow rate-equation variables of a class-B laser. A long-lived population reservoir and a much faster photon response make underdamped recovery and pronounced startup spiking common. The criterion and frequency follow from the linearization in part c; they are not universal consequences of the word “solid-state”.

Introduce $v=\dot{x}$. The phase plane system for the damped pendulum isAt an equilibrium point, $v=0$ and $\sin x=0$, so all equilibrium points are $\boxed{(x,v)=(n\pi ,0),n\in \mathbb{Z}}$. Angles identified modulo $2\pi$ give one downward and one upward equilibrium point.

The linearization of a dynamical system at $(n\pi ,0)$ has Jacobian matrix and eigenvaluesWhen $n$ is even and $k>\omega$, both eigenvalues are real and strictly negative, since $0<\sqrt{k^2-{\omega }^2}<k$. These downward equilibrium points are stable nodes. When $n$ is even and $k<\omega$, the eigenvalues are $-k\pm i\sqrt{{\omega }^2-k^2}$, so the downward equilibrium points are stable foci, approached with decaying oscillations.

When $n$ is odd, $\sqrt{k^2+{\omega }^2}>k$, so one eigenvalue is positive and one is negative in either parameter range. Every upward equilibrium point is consequently an unstable saddle equilibrium. All these equilibrium points are hyperbolic equilibria in the two requested cases, so their linearization of a dynamical system gives their local nonlinear classification.

The phase plane system is $\dot{x}=y$, $\dot{y}=-cy-\sin x$. Its equilibrium points are $(k\pi ,0)$ for every integer $k$. At such a point the Jacobian matrix and characteristic equation areFor even $k$, the equilibrium point is a stable focus for $0<c<2$, with eigenvalues $-c/2\pm i\sqrt{1-c^2/4}$; it is a stable node for $c>2$, with two distinct negative real eigenvalues. The excluded value $c=2$ gives the repeated critical case.

For odd $k$, the eigenvalues are $(-c\pm \sqrt{c^2+4})/2$. One is positive and one negative, so each is a saddle equilibrium for every $c>0$. These classifications follow from the linearization of a dynamical system, since all the relevant eigenvalues have nonzero real part for $c>0$. The even equilibria remain hyperbolic at $c=2$ too, although their linearization then has a repeated eigenvalue; that critical case is excluded from the requested classification.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

Write the deterministic drift of the Adler phase equation as $f(\theta )=\omega -ϵ\sin \theta$. Its minimum is $\omega -ϵ$ and its maximum is $\omega +ϵ$. For $0<\omega <ϵ$, the zero condition has exactly two solutions in the specified interval:The linearization of a dynamical system at an equilibrium point gives $\delta \dot{\theta }=f^{\prime }({\theta }_{\ast })\delta \theta$, with $f^{\prime }=-ϵ\cos \theta$. Defining $\kappa =\sqrt{{ϵ}^2-{\omega }^2}$ gives $f^{\prime }({\theta }_s)=-\kappa <0$, so ${\theta }_s$ is stable; $f^{\prime }({\theta }_u)=+\kappa >0$, so ${\theta }_u$ is unstable. For $\omega >ϵ$, $f$ is positive everywhere and there are no equilibrium points: the phase runs continuously. This is the distinction between phase locking and running phase dynamics.

With mobility scaled to one, the effective force is $-V^{\prime }$. Therefore the tilted washboard potential isFor $\omega <ϵ$, its alternating local minima and maxima trap noise-free trajectories in wells. Minima coincide with ${\theta }_s+2\pi j$, and maxima with ${\theta }_u+2\pi j$. For $\omega >ϵ$, $V^{\prime }=ϵ\sin \theta -\omega <0$ everywhere: there are no wells and the particle slides down the tilt. The potential is defined on the unwrapped phase and obeys $V(\theta +2\pi )=V(\theta )-2\pi \omega$; it is not a single-valued periodic equilibrium potential on the circle.

At the transition $\omega =ϵ$, the two equilibrium points merge at $\theta =\pi /2$ in a saddle-node bifurcation. There $f\simeq ϵ(\theta -\pi /2{)}^2/2$, so the point is attracting from the left and repelling from the right; a zero linear derivative alone does not establish stable trapping.

At the disease-free equilibrium point $(1,0)$, the linearization of a dynamical system of the infected equation is $\dot{y}=(\beta -d)y$. Thus exponential invasion occurs exactly when $\boxed{\beta >d}$. At equality there is no positive linear growth rate; the nonlinear boundary dynamics described below do not give sustained invasion.

There are $k{x}_n$ seeds and their germinating proportion is $e^{-\gamma x_n}$, giving the survival-augmented Ricker map with zero survival. A positive equilibrium point satisfies $e^{-\gamma x_{\ast }}=1/k$, soIt exists when $k>1$ and is strictly linearly stable when $|1-\log k|<1$. Thus the usual strict linear-stability interval is $\boxed{1<k<e^2}$.

With winter survival, surviving adults and new plants add:Writing $L=\log [k/(1-s)]$, the linearization of a dynamical system has multiplier $1-(1-s)L$. Hence the strict linear-stability interval becomesIts lower endpoint is less than $1$. Its upper endpoint exceeds $e^2$: for $t=1-s\in (0,1)$, $\log t+2/t$ is decreasing in $t$ and equals $2$ at $t=1$.

The nonlinear stability at the Ricker flip threshold needs a separate check. The upper endpoints have multiplier $-1$, so the strict linear test alone does not decide stability there. They are in fact locally asymptotically stable, with algebraic rather than geometric convergence. To see this, put $u=\gamma (x-x_{\ast })$ at $L=2/(1-s)$. The local map isand its second iterate is $u-2(s^2-s/2+1/6)u^3+O(u^4)$. The cubic coefficient in parentheses is positive for every $s\in [0,1)$. Thus, if “stable” includes nonlinear endpoint stability, the upper inequalities can be replaced by $\le$; the lower endpoints have no positive equilibrium.

Let $X=x-x_{L_3}$ and $Y=y$ be the displacement from the L3 Lagrange point, and put $U=\dot{X}$, $W=\dot{Y}$. The first-order linearization of a dynamical system iswhere every derivative is evaluated at $L_3$. For the state vector $\mathbf{\xi }=(X,Y,U,W{)}^{\mathsf{T}}$, this becomesThe question's reuse of $\mathbf{X}$ for the state vector is only notation; its first two entries are the small displacements $X,Y$, not the absolute coordinates.

Writing $\xi =x-1$, $\eta =y$, the linearization of a dynamical system is $\dot{\xi }=-6\xi$, $\dot{\eta }=-4\eta$. Eliminating $t$ gives $\eta =C{\xi }^{2/3}$.

A resistive velocity term proportional to the speed squared removes energy at rate proportional to $|\omega {|}^3$. Its derivative at zero velocity vanishes, so it does not appear in the linearization of a dynamical system at rest. Nonlinear energy arguments can still prove asymptotic stability.