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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.