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A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

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.

A planar equilibrium with real nonzero eigenvalues of the same sign. Negative eigenvalues give a stable node; positive eigenvalues give an unstable node. Repeated real eigenvalues may give a degenerate or improper node, depending on the eigenspace.

Both eigenvalues are real and negative: $\boxed{{\lambda }_1<0,{\lambda }_2<0}$, with distinctness as assumed. The two independent real eigendirections carry exponentially decaying factors $e^{{\lambda }_{j}t}$, giving a stable node.

The equilibrium equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian isAt $(0,0)$ it is $\mathrm{diag}(1,-1)$: a saddle. At $(-1,0)$ it is $\left(\begin{array}{cc}-1& 2\\ 0& -2\end{array}\right)$, with eigenvalues $-1,-2$: a stable node. At $(1,1)$ it is $\left(\begin{array}{cc}1& -2\\ 1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2-\lambda +2$ and eigenvalues $(1\pm i\sqrt{7})/2$: an unstable spiral. All three are hyperbolic, so the nonlinear local classifications follow from linearization.

Both eigenvalues are real and negative: $\boxed{{\lambda }_1<0,{\lambda }_2<0}$, with distinctness as assumed. The two independent real eigendirections carry exponentially decaying factors $e^{{\lambda }_{j}t}$, giving a stable node.

The equilibrium equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian isAt $(0,0)$ it is $\mathrm{diag}(1,-1)$: a saddle. At $(-1,0)$ it is $\left(\begin{array}{cc}-1& 2\\ 0& -2\end{array}\right)$, with eigenvalues $-1,-2$: a stable node. At $(1,1)$ it is $\left(\begin{array}{cc}1& -2\\ 1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2-\lambda +2$ and eigenvalues $(1\pm i\sqrt{7})/2$: an unstable spiral. All three are hyperbolic, so the nonlinear local classifications follow from linearization.

A stable node hasBoth exponential modes decay as $t\to \infty$, so the equilibrium is asymptotically stable. Generic linear orbits approach tangent to the eigenvector for the less negative eigenvalue; the eigenvector directions themselves are invariant straight lines. There is no rotation because both eigenvalues are real.

The first component of the vector field vanishes only when $y=0$ or $x=\pm 1$. The second vanishes only when $y=1$ or $x+y=3$. Intersecting these alternatives gives exactly five equilibria. Their Jacobian matrix isAt $(3,0)$,The real parts are negative and the imaginary parts are nonzero, so $(3,0)$ is a stable spiral.

At $(1,1)$ the Jacobian matrix is $\mathrm{diag}(2,-1)$. Its real eigenvalues have opposite signs, so $(1,1)$ is a saddle equilibrium.

At $(1,2)$,Both eigenvalues are positive, so $(1,2)$ is an unstable node.

At $(-1,1)$ the Jacobian matrix is $\mathrm{diag}(-2,-3)$, giving a stable node.

At $(-1,4)$,so $(-1,4)$ is a saddle equilibrium. All five equilibria are hyperbolic, so these linear classifications determine their nonlinear local stability as well.

Assume $\epsilon >0$. At an equilibrium, $w=\gamma v=f(v)$, soThe nonzero roots areConsequently, with ${\gamma }_c=(1-a{)}^2/4$, there are three distinct equilibria for $0<\gamma <{\gamma }_c$, two distinct equilibria for $\gamma ={\gamma }_c$, and only the origin for $\gamma >{\gamma }_c$. At equality the two nonzero equilibria coincide; for $0<\gamma <{\gamma }_c$ they both have $a<v<1$. Geometrically, ${\gamma }_c$ is the maximum of the concave quadratic $(a-v)(v-1)$.

Since $f^{\prime }(0)=-a$, the Jacobian matrix at the origin isThus the origin is always asymptotically stable. Its eigenvalues areWriting ${\gamma }_N=(a-\epsilon {)}^2/(4\epsilon )$, the origin is a stable node for $0<\gamma <{\gamma }_N$ and a stable focus for $\gamma >{\gamma }_N$. At $\gamma ={\gamma }_N>0$ it is a degenerate node: the repeated negative eigenvalue has only one independent eigenvector, since the nonzero off-diagonal entry prevents $J_0$ being a scalar matrix. If $a=\epsilon$, then ${\gamma }_N=0$ and every permitted $\gamma >0$ gives a stable focus. This change from node to focus does not change stability and is not a Hopf bifurcation.

The term $p$ represents prey reproduction and $-p^2/a$ limits growth by available food, giving carrying capacity $a$ in the absence of hunters. The loss $-ph/a$ represents encounters between prey and hunters. For hunters, $hp/(8b)$ is prey-supported growth and $-h/8$ is mortality; their per-capita growth changes sign at prey population $b$. The coordinate axes are invariant, and positive initial populations remain positive.

For $a=1$, the physical equilibrium points are the origin, $(1,0)$, and $(b,1-b)$ when $b<1$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1$ and $-1/8$, so the origin is always a saddle equilibrium, attracting along the hunter-only axis and repelling along the prey-only axis. At $(1,0)$ the eigenvalues areIt is a saddle equilibrium for $0<b<1/2$ and an asymptotically stable node for $b>1$.

For $0<b<1/2$, the coexistence Jacobian matrix isIts trace is negative and determinant positive. Its discriminant $b^2-(1-b)/2=(2b-1)(b+1)/2$ is negative, so $(b,1-b)$ is an asymptotically stable focus. Interior trajectories approach it in damped, counterclockwise oscillations in the $(p,h)$ plane. Both populations ultimately coexist.

To justify the global interior fate for this logistic predator-prey model, let $h_{\ast }=1-b$ and use the Lyapunov functionIt is nonnegative and its sublevel sets are compact inside the positive quadrant. Differentiation gives the explicit cancellationThe only invariant subset on which this vanishes is $p=b,h=h_{\ast }$: remaining at $p=b$ requires $\dot{p}=b(1-b-h)=0$. LaSalle's invariance principle therefore proves that every strictly positive initial state converges to this coexistence point.

For $b>1$, the formal coexistence point has negative $h$ and is outside the physical state space. The only physical equilibria are the saddle equilibrium at the origin and the stable prey-only node. Since $\dot{p}\le p(1-p)$, positive prey has $\mathrm{limsup}p\le 1$. Eventually $p<b$ with a fixed margin, so hunter population decays exponentially. When $h$ is sufficiently small, comparison with $\dot{p}=p(1-\epsilon -p)$ gives $\mathrm{liminf}p\ge 1-\epsilon$ for every $\epsilon >0$. Thus $h\to 0$ and $p\to 1$ for positive initial prey: hunters die out and prey reaches carrying capacity.

On the boundary, hunter-only initial data approach the origin; prey-only positive data approach $(1,0)$ in either parameter range. The sketches show representative values in the two requested ranges. Dashed and dotted lines are the prey and hunter nullclines; filled dots are attractors and open dots are saddle equilibria.

Write $k=\sqrt{\mu }$. The equilibrium points are a stable node $(-k,0)$ and a saddle equilibrium $(k,0)$, whose unstable and stable eigenvalues are $2k$ and $-\lambda$. The saddle equilibrium's local stable manifold is $x=k$. When the right-going unstable separatrix returns at $(\nu ,h)$ with $\nu =k$, it lands exactly on that stable manifold, producing a homoclinic orbit of the saddle. Crossing this condition changes whether the returning trajectory is captured by the node or lies on the outgoing side of the saddle equilibrium.

Use an incoming section $y=h$ and outgoing section $x=h$. For entry $(x,h)$ with $k<x<h$, integrate the local flow:so the outgoing height isThe global excursion is smooth and transverse. Its return coordinate has expansion $x_{\mathrm{next}}=\nu +cY+O(Y^2)$, with $c>0$ in these section orientations. Thus the local passage map near a dissipative saddle-node givesFor small $\mu >0$, $\lambda >2k$, hence $P(k^{+})=\nu$ and $P^{\prime }(k^{+})=0$. If $\nu -k>0$ is small, the graph starts above the diagonal and, with slope less than one nearby, crosses it close to $x=\nu$. More explicitly, $P(x)-x=(\nu -k)-(x-k)+O((x-k{)}^{\beta })$, giving a unique small fixed point with Floquet multiplier $0<P^{\prime }<1$. It corresponds to an attracting periodic orbit. For $\nu <k$ no such local crossing exists on $x>k$; the unstable branch returns to the node's side instead. The Poincaré return map panel shows this geometric fixed-point test.

For $\mu >0$, the left equilibrium point $(-\sqrt{\mu },0)$ is a stable node, the right equilibrium point $(\sqrt{\mu },0)$ a saddle equilibrium, and the saddle equilibrium's left unstable branch flows directly towards the node. The right branch takes the prescribed global excursion. Its landing relative to the two equilibrium points gives the six local portraits in the figure.

If $\mu >{\nu }^2$, then $-\sqrt{\mu }<\nu <\sqrt{\mu }$ for either sign of $\nu$. The global branch lands between the equilibrium points and flows left into the node. There is no nearby Poincaré return map cycle, and the landing approaches the node from its right.

If $0<\mu <{\nu }^2$ and $\nu >0$, the branch lands to the right of the saddle equilibrium. The contracting Poincaré return map has an attracting fixed point, giving a stable periodic orbit; the node also remains as a separate attractor with its own basin. If instead $\nu <0$, the branch lands to the left of the node and approaches it from the left. There is no nearby periodic orbit in that case.

If $\mu <0$, the two equilibrium points have vanished and a stable returning cycle exists for either sign of $\nu$. For negative $\nu$ the cycle traverses the full bottleneck and has the divergent period in part (c). For fixed positive $\nu$ it enters to the right of the bottleneck; its local period approaches $1/\nu -1/h$, so it need not diverge when $\mu \to 0^{-}$. The outer arcs in the figure encode only the given return; the local equations do not specify the full vector field away from the origin.

The boundaries explain the codimension-two organization. On $\nu =\sqrt{\mu }>0$ the saddle equilibrium has a homoclinic orbit. On $\mu =0$, $\nu <0$, the return passes through the semistable equilibrium point at a saddle-node bifurcation, producing a saddle-node bifurcation on an invariant circle. On $\mu =0$, $\nu >0$, the node and saddle equilibrium instead collide away from the persistent cycle. At $\mu =\nu =0$, the returning separatrix hits the equilibrium point's strong stable manifold at the saddle-node bifurcation: this is the saddle-node separatrix-loop bifurcation. The curve $\nu =-\sqrt{\mu }$ changes which side of the node receives the returning orbit; unlike the positive branch, it is not a global bifurcation through a saddle homoclinic orbit.

The Jacobian matrix at a diagonal steady state isIts eigenvectors $(1,1)$ and $(1,-1)$ represent common and opposing perturbations, with respective eigenvalues $(-1+g^{\prime })/\tau$ and $(-1-g^{\prime })/\tau$.

For a nonzero equilibrium, ${\sigma }^2+a^2{z}^2=m{a}^{2}z$ and ${\sigma }^2/a^2=z(m-z)$. Consequently $g^{\prime }(z)=2(m-z)/m$. The lower root has $g^{\prime }>1$ and is a saddle equilibrium; the upper root has $0<g^{\prime }<1$ and is a stable node. At the origin $g^{\prime }(0)=0$, so both eigenvalues are $-1/\tau$.

The supplied parameters give $\sigma /a=4$, and the nonzero roots solve $z^2-10z+16=0$, giving $2$ and $8$. The numerical stability results areBiologically, the origin is a quiescent state and the upper equilibrium a sustained mutually reinforced activity state. The saddle represents a collective activation threshold: a perturbation crossing its basin boundary can switch the network from low to high activity. This persistent-state mechanism can model a memory or decision variable, but the simplified rate equations do not establish that a particular biological circuit uses it.

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.

Set $Y=y+s$ and $\mu =2s-a$, and include $\dot{\mu }=0$ for the extended center manifold. The equations near $P_{-}$ becomeThe stable direction has eigenvalue $-2s$ and the center directions are $(x,\mu )$. Choose a symmetry-preserving center manifold $Y=h(x,\mu )$, even in $x$, containing the equilibrium line $x=0,Y=0$. Its invariance equation isWrite $h=A{x}^2+B\mu x^2+\cdots$. At degree two, $0=(1-2sA)x^2$, so $A=1/(2s)$. At degree three, $2A\mu x^2=-2sB\mu x^2$, so $B=-1/(2s^2)$. ConsequentlyWith $x=\sqrt{s}X$ this is $\dot{X}=\mu X-X^3+O(\mu X^3,X^5)$, the supercritical pitchfork bifurcation. For $\mu <0$, $P_{-}$ is attracting; for $\mu >0$ it becomes a saddle equilibrium and two attracting branches appear, consistent with the exact $Q_{\pm }$.

For $0<a^2<4b$, the vertical invariant line flows from $P_{+}$ towards $P_{-}$ between those points. The two off-axis stable equilibria lie below the horizontal axis, one in each invariant half-plane $x>0$ and $x<0$. For $a^2>4b$, these stable equilibria have merged into the stable node $P_{-}$, while $P_{+}$ remains a saddle equilibrium. Filled green points mark attracting equilibria, green rings mark centers, and red crosses mark saddle equilibria. The plotted arrows and nullclines show the corresponding phase portraits; a focus can become a node within the first parameter range without changing the pitchfork classification.

Put $s=\sqrt{b}$. The two fixed points on the invariant line $x=0$ are $P_{+}=(0,s)$ and $P_{-}=(0,-s)$. There are also $Q_{\pm }=(\pm \sqrt{b-a^2/4},-a/2)$ when $a^2<4b$. At equality they coalesce with $P_{-}$.

The Jacobian matrix isAt $P_{+}$ its eigenvalues are $-a-2s$ and $2s$, so it is always a saddle equilibrium. At $P_{-}$ they are $2s-a$ and $-2s$, so it is a saddle equilibrium for $a<2s$ and a stable node for $a>2s$, with a zero eigenvalue at $a=2s$.

At $Q_{\pm }$ the characteristic equation is ${\lambda }^2+a\lambda +4b-a^2=0$, givingFor $0<a^2<16b/5$ these points are stable spirals; for $16b/5<a^2<4b$ they are stable nodes. At $a^2=16b/5$ the double negative eigenvalue has only one independent eigenvector, so these are improper stable nodes. When $a=0$ they are center equilibria, as established in part (iii). Thus the collision at $\boxed{a^2=4b>0}$ changes both the number and stability of fixed points; it is the bifurcation classified below.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

The nullclines are $y=\mu x$ and $y=x^2/[\nu (1+x^2)]$. At a nonzero fixed point, cancellation of $x$ yields $\mu \nu =x/(1+x^2)$. The latter function attains its maximum $1/2$ at $x=1$. HenceFor $0<\mu <{\mu }_c$, $x_{-}<1<x_{+}$ and $x_{-}{x}_{+}=1$.

The Jacobian matrix isAt $O$ the distinct eigenvalues are $-\mu ,-\nu$, so the origin is a stable node. At a nonzero fixed point its trace is $-(\mu +\nu )<0$ andThe discriminant is $(\mu -\nu {)}^2+8x/(1+x^2{)}^2>0$. Thus $P_{-}$ is a saddle and $P_{+}$ is a stable node, never a focus.

Above the straight nullcline $\dot{x}>0$; below it $\dot{x}<0$. Below the curved nullcline $\dot{y}>0$; above it $\dot{y}<0$. At the saddle, an eigenvector can be chosen as $(1,\mu +\lambda )$: the stable eigenvalue gives negative slope, and the unstable eigenvalue gives positive slope. The stable manifold is the basin boundary. The two branches of the unstable manifold lead respectively to the origin and the positive stable node, as illustrated for representative parameters.

The nonnegative quadrant is forward invariant. Since $\dot{y}\le 1-\nu y$, $y$ stays bounded, and then $\dot{x}=-\mu x+y$ also bounds $x$. Moreover the divergence is the negative constant $-(\mu +\nu )$, excluding periodic or homoclinic loops by the Bendixson-Dulac criterion. The Poincaré-Bendixson theorem then leaves convergence to a fixed point. Initial data on the saddle's stable manifold approach the saddle; data in the two open basins approach extinction at $O$ or persistence at $P_{+}$, respectively.

The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. The Jacobian matrix isAt the origin its eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle equilibrium for every allowed $\mu$. At either nonzero fixed point the characteristic polynomial is ${\lambda }^2+\lambda +2{\mu }^2$, givingThus both are stable nodes for $0<|\mu |<1/(2\sqrt{2})$ and stable focuses for $|\mu |>1/(2\sqrt{2})$. At equality the repeated eigenvalue is $-1/2$ and the matrix is not scalar, so there is only one independent eigenvector: the fixed point is a stable improper node.

The stable manifold of the origin is $x=0$, exactly: there $\dot{y}=-y$. Conversely, a trajectory converging to the origin with $x\ne 0$ would eventually have $\dot{x}/x={\mu }^2-y>{\mu }^2/2$, which precludes convergence of $x$ to zero.

For the local unstable manifold, write $y=h(x)=a{x}^2+b{x}^4+O(x^6)$. Reflection symmetry makes $h$ even, and the invariance equation for a graph requiresComparing quadratic and quartic terms gives $2a{\mu }^2=1-a$ and $4b{\mu }^2-2a^2=-b$. Therefore