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Take an isothermal hydrostatic reference atmosphere, ${\rho }_0(z)={\rho }_s{e}^{-z/H}$, and linearize about rest. The horizontal acceleration and pressure-gradient force retain the equatorial Coriolis parameter $f\simeq \beta y$, with $\beta =2\Omega /a$. The traditional approximation retains only its vertical component. Quadratic perturbation advection is neglected. The vertical acceleration is balanced by perturbation buoyancy and pressure gradient after removing the reference hydrostatic balance.

The continuity constraint is the anelastic approximation, $\nabla \cdot ({\rho }_0\mathbf{u})=0$, giving $u_x+v_y+w_z-w/H=0$. It retains stratification-driven expansion while filtering acoustic density fluctuations. It is therefore not the full compressible continuity equation. Adiabatic displacement in a stably stratified background gives $b_t+N^{2}w=0$, with $N$ the Brunt–Väisälä frequency; upward displacement develops restoring negative buoyancy.

Pressure convention matters. With the entropy-buoyancy equation just described, a consistent reduced-pressure form writes the pressure acceleration as $-\nabla \Phi$, where $\Phi =p^{\prime }/{\rho }_0$. Its vertical term is $-{\partial }_z(p^{\prime }/{\rho }_0)$, not simply $-{\rho }_0^{-1}{p}_z^{\prime }$. The original source explicitly specifies variable ${\rho }_0(z)$ but prints the latter. Those differ by $p^{\prime }/({\rho }_{0}H)$, so the two cannot silently be equated. Parts b and c display the consequence and then solve the explicitly corrected reduced-pressure model that yields the requested wave equation. This is a mathematical inconsistency in the printed set, not missing source material.

Expand the vector field about the reference trajectory:Over a sufficiently short interval, the tangent-linear propagator isConsequentlyIf $S=J+J^T$ has $S\mathbf{e}=\lambda \mathbf{e}$ and the perturbation is initially parallel to $\mathbf{e}$, its squared Euclidean norm obeysPositive $\lambda$ means instantaneous norm-squared growth, negative $\lambda$ decay, and zero $\lambda$ no first-order change. The corresponding norm growth rate is $\lambda /2$, not $\lambda$. These are eigenvalues of the symmetric part governing the chosen error metric, not generally the eigenvalues of the nonsymmetric Jacobian matrix or the long-time Lyapunov exponents. The direction and the reference Jacobian matrix evolve during a finite forecast.

Earthquake location. The hypocentre is rupture initiation; its surface projection is the epicentre. Record calibrated arrival times at stations of known coordinates, identify P wave and S wave phases, and predict their travel times through a stated velocity model. Earthquake location by arrival times fitsfor the three source coordinates and origin time $t_0$. Linearization supplies residual equations $\delta t_i=\delta t_0+\nabla T_i\cdot \delta \mathbf{x}$. A grid search or iterative weighted least squares inversion can use pick uncertainties, while station geometry controls whether those parameters are well constrained. Four independent arrival constraints are the minimum for four unknowns, not a guarantee of a reliable solution.

In a homogeneous medium, an arrival-time difference between an S wave and a P wave cancels origin time and estimates source-to-station distance,Several distances intersect to locate the event; real layered or three-dimensional Earth structure requires curved-ray travel times. Nearby stations and diverse takeoff angles help resolve depth. Clock errors, phase misidentification, poor azimuthal coverage and incorrect velocities introduce uncertainty or systematic bias. Closely spaced earthquakes can be relocated precisely relative to each other because common-path errors partly cancel, even when their absolute locations remain uncertain. Report a location uncertainty and model assumptions, not merely a best-fit point.

H. F. Reid's earthquake-cycle model. A fault can remain locked while long-term tectonic motion slowly deforms the surrounding rocks. Elastic strain energy and shear stress accumulate. When resistance is overcome, rapid fault slip releases part of that stored energy as seismic waves, frictional heat and other irreversible work. The adjoining rocks partly recover their elastic distortion while retaining a permanent displacement discontinuity across the fault: this is elastic rebound. Surface markers initially crossing the fault become bent during loading and offset after slip.^[ref]

A simple sketch is a slow rise of stress followed by a rapid stress drop, then renewed loading; the plate displacement continues throughout. Geodetic deformation, observed fault offsets and earthquake stress changes can test the picture. Actual cycles need not repeat with identical slips or times: creep, afterslip, stress transfer and heterogeneous friction complicate the idealization. The release need not remove all pre-existing stress. Elastic rebound explains storage and release; it is not a deterministic earthquake clock.

A dense array does not correct an incorrect velocity model. The faster southwest block permits shorter travel times than an assumed uniform medium, with refraction favouring paths through fast rock. First arrivals may include refracted or seismic head wave paths rather than the assumed direct straight paths. Fitting those arrivals with one constant speed can make an earthquake on the fault appear closer to southwest stations and farther from northeast stations.

The sign follows even from a two-station linearization. Put southwest and northeast stations at $x=-L,+L$, and the true event at $x=0$ with depth $h$. In a constant-speed model,A negative observed differential residual therefore produces $\delta x<0$, an apparent southwest displacement. A common origin-time adjustment cannot remove this directional residual. The stated kilometre-scale offset is a model bias, not proof that the fault dips towards the apparent epicentres. Its exact magnitude cannot be predicted without the actual velocities and station geometry; the question supplies that magnitude as an observation.

The relevant contrast is independently documented at Parkfield: higher velocities lie southwest of the fault, with lower velocities northeast.^[ref] Account for the lateral structure and correct phases, then relocate and test whether the bias remains.

For $r>s$, $a>0$. When $0<a\le 1/4$, the roots of ${\mu }^2-\mu +a$ are real and lie strictly between zero and one. When $a>1/4$, they are a complex-conjugate pair of modulus $\sqrt{a}$. Thus the positive steady state is linearly asymptotically stable exactly when $0<a<1$, and unstable when $a>1$. Since $a=2s-2s^2/r$ increases strictly with $r$, the crossing occurs atAt the crossing,Every nonzero real linearized perturbation has period six. This establishes the intended linear stability threshold and linearized period six.

The nonlinear period-six claim in the printed paper does not follow, and is false as a local bifurcating-branch claim for this map. One can check this directly rather than infer a nonlinear period from the eigenvalues. At $r=r_c$ normalize $N_t=N^{\ast }(1+u)$, $N_{t-1}=N^{\ast }(1+v)$. The exact planar map isIts cubic Taylor expansion isHere is a concrete check of the sixfold return. Put $z=(u,v{)}^T$, $Q=u^2-uv+v^2$, andComposing the displayed cubic expansion six times cancels all quadratic terms and givesIn particularThe quartic coefficient is strictly negative. Hence, sufficiently near the equilibrium, every nonzero point has smaller $Q$ after six iterations: the critical nonlinear map has no nearby nontrivial period-six orbit, and the equilibrium itself is locally attracting despite its neutral linearization.

One can also rule out a branch of small period-six orbits on either side of the crossing. Normalize about the positive equilibrium at each nearby $r$, and put $\eta =a-1$. The derivative of the sixth power of $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ at $a=1$ is $3I+K$. Smooth dependence on $r$ therefore givesThe leading matrix has determinant $(3\eta +\alpha Q{)}^2+3(\eta +\beta Q{)}^2$. Both terms could vanish with $Q>0$ only if $\alpha =3\beta$, butThus its norm on nonzero $z$ is bounded below by a positive constant times $(|\eta |+\Vert z{\Vert }^2)\Vert z\Vert$, and the smaller remainder cannot cancel it. There is no nearby nonzero solution of $F_{\eta }^6(z)=z$. The eigenvalues satisfy the nonresonance conditions through order four, their moduli cross outward with increasing $r$, and the cubic radial coefficient is negative. The standard Neimark–Sacker bifurcation theorem therefore gives a supercritical invariant closed curve. Its rotation is near the linear resonance $1/6$; it is not a local branch of exact period-six orbits.

Finally, the discrete population map does not preserve nonnegativity for arbitrary large lagged populations: $r/(1+b{N}_{t-1}^2)-\delta$ can be negative. The positive-equilibrium stability conclusions above are local and do not assume a global positivity property that the formula lacks.

Let $f(X,Y)=X[r/(1+b{Y}^2)-\delta ]$. At the positive steady state, $1+b(N^{\ast }{)}^2=r/s$, where $s=\delta +1$. The two partial derivatives areThe first-order Taylor expansion with $N_t=N^{\ast }+n_t$ therefore givesEquivalently, the linearization on $(n_t,n_{t-1})$ has matrix $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ and characteristic polynomial ${\mu }^2-\mu +a$.

Momentum-resolved photoemission and directional tunneling can map the magnitude and nodes of the cuprate gap. Low-temperature heat capacity, thermal transport and penetration-depth measurements constrain the low-energy quasiparticle spectrum. These probes do not all determine the relative sign of the gap. A phase-sensitive test of d-wave pairing uses Josephson interference between differently oriented crystal regions. In the tricrystal half-flux experiment, a superconducting film crosses three designed grain boundaries; a frustrated ring can produce a half-flux signal detected with a scanning SQUID. Unfrustrated geometries supply controls. This probes phase structure beyond merely finding a V-shaped tunneling spectrum.^[ref]

For the ideal nearest-neighbor square lattice at one electron per site including both spins, particle-hole symmetry puts the chemical potential at zero. The two requested cuts areThe first runs from $-4t$ at the zone center to zero at $k_{x}a=\pm \pi$; the diagonal cut crosses zero at $k_{x}a=\pm \pi /2$. The Fermi surface obeys $\cos k_{x}a+\cos k_{y}a=0$, giving the diamond $|k_x|+|k_y|=\pi /a$ in the first Brillouin zone.

Near the specified node define $\delta \mathbf{k}=\mathbf{k}-{\mathbf{k}}_0$ andExpanding $\cos (\pi /2+\delta ka)=-a\delta k+O((a\delta k{)}^3)$ giveswhere $\alpha =2\sqrt{2}ta$ and $\beta =\sqrt{2}{\Delta }_{0}a$. ThereforeThere are four such nodes. An energy contour about one node encloses momentum area $\pi E^2/(\alpha \beta )$. Counting all four nodes and two independent spin quasiparticles, with one state per $(2\pi {)}^2$ of momentum area per unit real-space area, gives the positive-excitation count and nodal quasiparticle density of states in a two-dimensional d-wave superconductorThis counts thermally excitable Bogoliubov modes, including particle-like and hole-like excitations, not the electron-addition spectral density alone. That distinction prevents a factor-of-two ambiguity. It is a per-layer result; dividing by the layer spacing gives the corresponding volume density in the independent-layer approximation.

The thermal energy density above its zero-temperature value follows from Fermi occupation:ThusThis clean nodal contribution assumes $k_{B}T$ small compared with the scale over which linearization fails; impurity states and other excitations can add further terms.

In the vortex state, each flux quantum occupies area of order $\pi R^2={\Phi }_0/B$, so $R\propto B^{-1/2}$. Outside a core of radius $\xi$, the phase flow has magnitude $v_s(r)\simeq \kappa /(2\pi r)$ in the simple circular-cell estimate, with $\kappa =h/M_{\mathrm{pair}}$ and condensed-pair mass $M_{\mathrm{pair}}$. Its cell average isThe signed vector average can vanish around a vortex; the relevant quantity is a typical magnitude or Doppler shift, not a net transport velocity. Flow shifts nodal energies by order $\hslash {\mathbf{v}}_s\cdot {\mathbf{k}}_F$. Because the zero-field nodal density of states is linear in energy, averaging the magnitude of these shifts produces a residual density of states proportional to $\sqrt{B}$. The resulting Volovik effect givesThis is the clean semiclassical nodal regime, at temperatures below the field-induced Doppler scale and in an appropriate mixed-state field range. Core states, disorder, Zeeman splitting and screening can modify it. Its replacement by a square root of applied field additionally assumes the mean induction tracks that field away from the lower-critical-field onset; it is not an exact all-field law.

Linearization about the uniform resting state givesFor irrotational motion write ${\mathbf{u}}^{\prime }=\nabla \varphi$. The momentum equation integrates to $p^{\prime }=-{\rho }_0{\varphi }_t$ after choosing the time-dependent potential gauge. Substituting into the pressure equation yieldsThe constant is the adiabatic sound speed, also given by $(\partial p/\partial \rho {)}_S^{1/2}$ at the base state.

For the Ricker population map $H(u)=ru{e}^{-u}$, fixed points satisfy $u=0$ or $r{e}^{-u}=1$. Thus zero exists for all $r>0$, while $u_{\ast }=\log r$ is positive exactly when $r>1$. Since $H^{\prime }(u)=r{e}^{-u}(1-u)$, the respective periodic-orbit multipliers are $r$ and $1-\log r$.

Consequently zero is linearly attracting for $0<r<1$ and repelling for $r>1$; the positive biological population is linearly attracting for $1<r<e^2$ and repelling for $r>e^2$. The periodic-orbit multiplier of the positive point is positive below $r=e$ and negative above it, distinguishing monotone from alternating local convergence. At $r=1$, the coincident zero state has unit periodic-orbit multiplier but $0<H(u)<u$ for every $u>0$, giving nonlinear attraction from the physical right side. At $r=e^2$ the periodic-orbit multiplier is $-1$ and linearization is inconclusive. Writing $u=2+\eta$, the second iterate has expansion $\eta \mapsto \eta -{\eta }^3/3+O({\eta }^4)$, so the critical point is weakly attracting; beyond it a period-doubling instability occurs.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

An equilibrium is asymptotically stable when it is Lyapunov stable and attracts every initial state in some neighbourhood. Lyapunov stability means that sufficiently small initial displacement stays small for all future times; attraction additionally requires convergence to the equilibrium. A Lyapunov function is a differentiable positive-definite scalar function near the equilibrium whose derivative along solutions is nonpositive. Strict negativity, or an appropriate invariant-set argument for a non-strict derivative, gives asymptotic stability.

For $\nu =-1$, equilibria satisfy $y=0$ and $x+x^3=0$, so only $(0,0)$ is real. Its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with eigenvalues $\pm i$ for every $\kappa$. It is a linear centre equilibrium, so linearization alone cannot decide nonlinear attraction.

Use the positive proper energyFor $\kappa <0$, energy sublevel sets are compact and invariant. The zero-derivative set is the union of the axes, but its only invariant subset is the origin: on $x=0,y\ne 0$ one immediately leaves that axis, and on $y=0,x\ne 0$ one has $\dot{y}=-x-x^3\ne 0$. LaSalle's invariance principle therefore proves global asymptotic stability. At $\kappa =0$ the conserved energy levels are closed periodic orbits, giving a nonlinear centre equilibrium. For $\kappa >0$, time reversal together with $y\mapsto -y$ converts the system to the attracting negative-$\kappa$ case, so the origin is a nonlinear repeller and is unstable. Nonzero forward trajectories cannot remain bounded; this does not assert existence for all time if finite-time escape occurs.

The cubic terms vanish in the linearization, leaving $J_0=\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)$. Its eigenvalues are $1\pm 2i$, so the origin is a repelling focus equilibrium. The nonzero imaginary part gives rotation, while the positive real part makes small amplitudes grow. In particular it cannot be the forward limit of nearby nonzero trajectories, which motivates looking for an attracting periodic orbit away from the origin.

A fixed point $z_{\ast }$ of a map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ satisfies $F(z_{\ast })=z_{\ast }$. Its linear perturbations evolve by the Jacobian matrix $DF(z_{\ast })$: both eigenvalues must lie strictly inside the unit disk for linear asymptotic stability. One inside and one outside gives a saddle; both outside give a repeller. Unit-modulus eigenvalues require nonlinear analysis, unlike the continuous-time test using signs of real parts.

Here a fixed point has $x=y=p$ and $2p=a(p-p^3/3)$. ThereforeAt zero the characteristic equation is ${\lambda }^2+\lambda -a=0$. Its negative root is less than $-1$ for every $a>0$, while its positive root crosses $+1$ at $a=2$. Thus zero is a saddle for $0<a<2$ and a real repeller for $a>2$, with a pitchfork of fixed-point branches at $a=2$.

At either nonzero point the characteristic equation is ${\lambda }^2+\lambda +2a-6=0$, whose discriminant is $25-8a$. For $2<a<3$ the fixed points are saddles. At $a=3$ their periodic-orbit multipliers are $0,-1$, giving a flip threshold. For $3<a<25/8$ they are attracting real nodes; at $25/8$ repeated-eigenvalue degenerate nodes; and for $25/8<a<7/2$ attracting foci. The Jury stability criterion conditions reduce to $a>3$ and $a<7/2$, confirming the full stable interval.

At $a=7/2$ the periodic-orbit multipliers are $e^{\pm 2\pi i/3}$. For larger $a$ they form a repelling focus equilibrium. This is a unit-circle crossing at a strong resonance of a planar map with ratio $1:3$; linearization alone does not justify asserting a generic nonresonant Neimark-Sacker bifurcation. The period-two branch at the flip threshold is derived next, including an additional family absent from the source's proposed quadratic.

A genuine two-cycle must have the form $(x,y)\mapsto (y,x)$ with $x\ne y$. Hence $a(x-x^3/3)=x+y=a(y-y^3/3)$. Subtracting givesLet $S=x+y$ and $P=xy$. Using $x^3=3x-PS$ in either original equation gives $S(aP/3-1)=0$. The cubic swap-map two-cycle factorization exhibits two cases, so cancellation of $S$ would lose an actual branch.

If $S\ne 0$, then $P=3/a$ and $S^2=3+3/a$. Thus the source's quadratic holds for this family, withThere are two reflected genuine cycles for $a>3$, coalescing with the fixed points $p=\pm 1$ at $a=3$. For their two-step Jacobian matrix, $\mathrm{tr}D{F}^2=4-a$ and $\det D{F}^2=-2a^2+3a+9$. The unit-disk necessary inequality $1-\mathrm{tr}D{F}^2+\det D{F}^2>0$ fails, since it is $-2(a-3)(a+1)<0$. Therefore these quadratic-family cycles are unstable for every $a>3$. For these unstable asymmetric cycles of a cubic swap map, their periodic-orbit multipliers are real: one exceeds $1$, while the other crosses $-1$ at $a=(1+\sqrt{29})/2$. They are saddles before that secondary threshold and repelling beyond it. Their presence on the stable side of the fixed-point flip identifies the unstable branch of a subcritical flip.

If $S=0$, then $y=-x$ and the preceding quadratic identity gives $x^2=3$. Hence there is also the exact cycleThis antipodal cycle of a cubic swap map does not satisfy the printed $P=3/a$ condition: its product is $-3$. Along it the one-step Jacobian matrix is the same matrix $B=\left(\begin{array}{cc}0& 1\\ -2a& -1\end{array}\right)$ at both points, so the cycle periodic-orbit multipliers are the squares of the roots of ${\lambda }^2+\lambda +2a=0$. The Jury stability criterion conditions giveAt $a=1/2$ its periodic-orbit multipliers are on the unit circle and linearization is inconclusive. Its real-to-complex transition occurs at $a=1/8$. Thus the printed assertion about all period-two components is incomplete: it describes only $S\ne 0$, whereas the omitted antipodal cycle supplies the actual stable period-two interval.

A steady population $N_{\ast }$ is a root $h(N_{\ast })=0$. To linearize one needs $h$ differentiable near that root, an extra local regularity assumption beyond the source's continuity. Setting $N=N_{\ast }+\eta$ gives $\dot{\eta }=h^{\prime }(N_{\ast })\eta +o(\eta )$, hence $\eta (t)=\eta (0)e^{h^{\prime }(N_{\ast })t}$ at linear order.

Thus $h^{\prime }(N_{\ast })<0$ gives linear asymptotic stability and $h^{\prime }(N_{\ast })>0$ gives instability. If $h^{\prime }(N_{\ast })=0$, linearization is inconclusive. In a one-dimensional phase line, the more general nonlinear test is whether the vector field points towards the root from both sides. At the physical boundary $N=0$, only the right-hand side matters. Continuity alone does not guarantee uniqueness of trajectories, so a standard well-defined flow also presumes a local uniqueness condition, such as local Lipschitz continuity.

The equal-diffusivity Fisher reduction uses equal diffusion coefficients and the invariant relation $v=1-u$. The second equation then requires $u_{\tau }=\delta u(1-u)+u_{\xi \xi }$, while the first becomes $u_{\tau }=(1-\beta )u(1-u)+u_{\xi \xi }$. They agree whenThis reduction requires compatible initial and boundary data on that invariant relation. Tuning a parameter alone does not force arbitrary solutions of the two-field system onto it. For positive physical $\beta$, the displayed normalization also requires $0<\delta <1$. More generally the line $v=C(1-u)$ works when $\beta C=1-\delta$; the same positivity restriction remains for $C>0$.

Let $z=\xi -c\tau$ and $u=U(z)$ with $U(-\infty )=0$, $U(+\infty )=1$. The profile equation is $U^{\prime \prime }+c{U}^{\prime }+\delta U(1-U)=0$. Multiplication by $U^{\prime }$ and integration givesConsequently $c<0$: the increasing front moves left, with the $u=1$ state invading $u=0$. Linearization at its leading edge gives real monotone exponents only when $|c|\ge 2\sqrt{\delta }$, yielding the usual Fisher front speed range. A convenient exact example isSubstitution verifies it and the figure shows its leftward displacement. The source uses $x$ for the end limits after defining $\xi$; they denote the same spatial coordinate here.

The equilibria satisfy $y=0$ and $x-x^3=0$, giving $(0,0)$ and $(\pm 1,0)$. At the origin,Their product is $-1$, so the origin is a saddle equilibrium for every real $\Omega$.

At either nonzero equilibrium,The determinant is positive and the trace is $-(\Omega +1)$. Thus both equilibria are attracting for $\Omega >-1$, repelling for $\Omega <-1$, and have purely imaginary eigenvalues $\pm i\sqrt{2}$ at $\Omega =-1$. More precisely they are stable foci for $-1<\Omega <-1+2\sqrt{2}$ and stable nodes for $\Omega >-1+2\sqrt{2}$; they are unstable foci for $-1-2\sqrt{2}<\Omega <-1$ and unstable nodes for $\Omega <-1-2\sqrt{2}$. At the two discriminant-zero endpoints they are repeated-eigenvalue degenerate nodes, with a nondiagonalizable Jacobian. At $\Omega =-1$, linearization alone does not determine nonlinear stability or prove a nonlinear centre.

For $f(x)=rx(1-x)$, the fixed points are $x=0$ and $x_{\ast }=1-1/r$ when the latter is defined. On the physical interval the nonzero branch exists for $r>1$, meeting zero at $r=1$. Their multipliers are $r$ and $2-r$, respectively. Thus zero is linearly attracting for $0\le r<1$ and repelling for $r>1$; the positive point is linearly attracting for $1<r<3$ and repelling for $r>3$. At $r=1$ and $r=3$ the relevant multiplier has magnitude one, so linearization is inconclusive. At $r=1$, $x_{n+1}=x_n-x_n^2$ attracts small positive points to zero nonlinearly.

Factoring $f^2(x)-x$ and removing its fixed-point factors gives $r^2{x}^2-r(r+1)x+(r+1)=0$. Its distinct roots areThey map to each other, giving a genuine two-cycle only beyond $r_0=3$. The cycle multiplier isSolving $|M|<1$ in $r>3$ givesAt $r=3$, the two roots coalesce with $x_{\ast }=2/3$, where the fixed-point multiplier is $-1$: a supercritical period-doubling bifurcation creates the attracting two-cycle. At the upper endpoint its multiplier reaches $-1$, marking the next loss of linear stability. The two-cycle is superstable at $r=1+\sqrt{5}$, where $M=0$. At exactly $r=3$ the fixed point remains weakly attracting through higher-order terms: for $\eta =x-2/3$, the second iterate is $\eta \mapsto \eta -18{\eta }^3-27{\eta }^4$, showing alternating perturbations decay slowly.

The first equilibrium equation gives $y=x$, and the third gives $z=x^2/b$. The second then becomes $x(r-1-x^2/b)=0$. Hence the equilibria areAt the origin, one eigenvalue is $-b$ and the remaining two solve ${\lambda }^2+(\sigma +1)\lambda +\sigma (1-r)=0$. Their discriminant is $(\sigma -1{)}^2+4\sigma r>0$. For $r<1$ both are negative, so the origin is linearly stable, with all eigenvalues real and negative. For $r>1$ one is positive and one negative, so it is a saddle equilibrium with one unstable and two stable directions. At $r=1$ its eigenvalues are $0,-(\sigma +1),-b$: linearization has a neutral direction and cannot alone prove nonlinear stability. Given the source's global attraction for $r<1$, there are no nontrivial limit cycles or persistent chaotic invariant dynamics in that regime; all initial states tend to the origin.

To establish the local pitchfork, one needs the simple eigenvalue to cross zero, the two symmetry-related branches to exist and acquire the expected local stability, and nondegeneracy of the nonlinear coefficient. The derivative of the crossing eigenvalue is $\sigma /(\sigma +1)>0$. At $C_{\pm }$, the characteristic polynomial isFor $r>1$ sufficiently close to one, all coefficients are positive and $(\sigma +b+1)b(\sigma +r)>2\sigma b(r-1)$, so the cubic Routh-Hurwitz criterion makes both new equilibria stable. Together with the symmetry $(x,y,z)\mapsto (-x,-y,z)$ and the square-root branches, this yields the supercritical pitchfork. A local centre-manifold coordinate proportional to $x$ has leading normal form $\dot{q}=\frac{\sigma }{\sigma +1}(r-1)q-\frac{\sigma }{b(\sigma +1)}{q}^3+\cdots$, explicitly displaying the nonzero stabilizing cubic coefficient. At $r=1$ that cubic also gives weak local nonlinear attraction along the centre direction.

The travelling-wave reduction of a reaction-diffusion system of Fisher's equation isMultiplying by $U^{\prime }$ and integrating gives $c{\int }_{\mathbb{R}}(U^{\prime }{)}^{2}dz=1/6$, so the decreasing invasion front has $c>0$. The linearization at $U=0$ has spatial eigenvaluesFor $0<c<2$ they are complex: approaching zero spirals in the $(U,U^{\prime })$ plane and changes the sign of $U$, which is inadmissible for a density. Thus $c\ge 2$ is necessary.

It is also sufficient. Write $V=U^{\prime }$ and choose $\ell =(c-\sqrt{c^2-4})/2>0$. The triangleis forward invariant for $U^{\prime }=V$, $V^{\prime }=-cV-U(1-U)$. On $V=0$, $V^{\prime }\le 0$. On $V=-\ell U$,The boundary $U=1$ points left, and the boundary at $U=0$ consists only of the origin. The unstable branch from the saddle $(1,0)$ enters this triangle, where $U$ decreases and remains nonnegative. Its limiting state must have $V=0$ and $U(1-U)=0$; since it has left $U=1$, it tends to $(0,0)$. Therefore a monotone positive front exists for every $c\ge 2$, including the double-root endpoint:This is the Fisher-KPP minimum wave speed, obtained with an explicit invariant-region existence argument.