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Let $g>0$ be the magnitude of the downward gravity and write ${\alpha }_T$ for the thermal expansion coefficient, to distinguish it from the dynamo coefficient of the previous question. The conductive state has $\mathbf{u}=0$, $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $T_0(z)=T_r-\Delta Tz/d$. Its gravitational term is absorbed in a background hydrostatic pressure.

For the dimensional two-dimensional fields, choose the stream function and Cartesian magnetic flux function conventionsThese automatically give zero divergence. Define the Jacobian determinant bracket $J(f,h)=f_x{h}_z-f_z{h}_x$, so $\mathbf{u}\cdot \nabla h=J(\psi ,h)$. The full magnetic flux function, including the imposed field, must be retained in the magnetic force; using the perturbation alone would lose its linear restoring term.

The scalar vorticity is $(\nabla \times \mathbf{u}{)}_y=-{\nabla }^2\psi$. Also $(\nabla \times \mathbf{B}{)}_y=-{\nabla }^{2}A$, and henceThe buoyancy perturbation is $g{\alpha }_T(T-T_0)\hat{\mathbf{z}}$, whose curl in the $y$ direction is $-g{\alpha }_T(T-T_0{)}_x$. Taking the curl of the momentum equation and multiplying by minus one eliminates pressure and therefore givesWe have restored ${\mu }_0$ in this dimensional expression. The printed magnetic force uses units in which ${\mu }_0=1$, or equivalently a correspondingly rescaled field.

Use length scale $d$, time scale $d^2/\kappa$, velocity scale $\kappa /d$, stream-function scale $\kappa$, magnetic flux-function scale $B_{0}d$, and temperature perturbation scale $\Delta T$. Thus the dimensionless total flux function isHere $x,z$ now denote coordinates divided by $d$, and the same $T$ notation in the second identity denotes the physical temperature. The dimensionless parameters for thermal-diffusion scaling of planar magnetoconvection areThese are the Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio. The nondimensional magnetic-force coefficient is $B_0^2{d}^2/({\mu }_0{\rho }_r{\kappa }^2)=\sigma \zeta Q$, so omitting the factor $\zeta$ would use a different magnetic parameter.

The resistive induction equation becomes $A_t+J(\psi ,A)=\zeta {\nabla }^{2}A$. Since $J(\psi ,x)=-{\psi }_z$, its perturbation equation has source $+{\psi }_z$. Similarly, advection of the conductive temperature gradient gives $J(\psi ,-z)=-{\psi }_x$, producing the source $+{\psi }_x$ in the perturbation heat equation. The resulting nonlinear equations are thereforeThis derives both forcing signs from the background fields, rather than selecting them by a stream-function convention after the fact.

For the specified roll mode put $K=\pi /\lambda$ and $p_L=K^2+{\pi }^2$, the positive eigenvalue of minus the Laplacian on each spatial factor. Writeand take $\psi =(p_L/K)a{f}_s$, $\Theta =b{f}_T$, $\chi =\lambda c{f}_A$, with amplitudes depending on $\tau =p_{L}t$. These factors satisfy the printed boundary conditions: sine factors make the normal velocity and $B_x=-{\chi }_z$ vanish at the required faces; the second derivatives of the velocity factors give the stress-free conditions; $\Theta$ vanishes at the top and bottom and ${\Theta }_x$ vanishes at the sides.

Discard quadratic perturbation terms. Since ${\nabla }^{2}A={\nabla }^2\chi$, the magnetic force linearizes to $J(x,{\nabla }^2\chi )=({\nabla }^2\chi {)}_z$. The needed derivatives areThe heat and induction equations give $b^{\prime }=a-b$ and $c^{\prime }=a-\zeta c$. In the vorticity equation, its left side is $-(p_L^3/K)a^{\prime }{f}_s$ and its three right-side contributions are $\sigma [(p_L^3/K)a-RKb+\zeta Q{p}_L\pi \lambda c]f_s$. Consequently, withthe three-amplitude vertical-field magnetoconvection system isPrimes now mean $d/d\tau$. They are genuine time derivatives, as the original PDF shows.

For its linear stability, seek amplitudes proportional to $e^{s\tau }$. Eliminating $b,c$ or taking the determinant giveswhereAssume the physical range $\sigma >0$, $\zeta >0$, $q\ge 0$. The Routh-Hurwitz stability criterion gives strict decay exactly when $A_2>0$, $A_3>0$ and $A_1{A}_2>A_3$. At $r=0$ these hold. As $r$ increases, the first loss of stability can occur through a zero eigenvalue or a conjugate imaginary pair.

The stationary threshold isAt this value, $P(s)=s(s^2+A_{1}s+C)$ withIf $C>0$, the other two roots have negative real parts and the zero root is simple. Its derivative with respect to $r$ is $ds/dr=\sigma \zeta /C>0$. This is stationary onset; the reflection symmetry of the roll permits a generic pitchfork bifurcation. The linear calculation does not determine whether the nonlinear branch is supercritical or subcritical.

For oscillatory onset set $s=i\omega$, $\omega \ne 0$. Separating real and imaginary parts gives ${\omega }^2=A_2$ and $A_1{A}_2=A_3$. Solving the latter for $r$ yieldsA genuine Hopf bifurcation onset requires ${\omega }_H^2>0$, soAt this threshold $P(s)=(s+A_1)(s^2+{\omega }_H^2)$, giving the third root $-A_1<0$. The pair crosses transversely: differentiating $P(s,r)=0$ at $i{\omega }_H$ gives

The first instability of the conductive state is therefore stationary at $r_s$ if $\zeta \ge 1$, or if $0<\zeta <1$ and $q<q_{\ast }$. It is oscillatory at $r_H<r_s$ if $0<\zeta <1$ and $q>q_{\ast }$. At $q=q_{\ast }$, the two thresholds meet and $P(s)=s^2(s+A_1)$: the frequency is zero and neither a simple pitchfork nor a nonzero-frequency Hopf test applies. The oscillatory marginality in vertical-field magnetoconvection formula must not be used where its frequency squared is negative.

For conversion back to physical parameters, $R_{s,H}={\lambda }^2{p}_L^3{r}_{s,H}/{\pi }^2$, while the dimensional Hopf frequency is $(\kappa p_L/d^2){\omega }_H$. These are onset conditions for the specified spatial mode. Identifying the globally first mode would additionally require comparison over the allowed spatial modes. Nonlinear saturation and the stability of the bifurcating finite-amplitude states are not determined by this linear three-amplitude system alone.

Write $s=R^2$. A steady complex amplitude obeys $(\mu -s+i\Lambda )A=-1$, soThus every positive intensity root gives exactly one steady phase, and $s=0$ is impossible. The cubic intensity polynomial is $F(s,\mu )=s^3-2\mu s^2+({\mu }^2+{\Lambda }^2)s-1$. Its positive steady branch can be parameterized completely asThe two formulae join smoothly as a single curve at the maximum radius $R={\Lambda }^{-1}$; this joining point is not a saddle-node bifurcation. A turning point with respect to $\mu$ satisfies $F_s=0$. Subtracting the equilibrium and derivative conditions givesThese points lie on ${\mu }_{+}$. Since $f^{\prime }(s)=(1-s^3)/s^5$, its maximum is $f(1)=3/4$. Therefore there is exactly one distinct steady state for every $\mu$ when $\Lambda \ge \sqrt{3}/2$. One can also see uniqueness directly: ${\mu }_{-}$ is increasing and ${\mu }_{+}$ is decreasing when no turning points exist, and their respective ranges meet only at their common endpoint. At the equality there is a cusp bifurcation with a triple intensity root at $s=1,\mu =3/2$.

When $0<\Lambda <\sqrt{3}/2$, the two turning intensities satisfy $s_{-}<1<s_{+}$. Define ${\mu }_L=s_{-}+1/(2s_{-}^2)$ and ${\mu }_U=s_{+}+1/(2s_{+}^2)$; the curve has a local minimum at the former and a local maximum at the latter, with ${\mu }_L<{\mu }_U$. There are three distinct steady intensities for ${\mu }_L<\mu <{\mu }_U$, two distinct ones at an endpoint, and one otherwise. The bifurcation diagrams below show the folded small-$\Lambda$ curves and the unfurled large-$\Lambda$ curve, together with the linear stability calculated next. Both ends of the full branch have $R\to 0$ as $|\mu |\to \infty$, while its middle reaches the finite maximum ${\Lambda }^{-1}$.

For a stationary finite-wavenumber instability, the relevant Landau–Ginzburg model is the real Ginzburg–Landau equation: its coefficients are real although its pattern amplitude is complex. The complex amplitude records both the strength and the position of the underlying pattern. Assume a simple critical pair of Fourier modes $\pm k_c$, a quadratic nondegenerate maximum of the linear growth rate, and no additional conserved or marginal mode. The last assumption is essential: the conserved field in the previous problem requires an additional evolution equation.

To derive the model, write a real evolution equation schematically as $u_t=L(r)u+N_2(u,u)+N_3(u,u,u)+\cdots$, with symmetric multilinear nonlinearities. Put $r={\epsilon }^2{r}_2$, $X=\epsilon x$, $T={\epsilon }^{2}t$ and use the method of multiple scales expansionHere $L(k_c;0)v=0$. At the next order the noncritical mean and second harmonic are slaved to the critical mode:where $L(0;0)w_0=-2N_2(v,\overline{v})$ and $L(2k_c;0)w_2=-N_2(v,v)$. These inverses exist under the no-other-critical-mode hypothesis. Let $v^{\ast }$ be an adjoint null vector normalized by $⟨v^{\ast },v⟩=1$. At order ${\epsilon }^3$, the Fredholm solvability condition requires the resonant forcing to be orthogonal to $v^{\ast }$. The cubic resonant contribution isThis shows why the cubic coefficient includes feedback from slaved harmonics as well as any original cubic nonlinearity. The resulting amplitude equation, after naming its linear control coefficient $\mu$, isHere $\lambda (k)$ is the critical linear dispersion relation for growth. The condition $g>0$ gives a supercritical bifurcation; $g<0$ calls for higher-order saturation, for example the quintic real Ginzburg-Landau equation.

A concrete check is the cubic Swift–Hohenberg equation $u_t={\epsilon }^2\mu u-(1+{\partial }_x^2{)}^{2}u-g_0{u}^3$. With $u=\epsilon (A{e}^{ix}+\overline{A}{e}^{-ix})+\cdots$, its linear operator contributes $4A_{XX}$ at resonant order, while the cubic resonance is $3|A{|}^{2}A$. Hence $A_T=\mu A+4A_{XX}-3g_0|A{|}^{2}A$. This directly realizes the perturbation-theory coefficients above.

Symmetry independently constrains the form. A fast spatial translation rotates $A$ by a constant phase, so every nondifferential monomial must transform like $A$. The leading allowed nonlinear term is $|A{|}^{2}A$, and a quadratic term is forbidden. A spatial reflection maps $A(X)$ to $\overline{A}(-X)$; it makes the leading coefficients real and rules out a drift term in a reflection-symmetric rest frame. The vanishing first derivative of the linear growth rate at $k_c$ is consistent with the absence of a leading first spatial derivative. These arguments constrain terms but do not calculate $\xi ,g$ or replace the solvability calculation. Breaking reflection or studying an oscillatory instability generally leads to the complex Ginzburg–Landau equation with genuinely complex coefficients.

The principal steady solutions are the zero state and the Ginzburg-Landau plane wavesTheir physical carrier wavenumber is $k_c+\epsilon q$. For $q=0$ the constant phase is arbitrary, giving a circle of uniform-amplitude states. The zero state has Fourier growth rate $\mu -\xi p^2$ and is stable for $\mu <0$. For $g>0$, each existing plane wave is stable to spatially homogeneous amplitude perturbations, with radial eigenvalue $-2g{R}^2$, and neutral to a constant phase shift.

Steady modulated states are also possible. Writing $A=a(X)e^{i\varphi (X)}$ gives the conserved spatial phase current $j=a^2{\varphi }_X$ andIts first integral is $\xi a_X^2/2+\mu a^2/2-g{a}^4/4+\xi j^2/(2a^2)=E$. Boundary conditions select nonuniform periodic amplitudes, fronts or defects from this spatial dynamics. In the invariant real subspace, for example,is a steady kink joining opposite real states. It is not stable in the unrestricted complex equation: the imaginary perturbation $v=\mathrm{sech}(\sqrt{\mu /(2\xi )}X)$ has eigenvalue $\mu /2>0$ under $v_T=\xi v_{XX}+(\mu -g{A}^2)v$. This illustrates why a real-amplitude calculation can miss a phase instability. With real coefficients the Lyapunov functionalsatisfies $A_T=-\delta \mathcal{F}/\delta \overline{A}$ and $d\mathcal{F}/dT=-2\int |A_T{|}^{2}dX$ under compatible boundary conditions. It supplies a variational structure absent from a generic complex-coefficient model.

To test full linear stability of a plane wave, set $A=e^{iqX}(R+u+iv)$ with real perturbations $u,v$. Let $b=g{R}^2=\mu -\xi q^2>0$. The linear equations areFor perturbation wavenumber $p$, the sideband spectrum of a real Ginzburg-Landau plane wave isThe phase branch at small $p$ isTherefore the robust stable band on the infinite line is $q^2<\mu /(3\xi )$, narrower than the existence band $q^2<\mu /\xi$. Crossing its boundary gives the longitudinal Eckhaus instability. At the boundary the quadratic phase-diffusion coefficient vanishes and the leading nonzero term is negative of order $p^4$. More exactly, a nonzero sideband grows only when $p^2<6q^2-2\mu /\xi$; a finite periodic domain for the wave envelope may exclude that interval. Constant-amplitude stability is understood modulo its neutral global phase, and the allowed perturbation wavenumbers must always be specified.

In a second spatial dimension, an anisotropic medium that selects one roll orientation permits an equation with ${\xi }_x{A}_{XX}+{\xi }_y{A}_{YY}$. For positive coefficients its plane-wave analysis is similar, with the second-dimensional contribution adding damping. But simply replacing $A_{XX}$ by an isotropic Laplacian is not a faithful extension for rotationally invariant stationary rolls. There the critical wavevectors form a circle. Near $(k_c,0)$,This forces the anisotropic transverse scaling of an isotropic roll envelope: $X=\epsilon x$, $Y={\epsilon }^{1/2}y$, $T={\epsilon }^{2}t$. For the positive carrier convention used here the Newell–Whitehead–Segel equation isThe opposite fast-carrier convention reverses the accompanying imaginary sign and detuning convention. For a transverse phase perturbation of $A=R{e}^{iqX}$ our convention gives $\sigma =-\xi q{p}^2/k_c-\xi p^4/(4k_c^2)$. Thus $q<0$ permits the transverse zigzag instability, in addition to the longitudinal Eckhaus instability. The fourth transverse derivative and mixed derivative in the square encode the critical-circle geometry.

One complex amplitude still describes only a narrow cone of nearby roll orientations. Competing well-separated wavevectors require coupled amplitude equations; resonant triples can admit quadratic interactions and produce the hexagonal convection amplitude equations when the underlying sign symmetry allows them. Strong disorder, defect cores with large gradients, far-from-onset patterns, boundaries, mean flows and conserved fields may all lie outside the single-amplitude approximation. The method of multiple scales gives controlled leading behavior for small amplitudes and slow modulations, rather than a universal model of all two-dimensional pattern selection.

Here equilibrium means fixed position relative to the rotating wire: $r=a$ and $z=f(a)$ are constant, while $\theta =\omega t$ continues to change. Substituting $\dot{r}=\ddot{r}=0$ into the equation givesFor the linear stability calculation, set $r=a+ϵ$ and retain terms of first order in $ϵ$ and $\dot{ϵ}$. The term ${\dot{r}}^2$ contributes only at second order, and variation of the factor multiplying $\ddot{ϵ}$ also contributes only at second order. ThusThe two strict sign cases are treated in the following parts. If $g{f}^{\prime \prime }(a)={\omega }^2$, this linearization has no restoring term and higher orders are needed to classify the equilibrium.

For the specified wire,Hence the complete set of positive equilibrium radii isAt $a_n$,Therefore odd $n$ give a stable equilibrium, and even $n$ give an unstable equilibrium. There is no degenerate sign case in this example. The corresponding heights are $f(a_n)={\omega }^2{n}^2{\pi }^2{b}^2/(2g)+b(-1{)}^n$.

At a fixed point, the first equation gives $y=0$, the second then gives $x=0$, and the third gives $z=-\mu$. Thus there is exactly one equilibrium for every real $\mu$:The Jacobian matrix there isOne eigenvalue is $-1$. The other two have product one and sum $5ϵ\mu$. For $\mu <0$ their real parts are negative, so the equilibrium is asymptotically stable; for $\mu >0$ both real parts are positive, leaving two unstable directions and one stable direction. At $\mu =0$ the eigenvalues are $-1,i,-i$, withTherefore the only local bifurcation of this equilibrium is a Hopf bifurcation at $\mu =0$. A repeated real eigenvalue away from zero does not itself cause a bifurcation: the equilibrium remains hyperbolic there.

The cubic centre manifold calculation below gives the Hopf normal form radial coefficient $-9ϵ/8<0$, so this is a supercritical Hopf bifurcation. Small attracting periodic orbits occur on the $\mu >0$ side. In particular, at $\mu =0$ the leading cubic radial drift is inward, although linear stability alone cannot decide that boundary case.

The Jacobian matrix at the origin and its characteristic polynomial areThe fixed-point multipliers areBy the Jury stability criterion, both lie strictly inside the unit circle exactly whenThe last condition always holds, since ${\mu }^2+\mu +2=(\mu +1/2{)}^2+7/4$. Therefore the origin is asymptotically stable exactly for $-1<\mu <0$ among the hyperbolic parameter values.

For completeness, its linear stability throughout the parameter line is as follows. When $\mu <-1$, the complex-conjugate Floquet multipliers have modulus $|\mu |>1$, so the origin repels. When $-1<\mu <-1/3$, their modulus is $|\mu |<1$, so it attracts with oscillation. At $\mu =-1/3$ the repeated Floquet multiplier is $1/3$, still attracting. For $-1/3<\mu <0$, both Floquet multipliers are real and in $(0,1)$. For $0<\mu <1$, $p(1)<0$ and $p(0)>0$, so one Floquet multiplier lies in $(0,1)$ and the other exceeds one: the origin is a saddle fixed point of a map. For $\mu >1$, the complex-conjugate Floquet multipliers have modulus $\mu >1$, so it repels.

Possible changes at the unit circle are exhausted by $p(1)=0$, $p(-1)=0$, or a nonreal pair with product one. They giveAt $\mu =0$ the Floquet multipliers are $0,1$; the nonzero fixed point branches in part (a) meet the origin, and part (c) identifies a subcritical pitchfork bifurcation. Indeed at $\mu =0$ the exact restricted map is $y\mapsto y+y^3$, which makes the origin unstable despite its neutral linear direction.

At $\mu =-1$ the Floquet multipliers are $i,-i$: this is a 1:4 strong resonance of a planar map, coinciding with the change from repelling to attracting as $\mu$ increases. A generic nonresonant Neimark–Sacker bifurcation conclusion would be unjustified. At $\mu =1$ there is a double Floquet multiplier $+1$; the matrix has a nontrivial Jordan block, and the two further nonzero fixed point branches meet the origin. Linear stability alone is inconclusive at these unit-modulus boundary values. The repeated Floquet multiplier at $\mu =-1/3$ is strictly inside the unit circle and gives no bifurcation.

Use $y$ vertically and $\mu$ horizontally in the bifurcation diagram; the other coordinate is recovered from $x=\mu y$. The origin is the horizontal branch $y=0$. Draw it solid on $-1<\mu <0$, where it is asymptotically stable, and dashed on the remaining hyperbolic intervals, where it is unstable. Its three nonhyperbolic fixed points in parameter space are marked at $\mu =-1,0,1$; open markers distinguish these boundary cases from the strict linear stability classifications on the intervening intervals.

The nonzero fixed point branches are $y=\pm \sqrt{\mu (\mu -1)}$ for $\mu <0$ and $\mu >1$. Each is dashed because it is a saddle fixed point of a map, as proved in part (c). Near zero their shape is $y\sim \pm \sqrt{-\mu }$: the unstable branches lie on the same parameter side as the attracting origin, displaying the subcritical pitchfork bifurcation. The branches issuing from $\mu =1$ have $y\sim \pm \sqrt{\mu -1}$. At $\mu =-1$ no additional fixed point branch appears; the Floquet multipliers cross the unit circle at $\pm i$.

Potential temperature and a parcel. For a dry ideal gas of fixed composition, the first law of thermodynamics per unit mass isA reversible adiabatic process has $\delta q=0$. Taking $c_p,R_d$ constant gives $d\ln T=\kappa d\ln p$, where $\kappa =R_d/c_p$. Integrating to a fixed reference pressure $p_0$ defines the potential temperatureIt is the temperature a parcel would have after reversible dry adiabatic process taking it to $p_0$. Equivalently $ds=c_{p}d\ln \Theta$ under these assumptions, making potential temperature a thermodynamic entropy label useful for diagnosing stratification and air-mass motion. Latent heating, radiation, mixing and variable composition require modifying the conservation statement.

An air parcel is a small material volume followed with the flow. The parcel model treats it as internally approximately uniform, large compared with molecular scales and small compared with environmental variation scales. Rapid acoustic wave adjustment makes its pressure approximately equal to the surrounding pressure, while slower heat and mass exchange can be neglected during an idealized displacement. It is not a rigid container or an isolated molecule.

At the parcel's height, environmental hydrostatic pressure supplies an upward pressure force ${\rho }_{e}g$ per unit volume. Dividing by parcel density gives the upward accelerationThe last steps use equal pressure, equal gas constants and the same reference pressure. Moisture-dependent density requires virtual temperature or its potential-temperature counterpart. Nonhydrostatic pressure perturbations and entrainment are excluded from this elementary buoyancy expression.

The convective available potential energy is the positive buoyancy work per unit mass between the level of free convection and the equilibrium level:One sums positive-buoyancy layers when there is more than one. Negative work below free convection contributes convective inhibition rather than positive CAPE. In a lossless parcel estimate, conversion of CAPE to vertical kinetic energy gives $w_{\max }\simeq \sqrt{2\mathrm{CAPE}}$; pressure work beyond this model, mixing and precipitation can change that speed.

Linear stability. Let $\xi$ denote displacement from a neutrally buoyant reference altitude. Dry motion through an adiabatic process retains ${\Theta }_p={\Theta }_e(0)$, while ${\Theta }_e(\xi )={\Theta }_e(0)+\xi {\Theta }_e^{\prime }(0)+O({\xi }^2)$. The linear vertical momentum equation is thereforeFor $N_0^2>0$, $\xi =C\cos N_{0}t+D\sin N_{0}t$: stable buoyancy oscillations with period $2\pi /N_0$. For $N_0^2=0$, $\xi =C+Dt$: neutral stratification with no restoring force. For $N_0^2<0$, growing and decaying exponentials occur, indicating convective instability. The growing solution is generic unless the initial data select only the decaying branch. These are local small-displacement results with an approximately constant environmental gradient.

Lapse-rate estimate. Define $\Gamma =-dT/dz$. Hydrostatic balance and the ideal gas law imply $p^{\prime }/p=-g/(R_{d}T)$, soIn the free-parcel-oscillator interpretation, identifying the ridge encounter interval with the buoyancy period gives $P=10^4/20=500\mathrm{s}$ and $N_0=2\pi /P=0.01257{\mathrm{s}}^{-1}$. With $T=293.15\mathrm{K}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $c_p\simeq 1004\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$,The frequency identification is an extra modeling assumption. A stationary mountain gravity wave generally obeys ${\hat{\omega }}^2=N_0^2{k}_h^2/(k_h^2+m_z^2)$ without rotation, with $\hat{\omega }=U{k}_h$. Ridge spacing fixes $\hat{\omega }$, not the vertical wavenumber $m_z$. For example, $m_z=k_h$ allows the same spacing and wind with $N_0=\sqrt{2}\hat{\omega }$, and a lapse rate about $0.3\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. Thus $5.1$ is the requested parcel-period estimate, not a uniquely identifiable lapse rate for every possible mountain-wave field; a resolved wave field needs its vertical structure as well.

TODO: biological interpretations and organism-specific propagation are user-deferred. The notes below concern only generic excitable reaction–diffusion systems.

A front solution joins different asymptotic states in space. In a reaction–diffusion system, local nonlinear dynamics select possible states while diffusion couples neighboring points. A travelling wave with profile $u(x,t)=U(\xi )$, $\xi =x-ct$, converts a partial differential equation into a boundary value problem for the shape and speed. A pulse instead returns to the same resting state on both sides and can contain leading and trailing fronts.

A useful nondimensional cubic fast-slow reaction-diffusion model, in a convention of the FitzHugh-Nagumo model, iswhere $D>0$, $\epsilon \ll 1$, and $\beta >0$. Here $u$ is a fast activation variable, $w$ a slow negative-feedback variable and $I$ a generic drive; no physiological variable is assigned. Without spatial coupling, nullclines are $w=f(u)+I$ and $w=u/\beta$. At $I=0$, the resting state $(0,0)$ has Jacobian matrixIt has linear stability. With appropriate parameter choices a sufficiently large perturbation makes a fast excursion to the other outer branch of the cubic, followed by slow feedback and return. Thus a stable rest state can be excitable without being linearly unstable. Other parameters or applied drive can yield multiple equilibria or oscillations; these behaviours should not be asserted for every parameter set.

The fast-subsystem front can be obtained explicitly by freezing $w=I=0$. For the bistable cubic reaction-diffusion equation, $u=0,1$ are stable and $u=\alpha$ is unstable. The decreasing logistic front of a bistable cubic equation isTo check it, $U^{\prime }=-U(1-U)/\sqrt{2D}$ and substitution into $D{U}^{\prime \prime }+c{U}^{\prime }+f(U)=0$ cancels every term. For $\alpha <1/2$, the high state advances to the right; for $\alpha >1/2$, the direction reverses. At $\alpha =1/2$ the front is stationary. More generally, multiplying the travelling-wave equation by $U^{\prime }$ and integrating yieldsThis links propagation direction to the bias between the two stable states and recovers Maxwell balance for a scalar reaction-diffusion front at zero speed. It is a bistable selected-speed problem, rather than a universal Fisher-type minimum-speed formula.

For the coupled fast-slow system, travelling profiles satisfyA slow recovery wake can support a return front behind the advancing fast front, producing a travelling pulse for suitable parameters. Its width reflects the slow recovery time and propagation speed, while each sharp edge has the faster diffusion–reaction scale. The frozen-recovery speed above is a controlled subsystem result, not an exact speed for every coupled pulse. A definite pulse speed requires parameters and connecting boundary condition; none are specified by this brief-note question.

The fixed-point multiplier decreases continuously with $\gamma$ and first reaches the unit circle at $-1$. Solving $1-{\gamma }_{\ast }(1-\beta /\alpha )=-1$ givesIt is stable for $1<\gamma <{\gamma }_{\ast }$ and unstable for $\gamma >{\gamma }_{\ast }$. At the critical value linear stability is inconclusive. The crossing at $-1$ is the local signature of a period-doubling bifurcation, rather than a crossing through $+1$.

For an autonomous scalar differential equation, linear stability of a simple equilibrium point $N_e$ is determined by the derivative $f_N(N_e,a)=P^{\prime }(N_e)$: a negative derivative attracts nearby solutions and a positive derivative gives repulsion in a dynamical system of nearby solutions.

Here $f(0,a)=1+a\ge 1/2>0$ throughout the prescribed mathematical model parameter range, while $f(N,a)\to -\infty$ as $N\to \infty$. If there is only one equilibrium point, the function graph must cross the axis on one of its decreasing branches, outside the interval between its two extrema. The crossing is simple and $P^{\prime }(N_e)<0$. Thus the unique equilibrium point is linearly asymptotically stable. At a saddle-node bifurcation there are two distinct equilibrium points, one of them double, so that case does not fall under the one-equilibrium hypothesis.

For the modal matrix $M(s)$, $s\ge 0$, its trace is always negative:Its determinant isThe proposed inequality is exactly $\beta \kappa \eta <\alpha \kappa +1$. It makes the coefficient of $s$ positive, while the leading and constant coefficients are already positive. Therefore $D(s)>0$ for every $s\ge 0$, independent of which discrete wave numbers the domain allows.

A real two-by-two matrix with negative trace and positive determinant has both eigenvalues with negative real part: if they are real they have positive product and negative sum, and if nonreal their common real part is half the trace. Every mode therefore decays, proving linear stability for all domain sizes under the stated sufficient condition.

A fixed point $x_{\ast }$ satisfies $f(x_{\ast },\mu )=x_{\ast }$. A perturbation obeys $u_{n+1}=m{u}_n+O(u_n^2)$, where the periodic-orbit multiplier is $m=f_x(x_{\ast },\mu )$. It is linearly attracting for $|m|<1$ and repelling for $|m|>1$; at $m=\pm 1$ nonlinear terms decide Lyapunov stability. A crossing through $+1$ can produce a saddle-node, transcritical or pitchfork bifurcation, depending on the mathematical model parameter dependence and symmetry. A crossing through $-1$ generically gives a period-doubling bifurcation. A real one-dimensional function cannot have the complex conjugate pair needed for a Neimark–Sacker bifurcation.

For a genuine period-two orbit, study the return map $f\circ f$ at either point. By the chain rule its periodic-orbit multiplier isIts modulus determines linear stability by the same criterion, now per two iterates.

For the real quadratic iteration map, the fixed-point equation givesThe derivative is $-2x$. The lower branch $x_{-}$ is repelling whenever $\mu >-1/4$. The upper branch satisfies $|-2x_{+}|<1$ exactly forAt $\mu =-1/4$ the two fixed points merge in a saddle-node bifurcation. With $u=x+1/2$ at that mathematical model parameter, $u^{\prime }=u-u^2$, so the point attracts from the right and is repelling from the left. At $\mu =3/4$, $x_{+}=1/2$ has periodic-orbit multiplier $-1$ and undergoes a period-doubling bifurcation. At the threshold itself $u^{\prime }=-u-u^2$ and the two-step return is $u^{\prime \prime }=u-2u^3-u^4$, proving weak asymptotic stability there despite the inconclusive linear test.

Factor the second-iterate equation:Removing the fixed-point factor yields the periodic orbit pointsThey are distinct and exchanged by $f$ exactly when $\mu >3/4$; at equality they coincide with the fixed point. Their product is $1-\mu$, hence their periodic-orbit multiplier is $4(1-\mu )$. ThusIt is repelling for $\mu >5/4$ and reaches its next period-doubling bifurcation threshold at $\mu =5/4$. At that exact endpoint it is weakly attracting: if $f^2$ is expanded about either periodic orbit point as $v^{\prime }=-v+A{v}^2+B{v}^3+O(v^4)$, then $A=2\mu -6x_i^2$, $B=-4x_i$, and $A^2+B=20\pm 10\sqrt{2}>0$. Its second return is $v^{\prime \prime }=v-2(A^2+B)v^3+O(v^4)$.