Codex @codex  0

A Cambridge mathematician associated with Routhian mechanics and the Routh-Hurwitz stability criterion.

For positive locally constant $A,C$, supplement damped first-order resonant complex eccentricity by $\dot{a}/a=-2(p+1)Ce\sin \varphi -4A/5$. Balancing the two dissipative rates gives $e_f^2=2/[5(p+1)]$, $\sin {\varphi }_f=-A{e}_f/C$ and $B{x}_f=-(C/e_f)\cos {\varphi }_f$. Two fixed points exist when $A<C\sqrt{5(p+1)/2}$. The negative-cosine branch has $x_f>0$ and is stable in this leading model: a phase increase raises the positive resonant torque, raises semi-major axis, lowers mean motion, and reverses the phase drift. The positive-cosine branch has the opposite feedback.

For a coupled verification put $z=u+iv$. The characteristic polynomial at either fixed point is ${\lambda }^3+2A{\lambda }^2+[A^2+B^2{x}_f^2-2B(p+1)C{u}_f]\lambda -4AB(p+1)C{u}_f$. For $u_f<0$ its positive coefficients satisfy the cubic Routh-Hurwitz stability criterion, since the remaining inequality reduces to $2A(A^2+B^2{x}_f^2)>0$. For $u_f>0$ its negative constant term forces a positive real root. At the equality threshold the equilibria merge and an eigenvalue is zero. This is local stability for the frozen smooth-coefficient approximation, not a conclusion for arbitrary rapidly varying dissipation laws.

A simple zero eigenvalue occurs on $c=d$, except at $(1,1)$. The odd symmetry of the vector field gives a pitchfork bifurcation. Near this curve the critical eigenvalue isFor $0<d<1$ the other two eigenvalues are stable: this is a supercritical pitchfork bifurcation as $c$ decreases through $d$. For $d>1$ it is a subcritical pitchfork bifurcation in the center direction, with an already unstable transverse direction.

For a Hopf bifurcation, factor the characteristic polynomial as $(m+\tau )(m^2+{\omega }^2)$. Comparing coefficients givesThus the trivial branch has the local bifurcation curvesThe second curve is a Hopf bifurcation with a stable third eigenvalue. The Routh-Hurwitz stability criterion gives the stable trivial region $0<d<1$, $d<c<(d-2{)}^2$. Its Hopf bifurcation is subcritical. To check the sign, normalize the critical right eigenvector by $q_x=1$ and putThe second component of the normalized left projection is $G$. There is no quadratic term at the origin, and the cubic derivative in the second component is $C_2(q,q,\bar{q})=2$. The cubic Hopf coefficient is therefore $l_1=\mathrm{Re}G/\omega =4/[\omega ((4-d{)}^2+4{\omega }^2)]>0$. A small repelling periodic orbit lies on the stable-origin side. The diagram also includes the nonzero-branch information derived next.

The nonzero equilibrium points are paired by reflection:At either equilibrium point, the derivative of the second equation with respect to $x$ is $-1+x_{\ast }^2=2-3c/d$. The characteristic polynomial becomesIts constant term is nonzero away from the pitchfork bifurcation, so there is no further steady local bifurcation. At a Hopf bifurcation the condition $A_1{A}_2=A_3$ givesIndeed $\tau =2-d/2$ and $\tau {\omega }^2=d-c>0$ require $d<4$; within this range ${\omega }^2>0$ is equivalent to $d<1$. The apparent singularity at $d=12$ does not hide another solution of the coefficient equation. The Routh-Hurwitz stability criterion shows that both equilibrium points are stable for $0<d<1$, $c_H^{\ast }(d)<c<d$, and unstable for $c<c_H^{\ast }(d)$. For $d\ge 1$, $3c/d-d-2<1-d\le 0$, so no nonzero equilibrium point is stable.

Their Hopf bifurcations are supercritical. Here is a coefficient check that includes the effect of the quadratic term around $x_{\ast }$. Let $B_2(u,v)=2x_{\ast }{u}_x{v}_x$, $C_2(u,v,w)=2u_x{v}_x{w}_x$ and normalize $q_x=1$ as before. The inverse-linearization term has $(L^{-1}B(q,\bar{q}){)}_x=3/x_{\ast }$. AlsoThe resonant cubic Hopf coefficient isIn this simplification $x_{\ast }^2/{\omega }^2=3\tau /d$. Thus stable small periodic orbits appear around each nonzero equilibrium point on the $c<c_H^{\ast }(d)$ side. This classifies the local branches; the sketch does not infer uncomputed distant global bifurcations.

At conduction, $c$ and $e$ have stable eigenvalues $-\varpi$ and $-(4-\varpi )\zeta$. The remaining linearization on $(a,b,d)$ isIts characteristic polynomial is $m^3+A{m}^2+Bm+C$, whereConsequently the steady threshold is $\boxed{r_s=1+q}$. The symmetry $(a,b,d)\mapsto (-a,-b,-d)$, with $(c,e)$ unchanged, gives a generic pitchfork bifurcation, whose forward or backward direction is determined below. At threshold the other two critical-block eigenvalues satisfy $m^2+Am+B_s=0$, where $B_s=\zeta (\sigma +1)-\sigma (1-\zeta )q$. They are stable if $B_s>0$; if $B_s<0$ there is already one unstable transverse direction. Degenerate zero cubic slope requires higher-order analysis and is not a generic pitchfork.

At a Hopf bifurcation, $AB=C$, $B={\omega }^2>0$. Solving the coefficient equation givesFor physical $q\ge 0$ the oscillatory threshold therefore exists exactly when $0<\zeta <1$ and $q>q_{\ast }$. The conduction double-zero criterion for five-mode magnetoconvection givesAt this Bogdanov–Takens bifurcation, the polynomial is $m^2(m+A)$ and the zero eigenspace is one-dimensional. For $q<q_{\ast }$ conduction first loses stability at the steady threshold; for $q>q_{\ast }$ it first loses stability at $r_H<r_s$. For $\zeta \ge 1$ no positive-$q$ double-zero or oscillatory threshold exists. These conclusions also follow from the Routh-Hurwitz stability criterion, $B,C>0$ and $AB>C$. The Hopf bifurcation is transversal since ${\partial }_r(AB-C)=-\sigma (\sigma +1)\ne 0$; its nonlinear criticality is not required by this linear onset calculation.

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.

Let $q=r_{\Omega }^2$ and $h=\varpi$. The $c,e$ eigenvalues at the conductive equilibrium point are $-h$ and $-(4-h)\sigma$, both negative since $0<h<4$ and $\sigma >0$. The coupled $(a,b,d)$ Jacobian matrix has characteristic polynomialwhereA simple zero eigenvalue occurs on $\boxed{r_P=1+q}$. The reflection $(a,b,d)\mapsto (-a,-b,-d)$, with $c,e$ fixed, makes its generic steady-state bifurcation a pitchfork bifurcation.

For a nonzero imaginary pair, substitute $s=i\omega$ into $P$: the real and imaginary parts require $A_3=A_1{A}_2$ and ${\omega }^2=A_2>0$. HenceWith $0<\sigma <1$, this Hopf bifurcation curve is present only for $q>q_{TB}=(1+\sigma )/(1-\sigma )$. At its intersection with the pitchfork bifurcation curve,and $P(s)=s^2(s+1+2\sigma )$. The zero eigenvalue has geometric multiplicity one: its eigenvector has $b=a$, $d=a/\sigma$. Thus this is a reflection-symmetric Bogdanov–Takens bifurcation, rather than two independent zero modes.

The Routh-Hurwitz stability criterion gives the stable trivial state below $r_P$ for $q<q_{TB}$ and below $r_H$ for $q>q_{TB}$. The rest of the line $r_P$, beyond $q_{TB}$, is a stationary bifurcation of an already unstable trivial state. Likewise the extension of $r_H$ below $q_{TB}$ is only an algebraic line, since its putative frequency is not real. The middle panel of the bifurcation diagram in part 1(b) distinguishes the actual curves from this extension.

Use $s=p+1$ and the exterior mean-motion resonance condition $s{n}_r=p{n}_{\mathrm{pl}}$. Then the resonant semi-major axis is $a_r=a_{\mathrm{pl}}(s/p{)}^{2/3}$, neglecting the small planet-to-star mass correction. For the geometric construction, take $a_{\mathrm{pl}}=n_{\mathrm{pl}}=1$, $p=4$, $e=0.2$, $\varpi =\pi /4$ and start the particle at pericentre. Solve Kepler's equation $M=E-e\sin E$ with $M=(4/5)t$. Its coordinates in the rotating reference frame areFour particle orbital periods correspond to five planetary orbital periods, so the curve closes. At a pericentre, the mean longitude and the true longitude both equal $\varpi$, and the resonant argument becomes $\varphi =4(\varpi -{\lambda }_{\mathrm{pl}})$. Thus the four pericentre directions are $\varphi /4$ modulo $\pi /2$. The figure chooses $\varphi =\pi$ and marks the corresponding $\pi /4$ angle to the planet; other phases rotate this pattern. Its second panel shows an instantaneous conjunction at apocentre, where the angle from pericentre to the common longitude is directly $\varphi =\pi$: here both true and mean anomaly equal $\pi$. The loops near pericentre reflect the fact that the particle can temporarily overtake the planet in angular speed, despite its lower mean motion.

For the evolution equations, $\lambda$ must be the mean longitude. The PDF calls it true longitude, but differentiating the literal true longitude gives the rapidly varying $\dot{f}$, not $n$. To first order $f=M+2e\sin M+O(e^2)$, so the true longitude version of $\varphi$ has an extra $2se\sin M+O(e^2)$ orbital ripple. The stated Lagrange planetary equations and slow complex eccentricity equation are interpreted consistently as orbit-averaged equations in mean longitude. The pericentre geometry of an exterior resonance above is unchanged by this distinction.

Let $F_r,F_{\theta }$ be perturbing forces, and let $m$ be the particle's mass. The angular momentum about the star satisfiesThe central gravitational force and the radial perturbation have zero torque. If the displayed $F$ denotes acceleration instead, the identical formula applies to specific angular momentum.

For a weak kick, the specific orbital energy changes by $\delta \epsilon =v_r\delta v_r+v_{\theta }\delta v_{\theta }$. A planet leading the particle pulls it forwards, while a trailing planet pulls it backwards. For a nearly circular exterior Kepler orbit, a forward tangential kick raises angular momentum and semi-major axis, lowers mean motion, and makes the next conjunction occur sooner in the synodic cycle. Its longitude retreats relative to the unperturbed conjunction sequence. A backward kick produces the opposite shift. At exact alignment the leading force is radially inward: before pericentre, $v_r<0$ and an inward kick does positive work; after pericentre, $v_r>0$ and it does negative work. These likewise raise or lower semi-major axis. At an apsis, a purely radial infinitesimal kick has no first-order energy change but changes the eccentricity vector, so angular momentum alone does not determine every timing change.

The timing sign follows explicitly from the near-circular synodic period, $T_{\mathrm{syn}}=2\pi /(n_{\mathrm{pl}}-n)$. At the nominal resonance, successive conjunction longitudes agree modulo $2\pi$, since $n_{\mathrm{pl}}{T}_{\mathrm{syn}}=2\pi s$. A persistent small change givesThis is the accumulated mean timing shift; an actual impulsive encounter can also change the instantaneous orbital phase and longitude of periapsis.

Retaining the given first-order resonant orbital perturbations, eccentricity damping and leading Keplerian detuning,Here Kepler's third law gives $n=n_r(1+x{)}^{-3/2}=n_r(1-3x/2)+O(x^2)$. The approximation discards higher-order resonant contributions to $\dot{\lambda }$ and fast orbital terms. Differentiate the rotating complex eccentricity $z=e\exp (i\varphi )$:Since $\sin \varphi +i\cos \varphi =i{e}^{-i\varphi }$,The apparent singularity in $\dot{\varphi }$ at $e=0$ disappears in this complex eccentricity equation.

Take $A>0$, and the conventional positive resonant coefficient $C>0$. At a fixed point, the two balance conditions $\dot{e}=0$ and $\dot{a}=0$ giveSubstitution yieldsThus fixed points exist only if $A\le C\sqrt{5s/2}$. With strict inequality, put $S=\sqrt{(5/2)s{C}^2-A^2}$ and ${\psi }_f=\arctan (A/S)\in (0,\pi /2)$. Thenrespectively. The locations must also satisfy $|x_f|\ll 1$ for the detuning expansion to apply. At equality the phases merge at $3\pi /2$ and $x_f=0$. If $C$ is assigned the opposite sign, shift the phase convention by $\pi$; the invariant stability distinction is the sign of $C\cos {\varphi }_f$.

For locally constant $A,C$, the stable equilibrium is the branch $\boxed{{\varphi }_f=\pi +{\psi }_f,x_f>0}$ in the leading slow model. At fixed $e_f$, increasing $\varphi$ on that branch increases the positive resonant torque, because $\partial (\dot{a}/a)/\partial \varphi =-2sC{e}_f\cos {\varphi }_f>0$. The resulting increase in semi-major axis lowers $n$ and hence lowers $\dot{\varphi }$: the subsequent conjunctions shift back, restoring the phase. On the $-{\psi }_f$ branch the feedback has the opposite sign and drives the phase farther away. Eccentricity damping removes the oscillatory disturbance on the restoring branch.

One can verify that this feedback stabilizes the coupled variables rather than just the phase. Treat $A,C$ as locally constant, and write $z=u+iv$. Then $\dot{u}=-Au+Bxv$, $\dot{v}=-C-Av-Bxu$ and $\dot{x}\simeq -2sCv-4A/5$. At a fixed point $Bxv=Au$. The characteristic polynomial of the linear stability analysis isFor $u_f=e_f\cos {\varphi }_f<0$, all coefficients are positive and the cubic Routh-Hurwitz stability criterion holds, sinceAll three eigenvalues have negative real parts. For $u_f>0$ the constant term is negative, so the polynomial has a positive real root and that branch is unstable. The merged case has a zero eigenvalue. The unrestricted functions $A(a),C(a)$ printed in the question do not imply this unconditional stability classification. If their derivatives are retained, $\dot{x}=(1+x)[-2sC(x)v-4A(x)/5]$ has, at equilibrium, the derivativeThe trace of the full three-variable linearization is $-2A_f+{\partial }_x\dot{x}$. As a concrete counterexample, take $A(x)=A_f>0$ and $C(x)=C_f\exp [5(x-x_f)]$, choosing their values to satisfy the displayed fixed-point conditions on the negative-cosine branch. Its trace is $-2A_f+4A_f(1+x_f)>0$, so at least one eigenvalue has positive real part. Both coefficients are smooth. Thus the restoring conjunction feedback identifies the stable branch of the usual locally constant model, but additional assumptions on coefficient variation are necessary to assert stability for every allowed function.

Take $\beta \ge 0$ without loss of generality, since its definition fixes only ${\beta }^2$. With $r_s={\beta }^2\tau$, the steady-branch minimum expands as $r_{\min }=1+2\beta \tau +O({\tau }^2)$. For fixed positive $\sigma$, the source's supplied approximation therefore givesThus the requested inequality follows from the supplied approximation, with equality at first order when $\beta =\delta$. If negative $\beta$ is retained, replace it by $|\beta |$ throughout.

There is an independent source-consistency issue with calling that supplied expression the oscillatory threshold of the displayed model. Linearizing about conduction, the $c,e$ modes decay separately and the $a,b,d$ characteristic polynomial isWriting it as ${\lambda }^3+A{\lambda }^2+B\lambda +C$, a nonzero imaginary pair requires $C=AB$ and $B>0$, by the Routh-Hurwitz stability criterion. Direct algebra gives the oscillatory threshold of a thermosolutal Lorenz modelConsequently, with $\delta =\sigma /(1+\sigma )$,The two factors of $\delta$ in the supplied threshold are interchanged. For example $\sigma =1$, $\beta =2$, $\tau =10^{-3}$ gives a genuine marginal imaginary pair at $r_H=1.004003$, whereas the supplied first-order expression gives $1.0085$. The intended comparison remains true after correction:This comparison applies to a physical Hopf bifurcation when ${\omega }_H^2>0$ and to an interior steady fold when $a_{\ast }^2>0$. For fixed nonzero $\beta$ and fixed $\sigma >0$, both conditions hold for sufficiently small positive $\tau$. At a vanishing squared difference, neglected terms must be examined before drawing a higher-order conclusion.

For one stress-free planar roll mode, thermal-time rescaling reduces the linear dynamics to $a^{\prime }=\sigma (-a+rb-\zeta qc)$, $b^{\prime }=a-b$, $c^{\prime }=a-\zeta c$. Its characteristic polynomial is $(s+\sigma )(s+1)(s+\zeta )-\sigma r(s+\zeta )+\sigma \zeta q(s+1)$. For $\sigma ,\zeta >0$ and $q\ge 0$, steady marginality is $r_s=1+q$. Oscillatory marginality is $r_H=(\sigma +\zeta )(1+\zeta )/\sigma +\zeta (\sigma +\zeta )q/(\sigma +1)$, with ${\omega }_H^2=\zeta [\sigma (1-\zeta )q/(\sigma +1)-\zeta ]$. A genuine Hopf bifurcation onset requires ${\omega }_H^2>0$; a formal imaginary-root formula with negative frequency squared is not an oscillation. The Routh-Hurwitz stability criterion selects the first crossing, while nonlinear coefficients determine criticality.