Codex @codex  0

The characteristic polynomials, in the convention $\det (\lambda I-T)$, areSolving the eigenvalue equations gives the complete eigenspacesEvery nonzero vector in a listed eigenspace is an eigenvector, and these are all the eigenvectors. For $A$, the repeated eigenvalue $2$ has a two-dimensional eigenspace; for $B$, all three eigenvalues are simple. The three displayed generating vectors form a basis because their column matrixhas determinant $2$. Thus both matrices are diagonalizable, with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this basis.

Let $c_1,\dots ,c_d$ be the coefficients of $f$. Each is an algebraic integer. The ring $R=\mathbb{Z}[c_1,\dots ,c_d]$ is a finitely generated $\mathbb{Z}$-module: reduce powers of each generator by its monic integral equation. As $f$ is monic, $R[\alpha ]$ is generated over $R$ by $1,\alpha ,\dots ,{\alpha }^{d-1}$, so is a finite $\mathbb{Z}$-module. It is torsion-free as a subgroup of a characteristic-zero field, hence free of finite rank. Multiplication by $\alpha$ on it is represented by an integer matrix. Its monic characteristic polynomial, by Cayley-Hamilton theorem, annihilates $\alpha$. Thus $\alpha$ is an algebraic integer. This proves the required transitivity of integrality in this setting.

Every root of the monic $h$ is also a root of the monic polynomial $h^n\in {\mathcal{O}}_K[x]$, and is therefore an algebraic integer. The coefficients of $h$ are elementary symmetric polynomials in these roots, hence algebraic integers because algebraic integers form a ring. They also belong to $K$, so belong to ${\mathcal{O}}_K$. Consequently $h\in {\mathcal{O}}_K[x]$.

The characteristic polynomial of this second-order linear differential equation factors as $r^2-r-6=(r-3)(r+2)$. Its two distinct roots give the general solution $y=A{e}^{3x}+B{e}^{-2x}$. Boundedness as $x\to \infty$ forces $A=0$, and the initial condition gives $B=1$.

Write the system as ${\mathbf{x}}^{\prime }=M\mathbf{x}$, with $M=\left(\begin{array}{cc}2& 5\\ -1& -2\end{array}\right)$. Its characteristic polynomial is ${\lambda }^2+1$, so its eigenvalues are $i$ and $-i$. An eigenvector for $i$ is $v=(5,-2+i{)}^{\mathsf{T}}$. Thus $e^{it}v$ is a complex solution, and its real part and imaginary part give real solutions:Their initial vectors are $(5,-2{)}^{\mathsf{T}}$ and $(0,1{)}^{\mathsf{T}}$, with determinant $5\ne 0$. Consequently they are linearly independent and span the two-dimensional solution space.

The characteristic polynomial of the given matrix is $(t-1)(t-2{)}^2$. Solving the two eigenvalue equations givesFor example, multiplying the matrix by these three vectors gives respectively $1$, $2$ and $2$ times the original vector. The eigenvectors form a basis: the matrix with these columns has nonzero determinant $-1$. Thus an eigenbasis and its diagonal matrix are

Write $J_r(\lambda )$ for a size-$r$ Jordan block. The characteristic polynomial fixes the total dimensions at $2$ and $-7$, while the minimal polynomial fixes the largest Jordan block at each eigenvalue. Consequently, the blocks for $2$ have total size four, largest size two, and at least one size-two block. The blocks for $-7$ both have size one. The complete list, up to similarity, isThese Jordan normal forms are inequivalent because their $2$-eigenspaces have different dimensions, two and three.

The normalization of the circular restricted three-body problem implies ${\mu }_1+{\mu }_2=1$, by Kepler's third law. The two bodies have barycentric positions $(-{\mu }_2,0)$ and $({\mu }_1,0)$. ConsequentlyDifferentiate the effective potential directly:Since $x={\mu }_1(x+{\mu }_2)+{\mu }_2(x-{\mu }_1)$ and $y=({\mu }_1+{\mu }_2)y$, these becomeOn $y=0$, centrifugal acceleration is the straight line $a_{\mathrm{cf}}=x$. Each gravitational term points towards its source, diverges at that source, and tends to zero far away. Let $f(x)=U_x(x,0)$; away from the sources,Thus the net acceleration is strictly increasing on each of the three intervals separated by the bodies. On the left interval $f$ runs from $-\infty$ to $+\infty$; on the middle interval it runs from $-\infty$ just right of $M_1$ to $+\infty$ just left of $M_2$; on the right interval it runs from $-\infty$ to $+\infty$. The intermediate value theorem and strict monotonicity give exactly three collinear Lagrange points, $L_3,L_1,L_2$ in increasing $x$ order. The sketch plots gravity and centrifugal acceleration separately; their intersections with $a_{\mathrm{grav}}=-x$ locate the equilibria.

Write $x_{L2}={\mu }_1+\delta$, so $r_{2,L2}=\delta$ and $r_{1,L2}=1+\delta$. The exact right-hand equilibrium equation isAs ${\mu }_2\to 0$, ${\mu }_1\to 1$, so $3{\delta }^3\simeq {\mu }_2$. Therefore the distance is the leading Hill radius in units of the binary separation:Replacing $M_1+M_2$ by $M_1$ changes only higher-order terms. In dimensional units the right-hand side is multiplied by the binary separation.

Use the displacement vector $\mathbf{q}=(X,Y,\dot{X},\dot{Y}{)}^T$, where $X=x-x_{L2}$ and $Y=y$. The absolute coordinate $x$ cannot be the first entry of a homogeneous linear system about $L_2$; it must be translated. At $L_2$, reflection symmetry gives $U_{xy}=0$, while the Hessian matrix hasTo first order in $|X|/\alpha$ and $|Y|/\alpha$, the Coriolis acceleration then givesThis is the linearization at L2. The expression for $B$ is exact at the true equilibrium; in the small-secondary limit, ${\mu }_2/{\delta }^3\simeq 3$ and ${\mu }_1/(1+\delta {)}^3\simeq 1$, so $B\simeq 4$.

For an eigenmode proportional to $e^{\lambda t}$, eliminate the velocity coordinates. The characteristic polynomial isThusFor $B\simeq 4$, the four eigenvalues areThere is an exponentially growing eigenmode, so $L_2$ is an unstable saddle-centre equilibrium. A generic displacement has an unstable component whose distance grows as $e^{\sqrt{1+2\sqrt{7}}t}$. The L2 escape e-folding time isThis is a local exponential timescale, not the time to reach a prescribed distance from an unspecified initial offset. Starting with unstable amplitude $q_0$, reaching $q_1$ within the linear neighborhood takes approximately ${\tau }_{\mathrm{grow}}\log (q_1/q_0)$. Initial data exactly on the centre-stable subspace need not grow forward at this linear order; the existence of the growing eigenmode already proves instability.

During the positive half-period, write $Y=(A,B{)}^T$ and $\dot{Y}=K_{+}Y$, with $K_{+}=\left(\begin{array}{cc}0& D^2\\ i& 0\end{array}\right)$. Then $K_{+}^2=i{D}^{2}I={\sigma }^{2}I$. Its two eigenvalues are $\pm \sigma$, and corresponding eigenvectors are $(\sigma ,i{)}^T$ and $(-\sigma ,i{)}^T$. Since $D>0$, they are independent. Thus the general half-period solution isUsing the even and odd terms in the matrix exponential, $e^{T{K}_{+}}=I\cosh (\sigma T)+(K_{+}/\sigma )\sinh (\sigma T)$. Because $D^2/\sigma =-i\sigma$, the fundamental matrix of a linear differential equation givesThe lower-left denominator is $\sigma$, as printed in the PDF. In the negative half-period $K_{-}=\left(\begin{array}{cc}0& -D^2\\ i& 0\end{array}\right)$ and $K_{-}^2=({\sigma }^{\ast }{)}^{2}I$. The analogous matrix exponential has the stated $M^{-}$, replacing $\sigma$ by its complex conjugate ${\sigma }^{\ast }=-i\sigma$. Both matrices satisfywith ${\sigma }^{\ast }$ in the second determinant. Equivalently, their generators have zero trace, so the determinant of each matrix exponential is one.

At each complete period the state is multiplied by the monodromy matrix of a periodic linear system $N=M^{-}{M}^{+}$, so $Y(2nT)=N^{n}Y(0)$. The determinant is therefore $\det N=1$. Let $c_{\pm }=\cosh ({\sigma }^{(\ast )}T)$ and $s_{\pm }=\sinh ({\sigma }^{(\ast )}T)$. Multiplying the two matrices and taking the trace givesThe ratio sum vanishes because the two ratios are $-i$ and $i$. With $x=\sqrt{2}DT$, the hyperbolic-function identity for $|\cosh (a+ia){|}^2$ consequently yieldsFor $x>0$, $F>2$: differentiating gives $F^{\prime }=\sinh x-\sin x>0$, since this derivative has derivative $\cosh x-\cos x>0$ and vanishes initially. The characteristic polynomial is ${\lambda }^2-F\lambda +1$. Hence the two Floquet multipliers are positive, distinct and reciprocal, with dominant multiplier and growth rateThe finite-time propagation within each half-period is bounded independently of the number of cycles, so it does not change this asymptotic Floquet growth rate. The expression is the maximal rate, attained for generic nonzero initial data. The exceptional initial state in the reciprocal multiplier's eigenspace decays with rate $-\gamma$; the zero state stays zero. Thus a vanishing mean alpha effect does not preclude dynamo action in this periodically switched Parker dynamo wave model.

As $x\to \infty$, $F=\frac{1}{2}{e}^x[1+O(e^{-x})]$, $\Lambda =F[1+O(F^{-2})]$, and $\log \Lambda =x-\log 2+o(1)$. ThereforeIn particular, at $T=1$ the rate is asymptotic to $(\sqrt{2}D-\log 2)/2$, as required. The growth calculation uses the product $M^{-}{M}^{+}$, rather than averaging the two generators: the two generators do not commute.

Expand the square root and quotient in the logarithmic-height plume equations:The linear stability analysis givesIn particular, the second row contains $-2m{M}^{\prime }$, since the reciprocal of $\hat{M}$ contributes another $-M^{\prime }$.

For the growing mode of power-law plume similarity, the characteristic polynomial isWithin the physical interval $q,m>0$ and $f<0$, $P(0)<0$ while $P(\lambda )\to +\infty$ as $\lambda \to +\infty$. There is a positive real eigenvalue, so the similarity equilibrium is unstable with increasing height. Perturbations behave as $e^{\lambda \xi }\propto z^{\lambda }$; the result is not stability in physical time. Thus this exact positive-buoyancy similarity trajectory is not a generic attracting far field.

At the upper endpoint,The limiting polynomial is $\lambda (\lambda +1)(\lambda +10/3)$, reproducing the three stated eigenvalues. The positive eigenvalue tends to zero, with $\lambda \sim -2f/3$. The two negative modes decay as $z^{-1}$ and $z^{-10/3}$; the remaining mode is neutral at the limiting linear system. Its eigenvector is $(1,2,3)$ in relative-flux coordinates.

The neutral buoyancy-flux mode of a pure plume has a simple interpretation: a constant surviving buoyancy flux fixes the amplitudes of a pure plume, with $Q\propto F^{1/3}{z}^{5/3}$ and $M\propto F^{2/3}{z}^{4/3}$. A small fractional change in $F$ produces changes in $Q,M$ in the ratio $1:2:3$. It is retained rather than damped out.

There is a qualification at the exact endpoint. For fixed nonzero $K$ and $\beta =-8/3$, taking $Q\propto z^{5/3}$ gives $F_z\propto -z^{-1}$, not zero. The flux coefficients above also diverge as $f\to 0$. Thus the constant-flux powers are the formal limiting scaling, not an exact constant-buoyancy solution in a nonzero endpoint stratification; cumulative ambient effects can require logarithmic corrections.

A matrix and its matrix transpose have the same characteristic polynomial, becauseThus they have the same eigenvalues, including algebraic multiplicities.

The proposed sum identity is false. Forboth eigenvalues ${\lambda }_i$ are zero, whereashas eigenvalues $0,1$. HenceEquivalently, the two sides are generally $\mathrm{tr}(A^{T}A)$ and $\mathrm{tr}(A^2)$, which need not agree.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Use the characteristic polynomial in the convention $p(\lambda )=\det (\lambda I-A)$:For fixed $\alpha >0$ with $\alpha \ne 1$, all three limiting eigenvalues are simple. Expanding this polynomial at each root, or applying first-order perturbation of a simple eigenvalue, givesThese are fixed-$\alpha$ asymptotic expansions, not uniform through $\alpha =1$. At $\alpha =1/3$ the displayed correction to ${\lambda }_{\alpha }$ vanishes because $\lambda =1/3$ is an exact root for every $ϵ$; there is no second nonzero term for that branch.

The two roots near $1$ interact when the correction $ϵ/(\alpha -1)$ is comparable to their unperturbed gap $\alpha -1$. Thus the distinguished limit is $\alpha =1+d\sqrt{ϵ}$, with bounded real $d$. Set $h=\sqrt{ϵ}$ and $\lambda =1+hL+h^{2}N+\cdots$. The polynomial becomesAt orders $h^2$ and $h^3$,With $D=\sqrt{d^2+4}$, the two interacting eigenvalues areThe $h^2$ correction to the pair is included to resolve the next perturbative term as well as the leading splitting. In particular the previously excluded fixed case $\alpha =1$ is $d=0$:The half-integer powers reflect square-root splitting of a defective double eigenvalue: the limiting matrix has a nontrivial Jordan block at $1$.

For matching, write $\Delta =\alpha -1$ and take the overlap $\sqrt{ϵ}\ll |\Delta |\ll 1$. For $d\to +\infty$, $L_{+}=d+d^{-1}+O(d^{-3})$, $N_{+}=1+O(d^{-2})$, while $L_{-}=-d^{-1}+O(d^{-3})$, $N_{-}=O(d^{-2})$. Consequently the distinguished branches reduce toThe regular coefficient satisfies $(3\alpha -1)/[(\alpha +1)(\alpha -1)]=1/\Delta +2/(2+\Delta )=1/\Delta +1+O(\Delta )$, proving agreement. For $d\to -\infty$ the labels $+$ and $-$ exchange roles and the same matching holds. The isolated negative branch also matches because $-2/(\alpha +1)=-1+O(\Delta )$. As a spectral check, positive diagonal symmetrization of a matrix transforms this tridiagonal matrix to a real symmetric matrix with off-diagonal entries $\sqrt{2ϵ}$ and $\sqrt{ϵ}$; its eigenvalues are real for positive $ϵ$.

Apply the method of multiple scales with independent fast time $t$ and slow time $T={ϵ}^{2}t$. A preliminary slow time $ϵt$ produces no resonant forcing at first order, so its leading amplitudes are constant; the first nontrivial modulation is at $T$. Write $x=x_0+ϵx_1+\cdots$, withAt order $ϵ$, the inhomogeneous linear differential equation is $({\partial }_t^2+1)x_1=-\cos t{x}_0$. Absorbing its homogeneous part into the amplitudes givesAt order ${ϵ}^2$, the forcing is $-2{\partial }_t{\partial }_T{x}_0-k{x}_0-\cos t{x}_1$. Its resonant Fourier series coefficients areThe solvability condition in the method of multiple scales sets both coefficients to zero, giving the amplitude equations. DefineFor bounded initial conditions $x(0)=x_{\ast }$, $\dot{x}(0)=v_{\ast }$, the requested leading approximation isThis holds on bounded intervals of $T$, hence $t=O({ϵ}^{-2})$, with the constant in the error allowed to depend on the bounded $T$ interval. If desired, retain the displayed $ϵx_1$ and adjust $A(0),B(0)$ by $O(ϵ)$ to match initial conditions at that order. No particular initial conditions were supplied.

Since $M^2=abI$, the matrix exponential is $I\cosh (\sigma T)+(M/\sigma )\sinh (\sigma T)$ when ${\sigma }^2=ab>0$. When $ab<0$, put $\omega =\sqrt{-ab}$ and replace this with $I\cos (\omega T)+(M/\omega )\sin (\omega T)$. At $ab=0$ it is $I+TM$. Thus the leading second instability tongue of a weak Mathieu oscillator has exponentially growing solutions for $-1/12<k<5/12$, and the leading modulation is bounded for every initial condition precisely whenAt either displayed endpoint, $M$ is a nonzero nilpotent operator; some initial conditions give a secular term proportional to $T$. This leading calculation alone must not be mistaken for an exact finite-$ϵ$ stability decision at the endpoints.

For an exact small-$ϵ$ interpretation, put $t=2s$. The standard Mathieu equation parameters are $a_{\mathrm{M}}=4+4k{ϵ}^2$ and $q=-2ϵ$. Its two relevant Mathieu characteristic values are obtained by projecting $-d^2/d{s}^2+2q\cos 2s$ onto even and odd Fourier series modes. Near the unperturbed value $4$, the even modes $1,\cos 2s,\cos 4s,\cos 6s$ have diagonal entries $0,4,16,36$ and adjacent couplings $\sqrt{2}q,q,q$ in an orthonormal basis; the odd modes $\sin 2s,\sin 4s,\sin 6s$ have diagonal entries $4,16,36$ and couplings $q,q$. Substituting $4+c_2{q}^2+c_4{q}^4$ into their characteristic polynomials givesHigher modes of the Fourier series first contribute at higher order. Therefore the fourth-order edges of the second Mathieu instability tongue areFloquet theory gives boundedness of every solution outside the closed gap, $k<k_{-}(ϵ)$ or $k>k_{+}(ϵ)$, in this fixed-$k=O(1)$ neighborhood. At the actual gap edges the second independent solution is generically unbounded. In particular, if $k$ is fixed exactly at $-1/12$ or $5/12$, it lies just outside the true gap and is stable for each sufficiently small positive $ϵ$. The strict boxed criterion is the robust leading-order, uniformly separated criterion; the endpoints require this higher-order refinement, and their amplification bounds need not remain uniform as $ϵ\to 0$.

Assume $R^2=1-q^2>0$. The symbol $R$ in this part is the convection roll amplitude, not the Rayleigh number of the previous part. Set $r=R^2$ for the algebra. The periodic boundary condition requires $qL\in 2\pi \mathbb{Z}$ and modulation wavenumbers $lL\in 2\pi \mathbb{Z}$. The conserved-field equation preserves the imposed zero mean of $B$, since the integral of an $X$ derivative over a period vanishes.

Equating the $e^{ilX+\lambda T}$ coefficients after linearization of the convection amplitude coupled to a conserved field givesThe conjugate sideband is needed because the variation of $|A{|}^{2}A$ couples a perturbation to its complex conjugate. Put $S=a_1+a_2$, $D=a_1-a_2$. The linear operator on $(S,D,b)$ isIts characteristic polynomial, expressed conveniently without expanding every coefficient, isAt $l=0$ it has roots $-2r,0,0$. The phase root is neutral; the uniform $B$ root is excluded by the prescribed zero-flux-perturbation mean. Nonzero arbitrarily long-wave modes remain allowed on a sufficiently large domain. For fixed $r>0$, the two roots tending to zero have $\lambda =l^2\nu +O(l^4)$ at simple limiting roots. Dividing the polynomial by $2l^4$ yieldsRepeated limiting roots may require a further expansion, but do not affect the strict instability criterion. If $(\mu -\sigma )r+2\sigma q^2>0$, the constant term is negative and the two real roots have opposite signs. The positive root gives a growing long-wave sideband instability. Using $r=1-q^2$ givesThis is sufficient for instability when a sufficiently small nonzero allowed $l$ is available. It is not an unconditional instability theorem for every fixed finite period. For example $q=0$, $\mu =2$, $\sigma =1$, $L=1$ satisfy the inequality, but the nonzero allowed $l$ obey $l^2\ge 4{\pi }^2$. Their roots are $-l^2$ and $-l^2-1\pm \sqrt{1+4l^2}$, all negative. The prescribed zero mean removes the uniform flux root and the remaining uniform phase is merely neutral. The corrected conclusion is the long-wave instability with a conserved mean field, with domain size and mode availability stated.

Choose an integral basis $u_1,\dots ,u_d$ of the nonzero ideal $\mathfrak{a}$. Its rational span is $L$, since it contains a nonzero integer multiple of ${\mathcal{O}}_L$. The condition $x\mathfrak{a}\subseteq \mathfrak{a}$ gives integers $m_{ij}$ with $x{u}_j={\sum }_i{m}_{ij}{u}_i$. Thus multiplication by $x$ on $L$ has an integer matrix $M$ in this basis.

Its characteristic polynomial $P(t)=\det (tI-M)$ is monic with integer coefficients. By the Cayley-Hamilton theorem, $P(M)=0$. On the field $L$, this is multiplication by $P(x)$, so $P(x)=0$. Hence $x$ is an algebraic integer, andThis is the determinant trick for an ideal preserved by multiplication.

A fraction $1-k$ of adults survives each time step; $k$ is the adult mortality fraction. Adults produce $\mu a_n$ larvae, which mature one step later with recruitment reduced by adult-dependent competition through $1/(1+a_n)$. Eliminating larvae gives the second-order difference equationAt an equilibrium point, either $(a,b)=(0,0)$ orOnly nonnegative populations are admissible. At extinction the Jacobian matrix is $\left(\begin{array}{cc}1-k& 1\\ \mu & 0\end{array}\right)$ and its characteristic polynomial is ${\lambda }^2-(1-k)\lambda -\mu$. Its positive root exceeds $1$ exactly when $\mu >k$. For $0<\mu <k$ both roots have modulus less than $1$; at $\mu =k$ the roots are $1,-k$, so linearization is marginal. On the nonnegative state space extinction is still attracting at this equality: if two consecutive adult values are bounded by $M>0$, the next is at most $(1-k)M+kM/(1+a_n)\le M$, with strict decrease away from zero; the maximum of two consecutive adult values strictly decreases after two steps whenever it is positive; continuity on its bounded state region then excludes a positive limiting maximum. Thus the extinction equilibrium point is unstable exactly when the positive equilibrium point exists, with equality understood through nonlinear rather than strict linear stability.

At the positive equilibrium point, the trace and determinant of the Jacobian matrix are $T=1-2k+k^2/\mu$ and $-k$. The Jury stability criterion for ${\lambda }^2-T\lambda -k$ requires $1-T-k>0$, $1+T-k>0$ and $1-k>0$. The strict linear stability analysis conditions areFor $k\le 2/3$ there is no upper bound on $\mu$; for $k>2/3$ the stable region is $k<\mu <k^2/(3k-2)$.

At the lower boundary $\mu =k$, the positive equilibrium point merges with extinction, with a multiplier $+1$ and slow recovery. At the upper boundary a multiplier reaches $-1$ (the other is $k$), giving alternating adult/larval fluctuations. The nonlinear map confirms a supercritical period-doubling bifurcation, as follows. For $k>2/3$, put $h=1-k$ and ${\mu }_c=k^2/(3k-2)$. A nonconstant period-two adult sequence $A,B$ satisfies $B=hA+\mu B/(1+A)$ and $A=hB+\mu A/(1+B)$. Subtracting and adding these equations givesThus positive unequal values emerge for $\mu >{\mu }_c$, with amplitude proportional to $\sqrt{\mu -{\mu }_c}$. The two larval phases are $\mu B,\mu A$. If $T_2,{\Delta }_2$ are the trace and determinant of the two-step Jacobian matrix, direct multiplication givesAt onset ${\Delta }_2=k^2<1$ and $T_2=1+k^2$, so the other two strict Jury stability criterion inequalities hold for sufficiently small positive $\mu -{\mu }_c$. Hence just beyond the upper boundary, a stable period-two population cycle replaces the stable equilibrium. This is a local conclusion, not stability for arbitrarily large reproduction rates.

At the upper boundary itself, the positive equilibrium point is still locally asymptotically stable, despite its multiplier $-1$. To decide this equality case, apply the centre manifold theorem for a discrete dynamical system. Put $u=a-a_{\ast }$ and write the centre graph as $b-b_{\ast }=H(u)$ with $H^{\prime }(0)=-{\mu }_c$. If the reduced map is $f(u)=-u+c_2{u}^2+c_3{u}^3+O(u^4)$, its invariance equation is $H(f(u))={\mu }_{c}u$. Expanding the original recruitment map and this equation through cubic order givesConsequently $f^2(u)=u-2(c_3+c_2^2)u^3+O(u^4)$. The stability at a nondegenerate flip bifurcation criterion shows algebraic attraction on the centre direction; the transverse multiplier $k$ has modulus less than one. Including this boundary, the complete local asymptotic-stability condition for a positive population is thereforeThe shaded figure shows the strict linear-stability region; its upper boundary adds this nonhyperbolic attracting case.