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At $X=P(u,v)$, orient the unit normal vector $n=(P_u\times P_v)/\Vert P_u\times P_v\Vert$ towards the eye. Define the unit directions from the reflecting point to the source and eye byThe incident propagation direction is $-a$. The law of specular reflection is $b=-a+2(a\cdot n)n$, so $a+b$ is normal to the surface. Conversely, if the tangential components of $a+b$ vanish and $a\cdot n,b\cdot n>0$, its unit-length summands have equal positive normal components and obey the reflection law. Therefore solve the two scalar equationsThe stated orientation condition gives $b\cdot n>0$. It removes the back-facing ambiguity, but not multiple images or caustics. Exclude $Q(t)=P(u,v)$ and $E=P(u,v)$, where these directions are undefined.

At a root where the $2\times 3$ Jacobian matrix $DF$ has rank two, the implicit function theorem makes the solution set a local curve in parameter space. Compute a unit tangent $w$ in its nullspace; equivalently take the normalized cross product of the two rows of $DF$. A continuation method predicts $x_p=x+hw$, where $x=(u,v,t)$, then corrects with the Newton method applied to the three equationsThe last equation fixes a transverse section and allows tracing through a turning point of $t$, where solving only for $(u,v)$ at successive fixed source parameters would fail. The corresponding space point is $R=P(u,v)$, with derivative $R^{\prime }=P_u{u}^{\prime }+P_v{v}^{\prime }$ along the traced parameter-space curve.

Find initial roots by subdividing the parameter cube, using interval arithmetic bounds on $F$ to discard impossible boxes and validated local solves in the remaining boxes. Trace in both directions, stop at the boundary or a recognized closed loop, and retain a record of traced boxes to avoid duplicates. Search the remaining boxes as well: starting from just one seed would miss disconnected closed reflection curves. Where $DF$ loses rank, reduce the step and subdivide locally to find the outgoing branches instead of continuing a single guessed tangent. If physical visibility is required, discard roots for which either open segment $XE$ or $XQ(t)$ is occluded; test these segments against the surface or any supplied scene geometry.

Control the geometric approximation rather than just the parameter step. For any regular traced arc $R(s)$ with a bound $\Vert R^{\prime \prime }(s)\Vert \le M$, the chord error bound gives an arc-to-chord Hausdorff distance at most $M(\Delta s{)}^2/8$. Budget, for example, half of $ϵ$ for corrected-vertex error and half for chord error; accept a segment only whenBound the vertex error through the inverse transverse Jacobian matrix and a bound on $DP$, or use validated root enclosures. If analytic second-derivative bounds are unavailable, recursively subdivide and enclose each arc in a geometric tube of radius at most the remaining tolerance around its chord. A midpoint test alone is a useful heuristic but cannot certify the requested error for an arbitrary oscillatory curve.

The returned vertices $P(u_k,v_k)$ and their ordered connecting segments give the reflection curve to the prescribed tolerance on every resolved branch. Finite certified termination requires regular evaluable patches, derivative or enclosure bounds, and finitely resolvable branches; eye-facing normals alone do not guarantee these properties. Singular components or an identically vanishing reflection constraint must be handled as such rather than silently reported as an ordinary curve.

Use the limit subdivision surface and limit subdivision curve, rather than treating their initial control meshes as the reflecting geometry. Build adaptive refinement hierarchies for both. Each surface cell supplies a limit-position evaluator, local tangent vectors and a normal vector; each curve interval supplies a limit-position evaluator and tangent. Regular regions may be evaluated as local splines, while extraordinary neighborhoods require the evaluation machinery of the specified subdivision rules. A piecewise smooth surface is handled patch by patch, with explicit boundary or crease joins.

A useful initial polygonal construction refines the surface to small triangles and the source curve to small line segments. For a candidate mirror triangle in a plane $n\cdot (X-X_0)=0$, reflect the eye across that plane:For a source edge $Y(t)=(1-t)Y_0+t{Y}_1$, a physical reflection point is the plane intersection of the line $E^{\ast }Y(t)$:Keep only $0\le t\le 1$, $0<\lambda <1$, points inside the mirror triangle, and eye/source points on the appropriate reflecting side. Unfolding the reflected path across the mirror proves the specular reflection condition. As $Y$ moves along a straight edge, $X$ lies on the intersection of the mirror plane with the plane through $E^{\ast },Y_0,Y_1$, so the retained locus is a straight segment, a point or the empty set. Plane and triangle clipping computes its endpoints. If these defining points are collinear, or a denominator vanishes, use the corresponding limiting geometric case rather than a generic plane formula. A bounding volume hierarchy accelerates candidate search where geometric bounds exclude pairs; refined facet segments provide seeds and approximate connectivity.

To obtain the reflection curve of the actual limits, correct these seeds. In a local surface chart write $S(u,v)$, and on a source interval write $C(t)$. SetThe exact limit reflection locus is $F=0$ together with the front-side conditions. For a two-sided mirror, orient $n$ locally so that $b\cdot n>0$ and require $a\cdot n>0$; for a one-sided mirror use its prescribed orientation and discard the wrong side. At grazing incidence this orientation condition ceases to be strict and the case needs separate handling. Unlike the eye-facing assumption in the parametric question, no global orientation condition is supplied here.

Apply a continuation method: at a rank-two root of $DF$, predict in a nullspace tangent direction and correct by the Newton method on a transverse section. Map corrected parameters to $S(u,v)$, crossing patch and curve-interval boundaries through their hierarchy adjacency. Trace all branches, recognizing closed components and endpoints. Supplement facet seeds by recursive searches of the surface-cell/source-interval product, with interval arithmetic or other certified enclosures of $F$; otherwise a small loop absent from the initial polygonal approximation could be missed. Split singular neighborhoods where the Jacobian matrix loses rank, and test the open source-to-mirror and mirror-to-eye segments for occlusion when selecting actually visible components.

Allocate a geometric error budget to limit evaluation, root correction and polygonal chords. For example, require combined position-evaluation and corrected-root error at most $ϵ/2$ and, on every regular arc, require the chord error bound $M(\Delta s{)}^2/8\le ϵ/2$. Use subdivision position and normal bounds for evaluations: a tiny position error alone is insufficient, since a wrong normal vector changes the reflected ray. Away from singularities an inverse transverse Jacobian matrix bound converts reflection residual into parameter error and a bound on $DS$ converts this to space error. Refine until these bounds hold, or use validated enclosures of entire arcs and their distances to the chords.

Near a caustic the inverse bound can become large, so the hierarchy must refine more strongly or isolate the singular branch geometry. Mesh flatness alone does not certify a reflection tolerance. With regular convergent evaluators and finite resolvable components, the resulting ordered vertices and chords satisfyThe problem supplies no particular subdivision rules or smoothness guarantees, so this construction specifies what their evaluators and bounds must provide; an arbitrary nonsmooth or degenerate limit need not have an ordinary finite reflection curve. Refine for seeds, enforce reflection on the actual limit geometry, then refine the traced arcs to the required spatial tolerance.

A Hamiltonian system with $n$ degrees of freedom is Liouville integrable if it has $n$ functionally independent first integrals, including the Hamiltonian, whose pairwise Poisson brackets vanish. Functional independence is required on a regular open set. The Arnold-Liouville theorem says that a connected compact regular common level of these integrals is an $n$-torus, with a neighborhood admitting action-angle coordinates; the Hamiltonian flow there has constant angular velocities and fixed actions. Compactness is essential to the torus conclusion.

Use $\{f,g\}={\sum }_j(f_{q_j}{g}_{p_j}-f_{p_j}{g}_{q_j})$. On $r>0$, for differentiable $F$, Hamilton's equations give $\dot{q}=p$, $\dot{p}=-F^{\prime }(r)q/r$. Hence the angular momentum $M=q\times p$ obeysIn particular $M_1$ and $M_2$ are first integrals. The Jacobi identity for the Poisson bracket isTaking $f=H,g=M_1,h=M_2$ shows $\{H,\{M_1,M_2\}\}=0$. Direct differentiation gives $M_3=\{M_1,M_2\}=q_1{p}_2-q_2{p}_1$ and, cyclically, $\{M_i,M_j\}={ϵ}_{ijk}{M}_k$.

Set $L^2=M_1^2+M_2^2+M_3^2$. Then $\{L^2,M_3\}=2M_1(-M_2)+2M_2{M}_1=0$, and the brackets of $H$ with both expressions vanish. ThusFor functional independence, at $q=(r,0,0)$ with $r>0$ the Jacobian matrix of $(H,L^2,M_3)$ with respect to $(p_1,p_2,p_3)$ iswhose determinant is $-2r^3{p}_1{p}_3$. It is nonzero when $p_1{p}_3\ne 0$. More generally this minor is a nonzero polynomial in $q,p$, independent of $F$, so its zero set has empty interior and the integrals are independent on a dense open regular set. This proves Liouville integrability on the regular set. The arbitrary potential in the question does not ensure compact common levels: $F=0$ has unbounded free trajectories. The full torus conclusion applies to those regular connected common levels which are compact, rather than to every choice of $F$ and every level.

The equilibrium points are $(0,0)$ and $(\pm 1,\pm (b-a))$, with correlated signs. The Jacobian matrix is $\left(\begin{array}{cc}a-3b{x}^2& 1\\ 3x^2-1& 0\end{array}\right)$. At the origin its eigenvalues satisfy ${\lambda }^2-a\lambda +1=0$. It is a focus for $0<|a|<2$, a repeated-eigenvalue node for $|a|=2$, and a node for $|a|>2$; stability is determined by the sign of $a$, with stability for $a<0$. Its Poincaré index is $+1$. At either other equilibrium the determinant is $-2$, so it is a saddle point with Poincaré index $-1$.

Let $A=Dy(x)$ be the Jacobian matrix and $K(y)=H(x(y))$. The chain rule gives ${\nabla }_{x}H=A^T{\nabla }_{y}K$. Therefore the Hamilton equations transform to $\dot{y}=AJ{A}^T{\nabla }_{y}K$, with $J=\left(\begin{array}{cc}0& -I\\ I& 0\end{array}\right)$. They have the required canonical form for every Hamiltonian exactly whenNecessity follows by choosing Hamiltonians whose gradients take arbitrary values at a point; sufficiency follows from the displayed transformed equation. This is the symplectic matrix condition for a canonical transformation, and it also forces $A$ to be invertible.

For the given transformation, $A=\left(\begin{array}{cc}1& 2\\ -1/4& 1/2\end{array}\right)$ has determinant one. For any $2\times 2$ matrix, $AJ{A}^T=(\det A)J$, so the proposed transformation is canonical.

The homogeneous fixed point must satisfy $v=u^2$. The singular point $u=v=0$ is outside the domain of the reaction term, so substitution in the other equation gives the unique regular spatially homogeneous equilibriumThis is the basally forced quadratic activator-inhibitor model. The linearization of its reaction terms gives the Jacobian matrixPut $a=(r-1)/(r+1)$, so $-1<a<1$. Its trace and determinant are $a-q$ and $q$, and its eigenvalues areSince $q>0$, both eigenvalues have negative real part exactly when $q>a$. HenceFor $r\le 1$ the equilibrium is always strictly linearly stable. For $r>1$, equality $q=a$ is an oscillatory threshold with ${\lambda }_{\pm }=\pm i\sqrt{a}$; the nonlinear qualification at this threshold is proved below.

Now insert a perturbation proportional to $e^{\lambda t+ikx}$. The diffusion terms subtract $k^2\mathrm{diag}(1,p)$ from $J$. With $s=k^2$ the mode trace and determinant areFor a strictly stable homogeneous state, $q>a$ makes $\tau (s)<0$ for all $s\ge 0$. A nonzero Fourier mode can therefore grow only if $\Delta (s)<0$, in which case its two real eigenvalues have opposite signs. The determinant parabola has its minimum at $s_{\ast }=(pa-q)/(2p)$, and that minimum is $q-(pa-q{)}^2/(4p)$. The two-species Turing criterion is consequently derived here asThese force $a>0$, hence $r>1$. Equivalently, solving $pa-q>2\sqrt{pq}$ for $\sqrt{p}$ givesThus the inhibitor in a reaction-diffusion system must diffuse sufficiently fast. Inside this region, the growing wavenumbers obeyAt onset, the minimum touches zero and the two determinant roots coalesce. ThereforeHere $p$ is evaluated on the onset surface. These are the Turing threshold of a basally forced quadratic activator-inhibitor model and its continuum critical wavelength. On a finite or domain with periodic boundary conditions an admissible nonzero spatial mode must additionally fall in the stated band. Equality in the diffusion criterion is marginal onset, while $q=a$ is the distinct homogeneous oscillatory threshold.

For completeness, the weak focus at the Hopf threshold of a basal activator-inhibitor model resolves the nonlinear homogeneous stability when $q=a>0$. Let $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and set $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $X=x$, $Y=(hy-ax)/\omega$. Expanding the rational reaction term to cubic order giveswhereWrite $X=\varrho \cos \theta$, $Y=\varrho \sin \theta$ and evaluate these homogeneous polynomials on the unit circle. If $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, and $C=\cos \theta G_2-\sin \theta F_2$, division of the radial and angular equations givesThe angular integral of $A$ is zero because it changes sign under $\theta \mapsto \theta +\pi$. Iterating its quadratic term once over a full cycle contributes the square of this zero integral, so the first nonzero return map term is cubic. Using $⟨{\cos }^4\theta ⟩=3/8$, $⟨{\cos }^2\theta {\sin }^2\theta ⟩=1/8$ and the corresponding sixth moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map isBecause $0<a<1$, sufficiently small positive radii decrease on every return. The equilibrium at $q=a$ is therefore nonlinearly asymptotically stable, with weak rather than exponential attraction, despite its neutral linear eigenvalues. The strict inequalities above describe the ordinary diffusion-driven instability from a strictly linearly stable homogeneous state.

By the law of mass action, the net changes in the intermediate concentrations areThe autocatalytic reaction increases $X$ by one and decreases $Y$ by one; its reaction-rate convention is $k_3{X}^{2}Y$. With the specified common concentration scale and time scale, chooseSubstitution gives the nondimensional Brusselator system. Adding its two steady equations gives $u=1$, and then the second gives $v=b/a$. Its Jacobian matrix there isThe equilibrium is asymptotically stable for $b<1+a$ and unstable for $b>1+a$. At $b_c=1+a$ the eigenvalues are $\pm i\sqrt{a}$, crossing the imaginary axis with nonzero speed $\frac{d}{db}\mathrm{Re}\lambda =1/2$.

The nonlinear Hopf bifurcation is nondegenerate. At the threshold, put $x=u-1$ and $y=-\sqrt{a}[(u-1)+(v-b/a)]$. The translated equations areThe Hopf coefficient of the Brusselator in the planar radial normalization is $-(a+2)/8<0$: the cubic contribution is $-6a/16$ and the quadratic contribution is $[-2\sqrt{a}2(1-a)]/(16\sqrt{a})$. Consequently the bifurcation is supercritical, producing small stable oscillations on the unstable-equilibrium side. Their limiting period is

The species $u$ is the activator, since increasing $u$ increases its own production $u^2/v$; $v$ is the inhibitor, since increasing $v$ reduces that production. The unique positive uniform steady state solves $v=u^2$ and $u^2/v=bu$, giving $\boxed{u_{\ast }=1/b,v_{\ast }=1/b^2}$.

The reaction Jacobian matrix at this state isIts eigenvalues solve ${\lambda }^2+(1-b)\lambda +b=0$. Since $b>0$, both have negative real part precisely for $0<b<1$. This proves stability of the spatially uniform reaction kinetics.

A perturbation proportional to $e^{\lambda t+ikx}$ replaces $J$ by $J-k^2\mathrm{diag}(1,d)$. Put $q=k^2$. Its trace is $b-1-(1+d)q<0$ in the kinetically stable range, and its determinant isA growing spatial mode exists exactly when this determinant is negative for some $q>0$. Its minimum occurs at $q=(bd-1)/(2d)$, which is positive only if $bd>1$, and the minimum is negative precisely if $(bd-1{)}^2>4bd$. Together these conditions reduce to the Turing threshold of the quadratic activator-inhibitor model,The unstable band isAt the Turing instability threshold the band shrinks to its double root, yielding $\boxed{k_c^2=(1+\sqrt{2})/d}$. On a bounded spatial interval, an allowed wavenumber must additionally lie in the unstable band; the displayed parameter condition assumes the continuous spatial spectrum.

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

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

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

First find the positive spatially homogeneous equilibrium. Setting the reactions to zero gives $v=u^2$ and $u/v=b$, and hence the unique positive equilibrium isThe zero pair is not an admissible positive state and is singular in the $u^2/v$ term. For homogeneous deviations, the reaction Jacobian matrix at this equilibrium isThus $\mathrm{tr}J=b-1$, $\det J=b$, and its eigenvalues areFor $0<b<1$, both real parts are negative: a real pair has positive product and negative sum, while a complex pair has real part $(b-1)/2$. The reaction equilibrium is therefore asymptotically stable, with no growing oscillatory mode. This does not exclude damped oscillations. Indeed, for $3-2\sqrt{2}<b<1$ the eigenvalues are complex; for example, $b=1/2$ gives $-1/4\pm i\sqrt{7}/4$, a stable focus with oscillatory transients. Thus a literal exclusion of every oscillatory solution is false; the intended exclusion of sustained or growing oscillations is the meaningful statement.

One can strengthen this to exclude nonlinear periodic oscillations in the positive quadrant without relying only on the local linear stability analysis. For the reactions $f=u^2/v-bu$, $g=u^2-v$, use the Dulac function $H=u^{-2}$. ThenWere there a periodic orbit wholly in the positive quadrant, the flux of $H(f,g)$ through it would vanish, since this vector is tangent to the orbit; Green's theorem would equate that flux to a strictly negative interior integral. This contradiction proves there is no sustained positive periodic oscillation when $0<b<1$, the positive-quadrant Dulac multiplier for quadratic activation argument. It leaves the damped transients entirely consistent with the conclusion.

Now introduce a perturbation $(\delta u,\delta v)=\mathbf{a}{e}^{\sigma t+ikx}$ in the quadratic activator-inhibitor model. The Fourier mode of squared wavenumber $q=k^2$ evolves underIts matrix trace and determinant areA Turing instability must start with reaction stability, requiring $0<b<1$. Then $\tau (q)<0$ for every $q\ge 0$, so a mode is unstable exactly when $D(q)<0$: its two real eigenvalues have opposite signs. This also proves the diffusion-driven onset is stationary rather than oscillatory.

The upward-opening quadratic $D(q)$ has its minimum at $q_m=(db-1)/(2d)$. To have a negative value at a physically allowed positive squared wavenumber, one needsWriting $r=db$, the second inequality is $r^2-6r+1>0$, with roots $3\pm 2\sqrt{2}$. The first inequality discards the lower branch, givingThese are the Turing threshold of the quadratic activator-inhibitor model conditions. Equivalently, in the $(b,d)$ plane the instability region lies strictly above $d=(3+2\sqrt{2})/b$ and strictly to the left of $b=1$, with $b>0$. The boundary diverges as $b\to 0^{+}$ and approaches $3+2\sqrt{2}$ as $b\to 1^{-}$.

At equality, the determinant has a double root and one growth rate is zero, while the other remains negative. Substituting $db=3+2\sqrt{2}$ into its minimum location givesBeyond onset, the unstable band isThe stated parameter region assumes continuously available wavenumbers, as on an infinite domain. With specified finite-domain boundary conditions, an allowed nonzero spatial eigenmode must additionally lie in this band; homogeneous stability alone does not ensure that a finite box admits the critical wavelength.

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

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

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

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

The only equilibrium point is $(0,0)$. Its Jacobian matrix is $\left(\begin{array}{cc}-\mu & 1\\ -1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2+\mu \lambda +1$. For $\mu >0$ both eigenvalues have negative real parts, so the origin is locally asymptotically stable.

Use the Lyapunov function $V=(x^2+y^2)/2$. Along a solution,A Lyapunov function is positive away from the equilibrium and nonincreasing along trajectories; its sufficiently small sublevel sets prove stability. Asymptotic stability additionally means nearby trajectories converge to the equilibrium. The basin of attraction is the set of initial states whose forward solutions exist for all positive time and converge to it. Every closed disc of radius at most $\sqrt{3}$ is positively invariant, since $\dot{V}\le 0$ there. The LaSalle invariance principle says that a solution in a compact positively invariant set where $\dot{V}\le 0$ approaches the largest invariant subset of $\{\dot{V}=0\}$. Here that subset is only the origin: on $x=0$, remaining on the line would require $y=0$; the isolated points $x=\pm \sqrt{3},y=0$ on the outer disc do not remain there because $\dot{y}=-x\ne 0$. Hence the closed disc $x^2+y^2\le 3$ lies in the basin of attraction.

This estimate does not assert that the basin is exactly that disc, or the whole plane. For example, choose $R$ so large that $R^2\ge 6(\mu +1)/\mu$ and $R^2>3(\mu +2)/\mu$. The region $x\ge R,y\ge -x$ is positively invariant: on $x=R$ one has $\dot{x}>0$, and on $y=-x$ one has $(x+y{)}^{\cdot }=\mu x^3/3-(\mu +2)x>0$. Within it $\dot{x}\ge \mu x^3/6$, which forces finite-time escape. Thus the attraction is genuinely local. The Lyapunov function gives a rigorous sufficient domain for attraction without identifying its entire boundary.

At a fixed point, the first equation gives $y=-x$. Substitution into the second gives $2x(x^2-1)=0$, so the equilibrium points are $(0,0),(1,-1),(-1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1\pm i$, so the origin is an unstable spiral. The local rotation is clockwise: on the positive $x$-axis the velocity points downward. At either nonzero equilibrium point,Thus both nonzero equilibria are saddles, with eigenvalues $-2,2$. The stable manifold is tangent to $(1,-3)$, and the unstable manifold to $(1,1)$, at each saddle equilibrium. Since all these equilibrium points are hyperbolic, linear stability of a planar equilibrium gives their actual local types.

For the phase portrait, the nullclines are $y=-x$ and $y=x/(1-2x^2)$, the latter defined away from $x=\pm 1/\sqrt{2}$. The horizontal direction has the sign of $x+y$; the vertical direction has the sign of $(1-2x^2)y-x$. The system is invariant under $(x,y)\mapsto (-x,-y)$, so trajectories occur in centrally symmetric pairs. The arrows below show these directions; the highlighted stable manifolds enter the saddle equilibria, whereas the highlighted unstable manifolds leave them.

In Cartesian coordinates, independent coordinate variables satisfywhere ${\delta }_{ij}$ is the Kronecker delta. Applying the product rule and chain rule to $u_i=r^{\alpha }{x}_i$ gives the Jacobian matrixThis tensor is symmetric in $i,j$. Hence its contraction with the antisymmetric Levi-Civita symbol vanishes:Similarly, $v_i={ϵ}_{ijk}{k}_j{u}_k$ and the constancy of $\mathbf{k}$ giveThe trace of the Jacobian matrix iswhich vanishes for $\alpha =-3$. The curl of a cross product identity, with $\mathbf{k}$ constant, now givesFor $\alpha =-3$,soAll identities for this singular case hold on $r>0$. In particular, zero divergence here is a pointwise assertion away from the origin, not an assertion that the flux through a sphere enclosing the singularity is zero.

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

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

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

A steady-state bifurcation is a change in the local equilibrium branches or their stability as a parameter varies. At such a bifurcation the Jacobian matrix at the equilibrium has a zero eigenvalue, so the implicit function argument for a unique persistent local equilibrium fails.

For this system,At $(0,\mu )$ its eigenvalues are $1-{\mu }^2,-\mu$; at $(1,0)$ they are $-2,\mu -1$. Both have one simple zero eigenvalue at $\mu =1$, with the other eigenvalue negative. Interior equilibrium branches, obtained below, meet these boundary equilibria there, establishing the actual bifurcations rather than merely a necessary eigenvalue test.

The physical state space is the positively invariant set $X,Y,Z\ge 0$, $X+Y+Z=N$: adding the equations gives zero, and the vector field points inward at each coordinate boundary. At an equilibrium point, either $Y=0$, giving $(X,Y,Z)=(N,0,0)$, or $\beta X=b+r$. ThereforeA positive endemic equilibrium exists exactly when $N>N_c$.

For $N<N_c$, put $d=b+r-\beta N>0$. Since $X\le N$,The last limit follows either by evaluating the exponential integral (with a $t{e}^{-bt}$ term when $d=b$), or by splitting it into an early decaying part and a late part with uniformly small $Y$. Hence $X\to N$ for every physical initial condition. Although immune numbers need not decrease monotonically initially, both infected and immune numbers tend to zero.

At the endemic equilibrium, the Jacobian matrix of the $X,Y$ subsystem isA disturbance proportional to $e^{pt}$ therefore obeysBoth coefficients are positive. Real roots have negative sum and positive product; complex conjugate roots have real part $-(b+\beta Y_s)/2<0$. Thus the endemic equilibrium is asymptotically stable by linear stability analysis.

Under the given rank hypothesis, the necessary and sufficient local condition isVanishing only at the single point would not suffice: an extension preserves the coordinate Poisson brackets throughout its domain. Necessity follows from the coordinate criterion just proved.

For sufficiency, put $Q_i=a_i$ and $Y_i=X_{Q_i}$. Because the symplectic form is a nondegenerate bilinear form at every point, the independent differentials $d{Q}_i$ give independent fields $Y_1,\dots ,Y_n$. They commute, sinceMoreover $d{Q}_j(Y_i)=\{Q_j,Q_i\}=0$, so they span the tangent spaces of the $n$-dimensional fibres of the submersion $Q$. Choose a local section $s(Q)$ transverse to those fibres. Its joint local flow defines coordinates $(Q,t)$ byThe inverse function theorem makes $\Psi$ a local diffeomorphism, and commuting flows give ${\Psi }_{\ast }{\partial }_{t_i}=Y_i$.