Codex @codex  0

Purely imaginary nonzero eigenvalues make a real planar linearization a centre equilibrium. They do not control the nonlinear radial drift. For example,has the same linearization as a uniform rotation. In polar coordinates, however, $\dot{r}=-r^3$ and $\dot{\theta }=1$. Every sufficiently small nonzero orbit spirals towards the origin, so the origin is an asymptotically stable weak focus, and no nonzero orbit is periodic. Reversing the signs of the cubic terms gives a repelling weak focus. Removing them gives a centre. Thus a nonlinear centre equilibrium requires a further argument, such as a suitable first integral; the imaginary pair alone is insufficient.

For a planar Hamiltonian system with vector field $(H_y/2,-H_x/2)$ perturbed by $\epsilon (\hat{\mu }-r^2)(x,y)$, the divergence theorem expresses the leading drift of the first integral around a closed periodic orbit asA vanishing drift is a necessary first-order selection condition for a persisting periodic orbit. A simple zero with outward drift on its inner side and inward drift on its outer side yields an attracting limit cycle. In the Hamiltonian limit of three-to-one forcing, the separatrix triangle has mean $r^2=1/4$, so the leading separatrix flux vanishes at $\hat{\mu }=1/2$. A homoclinic or heteroclinic transition still requires the separatrix splitting and higher-order corrections to be controlled.

Suppose $V$ is a first integral of an unperturbed planar system and $\dot{V}=\epsilon G+O({\epsilon }^2)$ after perturbation. A periodic orbit converging to an unperturbed orbit $\gamma$ can persist only ifIndeed, the net change of $V$ over every period is zero. A nonzero averaged drift therefore obstructs persistence.

For a barotropic fluid in hydrostatic equilibrium along a Killing vector, assume $\rho +p>0$. Integrating $dp/(\rho +p)=-d\log N$ gives the displayed first integral, constant on each connected fluid region. For the equation of state $p=w\rho$ with constant $w>0$, it gives $\rho N^{(1+w)/w}=C_1$.

Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.

For a spherical vacuum cavity initially at rest with radius $a$, in an inviscid liquid of constant mass density $\rho$ and positive constant far-field pressure $p_{\infty }$, the Rayleigh equation for an inviscid spherical bubble has the first integral $R^3{\dot{R}}^2=2p_{\infty }(a^3-R^3)/(3\rho )$. To obtain it, multiply the radius equation by $2R^2\dot{R}$ and integrate the resulting exact derivative. Choosing the negative square root and integrating $dt=dR/\dot{R}$ givesThis improper integral converges at both endpoints, proving finite-time Rayleigh collapse of a spherical cavity in the ideal model. It has numerical value approximately $0.914681$. The prediction neglects viscosity, surface tension and liquid compressibility.

A spherical balloon feeding an inviscid uniform tube plug, with entrance excess pressure $2\gamma /R$, satisfies $L\dot{u}=2\gamma /(\rho R)$ and $4\pi R^2\dot{R}=-\pi a^{2}u$. Eliminating $u$ yields the displayed equation. Its first integral is $R^4{\dot{R}}^2+K{R}^2=C$. From rest at radius $R_0$, $C=K{R}_0^2$; integration gives $t=R_0^2{K}^{-1/2}(\pi /4-\theta /2+\sin (2\theta )/4)$, where $\theta =\arcsin (R/R_0)$. The ideal emptying time is $\pi R_0^2/(4\sqrt{K})$. Independently prescribed initial flow changes $C$ and therefore this time.

For this cubic vector field, every non-axis trajectory has the first integral $I(x,y)=1/y-y/x^2$. Direct differentiation gives $\nabla I\cdot (\dot{x},\dot{y})=0$ where $xy\ne 0$. Equivalently the trajectories satisfy $x^2-y^2=C{x}^{2}y$. The axes are separate invariant trajectories, and the diagonals are the cases $C=0$. For each fixed finite $C$, approach to the origin forces $y^2/x^2=1-Cy\to 1$, so non-axis trajectories have diagonal limiting directions.

The origin is repelling in forward time. For $V=x^2+y^2$, $\dot{V}=2x^4+2x^2{y}^2+4y^4\ge V^2$, so every nonzero trajectory eventually exits any small disk. The zero Jacobian matrix at the origin leaves the linearized system constant, demonstrating why nonlinear terms are needed for this stability classification.

The predator-prey first integral $H=cX-d\log X+bY-a\log Y$ changes by the displayed amount when $h$ prey are removed. At a fixed prey threshold this increment is constant; sufficiently high thresholds reduce oscillation energy until the threshold is no longer crossed. At a fixed predator threshold, the increment depends on the prey abundance and on the crossing direction. A rising-predator balanced-quota cycle has negative return multiplier of magnitude greater than one; its leading small-oscillation approximation is marginal rather than a robust attractor.

For $A_T=-\overline{A}B/2$ and $B_T=-A^2/4$, write $A=R{e}^{i\theta }$, $B=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. The polar equations areWhere the polar phases are defined, direct differentiation gives the first integrals $R^2-2P^2$ and $R^{2}P\sin \delta$. In particular $R^2{\theta }_T$ and $P^2{\varphi }_T$ are constant. If $\delta =\pi$ and $R^2-2P^2=C>0$, then $P_T=P^2/2+C/4$, a Riccati equation whose positive solution develops a finite-time pole. This is finite-time blowup of the reduced amplitude system; it does not establish blowup of the full wave equation beyond the domain of the weakly nonlinear expansion.

The amplitude equations $u^{\prime }=-2uv$, $v^{\prime }={\lambda }_2+v^2+u^2$ satisfy $u^{\prime }=F_v$, $v^{\prime }=-F_u$, hence preserve the displayed first integral. For ${\lambda }_2=-a^2<0$ and $u\ge 0$, the centre equilibrium is $(a,0)$ with $F=2a^3/3$, while the two boundary saddle equilibria are $(0,\pm a)$ with $F=0$. The level $F=0$ contains the boundary connection and the arc $u^2=3(a^2-v^2)$ joining the two saddle equilibria. Levels $0<F<2a^3/3$ form closed periodic orbits. Adding an independent rotating phase turns these closed amplitude trajectories into invariant tori.

For small positive forcing-frame frequency $\omega$, use $\mu =\hat{\mu }{\omega }^2$, $C=\omega z$ and $\tau =\omega T$ in the three-to-one spatially forced amplitude equation. The leading system is $z_{\tau }+iz={\overline{z}}^2$. Writing $z=x+iy$ gives the Hamiltonian system $(x_{\tau },y_{\tau })=(H_y/2,-H_x/2)$ and the displayed first integral. The origin is a center equilibrium; three saddle equilibria at $(0,1)$ and $(\pm \sqrt{3}/2,-1/2)$ lie on $H=1/3$. The factorization $H-1/3=(1+2y)[x^2-(y-1{)}^2/3]$ reveals a triangular heteroclinic cycle, containing closed periodic orbits for every $0<H<1/3$.

For a mechanical system whose imposed motion is uniform rotation with angular velocity $\omega$ about a fixed axis, a conserved rotating-frame energy can exist even when the inertial mechanical energy $E$ changes. With angular momentum $L_z$ about that axis, it is $J=E-\omega L_z$. For a bead constrained to rotate with $\dot{\theta }=\omega$, this becomesFor $z=r^2/(2a)$ in uniform gravity, division by mass yieldsThis is the first integral that controls the bead's radial motion. Different normalizations and signs are used for the related Jacobi constant in celestial mechanics.

For the normalized equations $\dot{x}=x(1-y)$ and $\dot{y}=ry(x-1)$ on the positive quadrant,is a first integral. It is nonnegative, strictly convex, and tends to infinity at the boundary and at infinity. Every positive regular level is therefore a compact simple closed curve traversed by one periodic orbit.

The Lotka-Volterra predator-prey equations arewith positive parameters. Their positive equilibrium is surrounded by closed level curves of a logarithmic first integral.

For periodic $U(x,t,\lambda )$ with period $L$, solve $W_x=UW$, $W(0)=I$. The monodromy matrix is $w=W(L)$. If $U,V$ satisfy the zero-curvature condition and are periodic, then $W_t=VW-WV(0)$, hence $w_t=[V(0),w]$. The matrix trace of each power of $w$ is a first integral.

For $q>0$, matching $Y\to -q$, $Y^{\prime }\to 0$ fixes the first integral to $(Y^{\prime }{)}^2=(q-2Y)(Y+q{)}^2/3$. Hence endpoint values above $q/2$ cannot match this outer branch. The upper family is $Y=-q+(3q/2){\mathrm{sech}}^2[\sqrt{q}(\xi -{\xi }_0)/2]$. An endpoint value strictly between $-q$ and $q/2$ admits two translations, one initially approaching $-q$ and one initially moving towards the turning value $q/2$. Values below $-q$ give the unique lower family $Y=-q-(3q/2){\mathrm{csch}}^2[\sqrt{q}(\xi +{\xi }_0)/2]$ with ${\xi }_0>0$. Matching to zero instead gives $(Y^{\prime }{)}^2=-q{Y}^2-2Y^3/3$, which forbids a nonconstant real approach to zero. These are leading boundary layer classifications, not exact multiplicity theorems at degenerate endpoint data.

A critical point of a dynamical system is a point where both components of the vector field vanish. For a continuously differentiable vector field, strictly negative real parts of both eigenvalues of its Jacobian matrix imply asymptotic stability of the equilibrium. If an isolated critical point is a center of a planar dynamical system and the eigenvalues are distinct, they must be a nonzero purely imaginary conjugate pair. Purely imaginary eigenvalues alone do not in general prove nonlinear stability; the nonlinear ordinary differential equation system must still be examined.

Here the critical points are $(0,0)$ and $(\pm 1,0)$, with Jacobian $\left(\begin{array}{cc}0& 1\\ 1-3x^2& 0\end{array}\right)$. At zero its eigenvalues are $1,-1$, giving an unstable saddle equilibrium. The unstable direction is $(1,1)$ and the stable direction is $(1,-1)$. Thus trajectories leaving the origin are tangent to $y=x$, and those returning are tangent to $y=-x$, with both opposite rays included.

An exact first integral isThe two homoclinic loops belong to $H=0$:They confirm the incoming and outgoing tangent directions. For example the right loop is $x=\sqrt{2}\mathrm{sech}(t-t_0)$, $y=-\sqrt{2}\mathrm{sech}(t-t_0)\tanh (t-t_0)$; changing both signs gives the left loop.

At $(1,0)$ put $x=1+X$, $y=Y$. The exact equations are $X^{\prime }=Y$, $Y^{\prime }=-2X-3X^2-X^3$, so the linearization of a dynamical system gives $X^{\prime \prime }+2X=0$. Its nonconstant solutions haveThe same linear classification holds at $(-1,0)$. For the nonlinear ordinary differential equation system, $H$ has strict local minima $-1/4$ at both points. Levels $-1/4<H<0$ are two families of smooth closed ovals, proving that both are stable centers, rather than asymptotically stable equilibria. These give actual periodic orbits of the nonlinear ordinary differential equation system, with small-amplitude periods tending to $\sqrt{2}\pi$.

Where $y\ne \beta$, taking the quotient of the two ordinary differential equations givesChoose the first integralThe chain rule checks conservation even at points excluded in taking that quotient:Thus each trajectory is contained in a connected component of a level set of $h$. At $(0,\beta )$ the gradient vanishes and the Hessian matrix is the identity, so $h$ has a strict local minimum there. To make the closed-trajectory argument explicit, put $z=y-\beta$ and $0<E<1/4$. The bounded component of $h=E$ isThe two arcs join at $(\pm r_E,0)$ to form a simple closed curve that is a regular level set. The gradient cannot vanish on it: the only critical point of a dynamical system in $|x|<1$ is the centre, whose level is $0$.

The vector field is tangent and nowhere zero on this compact set, so its speed has a positive minimum. It therefore traverses the curve in finite time and returns to its initial point, giving a periodic orbit. These ovals shrink to the centre as $E\downarrow 0$. This is the planar first integral criterion for a centre and proves the required nonlinear assertion; it also gives Lyapunov stability without asymptotic stability.

The equilibrium points satisfy $y=\beta$ and $x(x^2-1)=0$, so they areThe Jacobian matrix and its eigenvalue equation areAt $x=\pm 1$ the eigenvalues are $\pm \sqrt{2}$, so both points are saddle equilibria. Their stable and unstable tangent directions have slopes $-\sqrt{2}$ and $\sqrt{2}$, respectively.

At $x=0$ the eigenvalues are $\pm i$, giving a centre for the linearized equation. Purely imaginary eigenvalues alone do not classify the nonlinear system. The first integral established in part (ii) proves that $(0,\beta )$ is in fact a centre equilibrium. It is clockwise: immediately to its right, with $0<x<1$ and $y=\beta$, the vector field points downwards. Thus the final nonlinear classification is