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Near two homoclinic orbits of a planar saddle equilibrium with contracting ratio $\delta >1$, the signed Poincaré return map has separate offsets $\mu ,\nu$. Two single-lobe periodic orbits exist for $\mu <0$ and $\nu <0$, respectively. A two-lobe periodic orbit is represented by positive $p,q$ solving the displayed equations. Its boundary consists of the compound homoclinic orbit curves $\mu =-A{\nu }^{\delta }$ for $\nu >0$ and $\nu =-B{\mu }^{\delta }$ for $\mu >0$, with $A,B>0$. In a sufficiently small return domain the derivatives tend to zero, so each admissible symbolic branch has at most one attracting periodic orbit. Restricting to $\mu =\nu$, $A=B$ gives the usual symmetric homoclinic gluing bifurcation.

For unit-mass motion in $V(x)=3x^2-2x^3$, the mechanical energy is $E=v^2/2+V(x)$. The origin is a center equilibrium, and $(1,0)$ is a saddle equilibrium. Energies $0<E<1$ have trapped periodic orbits around the origin as well as a separate escaping component to the right. The barrier energy has a homoclinic orbit returning to the saddle after a turn at $x=-1/2$, plus two saddle branches on $x>1$. Above the barrier, particles traverse the well and eventually escape to the right.

An eigendirection is the one-dimensional vector subspace spanned by an eigenvector. Multiplying the eigenvector by a nonzero scalar leaves this direction unchanged. In a planar linear dynamical system, a negative real eigenvalue gives a direction of exponential approach to the equilibrium point, while a positive one gives exponential departure. For a hyperbolic saddle equilibrium, the nonlinear stable manifold and unstable manifold are tangent to the corresponding eigendirections.

A global bifurcation changes an invariant set through a global orbit connection or collision that cannot be identified by a single equilibrium point's linearization alone. For example, a global bifurcation through a saddle homoclinic orbit occurs when an outgoing unstable separatrix returns to the same saddle equilibrium's stable manifold. A smooth global Poincaré return map then combines with a local passage map to determine which side contains a periodic orbit. A heteroclinic orbit instead joins distinct equilibrium points; an energy-balance or Melnikov integral can locate its breaking in a nearly Hamiltonian system.

A planar periodic orbit, oriented by the flow, has Poincaré index $+1$. The index theorem equates this with the sum of the indices of the isolated equilibria enclosed by the orbit. A hyperbolic equilibrium has index $+1$ when its Jacobian matrix has positive determinant and index $-1$ when it is a saddle equilibrium.

If every Jacobian eigenvalue has negative real part, the equilibrium is locally asymptotically stable.

More generally, if the equilibrium is hyperbolic, the Hartman-Grobman theorem transfers the local attracting, repelling or saddle behavior of the linearized vector field to the nonlinear flow. Two real eigenvalues with opposite signs give a saddle equilibrium; all real parts positive give a source. Purely imaginary or zero eigenvalues require a separate nonlinear analysis. Merely having distinct eigenvalues does not remove this exception.

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 $\sqrt{2}\pi$.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

A critical point is an equilibrium: $\boxed{f(a,b)=g(a,b)=0}$. Put $X=x-a$, $Y=y-b$. The Jacobian matrix $J$ gives the first-order linearized equation $(\dot{X},\dot{Y}{)}^{\mathsf{T}}=J(X,Y{)}^{\mathsf{T}}$. The following eigenvalue tests classify that linearized system. For a continuously differentiable nonlinear system, the cases with nonzero real parts persist as the usual local node, saddle or spiral classifications. The purely imaginary case needs extra nonlinear information, as explained in (iv).

The equilibrium equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian isAt $(0,0)$ it is $\mathrm{diag}(1,-1)$: a saddle. At $(-1,0)$ it is $\left(\begin{array}{cc}-1& 2\\ 0& -2\end{array}\right)$, with eigenvalues $-1,-2$: a stable node. At $(1,1)$ it is $\left(\begin{array}{cc}1& -2\\ 1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2-\lambda +2$ and eigenvalues $(1\pm i\sqrt{7})/2$: an unstable spiral. All three are hyperbolic, so the nonlinear local classifications follow from linearization.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

A critical point is an equilibrium: $\boxed{f(a,b)=g(a,b)=0}$. Put $X=x-a$, $Y=y-b$. The Jacobian matrix $J$ gives the first-order linearized equation $(\dot{X},\dot{Y}{)}^{\mathsf{T}}=J(X,Y{)}^{\mathsf{T}}$. The following eigenvalue tests classify that linearized system. For a continuously differentiable nonlinear system, the cases with nonzero real parts persist as the usual local node, saddle or spiral classifications. The purely imaginary case needs extra nonlinear information, as explained in (iv).

The equilibrium equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian isAt $(0,0)$ it is $\mathrm{diag}(1,-1)$: a saddle. At $(-1,0)$ it is $\left(\begin{array}{cc}-1& 2\\ 0& -2\end{array}\right)$, with eigenvalues $-1,-2$: a stable node. At $(1,1)$ it is $\left(\begin{array}{cc}1& -2\\ 1& 0\end{array}\right)$, with characteristic polynomial ${\lambda }^2-\lambda +2$ and eigenvalues $(1\pm i\sqrt{7})/2$: an unstable spiral. All three are hyperbolic, so the nonlinear local classifications follow from linearization.

A saddle equilibrium hasThe negative eigenvalue gives a stable manifold direction and the positive eigenvalue an unstable manifold direction. Perturbations along the latter grow, so the equilibrium is unstable even though there are special trajectories approaching it.

The first component of the vector field vanishes only when $y=0$ or $x=\pm 1$. The second vanishes only when $y=1$ or $x+y=3$. Intersecting these alternatives gives exactly five equilibria. Their Jacobian matrix isAt $(3,0)$,The real parts are negative and the imaginary parts are nonzero, so $(3,0)$ is a stable spiral.

At $(1,1)$ the Jacobian matrix is $\mathrm{diag}(2,-1)$. Its real eigenvalues have opposite signs, so $(1,1)$ is a saddle equilibrium.

At $(1,2)$,Both eigenvalues are positive, so $(1,2)$ is an unstable node.

At $(-1,1)$ the Jacobian matrix is $\mathrm{diag}(-2,-3)$, giving a stable node.

At $(-1,4)$,so $(-1,4)$ is a saddle equilibrium. All five equilibria are hyperbolic, so these linear classifications determine their nonlinear local stability as well.

Applying time reversal of an autonomous differential equation to the system from part (i), and renaming its coordinates $(x,y)$, givesThe equilibrium points remain $(1,-1)$ and $(-1,-1)$. Time reversal of an autonomous differential equation preserves a saddle equilibrium, interchanging its stable and unstable directions; it also reverses the direction around a centre equilibrium. Therefore this system has precisely the required counterclockwise centre at $(1,-1)$ and saddle at $(-1,-1)$.

Using $x=-u$, $y=-v$ in the original autonomous planar system givesThe map $(x,y)\mapsto (-x,-y)$ is a rotation by $\pi$, preserving time and the clockwise or counterclockwise orientation of trajectories. Hence the given centre equilibrium at $(-1,1)$ becomes a clockwise centre at $(1,-1)$, and the given saddle equilibrium at $(1,1)$ becomes a saddle at $(-1,-1)$. These are the only equilibrium points, because rotation is invertible.

At an equilibrium point, $x=-y^2$ and $x^2=a^{2}y$, so $y(y^3-a^2)=0$. With $d=|a{|}^{2/3}>0$, the two equilibrium points and their Jacobian matrix areAt $(-d^2,d)$, $\det J=-3a^2<0$. The trace-determinant stability criterion therefore gives an unstable saddle equilibrium. At the origin, the eigenvalues are $\lambda =\pm i|a|$: its linearization is a center equilibrium. The linearization stability theorem does not classify this nonhyperbolic equilibrium of the nonlinear system.

Nonlinear classification. In fact the answer depends on $|a|$:Here is an explicit derivation, rather than an inference from the imaginary eigenvalues. When $a^2=1$, set $L=1-x+y$. Near the origin $L>0$, and the first integralhasIts Hessian matrix at the origin is the identity. Small level curves are closed ovals on which the vector field is nonzero, proving the center equilibrium classification and Lyapunov stability.

For $a^2\ne 1$, a higher-order Lyapunov function calculation settles the issue. Put $k=|a|$, $U=x$, $V=ky$, $\tau =kt$, $p=k^{-1}$ and $q=k^{-2}$. ThenWrite $R^2=U^2+V^2$ and introduce the differential operatorsSuccessively cancel the terms of degrees three, four and five in the derivative of $R^2+H_3+H_4+H_5$, using these homogeneous polynomials:Direct differentiation verifies $\mathcal{L}{H}_3=-\mathcal{N}{R}^2$, $\mathcal{L}{H}_4=-\mathcal{N}{H}_3$ and $\mathcal{L}{H}_5=-\mathcal{N}{H}_4$. The remaining degree-six term has circular averageFor example, the only nonzero averages of degree-six monomials needed here are $⟨U^6⟩=⟨V^6⟩=5/16$ and $⟨U^4{V}^2⟩=⟨U^2{V}^4⟩=1/16$. Since $\mathcal{L}={\partial }_{\theta }$ in polar coordinates, the zero-average part can be cancelled by a degree-six homogeneous polynomial $H_6$ satisfyingIndeed its restriction to the unit circle is obtained by integrating a zero-average trigonometric polynomial; its Fourier modes have orders $2,4,6$, so multiplication by $R^6$ gives a polynomial. Consequently the positive Lyapunov function $W=R^2+H_3+H_4+H_5+H_6$ satisfiesFor $k>1$ this is strictly negative near the origin, giving asymptotic stability. For $k<1$ it is strictly positive, and $W_{\tau }\ge C{W}^3$ in a sufficiently small neighbourhood; every nonzero starting point must eventually leave that neighbourhood. Finally ${\theta }_{\tau }=1+O(R)>0$, so the nearby trajectories rotate: these are weak foci, with the stability stated above. Merely calling the nonlinear origin a center for every $a\ne 0$ would miss this distinction.

At an equilibrium point, $x(1-x-y)=0$ and $2-2x-y=0$. The first equation gives $x=0$ or $y=1-x$. Substituting in the second gives, respectively,These are all the equilibrium points. The Jacobian matrix of the vector field isAt $(0,2)$,The eigenvalue $-1$ has algebraic multiplicity two and eigenspace $\mathrm{span}\{(0,1{)}^T\}$. For the linearized coordinates $u=x$, $v=y-2$,Both components tend to zero. The repeated eigenvalue with one independent eigenvector makes this a stable improper degenerate node, a sink equilibrium. The term $t{e}^{-t}$ is the signature of the nontrivial Jordan block. Since it is a hyperbolic equilibrium point, the linearization stability theorem gives local asymptotic stability of the nonlinear system. Approaching trajectories are tangent to the vertical eigendirection.

At $(1,0)$,One eigenvalue is positive and one negative, so this point is a saddle equilibrium. Its stable manifold is tangent to $y=\sqrt{2}(x-1)$, and its unstable manifold is tangent to $y=-\sqrt{2}(x-1)$.

For the phase portrait, the nullclines areAbove the second line the vertical flow is downward, and below it upward. For $x>0$, horizontal flow is to the right below $y=1-x$ and to the left above it; for $x<0$ the directions reverse. The line $x=0$ is invariant, with $\dot{y}=2-y$, so it flows toward $(0,2)$. Together with the saddle's two invariant directions, these signs determine the sketch.

Define the auxiliary variable $y=ϵ\dot{x}-x+x^3/3$. Differentiating gives $\dot{y}=ϵ\ddot{x}+(x^2-1)\dot{x}=-x$, while rearrangement gives $ϵ\dot{x}=x-x^3/3+y$. Conversely eliminating $y$ recovers the second-order equation, so this phase plane system is equivalent. At an equilibrium $\dot{y}=0$ forces $x=0$, and the first equation then forces $y=0$. Its Jacobian matrix isThus $ϵ{\lambda }^2-\lambda +1=0$. Expanding $\sqrt{1-4ϵ}=1-2ϵ-2{ϵ}^2-4{ϵ}^3+O({ϵ}^4)$ givesFor sufficiently small positive $ϵ$ both are positive: the equilibrium is an unstable node. For negative $ϵ$ the fast root is negative and the slow root positive, so it becomes a saddle equilibrium.