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The autonomous planar system $\dot{x}=y-2y^3$, $\dot{y}=-x$ preserves $K$. Its origin is a center equilibrium and $(0,\pm 1/\sqrt{2})$ are saddle equilibria. The level $K=1/4$ is $x=\pm (y^2-1/2)$: its two inner arcs are heteroclinic orbits enclosing periodic motion, while its outer branches connect saddles to escaping trajectories. Since $|\dot{y}|\sim y^2$ on escaping levels, infinity is reached in finite time. Adding $\alpha x/2$ to $\dot{x}$ gives $\dot{K}=\alpha x^2$, so positive $\alpha$ produces outward crossing of central levels and breaks the equal-energy heteroclinic connections.

The autonomous planar system has two real state variables, so its phase space is a subset of ${\mathbb{R}}^2$. It is autonomous because its vector field $(X(x,y),Y(x,y))$ has no explicit time dependence. The nullclines are $X(x,y)=0$ and $Y(x,y)=0$, on which the respective coordinate's instantaneous rate of change vanishes. A critical point of a dynamical system is an intersection where $X=Y=0$; the constant trajectory at that point is an equilibrium of an autonomous differential equation. A nullcline alone need not consist of equilibria.

Set $z=y-\beta$. Then the autonomous planar system becomeswhich has no $\beta$ dependence. ThereforeMore precisely, replacing $\beta$ by $\beta +\delta$ sends every trajectory $(x(t),y(t))$ to $(x(t),y(t)+\delta )$, with the same time parametrization. The equilibrium points, nullclines, periodic orbits and separatrices all undergo this same translation. Their types and orientations are independent of $\beta$, so varying $\beta$ produces no bifurcation.

One suitable autonomous planar system isIts equilibrium points are exactly the four combinations $(x,y)=(\pm 1,\pm 1)$. Its Jacobian matrix isThus $(-1,-1)$ and $(1,1)$ have eigenvalues $\pm 2$ and are saddle equilibria. At the other two points the eigenvalues are $\pm 2i$. To establish the nonlinear centre equilibrium classifications, use the first integralIts Hessian matrix is $\mathrm{diag}(2x,-2y)$, a negative-definite matrix at $(-1,1)$ and a positive-definite matrix at $(1,-1)$. Thus each is a strict local extremum, and nearby regular level sets are closed curves supporting periodic orbits, by the planar first integral criterion for a centre.

Immediately to the right of $(-1,1)$, the vector field points down because $x^2-1<0$, so that centre is clockwise. Immediately to the right of $(1,-1)$ it points up because $x^2-1>0$, so that centre is counterclockwise. This verifies both the nonlinear classifications and the prescribed orientations, with exactly four equilibrium points.

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.

An equilibrium point $p=(x_0,y_0)$ of an autonomous planar system is a point where its vector field vanishes:For a locally well-posed ordinary differential equation, Lyapunov stability means that every prescribed neighbourhood can be protected by choosing a sufficiently small initial neighbourhood. More precisely, for every $\epsilon >0$ there exists $\delta >0$ such that every trajectory starting with $\Vert p(0)-p\Vert <\delta$ exists for all $t\ge 0$ and satisfiesThe number $\delta$ may depend on $\epsilon$ but not on the particular starting point. Asymptotic stability additionally requires convergence to the equilibrium point for all sufficiently close starting points.

For both parameters the autonomous planar system is $\dot{x}=v$, $\dot{v}=-x(x^2+\lambda x+1)$. It has the conserved energyIndeed $\dot{H}=v\dot{v}+V_{\lambda }^{\prime }(x)\dot{x}=0$. At any equilibrium point $(x_{\ast },0)$ the eigenvalues satisfy

For $\lambda =1$, the quadratic $x^2+x+1$ is strictly positive, so the only equilibrium point is $(0,0)$, with eigenvalues $\pm i$. Since $V_1^{\prime }(x)$ has the sign of $x$, $V_1$ has a unique strict minimum at zero and tends to infinity in both directions. All positive energy levels are closed curves surrounding this stable center equilibrium. They are clockwise: rightward when $v>0$ and leftward when $v<0$.

For $\lambda =5/2$, factor $x^2+(5/2)x+1=(x+2)(x+1/2)$. The three equilibrium points and their classifications areThe two center equilibria are stable but not asymptotically stable; the saddle equilibrium is unstable. At the saddle equilibrium the local unstable and stable directions are $v=(\sqrt{3}/2)(x+1/2)$ and $v=-(\sqrt{3}/2)(x+1/2)$.

The potential values are $V(-2)=-2/3$, $V(0)=0$, and $V(-1/2)=7/192$. Consequently there are closed loops around the left center for $-2/3<H<0$, closed loops in both wells for $0<H<7/192$, and a figure-eight pair of homoclinic orbits at $H=7/192$. At $H=0$ the right center is a stationary point while the left well still has a closed orbit. For $H>7/192$, each closed curve encircles both centers and the saddle.

The two homoclinic orbits satisfy $v=\pm \sqrt{2(7/192-V(x))}$. Their outer turning points can be located exactly fromEvery nonstationary energy curve is oriented to the right in the upper half-plane and to the left in the lower half-plane. The sketches show the closed curves, the separating orbits, the stable and unstable saddle directions and the direction of flow.

Here the autonomous planar system is $\dot{x}=v$, $\dot{v}=-\cos x$. Its equilibrium points areThe Jacobian matrix isFor even $n$, the eigenvalues are $1,-1$, so these points are unstable saddle equilibria. Writing $\xi =x-x_{\ast }$, their local unstable and stable directions are $v=\xi$ and $v=-\xi$ respectively. For odd $n$, the eigenvalues are $i,-i$.

To classify the latter points nonlinearly, differentiate the first integralAt $x_{\ast }=3\pi /2+2m\pi$ this energy has a strict local minimum. Its nearby level sets are closed curves, proving that the odd-$n$ points are stable center equilibria, with no attraction to the center. The phase portrait consists of closed oscillatory orbits for $-1<H<1$, heteroclinic orbits at $H=1$ joining consecutive saddle equilibria, and open rotating orbits for $H>1$. The separating curves haveAbove the horizontal axis $\dot{x}=v>0$, so flow is rightward; below it flow is leftward. The closed orbits are clockwise, and these same signs orient every separating and rotating orbit.

Introduce the velocity variable $v=\dot{x}$. The second-order differential equation becomes the autonomous planar systemA critical point of a dynamical system is a point where both components of the vector field vanish: here $v=0$ and $f(x,0)=0$. It is a stationary solution in the phase plane. The Jacobian matrix at each such point gives the local linear stability of a planar equilibrium. For the equations below, a conserved energy also determines the nonlinear phase portrait; this is important because purely imaginary eigenvalues alone do not generally prove that an equilibrium is a center equilibrium.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

The phase plane is the two-dimensional state space of an autonomous planar system. Its oriented trajectories, equilibria, nullclines, and invariant sets display the system's qualitative dynamics.