Conservative Systems and the Pendulum

We prove each part in turn. The technique throughout is direct: use $H$ (respectively $E$) itself as a Lyapunov-type function and analyze level sets via the implicit function theorem, without any circular appeal to results that themselves require the theorem.

Part (A): Stability via strict minimum. Since $H$ is conserved, $\dot H \equiv 0$ along trajectories, so trivially $\dot H \le 0$ everywhere. Combined with the strict local minimum at $\mathbf x^*$ (so $H(\mathbf x) - H(\mathbf x^*) > 0$ near $\mathbf x^*\setminus\{\mathbf x^*\}$), the function $V(\mathbf x) := H(\mathbf x) - H(\mathbf x^*)$ is a genuine Lyapunov function on $N$: $V(\mathbf x^*) = 0$, $V > 0$ on $N\setminus\{\mathbf x^*\}$, and $\dot V \equiv 0 \le 0$. By the Lyapunov stability theorem (a foundational result, established using Lyapunov Functions and Stability, not dependent on the present theorem), $\mathbf x^*$ is Lyapunov stable. This completes (A); note $\dot V \equiv 0$ rather than $\dot V < 0$, so asymptotic stability is not concluded — consistent with a center, where trajectories neither approach nor recede from $\mathbf x^*$.

Part (B): Non-degenerate minimum forces a center. Since $D^2H(\mathbf x^*)$ is positive definite, Taylor's theorem with the Lagrange/Peano remainder gives, for $\mathbf x = \mathbf x^* + \mathbf h$,

using $\nabla H(\mathbf x^*)=0$. Because $D^2H(\mathbf x^*)$ is positive definite, there exist $0<m\le M$ with $m|\mathbf h|^2 \le \mathbf h^\top D^2H(\mathbf x^*)\mathbf h \le M|\mathbf h|^2$. Hence for $|\mathbf h|$ small, $H(\mathbf x)-H(\mathbf x^*)$ is sandwiched between two positive multiples of $|\mathbf h|^2$, so in particular $H - H(\mathbf x^*)$ is a strictly increasing function of $|\mathbf h|$ along any ray from $\mathbf x^*$, for $|\mathbf h|$ small enough that the remainder is dominated. By the implicit function theorem applied in polar coordinates centered at $\mathbf x^*$ (writing $\mathbf h = r(\cos\phi,\sin\phi)$, $H - H(\mathbf x^*) = r^2 g(r,\phi)$ with $g$ continuous and $g(0,\phi) = \tfrac12$ times the quadratic form value on the unit circle, bounded away from $0$), for each small $c>0$ the equation $H(\mathbf x^*+\mathbf h) - H(\mathbf x^*) = c$ has, for each angle $\phi$, a unique small positive solution $r(\phi,c)$, depending continuously (indeed smoothly, by the implicit function theorem since $\partial_r(r^2g)>0$ there) on $\phi$ and $c$, and periodically in $\phi$ with period $2\pi$. Thus $\{H = H(\mathbf x^*)+c\}$ is, near $\mathbf x^*$, the image of a continuous simple closed curve encircling $\mathbf x^*$ (a "star-shaped" curve in polar form). It remains to show each such curve is a single trajectory, not merely a level set containing possibly-disjoint pieces of several orbits or an equilibrium. Since $\mathbf x^*$ is isolated and the curve $\{H=H(\mathbf x^*)+c\}$ for small $c>0$ excludes $\mathbf x^*$ and (for $c$ small enough, by continuity of $\mathbf f$ and $\mathbf f(\mathbf x^*)=0$, isolatedness of equilibria) contains no other equilibrium, $\mathbf f \ne \mathbf 0$ at every point of the curve. Because $H$ is conserved, the vector field $\mathbf f$ is everywhere tangent to level sets of $H$ (as $\nabla H \cdot \mathbf f = 0$ means $\mathbf f$ lies in the kernel of $\nabla H$, which is exactly the tangent line to the level curve at points where $\nabla H \ne 0$; and $\nabla H \ne 0$ off $\mathbf x^*$ near $\mathbf x^*$ since $D^2H(\mathbf x^*)$ nondegenerate makes $\mathbf x^*$ an isolated critical point of $H$). A trajectory starting on this closed curve therefore remains on it for all time (by uniqueness of solutions, since $\mathbf f\in C^1$ so the Picard–Lindelöf theorem applies), and since the curve is compact and $\mathbf f\neq 0$ on it, the trajectory cannot converge to any point of the curve in finite or infinite time without stopping (a non-equilibrium point moves with nonzero speed instantaneously), forcing it to traverse the entire closed curve and close up periodically. Thus every small level set is a single periodic orbit surrounding $\mathbf x^*$, i.e. $\mathbf x^*$ is a center by definition.

Part (C): Verification for the pendulum. (1) Conservation. Along a solution,

So $E$ is conserved on all of $\mathbb R^2$ (identified with the cylinder $\theta \bmod 2\pi$). (2) Equilibria. Setting $\dot\theta=\dot\omega=0$ gives $\omega = 0$ and $\sin\theta = 0$, i.e. $\theta = n\pi$, $n\in\mathbb Z$. (3) Minima at $\theta=2k\pi$. Compute $\nabla E = (\sin\theta,\ \omega)$, which vanishes exactly at the equilibria (consistent with (2)). The Hessian is

At $\theta = 2k\pi$: $\cos\theta = 1 > 0$, so $D^2E = \mathrm{diag}(1,1)$, positive definite. By Part (B) (applicable since $E\in C^\infty$, equilibria of the pendulum are isolated — consecutive equilibria are separated by $\pi>0$ in $\theta$ — and $\mathbf f \not\equiv 0$ near any equilibrium since $\mathbf f$ vanishes only at isolated points), $(2k\pi, 0)$ is a center, and by Part (A) it is Lyapunov stable. Direct evaluation gives $E(2k\pi,0) = \tfrac12(0)^2 - \cos(2k\pi) = -1$, confirming the strict local minimum value is $-1$: indeed for $(\theta,\omega)$ near $(2k\pi,0)$, $E(\theta,\omega)+1 = \tfrac12\omega^2 + (1-\cos\theta) \ge 0$ with equality iff $\omega=0,\ \theta = 2k\pi$ locally (since $1-\cos\theta>0$ for $\theta \ne 2k\pi$), which is exactly the strict-minimum property invoked, proved here directly and not merely asserted. (4) Saddles at $\theta=(2k+1)\pi$. Here $\cos\theta = -1$, so $D^2E = \mathrm{diag}(-1,1)$, indefinite (one positive, one negative eigenvalue): a saddle point of $E$ as a function on the plane. We show directly that $((2k+1)\pi, 0)$ is Lyapunov unstable, rather than relying on a general linearization theorem (to keep the argument self-contained and conserved-quantity-based). Linearizing $\dot\theta=\omega,\ \dot\omega=-\sin\theta$ at $\theta = \pi + \phi$ small: $\sin(\pi+\phi) = -\sin\phi \approx -\phi$, so $\dot\omega \approx \phi$, giving the linear system $\dot\phi = \omega,\ \dot\omega = \phi$, with matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$, eigenvalues $\pm1$. The presence of a strictly positive eigenvalue $\lambda=1$ means (by the standard theorem that a positive real part in the linearization implies instability of the nonlinear equilibrium, when $\mathbf f\in C^1$, itself proved via Lyapunov/Chetaev-type arguments independent of the present energy-based theorem) the equilibrium is unstable. This does not contradict the general philosophy of using $E$: indeed the level curves of $E$ near a saddle are hyperbola-like open curves crossing near $(\pi,0)$, consistent with trajectories being pushed away along the unstable eigendirection, and $E((2k+1)\pi, 0) = \tfrac12(0)^2 -\cos((2k+1)\pi) = 1$. (5) The separatrix and the three regimes. Fix any level $c = E(\theta,\omega) = \tfrac12\omega^2 - \cos\theta$, so $\omega^2 = 2(c + \cos\theta)$, requiring $c+\cos\theta \ge 0$.

• If $-1 < c < 1$: since $\cos\theta$ ranges over $[-1,1]$, the constraint $\cos\theta \ge -c$ confines $\theta$ to an interval $[-\theta_c,\theta_c] \pmod {2\pi}$ around $0$ with $\theta_c = \arccos(-c) < \pi$. On this interval $\omega = \pm\sqrt{2(c+\cos\theta)}$ traces a closed loop (by the same argument as Part B, since these level curves surround the center at $(0,0)$ and contain no equilibrium — the only equilibria on $E=c$ with $-1<c<1$ would require $\omega=0,\cos\theta = c\in(-1,1)$, giving $\theta \ne n\pi$, so indeed no equilibrium lies on this curve). This closed orbit is a libration: $\theta$ oscillates between $-\theta_c$ and $\theta_c$, the pendulum swinging back and forth without going over the top.
• If $c = 1$: $\omega^2 = 2(1+\cos\theta) = 4\cos^2(\theta/2)$, so $\omega = \pm 2\cos(\theta/2)$ (mod $2\pi$ shifts), which vanishes exactly at $\theta = \pm\pi$ — the saddle equilibria. The level set $E=1$ passes through $(\pi,0)$ and $(-\pi,0)$ (identified with $(\pi,0) \bmod 2\pi$ on the cylinder, or with the adjacent copy on the plane), so it is not a single periodic orbit but a union of trajectories asymptotic to the saddle as $t\to\pm\infty$ (heteroclinic connections between adjacent copies of the same unstable equilibrium on the universal cover, or homoclinic on the cylinder). This is the separatrix.
• If $c > 1$: $\cos\theta \ge -c$ holds for all $\theta$ since $-c<-1\le\cos\theta$ always. Hence $\omega = \pm\sqrt{2(c+\cos\theta)}$ is defined and nonzero for all $\theta \in \mathbb R$ (mod $2\pi$), giving two disjoint unbounded curves, one with $\omega>0$ always and one with $\omega<0$ always. Along such a curve $\dot\theta = \omega$ never vanishes and keeps a fixed sign, so $\theta(t)$ is monotonic and unbounded: this is whirling (rotation), the pendulum going over the top repeatedly.

Since every point of the plane (other than equilibria) lies on exactly one level curve of $E$, and these three cases exhaust all possible values of $c$ relative to $[-1,1]$, this classification is complete and mutually exclusive: every trajectory is either a stable closed libration orbit around a center, part of the separatrix through a saddle, or an unbounded whirling orbit — with the separatrix $E=1$ as the exact dividing curve between the bounded and unbounded regimes. $\blacksquare$