A stable orbit without neighboring periodic orbits
For canonical pairs $(q_j,p_j)$, put $I_j=(p_j^2+q_j^2)/2$. Consider the consistent oscillator model
$$H=\omega_1I_1+\omega_2I_2,\qquad\omega_j>0,\qquad\omega_2/\omega_1\notin\mathbb Q.$$
Each $I_j$ is constant. Both components rotate uniformly, with angular speeds of magnitudes $\omega_1,\omega_2$. If both $I_1,I_2$ are positive, closure would require $\omega_1\tau,\omega_2\tau\in2\pi\mathbb Z$, contradicting irrationality. On a fixed energy $h>0$, the only geometric periodic orbits are the two coordinate circles with one action zero.
Near the first circle, the return map on a transverse disk is a rigid irrational rotation by $2\pi\omega_2/\omega_1$ up to the angular convention. Every radius is preserved, so the fixed point is stable, but no other point is periodic. Its twist coefficient vanishes. Stability alone cannot replace the nonzero-twist assumption.
Frequency maps and rational targets
On an action-angle domain $U\times\mathbb T^m$, with angles of period $2\pi$, let
$$H(I)=c\cdot I+\tfrac12 I^{\mathsf T}DI,\qquad D=D^{\mathsf T}.$$
Hamilton’s equations give $\dot I=0$, $\dot\theta=\omega(I)=c+DI$. A full torus is periodic with return time $n\tau$ if
$$c+DI=\frac{2\pi}{\tau}\frac{k}{n},\qquad k\in\mathbb Z^m.$$
When $D$ is invertible, every nearby frequency target has the unique solution
$$I=D^{-1}\left(\frac{2\pi}{\tau}\frac{k}{n}-c\right).$$
Rational vectors $k/n$ are dense. In an open action domain this gives periodic tori near any given action, by choosing targets near its actual frequency. The ratios must approximate $\tau\omega(I)/(2\pi)$; merely making them small does not bring the actions to zero. Near an oscillator boundary $I_j\geq0$, the resulting actions must additionally lie in the admissible cone. The formula alone cannot remove that constraint.
Invertibility of $D$ is a sufficient local frequency-adjustment condition in this integrable model, not a necessary condition for every possible abundance of periodic trajectories. For example, $H(I)=I_1+I_2$ has $D=0$ but every torus is periodic. Conversely, a constant irrational frequency ratio gives none on the full-dimensional tori. Degeneracy permits different outcomes.
Fixing energy changes the nondegeneracy question
For an autonomous problem at fixed energy, solving for arbitrary frequencies ignores the constraint $H(I)=h$. A useful local test is invertibility of the bordered matrix
$$\begin{pmatrix}D&\omega\\\omega^{\mathsf T}&0\end{pmatrix}.$$
To see its meaning, allow a scalar frequency rescaling $a$. Linearizing $\omega(I)=a\nu$ and $H(I)=h$ at a solution produces the equations $D\,\delta I-\nu\,\delta a=\delta b$, $\omega^{\mathsf T}\delta I=\delta h$. Up to a nonzero column factor, the displayed matrix is their Jacobian. Its invertibility is the implicit-function condition for varying the frequency direction while fixing energy. This condition differs from $\det D\ne0$.
This is an explanatory local calculation, not a theorem asserting persistence of these tori under perturbation. No KAM or higher-dimensional periodic-orbit theorem is proved here.
What the historical questions establish
Birkhoff asks whether his local conclusion survives in broader completely stable differential systems and doubts it without preserved area. That is a historical question in Original and Modernized. The rigid-rotation model proves a specific failure when twist vanishes; it does not answer every version of the broader question. The source also distinguishes continuing a periodic orbit by changing conserved quantities from finding different periodic orbits on the same joint level. The distinction remains essential.