能否判定给定微分方程组对应的轨线为极限环？

Great question! Let's start by grounding this system in its physical context—it describes the motion of a test particle in Schwarzschild spacetime (the geometry around a spherical black hole/star). First, let's fill in the missing piece of the radial equation (it's a standard result in general relativity):
$$ \frac{dr}{d\lambda} = \pm \sqrt{E^{2} - \left(1-\frac{2M}{r}\right)\left(1 + \frac{L^2}{r2}\right)} $$
Where $L$ is the particle's specific angular momentum (positive real), $E$ is its specific energy, and $M$ is the mass of the central object.

Now let's work through why limit cycles aren't possible here:

First, what is a limit cycle? #

A limit cycle is a closed, isolated periodic orbit that only exists in 2D autonomous dynamical systems. Key traits:

• It's a closed trajectory in phase space;
• No other periodic orbits exist in its immediate neighborhood;
• Nearby trajectories either spiral toward it (stable) or away from it (unstable).

Breaking down the system #

Let's simplify the original $dx/d\lambda$ and $dy/d\lambda$ equations first. From $ \frac{d\theta}{d\lambda} = \frac{L}{r^2} $, we get $ \frac{d\lambda}{d\theta} = \frac{r^2}{L} $. Substitute this into the equations:
$$
\begin{align*}
\frac{dx}{d\lambda} &= \frac{L}{r^2}\left( -y + \frac{dr}{d\lambda} \cdot \frac{r^2}{L} \cdot \frac{x}{r} \right) = -\frac{L y}{r^2} + \frac{x}{r}\frac{dr}{d\lambda} \
\frac{dy}{d\lambda} &= \frac{L}{r^2}\left( x + \frac{dr}{d\lambda} \cdot \frac{r^2}{L} \cdot \frac{y}{r} \right) = \frac{L x}{r^2} + \frac{y}{r}\frac{dr}{d\lambda}
\end{align*}
$$
If you substitute $x = r\cos\theta$ and $y = r\sin\theta$, you'll recognize this is just the Cartesian decomposition of polar coordinates velocity—so we're really looking at a particle's motion split into radial ($dr/d\lambda$) and angular ($d\theta/d\lambda$) components.

Case 1: Unbound trajectories ($E^2 \geq 1$) #

These are particles that either:

• Fall into the central object (approaching $r=2M$, the event horizon), or
• Escape to infinity ($r \to \infty$)
  In both cases, the trajectory is open—no periodicity, so definitely not a limit cycle.

Case 2: Bound trajectories ($E^2 < 1$) #

Here, the particle is trapped between a minimum radius $r_{\text{min}}$ and maximum radius $r_{\text{max}}$:

• Non-circular bound orbits: The radial motion oscillates between $r_{\text{min}}$ and $r_{\text{max}}$, but the angular motion is always increasing ($d\theta/d\lambda > 0$ because $L>0$). Due to the curvature of Schwarzschild spacetime, these orbits exhibit perihelion precession—they never close into a perfect ellipse, instead tracing a "rose petal" pattern in the $x$-$y$ plane. No periodicity here, so no limit cycle.
• Circular orbits ($dr/d\lambda = 0$): These occur when $E$ and $L$ satisfy $ E^2 = \left(1-\frac{2M}{r}\right)\left(1 + \frac{L^2}{r2}\right) $, making $r$ constant. While this is a closed trajectory in the $x$-$y$ plane:
  1. Circular orbits are not isolated—they correspond to specific $(E,L)$ pairs, not a unique, isolated orbit in phase space.
  2. Critically, our system lives in a 4D phase space (state variables: $x, y, dx/d\lambda, dy/d\lambda$). Limit cycles only exist in 2D autonomous systems, so this 4D system can't support them by definition.

Bonus: Reducing the system dimension #

If we use angular momentum conservation to reduce the system to a 1D system (treating $\theta$ as the parameter, state variable $r$), we get:
$$ \frac{dr}{d\theta} = \pm \frac{r^2}{L}\sqrt{E{2} - \left(1-\frac{2M}{r}\right)\left(1 + \frac{L^2}{r2}\right)} $$
1D systems only have fixed points (circular orbits, where $dr/d\theta=0$) and monotonic trajectories—no periodic orbits at all, let alone limit cycles.

Final Verdict #

No, the trajectories of this system cannot be limit cycles:

• Unbound trajectories are open and non-periodic.
• Bound trajectories are either precessing (non-closed) or circular (closed but not meeting the 2D/isolated requirements for a limit cycle).
• The system doesn't fit the 2D autonomous framework where limit cycles can exist.

内容的提问来源于stack exchange，提问作者Herr Schrödinger