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Suppose we have a mechanical system with 1 degree of freedom, i.e. an ODE

$$(1)\quad \ddot{q}+V^\prime(q)=0, $$

where $V \colon \mathbb{R} \to \mathbb{R}$ is some smooth function (potential energy). We then easily see that any solution of this equation must satisfy

$$\frac{\dot{q}^2}{2} + V(q)=\text{constant}.$$

In other words, if we put

$$E(q, \dot{q})=\frac{\dot{q}^2}{2} + V(q)$$

(energy), then the image of every solution of (1) must lie in a level set of $E$.


Question Can it be that a solution of (1) is projected properly in a connected level set of the energy?

More formally, can there exist a function $u \colon I \to \mathbb{R}$ that solves (1) and such that the set

$$T_u=\{(u(t), \dot{u}(t)) \mid t \in I\}$$

is a proper subset of

$$E_u=\{(a, b) \in \mathbb{R}^2\mid E(a, b)=E(u_0, \dot{u}_0)\}?$$

Of course $I$ must be the maximal interval of existance for the solution $u$. Require also $E_u$ to be connected (thanks to AlbertH who pointed this out).


By physical considerations I guess that this cannot happen. Also, I found in some physics textbooks phrases like: "finding the level sets of the energy is equivalent to finding the trajectories of the system", which corroborate this conjecture.

I'm not necessarily after a fully rigorous answer. Even a good intuition will suffice. Thank you.

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4 Answers 4

up vote 4 down vote accepted

You are essentially asking whether the particle can stop at a point inside the level set. To stop, it needs to have both zero velocity and zero acceleration (these conditions are sufficient because the dynamics is of second order). First condition requires $\dot q = 0$ implying also $V(q) = E$, while the second holds when $V'(q) = 0$. So just choose any potential with a local maximum $V_{\rm max}$ and choose initial conditions such that $E = V_{\rm max}$.

Physically these correspond to a ball rolling down a hill and then again up a hill. As the ball approaches the top, its kinetic energy decreases as $E - V$ and asymptotically it stops. This interval is proper in the level set because you can get another solution approaching the hill from the other side of the hill.

Perhaps the answer is even more obvious in more dimensions (or more degrees of freedom). There the level sets are of dimension $2N - 1$ and the failure of inclusion is more apparent. Just consider a potential with a local maximum (a boulge at the origin) which you can approach from any direction and asymptotically stop there.

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Ok, I finally got it! The example with two hills was truly illuminating. In fact the main question here was: "Should we know all level sets, does this mean that we know all trajectories?". Now the answer is yes, we know all possible trajectories, but we are giving up some information, as your example shows. Thank you. –  Giuseppe Negro Jun 12 '11 at 1:31
    
@dissonance: you're welcome. –  Marek Jun 12 '11 at 14:55

Consider a free particle with speed $v_0$. We have $T_u = \mathbb R \times \{v_0\}$ and $E_u = \mathbb R \times \{v_0, -v_0\}$.

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Right. I should have added: suppose $E_u$ is connected. –  Giuseppe Negro Jun 11 '11 at 19:05

yes, this behavior occurs in the mathematical pendulum. The situation where the pendulum is standing upwards is a unstable equilibrium. If one pertubs this equilibrium slightly, it is possible to let the pendulum take one swing (in an infinite amount of time). It never goes back (this motion has the same energy).

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Of course. This is another good example. –  Giuseppe Negro Jun 12 '11 at 1:32

Another positive answer to your question is obtained when the potential energy is $V(x)=\{0,\ \mathrm{for}\ x\leq 0;\ \exp(-1/x^2),\ \mathrm{for}\ x>0\}$.

In such a case the zero level set of the total energy is $]-\infty,0]$, but the trajectories of motion with zero energy are each one of its sigleton.

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