Proof that oscillations in 1d potential well occur between certain points In >>this<< situation, a particle with energy $E$ will oscillate between the positions $x_1$ and $x_2$ indicated on the diagram. This simple fact is taught in many introductory courses however I am unable to find any proof of this. The elementary level texts I looked at don't give a proof, and the more advanced ones treat this as obvious. 
I would like to gain some insight into this by reading a proof.
 A: So, if we knew for sure that this was a harmonic potential ($V \propto x^2$) we could solve the actual equation itself. But I think we can establish the "oscillation" part without recourse to the differential equations. Let's look at this purely in terms of conservation of energy. 
First, conservation of energy tells us that the particle will not slow down, because forces derived from potentials are conservative--they do not remove kinetic energy from the system. So the particle will continue moving "forever." We therefore have two options: either the particle must stay bounded in some region, or it must eventually go off to infinity. Now, conservation of energy tells us that at all points, the particle obeys (where $E_0$ is its initial energy; marked in your diagram as $E$):
$$
E = \frac{1}{2} m v^2 + V(x) = E_0
$$
Note that we can say that both of these terms are always and everywhere positive--mass is never negative, velocity squared is always positive, and $V(x)$ is finite so we can always add a meaningless constant to it to make it positive. Now, we know that the particle has some particular $E = E_0$, and that $E_0 < V(\pm \infty)$. But if this is the case, it follows that the particle cannot go out to infinity, since it could not fulfill the energy equation I wrote above there. In fact, since the lowest possible value of kinetic energy is 0, we can see that the particle can never be at any location where $V(x) > E_0$. The points where this condition is fulfilled are at $x_1$ and $x_2$ in your diagram, and so we can say definitely that the particle stays within those bounds.
Now, I want to point out where this argument fails. First of all, I haven't shown you for certain that the particle actually reaches these points, only that it can't go past them. But since nothing else slows the particle down (there is only this one potential), it will keep moving until that potential stops it. In addition, I haven't necessarily shown that the particle motion is oscillatory. This follows from the fact that Newton's second law is a second-order differential equation, and therefore its solutions are specified by two parameters. If I tell you that the particle is in a particular position, with a particular velocity, then you know it has one unique behavior. So every time the particle reaches $x_1$ with velocity 0, it starts a new period identical to the last time it was at $x_1$ with velocity 0. And, in addition, note that this argument would need to be somewhat refined if you had another well a little ways down the line--I think you can see that the particle would never reach that well, but I haven't exactly "proved" that.
Finally, in quantum mechanics, note that it is an important (and fascinating) result that the particle has a small (exponentially-decaying) chance of being found in the "classically forbidden region" and this is the source of what's usually known as quantum tunneling.
A: Since there is no reference config for Potential Energy specifically mentioned, you can consider that the minimum potential (the lowest point on the graph) is zero, and hence the KE of the particle would be maximum at that point. For one half of that potential energy curve, this would be analogous to a very simple case of throwing a ball upwards. The ball will go up till its Kinetic Energy becomes zero because of the work done by Gravitation. (Hence its potential energy would be $E$, at that point, which is the maximum energy for the particle in this case.) It would be equal and opposite on the other half of the potential curve.
Mathematically, the negative derivative of the potential would be the force on the particle. (you can observe the sign of this quantity by seeing the inclination of tangent lines to the curve.) For displacements between $x_1$ and $x_2$, which are directed away from the minimum point on the potential energy curve, the force is always directed opposite to the displacement. This shows that force on the particle is a restoring force of sorts  and always tries to bring the particle to its equilibrium ($F=0$) state.

This can only happen between ($V$ coordinates) $V(x_1)$ and $V(x_2)$ in this case, because the particle's total energy is $E$.  For $x_1$ and $x_2$, $V(x_1)=V(x_2)=E$, and this implies that the KE of the particle is zero at those points. This permits the force on the particle to move it in the direction of decrease in potential, and hence the particle falls back into the 'well' and oscillates.
