1. 1. We integrate minus the force, from a reference point to x, to obtain the potential:

      

      where we have chosen as our zero potential reference point. To sketch this potential we note that it 1) vanishes at x = 0; 2) goes to as ; 3) is symmetric under ; and 4) has as its extrema values of x satisfying

      

      At these points -- given by -- the second derivative of V is

      

      which is positive (indicating a local minimum, or stable equilibrium point) at x=0, and negative (indicating local maxima, or unstable equilibrium points) at . Putting these four facts together, our potential must look like

      

      This potential gives several different possible particle motions, depending on the relationship of a particle's energy to .

      For : (Case )

      The particle is unbound and has no turning points; thus a particle initially traveling right travels right forever, although it does slow down as x increases to , speed up until x=0, slow down until , then speed up as it runs off to .

      For : (Case )

      There are four turning points, marked a, b,-a, and -b. Positions between a and b, or between -a and -b, are forbidden.

      Here there are two types of allowed motions, depending on :

      For , the particle oscillates between the turning points , forever. This is a bound motion. The particle's speed increases when approaching the origin and decreases when moving away from the origin, vanishing at the turning points.

      For , the particle slows down as it approaches from , and turns around at the turning point , speeding back up as it returns to . This is an unbound motion.

      For E;SPMlt; 0: (Case )

      There are two turning points, marked c and -c. Positions between -c and c are forbidden.

      For , the particle slows down as it approaches from , and turns around at the turning point , speeding back up as it returns to . This is an unbound motion.

      Note the ``special'' dividing line case , the boundary between bound and unbound motion. Here a particle on either equilibrium point (if unperturbed) remains stationary forever. A particle approaching either equilibrium point, from the left or from the right, slows down as it approaches and stops at the equilibrium point, to rest forever if unperturbed. This type of solution ends up being important in determining tunneling in quantum mechanics and quantum field theory.

   2. To escape, a particle must have enough energy to overcome the barrier . A particle at x=0 must then have

      

      which equals

      

      to just barely escape. This gives, for ,

      

      A particle at has escape energy

      

      Setting this equal to gives

      

2. 1. We have the potential

      

      with derivatives

      

      As given, both and a are positive. Thus we have two local extrema, when V' vanishes at . At ,

      

      V'' is thus positive at x=a, making it a local minimum (stable equilibrium point). At x =-a, V'' is negative, so x=-a is a local maximum (unstable equilibrium point).

      Note that this agrees with the behavior of V': for example, for 0 ;SPMlt; x ;SPMlt; a, V is decreasing (V' is negative); while for x ;SPMgt;a, V is increasing (V' is positive), making x=a a minimum. Thus the potential looks like
      

      

      which decreases from x=0 to the minimum at x=a, then increases for x;SPMgt;a, with asymptotes that go as for small x, and for large x. Note that V(-x) = -V(x), so its graph is the negative reflection of V(x) for x;SPMgt;0.

      Only the stable equilibrium point at x=a determines small oscillations. They have frequency

      

   2. Turning points occur when the energy is entirely potential, so that the kinetic energy and speed must vanish. For the case, that occurs when

      

      Letting q = a/x, the turning point occurs when . We could solve this with a quadratic equation, but it's easier to guess the solutions q = 2, 1/2. These turning points bound the only allowed motion for positive x, an eternal oscillation between the turning points x = 1/2 and x = 2. (Positive x below 1/2 and above 2 are forbidden.) Negative x is allowed. For negative x, there is an unbound motion, with a turning point at the infinite potential barrier at x = 0. Particles with negative x, moving to the right, abruptly stop and reverse course to the left at x=0. (This potential is rather untrustworthy at x=0, we imagine a smooth continuation between its negative x and positive x behavior.)

      For the case, we have turning points at

      

      Again letting q = a/x, the turning points occur when , solved by q = -2, -1/2. Here however, motion between the turning points is forbidden. Instead we have an unbound motion, where a particle travels to the right from , slows down and reverses course at the turning point x = -2, returning to . We also have a bound motion between the x = -1/2 turning point and the origin, which is a turning point due to the infinite potential barrier at x= 0. These two turning points bound an eternal oscillation between x = -1/2 and x = 0.

3. An equilibrium point is stable if any deflection from the equilibrium point leads to a restoring force, pushing the particle back toward the equilibrium point. Such a force must be negative for positive deflection , and positive for negative (that is, a rightward deflection leads to a leftward restoring force, a leftward deflection a rightward restoring force). Thus the force must be odd in , and with overall sign opposing the deflection . A positive quadratic term in the potential, does this because it leads to a force which is odd in , and of sign opposite to . If an equilibrium point has vanishing , first and second order terms in the Taylor expansion for V vanish. The first nonzero term in that expansion will be , where denotes the nth derivative of V, for some n ;SPMgt; 2. If is nonzero for n=3, then the leading term in the force, F = -V', will be even. Thus the force will push in the same direction for each deflection away from , in one case accelerating the particle away from , making the equilibrium point unstable. Thus the first nonzero derivative at must be . This gives leading contribution to the force which will have opposite sign to only if is positive. Thus for stable equilibrium we must have

   

   Actually, if , we can still have stable equilibrium through the same argument, if and . Or most generally, we will have stable equilibrium if

   

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