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x
U
E
Lecture 12 Anharmonic Motion
Up to now we have (for good reason) focused on motion near stable equilibriums that
correspond to harmonic motion with linear restoring forces. As we try to move
beyond the simplest kinds of motion we should talk about confined, periodic motion
that is not simple harmonic motion, i.e ., anharmonic motion. The harmonic 1-D case
is described by
2
2 2
0 0
linear restoring force
k
F kx U x x
k
x x
m
Now we want to consider a perturbation that maintains the confining nature of this
potential and its symmetry under x x , i.e ., a quartic perturbation
2 2
2 2
0
1
2 2
1 0.
k
U x x
x x x
For the case 0 thischangeservesto“stiffen”therestoringforceleadingtomore
rapid oscillation, while for 0 the force is softened for small amplitude oscillations
and there are no oscillations for large enough
amplitude oscillations, i.e .,
2 2 1 M
x x ,
M
E U k . However, we can still analyze
this situation (assuming we are in the regime of
the force being a restoring force) much as we did
for the central force problem. The energy is
conserved and there will be turning points where
the kinetic energy vanishes and the total energy
is all potential energy as indicated in the figure. Motion in such a potential will have
the following properties:
1) Periodic x t x t for some period = twice the time from turning point to
turning point;
2) Symmetric about the turning points,
0
x t 0 ,
0 0
x t t x t t.
In particular, if we start the clock at a turning point, 0
t 0 , x t (^) x (^) t ). These 2
features do not depend on the reflection symmetry of the potential.
- The symmetry of the potential leads to asymmetry about the half-period,
x t (^) x (^) 2 t (^) or x (^) 2 t (^) x (^) t (^) x t (^) , which means
x (^) 4 t (^) x (^) 4 t (^) , i.e ., if (^) (^0) M
x x , then x 4 (^) 0.
So with the choice (^) 0 1
x t ^ 0, x 0 A consider the specific from of the solution for
0. We have (^) 0 1 0
x t A cos t
. In the general periodic case with 2
0 , we can use our knowledge of Fourier series to write the general periodic
solution as
1
cos , n
n
x t A n t
(^12.^3 )
which satisfies conditions 1) and 2) above. Condition 3) arising from the reflection
symmetry of the potential means that we require
cos cos cos odd.
n t n t n t n n
^
Factoring out the amplitude of the fundamental term,
, which carries any
dimensions, we have (the second term includes the harmonics)
1 2 1
1
cos cos 2 1.
n
n
x t A t n t
(^12.^5 )
Our expectations is that for small values of , i.e .,
2
1
A 1 , the amplitudes of the
harmonics will also be small,
2
3 1
A 1. Recall that in the unperturbed,
harmonic case, we have a linear problem as in Eq. ( 12. 1 ), for which the form of the
solution is independent of the size of the amplitude, i.e ., the problem is scale
invariant. Once we move to the anharmonic problem, we have broken the scale
invariance, i.e ., the constant has units of inverse distance squared and the deviation
from harmonic behavior depends on the size of the amplitude. Define the scaled
(dimensionless) variable 1
y x A , whose equation of motion is
2 3 2 2 5 2
t
-0.
1
x
0
2 3 2 2 5 2
t
-0.
1
x
0.
2 3 2 2 5 2
t
-0.
1
x
0.
2 3 2 2 5 2
t
-0.
1
x
1.
The approximation is in keeping just the linear terms in . As expected a positive
( ) leads to faster oscillations, while the amplitude of the first harmonic is linear in
3
1 1
A A. Thus keeping just the lowest order perturbation to harmonic behavior
we see that both the frequency of oscillation and the amplitude of the harmonics is
dependent on the magnitude of the amplitude, i.e ., we see the nonlinearity explicitly,
2 2
0
2
1
1
2 2
0 1 0 1
cos cos 3
x x x
A
x A t t
A A
The following plots correspond to 0 1
1, A 1 for various values of . Note that,
for values of comparable or larger than 1, where the anharmonic behavior is
obvious, the current approximations of keeping only the first harmonic and only the
linear terms in are not valid. For appropriately small values of , only the
shrinking of the period is clear.
2 3 2 2 5 2
t
-0.
1
x
5.
-1 -0.5 0 0.5 1
-0.
0
1
0
-1 -0.5 0 0.5 1
-0.
0
1
0.
For a truly perturbed (and inaccurate) case we have 5 in the next figure. Both the
period and the shape are distorted from the harmonic limit.
Let us also consider how this system looks in phase space, i.e., what does the phase
portrait look like (recall the Appendix to Lecture 7). The Hamiltonian and
Hamilton’sequations have the form
2
2 2
2
p k
H q q
m
H p H
q p kq q
p m q
The streamlines of the flow in phase space correspond to E H a constant, i.e .,
this is a conservative system. Typical streamlines are indicated in the following
figures for various values of . The different lines correspond to different values of
E. (with m k 1 ).
x
U
0.
x
U
0
x
U
0
2 2
0 0
2 2
0
sin 0
g
l
In this case the nonlinearity serves to lengthen rather shorten the period of oscillation.
What happens if the first nonlinear term does not preserve the reflection symmetry?
Consider the asymmetric cubic potential
2
k
F kx x U x x x
The potentials look like
Clearly in this case only small amplitude (small energy) systems are bound (for either
sign). The potential has a local maximum at 1 M
x where
2 6 M M
U x U k . Initial conditions corresponding to a total energy larger than
this value will exhibit at most 1 turning point. For energies below this critical value
there will be 2 turning points and, like the earlier systems, the motion will be
Periodic x t (^) x t (^) (^) for some period ;
Symmetric about the turning points, (^) 0
x t 0 , (^) 0 0
x t t x t t.
In particular, if we start the clock at a turning point, 0
t 0 , x t (^) x (^) t . Note
again that these 2 features do not depend on the reflection symmetry of the potential.
However, since in this case we do not have reflection symmetry, condition 3) will not
hold. This means that the expansion as a Fourier series will include all frequencies,
including n 0,
0
cos. n
n
x t A n t
( 12. 14 )
Similarly to what we did above, here we can solve this problem to order . Keeping
only n 0,1,
0 1 1 2
x t A A cos t A cos 2 t
( 12. 15 )
and substituting in the equation of motion we have
-2 -1.5 -1 -0.5 0 0.
-0.
0
1
0.
-1 -0.5 0 0.
-0.
0
1 1.
shift) and we see that large energy states (or states that start to the left of the local
maximum in U) are just scattering trajectories.
