Perturbation treatment of the anharmonic oscillator
Suppose we have a third order non-linear term in the equation of motion of a particular
system such as an oscillating pendulum:
ψ̈ + ω02 ψ − cψ 3 = 0
g
1
ω02 = , c = ω02
l
6
If the anharmonic term ψ 3 were ignored the solution would be in the form
ψ ' sin ω0 t.
Since now the equation of motion involves the cube of the displacement we seek a trial
solution in an additive form written as a superposition of two different motions, one in
sin ωt, other in sin 3ωt.
It should be clear that sin 3ωt has been included as suggested by
sin3 x =
3
1
sin x − sin 3x.
4
4
In fact, the solution of the unperturbed part is in the form sin ωt the cube of which
generates sin 3ωt and the differential equation takes the form
A sin ωt + B sin 3ωt = 0
which cannot be satisfied for any t unless A and B are identically zero. Thus, so as to
cancel the term sin 3ωt generated by ψ 3 we are forced to add to sin ωt a term such as
sin 3ωt.
Going further, the new sin 3ωt in the trial solution generates a term in 3 sin 9ωt, and
so on. However, if 1 the series converges very rapidly. Therefore, we retain the first
order correction and ignore terms like 3 , 9 , etc.
We thus write an approximate solution of the form
ψ = ψ0 {sin ωt + sin 3ωt}
where is a dimensionless constant much less than unity for when | ψ0 | 1.
Substituting the trial solution into the equation of motion:
ψ̈ = −ω 2 ψ0 sin ωt − 9ω 2 ψ0 sin 3ωt
ω02 ψ = ω02 ψ0 sin ωt + ω02 ψ0 sin 3ωt
1
1
1
1
− ω02 ψ 3 = − ω02 ψ03 sin ωt + ω02 ψ03 sin 3ωt − ω02 ψ03 sin2 ωt sin 3ωt
6
8
24
2
Equating the respective coefficients of sin ωt and sin 3ωt we obtain
1
−ω 2 + ω02 − ω02 ψ02 = 0
8
or
1
ω 2 = ω02 {1 − ψ02 }
8
1
ω ' ω0 {1 − ψ02 }
16
and further
−9ω 2 + ω02 +
1 2 2
ω ψ =0
24 0 0
1
Setting ω ' ω0 , we obtain
=
1 2
ψ
192 0
which is a measure of the fractional contribution of the anharmonic effect to the solution
of the unperturbed equation of motion. For ψ0 = 0.3 rad, we have ' 0.001, which
indeed is very small.
The reason why a term in sin 2ωt were not included in the trial solution is that if a solution
of the form
ψ = ψ0 {sin ωt + η sin 2ωt}
were tried we would find η = 0. The pendulum generates chiefly third harmonics, i.e.,
terms in sin 3ωt, and no terms in sin 2ωt. The situation would be different for a device
for which the equation of motion included a term in ψ 2 . In such a case the solution will
have a term in sin 2ωt, and the same technique can be used.
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