Student Seminar: Classical and Quantum Integrable Systems

Thus, the combination ˙θ 2 − 2mg
L
cos θ is an integral of motion. In fact, this is nothing
else as the total energy. Indeed, the total energy is (up to an additive constant which
can be always added)
E = m⃗v2
2 + U = mL2 ˙θ 2 + mgL(1 − cos θ) .
2
We rewrite the conservation law in the form
L 2 ˙θ2 = 2gh − 4gL sin 2 θ 2 ,
where h is an integration constant. Making the change of variables y = sin θ 2
arrive at
ẏ 2 = g ( h
)
L (1 − y2 )
2L − y2 .
We have now several cases to consider
we
• Under the oscillatory motion the point does not reach the top of a circle. This
h
means that ẏ terns to zero for some y < 1. Thus, < 1. Denoting h = 2L 2Lk2 ,
where k is a positive constant less then one we obtain
) ( )
ẏ 2 =
(1 gk2 − k 2 y2
1 − y2
.
L k 2 k 2
Solution to this equation is
( √ g
)
y = k sn
L (t − t 0), k .
The integration constants are√t 0 and k, they are determined from the initial
L
conditions. the period is T = K(k). g
• Rotatory motion. Here h > 2L. Thus, taking 2L = hk 2 we will have k 2 < 1.
Equation becomes
ẏ 2 =
g
Lk (1 − 2 y2 )(1 − k 2 y 2 )
whose solution is
( √ g t − t
)
0
y = sn
, k .
L k
• The point reaches the top. Here h = 2L and we get
ẏ 2 = g L (1 − y2 ) 2 → ẏ =
√ g
L (1 − y2 ) .
Solution is
(√ ) g
y = tanh
L (t − t 0) .
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