1Integrability of ODE's

II Integrable Systems



1.4 The Arnold-Liouville theorem
We now get to the Arnold-Liouville theorem. This theorem says that if a
Hamiltonian system is integrable, then we can find a canonical transformation
(
q, p
)
7→
(
Q, P
) such that
˜
H
depends only on
P
. If this happened, then
Hamilton’s equations reduce to
˙
Q =
˜
H
P
,
˙
P =
˜
H
Q
= 0,
which is pretty easy to solve. We find that
P
(
t
) =
P
0
is a constant, and since
the right hand side of the first equation depends only on
P
, we find that
˙
Q
is
also constant! So Q = Q
0
+ t, where
Ω =
˜
H
P
(P
0
).
So the solution just falls out very easily.
Before we prove the Arnold-Liouville theorem in full generality, we first see
how the canonical transformation looks like in a very particular case. Here we
will just have to write down the canonical transformation and see that it works,
but we will later find that the Arnold-Liouville theorem give us a general method
to find the transformation.
Example. Consider the harmonic oscillator with Hamiltonian
H(q, p) =
1
2
p
2
+
1
2
ω
2
q
2
.
Since is a 2-dimensional system, so we only need a single first integral. Since
H
is a first integral for trivial reasons, this is an integrable Hamiltonian system.
We can actually draw the lines on which
H
is constant they are just
ellipses:
q
q
We note that the ellipses are each homeomorphic to
S
1
. Now we introduce the
coordinate transformation (q, p) 7→ (φ, I), defined by
q =
r
2I
ω
sin φ, p =
2Iω cos φ,
For the purpose of this example, we can suppose we obtained this formula
through divine inspiration. However, in the Arnold-Liouville theorem, we will
provide a general way of coming up with these formulas.
We can manually show that this transformation is canonical, but it is merely
a computation and we will not waste time doing that. IN these new coordinates,
the Hamiltonian looks like
˜
H(φ, I) = H(q(φ, I), p(φ, I)) = ωI.
This is really nice. There is no φ! Now Hamilton’s equations become
˙
φ =
˜
H
I
= ω,
˙
I =
˜
H
φ
= 0.
We can integrate up to obtain
φ(t) = φ
0
+ ωt, I(t) = I
0
.
For some unexplainable reason, we decide it is fun to consider the integral along
paths of constant H:
1
2π
I
p dq =
1
2π
Z
2π
0
p(φ, I)
q
φ
dφ +
q
I
dI
=
1
2π
Z
2π
0
p(φ, I)
q
φ
dφ
=
1
2π
Z
2π
0
r
2I
ω
2Iω cos
2
φ dφ
= I
This is interesting. We could always have performed the integral
1
2π
H
p
d
q
along
paths of constant
H
without knowing anything about
I
and
φ
, and this would
have magically gave us the new coordinate I.
There are two things to take away from this.
(i) The motion takes place in S
1
(ii) We got I by performing
1
2π
H
p dq.
These two ideas are essentially what we are going to prove for general Hamiltonian
system.
Theorem
(Arnold-Liouville theorem)
.
We let (
M, H
) be an integrable 2
n
-
dimensional Hamiltonian system with independent, involutive first integrals
f
1
, ··· , f
n
, where f
1
= H. For any fixed c R
n
, we set
M
c
= {(q, p) M : f
i
(q, p) = c
i
, i = 1, ··· , n}.
Then
(i) M
c
is a smooth
n
-dimensional surface in
M
. If
M
c
is compact and
connected, then it is diffeomorphic to
T
n
= S
1
× ··· × S
1
.
(ii)
If
M
c
is compact and connected, then locally, there exists canonical coor-
dinate transformations (
q, p
)
7→
(
φ, I
) called the action-angle coordinates
such that the angles
{φ
k
}
n
k=1
are coordinates on
M
c
; the actions
{I
k
}
n
k=1
are first integrals, and
H
(
q, p
) does not depend on
φ
. In particular,
Hamilton’s equations
˙
I = 0,
˙
φ =
˜
H
I
= constant.
Some parts of the proof will refer to certain results from rather pure courses,
which the applied people may be willing to just take on faith.
Proof sketch.
The first part is pure differential geometry. To show that
M
c
is
smooth and
n
-dimensional, we apply the preimage theorem you may or may not
have learnt from IID Differential Geometry (which is in turn an easy consequence
of the inverse function theorem from IB Analysis II). The key that makes this
work is that the constraints are independent, which is the condition that allows
the preimage theorem to apply.
We next show that
M
c
is diffeomorphic to the torus if it is compact and
connected. Consider the Hamiltonian vector fields defined by
V
f
i
= J
f
i
x
.
We claim that these are tangent to the surface
M
c
. By differential geometry, it
suffices to show that the derivative of the
{f
j
}
in the direction of
V
f
i
vanishes.
We can compute
V
f
i
·
x
f
j
=
f
j
x
J
f
i
x
= {f
j
, f
i
} = 0.
Since this vanishes, we know that
V
f
i
is a tangent to the surface. Again by
differential geometry, the flow maps
{g
i
}
must map
M
c
to itself. Also, we know
that the flow maps commute. Indeed, this follows from the fact that
[V
f
i
, V
f
j
] = V
{f
i
,f
j
}
= V
0
= 0.
So we have a whole bunch of commuting flow maps from M
c
to itself. We set
g
t
= g
t
1
1
g
t
2
2
···g
t
n
n
,
where t R
n
. Then because of commutativity, we have
g
t
1
+t
2
= g
t
1
g
t
2
.
So this is gives a group action of
R
n
on the surface
M
c
. We fix
x M
c
. We
define
stab(x) = {t R
n
: g
t
x = x}.
We introduce the map
φ :
R
n
stab(x)
M
c
given by
φ
(
t
) =
g
t
x
. By the orbit-stabilizer theorem, this gives a bijection
between
R
n
/ stab
(
x
) and the orbit of
x
. It can be shown that the orbit of
x
is
exactly the connected component of
x
. Now if
M
c
is connected, then this must
be the whole of
x
! By general differential geometry theory, we get that this map
is indeed a diffeomorphism.
We know that
stab
(
x
) is a subgroup of
R
n
, and if the
g
i
are non-trivial, it can
be seen (at least intuitively) that this is discrete. Thus, it must be isomorphic
to something of the form Z
k
with 1 k n.
So we have
M
c
=
R
n
/ stab(x)
=
R
n
/Z
k
=
R
k
/Z
k
× R
nk
=
T
k
× R
nk
.
Now if
M
c
is compact, we must have
n k
= 0, i.e.
n
=
k
, so that we have no
factors of R. So M
c
=
T
n
.
With all the differential geometry out of the way, we can now construct the
action-angle coordinates.
For simplicity of presentation, we only do it in the case when
n
= 2. The
proof for higher dimensions is entirely analogous, except that we need to use
a higher-dimensional analogue of Green’s theorem, which we do not currently
have.
We note that it is currently trivial to re-parameterize the phase space with
coordinates (
Q, P
) such that
P
is constant within the Hamiltonian flow, and
each coordinate of
Q
takes values in
S
1
. Indeed, we just put
P
=
c
and use the
diffeomorphism
T
n
=
M
c
to parameterize each
M
c
as a product of
n
copies of
S
n
. However, this is not good enough, because such an arbitrary transformation
will almost certainly not be canonical. So we shall try to find a more natural
and in fact canonical way of parametrizing our phase space.
We first work on the generalized momentum part. We want to replace
c
with something nicer. We will do something analogous to the simple harmonic
oscillator we’ve got.
So we fix a
c
, and try to come up with some numbers
I
that labels this
M
c
.
Recall that our surface M
c
looks like a torus:
Up to continuous deformation of loops, we see that there are two non-trivial
“single” loops in the torus, given by the red and blue loops:
More generally, for an n torus, we have n such distinct loops Γ
1
, ··· , Γ
n
. More
concretely, after identifying M
c
with S
n
, these are the loops given by
{0} × ··· × {0} × S
1
× {0} × ··· × {0} S
1
.
We now attempt to define:
I
j
=
1
2π
I
Γ
j
p · dq,
This is just like the formula we had for the simple harmonic oscillator.
We want to make sure this is well-defined — recall that Γ
i
actually represents
a class of loops identified under continuous deformation. What if we picked a
different loop?
Γ
0
2
Γ
2
On M
c
, we have the equation
f
i
(q, p) = c
i
.
We will have to assume that we can invert this equation for
p
locally, i.e. we can
write
p = p(q, c).
The condition for being able to do so is just
det
f
i
p
j
6= 0,
which is not hard.
Then by definition, the following holds identically:
f
i
(q, p(q, c)) = c
i
.
We an then differentiate this with respect to q
k
to obtain
f
i
q
k
+
f
i
p
`
p
`
q
k
= 0
on M
c
. Now recall that the {f
i
}’s are in involution. So on M
c
, we have
0 = {f
i
, f
j
}
=
f
i
q
k
f
j
p
k
f
i
p
k
f
j
q
k
=
f
i
p
`
p
`
q
k
f
j
p
k
f
i
p
k
f
j
p
`
p
`
q
k
=
f
i
p
k
p
k
q
`
f
j
p
`
f
i
p
k
f
j
p
`
p
`
q
k
=
f
i
p
k
p
`
q
k
p
k
q
`
f
j
p
`
.
Recall that the determinants of the matrices
f
i
p
k
and
f
j
p
`
are non-zero, i.e. the
matrices are invertible. So for this to hold, the middle matrix must vanish! So
we have
p
`
q
k
p
k
q
`
= 0.
In our particular case of
n
= 2, since
`, k
can only be 1
,
2, the only non-trivial
thing this says is
p
1
q
2
p
2
q
1
= 0.
Now suppose we have two “simple” loops Γ
2
and Γ
0
2
. Then they bound an area
A:
Γ
0
2
Γ
2
A
Then we have
I
Γ
2
I
Γ
0
2
!
p · dq =
I
A
p · dq
=
ZZ
A
p
2
q
1
p
1
q
2
dq
1
dq
2
= 0
by Green’s theorem.
So I
j
is well-defined, and
I = I(c)
is just a function of
c
. This will be our new “momentum” coordinates. To figure
out what the angles
φ
should be, we use generating functions. For now, we
assume that we can invert I(c), so that we can write
c = c(I).
We arbitrarily pick a point x
0
, and define the generating function
S(q, I) =
Z
x
x
0
p(q
0
, c(I)) · dq
0
,
where
x
= (
q, p
) = (
q, p
(
q, c
(
I
))). However, this is not a priori well-defined,
because we haven’t said how we are going to integrate from
x
0
to
x
. We are going
to pick paths arbitrarily, but we want to make sure it is well-defined. Suppose
we change from a path γ
1
to γ
2
by a little bit, and they enclose a surface B.
x
0
x
γ
2
γ
1
B
Then we have
S(q, I) 7→ S(q, I) +
I
B
p · dq.
Again, we are integrating p · dq around a boundary, so there is no change.
However, we don’t live in flat space. We live in a torus, and we can have a
crazy loop that does something like this:
Then what we have effectively got is that we added a loop (say) Γ
2
to our path,
and this contributes a factor of 2
πI
2
. In general, these transformations give
changes of the form
S(q, I) 7→ S(q, I) + 2πI
j
.
This is the only thing that can happen. So differentiating with respect to
I
, we
know that
φ =
S
I
is well-defined modulo 2
π
. These are the angles coordinates. Note that just like
angles, we can pick
φ
consistently locally without this ambiguity, as long as we
stay near some fixed point, but when we want to talk about the whole surface,
this ambiguity necessarily arises. Now also note that
S
q
= p.
Indeed, we can write
S =
Z
x
x
0
F · dx
0
,
where
F = (p, 0).
So by the fundamental theorem of calculus, we have
S
x
= F.
So we get that
S
q
= p.
In summary, we have constructed on M
c
the following: I = I(c), S(q, I), and
φ =
S
I
, p =
S
q
.
So
S
is a generator for the canonical transformation, and (
q, p
)
7→
(
φ, I
) is a
canonical transformation.
Note that at any point
x
, we know
c
=
f
(
x
). So
I
(
c
) =
I
(
f
) depends on the
first integrals only. So we have
˙
I = 0.
So Hamilton’s equations become
˙
φ =
˜
H
I
,
˙
I = 0 =
˜
H
φ
.
So the new Hamiltonian depends only on I. So we can integrate up and get
φ(t) = φ
0
+ t, I(t) = I
0
,
where
Ω =
˜
H
I
(I
0
).
To summarize, to integrate up an integrable Hamiltonian system, we identify
the different cycles Γ
1
, ··· , Γ
n
on M
c
. We then construct
I
j
=
1
2π
I
Γ
j
p · dq,
where p = p(q, c). We then invert this to say
c = c(I).
We then compute
φ =
S
I
,
where
S =
Z
x
x
0
p(q
0
, c(I)) · dq
0
.
Now we do this again with the Harmonic oscillator.
Example. In the harmonic oscillator, we have
H(q, p) =
1
2
p
2
+
1
2
ω
2
q
2
.
We then have
M
c
=
(q, p) :
1
2
p
2
+
1
2
ω
2
q
2
= c
.
The first part of the Arnold-Liouville theorem says this is diffeomorphic to
T
1
=
S
1
, which it is! The next step is to pick a loop, and there is an obvious
one the circle itself. We write
p = p(q, c) = ±
p
2c ω
2
q
2
on M
c
. Then we have
I =
1
2π
Z
p · dq =
c
ω
.
We can then write c as a function of I by
c = c(I) = ωI.
Now construct
S(q, I) =
Z
x
x
0
p(q
0
, c(I)) dq
0
.
We can pick x
0
to be the point corresponding to θ = 0. Then this is equal to
Z
q
0
p
2ωI ω
2
q
02
dq
0
.
To find φ, we need to differentiate this thing to get
φ =
S
I
= ω
Z
q
0
dq
0
p
2ωI ω
2
q
02
= sin
1
r
ω
2I
q
As expected, this is only well-defined up to a factor of 2
π
! Using the fact that
c = H, we have
q =
r
2π
ω
sin φ, p =
2Iω cos φ.
These are exactly the coordinates we obtained through divine inspiration last
time.