ME201 Advanced Dynamics (Fall 2007)
HW2 Solution (
100 pts
)
1.
50pts
Consider the system of two oscillating equal masses given by
m
¨
x
1
=

kx
1
+
c
1
(
x
1

x
2
) +
c
2
(
x
1

x
2
)
3
m
¨
x
2
=

kx
2

c
1
(
x
1

x
2
)

c
2
(
x
1

x
2
)
3
(a)
Is this a conservative system?
At first, you may want to integrate the right hand sides to obtain
potentials and see if the time derivative of the Hamiltonian is zero. This works for this system,
but what happens if it doesn’t.
Does that mean that there exists no invariant function for
the dynamics? So we want to analyze this in a way that doesn’t use this unknown potential
function. Another way to think of this is that all conservative systems do not necessarily have the
symmetries of a Hamiltonian system (or are not Hamiltonian). So trying to show that a system
is Hamiltonian, to show that it is conservative, may not be the best approach. From Arnold’s
book on mechanics we have that a system if conservative if there exists a function
U
:
E
2
→
R
such that
f
=

∂U
∂
x
where both
f
and
x
are vectors. This is not directly helpful as we don’t
have the function
U
that is being discussed here. However, we can do some manipulations to
obtains something useful. From the definition we have:
f
1
=

∂U
(
x
1
, x
2
)
∂x
1
f
2
=

∂U
(
x
1
, x
2
)
∂x
2
To simplify things, we can take the partial of the first equation with respect to
x
2
and the
second with respect to
x
1
. We then use the fact that for a continuously differential
U
(
x
1
, x
2
),
∂
2
U
(
x
1
,x
2
)
∂x
1
∂x
2
=
∂
2
U
(
x
1
,x
2
)
∂x
2
∂x
1
to get a simple relation that does not involve U.
∂f
1
∂x
2
=
∂
∂x
2

∂U
(
x
1
, x
2
)
∂x
1
¶
∂f
2
∂x
1
=
∂
∂x
1

∂U
(
x
1
, x
2
)
∂x
2
¶
⇒
∂f
1
∂x
2
=
∂f
2
∂x
1
The last equality is an equality that is easiest route to show that the system is conservative.
As an aside, a conservative system admits closed contours in its phase space.
Therefore, the
contour integral on this space should equal zero. That is:
I
(
f
1
dx
1
+
f
2
dx
2
) = 0
By Green’s theorem we have
I
(
f
1
dx
1
+
f
2
dx
2
) =
Z Z
∂f
2
∂x
1

∂f
1
∂x
2
¶
dx
1
dx
2
= 0
ME201 Advanced Dynamics
1
HW2 Solution
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This is the same result. Now plugging in our actual equations:
∂f
1
∂x
2
=

c
1

3(
x
1

x
2
)
2
∂f
2
∂x
1
=

c
1

3(
x
1

x
2
)
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 Fall '08
 Mezic,I
 Equations, Nonlinear system, phase space

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