Phase flow is the one-parameter group of transformations

In summary: Therefore, [itex]g^{t+s}=g^t\circ g^s[/tex] for all t,s, so [itex]\{g^t\}[/tex] is a group. In summary, the phase flow is a one-parameter group of transformations of phase space, denoted by [itex]\{g^t\}[/tex]. This group is defined by [itex]g^t(q_0,p_0)=(q(t),p(t))[/tex], where [itex]q(t)[/tex] and [itex]p(t)[/tex] are solutions of the Hamilton's system of equations corresponding to initial conditions [itex]q_0[/
  • #1
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The phase flow is the one-parameter group of transformations of phase space

[tex]g^t:({\bf{p}(0),{\bf{q}(0))\longmapsto({\bf{p}(t),{\bf{q}(t)) [/tex],

where [tex]{\bf{p}(t)[/tex] and [tex]{\bf{q}}(t)[/tex] are solutions of the Hamilton's system of equations corresponding to initial condition [tex]{\bf{p}}(0) [/tex]and [tex]{\bf{q}}(0)[/tex].

Show that [tex]\{g^t\}[/tex] is a group.


Can anyone help me prove the composition?

[tex]g^t\circ g^s=g^{t+s}[/tex]
 
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  • #2


On the one hand, [itex]g^{t+s}(p_0,q_0)=(p(t+s),q(t+s))[/itex] where p and q are solutions of the Hamilton's system of equations corresponding to initial condition p_0 and q_0. On the other hand [itex]g^t\circ g^s(p_0,q_0)=g^t(p(s),q(s))=(p'(t),q'(t))[/itex] where p' and q' are solutions of the Hamilton's system of equations corresponding to initial condition p(s) and q(s). To show that (p(t+s),q(t+s))=(p'(t),q'(t)) for all s and t, observe that for s fixed, both sides satisfy the same initial value problem (in the variable t). By uniqueness of the solution to an initial value problem, you get the desired equality.
 
  • #3


quasar987 said:
] To show that (p(t+s),q(t+s))=(p'(t),q'(t)) for all s and t, observe that for s fixed, both sides satisfy the same initial value problem (in the variable t). By uniqueness of the solution to an initial value problem, you get the desired equality.
Can you show that explicitly? I don't follow.

Associativity:

[itex]
( g^t\circ g^s) \circ g^r = g^{(t+s)} \circ g^r = g^{(t+s) +r}
[/itex]

[itex]
g^t\circ (g^s \circ g^r) = g^{t} \circ g^{(s+r)} = g^{t+(s +r)}
[/itex]

Identity,

at t = 0
[tex]
g^t:({\bf{p}(0),{\bf{q}(0))\longmapsto({\bf{p}(0), {\bf{q}(0))
[/tex]

Thus
[itex]
g^0 \circ g^s = g^{(0+s)} =g^s
[/itex] [itex]
g^s \circ g^0 = g^{(s+0)} =g^s
[/itex]Invertible

reverse the sign of Hamiltonian = reverse time, so [tex]g^t[/tex] is invertible and it follows that [tex](g^t)^{-1}=g^{-t}[/tex], so [tex]g^t\circ g^{-1}=g^{t+(-t)} = g^0[/tex] for all [tex]t,s\in\mathbb{R}[/tex].

Is that correct? I'm not sure of the quantifiers are correct for the composition. Should it be t,s > 0 or t,s in R?
 
Last edited:
  • #4


Your last post shows that all the group properties follow easily once you have

[tex]g^t\circ g^s=g^{t+s}[/tex].

I thought this is what you wer trying to show. Is it not?
 
  • #5


Yes.

quasar987 said:
] To show that (p(t+s),q(t+s))=(p'(t),q'(t)) for all s and t, observe that for s fixed, both sides satisfy the same initial value problem (in the variable t). By uniqueness of the solution to an initial value problem, you get the desired equality.
Can you show that explicitly? I don't follow.The remainder of the post was just to see if there was anything wrong.
 
  • #6


For invertibility, perhaps simpler is this:

[tex]g^t\circ g^{-t}=g^{-t}\circ g^{t}=g^0=\mathrm{id}[/tex]

so g^t is invertible with inverse g^{-t}.
 
  • #7


Fix s and set Q(t):=p(t+s), P(t):=p(t+s). Then, by the chain rule and the fact that q,p satisfy Hamilton's equations, we get

[tex]\frac{dQ}{dt}(t) =\frac{d}{dt}q(t+s)=\frac{dq}{dt}(t+s)=\frac{\partial H}{\partial p}(q(t+s),p(t+s))=\frac{\partial H}{\partial p}(Q(t),P(t))[/tex]
[tex]-\frac{dP}{dt}(t) =-\frac{d}{dt}p(t+s)=-\frac{dp}{dt}(t+s)=\frac{\partial H}{\partial q}(Q(t),P(t))[/tex]
[tex]Q(0)=q(0+s)=q(s), \ \ P(0)=p(0+s)=p(s)[/tex]

And by definition of q',p',

[tex]\frac{dq'}{dt}(t)=\frac{\partial H}{\partial p}(q'(t),p'(t))[/tex]
[tex]-\frac{dp'}{dt}(t)=\frac{\partial H}{\partial q}(q'(t),p'(t))[/tex]
[tex]q'(0)=q(s), \ \ p'(0)=p(s)[/tex]

So we see that Q,P satisfies the same initial value problem as q',p', so by uniqueness of the solution of such systems, it must be that Q=q' and P=p'. That is, [itex]g^{t+s}(q_0,p_0)=(g^t\circ g^s)(q_0,p_0)[/tex].
 

What is phase flow?

Phase flow is a mathematical concept used in the study of dynamical systems. It is the one-parameter group of transformations that describes the evolution of a system over time.

What is a one-parameter group of transformations?

A one-parameter group of transformations is a set of transformations that can be described by a single parameter, typically denoted as t. These transformations can be applied to a system or object to describe its changes over time.

How is phase flow related to dynamical systems?

Phase flow is a fundamental concept in the study of dynamical systems. It is used to describe the evolution of a system over time, taking into account factors such as initial conditions and external influences.

What does the one-parameter in one-parameter group of transformations refer to?

The one-parameter in one-parameter group of transformations refers to the single parameter, typically denoted as t, that is used to describe the transformations. This parameter represents time and is used to track the changes in a system over time.

How is phase flow used in scientific research?

Phase flow is used in a variety of scientific fields, including physics, engineering, and mathematics. It is a powerful tool for understanding the behavior of complex systems and is often used in mathematical modeling and simulation studies.

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