Is each path in a phase portrait unique?

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  • Thread starter crastinus
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  • #1
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As I understand it, the phase space is the space of all possible states, without reference to the rule of transformation from one state to another; whereas the phase portrait is the phase space with the paths from one state to another connected following the system's rule of transformation.

If that is so, then, for a classical system, is there only one path in the phase portrait between two states A and B over a fixed time interval from ti to tj?
 

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  • #2
wrobel
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I believe that the best way to grasp such things is to use theorems.
Consider an initial value problem $$\dot x=v(x),\quad x(0)=\hat x,\quad x=(x^1,\ldots,x^m)\in\mathbb{R}^m.\qquad (*)$$ We also assume for simplicity that the vector field ##v## is smooth in ##\mathbb{R}^m##. This assumption is sufficient to guarantee existence and uniqueness of the solution by the Cauchy theorem. The phase space is ##\mathbb{R}^m##.
Let ##g^t(\hat x)## stand for solution to problem (*)

Theorem. Let ##K\subset \mathbb{R}^m## be any bounded open domain. Then there exists a positive constant ##T## such that the solution ##g^t(\hat x)## is defined for all ##t\in[-T,T]## and for all ##\hat x\in K##. Moreover (this is the answer to your question) for any ##t\in[-T,T]## the mapping ##\hat x\mapsto g^t(\hat x)## is a diffeomorphism between ##K## and ##g^t(K)##.
If ##t,s,t+s\in[-T,T]##then ##g^{t+s}(\hat x)=g^t(g^s(\hat x))##
 
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  • #3
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This is great. Just what I need! I confess that I'm not sure I understand it yet, though. Let me work on it a bit before replying.

Thanks again!
 

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