# Is each path in a phase portrait unique?

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## Main Question or Discussion Point

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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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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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!