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Physics
Classical Physics
Is each path in a phase portrait unique?
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[QUOTE="wrobel, post: 6065343, member: 593228"] 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))## [/QUOTE]
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Is each path in a phase portrait unique?
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