- #1

- 78

- 9

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

*t*to

_{i}*t*?

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In summary, the conversation discusses the concept of phase space and phase portrait, with the former being the space of all possible states and the latter being the phase space with connected paths following the system's rule of transformation. For a classical system, there is only one path in the phase portrait between two states over a fixed time interval. To understand this concept, the conversation suggests using theorems and provides one for initial value problems with a smooth vector field. The theorem states that for any bounded open domain, there exists a constant time interval where the solution is defined and the mapping between the domain and its transformed state is a diffeomorphism.

- #1

- 78

- 9

If that is so, then, for a classical system, is there only one path in the phase portrait between two states

- #2

Science Advisor

- 1,104

- 945

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))##

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))##

Last edited:

- #3

- 78

- 9

Thanks again!

A phase portrait is a graphical representation of the qualitative behavior of a dynamical system. It shows the trajectories, or paths, that the system can take as it evolves over time.

A phase portrait provides a visual representation of the behavior of a dynamical system, making it easier to understand and analyze. It can reveal important features such as stability, periodicity, and attractors.

No, not all paths in a phase portrait are unique. Some paths may overlap or intersect with each other, while others may follow the same trajectory but in reverse direction.

The uniqueness of a path in a phase portrait depends on the specific conditions and parameters of the dynamical system. To determine if a path is unique, we can analyze the equations that govern the system and look for solutions that yield the same trajectory.

The uniqueness of paths in a phase portrait can be influenced by the initial conditions, parameters of the system, and the behavior of the system itself. Nonlinear systems, in particular, can exhibit complex and non-unique paths in their phase portraits.

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