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From Wikipedia

Given an initial value problem $$ \dot{\boldsymbol{x}}=\boldsymbol{v}(t,\boldsymbol{x}) $$ $$ \boldsymbol{x}|_{{t=0}}=\boldsymbol{x}_0 $$

The solution is an evolution function $$ \boldsymbol{{x}}(t)=\Phi(t,\boldsymbol{{x}}_0) $$ Some formal manipulation of the system of differential equations shown above gives a more general form of equations a dynamical system must satisfy $$ \dot{\boldsymbol{x}}-\boldsymbol{v}(t,\boldsymbol{x})=0 \qquad\Leftrightarrow\qquad \mathfrak{{G}}\left(t,\Phi(t,\boldsymbol{{x}}_0)\right)=0 $$ where $\scriptstyle\mathfrak{G}:{{(T\times M)}^M}\to\mathbf{C}$ is a functional from the set of evolution functions to the field of the complex numbers.

I was wondering why the domain of $\scriptstyle\mathfrak{G}$ is ${(T\times M)}^M$? I thought it was $T\times M$, since $\Phi(t, x_0) \in M$.

What is $\scriptstyle\mathfrak{G}$? How can it be determined from the initial value problem?

Thanks!

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Each flow $\Phi: T \times M \rightarrow M$, and the domain of that functional is all such flows, and so the set of such flows can be regarded as $(T \times M)^M$. – Christopher A. Wong Oct 18 '12 at 1:17
(1) Isn't the set of such flows $M^{T \times M}$ instead? (2) The domain of $\scriptstyle\mathfrak{G}$ doesn't look like the set of such flows, by looking at $ \mathfrak{{G}}\left(t,\Phi(t,\boldsymbol{{x}}_0)\right)=0 $. – Tim Oct 18 '12 at 1:22
Gosh, sorry, I was confusing myself with something else. – Christopher A. Wong Oct 18 '12 at 1:36

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