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definition.

The phase flow of the differential equation $\dot{x}=\vec{v}\ (x)$ is the one-parameter diffeomorphism group for which $\vec{v}$ is the phase velocity vector field, namely, $$ \vec{v}=\frac{d}{dt} \Big|_{t=0} (g^tx) $$

In the book, for the problem to find the phase flow of $\dot{x}=x-1$, the provided answer, $g^tx=(x-1)e^t+1$, is easy to verify. However, I have not idea to solve this problem. Any helps?

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    $\begingroup$ $\dot x=x-1$ makes $x$ look like a scalar. $\dot x=v(x)$ makes $x$ look like a vector. Can you clarify? $\endgroup$
    – robjohn
    Aug 16, 2019 at 8:05
  • $\begingroup$ @Mattos Yeah, since in the book it is defined as the $phase\ velocity\ vector\ field$, I use what the book indicates. $\endgroup$
    – Tab1e
    Aug 16, 2019 at 13:03
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    $\begingroup$ If $x$ is a vector, how can $\dot x=x-1$ since $1$ is not a vector? $\endgroup$
    – robjohn
    Aug 16, 2019 at 15:32
  • $\begingroup$ @robjohn It is just a notation in the definition, and the question is stated separately. You may assume it is a scalar and assume $v$ is also a scalar too. $\endgroup$
    – Tab1e
    Aug 17, 2019 at 2:34

3 Answers 3

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You seem to have a notation ambiguity. In $g^tx$ the $g^t$ is the group or semi-group action on the real line, not a power of some number $g$ with exponent $t$. What it means is $g^t(x)$ or $g(t,x)$. There is nothing more to do.

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  • $\begingroup$ Exactly, in the book, it says that it is a one-parameter group of transformation of a set, but how can I find this mapping tho? $\endgroup$
    – Tab1e
    Aug 16, 2019 at 13:21
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It is not that simple, $x$ is a function of $t$, so just integrating on both sides won't work. To solve $\dot x=x-1$ you could use separation of variables: $\int \dot x/(x-1)dt=t$ which, after substituting $u$ for $x-1$ becomes $$t=\int_{x_0}^{x(t)}\frac{1}{u}dt=\ln\left(x(t)-1\right)-\ln(x_0)$$ Therefore $$x(t)=x_0 e^t+1$$

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    $\begingroup$ Why did you leave out the integration constant? It is quite important for a flow function to have the initial value as a variable. $\endgroup$ Aug 16, 2019 at 7:28
  • $\begingroup$ I am sorry but your method for solving ordinary differential equation $\dot{x}-x=-1 (1)$ misses an arbitrary constant (moreover it does not necessitate integration).This ODE has associated homogeneous equation $\dot{x}-x=0$ whose evident general solution is $x=Ae^t$ ; therefore, as a particular solution of (1) is $1$, one adds it to the general solution found above, giving $x(t)=1+Ae^t$ $\endgroup$
    – Jean Marie
    Aug 16, 2019 at 7:37
  • $\begingroup$ Thanks for answering, can you please indicate what I should do next to find the phase flow? $\endgroup$
    – Tab1e
    Aug 16, 2019 at 7:50
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The proceeding is the following (from Example 2, pag 65, Ordinary Differential Equations (3ed) - Arnol'd):

The phase velocity vector field of the ODE is: $\vec{v}(x)= x-1$, then the solution of the equation $\dot{x} = \vec{v}(x)$ with initial condition $x_0$ for $t=0$ can easily be found explicitly:

\begin{align} \frac{dx}{dt} &= x -1 \\ \frac{dx}{x-1}&= dt \\ \int \frac{1}{x-1} dx &= \int dt \\ \ln(x-1) &= t+ C \end{align}

We left the integration constant of both sides on the right side.

Then we know that for $t=0$ we have $x(t=0)=x_0$. Thus, evaluating this in the last expression, we obtain:

\begin{align} C = \ln(x_0 -1) \end{align}

Now we have:

\begin{align} \ln(x-1) &= t+ \ln(x_0 -1) \end{align}

rearranging

\begin{align} \ln(x-1)-\ln(x_0 -1) &= t \\ \ln\left(\frac{x-1}{x_0-1} \right) &= t\\ \frac{x-1}{x_0-1} &= e^t \\ x &= (x_0-1)e^t + 1 \end{align}

Now, the crucial final step is to identify correctly in the last expression the phase flow $g^t$. For this, we can think the right side as the motion under the action of the phase flow $g^t$ of a fixed point $x_0$ of the phase space, i.e, $g^tx_0$. Then¸ we can say that:

\begin{align} g^tx = (x-1)e^t + 1 \end{align}

(pag 64, Ordinary Differential Equations (3ed) - Arnol'd):

In other words, under the action of the phase flow ($g^t$) the phase point ($x$) moves so that its velocity vector at any instant equals the phase velocity vector at the point of the phase space at which the moving point is located.

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