0
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ $\dot x=x-1$ 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
  • 2
    $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

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 $$x(t)=x_0 e^t+1$$

$\endgroup$
3
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.