First of all, I must say I have a physics background. I am beggining the read of the Dynamical Systems series by Arason, Arnold et al. (Springer) and, although I can understand seemingly more complex concepts I am stuck with this one.


In the first chapter (definitions), a couple of definitions go like this, and a theorem follows immidately:

A one-parameter group $\{g^t\}$ of diffeomorphisms of a domain $U$ is a map of the direct product $\mathbb{R}\times U$ into $U$ (here $U$ is the phase space and $\mathbb{R}\times U$ is the extended phase space):

$$\begin{matrix} g:\mathbb{R}\times U \to U \\ g(t, x) = g^t(x), \space t\in\mathbb{R}, \space x\in U \end{matrix}$$

such that

  1. g is differentiable (here they mean continuous derivatives of all orders)
  2. the family $\{g^t, t\in\mathbb{R}\}$ is a one-parameter group of transformations; that is, $g^t \circ g^s = g^{t+s}, \space \forall t,s\in \mathbb{R}$ and $g^0$ is the identity.
  3. (implied by 1&2) $\forall t \in \mathbb{R}$, $g^t$ is a diffeomorphism.

$\{g^t\}$ is also called a phase flow.

The phase velocity field of a flow $\{g^t\}$ is the vector field

$$v(x) = \frac{d(g^tx)}{dt} \mid_{t=0}$$

which is also called the generating field of $\{g^t\}$.

Theorem. The motion of a point under the action of a phase flow $\varphi(t) = g^t(x)$ is a solution of the autonomous system $\dot{x} = v(x)$ defined by $v(x)$.

So far, so good. Now, they note that it may be possible to construct $\{g^t\}$ for which $v(x)$ is the phase-velocity field. However, it is not always the case, and they give the example in 1D: $\dot{x} = -x^2$. Apparently, it is easy to compute that the only possible field would be $g^tx = \frac{x}{1-tx}$, and it satisfies the group property $g^t \circ g^s = g^{t+s}$ but should be defined in $\mathbb{R}\cup\{\infty\}$ instead of $\mathbb{R}$.

My problem

Okay, I get that if this is the field, it should be defined as such. My problem is: I really don't know how to compute this field from the differential equation and see that it is the only possibility of a phase flow for this system. Which makes me think about how well I am understanding the concept on itself. If someone can give insight in how to prove that, or further understaning (more intuitive maybe) of this concept it would be much appreciated.

(Apologies if it was such a long text for such a small question). And thanks in advance!

  • $\begingroup$ It's not clear what the question is. You need to distinguish between the velocity field $v(x)$ and the trajectories of the solutions $g^tx$ (whose derivative matches the velocity field). Which field do you mean in the question? The $x$ in $g^tx$ is the starting point. $\endgroup$ Commented Mar 6, 2019 at 18:51
  • $\begingroup$ So, my exact problem is to compute the phase flow $g^tx$ for the example that they give (which is actually not a phase flow as discussed afterwards). And you say that the x in $g^tx$ denotes the initial condition? $\endgroup$ Commented Mar 6, 2019 at 19:08

1 Answer 1


I don't know if I am too late here. Let me make a few comments. Yours is not a question, but, perhaps, about how to think of the system.

For me, the best way to see this is to first solve the equation $\frac{dx}{dt}=x^2$ (the sign you gave is not correct!!)with $x_0=x(0)$. This gives $x(t)=\frac{x_0}{1-tx_0}$ . Now think of $x(t)$ as $g^tx_0$, that is $x(t)=g^tx_0$. Then, $g^tx=\frac{x}{1-tx}$. This is simply a function $g$ with parameter $t$ on $x\in\mathbb{R}$, denoted $\phi(t)$ there.

In order to qualify as a flow, you need $\phi(t)$ to be a diffeomorphism over the whole $\mathbb{R}\ni t$. They show that $g^t$ fails here.

Hope this helps.


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