This seems to be a somewhat common theme among users on here trying to start reading Arnold's ODE book. I have found a few similar questions but none have sufficiently answered my problems.

Specifically I am having a hard time understanding the phase velocity,

$v(x) = \dot{x} = \frac{d}{dt}|_{t=0}g^tx$.

According to this post, Arnold's ODE computation of phase velocity, it is suggested that in the notation of $g^tx$, $x$ is considered a constant or equal to the initial value $x_0$.

But how do I then reconcile this fact with the expression $\dot{x}=\frac{d}{dt}|_{t=0}g^tx$? In my simple world $\dot{x}=\frac{d}{dt}g^tx$ would make a lot more sense. I find this alteration to be especially fitting considering the examples put forth later, e.g. draw the integral curves of $v(x) = \dot{x} = -kx$.

What am I missing here? Help would be greatly appreciated as I am reading this in my spare time and have no access to feedback other than from sources such as this site. Thank you!

Edit by request: I have the Richard A. Silverman translated edition and the definition is found in subsection 1.4 on Vector Fields in Sec. 1 Phase Spaces and Phase Flows on page 7.

  • 1
    $\begingroup$ It would help if you gave the chapter and section in Arnold where this is. I agree with what you wrote as an interpretation. $g^t x = g(t,x)$ is a function which maps out a curve for each $x$, the curve being a mapping of $t \in [0,a]$ a real interval. It's the same as saying $g$ is a one parameter family of maps on wherever $x$ lives -- i.e. a flow. $\endgroup$ – hkr Mar 29 '17 at 4:43

The answer lies in the one-parameter family of diffeomorphisms $\{g^{t}\}$ and that $x$ denotes a point in a manifold $M$. The action of $g^{t}$ on $x$ just moves the point around. Consider the following example: let $M=\mathbb{R}^{2}$ and define

$$\begin{align} g^{t}:M &\,\longrightarrow\, M \\ (x,y) &\,\longmapsto g^{t}(x,y) \,=\, (x+t,y) \end{align}$$

Given any point in $M$, $g^{t}$ just translates it to another point. For example, the point $(1,2)\in M$ gets translated to the point $(1+t,2)$ for some value of $t\in\mathbb{R}$. Since $t$ can actually vary over any value of $t$, this actually describes a curve: the horizontal line at $y=2$. Importantly, notice that when $t=0$ we return the point we started with: $g^{0}(x,y)=(x,y)$ and hence $g^{0}$ is the identity transformation. This is an important property of these groups of transformations (and by group I really mean group in the abstract algebraic sense!).

Now, back to the velocity. The application of $g^{t}$ on a point $x\in M$ produces a curve; we now wish to find the velocity of the curve at the point $x$. To do this, is is natural to differentiate the curve with respect to the curve parameter, which in this case is $t$:

$$v \,=\, \frac{d}{dt} \;g^{t}x$$

But this gives the velocity field on all points of the curve. To find the velocity field at the point $x\in M$ on the curve, we must evaluate the derivative at $t\in\mathbb{R}$ which returns the point $x\in M$ which we are interested in. But we know this occurs at $t=0$ as $g^{0}$ induces the identity ($g^{0}x=x$), so we then evaluate the above derivative at $t=0$:

$$v(x) \,=\, \left.\frac{d}{dt}\right|_{t=0} \;g^{t}x.$$

The notation means the velocity $v(x)$ at the point $x\in M$ is given by the derivative of the curve $g^{t}x\in M$ evaluated at $t=0$.

Hope this helps!

  • $\begingroup$ Thank you very much! It makes much more sense to me now. But for a differential equation $\dot{x}=kx$, the t advance mapping must still be the solution, ie $g^tx=Ce^{kt}$? $\endgroup$ – plebmatician Jun 8 '17 at 8:27

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