Here are the preliminaries of my question:

Let $\Omega$ be a compact metric space, and suppose a flow $(\Omega,\mathbb{R})$ is given. Let $A\colon\Omega\to M^2$ be continuous. Here $M^2$ is the set of the real $2\times 2$-matrices. If $\omega\in\Omega$ and $x_0\in\mathbb{R}^2$, define $(\omega,x_0)\cdot t=(\omega\cdot t,x(t))$, where $x(t)$ is the solution to $$ E(\omega):~~~\dot{x}=A(\omega\cdot t)x, $$ satisfying $x(0)=x_0$. (We read $E(\omega)$ as "the equation corresponding to $\omega$.") Now write $A\colon\Omega\to M^2$ $$ A(\omega)=\begin{bmatrix}a(\omega) & -b(\omega)\\b(\omega) & a(\omega)\end{bmatrix}+\begin{bmatrix}\delta(\omega) & \epsilon(\omega)\\\epsilon(\omega) & -\delta(\omega).\end{bmatrix} $$ In polar coordinates $r,\theta$, equation $E(\omega)$ becomes $$ \frac{d}{dt}\theta(t)=b(\omega t)+\epsilon(\omega t)\cos 2\theta(t)-\delta(\omega t)\sin 2\theta(t),~~~~~(3)\\ \frac{d}{dt}\ln r(t)=a(\omega t)+\delta(\omega t)\cos 2\theta(t)+\epsilon(\omega t)\sin 2\theta(t).~~~(4) $$ Moreover define $$ \sigma\colon\Omega\times S^1\to\Sigma, $$ whereat $S^1=\left\{x\in\mathbb{R}^2 : \lVert x\rVert =1\right\}$ and $\Sigma=\Omega\times P^1$. Let $P^1$ be obtained from $S^1$ by identifying antipodal points. Let $\Psi$ be a flow on $\Sigma$; letting $\phi=\exp(i\cdot 2\theta)$ be a coordinate on $P^1$, we will write $(\omega,\phi)\cdot t=(\omega t,\phi(t))$ when referring to this flow.

Now consider the following:

Let $x(t)$ be a solution to equation $E(\omega)$. For fixed $\tau$, let $(\omega,\Phi_0)=\sigma(\omega,x(\tau)/\lVert x(\tau)\rVert)$. Assume that the trace of $A(\omega)$ is $0$, i.e. $\text{trace}(A(\omega))=0$ $(\omega\in\Omega$).

Now the text says:

By the assumption $\text{trace}(A(\omega))=0$ $(\omega\in\Omega$) and equation (4), we have $$ \frac{1}{t-\tau}\ln\frac{\lVert x(t)\rVert}{\lVert x(\tau)\rVert}=\frac{1}{t-\tau}\int_{\tau}^t f(\omega s,\phi_0(s))\, ds, $$ where $$ f(\omega,\phi)=\delta(\omega)\cos 2\theta + \epsilon(\omega)\sin 2\theta,~~\phi=\exp(i\cdot 2\theta). $$

My question is how we get the equation $$ \frac{1}{t-\tau}\ln\frac{\lVert x(t)\rVert}{\lVert x(\tau)\rVert}=\frac{1}{t-\tau}\int_{\tau}^t f(\omega s,\phi_0(s))\, ds. $$

Could you pls explain that to me?


You have $$ \dot z = f(t) $$ where $z(t)=\ln(r(t)$ and $f(t)=δ(ωt)\cos(2θ(t))+ϵ(ωt)\sin(2θ(t))$ So obviously $$ z(t)=z(\tau)+\int_\tau^t f(s) ds $$ Namely $$ z(t)-z(\tau)=\ln(r(t))-\ln(r(\tau)=\ln(\frac{r(t)}{r(\tau)})=\int_\tau^t f(s) ds $$ but $r(t)=\sqrt{x_1^2(t)+x_2^2(t)}=||x(t)||$

The fact that $f(s)=f(\omega(s),\phi(s))$ follows from the condition $a(\omega)=0$ (trace of $A$ is equal to zero)

  • $\begingroup$ And where does the factor $\frac{1}{t-\tau}$ come from and which role does $\sigma$ play? $\endgroup$ – mathfemi Oct 22 '14 at 19:52
  • $\begingroup$ This the same factor on the both sides so it is not relevant. What $\sigma$ you are talking about? $\endgroup$ – Alexander Vigodner Oct 22 '14 at 19:54
  • $\begingroup$ The $\sigma$ I mentioned in the preliminaries, i.e. the fact that $f(\omega s,\Phi_0(s))=f((\sigma(\omega,x(\tau)/\lVert x(\tau)\rVert)\cdot s)=f(\Psi(\sigma(\omega,x(\tau)/\lVert x(\tau)\rVert),s)$ $\endgroup$ – mathfemi Oct 22 '14 at 19:58
  • $\begingroup$ The result is not connected to $\sigma$. $\sigma$ is your internal way to describe the system. $\endgroup$ – Alexander Vigodner Oct 22 '14 at 20:09
  • $\begingroup$ Another and last question: What exactly is $P^1$? I cannot imagaine it. $\endgroup$ – mathfemi Oct 22 '14 at 20:28

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