# Get a special form of an linear System of ODE (using polar form)

In this post Converting an ODE in polar form it is shown that a linear system of ODE $$x'=\begin{pmatrix}a(t) & b(t)\\c(t) & d(t)\end{pmatrix}x$$ can be written in polar coordinates $r,\Phi$ as $$\frac{d}{dt}\Phi(t)=(d-a)\cdot\cos\Phi(t)\sin\Phi(t)-b(t)\sin^2\Phi(t)+c(t)\cos^2\Phi(t)\\\frac{d}{dt}\ln r(t)=a(t)\cos^2\Phi(t)+(b(t)-c(t))\sin\Phi(t)\cos\Phi(t)-d(t)\sin^2\Phi(t).$$

Now in an article (http://www.sciencedirect.com/science/article/pii/0022039678900773) I found the following concerning this converting on page 26, 1.14:

Write $$A(t)=\begin{bmatrix}\alpha(t) & -\beta(t)\\\beta(t) & \alpha(t)\end{bmatrix}+\begin{bmatrix}\delta(t) & \epsilon(t)\\\epsilon(t) & -\delta(t)\end{bmatrix}.$$ In polar coordinates $r,\theta$, the equation $x'=A(t)x$ becomes $$\frac{d}{dt}\theta(t)=\beta(t)+\epsilon(t)\cos 2\theta(t)-\delta(t)\sin 2\theta(t)\\\frac{d}{dt}\ln r(t)=\alpha(t)+\delta(t)\cos 2\theta(t)+\epsilon(t)\sin 2\theta(t).$$

Ok, when setting $$a(t):=\alpha(t)+\delta(t)\\ b(t):=-\beta(t)+\epsilon(t)\\ c(t):=\beta(t)+\epsilon (t)\\ d(t):=\alpha(t)-\delta(t)$$ in the above expression of the ODE system in polar form and using $$\cos 2x=\cos^2 x - \sin^2 x,\\ \sin 2x=2\sin x\cos x,$$ I get these two equations.

But what does this way of writing $A(t)$ mean resp. where does it come from resp. where are the functions $\alpha, \beta, \delta$ and $\epsilon$ from?

Can any matrix $A(t)$ be written like this?

I do not understand the motivation of this way of writing $A(t)$.

Maybe you can clarify it.

Hope to hear from you.

With greetings