# Converting an ODE in polar form

Convert the ODE system $$\dot{x}=\begin{pmatrix}a(t) & b(t)\\c(t) & d(t)\end{pmatrix}x$$ into polar form. You should get two equations $$\frac{d}{dt}\Phi(t)=...\\ \frac{d}{dt}\ln r(t)=....$$

I set $$x_1:=r(t)\cos\Phi(t)\\ x_2:=r(t)\sin\Phi(t)$$ and got $$\frac{d}{dt}\Phi(t)=b(t)+\frac{\frac{d}{dt}r(t)\cos\Phi(t)}{r(t)\sin\Phi(t)}-\frac{a(t)\cos\Phi(t)}{\sin\Phi(t)}\\ \frac{d}{dt}\ln r(t)=d(t)+\frac{c(t)\cos\Phi(t)}{\sin\Phi(t)}-\frac{\cos\Phi(t)\frac{d}{dt}\Phi(t)}{\sin\Phi(t)}$$

Would like to know if this is right.

With greetings

• The conversion is not over, you want to express $d\Phi(t)/dt$ as a function of $r(t)$, $\Phi(t)$, and the matrix at time $t$, only, and likewise for $dr(t)/dt$.
– Did
Oct 21, 2014 at 18:50

I have the following:

$$x_1:=r(t) \cos \Phi(t),$$ $$x_2:=r(t) \sin \Phi(t)$$

and the derivatives are:

$$\dot{x_1}=\dot r \cos \Phi - r \sin(\Phi)\, \dot \Phi,$$ $$\dot{x_2}=\dot r \sin \Phi + r \cos(\Phi)\, \dot \Phi.$$

with $\dot x \equiv \dfrac{d x}{dt}.$

So you get:

$$\dot r \cos \Phi - r \sin(\Phi)\, \dot \Phi =a r\cos \Phi+br\sin\Phi,$$ $$\dot r \sin \Phi + r \cos(\Phi)\, \dot \Phi =c r\cos \Phi+r d\sin\Phi.$$

The equation for $\dot\Phi$ is:

$$\dot\Phi =\dfrac{(a r\cos \Phi+br\sin\Phi)\sin\Phi-(c r\cos \Phi+rd\sin\Phi)\cos\Phi}{-r},$$

and for $\dot r$ is :

$$\dot r = (a r\cos \Phi+br\sin\Phi)\cos\Phi - (c r\cos \Phi+rd\sin\Phi)\sin\Phi,$$

but:

$$\dfrac{\dot r}{r(t)} = \dfrac{d}{dt}\ln[r(t)],$$

so:

$$\dfrac{d}{dt}\ln[r(t)]=(a \cos \Phi+b\sin\Phi)\cos\Phi - (c\cos \Phi+d\sin\Phi)\sin\Phi.$$

• How do you get the equations for $\dot{\Phi}$ and $\dot{r}$? Oct 21, 2014 at 19:07
• @mathfemi with some algebra... you have two equations and $\dot \Phi$ and $\dot r$ in both equations. Oct 21, 2014 at 19:10