differential equations in polar coordinates

[robster]

Hi, i have the following ODE system in x und y:

x'(t) = x - y - x*(x²+y²)
y'(t) = x + y - y*(x²+y²)

By use of polar coordinates

x(t)=r(t) * cos ( @(t) )
y(t)=r(t) * sin ( @(t) )

[@ = theta]

it is supposed to be transformed into the system

r'(t) = r - r³
@'(t) = 1

how is that done?

My ideas so far:

I derive x(t)=r(t) * cos ( @(t) ) and obtain

x'(t) = r'(t)*cos(@(t)) + r(t)*(-sin(@(t)))*@'(t)
y'(t) = r'(t)*sin(@(t)) + r(t)*cos(@(t))*@'(t) (1)

Now I replace x(t) by r(t)*cos (@(t)) in the original ode, i.e.

x - y - x*(x²+y²) -> r*cos@ - r*sin@ - r*cos@*r²
x + y - y*(x²+y²) -> r*cos@ + r*sin@ - r*sin@*r² (2)

now identify the expessions in (1) with those in (2) leads to

r'(t)*cos(@(t)) + r(t)*(-sin(@(t)))*@'(t) = r*cos@ - r*sin@ - r*cos@*r²
r'(t)*sin(@(t)) + r(t)*cos(@(t))*@'(t) = r*cos@ + r*sin@ - r*sin@*r²

now this should be equal to the given solution

r'(t) = r - r³
@'(t) = 1

but so far i failed to find an adequate transformation...

thanks for your help

 

[galactus]

\(\displaystyle x'(t)=x-y-x(x^{2}+y^{2})\)

Using polar subs \(\displaystyle x(t)=rcos(t), \;\ y(t)=rsin(t)\)

\(\displaystyle -rsin(t)+r'cos(t)=rcos(t)(1-r^{2})-rsin(t)\)

\(\displaystyle r'cos(t)=rcos(t)(1-r^{2})\Leftarrow \text{Of course, here we divide by cos(t). This is assuming cos(t) is not equal to 0}\)

\(\displaystyle r'=r(1-r^{2})=r-r^{3}\)


What most overlook is the use of the product rule on the left side and just write -rsin(t).

 

[robster]

no, you forgot the second variable @=theta, which is a function in t, @ = @(t)

so the derivate of x is as i said

x'(t) = r'(t)*cos(@(t)) + r(t)*(-sin(@(t)))*@'(t) , (note @'(t) which is necessary for an ode in @)

i found a different way to solve it, thanks anyway ;-)

 

[galactus]

My goodness, DUH, you're correct. I didn't even notice that.