Poirot1
- 243
- 0
system is $x'=y+xf(r^2)$ and $y'=-x+yf(r^2)$. where $r^2=x^2+y^2$
(i) prove that $\frac{dr^2}/{dt}=2r^2f(r^2)$. My solution ( I won't write out details): use chain rule and the fact that rr'=xx'+yy'.
(ii)assume $f(r^2)$ has N zeroes. determine the number of fixed points and periodic solutions the system has and write about the stability of fixed points.
This one I think I did the first (I will give details) but can't do the others
Solution: $r^2{\theta}'=xy'-yx'$and if (x,y) is fixed point, then xy'-yx'=0. If you work this out you get fixed point implies -r^2=0 so x=y=0 is only fixed point. In the solution however it has a different justification, namely that ${\theta}'=-1$ but I don't understand how this means (0,0) is only fixed point.
(i) prove that $\frac{dr^2}/{dt}=2r^2f(r^2)$. My solution ( I won't write out details): use chain rule and the fact that rr'=xx'+yy'.
(ii)assume $f(r^2)$ has N zeroes. determine the number of fixed points and periodic solutions the system has and write about the stability of fixed points.
This one I think I did the first (I will give details) but can't do the others
Solution: $r^2{\theta}'=xy'-yx'$and if (x,y) is fixed point, then xy'-yx'=0. If you work this out you get fixed point implies -r^2=0 so x=y=0 is only fixed point. In the solution however it has a different justification, namely that ${\theta}'=-1$ but I don't understand how this means (0,0) is only fixed point.