Non-linear differential equation

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SUMMARY

The discussion centers on the non-linear differential equation system defined by $x' = y + x f(r^2)$ and $y' = -x + y f(r^2)$, where $r^2 = x^2 + y^2$. The first part establishes that $\frac{dr^2}{dt} = 2r^2 f(r^2)$ using the chain rule. The second part explores the number of fixed points and periodic solutions, concluding that the only fixed point is at (0,0) due to the condition $-r^2 = 0$. The discussion also touches on the stability of fixed points and the implications of the angular velocity ${\theta}' = -1$.

PREREQUISITES
  • Understanding of non-linear differential equations
  • Familiarity with the chain rule in calculus
  • Knowledge of fixed points and their stability analysis
  • Concept of periodic solutions in dynamical systems
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  • Study the stability of fixed points in non-linear systems
  • Learn about periodic solutions in dynamical systems
  • Explore the implications of angular velocity in phase plane analysis
  • Investigate the role of zeroes of functions in determining system behavior
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Mathematicians, physicists, and engineers working with dynamical systems, particularly those focusing on non-linear differential equations and stability analysis.

Poirot1
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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.
 
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You already know that x'= y+ xf(r^2) and y'= -x+ yf(r^2) so that xy'- yx'= x(-x+ yf(r^2))- y(y+ xf(r^2)= -x^2+ xyf(r^2)- (y^2+ xyf(r^2)= -(x^2+y^2)= -r^2= 0 if and only if r= 0.
 
ok what about the question about periodic solutions
 

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