I have a dynamical system in three dimensions given by:

$\dot x = (1-x^2-y^2-z^2)x+xz-y$

$\dot y = (1-x^2-y^2-z^2)y+yz+x$

$\dot z = (1-x^2-y^2-z^2)z-x^2-y^2$

I analyzed the system by first finding the fixed points which are at (0,0,0) , (0,0,1) and (0,0,-1).

Next, I found the Jacobian of the system and for each fixed point, I computed the eigenvalues of the Jacobian matrix.

For the fixed point (0,0,0), $\lambda = 1, 1+i, 1-i$

For the fixed point (0,0,1), $\lambda = -2, 1+i, 1-i$

For the fixed point (0,0,-1), $\lambda = -2, -1+i, -1-i$

Hence, (0,0,0) is unstable since the eigenvalues are positive. (0,0,1) is unstable since the real part of the complex eigenvalue is positive. (0,0,-1) is negative since the eigenvalues are negative.(i.e. real part of complex eigenvalues are negative).

Is there anything else I can say about the long term behavior of this system?


A few hints that should let you complete the picture:

  • $r'=(1-r^2)r$, where $r=(x^2+y^2+z^2)^{1/2}$;

  • $\theta'=1$ for the angle in the $xy$-plane;

  • the $z$-axis is invariant, satisfying the equation $z'=(1-z^2)z$.

In particular, the sphere of radius $1$ centered at the origin is invariant.

  • $\begingroup$ How are you getting r' = (1-r^2)*r ? Are you applying rr'=xx'+yy'+zz' ? Because the sphere is invariant, does that also mean that trajectories are periodic around the z axis? $\endgroup$ – Jay Apr 18 '16 at 17:04
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    $\begingroup$ Sometimes we really need to make an effort. $\endgroup$ – John B Apr 18 '16 at 18:28
  • $\begingroup$ I am in the process of working out a solution. I'm asking these questions because I have never dealt with a 3d dynamical system . I'm familiar with the 2d form of converting to polar coordinates via rr'=xx'+yy' so I wanted to see how the transformation works in 3d. $\endgroup$ – Jay Apr 18 '16 at 19:30

From $r \dot r =x\dot x + y\dot y +z\dot z$

$r \dot r =x\dot x + y\dot y +z\dot z$

$r\dot r = -x^4-2x^2y^2-2x^2z^2-y^4-2y^2z^2-z^4+x^2+y^2+z^2$

$ r\dot r = (x^2+y^2+z^2) (1 - (x^2+y^2+z^2)) = r^2(1-r^2) $

$\dot r = r(1-r^2) $

In terms of stability, r=0 is a repeller and r=1 is an attractor.

Looking at $\dot \theta = (x\dot y-y \dot x)/r^2$

$ \dot \theta = (x^2+y^2)/r^2 = 1 $

Additionally, the ODEs are invariant under the transformation (x,y,z,t) ==> (-x,-y,z,t)

which shows that solutions are symmetric about the z axis.

Thus, based on this analysis, trajectories for the system will go out to the sphere of radius 1 and rotate counterclockwise with respect to the xy plane. There is no periodicity as there is no time reversal symmetry about the z axis.


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