1
$\begingroup$

Consider the ODEs for $x(t),y(t),z(t)$

$\dot{x} = - y z - \alpha x$

$\dot{y} = 1 + 2 x z - \alpha y$

$\dot{z} = - x y - \alpha z$

with initial conditions $x_0,y_0,z_0\in \mathbb{R}$.

Numerically investigating these ODEs, it seems that where $\alpha\ll 1$ and $t\to \infty$, these ODEs have a universal behavior which is

$(x(t),y(t),z(t))\to 1/\sqrt{2}(1,0,-1)$ or $1/\sqrt{2}(-1,0,1)$ as $t\to \infty$ then $\alpha\to 0$.

Is there a clear way to see this analytically?

$\endgroup$
  • $\begingroup$ Are you sure about the differential equations and the equilibria? Namely the points you gave are not equilibria of the differential equations; maybe it should be $\dot{y}=1+xz-\alpha y$ or the equilibria have to be divided by $\sqrt{2}$? $\endgroup$ – Kwin van der Veen Jul 21 '19 at 0:02
  • $\begingroup$ you're absolutely right, the equilibria should be scaled by $1/\sqrt{2}$. I will edit. $\endgroup$ – J. Doee Jul 21 '19 at 12:13
1
$\begingroup$

For nonlinear systems it is often hard to say something about global system behavior. Instead one can look at local behavior by analyzing the stability of each of the equilibrium points by linearizing the system. Solving for $\dot{x}=\dot{y}=\dot{z}=0$ yields the following five equilibria

\begin{align} (x_1,y_1,z_1) &= \left(0, \frac{1}{\alpha}, 0\right), \\ (x_2,y_2,z_2) &= \left(-\sqrt{\frac{1-\alpha^2}{2}}, \alpha, \sqrt{\frac{1-\alpha^2}{2}}\right), \\ (x_3,y_3,z_3) &= \left(\sqrt{\frac{1-\alpha^2}{2}}, \alpha, -\sqrt{\frac{1-\alpha^2}{2}}\right), \\ (x_4,y_4,z_4) &= \left(i\sqrt{\frac{1+\alpha^2}{2}}, -\alpha, i\sqrt{\frac{1+\alpha^2}{2}}\right), \\ (x_5,y_5,z_5) &= \left(-i\sqrt{\frac{1+\alpha^2}{2}}, -\alpha, -i\sqrt{\frac{1+\alpha^2}{2}}\right). \end{align}

The stability of these equilibria can be checked by evaluating the eigenvalues of the Jacobi matrix of the system,

$$ J = \begin{bmatrix} -\alpha & -z & -y \\ 2\,z & -\alpha & 2\,x \\ -y & -x & -\alpha \end{bmatrix}, $$

at each of the equilibria. Doing so yields the following eigenvalues

\begin{align} \lambda_1 &= \left\{-\alpha, \frac{1 - \alpha^2}{\alpha}, \frac{-1 - \alpha^2}{\alpha}\right\}, \\ \lambda_2=\lambda_3 &= \left\{-\frac{\alpha+\sqrt{9\,\alpha^2-8}}{2}, -\frac{\alpha-\sqrt{9\,\alpha^2-8}}{2}, -2\,\alpha\right\}, \\ \lambda_4=\lambda_5 &= \left\{-\frac{\alpha+\sqrt{8-7\,\alpha^2}}{2}, -\frac{\alpha-\sqrt{8-7\,\alpha^2}}{2}, -2\,\alpha\right\}. \end{align}

When using that you consider $0 < \alpha \ll 1$ only the second and third equilibria are stable (all eigenvalues have negative real part). In the limit of $\alpha\to0$ does indeed yield that these two equilibria are located at $1/\sqrt{2}\left(-1, 0, 1\right)$ and $1/\sqrt{2}\left(1, 0, -1\right)$ respectively. However, the fact that these are the only stable equilibria does not mean that the union of their basins of attraction has to span the entire $\mathbb{R}^3$. For example any solution that starts at $z=x$ doesn't converge to either of the two stable equilibria. In order to get a lower bound on the size of these basins of attraction one could use a quadratic Lyapunov function based on the linearization at the equilibrium point and see for which closed set of points around the equilibrium the time derivative of the Lyapunov function remains negative.

It can also be noted that when $\alpha = 0$ there are infinitely many equilibria, namely when $y=0$ and $2\,x\,z = -1$. Stability analysis of these points is inconclusive, since all eigenvalues have a zero real part at every point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.