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I consider a discrete dynamical system.

We define the current state of the system as follows: $$s(t) = \left(x_1(t), x_2(t), x_3(t), y_1(t), y_2(t), y_3(t)\right)$$

\begin{align*} x_1(t+1) &= f_1(s(t)) = \frac{1}{2 \cdot \sqrt{x_1(t)^2 + y_1(t)^2}}x_1(t) + \frac{1}{2} x_2(t) \\ x_2(t+1) &= f_2(s(t)) = \frac{1}{2} x_1(t) + \frac{1}{2} x_3(t) \\ x_3(t+1) &= f_3(s(t)) = \frac{1}{2} x_2(t) + \frac{1}{2 \cdot \sqrt{x_3(t)^2 + y_3(t)^2}}x_3(t) \\ y_1(t+1) &= f_4(s(t)) = \frac{1}{2 \cdot \sqrt{x_1(t)^2 + y_1(t)^2}}y_1(t) + \frac{1}{2} y_2(t) \\ y_2(t+1) &= f_5(s(t)) = \frac{1}{2} y_1(t) + \frac{1}{2} y_3(t) \\ y_3(t+1) &= f_6(s(t)) = \frac{1}{2} y_2(t) + \frac{1}{2 \cdot \sqrt{x_3(t)^2 + y_3(t)^2}}y_3(t) \\ \end{align*}

This is the Jacobi matrix of the system (I omitted the time parameter t for better readability):

\begin{align*} J &=\begin{bmatrix} \frac{y_1^2}{2(x_1^2 +y_1^2)^{3/2}} & \frac{1}{2} & 0 & - \frac{x_1 \cdot y_1 }{2(x_1^2 + y_1^2)^{3/2}} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{y_3^2}{2(x_3^2 +y_3^2)^{3/2}} & 0 & 0 & - \frac{x_3 \cdot y_3}{2(x_3^2 + y_3^2)^{3/2}} \\ - \frac{x_1 \cdot y_1}{2(x_1^2 + y_1^2)^{3/2}}& 0 & 0 & \frac{x_1^2}{2(x_1^2 +y_1^2)^{3/2}} & \frac{1}{2} & 0 \\ 0& 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ 0 & 0 & - \frac{x_3 \cdot y_4}{2(x_3^2 + y_3^2)^{3/2}} & 0 & \frac{1}{2} & \frac{x_3^2}{2(x_3^2 +y_3^2)^{3/2}} \end{bmatrix} \end{align*}

In general, $x_1 = x_2 = x_3$ and $y_1 = y_2 = y_3$is a fix point of the system in case $(x_1,y_1)$ is a unit vector, but there does not seem to be any attracting fix point. Is there any way to determine the fix point this system converges to? (it does so in simulations).

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1 Answer 1

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Showing whether a certain fixed point has global convergence is difficult. However, local convergence can be shown by linearizing the system at each fixed point, namely if all the eigenvalues of the Jacobi matrix evaluated at that fixed point lie inside the unit circle would mean that all initial conditions sufficiently close to that fixed point will converge to that fixed point. How close "sufficiently close" would have to be is hard to say though.

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