# Attractors in random dynamics

Let $$\Delta$$ be the interval $$[-1,1]$$, then we can consider the probability space $$(\Delta , \mathcal{B}(\Delta),\nu)$$, where $$\mathcal{B}(\Delta)$$ is the Borel $$\sigma$$-algebra and $$\nu$$ is equal half of the Lebesgue measure.

Then we can endow the space $$\Delta^{\mathbb{N}}:= \{ (\omega_n)_{n\in \mathbb{N}};\ \omega_n \in \Delta, \ \forall \ n\in \mathbb{N}\}$$ with the $$\sigma$$-algebra $$\mathcal{B}(\Delta^{\mathbb{N}})$$ (Borel $$\sigma$$-algebra of $$\Delta^{\mathbb{N}}$$ induced by the product topology) and the probability measuare $$\nu^{\mathbb{N}}$$ in the measurable space$$(\Delta^{\mathbb{N}},\mathcal{B}(\Delta^{\mathbb{N}}))$$, such that $$\nu^{\mathbb{N}} \left(A_1\times A_2\times \ldots \times A_n \times \prod_{i=n+1}^{\infty} \Delta\right)=\nu(A_1) \cdot \ldots\cdot \nu(A_n).$$

Now, let $$\sigma>2/(3\sqrt{3})$$ be a real number, and define $$x_-^* = \text{The unique real root of the polynomial }x^ 3+ \sigma = x,$$ $$x_+^* = \text{The unique real root of the polynomial }x^ 3- \sigma = x,$$ it is easy to see that $$x_+^* = -x_-^*$$.

We can then define the function $$h:\mathbb{N}\times \Delta^ \mathbb{N}\times \mathbb{R}\to \mathbb{R}$$ in the following recursive way,

• $$h(0,(\omega_n)_{n},x) = x$$, $$\forall\ (\omega_n)_n\in \mathbb{N}$$ and $$\forall\ x\in \mathbb{R}$$;
• $$h(i+1,(\omega_n)_{n},x) = \sqrt[3]{h(i,(\omega_n)_{n},x) + \sigma \omega_i}.$$

This way we are for, every $$x \in \mathbb{R}$$ and $$(\omega_n)_n\in\Delta^\mathbb {N}$$, defining the following sequence $$\left\{x, \sqrt[3]{x + \sigma \omega_1},\sqrt[3]{\sqrt[3]{x + \sigma \omega_1}+\sigma w_2},\sqrt[3]{\sqrt[3]{\sqrt[3]{x + \sigma \omega_1}+\sigma w_2} + \sigma w_3}, \ldots \right\}.$$

Using Mathematica I did some numerical experiments and seems like that $$\lim_{n\to\infty}\left| h(n,(\omega_n)_n, x^*_+) - h(n,(\omega_n)_n, x^*_-) \right| = 0,\ \text{for \nu^{\mathbb{N}}-almost surely (\omega_n)_n} \in \Delta^{\mathbb{N}}.$$

Can anyone tell me, or present me a good reason why this should happen?

## Numerical evidences

Below this paragraph you can find my Mathematica code, imputing a value of $$\sigma$$ and a number of iterations ($$b$$) this program has as output two discrete plot, the first one is the discrete plot of the sequences $$h(\cdot, (\omega_n)_n, x_+^ *)$$ and $$h(\cdot, (\omega_n)_n, x_-^ *)$$ in the same plot, the second one is a discrete plot of the difference $$h(\cdot , (\omega_n)_n, x_+^ *) - h(\cdot , (\omega_n)_n, x_-^ *)$$, for some random sequence $$(\omega_n)_n$$.

\[Sigma] = 2/(3*Sqrt[3]) + 0.15; *Insert here your value of sigma*
b = 10000; *Insert here your the number of iterations*
A = Solve[x^3 + \[Alpha] == x , x, Reals];
B = Solve[x^3 - \[Alpha] == x , x, Reals];
y = x /. A[[1]];
z = x /. B[[1]];
W = Table[RandomReal[{-\[Sigma], \[Sigma]}], {i, 1, b}];
P = RecurrenceTable[{x[n + 1] == CubeRoot[x[n] + W[[n]]], x[1] == y},
x, {n, 1, b} ];
Q = RecurrenceTable[{x[n + 1] == CubeRoot[x[n] + W[[n]]], x[1] == z},
x, {n, 1, b} ];
DiscretePlot[{P[[i]], Q[[i]]}, {i, 1, b}, PlotRange -> {{1, b}, {-1.5, 1.5}}]
DiscretePlot[Q[[i]] - P[[i]], {i, 1, b},PlotRange -> {{1, b}, {-2, 2}}, PlotStyle -> Brown]


Example 1: $$\sigma = 1$$ and $$b= 50$$

Example 2: $$\sigma = \frac{2}{3 \sqrt{3}} +0.15$$ and $$b= 100000.$$

for some reason when $$\sigma$$ is big, the convergence of $$\left| h(n,(\omega_n)_n, x^*_+) - h(n,(\omega_n)_n, x^*_-) \right|$$ seems to occur "faster", and I do not have idea why this happens.

Can anyone help me?

Since this is a monotone continuous random dynamical system such that the underlying Markov process is uniquely ergodic, it admits a unique random fixed point. The standard proof goes as follows: by monotonicity and continuity, the bounded set attractor $$A$$ is such that $$A_+ = \sup A$$ and $$A_- = \inf A$$ are random fixed points. Since the law of $$A_\pm$$ is a (and therefore the by unique ergodicity) invariant measure, one must have $$A_+ = A_-$$ almost surely.
• I think that is the family of process $$\{\Phi_{n}^{x}(\omega) := h(n,x,\omega)\},$$ with transition probability $$P_{n}(x,\Gamma) = \int_{\Omega} 1_{\Gamma}\circ h(n,\omega,x)\ \text{d} \nu^{\mathbb{N}}(\omega),$$ but I am not sure. – Matheus Manzatto Aug 13 at 21:55