## Homework 13

For this homework, you can submit up to 3 times.

You need to write the code by yourself.

You need to write the code by yourself.

#### Q13.1 \(\quad\) Chains of piecewise linear maps (Example 24.1)

Let us simulate the diffusion process in the chains of piecewise linear maps defined as \[ \hat{f}(\hat{x}) = \begin{cases} \Lambda \hat{x} & \quad \hat{x} \in [0, 1/2) \\ \Lambda \hat{x} + 1 - \Lambda & \quad \hat{x} \in (1/2, 1] \end{cases} \] extended to the entire real line by \(\hat{f}(\hat{x} + n) = \hat{f}(\hat{x}) + n\) for \(n \in \mathbb{z}\). Please write a program to simulate this system. Choose a large number of equally spaced points \(x_i\) in range [-1, 1] as the initial conditions, and iterate them for a certain number of steps. Since this map is anti-symmetric, we expect the mean drift is zero, so we just record \(x_i^2\). Figure (a) below is what I got for slope \(\Lambda = 4 \). The y-axis is the mean value of \(x_i^2\) at each iteration step, and the x-axis is the iteration index. As is shown, the slope of the straight line is approximate 0.499, so the diffusion constant is about 0.2495 in this case.Set \(\Lambda = 5\), what is the approximate diffusion constant in your simulation ?

#### Q13.2 \(\quad\) modification of Example 24.4

In example 24.4, we choose an appropriate slope of the linear map \(\Lambda = 2(\sqrt{2}+1)\) such that the critical point is mapped onto the right border of \(\mathcal{M}_{1_+}\), therefore region \(\mathcal{M}_{2_+}\) is mapped to \(\mathcal{M}_{0_+}\) and \(\mathcal{M}_{1_+}\). Now try to find the appropriate slope \(\Lambda\) in range [4, 6] such that the critical point is mapped onto the right border of \(\mathcal{M}_{0_+}\). Now follow the same procedure in example 24.4 to find the dynamical zeta function and try to get the diffusion constant \(D\) in this case. In order to validate your result, please check \(1/\zeta(0,1) = 0\). Also, you can simulate diffusion for this particular slope and obtain the numerical diffusion constant to double check your analytical result.What is the diffusion coefficient in this case ? ( Note, at least 5 significant digits should be accurate. )

#### Q13.3 \(\quad\) Diffusion for odd integer \(\Lambda\) (Exercise 24.1)

Example 24.2 illustrates the process of applying cycle expansion to the piecewise linear map to calculate the diffusion constant for the case \(\Lambda\) being even. Now let us turn to the case that \(\Lambda\) is odd. For example, when \(\Lambda = 3\), interval [0, 1] can be divided into 4 pieces: \(\{[0,1/3), [1/3, 1/2), (1/2, 2/3), [2/3,1]\}\), with the corresponding symbols \(\{ 0^+, 1^+, 1^-, 0^-\}\). You can easily verify that \(f(\mathcal{M}_{1^+}) = \mathcal{M}_{0^+} \cup \mathcal{M}_{1^+}\), so sequences \(1^+1^-\) and \(1^+0^-\) are pruned. Similar argument goes for region \(1^-\). In this sense, we can treat sequences \(\{1^+0^+, 1^-0^-, 1^+1^+0^+, 1^-1^-0^-, \cdots\}\) as new symbols, and we get the dynamical zeta function as \[ 1/\zeta = 1 - t_{0^+} - t_{0^-} - t_{1^+0^+} - t_{1^-0^-} - t_{1^+1^+0^-} - t_{1^-1^-0^-} - \cdots \] You may consult example 24.4 to help you understand the above process. The difference between \(\Lambda\) even and \(\Lambda\) odd case is that symbolic dynamics in the latter case has infinite number of symbols, as opposed to the finite expression in the \(\Lambda\) even case: \( 1/\zeta = 1 - t_{0^+} - t_{0^-} - \cdots - t_{(a-1)^+} - t_{(a-1)^-} \,. \) After you get the dynamical function, you can calculate the mean cycle length \(\langle n \rangle_\zeta\) and mean cycle displacement \(\langle n^2 \rangle_\zeta\) to obtain the diffusion constant.We can generalize the above analysis to any odd \(\Lambda\) case. What is the mean cycle length for \(\Lambda=11\) ?

#### Q13.4 \(\quad\) Dependence of diffusion constant on slope -- Q13.1 continued

If the slope \(\Lambda\) is changed continuously, you may expect that the diffusion constant changes continuously too in a similar way. However, as shown in Chaosbook figure 24.5, the dependence is not a smooth function. This is not the first time we encounter such a situation. Remember that in homework 10, we observe that a small change of parameter in Henon map results in a totally different natural measure. In this sense, structural stability should not be an easy assumption in dynamical systems.Figure (b) below shows the numerical diffusion constant I got for \(5<\Lambda<6\). Although only 100 points are chosen in this range, we still can see the non-smoothness of this curve. Try to plot the diffusion constant for range \(3<\Lambda<4\). What is the approximate maximal diffusion coefficient in this range ?

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