Homework 12

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We provide a template python code for you to start with. Download it from here.

Q12.1 \(\quad\) periodic orbits in Logistic map (Exercise 23.2)

In homework 10, we use a brute force way to estimate the escape rate in Logistic map: \[ f(x) = Ax(1 - x) \] and we promised to use a more elegant way to solve the same problem. Now, after studying chapter 23, it is time to apply the idea of cycle expansion to the escape rate problem inside this system.

The prerequisite of cycle expansion is the knowledge of the Floquet multipliers of the first few periodic orbits up to a certain truncation length \( N\). Here we set \(N=4\). The symbolic representations of the periodic orbit up to length 4 are : \(\{ 0, 1, 01, 001, 011, 0001, 0011, 0111\}\) (table 18.1). In template code "logisticMap.py", please set "case=1" to find all these orbits and their Floquet multipliers for parameter \(A=6.0\). For example, the multiplier of orbit \(01\) is -20.0 and the multiplier of orbit \(0111\) is -328.669578465.

What is the Floquet multiplier of orbit \(0011\) ? ( at least 2 decimal digits should be accurate )

Q12.2 \(\quad\) dynamical zeta function in Logistic map

From previous chapters, we know that the escape rate in a dynamical system is \(\gamma = -\ln s(0)\). Here \(s(0)\) is the leading eigenvalue of evolution operator, which turns out to be the inverse of the smallest zero point of dynamical zeta funcion or spectral determinant. On the other hand, Formula (23.9) and (23.10) describe the numerical procedure of evaluating dynamical zeta function up to some truncation length. Since we only have periodic orbits up to length 4 from Q12.1. We try to truncate dynamizcal zeta function to \(N=2,3,4\) respectively and observe its convergence.

Please set "case=2", and try to obtain dynamical zeta function for \(A=6.0\) up to length 2, 3 an 4. For example, when \(N=2\), \[ 1/\zeta = -0.00834z^2 - 0.4167z + 1 \] Once you have dynamical function, you can get its zero points and find the escape rate in this system. Also observe the convergence rate of escape rate.

When \(N=4\), the coefficient of the leading term \(z^4\) is a very small number ( \(a_4\) is around \(10^{-7} \) ), What is \(10^7 \times a_4\) ? ( at least three significant digits should be accurate )

Q12.3 \(\quad\) spectral determinant in Logistic map

The cycle expansion of spectral determinant is more convolved than that of dynamical zeta function. Actually, we first try to get the expansion of trace formula as shown in the formula below (23.13), then we use the relation between spectral determinant and the trace (formula below 23.15) to get the relation between coefficients of trace expansion series and coefficients of spectral determinant expansion series ( formula 23.16 ).

Now, set "case=3", and try to obtain spectral determinant for \(A=6.0\) up to length 2, 3 an 4. For example, when \(N=2\), \[ \det(1-z\mathcal{L}) = 1 - 0.4z -0.015238z^2 \] See how quickly the escape rate converges. When \(N=4\), the coefficient of the leading term \(z^4\) is a very small number ( \(a_4\) is around \(10^{-9} \) ), What is \(10^9 \times a_4\) ? ( at least three significant digits should be accurate )

Q12.4 \(\quad\) spectral determinant in Logistic map -- continued

Now set \(A=5.0\), and try to find the periodic orbits up to length \(N=5\). Obtain the cycle expansion of spectral determinant up to truncation number 5. What is the escape rate? So far, you understand that cycle expansion is far superior to the numerical simulation we did before. (Note: at least 5 digits should be accurate. )

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