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Rigorous Computation of Linear Response for Intermittent Maps

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Abstract

We present a rigorous numerical scheme for the approximation of the linear response of the invariant density of a map with an indifferent fixed point with respect to the order of the fixed point, with explicit and computed estimates for the error and all the involved constants.

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The code implimented for this work will be published online as soon as possible and is available upon request.

Notes

  1. See also earlier related work [33]. See also [27] for a comprehensive historical account including literature from physics.

  2. See also [32] for related work on dynamical determinants.

  3. Note that in the finite measure case, \(h^*\) is the derivative of the non-normalized density \(h_\epsilon \). The advantage in working with \(h_{\epsilon }\) is reflected in keeping the operator \(F_{\epsilon }\) linear and to accommodate the infinite measure preserving case. In the finite measure case, once the derivative of \(h_{\epsilon }\) is obtained, the derivative of the normalized density can be easily computed. Indeed, \(h_\epsilon =h +\epsilon h^*+o(\epsilon )\). Consequently, \(\int h_\epsilon =\int h +\epsilon \int h^*+o(\epsilon )\). Hence, \(\partial _{\epsilon }(\frac{h_\epsilon }{\int h_\epsilon }){|}_{\epsilon =0}=h^*-h\int h^*\).

  4. The existence of a uniform constant \(D>0\) is implied by condition (3).

  5. it is straightforward to see that these inequalities imply Lasota–Yorke inequalities on \(W^{k,1}\) with weak norm \(W^{k-1,1}\).

  6. It should be noted that finding the integral of \(A_0{\hat{h}}_n'\) and \(B_0{\hat{h}}_n\) is not easy so in our calculations instead of a true Ulam approximation where \(\zeta \) corresponds to the value that gives the integral we simply use the midpoint; this error is taken into account explicitly and depends on the regularity estimates we have on \(h_{\eta }\).

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Acknowledgements

The authors would like to thank Prof. Bahsoun and Prof. Galatolo for their guidance and assistance, their patience and attention, and also the reviewers and editor for their helpful remarks. Isaia Nisoli is partially supported by CNPq, CAPES (through the programs PROEX and the CAPES-STINT project “Contemporary topics in non uniformly hyperbolic dynamics”). The author is currently under “Afastamento do país para qualificação profissional, apresentação de trabalhos técnico-cientìficos e colaboração institucional do pessoal docente e técnico-administrativo” from UFRJ and is currently a Specially Appointed Associate Professor at Hokkaido University, and would like to thank Prof. Yuzuru Sato for the hospitality and scientific discussions.The research of Toby Taylor-Crush was supported by EPSRC.

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  1. Instituto de Matematica - UFRJ, Av. Athos da Silveira Ramos 149, Centro de Tecnologia - Bloco C Cidade Universitaria - Ilha do Fundão, Caixa Postal 68530, Rio de Janeiro, RJ, 21941-909, Brazil

    Isaia Nisoli

  2. Department of Mathematical Sciences, Loughborough University, Loughborough, Leicestershire, LE11 3TU, UK

    Toby Taylor-Crush

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Appendix A: Computing Derivatives

Appendix A: Computing Derivatives

In order to calculate \(A_0\), \(B_0\), \(a_\omega \) and \(b_\omega \) we use an iterative formula. We start with

$$\begin{aligned} g_\omega \circ T_\omega (x)=x \end{aligned}$$

from which we use the chain rule to get

$$\begin{aligned} (g_\omega \circ T_\omega )'(x)&=g_\omega '\circ T_\omega (x)\cdot T_\omega '(x)=1\\ \implies g_\omega '(x)&=\frac{1}{T_\omega '\circ g_\omega (x)} \end{aligned}$$

and

$$\begin{aligned} \partial _\alpha (g_\omega \circ T_\omega )(x)&=\partial _\alpha g_\omega \circ T_\omega (x)+ g_\omega '\circ T_\omega (x)\cdot \partial _\alpha T_\omega (x)=0\\ \implies \partial _\alpha g_\omega (x)&=-\frac{\partial _\alpha T_\omega \circ g_\omega (x)}{T_\omega '\circ g_\omega (x)}. \end{aligned}$$

We then use further chain rules to get

$$\begin{aligned} (g_\omega '\circ T_\omega )'(x)&=g_\omega ''\circ T_\omega (x)\cdot T_\omega '(x)=-\frac{T_\omega ''(x)}{(T_\omega '(x))^2}\\ \implies g_\omega ''(x)&=-\frac{T_\omega ''\circ g_\omega (x)}{(T_\omega '\circ g_\omega (x))^3} \end{aligned}$$

and

$$\begin{aligned} \partial _\alpha (g_\omega '\circ T_\omega )(x)&=\partial _\alpha g_\omega ' \circ T_\omega (x)+ g_\omega ''\circ T_\omega (x)\cdot \partial _\alpha T_\omega (x)=\partial _\alpha \frac{1}{T_\omega '(x)}\\ \implies \partial _\alpha g_\omega '(x)&=\frac{T_\omega ''\circ g_\omega (x)\cdot \partial _\alpha T_\omega \circ g_\omega (x)}{(T_\omega '\circ g_\omega )^3}-\frac{\partial _\alpha T_\omega '\circ g_\omega (x)}{(T_\omega '\circ g_\omega (x))^2}. \end{aligned}$$

We already can calculate \(g_\omega \) so we need to calculate \(\partial _\alpha T_\omega \), \(T_\omega '\), \(\partial _\alpha T_\omega '\) and \(T_\omega ''\). Note that \(T_\omega = T_0^{n}\circ T_1\) where \(|\omega |=n\), so

$$\begin{aligned} (T_0^{n}\circ T_1)'&=(T_0^{n})'\circ T_1\cdot T_1'\\ \partial _\alpha (T_0^{n}\circ T_1)&=\partial _\alpha (T_0^{n})\circ T_1 + (T_0^{n})'\circ T_1\cdot \partial _\alpha T_1\\ (T_0^{n}\circ T_1)''&=(T_0^{n})''\circ T_1\cdot (T_1')^2+(T_0^{n})'\circ T_1\cdot T_1''\\ \partial _\alpha (T_0^{n}\circ T_1)'&=\partial _\alpha (T_0^{n})'\circ T_1\cdot T_1' + (T_0^{n})''\circ T_1\cdot T_1'\cdot \partial _\alpha T_1+(T_0^{n})'\circ T_1\cdot \partial _\alpha T_1' \end{aligned}$$

which we may write as a matrix

$$\begin{aligned} \begin{pmatrix} T_\omega '\\ \partial _\alpha T_\omega \\ T_\omega ''\\ \partial _\alpha T_\omega ' \end{pmatrix}= \begin{pmatrix} T_1' &{} 0 &{} 0 &{} 0\\ \partial _\alpha T_1 &{} 1 &{} 0 &{} 0\\ T_1''&{} 0 &{} (T_1')^2 &{} 0\\ \partial _\alpha T_1' &{} 0 &{} T_1'\partial _\alpha T_1 &{} T_1' \end{pmatrix}\cdot \begin{pmatrix} (T_0^{n})'\circ T_1\\ \partial _\alpha (T_0^{n})\circ T_1\\ (T_0^{n})''\circ T_1\\ \partial _\alpha (T_0^{n})'\circ T_1 \end{pmatrix}. \end{aligned}$$

By the same logic we may write

$$\begin{aligned}{} & {} \begin{pmatrix} T_\omega '\\ \partial _\alpha T_\omega \\ T_\omega ''\\ \partial _\alpha T_\omega ' \end{pmatrix}= \begin{pmatrix} T_1' &{}\quad 0 &{}\quad 0 &{}\quad 0\\ \partial _\alpha T_1 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ T_1''&{}\quad 0 &{}\quad (T_1')^2 &{}\quad 0\\ \partial _\alpha T_1' &{}\quad 0 &{}\quad T_1'\partial _\alpha T_1 &{}\quad T_1' \end{pmatrix}\\{} & {} \cdot \begin{pmatrix} T_0'\circ T_1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ \partial _\alpha T_0\circ T_1 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ T_0''\circ T_1&{}\quad 0 &{}\quad (T_0'\circ T_1)^2 &{}\quad 0\\ \partial _\alpha T_0'\circ T_1 &{}\quad 0 &{}\quad T_0'\circ T_1\partial _\alpha T_0\circ T_1 &{}\quad T_0'\circ T_1 \end{pmatrix} \begin{pmatrix} (T_0^{n-1})'\circ T_0\circ T_1\\ \partial _\alpha (T_0^{n-1})\circ T_0\circ T_1\\ (T_0^{n-1})''\circ T_0\circ T_1\\ \partial _\alpha (T_0^{n-1})'\circ T_0\circ T_1 \end{pmatrix}. \end{aligned}$$

and use induction to give a series of matrices such that

$$\begin{aligned}{} & {} \begin{pmatrix} T_\omega '\\ \partial _\alpha T_\omega \\ T_\omega ''\\ \partial _\alpha T_\omega ' \end{pmatrix}= \begin{pmatrix} T_1' &{}\quad 0 &{}\quad 0 &{}\quad 0\\ \partial _\alpha T_1 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ T_1''&{}\quad 0 &{}\quad (T_1')^2 &{}\quad 0\\ \partial _\alpha T_1' &{}\quad 0 &{}\quad T_1'\partial _\alpha T_1 &{}\quad T_1' \end{pmatrix}\\ {}{} & {} \cdot \begin{pmatrix} T_0'\circ T_1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ \partial _\alpha T_0\circ T_1 &{}\quad 1 &{}\quad 0 &{}\quad 0\\ T_0''\circ T_1&{}\quad 0 &{}\quad (T_0'\circ T_1)^2 &{}\quad 0\\ \partial _\alpha T_0'\circ T_1 &{}\quad 0 &{}\quad T_0'\circ T_1\partial _\alpha T_0\circ T_1 &{}\quad T_0'\circ T_1 \end{pmatrix}\dots \begin{pmatrix} T_0'\circ T_0^{n-1}\circ T_1\\ \partial _\alpha T_0\circ T_0^{n-1}\circ T_1\\ T_0''\circ T_0^{n-1}\circ T_1\\ \partial _\alpha T_0'\circ T_0^{n-1}\circ T_1 \end{pmatrix}. \end{aligned}$$

Using the explicit formulas for our branches we have

$$\begin{aligned}&T_0 = x(1+(2x)^\alpha )\\&T_1 = 2x-1\\&T_0' = 1+(1+\alpha )(2x)^\alpha \\&T_1' = 2\\&\partial _\alpha T_0 = (\log (x)+\log (2))2^\alpha x^{\alpha +1}\\&\partial _\alpha T_1 = 0\\&T_0'' = \alpha (1+\alpha ) 2^\alpha x^{\alpha -1}\\&T_1'' = 0\\&\partial _\alpha T_0' = (2x)^\alpha ((\alpha +1)(\log (x)+\log (2))+1)\\&\partial _\alpha T_1' = 0 \end{aligned}$$

and we are able to calculate explicitly the values \(A_0\), \(B_0\), \(a_\omega \) and \(b_\omega \). To calculate \(a_\omega \) and \(b_\omega \) we use \(g_\omega =g_1\circ g_0^{n-1}\) and we use \(T_0^{m}\circ T_1\circ g_1\circ g_0^{n-1}= g_0^{n-1-m}\) to calculate

$$\begin{aligned} \begin{pmatrix} T_\omega '\circ g_\omega \\ \partial _\alpha T_\omega \circ g_\omega \\ T_\omega ''\circ g_\omega \\ \partial _\alpha T_\omega '\circ g_\omega \end{pmatrix}= \begin{pmatrix} T_1'\circ g_\omega &{}\quad 0 &{}\quad 0 &{}\quad 0\\ \partial _\alpha T_1\circ g_\omega &{}\quad 1 &{}\quad 0 &{}\quad 0\\ T_1''\circ g_\omega &{}\quad 0 &{}\quad (T_1')^2\circ g_\omega &{}\quad 0\\ \partial _\alpha T_1'\circ g_\omega &{}\quad 0 &{}\quad T_1'\circ g_\omega \partial _\alpha T_1\circ g_\omega &{}\quad T_1'\circ g_\omega \end{pmatrix}\\ \\ \cdot \begin{pmatrix} T_0'\circ g_0^{n} &{}\quad 0 &{}\quad 0 &{}\quad 0\\ \partial _\alpha T_0\circ g_0^{n} &{}\quad 1 &{}\quad 0 &{}\quad 0\\ T_0''\circ g_0^{n}&{}\quad 0 &{}\quad (T_0'\circ g_0^{n})^2 &{}\quad 0\\ \partial _\alpha T_0'\circ g_0^{n} &{}\quad 0 &{}\quad T_0'\circ g_0^{n}\partial _\alpha T_0\circ g_0^{n} &{}\quad T_0'\circ g_0^{n} \end{pmatrix}\dots \begin{pmatrix} T_0'\circ g_0\\ \partial _\alpha T_0\circ g_0\\ T_0''\circ g_0\\ \partial _\alpha T_0'\circ g_0 \end{pmatrix} \end{aligned}$$

where we calculate \(g_0^{m}\) using the shooting method from section 7.1.1.

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Nisoli, I., Taylor-Crush, T. Rigorous Computation of Linear Response for Intermittent Maps. J Stat Phys 190, 192 (2023). https://doi.org/10.1007/s10955-023-03174-8

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