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dynamical system, a.c.i.m., response function, susceptibility, causality
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\def\qed{\unskip\kern 6pt\penalty 500\raise -2pt\hbox
{\vrule\vbox to 10pt{\hrule width 4pt\vfill\hrule}\vrule}}
\centerline{DIFFERENTIATING THE ABSOLUTELY CONTINUOUS}
\centerline{INVARIANT MEASURE OF AN INTERVAL MAP $f$}
\centerline{WITH RESPECT TO $f$.}
\bigskip
\centerline{by David Ruelle\footnote{*}{Mathematics Dept., Rutgers University, and IHES. 91440 Bures sur Yvette, France.\break $<$ruelle@ihes.fr$>$}.}
\bigskip\bigskip\noindent
{\leftskip=2cm\rightskip=2cm\sl {\bf Abstract.} Let the map $f:[-1,1]\to[-1,1]$ have a.c.i.m. $\rho$ (absolutely continuous $f$-invariant measure with respect to Lebesgue). Let $\delta\rho$ be the change of $\rho$ corresponding to a perturbation $X=\delta f\circ f^{-1}$ of $f$. Formally we have, for differentiable $A$,
$$ \delta\rho(A)=\sum_{n=0}^\infty\int\rho(dx)\,X(x){d\over dx}A(f^nx) $$
but this expression does not converge in general. For $f$ real-analytic and Markovian in the sense of covering $(-1,1)$ $m$ times, and assuming an {\it analytic expanding} condition, we show that
$$ \lambda\mapsto\Psi(\lambda)=\sum_{n=0}^\infty\lambda^n
\int\rho(dx)\,X(x){d\over dx}A(f^nx) $$
is meromorphic in ${\bf C}$, and has no pole at $\lambda=1$. We can thus formally write $\delta\rho(A)=\Psi(1)$.
\par}
\vfill\eject
We postpone a discussion of the significance of our result, and start to describe the conditions under which we prove it. Note that these conditions are certainly too strong: suitable differentiability should replace analyticity, and a weaker Markov property should be sufficient. But the point of the present note is to show how it is that $\Psi(\lambda)$ has no pole at $\lambda=1$, rather than deriving a very general theorem.
\medskip\noindent
{\bf Setup.}
\medskip
We assume that $f:[-1,1]\to[-1,1]$ is real analytic and piecewise monotone on $[-1,1]$ in the following sense: there are points $c_j$ ($j=0,\ldots,m$, with $m\ge2$) such that $-1=c_00$ and, if $a>0$,
$$ \varpi(\pm(1-a\xi^2+b\xi^3\ldots))
=\pm(1-\sqrt{{a\over C}}\xi+{b\over2\sqrt{aC}}\xi^2\ldots) $$
[We may for instance take
$$ \omega(x)=\sin{\pi x\over2}\qquad,
\qquad\varpi(x)={2\over\pi}\arcsin x $$
or
$$ \qquad\omega(x)={1\over16}(25x-10x^3+x^5)\qquad,
\qquad\varpi(x)={16\over25}x\ldots\qquad\qquad] $$
\indent
The function $g:\varpi\circ f\circ\omega$ from $[-1,1]$ to $[-1,1]$ has monotone restrictions to the intervals $\varpi[c_{j-1},c_j]=[d_{j-1},d_j]$. It is readily seen that $g_j$ extends to a holomorphic function in a neighborhood of $[d_{j-1},d_j]$, and that
$$ g_1(-1+\xi)=-1+\sqrt{f'(-1)}\xi+\alpha_-\xi^3\ldots $$
$$ g_m(1-\xi)=(-1)^{m+1}(1-\sqrt{|f'(1)|}\xi-\alpha_+\xi^3\ldots) $$
with no $\xi^2$ terms in the right-hand sides [this follows from our choice of $\omega$, which has no $\xi^3$ term]. One also finds that, for $j=1,\ldots,m-1$
$$ g_j(d_j-\xi)
=(-1)^{j+1}(1-\sqrt{|f''(c_j)|\over2C}\omega'(d_j)\xi+\gamma_j\xi^2\ldots) $$
$$ g_{j+1}(d_j+\xi)
=(-1)^{j+1}(1-\sqrt{|f''(c_j)|\over2C}\omega'(d_j)\xi-\gamma_j\xi^2\ldots) $$
where $\gamma_j$ is the same in the two relations. We note the following easy consequences of the above developments:
\medskip\noindent
{\bf Lemma 1.} {\sl Let $\psi_j:[-1,1]\to[d_{j-1},d_j]$ be the inverse of $g_j$ for $j=1,\ldots,m$ (increasing for $j$ odd, decreasing for $j$ even). Then
$$ \psi_1(-1+\xi)=-1+{1\over\sqrt{f'(-1)}}\xi+\beta_-\xi^3 $$
$$ \psi_m((-1)^{m+1}(1-\xi))=1-{1\over\sqrt{|f'(1)|}}\xi+\beta_+\xi^3 $$
(there are no $\xi^2$ terms in the right-hand sides). If $j0$ we can thus, for each ${\bf j}$, find $\ell$ such that ${\rm diam}\psi_{j_1}\circ\cdots\circ\psi_{j_\ell}\bar U<\epsilon$. Hence (using the compactness of the Cantor set of sequences ${\bf j}$) one can choose $L$ so that the $m^L$ sets
$$ \psi_{j_1}\circ\cdots\psi_{j_L}\bar U $$
have diameter $<\epsilon$. The open connected set
$$ V=\cup_{j_1,\ldots,j_L}\psi_{j_1}\circ\cdots\psi_{j_L}U $$
satisfies $[-1,1]\subset V\subset U$, and $\psi_j\bar V=\cup_{j_1,\ldots,j_L}\psi_j\circ\psi_{j_1}\circ\cdots\circ\psi_{j_L}\bar U\subset\cup_{j_0,j_1,\ldots,j_{L-1}}\psi_{j_0}\circ\psi_{j_1}\circ\psi_{i_{L-1}}U=V$. This shows that $U$ can be replaced in Assumption A by a set $V$ contained in an $\epsilon$-neighborhood of $[-1,1]$.
\medskip
Since we have shown above that ${\rm diam}\psi_{j_1}\circ\cdots\psi_{j_L}\bar U<\epsilon$, we see that $\psi_1^L$ maps a small circle around $-1$ strictly inside itself. We have thus $\psi'_1(-1)<1$ ({\it i.e.}, $f'(-1)>1$) and similarly, if $m$ is odd, $\psi'_m(1)<1$ ({\it i.e.}, $f'(1)>1$).
\medskip
The following two lemmas state some easy facts to be used later.
\medskip\noindent
{\bf Lemma 2.} {\sl Let $H$ be the Hilbert space of functions $\bar U\to{\bf C}$ which are square integrable (with respect to Lebesgue) and holomorphic in $U$. The operator ${\cal L}$ on $H$ defined by
$$ ({\cal L}\Phi)(z)=\sum_{j=1}^m(-1)^{j+1}\psi'_j(z)\Phi(\psi_j(z)) $$
is holomorphy improving. In particular ${\cal L}$ is compact and trace-class.}\qed
\medskip\noindent
{\bf Lemma 3.} {\sl On $[-1,1]$ we have
$$ ({\cal L}\Phi)(x)=\sum_j|\psi'_j(x)|\Phi(\psi_j(x)) $$
hence $\Phi\ge0$ implies ${\cal L}\Phi\ge0$ (${\cal L}$ preserves positivity) and
$$ \int_{-1}^1dx\,({\cal L}\Phi)(x)=\int_{-1}^1dx\,\Phi(x) $$
(${\cal L}$ preserves total mass).}\qed
\medskip\noindent
{\bf Lemma 4.} {\sl ${\cal L}$ has a simple eigenvalue $\mu_0=1$ corresponding to an eigenfunction $\sigma_0>0$. The other eigenvalues $\mu_k$ ($k\ge1$) satisfy $|\mu_k|<1$, and their (generalized) eigenfunctions $\sigma_k$ satisfy $\int_{-1}^1dx\,\sigma_k(x)=0$.}
\medskip
Let $(\mu_k,\sigma_k)$ be a listing of the eigenvalues and generalized eigenfunctions of the trace-class operator ${\cal L}$. For each $\mu_k$ there is some $\sigma_k$ such that ${\cal L}\sigma_k=\mu_k\sigma_k$, hence
$$ |\mu_k|\int_{-1}^1dx\,|\sigma_k(x)|=\int_{-1}^1dx\,|\mu_k\sigma_k(x)|
=\int_{-1}^1dx\,|({\cal L}\sigma_k)(x)| $$
$$ \le\int_{-1}^1dx\,({\cal L}|\sigma_k|)(x)=\int_{-1}^1dx\,|\sigma_k(x)| $$
hence $|\mu_k|\le1$. Denote by $S_<$ and $S_1$ the spectral spaces of ${\cal L}$ corresponding to eigenvalues $\mu_k$ with $|\mu_k|<1$, and $|\mu_k|=1$ respectively. If $\sigma_k\in S_<$ then, for some $n\ge1$,
$$ 0=\int_{-1}^1dx\,(({\cal L}-\mu_k)^n\sigma_k)(x)
=\int_{-1}^1dx\,(1-\mu_k)^n\sigma_k(x) $$
hence $\int_{-1}^1dx\,\sigma_k(x)=0$.
\medskip
On the finite dimensional space $S_1$, there is a basis of eigenvectors $\sigma_k$ diagonalizing ${\cal L}$ (if ${\cal L}|S_1$ had non-diagonal normal form, $||{\cal L}^n|S_1||$ would tend to infinity with $n$, in contradiction with $\int_{-1}^1dx\,|({\cal L}^n\Phi)(x)|\le\int_{-1}^1dx\,|\Phi(x)|$). We shall now show that, up to multiplication by a constant $\ne0$, we may assume $\sigma_k\ge0$. If not, because $\sigma_k$ is continuous and the intervals $\psi_{j_1}\circ\cdots\circ\psi_{j_n}[-1,1]$ are small for large $n$ (mixing), we would have $|({\cal L}^n\sigma_k)(x)|<({\cal L}^n|\sigma_k|)(x)$ for some $n$ and $x$. This would imply $\int_{-1}^1dx\,|({\cal L}^n\sigma_k)(x)|<\int_{-1}^1dx\,|\sigma_k(x)|$ in contradiction with ${\cal L}\sigma_k=\mu_k\sigma_k$ and $|\mu_k|=1$. From $\sigma_k\ge0$ we get $\mu_k=1$, and the corresponding eigenspace is at most one dimensional (otherwise it would contain functions not $\ge0$). But we have $1\notin S_<$ because $\int_{-1}^1dx
\!
,1\ne0$, so that $S_1\ne\{0\}$. Thus $S_1$ is spanned by an eigenfunction, which we call $\sigma_0$, to the eigenvalue $\mu_0=1$. Finally, $\sigma_0>0$ because if $\sigma_0(x)=0$ we would have also $\sigma_0(y)=0$ whenever $g^n(y)=x$, which is not compatible with $\sigma_0$ continuous $\ne0$.\qed
\medskip\noindent
{\bf Lemma 5.} {\sl If we normalize $\sigma_0$ by $\int_{-1}^1dx\,\sigma_0(x)=1$, then $\sigma_0(dx)=\sigma_0(x)dx$ is the unique $g$-invariant probability measure absolutely continuous with respect to Lebesgue on $[-1,1]$. In particular, $\sigma_0(dx)$ is ergodic.}
\medskip
For continuous $A$ on $[-1,1]$ we have
$$ \int_{-1}^1\sigma_0(dx)(A\circ g)(x)=\int_{-1}^1dx\,\sigma_0(x)A(g(x))
=\int_{-1}^1dx\,({\cal L}\sigma_0)(x)A(x)=\int_{-1}^1\sigma_0(dx)A(x) $$
so that $\sigma_0(dx)$ is $g$-invariant. Let $\tilde\sigma(x)dx$ be another $g$-invariant probability measure absolutely invariant with respect to Lebesgue. Then, if $\tilde\sigma\ne\sigma_0$
$$ \int_{-1}^1dx\,|\sigma_0(x)-\tilde\sigma(x)|
=\int_{-1}^1dx\,|({\cal L}(\sigma_0-\tilde\sigma))(x)| $$
$$ <\int_{-1}^1dx\,({\cal L}|\sigma_0-\tilde\sigma|)(x)
=\int_{-1}^1dx\,|\sigma_0(x)-\tilde\sigma(x)| $$
by mixing: contradiction.\qed
\medskip\noindent
{\bf Lemma 6.} {\sl Let $H_1\subset H$ consist of those functions $\Phi$ with derivatives vanishing at $\pm1$: $\Phi'(-1)=\Phi'(1)=0$. Then ${\cal L}H_1\subset H_1$ and $\sigma_0\in H_1$}.
\medskip
${\cal L}H_1\subset H_1$ is an easy calculation using Lemma 1. Furthermore, by Lemma 4, $\sigma_0=\lim_{n\to\infty}{\cal L}^n{1\over2}$, and ${1\over2}\in H_1$ implies $\sigma_0\in H_1$.\qed
\medskip
The image $\rho(dx)=\rho(x)dx$ of $\sigma_0(x)dx$ by $\omega$ is the unique $f$-invariant probability measure absolutely continuous with respect to Lebesgue on $[-1,1]$. We have
$$ \rho(x)=\sigma_0(\varpi x)\varpi'(x) $$
Consider now the expression
$$ \Psi(\lambda)
=\sum_{n=0}^\infty\lambda^n\int_{-1}^1\rho(dx)\,X(x){d\over dx}A(f^nx) $$
where we assume that $X$ extends to a holomorphic function in a neighborhood of $[-1,1]$ and $A\in{\cal C}^1[-1,1]$. For sufficiently small $|\lambda|$, the series defining $\Psi(\lambda)$ converges. Writing $B=A\circ\omega$ and $x=\omega y$ we have
$$ X(x){d\over dx}A(f^nx)=X(\omega y){1\over\omega'(y)}{d\over dy}B(g^ny) $$
hence
$$ \Psi(\lambda)=\sum_{n=0}^\infty\lambda^n\int_{-1}^1dy\,
\sigma_0(y){X(\omega y)\over\omega'(y)}{d\over dy}B(g^ny) $$
Defining $Y(y)=\sigma_0(y)X(\omega y)/\omega'(y)$, we see that $Y$ extends to a function holomorphic in a neighborhood of $[-1,1]$, which we may take to be $U$, except for simple poles at $-1$ and $1$. We may write
$$ \int_{-1}^1dy\,\sigma_0(y){X(\omega y)\over\omega'(y)}{d\over dy}B(g^ny)
=\int_{-1}^1dy\,Y(y)g'(y)\cdots g'(g^{n-1}y)B'(g^ny) $$
$$ =\int_{-1}^1ds\,({\cal L}_0^nY)(s)B'(s) $$
where
$$ ({\cal L}_0\Phi)(s)=\sum_{j=1}^m(-1)^{j+1}\Phi(\psi_js) $$
and we have thus
$$ \Psi(\lambda)=\sum_{n=0}^\infty\lambda^n\int_{-1}^1ds\,
({\cal L}_0^nY)(s)B'(s) $$
\noindent
{\bf Lemma 7.} {\sl Let $H_0\subset H$ be the space of functions vanishing at $-1$ and $1$. Then ${\cal L}_0H_0\subset H_0$.}
\medskip
This follows readily from Lemma 1.\qed
\medskip\noindent
{\bf Lemma 8.} {\sl There are meromorphic functions $\Phi_\pm$ with Laurent series
$$ \Phi_\pm(z)={1\over z\mp1}+O(z\mp1) $$
at $\pm1$ and $\Phi_\pm(\mp1)=0$ such that
$$ {\cal L}_0\Phi_-=\sqrt{f'(-1)}\Phi_- $$
$$ \left\{
\matrix{{\cal L}_0\Phi_+=\sqrt{f'(1)}\Phi_+\qquad\hbox{if $m$ is odd}\cr
{\cal L}_0(\Phi_+/\sqrt{|f'(1)|}+\Phi_-/\sqrt{f'(-1)})=\tilde Y\in H_0
\qquad\hbox{if $m$ is even}\cr}\right. $$}
\indent
Define
$$ p_\pm(z)={1\over z\mp1}-{1\over4}(z\mp1) $$
then Lemma 1 yields
$$ ({\cal L}_0-\sqrt{f'(-1)})p_-=u_-\in H_0 $$
$$ \left\{
\matrix{({\cal L}_0-\sqrt{f'(1)})p_+=u_+\in H_0\qquad\hbox{if $m$ is odd}\cr
{\cal L}_0p_++\sqrt{|f'(1)|}p_-=u_0\in H_0\qquad\hbox{if $m$ is even}\cr}
\right. $$
Since $f'(-1)>1$, Lemma 4 shows that ${\cal L}-\sqrt{f'(-1)}$ is invertible on $H$, hence there is $v_-$ such that
$$ ({\cal L}-\sqrt{f'(-1)})v_-=u'_- $$
and since $\int_{-1}^1dx\,u'_-(x)=0$, also $\int_{-1}^1dx\,v_-(x)=0$ and we can take $w_-\in H_0$ such that $w'_-=v_-$. Then
$$ (({\cal L}_0-\sqrt{f'(-1)})w_-)'=({\cal L}-\sqrt{f'(-1)})w'_-
=({\cal L}-\sqrt{f'(-1)})v_-=u'_- $$
so that
$$ ({\cal L}_0-\sqrt{f'(-1)})w_-=u_- $$
without additive constant because the left-hand side is in $H_0$ by Lemma 7. In conclusion
$$ ({\cal L}_0-\sqrt{f'(-1)})(p_--w_-)=0 $$
and we may take $\Phi_-=p_--w_-$.
\medskip
If $m$ is odd, $\Phi_+$ is handled similarly. If $m$ is even, taking $\Phi_+=p_+$ and writing $\tilde Y=u_0/\sqrt{|f'(1)|}-w_-$ we obtain
$$ {\cal L}_0({\Phi_+\over\sqrt{|f'(1)|}}+{\Phi_-\over\sqrt{f'(-1)}})
=\tilde Y\in H_0 $$
which completes the proof.\qed
\medskip
We have $\sigma_0\in H_1$ (Lemma 6), and $X\circ\omega\in H_1$ by our choice of $\omega$. Also
$$ \omega'(\pm(1-\xi))=2C\xi-4D\xi^3\ldots $$
so that
$$ Y={\bf C}\Phi_-+{\bf C}\Phi_++H_0 $$
\indent
If $m$ is odd let $Y=c_-\Phi_-+c_+\Phi_++Y_0$, with $Y_0\in H_0$. Then
$$\Psi(\lambda)={c_-\over{1-\lambda\sqrt{f'(-1)}}}\int_{-1}^1ds\,\Phi_-(s)B'(s)
+{c_+\over{1-\lambda\sqrt{f'(1)}}}\int_{-1}^1ds\,\Phi_+(s)B'(s)
+\Psi_0(\lambda) $$
where $\Psi_0$ is obtained from $\Psi$ when $Y$ is replaced by $Y_0$.
\medskip
If $m$ is even let $Y=c_-\Phi_-+\tilde c({\Phi_+/\sqrt{|f'(1)|}}+{\Phi_-/\sqrt{f'(-1)}})+Y_0$, with $Y_0\in H_0$. Then
$$ \Psi(\lambda)
={c_-\over{1-\lambda\sqrt{f'(-1)}}}\int_{-1}^1ds\,\Phi_-(s)B'(s)
+\tilde c\int_{-1}^1ds\,({\Phi_+\over\sqrt{|f'(1)|}}
+{\Phi_-\over\sqrt{f'(-1)}})B'(s) $$
$$ +\lambda\tilde\Psi(\lambda)+\Psi_0(\lambda) $$
where $\tilde\Psi(\lambda)$ is obtained from $\Psi$ when $Y$ is replaced by $\tilde Y$.
\medskip
Writing $\mu_\pm=\sqrt{f'(\pm1)}$ we see that $\Psi(\lambda)$ has two poles at $\mu_\pm^{-1}$ if $m$ is odd, and one pole at $\mu_-^{-1}$ if $m$ is even; the other poles are those of $\Psi_0(\lambda)$ and possibly $\tilde\Psi(\lambda)$. Since $Y_0\in H_0$ and ${\cal L}_0H_0\subset H_0$, we have
$$ \Psi_0(\lambda)
=\sum_{n=0}^\infty\lambda^n\int_{-1}^1ds\,({\cal L}_0^nY_0)(s)B'(s)
=-\sum_{n=0}^\infty\lambda^n\int_{-1}^1ds\,({\cal L}_0^nY_0)'(s)B(s) $$
$$ =-\sum_{n=0}^\infty\lambda^n\int_{-1}^1ds\,({\cal L}^nY'_0)(s)B(s) $$
It follows that $\Psi_0(\lambda)$ extends meromorphically to ${\bf C}$ with poles at the $\mu_k^{-1}$. We want to show that the residue of the pole at $\mu_0^{-1}=1$ vanishes . By Lemma 4, $\int_{-1}^1dx\,\sigma_k(x)=0$ for $k\ge1$. Thus, up to normalization, the coefficient of $\sigma_0$ in the expansion of $Y'_0$ is
$$ \int_{-1}^1dx\,Y'_0(x)=Y_0(1)-Y_0(-1)=0 $$
because $Y_0\in H_0$. Therefore $\Psi_0(z)$ is holomorphic at $z=1$, and the same argument applies to $\tilde\Psi(z)$, concluding the proof of the theorem.\qed
\medskip\noindent
{\bf Discussion.}
\medskip
It can be argued that the {\it physical measure} describing a physical dynamical system is an SRB (Sinai-Ruelle-Bowen) measure $\rho$ (see the recent reviews [11], [2] which contain a number of references), or an a.c.i.m. $\rho$ in the case of a map of the interval. But, typically, physical systems depend on parameters, and it is desirable to know how $\rho$ depends on the parameters ({\it i.e.}, on the dynamical system). The dependence is smooth for uniformly hyperbolic dynamical systems (see [5], [6] and references given there), but discontinuous in general.
\medskip
The present note is devoted to an example in support of an idea put forward in [8]: that derivatives of $\rho(A)$ with respect to parameters can be meaningfully defined in spite of discontinuities. An ambitious project would be to have Taylor expansions on a large set $\Sigma$ of parameter values and, using a theorem of Whitney [10], to connect these expansions by a function extrapolating $\rho(A)$ smoothly outside of $\Sigma$. In a different dynamical situation, that of KAM tori, a smooth extension \`a la Whitney has been achieved by Chierchia and Gallavotti [3], and P\"oschel [4].
\medskip
In our study we have considered only a rather special set $\Sigma$ consisting of maps satisfying a Markov property. (Reference [1] should be consulted for a discussion of the poles encountered in the study of a Markovian map $f$). Note that the studies of a.c.i.m. for maps of the interval, and of SRB measures for H\'enon-like maps, are typically based on perturbations of a map satisfying a Markov property (for the use of slightly more general Misiurewicz-type maps see [9], which also gives references to earlier work).
\medskip
The function $\Psi(\lambda)$ that we have encountered is related to the {\it susceptibility} $\omega\mapsto\Psi(e^{i\omega})$ giving the response of a system to a periodic perturbation. The existence of a holomorphic extension of the susceptibility to the upper half complex plane is expected to follow from {\it causality} (causality says that cause preceeds effect, resulting in a {\it response function} $\kappa$ having support on the positive half real axis, and its Fourier transform $\hat\kappa$ extending holomorphically to the upper half complex plane). A discussion of nonequilibrium statistical mechanics [7] shows that the expected support and holomorphy properties hold close to equilibrium, or if uniform hyperbolicity holds. In the example discussed in this note, $\kappa$ has the right support property, but increases exponentially at infinity, and holomorphy in the upper half plane fails, corresponding the existence of a pole of $\Psi$ at $\lambda=1/\sqrt{f'(-1)}$. This might be expressed by saying that $\rho$ is {\it not linearly stable}. The physically interesting situation of {\it large systems} (thermodynamic limit) remains quite unclear at this point.
\medskip\noindent
{\bf Acknowledgments.}
For many discussions on the subject of this note, I am indebted to V. Baladi, M. Benedicks, G. Gallavotti, M. Viana, and L.-S. Young.
\medskip\noindent
{\bf References.}
[1] V. Baladi, Y. Jiang, H.H. Rugh. ``Dynamical determinants via dynamical conjugacies for postcritically finite polynomials.'' J. Statist. Phys. {\bf 108},973-993(2002).
[2] C. Bonatti, L. Diaz, and M. Viana. {\it Dynamics beyond uniform hyperbolicity: a global geometric and probabilistic approach.} Springer, to appear.
[3] L. Chierchia and G. Gallavotti. ``Smooth prime integrals for quasi-integrable Hamiltonian systems.'' Nuovo Cim. {\bf 67B},277-295(1982).
[4] J. P\"oschel. ``Integrability of Hamiltonian systems on Cantor sets.'' Commun. in Pure and Applied Math. {\bf 35},653-696(1982).
[5] D. Ruelle. ``Differentiation of SRB states.'' Commun. Math. Phys. {\bf 187},227-241(1997); ``Correction and complements.'' Commun. Math. Phys. {\bf 234},185-190(2003).
[6] D. Ruelle. ``Differentiation of SRB states for hyperbolic flows.'' In preparation.
[7] D. Ruelle. ``Smooth dynamics and new theoretical ideas in nonequilibrium statistical mechanics.'' J. Statist. Phys. {\bf 95},393-468(1999).
[8] D. Ruelle. ``Application of hyperbolic dynamics to physics: some problems and conjectures.'' Bull. Amer. Math. Soc. (N.S.) {\bf 41},275-278(2004).
[9] Q. Wang and L.-S. Young. ``Towards a theory of rank one attractors.'' Preprint
[10] H. Whitney. ``Analytic expansions of differentiable functions defined in closed sets.'' Trans. Amer. Math. Soc. {\bf 36},63-89(1934).
[11] L.-S. Young. ``What are SRB measures, and which dynamical systems have them?'' J. Statist. Phys. {\bf 108},733-754(2002).
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