Physical entropy production

A simple fact: let T:X\to X be a homeomorphism preserving the measure-class of \omega, J(T^k,x) be the Jacobian of T^k at x. Then for any sequence a(k)\to\infty with \sum_k \frac{1}{a(k)}<+\infty we have \limsup_k\frac{J(T^k,x)}{a(k)}=0 for \omega-a.e. x\in X. For example a(k)=e^{t k} with t>0 or a(k)=k^{\alpha} with \alpha>1. This is a direct corollary of Borel-Cantelli property:

Consider the set E_{k,\delta}=\{x\in X: \frac{J(T^k,x)}{a(k)}\ge\delta\}. It is easy to see \omega(E_{k,\delta})\le \frac{1}{\delta\cdot a(k)} and hence \sum\omega(E_{k,\delta})\le \sum\frac{1}{\delta\cdot a(k)}<+\infty. So \omega(x\in E_{k,\delta} \text{infinitely often})=0 for all \delta>0

Notes from papers by Jaksic, Pillet, Rey-Bellet, Ruelle and Young.

4. Let \nu\ll \mu be two probability measures on X and \phi=\frac{d\nu}{d\mu} be its density. Then 1=\nu(X)=\mu(\phi). Moreover 1/\phi is well-defined with respect to \nu and \nu(1/\phi)=\mu(1)=1, too.

The relative entropy can be defined as E(\nu|\mu)=\nu(\log\phi) when \nu\ll\mu, +\infty otherwise.
Note that E(\nu|\mu)=\nu(-\log\frac{1}{\phi})\ge-\log\nu(1/\phi)=0. So the relative entropy is nonnegative.

Convexity: assume \nu_1,\nu_2\ll\mu and p+q=1, then \phi=p\phi_1+q\phi_2 and E(p\nu_1+q\nu_2|\mu)\le p\nu_1(\log\phi_1)+q\nu_2(\log\phi_2).

3. Let f:X\to X be a homeomorphism, and \phi be a continuous function on X. Then a probability \mu on X is said to be a Gibbs measure with respect to (f,\phi), if there exist C_n(\epsilon) with \displaystyle \lim_{n\to\infty}\frac{\log C_n(\epsilon)}{n}=0 for all \epsilon>0, such that 1/C_n(\epsilon)\le \frac{\mu(B_n(x,\epsilon))}{e^{\phi_n(x)}}\le C_n(\epsilon) for all x\in X.

Prop. Assume there exists some Gibbs measure \mu for (f,\phi). Then for any continuous function g, the moment generating function \displaystyle e(\theta)=\lim_{n\to\infty}\frac{1}{n}\log\mu(e^{\theta g_n})=P(f,\phi+\theta g).
Proof. Let \epsilon>0 and \delta=\text{var}_\epsilon g. Then \displaystyle e^{\phi_n(x)+\theta g_n(x)}\overset{C}{\sim} e^{\theta g_n(x)}\cdot\mu(B_n(x,\epsilon))\overset{\delta}{\sim}\int_{B_n(x,\epsilon)}e^{\theta g_n}d\mu.

Assume f is expansive and (f,\phi) admits a unique equilibrium state, say \mu_\phi. Then the pressure function is differentiable at \phi and \frac{d}{d\theta}|_{\theta=0}P(f,\phi+\theta g)=\mu_\phi(g). Then according to Prop., \displaystyle \mu_\phi(g)=e'(0)=\lim_{n\to\infty}\frac{1}{n}\mu(g_n)=\lim_{n\to\infty}\mu_n(g) for all g. In particular we see \mu_n\to\mu_\phi for every (f,\phi)-Gibbs measure \mu.

Remark. The Gartner-Ellis Theorem actually says that for every (f,\phi)-Gibbs measure \mu, \nu_{n,x}\to\mu_\phi for \mu-a.e. x\in X.

2. Let f:(X,\mu)\to (X,\mu) be a local homeomorphism preserving the measure-class of \mu. Suppose there is an decomposition X=\bigcup_k X_k, and the branch inverse f_k^{-1}:X\to X_k. Let J(f) be the Jacobian of f, and L=L_0 be the induced transfer operator: \displaystyle (L\phi)(x)=\sum_{fy=x}\frac{\phi(y)}{J(f,y)}. Then for any measurable function \phi: \int L(\phi) d\mu=\int\phi d\mu.
Proof. \int L(\phi) d\mu=\sum_k \int\frac{\phi(x_k)}{J(f,x_k)}d\mu(x)=\sum_k \int \phi(x_k) J(f_k^{-1},x)d\mu(x)
=\sum_k \int_{X_k}\phi d\mu(x_k)=\int\phi d\mu.

Moreover for any two functions \phi,\psi, we have
\int (L\phi)\cdot\psi d\mu=\int L(\phi\cdot\psi\circ f)d\mu=\int \phi\cdot\psi\circ f d\mu.

An \mu-abs. cts measure \nu=h\mu is f-invariant if and only if Lh=h.
Proof. \int \phi\circ\!f\, d\nu=\int \phi\circ\!f\, h d\mu=\int \phi\cdot (Lh) d\mu=\int\phi\, hd\mu=\int\phi d\nu.

Then we generalize the transfer operator L_g:\phi\mapsto (x\mapsto \sum_{fy=x}\frac{e^{g(y)\phi(y)}}{J(f,y)}). The moment generating function of g with respect to an f-invariant measure \nu=h\mu is e(\theta)=\lim_{n\to\infty}\frac{1}{n}\log\nu(e^{\theta g_n}). This is related to the transfer operator L_g:
\nu(e^{\theta g_n})=\int e^{\theta g_n}\cdot hd\mu=\int L^n(e^{\theta g_n}\cdot h)d\mu=\mu(L_{\theta g}^n h).

In particular, if L_{\theta g} has a spectral gap, then e(\theta) is analytic and leads to large deviation results of (\nu,g).

1. Let f:M\to M be a C^2 diffeomorphism, m a pre-chosen smooth probability volume, J(f,\cdot) the Jacobian of f with respect to m. The entropy production of (f,m) is e_f(m)=\int-\log J(f,x)dm(x)\ge-\log\int J(f,x)dm(x)=0 (that is, entropy is non-decreasing). Moreover, e_f(m)=0 if and only if m is f-invariant (that is, an equilibrium state).

Ruelle introduced the entropy production for general  f-invariant measure \nu as e_f(\nu)=\int-\log J(f,x)d\nu(x). Clearly e_{f^{-1}}(\mu)=\int-\log J(f^{-1},x)d\nu(x)=\int\log J(f,f^{-1}x)d\nu(x) =\int\log J(f,x)d\nu(x)=-e_f(f), since \mu is f-invariant. Compare with e_{f^{-1}}(m)=\int-\log J(f^{-1},x)dm(x)=\int\log J(f,f^{-1}x)dm(x) =\int\log J(f,x)dm(fx)=\int J(f,x)\log J(f,x)dm(x).

It is observed in his paper that e_f(\nu)=-\int\sum_i\lambda_i(x)d\nu(x)=-\sum_i\lambda_i(\nu) be Oseledec Multiplicative Ergodic Theorem, and hence independent of the choice of volumes.

Theorem 1.2. (1). Suppose h_\mu(f)=\sum_i\lambda^+_i(\mu), then e_f(\mu)\ge 0. (He called such measures SRB there)
(2). Assume \mu is hyperbolic SRB. Then e_f(\mu)>0 if and only if \mu is singular.
(3). m(B(\mu,f))=0 if e_f(\mu)<0.

Proof. (1). e_f(\mu)=-\Lambda^u(\mu)-\Lambda^s(\mu)=-(h_\mu(f)+\Lambda^s(\mu))\ge0 by Ruelle entropy inequality.
(2). Note that SRB is equivalent to \mu^u_x\ll m^u_x. So \mu\ll m if and only if \mu^s_x\ll m^s_x, which also equivalent to (by Ledrappier-Young) h_\mu(f)=-\Lambda^s(\mu) and hence e_f(\mu)=-\Lambda^u(\mu)-\Lambda^s(\mu)=-h_\mu(f)-\Lambda^s(\mu)=0.
(3). Pick N\ge1. For each x\in B(\mu,f), there exists n_x\ge N with \displaystyle \frac{1}{n}\sum_{0\le k\le n-1}-\log J(f,f^kx)\le -c=:e_f(\mu)/2<0. So B(\mu,f)\subset\bigcup_{n\ge N}A_n, where A_n=\{x:J(f^n,x)\ge e^{nc}\}. Clearly m(A_n)\le e^{-nc} and hence m(B(\mu,f))\le\sum_{n\ge N}e^{-nc}\to 0 as N\to\infty. In particular m(B(\mu,f))=0.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Please log in using one of these methods to post your comment: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: