We have a family of potentials Vt on Rn, evolving with time, and their associated Gibbs measures pt, ie
pt(x)=Zte−Vt(x)whereZt=∫Rne−Vt(x)dx.
The goal is to compute, at least numerically, the partition functions Zt or equivalently the free energy Ft=−logZt. In many problems, the family of potentials (Vt) connects an unknown potential V0, for which we would like to compute F0, and another potential V1 for which F1 is known.
An elegant way of estimating these free energies uses the Jarzynski identity, discovered in 1996. It is based on the Langevin dynamics associated with the evolving potential Vt:
dXt=−∇Vt(Xt)dt+2dWt
with (Wt) a Brownian motion. Upon standard conditions on Vt, this system – a standard Stochastic Differential Equation – has a solution. Note that the distribution of Xt is notpt in general. However, there is a remarkable identity relating the distribution of Xt with pt.
Jarzynski's identity states that
E[e−∫0tV˙s(Xs)−F˙sds]=1.
Here and after, the dot in V˙t or F˙t always represents a derivative with respect to time.
The consequence of this identity is that
logE[e−∫0tV˙s(Xs)ds]=Ft−F0,
giving rise to a natural estimator for the free energy differenceFt−F0: one samples many paths from the dynamics (2), then computes the path integral wt=∫0tV˙s(Xs)ds along each path, and then average the values of e−wt and take the log.
In general, exponential of paths integrals are not easy to study. The classical way is to use the Feynman-Kac representation of PDEs, but this proof seems quite involved; instead, there is a nice trick shown to me by Eric Vanden-Eijnden. The trick is as follows: if ρt(x,y) is the probability density of the system (Xt,e−∫0tV˙s(Xs)−F˙sds) at t, then
clearly, we want to compute ∫uρt(x,u)dudx,
the quantity at(x)=∫uρt(x,u)du solves an explicit equation,
and from this equation we can see that at(x)=e−Vt(x)/Zt=pt(x).
The proof is then finished, since ∫at=∫pt=1. From now on, we'll set Ut=Vt−Ft, so that pt(x)=e−Ut(x). Clearly, ∇Ut=∇Vt.
subject to the starting condition I0=1 and X0∼p0. Then clearly,
It=e−∫0tU˙s(Xs)ds.
We consider the system Zt=(Xt,It) as a single SDE dZt=f(t,Zt)dt+σdWt with
ft(x,y)=(−∇Ut(x)−U˙t(x)y) and σ=diag(2,…,2,0).
Let us note ρt(x,y) the density of (Xt,It), and write the associated Fokker-Planck equation; here and after, ∇ is a gradient and ∇⋅ is the divergence, ie ∇⋅φ=∑i∂iφ. Subscripts denote partial differentiation with respect to some variables.
An integration by parts shows that ∫y2∂yρt(x,y)dy=−2∫yρt(x,y)dy=−2at, providing a sufficient decay of y→y2ρ(x,y) at infinity. We obtain the following equation:
a˙t=−(∇⋅∇Ut)at−U˙tat+∇Ut⋅∇at+Δat.
It turns out that the density pt(x)=e−Ut(x) is the solution of this equation. Indeed, pt=−U˙tpt and ∇pt=−∇Utpt, so that it is easy to check that (11) is satisfied. Moreover, a0(x)=∫yρ0(x,y)dy=e−U0(x)=p0(x), so the initial conditions are identical.