Let Vt be a family of smooth potentials on Rn, continuously evolving with time t∈[0,1]. Let pt be their associated Gibbs measure:
It is not easy to find a continuous dynamic Xt, such that at each time t the distribution of Xt matches pt. One might be tempted to use the overderdamped Langevin dynamics,
with (Wt) a Brownian motion. But the density ρt of Xtis not equal to pt in general. Indeed, ρt solves the Fokker-Planck equation
If ρt was equal to pt, then pt should also satisfy this equation; but for pt one easily checks that the RHS is zero, resulting in p˙t=0 which is false in general if we choose a family of evolving potentials; it would only be true at equilibrium when Vt is constant.
The Jarczynski trick allows to reconnect the distribution of Xt with pt. It introduces a weight, the solution of an auxiliary ODE, such that its conditional expectation given Xt=x gives pt(x), hence correcting the mismatch between ρt and pt.
(In the sequel we note Ft=logZt, and the dot in V˙t or F˙t always represents a derivative with respect to time. )
Let (Xt,Wt) be the solution of the following augmented system: dXt=−∇Vt(Xt)dt+2dWtdWt=(−V˙t(Xt)−Ft˙)WtdtX0∼p0W0=1. Then, for any test function φ, E[φ(Xt)Wt]=∫φ(x)Zte−Vt(x)dx. In other words, the probability measure P(A)=E[1AWt] is exactly pt.
Jarczynski's goal was to compute, at least numerically, the partition functions Zt or equivalently the free energy Ft=−logZt. The Jarczynski estimator, discovered in 1996, simply consists in applying (5) to φ=1, and noting that Wt is a simple ODE solved by
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 to prove Jarczynski's identity directly is to use the Feynman-Kac representation of PDEs, but this proof seems quite involved; instead, the trick above with the augmented system seems to be easier. It was shown to me by Eric Vanden-Eijnden.
Equation (5) can be reformulated as follows: if we note ϱt(x,w) the density of (Xt,Wt), then
Note that the integral is only on the positive axis because Wt is always positive from (6). We note qt(x) the term on the right; we are going to prove pt=qt by proving that both satisfy the same parabolic equation, then invoke a uniqueness result.
From now on, we'll set Ut=Vt−Ft, so that pt(x)=e−Ut(x) is normalized. Clearly, ∇Ut=∇Vt.
We consider the system Zt=(Xt,Wt) in (4) as a single SDE
ft(x,w)=(−∇Ut(x)−U˙t(x)w) and σ=diag(2,…,2,0).
Let us note ϱt(x,y) the density of (Xt,Wt), and write the associated Fokker-Planck equation; here and after, ∇ is a gradient and ∇⋅ is the divergence. For clarity in this section, subscripts denote partial differentiation with respect to the corresponding variables.
We're now left with checking that the solutions of
are essentially unique if they start with the same initial conditions, which is the case here. In the equation Bt=∇Ut and Ct=−U˙t. Uniqueness for solutions of parabolic equations is well-known to PDE theorists and can be found in Evans' book, chapter 7. We provide a proof for completeness. Indeed, by linearity, it is sufficient to check that the only solution of (17) started at u0(x)=0 is identically zero. Set γ(t):=∫∣ut(x)∣2dx; we're going to bound its derivative using "energy estimates". We have
The first term is −∫∣∇ut∣2. The second term is equal to +∫∇ut⋅utBt by integration by parts. Young's inequality a⋅b⩽λ∣a∣2+∣b∣2/4λ, valid for any λ>0, yields
In particular with λ=1 we cancel the first term and we obtain
Now, if Bt,Ct are bounded by some constant c, we get γ˙(t)⩽cγ(t). Grönwall's lemma yields γ(t)⩽γ(0)ect=0 and since γ(t) is always positive, we get γ(t)=0 and ut=0 for all t.