Let be a family of smooth potentials on , continuously evolving with time . Let be their associated Gibbs measure:
It is not easy to find a continuous dynamic , such that at each time the distribution of matches . One might be tempted to use the overderdamped Langevin dynamics,
with a Brownian motion. But the density of is not equal to in general. Indeed, solves the Fokker-Planck equation
If was equal to , then should also satisfy this equation; but for one easily checks that the RHS is zero, resulting in which is false in general if we choose a family of evolving potentials; it would only be true at equilibrium when is constant.
The Jarczynski trick allows to reconnect the distribution of with . It introduces a weight, the solution of an auxiliary ODE, such that its conditional expectation given gives , hence correcting the mismatch between and .
(In the sequel we note , and the dot in or always represents a derivative with respect to time. )
be the solution of the following augmented system
Then, for any test function
In other words, the probability measure
Jarczynski's goal was to compute, at least numerically, the partition functions or equivalently the free energy . The Jarczynski estimator, discovered in 1996, simply consists in applying (5) to , and noting that is a simple ODE solved by
Jarzynski's identity states that
The consequence of this identity is that
giving rise to a natural estimator for the free energy difference : one samples many paths from the dynamics (2), then computes the path integral along each path, and then average the values of 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 the density of , then
Note that the integral is only on the positive axis because is always positive from (6). We note the term on the right; we are going to prove by proving that both satisfy the same parabolic equation, then invoke a uniqueness result.
From now on, we'll set , so that is normalized. Clearly, .
We consider the system in (4) as a single SDE
Let us note the density of , 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.
Now by the definition of , we have , hence finally
We already set , so that for example. Multiplying the last equation by , integrating, and swapping integrals and derivatives, we get
We have , hence
An integration by parts shows that , provided a sufficient decay of at infinity. We obtain the following equation:
where I removed the subscript for clarity. This equation is a Fokker-Planck equation (first terms) but with a birth-death term added at the end.
It turns out that the density is a solution of this equation. Indeed, and , so that it is easy to check that (16) is satisfied.
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 and . 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 is identically zero. Set ; we're going to bound its derivative using "energy estimates". We have
The first term is . The second term is equal to by integration by parts. Young's inequality , valid for any , yields
In particular with we cancel the first term and we obtain
Now, if are bounded by some constant , we get . Grönwall's lemma yields and since is always positive, we get and for all .