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In a deterministic system, given by the system of differential equation

$$\frac{dx_n}{dt}=F_n(x)$$

Which is ergodi, and mixing with respect to a $\rho^{inv}(x)$, in a limited subspace of $R^N$,show that the correlation functions decays, for $t$ small and positive, as:

$$\left<x_n(t)x_n(0)\right>=\left<x_n^2\right>-Ct^2+ O(t^3)\qquad C>0$$

Instead, if we add a noise:

$$\frac{dx_n}{dt}=F_n(x)+\sqrt{2\epsilon}\eta_n$$

with $\eta_n$ white noises, delta correlated $$\left<\eta_n(t)\eta_m(t’)\right>=\delta_{nm}\delta(t-t’)$$

we have that the correlation function decays linearly:

$$\left<x_n(t)x_n(0)\right>=\left<x_n^2\right>-C(\epsilon)t+ O(t^2)\qquad C>0$$

  • How do I show these two results: in particular, how do I extract the $t^2$ dependence in the first case?
  • Furthermore, how do we reconcile the two results when $\epsilon<<1$?
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  • $\begingroup$ For the first part, I tried to Taylor expand the F_n to get the linear evolution of the (stai), but to have a second-order dependance I should bo on a maxiomium/minimum of F, but I don’t see why this should be the case. For the latter: is the fact that I’m approximating a deterministic process with a stochastic one, hence the second description is no longer valid? $\endgroup$
    – Drebin J.
    Jul 15, 2018 at 13:37
  • $\begingroup$ @Derbin J. Can you include the contents of your previous comment in the question? $\endgroup$
    – user191954
    Jul 15, 2018 at 13:42
  • $\begingroup$ @DrebinJ. Can you tell us the source of this problem? $\endgroup$
    – stafusa
    Jul 15, 2018 at 14:48
  • $\begingroup$ @stafusa it’s an exercise, no more context sorry Sure: most easily, if you Taylor expand around 0 we get: $x(t) = x(0) + F_n(0) t$ for small t Hence a linear decay (defining the average of F(0) and x(o) to be C). Apparently thought I’doing something very wrong $\endgroup$
    – Drebin J.
    Jul 15, 2018 at 14:59

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For the deterministic case, I think we need to assume that the dynamics generates a stationary ensemble, in which case we can write (time origin invariance) $$ \frac{d}{d t_0} \langle A(t_0+t) B(t_0) \rangle = \langle \dot{A}(t_0+t) B(t_0) \rangle + \langle A(t_0+t) \dot{B}(t_0) \rangle = 0 $$ where $A$ and $B$ are any functions of $x$, and the dot denotes the time derivative. Then we may set $t_0=0$ and, furthermore, $t=0$: $$ \langle \dot{A} B \rangle = - \langle A \dot{B} \rangle $$ the so-called "dot shifting" formula. Setting $A=B=x_n$ shows us that $$ \langle \dot{x}_n x_n \rangle = - \langle x_n \dot{x}_n \rangle = 0 $$ i.e the linear term in the Taylor expansion of $\langle x_n(t) x_n \rangle$ is equal to minus itself, and therefore vanishes. Of course this also seems to give us a restriction on the form of $F_n(x)$ since it implies $\langle x_n F_n(x)\rangle=0$. Setting $A=\dot{x}_n$, $B=x_n$ gives us $$ \langle \ddot{x}_n x_n \rangle = - \langle \dot{x}_n \dot{x}_n \rangle = -\langle F_n(x)^2\rangle $$ showing that the quadratic term in the Taylor expansion of $\langle x_n(t) x_n \rangle$ indeed has a negative sign, and giving the expression for $C$.

The situation with the stochastic equation is discussed in many textbooks on the general background to the Langevin equation, and on stochastic processes, and I will only give a few comments. Firstly, the extra linear term in the Taylor expansion of $\langle x_n(t) x_n \rangle$ is entirely the result of the noise term, which is why the coefficient $C(\epsilon)$ is written simply as a function of $\epsilon$. Therefore, when you let $\epsilon\rightarrow 0$, this term simply goes away, and there is no significant problem reconciling with the deterministic equation. Secondly, the stochastic differential equation, as written, is formally not correct: you need to separate the differentials $d x_n$ and $F_n(x) dt$ from the time integrated random force, which is not proportional to $dt$: it is a Wiener process. If you look at the simple Langevin equation, you'll see why this gives rise to an initial linear decay of the time correlation function (which is, incidentally, also related to the form of $F_n$ through the fluctuation-dissipation theorem).

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  • $\begingroup$ Thank you very much! Are there any assumptions we need to make to ensure we have a stationary ensemble? Which ones, in case? $\endgroup$
    – Drebin J.
    Jul 15, 2018 at 16:27
  • $\begingroup$ I'm afraid I can't say for sure. My worry is that an arbitrary right hand side $F_n$ might generate a nonstationary distribution; in that case, though, what are we to understand by $\langle \ldots \rangle$? I think it's cleanest to specify that this is a stationary ergodic process, but maybe others with more expertise will give a more satisfactory answer on this point. $\endgroup$
    – user197851
    Jul 15, 2018 at 17:06

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