I have the following:

$ \dot{x} = \frac{dx}{dt}= A\left( x\right) + \sqrt{B\left( x\right)}\eta\left( t\right) $

where $ A\left( x\right)=a_0 - a_1x $ and $ B\left( x\right)=b_0-b_1x+b_2x^2 $. All $ a_k,b_k \geq 0 $. $ \eta $ is related to a gaussian with null mean and unit variance.

Defining $ G\left( \tau\right)=\langle x\left( t\right)x\left( t+\tau\right)\rangle $ and supposing $ a_0 = 0 $ we have to prove that:

$ G\left( \tau\right) = G\left( 0\right)\ e^{a_1\tau} $.

I tried this:

1) Considering $ \tau $ small enough to allow the use of approximation $ x\left( t+\tau\right)=x\left( t\right)+\frac{1}{2}\tau\ \dot{x}\left( t\right) \ $, I do:

\begin{align*} G\left( \tau\right) &= \langle x\left( t\right)x\left( t+\tau\right) \rangle \\ &= \int x\left( t\right)\left[ x\left( t\right) + \frac{1}{2}\tau \dot{x}\left( t\right) \right]\rho\left( x\right)dx \\ &= \int x\left( t\right)^2\rho\left( x\right)dx + \frac{1}{2}\tau\int x\left( t\right)\dot{x}\left( t\right)\rho\left( x\right)dx \\ &= \int x\left( t\right)^2\rho\left( x\right)dx + \frac{1}{4}\tau\int \frac{dx^2}{dt}\rho\left( x\right)dx \\ &= \langle x\left( t\right)^2\rangle + \frac{\tau}{4}\langle\frac{dx^2}{dt}\rangle\ . \end{align*}

Since the system is in thermodynamic equilibrium, $\frac{d\rho}{dt} = 0 $ and then:

$ G\left( \tau\right) = \langle x^2\rangle + \frac{\tau}{4}\frac{d}{dt}\langle x^2 \rangle $

I don't see how this result can help me to get the proof. In this way the $ a_0=0 $ hypothesis was not required, which makes me think I'm in a way won't help me. The only thing I can see from here is something like:

$$ G\left( \tau\right) = \langle x^2\rangle\left( 1 + \frac{\tau}{4}\frac{d}{dt}\right) \Rightarrow G\left(\tau^\prime\right)=\langle x^2\rangle e^{\frac{\tau^\prime}{4}} = G\left( 0\right) e^{\frac{\tau^\prime}{4}} \neq G\left( 0\right)\ e^{a_1\tau}\ ,$$

where $ \tau $ is small and $ \tau^\prime $ arbitrary.

2) Doing the same approximation of "1)" I decided to use the $ \dot{x} $ equation:

\begin{align*} G\left( \tau\right) &= \langle x\left( t\right)^2\rangle + \frac{1}{2}\tau\int x\left( t\right)\dot{x}\left( t\right)\rho\left( x\right)dx \\ &= \langle x^2\rangle + \frac{\tau}{2}\int x\left( t\right)\left[ A\left( x\right) + \sqrt{B\left( x\right)}\eta\left( t\right)\right]\rho\left( x\right)dx \\ &= \langle x^2\rangle\left( 1 - \tau\frac{a_1}{2}\right) + \frac{\tau}{2}\eta\left( t\right)\int x\left( t\right)\sqrt{B\left( x\right)}\rho\left( x\right)dx \ . \end{align*}

I stopped here because the coefficients of $ B $ don't appear at the expression what I want to get. If I neglect the last integral making $ \eta\rightarrow 0 $ I have something like:

$ G\left( \tau\right) = \langle x^2 \rangle\left( 1 - \tau\frac{a_1}{2}\right)^1 \approx \langle x^2 \rangle\left( 1 - \tau\frac{a_1}{2}\right)^{1+\tau} = \langle x^2 \rangle\left( 1 - \frac{1}{n}\frac{a_1}{2}\right)^{1+\frac{1}{n}} \ .$

The last step was based on the arquimedian property of real set, $ n $ is a natural number. I almost can see the $ n\rightarrow \infty $ making $ G\left( \tau\right) = \langle x^2 \rangle e^{-\frac{a_1}{2}} = G\left( 0\right) e^{-\frac{a_1}{2}} \neq G\left( 0\right)\ e^{a_1\tau} $ that's what I want.

This problem comes from Statistical Mechanics discipline of Mastering program on physics. As I assume $ \tau $ very small to make these approximations, I think the $ G\left(\tau\right) $ is something like infinitesimal generator of something in the system.

I appreciate some guidance to solve this. I appreciate most some guidance with mathematical rigor, telling why some step can (or cannot) be taken.


1 Answer 1


First of all, there a couple of errors in your computations. For example, the average you are taking are over time so you should use $\rho(t)dt$, not $\rho(x)dx$!. Also the Taylor approximation should be $$x(t+\tau)\sim x(t)+\tau \dot{x}$$

Moreover, approximating $G(\tau)$ for small $\tau$ would just give you an hint of what would happen at small $\tau$, you would not be able to recover the full $G(\tau)$. If you had not made the mistakes you did, indeed, following your computations but slightly corrected and using $\left< * \right>$ for the average of $*$ (i.e. $\left< * \right> = \int_t * \rho(t)dt$ ):

$$G(\tau)\sim \left< x(t)(x(t) +\tau\dot{x} ) \right>=\left<x^2(t)+\tau x(t)\dot{x} \right>=\left< x^2(t) \right> +\tau\left<x(t)\dot{x} \right> $$

now, using the expression you have for $\dot{x}$ and the fact that $\left< x^2(t) \right>=G(0)$:

$$G(\tau)\sim G(0) + \tau \left < x\left(-a_1 x+\sqrt{B(x)}\eta(t)\right) \right>$$ i.e.

$$G(\tau)\sim G(0)-\tau a_1\left<x^2\right>+\tau\left< x\sqrt{B(x)}\eta(t)\right>$$

now we make the assumption that $\eta(t)$ is not correlated with the $x$-terms [notice that this is the only step in which I actually have to assume. I think it is right or that any similar assumption applies, but maybe think about it), i.e. that we can write:

$$G(\tau)\sim G(0)-a_1\tau \left<x^2\right>+\tau \left<x\sqrt{B(x)}\right> \left< \eta(t)\right>$$ and now because $\left< \eta(t)\right>=0$ we get, again because $\left <x^2(t)\right>=G(0)$: $$G(\tau)\sim G(0)(1-a_1\tau)$$ which is the small $\tau$ expansion of the solution you need: $$G(\tau)=G(0)e^{-a_1\tau}\sim G(0)(1-a_1\tau)$$

(I get a minus sign which you don't have, which I think is also right as otherwise the correlation would increase over time, which is weird... who of us made the mistake..?)

Anyways this procedure could have given you a hint, and a small-$\tau$ proof of the result, but not the final solution.

What instead if try to compute

$${d G(\tau)\over d\tau} = \left< x(t){dx(t+\tau)\over d \tau}\right>$$ (where I only take the derivative of the second one because the first on has no $\tau$ dependence)?. So as ${dx(t+\tau)\over d \tau}={dx(\tau)\over d \tau}|_{t+\tau}=\dot{x}|_{t+\tau}$:

$${d G(\tau)\over d\tau} = \left< x(t)\left(-a_1x(t+\tau)+\sqrt{B(x)}\eta(t+\tau)\right)\right>$$ for the exact same reasons as before $$\left<\eta(t+\tau)\right>=0$$ and we are left with $${d G(\tau)\over d\tau} = -a_1\left< x(t)x(t+\tau)\right>=-a_1G(\tau)$$ so that our solution is, solving the easy $\dot{y}=-Ay\rightarrow y(t)=y(0)e^{-At}$ differential equation

$$G(\tau)=G(0)e^{-a_1\tau}$$ (again with a minus sign which I trust - but I am open to discussion!)

Hope this helps not only solving it, but also showing some of your mistakes and wrong (but still not trivial!) reasoning.

  • $\begingroup$ In fact your proof (with $-a_1$ signal) is right. Probably some typo from who build the exercise. This help me to clarify some points also, as you point out my mistakes. Thanks a lot! $\endgroup$ Commented Dec 8, 2018 at 22:30

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