# Stochastic differential equations with null mean and unit variance

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.

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. $$\left< * \right> = \int_t * \rho(t)dt$$ ):

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

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.

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

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:

$$G(\tau)\sim G(0)-a_1\tau \left+\tau \left \left< \eta(t)\right>$$ and now because $$\left< \eta(t)\right>=0$$ we get, again because $$\left =G(0)$$: $$G(\tau)\sim G(0)(1-a_1\tau)$$ 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.

• 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! – Enrique René Dec 8 at 22:30