I have a very basic question regarding the mean of the Langevin equation. So we have an equation of the form:
$$\dot{v}(t)=-\beta v(t)+ \xi (t)$$
Where $\xi (t)$ is a Gaussian white noise with an average zero and a $\delta$ correlation in time.
As known, this equation has the following solution:
$$v(t)=v(0) e^{-\beta t}+\int_0^t dt' e^{-\beta (t-t')} \xi(t') $$ and I want to take the mean of this equation, i.e., $\langle v(t)\rangle$
The second term goes is zero since $\langle\xi (t)\rangle=0$, which leaves the first term. This, according to some books I have been reading, should be: $$\langle v(t)\rangle=v(0) e^{-\beta t},$$ but I really don't get how we got this result? It's a bit confusing to me.

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    $\begingroup$ Doesn’t it just follow from your equation? $\endgroup$ – CAF Dec 30 '18 at 17:13
  • $\begingroup$ Are you asking how the solution $v(t)$ is derived? Or the mean of the solution reduces to the first term of the solution? $\endgroup$ – Kyle Kanos Dec 30 '18 at 17:29
  • $\begingroup$ How does it follow from my equation? it's just not my brightest day today lol $\endgroup$ – George Farah Dec 30 '18 at 17:31
  • $\begingroup$ How the mean of the solution reduces to the first term.. $\endgroup$ – George Farah Dec 30 '18 at 17:32
  • $\begingroup$ Well you state it in the post itself: $\xi$ is a Gaussian random variable with mean zero, so the mean of $v(t)$ reduces to what's left: $v(0)\exp(-\beta t)$. $\endgroup$ – Kyle Kanos Dec 30 '18 at 17:39

Let me answer the original question as I understood it literally. It has a methodological value.

I understand taking a mean value as some integration over a limited period $T$: $$\langle f(t)\rangle=\frac{1}{T}\int_t^{t+T}f(\tau)d\tau$$ Only doing this we may "eliminate" the fluctuating force $\xi(t)$. Now, applying this averaging to the solution $v(t)$ we obtain: $$\langle v(t)\rangle=\frac{1}{T}\int_t^{t+T}v(0)\text{e}^{-\beta \tau}d\tau + \frac{1}{T}\int_t^{t+T}\text{e}^{-\beta \tau}d\tau\int_0^{\tau} dt' \text{e}^{t'} \xi(t')$$ Now let us take the first integral: $$\frac{1}{T}\int_t^{t+T}v(0)\text{e}^{-\beta \tau}d\tau=v(0)\frac{\text{e}^{-\beta t}(1-\text{e}^{-\beta T})}{\beta T}$$ This is an exact result of our averaging over some period $T$. Now, if inequality $\beta T\ll 1$ holds, in the first approximation the difference in the nominator is equal to $\beta T$ with a good precision, thus you obtain the answer to the original question.

In order to make sure that the second term in the exact solution vanishes after averaging, you must respect - in addition! - another inequality, namely $T/\delta\gg 1$. Otherwise some "long-time" fluctuations will still be present in the mean solution. I leave the proof of this as an exercise to those who downvoted my answer without explanation.

Finally, let me note that although the noise force mean value is zero, it does not mean that this force does not displace the particle in the space. Originally still particle ($v(0)=0$) and without friction ($\beta=0$) may be found elsewhere: $x(t)=\int_0^t dt'\int_0^{t'}\xi(t'')dt''\ne x(0)$, remember the Brownian motion, for example. Another (funny) example: a periodic external force $F(t)=F_0\cdot \sin(\omega t)$. On average this force is zero, but it displaces the particle with time far far away ;-) One cannot obtain this result by just dropping out this force from the exact equation of motion.

EDIT: As pointed out Alexander in his comment, "Langevin equation is about stochastic processes. Averages of stochastic processes are (usually) with respect to different noise realizations (all taken from same distribution). In different language - tracing out the ensemble of stochastic environment." In other words, averaging is done in fact over some parameter independent of time. It is like averaging over a random initial phase $\varphi_0$ in my last example with a periodic external force $F_0\cdot\sin(\omega t + \varphi_0)$. Thus, it is not averaging the equation or its solution over $t\le\tau\le (t+T)$. Although mathematically it is possible, I do not fully understand how it may naturally arise in physical calculations and what meaningful conclusion one can draw from such averaged things.

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  • $\begingroup$ I don't see how this answers the question. $\endgroup$ – Kyle Kanos Dec 30 '18 at 18:25
  • $\begingroup$ Please, define what is meant by taking the mean value explicitly. Is it an integration over a finite period? $\endgroup$ – Vladimir Kalitvianski Dec 30 '18 at 18:27
  • $\begingroup$ Langevin equation is about stochastic processes. Averages of stochastic processes are (usually) with respect to different noise realizations (all taken from same distribution).In different language - tracing out the ensemble of stochastic environment. $\endgroup$ – Alexander Dec 31 '18 at 13:37
  • $\begingroup$ @Alexander: Thus, averaging is done in fact over some parameter independent of time. I was mislead by the $\delta$ as a time constant in the time-dependent force $\xi(t)$. $\endgroup$ – Vladimir Kalitvianski Dec 31 '18 at 13:47

To see why the first is left when taking the mean, re-write the SDE as, $$ \mathrm dv=-\beta v\mathop{}\!\mathrm dt+g(x,\,t)\mathop{}\!\mathrm dW, $$ where $g(x,\,t)$ is some function and $\mathrm dW$ the stochastic noise. Then we can obtain a differential equations for the mean of $v$ by taking the mean of both sides: $$ \mathrm d\langle\mathrm v\rangle=\langle\mathrm d v\rangle =\langle -\beta v\rangle\mathop{}\!\mathrm dt $$ since $\beta$ is a constant and $\langle\mathrm dW\rangle=0$. This can be rearranged to, $$ \frac{\mathrm d\langle v\rangle}{\mathrm dt}=-\beta\langle v\rangle, $$ the solution of this differential equation should be immediately seen: $$ \langle v(t)\rangle=e^{-\beta t}\langle v(0)\rangle $$ which matches the sources you've seen, aside from the lack of $\langle \cdot\rangle$ wrapping $v(0)$, which I suspect is some level of your confusion (the other bit being the exponential being dependent on $t$, though this confusion should be cleared up from the derivation above).

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  • $\begingroup$ Actually you solve another equation- your second one, which you postulate rather than derive; thus it has a limited relation to OP question. $\endgroup$ – Vladimir Kalitvianski Dec 31 '18 at 8:09
  • $\begingroup$ @VladimirKalitvianski my Eq 2 is derived from Eq 1, which isn't exactly the same as OPs but has the same deterministic part which is the important part in the mean. Don't be so upset that your answer isn't well received that you "take revenge" on a better answer. $\endgroup$ – Kyle Kanos Dec 31 '18 at 12:16
  • $\begingroup$ You too, don't be upset that your answer is not the right one. You do not derive the second equation! $\endgroup$ – Vladimir Kalitvianski Dec 31 '18 at 13:08
  • $\begingroup$ @vlad I suppose you're right in that I don't derive it, it just follows from the text (i.e., applying averages). I think your answer still overcomplicates the matter compared to mine, but it is significantly better now that you've fleshed out why the decaying matters (instead of your simply asserting that's the reason). I'm going to remove the "edit" market in your post, though, because SE tracks edits and those interested can just click the edit history to see what changed. $\endgroup$ – Kyle Kanos Dec 31 '18 at 13:26
  • $\begingroup$ With removing some parts of my text, you violate my authorship. In addition it makes the paragraphs somewhat inconsistent. I disagree with such a practice. $\endgroup$ – Vladimir Kalitvianski Dec 31 '18 at 13:32

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