Langevin equation - derivative of random force?

[SchroedingersLion]

TL;DR Summary

calculation exercise, problems with Ito calculus?

Greetings,

I am struggling with an exercise to the Langevin equation.

Suppose we are given the following differential equation for a particle's 1D time-dependent momentum $p(t)$:
$$\text{d}p = -\gamma p \text{d}t + F(r)\text{d}t + \sqrt{C\gamma}\text{d}W $$
with a constant $C$, a position-dependent function $F(r)$, a friction parameter $\gamma$ and a Wiener process $W$.
Moreover:
$$ \text{d}r = \frac {p}{m}\text{d}t $$

I am interested in the solution in the limit $\gamma \rightarrow \infty$.

In a first step, I am to show that $p(t)$ is given by:
$$p(t)=p(0)e^{-\gamma t} + \int_0^t e^{-\gamma(t-s)}(F(r)+\sqrt{C\gamma}f(t)) \, ds $$

The hint was to first rewrite the differential equation (with a lack of rigor) as $\frac {dp}{dt} + \gamma p = F(r) + \sqrt{C\gamma}f(t)$, so apparently $f(t) = \frac{dW}{dt}$, a random force.

My naive try was now to simply take the time derivative of the given $p(t)$ and show that it suffices the differential equation of the hint. The integral can be calculated analytically, but I don't know what I have to do about the time derivatives of the function $f(t)$. I get the correct result, aside from terms ~$\frac {d}{dt} f(t)$. Can I assume that these are zero? If the random force is 0 aside from discrete points in time, this might or might not be legit. I might be missing some Ito calculus basics here, so would be happy for a hint.
SL

 

[vanhees71]

Usually you take the "Green's function" of the non-stochastic part and express the solution as an integral over the fluctuating force. Then you never need to take the derivative of the stochastic force, which is ill-defined.

 

[SchroedingersLion]

Thank you Vanhees.

Is there a way to do it without Green's function? We did not cover it in that lecture. The exercise also says that the proof does not need to be entirely rigorous.

 

[vanhees71]

Maybe, there is another way, but Green's functions are always good to learn about. Another way is of course to derive the Fokker-Planck equation from the Langevin equation and analyse the phase-space distribution function describing the stochastic motion of the Brownian particle statistically. You find a simple (non-rigorous) derivation for the most simple relativistic case in Sect. 3.1 of

https://arxiv.org/abs/0903.1096

 

[SchroedingersLion]

Thank you Vanhees, but I would like to follow the lined out hints. Maybe you can say something to the stuff below?

I made progress:
If I take the equation from the hint
$\frac {dp}{dt} + \gamma p = F(r) + \sqrt{C\gamma}f(t)$, replace $t$ with dummy variable $s$, multiply both sides with $e^{-\gamma(t-s)}$ and integrate over $s$ from $0$ to $t$, I obtain
$$p(t)=p(0)e^{-\gamma t} + \int_0^t e^{-\gamma(t-s)}(F(r)+\sqrt{C\gamma}f(s)) \, ds $$
This is what I need (aside from the fact that the $f(s)$ was an $f(t)$ in the exercise description in my first post, so I assume that was a misprint - otherwise one could calculate the integral analytically immediately).

In the next step, I should show that
$$ r(t)=r(0) + \frac{1}{m\gamma}p(0)(1-e^{-\gamma t}) +\frac 1 m \int_0^t \int_0^{s'}e^{-\gamma (s'-s)}(F(r)+\sqrt{C\gamma}f(t))\, ds \, ds' $$
This can be done by writing $\frac{dr(s')}{ds'} = \frac {p(s')}{m}$ and integrating over $s$ from $0$ to $t$ (only that I have $f(s)$ under the integral again).

The next step is to make use of Leibniz integral rule to rewrite $r(t)$ as
$$ r(t)=r(0) + \frac{1}{m\gamma}p(0)(1-e^{-\gamma t}) +\frac 1 m \int_0^t (1-e^{-\gamma(t-s)})(F(r)+\sqrt{C\gamma}f(t))\, ds $$
where I don't know whether it really is an $f(t)$ this time or not.

I have no idea how to use Leibniz rule here to obtain the desired equation.
If one tries to take the the time derivative of the double integral (assuming $f$ to be $f(s)$) and uses Leibniz rule, one obtains the correct exponential $e^{-\gamma(t-s)}$ under the integral, but the $1-$ is missing.

 

[vanhees71]

The problem is that $F(r)$ is not known but contains the variable $r$. The technique of the Green's function only works for the harmonic oscillator, of course, i.e., for $F(r)=-k r$. Then you have to look at the complete system of equations of motion,
$$p(t)=m \dot{r}(t), \quad \dot{p}(t)=-\gamma p + F(r) +\sqrt{C \gamma} f(t).$$
Now you can of cours write
$$m \ddot{r}+m \gamma \dot{r} + k r =\sqrt{C \gamma} f(t),$$
which you can rewrite using the Green's function of the operator on the left-hand-side, i.e.,
$$m \partial_t^2 G(t,t') + m \gamma \partial_t G(t,t') + k G(t,t')=\delta(t-t').$$
Using the retarded Green's function $G(t,t') \propto \Theta(t-t')$ leads to
$$r(t)=r_0 (t) + \int_0^t \mathrm{d} t G(t,t') \sqrt{C \gamma} f(t'),$$
where $r_0(t)$ is an arbitrary solution of the homogeneous equation,
$$m \ddot{r}+m \gamma \dot{r} + k r=0.$$

 

[SchroedingersLion]

You keep speaking about Green's function, but as it was not part of the lecture, I do not intend to make use of it.
The function $F(r)$ is not specified in the exercise.

There must be some calculational trick to go from

$$ r(t)=r(0) + \frac{1}{m\gamma}p(0)(1-e^{-\gamma t}) +\frac 1 m \int_0^t \int_0^{s'}e^{-\gamma (s'-s)}(F(r)+\sqrt{C\gamma}f(s))\, ds \, ds' $$
over to
$$ r(t)=r(0) + \frac{1}{m\gamma}p(0)(1-e^{-\gamma t}) +\frac 1 m \int_0^t (1-e^{-\gamma(t-s)})(F(r)+\sqrt{C\gamma}f(t))\, ds $$
using Leibniz rule.

 

[vanhees71]

Just flip the integration order (be careful with the boundaries). The expression you are using is still implicit though. BTW: What you are using is nothing else than the "free" Green's function.

 

[SchroedingersLion]

Thanks Vanhees, I got it now. Flipping order of integration leads to the result even without Leibniz rule. Don't get why they gave that rule as a hint.