Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone tell me where I am going wrong here? (I am leaving out any random fluctuation forcings, because I don't think they are relevant to my problem.)

1: $\displaystyle \frac{dv(t)}{dt}=-\eta v(t)$

2: $\displaystyle \frac{dv(t+\tau)}{dt}\frac{dv(t)}{dt}=\eta^2 v(t+\tau)v(t)$

3: $\displaystyle \frac{d}{d\tau}\left(v(t+\tau)\frac{dv(t)}{dt}\right)=\eta^2 v(t+\tau)v(t)$

4: $\displaystyle \frac{d}{d\tau}\left(\frac{d}{dt}(v(t+\tau)v(t))-\frac{dv(t+\tau)}{dt}v(t)\right)=\eta^2 v(t+\tau)v(t)$

5: $\displaystyle -\frac{d^2}{d\tau^2}\left(v(t+\tau)v(t)\right)=\eta^2 v(t+\tau)v(t)$

6: $\displaystyle \frac{d^2}{d\tau^2}\langle v(t+\tau)v(t)\rangle =-\eta^2 \langle v(t+\tau)v(t)\rangle$

That minus sign on the RHS can't be right, but I don't see where I am going wrong. Thanks.

share|improve this question
Where's Brownian motion here? –  Ilya Jun 6 '13 at 17:14
In this simple (and common) model, the effects of Brownian acceleration are divided into two pieces -- a deterministic exponential decay (since the particle in question will be getting more collisions opposing its motion than accelerating it) and a random part. I have, as I said in the question, left out the random part because I don't think it is the source of my problem. –  bob.sacamento Jun 6 '13 at 19:29
what I see in your question is just an ODE, lacking a random perturbation part. I recommend you to be more specific. –  Sangchul Lee Jun 6 '13 at 23:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.