Solutions of a Focker-Planck Equation

Transcription

1 Solutions of a Focker-Planck Equation W. C. Troy March 31, Overview.. The Langevin Equation for Brownian Motion. 3. The Focker-Planck equation. 4. The steady state equation. 1 Overview. In these notes we go over the analysis of a Focker-Planck equation which models the behavior of a rachet engine problem Our goal is show how the authors answer the following questions: i What is the Langevin equation for Brownian motion? ii What is the associated Focker-Planck equation? How do we find the energy equation? iii How to analyze the steady state equation and find its solution in closed form? How to find the equation for energy for the steady state equation? The Langevin equation for Brownian motion. The position of a Brownian motion particle at tmeperature T in a potential V x is described by the Langevin equation m d x = dv dx λdx + ζt,.1 where λ is the friction coefficient, and ζt is Gaussian white noise with zero mean and autocorellation E ζt, ζs =< ζtζt >= λk B T δt t.. Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 1560, U.S.A. 1

2 3 The Focker-Planck equation. The Focker-Planck equation associated with.1 is ρx, p, t = p t x m + V x + λm ] p p + λk B T p ρx, p, t, 3.1 where ρx, p, t is the probability density function for position x and momentum p = m dx particle at time t; T is temperature and k B is Boltzmann s constant. of the Our goal here is to show how to derive the energy evolution equation d E p m + V x = λ ] k B T E p m m. 3. The first step in the derivation of 3. is to multiply both sides of 3.1 by p. This gives t p ρx, p, t = p3 m x ρx, p, t + V p x + λm ] p p ρx, p, t + λk B T p ρx, p, t 3.3 p Integrate both sides of 3.3 with respect to p. The left side becomes t p ρx, p, tdp = t Ep. 3.4 To integrate the first term on the right side of 3.3 I m guessing that xρx, p, t is an even function of p or possibly p3 ρx, p, tdp is independent of x. If xρx, p, t is even then we conclude, since p 3 is odd, that p3 ρx, p, tdp = m x Integrating the second term on the right side of 3.3 gives p V x + λm ] p p ρx, p, t dp = V x p p ρx, p, tdp + λ m p pρx, p, tdp. p 3.6 Integration by parts reduces 3.6 to p V x + λm ] p p ρx, p, t dp = V x pρx, p, tdp λ p ρx, p, tdp. 3.7 m Recall that p = m dx. The the first integral on the right side of 3.7 can be written as V x Substituting pρx, p, tdp = V x ρx, p, tdp = 1 into 3.8 gives V x m dx ρx, p, tdp = mv x dx ρx, p, tdp. 3.8 pρx, p, tdp = m d V x. 3.9 From 3.9 and the identity Ep = p ρx, p, tdp it follows that 3.7 becomes p V x + λm ] p p ρx, p, t dp = m d V x λ m Ep. 3.10

3 Finally, integrating the last term on the right side of 3.3 twice by parts, we obtain Now combine 3.4, 3.5, 3.10 and 3.11 with 3.3 and obtain λk B T p p ρx, p, t = λk BT 3.11 t Ep = λk B T m d V x λ m Ep. 3.1 Dividing both sides of 3.1 by m and rearranging terms gives E p t m + V x = λ k B T E p m m This completes the derivation of The steady state equation. The time-independent steady state steady state equation is p x m + V x + λm ] p p + λk B T p ρx, p = 0, 4.1 or equvivalently, p m x ρx, p + p V x + λm ] p ρx, p + λk B T ρx, p = p We have two goals here. The first is to solve 4. and derive the steady state solution A ρx, p = πmkb T exp 1 ] p k B T m + V x, 4.3 where A is a constant. Our second goal is to show that the energy E p m = p m hpdp satisfies E p m = k BT. 4.4 To solve 4. we assume that ρx, p = wxhp, which reduces 4. to p m w xhp + λ m wxhp + V x + λ m p ] wxh p + λk B T wxh p = Divide both sides of 4.5 by pwxhp and get 1 w x m wx + λ m + Now break 4.6 into the system of two equations V x + λ m p ] h p php + λk BT h p php = λ pm + λ h p m hp + λk BT h p = 0, 4.7 php 3

4 1 w x m wx + V x h p = php To solve 4.7, divide both sides by λ, multiply by mphp, and obtain The solution of 4.9 which is important here is mk B T h p + ph p + hp = hp = πmkb T exp ρ mk B T since this solution saitsifies the normalization condition Next, we solve 4.8. First, it follows from 4.10 that Substituting 4.1 into 4.8, and manipulating the algebera results in The general solution of 4.13 is hpdp = h p php = 1 mk B T. 4.1 w x wx + V x = k B T wx = Aexp V x, 4.14 k B T where A is a constant. Multiplying 4.10 and 4.14, we obtain the required steady state solution A ρx, p = πmkb T exp 1 ] p k B T m + V x Finally, we show how to derive 4.4. The first step here is to multiply boths sides of 4.10 by and get Integration of both sides of 4.16 gives mk B T mk B T p h p + p 3 h p + p hp = p h pdp + Integrating the first integrals twice by parts, we conclude that p h pdp = p 3 h pdp + p hpdp = hpdp = 4.18 since hp the solution given in 4.10 satisfies Integration of the second integral by parts shows that p 3 h pdp = 3 p hpdp

5 Now substitute 4.17 and 4.19 into 4.17 and get Since p hpdp = Ep, we can write 4.0 in the form This completes the derivation of 4.4. p hpdp = mk B T. 4.0 E p m = k BT