5 Applying the Fokker-Planck equation

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1 5 Applying the Fokker-Planck equation We begin with one-dimensional examples, keeping g = constant. Recall: the FPE for the Langevin equation with η(t 1 )η(t ) = κδ(t 1 t ) is = f(x) + g(x)η(t) t = x [f(x)p ] + κ x with appropriate initial and boundary conditions. Ex 1: Overdamped Brownian motion Langevin equation: (f, g 1) with x() = x. The FPE is with initial condition t = η(t) [ g(x) ] x (g(x)p ), = [P ] + κ x x P = κ P x, P (x, ) = δ(x x ) and boundary conditions P as x. This is the well-known heat or diffusion equation, with solution on x (, ) given by 1 P (x, t) = e (x x ) 4κt. 4πκt We can immediately find e.g. moments x(t) n = x n P (x, t). 1

2 Thus the mean is x(t) = x and the variance is x(t) x = κt. Note x(t) is a non-stationary process. Large-time behaviour: at any fixed point x, P (x, t) as t. Ex : The Ornstein-Uhlenbeck process Langevin equation for f = αx, g 1, with α a positive constant: with x() =. = αx + η(t), This is a linear equation for x, forced by a Gaussian white noise: we therefore expect x to be a Gaussian process also. The FPE is t + x [ αxp ] P κ x =, with initial condition P (x, ) = δ(x) and boundary conditions as in the previous example. This may be solved explicitly for P (x, t), see e.g. page 1 of [Risken], but an easier question is to find the stationary distribution if it exists. P (x) = lim t P (x, t), In steady-state, t, so we solve the ODE d (αxp ) κ d P = to determine P (x). Integrating: αxp + κ d P = S, where the constant S is called the probability flux.

3 As x we expect P, faster than x 1 (in order to have unit integral): S =. Then d P = α κ xp P (x) = C exp [ α ] κ x. The constant C is found by the normalization requirement α P (x) = 1 C = πκ. Note: if a stationary solution exists for a one-variable (n = 1) problem then this approach (setting / t to zero) yields an ODE for P (x) which can often be solved relatively easily. If no stationary solution exists, this method gives unnormalizable solutions. Ex 3: Example with no stationary solution One-dimensional overdamped Brownian motion (Ex 1), with FPE t = κ P x. We already know the time-dependent solution P (x, t), but suppose we try to find a stationary solution P (x) by setting / t to zero: d P = P (x) = Ax + B. The boundary conditions at ± require A =, giving P = B, a constant. But this solution cannot be normalized, since does not exist. P (x) = B Conclude: no stationary solution. This agrees with our knowledge of the Gaussian hump with always-growing wih and shrinking height. 3

4 Ex 4: A bistable potential The Langevin equation with f(x) = U (x), g 1, and potential is U(x) = x4 4 x = x3 + x + η(t). The deterministic system has equilibrium points at x = 1, x = 1 (both stable), and an unstable point at x =. Look for a stationary solution to the case with white noise: η(t)η(t ) = κδ(t t ). The FPE is t x [U (x)p ] κ P x =, and we look for a stationary solution P (x): κ d P + d [U P ] =. Integrating: dp = U κ P + C 1, and C 1 = as before because of boundary conditions as x ±. So P (x) = C exp [ 1κ ] U(x), provided this is normalizable. In this example we get [ P (x) = c exp 1 ( )] x 4 κ 4 x. This is a bimodal distribution for small κ: most of time is spent near the stable points of the deterministic system. A two-dimensional example 4

5 The D deterministic system (with positive constants ν and Ω) 1 = ν ( 1 x 1 x ) x1 Ωx = ν ( 1 x 1 x ) x + Ωx 1 (1) can be expressed in polar coordinates r = x 1 + x θ = tan 1 x x 1 as dr dθ = ν(1 r )r = Ω. [To see this, set x 1 = r cos θ and x = r sin θ]. This represents a nonlinear ( self-sustained ) oscillator, with an asymptotically stable limit cycle at r = 1. The rotation frequency is Ω, a constant. The steady-state deterministic solution in the original (rectangular Cartesian) variables x 1 and x is where θ is the initial phase. x 1 (t) = cos(ωt + θ ) x (t) = sin(ωt + θ ), This is an asymptotically stable oscillating system with frequency Ω. This mathematical model of nonlinear oscillation arises in many applications, e.g. semiclassical laser equations (Risken chapter 1), with x 1 (t) and x (t) being the real and imaginary components of the electric field [Risken equation (1.39)]. Electronic oscillators, e.g. Van der Pol s circuit. Dynamical systems near a Hopf bifurcation point. It is therefore of considerable interest to understand the effects of noise upon the system (1). For now, we consider additive white noise, with independent equalintensity noise sources η 1 (t) and η (t) in the x 1 and x directions, respectively. 5

6 The Langevin equations are with 1 = f 1 (x 1, x ) = ν ( 1 x 1 x ) x1 Ωx + η 1 (t) = f (x 1, x ) = ν ( 1 x 1 x ) x + Ωx 1 + η (t), η i (t)η j (t ) = κδ ij δ(t t ). The Fokker-Planck equation for the joint PDF P (x 1, x, t) is (in rectangular Cartesian variables): t + [f 1 P ] + ( ) P [f P ] κ + P =. x 1 x x 1 x This is more usefully written in cylindrical polar coordinates [Risken equation (1.39)] for P (r, θ, t): t + 1 [ ν(1 r )r P ] + ( ( 1 r r θ [ΩP ] κ r ) + 1 ) P =. r r r r θ Let s seek a stationary solution of this D problem, i.e. find P (r, θ) satisfying ( ( 1 [ ] ν(1 r )r 1 P + r r θ [ΩP ] κ r ) + 1 ) P =. r r r r θ Physical intuition: expect probability to be spread around the limit cycle. We try looking for a θ-independent solution, so set θ =. This reduces the equation to a one-variable problem, i.e., an ODE: ( 1 d [ ] ν(1 r )r 1 d P κ r dp ) =. r dr r dr dr Solve this ODE: [ ( )] ν r P = B exp κ r4, 4 6

7 with B given by the normalization condition rdr π dθp (r) = 1 rp (r)dr = 1 π ( B = π B = Here we have used the error function: ν κ π 3 erf(z) = π z [ ( )] ) ν r 1 r exp κ r4 dr 4 e ν 4κ 1 + erf ( ). ν 4κ e t. To understand the form of P, examine P (r) near the deterministic limit cycle r = 1. Let r = 1 + ρ, with ρ 1. Then and so r r ρ, P Be ν κ( 1 4 ρ ) = ν κ π erf ( )e ν κ ρ ν 4κ This is a Gaussian hump centered at ρ = (r = 1) and of variance σ = κ ν, so the wih (standard deviation) of the hump is proportional to κ and inversely proportional to ν. Phase dynamics Often (especially in electronic circuit applications) we are concerned only with noise effects on the phase angle θ(t). In this example the phase dynamics are very simple in the deterministic case: θ = Ω, and when the noise terms are added, we get θ = Ω η 1(t) r sin θ + η (t) r cos θ. 7

8 If we assume r 1 (amplitude strongly controlled), then we have a 1D Langevin equation (with multiplicative noises) θ = Ω η 1 (t) sin θ + η (t) cos θ. Note θ is an angle here so P (θ) must be periodic in θ. This leads us to the question of boundary conditions for the Fokker-Planck equation. 8