Exercise sheet 3: Random walk and diffusion equation, the Wiener-Khinchin theorem, correlation function and power spectral density

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1 AMSI (217) Stochastic Equations and Processes in Physics and Biology Exercise sheet 3: Random walk and diffusion equation, the Wiener-Khinchin theorem, correlation function and power spectral density 1. Q1: Inhomogeneous biased random walk. A walker steps a distance with probabilities p(x) and q(x) to the right and to the left, respectively every δt seconds. Show that the Fokker-Planck equation is given by with the drift force f(x) = p(x) q(x) δt t P(x,t) = x (f(x)p(x,t))+d 2 xp(x,t), and the diffusion coefficient D = 2 2δt. 2. Q2: Diffusion equation. A one-dimensional domain is bounded by a wall at x = and has a sink at x = a. You are releasing random walkers at < x = x < a with the rate five walkers every second. Determine the average number of walkers in the domain in the stationary state. Solve the problem in the continuous limit, assuming that the random walkers are symmetric and the diffusion coefficient is D. 3. Q3: The Wiener-Khinchin theorem. The power spectral density S(ω) of the process s(t) is known. Find the corresponding stationary ACF for (a) (b) (c) S(ω) = S(ω) = S(ω) = 1 { 1 ω a, otherwise { 1 ω ω 1, otherwise 4. Q4: Power spectral density of a linear system. Compute the power spectral density S(ω) for the following stochastic processes (a) ẋ = αx+βx(t τ)+dξ(t) (b) The Van der Pol oscillator: ẋ = y, ẏ = x αy +βy(t τ)+dξ(t), where ξ(t) is the Gaussian white noise with ξ(t)ξ(t ) = δ(t t ). (Andrey Pototsky, Swinburne University of Technology, apototskyy@swin.edu.au)

2 Q1: Solution The master equation is given by Using Taylor series expansion We obtain P(x,t+δt) = p(x )P(x,t)+q(x+ )P(x+,t) P(x,t+δt) p(x,t)+ t p(x,t)δt+..., p(x )P(x,t) p(x)p(x,t) x (p(x)p(x,t)) x(p(x)p(x,t))+..., q(x+ )P(x+,t) q(x)p(x,t)+ x (q(x)p(x,t)) x(q(x)p(x,t))+... t P(x,t) 1 [p(x)p(x,t) x (p(x)p(x,t)) + 2 δt 2 2 x(p(x)p(x,t)) ] + q(x)p(x,t)+ x (q(x)p(x,t)) x(q(x)p(x,t)) P(x,t) [ = x p(x) q(x) ] P(x,t)+ 2 δt 2δt xp(x,t).

3 Q2: Solution The entire domain [,a] is divided into two parts: D 1 : [,x ] and D 2 : [x,a]. In the stationary regime, the walkers that move to the left from x into the domain D 1 will eventually return to x, after bouncing off the wall at x =. This implies that the probability current J 1 = D x P 1 (x) is zero in D 1. Therefore, the stationary density P 1 (x) in D 1 is constant P 1 (x) = C 1. In the domain D 2 the current J in the stationary regime is constant J = D x P 2 (x). Consequently, P 2 (x) = J D x+c 2, where the constants C 2 can be determined from the boundary conditions. Namely, we require that the density is continuous at x = x and that P 2 (x = a) = (absorbing boundary). This yields Solving for C 1 and C 2, we obtan C 1 = J D x +C 2, J D a+c 2 =. P 1 = J D (a x ), P 2 = J D (a x). The total number of walkers N in the domain is found as x a N = P 1 (x)dx+ P 2 (x)dx = J ] [(a x )x +a(a x ) a2 x D 2 + x2 2 = J [ a 2 x 2 ]. 2D The current J gives the number of walkers that pass through the system per unit of time (per one second) ( ) walkers J = 5. second Under this condition, the number of walkers in the domain is N = 5 2D (a2 x 2 ).

4 Q3: Solution (a) (b) G(τ) = ( a G(τ) = 2Re ( e iaτ ) 1 = 2Re iτ e ±iωτ dω = (2π)δ(τ) ) ( e e ±iωτ iωτ dω = 2Re iτ = 2sin(aτ). τ ) a Note that Dirac s delta function can be represented as sin(ax) lim = δ(x). a πx The ACF is shown in Fig.1 for three different values of a 4 sin(x)/x sin(2x)/x sin(5x)/x ACF(x) x -3 3 x Figure 1: ACF from part (b) 2sin(aτ) τ for a = 1,2, x

5 (c) ( 1 ) ( 1 e G(τ) = 2Re e ±iωτ iτ ) +τi (1 ω)dω = 2Re τ 2 The ACF is shown in Fig.2 = 2 1 cosτ τ 2. ACF(x) (1-cos(x))/x x Figure 2: ACF from part (c) 1 cosτ τ 2.

6 Q4: Solution (a) In the Fourier space Solving for ˆx Taking the modulus iωˆx = αˆx+βe iωτˆx+dˆξ ˆx = Dˆξ iω +α βe iωτ. S x (ω) = 1 D 2 2π (α βcos(ωτ)) 2 +(ω +βsin(ωτ)) 2 S(ω) α=.5 α= ω Figure 3: Power spectrum S(ω) from part (a) for β = 1, τ = 1 and two different values of α as in the legend.

7 (b) Van der Pol oscillator In the Fourier space iωˆx = ŷ, iωŷ = ˆx αŷ +e iωτ βŷ +Dˆξ. Solving for ˆx and ŷ ˆx = The power spectrum of the x coordinate Dˆξ 1 ω 2 +iωα iβωe ( iωτ) ŷ = iωˆx. S x (ω) = 1 D 2 2π (1 ω 2 βωsinωτ) 2 +ω 2 (α βcosωτ) 2. 1 S x τ= S x τ=1 S 1 S 1, S x 1 - ω m ω Figure4: PowerspectrumS x (ω)forτ = (solidline), τ = 1(dottedline)andthebackgroundspectrum S 1 (ω) (dashed line). Other parameters are β =.5 α = 1. The background function S 1 (ω) can be determined as the limit of the running average of S x (ω) over 2π/τ This yields S 1 (ω) = τ S 1 (ω) = lim τ 2π ω+2π/τ ω S x (ω )dω. D 2 [ω 2 1] 2 +ω 2 [(α+β) 2 +2(α+β)β].

8 ACF: A. Pototsky and N. Janson, Correlation theory of delayed feedback in stochastic systems below Andronov-Hopf bifurcation, Phys. Rev. E 76, 5628 (27) Valentin Flunkert and Eckehard Schöll Suppressing noise-induced intensity pulsations in semiconductor lasers by means of time-delayed feedback, Phys. Rev. E 76, 6622 (27) 6 3 (a) 4 (b) -3-4 ψ xx /D (c) (d) (e) (f) t/τ t/τ Figure 5: ACF of x in the units of D 2 at different values of the delay time τ: (a) and (b) τ = 1, (c) and (d) τ = 1, (e) and (f) τ = 3. Panels (a), (c) and (e) show the behaviour of the ACF on the scale of t = 25τ. Panels (b), (d) and (f) reveal the behaviour on the scale of t = 2τ.