Lecture 10 Notes: Persistent Random Walks and the Telegrapher’s Equation

Lecture 10: Persistent Random Walks and the Telegrapher’s Equation March 11, 2003 1 Review from last lecture To review from last lecture, we’ve considered a random walk with correlated displacements: XN = N ∑Δ x n < Δx n ⋅ Δx n +m >= C(m) , n =1 where C(m) is the correlation function. As an example, we considered an exponentially decaying correlation function: C(m) = m =e −2 m nc where nc = 2 −log We found: < XN2 > 2 = 1+ 1− N +2 −1 ( − 1)2 N We saw the transition from ballistic to diffusive behavior by scaling and taking the limit ρ�1: < XN2 > 2 = nc 1 N 1 −2N nc + (e −1) nc 2 ~ ~ N nc for N << 1 nc ballistic N nc for N >> 1 nc diffusive Lecture #10 2 2 Persistent Random Walk: Definition Here we consider the Persistent Random Walk example, a correlated random walk on a hypercubic lattice in which the walker has probability α of continuing in the same direction as the previous step. In d dimensions, the walker has probability (1-α)/(2d-1) of going any other direction. Today, we specifically consider the d=1 example (note that it is non-trivial to scale up to higher dimensions). Figure 1 shows a diagram of the one-dimensional persistent random walk. The walker (at position 0), is about to take step n after taking step n-1. step n-1 −2 σ α −σ step n σ 0 β=1−α 2σ Figure 1. The persistent random walk (d=1) The walker’s displacements are correlated as discussed last lecture. We determine the correlation coefficient, ρ: = < Δx n+ m ⋅ Δxn > 2 = − (1− ) = 2 −1 Note that if α=1/2 then ρ=0 and we have an uncorrelated iid random walk. If α=1 then ρ=1 and we have a perfectly correlated random walk (always moving same direction). If α=-1 then ρ=-1 and we have an anti-correlated random walk (hopping between two sites). We can already predict from the exponentially decaying correlation example that we will eventually get diffusive scaling:1 D( ) 1 + = Do 1− = where Do = 2 2 Our goal will be to solve for the probability distribution function Pn(m) of the displacement m after n steps for the persistent random walk. 1 Note that in higher dimensions (d>1) Do=σ2/2dτ. Lecture #10 3 3 Difference Equations We will solve for Pn(m) by use of difference equations, a technique introduced by M. Kai. First, we decompose Pn(m) into two probabilities: An(m) for the walker to end up at m after n steps coming from the left of m and Bn(m) for the walker to end up at m after n steps coming from the right of m. An(m) = Pr ob(Xn = m | Xn −1 = m − 1) Prob(X n−1 = m −1) Bn (m) = Prob( Xn = m | Xn −1 = m +1) Prob(Xn − 1 = m +1) Pn (m) = An(m) + Bn (m) Let α be the transition probability for the random walker to move in the same direction as the previous step. Let β (=1-α) be the transition probability for reversing direction. We can decompose the system into two situations (coming from left vs. right at step n), and for each of these possibilities, there are two ways to arrive at m (depends on direction of step n-1). α A m-1 m β β B m-1 m+1 m m+1 α Figure 2. A split probability diagram of the persistent random walk. Continuing in the same direction maps to horizontal movement on this diagram, whereas reversing direction corresponds to vertically jumping from A to B (or vice versa). Thus the difference equations for the probabilities An+1(m) and Bn+1(m) are: An+1 (m) = An (m −1)+ Bn (m − 1) Bn +1 (m) = An (m + 1) + Bn (m +1) The most useful property of decomposing Pn(m) and using the difference equations is that if the previous state [An(m),Bn(m)] is known, then the next state can be determined. Thus the evolution of [An(m), Bn(m)] is a Markov chain. Conversely, directly using the probability Pn(m) requires knowledge of two previous time steps. To solve for [An(m), Bn(m)] we can either solve the problem exactly with a DFT or we can find an asymptotic solution by taking the continuum limit. We will discuss the latter, which will yield a partial differential equation in the limit n�∞. Lecture #10 4 3.1 Continuum limit approximation Let’s consider continuous variables as n�∞ : a(m b(m p(m q(m ,n ,n ,n ,n ) ≡ An(m) ) ≡ Bn (m) ) ≡ Pn (m) = An (m) + Bn (m) ) ≡ Qn (m) = An (m) − Bn(m) The function q has been introduced which encodes the effect of correlations. Note that q=0 when α=1/2 (uncorrelated) and q is non-zero otherwise. Taylor expand the difference equations around x=mσ and t=nτ: 2 2 ⎛⎜ ⎛ ⎞ ⎞ a − ax + axx − ... + ⎜ b − bx + bxx −... 2 ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 2 ⎛ ⎛ ⎞ ⎞ b + bt + btt + ... = ⎜ a + a x + a xx +... + ⎜ b + bx + b +... ⎝ ⎠ ⎝ ⎠ 2 2 2 xx a + at + 2 att + ... = Then add these equations, using α+β=1, a+b=p, a-b=q, to get: pt + 2 2 ptt + ... = ( − ) q x + 2 2 pxx +... (1) Also subtract the two expansions to get: q + qt + 2 2 qtt + ... = ( − )q − px + ( − ) 2 2 qxx + ... (2) Take the x-derivative of Equation 2 : (1− + )q x + qtx + ... = − pxx + ... (3) and substitute this expression for qx into Equation 1 to arrive at: pt + 2 ptt + ... = ( − ) ⎛⎜ − pxx ⎝1 − + − 2 qtx ⎞ + p + ... 1 − + ⎠ 2 xx It will matter how we take the continuum limit to see which terms are important. (4) Lecture #10 5 3.2 Diffusive scaling Here we take the continuum limit (as n�∞, σ�0 & τ� 0) with Do=σ2/2τ remaining constant. Going back to the x-derivative equation (Eq. 3), in this limit and using β=1-α: qx ~ − p 2 xx (5) Substituting this back into Equation 1 and taking the limits σ�0 and τ�0 yields a diffusion equation with a modified diffusion coefficient: ⎛ pt = Do ⎜1 + ⎝ where D =1+ Do − − ⎞ p + O( ) ⎠ xx = = = 1− (6) 1+ 1− as expected for an exponentially decaying correlation function. 3.3 Ballistic scaling Here we will take the continuum limit in a different way in order to probe the transition from persistence (“ballistic scaling”) to diffusive scaling. We again take the limit as n�∞, σ�0 & τ� 0, however we now require v=σ/τ to remain fixed. We further take the strong persistence limit with α�1 and ρ�1: = 1− , 2 = 2 = − = 1− The scaling of ε with σ and τ is still to be determined. Returning to Equation 3 we see it simplifies to: q x ~ − qtx − pxx Now we introduce a characteristic timescale for ballistic motion τc=τ/ε. Applying this to Equation 7 gives: q x ~ − cqtx − v c pxx Now we return to Equation 1, and in this limit we find: (7) (8) Lecture #10 6 pt ~ ( − ) qx = −(1− )vq x (9) Substitute for the qtx term in Equation 8 by taking the time derivative of Equation 9: ptt ~ −(1− )vqtx ~ −vqtx (10) Inserting Equations 9 & 10 into Equation 8 yields the telegrapher’s equation: ptt + 1 pt = v 2 pxx c Which is at leading order for n�∞, (σ�0 & τ�0 approaching with σ/τ=v=constant), and α�1 (approaching as α=1-τ/2τc.) For times much smaller than τc, the telegrapher’s equation reduces to the wave equation; at times much longer than τc, it reduces to the diffusion equation. Thus it correctly models a signal which moves initially as a wave (Fig. 3A), but over time decays due to noise (Fig. 3B). A B Figure 3. A) Wave motion of a signal modeled by the telegrapher’s equation B) Diffusive motion of a signal modeled by the telegrapher’s equation. (11)