Chapter 3 Random Process & Partial Differential Equations
3.0 Introduction Deterministic model ( repeatable results ) Consider particles with coordinates rr 1, rr, rr the interactions between particles modeled by potential VV rr 1, rr, rr the dynamics of the system represented by ordinary differential equations d ri mi = f, 1,,3 i i = dt V V V fi = rv ( r1, r, r ),, i = xi yi zi These coupled equations can be solved using numerical method
3.0 Introduction Verlet algorithm Taylor expansion 相加 相减 ( t ) + O Verlet integrator is an order more accurate than integration by simple Taylor expansion alone, with the same term ΔΔtt.
3.0 Introduction Verlet algorithm is not self-starting,we will use velocity Verlet algorithm in molecular dynamics simulations. v ( ) v( ) velocity Verlet algorithm ( t) + a( t+ t) a t+ t = t + t+ O t ( ) ( 3 t ) + O
3.0 Introduction Probabilistic model ( unrepeatable results ) Consider the same system of particles as before when there is uncetainty, say rrrrrrrrrrrr ffffffffff RR ii, which has a Gaussian probability distribution with correlation function ( t) ( t' ) = 6 γktδ ( t t' ) R R γγ: friction coefficient i i B Langevin Equation: d ri dri m ( ) i + γ = f, 1,,3 i + Ri t i = dt dt V V V f ( ) i = rv r1, r, r,, i = xi yi zi Each particle's motion can be described by certain probabilities, derived from Fokker- Planck Equation.
3.0 Introduction (a) Wingless Locusts marching in the field. (b) A rotating colony of army ants. (c) A three-dimensional array of golden rays. (d) Fish are known to produce such vortices. (e) Before roosting, thousands of starlings producing a fascinating aerial display. (f) A herd of zebra. (g) People spontaneously ordered into traffic lanes as they cross a pedestrian bridge in large numbers. (h) Although sheep are known to move very coherently, just as the corresponding theory predicts, when simply hanging around (no motion), well developed orientational patterns cannot emerge. Tamás Vicsek, Collective motion, Physics Reports, Vol 517, 01, Pages 71-140
3.0 Introduction
our primary goal: 3.0 Introduction to investigate the connection between probabilistic and deterministic models of the same phenomenon. Connection? probabilistic deterministic Micro view: single random process Macro view: (ensemble average) definite distribution function described by partial differential equation Looks paradoxical that a random process can be characterized by a definite equation But we know it is true from a lot of daily experience, such as coin tossing.
3.0 Introduction coin tossing many parameters unknown the initial orientation, velocity, and spin; the properties of the table surface; Various atomic defects, dislocations, grains, voids ~50% ~50% For single toss: no idea whether head or tail turns up. After a large number of tosses, proportion of heads or tails is ~0.5. With this example, it is not so surprising that there is a determinable distribution of probabilities which characterizes a random process.
3.0 Introduction Probabilistic model originates from Incompleteness of information e.g. coin tossing Parameters sensitivity --- tiny perturbation in input induces huge variation in output. e.g. In kinetics of gases, a slight change in the initial conditions would result in a tremendous change after many collisions. By an averaging of the solutions with varying initial conditions random processes can be modeled successfully.
3.0 Introduction in the following sections, Section 3.1 1-D Brownian motion. An explicit expression of the probability w(m, ). Section 3. Simplified expression of w(m, ). Asymptotic Series, Laplace s Method Section 3.3 a difference equation for w(m, ) leads to a partial differential equation. Section 3.4 The connection between probability and differential equations.
3.1 Random Walk in One Dimension; Langevin s Equation In Brownian motion, small particles move about in liquid or gas. no possibility and no interest of computing the trajectory of each molecule. one wishes to have an average understanding of the phenomenon. The Roman Lucretius's ( 卢克莱修 ) described Brownian motion of dust in his scientific poem "On the ature of Things" (60 BC) Botanist Robert Brown in 187 studied pollen grains suspended in water. Albert Einstein in 1905 solved this problem.
3.1.1 An one dimension random walk model The minimum model of random walk: 1-D lattice model b = 1 a a Particle moves according to the following rules: Move in steps of a fixed length dx in a fixed time interval dt. The probability to the right p and to the left q=1-p. Goal: to obtain the probability w(m, ) m steps to the right of the origin the total steps Probability = number of observed events total number of events
3.1.1 An one dimension random walk model random walk model can be found in various situations a drunk staggering down a street a gambling game in which a coin is tossed Polymer Physics: Freely jointed chain model
3.1.1 An one dimension random walk model Fair coin tossing Head ~50% Tail ~50% 0.5 0.5
Random walk 3.1.1 An one dimension random walk model
3.1. Explicit solution To find w(m, ), the probability that a particle at a point m [-,] steps to the right of its origin after total steps. Suppose that the particle p steps to the right, p>0 -p steps to the left Displacment m m = p - (-p) = p p = ( + m)/ m p p e.g. =1 -p=5 p=7 m= is even m is even. is odd m is odd For example, if =3, the possible values of m = -3, -1, 1, 3. if =4, the possible values of m = -4, -, 0,, 4.
3.1. Explicit solution p p To find out the number of paths with p steps to the right and - p to the left. left right equivalent m Path 1: Step: 1 3 4 5 6 7 8 9 10 11 1 Path : equivalent Step: 1 3 4 5 6 7 8 9 10 11 1 The number of choices with p indistinguishable pink ball in boxes
3.1. Explicit solution Consider p distinguishable balls, which can be placed in boxes in the following number of ways: 1 4 3 5 7 8 ( 1)( ) ( 1) p+ =! ( p)! 1 4 7 3 8 5 Interchanging distinguishable balls does not change the pattern. There are p! permutations of p balls. 排列
3.1. Explicit solution number of ways p distinguishable balls = can be placed in boxes! p ( )! number of full box empty box patterns = C p number of full permutations of distinguishable balls within a pattern p! binomial coefficient C p =! p! p! ( ) ( ) p p p p= 0 x y C x y + =
3.1. Explicit solution The total number of possible path is the probability that a particle at a point m steps to the right of its origin after total steps. wm, = ( ) C p where p = ( + m)/ The sum of all probabilities is unity (, ) wm m= p= 0 1 = Cp 1 1 = C p p= 0 1 1 = + = 1 p p wm, ( 0) m
3.1. Explicit solution Characteristic functions the characteristic function of any real-valued random variable defines its probability distribution. Let iθ i ( ) = ae + be λθ θ b = 1 a a λ ( ) ( θ ) = ae + be = a e + abe + b e iθ iθ iθ i0 iθ P ( m) m= m=0 m=- P ( ) = a a P ( ) 0 = ab ( ) P = bb these coefficients are the probability λθ = ae + be = e x = ± 1 iθ iθ iθx characteristic function ( ) mean
To extract the coefficient analytically, for example as m= λ ( ) ( θ ) 3.1. Explicit solution = ae + be = a e + abe + b e iθ iθ iθ i0 iθ P ( ) 1 = π ( ) iθ λ θ θ 1 = + + π e d ( ) iθ i0 iθ iθ a e abe b e e d 1 ( i0θ iθ i4θ = a e + abe + b e ) dθ π π π π 1 iθ i4θ = a dθ + ab e dθ + b e dθ π π π π = a π π π π π π Fourier transform θ
3.1. Explicit solution Generally, we extract the coefficient via Fourier transform π 1 i m P ( m) = θ λ ( θ) e dθ π π 1 ( iθ iθ) iθm = ae + be e dθ π 1 = π 1 = π 1 = π = 1 π π π π π k = 0 π π k = 0 π π p= 0 ( ) C a e be e dθ k iθ k k i θ k i θ m k ( ) C a e be e dθ k iθ k k iθk iθm k ( ) C ae b e e dθ p i θ p p iθ p i θ m p π p p iθ ( p m) Cp ab e d p= 0 π θ C p = C p p= k
we notice 1 π π π iθ p m 0, if p m / e 1, / Thus, we have 3.1. Explicit solution ( ) ( + ) dθ = if p = ( + m) p p ( m) = = ( + ) P Cp ab, p m/ Specifically, ( ) (, ) P m = wm = when a = b= 1/ C p
3.1. Explicit solution General Characteristic function probability density function (PDF) p(x). displacement x is continuous. the characteristic function is given by λ ikx k e p x e dx = = ( ) ikx ( ) Fourier transform of p(x). i i i x ( ) ae θ = + be θ = e θ λθ One important property of the characteristic function ( ) ( ) λ 0 = p x dx = 1
General random walk model Consider a continuous 1-D random walk process of n steps we have recursion relation: convolution n = n 1 Pn 1 ( x) * p( x) ( ) ( ) ( ) P x P y p x y dy This means that the probability PP nn xx of a particle at x after n steps is PP nn 1 yy pp xx yy = 3.1. Explicit solution the probability of arriving at y in n 1 steps the probability of displacements x-y in one step. ( )* ( ) = ( ) ( ) f x g x f y g x y dy Pn p( x y 1 ) ( ) x p x y y1 y 1 ( y) Pn ( x) Pn 1 ( y) ( ) = ( ) ( ) P x P y p x y n n 1 i i i n ( ) ( ) ( ) P x = P y p x y dy n 1
3.1. Explicit solution Let us define P ( ) ( ) ikx n k = Pn x e dx n ( ) = ( ) λ ( ) P k P k k n 1 ( )* ( ) = ( ) ( ) f x g x f y g x y dy 1 F f x g x f k g k π ( )* ( ) = ( ) ( ) ( ) = ( ) ( ) P x P x p x λ n convolution theorem n 1 * ( k ) ( ) ikx = p x e dx P k = P k λ k = P k λ k = = λ k ( ) ( ) ( ) ( ) ( ) n ( ) n n 1 n 1 ikx 1 n ikx Pn( x) = Pn( k ) e dx = λ ( k ) e dx π π
3.1.3 Mean, Variance, and the Generating function The expected value of function f is defined by f = m= ( ) (, ) f mwm 1 p = p ( m) w( m, ) = pc p m= p= 0 n-th moment n n n 1 m = mw m = mc m= m= (, ) where p = ( + m)/ p m m mean displacement mean square displacement or variance
3.1.3 Mean, Variance, and the Generating function In order to evaluate the various moments, we introduce the generating function p ( ) uwm (, ) Gu = p= 0 or p p p 1 p 1 1 1 Gu ( ) = ucp = Cu p = ( 1+ u) p= 0 p= 0 Example: to calculate mm p 1 ( ) ( ) ( ) ( ) G ' u = pu w m, G '1 = pw m, = p p= 0 p= 0 1 G( u) ( 1 u) = + G' ( 1 ) = / since p = ( + m)/ p = / 1 p = + mwm p= 0 ( ) (, ) 1 1 m = wm (, ) + mwm (, ) = + p= 0 p= 0 m = 0
3.1.3 Mean, Variance, and the Generating function Example: 1/ m =? p 1 ( ) ( ) G ' u = pu w m, p= 0 p ( ) = ( ) ( ) G'' u p p 1 u wm, p= 0 G'' 1 = p p 1 w m, = p p 1 = p p ( ) ( ) ( ) ( ) p= 0 1 = + ( ) ( 1 u) G u ( ) G '' 1 = ( 1) ( ) 4 G'' 1 = p p G'' ( u) = ( 1)( 1+ u) p 1 ( ) 1 = p + = + 4 4 4 m = p ( ) 4 4 4 m = p = p + p = + + = m 1/ 1/ =
3.1.3 Mean, Variance, and the Generating function Generally, Its characteristic function 3 3 ikx ikx kx kx λ ( k ) = e = p ( x) e dx = dx p ( x) 1+ ikx + i +! 3! = n= 0 We obtain n-th moments using characteristic function x n n ik n! n ( i) = x n n d n λ dk ( k) n k = 0 ( ) n n the nth- moment x = p x x dx
3.1.4 To determine Boltzmann s constant from Brownian Motion Theory by Einstein, experiment by Perrin Einstein Assume that the macroscopic resistance on the particle is proportional to the velocity - by classical hydrodynamics showed diffusion obey the statistical law 1 1 3 3 x = x + y + z = r = Dt the diffusion coefficient D is given by D = kt / f T : absolute temperature; K : Boltzmann s constant f : the coefficient of resistance f = 6πµ a (Stokes law) μ : viscosity coefficient; a : particle size Verified by Perrin
3.1.4 To determine Boltzmann s constant from Brownian Motion The modern theory of the Brownian motion dv dt = + Langevin s equation m fv F( t) where v the velocity of the particle and m mass. The random force follows Fluctuation-dissipation relation F ( t) F ( t' ) = 6 fk Tδ ( t t' ) i i B
3.1.4 To determine Boltzmann s constant from Brownian Motion To solve dv m f t dt = v + F ( ) multiply with x, and take the ensemble average dv m x = f xv + xf t dt ( ) d x v dt m v = f xv + xf t d x v dt f m + x v v = ( ) 0 xf ( t) = 0 ot correlated f t m m m x v = ce + v v f f stationary solution
3.1.4 To determine Boltzmann s constant from Brownian Motion 1 x = r = Dt 3 1 d xx 1 d r 1 d ( 6Dt) x v = = = = dt dt dt 3D x v = m v f 1 1 3 1 3 m v = kt m v = f x v = fd energy equipartition principle Boltzmann constant f k = D = D T 6πµ a T f = 6πµ a
作业 : Ex. 4 Page 90 修正 :Eq.(4) 应该为 J p ( x) k ( 1) ( ) x = k = 0 k! Γ k+ p+ 1 k+ p