17 brownian Motion and diffusion equation in one spatial dimension

Question 1

We consider partial derivatives for example because

Answer 1

e.g. we are considering interactions wrt space F(x,t)
deriv wrt to one variable

Question 2

Standard brownian motion

Answer 2

is a set of positions {X_t}t≥0 that form a continuous path
(continuous time and space limit of a randon walk)

The increment ∆X = X_{t+∆t} − Xₜ
is a Gaussian random
variable with mean zero and variance proportional to ∆t,

(larger time larger variance further you can get)
(random walk equally likely)
and specifically

E(∆X) = 0,
var(∆X) = 2D∆t,
where D is the diffusivity or diffusion coefficient

Question 3

Why is diffusion equation nice? for the real line

Answer 3

smoothing property- even if start with an irregular initial function where all of the mass is concentrated on x_0 as soon as you let the system run (t>0) we get a nice guassian distribution as a normal around x_0

left and right for individuals

for any time t we find individuals anywhere on the line (positive probability)

no loss of mas:
δ means

∫-∞, ∞ δ(y − y₀).dy=1 means
∫-∞, ∞p(x,t).dt=1
mass distributes on the real line

Question 4

Random walk

Answer 4

One dim random walk, Markov chain state space is the integers and only transitions to neighbouring states
if X_n = k
then X_{n+1)=k
X_{n+1} = k+1
or X_{n+1}=k-1

consider one where the prob of making a jump to the right is p and a jump to the left is 1-p

random walk with prob p=0.5

Question 5

Brownian motion using general

Answer 5

Example of a continuous Markov process (MP if we know probability at current time step the next only dep on this not prev history)
Consider a random walk, left or right by ∆x.

P(of finding X(t+∆t)∈ [x, x + ∆x]]
= p( x, t+∆t)∆x = sum of

1) The probability that X(t) ∈ [x − ∆x, x] and there is a jump to the right at time t occurs with probability p(x − ∆x, t)∆x/2;

(2) the probability that X(t) ∈ [x + ∆x, x + 2∆x] and there is a jump to the left at time t. This occurs with a probability p(x + ∆x, t)∆x/2.

Question 6

pdf of the wiener process/ brownian motion on R

pdf of Xₜ for an initial concentration:

Answer 6

we require an initial bc p(x,0)=p_0(x)

for initial concentration p(x,0)=δ(x − x₀)
solution is
p(x,t) =
( 1/√{4πDt}) * exp (-[(x−x₀)²]/[4Dt]

We see that the pdf of this satisfies the diffusion equation of partial derivatives

and we have seen the sol on the real line, and finite lin [0,L]

Question 7

pdf of the wiener process/ brownian motion on R

How to find
solution p(x,t) on the real line (infinite(

Answer 7

use the fourier transform of the function/variable p

p~(k,t) = F(p)(k) =
(1/√{2π}) ∫₀ ∞ p(x,t) exp(ikx) . dx

Fourier transform both sides of the diffusion partal derivs eq of this

(∂/∂t)p~(k,t) = -k²Dp~(k,t)
(linear PDE in one var)

giving
p~(k,t) =p~(k,o)exp(-k²Dt)

then inverse fourier transform this to find

p(x,t) =
( 1/√{4πDt}) *
∫-∞, ∞ p₀(y)* exp (-[(x−y²]/[4Dt] .dy

(p₀(y) is initial condition)

if we assume all particles at x_0
then δ (x-x
then we get the fundamenental sol given (fundamental sol means all particles iniitally at some point)

Question 8

pdf of the wiener process/ brownian motion on R

what is the difference between sol on R line and on interval [0,L]

Answer 8

can use fourier as can int from -inf to inf for real line,

require BCs absorbing/reflecting for the interval
that one is solved by assuming p is seperable we then need to use the diffusion eq and solve the two odes

Question 9

pdf of the wiener process/ brownian motion on R
solution p(x,t) on the finite interval [0,L]

What kind of conditions?

Answer 9

different as we need to prescribe boundary conditions

we can have two possibilities
Absorbing or reflecting boundaries (L5 mixed)

absorbing:
p(0,t)=p(L,t) = 0 (DIRICHLET HOMOGENEOUS) (prescribing value of funct at end point, homogeneous as value is 0)

reflecting:
∂p(0,t)/∂x = ∂P(L,t)/∂x (NEUMANN HOMOGENEOUS) (prescribing derivative of the two end points )

Question 10

pdf of the wiener process/ brownian motion on R

solution p(x,t) on the finite interval [0,L] with given condition

How do we go about it?
also reviews in L5

Answer 10

we don’t use fourier transform as cant integrate from -inf to inf
so we solve by
step 1 Assume function is seperable
p(x,t)=F(x)G(t)
use in diffusion equation
FG* = DGF’’
G*/DG = F’‘/F =-λ constant λ>0
step 2 solve seperately the differential eq for G and for F
G gives exp sol
F is 2nd order and thus sol is linear combo of sin and cos
G(t) = G_0(0)exp(-λtD)
F(t)=asin(√λ x) + bcos (√λ x)
a and b determined by BCs
infinite values of:
√λₙ = (nπ/L)
giving a family of values

Step 3
superposition principle gives the general solution(is given by the sum of all from the family)
p(t,x) = Σ_[n=0,..∞]
aₙ sin(√λₙ x) + bₙ cos(√λₙ x)]C_n

C_n determined by initial condition
boundary conditions determine a and b

Question 11

fundamental solution of the n-dimensional diffusion
equation:

Answer 11

∂p(x, t)/
∂t

= D∇²p(x, t)
= Σᵢ ∂²p(x, t)/∂xᵢ²
i=1,..,n

Question 12

pdf of the wiener process/ brownian motion on R

solution p(x,t) on the finite interval [0,L] with given condition

G(t) = G_0(0)exp(-λtD)
F(t)=asin(√λ x) + bcos (√λ x)

how do the different absorbing reflecting conditions change?

Answer 12

a and b are determined by BCs

absorbing:p(0,t)=p(L,t) = 0
means b=0 for all family of sines from a_n

reflecting:
∂p(0,t)/∂x = ∂P(L,t)/∂x
means a=0 left with family of cosines

family as infinite values of lambda: SHOULD ALWAYS START FROM 0 for both

Question 13

how do the different absorbing reflecting conditions affect over a long period of time

Answer 13

absorbing:p(0,t)=p(L,t) = 0
means b=0 for all family of sines from a_n

removing ants
(as we remove walkers the solution tends to 0 as t goes to infinity, as we are losing mass in the system)

can be seen as all terms have exp(-) so tends to 0

reflecting:
∂p(0,t)/∂x = ∂P(L,t)/∂x
means a=0 left with family of cosines

we still have exp(-) but we have a leading term not dep on time b_0/2 so no loss of mass

(mass remains in this system)

Question 14

Brownian motion using p= 1/2

Answer 14

Example of a continuous Markov process (MP if we know probability at current time step the next only dep on this not prev history)
Consider a random walk, left or right by ∆x.

P(of finding X(t+∆t)∈ [x, x + ∆x]]
= p( x, t+∆t)∆x =

p( x, t+∆t)=
0.5p(x-∆x ,t) + 0.5p(x+∆x,t)
(transition to x from x-∆x with prob 0.5) +(transition to x from x+∆x with prob 0.5)

Question 15

To consider the continuous time and space limit of the random walk, we let ∆x and ∆t → 0,
and Taylor-expand both sides of Eq.

p( x, t+∆t)=
0.5p(x-∆x ,t) + 0.5p(x+∆x,t)

** important result about the
diffusion equation of Brownian motion as a suitable continuum limit (∆t → 0 and ∆x → 0)
of the random walk**

Answer 15

Using Taylor series on both sides
p(x,t) + ∆t∂/∂t p(x,t) +(1/2∂^2/∂t^2)p(x,t) +…

=
0.5[p(x,t) -∆x∂/∂xp(x,t) + 0.5∆t^2∂^2/∂x^2p(x,t) ]

+ 0.5[p(x,t) + ∆x∂/∂xp(x,t) + 0.5∆t^2∂^2/∂x^2p(x,t) ]
so we have
(∂/∂t) p(x,t)+…. =(∆x²/2∆t)* (∂²/∂x²) p(x,t)+…..

Question 16

if both ∆x and ∆t are small but ∆t proportional to ∆x² we can write

Answer 16

D= ∆x²/2∆t
and
(∂/∂t) p(x,t)+…. =D(∂²/∂x²) p(x,t)

a solution is p(x,t)=
(4piDt)^-0.5 * EXP(- (x-x_0)^2/4Dt)

Question 17

thinking about how far a particle moves in an interval
∆x not discrete

P {∆x1 ≤ ∆x ≤ ∆x2}

Answer 17

we describe in terms of probability density function

P {∆x_1 ≤ ∆x ≤ ∆x_2} = ∫[∆x_1,∆x_2] w(y).dy
The function satisfies as pdf:
∫[-∞,∞] w(y).dy = 1

assumption equally likely left and right w(-y) = w(y) is used also to cancel out

Question 18

Einstein and the diffusion equation

Answer 18

With the defn of the diffusion coefficient
D= (1/2∆t)* ∫_[-∞,∞] y²w.dy

diffusion equation
∂f/∂t= D ∂²f/∂x²

found by:
f (x, t + ∆t) = ∫_[-∞,∞] f(x-y,t)w(y).dy
using taylor expansion on both sides with partial deriv:
f (x, t + ∆t) = f(x,t) + (∂f/∂t)(x,t)∆t+…
f(x-y,t) = f(x,t) - (∂f/∂t)(x,t)y + (1/2)(∂²f/∂x²)y+…..

then using ∫[-∞,∞] w(y).dy = 1 and ∫[-∞,∞] yw(y).dy = 0

Question 19

Stochastic processes

Answer 19

A stochastic process is a series of random variables labelledby time. The value of the process at time t is a random variable thatwe write Xt

When time is continuous (t is a real number) we say that X is a continuous-time stochastic process. One realization of a stochastic process is a sample path. When we can draw these paths without lifting the pen from the paper, we say that the stochastic process has continuous paths

Question 20

W_t

Answer 20

The fundamental stochastic process in continuous space andtime is standard Brownian motion, or the Wiener process.The position at time t is the random variable denoted Wt

It is a Markov process with the property that, no matter how small ∆t is,W_{t+∆t} − W_t has a Gaussian distribution with mean zero and variance ∆t,and is independent of the value of W_t

if f(x,t) is the probability density function of W_t then
∂/∂t (f (x, t)) = (1/2)∂^2/∂x^2) f (x, t)
mean is always 0?