20 brownian motion in space Flashcards
extending brownian motion to a plane
in each component we get brownian motion
A Brownian motion with diffusivity D in the n-dimensional
space Rⁿ, denoted by B_t
, is such that each Cartesian coordinate describes the position of an independent one-dimensional Brownian motion in R
extending brownian motion to a plane notation pdf? in general
in Rⁿ label (X_1,…X_n) in each component indep brownian motion in R³ = (X,Y,Z)
Probability density function in Rⁿ
p(x,t) = product of PDF for each component in R as they are independent
x in rⁿ t=>0
p⁽ⁿ⁾(x,t) =p⁽¹⁾(x₁,t)…p⁽¹⁾(xₙ,t)
extending brownian motion to a plane pdf?
fundamental solution
of the
n-dimensional diffusion
equation:
∂p(x, t)∂t = D∇² p(x, t)
= D Σ ∂²p(x, t)/∂x² i=1,..,n
so now we assume x_0=0 with no boundaries or barriers
p⁽ⁿ⁾(x,t) =p⁽¹⁾(x₁,t)…p⁽¹⁾(xₙ,t)
= ( 1/√{4πDt}) * exp (-[(x₁−x₀,₁)²]/[4Dt])…( 1/√{4πDt}) * exp (-[(xₙ−x₀,ₙ)²]/[4Dt])
now if x₀=0 origin
then
thus the pdf of Bₜ at t>0 is
p⁽ⁿ⁾(x,t)
= ((4πDt)⁻ⁿ/² exp(−|x|²/4Dt)
where x=(x_1,..x_n)
and |x|² = Σxᵢ² i=1,..,n
and this is the fundamental solution of the
n-dimensional diffusion
equation:
∂p⁽ⁿ⁾(x, t)/∂t = D∇² p⁽ⁿ⁾(x, t)
= D Σ ∂²p(x, t)/∂x² i=1,..,n
what is where x=(x_1,..x_n)
and |x|² = Σxᵢ² i=1,..,n
r^2 =|x|² = Σxᵢ² i=1,..,n
distance function from the origin
∇²
laplacian
∇²φ
=∂²φ/∂x₁² + ∂²φ/∂x₂²+…+∂²φ/∂xₙ²
this is the sum of the second derivatives in each component
consider R^2
p⁽ⁿ⁾(x,t)
= ((4πDt)⁻ⁿ/² exp(−|x|²/4Dt)
verify it solves the diffusion eq
∂p⁽ⁿ⁾(x, t)/∂t = D∇² p⁽ⁿ⁾(x, t)
= D Σ ∂²p(x, t)/∂x² i=1,..,n
p⁽²⁾(x,t)
= ((4πDt)⁻²/² exp(−|x|²/4Dt)=
1/(4πDt) * exp(−|x|²/4Dt)
let x= (x,y)
|x|² =x² + y²
∂p⁽²⁾(x, t)/∂t =∂/∂t[
1/(4πDt) * exp(−(x² + y²)/4Dt]
=
∂/∂t[
(1/√{4πDt})exp(-x²/4Dt) *(1/√{4πDt})exp(-y²/4Dt)]
=
∂/∂t[ p⁽¹⁾(x,t) * p⁽¹⁾(y,t)]
=∂/∂t[ p⁽¹⁾(x,t)]p⁽¹⁾(x,t) + p⁽¹⁾(x,t)∂/∂t[ p⁽¹⁾(x,t)] X
(note that:.p⁽¹⁾(x,t) = ( 1/√{4πDt}) * exp (-[(x)²]/[4Dt])
verify laplacian of
D∇²p⁽²⁾(x, t)=
D(∂²p⁽¹⁾(x,t)⁽²/∂x²) p⁽¹⁾(y,t)
+ D(∂²p⁽¹⁾(y,t)⁽²/∂y²) p⁽¹⁾(x,t)
because each p⁽¹⁾ is the sol of diffusion in 1D
=(∂p⁽¹⁾(x,t)/∂t)p⁽¹⁾(y,t) + (∂p⁽¹⁾(y,t)/∂t)p⁽¹⁾(x,t)
comparing to what we have here X
=∂p⁽²⁾(x, t)/∂t
THIS GENERALISES TO Rⁿ
brownian motion to
What happens to the distribution of the particulas as t goes to infinity?
1D
in R^n?
on average the mean =x_0 starting point
if x_0=0 starting point
in one dimension:
Var(Xₜ)= 2D∆t
in R^2:
What is the variance of the distance from the origin? we work this out
r=sqrt(x₁²+…+xₙ²)
exit times in brownian motion
for Rₙ
what do they satisfy?
extends to a balls
imagine you start at origin in R^3
more natural to look at spheres as distance from the origin
due to radial symmetry the hitting time/probabilities dep on r
from the 1d conditions we get the R^n conditions for exit times:
∇²P(r) = 0 and D∇²T(r)=-1
(we then provide boundary conditions)
brownian motion
Expectation and variance
in R²
in R³
In Rₙ
in one dimension:
Var(Xₜ)= 2D∆t for each component
Var(Xₜ) =E( X²) as E(X)=0
as brownian motion is independent
Bₜ = (Xₜ, Yₜ),
E(Rₜ²) =
E( X²+Y²)
=E( X²)+E(Y²)
=4Dt
in R³ :
Bₜ = (Xₜ, Yₜ,Zₜ),
E(Rₜ²) =
E( X²+Y²+Z²)
=6Dt
In Rₙ:
E(Rₜ²) = n(2Dt)
polar coords in R^2
r distance from origin
θ angle between vector from origin to point and x axis
(x, y) = (r cos θ, r sin θ)
∇²ψ =∂²/∂x²+∂²/∂y²
=(1/r)(d/dr)(rdψ/dr)
θ wont play a role in the probabilities of brownian motion
don’t need to learn by heart, whenever we need these they will be given in exam
(1/r) captures the fact that it has radial symmetry
spherical coordinates in R^3
(x, y, z) = (r cos ϕ cos θ, r sin ϕ cos θ, r sin θ)
θ angle between vector and z axis
ϕ angle between projection of vector onto xy plane and x axis
∇²ψ =∂²/∂x²+∂²/∂y² +∂²/∂z²
=(1/r²) d/dr(r²dψ/dr)
don’t need to learn by heart, whenever we need these they will be given in exam
Example:
Hitting probability for circles in IR^2
same BC as R₃:
P(a)=1 (probability of hitting a first given you start at a)
P(R)=0 (probability of hitting R first given you start at a)
∇²P(r) = 0 in general:
this example therefore
∇²P(r) =
(1/r) d/dr(rdP(r)/dr) = 0
rdP(r)/dr =C₁ constant
using BCs:
P(r) =
[ln(r/R)]/ln(a/R)
Note that:
P(r) → 1 as R → ∞
independently of a: in 2D, the Brownian motion does not escape at ∞
Example:
Hitting probability for two spheres in IR^3
Consider two concentric spherical shells of radius
0 < a < R
and let
P(r) = probability of hitting the smaller sphere first, starting
from distance r, with a < r < R
Then P(r) satisfies
..
what BCs shall we impose?
∇²P(r) = 0 in general:
this example therefore
∇²P(r) =
(1/r²) d/dr(r²dP(r)/dr)
imagine two spheres smaller radius a contained in larger sphere radius R
suppose initially at r distance
a<r<R
which sphere would be hit first?
P(r) = probability of hitting the smaller sphere first, starting
from distance r, with a < r < R
suitable BC:
P(a)=1 (probability of hitting a first given you start at a)
P(R)=0 (probability of hitting R first given you start at a)
(1/r²) d/dr(r²dP(r)/dr) = 0
means
r² dP(r)/dr)=C₁ constant
dP(r)/dr)=C₁/r²
P(r)= -C₁/r + C₂
P(a)=1: C₁ = aR/(a-R)
P(R)=0 C₂= C₁/r = a/(a-R)
P(r)= (a/(a-r)) [r-R]/[r]
=(a/(R-a)) (R/r -1)
(in 1D P(x) linear but in 3D P(r) decreases from p(a)=1 to P(R)=0) not linearly, many more directions considered)
as R → ∞ P(r)→ 1/r > 0
(this is comparing starting outside the spheres, will we end up closer to the sphere?)
i.e., there is a positive probability 1 − a/r of escaping at ∞ for every r > a.
but also probability a/r of hitting also positive
Mean time to hit boundary
Exit time from a sphere
in R^2
In general, the mean time T to hit the boundary of a region is a
function of the starting point (distance r from the origin) and
satisfies the PDE
D∇2T(r) = −1.
with boundary conditions
T(R) = 0, T′(0) = 0 (reflecting)
Mean time to hit boundary
Exit time from a sphere
Start inside the two spheres, mean time of hitting either
what are the BCs
Mean time of hitting boundary if starting at boundary (either one)
T(a)=0
T(R)=0 used
