18 Solutions of the one-dimensional diffusion equation Flashcards
The one-dimensional diffusion equation
∂p(x, t)/∂t = D(∂^2p(x, t)/∂x^2)
where D is the diffusivity or diffusion coefficient/constant
solving the one-dimensional diffusion eq on R line
A solution is
N(x, t) = (4πDt)^{− 1/2} *
exp(−[(x − x_0)^2]/[4Dt]).
bell curve, max not at 0, can positive for some neg x values
whatever value of x_0 we choose, the solution that satisfies
integral from -inf to inf of N(x,t) .dx = 1 is as above
peak at x_0
we need this to be able to interpret function N as probability density
checking the solution of the one-dimensional diffusion eq on R line
using N(x,t)
∂N/∂x = -x/2Dt (N)
∂²N/∂x² = -(1/2Dt)N + (-x/2Dt)(-x/2Dt)N
= (-1/2Dt)N + (x²/4D²t²)*N
(treat t as constant)
∂N/∂t =[2piD/((4piDt)^(3/2))]exp(-x^2/4Dt)+1/(4piDt)^0.5exp(-x²/4Dt)
=-(1/2t)N +N(x²/4Dt²)
issue is that i can’t have t=0
on denominator width of the curve =t
The Fourier transform
helps to describe the motion in more than one way.
We can solve the diffusion equation on x ∈ IR by a Fourier
transform.
We take our function in terms of x and t
multiply by exp(isx)
and integrate wrt x
F(N) = N˜(s, t)
= (1/√2π) ∫_[-∞,∞] of
N(x, t)e^{isx} dx.
Linear solution multiplied still satisfies the linear ODE, we transformed using that particular constant we want to think of it as a pdf
We think of the function N(x, t) at at time t as being
decomposed into Fourier components in x
…
“What is the fourier transform of the derivative of the function”
solving in Fourier space
transform the initial conditions
F(∂N/∂t) = -isN~
F(∂²N∂t²) = -s²N~
∂N~/∂t =-Ds~N~
N~(s,t)= N~(s,0)exp(-Ds²t)
N~(s,0)= (1/√2π) ∫_[-∞,∞] of
N(x, 0)e^{isx} dx.
for every s (each value of x treated seperately)
inverse fourier transform
we start of thinking in x,t would like it back in x and t
N(x,t) = (1/√2π) ∫_[-∞,∞] of
N~(x, t)e^{-isx} dx.
(negative in exp)
inverse fourier transform of 2 products
f
inverse fourier transform of
(e^{−s/2Dt})
f
approx bell shape with
sin functions
approx
I= ∫_[-∞,∞] of exp(-x²).dx integrate
I²
I² = ∫[-∞,∞] of exp(-x²).dx * ∫[-∞,∞] of exp(-y²).dy
= ∫[-∞,∞] ∫[-∞,∞] exp(-y²) exp(-x²).dx.dy
x=rcosθ y=rsinθ
x²+y² = r² dxdy=r dθdr
18.3 L5 (The diffusion equation on the semi-infinite line and the method
of images)
the diffusion equation on the semi-infinite line
subject to
absorbing (Dirichlet) and reflecting (Neumann)
BCs at one end of the semi-infinite line. The
other boundary condition simply requires the density to vanish at the other end of the semiinfinite line (e.g. x → −∞)
Consider the diffusion equation which gives solutions to be considered as a sum of N’s
N_1 +N_2 +… etc
for each N we could choose a different x_0
this would give rise to a series of bell curves all translated, e.g. their x_0 giving the peaks differ
you would also have to change the prefactor so that the sum of integrals from -inf to inf =1 to interpret as a pdf HOWEVER
(linear partial differential equation)
L5: Brownian motion in one dimension absorbed at 0
condition?
Suppose that particles are absorbed whenever they reach zero.
Then the probability density, which we now denote pₖ(x, t), still obeys the diffusion equation when x > 0, but
with the extra condition
pₖ(0, t) = 0 ∀t
L5: Brownian motion in one dimension absorbed at 0
solution
The solution is now
pₖ(x, t)
= (4πDt)^{− 1/2} *
[
exp(−[(x − x₀)²]/[4Dt]).
-
exp(-(x+x₀)²/4Dt)]
the difference of two terms
one gaussian shape centred at x₀
subtract
gaussian shape centred at -x₀
L5: Brownian motion in one dimension absorbed at 0
density pₖ(x, t) is
defined for x ≥ 0
(not valid for negative values of x)
brownian walker wanders for large values of x,
if gets to 0 then absorbs
property:
pₖ(0, t) = 0 ∀t
curve that follows the subtraction of two bell curves (one upside down)
is the difference of two Gaussian density functions, one centred at x₀ at the other at −x₀.
however now we dont have the condition of intgeral =1 as brownian walkers get absorbed, so if they are absorbed we have less and less and probability integrates to <1
L5: Brownian motion in one dimension absorbed at 0
pₖ(x, t)
= (4πDt)^{− 1/2} *
[
exp(−[(x − x₀)^2]/[4Dt]).
-
exp(-(x+x₀)²/4Dt)]
also written
pₖ(x, t)
= (4πDt)^{− 1/2} *
exp(−[(x − x₀)^2]/[4Dt])*
[1- exp(-(x*x₀)/Dt)]
written otherwise
as
L5: Brownian motion in one dimension absorbed at 0
**
obtaining solution**
pₖ(x, t)
= (4πDt)^{− 1/2} *
[
exp(−[(x − x₀)^2]/[4Dt]).
-
exp(-(x+x₀)²/4Dt)]
also written alt how?
(x+x₀)² = x²+2xx₀ +x₀²
=x²-2xx₀ +x₀² +4xx₀
=(x − x₀)² + 4xx₀
(x+x₀)² /4Dt = (x − x₀)²/4Dt + xx₀/Dt
exp(-(x+x₀)² /4Dt )
= exp(-(x − x₀)²/4Dt - xx₀/Dt)
= exp(-(x − x₀)²/4Dt) *exp( xx₀/Dt))
