Diffusion Equation and Mean Free Path Speaker: Xiaolei Chen Advisor: Prof. Xiaolin Li Department of Applied Mathematics and Statistics Stony Brook University (SUNY)
Content General Introduction Analytic Solution of Diffusion Equation Numerical Schemes Mean Free Path
General Introduction
General Introduction 1D diffusion equation u t = νu xx Parabolic partial differential equation ν: thermal conductivity, or diffusion coefficient In physics, it is the transport of mass, heat, or momentum within a system In connection with Probability, Brownian motion, Black- Scholes equation, etc
Analytic Solution For the parabolic diffusion equation u t = νu xx and initial condition u x, 0 = u 0 x, use Fourier Transform to obtain the analytic solution. u k, t = 1 2π + u(x, t)e ikx dx Apply the Fourier Transform to the diffusion equation. u t = νk 2 u and initial condition u k, 0 = u 0 k The solution to the above equation is given by u k, t = u 0 k e νk2 t where u 0 k = 1 + u0 x e ikx dx. 2π
Analytic Solution Then, apply inverse Fourier Transform to u k, t. u x, t = 1 2π + u k, t e ikx dk u x, t = 1 + 2π + u 0 y e νk2 t+ik(x y) dk dy Consider the integral of k. I(β) = + e νk2 t+ik(x y) dk = Where α = νt and β = i(x y). + e α2 k 2 +βk dk
Analytic Solution Easy to verify that di(β) dβ = β 2α 2 I I(β) = Ceβ2 /4α 2 + The constant C = I 0 = e α 2 k 2 dk = π α2. So, I = π α 2 eβ2 /4α 2 = π νt e x y 2 /4νt Therefore, the analytic solution of the diffusion equation is u x, t = 1 + u 0 y e x y 2 /4νt dy 4πνt
Analytic Solution ---initial condition is the delta function Example 1: u t = νu xx and initial condition u x, 0 = u 0 x = δ(x) The solution is given by u x, t = 1 + u 0 y e x y 2 /4νt dy 4πνt u x, t = 1 + δ y e x y 2 /4νt dy 4πνt u x, t = 1 4πνt e x2 /4νt
Analytic Solution ---initial condition is a step function Example 2: u t = νu xx and initial condition u x, 0 = u 0 x = u l, if x < 0 u r, if x > 0 The solution is given by u x, t = 1 u x, t = 1 + 4πνt 0 4πνt ( u 0 y e x y 2 /4νt dy u l e x y 2 4νt dy + u x, t = u l + 1 π (u r u l ) 0 + u r e x y 2 x 4νt e y 2 dy 4νt dy)
Numerical Schemes Central Explicit Scheme u n+1 n j u j = ν u n j+1 t Consistency: O Δx 2, Δt 2u n n j + u j 1 x 2 Stability: νδt Δx 2 < 1 2 Central Implicit Scheme u n+1 n j u j = ν u j+1 n+1 2u n+1 n+1 j + u j 1 t x 2 Consistency: O Δx 2, Δt Stability: unconditionally stable
Numerical Schemes Crank-Nicolson Scheme u n+1 n j u j = 1 n t 2 ν(u j+1 2u n n j + u j 1 x 2 + u j+1 n+1 2u n+1 n+1 j + u j 1 x 2 ) Consistency: O Δx 2, Δt 2 Stability: unconditionally stable Leap Frog Scheme u j n+1 u j n 1 2 t Consistency: O Δx 2, Δt 2 = ν u n j+1 2u n n j + u j 1 x 2 Stability: unconditionally unstable
Numerical Schemes Du Fort-Frankel Scheme u n+1 n 1 j u j = ν u n j+1 (u n+1 j + u n 1 n j ) + u j 1 2 t x 2 Consistency: O Δt 2 /Δx 2, Δt 2 conditionally consistent Stability: unconditionally stable
Mean Free Path In physics, mean free path is the average distance travelled by a moving particle between successive collisions, which modify its direction or energy or other particle properties. Relation to diffusion coefficient ν λ 2 ν = 1 2 Δτ = 1 2 λu ave where λ is the mean free path, Δτ is the average time between collisions, and u ave is the average molecular speed.
Mean Free Path --- 1D Brownian Motion Consider a 1D random walk: during each time step size Δτ, a particle can move by +λ or λ so that a collision happens.
Mean Free Path --- 1D Brownian Motion The displacement from the original location after n time steps (or n collisions) is X (n) = where x i = ±λ with equal probability. Then, we have n i=1 E x i = 0, E X n = E x i n i=0 x i = 0 Var x i = λ 2, Var X n = E X n E X n 2 = nλ 2 Note: Brownian motion is a markov process, which means the movement at each time step is independent of the previous ones.
Mean Free Path --- 1D Brownian Motion According to the Central Limit Theorem, as n n i=1 n x i n. This is equivalent to X n d Ν 0, nλ 2 where, t is the total time. Then, the distribution of X n p x, t = E x i d Ν 0, Var x i 1 = Ν(0, tλ2 Δτ ) 2πtλ 2 /Δτ e x2 /(2tλ2 /Δτ)
Mean Free Path --- 1D Brownian Motion Now, consider the diffusion process with initial condition u x, 0 = u 0 x = δ(x) Its solution is given in example 1. u x, t = 1 4πνt e x2 /4νt Particle Movement Diffusion Process So, p x, t = u(x, t) and this leads to tλ 2 Δτ = 2νt ν = 1 λ 2 2 Δτ = 1 2 λu ave
Mean Free Path --- Kinematic Viscosity Molecular Diffusion
Mean Free Path --- Kinematic Viscosity Molecular Diffusion For typical air at room conditions, the average speed of molecular is about 500 m/s.and the mean free path of the air at the same condition is about 68nm. So, ν = 1 2 λu ave 1 2 500 68 10 9 = 1.7 10 5 m 2 /s This is close to the ratio of dynamic viscosity (1.81 10 5 kg/(m s)) to the density (1.205kg/m 3 ) ν = μ ρ 1.81 10 5 1.205 1.502 10 5 m 2 /s
Mean Free Path --- Kinematic Viscosity Eddy Diffusion It is mixing that is caused by eddies with various sizes. The mean free path is related to the size of the vortices. And it is usually much larger than that of the molecular diffusion process. Larger Mean Free Path Large Kinematic Viscosity Use RANS (Reynolds-Averaged Navier Stokes), LES (Large Eddy Simulation) to modify the viscosity
References Notes by Prof. XiaolinLi Wikipedia: diffusion, viscosity, mean free path, turbulence, Brownian motion, molecular diffusion, eddy diffusion