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 푢푡 = 휈푢푥푥

• Parabolic partial differential equation • 휈: 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

푢푡 = 휈푢푥푥 and initial condition 푢 푥, 0 = 푢0 푥 , use Fourier Transform to obtain the analytic solution. 1 +∞ 푢 푘, 푡 = 푢(푥, 푡)푒−푖푘푥푑푥 2휋 −∞ Apply the Fourier Transform to the diffusion equation. 2 푢 푡 = −휈푘 푢 and initial condition 푢 푘, 0 = 푢 0 푘 The solution to the above equation is given by −휈푘2푡 푢 푘, 푡 =푢 0 푘 푒 1 +∞ where 푢 푘 = 푢 푥 푒−푖푘푥푑푥. 0 2휋 −∞ 0 Analytic Solution

Then, apply inverse Fourier Transform to 푢 푘, 푡 . 1 +∞ 푢 푥, 푡 = 푢 푘, 푡 푒푖푘푥푑푘 2휋 −∞ +∞ +∞ 1 −휈푘2푡+푖푘(푥−푦) 푢 푥, 푡 = 푢0 푦 푒 푑푘 푑푦 2휋 −∞ −∞

Consider the integral of 푘. +∞ +∞ 2 2 2 퐼(훽) = 푒−휈푘 푡+푖푘(푥−푦)푑푘 = 푒−훼 푘 +훽푘 푑푘 −∞ −∞ Where 훼 = 휈푡 and 훽 = 푖(푥 − 푦). Analytic Solution

Easy to verify that 푑퐼(훽) 훽 2 2 = 퐼 퐼(훽) = 퐶푒훽 /4훼 푑훽 2훼2

+∞ 2 2 휋 The constant 퐶 = 퐼 0 = 푒−훼 푘 푑푘 = . So, −∞ 훼2

휋 2 2 휋 2 퐼 = 푒훽 /4훼 = 푒− 푥−푦 /4휈푡 훼2 휈푡 Therefore, the analytic solution of the diffusion equation is +∞ 1 − 푥−푦 2/4휈푡 푢 푥, 푡 = 푢0 푦 푒 푑푦 4휋휈푡 −∞ Analytic Solution ---initial condition is the delta function

Example 1: 푢푡 = 휈푢푥푥 and initial condition 푢 푥, 0 = 푢0 푥 = 훿(푥) The solution is given by +∞ 1 − 푥−푦 2/4휈푡 푢 푥, 푡 = 푢0 푦 푒 푑푦 4휋휈푡 −∞ +∞ 1 2 푢 푥, 푡 = 훿 푦 푒− 푥−푦 /4휈푡 푑푦 4휋휈푡 −∞ 1 2 푢 푥, 푡 = 푒−푥 /4휈푡 4휋휈푡 Analytic Solution ---initial condition is a step function

Example 2: 푢푡 = 휈푢푥푥 and initial condition 푢푙, 푖푓 푥 < 0 푢 푥, 0 = 푢0 푥 = 푢푟, 푖푓 푥 > 0 The solution is given by +∞ 1 − 푥−푦 2/4휈푡 푢 푥, 푡 = 푢0 푦 푒 푑푦 4휋휈푡 −∞ 0 푥−푦 2 +∞ 푥−푦 2 1 − − 푢 푥, 푡 = ( 푢푙푒 4휈푡 푑푦 + 푢푟푒 4휈푡 푑푦) 4휋휈푡 −∞ 0 푥 1 4휈푡 −푦2 푢 푥, 푡 = 푢푙 + (푢푟 − 푢푙) 푒 푑푦 휋 −∞ Numerical Schemes

 Central Explicit Scheme 푛+1 푛 푛 푛 푛 푢푗 − 푢푗 푢푗+1 − 2푢푗 + 푢푗−1 = 휈 ∆푡 ∆푥2 휈Δ푡 1 Consistency: 푂 Δ푥2, Δ푡 Stability: < Δ푥2 2

 Central Implicit Scheme 푛+1 푛 푛+1 푛+1 푛+1 푢푗 − 푢푗 푢푗+1 − 2푢푗 + 푢푗−1 = 휈 ∆푡 ∆푥2 Consistency: 푂 Δ푥2, Δ푡 Stability: unconditionally stable Numerical Schemes

 Crank-Nicolson Scheme 푛+1 푛 푛 푛 푛 푛+1 푛+1 푛+1 푢푗 − 푢푗 1 푢푗+1 − 2푢푗 + 푢푗−1 푢푗+1 − 2푢푗 + 푢푗−1 = 휈( + ) ∆푡 2 ∆푥2 ∆푥2 Consistency: 푂 Δ푥2, Δ푡2 Stability: unconditionally stable

 Leap Frog Scheme 푛+1 푛−1 푛 푛 푛 푢푗 − 푢푗 푢푗+1 − 2푢푗 + 푢푗−1 = 휈 2∆푡 ∆푥2 Consistency: 푂 Δ푥2, Δ푡2 Stability: unconditionally unstable Numerical Schemes

 Du Fort-Frankel Scheme 푛+1 푛−1 푛 푛+1 푛−1 푛 푢푗 − 푢푗 푢푗+1 − (푢푗 + 푢푗 ) + 푢푗−1 = 휈 2∆푡 ∆푥2 Consistency: 푂 Δ푡2/Δ푥2, Δ푡2 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 휈 1 휆2 1 휈 = = 휆푢 2 Δ휏 2 푎푣푒 where 휆 is the mean free path, Δ휏 is the average time between collisions, and 푢푎푣푒 is the average molecular speed. Mean Free Path --- 1D Brownian Motion

Consider a 1D random walk: during each time step size Δ휏, a particle can move by +휆 or −휆 so that a collision happens. Mean Free Path --- 1D Brownian Motion

The displacement from the original location after 푛 time steps (or 푛 collisions) is 푛 (푛) 푋 = 푥푖 푖=1 where 푥푖 = ±휆 with equal probability. Then, we have 푛 푛 퐸 푥푖 = 0, 퐸 푋 = 퐸 푥푖 = 0 푖=0 2 푛 푛 푛 2 2 푉푎푟 푥푖 = 휆 , 푉푎푟 푋 = 퐸 푋 − 퐸 푋 = 푛휆 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, 푛 푥 푑 푛 푖=1 푖 − 퐸 푥 Ν 0, 푉푎푟 푥 푛 푖 푖 as 푛 ∞. This is equivalent to 푑 푡휆2 푋 푛 Ν 0, 푛휆2 = Ν(0, ) Δ휏 where, 푡 is the total time. Then, the distribution of 푋 푛 1 2 2 푝 푥, 푡 = 푒−푥 /(2푡휆 /Δ휏) 2휋푡휆2/Δ휏 Mean Free Path --- 1D Brownian Motion

Now, consider the diffusion process with initial condition 푢 푥, 0 = 푢0 푥 = 훿(푥) Its solution is given in example 1. 1 2 푢 푥, 푡 = 푒−푥 /4휈푡 4휋휈푡

Particle Movement Diffusion Process So, 푝 푥, 푡 = 푢(푥, 푡) and this leads to 푡휆2 1 휆2 1 = 2휈푡 휈 = = 휆푢 Δ휏 2 Δ휏 2 푎푣푒 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 푚/푠.And the mean free path of the air at the same condition is about 68푛푚. So, 1 1 휈 = 휆푢 ≈ × 500 × 68 × 10−9 = 1.7 × 10−5푚2/푠 2 푎푣푒 2 This is close to the ratio of dynamic viscosity (1.81× 10−5푘푔/(푚 ∙ 푠)) to the density (1.205푘푔/푚3) 휇 1.81 × 10−5 휈 = ≈ ≅ 1.502 × 10−5푚2/푠 휌 1.205 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