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

𝑢𝑡 = 𝜈𝑢𝑥𝑥

• 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 SolutionFor the parabolic diffusion equation

𝑢𝑡 = 𝜈𝑢𝑥𝑥 and initial condition 𝑢 𝑥, 0 = 𝑢0 𝑥 ,

use Fourier Transform to obtain the analytic solution.

𝑢 𝑘, 𝑡 =1

2𝜋 −∞

+∞

𝑢(𝑥, 𝑡)𝑒−𝑖𝑘𝑥𝑑𝑥

Apply the Fourier Transform to the diffusion equation.

𝑢𝑡 = −𝜈𝑘2 𝑢 and initial condition 𝑢 𝑘, 0 = 𝑢0 𝑘

The solution to the above equation is given by

𝑢 𝑘, 𝑡 = 𝑢0 𝑘 𝑒−𝜈𝑘2𝑡

where 𝑢0 𝑘 =1

2𝜋 −∞+∞

𝑢0 𝑥 𝑒−𝑖𝑘𝑥𝑑𝑥.

Analytic SolutionThen, apply inverse Fourier Transform to 𝑢 𝑘, 𝑡 .

𝑢 𝑥, 𝑡 =1

2𝜋 −∞

+∞

𝑢 𝑘, 𝑡 𝑒𝑖𝑘𝑥𝑑𝑘

𝑢 𝑥, 𝑡 =1

2𝜋 −∞

+∞

−∞

+∞

𝑢0 𝑦 𝑒−𝜈𝑘2𝑡+𝑖𝑘(𝑥−𝑦)𝑑𝑘 𝑑𝑦

Consider the integral of 𝑘.

𝐼(𝛽) = −∞

+∞

𝑒−𝜈𝑘2𝑡+𝑖𝑘(𝑥−𝑦)𝑑𝑘 =

−∞

+∞

𝑒−𝛼2𝑘2+𝛽𝑘 𝑑𝑘

Where 𝛼 = 𝜈𝑡 and 𝛽 = 𝑖(𝑥 − 𝑦).

Analytic Solution

Easy to verify that 𝑑𝐼(𝛽)

𝑑𝛽=

𝛽

2𝛼2𝐼 𝐼(𝛽) = 𝐶𝑒𝛽

2/4𝛼2

The constant 𝐶 = 𝐼 0 = −∞+∞

𝑒−𝛼2𝑘2

𝑑𝑘 =𝜋

𝛼2. So,

𝐼 =𝜋

𝛼2𝑒𝛽

2/4𝛼2=

𝜋

𝜈𝑡𝑒− 𝑥−𝑦 2/4𝜈𝑡

Therefore, the analytic solution of the diffusion equation is

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡 −∞

+∞

𝑢0 𝑦 𝑒− 𝑥−𝑦 2/4𝜈𝑡 𝑑𝑦

Analytic Solution---initial condition is the delta function

Example 1: 𝑢𝑡 = 𝜈𝑢𝑥𝑥 and initial condition

𝑢 𝑥, 0 = 𝑢0 𝑥 = 𝛿(𝑥)

The solution is given by

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡 −∞

+∞

𝑢0 𝑦 𝑒− 𝑥−𝑦 2/4𝜈𝑡𝑑𝑦

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡 −∞

+∞

𝛿 𝑦 𝑒− 𝑥−𝑦 2/4𝜈𝑡 𝑑𝑦

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡𝑒−𝑥

2/4𝜈𝑡

Analytic Solution---initial condition is a step function

Example 2: 𝑢𝑡 = 𝜈𝑢𝑥𝑥 and initial condition

𝑢 𝑥, 0 = 𝑢0 𝑥 = 𝑢𝑙 , 𝑖𝑓 𝑥 < 0𝑢𝑟 , 𝑖𝑓 𝑥 > 0

The solution is given by

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡 −∞

+∞

𝑢0 𝑦 𝑒− 𝑥−𝑦 2/4𝜈𝑡𝑑𝑦

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡(

−∞

0

𝑢𝑙𝑒−

𝑥−𝑦 2

4𝜈𝑡 𝑑𝑦 + 0

+∞

𝑢𝑟𝑒−

𝑥−𝑦 2

4𝜈𝑡 𝑑𝑦)

𝑢 𝑥, 𝑡 = 𝑢𝑙 +1

𝜋(𝑢𝑟 − 𝑢𝑙)

−∞

𝑥

4𝜈𝑡𝑒−𝑦

2𝑑𝑦

Numerical Schemes

Central Explicit Scheme

𝑢𝑗𝑛+1 − 𝑢𝑗

𝑛

∆𝑡= 𝜈

𝑢𝑗+1𝑛 − 2𝑢𝑗

𝑛 + 𝑢𝑗−1𝑛

∆𝑥2

Consistency: 𝑂 Δ𝑥2, Δ𝑡 Stability: 𝜈Δ𝑡

Δ𝑥2 <1

2

Central Implicit Scheme

𝑢𝑗𝑛+1 − 𝑢𝑗

𝑛

∆𝑡= 𝜈

𝑢𝑗+1𝑛+1 − 2𝑢𝑗

𝑛+1 + 𝑢𝑗−1𝑛+1

∆𝑥2

Consistency: 𝑂 Δ𝑥2, Δ𝑡 Stability: unconditionally stable

Numerical Schemes

Crank-Nicolson Scheme

𝑢𝑗𝑛+1 − 𝑢𝑗

𝑛

∆𝑡=

1

2𝜈(𝑢𝑗+1𝑛 − 2𝑢𝑗

𝑛 + 𝑢𝑗−1𝑛

∆𝑥2+𝑢𝑗+1𝑛+1 − 2𝑢𝑗

𝑛+1 + 𝑢𝑗−1𝑛+1

∆𝑥2)

Consistency: 𝑂 Δ𝑥2, Δ𝑡2 Stability: unconditionally stable

Leap Frog Scheme

𝑢𝑗𝑛+1 − 𝑢𝑗

𝑛−1

2∆𝑡= 𝜈

𝑢𝑗+1𝑛 − 2𝑢𝑗

𝑛 + 𝑢𝑗−1𝑛

∆𝑥2

Consistency: 𝑂 Δ𝑥2, Δ𝑡2 Stability: unconditionally unstable

Numerical Schemes

Du Fort-Frankel Scheme

𝑢𝑗𝑛+1 − 𝑢𝑗

𝑛−1

2∆𝑡= 𝜈

𝑢𝑗+1𝑛 − (𝑢𝑗

𝑛+1 + 𝑢𝑗𝑛−1) + 𝑢𝑗−1

𝑛

∆𝑥2

Consistency: 𝑂 Δ𝑡2/Δ𝑥2, Δ𝑡2 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

𝜈 =1

2

𝜆2

Δ𝜏=

1

2𝜆𝑢𝑎𝑣𝑒

where 𝜆 is the mean free path, Δ𝜏 is the average time

between collisions, and 𝑢𝑎𝑣𝑒 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 +𝜆or −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) =

𝑖=1

𝑛

𝑥𝑖

where 𝑥𝑖 = ±𝜆 with equal probability. Then, we have

𝐸 𝑥𝑖 = 0, 𝐸 𝑋 𝑛 = 𝐸 𝑖=0

𝑛

𝑥𝑖 = 0

𝑉𝑎𝑟 𝑥𝑖 = 𝜆2, 𝑉𝑎𝑟 𝑋 𝑛 = 𝐸 𝑋 𝑛 − 𝐸 𝑋 𝑛 2= 𝑛𝜆2

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,

𝑛 𝑖=1𝑛 𝑥𝑖𝑛

− 𝐸 𝑥𝑖 𝑑Ν 0, 𝑉𝑎𝑟 𝑥𝑖

as 𝑛 ∞. This is equivalent to

𝑋 𝑛 𝑑Ν 0, 𝑛𝜆2 = Ν(0,

𝑡𝜆2

Δ𝜏)

where, 𝑡 is the total time.

Then, the distribution of 𝑋 𝑛

𝑝 𝑥, 𝑡 =1

2𝜋𝑡𝜆2/Δ𝜏𝑒−𝑥

2/(2𝑡𝜆2/Δ𝜏)

Mean Free Path--- 1D Brownian Motion

Now, consider the diffusion process with initial condition

𝑢 𝑥, 0 = 𝑢0 𝑥 = 𝛿(𝑥)

Its solution is given in example 1.

𝑢 𝑥, 𝑡 =1

4𝜋𝜈𝑡𝑒−𝑥

2/4𝜈𝑡

Particle Movement Diffusion Process

So, 𝑝 𝑥, 𝑡 = 𝑢(𝑥, 𝑡) and this leads to 𝑡𝜆2

Δ𝜏= 2𝜈𝑡 𝜈 =

1

2

𝜆2

Δ𝜏=

1

2𝜆𝑢𝑎𝑣𝑒

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

2𝜆𝑢𝑎𝑣𝑒 ≈

1

2× 500 × 68 × 10−9 = 1.7 × 10−5𝑚2/𝑠

This is close to the ratio of dynamic viscosity (1.81×10−5𝑘𝑔/(𝑚 ∙ 𝑠)) to the density (1.205𝑘𝑔/𝑚3)

𝜈 =𝜇

𝜌≈

1.81 × 10−5

1.205≅ 1.502 × 10−5𝑚2/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