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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