# 1.2.1.7. 2D Second-order Linear Diffusion¶

## 1.2.1.7.1. Understand the Problem¶

• What is the final velocity profile for 2D linear diffusion when the initial conditions are a square wave and the boundary conditions are unity?

• 2D diffusion is described as follows:

${\partial u \over \partial t} = \nu \left ( {\partial^2 u \over \partial x^2}+ {\partial^2 u \over \partial y^2} \right )$
${\partial v \over \partial t} = \nu \left ( {\partial^2 v \over \partial x^2}+ {\partial^2 v \over \partial y^2} \right )$

## 1.2.1.7.2. Formulate the Problem¶

### 1.2.1.7.2.1. Input Data¶

• nt = 51 (number of temporal points)

• tmax = 0.5

• nx = 21 (number of x spatial points)

• nj = 21 (number of y spatial points)

• xmax = 2

• ymax = 2

• nu = 0.1

Initial Conditions: $$t=0$$

x

i

y

j

u(x,y,t), v(x,y,t)

$$0$$

$$0$$

$$0$$

$$0$$

$$1$$

$$0 < x \le 0.5$$

$$0 < i \le 5$$

$$0 < y \le 0.5$$

$$0 < j \le 5$$

$$1$$

$$0.5 < x \le 1$$

$$5 < i \le 10$$

$$0.5 < y \le 1$$

$$5 < j \le 10$$

$$2$$

$$1 < x < 2$$

$$10 < i < 20$$

$$1 < y < 2$$

$$10 < j < 20$$

$$1$$

$$2$$

$$20$$

$$2$$

$$20$$

$$1$$

Boundary Conditions: $$x=0$$ and $$x=2$$, $$y=0$$ and $$y=2$$

t

n

u(x,y,t), v(x,y,t)

$$0 \le t \le 0.5$$

$$0 \le n \le 50$$

$$1$$

### 1.2.1.7.2.2. Output Data¶

x

y

t

u(x,y,t), v(x,y,t)

$$0 \le x \le 2$$

$$0 \le y \le 2$$

$$0 \le t \le 0.5$$

$$?$$

## 1.2.1.7.3. Design Algorithm to Solve Problem¶

### 1.2.1.7.3.1. Space-time discretisation¶

• i $$\rightarrow$$ index of grid in x

• j $$\rightarrow$$ index of grid in y

• n $$\rightarrow$$ index of grid in t

• FD in time

• CD in space

### 1.2.1.7.3.3. Discrete equation¶

${{u_{i,j}^{n+1} - u_{i,j}^n} \over {\Delta t}} = \nu {{u_{i-1,j}^n - 2u_{i,j}^n + u_{i+1,j}^n} \over \Delta x^2} + \nu {{u_{i,j-1}^n - 2u_{i,j}^n + u_{i,j+1}^n} \over \Delta y^2}$
${{v_{i,j}^{n+1} - v_{i,j}^n} \over {\Delta t}} = \nu {{v_{i-1,j}^n - 2v_{i,j}^n + v_{i+1,j}^n} \over \Delta x^2} + \nu {{v_{i,j-1}^n - 2v_{i,j}^n + v_{i,j+1}^n} \over \Delta y^2}$

### 1.2.1.7.3.4. Transpose¶

$u_{i,j}^{n+1} = u_{i,j}^n + \nu \Delta t \left ( {{u_{i-1,j}^n - 2u_{i,j}^n + u_{i+1,j}^n} \over \Delta x^2} + {{u_{i,j-1}^n - 2u_{i,j}^n + u_{i,j+1}^n} \over \Delta y^2} \right )$
$v_{i,j}^{n+1} = v_{i,j}^n + \nu \Delta t \left ( {{v_{i-1,j}^n - 2v_{i,j}^n + v_{i+1,j}^n} \over \Delta x^2} + {{v_{i,j-1}^n - 2v_{i,j}^n + v_{i,j+1}^n} \over \Delta y^2} \right )$

### 1.2.1.7.3.5. Pseudo-code¶

#Constants
nt = 51
tmax = 0.5
dt =  tmax/(nt-1)
nx =  21
xmax = 2
ny = 21
ymax = 2
dx = xmax/(nx-1)
dy = ymax/(ny-1)
nu = 0.1

#Boundary Conditions
#u boundary, bottom and top
u(0:20,0,0:50)=u(0:20,20,0:50)=1
#v boundary, bottom and top
v(0:20,0,0:50)=v(0:20,20,0:50)=1
#u boundary left and right
u(0,0:20,0:50)=u(20,0:20,0:50)=1
#v boundary left and right
v(0,0:20,0:50)=v(20,0:20,0:50)=1

#Initial Conditions
u(:,:,0) = 1
v(:,:,0) = 1
u(5:10,5:10,0) = 2
v(5:10,5:10,0) = 2

#Iteration
for n between 0 and 49
for i between 1 and 19
for j between 1 and 19
u(i,j,n+1) = u(i,j,n)-dt*nu*[ ( u(i-1,j,n)-2*u(i,j,n)+u(i+1,j,n) ) /dx^2 +
( u(i,j-1,n)-2*u(i,j,n)+u(i,j+1,n) ) /dy^2 ]
v(i,j,n+1) = v(i,j,n)-dt*nu*[ ( v(i-1,j,n)-2*v(i,j,n)+v(i+1,j,n) ) /dx^2 +
( v(i,j-1,n)-2*v(i,j,n)+v(i,j+1,n) ) /dy^2 ]


## 1.2.1.7.4. Implement Algorithm in Python¶

def diffusion(nt, nx, ny, tmax, xmax, ymax, nu):
"""
Returns the velocity field and distance for 2D diffusion
"""
# Increments
dt = tmax/(nt-1)
dx = xmax/(nx-1)
dy = ymax/(ny-1)

# Initialise data structures
import numpy as np
u = np.zeros(((nx,ny,nt)))
v = np.zeros(((nx,ny,nt)))
x = np.zeros(nx)
y = np.zeros(ny)

# Boundary conditions
u[0,:,:] = u[nx-1,:,:] = u[:,0,:] = u[:,ny-1,:] = 1
v[0,:,:] = v[nx-1,:,:] = v[:,0,:] = v[:,ny-1,:] = 1

# Initial conditions
u[:,:,:] = v[:,:,:] = 1
u[int((nx-1)/4):int((nx-1)/2),int((ny-1)/4):int((ny-1)/2),0] = 2
v[int((nx-1)/4):int((nx-1)/2),int((ny-1)/4):int((ny-1)/2),0] = 2

# Loop
for n in range(0,nt-1):
for i in range(1,nx-1):
for j in range(1,ny-1):
u[i,j,n+1] = ( u[i,j,n]+dt*nu*( ( u[i-1,j,n]-2*u[i,j,n]+u[i+1,j,n] ) /dx**2 +
( u[i,j-1,n]-2*u[i,j,n]+u[i,j+1,n] ) /dy**2 ) )
v[i,j,n+1] = ( v[i,j,n]+dt*nu*( ( v[i-1,j,n]-2*v[i,j,n]+v[i+1,j,n] ) /dx**2 +
( v[i,j-1,n]-2*v[i,j,n]+v[i,j+1,n] ) /dy**2 ) )

# X Loop
for i in range(0,nx):
x[i] = i*dx

# Y Loop
for j in range(0,ny):
y[j] = j*dy

return u, v, x, y

def plot_3D(u,x,y,time,title,label):
"""
Plots the 2D velocity field
"""

import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
fig=plt.figure(figsize=(11,7),dpi=100)
ax=fig.gca(projection='3d')
ax.set_xlabel('x (m)')
ax.set_ylabel('y (m)')
ax.set_zlabel(label)
X,Y=np.meshgrid(x,y)
surf=ax.plot_surface(X,Y,u[:,:,time],rstride=2,cstride=2)
plt.title(title)
plt.show()

u,v,x,y = diffusion(151, 51, 51, 0.5, 2.0, 2.0, 0.1)

plot_3D(u,x,y,0,'Figure 1: nu=0.1, nt=151, nx=51, ny=51, t=0sec','u (m/s)')
plot_3D(u,x,y,100,'Figure 2: nu=0.1, nt=151, nx=51, ny=51, t=0.5sec','u (m/s)')
plot_3D(v,x,y,0,'Figure 3: nu=0.1, nt=151, nx=51, ny=51, t=0sec','v (m/s)')
plot_3D(v,x,y,100,'Figure 4: nu=0.1, nt=151, nx=51, ny=51, t=0.5sec','v (m/s)') (png, hires.png, pdf) (png, hires.png, pdf) (png, hires.png, pdf) (png, hires.png, pdf)

## 1.2.1.7.5. Conclusions¶

### 1.2.1.7.5.1. Why isn’t the square wave maintained?¶

• As with 1D, the square wave isn’t maintained because the system is attempting to reach equilibrium - the rate of change of velocity being equal to the shear force per unit mass. There are no external forces and no convective acceleration terms.