# 1.2.1.5. 2D First-order Linear Convection¶

## 1.2.1.5.1. Understand the Problem¶

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

• 2D linear convection is described as follows:

${\partial u \over \partial t} + c {\partial u \over \partial x} + c {\partial u \over \partial y} = 0$
${\partial v \over \partial t} + c {\partial v \over \partial x} + c {\partial v \over \partial y} = 0$

## 1.2.1.5.2. Formulate the Problem¶

### 1.2.1.5.2.1. 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

• c = 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)

$$0 \le t \le 0.5$$

$$0 \le n \le 50$$

$$1$$

### 1.2.1.5.2.2. Output Data¶

x

y

t

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

$$0 \le x \le 2$$

$$0 \le y \le 2$$

$$0 \le t \le 0.5$$

$$?$$

## 1.2.1.5.3. Design Algorithm to Solve Problem¶

### 1.2.1.5.3.1. 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

• BD in space

### 1.2.1.5.3.3. Discrete equation¶

${{u_{i,j}^{n+1} - u_{i,j}^n} \over {\Delta t}} + c {{u_{i,j}^n - u_{i-1,j}^n} \over \Delta x} + c {{u_{i,j}^n - u_{i,j-1}^n} \over \Delta y} = 0$
${{v_{i,j}^{n+1} - v_{i,j}^n} \over {\Delta t}} + c {{v_{i,j}^n - v_{i-1,j}^n} \over \Delta x} + c {{v_{i,j}^n - v_{i,j-1}^n} \over \Delta y} = 0$

### 1.2.1.5.3.4. Transpose¶

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

### 1.2.1.5.3.5. 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)

#Boundary Conditions
for n between 0 and 50
for i between 0 and 20
u(i,0,n)=v(i,0,n)=1
u(i,20,n)=v(i,20,n)=1
for j between 0 and 20
u(0,j,n)=v(0,j,n)=1
u(20,j,n)=v(20,j,n)=1

#Initial Conditions
for i between 1 and 19
for j between 1 and 19
if(5 < i < 10 OR 5 < j < 10)
u(i,j,0) = v(i,j,0)= 2
else
u(i,j,0) = v(i,j,0)= 1

#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)-c*dt*[(1/dx)*(u(i,j,n)-u(i-1,j,n))+
(1/dy)*(u(i,j,n)-u(i,j-1,n))]
v(i,j,n+1) = v(i,j,n)-c*dt*[(1/dx)*(v(i,j,n)-v(i-1,j,n))+
(1/dy)*(v(i,j,n)-v(i,j-1,n))]


## 1.2.1.5.4. Implement Algorithm in Python¶

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

# Initialise data structures
import numpy as np
u = np.ones(((nx,ny,nt)))
v = np.ones(((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[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]-c*dt*((1/dx)*(u[i,j,n]-u[i-1,j,n])+
(1/dy)*(u[i,j,n]-u[i,j-1,n])))
v[i,j,n+1] = (v[i,j,n]-c*dt*((1/dx)*(v[i,j,n]-v[i-1,j,n])+
(1/dy)*(v[i,j,n]-v[i,j-1,n])))

# 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
import numpy as np
from mpl_toolkits.mplot3d import Axes3D
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 = convection(101, 81, 81, 0.5, 2.0, 2.0, 0.5)
plot_3D(u,x,y,0,'Figure 1: c=0.5m/s, nt=101, nx=81, ny=81, t=0sec','u (m/s)')
plot_3D(u,x,y,100,'Figure 2: c=0.5m/s, nt=101, nx=81, ny=81, t=0.5sec','u (m/s)')
plot_3D(v,x,y,0,'Figure 3: c=0.5m/s, nt=101, nx=81, ny=81, t=0sec','v (m/s)')
plot_3D(v,x,y,100,'Figure 4: c=0.5m/s, nt=101, nx=81, ny=81, t=0.5sec','v (m/s)')


## 1.2.1.5.5. Conclusions¶

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

• As with 1D, the first order backward differencing scheme in space creates false diffusion.

### 1.2.1.5.5.2. Why does the wave shift?¶

• The square wave is being convected by the constant linear wave speed c in 2 dimensions

• For $$c > 0$$ profiles shift by $$c \Delta t$$ - compare Figure 1, 2, 3 and 4