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

1.2.1.5.3.2. Numerical scheme

  • 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()

(Source code)

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