Methods for 1D Advection

Copyright (C) 2010-2020 Luke Olson
Copyright (C) 2020 Andreas Kloeckner

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In [6]:
import numpy as np
import scipy as sp
import matplotlib.pyplot as plt

Problem Description

Here you will set up the problem for $$ u_t + c u_x = 0$$ with periodic BC on the interval [0,1]

In [7]:
c = 1.0
T = 1.0 / c # end time

Set up the Grid

  • dx will be the grid spacing in the $x$-direction
  • x will be the grid coordinates
  • xx will be really fine grid coordinates
In [8]:
nx = 82
x = np.linspace(0, 1, nx, endpoint=False)
dx = x[1] - x[0]
xx = np.linspace(0, 1, 1000, endpoint=False)

Now define an initial condition:

In [9]:
def f(x):
    u = np.zeros(x.shape)
    u[np.intersect1d(np.where(x>0.4), np.where(x<0.6))] = 1.0
    return u
In [10]:
plt.plot(x, f(x), lw=3, clip_on=False)
Out[10]:
[<matplotlib.lines.Line2D at 0x7f679cd039d0>]

Setting the Time Step

Have spatial grid. Now we need a time step. So define a ratio parameter $\lambda$. Let $$ \Delta t = \Delta x \frac{\lambda}{c}$$

In [46]:
lmbda = 0.93
dt = dx * lmbda / c
nt = int(T/dt)
print('T = %g' % T)
print('tsteps = %d' % nt)
print('    dx = %g' % dx)
print('    dt = %g' % dt)
print('lambda = %g' % lmbda)

T = 1
tsteps = 88
    dx = 0.0121951
    dt = 0.0113415
lambda = 0.93

Now make an index list, called $J$, so that we can access $J+1$ and $J-1$ easily

In [47]:
J = np.arange(0, nx - 1)  # all vertices
Jm1 = np.roll(J, 1)
Jp1 = np.roll(J, -1)

Run and Animate

Experiments:

  • Try different values of $\lambda$.
  • Try all the methods.
In [48]:
import time

method = 'ETBS'
plotit = True
u = f(x)

fig = plt.figure(figsize=(10,10))
plt.title('u vs x')
line1, = plt.plot(x, u, lw=3, clip_on=False)
line2, = plt.plot(x, u, lw=3, clip_on=False)
        
def timestepper(n):
    if method == 'ETBS':
        u[J] = u[J] - lmbda * (u[J] - u[Jm1])
            
    if method == "ETFS":
        u[J] = u[J] - lmbda * (u[Jp1] - u[J])
        
    if method == "ETCS":
        u[J] = u[J] - lmbda * (1.0 / 2.0) * (u[Jp1] - u[Jm1])
    
    uex = f((xx - c * (n+1) * dt) % 1.0)
        
    line1.set_data(xx, uex)
    line2.set_data(x, u)
    
    return line1, line2

from matplotlib import animation
from IPython.display import HTML

ani = animation.FuncAnimation(fig, timestepper, frames=nt, interval=30)
html = HTML(ani.to_jshtml())
plt.clf()
html
Out[48]:
<Figure size 720x720 with 0 Axes>

Check the Error

In [49]:
uex = f((x - c * (nt+1) * dt) % 1.0)
error = u - uex
l2err = np.sqrt(dx * np.sum(error**2))
print(l2err)

0.1847348715534358
In [ ]: