Experimenting with Dispersion and Dissipation

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

MIT License Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
In [1]:
import numpy as np
import scipy as sp
import matplotlib.pyplot as plt

Problem Description

Consider $$ u_t + a u_x = 0$$ with periodic BC on the interval $[0,1]$.

In [2]:
a = 1.0
T = 4.0 / a # end time for four cycles

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 [27]:
nx = 90
k = 10
x = np.linspace(0, 2*np.pi, nx, endpoint=False)
dx = x[1] - x[0]
xx = np.linspace(0, 2*np.pi, 1000, endpoint=False)

Now define an initial condition:

In [28]:
def f(x):
    u = np.sin(k * x)
    return u
In [29]:
plt.plot(xx, f(xx), lw=3, clip_on=False)
Out[29]:
[<matplotlib.lines.Line2D at 0x7feaed778190>]

Setting the Time Step

Now we need a time step. Let $$ \Delta t = \Delta x \frac{\lambda}{a}$$ with CFL number $\lambda$.

In [30]:
lmbda = 0.9
dt = dx * lmbda / a
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 = 4
tsteps = 63
    dx = 0.0698132
    dt = 0.0628319
lambda = 0.9

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

In [31]:
J = np.arange(0, nx)  # all vertices
Jm1 = np.roll(J, 1)
Jp1 = np.roll(J, -1)

Run and Plot Final Solution

In [32]:
plotit = True

uETBS = f(x)
uLW = f(x)
if plotit:
    fig = plt.figure(figsize=(10,10))
    ax = fig.add_subplot(111)
    ax.set_title('u vs x')

for n in range(0, nt):
    uETBS[J] = uETBS[J] - lmbda * (uETBS[J] - uETBS[Jm1])
    
    uLW[J] = uLW[J] - lmbda * (1.0 / 2.0) * (uLW[Jp1] - uLW[Jm1]) \
           + (lmbda**2 / 2.0) * (uLW[Jp1] - 2 * uLW[J] + uLW[Jm1])
    
    uex = f((xx - a * (n+1) * dt) % 1.0)

if plotit:
    ax.plot(x, uETBS, '-', clip_on=False, label="ETBS")
    ax.plot(x, uLW, '-', clip_on=False, label="Lax-Wendroff")
    ax.plot(xx, uex, 'k-', clip_on=False, label='exact')
    ax.legend(frameon=False)
    plt.show()
In [ ]:
 
In [ ]: