Finite Volume Burgers

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

from matplotlib import animation
from IPython.display import HTML
In [2]:
def gaussian(x):
    u = sp.exp(-100 * (x - 0.25)**2)
    return u

def step(x):
    u = np.zeros(x.shape)
    for j in range(len(x)):
        if (x[j] >= 0.6) and (x[j] <= 0.8):
            u[j] = 1.0

    return u

def g1(x):
    return 1+gaussian(x)

def g2(x):
    return 1+gaussian(x) + step(x)

g = g1
In [20]:
nx = 164
x = np.linspace(0, 1, nx, endpoint=False)
dx = x[1] - x[0]
xx = np.linspace(0, 1, 1000, endpoint=False)

lmbda = 0.95
nt = 250
print('tsteps = %d' % nt)
print('    dx = %g' % dx)
print('lambda = %g' % lmbda)

J = np.arange(0, nx)  # all vertices
Jm1 = np.roll(J, 1)
Jp1 = np.roll(J, -1)

plt.plot(x, g(x))
tsteps = 250
    dx = 0.00609756
lambda = 0.95
[<matplotlib.lines.Line2D at 0x7f1ec5d4ccd0>]

Plot the solution:

In [21]:
if 1:
    # Burgers
    def f(u):
        return u**2/2
    def fprime(u):
        return u
    # advection
    def f(u):
        return u
    def fprime(u):
        return 1+0*u

steps_per_frame = 2

Part I: Lax-Friedrichs

Implement rhs for a Lax-Friedrichs flux:

Recall (local) Lax-Friedrichs: $$ f^{\ast} (u^{_-}, u^+) = \frac{f (u^-) + f (u^+)}{2} - \frac{\alpha}{2} (u^+ - u^-) \quad\text{with}\quad \alpha = \max \left( |f' (u^-)|, |f' (u^+)| \right).$$ Recall FV: $$ \bar{u}_{j,\ell+1} = \bar{u}_{j,\ell} - \frac{h_t}{h_x} (f (u_{j + 1 / 2,\ell}) - f (u_{j - 1 / 2,\ell})) . $$

In [29]:
def rhs(u):
    uplus = u[Jp1]
    uminus = u[J]
    alpha = np.maximum(np.abs(fprime(uplus)), np.abs(fprime(uminus)))
    # right-looking, between J and Jp1
    fluxes = (
        - alpha/2*(uplus-uminus)
    return - 1/dx*(fluxes[J]-fluxes[Jm1])
In [30]:
u = g(x)

def timestepper(n):
    for i in range(steps_per_frame):
        dt = dx*lmbda/np.max(np.abs(u))
        u[:] = u + dt*rhs(u)
    line.set_data(x, u)
    return line

fig = plt.figure(figsize=(5,5))
line, = plt.plot(x, u)

ani = animation.FuncAnimation(
    fig, timestepper,
html = HTML(ani.to_jshtml())