Finite Volume Burgers¶

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

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In [2]:
import numpy as np
import scipy as sp
import matplotlib.pyplot as plt

from matplotlib import animation
from IPython.display import HTML
In [3]:
def gaussian(x):
    u = np.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 [4]:
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
Out[4]:
[<matplotlib.lines.Line2D at 0x7ffb205e7fa0>]

Plot the solution:

In [5]:
if 1:
    # Burgers
    def f(u):
        return u**2/2
    def fprime(u):
        return u
else:
    # 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 [6]:
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 = (
        (f(uplus)+f(uminus))/2
        - alpha/2*(uplus-uminus)
        )
    return - 1/dx*(fluxes[J]-fluxes[Jm1])
In [7]:
u = g(x)

def timestepper(n):
    for i in range(steps_per_frame):
        dt = dx*lmbda/np.max(fprime(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,
    frames=nt//steps_per_frame,
    interval=30)
html = HTML(ani.to_jshtml())
plt.clf()
html
Out[7]:
<Figure size 360x360 with 0 Axes>

Part II: Second-Order Reconstruction¶

First, need a second-order time integrator:

In [8]:
def rk2_step(dt, u, rhs):
    k1 = rhs(u)
    k2 = rhs(u+dt*k1)
    return u+0.5*dt*(k1+k2)

Now upgrade the reconstruction to second order.

NOTE: It's OK to assume (here!) that the wind blows from the left (/to the right) to simplify upwinding.

In [9]:
def rhs(u):
    # right-looking, between J and Jp1
    # 2nd order central
    fluxes = f(u[J] + 1/2*(u[Jp1]-u[J]))
    
    # 2nd order upwind
    #fluxes = f(u[J] + 1/2*(u[J]-u[Jm1]))
    
    return - 1/dx*(fluxes[J]-fluxes[Jm1])
In [10]:
u = g(x)

def timestepper(n):
    # to simplify upwinding
    assert np.min(u) >= 0
    
    for i in range(steps_per_frame):
        dt = 0.7*dx*lmbda/np.max(fprime(np.abs(u)))
        u[:] = rk2_step(dt, u, rhs)
                       
    line.set_data(x, u)
    return line

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

ani = animation.FuncAnimation(
    fig, timestepper,
    frames=nt//steps_per_frame,
    interval=30)
html = HTML(ani.to_jshtml())
plt.clf()
html
Out[10]:
<Figure size 360x360 with 0 Axes>
In [ ]: