

In [1]:

    
import numpy
from matplotlib import pyplot
from matplotlib import colors
% matplotlib inline



In [2]:

    
import os, sys
sys.path.append(os.path.split(os.getcwd())[0])



In [3]:

    
import utils.poly as poly
import utils.quadrature as quad

The coordinates of solution points using Gauss-Legendre quadrature points.



In [4]:

    
xi = quad.GaussJacobi(4).nodes

The Lagrange basis using the Gauss-Legendre quadrature points.



In [5]:

    
Lk = poly.LagrangeBasis(xi)

The exact solution of $u(x)$.



In [6]:

    
def u_exact(x):
    '''exact solution of u'''
    return numpy.sin(x)

Define a function to calculate flux at a location given a known $u$.



In [7]:

    
def f(ui):
    '''flux at spificic location using known velocity at the same location'''
    return 0.5 * ui * ui

Calculate the $u$ at solution points.



In [8]:

    
ui = numpy.dot(Lk(xi), u_exact(xi))

Calculate the flux at solution points using $u$ at those locations.



In [9]:

    
fi = f(ui)

Use $u$ at solution points and Lagrange interpolation to calculate $u$ at left boundary.



In [10]:

    
uL = numpy.dot(Lk(-1), u_exact(xi))

Use $u$ at the left boundary to calculate flux at that boundary (that is, the $f(u_L)$ in the paper).



In [11]:

    
f(uL)









    Out[11]:





0.35250512546198559

Use $f$ at the solution points and Lagrange interpolation to calculate the flux at that boundary (i.e., the $f_L$ in the paper).



In [12]:

    
fL = numpy.dot(Lk(-1), fi)
print(fL)









    



0.383561492566

We can now see that $$f_L \ne f(u_L)$$