#!/usr/bin/env python
# coding: utf-8
# # Rates of Convergence
#
# 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
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#
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# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
# THE SOFTWARE.
#
# In[1]:
from firedrake import *
import numpy as np
import matplotlib.pyplot as plt
# ## A Boundary Value Problem
#
# Consider
# $$u_{*} = \sin(\omega \pi x) \sin(\omega \pi y)$$
# on the unit square with $\omega = 2$ as a start.
# In[2]:
omega = 2
# ## Refine the Mesh and Check the Error
# In[12]:
errsH0 = []
errsH1 = []
hs = []
for nx in [4, 8, 16, 32, 64, 128]: # , 256, 512]:
print(f"Now solving {nx}x{nx}...")
mesh = UnitSquareMesh(nx, nx)
V = FunctionSpace(mesh, "Lagrange", 1)
Vexact = FunctionSpace(mesh, "Lagrange", 7)
x = SpatialCoordinate(V.mesh())
u_exact = interpolate(sin(omega*pi*x[0])*sin(omega*pi*x[1]), Vexact)
f = 2*pi**2*omega**2*u_exact
# all four sides of the square
bc = DirichletBC(V, 0.0, [1, 2, 3, 4])
u = TrialFunction(V)
v = TestFunction(V)
a = inner(grad(u), grad(v))*dx
L = f*v*dx
u = Function(V)
solve(a == L, u, bc)
EH0 = errornorm(u_exact, u, norm_type='L2')
EH1 = errornorm(u_exact, u, norm_type='H1')
errsH0.append(EH0)
errsH1.append(EH1)
hs.append(1/nx)
# In[13]:
errsH0 = np.array(errsH0)
errsH1 = np.array(errsH1)
hs = np.array(hs)
rH0 = np.log(errsH0[1:] / errsH0[0:-1]) / np.log(hs[1:] / hs[0:-1])
rH1 = np.log(errsH1[1:] / errsH1[0:-1]) / np.log(hs[1:] / hs[0:-1])
print(rH0)
print(rH1)
# In[15]:
plt.loglog(hs, errsH0, "o-", label="$H^0$ error")
plt.loglog(hs, errsH1, "o-", label="$H^1$ error")
plt.loglog(hs, hs**1, "o-", label="$h^1$")
plt.loglog(hs, hs**2, "o-", label="$h^2$")
plt.legend()
# - Now play with $\omega$ and the element order.
# In[ ]: