# Rates of Convergence¶

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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)

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)

Now solving 4x4...
Now solving 8x8...
Now solving 16x16...
Now solving 32x32...
Now solving 64x64...
Now solving 128x128...

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)

[1.63571727 1.8993806  1.97405822 1.9934556  1.99840339]
[0.8332833  0.95536396 0.98862547 0.99714187 0.99928454]

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()

Out[15]:
<matplotlib.legend.Legend at 0x7fcc02d964d0>
• Now play with $\omega$ and the element order.
In [ ]: