A FEniCS Tutorial - FEniCS Project

All mesh objects are of type Mesh so typing the command pydoc dolfin.Mesh
in a terminal window will give a list of methods (that is, functions in a class)
that can be called through any Mesh object. In fact, pydoc dolfin.X shows
the documentation of any DOLFIN name X.
Writing out the solution on the screen can now be done by a simple loop:
coor = mesh.coordinates()
if mesh.num_vertices() == len(u_array):
for i in range(mesh.num_vertices()):
print ’u(%8g,%8g) = %g’ % (coor[i][0], coor[i][1], u_array[i])
The beginning of the output looks like this:
u( 0, 0) = 1
u(0.166667, 0) = 1.02778
u(0.333333, 0) = 1.11111
u( 0.5, 0) = 1.25
u(0.666667, 0) = 1.44444
u(0.833333, 0) = 1.69444
u( 1, 0) = 2
For Lagrange elements of degree higher than one, the vertices do not correspond
to all the nodal points and the ‘if‘-test fails.
For verification purposes we want to compare the values of the computed u
at the nodes (given by u_array) with the exact solution u0 evaluated at the
nodes. The difference between the computed and exact solution should be less
than a small tolerance at all the nodes. The Expression object u0 can be
evaluated at any point x by calling u0(x). Specifically, u0(coor[i]) returns
the value of u0 at the vertex or node with global number i.
Alternatively, we can make a finite element field u_e, representing the exact
solution, whose values at the nodes are given by the u0 function. With
mathematics, ue = ∑ N
j=1 E jφ j , where E j = u 0 (x j ,y j ), (x j ,y j ) being the coordinates
of node number j. This process is known as interpolation. FEniCS has
a function for performing the operation:
u_e = interpolate(u0, V)
The maximum error can now be computed as
u_e_array = u_e.vector().array()
print ’Max error:’, numpy.abs(u_e_array - u_array).max()
The value of the error should be at the level of the machine precision (10 −16 ).
To demonstrate the use of point evaluations of Function objects, we write
out the computed u at the center point of the domain and compare it with the
exact solution:
center = (0.5, 0.5)
print ’numerical u at the center point:’, u(center)
print ’exact u at the center point:’, u0(center)
Trying a 3×3 mesh, the output from the previous snippet becomes
19