A FEniCS Tutorial - FEniCS Project

Here, Γ 0 is the boundary x = 0, while Γ 1 corresponds to the boundary x = 1.
We have that u L = 1+2y 2 , u R = 2+2y 2 , and g = −4y. For the left boundary
Γ 0 we define the usual triple of a function for the boundary value, a function
for defining the boundary of interest, and a DirichletBC object:
u_L = Expression(’1 + 2*x[1]*x[1]’)
def left_boundary(x, on_boundary):
tol = 1E-14 # tolerance for coordinate comparisons
return on_boundary and abs(x[0]) < tol
Gamma_0 = DirichletBC(V, u_L, left_boundary)
For the boundary x = 1 we write a similar code snippet:
u_R = Expression(’2 + 2*x[1]*x[1]’)
def right_boundary(x, on_boundary):
tol = 1E-14 # tolerance for coordinate comparisons
return on_boundary and abs(x[0] - 1) < tol
Gamma_1 = DirichletBC(V, u_R, right_boundary)
The various essential conditions are then collected in a list and used in the
solution process:
bcs = [Gamma_0, Gamma_1]
...
solve(a == L, u, bcs)
# or
problem = LinearVariationalProblem(a, L, u, bcs)
solver = LinearVariationalSolver(problem)
solver.solve()
In other problems, where the u values are constant at a part of the boundary,
we may use a simple Constant object instead of an Expression object.
The file dn2_p2D.py contains a complete program which demonstrates the
constructions above. An extended example with multiple Neumann conditions
would have been quite natural now, but this requires marking various parts of
theboundaryusingthemeshfunctionconceptandisthereforelefttoSection5.3.
1.15 A Linear Algebra Formulation
Given a(u,v) = L(v), the discrete solution u is computed by inserting u =
∑ N
j=1 U jφ j into a(u,v) and demanding a(u,v) = L(v) to be fulfilled for N test
functions ˆφ 1 ,..., ˆφ N . This implies
N∑
a(φ j , ˆφ i )U j = L(ˆφ i ), i = 1,...,N,
j=1
43