A FEniCS Tutorial - FEniCS Project

from the weak form. We then only need to take care of Dirichlet conditions at
two sides:
tol = 1E-14 # tolerance for coordinate comparisons
def Dirichlet_boundary0(x, on_boundary):
return on_boundary and abs(x[0]) < tol
def Dirichlet_boundary1(x, on_boundary):
return on_boundary and abs(x[0] - 1) < tol
bc0 = DirichletBC(V, Constant(0), Dirichlet_boundary0)
bc1 = DirichletBC(V, Constant(1), Dirichlet_boundary1)
bcs = [bc0, bc1]
Note that this code is independent of the number of space dimensions. So are
the statements defining and solving the variational problem:
u = TrialFunction(V)
v = TestFunction(V)
f = Constant(-2)
a = inner(nabla_grad(u), nabla_grad(v))*dx
L = f*v*dx
u = Function(V)
solve(a == L, u, bcs)
The complete code is found in the file paD.py (Poisson problem in ”anyD”).
If we want to parameterize the direction in which u varies, say by the space
direction number e, we only need to replace x[0] in the code by x[e]. The
parameter e could be given as a second command-line argument. The reader is
encouraged to perform this modification.
2 Nonlinear Problems
Now we shall address how to solve nonlinear PDEs in FEniCS. Our sample PDE
for implementation is taken as a nonlinear Poisson equation:
−∇·(q(u)∇u) = f . (38)
The coefficient q(u) makes the equation nonlinear (unless q(u) is constant in u).
To be able to easily verify our implementation, we choose the domain, q(u),
f, and the boundary conditions such that we have a simple, exact solution u.
Let Ω be the unit hypercube [0,1] d in d dimensions, q(u) = (1 + u) m , f = 0,
u = 0 for x 0 = 0, u = 1 for x 0 = 1, and ∂u/∂n = 0 at all other boundaries
x i = 0 and x i = 1, i = 1,...,d−1. The coordinates are now represented by the
symbols x 0 ,...,x d−1 . The exact solution is then
u(x 0 ,...,x d ) = ( (2 m+1 −1)x 0 +1 ) 1/(m+1)
−1. (39)
We refer to Section 1.16 for details on formulating a PDE problem in d space
dimensions.
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