A FEniCS Tutorial - FEniCS Project

seek a new (hopefully improved) solution u k+1 in iteration k+1 such that u k+1
solves the linear problem,
∇·(q(u k )∇u k+1) = 0, k = 0,1,... (43)
The iterations require an initial guess u 0 . The hope is that u k → u as k → ∞,
and that u k+1 is sufficiently close to the exact solution u of the discrete problem
after just a few iterations.
We can easily formulate a variational problem for u k+1 from (43). Equivalently,
we can approximate q(u) by q(u k ) in (41) to obtain the same linear
variational problem. In both cases, the problem consists of seeking u k+1 ∈ V
such that
˜F(u k+1 ;v) = 0 ∀v ∈ ˆV, k = 0,1,..., (44)
with
∫
˜F(u k+1 ;v) =
Ω
q(u k )∇u k+1 ·∇vdx. (45)
Since this is a linear problem in the unknown u k+1 , we can equivalently use the
formulation
a(u k+1 ,v) = L(v), (46)
with
∫
a(u,v) =
Ω
q(u k )∇u·∇vdx (47)
L(v) = 0. (48)
The iterations can be stopped when ǫ ≡ ||u k+1 −u k || < tol, where tol is a
small tolerance, say 10 −5 , or when the number of iterations exceed some critical
limit. The latter case will pick up divergence of the method or unacceptable
slow convergence.
In the solution algorithm we only need to store u k and u k+1 , called u_k and
u in the code below. The algorithm can then be expressed as follows:
def q(u):
return (1+u)**m
# Define variational problem for Picard iteration
u = TrialFunction(V)
v = TestFunction(V)
u_k = interpolate(Constant(0.0), V) # previous (known) u
a = inner(q(u_k)*nabla_grad(u), nabla_grad(v))*dx
f = Constant(0.0)
L = f*v*dx
# Picard iterations
u = Function(V) # new unknown function
eps = 1.0 # error measure ||u-u_k||
tol = 1.0E-5 # tolerance
iter = 0 # iteration counter
maxiter = 25 # max no of iterations allowed
while eps > tol and iter < maxiter:
50