A FEniCS Tutorial - FEniCS Project

iter += 1
solve(a == L, u, bcs)
diff = u.vector().array() - u_k.vector().array()
eps = numpy.linalg.norm(diff, ord=numpy.Inf)
print ’iter=%d: norm=%g’ % (iter, eps)
u_k.assign(u) # update for next iteration
We need to define the previous solution in the iterations, u_k, as a finite element
function so that u_k can be updated with u at the end of the loop. We may
create the initial Function u_k by interpolating an Expression or a Constant
to the same vector space as u lives in (V).
Inthecodeabovewedemonstratehowtousenumpyfunctionalitytocompute
the norm of the difference between the two most recent solutions. Here we apply
the maximum norm (l ∞ norm) on the difference of the solution vectors (ord=1
and ord=2 give the l 1 and l 2 vector norms – other norms are possible for numpy
arrays, see pydoc numpy.linalg.norm).
The file picard_np.py contains the complete code for this nonlinear Poisson
problem. The implementation is d dimensional, with mesh construction and
setting of Dirichlet conditions as explained in Section 1.16. For a 33×33 grid
with m = 2 we need 9 iterations for convergence when the tolerance is 10 −5 .
2.2 A Newton Method at the Algebraic Level
After having discretized our nonlinear PDE problem, we may use Newton’s
method to solve the system of nonlinear algebraic equations. From the continuous
variational problem (40), the discrete version (42) results in a system of
equations for the unknown parameters U 1 ,...,U N (by inserting u = ∑ N
j=1 U jφ j
and v = ˆφ i in (42)):
F i (U 1 ,...,U N ) ≡
N∑
∫
j=1
Ω
( ( N
) )
∑
q U l φ l ∇φ j U j ·∇ˆφ i dx = 0, i = 1,...,N .
l=1
(49)
Newton’s method for the system F i (U 1 ,...,U j ) = 0, i = 1,...,N can be formulated
as
N∑
j=1
∂
∂U j
F i (U k 1,...,U k N)δU j = −F i (U k 1,...,U k N), i = 1,...,N, (50)
U k+1
j = U k j +ωδU j , j = 1,...,N, (51)
where ω ∈ [0,1] is a relaxation parameter, and k is an iteration index. An initial
guess u 0 must be provided to start the algorithm.
The original Newton method has ω = 1, but in problems where it is difficult
toobtainconvergence, so-calledunder-relaxation withω < 1mayhelp. Itmeans
that one takes a smaller step than what is suggested by Newton’s method.
We need, in a program, to compute the Jacobian matrix ∂F i /∂U j and the
right-hand side vector −F i . Our present problem has F i given by (49). The
51