A FEniCS Tutorial - FEniCS Project

derivative ∂F i /∂U j becomes
⎡
∫
(
N∑ N∑
N
) ⎤
∑
⎣q ′ ( Ul k φ l )φ j ∇( Uj k φ j )·∇ˆφ i +q Ul k φ l ∇φ j ·∇ˆφ i
⎦ dx. (52)
Ω
l=1 j=1
l=1
The following results were used to obtain (52):
∂u
= ∂ ∑ N
U j φ j = φ j ,
∂U j ∂U j
j=1
∂
∂U j
∇u = ∇φ j ,
∂
∂U j
q(u) = q ′ (u)φ j . (53)
We can reformulate the Jacobian matrix in (52) by introducing the short notation
u k = ∑ N
j=1 Uk j φ j:
∫
∂F
]
i
=
[q ′ (u k )φ j ∇u k ·∇ˆφ i +q(u k )∇φ j ·∇ˆφ i dx. (54)
∂U j
Ω
In order to make FEniCS compute this matrix, we need to formulate a corresponding
variational problem. Looking at the linear system of equations in
Newton’s method,
N∑
j=1
∂F i
∂U j
δU j = −F i , i = 1,...,N,
wecanintroducev asageneraltestfunctionreplacing ˆφ i ,andwecanidentifythe
unknownδu = ∑ N
j=1 δU jφ j . Fromthelinearsystemwecannowgo”backwards”
to construct the corresponding linear discrete weak form to be solved in each
Newton iteration:
∫
[
q ′ (u k )δu∇u k ·∇v +q(u k )∇δu·∇v ] ∫
dx = − q(u k )∇u k ·∇vdx. (55)
Ω
This variational form fits the standard notation a(δu,v) = L(v) with
∫
[
a(δu,v) = q ′ (u k )δu∇u k ·∇v +q(u k )∇δu·∇v ] dx
Ω
∫
L(v) = − q(u k )∇u k ·∇vdx.
Ω
Note the important feature in Newton’s method that the previous solution u k
replaces u in the formulas when computing the matrix ∂F i /∂U j and vector F i
for the linear system in each Newton iteration.
We now turn to the implementation. To obtain a good initial guess u 0 , we
can solve a simplified, linear problem, typically with q(u) = 1, which yields
the standard Laplace equation ∇ 2 u 0 = 0. The recipe for solving this problem
appears in Sections 1.2, 1.3, and 1.13. The code for computing u 0 becomes as
follows:
Ω
52