A FEniCS Tutorial - FEniCS Project

To see how repeated assembly can be avoided, we look at the L(v) form in
(75), which in general varies with time through u k−1 , f k , and possibly also with
∆t if the time step is adjusted during the simulation. The technique for avoiding
repeated assembly consists in expanding the finite element functions in sums
overthebasisfunctionsφ i , asexplainedinSection1.15, toidentifymatrix-vector
products that build up the complete system. We have u k−1 = ∑ N
j=1 Uk−1 j φ j ,
and we can expand f k as f k = ∑ N
j=1 Fk j φ j. Inserting these expressions in L(v)
and using v = ˆφ i result in
∫
Ω
(
u k−1 +∆tf k) ∫
vdx =
=
Ω
⎛
N∑
⎝
j=1
N∑
(∫
j=1
Ω
U k−1
j
φ j +∆t
)
ˆφ i φ j dx U k−1
j
N∑
j=1
F k j φ j
⎞
⎠ ˆφi dx,
+∆t
N∑
(∫
j=1
Ω
)
ˆφ i φ j dx Fj k .
Introducing M ij = ∫ Ω ˆφ i φ j dx, we see that the last expression can be written
N∑
j=1
M ij U k−1
j
+∆t
N∑
M ij Fj k ,
j=1
which is nothing but two matrix-vector products,
MU k−1 +∆tMF k ,
if M is the matrix with entries M ij and
and
U k−1 = (U k−1
1 ,...,U k−1
N )T ,
F k = (F k 1,...,F k N) T .
We have immediate access to U k−1 in the program since that is the vector
in the u_1 function. The F k vector can easily be computed by interpolating
the prescribed f function (at each time level if f varies with time). Given M,
U k−1 , and F k , the right-hand side b can be calculated as
b = MU k−1 +∆tMF k .
That is, no assembly is necessary to compute b.
The coefficient matrix A can also be split into two terms. We insert v = ˆφ i
and u k = ∑ N
j=1 Uk j φ j in the expression (74) to get
N∑
(∫
j=1
Ω
)
ˆφ i φ j dx Uj k +∆t
N∑
(∫
j=1
∇ˆφ i ·∇φ j dx
Ω
which can be written as a sum of matrix-vector products,
MU k +∆tKU k = (M +∆tK)U k ,
)
Uj k ,
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