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We now calculate the stiffness matrix
A
BS
N
×
N
∈ R
using the finite element
space
V
N
. We have, by (
8.17
),
A
BS
j,j
=
a
BS
(ϕ
j
,ϕ
j
)
=
a
BS
(ϕ
N
2
(i
1
−
1
)
+
i
2
,ϕ
N
2
(i
1
−
1
)
+
i
2
)
=
a
BS
(b
i
1
b
i
2
,b
i
1
b
i
2
)
=
Q
11
2
+
Q
12
2
∂
x
1
(b
i
1
b
i
2
)∂
x
1
(b
i
1
b
i
2
)
d
x
∂
x
1
(b
i
1
b
i
2
)∂
x
2
(b
i
1
b
i
2
)
d
x
G
G
+
Q
21
2
∂
x
2
(b
i
1
b
i
2
)∂
x
1
(b
i
1
b
i
2
)
d
x
+
Q
22
∂
x
2
(b
i
1
b
i
2
)∂
x
2
(b
i
1
b
i
2
)
d
x
2
G
G
μ
1
μ
2
+
∂
x
1
(b
i
1
b
i
2
)b
i
1
b
i
2
d
x
+
∂
x
2
(b
i
1
b
i
2
)b
i
1
b
i
2
d
x
G
G
r
+
b
i
1
b
i
2
b
i
1
b
i
2
d
x
G
b
i
1
b
i
1
d
x
1
b
i
1
b
i
1
d
x
1
R
R
R
R
=
Q
11
2
b
i
2
b
i
2
d
x
2
+
Q
12
b
i
2
b
i
2
d
x
2
2
−
R
−
R
−
R
−
R
b
i
1
b
i
1
d
x
1
b
i
1
b
i
1
d
x
1
R
R
R
R
+
Q
21
2
b
i
2
b
i
2
d
x
2
+
Q
22
b
i
2
b
i
2
d
x
2
2
−
R
−
R
−
R
−
R
+
μ
1
b
i
1
b
i
1
d
x
1
b
i
2
b
i
2
d
x
2
+
μ
2
b
i
1
b
i
1
d
x
1
R
R
R
R
b
i
2
b
i
2
d
x
2
−
R
−
R
−
R
−
R
+
r
b
i
1
b
i
1
d
x
1
R
R
b
i
2
b
i
2
d
x
2
.
Using that the matrices
S
,
B
and
M
are given by
S
i,i
=
b
i
b
i
,
B
i,i
=
b
i
b
i
and
−
R
−
R
M
i,i
=
b
i
b
i
, see, e.g. (3.22), and noting that
b
i
b
i
=−
b
i
b
i
=−
B
i,i
,we
obtain
j,j
=
Q
11
S
i
1
,i
1
M
i
2
,i
2
+
Q
12
B
)
i
2
,i
2
+
Q
21
A
BS
B
i
1
,i
1
(
−
(
−
B
)
i
1
,i
1
B
i
2
,i
2
2
2
2
+
Q
22
2
M
i
1
,i
1
S
i
2
,i
2
+
μ
1
B
i
1
,i
1
M
i
2
,i
2
+
μ
2
M
i
1
,i
1
B
i
2
,i
2
+
r
M
i
1
,i
1
M
i
2
,i
2
.
Denote by
S
k
1
,
2, and analogously for
B
k
,
M
k
.
the matrix with entries
S
i
k
,i
k
,
k
=
Using that the covariance matrix
Q
is symmetric and Eq. (
8.14
), we have shown
Proposition 8.4.2
Assume d
2
in
(
8.9
)
and assume that the finite element space
V
N
is as in
(
8.18
).
Then
,
the stiffness matrix
A
BS
=
to the bilinear form a
BS
(
·
,
·
) is
given by
=
Q
11
2
+
Q
22
2
A
BS
S
1
M
2
−
Q
12
B
1
B
2
M
1
S
2
⊗
⊗
⊗
μ
1
B
1
M
2
μ
2
M
1
B
2
r
M
1
M
2
.
+
⊗
+
⊗
+
⊗
Since the Kronecker product is distributive, we can reduce the number of prod-
ucts by
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