Biomedical Engineering Reference
In-Depth Information
Then in every discrete time step and for every
i, j, k
we construct (west, east,
south, north, bottom and top) coefficients
G
σ,w,l
G
σ,e,l
4
i,j,k
)
ε
2
+
G
w,l
i,j,k
g
(
4
i,j,k
)
ε
2
+
G
e,l
i,j,k
g
(
a
i,j,k
=
τ
1
4
,a
i,j,k
=
τ
1
4
,
l
=1
l
=1
G
σ,s,l
G
σ,n,l
4
i,j,k
)
ε
2
+
G
s,l
i,j,k
g
(
4
i,j,k
)
ε
2
+
G
n,l
i,j,k
g
(
a
i,j,k
=
τ
1
,a
i,j,k
=
τ
1
,
(25)
4
4
l
=1
l
=1
G
σ,b,l
G
σ,t,l
4
i,j,k
)
ε
2
+
G
b,l
i,j,k
g
(
4
i,j,k
)
ε
2
+
G
t,l
i,j,k
g
(
a
i,j,k
=
τ
1
4
,a
i,j,k
=
τ
1
4
,
l
=1
l
=1
and we use, cf. (14),
1
m
i,j,k
=
ε
2
+
2
G
z,l
i,j,k
4
1
24
z∈C
p
l
=1
to define diagonal coefficients
a
i,j,k
=
a
i,j,k
+
a
i,j,k
+
a
i,j,k
+
a
i,j,k
+
a
i,j,k
+
a
i,j,k
+
m
i,j,k
h
2
.
If we define the right-hand sides at the
n
th discrete time step by
b
i,j,k
=
m
i,j,k
h
2
u
n−
1
i,j,k
,
then for the DF node corresponding to triple (
i, j, k
) we get the equation
a
i,j,k
u
i,j,k
−
a
i,j,k
u
i−
1
,j,k
−
a
i,j,k
u
i
+1
,j,k
−
a
i,j,k
u
i,j−
1
,k
−
(26)
a
i,j,k
u
i,j
+1
,k
−
a
i,j,k
u
i,j,k−
1
−
a
i,j,k
u
i,j,k
+1
=
b
i,j,k
.
Collecting these equations for all DF nodes and taking into account Dirichlet
boundary conditions we get the linear system to be solved.
3.4. Solution of Linear Systems
We can solve system (26) by any efficient preconditioned linear iterative solver
suitable for sparse, symmetric, diagonally dominant M-matrices [66]. For exam-
ple, the so-called SOR (Successive Over Relaxation) method can be used. Then,
