Biomedical Engineering Reference
In-Depth Information
1
0.8
0.6
0.4
0.2
0
−
0.2
−
0.4
−
0.6
−
0.8
−
1
−
1
−
0.8
−
0.6
−
0.4
−
0.2
0
0.2
0.4
0.6
0.8
1
Figure 4.13:
Example of two circles expanding using the current velocity ex-
tension method.
of equations results in a quartic polynomial in
F
i
,
j
which can be solved using a
Newton solver or by a direct quartic polynomial solver. Once
F
i
,
j
is computed,
the value of
φ
i
,
j
is easily computed from Eq. 4.43.
The initialization of this method uses a similar bicubic representation as was
discussed in Section 4.2.3. The initialization process is based upon the following
theorem from [23], and illustrated in Fig. 4.14:
Theorem 1.
Suppose
={
(
x
,
y
):
ax
+
by
=
c
}
and F
0
(
x
,
y
)
=
dx
+
ey
+
f
for
(
x
,
y
)
∈
with F
0
not identically zero on
, then the equations
F
∇
φ
=
1
,
(4.44)
∇
F
·∇
φ
=
0
,
(4.45)
