Biomedical Engineering Reference
In-Depth Information
2.6 Collapse of vessels
Veins collapse under normal physiological conditions. For example, in normal indi-
viduals venous blood is drained from the brain mainly through the jugular veins in
the supine position, whereas in the upright position these veins collapse and most of
the venous flow takes place through the vertebral veins [25]. Arteries do collapse,
but rarely do so under normal physiological conditions. In this section we construct
exact solutions for tube collapse that can be used as reference solutions for validating
numerical methods. To be consistent with the main body of the paper we continue
the analysis of the equations for a tube law intended for arteries, and thus the solution
have a limited physical meaning.
The Riemann problem (2.24) is considered, neglecting variations of
K
, leading
to the reduced 2
×
2 system (2.25) with
m
=
1
/
2. Consider the physical variables
T
. There are three cases of interest. Case 1 is that in which the left hand
portion of the collapsible tube
W
=[
A
,
u
]
(
−
∞
,
0
]
is not collapsed but the right hand portion
[
is collapsed and is therefore
dry
. Case 2 is the opposite case, the left portion is
collapsed but the right portion is not. In case 3 the initial condition does not include
collapse but this arises as time evolves. Below we provide a complete solution to all
three possible cases.
0
,
∞
)
Proposition 6.1.
The solution of the Riemann problem (2.24) with data
A
L
>
0,
u
L
arbitrary;
A
R
=
0 has exact solution consisting of a single rarefaction
wave associated with the eigenvalue
0and
u
R
=
λ
1
=
u
−
c
and is given as
⎧
⎨
W
L
if
x
/
t
≤
u
L
−
c
L
,
W
LO
(
x
,
t
)=
W
Lfan
if
u
L
−
c
L
≤
x
/
t
≤
S
∗
L
,
(2.52)
⎩
W
0
if
S
∗
L
≤
x
/
t
,
where
W
L
is the left data state,
W
R
=
W
0
is the collapsed right data state and
=
+
S
∗
L
u
L
4
c
L
(2.53)
is the speed of the blood/no blood front separating the collapsed state (right) from
the uncollapsed new state (left) given as
⎧
⎨
u
L
+
1
5
4
x
t
u
=
4
c
L
+
,
W
Lfan
≡
(2.54)
u
L
⎩
1
5
x
t
c
=
+
4
c
L
−
.
Proof.
First it is easy to show that a shock solution of the Riemann problem with
initial condition as for case 1 is not possible [21]. Therefore the only possible way for
connecting the left state
W
L
and the collapsed state
W
0
is through a rarefaction wave
associated with the left eigenvalue
λ
1
=
u
−
c
. This wave, denoted by
W
Lfan
,has
