Biomedical Engineering Reference
In-Depth Information
Proposition 6.3.
The solution of the Riemann problem (2.24) with data
A
L
,
u
L
;
A
R
,
u
R
arbitrary but subject to the condition
4
(
c
L
+
c
R
)
≤
u
R
−
u
L
≡
Δ
u
crit
(2.58)
has the form
⎧
⎨
⎩
W
L
0
(
x
,
t
)
if
x
/
t
≤
S
∗
L
,
W
(
x
,
t
)=
W
0
(collapsed state) if
S
∗
L
≤
x
/
t
≤
S
∗
R
,
(2.59)
W
R
0
(
x
,
t
)
if
S
∗
R
≤
x
/
t
,
where
W
L
0
and
W
R
0
are given by (2.52) and (2.55) respectively.
Proof.
For this case to apply the condition
S
∗
L
≤
S
∗
R
must hold, from which (2.58)
follows. The rest of the proof follows from Propositions 6.1 and 6.2.
We now show some examples of tube collapse with uniform data: Young's
modulus
E
10
5
N/m
2
, exponent in tube law
m
=
3
.
0
×
=
1
/
2, wall thickness
h
0
=
10
−
3
m. Fig. 2.5 illus-
trates a solution of the type constructed in Proposition 6.1. The solution consists of
a single left rarefaction wave associated with the eigenvalue
10
−
4
m, radius at zero transmural pressure
R
0
=
5
.
0
×
6
.
5
×
λ
1
=
u
−
c
,inwhich
0.02
20
0.01
15
0
10
−0.01
5
−0.02
0
−0.2
0
0.2
−0.2
0
0.2
x [m]
x [m]
0.01
100
0
0.005
−100
−200
0
−0.2
0
0.2
−0.2
0
0.2
x [m]
x [m]
Fig. 2.5.
Exact solution dry-bed Riemann problem at time 0
.
01
s
. Initial discontinuity is positioned
10
−
4
m
2
,
u
L
=
0m
2
,
u
R
=
at
x
0
=
0
.
0 m. Initial conditions are:
A
L
=
3
.
5
×
1
.
0m/s,
A
R
=
0
.
0
.
0m/s
