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Without loss of generality, we take the simple case: two reaches in cascade. These
reaches are assumed to have the same length
L
. As depicted in Fig. 2,
x
=0and
x
=
L
corresponds to the left and right side of each reach respectively.
Fig. 2.
A canal system consisting of 2 reaches
The Saint-Venant equations are written for each reach
i
,(
i
=1
,
2)
a
i
v
i
+
∂
∂x
a
i
v
i
=
0
0
,
V
i
+
v
i
A
i
+
a
i
g/b
i
(
A
i
+
a
i
)
V
i
+
v
i
∂
∂t
(11)
(
i
=1
,
2)
,
(
x,t
)
∈
[0
,L
]
×
[0
,
+
∞
)
.
Equation (11) has to be complemented by the initial conditions
a
(
x,
0) =
a
0
(
x
)
,
v
(
x,
0) =
v
0
(
x
)
(12)
where
a
(
x,t
)=(
a
1
(
x,t
)
,a
2
(
x,t
))
T
,
v
(
x,t
)=(
v
1
(
x,t
)
,v
2
(
x,t
))
T
.
The boundary conditions are as follows.
F
1
(
v
1
(0
,t
)
,a
1
(0
,t
)) =
v
1
(0
,t
)
−f
1
(
a
1
(0
,t
)) = 0
,
F
2
(
v
1
(
L,t
)
,a
1
(
L,t
)) =
v
1
(
L,t
)
−f
1
(
a
1
(
L,t
)) = 0
,
F
4
(
v
2
(
L,t
)
,a
2
(
L,t
)) =
v
2
(
L,t
)
−f
2
(
a
2
(
L,t
)) = 0
,
F
3
(
v
1
(
L,t
)
,a
1
(
L,t
)
,v
2
(0
,t
)
,a
2
(0
,t
)) = (
A
1
+
a
1
(
L,t
))
(
V
1
+
v
1
(
L,t
))
(13)
(
A
2
+
a
2
(0
,t
))(
V
2
+
v
2
(0
,t
)) = 0
,
−
where
f
i
(
u
)and
f
i
(
u
), (
i
=1
,
2) are continuously differentiable about
u
in a
neighborhood of 0
R
.Theysatisfy
f
i
(0) = 0 and
f
i
(0) = 0.
The Riemann invariants are computed as
ξ
i
±
∈
(
a
i
,v
i
)=
2
(
v
i
+
a
i
p
i
(
α
)
dα
),
where
p
i
(
α
)=
g
0
(
A
i
+
α
)
b
i
(
A
i
+
α
)
. For the sake of simplicity, we denote the fol-
lowing vectors
ξ
−
(
x,t
),
ξ
+
(
x,t
)and
ξ
(
x,t
)by
ξ
−
(
x,t
) =(
ξ
1
(
x,t
)
,ξ
2
−
(
x,t
))
T
,
−
ξ
+
(
x,t
) =(
ξ
+
(
x,t
)
,ξ
+
(
x,t
))
T
,
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