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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 ) ),
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 ) 2
( x,t )) T ,
ξ + ( x,t ) =( ξ + ( x,t ) + ( x,t )) T ,
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