Notes on RKDG methods for shallow-water equations in canal networks

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1 Notes on RKDG methods for shallow-water equations in canal networks Maya Briani, Benedetto Piccoli, Jing-Mei Qiu December 9, 015 Abstract PDE models for network flows are used in a number of different applications, including modeling of water channel networks. While the theory and first-order numerics are well developed, high-order schemes are not well developed. We propose a Runge-Kutta discontinuous Galerkin method as an efficient, effective and compact numerical approach for numerical simulations of 1-D models for water flow in open canals. Our numerical tests show the advantages of RKDG over firstorder schemes. 1 Introduction The interest for Partial Differential Equation (PDE) models for networks flows is nowadays well established. One reason for such interest can be found in rich set of applications ranging from vehicular traffic to supply chain and blood flow (see [8, ] for a fairly complete account of recent results). Within this research area, water channels networks (for irrigation or water distribution) is a quite richly investigated theme, see [11, 13, 14, 16, 9, 17, 13, 10]. Both for general network flow PDE models and specific water channels ones, the modeling and control problems were treated in a number of papers [1, ]. On one side, a wealth of scientific works dealt with the problem of numerically simulating water flow in a single channel, see [6, 3, 19]. On the other side, high order numerics for network flows is a relatively newly developing one [17, 13, 10]. In [17], the authors apply RKDE method to treat flows in individual channels as a 1 D flows and use a D junction Istituto per le Applicazioni del Calcolo M. Picone, Consiglio Nazionale delle Ricerche, Via dei Taurini 19, Rome, Italy (m.briani@iac.cnr.it) Rutgers University Camden 311 N. 5th Street Camden, NJ 0810 (piccoli@camden.rutgers.edu) Department of Mathematics, University of Houston, Houston, TX, 7704 (jingqiu@math.uh.edu) 1

2 region to couple them. In [13] the authors consider a local zooming of a T-junction, resulting in a two-dimensional flow problem at the canals intersection, for which a high-order non-oscillatory method is applied. Then the water flow solutions are space-averaged over the junction areas. [10] presents a comparison of 1 D and D network simulations, which are validated by experimental results. In this paper, we focus on Runge-Kutta (RK) Discontinuous Galerkin (DG) methods to deal with shallow water equations on a channel network. The RKDG method is an efficient, effective and compact numerical approach for numerical simulations of water flow in open canals. The shallow water equations are given by a system of balance (or conservation) laws. One of the difficulty of such system is the presence of source terms due to friction of channel bed varying slope. Since we focus on the network problem, we assume that no source term is present and deal with the delicate problem of defining and numerically simulating dynamics at channel junctions. For the latter, we restrict to two of the main used approaches, which impose at junctions equality of water height [16], continuity of energy [14, 13], between incoming and outgoing channels (besides the conservation of flow). In our proposed framework, both for the simple 1-to-1 case and for the 1-to- or -to-1 junctions, the junction condition do not depend on the angle between canals. There are papers which take into account the angle between canals, most of them are on D models [10, 18, 5]. We refer readers to Goudiaby-Kreiss [9] and Marigo [16] for more details on the existence of the solution in the case of a network with three canals with a 1-to- junction. Specifically, in [9] the authors prove that under certain condition, there is a unique solution to the junction Riemann problem assuming the conservation of mass (see eq.(1)) and energy continuity (see eq.(14)) at the junction, while in [16] the authors prove the existence and the uniqueness of the solution to the junction Riemann problem assuming the conservation of mass (see eq.(1)) and equal heights (see eq.(13)) at the junction. Both works focus on the case of a network with three canals. Other results for the existence and uniqueness of conservation laws at a junction can be found in Colombo, Herty, Sachers [5] and references therein, where the authors give results on the well posedness for a junction of n open canals under the model proposed by Leugering and Schmidt [14](see formulae (.9)-(.16)) with energy continuity at the junction. Physical reasons motivate different choices of conditions that are originally derived by engineers. Which conditions are used often depends on if the flow is subcritical or supercritical. See for instance [1, 14] and references therein for a subcritical case. First we recall the definition of shallow water equations and explain in details the solution to Riemann problems, i.e. Cauchy problems with Heaviside-type initial conditions. The system has two regimes: subcritical where the two eigenvalues have different signs and supercritical where both

3 eigenvalues are positive. The first case is also called the fluvial condition opposed to the second one called torrential. The fluvial/subcritical condition is more natural for large channels (or rivers) and we restrict most of the paper to this case. Then we revise the definition of solution at junctions using the two methods: equal water height and equal energy. For the simple case of a one-to-one junction (which could represent a bottleneck or increase in channel size), we explain in details the solution, compare the two methods and provide conditions to have solutions in the subcritical regime. Notice that the subcritical condition is necessary to provide a good definition of solution on the network and thus is a basic step to construct numerical methods. We describe solutions for the one-to-two and two-to-one junctions. After revisiting the basics of RKDG on a single channel, we describe the numerical coupling condition at junctions for the two proposed methods. The obtained scheme is then tested in terms of accuracy and on one-to-one junctions. Tests on more complicate junctions (one-to-two and two-to-one) show how the use of high order schemes is necessary to obtain an accurate approximation of the true solution by comparing our methods with firstorder ones. The Saint-Venant equation The Saint-Venant equation (or shallow water) describes the water propagation in a canal with rectangular cross-section and constant slope as follows: t h + x (hv) = 0, (1) t (hv) + x (hv + 1 gh ) = gh(s 0 S f ), with h(x, t) the water height, v(x, t) the water velocity at time t and location x along the canal, g the gravity constant, S 0 the bed slope function and S f the friction slope function. For the purpose of this work, we assume a steady state friction on all canals and we assume horizontal canals with zero slope. We set q = hv (The quantity hv is often called the discharge in shallow water theory, since it measures the flow rate of water past a point) and we reformulate system (1) in a vector form as where u = ( h q t u + x f(u) = 0, () ) (, f(u) = hv hv + 1 gh ). (3) For a smooth solution, system () can equivalently be written in the quasilinear form t u + A(u) x u = 0, (4) 3

4 where the Jacobian matrix A(u) = f (u) is ( 0 1 A(u) = v + gh v The eigenvalues of the matrix A(u) are ). (5) λ 1 (u) = v gh, λ (u) = v + gh, with the corresponding eigenvectors ( ) ( 1 r 1 (u) = v, r gh (u) = 1 v + gh ). (6) Note that in general λ 1 and λ can be of either sign. When the velocity v = q/h of the fluid is smaller than the speed gh of the gravity waves, that is v < gh (7) the fluid is said to be fluvial or subcritical and then one has λ 1 < 0, λ > 0. Hence, under subcritical condition (7), there are two waves propagating in opposite directions. The left and right characteristics are associated to λ 1 and λ respectively. The ratio F r = v / gh is called the Froude number..1 The Riemann problem on a flat bottom Assuming that the river bed is flat, the system () becomes homogeneous. Hence, we obtain a conservative system, which leads to the following Riemann problem: t u + x f(u) = 0, { (8) ul if x < 0, u(x, 0) = u r if x > 0. Here u(x, 0) = (h(x, 0), q(x, 0)) and u l = (h l, q l ) and u r = (h r, q r ). The shallow water equations are genuinely nonlinear and so the Riemann problem always consists of two waves, each of which is a shock or rarefaction. Due to the subcritical flow condition (7), there will be one left (with negative speed) and one right (with positive speed) going wave. In the sequel the left and right going waves are denoted by l-wave and r-wave, respectively. The solution to this Riemann problem consists of the l-wave and the r-wave separated by an intermediate state u = (h, q ). This intermediate state is connected to u l = (h l, q l ) through a physically correct l-waves, and to u r = (h r, q r ) through a physically correct r-wave, see Figure 1. To find the 4

5 t l-wave u l u u r r-wave 0 x Figure 1: The solution to the Riemann problem (8). The intermediate state u is constant in the region delimited by l-wave and r-wave. l- and r-waves are shocks or rarefactions. intermediate state u in general we can define two functions φ l and φ r by { vl ( gh gh l ) if h < h l (rarefaction) φ l (h) = v l (h h l ) if h > h l (shock wave), g h+h l hh l (9) and { vr + ( gh gh r ) if h < h r (rarefaction) φ r (h) = v r + (h h r ) if h > h r (shock wave). g h+hr hh r (10) For a given state h, the function φ l (h) returns the value of v such that (h, hv) can be connected to u l by a physically correct l-wave (or 1-wave), while φ r (h) returns the value of v such that (h, hv) can be connected to u r by a physically correct r-wave (or -wave). We want to determine h so that φ l (h ) = φ r (h ) and such that the subcritical condition is still verified. If there exists a unique solution to the non-linear system φ l (h) φ r (h) = 0, it can be computed by applying a nonlinear root finder to the function φ(h) φ l (h) φ r (h). Example.1 Consider the Riemann problem (8), where ( ) ( ) ( ) ( hl 1 hr 0.5 u l = =, u 0 r = = q r 0 q l ). (11) One has h l > h r and q l = q r = 0, then this Riemann problem models what happens in a dam separating two levels of water breaks at time t = 0. In this case the solution consists of a l-rarefaction and a r-shock. 3 Water flow in canal network A canal network is described by a topological graph, i.e. a couple (I, J ), where I is a collection of intervals representing canals, and J is a collection of vertices representing junctions. We distinguish between multiple nodes, indexed by j J M, at which various canals come together, and the single 5

6 nodes, indexed by j J S, which are ending points of a single canal. On each canal I k = [a k, b k ] system () has to be solved for the couple u k = (h k, h k v k ). To simulate a canal network over a time interval [0, T ], the system () has to be complemented by initial data u 0 k = (h0 k, v0 k )T and appropriate boundary conditions at x = a k or x = b k for k J M or k J S. We restrict the discussion to the case of a single junction and we assume that different canals are connected at x = 0. Typically, at x = 0, algebraic conditions are prescribed, which couple the dynamics on different canals k and l. The conservation of mass, i.e. h k v k = 0, t > 0, (1) is usually coupled with the following: Equal water pressure (equal water heights) Energy continuity (equal energy levels) k 1 gh k = 1 gh l, t > 0. (13) h k + v k g = h l + v l, t > 0. (14) g Note that because of the parametrization, not all canals are pointing away from the junction. Then condition (1) is equivalent to (hv) i = (hv) j, t > 0. (15) i:incoming j:outgoing Not that the set of conditions (1) and (13) or (1) and (14) are imposed a priori. Physical reasons motivate different choices of conditions that are originally derived by engineers. Which conditions are used often depends on if the flow is subcritical or supercritical. See for instance [1, 14] and references therein for a subcritical case. However, to get a well-posed problem we need additional conditions at junctions. 3.1 The Junction Riemann problem For a fixed junction J M, a Riemann Problem is a Cauchy Problem with initial data which are constant on each canal incident at the junction. The evolution on the whole network of the solution to () is determined once one assigns a Riemann Solver at each junction. Considering only subcritical states, given constant initial conditions (u 0 i, u0 j ), where i ranges over incoming canals and j over outgoing ones with respect to the given parametrization, the Riemann solution consists of intermediate states (u i, u j ) satisfying 6

7 t l-wave u 0 i junction r-wave u conditions i u j u 0 j 0 x Figure : The states in the Junction Riemann solution. Left state u 0 i, right state u0 j intermediate states u i and u j. Note that l- and r-waves are shocks or rarefactions. and the some junction conditions (as (1)-(13) or (1)-(14)) with h i > 0 and h j > 0, and such that, for each i and j, the state u i is connected to u 0 i through a physically correct l-wave and the state u j is connected to u0 j through a physically correct r-wave, i.e. v i = φ l (h i ) and v j = φ r (h j), (16) where φ l and φ r are given by (9) and (10) respectively. See Figure. We then look for solutions (h i, h j, q i, q j ) satisfying (1)-(13) (or (1)- (14)) and (16). The global system shows different combinations given by the fact that we have two waves which can be shocks or/and rarefactions. In the next two sections we shall set up the algebraic systems to be solved to compute the water flow in a simple junction and a T-network defined by a 1-to- junction and a -to-1 junction respectively The simple junction As a first example, we assume to have two canals intersecting at one single point. We parametrize the first canal by (, 0) and the second one by (0, + ); they then connect at x = 0. We assume the conservation of mass (1) and equal heights (13) at the junction. We then have to solve the two following equations on each canal respectively, t u 1 + x f(u 1 ) = 0, for x < 0, t u + x f(u ) = 0, for x > 0, (17) where f(u) is defined in (3), coupled with junction conditions (1)-(13), i.e. for all t > 0 q 1 (0, t) = q (0 +, t), (18) h 1 (0, t) = h (0 +, t). The junction Riemann problem at x = 0 is then given by (17)-(18) together with Riemann data u 1 (0, t ) = u 0 1 and u (0 +, t ) = u 0. (19) 7

8 At time t, the solution at the junction consists of two waves separated by intermediate states u 1 and u satisfying the subcritical assumption v k < gh k, k = 1,, and the non-linear system h 1 v 1 = h v, h 1 = h = h, v1 = φ l(h ), v = φ r(h ), (0) with h > 0 and where φ l ( ) and φ r ( ) are defined in (9) and (10) respectively. System (0) extends easily to the case in which the two canals have different cross sections. Let a i be the (constant) width of the i-th canal, the flow rate is then q i = a i v i h i and the first equation in (0) becomes a 1 h 1v 1 = a h v. A system similar to (0) can be set up by assuming the energy continuity at the junction. In this case we should solve h 1 v 1 = h v or (a 1h 1 v 1 = a h v ), gh v 1 = gh + 1 v, v 1 = φ l(h 1 ), v = φ r(h ). (1) Remark 3.1 In general, system (1) does not admit a unique solution. Rewriting it in terms of (h, q), for the first two equalities, we get the following relations q1 = q = q, gh q h 1, = gh + 1 q h, Let us assume a subcritical state u 1 = (h 1, q ) be given. From the energy relation, to derive u such that h > 0 and q < h gh we have to compute the solutions of the following third-order polynomial ( ) P(x) = x 3 h 1 + q gh 1,. x + q g, (x = h ). It is trivial that x = h 1 is a solution, then P(x) = (x h 1, )(x q gh 1, x q gh 1, ) 8

9 and the two other solutions are x ± = q 4gh 1, ± 1 q 4 4g h 4 1, + q gh. 1 It is easy to check that x < 0 and then it is excluded as a possible solution since it is non-positive. Moreover, it is possible to check that the couple (x +, q ) does not satisfy the subcritical condition. Specifically, setting ξ = q, by the fluvial assumption on (h gh 3 1, q ), ξ (0, 1), we then get, for the 1, couple (x +, q ), q gx 3 + = 64ξ (ξ + > 1 ξ (0, 1). ξ + 8ξ) 3 Therefore, the only admissible solution is h = h 1. We can conclude this remark by saying that, for a simple junction in a fluvial regime, assuming energy continuity at the junction is equivalent to assume equal heights. Further, in Section 3.1. we shall give some more comments on the subcritical flow condition. The fluvial regime allows us to select a unique solution when starting from constant initial data on each canal. Assuming that u 1 (x, 0) = u 0 1 x (, 0) and u (x, 0) = u 0 x (0, + ), for u0 1 and u0 constant values, we define the solution on the network as follows: let (u 1, u ) be the intermediate state satisfying the system (0) and verifying the subcritical assumption; fixing x 0 = 0, we get on the first canal: (a) if h 0 1 < h 1 (shock), for s = (q 1 q0 1 )/(h 1 h0 1 ), { u 0 u 1 (x, t) = 1 for x < x 0 + st, for x x 0 + st. u 1 () (b) if h 0 1 > h 1 (rarefaction), for a = (v0 1 gh 0 1 )t + x 0 and b = (v 1 gh 1 )t + x 0, u 0 1 for x a, u 1 (x, t) = u r 1 (x, t) for a < x < b, u 1 for x b, (3) where u r 1 = (hr 1, hr 1 vr 1 ) with h r 1(x, t) = (v 1 + gh 1 (x x 0)/t) 9g and v r 1(x, t) = v 1+( gh 1 gh r 1 ). On the other hand, for the second canal we get: 9

10 if h > h0 (shock) and s = (q q0 )/(h h0 ), { u u (x, t) = for x < x 0 + st, for x x 0 + st. u 0 (4) if h < h0 (rarefaction), for a = (v + gh )t + x 0 and b = (v 0 + gh 0 )t + x 0, u for x a, u (x, t) = u r (x, t) for a < x < b, u 0 for x b, (5) where u r = (hr, hr vr ) with h r (x, t) = ((x x 0)/t (v gh )) 9g and v r (x, t) = v ( gh gh r ). This procedure extends in a straightforward manner to more general networks Comments on the subcritical flow condition As explained in detail in previous sections, the subcritical flow condition is necessary to select a physically reasonable solution at the junction. In general the subcritical condition may be violated by junction solutions. For instance, even if u i (0, t ) and u j (0 +, t ) are subcritical, the resulting new states may be not. Here we focus on this issue and determine, for a simple 1-1 junction, some local constraints on u i (0, t ) and u j (0 +, t ) ensuring that all new states at time t satisfy the subcritical flow condition. Sufficient conditions for subcriticality in case of more complicate junctions or for channels with different cross-sections appear to be challenging. As done before, we focus on a 1-1 junction. Let the values of the solutions u 1 = u 1 (0, t) and u = u (0 +, t) at the junction (x = 0) be given. Assuming that u 1 and u are such that v 1 < gh 1 and v < gh, the junction values are then given by solving the non-linear system (0) (in the case of a simple junction, system (1) is equivalent to (0), see Remark 3.1). We want to find a priori constraints on u 1 and u such that u 1 = (h, h v1 ) and u = (h, h v ) verify the subcritical flow condition, i.e. v1 < gh and v < gh. For a given point (h 0, v 0 ), we set R 1 (h 0, v 0 ; h) = {(h, v) : v = v 0 ( gh gh 0 ), h < h 0 }, S 1 (h 0, v 0 ; h) = {(h, v) : v = v 0 (h h 0 ) g h+h 0 hh 0, h > h 0 }, R (h 0, v 0 ; h) = {(h, v) : v = v 0 ( gh 0 gh), h > h 0 }, S (h 0, v 0 ; h) = {(h, v) : v = v 0 (h 0 h) g h+h 0 hh 0, h < h 0 }. (6) 10

11 With these notations we get that φ l (h) = R 1 (h l, v l ; h) S 1 (h l, v l ; h) and φ r (h) = R 1 (h r, v r ; h) S 1 (h r, v r ; h), where R 1 (h 0, v 0 ; h) = {(h, v) : v = v 0 + ( gh gh 0 ), h < h 0 }, S 1 (h 0, v 0 ; h) = {(h, v) : v = v 0 + (h h 0 ) g h+h 0 hh 0, h > h 0 }. (7) Similarly we set R 1 1 (h 0, v 0 ; h) = {(h, v) : v = v 0 + ( gh 0 gh), h > h 0 }, S1 1 (h 0, v 0 ; h) = {(h, v) : v = v 0 + (h 0 h) g h+h 0 hh 0, h < h 0 }. (8) Following [9], we will consider the 1-critical curve and the -critical curve, C 1 (h) = {(h, v) : v = gh} and C (h) = {(h, v) : v = gh}, (9) respectively. In [9], the authors show that the states (h 1, v 1 ) such that v1 = φ l(h 1 ) and h 1 < ĥ1 are not subcritical, where û 1 = (ĥ1, ˆv 1 ) given by ĥ 1 = 1 9g (v 1 + gh 1 ) and ˆv 1 = gĥ1, (30) is the intersection point between C 1 (h) and R 1 (h 1, v 1 ; h). On the other hand, the states (h, v ) such that v = φ r(h ) and h < ĥ are not subcritical, where û = (ĥ, ˆv ) given by ĥ = 1 9g ( v + gh ) and ˆv = gĥ, (31) is the intersection point between C (h) and R 1 (h, v ; h). We now want to determine constraints on (h 1, v 1, h, v ) such that the solution of system (0) verifies the subcritical flow condition. Proposition 3. Assuming v 1 < gh 1 and v < gh, the solution (h, v ) of system (0), verifies the subcritical flow condition if one of the following situation occurs: (i) h 1 h and v S (h 1, v 1 ; h ), S given in (6). (ii) h 1 h and h > ĥ1, where ĥ1 is given by (30). (iii) h 1 ĥ1 > h and v < S (ĥ1, ˆv 1 ; h ), S given in (6). (iv) h 1 < h and v 1 S 1 1 (h, v ; h 1 ), S 1 1 given in (7). (v) h 1 < h and h 1 ĥ, where ĥ is given by (31). (vi) h 1 < ĥ < h and v 1 > S 1 1 (ĥ, ˆv ; h 1 ), S 1 1 given in (7). 11

12 R 1 C 1 u u 1 ū 1 S S 1 u S 1 R 1 u C 1 û 1 u S 1 u u R 1 u 1 S 1 v 1 = S (h 1, v 1 ; h ) C C (i) (ii) R 1 ū 1 û 1 u S u u 1 S 1 S1 C1 ū 1 = S (ĥ1, ˆv 1 ; h ) C (iii) Figure 3: Admissible subcritical states for system (0). Proof: At the junction we have that admissible states for the incoming canal lie on the curves of the first family R 1 (h 1, v 1 ; h) and S 1 (h 1, v 1 ; h), while admissible states for the outgoing canal lie on the curves of the second family R 1 (h, v ; h) and S 1 (h, v ; h). Let study the following different cases: Case 1. For h 1 h we may have the following combinations S 1 S 1, R 1 R 1 or R 1 S 1. (i) If h < h 1 and v S (h 1, v 1 ; h ), the case S 1 S 1 occurs (see Figure 3-(i)) and the solution (h, v ) is such that h > h 1 > h and v < v < v 1. In this case the subcritical flow condition v < gh is trivially achieved: gh < gh < v < v < v 1 < gh 1 < gh. (ii) Let us now assume v > S (h 1, v 1 ; h ). In this case it may occur the combination R 1 R 1 or R 1 S 1. In both cases the point (h, v ) lies on R 1 (h 1, v 1 ; h) and we have that h < h 1 and v > v 1 ; the v value may then exceed the subcritical limit. So, as previously observed, computing the intersection point between R 1 (h 1, v 1 ; h) and the 1-critical curve C 1 (h), given by (30), we get that for all ĥ1 h h 1 the corresponding v = R 1 (h 1, v 1 ; h ) verifies the subcritical flow condition. Then, for all (h, v ) such that ĥ1 h h 1 the solution is fluvial (see Figure 3-(ii)). (iii) On the other hand, if h < ĥ1 we may have a supercritical solution. This does not occur if v < S (ĥ1, ˆv 1 ; h ), see Figure 3-(iii). Case. For h 1 < h we may have the following combinations S 1 S 1, S 1R 1 or R 1 R 1. 1

13 (iv) If S 1 S 1 occurs, (h, v ) is such that h > h > h 1 and v < v < v 1 and the subcritical flow condition is trivially achieved. This situation arises for all (h 1, v 1 ) such that h 1 < h and v 1 S 1 1 (h, v ; h 1 ). (v) Let us now assume v 1 < S1 1 (h, v ; h 1 ). In this case it may occur only S 1 R 1 or R 1 R 1. In both cases, it is only the rarefaction R 1 (h, v ; h) which can intersect with the -critical curve C. That intersection denoted by (ĥ, ˆv ) is given by (31). Hence, if ĥ h 1 < h the solution is fluvial for all v 1 < gh 1 and v < gh. (vi) On the other hand, if h 1 < ĥ < h, we need v 1 > S 1 1 (ĥ, ˆv ; h 1 ). Example 3.3 For instance, Proposition 3. can be useful when one wants to model what happens in a dam separating two levels of water breaks. In this particular case v 1 = v = 0 and h 1 > h > 0 and one can a priori define a constrain on the ratio h /h 1 to get a fluvial (subcritical) or torrential (supercritical) behaviour. In fact, by applying Proposition 3. one falls in the case (ii) or (iii). For v 1 = v = 0 we then get ĥ1 = 4 9 h 1 and h > 4 9 h 1 or, for h < 4 9 h h 1 + (h 4 9 h 9h + 4h 1 1) > 0 h > c, c h 1 h h 1 Therefore, we know a priori that if h /h 1 > the water behaviour after the dam break will be fluvial. In the sequel, we shall always assume to stay in the fluvial regime and all numerical tests will be selected to verify this assumption One incoming canal and two outgoing canals, 1-to- junction In this section, we shall consider the case of a network composed of three canals. We parametrize the first canal by (, 0) and the other two canals by (0, + ) assuming that the three canals are connected at x = 0. Then, the Saint-Venant equations are t u 1 + x f(u 1 ) = 0, for x < 0, t u + x f(u ) = 0, for x > 0, t u 3 + x f(u 3 ) = 0, for x > 0, (3) where f(u) is defined in (3). 13

14 Assuming the conservation of mass (1) and equal heights (13) at the junction, conditions (1)-(13) are reduced to the following: for all t > 0 q 1 (0, t) = q (0 +, t) + q 3 (0 +, t), h 1 (0, t) = h (0 +, t) = h 3 (0 +, t). (33) The junction Riemann problem at x = 0 is then given by (3)-(33) together with Riemann data u 1 (0, t ) = u 0 1, u (0 +, t ) = u 0, u 3(0 +, t ) = u 0 3. (34) The Riemann solution at the junction consists of three waves separated by intermediate states u 1, u and u 3 satisfying the junction conditions (33) and relations (16), i.e. the non-linear system h 1 = h = h 3 = h, v 1 = v + v 3, v 1 = φ l(h ), v = φ r(h ), v 3 = φ r(h ), (35) with h > 0 and the subcritical assumption vk < gh, k = 1,, 3 and φ l ( ) and φ r ( ) are defined in (9) and (10) respectively. To get the solution (h, v1, v, v 3 ) at each time, system (35) shows different possible combinations given by the fact that we have three waves which can be shocks or/and rarefactions. In the numerical tests presented in this work, we will solve system (35) by a numerical root-finding algorithm, see Section 5 for more details. Notice that system (35) admits a unique solution (see for instance [16]), but the solution not always verifies the subcritical condition. Then, suitable initial data have to be given to ensure the fluvial regime to the problem. If we assume at the junction the conservation of mass coupled with energy continuity (14), the junction conditions reduce to q 1 (0, t) = q (0 +, t) + q 3 (0 +, t), gh 1 (0, t) + 1 q1 (0, t) h 1 (0, t) = gh (0 +, t) + 1 q (0+, t) h (0+, t) = gh 3(0 +, t) + 1 q3 (0+, t) h 3 (0+, t) (36) and the analogous of system (35) is now gh v 1 = gh + 1 v = gh v 3, h 1 v 1 = h v + h 3 v 3, v 1 = φ l(h 1 ), v = φ r(h ), v 3 = φ r(h 3 ), 14 (37)

15 with h k > 0, k = 1,, 3 and the subcritical assumption v k < gh, k = 1,, 3. In this case the existence of the solution (h 1, h, h 3, v 1, v, v 3 ) is not always assured. In [9] the authors prove the existence of the solution in the particular case of two identical outgoing canals and for an initial data as a perturbation of a steady state. A more general result may be found in [5] Two incoming canals and one outgoing canal, -to-1 junction As in the previous section, here we would like to set up the algebraic system at a -to-1 junction. The system of equations becomes now t u 1 + x f(u 1 ) = 0, for x < 0, t u + x f(u ) = 0, for x < 0, t u 3 + x f(u 3 ) = 0, for x > 0. (38) Assuming first the conservation of mass (1) and equal heights (13) at the junction, we get that the Riemann solution at the junction consists of three waves separated by intermediate states u 1, u and u 3 satisfying the nonlinear system h 1 = h = h 3 = h, v1 + v = v 3, v1 = φ l(h ), (39) v = φ l(h ), v 3 = φ r(h ), with h > 0 and the subcritical assumption v k < gh, k = 1,, 3 and where φ l ( ) and φ r ( ) are again defined in (9) and (10) respectively. If, however, we assume the conservation of mass coupled with energy continuity (14) at the junction, we get gh v 1 = gh + 1 v = gh v 3, h 1 v 1 + h v = h 3 v 3, v 1 = φ l(h 1 ), v = φ l(h ), v 3 = φ r(h 3 ). (40) 4 High Order Schemes for the Shallow Water Equations High order schemes have been developed over years for shallow water equations. For example, there are the finite difference, finite volume WENO 15

16 methods and recently developed Runge-Kutta (RK) discontinuous Galerkin (DG) methods [19]-[7]. In our simulations, we use the RK DG approach, taking advantage of its compactness, especially for setting the numerical boundary condition at junctions. 4.1 RKDG for Shallow Water Equations In our description below, the 1D computational domain is discretized into cells C m = [x m 1, x m+ 1 ] with x m = (x m 1 + x m+ 1 )/, m = 1,..., M, where M is the total number of computational cells. We let the size of the m-th cell be x m and the maximum mesh size x = max m x m. Given a non-negative integer k, we define a finite dimensional discrete space, W x = W k x = {w : w Cm P k (C m ), m = 1,..., M}, (41) where P k (C m ) denotes the space of polynomials of degree at most k on cell C m. We let W x to define its vector version. Note that functions in W x is piecewise defined and is discontinuous at the cell boundaries x m± 1. For a function w W x, we let w ± to denote left and right limits of the m+ 1 function values. The DG method for () is formulated by multiplying the equation system by some test functions w, integrating over each computational cell, and performing integration by parts. Specifically, we seek U = (u 1, u ) W x such that, w(x) t Udx = f(u) x w(x)dx C m C m where the numerical fluxes ˆf m± 1 ( ) ˆf m+ 1 w(x ) ˆf m+ 1 m 1 w(x + ), m 1 w W x (4) = f(u, U + ) are approximate Rie- m± 1 m± 1 mann solvers. For example, in our simulations, we use the Lax-Friedrich flux with f(u, U + ) = 1 (f(u ) + f(u + )) + α (U U + ) where α being the largest eigenvalue of the Jacobian matrix (5) over the entire domain. For implementation, each component of the approximate solution U, e.g. u 1, on mesh C m can be expressed as u 1 (x, t) = k l=0 û 1,l m (t)ψ l m(x), (43) where {ψm(x)} l k l=0 is the set of basis functions of P k (C m ). In our simulations, we choose the Legendre polynomials as local orthogonal basis of P k (C m ) with ψ 0 m = 1, ψ 1 m = ( ) x xm, ψm = x m / 16 ( ) x xm 1 x m / 3,...

17 The test function w(x) in eq. (4) can be taken as the set of basis functions ψ l m(x), l = 0 k. More implementation details can be found in the original paper [6] and the review article [7]. The equation system (4) can be evolved in time via the method of lines approach by a TVD RK method [6] in the following form, U (1) = U n + t n L(U n ), U () = 3 4 U n U (1) tn L(U (1) ), (44) U n+1 = 1 3 U n + 3 U () + 3 tn L(U () ), where L is the spatial operator, which denotes the R.H.S. of eq. (4), and t n is the numerical time step. Another important ingredient for the DG methods is that a slope limiter procedure might be needed after each inner stage in the RK time stepping, when the solution contains discontinuities. We use the characteristic-wise total variation bounded (TVB) limiter in [6]. 4. Numerical coupling conditions at junctions In this subsection, we briefly discuss how to set up numerical boundary conditions at junctions in a network of open canals. The purpose is indeed to use the RKDG method as a general approach for simulating hyperbolic network problems. The RKDG method is known to be very compact in the sense that the solution on one computational cell depends only on direct neighboring cells via numerical fluxes, thus allowing for easy handling of various boundary conditions. Below, we take a 1-to- junction as an example to illustrate our proposed procedure in setting up numerical boundary conditions at junctions. The approach can be directly generalized to other cases. We consider the Riemann problem at the junction from one incoming canal (denoted by a) to two outgoing canals (denoted by b and c). We assume DG solutions from canal a at the left limit of the junction being U a = (h a, q a ) = lim x 0 U a(x) with 0 being the right-most point of the incoming canal a, and assume DG solutions from canal b at the right limits of the junction being U b = (h b, q b ) = lim x 0 + U b(x), with 0 being the left-most point of the outgoing canal b. Similar notations are introduced for the canal c. From the discussions in Section 3.1, one can find intermedia states U a = (h a, h av a), U b = (h b, h b v b ), U c = (h c, h cv c ) by 17

18 a root-finding algorithm from a nonlinear system, which is formulated such that Ua (or Ub, U c ) can be connected to U a (or U b, U c ) via a physically correct wave (shock or rarefactions) and junction conditions are satisfied. For example, the equal heights assumption gives h a = h b = h c, h av a = h b v b + h cv c, and the energy continuity assumption gives gh a + 1 v a = gh b + 1 v b = gh c + 1 v c, h av a = h b v b + h cv c. With these numerically computed intermedia states, one can directly assign the numerical fluxes at the junction as ˆf Ma+ 1. = f(u a ) for the canal a, ˆf 1. = f(ub ) for the canal b, The flux at the junction for the canal c is set similar to that for the canal b. Such boundary condition is imposed at each RK stage via the method-ofline approach. Finally, we emphasize that the proposed boundary treatment, via evaluating fluxes at intermediate states, exactly preserves the mass in a hyperbolic network as h av a = h b v b + h cv c. 5 Numerical tests In this Section, we examine the accuracy of the proposed method for several different situations. We shall refer to the high order scheme (4) with polynomial degree p = and RK scheme (44) in time, as RKDG and to the first order approximation, defined by an Euler in time approximation with polynomial degree p = 0, as EDG0. In both approaches, the numerical flux in (4) is the Lax Friedrich flux, unless otherwise specified. In all tests, we use x = 0.05 in simulations for all figures in this section. we set the calculation times so as to avoid interaction with the boundaries of channels. Moreover, all tests are set up to verify the subcritical condition. 5.1 Testing the accuracy In this example we test the third order accuracy of the proposed scheme for a smooth solution, when applied to a simple one-to-one junction. Starting from a single canal problem with the following initial conditions: h(x, 0) = 10 + e cos(πx), (hu)(x, 0) = cos(sin(πx)), x (0, 1), (45) we split in half the canal adding a virtual junction at x = 1. At each time step, to compute the two junction values (u 1, u ), we solve numerically 18

19 system (0) with the Riemann data given by the left and right limits U J+ 1 and U + obtained by the DG polynomials in (43), where J denotes the J+ 1 index such that J x = 0.5. The numerical fluxes ˆf + and ˆf at the J+ 1 J+ 1 junction are given by ˆf J+ 1 = f(u, u + J+ 1) and ˆf 1 J+ 1 = f(u, U + J+ 1 ). Since the exact solution is not known explicitly for this case, we use the same scheme with M = 0, 480 points to compute a reference solution, and treat this reference solution as the exact solution in computing the numerical errors. We compute up to t = 0.1 when the solution is still smooth. Table 1 contains the l 1 errors and numerical orders of accuracy. We can clearly see that third order accuracy is achieved. x RKDG error on (h, q) order on (h, q) 0.05 (0.9E-0, 0.3E-01) (3.16, 3.17) 0.05 (0.4E-03, 0.48E-0) (.4,.7) (0.51E-04, 0.58E-03) (3.04, 3.04) (0.66E-05, 0.73E-04) (.96,.98) (0.84E-06, 0.9E-05) (.97,.99) (0.10E-06, 0.11E-05) (.99, 3.00) Table 1: l 1 -errors and order accuracy for a smooth solution on a simple junction. 5. The simple junction In this section we shall assume to have two canals of different cross sections connected at one point, x = 0. We then solve numerically system (0) at the junction and we compare the numerical solution obtained on each canal by applying RKDG and EDG0 methods. In all tests we set a 1 = 1 and a = 0.5, where a 1 and a are the constant widths of the two canals respectively. In Figures 4 we assume a constant initial data on both canals and we compare the numerical solution with the exact one (see Section 3.1.1). Specifically, at t = 0, we assume water on canal at rest and all along canal 1 water that moves with positive speed towards canal, i.e. x canal 1 : h 1 (x, 0) = 1, q 1 (x, 0) = 0.1; canal : h (x, 0) = 1, q (x, 0) = 0. (46) In Figure 4 we assume equal heights at x = 0. It is clear as RKDG captures better and well the slope of the solution. 19

20 In Figure 5 we assume on canal 1 a non-constant initial state and we compare the numerical solution with a reference solution obtained by applying RKDG scheme with a fine mesh ( x = 10 4 ). The initial condition on the first canal is now given by canal 1 : h 1 (x, 0) = 1 + e 0 x, q 1 (x, 0) = 0. h 1 (x, 0). (47) It appears that RKDG captures well the oscillations of the solution and it is very accurate around the junction, see Figure 5. Figure 4: A 1-to-1 junction with constant cross-sections a 1 = 1 and a = 0.5; RKDG in red line (- -), EDG0 in blue line (- ) and the exact solution in black ( ). Initial data (46) and final time T = The T-network In this Section we propose some numerical tests for two types of T-network which are defined by a 1-to- junction and by a -to-1 junction respectively. In all tests, we assume that each canal has equal length and we compare the solutions obtained by applying the RKDG procedure with the solution obtained by the first-order approximation EDG0. Since the exact solution is not known explicitly for almost all cases under consideration, we use RKDG with an extra fine grid ( x = 10 4 ) to compute a reference solution. 0

21 Figure 5: A 1-to-1 junction with constant cross-sections a 1 = 1 and a = 0.5; RKDG in red line (- -), EDG0 in blue line (- ) and the exact solution in black ( ). Initial data (47) and final time T = We test first the schemes with the following constant initial data: x [0, 1], canal 1 : h 1 (x, 0) = 0.5, q 1 (x, 0) = 0.1; canal : h (x, 0) = 0.5, q (x, 0) = 0; (48) canal 3 : h 3 (x, 0) = 1, q 3 (x, 0) = 0. As expected, we observe in Figure 6 that RKDG has better performance and less numerical diffusion than EDG0. More interesting is to evaluate how the numerical solution within the canals affects the value at the junction. In this particular test case, for each canal k, k = 1,, 3, the correct value u k at the junction is given by solving only once system (1) with Riemann data (48) respectively. The numerical procedure instead solves system (1) at each time step, by using the numerical values computed on each canal around the junction. This affects the values at the junction and therefore the solution within canals. In Table we provide the l 1 error at the junction as a function of time. Specifically, we compute for each canal the value e k = t N u k U k (t n), k = 1,, 3, (49) n=0 1

22 where u k is the solution of (1) computed only once at first and U k (t n) are the solutions computed at each time step which are influenced by the numerical procedure applied within canals. Table shows in fact that the numerical solution within canals affects the values at the junction, but by applying RKDG as solver inside canals, the error (49) at the junction is of one order smaller. final time T = 0.1 Canal 1 Canal Canal 3 data (h, q) (0.5, 0.1) (0.5, 0) (1, 0) RKDG (0.13E-03, 0.7E-03) (0.15E-03, 0.7E-03) (0.19E-03, 0.43E-03) EDG0 (0.94E-03, 0.1E-0) (0.78E-03, 0.6E-0) (0.1E-03, 0.37E-0) data (h, q) (1.5, 0.5) (0., 0) (0.4, 0) RKDG (0.45E-03, 0.9E-03) (0.45E-03, 0.91E-03) (0.45E-03, 0.99E-03) EDG0 (0.17E-0, 0.15E-0) (0.17E-0, 0.39E-0) (0.17E-0, 0.54E-0) Table : l 1 -errors at the junction as defined in (49). We consider now the following initial data: x [, 1], canal 1 : h 1 (x, 0) = 1 + e 0 x, q 1 (x, 0) = 1 h 1(x, 0); canal : h (x, 0) = 1, q (x, 0) = 0; canal 3 : h 3 (x, 0) = 0.5, q 3 (x, 0) = 0. (50) The numerical results are presented in Figure 7. It is clear as the higher order RKDG scheme has better performance in resolving solution structures than the lower order scheme. Similar results are given by considering the following initial data on the first canal, see Figure 8: { 1.5 if x [0, 0.] [0.4, 0.6] [0.8, 1], canal 1 : h 1 (x, 0) = 1 otherwise q 1 (x, 0) = 0. h 1 (x, 0). (51) Again, assuming a T-network defined by a -to-1 junction and data (51) on both two incoming canals, we get that RKDG captures better all the structures produced by the solution, see Figure 9.

23 Figure 6: Equal energy at the 1-to- junction. RKDG in red line, EDG0 in blue line and the exact solution in black. References [1] Bastin, G.; Bayen, A. M.; D Apice, C.; Litrico, X.; Piccoli, B.. Open problems and research perspectives for irrigation channels. Netw. Heterog. Media 4, 009, Vol., pp.i v. [] Bressan, A.; Canic, S.; Garavello, M.; Herty, M.; Piccoli, B. Flows on networks: recent results and perspectives. EMS Surv. Math. Sci., 014, Vol.1, No.. 1: pp [3] Bretti, G.; Natalini, R.; Piccoli, B. Numerical approximations of a traffic flow model on networks. Netw. Heterog. Media, 006, Vol.1 no. 1: pp [4] Canic, S.; Piccoli, B.; Qiu, J.-M.; Ren, T. Runge-Kutta discontinuous Galerkin method for traffic flow model on networks. J. Sci. Comput., 015, Vol. 63 no. 1: pp [5] Colombo, R.M.; Herty, M.;Sachers, V. On Conservation Laws at a Junction. SIAM Journal on Mathematical Analysis, 008, Vol. 40, No. : pp

24 (a) Final time T = 0.1 (b) Final time T = 0.4 Figure 7: Equal heigths at the junction. A 1-to- junction with initial data (50). RKDG in red line (- -), EDG0 in blue line (- ) and the exact solution in black ( ). In Figure (a) final time T = 0.1 and in (b) final time T =

25 (a) Final time T = 0.1 (b) Final time T = 0.4 Figure 8: Equal heigths at the junction. A 1-to- junction with initial data (51). RKDG in red line (- -), EDG0 in blue line (- ) and the exact solution in black ( ). In Figure (a) final time T = 0.1 and in (b) final time T =

26 (a) Final time T = 0.1 (b) Final time T = 0.4 Figure 9: Equal heigths at the junction. A -to-1 junction with initial data (51). RKDG in red line (- -), EDG0 in blue line (- ) and the exact solution in black ( ). In Figure (a) final time T = 0.1 and in (b) final time T =

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