Admissibility conditions for Riemann data in shallow water theory

1 Introduction

Many physical systems can be described by systems of conservation laws. Such systems are first-order quasilinear partial differential equations, and for one-dimensional problems they can be written in the general form

where u is a vector of unknowns, x is the one-dimensional spatial coordinate, and t is the time. The flux function f is a nonlinear vector function often satisfying certain mild assumptions, such as that the function be twice continuously differentiable, the flux Jacobian ∇ f have a full set of distinct eigenvalues and the wave families be either genuinely nonlinear or linearly degenerate [12], [17], [24].

Due to the special nonlinear structure of such systems, solutions naturally develop discontinuities in time, even if the original state of the system is given by a smooth function of x. Once a discontinuity has developed, solutions of the system need to be interpreted in a weak sense. If the solution is to include a jump between a left state u L and a right state u R , the weak formulation leads to the well-known Rankine-Hugoniot condition

The Riemann problem distills the essence of the problem of singularity formation into a simple initial-value problem where the initial data have a prescribed discontinuity. For the system above, the Riemann problem would prescribe initial data of the form

where u L and u R are given left and right states. Under the conditions on f mentioned above, the Riemann problem can always be solved as long as the left and right states are close enough (see [12], [17], [19], [20]). However, if the left and right state are not close, then there is no general theory guaranteeing the existence of a solution to the Riemann problem (see [8]). Indeed, it can be shown explicitly, that there is no solution using the standard theory in some cases because the solution becomes unbounded [27]. On the other hand, for a large number of systems, solutions of the Riemann problem can be shown to exist by elementary methods.

In the present work, our focus is on the well-known shallow-water system which appears in the general form (1) when defining the principal unknown vector u and the flux function f , respectively by

In physical terms, the unknown h ( x ,   t ) represents the local flow depth at a point x in space and at a time t. The unknown u ( x ,   t ) represents the horizontal fluid velocity at x and t, averaged over the fluid column.

The solution of the Riemann problem for the shallow-water equations is well known, and can be found in many texts on conservation laws (cf. [2], [12]). One way to normalize the problem is to consider the left state u L given, and look at all possible right states. Since h represents the total flow depth of the fluid, an additional admissibility condition is usually imposed, requiring both u L and u R to feature non-negative flow depth. Indeed, if imposing the requirement that h L ≥ 0 and h R ≥ 0 , then it can be shown that the Riemann problem can be solved for all right states u R satisfying this admissibility condition.

As can be gleaned from Figure 1, the condition that h be non-negative restricts the analysis to the right half-plane in the ( h ,   u ) phase space. However, if one looks closely at the solution of the Riemann problem, it appears that the solution features dry states for many possible right states (see Figure 1). In particular, in order to resolve the Riemann problem with a right state in the shaded region in Figure 1, one needs to incorporate a dry region ( h = 0 ) into the solution. Even though the solution is well defined mathematically, from a physical point of view, it does not seem reasonable for a dry region to develop from initial conditions which otherwise seem perfectly normal (just as it does not seem reasonable to include states with h < 0 ). Note that one obvious admissibility condition would be to simply specify that any right state in the shaded region of Figure 1 should be disregarded. To set such a criterion we would set the threshold by u R ≤ u * + 2 g h , representing the boundary of the shaded region with u * given by the R 1 curve evaluated at h = 0 (see general solution in Section 4 for context). However, there does not appear to exist any physical motivation for this condition. Therefore, in the present work, we aim to identify conditions on the right state u R which are rooted in the physical interpretation of the equations, and also guarantee that the solution of the Riemann problem does not include a dry state.

Figure 1:

Phase space for a particular left state ( h L ,   u L ) . The red curves denoted by S 1 and S 2 indicate possible right states which can be reached through a single discontinuity. The blue curves denoted by R 1 and R 2 show right states which can be reached through a continuous solution.

Recall that the shallow-water assumption applies to surface waves that are slowly varying, and a shock represents a bore (i. e., a traveling hydraulic jump) where the shock structure which may feature oscillations or turbulent structures) has been neglected [7], [22], [28]. If this physical interpretation is taken as a point of departure, then it appears that a Riemann problem may develop from the collision of two bores, and a natural admissibility condition would be to consider only such Riemann problems. Thus we will study the history of the Riemann problem, or more succinctly the backwards problem for t < 0 . Examining possible solutions of the backwards problem will lead to clear conditions on whether or not a Riemann problem can develop. As it turns out, these conditions will exclude Riemann problems whose solution involves a dry state.

Note that a Riemann problem could also develop from certain initial data which are arranged in such a way that the solution lines up at some time so as to give a perfect Riemann problem. While this is possible, it would clearly be unstable to even the smallest perturbations. On the other hand, one might also consider the collision of three or more traveling hydraulic jumps, or the collision between rarefaction waves and shocks, but these situations are so unlikely to happen that they would constitute a set of measure zero in the configuration space. In the current work, we focus on the origin of the Riemann problem which can be represented by a set of non-zero measure in the configuration space given by the phase plane.

The outline of this paper is as follows. In sections 2 and 3, a short discussion of the properties of basic admissible waves for the shallow-water system is given, and the standard solution of the Riemann problem is explained in section 4. This is standard fare, but we need various formulas in order to set up the problem to be attacked. In sections 5, 6, and 7, Riemann problems originating from various configurations are investigated. Some ramifications of our results are discussed in the Conclusion.

2 Shock waves and bore properties

As already mentioned above, the shallow-water system can be written in terms of mass and momentum conservation in the form

A derivation of this system from first principles can be found in [25], where it is also shown that the conservation of energy is formulated as

Discontinuous solutions develop naturally in this system even in the case of flat bathymetry which is under study here. In the case when the solution features jumps, the imposition of mass and momentum conservation leads to an energy loss (see [23]) which has been the subject of a number of studies [3], [4], [5], [6], [15], [26]. In the context of the conservation laws, the energy loss means that (6) becomes an inequality, which is then taken as the mathematical entropy in order to pick out physically reasonable discontinuous solutions.

In the context of the shallow-water equations (4) and (5) the Rankine-Hugoniot condition (2) yields the following relations:

Combining these two equations enables us to find an expression for u R in terms of h ,   h L and u L as shown in [2], [12]. Indeed, one may define the Hugoniot locus of all possible right states ( h ,   u ) for a given left state ( h L ,   u L ) in terms of the shock curves S 1 and S 2 as follows.

A useful observation to be used later is that the fluid velocity of u on S 1 is strictly decreasing, while the velocity on S 2 is strictly decreasing. In fact taking the derivative yields the expression

where the minus sign refers to the S 1 curve and the plus sign to the S 2 curve. Inspecting the term on the right in the above relation confirms that the sign of the derivative u ′ ( h ) depends only on whether the derivative is taken on S 1 or on S 2 .

The Hugoniot loci may also be described in terms of the momentum q = h u . Indeed, for a given left ( h L ,   q L ) , the possible right states must satisfy one of the following relations

Taking the second derivative of these expressions shows that these curves are strictly concave and convex, respectively:

Finally, the speed of the discontinuity may be found from the Rankine-Hugoniot condition as

Next, let us discuss the entropy condition for shock waves. It is well known [12], [24] that it is necessary to impose both the Rankine-Hugoniot and the entropy condition to ensure uniqueness of a solution. In the context of the shallow-water theory, the mechanical energy serves as a mathematical entropy. In fact, it is well known that energy is lost in a shock either due to turbulence or the continuous creation of surface oscillations [3], [6], [11], [13], [14], [25]. Similar considerations can be used in various other applications, such as for example in the context of porous media [1].

In the present case, the expected loss of mechanical energy is enforced by imposing the inequality

for discontinuous solutions. It is also convenient to introduce the relative mass flux m by

Using m, we can express the rate at which energy is lost at the shock by

Note that since we always require Δ E < 0 for discontinuous solutions, if h L < h R , then we must have m > 0 from the previous relation. Invoking (13) then shows that u R > σ and u L > σ . On the other hand, similar considerations show that if h L > h R , then (12) requires that u R < σ and u L < σ . These relations show that fluid particles always move across the shock from the region of lower depth to the region of higher depth, a fact already noted in [25]. Moreover, combining equations (7), (8) and (13) and using the condition (12) shows that we must have u R < u L for all discontinuous solutions. The most important properties of the shock curves are summarized in Tables 1 and 2 and Figure 2.

Figure 2:

Left panel: bore with flow depth h L < h R which corresponds to the right state being on the shock curve S 1 . Right panel: bore with flow depth h R < h L corresponding to the right state being on the shock curve S 2 . In both cases the bore front may feature positive (right-moving), zero, or negative (left-moving) propagation velocity.

One should also remark that both S 1 and S 2 shocks satisfy the Lax entropy condition (cf. [12]). This condition states that the speed σ i of an S i shock must satisfy the relation

where λ i are the eigenvalues of the flux Jacobian matrix ∇ f . For the shallow water equations this eigenvalues are given by

A geometrical representation of the Lax entropy condition in the ( x ,   t ) − plane is shown in Figures 3 and 4.

Figure 3:

Left moving bore with speed σ 1 , λ i ( L ) and λ i ( R ) for a S 1 shock.

Figure 4:

Right moving bore with speed σ 2 , λ i ( L ) and λ i ( R ) for a S 2 shock.

3 Rarefaction waves

Following the classical theory (presented for example in [10], [28], [12]) we seek traveling wave solutions of the form u ( x ,   t ) = v ( ξ ) with ξ = x t . Substituting this term the into the conservation law one can easily verify that the system reduces to a system of ODEs of the form

The solution is then given by the integral of (16). We may now exploit this insight using the theory of Riemann invariants w : R 2 → R , which is a smooth function that is constant along the integral curves [10]. For the shallow-water equations the Riemann invariants are given by

Along a Riemann invariant, the solution must therefore satisfy

Hence, for a given left state we may write the rarefaction wave solution as follows

By comparison, one can also show that v ˙ is the right eigenvector r ( v ) , and ξ is the corresponding eigenvalue λ ( v ) belonging to the Jacobi matrix of the flux function. Having λ = ξ would mean that the eigenvalues must be increasing from left to right. This implies λ i ( L ) < λ i ( R ) and by equation (15) that u L < u R whenever there is a rarefaction wave. Figures 5 and 6 depicts two rarefaction waves propagating left and right in the ( x ,   t ) − plane. Following the characteristics one can see how the waves moves forward in time.

Figure 5:

Left moving rarefaction wave smoothly varying between λ 1 ( L ) and λ 1 ( R ) .

Figure 6:

Right moving rarefaction wave providing a smooth transition between λ 2 ( L ) and λ 2 ( R ) .

In fluid mechanics, some authors refer to these waves as negative surges resulting from a decrease in flow depth [9]. Interestingly, Peregrine was able to show that a negative surge together with a bore advancing in positive direction originates from the collision of two fast shocks [21]. Therefore, we will discuss the development of the Riemann problem from a collision of two S 2 shocks in Section 6.

4 General solution of the Riemann problem

Using the results from sections 3 and 4, the general solution of the Riemann problem can be fond using the rarefaction curves defined by (19) and (20)

and the shock curves (7) and (8)

In the following, it will be convenient to plot the integral curves and shock curves for a particular left state ( h L ,   u L ) plotted in two different coordinate systems. Figure 7 shows the integral curves in ( h ,   q ) − coordinates, where q = h u is the momentum, while Figure 8 shows the integral curves in ( h ,   u ) − coordinates. The benefit of the former representation lies in the fact that the shock speed between q 1 and q 2 is given by

which is simply the secant line joining the two states. For example, any right state given on the S 2 ( L ) curve in Figure 8 would give rise to a right- moving bore since the slope of the secant line joining left and right states would be positive, i. e., σ R > 0 . Of course, if the right state is not given on any of the integral or shock curves, S 1 ,   S 2 ,   R 1 and R 2 need to be combined to give a solution of the Riemann problem. Indeed as explained in [12], [18], given a left state we may consider all possible right states and then find the solution depending on whether the right state is in region I ,   I I ,   I I I or I V . For instance, let us say we have a right state somewhere in region four. Then in order to find an entropy solution we must define a middle state at some point ( h M ,   u M ) on the shock curve S 1 ( L ) such that it is connected to a rarefaction curve R 2 ( M ) . Similarly, the entropy solution for each region is found by two elementary waves going through some middle state. For region I, we follow R 1 ( L ) connecting the right state with R 2 ( M ) for a middle state. In region I I , we first go along R 1 ( L ) then S 2 ( M ) . Finally, in region I I I we connect S 1 ( L ) with S 2 ( M ) for some middle state (see example in Figures 9 and 10). Concluding this section we remark that the solution is in fact unique since all of these solutions satisfy the admissibility conditions, and it has been shown that there is only one middle state M for each region.

Figure 7:

Phase-space in ( h ,   q ) − coordinates.

Figure 8:

Phase-space in ( h ,   u ) − coordinates.

Figure 9:

Solution in region I I I using ( h ,   q ) − coordinates.

Figure 10:

Solution in region I I I using ( h ,   u ) − coordinates.

5 Development of the Riemann problem from a collision of S 2 and S 1 shocks

In this section, we will discuss the origin of the Riemann problem from a collision of two bores. It is most convenient to focus the discussion by assuming that a left state is given. With this proviso, we will prove that the Riemann problem associated to certain right states in region I I I arises from the head-on collision of two counter-propagating bores, while other right states are connected to an overtaking collision of co-propagating bores (see Figure 11). Indeed, we will show that these two scenarios cover all possible right states in region I I I . Finally, we show that Riemann problems with right states in regions I ,   I I and I V cannot develop from either head-on or overtaking collisions of an S 1 and an S 2 shock.

Figure 11:

Backwards problem in ( h ,   u ) − coordinates.

In order to understand how a given Riemann problem develops, we consider the backwards problem for t < 0 . In order to solve the backwards problem, the usual disposition of a slow shock on the left left and a fast shock on the right has to be reversed. Indeed, to solve the backwards problem, the left state is connected to a center state lying on the locus S 2 ( L ) . The center state is then connected to the right state by ensuring that the right state lies on the locus S 1 ( C ) . This configuration then leads to the collision of the two shocks at time t = 0 . Note that we have chosen to use the term center state for the backwards problem versus middle state for the forward problem.

As indicated in Figure 12, the solution of the Riemann problem for a right state in region I I I consists of a 1 − shock and a 2 − shock connected by a middle state on S 1 ( L ) . Note that the flow depth of the middle state will always be higher than for both the left and the right state. Specifically we always have h M > h L and h M > h L in region III. In fact, it can be observed that fluid particles from both sides will move back towards the middle, thus contributing to the raised flow depth of the middle state. In that respect, it seems natural that the Riemann problem should result from two colliding bores. Figure 13 depicts the case of a head-on collision of a left-moving shock and a right-moving shock. Note that the backwards solution has the two shocks connected by a center state ( h C ,   u C ) , then moving towards each other resulting in a Riemann problem at time t = 0 .

Figure 12:

Forward problem in ( h ,   u ) − coordinates.

Figure 13:

The Riemann problem at t = 0 originates from two counter-propagating shocks ( t < 0 ). The solution is given for t > 0 .

Figure 14:

Development of the Riemann problem in phase space.

Figure 15:

Development of the Riemann problem in ( x ,   t ) − coordinates.

6 Development of the Riemann problem from a collision of two S 2 shocks

Consideration will now be given to the Riemann problem arising from a S 2 − S 2 connection. As it will turn out, the resulting Riemann problem will have a right state in region I I . As before, we consider the left state given. It is then straightforward to see that the center state in the backwards problem must lie in the Rankine-Hugoniot locus S 2 ( L ) . Thus the center state is given by the formula

On the other hand, if the center state is to be connected to the right state by an S 2 -shock, then the right state must lie on the S 2 ( C ) shock curve and therefore satisfy the relation

Putting these two formulas together defines the region of all possible right states as

We have the following theorem.

Figure 16:

Backwards problem in ( h ,   q ) − coordinates.

Figure 17:

Backwards problem in ( h ,   u ) − coordinates.

Figure 18:

Two colliding bores forming the Riemann Problem.

7 Development of the Riemann problem from a collision of two S 1 shocks

The final case to be considered is when a Riemann problem develops from the collision between two S 1 shocks. The situation is similar, and the arguments in the proofs are virtually the same as in the previous section. Finding a center center state turns out to only be possible for some right states in region I V .

As before, we consider the left state given. It is then straightforward to see that the center state in the backwards problem must lie in the Rankine-Hugoniot locus S 2 ( L ) . Thus the center state is given by the formula

On the other hand, if the center state is to be connected to the right state by an S 2 -shock, then the right state must lie on the S 2 ( C ) shock curve and therefore satisfy the relation

Putting these two formulas together defines the region of all possible right states as

We have the following theorem.

Figure 19:

Backwards problem in ( h ,   q ) − coordinates.

Figure 20:

Backwards problem in ( h ,   u ) − coordinates.

8 Conclusion

In this article, we have considered the Riemann problem associated to the shallow-water equations. The study of the Riemann problem is important when trying to understand the behavior of solutions of a system of conservation laws. For example the Riemann problem can used as a tool in the front-tracking method where general initial data are decomposed into piecewise constant functions which gives rise to a series of Riemann problems [12]. This approach is used in existence proofs and numerical schemes, but one may face difficulties interpreting solutions of the Riemann problem for the shallow-water equations in the case when the solution includes a dry region ( h = 0 ). In gas dynamics, this situation is known as cavitation and is a well-defined concept, but the creation of a dry zone between two propagating waves does not seem reasonable from a physical point of view in the case of shallow water theory.

In the present work, we have imposed the condition that the Riemann problem should arise from the collision of two bores. With this condition in place, we were able to show that solutions of the Riemann problem do not feature cavitation. In summary, for a given left state, the collision of an S 1 ( L ) and an S 2 ( L ) shock gives rise to a Riemann problem in Region I I I (Theorem 1). The collision of two S 2 ( L ) shocks gives rise to a Riemann problem in Region I I (Theorem 2), and the collision of two S 1 ( L ) shocks gives rise to a Riemann problem in Region I V (Theorem 3). It is clear that a right state in region I is not permitted if these admissibility conditions are used. In particular, we avoid a right state in the shaded region of Figure 1 which is the region where the resolution of the Riemann problem features a dry state.