Positivity-Preserving Well-Balanced Arbitrary Lagrangian–Eulerian Discontinuous Galerkin Methods for the Shallow Water Equations

Abstract

In this paper, we develop well-balanced arbitrary Lagrangian–Eulerian discontinuous Galerkin (ALE-DG) methods for the shallow water equations, which preserve not only the still water equilibrium but also the moving water equilibrium. Based on the time-dependent linear affine mapping, the ALE-DG method for conservation laws maintains almost all mathematical properties of DG methods on static grids, such as conservation, geometric conservation law (GCL), entropy stability. The main difficulty to obtain the well-balanced property of the ALE-DG method for shallow water equations is that the grid movement and usual time discretization may destroy the equilibrium. By adopting the GCL preserving Runge–Kutta methods and the techniques of well-balanced DG schemes on static grids, we successfully construct the high order well-balanced ALE-DG schemes for the shallow water equations with still and moving water equilibria. Meanwhile, the ALE-DG schemes can also preserve the positivity property at the same time. Numerical experiments in different circumstances are provided to illustrate the well-balanced property, positivity preservation and high order accuracy of these schemes.

Appendices

Proofs for the Still Water Equilibrium Schemes

In order to show the main idea of the proof, we only show the proofs of the properties in Sect. 3 in one dimensional case. The two-dimensional proofs are similar and we omit the detail for simplicity. In one dimensional case, the flux term is defined by

$$\begin{aligned} {\varvec{G}}(\omega ,{\varvec{v}})={\varvec{f}}({\varvec{v}})-\omega {\varvec{v}},\text { with }{\varvec{f}}({\varvec{v}})=\left( \begin{matrix} hu\\ \displaystyle \frac{(hu)^2}{\eta -b}+\frac{1}{2}g(\eta -b)^2 \end{matrix}\right) . \end{aligned}$$

1.1 Proof of Lemma 3.1 for Scheme (3.12)

We first simplify the numerical flux $\hat{{\varvec{G}}}^{l,r}$. We denote the two components of the numerical flux $\hat{{\varvec{G}}}$ by $\hat{{\varvec{G}}}^{[1]}$ and $\hat{{\varvec{G}}}^{[2]}$.

We write the specific form of $\hat{{\varvec{G}}}^{[1],l}$ as follows

$$\begin{aligned} \hat{{\varvec{G}}}^{[1],l}&=\hat{{\varvec{G}}}^{[1]}(\omega ,{\varvec{v}}^{*,-},{\varvec{v}}^{*,+})+\frac{1}{2}\left( (\omega +\alpha )(\eta ^{*,+}-\eta ^+)+(\omega -\alpha )(\eta ^{*,-}-\eta ^-)\right) \end{aligned}$$

(A.1)

$$\begin{aligned}&=\frac{1}{2}\left( hu^--\omega \eta ^{*,-}+hu^--\omega \eta ^{*,+}-\alpha (\eta ^{*,+}-\eta ^{*,-})\right. \nonumber \\&\left. \quad +(\omega +\alpha )(\eta ^{*,+}-\eta ^+)+(\omega -\alpha )(\eta ^{*,-}-\eta ^-)\right) \nonumber \\&=\frac{1}{2}\left( hu^--\omega \eta ^{-}+hu^--\omega \eta ^{+}-\alpha (\eta ^{+}-\eta ^{-}\right) =\hat{{\varvec{G}}}^{[1]}(\omega ,{\varvec{v}}^-,{\varvec{v}}^+). \end{aligned}$$

(A.2)

Similarly we have

$$\begin{aligned} \hat{{\varvec{G}}}^{[1],r}=\hat{{\varvec{G}}}^{[1]}(\omega ,{\varvec{v}}^-,{\varvec{v}}^+)=\hat{{\varvec{G}}}^{[1],r}. \end{aligned}$$

(A.3)

Here without causing conflict, we have omitted the subscript $j+\frac{1}{2}$ and we note that we don’t use the assumption of still water equilibrium.

Next, since ${\varvec{v}}$ equals the still water equilibrium, i.e. $h(x)+b(x)=\text{ constant }$ and $hu(x)=0$. We have $\eta _{j+\frac{1}{2}}^+=\eta _{j+\frac{1}{2}}^-$ and $(hu)_{j+\frac{1}{2}}^\pm =0$. For the numerical flux of the first equation, we use (A.3) to get

$$\begin{aligned} \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1],l}=\hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1],r}=\hat{{\varvec{G}}}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^-,{\varvec{v}}_{j+\frac{1}{2}}^+\right) ={\varvec{G}}^{[1]}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^-\right) . \end{aligned}$$

(A.4)

For the numerical flux of the second equation, we have

$$\begin{aligned} \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[2],l}&=\hat{{\varvec{G}}}^{[2]}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^{*,-},{\varvec{v}}_{j+\frac{1}{2}}^{*,+}\right) +\frac{g}{2}\left( \eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-\right) ^2-\frac{g}{2}\left( h_{j+\frac{1}{2}}^{*,-}\right) ^2\nonumber \\&=\frac{1}{2}\left( {\varvec{G}}(\omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^{*,-})+{\varvec{G}}(\omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^{*,+})\right) +\frac{g}{2}\left( \eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-\right) ^2-\frac{g}{2}\left( h_{j+\frac{1}{2}}^{*,-}\right) ^2\nonumber \\&=\frac{1}{2}\left( \frac{g}{2}\left( \eta _{j+\frac{1}{2}}^{*,-}-b_{j+\frac{1}{2}}^-\right) ^2+\frac{g}{2}\left( \eta _{j+\frac{1}{2}}^{*,+}-b_{j+\frac{1}{2}}^+\right) ^2\right) +\frac{g}{2}\left( \eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-\right) ^2-\frac{g}{2}\left( h_{j+\frac{1}{2}}^{*,-}\right) ^2\nonumber \\&=\frac{g}{2}\left( h_{j+\frac{1}{2}}^{*,-}\right) ^2+\frac{g}{2}\left( \eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-\right) ^2-\frac{g}{2}\left( h_{j+\frac{1}{2}}^{*,-}\right) ^2\nonumber \\&=\frac{g}{2}\left( \eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-\right) ^2={\varvec{G}}^{[2]}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^-\right) . \end{aligned}$$

(A.5)

Here we use the fact that $hu_{j+\frac{1}{2}}^-=hu_{j+\frac{1}{2}}^+=0$. Similarly, we have

$$\begin{aligned} \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[2],r}={\varvec{G}}^{[2]}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^+\right) . \end{aligned}$$

(A.6)

So we combine the results (A.2)–(A.6) and have

$$\begin{aligned} \begin{aligned} \hat{{\varvec{G}}}_{j+\frac{1}{2}}^l={\varvec{G}}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^-\right) =\left( \begin{matrix} \omega _{j+\frac{1}{2}}\eta _{j+\frac{1}{2}}^-\\ \frac{g}{2}\left( \eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-\right) ^2 \end{matrix} \right) ,\\ \hat{{\varvec{G}}}_{j-\frac{1}{2}}^r={\varvec{G}}\left( \omega _{j-\frac{1}{2}},{\varvec{v}}_{j-\frac{1}{2}}^+\right) =\left( \begin{matrix} \omega _{j-\frac{1}{2}}\eta _{j-\frac{1}{2}}^+\\ \frac{g}{2}\left( \eta _{j-\frac{1}{2}}^+-b_{j-\frac{1}{2}}^+\right) ^2 \end{matrix} \right) . \end{aligned} \end{aligned}$$

Thus we have

$$\begin{aligned} \begin{aligned} {\mathcal {L}}^h_j\left( \omega ,{\varvec{v}},\varvec{\phi },t\right) =&\int \limits _{K_j(t)}\left( \begin{matrix} -\omega \eta \\ \frac{g}{2}(\eta -b)^2 \end{matrix} \right) \cdot \varvec{\phi }_x\mathrm {d} x-\left( \begin{matrix} \omega _{j+\frac{1}{2}}\eta _{j+\frac{1}{2}}^-\\ \frac{g}{2}(\eta _{j+\frac{1}{2}}^--b_{j+\frac{1}{2}}^-)^2 \end{matrix} \right) \cdot \varvec{\phi }_{j+\frac{1}{2}}^-\\&+\left( \begin{matrix} \omega _{j-\frac{1}{2}}\eta _{j-\frac{1}{2}}^+\\ \frac{g}{2}(\eta _{j-\frac{1}{2}}^+-b_{j-\frac{1}{2}}^+)^2 \end{matrix} \right) \cdot \varvec{\phi }_{j-\frac{1}{2}}^++\int \limits _{K_j(t)}\left( \begin{matrix} 0\\ -g(\eta -b)b_x \end{matrix} \right) \cdot \varvec{\phi }\mathrm {d} x\\ =&-\int \limits _{K_j(t)}\partial _x\left( \begin{matrix} -\omega \eta \\ \frac{g}{2}(\eta -b)^2 \end{matrix} \right) \cdot \varvec{\phi }\mathrm {d} x+\int \limits _{K_j(t)}\left( \begin{matrix} 0\\ -g(\eta -b)b_x \end{matrix} \right) \cdot \varvec{\phi }\mathrm {d} x\\ =&\int \limits _{K_j(t)}\partial _x\left( \begin{matrix} \omega \eta \\ 0 \end{matrix} \right) \cdot \varvec{\phi }\mathrm {d} x\\ =&\int \limits _{K_j(t)}(\omega {\varvec{v}})_x \cdot \varvec{\phi }\mathrm {d} x. \end{aligned} \end{aligned}$$

where we use the integration by parts to cancel the boundary interface terms. This finishes the proof.

1.2 Proof of Lemma 3.1 for Scheme (3.15)

Since ${\varvec{v}}$ equals the still water equilibrium, i.e. ${\varvec{v}}$ is constant vector, we have

$$\begin{aligned} {\varvec{v}}_{j+\frac{1}{2}}^-={\varvec{v}}_{j+\frac{1}{2}}^+. \end{aligned}$$

Because $\widehat{{\varvec{G}}}$ is a monotone flux, then we have

$$\begin{aligned} \widehat{{\varvec{G}}}_{j+\frac{1}{2}}&=\frac{1}{2}\left( {\varvec{G}}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^-\right) +{\varvec{G}}\left( \omega _{j+\frac{1}{2}},{\varvec{v}}_{j+\frac{1}{2}}^+\right) \right) \nonumber \\&=\frac{1}{2}\left( {\varvec{f}}\left( {\varvec{v}}_{j+\frac{1}{2}}^-\right) +{\varvec{f}}\left( {\varvec{v}}_{j+\frac{1}{2}}^+\right) \right) -\omega _{j+\frac{1}{2}}{\varvec{v}}_{j+\frac{1}{2}}^-. \end{aligned}$$

(A.7)

Similarly we have

$$\begin{aligned} \widehat{{\varvec{G}}}_{j-\frac{1}{2}}=\frac{1}{2}\left( {\varvec{f}}\left( {\varvec{v}}_{j-\frac{1}{2}}^-\right) +{\varvec{f}}\left( {\varvec{v}}_{j-\frac{1}{2}}^+\right) \right) -\omega _{j-\frac{1}{2}}{\varvec{v}}_{j-\frac{1}{2}}^+. \end{aligned}$$

(A.8)

Then we can compute that

$$\begin{aligned}&{\mathcal {L}}^s_j\left( \omega ,{\varvec{v}},\varvec{\phi },t\right) \\&\quad =\int \limits _{K_j(t)}-\omega {\varvec{v}} \cdot \varvec{\phi }_x\mathrm {d} x+\omega _{j+\frac{1}{2}}{\varvec{v}}_{j+\frac{1}{2}}^- \cdot \varvec{\phi }_{j+\frac{1}{2}}^--\omega _{j-\frac{1}{2}}{\varvec{v}}_{j-\frac{1}{2}}^+ \cdot \varvec{\phi }_{j-\frac{1}{2}}^+\\&\qquad +\int \limits _{K_j(t)}{\varvec{f}}({\varvec{v}}) \cdot \varvec{\phi }_x\mathrm {d} x -\frac{1}{2}\left( {\varvec{f}}\left( {\varvec{v}}_{j-\frac{1}{2}}^-\right) \right. \\&\left. \qquad +{\varvec{f}}\left( {\varvec{v}}_{j-\frac{1}{2}}^+\right) \right) \cdot \varvec{\phi }_{j+\frac{1}{2}}^-+\frac{1}{2}\left( {\varvec{f}}\left( {\varvec{v}}_{j-\frac{1}{2}}^-\right) +{\varvec{f}}\left( {\varvec{v}}_{j-\frac{1}{2}}^+\right) \right) \cdot \varvec{\phi }_{j-\frac{1}{2}}^++s_j\\&\quad =\int \limits _{K_j(t)}(\omega {\varvec{v}})_x \cdot \varvec{\phi }\mathrm {d} x+\int \limits _{K_j(t)}\frac{1}{2}g(\eta -b)^2\,\psi _x\mathrm {d} x\\&\qquad -\frac{g(\eta -b_{j+\frac{1}{2}}^-)^2+g\left( \eta -b_{j+\frac{1}{2}}^+\right) ^2}{4} \psi _{j+\frac{1}{2}}^-+\frac{g\left( \eta -b_{j+\frac{1}{2}}^-\right) ^2+g\left( \eta -b_{j+\frac{1}{2}}^+\right) ^2}{4} \psi _{j-\frac{1}{2}}^++s_j\\&\quad =\int \limits _{K_j(t)}(\omega {\varvec{v}})_x \cdot \varvec{\phi }\mathrm {d} x+\int \limits _{K_j(t)}\left( \frac{1}{2}g\eta ^2+\frac{1}{2}gb^2-g\eta b\right) \psi _x\mathrm {d} x\\&\qquad -\left( \frac{1}{2}g\eta ^2+\frac{1}{4}g\left( b_{j+\frac{1}{2}}^-\right) ^2+\frac{1}{4}g\left( b_{j+\frac{1}{2}}^+\right) ^2-\frac{1}{2}g\eta \,b_{j+\frac{1}{2}}^--\frac{1}{2}g\eta \,b_{j+\frac{1}{2}}^+\right) \psi _{j+\frac{1}{2}}^-\\&\qquad +\left( \frac{1}{2}g\eta ^2+\frac{1}{4}g\left( b_{j-\frac{1}{2}}^-\right) ^2+\frac{1}{4}g\left( b_{j-\frac{1}{2}}^+\right) ^2-\frac{1}{2}g\eta \,b_{j-\frac{1}{2}}^--\frac{1}{2}g\eta \,b_{j-\frac{1}{2}}^+\right) \psi _{j-\frac{1}{2}}^+\\&\qquad +\int \limits _{K_j(t)}\left( -\frac{1}{2}gb^2+g\eta b\right) \psi _x\mathrm {d} x\\&\qquad -\left( -\frac{1}{4}g\left( b_{j+\frac{1}{2}}^-\right) ^2-\frac{1}{4}g\left( b_{j+\frac{1}{2}}^+\right) ^2+\frac{1}{2}g\eta \,b_{j+\frac{1}{2}}^-+\frac{1}{2}g\eta \,b_{j+\frac{1}{2}}^+\right) \psi _{j+\frac{1}{2}}^-\\&\qquad +\left( -\frac{1}{4}g\left( b_{j-\frac{1}{2}}^-\right) ^2-\frac{1}{4}g\left( b_{j-\frac{1}{2}}^+\right) ^2+\frac{1}{2}g\eta \,b_{j-\frac{1}{2}}^-+\frac{1}{2}g\eta \,b_{j-\frac{1}{2}}^+\right) \psi _{j-\frac{1}{2}}^+\\&\quad =\int \limits _{K_j(t)}(\omega {\varvec{v}})_x \cdot \varvec{\phi }\mathrm {d}x+\int \limits _{K_j(t)}\frac{1}{2}g\eta ^2 \psi _x\mathrm {d} x-\frac{1}{2}g\eta ^2\psi _{j+\frac{1}{2}}^-+\frac{1}{2}g\eta ^2\psi _{j-\frac{1}{2}}^+\\&\quad =\int \limits _{K_j(t)}(\omega {\varvec{v}})_x \cdot \varvec{\phi }\mathrm {d}x+\int \limits _{K_j(t)}\left( \frac{1}{2}g\eta ^2\right) _x\psi \mathrm {d}x=\int \limits _{K_j(t)}(\omega {\varvec{v}})_x \cdot \varvec{\phi }\mathrm {d} x. \end{aligned}$$

The second equality is due to $hu=0$ and $\eta =\text{ constant }$. The sixth equality uses the fact $\eta =\text{ constant }$.

1.3 Proof of Proposition 3.2

It is sufficient and necessary to prove that if ${\varvec{v}}$ are the still water equilibrium at time $t=0$, i.e.

$$\begin{aligned} \eta ^0(x)=\text{ constant },\quad \text{ and }\quad (hu)^0(x)=0, \end{aligned}$$

(A.9)

then the numerical solution at any time $t^n$ has to be

$$\begin{aligned} \eta ^n(x)=\text{ constant },\quad \text{ and }\quad (hu)^n(x)=0. \end{aligned}$$

(A.10)

Now we will use induction to prove the proposition.

Basic Step: Suppose that the initial data ${\varvec{v}}^0$ are the still water equilibrium, i.e. $\eta ^0(x)=\text{ constant }$ and $(hu)^0(x)=0$. So ${\varvec{v}}^0$ is true.

Inductive Step: Now we assume the truth of ${\varvec{v}}^k$, for some $k\in {\mathbb {N}}$ —that is, we assume that

$$\begin{aligned} \eta ^k(x)=\text{ constant },\quad (hu)^k(x)=0, \end{aligned}$$

(A.11)

is a true state. From this assumption we want to deduce that

$$\begin{aligned} \eta ^{k+1}(x)=\text{ constant },\quad (hu)^{k+1}(x)=0. \end{aligned}$$

(A.12)

In each time step, we use the Euler forward scheme (3.20). For both two schemes, we can use Lemma 3.1 to simplify the equation in (3.20) and get

$$\begin{aligned} \begin{aligned} \int \limits _{K_j^{k+1}}{\varvec{v}}^{k+1}\cdot \varvec{\phi }^{k+1}\mathrm {d} x=&\int \limits _{K_j^k}{\varvec{v}}^k\cdot \varvec{\phi }^k\mathrm {d} x+\Delta t\int \limits _{K_j^k}(\omega ^k{\varvec{v}}^k)_x \cdot \varvec{\phi }^k\mathrm {d} x\\ =&{\varvec{v}}^k\cdot \,\left( \int \limits _{K_j^k}\varvec{\phi }^k\mathrm {d} x+\Delta t\int \limits _{K_j^k}(\omega ^k)_x\varvec{\phi }^k\mathrm {d} x\right) \\ =&{\varvec{v}}^k\cdot \,\int \limits _{K_j^{k+1}}\varvec{\phi }^{k+1}\mathrm {d} x\\ =&\int \limits _{K_j^{k+1}}{\varvec{v}}^k\cdot \,\varvec{\phi }^{k+1}\mathrm {d} x, \end{aligned} \end{aligned}$$

(A.13)

where the second equality holds since ${\varvec{v}}^k$ is a constant vector and the third equality is due to the dGCL in [23].

So we prove that ${\varvec{v}}^{k+1}={\varvec{v}}^k$, which is also the still water equilibrium, i.e.

$$\begin{aligned} \eta ^{k+1}=\text{ constant }, \quad (hu)^{k+1}=0. \end{aligned}$$

This claim means that the numerical solution ${\varvec{v}}^n$ for $n\in {\mathbb {N}}$ is in the case of still water equilibrium, thus the fully discrete schemes are well-balanced.

1.4 Proof of Theorem 3.4

According to the results in (A.2) and (A.3), we simplify the numerical flux in (3.12) to get that the numerical scheme for the first equation in (3.12) and (3.15) are the same. Then according to the first equation of (3.12) or (3.15), we write the equation satisfied by the cell average of surface level $\eta$:

$$\begin{aligned} \frac{\Delta _j^{n+1}}{\Delta _j^n}{\bar{\eta }}_j^{n+1}={\bar{\eta }}_j^n-\lambda \left( \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}+\hat{{\varvec{G}}}_{j-\frac{1}{2}}^{[1]}\right) . \end{aligned}$$

(A.14)

where $\hat{{\varvec{G}}}^{[1]}$ denote the first component of $\hat{{\varvec{G}}}$:

$$\begin{aligned} \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}=\frac{1}{2}\left( (hu)_{j+\frac{1}{2}}^{n,+}-\omega _{j+\frac{1}{2}}\eta _{j+\frac{1}{2}}^{n,+}+(hu)_{j+\frac{1}{2}}^{n,-}-\omega _{j+\frac{1}{2}}\eta _{j+\frac{1}{2}}^{n,-}-\alpha (\eta _{j+\frac{1}{2}}^{n,+}-\eta _{j+\frac{1}{2}}^{n,-})\right) ,\nonumber \\ \end{aligned}$$

(A.15)

and $\lambda =\frac{\Delta t}{\Delta _j^n}$. Moreover, we write out the equation satisfied by the cell average of bottom topography b:

$$\begin{aligned} \frac{\Delta _j^{n+1}}{\Delta _j^n}{\bar{b}}_j^{n+1}={\bar{b}}_j^n+\lambda \left( \widehat{\omega b}_{j+\frac{1}{2}}-\widehat{\omega b}_{j-\frac{1}{2}}\right) . \end{aligned}$$

(A.16)

We denote $\eta _j-b_j$ by $h_j$. We firstly decompose ${\bar{h}}_j^{n+1}$ in (A.14) into two parts:

$$\begin{aligned} \frac{\Delta _j^{n+1}}{\Delta _j^n}{\bar{h}}_j^{n+1}=W_1+W_2, \end{aligned}$$

(A.17)

with

$$\begin{aligned} W_1=&{\bar{b}}_j^n-\frac{\Delta _j^{n+1}}{\Delta _j^n}{\bar{b}}_j^{n+1}+\lambda \left( \widehat{\omega b}_{j+\frac{1}{2}}-\widehat{\omega b}_{j-\frac{1}{2}}\right) ,\\ W_2=&{\bar{h}}_j^n-\lambda \left( \left( \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j+\frac{1}{2}}\right) -\left( \hat{{\varvec{G}}}_{j-\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j-\frac{1}{2}}\right) \right) . \end{aligned}$$

We note that $W_1=0$. This is because the definition of b in (A.16). So we only need to prove $W_2>0$.

Since we use the Lax-Friedrichs flux in this paper, we first simplify $\hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j+\frac{1}{2}}$ by combining (A.15) and (3.22):

$$\begin{aligned} \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j+\frac{1}{2}}&=\frac{1}{2}\left( (hu)_{j+\frac{1}{2}}^{n,+}-\omega _{j+\frac{1}{2}}h_{j+\frac{1}{2}}^{n,+}+(hu)_{j+\frac{1}{2}}^{n,-}-\omega _{j+\frac{1}{2}}h_{j+\frac{1}{2}}^{n,-}-\alpha (h_{j+\frac{1}{2}}^{n,+}-h_{j+\frac{1}{2}}^{n,-})\right) \\&=\frac{1}{2}\left( \left( u_{j+\frac{1}{2}}^{n,+}-\omega _{j+\frac{1}{2}}-\alpha \right) h_{j+\frac{1}{2}}^{n,+}+\left( u_{j+\frac{1}{2}}^{n,-}-\omega _{j+\frac{1}{2}}+\alpha \right) h_{j+\frac{1}{2}}^{n,-}\right) \\&<\frac{1}{2}\left( \left( \alpha -\alpha \right) h_{j+\frac{1}{2}}^{n,+}+\left( \alpha +\alpha \right) h_{j+\frac{1}{2}}^{n,-}\right) =\alpha h_{j+\frac{1}{2}}^{n,-}. \end{aligned}$$

The first equality is due to $\hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}$ and $\widehat{\omega b}_{j+\frac{1}{2}}$ use the same $\alpha$. The inequality is due to the definition of $\alpha$:

$$\begin{aligned} -\alpha<u_{j+\frac{1}{2}}^{n,+}-\omega _{j+\frac{1}{2}}<\alpha . \end{aligned}$$

Similarly we have

$$\begin{aligned} \hat{{\varvec{G}}}_{j-\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j-\frac{1}{2}}>-\alpha h_{j-\frac{1}{2}}^{n,+}. \end{aligned}$$

So we use the Gauss quadrature rule and above inequalities to get

$$\begin{aligned} W_2=&\sum _{\nu =1}^L{\hat{\omega }}_\nu h_j^n({\hat{x}}_j^{(\nu )})-\lambda \left( \left( \hat{{\varvec{G}}}_{j+\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j+\frac{1}{2}}\right) -\left( \hat{{\varvec{G}}}_{j-\frac{1}{2}}^{[1]}+\widehat{\omega b}_{j-\frac{1}{2}}\right) \right) \\>&\sum _{\nu =1}^{L}{\hat{\omega }}_\nu h_j^n({\hat{x}}_j^{(\nu )})-\alpha \lambda h_{j-\frac{1}{2}}^{n,+}-\alpha \lambda h_{j+\frac{1}{2}}^{n,-}\\ =&\sum _{\nu =2}^{L-1}{\hat{\omega }}_\nu h_j^n({\hat{x}}_j^{(\nu )})+\left( {\hat{\omega }}_1-\alpha \lambda \right) h_{j-\frac{1}{2}}^{n,+}-\left( {\hat{\omega }}_L-\alpha \lambda \right) h_{j+\frac{1}{2}}^{n,-}\\ >&0. \end{aligned}$$

Notice that ${\hat{\omega }}_1={\hat{\omega }}_L$ and $\lambda <\frac{1}{{\hat{\alpha }}_0}=\frac{{\hat{\omega }}_1}{\alpha }$. Combined with the assumption $h(x)>0$, for $x\in {\mathbb {S}}_{K_j^n}$, we proved that $W_2>0$.

Proofs for the Moving Water Equilibrium Scheme

1.1 Proof of Lemma 4.2

We first prove that $\hat{{\varvec{g}}}(\omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{*,-},{\varvec{u}}_{j+\frac{1}{2}}^{*,+})={\varvec{g}}(\omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{d,-})={\varvec{g}}(\omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{d,-})$. Under such assumption, we have

$$\begin{aligned} {\varvec{u}}^r=0,\quad {\varvec{u}}^*={\varvec{u}}^d. \end{aligned}$$

(B.1)

With stationary shocks of ${\varvec{u}}^d$ and the assumption $\omega _{j+\frac{1}{2}}=0$ at the shock, we can calculate that

$$\begin{aligned} \hat{{\varvec{g}}}\left( \omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{*,-},{\varvec{u}}_{j+\frac{1}{2}}^{*,+}\right) =\hat{{\varvec{g}}}\left( \omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{d,-},{\varvec{u}}_{j+\frac{1}{2}}^{d,+}\right) ={\varvec{g}}\left( \omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{d,-}\right) ={\varvec{g}}\left( \omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{d,-}\right) .\nonumber \\ \end{aligned}$$

(B.2)

The first equality is due to (B.1). The second and the third equality is due to $\omega _{j+\frac{1}{2}}=0$, the property of Roe’s flux that it can exactly solve the Riemann problem [35], and the shock at point $x_{j+\frac{1}{2}}$ is a stationary shock.

With the continuity of ${\varvec{u}}^d$, we have ${\varvec{u}}_{j+\frac{1}{2}}^{*,\pm }={\varvec{u}}_{j+\frac{1}{2}}^{d,\pm }$ and we have the same conclusion with (B.2) according to the property of the Lax-Friedrichs flux. This means that for all the numerical fluxes,

$$\begin{aligned} \hat{{\varvec{g}}}_{j+\frac{1}{2}}^l={\varvec{f}}({\varvec{u}}_{j+\frac{1}{2}}^-)-\omega _{j+\frac{1}{2}}{\varvec{u}}_{j+\frac{1}{2}}^{d,-}\quad \text{ and }\quad \hat{{\varvec{g}}}_{j-\frac{1}{2}}^r={\varvec{f}}({\varvec{u}}_{j-\frac{1}{2}}^+)-\omega _{j-\frac{1}{2}}{\varvec{u}}_{j-\frac{1}{2}}^{d,+}. \end{aligned}$$

Now, we have

$$\begin{aligned} \begin{aligned} {\mathcal {L}}^m_j(\omega ,{\varvec{u}},\varvec{\phi },t)=&\int \limits _{K_j(t)}{\varvec{g}}(\omega ,{\varvec{u}})\cdot \varvec{\phi }_x\mathrm {d}x-\left( {\varvec{f}}({\varvec{u}}_{j+\frac{1}{2}}^-)-\omega _{j+\frac{1}{2}}{\varvec{u}}_{j+\frac{1}{2}}^{d,-}\right) \cdot \varvec{\phi }_{j+\frac{1}{2}}^-\\&+\left( {\varvec{f}}({\varvec{u}}_{j-\frac{1}{2}}^+)-\omega _{j-\frac{1}{2}}{\varvec{u}}_{j-\frac{1}{2}}^{d,+}\right) \cdot \varvec{\phi }_{j-\frac{1}{2}}^+-\int \limits _{K_j(t)}{\varvec{f}}({\varvec{u}}^e)\cdot \varvec{\phi }_x\mathrm {d}x\\&+f({\varvec{u}}_{j+\frac{1}{2}}^{e,-})\cdot \varvec{\phi }_{j+\frac{1}{2}}^--{\varvec{f}}({\varvec{u}}_{j-\frac{1}{2}}^{e,+})\cdot \varvec{\phi }_{j-\frac{1}{2}}^+\\ =&-\int \limits _{K_j(t)}\omega {\varvec{u}}\cdot \varvec{\phi }_x\mathrm {d}x +\omega _{j+\frac{1}{2}} {\varvec{u}}_{j+\frac{1}{2}}^{d,-} \cdot \varvec{\phi }_{j+\frac{1}{2}}^--\omega _{j-\frac{1}{2}} {\varvec{u}}_{j-\frac{1}{2}}^{d,+} \cdot \varvec{\phi }_{j-\frac{1}{2}}^+. \end{aligned} \end{aligned}$$

This is because of the assumption ${\varvec{u}}={\varvec{u}}^e$ and we finish the proof.

1.2 Proof of Proposition 4.4

Similar to previous proof in Proposition 3.2, we will use induction to prove the proposition.

Basic Step: Suppose that ${\varvec{u}}^d$ is the initial condition. Then we have ${\varvec{u}}^0=P{\varvec{u}}^d$ and ${\varvec{u}}^{e,0}=P{\varvec{u}}^d$, which will lead to ${\varvec{u}}^0={\varvec{u}}^{e,0}$. So ${\varvec{u}}^0$ is true.

Inductive Step: Now we assume the truth of ${\varvec{u}}^k$, for some $k\in {\mathbb {N}}$ —that is, we assume that ${\varvec{u}}^k={\varvec{u}}^{e,k}$ is a true state. From this assumption we want to deduce that ${\varvec{u}}^{k+1}={\varvec{u}}^{e,k+1}$.

Since ${\varvec{u}}^k={\varvec{u}}^{e,k}$, we have ${\varvec{u}}^{r,k}=0$. Use Lemma 4.2, we have

$$\begin{aligned} {\mathcal {L}}^m_j(\omega ^k,{\varvec{u}}^k,\varvec{\phi }^k,t^k)=\omega ^k_{j+\frac{1}{2}}{\varvec{u}}_{j+\frac{1}{2}}^{d,-}\cdot \varvec{\phi }_{j+\frac{1}{2}}^--\omega ^k_{j-\frac{1}{2}}{\varvec{u}}_{j-\frac{1}{2}}^{d,+}\cdot \varvec{\phi }_{j-\frac{1}{2}}^+-\int \limits _{K_j^k}\omega ^k {\varvec{u}}^{e,k}\cdot (\varvec{\phi }^k)_x\mathrm {d}x.\nonumber \\ \end{aligned}$$

(B.3)

By comparing (4.19) and (4.20), we have

$$\begin{aligned} \int \limits _{K_j^{k+1}}({\varvec{u}}^{k+1}-{\varvec{u}}^{e,k+1})\cdot \varvec{\phi }^{k+1}\mathrm {d}x=0. \end{aligned}$$

(B.4)

This means ${\varvec{u}}^{k+1}={\varvec{u}}^{e,k+1}$. This finishes the proof that for $n\in {\mathbb {N}}$ the numerical solution ${\varvec{u}}^n$ is in the case of moving water equilibrium, which means our scheme do preserve the moving water equilibrium.

1.3 Proof of Theorem 4.6

Since we use the Lax-Friedrichs numerical flux, we can write the specific form of $\hat{{\varvec{g}}}_{j+\frac{1}{2}}^{l,[1]}$:

$$\begin{aligned} \hat{{\varvec{g}}}_{j+\frac{1}{2}}^{l,[1]}= & {} \hat{{\varvec{g}}}^{[1]}(\omega _{j+\frac{1}{2}},{\varvec{u}}_{j+\frac{1}{2}}^{*,-},{\varvec{u}}_{j+\frac{1}{2}}^{*,+})\nonumber \\= & {} \frac{1}{2}\left( m_{j+\frac{1}{2}}^{*,+}-\omega _{j+\frac{1}{2}} h_{j+\frac{1}{2}}^{*,+}-\alpha h_{j+\frac{1}{2}}^{*,+}+m_{j+\frac{1}{2}}^{*,-}-\omega _{j+\frac{1}{2}}h_{j+\frac{1}{2}}^{*,-}+\alpha h_{j+\frac{1}{2}}^{*,-}\right) \nonumber \\= & {} \frac{1}{2}\left( \left( u_{j+\frac{1}{2}}^{*,+}-\omega _{j+\frac{1}{2}}-\alpha \right) h_{j+\frac{1}{2}}^{*,+}+\left( u_{j+\frac{1}{2}}^{*,-}-\omega _{j+\frac{1}{2}}+\alpha \right) h_{j+\frac{1}{2}}^{*,-}\right) \nonumber \\< & {} \frac{1}{2}\left( \left( \alpha -\alpha \right) h_{j+\frac{1}{2}}^{*,+}+\left( \alpha +\alpha \right) h_{j+\frac{1}{2}}^{*,-}\right) =\alpha h_{j+\frac{1}{2}}^{*,-}. \end{aligned}$$

(B.5)

The first equality is due to $m^d=m^e$ is constant, which is same as the proof of mass conservation and the inequality is due the definition of $\alpha$ in (4.17). Similar we simplify $\hat{{\varvec{g}}}_{j-\frac{1}{2}}^{r,[1]}$ to get

$$\begin{aligned} \hat{{\varvec{g}}}_{j-\frac{1}{2}}^{r,[1]}>-\alpha h_{j-\frac{1}{2}}^{*,+}. \end{aligned}$$

(B.6)

Now we combine (B.5), (B.6) and Gauss-Lobatto quadrature rule, which introduced in (3.6) to simplify (4.22):

$$\begin{aligned} \frac{\Delta _j^{n+1}}{\Delta _j^n}{\bar{h}}_j^{n+1}&={\bar{h}}_j^n-\lambda \left( \hat{{\varvec{g}}}_{j+\frac{1}{2}}^{l,[1]}-\hat{{\varvec{g}}}_{j-\frac{1}{2}}^{r,[1]}\right)>\sum _{\nu =1}^L{\hat{\omega }}_\nu h_j^n({\hat{x}}_j^{(\nu )})-\alpha \lambda \left( h_{j+\frac{1}{2}}^{*,-}+h_{j-\frac{1}{2}}^{*,+}\right) \\&=\sum _{\nu =2}^{L-1}{\hat{\omega }}_\nu h_j^n({\hat{x}}_j^{(\nu )})+\left( {\hat{\omega }}_1h_{j-\frac{1}{2}}^{n,+}-\alpha \lambda h_{j-\frac{1}{2}}^{*,+}\right) +\left( {\hat{\omega }}_Lh_{j+\frac{1}{2}}^{n,-}-\alpha \lambda h_{j+\frac{1}{2}}^{*,-}\right) >0. \end{aligned}$$

This is because the assumption of the theorem that $h_j^n({\hat{x}}_j^{(\nu )})>0$ and (4.25):

$$\begin{aligned} 1>{\hat{\alpha }}_0\lambda =\frac{\alpha \lambda }{{\hat{\omega }}_1}\max _{x\in \{x_{j-\frac{1}{2}}^+,x_{j+\frac{1}{2}}^-\}}\frac{h_j^*(x)}{h_j^n(x)}\rightarrow {\hat{\omega }}_Lh_j^n(x)>\alpha \lambda h_j^*(x),\,\text {for }x\in \{x_{j-\frac{1}{2}}^+,x_{j+\frac{1}{2}}^-\},\nonumber \\ \end{aligned}$$

(B.7)

and the assumption of Gauss-Lobatto quadrature rule: ${\hat{x}}_1^{(\nu )}$ and ${\hat{x}}_L^{(\nu )}$ are the cell boundaries of $K_j^n$ and ${\hat{\omega }}_1={\hat{\omega }}_L$. Now we have proved ${\bar{h}}_j^{n+1}>0$ and finish the proof.