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A Class of Positive Semi-discrete Lagrangian–Eulerian Schemes for Multidimensional Systems of Hyperbolic Conservation Laws

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Abstract

In this paper, we design and analyze a new class of positive Semi-Discrete Lagrangian–Eulerian (SDLE) schemes for solving multidimensional initial value problems for scalar models and systems of conservation laws. The construction of the schemes is based on the space–time no flow surface region, previously presented and analyzed by the authors for fully-discrete schemes. The implementation of the scheme in the case of systems is a straightforward componentwise application of the multidimensional scalar case, but, importantly, the semi-discrete approach does not require dimensional splitting strategies. Entropy-convergence proof for the multidimensional scalar case is provided via weak asymptotic analysis. We also prove that the new two-dimensional Lagrangian–Eulerian scheme satisfies the scalar maximum principle along with relevant estimates, which also implies the uniqueness of the weak solution satisfying Kruzhkov entropy condition. Moreover, we show that the new semi-discrete Lagrangian–Eulerian scheme, in the more general context of multidimensional hyperbolic systems of conservation laws, also satisfies the positivity principle. Indeed, by using the no flow properties, it is not necessary to obtain the eigenvalues associated with the hyperbolic flux to guarantee the positivity of our numerical scheme. We also use the no flow estimates for numerically stable computations—for scalar equations and multidimensional systems—in a similar fashion to that of the well-known stability condition by Courant–Friedrichs–Lewy (CFL), but without the need to employ the eigenvalues (exact and approximate values) of the relevant Jacobian of the numerical flux functions. Another interesting feature of the no flow Lagrangian–Eulerian construction is that the matrices are symmetric for free (actually, they are diagonal), which is independent for a general class of hyperbolic flux for scalar problems and systems as well. We provide robust numerical examples to verify the theory and illustrate the capabilities of the semi-discrete approach in three cases (1) a 4 \(\times \) 4 system of compressible Euler flows (Double Mach reflection and Wind tunnel problems), (2) a 3 \(\times \) 3 shallow-water system of equations with and without bottom discontinuous topography, and (3) 2 \(\times \) 2 nonstrictly hyperbolic three-phase flows with a resonance point. We also demonstrate the application of our semi-discrete scheme to nontrivial prototype two-dimensional scalar problems that display intricate wave interactions (e.g., 2D inviscid Burgers’ equation for an oblique Riemann problem, Buckley–Leverett equation with gravity, and nonlinear equation with non-convex fluxes). We present satisfactory qualitative results for the Orszag–Tang vortex system in a prototype ideal magnetohydrodynamic model with no need of an additional constrained-transport enforcement to preserve the magnetic field. We also verify non-oscillatory and sufficiently accurate numerical solutions for the 2D Euler system, isentropic smooth vortex convection problem. The multidimensional SDLE scheme retains simplicity, with a very good resolution, is efficient in terms of computational and memory cost, and is simple to implement as well since no (local) Riemann problems are solved; hence, time-consuming field-by-field type decompositions are avoided in the case of systems. These features are significant and ensure the simplicity and power of this class of positive semi-discrete schemes.

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Acknowledgements

E. Abreu gratefully acknowledges the financial support of the São Paulo Research Foundation (FAPESP) (Grant 2019/20991-8), the National Council for Scientific and Technological Development - Brazil (CNPq) (Grant 306385/2019-8), PETROBRAS - Brazil (Grant 2015/00398-0), and the MATHDATA - AUIP Network (Red Iberoamericana de Investigación en Matemáticas Aplicadas a Datos) (https://www.mathdata.science/). This research is part of a doctoral thesis funded by CAPES (Coordination for the Improvement of Higher Education Personnel), which J. François would like to point out and thank the institutional grant provided by CAPES and IMECC/Unicamp as well. W. Lambert gratefully acknowledges the financial support of FAPESP (Grant 2019/20991-8).

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Appendices

Appendix A. The Maximum Principle and the Entropy Solution

Here, we obtain an abstract proposition that can be used for any numerical method satisfying the hypotheses of Theorem 6.1. In this case, we show that our numerical scheme satisfies the maximum principle. Since the initial data in (3.1) satisfies \(u_0(x,y)\in L^\infty ({\mathbb {T}})\), we are able to obtain a smooth approximation to \(u_0(x,y)\), denoted as \(u_0(x,y,\epsilon )\in {\mathcal {C}}^\infty ({\mathbb {T}})\), such that \(u_0(x,y,\epsilon )\) uniformly converges to \(u_0(x,y)\). In the following proposition, we use this approximation and state our result as follows:

PropositionAppendix A.1

Let us assume that numerical method (5.6) satisfies the hypothesis of Theorem 6.1. Then, any local solution on [0, T), for \(T>0\), of (3.1), using the numerical scheme (5.6), takes its values between range \([\min _{(x,y)\in {{\mathbb {T}}}}u_0(x,y),\max _{(x,y)\in {{\mathbb {T}}}}u_0(x,y)]\).

Proof

We consider \((x,y)\in {{\mathbb {T}}}\). First, we take values \(\epsilon \) (not fixed, for example \(\epsilon \) can be rational values) so that \(\{n\epsilon \}_{n\in {\mathbb {Z}}}\) forms a dense set in \({{\mathbb {T}}}\). By absurd, we assume that there exists a \(\epsilon _0>0\) satisfying, for \(T>0\),

$$\begin{aligned} \sup _{x\in {{\mathbb {T}}}}u(x,y,t,\epsilon _0)>\sup _{(x,y)\in {{\mathbb {T}}}}u_0(x,y,\epsilon _0) \quad \text { for some } t\in [0,T]. \end{aligned}$$
(A.1)

Since \(u_0(x,y,\epsilon )\) is continuous, we can choose a small enough \(\epsilon _0\) and \(\eta >0\) so that \(\{u(x,y,t,\epsilon _0)\}\subset [\min _{(x,y)\in {{\mathbb {T}}}}u_0(x,y,\epsilon ) -\eta ,\max _{(x,y)\in {{\mathbb {T}}}}u_0(x,y,\epsilon )+\eta ]\). Given that \(u_0(x,y,\epsilon _0)\) is smooth, then solution \(u(x,y,t,\epsilon _0)\) from Eq. (5.6) is also smooth because this space can be considered a Banach space using the \(L^\infty \) norm. Thus, there exists \(x_0\), \(y_0\), and \(t_0\) such that \(\sup _{(x,y)\in {{\mathbb {T}}}}u(x,y,t,\epsilon _0)= u(x_0,y_0,t_0,\epsilon _0)\). Since \((x_0,t_0)\) is a maximum, solution \(u(x,y,t,\epsilon _0)\) satisfies

$$\begin{aligned} \partial _tu(x_0,t_0,\epsilon _0)\ge 0. \end{aligned}$$
(A.2)

Moreover, if scheme (5.6) satisfies the hypothesis of Theorem 6.1, we obtain

$$\begin{aligned} \partial _t (u(x_0,y_0,t_0,\epsilon _0)&=-\frac{1}{(\epsilon _1)_0} \big [{\mathcal {F}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0+(\epsilon _1)_0,y_0,t_0,\epsilon _0)\nonumber \\&\qquad -{\mathcal {F}}(u(x_0-(\epsilon _1)_0,y_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)) \big ]\nonumber \\&\quad -\frac{1}{(\epsilon _2)_0} \big [{\mathcal {G}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0,y_0+(\epsilon _2)_0,t_0,\epsilon _0)\nonumber \\&\quad -{\mathcal {G}}(u(x_0,y_0-(\epsilon _2)_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)) \big ]. \end{aligned}$$
(A.3)

We can rewrite (A.3) as

$$\begin{aligned} \partial _t (u(x_0,y_0,t_0,\epsilon _0)&= -\frac{1}{(\epsilon _1)_0} \big [{\mathcal {F}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0\nonumber \\&\quad +(\epsilon _1)_0,y_0,t_0,\epsilon _0) -{\mathcal {F}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)\big ]\nonumber \\&\quad -\frac{1}{(\epsilon _1)_0} \big [{\mathcal {F}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)\nonumber \\&\quad -{\mathcal {F}}(u(x_0-(\epsilon _1)_0,y_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)) \big ]\nonumber \\&\quad -\frac{1}{(\epsilon _2)_0} \big [{\mathcal {G}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0,y_0+(\epsilon _2)_0,t_0,\epsilon _0)\nonumber \\&\quad -{\mathcal {G}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)\big ]\nonumber \\&\quad -\frac{1}{(\epsilon _2)_0} \big [{\mathcal {G}}(u(x_0,y_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)\nonumber \\&\quad -{\mathcal {G}}(u(x_0,y_0-(\epsilon _2)_0,t_0,\epsilon _0),u(x_0,y_0,t_0,\epsilon _0)) \big ] . \end{aligned}$$
(A.4)

From the mean value theorem, (A.4) becomes

$$\begin{aligned}&\partial _t (u(x_0,y_0,t_0,\epsilon _0)\nonumber \\&\quad =-\frac{1}{(\epsilon _1)_0} \left[ \frac{\partial {\mathcal {F}}}{\partial y}(u(x_0,y_0,t_0,\epsilon _0),\theta _{1})\right] (u(x_0+(\epsilon _1)_0,y_0,t_0,\epsilon _0)-u(x_0,y_0,t_0,\epsilon _0))\nonumber \\&\qquad -\frac{1}{(\epsilon _1)_0} \left[ \frac{\partial {\mathcal {F}}}{\partial x}(\theta _2,u(x_0,y_0,t_0,\epsilon _0)) \right] (u(x_0,y_0,t_0,\epsilon _0)-u(x_0-(\epsilon _1)_0,y_0,t_0,\epsilon _0))\nonumber \\&\qquad -\frac{1}{(\epsilon _2)_0} \left[ \frac{\partial {\mathcal {G}}}{\partial y} (u(x_0,y_0,t_0,\epsilon _0),\eta _1))\right] (u(x_0,y_0+(\epsilon _2)_0,t_0,\epsilon _0)-(u(x_0,y_0,t_0,\epsilon _0))\nonumber \\&\qquad -\frac{1}{(\epsilon _2)_0} \left[ \frac{\partial {\mathcal {G}}}{\partial x} (\eta _2,u(x_0,y_0,t_0,\epsilon _0)) \right] (u(x_0,y_0,t_0,\epsilon _0)-(u(x_0,y_0-(\epsilon _2)_0,t_0,\epsilon _0)) , \end{aligned}$$
(A.5)

where \(\theta _1\) is a state between \(u(x_0+(\epsilon _1)_0,y_0,t_0,\epsilon _0)\) and \(u(x_0,y_0,t_0,\epsilon _0)\); \(\theta _2\), a state between \(u(x_0,y_0-(\epsilon _1)_0,t_0,\epsilon _0)\) and \(u(x_0,y_0,t_0,\epsilon _0)\); \(\eta _1\), a state between \(u(x_0,y_0,t_0,\epsilon _0)\) and \(u(x_0,y_0+(\epsilon _2)_0,t_0,\epsilon _0)\); and \(\eta _2\), a state between \(u(x_0,y_0-(\epsilon _2)_0,t_0,\epsilon _0)\) and \(u(x_0,y_0,t_0,\epsilon _0)\). Since \(u(x_0,y_0,t_0,\epsilon _0)\) is a maximum and conditions (6.8) are valid, from Eq. (A.5), we have that

$$\begin{aligned} \partial _tu(x_0,y_0,t_0,\epsilon _0)\le 0. \end{aligned}$$
(A.6)

From inequalities (A.2) and (A.6), we obtain that \(\partial _tu(x_0,y_0,t_0,\epsilon _0)=0\). Thus, the right hand side of (A.5) is null. This leads to \(u(x_0-(\epsilon _1)_0,y_0,t_0,\epsilon _0)= u(x_0+(\epsilon _1)_0,y_0,t_0,\epsilon _0) =u(x_0,y_0,t_0,\epsilon _0)\) and \(u(x_0,y_0-(\epsilon _2)_0,t_0,\epsilon _0)= u(x_0,y_0+(\epsilon _2)_0,t_0,\epsilon _0) =u(x_0,y_0,t_0,\epsilon _0)\). By recursion, we obtain that \(u(x_0+n(\epsilon _1)_0,y_0,t_0,\epsilon _0)=u(x_0,y_0+m(\epsilon _2)_0,t_0,\epsilon _0)= u(x_0,t_0,\epsilon _0)\) for all n and m, i.e., u is constant because u is (at least) continuous and \({\mathbb {N}}(\epsilon _1)_0\times {\mathbb {N}}(\epsilon _2)_0\) is dense in \({{\mathbb {T}}}\) modulus 1 (because \((\epsilon _1)_0\) and \((\epsilon _2)_0\) are taken as irrational numbers). From ODE (5.6), u is constant, and the solution is trivial, which leads to a contradiction by the assumption. The same argument can be used by substituting \(\sup \) by \(\inf \) in Eq. (A.1), and the proof is completed. \(\square \) .

Proposition Appendix A.1 yields the following Corollary:

CorollaryAppendix A.2

Let us assume that numerical method (5.6) satisfies the conditions of Proposition 6.2. Then, it also satisfies the maximum principle, i.e., the solution satisfies \(u\in [\min _{(x,y)\in {{\mathbb {T}}}}u_0(x,y),\max _{(x,y)\in {{\mathbb {T}}}} u_0(x,y)]\).

In Sect. 5, we make the hypothesis that there exists a \(a>0\) such that \(u>a>0\). From the Corollary (Appendix A.2) we can verify that this hypothesis is not restrictive to our method.

RemarkAppendix A.3

If initial data \(u_0(x,y)\in L^\infty ({\mathbb {T}})\) in Eq. (3.1) assumes negative, positive, and null values, we consider \(N=\sup _{(x,y)\in {\mathbb {T}}}|u_0(x,y)|+\beta \) for \(\beta \) a positive constant. Then, we consider the following auxiliary problem:

$$\begin{aligned}&\frac{\partial u}{\partial t}+ \frac{\partial H(u-N)}{\partial x} +\frac{\partial G(u-N)}{\partial y} = 0, \quad (x,y) \in {\mathbb {T}}, \quad t > 0,\nonumber \\&\quad \quad \quad u(x,y,0) = u_0(x,y)+N, \end{aligned}$$
(A.7)

Notice that the new initial data for (A.7) assume only positive values. Under a suitable hypothesis (see Proposition Appendix A.1), we demonstrate that the numerical method satisfies the maximum principle, that is, solution u(xyt) takes its values between the maximum and minimum values of the initial data. Then, the solution to (A.7) assumes only positive values; hence, the \(u>a>0\) assumption is valid. In Appendix C, we prove that if u(xyt) is the (weak and entropic) solution to (A.7), then \(u(x,y,t)-N\) is the solution to (3.1).

RemarkAppendix A.4

From Remark Appendix A.3, given any Cauchy problem for (3.1), we can define auxiliary problem (A.7), which assumes only positive values. This auxiliary problem satisfies the \(u>a>0\) hypothesis, and the proof of convergence is valid for the solution to (A.7), which is u(xyt), and for that to (3.1), which is \(u(x,y,t)-N\). The \(u>a>0\) assumption helps to avoid several technical details in such proof. In addition, by means of Remark Appendix A.3, we guarantee the convergence of the numerical method for (3.1), without using any technical details. This is a more elegant strategy to deal with problems of the Lagrangian–Eulerian type using the no flow curve, which requires H(u)/u and G(u)/u to be defined. In summary, for the sake of concreteness and simplicity, and without loss of generality, the proof of convergence of the SDLE scheme via the weak asymptotic analysis covers all initial data \(u_0(x,y)\in L^\infty ({\mathbb {R}}\times {\mathbb {R}})\) in Eq. (3.1) for negative, positive, and null values, which is necessary for industrial and real world problems.

The next step of our construction is to prove that the proposed scheme satisfies some kind of entropy condition. In this paper, we use the Kruzhkov entropy solution. We say that a solution \(u(x,y,t)\in L^\infty ({\mathbb {T}}\times [0,T))\) satisfies the Kruzhkov entropy solution if

$$\begin{aligned}&\int _{0}^T\int _{{{\mathbb {T}}}}\Big (\left| u(x,y,t)-K\right| \varphi _t(x,y,t) +sign(u(x,y,t)-K)\nonumber \\&\quad [u(x,y,t)f(u(x,y,t))-Kf(K))]\varphi _x(x,y,t)\nonumber \\&\qquad +sign(u(x,y,t)-K)[u(x,y,t)g(u(x,y,t))-Kg(K))]\varphi _y(x,y,t)\Big )dxdydt \nonumber \\&\qquad + \int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy\ge 0 \end{aligned}$$
(A.8)

for all \(\varphi (x,y,t)\in {\mathcal {C}}^\infty _0({\mathbb {T}}\times [0,T))\). In eq. (A.8), the subscripts in function \(\varphi \) represent the partial derivatives.

In this proof, we use that the sequence generated by scheme (5.6) is pre-compact. Here, we assume that

$$\begin{aligned} \mu =||u_0(x,y)||_{\infty }. \end{aligned}$$
(A.9)

PropositionAppendix A.5

(Kruzhkov entropy) Let us assume that the conditions of Proposition 6.2 are satisfied. Then, \(u(x,y,t,\epsilon )\longrightarrow u(x,y,t)\) when \(\epsilon \longrightarrow 0\) in \(L_{loc}^1({{\mathbb {T}}}\times [0,\infty ))\), when u(xyt) is the unique entropy solution to (3.1).

Proof

We consider a (fixed but generic) constant \(K\in [-\mu ,\mu ]\) , \(b^x_{\epsilon _1+1/2;\epsilon _2}={\hat{M}}_1\), and \(b^y_{\epsilon _1;\epsilon _2+1/2}={\hat{M}}_2\) for all \(\epsilon \). For almost \((x,y,t)\in {{\mathbb {T}}}\times (0,\infty )\) and fixed x and y, and then using (5.6),

$$\begin{aligned}&\frac{d}{dt}|u(x,y,t,\epsilon )-K|\nonumber \\&\quad = sign(u_{_{\epsilon _1;\epsilon _2}}-K)\frac{d}{dt}u(x,y,t,\epsilon )\nonumber \\&\quad =-\frac{1}{4\epsilon _1}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ {\hat{M}}_1\left( u_{_{\epsilon _1;\epsilon _2}} +\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}-\left( u_{_{\epsilon _1+1;\epsilon _2}} -\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1+1;\epsilon _2}}\right) \right) +\right. \nonumber \\&\qquad +\left( (f(u_{_{\epsilon _1;\epsilon _2}})+f(u_{_{\epsilon _1+1;\epsilon _2}})) \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}} +\left( u_{_{\epsilon _1+1;\epsilon _2}}-\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1+1;\epsilon _2}}\right) \right) \right) \nonumber \\&\qquad -\left( {\hat{M}}_1\left( u_{_{\epsilon _1-1;\epsilon _2}}+\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1-1;\epsilon _2}} -\left( u_{_{\epsilon _1;\epsilon _2}}-\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) +\nonumber \\&\qquad \left. -\left( (f(u_{_{\epsilon _1-1;\epsilon _2}})+f(u_{_{\epsilon _1;\epsilon _2}})) \left( u_{_{\epsilon _1-1;\epsilon _2}}+\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1-1;\epsilon _2}} +\left( u_{_{\epsilon _1;\epsilon _2}}-\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) \right] \nonumber \\&\qquad -\frac{1}{4\epsilon _1}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ {\hat{M}}_2\left( u_{_{\epsilon _1;\epsilon _2}} +\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}-\left( u_{_{\epsilon _1;\epsilon _2+1}} -\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2+1}}\right) \right) +\right. \nonumber \\&\qquad +\left( (g(u_{_{\epsilon _1;\epsilon _2}})+g(u_{_{\epsilon _1;\epsilon _2+1}})) \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}} +\left( u_{_{\epsilon _1;\epsilon _2+1}}-\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2+1}}\right) \right) \right) \nonumber \\&\qquad -\left( {\hat{M}}_2\left( u_{_{\epsilon _1;\epsilon _2-1}}+\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2-1}} -\left( u_{_{\epsilon _1;\epsilon _2}}-\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) +\nonumber \\&\qquad \left. -\left( (g(u_{_{\epsilon _1;\epsilon _2-1}})+g(u_{_{\epsilon _1;\epsilon _2}})) \left( u_{_{\epsilon _1;\epsilon _2-1}}+\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2-1}} +\left( u_{_{\epsilon _1;\epsilon _2}}-\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) \right] . \end{aligned}$$
(A.10)

Rearranging the terms, we obtain

$$\begin{aligned}&\frac{d}{dt}|u(x,y,t,\epsilon )-K|\nonumber \\&\quad = sign(u_{_{\epsilon _1;\epsilon _2}}-K)\frac{d}{dt}u(x,y,t,\epsilon )\nonumber \\&\quad =\frac{1}{4\epsilon _1}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ u_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}+u_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- u_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}+{\hat{A}}_{\epsilon _1;\epsilon _2})\right] + \nonumber \\&\qquad -\frac{1}{16}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] +\nonumber \\&\qquad +\frac{1}{4\epsilon _2}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ u_{_{\epsilon _1;\epsilon _2+1}}{\hat{D}}_{\epsilon _1;\epsilon _2+1}+u_{_{\epsilon _1;\epsilon _2-1}}{\hat{C}}_{\epsilon _1;\epsilon _2-1}- u_{_{\epsilon _1;\epsilon _2}}({\hat{D}}_{\epsilon _1;\epsilon _2}+{\hat{C}}_{\epsilon _1;\epsilon _2})\right] + \nonumber \\&\qquad -\frac{1}{16}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ (u_y)_{_{\epsilon _1;\epsilon _2+1}}{\hat{D}}_{\epsilon _1;\epsilon _2+1}-(u_y)_{_{\epsilon _1;\epsilon _2-1}}{\hat{C}}_{\epsilon _1;\epsilon _2-1}- (u_y)_{_{\epsilon _1;\epsilon _2}}({\hat{D}}_{\epsilon _1;\epsilon _2}-{\hat{C}}_{\epsilon _1;\epsilon _2})\right] , \end{aligned}$$
(A.11)

where

$$\begin{aligned}&{\hat{A}}_{\epsilon _1;\epsilon _2}={\hat{M}}_1+f({u}_{_{\epsilon _1;\epsilon _2}})+f({u}_{_{\epsilon _1+1;\epsilon _2}})\quad \text { and } \quad {\hat{B}}_{\epsilon _1+1;\epsilon _2}={\hat{M}}_1-( f({u}_{_{\epsilon _1;\epsilon _2}})+f({u}_{_{\epsilon _1+1;\epsilon _2}})), \end{aligned}$$
(A.12)
$$\begin{aligned}&{\hat{C}}_{\epsilon _1;\epsilon _2}={\hat{M}}_2+g({u}_{_{\epsilon _1;\epsilon _2}})+g({u}_{_{\epsilon _1;\epsilon _2+1}}) \quad \text { and } \quad {\hat{D}}_{\epsilon _1;\epsilon _2+1}={\hat{M}}_2-( g({u}_{_{\epsilon _1;\epsilon _2}})+g({u}_{_{\epsilon _1;\epsilon _2+1}})). \end{aligned}$$
(A.13)

To prove that the method satisfies the Kruzhkov entropy solution, we rewrite (A.11) as

$$\begin{aligned}&\frac{d}{dt}|u(x,y,t,\epsilon )-K|=sign(u_{_{\epsilon _1;\epsilon _2}}-K)\frac{d}{dt}u(x,y,t,\epsilon )=\nonumber \\&\qquad =sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left\{ \frac{1}{4\epsilon _1}\left[ u_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}+u_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}\right. \right. \nonumber \\&\quad -K({\hat{M}}_1-2f(K)+{\hat{M}}_1+2f(K))\nonumber \\&\qquad \left. \left. -\Big (u_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}+{\hat{A}}_{\epsilon _1;\epsilon _2})-K({\hat{M}}_1-2f(K)+{\hat{M}}_1+2f(K))\Big ) \right] \right. \nonumber \\&\qquad -\frac{1}{16}\left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2} \right. \nonumber \\&\qquad \left. \left. -(u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \right\} \nonumber \\&\qquad +sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left\{ \frac{1}{4\epsilon _2}\left[ u_{_{\epsilon _1;\epsilon _2+1}}{\hat{D}}_{\epsilon _1;\epsilon _2+1}+u_{_{\epsilon _1;\epsilon _2-1}}{\hat{C}}_{\epsilon _1;\epsilon _2-1}\right. \right. \nonumber \\&\quad -K({\hat{M}}_2-2g(K)+{\hat{M}}_2+2g(K))\nonumber \\&\qquad \left. -\Big (u_{_{\epsilon _1;\epsilon _2}}({\hat{D}}_{\epsilon _1;\epsilon _2}+{\hat{C}}_{\epsilon _1;\epsilon _2})-K({\hat{M}}_2-2g(K)+{\hat{M}}_2+2g(K))\Big ) \right] \nonumber \\&\qquad \left. -\frac{1}{16}\left[ (u_y)_{_{\epsilon _1;\epsilon _2+1}}{\hat{D}}_{\epsilon _1;\epsilon _2+1}-(u_y)_{_{\epsilon _1;\epsilon _2-1}}{\hat{C}}_{\epsilon _1;\epsilon _2-1}- (u_y)_{_{\epsilon _1;\epsilon _2}}({\hat{D}}_{\epsilon _1;\epsilon _2}-{\hat{C}}_{\epsilon _1;\epsilon _2})\right] \right\} . \end{aligned}$$
(A.14)

Now, we analyze

$$\begin{aligned}&sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left( u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right) \nonumber \\&\quad =sign(u_{_{\epsilon _1;\epsilon _2}}-K)( (u_{_{\epsilon _1;\epsilon _2}}-K){\hat{M}}_1+(Kf(K)-u_{_{\epsilon _1;\epsilon _2}}f({u}_{_{\epsilon _1-1;\epsilon _2}})+(Kf(K)\nonumber \\&\quad -u_{_{\epsilon _1;\epsilon _2}}f({u}_{_{\epsilon _1;\epsilon _2}}))). \end{aligned}$$
(A.15)

We are interested in studying (A.15). We have the following possibilities between K and \(u_{_{\epsilon _1;\epsilon _2}}\):

  1. 1.

    \(K=u_{_{\epsilon _1;\epsilon _2}}\). In this case, since the numerical method is \(TVNI_\epsilon \) according to the hypothesis of the numerical scheme, the set for which \(K=u_{_{\epsilon _1;\epsilon _2}}\ne u_{_{\epsilon _1-1;\epsilon _2}}\) has null measure in \({{\mathbb {T}}}\times {\mathbb {R}}^+\). Thus, for \(K=u_{_{\epsilon _1;\epsilon _2}}\), we obtain

    $$\begin{aligned}&( (u_{_{\epsilon _1;\epsilon _2}}-K){\hat{M}}_1+(Kf(K)-u_{_{\epsilon _1;\epsilon _2}}f({u}_{{\epsilon -1}})+(Kf(K) -u_{_{\epsilon _1;\epsilon _2}}f({u}_{{\epsilon }}))\\&\quad =(Kf(K)-u_{_{\epsilon _1;\epsilon _2}}f({u}_{{\epsilon -1}})\ne 0 \end{aligned}$$

    only in a null measure set.

  2. 2.

    \(K\ne u_{_{\epsilon _1;\epsilon _2}}\). Since f is a Lipschitzian function, we have that

    $$\begin{aligned}&|Kf(K)-u_{_{\epsilon _1;\epsilon _2}}f({u}_{_{\epsilon _1;\epsilon _2}})+Kf(K)-u_{_{\epsilon _1-1;\epsilon _2}} f({u}_{_{\epsilon _1;\epsilon _2}})|\nonumber \\&\quad \le |Kf(K)-u_{_{\epsilon _1;\epsilon _2}}f({u}_{_{\epsilon _1;\epsilon _2}})|+|Kf(K)-u_{_{\epsilon _1-1;\epsilon _2}}f({u}_{_{\epsilon _1;\epsilon _2}})|\nonumber \\&\quad \le (2(M+Lu_{_{\epsilon _1;\epsilon _2}})|K-{u}_{_{\epsilon _1;\epsilon _2}}|+u_{_{\epsilon _1;\epsilon _2}}L|{u}_{_{\epsilon _1;\epsilon _2}}-{u}_{_{\epsilon _1;\epsilon _2}}|\nonumber \\&\quad \le (2(M+L\mu )|K-{u}_{_{\epsilon _1;\epsilon _2}}|+\mu L|{u}_{_{\epsilon _1;\epsilon _2}}-{u}_{_{\epsilon _1;\epsilon _2}}|. \end{aligned}$$
    (A.16)

    Here, we use that

    $$\begin{aligned}&Kf(K)-{u}_{_{\epsilon _1;\epsilon _2}}f(u_{_{\epsilon _1;\epsilon _2}})=(K-{u}_{_{\epsilon _1;\epsilon _2}})f(K)+{u}_{_{\epsilon _1;\epsilon _2}}(f(K)-f({u}_{_{\epsilon _1;\epsilon _2}})); \\&\qquad Kf(K)-{u}_{_{\epsilon _1;\epsilon _2}}f(u_{_{\epsilon _1-1;\epsilon _2}})=(K-{u}_{_{\epsilon _1;\epsilon _2}})f(K)\\&\qquad +{u}_{_{\epsilon _1;\epsilon _2}}(f(K)-f({u}_{_{\epsilon _1;\epsilon _2}})+(f({u}_{_{\epsilon _1;\epsilon _2}})-f({u}_{_{\epsilon _1-1;\epsilon _2}}))); \end{aligned}$$

    L is the Lipschitzian constant of f; and \(M_1\) denotes the sup of the modulus of f given by Eq. (5.5) and of \(\mu \) given by Eq. (A.9). Here, we also use that the solution satisfies the maximum principle. The states for which \(u(x,y,t,\epsilon )\) is discontinuous is a set of null measure. In the case of the states for which \(u(x,y,t,\epsilon )\) is continuous, we can choose a small enough \(\epsilon \) such that \(|{u}_{_{\epsilon _1;\epsilon _2}}-{u}_{_{\epsilon _1-1;\epsilon _2}}|\le |K-{u}_{_{\epsilon _1;\epsilon _2}}|\).

Thus, up to a null measure set, and for a small enough \(\epsilon \) and taking \({\hat{M}}_1>2M_1+3L_1\mu \), we have

$$\begin{aligned}&sign({u}_{_{\epsilon _1;\epsilon _2}}-K)({\hat{M}}_1({u}_{_{\epsilon _1;\epsilon _2}}-K))={\hat{M}}_1|{u}_{_{\epsilon _1;\epsilon _2}}-K|\nonumber \\&\quad \ge 2Kf(K)-{u}_{_{\epsilon _1;\epsilon _2}}(f({u}_{_{\epsilon _1-1;\epsilon _2}})+f({u}_{_{\epsilon _1;\epsilon _2}})). \end{aligned}$$
(A.17)

This leads to

$$\begin{aligned}&sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left( u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right) = \nonumber \\&\quad =sign(u_{_{\epsilon _1;\epsilon _2}}-K)( (u_{_{\epsilon _1;\epsilon _2}}-K){\hat{M}}_1+(2Kf(K)-u_{{\epsilon _1;\epsilon _2}}(f({u}_{_{\epsilon _1-1;\epsilon _2}})+f({u}_{_{\epsilon _1;\epsilon _2}})))\nonumber \\&\quad =|(u_{_{\epsilon _1;\epsilon _2}}-K){\hat{M}}_1+(2Kf(K)-u_{_{\epsilon _1;\epsilon _2}}(f({u}_{_{\epsilon _1-1;\epsilon _2}}) +f(u_{_{\epsilon _1;\epsilon _2}}))|\nonumber \\&\quad =|u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))|. \end{aligned}$$
(A.18)

The same argument can be used for

$$\begin{aligned}&sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left( u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right) =\nonumber \\&\qquad =sign(u_{_{\epsilon _1;\epsilon _2}}-K)( (u_{_{\epsilon _1;\epsilon _2}}-K){\hat{M}}_1-(f(K)\nonumber \\&\qquad -u_{_{\epsilon _1;\epsilon _2}}f({u}_{_{\epsilon _1-1;\epsilon _2}})-(f(K)-u_{_{\epsilon _1;\epsilon _2}}f({u}_{_{\epsilon _1;\epsilon _2}}))), \end{aligned}$$
(A.19)

and, for a small enough \(\epsilon \) and up to a null measure set, we can write

$$\begin{aligned}&sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left( u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right) =|u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))|. \end{aligned}$$
(A.20)

Performing similar calculations and taking \({\hat{M}}_2>2M_2+3L_2\mu \), where \(M_2\) is given by (5.5), for a small enough \(\epsilon =(\epsilon _1,\epsilon _2)\), we also have

$$\begin{aligned}&sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left( u_{_{\epsilon _1;\epsilon _2}}{\hat{C}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_2+2g(K))\right) =|u_{_{\epsilon _1;\epsilon _2}}{\hat{C}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_2+2g(K))| \end{aligned}$$
(A.21)

and

$$\begin{aligned} sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left( u_{_{\epsilon _1;\epsilon _2}}{\hat{D}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_2+2g(K))\right) =|u_{_{\epsilon _1;\epsilon _2}}{\hat{D}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_2+2g(K))|.\nonumber \\ \end{aligned}$$
(A.22)

Now, consider a small enough \(\epsilon \), and except on a null set. Thus, we use equations (A.18), (A.20), (A.21) and (A.22) in (A.14) to obtain

$$\begin{aligned} \frac{d}{dt}|u(x,y,t,\epsilon )&-K|\le \Psi _{\epsilon _1;\epsilon _2}+\Theta _{\epsilon _1;\epsilon _2}, \end{aligned}$$
(A.23)

where

$$\begin{aligned} \Psi _{\epsilon _1;\epsilon _2}&=\frac{1}{4\epsilon _1} \Big (\left| u_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right| -\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right| \Big ) \nonumber \\&\quad -\frac{1}{4\epsilon _1}\Big (\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right| -\left| u_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right| \Big ) \nonumber \\&\quad -\frac{1}{16}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \end{aligned}$$
(A.24)

and

$$\begin{aligned} \Theta _{\epsilon _1;\epsilon _2}&=\frac{1}{4\epsilon _2} \Big (\left| u_{_{\epsilon _1;\epsilon _2+1}}{\hat{D}}_{\epsilon _1;\epsilon _2+1}-K({\hat{M}}_2-2g(K))\right| -\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{D}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_2-2g(K))\right| \Big ) \nonumber \\&\quad -\frac{1}{4\epsilon _2}\Big (\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{C}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_2+2g(K))\right| -\left| u_{_{\epsilon _1;\epsilon _2-1}}{\hat{C}}_{\epsilon _1;\epsilon _2-1}-K({\hat{M}}_2+2g(K))\right| \Big ) \nonumber \\&\quad -\frac{1}{16}sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ (u_y)_{_{\epsilon _1;\epsilon _2+1}}{\hat{D}}_{\epsilon _1;\epsilon _2+1}-(u_y)_{_{\epsilon _1;\epsilon _2-1}}{\hat{C}}_{\epsilon _1;\epsilon _2-1}- (u_y)_{_{\epsilon _1;\epsilon _2}}({\hat{D}}_{\epsilon _1;\epsilon _2}-{\hat{C}}_{\epsilon _1;\epsilon _2})\right] . \end{aligned}$$
(A.25)

In other to simplify our calculations, we prove our next result for

$$\begin{aligned} \frac{d}{dt}|u(x,y,t,\epsilon )&-K|\le \Psi _{\epsilon _1;\epsilon _2}. \end{aligned}$$
(A.26)

Nevertheless, the result for (A.26) is very similar.

Multiplying inequality (A.26) by the non-negative test function \(\varphi =\varphi (x,y,t)\in C^\infty _0({{\mathbb {T}}}\times [0,T)),T>0\), and integrating by parts, we obtain (notice that the null measure set does not modify the value of the integral)

$$\begin{aligned}&-\int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy-\int _{0}^T\int _{{{\mathbb {T}}}}\left| u(x,y,t,\epsilon )-K\right| \varphi _t(x,y,t)dxdydt\le \nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4\epsilon _1} \Big (|u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-K({\hat{M}}_1-2f(K))|-|u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))|\Big ) \nonumber \\&\quad -\frac{1}{4\epsilon _1}\Big (|u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))|-|u_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}-K({\hat{M}}_1+2f(K))|\Big ) \nonumber \\&\quad -\frac{1}{16}sign(u_{_{\epsilon _1;\epsilon _2}}{-}K) \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}{-}(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}{-} (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}{-}{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\, g\,dx\, dy\,dt. \end{aligned}$$
(A.27)

Note that if we take \(\epsilon _1 \longrightarrow \epsilon _1-1\) in the index, we have that

\(\displaystyle {\left| u_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right| }\) \(\varphi (x,y,t)\) \(\longrightarrow \) \(\displaystyle {\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right| }\) \(\varphi (x-\epsilon _1,y,t)\) and that if we take \(\epsilon _1 \longrightarrow \epsilon _1+1\) in the index, we have that \(\displaystyle {\left| u_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right| }\) \(\varphi (x,y,t)\) \(\longrightarrow \) \(\displaystyle {\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right| }\) \(\varphi (x+\epsilon _1,y,t)\). Thus, inequality (A.27) is written as

$$\begin{aligned}&-\int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy-\int _{0}^T\int _{{{\mathbb {T}}}}\left| u(x,y,t,\epsilon )-K\right| \varphi _t(x,y,t)dxdydt\le \nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4} \Big (-\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K))\right| \Big )\left( \frac{\varphi (x,y,t)-\varphi (x-\epsilon _1,y,t)}{\epsilon _1}\right) dxdydt+ \nonumber \\&\quad +\int _0^T\int _{{{\mathbb {T}}}}\frac{1}{4}\Big (\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}-K({\hat{M}}_1+2f(K))\right| \Big )\left( \frac{\varphi (x+\epsilon _1,y,t)-\varphi (x,y,t)}{\epsilon _1} \right) dxdydt\nonumber \\&\quad -\frac{1}{16}\int _0^T\int _{{{\mathbb {T}}}}\Big \{sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}\right. \nonumber \\&\left. \quad - (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t) dxdydt. \end{aligned}$$
(A.28)

Since \(\varphi \in C^\infty _0({{\mathbb {T}}}\times [0,T))\), Eq. (A.28) becomes

$$\begin{aligned}&-\int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy-\int _{0}^T\int _{{{\mathbb {T}}}}\left| u(x,y,t,\epsilon )-K\right| \varphi _t(x,y,t)dxdydt\le \nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4} \Big (\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2} -K({\hat{M}}_1+2f(K))\right| \nonumber \\&\quad -\left| u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2} -K({\hat{M}}_1-2f(K))\right| \Big )\varphi _x(x,y,t)\Big )dxdydt+ I(\epsilon _1) \nonumber \\&\quad -\frac{1}{16}\int _0^T\int _{{{\mathbb {T}}}}\Big \{sign(u_{_{\epsilon _1;\epsilon _2}}-K)\nonumber \\&\quad \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t)dxdydt. \end{aligned}$$
(A.29)

Here, \(I(\epsilon )\longrightarrow 0\) when \(\epsilon \longrightarrow 0\). Considering that (A.18) and (A.20) are valid, we obtain

$$\begin{aligned}&|u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))|-|u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}-2f(K))|=\nonumber \\&\quad =sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left[ u_{_{\epsilon _1;\epsilon _2}}{\hat{A}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1+2f(K))-(u_{_{\epsilon _1;\epsilon _2}}{\hat{B}}_{\epsilon _1;\epsilon _2}-K({\hat{M}}_1-2f(K)))\right] =\nonumber \\&\quad =sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left[ u_{_{\epsilon _1;\epsilon _2}}(2f({u}_{_{\epsilon _1;\epsilon _2}})+f({u}_{_{\epsilon _1+1;\epsilon _2}})+f({u}_{_{\epsilon _1-1;\epsilon _2}})-4Kf(K)\right] . \end{aligned}$$
(A.30)

By substituting (A.30) in (A.29), we have that

$$\begin{aligned}&-\int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy-\int _{0}^T\int _{{{\mathbb {T}}}}\left| u(x,y,t,\epsilon )-K\right| \varphi _t(x,y,t)dxdydt\le \nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4} \Big (sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left[ u_{_{\epsilon _1;\epsilon _2}}(2f({u}_{_{\epsilon _1;\epsilon _2}})\right. \nonumber \\&\left. \qquad +f({u}_{_{\epsilon _1+1;\epsilon _2}})+f({u}_{_{\epsilon _1-1;\epsilon _2}})-4Kf(K)\right] \Big )\varphi _x(x,y,t)\Big )dxdydt+I(\epsilon _1) \nonumber \\&\qquad -\frac{1}{16}\int _0^T\int _{{{\mathbb {T}}}}\Big \{sign(u_{_{\epsilon _1;\epsilon _2}}-K) \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}\right. \nonumber \\&\left. \qquad - (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t)dxdydt. \end{aligned}$$
(A.31)

To complete our proof, we must analyze

$$\begin{aligned}&\frac{1}{16}\int _0^T\int _{{{\mathbb {T}}}}\Big \{sign(u_{_{\epsilon _1;\epsilon _2}}-K)\nonumber \\&\quad \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t)dxdydt.\nonumber \\ \end{aligned}$$
(A.32)

Considering previous arguments, we know that, since the proposed scheme is \(TVNI_\epsilon \), for each fixed t, there exists only a finite number of x such that \(u_{\epsilon }(x,t,\epsilon )-K\) changes the signal. Then, we split \({{\mathbb {T}}}={\mathbb {S}}_1\cup {\mathbb {S}}_2\cup {\mathbb {S}}_3\) such that

$$\begin{aligned}&x\in {\mathbb {S}}_1\longrightarrow sign(u_{\epsilon }(x,t,\epsilon )-K)<0 \quad \text { and } \quad \end{aligned}$$
(A.33)
$$\begin{aligned}&x\in {\mathbb {S}}_2\longrightarrow sign(u_{\epsilon }(x,t,\epsilon )-K)>0, \end{aligned}$$
(A.34)

and \({\mathbb {S}}_3\) has null measure. Here, \( {\mathbb {S}}_1\) and \( {\mathbb {S}}_2\) consist (each) of a finite number of subintervals. We rewrite the spatial integral on \({{\mathbb {T}}}\) in (A.32) as

$$\begin{aligned}&-\int _{{\mathbb {S}}_1}\Big \{ \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t)dxdydt+\nonumber \\&\quad +\int _{{\mathbb {S}}_2}\Big \{ \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}- (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t)dxdydt. \end{aligned}$$
(A.35)

Now, notice that, due to translations of \(\pm \epsilon \), we have that

$$\begin{aligned}&\int _{{\mathbb {S}}_1}\Big \{ \left[ (u_x)_{_{\epsilon _1+1;\epsilon _2}}{\hat{B}}_{\epsilon _1+1;\epsilon _2}-(u_x)_{_{\epsilon _1-1;\epsilon _2}}{\hat{A}}_{\epsilon _1-1;\epsilon _2}\right. \nonumber \\&\left. \quad - (u_x)_{_{\epsilon _1;\epsilon _2}}({\hat{B}}_{\epsilon _1;\epsilon _2}-{\hat{A}}_{\epsilon _1;\epsilon _2})\right] \Big \}\varphi (x,y,t)dxdydt={\mathbb {O}}(\epsilon _1) \end{aligned}$$
(A.36)

because the terms inside the integral cancel out (actually, they do not cancel out only at the extremes of the subintervals of \( {{\mathbb {S}}_1}\)). The same argument is valid for the integral on \({\mathbb {S}}_2\). Given that the argument is valid for any \(t\in [0,T]\), we have that (A.32) is \({\mathbb {O}}(\epsilon _1)\).

Then, Eq. (A.31) is written as

$$\begin{aligned}&\int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy +\int _{0}^T\int _{{{\mathbb {T}}}}\left| u(x,y,t,\epsilon ) -K\right| \varphi _t(x,y,t)dxdydt+\nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4} \Big (sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left[ u_{_{\epsilon _1;\epsilon _2}}(2f({u}_{_{\epsilon _1;\epsilon _2}}) +f({u}_{_{\epsilon _1+1;\epsilon _2}})\right. \nonumber \\&\left. \quad +f({u}_{_{\epsilon _1-1;\epsilon _2}}) -4Kf(K)\right] \Big )\varphi _x(x,y,t)\Big )dxdydt\ge -I(\epsilon _1)+{\mathbb {O}}(\epsilon _1). \end{aligned}$$
(A.37)

Using (A.23) and performing similar calculations, we obtain

$$\begin{aligned}&\int _{{{\mathbb {T}}}}\left| u_0(x,y)-K\right| \varphi (x,y,0)dxdy +\int _{0}^T\int _{{{\mathbb {T}}}}\left| u(x,y,t,\epsilon ) -K\right| \varphi _t(x,y,t)dxdydt+\nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4} \Big (sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left[ u_{_{\epsilon _1;\epsilon _2}}(2f({u}_{_{\epsilon _1;\epsilon _2}}) +f({u}_{_{\epsilon _1+1;\epsilon _2}})\right. \nonumber \\&\left. \qquad +f({u}_{_{\epsilon _1-1;\epsilon _2}}) -4Kf(K)\right] \Big )\varphi _x(x,y,t)\Big )dxdydt+\nonumber \\&\quad \int _0^T\int _{{{\mathbb {T}}}}\Big \{\frac{1}{4} \Big (sign(u_{_{\epsilon _1;\epsilon _2}}-K)\left[ u_{_{\epsilon _1;\epsilon _2}}(2g({u}_{_{\epsilon _1;\epsilon _2}}) +g({u}_{_{\epsilon _1;\epsilon _2+1}})\right. \nonumber \\&\left. \qquad +g({u}_{_{\epsilon _1;\epsilon _2-1}}) -4Kg(K)\right] \Big )\varphi _y(x,y,t)\Big )dxdydt\ge -I(\epsilon )+{\mathbb {O}}(\epsilon ). \end{aligned}$$
(A.38)

Here, \(u_{_{\epsilon _1;\epsilon _2}}=u(x,y,t,\epsilon )\) and \(I(\epsilon )=I(\epsilon _1)+I(\epsilon _2)\). In  Appendix B, we show that family \(u(x,y,t,\epsilon )\) for \(\epsilon _1>0\) and \(\epsilon _2>0\) and \(\epsilon =(\epsilon _1,\epsilon _2)\) is a pre-compact sequence in \(L^1({{\mathbb {T}}}\times [0,T])\). Let u(xyt) be an accumulation point of family \(u(x,y,t,\epsilon )\). Thus, for a subsequence \(\epsilon _r\), we have that \(u(x,y,t,\epsilon _r)\longrightarrow u(x,y,t)\) when \(r\longrightarrow \infty \) in \(L^1({{\mathbb {T}}}\times [0,T])\). In other words, \(u(x,y,t,\epsilon _r+1)\longrightarrow u(x,y,t)\) and \(u(x,y,t,\epsilon _r-1)\longrightarrow u(x,y,t)\) when point-wise up to a set of null measure. Taking the limit as \(\epsilon =\epsilon _r\longrightarrow 0\) in (A.37), we obtain the entropy relation, remembering that \(I(\epsilon )+{\mathbb {O}}(\epsilon )\longrightarrow 0\) and noticing that \(f({u}_{_{\epsilon _1;\epsilon _2}})\longrightarrow f(u)\), \(f({u}_{_{\epsilon _1\pm 1;\epsilon _2}})\longrightarrow f(u)\), \(g({u}_{_{\epsilon _1;\epsilon _2}})\longrightarrow g(u)\), and \(g({u}_{_{\epsilon _1;\epsilon _2\pm 1}})\longrightarrow g(u)\). Thus, u(xyt) satisfies Eq. (A.8).

In Eq. (A.8), \( K \in [-\mu ,\mu ]\). However, for \(|K|\ge \mu \), note that the inequality that is Eq. (A.8) reduces to the equality (weak solution),

$$\begin{aligned}&\int _{0}^T\int _{{{\mathbb {T}}}}\Big (u(x,y,t)\varphi _t(x,y,t) +u(x,y,t)f(u)\varphi _x(x,y,t)+u(x,y,t)g(u)\varphi _y(x,y,t)\Big )dxdydt\nonumber \\&\quad + \int _{{{\mathbb {T}}}}u_0(x,y)g(x,y,0)dx= 0. \end{aligned}$$

From these results, we obtain that (A.8) holds for all \(K\in {\mathbb {R}}\). Since \(T>0\) and \(\varphi =\varphi (x,y,t)\in C^\infty _0({{\mathbb {T}}}\times [0,T))\) are arbitrary functions, inequality (A.8) leads to solution u(xyt), which is the (unique) entropy solution to (3.1). From the above, it is easy to see from (5.6) that an accumulation point u(xt) of \(u(x,t,\epsilon )\) is unique. This implies that \(u(x,y,t,\epsilon )\) converges to u(xyt) as \(\epsilon \rightarrow 0\) in \(L^1_{loc}({{\mathbb {T}}}\times [0,\infty ))\) because T is arbitrary, which completes the proof. \(\square \) .

Therefore, we prove that the SDLE method (3.3)–(3.7) obtained from (5.6)–(5.7) converges to the entropy solution to (3.1).

Appendix B. The Pre-compactness of Sequence \(u(x,t,\epsilon )\)

Here, for Lemma 1 and Corollary Appendix B.1 below, we use the same notation employed in [13], i.e., \(x=(x,y)\) for the 2D case.

To prove that sequence \(u(x,t,\epsilon )\) is pre-compact, we used some of the results reported in [13]. The first result we need is Lemma 1 in [13].

Lemma 1. Suppose that \(u(x)\in L^1({\mathbb {T}}^n)\), \(h>0\). Then

$$\begin{aligned} \int _{{\mathbb {T}}^n}|u(x)(signu)^h(x)-|u(x)||dx\le 2\omega ^x(h), \end{aligned}$$

where

$$\begin{aligned} \omega ^x(h)=\sup _{|\Delta x|\le h}\int _{{\mathbb {T}}^n}|u(x+\Delta x)-u(x)|dx \end{aligned}$$

is the continuity modulus of u(x) in \(L^1({\mathbb {T}}^n)\).

Here, \({\mathbb {T}}^n\) is the n-dimensional torus. In this study, we are interested in a one-dimensional problem. For \(n=1\), \({\mathbb {T}}^n\) reduces to \({{\mathbb {T}}}\). Since the proof of the previous Lemma does not depend on the scheme, we refer to [13].

Notice that \(\omega ^x(h)\) is a measure of \(TVNI_\epsilon \), as described in Eq. (6.2). Thus, under the same hypothesis of Proposition 6.2, we can prove the following Corollary:

CorollaryAppendix B.1

Let us assume that \(u(x,t,\epsilon )\) is given by scheme (5.6) and satisfies the hypothesis of Proposition 6.2. Then, for all \(t>0\), \(\Delta x \in {\mathbb {R}}\), we have that

$$\begin{aligned} \int _{{\mathbb {T}}}|u(x+\Delta x,t,\epsilon ) -u(x,t,\epsilon )|dx\le \int _{{\mathbb {T}}}| u_0(x+\Delta x,t,\epsilon )-u_0(x,t,\epsilon )|dx\le \omega ^x(|\Delta x|),\nonumber \\ \end{aligned}$$
(B.1)

where

$$\begin{aligned} \omega ^x(|\Delta x|)\le \sup _{|\Delta x|\le h} \int _{{\mathbb {T}}}|u_0(x+\Delta x,t,\epsilon ) -u_0(x,t,\epsilon )|dx \end{aligned}$$

is the continuity modulus of initial data \(u_0(x)\) in \({{\mathbb {T}}}\).

The proof of Corollary Appendix B.1 follows from Proposition 6.2 and the supremum properties of a function. Now, we prove the result to obtain the pre-compactness of sequence \(u(x,y,t,\epsilon )\). The first useful result, similar to that obtained in [13], is

LemmaAppendix B.2

Let us assume that \(\phi (x,y)\in C^1({{\mathbb {T}}})\). Then \(\forall \Delta t>0\),

$$\begin{aligned} \int _{{\mathbb {T}}} (u(x,t+\Delta t,\epsilon ) -u(x,t,\epsilon )\phi (x)dx\le N ||\nabla \phi ||_{\infty } \zeta ({{\mathbb {T}}}) \Delta t. \end{aligned}$$
(B.2)

Here, \(\zeta ({\mathbb {T}})\) is the measure of \({\mathbb {T}}\) and

$$\begin{aligned} N=\left( \frac{{\hat{M}}_1}{2}+M_1\right) \left( \mu +\frac{\epsilon _1}{8}\upsilon _1\right) +\left( \frac{{\hat{M}}_2}{2}+M_2\right) \left( \mu +\frac{\epsilon _2}{8}\upsilon _2\right) \quad \text { and }\quad \mu =||u_0||_\infty . \end{aligned}$$

Proof

Let us denote \(I(t)=\int _{{{\mathbb {T}}}} u(x,t,y,\epsilon )\phi (x)dx \). Differentiating I(t) from t and using (5.6), we have that

$$\begin{aligned} I^\prime (t)=&\int _{{{\mathbb {T}}} }\left\{ -\frac{1}{4\epsilon _1}\left[ \left( b^x_{_{\epsilon _1+1/2;\epsilon _2}} \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}-\left( u_{_{\epsilon _1+1;\epsilon _2}}-\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) +\right. \right. \nonumber \\&+\left( (f(u_{_{\epsilon _1;\epsilon _2}})+f(u_{_{\epsilon _1+1;\epsilon _2}})) \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}} +\left( u_{_{\epsilon _1+1;\epsilon _2}}-\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1+1;\epsilon _2}}\right) \right) \right) \nonumber \\&-\left( b^x_{_{\epsilon _1-1/2;\epsilon _2}}\left( u_{_{\epsilon _1-1;\epsilon _2}} +\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1-1;\epsilon _2}}-\left( u_{_{\epsilon _1;\epsilon _2}}-\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) +\nonumber \\&\left. -\left( (f(u_{_{\epsilon _1-1;\epsilon _2}}) +f(u_{_{\epsilon _1;\epsilon _2}}))\left( u_{_{\epsilon _1-1;\epsilon _2}} +\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1-1;\epsilon _2}} +\left( u_{_{\epsilon _1;\epsilon _2}} -\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) \right] \nonumber \\&-\frac{1}{4\epsilon _2}\left[ \left( b^y_{_{\epsilon _1;\epsilon _2+1/2}} \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}-\left( u_{_{\epsilon _1;\epsilon _2+1}}-\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) +\right. \nonumber \\&+\left( (g(u_{_{\epsilon _1;\epsilon _2}})+g(u_{_{\epsilon _1;\epsilon _2+1}})) \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}} +\left( u_{_{\epsilon _1;\epsilon _2+1}}-\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2+1}}\right) \right) \right) \nonumber \\&-\left( b^y_{_{\epsilon _1;\epsilon _2-1/2}}\left( u_{_{\epsilon _1;\epsilon _2-1}} +\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2-1}}-\left( u_{_{\epsilon _1;\epsilon _2}}-\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) +\nonumber \\&\left. \left. -\left( (g(u_{_{\epsilon _1;\epsilon _2-1}}) +g(u_{_{\epsilon _1;\epsilon _2}}))\left( u_{_{\epsilon _1;\epsilon _2-1}} +\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2-1}} +\left( u_{_{\epsilon _1;\epsilon _2}} -\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}\right) \right) \right) \right] \right\} \nonumber \\&\phi (x,y)dxdy. \end{aligned}$$
(B.3)

Changing the order in the integration variable, we obtain

$$\begin{aligned} I^\prime (t)=&\int _{{{\mathbb {T}}}} \bigg \{\frac{1}{4}\left[ \left( b^x_{_{\epsilon _1+1/2;\epsilon _2}} \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}} -\left( u_{_{\epsilon _1+1;\epsilon _2}} -\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1+1;\epsilon _2}}\right) \right) \right) +\right. \nonumber \\&\left. +\left( (f(u_{_{\epsilon _1;\epsilon _2}}) +f(u_{_{\epsilon _1+1;\epsilon _2}}))\left( u_{_{\epsilon _1;\epsilon _2}} +\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1;\epsilon _2}}+\left( u_{_{\epsilon _1+1;\epsilon _2}} -\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1+1;\epsilon _2}}\right) \right) \right) \right] \nonumber \\&\frac{\phi (x+\epsilon _1,y)-\phi (x,y)}{\epsilon _1}\nonumber \\&+\frac{1}{4}\left[ \left( b^y_{_{\epsilon _1;\epsilon _2+1/2}} \left( u_{_{\epsilon _1;\epsilon _2}}+\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}} -\left( u_{_{\epsilon _1;\epsilon _2+1}} -\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2+1}}\right) \right) \right) +\right. \nonumber \\&\left. \left. +\left( (g(u_{_{\epsilon _1;\epsilon _2}}) +g(u_{_{\epsilon _1;\epsilon _2+1}}))\left( u_{_{\epsilon _1;\epsilon _2}} +\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2}}+\left( u_{_{\epsilon _1;\epsilon _2+1}} -\frac{\epsilon _2}{4}(u_y)_{_{\epsilon _1;\epsilon _2+1}}\right) \right) \right) \right] \right. \nonumber \\&\left. \frac{\phi (x,y+\epsilon _2)-\phi (x,y)}{\epsilon _2}\right\} dxdy. \end{aligned}$$
(B.4)

Since \(I^\prime (t)=G(t)\) implies that \(|I(t+\Delta t)-I(t)|\le max G(t)\Delta t\), we can estimate the right hand side inside of the integral of Eq. (B.4) as

$$\begin{aligned}&|RHS|\le \bigg \{\frac{1}{4}\left[ \left( \left| b^x_{_{\epsilon _1+1/2;\epsilon _2}}\right| \left( \left| u_{_{\epsilon _1;\epsilon _2}}\right| +\frac{\epsilon _1}{4}\left| (u_x)_{_{\epsilon _1;\epsilon _2}}\right| -\left( u_{_{\epsilon _1+1;\epsilon _2}} -\frac{\epsilon _1}{4}(u_x)_{_{\epsilon _1+1;\epsilon _2}}\right) \right) \right) +\right. \nonumber \\&\quad \left. +\left( (f(u_{_{\epsilon _1;\epsilon _2}}) +f(u_{_{\epsilon _1+1;\epsilon _2}}))\left( \left| u_{_{\epsilon _1;\epsilon _2}}\right| +\frac{\epsilon _1}{4}\left| (u_x)_{_{\epsilon _1;\epsilon _2}}\right| +\left( \left| u_{_{\epsilon _1+1;\epsilon _2}}\right| +\frac{\epsilon _1}{4}\left| (u_x)_{_{\epsilon _1+1;\epsilon _2}}\right| \right) \right) \right) \right] \nonumber \\&\left| \frac{\phi (x+\epsilon _1,y)-\phi (x,y)}{\epsilon _1}\right| \nonumber \\&\quad +\frac{1}{4}\left[ \left( \left| b^y_{_{\epsilon _1;\epsilon _2+1/2}}\right| \left( \left| u_{_{\epsilon _1;\epsilon _2}}\right| +\frac{\epsilon _2}{4}\left| (u_y)_{_{\epsilon _1;\epsilon _2}}\right| +\left( \left| u_{_{\epsilon _1;\epsilon _2+1}}\right| +\frac{\epsilon _2}{4}\left| (u_y)_{_{\epsilon _1;\epsilon _2+1}}\right| \right) \right) \right) +\right. \nonumber \\&\quad \left. +\left( (\left| g(u_{_{\epsilon _1;\epsilon _2}})\right| +\left| g(u_{_{\epsilon _1;\epsilon _2+1}})\right| )\left( \left| u_{_{\epsilon _1;\epsilon _2}}\right| +\frac{\epsilon _2}{4}\left| (u_y)_{_{\epsilon _1;\epsilon _2}}\right| +\left( \left| u_{_{\epsilon _1;\epsilon _2+1}}\right| +\frac{\epsilon _2}{4}\left| (u_y)_{_{\epsilon _1;\epsilon _2+1}}\right| \right) \right) \right) \right] \nonumber \\&\left| \frac{\phi (x,y+\epsilon _2)-\phi (x,y)}{\epsilon _2}\right| . \end{aligned}$$
(B.5)

Using that

$$\begin{aligned} \left| \frac{\phi (x\pm \epsilon ,y)-\phi (x)}{\epsilon _1}\right| \le ||\nabla \phi ||_\infty \quad \text { and }\quad \left| \frac{\phi (x,y\pm \epsilon )-\phi (x)}{\epsilon _2}\right| \le ||\nabla \phi ||_\infty , \end{aligned}$$
(B.6)

\(\displaystyle {b^x_{\epsilon _1+1/2;\epsilon _2}={\hat{M}}_1}\), \(\displaystyle {b^y_{\epsilon _1;\epsilon _2+1/2}={\hat{M}}_2}\), \(||(u_x)||_{\infty } \le \upsilon _1\), \(||(u_y)||_{\infty } \le \upsilon _2\), \(||u(x,y,t,\epsilon )||_\infty = \mu \), and \(M_1\) and \(M_2\) are given by (5.5), condition (B.2) is satisfied. \(\square \)

Since we obtained similar estimates in [13], we used Lemma 3 in [13].

Lemma 3. For every \(t\ge 0\), \(\Delta t>0\)

$$\begin{aligned} \int _{{{\mathbb {T}}}}|u(x,t+\Delta t,\epsilon ) -u(x,t,\epsilon )|dx \le \omega ^t(\Delta t), \end{aligned}$$
(B.7)

where \(\omega ^t(\Delta t)=\inf _{h>0}(4\omega ^x(h)+cN\Delta t/h)\), and c is a universal constant.

Note that in \(\omega ^t(\Delta t)\), since this parameter is the infimum, \(\omega ^t(\Delta t)\), for fixed \(\Delta t\), reduces to \(\inf _{h>0}(4\omega ^x(h))\).

Moreover, since \(\omega ^x(h)\longrightarrow 0\) as \(h\longrightarrow 0\) and does not depend on \(\epsilon \) (based on previous results), family \(u(x,t,\epsilon )\) is uniformly bounded and equicontinuous in \(L^1({{\mathbb {T}}}\times [0,T])\) for every \(T>0\). Thus, \(u(x,t,\epsilon )\) is a pre-compact sequence in \(L^1({{\mathbb {T}}}\times [0,T])\), which implies that we can extract a sequence \(\epsilon _k\longrightarrow 0\) such that \(u_k(x,y,t)=u(x,y,t,\epsilon _k)\longrightarrow u(x,y,t)\) as \(k\longrightarrow \infty \) in \(L_{loc}^1({{\mathbb {T}}}\times [0,\infty ])\).

Appendix C. Proof that if u(xyt) is the Solution to Eq. (A.7), then \(u(x,y,t)-N\) is the Solution to Eq. (3.1)

Here, we give the proof considering the classical weak solution defined in \({\mathbb {R}}^2\times {\mathbb {R}}^+\). The proof is similar for the domain \({\mathbb {T}}\times {\mathbb {R}}^+\). First, we consider the weak solution to (A.7) to be

$$\begin{aligned}&\int _{{\mathbb {R}}^+}\int _{{\mathbb {R}}\times {\mathbb {R}}} \left( u\varphi _t+H(u-N)\varphi _x+G(u-N)\varphi _y\right) dxdydt\nonumber \\&\quad +\int _{{\mathbb {R}}\times {\mathbb {R}}} (u_0(x,y)+N)\varphi (x,y,0)dxdy=0, \end{aligned}$$
(C.1)

where \(\varphi =\varphi (x,y,t)\) is a test function with support in \({\mathbb {R}}^2\times {\mathbb {R}}^+\). If we consider the domain \({\mathbb {T}}\times {\mathbb {R}}^+\), the support of \(\varphi \) lies in domain \({\mathbb {T}}\times {\mathbb {R}}^+\).

By substituting \(u-N=U\) in (C.1), we obtain

$$\begin{aligned}&\int _{{\mathbb {R}}^+}\int _{{\mathbb {R}}\times {\mathbb {R}}} \left( (U+N)\varphi _t+H(U)\varphi _x+G(U)\varphi _y\right) dxdydt\nonumber \\&\quad +\int _{{\mathbb {R}}\times {\mathbb {R}}} (u_0(x,y)+N)\varphi (x,y,0)dxdy\nonumber \\ \quad&= \int _{{\mathbb {R}}^+}\int _{{\mathbb {R}}\times {\mathbb {R}}} \left( U\varphi _t+H(U)\varphi _x+G(U)\varphi _y\right) dxdydt\nonumber \\&\quad +\int _{{\mathbb {R}}\times {\mathbb {R}}} (u_0(x,y)+N)\varphi (x,y,0)dxdy\nonumber \\&\qquad +\int _{{\mathbb {R}}^+}\int _{{\mathbb {R}}\times {\mathbb {R}}}N\varphi _t dxdydt =0. \end{aligned}$$
(C.2)

Using Fubini’s theorem for the integral and using that \(\varphi (x,y,t)\) has compact support, then

$$\begin{aligned} \int _{{\mathbb {R}}^+}\int _{{\mathbb {R}}\times {\mathbb {R}}}N\varphi _t dxdydt= -\int _{{\mathbb {R}}\times {\mathbb {R}}}N\varphi (x,y,0)dxdy. \end{aligned}$$
(C.3)

By substituting (C.3) in (C.2), we get

$$\begin{aligned} \int _{{\mathbb {R}}^+}\int _{{\mathbb {R}}\times {\mathbb {R}}} \left( U\varphi _t+H(U)\varphi _x+G(U)\varphi _y\right) dxdt+\int _{{\mathbb {R}}\times {\mathbb {R}}} u_0(x,y)\varphi (x,y,0)dx=0,\qquad \end{aligned}$$
(C.4)

which is the solution to (3.1). The previous calculations are reversible, i.e., (C.4) in (C.1). Thus, if u(xyt) is the solution to (3.1), then \(u(x,y,t)+N\) is the solution to (A.7), and the reciprocal is true.

A similar calculation can be performed to prove that the entropic solutions are the same.

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Abreu, E., François, J., Lambert, W. et al. A Class of Positive Semi-discrete Lagrangian–Eulerian Schemes for Multidimensional Systems of Hyperbolic Conservation Laws. J Sci Comput 90, 40 (2022). https://doi.org/10.1007/s10915-021-01712-8

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