Convergence via OSLC of the Godunov scheme for a scalar conservation law with time and space flux discontinuities

Abstract

This paper deals with a Godunov scheme as applied to a scalar conservation law whose flux has discontinuities in both space and time. We extend the definition of vanishing viscosity solution of Karlsen and Towers (J Hyperbolic Differ Equ 14:671–702, 2017) (which applies to a flux with a spatial discontinuity) in order to accommodate the addition of temporal flux discontinuities, and prove that this extended definition implies uniqueness. We prove convergence of the Godunov approximations to the unique vanishing viscosity solution as the mesh size converges to zero, thus establishing well-posedness for the problem. The novel aspect of this paper is the use of a discrete one-sided Lipschitz condition (OSLC) in the discontinuous flux setting. In the classical setting where flux discontinuities are not present, the OSLC is well known to produce an immediate regularizing effect, with a local spatial variation bound resulting at any positive time. We show that the OSLC also produces a regularizing effect at any finite distance from the spatial flux discontinuity. This regularizing effect is not materially affected by temporal flux discontinuities. When combined with a Cantor diagonal argument, these regularizing effects imply convergence of the Godunov approximations. With this new method it is possible to forgo certain assumptions about the flux that seem to be required when using two commonly used convergence methods.

Proof of Lemma 4.2

In this section we prove Lemma 4.2. In the process we prove that the lemma also holds if the Engquist–Osher flux is substituted for the Godunov flux. Our proof is an adaptation of a more global version of the lemma that is found in [31]. We require a local version in order to accommodate the spatial flux discontinuity. Reference [46] also contains a global version of this result applied to several numerical fluxes in addition to the Godunov flux, and [49] has a global version for the Lax–Friedrichs scheme.

Proof

Part 1. Preliminaries We start by presenting a calculation appearing in [31] showing that

$$\begin{aligned} \varDelta _+U_j^n, D_j^n \le {{1} \over {\lambda \mu }}, \quad j \in \mathbb {Z}, \quad n=0,1, \ldots , N. \end{aligned}$$

(A.1)

To this end, let q denote any of the flux functions $f^m(u), g^m(u)$, $m=0,\ldots ,M$. In what follows we will use the prime notation $q'(u)=q_u(u)$, $q''(u)=q_{uu}(u)$. By Lemma 4.1, we can assume that $U_j^n \in [\underline{u},\overline{u}]$, and so we can apply the convexity assumption (1.3) and the CFL condition (3.8). This yields

$$\begin{aligned} \begin{aligned} \lambda \mu \left( U_{j+1}^n - U_j^n \right)&\le \lambda \int _{U_j^n}^{U_{j+1}^n} q''(\xi ) \, d \xi \\&= \lambda \left( q'(U_{j+1}^n) - q'(U_{j}^n) \right) \le 2\lambda L_q \le 1, \end{aligned} \end{aligned}$$

and (A.1) follows directly from this.

For the rest of the proof, $j \in {\hat{\mathbb {Z}}}$, $n\in \{0,\ldots ,N\}$ are fixed. Due to the assumption $j \in {\hat{\mathbb {Z}}}$, the interface at $x=0$ is not involved. Thus we may assume the numerical flux is either ${\bar{f}}^m$ for ${\bar{g}}^m$ for some fixed m. We denote this numerical flux by ${\bar{q}}$, and the underlying flux ($f^m$ or $g^m$) by q. By the convexity assumption, there is a unique $u^* \in [\underline{u},\overline{u}]$ such that $u^* = \text {argmin}_{w \in [\underline{u},\overline{u}]} q(w)$. It follows that $\mathrm {sign}(q'(u)) = \mathrm {sign}(u-u^*)$. Without loss of generality, take $q(u^*) = 0$. Define $\psi (z;a) = z - a z^2$, where $a = \lambda \mu /4>0$. We have

$$\begin{aligned}&\psi (z;a) \ge 0 \quad \hbox {for } z \in [0, 1/a], \nonumber \\&\quad z\mapsto \psi (z;a)\text { is nondecreasing for }z \le 1/(2a) = 2/(\lambda \mu ). \end{aligned}$$

(A.2)

We must prove (4.3), which can be stated in terms of $\psi$:

$$\begin{aligned} D_j^{n+1} \le \psi (\max (D_{j-1}^n,D_j^n,D_{j+1}^n); a). \end{aligned}$$

(A.3)

Note that by (A.1) $\max (D_{j-1}^n,D_j^n,D_{j+1}^n) \le 1/(4a)$, allowing us to use (A.2).

We claim that we can assume that $\varDelta _+ U^n_{j+1}\ge 0$, $\varDelta _+ U^n_{j-1} \ge 0$. To prove the claim suppose that

$$\begin{aligned} \text {inequality }(A.3)\text { holds when } \varDelta _+U_{j-1}^n, \varDelta _+U_{j+1}^n \ge 0. \end{aligned}$$

(A.4)

The marching formula (3.4) can be expressed as

$$\begin{aligned} U_j^{n+1} = G(U_{j-1}^n,U_j^n,U_{j+1}^n), \end{aligned}$$

where G is nondecreasing in all three arguments; this follows in a straightforward manner from the monotonicity of the numerical flux ${\bar{q}}$ along with the CFL condition (3.8). Define

$$\begin{aligned} {\tilde{U}}_{j-1}^{n} = U_{j-1}^n \wedge U_j^n, \quad {\tilde{U}}_{j}^{n} = U_j^n, \quad {\tilde{U}}_{j+1}^{n} = U_{j+1}^n, \quad {\tilde{U}}_{j+2}^{n} = U_{j+2}^n \vee U_{j+1}^n. \end{aligned}$$

and let

$$\begin{aligned} {\tilde{U}}_j^{n+1} = G({\tilde{U}}_{j-1}^n,{\tilde{U}}_j^n,{\tilde{U}}_{j+1}^n), \quad {\tilde{U}}_{j+1}^{n+1} = G({\tilde{U}}_{j}^n,{\tilde{U}}_{j+1}^n,{\tilde{U}}_{j+2}^n). \end{aligned}$$

By the monotonicity property of G we have

$$\begin{aligned} \begin{aligned} \varDelta _+U_j^{n+1}&= G(U_{j}^n,U_{j+1}^n,U_{j+2}^n) - G(U_{j-1}^n,U_j^n,U_{j+1}^n)\\&\le G({\tilde{U}}_{j}^n,{\tilde{U}}_{j+1}^n,{\tilde{U}}_{j+2}^n ) - G({\tilde{U}}_{j-1}^n,{\tilde{U}}_j^n,{\tilde{U}}_{j+1}^n)\\&= \varDelta _+{\tilde{U}}_j^{n+1}. \end{aligned} \end{aligned}$$

(A.5)

Let ${\tilde{D}}_{j+i}^n = (\varDelta _+ {\tilde{U}}_{j+i}^n )_+$, $i=-1,0,1$. We have $0 \le {\tilde{D}}_{j+i}^n = D_{j+i}^n \le 1/(2a)$, $i=-1,0,1$, so (A.2) still applies. We also have $\varDelta _+ {\tilde{U}}_{j+1}^n \ge 0$, $\varDelta _+ {\tilde{U}}_{j-1}^n \ge 0$, so (A.4) implies that

$$\begin{aligned} {\tilde{D}}_j^{n+1} \le \psi \left( \max ({\tilde{D}}_{j-1}^n,{\tilde{D}}_j^n,{\tilde{D}}_{j+1}^n );a\right) . \end{aligned}$$

(A.6)

By (A.5) we have

$$\begin{aligned} D_j^{n+1} \le {\tilde{D}}_j^{n+1}, \end{aligned}$$

(A.7)

and since ${\tilde{D}}_{j+i}^n = D_{j+i}^n$, $i=-1,0,1$,

$$\begin{aligned} \psi \left( \max ({D}_{j-1}^n,{D}_j^n,{D}_{j+1}^n) ;a\right) =\psi \left( \max ({\tilde{D}}_{j-1}^n,{\tilde{D}}_j^n,{\tilde{D}}_{j+1}^n) ;a\right) . \end{aligned}$$

(A.8)

The proof of the claim is completed by combining (A.6), (A.7), and (A.8). We now continue the proof under the assumption that $\varDelta _+ U_{j-1}^n \ge 0$, $\varDelta _+ U_{j+1}^n \ge 0$.

Part 2. Proof for the Engquist–Osher (EO) flux It is convenient to first prove the lemma when the EO flux ${\bar{p}}$ is substituted for the Godunov flux ${\bar{q}}$. After that the proof for the Godunov flux requires only a slight modification. Under the assumptions stated about the flux q, the EO flux has the form

$$\begin{aligned} {\bar{p}}(v,u)= & {} q_-(v) + q_+(u),\text { where }\nonumber \\&\quad q_-(v):= q(v \wedge u^*), \quad q_+(u):= q(u \vee u^*). \end{aligned}$$

(A.9)

We will use the following formulas which follow readily from the definitions of $q_{\pm }$ in (A.9):

$$\begin{aligned}&q_-(b) - q_-(a) \ge q_-'(a)\left( b\wedge u^* - a\wedge u^* \right) \nonumber \\&\quad +\,{\mu \over 2} \left( b\wedge u^* - a\wedge u^* \right) ^2,\nonumber \\&q_+(b) - q_+(a) \ge q_+'(a)\left( b\vee u^* - a\vee u^* \right) \nonumber \\&\quad +\,{\mu \over 2} \left( b\vee u^* - a\vee u^* \right) ^2. \end{aligned}$$

(A.10)

Starting from

$$\begin{aligned} U_j^{n+1} = U_j^n - \lambda \varDelta _- {\bar{p}}(U_{j+1}^n,U_j^n), \end{aligned}$$

then differencing and applying (A.9) the result is

$$\begin{aligned} \varDelta _+ U_j^{n+1} = \varDelta _+ U_j^n -\lambda \varDelta _+\varDelta _- q_-(U_{j+1}^n) - \lambda \varDelta _+\varDelta _- q_+(U_{j}^n). \end{aligned}$$

(A.11)

The second differences appearing in (A.11) can be expressed in the following form:

$$\begin{aligned} \varDelta _+\varDelta _- q_-(U_{j+1}^n)= & {} \left( q_-(U_{j+2}^n) - q_-(U_{j+1}^n) \right) \nonumber \\&\quad + \left( q_-(U_{j}^n) - q_-(U_{j+1}^n) \right) ,\nonumber \\ \varDelta _+\varDelta _- q_+(U_{j}^n)= & {} \left( q_+(U_{j+1}^n) - q_+(U_{j}^n) \right) \nonumber \\&\quad + \left( q_+(U_{j-1}^n) - q_+(U_{j}^n) \right) . \end{aligned}$$

(A.12)

Let

$$\begin{aligned} \varDelta _+ U^-_{j} = U_{j+1}^n \wedge u^* - U_{j}^n \wedge u^*, \quad \varDelta _+ U^+_{j} = U_{j+1}^n \vee u^* - U_{j}^n \vee u^*. \end{aligned}$$

Starting from (A.12) and applying (A.10) we find that

$$\begin{aligned} \varDelta _+\varDelta _- q_-(U_{j+1}^n)\ge & {} q'_-(U_{j+1}^n) \varDelta _+ U^-_{j+1} + {\mu \over 2} ( \varDelta _+ U^-_{j+1})^2\nonumber \\&\quad - \,q'_-(U_{j+1}^n) \varDelta _+ U^-_{j} + {\mu \over 2} ( \varDelta _+ U^-_{j})^2,\nonumber \\ \varDelta _+\varDelta _- q_+(U_{j}^n)\ge & {} q'_+(U_{j}^n) \varDelta _+ U^+_{j} + {\mu \over 2} ( \varDelta _+ U^+_{j})^2 \nonumber \\&\quad -\, q'_+(U_{j}^n) \varDelta _+ U^+_{j-1} + {\mu \over 2} ( \varDelta _+ U^+_{j-1})^2. \end{aligned}$$

(A.13)

Substituting (A.13) into (A.11) yields

$$\begin{aligned} \varDelta _+ U_j^{n+1}\le & {} \varDelta _+U_j^n - \lambda q'_-(U_{j+1}^n) \varDelta _+ U^-_{j+1} - {{\lambda \mu } \over 2} ( \varDelta _+ U^-_{j+1})^2 \nonumber \\&\quad +\, \lambda q'_-(U_{j+1}^n) \varDelta _+ U^-_{j} - {{\lambda \mu } \over 2} ( \varDelta _+ U^-_{j})^2\nonumber \\&\quad -\,\lambda q'_+(U_{j}^n) \varDelta _+ U^+_{j} - {{\lambda \mu } \over 2} ( \varDelta _+ U^+_{j})^2 \nonumber \\&\quad +\,\lambda q'_+(U_{j}^n) \varDelta _+ U^+_{j-1} - {{\lambda \mu } \over 2} ( \varDelta _+ U^+_{j-1})^2. \end{aligned}$$

(A.14)

From here the proof reduces to four cases, depending on the ordering of $u^*, U_j^n, U_{j+1}^n$.

Case 1 $u^* \le U_j^n, U_{j+1}^n$. We are assuming that $U^n_{j+2} \ge U^n_{j+1}$, and so we also have $u^* \le U_j^n, U_{j+1}^n, U_{j+2}^n$. Inequality (A.14) becomes

$$\begin{aligned} \varDelta _+ U_j^{n+1}\le & {} \varDelta _+U_j^n -\lambda q'(U_{j}^n) \varDelta _+ U^n_{j} - {{\lambda \mu } \over 2} ( \varDelta _+ U^n_{j})^2 \nonumber \\&\quad +\,\lambda q'(U_{j}^n) \varDelta _+ U^+_{j-1} - {{\lambda \mu } \over 2} ( \varDelta _+ U^+_{j-1})^2 \nonumber \\\le & {} \left( 1 - \lambda q'(U_{j}^n)\right) \varDelta _+ U^n_{j} + \lambda q'(U_{j}^n) \varDelta _+ U^+_{j-1} \nonumber \\&\quad -\, {{\lambda \mu } \over 4} \left( ( \varDelta _+ U^n_{j})^2 + ( \varDelta _+ U^+_{j-1})^2\right) . \end{aligned}$$

(A.15)

Due to the CFL condition, the first two terms are a convex combination of $\varDelta _+ U^n_{j}, \varDelta _+ U^+_{j-1}$, and

$$\begin{aligned} ( \varDelta _+ U^n_{j})^2 + ( \varDelta _+ U^+_{j-1})^2 \ge \left( \max \left( \varDelta _+ U^n_{j} , \varDelta _+ U^+_{j-1}\right) \right) ^2. \end{aligned}$$

Combining these observations with (A.15), and recalling $a= \lambda \mu /4$, results in

$$\begin{aligned} \begin{aligned} \varDelta _+ U_j^{n+1}&\le \max \left( \varDelta _+ U^n_{j}, \varDelta _+ U^+_{j-1}\right) - a \left( \max \left( \varDelta _+ U^n_{j} , \varDelta _+ U^+_{j-1}\right) \right) ^2.\quad \end{aligned} \end{aligned}$$

(A.16)

It follows from $U_{j-1}^n \le U_j^n$ that $0 \le \varDelta _+ U^+_{j-1} \le \varDelta _+U_{j-1}^n$, and then by monotonicity of $\psi (\cdot ; a)$ the inequality (A.16) becomes

$$\begin{aligned} \begin{aligned} \varDelta _+ U_j^{n+1}&\le \max \left( \varDelta _+ U^n_{j}, \varDelta _+ U^n_{j-1}\right) - a \left( \max \left( \varDelta _+ U^n_{j} , \varDelta _+ U^n_{j-1}\right) \right) ^2. \end{aligned} \end{aligned}$$

Two more applications of monotonicity of $\psi$ give

$$\begin{aligned} \varDelta _+ U_j^{n+1}&\le \max \left( D_{j-1}^n,D_j^n,D_{j+1}^n\right) - a \left( \max \left( D_{j-1}^n,D_j^n,D_{j+1}^n \right) \right) ^2\nonumber \\&= \psi \left( \max \left( D_{j-1}^n,D_j^n,D_{j+1}^n\right) ;a \right) . \end{aligned}$$

(A.17)

Finally, we are guaranteed that the right side of (A.17) is nonnegative, so the left side of (A.17) can be replaced by $D_j^{n+1}$, and the proof of Case 1 is complete.

Case 2 $U_j^n, U_{j+1}^n \le u^*$. The proof of this case is similar to Case 1 and we omit it.

Case 3 $U_j^n<u^* < U_{j+1}^n$. With the assumption $\varDelta _+ U_{j-1}^n \ge 0$, $\varDelta _+ U_{j+1}^n \ge 0$, we also have $U_{j-1}^n \le U_j^n<u^* < U_{j+1}^n \le U_{j+2}^n$. Using this ordering, (A.14) becomes

$$\begin{aligned} \begin{aligned} \varDelta _+ U_j^{n+1}&\le \varDelta _+ U_j^n - {{\lambda \mu }\over 2} \left( (U_{j+1}^n - u^*)^2 + (u^* - U_{j}^n)^2\right) \\&\le \varDelta _+ U_j^n - {{\lambda \mu } \over 4} (\varDelta _+U_j^n)^2\\&= \psi (\varDelta _+ U_j^n;a), \end{aligned} \end{aligned}$$

where we have used the inequality $(\alpha +\beta )^2 \le 2(\alpha ^2+\beta ^2)$. The proof of this case is completed by using the monotonicity property of $\psi$ as in the proof of Case 1.

Case 4 $U_{j+1}^n<u^* < U_{j}^n$. Using the identity $\varDelta _+ U_j^n = \varDelta _+U_j^+ + \varDelta _+U_j^-$, the inequality (A.14) becomes

$$\begin{aligned}&\varDelta _+ U_j^{n+1} \le \left( 1 + \lambda q'_-(U_{j+1}^n)\right) \varDelta _+ U^-_{j} - \lambda q'_-(U_{j+1}^n) \varDelta _+ U^-_{j+1} \nonumber \\&\quad - {{\lambda \mu } \over 2}\left( ( \varDelta _+ U^-_{j})^2 + \varDelta _+ U^-_{j+1})^2 \right) \nonumber \\&\quad +\left( 1 - \lambda q'_+(U_{j}^n)\right) \varDelta _+ U^+_{j} + \lambda q'_+(U_{j}^n) \varDelta _+ U^+_{j-1} \nonumber \\&\quad - {{\lambda \mu } \over 2}\left( ( \varDelta _+ U^+_{j})^2 + ( \varDelta _+ U^+_{j-1})^2\right) . \end{aligned}$$

(A.18)

The CFL condition, along with $U_{j+1}^n<u^* < U_{j}^n$, results in

$$\begin{aligned} \left( 1 + \lambda q'_-(U_{j+1}^n)\right) \varDelta _+ U^-_{j} \le 0, \left( 1 - \lambda q'_+(U_{j}^n)\right) \varDelta _+ U^+_{j} \le 0, \end{aligned}$$

and so we can replace (A.18):

$$\begin{aligned} \varDelta _+ U_j^{n+1}\le & {} - \lambda q'_-(U_{j+1}^n) \varDelta _+ U^-_{j+1} - {{\lambda \mu } \over 2}\left( ( \varDelta _+ U^-_{j})^2 + \varDelta _+ U^-_{j+1})^2 \right) \nonumber \\&\quad +\, \lambda q'_+(U_{j}^n) \varDelta _+ U^+_{j-1} - {{\lambda \mu } \over 2}\left( ( \varDelta _+ U^+_{j})^2 + ( \varDelta _+ U^+_{j-1})^2\right) \nonumber \\\le & {} - \lambda q'_-(U_{j+1}^n) \varDelta _+ U^-_{j+1} + \lambda q'_+(U_{j}^n) \varDelta _+ U^+_{j-1} \nonumber \\&\quad -{{\lambda \mu } \over 4}\left( (\varDelta _+ U^-_{j+1})^2 + ( \varDelta _+ U^+_{j-1})^2 \right) . \end{aligned}$$

(A.19)

As a result of our assumption that $\varDelta _+ U_{j-1}^n, \varDelta _+ U_{j+1}^n \ge 0$, we have

$$\begin{aligned} 0 \le \varDelta _+ U^-_{j+1} \le \varDelta _+ U_{j+1}^n, \quad 0 \le \varDelta _+ U^+_{j-1} \le \varDelta _+ U_{j-1}^n. \end{aligned}$$

(A.20)

Invoking the CFL condition again, from (A.19) and (A.20), along with $a=\lambda \mu /4$, we get

$$\begin{aligned} \begin{aligned} \varDelta _+ U_j^{n+1}&\le {1 \over 2} \left( \varDelta _+ U^-_{j+1} + \varDelta _+ U^+_{j-1} \right) - a\left( (\varDelta _+ U^-_{j+1})^2 + ( \varDelta _+ U^+_{j-1})^2 \right) \\&\le \max (\varDelta _+ U^-_{j+1}, \varDelta _+ U^+_{j-1}) - a\left( \max (\varDelta _+ U^-_{j+1} , \varDelta _+ U^+_{j-1})\right) ^2 .\\ \end{aligned} \end{aligned}$$

The proof of this case is completed by using the monotonicity of $\psi (\cdot ;a)$.

Part 3. Proof for the Godunov flux Under the assumptions stated at the beginning of this appendix, the Godunov flux ${\bar{q}}(v,u)$ is identical to the EO flux ${\bar{p}}(v,u)$, except for the case where $v<u^*<u$, and in that case

$$\begin{aligned} {\bar{q}}(v,u) = \max (q(v),q(u)) \le q(v)+q(u) = {\bar{p}}(v,u). \end{aligned}$$

Thus in all cases we have

$$\begin{aligned} {\bar{q}}(v,u) \le {\bar{p}}(v,u). \end{aligned}$$

(A.21)

For the Godunov scheme the differences evolve according to

$$\begin{aligned} \varDelta _+U_j^{n+1} = \varDelta _+U_j^n - \lambda \left( {\bar{q}}(U_{j+2}^n,U_{j+1}^n) -2{\bar{q}}(U_{j+1}^n,U_j^n) +{\bar{q}}(U_{j}^n,U_{j-1}^n) \right) . \end{aligned}$$

We assumed that $U_{j+1}^n \le U_{j+2}^n$, $U_{j-1}^n \le U_{j}^n$, and so

$$\begin{aligned} {\bar{q}}(U_{j+2}^n,U_{j+1}^n) = {\bar{p}}(U_{j+2}^n,U_{j+1}^n), \quad {\bar{q}}(U_{j}^n,U_{j-1}^n) = {\bar{p}}(U_{j}^n,U_{j-1}^n). \end{aligned}$$

Applying these identities, along with (A.21) for the $2{\bar{q}}(U_{j+1}^n,U_j^n)$ term, we have

$$\begin{aligned} \varDelta _+U_j^{n+1} \le \varDelta _+U_j^n - \lambda \left( {\bar{p}}(U_{j+2}^n,U_{j+1}^n) -2{\bar{p}}(U_{j+1}^n,U_j^n) +{\bar{p}}(U_{j}^n,U_{j-1}^n) \right) . \end{aligned}$$

The conclusion now follows directly from Part 2. $\square$