Convergence of a Godunov scheme to an Audusse–Perthame adapted entropy solution for conservation laws with BV spatial flux

Abstract

In this article we consider the initial value problem for a scalar conservation law in one space dimension with a spatially discontinuous flux. There may be infinitely many flux discontinuities, and the set of discontinuities may have accumulation points. Thus the existence of traces cannot be assumed. In Audusse and Perthame (Proc R Soc Edinb Sect A 135:253–265, 2005) proved a uniqueness result that does not require the existence of traces, using adapted entropies. We generalize the Godunov-type scheme of Adimurthi et al. (SIAM J Numer Anal 42(1):179–208, 2004) for this problem with the following assumptions on the flux function, (i) the flux is BV in the spatial variable and (ii) the critical point of the flux is BV as a function of the space variable. We prove that the Godunov approximations converge to an adapted entropy solution, thus providing an existence result, and extending the convergence result of Adimurthi, Jaffré and Gowda.

Appendix

Let $g:\mathbb {R}\rightarrow \mathbb {R}$ be defined as $g(x)=\left| x\right|$. Suppose u satisfies entropy condition (12) then set $v(x,t)=\Psi (u(x,t),x)$ where $\Psi$ is as in (10). We denote inverse of the map $\zeta \mapsto \Psi (\zeta ,x)$ by $\alpha (\cdot ,x)$. Then we have

$$\begin{aligned} \int \limits _{0}^{\infty }\int \limits _{\mathbb {R}}\left[ \left| \alpha (v(x,t),x)-\alpha (k,x)\right| \frac{\partial \phi }{\partial t}+sgn(v-k)(g(v)-g(k))\frac{\partial \phi }{\partial x}\right] \,dxdt\ge 0\nonumber \\ \end{aligned}$$

(93)

for any $k\in \mathbb {R}$ and $0\le \phi \in C_0^{\infty }(\mathbb {R}\times \mathbb {R}_+)$.

Lemma 5.1

[18] Let $v_1,v_2\in L^{\infty }(\mathbb {R}\times \mathbb {R}_+)$ be two functions satisfying (93). Then we have

$$\begin{aligned}&\frac{\partial }{\partial t}\left| \alpha (v_1(x,t),x)-\alpha (v_2(x,t),x)\right| +\frac{\partial }{\partial x}sgn(v_1-v_2)(g(v_1)-g(v_2))\nonumber \\&\quad \le 0 \text{ in } \mathcal {D}^{\prime }(\mathbb {R}\times \mathbb {R}_+). \end{aligned}$$

(94)

Proof

For $0\le \phi \in C_c^{\infty }(\mathbb {R}_+\times \mathbb {R})$ and $0\le \psi \in C_c^{\infty }(\mathbb {R}_+\times \mathbb {R})$ we have

$$\begin{aligned}&\int \limits _{0}^{\infty }\int \limits _{\mathbb {R}}\left[ \left| \alpha (v_1(x,t),x)-\alpha (k,x)\right| \frac{\partial \phi }{\partial t}\right. \nonumber \\&\quad \left. +sgn(v_1-k)(g(v_1)-g(k))\frac{\partial \phi }{\partial x}\right] \,dxdt\ge 0 \end{aligned}$$

(95)

and

$$\begin{aligned}&\int \limits _{0}^{\infty }\int \limits _{\mathbb {R}}\left[ \left| \alpha (v_2(y,s),y)-\alpha (l,y)\right| \frac{\partial \psi }{\partial s}\right. \nonumber \\&\left. +sgn(v_2-l)(g(v_2)-g(l))\frac{\partial \psi }{\partial y}\right] \,dyds\ge 0. \end{aligned}$$

(96)

Fix a $\Phi \in C_c^{\infty }(\mathbb {R}\times \mathbb {R}_+)$. Let $\eta _\epsilon$ be Friedrichs mollifiers. Consider

$$\begin{aligned} \phi (x,t)= & {} \Phi \left( x,t\right) \eta _\epsilon \left( {y-x}\right) \eta _\delta \left( {s-t}\right) , \end{aligned}$$

(97)

$$\begin{aligned} \psi (y,s)= & {} \Phi \left( x,t\right) \eta _\epsilon \left( {y-x}\right) \eta _\delta \left( {s-t}\right) . \end{aligned}$$

(98)

Putting $k=v_2(y,s)$ and $l=v_1(x,t)$ in (95) and (96) respectively and adding the resultants we get

$$\begin{aligned}&\int \limits _{\mathbb {R}\times \mathbb {R}_+}\int \limits _{\mathbb {R}\times \mathbb {R}_+}P_1(x,t,y,s)\frac{\partial }{\partial t}(\Phi (x,t)\eta _\epsilon (y-x)\eta _\delta (s-t))\,dtdxdsdy\nonumber \\&\quad +\int \limits _{\mathbb {R}\times \mathbb {R}_+}\int \limits _{\mathbb {R}\times \mathbb {R}_+}P_2(x,t,y,s)\frac{\partial }{\partial s}(\Phi (x,t)\eta _\epsilon (y-x)\eta _\delta (s-t))\,dtdxdsdy\nonumber \\&\quad +\int \limits _{\mathbb {R}\times \mathbb {R}_+}\int \limits _{\mathbb {R}\times \mathbb {R}_+}Q(x,t,y,s)\eta _\epsilon (y-x)\eta _\delta (s-t)\frac{\partial }{\partial x}\Phi (x,t),dtdxdsdy\ge 0\nonumber \\ \end{aligned}$$

(99)

where

$$\begin{aligned} P_1(x,t,y,s):= & {} \left| \alpha (v_1(x,t),x)-\alpha (v_2(y,s),x)\right| ,\\ P_2(x,t,y,s):= & {} \left| \alpha (v_1(x,t),y)-\alpha (v_2(y,s),y)\right| ,\\ Q(x,t,y,s):= & {} sgn(v_1(x,t)-v_2(y,s))(g(v_1(x,t))-g(v_2(y,s))). \end{aligned}$$

Let $E_0,E_1,E_2\subset \mathbb {R}$ be three sets such that

$$\begin{aligned} E_0:= & {} \left\{ t\in \mathbb {R}_+; t \text{ is } \text{ a } \text{ Lebesgue } \text{ point } \text{ of } v_2(x,t) \text{ for } \text{ a.e. } x\in \mathbb {R}\right\} , \end{aligned}$$

(100)

$$\begin{aligned} E_1:= & {} \left\{ x\in \mathbb {R}; x \text{ is } \text{ a } \text{ Lebesgue } \text{ point } \text{ of } v_2(x,t) \text{ for } \text{ a.e. } t\in \mathbb {R}_+\right\} , \end{aligned}$$

(101)

$$\begin{aligned} E_2:= & {} \left\{ x;\lim \limits _{\epsilon \rightarrow 0}\int \eta _\epsilon \left( {x-y}\right) \max \limits _{\left| u\right| \le r}\left| \alpha (u,x)-\alpha (u,y)\right| =0\right\} , \end{aligned}$$

(102)

where $r=\max \{\Vert v_1\Vert _{L^{\infty }(\mathbb {R}\times \mathbb {R}_+)},\Vert v_2\Vert _{L^\infty (\mathbb {R}\times \mathbb {R}_+)}\}$. Since $v_2\in L^{\infty }(\mathbb {R}\times \mathbb {R}_+)$, $E_0,E_1$ are measurable sets and $meas(\mathbb {R}_+\setminus E_0)=meas(\mathbb {R}\setminus E_1)=0$. By our assumption, for a fixed $x\in \mathbb {R}$, $\Psi (x,\cdot )$ is Lipschitz on $[-r,r]$. Since $C([-r,r])$ is separable, by Pettis Theorem we have measurability of $E_2$ and $meas(\mathbb {R}\setminus E_2)=0$. Therefore we can get

$$\begin{aligned}&\left| \int _{\mathbb {R}}P_1(x,t,y,s)\eta _{\epsilon }(y-x)\,dy-P_1(x,t,x,s)\right| \nonumber \\&\quad \le \int \limits _{\mathbb {R}}\left| \alpha (v_2(y,s),x)-\alpha (v_2(x,s),x)\right| \eta _\epsilon (y-x)\,dy\rightarrow 0 \end{aligned}$$

(103)

as $\epsilon \rightarrow 0$ for $x\in E_1$ and a.e. $t,s\in \mathbb {R}_+$. We can also obtain

$$\begin{aligned}&\left| \int \limits _{\mathbb {R}}P_2(x,t,y,s)\eta _\epsilon (y-x)dy-P_2(x,t,x,s)\right| \nonumber \\&\quad \le \int \limits _{\mathbb {R}}\left| \alpha (v_2(y,s),x)-\alpha (v_2(x,s),x)\right| \eta _\epsilon (y-x)\,dy\nonumber \\&\qquad +2\int \limits _\mathbb {R}\eta _\epsilon \left( {x-y}\right) \max \limits _{\left| u\right| \le r}\left| \alpha (u,x)-\alpha (u,y)\right| \nonumber \\&\qquad \rightarrow 0 \end{aligned}$$

(104)

as $\epsilon \rightarrow 0$ for $x\in E_2$ and a.e. $t,s\in \mathbb {R}_+$. With the help of (103) and (104) and Lebesgue Dominated Convergence Theorem we have

$$\begin{aligned}&\lim \limits _{\epsilon \rightarrow 0}\int \limits _{\mathbb {R}^2_+}\int \limits _{\mathbb {R}^2_+}\left( P_1(x,t,y,s)\frac{\partial }{\partial t}(\Phi (x,t)\eta _{\delta }(s-t))\right. \nonumber \\&\quad \left. + P_2(x,t,y,s)\frac{\partial }{\partial s}(\Phi (x,t)\eta _{\delta }(s-t))\right) \end{aligned}$$

(105)

$$\begin{aligned}&\eta _\epsilon (y-x)\,dtdxdsdy =\int \limits _{\mathbb {R}^2_+}\int \limits _{\mathbb {R}_+}P_1(x,t,x,s)\eta _{\delta }(s-t)\frac{\partial }{\partial t}\Phi (x,t)\,dtdxds.\qquad \end{aligned}$$

(106)

In a similar way we can show

$$\begin{aligned}&\lim \limits _{\delta \rightarrow 0}\int \limits _{\mathbb {R}^2_+}\int \limits _{\mathbb {R}_+}P_1(x,t,x,s)\eta _{\delta }(s-t)\frac{\partial }{\partial t}\Phi (x,t)\,dtdxds\nonumber \\&\quad =\int \limits _{\mathbb {R}^2_+}P_1(x,t,x,t)\frac{\partial }{\partial t}\Phi (x,t)\,dtdx. \end{aligned}$$

(107)

Similarly we have

$$\begin{aligned}&\left| \int \limits _{\mathbb {R}}Q(x,t,y,s)\eta _\epsilon (y-x)\,dy-Q(x,t,x,s)\right| \nonumber \\&\quad \le \int \limits _{\mathbb {R}}\left| v_2(y,s)-v_2(x,s)\right| \eta _{\epsilon }(y-x)\,dy\rightarrow 0 \end{aligned}$$

(108)

as $\epsilon \rightarrow 0$ for $x\in E_1$ and a.e. $t,s\in \mathbb {R}_+$. Then by Lebesgue Dominated Convergence Theorem we have

$$\begin{aligned}&\lim \limits _{\epsilon \rightarrow 0}\int \limits _{\mathbb {R}^2_+}\int \limits _{\mathbb {R}^2_+}Q(x,t,y,s)\eta _\epsilon (y-x)\eta _\delta (s-t)\frac{\partial }{\partial x}\Phi \,dtdxdsdy\nonumber \\&\quad =\int \limits _{\mathbb {R}^2_+}\int \limits _{\mathbb {R}_+}Q(x,t,x,s)\eta _\delta (s-t)\frac{\partial }{\partial x}\Phi \,dtdxds. \end{aligned}$$

(109)

We also have for a.e. $x\in \mathbb {R}$ and $t\in E_0$

$$\begin{aligned}&\left| \int \limits _{\mathbb {R}_+}Q(x,t,x,s)\eta _\delta (s-t)\,ds-Q(x,t,x,t)\right| \nonumber \\&\quad \le \int \limits _{\mathbb {R}_+}\left| v_2(x,t)-v_2(x,s)\right| \eta _{\delta }(s-t)\,ds\rightarrow 0 \end{aligned}$$

(110)

as $\delta \rightarrow 0$. This yields

$$\begin{aligned}&\lim \limits _{\epsilon \rightarrow 0,\delta \rightarrow 0}\int \limits _{\mathbb {R}^2_+}\int \limits _{\mathbb {R}^2_+}Q(x,t,y,s)\eta _\epsilon (y-x)\eta _\delta (s-t)\frac{\partial }{\partial x}\Phi \,dtdxdsdy\nonumber \\&\quad =\int \limits _{\mathbb {R}^2_+}Q(x,t,x,t)\frac{\partial }{\partial x}\Phi \,dtdx. \end{aligned}$$

(111)

This completes the proof. $\square$

Observe the following

$$\begin{aligned} g(v(x,t))=g(\Psi (u(x,t),x))=A(u(x,t),x)=A(\alpha (v(x,t),x),x). \end{aligned}$$

(112)

From Lemma 5.1 we can prove the following by a similar argument as in [15].

Lemma 5.2

Let $v_1,v_2\in C([0,T],L^{1}_{loc}(\mathbb {R}))\cap L^{\infty }(\mathbb {R}\times \mathbb {R}_+)$ be two function satisfying (93). Then for a.e. $t\in [0,T]$ and any $r>0$ we have

$$\begin{aligned}&\int \limits _{\left| x\right| \le r}\left| \alpha (v_1(x,t),x)-\alpha (v_2(x,t),x)\right| \,dx\nonumber \\&\quad \le \int \limits _{\left| x\right| \le r+L_1t}\left| \alpha (v_1(x,0),x)-\alpha (v_2(x,0),x)\right| \,dx \end{aligned}$$

(113)

swhere $L_1:=\sup \{\partial _u A(u,x);\,x\in \mathbb {R},\left| u\right| \le \max (\Vert v_1(x,0)\Vert _{L^{\infty }},\Vert v_2(x,0)\Vert _{L^{\infty }})\}$.