Monotone finite-volume entropy convergence — Examples

The polynomial $F$ is continuous. On the diagonal, its correction term vanishes and $F(s,s)=s^2\mathrm{/}2=f_B(s)$, so it is consistent. Its first partial derivative is ${\mathrm{\partial }}_{a}F(a,b)=\frac{a}{2}+\frac{1}{16}((a^2-1)(b^2-1)+2a(a-b)(b^2-1))\mathrm{.}$ At $(-1,0)$ this equals $-1\mathrm{/}2-2\mathrm{/}16=-5\mathrm{/}8<0$, so it is not nondecreasing in its first coordinate.

On the uniform cells $I_j=[jh,(j+1)h)$ with $h>0$, the initial averages are $U_j^0=-1$ for $j<0$ and $U_j^0=1$ for $j\ge 0$. Every same-state face flux is $F(-1,-1)=F(1,1)=1\mathrm{/}2$. The only mixed pair is $(-1,1)$; both correction factors vanish and $F(-1,1)=1\mathrm{/}2$. Thus every face flux equals $1\mathrm{/}2$, and the recurrence leaves every cell average unchanged at every step.

The datum has $\mathrm{TV}(u_0)=2$ by [F5]. For any sequence $h_m,{\tau }_m\downarrow 0$ with this aligned grid convention, the piecewise-constant reconstruction is exactly the stationary step function almost everywhere. It therefore converges strongly in local space-time $L^1$ to $u(x,t)=-1$ for $x<0$ and $u(x,t)=1$ for $x>0$. Its initial trace is $u_0$.

Since $f_B(u)=1\mathrm{/}2$ almost everywhere, for any $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$ one has ${\int }_0^{T}u(x,t){\phi }_t(x,t)dt=-u_0(x)\phi (x,0)$ for almost every $x$, because $u(x,t)=u_0(x)$ is stationary and $\phi$ vanishes near $t=T$. Integrating in $x$ cancels the initial term in [F2]. The flux contribution is $\frac{1}{2}{\int }_{\mathbb{R}}{\phi }_x(x,t)dx=0$ for each $t$, since $\phi$ has compact spatial support. Thus $u$ satisfies the weak identity and is a locally bounded distributional weak solution.

Let $R:=T\mathrm{/}8$ and choose the cutoff $\chi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ of [F4] with $\chi =1$ on the unit ball and support in the radius-two ball. The function $\phi (x,t):=\chi (x\mathrm{/}R,(t-T\mathrm{/}2)\mathrm{/}R)$ is nonnegative, smooth, supported in $\mathbb{R}\times (0,T)$, and satisfies ${\int }_0^T\phi (0,t)dt>0$. For $k=0$, $\mathrm{\mid }u\mathrm{\mid }=1$ is stationary, so $\int \mathrm{\mid }u\mathrm{\mid }{\phi }_t=0$, while $q_0(-1)=-1\mathrm{/}2$ and $q_0(1)=1\mathrm{/}2$ give ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}{q}_0(u){\phi }_{x}dxdt=-{\int }_0^T\phi (0,t)dt<0.$ This contradicts the required nonnegative entropy pairing in [F3].

Fix $b\in [-M,M]$. On each branch in [F1], the function $a\mapsto G(a,b)$ is either constant, $a^2\mathrm{/}2$, or $\max \{a^2,b^2\}\mathrm{/}2$. It is nondecreasing: when $a\le b$, the active values are $b^2\mathrm{/}2$ on the negative-negative branch, $0$ on the crossing branch, and $a^2\mathrm{/}2$ on the nonnegative branch; when $b<a$, the maximum is constant while $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }b\mathrm{\mid }$ and increases as $a^2\mathrm{/}2$ once $a$ is the larger positive magnitude. The branch values agree at their boundaries by [F2], so the monotonicity persists across each switch.

Fix $a\in [-M,M]$. On the branches of [F1], $b\mapsto G(a,b)$ is either constant, $b^2\mathrm{/}2$ for $b\le 0$, or $\max \{a^2,b^2\}\mathrm{/}2$ on $b<a$. As $b$ increases, $b^2\mathrm{/}2$ decreases on the negative axis; on $b<a$, the maximum decreases while $\mathrm{\mid }b\mathrm{\mid }>\mathrm{\mid }a\mathrm{\mid }$ and is constant after $\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }$; the remaining $a\le b$ branches are constant or decrease to zero. Boundary values agree by [F2]. Hence $G(a,\cdot )$ is nonincreasing.

On every smooth branch the derivative in the varying coordinate has absolute value at most $R$ on $[-R,R{]}^2$. The maximum branch has the same bound because $\max \{a^2,b^2\}\mathrm{/}2$ is piecewise one of the two square branches. The finite branch partition and continuity from [F2] therefore give coordinate Lipschitz bounds $R$ on every compact square $[-R,R{]}^2$, proving that $G$ is locally Lipschitz on ${\mathbb{R}}^2$. Applying the same estimate with $R=M$ gives coordinate Lipschitz bounds $L_1=L_2=M$ on the target square. Consistency is [F2], so the full CFL bound is $\lambda (L_1+L_2)=2M\lambda \le 1$.

The hypotheses of [F4] now hold for every stated family: the data are bounded BV by [F3], the flux is locally Lipschitz, monotone and consistent by steps 1.1–1.3, the full CFL bound holds, and the mesh ratio has the required positive lower bound. Thus the reconstructions converge strongly in local $L^1$ to the unique entropy solution.

On the diagonal, $F_{\mathrm{L}\mathrm{F}}(s,s)=f(s)$, so the flux is consistent. If $a_2=a_1$, the first-coordinate difference is zero; if $a_2>a_1$, the mean-value theorem gives $F_{\mathrm{L}\mathrm{F}}(a_2,b)-F_{\mathrm{L}\mathrm{F}}(a_1,b)=\frac{f(a_2)-f(a_1)+\alpha (a_2-a_1)}{2}\ge \frac{\alpha -L}{2}(a_2-a_1)\ge 0.$ Thus it is nondecreasing in its first coordinate.

If $b_2=b_1$, the second-coordinate difference is zero; if $b_2>b_1$, the mean-value theorem gives $F_{\mathrm{L}\mathrm{F}}(a,b_2)-F_{\mathrm{L}\mathrm{F}}(a,b_1)=\frac{f(b_2)-f(b_1)-\alpha (b_2-b_1)}{2}\le -\frac{\alpha -L}{2}(b_2-b_1)\le 0,$ so it is nonincreasing in its second coordinate. The mean-value bounds on each variable also give $L_1=L_2=(L+\alpha )\mathrm{/}2$.

Since $f$ is locally Lipschitz and the flux adds a linear term, $F_{\mathrm{L}\mathrm{F}}$ is locally Lipschitz. The displayed flux is consistent by its diagonal formula, and steps 1.1 and 1.2 prove monotonicity with coordinate bounds whose sum is $L+\alpha$. Hence the displayed upper ratio gives the full CFL condition, while the lower ratio and bounded BV data satisfy the remaining hypotheses of [F3]. The convergence theorem applies.

The initial cell averages are $U_j^0=1$ for $j<0$ and $U_j^0=-1$ for $j\ge 0$. Same-state faces have flux $1\mathrm{/}2$ by [F1], and the only mixed face has pair $(1,-1)$ and flux $\max \{1,1\}\mathrm{/}2=1\mathrm{/}2$. Thus every face flux equals $1\mathrm{/}2$.

Every cell's left and right fluxes are both $1\mathrm{/}2$, so the conservative update leaves every $U_j^n$ unchanged. Induction in $n$ proves that both reconstructions equal the stationary step function almost everywhere for every $m$; in particular they converge strongly locally to it.

The stationary profile is a bounded weak solution: for every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$, its time term satisfies ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}u{\phi }_{t}dxdt=-{\int }_{\mathbb{R}}{u}_0(x)\phi (x,0)dx$, while $f_B(u)=1\mathrm{/}2$ almost everywhere and hence ${\int }_{\mathbb{R}}{f}_B(u){\phi }_{x}dx=0$ at each time. This is exactly the weak identity of [F3].

Fix $k\in \mathbb{R}$. The entropy flux traces are $q_k(1)$ on the left and $q_k(-1)$ on the right. If $k\le -1$ or $k\ge 1$, both states lie on the same side of $k$ and these fluxes are equal because $f_B(1)=f_B(-1)$. If $-1<k<1$, then $q_k(1)=\frac{1-k^2}{2},q_k(-1)=-\frac{1-k^2}{2},q_k(1)-q_k(-1)=1-k^2\ge 0.$ For any nonnegative interior test $\phi$, the time term $\int \mathrm{\mid }u-k\mathrm{\mid }{\phi }_t$ vanishes because $u$ is stationary. Integration on the two half-lines gives ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}{q}_k(u){\phi }_{x}dxdt=(q_k(1)-q_k(-1)){\int }_0^T\phi (0,t)dt\ge 0.$ The initial trace is exactly $u_0$, so [F5] shows the stationary shock is a Kruzhkov entropy solution.

The left and right traces are $u_L=1$ and $u_R=-1$. By [F4], the shock speed is $s=\frac{f_B(-1)-f_B(1)}{-1-1}=\frac{1\mathrm{/}2-1\mathrm{/}2}{-2}=0,$ which agrees with the stationary interface. The Rankine–Hugoniot supplier applies because step 3.1 verified the weak identity and the traces are distinct.

The translation variation of the datum is $2$: for $h>0$, $u_0(x+h)-u_0(x)$ is nonzero only on $(-h,0)$, where its absolute value is $2$, so $\int \mathrm{\mid }{u}_0(x+h)-u_0(x)\mathrm{\mid }dx=2h$. The ratio is ${\tau }_m\mathrm{/}{h}_m=1\mathrm{/}4$, and the flux bounds $L_1=L_2=1$ from [F2] give $\lambda (L_1+L_2)=1\mathrm{/}2\le 1$. Thus the hypotheses of the main convergence theorem in [F2] hold; since the exact reconstruction is the entropy solution from step 3.2, it is the unique limit. No global $L^1$ assertion is made because the two tails are nonzero.