Monotone finite-volume entropy convergence

For a finite set $J\subset \mathbb{Z}$, the cell-average formula gives ${\sum }_{j\in J}\mathrm{\mid }{U}_{j+1}^0-U_j^0\mathrm{\mid }\le \frac{1}{h}{\int }_{{\cup }_{j\in J}{I}_j}\mathrm{\mid }{u}_0(x+h)-u_0(x)\mathrm{\mid }dx\le V\mathrm{.}$ Taking the supremum over finite $J$ proves the initial discrete variation bound. Every initial cell average lies in $I$, since $a_{-}\le u_0\le a_{+}$ almost everywhere on its cell.

If $a$ or $c$ increases, monotonicity of $F$ in its first or second coordinate makes $H$ nondecreasing in that variable. If $b_2\ge b_1$, coordinate Lipschitz bounds give $H(a,b_2,c)-H(a,b_1,c)\ge (b_2-b_1)(1-\lambda (L_1+L_2))\ge 0.$ Thus $H$ is nondecreasing in all coordinates. Consistency gives $H(s,s,s)=s$.

Suppose $U^n$ takes values in $I$. Coordinatewise comparison with the constant triples $(a_{-},a_{-},a_{-})$ and $(a_{+},a_{+},a_{+})$, followed by step 1.2, yields $a_{-}=H(a_{-},a_{-},a_{-})\le U_j^{n+1}\le H(a_{+},a_{+},a_{+})=a_{+}$. Induction from step 1.1 proves the range bound at every time.

Let $S$ denote the sequence update. If $x,y\in I^{\mathbb{Z}}$ and ${\sum }_j\mathrm{\mid }{x}_j-y_j\mathrm{\mid }<\mathrm{\infty }$, monotonicity gives $\mathrm{\mid }Sx-Sy\mathrm{\mid }\le S(x\vee y)-S(x\wedge y)$ coordinatewise. The difference on the right is summable: the update formula and the coordinate Lipschitz bounds make its flux differences summable. For any $z,w$ with ${\sum }_j\mathrm{\mid }{z}_j-w_j\mathrm{\mid }<\mathrm{\infty }$, summing the update over $-N\le j\le N$ yields ${\sum }_{j=-N}^N(S{z}_j-S{w}_j)={\sum }_{j=-N}^N(z_j-w_j)-\lambda (d_N-d_{-N-1}),d_j:=F(z_j,z_{j+1})-F(w_j,w_{j+1})\mathrm{.}$ The coordinate Lipschitz bounds make ${\sum }_j\mathrm{\mid }{d}_j\mathrm{\mid }<\mathrm{\infty }$, so both boundary terms tend to zero. The update difference is absolutely summable, and hence ${\sum }_j(S{z}_j-S{w}_j)={\sum }_j(z_j-w_j)$. Applying this to $z=x\vee y$ and $w=x\wedge y$ proves $\mathrm{\parallel }Sx-Sy{\mathrm{\parallel }}_1\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_1$.

The initial neighbor difference is summable by step 1.1. The update commutes with integer shifts because the same $H$ is applied at every index. Apply step 2.2 to $x=U^n$ and its shift $y_j=U_{j+1}^n$ to obtain ${\sum }_j\mathrm{\mid }{U}_{j+1}^{n+1}-U_j^{n+1}\mathrm{\mid }\le {\sum }_j\mathrm{\mid }{U}_{j+1}^n-U_j^n\mathrm{\mid }\mathrm{.}$ Induction proves the stated TVD bound.

Adding and subtracting $F(U_j^n,U_j^n)$ in the update and using the two coordinate Lipschitz bounds gives $\mathrm{\mid }{U}_j^{n+1}-U_j^n\mathrm{\mid }\le \lambda (L_2\mathrm{\mid }{U}_{j+1}^n-U_j^n\mathrm{\mid }+L_1\mathrm{\mid }{U}_j^n-U_{j-1}^n\mathrm{\mid })\mathrm{.}$ Multiply by $h\mathrm{/}\tau =1\mathrm{/}\lambda$, sum over $j$, and use step 3.1 to obtain the time-increment estimate.

If $s\ge k$, then $s\vee k=s$ and $s\wedge k=k$, so $G_k(s,s)=f(s)-f(k)=q_k(s)$. If $s<k$, then $s\vee k=k$ and $s\wedge k=s$, so $G_k(s,s)=f(k)-f(s)=q_k(s)$.

First suppose $k\in I$. Put $X=(U_{j-1}^n,U_j^n,U_{j+1}^n)$ and let $X^{+}$ and $X^{-}$ be its coordinatewise maximum and minimum with $(k,k,k)$. Both triples lie in $I^3$. By [F1], the two numbers $H(X)$ and $H(k,k,k)=k$ lie between $H(X^{-})$ and $H(X^{+})$; therefore $\mathrm{\mid }H(X)-k\mathrm{\mid }\le H(X^{+})-H(X^{-})\mathrm{.}$ Expanding the definition of $H$, the right side is $\mathrm{\mid }{U}_j^n-k\mathrm{\mid }-\lambda (G_k(U_j^n,U_{j+1}^n)-G_k(U_{j-1}^n,U_j^n)),$ which gives the claimed inequality after rearrangement.

If $k<a_{-}$, every state is above $k$ by the range assertion in [F1]. Then $\mathrm{\mid }{U}_j^n-k\mathrm{\mid }=U_j^n-k$ and $G_k(a,b)=F(a,b)-F(k,k)$ for state pairs; substituting the scheme recurrence makes the displayed left side zero. If $k>a_{+}$, every state is below $k$, so $\mathrm{\mid }{U}_j^n-k\mathrm{\mid }=k-U_j^n$ and $G_k(a,b)=F(k,k)-F(a,b)$; substitution again gives zero. The cases $k=a_{-}$ and $k=a_{+}$ are already covered by step 2.1.

For every $t$, $W_m(\cdot ,t)$ is piecewise constant with values in $I$ and spatial variation at most $V$: on each time slab it is a convex combination of two states whose variations are at most $V$ by [F2]. Here the variation of a piecewise-constant section on an interval is the sum of the absolute jump sizes inside that interval. Fix an integer $R\ge 1$ and consider any sequence $v_m$ of such sections on $[-R,R]$. At each rational point in $[-R,R]$, the values lie in the compact interval $I$. The bounded-value subsequence at one rational point exists by [F6]; explicitly, partition $I$ into finitely many closed intervals of length at most $2^{-1}$, take the first interval containing infinitely many values, and repeat inside it with successive meshes $2^{-2},2^{-3},\dots$, always selecting the first subinterval with infinitely many retained indices. The nested intervals have diameters tending to zero, so the least increasing indices in the nested sets give a convergent subsequence. Use the enumeration fixed by [F5], and repeat this deterministic extraction successively on the current subsequence at each enumerated rational that lies in $[-R,R]$, skipping the others. The nested subsequences yield a diagonal subsequence converging at every rational point; each interval and index is selected by a least-index rule, so this uses no choice axiom. Given $\delta >0$, choose a finite rational partition of $[-R,R]$ with mesh at most $\delta$, possible by [F4]. For each subinterval $J=[r_i,r_{i+1}]$ and any two selected functions $v_p,v_q$, variation gives $\mathrm{\mid }{v}_p(x)-v_q(x)\mathrm{\mid }\le \mathrm{\mid }{v}_p(r_i)-v_q(r_i)\mathrm{\mid }+{\mathrm{Var}}_J(v_p)+{\mathrm{Var}}_J(v_q)$ for almost every $x\in J$. Integrating and summing yields $\mathrm{\parallel }{v}_p-v_q{\mathrm{\parallel }}_{L^1([-R,R])}\le 2\delta V+2R{\max }_i\mathrm{\mid }{v}_p(r_i)-v_q(r_i)\mathrm{\mid }\mathrm{.}$ For the fixed partition the maximum tends to zero as $p,q\to \mathrm{\infty }$; then taking $\delta \downarrow 0$ proves the selected sections are Cauchy in local $L^1$.

On each slab, ${\mathrm{\partial }}_t{W}_m=(U_j^{n+1}-U_j^n)\mathrm{/}{\tau }_m$ on cell $I_j$. By [F2], its $L^1(\mathbb{R})$ norm is at most $C:=(L_1+L_2)V$. Integrating over any time interval gives $\mathrm{\parallel }{W}_m(\cdot ,t)-W_m(\cdot ,s){\mathrm{\parallel }}_{L^1(\mathbb{R})}\le C\mathrm{\mid }t-s\mathrm{\mid }\mathrm{.}$ Use the enumeration fixed by [F5] and process its indices together with integers $R\ge 1$ in order of increasing sum. For each pair $(\mathrm{\ell },R)$, if $q_{\mathrm{\ell }}\in [0,T]$, apply step 1.1 at time $q_{\mathrm{\ell }}$ on $[-R,R]$; otherwise skip that pair. The resulting diagonal extraction covers every rational time in $[0,T]$ and every integer interval. The subsequence converges in $L^1([-R,R])$ at every rational time for every integer $R$; this recursive construction uses no choice axiom.

Fix $R$ and $\epsilon >0$, and put $\delta :=\epsilon \mathrm{/}(4(C+1))$. Choose a finite set of rational times in $[0,T]$ that is a $\delta$-net: include $0$, choose $q\in \mathbb{Q}\cap [0,T]$ with $T-q<\delta$ (using $q=0$ if $T<\delta$), and subdivide $[0,q]$ by rational points with gaps below $\delta$, possible by [F4]. At those finitely many times the subsequence in step 2.1 is Cauchy in $L^1([-R,R])$. For any $t\in [0,T]$, choose a net time $s$ within the gap bound; the triangle inequality and the uniform time estimate in step 2.1 give $\mathrm{\parallel }{W}_p(\cdot ,t)-W_q(\cdot ,t){\mathrm{\parallel }}_{L^1([-R,R])}\le 2C\mathrm{\mid }t-s\mathrm{\mid }+\mathrm{\parallel }{W}_p(\cdot ,s)-W_q(\cdot ,s){\mathrm{\parallel }}_{L^1([-R,R])}\mathrm{.}$ Thus the subsequence is uniformly Cauchy in time in local $L^1$. For fixed $t,R$, its bounded measurable sections have an $L^1([-R,R])$ limit: for each $j$, let $N_j$ be the least integer such that all $p,q\ge N_j$ have $L^1([-R,R])$ difference below $2^{-j}$, which exists by Cauchyness. Put ${\widehat{N}}_j:={\max }_{1\le r\le j}{N}_r$ and recursively choose the least increasing indices $n_j\ge {\widehat{N}}_j$. Then both $n_j,n_{j+1}\ge N_j$, so successive differences have norms below $2^{-j}$. By [F8] their pointwise absolute sum is finite almost everywhere, and dominated convergence [F7] gives the $L^1$ limit. Uniqueness makes these limits consistent as $R$ increases, and the uniform Cauchy estimate makes $t\mapsto u(t)$ continuous in each local $L^1$ norm. The subsequence is also Cauchy in $L^1([-R,R]\times (0,T))$ for every integer $R$, by integrating its uniform-in-time Cauchy bound. For each $j$, let $M_j$ be the least integer such that all $p,q\ge M_j$ have difference below $2^{-j}$ on every cylinder with integer $R\le j$; this exists because there are finitely many such cylinders and the sequence is Cauchy on each. Put ${\widehat{M}}_j:={\max }_{1\le r\le j}{M}_r$ and recursively choose the least increasing indices $m_j\ge {\widehat{M}}_j$. For every fixed integer $R$ and every $j\ge R$, both $m_j,m_{j+1}\ge M_j$, so their successive differences have norm below $2^{-j}$ on that cylinder. [F8] then gives a finite pointwise sum of differences almost everywhere on each cylinder, and [F7] yields a measurable space-time limit $v$ with strong local $L^1$ convergence. For each integer $R$, the sum of the $L^1([-R,R]\times (0,T))$ norms of these successive differences is finite. By [F9], for almost every $t$ their spatial absolute sum is integrable on $[-R,R]$; hence $W_{m_j}(\cdot ,t)$ converges almost everywhere in $x$ there to $v(\cdot ,t)$. Since all these sections and their limit take values in the bounded interval $I$, dominated convergence [F7] gives convergence to $v(\cdot ,t)$ in $L^1([-R,R])$. But this is a subsequence of the sequence that converges in $L^1([-R,R])$ to $u(t)$ for every $t$, so uniqueness of the $L^1$ limit gives $v(\cdot ,t)=u(t)$ for almost every $t$. Intersecting these full-measure time sets over integer $R$ shows that the space-time limit and the continuous local-$L^1$ map are the same limit on all bounded intervals.

Every $W_m$ takes values in $I$, so its local $L^1$ limit also takes values in $I$ almost everywhere: if, for example, $u(t)>a_{+}$ on a set of positive measure, then some set where $u(t)\ge a_{+}+1\mathrm{/}r$ has positive measure, contradicting $L^1$ convergence from functions bounded by $a_{+}$. The same argument applies to the lower bound.

For $r<0$, translation invariance of Lebesgue measure gives the same bound as for $-r$, so $\mathrm{\parallel }{u}_0(\cdot +r)-u_0{\mathrm{\parallel }}_{L^1(\mathbb{R})}\le \mathrm{\mid }r\mathrm{\mid }V$ for every $r\in \mathbb{R}$. On each cell, the initial cell-average error is controlled by translations: ${\sum }_j{\int }_{I_j}\mathrm{\mid }{u}_0(x)-U_j^0\mathrm{\mid }dx\le \frac{1}{h}{\int }_{-h}^h{\int }_{\mathbb{R}}\mathrm{\mid }{u}_0(x+r)-u_0(x)\mathrm{\mid }dxdr\le hV\mathrm{.}$ The first inequality follows by writing each cell average as an integral over that same cell, applying the triangle inequality, and changing variables to $r=y-x$; the restriction that $x$ and $x+r$ lie in the same cell can only decrease the integral. Since $W_m(\cdot ,0)=U^0$, step 3.1 gives $u(0)=u_0$ in local $L^1$.

On each slab, $\mathrm{\mid }{W}_m-U_j^n\mathrm{\mid }=\theta \mathrm{\mid }{U}_j^{n+1}-U_j^n\mathrm{\mid }$. Integrating in time contributes the factor ${\tau }_m\mathrm{/}2$; the time-increment bound in [F2] gives, on $[0,T]$, $\mathrm{\parallel }{W}_m-{\mathcal{U}}_m{\mathrm{\parallel }}_{L^1(\mathbb{R}\times (0,T))}\le \frac{(T+{\tau }_m){\tau }_m}{2}(L_1+L_2)V⟶0.$ Consequently ${\mathcal{U}}_m$ converges strongly in local space-time $L^1$ to the same limit as $W_m$.

Start with any subsequence of meshes. By [F4] it has a further subsequence along which $W_m$ converges in $C([0,T];L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1)$ to a map $u(t)$ with $u(0)=u_0$, values in $I$, and ${\mathcal{U}}_m\to u$ strongly in space-time $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. The range condition gives the uniform local state bound required below.

Fix $k\in \mathbb{R}$ and a nonnegative $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (0,T))$. Let ${\mathrm{\Phi }}_j^n$ be the average of $\phi$ on $I_j\times [t_n,t_{n+1}]$, and choose $N$ so that $\phi$ vanishes on and after the last time slab. Multiply the one-step inequality [F9] by $h{\mathrm{\Phi }}_j^n$, sum over the finitely many relevant $j,n$, telescope in time, and reindex the shared entropy fluxes in space. The resulting exact inequality is ${\sum }_{n=1}^{N-1}{\sum }_{j}h\mathrm{\mid }{U}_j^n-k\mathrm{\mid }({\mathrm{\Phi }}_j^n-{\mathrm{\Phi }}_j^{n-1})+{\sum }_{n=0}^{N-1}{\sum }_j\tau G_k(U_j^n,U_{j+1}^n)({\mathrm{\Phi }}_{j+1}^n-{\mathrm{\Phi }}_j^n)\ge 0.$ There is no initial term because $\phi$ is supported away from $t=0$.

The hypotheses of [F5] hold along this further subsequence: $f$ is continuous, the cell data and shared-flux recurrence are those of [F1], the mesh widths vanish, the states are bounded by $I$, and the strong local convergence is from step 1.1. Therefore $u$ satisfies the distributional weak equation with the prescribed initial boundary term.

On the compact state interval $J$ spanned by $I\cup \{k\}$, local Lipschitz continuity of $F$ gives a finite constant $C_k$ for which $\mathrm{\mid }{G}_k(a,b)-G_k(a,a)\mathrm{\mid }\le C_k\mathrm{\mid }b-a\mathrm{\mid }$. By [F3], the local space-time $L^1$ difference between the face entropy flux $G_k(U_j^n,U_{j+1}^n)$ and $q_k(U_j^n)=G_k(U_j^n,U_j^n)$ is at most $C_{k}T{h}_m\mathrm{TV}(u_0)$, which tends to zero. Also $q_k$ is Lipschitz on $I$ because $f\in C^1$, so strong convergence in step 1.1 gives $q_k({\mathcal{U}}_m)\to q_k(u)$ in local $L^1$.

The finite-sum expression in step 1.2 converges to ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}(\mathrm{\mid }u-k\mathrm{\mid }{\phi }_t+q_k(u){\phi }_x)dxdt\mathrm{.}$ Writing $\overline{{\phi }_t}{}_j^n$ for the average of ${\phi }_t$ on $I_j\times [t_n,t_{n+1}]$, Taylor's formula in time gives $\mathrm{\mid }{\mathrm{\Phi }}_j^n-{\mathrm{\Phi }}_j^{n-1}-{\tau }_m\overline{{\phi }_t}{}_j^n\mathrm{\mid }\le C_{\phi }{\tau }_m^2$. After multiplication by $h_m\mathrm{\mid }{U}_j^n-k\mathrm{\mid }$ and summing the $O(h_m^{-1})$ cells and $O({\tau }_m^{-1})$ slabs in the compact support, the error is $O({\tau }_m)$. For the flux term, translation of adjacent cell averages gives ${\tau }_m({\mathrm{\Phi }}_{j+1}^n-{\mathrm{\Phi }}_j^n)={\int }_{t_n}^{t_{n+1}}{\int }_{I_j}{h}_m^{-1}{\int }_0^{h_m}{\phi }_x(x+s,t)dsdxdt$; replacing the averaged derivative by ${\phi }_x$ has total error $O(h_m)$ because ${\phi }_{xx}$ and the numerical entropy fluxes are uniformly bounded on the compact support. The face-flux replacement in step 2.2 and strong state convergence supply the remaining limit. Passing to the limit in step 1.2 proves the Kruzhkov inequality for this $k,\phi$. Since both were arbitrary, all interior Kruzhkov inequalities hold.

The initial value and continuity from [F4] imply the strong local $L^1$ trace: for every compact interval $K$, $\mathrm{\parallel }u(t)-u_0{\mathrm{\parallel }}_{L^1(K)}\to 0$ as $t\downarrow 0$, hence the essential supremum over $0<t<\delta$ also tends to zero. Together with the entropy inequalities of step 3.1, [F6] makes $u$ a bounded Kruzhkov entropy solution.

Put $M_0:=\max \{\mathrm{\mid }{a}_{-}\mathrm{\mid },\mathrm{\mid }{a}_{+}\mathrm{\mid }\}$. By [F10], $f$ is Lipschitz on $[-M_0,M_0]$, which contains the range $I$ of every limit from step 4.1. Thus the range and flux hypotheses of [F7] hold for any pair of such limits. If $u_0\in L^1(\mathbb{R})$, [A2] and [F8] supply a bounded Kruzhkov solution, and [F7] makes it unique and identifies it with the limit of step 4.1. If $u_0\mathrm{\notin }{L}^1(\mathbb{R})$, step 4.1 itself supplies existence and [F7] supplies uniqueness. Thus in either branch there is a unique target entropy solution, without applying the global existence theorem outside its $L^1$ domain.

Every convergent subsequence from step 1.1 has an entropy-solution limit with the same initial data; by [F7] all such limits agree with the unique target from step 5.1 as space-time $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ classes. If the full sequence failed to converge strongly on some compact space-time set, a subsequence would stay at distance at least $\epsilon >0$ from that target; [F4] would give a further subsequence converging strongly there to an entropy solution, contradicting uniqueness. Therefore ${\mathcal{U}}_m\to u$ strongly in local space-time $L^1$. For the uniform-in-time conclusion, Fubini in [F6] turns this space-time equality into equality of the $L^1(K)$ sections for almost every $t$, for each compact interval $K$. The two local-$L^1$ limit maps are continuous, so they agree for every $t$: if their distance were positive at one time, continuity would make it positive on an interval of positive measure. Uniform-in-time local convergence of $W_m$ then follows by the same subsequence contradiction using [F4].