Well-balanced balance laws and wet/dry fronts

Write $r_{-}:=\max (-r,0)$. On each compact rectangle $Q\subset \mathbb{R}\times (0,T)$, $h_k\ge 0$ implies $h_{-}\le \mathrm{\mid }h-h_k\mathrm{\mid }$, so ${\int }_Q{h}_{-}\le \mathrm{\parallel }h-h_k{\mathrm{\parallel }}_{L^1(Q)}\to 0$. On each compact interval $I\subset \mathbb{R}$, $h_k^0\ge 0$ gives $(h_0{)}_{-}\le \mathrm{\mid }{h}_0-h_k^0\mathrm{\mid }$, so ${\int }_I(h_0{)}_{-}\le \mathrm{\parallel }{h}_0-h_k^0{\mathrm{\parallel }}_{L^1(I)}\to 0$. Countable compact exhaustions of $\mathbb{R}\times (0,T)$ and $\mathbb{R}$ now give $h\ge 0$ and $h_0\ge 0$ almost everywhere on their domains.

Where $h>0$, [F1] allows arbitrary momentum $m$; where $h=0$, $m=hu=0$, so $U\in K$. For the initial data, on each compact interval $I$ the bound on $u_k^0$ and $U_k^0\in K$ imply $\mathrm{\mid }{m}_k^0\mathrm{\mid }\le M_I{h}_k^0$. Hence $(\mathrm{\mid }{m}_0\mathrm{\mid }-M_I{h}_0{)}_{+}\le \mathrm{\mid }{m}_0-m_k^0\mathrm{\mid }+M_I\mathrm{\mid }{h}_k^0-h_0\mathrm{\mid }$ pointwise, and integration followed by strong $L^1(I)$ convergence gives $\mathrm{\mid }{m}_0\mathrm{\mid }\le M_I{h}_0$ almost everywhere. Thus $m_0=0$ wherever $h_0=0$; step 1.1 gives $h_0\ge 0$, and [F1] yields $U_0\in K$. A countable compact exhaustion proves the bulk and initial claims almost everywhere on their domains.

Let $(h_1,m_1),(h_2,m_2)\in K$ and $\theta \in [0,1]$. The convex combination has depth $h_{\theta }=\theta h_1+(1-\theta )h_2\ge 0$. If $\theta =0$ or $\theta =1$, it is an endpoint in $K$ by [F1]. If $0<\theta <1$ and $h_{\theta }=0$, nonnegativity forces $h_1=h_2=0$; [F1] then gives $m_1=m_2=0$, so the combination is $(0,0)\in K$. If $h_{\theta }>0$, [F1] places the combination in $K$ for any momentum.

Thus every segment between two states in $K$ lies in $K$, which is exactly the convexity condition in [F2].

In the middle branch $\alpha ,\beta >0$. Define $p=\alpha (u_L^{\ast }+\beta )\mathrm{/}(\alpha +\beta )$ and $q=\beta (\alpha -u_R^{\ast })\mathrm{/}(\alpha +\beta )$. Substituting $a_L=-\beta$ and $a_R=\alpha$ into [F1], then using [F2], gives ${\widehat{F}}^h=(\alpha h_L^{\ast }{u}_L^{\ast }+\beta h_R^{\ast }{u}_R^{\ast }-\alpha \beta (h_R^{\ast }-h_L^{\ast }))\mathrm{/}(\alpha +\beta )=p{h}_L^{\ast }-q{h}_R^{\ast }$.

The speed definitions give $u_L^{\ast }+\beta \ge c_L^{\ast }\ge 0$, $\alpha -u_L^{\ast }\ge c_L^{\ast }\ge 0$, $\alpha -u_R^{\ast }\ge c_R^{\ast }\ge 0$, and $\beta +u_R^{\ast }\ge c_R^{\ast }\ge 0$. Hence the coefficients in step 1.1 satisfy $p,q\ge 0$, $p\le \alpha$, and $q\le \beta$. When $h_L^{\ast }>0$, $c_L^{\ast }>0$ and $\alpha -u_L^{\ast }>0$, so $p<\alpha$; when $h_R^{\ast }>0$, $c_R^{\ast }>0$ and $\beta +u_R^{\ast }>0$, so $q<\beta$.

If $a_L=0<a_R$, the first HLL branch gives ${\widehat{F}}^h=h_L^{\ast }{u}_L^{\ast }$; the speed definitions imply $u_L^{\ast }\ge c_L^{\ast }\ge 0$ and $\alpha \ge u_L^{\ast }+c_L^{\ast }$, so take $p=u_L^{\ast }$ and $q=0$. If $a_L<0=a_R$, the second branch gives ${\widehat{F}}^h=h_R^{\ast }{u}_R^{\ast }$, with $u_R^{\ast }\le -c_R^{\ast }\le 0$ and $\beta \ge c_R^{\ast }-u_R^{\ast }$, so take $p=0$ and $q=-u_R^{\ast }$. If both speeds vanish, their definitions force $u_s^{\ast }-c_s^{\ast }\ge 0$ and $u_s^{\ast }+c_s^{\ast }\le 0$ on each side, hence $c_s^{\ast }=u_s^{\ast }=h_s^{\ast }=0$ and the first branch returns zero. These choices satisfy the bounds in step 2.1 and cover every equality-speed branch.

Fix a face and set $b^{\ast }=\max (b_L,b_R)$. If a side cell is wet, its free surface is ${\eta }_0$, so its reconstructed depth is $({\eta }_0-b^{\ast }{)}_{+}$. If a side cell is dry, then $b_s\ge {\eta }_0$ and $b^{\ast }\ge b_s$, so its reconstructed depth is $(b_s-b^{\ast }{)}_{+}=0=({\eta }_0-b^{\ast }{)}_{+}$. Thus both reconstructed depths equal $H=({\eta }_0-b^{\ast }{)}_{+}$, and both reconstructed velocities are zero.

If $H>0$, the states on the face are equal to $(H,0)$ and the middle HLL formula has speeds $-\sqrt{gH}$ and $\sqrt{gH}$, so its flux is $F(H,0)=(0,g{H}^2\mathrm{/}2)$. If $H=0$, both states and speeds are zero and the dry branch returns $(0,0)$. Thus in either case the common flux is $(0,g{H}^2\mathrm{/}2)$.

On the side belonging to a cell of depth $h_s$, the correction in [F2] changes this common flux to $(0,g{h}_s^2\mathrm{/}2)$. Therefore the right corrected flux of cell $i$ and the left corrected flux of that same cell are both $(0,g{h}_i^2\mathrm{/}2)$. Their difference is zero, so the update leaves every $U_i$ unchanged.

Fix a compact spatial interval $I$ containing the cell centers under consideration and let $I^{+}$ be its closed unit enlargement. For $0<d\le 1$, Taylor's formula at $x_i\in I$ gives $b(x_i+d)-b(x_i)=a+r_{+}$ and $b(x_i-d)-b(x_i)=-a+r_{-}$, where $a=b^{\mathrm{\prime }}(x_i)d$ and $\mathrm{\mid }{r}_{\pm }\mathrm{\mid }\le \frac{1}{2}{\sup }_{I^{+}}\mathrm{\mid }{b}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }{d}^2$. Since $s\mapsto s_{+}$ is $1$-Lipschitz and $a_{+}-(-a{)}_{+}=a$, this yields $\mathrm{\mid }{\delta }_{i,+}-{\delta }_{i,-}-b^{\mathrm{\prime }}(x_i)d\mathrm{\mid }\le C_0{d}^2$. The same local derivative bounds give $0\le {\delta }_{i,\pm }\le C_{1}d$, with constants depending only on $I$ and $b$.

For $h\ge 0$ and $\delta \ge 0$, write $(h-\delta {)}_{+}^2=h^2-2h\delta +R(h,\delta )$. If $\delta \le h$, then $R={\delta }^2$; if $\delta >h$, then $R=2h\delta -h^2$, which lies between $0$ and ${\delta }^2$ because ${\delta }^2-(2h\delta -h^2)=(\delta -h{)}^2$. For $0<d\le 1$, steps 1.1 and [F1] therefore give $\mid (S_i^n{)}_2+g{h}_i^n{b}^{\mathrm{\prime }}(x_i)\mid \le gH{C}_{0}d+\frac{g}{2d}({\delta }_{i,+}^2+{\delta }_{i,-}^2)\le C_{2}d\mathrm{.}$ For $d\ge 1$, both squared positive parts in [F1] lie in $[0,H^2]$, so $\mathrm{\mid }(S_i^n{)}_2\mathrm{\mid }\le g{H}^2\mathrm{/}(2d)$. Since $\mathrm{\mid }{b}^{\mathrm{\prime }}(x_i)\mathrm{\mid }\le {\sup }_I\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }$, it follows that $\mid (S_i^n{)}_2+g{h}_i^n{b}^{\mathrm{\prime }}(x_i)\mid \le \frac{g{H}^2}{2d}+gH{\sup }_I\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }\le C_{3}d\mathrm{.}$ Taking $C=\max (C_2,C_3)$ proves the stated estimate for every $d>0$.

Now let $d\to 0$ and fix a compact space-time rectangle. For all sufficiently small $d$, the centers of cells meeting this rectangle lie in a fixed compact interval, so the estimate in step 2.1 and $\mathrm{\mid }{b}^{\mathrm{\prime }}(x)-b^{\mathrm{\prime }}(x_i)\mathrm{\mid }\le \frac{1}{2}d\sup \mathrm{\mid }{b}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }$ on a fixed enlargement show that ${\mathcal{S}}_d-(0,-g{h}_d{b}^{\mathrm{\prime }}(x))\to 0$ uniformly there. Since $b^{\mathrm{\prime }}$ is bounded on compact intervals and $h_d\to h$ strongly in local $L^1$, $(0,-g{h}_d{b}^{\mathrm{\prime }}(x))\to (0,-gh{b}^{\mathrm{\prime }}(x))$ strongly in local $L^1$, proving the asserted source convergence.

Multiply the recurrence in [F1] by ${\mathrm{\Phi }}_i^n$ using the Euclidean dot product and sum over cells $i$ and time levels $0\le n<N$. Compact support leaves only finitely many nonzero terms, and the resulting exact sum is zero.

For each cell, summing in time and using ${\mathrm{\Phi }}_i^{N-1}=0$ gives ${\sum }_{n=0}^{N-1}(U_i^{n+1}-U_i^n)\cdot {\mathrm{\Phi }}_i^n=-U_i^0\cdot {\mathrm{\Phi }}_i^0+{\sum }_{n=1}^{N-1}{U}_i^n\cdot ({\mathrm{\Phi }}_i^{n-1}-{\mathrm{\Phi }}_i^n)$. This is finite summation by parts; no limit or infinite rearrangement is used.

Reindexing the spatial face sum gives ${\sum }_i({\widehat{F}}_{i+1\mathrm{/}2}^n-{\widehat{F}}_{i-1\mathrm{/}2}^n)\cdot {\mathrm{\Phi }}_i^n={\sum }_i{\widehat{F}}_{i+1\mathrm{/}2}^n\cdot ({\mathrm{\Phi }}_i^n-{\mathrm{\Phi }}_{i+1}^n)$. Insert this and step 2.1 into step 1.1, retain the source sum with its minus sign from [F1], and multiply the zero identity by $-1$. The displayed identity follows.

Fix a compact space-time rectangle $Q$ and a fixed compact enlargement $Q^{\mathrm{\prime }}$ containing $Q$ and every spatial translate $Q+d_k$ for all sufficiently large $k$. Choose local bounds $H,M$ so that $0\le h_k\le H$ and $\mathrm{\mid }{u}_k\mathrm{\mid }\le M$ on $Q^{\mathrm{\prime }}$. Strong local $L^1$ convergence implies the same bounds for $h,u$ almost everywhere on $Q^{\mathrm{\prime }}$: the sum $\mathrm{\mid }{h}_k-h\mathrm{\mid }+\mathrm{\mid }{u}_k-u\mathrm{\mid }$ tends to zero in $L^1(Q^{\mathrm{\prime }})$, so [F4] gives a subsequence on which both converge almost everywhere and inherit the bounds. By [F3], multiplying $h,u$ by the indicator of $Q^{\mathrm{\prime }}$ therefore gives $L^1({\mathbb{R}}^2)$ functions. The given strong convergence gives $h_k\to h$ and $u_k\to u$ in $L^1(Q)$. Applying [F2] to the cutoffs shows $h(\cdot +d_k,\cdot )\to h$ and $u(\cdot +d_k,\cdot )\to u$ in $L^1(Q)$. Strong local convergence on $Q^{\mathrm{\prime }}$ and the triangle inequality then show $h_k(\cdot +d_k,\cdot )\to h$ and $u_k(\cdot +d_k,\cdot )\to u$ in $L^1(Q)$ as well.

Let $L$ be a Lipschitz constant for $b$ on the spatial enlargement, and write $h_k^{+}(x,t)=h_k(x+d_k,t)$ and $u_k^{+}(x,t)=u_k(x+d_k,t)$. Hydrostatic clipping changes either adjacent depth by at most $L{d}_k$. Take any subsequence of $k$. By step 1.1, the sum of the four errors $\mathrm{\mid }{h}_k-h\mathrm{\mid }+\mathrm{\mid }{u}_k-u\mathrm{\mid }+\mathrm{\mid }{h}_k^{+}-h\mathrm{\mid }+\mathrm{\mid }{u}_k^{+}-u\mathrm{\mid }$ converges to zero in measure on $Q$; [F4] gives a further subsequence on which all four errors tend to zero almost everywhere. At a point where $h>0$, both cell depths are eventually larger than $L{d}_k$, so neither reconstructed state is clipped to zero and both reconstructed velocities converge to $u$. At a point where $h=0$, the hypothesis gives $u=0$ and both cell velocities converge to zero; resetting either velocity to zero when its reconstructed depth vanishes therefore still gives convergence to $u$. The reconstructed depths in both states converge to $h$ as well. Thus along this common further subsequence both reconstructed primitive states converge almost everywhere to $(h,u)$. Since the starting subsequence was arbitrary, every subsequence has such a further subsequence.

At a wet limit state $h>0$, the two face inputs tend to the same $(h,hu)$; the left, middle and right HLL branches all tend to $F(h,hu)$, and the formulas agree at the branch boundaries. At a dry limit state, $h=0$ and $u=0$, so both reconstructed fluxes and both speeds tend to zero. In the middle branch write $\alpha =a_R$, $\beta =-a_L$; its equivalent formula is $\widehat{F}=(\alpha F_L^{\ast }+\beta F_R^{\ast })\mathrm{/}(\alpha +\beta )-[\alpha \beta \mathrm{/}(\alpha +\beta )](U_R^{\ast }-U_L^{\ast })$. Here $\alpha \beta \mathrm{/}(\alpha +\beta )\le \max (\alpha ,\beta )$, which tends to zero at the dry state, while $F_L^{\ast },F_R^{\ast },U_L^{\ast },U_R^{\ast }$ tend to zero. The outer branches also tend to zero. Thus along the almost-everywhere convergent further subsequence from step 2.1, the HLL flux tends almost everywhere to $F(h,hu)$. The same equivalent formula and $\alpha ,\beta \le M+\sqrt{gH}$ give a uniform bound on the fluxes on $Q$.

Starting from any subsequence, step 2.1 supplies a further subsequence with almost-everywhere convergence of both face inputs; step 3.1 gives almost-everywhere convergence of its fluxes and a uniform bound. Since $Q$ has finite measure by [F3], [F5] yields $L^1(Q)$ convergence along that further subsequence. If the full flux sequence failed to converge in $L^1(Q)$, some subsequence would stay a fixed positive distance from $F(h,hu)$, contradicting this property. Therefore ${\mathcal{F}}_k\to F(h,hu)$ in $L^1(Q)$. As $Q$ was arbitrary, the convergence is local.

Suppose $h_R^{\ast }=0$. Then $u_R^{\ast }=c_R^{\ast }=0$. If both speeds are zero, [F1] makes the full flux zero. If $a_L<0=a_R$, the second branch is $F_R^{\ast }=(0,0)$. If $a_L=0<a_R$, the speed definition forces $u_L^{\ast }\ge c_L^{\ast }$. When $h_L^{\ast }>0$, this gives ${\widehat{F}}^h=h_L^{\ast }{u}_L^{\ast }>0$; when $h_L^{\ast }=0$, both states and the flux are dry and zero. In the middle branch, [F2] gives ${\widehat{F}}^h=p{h}_L^{\ast }\ge 0$; it is strictly positive for a wet left state and otherwise both states are dry and the full flux is zero.

Suppose instead $h_L^{\ast }=0$. Then $u_L^{\ast }=c_L^{\ast }=0$. If both speeds vanish, the full flux is zero. If $a_L=0<a_R$, the first branch is $F_L^{\ast }=(0,0)$. If $a_L<0=a_R$, the speed definition forces $u_R^{\ast }\le -c_R^{\ast }$. For $h_R^{\ast }>0$ this makes ${\widehat{F}}^h=h_R^{\ast }{u}_R^{\ast }<0$; if $h_R^{\ast }=0$, the full flux is zero. In the middle branch, [F2] gives ${\widehat{F}}^h=-q{h}_R^{\ast }\le 0$, strictly negative for a wet right state and otherwise zero with both states dry.