Hydrostatic HLL face fluxes converge under strong state and velocity convergence

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.