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For a strictly hyperbolic two-component system of conservation laws with ordered speeds ${\lambda }_{-}<{\lambda }_{+}$, a compressive family-$j$ shock has its family characteristics entering from both sides: ${\lambda }_j(U_R)<s<{\lambda }_j(U_L)$. The other family crosses the interface with the same sign of relative speed on both sides. These are the Lax shock admissibility conditions and give three incoming characteristic directions for two conservation laws.

In the present problem,Since $u_1=s-s/h_1$, we always have $u_1-\sqrt{h_1}<s$. Hence a positive-speed shock of this form cannot be a family-minus shock. For a family-plus shock the remaining strict inequalities areThe first is equivalent to $h_1(h_1+1)>2$, or $h_1>1$. The second is equivalent to $s<h_1^{3/2}$, whose square reduces to $h_1+1<2h_1^2$, again equivalent to $h_1>1$ for positive $h_1$. The minus speed on the right is $-1<s$ automatically. Thus all causal inequalities hold exactly whenDepth increases behind the right-moving shock. For $0<h_1<1$, the algebraic jump solution has the wrong characteristic orientation and is an inadmissible expansion shock; equality is the no-jump limit.

For the dimensionless shallow water equations with gravity coefficient one, let the left state be $(h,u)$, $h>0$, and the right state $(1,0)$. The Rankine-Hugoniot conditions are $s(h-1)=hu$ and $shu=h{u}^2+(h^2-1)/2$. For $h\ne 1$ and $s>0$, eliminating $u$ gives the displayed formulas. The characteristic speeds are $u\pm \sqrt{h}$ and $\pm 1$. Because $u-\sqrt{h}<s$ always, the positive branch can only be an admissible family-plus shock. Its remaining Lax shock admissibility conditions, $1<s<u+\sqrt{h}$, both reduce to $h>1$. At $h=1$ there is no jump and the labelled interface has no determined shock speed.