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For bounded entropy solutions of the same scalar law, $Q(u,v)=\mathrm{sgn}(u-v)(f(u)-f(v))$ satisfies the displayed distributional inequality, with initial value $|u_0-v_0|$ in its test-function form. The doubling of variables for scalar conservation laws, local translation continuity and the averaged initial trace of an entropy solution prove it. Its significance is comparison between two solutions rather than an inequality against a constant state.

Set $H(a,b)=|a-b|$ and $Q(a,b)=\mathrm{sgn}(a-b)(f(a)-f(b))$. Both are symmetric in $a,b$. For fixed $(s,y)$ apply the entropy inequality for $u$ at the constant level $k=v(s,y)$ with test $\Phi (t,x,s,y)$, then integrate over $(s,y)$. This producesHere the first integral is over both time-space pairs and the second over the three remaining variables. Repeat with $v$, level $u(t,x)$, integrating over $(t,x)$, to obtain the analogous terms with ${\Phi }_s,{\Phi }_y$ and initial value $v_0$.

Adding and using symmetry giveswhere $B_u=\int H(u_0(x),v(s,y))\Phi (0,x,s,y)dsdxdy$ and $B_v=\int H(u(t,x),v_0(y))\Phi (t,x,0,y)dtdxdy$. All integrals are finite on the test's compact support because the states are bounded. The levels being inserted are constants relative to the variables in each entropy inequality; there is no differentiation of the merely measurable other solution. Fubini's theorem and continuity of $H,Q$ justify the integration. This is doubling of variables for scalar conservation laws.