An additive constant in an entropy flux does not change the entropy inequality

For a test function $\phi$, [F1] and the definition of $q^C$ give $⟨{\mathrm{div}}_x{q}^C(u),\phi ⟩=-\int q^C(u)\cdot \mathrm{\nabla }\phi =-\int q(u)\cdot \mathrm{\nabla }\phi -{\sum }_{i=1}^n{C}_i\int {\mathrm{\partial }}_{x_i}\phi$.

Each ${\int }_{{\mathrm{\Pi }}_T}{\mathrm{\partial }}_{x_i}\phi dxdt=0$: the inner spatial integral vanishes because $\phi$ is compactly supported in $x$, so the function $x_i\mapsto \phi (t,x_i,x^{\mathrm{\prime }})$ is smooth compactly supported and the fundamental theorem of calculus applies, and the remaining integral over the other variables is finite as $\phi$ has compact support.

By steps 1.1 and 1.2, $⟨{\mathrm{div}}_x{q}^C(u),\phi ⟩=-\int q(u)\cdot \mathrm{\nabla }\phi =⟨{\mathrm{div}}_{x}q(u),\phi ⟩$ for every test function, hence ${\mathrm{div}}_x{q}^C(u)={\mathrm{div}}_{x}q(u)$ in ${\mathcal{D}}^{\mathrm{\prime }}({\mathrm{\Pi }}_T)$.

Adding the common distribution ${\mathrm{\partial }}_t\eta (u)$ to both sides of step 2.1, the two inequalities ${\mathrm{\partial }}_t\eta (u)+{\mathrm{div}}_{x}q(u)\le 0$ and ${\mathrm{\partial }}_t\eta (u)+{\mathrm{div}}_x{q}^C(u)\le 0$ are literally the same distributional inequality, so they are equivalent; in particular the entropy condition does not depend on the additive normalisation of the entropy flux.