Conditional ideal-gas Euler kinetic interface

Use the collision-invariant multipliers $\psi =m$, $\psi =m{v}_i$ for $i=1,2,3$, and $\psi =m\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2$. Their weak collision brackets vanish by the elastic constraints, and the stated absolute convergence makes $\int \psi Q_{\nu }(f^{\epsilon })d^{3}v=0$ for each of them. Since $\epsilon >0$, the scaled equation therefore gives $\int \psi ({\mathrm{\partial }}_t{f}^{\epsilon }+v\cdot {\mathrm{\nabla }}_x{f}^{\epsilon })d^{3}v=0$. The assumed ${\omega }_3\mathrm{\mid }{\mathrm{\partial }}_t{f}^{\epsilon }\mathrm{\mid }$ and ${\omega }_3\mathrm{\mid }v\cdot {\mathrm{\nabla }}_x{f}^{\epsilon }\mathrm{\mid }$ majorants justify these integrals. This uses the summed transport derivative as supplied; separate majorants for $\psi v_j{\mathrm{\partial }}_{x_j}{f}^{\epsilon }$ are not needed.

Let $\chi \in C^1([0,T_{\ast }]\times \mathrm{\Omega })$ be periodic in space and vanish near $T_{\ast }$, allowing $\chi (0,x)\mathrm{\ne }0$. For each fixed velocity, integrate $\psi f^{\epsilon }$ against ${\mathrm{\partial }}_t\chi +v\cdot {\mathrm{\nabla }}_x\chi$ and use [F4] in time and in each spatial coordinate. The time boundary is $-{\int }_{\mathrm{\Omega }}\psi f^{\epsilon }(0,x,v)\chi (0,x)d^{3}x$, and opposite periodic spatial boundaries cancel. Integrate this identity in velocity using [F3]. Indeed, $\psi$ has degree at most two, so $\mathrm{\mid }\psi \mathrm{\mid }$ and $\mathrm{\mid }\psi \mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }$ are bounded by constants times ${\omega }_3$; the displayed moment, time-derivative, and summed-transport envelopes make the grouped terms absolutely integrable. The initial term is integrable by the same moment bound at $t=0$. Thus step 1.1 gives the weak moment identity ${\int }_0^{T_{\ast }}{\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^3}\psi f^{\epsilon }({\mathrm{\partial }}_t\chi +v\cdot {\mathrm{\nabla }}_x\chi )d^{3}v{d}^{3}xdt+{\int }_{\mathrm{\Omega }}{\int }_{{\mathbb{R}}^3}\psi f^{\epsilon }(0,x,v)\chi (0,x)d^{3}v{d}^{3}x=0.$ The spatial integration by parts was performed at each fixed velocity before applying Fubini to the summed transport and final test-gradient terms, so no separate weighted spatial derivative bound has been assumed.

Replace $f^{\epsilon }$ in step 2.1 by $M=M_{n,u,T}$, including its value at $t=0$. The absolute difference of the spacetime terms is at most $C_{\chi }{\int }_0^{T_{\ast }}{\int }_{\mathrm{\Omega }}\int {\omega }_3\mathrm{\mid }{f}^{\epsilon }-M\mathrm{\mid }$, since the coefficients $\psi {\mathrm{\partial }}_t\chi +\psi v\cdot {\mathrm{\nabla }}_x\chi$ have velocity degree at most three and the test derivatives are bounded. The initial difference is at most $C_{\chi }{\int }_{\mathrm{\Omega }}\int {\omega }_2\mathrm{\mid }{f}^{\epsilon }(0)-M(0)\mathrm{\mid }$, since $\psi$ has degree at most two. Both tend to zero by the two stated convergence hypotheses. Hence the displayed weak moment identity holds with $M$ and its initial value.

Apply [F1] to the Gaussian moments in step 3.1. The multiplier $m$ gives density $\rho$ and flux $\rho u$; the three multipliers $m{v}_i$ give momentum $\rho u$ and flux tensor $\rho u\otimes u+p{I}_3$; the multiplier $m\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}2$ gives energy $E$ and flux $(E+p)u$. Applying the scalar identity componentwise to a vector test gives the momentum weak identity. Thus all three Euler equations hold in distributions, and their weak identities include exactly the initial values of the displayed Gaussian moments, identified by the ${\omega }_2$ initial convergence. These are the stated initial traces of the $C^1$ limiting fields.

The residuals of the three displayed Euler equations are continuous because the limiting fields are $C^1$. If a residual component were nonzero at an interior point, it would have one strict sign on a sufficiently small coordinate ball in $(0,T_{\ast })\times \mathrm{\Omega }$. Choose concentric balls within that chart and a nonnegative smooth bump from [F5], equal to one on the inner ball and supported in the outer ball; extend it by zero to the periodic cylinder. Its pairing with the residual would have strict sign, contradicting the distributional identity from step 4.1. Every residual therefore vanishes pointwise for $0<t<T_{\ast }$.

The observables [F2] identify $\rho =mn$ and the physical kinetic-energy density. The Gaussian moments in [F1] give $E=\rho \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}2+3n{k}_{B}T\mathrm{/}2$ and $p=n{k}_{B}T$, with precisely the stated SI units. The argument has used weighted strong convergence and its initial counterpart as assumptions, not derived them from small $\epsilon$ or entropy. The full conclusion is conditional on the adopted physical model [A1] and convergence hypotheses [A2], with no empirical validity claim.