Mass conservation for compactly supported entropy solutions

The cone support holds at every time for the chosen representative. By [F2] there is a full-measure set $E\subseteq (0,\mathrm{\infty })$ with $u(t)=0$ almost everywhere outside $B(x_0,R+Lt)$ for $t\in E$. Fix $T>0$, $t_0\in [0,T]$ and a compact set $K\subseteq \{x:\mathrm{\mid }x-x_0\mathrm{\mid }>R+L{t}_0\}$. The positive distance of $K$ from $\bar{B}(x_0,R+L{t}_0)$ lets us choose $t_j\in E\cap (0,T)$ with $t_j\to t_0$ and $K\subseteq \{x:\mathrm{\mid }x-x_0\mathrm{\mid }>R+L{t}_j\}$ for all $j$; then $u(t_j)=0$ in $L^1(K)$, and the $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$-continuity of the representative [F1] gives $u(t_0)=0$ in $L^1(K)$. A countable exhaustion of the strict exterior of $B(x_0,R+L{t}_0)$ by compact sets gives $u(t_0)=0$ almost everywhere there; hence for every $t_0\in [0,T]$, $u(t_0)$ is supported in $\bar{B}(x_0,R+L{t}_0)\subseteq \bar{B}(x_0,R+LT)$.

Truncated mass balance. By step 1.1, for every $t\in [0,T]$ the function $u(t,\cdot )$ vanishes almost everywhere outside the fixed ball $B(x_0,R+LT)$; in particular $u\in L^1$ at every time with $\mathrm{\parallel }u(t){\mathrm{\parallel }}_1\le M\mathrm{v}\mathrm{o}\mathrm{l}(B(x_0,R+LT))$, and the $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ continuity of [F1] is continuity in $L^1({\mathbb{R}}^n)$. Choose $\beta$ as in [F4] and, for $\psi \in C_c^{\mathrm{\infty }}((0,T))$, test [F3] with $\phi (t,x)=\psi (t)\beta (x)$: since $u(t,\cdot )$ is supported where $\beta =1$ and $f(0)=0$ implies $f(u)=0$ wherever $u=0$, the flux term vanishes and only flat boundary terms contribute. The divergence theorem in the form of the weak identity then gives ${\int }_0^T{\psi }^{\mathrm{\prime }}(t)m(t)dt=0$ for the function $m(t)={\int }_{{\mathbb{R}}^n}u(t,x)dx$, that is, $m^{\mathrm{\prime }}=0$ in the sense of distributions on $(0,T)$.

Conclusion. Since $u$ is continuous in $L^1$ on $[0,T]$ and the support lies in the fixed ball, $m$ is continuous on $[0,T]$; its distributional derivative vanishes on $(0,T)$ by step 2.1 and $m(0)=\int u_0$ by the strong $L^1$ trace. To see constancy directly, convolve $m$ locally in time with a smooth unit-mass bump: its derivative is zero by $m^{\mathrm{\prime }}=0$ tested against translated kernels, so FTC makes each convolution constant on every interior compact interval. Uniform continuity of $m$ on such intervals makes the convolutions converge uniformly to $m$, hence $m$ is constant on $(0,T)$ and by continuity at its endpoints. Therefore $m(t)=\int u_0$ for every $t\in [0,T]$; in particular the equality holds for almost every $t$ and the chosen representative makes $t\mapsto \int u(t,x)dx$ constant on every $[0,T]$. As $T>0$ was arbitrary, the claims follow on $[0,\mathrm{\infty })$.