If $u \in L^{\infty} (\mathbb{R} \times \mathbb{R}^{+})$ is the weak solutions of \begin{equation} u_t +f(u)_x=0 \end{equation} then, define $v$ by $v(x,t)=u(x-st,t)$, then v is the weak solution of \begin{equation} v_t + (f(v)-sv)_x=0 \end{equation} How to prove this? For smooth solution the proof is easy. How to justify it for any $u \in L^{\infty}$
Definition: $u \in L^{\infty}(\mathbb{R} \times \mathbb{R}^{+})$ is said to be a weak solution of a conservation law if for all $\phi \in C_c^{\infty} (\mathbb{R} \times \mathbb{R}^{+})$ u satisfies $$\int\limits_{\mathbb{R \times \mathbb{R}^+}} u \phi_t+f(u)\phi_x=0$$
A natural weak formulation of the conservation law $u_t + f(u)_x = 0$ is $$ \iint_{\Bbb R^+\times\Bbb R} [\phi u_t + \phi f(u)_x]\,\text dx \,\text dt = 0 , $$ for all $\phi \in C_c^\infty(\Bbb R\times \Bbb R^+)$, or $$ \iint_{\Bbb R^+\times\Bbb R} [\phi_t u + \phi_x f(u)]\,\text dx \,\text dt = 0 $$ after integration by parts and application of Fubini's theorem. Let us start from the latter equation. Similarly to this post, we make the change of variable $(\xi, \tau) = (x-st,t)$. Thus, we have $$ \iint_{\Bbb R^+\times\Bbb R} [(\phi_\tau - s \phi_\xi) u + \phi_\xi f(u)]\,\text d\xi \,\text d\tau = 0 , $$ which is nothing else but the weak formulation of the conservation law $u_\tau + (f(u)-su)_\xi = 0$.