Problem Let $u(x, t)$ be a weak solution to the conservation law $$u_t + [f(u)]_x = 0$$ Prove that, for every fixed $(x_0, t_0)$ and each $ε > 0$, the function $u_ε(x, t) = u(x_0 + εx, t_0 + εt)$ is another weak solution of the given conservation law.
My Attempt I plugged in $u_ε(x, t)$ into the conservation law and after simplifying a bit, I got: $$u_t(x_0 + εx, t_0 + εt)+[f(u(x_0 + εx, t_0 + εt))]_x=0$$ From here, how do I take the limit as $ε$ goes to zero? Or did I do something wrong? Thanks for the help
Generally, any PDE that does not involve the variables explicitly is invariant under shifts of variables. Indeed, the shift commutes with derivatives; hence, any solution of the equation $F(u,u_x,u_t,u_{xx},u_{tt},u_{xt},\dots)=0$ remains a solution after shift.
But perhaps one should be more careful to make the above rigorous when the PDE is understood in a weak sense. The weak form amounts to saying that for every test function $\phi(x,t)$ we have $$\int \bigg(u(x,t) \phi_t(x,t) + f(u(x,t)) \phi_x(x,t)\bigg)\,dx\,dt=0$$ If we let $x=X+h$, $t=T+k$, and introduce notation $v(X,T)=u(X+h,T+k)$ and $\psi(X,T)=\phi_X(X+h,T+k)$, the above becomes $$\int \bigg(v(X,T) \psi_T (X ,T ) + f(v(X , T )) \psi_X(X ,T )\bigg)\,dX \,dT=0$$ Which says precisely that $v$ is a weak solution of the same equation, because $\psi$ can be any test function here.