Let $d\in\mathbb N$, $b\in\mathbb R^d$ and $u_0:\mathbb R^d\to\mathbb R$ be differentiable.
We can easily show that the unique (classical) solution of \begin{align}\left(\frac\partial{\partial t}+b\cdot\nabla_x\right)u&=0&\text{in }(0,\infty)\times\mathbb R^d;\tag1\\ u&=u_0&\text{on }\{0\}\times\mathbb R^d\tag2\end{align} is given by $$u(t,x):=u_0(x-tb)\;\;\;\text{for }(t,x)\in[0,\infty)\times\mathbb R^d.\tag3$$
If $u_0$ is not differentiable, will a "weak" solution of $(1)$ and $(2)$ still be given by $(3)$? And is this weak solution still unique?
To be precise, let $\Omega\subseteq\mathbb R^d$ be bounded and open, $H:=L^2(\Omega)$, $V:=H_0^1(\Omega)$ and $u_0\in V'$. Say that $u\in\mathcal L^2((0,\infty),V)$ is a weak solution of $(1)$ and $(2)$, with $\mathbb R^d$ replaced by $\Omega$, if $$\frac{\rm d}{{\rm d}t}\langle\varphi,u(t)\rangle_H=-\langle b\varphi,\nabla u(t)\rangle_{L^2(\Omega,\:\mathbb R^d)}\tag4$$ and $$\langle\varphi,u(0)\rangle=u_0\tag5$$ for all $\varphi\in V$.
Since this definition requires $u$ to take values in $V$ (and not only in $V'$), I guess we need to assume that $u_0\in V$ (but I would also like to know if there is also a sensible notion of solution which takes values in $V'$) such that we can at least ask whether such a weak solution could be given by $(3)$. However, there is still the problem that $u_0$ is only an equivalence class and hence we cannot make sense of the evaluation $u_0(x-tb)$ of $u_0$ at $x-tb$ ...