[proofplan] We verify the displayed formula by differentiating it in time and space. For uniqueness, we restrict an arbitrary classical solution to the characteristic line through the point under consideration and use the differential equation to show that the restricted one-variable function is constant. [/proofplan] [step: Verify the displayed formula] Set $u(x,t)=u_0(x-bt)$. Since $u_0\in C^1(\mathbb R^n)$ and the map $(x,t)\mapsto x-bt$ is smooth, the function $u$ is $C^1$ on $\mathbb R^n\times\mathbb R$. The chain rule gives \begin{align*} \partial_t u(x,t)=-b\cdot \nabla u_0(x-bt) \end{align*} and \begin{align*} \nabla u(x,t)=\nabla u_0(x-bt). \end{align*} Therefore \begin{align*} \partial_t u(x,t)+b\cdot\nabla u(x,t)=0. \end{align*} At $t=0$ one has $u(x,0)=u_0(x)$, so the formula gives a classical solution with the prescribed initial data. [/step] [step: Prove uniqueness along characteristics] Let $v:\mathbb R^n\times\mathbb R\to\mathbb R$ be another $C^1$ solution with $v(x,0)=u_0(x)$. Fix $(x,t)\in\mathbb R^n\times\mathbb R$ and define the curve $\gamma:\mathbb R\to\mathbb R^n$ by \begin{align*} \gamma(s)=x-bt+bs. \end{align*} The curve satisfies $\gamma(0)=x-bt$ and $\gamma(t)=x$. Define $w:\mathbb R\to\mathbb R$ by $w(s)=v(\gamma(s),s)$. The chain rule and the equation for $v$ give \begin{align*} w'(s)=b\cdot\nabla v(\gamma(s),s)+\partial_t v(\gamma(s),s)=0. \end{align*} Thus $w$ is constant, and hence \begin{align*} v(x,t)=w(t)=w(0)=v(x-bt,0)=u_0(x-bt). \end{align*} This equals the displayed formula for $u$, so the classical solution is unique. [guided] The uniqueness argument follows the geometry of the transport equation. Let $v:\mathbb R^n\times\mathbb R\to\mathbb R$ be any $C^1$ solution of $\partial_t v+b\cdot\nabla v=0$ with the same initial data $v(x,0)=u_0(x)$. Through a fixed point $(x,t)\in\mathbb R^n\times\mathbb R$, the curve $\gamma:\mathbb R\to\mathbb R^n$ given by $\gamma(s)=x-bt+bs$ is the characteristic that starts at $x-bt$ when $s=0$ and reaches $x$ when $s=t$. For $w:\mathbb R\to\mathbb R$ defined by $w(s)=v(\gamma(s),s)$, the chain rule gives $w'(s)=b\cdot\nabla v(\gamma(s),s)+\partial_t v(\gamma(s),s)=0$. Thus $w$ is constant, so $v(x,t)=w(t)=w(0)=v(x-bt,0)=u_0(x-bt)$. Every classical solution therefore agrees with the displayed formula. [/guided] [/step]