> [!NOTE] > The solution to the equation $\partial_{t}u(x,t)+v(x,t)\partial_{x}u(x,t)=0, \quad (x,t)\in \mathbb{R}\times(0,T)\tag{1}$with the initial condition $u(x,0)=\Phi(x)$ is given by implicitly by $u(x,t)=\Phi(x_{0})$ > where $x_{0}$ is the such that $\xi'(t)=v(\xi(t),t), \quad \xi(0)=x_{0}$ **Proof**: Suppose $u$ is indeed a solution to $(1).$ Consider the function $t \mapsto(\xi(t),t).$ Using the [[Chain rule for derivative|chain rule]], $\begin{align} \frac{d}{dt}u(\xi(t),t) &= \partial_{x}u(\xi(t),t)\xi'(t)+\partial_{t}u(\xi(t),t) \\ &= \partial_{x}u(\xi(t),t)v(\xi(t),t) + \partial_{t}u(\xi(t),t)= 0. \end{align}$ It follows from [[Real Function with Zero Derivative is Constant|mean value theorem]] that the value of $u$ along the curve $(\xi(t),t)$ is constant.