Simple Transport $u_t+cu_x=0$, why is the substance transported to the right at a fixed speed c? In Partial Differential Equations by Walter Strauss, Ch 1.3, Example 1, they present the transport equation $u_t+cu_x=0$ whose solution is $u(x,t)=f(x-ct)$. They claim that this solution means the substance is transported to the right at a fixed speed c, why does this conclusion follow from the solution?
They also state that each particle moves in the $xt$ plane along a characteristic line. Why is this so?
 A: The transport equation models the concentration of a substance flowing at a constant rate.
Formally, for a parameter $v\in \mathbb{R}$, the transport equation on $\mathbb{R}\times \mathbb{R}^{+}$ is given by $$\frac{\partial u}{\partial t}(x,t)+v\frac{\partial u}{\partial x}(x,t)=0 \quad [1],$$
where $v$ is the velocity of the fluid. Now, the general solution by $[1]$ is given by $$u(x,t)=\phi(x-vt),$$where $\phi$ is an arbitrary function. But, if you have the initial value problem for the transfport equation given by $$(\operatorname{IVP}):\begin{cases}u_{t}+vu_{x}=0, \quad x\in \mathbb{R},t\in \mathbb{R}^{+}\\\left.u\right|_{t=0}=f(x), \quad x\in \mathbb{R}.\end{cases} \quad [2]$$
Then, the particular solution to $[2]$ is given by $$u(x,t)=f(x-vt).$$

They claim that this solution means the substance is transported to the right at a fixed speed $c$, why does this conclusion follow from the solution?


They also state that each particle moves in the $x-t$ plane along a characteristic line. Why is this so?

First, note that here I'm using $c=v$. Using the gradient operator $$\nabla:= \left( \frac{\partial }{\partial t}, \frac{\partial}{\partial x}\right),$$we can re-write $[1]$ as $$\nabla_{\vec{c}}u:=\vec{c}\cdot \nabla u=\begin{pmatrix} 1\\ v \end{pmatrix}\cdot \nabla u(x,t)=0. \quad [3]$$
Note that if there's not gradient, so there's not transport. Now, note that $[3]$ says that the directional derivative in the $\begin{pmatrix} 1\\ v \end{pmatrix}$ in the $x-t$ plane is zero. So, the solution of $[1]$ it's to say $u(x,t)$ must be constant in this direction. In the plane $x-t$, the $\begin{pmatrix} 1 \\ v \end{pmatrix}-$direction is along lines parralel $x=vt$ (characteristics of $[1]$).
I think this answers both questions. A tip if you are interested in the physical interpretation of the solution is to solve problems that involve the transport equation and you can observe some basic models.
Here, you can read a JJacquelin's answer about the "Understanding the solution to the transport equation $u_{t}+au_{x}=0$". Maybe it will be useful for you.
