Understanding the solution to the transport equation $u_t + au_x = 0$

Question

I am reading the first two pages of this pdf and am confused about why they let $z(x,t) = f(x-at)$ at the bottom of page $2$. They say that since $z(x,t)$ is constant along the lines $x-at = x_0$, then $z(x,t) = f(x-at)$. But, I don't see the connection between those two statements.

Can someone please explain what they are doing?

Answer 1

Citation : They say that since $z(x,t)$ is constant along the lines $x-at = x_0$, then $z(x,t) = f(x-at)$.

$z(x,t)$ is constant along the line $\quad x-at = x_0\quad$ means that for the given value $x_0$ we have $\quad z=z_0=$constant.

The same, on another line $\quad x-at = x_1\quad$ , we have $\quad z=z_1=$another constant. Again $x-at = x_1$ , of course with different $x$ and $t$ than previously.

On another line $\quad x-at = x_2\quad$ , we have $\quad z=z_2=$another constant. Again $x-at = x_2$ , of course with different $x$ and $t$ than previously.

And so on.

One can understand that, at each constant $x_0$ , $x_1$ , $x_2$ , etc. corresponds a value of $z$, respectively $z_0$ , $z_1$ , $z_2$ , etc.

In other words, $z$ is a function $f$ of these constants, each one $=x-at$ , of course with different $x$ and $t$, that is: $$z=f(x-at).$$

Answer 2

Let's consider the equation:

$$u_t+au_x=0$$

The form of the equation suggests it might be a full time derivative:

$$\frac{du}{dt}=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} \frac{dx}{dt}=0$$

The fact that the derivative is $0$ means the function is constant. But how can it be constant?

Looking at the original equation we can see that:

$$\frac{dx}{dt}=a$$

So, in other words:

$$x=x_0+at$$

$$x-at=x_0$$

So, if this relation between the two variables is satisfied, then $u=u_0$, the function is constant. Which is what the book means.

As the function is constant, and $x-at$ is constant, you can write:

$$u_0=f(x_0)$$

Or:

$$u=f(x-at)$$

Here $f$ is a function of single variable (why did $x_0$ suddenly become a variable? Just because, it's an arbitrary constant. I'm sure the book explains it better).

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For the latter form of the solution, we see it clearly by substitution:

$$u_x=f'$$

$$u_t=-a f'$$

$$u_t+au_x=-af'+af'=0$$