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I have a question about the following paper:

https://web.stanford.edu/class/math220a/handouts/firstorder.pdf

My question is on the end of the second page, when the author solves the transport equation.

"Now let $S$ be the integral of surface formed from a union of these characteristic curves. In doing so, we see that $z(x,t)$ is constant along the lines $x - at = x_{0}$. That is, $z(x,t) = f(x - at)$. Letting $u(x,t) = z(x,t) = f(x - at)$ for any (smooth) function $f$..."

I don't understand why

"$z(x,t)$ is constant along the lines $x - at = x_{0}$" implies "$z(x,t) = f(x - at)$".

Maybe it's a dumb question, but could anyone help me?

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The intuition is as follows: $z(x,t)$ is constant along the lines $x-at=x_0$, so $z$ only "sees" which of these lines you are on. This means that we can write it as $z(x,t)=f(x-at)$ in a well defined way. Another common example in pde is functions from $u:\mathbb{R}^2\to \mathbb{R}$ invariant under rotations are those that may be written as $$ u(r,\theta)=f(r) $$ for a function $f$ depending only on the radius, the only thing the original function $u$ cared about.

To see this more algebraically, note that for $(x,t)$ satisfying $x-at=x_0$, we have $$ z(x,t)=z(x_0,0) $$ by plugging in $t=0$. So, $$ z(x,t)=z(x-at,0) $$ Name $$ f(x-at)=z(x-at,0)=z(x,t) $$ This process works for any given initial position $x_0$ and has the nice visualization of tracing back along each of these lines $x-at=x_0$ in the $(x,t)$ plane to find the value of $z$ at a point $(x_0,0)$ where a value has been prescribed by initial data.

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    $\begingroup$ Great explanation! $\endgroup$ – Greg Oct 6 '18 at 16:54

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