# Trouble with solution for transport equation.

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?

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.