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I am a little confused about the connection between variables for the plain advection pde: $$u_t+au_x=0$$ So initially I thought $x$ and $t$ are independent and $u$ is a function of those, but then we can write the same PDE and say that there are characteristics and then we have ODE for them: $$\frac{dx}{dt}=a$$ So, now $x$ is a function of $t$. So are they dependent or not?

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    $\begingroup$ The notation in some textbooks can be confusing; it's better to distinguish the characteristic curves by writing them with a different symbol, e.g. $X(\xi, t)$ for the curve such that $X(\xi, 0) = \xi$, $\dot X = a(X)$. $\endgroup$ – Daniel Shapero Jun 24 '16 at 16:15
  • $\begingroup$ The characteristic curves are not solutions to the PDE, instead they represent the paths in the $x-t$ plane (i.e. the domain of $u$ -- the actual solution of the PDE) along which information travels/propagates. So yes $x$ and $t$ are independent variables of the solution $u(x,t)$, but along characteristics of the PDE they are not independent. $\endgroup$ – okrzysik Jun 27 '16 at 9:29
  • $\begingroup$ @okrzysik: thanks for answering my questions exactly. $\endgroup$ – Kamil Jun 27 '16 at 14:00
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You are correct in your initial assumption. x and t are independent and u is the dependent variable.

Your confusion is because you are assuming that $\frac{dx}{dt} = a$ is the solution to the equation. Notice that $u$ is not in this ODE. Basically this is an equation for the symmetry of the PDE. After getting this equation, you can write

u = F(x-at)

which is the actual solution since you will know u at t = 0 by specifying the initial conditions.

x and t are not dependent on each other. $\frac{dx}{dt} = a$ merely states that the solution is fully specified on lines that are solutions of this ODE, and thus are called characteristics.

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