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?

  • 2
    $\begingroup$ 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 $X(\xi, 0) = \xi$, $\dot X = a(X)$. $\endgroup$ Commented Jun 24, 2016 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
    Commented Jun 27, 2016 at 9:29
  • $\begingroup$ @okrzysik: thanks for answering my questions exactly. $\endgroup$
    – Kamil
    Commented Jun 27, 2016 at 14:00

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.