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I don't quite understand why this advection equation, with some initial condition $u(0,x) = u_{0}(x)$, is considered hyperbolic (as for instance here). If I apply the test mentioned in Farlow, comparing it to $$Au_{xx}+Bu_{xy} + Cu_{yy} + D u_{x} + Eu_{y} + Fu = G$$ then both $B^{2}$ and $AC$ are zero, and thus $B^{2} - 4AC = 0 \Rightarrow \text{parabolic}$. Or does a different criterion apply when there aren't any second-order terms?

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  • $\begingroup$ The criterion you mentioned is probably for 2nd order linear PDEs, but the PDE you have is 1st order. There is probably some other definition for 1st order linear PDEs (I am not expert though, but I can try to look it up). $\endgroup$
    – Malkoun
    Dec 31, 2020 at 5:05
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    $\begingroup$ Look at p. 45 in the book you have mentioned <www2.math.umd.edu/~petersd/462/charact.pdf>. From what I understand, the system of PDEs $u_t + A u_x = 0$ is said to be hyperbolic if $A$ has only real eigenvalues. In the case you are interested in, $a$ is a real number (a $1$ by $1$ matrix), so the PDE is hyperbolic. $\endgroup$
    – Malkoun
    Dec 31, 2020 at 5:16
  • $\begingroup$ Thanks, @Malkoun, that makes sense. But doesn't that imply that essentially all first-order PDEs are hyperbolic? $\endgroup$
    – Durden
    Dec 31, 2020 at 5:56
  • $\begingroup$ Well, even if we restricted to the class of real $n$ by $n$ matrices $A$, the characteristic polynomial $p_A$ of $A$ is then a polynomial with real coefficients. If $p_A$ has all its roots real and distinct, then I think there is an open neighborhood of $p_A$ consisting of polynomials with real coefficients and having also all their roots real and distinct. But if $p_A$ has at least one root with nonzero imaginary part, then there is an open neighborhood of $p_A$ which also has at least one root with nonzero imaginary part. $\endgroup$
    – Malkoun
    Dec 31, 2020 at 6:08
  • $\begingroup$ So the set of polynomials with real coefficients with all their roots being real and distinct is open but not dense in the set of polynomials with real coefficients. Similar statements should hold for real $n$ by $n$ matrices and their eigenvalues. The strictly hyperbolic such matrices form (I think) and open subset $U$ of the set of all real $n$ by $n$ matrices, but $U$ is not dense. $\endgroup$
    – Malkoun
    Dec 31, 2020 at 6:12

1 Answer 1

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You have to look at the principal symbol $$ a_1(x,t,\xi,\tau)=i\tau+ai\xi. $$ For every fixed $\xi$ the equation $a_1(x,t,\xi,\tau)=0$ has exactly one real solution in $\tau$. Therefore the equation is hyperbolic with respect to $t$. Basically, this means that there is a real characteristic passing through each point. You can read more on the topic here: https://people.sissa.it/~dabrow/IFM12small1.pdf

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