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In the $n \times n$ hyperbolic system $$u_t + A(u)u_x = F(u)$$ what's the name of the assumption that $A$ has no zero eigenvalues?

Note that if the eigenvalues are all real and distinct the system is called strictly hyperbolic.

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    $\begingroup$ What's the significance of having no zero eigenvalues? $\endgroup$
    – Deane Yang
    Commented Mar 20, 2020 at 23:04

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I believe that a zero eigenvalue of $A(u)$ implies that the vector field $\partial_t$ is a characteristic direction. For instance, the equation $u_s + u_y = 0$ takes the form $u_t=0$ after switching to the coordinates $(t,x)=(s,y-s)$, where the spatial surfaces $t=$const and $s=$const are the same, while $\partial_t = \partial_s + \partial_y$. So perhaps you could say that when $A(u)$ has only real non-zero eigenvalues the system is "weakly hyperbolic with non-characteristic time".

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  • $\begingroup$ Thanks. Could you just clarify what's the definition of a characteristic direction? $\endgroup$
    – user140746
    Commented Mar 21, 2020 at 0:16
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    $\begingroup$ This doesn't agree with what I think of as a characteristic direction. What I think of as a noncharacteristic direction is one such that any initial data specified on a surface (here, a curve) transversal to the direction is well posed. Here, I believe $\partial_t$ is noncharacteristic in that sense. However, if the eigenvalue of $A$ is zero, then the initial value problem with initial data on the curve $x = 0$ is ill-posed. It is, however, well-posed, if $A$ is invertible, i.e., has no zero eigenvalues. $\endgroup$
    – Deane Yang
    Commented Mar 21, 2020 at 1:40
  • $\begingroup$ I more or less agree with Deane. The eigenvalues of $A$ are the characteristic velocities (wave speeds). I don't really see any significance of $A$ having no zero eigenvalues. $\endgroup$
    – Willie Wong
    Commented Mar 21, 2020 at 2:02
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    $\begingroup$ @DeaneYang I distinguish between characteristic directions (elements of the characteristic cone in the tangent bundle, sometimes also called ray cone) and characteristic surfaces (conormals are elements of the characteristic cone in the cotangent bundle). These two distinct characteristic cones are duals of each other, of course. In my example, $t=0$ is a non-characteristic surface, but $\partial_t$ is a characteristic direction. On the other hand, $s-y=0$, which is the same as $x=0$, is a characteristic surface. Also notice that $\partial_t$ is tangent to $x=0$, as you would expect! :-) $\endgroup$
    – Igor Khavkine
    Commented Mar 21, 2020 at 2:35
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    $\begingroup$ @Zyl In your case, for fixed $u$, if $v$ is an eigenvalue of $A(u)$, then $\partial_t - v\partial_x$ is a characteristic direction or a ray vector. These vectors are the directions along which wave packets propagate in the geometric optics approximation. A characteristic (codimension-1) surface is any surface whose tangent plane contains at least one characteristic direction. These surfaces are wave fronts in the plane wave approximation. $\endgroup$
    – Igor Khavkine
    Commented Mar 21, 2020 at 2:47

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