# Characteristic surface for systems of PDE

Despite the title, this is probably actually a question in linear algebra or algebraic geometry. Let me write the question(s) first, before I explain the background.

Problems

Let $h^{\mu\nu}_{ij}$ represent a map from $\mathbb{R}^4\otimes\mathbb{R}^4$ to $\mathbb{R}^2\otimes\mathbb{R}^2$ (here $\mu\nu$ are indices in the $\mathbb{R}^4$ directions, and $ij$ are in $\mathbb{R}^2$ directions). We shall assume that $h$ is symmetric swapping the $\mu,\nu$ indices and also symmetric swapping the $i,j$ indices. Then for any $\xi_\mu$ in $\mathbb{R}^4$, the object $H(\xi):= h^{\mu\nu}_{ij}\xi_\mu\xi_\nu$ is a symmetric bilinear form on $\mathbb{R}^2$. We say that $\xi$ is characteristic if $H(\xi)$ is degenerate. In other words, $\xi$ is characteristic if $\det(H(\xi)) = 0$.

Since $H(\xi)$ is quadratic in $\xi$, the determinant is an 8th degree homogeneous polynomial in $\xi$. Furthermore, by definition if $\xi$ is characteristic, so is $-\xi$. Observe also that in general the characteristic set will have multiple sheets.

Question 1, very specific

Does there exist an $h$ such that the characteristic surface is given by $\xi_1^4 + \xi_2^4 + \xi_3^4 - \xi_4^4 = 0$?

Question 2, slightly more general

In general are there any obstructions to having a sheet of the characteristic surface described by the zero set of an irreducible (over the reals) polynomial of degree strictly higher than 2?

Question 3, even more general

What if we relax the condition on $h$ so that it is a map from $\mathbb{R}^m\otimes\mathbb{R}^m$ to $\mathbb{R}^d\otimes\mathbb{R}^d$ with the same symmetric properties. Define $H(\xi)$ analogously. Can a sheet of the characteristic surface have algebraic degree more than 2?

I'm particularly interested in concrete examples.

Motivation

This comes from the study of hyperbolic systems partial differential equations. Recall that a second degree partial differential equation $$h^{\mu\nu}_{ij} \partial_\mu\partial_\nu u^i = 0$$ is said to be strictly hyperbolic in the direction of $e_\mu$ if the characteristic polynomial (a polynomial in $t$) $\det(H(x_\mu - te_\mu))$ is hyperbolic for any fixed $x_\mu$ linearly independent from $e_\mu$ and that the roots are distinct (it is enough that the second condition only holds for all by finitely many $x_\mu$ modulo $e_\mu$).

The classical examples for strictly hyperbolic systems (wave equation, crystal optics, etc) all have the sheets of the characteristic surfaces being linearly transformed versions of the standard quadratic double cone: in other words there exists a basis of $\mathbb{R}^m$ such that a sheet is given by $\sum_{i = 1}^{m-1} e_i^2 - e_m^2 = 0$.

I am guessing that for strictly hyperbolic systems in fact all sheets must be of this form due to homogeneity (though please let me know if I am wrong).

So my question is: is it possible for a non-strictly hyperbolic system (but one still hyperbolic) where some of the sheets have higher multiplicity to not come from "the square of a quadratic sheet" but from a genuinely quartic or higher polynomial?

Postscript

Please do let me know if you need any clarification on my question. Thanks.

Update

I struck out question 1 for the following reason: in view of my motivation from hyperbolic polynomials arising from second order PDEs, the answer is negative. The argument is thus: for a hyperbolic system of PDEs, the time-like direction $\xi_4$ should have its corresponding $h^{44}_{ij}$ negative definite, whereas the space-like directions $\xi_1,\xi_2,\xi_3$ should have their corresponding $h^{aa}_{ij}$ positive definite. A simple computation shows that the coefficient to the $\xi_a^4$ term in $\det H(\xi)$ must be $\det h^{aa}_{ij}$. If the target space is two dimensional, both positive definite and negative definite matrices have positive determinants. So for any hyperbolic polynomial arising from a second order system of PDEs, the coefficients for $\xi_a^4$ must be positive.

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just a question: by sheet you mean an irreducible component? As to question 1: have you tried to write down the equation $\det H(\xi)=p^2$ and solve for the $h$ (where $p$ is your polynomial of degree 4)? – Michael Bächtold Jul 14 '10 at 14:45
By sheet I mean the zero set of a irreducible component of the characteristic polynomial. Yes. Let's just say that I started writing down the equation for $\det H(\xi) = p^2$ and looked at the horrible mess and decide to ask to see if anyone else know of a quick example. So I am indeed in the middle of trying to see if it is explicitly computable. – Willie Wong Jul 14 '10 at 14:56
So question 1 is a no, at least for the applications I care about. See above. – Willie Wong Jul 15 '10 at 13:41
Hum, for a hyperbolic map from $\mathbb{R}^m\to\mathbb{R}^n$, the number of free variables is essentially $(\frac{m(m+1)}{2}-1)(\frac{n(n+1)}{2}) - \frac{n(n-1)}{2}$, while the number of equations for prescribing the characteristic polynomial is $\frac{(2n+m-1)!}{(m-1)! 2n !}$. The latter tend to be rather bigger than the former. – Willie Wong Jul 15 '10 at 16:15

There is a natural example where an irreducible component of the symbol, a hyperbolic polynomial, is of degree higher than $2$. It occurs in compressible magnetohydrodynamics (MHD). It is a coupling of the Euler system of compressible, inviscid gas dynamics, with Maxwell's equations, when the magnetic part dominates the EM field. It consists of $8$ conservation laws, governing mass density ($1$), momentum ($3$), energy ($1$) and magnetic field ($3$). When linearizing around a uniform state, we obtain a linear, first-order hyperbolic system. The determinant of the symbol splits into the square of a linear factor (pure transport), a quadratic factor (Alfvén mode) and an irreducible factor of degree $4$. The roots of the latter are the velocities in direction $\xi$ of the fast/slow backward/forward waves.

You can also get a complex structure in nonlinear models of EM fields. See here.

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Can it actually happen in nonlinear electrodynamics? Somehow earlier I thought the desired behaviour maybe in contradiction with Boillat's characterization of non-birefringent models. But thinking about it again now I am not too sure. I'll have to spend some time to study the models you gave in the answer. Thanks! – Willie Wong Oct 11 '10 at 15:39

I dont't see why "for strictly hyperbolic systems in fact all sheets must be of this form due to homogeneity" should hold. Your questions appear to be about the concept of second order symmetric hyperbolic systems ("symmetric" here refers to the symmetry in $\mu$ and $\nu$). First order symmetric hyperbolic systems definitely don't necessarily have a characteristic surface that consists only of quadratic sheets. Since there are even more degrees of freedom for second order systems, I don't see why they would have more constraints (but maybe I missed something). The simplest non-quadratic example is a diagonal system where each diagonal entry is a first order operator squared (instead of a scalar wave operator).

As always, I am extremely curious about why you want to study nonstandard examples of hyperbolic systems. The only place I know where such things have arisen in nature is in the isoemetric embedding problem.

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Right, that's why I said it is a guess. My thinking goes something like this: if a system is strictly hyperbolic, then by rescaling we can fix $t = 1$ and now we study the zero set of some polynomial on the projective plane, which has several connected components each of multiplicity 2. Now my algebra is really rusty, so my guess that this implies each connected component is quadratic may be completely off. – Willie Wong Jul 14 '10 at 14:16
And yes, I am focused only on second order "symmetric" systems. The motivation for that is because they are the natural objects arising from Lagrangian field theories. But believe it or not, until 2001 there has been no formal treatment of wellposedness properties of the second order symmetric hyperbolic systems (as you call them). The theory for those reducible to a first order symmetric hyperbolic system is well known, as that for second order strictly/Leray hyperbolic systems. – Willie Wong Jul 14 '10 at 14:20
For systems where the characteristic sets are quadratic surfaces, one can adapt a lot of pseudo-Riemannian techniques to study domains of dependence properties and formation of singularities. For systems where the characteristic sets are higher order surfaces, we lose the inner product structure and it is not clear whether there is a Finsler geometry approach. – Willie Wong Jul 14 '10 at 14:28
I don't know anything about Lagrangian field theories, so I might be totally off the mark. But my advice is to steer clear of anything that leads you to a weird PDE with weird null bicharacteristics. This seems to me to be very unphysical, if only because it will have weird invariance properties. – Deane Yang Jul 14 '10 at 14:30
As for your "strictly hyperbolic" implies "quadratic" argument, I don't see why the components have to have multiplicity 2. A simple example (slightly different from the one I gave before) is a diagonal system, where each diagonal element is the composition of two different first order operators. – Deane Yang Jul 14 '10 at 14:32