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Consider the $n \times n$ system $$u_t + A(u)u_x = F(u).$$

If the eigenvalues of $A$ are all real and distinct the system is called strictly hyperbolic.

What is the relationship between this definition and the existence of a strictly convex entropy? Are they equivalent?

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They are not. First of all, the existence of a convex entropy is not meaningful for a system given in this quasi-linear form. The reason is that you might make a change $v=\phi(u)$ of unknown, but the convexity is not preserved by composition by the diffeomorphism $\phi$.

In addition, if $n\ge3$, a generic quasi-linear system does not admit conservation laws $\partial_t\eta(u)+\partial_q(u)=g(u)$, because the compatibility condition $\nabla\eta A=\nabla q$ is over-determined.

Now, if you give yourself a system of balance laws $u_t+f(u)_x=F(u)$, whose principal part is in conservation form, then the notion of convex entropy becomes meaningful, because you authorize only linear change of variables. Once again, a generic system with $n\ge3$ does not admit an entropy. So the system can be hyperbolic without having this convex entropy.

Of course, systems coming from thermodynamics are not generic. They were characterized by Godunov as those for which there are two functions $E(w),M(w)$ with $E$ strictly convex, such that $u=\nabla E(w)$ and $f(u)=\nabla M(w)$. Then the system is symmetric hyperbolic, in the sense that $$\nabla^2Ew_t+\nabla^2Mw_x=F(u)$$ where the matrices are symmetric (Hessians), and the first one is positive definite. Such systems do have a convex entropy, namely the Legendre transform $E^*(u)$. Conversely, a system of balance laws equipped with a strictly convex entropy can be written that way.

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No, entropy convexity and hyperbolicity are not equivalent conditions. A necessary and sufficient condition for the system of differential equations to possess a strictly convex entropy is that the system is symmetrizable and hence hyperbolic. The symmetrizability condition is stronger than the condition of hyperbolicity, a system may have real eigenvalues and be therefore hyperbolic without being symmetrizable, and therefore without having a strictly convex entropy.

See for example these notes, or theorem 3.2 of this book.

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    $\begingroup$ Dear Carlo. These notes speak of a system (1) which is never written. And the statement that a convex entropy exists iff the system is symmetrizable is given without proof ; only a reference to G. & R.'s book is given. The point is that (1) is certainly already in conservation form. Actually the theorem does not apply to a more general system in quasi-linear form. Just because a strictly hyperbolic system is always symmetrizable, even if it has no entropy at all. $\endgroup$
    – Denis Serre
    Commented Mar 21, 2020 at 15:12
  • $\begingroup$ thanks, Denis, I have added a reference to a book where the proof is written out. $\endgroup$
    – Carlo Beenakker
    Commented Mar 21, 2020 at 15:38
  • $\begingroup$ But isn't strict hyperbolicity (i.e. having real and distinct eigenvalues) a sufficient condition for a matrix to be symmetrizable? $\endgroup$
    – user140746
    Commented Mar 22, 2020 at 21:31
  • $\begingroup$ I don't think so : a matrix can have real distinct eigenvalues and not be symmetrizable. $\endgroup$
    – Carlo Beenakker
    Commented Mar 22, 2020 at 21:39
  • $\begingroup$ @CarloBeenakker But isn't in that case the matrix diagonalizable (and therefore simmetrizable)? $\endgroup$
    – user140746
    Commented Mar 26, 2020 at 20:05

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