Why is a first order PDE hyperbolic?

Question

I am learning about classification of PDE's into hyperbolic, parabolic and elliptic PDEs and I was reading this post which if I understand correctly says that first order PDE are hyperbolic. However if we have a second-order PDE of the form $a\partial_{xx}u+b\partial_{xy}u+c\partial_{yy}u+d\partial_xu+e\partial_yu+fu=0$ then we can classify it as follows :

$b^2-4ac>0$ : hyperbolic
$b^2-4ac<0$ : elliptic
$b^2-4ac=0$ : parabolic

However, for a first order PDE we have $a=b=c=0$ so shouldn't the first order PDE be parabolic?

It could be derivated wrt x and then the criterion is about $e^2-4d$ ? — QuantumPotatoïd, Feb 15 '22 at 17:42
$a=b=c=0\Rightarrow d\partial_x u+e\partial_y u +fu=0\Rightarrow d\partial_{xx}u+e\partial_{xy}+f\partial_x u=0,d\partial_{xy}u+e\partial_{yy}u+f\partial_y u=0$. Thus we could take your classification with $e^2-4d$ or $d^2-4e$ ? — QuantumPotatoïd, Feb 16 '22 at 03:00
We can see too that : The solutions of $u'=fu$ are expressible as functions of sinh and cosh : hyperbolic functions. — QuantumPotatoïd, Feb 16 '22 at 17:40
@Cretin2, could you explain why you consider the EDO $u'=fu$? — edamondo, Feb 16 '22 at 23:07

Artem · Answer 1 · 2022-02-15T21:41:02.353

2

There is much more general definition of a PDE being hyperbolic (or even a system of PDE being hyperbolic). You can find it in many (usually graduate level and above) textbooks.

Non-technically speaking a PDE of order $n$ is called hyperbolic if an initial value problem for $n-1$ derivatives is well-posed, i.e., its solution exists (locally), unique, and depends continuously on initial data.

So, for instance, if you take a first order PDE (transport equation) with initial condition $$ u_t+u_x=0,\quad u(0,x)=f(x), $$ then it can be shown that this problem is well-posed and hence hyperbolic (as for any other first order PDE under some technical conditions about characteristics and the curve of initial condition).

If you take the usual wave equation with two initial conditions $$ u_{tt}-u_{xx}=0\quad u(0,x)=f(x), u'_t(0,x)=g(x), $$ then it can be shown that this problem is well-posed and hence the wave equation is hyperbolic (as well as any other second order PDE with $b^2-4ac>0$ according to your classification).

Finally, if you take the Laplace equation with two initial conditions $$ u_{tt}+u_{xx}=0\quad u(0,x)=f(x), u'_t(0,x)=g(x), $$ then it can be shown that this problem is not well-posed and hence not hyperbolic (it is actually elliptic).

edited Feb 15 '22 at 21:41

answered Feb 15 '22 at 21:35

Artem

14,414

Thank you. But then how is parabolic and elliptic defined in general? Also is is true that in the case of second order PDE, the problem is well posed iff $b^2-4ac>0$? – edamondo Feb 16 '22 at 11:17
1

@edamondo For your first question, please read this post, it is difficult to explain it better. For the second question: note that I said well-posed for the initial value problem. It is possible to pose different problems. For instance for Laplace equation it is boundary value problem that is well-posed, etc. This is an interesting mathematical question: how to set the additional conditions for the given PDE so that the problem you obtain is well posed. – Artem Feb 16 '22 at 12:37
Okay, so hyperbolic problems are necessarily initial value problems? I guess I then should have rather asked : is is true that in the case of second order PDE with initial value, the problem is well posed iff $b^2-4ac>0$? – edamondo Feb 16 '22 at 23:05
First question: not necessarily; second question: no. – Artem Feb 17 '22 at 01:11

Why is a first order PDE hyperbolic?

1 Answers1