Why is a first order PDE hyperbolic? 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?
 A: 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).
