# Why are second order linear PDEs classified as either elliptic, hyperbolic or parabolic?

Is there a geometric interpretation of second order linear partial differential equations which explains why they are classified as either elliptic, hyperbolic or parabolic, or is this just a naming convention? That is, do they have any relation with actual ellipses, hyperbolas and parabolas?

• Feels related to my own question on the term “elliptic”, currently without an answer. I realize that I failed to mention PDEs in that list, and probably forgot some other uses as well. – MvG May 8 '16 at 20:11

Yes, there is. In elliptic case, we have that $A:=(a_{ij})$, the matrix corresponding to second derivatives, is positive definite. Psitive definiteness implies that $$v^t.A.v$$ is positive for all $v\neq 0$ and zero for $v=0$. If one draws all the $v$'s which satisfy $v^t.A.v =c$ for a constant $c$, it will be an oval-like shape, an ellipsis (in n-dimension of course.) So, the term elliptic.

Case of parabolic: The are defined as having their "symbol" of the form $\frac{d}{dx_n}-\{an \ elliptic \ part\}$. In view of above, think of the elliptic part as ($x_1^2+x_2^2+x_3^2+...+x_{n-1}^2$). The general case will be same but with positive weights behind each squared term. Anyway, what will shapes defined as $$x_n-(x_1^2+x_2^2+x_3^2+...+x_{n-1}^2)=c,$$ $c$ a constant look like? A parabola.

The hyperbolic case: The symbol is of the form $$\frac{d^2}{dx_n^2}-\{an\ elliptic \ part\}$$

Solutions to $$x_n^2-(x_1^2+x_2^2+x_3^2+...+x_{n-1}^2)=c,$$ are hyperbolas.

BUT, they don't classify them simply because of this. They do, because all elliptic operators share many qualitative behavior. All parabolic ones, although similar to elliptic, have some other common features. And finally, the hyperbolic ones are totally different in behavior. It is interesting that changing a + in an elliptic to a - (making it hyperbolic) does such a drastic change to the PDE.

BUT, again! Many important PDE's do not fall within any class, in fact. However, understanding the basic cases is always the first step in math. That is why people spent so much time investigating these cases in a kind of isolated fashion.

• Interesting, thank you! Just out of curiosity, do hyperbolic PDEs have anything to do with hyperbolic geometry? (I'm wondering about this because, in physics, some hyperbolic PDEs, like the electromagnetic wave equation, are invariant under Lorentz transformations, which are geometrically hyperbolic rotations of the 4-dimensional Minkowski space. Is this just coincidence?) – David Herrero Martí May 8 '16 at 21:56
• I have no idea except that I know in differential geometry hyperbolic means negative curvature. I don't know if from this we get a hyperbolic PDE, but I think not. – Behnam Esmayli May 8 '16 at 23:07