Why are certain PDE called "elliptic", "hyperbolic", or "parabolic"?

A general 2nd order linear PDE in two variables is written

$$Au_{xx} + 2Bu_{xy} + Cu_{yy} + Du_x + Eu_y + F = 0$$

and $A,B,C,D,E,F$ can be functions depending on $x$ and $y$. We say a PDE is elliptic, hyperbolic or parabolic if \begin{align} B^2 - AC &= 0, &\text{parabolic} \\ B^2 - AC &>0, &\text{hyperbolic} \\ B^2 - AC &<0, &\text{elliptic} \end{align} Note that if $A,B,C,D,E,F$ depend on $x$ or $y$, there can be regions where the PDE is elliptic, hyperbolic or parabolic and different techniques are used to solve each type. If the coefficients are constant the naming comes form considering the polynomial equation $$Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F = 0$$ depending on the sign of $B^2 - AC$, this forms an ellipse, hyperbola or parabola in $\mathbb{R}^2$. This can be extended to higher dimensions as well with hyperboloids, paraboloids, or ellipsoids.


All quadratic curves can be studied using the equation $Ax^2+2Bxy+Cy^2 + Dx + Ey + F=0$ the discriminant of which is $B^2-AC$ and the solution curve will be a ellipse, hyperbola, or parabola depending on whether the discriminant is positive, negative, or zero.

Partial second-order differential equations take a much similar form with $Au_{xx}+2Bu_{xy}+Cu_{yy} + Du_x + Eu_y + F = 0$ and the discriminant plays a similar role in classifying the solutions so they are named after the algebraic curves that resemble this expression.


I'd like to address this comment of the OP to one of the answers:

But I think the solutions of the corresponding differential equations have nothing to do with their name? Or do they have?.

Actually, we can relate the names to the corresponding geometrical curves. To do this, consider how the following homogeneous PDE will look in the Fourier-transformed form.

Original PDE (with $u^{(n,m)}(x,y)$ denoting $n$th partial derivative of $u$ in $x$ and $m$th in $y$):

$$Au^{(2,0)}(x,y) + 2Bu^{(1,1)}(x,y) + Cu^{(0,2)}(x,y) + Du^{(1,0)}(x,y) + Eu^{(0,1)}(x,y) = 0.\tag1$$

Fourier-transformed one (with $\hat u(k_x,k_y)$ denoting the Fourier transform of $u(x,y)$):

$$\mathcal L\hat u(k_x,k_y) = 0,\tag2$$

where

$$\mathcal L=Ak_x^2 + 2Bk_xk_y + Ck_y^2 + Dk_x + Ek_y.\tag3$$

This multiplication by $\mathcal L$ is the Fourier-space version of the differential operator from $(1)$. Notice that $\mathcal L$ is just a second-degree polynomial in $k_x$, $k_y$ — a quadratic form. Now consider the nodes of $\mathcal L$ in the $(k_x,k_y)$ plane: they will be defined by the equation

$\mathcal L=0.\tag4$

Geometrically, these nodes will be exactly the kinds of curves with the names corresponding to the name of the kind of PDE, i.e. ellipses, hyperbolas or parabolas.