Consider the following 2D linear PDE: $$ a_1 \frac{\partial^2 \phi}{\partial x^2} + 2 a_2 \frac{\partial^2 \phi}{\partial x \partial y} + a_3 \frac{\partial^2 \phi}{\partial y^2} + a_4 \frac{\partial \phi}{\partial x} + a_5 \frac{\partial \phi}{\partial y} + a_6 \phi = b $$ where the coefficients $a_1,\dots,a_6,b$ are all functions of $x,y$ (i.e. this is the most general second-order 2D linear PDE). A lot of books on PDEs seem to make a big deal about what type the equation is, i.e. elliptic, hyperbolic, or parabolic. For example an elliptic PDE is defined as one in which $$ a_2^2 - a_1 a_3 < 0 $$ From the point of view of actually wanting to solve the equation numerically (given appropriate boundary conditions), I don't see why knowing the type of equation it is has any relevance.
I would use a 2nd order Taylor series expansion to discretize all the derivatives into a finite difference approximation $$ A \boldsymbol{\phi} = \mathbf{b} $$ and then just solve the resulting sparse matrix system. It seems to me this would work just fine no matter what the value of $a_2^2 - a_1 a_3$ is. I suspect a similar statement could be made if using Finite Elements or Finite Volume methods. So my question is why do people care so much about the type of PDE when a numerical solution is desired?
A related question is related to Multigrid methods. Multigrid methods seem to be applied primarily to elliptic PDEs. But for the example above, if I choose appropriate mesh sizes, why can't I apply the same multigrid method regardless of the value of $a_2^2 - a_1 a_3$? Do multigrid methods work only when this discriminent is negative for some reason?