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The title of the question says it all. Why can we only apply the method to parabolic PDEs like the heat equation, and not to ordinary PDEs?

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Who gave you that idea?

You absolutely can use Finite Differences for other PDEs. They are routinely used to solve hyperbolic PDEs (wave equation, both first and second order) and elliptic PDEs (steady state diffusion/heat equation). You can even mix and match the equation types and create PDEs that have characteristic of both hyperbolic and parabolic equations, such as the Navier-Stokes equations.

If you're interested in learning how to implement solvers for these, most numerical fluid mechanics textbooks have a pretty thorough treatment on discretizing PDEs of many types with finite differences.

For reference:

First order wave equation: $\frac{\partial u}{\partial t} + \nabla \cdot(c\,u) = 0$

Second order wave equation: $\frac{\partial^2 u}{\partial^2 t} - c^2 \nabla^2 u = 0$

Elliptic diffusion: $\nabla^2 u + f = 0$

Navier Stokes:

$$ \rho\left(\frac{\partial\mathbf{u}}{\partial t} + (\mathbf{v}\cdot\nabla)\mathbf{v}\right) = -\nabla{P} + \mu\nabla^2 \mathbf{u} $$

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  • $\begingroup$ thanks for your answer, it makes perfect sense. So that would mean that even if we have a PDE of first whose partial derivatives are respect to a variable $x$, we could safely use the method of finite differences? $\endgroup$ – Adam Sep 16 '15 at 12:34
  • $\begingroup$ Most likely. What PDE are you trying to solve? $\endgroup$ – Tyler Olsen Sep 16 '15 at 12:35
  • $\begingroup$ I was just simply thinking of a generic first order PDE. I got asked this question at an interview for a quant role and it took me a bit aback. $\endgroup$ – Adam Sep 16 '15 at 12:38
  • $\begingroup$ Gotcha. Yes, you should be able to do it. You have to be careful, though, to ensure that your discretization (forward difference, backward difference, central difference,...) is stable. There are rigorous ways to do this (eg, Von Neumann analysis for linear PDEs). It can require some experience and intuition to get it right on the first try, as a discretization that works for one PDE can often fail for another. (try coding up a few different discretizations of the linear 1st-order wave equation, as an illustrative example). $\endgroup$ – Tyler Olsen Sep 16 '15 at 12:45

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