Characteristics in the theory of PDE's - What's Going On?

Warning: the following is by no means rigorous or even remotely constitute as proof of anything. This is just an intuition about characteristic lines to understand what purpose they really serve and how they relate to discontinuities.

TL;DR: Characteristic curves are curves where values propagate as in the transport equation. This is why derivative = $0$. Only characteristic curves can propagate discontinuities, as in the discontinuities don't survive the next timestep outside those curves. Therefore, looking for discontinuities (or persistent discontinuities) amount to looking for characteristic curves. Hence, defining characteristic curves as where the higher order terms are discontinuous.

The way I understand characteristic curves is that I think of them as some parametric curves where things don't change much. (As in some derivatives/partial derivatives/gradient = $0$)

The reason these things are useful is because these curves allow us to simplify PDE's into simpler differential equations that are easier to analyze. (Pro-tip: it's always about $\frac{dy}{dx}$, at least in 2d)

In your first example, you showed it when you said that you want $$ \vec{v} = \left(\frac{\partial^2 \phi}{\partial x^2}, \frac{\partial^2 \phi}{\partial x \partial y}, \frac{\partial^2 \phi}{\partial y^2}\right) = \vec{0}. $$ This is intuitively the second order equivalent of saying that $\frac{du}{dt} = 0 $ in first order equations.

More on intuition:

Think back to the transport equation (I know... basics). What does the transport equation do along characteristic curves? It just propagates some value forward in time in a very clean and simply describable way. The same thing is true even with second order equations except, it might not be as simple as in the transport equation case.

Example:

Say $u$ is a function of $x$ and $t$ and $x$ is a function of $t$, i.e: $u(x(t), t)$

Now, suppose you wanted to see how u changed with time: i.e: $\frac{du}{dt}$

$\frac{du}{dt} = u_x \frac{dx}{dt} + u_t$ by chain rule

But if you look closely, this is just the transport equation in disguise:

recall: transport eq: $0 = cu_x + u_t$. So the transport equation has the peculiarity of $\frac{dx}{dt} = c$ and

$\frac{du}{dt} = 0$ <----- Our famous zero derivative

(helpful source: http://www.youtube.com/watch?v=tNP286WZw3o )

So, if we apply the "pro-tip" with x as y and t as x, we have that the characteristic curves are described by $\frac{dx}{dt} = c$, then $x = ct + x_0$ defines our characteristic curves given that $x_0$ is the initial value for $x$

Then, since $\frac{du}{dt} = 0$, u is constant along the characteristic lines.

And in reality, this is all characteristic curves are. Think of them as little rivers of simple behavior in an otherwise complex topological landscape filled with scary discontinuities and shocks and other horrible monsters that come out of PDE's.

Now, why did the first source you cited focus on discontinuities on the higher order terms? Turns out it's a nice trick. Basically, only characteristics can handle discontinuities because there's no chance for characteristic curves to do crazy stuff. (their behavior is in some way loosely constant-ish...)

Check out this source at § 1.5.2 Characteristics as the carriers of discontinuity

Hopefully the source will be enlightening a bit because you'll see some of the talk of the whole vanishing stuff.

So, if you know that the discontinuities will only be carried through by characteristic curves, then you can "define" that characteristic curves are the curves such that the higher order terms are discontinuous.

If you have a hard time seeing this whole characteristic curves as carriers of discontinuity, think of Burger's equation. (I know, it's non-linear, but bear with me).

Look at where in Burger's equation you are forced to insert the shocks: where characteristic lines meet. Burger's equation, as non-linear as it looks, doesn't get any other discontinuities but those due to the characteristic lines.

This is why your first source goes for the discontinuities at the higher order terms. Any discontinuities that may occur, as the professor in the video says, will just vanish and not be there at the next timestep. Especially in cases like the heat equation with infinite propagation speed.

Hope this helps.