Waves - determining whether a given formula represents a wave Well the basic formula of a wave needs to contain $$y(x,t) = f(x \pm vt)$$ where the sign depends upon the direction of propagation of the wave. 
However, not every function in the form $y(x,t) = f(x \pm vt)$ will represent a wave. Why is this so? An example given to me was $(x - vt)^2$. Why does this not represent a wave? It is defined for all values of $x$, so what is the problem? And what is the best way to recognize whether a given function represents a wave or not?
 A: 
In physics, a wave is disturbance or oscillation (of a physical quantity), that travels through matter or space, accompanied by a transfer of energy.

This is the description of a wave in Wikipedia. If you plot the function you have given in your question, you would see that the the function $x^2$ travelling in space. I don't see a reason why this function cannot be considered as a wave, since it also obeys the wave equation. It is not an ordinary wave and the values of the function stretch to infinity (which means an infinite amplitude) but in principle if you limit this function to an interval $[a,b]$ it is a perfect wave and you can find its Fourier components.
In my opinion the best way to see whether a function is a wave or not, would be to plug the function in the wave equation. If it holds, it is a wave (or at least it describes a wave for a given interval). If you don't want to do the algebra, you can also plot it using GeoGebra or a similar software and fiddle around with $t$ in order to see if the function actually translates when you change $t$. 
A: Weakly speaking, your proposed function is indeed a propagating disturbance (a wave) which obeys the wave equation, but it carries infinite energy and goes to infinity on both sides (so it is not physically meaningful situation). You can restrict this in a more quantitative fashion: for a wave packet to have a finite energy, it has to be square integrable ($L^2$ space), that is, $\int y^2\,dx<\infty$, which also as a consequence, means, that it has a well defined frequency spectrum. This even excludes an infinite sine wave, because it carries an infinite energy. A weaker class of signals (very physically significant) are signals of finite power, so: $\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T f^2 \,dx<\infty$. This allows neverending oscillatory waves but excludes nonphysical nonsense as ever-growing signals. This is usually a reasonable restriction. However, infinite spikes in amplitude are still allowed, so I'd say that just restriction in amplitude ($|f(x,t)|<M<\infty$ for all $x,t$) is a plain and simple common sense rule of what is a valid wave.
Also, instead of trying to find the form $f(x\pm vt)$, you can always just put the function into the wave equation and check if it holds:
$$\frac{\partial^2 f }{\partial x^2}-\frac1{v^2}\frac{\partial^2 f }{\partial t^2}=0$$
where I used $v$ for wave speed, following notation in the question.
To sum up: every function that satisfies the wave equation is a wave. However, every physical model is composed of the differential equation, its boundary and initial conditions, and its domain where it's defined. The boundary conditions exclude infinitely growing functions and domain excludes spikes/poles/gaps. Everything else is ok.
A: At $t \to -\infty$ your function differs from zero in all the space, at any finite point. So, at any point in the universe you have a source. As $t$ increases, there appears an quenching of the source at the location $x = vt$, and this quenching propagates in space with the velocity $v$ in the positive direction of the $x$ axis. In this interpretation, your function can be a wave.
However, there are problems: at any given time the intensity of your function over all the space is infinite
$lim_{A \to \infty} \int_{-A}^A [(x - vt)^2]^2 dx = lim_{A \to \infty}\left( \frac {2A^5}{5} + \frac {4A^3}{3} + 2v^4t^4A \right)$ .
No physical wave can comprise infinite energy. We use frequently in our calculi Fourier waves ~ $C_0 e^{i(kx - \omega t)}$, and the intensity of such a wave integrated over all the space is infinite, but in practice any wave is limited in space. The Fourier waves are only an ideal approximation, which we take in consideration in our calculi.
A: A wave is usually just a finite disturbance (usually periodic as well). Just like the ocean waves are disturbances in the flat water level, electromagnetic waves, for example, are disturbances in the electric and magnetic fields in the flat level of empty space, in which no fields are present.
Usually when talking about waves, one thinks of a periodic function moving at a constant speed, like in your case. But, unlike what I described previously, in your case the disturbance is not finite like a sine or a square wave, but infinite. Your function grows bigger as you get farther away from x=vt (or x=-vt), which means it is not a physical wave describing a physical phenomenon like the water level or an electromagnetic field. Therefore, you might not call this function a "wave", even though it is a solution of the wave equation.
