This is simply a rescaling of the axes in $k$-space. Since in your 1D-example the first reciprocal lattice point is at $2\pi/a$, dividing the point at the brillouin zone boundary by this value gives $1/2$, as is stated in the text. So the point $a^*/2$ is not, as you assumed, the position of the brillouin zone boundary in reduced units, but the boundary in the not-reduced units.
I assume $a^* \equiv 2\pi/a$ is the length of the reciprocal lattice vector, since it would make sense in this context.
Edit 1
For phonons, the reason why the brillouin zone boundary is halfway to the first reciprocal lattice point is that the shortest wavelength you can have is a sign change from one atom to the other. Picture a chain of atoms with the first up, the second down, the third up again. There is no shorter wavelength than this. We also know that the solutions are plane waves (in the simplest case), which means (1D) $s(x) = \mathfrak{Re}(A\cdot e^{ikx})$, where $s(x)$ is the amplitude of the atom at position $x$ and $A$ is the maximum amplitude of the oscillation. For this to change sign from site to site, $k\overset{!}{=}\pi/a$, which you can easily verify.
As to how to construct the first brillouin zone, have a look at any solid state physics book. You just draw lines from the origin to every reciprocal lattice point and bisect them with a plane perpendicular to the line. Every point you can get to without crossing any of these planes is in the first brillouin zone, and the planes themselves are the boundaries.
Edit 2
The brillouin zone is constructed in such a way that it is sufficient to consider all $k$-points inside it, as it can be shown that they are equivalent to points outside. We know the waves to have bloch form
$$s(x) = e^{ika} u(x)$$
where $u(x)$ has the periodicity of the lattice. From this expression we can see, that $ka$ gives you the phase change from one lattice site to the next. If now $ka$ is bigger than $\pi$, say $\pi+\Delta$, this point on the $k$-axis is equivalent to $-\pi+\Delta$, because $e^{i(\pi+\Delta)}=e^{i(\pi+\Delta -2\pi)}=e^{i(-\pi+\Delta)}$. So we see that looking at $k$-points up to $\pi/a$ is sufficient for all properties, because the points outside have an equivalent point inside. And by construction of the reciprocal lattice (its first point in the positive direction is at $2\pi/a$), this is precisely at $a^*/2$.
Wave equation
I think the question is vaguely posed, since the answer depends on what we define as waves and wave equations. In the question cited in the OP many answers simply assumed that waves mean electromagnetic waves and wave equations means
$$
\partial_t^2u(\mathbf{x},t)=c^2\nabla^2u(\mathbf{x},t).$$
The dispersion relation in this case is obvious:
$$\omega^2-c^2\mathbf{k}^2=0.$$
Linear equations
One could talk about waves in more general sense, as solutions to any linear equation, solvable via Fourier transform, i.e., having solutions
$$
u(\mathbf{x},t) =\int d\mathbf{k}\int d\omega \tilde{u}(\mathbf{k},\omega)e^{i(\mathbf{k}\mathbf{x}-\omega t)},$$
in which case any linear operator would suffice
$$F(\partial_t, \nabla)u(\mathbf{x},t)=0.$$
By choosing function $F(\partial_t, \nabla)$ one could get almost anything. E.g.,
$$\partial_t^4u(\mathbf{x},t)=a\nabla^8u(\mathbf{x},t) + \nabla^4u(\mathbf{x},t) + cu(\mathbf{x},t)$$
has several dispersion branches.
Among more basic equations with several branches one could cite Dirac equation and Klein-Gordon equation (the latter being simply the wave equation with a constant term added).
Non-linear equations
One could go even further and consider non-linear equations that allow running solutions of the type $$f(\mathbf{k}\mathbf{x}-\omega t),$$ such as, e.g., Korteveg-de Vries equation or Sine-Gordon equation.
Which of these equations do happen?
In university physics courses one typically deals with linear theories, because the fundamental phsyics is described (mainly?) by linear theories. In more domain-specific courses one however quickly encounters equations that have higher derivatives or non-linear terms. The domains to look for more complex equations are:
- hydrodynamics
- elasticity theory
- electrodynamics of non-linear media
- non-linear theory (which deals more specifically with the equations rather than their physical content).
Remarks
First-order equations One can have also first-order wave equations, e.g.,
$$\partial_t u(x,t)\pm v\partial u(x,t)=0,$$
which give actial travelling wave solutions of type $f(x-vt)$. In more dimensions:
$$\partial_t u(\mathbf{x},t)-\mathbf{v}\cdot\nabla u(\mathbf{x},t)=0.$$
The nuance of these equations is taht they have a preferred direction for the wave propagation (even in 1D we have either right- or left-moving wave, depending on the sign). This is why the physical theories that are symmetric in space and/or time usually have second (or generally even) partial derivatives.
One example of an equation with such a first-order term is the Navier-Stokes equation, although it is a non-linear one (but it can be linearized to give simple wave solutions).
Waves vs. running waves When dealing with general form of equation $F(\partial_t, \nabla)u(\mathbf{x},t)=0$, it is necessary to keep in mind that, although it is solvable by Fourier transform, its solutions are not necessarily running waves of the form $f(\omega t-\mathbf{k}\mathbf{x})$. Requiring that solutions have this form would restrict the type of the differential operators that can be used, e.g., excluding diffusion equation.
Schrödinger equation On the other hand, Schrödinger equation (which can be viewed as a diffusion equation with complex coefficients) is certainly considered a wave equation and its solutions are often referred to as matter waves, even though they are not running waves in the restricted sense mentioned above.
Broad/flat band limit in some solid state phsyics problems one considers a broad-band limit where all electrons are assumed to have the same wave function (or wave number), while possibly ahving different energies - this can be interpreted as a continuum of frequencies corresponding to the same wavelength. The opposite and also used is the flat-band limit, where one assumes that all the wave numbers correspond to the same energy/frequency.
Best Answer
I think the short version is that dispersion relations only tell a small part of the story of a phonon. They say nothing about the motion of the corresponding wave; all they tell you is how the oscillatory frequency is related to the wavevector.
As for how to read the curves: you read them just like any function. There is no inherent cause and effect, just as there is no inherent cause and effect for $y = x^2$. $x=2$ does not cause $y=4$ any more than $y=4$ causes $x=2$. For phonons, an oscillatory source with a given frequency would cause phonon(s) with specific wavelengths, and an excitation with a given wavelength would cause phonons(s) with specific frequencies. The cause depends on what you do --- not on a curve.
Let's start with a simpler system: an infinite, isotropic, continuum solid (e.g. jello). This systems supports two kinds of waves: longitudinal (also known as pressure or p-waves) and transverse (also known as shear or s-waves). The two types of waves will, in general, have different velocities --- say $c_l$ and $c_t$, respectively. The dispersion relation for "phonons" in this system will then have two branches: $\omega_l = c_l \left|\vec{k}\right|$ and $\omega_t = c_t \left|\vec{k}\right|$.
The real world meaning is that, at any given $\vec{k}$, both longitudinal and transverse waves can exist. They will, however, look quite different (see below --- taken from Wikipedia). In this system, you can find longitudinal and transverse waves to match any frequency.
When you're working with atoms (or meta materials), things become more complicated, but the long-wavelength limit of acoustic phonons converges to elastic waves in a continuum system. To answer your questions: