Are there rigorous mathematical definitions for these waves? My friend linked this .gif to me tonight, and asked me if I knew of any equations that might model these bottom two waves (the blue and green waves). Unfortunately, I am not far enough in my education to recognize if any such model exists. Are these waves modeled after some equation, or is this just some piece of eye candy?

 A: All can be modeled by some $2 \pi$ periodic function $r:\mathbb{R} \to \mathbb{R}$, then the equation is $t \mapsto r(t) \sin t$.
For the red wave, use $r(t) = 1$.
For the blue wave, use $r(t) = \sqrt{1+\sin^2t}$, for $t \in [-\frac{\pi}{4}, \frac{\pi}{4})$, and let $r$ by $\frac{\pi}{2}$ periodic.
For the green wave, use the same formula as for the blue wave, except for the domain $[-\frac{\pi}{6}, \frac{\pi}{6})$, and let $r$ by $\frac{\pi}{3}$ periodic.
A: So, for the fist one obviously, it's sine function.
$$
h(t) = r \sin \omega t
$$
As for the second one,
$$
h(t) = \frac a2 \cdot \left \{ \begin{array}{lcc}
\tan \left (-\frac \pi 4 + \omega t \right )& \text{if} & 0 \le t \le \frac T4 \\
1 & \text{if} & \frac T4 \le t \le \frac T2 \\
\tan \left ( \frac {3\pi}4 + \omega t \right ) & \text{if} & \frac T2 \le t \le \frac {3T}4 \\
-1 & \text{if} & \frac {3T}4 \le t \le T
\end{array}\right .
$$
For the last one,
$$
h(t) = a \cdot \left \{ \begin{array}{lcc}
\frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t\right) & \text{if} & 0 \le t \le \frac T6 \\
\frac 12 + \frac 12 \left [\frac 12 + \frac {\sqrt 3}2 \tan \left( -\frac \pi 6 + \omega t - \frac \pi 3\right) \right ] & \text{if} & \frac T6 \le t \le \frac T3 \\
1 - \frac 12 \left [\frac 12 + \frac {\sqrt 3}2 \tan \left( -\frac \pi 6 + \omega t - \frac {2\pi} 3\right) \right ] & \text{if} & \frac T3 \le t \le \frac T2 \\
\frac {\sqrt 3}2 \tan \left( \pi - \omega t + \frac \pi 6 \right) & \text{if} & \frac T2 \le t \le \frac {2T}3 \\
-\frac 12 - \frac 12 \left [ \frac 12 + \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t - \frac {4 \pi}3 \right )\right ] & \text{if} & \frac {2T}3 \le t \le \frac {5T}6 \\
-1 + \frac 12 \left [ \frac 12 + \frac {\sqrt 3}2 \tan \left ( -\frac \pi 6 + \omega t - \frac {5 \pi}3\right )\right ] & \text{if} & \frac {5T}6 \le t \le T
\end{array}\right .
$$
PS: For last two cases $a$ is a square or hexagon side. $T = \frac {2\pi}\omega$
Initial time angles are chosen according to this

