The wave equation in physics

The wave equation describes physical processes which follow the same pattern in space and time. For example, if you plot the height of the surface of the sea vs. horizontal position, it's a sine (or cosine) wave. On the other hand, if you plot the height of the surface of the sea at a given point vs. time, it's also a sine (or cosine) wave. Space and time are symmetric as far as oscillations are concerned. The wave equation has second derivatives with respect to both space and time.

We'll only solve the wave equation in one dimension here. That corresponds to the motion of a vibrating string, $y(x,t)$. In two dimensions, e.g. in the case of a drum, $z(x,y,t)$, an additional separation of variables step is needed after the time part has been separated.

General solution

The wave equation is, generally: $\nabla^2z=\frac{1}{v^2}\frac{\partial^2z}{\partial t^2}$,
or, to keep things simple, in 1D: $\frac{\partial^2y}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2y}{\partial t^2}$.
Assume product solution: $y(x,t)=X(x)\cdot T(t)$.
Substitute product: $T\frac{{\rm d}^2X}{{\rm d}x^2}=\frac{X}{v^2}\frac{{\rm d}^2T}{{\rm d}t^2}$,
separate terms, and introduce separation constant: $\frac{1}{X}\frac{{\rm d}^2X}{{\rm d}x^2}=\frac{1}{v^2T}\frac{{\rm d}^2T}{{\rm d}t^2}=-k^2$.
This leaves us with two ODEs with constant coefficients: $\frac{{\rm d}^2X}{{\rm d}x^2}+k^2X=0$, $\frac{{\rm d}^2T}{{\rm d}t^2}+k^2v^2T=0$
The 2nd-order coefficients are: $a_2=1$, $a_2=1$,
the 1st-order coefficients are: $a_1=0$, $a_1=0$,
and the zero-order coefficients are: $a_0=k^2$, $a_0=k^2v^2$.
The roots of the characteristic polynomials are: $-\frac{0}{2}\pm\sqrt{\frac{0^2}{4}-k^2}=\pm {\rm i}k$, $-\frac{0}{2}\pm\sqrt{\frac{0^2}{4}-k^2v^2}=\pm {\rm i}kv$,
and the solutions are: $X(x)={\rm e}^{{\rm i}kx}=\begin{cases}\sin{kx}&\\\cos{kx}\end{cases}$, $T(t)={\rm e}^{{\rm i}kvt}=\begin{cases}\sin{kvt}&\\\cos{kvt}\end{cases}$.

Note that the constant $k$ in both solutions is the same. Both are complex exponentials, i.e. oscillations, which is consistent with the observation of waves on a lake as indicated above. As $\sin{kx}$ describes the spatial oscillation, $k$ must be measured in m-1. It is the wave number of the oscillation. To balance the units in $\cos{kvt}$, $v$ must be in ms-1; it is the velocity of the wave. In physics, it is more common to join the two constants together and specify the angular frequency, $\omega=2\pi kv$, instead.

Finally, we need to put the two partial solutions back together. Therefore the general solutions of the wave equation are:

$$y(x,t)=X(x)\cdot T(t)=\begin{cases}\cos{(kx)}\cos{(\omega t)}&\sin{(kx)}\cos{(\omega t)}\\\cos{(kx)}\sin{(\omega t)}&\sin{(kx)}\sin{(\omega t)}\end{cases}$$

To conclude the section on PDEs, we'll have another example of how to use boundary conditions to find the physically sensible solutions of the wave equation for a given problem.