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.

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.