21.1 Introduction

The wave equation in two dimensions,

\begin{displaymath}
u_{tt} = a^2 u_{xx}
\end{displaymath}

has the general solution

\begin{displaymath}
u(x,t) = f_1(x-at) + f_2(x+at)
\end{displaymath}

Here $f_1(x-at)$ is a function that moves to the right with speed $a$; a 'right-going wave'. And $f_2(x-at)$ is a function that moves to the left with speed $a$; a 'left-going wave'.

This solution was derived earlier. The functions $f_1$ and $f_2$ must be found from whatever initial and boundary conditions are given. One special case of that is the D'Alembert solution, which is the subject of this brief chapter.

In its simplest form, the D'Alembert solution assumes that there are no boundaries. That means that the $x$-range is doubly infinite:

\begin{displaymath}
-\infty<x<\infty
\end{displaymath}

Therefore, only initial conditions are needed at the starting time $t$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0:

\begin{displaymath}
u(x,0) = f(x) \qquad u_t(x,0) = g(x)
\end{displaymath}

Here $f$ and $g$ are assumed to be given functions.

Note that since the wave equation is second order in time, it needs two initial conditions. Physically, the wave equation might describe the vibrations of a string. The two initial conditions are then that both the initial position and the initial velocity of the string must be given at each point.

The two given initial conditions allow the unknown functions $f_1$ and $f_2$ in the solution of the wave equation to be determined. That gives the solution directly in terms of the given functions $f$ and $g$:

\begin{displaymath}
\fbox{$\displaystyle
u(x,t) = \frac{f(x-at) + f(x+at)}{2}
+ \frac{1}{2a} \int_{x-at}^{x+at} g(\xi) { \rm d}\xi
$}
\end{displaymath} (21.1)

This is derived in example 4.10 in the book.

The solution can be understood more physically from looking at the $x,t$-plane:

\begin{displaymath}
\hbox{\epsffile{dal.eps}}
\end{displaymath}

To get the solution $u$ at a position and time P, we need to average the $f$, (i.e. $u$), values at points Q and R. To that we need to add an integral of $g$, (i.e. the velocity $u_t$), between points Q and R:

\begin{displaymath}
u_P = \frac{u_Q + u_R}{2}
+ \frac{1}{2a} \int_Q^R u_t { \rm d}\xi
\end{displaymath}

Note that the solution at point P depends on the initial conditions in the interval QR. In other words, the ``region of dependence'' of point P is the triangle QPR. Whatever is outside that triangle does not affect the solution at P at all.

Conversely, point P is inside the region of influence of all points inside the triangle. A change in the initial conditions at any initial point inside the triangle will influence the solution at P.