Chapter 12.4, Problem 20E

The vertical displacement u(x, t) of an infinitely long string is determined from the initial-value problem

$\begin{array}{l}{a}^2\frac{{\partial }^{2}u}{\partial x^2}=\frac{{\partial }^{2}u}{\partial t^2},-\infty <x<\infty ,t>0\\ u(x,0)=f(x),{\frac{\partial u}{\partial t}|}_{t=0}=g(x).\end{array}$

This problem can be solved without separating variables.

1. (a) Show that the wave equation can be put into the form ${\partial }^{2}u/\partial \eta \partial \xi =0$ by means of the substitutions ξ = x + at and η = x – at.
2. (b) Integrate the partial differential equation in part (a), first with respect to η and then with respect to ξ, to show that u(x, t) = F(x + at) + G(x – at), where F and G are arbitrary twice differentiable functions, is a solution of the wave equation. Use this solution and the given initial conditions to show that

   $\begin{array}{l}F(x)=\frac{1}{2}f(x)+\frac{1}{2a}{\displaystyle {\int }_{x_0}^{x}g(s)}ds+c\\ \text{and}G(x)=\frac{1}{2}f(x)-\frac{1}{2a}{\displaystyle {\int }_{x_0}^{x}g(s)}ds-c,\end{array}$

   where x₀ is arbitrary and c is a constant of integration.

3. (c) Use the results in part (b) to show that $u(x,t)=\frac{1}{2}[f(x+at)+f(x-at)]+\frac{1}{2a}{\displaystyle {\int }_{x-at}^{x+at}g(s)}ds$. (14)

   Note that when the initial velocity g(x) = 0, we obtain $u(x,t)=\frac{1}{2}[f(x+at)+f(x-at)],-\infty <x<\infty .$

   This last solution can be interpreted as a super position of two traveling waves, one moving to the right (that is, $\frac{1}{2}f(x-at)$) and one moving to the left ( $\frac{1}{2}f(x+at)$). Both waves travel with speed a and have the same basic shape as the initial displacement f (x). The form of u(x, t) given in (14) is called d’Alembert’s solution.