4.1: One-Dimensional Wave Equation

The one-dimensional wave equation is given by

$$\begin{equation} \label{waveone} \dfrac{1}{c^2}u_{tt}-u_{xx}=0, \end{equation}$$
where $u=u(x,t)$ is a scalar function of two variables and $c$ is a positive constant. According to previous considerations, all $C^2$-solutions of the wave equation are

$$\begin{equation} \label{wavegen} u(x,t)=f(x+ct)+g(x-ct), \end{equation}$$

with arbitrary $C^2$-functions $f$ and $g$

The Cauchy initial value problem for the wave equation is to find a $C^2$-solution of

$$\begin{eqnarray*} \dfrac{1}{c^2}u_{tt}-u_{xx}&=&0\\ u(x,0)&=&\alpha(x)\\ u_t(x,0)&=&\beta(x), \end{eqnarray*}$$

where $\alpha,\ \beta\in C^2(-\infty,\infty)$ are given.

Theorem 4.1. There exists a unique $C^2(\mathbb{R}^1\times\mathbb{R}^1)$-solution of the Cauchy initial value problem, and this solution is given by d'Alembert's¹ formula

$$\begin{equation} \label{waveform} u(x,t)=\dfrac{\alpha(x+ct)+\alpha(x-ct)}{2}+\dfrac{1}{2c}\int_{x-ct}^{x+ct}\ \beta(s)\ ds. \end{equation}$$

Proof. Assume there is a solution $u(x,t)$ of the Cauchy initial value problem, then it follows from ($\ref{wavegen}$) that

$$\begin{eqnarray} \label{ini1} u(x,0)&=&f(x)+g(x)=\alpha(x)\\ \label{ini2} u_t(x,0)&=&cf'(x)-cg'(x)=\beta(x). \end{eqnarray}$$

From ($\ref{ini1}$) we obtain

$$f'(x)+g'(x)=\alpha'(x),$$

which implies, together with ($\ref{ini2}$), that

$$\begin{eqnarray*} \label{12a} f'(x)&=&\dfrac{\alpha'(x)+\beta(x)/c}{2}\\ \label{12b}g'(x)&=&\dfrac{\alpha'(x)-\beta(x)/c}{2}. \end{eqnarray*}$$

Then

$$\begin{eqnarray*} f(x)&=&\dfrac{\alpha(x)}{2}+\dfrac{1}{2c}\int_0^x\ \beta(s)\ ds +C_1\\ g(x)&=&\dfrac{\alpha(x)}{2}-\dfrac{1}{2c}\int_0^x\ \beta(s)\ ds +C_2. \end{eqnarray*}$$

The constants $C_1$, $C_2$ satisfy

$$C_1+C_2=f(x)+g(x)-\alpha(x)=0,$$ 

see ($\ref{ini1}$). Thus each $C^2$-solution of the Cauchy initial value problem is given by d'Alembert's formula. On the other hand, the function $u(x,t)$ defined by the right hand side of ($\ref{waveform}$) is a solution of the initial value problem.

$\Box$

Corollaries. 1. The solution $u(x,t)$ of the initial value problem depends on the values of $\alpha$ at the endpoints of the interval $[x-ct,x+ct]$ and on the values of $\beta$ on this interval only, see Figure 4.1.1. The interval $[x-ct,x+ct]$ is called {\it domain of dependence}.

Figure 4.1.1: Interval of dependence

2. Let $P$ be a point on the $x$-axis. Then we ask which points $(x,t)$ need values of $\alpha$ or $\beta$ at $P$ in order to calculate $u(x,t)$? From the d'Alembert formula it follows that this domain is a cone, see Figure 4.2.1. This set is called domain of influence.

Figure 4.2.1: Domain of influence

¹ d'Alembert, Jean Babtiste le Rond, 1717-1783

Contributors and Attributions

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.