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4.1: One-Dimensional Wave Equation

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    2148
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    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}.

     Interval 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.

    Domain of influence

    Figure 4.2.1: Domain of influence

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

    Contributors and Attributions


    This page titled 4.1: One-Dimensional Wave Equation is shared under a not declared license and was authored, remixed, and/or curated by Erich Miersemann.

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