Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.2.1: Case n=3

[ "article:topic", "showtoc:no" ]
  • Page ID
    2177
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    The Euler-Poisson-Darboux equation in this case is

    $$(rM)_{rr}=c^{-2}(rM)_{tt}.$$

    Thus \(rM\) is the solution of the one-dimensional wave equation with initial data

    \begin{equation}
    \label{initialn3}\tag{4.2.1.1}
    (rM)(r,0)=rF(r)\ \ \ (rM)_t(r,0)=rG(r).
    \end{equation}
    From the d'Alembert formula we get formally
    \begin{eqnarray}
    \label{meansol1}
    M(r,t)&=&\dfrac{(r+ct)F(r+ct)+(r-ct)F(r-ct)}{2r}\\ \tag{4.2.1.2}
    &&+\dfrac{1}{2cr}\int_{r-ct}^{r+ct}\ \xi G(\xi)\ d\xi.
    \end{eqnarray}

    The right hand side of the previous formula is well defined if the domain of dependence \([x-ct,x+ct]\) is a subset of \((0,\infty)\). We can extend \(F\) and \(G\)  to \(F_0\) and \(G_0\) which are defined on \((-\infty,\infty)\) such that \(rF_0\) and \(rG_0\) are \(C^2(\mathbb{R}^1)\)-functions as follows.
    Set

    $$
    F_0(r)=\left\{\begin{array}{r@{\quad:\quad}l}
                    F(r)&r>0\\
                    f(x)&r=0\\
                    F(-r)&r<0
                    \end{array}\right.\
    $$

    The function \(G_0(r)\) is given by the same definition where \(F\) and \(f\) are replaced by \(G\) and \(g\), respectively.


    Lemma. \(rF_0(r),\ rG_0(r)\in C^2(\mathbb{R}^2)\).

    Proof. From definition of \(F(r)\) and \(G(r)\), \(r>0\), it follows from the mean value theorem

    $$\lim_{r\to+0} F(r)=f(x),\ \ \    \lim_{r\to+0} G(r)=g(x).$$

    Thus \(rF_0(r)\) and \(rG_0(r)\) are \(C(\mathbb{R}^1)\)-functions. These functions are also in \(C^1(\mathbb{R}^1)\). This follows since \(F_0\) and \(G_0\) are in \(C^1(\mathbb{R}^1)\). We have, for example,

    \begin{eqnarray*}
    F'(r)&=&\dfrac{1}{\omega_n}\int_{\partial B_1(0)}\ \sum_{j=1}^n f_{y_j}(x+r\xi)\xi_j\ dS_\xi\\
    F'(+0)&=&\dfrac{1}{\omega_n}\int_{\partial B_1(0)}\ \sum_{j=1}^n f_{y_j}(x)\xi_j\ dS_\xi\\
    &=&\dfrac{1}{\omega_n}\sum_{j=1}^n f_{y_j}(x)\int_{\partial B_1(0)}\ n_j\ dS_\xi\\
    &=&0.
    \end{eqnarray*}

    Then, \(rF_0(r)\) and \(rG_0(r)\) are in \(C^2(\mathbb{R}^1)\), provided  \(F''\) and \(G''\)  are bounded as \(r\to+0\).  This property follows from        

    $$F''(r)=\dfrac{1}{\omega_n}\int_{\partial B_1(0)}\ \sum_{i,j=1}^n f_{y_iy_j}(x+r\xi)\xi_i\xi_j\ dS_\xi.$$

    Thus

    $$F''(+0)=\dfrac{1}{\omega_n}\sum_{i,j=1}^n f_{y_iy_j}(x)\int_{\partial B_1(0)}\ n_in_j\ dS_\xi.$$

    We recall that \(f,g\in C^2(\mathbb{R}^2)\) by assumption.

    \(\Box\)

    The solution of the above initial value problem, where \(F\) and \(G\) are replaced by \(F_0\) and \(G_0\), respectively, is

    \begin{eqnarray*}
    M_0(r,t)&=&\dfrac{(r+ct)F_0(r+ct)+(r-ct)F_0(r-ct)}{2r}\\
    &&+\dfrac{1}{2cr}\int_{r-ct}^{r+ct}\ \xi G_0(\xi)\ d\xi.
    \end{eqnarray*}

    Since \(F_0\) and \(G_0\) are even functions, we have

    $$\int_{r-ct}^{ct-r}\ \xi G_0(\xi)\ d\xi=0.$$

    Thus

    \begin{eqnarray}
    M_0(r,t)&=&\dfrac{(r+ct)F_0(r+ct)-(ct-r)F_0(ct-r)}{2r}\nonumber\\
    \label{meansol2} \tag{4.2.1.3}
    &&+\dfrac{1}{2cr}\int_{ct-r}^{ct+r}\ \xi G_0(\xi)\ d\xi,
    \end{eqnarray}

    see Figure 4.2.1.1.

    Changed domain of integration


    Figure 4.2.1.1: Changed domain of integration

    For fixed \(t>0\) and \(0<r<ct\) it follows that \(M_0(r,t)\) is the solution of the initial value problem with given initially data (\ref{initialn3}) since \(F_0(s)=F(s)\), \(G_0(s)=G(s)\) if \(s>0\).

    Since for fixed \(t>0\)

    $$u(x,t)=\lim_{r\to 0} M_0(r,t),$$

    it follows from d'Hospital's rule that

    \begin{eqnarray*}
    u(x,t)&=&ctF'(ct)+F(ct)+tG(ct)\\
    &=&\dfrac{d}{dt}\left(tF(ct)\right)+tG(ct).
    \end{eqnarray*}

    Theorem 4.2. Assume \(f\in C^3(\mathbb{R}^3)\) and \(g\in C^2(\mathbb{R}^3)\) are given. Then there exists a unique solution \(u\in C^2(\mathbb{R}^3\times [0,\infty))\) of the initial value problem (4.2.2)-(4.2.3), where \(n=3\), and the solution is given by the Poisson's formula

    \begin{eqnarray}
    u(x,t)&=&\dfrac{1}{4\pi c^2}\dfrac{\partial}{\partial t}\left(\dfrac{1}{t}\int_{\partial B_{ct}(x)}\ f(y)\ dS_y\right)\\ \tag{4.2.1.4}
    &&+\dfrac{1}{4\pi c^2 t}\int_{\partial B_{ct}(x)}\ g(y)\ dS_y.
    \end{eqnarray}

    Proof. Above we have shown that a \(C^2\)-solution is given by Poisson's formula. Under the additional assumption \(f\in C^3\) it follows from Poisson's formula that this formula defines a solution which is in \(C^2\), see F. John [10], p. 129.

    \(\Box\)

    Corollary. From Poisson's formula we see that the domain of dependence for \(u(x,t_0)\) is the intersection of the cone defined by \(|y-x|=c|t-t_0|\) with the hyperplane defined by \(t=0\), see Figure 4.2.1.2.

    Domain of dependence, case \(n=3\).

    Figure 4.2.1.2: Domain of dependence, case \(n=3\).

    Contributors