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Mathematics LibreTexts

4.2.1: Case n=3

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    The Euler-Poisson-Darboux equation in this case is


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

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

    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.


    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,

    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\\

    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

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


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

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


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

    &&+\dfrac{1}{2cr}\int_{r-ct}^{r+ct}\ \xi G_0(\xi)\ d\xi.

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

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


    \label{meansol2} \tag{}
    &&+\dfrac{1}{2cr}\int_{ct-r}^{ct+r}\ \xi G_0(\xi)\ d\xi,

    see Figure

    Changed domain of integration

    Figure 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


    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

    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{}
    &&+\dfrac{1}{4\pi c^2 t}\int_{\partial B_{ct}(x)}\ g(y)\ dS_y.

    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.


    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

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

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