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Mathematics LibreTexts

4.2.1: Case n=3

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 \(n=3\).