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# 4.2.1: Case n=3

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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. 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 , 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. Figure 4.2.1.2: Domain of dependence, case $$n=3$$.

## Contributors

• Integrated by Justin Marshall.