\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 103, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/103\hfil A classical solution weakened on the axis]
{A classical solution weakened on the axis for a mixed problem of
inhomogeneous hyperbolic equations}

\author[Raid Al-Momani \hfil EJDE--2003/103\hfilneg]{Raid Al-Momani} 

\address{Raid Al-Momani \hfill\break 
Department of Mathematics,
Yarmouk University, Irbid, Jordan}
\email{raid@yu.edu.jo}

\date{}
\thanks{Submitted December 20, 2002. Published October 9, 2003.}
\subjclass[2000]{35L70, 58J45}
\keywords{Weakened classical solutions, hyperbolic equations}

\begin{abstract}
  The main purpose of this paper is to present sufficient conditions, on the
  forcing term of a mixed problem for three dimensional hyperbolic equations
  of any even order, for the existence of axially weakened classical solutions.
\end{abstract}

\maketitle

\numberwithin{equation}{section} 
\newtheorem{theorem}{Theorem}[section]

\section{introduction}

Let $r=\sqrt{x_1^2+x_2^2+x_3^2}$ and $G=\{x\in \mathbb{R}^3: 0<r<R<\infty \}$.
 On the cylinder $\overline{Q}=\overline{G}\times [0,T]$, we consider the
mixed problem
\begin{gather}
\prod_{k=1}^m\big( \frac{\partial ^2}{\partial t^2}-a_k^2\Delta \big) 
u+F(r,t;u)=f(r,t)  \label{1} \\
\frac{\partial ^pu}{\partial t^p}\big|_{t=0}=\phi _p(r),\quad 0\leq p\leq
2m-1  \label{2} \\
\big( \frac{\partial \Delta ^ku}{\partial \nu }+\frac 1{r}\Delta ^ku\big) 
\big| _\Gamma =0,\quad 0\leq k\leq m-1 \,,  \label{3}
\end{gather}
where $\Gamma $ denotes the lateral surface of the cylinder $Q,\frac
\partial {\partial \nu }$ is the partial derivative with respect to the
normal $\nu $ which is outward to the lateral surface $\Gamma $ of the
cylinder $Q$, the reals $a_1,\dots ,a_n$ are different, 
$\Delta =\frac{\partial ^2}{\partial x_1^2}+\frac{\partial ^2}{\partial x_2^2} 
+\frac{\partial ^2}{\partial x_3^2}$ is the Laplacian, $\Delta_r \varphi 
=\frac{\partial ^2 \varphi}{\partial r^2}+ \frac{2}{r} 
+ \frac{\partial \varphi}{\partial r}$, and
\begin{equation}
\begin{aligned} F(r,t;u)
&=\sum_{p+q<m}\big[b_{p,q}(r,t)\sum_{i=1}^3x_i\frac{\partial ^{2p+1}\Delta
^qu}{\partial t^{2p}\partial x_i} \\ &\quad +C_{p+q}(r,t)\frac{\partial
^{2p+1}\Delta ^qu}{\partial t^{2p+1}}+d_{p,q}(r,t)\frac{\partial ^{2p}\Delta
^qu}{\partial t^{2p}}\big]. \end{aligned}  \label{4}
\end{equation}
We assume that
\begin{equation}
b_{p,q}(R,t)=0  \label{5}
\end{equation}
which is the consistency condition. We also assume that the coefficients 
$b_{p,q}$, $C_{p,q}$, $d_{p,q} $ are continuous in $\overline{Q}$ and their
derivatives $\frac{\partial b_{p,q}}{\partial r}, \frac{\partial C_{p,q}}
{\partial r},\frac{\partial d_{p,q}}{\partial r}$ are bounded in 
$\overline{Q}$.

The central symmetry of the problem with respect to the variable
$x$ is reflected by the dependence of the data on it, in the
invariance of the Laplacian with respect to rotation, and in the
geometry of $G$. Before we state our results, let us dwell a
moment on the existing literature concerning similar problems.
First, we recall known facts concerning hyperbolic equations; as
it is well known, hyperbolic equations do not keep smoothness. So,
the space of continuous functions is badly adapted for
investigating them; it is not suitable in the sense that for more
than one space variable, we can not establish the agreement of
necessary and sufficient conditions imposed on the initial
functions for the existence of the classical solutions as it was
shown by Sobolev [10] for the Cauchy problem of the wave equation.
In contrast, Sobolev spaces are more adapted for investigating
hyperbolic equations. But if the dimension is greater than the
order of the operator, then solutions in Sobolev spaces may be
discontinuous even at any point. Therefore, it arises the question
of how to use the good quality of the spaces of continuous
functions. Naturally, arises the question to localize the set of
irregular points of the solutions. We will then make some
compromise, we weaken the requirements on the sought for
solutions; As, in our situation, they are classical except on the
axis of the cylinder, we will impose some conditions there, and
call the new defined solutions, \textit{weakened on the axis}.

Classical solution of mixed problems for hyperbolic equations has been the
subject of extensive research. However, their common drawback, is that the
sufficient conditions, imposed on the initial functions for the existence of
a classical solution, disagree with the necessary ones, see, for example,
the works of Steklov [11] and Petrovsky [9].

This limitation has been eliminated for one-dimensional hyperbolic
equations; an agreement between necessary and sufficient conditions for the
initial functions has been established in the case of mixed problems for the
telegraph equation by Chernyatin [5,6] and Egorov [7]. Baranowskaya [2]
extended this result to more general one-dimensional hyperbolic equations of
second order. Also, it has been extended to one-dimensional hyperbolic
equations of any even order in the work of Koku and Yurchuk [4].

For multidimensional hyperbolic equations, the situation is much more
delicate. All the known sufficient conditions for the existence of classical
solutions contain additional conditions for the initial functions. Moreover,
Sobolev [10] proved that the requirements for the initial-value (Cauchy)
problem for the wave equation could not be reduced by one derivative. The
solution at the point $r=0$ requires that $\Phi \in C^3$ and $\Psi \in C^2$
in the case of central symmetry. Besov [3] arrived at similar results in
terms of what is called now Besov spaces. An agreement between necessary and
sufficient conditions for the initial functions has been established in the
case of mixed problems for homogeneous three-dimensional hyperbolic
equations of any even order in the case of central symmetry by Al-Momani [1]
and Al-Momani and Yurchuk [12].

By $C_{\{|x|=0\}}^{2m}(\overline{Q})$ we denote the set of functions $u\in
C^1(\overline{Q})\cap C^{2m}(\overline{Q}\backslash \{0\}\times [0,T])$
satisfying the conditions
\begin{gather}
\frac{\partial ^p\Delta ^nu}{\partial t^p}\in C^1(\overline{Q}),\quad
p+2n<2m-1  \label{1.6} \\
\frac{\partial ^{p+2}\Delta ^nu}{\partial t^p\partial r^2}\in C(\overline{Q}%
),\quad p+2n<2m-2  \label{1.7} \\
\lim_{r\to 0}\sum_{i=1}^3x_i\frac{\partial ^{p+1}\Delta ^nu(x,t)}{\partial
t^p\partial x_i}=0,\quad p+2n=2m-1  \label{1.8} \\
\lim_{r\to 0}r\frac{\partial ^p\Delta ^nu(x,t)}{\partial t^p},\quad
p+2n=2m\,.  \label{1.9}
\end{gather}


\noindent\textbf{Definition.} The \textit{weakened on the axis}
$r=0$ classical solution in $\overline{Q}$ of the problem
(1.1)-(1.3) is a function in $C_{\{|x|=0\}}^{2m}(\overline{Q})$,
which satisfies equation (\ref{1}) pointwise in
$\overline{Q}\backslash \{0\}\times [0,T]$ and satisfies the
conditions (\ref{2}), (\ref{3}) in the usual sense.
\smallskip

Next, we state a result from [1] which provides the existence of the \textit{%
weakened on the axis} classical solution in $\overline{Q}$ of our problem (%
\ref{1})-(\ref{3}) with $f(r,t)=0$ in (\ref{1}).

\begin{theorem}
\label{thm1} Let the coefficients $b_{p,q},\;C_{p,q},\;d_{p,q}$ be
continuous in $\overline{Q}$ and have bounded first partial derivatives in $%
\overline{Q}$ with respect to $r$ and the condition (\ref{5}) holds. The
necessary and sufficient conditions for the existence of the \textit{%
weakened on the axis} classical solution in $\overline{Q}$ of the problem
\[
\prod_{k=1}^m\big( \frac{\partial ^2}{\partial t^2}-a_k^2\Delta \big)
u+F(r,t;u)=0
\]
with (\ref{2}) and (\ref{3}) are the following conditions on the
initial functions $\phi _p$:
\begin{gather}
\phi _p\in C^{2m-p}(0,R]  \label{6} \\
\Delta _r^n\phi _p\in C^2[0,R],\quad p+2n<2m-2  \label{7} \\
\Delta _r^n\phi _p\in C^1(0,R],\quad p+2n<2m-1  \label{8} \\
\lim_{r\to 0}r\frac d{dr}\Delta _r^n\phi _p(r)=0,\quad p+2n<2m-1  \label{9}
\\
\lim_{r\to 0}r\Delta _r^n\phi _p(r)=0,\quad p+2n=2m  \label{10} \\
\big[ \frac d{dr}\Delta _r^n\phi _p(r)+\frac 1r\Delta _r^n\phi _p(r)\big] %
\big|_{r=R}=0,\quad p+2n<2m  \label{11}
\end{gather}
\end{theorem}

\section{Main result}

\begin{theorem}
\label{thm2} Sufficient conditions for the existence of the weakened on the
axis classical solution in $\overline{Q}$ of problem \eqref{1}-\eqref{3} are
given by \eqref{6}-\eqref{11} on the initial functions $\phi _p$, and the
following conditions on the function $f$:
\begin{gather}
f\in C(\overline{Q}\backslash \{0\}\times [0,T]),  \label{12} \\
\lim_{r\to 0}rf(r,t)=0,\quad 0\leq t\leq T  \label{13} \\
\int_0^Rr^2|\frac{\partial f(r,t)}{\partial r}|^2dr\in C[0,T].  \label{14}
\end{gather}
\end{theorem}

\begin{proof}
 We transform \eqref{1}-\eqref{3} to the spherical coordinates
 and set $\nu (r,t)=ru(r,t)$ and use the equality
\[
\Delta _r\frac{\nu (r,t)}r=\frac 1r\frac{\partial ^2\nu (r,t)}{\partial r^2}.
\]
Then the function $\nu (r,t)$ satisfies
\begin{equation}
\prod_{k=1}^m\big( \frac{\partial ^2}{\partial t^2}-a_k^2\frac{\partial ^2}
{\partial r^2}\big) \nu +F(r,t;\nu )=f_1(r,t)  \label{15}
\end{equation}
with the initial and boundary conditions
\begin{gather}
\frac{\partial ^pu}{\partial t^p}|_{t=0} =\Phi _p(r),\Phi _p(r)=r\phi
_p(r),\quad 0\leq p\leq 2m-1  \label{16} \\
\frac{\partial ^{2k}\nu }{\partial r^{2k}}|_{r=0} =\frac{\partial
^{2k+1}\nu }{\partial r^{2k+1}}|_{r=R}=0,\quad 0\leq k\leq m-1\,,  \label{17}
\end{gather}
where
\begin{align*}
 F(r,t;\nu ) 
 &=\sum_{p+q<m} b_{p,q}(r,t)r\frac{\partial ^{2p+2q+1}\nu }{\partial t^{2p}
\partial r^{2q+1}} \\
&\quad +C_{p,q}(r,t)\frac{\partial^{2p+2q+1}\nu }{\partial t^{2p+1}\partial r^{2q}}
+\big(d_{p,q}(r,t)-b_{p,q}(r,t)\big) 
\frac{\partial ^{2p+2q}\nu}{\partial t^{2p}\partial r^{2q}}\big]
\end{align*}
and the functions $\Phi _p(r)=r\phi _p(r)$ satisfy the conditions
\begin{gather*}
\Phi _p\in C^{2m-p}[0,R],\quad \Phi _p^{(2n)}(0)=0,\quad 0\leq 2n\leq 2m-p\\
\Phi _p^{(2k+1)}(R)=0,\quad 0\leq 2k+1\leq 2m-p\,.
\end{gather*}
The function $f_1(r,t)=rf(r,t)$ satisfies the conditions
\begin{gather*}
f_1\in C\left( [0,R]\times [0,T]\right) ,\quad f_1(0,t)=0  \label{19} \\
\int_0^R|\frac{\partial f_1(r,t)}{\partial r}|^2dr\in C[0,T]\,.  \label{20}
\end{gather*}
As is shown in [4], under the conditions (\ref{12})-(\ref{14}), problem
(\ref{15})-(\ref{17}) has a classical solution $\nu \in C^{2m}(\overline{Q})$
which satisfies
\begin{equation}
\begin{aligned}
&\nu (r,t)\\
&=\frac 12\sum_{1\leq k\leq m}a_k^p \Big\{\sum_{0\leq i\leq
2m-1}\alpha _{i,k}\big[\overline{\Phi }_i^{(-i)}(r+a_kt)
 +(-1)^i\overline{\Phi }_i^{(-i)}(r-a_kt)\big]\\
&\quad -\alpha _{2m-1,k}\int_0^t\big[\overline{F}^{(1-2m)}(r+a_k(t-\tau ),\tau ;\nu )
-\overline{F}^{(1-2m)}(r+a_k(t-\tau ),\tau ;\nu )\big]d\tau  \\
&\quad +\alpha _{2m-1,k}\int_0^t\big[ \overline{f}^{(1-2m)}(r+a_k(t-\tau ),\tau)
-\overline{f}_1^{(1-2m)}(r-a_k(t-\tau ),\tau )\big] d\tau  \Big\}\,.
\end{aligned} \label{21}
\end{equation}
Here,
\begin{gather}
\overline{f}^{(-k)}(r,\tau ) =\int_0^r\frac{(r-\zeta )^{k-1}}{(k-1)!}
\overline{f}_1(\zeta ,\tau )d\tau  \label{22} \\
\overline{f}_1^{(k)}(r,F) =\frac{\partial ^k}{\partial r^k}\overline{f}_1(r,\tau ),
\quad k\geq 0  \label{23}
\end{gather}
we denote by $\overline{f}_1$ the $4R$-periodic continuation with respect to
$r$ of the function $f_1(r,t)$ from $\overline{Q}$ to $R^1\times [0,T]$
which is constructed in the same way as $\overline{\Phi }$ and $\overline{F}$
in [1]. Let
$$
\alpha _{i,k}=\begin{cases}
1, & \mbox{if } i=k  \\
0, & \mbox{if } i\neq k\,. \end{cases}
$$
Using (\ref{21}), we will show that the function $u(x,t)=\frac{\nu (r,t)}r$
is the {\it weakened on the axis} classical solution in $\overline{Q}$ of
the problem (\ref{1})-(\ref{3}). By substitution we can check that
the function $u$ satisfies at $r=0$ the equation (\ref{1}), initial
conditions (\ref{2}) and boundary conditions (\ref{3}). Therefore, to
complete the proof of our theorem it remains to prove that
\[
u(x,t)=\frac{\nu (r,t)}r,\quad u\in C_{\{r=0\}}^{2m}(\overline{Q})
\]
since $\nu \in C^{2m}(\overline{Q})$ the function $u=\frac \nu
r\in C^{2m}(\overline{Q}\backslash \{0\}\times [0,T])$. Now we
must show that the function $u$ satisfies the conditions
(1.6)-(1.9).

 Equation (\ref{21}) implies that $\frac{\partial
^{p+q}\nu }{\partial t^p\partial r^q}$ satisfies
\begin{equation*}
\begin{aligned}
\frac{\partial ^{p+q}\nu }{\partial t^p\partial r^q}
&=\frac 12\sum_{1\leq k\leq m}a_k^p\Big\{\sum_{0\leq i\leq 2m-1}\alpha _{i,k}
\big[\overline{\Phi }_i^{(p+q-i)}(r+a_kt)\\
&\quad +(-1)^{p+i}\overline{\Phi }_i^{(p+q-i)}(r-a_kt)\big]\\
&\quad -\alpha _{2m-1,k}\int_0^t\big[
\overline{F}^{(p+q+1-2m)}(r+a_k(t-\tau ),\tau ;\overline{\nu })\\
&\quad +(-1)^{p+1}\overline{F}^{(p+q+1-2m)}(r-a_k(t-\tau ),\tau ;\overline{\nu })\big]
d\tau    \\
&\quad+\alpha _{2m-1,k}\int_0^t[\overline{f}_1^{(p+q+1-2m)}(r+a_k(t-\tau ),\tau ) \\
&\quad +(-1)^{p+1}\overline{f}_1^{(p+q+1-2m)}(r-a_k(t-\tau ),\tau )]d\tau \Big\}\,.
\end{aligned}\label{28}
\end{equation*}
The continuations $\overline{\Phi }_i$, $\overline{\nu }$, $\overline{F}$
and $\overline{f}_1$ satisfy the following conditions:
\begin{equation}
\overline{\Phi }_i^{(p+q-i)}(r\pm a_kt)=\sum_{j=0}^{2m-p-q}\overline{\Phi }%
_i^{(p+q-i+j)}(\pm a_kt)\frac{r^j}{j!}+\theta (r^{2m-p-q})  \label{29}
\end{equation}
where $\theta (r^{2m-p-q})$ means a function $g(r)$ such that
$\lim_{r\to 0} g(r)/(r^{2m-p-q})=0$,  %\label{30}
\begin{equation}
\begin{aligned}
&\int_0^t\overline{F}^{(p+q+1-2m)}(r\pm a_k(t-\tau ),\tau ;\overline{\nu })d\tau \\
&=\sum_{j=0}^{2m-p-q}\int_0^t\overline{F}^{p+q+1-2m+j}(\pm a_k(t-\tau
),\tau ;\overline{\nu })d\tau \frac{r^j}{j!}+\theta (r^{2m-p-q})\,,
\end{aligned}  \label{31}
\end{equation}
and
\begin{equation}
\overline{f}_1\in C(R^1\times [0,T]),  \quad
\int_0^t\overline{f}_1(r\pm a_k(t-\tau ),\tau )d\tau \in C^1(R^1\times
[0,T])\,. \label{32}
\end{equation}
By the Taylor formula with remainder in Peano form [8],
\begin{equation}
\begin{aligned}
&\int_0^t[\overline{f}_1^{(p+q+1-2m)}(r\pm a_k(t-\tau ),\tau )d\tau \\
&=\sum_{j=0}^{2m-p-q}\int_0^t\overline{f}_1^{(p+q+1-2m)}(\pm a_k(t-\tau ),\tau )
\frac{r^j}{j!}  +\theta (r^{2m-p-q})
\end{aligned} \label{33}
\end{equation}
and
\begin{equation}
\overline{f}_1^{(-k)}(r,\tau )=(-1)^{1+k}\overline{f}_1^{(-k)}(r,t). \label{34}
\end{equation}
 From (\ref{14}) and
\begin{equation*}
\frac{\partial u(x,t)}{\partial x_i}=\frac{x_i}r
\frac{\partial u(r,t)}{\partial r}  \label{35}
\end{equation*}
and on the basis of (\ref{29})-(\ref{34}), we have
\begin{equation}
\begin{aligned}
\frac{\partial ^p\Delta ^nu}{\partial t^p}
&=\frac 1{2r}\sum_{1\leq k\leq m}a_k^p\Big\{\sum_{j=0}^{2m-2n-p}(1-(-1)^j)
\Big[\sum_{0\leq i\leq 2m-1}\alpha _{i,k}\overline{\Phi }_i^{(2n+p-i+j)}(a_kt)   \\
&\quad -\alpha _{2m-1,k}\int_0^t\overline{F}^{(2n+p+1-2m+j)}(a_k(t-\tau ),\tau ;
\overline{\nu }) d\tau  \\
&\quad +\alpha _{2m-1,k}\int_0^t\overline{f}_1^{(2n+p+1-2m+j)}(a_k(t-\tau ),\tau
)d\tau \Big]\frac{r^j}{j!}\Big\}+\theta (r^{2m-2n-p})\,,
\end{aligned} \label{36}
\end{equation}
\begin{equation}
\begin{aligned}
\frac{\partial ^{p+1}\Delta ^nu}{\partial t^p\partial x_i}
&=\frac{x_i}{2r^2}\sum_{1\leq k\leq m}^pa_k^p
\Big\{\sum_{j=0}^{2m-2n-p-1}(1+(-1)^j)(\frac 1{j!}-\frac 1{(j+1)!})\\
&\quad\times \Big[\sum_{0\leq i\leq 2m-1}\alpha _{i,k}\overline{\Phi }_i^{(2n+p+1-i+j)}
(a_kt)\\
&\quad -\alpha _{2m-1,k}\int_0^t\overline{F}^{(2n+p+2-2m+j)}(a_k(t-\tau ),\tau ;
\overline{\nu })d\tau  \\
&\quad +\alpha _{2m-1,k}\int_0^t\overline{f}_1^{(2n+p+2-2m+j)}(a_k(t-\tau ),\tau
)d\tau \Big]\frac{r^j}{j!}\Big\}+\theta (r^{2m-2n-p-1})
\end{aligned} \label{37}
\end{equation}
Equalities (\ref{36}) and (\ref{37}) show that
\begin{equation}
\begin{aligned}
\lim_{r\to 0}\frac{\partial ^p\Delta ^nu}{\partial t^p}
&=\sum_{1\leq k\leq m}a_k^p\Big[\sum_{0\leq i\leq 2m-1}\alpha _{i,k}
\overline{\Phi }_i^{(2n+p-i+1)}(a_kt)  \\
&\quad -\alpha _{2m-1,k}\int_0^t\overline{F}^{(2n+p+2-2m)}(a_k(t-\tau ),\tau ,
\overline{\nu })d\tau    \\
&\quad +\alpha _{2m-1,k}\int_0^t\overline{f}_1^{(2n+p+2-2m)}(a_k(t-\tau ),\tau
)d\tau \Big]
\end{aligned}\label{38}
\end{equation}
and
\begin{equation*}
\lim_{r\to 0}\frac{\partial ^{p+1}\Delta ^nu}{\partial t^p\partial
x_i} =0\,.  \label{39}
\end{equation*}
Therefore, condition (\ref{1.6}) holds. Equation (\ref{38}) also
implies that
\begin{equation*}
\begin{aligned}
\lim_{r\to 0}\frac 1r\frac{\partial ^{p+1}\Delta ^nu}{\partial t^p\partial r}
&=\lim_{r\to 0}\frac 1{r^2}\sum_{i=1}^3x_i\frac{\partial ^{p+1}
\Delta ^nu}{\partial t^p\partial x_i}  \\
&= \frac 13\sum_{1\leq k\leq m}a_k^p[\sum_{0\leq i\leq 2m-1}\alpha _{i,k}
\overline{\Phi }_i^{(2n+p+3-i)}(a_kt)   \\
&\quad -\alpha _{2m-1,k}\int_0^t\overline{F}^{(2n+p+4-2m)}(a_k(t-\tau ),\tau ,
\overline{\nu })d\tau    \\
&\quad +\alpha _{2m-1,k}\int_0^t\overline{f}_1^{(2n+p+4-2m)}(a_k(t-\tau ),\tau
)d\tau \Big]\,.
\end{aligned}  \label{40}
\end{equation*}
Replacing {\em n} by $n+1$ in equations (\ref{38}) and (\ref{4}) and taking
into account that
\[
\frac{\partial ^{p+2}\Delta ^nu}{\partial t^p\partial r^2}=\frac{\partial
^p\Delta ^{n+1}u}{\partial t^p}-\frac 2r\frac{\partial ^{p+1}\Delta ^nu}{%
\partial t^p\partial r}
\]
we obtain
\begin{align*}
\lim_{r\to 0}\frac{\partial ^{p+2}\Delta ^nu}{\partial t^p\partial r^2}
&=\frac 13\sum_{1\leq k\leq m}a_k^p[\sum_{0\leq i\leq 2m-1}\alpha
_{i,k}\overline{\Phi }_i^{(2n+p+3-i)}(a_kt) \\
&\quad -\alpha _{2m-1,k}\int_0^t\overline{F}^{(2n+p+4-2m)}(a_k(t-\tau ),\tau ,
\overline{\nu })d\tau \\
&\quad +\alpha _{2m-1,k}\int_0^t\overline{f}_1^{(2n+p+4-2m)}(a_k(t-\tau ),\tau
)d\tau \Big]
\end{align*}
and consequently condition (\ref{1.7}) holds.
 From (\ref{37}) with $p+2n=2m-1$, we have
\begin{equation}
\sum_{i=1}^3x_i\frac{\partial ^{p+1}\Delta ^nu}{\partial t^p\partial x_i}%
=\theta (r^o)  \label{41}
\end{equation}
and from (\ref{36}) with $p+2n=2m$, we have
\begin{equation}
r\frac{\partial ^p\Delta ^nu}{\partial t^p}=\theta (r^o)\,.  \label{42}
\end{equation}
Since $\lim_{r\to 0}\theta (r^o)=0$, so passing in (\ref{41}) and
(\ref{42}) to the limit as $r\to 0$, we show that the conditions
(\ref{1.8}) and (\ref{1.9}) consequently hold.
\end{proof}

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