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\def\rightheadline{\hfil $n$-dimensional Laplace transforms
\hfil\folio}
\def\leftheadline{\folio\hfil R. S. Dahiya \& Jafar Saberi-Nadjafi
 \hfil}

\def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt %
15th Annual Conference of Applied Mathematics, Univ. of Central Oklahoma, 
\hfill\break
Electronic Journal of Differential Equations, Conference~02, 1999, pp. 61--74.
\hfill\break
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfill\break 
ftp ejde.math.swt.edu  or ejde.math.unt.edu (login: ftp)\bigskip} }

\topmatter
\title
Theorems on $n$-dimensional Laplace transforms and their applications
\endtitle

\thanks 
{\it 1991 Mathematics Subject Classifications:} 44A30.\hfil\break\indent
{\it Key words and phrases:} multi-dimensional Laplace transforms.
\hfil\break\indent
\copyright 1999 Southwest Texas State University  and
University of North Texas. \hfil\break\indent
Published November 24, 1999.
\endthanks
\author R. S. Dahiya \& Jafar Saberi-Nadjafi \endauthor

\address R. S. Dahiya \hfill\break
Department of Mathematics \hfill\break
Iowa State University \hfill\break 
Ames, IA  50011, USA
\endaddress
\email dahiya\@iastate.edu
\endemail

\address Jaffar Saberi-Nadjafi \hfill\break
Department of Mathematics,\hfill\break
Ferdowsi, University of Mashhad \hfill\break
 Mashhad, Iran 
\endaddress


\abstract
In the present paper we prove certain theorems involving the classical 
Laplace transform of $n$-variables.  The theorems are 
then shown to yield a nice algorithm for evaluating $n$-dimensional Laplace 
transform pairs.  In the second part, boundary value problems are solved by 
using the double Laplace transformation.
\endabstract
\endtopmatter

\document
\head 1. Introduction and Notation\endhead

The generalization of the well-known Laplace transform\newline
$L[f(t); s]=\int^{\infty}_0 e^{-st}f(t)dt$ to $n$-dimensional is given by
$$
L_n[f(\bar t); \bar s]=\int^{\infty}_0\int^{\infty}_0\dots\int^{\infty}_0
\exp(-\bar s\cdot\bar t)f(\bar t) P_n(d\bar t)
$$
where $\bar t=(t_1,t_2,\dots, t_n),\ \bar s=(s_1,s_2,\dots, s_n),\ 
\bar s\cdot\bar t=\sum\limits^n_{i=1}s_it_i$ and 
$P_n(d\bar t)=\prod\limits^n_{k=1}dt_k$.

In addition to the notations introduced above, we will use the following 
throughout this article. Let $\overline{t^v}=(t^v_1,t^v_2,\dots, t^v_n)$ for 
any real exponent $v$ and let $p_k(\overline t)$ be the $k$-th symmetric 
polynomial in the components $t_k$ of $\overline t$.  Then
\roster
\item"{(i)}" $p_1(\overline{t^v})=t^v_1+t^v_2+\dots+t^v_n$
\item"{(ii)}" $p_n(\overline{t^v})=t^v_1\cdot t^v_2\dots t^v_n$.\endroster


\head{2.\ Main Results}\endhead

\proclaim{Theorem 2.1}  Suppose that $f(x)$ and $f(x^2)$ are functions of 
class $\Omega$.  Let
\roster
\item"{(i)}" $\Cal L\{f(x); s\}=\phi(s)$,
\item"{(ii)}" $\Cal L\{x^{-\frac 32}\phi\left(\frac 1x\right); s\}=\xi(s)$,
\item"{(iii)}" $\Cal L\{x^{-\frac 12}\xi\left(\frac 1{x^2}\right); s\}=\zeta(s)$, and
\item"{(iv)}" $\Cal L\{f(x^2); s\}=G(s)$,\endroster
where $x^{-\frac 32}\phi\left(\frac 1x\right)$ and $x^{-\frac 12}\xi
\left(\frac 1{x^2}\right)$ are also functions of class $\Omega, x^{-\frac 32}
\exp\left(-sx-\frac ux\right)f(u)$ and 
$u^{-\frac 12}x^{-\frac 32}\exp\left(-sx-\frac{2u^{\frac 12}}x\right)f(u)$ 
belong to $L_1\left[\left(0,\infty)\times (0,\infty\right)\right]$.

\flushpar Then
$$
\Cal L_n\left\{\frac 1{p_n\left(\overline{x^{\frac 12}}\right)}G[2p_1(
\overline{x^{-1}})];\overline s\right\}=\frac{\pi^{\frac{n-2}2}}2\cdot
\frac{p_1\left(\overline{s^{\frac 12}}\right)}{p_n
\left(\overline{s^{\frac 12}}\right)}\zeta\left[\left(P_1
\left(\overline{s^{\frac 12}}\right)\right)^2\right],\tag 2.1
$$
where $\Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>d$, a 
constant, provided the integrals involved exist for $n=2,3,\dots, N$. 
The existence conditions for two-dimensions are given in Ditkin and Prudnikov 
\cite{11; p.4} and similar conditions hold for $N$-dimensions, we refer to 
Brychkov et al. \cite{2; Ch.2}.\endproclaim

\demo{Proof}  Using (i) and (ii), we obtain
$$
\xi(s)=\int^{\infty}_0\left[\int^{\infty}_0x^{-\frac 32}\exp
\left(-sx-\frac ux\right)f(u)du\right]dx.\tag 2.2
$$
Next we wish to interchange the order of the integral on the right side of 
(2.2).  The integrand $x^{-\frac 32}\exp\left(-sx-\frac ux\right)f(u)$ belongs 
to $L_1[(0,\infty)\times (0,\infty)]$, that, by Fubini's Theorem, 
interchanging the order of the integral on the right of (2.2) is permissible.

\flushpar Therefore,
$$
\xi(s)=\int^{\infty}_0 f(u)\left[\int^{\infty}_0 x^{-\frac 32}\exp
\left[-sx-\frac ux\right] dx\right]du\tag 2.3
$$
We then use a well-known result in Robert and Kaufman \cite{15} on the right 
side of (2.3) to obtain
$$
\xi(s)=\pi^{\frac 12}\int^{\infty}_0 u^{-\frac 12}\exp\left(-2u^{
\frac 12}s^{\frac 12}\right)f(u)du.\tag 2.4
$$
Using (2.4) and (iii), it follows that
$$
\zeta (s)=\pi^{\frac 12}\int^{\infty}_0\left[\int^{\infty}_0 x^{-\frac 12}
u^{-\frac 12}\exp\left(-sx-\frac{2u{\frac 12}}x\right)f(u)du\right]dx.,\ 
\text{ where }\ \Cal Re\ s>\lambda_1.\tag 2.5
$$
Since $x^{-\frac 12}u^{-\frac 12}\exp\left(-sx-\frac{2u^{\frac 12}}x\right)$ 
belongs to $L_1[(0,\infty)\times (0,\infty)]$; therefore, according to 
Fubini's Theorem, (2.5) can be rewritten as 
$$
\zeta(s)=\pi^{\frac 12}\int^{\infty}_0u^{-\frac 12}f(u)\left[\int^{\infty}_0
x^{-\frac 12}\exp\left(-sx-\frac{\left(2^{\frac 32}u^{\frac 14}\right)^2}{4x}
\right)dx\right]du,\ \text{ where }\ \Cal Re\ s>\lambda_1.
$$
 From the tables of Roberts and Kaufman \cite{15}, we obtain
$$
s^{\frac 12}\zeta(s)=\pi\int^{\infty}_0u^{-\frac 12}f(u)\exp\left(-2^{\frac 32}
u^{\frac 14}s^{\frac 12}\right)du.\tag 2.6
$$
Next, we substitute $u=v^2$ in (2.6) to obtain
$$
s^{\frac 12}\zeta(s)=2\pi\int^{\infty}_0\exp\left(-2^{\frac 32}s^{\frac 12}
v^{\frac 12}\right)f(v^2)dv.\tag 2.7
$$
Replacing $s$ by $\left[p_1\left(\overline{s^{\frac 12}}\right)\right]^2$, 
multiplying both sides of (2.7) by $p_n\left(\overline{s^{\frac 12}}\right)$, 
we obtain 
$$
p_1\left(\overline{s^{\frac 12}}\right)p_n\left(\overline{s^{\frac 12}}\right)
\zeta\left[p_1\left(\overline{s^{\frac 12}}\right)\right] =2\pi
\int^{\infty}_0p_n\left(\overline{s^{\frac 12}}\right)\exp\left[-2^{\frac 32} 
p_1\left(\overline{s^{\frac 12}}\right)
v^{\frac 12}\right]f(v^2)dv\tag 2.8
$$
Now we use the operational relation given in Ditkin and Prudnikov \cite{11}
$$
s^{\frac 12}_i\exp\left(-as_i^{\frac 12}\right)\Doteq (\pi x_i)^{-\frac 12}
\exp\left(-\frac{a^2}{4x_i}\right)\text{ for } i=1,2,\dots, n\tag 2.9
$$
Equation (2.8) reads as follows
$$
p_n\left(\overline{s^{\frac 12}}\right)p_1\left(\overline{s^{\frac 12}}
\right)\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^2\right]
\overset n\to{\underset n\to =}\frac 2{\pi^{\frac{n-2}2}p_n
\left(\overline{x^{\frac 12}}\right)}\int^{\infty}_0\exp\left(-2 vp_1
\left(\overline{x^{-1}}\right)\right)f(v^2)dv\tag 2.10
$$
Applying (iv) in (2.10), we arrive at
$$
p_n\left(\overline{s^{\frac 12}}\right)p_1\left(\overline{s^{\frac 12}}\right)\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^2\right]
\overset n\to{\underset n\to=}\frac 2{\pi^{\frac{n-2}2}p_n\left(\overline{x^{\frac 12}}\right)}\ G\left[2p_1\left(\overline{x^{-1}}\right)\right].
$$
Therefore,
$$
\Cal L_n\left\{\frac 1{p_n\left(\overline{x^{\frac 12}}\right)}\ G\left[2p_1\left(\overline{x^{\frac 12}}\right)\right]; \overline s\right\}=\frac{\pi^{\frac{n-2}2}}2\cdot\frac{p_1\left(\overline{s^{\frac 12}}\right)}{p_n\left(\overline{s^{\frac 12}}\right)}\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^2\right],
$$
where $n=2,3,\dots, N$.\enddemo

To show the applicability of Theorem 2.1, we will construct certain functions with $n$ variables and calculate their Laplace transformation.

\example{Example 2.1}  Assume that $f(x)=x^{\frac{\tau-1}4}$.  Then
$$\align
\phi(s)&=\frac{\Gamma\left(\frac{\tau+3}4\right)}{s^{\frac{\tau+3}4}},\Cal Re\ s>0,\ \Cal Re\ v>-3;\\
\xi(s) &=\frac{\Gamma\left(\frac{\tau+3}4\right)\Gamma\left(\frac{\tau+1}4\right)}{s^{\frac{\tau+1}4}},\ \Cal Re\ \tau>-1,\Cal Re\ s>0,\ \text{and}\\
\zeta(s)&=\frac{\Gamma\left(\frac{\tau+3}4\right)\Gamma\left(\frac{\tau+1}4\right)\Gamma\left(\frac{\tau}2+1\right)}
{s^{\frac{\tau}2+1}},\ \Cal Re\ s>0,\ \Cal Re\ \tau >-1.\\
G(s) &=\frac{\Gamma\left(\frac{\tau +1}2\right)}{s^{\frac{\tau+1}2}},\ \Cal Re\ \tau >-1.\endalign
$$  
Therefore,
$$
\Cal L_n\left\{\frac 1{p_n\left(\overline{x^{\frac 12}}\right)\left[p_1\left(\overline{x^{-1}}\right)\right]^{\frac{\tau+2}2}};\ 
\overline s\right\}=\pi^{\frac{n-1}2}\Gamma\left(\frac{\tau}2+\frac 32\right)\cdot\frac 1{p_n\left(\overline{s^{\frac 12}}\right)\left[p_1\left(\overline{s^{\frac 12}}\right)\right]^{\tau+2}}\tag 2.11
$$
where $\Cal Re\ \tau >-1,\ \Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>0$, and $n=2,3,\dots, N$.
\endexample

\example{Example 2.2}  Suppose that $f(x)=I_0(ax^{\frac 12})$.  Then
$$\align
\phi(s) &=\frac 1s\exp \left(\frac{a^2}{4s}\right),\ \Cal Re\ s>0,\\
\xi(s) &=\frac{\pi^{\frac 12}}{\left(s-\frac{a^2}4\right)^{\frac 12}},\ \Cal Re\ s>\Cal Re\ \frac{a^2}4,\\
\zeta(s) &=\frac{4\pi}{a^{\frac 52}}\left\{\frac{4\pi}{\left[\Gamma\left(\frac 14\right)\right]^2}{}_1F_2\left[\matrix \frac 34\phantom{M};\\ \vspace{2\jot}\frac 12,\frac 54;\endmatrix \frac{s^2}{a^2}\right]-\frac{\left[\Gamma\left(\frac 14\right)\right]^2s}{6\pi}{}_1F_2\left[\matrix \frac 34\phantom{M};\\ \vspace{2\jot}
\frac 32,\frac 74;\endmatrix \frac{s^2}{a^2}\right]\right\}.\\
G(s) &=\frac 1{(s^2-a^2)^{\frac 12}},\ \Cal Re\ s>|\Cal Re\ a|.\endalign
$$
So that
$$\aligned
&\Cal L_n\left\{\frac 1{p_n(x^{\frac 12})}\cdot\frac 1{\left[4p^2_1\left(\overline{x^{-1}}\right)-a^2\right]^{\frac 12}}; \overline s\right\}\\
=\frac{2\pi^{\frac n2}p_1\left(\overline{s^{\frac 12}}\right)}{a^{\frac 52}p_n\left(\overline{s^{\frac 12}}\right)}&\left\{\frac{4\pi}{\left(\Gamma\left(\frac 14\right)\right]^2}{}_1F_2\left[\matrix \frac 34\phantom{M};\\ \vspace{2\jot}
\frac 12,\frac 54;\endmatrix \frac{p_1^2\left(\overline{s^{\frac 12}}\right)}{a^2}\right]-\frac{\left[\Gamma\left(\frac 14\right)\right]^2 p^2_1\left(\overline{s^{\frac 12}}\right)}{6\pi}\right.\\
&\qquad \left.{}_1F_2\left[\matrix\frac 54\phantom{M};\\ \vspace{2\jot}
\frac 32,\frac 74;\endmatrix \frac{p^2_1\left(\overline{s^{\frac 12}}\right)}{a^2}\right]; \overline s\right\},\endaligned\tag 2.12
$$
where $\Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>|\Cal Re\ a|$.
\endexample

\example{Example 2.3}  Consider $f(x)={}_pF_q\left[\matrix (a)_p;\\ \vspace{2\jot}
(b)_q;\endmatrix cx\right]$.  Then\newline
\phantom{Example 2.3.  Consider M}$\phi(s)=\frac 1s{}_{p+1}F_q\left[\matrix (a)_p;\\ \vspace{2\jot}
(b)_q;\endmatrix cx\right]$,\newline
where $p\le q,\ \Cal Re\ s>|\Cal Re\ c|$.
$$
\xi(s)=\frac{\pi^{\frac 12}}{s^{\frac 12}}{}_{p+2}F_q\left[\matrix (a)_p,1,\frac 12;\\ \vspace{2\jot}
(b)_q\phantom{MM};\endmatrix \frac cs\right],
$$
where $p\le q-1, \Cal Re\ \tau >0$ if $p+1<q, \Cal Re\ s>\Cal Re\ c$ if $p+1=q$,
$$
\zeta(s)=\frac{\pi}{2s^{\frac 32}}{}_{p+4}F_q\left[\matrix (a)_p,1,\frac 12,\frac 34,\frac 54 ;\\ \vspace{2\jot}
(b)_q\phantom{MM};\endmatrix\frac{4c}{s^2}\right],
$$
where $p\le q-3; \Cal Re\ s>0$ if $p\le q-4$; and $\Cal Re(s+2c^{\frac 12}\cos\pi r)>0$\newline
$(r=0,1)$ if $p=q-3$.\newline
Hence
$$\align
&\Cal L_n\left\{\frac 1{p_n\left(\overline{x^{\frac 12}}\right)\left[p_1\left(\overline{x^{-1}}\right)\right]}{}_{p+2}F_q\left[\matrix (a)_p,1,\frac 12;\\ \vspace{2\jot}
(b)_q\phantom{MM};\endmatrix \frac c{p^2_1\left(\overline{x^{-1}}\right)}\right];s\right\}\\
&\phantom{M}=\frac{\pi^{\frac n2}}{2p_n\left(\overline{s^{\frac 12}}\right)p^2_1\left(\overline{s^{\frac 12}}\right)}{}_{p+4}F_q\left[\matrix (a)_p,1,\frac 12,\frac 34,\frac 54;\\ \vspace{2\jot}
(b)_q\phantom{MMMm};\endmatrix \frac{4c}{p^4_1\left(\overline{s^{\frac 12}}\right)}\right],\tag 2.13\endalign
$$
where $p\le q-3, \Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>0$ if $p\le q-4$; and $\Cal Re \left[p_1\left(\overline{s^{\frac 12}}\right)+2c^{\frac 12}\cos \pi r\right]$\newline
$>0 (r=0,1)$ if $p=q-3$.
\endexample

\example{Example 2.4}  Assume that $f(x)=x^{\frac 12}J_0(ax^{\frac 12})$.  Then
$$\align
&\phi(s)=\frac{\pi^{\frac 12}}{2s^{\frac 32}}{}_1F_1\left[\matrix \frac 32;\\ \vspace{2\jot}
1;\endmatrix -\frac{a^2}{4s}\right], \Cal Re\ s>0,\\
&\xi(s)=\frac{\pi^{\frac 12}}2{}_2F_1\left[\matrix \frac 32,1;\\ \vspace{2\jot}
1\phantom{M};\endmatrix -\frac{a^2}{4s}\right],\Cal Re\ s>-\Cal Re-\frac{a^2}4,\\
&\zeta(s)=\frac 1{\pi^{\frac 12}s^{\frac 12}}\bold G^{1,4}_{4,2}
\left(\frac {a^2}{s^2}\bigg\vert\matrix \frac 14, \frac 34,0,-\frac 12\\ \vspace{2\jot}
0,0\endmatrix\right),\endalign
$$
where $\Cal Re\ s>0,|\arg\ a|<2\pi$.
$$
G(s)=\frac{\pi^{\frac 12}}2{}_2F_1\left[\matrix \frac 32,1;\\ \vspace{2\jot}
1\phantom{M};\endmatrix -\frac{a^2}{4s}\right], \Cal Re\ s>\Cal Re-\frac {a^2}4.
$$
Hence
$$\align
&\Cal L_n\left\{\frac{p_1\left(\overline{x^{-1}}\right)}{p_n\left(\overline{x^{\frac 12}}\right)\left[4p^2_1\left(\overline{x^{-1}}\right)+a^2\right]^{\frac 32}}, \overline s\right\}\\
= &\frac{\pi^{\frac{n-3}2}p_1\left(\overline{s^{\frac 12}}\right)}{p_n\left(\overline{s^{\frac 12}}\right)}\bold G^{1,4}_{4,2}\left(\frac{a^2}{p^4_1\left(\overline{s^{\frac 12}}\right)}\bigg\vert \matrix\frac 12,\frac 34,0,-\frac 12\\ \vspace{2\jot}
0,0\phantom{MM}\endmatrix\right),\tag 2.14\endalign
$$
where $\Cal Re \left[p_1\left(\overline{s^{\frac 12}}\right)\right]>0,|\arg\ a|<2\pi$ and $\bold G^{1,4}_{4,2}$ is a Meijer's $G$-function.\endexample

\example{Example 2.5 (Two-Dimensions)}

Upon substituting $n=2$ in Examples 2.1, 2.2, 2.3, and 2.5 we arrive at the following results, respectively
$$
\Cal L_2\left\{\frac{(xy)^{\frac{\tau}2}}{(x+y)^{\frac{\tau+1}2}}; s_1, s_2\right\}=\pi^{\frac 12}\Gamma\left(\frac{\tau}2+1\right)\cdot\frac 1{(s_1s_2)^{\frac 12}\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^{\tau+1}}\tag 2.15
$$
where $\Cal Re\ \tau >-1,\ \Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0$.
$$\aligned
&\qquad\qquad\qquad\Cal L_2\left\{\frac{(xy)^{\frac 12}}{[4(x+y)^2-(axy)^2]^{\frac 12}}; s_1, s_2\right\}\\
&=\frac{2\pi\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}{a^{\frac 52}(s_1 s_2)^{\frac 12}}\left\{\frac{4\pi}{\left[\Gamma\left(\frac 14\right)\right]^2}{}_1F_2\left[\matrix \frac 34\phantom{M};\\ \vspace{2\jot}
\frac 12,\frac 54;\endmatrix\ \frac{\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2}{a^2}\right]\right.\\
&\qquad\qquad\left.-\frac{\left[\Gamma\left(\frac 14\right)\right]^2\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2}{6\pi}{}_1F_2\left[\matrix \frac 54\phantom{M};\\ \vspace{2\jot}
\frac 32,\frac 74;\endmatrix\ \frac{\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2}{a^2}\right]\right\}\endaligned\tag 2.16
$$
where $\Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0, |\arg\ a|<2\pi$.
$$\aligned
&\qquad \Cal L_2\left\{\frac{(xy)^{\frac 12}}{x+y}{}_{p+2}F_q\left[\matrix (a)_p,1,\frac 12;\\ \vspace{2\jot}
(b)_q\phantom{MM};\endmatrix\ \frac{c(xy)^2}{(x+y)^2}\right]; s_1, s_2\right\}\\
=&\frac{\pi}{2(s_1s_2)^{\frac 12}\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2}{}_{p+4}F_q\left[\matrix (a)_p,\frac 12,\frac 34,\frac 54;\\ \vspace{2\jot}
(b)_q\phantom{MM};\endmatrix\ \frac{4c}{\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^4}\right],\endaligned\tag 2.17
$$
where $p\le q-3,\ \Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0$ if $p\le q-4$; and $\Cal Re\left[\left(s^{\frac 12}_1+s^{\frac 12}_2\right)+2c^{\frac 12}\cos\pi r\right]$\newline
$>0(r=0,1)$ if $p=q-3$.
$$
\Cal L_2\left\{\frac{(xy)^{\frac 32}(x+y)}{\left[4(x+y)^2+(axy)^2\right]^{\frac 32}}; s_1, s_2\right\}=\frac{s^{\frac 12}_1+s^{\frac 12}_2}{(\pi s_1s_2)^{\frac 12}}\bold G^{1,4}_{4,2}\left(\frac{a^2}{\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^4}\bigg\vert\matrix \frac 12,\frac 32,0,-\frac 12\\ \vspace{2\jot}
0,0\phantom{MMM}\endmatrix\right),\tag 2.18
$$
where $\Cal R\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>|\Cal Re\ a|$ and $\bold G^{1,4}_{4,2}$ is a Meijer's $G$-function.
\endexample

\remark{Remark 2.1}  If we let $\tau=0$ in Relation (2.15), and then using Relation (1) in \cite{24}.  We deduced that
$$
\Cal L_2\left\{(x+y)^{\frac 12}; s_1,s_2\right\}=\frac{\pi^{\frac 12}\left(s_1+s_2+s_1^{\frac 12}s^{\frac 12}_2\right)}{2(s_1s_2)^{\frac 32}\left(s_1^{\frac 12}+s^{\frac 12}_2\right)}\tag 2.19
$$\endremark

\proclaim{Theorem 2.2}  Assume that $f$ belongs to class $\Omega$ and $\phi$ be the one-dimensional Laplace transformation of $f$.  Let\roster
\item"{(i)}" $\Cal L\left\{x^{-\frac 12}\phi\left(\frac 1x\right); s\right\}=\gamma(s)$,
\item"{(ii)}" $-\frac d{ds}\left\{s^{-v}\gamma\left(\frac 1{s^2}\right)\right\}=\zeta(s)$,
\item"{(iii)}" $\Cal L\left\{xf(x);s\right\}=H(s)$,\endroster
and suppose that $x^{-\frac 12}\phi\left(\frac 1x\right)$ belongs to $\Omega$ and $\frac d{ds}
\left\{s^{-v}\gamma\left(\frac 1{s^2}\right)\right\}$ exists for $\Cal Re\ s>c_1$ where $c_1$ is a constant.  Then
$$\aligned
&\Cal L_n\left\{\frac{(v-1)\phi\left[p_1\left(\overline{x^{-1}}\right)\right]-2p_1\left(\overline{x^{-1}}\right)H\left[p_1\left(\overline{x^{-1}}\right)\right]}{p_n\left(\overline{x^{\frac 12}}\right)}; \overline s\right\}=\\
&\qquad\qquad\frac {\pi^{\frac{n-1}2}}{p_n\left(\overline{s^{\frac 12}}\right)\left[p_1\left(\overline{s^{\frac 12}}\right)\right]^v}\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^{-1}\right],\endaligned\tag 2.20
$$
where $\Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>d$, a constant $n=2,3,\dots,N$.  It is asssumed that the integral on the left exists.
\endproclaim

\demo{Proof}  A similar method to Theorem 2.2 can be used to prove this theorem.  The outline of the proof is as follows.

Making use of our hpothesis and (i) yields
$$
\gamma(s)=\pi^{\frac 12}\int^{\infty}_0\left[\int^{\infty}_0x^{-\frac 12}\exp\left(-sx-\frac ux\right)f(u)du\right]dx,\text{ where }\Cal Re\ s>c_1.\tag 2.21
$$
Using Fubini's Theorem to interchange the order of the integral on the right side of (2.21) and applying a result in Roberts and Kaufman \cite{15} we obtain
$$
\gamma(s)=\pi^{\frac 12}\int^{\infty}_0f(u)s^{-\frac 12}\exp\left(-2u^{\frac 12}s^{\frac 12}\right)du.\tag 2.22
$$
Taking into account the condition (ii) we see that the equation (2.22) implies that
$$\align
&s^v\zeta(s)=\pi^{\frac 12}(v-1)\int^{\infty}_0f(u)\exp\left(-\frac{2u^{\frac 12}}s\right)du-2\pi^{\frac 12}s^{-1}\int^{\infty}_0u^{\frac 12}f(u)\\
&\qquad\qquad\exp\left(\frac{-2u^{\frac 12}}s\right)du,\text{ where } \Cal Re\ s>c_1.\tag 2.23\endalign
$$
Now. replacing $s$ by $\left[p_1\left(\overline{s^{\frac 12}}\right)\right]^{-1}$ and then multiplying each side of (2.23) by $p_n\left(\overline{s^{\frac 12}}\right)$ and then making use of operational relations (2.9) and (2.24)
$$
s_i\exp\left(-as_i^{\frac 12}\right)\Doteq \frac a2\pi^{-\frac 12}x_i^{-\frac 32}\exp\left(-\frac{a^2}{4x_i}\right)\text{ for } i=1,2,\dots, n,\tag 2.24
$$
equation (2.23) reads
$$\align
&p_n\left(\overline{s^{\frac 12}}\right)\left[p_1\left(\overline{s^{\frac 12}}\right)\right]^{-v}\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^{-1}\right]\overset n\to{\underset n\to =}\frac 1{\pi^{\frac{n-1}2}p_n\left(\overline{x^{\frac 12}}\right)}\left[(v-1)\int^{\infty}_0f(u)\right.\\
&\exp\left[-up_1\left(\overline{x^{-1}}\right)\right]dx -2p_1\left(\overline{x^{-1}}\right)\int^{\infty}_0uf(u)\exp\left[-up_1\left(\overline{x^{-1}}\right)\right]du.\tag 2.25\endalign
$$
Equation (2.25) with (iii) and the definition of the one-dimensional Laplace transform, leads to
$$\align
&p_n\left(\overline{s^{\frac 12}}\right)\left[p_1\left(\overline{s^{\frac 12}}\right)\right]^{-v}\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^{-1}\right]\\
&\qquad\qquad \overset n\to{\underset n\to =}\frac 1{\pi^{\frac{n-1}2}p_n\left(\overline{x^{\frac 12}}\right)}\left\{(v-1)\phi\left[p_1\left(\overline{x^{-1}}\right)\right]-2p_1\left(\overline{x^{-1}}\right)H\left[p_1\left(\overline{x^{-1}}\right)\right]\right\}.\endalign
$$
Therefore,
$$\align
&\Cal L_n\left\{\frac{(v-1)\phi\left[p_1\left(\overline{x^{-1}}\right)\right]-2p_1\left(\overline{x^{-1}}\right)H\left[p_1\left(\overline{x^{-1}}\right)\right]}{p_n\left(\overline{x^{\frac 12}}\right)}; \overline s\right\}\\
&\qquad\qquad\qquad =\frac{\pi^{\frac{n-1}2}}{p_n\left(\overline{s^{\frac 12}}\right)\left[p_n\left(\overline{s^{\frac 12}}\right)\right]^v}\zeta\left[\left(p_1\left(\overline{s^{\frac 12}}\right)\right)^{-1}\right],\endalign
$$
where $\Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>d$ for some constant $d, n=2,3,\dots, N$.\enddemo

\example{Example 2.6}  Let $f(x)=J_0\left(2x^{\frac 12}\right)$.  Then
$$\align
&\phi(s)=\frac 1s\exp\left(-\frac 1s\right),\ \Cal Re\ s>0,\\
&\gamma(s)=\frac{\pi^{\frac 12}}{2(s+1)^{\frac 32}},\Cal Re\ s>-1.\endalign
$$
Thus
$$
\zeta(s)=\frac{\pi^{\frac 12}(vs^2+v-3)}{2s^{v-2}(1+s^2)^{\frac 52}},\ \Cal Re\ s>-1.
$$
Therefore
$$\align
&\Cal L_n\left\{\frac 1{p_1\left(\overline{x^{-1}}\right)p_n\left(\overline{x^{\frac 12}}\right)}\exp\left(-\frac 1{p_1\left(\overline{x^{-1}}\right)}\right)\left\{ (v-1)-2_1F_1\left[\matrix -1;\\ \phantom{-}1;\endmatrix\ \frac 1{p_1\left(\overline{x^{-1}}\right)}\right]\right\};\overline s\right\}\\ \vspace{2\jot}
&=\frac{\pi^{\frac n2}p_1\left(\overline{s^{\frac 12}}\right)\left[v+(v-3)p^2_1\left(\overline{s^{\frac 12}}\right)\right]}{2p_n\left(\overline{s^{\frac 12}}\right)\left[1+p^2_1\left(\overline{s^{\frac 12}}\right)\right]^{\frac 52}},\ \text{where } \Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>-1.\tag 2.26\endalign
$$
\endexample

\remark{Remark 2.2}  If we let $n=2$ and $v=1$ or $v=3$, from the equation (2.26) we deduce the following results, respectively
$$\align
(i)\quad &\Cal L_2\left\{\frac{(xy)^{\frac 12}}{(x+y)} \exp\left(-\frac{xy}{x+y}\right){}_1F_1\left[\matrix -1;\\ \phantom{-}1;\endmatrix\frac{xy}{x+y}\right]; s_1, s_2\right\}=\phantom{MMMMMMMMM}\\
\qquad &\frac{\pi\left(s^{\frac 12}_1+s^{\frac 12}_2\right)\left[2\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2-1\right]}{4(s_1s_2)^{\frac 12}\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2\right]^{\frac 52}},\tag 2.26{$'$}\endalign
$$
where $\Cal Re \left[s^{\frac 12}_1+s^{\frac 12}_2\right]>-1$.
$$\align
(ii)\quad &\Cal L_2\left\{\frac{(xy)^{\frac 12}}{(x+y)}\exp\left(-\frac{xy}{x+y}\right)\left\{1-{}_1F_1\left[\matrix -1;\\ \phantom{-}1;\endmatrix\ \frac{xy}{x+y}\right]\right\}; s_1, s_2\right\}=\phantom{MMMMM}\\
\qquad &\frac{3\pi\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}{2(s_1s_2)^{\frac 12}\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2\right]^{\frac 52}},\tag 2.26{$''$}\endalign
$$
where $\Cal Re \left[s^{\frac 12}_1+s^{\frac 12}_2\right]>1$.

\flushpar Notice, with the help of (2.26$'$) from (2.26$''$) we arrive at the following result
$$
\Cal L_2\left\{\frac{(xy)^{\frac 12}}{(x+y)}\exp\left(-\frac{xy}{x+y}\right); s_1, s_2\right\}=\frac{\pi}2\frac{\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}{(s_1s_2)^{\frac 12}\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2\right]^{\frac 23}}.\tag 2.26{$'''$}
$$
This is the same as the result (2.109) in Ditkin and Prudnokov \cite{11; p. 140}.

Furthermore, with the help of (2.26$'''$) and the operational relation (47) in Voelker and Doetsch \cite{24; p. 159}, we derive the following new results.
$$\align
&\Cal L_2\left\{\left(\frac yx\right)^{\frac 12}\cdot \frac 1{x+y}\exp\left(-\frac{xy}{x+y}\right); s_1, s_2\right\}=\frac{\pi}{s^{\frac 12}_2\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2\right]^{\frac 12}},\phantom{MMMM}\\
&\Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>1.\tag 2.26{$^{iv}$}\\
&\Cal L_2\left\{\left(\frac xy\right)^{\frac 12}\cdot\frac 1{x+y}\exp\left(-\frac{xy}{x+y}\right); s_1, s_2\right\}=\frac{\pi}{s^{\frac 12}_1\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2\right]^{\frac 12}},\tag 2.26{$^v$}\endalign
$$
where $\Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>1$.\endremark

\example{Example 2.7}  Assuming $x^{-\frac 12}\cos 2x^{\frac 12}$, we obtain
$$\align
\phi(x) &=\left(\frac{\pi}s\right)^{\frac 12}\exp\left(-\frac 1x\right),\ \Cal Re\ s>0.\\
\gamma(s) &=\frac{\pi}{s+1},\ \Cal Re\ s>-1.\\
\zeta(s) &=\left(vs^2+v-2\right)\frac{\pi^{\frac 12}s^{-v+1}}{(s^2+1)^2},\ \Cal Re\ s>-1.\\
H(s) &= \frac{\pi^{\frac 12}}{s^{\frac 52}}\left(\frac s2-1\right)\exp\left(-\frac 1s\right),\ \Cal Re(s)>0.\endalign
$$
Therefore,
$$\align
&\Cal L_n\left\{\frac{(v-1)p_1\left(\overline{x^{-1}}\right)-\left(p_1\left(\overline{x^{-1}}\right)-2\right)\exp\left(-\frac 1{p_1\left((\overline{x^{-1}}\right)}\right)}{\left[p_1\left(\overline{x^{-1}}\right)\right]^{\frac 32}p_n\left(\overline{x^{-1}}\right)};\overline s\right\}\\
&\ =\frac{\pi^{\frac{n-1}2}p_1\left(\overline{s^{\frac 12}}\right)}{p_n\left(\overline{s^{\frac 12}}\right)\left[1+p^2_1\left(\overline{s^{\frac 12}}\right)\right]^2}\left\{(v-2)p^2_1\left(\overline{s^{\frac 12}}\right)+v\right\}, \Cal Re\left[p_1\left(\overline{s^{\frac 12}}\right)\right]>0.\tag 2.27\endalign
$$
\endexample

\remark{Remark 2.3}  If we let $n=2$ and $v=1$ in (2.27) by using the following well-known formula
$$\align
&\Cal L_2\left\{\frac{xy}{(x+y)^{\frac 32}}\exp\left(-\frac{xy}{x+y}\right); s_1, s_2\right\}=\frac{\pi^{\frac 12}\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}{(s_1s_2)^{\frac 12}\left[1+\left(s_1^{\frac 12}+s^{\frac 12}_2\right)^2\right]^2},\\
&\qquad \Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0.\endalign
$$
we derive that
$$\aligned
&\Cal L_2\left\{\frac 1{(x+y)^{\frac 12}}\exp\left(-\frac{xy}{x+y}\right); s_1, s_2\right\}=\frac{\pi^{\frac 12}\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}{(s_1s_2)^{\frac 12}\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^2\right]},\\
&\Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0.\endaligned\tag 2.27{$'$}
$$
\endremark
\vskip .3in
\head{3.\ Non-Homogeneous Second Order Partial Differential Equations of Parabolic type}\endhead

In this section we solved a few partial differential equations of the type
$$
u_{xx}+2u_{xy}+u_{yy}+u=f(x,y), 0<x<\infty, 0<y<\infty,\tag 3.1
$$
under the following initial and boundary conditions
$$
u(x,0)=u(0,y)=u_y(x,0)=u_x(0,y)=u(0,0)=0\tag 3.2
$$
by means of some of our results established in Section 2 using the double Laplace transformation.

\example{Example 3.1}  Determination of a solution $u=u(x,y)$ of (3.1) and (3.2) for\roster
\item"{(a)}" $f(x,y)=(x+y)^{\frac 12}$\newline
\item"{(b)}" $f(x,y)=\frac{(xy)^{\frac{\tau}2}}{(x+y)^{\frac{\tau+1}2}}$\newline
\item"{(c)}" $f(x,y)=\frac{(xy)^{\frac {\tau}2}}{(x+y)^{\frac{\tau+2}2}}$\endroster

We will use the following for the rest of this section.  If
$$\align
&u(x,0)=f(x), u(0,y)=g(y),\\
&u_y(x,y)|_{y=0}=u_y (x,0)=f_1(x), u_x(x,y)|_{x=0}=u_x(0,y)=g_1(y)\endalign
$$
and if their one-dimensional Laplace transformations are $F(s_1), G(s_2), F_1(s_1)$ and $G_1(s_2)$, respectively, then
$$\align
&\Cal L_2\{u(x,y); s_1, s_2\}=\int^{\infty}_0\int^{\infty}_0\exp(-s_1x-s_2y)u(x,y)dxdy=U(s_1,s_2)\tag 3.3\\
&\qquad \Cal L_2\{u_{xx}; s_1,s_2\}=s^2_1U(s_1,s_2)-s_1G(s_2)-G_1(s_2)\tag 3.4\\
&\qquad \Cal L_2\{u_{yy}; s_1, s_2\}=s^2_2U(s_1,s_2)-s_2F(s_1)-F_1(s_1)\tag 3.5\\
&\quad \Cal L_2\{u_{xy}; s_1, s_2\}=s_1s_2U(s_1,s_2)-s_1F(s_1)-s_2G(s_2)+u(0,0)\tag 3.6\endalign
$$

(a)\ By applying the double Laplace transformation termwise to partial differential equation and the initial-boundary condition in (3.1) and (3.2), using (3.3)--(3.6), and with the aid of Relation (2.19) in Remark 2.1, we obtain the transformed problem
$$
U(s_1, s_2)=\frac 1{(s_1+s_2)^2+1}\cdot\frac{\pi^{\frac 12}\left[s_1+s_2+(s_1s_2)^{\frac 12}\right]}{2(s_1s_2)^{\frac 32}\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}, \Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0.\tag 3.7
$$

The inversion of (3.7) will provide us with the solution of (3.1) and (3.2).  So that, the inverse transform of (3.7) can be obtained using formula (133) in \cite{24}
$$
u(x,y)=\int^x_0(x+y-2\xi)^{\frac 12}\sin\xi d\xi\tag 3.8
$$
By a simple change of variable $x+y-2\xi=2t^2$ in (3.8), we obtain
$$
u(x,y)=2^{+\frac 32}\int^{\left(\frac{y-x}2\right)^{\frac 12}}_{\left(\frac{x+y}2\right)^{\frac 12}}t^2\sin\left(t^2-\frac{x+y}2\right)dt\ \text{ if }\ y>x.
$$
Expanding the sine and making some simplification, we deduce that
$$\aligned
&u(x,y)=2^{+\frac 32}\left[\cos\left(\frac{x+y}2\right)\int^{\left(\frac{y-x}2\right)^{\frac 12}}_{\left(\frac{x+y}2\right)^{\frac 12}}t^2\sin t^2dt-\sin\left(\frac{x+y}2\right)\right.\\
&\left.\int^{\left(\frac{x-y}2\right)^{\frac 12}}_{\left(\frac{x+y}2\right)^{\frac 12}}t^2\cos t^2dt\right]\endaligned\tag 3.9
$$
Calculating the integrals involved in (3.9), we arrive at
$$\align
&u(x,y)=\frac 14\left\{\matrix \left(\frac{x+y}2\right)^{\frac 12}-\left(\frac{y-x}2\right)^{\frac 12}\cos x+\pi^{\frac 12}\cos\left(\frac{x+y}2\right)\left[C\left(\frac{y-x}2\right)-D\left(\frac{x+y}2\right)\right]\\ \vspace{2\jot}
\phantom{MMMMMMMMM}+\pi^{\frac 12}\sin\left(\frac{x+y}2\right)\left[S\left(\frac{y-x}2\right)-S\left(\frac{x+y}2\right)\right]\endmatrix\right\}\\
&\qquad \text{ if }y>x,\endalign
$$
where $C(\cdot)$ and $S(\cdot)$ are Fresnel integrals.

Similarly, to obtain the transform equations for parts (b) and (c).  We replace Relation (2.19) in Example (2.5) for part (b) and (2.15) in the same example with Formula 181 in Brychkov et al.  \cite{2; p.300} for part (c) to arrive at
$$
U(s_1,s_2)=\frac{\pi^{\frac 12}\Gamma\left(\frac{\tau}2+1\right)}{\left[(s_1+s_2)^2+1\right](s_1s_2)^{\frac {\tau}2}\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^{\tau+1}}\tag 3.10
$$
$$
U(s_1,s_2)=\frac{\pi\Gamma(\tau+1)}{2^{\tau}\Gamma\left(\frac{\tau+3}2\right)\left[(s_1+s_2)^2+1\right]\left(s^{\frac 12}_1+s^{\frac 12}_2\right)^{\tau+1}},\tag 3.11
$$
where $\Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0$.  Thus, we obtain the following solutions, respectively
$$\align
&u(x,y)=\frac 1{2^{\tau+1}}\int^{y-x}_{x+y}t^{-\frac{\tau+1}2}
\left[t^2-(x-y)^2\right]^{\frac{\tau}2}\sin\left(\frac{t-(x+y)}2\right)dt\\
&\qquad \text{if}\ y>x\Cal Re\ v>-1.\tag 3.12\\
&u(x,y)=\frac 1{2^{\tau+1}}\int^{y-x}_{x+y}t^{-\frac{\tau+3}2}\left[t^2-(x-y)^2\right]^{\frac{\tau}2}\sin\left(\frac{t-(x+y)}2\right)dt\\
&\qquad \text{if}\ y>x,\Cal Re\ v>-1.\tag 3.13\endalign
$$
\endexample

\remark{Remark 2.4}  Substituting $\tau=0$ in parts (b) and (c), lead to
$$\align
&u_{xx}+2u_{xy}+u_{yy}+u=\frac 1{(x+y)^{\frac 12}}\tag 3.14\\
&u_{xx}+2u_{xy}+u_{yy}+u=\frac 1{(x+y)^{\frac 32}}\tag 3.15\endalign
$$
Next, using (3.14) and (3.15), we arrive at the following explicit solutions for the equations (3.14) and (3.15) respectively
$$\align
&u(x,y)=\left(\frac{\pi}2\right)^{\frac 12}\left\{\cos\left(\frac{x+y}2\right)\left[S\left(\frac{y-x}2\right)-S\left(\frac{x+y}2\right)\right]-\right.\\
&\qquad\left. \sin\left(\frac{x+y}2\right)\left[C\left(\frac{y-x}2\right)-C\left(\frac{x+y}2\right)\right]\right\}\ \text{if}\ y>x.\endalign
$$
$$\align
&u(x,y)=\frac 1{(x+y)^{\frac 12}}\cos(x+y)-\frac 1{(y-x)^{\frac 12}}\cos y-\pi^{\frac 12}\left\{\cos\left(\frac{x+y}2\right)\left[S\left(\frac{y-x}2\right)
\right.\right.\\
&\qquad \left.\left. -S\left(\frac{x+y}2\right)\right]-\sin\left(\frac{x+y}2\right)\left[C\left(\frac{y-x}2\right)-C\left(\frac{x+y}2\right)\right]\right\}\ \text{if}\ y>x.\endalign
$$                 
\endremark

\example{Example 3.2}  Solve the following Parabolic differential equation described by
$$
u_{xx}+2u_{xy}+u_{yy}+u=\frac 1{(x+y)^{\frac 12}}\exp\left(-\frac{xy}{x+y}\right),\ 0<x<\infty,\ 0<y<\infty,\tag 3.16
$$
under the initial and boundary conditions
$$
u(x,0)=u(0,y)=u_y(x,0)=u_x(0,y)=u(0,0)=0.\tag 3.17
$$
With the aid of (2.27') in Remark 2.3 and the similar procedure we have followed for Example 3.1, the transformed problems reads
$$
U(s_1,s_2)=\frac{\pi\left(s^{\frac 12}_1+s^{\frac 12}_2\right)}{\left[(s_1+s_2)^2+1\right]\left[1+\left(s^{\frac 12}_1+s^{\frac 12}_2\right)\right]^{\frac 12}},\tag 3.18
$$
where $\Cal Re\left[s^{\frac 12}_1+s^{\frac 12}_2\right]>0$.  Using formula (133) in \cite{24} the inverse transform of (3.18) leads to the following integral representation.
$$
u(x,y)=\int^{y-x}_{x+y}t^{-\frac 12}\exp\left[\frac{t^2-(x+y)^2}{4t}\right]\sin\left(\frac{t-(x+y)}2\right)dt\ \text{if}\ y>x.
$$
\endexample

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\enddocument