\input amstex \documentstyle{amsppt} \loadmsbm \hcorrection{1cm} \vcorrection{-6mm} \nologo \TagsOnRight \NoBlackBoxes \pageno=25 \headline={\ifnum\pageno=25 \hfill\else% {\tenrm\ifodd\pageno\rightheadline \else \leftheadline\fi}\fi} \def\rightheadline{\hfil Systems of multi-dimensional Laplace transforms \hfil\folio} \def\leftheadline{\folio\hfil Ali Babakhani \& R. S. Dahiya \hfil} \def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt % 16th Conference on Apllied Mathematics, Univ. of Central Oklahoma,\hfill\break Electronic Journal of Differential Equations, Conf. 07, 2001, pp. 25--36.\hfill\break ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \hfill\break ftp ejde.math.swt.edu (login: ftp)\bigskip} } \topmatter \title Systems of multi-dimensional Laplace transforms and a heat equation \endtitle \thanks {\it Mathematics Subject Classifications:} 44A30.\hfil\break\indent {\it Key words and phrases:} Two-dimensional Laplace transforms. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Published July 20, 2001. \endthanks \author Ali Babakhani \& R. S. Dahiya \endauthor \address Ali Babakhani \hfill\break Department of Mathematics, College of Engineering, University of Tehran, Tehran, Iran \endaddress \address R. S. Dahiya\hfill\break Department of Mathematics, Iowa State University, Ames, IA 50011, USA \endaddress \email dahiya\@math.iastate.edu \endemail \abstract The object of this paper is to establish several new theorems involving systems of two-dimensional Laplace transforms containing five to seven equations. These systems can be used to calculate new Laplace transform pairs. In the second part, a boundary value problem is solved by using the double Laplace transformation. \endabstract \endtopmatter \define\({\left(} \define\){\right)} \define\[{\left[} \define\]{\right]} \def\dotseq{\underset{\raise5pt\hbox{..}}\to{\overset{\raise1pt\hbox{..}}\to =}} \document \head 1. Introduction \endhead The two-dimensional Laplace transform of function $f(x,y)$ is defined by Ditkin and Prudnikov [5] as follows: $$ F(p,q)=pq\int^{\infty}_0\int^{\infty}_0e^{-px-qy} f(x,y)dx\ dy $$ and symbolically is denoted by $F(p,q)\dotseq f(x,y)$ where the symbol $\dotseq $ is called ``operational''. The correspondence between $f(x,y)$ and $F(p,q)$ may be interpreted as a transformation which transforms the function $f(x,y)$ into the function $F(p,q)$. Thus we call $F(p,q)$ the image of $f(x,y)$ and $f(x,y)$ is the original of $F(p,q)$. In this paper we derive new rules on systems involving double Laplace transformations. We also solve a boundary value problem. \subhead{2.\ Systems of two-dimensional Laplace transforms}\endsubhead \subhead{2.1.\ The Image of $G(\frac 1{4x}, \frac 1{4y})$}\endsubhead The six systems of two dimensional Laplace transforms are obtained in this s ection. Each system contains five to seven equations. These systems can be used to calculate one of the functions, when the others are known, especially to compute the image of $x^{i/2}y^{j/2}G(\frac 1{4x},\frac 1{4y})$ when $i=-3,\pm 5$ and $j=\pm 1,\pm 3$. They are further used to obtain new Laplace transform pairs. These systems are proved in six theorems and some typical examples are given after each theorem. The image of $x^{i/2}y^{j/2}G(\frac 1{4x},\frac 1{4y})$ when $i=3$ and $j=\pm 1,3$ is given by Dahiya \cite{3}. \proclaim{Theorem 2.1} Let \roster \item"{(I)}" $F(p,q)\dotseq f(x,y)$ \item"{(II)}" $\psi(p,q)=-pq\frac{\partial}{\partial p}\(\frac{F(p,q)}{pq}\)\dotseq xf(x,y)$ \item"{(III)}" $\sigma(p,q)=-pq\frac{\partial}{\partial q}\(\frac{F(p,q)}{pq}\)\dotseq yf(x,y)$ \item"{(IV)}" $\varphi(p,q)=pq\frac{\partial^2}{\partial p\partial q}\(\frac{F(p,q)}{pq}\)\dotseq xyf(x,y)$ \item"{(V)}" $G(p,q)\dotseq xyf\(\sqrt x,\sqrt y\)$.\endroster Then $$\aligned (pq)^{1/2}&F\(\sqrt p,\sqrt q\)+pq^{1/2}\psi\(\sqrt p,\sqrt q\)+p^{1/2}q\sigma \(\sqrt p,\sqrt q\)+pq\varphi\(\sqrt p,\sqrt q\)\\ &\dotseq \frac 1{4\pi}(xy)^{-3/2}G\(\frac 1{4x},\frac 1{4y}\)\endaligned\tag 2.1.1 $$ \endproclaim \demo{Proof} We start from two operational relations $$\align &p(1+s\sqrt p)e^{-s\sqrt p}\doteqdot \frac 1{4\sqrt{\pi}} s^3x^{-5/2}e^{-s^2/4x}\\ &q(1+t\sqrt q)e^{-t\sqrt q}\doteqdot \frac 1{4\sqrt{\pi}} t^3y^{-5/2}e^{-t^2/4y}\endalign $$ We multiply together the above equations to get $$ pq(1+\sqrt p)(1+t\sqrt q)e^{-s\sqrt p-t\sqrt q}\dotseq \frac 1{16\pi}(xy)^{-5/2}(st)^3e^{-\frac{s^2}{4x}-\frac{t^2}{4y}}. $$ Now, we multiply both sides by $f(s,t)$ and integrate with respect to $s$ and $t$ over the positive quarter plane. $$\align &pq\int^{\infty}_0\int^{\infty}_0(1+s\sqrt p)(1+t\sqrt q)e^{-s\sqrt p-t\sqrt q}f(s,t)ds dt\\ &\quad \dotseq \frac 1{16\pi}(xy)^{-5/2}\int^{\infty}_0\int^{\infty}_0e^{-\frac{s^2}{4x}-\frac{t^2}{4y}}(st)^3f(s,t)ds dt.\endalign $$ We make the change of variables $s=\sqrt u$ and $t=\sqrt v$ on the right hand side to obtain $$\align &pq\int^{\infty}_0\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}f(s,t)dsdt+p^{3/2}q\int^{\infty}_0\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}sf(s,t)dsdt\\ &+pq^{3/2}\int^{\infty}_0\!\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}tf(s,t)dsdt +(pq)^{3/2}\int^{\infty}_0\!\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}(st)f(s,t)dsdt\\ &\qquad \dotseq \frac 1{64\pi}(xy)^{-5/2}\int^{\infty}_0\int^{\infty}_0e^{-\frac u{4x}-\frac v{4y}}(uv)f(\sqrt u,\sqrt v)dudv.\endalign $$ Finally by using (V) on the right hand side and (I), (II), (III), (IV) on the left hand side we get $$\align &(pq)^{1/2}F(\sqrt p,\sqrt q)+pq^{1/2}\psi(\sqrt p,\sqrt q)+p^{1/2}q\sigma(\sqrt p,\sqrt q)+pq\varphi(\sqrt p,\sqrt q)\\ &\quad \dotseq \frac 1{4\pi}(xy)^{-3/2}G\(\frac 1{4x},\frac 1{4y}\).\qed\endalign $$ \enddemo Thus we have the following system of operational relations $$\alignedat2 &F(p,q)\dotseq f(x,y)&\quad &\varphi(p,q)\dotseq xyf(x,y)\\ &\psi(p,q)\dotseq xf(x,y)&\quad &\sigma(p,q)\dotseq yf(x,y)\\ &G(p,q)\dotseq xyf(\sqrt x,\sqrt y)& &\endalignedat $$ $$\align K(p,q)&=(pq)^{1/2}F(\sqrt p,\sqrt q)+pq^{1/2}\psi(\sqrt p,\sqrt q)+p^{1/2}q\sigma(\sqrt p,\sqrt q)+pq\psi(\sqrt p,\sqrt q)\\ &\dotseq \frac 1{4\pi}(xy)^{-3/2}G\(\frac 1{4x},\frac 1{4y}\)\endalign $$ and it is always possible to calculate one of the twelve functions, when the others are known. Hence by using the above system we can derive twelve rules. For example, we obtain:\roster \item"{1)}" the original of the function $G(p,q)$ from the original of the function $F(p,q)$ by replacing $x$ and $y$ by $\sqrt x$ and $\sqrt y$ respectively, and finally multiplying by $xy$. \item"{2)}" the original of the function $F(p,q)$ from the original of the function $\varphi(p,q)$ by multiplying by $\frac 1{xy}$. \item"{3)}" the image of the function $xyf(\sqrt x,\sqrt y)$ from the original of the function $K(p,q)$ by replacing $x$ and $y$ by $\frac 1{4p}$ and $\frac 1{4q}$ respectively, and finally multiplying by $\frac{\pi}{16}(pq)^{-3/2}$ \item"{4)}" the image of the function $xf(x,y)$ from the image of the function \break $\frac 1{4\pi}(xy)^{-3/2}G(\frac 1{4x},\frac 1{4y})$ by replacing $p$ and $q$ by $p^2$ and $q^2$ respectively, then subtracting $pqF(p,q)+pq^2\sigma(p,q)+(pq)^2\varphi(p,q)$, and finally multiplying by $\frac 1{p^2q}$.\endroster \example{Example 2.1} Let $f(x,y)=yJ_2(2\sqrt{xy})$, then $$\alignedat2 &F(p,q)=\frac p{(pq+1)^2}, &\quad &\sigma(p,q)=\frac{p(3pq+1)}{q(pq+1)^3}\quad \text{[5; p.137]}\\ &\psi(p,q)=\frac{2pq}{(pq+1)^3}, &\quad &\varphi(p,q)=\frac{6p^2q}{(pq+1)^4}\\ &G(p,q)=\frac{\sqrt{\pi}}{64}\cdot\frac{48pq-16\sqrt{pq}+1}{p^{5/2}q^3}e^{-\frac 1{2\sqrt{pq}}}&& \endalignedat $$ Using (2.1.1) and simplifying a bit, results in $$ \frac{p\sqrt q(6pq+4\sqrt{pq}+1)}{(\sqrt{pq}+1)^4}\dotseq \frac 1{\sqrt{\pi}}(4xy-16\sqrt{xy}+3)\sqrt y e^{-2\sqrt{xy}}.\qed $$ \endexample \example{Example 2.2} Let $f(x,y)=\frac 1{\sqrt{x^2+y^2}}$, then $$\align &F(p,q)=\frac{pq}{\sqrt{p^2+q^2}}\text{Ln}\frac{p+q+\sqrt{p^2+q^2}}{p+q-\sqrt{p^2+q^2}}\qquad \text{[5; p.147]}\\ &\psi(p,q)=\frac{p^2q}{\sqrt{(p^2+q^2)^3}}\text{Ln}\frac{p+q+\sqrt{p^2+q^2}}{p+q-\sqrt{p^2+q^2}}+\frac{q(q-p)}{p^2+q^2}\\ &\sigma(p,q)=\frac{pq^2}{\sqrt{(p^2+q^2)^3}}\text{Ln}\frac{p+q+\sqrt{p^2+q^2}}{p+q-\sqrt{p^2+q^2}}+\frac{p(p-q)}{p^2+q^2}\\ &\varphi(p,q)=\frac{2p^2q^2}{\sqrt{(p^2+q^2)^5}}\text{Ln}\frac{p+q+\sqrt{p^2+q^2}}{p+q-\sqrt{p^2+q^2}}+\frac{(p+q)(p^2-3pq+q^2)}{(p^2+q^2)^2}\\ &G(p,q)=\frac{\sqrt{\pi}}2\cdot\frac{(p+q+3\sqrt{pq})}{2\sqrt{pq}(\sqrt p+\sqrt q)^3}\qquad\qquad\qquad \text{[5; p.127]}\endalign $$ Using (2.1.1) and simplifying a bit, we obtain $$\align &\frac{pq(2p^2+2q^2+7pq)}{\sqrt{(p+q)^5}}\text{ Ln}\frac{\sqrt p+\sqrt q+\sqrt{p+q}}{\sqrt p+\sqrt q-\sqrt{p+q}}+\frac{pq(\sqrt p+\sqrt q)(p-3\sqrt{pq}+q)}{(p+q)^2}\\ &\qquad \dotseq \frac 1{2\sqrt{\pi}}\cdot\frac{x+y+3\sqrt{xy}}{\sqrt{xy}(\sqrt x+\sqrt y)^3}.\qed\endalign $$ \endexample \example{Example 2.3} Let $f(x,y)=\text{Ln }(x^2+y^2)$, then $$\align F(p,q)=&2\Big(\Gamma'(1)+\frac{\frac{\pi}2 pq-p^2\text{Ln } q-q^2 \text{Ln }p}{p^2+q^2}\Big)\qquad \text{[5; p. 147]}\\ \psi(p,q)=&2\Big(\frac{\Gamma'(1)}p+\frac {p(q^2-p^2) \text{Ln }q-q^2(p^{-1}q^2+3p) \text{ Ln } p+pq^2+p^{-1}q^4+\pi p^2q}{(p^2+q^2)^2}\Big)\\ \sigma(p,q)=&2\Big(\frac{\Gamma'(1)}q+\frac{q(p^2-q^2) \text{Ln } p-p^2(p^2q^{-1}+3q)\text{Ln } p+pq^2+p^4q^{-1} +\pi pq^2}{(p^2+q^2)^2}\Big)\\ \psi(p,q)=&2\Big(\frac{\Gamma'(1)}{pq}+\frac{pq(3p^2-6q^2-p^{-2}q^4) \text{ Ln } p+pq(3q^2-6p^2-p^4q^{-2}) \text{ Ln } q}{(p^2+q^2)^3}\\ & +\frac{-pq(p^2+q^2)+p^5q^{-1}+p^{-1}q^5 +4\pi(pq)^2}{(p^2+q^2)^3}\Big)\\ G(p,q)=&\frac{q(3-p^{-1}q))\text{Ln } p-p(3-pq^{-1}) \text{Ln } q+p-q+p^{-1}q^2-p^2q^{-1}}{(q-p)^3}+\frac{\Gamma(1)}{pq}. \endalign $$ Using (2.1.1) and manipulating a bit, we get $$\align &\frac{2\sqrt{pq}(2p^3+2q^3+3p^2q+3pq^2-pq)+\pi pq(3p^2+3q^2+14pq)}{(p+q)^3}\\ & -\frac{4q^2\sqrt{pq}(3p+q)\text{Ln }p+4p^2\sqrt{pq}(3p+q)\text{Ln }q}{(p+q)^3}+8\Gamma'(1)\sqrt{pq}\\ &\dotseq \frac 4{\pi}\(\frac{x^2(x-3y)\text{Ln }(4x)+y^2(3x-y)\text{Ln }(4x) +(x-y)(x^2+y^2)}{\sqrt{xy}(x-y)^3}+\frac{\Gamma'(1)}{\sqrt{xy}}\).\qed\endalign $$ \endexample \proclaim{Theorem 2.2} Let\newline (I)\quad $F(p,q)\dotseq f(x,y)$ \qquad (II)\quad $\varphi(p,q)\dotseq xyf(x,y)$\qquad (III)\quad $\psi(p,q)\dotseq xf(x,y)$ \newline (IV)\quad $\sigma(p,q)\dotseq yf(x,y)$\qquad (V)\quad $G(p,q)\dotseq xy^{-1/2}f(\sqrt x,\sqrt y)$ \newline Then $$\aligned K(p,q)=&p^{1/2}q^{-1}F(\sqrt p,\sqrt q)+pq^{-1}\psi(\sqrt p,\sqrt q)\\ &+p^{1/2}q^{-1/2}\sigma(\sqrt p,\sqrt q)+pq^{-1/2}\varphi(\sqrt p,\sqrt q)\\ \dotseq & \frac 2{\pi}x^{-3/2}y^{3/2}G\(\frac 1{4x},\frac 1{4y}\).\endaligned\tag 2.1.2 $$ \endproclaim \demo{Proof} We know that $$ p(1+s\sqrt p)e^{-s\sqrt p}\doteqdot \frac 1{4\pi} s^3x^{-5/2}e^{-\frac{s^2}{4x}}, q^{-1/2}(1+t\sqrt q)e^{-t\sqrt q}\doteqdot \frac 2{\sqrt{\pi}}y^{1/2}e^{-\frac{t^2}{4y}} $$ Multiplying together the above equations, results in $$ pq^{-1/2}(1+s\sqrt p)(1+t\sqrt q)e^{-s\sqrt p-t\sqrt q}\dotseq \frac 1{2\pi}x^{-5/2}y^{1/2}s^3e^{-\frac{s^2}{4x}-\frac{t^2}{4y}}. $$ If we multiply both sides by $f(s,t)$ and integrate with respect to $s$ and $t$ over the positive quarter plane, we obtain $$\align pq^{-1/2}&\int^{\infty}_0\int^{\infty}_0(1+s\sqrt p)(1+t\sqrt q)e^{-s\sqrt p-t\sqrt q}f(s,t)dsdt\\ &\dotseq \frac 1{2\pi}x^{-5/2}y^{1/2}\int^{\infty}_0\int^{\infty}_0e^{-\frac{s^2}{4x}-\frac{t^2}{4y}}s^3f(s,t)dsdt\endalign $$ which is $$\align &pq^{-1/2}\int^{\infty}_0\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}f(s,t)dsdt+p^{3/2}q^{-1/2}\int^{\infty}_0\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}sf(s,t)dsdt\\ &+p\int^{\infty}_0\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}tf(s,t)dsdt+p^{3/2}\int^{\infty}_0\int^{\infty}_0e^{-s\sqrt p-t\sqrt q}(st)f(s,t)dsdt\\ &\qquad\qquad\dotseq \frac 1{8\pi}x^{-5/2}y^{1/2}\int^{\infty}_0\int^{\infty}_0e^{-\frac u{4x}-\frac v{4y}}uv^{-1/2}f(\sqrt u,\sqrt v)dudv.\endalign $$ By using (V) on the right hand side and (I), (II), (III), (IV) on the left hand side, we get the desired result.\qed\enddemo Theorem 2.2 gives a system of six equations (I)--(V) and (2.1.2). For this system we will formulate twelve rules analogous to those mentioned for the system obtained from Theorem 2.1. For example, we can derive:\roster \item"{1)}" The original of the function $K(p,q)$ from the image of the function \break $xy^{-1/2}f(\sqrt x,\sqrt y)$ by replacing $p$ and $q$ by $\frac 1{4x}$ and $\frac 1{4y}$ respectively, and finally multiplying by $\frac 2{\pi}x^{-3/2}y^{3/2}$. \item"{2)}" The image of the function $\frac 2{\pi}x^{-3/2}y^{3/2}G(\frac 1{4x},\frac 1{4y})$ from the images of the functions $f(x,y), xf(x,y), yf(x,y)$ and $xyf(x,y)$ by multiplying by $pq^{-2}$, $p^2q^{-2}$, $pq^{-1}$ and $p^2q^{-1}$ respectively, then adding them, and finally replacing $p$ and $q$ by $\sqrt p$ and $\sqrt q$ respectively. \endroster \example{Example 2.4} Let $f(x,y)=\frac{(xy)^a}{\Gamma(2a+1)}J_{2a}(2\sqrt{xy})$ where Re$(2a+1)>0$, then $$\align &F(p,q)=\frac{pq}{(pq+1)^{2a+1}},\qquad \sigma(p,q)=\frac{(2a+1)p^2q}{(pq+1)^{2a+2}}\qquad \quad\text{[5; p.137]}\\ &\psi(p,q)=\frac{(2a+1)pq^2}{(pq+1)^{2a+2}},\qquad \psi(p,q)=(2a+1)\frac{((2a+1)pq-1)pq}{(pq+1)^{2a+3}}\\ &\qquad G(p,q)=\frac{\sqrt{\pi}}{4^{a+1}}\cdot\frac{4(a+1)\sqrt{pq}-1}{\Gamma(2a+1)p^{a+\frac 32}q^a} e^{-\frac 1{2\sqrt{pq}}}\qquad\qquad \text{[5; p.144]}\endalign $$ Using (2.1.2) and simplifying a bit, we obtain $$ \frac{p((a+1)^2pq+(2a+3)\sqrt{pq}+1)}{\sqrt q(\sqrt{pq}+1)^{2a+3}} \dotseq \frac{4^{a+1}}{\Gamma(2a+1)\sqrt{\pi}}(a+1-\sqrt{xy}) x^{a-\frac 12}y^{a+1}e^{-2\sqrt{xy}}. $$\endexample \proclaim{Theorem 2.3} Let\newline (I)\quad $F(p,q)\dotseq f(x,y)$\quad (II)\quad $\psi(p,q)\dotseq xf(x,y)$\quad (III)\quad $G(p,q)\dotseq xf(\sqrt x,\sqrt y)$ Then $$ (pq)^{1/2}F(\sqrt p,\sqrt q)+pq^{1/2}\psi(\sqrt p,\sqrt q)\dotseq \frac 1{2\pi}x^{-3/2}y^{-1/2}G\(\frac 1{4x},\frac 1{4y}\)\tag a $$ Moreover if we let\newline (IV)\qquad $H(p,q)\dotseq xy^{-1/2}f(\sqrt x,\sqrt y) $\newline then $$ p^{1/2}F(\sqrt p,\sqrt q)=p\psi(\sqrt p,\sqrt q)\dotseq \frac 1{\pi}x^{-3/2}y^{1/2}H\(\frac 1{4x},\frac 1{4y}\).\tag b $$\endproclaim \demo{Proof} We have the following operational relations $$\align &p(1+s\sqrt p)e^{-s\sqrt p}\doteqdot\frac 1{4\sqrt{\pi}}s^3x^{-5/2}e^{-\frac{s^2}{4x}}\tag 2.1.3\\ &qe^{-t\sqrt q}\doteqdot\frac 1{2\sqrt{\pi}}t y^{-3/2}e^{-\frac{t^2}{4y}}\tag 2.1.4\\ &q^{1/2}e^{-t\sqrt q}\doteqdot\frac 1{\sqrt{\pi}}y^{-1/2}e^{-\frac{t^2}{4y}}\tag 2.1.5\endalign $$ To get (a), we start with (2.1.3) and (2.1.4) and to get (b) we start with (2.1.3) and (2.1.5). The rest of the proof is similar to the proof of Theorem 2.2.\qed\enddemo We can derive two systems of operational relations in a similar way we did for the preceding theorem. For each system we can formulate eight rules analogous to those mentioned for the preceding systems. \example{Example 2.5} Let $f(x,y)=\frac 1{\sqrt{x^2+y^2}}$, then similar to what we performed in Example 2.2, from (a) we obtain $$\align &\frac{pq(2p+q)}{(p+q)^{3/2}}\text{Ln}\frac{\sqrt p+\sqrt q+\sqrt{p+q}}{\sqrt p+\sqrt q-\sqrt{p+q}}+\frac{pq(\sqrt q-\sqrt p)}{p+q}\\ &\qquad \dotseq \frac 1{2\sqrt{\pi}}\cdot\frac{\sqrt x+2\sqrt y}{\sqrt{xy}(\sqrt x+\sqrt y)^2}.\qed\endalign $$\endexample \example{Example 2.6} Let $f(x,y)=\text{Ln} (x^2+y^2)$, then as we did in Example 2.3, (a) results in $$\align &4\Gamma'(1) \sqrt{pq}+\frac{2pq\sqrt{pq}+2q^2\sqrt{pq}-2p^2\sqrt{pq}\ \text{Ln} q-2q\sqrt{pq}(2p+q)\text{Ln}p+\pi pq(3p+q)}{(p+q)^2}\\ &\qquad \dotseq \frac 2{\pi}\(\frac{x(x-2y)\text{Ln}(4x)+y^2/\ln(4y) +x(x-y)-\Gamma'(1)(x-y)^2}{\sqrt{xy}(x-y)^2}\)\qed\endalign $$\endexample \example{Example 2.7} Let $f(x,y)=y J_2(2\sqrt{xy})$, then similar to what we did in Example 2.1, from (a) we get $$ \frac{p\sqrt q(3\sqrt{pq}+1)}{(\sqrt{pq}+1)^3}\dotseq \frac 2{\sqrt{\pi}}(3-2\sqrt{xy})\sqrt y e^{-2\sqrt{xy}}\qed $$\endexample \example{Example 2.8} Let $f(x,y)=\frac{(xy)^a}{\Gamma(2a+1)}J_{2a}(2\sqrt{xy})$ where Re$(2a+1)>0$, then similar to what we did in Example 2.4, (b) results in $$ \frac{p\sqrt q+2(a+1)p^{3/2}q}{(\sqrt{pq}+1)^{2a+2}}\dotseq \frac{2^{2a+1}}{\sqrt{\pi}\Gamma(2a+1)} (a+1-\sqrt{xy})x^{a-\frac 12}y^a e^{-2\sqrt{xy}}.\qed $$\endexample We now state four more thoerems. The proofs of these theorems are omitted because they are similar to the proofs of the preceding theorems. Some examples are given after each theorem. \proclaim{Theorem 2.4} Let\newline (I)\quad\ \ $F(p,q)\dotseq f(x,y)$\qquad\qquad\quad (II)\quad $\eta(p,q)\dotseq x^2f(x,y)$\newline (III)\quad $\psi(p,q)\dotseq xf(x,y)$\qquad\qquad (IV)\quad $G(p,q)\dotseq x^{-1/2} f(\sqrt x, \sqrt y)$.\newline Then $$\align &3p^{-2} q^{1/2}F(\sqrt p,\sqrt q)+3p^{-\frac 32}q^{1/2}\psi(\sqrt p,\sqrt q)+p^{-1}q^{1/2}\eta(\sqrt p,\sqrt q)\tag a\\ &\qquad \dotseq \frac 8{\pi}x^{5/2}y^{-1/2}G\(\frac 1{4x}, \frac 1{4y}\).\endalign $$ Moreover if we assume\newline (V)\quad $H(p,q)\dotseq (xy)^{-1/2}f(\sqrt x,\sqrt y)$.\newline Then $$\align &3p^{-2}F(\sqrt p,\sqrt q)+3p^{-3/2}\psi(\sqrt p,\sqrt q)+p^{-1}\eta(\sqrt p,\sqrt q)\tag b\\ &\qquad \dotseq \frac{16}{\pi}x^{5/2}y^{1/2}H\(\frac 1{4x},\frac 1{4y}\)\qed\endalign $$\endproclaim \example{Example 2.9} Let $f(x,y)=\frac{(xy)^a}{\Gamma(2a+1)} J_{2a}(2\sqrt{xy})$ where Re$(2a+1)>0$, then (a) gives $$\align &\frac{q(4(a+1)(a+2)pq+3(2a+3)\sqrt{pq}+3)}{p^{3/2}(\sqrt{pq}+1)^{2a+3}}\\ &\qquad \dotseq \frac{4^{a+1}}{\sqrt{\pi}\Gamma(2a+1)}x^{a+2}y^{a-\frac 12}e^{-2\sqrt{xy}}.\qed\endalign $$ \endexample \example{Example 2.10} Let $$ f(x,y)=x^{2c_1}y^{2c_2}G^{i,j}_{u,v}\(c(xy)^2\bigg\vert \matrix a_1,\ a_2,\ \dots,\ a_u\\ b_1,\ b_2,\ \dots,\ b_v\endmatrix\) $$ where $(u+v)<2(i+j),|arg c|<\frac{\pi}2(i+j-\frac u2-\frac v2)$ and Re$(2c_k+b_{\ell}+1)>0, k=1,2;\ \ell=1,2,\dots, i$. Then $$\align F(p,q)=&\frac{2^{2c_1+2c_2}}{\pi p^{2c_1}q^{2c_2}} G^{i,j}_{u+4, v+4}\(\frac{16c}{(pq)^2}\bigg\vert\matrix &-c_1,\ -c_1+\frac 12,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ &(b_v)\phantom{MMMMMMMMMMMM}\endmatrix\)\\ \psi(p,q)=&\frac{2^{2c_1+2c_2+1}}{\pi p^{2c_1+1}q^{2c_2}} G^{i,j}_{u+4, v+4} \(\frac{16c}{(pq)^2}\bigg\vert\matrix &-c_1,\ -c_1-\frac 12,\ -c_2,\ -c_2 +\frac 12,(a_u)\\ &(b_v)\phantom{MMMMMMMMMMMMM}\endmatrix\)\\ \eta(p,q)=&\frac{2^{2c_1+2c_2+2}}{\pi p^{2c_1+2}q^{2c_2}}G^{i,j}_{u+4, v+4} \Big(\frac{16c}{(pq)^2}\bigg\vert\matrix &-c_1-\frac 12,\ -c_1-1,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ &(b_v)\phantom{MMMMMMMMMMMMMM}\endmatrix\Big)\\ H(p,q)=&p^{-2c_1+\frac 12}q^{-2c_2+\frac 12}G^{i,j}_{u+2,v+2} \(\frac c{pq}\bigg\vert\matrix &-2c_1+\frac 12,\ -2c_2+\frac 12,\ (a_u)\\ &(b_v)\phantom{MMMMMMMMM}\endmatrix\)\endalign $$ where $(a_u)=a_1,a_2,\dots, a_u$ and $(b_v)$ is defined similarly. Hence (b) gives $$\align &\frac{(3)2^{2c_1+2c_2}}{\pi p^{c_1+2}q^{c_2}}G^{i,j}_{u+4,v+4}\(\frac{16c}{pq}\bigg\vert\matrix -c_1,\ -c_1+\frac 12,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMMMMM}\endmatrix\)\\ &+\frac{(3)2^{2c_1+2c_2+1}}{\pi p^{c_1+2}q^{c_2}} G^{i,j}_{u+4, v+4}\(\frac{16c}{pq}\bigg\vert\matrix -c_1,\ -c_1-\frac 12,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMMMMM}\endmatrix\)\\ &+\frac{2^{2c_1+2c_2+2}}{\pi p^{c_1+2}q^{c_2}}G^{i,j}_{u+4, v+4}\(\frac{16c}{pq}\bigg\vert\matrix -c_1-\frac 12,\ -c_1-1,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMMMMMM}\endmatrix\)\\ &\qquad\dotseq \frac{2^{4c_1+4c_2+2}}{\pi}x^{2c_1+2}y^{2c_2}G^{i,j}_{u+2, v+2}\(16c(xy)\bigg\vert\matrix -2c_1+\frac 12,\ -2c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMM}\endmatrix\).\endalign $$ Recall $$ F(ap,bq)\dotseq f\(\frac xa,\frac yb\)\tag 2.1.6 $$ where $F(p,q)$ is the Laplace Carson transform of $f(x,y)$. If we use (2.1.6) with $a=b=4$, we get $$\align &p^{-c_1-2}q^{-c_2}\[3G^{i,j}_{u+4,v+4}\(\frac c{pq}\bigg\vert\matrix -c_1,\ -c_1+\frac 12,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMMMM}\endmatrix\)\right.\\ &+6G^{i,j}_{u+4,v+4}\(\frac c{pq}\bigg\vert\matrix -c_1,\ -c_1-\frac 12,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMMM}\endmatrix\)\\ &\left. +4G^{i,j}_{u+4,v+4}\(\frac c{pq}\bigg\vert\matrix -c_1-\frac 12,\ -c_1-1,\ -c_2,\ -c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMMMMMMM}\endmatrix\)\]\\ &\qquad \dotseq 4x^{2c_1+2}y^{2c_2}G^{i,j}_{u+2,v+2}\(c(xy) \bigg\vert\matrix -2c_1+\frac 12,\ -2c_2+\frac 12,\ (a_u)\\ (b_v)\phantom{MMMMMMMM}\endmatrix\).\qed\endalign $$ \endexample \proclaim{Theorem 2.5} Let\newline (I)\qquad $F(p,q)\dotseq f(x,y)$\qquad\qquad (II)\quad $\eta(p,q)\dotseq x^2f(x,y)$\newline (III)\quad $\psi(p,q)\dotseq xf(x,y)$\qquad\qquad (IV)\quad $G(p,q)\dotseq x^2f(\sqrt x,\sqrt y)$.\newline Then $$\align &3(pq)^{1/2}F(\sqrt p,\sqrt q)+3pq^{1/2}\psi(\sqrt p,\sqrt q)+p^{3/2}q^{1/2}\eta (\sqrt p,\sqrt q)\tag a\\ &\qquad \dotseq \frac 1{4\pi}x^{-5/2}y^{-1/2}G\(\frac 1{4x},\frac 1{4y}\).\endalign $$ Moreover if we assume\newline (V)\quad $H(p,q)\dotseq x^2y^{-1/2}f(\sqrt x,\sqrt y)$.\newline Then $$\align &3p^{1/2}F(\sqrt p, \sqrt q)+3p\psi(\sqrt p,\sqrt q)+p^{3/2}\eta(\sqrt p,\sqrt q)\tag b\\ &\qquad \dotseq \frac 1{2\pi} x^{-5/2}y^{1/2}H\(\frac 1{4x},\frac 1{4y}\)\qed\endalign $$ \endproclaim \example{Example 2.11} Let $f(x,y)=\frac 1{\sqrt{x^2+y^2}}$, then from (a), we get $$\align &\frac{8p^2+8pq+3q^2)}{(p+q)^{5/2}}\text{Ln}\frac{\sqrt p+\sqrt q+\sqrt{p+q}}{\sqrt p+\sqrt q-\sqrt{p+q}}+\frac{pq(5p\sqrt q-q\sqrt p-4p^{3/2}+2q^{3/2}}{(p+q)^2}\\ &\qquad \dotseq 2\sqrt{\pi}\ \frac{x^{5/2}(6\sqrt x+16\sqrt y+9)}{(\sqrt x+\sqrt y)^3}.\qed\endalign $$ \endexample \example{Example 2.12} Let $f(x,y)=yJ_2(2\sqrt{xy})$, then (a) gives $$ \frac{3p\sqrt q(5pq+4\sqrt{pq}+1)}{(\sqrt{pq}+1)^4}\dotseq \frac 2{\sqrt{\pi}}(4xy-18\sqrt{xy}+15)\sqrt y\ e^{-2\sqrt{xy}}.\qed $$\endexample \example{Example 2.13} Let $f(x,y)=\frac{(xy)^a}{\Gamma(2a+1)} J_{2a}(2\sqrt{xy})$ where Re$(2a+1)>0$, then (b) gives $$\align &\frac{p\sqrt q(4(a+1)(a+2)pq+3(2a+3)\sqrt{pq}+3)}{(\sqrt{pq}+1)^{2a+3}}\\ &\qquad \dotseq \frac{4^{a+\frac 12}}{\sqrt{\pi}}(2xy-(4a+7)\sqrt{xy}+2(a+1)(a+2))x^{a-\frac 12}y^ae^{-2\sqrt{xy}}.\qed\endalign $$ \endexample \proclaim{Theorem 2.6} Let\newline (I)\quad $F(p,q)\dotseq f(x,y)$\qquad (II)\quad $\eta(p,q)\dotseq x^2f(x,y)$\newline (III)\quad $\psi(p,q)\dotseq xf(x,y)$\qquad (IV)\quad $\xi(p,q)\dotseq x^2yf(x,y)$\newline (V)\quad $\sigma(p,q)\dotseq yf(x,y)$\qquad (VI)\quad $\varphi(p,q)\dotseq xyf(x,y)$\newline (VII)\quad $G(p,q)\dotseq x^{-1/2} yf(\sqrt x, \sqrt y)$.\newline Then $$\align &3p^{-2}q^{1/2}F(\sqrt p,\sqrt q)+2p^{-3/2}q^{1/2}\psi(\sqrt p,\sqrt q)+p^{-1}q^{1/2}\eta(\sqrt p,\sqrt q)\tag a\\ &+ 3p^{-2}q\sigma(\sqrt p,\sqrt q)+3p^{-3/2}q\psi(\sqrt p,\sqrt q)+p^{-1}q\xi(\sqrt p,\sqrt q)\\ &\qquad \dotseq \frac 4{\pi}x^{5/2}y^{-3/2}G\(\frac 1{4x},\frac 1{4y}\).\endalign $$ Moreover if we define\newline (VII)\quad $H(p,q)\dotseq (x,y)^{-1/2}f(\sqrt x,\sqrt y)$\newline then $$\align &3p^{-2}q^{-1}F(\sqrt p,\sqrt q)+3p^{-3/2}q^{-1}\psi(\sqrt p,\sqrt q)+p^{-1}q^{-1}\eta(\sqrt p,\sqrt q)\tag b\\ &+3p^{-2}q^{-1/2}\sigma(\sqrt p,\sqrt q)+3p^{-3/2}q^{-1/2}\varphi(\sqrt p,\sqrt q)+p^{-1}q^{-1/2}\xi(\sqrt p,\sqrt q)\\ &\qquad \dotseq \frac{32}{\pi}x^{5/2}y^{3/2}H\(\frac 1{4x},\frac 1{4y}\).\qed\endalign $$ \endproclaim \example{Example 2.14} Let $f(x,y)\frac{(xy)^a}{\Gamma(2a+1)} J_{2a}(2\sqrt{xy})$ where Re$(2a+1)>0$, then from (a) we obtain $$\align &\frac{q(8(a+1)^2(a+2)(pq)^{3/2}+(8a^2+20a+17)pq+6(a+2)\sqrt{pq}+3)}{p^{3/2}(\sqrt{pq}+1)^{2a+4}}\\ &\qquad\dotseq \frac{4^{a+\frac 32}}{\sqrt{\pi}\Gamma(2a+1)}\ (a+1-\sqrt{xy})x^{a+2}y^{a-\frac 12}e^{-2\sqrt{xy}}.\qed\endalign $$ \endexample \subhead{3.\ Second Order Parabolic Equation}\endsubhead Consider $$\gathered u_{xx}-u_y=f(x,y)\quad 0-1\\ \lim_{x\rightarrow\infty}u(x,y) \text{ bounded.} \endgathered\tag 3.1 $$ In order to apply the operational calculus and find the image of $u_{xx}$, we need the Laplace transform of $u_x(0,y)$. Let us assume $$ U(p,q)\dotseq u(x,y),\quad F(p,q)\dotseq f(x,y),\quad H(q)\doteqdot u_x(0,y). $$ Then the transformed equation (3.1) takes the form $$ p^2\[U(p,q)-\frac{\Gamma(a+1)}{q^a}\]-pH(q)-q\[U(p,q)-\frac p{p^2+1}\]= F(p,q) $$ from which we derive $$ U(p,q)=\frac{F(p,q)}{p^2-q}+\frac{pH(q)}{p^2-q}+\frac{\Gamma(a+1)p^2}{q^a(p^2-q)}-\frac{pq}{(p^2+1)(p^2-q)}. $$ First we assume $f(x,y)=0$. Then we have $$ U(p,q)=\frac{pH(q)}{p^2-q}+\frac{\Gamma(a+1)p^2}{q^a(p^2-q)}-\frac{pq}{(p^2+1)(p^2-q)}. $$ To find $u(x,y)$, we take the inverse of each term with respect to $p$ only, i.e., $$\align &\frac p{p^2-q}\doteqdot\frac{\sin h \sqrt q\ x}{\sqrt q},\quad \frac{p^2}{p^2-q}\doteqdot\cosh\sqrt q\ x\\ &\frac p{(p^2+1)(p^2-q)}\doteqdot\frac{\sin h\sqrt q\ x-\sqrt q\ \sin x}{\sqrt q(q+1)}\qquad\text{[6; p.197]}.\endalign $$ Then we obtain $$\align U(p,q)\doteqdot&\frac{H(q)}{\sqrt q}\sin h\sqrt q\ x+\frac{\Gamma(a+1)}{q^a}\cosh(\sqrt q\ x)-\frac{\sqrt q}{q+1}(\sin h\sqrt q\ x-\sqrt q\sin x)\\ U(p,q)\doteqdot& \frac{H(q)}{2\sqrt q}(e^{\sqrt q\ x}-e^{-\sqrt q\ x})+\frac{\Gamma(a+1)}{2q^a}(e^{\sqrt q\ x}+e^{-\sqrt q\ x})\\ &-\frac{\sqrt q}{2(q+1)}(e^{\sqrt q\ x}-e^{-\sqrt q\ x})+\frac q{q+1}\sin x\endalign $$ $$ u(x,q)=\frac{q^{a-\frac 12}(q+1)H(q)+\Gamma(a+1)(q+1)-q^{a+1}}{2q^a(q+1)}\ e^{\sqrt q\ x}\tag 3.3 $$ $$ +\frac{-q^{a-\frac 12}(q+1)H(q)+\Gamma(a+1)(q+1)+q^{a+1}}{2q^a(q+1)} e^{-\sqrt q\ x}+\frac q{q+1}\sin x. $$ Since the limit of $u(x,q)$ is bounded as $x$ approaches infinity, we must have $$ q^{a-\frac 12}(q+1)H(q)+\Gamma(a+1)(q+1)-q^{a+1}=0 $$ which implies $$ H(q)=\frac{q^{a+1}-\Gamma(a+1)(q+1)}{q^{a-\frac 12}(q+1)}. $$ Replacing $H(q)$ in (3.3) results in $$ u(x,q)=\frac{\Gamma(a+1)}{q^a}e^{-\sqrt q\ x}+\frac q{q+1}\sin x. $$ Inverting the above function with respect to $q$ and using $$ q^{-a}e^{-\sqrt q\ x}\doteqdot \frac{2^{a+\frac 12}}{\sqrt{\pi}}\ y^ae^{-\frac{x^2}{8y}}D_{-2a-1}\(\frac x{\sqrt{2y}}\),\qquad \text{[6; p. 246]} $$ we obtain $$ u(x,y)=\frac{2^{a+\frac 12}\Gamma(a+1)}{\sqrt{\pi}}\ y^ae^{-\frac{x^2}{8y}}D_{-2a-1}\(\frac x{\sqrt{2y}}\)+e^{-y}\sin x.\tag 3.4 $$ \Refs \ref\no 1\by R.S. Dahiya \paper Computation of two-dimensional Laplace transforms \jour Rendiconti di Matimatica, Rome\vol 8 \finalinfo ser.vi (1975), 805--813\endref \ref\no 2\by R.S. Dahiya \paper Calculation of two-dimensional Laplace transform pairs-1 \jour Simon Stevin, A quarterly journal of Pure and Applied Mathematics\vol 56 \issue 1-2\yr 1982\pages 97--107\endref \ref\no 3\by R.S. Dahiya \paper Systems of two-dimensional Laplace transforms and their applications \jour Simon Stevin, A quarterly journal of Pure and Applied Mathematics\vol 59 \issue 4\yr 1985\pages 373--384\endref \ref\no 4\by R.S. Dahiya and J. Debnath \paper Theorems on multidimensional Laplace transform for solution of boundary value problems \jour International Journal of computers and Mathematics with Applications \vol 28 (12)\yr 1989\pages 1033--1056\endref \ref\no 5\by V.A. Ditkin and A.P. Prudnikov \paper Operational calculus in two variables and its applications \finalinfo (English translation from Russian), Pergaman Press (1962)\endref \ref\no 6\by Roberts, G.E. and Kaufman, H. \book Table of Laplace transforms\publ Philadelphia: W.B. Saunders Co. \yr 1966\endref \ref\no 7\by N.A. Sastri \paper On simultaneous operational calculus. Jour. of Indian Math. Soc. \issue 1\yr 1934\pages 235\endref \ref\no 8\by D. Voelker and G. Doetsch \book Die zweidimensional Laplace transformation\publ Birkhauser Verlag \publaddr Basel\yr 1950\endref \endRefs \enddocument