16th Conference on Apllied Mathematics, Univ. of Central Oklahoma, Electronic Journal of Differential Equations, Conf. 07, 2001, pp. 25–36. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

SYSTEMS OF MULTI-DIMENSIONAL LAPLACE

TRANSFORMS AND A HEAT EQUATION

Ali Babakhani & R. S. Dahiya

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.

1. Introduction

The two-dimensional Laplace transform of function f(x, y) is defined by Ditkin and Prudnikov [5] as follows:

F (p, q) = pq ∫_{0}^{∞} ∫_{0}^{∞} e^{−px−qy}f(x, y)dx dy

and symbolically is denoted by F (p, q) {.. \atop =} .. f(x, y) where the symbol {.. \atop =} .. 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.

2. Systems of two-dimensional Laplace transforms.

2.1. The Image of G(\frac{1}{4x}, \frac{1}{4y}).

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, ±5 and j = ±1, ±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 = ±1, 3 is given by Dahiya [3].

Mathematics Subject Classifications: 44A30.

Key words and phrases: Two-dimensional Laplace transforms.

\begin{matrix}c \\ ◯ \end{matrix}2001 Southwest Texas State University.

Published July 20, 2001.

Theorem 2.1. Let

(I) F (p, q) {.. \atop =} .. f(x, y)

(II) ψ(p, q) = −pq \frac{∂}{∂p} (\frac{F (p,q)}{pq}) {.. \atop =} .. xf(x, y)

(III) σ(p, q) = −pq \frac{∂}{∂q} (\frac{F (p,q)}{pq}) {.. \atop =} .. yf(x, y)

(IV) φ(p, q) = pq \frac{∂^{2}}{∂p∂q} (\frac{F (p,q)}{pq}) {.. \atop =} .. xyf(x, y)

(V) G(p, q) {.. \atop =} .. xyf (\sqrt{x}, \sqrt{y}).

Then

\begin{aligned}(pq)^{1/2} & F (\sqrt{p}, \sqrt{q}) + pq^{1/2}ψ (\sqrt{p}, \sqrt{q}) + p^{1/2}qσ (\sqrt{p}, \sqrt{q}) + pqφ (\sqrt{p}, \sqrt{q}) \\ & {.. \atop =} .. \frac{1}{4π} (xy)^{−3/2}G (\frac{1}{4x}, \frac{1}{4y} ) \end{aligned} (2.1.1)

Proof. We start from two operational relations

p(1 + s\sqrt{p})e^{−s\sqrt{p}} ≑ \frac{1}{4\sqrt{π}} s^{3}x^{−5/2}e^{−s^{2}/4x}

q(1 + t\sqrt{q})e^{−t\sqrt{q}} ≑ \frac{1}{4\sqrt{π}} t^{3}y^{−5/2}e^{−t^{2}/4y}

We multiply together the above equations to get

pq(1 + \sqrt{p})(1 + t\sqrt{q})e^{−s\sqrt{p}−t\sqrt{q}} {.. \atop =} .. \frac{1}{16π} (xy)^{−5/2}(st)^{3}e^{− \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.

pq ∫_{0}^{∞} ∫_{0}^{∞}(1 + s\sqrt{p})(1 + t\sqrt{q})e^{−s\sqrt{p}−t\sqrt{q}}f(s, t)dsdt

{.. \atop =} .. \frac{1}{16π} (xy)^{−5/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{− \frac{s^{2}}{4x} − \frac{t^{2}}{4y}} (st)^{3}f(s, t)dsdt.

We make the change of variables s = \sqrt{u} and t = \sqrt{v} on the right hand side to obtain

pq ∫_{0}^{∞} ∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}f(s, t)dsdt + p^{3/2}q ∫_{0}^{∞} ∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}sf(s, t)dsdt

+ pq^{3/2} ∫_{0}^{∞}∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}tf(s, t)dsdt + (pq)^{3/2} ∫_{0}^{∞}∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}(st)f(s, t)dsdt

{.. \atop =} .. \frac{1}{64π} (xy)^{−5/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{− \frac{u}{4x} − \frac{v}{4y}} (uv)f(\sqrt{u}, \sqrt{v})dudv.

Finally by using (V) on the right hand side and (I), (II), (III), (IV) on the left hand side we get

(pq)^{1/2}F (\sqrt{p}, \sqrt{q}) + pq^{1/2}ψ(\sqrt{p}, \sqrt{q}) + p^{1/2}qσ(\sqrt{p}, \sqrt{q}) + pqφ(\sqrt{p}, \sqrt{q})

{.. \atop =} .. \frac{1}{4π} (xy)^{−3/2}G (\frac{1}{4x}, \frac{1}{4y} ) . □

Thus we have the following system of operational relations

F (p, q) {.. \atop =} .. f(x, y)  φ(p, q) {.. \atop =} .. xyf(x, y)
ψ(p, q) {.. \atop =} .. xf(x, y)  σ(p, q) {.. \atop =} .. yf(x, y)

G(p, q) {.. \atop =} .. xyf(\sqrt{x}, \sqrt{y})

K(p, q) = (pq)^{1/2}F (\sqrt{p}, \sqrt{q}) + pq^{1/2}ψ(\sqrt{p}, \sqrt{q}) + p^{1/2}qσ(\sqrt{p}, \sqrt{q}) + pqψ(\sqrt{p}, \sqrt{q})

{.. \atop =} .. \frac{1}{4π} (xy)^{−3/2}G (\frac{1}{4x}, \frac{1}{4y} )

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:

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.

2) the original of the function F (p, q) from the original of the function φ(p, q) by multiplying by \frac{1}{xy}.

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{π}{16}(pq)^{−3/2}

4) the image of the function xf(x, y) from the image of the function \frac{1}{4π} (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}σ(p, q) + (pq)^{2}φ(p, q), and finally multiplying by \frac{1}{p^{2}q}.

Example 2.1. Let f(x, y) = yJ_{2}(2\sqrt{xy}), then

F (p, q) = \frac{p}{(pq + 1)^{2}} ,  σ(p, q) = \frac{p(3pq + 1)}{q(pq + 1)^{3}} [5; p.137]
ψ(p, q) = \frac{2pq}{(pq + 1)^{3}} ,  φ(p, q) = \frac{6p^{2}q}{(pq + 1)^{4}}

G(p, q) = \frac{\sqrt{π}}{64} · \frac{48pq − 16\sqrt{pq} + 1}{p^{5/2}q^{3}}e^{− \frac{1}{2\sqrt{pq}}}

Using (2.1.1) and simplifying a bit, results in

\frac{p\sqrt{q}(6pq + 4\sqrt{pq} + 1)}{(\sqrt{pq} + 1)^{4}} {.. \atop =} .. \frac{1}{\sqrt{π}} (4xy − 16\sqrt{xy} + 3)\sqrt{y}e^{−2\sqrt{xy}}. □

28 Ali Babakhani & R. S. Dahiya Example 2.2. Let f(x, y) = \frac{1}{\sqrt{x^{2}+y^{2}}} , then

F (p, q) = \frac{pq}{\sqrt{p^{2} + q^{2}}} Ln\frac{p + q + \sqrt{p^{2} + q^{2}}}{p + q − \sqrt{p^{2} + q^{2}}} [5; p.147]

ψ(p, q) = \frac{p^{2}q}{\sqrt{(p^{2} + q^{2})^{3}}} Ln\frac{p + q + \sqrt{p^{2} + q^{2}}}{p + q − \sqrt{p^{2} + q^{2}}} + \frac{q(q − p)}{p^{2} + q^{2}}

σ(p, q) = \frac{pq^{2}}{\sqrt{(p^{2} + q^{2})^{3}}} Ln\frac{p + q + \sqrt{p^{2} + q^{2}}}{p + q − \sqrt{p^{2} + q^{2}}} + \frac{p(p − q)}{p^{2} + q^{2}}

φ(p, q) = \frac{2p^{2}q^{2}}{\sqrt{(p^{2} + q^{2})^{5}}} 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{π}}{2} · \frac{(p + q + 3\sqrt{pq})}{2\sqrt{pq}(\sqrt{p} + \sqrt{q})^{3}} [5; p.127]

Using (2.1.1) and simplifying a bit, we obtain

\frac{pq(2p^{2} + 2q^{2} + 7pq)}{\sqrt{(p + q)^{5}}} 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}}

{.. \atop =} .. \frac{1}{2\sqrt{π}} · \frac{x + y + 3\sqrt{xy}}{\sqrt{xy}(\sqrt{x} + \sqrt{y})^{3}} . □

Example 2.3. Let f(x, y) = Ln (x^{2} + y^{2}), then

F (p, q) =2(Γ′(1) + \frac{\frac{π}{2} pq − p^{2}Ln q − q^{2}Ln p}{p^{2} + q^{2}})  [5; p. 147]
ψ(p, q) =2(\frac{Γ′(1)}{p} + \frac{p(q^{2} − p^{2})Ln q − q^{2}(p^{−1}q^{2} + 3p) Ln p + pq^{2} + p^{−1}q^{4} + πp^{2}q}{(p^{2} + q^{2})^{2}} )
σ(p, q) =2(\frac{Γ′(1)}{q} + \frac{q(p^{2} − q^{2})Ln p − p^{2}(p^{2}q^{−1} + 3q)Ln p + pq^{2} + p^{4}q^{−1} + πpq^{2}}{(p^{2} + q^{2})^{2}} )
ψ(p, q) =2(\frac{Γ′(1)}{pq} + \frac{pq(3p^{2} − 6q^{2} − p^{−2}q^{4}) Ln p + pq(3q^{2} − 6p^{2} − p^{4}q^{−2}) Ln q}{(p^{2} + q^{2})^{3}}

+ \frac{−pq(p^{2} + q^{2}) + p^{5}q^{−1} + p^{−1}q^{5} + 4π(pq)^{2}}{(p^{2} + q^{2})^{3}} )

G(p, q) =\frac{q(3 − p^{−1}q))Ln p − p(3 − pq^{−1})Ln q + p − q + p^{−1}q^{2} − p^{2}q^{−1}}{(q − p)^{3}} + \frac{Γ(1)}{pq}.

Using (2.1.1) and manipulating a bit, we get

\frac{2\sqrt{pq}(2p^{3} + 2q^{3} + 3p^{2}q + 3pq^{2} − pq) + πpq(3p^{2} + 3q^{2} + 14pq)}{(p + q)^{3}}

− \frac{4q^{2}\sqrt{pq}(3p + q)Ln p + 4p^{2}\sqrt{pq}(3p + q)Ln q}{(p + q)^{3}} + 8Γ′(1)\sqrt{pq}

{.. \atop =} .. \frac{4}{π} (\frac{x^{2}(x − 3y)Ln (4x) + y^{2}(3x − y)Ln (4x) + (x − y)(x^{2} + y^{2})}{\sqrt{xy}(x − y)^{3}} + \frac{Γ′(1)}{\sqrt{xy}}) . □

Theorem 2.2. Let

(I) F (p, q) {.. \atop =} .. f(x, y)  (II) φ(p, q) {.. \atop =} .. xyf(x, y)  (III) ψ(p, q) {.. \atop =} .. xf(x, y)

(IV) σ(p, q) {.. \atop =} .. yf(x, y) (V) G(p, q) {.. \atop =} .. xy^{−1/2}f(\sqrt{x}, \sqrt{y})

Then

\begin{matrix}K(p, q) = & p^{1/2}q^{−1}F (\sqrt{p}, \sqrt{q}) + pq^{−1}ψ(\sqrt{p}, \sqrt{q}) \\ + p^{1/2}q^{−1/2}σ(\sqrt{p}, \sqrt{q}) + pq^{−1/2}φ(\sqrt{p}, \sqrt{q}) \\ {.. \atop =} .. & \frac{2}{π} x^{−3/2}y^{3/2}G (\frac{1}{4x}, \frac{1}{4y} ) . \end{matrix}  (2.1.2)

Proof. We know that

p(1 + s\sqrt{p})e^{−s\sqrt{p}} ≑ \frac{1}{4π} s^{3}x^{−5/2}e^{− \frac{s^{2}}{4x}} , q^{−1/2}(1 + t\sqrt{q})e^{−t\sqrt{q}} ≑ \frac{2}{\sqrt{π}} 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}} {.. \atop =} .. \frac{1}{2π} x^{−5/2}y^{1/2}s^{3}e^{− \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

pq^{−1/2} ∫_{0}^{∞} ∫_{0}^{∞}(1 + s\sqrt{p})(1 + t\sqrt{q})e^{−s\sqrt{p}−t\sqrt{q}}f(s, t)dsdt

{.. \atop =} .. \frac{1}{2π} x^{−5/2}y^{1/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{− \frac{s^{2}}{4x} − \frac{t^{2}}{4y}} s^{3}f(s, t)dsdt

which is

pq^{−1/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}f(s, t)dsdt + p^{3/2}q^{−1/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}sf(s, t)dsdt

+ p ∫_{0}^{∞} ∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}tf(s, t)dsdt + p^{3/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{−s\sqrt{p}−t\sqrt{q}}(st)f(s, t)dsdt

{.. \atop =} .. \frac{1}{8π} x^{−5/2}y^{1/2} ∫_{0}^{∞} ∫_{0}^{∞} e^{− \frac{u}{4x} − \frac{v}{4y}} uv^{−1/2}f(\sqrt{u}, \sqrt{v})dudv.

By using (V) on the right hand side and (I), (II), (III), (IV) on the left hand side, we get the desired result. □

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:

1) The original of the function K(p, q) from the image of the function 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}{π} x^{−3/2}y^{3/2}.

2) The image of the function \frac{2}{π} 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^{2}q^{−2}, pq^{−1} and p^{2}q^{−1} respectively, then adding them, and finally replacing p and q by \sqrt{p} and \sqrt{q} respectively.

Example 2.4. Let f(x, y) = \frac{(xy)^{a}}{Γ(2a+1)}J_{2a}(2\sqrt{xy}) where Re(2a + 1) > 0, then

F (p, q) = \frac{pq}{(pq + 1)^{2a+1}} ,  σ(p, q) = \frac{(2a + 1)p^{2}q}{(pq + 1)^{2a+2}}  [5; p.137]

ψ(p, q) = \frac{(2a + 1)pq^{2}}{(pq + 1)^{2a+2}} , ψ(p, q) = (2a + 1)\frac{((2a + 1)pq − 1)pq}{(pq + 1)^{2a+3}}

G(p, q) = \frac{\sqrt{π}}{4^{a+1}} · \frac{4(a + 1)\sqrt{pq} − 1}{Γ(2a + 1)p^{a+ \frac{3}{2}} q^{a}} e^{− \frac{1}{2\sqrt{pq}}} [5; p.144]

Using (2.1.2) and simplifying a bit, we obtain

\frac{p((a + 1)^{2}pq + (2a + 3)\sqrt{pq} + 1)}{\sqrt{q}(\sqrt{pq} + 1)^{2a+3}} {.. \atop =} .. \frac{4^{a+1}}{Γ(2a + 1)\sqrt{π}} (a + 1 − \sqrt{xy})x^{a− \frac{1}{2}} y^{a+1}e^{−2\sqrt{xy}}.

Theorem 2.3. Let

(I) F (p, q) {.. \atop =} .. f(x, y) (II) ψ(p, q) {.. \atop =} .. xf(x, y) (III) G(p, q) {.. \atop =} .. xf(\sqrt{x}, \sqrt{y})

Then

(pq)^{1/2}F (\sqrt{p}, \sqrt{q}) + pq^{1/2}ψ(\sqrt{p}, \sqrt{q}) {.. \atop =} .. \frac{1}{2π} x^{−3/2}y^{−1/2}G (\frac{1}{4x}, \frac{1}{4y} )  (a)

Moreover if we let

(IV) H(p, q) {.. \atop =} .. xy^{−1/2}f(\sqrt{x}, \sqrt{y})

then

p^{1/2}F (\sqrt{p}, \sqrt{q}) = pψ(\sqrt{p}, \sqrt{q}) {.. \atop =} .. \frac{1}{π} x^{−3/2}y^{1/2}H (\frac{1}{4x}, \frac{1}{4y} ) .  (b)

Proof. We have the following operational relations

p(1 + s\sqrt{p})e^{−s\sqrt{p}} ≑ \frac{1}{4\sqrt{π}} s^{3}x^{−5/2}e^{− \frac{s^{2}}{4x}}  (2.1.3)
qe^{−t\sqrt{q}} ≑ \frac{1}{2\sqrt{π}} ty^{−3/2}e^{− \frac{t^{2}}{4y}}  (2.1.4)
q^{1/2}e^{−t\sqrt{q}} ≑ \frac{1}{\sqrt{π}} y^{−1/2}e^{− \frac{t^{2}}{4y}}  (2.1.5)

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. □

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 2.5. Let f(x, y) = \frac{1}{\sqrt{x^{2}+y^{2}}} , then similar to what we performed in Ex-

ample 2.2, from (a) we obtain

\frac{pq(2p + q)}{(p + q)^{3/2}}Ln\frac{\sqrt{p} + \sqrt{q} + \sqrt{p + q}}{\sqrt{p} + \sqrt{q} − \sqrt{p + q}} + \frac{pq(\sqrt{q} − \sqrt{p})}{p + q}

{.. \atop =} .. \frac{1}{2\sqrt{π}} · \frac{\sqrt{x} + 2\sqrt{y}}{\sqrt{xy}(\sqrt{x} + \sqrt{y})^{2}} . □

Example 2.6. Let f(x, y) = Ln(x^{2} + y^{2}), then as we did in Example 2.3, (a) results in

4Γ′(1)\sqrt{pq} + \frac{2pq\sqrt{pq} + 2q^{2}\sqrt{pq} − 2p^{2}\sqrt{pq} Lnq − 2q\sqrt{pq}(2p + q)Lnp + πpq(3p + q)}{(p + q)^{2}}

{.. \atop =} .. \frac{2}{π} (\frac{x(x − 2y)Ln(4x) + y^{2}/ ln(4y) + x(x − y) − Γ′(1)(x − y)^{2}}{\sqrt{xy}(x − y)^{2}} ) □

Example 2.7. Let f(x, y) = yJ_{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}} {.. \atop =} .. \frac{2}{\sqrt{π}} (3 − 2\sqrt{xy})\sqrt{y}e^{−2\sqrt{xy}} □

Example 2.8. Let f(x, y) = \frac{(xy)^{a}}{Γ(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}} {.. \atop =} .. \frac{2^{2a+1}}{\sqrt{π}Γ(2a + 1)}(a + 1 − \sqrt{xy})x^{a− \frac{1}{2}} y^{a}e^{−2\sqrt{xy}}. □

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.

Theorem 2.4. Let

(I) F (p, q) {.. \atop =} .. f(x, y) (II) η(p, q) {.. \atop =} .. x^{2}f(x, y)

(III) ψ(p, q) {.. \atop =} .. xf(x, y) (IV) G(p, q) {.. \atop =} .. x^{−1/2}f(\sqrt{x}, \sqrt{y}).

Then

3p^{−2}q^{1/2}F (\sqrt{p}, \sqrt{q}) + 3p^{− \frac{3}{2}} q^{1/2}ψ(\sqrt{p}, \sqrt{q}) + p^{−1}q^{1/2}η(\sqrt{p}, \sqrt{q})  (a)

{.. \atop =} .. \frac{8}{π} x^{5/2}y^{−1/2}G (\frac{1}{4x}, \frac{1}{4y} ) .

Moreover if we assume

(V) H(p, q) {.. \atop =} .. (xy)^{−1/2}f(\sqrt{x}, \sqrt{y}).

Then

3p^{−2}F (\sqrt{p}, \sqrt{q}) + 3p^{−3/2}ψ(\sqrt{p}, \sqrt{q}) + p^{−1}η(\sqrt{p}, \sqrt{q})  (b)

{.. \atop =} .. \frac{16}{π}x^{5/2}y^{1/2}H (\frac{1}{4x}, \frac{1}{4y} ) □

Example 2.9. Let f(x, y) = \frac{(xy)^{a}}{Γ(2a+1)}J_{2a}(2\sqrt{xy}) where Re(2a + 1) > 0, then (a) gives

\frac{q(4(a + 1)(a + 2)pq + 3(2a + 3)\sqrt{pq} + 3)}{p^{3/2}(\sqrt{pq} + 1)^{2a+3}}

{.. \atop =} .. \frac{4^{a+1}}{\sqrt{π}Γ(2a + 1)}x^{a+2}y^{a− \frac{1}{2}} e^{−2\sqrt{xy}}. □

Example 2.10. Let

f(x, y) = x^{2c_{1}}y^{2c_{2}}G_{u,v}^{i,j} (c(xy)^{2}| \begin{matrix}a_{1}, a_{2}, …, a_{u} \\ b_{1}, b_{2}, …, b_{v} \end{matrix} )

where (u + v) < 2(i + j), |argc| < \frac{π}{2} (i + j − \frac{u}{2} − \frac{v}{2} ) and Re(2c_{k} + b_{ℓ} + 1) > 0, k = 1, 2; ℓ = 1, 2, …, i. Then

F (p, q) =\frac{2^{2c_{1}+2c_{2}}}{πp^{2c_{1}}q^{2c_{2}}} G_{u+4,v+4}^{i,j} (\frac{16c}{(pq)^{2}} | \begin{matrix}−c_{1}, −c_{1} + \frac{1}{2}, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )
ψ(p, q) =\frac{2^{2c_{1}+2c_{2}+1}}{πp^{2c_{1}+1}q^{2c_{2}}} G_{u+4,v+4}^{i,j} (\frac{16c}{(pq)^{2}} | \begin{matrix}−c_{1}, −c_{1} − \frac{1}{2}, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )
η(p, q) =\frac{2^{2c_{1}+2c_{2}+2}}{πp^{2c_{1}+2}q^{2c_{2}}} G_{u+4,v+4}^{i,j}(\frac{16c}{(pq)^{2}} | \begin{matrix}−c_{1} − \frac{1}{2}, −c_{1} − 1, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )
H(p, q) =p^{−2c_{1}+ \frac{1}{2}} q^{−2c_{2}+ \frac{1}{2}} G_{u+2,v+2}^{i,j} (\frac{c}{pq} | \begin{matrix}−2c_{1} + \frac{1}{2}, −2c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )

where (a_{u}) = a_{1}, a_{2}, …, a_{u} and (b_{v}) is defined similarly. Hence (b) gives

\frac{(3)2^{2c_{1}+2c_{2}}}{πp^{c_{1}+2}q^{c_{2}}} G_{u+4,v+4}^{i,j} (\frac{16c}{pq}| \begin{matrix}−c_{1}, −c_{1} + \frac{1}{2}, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )

+ \frac{(3)2^{2c_{1}+2c_{2}+1}}{πp^{c_{1}+2}q^{c_{2}}} G_{u+4,v+4}^{i,j} (\frac{16c}{pq}| \begin{matrix}−c_{1}, −c_{1} − \frac{1}{2}, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )

+ \frac{2^{2c_{1}+2c_{2}+2}}{πp^{c_{1}+2}q^{c_{2}}} G_{u+4,v+4}^{i,j} (\frac{16c}{pq}| \begin{matrix}−c_{1} − \frac{1}{2}, −c_{1} − 1, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )

{.. \atop =} .. \frac{2^{4c_{1}+4c_{2}+2}}{π} x^{2c_{1}+2}y^{2c_{2}}G_{u+2,v+2}^{i,j} (16c(xy)| \begin{matrix}−2c_{1} + \frac{1}{2}, −2c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} ) .

Recall

F (ap, bq) {.. \atop =} .. f (\frac{x}{a}, \frac{y}{b} )  (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

p^{−c_{1}−2}q^{−c_{2}} [3G_{u+4,v+4}^{i,j} (\frac{c}{pq} | \begin{matrix}−c_{1}, −c_{1} + \frac{1}{2}, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )

+ 6G_{u+4,v+4}^{i,j} (\frac{c}{pq} | \begin{matrix}−c_{1}, −c_{1} − \frac{1}{2}, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )

+4G_{u+4,v+4}^{i,j} (\frac{c}{pq} | \begin{matrix}−c_{1} − \frac{1}{2}, −c_{1} − 1, −c_{2}, −c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} )]

{.. \atop =} .. 4x^{2c_{1}+2}y^{2c_{2}}G_{u+2,v+2}^{i,j} (c(xy)| \begin{matrix}−2c_{1} + \frac{1}{2}, −2c_{2} + \frac{1}{2}, (a_{u}) \\ (b_{v}) \end{matrix} ) . □

Systems of multi-dimensional Laplace transforms 33

Theorem 2.5. Let

(I) F (p, q) {.. \atop =} .. f(x, y) (II) η(p, q) {.. \atop =} .. x^{2}f(x, y)

(III) ψ(p, q) {.. \atop =} .. xf(x, y) (IV) G(p, q) {.. \atop =} .. x^{2}f(\sqrt{x}, \sqrt{y}).

Then

3(pq)^{1/2}F (\sqrt{p}, \sqrt{q}) + 3pq^{1/2}ψ(\sqrt{p}, \sqrt{q}) + p^{3/2}q^{1/2}η(\sqrt{p}, \sqrt{q})  (a)

{.. \atop =} .. \frac{1}{4π} x^{−5/2}y^{−1/2}G (\frac{1}{4x}, \frac{1}{4y} ) .

Moreover if we assume

(V) H(p, q) {.. \atop =} .. x^{2}y^{−1/2}f(\sqrt{x}, \sqrt{y}).

Then

3p^{1/2}F (\sqrt{p}, \sqrt{q}) + 3pψ(\sqrt{p}, \sqrt{q}) + p^{3/2}η(\sqrt{p}, \sqrt{q})  (b)

{.. \atop =} .. \frac{1}{2π} x^{−5/2}y^{1/2}H (\frac{1}{4x}, \frac{1}{4y} ) □

Example 2.11. Let f(x, y) = \frac{1}{\sqrt{x^{2}+y^{2}}} , then from (a), we get

\frac{8p^{2} + 8pq + 3q^{2})}{(p + q)^{5/2}}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}}

{.. \atop =} .. 2\sqrt{π} \frac{x^{5/2}(6\sqrt{x} + 16\sqrt{y} + 9)}{(\sqrt{x} + \sqrt{y})^{3}}. □

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}} {.. \atop =} .. \frac{2}{\sqrt{π}} (4xy − 18\sqrt{xy} + 15)\sqrt{y} e^{−2\sqrt{xy}}. □

Example 2.13. Let f(x, y) = \frac{(xy)^{a}}{Γ(2a+1)}J_{2a}(2\sqrt{xy}) where Re(2a + 1) > 0, then (b) gives

\frac{p\sqrt{q}(4(a + 1)(a + 2)pq + 3(2a + 3)\sqrt{pq} + 3)}{(\sqrt{pq} + 1)^{2a+3}}

{.. \atop =} .. \frac{4^{a+ \frac{1}{2}}}{\sqrt{π}} (2xy − (4a + 7)\sqrt{xy} + 2(a + 1)(a + 2))x^{a− \frac{1}{2}} y^{a}e^{−2\sqrt{xy}}. □

Theorem 2.6. Let

(I) F (p, q) {.. \atop =} .. f(x, y) (II) η(p, q) {.. \atop =} .. x^{2}f(x, y)

(III) ψ(p, q) {.. \atop =} .. xf(x, y) (IV) ξ(p, q) {.. \atop =} .. x^{2}yf(x, y)

(V) σ(p, q) {.. \atop =} .. yf(x, y) (VI) φ(p, q) {.. \atop =} .. xyf(x, y)

(VII) G(p, q) {.. \atop =} .. x^{−1/2}yf(\sqrt{x}, \sqrt{y}).

Then

3p^{−2}q^{1/2}F (\sqrt{p}, \sqrt{q}) + 2p^{−3/2}q^{1/2}ψ(\sqrt{p}, \sqrt{q}) + p^{−1}q^{1/2}η(\sqrt{p}, \sqrt{q})  (a)
+ 3p^{−2}qσ(\sqrt{p}, \sqrt{q}) + 3p^{−3/2}qψ(\sqrt{p}, \sqrt{q}) + p^{−1}qξ(\sqrt{p}, \sqrt{q})

{.. \atop =} .. \frac{4}{π} x^{5/2}y^{−3/2}G (\frac{1}{4x}, \frac{1}{4y} ) .

Moreover if we define

(VII) H(p, q) {.. \atop =} .. (x, y)^{−1/2}f(\sqrt{x}, \sqrt{y})

then

3p^{−2}q^{−1}F (\sqrt{p}, \sqrt{q}) + 3p^{−3/2}q^{−1}ψ(\sqrt{p}, \sqrt{q}) + p^{−1}q^{−1}η(\sqrt{p}, \sqrt{q})  (b)
+ 3p^{−2}q^{−1/2}σ(\sqrt{p}, \sqrt{q}) + 3p^{−3/2}q^{−1/2}φ(\sqrt{p}, \sqrt{q}) + p^{−1}q^{−1/2}ξ(\sqrt{p}, \sqrt{q})

{.. \atop =} .. \frac{32}{π}x^{5/2}y^{3/2}H (\frac{1}{4x}, \frac{1}{4y} ) . □

Example 2.14. Let f(x, y)\frac{(xy)^{a}}{Γ(2a+1)}J_{2a}(2\sqrt{xy}) where Re(2a+1) > 0, then from (a) we obtain

\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}}

{.. \atop =} .. \frac{4^{a+ \frac{3}{2}}}{\sqrt{π}Γ(2a + 1)} (a + 1 − \sqrt{xy})x^{a+2}y^{a− \frac{1}{2}} e^{−2\sqrt{xy}}. □

3. Second Order Parabolic Equation.

Consider

u_{xx} − u_{y} = f(x, y) 0 < x < ∞, 0 < y < ∞
u(x, 0) = sin x
u(0, y) = y^{a} Re(a) > −1
lim_{x→∞} u(x, y) bounded.

(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) {.. \atop =} .. u(x, y), F (p, q) {.. \atop =} .. f(x, y), H(q) ≑ u_{x}(0, y).

Then the transformed equation (3.1) takes the form

p^{2} [U(p, q) − \frac{Γ(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{Γ(a + 1)p^{2}}{q^{a}(p^{2} − q)} − \frac{pq}{(p^{2} + 1)(p^{2} − q)}.

Systems of multi-dimensional Laplace transforms 35

First we assume f(x, y) = 0. Then we have

U(p, q) = \frac{pH(q)}{p^{2} − q} + \frac{Γ(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.,

\frac{p}{p^{2} − q} ≑ \frac{sin h\sqrt{q} x}{\sqrt{q}}, \frac{p^{2}}{p^{2} − q} ≑ cosh \sqrt{q} x

\frac{p}{(p^{2} + 1)(p^{2} − q)} ≑ \frac{sin h\sqrt{q} x − \sqrt{q} sin x}{\sqrt{q}(q + 1)} [6; p.197].

Then we obtain

U(p, q) ≑\frac{H(q)}{\sqrt{q}} sin h\sqrt{q} x + \frac{Γ(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) ≑\frac{H(q)}{2\sqrt{q}}(e^{\sqrt{q} x} − e^{−\sqrt{q} x}) + \frac{Γ(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

u(x, q) = \frac{q^{a− \frac{1}{2}} (q + 1)H(q) + Γ(a + 1)(q + 1) − q^{a+1}}{2q^{a}(q + 1)} e^{\sqrt{q} x}  (3.3)

+\frac{−q^{a− \frac{1}{2}} (q + 1)H(q) + Γ(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{1}{2}} (q + 1)H(q) + Γ(a + 1)(q + 1) − q^{a+1} = 0

which implies

H(q) = \frac{q^{a+1} − Γ(a + 1)(q + 1)}{q^{a− \frac{1}{2}} (q + 1)}.

Replacing H(q) in (3.3) results in

u(x, q) = \frac{Γ(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} ≑ \frac{2^{a+ \frac{1}{2}}}{\sqrt{π}} y^{a}e^{− \frac{x^{2}}{8y}} D_{−2a−1} (\frac{x}{\sqrt{2y}} ) , [6; p. 246]

we obtain

u(x, y) = \frac{2^{a+ \frac{1}{2}} Γ(a + 1)}{\sqrt{π}} y^{a}e^{− \frac{x^{2}}{8y}} D_{−2a−1} (\frac{x}{\sqrt{2y}} ) + e^{−y} sin x.  (3.4)

36 Ali Babakhani & R. S. Dahiya

References

1. R.S. Dahiya, Computation of two-dimensional Laplace transforms, Rendiconti di Matimatica, Rome 8, ser.vi (1975), 805–813.

2. R.S. Dahiya, Calculation of two-dimensional Laplace transform pairs-1, Simon Stevin, A quarterly journal of Pure and Applied Mathematics 56 (1982), no. 1-2, 97–107.

3. R.S. Dahiya, Systems of two-dimensional Laplace transforms and their applications, Simon Stevin, A quarterly journal of Pure and Applied Mathematics 59 (1985), no. 4, 373–384.

4. R.S. Dahiya and J. Debnath, Theorems on multidimensional Laplace transform for solution of boundary value problems, International Journal of computers and Mathematics with Applications 28 (12) (1989), 1033–1056.

5. V.A. Ditkin and A.P. Prudnikov, Operational calculus in two variables and its applications, (English translation from Russian), Pergaman Press (1962).

6. Roberts, G.E. and Kaufman, H., Table of Laplace transforms, Philadelphia: W.B. Saunders Co., 1966.

7. N.A. Sastri, On simultaneous operational calculus. Jour. of Indian Math. Soc. (1934), no. 1, 235.

8. D. Voelker and G. Doetsch, Die zweidimensional Laplace transformation, Birkhauser Verlag, Basel, 1950.

Ali Babakhani Department of Mathematics, College of Engineering, University of Tehran, Tehran, Iran

R. S. Dahiya Department of Mathematics, Iowa State University, Ames, IA 50011, USA

E-mail address: dahiya@math.iastate.edu