15th Annual Conference of Applied Mathematics, Univ. of Central Oklahoma, Electronic Journal of Differential Equations, Conference 02, 1999, pp. 61–74. ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)

THEOREMS ON n-DIMENSIONAL LAPLACE

TRANSFORMS AND THEIR APPLICATIONS

R. S. Dahiya & Jafar Saberi-Nadjafi

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.

1. Introduction and Notation

The generalization of the well-known Laplace transform L[f(t); s] = ∫_{0}^{∞} e^{−st}f(t)dt to n-dimensional is given by

L_{n}[f(t̄); s̄] = ∫_{0}^{∞} ∫_{0}^{∞}⋯ ∫_{0}^{∞} exp(−s̄ · t̄)f(t̄)P_{n}(dt̄)

where t̄= (t_{1}, t_{2}, …, t_{n}), s̄ = (s_{1}, s_{2}, …, s_{n}), s̄ · t̄= ∑_{i=1}^{n} s_{i}t_{i} and P_{n}(dt̄) = ∏_{k=1}^{n} dt_{k}.

In addition to the notations introduced above, we will use the following throughout this article. Let \overline{t^{v}} = (t_{1}^{v}, t_{2}^{v}, …, t_{n}^{v}) 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

(i) p_{1}(\overline{t^{v}}) = t_{1}^{v} + t_{2}^{v} + ⋯ + t_{n}^{v} (ii) p_{n}(\overline{t^{v}}) = t_{1}^{v} · t_{2}^{v} … t_{n}^{v}.

2. Main Results

Theorem 2.1. Suppose that f(x) and f(x^{2}) are functions of class Ω. Let (i) L{f(x); s} = ϕ(s), (ii) L\{x^{− \frac{3}{2}} ϕ (\frac{1}{x}); s\} = ξ(s), (iii) L\{x^{− \frac{1}{2}} ξ (\frac{1}{x^{2}} ); s\} = ζ(s), and (iv) L{f(x^{2}); s} = G(s),

where x^{− \frac{3}{2}} ϕ (\frac{1}{x}) and x^{− \frac{1}{2}} ξ (\frac{1}{x^{2}} ) are also functions of class Ω, x^{− \frac{3}{2}} exp (−sx − \frac{u}{x}) f(u)

and u^{− \frac{1}{2}} x^{− \frac{3}{2}} exp (−sx − \frac{2u^{\frac{1}{2}}}{x} ) f(u) belong to L_{1} [(0, ∞) × (0, ∞)].

1991 Mathematics Subject Classifications: 44A30.

Key words and phrases: multi-dimensional Laplace transforms.

\begin{matrix}c \\ ◯ \end{matrix}1999 Southwest Texas State University and University of North Texas.

Published November 24, 1999.

Then

L_{n} \{\frac{1}{p_{n} (\overline{x^{\frac{1}{2}}} )}G[2p_{1}(\overline{x^{−1}})]; \overline{s}\} = \frac{π^{\frac{n−2}{2}}}{2} · \frac{p_{1} (\overline{s^{\frac{1}{2}}} )}{p_{n} (\overline{s^{\frac{1}{2}}} )}ζ [(P_{1} (\overline{s^{\frac{1}{2}}} ))^{2}] ,  (2.1)

where Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > d, a constant, provided the integrals involved exist for n = 2, 3, …, N. The existence conditions for two-dimensions are given in Ditkin and Prudnikov [11; p.4] and similar conditions hold for N-dimensions, we refer to Brychkov et al. [2; Ch.2].

Proof. Using (i) and (ii), we obtain

ξ(s) = ∫_{0}^{∞} [∫_{0}^{∞} x^{− \frac{3}{2}} exp (−sx − \frac{u}{x}) f(u)du] dx.  (2.2)

Next we wish to interchange the order of the integral on the right side of (2.2). The integrand x^{− \frac{3}{2}} exp (−sx − \frac{u}{x}) f(u) belongs to L_{1}[(0, ∞) × (0, ∞)], that, by Fubini’s Theorem, interchanging the order of the integral on the right of (2.2) is permissible. Therefore,

ξ(s) = ∫_{0}^{∞} f(u) [∫_{0}^{∞} x^{− \frac{3}{2}} exp [−sx − \frac{u}{x}] dx] du  (2.3)

We then use a well-known result in Robert and Kaufman [15] on the right side of (2.3) to obtain

ξ(s) = π^{\frac{1}{2}} ∫_{0}^{∞} u^{− \frac{1}{2}} exp (−2u^{\frac{1}{2}} s^{\frac{1}{2}} ) f(u)du.  (2.4)

Using (2.4) and (iii), it follows that

ζ(s) = π^{\frac{1}{2}} ∫_{0}^{∞} [∫_{0}^{∞} x^{− \frac{1}{2}} u^{− \frac{1}{2}} exp (−sx − \frac{2u\frac{1}{2}}{x} ) f(u)du] dx., where Re s > λ_{1}.

(2.5)

Since x^{− \frac{1}{2}} u^{− \frac{1}{2}} exp (−sx − \frac{2u^{\frac{1}{2}}}{x} ) belongs to L_{1}[(0, ∞) × (0, ∞)]; therefore, accord-

ing to Fubini’s Theorem, (2.5) can be rewritten as

ζ(s) = π^{\frac{1}{2}} ∫_{0}^{∞} u^{− \frac{1}{2}} f(u) [∫_{0}^{∞} x^{− \frac{1}{2}} exp (−sx − \frac{(2^{\frac{3}{2}} u^{\frac{1}{4}} )^{2}}{4x} ) dx] du, where Re s > λ_{1}.

From the tables of Roberts and Kaufman [15], we obtain

s^{\frac{1}{2}} ζ(s) = π ∫_{0}^{∞} u^{− \frac{1}{2}} f(u) exp (−2^{\frac{3}{2}} u^{\frac{1}{4}} s^{\frac{1}{2}} ) du.  (2.6)

Next, we substitute u = v^{2} in (2.6) to obtain

s^{\frac{1}{2}} ζ(s) = 2π ∫_{0}^{∞} exp (−2^{\frac{3}{2}} s^{\frac{1}{2}} v^{\frac{1}{2}} ) f(v^{2})dv.  (2.7)

n-dimensional Laplace transforms 63

Replacing s by [p_{1} (\overline{s^{\frac{1}{2}}} )]^{2}, multiplying both sides of (2.7) by p_{n} (\overline{s^{\frac{1}{2}}} ), we obtain

p_{1} (\overline{s^{\frac{1}{2}}} ) p_{n} (\overline{s^{\frac{1}{2}}} ) ζ [p_{1} (\overline{s^{\frac{1}{2}}} )] = 2π ∫_{0}^{∞} p_{n} (\overline{s^{\frac{1}{2}}} ) exp [−2^{\frac{3}{2}} p_{1} (\overline{s^{\frac{1}{2}}} ) v^{\frac{1}{2}} ] f(v^{2})dv

(2.8)

Now we use the operational relation given in Ditkin and Prudnikov [11]

si^{\frac{1}{2}} exp (−asi^{\frac{1}{2}} ) ≑ (πx_{i})^{− \frac{1}{2}} exp (−\frac{a^{2}}{4x_{i}} ) for i = 1, 2, …, n  (2.9)

Equation (2.8) reads as follows

p_{n} (\overline{s^{\frac{1}{2}}} ) p_{1} (\overline{s^{\frac{1}{2}}} ) ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{2}] =_{n}^{n} \frac{2}{π^{\frac{n−2}{2}} p_{n} (\overline{x^{\frac{1}{2}}} )} ∫_{0}^{∞} exp (−2vp_{1} (\overline{x^{−1}})) f(v^{2})dv

(2.10)

Applying (iv) in (2.10), we arrive at

p_{n} (\overline{s^{\frac{1}{2}}} ) p_{1} (\overline{s^{\frac{1}{2}}} ) ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{2}] =_{n}^{n} \frac{2}{π^{\frac{n−2}{2}} p_{n} (\overline{x^{\frac{1}{2}}} )} G [2p_{1} (\overline{x^{−1}})] .

Therefore,

L_{n} \{\frac{1}{p_{n} (\overline{x^{\frac{1}{2}}} )} G [2p_{1} (\overline{x^{\frac{1}{2}}} )]; \overline{s}\} = \frac{π^{\frac{n−2}{2}}}{2} · \frac{p_{1} (\overline{s^{\frac{1}{2}}} )}{p_{n} (\overline{s^{\frac{1}{2}}} )}ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{2}] ,

where n = 2, 3, …, N.

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

Example 2.1. Assume that f(x) = x^{\frac{τ−1}{4}} . Then

ϕ(s) = \frac{Γ (\frac{τ+3}{4})}{s^{\frac{τ+3}{4}}}, Re s > 0, Re v > −3;
ξ(s) = \frac{Γ (\frac{τ+3}{4}) Γ (\frac{τ+1}{4})}{s^{\frac{τ+1}{4}}}, Re τ > −1, Re s > 0, and
ζ(s) = \frac{Γ (\frac{τ+3}{4}) Γ (\frac{τ+1}{4}) Γ (\frac{τ}{2} + 1)}{s^{\frac{τ}{2} +1}}, Re s > 0, Re τ > −1.
G(s) = \frac{Γ (\frac{τ+1}{2})}{s^{\frac{τ+1}{2}}}, Re τ > −1.

Therefore,

L_{n} \{\frac{1}{p_{n} (\overline{x^{\frac{1}{2}}} ) [p_{1} (\overline{x^{−1}})]^{\frac{τ+2}{2}}} ; \overline{s}\} = π^{\frac{n−1}{2}} Γ (\frac{τ}{2} + \frac{3}{2}) · \frac{1}{p_{n} (\overline{s^{\frac{1}{2}}} ) [p_{1} (\overline{s^{\frac{1}{2}}} )]^{τ+2}}

(2.11)

where Re τ > −1, Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > 0, and n = 2, 3, …, N.

Example 2.2. Suppose that f(x) = I_{0}(ax^{\frac{1}{2}} ). Then

ϕ(s) = \frac{1}{s} exp (\frac{a^{2}}{4s}) , Re s > 0,
ξ(s) = \frac{π^{\frac{1}{2}}}{(s − \frac{a^{2}}{4} )^{\frac{1}{2}}} , Re s > Re \frac{a^{2}}{4} ,

ζ(s) = \frac{4π}{a^{\frac{5}{2}}} \{\frac{4π}{[Γ (\frac{1}{4})]^{2}}_{1}F_{2} [ \begin{matrix}\frac{3}{4} ; \\ \frac{1}{2}, \frac{5}{4}; \end{matrix} \frac{s^{2}}{a^{2}} ] − \frac{[Γ (\frac{1}{4})]^{2} s}{6π}_{1}F_{2} [ \begin{matrix}\frac{3}{4} ; \\ \frac{3}{2}, \frac{7}{4}; \end{matrix} \frac{s^{2}}{a^{2}} ]\} .

G(s) = \frac{1}{(s^{2} − a^{2})^{\frac{1}{2}}} , Re s > |Re a|.

So that

\begin{aligned}& L_{n} \{\frac{1}{p_{n}(x^{\frac{1}{2}} )} · \frac{1}{[4p_{1}^{2} (\overline{x^{−1}}) − a^{2}]^{\frac{1}{2}}} ; \overline{s}\} \\ = \frac{2π^{\frac{n}{2}} p_{1} (\overline{s^{\frac{1}{2}}} )}{a^{\frac{5}{2}} p_{n} (\overline{s^{\frac{1}{2}}} )} & \{\frac{4π}{(Γ (\frac{1}{4})]^{2}}_{1}F_{2} [ \begin{matrix}\frac{3}{4} ; \\ \frac{1}{2}, \frac{5}{4}; \end{matrix} \frac{p_{1}^{2} (\overline{s^{\frac{1}{2}}} )}{a^{2}} ] − \frac{[Γ (\frac{1}{4})]^{2} p_{1}^{2} (\overline{s^{\frac{1}{2}}} )}{6π} \\ & _{1}F_{2} [ \begin{matrix}\frac{5}{4} ; \\ \frac{3}{2}, \frac{7}{4}; \end{matrix} \frac{p_{1}^{2} (\overline{s^{\frac{1}{2}}} )}{a^{2}} ]; \overline{s}\} , \end{aligned}  (2.12)

where Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > |Re a|.

Example 2.3. Consider f(x) = _{p}F_{q} [ \begin{matrix}(a)_{p}; \\ (b)_{q}; \end{matrix} cx]. Then

ϕ(s) = \frac{1}{s}_{p+1}F_{q} [ \begin{matrix}(a)_{p}; \\ (b)_{q}; \end{matrix} cx],

where p ≤ q, Re s > |Re c|.

ξ(s) = \frac{π^{\frac{1}{2}}}{s^{\frac{1}{2}}} _{p+2}F_{q} [ \begin{matrix}(a)_{p}, 1, \frac{1}{2}; \\ (b)_{q} ; \end{matrix} \frac{c}{s}] ,

where p ≤ q − 1, Re τ > 0 if p + 1 < q, Re s > Re c if p + 1 = q,

ζ(s) = \frac{π}{2s^{\frac{3}{2}}} _{p+4}F_{q} [ \begin{matrix}(a)_{p}, 1, \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \\ (b)_{q} ; \end{matrix} \frac{4c}{s^{2}}] ,

where p ≤ q − 3; Re s > 0 if p ≤ q − 4; and Re(s + 2c^{\frac{1}{2}} cos πr) > 0 (r = 0, 1) if p = q − 3. Hence

L_{n} \{\frac{1}{p_{n} (\overline{x^{\frac{1}{2}}} ) [p_{1} (\overline{x^{−1}})]}_{p+2}F_{q} [ \begin{matrix}(a)_{p}, 1, \frac{1}{2}; \\ (b)_{q} ; \end{matrix} \frac{c}{p_{1}^{2} (\overline{x^{−1}})}]; s\}

= \frac{π^{\frac{n}{2}}}{2p_{n} (\overline{s^{\frac{1}{2}}} ) p_{1}^{2} (\overline{s^{\frac{1}{2}}} )}_{p+4}F_{q} [ \begin{matrix}(a)_{p}, 1, \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \\ (b)_{q}  ; \end{matrix} \frac{4c}{p_{1}^{4} (\overline{s^{\frac{1}{2}}} )}] ,  (2.13)

where p ≤ q − 3, Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > 0 if p ≤ q − 4; and Re [p_{1} (\overline{s^{\frac{1}{2}}} ) + 2c^{\frac{1}{2}} cos πr] > 0(r = 0, 1) if p = q − 3.

Example 2.4. Assume that f(x) = x^{\frac{1}{2}} J_{0}(ax^{\frac{1}{2}} ). Then

ϕ(s) = \frac{π^{\frac{1}{2}}}{2s^{\frac{3}{2}}} _{1}F_{1} [ \begin{matrix}\frac{3}{2}; \\ 1; \end{matrix} − \frac{a^{2}}{4s}] , Re s > 0,

ξ(s) = \frac{π^{\frac{1}{2}}}{2} _{2}F_{1} [ \begin{matrix}\frac{3}{2}, 1; \\ 1 ; \end{matrix} − \frac{a^{2}}{4s}] , Re s > −Re − \frac{a^{2}}{4} ,

ζ(s) = \frac{1}{π^{\frac{1}{2}} s^{\frac{1}{2}}} G_{4,2}^{1,4} (\frac{a^{2}}{s^{2}} | \begin{matrix}\frac{1}{4}, \frac{3}{4}, 0, −\frac{1}{2} \\ 0, 0 \end{matrix} ) ,

where Re s > 0, | arg a| < 2π.

G(s) = \frac{π^{\frac{1}{2}}}{2} _{2}F_{1} [ \begin{matrix}\frac{3}{2}, 1; \\ 1 ; \end{matrix} − \frac{a^{2}}{4s}] , Re s > Re − \frac{a^{2}}{4} .

Hence

L_{n} \{\frac{p_{1} (\overline{x^{−1}})}{p_{n} (\overline{x^{\frac{1}{2}}} ) [4p_{1}^{2} (\overline{x^{−1}}) + a^{2}]^{\frac{3}{2}}} , \overline{s}\}

=\frac{π^{\frac{n−3}{2}} p_{1} (\overline{s^{\frac{1}{2}}} )}{p_{n} (\overline{s^{\frac{1}{2}}} )}G_{4,2}^{1,4} (\frac{a^{2}}{p_{1}^{4} (\overline{s^{\frac{1}{2}}} )}| \begin{matrix}\frac{1}{2}, \frac{3}{4}, 0, −\frac{1}{2} \\ 0, 0 \end{matrix} ) ,  (2.14)

where Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > 0, | arg a| < 2π and G_{4,2}^{1,4} is a Meijer’s G-function.

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

L_{2} \{\frac{(xy)^{\frac{τ}{2}}}{(x + y)^{\frac{τ+1}{2}}} ; s_{1}, s_{2}\} = π^{\frac{1}{2}} Γ (\frac{τ}{2} + 1) · \frac{1}{(s_{1}s_{2})^{\frac{1}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{τ+1}}  (2.15)

where Re τ > −1, Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0.

\begin{matrix}L_{2} \{\frac{(xy)^{\frac{1}{2}}}{[4(x + y)^{2} − (axy)^{2}]^{\frac{1}{2}}} ; s_{1}, s_{2}\} \\ = \frac{2π (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}{a^{\frac{5}{2}} (s_{1}s_{2})^{\frac{1}{2}}} \{\frac{4π}{[Γ (\frac{1}{4})]^{2}}_{1}F_{2} [ \begin{matrix}\frac{3}{4} ; \\ \frac{1}{2}, \frac{5}{4}; \end{matrix} \frac{(s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}}{a^{2}} ] \\ −\frac{[Γ (\frac{1}{4})]^{2} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}}{6π}_{1}F_{2} [ \begin{matrix}\frac{5}{4} ; \\ \frac{3}{2}, \frac{7}{4}; \end{matrix} \frac{(s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}}{a^{2}} ]\} \end{matrix}  (2.16)

where Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0, | arg a| < 2π.

\begin{aligned}& L_{2} \{\frac{(xy)^{\frac{1}{2}}}{x + y}_{p+2}F_{q} [ \begin{matrix}(a)_{p}, 1, \frac{1}{2}; \\ (b)_{q}  ; \end{matrix} \frac{c(xy)^{2}}{(x + y)^{2}} ]; s_{1}, s_{2}\} \\ = & \frac{π}{2(s_{1}s_{2})^{\frac{1}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}}_{p+4}F_{q} [ \begin{matrix}(a)_{p}, \frac{1}{2}, \frac{3}{4}, \frac{5}{4}; \\ (b)_{q}  ; \end{matrix} \frac{4c}{(s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{4}} ] , \end{aligned}  (2.17)

where p ≤ q − 3, Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0 if p ≤ q − 4; and Re [(s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ) + 2c^{\frac{1}{2}} cos πr]

> 0(r = 0, 1) if p = q − 3.

L_{2} \{\frac{(xy)^{\frac{3}{2}} (x + y)}{[4(x + y)^{2} + (axy)^{2}]^{\frac{3}{2}}} ; s_{1}, s_{2}\} = \frac{s1^{\frac{1}{2}} + s2^{\frac{1}{2}}}{(πs_{1}s_{2})^{\frac{1}{2}}} G_{4,2}^{1,4} (\frac{a^{2}}{(s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{4}} | \begin{matrix}\frac{1}{2}, \frac{3}{2}, 0, −\frac{1}{2} \\ 0, 0 \end{matrix} ) ,

(2.18)

where R [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > |Re a| and G_{4,2}^{1,4} is a Meijer’s G-function.

Remark 2.1. If we let τ = 0 in Relation (2.15), and then using Relation (1) in [24]. We deduced that

L_{2} \{(x + y)^{\frac{1}{2}} ; s_{1}, s_{2}\} = \frac{π^{\frac{1}{2}} (s_{1} + s_{2} + s1^{\frac{1}{2}} s2^{\frac{1}{2}} )}{2(s_{1}s_{2})^{\frac{3}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}  (2.19)

Theorem 2.2. Assume that f belongs to class Ω and ϕ be the one-dimensional Laplace transformation of f. Let

(i) L \{x^{− \frac{1}{2}} ϕ (\frac{1}{x}); s\} = γ(s),

(ii) −\frac{d}{ds} \{s^{−v}γ (\frac{1}{s^{2}} )\} = ζ(s), (iii) L {xf(x); s} = H(s),

and suppose that x^{− \frac{1}{2}} ϕ (\frac{1}{x}) belongs to Ω and \frac{d}{ds} \{s^{−v}γ (\frac{1}{s^{2}} )\} exists for Re s > c_{1} where c_{1} is a constant. Then

\begin{matrix}L_{n} \{\frac{(v − 1)ϕ [p_{1} (\overline{x^{−1}})] − 2p_{1} (\overline{x^{−1}}) H [p_{1} (\overline{x^{−1}})]}{p_{n} (\overline{x^{\frac{1}{2}}} )}; \overline{s}\} = \\ \frac{π^{\frac{n−1}{2}}}{p_{n} (\overline{s^{\frac{1}{2}}} ) [p_{1} (\overline{s^{\frac{1}{2}}} )]^{v}} ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{−1}] , \end{matrix}  (2.20)

where Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > d, a constant n = 2, 3, …, N. It is asssumed that the integral on the left exists.

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

n-dimensional Laplace transforms 67

Making use of our hpothesis and (i) yields

γ(s) = π^{\frac{1}{2}} ∫_{0}^{∞} [∫_{0}^{∞} x^{− \frac{1}{2}} exp (−sx − \frac{u}{x}) f(u)du] dx, where Re s > c_{1}. (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 [15] we obtain

γ(s) = π^{\frac{1}{2}} ∫_{0}^{∞} f(u)s^{− \frac{1}{2}} exp (−2u^{\frac{1}{2}} s^{\frac{1}{2}} ) du.  (2.22)

Taking into account the condition (ii) we see that the equation (2.22) implies that

s^{v}ζ(s) = π^{\frac{1}{2}} (v − 1) ∫_{0}^{∞} f(u) exp (−\frac{2u^{\frac{1}{2}}}{s} ) du − 2π^{\frac{1}{2}} s^{−1} ∫_{0}^{∞} u^{\frac{1}{2}} f(u)

exp (\frac{−2u^{\frac{1}{2}}}{s} ) du, where Re s > c_{1}.  (2.23)

Now. replacing s by [p_{1} (\overline{s^{\frac{1}{2}}} )]^{−1} and then multiplying each side of (2.23) by p_{n} (\overline{s^{\frac{1}{2}}} ) and then making use of operational relations (2.9) and (2.24)

s_{i} exp (−asi^{\frac{1}{2}} ) ≑ \frac{a}{2}π^{− \frac{1}{2}} x_{i}^{− \frac{3}{2}} exp (−\frac{a^{2}}{4x_{i}} ) for i = 1, 2, …, n,  (2.24)

equation (2.23) reads

p_{n} (\overline{s^{\frac{1}{2}}} ) [p_{1} (\overline{s^{\frac{1}{2}}} )]^{−v} ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{−1}] =_{n}^{n} \frac{1}{π^{\frac{n−1}{2}} p_{n} (\overline{x^{\frac{1}{2}}} )} [(v − 1) ∫_{0}^{∞} f(u)

exp [−up_{1} (\overline{x^{−1}})] dx − 2p_{1} (\overline{x^{−1}}) ∫_{0}^{∞} uf(u) exp [−up_{1} (\overline{x^{−1}})] du.  (2.25)

Equation (2.25) with (iii) and the definition of the one-dimensional Laplace transform, leads to

p_{n} (\overline{s^{\frac{1}{2}}} ) [p_{1} (\overline{s^{\frac{1}{2}}} )]^{−v} ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{−1}]

=_{n}^{n} \frac{1}{π^{\frac{n−1}{2}} p_{n} (\overline{x^{\frac{1}{2}}} )} \{(v − 1)ϕ [p_{1} (\overline{x^{−1}})] − 2p_{1} (\overline{x^{−1}}) H [p_{1} (\overline{x^{−1}})]\} .

Therefore,

L_{n} \{\frac{(v − 1)ϕ [p_{1} (\overline{x^{−1}})] − 2p_{1} (\overline{x^{−1}}) H [p_{1} (\overline{x^{−1}})]}{p_{n} (\overline{x^{\frac{1}{2}}} )}; \overline{s}\}

= \frac{π^{\frac{n−1}{2}}}{p_{n} (\overline{s^{\frac{1}{2}}} ) [p_{n} (\overline{s^{\frac{1}{2}}} )]^{v}} ζ [(p_{1} (\overline{s^{\frac{1}{2}}} ))^{−1}] ,

where Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > d for some constant d, n = 2, 3, …, N.

68 R. S. Dahiya & Jafar Saberi-Nadjafi Example 2.6. Let f(x) = J_{0} (2x^{\frac{1}{2}} ). Then

ϕ(s) = \frac{1}{s} exp (−\frac{1}{s}) , Re s > 0,

γ(s) = \frac{π^{\frac{1}{2}}}{2(s + 1)^{\frac{3}{2}}} , Re s > −1.

Thus

ζ(s) = \frac{π^{\frac{1}{2}} (vs^{2} + v − 3)}{2s^{v−2}(1 + s^{2})^{\frac{5}{2}}}, Re s > −1.

Therefore

L_{n} \{\frac{1}{p_{1} (\overline{x^{−1}}) p_{n} (\overline{x^{\frac{1}{2}}} )} exp (−\frac{1}{p_{1} (\overline{x^{−1}})}) \{(v − 1) − 2_{1}F_{1} [ \begin{matrix}−1; \\ 1; \end{matrix} \frac{1}{p_{1} (\overline{x^{−1}})}]\}; \overline{s}\}

= \frac{π^{\frac{n}{2}} p_{1} (\overline{s^{\frac{1}{2}}} ) [v + (v − 3)p_{1}^{2} (\overline{s^{\frac{1}{2}}} )]}{2p_{n} (\overline{s^{\frac{1}{2}}} ) [1 + p_{1}^{2} (\overline{s^{\frac{1}{2}}} )]^{\frac{5}{2}}}, where Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > −1.  (2.26)

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

(i) L_{2} \{\frac{(xy)^{\frac{1}{2}}}{(x + y)} exp (−\frac{xy}{x + y} )_{1}F_{1} [ \begin{matrix}−1; \\ 1; \end{matrix} \frac{xy}{x + y} ]; s_{1}, s_{2}\} =
\frac{π (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ) [2 (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2} − 1]}{4(s_{1}s_{2})^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]^{\frac{5}{2}}},  (2.26′)

where Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > −1.

(ii) L_{2} \{\frac{(xy)^{\frac{1}{2}}}{(x + y)} exp (−\frac{xy}{x + y} ) \{1 −_{1}F_{1} [ \begin{matrix}−1; \\ 1; \end{matrix} \frac{xy}{x + y} ]\}; s_{1}, s_{2}\} =
\frac{3π (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}{2(s_{1}s_{2})^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]^{\frac{5}{2}}} ,  (2.26′′)

where Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 1.

Notice, with the help of (2.26′) from (2.26′′) we arrive at the following result

L_{2} \{\frac{(xy)^{\frac{1}{2}}}{(x + y)} exp (−\frac{xy}{x + y} ); s_{1}, s_{2}\} = \frac{π}{2} \frac{(s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}{(s_{1}s_{2})^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]^{\frac{2}{3}}} . (2.26′′′)

This is the same as the result (2.109) in Ditkin and Prudnokov [11; p. 140].

Furthermore, with the help of (2.26′′′) and the operational relation (47) in Voelker and Doetsch [24; p. 159], we derive the following new results.

L_{2} \{(\frac{y}{x})^{\frac{1}{2}} · \frac{1}{x + y} exp (−\frac{xy}{x + y} ); s_{1}, s_{2}\} = \frac{π}{s2^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]^{\frac{1}{2}}} ,

Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 1.  (2.26^{iv})
L_{2} \{(\frac{x}{y})^{\frac{1}{2}} · \frac{1}{x + y} exp (−\frac{xy}{x + y} ); s_{1}, s_{2}\} = \frac{π}{s1^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]^{\frac{1}{2}}} , (2.26^{v})

where Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 1.

Example 2.7. Assuming x^{− \frac{1}{2}} cos 2x^{\frac{1}{2}} , we obtain

ϕ(x) = (\frac{π}{s} )^{\frac{1}{2}} exp (−\frac{1}{x}) , Re s > 0.
γ(s) = \frac{π}{s + 1}, Re s > −1.
ζ(s) = (vs^{2} + v − 2) \frac{π^{\frac{1}{2}} s^{−v+1}}{(s^{2} + 1)^{2}} , Re s > −1.
H(s) = \frac{π^{\frac{1}{2}}}{s^{\frac{5}{2}}} (\frac{s}{2} − 1) exp (−\frac{1}{s}) , Re(s) > 0.

Therefore,

L_{n} \{\frac{(v − 1)p_{1} (\overline{x^{−1}}) − (p_{1} (\overline{x^{−1}}) − 2) exp (−\frac{1}{p_{1}((\overline{x^{−1}})})}{[p_{1} (\overline{x^{−1}})]^{\frac{3}{2}} p_{n} (\overline{x^{−1}})}; \overline{s}\}

= \frac{π^{\frac{n−1}{2}} p_{1} (\overline{s^{\frac{1}{2}}} )}{p_{n} (\overline{s^{\frac{1}{2}}} ) [1 + p_{1}^{2} (\overline{s^{\frac{1}{2}}} )]^{2}} \{(v − 2)p_{1}^{2} (\overline{s^{\frac{1}{2}}} ) + v\} , Re [p_{1} (\overline{s^{\frac{1}{2}}} )] > 0. (2.27)

Remark 2.3. If we let n = 2 and v = 1 in (2.27) by using the following well-known formula

L_{2} \{\frac{xy}{(x + y)^{\frac{3}{2}}} exp (−\frac{xy}{x + y} ); s_{1}, s_{2}\} = \frac{π^{\frac{1}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}{(s_{1}s_{2})^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]^{2}} ,

Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0.

we derive that

\begin{matrix}L_{2} \{\frac{1}{(x + y)^{\frac{1}{2}}} exp (−\frac{xy}{x + y} ); s_{1}, s_{2}\} = \frac{π^{\frac{1}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}{(s_{1}s_{2})^{\frac{1}{2}} [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{2}]}, \\ Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0. \end{matrix} (2.27′)

3. Non-Homogeneous Second Order Partial Differential Equations of Parabolic type

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 < ∞, 0 < y < ∞,  (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  (3.2)

by means of some of our results established in Section 2 using the double Laplace transformation.

Example 3.1. Determination of a solution u = u(x, y) of (3.1) and (3.2) for

(a) f(x, y) = (x + y)^{\frac{1}{2}}

(b) f(x, y) = \frac{(xy)^{\frac{τ}{2}}}{(x+y)^{\frac{τ+1}{2}}}

(c) f(x, y) = \frac{(xy)^{\frac{τ}{2}}}{(x+y)^{\frac{τ+2}{2}}}

We will use the following for the rest of this section. If

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)

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

L_{2}{u(x, y); s_{1}, s_{2}} = ∫_{0}^{∞} ∫_{0}^{∞} exp(−s_{1}x − s_{2}y)u(x, y)dxdy = U(s_{1}, s_{2}) (3.3)
L_{2}{u_{xx}; s_{1}, s_{2}} = s_{1}^{2}U(s_{1}, s_{2}) − s_{1}G(s_{2}) − G_{1}(s_{2})  (3.4)
L_{2}{u_{yy}; s_{1}, s_{2}} = s_{2}^{2}U(s_{1}, s_{2}) − s_{2}F (s_{1}) − F_{1}(s_{1})  (3.5)
L_{2}{u_{xy}; s_{1}, s_{2}} = s_{1}s_{2}U(s_{1}, s_{2}) − s_{1}F (s_{1}) − s_{2}G(s_{2}) + u(0, 0)  (3.6)

(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} · \frac{π^{\frac{1}{2}} [s_{1} + s_{2} + (s_{1}s_{2})^{\frac{1}{2}} ]}{2(s_{1}s_{2})^{\frac{3}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}, Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0. (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 [24]

u(x, y) = ∫_{0}^{x}(x + y − 2ξ)^{\frac{1}{2}} sin ξdξ  (3.8)

By a simple change of variable x + y − 2ξ = 2t^{2} in (3.8), we obtain

u(x, y) = 2^{+ \frac{3}{2}} ∫\begin{matrix}(\frac{y−x}{2}) & \frac{1}{2} \\ (\frac{x+y}{2} ) & \frac{1}{2} \end{matrix} t^{2} sin (t^{2} − \frac{x + y}{2} ) dt if y > x.

Expanding the sine and making some simplification, we deduce that

u(x, y) = 2^{+ \frac{3}{2}} [cos (\frac{x + y}{2} ) ∫\begin{matrix}(\frac{y−x}{2}) & \frac{1}{2} \\ (\frac{x+y}{2} ) & \frac{1}{2} \end{matrix} t^{2} sin t^{2}dt − sin (\frac{x + y}{2} )

∫\begin{matrix}(\frac{x−y}{2} ) & \frac{1}{2} \\ (\frac{x+y}{2} ) & \frac{1}{2} \end{matrix} t^{2} cos t^{2}dt]

(3.9)

Calculating the integrals involved in (3.9), we arrive at

u(x, y) = \frac{1}{4} \{ \begin{matrix}(\frac{x+y}{2} )^{\frac{1}{2}} − (\frac{y−x}{2})^{\frac{1}{2}} cos x + π^{\frac{1}{2}} cos (\frac{x+y}{2} ) [C (\frac{y−x}{2}) − D (\frac{x+y}{2} )] \\ + π^{\frac{1}{2}} sin (\frac{x+y}{2} ) [S (\frac{y−x}{2}) − S (\frac{x+y}{2} )] \end{matrix} \}

if y > x,

where C(·) and S(·) 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. [2; p.300] for part (c) to arrive at

U(s_{1}, s_{2}) = \frac{π^{\frac{1}{2}} Γ (\frac{τ}{2} + 1)}{[(s_{1} + s_{2})^{2} + 1] (s_{1}s_{2})^{\frac{τ}{2}} (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{τ+1}}  (3.10)

U(s_{1}, s_{2}) = \frac{πΓ(τ + 1)}{2^{τ} Γ (\frac{τ+3}{2}) [(s_{1} + s_{2})^{2} + 1] (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )^{τ+1}} ,  (3.11)

where Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0. Thus, we obtain the following solutions, respectively

u(x, y) = \frac{1}{2^{τ+1}} ∫_{x+y}^{y−x} t^{− \frac{τ+1}{2}} [t^{2} − (x − y)^{2}]^{\frac{τ}{2}} sin (\frac{t − (x + y)}{2}) dt

if y > xRe v > −1.  (3.12)

u(x, y) = \frac{1}{2^{τ+1}} ∫_{x+y}^{y−x} t^{− \frac{τ+3}{2}} [t^{2} − (x − y)^{2}]^{\frac{τ}{2}} sin (\frac{t − (x + y)}{2}) dt

if y > x, Re v > −1.  (3.13)

Remark 2.4. Substituting τ = 0 in parts (b) and (c), lead to

u_{xx} + 2u_{xy} + u_{yy} + u = \frac{1}{(x + y)^{\frac{1}{2}}}  (3.14)
u_{xx} + 2u_{xy} + u_{yy} + u = \frac{1}{(x + y)^{\frac{3}{2}}}  (3.15)

Next, using (3.14) and (3.15), we arrive at the following explicit solutions for the equations (3.14) and (3.15) respectively

u(x, y) = (\frac{π}{2} )^{\frac{1}{2}} \{cos (\frac{x + y}{2} ) [S (\frac{y − x}{2}) − S (\frac{x + y}{2} )] −

sin (\frac{x + y}{2} ) [C (\frac{y − x}{2}) − C (\frac{x + y}{2} )]\} if y > x.

u(x, y) = \frac{1}{(x + y)^{\frac{1}{2}}} cos(x + y) − \frac{1}{(y − x)^{\frac{1}{2}}} cos y − π^{\frac{1}{2}} \{cos (\frac{x + y}{2} ) [S (\frac{y − x}{2})

−S (\frac{x + y}{2} )] − sin (\frac{x + y}{2} ) [C (\frac{y − x}{2}) − C (\frac{x + y}{2} )]\} if y > x.

Example 3.2. Solve the following Parabolic differential equation described by

u_{xx} + 2u_{xy} + u_{yy} + u = \frac{1}{(x + y)^{\frac{1}{2}}} exp (−\frac{xy}{x + y} ) , 0 < x < ∞, 0 < y < ∞, (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.  (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{π (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )}{[(s_{1} + s_{2})^{2} + 1] [1 + (s1^{\frac{1}{2}} + s2^{\frac{1}{2}} )]^{\frac{1}{2}}} ,  (3.18)

where Re [s1^{\frac{1}{2}} + s2^{\frac{1}{2}} ] > 0. Using formula (133) in [24] the inverse transform of (3.18) leads to the following integral representation.

u(x, y) = ∫_{x+y}^{y−x} t^{− \frac{1}{2}} exp [\frac{t^{2} − (x + y)^{2}}{4t} ] sin (\frac{t − (x + y)}{2}) dt if y > x.

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

   E-mail address: dahiya@iastate.edu

Jaffar Saberi-Nadjafi Department of Mathematics, Ferdowsi, University of Mashhad Mashhad, Iran