integral transforms and special functions theorems on n

PLEASE SCROLL DOWN FOR ARTICLE

This article was downloaded by: [Ferdowsi University of Mashhad]On: 27 August 2010Access details: Access Details: [subscription number 912974449]Publisher Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registered office: Mortimer House, 37-41 Mortimer Street, London W1T 3JH, UK

Integral Transforms and Special FunctionsPublication details, including instructions for authors and subscription information:http://www.informaworld.com/smpp/title~content=t713643686

Theorems on n-dimensional laplace transformations involving the solutionof wave equationsJafar Saberi-nadjafia; R. S. Dahiyab

a Department of Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran b Department ofMathematics, Iowa State University, Ames, Iowa, USA

To cite this Article Saberi-nadjafi, Jafar and Dahiya, R. S.(2004) 'Theorems on n-dimensional laplace transformationsinvolving the solution of wave equations', Integral Transforms and Special Functions, 15: 4, 337 — 358To link to this Article: DOI: 10.1080/10652460410001672933URL: http://dx.doi.org/10.1080/10652460410001672933

Full terms and conditions of use: http://www.informaworld.com/terms-and-conditions-of-access.pdf

This article may be used for research, teaching and private study purposes. Any substantial orsystematic reproduction, re-distribution, re-selling, loan or sub-licensing, systematic supply ordistribution in any form to anyone is expressly forbidden.

The publisher does not give any warranty express or implied or make any representation that the contentswill be complete or accurate or up to date. The accuracy of any instructions, formulae and drug dosesshould be independently verified with primary sources. The publisher shall not be liable for any loss,actions, claims, proceedings, demand or costs or damages whatsoever or howsoever caused arising directlyor indirectly in connection with or arising out of the use of this material.

http://www.informaworld.com/smpp/title~content=t713643686

http://dx.doi.org/10.1080/10652460410001672933

http://www.informaworld.com/terms-and-conditions-of-access.pdf

Integral Transforms and Special FunctionsVol. 15, No. 4, August 2004, pp. 337–358

THEOREMS ON N-DIMENSIONAL LAPLACETRANSFORMATIONS INVOLVING THE SOLUTION OF

WAVE EQUATIONS

JAFAR SABERI-NADJAFIa and R. S. DAHIYAb,∗

aDepartment of Mathematics, Ferdowsi University of Mashhad, Mashhad, Iran; bDepartment ofMathematics, Iowa State University, Ames, Iowa 50011, USA

(Received 28 January 2003)

The primary objective of this article is to establish several new results for calculating Laplace transformation pairs ofN -dimensions from known one-dimensional Laplace transformations. Next, we applied these results to a number ofcommonly used special functions to obtain a new Laplace transformation in two and N -dimensions. Finally, severalone-dimensional wave equations involving special functions are explicitly solved by using two-dimensional Laplacetransformations and applying some of the developed results in Section 2.

Keywords: Multi-dimensional transforms; Laplace transforms; Wave equation

MSC 1991: 44A30; 44A10; 35F05

1 INTRODUCTION AND NOTATION

The Laplace transform, it can be fairly said, stands first in importance among all integraltransforms for which there are many specific examples in which other transforms prove moreexpedient. The Laplace transform is the most powerful in dealing with both initial-boundary-value problems (IBVPs) and transforms.

By the use of N -dimensional Laplace transformations, a partial differential equation (PDE)and its associated boundary conditions can be transformed into an algebraic equation in n

independent complex variables. This algebraic equation can be solved for multiple transformof the solution of the original PDE. Multiple inversion of this transform then gives the desiredsolution [8], [9], [11], and [12]. This technique is used in Section 3 of this article to solveseveral one-dimensional Wave equations. For more results, see [2–7], [13], [15–20].

In what follows, we will use the following notations.For any real n-dimensional variable x = (x1, x2, . . . , xn), and for any complex n-

dimensional variable s = (s1, s2, . . . , sn), we denote xv = (xv1 , xv

2 , . . . , xvn) and sv =

(sv1 , sv

2 , . . . , svn) where v is any real number.

Let pk(x) or pk(s) be the kth symmetric polynomial in the components xk or sk , of x or s,respectively. Then for s = (s1, s2, . . . , sn), we denote

∗ Corresponding author. E-mail: [email protected]

ISSN 1065-2469 print; ISSN 1476-8291 online c© 2004 Taylor & Francis LtdDOI: 10.1080/10652460410001672933

Downloaded By: [Ferdowsi University of Mashhad] At: 11:48 27 August 2010

338 J. SABERI-NADJAFI AND R. S. DAHIYA

(i) p1(sv) = sv1 + sv

2 + · · · + svn =

n∑j=1

svj .

(ii) pn(sv) = sv1 · sv

2 · · · svn = ∏n

j=1 svj .

In addition, we shall write

(iii) x · s =n∑

j=1xj sj .

(iv) Re (s) = Real part of a complex number s.(v) Re [p1(sv)] = Real part of a complex number (sv

1 + sv2 + · · · + sv

n).

As usual, we denote by N the set of natural numbers, N = {1, 2, . . .} and N0 = N ∪ {0}. By Rn

we mean the n-dimensional Euclidean space, n ∈ N . Analogously, we denote unitary spaceby Cn. By subsets Rn+ and R

n

+ of Rn, we mean the following

Rn+ = {x: x ∈ Rn, x > 0},

Rn

+ = {x: x ∈ Rn, x ≥ 0}.

By L1[Rn+; exp(−a · x)], we mean the linear space of all measurable function f defined onRn+ for which the following conditions hold:

(i)∫ω

|f (x) dx are finite.(ii) f (x) satisfies the condition

|f (x)| ≤ M exp(a · x),

for all measurable functions f on Rn+, where M and a are positive constants.

By Ea , we mean the set of all functions from Rn into C with the following properties:There exists a point a ∈ Rn such that f ∈ L1[Rn+; exp(−a · x)] and

f (x) = 0, x ∈ Rn+\Rn

+,

Ea is equipped with the norm ‖f ‖Ea= ‖f ‖L1[Rn+;exp(−a·x)].

In the present work, the class � is taken to consist of all complex valued functions f thatare piecewise continuous in the range x ≥ 0 and of exponential order as x → ∞.

If u(x, y) be a function of two variables x and y, we denote

∂u(x, y)

∂x= ux,

∂u(x, y)

∂y= uy,

∂2(x, y)

∂x∂y= uxy,

∂2u(x, y)

∂x2= uxx,

∂2u(x, y)

∂y2= uyy.

We denote

u(x, 0) = α(x), u(0, y) = β(y),

uy(x, 0) = θ(x), ux(0, y) = δ(y)

and assume α(x), β(y), θ(x), and δ(y) are Laplace transformable.


THEOREMS ON N -DIMENSIONAL LAPLACE TRANSFORMATIONS 339

The n-dimensional Laplace transform of a function from Rn+ into C is defined by means of

F(s) = Ln{f (x); s} =∞∫

0

∞∫0

· · ·∞∫

0

exp[−s · x]f (x) dx1 dx2 . . . dxn

=∫

Rn+

exp[−s · x]f (x) dx. (1)

The domain of definition of F is the set of all points s ∈ Cn such that the integral in Eq. (1) isconvergent.

Instead of the n-dimensional Laplace transform (1), one sometimes calculates the so-calledn-dimensional Laplace–Carson transform:

F(s) = pn(s)

∫Rn+

exp(−s · x)f (x) dx. (2)

Symbolically, we denote the pair F(s) and f (x) with the operational relation

F(s)n=n

f (x) or f (x)n=n

F(s). (3)

In this notation, some formulas become more simple.We denote Eq. (3) in one-dimensional case by the following

F(s) .=. f (x).

Let f ∈ Ea, a ∈ Rn+ and let there exist (∂/∂xj )f = Djf, j = 1, 2, . . . n, D1f =∂nf/(∂x1∂x2 . . . ∂xn) and Djf ∈ C(Rn+), D1f ∈ Ea . Then at each point of continuity of f ,the so-called complex inversion formula holds

L−1n {F(s); x} = f (x) = (2πi)−n

∫(α)

exp(s · x)F (s) ds, α ∈ Rn+, α > α. (4)

Here, the integral has to be understood in the sense of the principal value of Cauchy, i.e.,

∫(α)

. . . ds = limβj→∞

j=1,2,...,n

α1+iβ1∫α1−iβ1

α2+iβ2∫α2−iβ2

· · ·αn+iβn∫

αn−iβn

· · · ds1 ds2 . . . dsn.

The integral at the right side of Eq. (4) is called the n-dimensional inverse Laplace trans-formation of F .

If we denote

u(x, 0) = f (x), u(0, y) = g(y),

uy(x, y)|y=0 = uy(x, 0) = f1(x), ux(x, y)|x=0 = ux(0, y) = g1(y),



and their one-dimensional Laplace transformations are denoted by G(s2), F1(s1), and G1(s2),respectively, then

L2{u(x, y); s1, s2} =∞∫

0

∞∫0

exp(−s1x − s2y)u(x, y) dx dy = U(s1, s2), (5)

L2{uxx; s1, s2} = s21U(s1, s2) − s1G(s2) − G1(s2), (6)

L2{uyy; s1, s2} = s22U(s1, s2) − s2F(s1) − F1(s1). (7)

It is assumed that the integrals involved exist. The existence conditions for two-dimensionsare given in Ditkin and Prudnikov [10, p. 4] and for similar conditions hold for N -dimensions,we refer to Brychkov et al. [1; ch. 2].

2 THEOREMS ON N-DIMENSIONAL LAPLACE TRANSFORMATION

THEOREM 2.1 Let

(i) L{f (x); s} = φ(s), and(ii) L{x−1/2f (x2); s} = F(s), assuming that f (x2) is a function of class �.

(iii) L{x−3/2φ(

1x

); s} = ξ(s),

(iv) L{x−3/2ξ(

1x2

); s} = η(s),

where x−3/2φ(1/x) and x−3/2ξ(1/x2) are also functions of class �, and x−3/2 exp(−sx −u/x)f (u) and u−1/2x−3/2 exp(−sx − 2u1/2/x)f (u) belong to L1[(0, ∞) × (0, ∞)].Then

Ln

1

p(x1/2)n

F [2p1(x−1)]; s

= π(n−2)/2

21/2pn(s1/2)η[(p1(s1/2))2], (8)

where n = 2, 3, . . . , N and Re [p1(s1/2)] > c and c is a constant.

Proof From (i), we obtain φ(s) =∞∫0

exp(−st)f (t) dt for Re s > c0, where c0 is a constant,

so that

x−3/2φ

(1

x

)= x−3/2

∞∫0

exp(−u

x

)f (u) du. (9)

Multiplying both sides of Eq. (9) by exp(−sx) and integrating from 0 to ∞, we obtain

∞∫0

exp(−sx)x−3/2φ

(1

x

)dx =

∞∫0

∞∫

0

x−3/2 exp(−sx − u

x

)f (u) du

dx.

The integrand x−3/2 exp(−sx − u/x)f (u) belongs to L1[(0, ∞) × (0, ∞)], that, by Fubini’sTheorem, interchanging the order of the integral on the right side of Eq. (9) is permissible. By



using (iii) on the left and interchanging the order of integration on the right side of Eq. (9), weobtain

ξ(s) =∞∫

0

f (u)

∞∫

0


x

)dx

du, (10)

where Re s > λ1 and λ1 is a constant.A result in Roberts and Kaufman [14] regarding the inner integral in Eq. (10) be used to evaluatethis integral as

∞∫0


x

)dx = π1/2u−1/2 exp

(−2u1/2s1/2).

Therefore, Eq. (10) can be written as

ξ(s) = π1/2

∞∫0

u−1/2 exp(−2u1/2s1/2

)f (u) du. (11)

Using Eq. (2.4) in (iv), we arrive at

η(s) = π1/2

∞∫0

∞∫

0

s−3/2u−1/2 exp

(−sx − 2u1/2

x

)f (u) du

dx, (12)

where Re s > λ2 for some constant λ2.Again, because x−3/2u1/2 exp(−sx − 2u1/2/x)f (u) belongs to L1[(0, ∞) × (0, ∞)], by

Fubini’s Theorem, we can interchange the order of integration on the right side of Eq. (12) toobtain

η(s) = π1/2

∞∫0

u−1/2f (u)

∞∫

0

x−3/2 exp

(−sx − (23/2u1/4)2

4x

)dx

du. (13)

Using the previously mentioned result in Roberts and Kaufman [14] on the inner integral inEq. (10), we can evaluate the integral inside the brackets.

Consequently, Eq. (13) becomes

η(s) = 2−1/2π

∞∫0

u−3/4f (u)exp(−23/2u1/4s1/2) du. (14)

By substituting u = v2 in Eq. (14), we obtain

η(s) = 21/2π

∞∫0

v−1/2f (v2) exp(−23/2v1/2s1/2) dv. (15)



Replacing s by [p1(s1/2)]2 and multiplying both sides of Eq. (15) by pn(s1/2), we arrive at

pn(s1/2)η[(p1(s1/2))2] = 21/2x

∞∫0

v−1/2f (v2)pn(s1/2) exp(−23/2v1/2p1(s1/2)) dv. (16)

Now, we use the operational relation given in Ditkin and Prudnikov [10]:

s1/2i exp(−as

1/2i ) .=. (πxi)

−1/2 exp

(− a2

4xi

)for i = 1, 2, . . . , n. (17)

Equation (16) can be rewritten as

pn(s1/2)η[(p1(s1/2))2]n=n

21/2

π(n−2)/2pn(x1/2)

∞∫0

v−1/2f (v2) exp(−2vp1(x−1)) dv (18)

Using (ii) and Eq. (18), we obtain

pn(s1/2)η[(p1(s1/2))2] n=n21/2

π(n−2)/2· 1

pn(x1/2)F [2p1(x−1)].

Hence,

Ln

{1

pn(x1/2)F [2p1(x−1)]; s

}= π(n−2)/2

21/2pn(s1/2)η[(p1(s1/2))2],

where n = 2, 3, . . . , N .

Example 2.1′ Let f (x) = xv/4. Then

φ(s) = ((v/4) + 1)

s((v/4)+1), Re s > 0, Re v > −4,

ξ(s) = ((v/4) + 1) ((v/4) + (1/2))

s(v/4)+(1/2), Re v > −2, Re s > 0

and

η(s) = ((v/4) + 1) ((v/4) + (1/2)) ((v + 1)/2)

s(v+1)/2, Re v > −1, Re s > 0.

F (s) = ((v + 1)/2)

s(v+1)/2.

Therefore,

Ln

{1

pn(x1/2)[p1(x−1)](v+1)/2; s

}= π(n−1)/2

(v

2+ 1

)· 1

pn(s1/2)[p1(s1/2)]v+1, (19)

where Re v > −1, Re [p1(s1/2)] > 0, and n = 2, 3, . . . , N .



Example 2.2 Assume that f (x) = sin(ax1/2). Then

φ(s) = απ1/2

2s−3/2 exp

(a2

4s

), Re s > 0

and

ξ(s) = 2απ1/2

4s + α2, Re s > −Re

a2

4,

η(s) = π

αs1/2S−1,1/2

(2s

α

), Re a > 0, Re s > 0,

F (s) = π1/2

(s2 + α2)1/4sin

[1

4tan−1

(α

s

)], Re s > |Im a|.

where Re a > 0, Re s > 0.Hence, we obtain

Ln

{sin[(1/4) tan−1(α/(2p1(x−1)))]

pn(x1/2)[4p21(x

−1) + α2]1/4; s

}=(

πn+1

2

)1/2

· p1(s1/2)

αpn(s1/2)S−1,1/2

[2

αp2

1(s1/2)

],

(20)

where Re α > 0, Re[p1(s1/2)] > 0.

Example 2.3 Consider f (x) = xτpF q

[(a)p;(b)q;

kx

]. Then

φ(s) = (τ + 1)

sτ+1p + 1F q

[(a)p, τ + 1;(b)q;

ks

],

where p ≤ q, Re τ > −1 and Re s >

{0 if p < q

Re k if p = q.

F (s) = ((4τ + 1)/2)

s(4τ+1)/2p + 2F q

[(a)p, τ + 1/4, τ + 3/4;(b)q;

4k

s2

],

where p ≤ q − 1, Re τ > −1/4; Re s > 0 if p ≤ q − 1; and Re (s + 2k cos πr) > 0(r =0, 1) if p = q − 1.

ξ(s) = (τ + 1) (τ + 1/2)

sτ+1/2p + 2F q

(a)p, τ + 1, τ + 1

2;

(b)q;ks

,

where p ≤ q − 1, Re τ > −1/2 and Re s >

{0 if p < q

Re k if p = q.

η(s) = (τ + 1) (τ + 1/2) (2τ + 1/2)

s2τ+1/2p + 4F q



(a)p, τ + 1, τ + 1

2, τ + 1

4, τ + 3

4;

(b)q;4k2

s2

,

where p ≤ q − 3, Re τ > −1/4; Re s > 0 if p ≤ q − 4; and Re (s + 2k cos πr) > 0(r =0, 1) if p = q − 3.

Therefore, we have

Ln

1

pn(x1/2)[p1(x−1)]2τ+1/2p + 2F q

(a)p,

4τ + 1

4,

4τ + 3

4;

(b)q;k

p21(x

−1)

; s

= π(n−1)/2 (2τ + 1)

pn(s1/2)[p1(s1/2)]4τ+1p + 4F q

(a)p, τ + 1, τ + 1

2, τ + 1

4, τ + 3

4;

(b)q;4k

p41(s

−1)

(21)

where p ≤ q − 3, Re τ > −1/4; Re [p1(s1/2)] > 0 if p ≤ q − 4; and Re [p1(s1/2) +2k cos πr] > 0 (r = 0, 1) if p = q − 3.

Example 2.4 Suppose that

f (x) =

1 if 0 < x <1

64

0 if x >1

64.

Then,

φ(s) = 1 − exp(−s/64)

s,

In addition, we obtain

ξ(s) =(π

s

)1/2[

1 − exp

(− s1/2

4

)], Re s > 0,

η(s) = π

s1/2[1 − exp(−s1/2)], Re s > 0,

F (s) = γ (1/2, s/8)

s1/2, Re s > −∞.

Therefore, using Theorem 2.1, we obtain the following result:

Ln

1

pn(x1/2)[p1(x−1)]

1

2

Erf

(p1(x−1)

2

)1/2 ; s

= π(n−1)/2

pn(s1/2)p1(s1/2)[1 − exp(−p1(s1/2))]. (22)

where Re [p1(s1/2)] > 0, n = 2, 3, . . . , N .



Example 2.1′ Let n = 2, from Example 2.1, we arrive at the following well-known result intwo-dimensions.

L2

{(xy)v/2

(x + y)(v+1)/2; s1, s2

}= π1/2

(v

2+ 1

)· 2

(s1s2)1/2(s1/21 + s

1/22 )v+1

, (23)

where Re v > −1, Re [s1/21 + s

1/22 ] > 0.

If we let v = 0 in Eq. (23), we obtain

L2

{1

(x + y)1/2, s1, s2

}= π1/2

(s1s2)1/2(s1/21 + s

1/21 )

. (24)

From Eq. (24) and the following relations

L2{xF(x, y); s1, s2} = − ∂

∂s1f (s1, s2),

L2{yF(x, y); s1, s2} = ∂

∂s2f (s1, s2),

where f (s1, s2) = L2{F(x, y); s1, s2}.We deduce that

L2{(x + y)1/2; s1, s2

} = π1/2(s1 + s2 + s1/21 s

1/22 )

2(s1s2)3/2(s1/21 + s

1/22 )

. (25)

THEOREM 2.2 Suppose f and x−3/2φ(1/x) belong to class �, where φ is the one-dimensionalLaplace transform of f .Let

(i) L{xj−1/2f (x); s} = Fj (s) for j = 0, 1, and let d/ds{s−vζ(1/s2)} exist for Re s > c0,where c0 is a constant.If x−3/2 exp(−sx − u/x) belongs to L1[(0, ∞) × (0, ∞)] and the following conditionshold:

(ii) L{x−3/2φ(1/x); s} = ζ(s),(iii) −d/ds{s−vζ(1/s2)} = η(s).

Then

Ln

{vF0[p1(x−1)] − 2p1(x−1)F1[p1(x−1)]

pn(x1/2); s

}= π(n−1)/2

pn(s1/2)[p1(s1/2)]v+1η[(p1(s1/2))−1],

(26)

where Re [p1(s1/2)] > c, c is a constant. It is assumed that the integrals involved exist forn = 2, 3, . . . , N .

Proof By the hypothesis and (ii), we get

ζ(s) =∞∫

0

x−3/2φ

(1

x

)exp(−sx) dx =

∞∫0

x−3/2 exp(−sx)

∞∫

0

f (u) exp(−u

x

)du

dx,



where Re s > c0 for some constant c0 which leads to

ζ(s) =∞∫

0

∞∫

0


x

)f (u) du

dx. (27)

Next we wish to interchange the order of the integral on the right side of Eq. (27). Theinterchange is permissible, according to Fubini’s Theorem.Therefore,

ζ(s) =∞∫

0

f (u)

∞∫

0


x

)dx

du, where Re s > c0. (28)

A result in Roberts and Kaufman [14] regarding the inner integral in Eq. (28) is used to evaluatethe integral as

ζ(s) = π1/2

∞∫0

u−1/2f (u)exp(−2u1/2s1/2) du. (29)

Equation (29) together with (iii) shows that

η(s) = π1/2s−v−1

v

∞∫0

u−1/2f (u) exp

(−2u1/2

s

)du − 2s−1

∞∫0

f (u) exp

(−2u1/2

s

)du

,

so that

sv+1η(s) = π1/2

v

∞∫0

u−1/2f (u) exp

(−2u1/2

s

)du − 2s−1

∞∫0

f (u) exp

(−2u1/2

s

)du

,

(30)

where Re s > c0.Next, we replace s by [p1(s1/2)]−1 and, then, multiply both sides of Eq. (30) by pn(s1/2),

to get

pn(s1/2)[p1(s1/2)]−v−1η[(p1(s1/2))−1]

= π1/2v

∞∫0

u−1/2pn(s1/2)f (u) exp[−2u1/2p1(s1/2)] du

− 2π1/2

∞∫0

pn(s1/2)p1(s1/2)f (u) exp[−2u1/2p1(s1/2)] du. (31)

Now, we use the following operational relations given in Ditkin and Prudnikov [10]:

s1/2i exp(−αs

1/2i ) .=. (πxi)

−1/2 exp

(− α2

4xi

)for i = 1, 2, . . . , n,

si exp(−αs1/2i ) .=.

α

2π−1/2x

−3/2i exp

(− α2

4xi

)for i = 1, 2, . . . , n.



Equation (31) can be rewritten as

pn(s1/2)[p1(s1/2)]−v−1η[(p1(s1/2))−1]

n=n

v

π(n−1)/2pn(x1/2)

∞∫0

u−1/2f (u) exp[−2u1/2p1(x−1)] du

− 2p1(x−1)

π(n−1)/2pn(x1/2)

∞∫0

u1/2f (u) exp[−2u1/2p1(x−1)] du. (32)

Equation (32) with (i) for j = 0, 1, yields to

pn(s1/2)[p1(s1/2)]−v−1η[(p1(s1/2))−1]

n=n1

π(n−1)/2pn(x1/2){vF0[p1(x−1)] − 2(p1(x−1)F1[p1(x−1)])}.

Thus,

Ln

{vF0[p1(x−1)] − 2p1(x−1)F1[p1(x−1)]

pn(x1/2); s

}= π(n−1)/2

pn(s1/2)[pn(s1/2)]v+1η[(p1(s1/2))−1],

where Re[p1(s1/2)] > c and n = 2, 3, . . . , N . �

The following examples will illustrate the applications of Theorem 2.2. We shall considerthe function f to be an elementary or some special function to construct certain functions withn variables, and we calculate their Laplace transforms using Theorem 2.2. We will use thetwo-dimensional case for some of these examples in Section 3 to solve the Wave equations.

Example 2.5 Let f (x) = J0(2x1/2). The

φ(s) = 1

sexp

(−1

s

), Re s > 0,

ζ(s) = π1/2

(s + 1)1/2, Re s > 0,

so that

η(s) = π1/2(vs2 + v − 1)

sv(1 + s2)3/2, Re s > 0.

Fj (s) = (j − 1/2)

sj−1/2exp

(−1

s

)1F1

j − 1

2;

1;1s

for j = 0, 1, Re s > 0.

Hence,

Ln

π1/2v exp[−1/p1(x−1)]1F1

[1/2;1; 1/p1(x−1)

]

−2π1/2/2 exp[−1/p1(x−1)]1F1

−1

2;

1;1/p1(x−1)

[p1(x−1)]1/2pn(x1/2)



= πn−1

2 · π1/2[v[p1(s1/2)]−2 + v − 1][p1(s1/2)]v

pn(s1/2)[p1(s1/2)]v+1[1 + (p1(s1/2))−2]3/2.

Thus,

Ln

exp[−1/p1(x−1)]

{v1 F1

[1/2;

1; 1p1(x−1)

]− 1F1

[−1/2;1; 1p1(x−1)

]

pn(x1/2)[p1(x−1)]1/2; s

= π(n−1)/2[v + (v − 1)p21(s

1/2)]

pn(s1/2)[1 + p21(s

1/2)]3/2, (33)

where Re pn(s1/2) > 0.

Remark 2.1 Equation (33) in two-dimensional case for v = 0 or v = 1 reduces to thefollowing new results, respectively.

(i)

L2

exp[−xy/(x + y)]1F1

[−1/2;1; xy/(x + y)

]

(x + y)1/2; s1, s2

= π1/2(s1/21 + s

1/22 )2

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]3/2

, (34)

where Re [s1/21 + s

1/22 ] > 0.

L2

exp[−xy/(x + y)]

{1F1

[1/2;

1; xy/(x + y)

]− 1F1

[−1/2;1; xy/(x + y)

]}

(x + y)1/2; s1, s2

= π1/2

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]3/2

using the following result

1F1

[1/2;

1; xyx + y

]− 1F1

[−1/2;1; xyx + y

]= xy

x + y1F1

[1/2;

2; xyx + y

].



The last formula can be simplified as

L2

{exp

[− xy

x + y

]xy(x + y)1/2

1F1

[1/2;2; xyx + y

]; s1, s2

}

= π1/2

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]3/2

, (35)

where Re [s1/21 + s

1/22 ] > 0.

Using the following version of Fj

Fj (s) = (j − 1/2)

sj−1/2 1F1

[j − 1/2;

1; −1s

]for j = 0, 1.

The Formula (33) turns out to be

Ln

v 1F1

[1/2;

1; −1p1

]− 1F1

[3/2;

1; −1p1(x−1)

]

p1(x1/2)[p1(x−1)]1/2; s

= π(n−1)/2[v + (v − 1)p21(s

1/2)]

pn(s1/2)[1 + p21(s

1/2)]3/2, (36)

where Re [p1(s1/2)] > 0.Special cases of Eq. (36) for n = 2, v = 0 or v = 1 turn out to be

L2

1

(x + y)1/2 1F1

3

2;

1;−xyx + y

;s1, s2

= π1/2

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]3/2

, Re [s1/21 + s

1/22 ] > 0. (37)

L2

xy(x + y)1/2

1F1

1

2;

2;−xyx + y

;s1, s2

= π1/2

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]3/2

, Re [s1/21 + s

1/22 ] > 0. (38)

Example 2.6 If we let f (x) to be 0F1

[ ·;1; x

], x1/4J1/2(2x1/2) or cos 2x1/2, then using

Theorem 2.2 we derive the following results

(i)

Ln

1

p1(x−1)pn(x1/2)

v 1F1

1

2;

1;1p1(x−1)



− 1F1

3

2;

1;1p1(x−1)

; s

= π(n−1)/2

p1(s1/2)pn(s1/2)

(v − 1)2F1

1,

1

2;

1;1p2

1(s12)

× 1

p1(s1/2)2F1

2,

3

2;

2;1p2

1(s1/2)

, (39)

where Re [p1(s1/2)] > 1 and n = 2, 3, . . . , N .(ii)

Ln

1

p1(x−1)pn(x1/2)

v 1F1

1;

3

2; − 1p1(x−1)

− 21F1

2;

3

2; − 1p1(x−1)

;s

= πn/2[v + (v − 2)p21(s

1/2)]

2pn(s1/2)[1 + p21(s

1/2)]2, Re [p1(s1/2)] > 0, n = 2, 3, . . . , N. (40)

(iii)

Ln

{1

p3/21 (x−1)pn(x1/2)

[2 + (v − 1)p1(x−1) exp

[−1p1(x−1)

]]; s

}

= π(n−1)/2p1(s1/2)

pn(s1/2)

[(v + 1) + (v − 1)p2

1(s1/2)

(1 + p21(s

1/2))2

],

Re [p1(s1/2)] > 0, n = 2, 3, . . . , N. (41)

Example 2.6′ (Two-dimensions)

(i) Upon substituting n = 2 and v = 1 in Eq. (39) we arrive at a new result as follows

L2

(xy)

(x + y)3/2 1F1

5

2;

2;xyx + y

; s1, s2

= π1/2

(s1s2)1/2(s1/21 + s

1/22 )3

2F1

2,

3

2;

2;1(s12

1 + s1/22 )2

, (42)

where Re [s1/21 + s

1/22 ] > 1.

(ii) On substitution of n = 2 and v = 2 or v = 0 in Eq. (40), we obtain the following results,respectively

L2

(xy)3/2

(x + y)2 1F1

2;

5

2; −xy x + y

; s1, s2



= 3π

4(s1s2)1/2[1 + (s1/21 + s

1/22 )]2

, Re [s1/21 + s

1/22 ] > 0. (43)

L2

(xy)1/2

(x + y)1F1

2;

3

2; −xyx + y

; s1, s2

= π(s1/21 + s

1/22 )2

2(s1s2)1/2[1 + (s1/21 + s

1/22 )]2

, Re[s1/21 + s

1/22 ] > 0. (44)

(iii) If we let n = 2, v = 1 in Eq. (41), we obtain

L2

{xy

(x + y)3/2exp

(− xy

x + y

);s1, s2

}

= π1/2(s1/21 + s

1/22 )

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]2

, Re[s1/21 + s

1/22 ] > 0. (45)

Furthermore, if we substitute v = 3 and with the help of Eq. (45), we arrive at

L2

{1

x + yexp

(− xy

x + y

); s1, s2

}

= π1/2(s1/21 + s

1/22 )

(s1s2)1/2[1 + (s1/21 + s

1/22 )2]

, Re[s1/21 + s

1/22 ] > 0. (46)

Both the operational relations (45) and (46) are well-known results.

Similarly, many more double Laplace transforms can be derived by taking different valuesof v in Eqs. (39)–(41).

Example 2.7 Suppose f (x) = xα exp(−bx). Then

φ(s) = (α + 1)

(s + b)α+1, Re α > −1, Re s > −Re b,

ξ(s) =(

2

b

)α+1

(α + 1)

(α + 1

2

)exp

( s

2b

)D−2α−1

[(2s

b

)], Re α > −1

2.

For v = 0

η(s) = −(2

b)a+3/2 (α + 1)

(α + 1

2

)(2α + 1) · exp(1/(2bs2))

s2D−2α−2

[(2

b

)1/2 1

s

],

where Re α > −1/2, Re s > 0, and |arg 1/b| < π .Next,

F1(s) = (α + 3/2)

(s + b)α+3/2, Re s > −Re b, Re α > −3

2.



Therefore, from Eq. (26) for v = 0, we obtain

Ln

{p1(x−1)

pn(x1/2)[b + p1(x−1)]a+3/2; s

}

= π(n−1)/2

(2

b

)α+3/2 (α + 1)p1(s1/2)

pn(s1/2)

× exp

[1

2bp2

1(s1/2)

]D−2α−2

[(2

b

)1/2

p1(s1/2)

], (47)

where Re α > −1/2, Re[p1(s1/2)] > 0 and n = 2, 3, . . . , N .

Example 2.7′ If we let n = 2 in Eq. (47) we obtain the following new result

L2

{(x + y)(xy)α

(x + y + bxy)α+3/2; s1, s2

}

= (2/b)α+3/2π1/2 (α + 1)(s1/21 + s

1/22 )

(s1s2)1/2

× exp

[1

2b(s

1/21 + s

1/22 )2

]D−2α−2

[(2

b

)1/2

(s1/21 + s

1/22 )

]. (48)

One can develop many other operational relations from given function f in Example 2.7 viausing Eq. (26) for different values of v.

3 THE WAVE EQUATIONS

To illustrate the use of some of our results which are derived in Section 2 in wave mechanics,we shall consider the one-dimensional wave equation in a normalized form

uxx − uyy = f (x, y), 0 < x < ∞, 0 < y < ∞. (49)

Under the initial and boundary conditions (necessary in the first view of the application of theLaplace transformation)

u(x, 0) = α(x), u(0, y) = β(y)

uy(x, 0) = θ(x), ux(0, y) = δ(y),

α(0) = β(0), (50)

subject to the following condition of compatibility:

x∫0

f (x − τ, τ ) dτ + δ(x) − θ(x) + ∂

∂x[β(x) − α(x)] = 0. (51)

Example 3.1 Determination of a solution u = u(x, y) of Eqs. (49) and (50) for



(a)

f (x, y) = 0,

α(x) = x1/2, β(y) = y1/2,

θ(x) = 1

2x−1/2 and δ(y) = 1

2y−1/2.

(b)

f (x, y) = yn − xn,

α(x) = x1/2, β(y) = y1/2,

θ(x) = 1

2x−1/2 and δ(y) = 1

2y−1/2.

(c)

f (x, y) = xv−2yv − xvyv−2, Re v > 1,

α(x) = x1/2, β(y) = y1/2,

θ(x) = 1

2x−1/2 and δ(y) = 1

2y−1/2.

We provide the solution for part (c). By an argument similar to that employed in part (c) thesolution for the two other parts are established in brief.

(c) Taking the two-dimensional Laplace transformation from each term of Examples 3.1(a)and 3.2(a) with the aid of Eqs. (6) and (7), yield the following transform equation

(s21 − s2

2 )U(s1, s2) − (3/2)s1

s3/22

− (3/2)

s1/22

+ (3/2)s2

s3/21

+ (3/2)

s1/21

= (v − 1) (v + 1)

sv−11 sv+1

2

− (v − 1) (v + 1)

sv+11 sv−1

2

or, equivalently

U(s1, s2) =

(3

2

)[s

1/21 + s

1/22

(s1s2)3/2− 1

s1s2(s1/21 + s

1/22 )

]

+ (v − 1) (v + 1)

sv+11 sv+1

2

, Re[s1/21 + s

1/22 ] > 0. (52)

For the inversion of Eqs. (52), we combine Example 2.1′ with Formula 87 in Voelker andDoetsch [21, p. 236]. Finally, we obtain the solution in the form

u(x, y) = (x + y)1/2 + 1

v(v − 1)xvyv, Re v > 1. (53)



Proceeding in the same way as in the establishment of solution (53), we can then show thatthe transform equations for parts (a) and (b) are as follows, respectively

U(s1, s2) = (3/2)(s1/21 + s

1/22 )

(s1s2)3/2− 1

s1s2(s1/21 + s

1/22 )

,

U(s1, s2) = (3/2)(s1/21 + s

1/22 )

(s1s2)3/2− 1

s1s2(s1/21 + s

1/22 )

+∑n−1

i,j=1i<j

sisj

(s1s2)n+1(s1 + s2).

Therefore,

u(x, y) = (x + y)1/2,

u(x, y) = (x + y)1/2

+

xyn+1

(1) (n)2F1

[1, −1;n + 2; yx

]+ x2yn

(2) (n − 1)2F1

[1, −2;n + 1; yx

]

+ · · · + xn+1y

(n) (1)2F1

[1, −n;

3; yx

]if y > x

xn+1y

(n) (1)2F1

[1, −n;3; xy

]+ xny2

(n − 1) (2)2F1

[1, 1 − n;4; xy

]

+ · · · + xyn+1

(n) (1)2F1

[1, −1;n + 2; xy

]if y < x,

by virtue of Example 2.1 and the following inversion formula given in the unpublishedmonograph by Dahiya is

L−12 {sk−µ−1

2 sk−v−12 (s1 + s2)

−k; x, y}

=

xµ−kyv

(µ − k) (v + 1)2F1

[k, k − µ;

v + 1; yx

]if y > x µ, v > −1.

xµyv−k

(µ + 1) (v − k)2F1

[k, k − v;

µ + 1; xy

]if y < x.


(a)

f (x, y) = xv−1yv − xvyv−1

(x + y)v+1/2and

α(x) = 0 = β(y),

θ(x) = x−1/2 and δ(y) = y−1/2.



(b)

f (x, y) = xv−1yv − xvyv−1

(x + y)v+(3/2)and

α(x) = 0 = β(y),

θ(x) = x−(1/2) and δ(y) = y−1/2.

We only present the solution of part (b) completely, the solution of part (a) is similar to thatof part (b), hence the details of part (a) have been omitted.

(b) Applying the double Laplace transform to the terms of Examples 3.1(b) and 3.2(b) withthe aid of Eqs. (6) and (7), Formula 47 in Voelker and Doetsch [21, p. 159] and Formula 181in Brychkov et al. [1, p. 300], finally we arrive at

(s1 − s2)U(s1, s2) − π1/2

s1/22

+ π1/2

s1/21

= π (2v + 1)

22v (v + 3/2)

{∫ ∞

s1

dλ

(λ1/2 + s1/22 )2v+1

−∫ ∞

s2

dλ

(s1/21 + λ1/2)2v+1

}.

Evaluating the integrals and simplifying, we get

U(s1, s2) = π1/2

(s1s2)1/2(s1/21 + s

1/22 )(s1 + s2)

+ π (2v + 1)

v22v (v + 3/2)· s

1/21 − s

1/22

(s1/21 + s

1/22 )(s1 + s2)

,

(54)

where Re[s1/21 + s

1/22 ] > 0.

The inversion of Eq. (54) will be accomplished from Formula 42 inVoelker and Doetsch [21,p. 186], Example 2.1 and Formula 181 in Brychkov et al. [1]

u(x, y) = 1

v

∫ x

0

v(x + y − 2ξ)v+1 + [(x − ξ)(y − ξ)]v

(x + y − 2ξ)v+(3/2)dξ if y > x

∫ y

0

v(x + y − 2ξ)v+1 + [(x − ξ)(y − ξ)]v

(x + y − 2ξ)v+(3/2)dξ if y < x.

(55)

Substituting x + y − 2ξ = t in Eq. (5), we obtain

u(x, y) =

(x + y)1/2 − (y − x)1/2

− 1

v22v+1

∫ y−x

x+y[t2 − (x − y)2]vt−v−3/2 dt if y > x,

(x + y)1/2 − (x − y)1/2

− 1

v22v+1

∫ x−y

x+y[t2 − (x − y)2]vt−v−3/2 dt if y < x, Re v > 0.

(56)

(a) Similarly, the transform equation for part (a) is as follows:

U(s1, s2) = π1/2

(s1s2)1/2(s1/21 + s

1/22 )(s1 + s2)

+ π1/2 (v + 1)

v(s1s2)1/2(s1/21 + s

1/22 )(s1 + s2)

, Re [s1/21 + s

1/22 ] > 0.



The inversion of u(s1, s2) will be accomplished from Formula 42 in Voelker and Doetsch [21,p. 186] and Relation (2.12′) in Example 2.1′

u(x, y) = 1

v

∫ x

0

v(x + y) − 2ξ)v + [(x − ξ)(y − ξ)]v

(x + y − 2ξ)v+(1/2)dξ if y > x

∫ y

0

v(x + y − 2ξ)v + [(x − ξ)(y − ξ)]v

(x + y − 2ξ)v+(1/2)dξ if y < x.

Remark 3.1 One may easily check that the condition of compatibility (51) holds true for allIBVPs given in Example 3.1.Remark 3.2 If we let v = 1, in part (a) and part (b) of Example 3.2, we obtain the followingsolutions, respectively

u(x, y) =

(x + y)1/2 − (y − x)1/2 − 1

12[(y − x)3/2 − (x + y)3/2]

−1

4(x − y)2[(y − x)−1/2 − (x + y)−1/2] if y > x,

(x + y)1/2 − (x − y)1/2 − 1

12[(x − y)3/2 − (x + y)3/2]

−1

4(x − y)2[(x − y)−1/2 − (x + y)−1/2] if y < x.

u(x, y) =

5

4[(x + y)1/2 − (y − x)1/2]

− 1

12(x − y)2[(y − x)−3/2 − (x + y)−3/2] if y > x,

5

4[(x + y)1/2 − (x − y)1/2]

− 1

12(x − y)2[(x − y)−3/2 − (x + y)−3/2] if y < x.


f (x, y) = (xy)(x − y)

(x + y)3/2exp(− xy

x + y) and

α(x) = exp(−x), β(y) = exp(−y)

θ(x) = 0 = δ(y).

We begin to solve the IBVP by taking the two-dimensional Laplace transformation of eachterm of Eqs. (49) and (50), with the help of Eqs. (6) and (7), and also using the Relation (45′) inExample 2.6′ and with the aid of Formula 47 in Voelker and Doetsch [21, p. 159], we arrive at

(s1 − s2)U(s1, s2) = (s21 − s2

2 ) + (s1 − s2)

(s1 + 1)(s2 + 1)+ π(s

1/21 − s

1/22 )

(s1s2)1/2

∫ ∞

s1/21 +s

1/22

t

(1 + t2)2dt

or, equivalently

U(s1, s2) = 1

(s1 + 1)(s2 + 1)+ 1

(s1 + 1)(s2 + 1)(s1 + s2)

+ π

2(s1s2)1/2(s1/21 + s

1/22 )[1 + (s

1/21 + s

1/22 )](s1 + s2)

.



Making use of Relations (1.2) in Ditkin and Prudnikov [10, p. 100] together with Relation(201) in Brychkov et al. [1, p. 303] and finally with the aid of Formula 42 in Voelker andDoetsch [21; p. 186], we obtain the following solution

u(x, y) = exp[−(x − y)]

+

∫ x

0 exp[−(x + y − 2ξ)] dξ

+ ∫ x

0

1

(x + y − 2ξ)1/2γ

(1,

(x − ξ)(y − ξ)

(x + y − 2ξ)

)dξ if y > x∫ y

0 exp[−(x + y − 2ξ)] dξ

+ ∫ y

0

1

(x + y − 2ξ)1/2γ

(1,

(x − ξ)(y − ξ)

(x + y − 2ξ)

)dξ, if y < x.

Hence,

u(x, y) = 1

2

exp[−(y − x)] + [exp −(x + y)]

− ∫ (y−x)

(x+y)t−1/2γ

(1,

t2 − (x − y)2

4t

)dt if y > x

exp[−(x − y)] + exp[−(x + y)]

− ∫ (x−y)

(x+y)t−1/2γ

(1,

t2 − (x − y)2

4t

)dt if y < x.

Remark 3.3 It is easy to check that the condition of compatibility (51) holds true for theIBVPs in Example 3.3.

References

[1] Brychkov, Y. A., Glaeske, H. J., Prudnikov, A. P. and Tuan, V. K. (1992). Multidimensional IntegralTransformations. Gordon and Breach, Philadelphia.

[2] Churchill, R. V. (1972). Operational Mathematics, 3rd ed., McGraw-Hill Book Company, New York.[3] Dahiya, R. S. (1967). Certain theorems on n-dimensional operational calculus. Compositio Mathematica.

Amsterdam, Holland, 18 Fasc, 1,2, pp. 17–24.[4] Dahiya, R. S. (1977). Computation of n-dimensional Laplace transforms. Journal of Computational and Applied

Mathematics, 3(3), 185–188.[5] Dahiya, R. S. (1981). Calculation of two-dimensional Laplace transforms pairs-I, Simon Stevin. A Quarterly

Journal of Pure and Applied Mathematics, 56 , 97–108.[6] Dahiya, R. S. (1983). Calculation of two-dimensional Laplace transforms pairs-II, Simon Steven. A Quarterly

Journal of Pure and Applied Mathematics (Belgium), 57, 163–172.[7] Dahiya, R. S. (1985). Laplace transform pairs of n-dimensions. Internat. J. Math. and Math. Sci., 9, 449–454.[8] Dahiya, R. S. and Debenath, J. C. (1989). Theorems on multidimensional Laplace transforms for solution of

boundary value problems. Computers Math. Applications, 18(12), 1033–1056.[9] Dahiya, R. S. and Vinayagamorthy, M. (1990). Laplace transform pairs of dimensions and heat conduction

problem. Math. Comput. Modeling, 13(10), 35–50.[10] Ditkin, V. A. and Prudnikov, A. P. (1962). Operational Calculus In Two Variables and its Applications. Pergamon

Press, New York.[11] Estrin, T. A. and Higgins, T. J. (1951). The solution of boundary value problems by multiple Laplace

transformations. Journal of the Franklin Institute, 252(2), 153–167.[12] Jaeger, J. C. (1940). The solution of boundary value problems by a double Laplace transforms. Bull. Amer. Math.

Soc., 46, 687–693.[13] Magnus, W. and Oberhettinger. (1966). Formulas and Theorems for Special Functions of Mathematical Physics.

Springer-Verlag, New York.[14] Roberts, G. E. and Kaufman. (1966). Tables of Laplace Transforms. W.B. Saunders Company, Philadelphia and

London.[15] Pipes, L. S. (1939). The operational calculus. (I), (II), (III), Journal of Applied Physics, 10, 172–180, 258–264

and 301–311.



[16] Royden, H. L. (1988). Real Analysis, 3rd ed., Macmillan Publishing Company, New York.[17] Saberi-Nadjafi, J. (1992). Theorems on N-dimensional Inverse Laplace transformations. Proc. of the Eighth

Annual Conference On Applied Mathematics, Oklahoma, pp. 317–330.[18] Saberi-Nadjafi, J. and Dahiya, R. S. (1992). Certain theorems on N -dimensional laplace transformations and

their applications. Proc. of Eighth annual Conference On Applied Mathematics, Oklahoma, pp. 245–258.[19] Snedden, I. N. (1972). The Use of Integral Transforms. McGraw-Hill Book Company, New York.[20] Stromberg, K. R. (1981). Introduction to Classical Real Analysis. Wadsworth International Group, Belmount,

California.[21] Voelker, D. and Und Doetsch, G. (1950). Die Zweidimensionale Laplace Transformation. Verlag Birkuauser

Basel.


integral transforms and special functions theorems on n

Documents