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
R∞
L[f (t); s] = 0 e−st f (t)dt to n-dimensional is given by
Z
Ln [f (t̄); s̄] =
∞
0
Z
∞
0
Z
···
0
∞
exp(−s̄ · t̄)f (t̄)Pn (dt̄)
where t̄ = (t1 , t2 , . . . , tn ), s̄ = (s1 , s2 , . . . , sn ), s̄ · t̄ =
n
P
i=1
n
Q
si ti and Pn (dt̄) =
dtk .
k=1
In addition to the notations introduced above, we will use the following throughout this article. Let tv = (tv1 , tv2 , . . . , tvn ) for any real exponent v and let pk (t) be
the k-th symmetric polynomial in the components tk of t. Then
(i) p1 (tv ) = tv1 + tv2 + · · · + tvn
(ii) pn (tv ) = tv1 · tv2 . . . tvn .
2. Main Results
Theorem 2.1. Suppose that f (x) and f (x2 ) are functions of class Ω. Let
L{f (x); s} =
φ(s),
1
− 32
L{x φ x ; s} = ξ(s),
1
L{x− 2 ξ x12 ; s} = ζ(s), and
L{f (x2 ); s} = G(s),
3
1
u
− 12
− 32
where x− 2 φ x1 and
x
are
also
functions
of
class
Ω,
x
f (u)
ξ
exp
−sx
−
2
x
x
1
1
3
2
and u− 2 x− 2 exp −sx − 2ux
f (u) belong to L1 [(0, ∞) × (0, ∞)].
(i)
(ii)
(iii)
(iv)
1991 Mathematics Subject Classifications: 44A30.
Key words and phrases: multi-dimensional Laplace transforms.
c
1999
Southwest Texas State University and University of North Texas.
Published November 24, 1999.
61
R. S. Dahiya & Jafar Saberi-Nadjafi
62
Then




n−2
2
1
π
1 G[2p1 (x−1 )]; s =
p x2

2
n
Ln
1
p1 s 2
2
1
1 ζ P1 s 2
,
·
pn s 2
(2.1)
h 1 i
where Re p1 s 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
Z
∞
Z
∞
ξ(s) =
x
0
− 32
0
u
exp −sx −
f (u)du dx.
x
(2.2)
Next we wish to interchange the order of the integral on the right side of (2.2). The
3
integrand x− 2 exp −sx − ux f (u) belongs to L1 [(0, ∞) × (0, ∞)], that, by Fubini’s
Theorem, interchanging the order of the integral on the right of (2.2) is permissible.
Therefore,
Z ∞
Z ∞
h
3
ui
ξ(s) =
f (u)
x− 2 exp −sx −
dx du
(2.3)
x
0
0
We then use a well-known result in Robert and Kaufman [15] on the right side of
(2.3) to obtain
Z ∞
1
1
1
1
ξ(s) = π 2
(2.4)
u− 2 exp −2u 2 s 2 f (u)du.
0
Using (2.4) and (iii), it follows that
ζ(s) = π
1
2
Z
∞
Since x
∞
x
0
− 12
Z
u
− 12
− 12
u
− 12
0
exp −sx −
2u 12
exp −sx −
x
1
2u 2
f (u)du dx., where Re s > λ1 .
(2.5)
belongs to L1 [(0, ∞) × (0, ∞)]; therefore, accord-
x
ing to Fubini’s Theorem, (2.5) can be rewritten as
1

Z
∞
ζ(s) = π 2
0
1

u− 2 f (u) 

Z
∞
0
3
2
1
4
2 u
1

x− 2 exp −sx −
4x
2 

 
 dx du, where Re s > λ1 .
From the tables of Roberts and Kaufman [15], we obtain
Z
1
2
∞
s ζ(s) = π
0
3 1 1
1
u− 2 f (u) exp −2 2 u 4 s 2 du.
(2.6)
Next, we substitute u = v 2 in (2.6) to obtain
1
2
Z
∞
s ζ(s) = 2π
0
3 1 1
exp −2 2 s 2 v 2 f (v 2 )dv.
(2.7)
n-dimensional Laplace transforms
63
1
h 1 i2
Replacing s by p1 s 2
, multiplying both sides of (2.7) by pn s 2 , we obtain
Z
1 1 h 1 i
2
2
2
p1 s pn s ζ p1 s
= 2π
∞
0
1
h 3 1 1i
pn s 2 exp −2 2 p1 s 2 v 2 f (v 2 )dv
(2.8)
Now we use the operational relation given in Ditkin and Prudnikov [11]
1
1
a2
− 12
2
2
si exp −asi + (πxi ) exp −
for i = 1, 2, . . . , n
4xi
Equation (2.8) reads as follows
1 1 1 2 n
pn s 2 p1 s 2 ζ p1 s 2
=
n
2
π
n−2
2
1
pn x 2
Z
∞
0
(2.9)
exp −2vp1 x−1 f (v 2 )dv
(2.10)
Applying (iv) in (2.10), we arrive at
1 1 1 2 n
pn s 2 p1 s 2 ζ p1 s 2
=
n
i
h
−1
.
G
2p
x
1
n−2
1
π 2 pn x 2
2
Therefore,
1


n−2


i
h
s2
p
2
1
1
π 2
1
1
1 G 2p1 x 2 ; s =
1 ζ p1 s 2
Ln
,
·
p x2

2
pn s 2
n
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.
τ −1
Example 2.1. Assume that f (x) = x 4 . Then
Γ τ +3
4
φ(s) =
, Re s > 0, Re v > −3;
τ +3
s 4 τ +1
Γ τ +3
Γ
4
4
ξ(s) =
, Re τ > −1, Re s > 0, and
τ +1
s 4
Γ τ +3
Γ τ +1
Γ τ2 + 1
4
4
ζ(s) =
, Re s > 0, Re τ > −1.
τ
s 2 +1
Γ τ +1
2
G(s) =
, Re τ > −1.
τ +1
s 2
Therefore,




1
Ln
;
1h i τ +2

2

 pn x 2 p1 x−1
h
where Re τ > −1, Re p1 s
1
2




s



=π
n−1
2
Γ
τ
3
+
2 2
·
1
1 h 1 iτ +2
pn s 2 p1 s 2
(2.11)
i
> 0, and n = 2, 3, . . . , N .
R. S. Dahiya & Jafar Saberi-Nadjafi
64
1
Example 2.2. Suppose that f (x) = I0 (ax 2 ). Then
2
1
a
φ(s) = exp
, Re s > 0,
s
4s
1
ξ(s) =
π2
2
1 , Re s > Re
a2
,
4
s − a4 2
(
"3
# "3
#)
2
; s2
; s2
Γ 14
s
4π
4π
4
4
ζ(s) = 5 .
2 1 F2 1 5 2 −
1 F2
3 7 a2
a
6π
a2
Γ 14
,
;
,
;
2 4
2 4
1
G(s) =
1 , Re s > |Re a|.
(s2 − a2 ) 2
So that



1



1
i1 ; s

 pn (x ) 4p2 x−1 − a2 2 

1
1 
1 
n
2
3
2
2π 2 p1 s 2  4π
Γ
p
;
1 s
4
−
= 5 1
F
2
1
2
1 5
a2
a 2 pn s 2  Γ 14
2, 4;
1  

2
5

s2
p
;
1
4


;s ,
1 F2
3 7

a2
2, 4;
h 1 i
where Re p1 s 2
> |Re a|.
"
#
(a)p ;
Example 2.3. Consider f (x) = p Fq
cx . Then
(b)
;
q
"
#
(a)p ;
φ(s) = 1s p+1 Fq
cx ,
(b)q ;
where p ≤ q, Re s > |Re c|.
"
#
1
(a)p , 1, 12 ; c
π2
ξ(s) = 1 p+2 Fq
,
s2
(b)q
;s
Ln
1
2
·h
1
4
2
1
p21 s 2
(2.12)
6π
where p ≤ q − 1, Re τ > 0 if p + 1 < q, Re s > Re c if p + 1 = q,
"
#
(a)p , 1, 12 , 34 , 54 ; 4c
π
ζ(s) = 3 p+4 Fq
,
2s 2
(b)q
; s2
1
where p ≤ q − 3; Re s > 0 if p ≤ q − 4; and Re(s + 2c 2 cos πr) > 0
(r = 0, 1) if p = q − 3.
Hence


 


(a)p , 1, 12 ;
1
c
1h i p+2 Fq 
;s
Ln
 p x 2 p x−1

(b)q
; p21 x−1
n
1


n
1 3 5
(a)
,
1,
,
,
;
2
p
π
4c
2 4 4
1 1 p+4 Fq 
1,
=
(b)q
; p41 s 2
2pn s 2 p21 s 2
(2.13)
n-dimensional Laplace transforms
h 1
h 1 i
i
1
> 0 if p ≤ q − 4; and Re p1 s 2 + 2c 2 cos πr
where p ≤ q − 3, Re p1 s 2
> 0(r = 0, 1) if p = q − 3.
1
65
1
Example 2.4. Assume that f (x) = x 2 J0 (ax 2 ). Then
"3
#
1
π2
a2
2;
φ(s) = 3 1 F1
−
, Re s > 0,
4s
2s 2
1;
"3
#
1
π2
a2
a2
2 , 1;
ξ(s) =
−
F
,
Re
s
>
−Re
−
,
2 1
2
4s
4
1 ;
!
1 3
1
1
a2 4 , 4 , 0, − 2
1,4
ζ(s) = 1 1 G4,2
,
s2 π 2 s2
0, 0
where Re s > 0, | arg a| < 2π.
"3
#
1
2
,
1;
π2
a
a2
2
G(s) =
−
, Re s > Re − .
2 F1
2
4s
4
1 ;
Hence



x−1



p1
h
i3 , s

 p x 12 4p2 x−1 + a2 2 

n
1
1


n−3
1 3
1
π 2 p1 s 2
2
,
,
0,
−
a
2 
 2 4
1 G1,4
=
,
4,2
1
4
0,
0
pn s 2
p1 s 2
Ln
(2.14)
h 1 i
where Re p1 s 2
> 0, | arg a| < 2π and G1,4
4,2 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
(
)
τ
τ
(xy) 2
1
1
L2
= π2Γ
(2.15)
+1 ·
1
τ +1
τ +1 ; s1 , s2
1
1
2
(x + y) 2
(s1 s2 ) 2 s12 + s22
h 1
i
1
where Re τ > −1, Re s12 + s22 > 0.
(
L2
1
(xy) 2
)
1 ; s1 , s2
[4(x + y)2 − (axy)2 ] 2

1

2 
1
1
1

2
2
2
2
3

2π s1 + s2
4π
 4 ; s1 + s2 
=

1 2 1 F2  1 5
5
1
a2
a 2 (s1 s2 ) 2 
, ;
 Γ 4
2 4

2
2 
1
1
1
1 2 12

2
2
5

Γ 4
s1 + s22
+
s
s
1
2
4 ;

−
F


1 2
3 7

6π
a2

2, 4;
(2.16)
R. S. Dahiya & Jafar Saberi-Nadjafi
66
h 1
i
1
where Re s12 + s22 > 0, | arg a| < 2π.
(
"
#
)
1
(a)p , 1, 12 ; c(xy)2
(xy) 2
L2
; s1 , s2
p+2 Fq
2
x+y
(b)q
; (x + y)


1 3 5
π
4c
 (a)p , 2 , 4 , 4 ;

=
F

q
1
4  ,
p+4
2
1
1
1
1
(b)q
;
s12 + s22
2(s1 s2 ) 2 s12 + s22
(2.17)
h 1
i
h 1
i
1
1
1
where p ≤ q − 3, Re s12 + s22 > 0 if p ≤ q − 4; and Re s12 + s22 + 2c 2 cos πr
> 0(r = 0, 1) if p = q − 3.
(
L2
)
3
2
(xy) (x + y)
3
[4(x + y)2 + (axy)2 ] 2
h
1
2
1
2
i
; s1 , s2
where R s1 + s2 > |Re a| and


1
3
1
2 , 2 , 0, − 2 
a2
s + s2 1,4 
= 1
,
1 G4,2  1
4
1
(πs1 s2 ) 2
0,
0
2
2
s1 + s2
1
2
1
2
(2.18)
G1,4
4,2
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
1
1
1
2
2
2
n
o
s
π
+
s
+
s
s
1
2
1 2
1
1
L2 (x + y) 2 ; s1 , s2 =
(2.19)
1
3
2(s1 s2 ) 2 s12 + s22
Theorem 2.2. Assume that f belongs to class Ω and φ be the one-dimensional
Laplace transformation of f . Let
n 1
o
(i) L x− 2 φ x1 ; s = γ(s),
−v
d
(ii) − ds
s γ s12 = ζ(s),
(iii) L {xf (x); s} = H(s),
−v
1
d
and suppose that x− 2 φ x1 belongs to Ω and ds
s γ s12
exists for Re s > c1
where c1 is a constant. Then
i
h i 
h 
 (v − 1)φ p1 x−1 − 2p1 x−1 H p1 x−1

1
Ln
;s =


pn x 2
n−1
−1
π 2
1
1 h 1 iv ζ p1 s 2
,
pn s 2 p1 s 2
(2.20)
h 1 i
where Re p1 s 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
Z ∞ Z ∞
1
1
u
γ(s) = π 2
x− 2 exp −sx −
f (u)du dx, where Re s > c1 .
x
0
0
(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
Z ∞
1
1
1
1
2
γ(s) = π
(2.22)
f (u)s− 2 exp −2u 2 s 2 du.
0
Taking into account the condition (ii) we see that the equation (2.22) implies that
!
Z ∞
Z ∞
1
2
1
1
1
2u
v
−1
2
2
s ζ(s) = π (v − 1)
f (u) exp −
u 2 f (u)
du − 2π s
s
0
0
!
1
−2u 2
exp
(2.23)
du, where Re s > c1 .
s
h 1 i−1
Now. replacing s by p1 s 2
and then multiplying each side of (2.23) by
1
pn s 2 and then making use of operational relations (2.9) and (2.24)
a 1 3
1
a2
−2 −2
2
si exp −asi + π xi exp −
for i = 1, 2, . . . , n,
(2.24)
2
4xi
equation (2.23) reads
1 h 1 i−v 1 −1 n
pn s 2 p1 s 2
ζ p1 s 2
=
n
1
n−1
2
Z
(v − 1)
1
∞
f (u)
0
pn x 2
h
i
Z ∞
i
h
exp −up1 x−1 dx − 2p1 x−1
uf (u) exp −up1 x−1 du.
π
0
(2.25)
Equation (2.25) with (iii) and the definition of the one-dimensional Laplace transform, leads to
1 h 1 i−v 1 −1 pn s 2 p1 s 2
ζ p1 s 2
n
=
n
Therefore,
i
h io
n
h −1
−1 H p
−1
−
2p
.
(v
−
1)φ
p
x
x
x
1
1
1
n−1
1
π 2 pn x 2
1
i
h i 
h 
 (v − 1)φ p1 x−1 − 2p1 x−1 H p1 x−1

1
Ln
;s


pn x 2
n−1
−1
π 2
1
1 h 1 iv ζ p1 s 2
=
,
pn s 2 pn s 2
h 1 i
where Re p1 s 2
> d for some constant d, n = 2, 3, . . . , N .
R. S. Dahiya & Jafar Saberi-Nadjafi
1
Example 2.6. Let f (x) = J0 2x 2 . Then
68
1
1
φ(s) = exp −
, Re s > 0,
s
s
1
π2
γ(s) =
Thus
3
2(s + 1) 2
, Re s > −1.
1
ζ(s) =
π 2 (vs2 + v − 3)
5
2sv−2 (1 + s2 ) 2
, Re s > −1.
Therefore




 


 
1
1
1
−1;




Ln
exp −
(v − 1) − 21 F1
;s
1; p x−1  
 p x−1 p x 12

p1 x−1
1
n
1
1h
1 i
n
h 1 i
π 2 p1 s 2 v + (v − 3)p21 s 2
=
> −1.
,
where
Re
p1 s 2
1h
1 i 52
2
2
2
1 + p1 s
2pn s
(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
(
)
1
(xy) 2
xy
−1; xy
(i) L2
exp −
; s1 , s2 =
1 F1
1; x + y
(x + y)
x+y
1
1
2
1
1
2
2
2
2
π s1 + s2
2 s1 + s2
−1
(2.260 )
52 ,
2
1
1
1
4(s1 s2 ) 2 1 + s12 + s22
h 1
i
1
where Re s12 + s22 > −1.
(
(ii)
)
1
(xy) 2
xy
−1; xy
L2
exp −
1 − 1 F1
; s1 , s2 =
1; x + y
(x + y)
x+y
1
1
3π s12 + s22
,
1
2 52
1
1
2
2
2(s1 s2 ) 2 1 + s1 + s2
(2.2600 )
h 1
i
1
where Re s12 + s22 > 1.
Notice, with the help of (2.260 ) from (2.2600 ) we arrive at the following result
1
1
(
)
2
2
1
+
s
s
1
2
(xy) 2
π
xy
000
L2
exp −
; s1 , s2 =
23 . (2.26 )
(x + y)
x+y
2
2
1
1
1
(s1 s2 ) 2 1 + s12 + s22
n-dimensional Laplace transforms
69
This is the same as the result (2.109) in Ditkin and Prudnokov [11; p. 140].
Furthermore, with the help of (2.26000 ) and the operational relation (47) in
Voelker and Doetsch [24; p. 159], we derive the following new results.
L2
1
y 2
x
h
1
xy
·
exp −
; s1 , s2 =
x+y
x+y
1
2
Re s12 + s22 > 1.
( 1
)
x 2
1
xy
L2
·
exp −
; s1 , s2 =
y
x+y
x+y
1
2
1
2
1
2
1
2
2 12
,
(2.26iv )
1
2
s1
h
s2 1 + s1 + s2
i
1
1
π
π
1
2 12
1
2
2
1 + s1 + s2
,
(2.26v )
i
where Re s1 + s2 > 1.
1
1
Example 2.7. Assuming x− 2 cos 2x 2 , we obtain
π 12
1
exp −
x
, Re s > 0.
s
π
γ(s) =
, Re s > −1.
s+1
π 12 s−v+1
2
ζ(s) = vs + v − 2
, Re s > −1.
(s2 + 1)2
1
π2 s
1
H(s) = 5
− 1 exp −
, Re(s) > 0.
s
s2 2
φ(x) =
Therefore,
Ln


−1
−1

− p1 x
− 2 exp −
 (v − 1)p1 x



h
i 32
pn x−1
1
p1 ((x−1 )




;s



p1
n−1
1
n
1
o
h 1 i
π 2 p1 s 2
2
2
=
(v
−
2)p
+
v
,
Re
p1 s 2
> 0.
s
1h
1 i2
1
2
2
2
(2.27)
1 + p1 s
pn s
x−1
Remark 2.3. If we let n = 2 and v = 1 in (2.27) by using the following well-known
formula
1
1
1
2
2
2
π s1 + s2
xy
xy
L2
−
; s1 , s2 =
3 exp
1
2 2 ,
x+y
1
(x + y) 2
1
(s1 s2 ) 2 1 + s12 + s22
h 1
i
1
Re s12 + s22 > 0.
R. S. Dahiya & Jafar Saberi-Nadjafi
70
we derive that
L2
1
(x + y)
1
2
exp −
xy
x+y
; s1 , s2
h 1
i
1
Re s12 + s22 > 0.
1
1
1
π 2 s12 + s22
=
1
2 ,
1
1
2
2
(s1 s2 ) 2 1 + s1 + s2
(2.270 )
3. Non-Homogeneous Second Order Partial
Differential Equations of Parabolic type
In this section we solved a few partial differential equations of the type
uxx + 2uxy + uyy + u = f (x, y), 0 < x < ∞, 0 < y < ∞,
(3.1)
under the following initial and boundary conditions
u(x, 0) = u(0, y) = uy (x, 0) = ux (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
1
(a) f (x, y) = (x + y) 2
τ
(b) f (x, y) =
(xy) 2
τ +1
(x+y) 2
τ
(c) f (x, y) =
(xy) 2
τ +2
(x+y) 2
We will use the following for the rest of this section. If
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 if their one-dimensional Laplace transformations are F (s1 ), G(s2 ), F1 (s1 ) and
G1 (s2 ), respectively, then
Z
L2 {u(x, y); s1 , s2 } =
L2 {uxx ; s1 , s2 } =
L2 {uyy ; s1 , s2 } =
∞
0
Z
0
∞
exp(−s1 x − s2 y)u(x, y)dxdy = U (s1 , s2 )
s21 U (s1 , s2 ) − s1 G(s2 ) − G1 (s2 )
s22 U (s1 , s2 ) − s2 F (s1 ) − F1 (s1 )
L2 {uxy ; s1 , s2 } = s1 s2 U (s1 , s2 ) − s1 F (s1 ) − s2 G(s2 ) + u(0, 0)
(3.3)
(3.4)
(3.5)
(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),
n-dimensional Laplace transforms
71
and with the aid of Relation (2.19) in Remark 2.1, we obtain the transformed problem
h
i
1
1
2
2
i
h 1
s
+
s
+
(s
s
)
π
1
2
1 2
1
1
2
2
U (s1 , s2 ) =
,
Re
s
+
s
·
1
1
1
2 > 0.
3
(s1 + s2 )2 + 1
2(s1 s2 ) 2 s12 + s22
(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]
Z
x
u(x, y) =
0
1
(x + y − 2ξ) 2 sin ξdξ
(3.8)
By a simple change of variable x + y − 2ξ = 2t2 in (3.8), we obtain
+ 32
u(x, y) = 2
1
Z ( y−x
2
2 )
1
2
( x+y
2 )
x+y
t sin t −
2
2
2
dt if y > x.
Expanding the sine and making some simplification, we deduce that

+ 32
u(x, y) = 2
1
Z ( x−y
2
2 )
(
x+y
2
)
1
2
cos x + y
2

Z ( y−x ) 12
2
1
2
( x+y
2 )
2
2
t sin t dt − sin
x+y
2
(3.9)
t2 cos t2 dt
Calculating the integrals involved in (3.9), we arrive at
u(x, y) =

1
4
x+y
2
12
−
y−x
2
12
1
x+y
2
1
x+y
2
cos x + π 2 cos
+ π 2 sin
C
y−x
2
S
y−x
2
−D
−S
x+y
2
x+y
2



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
+1
U (s1 , s2 ) =
1
τ +1
1
τ
[(s1 + s2 )2 + 1] (s1 s2 ) 2 s12 + s22
1
π2Γ
U (s1 , s2 ) =
2τ Γ
τ +3
2
τ
2
πΓ(τ + 1)
τ +1 ,
1
1
[(s1 + s2 )2 + 1] s12 + s22
(3.10)
(3.11)
R. S. Dahiya & Jafar Saberi-Nadjafi
72
h 1
i
1
where Re s12 + s22 > 0. Thus, we obtain the following solutions, respectively
u(x, y) =
1
2τ +1
Z
y−x
t
− τ +1
2
2
τ
t − (x − y)2 2 sin
x+y
t − (x + y)
2
dt
if y > xRe v > −1.
Z y−x
2
τ
1
t − (x + y)
− τ +3
2 2
2
u(x, y) = τ +1
t − (x − y)
t
sin
dt
2
2
x+y
if y > x, Re v > −1.
(3.12)
(3.13)
Remark 2.4. Substituting τ = 0 in parts (b) and (c), lead to
uxx + 2uxy + uyy + u =
uxx + 2uxy + uyy + u =
1
1
(3.14)
3
(3.15)
(x + y) 2
1
(x + y) 2
Next, using (3.14) and (3.15), we arrive at the following explicit solutions for the
equations (3.14) and (3.15) respectively
π 12 x+y
y−x
x+y
u(x, y) =
cos
S
−S
−
2
2
2
2
x+y
y−x
x+y
sin
C
−C
if y > x.
2
2
2
x+y
y−x
u(x, y) =
cos
S
1 cos(x + y) −
1 cos y − π
2
2
(x + y) 2
(y − x) 2
x+y
x+y
y−x
x+y
−S
− sin
C
−C
if y > x.
2
2
2
2
1
1
1
2
Example 3.2. Solve the following Parabolic differential equation described by
1
xy
uxx + 2uxy + uyy + u =
−
, 0 < x < ∞, 0 < y < ∞, (3.16)
1 exp
x+y
(x + y) 2
under the initial and boundary conditions
u(x, 0) = u(0, y) = uy (x, 0) = ux (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
1
1
π s12 + s22
U (s1 , s2 ) =
(3.18)
i 12 ,
h
1
1
2
2
2
[(s1 + s2 ) + 1] 1 + s1 + s2
h 1
i
1
where Re s12 + s22 > 0. Using formula (133) in [24] the inverse transform of (3.18)
leads to the following integral representation.
2
Z y−x
t − (x + y)2
t − (x + y)
− 12
u(x, y) =
t exp
sin
dt if y > x.
4t
2
x+y
n-dimensional Laplace transforms
73
References
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R. S. Dahiya & Jafar Saberi-Nadjafi
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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