Gen. Math. Notes, Vol. 14, No. 1, January 2013, pp. 6-20
ISSN 2219-7184; Copyright © ICSRS Publication, 2013 www.i-csrs.org
Available free online at http://www.geman.in
A. Aghili
1
and H. Zeinali
2
1,2
Department of Applied Mathematics
Faculty of Mathematical Sciences,
University of Guilan,
P.O. Box- 1841, Rasht – Iran
1
E-mail: arman.aghili@gmail.com
2
E-mail: homa_zeinaly@yahoo.com
(Received: 8-10-12 / Accepted: 19-11-12)
Abstract
In this work, the authors implemented Laplace transform method for solving certain partial fractional differential equations and Volterra singular integral equations. Constructive examples are also provided to illustrate the ideas. The result reveals that the transform method is very convenient and effective.
Keywords: Non-homogeneous time fractional heat equations; Laplace transform; Volterra singular integral equations.
In this work, the authors used Laplace transform for solving Volterra singular integral equations and PFDEs.

Solution to Volterra Singular Integral… 7
The Laplace transform is an alternative method for solving different types of
PDEs. Also it is commonly used to solve electrical circuit and systems problems.
In this work, the authors implemented transform method for solving the partial fractional heat equation which arise in applications. Several methods have been introduced to solve fractional differential equations, the popular Laplace transform method, [ 1 ] , [ 2 ] , [ 3 ], [ 4 ] , and operational method [ 10].
However, most of these methods are suitable for special types of fractional differential equations, mainly the linear with constant coefficients. More detailed information about some of these results can be found in a survey paper by Kilbas and Trujillo [10]. Atanackovic and Stankovic [5],[6]and Stankovic [20] used the
Laplace transform in a certain space of distributions to solve a system of partial differential equations with fractional derivatives, and indicated that such a system may serve as a certain model for a visco elastic rod. Oldham and Spanier I, [13] and [14] , respectively, by reducing a boundary value problem involving Fick’s second low in electro analytic chemistry to a formulation based on the partial
Riemann – Liouville fractional with half derivative. Oldham and Spanier [14] gave other application of such equations for diffusion problems.
K.Sharma et al, in [18],derive a solution of a generalized fractional Volterra integral equation involving K
4
−
function with the help of the Sumudo transform.
Wyss [22] considered the time fractional diffusion and wave equations and obtained the solution in terms of Fox functions.
Laplace transform of function f ( ) is as follows
If
{ ( )}
= ∫
0
∞ e
− s t f ( ) :
=
( ).
{ ( )}
=
F s , then L
−
1
{ ( )} is given by
=
1
2
π i
∫ st ( ) ,
Where F(s) is analytic in the half- plane Re ( )
> c . For n
α n ,one gets
[15],[16]
L {
C
0
D t
α f ( )}
= − n k
−
1
∑
=
0 s
α
1 f ( 0 ).

8 A. Aghili et al.
Theorem 1.1 (Effros’s Theorem [ 9 ])
L
Let
φ
(
0
∞
{ ( )}
=
F s f
and
τ =
U s
are analytic, then one has
τ τ d
τ
)
=
( ) ( ( ) ).
− τ
and assuming
Example 1.1 Let us assume that, ( )
=
1
and ( ) s
α q s
= s
α
, then one has
τ = exp( s
−
α
τ s
α
)
, which leads to
τ = t
α −
1
W (
− α α τ t
− α
).
Then, we obtain
L

 t
1
1
− α
0
∞ f
τ
W (
−
, ; t
− α
) d
τ


=
(
α
) s
α
.
Provided that the integral in bracket converges absolutely.
Like the Fourier transform, the Laplace transform is used in a variety of applications.
Perhaps the most common usage of the Laplace transform is in the solution of initial value problems. However, there are other situations for which the properties of the Laplace transform are also very useful, such as in the evaluation of certain integrals and in the solution of fractional singular integral equations of
Volterra type.
In the following, we may show some applications of integral transform in evaluating certain integrals.
Lemma 1.1 The following relations hold true
1 -
∫
2
π
0
=
2
π α
− α
∫ ϕ ϕ =
2
π
I x
2 -
2
− α
∫ ϕ + y ϕ ϕ =
2
π
I
0
( x
2 + y
2
),

Solution to Volterra Singular Integral… 9
3 -
2
π
∫
0
I
0
( cos ϕ
) d ϕ = k
∞
∑
=
0
1 k k
3 k
.
Proof. 1 – In order to show the above relation, let us introduce the function
=
2
π
∫
,
0 we first calculate the Laplace transform of
{ ( )}
=
+∞
∫ exp(
− sx dx
2
π
∫
0 0
( ) as following, ϕ ϕ
, changing the order of integration and simplifying to obtain
=
2
π d ϕ
+∞
∫ ∫ exp( sx x dx
0 0
At this point we introduce the change of variable e i ϕ = z
=
2
π
∫
0 d ϕ cos ϕ − s and simplifying to get
=
2
π
∫
0 cos d ϕ ϕ − s
= z
∫
=
1 iz
2 dz
−
2 sz
+ i
The value of the complex integral after using Cauchy integral formula is
= z
∫
=
1 iz
2 dz
−
2 sz
+ i
=
2
π s
2 −
1
=
2
π
L I x
0
2- We can rewrite the left side of the equation in the following form
I
=
2
∫
− α e x
2 + y
2
( x x
2 + y
2 cos
θ + y x
2 + y
2 d
θ
, we introduce a new variable
α such that x
= y x
2 + y
2 x
2 + y
2
= cos
α then
I
=
2
− α
e x
2 + y
2 sin(
α θ
) d
, and again introducing the new variable
I
= ∫
0
2
π e x
2 + y
2 sin ϕ d ϕ =
2
π
I
0
( x
2 + y
2
).
leads to

10 A. Aghili et al.
It is obvious that if we set y
=
0 , then we get the relationships 1 and 2.
3- Let us define a new function by the integral
= ∫
2
π
I
0
( 2 x cos
θ θ
,
0 taking Laplace transform of the above relation ,we get
{ ( )}
=
2
π
∫
(
0 e cos
θ s s
)
= s
1
2
π
∫
0 cos
θ e s d
θ =
1 1
I
0
( ).
s s
On the other hand, one has the following expansion for modified Bessel’s function of order zero
I
0
( )
= k
∞
∑
=
0 y
2 k k
2 k
, so we have
∫
0
2
π
I
0
( 2 x cos
θ θ = k
∞
∑
=
0 x
2 k k k
2 k
.
If we set x
=
1 in the above relation, we get the desired equation.
π
Lemma 1.2 Let us assume that ( ( ))
=
( )
= ∫
4 exp{
− s (sec
2
α
0
+ csc
2
α
)} d
α
.
Show that the following relations hold true
1 -
2 -
=
1
−
4
,
=
2 t t
π erf ( 2 s ),
2
3 -
∫
2
0
π tan
α −
2 )
= − e
2
Ei
− where for x
>
0
( )
= ∫ x
−∞ e t t
Γ dt is exponential integral and the relation z x is incomplete Gamma function.
Ei (
− x )
= −Γ
(0, )
Proof: 1- We have
=
π
4
0 e
−
(
4 s
θ
) d
,

Solution to Volterra Singular Integral… 11 and consequently applying Bromwich's integral, we get
π
=
1
2
π i
4
0 e
−
1 sin 2
θ
) d
θ
) e ds , changing the order of integration, leads to
=
π
0
∫
4
δ
( t
−
4
θ
) d
θ
.
Now, we introduce the new variable t
−
4
θ
= w , then we get
=
1
2 t t
−
4
.
2- We take Laplace transform of the above equation to obtain
=
+∞
∫
4 e
− st
2 t t
−
4 dt .
Now we introduce the new variable t
{ ( )}
= e
−
4 s
∫
+∞
0 e
− su
2 u
2 +
4 du .
At this point, let us assume that
= ∫
+∞
0
4 u
2
to get u e
− su
2
2 +
4 du , (0)
=
′ = − ∫
+∞
0
= −
1
2
2 u e
− su
2 u
2 +
4
π s
+ du
= −
+∞
∫
0
− su
2 e du
+
4
∫
+∞
0 u e
− su
2
2 +
4
Solving the above ODE we obtain
( )
= e
4 s
π
4 erf (2 s ), and consequently
=
π
4 erf (2 s ).
π
4 then

12 A. Aghili et al.
Lemma 1.3 If ( ) and ϕ x are Laplace transformable functions then we have the following relationship where K s
Φ s
L 

∫
+∞ x
(
− ϕ


=
K ( s ) ( ), are Laplace transforms of functions k (
− x ϕ x respectively.
Proof: See [ 9 ][ 17 ].
Laplace transform can be used to solve certain types of Volterra singular integral equations.
Problem 2.1 Let us consider fractional Volterra singular integral equation of the form, c
α
( )
=
( )
+ in which
λ ∫
+∞
(
−
) ( ) , f ( 0 )
=
0, (2.1)
( , )
=
( x
−
) is the kernel and g x is assumed to be a Laplace transformable function.
Then (2.1) has the formal solution
=
1
2
π i
∫
{
λ
K (
− − s
α
} e ds
Solution: Let transforms of
( ( ))
=
( ) , ( ( ))
=
( ), ( (
− x ))
=
( ) be the Laplace
( ), ( ), (
− x ) , respectively, then by using Lemma 1.3 one gets the following relationship,
α
( )
=
( )
+ λ
K (
−
) ( ).
(2.2)
So one can write,
=
λ
−
K (
− − s
α
,
(2.3) and consequently by Bromwich's integral we get the following relation,
=
1
2
π i
∫
{
λ
−
K (
− − s
α
} st e ds ,
(2.4) which can be solved by the use of Residue theorem.
Note that F(s) is analytic in the half plane Res
> c .
Example 2.1: Solve the following singular integral equation c
α
( )
= exp(
− ax )
+ λ ∫
+∞
J
0
( 2 (
−
) ) ( ) , f ( 0 )
=
0.
(2.5) x

Solution to Volterra Singular Integral… 13
Solution: Laplace-transform of the above integral equation leads to
1
= s
1
+ a
λ e s s and consequently
(2.6)
=
1 s
+ a
1
λ e s
+ s s
α
. (2.7)
Using Bromwich's integral yields
=
2
1
π i
∫ se sx
( s
+ a )(
λ e
1 s
+ s
α +
1
) ds .
(2.8)
Now let us consider the case:
α =
0.5
, then relation (2.8) becomes
=
1
2
π i
∫ e sx
(
+ a )(
λ
1
−
3 e s 2
+
1 ) ds .
(2.9)
So we may apply Laplace transform of convolution of functions and using the fact that
L
−
1
{
1
(
+ a )
; s
→ =
0
x e a (
η − x )
π x d
η
, (2.10) and also the following relationship
L
−
1
{
1
+ λ
1
1
−
3
2
}
=
L
−
1
λ
1
−
3
2
=
L
−
1
{1
+
∞
( 1) k
λ k k
∑
=
1
λ
and,
L
−
1
{1
∞
+
( 1) k
∑
=
1 k
λ k k
−
3 k
2 k
1
−
3
2 )
2 −
...}
=
−
3
2 k
},
}
= δ
( )
+ k
∞
∑
=
1
− k
λ k x
( ) k
3 k
−
2
4 I
3 k
−
2
(2
2 kx ).
(2.11)

14 A. Aghili et al.
From relationships (2.9)-(2.11), one gets the formal solution as follows
( )
=
{
∫ x
0 e a (
η − x )
π x d
η δ + k
∞
∑
=
1
− k
λ k x
( ) k
3 k
−
2
4 I
3 k
−
2
2
(2 kx )}, which can be calculated as bellow
( )
= x
∫ ∫ w
0 0 e a (
η − w )
π w d x
− w )
+
∞
( 1) k
λ k
( k
∑
=
1 x
− w
) k
3 k
−
2
4 I
3 k
−
2
(2 (
− w ) )} dw , (2.12)
2
Problem 2.2 Solving the system of fractional singular integral equations of the form,





 c c
D
α φ
( )
=
( )
D
α ψ
( )
=
( )
−
+
λ
λ
+∞
∫ x x
+∞
∫
(
− ψ
(
− φ
, (2.13) with conditions
φ ψ
Solution: Multiplying second equation by i and adding to the first equation leads to c
D
α
(
φ ψ
)
=
( g
+
)( )
+ i
λ ∫
+∞
(
−
)(
φ ψ
)( ) .
x
(2.14)
Now let ( i )( )
= ξ
( ), (
+
)( )
=
( ),
λ γ
, then one can rewrite the above equation in the form c
D
α ξ
( )
=
( )
+ γ
(
− ξ ∫
+∞ x
At this point, we can apply previous example to this one as bellow.
Taking Laplace transform of equation (2.15) leads to s
α Φ
( )
=
( )
+ γ
K (
− Φ where
(2.15)
Φ
( ), ( ), ( ) are Laplace transforms of functions
ξ
( ), ( ), (
− x ) respectively. Hence, one gets the following relationship

Solution to Volterra Singular Integral… 15
Φ
( )
=
α
λ
2
( K
+
(
λ
−
K s
(
))
2
−
+
) ( ) s
2
α
+ i
α
λ
2
( )
( K
+
(
λ
− s
K (
−
) ( )
))
2 + s
2
α
, where ( ), ( ) are Laplace transforms of g x h x respectively.
So one gets
φ ɶ =
α
λ
2
( )
+
( (
λ
− s
K
))
(
2
− s H s
+ s
2
α
( )
,
ψ ɶ ( )
=
α
λ
2
( )
+
( (
−
λ s
K
))
2
(
−
+ s G s s
2
α
And finally by using inversion formula, the solution will be
φ =
1
2
π i
∫
α
λ
( )
2
( K
+
(
λ
−
K s
(
))
2
− s H s
+ s
2
α e ds ,
.
(2.16)
ψ =
1
2
π i
∫
α
λ
2
( K
+
(
λ
− s
K (
−
) ( )
))
2 + s
2
α e ds .
Example 2.2 consider the following system






 c c
1
D 2
φ
( )
D
1
2
ψ
( )
= −
=
2 x
−
3
2
1
π x
π
−
+ ∫
+∞ x
∫
+∞ x
π
1
( x
− t )
ψ
1
π
( x
− t )
φ
, then we have
=
1 s
, ( )
=
, (
− =
1 s
, so by using relationships (2.16) one gets



φ
ψ
( )
= δ x
( )
=
2 cos x
,
Problem 2.3 Let us consider fractional Volterra singular integral equation of the form, c
D
α φ
( )
=
( )
+ λ ∫ x ln(
− φ
,
φ
(0)
=
0, 0 1.
(2.17)
0
Solution: After taking Laplace-transform of the above integral equation and simplifying, one gets
Φ = s
α +
1 + + ln s )
(2.18)

16 A. Aghili et al.
φ in
γ ≈
0.577
is Euler constant. Using complex inversion formula for the above relation leads to
=
1
2
π
∫ i s
α +
1 sx
+ λ γ + ln ) ds .
(2.19)
Example 2.3 Consider the following fractional singular integral equation c
1
D 2
φ
( )
= x
+ ∫ x
0 ln(
− φ
,
φ
(0)
=
0, .
By using equation (2.18), we get the solution as bellow
Φ =
=
= s
3/2
π
γ s ln s
=
π s s 2
3
{
1
+
γ
1
+ ln s
3/2 s
}
=
π s
{1
−
γ + ln s
+
(
γ + ln
3 s
3/2 s
3/2 s 2
π
{
1
−
γ + s s ln s
+
1
5/2
γ
( s s
+ ln
3/2 s
)
2 −
....
}
= s
)
2 −
.... }
= π
∞
n
=
0
1 s
(
−
γ + ln s s
3/2
At this point, we may invert
Φ
( ) easily by using convolution. Therefore, one can find that ln x
∗ x
= ∫ x
0
( where ln x
φ = π
{1
+
γ + ln s
, s
1
Γ
(3 / 2)
∫ x
0 x
−
)(ln )
=
2 x t t dt x
3
3/ 2
Γ
(3 / 2) s
3/ 2
,so we can write
(
−
) ln tdt
+ ∫ x
( (
−
) ln tdt x
0
−
+
Engineering and other areas of sciences can be successfully modeled by the use of fractional derivatives. That is because of the fact that, a realistic modeling of physical phenomenon having dependence not only at the time instant, but also the previous time history.
In this section, the authors consider certain non-homogeneous time fractional heat equation in a spherical domain that is a generalization to the problem which is

Solution to Volterra Singular Integral… 17 studied by Jordan and Puri [21]. In this work, only the Laplace transformation is considered as a powerful tool for solving the above mentioned problem. This goal has been achieved by formally deriving exact analytical solution.
Problem 3.1 Solve the non-homogeneous time fractional heat equation c t
α =
∂ 2
∂
( , ) r 2
+
2 r
∂
( , )
∂ r
0
− λ
( , )
− r 1, t
>
0,
( ),
0
α
1, (3.1) with the boundary conditions: lim r
→
0
< ∞
, u r t
=
1 and the initial condition ( , 0)
=
0 0 r 1 . Let ( ) be Laplace transformable function.
Solution: Let us introduce a new variable ( , )
=
( , ) . Then equation (3.1) becomes t
α =
∂ 2
∂ r
2
− λ c
( , )
−
( ).
(3.2)
By taking the Laplace transform with respect to variable t of equation (3.2) and boundary conditions we get
α =
∂ 2
V
∂ r
2
− λ
( , )
−
( ), (3.3) or
∂ 2
V
∂ r
2
−
(
λ + s
α
) ( , )
=
( ),
(3.4) with the boundary conditions r lim
→
0
V r s
=
0,
Solving the above equation (3.4) leads to
V (1, ) V (1, ) r
− =
1 s
( , )
=
A cosh( r
λ + s
α
)
+
B sinh( r
λ + s
α
)
−
λ
Now we apply the boundary conditions to get
+ s
α r .
= s (
λ + s
α sinh( r cosh(
λ
λ + s
α
)
+ s
α
) sinh(
λ + s
α
))
−
λ + s
α r .
(3.5)
So by using Bromwich's integral we have the following relationship

18 A. Aghili et al.
=
1
2
π i
∫

 s (
λ + s
α sinh( r cosh(
λ
λ + s
α
)
+ s
α
) sinh(
λ + s
α
))
−
λ + s
α

 st e ds .
(3.6)
To use the residue theorem let us assume that 0.5
α =
, so relationship (3.6) will be changed to
=
1
2
π i
∫

 s (
λ + s sinh( r cosh(
λ +
λ s
+ s )
λ + s ))
−
λ + s

 st e ds .
(3.7) st s
=
0, s
= λ
2 and also simple poles at
λ + s n
= i
β n or s n
=
(
λ β n
2 2
) where tan
β n
= β n for n
=
1,2,....
By using residue theorem, one gets
( , )
= r
2
λ n
∞
=
1
t
0
(
− η
) e d
η s in (
λ + β n
2 r
β n
) s in
)
β n
+ e
(
λ s in h ( c o s h
λ r
−
λ
) s in h n
2
)
2 t
,
λ
− and consequently the final solution is as bellow
= r
2
λ t
0
− η
) e d
η + sinh( r r (
λ cosh
λ −
λ
) sinh
λ
)
− r
1 n
∞
=
1
( sin(
λ β
2 n r
β n
)
) sin
β n e
(
λ β n
2
)
2 t
.
Note that if we set
α =
1 , the problem becomes the non-homogeneous heat equation.
Case 3.1. For c t
1
( , )
=
∂ 2
( ) 0 (homogeneous equation),
∂
( , ) r 2
+
2 r
∂
( , )
∂ r
−
α
( , ),
>
0,0 r
=
0.5,
λ =
1 we have
1,
Then the solution is
( , )
= e
−
1 sinh( )
−
1 r r
∞
n
=
1
( 1 sin(
+ β
2 n r
β n
) sin
)
β n e ( 1
+ β n in which tan
β n
= β n for n
=
1,2,...
(figure1). t ,

Solution to Volterra Singular Integral… 19
Figure1.
In the present paper the authors implemented the Laplace transform method for solving fractional singular integral equations. They also considered certain time fractional heat equation which is a generalization to the problem investigated in
[21], the problem of dynamic thermo -elastic stresses in a spherical shell with fixed boundaries whose inner surface is subjected to a step jump in temperature.
We hope that it will also benefit many researchers in the disciplines of applied mathematics, mathematical physics and engineering.
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