EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 2, 2023, 919-933
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Solution of Integral Equations Via Laplace ARA
Transform

Ahmad Qazza

Department of Mathematics, Zarqa University, Zarqa 13110, Jordan.

Abstract. This research article demonstrates an efficient method for solving partial integro-
differential equations. The intention of this research is to establish the solution of some different
classes of integral equations, by utilizing the double Laplace ARA transform. We present some
definitions and basic concepts related to the double Laplace ARA transform. The results of the
examples support the theoretical results and show the accuracy and applicability of the presented
approach.

2020 Mathematics Subject Classifications: 44A05, 34A25

Key Words and Phrases: Double Laplace-ARA transform, ARA transform, Laplace transform,
Partial integro-differential equations

1. Introduction

Integral transforms play a vital role in solving integral equations and partial integro-
differential equations. For this reason, many phenomena in the field of engineering, sci-
ence, and mathematical physics can be represented by integral equations of different types
[6, 8, 9, 12, 15, 16, 26]. Using integral transformations, we can transform integral equa-
tions into algebraic or differential equations and get the exact solution of the target integral
equations. Developed through the hard work of many scientists and researchers, these tech-
niques are used today to tackle challenging problems in contemporary arithmetic. These
transformations enable us to get the exact solutions of the objective equations without
the need for linearization or discretization, like Laplace, Fourier, Elzaki, Natural, Sumudu,
and ARA transformations [14, 17, 18, 24, 25, 27]. They are used in transforming the par-
tial differential equations into ordinary equations using a simple transformation, or into
algebraic equations using a double integral transformation.
The double transformations have also widespread applied to solve partial differential equa-
tions with unknown two variable functions, and as a result, double transformations have
been considered to be very effective in handling partial differential equations compared to
other numerical approaches [3, 7, 11]. In addition, extensions of the double transformation

DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4745

Email address: aqazza@zu.edu.jo (A. Qazza)

https://www.ejpam.com 919 © 2023 EJPAM All rights reserved.



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 920

have been developed in the relevant literature, such as double Laplace transform, double
Shehu transform [10], double Sumudu transform [21, 28], double Elzaki transform [13],
double Laplace-Sumudu transform [20], ARA–Sumudu transform [5, 19] and double ARA
transform [22, 23].
The ARA transformation is introduced in 2020 [18]. It is defined by the improper integral.

Gn [φ (u)] = v

∫ ∞

0
un−1e−uvφ (u) du, v > 0.

This transformation has attracted much attention from researchers due to its ability to
generate multiple transformations of index n, and it could also easily overcome the chal-
lenges of having singular points in differential equations. Despite all these merits it could
be used to solve different kinds of problems [1, 4].
In this research, we introduce a new Laplace and ARA combination, so that we can take
advantage of these two powerful transforms. This combination is called the double Laplace-
ARA transform (DL-ARAT) [2]. Basic properties and concepts related to DL-ARAT are
obtained and proven, also we process the values of some functions by DL-ARAT. To help
us in solving integral equations new relations related to the double convolution theorem
and partial derivatives are implemented and established. The novelty of this research is
evident in these combinations between Laplace and ARA transforms, in which the new
DL-ARAT have the advantages of the two transforms, the applicability of ARA in han-
dling some singular points found in the equations and the simplicity of Laplace.
In this work we use the first order ARA transform G1 [φ(u)], which we denote by G [φ(u)]
for the sake of simplicity.
The motivation of this work is to present a novel double integral transform, that combines
two powerful transforms, Laplace and ARA transforms. The new approach has the merits
of the two transforms and can solve different kinds of problems.
The remaining part of the paper is set up as follows. Section 2 defines the basic definitions
and properties of the ARA transform and Laplace transform. In Section 3 basic properties
and theorems of DL-ARAT are presented and proved, and we apply DL-ARAT to some
functions. By applying the integral transform DL-ARAT to solve the second type nonlin-
ear VIE and solving significant examples in Section 4, the effectiveness and efficiency of
the proposed method are illustrated. Finally, in Section 5, the conclusion of the work is
presented.

2. Preliminaries and Notations

In this part, we will provide the basic definitions and some properties of the Laplace
and the ARA transforms that will be needed in later sections.

Definition 1. [27] The Laplace transform of the function φ(t) of t > 0 is the function
Φ (s) = L[φ(t)], defined by

L [φ (t)] =

∫ ∞

0
e−stφ (t)dt, Re(s) > 0, (1)



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 921

inverse Laplace transform of Φ (s) is given by

L−1 [Φ (s)] =
1

2πi

∫ c+i∞

c−i∞
estΦ (s) ds = φ (t) , t > 0. (2)

Theorem 1. [27] If the piecewise continuous function φ (t) and of exponential order k on
the interval 0 ≤ t <∞. Then L [φ(t)] exists for Re(s) > k and satisfies

|φ (t)| ≤Mekt, M > 0,

where M is a constant. Then Laplace transform integral converges absolutely for Re (s) >
k.

Proof. Using the definition of Laplace transform, we get

|Φ (s)| =
∣∣∣∣∫ ∞

0
e−stφ (t)dt

∣∣∣∣ ≤ ∫ ∞

0
e−st |φ (t)|dt ≤M

∫ ∞

0
e−(s−k)tdt =

M

s− k
,

where Re (s) > k.
Thus, Laplace transform integral converges absolutely for Re (s) > k.

Definition 2. [18] The first order ARA integral transform of a continuous function φ(u)
on the interval (0,∞) is introduced as

G [φ (u)] (v) = Φ (v) = v

∫ ∞

0
e−uvφ (u)du, Re(v) > 0. (3)

The inverse ARA transform is defined by

G−1 [G [φ (u)]] =
1

2πi

∫ c+i∞

c−i∞
euvΦ (v) dv = φ (u) , (4)

where

Φ (v) =

∫ ∞

0
e−uvφ (u)du.

Theorem 2. [18] If φ(u) is piecewise continuous in 0 ≤ u ≤ k and satisfies

|uφ (u)| ≤Meku, M > 0,

where M is a constant, then the ARA transform exists for all Re(v) > k.

Proof. Using the definition of the ARA transform, we have

|Φ (n, v)| =
∣∣∣∣v ∫ ∞

0
ue−uvφ (u)du

∣∣∣∣ = ∣∣∣∣v ∫ α

0
ue−uvφ (u)du+ v

∫ ∞

α
ue−uvφ (u)du

∣∣∣∣
≤ v

∣∣∣∣∫ ∞

α
ue−uvφ (u)du

∣∣∣∣ ≤ v

∫ ∞

α
e−uv |u φ (u)|du

≤ v

∫ ∞

α
e−uvMekudu = vM

∫ ∞

α
e−(v−k)udu =

vM

v − k
e−α(v−k).



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 922

This integral converges for all Re(v) > k.
Thus, G [φ (u)] exists.

In the table below (Table 1) we introduce the Laplace transform and the ARA trans-
form for some functions and give some basic properties of both transforms, where φ (t)
and ψ(t) are continuous functions, α, β ∈ R.

Table 1: Laplace transform and ARA transform for some basic functions.

Laplace transform ARA transform

αfφ (t) + β ψ(t) α L [φ (t)] + β L [ψ(t)] α G [φ (t)] + β G [ψ(t)]

tα Γ(α+1)
sα+1 , α ≥ 0 Γ(α+1)

vα , α > 0
eαx 1

s−α , α ∈ R v
v−α

φ′ (t) sL [φ (t)]− φ (0) vG [φ (t)]− vφ(0)

φ(m) (t) smL [φ (t)]−
∑m

l=1 s
m−lφ(l−1) (0) vmG [φ (t)]−

∑m
l=1 v

m−l+1φ(l−1) (0)
sin (αt) α

s2+α2
αv

v2+α2

cos (αt) s
s2+α2

v2

v2+α2

sinh (αt) α
s2−α2

αv
v2−α2

cosh (αt) s
s2−α2

v2

v2−α2 ,

(φ ∗ ψ)(t) L [φ (t)] L [ψ (t)] G[φ(t)] G[ψ(t)]
v

3. Double Laplace-ARA Transform of First Order (DL-ARAT)

The integral transform DL-ARAT is introduced in this section that combines the
Laplace transform and the ARA transform of first order. A fundamental properties and
theorems for DL-ARA are presented.

Definition 3. The DL-ARAT of a continuous function φ(t, u) is defined as

LtGu [φ (t, u)] = Φ (s, v) = v

∫ ∞

0

∫ ∞

0
e−st−uvφ (t, u)dt du, s, v > 0. (5)

Clearly that DL-ARAT is a linear

LtGu [α φ (t, u) + β ψ(t, u)] = α LtGu [ φ (t, u)] + β LtGu [ ψ(t, u)] , (6)

where α and β are constants and the functions LtGu [φ (t, u)] ,LtGu [ψ(t, u)] are exists.
The inverse of the DL-ARAT is provided by

L−1
t G−1

u [Φ (s, v)] =
1

2πi

∫ c+i∞

c−i∞
etsds

1

2πi

∫ r+i∞

r−i∞

euv

v
Φ (s, v) dv = φ (t, u) . (7)



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 923

Properties

(i) Suppose that φ (t, u) = φ1 (t)φ2 (u), t, u > 0. Then

LtGu [φ(t, u)] = Lt [φ1 (t)]Gu [φ2(u)] .

Proof.

LtGu [φ(t, u)] = LtGu [φ1 (t)φ2(u)] = v

∫ ∞

0

∫ ∞

0
e−ts−uvφ1 (t)φ2(u)dt du

=

∫ ∞

0
φ1 (t) e

−tsdt · v
∫ ∞

0
φ2 (u) e

−uvdu = Lt [φ1 (t)]Gu [φ2(u)] .

(ii) DL-ARAT of basic functions

• Suppose that φ(t, u) = 1, t, u > 0. Then

LtGu [1] = v

∫ ∞

0

∫ ∞

0
e−ts−uvdt du =

∫ ∞

0
e−tsdt v

∫ ∞

0
e−uvdu =

1

s
,

where Re(t) > 0.

• Suppose that φ (t, u) = tαuβ, α, β are constants, and t, u > 0. Then

LtGu
[
tα uβ

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uvtαuβdt du =

∫ ∞

0
e−tstαdt v

∫ ∞

0
e−uvuβdu

=
Γ (α+ 1) Γ (β + 1)

sα+1vβ
, Re (α) > −1 , Re (β) > −1.

• Suppose that φ(t, u) = eαt+βu, α, β are constants, and t, u > 0. Then

LtGu
[
eαt+βu

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uveαt+βudt du

=

∫ ∞

0
e−tseαtdt v

∫ ∞

0
e−uveuvdu

=
v

(s− a) (v − b)
.

Likewise,

LtGu
[
ei(αt+βu)

]
=

v

(s− iα) (v − iβ)
=
v (sv − αβ) + iv(sβ + vα)

(s2 + α2) (v2 + β2)
.

Consequently,

LtGu [sin (αt+ βu)] =
v(βs+ αv)

(s2 + α2) (sv2 + β2)
,

LtGu [cos (αt+ βu)] =
v (sv − αβ)

(s2 + α2) (v2 + β2)
.



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 924

• Suppose that φ(t, u) = sinh (αt+ βu) or φ(t, u) = cosh (αt+ βu).
Recall that

LtGu [sinh (αt+ βu)] =
v(sβ + vα)

(s2 − α2) (v2 − β2)
,

LtGu [cosh (αt+ βu)] =
v (sv + αβ)

(s2 − α2) (v2 − β2)
.

• Suppose that φ(t, u) = J0
(
c
√
tu
)

LtGu
[
J0

(
c
√
tu
)]

= v

∫ ∞

0

∫ ∞

0
e−ts−uvJ0

(
c
√
tu
)
dt du

=

∫ ∞

0
e−tsJ0

(
c
√
tu
)
dt s

∫ ∞

0
e−uvdu

= v

∫ ∞

0
e−

c2

4v
ue

−uv
du =

4v

4sv + c2
.

• Suppose the function φ(t, u) of exponential order α and β as t and u goto ∞.
If ∃N > 0 such that ∀t > T and u > U , we have

|φ(t, u)| ≤ Neαt+βu.

We can write φ(t, u) = O(eαt+βu) as t and u goes to ∞, s > α and v > β.

Theorem 3. Suppose that φ(t, u) of exponential orders α and β is a continuous function
on [0, T )× [0, U). Then LtGu [φ(t, u)] exists for s and v gave Re(s) > α and Re(v) > β.

Proof. By the definition of DL-ARAT, we get

|Φ(s, v)| =
∣∣∣∣v ∫ ∞

0

∫ ∞

0
e−ts−uvφ(t, u)dt du

∣∣∣∣ ≤ v

∫ ∞

0

∫ ∞

0
e−ts−uv |φ (t, u)|dt du

≤ K

∫ ∞

0
e−(s−α)tdt v

∫ ∞

0
e−(v−β)udu =

Kv

(u− α) (v − β)
,

Re (s) > α, Re (v) > β.

Thus, LtGu [φ(t, u)] exists.

Theorem 4. Suppose that LtGu [φ(t, u)] and LtGu [ψ(t, u)] are exists and LtGu [φ(t, u)] =
Φ(s, v), LtGu [ψ(t, u)] = Ψ(s, v), then the double convolution

LtGu [φ(t, u) ∗ ∗ψ(t, u)] =
1

v
Φ(s, v) Ψ(s, v), (8)

where

φ(t, u) ∗ ∗ψ(t, u) =
∫ t

0

∫ u

0
φ(t− τ, u− ρ)ψ(τ, ρ)dτ dρ. (9)



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 925

Proof. By the definition of DL-ARAT, we get

LtGu [φ(t, u) ∗ ∗ψ(t, u)] = v

∫ ∞

0

∫ ∞

0
e−ts−uv (φ(t, u) ∗ ∗ψ(t, u))dt du

= v

∫ ∞

0

∫ ∞

0
e−ts−uv

(∫ t

0

∫ u

0
φ(t− τ, u− ρ)ψ(τ, ρ)dτ dρ

)
dt du.

(10)

Equation (10) can be written as

LtGu [φ(t, u) ∗ ∗ψ(t, u)]

= v

∫ ∞

0

∫ ∞

0
e−ts−uv

(∫ ∞

0

∫ ∞

0
φ(t− τ, u− ρ)H (t− τ, u− ρ)ψ(τ, ρ)dτ dρ

)
dt du

=

∫ ∞

0

∫ ∞

0
ψ (τ, ρ) dτ dρ

(
v

∫ ∞

0

∫ ∞

0
e−s(t+τ)−v(u+ρ)φ(t− τ, u− ρ)H (t− τ, u− ρ)

)
dt du

= Ψ(s, v)

∫ ∞

0

∫ ∞

0
e−sτ−vρφ (τ, ρ) dτ dρ =

1

v
Ψ(s, v)Φ (s, v).

where H (t− τ, u− ρ) is the Heaviside unit step function.

Theorem 5. Suppose that φ(t, u) is a continuous function and LtGu [φ(t, u)] = Φ (s, v).
Then, we have the following properties of derivatives

(i)

LtGu
[
∂φ(t, u)

∂u

]
= vΦ (s, v)− vLt [φ(t, 0)] ,

(ii)

LtGu
[
∂φ(t, u)

∂t

]
= sΦ (s, v)− Gu [φ(0, u)] ,

(iii)

LtGu
[
∂2φ(t, u)

∂u2

]
= v2Φ (s, v)− v2Lt [φ(t, 0)]− vLt

[
∂φ(t, 0)

∂u

]
,

(iv)

LtGu
[
∂2φ(t, u)

∂t2

]
= s2Φ (s, v)− sGu [φ(0, u)]− Gu

[
∂φ(0, u)

∂u

]
,

(v)

LtGu
[
∂2φ(t, u)

∂t∂u

]
= svΦ (s, v)− svLt [φ(t, 0)]− vGu [φ(0, u)] + vφ(0, 0).

Proof.



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 926

(i)

LtGu
[
∂φ(t, u)

∂u

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uv

∂φ(t, u)

∂u
dt du

=

∫ ∞

0
e−tsdt v

∫ ∞

0
e−uv

∂φ (t, u)

∂u
du

=

∫ ∞

0
e−ts (vΦ (t, v)− vφ (t, 0)) dt

= vΦ (s, v)− vLt [φ(t, 0)] .

(ii)

LtGu
[
∂φ(t, u)

∂t

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uv

∂φ(t, u)

∂t
dt du

= v

∫ ∞

0
e−uvdu

∫ ∞

0
e−ts

∂φ(t, u)

∂t
dt

= v

∫ ∞

0
e−uv (Φ(s, u)− φ (0, u)) du

= sΦ (s, v)− Gu [φ(0, u)] .

(iii)

LtGu
[
∂2φ(t, u)

∂u2

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uv

∂2φ(t, u)

∂u2
dt du

=

∫ ∞

0
e−tsdt v

∫ ∞

0
e−uv

∂2φ(t, u)

∂u2
du

=

∫ ∞

0
e−ts

(
v2Φ (t, v)− v2φ(t, 0)− v

∂φ(t, 0)

∂u

)
dt

= v2Φ (s, v)− v2Lt [φ(t, 0)]− vLt
[
∂φ(t, 0)

∂u

]
.

(iv)

LtGu
[
∂2φ(t, u)

∂t2

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uv

∂2φ(t, u)

∂t2
dt du

= v

∫ ∞

0
e−uvdu

∫ ∞

0
e−ts

∂2φ(t, u)

∂t2
dt

= v

∫ ∞

0
e−uv

(
s2Φ (s, u)− sφ(0, u)− ∂φ(0, u)

∂t

)
du

= s2Φ (s, v)− sGu [φ(0, u)]− Gu
[
∂φ(0, u)

∂t

]
.



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 927

(v)

LtGu
[
∂2φ(t, u)

∂t∂u

]
= v

∫ ∞

0

∫ ∞

0
e−ts−uv

∂2φ(t, u)

∂t∂u
dt du

= v

∫ ∞

0
e−uvdu

∫ ∞

0
e−ts

∂2φ(t, u)

∂t∂u
dt

= sv

∫ ∞

0

∫ ∞

0
e−ts−uv

∂φ(t, u)

∂u
dt du− v

∫ ∞

0
e−st

∂φ(0, u)

∂u
du

= svΦ (s, v)− v Gu [φ(0, u)]− sv Lt [φ(t, 0)] + v φ(0, 0).

4. Applications of DL-ARAT to Solve Volterra Integral Differential
Equations

In this part, we apply DL-ARAT to the following classes of Volterra integral equations
(VIEs) and Volterra partial integro-differential equations (VPIDEs) first and second order.

4.1. VIEs of two variables

Considering the following VIE

φ (t, u) = ω (t, u) + a

∫ t

0

∫ u

0
φ (t−m,u− n)ψ (m,n) dm dn, (11)

where a is constant, ω (t, u) and ψ (t, u) are two known functions, and φ (t, u) is an unknown
function.
Running DARA-ST on equation (1)

LtGu [φ (t, u)] = LtGu [ω (t, u)] + LtGu
[
a

∫ t

0

∫ u

0
φ (t−m,u− n)ψ (m,n) dm dn

]
. (12)

According to Theorem 5 and the linearity property (6), equation (12) can be written

Φ (s, v) = Ω (s, v) + a
1

v
Φ (s, v)Ψ (s, v) , (13)

where Φ (s, v) = LtGu [φ (t, u)], Ω (s, v) = LtGu [ω (t, u)] and Ψ (s, v) = LtGu [ψ (x, u)].
Consequently,

Φ (s, v) =
v Ω (s, v)

v − a Ψ(s, v)
. (14)

Using the inverse transform L−1
t G−1

u , we get the exact solution of (11)

φ (t, u) = L−1
t G−1

u

[
v Ω (s, v)

v − a Φ (s, v)

]
. (15)

Now, we give three illustrative problems to above technique.



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 928

Problem 1. Consider the following VIE

φ (t, u) = b− a

∫ t

0

∫ u

0
φ (m,n) dm dn, (16)

where a and b are constant.

Solution. By implementing DL-ARA in equations (16) and using the linearity prop-
erty and convolution theorem, we get

Φ (s, v) =
b

s
− a

1

vs
Φ (s, v) . (17)

As a result,

Φ (s, v) =
b v

sv + a
. (18)

Using the inverse transform L−1
t G−1

u , we get the exact solution of (18)

φ (t, u) = L−1
t G−1

u

[
b v

sv + a

]
= b J0

(
2
√
atu

)
.

Problem 2. Consider the following VIE

a2u =

∫ t

0

∫ u

0
φ (t−m,u− n)φ (m,n) dm dn, (19)

where a is a constant.

Solution. By implementing DL-ARA in equations (19) and using Theorem 4 on
equation (19), we get

a2

sv
=

1

v
Φ2(s, v). (20)

Thus,

Φ (s, v) =
a√
s
. (21)

Using the inverse transform L−1
t G−1

u , we obtain the solution exact equation of (19) as
following

φ (t, u) = L−1
t G−1

u

[
a√
s

]
=

a√
π

1√
t
. (22)

Problem 3. Consider the following VIE∫ t

0

∫ u

0
em−nφ (t−m,u− n)dm dn = tet−u − tet. (23)



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 929

Solution. By implementing DL-ARA in equations (23) and using Theorem 4 on (23),
we get

v Φ(s, v)

(s− 1) (1 + v)
=

v

(v + 1) (s− 1)2
− 1

(s− 1)2
. (24)

Thus,

Φ(s, v) =
−1

v (s− 1)
. (25)

Taking the inverse transform L−1
t G−1

u , we have the exact solution of (23) as follows

φ (t, u) = L−1
t G−1

u

[
−1

v (s− 1)

]
= −u et. (26)

4.2. VPIDEs of first order

Considering the following VPIDE

∂φ (t, u)

∂t
+

∂φ (t, u)

∂u
= ω (t, u) + a

∫ t

0

∫ u

0
φ (t−m,u− n)ψ (m,n) dm dn, (27)

with the conditions
φ (t, 0) = f0 (t) , φ (0, u) = h0 (u) , (28)

where a is a constant, φ (t, u) is an unknown function, ω (t, u) and ψ (t, u) are known
functions.
Implement DL-ARAT on Equation (27), we achieve

sΦ (s, v)− Gu [φ (0, u)] + sΦ(s, v)− sLt [φ (t, 0)] = Ω (s, v) + a
1

v
Φ (s, v)Ψ (s, v) .

Substituting the values of the transformed condition (28)

Φ (s, v) =
vΩ (s, v) + vH0(v) + v2F0(s)

sv + v2 − aΨ(v, s)
, (29)

where F0 (s) = Lt [ψ (t, 0)] and H0 (v) = Gu [ψ (0, u)].
Running the inverse transform L−1

t G−1
u , we get the exact solution of (27) as follows

φ (t, u) = L−1
t G−1

u

[
vΩ (s, v) + vH0(v) + v2F0(s)

sv + v2 − aΨ(s, v)

]
. (30)

Now, we give illustrative problem to above technique.

Problem 4. Consider the following VPIDE

∂φ (t, u)

∂t
+
∂φ (t, u)

∂u
= −1 + et + eu + et+u +

∫ t

0

∫ u

0
φ (t−m,u− n) dm dn, (31)

with the conditions
φ (t, 0) = et, φ (0, u) = eu. (32)



A. Qazza / Eur. J. Pure Appl. Math, 16 (2) (2023), 919-933 930

Solution. By implementing DL-ARA in equation (32) and the source function ω (t, u)
F0 (s) = Lt

[
et
]
= 1

s−1 ,

H0 (v) = Gu [eu] = v
v−1 ,

ω (t, u) = LtGu
[
−1 + et + eu + et+u

]
= v(1−2sv)

(v−v2)(−1+s)s
.

(33)

Now, putting values in (33), into (30), we get the solution of (31) as follows

φ (t, u) = L−1
t G−1

u

v
(

v(1−2sv)
(v−v2)(−1+s)s

)
s v + v2 − 1

s

+
v v
v−1 + v2 1

s−1

s v + v2 − 1
s


= L−1

x G−1
u

[
v

(−1 + v) (−1 + s)

]
= et+u.

(34)

4.3. VPIDE of Second order

Given the following VPIDE

∂2φ (t, u)

∂u2
− ∂2φ (t, u)

∂t2
+ φ (t, u) +

∫ t

0

∫ u

0
ψ (t− δ, u− ε) φ (δ, ε)dδ dε = ω (t, u) , (35)

with the conditions
φ (t, 0) = f0 (t) , φu (t, 0) = f1 (t) ,
φ (0, u) = h0 (u) , φt (0, u) = h1 (u) .

(36)

Applying DL-ARAT to (36), we have

v2Φ (s, v)− v2Lt [φ (t, 0)]− vLt [φu (t, 0)]−
(
s2Φ (s, v)− sGu [φ (0, u)]− Gu [φt (0, u)]

)
+Φ(s, v) +

1

v
Φ (s, v)Ψ (s, v) = Ω (s, v) .

After simple calculations, one can obtain

Φ (s, v) =
v3F0(s) + v2F1(s)− svH0(v)

v3 − s2v + v +Ψ(s, v)
+

−vH1(v) + vΩ (s, v)

v3 − s2v + v +Ψ(s, v)
, (37)

where F0 (s) = Lt [φ (t, 0)], F1 (s) = Lt [φu (§, 0)], H0 (v) = Gu [φ (0, u)] and H1 (v) =
Gu [φt (0, u)].
Running the inverse transform L−1

t G−1
u , we get the solution of (35) as follows

φ (t, u) = L−1
t G−1

u

[
v3F0(s) + v2F1(s)

v3 − s2v + v +Φ(s, v)
+

−svH0(v)− vH1(v) + vΩ (s, v)

v3 − s2v + v +Ψ(s, v)

]
. (38)

Now, we give illustrative problem to above technique.



REFERENCES 931

Problem 5. Consider the following VPIDE

∂2φ (t, u)

∂u2
− ∂2φ (t, u)

∂t2
+ φ (t, u) +

∫ t

0

∫ u

0
et−m+u−nφ (m,n)dm dn = et+u + tuet+u, (39)

with conditions
φ (t, 0) = et, φu (t, 0) = et,
φ (0, u) = eu, φt (0, u) = eu.

(40)

Solution.By implementing DL-ARA in equation (40) and the functions Ψ (s, v) and
ω(t, u), we achieve 

F0 (s) = F1 (s) =
1
s−1 ,

H0 (u) = H1 (u) =
u
u−1 ,

Ψ(s, v) = v
(v−1)(s−1) ,

Ω (s, v) = v(2−s+v(s−1))

(−1+s)2(−1+v)
2 .

(41)

Now, putting values in (41), into (38), we obtain the solution of equation (39) as follows

φ (t, u) = L−1
t G−1

u

v3 1
s−1 + v2 1

s−1 − sv v
v−1 − v v

v−1

v3 − s2v + v + v
(v−1)(s−1)

+
v v(2−s+v(−1+s))

(v−1)2(s−1)2

v3 − s2v + v + v
(v−1)(s−1)


= L−1

t G−1
u

[
v

(v − 1)(s− 1)

]
= et+u.

5. Conclusion

In this research manuscript, we propose DL-ARAT approach to solve IDEs. Theorems
and basic properties of the new DL-ARAT are presented in detail. Two types of integral
equations have been discussed: partial integral and PIDEs. The solutions for IDEs are
examined and found to best represent the true dynamics of the problem. The method
offers a useful way to develop an analytical treatment for these equations. In a future
work we will use the proposed scheme to solve other nonlinear equations.

Acknowledgements

The author is grateful to the reviewers for their valuable remarks, comments and
advice, that help to improve the quality of the manuscript.

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