429 
 

IHJPAS. 37 (1) 2024 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

PISSN: 1609-4042, EISSN: 2521-3407 

  

 

 

 

 

   

1Ahmad Issa*       2Emad A. Kuffi   
1Department of Mathematics, Faculty of Science, Karabuk University, Karabuk, Turkey. 

2 Department of Mathematics, College of Basic Education, Mustansiriyah University, Baghdad, Iraq. 

 

 

*Corresponding Author: ahmad93.issa18@gmail.com 

 

 

 

 

Abstract  

Due to the importance of solutions of partial differential equations, linear, nonlinear, 

homogeneous, and non-homogeneous, in important life applications, including engineering 

applications, physics and astronomy, medical sciences, and life technology, and their 

importance in solutions to heat transfer equations, wave, Laplace equation, telegraph, etc. In 

this paper, a new double integral transform has been proposed. 

In this work, we have introduced a new double transform ( Double Complex EE Transform ). 

In addition, we presented the convolution theorem and proved the properties of the proposed 

transform, which has an effective and useful role in dealing with the solution of two-

dimensional partial differential equations. Moreover, two examples of important mathematical 

equations are solved to illustrate method. This double  integral transformation has a complex 

kernel. 

 

Keywords Complex EE Transform, Double Complex EE Transform, partial differential 

equations. 

 

1. Introduction 

The integral transformation is one of the important and famous topics of applied mathematics 

[1-10]. In recent years many researchers have discovered new transformations to solve several 

problems in mathematics, physics, and engineering [11-27]. [28] have implemented a new 

transformation (Double Aboodh Transform) for solving Telegraph Equation.  [29] have solved 

ordinary differential equations by a new transformation (Complex EE Transform). In [30], 

Meddahi and Jafari, introduced the new transformation (General Double Integral Transform) to 

solve telegraph and Klein-Gordon equations. 

 Received 13 March 2023, Received 23 March 2023,  Accepted 29 May 2023, Published 20 January 2024 

 

On The Double Integral Transform (Complex EE Transform) and Their 

Applications 

 

 

doi.org/10.30526/37.1.3329 

https://jih.uobaghdad.edu.iq/index.php/j/index#1609-4042
https://jih.uobaghdad.edu.iq/index.php/j/index#2521-3407
https://orcid.org/0000-0001-7495-3443
mailto:ahmad93.issa18@gmail.com
https://orcid.org/0009-0004-5254-674X
mailto:emad.kuffi@uomustansiriyah.edu.iq


 IHJPAS. 37(1)2024 

430 
 

The relationship between this transform and the two-dimensional Laplace transform is the change 

in the definition of the kernel, the kernel in the proposed transform is wider than the kernel in the 

two-dimensional Laplace transform. 

 

2.  Definitions and Properties of Double Complex EE Transform : 

Definition 2.1. [29] The Complex EE (Emad-Elaf ) Transform of 𝑓(𝑡) denoted by the operator 𝐸 

is given by: 

𝐸[𝑓(𝑡)] = 𝐸(𝑖𝑣) = ∫ 𝑓(𝑡)𝑒−𝑖𝑣𝑛𝑡   𝑑𝑡, 𝑛 ∈  𝑍, 𝑡 ≥  0, 𝑖 ∈  𝐶

∞

𝑡=0

, 

where 𝑣 is a complex parameter and 𝐼𝑚 (𝑣𝑛) < 0. 

Now, we will introduce the definition of the double integral transform of the complex EE transform 

and its properties. 

 

Definition 2.2. The Double Complex EE Transform of 𝑓(𝑥, 𝑡) denoted by the operator 𝐷𝐸𝐸  is 

given by: 

𝐷𝐸𝐸[𝑓(𝑥, 𝑡)] = 𝐴(𝑖𝑢, 𝑖𝑣) = ∫ ∫ 𝑓(𝑥, 𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥, 𝑛 ∈  𝑍, 𝑡, 𝑥 ≥  0, 𝑖 ∈  𝐶

∞

0

∞

0

, 

where 𝑣 and 𝑢 are complex parameters and every 𝐼𝑚 (𝑣𝑛) < 0 and 𝐼𝑚 (𝑢𝑛) < 0. 

Some properties of the Double Complex EE Transform are as follows 

 

I. If 𝑓(𝑥, 𝑡) = 1, then 

𝐷𝐸𝐸[1] = ∫ ∫ 1. 𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 1. 𝑒−𝑖𝑢𝑛𝑥  𝑑𝑥

∞

0

∫ 1. 𝑒−𝑖𝑣𝑛𝑡  𝑑𝑡

∞

0

 

= 𝐸𝐸[1]. 𝐸𝐸[1] 

=
−𝑖

𝑢𝑛
.
−𝑖

𝑣𝑛
 

 

=
−1

(𝑢𝑣)𝑛
 . 

II. If 𝑓(𝑥, 𝑡) = 𝑥𝑟𝑡𝑚, then 

 

𝐷𝐸𝐸[𝑥𝑟𝑡𝑚] = ∫ ∫ 𝑥𝑟𝑡𝑚𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑥𝑟𝑒−𝑖𝑢𝑛𝑥  𝑑𝑥

∞

0

∫ 𝑡𝑚  𝑒−𝑖𝑣𝑛𝑡  𝑑𝑡

∞

0

 

= 𝐸𝐸[𝑥𝑟]. 𝐸𝐸[𝑡𝑚] 



IHJPAS. 37 (1) 2024 

431 
 

=
𝑟!

(𝑖𝑢)(𝑟+1)𝑛
.

𝑚!

(𝑖𝑣)(𝑚+1)𝑛
. 

 

III. If 𝑓(𝑥, 𝑡) = 𝑒𝑎𝑥+𝑏𝑡, then 

 

𝐷𝐸𝐸[𝑒𝑎𝑥+𝑏𝑡] = ∫ ∫ 𝑒𝑎𝑥+𝑏𝑡𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒𝑎𝑥𝑒−𝑖𝑢𝑛𝑥  𝑑𝑥

∞

0

∫ 𝑒𝑏𝑡 𝑒−𝑖𝑣𝑛𝑡   𝑑𝑡

∞

0

 

= 𝐸𝐸[𝑒𝑎𝑥]. 𝐸𝐸[𝑒𝑏𝑡] 

= (−1) [
𝑎 + 𝑖𝑢𝑛

𝑎2 + 𝑢2𝑛
] . (−1) [

𝑏 + 𝑖𝑣𝑛

𝑏2 + 𝑣2𝑛
] 

 

=
(𝑎 + 𝑖𝑢𝑛)(𝑏 + 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
. 

IV. If  𝑓(𝑥, 𝑡) = 𝑒−(𝑎𝑥+𝑏𝑡), then 

 

𝐷𝐸𝐸[𝑒−(𝑎𝑥+𝑏𝑡)] = ∫ ∫ 𝑒−(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒−𝑎𝑥𝑒−𝑖𝑢𝑛𝑥  𝑑𝑥

∞

0

∫ 𝑒−𝑏𝑡 𝑒−𝑖𝑣𝑛𝑡  𝑑𝑡

∞

0

 

= 𝐸𝐸[𝑒−𝑎𝑥]. 𝐸𝐸[𝑒−𝑏𝑡] 

=
(−𝑎 + 𝑖𝑢𝑛)(−𝑏 + 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
 

 

=
(𝑎 − 𝑖𝑢𝑛)(𝑏 − 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
. 

V. If 𝑓(𝑥, 𝑡) = 𝑒𝑖(𝑎𝑥+𝑏𝑡), then 

 

𝐷𝐸𝐸[𝑒𝑖(𝑎𝑥+𝑏𝑡)] = ∫ ∫ 𝑒𝑖(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒𝑖𝑎𝑥𝑒−𝑖𝑢𝑛𝑥  𝑑𝑥

∞

0

∫ 𝑒𝑖𝑏𝑡 𝑒−𝑖𝑣𝑛𝑡  𝑑𝑡

∞

0

 

= 𝐸𝐸[𝑒𝑖𝑎𝑥]. 𝐸𝐸[𝑒𝑖𝑏𝑡] 

=
1

𝑖(𝑢𝑛 − 𝑎)
.

1

𝑖(𝑣𝑛 − 𝑏)
 

 

=
−1

(𝑢𝑛 − 𝑎)(𝑣𝑛 − 𝑏)
. 



IHJPAS. 37 (1) 2024 

432 
 

 

 

 

VI. If 𝑓(𝑥, 𝑡) = 𝑒−𝑖(𝑎𝑥+𝑏𝑡), then 

 

𝐷𝐸𝐸[𝑒−𝑖(𝑎𝑥+𝑏𝑡)] = ∫ ∫ 𝑒−𝑖(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑎𝑥𝑒−𝑖𝑢𝑛𝑥  𝑑𝑥

∞

0

∫ 𝑒−𝑖𝑏𝑡 𝑒−𝑖𝑣𝑛𝑡  𝑑𝑡

∞

0

 

= 𝐸𝐸[𝑒−𝑖𝑎𝑥]. 𝐸𝐸[𝑒−𝑖𝑏𝑡] 

=
1

𝑖(𝑢𝑛 + 𝑎)
.

1

𝑖(𝑣𝑛 + 𝑏)
 

 

=
−1

(𝑢𝑛 + 𝑎)(𝑣𝑛 + 𝑏)
. 

VII. If 𝑓(𝑥, 𝑡) = sin (𝑎𝑥 + 𝑏𝑡), then 

 

𝐷𝐸𝐸[sin (𝑎𝑥 + 𝑏𝑡)] = ∫ ∫ sin (𝑎𝑥 + 𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ ∫(
𝑒𝑖(𝑎𝑥+𝑏𝑡) − 𝑒−𝑖(𝑎𝑥+𝑏𝑡)

2𝑖
)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

=
1

2𝑖
[∫ ∫ 𝑒𝑖(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

−
1

2𝑖
[∫ ∫ 𝑒−𝑖(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

=
1

2𝑖
[𝐷𝐸𝐸[𝑒𝑖(𝑎𝑥+𝑏𝑡)] − 𝐷𝐸𝐸[𝑒−𝑖(𝑎𝑥+𝑏𝑡)]] 

=
1

2𝑖
[

−1

(𝑢𝑛 − 𝑎)(𝑣𝑛 − 𝑏)
+

1

(𝑢𝑛 + 𝑎)(𝑣𝑛 + 𝑏)
] 

 

=
𝑖(𝑏𝑢𝑛 + 𝑎𝑣𝑛)

(𝑢2𝑛 − 𝑎2)(𝑣2𝑛 − 𝑏2)
. 

VIII. If 𝑓(𝑥, 𝑡) = cos (𝑎𝑥 + 𝑏𝑡), then 

 

𝐷𝐸𝐸[cos (𝑎𝑥 + 𝑏𝑡)] = ∫ ∫ cos (𝑎𝑥 + 𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 



IHJPAS. 37 (1) 2024 

433 
 

= ∫ ∫(
𝑒𝑖(𝑎𝑥+𝑏𝑡) + 𝑒−𝑖(𝑎𝑥+𝑏𝑡)

2
)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

=
1

2
[∫ ∫ 𝑒𝑖(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

+
1

2
[∫ ∫ 𝑒−𝑖(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

=
1

2
[𝐷𝐸𝐸[𝑒𝑖(𝑎𝑥+𝑏𝑡)] + 𝐷𝐸𝐸[𝑒−𝑖(𝑎𝑥+𝑏𝑡)]] 

=
1

2
[

−1

(𝑢𝑛 − 𝑎)(𝑣𝑛 − 𝑏)
+

−1

(𝑢𝑛 + 𝑎)(𝑣𝑛 + 𝑏)
] 

 

=
−((𝑢𝑣)𝑛 + 𝑎𝑏)

(𝑢2𝑛 − 𝑎2)(𝑣2𝑛 − 𝑏2)
. 

IX. If 𝑓(𝑥, 𝑡) = sinh (𝑎𝑥 + 𝑏𝑡), then 

 

𝐷𝐸𝐸[sinh (𝑎𝑥 + 𝑏𝑡)] = ∫ ∫ sinh (𝑎𝑥 + 𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ ∫ (
𝑒(𝑎𝑥+𝑏𝑡) − 𝑒−(𝑎𝑥+𝑏𝑡)

2
)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

=
1

2
[∫ ∫ 𝑒(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

−
1

2
[∫ ∫ 𝑒−(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

=
1

2
[𝐷𝐸𝐸[𝑒(𝑎𝑥+𝑏𝑡)] − 𝐷𝐸𝐸[𝑒−(𝑎𝑥+𝑏𝑡)]] 

=
1

2
[

(𝑎 + 𝑖𝑢𝑛)(𝑏 + 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
−

(𝑎 − 𝑖𝑢𝑛)(𝑏 − 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
] 

 

=
𝑖(𝑎𝑣𝑛 + 𝑏𝑢𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
. 

X. If 𝑓(𝑥, 𝑡) = cosh (𝑎𝑥 + 𝑏𝑡), then 

 

𝐷𝐸𝐸[cosh (𝑎𝑥 + 𝑏𝑡)] = ∫ ∫ cosh (𝑎𝑥 + 𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ ∫ (
𝑒(𝑎𝑥+𝑏𝑡) + 𝑒−(𝑎𝑥+𝑏𝑡)

2
)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 



IHJPAS. 37 (1) 2024 

434 
 

=
1

2
[∫ ∫ 𝑒(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

+
1

2
[∫ ∫ 𝑒−(𝑎𝑥+𝑏𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

] 

=
1

2
[𝐷𝐸𝐸[𝑒(𝑎𝑥+𝑏𝑡)] + 𝐷𝐸𝐸[𝑒−(𝑎𝑥+𝑏𝑡)]] 

=
1

2
[

(𝑎 + 𝑖𝑢𝑛)(𝑏 + 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
+

(𝑎 − 𝑖𝑢𝑛)(𝑏 − 𝑖𝑣𝑛)

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
] 

 

=
𝑎𝑏 − (𝑢𝑣)𝑛

(𝑎2 + 𝑢2𝑛)(𝑏2 + 𝑣2𝑛)
. 

 

3.  (Convolution Theorem) :  

If 𝐷𝐸𝐸[𝑓(𝑥, 𝑡)] = 𝐴(𝑖𝑢, 𝑖𝑣) , 𝐷𝐸𝐸[𝑔(𝑥, 𝑡)] = 𝐵(𝑖𝑢, 𝑖𝑣), and  

𝐷𝐸𝐸[𝑓 ∗∗ 𝑔(𝑥, 𝑡)] = ∫ ∫ 𝑓(𝑥 − 𝛽, 𝑡 − 𝛼)𝑔(𝑥 − 𝛽, 𝑡 − 𝛼) 𝑑𝑡𝑑𝑥,

∞

0

∞

0

 

then  

𝐷𝐸𝐸[𝑓 ∗∗ 𝑔(𝑥, 𝑡)] = 𝐷𝐸𝐸[𝑓(𝑥, 𝑡)] . 𝐷𝐸𝐸[𝑔(𝑥, 𝑡)] = 𝐴(𝑖𝑢, 𝑖𝑣) . 𝐵(𝑖𝑢, 𝑖𝑣). 

 

4.  Double EE Integral Transform for the First and Second Derivatives : 

1.  

𝐷𝐸𝐸 [
𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
] = ∫ ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡(∫
𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
𝑒−𝑖𝑢𝑛𝑥 𝑑𝑥)𝑑𝑡

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡 (𝑒−𝑖𝑢𝑛𝑥𝑓(𝑥, 𝑡)|0
∞ + 𝑖𝑢𝑛 ∫ 𝑓(𝑥, 𝑡)

∞

0

𝑒−𝑖𝑢𝑛𝑥𝑑𝑥) 𝑑𝑡

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡 (−𝑓(0, 𝑡) + 𝑖𝑢𝑛 ∫ 𝑓(𝑥, 𝑡)

∞

0

𝑒−𝑖𝑢𝑛𝑥𝑑𝑥) 𝑑𝑡

∞

0

 

 

= − ∫ 𝑓(0, 𝑡)𝑒−𝑖𝑣𝑛𝑡𝑑𝑡 + 𝑖𝑢𝑛 ∫ ∫ 𝑓(𝑥, 𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

∞

0

 

= −𝐴(0, 𝑖𝑣) + 𝑖𝑢𝑛𝐴(𝑖𝑢, 𝑖𝑣). 

 

2.  

𝐷𝐸𝐸 [
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑥2
] = ∫ ∫

𝜕2𝑓(𝑥, 𝑡)

𝜕𝑥2
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 



IHJPAS. 37 (1) 2024 

435 
 

= ∫ 𝑒−𝑖𝑣𝑛𝑡(∫
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑥2
𝑒−𝑖𝑢𝑛𝑥 𝑑𝑥)𝑑𝑡

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡 (𝑒−𝑖𝑢𝑛𝑥
𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
|0

∞ + 𝑖𝑢𝑛 ∫
𝜕𝑓(𝑥, 𝑡)

𝜕𝑥

∞

0

𝑒−𝑖𝑢𝑛𝑥𝑑𝑥) 𝑑𝑡

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡 (−
𝜕𝑓(0, 𝑡)

𝜕𝑥
+ 𝑖𝑢𝑛 ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥

∞

0

𝑒−𝑖𝑢𝑛𝑥𝑑𝑥) 𝑑𝑡

∞

0

 

 

= − ∫
𝜕𝑓(0, 𝑡)

𝜕𝑥
𝑒−𝑖𝑣𝑛𝑡𝑑𝑡 + 𝑖𝑢𝑛 ∫ ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

∞

0

 

= −
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑥
+ 𝑖𝑢𝑛𝐷𝐸𝐸 [

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
] 

 

= −
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑥
− 𝑖𝑢𝑛𝐴(0, 𝑖𝑣) − 𝑢2𝑛𝐴(𝑖𝑢, 𝑖𝑣). 

3.  

 

𝐷𝐸𝐸 [
𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
] = ∫ ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥(∫
𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
𝑒−𝑖𝑣𝑛𝑡 𝑑𝑡)𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥 (𝑒−𝑖𝑣𝑛𝑡𝑓(𝑥, 𝑡)|0
∞ + 𝑖𝑣𝑛 ∫ 𝑓(𝑥, 𝑡)

∞

0

𝑒−𝑖𝑣𝑛𝑡𝑑𝑡) 𝑑𝑥

∞

0

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥 (−𝑓(𝑥, 0) + 𝑖𝑣𝑛 ∫ 𝑓(𝑥, 𝑡)

∞

0

𝑒−𝑖𝑣𝑛𝑡𝑑𝑡) 𝑑𝑥

∞

0

 

, 

= − ∫ 𝑓(𝑥, 0)𝑒−𝑖𝑢𝑛𝑥𝑑𝑥 + 𝑖𝑣𝑛 ∫ ∫ 𝑓(𝑥, 𝑡)𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

∞

0

 

= −𝐴(𝑖𝑢, 0) + 𝑖𝑣𝑛𝐴(𝑖𝑢, 𝑖𝑣). 

4.  

 

𝐷𝐸𝐸 [
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑡2
] = ∫ ∫

𝜕2𝑓(𝑥, 𝑡)

𝜕𝑡2
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥(∫
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑡2
𝑒−𝑖𝑣𝑛𝑡 𝑑𝑡)𝑑𝑥

∞

0

∞

0

 



IHJPAS. 37 (1) 2024 

436 
 

= ∫ 𝑒−𝑖𝑢𝑛𝑥 (𝑒−𝑖𝑣𝑛𝑡
𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
|0

∞ + 𝑖𝑣𝑛 ∫
𝜕𝑓(𝑥, 𝑡)

𝜕𝑡

∞

0

𝑒−𝑖𝑣𝑛𝑡𝑑𝑡) 𝑑𝑥

∞

0

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥 (−
𝜕𝑓(𝑥, 0)

𝜕𝑡
+ 𝑖𝑣𝑛 ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡

∞

0

𝑒−𝑖𝑣𝑛𝑡𝑑𝑡) 𝑑𝑥

∞

0

 

 

= − ∫
𝜕𝑓(𝑥, 0)

𝜕𝑡
𝑒−𝑖𝑢𝑛𝑥𝑑𝑥 + 𝑖𝑣𝑛 ∫ ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

∞

0

 

= −
𝜕𝐴(𝑖𝑢, 0)

𝜕𝑡
+ 𝑖𝑣𝑛𝐷𝐸𝐸 [

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
] 

 

= −
𝜕𝐴(𝑖𝑢, 0)

𝜕𝑡
− 𝑖𝑣𝑛𝐴(𝑖𝑢, 0) − 𝑣2𝑛𝐴(𝑖𝑢, 𝑖𝑣). 

5.  

 

𝐷𝐸𝐸 [
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑡𝜕𝑥
] = ∫ ∫

𝜕2𝑓(𝑥, 𝑡)

𝜕𝑡𝜕𝑥
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥(∫
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑡𝜕𝑥
𝑒−𝑖𝑣𝑛𝑡 𝑑𝑡)𝑑𝑥

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥 (𝑒−𝑖𝑣𝑛𝑡
𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
|0

∞ + 𝑖𝑣𝑛 ∫
𝜕𝑓(𝑥, 𝑡)

𝜕𝑥

∞

0

𝑒−𝑖𝑣𝑛𝑡𝑑𝑡) 𝑑𝑥

∞

0

 

= ∫ 𝑒−𝑖𝑢𝑛𝑥 (−
𝜕𝑓(𝑥, 0)

𝜕𝑥
+ 𝑖𝑣𝑛 ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥

∞

0

𝑒−𝑖𝑣𝑛𝑡𝑑𝑡) 𝑑𝑥

∞

0

 

 

= − ∫
𝜕𝑓(𝑥, 0)

𝜕𝑥
𝑒−𝑖𝑢𝑛𝑥𝑑𝑥 + 𝑖𝑣𝑛 ∫ ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

∞

0

 

= −
𝜕𝐴(𝑖𝑢, 0)

𝜕𝑥
+ 𝑖𝑣𝑛𝐷𝐸𝐸 [

𝜕𝑓(𝑥, 𝑡)

𝜕𝑥
] 

 

= −
𝜕𝐴(𝑖𝑢, 0)

𝜕𝑥
− 𝑖𝑣𝑛𝐴(0, 𝑖𝑣) − (𝑢𝑣)𝑛𝐴(𝑖𝑢, 𝑖𝑣). 

6.  

 

𝐷𝐸𝐸 [
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑥𝜕𝑡
] = ∫ ∫

𝜕2𝑓(𝑥, 𝑡)

𝜕𝑥𝜕𝑡
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

 



IHJPAS. 37 (1) 2024 

437 
 

= ∫ 𝑒−𝑖𝑣𝑛𝑡(∫
𝜕2𝑓(𝑥, 𝑡)

𝜕𝑥𝜕𝑡
𝑒−𝑖𝑢𝑛𝑥 𝑑𝑥)𝑑𝑡

∞

0

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡 (𝑒−𝑖𝑢𝑛𝑥
𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
|0

∞ + 𝑖𝑢𝑛 ∫
𝜕𝑓(𝑥, 𝑡)

𝜕𝑡

∞

0

𝑒−𝑖𝑢𝑛𝑥𝑑𝑥) 𝑑𝑡

∞

0

 

= ∫ 𝑒−𝑖𝑣𝑛𝑡 (−
𝜕𝑓(0, 𝑡)

𝜕𝑡
+ 𝑖𝑢𝑛 ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡

∞

0

𝑒−𝑖𝑢𝑛𝑥𝑑𝑥) 𝑑𝑡

∞

0

 

 

= − ∫
𝜕𝑓(0, 𝑡)

𝜕𝑡
𝑒−𝑖𝑣𝑛𝑡𝑑𝑡 + 𝑖𝑢𝑛 ∫ ∫

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
𝑒−𝑖(𝑢𝑛𝑥+𝑣𝑛𝑡) 𝑑𝑡𝑑𝑥

∞

0

∞

0

∞

0

 

= −
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑡
+ 𝑖𝑢𝑛𝐷𝐸𝐸 [

𝜕𝑓(𝑥, 𝑡)

𝜕𝑡
] 

 

= −
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑡
− 𝑖𝑢𝑛𝐴(𝑖𝑢, 0) − (𝑢𝑣)𝑛𝐴(𝑖𝑢, 𝑖𝑣). 

 

5. Examples of Applying The Double EE Integral Transform on Partial Differential 

Equations : 

 

Example 1.  consider the telegraph equation 

 

 

𝑢𝑥𝑥 = 𝑢𝑡𝑡 + 2𝑢𝑡 + 𝑢,                                                                                                                                (1) 

with ICs and BCs 

 

𝑢(0, 𝑡) = 𝑒−2𝑡, 𝑢𝑥(0, 𝑡) = 𝑒−2𝑡 , 𝑢(𝑥, 0) = 𝑒𝑥, 𝑢𝑡(𝑥, 0) = −2𝑒𝑥.                                                 (2) 

The exact solution of Eq (1) is 𝑢(𝑥, 𝑡) = 𝑒𝑥−2𝑡. 

 

Using the Double Complex EE Transform of both sides to Eq (1), we get: 

 

 

𝐷𝐸𝐸[𝑢𝑥𝑥] = 𝐷𝐸𝐸[𝑢𝑡𝑡] + 𝐷𝐸𝐸[2𝑢𝑡] + 𝐷𝐸𝐸[𝑢 ].                                                                                     (3) 

 

Eq (3) can be written as : 

 

 

−
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑥
− 𝑖𝑢𝑛𝐴(0, 𝑖𝑣) − 𝑢2𝑛𝐴(𝑖𝑢, 𝑖𝑣) = −

𝜕𝐴(𝑖𝑢, 0)

𝜕𝑡
− 𝑖𝑣𝑛𝐴(𝑖𝑢, 0) − 𝑣2𝑛𝐴(𝑖𝑢, 𝑖𝑣) 

                                        

                                                                                     −2𝐴(𝑖𝑢, 0) + 2𝑖𝑣𝑛𝐴(𝑖𝑢, 𝑖𝑣) + 𝐴(𝑖𝑢, 𝑖𝑣).       (4) 

 



IHJPAS. 37 (1) 2024 

438 
 

 

Besides by using the Double Complex EE Transform of ICs and BCs, we have: 

 

 

𝐴(0, 𝑖𝑣) =
−(−2 + 𝑖𝑣𝑛)

4 + 𝑣2𝑛
, 𝐴(𝑖𝑢, 0) =

−(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛
, 

 

                                       
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑥
=

−(−2 + 𝑖𝑣𝑛)

4 + 𝑣2𝑛
,
𝜕𝐴(𝑖𝑢, 0)

𝜕𝑡
=

2(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛
 ,                              (5) 

 

set Eq (5) into Eq (4), we obtain : 

 

(−2 + 𝑖𝑣𝑛)

4 + 𝑣2𝑛
+ 𝑖𝑢𝑛

(−2 + 𝑖𝑣𝑛)

4 + 𝑣2𝑛
− 𝑢2𝑛𝐴(𝑖𝑢, 𝑖𝑣)

=
−2(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛
+ 𝑖𝑣𝑛

(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛
− 𝑣2𝑛𝐴(𝑖𝑢, 𝑖𝑣) +

2(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛
+ 2𝑖𝑣𝑛𝐴(𝑖𝑢, 𝑖𝑣)

+ 𝐴(𝑖𝑢, 𝑖𝑣),      

                                        

𝐴(𝑖𝑢, 𝑖𝑣)(𝑣2𝑛 − 2𝑖𝑣𝑛 − 1 − 𝑢2𝑛) = 𝑖𝑣𝑛
(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛
+

𝑢𝑛(𝑣𝑛 + 2𝑖) − 𝑖(𝑣𝑛 + 2𝑖)

4 + 𝑣2𝑛
, 

 

 

𝐴(𝑖𝑢, 𝑖𝑣) =
𝑖𝑣𝑛(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛((𝑣𝑛 − 𝑖)2 − 𝑢2𝑛)
+

(𝑣𝑛 + 2𝑖)(𝑢𝑛 − 𝑖)

(4 + 𝑣2𝑛)((𝑣𝑛 − 𝑖)2 − 𝑢2𝑛)
.

=
(1 + 𝑖𝑢𝑛)

1 + 𝑢2𝑛

(−2 + 𝑖𝑣𝑛)

4 + 𝑣2𝑛
 .                                                                                          (6) 

 

To find the exact solution of Eq (1), we use the inverse Double Complex EE Transform to Eq (6), 

to get: 

𝑢(𝑥, 𝑡) = 𝑒𝑥−2𝑡. 

 

Example 2. consider the wave equation 

 

 

                                                     𝑢𝑥𝑥 = 𝑢𝑡𝑡 ,                                                                                               (7)   

with ICs and BCs 

 

𝑢(0, 𝑡) = 2𝑡, 𝑢𝑥(0, 𝑡) = cos 𝑡  , 𝑢(𝑥, 0) = sin 𝑥 , 𝑢𝑡(𝑥, 0) = 2.                                                      (8) 

The exact solution of Eq (7) is 𝑢(𝑥, 𝑡) = sin 𝑥 cos 𝑡 + 2𝑡. 

 

Using the Double Complex EE Transform of both sides to Eq (7), we get: 

 

 



IHJPAS. 37 (1) 2024 

439 
 

                                                 𝐷𝐸𝐸[𝑢𝑥𝑥] = 𝐷𝐸𝐸[𝑢𝑡𝑡].                                                                              (9) 

 

Eq (9) can be written as : 

 

 

−
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑥
− 𝑖𝑢𝑛𝐴(0, 𝑖𝑣) − 𝑢2𝑛𝐴(𝑖𝑢, 𝑖𝑣) = −

𝜕𝐴(𝑖𝑢, 0)

𝜕𝑡
− 𝑖𝑣𝑛𝐴(𝑖𝑢, 0) − 𝑣2𝑛𝐴(𝑖𝑢, 𝑖𝑣).   (10) 

                                        

 

Besides by using the Double Complex EE Transform of ICs and BCs, we have: 

 

 

𝐴(0, 𝑖𝑣) =
−2

𝑣2𝑛
, 𝐴(𝑖𝑢, 0) =

−1

𝑢2𝑛 − 1
,
𝜕𝐴(0, 𝑖𝑣)

𝜕𝑥
=

−𝑖𝑣𝑛

𝑣2𝑛 − 1
,
𝜕𝐴(𝑖𝑢, 0)

𝜕𝑡
=

−2𝑖

𝑢𝑛
,                       (11) 

 

set Eq (11) into Eq (10), we obtain : 

 

𝑖𝑣𝑛

𝑣2𝑛 − 1
+

2𝑖𝑢𝑛

𝑣2𝑛
− 𝑢2𝑛𝐴(𝑖𝑢, 𝑖𝑣) =

2𝑖

𝑢𝑛
+

𝑖𝑣𝑛

𝑢2𝑛 − 1
− 𝑣2𝑛𝐴(𝑖𝑢, 𝑖𝑣),      

                                        

𝐴(𝑖𝑢, 𝑖𝑣)(𝑣2𝑛 − 𝑢2𝑛) =
2𝑖

𝑢𝑛
+

𝑖𝑣𝑛

𝑢2𝑛 − 1
−

𝑖𝑣𝑛

𝑣2𝑛 − 1
−

2𝑖𝑢𝑛

𝑣2𝑛
, 

 

 

𝐴(𝑖𝑢, 𝑖𝑣) =
2𝑖𝑣2𝑛 − 2𝑖𝑢2𝑛

𝑢𝑛𝑣2𝑛(𝑣2𝑛 − 𝑢2𝑛)
+

𝑖𝑣𝑛(𝑣2𝑛 − 1) − 𝑖𝑣𝑛(𝑢2𝑛 − 1)

(𝑢2𝑛 − 1)(𝑣2𝑛 − 1)(𝑣2𝑛 − 𝑢2𝑛)

=
2𝑖

𝑢𝑛𝑣2𝑛
+

𝑖𝑣𝑛

(𝑢2𝑛 − 1)(𝑣2𝑛 − 1)
 .                                                                          (12) 

 

To find the exact solution of Eq (7), we use the inverse Double Complex EE Transform to Eq (12), 

to get: 

𝑢(𝑥, 𝑡) = 2𝑡 + sin 𝑥 cos 𝑡. 

5. Conclusion 

In this paper, we have dealt with the new double transform ( Double Complex EE Transform ) and 

some important properties and definitions to it, then after that we applied this transform to solve 

the telegraph equation, and wave equation, which are important partial differential equations in 

applied mathematics. 

Acknowledgment 

The authors are greatly appreciated the referees for their valuable comments and suggestions for 

improving the paper 

Conflict of Interest  

The authors declare that they have no conflicts of interest. 

Funding 

There is no financial support in preparation for the publication. 



IHJPAS. 37 (1) 2024 

440 
 

 

References 

 

1. Turq S. M.; Kuffi E. A. On the Double of the Emad-Falih Transformation and Its Properties with 

Applications. Ibn Al-Haitham Journal for Pure and Applied Sciences 2022, 35(4), 220–234. 

2. Saadeh, R.; Ghazal, B.; Burqan, A. A study of double general transform for solving fractional 

partial differential equations. Mathematical Methods in the Applied Sciences 2023, 46(16), 

17158-17176. 

3. Issa, A.; Kuffi, E. A.; Düz, M. A Further Generalization of the General Polynomial Transform 

and Its Basic Characteristics and Applications. Journal of University of Anbar for Pure Science 

2023, 17(2), 338–342. 

4. Kilicman, A.; Gadain, H. An application of double Laplace transform and double Sumudu 

transform. Lobachevskii Journal of Mathematics 2009, 30(3), 214-223. 

5. Ramadan, M. A.; Hadhoud, A. R. Double Ramadan group integral transform: definition and 

properties with applications to partial differential equations. Applied Mathematics and 

Information Sciences 2018, 12, 389-396. 

6. Yang, X. J. A new integral transform operator for solving the heat-diffusion problem. Applied 

Mathematics Letters 2017, 64, 193-197. 

7. Dhunde, R. R.; Waghmare, G. L. Double Laplace transform method in mathematical physics. 

International Journal of Theoretical and Mathematical Physics 2017, 7, 14-20. 

8. Tchuenche, J. M.; Mbare, N.S. An application of the double Sumudu transform. Applied 

Mathematical Sciences 2007, 1(1), 31–39. 

9. Eltayeb, H.; Kilicman, A. On Double Sumudu Transform and Double Laplace Transform. 

Malaysian journal of mathematical sciences 2010, 4, 17–30.  

10. Ali, A.; Suwan, I.; Abdeljawad, T. Numerical simulation of time partial fractional diffusion 

model by Laplace transform. AIMS Mathematics 2022, 7(2), 2878-2890.  

11. Mohand, M.; Mahgoub, A. The New Integral Transform (Mohand Transform), Advances in 

Theoretical and Applied Mathematics 2017, 12, 113–120.  

12. Kamal, A.; Sedeeg, H. The New Integral Transform Kamal Transform, Advances in 

Theoretical and Applied Mathematics 2016, 11, 451–458. 

13. Verma, D.; Alam, A.; Singh, A. P. Applications of Aboodh Transform. International Research 

Journal of Innovations in Engineering and Technology 2020, 4, 47–51.  

14. Jaabar, S.M.; Hussain, A.H. Solving Linear System of Third order of PDEs by Using Al-

Zughair Transform, Journal of Discrete Mathematical Sciences and Cryptography 2021, 24, 

1683–1688.  

15. Ahmed, S.A.; Elzaki, T.M.; Elbadri, M.; Mohamed, M.Z. Solution of Partial Differential 

Equations by New Double Integral Transform (Laplace-Sumudu Transform), Ain Shams 

Engineering Journal 2021, 12, 4045–4049. 

16. Georgieva, A. Double Fuzzy Sumudu Transform to Solve Partial Volterra Fuzzy Integro-

Differential Equations. Mathematics 2020, 8(5), 692. 

17. Debnath, L. The Double Laplace Transforms and Their Properties with Applications to 

Functional, Integral and Partial Differential Equations. International Journal of Applied and 

Computational Mathematics 2016, 2, 223–241. 

18. Kiliçman, A.; Omran, M. On Double Natural Transform and Its Applications. Journal of 

Nonlinear Sciences & Applications (JNSA) 2017, 10 (4), 1744–1754. 



IHJPAS. 37 (1) 2024 

441 
 

19. Aboodh, K.S.; Farah, R.A.; Almardy, I.A.; Almostafa, F.A. Solution of Partial Integro-

Differential Equations by Using Aboodh and Double Aboodh Transforms Methods. Global 

Journal of Pure and Applied Mathematics 2016, 13(8), 4347-4360. 

20. Sonawane, S.M.; Kiwne, S.B. Double Kamal transforms: Properties and Applications. J. Appl. 

Sci. Comput 2019, 4, 1727-1739. 

21. Jadhav, S.; Basotia, V.; Hiwarekar, A. An Application of The Double Elzaki Transform in 

Partial Differential Equations. International Journal of Advanced Research in Science, 

Communication and Technology (IJARSCT) 2022, 2(1), 181-186. 

22. Alfaqeih, S.; Misirli, E. On Double Shehu Transform and Its Properties with Applications. 

International Journal of Analysis and Applications 2020, 18(3), 381-395. 

23. Patil, D.P. Double Mahgoub Transform for The Solution of Parabolic Boundary Value 

Problems. Journal of Engineering Mathematics and Stat 2020, 4(2), 28-36. 

24. Gupta, R. Solving Electric Transmission Line Wave Equations via Double Gupta Transform. 

International Journal of Advanced Research in Electrical, Electronics and Instrumentation 

Engineering (IJAREEIE) 2022, 11(6), 2533-2538. 

25. Saadeh, R. Applications of Double ARA Integral Transform. Computation 2022, 10(12), 216. 

26. Patil, D.P.; Patil, D.S.; Kanchan, S.M. New Integral Transform,“Double Kushare Transform”. 

IRE Journals 2022, 6(1), 45-52. 

27. Ahmed, S.A.; Elzaki, T.M.; Elbadri, M.; Mohamed, M.Z. Solution of Partial Differential 

Equations by New Double Integral Transform (Laplace-Sumudu Transform). Ain Shams 

Engineering Journal 2021, 12(4), 4045-4049. 

28. Aboodh, K.S.; Farah, R.A.; Almardy, I.A.; ALmostafa, F.A. Solution of Telegraph Equation 

by Using Double Aboodh Transform, Elixir International Journal 2017, 110, 48213–48217. 

29. Kuffi, E. A.; Abbas, E. S. A Complex Integral Transform (Complex EE Transform) and Its 

Applications, Mathematical Statistician and Engineering Applications 2022, 71, 263–266.  

30. Meddahi, M.; Jafari, H.; Yang, X.J. Towards New General Double Integral Transform and Its 

Applications to Differential Equations. Mathematical Methods in the Applied Sciences 2022, 

45(4), 1916-1933.