# Solving Partial Integro-Differential Equations via Double Formable Transform

Authors: Bayan Ghazal; Rania Saadeh; Abdelilah K. Sedeeg
Journal: Applied Computational Intelligence and Soft Computing
 (2022)

---

## Abstract

In this study, we present a new double integral transform called the double formable transform. Several properties and theorems related to existing conditions, partial derivatives, the double convolution theorem, and others are presented. Additionally, we use a convolution kernel to solve linear partial integro-differential equations (PIDE) by using the double formable transform. By solving numerous cases, the double formable transform’s ability to turn the PIDE into an algebraic equation that is simple to solve is demonstrated.


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## Body

## 1. Introduction

One of the most significant subjects that might be interesting in life issues is mathematics; it helps to improve mental abilities and foster logical and sound thinking. Additionally, it aids in the understanding and resolution of problems arising from a variety of natural phenomena as well as those pertaining to physical, engineering, biological, and computer science [1–5]. Recent studies have demonstrated the value of integral transforms in the solution of partial equations relating to, for example, sound, heat, static electricity, elasticity, fluid flow, and other physical and engineering phenomena [6, 7].In addition, the integral transform helped to convert the partial and integral equations into algebraic equations that provide the problems’ precise solutions in the most straightforward manner [8–11]. The double integral transforms have drawn researchers’ attention recently because of their contribution to the solution of equations involving two variables and finding precise solutions in the simplest ways possible. However, with the development of science and the urgent need for mathematics to solve newly emerged problems, it was necessary to pay attention to scientists and researchers to obtain new and advanced methods to keep speed with these problems [12–14]. The most popular double integral transforms that are used to solve partial differential equations are the double Laplace transform [15, 16], the double Laplace-Sumudu transform [17], the double Sumudu transform [18], the ARA-Sumudu transform [19, 20], the double Shehu transform [21], the double Elzaki transform [12], and others [22]. Integral equations of various types played a significant role in explaining many phenomena and coming up with exact solutions when various sciences became complex as a result of their interactions with one another and scientists started studying all physical, chemical, biological, and engineering phenomena. As a result, integral equations are crucial for representing the many disciplines, which prompted scientists and researchers to look for a solution to make it possible for them to solve the integral equations.There are several types of integral equations, including Volterra integral equations and integro-differential equations; in this paper, we will focus on these two types of integral equations and solve many examples of each [23].In 2021, with the advent of the formable integral transform (FT) that was presented by Saadah and Ghazal [24], its ability to solve ordinary, partial differential equations, and integral equations, and its properties that could enable us to solve a wide kind of differential and integral problems, we intend to define the new double formable transform. Moreover, it is worth mentioning here that the formable transform might be considered a special case of the new integral transform presented by Jafari in [25], and the idea of double integral transforms is also generalized by the authors in [14]. The main advantage of using the double formable transform is that it preserves the values of constants, which simplifies the calculations during the solving of problems, and it transforms the target problem from two-dimensional space into four-dimensional space. In this study, we focus on the double formable transform and study the main properties and the application of solving integral equations.Our goal in this paper is to provide a solution to two types of integral equations, the Volterra integral equations, and the partial integro-differential equations, by using the double formable transform (DFT). We have also clarified and explained some theorems and properties with their proofs, and among the most important theorems that will help us solve the examples are the double convolution, partial derivatives, and many more, as we explain in the subsequent sections.This article is organized as follows: in Section2, the fundamental facts of formable transform are introduced. In Section 3, we introduce a new double formable transform and present some properties and theorems. In Section 4, we apply the DFT to different types of IDEs and solve some examples. Lastly, in Section 5, the conclusion section is presented.


## 2. Fundamental Facts of Formable Transform [24]

A new integral transform known as the formable transform of the continuous functionht on the interval 0,∞ is defined by(1)ℛht=Bs,u=su∫0∞e−suthtdt.The inverse formable transform is defined as(2)ℛ−1ℛht=12πi∫c−i∞c+i∞es/utBs,uds=ht.


### 2.1. Formable Transform of Some Basic Functions

(3)
i
.
ℛ
1
=
1
.
(4)
ii
.
ℛ
t
α
=
u
α
s
α
Γ
α
+
1
,
α
>
0
,
(5)
iii
.
ℛ
e
at
=
s
s
−
au
,
(6)
iv
.
ℛ
sin
at
=
asu
s
2
+
a
2
u
2
,
(7)
v
.
ℛ
sinh
at
=
asu
s
2
−
a
2
u
2
,
(8)
vi
.
ℛ
cos
at
=
s
2
s
2
+
a
2
u
2
,
(9)
vii
.
ℛ
cosh
at
=
s
2
s
2
−
a
2
u
2
,
(10)
viii
.
ℛ
th
t
=
u
2
s
d
B
s
,
u
du
+
u
s
B
s
,
u
.


### 2.2. Formable Transform of Derivatives

(11)
i
.
ℛ
h
′
t
=
s
u
B
s
,
u
−
s
u
h
0
,
(12)
ii
.
ℛ
h
m
+
1
t
=
s
u
m
+
1
B
s
,
u
−
∑
j
=
0
m
s
u
m
+
1
−
j
h
j
0
,
m
=
0,1,2
,
…
.The above results can be obtained from the definition of FT with simple calculations.


## 2.1. Formable Transform of Some Basic Functions

(3)
i
.
ℛ
1
=
1
.
(4)
ii
.
ℛ
t
α
=
u
α
s
α
Γ
α
+
1
,
α
>
0
,
(5)
iii
.
ℛ
e
at
=
s
s
−
au
,
(6)
iv
.
ℛ
sin
at
=
asu
s
2
+
a
2
u
2
,
(7)
v
.
ℛ
sinh
at
=
asu
s
2
−
a
2
u
2
,
(8)
vi
.
ℛ
cos
at
=
s
2
s
2
+
a
2
u
2
,
(9)
vii
.
ℛ
cosh
at
=
s
2
s
2
−
a
2
u
2
,
(10)
viii
.
ℛ
th
t
=
u
2
s
d
B
s
,
u
du
+
u
s
B
s
,
u
.


## 2.2. Formable Transform of Derivatives

(11)
i
.
ℛ
h
′
t
=
s
u
B
s
,
u
−
s
u
h
0
,
(12)
ii
.
ℛ
h
m
+
1
t
=
s
u
m
+
1
B
s
,
u
−
∑
j
=
0
m
s
u
m
+
1
−
j
h
j
0
,
m
=
0,1,2
,
…
.The above results can be obtained from the definition of FT with simple calculations.


## 3. Double Formable Transform (DFT)

Definition 1.
Lethx,t be a continuous function of two variables x>0 and t>0. Then, the double formable transform (DFT) of a function hx,t is defined as(13)ℛxℛthx,t=ℛxℛthx,t;v,r,s,u=ℛxℛthx,t;t⟶s,u;x⟶v,r=vr∫0∞e−v/rxsu∫0∞e−s/uthx,tdtdx,=suvr∫0∞∫0∞e−v/rx+s/uthx,tdxdt,which is equivalent to(14)ℛxℛthx,t=Hv,r,s,u=sv∫0∞∫0∞e−vx+sthrx,utdxdt.
The inverse of DFT is given by(15)ℛx−1ℛt−1ℛxℛthx,t=12πi∫c−i∞c+i∞12πi∫d−i∞d+i∞1s1resut+vrtHv,r,s,udsdr=hx,t.
We denote the single FT as follows:(i)
With respect tox: ℛxhx,t=Hv,r,t.(ii)
With respect tot: ℛthx,t=Hx,s,u.
Clearly, the DFT is a linear transformation as shown below:(16)ℛxℛtahx,t+bmx,t=suvr∫0∞∫0∞e−v/rx+s/utahx,t+bmx,tdxdt=asuvr∫0∞∫0∞e−v/rx+s/uthx,tdxdt+bsuvr∫0∞∫0∞e−v/rx+s/utmx,tdxdt=aℛxℛthx,t+bℛxℛtmx,t,where α and β are constants and ℛxℛthx,t,ℛxℛtmx,t are exists.Property 1.
Lethx,t=fxgt,x>0,t>0. Then,(17)ℛxℛthx,t=ℛxfxℛtgt.Proof.
(18)
ℛ
x
ℛ
t
h
x
,
t
=
ℛ
x
ℛ
t
f
x
g
t
=
s
u
v
r
∫
0
∞
∫
0
∞
e
−
v
/
r
x
+
s
/
u
t
f
x
g
t
d
x
d
t
=
v
r
∫
0
∞
e
−
v
/
r
x
f
x
d
x
s
u
∫
0
∞
e
−
s
/
u
t
g
t
d
t
=
ℛ
x
f
x
ℛ
t
g
t
.


### 3.1. DFT of Some Basic Functions

(i)
Lethx,t=1,x>0,t>0. Then,(19)ℛxℛt1=suvr∫0∞∫0∞e−v/rx+s/utdxdt=vr∫0∞e−v/rxdxsu∫0∞e−s/utdt=ℛx1ℛt1.
From equation(3), we get ℛxℛt1=1.
(ii)
Lethx,t=xαtβ,x>0,t>0 and α,β are constants. Then,(20)ℛxℛtxαtβ=suvr∫0∞∫0∞e−v/rx+s/utxαtβdxdt=vr∫0∞e−v/rxxαdxsu∫0∞e−s/uttβdt=ℛxxαℛttβ.
From equation (4), we get ℛxℛtxαtβ=u/sαr/vβΓα+1Γβ+1.
(iii)
Lethx,t=eαx+βt,x>0,t>0 and α,β are positive constants. Then,(21)ℛxℛteαx+βt=suvr∫0∞∫0∞e−v/rx+s/uteαx+βtdxdt=vr∫0∞e−v/rxeαxdxsu∫0∞e−s/uteβtdt=ℛxeαxℛteβt.From equation (5), we get ℛxℛteαx+βt=sv/s−αuv−βr.Similarly,ℛxℛteiαx+βt=sv/s−iαuv−iβr.Using the property of complex analysis, we have(22)ℛxℛteiαx+βt=svsv−αβur+isvsrβ+uvαs2+α2u2v2+β2r2.Using Euler’s formula,sinx=eix−e−ix/2i,cosx=eix+e−ix/2,and the formula,sinhx=ex−e−x/2i,coshx=ex+e−x/2.Thus, we find the DFT of the following functions:(23)ℛxℛtsinαx+βt=svsrβ+uvαs2+α2u2v2+β2r2,ℛxℛtcosαx+βt=svsv−αβurs2+α2u2v2+β2r2,ℛxℛtsinhαx+βt=svsrβ+uvαs2−α2u2v2−β2r2,ℛxℛtcoshαx+βt=svsv+αβurs2−α2u2v2−β2r2,ℛxℛtJ0ℷxt==suvr∫0∞∫0∞e−v/rx+s/utJ0ℷxtdxdt=vr∫0∞J0ℷxte−v/rxdxsu∫0∞e−s/utdt=su∫0∞e−ℷ2r/4vte−s/utdt.Thus, we getℛxℛtJ0ℷxt=4sv/4vs+ℷ2ru.(24)ℛxℛtxtex+t=suvr∫0∞∫0∞e−v/rx+s/utxtex+tdxdt=vr∫0∞xexe−v/rxdtsu∫0∞tete−s/utdt.From equations (5) and (6), we get(25)ℛxℛtxtex+t=u2sds/s−udu+usss−ur2vdv/v−rdr+rvvv−r=suvrv−r2s−u2.


### 3.2. Existence Conditions for DFT

Ifhx,t is a function of exponential orders α and β as x⟶∞ and t⟶∞, and if there exists a positive constant M such that ∀x>X and t>T, we have(26)hx,t≤Meαx+βt,we can write hx,t=Oeαx+βt as x⟶∞ and t⟶∞, v/r>α and s/u>β.Theorem 1.
Lethx,t be a continuous function on the region 0,X×0,T of exponential orders α and β. Then, ℛxℛthx,t exists for v/r and s/u provided Rev/r>α and Res/u>β.Proof.
Using the definition of DFT, we get(27)ℛxℛthx,t=suvr∫0∞∫0∞e−v/rx+s/uthx,tdxdt≤suvr∫0∞∫0∞e−v/rx+s/uthx,tdxdt≤Mvr∫0∞e−v/r−αxdxsu∫0∞e−s/u−βtdx=Mvsv−αrs−βu,Revr>αResu>β.


### 3.3. Some Basic Theorems of DFT

Theorem 2 (Shifting Property).
Lethx,t be a continuous function, and ℛ2hx,t=Hv,r,s,u. Then,(28)ℛxℛteαx+βthx,t=svv−αrs−βuHv,vrv−αr,s,sus−βu.Proof.
Using the definition in equation (14),(29)ℛxℛteαx+βthx,t=sv∫0∞∫0∞e−vx+steαrx+βuthrx,utdxdt=sv∫0∞∫0∞e−v−αrx+s−βuthrx,utdxdt.
By lettingv−αrx=vy, and dx=v/v−αrdy and s−βut=sτ, and dt=s/s−βudτ in (29), we have(30)ℛxℛteαx+βthx,t=svv−αrs−βu∫0∞∫0∞e−vy+sτhvrv−αry,sus−βuτdydτ=svv−αrs−βuHv,vrv−αr,s,sus−βu.Theorem 3 (Periodic function).
Letℛxℛthx,t be exists, where hx,t is a periodic function of periods α and β such that,(31)hx+α,t+β=hx,t,∀x,y.Then,(32)ℛxℛthx,t=1−e−vrβ+suα−1suvr∫0α∫0βe−vrx+suthx,tdxdt.Proof.
Using the definition of DFT, we get(33)ℛxℛthx,t=suvr∫0∞∫0∞e−vrx+suthx,tdxdt.
Using the property of improper integral, equation (33) can be written as(34)ℛxℛthx,t=suvr∫0α∫0βe−v/rx+s/uthx,tdxdt+suvr∫α∞∫β∞e−v/rx+s/uthx,tdxdt.
Puttingx=ρ+β and t=τ+α on the second integral in equation (34). We obtain(35)ℛxℛthx,t=suvr∫0α∫0βe−vrx+suthx,tdxdt+suvr∫α∞∫β∞e−vrρ+β+suτ+αhρ+β,τ+αdρdτ.
Using the periodicity of the functionhx,t, equation (35) can be written by(36)ℛxℛthx,t=suvr∫0α∫0βe−vrx+suthx,tdxdt+e−vrβ+suαsuvr∫0∞∫0∞e−vrρ+suτhρ,τdρdτ.
Using the definition of DFT, we get(37)ℛxℛthx,t=suvr∫0α∫0βe−vrx+suthx,tdxdt+e−vrβ+suαℛxℛthx,t.
Thus, equation (37) can be simplified into(38)ℛxℛthx,t=11−e−v/rβ+s/uαsuvr∫0α∫0βe−vrx+suthx,tdxdt.Theorem 4 (Heaviside function).
Letℛxℛthx,t exists, then(39)ℛxℛthx−δ,t−σHx−δ,t−σ=e−vrδ−suσℛxℛthx,t,where Hx−δ,t−σ is the Heaviside unit step function defined as(40)Hx−δ,t−σ=1,x>δ,t>σ,0,Ohtherwise.Proof.
Using the definition of DFT, we get(41)ℛxℛthx−δ,t−σHx−δ,t−σ=suvr∫0∞∫0∞e−vrx+suthx−δ,t−σHx−δ,t−σdxdt=suvr∫0∞∫0∞e−vrx+suthx−δ,t−σdxdt
Puttingx−δ=ρ and t−σ=τ in equation (41). We obtain(42)ℛxℛthx−δ,t−σHx−δ,t−σ=suvr∫0∞∫0∞e−vrρ+δ+suτ+σhρ,τdρdτ.Thus, equation (42) can be simplified into(43)ℛxℛthx−δ,t−σHx−δ,t−σ=e−vrδ−suσsuvr∫0∞∫0∞e−vrρ−suτhρ,τdρdτ=e−vrδ−suσℛxℛthx,t.Theorem 5 (Convolution theorem).
Letℛxℛthx,t and ℛxℛtkx,t are exists, then(44)ℛxℛthx,t∗∗kx,t=ursvℛxℛthx,tℛxℛtkx,t,where hx,t∗∗kx,t=∫0x∫0thx−ρ,t−τkρ,τdρdτ and the symbol ∗∗ denotes the double convolution with respect to x and t.Proof.
Using the definition of DFT, we get(45)ℛxℛthx,t∗∗kx,t=suvr∫0∞∫0∞e−vrx+sut∫0x∫0thx−ρ,t−τkρ,τdρdτdxdt.
Using the Heaviside unit step function, equation (45) can be written as(46)ℛxℛthx,t∗∗kx,t=suvr∫0∞∫0∞e−vrx+sut∫0∞∫0∞hx−ρ,t−τHx−ρ,t−τkρ,τdρdτdxdt.
Thus, equation (46) can be written as(47)ℛxℛthx,t∗∗kx,t=∫0∞∫0∞kρ,τdρdτsuvr∫0∞∫0∞e−vrx+suthx−ρ,t−τHx−ρ,t−τdxdt=∫0∞∫0∞kρ,τdρdτe−vrρ−suτℛxℛthx,t=ℛxℛthx,t∫0∞∫0∞e−vrρ−suτkρ,τdρdτ=ursvℛxℛthx,tℛxℛtkx,t.Theorem 6 (Derivatives properties).
Lethx,t be a continuous function. Then, we get the following derivatives properties:(a)
ℛxℛt∂hx,t/∂t=s/uℛxℛthx,t−s/uℛxhx,0.(b)
ℛxℛt∂hx,t/∂x=v/rℛxℛthx,t−v/rℛth0,t.(c)
ℛxℛt∂2hx,t/∂t2=s2/u2ℛxℛthx,t−s2/u2ℛxhx,0−s/uℛx∂hx,0/∂t.(d)
ℛxℛt∂2hx,t/∂x2=v2/r2ℛxℛthx,t−v2/r2ℛth0,t−v/rℛt∂h0,t/∂x(48)eℛxℛt∂2hx,t∂x∂t=vsruℛxℛthx,t−ℛxhx,0−ℛth0,t+h0,0.We only present the proof of (a), (c), and (e), the proof of (b) and (d) are the same.Proof.
(a)ℛxℛt∂hx,t/∂t=s/uv/r∫0∞∫0∞e−v/rx+s/ut∂hx,t/∂tdxdt=v/r∫0∞e−v/rxdxs/u∫0∞e−s/ut∂hx,t/∂tdt
Using the integration by part, we obtain.
Letu=e−s/ut⇒du=−s/ue−s/utdt, dv=∂hx,t/∂tdt⇒v=hx,t.
Thus,s/u∫0∞e−s/ut∂hx,t/∂tdt=s/u−hx,0+s/u∫0∞e−s/uthx,tdt,
So, we get,(49)ℛxℛt∂hx,t∂t=suℛxℛthx,t−suℛxhx,0.Proof.
(c).ℛxℛt∂2hx,t/∂t2=s/uv/r∫0∞∫0∞e−v/rx+s/ut∂2hx,t/∂t2dxdt=v/r∫0∞e−v/rxdxs/u∫0∞e−s/ut∂2hx,t/∂t2dt.
Using the integration by part, we obtain.
Letu=e−s/ut⇒du=−s/ue−s/utdt,dv=∂2hx,t/∂t2dt⇒v=∂hx,t/∂t.
Thus,s/u∫0∞e−s/ut∂2hx,t/∂t2dt=s/u−∂hx,0/∂t+s/u∫0∞e−s/ust∂hx,t/∂tdt.
Using equation (49), we have(50)ℛxℛt∂2hx,t∂t2=s2u2ℛxℛthx,t−s2u2ℛxhx,0−suℛx∂hx,0∂t.
proof of (e) by similar (a) and (c) arguments, one can get(51)ℛxℛt∂2hx,t∂x∂t=suvr∫0∞∫0∞e−vrx+sut∂2hx,t∂x∂tdxdt=vr∫0∞e−vrx∂2hx,t∂x∂tdxsu∫0∞e−sutdt.
Using the integration by part, we obtain(52)vr∫0∞e−vrx∂2hx,t∂x∂tdx=−vr∫0∞e−vrx∂h0,t∂tdt+vr2su∫0∞∫0∞e−vrx+sut∂hx,t∂tdxdt.
Using equations (11) and (49), we have(53)ℛxℛt∂2hx,t∂x∂t=vsruℛxℛthx,t−ℛxhx,0−ℛth0,t+h0,0.
We summarize the previous results in the following table below Table1:Table 1
DFT for some functions.
h
x
,
t
ℛ
x
ℛ
t
h
x
,
t
1
1
x
α
t
β
u
/
s
α
r
/
v
β
Γ
α
+
1
Γ
β
+
1
.
e
α
x
+
β
t
s
v
/
s
−
α
u
v
−
β
r
sin
α
x
+
β
t
s
v
v
u
β
+
s
r
α
/
s
2
+
α
2
u
2
v
2
+
β
2
r
2
cos
α
x
+
β
t
s
v
s
v
−
α
β
u
r
/
s
2
+
α
2
u
2
v
2
+
β
2
r
2
sinh
α
x
+
β
t
s
v
v
u
β
+
s
r
α
/
s
2
−
α
2
u
2
v
2
−
β
2
r
2
cosh
α
x
+
β
t
s
v
s
v
+
α
β
u
r
/
s
2
−
α
2
u
2
v
2
−
β
2
r
2
xte
x
+
t
suvr
/
v
−
r
2
s
−
u
2
J
0
ℷ
x
t
4
v
s
/
4
v
s
+
ℷ
2
ru
e
α
x
+
β
t
h
x
,
t
s
v
/
v
−
α
r
s
−
β
u
H
v
,
vr
/
v
−
α
r
,
s
,
s
u
/
s
−
β
u
h
x
−
δ
,
t
−
σ
H
x
−
δ
,
t
−
σ
e
−
v
/
r
δ
−
s
/
u
σ
ℛ
x
ℛ
t
h
x
,
t
h
x
,
t
∗
∗
k
x
,
t
ur
/
sv
ℛ
x
ℛ
t
h
x
,
t
ℛ
x
ℛ
t
k
x
,
t
∂
h
x
,
t
/
∂
t
s
/
u
ℛ
x
ℛ
t
h
x
,
t
−
s
/
u
ℛ
x
h
x
,
0
∂
h
x
,
t
/
∂
x
v
/
r
ℛ
x
ℛ
t
h
x
,
t
−
v
/
r
ℛ
t
h
0
,
t
∂
2
h
x
,
t
/
∂
t
2
s
2
/
u
2
ℛ
x
ℛ
t
h
x
,
t
−
s
2
/
u
2
ℛ
x
h
x
,
0
−
s
/
u
ℛ
x
∂
h
x
,
0
/
∂
t
∂
2
h
x
,
t
/
∂
x
2
v
2
/
r
2
ℛ
x
ℛ
t
h
x
,
t
−
v
2
/
r
2
ℛ
t
h
0
,
t
−
v
/
r
ℛ
t
∂
h
0
,
t
/
∂
x
∂
h
x
,
t
/
∂
x
∂
t
v
s
/
r
u
ℛ
x
ℛ
t
h
x
,
t
−
ℛ
x
h
x
,
0
−
ℛ
t
h
0
,
t
+
h
0,0


## 3.1. DFT of Some Basic Functions

(i)
Lethx,t=1,x>0,t>0. Then,(19)ℛxℛt1=suvr∫0∞∫0∞e−v/rx+s/utdxdt=vr∫0∞e−v/rxdxsu∫0∞e−s/utdt=ℛx1ℛt1.
From equation(3), we get ℛxℛt1=1.
(ii)
Lethx,t=xαtβ,x>0,t>0 and α,β are constants. Then,(20)ℛxℛtxαtβ=suvr∫0∞∫0∞e−v/rx+s/utxαtβdxdt=vr∫0∞e−v/rxxαdxsu∫0∞e−s/uttβdt=ℛxxαℛttβ.
From equation (4), we get ℛxℛtxαtβ=u/sαr/vβΓα+1Γβ+1.
(iii)
Lethx,t=eαx+βt,x>0,t>0 and α,β are positive constants. Then,(21)ℛxℛteαx+βt=suvr∫0∞∫0∞e−v/rx+s/uteαx+βtdxdt=vr∫0∞e−v/rxeαxdxsu∫0∞e−s/uteβtdt=ℛxeαxℛteβt.From equation (5), we get ℛxℛteαx+βt=sv/s−αuv−βr.Similarly,ℛxℛteiαx+βt=sv/s−iαuv−iβr.Using the property of complex analysis, we have(22)ℛxℛteiαx+βt=svsv−αβur+isvsrβ+uvαs2+α2u2v2+β2r2.Using Euler’s formula,sinx=eix−e−ix/2i,cosx=eix+e−ix/2,and the formula,sinhx=ex−e−x/2i,coshx=ex+e−x/2.Thus, we find the DFT of the following functions:(23)ℛxℛtsinαx+βt=svsrβ+uvαs2+α2u2v2+β2r2,ℛxℛtcosαx+βt=svsv−αβurs2+α2u2v2+β2r2,ℛxℛtsinhαx+βt=svsrβ+uvαs2−α2u2v2−β2r2,ℛxℛtcoshαx+βt=svsv+αβurs2−α2u2v2−β2r2,ℛxℛtJ0ℷxt==suvr∫0∞∫0∞e−v/rx+s/utJ0ℷxtdxdt=vr∫0∞J0ℷxte−v/rxdxsu∫0∞e−s/utdt=su∫0∞e−ℷ2r/4vte−s/utdt.Thus, we getℛxℛtJ0ℷxt=4sv/4vs+ℷ2ru.(24)ℛxℛtxtex+t=suvr∫0∞∫0∞e−v/rx+s/utxtex+tdxdt=vr∫0∞xexe−v/rxdtsu∫0∞tete−s/utdt.From equations (5) and (6), we get(25)ℛxℛtxtex+t=u2sds/s−udu+usss−ur2vdv/v−rdr+rvvv−r=suvrv−r2s−u2.


## 3.2. Existence Conditions for DFT

Ifhx,t is a function of exponential orders α and β as x⟶∞ and t⟶∞, and if there exists a positive constant M such that ∀x>X and t>T, we have(26)hx,t≤Meαx+βt,we can write hx,t=Oeαx+βt as x⟶∞ and t⟶∞, v/r>α and s/u>β.Theorem 1.
Lethx,t be a continuous function on the region 0,X×0,T of exponential orders α and β. Then, ℛxℛthx,t exists for v/r and s/u provided Rev/r>α and Res/u>β.Proof.
Using the definition of DFT, we get(27)ℛxℛthx,t=suvr∫0∞∫0∞e−v/rx+s/uthx,tdxdt≤suvr∫0∞∫0∞e−v/rx+s/uthx,tdxdt≤Mvr∫0∞e−v/r−αxdxsu∫0∞e−s/u−βtdx=Mvsv−αrs−βu,Revr>αResu>β.


## 3.3. Some Basic Theorems of DFT

Theorem 2 (Shifting Property).
Lethx,t be a continuous function, and ℛ2hx,t=Hv,r,s,u. Then,(28)ℛxℛteαx+βthx,t=svv−αrs−βuHv,vrv−αr,s,sus−βu.Proof.
Using the definition in equation (14),(29)ℛxℛteαx+βthx,t=sv∫0∞∫0∞e−vx+steαrx+βuthrx,utdxdt=sv∫0∞∫0∞e−v−αrx+s−βuthrx,utdxdt.
By lettingv−αrx=vy, and dx=v/v−αrdy and s−βut=sτ, and dt=s/s−βudτ in (29), we have(30)ℛxℛteαx+βthx,t=svv−αrs−βu∫0∞∫0∞e−vy+sτhvrv−αry,sus−βuτdydτ=svv−αrs−βuHv,vrv−αr,s,sus−βu.Theorem 3 (Periodic function).
Letℛxℛthx,t be exists, where hx,t is a periodic function of periods α and β such that,(31)hx+α,t+β=hx,t,∀x,y.Then,(32)ℛxℛthx,t=1−e−vrβ+suα−1suvr∫0α∫0βe−vrx+suthx,tdxdt.Proof.
Using the definition of DFT, we get(33)ℛxℛthx,t=suvr∫0∞∫0∞e−vrx+suthx,tdxdt.
Using the property of improper integral, equation (33) can be written as(34)ℛxℛthx,t=suvr∫0α∫0βe−v/rx+s/uthx,tdxdt+suvr∫α∞∫β∞e−v/rx+s/uthx,tdxdt.
Puttingx=ρ+β and t=τ+α on the second integral in equation (34). We obtain(35)ℛxℛthx,t=suvr∫0α∫0βe−vrx+suthx,tdxdt+suvr∫α∞∫β∞e−vrρ+β+suτ+αhρ+β,τ+αdρdτ.
Using the periodicity of the functionhx,t, equation (35) can be written by(36)ℛxℛthx,t=suvr∫0α∫0βe−vrx+suthx,tdxdt+e−vrβ+suαsuvr∫0∞∫0∞e−vrρ+suτhρ,τdρdτ.
Using the definition of DFT, we get(37)ℛxℛthx,t=suvr∫0α∫0βe−vrx+suthx,tdxdt+e−vrβ+suαℛxℛthx,t.
Thus, equation (37) can be simplified into(38)ℛxℛthx,t=11−e−v/rβ+s/uαsuvr∫0α∫0βe−vrx+suthx,tdxdt.Theorem 4 (Heaviside function).
Letℛxℛthx,t exists, then(39)ℛxℛthx−δ,t−σHx−δ,t−σ=e−vrδ−suσℛxℛthx,t,where Hx−δ,t−σ is the Heaviside unit step function defined as(40)Hx−δ,t−σ=1,x>δ,t>σ,0,Ohtherwise.Proof.
Using the definition of DFT, we get(41)ℛxℛthx−δ,t−σHx−δ,t−σ=suvr∫0∞∫0∞e−vrx+suthx−δ,t−σHx−δ,t−σdxdt=suvr∫0∞∫0∞e−vrx+suthx−δ,t−σdxdt
Puttingx−δ=ρ and t−σ=τ in equation (41). We obtain(42)ℛxℛthx−δ,t−σHx−δ,t−σ=suvr∫0∞∫0∞e−vrρ+δ+suτ+σhρ,τdρdτ.Thus, equation (42) can be simplified into(43)ℛxℛthx−δ,t−σHx−δ,t−σ=e−vrδ−suσsuvr∫0∞∫0∞e−vrρ−suτhρ,τdρdτ=e−vrδ−suσℛxℛthx,t.Theorem 5 (Convolution theorem).
Letℛxℛthx,t and ℛxℛtkx,t are exists, then(44)ℛxℛthx,t∗∗kx,t=ursvℛxℛthx,tℛxℛtkx,t,where hx,t∗∗kx,t=∫0x∫0thx−ρ,t−τkρ,τdρdτ and the symbol ∗∗ denotes the double convolution with respect to x and t.Proof.
Using the definition of DFT, we get(45)ℛxℛthx,t∗∗kx,t=suvr∫0∞∫0∞e−vrx+sut∫0x∫0thx−ρ,t−τkρ,τdρdτdxdt.
Using the Heaviside unit step function, equation (45) can be written as(46)ℛxℛthx,t∗∗kx,t=suvr∫0∞∫0∞e−vrx+sut∫0∞∫0∞hx−ρ,t−τHx−ρ,t−τkρ,τdρdτdxdt.
Thus, equation (46) can be written as(47)ℛxℛthx,t∗∗kx,t=∫0∞∫0∞kρ,τdρdτsuvr∫0∞∫0∞e−vrx+suthx−ρ,t−τHx−ρ,t−τdxdt=∫0∞∫0∞kρ,τdρdτe−vrρ−suτℛxℛthx,t=ℛxℛthx,t∫0∞∫0∞e−vrρ−suτkρ,τdρdτ=ursvℛxℛthx,tℛxℛtkx,t.Theorem 6 (Derivatives properties).
Lethx,t be a continuous function. Then, we get the following derivatives properties:(a)
ℛxℛt∂hx,t/∂t=s/uℛxℛthx,t−s/uℛxhx,0.(b)
ℛxℛt∂hx,t/∂x=v/rℛxℛthx,t−v/rℛth0,t.(c)
ℛxℛt∂2hx,t/∂t2=s2/u2ℛxℛthx,t−s2/u2ℛxhx,0−s/uℛx∂hx,0/∂t.(d)
ℛxℛt∂2hx,t/∂x2=v2/r2ℛxℛthx,t−v2/r2ℛth0,t−v/rℛt∂h0,t/∂x(48)eℛxℛt∂2hx,t∂x∂t=vsruℛxℛthx,t−ℛxhx,0−ℛth0,t+h0,0.We only present the proof of (a), (c), and (e), the proof of (b) and (d) are the same.Proof.
(a)ℛxℛt∂hx,t/∂t=s/uv/r∫0∞∫0∞e−v/rx+s/ut∂hx,t/∂tdxdt=v/r∫0∞e−v/rxdxs/u∫0∞e−s/ut∂hx,t/∂tdt
Using the integration by part, we obtain.
Letu=e−s/ut⇒du=−s/ue−s/utdt, dv=∂hx,t/∂tdt⇒v=hx,t.
Thus,s/u∫0∞e−s/ut∂hx,t/∂tdt=s/u−hx,0+s/u∫0∞e−s/uthx,tdt,
So, we get,(49)ℛxℛt∂hx,t∂t=suℛxℛthx,t−suℛxhx,0.Proof.
(c).ℛxℛt∂2hx,t/∂t2=s/uv/r∫0∞∫0∞e−v/rx+s/ut∂2hx,t/∂t2dxdt=v/r∫0∞e−v/rxdxs/u∫0∞e−s/ut∂2hx,t/∂t2dt.
Using the integration by part, we obtain.
Letu=e−s/ut⇒du=−s/ue−s/utdt,dv=∂2hx,t/∂t2dt⇒v=∂hx,t/∂t.
Thus,s/u∫0∞e−s/ut∂2hx,t/∂t2dt=s/u−∂hx,0/∂t+s/u∫0∞e−s/ust∂hx,t/∂tdt.
Using equation (49), we have(50)ℛxℛt∂2hx,t∂t2=s2u2ℛxℛthx,t−s2u2ℛxhx,0−suℛx∂hx,0∂t.
proof of (e) by similar (a) and (c) arguments, one can get(51)ℛxℛt∂2hx,t∂x∂t=suvr∫0∞∫0∞e−vrx+sut∂2hx,t∂x∂tdxdt=vr∫0∞e−vrx∂2hx,t∂x∂tdxsu∫0∞e−sutdt.
Using the integration by part, we obtain(52)vr∫0∞e−vrx∂2hx,t∂x∂tdx=−vr∫0∞e−vrx∂h0,t∂tdt+vr2su∫0∞∫0∞e−vrx+sut∂hx,t∂tdxdt.
Using equations (11) and (49), we have(53)ℛxℛt∂2hx,t∂x∂t=vsruℛxℛthx,t−ℛxhx,0−ℛth0,t+h0,0.
We summarize the previous results in the following table below Table1:Table 1
DFT for some functions.
h
x
,
t
ℛ
x
ℛ
t
h
x
,
t
1
1
x
α
t
β
u
/
s
α
r
/
v
β
Γ
α
+
1
Γ
β
+
1
.
e
α
x
+
β
t
s
v
/
s
−
α
u
v
−
β
r
sin
α
x
+
β
t
s
v
v
u
β
+
s
r
α
/
s
2
+
α
2
u
2
v
2
+
β
2
r
2
cos
α
x
+
β
t
s
v
s
v
−
α
β
u
r
/
s
2
+
α
2
u
2
v
2
+
β
2
r
2
sinh
α
x
+
β
t
s
v
v
u
β
+
s
r
α
/
s
2
−
α
2
u
2
v
2
−
β
2
r
2
cosh
α
x
+
β
t
s
v
s
v
+
α
β
u
r
/
s
2
−
α
2
u
2
v
2
−
β
2
r
2
xte
x
+
t
suvr
/
v
−
r
2
s
−
u
2
J
0
ℷ
x
t
4
v
s
/
4
v
s
+
ℷ
2
ru
e
α
x
+
β
t
h
x
,
t
s
v
/
v
−
α
r
s
−
β
u
H
v
,
vr
/
v
−
α
r
,
s
,
s
u
/
s
−
β
u
h
x
−
δ
,
t
−
σ
H
x
−
δ
,
t
−
σ
e
−
v
/
r
δ
−
s
/
u
σ
ℛ
x
ℛ
t
h
x
,
t
h
x
,
t
∗
∗
k
x
,
t
ur
/
sv
ℛ
x
ℛ
t
h
x
,
t
ℛ
x
ℛ
t
k
x
,
t
∂
h
x
,
t
/
∂
t
s
/
u
ℛ
x
ℛ
t
h
x
,
t
−
s
/
u
ℛ
x
h
x
,
0
∂
h
x
,
t
/
∂
x
v
/
r
ℛ
x
ℛ
t
h
x
,
t
−
v
/
r
ℛ
t
h
0
,
t
∂
2
h
x
,
t
/
∂
t
2
s
2
/
u
2
ℛ
x
ℛ
t
h
x
,
t
−
s
2
/
u
2
ℛ
x
h
x
,
0
−
s
/
u
ℛ
x
∂
h
x
,
0
/
∂
t
∂
2
h
x
,
t
/
∂
x
2
v
2
/
r
2
ℛ
x
ℛ
t
h
x
,
t
−
v
2
/
r
2
ℛ
t
h
0
,
t
−
v
/
r
ℛ
t
∂
h
0
,
t
/
∂
x
∂
h
x
,
t
/
∂
x
∂
t
v
s
/
r
u
ℛ
x
ℛ
t
h
x
,
t
−
ℛ
x
h
x
,
0
−
ℛ
t
h
0
,
t
+
h
0,0


## 4. Presentation of DFT Method in Solving Integral Equations and Applications

To show the applicability and strength of DFT in solving integral equations, we use it to solve some kinds of integral equations and illustrate the method by solving several examples.


### 4.1. Volterra Integral Equation

Consider the following Volterra-integral equation:(54)hx,t=fx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ,where hx,t is the unknown function, fx,t and wx,t are two known functions and μ is a constant. The main idea of the method is to apply the DFT to equation (54) as the following:(55)ℛxℛthx,t=ℛxℛtfx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ.Now, using the differentiation property of the DFT on equation (55) and Theorem 5, we have(56)ℛxℛthx,t=Fv,r,s,u+μrusvℛxℛthx,tℛxℛtwx,t.Equation (56) can be simplified as follows:(57)ℛxℛthx,t=Fv,r,s,u1−μrusvℛxℛtwx,t−1,Operating with the inverse of DFT on both sides of equation (57) gives(58)hx,t=ℛx−1ℛt−1Fv,r,s,u1−μrusvℛxℛtwx,t−1,where hx,t represents the term arising from the known functions fx,t and wx,t.Example 1.
Let us consider the following Volterra integral equation:(59)hx,t=α−μ∫0x∫0thρ,τdρdτ.
By substituting the values of functions
F
=
ℛ
x
ℛ
t
α
=
α
,
W
=
ℛ
x
ℛ
t
−
1
=
−
1 in the general formula in equation (58), we obtain(60)ℛxℛthx,t=α1+μrusv−1=αsvsv+μru.
Now, applying inverse DFT to equation (60), then the solution of equation (59) is given by(61)hx,t=ℛx−1ℛt−1αsvsv+μru=ℛx−1ℛt−14αsv4sv+2μ2ru=αJ02μxt.Example 2.
Let us consider the following Volterra integral equation:(62)4t=∫0x∫0thx−ρ,t−τhρ,τdρdτ.
Appling the DFT to equation (62), we have(63)4us=rusvℛxℛthx,tℛxℛthx,t=rusvℛxℛthx,t2.
Equation (63) can be simplified as follows:(64)ℛxℛthx,t=2vr.
Taking the inverse DFT to equation (64), we get the solution of equation (62) as follows:(65)hx,t=ℛx−1ℛt−12vr=2πx.Example 3.
Let us consider the following Volterra integral equation:(66)xexe−t−1=∫0x∫0teρ−τhx−ρ,t−τdρdτ.
Appling the DFT to equation (66), we have(67)r2vddrvv−r+rvvv−rss+u−1=rusvsvℛxℛthx,tv−rs+u.
Equation (67) can be simplified as follows:(68)ℛxℛthx,t=rvv−r2−us+uv−rs+uru=−vv−r.
Applying the inverse DFT to equation (68), we get the solution of equation (66), we get(69)hx,t=ℛx−1ℛt−1−vv−r=−ex.


### 4.2. Volterra Integro-Partial Differential Equations

Consider the following Volterra integro-partial differential equations:(70)∂hx,t∂x+∂hx,t∂t=fx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ,with the conditions(71)hx,0=kx,h0,t=gt,where hx,t is the unknown function, fx,t and wx,t are two known functions, and μ is a constant.To get the solution of equation (70), we firstly apply the single formable transform to all conditions in equation (71) and the DFT to equation(70), as follows:(72)ℛxhx,0=ℛxkx=K,ℛth0,t=ℛtgt=G.Thus, we have(73)ℛxℛt∂hx,t∂x+∂hx,t∂t=ℛxℛtfx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ.Using the differentiation property of the DFT to equation (73) and the double convolution theorem, we have(74)vrℛxℛthx,t−vrG+suℛxℛthx,t−suK=Fv,r,s,u+μrusvℛxℛthx,tℛxℛtwx,t.Equation (44) can be simplified as follows:(75)ℛxℛthx,t=Fv,r,s,u+vrG+suKvr+su−μursvℛxℛtwx,t−1.Operating the inverse of DFT to both sides of equation (75), we get(76)hx,t=ℛx−1ℛt−1F+vrG+suKvr+su−μursvℛxℛtwx,t−1,where hx,t represents the term arising from the known functions fx,t,wx,t,kx, and gt.Example 4.
Let us consider the following Volterra integro-partial differential equations:(77)∂hx,t∂x+∂hx,t∂t=−1+ex+et+ex+t+∫0x∫0thx−ρ,t−τdρdτ.subject to the conditions(78)hx,0=ex,h0,t=et.
By substituting the values of functions(79)K=ℛxex=vv−r,G=ℛtet=ss−u,ℛxℛtwx,t=ℛxℛt1=1,F=ℛxℛt−1+ex+et+ex+t=−1+vv−r+ss−u+svs−uv−r,in the general formula in equation (76), we obtain(80)ℛxℛthx,tvr+su−ursv=−1+vv−r+ss−u+svs−uv−r+vrss−u+suvv−r.
Equation (80) can be simplified as follows:(81)ℛxℛthx,t=rv−r+ss−u+svs−uv−r+vrss−u+suvv−rvr+su−ursv−1=suvr−ur/sv+v/r+s/uurs−uv−rvr+su−ursv−1.
Thus,(82)ℛxℛthx,t=svs−uv−r.
Taking the inverse of DFT to equation (82), then the solution of equation (77) is(83)hx,t=ℛx−1ℛt−1svs−uv−r=ex+t.


### 4.3. Integro-Partial Differential Equations

Consider the following integro-partial differential equation:(84)∂2hx,t∂t2−∂2hx,t∂x2+hx,t+∫0x∫0twx−ρ,t−τhρ,τdρdτ=fx,t,with the conditions(85)hx,0=k1x,∂hx,0∂t=k2x,h0,t=g1t,∂h0,t∂x=g2t,where hx,t is unknown function, fx,t and wx,t are two known functions.The main idea of this method is to apply formable transform to all conditions in equation (85) and the DFT to equation (84), as the following:The single formable transform to conditions in equation (85), we get(86)ℛxhx,0=ℛxk1x=K1,ℛx∂hx,0∂t=ℛxk2x=K2,ℛth0,t=ℛtg1t=G1,ℛt∂h0,t∂x=ℛtg2t=G2.The DFT on both sides of equation (84), to get(87)ℛxℛt∂2hx,t∂t2−∂2hx,t∂x2+hx,t+∫0x∫0twx−ρ,t−τhρ,τdρdτ=ℛxℛtfx,t.Using the differentiation property of the DFT of equation (87), above conditions and Theorem 5, we have(88)s2u2ℛxℛthx,t−s2u2K1−suK2−v2r2ℛxℛthx,t−v2r2G1−vrG2+ℛxℛthx,t+rusvℛxℛthx,tℛxℛtwx,t=Fv,r,s,u.Equation (88) can be simplifying as follows:(89)ℛxℛthx,t=Fv,r,s,u+s2u2K1+suK2−v2r2G1−vrG21s2/u2−v2/r2+1+ru/svℛxℛtwx,t.Operating with the inverse of DFT on both sides of equation (89), gives(90)hx,t=ℛx−1ℛt−1F+s2u2K1+suK2−v2r2G1−vrG21s2/u2−v2/r2+1+ru/svℛxℛtwx,t,where hx,t represents the term arising from the known functions fx,t,k1x,g1t,k2x,g2t, and wx,t.Example 5.
Let us consider the following integro-partial differential equation:(91)∂2hx,t∂t2−∂2hx,t∂x2+hx,t+∫0X∫0tex−ρ+t−τhρ,τdρdτ=ex+t+xtex+t,with the conditions(92)hx,0=∂hx,0∂t=ex,h0,t=∂h0,t∂x=et.
By substituting the values of functions(93)K1=K2=ℛxex=vv−r,G1=G2=ℛtet=ss−u,F=ℛxℛtex+t+xtex+t=svv−rs−u+suvrv−r2s−u2,ℛxℛtwx,t=ℛxℛtex+t=sv/v−rs−u, in the general formula in equation (89), we obtain(94)ℛxℛthx,t=svs−uv−r.
Taking the inverse DFT to equation (94), then the solution of equation (91) is(95)hx,t=ℛx−1ℛt−1svv−rs−u=ex+t.


## 4.1. Volterra Integral Equation

Consider the following Volterra-integral equation:(54)hx,t=fx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ,where hx,t is the unknown function, fx,t and wx,t are two known functions and μ is a constant. The main idea of the method is to apply the DFT to equation (54) as the following:(55)ℛxℛthx,t=ℛxℛtfx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ.Now, using the differentiation property of the DFT on equation (55) and Theorem 5, we have(56)ℛxℛthx,t=Fv,r,s,u+μrusvℛxℛthx,tℛxℛtwx,t.Equation (56) can be simplified as follows:(57)ℛxℛthx,t=Fv,r,s,u1−μrusvℛxℛtwx,t−1,Operating with the inverse of DFT on both sides of equation (57) gives(58)hx,t=ℛx−1ℛt−1Fv,r,s,u1−μrusvℛxℛtwx,t−1,where hx,t represents the term arising from the known functions fx,t and wx,t.Example 1.
Let us consider the following Volterra integral equation:(59)hx,t=α−μ∫0x∫0thρ,τdρdτ.
By substituting the values of functions
F
=
ℛ
x
ℛ
t
α
=
α
,
W
=
ℛ
x
ℛ
t
−
1
=
−
1 in the general formula in equation (58), we obtain(60)ℛxℛthx,t=α1+μrusv−1=αsvsv+μru.
Now, applying inverse DFT to equation (60), then the solution of equation (59) is given by(61)hx,t=ℛx−1ℛt−1αsvsv+μru=ℛx−1ℛt−14αsv4sv+2μ2ru=αJ02μxt.Example 2.
Let us consider the following Volterra integral equation:(62)4t=∫0x∫0thx−ρ,t−τhρ,τdρdτ.
Appling the DFT to equation (62), we have(63)4us=rusvℛxℛthx,tℛxℛthx,t=rusvℛxℛthx,t2.
Equation (63) can be simplified as follows:(64)ℛxℛthx,t=2vr.
Taking the inverse DFT to equation (64), we get the solution of equation (62) as follows:(65)hx,t=ℛx−1ℛt−12vr=2πx.Example 3.
Let us consider the following Volterra integral equation:(66)xexe−t−1=∫0x∫0teρ−τhx−ρ,t−τdρdτ.
Appling the DFT to equation (66), we have(67)r2vddrvv−r+rvvv−rss+u−1=rusvsvℛxℛthx,tv−rs+u.
Equation (67) can be simplified as follows:(68)ℛxℛthx,t=rvv−r2−us+uv−rs+uru=−vv−r.
Applying the inverse DFT to equation (68), we get the solution of equation (66), we get(69)hx,t=ℛx−1ℛt−1−vv−r=−ex.


## 4.2. Volterra Integro-Partial Differential Equations

Consider the following Volterra integro-partial differential equations:(70)∂hx,t∂x+∂hx,t∂t=fx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ,with the conditions(71)hx,0=kx,h0,t=gt,where hx,t is the unknown function, fx,t and wx,t are two known functions, and μ is a constant.To get the solution of equation (70), we firstly apply the single formable transform to all conditions in equation (71) and the DFT to equation(70), as follows:(72)ℛxhx,0=ℛxkx=K,ℛth0,t=ℛtgt=G.Thus, we have(73)ℛxℛt∂hx,t∂x+∂hx,t∂t=ℛxℛtfx,t+μ∫0x∫0thx−ρ,t−τwρ,τdρdτ.Using the differentiation property of the DFT to equation (73) and the double convolution theorem, we have(74)vrℛxℛthx,t−vrG+suℛxℛthx,t−suK=Fv,r,s,u+μrusvℛxℛthx,tℛxℛtwx,t.Equation (44) can be simplified as follows:(75)ℛxℛthx,t=Fv,r,s,u+vrG+suKvr+su−μursvℛxℛtwx,t−1.Operating the inverse of DFT to both sides of equation (75), we get(76)hx,t=ℛx−1ℛt−1F+vrG+suKvr+su−μursvℛxℛtwx,t−1,where hx,t represents the term arising from the known functions fx,t,wx,t,kx, and gt.Example 4.
Let us consider the following Volterra integro-partial differential equations:(77)∂hx,t∂x+∂hx,t∂t=−1+ex+et+ex+t+∫0x∫0thx−ρ,t−τdρdτ.subject to the conditions(78)hx,0=ex,h0,t=et.
By substituting the values of functions(79)K=ℛxex=vv−r,G=ℛtet=ss−u,ℛxℛtwx,t=ℛxℛt1=1,F=ℛxℛt−1+ex+et+ex+t=−1+vv−r+ss−u+svs−uv−r,in the general formula in equation (76), we obtain(80)ℛxℛthx,tvr+su−ursv=−1+vv−r+ss−u+svs−uv−r+vrss−u+suvv−r.
Equation (80) can be simplified as follows:(81)ℛxℛthx,t=rv−r+ss−u+svs−uv−r+vrss−u+suvv−rvr+su−ursv−1=suvr−ur/sv+v/r+s/uurs−uv−rvr+su−ursv−1.
Thus,(82)ℛxℛthx,t=svs−uv−r.
Taking the inverse of DFT to equation (82), then the solution of equation (77) is(83)hx,t=ℛx−1ℛt−1svs−uv−r=ex+t.


## 4.3. Integro-Partial Differential Equations

Consider the following integro-partial differential equation:(84)∂2hx,t∂t2−∂2hx,t∂x2+hx,t+∫0x∫0twx−ρ,t−τhρ,τdρdτ=fx,t,with the conditions(85)hx,0=k1x,∂hx,0∂t=k2x,h0,t=g1t,∂h0,t∂x=g2t,where hx,t is unknown function, fx,t and wx,t are two known functions.The main idea of this method is to apply formable transform to all conditions in equation (85) and the DFT to equation (84), as the following:The single formable transform to conditions in equation (85), we get(86)ℛxhx,0=ℛxk1x=K1,ℛx∂hx,0∂t=ℛxk2x=K2,ℛth0,t=ℛtg1t=G1,ℛt∂h0,t∂x=ℛtg2t=G2.The DFT on both sides of equation (84), to get(87)ℛxℛt∂2hx,t∂t2−∂2hx,t∂x2+hx,t+∫0x∫0twx−ρ,t−τhρ,τdρdτ=ℛxℛtfx,t.Using the differentiation property of the DFT of equation (87), above conditions and Theorem 5, we have(88)s2u2ℛxℛthx,t−s2u2K1−suK2−v2r2ℛxℛthx,t−v2r2G1−vrG2+ℛxℛthx,t+rusvℛxℛthx,tℛxℛtwx,t=Fv,r,s,u.Equation (88) can be simplifying as follows:(89)ℛxℛthx,t=Fv,r,s,u+s2u2K1+suK2−v2r2G1−vrG21s2/u2−v2/r2+1+ru/svℛxℛtwx,t.Operating with the inverse of DFT on both sides of equation (89), gives(90)hx,t=ℛx−1ℛt−1F+s2u2K1+suK2−v2r2G1−vrG21s2/u2−v2/r2+1+ru/svℛxℛtwx,t,where hx,t represents the term arising from the known functions fx,t,k1x,g1t,k2x,g2t, and wx,t.Example 5.
Let us consider the following integro-partial differential equation:(91)∂2hx,t∂t2−∂2hx,t∂x2+hx,t+∫0X∫0tex−ρ+t−τhρ,τdρdτ=ex+t+xtex+t,with the conditions(92)hx,0=∂hx,0∂t=ex,h0,t=∂h0,t∂x=et.
By substituting the values of functions(93)K1=K2=ℛxex=vv−r,G1=G2=ℛtet=ss−u,F=ℛxℛtex+t+xtex+t=svv−rs−u+suvrv−r2s−u2,ℛxℛtwx,t=ℛxℛtex+t=sv/v−rs−u, in the general formula in equation (89), we obtain(94)ℛxℛthx,t=svs−uv−r.
Taking the inverse DFT to equation (94), then the solution of equation (91) is(95)hx,t=ℛx−1ℛt−1svv−rs−u=ex+t.


## 5. Conclusions

Different phenomena are modeled using partial integro-differential equations in sciences, engineering, and social sciences. Solving partial integral equations using the double transform technique is effective. Convolution kernels are used to solve similar problems. In this study, we use the DFT technique to solve integral equations of different kinds. We present the definition of DFT and prove some basic properties and theorems related to its existence and derivatives. Finally, we apply DFT to solve partial integral equations and partial integro-differential equations. The outcomes show the simplicity and applicability of DFT in solving such problems.



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