# Solution of Integral Differential Equations by New Double Integral Transform (Laplace–Sumudu Transform)

Authors: Shams A. Ahmed; Tarig M. Elzaki; Abdelgabar Adam Hassan
Journal: Abstract and Applied Analysis
 (2020)

---

## Abstract

The primary purpose of this research is to demonstrate an efficient replacement double transform named the Laplace–Sumudu transform (DLST) to unravel integral differential equations. The theorems handling fashionable properties of the Laplace–Sumudu transform are proved; the convolution theorem with an evidence is mentioned; then, via the usage of these outcomes, the solution of integral differential equations is built.


---

## Body

## 1. Introduction

Double integral transform and their characteristics and theories are nevertheless new and below studies [1–3], in which the preceding research treated some components of them along with definitions, simple theories, and the answer of normal and partial differential equations [4–16]; additionally, some researchers addressed these transforms and combine them with exclusive mathematical method such as differential transform approach, homotopy perturbation technique, Adomian decomposition method, and variational iteration method [7–16] so that we can solve the linear and nonlinear fractional differential equations.In this paper, we are ready to spotlight the way during which the Laplace–Sumudu transform is blend to solve the integral differential equations.A wide range of linear integral differential equations are considered which include the Volterra integral equation (Section3.1), the Volterra integro-partial differential equation (Section 3.2), and the partial integro-differential equation (Section 3.3).Definition 1.
The double Laplace–Sumudu transform of the functionϕx,t of two variables x>0 and t>0 is denoted by LxStϕx,t=ϕ¯ρ,σ and defined as
(1)LxStϕx,t=ϕ¯ρ,σ=1σ∫0∞∫0∞e−ρx−t/σϕx,tdxdt.
Clearly, double Laplace–Sumudu transform is a linear integral transformation as shown below:(2)LxStγϕx,t+ηφx,t=1σ∫0∞∫0∞e−ρx−t/σγϕx,t+ηφx,tdxdt=1σ∫0∞∫0∞e−ρx−t/σγϕx,tdxdt+1σ∫0∞∫0∞e−ρx−t/σηφx,tdxdt=γσ∫0∞∫0∞e−ρx−t/σϕx,tdxdt+ησ∫0∞∫0∞e−ρx−t/σφx,tdxdt=γLxStϕx,t+ηLxStφx,t,where γ and η are constants.Definition 2.
The inverse double Laplace–Sumudu transformLx−1St−1ϕ¯ρ,σ=ϕx,t is defined by the following form:
(3)Lx−1St−1ϕ¯ρ,σ=ϕx,t=12πi∫γ−i∞γ+i∞eρxdρ12πi∫ω−i∞ω+i∞1σet/σϕ¯ρ,σdσ.


## 2. Double Laplace–Sumudu Transform of Basic Functions

(1)
Letϕx,t=1,x>0,t>0, then
(4)ϕ¯ρ,σ=LxSt1=1σ∫0∞∫0∞e−ρx−t/σdxdt=∫0∞e−ρxdx∫0∞1σe−t/σdt=1ρ.(2)
Letϕx,t=xctd,x>0,t>0, then
(5)ϕ¯ρ,σ=LxStxctd=1σ∫0∞∫0∞e−ρx−t/σxctddxdt=∫0∞e−ρxxcdx∫0∞1σe−t/σtddt=Γc+1ρc+1σdΓd+1,Rec>−1,Red>−1.Ifc and d are positive integral, then
(6)LxStxctd=c!d!ρc+1σd.(3)
Letϕx,t=ecx+dt, then
(7)ϕ¯ρ,σ=LxStecx+dt=1σ∫0∞∫0∞e−ρx−t/σecx+dtdxdt=dx∫0∞1σe−t/σ−dtdt=1ρ−c1−dσ.Similarly,(8)LxSteicx+dt=1ρ−ic1−idσ=ρ−cdσ+ic+dσρρ2+c21+d2σ2.Consequently,(9)LxStsincx+dt=c+dσρρ2+c21+d2σ2,LxStcoscx+dt=ρ−cdσρ2+c21+d2σ2.(4) Letϕx,t=sinhcx+dtorcoshcx+dt.Recall that(10)sinhcx+dt=ecx+dt−e−cx+dt2,coshcx+dt=ecx+dt+e−cx+dt2.Therefore,(11)LxStsinhcx+dt=c+dσρρ2−c21−d2σ2,LxStcoshcx+dt=ρ+cdσρ2−c21−d2σ2.(5) Letϕx,t=J0cxt, then
(12)LxStJ0cxt=1σ∫0∞∫0∞e−ρx−t/σJ0cxtdxdt=∫0∞1σe−t/σdt∫0∞e−ρxJ0cxtdx=1ρSe−c2t/4ρ=44ρ+σc2,whereJ0x is the modified Bessel function of order zero.(6) Letϕx,t=fxgt, then
(13)ϕ¯ρ,σ=LxStfxgt=1σ∫0∞∫0∞e−ρx−t/σfxgtdxdt=∫0∞e−ρxfxdx∫0∞1σe−t/σgtdt=LxfxStgt.


### 2.1. Existence Condition for the Double Laplace–Sumudu Transform

Ifϕx,t is an exponential order, then c and d as x→∞,t→∞., and if ∃ a positive constant K such that ∀x>X,t>T, then
(14)ϕx,t=Kecx+dt,and we write ϕx,t=Oecx+dt as x→∞,t→∞. Or, equivalently,
(15)limx→∞,t→∞e−ρx−t/σϕx,t=Klimx→∞,t→∞e−ρ−cx−t/σ−dt=0,ρ>c,1σ>d.The function ϕx,t is called an exponential order as x→∞,t→∞, and clearly, it does not grow faster than Kecx+dt as x→∞,t→∞.Theorem 3.
If a functionϕx,t is a continuous function in every finite interval 0,X and 0,T of exponential order ecx+dt, then the double Laplace–Sumudu transform of ϕx,t exists for all ρ and 1/σ provided Reρ>c and Re1/σ>d.Proof.
From the Definition1., we have
(16)ϕ¯ρ,σ=1σ∫0∞∫0∞e−ρx−t/σϕx,tdxdt≤K∫0∞e−ρ−cxdx∫0∞1σe−t/σ−dtdt=Kρ−c1−dσ,Reρ>c,Re1σ>d.Then, from Eq. (16) we have limx→∞,t→∞ϕ¯ρ,σ=0, or limx→∞,t→∞ϕ¯ρ,σ=0.


### 2.2. Basic Derivative Properties of the Double Laplace–Sumudu Transform

Ifϕ¯ρ,σ=LxStϕx,t, then
(17)(I)LxSt∂ϕx,t∂x=ρϕ¯ρ,σ−Sϕ0,t.Proof.
(18)LxSt∂ϕx,t∂x=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂xdxdt=1σ∫0∞e−t/σdt∫0∞e−ρx∂ϕx,t∂xdx.Using integration by parts, letu=e−ρx,dv=∂ϕx,t/∂xdx, then
(19)LxSt∂ϕx,t∂x=1σ∫0∞e−t/σdt−ϕ0,t+ρ∫0∞e−ρxϕx,tdx=ρϕ¯ρ,σ−Sϕ0,t.(20)(II)LxSt∂ϕx,t∂t=1σϕ¯ρ,σ−1σLϕx,0.Proof.
(21)LxSt∂ϕx,t∂t=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂tdxdt=1σ∫0∞e−ρxdx∫0∞e−tσ∂ϕx,t∂tdt.Using integration by parts, letu=e−t/σ,dv=∂ϕx,t/∂tdt, then LxSt∂ϕx,t/∂t=1/σ∫0∞e−ρxdx−ϕx,0+1/σ∫0∞e−t/σϕx,tdt=1/σϕ¯ρ,σ−1/σLϕx,0.Similarly, we can prove(22)LxSt∂2ϕx,t∂x2=ρ2ϕ¯ρ,σ−ρSϕ0,t−Sϕx0,t.(III)LxSt∂2ϕx,t∂t2=1σ2ϕ¯ρ,σ−1σ2Lϕx,0−1σLϕtx,0.LxSt∂2ϕx,t∂x∂t=ρσϕ¯ρ,σ−ρσLϕx,0−Sϕt0,t.Theorem 4.
Ifϕ¯ρ,σ=LxStϕx,t, then
(23)LxStϕx−δ,t−εHx−δ,t−ε=e−ρδ−ε/σϕ¯ρ,σ,whereHx,t is the Heaviside unit step function defined by
(24)Hx−δ,t−ε=1,x>δ,t>ε0,otherwise.Proof.
We have, by Definition1.,
(25)LxStϕx−δ,t−εHx−δ,t−ε=1σ∫0∞∫0∞e−ρx−t/σϕx−δ,t−εHx−δ,t−εdxdt=1σ∫δ∞∫ε∞e−ρx−t/σϕx−δ,t−εdxd,that is, by puttingx−δ=q,t−ε=w(26)=e−ρδ−ε/σ1σ∫0∞∫0∞e−ρq−w/σϕq,wdqdw=e−ρδ−ε/σϕ¯ρ,σ.


### 2.3. Convolution Theorem of Double Laplace–Sumudu Transform

Definition 5.
The convolution ofϕx,t and ψx,t is denoted by ϕ∗∗ψx,t and defined by
(27)ϕ∗∗ψx,t=∫0x∫0tϕx−δ,t−εψδ,εdδdε.Theorem 6.
(convolution theorem) IfLxStϕx,t=ϕ¯ρ,σ and LxStψx,t=ψ¯ρ,σ, then
(28)LxStϕ∗∗ψx,t=σϕ¯ρ,σψ¯ρ,σ.Proof.
From the definition1., we have
(29)LxStϕ∗∗ψx,t=1σ∫0∞∫0∞e−ρx−t/σϕ∗∗ψx,tdxdt=1σ∫0∞∫0∞e−ρx−t/σ∫0x∫0tϕx−δ,t−εψδ,εdδdεdxdt,which is, using the Heaviside unit step function,
(30)=1σ∫0∞∫0∞e−ρx−t/σ∫0∞∫0∞ϕx−δ,t−εHx−δ,t−εψδ,εdδdεdxdt=∫0∞∫0∞ψδ,εdδdε1σ∫0∞∫0∞e−ρx−t/σϕx−δ,t−εHx−δ,t−εdxdt,that is, by Theorem 4 gives
(31)=∫0∞∫0∞ψδ,εdδdεe−ρδ−ε/σϕ¯ρ,σ=ϕ¯ρ,σ∫0∞∫0∞e−ρδ−ε/σψδ,εdδdε=σϕ¯ρ,σψ¯ρ,σ.


## 2.1. Existence Condition for the Double Laplace–Sumudu Transform

Ifϕx,t is an exponential order, then c and d as x→∞,t→∞., and if ∃ a positive constant K such that ∀x>X,t>T, then
(14)ϕx,t=Kecx+dt,and we write ϕx,t=Oecx+dt as x→∞,t→∞. Or, equivalently,
(15)limx→∞,t→∞e−ρx−t/σϕx,t=Klimx→∞,t→∞e−ρ−cx−t/σ−dt=0,ρ>c,1σ>d.The function ϕx,t is called an exponential order as x→∞,t→∞, and clearly, it does not grow faster than Kecx+dt as x→∞,t→∞.Theorem 3.
If a functionϕx,t is a continuous function in every finite interval 0,X and 0,T of exponential order ecx+dt, then the double Laplace–Sumudu transform of ϕx,t exists for all ρ and 1/σ provided Reρ>c and Re1/σ>d.Proof.
From the Definition1., we have
(16)ϕ¯ρ,σ=1σ∫0∞∫0∞e−ρx−t/σϕx,tdxdt≤K∫0∞e−ρ−cxdx∫0∞1σe−t/σ−dtdt=Kρ−c1−dσ,Reρ>c,Re1σ>d.Then, from Eq. (16) we have limx→∞,t→∞ϕ¯ρ,σ=0, or limx→∞,t→∞ϕ¯ρ,σ=0.


## 2.2. Basic Derivative Properties of the Double Laplace–Sumudu Transform

Ifϕ¯ρ,σ=LxStϕx,t, then
(17)(I)LxSt∂ϕx,t∂x=ρϕ¯ρ,σ−Sϕ0,t.Proof.
(18)LxSt∂ϕx,t∂x=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂xdxdt=1σ∫0∞e−t/σdt∫0∞e−ρx∂ϕx,t∂xdx.Using integration by parts, letu=e−ρx,dv=∂ϕx,t/∂xdx, then
(19)LxSt∂ϕx,t∂x=1σ∫0∞e−t/σdt−ϕ0,t+ρ∫0∞e−ρxϕx,tdx=ρϕ¯ρ,σ−Sϕ0,t.(20)(II)LxSt∂ϕx,t∂t=1σϕ¯ρ,σ−1σLϕx,0.Proof.
(21)LxSt∂ϕx,t∂t=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂tdxdt=1σ∫0∞e−ρxdx∫0∞e−tσ∂ϕx,t∂tdt.Using integration by parts, letu=e−t/σ,dv=∂ϕx,t/∂tdt, then LxSt∂ϕx,t/∂t=1/σ∫0∞e−ρxdx−ϕx,0+1/σ∫0∞e−t/σϕx,tdt=1/σϕ¯ρ,σ−1/σLϕx,0.Similarly, we can prove(22)LxSt∂2ϕx,t∂x2=ρ2ϕ¯ρ,σ−ρSϕ0,t−Sϕx0,t.(III)LxSt∂2ϕx,t∂t2=1σ2ϕ¯ρ,σ−1σ2Lϕx,0−1σLϕtx,0.LxSt∂2ϕx,t∂x∂t=ρσϕ¯ρ,σ−ρσLϕx,0−Sϕt0,t.Theorem 4.
Ifϕ¯ρ,σ=LxStϕx,t, then
(23)LxStϕx−δ,t−εHx−δ,t−ε=e−ρδ−ε/σϕ¯ρ,σ,whereHx,t is the Heaviside unit step function defined by
(24)Hx−δ,t−ε=1,x>δ,t>ε0,otherwise.Proof.
We have, by Definition1.,
(25)LxStϕx−δ,t−εHx−δ,t−ε=1σ∫0∞∫0∞e−ρx−t/σϕx−δ,t−εHx−δ,t−εdxdt=1σ∫δ∞∫ε∞e−ρx−t/σϕx−δ,t−εdxd,that is, by puttingx−δ=q,t−ε=w(26)=e−ρδ−ε/σ1σ∫0∞∫0∞e−ρq−w/σϕq,wdqdw=e−ρδ−ε/σϕ¯ρ,σ.


## 2.3. Convolution Theorem of Double Laplace–Sumudu Transform

Definition 5.
The convolution ofϕx,t and ψx,t is denoted by ϕ∗∗ψx,t and defined by
(27)ϕ∗∗ψx,t=∫0x∫0tϕx−δ,t−εψδ,εdδdε.Theorem 6.
(convolution theorem) IfLxStϕx,t=ϕ¯ρ,σ and LxStψx,t=ψ¯ρ,σ, then
(28)LxStϕ∗∗ψx,t=σϕ¯ρ,σψ¯ρ,σ.Proof.
From the definition1., we have
(29)LxStϕ∗∗ψx,t=1σ∫0∞∫0∞e−ρx−t/σϕ∗∗ψx,tdxdt=1σ∫0∞∫0∞e−ρx−t/σ∫0x∫0tϕx−δ,t−εψδ,εdδdεdxdt,which is, using the Heaviside unit step function,
(30)=1σ∫0∞∫0∞e−ρx−t/σ∫0∞∫0∞ϕx−δ,t−εHx−δ,t−εψδ,εdδdεdxdt=∫0∞∫0∞ψδ,εdδdε1σ∫0∞∫0∞e−ρx−t/σϕx−δ,t−εHx−δ,t−εdxdt,that is, by Theorem 4 gives
(31)=∫0∞∫0∞ψδ,εdδdεe−ρδ−ε/σϕ¯ρ,σ=ϕ¯ρ,σ∫0∞∫0∞e−ρδ−ε/σψδ,εdδdε=σϕ¯ρ,σψ¯ρ,σ.


## 3. Application of Laplace–Sumudu Transform (DLST) of Integral Differential Equations

In this section, we apply the double Laplace–Sumudu transform (DLST) method to linear integral differential equations.


### 3.1. Volterra Integral Equation

Consider the linear Volterra integral equation as form(32)ϕx,t=gx,t+λ∫0x∫0tϕx−δ,t−εψδ,εdδdε,where ϕx,t is the unknown function, λ is a constant, and gx,t and ψx,t are two known functions. Applying the double Laplace–Sumudu transform (DLST) with linearity to both sides of equation (32) and using Theorem 6 (convolution theorem), we get
(33)ϕ¯ρ,σ=g¯ρ,σ+λσϕ¯ρ,σψ¯ρ,σ.Consequently,(34)ϕ¯ρ,σ=g¯ρ,σ1−λσψ¯ρ,σ.TakingLx−1St−1ϕ¯ρ,σ for equation (34), we obtain the solution ϕx,t of equation (32).
(35)ϕx,t=Lx−1St−1g¯ρ,σ1−λσψ¯ρ,σ.We illustrate the above method by simple examples.(a)
Solve the equation(36)ϕx,t=a−λ∫0x∫0tϕδ,εdδdε.where a and λ are constant.Applying the double Laplace–Sumudu transform (DLST) of equation (36), we get
(37)ϕ¯ρ,σ=aρ−λσρϕ¯ρ,σ.Consequently,(38)ϕ¯ρ,σ=aρ+λσ.TakingLx−1St−1 for equation (38), we obtain the solution ϕx,t of equation (36).
(39)ϕx,t=Lx−1St−1aρ+λσ=aJ02λxt.(b)
Solve the equation(40)a2t=∫0x∫0tϕx−δ,t−εϕδ,εdδdε,where a is a constant.Applying (DLST) of equation (40), we get
(41)a2σρ=σϕ2¯ρ,σ.Or
(42)ϕ¯ρ,σ=aρ.TakingLx−1St−1 for equation (42), we obtain the solution ϕx,t of equation (40).
(43)ϕx,t=Lx−1St−1aρ=aπ.1x.(c)
Solve the equation(44)∫0x∫0teδ−εϕx−δ,t−εdδdε=xex−t−xex.Applying (DLST) of equation (44), we get
(45)σϕ¯ρ,σρ−11+σ=1ρ−121+σ−1ρ−12.Simplifying and takingLx−1St−1 for equation (45), we obtain
(46)ϕx,t=Lx−1St−1−1ρ−1=−ex.


### 3.2. Volterra Integro-Partial Differential Equations

Consider the linear Volterra integro-partial differential equation as form(47)∂ϕx,t∂x+∂ϕx,t∂t=gx,t+λ∫0x∫0tϕx−δ,t−εψδ,εdδdε,with the conditions:
(48)ϕx,0=f0x,ϕ0,t=h0t,where ϕx,t is the unknown function, λ is a constant, and gx,t and ψx,t are two known functions.Applying (DLST) to both sides of (47) and single (LT) and (ST) for equation (48) and simplification, we get
(49)ϕ¯ρ,σ=f0¯ρ+σh0¯σ+σg¯ρ,σ1+σρ−λσ2ψ¯ρ,σ.ApplyingLx−1St−1 to (49), we obtain the solution of (47) in the form
(50)ϕx,t=Lx−1St−111+σρ−λσ2ψ¯ρ,σf0¯ρ+σh0¯σ+σg¯ρ,σ.We illustrate the above method by a simple example.(d)
Solve the equationBy substitutingψδ,ε=1,λ=1,gx,t=−1+ex+et+ex+t in (47), we have got
(51)∂ϕx,t∂x+∂ϕx,t∂t=−1+ex+et+ex+t+∫0x∫0tϕx−δ,t−εdδdε,with the conditions:
(52)ϕx,0=ex=f0x,ϕ0,t=et=h0t.Substituting(53)f0¯ρ=1ρ−1,h0¯σ=11−σ,g¯ρ,σ=−1ρ+1ρ−1+1ρ1−σ+1ρ−11−σin (50) and simplifying, we get the solution of (51)
(54)ϕx,t=Lx−1St−11ρ−11−σ=ex+t.


### 3.3. Partial Integro-Differential Equation

Consider the linear partial integro-differential equation as form(55)∂2ϕx,t∂t2−∂2ϕx,t∂x2+ϕx,t+∫0x∫0tψx−δ,t−εϕδ,εdδdε=gx,t,with the conditions:
(56)ϕx,0=f0x,∂ϕx,0∂t=f1x,ϕ0,t=h0t,∂ϕ0,t∂x=h1t.Applying (DLST) to both sides of (55) and single (LT) and (ST) for equation (56) and simplification, we get
(57)ϕ¯ρ,σ=f0¯ρ+σf1¯ρ−σ2ρh0¯σ−σ2h1¯σ+σ2g¯ρ,σ1−σ2ρ2+σ2+σ3ψ¯ρ,σ.ApplyingLx−1St−1 to (57), we obtain the solution of (55) in the form
(58)ϕx,t=Lx−1St−1f0¯ρ+σf1¯ρ−σ2ρh0¯σ−σ2h1¯σ+σ2g¯ρ,σ1−σ2ρ2+σ2+σ3ψ¯ρ,σ.We illustrate the above method by a simple example.(e)
Solve the equation:By substitutingψx−δ,t−ε=ex−δ+t−ε,gx,t=ex+t+xtex+t in (55), we have got
(59)∂2ϕx,t∂t2−∂2ϕx,t∂x2+ϕx,t+∫0x∫0tex−δ+t−εϕδ,εdδdε=ex+t+xtex+t,with the conditions:
(60)ϕx,0=ex=f0x,∂ϕx,0∂t=ex=f1x,ϕ0,t=et=h0t,∂ϕ0,t∂x=et=h1t.Substituting(61)f0¯ρ=f1¯ρ=1ρ−1,h0¯σ=h1¯σ=11−σ,g¯ρ,σ=1ρ−11−σ+σρ−121−σ2in (58) and simplifying, we get a solution of (59)
(62)ϕx,t=Lx−1St−11ρ−11−σ=ex+t.


## 3.1. Volterra Integral Equation

Consider the linear Volterra integral equation as form(32)ϕx,t=gx,t+λ∫0x∫0tϕx−δ,t−εψδ,εdδdε,where ϕx,t is the unknown function, λ is a constant, and gx,t and ψx,t are two known functions. Applying the double Laplace–Sumudu transform (DLST) with linearity to both sides of equation (32) and using Theorem 6 (convolution theorem), we get
(33)ϕ¯ρ,σ=g¯ρ,σ+λσϕ¯ρ,σψ¯ρ,σ.Consequently,(34)ϕ¯ρ,σ=g¯ρ,σ1−λσψ¯ρ,σ.TakingLx−1St−1ϕ¯ρ,σ for equation (34), we obtain the solution ϕx,t of equation (32).
(35)ϕx,t=Lx−1St−1g¯ρ,σ1−λσψ¯ρ,σ.We illustrate the above method by simple examples.(a)
Solve the equation(36)ϕx,t=a−λ∫0x∫0tϕδ,εdδdε.where a and λ are constant.Applying the double Laplace–Sumudu transform (DLST) of equation (36), we get
(37)ϕ¯ρ,σ=aρ−λσρϕ¯ρ,σ.Consequently,(38)ϕ¯ρ,σ=aρ+λσ.TakingLx−1St−1 for equation (38), we obtain the solution ϕx,t of equation (36).
(39)ϕx,t=Lx−1St−1aρ+λσ=aJ02λxt.(b)
Solve the equation(40)a2t=∫0x∫0tϕx−δ,t−εϕδ,εdδdε,where a is a constant.Applying (DLST) of equation (40), we get
(41)a2σρ=σϕ2¯ρ,σ.Or
(42)ϕ¯ρ,σ=aρ.TakingLx−1St−1 for equation (42), we obtain the solution ϕx,t of equation (40).
(43)ϕx,t=Lx−1St−1aρ=aπ.1x.(c)
Solve the equation(44)∫0x∫0teδ−εϕx−δ,t−εdδdε=xex−t−xex.Applying (DLST) of equation (44), we get
(45)σϕ¯ρ,σρ−11+σ=1ρ−121+σ−1ρ−12.Simplifying and takingLx−1St−1 for equation (45), we obtain
(46)ϕx,t=Lx−1St−1−1ρ−1=−ex.


## 3.2. Volterra Integro-Partial Differential Equations

Consider the linear Volterra integro-partial differential equation as form(47)∂ϕx,t∂x+∂ϕx,t∂t=gx,t+λ∫0x∫0tϕx−δ,t−εψδ,εdδdε,with the conditions:
(48)ϕx,0=f0x,ϕ0,t=h0t,where ϕx,t is the unknown function, λ is a constant, and gx,t and ψx,t are two known functions.Applying (DLST) to both sides of (47) and single (LT) and (ST) for equation (48) and simplification, we get
(49)ϕ¯ρ,σ=f0¯ρ+σh0¯σ+σg¯ρ,σ1+σρ−λσ2ψ¯ρ,σ.ApplyingLx−1St−1 to (49), we obtain the solution of (47) in the form
(50)ϕx,t=Lx−1St−111+σρ−λσ2ψ¯ρ,σf0¯ρ+σh0¯σ+σg¯ρ,σ.We illustrate the above method by a simple example.(d)
Solve the equationBy substitutingψδ,ε=1,λ=1,gx,t=−1+ex+et+ex+t in (47), we have got
(51)∂ϕx,t∂x+∂ϕx,t∂t=−1+ex+et+ex+t+∫0x∫0tϕx−δ,t−εdδdε,with the conditions:
(52)ϕx,0=ex=f0x,ϕ0,t=et=h0t.Substituting(53)f0¯ρ=1ρ−1,h0¯σ=11−σ,g¯ρ,σ=−1ρ+1ρ−1+1ρ1−σ+1ρ−11−σin (50) and simplifying, we get the solution of (51)
(54)ϕx,t=Lx−1St−11ρ−11−σ=ex+t.


## 3.3. Partial Integro-Differential Equation

Consider the linear partial integro-differential equation as form(55)∂2ϕx,t∂t2−∂2ϕx,t∂x2+ϕx,t+∫0x∫0tψx−δ,t−εϕδ,εdδdε=gx,t,with the conditions:
(56)ϕx,0=f0x,∂ϕx,0∂t=f1x,ϕ0,t=h0t,∂ϕ0,t∂x=h1t.Applying (DLST) to both sides of (55) and single (LT) and (ST) for equation (56) and simplification, we get
(57)ϕ¯ρ,σ=f0¯ρ+σf1¯ρ−σ2ρh0¯σ−σ2h1¯σ+σ2g¯ρ,σ1−σ2ρ2+σ2+σ3ψ¯ρ,σ.ApplyingLx−1St−1 to (57), we obtain the solution of (55) in the form
(58)ϕx,t=Lx−1St−1f0¯ρ+σf1¯ρ−σ2ρh0¯σ−σ2h1¯σ+σ2g¯ρ,σ1−σ2ρ2+σ2+σ3ψ¯ρ,σ.We illustrate the above method by a simple example.(e)
Solve the equation:By substitutingψx−δ,t−ε=ex−δ+t−ε,gx,t=ex+t+xtex+t in (55), we have got
(59)∂2ϕx,t∂t2−∂2ϕx,t∂x2+ϕx,t+∫0x∫0tex−δ+t−εϕδ,εdδdε=ex+t+xtex+t,with the conditions:
(60)ϕx,0=ex=f0x,∂ϕx,0∂t=ex=f1x,ϕ0,t=et=h0t,∂ϕ0,t∂x=et=h1t.Substituting(61)f0¯ρ=f1¯ρ=1ρ−1,h0¯σ=h1¯σ=11−σ,g¯ρ,σ=1ρ−11−σ+σρ−121−σ2in (58) and simplifying, we get a solution of (59)
(62)ϕx,t=Lx−1St−11ρ−11−σ=ex+t.


## 4. Conclusion

In this paper, the Laplace–Sumudu transform approach for solving integral differential equations is studied. We provided the theorems and popular properties for this new double transform and furnished some examples. The examples show that the Laplace–Sumudu transform approach is powerful in solving the equations of taken into consideration type, and a couple of advanced problems in linear and nonlinear partial differential equations and nonlinear integral differential equations could be discussed during a later paper.



---

*Source: 4725150-2020-10-19.xml*