# Applications of New Double Integral Transform (Laplace–Sumudu Transform) in Mathematical Physics

Authors: Shams A. Ahmed
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

The primary purpose of this research is to demonstrate an efficient replacement of double transform called the double Laplace–Sumudu transform (DLST) and prove some related theorems of the new double transform. Also, we will discuss the fundamental properties of the double Laplace–Sumudu transform of some basic functions. Then, by utilizing those outcomes, we will apply it to the partial differential equations to show its simplicity, efficiency, and high accuracy.


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## Body

## 1. Introduction

Partial differential equations are used to describe many real-world problems arising in all the fields of applied science and issues associated with engineering. Due to the rapid advancement in science and engineering, various integral transforms have been used to solve the differential and integral equations, for instance, the Laplace transform [1], the Sumudu transform [2], the Elzaki transform [3], the natural transform [4], and many other double integral transforms [5, 6]. However, most of the existing integral transforms have some limitations and cannot be used directly to solve nonlinear problems or many complex mathematical models. As a result, some researchers have combined these integral transforms with other methods such as the differential transform method, homotophy perturbation method, Adomian decomposition method and variational iteration method [7–13] for solving many nonlinear differential equations.Table 1
The double Laplace–Sumudu transform of some basic functions.
Sr. no.ϕx,tLxStϕx,t=ϕ¯ρ,σ111/ρ2xctd, if c and d are positive integralc!d!/ρc+1σd3ecx+dt1/ρ−c1−dσ4sincx+dtc+dσρ/ρ2+c21+d2σ25coscx+dtρ−cdσ/ρ2+c21+d2σ26sinhcx+dtc+dσρ/ρ2−c21−d2σ27coshcx+dtρ+cdσ/ρ2−c21−d2σ28J0cxt, where J0x is the modified Bessel function of order zero4/4ρ+σc29ϕx−δ,t−εHx−δ,t−ε, where Hx,t is the Heaviside unit step functione−ρδ−ε/σϕ¯ρ,σ10ϕ∗∗ψx,t (convolution theorem)σϕ¯ρ,σψ¯ρ,σ11fxgtLxfxStgtThe usefulness of these equations has attracted the attention of many scholars throughout the history of applied sciences. One of the problems with these equations is that they are very difficult to solve equations with unknown functions of two variables sometimes. For this reason, this problem was solved by combining the Laplace transforms and Sumudu transform to give another double transform which is called the double Laplace–Sumudu transform.In the present research, we consider linear, one-dimensional, time-dependent partial differential equation (PDE) of the form(1)∑n=0Ncn∂nϕx,t∂tn=∑m=1Mdm∂mϕx,t∂xm+gx,t,x,t∈ℝ+2,with the initial conditions:
(2)∂nϕx,0∂tn=fnx,n=0,1,⋯,N−1,x∈ℝ+,and boundary conditions:
(3)∂mϕ0,t∂xm=hmt,m=0,1,⋯,M−1,t∈ℝ+,where cn,0≤n≤N; dm,1≤m≤M are given coefficients and N, M are positive integers and gx,t is the source term.The main objective of this paper is to develop new applications of the double Laplace–Sumudu transform for solving linear partial differential equations of the type (1) subject to the initial conditions (2) and boundary conditions (3). The proposed integral transform is successfully applied to a wide range of linear partial differential equations in mathematical physics.


## 2. Preliminaries

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
(4)LxStϕx,t=ϕ¯ρ,σ=1σ∫0∞∫0∞e−ρx−t/σϕx,tdxdt,whenever that integral exists. Here, ρ and σ are complex numbers.Clearly, the double Laplace–Sumudu transform is a linear integral transformation as shown below:(5)LxStγϕx,t+ηψx,t=1σ∫0∞∫0∞e−ρx−t/σγϕx,t+ηψx,tdxdt=1σ∫0∞∫0∞e−ρx−t/σγϕx,tdxdt+1σ∫0∞∫0∞e−ρx−t/σηψx,tdxdt=γσ∫0∞∫0∞e−ρx−t/σϕx,tdxdt+ησ∫0∞∫0∞e−ρx−t/σψx,tdxdt=γLxStϕx,t+ηLxStψx,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:
(6)Lx−1St−1ϕ¯ρ,σ=ϕx,t=12πi∫γ−i∞γ+i∞eρxdρ12πi∫η−i∞η+i∞1σet/σϕ¯ρ,σdσ.Definition 3.
The double Laplace–Sumudu transform formulas for the partial derivatives of an arbitrary integer order are given by (see the proof in Section3.2)
(7)LxSt∂nϕx,t∂tn=σ−nϕ¯ρ,σ−∑j=0n−1σ−n+jLx∂jϕx,0∂tj,LxSt∂mϕx,t∂xm=ρmϕ¯ρ,σ−∑k=0m−1ρm−1−kSt∂kϕ0,t∂xk.


## 3. Fundamental Properties of Double Laplace–Sumudu Transform

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

Ifϕx,t is an exponential order c and d as x→∞,t→∞., if there exists a positive constant K such that ∀x>X,t>T(8)ϕx,t=Kecx+dt,and we write
(9)ϕx,t=Oecx+dtasx→∞,t→∞.Or, equivalently,(10)limx→∞,t→∞e−ρx−t/σϕx,t=Klimx→∞,t→∞e−ρ−cx−1/σ−dt=0,ρ>c,1σ>d.The functionϕx,t is called an exponential order as x→∞,t→∞, and clearly, it does not grow faster than Kecx+dt as x→∞,t→∞.Theorem 4.
If a functionϕx,tis a continuous function in every finite interval0,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 Definition1, we have
(11)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 Equation (11), we have
(12)limx→∞,t→∞ϕ¯ρ,σ=0,orlimx→∞,t→∞ϕ¯ρ,σ=0.


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

Ifϕ¯ρ,σ=LxStϕx,t, then
(13)ILxSt∂ϕx,t∂x=ρϕ¯ρ,σ−Sϕ0,t,IILxSt∂ϕx,t∂t=1σϕ¯ρ,σ−1σLϕx,0,IIILxSt∂2ϕx,t∂x2=ρ2ϕ¯ρ,σ−ρSϕ0,t−Sϕx0,t,IVLxSt∂2ϕx,t∂t2=1σ2ϕ¯ρ,σ−1σ2Lϕx,0−1σLϕtx,0,VLxSt∂2ϕx,t∂x∂t=ρσϕ¯ρ,σ−ρσLϕx,0−Sϕt0,t.Proof.
(14)ILxSt∂ϕx,t∂x=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂xdxdt=1σ∫0∞e−t/σdt∫0∞e−ρx∂ϕx,t∂xdxlet u=e−ρx,dv=∂ϕx,t/∂xdx, thus
(15)LxSt∂ϕx,t∂x=1σ∫0∞e−t/σdt−ϕ0,t+ρ∫0∞e−ρxϕx,tdx=ρϕ¯ρ,σ−Sϕ0,t.IILxSt∂ϕx,t∂t=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂tdxdt=1σ∫0∞e−ρxdx∫0∞e−t/σ∂ϕx,t∂tdtlet u=e−t/σ,dv=∂ϕx,t/∂tdt, then
(16)LxSt∂ϕx,t∂t=1σ∫0∞e−ρxdx−ϕx,0+1σ∫0∞e−t/σϕx,tdt=1σϕ¯ρ,σ−1σLϕx,0.
Similarly, we can prove that (III), (IV), and (V).


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

Ifϕx,t is an exponential order c and d as x→∞,t→∞., if there exists a positive constant K such that ∀x>X,t>T(8)ϕx,t=Kecx+dt,and we write
(9)ϕx,t=Oecx+dtasx→∞,t→∞.Or, equivalently,(10)limx→∞,t→∞e−ρx−t/σϕx,t=Klimx→∞,t→∞e−ρ−cx−1/σ−dt=0,ρ>c,1σ>d.The functionϕx,t is called an exponential order as x→∞,t→∞, and clearly, it does not grow faster than Kecx+dt as x→∞,t→∞.Theorem 4.
If a functionϕx,tis a continuous function in every finite interval0,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 Definition1, we have
(11)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 Equation (11), we have
(12)limx→∞,t→∞ϕ¯ρ,σ=0,orlimx→∞,t→∞ϕ¯ρ,σ=0.


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

Ifϕ¯ρ,σ=LxStϕx,t, then
(13)ILxSt∂ϕx,t∂x=ρϕ¯ρ,σ−Sϕ0,t,IILxSt∂ϕx,t∂t=1σϕ¯ρ,σ−1σLϕx,0,IIILxSt∂2ϕx,t∂x2=ρ2ϕ¯ρ,σ−ρSϕ0,t−Sϕx0,t,IVLxSt∂2ϕx,t∂t2=1σ2ϕ¯ρ,σ−1σ2Lϕx,0−1σLϕtx,0,VLxSt∂2ϕx,t∂x∂t=ρσϕ¯ρ,σ−ρσLϕx,0−Sϕt0,t.Proof.
(14)ILxSt∂ϕx,t∂x=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂xdxdt=1σ∫0∞e−t/σdt∫0∞e−ρx∂ϕx,t∂xdxlet u=e−ρx,dv=∂ϕx,t/∂xdx, thus
(15)LxSt∂ϕx,t∂x=1σ∫0∞e−t/σdt−ϕ0,t+ρ∫0∞e−ρxϕx,tdx=ρϕ¯ρ,σ−Sϕ0,t.IILxSt∂ϕx,t∂t=1σ∫0∞∫0∞e−ρx−t/σ∂ϕx,t∂tdxdt=1σ∫0∞e−ρxdx∫0∞e−t/σ∂ϕx,t∂tdtlet u=e−t/σ,dv=∂ϕx,t/∂tdt, then
(16)LxSt∂ϕx,t∂t=1σ∫0∞e−ρxdx−ϕx,0+1σ∫0∞e−t/σϕx,tdt=1σϕ¯ρ,σ−1σLϕx,0.
Similarly, we can prove that (III), (IV), and (V).


## 4. Double Laplace–Sumudu Transform (DLST)

Applying the double Laplace–Sumudu transform on both sides of (1), we get
(17)∑n=0Ncnσ−nϕ¯ρ,σ−∑j=0n−1σ−n+jLx∂jϕx,0∂tj=∑m=1Mdmρmϕ¯ρ,σ−∑k=0m−1ρm−1−kSt∂kϕ0,t∂xk+g¯ρ,σ.Further, applying single Laplace transform (LT) to initial (2) and single Sumudu transform (ST) to boundary (3), we get
(18)Lx∂nϕx,0∂tn=fn¯ρ,n=0,1,⋯,N−1,St∂mϕ0,t∂xm=hm¯σ,m=0,1,⋯,M−1,substituting (18) in (17) and simplifying, we obtain
(19)∑n=0Ncnσ−nϕ¯ρ,σ−∑j=0n−1σ−n+jfn¯ρ=∑m=1Mdmρmϕ¯ρ,σ−∑k=0m−1ρm−1−khm¯σ+g¯ρ,σ.Equation (19) is an algebraic equation in ϕ¯ρ,σ. Solving this algebraic equation and taking Lx−1St−1ϕ¯ρ,σ, we obtain an exact solution ϕx,t of Equation (1).


## 5. Elucidative Examples

In this section, the applications of the proposed transform are presented. The simplicity, efficiency, and high accuracy of the double Laplace–Sumudu transform are clearly illustrated.


### 5.1. The Advection-Diffusion Equation

By substitutingN=1,M=2,c0=g=0,c1=1 in (1), we have got the advection-diffusion equation:
(20)∂ϕx,t∂t=d2∂2ϕx,t∂x2+d1∂ϕx,t∂x,with ICs and BCs
(21)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (20) as
(22)ϕx,t=Lx−1St−111−d2σρ2−d1σρf0¯ρ−d2σρh0¯σ−d2σh1¯σ−d1σh0¯σ.Example 5.
Puttingd1=−1,d2=1 in (20), we got
(23)∂ϕx,t∂t=∂2ϕx,t∂x2−∂ϕx,t∂x,with the conditions,
(24)ϕx,0=ex−x=f0x,ϕ0,t=1+t=h0t,ϕx0,t=0=h1t.Substituting(25)f0¯ρ=1ρ−1−1ρ2,h0¯σ=1+σ,h1¯σ=0,in (22) and simplifying, we get a solution of (23)
(26)ϕx,t=Lx−1St−11ρ−1−1ρ2+σρ=ex−x+t.


### 5.2. The Reaction-Diffusion Equation

By substitutingN=1, M=2, g=0, c1=1, d1=0, and d2>0 in (1), we have the reaction-diffusion equation:
(27)c0ϕx,t+∂ϕx,t∂t=d2∂2ϕx,t∂x2,with ICs and BCs:
(28)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (27) as
(29)ϕx,t=Lx−1St−111+c0σ−d2σρ2f0¯ρ−d2σρh0¯σ−d2σh1¯σ.


#### 5.2.1. The Heat (Diffusion) Equation

Takingc0=0 in (27), we have the linear heat equation:
(30)∂ϕx,t∂t=d2∂2ϕx,t∂x2,with ICs and BCs:
(31)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (29) gives the solution of (30) as
(32)ϕx,t=Lx−1St−111−d2σρ2f0¯ρ−d2σρh0¯σ−d2σh1¯σ.Example 6.
Puttingd2=1 in (30) to yield
(33)∂ϕx,t∂t=∂2ϕx,t∂x2,with the conditions,
(34)ϕx,0=sinx=f0x,ϕ0,t=0=h0t,ϕx0,t=e−t=h1t.Substituting(35)f0¯ρ=1ρ2+1,h0¯σ=0,h1¯σ=11+σ,in (32) and simplifying, we get a solution of (33)
(36)ϕx,t=Lx−1St−11ρ2+11+σ=e−tsinx.


### 5.3. The Telegraph Equation

By substitutingN=M=2,d1=g=0 in (1), we have got the linear telegraph equation
(37)c0ϕx,t+c1∂ϕx,t∂t+c2∂2ϕx,t∂t2=d2∂2ϕx,t∂x2,with ICs and BCs:
(38)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (37) as
(39)ϕx,t=Lx−1St−11c0σ2+c1σ+c2−d2σ2ρ2c1f0¯ρ+c2f0¯ρ+c2σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ.


#### 5.3.1. The Wave Equation

Takingc0=c1=0,c2=1 in (37), we obtain the linear wave equation:
(40)∂2ϕx,t∂t2=d2∂2ϕx,t∂x2,with ICs and BCs
(41)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (39) gives the solution of (40) as
(42)ϕx,t=Lx−1St−111−d2σ2ρ2f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ.


### 5.4. The Klein-Gordon Equation

By substitutingN=M=2,c1=d1=0,c2=1 in (1), we have the Klein-Gordon equation:
(43)c0ϕx,t+∂2ϕx,t∂t2=d2∂2ϕx,t∂x2+gx,t,with ICs and BCs
(44)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (43) as
(45)ϕx,t=Lx−1St−111+c0σ2−d2σ2ρ2f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ+σ2g¯ρ,σ.Example 7.
Puttingc0=−2,d2=1, and gx,t=−2sinxsint in (43) to yield
(46)∂2ϕx,t∂t2−2ϕx,t=∂2ϕx,t∂x2−2sinxsint,with the conditions,
(47)ϕx,0=0=f0x,ϕtx,0=sinx=f1x,ϕ0,t=0=h0t,ϕx0,t=sint=h1t.Substituting(48)f0¯ρ=0,f1¯ρ=1ρ2+1,h0¯σ=0,h1¯σ=σ1+σ2,g¯ρ,σ=−2σρ2+11+σ2,in (45) and simplifying, we get a solution of (46)
(49)ϕx,t=Lx−1St−1σρ2+11+σ2=sinxsint.


### 5.5. The Dissipative Wave Equation

By substitutingN=M=2,c0=0,c2=1 in (1), we have the linear dissipative wave equation:
(50)c1∂ϕx,t∂t+∂2ϕx,t∂t2=d2∂2ϕx,t∂x2+d1∂ϕx,t∂x+gx,t,with ICs and BCs:
(51)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (50) as
(52)ϕx,t=Lx−1St−111+c1σ−d2σ2ρ2−d1σ2ρc1σf0¯ρ+f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ−d1σ2h0¯σ+σ2g¯ρ,σ.Example 8.
Puttingc1=d1=d2=1 and gx,t=2t−x in (50) to yield
(53)∂ϕx,t∂t+∂2ϕx,t∂t2=∂2ϕx,t∂x2+∂ϕx,t∂x+2t−x,with the conditions,
(54)ϕx,0=x2=f0x,ϕtx,0=0=f1x,ϕ0,t=t2=h0t,ϕx0,t=0=h1t.Substituting(55)f0¯ρ=2ρ3,f1¯ρ=0,h0¯σ=2σ2,h1¯σ=0,g¯ρ,σ=2σρ−1ρ2,in (52) and simplifying, we get a solution of (53)
(56)ϕx,t=Lx−1St−12ρ3+2σ2ρ=x2+t2.


### 5.6. The Korteweg-de Vries (KdV) Equation

By substitutingN=1,M=3,c0=d2=0,c1=1 in (1), we have the linear Korteweg-de Vries (KdV) equation:
(57)∂ϕx,t∂t=d3∂3ϕx,t∂x3+d1∂ϕx,t∂x+gx,t,with ICs and BCs
(58)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,ϕxx0,t=h2t,then (19) gives the solution of (57) as
(59)ϕx,t=Lx−1St−111−d3σρ3−d1σρf0¯ρ−d3σρ2h0¯σ−d3σρh1¯σ−d3σh2¯σ−d1σh0¯σ+σg¯ρ,σ.Example 9.
Puttingd3=d1=−1 and gx,t=0 in (57) to yield
(60)∂ϕx,t∂t=−∂3ϕx,t∂x3−∂ϕx,t∂x,with the conditions,
(61)ϕx,0=e−x=f0x,ϕ0,t=e2t=h0t,ϕx0,t=−e2t=h1t,ϕxx0,t=e2t=h2t.Substituting(62)f0¯ρ=1ρ+1,h0¯σ=h2¯σ=11−2σ,h1¯σ=−11−2σ,in (59) and simplifying, we get a solution of (60)
(63)ϕx,t=Lx−1St−11ρ+11−2σ=e2t−x.Example 10.
Puttingd3=1, d1=0, and gx,t=2et−x in (57) to obtain the linear third-order dispersive, inhomogeneous equation:
(64)∂ϕx,t∂t=∂3ϕx,t∂x3+2et−x,with the conditions,
(65)ϕx,0=1+e−x=f0x,ϕ0,t=1+et=h0t,ϕx0,t=−et=h1t,ϕxx0,t=et=h2t.Substituting(66)f0¯ρ=1ρ+1ρ+1,h0¯σ=1+11−σ,h1¯σ=−11−σ,h2¯σ=11−σ,g¯ρ,σ=2ρ+11−σ,in (59) and simplifying, we get a solution of (64)
(67)ϕx,t=Lx−1St−11ρ+1ρ+11−σ=1+et−x.


### 5.7. The Euler-Bernoulli Equation

By substitutingN=2,M=4,c0=c1=d1=d2=d3=0,c2=−1 in (1), we have the Euler-Bernoulli equation:
(68)−∂2ϕx,t∂t2=d4∂4ϕx,t∂x4+gx,t,with ICs and BCs:
(69)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,ϕxx0,t=h2t,ϕxxx0,t=h3t,then (19) gives the solution of (68) as
(70)ϕx,t=Lx−1St−111+d4σ2ρ4f0¯ρ+σf1¯ρ+d4σ2ρ3h0¯σ+d4σ2ρ2h1¯σ+d4σ2ρh2¯σ+d4σ2h3¯σ−σ2g¯ρ,σ.Example 11.
Puttingd4=1 and gx,t=−xt−t2 in (68) to yield
(71)−∂2ϕx,t∂t2=∂4ϕx,t∂x4−xt−t2,with the conditions,
(72)ϕx,0=0=f0x,ϕtx,0=x55!=f1x,ϕ0,t=t412=h0t,ϕx0,t=0=h1t,ϕxx0,t=0=h2t,ϕxxx0,t=0=h3t,Substituting(73)f0¯ρ=0,f1¯ρ=1ρ6,h0¯σ=2σ4,h1¯σ=h2¯σ=h3¯σ=0,g¯ρ,σ=−σρ2−2σ2ρ,in (70) and simplifying, we get a solution of (71)
(74)ϕx,t=Lx−1St−1σρ6+2σ4ρ=x5t5!+t412.


## 5.1. The Advection-Diffusion Equation

By substitutingN=1,M=2,c0=g=0,c1=1 in (1), we have got the advection-diffusion equation:
(20)∂ϕx,t∂t=d2∂2ϕx,t∂x2+d1∂ϕx,t∂x,with ICs and BCs
(21)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (20) as
(22)ϕx,t=Lx−1St−111−d2σρ2−d1σρf0¯ρ−d2σρh0¯σ−d2σh1¯σ−d1σh0¯σ.Example 5.
Puttingd1=−1,d2=1 in (20), we got
(23)∂ϕx,t∂t=∂2ϕx,t∂x2−∂ϕx,t∂x,with the conditions,
(24)ϕx,0=ex−x=f0x,ϕ0,t=1+t=h0t,ϕx0,t=0=h1t.Substituting(25)f0¯ρ=1ρ−1−1ρ2,h0¯σ=1+σ,h1¯σ=0,in (22) and simplifying, we get a solution of (23)
(26)ϕx,t=Lx−1St−11ρ−1−1ρ2+σρ=ex−x+t.


## 5.2. The Reaction-Diffusion Equation

By substitutingN=1, M=2, g=0, c1=1, d1=0, and d2>0 in (1), we have the reaction-diffusion equation:
(27)c0ϕx,t+∂ϕx,t∂t=d2∂2ϕx,t∂x2,with ICs and BCs:
(28)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (27) as
(29)ϕx,t=Lx−1St−111+c0σ−d2σρ2f0¯ρ−d2σρh0¯σ−d2σh1¯σ.


### 5.2.1. The Heat (Diffusion) Equation

Takingc0=0 in (27), we have the linear heat equation:
(30)∂ϕx,t∂t=d2∂2ϕx,t∂x2,with ICs and BCs:
(31)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (29) gives the solution of (30) as
(32)ϕx,t=Lx−1St−111−d2σρ2f0¯ρ−d2σρh0¯σ−d2σh1¯σ.Example 6.
Puttingd2=1 in (30) to yield
(33)∂ϕx,t∂t=∂2ϕx,t∂x2,with the conditions,
(34)ϕx,0=sinx=f0x,ϕ0,t=0=h0t,ϕx0,t=e−t=h1t.Substituting(35)f0¯ρ=1ρ2+1,h0¯σ=0,h1¯σ=11+σ,in (32) and simplifying, we get a solution of (33)
(36)ϕx,t=Lx−1St−11ρ2+11+σ=e−tsinx.


## 5.2.1. The Heat (Diffusion) Equation

Takingc0=0 in (27), we have the linear heat equation:
(30)∂ϕx,t∂t=d2∂2ϕx,t∂x2,with ICs and BCs:
(31)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,then (29) gives the solution of (30) as
(32)ϕx,t=Lx−1St−111−d2σρ2f0¯ρ−d2σρh0¯σ−d2σh1¯σ.Example 6.
Puttingd2=1 in (30) to yield
(33)∂ϕx,t∂t=∂2ϕx,t∂x2,with the conditions,
(34)ϕx,0=sinx=f0x,ϕ0,t=0=h0t,ϕx0,t=e−t=h1t.Substituting(35)f0¯ρ=1ρ2+1,h0¯σ=0,h1¯σ=11+σ,in (32) and simplifying, we get a solution of (33)
(36)ϕx,t=Lx−1St−11ρ2+11+σ=e−tsinx.


## 5.3. The Telegraph Equation

By substitutingN=M=2,d1=g=0 in (1), we have got the linear telegraph equation
(37)c0ϕx,t+c1∂ϕx,t∂t+c2∂2ϕx,t∂t2=d2∂2ϕx,t∂x2,with ICs and BCs:
(38)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (37) as
(39)ϕx,t=Lx−1St−11c0σ2+c1σ+c2−d2σ2ρ2c1f0¯ρ+c2f0¯ρ+c2σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ.


### 5.3.1. The Wave Equation

Takingc0=c1=0,c2=1 in (37), we obtain the linear wave equation:
(40)∂2ϕx,t∂t2=d2∂2ϕx,t∂x2,with ICs and BCs
(41)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (39) gives the solution of (40) as
(42)ϕx,t=Lx−1St−111−d2σ2ρ2f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ.


## 5.3.1. The Wave Equation

Takingc0=c1=0,c2=1 in (37), we obtain the linear wave equation:
(40)∂2ϕx,t∂t2=d2∂2ϕx,t∂x2,with ICs and BCs
(41)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (39) gives the solution of (40) as
(42)ϕx,t=Lx−1St−111−d2σ2ρ2f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ.


## 5.4. The Klein-Gordon Equation

By substitutingN=M=2,c1=d1=0,c2=1 in (1), we have the Klein-Gordon equation:
(43)c0ϕx,t+∂2ϕx,t∂t2=d2∂2ϕx,t∂x2+gx,t,with ICs and BCs
(44)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (43) as
(45)ϕx,t=Lx−1St−111+c0σ2−d2σ2ρ2f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ+σ2g¯ρ,σ.Example 7.
Puttingc0=−2,d2=1, and gx,t=−2sinxsint in (43) to yield
(46)∂2ϕx,t∂t2−2ϕx,t=∂2ϕx,t∂x2−2sinxsint,with the conditions,
(47)ϕx,0=0=f0x,ϕtx,0=sinx=f1x,ϕ0,t=0=h0t,ϕx0,t=sint=h1t.Substituting(48)f0¯ρ=0,f1¯ρ=1ρ2+1,h0¯σ=0,h1¯σ=σ1+σ2,g¯ρ,σ=−2σρ2+11+σ2,in (45) and simplifying, we get a solution of (46)
(49)ϕx,t=Lx−1St−1σρ2+11+σ2=sinxsint.


## 5.5. The Dissipative Wave Equation

By substitutingN=M=2,c0=0,c2=1 in (1), we have the linear dissipative wave equation:
(50)c1∂ϕx,t∂t+∂2ϕx,t∂t2=d2∂2ϕx,t∂x2+d1∂ϕx,t∂x+gx,t,with ICs and BCs:
(51)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,then (19) gives the solution of (50) as
(52)ϕx,t=Lx−1St−111+c1σ−d2σ2ρ2−d1σ2ρc1σf0¯ρ+f0¯ρ+σf1¯ρ−d2σ2ρh0¯σ−d2σ2h1¯σ−d1σ2h0¯σ+σ2g¯ρ,σ.Example 8.
Puttingc1=d1=d2=1 and gx,t=2t−x in (50) to yield
(53)∂ϕx,t∂t+∂2ϕx,t∂t2=∂2ϕx,t∂x2+∂ϕx,t∂x+2t−x,with the conditions,
(54)ϕx,0=x2=f0x,ϕtx,0=0=f1x,ϕ0,t=t2=h0t,ϕx0,t=0=h1t.Substituting(55)f0¯ρ=2ρ3,f1¯ρ=0,h0¯σ=2σ2,h1¯σ=0,g¯ρ,σ=2σρ−1ρ2,in (52) and simplifying, we get a solution of (53)
(56)ϕx,t=Lx−1St−12ρ3+2σ2ρ=x2+t2.


## 5.6. The Korteweg-de Vries (KdV) Equation

By substitutingN=1,M=3,c0=d2=0,c1=1 in (1), we have the linear Korteweg-de Vries (KdV) equation:
(57)∂ϕx,t∂t=d3∂3ϕx,t∂x3+d1∂ϕx,t∂x+gx,t,with ICs and BCs
(58)ϕx,0=f0x,ϕ0,t=h0t,ϕx0,t=h1t,ϕxx0,t=h2t,then (19) gives the solution of (57) as
(59)ϕx,t=Lx−1St−111−d3σρ3−d1σρf0¯ρ−d3σρ2h0¯σ−d3σρh1¯σ−d3σh2¯σ−d1σh0¯σ+σg¯ρ,σ.Example 9.
Puttingd3=d1=−1 and gx,t=0 in (57) to yield
(60)∂ϕx,t∂t=−∂3ϕx,t∂x3−∂ϕx,t∂x,with the conditions,
(61)ϕx,0=e−x=f0x,ϕ0,t=e2t=h0t,ϕx0,t=−e2t=h1t,ϕxx0,t=e2t=h2t.Substituting(62)f0¯ρ=1ρ+1,h0¯σ=h2¯σ=11−2σ,h1¯σ=−11−2σ,in (59) and simplifying, we get a solution of (60)
(63)ϕx,t=Lx−1St−11ρ+11−2σ=e2t−x.Example 10.
Puttingd3=1, d1=0, and gx,t=2et−x in (57) to obtain the linear third-order dispersive, inhomogeneous equation:
(64)∂ϕx,t∂t=∂3ϕx,t∂x3+2et−x,with the conditions,
(65)ϕx,0=1+e−x=f0x,ϕ0,t=1+et=h0t,ϕx0,t=−et=h1t,ϕxx0,t=et=h2t.Substituting(66)f0¯ρ=1ρ+1ρ+1,h0¯σ=1+11−σ,h1¯σ=−11−σ,h2¯σ=11−σ,g¯ρ,σ=2ρ+11−σ,in (59) and simplifying, we get a solution of (64)
(67)ϕx,t=Lx−1St−11ρ+1ρ+11−σ=1+et−x.


## 5.7. The Euler-Bernoulli Equation

By substitutingN=2,M=4,c0=c1=d1=d2=d3=0,c2=−1 in (1), we have the Euler-Bernoulli equation:
(68)−∂2ϕx,t∂t2=d4∂4ϕx,t∂x4+gx,t,with ICs and BCs:
(69)ϕx,0=f0x,ϕtx,0=f1x,ϕ0,t=h0t,ϕx0,t=h1t,ϕxx0,t=h2t,ϕxxx0,t=h3t,then (19) gives the solution of (68) as
(70)ϕx,t=Lx−1St−111+d4σ2ρ4f0¯ρ+σf1¯ρ+d4σ2ρ3h0¯σ+d4σ2ρ2h1¯σ+d4σ2ρh2¯σ+d4σ2h3¯σ−σ2g¯ρ,σ.Example 11.
Puttingd4=1 and gx,t=−xt−t2 in (68) to yield
(71)−∂2ϕx,t∂t2=∂4ϕx,t∂x4−xt−t2,with the conditions,
(72)ϕx,0=0=f0x,ϕtx,0=x55!=f1x,ϕ0,t=t412=h0t,ϕx0,t=0=h1t,ϕxx0,t=0=h2t,ϕxxx0,t=0=h3t,Substituting(73)f0¯ρ=0,f1¯ρ=1ρ6,h0¯σ=2σ4,h1¯σ=h2¯σ=h3¯σ=0,g¯ρ,σ=−σρ2−2σ2ρ,in (70) and simplifying, we get a solution of (71)
(74)ϕx,t=Lx−1St−1σρ6+2σ4ρ=x5t5!+t412.


## 6. Conclusion

We introduced an efficient integral transform called the double Laplace–Sumudu transform to obtain exact solutions of linear partial differential equations. We presented its existence and inverse transform. Also, we presented some useful properties and theorems to seek out the solution of a wide range of linear partial differential equations. Finally, based on the mathematical formulations, simplicity, and the findings of the proposed integral transform, we conclude that it is highly efficient because of the following advantages:(i)
The single Laplace and Sumudu transform to solve equations with unknown functions of two variables were hard and useless sometimes. For this reason, this problem was solved by combining the Laplace transforms and Sumudu transform to give another double transform which is called the double Laplace–Sumudu transform(ii)
It can easily be applied directly to solve a wide range of linear partial differential equations in mathematical physics by turning these equations into algebraic ones(iii)
For advanced research in physical science and engineering, the proposed integral transform can be considered a stepping stone to solve a wide range of linear partial differential equations.



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*Source: 6625247-2021-03-01.xml*