# A Note on Conformable Double Laplace Transform and Singular Conformable Pseudoparabolic Equations

Authors: Hassan Eltayeb; Said Mesloub
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this work, we combine conformable double Laplace transform and Adomian decomposition method and present a new approach for solving singular one-dimensional conformable pseudoparabolic equation and conformable coupled pseudoparabolic equation. Furthermore, some examples are given to show the performance of the proposed method.


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## Body

## 1. Introduction

Fractional partial differential equations have attracted much attention in applied sciences and engineering such as acoustics, control, and viscoelasticity. The parabolic equation appeared in different fields of applied mathematics, such as heat conduction and fluid mechanics (for instance, see [1–4]). The authors in [5, 6] studied the fractional diffusion equations problems by using the Adomian decomposition method and series expansion method. Many papers exist in the literature, which are related to conformable fractional derivative with its properties and applications [7, 8]. This new method was quickly generalized by Katugampola [9, 10]. The authors in [11] investigated existence and uniqueness theorems for sequential linear conformable fractional differential equations. The authors in [12] revisited the Grünwald Letnikov, Riemann–Liouville, and Caputo fractional derivatives and analysed under the light of the proposed criteria. The nonhomogeneous nonlocal theory has been presented based on conformable derivatives (CD) to study the critical point instability of micro/nanobeams under a distributed variable-pressure force (see [13]). The authors in [14] proposed a new fractional nonlocal model and its application in free vibration of Timoshenko and Euler–Bernoulli beams. Recently, several researchers applied the conformable Laplace transform method to solve different types of fractional differential equation (see [15, 16]). Many exact solutions in various wave forms for the nonlinear conformable time-fractional parabolic equation with exponential nonlinearity are formally constructed in [17]. The goal of this paper is to investigate the solution of singular conformable fractional pseudoparabolic equation and conformable coupled pseudoparabolic equation by conformable double Laplace transform decomposition methods (CDLTDMs). Moreover, we are able to prove some theorems related to this work.


### 1.1. Conformable Partial Derivatives

Definition 1.
(see [18]). Given a function fx,t:R×0,∞⟶R, the conformable space fractional partial derivative of order α of the function fx,t is denoted by(1)∂α∂xαfx,t=limε⟶0fx+εx1−α,t−fx,tε,x,t>0,0<α≤1.Definition 2.
(see [18]). Given a function fx,t:R×0,∞⟶R, the conformable time partial derivative of order β of the function fx,t is defined as(2)∂β∂tβfx,t=limσ⟶0fx,t+σt1−β−fx,tε,x,t>0,0<β≤1,where ∂α/∂xα and ∂β/∂tβ are called the fractional derivatives of order α and β, respectively.
In Theorem 1, the connection between the conformable derivatives and the first derivative can be represented as follows.Theorem 1.
Letα,β∈0,1 and fx,t be αandβ differentiable at a point x,t>0.
Then,(3)∂αfx,t∂xα=x1−α∂fx,t∂x,∂βfx,t∂tβ=t1−β∂fx,t∂x.Proof.
By using definitions 1 and 2 andh=ϵx1−α in equation (1), we have(4)∂αfx,t∂xα=limε⟶0fx+εx1−α,t−fx,tε=limh⟶0fx+h,t−fx,thxα−1=x1−αlimh⟶0fx+h,t−fx,th=x1−α∂fx,t∂x.
Similarly, we prove equation (2).
In the next example, we introduce the conformable derivative of specific functions, by using Theorem1 as follows.Example 1.
Letα,β∈0,1 and a,b,m,n,λ,andμ∈ℝ; then(1)
∂α/∂xαaux,t+bvx,t=a∂αux,t/∂xα+b∂αvx,t/∂xα(2)
∂α+β/∂xα∂tβxμtλ=μλxκ−αtλ−β, λ,μ∈ℝ(3)
∂α/∂xαeλxα/α+τtβ/β=λeλxα/α+τtβ/β, ∂β/∂tβeλxα/α+τtβ/β=τeλxα/α+τtβ/β(4)
∂β/∂tβxα/αtβ/β=xα/α, ∂β/∂tβxα/αntβ/βm=mxα/αntβ/βm−1(5)
∂α/∂xαxα/αntβ/β=nxα/αn−1tβ/β, ∂β/∂tβxα/αntβ/βm=mxα/αntβ/βm−1(6)
∂β/∂tβsinxα/αsintβ/β=sinxα/αcostβ/β,∂α/∂xαsinaxα/αsintβ/β=sinxα/αcostβ/β


## 1.1. Conformable Partial Derivatives

Definition 1.
(see [18]). Given a function fx,t:R×0,∞⟶R, the conformable space fractional partial derivative of order α of the function fx,t is denoted by(1)∂α∂xαfx,t=limε⟶0fx+εx1−α,t−fx,tε,x,t>0,0<α≤1.Definition 2.
(see [18]). Given a function fx,t:R×0,∞⟶R, the conformable time partial derivative of order β of the function fx,t is defined as(2)∂β∂tβfx,t=limσ⟶0fx,t+σt1−β−fx,tε,x,t>0,0<β≤1,where ∂α/∂xα and ∂β/∂tβ are called the fractional derivatives of order α and β, respectively.
In Theorem 1, the connection between the conformable derivatives and the first derivative can be represented as follows.Theorem 1.
Letα,β∈0,1 and fx,t be αandβ differentiable at a point x,t>0.
Then,(3)∂αfx,t∂xα=x1−α∂fx,t∂x,∂βfx,t∂tβ=t1−β∂fx,t∂x.Proof.
By using definitions 1 and 2 andh=ϵx1−α in equation (1), we have(4)∂αfx,t∂xα=limε⟶0fx+εx1−α,t−fx,tε=limh⟶0fx+h,t−fx,thxα−1=x1−αlimh⟶0fx+h,t−fx,th=x1−α∂fx,t∂x.
Similarly, we prove equation (2).
In the next example, we introduce the conformable derivative of specific functions, by using Theorem1 as follows.Example 1.
Letα,β∈0,1 and a,b,m,n,λ,andμ∈ℝ; then(1)
∂α/∂xαaux,t+bvx,t=a∂αux,t/∂xα+b∂αvx,t/∂xα(2)
∂α+β/∂xα∂tβxμtλ=μλxκ−αtλ−β, λ,μ∈ℝ(3)
∂α/∂xαeλxα/α+τtβ/β=λeλxα/α+τtβ/β, ∂β/∂tβeλxα/α+τtβ/β=τeλxα/α+τtβ/β(4)
∂β/∂tβxα/αtβ/β=xα/α, ∂β/∂tβxα/αntβ/βm=mxα/αntβ/βm−1(5)
∂α/∂xαxα/αntβ/β=nxα/αn−1tβ/β, ∂β/∂tβxα/αntβ/βm=mxα/αntβ/βm−1(6)
∂β/∂tβsinxα/αsintβ/β=sinxα/αcostβ/β,∂α/∂xαsinaxα/αsintβ/β=sinxα/αcostβ/β


## 2. Some Properties of the Conformable Laplace Transform

Here, we work with the single conformable Laplace transform and conformable double Laplace transform (CDLT) which are defined, respectively as follows.Definition 3.
(see [7, 19, 20]). Let f:a,∞⟶ℝ and 0<β≤1. Then, the fractional Laplace transform of order β is defined by(5)Ltβft=∫0∞e−stβ/βfttβ−1dt.Definition 4.
(see [21]). Let ux,t be a piecewise continuous function on the interval a,∞×a,∞ of exponential order. Consider for some a,b∈ℝ sup x,t>0, ux,t/eaxα/α+btβ/β in these conditions (CDLT) is defined by(6)LxαLtβux,t=∫0∞∫0∞e−pxα/α−stβ/βux,ttβ−1xα−1dtdx,where p,s∈ℂ,0<α,β≤1, and the integrals are by means of conformable integral with respect to t and x, respectively.Theorem 2.
IfLxαLtβfxα/α,tβ/β=Fα,βp,s, then(7)LxαLtβfxαα−ζαα,tββ−ηββHxαα−ζαα,tββ−ηββ=e−pζα/α−sηβ/βFα,βp,s,where Hx,t is the Heaviside unit step function defined by Hxα/α−a,tβ/β−b=1 when x>a and t>b and Hxα/α−a,tβ/β−b=0 when x<a and t<b.Proof.
By applying the definition of double conformable Laplace transform,(8)LxαLtβfxαα−ζαα,tββ−ηββHxαα−ζαα,tββ−ηββ=∫0∞∫0∞e−pxα/α−stβ/βfxαα−ζαα,tββ−ηββHxαα−ζαα,tββ−ηββtβ−1xα−1dtdx=∫ζα/α∞∫ηβ/β∞e−pxα/α−stβ/βfxαα−ζαα,tββ−ηββtβ−1xα−1dtdx,which is, by putting xα/α−ζα/α=τα/α,tβ/β−ηβ/β=νβ/β, we have(9)LxαLtβfxαα−ζαα,tββ−ηββHxαα−ζαα,tββ−ηββ=e−pζα/α−sηβ/β∫0∞∫0∞e−pτα/α−sυβ/βfταα,υββυβ−1τα−1dυdτ=e−pζα/α−sηβ/βFα,βp,s.
In the next example, we reported that some conformable Laplace transforms of definite functions are important in this study.Example 2.
(1)
LxαLtβxα/αntβ/βm=LxLtxntm=n!m!/pn+1sm+1, where m and n are positive integers(2)
LxαLtβeλxα/α+τtβ/β=LxLteλx+τt=1/p−λs−τ(3)
LxαLtβsinλxα/αsinτtβ/β=LxLtsinxcost=1/p2+λ2s2+τ2Theorem 3.
Letf be piecewise continuous on a,∞×a,∞; the (CDLT) of the conformable partial derivatives of orders α-th and β-th, ∂αu/∂xα,∂βu/∂tβ,∂2αu/∂x2α, and ∂2βu/∂t2β is given by(10)LxαLtβ∂αu∂xα=pUp,s−U0,s,(11)LxαLtβ∂βu∂tβ=sUp,s−Up,0,(12)LxαLtβ∂2αu∂x2α=p2Up,s−pU0,s−Ux0,s,LxαLtβ∂βu∂tβ=s2Up,s−sUp,0−Utp,0.Proof.
By using definition (CDLT) for∂αu/∂xα, we have(13)LxαLtβ∂αu∂xα=∫0∞∫0∞e−pxα/α−stβ/β∂αu∂xαtβ−1xα−1dtdx=∫0∞e−stβ/βtβ−1∫0∞e−pxα/α∂αu∂xαxα−1dxdt.
By applying Theorem1,∂αu/∂xα=x1−α∂ux,t/∂x equation (13) becomes(14)LxαLtβ∂αu∂xα=∫0∞e−stβ/βtβ−1∫0∞e−pxα/α∂ux,t∂xdxdt.
The integral inside bracket given by(15)∫0∞e−pxα/α∂ux,t∂xdx=pUp,t−U0,t.
By substituting equation (15) into equation (14), we obtain(16)LxαLtβ∂αu∂xα=pUp,s−U0,s.
In the same manner, the (CDLT) of∂βu/∂tβ,∂2αu/∂x2α, and ∂2βu/∂t2β can be obtained.
Double Laplace transform of the functionxα/αn∂βf/∂tβ and xα/αfx,t are studied in the next theorem.Theorem 4.
If the (CDLT) of the conformable partial derivatives∂β/∂tβu is given by equation (11), then double Laplace transform of xα/αn∂β/∂tβux,t and xα/αux,t are given by(17)−1ndndpnLxαLtβux,t=LxαLtβxααnux,t,(18)−1ndndpnLxαLtβ∂βu∂tβ=LxαLtβxααn∂βu∂tβ,where n=1,2,3,….Proof.
By applying the nth derivative with respect top for both sides of equation (6), we get equation (17) as follows:(19)dndpnLxαLtβux,t=∫0∞∫0∞dndpne−pxα/α−stβ/βux,ttβ−1xα−1dtdx=−1n∫0∞∫0∞xααne−pxα/α−stβ/βtβ−1xα−1ux,tdtdx=−1nLxαLtβxααnux,t.
We obtain(20)−1ndndpnLxαLtβux,t=LxαLtβxααnux,t.
Similarly, we can prove equation (18).


## 3. Singular One-Dimensional Conformable Fractional Pseudoparabolic Equation

The conformable double Laplace decomposition methods (CDLTDMs) are an efficient technique which is used to obtain the solution linear and nonlinear singular pseudoparabolic equation.Problem.
We consider0<α≤1 and 0<β≤1 as singular one-dimensional pseudoparabolic equations with initial conditions in the form(21)∂βu∂tβ−αxα∂α∂xαxαα∂α∂xαu−αxα∂α+β∂xα∂tβxαα∂α∂xαu=fx,t,subject to(22)ux,0=f1x,where, the term, α/xα∂αu/∂xαxα/α∂α/∂xα is called conformable Bessel’s operator and fx,t and f1x are known functions. In order to solve equation (21), we apply the following steps:Step 1: multiplying equation (21) by xα/α:(23)xαα∂βu∂tβ−∂α∂xαxαα∂α∂xαu−∂α+β∂xα∂tβxαα∂α∂xαu=xααfx,t.Step 2: using Lemma 1 and equation (18) for equations in step 1 and single conformable Laplace transform for equation (22), we obtain(24)ddpUp,s=1sddpLxαf1x+1sddpLxαLtβfx,t−1sLxαLtβ∂α∂xαxαα∂α∂xαu+∂α+β∂xα∂tβxαα∂α∂xαu,where the symbolLxαLtβ indicates (CDLT) with respect to xandt.Step 3: applying the integral for both sides of equation (24), from 0 to p with respect to p, we have(25)Up,s=1s∫0pddpLxαf1xdp−1s∫0pLxαLtβ∂α∂xαxαα∂α∂xαu+∂α+β∂xα∂tβxαα∂α∂xαudp+1s∫0pddpLxαLtβfx,tdp.Step 4: next, the (CDLTDM) consists of representing the solution of the singular pseudoparabolic equation asux,t by the infinite series(26)ux,t=∑n=0∞unx,t.Step 5: working with the double Laplace transform on both sides of equation (25) and using equation (26), we receive(27)∑n=0∞unx,t=f1x+Lp−1Ls−11s∫0pddpLxαLtβfx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂α∂xα∑n=0∞unx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂α∂xα∑n=0∞unx,tdp.
We define the following recursive formula:(28)u0x,t=f1x+Lp−1Ls−11s∫0pddpLxαLtβfx,tdp.
The rest of the terms can be written as follows:(29)un+1x,t=−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂α∂xα∑n=0∞unx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂α∂xα∑n=0∞unx,tdp,where Lp−1Ls−1 indicates double inverse Laplace transform with respect to p and s.
Here, we assume that double inverse Laplace transform with respect top and s exists for each terms in equations (28) and (29). To confirm our method, we solve the next example.Example 3.
Consider the following nonhomogeneous form of a singular one-dimensional pseudoparabolic equation:(30)∂βu∂tβ−αxα∂α∂xαxαα∂α∂xαu−αxα∂α+βu∂xα∂tβxαα∂α∂xαu=−xαα2sintββ+4sintββ−4costββ,0≤x<∞,0≤t<∞,0<α≤1,0<β≤1,with the condition(31)ux,t=x2.
By applying the above steps and Theorem1, we obtain(32)u0x,t=xαα2costββ−4costββ−4sintββ+4,un+1x,t=−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αun∂xαdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂αun∂xαdp.
Based on the (CDLTDM), we obtain(33)u1x,t=−Lp−1Ls−11s∫0pLxLt∂α∂xαxαα∂α∂xαu0+∂α+β∂xα∂tβxαα∂α∂xαu0dp=−4+4costββ+4sintββ.
In a similar manner, we obtain that(34)u2x,t=−Lp−1Ls−11s∫0pLxLt∂α∂xαxαα∂α∂xαu1+∂α+β∂xα∂tβxαα∂α∂xαu1dp=0,u3=0,u4=0,….
By adding all the terms, we get(35)ux,t=u0x,t+u1x,t+u2x,t+⋯.
Thus, the exact solution is obtained as follows:(36)ux,t=xαα2costββ.
By takingα=1 and β=1, the fractional solution becomes(37)ux,t=x2cost.Problem.
Consider the following nonlinear singular one-dimensional pseudoparabolic equation:(38)∂βu∂tβ−αxα∂α∂xαxαα∂αu∂xα−αxα∂α+β∂xα∂tβxαα∂αu∂xα=−2u∂αu∂xα+xαα∂αu∂xα2+fx,t,0<α≤1,0<β≤1,0≤x<∞,0≤t<∞,subject to(39)ux,0=f1x.
Using our method, we get(40)u0x,t=f1x+Lp−1Ls−11s∫0pddpLxαLtβfx,tdp.The rest of the terms are given by(41)un+1x,t=−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αun∂xαdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂αun∂xαdp+Lp−1Ls−11s∫0pLxαLtβ2xααAndp−Lp−1Ls−11s∫0pLxαLtβxαα2Bndp,where An and Bn are the so-called Adomian polynomials, given by(42)An=∑n=0∞ununx,Bn=∑n=0∞unx2.
The nonlinear termsuux and ux2 are represented as(43)A0=u0u0x,A1=u0u1x+u1u0x,A2=u0u2x+u1u1x+u2u0x,A3=u0u3x+u1u2x+u2u1x+u3u0x,B0=u0x2,B1=2u0xu1x,B2=2u0xu2x+u1x2,B3=2u0xu3x+2u1xu2x.
To illustrate this method for nonlinear problem, we consider the following example.Example 4.
Consider the following nonlinear pseudoparabolic equation:(44)∂βu∂tβ−αxα∂α∂xαxαα∂αu∂xα−αxα∂α+β∂xα∂tβxαα∂αu∂xα=−2u∂αu∂xα+xαα∂αu∂xα2−xαα2e−tβ/β,0≤x<∞,0≤t<∞,0<α≤1,0<β≤1,subject to(45)ux,0=xαα2.
By applying the aforesaid conformable double Laplace decomposition method and Theorem1, we have(46)u0x,t=xαα2e−tβ/β,u1x,t=−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αun∂xα+∂α+β∂xα∂tβxαα∂αun∂xαdp+Lp−1Ls−11s∫0pLxαLtβ2xααAn−xαα2Bndp=0.
Proceeding in a similar manner, we have(47)u2=0,u3=0,u4=0,….
So according to equation (26), we have(48)ux,t=xαα2e−tβ/β,which is the exact solution of equation (44).


## 4. Conformable Double Laplace Transform Method and Singular Conformable Coupled Pseudoparabolic Equation

The purpose of this part is to examine the use of the (CDLTDM) for the linear one-dimensional conformable coupled pseudoparabolic equation. We consider the following conformable coupled pseudoparabolic equations:(49)∂βu∂tβ−αxα∂α∂xαxαα∂αu∂xα−αxα∂α+β∂xα∂tβxαα∂αu∂xα+ζv=fx,t,∂βv∂tβ−αxα∂α∂xαxαα∂αv∂xα−αxα∂α+β∂xα∂tβxαα∂αv∂xα+ζu=gx,t,with conditions(50)ux,0=f1x,vx,0=g1x,where fx,t,gx,t,f1x, and g1x are the known functions and ζ is the coupling parameter. One can get the solution of equation (49), by using (CDLTDM); this method consists of the following steps:(1)
Multiply both sides of equation (49) by xα/α leading to the following equation:(51)xαα∂βu∂tβ−∂α∂xαxαα∂αu∂xα−∂α+β∂xα∂tβxαα∂αu∂xα+ζxααv=xααfx,t,xαα∂βv∂tβ−∂α∂xαxαα∂αv∂xα−∂α+β∂xα∂tβxαα∂αv∂xα+ζxααu=xααgx,t.(2)
Applying (CDLT) on both sides of equation (51) and single conformable Laplace transform for equation (50), we get(52)LxαLtβxαα∂βu∂tβ=LxαLtβ∂α∂xαxαα∂αu∂xα+∂α+β∂xα∂tβxαα∂αu∂xα−ζxααv+xααfx,t,LxαLtβxαα∂βv∂tβ=LxαLtβ∂α∂xαxαα∂αv∂xα+∂α+β∂xα∂tβxαα∂αv∂xα−ζxααu+xααgx,t.On using Theorem1 and Theorem 2, we obtain(53)ddpUp,s=1sddpLxαf1x−1sLxαLtβ∂α∂xαxαα∂α∂xαu+∂α+β∂xα∂tβxαα∂α∂xαu−ζxααv+1sddpLxαLtβfx,t,ddpVp,s=1sddpLxαg1x−1sLxαLtβ∂α∂xαxαα∂α∂xαv+∂α+β∂xα∂tβxαα∂α∂xαv−ζxααu+1sddpLxαLtβgx,t.(3)
By integrating both sides of equation (53) from 0 to p with respect to p, we have(54)Up,s=1s∫0pddpLxαf1xdp−1s∫0pLxαLtβ∂α∂xαxαα∂α∂xαu+∂α+β∂xα∂tβxαα∂α∂xαu−ζxααvdp+1s∫0pddpLxαLtβfx,tdp,Vp,s=1s∫0pddpLxαg1xdp−1s∫0pLxαLtβ∂α∂xαxαα∂α∂xαv+∂α+β∂xα∂tβxαα∂α∂xαv−ζxααudp+1s∫0pddpLxαLtβgx,tdp.By applying double inverse Laplace transform for equation (54), we have(55)ux,t=f1x+Lp−1Ls−11s∫0pddpLxαLtβfx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αu∂xαdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂αu∂xα−ζxααvdp,(56)vx,t=g1x+Lp−1Ls−11s∫0pddpLxαLtβgx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αv∂xαdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂αv∂xα−ζxααudp.The (CDLTDM) defines the solutions of conformable coupled pseudoparabolic equations asux,t and vx,t by the infinite series:(57)ux,t=∑n=0∞unx,t,vx,t=∑n=0∞vnx,t.By substituting equation (57) into equations (55) and (56), we get(4)
Working with the double Laplace transform on both sides of equation (25) and using equation (26), we receive(58)∑n=0∞unx,t=f1x+Lp−1Ls−11s∫0pddpLxαLtβfx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂α∂xα∑n=0∞undp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂α∂xα∑n=0∞undp+Lp−1Ls−11sLxLtζxαα∑n=0∞vndp,∑n=0∞vnx,t=g1x+Lp−1Ls−11s∫0pddpLxαLtβgx,tdp−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂α∂xα∑n=0∞vndp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂α∂xα∑n=0∞vndp+Lp−1Ls−11sLxLtζxαα∑n=0∞undp.This technique suggests that the zeroth componentsu0 and v0 are identified by the initial conditions and from source terms as follows:(59)u0=f1x+Lp−1Ls−11s∫0pddpLxαLtβfx,tdp,v0=g1x+Lp−1Ls−11s∫0pddpLxαLtβgx,tdp.The rest of the terms are given by(60)un+1=−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αun∂xαdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂αun∂xαdp+Lp−1Ls−11sLxLtζxααvndp,(61)vn+1=−Lp−1Ls−11s∫0pLxαLtβ∂α∂xαxαα∂αvn∂xαdp−Lp−1Ls−11s∫0pLxαLtβ∂α+β∂xα∂tβxαα∂αvn∂xαdp+Lp−1Ls−11sLxLtζxααundp.In order to ensure the four techniques for solving the conformable fractional coupled pseudoparabolic equation, we will consider the following example.Example 5.
Consider the following homogeneous form of conformable coupled pseudoparabolic equations:(62)∂βu∂tβ−2αxα∂α∂xαxαα∂αu∂xα−αxα∂α+β∂xα∂tβxαα∂αu∂xα+2v=0,∂βv∂tβ−2αxα∂α∂xαxαα∂αv∂xα−αxα∂α+β∂xα∂tβxαα∂αv∂xα+2u=0,where(63)0≤x<∞,0≤t<∞,0<α≤1,0<β≤1,with conditions(64)ux,0=xαα2,vx,0=xαα2.
By applying the above method and Theorem1 for equation (62) and (64) and using equations (59), (60), and (61), we obtain(65)u0=xαα2,v0=xαα2,u1=−Lp−1Ls−11sLxLt2∂α∂xαxαα∂α∂xαu0+∂α+β∂xα∂tβxαα∂α∂xαu0−2xααv0=8tββ−2xαα2tββ,v1=−Lp−1Ls−11sLxLt2∂α∂xαxαα∂α∂xαv0+∂α+β∂xα∂tβxαα∂α∂xαv0−2xααu0=8tββ−2xαα2tββ,u2=−Lp−1Ls−11sLxLt2∂α∂xαxαα∂α∂xαu1+∂α+β∂xα∂tβxαα∂α∂xαu1−2xααv1=−16tββ2−8tββ+2xαα2tββ2,v2=−Lp−1Ls−11sLxLt2∂α∂xαxαα∂α∂xαv1+∂α+β∂xα∂tβxαα∂α∂xαv1−2xααu1=−16tββ2−8tββ+2xαα2tββ2,u3=16tββ3+16tββ2−43xαα2tββ3,v3=16tββ3+16tββ2−43xαα2tββ3,u4=−323tββ4−16tββ3+23xαα2tββ4,v4=−323tββ4−16tββ3+23xαα2tββ4,and similarly for the rest components. Using equation (57), the series solutions are therefore given by(66)u=u0+u1+u2+⋯=1−2tββ+2tβ/β22!−2tβ/β33!+2tβ/β44!−⋯xαα2,v=v0+v1+v2+⋯=1−2tββ+2tβ/β22!−2tβ/β33!+2tβ/β44!−⋯xαα2.Hence,(67)ux,t=xαα2e−2tβ/β,vx,t=xαα2e−2tβ/β.
The exact solution is obtained by takingα=1 and β=1, as follows:(68)ux,t=x2e−2t,vx,t=x2e−2t.


## 5. Numerical Result

In this section, we discuss the precision and efficiency of the (CDLTDM) by numerical results ofux,t for the exact solution when α=β=1 and approximate solutions at α and β taking different fractional values for conformable pseudoparabolic equation and conformable coupled pseudoparabolic equation. The solutions of equation (30) are depicted in Figures 1 and 2, respectively.Figure 1
The exact and approximate solutions ofux,t for Example 3, when α=β.Figure 2
The exact and approximate solutions ofux,t, for Example 3, when we take different values of fractional order β and α=0.99.In Figure1, the approximate solutions of equation (30) at t=1 and α=β, taking different fractional values, are compared and we found that the numerical solution becomes close to the exact solution when the fractional value increases:(69)α=β=1.Figure2 indicates that the exact solution at αβ=1 of equation (30) and the approximate solution of equation (30) decrease at the fractional derivative values α0.99 and β0.98,0.96,0.94. Similarly, the exact solution and approximate solution of equation (44) are demonstrated in Figures 3 and 4. Figure 3 give the plots of the behaviour of equation (44) when t=1 and α=β with different fractional values taken in this case; the solution ux,t becomes close to the exact solution at α=β close to one.Figure 3
The exact and approximate solutions ofux,t for Example 4, when α=β.Figure 4
The exact and approximate solutions ofux,t, for Example 4, when we take different values of fractional order α and β=0.99.Figure4 shows the approximate solution of equation (44) with 0<α≤1, β=0.99, and t=1; in such a case, the function ux,t gradually decreases. Finally, Figure 5 suggests that in the solutions of equation (62) at t=1 and 0<α=β≤1, we find that the numerical solution becomes close to the exact solution when the fractional value increases.Figure 5
The exact and approximate solutions ofux,t for Example 5, when α=β.Figure6 demonstrates that the exact solution at α=β=1 of equation (62) and the approximate solution of equation (62) are concave upward at the fractional derivative increasing when 0<α≤1 and β fixed.Figure 6
The exact and approximate solutions ofux,t, for Example 5, when we take different values of fractional order α and β=0.99.


## 6. Conclusion

In this work, singular one-dimensional conformable pseudoparabolic equation and conformable coupled pseudoparabolic equation have been considered. Then, new conformable double Laplace transform decomposition methods have been applied to the problems. Finally, we gave three differential examples to show that this method is applicable and valid. The suggested method can also be applied for systems with more than two linear and nonlinear partial differential equations. In addition, if we letα=1 and β=1 in Examples 3 and 4, we get the solution which is considered in [22]. All figure results are obtained by using Matlab.



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