# Solving Some Partial Differential Equations by Using Double Laplace Transform in the Sense of Nonconformable Fractional Calculus

Authors: Sami Injrou; Iyad Hatem
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

In this paper, we introduce the non-conformable double Laplace transform. Its properties are studied, and it is applied to solve some fractional PDEs involving the nonconformable fractional derivative. Graphical representations of the obtained solutions are shown in figures. The study shows that this transform is effective and easy to apply to create an exact solution for types of fractional PDEs.


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## Body

## 1. Introduction

Throughout the decades, mathematics has played an influential role in the developed civilization. Through mathematical modelling, it has allowed to describe and predict phenomena in real world. From this viewpoint, it is essential to assure the importance of calculus to study many of the laws of nature.On September 30th, 1695, a generalization of calculus was born in a letter between L’Hopltal and Leibniz in which a question got raised about the meaning of taking a fractional derivative such as d1/2y/dx1/2 [1]. This generalization is called fractional calculus. In the last and present centuries, the importance of the fractional calculus has been growing because of its deep applications in all related fields of science and engineering [2–5].Actually, many definitions of fractional derivative and fractional integral have been proposed. The most popular ones are Riemann–Liouville, Caputo, Grunwald–Letnikov, Hadamard, Erdely, Kober, Marchaud, and Riesz [6–9]. Most of these fractional derivatives do not satisfy the classical formulas of derivatives such as product, quotient, and chain rules except Caputo [10, 11].Recently, Khalil et al. [12] introduced an extension of the ordinary limit definition for the derivative of a function, namely, the conformable derivative. More recently, Martinez proposed a new nonconformable local derivative in [13]. One can see that the last two definitions satisfy the classical properties mentioned above. Many researchers solved nonlinear fractional partial differential equations in the sense of conformable derivative and Caputo derivative [14–31]. There is an important study [32], where the authors treat the price adjustment equation in many senses of fractional derivatives, such as truncated M-derivative including the Mittag–Leffer function, beta-derivative, and conformable derivative defined in the form of limit for α-differentiable functions.In [13], Martinez introduced the nonconformable Laplace transform local fractional derivative, and they proved its existence beside main properties. Ozarslan et al. in [33] presented Atangana–Baleanu fractional derivative in the Caputo sense with the n+αth order and its Laplace transform, and they studied that the wind influenced projectile motion equations involving this sense of derivative. In [34], Ozkan and Kurt introduced the conformable double Laplace transform and used it to solve some fractional partial differential equations that represent many physical and engineering models.In this manuscript, we establish the nonconformable double Laplace transform and apply this new definition to solve a nonconformable fractional heat, wave, and telegraph equations with sense of nonconformable fractional derivative.


## 2. Preliminary

In this section, we present the definition of nonconformable fractional derivative and some important definitions.Definition 1.
(see [13]). Given a function h:0,∞⟶R, then the nonconformable fractional derivative N3αht of order α of h at t is defined by(1)N3αht=limε⟶0ht+εt−α−htε,α∈0,1,t>0.
In addition, if the nonconformable fractional derivativeN3α of h of order α exists, then h is N-differentiable.Definition 2.
(see [13]). Let α∈0,1 and c be a real number, then the fractional exponential is defined in the following way:(2)EαN3c,t=expctα+1α+1.Definition 3.
(see [13]). Let α∈0,1 and 0≤a≤b, then we can say that a function h:a,b⟶R is α− fractional integrable on a,b, if the integral(3)JaαN3hx=∫axhtt−αdt=∫axhtdαt,exists and is finite.
In the next section, we present the definitions of the nonconformable double and single Laplace transform of a function with tow variables. We prove some properties. We claim that the results presented here are new in its domain.


## 3. Nonconformable Double Laplace Transform

Definition 4.
Letux,t be a piecewise continuous function on 0,∞×0,∞ of generalized exponential order; that is, there exist constants M,a and b such that ux,t≤MEαN3a,tEβN3b,t for sufficiently large t. The nonconformable double Laplace transform of ux,t is defined by(4)LtαLxβux,t=Up,s=∫0∞∫0∞EαN3−s,tEβN3−p,xux,tdαtdβx,where p,s∈C, 0<α,β≤1, and the integrals are in sense of nonconformable fractional integral with respect to t and x.
Now, we define the single nonconformable Laplace transform of a function with two variables.Definition 5.
Letux,t be a piecewise continuous function on 0,∞×0,∞ of the generalized exponential order. The nonconformable Laplace transform with respect to x of ux,t is defined by(5)Lxβux,t=Up,t=∫0∞EβN3−p,xux,tdβx,and the nonconformable Laplace transform with respect to t of ux,t is defined by(6)Ltαux,t=Ux,s=∫0∞EαN3−s,tux,tdαt,where the integrals are in nonconformable sense.


### 3.1. Some Properties of Nonconformable Double Laplace Transform

Here, we consider some of the properties and theorems of the nonconformable double Laplace Transform with their verification.If we assume that the functionux,t provides the sufficient conditions [35] and the order of transformation can be changed, then(7)∫0∞∫0∞EαN3−s,tEβN3−p,xux,tdαtdβx=∫0∞∫0∞EβN3−p,xEαN3−s,tux,tdβxdαt.Therefore, we get(8)LtαLxβux,t=LxβLtαux,t,=Up,s.Theorem 1.
Letux,t and wx,t be two functions which have the nonconformable double Laplace transform. Then, we get(1)
LtαLxβc1ux,t+c2wx,t=c1LtαLxβux,t+c2LtαLxβwx,t, where c1 and c2 are real constants(2)
LtαLxβEβN3d,xEαN3c,tux,t=Up−d,s−c, where c and d are any real constants, s−d>0 and p−c>0(3)
LtαLxβuηx,μt=1/ηβ+1μα+1Up/ηβ+1,s/μα+1, where μ and η are two nonzero real numbers(4)
Form,n∈N, −1m+nLtαLxβxmβ+1/β+1m.tnα+1/α+1nux,t=∂m+nUp,s/∂pm∂snProof.
(1)
We obtain the proof from the definition directly.(2)
By using nonconformable Laplace transform definition, one can find(9)LtαLxβEβN3d,xEαN3c,tux,t=∫0∞∫0∞EβN3d,xEαN3c,tEβN3−p,xEαN3−s,tux,tdβxdαt=∫0∞∫0∞EβN3−p−d,xEαN3−s−c,tux,tdβxdαt=Up−d,s−c,LtαLxβuηx,μt=∫0∞∫0∞EαN3−s,tEβN3−p,xuηx,μtdαtdβx=∫0∞∫0∞e−pxβ+1/β+1e−stα+1/α+1uηx,μtdαtdβx=∫0∞e−pxβ+1/β+1∫0∞e−stα+1/α+1uηx,μtdαtdβx,but(10)∫0∞e−stα+1/α+1uηx,μtdαt=∫0∞e−stα+1/α+1uηx,μtt−αdt=t=q/μ∫0∞e−sq/μα+1/α+1uηx,qq−α/μ−αdqμ=1μα+1∫0∞e−s/μα+1qα+1/α+1uηx,qdαq=1μα+1Uηx,sμα+1.Therefore,(11)LtαLxβuηx,μt=∫0∞e−pxβ+1/β+11μα+1Uηx,sμα+1dβx=1ηβ+1μα+1Upηβ+1,sμα+1.(3)
Taking into account the convergence properties of improper integral, one can change the order of the operation of differentiation and integration. Therefore, one can differentiate with respect top, s under the sign of integral. Thus,(12)∂m+nUp,s∂pm∂sn=∫0∞∂m∂pme−pxβ+1/β+1∫0∞∂n∂sne−stα+1/α+1ux,tdαtdβx,=−1m+nLtαLxβxmβ+1β+1m.tnα+1α+1nux,t,where we repeated differentiation with respect top, s, m and n times, respectively.Lemma 1.
Under the assumptions of the definition of the nonconformable double Laplace transform, we have(13)LtαLxβxN3βux,t=pUp,s−U0,s,(14)LtαLxβtN3αux,t=sUp,s−Up,0(15)LtαLxβxN3βtN3αux,t=psUp,s−pUp,0−sU0,s+U0,0,where xN3βux,t and tN3αux,t are the β-th and α-th order nonconformable fractional partial derivatives, respectively, and xN3βtN3αux,t are the mixed β-th and α-th order nonconformable fractional partial derivatives.Proof.
(1)
LtαLxβxN3βux,t=pUp,s−U0,s.Using the definition of non-comformable double Laplace transform and the Proposition 2.3 in [13], we get(16)LtαLxβxN3βux,t=∫0∞∫0∞EαN3−s,tEβN3−p,xxN3βux,tdαtdβx,=∫0∞EαN3−s,t∫0∞EβN3−p,xxN3βux,tdβxdαt,=∫0∞EαN3−s,tLxβxN3βux,tdαt,=∫0∞EαN3−s,tLxβxux,t−u0,tdαt,=pUp,s−U0,s.(2)
In the same manner, one can prove(17)LtαLxβtN3αux,t=sUp,s−Up,0.(3)
LtαLxβxN3βtN3αux,t=∫0∞∫0∞EαN3−s,tEβN3−p,xxN3βtN3αux,tdαtdβx.
Settinggx,t=tN3αux,t and using (13), we have(18)LtαLxβxN3βgx,t=pGp,s−G0,s,where(19)Gp,s=LtαLxβgx,t=LtαLxβtN3αux,t=sUp,s−Up,0,and(20)G0,s=LtαLxβg0,t=LtαLxβtN3αu0,t=sU0,s−U0,0.
Now, substituting equations (18) and (19) into equation (17) yields(21)LtαLxβxN3βtN3αux,t=psUp,s−pUp,0−sU0,s+U0,0.
Analogously, one can prove the following theorem.Theorem 2.
Letα,β∈0,1 and m,n∈N such that ux,t∈ClR+×R+, where l=maxm,n. Also, let ux,t,xiN3βux,t, and tjN3αux,t for i=1,...,m,j=1,...,n are N-transformable; then, we have(22)LtαLxβxmN3βux,t=pmUp,s−pm−1U0,s−∑i=1m−1pm−1−iLtαxiN3βU0,t,LtαLxβtnN3βux,t=snUp,s−sn−1Up,0−∑j=1n−1sn−1−jLxβtjN3αUx,0,LtαLxβxmN3βtnN3βux,t=pmsnUp,s−s−1Up,0−p−1U0,s−∑j=1n−1s−1−jLxβtjN3αUx,0−∑i=1m−1p−1−iLtαxiN3βU0,t+∑i=1m−1∑j=1n−1s−1−jp−1−ixiN3βtjN3αU0,0+p−ms−nU0,0,where xmN3βux,t and tnN3αux,t are m, n times nonconformable fractional derivative of function ux,t, with order β and α, respectively, and xmN3βtnN3αux,t are the mixed β-th and α-th order nonconformable fractional partial derivatives of function ux,t.


## 3.1. Some Properties of Nonconformable Double Laplace Transform

Here, we consider some of the properties and theorems of the nonconformable double Laplace Transform with their verification.If we assume that the functionux,t provides the sufficient conditions [35] and the order of transformation can be changed, then(7)∫0∞∫0∞EαN3−s,tEβN3−p,xux,tdαtdβx=∫0∞∫0∞EβN3−p,xEαN3−s,tux,tdβxdαt.Therefore, we get(8)LtαLxβux,t=LxβLtαux,t,=Up,s.Theorem 1.
Letux,t and wx,t be two functions which have the nonconformable double Laplace transform. Then, we get(1)
LtαLxβc1ux,t+c2wx,t=c1LtαLxβux,t+c2LtαLxβwx,t, where c1 and c2 are real constants(2)
LtαLxβEβN3d,xEαN3c,tux,t=Up−d,s−c, where c and d are any real constants, s−d>0 and p−c>0(3)
LtαLxβuηx,μt=1/ηβ+1μα+1Up/ηβ+1,s/μα+1, where μ and η are two nonzero real numbers(4)
Form,n∈N, −1m+nLtαLxβxmβ+1/β+1m.tnα+1/α+1nux,t=∂m+nUp,s/∂pm∂snProof.
(1)
We obtain the proof from the definition directly.(2)
By using nonconformable Laplace transform definition, one can find(9)LtαLxβEβN3d,xEαN3c,tux,t=∫0∞∫0∞EβN3d,xEαN3c,tEβN3−p,xEαN3−s,tux,tdβxdαt=∫0∞∫0∞EβN3−p−d,xEαN3−s−c,tux,tdβxdαt=Up−d,s−c,LtαLxβuηx,μt=∫0∞∫0∞EαN3−s,tEβN3−p,xuηx,μtdαtdβx=∫0∞∫0∞e−pxβ+1/β+1e−stα+1/α+1uηx,μtdαtdβx=∫0∞e−pxβ+1/β+1∫0∞e−stα+1/α+1uηx,μtdαtdβx,but(10)∫0∞e−stα+1/α+1uηx,μtdαt=∫0∞e−stα+1/α+1uηx,μtt−αdt=t=q/μ∫0∞e−sq/μα+1/α+1uηx,qq−α/μ−αdqμ=1μα+1∫0∞e−s/μα+1qα+1/α+1uηx,qdαq=1μα+1Uηx,sμα+1.Therefore,(11)LtαLxβuηx,μt=∫0∞e−pxβ+1/β+11μα+1Uηx,sμα+1dβx=1ηβ+1μα+1Upηβ+1,sμα+1.(3)
Taking into account the convergence properties of improper integral, one can change the order of the operation of differentiation and integration. Therefore, one can differentiate with respect top, s under the sign of integral. Thus,(12)∂m+nUp,s∂pm∂sn=∫0∞∂m∂pme−pxβ+1/β+1∫0∞∂n∂sne−stα+1/α+1ux,tdαtdβx,=−1m+nLtαLxβxmβ+1β+1m.tnα+1α+1nux,t,where we repeated differentiation with respect top, s, m and n times, respectively.Lemma 1.
Under the assumptions of the definition of the nonconformable double Laplace transform, we have(13)LtαLxβxN3βux,t=pUp,s−U0,s,(14)LtαLxβtN3αux,t=sUp,s−Up,0(15)LtαLxβxN3βtN3αux,t=psUp,s−pUp,0−sU0,s+U0,0,where xN3βux,t and tN3αux,t are the β-th and α-th order nonconformable fractional partial derivatives, respectively, and xN3βtN3αux,t are the mixed β-th and α-th order nonconformable fractional partial derivatives.Proof.
(1)
LtαLxβxN3βux,t=pUp,s−U0,s.Using the definition of non-comformable double Laplace transform and the Proposition 2.3 in [13], we get(16)LtαLxβxN3βux,t=∫0∞∫0∞EαN3−s,tEβN3−p,xxN3βux,tdαtdβx,=∫0∞EαN3−s,t∫0∞EβN3−p,xxN3βux,tdβxdαt,=∫0∞EαN3−s,tLxβxN3βux,tdαt,=∫0∞EαN3−s,tLxβxux,t−u0,tdαt,=pUp,s−U0,s.(2)
In the same manner, one can prove(17)LtαLxβtN3αux,t=sUp,s−Up,0.(3)
LtαLxβxN3βtN3αux,t=∫0∞∫0∞EαN3−s,tEβN3−p,xxN3βtN3αux,tdαtdβx.
Settinggx,t=tN3αux,t and using (13), we have(18)LtαLxβxN3βgx,t=pGp,s−G0,s,where(19)Gp,s=LtαLxβgx,t=LtαLxβtN3αux,t=sUp,s−Up,0,and(20)G0,s=LtαLxβg0,t=LtαLxβtN3αu0,t=sU0,s−U0,0.
Now, substituting equations (18) and (19) into equation (17) yields(21)LtαLxβxN3βtN3αux,t=psUp,s−pUp,0−sU0,s+U0,0.
Analogously, one can prove the following theorem.Theorem 2.
Letα,β∈0,1 and m,n∈N such that ux,t∈ClR+×R+, where l=maxm,n. Also, let ux,t,xiN3βux,t, and tjN3αux,t for i=1,...,m,j=1,...,n are N-transformable; then, we have(22)LtαLxβxmN3βux,t=pmUp,s−pm−1U0,s−∑i=1m−1pm−1−iLtαxiN3βU0,t,LtαLxβtnN3βux,t=snUp,s−sn−1Up,0−∑j=1n−1sn−1−jLxβtjN3αUx,0,LtαLxβxmN3βtnN3βux,t=pmsnUp,s−s−1Up,0−p−1U0,s−∑j=1n−1s−1−jLxβtjN3αUx,0−∑i=1m−1p−1−iLtαxiN3βU0,t+∑i=1m−1∑j=1n−1s−1−jp−1−ixiN3βtjN3αU0,0+p−ms−nU0,0,where xmN3βux,t and tnN3αux,t are m, n times nonconformable fractional derivative of function ux,t, with order β and α, respectively, and xmN3βtnN3αux,t are the mixed β-th and α-th order nonconformable fractional partial derivatives of function ux,t.


## 4. Examples and Applications

In this section, we use the proposed transform to solve some fractional PDEs in sense of nonconformable fractional derivative arising from several physical and engineering problems.


### 4.1. Homogeneous Space-Time Nonconformable Fractional Heat Equation

Consider the following homogeneous space-time nonconformable fractional heat equation in one dimension:(23)tN3αux,t=x2N3βux,t,0<x,0<t,subject to the initial and boundary conditions(24)u0,t=EαN31,t,(25)ux,0=EβN31,x,(26)xN3βu0,t=EαN31,t.Applying nonconformable double Laplace transform (4) to equation (22), we get(27)s−p2Up,s=Up,0−pU0,s−xN3βU0,s.Then, applying the nonconformable single Laplace transform to the conditions (23)–(25) by using Theorem 1.2 in [13], we get(28)U0,s=Ltαu0,t,=LtαEαN31,t,=1s−1,Up,0=Lxβux,0,=LxβEβN31,x,=1p−1,xN3βU0,s=LtαxN3βu0,t,=LtαEαN31,t,=1s−1.Substituting (27) into (26), one can obtain(29)Up,s=1p−1s−1.Therefore, the solution of problem (22)–(25) is(30)ux,t=EαN31,tEβN31,x.


### 4.2. Homogeneous Space-Time Nonconformable Fractional Wave Equation

Consider the following homogeneous space-time nonconformable fractional wave equation in one dimension:(31)t2N3αux,t=x2N3βux,t,0<x,0<t,subject to the initial and boundary conditions:(32)u0,t=0,(33)ux,0=sinxβ+1β+1,(34)xN3βu0,t=costα+1α+1,(35)tN3αux,0=0.Applying nonconformable double Laplace transform (4) to equation (30), we get(36)s2−p2Up,s+pU0,s+xN3βU0,s−sUp,0−tN3αUp,0=0.Then, applying the nonconformable single Laplace transform to the conditions (31)–(34) by using Theorem 1.2 in [13].(37)U0,s=Ltαu0,t,=0,Up,0=Lxβux,0,=Lxβsinxβ+1β+1,=1p2+1,xN3βU0,s=LtαxN3βu0,t,=Ltαcostα+1α+1,=ss2+1,tN3αUp,0=LxβtN3αux,0,=0.Substituting (36) into (35), one can get(38)Up,s=sp2+1s2+1.Hence, the solution of problem (30)–(34) is(39)ux,t=sinxβ+1β+1costα+1α+1.


### 4.3. Nonhomogeneous Space-Time Nonconformable Fractional Wave Equation

Consider the following nonhomogeneous space-time nonconformable fractional wave equation in one dimension:(40)t2N3αux,t=x2N3βux,t+2EαN31,tcosxβ+1β+1,0<x,0<t,subject to the initial and boundary conditions(41)u0,t=EαN31,t,(42)ux,0=cosxβ+1β+1,(43)xN3βu0,t=0,(44)tN3αux,0=cosxβ+1β+1.Applying nonconformable double Laplace transform (4) to equation (39), we get(45)s2−p2Up,s+pU0,s+xN3βU0,s−sUp,0−tN3αUp,0=2psp2+1.Then, applying the nonconformable single Laplace transform to the conditions (40)–(43) by using Theorem 1.2 in [13], we get(46)U0,s=LtαEαN31,t=1s−1,Up,0=Lxβcosxβ+1β+1=pp2+1,xN3βU0,s=LtαxN3βu0,t,=0,tN3αUp,0=Lxβcosxβ+1β+1,=pp2+1.Substituting (45) into (44), one can obtain(47)Up,s=pp2+1s−1.Thus, the solution of problem (39)–(43) is(48)ux,t=EαN31,tcosxβ+1β+1.


### 4.4. Nonhomogeneous Space-Time Nonconformable Fractional Telegraph Equation

Consider the following nonhomogeneous space-time nonconformable fractional telegraph equation in one dimension:(49)x2N3βux,t−t2N3αux,t−tN3αux,t−ux,t=−4EαN31,tsinxβ+1β+1,0<x,0<t,subject to the initial and boundary conditions(50)u0,t=0,(51)ux,0=sinxβ+1β+1,(52)xN3βu0,t=EαN31,t,(53)tN3αux,0=sinxβ+1β+1.Applying nonconformable double Laplace transform (4) to equation (48), we get(54)p2−s2−s−1Up,s−pU0,s−xN3βU0,s+s+1Up,0+tN3αUp,0=−4s−1p2+1.Then, applying the nonconformable single Laplace transform to the conditions (49)–(52) by using Theorem 1.2 in [13], we get(55)U0,s=0,Up,0=Lxβsinxβ+1β+1,=1p2+1,xN3βU0,s=LtαEαN31,t=1s−1,tN3αUp,0=Lxβsinxβ+1β+1,=1p2+1.Substituting (54) into (53), one can obtain(56)Up,s=1p2+1s−1.Therefore, the solution of problem (48)–(52) is(57)ux,t=EαN31,tsinxβ+1β+1.


## 4.1. Homogeneous Space-Time Nonconformable Fractional Heat Equation

Consider the following homogeneous space-time nonconformable fractional heat equation in one dimension:(23)tN3αux,t=x2N3βux,t,0<x,0<t,subject to the initial and boundary conditions(24)u0,t=EαN31,t,(25)ux,0=EβN31,x,(26)xN3βu0,t=EαN31,t.Applying nonconformable double Laplace transform (4) to equation (22), we get(27)s−p2Up,s=Up,0−pU0,s−xN3βU0,s.Then, applying the nonconformable single Laplace transform to the conditions (23)–(25) by using Theorem 1.2 in [13], we get(28)U0,s=Ltαu0,t,=LtαEαN31,t,=1s−1,Up,0=Lxβux,0,=LxβEβN31,x,=1p−1,xN3βU0,s=LtαxN3βu0,t,=LtαEαN31,t,=1s−1.Substituting (27) into (26), one can obtain(29)Up,s=1p−1s−1.Therefore, the solution of problem (22)–(25) is(30)ux,t=EαN31,tEβN31,x.


## 4.2. Homogeneous Space-Time Nonconformable Fractional Wave Equation

Consider the following homogeneous space-time nonconformable fractional wave equation in one dimension:(31)t2N3αux,t=x2N3βux,t,0<x,0<t,subject to the initial and boundary conditions:(32)u0,t=0,(33)ux,0=sinxβ+1β+1,(34)xN3βu0,t=costα+1α+1,(35)tN3αux,0=0.Applying nonconformable double Laplace transform (4) to equation (30), we get(36)s2−p2Up,s+pU0,s+xN3βU0,s−sUp,0−tN3αUp,0=0.Then, applying the nonconformable single Laplace transform to the conditions (31)–(34) by using Theorem 1.2 in [13].(37)U0,s=Ltαu0,t,=0,Up,0=Lxβux,0,=Lxβsinxβ+1β+1,=1p2+1,xN3βU0,s=LtαxN3βu0,t,=Ltαcostα+1α+1,=ss2+1,tN3αUp,0=LxβtN3αux,0,=0.Substituting (36) into (35), one can get(38)Up,s=sp2+1s2+1.Hence, the solution of problem (30)–(34) is(39)ux,t=sinxβ+1β+1costα+1α+1.


## 4.3. Nonhomogeneous Space-Time Nonconformable Fractional Wave Equation

Consider the following nonhomogeneous space-time nonconformable fractional wave equation in one dimension:(40)t2N3αux,t=x2N3βux,t+2EαN31,tcosxβ+1β+1,0<x,0<t,subject to the initial and boundary conditions(41)u0,t=EαN31,t,(42)ux,0=cosxβ+1β+1,(43)xN3βu0,t=0,(44)tN3αux,0=cosxβ+1β+1.Applying nonconformable double Laplace transform (4) to equation (39), we get(45)s2−p2Up,s+pU0,s+xN3βU0,s−sUp,0−tN3αUp,0=2psp2+1.Then, applying the nonconformable single Laplace transform to the conditions (40)–(43) by using Theorem 1.2 in [13], we get(46)U0,s=LtαEαN31,t=1s−1,Up,0=Lxβcosxβ+1β+1=pp2+1,xN3βU0,s=LtαxN3βu0,t,=0,tN3αUp,0=Lxβcosxβ+1β+1,=pp2+1.Substituting (45) into (44), one can obtain(47)Up,s=pp2+1s−1.Thus, the solution of problem (39)–(43) is(48)ux,t=EαN31,tcosxβ+1β+1.


## 4.4. Nonhomogeneous Space-Time Nonconformable Fractional Telegraph Equation

Consider the following nonhomogeneous space-time nonconformable fractional telegraph equation in one dimension:(49)x2N3βux,t−t2N3αux,t−tN3αux,t−ux,t=−4EαN31,tsinxβ+1β+1,0<x,0<t,subject to the initial and boundary conditions(50)u0,t=0,(51)ux,0=sinxβ+1β+1,(52)xN3βu0,t=EαN31,t,(53)tN3αux,0=sinxβ+1β+1.Applying nonconformable double Laplace transform (4) to equation (48), we get(54)p2−s2−s−1Up,s−pU0,s−xN3βU0,s+s+1Up,0+tN3αUp,0=−4s−1p2+1.Then, applying the nonconformable single Laplace transform to the conditions (49)–(52) by using Theorem 1.2 in [13], we get(55)U0,s=0,Up,0=Lxβsinxβ+1β+1,=1p2+1,xN3βU0,s=LtαEαN31,t=1s−1,tN3αUp,0=Lxβsinxβ+1β+1,=1p2+1.Substituting (54) into (53), one can obtain(56)Up,s=1p2+1s−1.Therefore, the solution of problem (48)–(52) is(57)ux,t=EαN31,tsinxβ+1β+1.


## 5. Discussion

In this section, we illustrate the obtained solutions in the four examples for different values ofα and β in 2D and 3D graphs by using Maple 13 software.In Figure1(a), the solutions of problem (22)–(25) obtained from the nonconformable double Laplace transform method, are shown for different values of α and β when t=1 and x∈0,2.5, Figure 1(b) shows the solution of problem (22)–(25) in x − t plane when α=β=0.05, x∈0,5, and t∈0,5.Figure 1
(a) Solutions of problems (22)–(25) for different values of α and β when t=1. (b) 3D graph of the solution of problem (22)–(25) when α=β=0.05.
(a)(b)Figure2(a) illustrates the nonconformable double Laplace transform solutions of problem (30)–(34) with different values of α and β when t=1 and x∈0,7. Figure 2(b) shows the 3D graph of the solution, where α=β=0.1, x∈0,7, and t∈0,7.Figure 2
(a) Solutions of problems (30)–(34) for different values of α and β when t=1. (b) 3D graph of the solution of problem (30)–(34) when α=β=0.1.
(a)(b)With different values ofα and β, the nonconformable double Laplace transform solutions of problem (39)–(43) are shown in Figure 3(a) with t=1 and x∈0,6. In Figure 3(b), the solution is illustrated in the x − t plane, where α=β=0.01, x∈0,7, and t∈0,7.Figure 3
(a) Solutions of problems (39)–(43) for different values of α and β when t=1. (b) 3D graph of the solution of problem (39)–(43) when α=β=0.01.
(a)(b)The solutions of problem (48)–(52) are shown in Figure 4(a) for different values of α and β when t=1 and x∈0,7, and the 3D graph of the solution is shown in Figure 4(b) where α=β=0.1, x∈0,7, and t∈0,7.Figure 4
(a) Solutions of problems (48)–(52) for different values of α and β when t=1. (b) 3D graph of the solution of problem (48)–(52) when α=β=0.5.
(a)(b)In Figures1–4, one can observe that the solutions in the case of nonconformable fractional derivative approach values of the classical case, i.e., α=β=1, whenever α and β approach one.In Figures2 and 3, one can see that the wavelength increases as α and β approach zero. Thus, the number of crests and troughs increases as α and β approach one, the wave amplitude increases in Figure 1 and decreases in Figures 2 and 3 as α and β approach one.


## 6. Conclusion

In this work, we introduce the nonconformable double Laplace transform with proof of its main properties. Then, we use the transform to solve selected fractional PDEs in sense of nonconformable derivative, such as homogeneous space-time nonconformable fractional heat equation, homogeneous and nonhomogeneous space-time nonconformable fractional wave equations, and nonhomogeneous space-time nonconformable fractional telegraph equation. In addition, 3D graphical representations are offered for obtained solutions with illustrations for different values ofα and β with t=1 fixed. The 2D graphs show that the obtained solutions, by using the nonconformable double Laplace transform, are close to that of classical derivatives as α⟶1 and β⟶1 with t=1 fixed. We observe that the wavelength increases and the amplitude decreases in the case of the homogeneous space-time nonconformable fractional wave equation as α⟶0 and β⟶0 with t=1 fixed. It increases in the last two cases as α and β approaches zero with t=1 fixed. We may conclude that this transform can be a very powerful technique to solve many fractional PDEs involving nonconformable derivative arising in physics, chemistry, and engineering.



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