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Results Involving Partial Differential Equations and Their Solution by Certain Integral Transform

Rania Saadah1, Mohammed Amleh1, Ahmad Qazza1, Shrideh Al-Omari2,*, Ahmet Ocak Akdemir3

1 Department of Mathematics, Faculty of Science, Zarqa University, Zarqa, 13110, Jordan
2 Department of Mathematics, Faculty of Science, Al-Balqa Applied University, Amman, 11134, Jordan
3 Department of Mathematics, Faculty of Science and Letters, Agri Ibrahim Çeçen University, Agri, 04100, Turkey

* Corresponding Author: Shrideh Al-Omari. Email: email

(This article belongs to the Special Issue: On Innovative Ideas in Pure and Applied Mathematics with Applications)

Computer Modeling in Engineering & Sciences 2024, 138(2), 1593-1616. https://doi.org/10.32604/cmes.2023.029180

Abstract

In this study, we aim to investigate certain triple integral transform and its application to a class of partial differential equations. We discuss various properties of the new transform including inversion, linearity, existence, scaling and shifting, etc. Then, we derive several results enfolding partial derivatives and establish a multi-convolution theorem. Further, we apply the aforementioned transform to some classical functions and many types of partial differential equations involving heat equations, wave equations, Laplace equations, and Poisson equations as well. Moreover, we draw some figures to illustrate 3-D contour plots for exact solutions of some selected examples involving different values in their variables.

Keywords


1  Introduction

Mathematical modeling is one of the main aspects of mathematics that deals with the real-life problems involved in various fields of science. Mathematical modeling transforms life problems into mathematical models and analyzes models mathematically. It is generally used in the computational theory of the field of natural sciences, computer sciences, engineering and social sciences as well. Mathematical models are the cornerstone of the modern scientific understanding of sound, heat, diffusion, static electricity, thermodynamics, fluid dynamics, elasticity, quantum mechanics and general relativity. They can be generated from various mathematical considerations comprising differential geometry and calculus of variation. Partial differential equations (PDEs) occupy a large sector of oriented scientific fields in physics and engineering. Various methods have been developed over the years to find solutions of partial differential equations, some of which reduce partial differential equations to one or more than one ordinary differential equation (ODE). However, a number of approaches have been proposed to deal with partial differential equations, such as the method of group static solutions [1,2], the non-classical method of group static [3], the method of partially invariant solutions [1] and the general method of variable separation for both linear systems. In the non-linear equations we recall the Hamilton-Jacobi equation, residual power series [4], reproducing kernel reproduction [5,6], Laplace residual power series and others [79].

Partial differential equations can also be handled using integral transformations to determine the exact solution to the target equations without the need for linearization or discretization. The transformation techniques are powerful and straightforward. Therefore, scientists have put a lot of efforts into understanding and improving techniques. Many integral transforms have been developed such as the Laplace transform [10], Novel transform [11], Sumudu transform [12], Elzaki transform [13], Natural transform [14], ARA transform [15], formable transform [16] and others [1719].

In the sequel, extensions of double transformations are extensively developed for solving PDEs and obtaining results by performing solutions based on numerical approaches [20]. Double transforms such as the double Laplace transform [2123], double Shehu transform [24], double Kamal transform [25], double Sumudu transform [2631], double Elzaki transform [32], double Laplace-Sumudu transform [3335] and ARA–Sumudu transform [36,37] have been considered in the literature. The ARA transform [15] was used for solving ordinary fractional differential equations [38] and some nonlinear PDEs [39]. The double ARA transform was employed for some kinds of partial and integral equations in [4047].

In this work, we extend the idea of the single and double ARA transforms to a triple ARA transform which can be used in solving larger classes of PDEs. In this context, we define the new transform and present its main properties. More results regarding the existence, partial derivatives, and convolution theorems are also presented. However, the novelty of this study fulfills the generalization of the double ARA transform and its application to wider classes of PDEs.

Motivation in this work emerges from introducing a new approach that solves partial differential equations of multi-variables. We present characteristics of the transform and compute its values for some basic functions. Further, we show its applicability to various types of PDEs of great importance in physics. However, we organize this paper as follows. In Sections 2 and 3, we discuss the single and double ARA transforms and obtain their main properties. In Section 4, we introduce the triple ARA transform and prove some basic results on existence, elementary functions, convolutions, and partial derivatives. In Section 5, we apply the TARAT in solving PDEs and consider some examples and their 3D graphs. In Section 6, we give a conclusion section.

2  Basic Definitions and Preliminary Results

In this section, we recall some definitions and notations from [15] and establish several basic properties of the ARA transform.

Definition 2.1. The nth order ARA transform of a continuous function ψ is defined over an infinite interval (0,∞) as

𝒢n[ψ(x)](s)=Ψ(n,s)=s0xn1esxψ(x)dx,s>0,(1)

provided the integral exists. In particular, for n=1, the ARA integral transform of a continuous function ψ is given by

𝒢1[ψ(x)](s)=Ψ(s)=s0esxψ(x)dx,s>0,(2)

when the integral exists. For simplicity, we make use of the notation 𝒢[ψ]of the ARA transform to replacce  the  notation 𝒢1[ψ].

The inverse ARA transform is defined by [15]

𝒢x1[Ψ(s)]=ψ(x)=12πirir+iesxsΨ(s)ds.(3)

Following is a result which is very needful in the sequel.

Theorem 2.1. Let the function ψ be piecewise continuous in every finite closed interval 0xα [15], and satisfies the condition

|xn1ψ(x)|Reαx,

where R is a positive constant. Then, the ARA transform exists for all s>α,α being real constant.

Proof. By employing the definition of the ARA transform we obtain

|Ψ(n,s)|=|s0xn1esxψ(x)dx|,

which reveals

|Ψ(n,s)|s|0xn1esxψ(x)dx|s0esx|xn1ψ(x)|dxs0esxReαxdx=sR0e(sα)xdx=sRsαe(sα), for  all s>α.

It is clear from the above equation that the improper integral converges for all s>α. Therefore, the 𝒢n+1[ψ] exists for all s>α,α being a real constant.

This completes the proof of the theorem.

Note that, the existence condition of the ARA transform fulfils when the function ψ is continuous and possessing the property.

|ψ(x)|Reαx,for  some  real  constant R.

Let Ψ(s)=𝒢[ψ], Q(s)=𝒢[q] and a,bR. Then, the following are some basic properties of the ARA transform:

•   𝒢[aq(x)+bψ(x)]=a𝒢[q(x)]+b𝒢[ψ(x)].

•   𝒢1[aQ(s)+bΨ(s)]=a𝒢1[Q(s)]+b𝒢1[Ψ(s)].

•   𝒢[xα]=Γ(α+1)sα,α>0.

•   𝒢[eax]=ssa,aR.

•   𝒢[ψ(n)(x)]=sn𝒢[ψ(x)]k=1nsnk+1ψ(k1)(0).

•   𝒢[xnψ(x)]=Ψ(1+n,s),nN.

3  Double ARA Transform

This section discusses the new double ARA transform DARAT. It provides fundamental properties and characteristics of the transform including existence conditions, linearity and inversion formula for the proposed new double transform. Moreover, it provides some important properties and results and computes the double ARA transform of some elementary functions.

Definition 3.1. Let ψ be a continuous function of two positive variables x and y. Then, the DARAT of ψ is defined by

𝒢x,y[ψ(x,y)]=Ψ(s,p)=sp00e(sx+py)ψ(x,y)dxdy,s,p>0,(4)

provided the integral exists. Clearly, the double ARA transform is linear, which can be shown from the equation

𝒢x,y[Aq(x,y)+Bψ(x,y)]=sp00e(sx+py)(Aq(x,y)+Bψ(x,y))dxdy=Asp00e(sx+py)q(x,y)dxdy+Bsp00e(sx+py)ψ(x,y)dxdy=A𝒢x,y[q(x,y)]+B𝒢x,y[ψ(x,y)],

provided A and B are constants and, 𝒢x,y[q] and 𝒢x,y[ψ] exist. The inverse DARAT is defined by

𝒢x,y1[Ψ(s,p)]=𝒢x1[𝒢y1[Ψ(s,p)]]=(12πi)cic+iesxsds(12πi)rir+iepypΨ(s,p)dp=ψ(x,y).

Now, we state without proof the following theorems [40,41].

Theorem 3.1 [40] Let ψ be a continuous function of exponential orders α and β, defined on the region [0,X)×[0,Y). Then Ψ(s,p) exists for s and p, provided Re(s)>α and Re(p)>β.

Theorem 3.2 (Periodic Function) [40] Let 𝒢x,y[ψ] exist, where ψ is a periodic function of periods α>0 and β>0, and ψ(x+α,y+β)=ψ(x,y),x,y. Then,

𝒢x,y[ψ(x,y)]=1(1e(sα+pβ))(sp0α0βe(sx+py)(ψ(x,y))dxdy).

Theorem 3.3 (Heaviside Function) [40] Let 𝒢x,y[ψ(x,y)]=Ψ(s,p)be  exist. Then, we have

𝒢x,y[ψ(xδ,yε)H(xδ,yε)]=esδpεΨ(s,p),

where H(xδ,yε) is the Heaviside function defined by

H(xδ,yε)={1,x>δ,y>ε,0,otherwise.

Theorem 3.4 (Convolution Theorem). [40] Let 𝒢x,y[q] and 𝒢x,y[ψ] be given such that 𝒢x,y[q(x,y)]=Q(s,p) and 𝒢x,y[ψ(x,y)]=Ψ(s,p). Then, we have

𝒢x,y[q(x,y)ψ(x,y)]=1spQ(s,p)Ψ(s,p),(5)

where q(x,y)ψ(x,y)=0x0yq(xδ,yε)ψ(δ,ε) dδdε and the symbol denotes the double convolution with respect to x and y.

In the following, we present Table 1 to illustrate the values of the double ARA transform of some elementary functions and give main properties of the transform.

images

4  The Triple ARA Transform

In this section, we introduce a new triple transform, called the triple ARA transform (TARAT), and derive some basic properties and establish certain results involving partial derivatives.

Definition 4.1. Let ψ be a continuous function. Then, the triple ARA transform of ψ is defined by

𝒢x,y,t[ψ(x,y,t)]=Ψ(s,p,k)=𝒢x{𝒢y{𝒢t{ψ(x,y,t);tk};yp};xs}=spk000esxpyktψ(x,y,t)dxdydt,

where x,y,t>0 and, s,p and k are complex variables of the transform. The inverse triple ARA transform is given by

ψ(x,y,t)=𝒢x,y,t1[Ψ(s,p,k)]=12πaia+iesxs×12πibib+iepyp×12πicic+iektkΨ(s,p,k)dk dp ds.

Note that, the triple ARA transform is a linear transform, due to the fact that if ψ and q are two continuous functions, then

𝒢x,y,t[αψ(x,y,t)+βq(x,y,t)]=spk000[αψ(x,y,t)+βq(x,y,t)]dxdydt=αspk000ψ(x,y,t)dxdydt+βspk000q(x,y,t)dxdydt=αΨ(s,p,k)+βQ(s,p,k).

To simplify the application of the triple ARA transform, we need to recall the following notations:

𝒢x[ψ(x,y,t)]=Ψ¯(s,y,t) is the single ARA transform of the function ψ with respect to x, and

𝒢xy[ψ(x,y,t)]=Ψ¯¯(s,p,t) is the double ARA transform of the function ψ with respect to x and y.

By a similarity in the variables x,y and t, we may consider the following notations:

𝒢y[ψ(x,y,t)]=Ψ¯(x,p,t),𝒢t[ψ(x,y,t)]=Ψ¯(x,y,k),𝒢x,t[ψ(x,y,t)]=Ψ¯¯(s,y,k),𝒢yt[ψ(x,y,t)]=Ψ¯¯(x,p,k).

The existence conditions of the triple ARA transform are presented in the following theorem.

Theorem 4.1. Let ψ be a continuous function on I3, where I=[0,), and satisfy the following condition

|ψ(x,y,t)|Meαx+βy+γt,

where α,β,γ and M>0. Then, the triple ARA transform exists for all Re(s)>α, Re(p)>β and Re(k)>γ. In addition, the triple ARA transform is unique, i.e., if Ψ=Q, then ψ=q.

Proof. Assume that α,β,γ>0 such that Re(s)>α, Re(p)>β and Re(k)>γ. Then, we have

Ψ(s,p,k)=𝒢x,y,t[ψ(x,y,t),(s,p,k)]|spk|000|ψ(x,y,t)|eRe(s)xRe(p)yRe(k)tdxdydtM|s|αRe(s)×|p|βRe(p)×|k|γRe(k).

Also, the inverse triple ARA transform implies that

ψ(x,y,t)=12πiaia+iesxs×12πibib+iepyp×12πicic+iektkΨ(s,p,k)dk dp ds=12πiaia+iesxs×12πibib+iepyp×12πicic+iektkQ(s,p,k)dk dp ds=q(x,y,t).

This finishes the proof of the theorem.

In Table 2, we provide some values of the triple ARA transform for some basic functions.

images

Now, we introduce some properties of the new approach.

Property 4.1. (Change of scale property) Let ψ be a continuous function on which the triple ARA transform exists. Then, we have

𝒢x,y,t[ψ(ax,by,ct)]=Ψ(sa,pb,kc),(6)

where a,b and c are real constants.

Proof. Based on the definition of the triple ARA transform, we have

𝒢x,y,t[ψ(ax,by,ct)]=spk000esxpykt.ψ(ax,by,ct) dxdydt.(7)

Putting ax=u,by=v and ct=w ensures that Eq. (7) becomes

𝒢x,y,t[ψ(ax,by,ct)]=spk000esaupbvkcwψ(u,v,w)duadvbdwc=sapbkc000esaupbvkcwψ(u,v,w)du dv dw=Ψ(sa,pb,kc).

This finishes the proof of our result.

Property 4.2. (First shifting property) Let ψ be a continuous function on which the triple ARA transform exists. Then, we have

𝒢x,y,t[eax+by+ctψ(x,y,t)]=spk(sa)(pb)(kc)Ψ(sa,pb,kc),

where a,b and c are real constants.

Proof. Based on the definition of the triple ARA transform, we obtain

𝒢x,y,t[eax+by+ctψ(x,y,t)]=spk000eax+by+ct.esxpykt.ψ(x,y,t)dxdydt

=spk000e(sa)x.e(pb)ye(kc)t.ψ(x,y,t)dxdydt=spk.(sa)(pb)(kc)(sa)(pb)(kc)000e(sa)x.e(pb)ye(kc)t.ψ(x,y,t)dx dy dt=spk(sa)(pb)(kc)Ψ(sa,pb,kc).

This finishes the proof of our result.

The following result is very useful in solving some kinds of partial differential equations.

Theorem 4.2. If 𝒢x,y,t[ψ(x,y,t)]=Ψ(s,p,k) be given. Then, we have

𝒢x,y,t[xnymtrψ(x,y,t)]=(1)n+m+r×spkn+m+rsnpmkr(Ψ(s,p,k)spk),(8)

where n,m and rN.

Proof: The definition of the triple ARA transform implies

𝒢x,y,t[xnymtrψ(x,y,t)]=spk000esxpyktxnymtrψ(x,y,t)dx dy dt=pk00epyktymtr(s0esxxnψ(x,y,t)dx)dy dt.

From the properties of the single ARA, we get

𝒢x,y,t[xnymtrψ(x,y,t)]=pk00ymtrepykts(1)nnsn[𝒢x[ψ(x,y,t)]s]dy dt.

Again, using some properties of the ARA transform to both variables y and t leads to write

𝒢x,y,t[xnymtrψ(x,y,t)]=(1)n(1)m(1)rspkn+m+rsnpmkr(𝒢x,y,t[ψ(x,y,t)]spk).

This finishes the proof of the theorem.

The application of triple ARA on partial derivatives of different types is illustrated on the following theorem.

Theorem 4.3. Let Ψ be the triple ARA transform of the function ψ. Then, the following hold true:

i)   𝒢x,y,t[ψ(x,y,t)x]=sΨ(s,p,k)sΨ¯¯(0,p,k).

ii)   𝒢x,y,t[2ψ(x,y,t)x2]=s2Ψ(s,p,k)s2Ψ¯¯(0,p,k)sΨ¯¯(0,p,k)x.

iii)   𝒢x,y,t[3ψ(x,y,t)x3]=s3Ψ(s,p,k)s3Ψ¯¯(0,p,k)s2Ψ¯¯(0,p,k)xs2Ψ¯¯(0,p,k)x2.

iv)   𝒢x,y,t[3ψ(x,y,t)xyt]=spkΨ(s,p,k)spkΨ¯¯(0,p,k)spkΨ¯¯(s,0,k)spkΨ¯¯(s,p,0)+spkΨ¯(s,0,0)+spkΨ¯(0,p,0)+spkΨ¯(0,0,k)spkψ(0,0,0).

Proof (i). From the definition of the triple ARA transform, we write

𝒢x,y,t[ψ(x,y,t)x]=spk000esxpyktψ(x,y,t)xdx dy dt=pk0ektdt0epydy(s0esxψ(x,y,t)xdx).

To solve the integration inside the brackets, we use the idea of integration by parts with the assumptions that

u=esx and dv=ψ(x,y,t)x to get

𝒢x,y,t[ψ(x,y,t)x]=pk0ektdt0epydt(sψ(0,y,t)+s20ψ(x,y,t)dx)=sΨ(s,p,k)sΨ¯¯(0,p,k).

(ii). By using part (i) of the theorem, it follows that:

𝒢x,y,t[2ψ(x,y,t)x2]=𝒢x,y,t[x(ψ(x,y,t)x)]=s𝒢x,y,t[ψ(x,y,t)x]sΨ¯¯(0,p,k)x=s2Ψ(s,p,k)s2Ψ¯¯(0,p,k)sΨ¯¯(0,p,k)x,

where,

Ψ¯¯(0,p,k)x=𝒢x,y,t[ψ(0,y,t)x].

Similarly, one can follow the proof of (iii) and (iv).

This proves the theorem.

In view of the similarity between x,y and t, one can conclude the rest of derivative rules which can enlisted as follows.

The following property is a basic result from Theorem 4.3, and the proof can be obtained by similar arguments to the above theorem.

Corollary 4.1. Let Ψ(s,p,k)=𝒢x,y,t[ψ(x,y,t)]. Then, we have the following results hold:

•   𝒢x,y,t[ψ(x,y,t)y]=pΨ(s,p,k)pΨ¯(s,0,k).

•   𝒢x,y,t[ψ(x,y,t)t]=kΨ(s,p,k)kΨ¯(s,p,0).

•   𝒢x,y,t[2ψ(x,y,t)xy]=spΨ(s,p,k)spΨ¯¯(s,0,k)spΨ¯¯(0,p,k)+spΨ¯(0,0,k).

•   𝒢x,y,t[2ψ(x,y,t)xt]=skΨ(s,p,k)skΨ¯¯(s,p,0)skΨ¯¯(0,p,k)+skΨ¯(0,p,0).

•   𝒢x,y,t[2ψ(x,y,t)yt]=pkΨ(s,p,k)pkΨ¯¯(s,0,k)pkΨ¯¯(s,p,0)+pkΨ¯(s,0,0).

The following corollary presents a new application of the triple ARA transform, that is useful in solving some kinds of integral equations.

Corollary 4.2. If 𝒢x,y,t[ψ(x,y,t)]=Ψ(s,p,k), then we have

𝒢x,y,t[0t0y0xψ(u,v,w)du dv dw]=Ψ(s,p,k)spk,s,p,k>0.

Proof. Suppose that

g(x,y,t)=0t0y0xψ(u,v,w)du dv dw,

and g(0,0,0)=0. Now, consider

3g(x,y,t)xyt=ψ(x,y,t).(9)

Applying the triple ARA transform to both sides of Eq. (4), and using Theorem 4.3 suggest to write

𝒢x,y,t[3g(x,y,t)xyt]=𝒢x,y,t[ψ(x,y,t)]=Ψ(s,p,k).

Thus, we write

Ψ(s,p,k)=𝒢x,y,t[3g(x,y,t)xyt]=spk G(s,p,k)spk G¯¯(s,p,0)spk G¯¯(S,0,K)spk G¯¯(0,p,k)+psk G¯(0,p,0)+ksp G¯(0,0,k)+spk G¯(s,0,0)spk g(0,0,0),=spk G(s,p,k),

which implies that

G(s,p,k)=1spkΨ(s,p,k).

This ends the proof of our corollary.

Now we study the effect of the triple ARA on the Heaviside function.

Theorem 4.4. Let 𝒢x,y,t[ψ(x,y,t)]=Ψ(s,p,k). Then, we have

𝒢x,y,t[ψ(xδ,yε,tσ)H(xδ,yε,tσ)]=eδsϵpσkΨ(s,p,k),

where H(x,y,t) denotes the Heaviside function defined by

H(xδ,yε,tσ)={1,x>δ,y>ϵ,t>σ0,otherwise.

Proof. From the definition of the triple ARA transform, we have

𝒢x,y,t[ψ(xδ,yε,tσ)H(xδ,yε,tσ)]=spk000esxpykt[ψ(xδ,yε,tσ)H(xδ,yε,tσ)]dxdydt=spkσεδesxpykt[ψ(xδ,yε,tσ)]dxdydt.

Putting xδ=ρ, yϵ=β and tσ=λ in Eq. (32), we obtain

𝒢x,y,t[ψ(xδ,yε,tσ)H(xδ,yε,tσ)]=spk000es(ρ+δ)p(β+ϵ)k(λ+σ)[ψ(ρ,β,λ)]dρdβdλ.

Thus, we infer

𝒢x,y,t[ψ(xδ,yε,tσ)H(xδ,yε,tσ)]=eδsϵpσk(spk000esρpβσλ[ψ(ρ,β,λ)]dρdβdλ)=eδsϵpσkΨ(s,p,k).

This ends the proof of our theorem.

The following theorem presents the application of the triple ARA on the multi convolution property.

Theorem 4.5.(Convolution Theorem) If 𝒢x,y,t[ψ]=Ψ and 𝒢x,y,t[q]=Q, then we have

𝒢x,y,t[(ψq)(x,y,t)]=1spkΨ(s,p,k)Q(s,p,k),

where

(ψq)(x,y,t)=0x0y0tψ(xδ,yε,tσ)q(δ,ε,σ)dδdεdσ.

Proof. Following the definition of the triple ARA transform, we write

𝒢x,y,t[(ψq)(x,y,t)]=spk000esxpykt(0x0y0tψ(xδ,yε,tσ)q(δ,ε,σ)dδdεdσ)dxdydt.

The definition of the Heaviside function implies

𝒢x,y,t[(ψq)(x,y,t)]=spk000esxpykt(000ψ(xδ,yε,tσ)H(xδ,yε,tσ)×0q(δ,ε,σ)dδdεdσ)dxdydt.(10)

Thus, we get

𝒢x,y,t[(ψq)(x,y,t)]=000q(δ,ε,σ)dδdεdσ(spk000esxpyktψ(xδ,yε,tσ)×0H(xδ,yε,tσ))dxdydt.

Using Theorem 4.4 reveals

𝒢x,y,t[(ψq)(x,y,t)]=000q(δ,ε,σ)dδdεdσeδsϵpσkΨ(s,p,k)=Ψ(s,p,k)000eδsϵpσkq(δ,ε,σ)dδdεdσ=1spkΨ(s,p,k)Q(s,p,k).

This completes the proof of the theorem.

5  Algorithm and Applications

In this section, we present an algorithm for using the triple ARA transform in solving partial differential equations and provide solutions of some examples including partial differential equations.

5.1 Algorithm of the TARAT Method

To illustrate the method of using the triple ARA transform in solving partial differential equations, let us consider the following partial differential equation:

L(ψ(x,y,t))+A(ψ(x,y,t))=g(x,y,t),(11)

where L is any linear differential operator, with respect to x,y or t, A is any constant, ψ is the unknown function and g is the source term. To apply the triple ARA transform method, we operate the TARAT on both sides of Eq. (7) to get

𝒢x,y,t[L(ψ(x,y,t))+𝒢x,y,t[Aψ(x,y,t)]]=𝒢x,y,t[g(x,y,t)],

which yields

𝒢x,y,t[L(ψ(x,y,t))]+AΨ(s,p,k)=G(s,p,k),(12)

where Ψ=𝒢x,y,t[ψ] and G=𝒢x,y,t[g]. Now, since L and 𝒢x,y,t are linear operators, then by running the triple ARA transform on the operator L and doing some algebraic simplifications lead to the following form

Ψ(s,p,k)H1(s,p,k)+H2(s,p,k)+AΨ(s,p,k)=G(s,p,k),(13)

where H1 and H2 are functions depending on the discussed equation to be determined. Simplifying Eq. (13) suggests to have

Ψ(s,p,k)=G(s,p,k)H2(s,p,k)H1(s,p,k)+A.(14)

To obtain the solution in the original space, we apply the inverse triple ARA transform to Eq. (14) to have

ψ(x,y,t)=𝒢x,y,t1[G(s,p,k)H2(s,p,k)H1(s,p,k)+A].

Hence, the desired result is therefore established.

5.2 Applications

In this section, we apply the triple ARA transform operator to solve the wave equation, heat equation, Poisson equation and some few others.

Example 1: Consider the following wave equation

22ψ(x,y,t)t22ψ(x,y,t)x22ψ(x,y,t)y2=24t2+4y,(15)

where x,y ϵ R and t>0 subject to the conditions:

ψ(0,y,t)=t4+yt2,ψ(x,0,t)=t4,ψ(x,y,0)=0,

ψ(0,y,t)x=sinysint,ψ(x,0,t)y=t2+sinxsint,

ψ(x,y,0)t=sinxsiny.(16)

Solution: Operating the triple ARA transform to both sides of Eq. (5) yields

2𝒢x,y,t[2ψ(x,y,t)t2]𝒢x,y,t[2ψ(x,y,t)x2]𝒢x,y,t[2ψ(x,y,t)y2]=𝒢x,y,t[24t2+4y].

Hence, using the derivative properties implies

2(k2ΨΨ¯¯(s,p,0)t(s,p,k)k2Ψ¯¯(s,p,0)kΨ¯¯(s,p,0)t)(s2Ψ(s,p,k)s2Ψ¯¯(0,p,k)sΨ¯¯(0,p,k)x)(p2Ψ(s,p,k)p2Ψ¯¯(s,0,k)pΨ¯¯(s,0,k)y)=48k2+4p.(17)

To simplify Eq. (17), we find the transformed values of the conditions (16) as follows:

Ψ¯¯(0,p,k)=𝒢y,t[ψ(0,y,t)]=𝒢y,t[t4+yt2]=24k4+2pk2,

Ψ¯¯(s,0,k)=𝒢x,t[ψ(x,0,t)]=𝒢x,t[t4]=24k4,

Ψ¯¯(s,p,0)=Gx,y[ψ(x,y,0)]=Gx,y[0]=0 ,

Ψ¯¯(0,p,k)x=𝒢y,t[ψ(0,y,t)x]=𝒢y,t[sinysint]=pk(p2+1)(k2+1),

Ψ¯¯(s,0,k)y=𝒢x,t[ψ(x,0,t)y]=𝒢x,t[t2+sinxsint]=2k2+k(s2+1)(k2+1),

Ψ¯¯(s,p,0)t=𝒢x,y[ψ(x,y,0)t]=𝒢x,y[sinxsiny]=sp(s2+1)(p2+1).

By substituting the above values and using simple calculations, we can derive

Ψ(s,p,k)=24k4+2pk2+spk(s2+1)(p2+1)(k2+1).(18)

Applying the inverse triple ARA transform 𝒢x,y,t1 to Eq. (18), we obtain a solution to Example 1 in the form

ψ(x,y,t)=t4+yt2+sinxsinysint.

The following Fig. 1 illustrates the contours 3D plot of the exact solution to Example 1, with different values of the variables as follows.

images

Figure 1: Contour 3D plot of the solution to Example 1, ψ(x,y,t) with the values x[2,2],y[2,2] and three values of t in (a) t[1,2], (b) t=1,2,3, (c) t=1,2,3

Example 2: Consider the heat partial differential equation

ψ(x,y,t)t=2ψ(x,y,t)x2+2ψ(x,y,t)y2+2cos(x+y),(19)

where x,yϵR and t>0 subject to the conditions:

ψ(0,y,t)=e2tsiny+cosy,ψ(x,0,t)=e2tsinx+cosx,

ψ(x,y,0)=sin(x+y)+cos(x+y),ψ(0,y,t)x=e2tcosysiny,

ψ(x,0,t)y=e2tcosxsinx.(20)

Solution: Operating the triple ARA transform to both sides of Eq. (19) gives rise to

𝒢x,y,t[ψ(x,y,t)t]=𝒢x,y,t[2ψ(x,y,t)x2]+𝒢x,y,t[2ψ(x,y,t)y2].+𝒢x,y,t[2cos(x+y)].(21)

Employing properties of the derivative infer

kΨ(s,p,k)kΨ¯¯(s,p,0)=s2Ψ¯¯(s,p,k)s2Ψ¯¯(0,p,k)sΨ¯¯(0,p,k)x+p2Ψ(s,p,k)p2Ψ¯¯(s,0,k)pΨ¯¯(s,0,k)+sp(sp1)s2+1.(22)

To simplify Eq. (22), we compute the transformed values of the conditions as follows:

Ψ¯¯(0,p,k)=𝒢y,t[e2tsiny+cosy]=pk(p2+1)(k+2)+p2p2+1,

Ψ¯¯(s,0,k)=𝒢x,t[e2tsinx+cosx]=sk(s2+1)(k+2)+s2s2+1,

Ψ¯¯(s,p,0)=𝒢x,y[sin(x+y)+cos(x+y)]=sp(s+p)p2+1+sp(sp1)s2+1,

Ψ¯¯(0,p,k)x=𝒢y,t[e2tcosysiny]=p2k(p2+1)(k+2)pp2+1,

Ψ¯¯(s,0,k)y=𝒢x,t[e2tcosxsinx]=s2k(s2+1)(k+2)ss2+1.

Substituting the above values included in Eq. (22) and applying some calculations we reach the conclusion

Ψ(s,p,k)=ksp(s+p)(k+2)(p2+1)+sp(sp1)s2+1.(23)

Applying the inverse 𝒢x,y,t1 to Eq. (23), we get a solution in the original space as

ψ(x,y,t)=e2tsin(x+y)+cos(x+y).

Our example has, therefore, been solved.

The following Fig. 2 illustrates the contours 3D plot of the exact solution to Example 2, with different values of the variables as follows.

images

Figure 2: Contour the 3D plot of the solution to Example 2, ψ(x,y,t) with the values x[2,2],y[2,2] and three values of t in (a) t[1,2], (b) t=1,2,3, (c) t=1,2,3

Example 3: Consider the following Poisson partial differential equation:

2ψ(x,y,t)x2+2ψ(x,y,t)y2+2ψ(x,y,t)t2=2sinxcosysinh2t,(24)

where x,y and t are positive real numbers subject to the initial conditions:

ψ(0,y,t)=0,ψ(x,0,t)=sinxsinh2t,ψ(x,y,0)=0,

ψ(0,y,t)x=cosysinh2t,ψ(x,0,t)y=0,ψ(x,y,0)t=2sinxcosy.(25)

Solution: By arguments alike to those employed for Examples 1 and 2, we apply the TARAT to Eq. (24) and compute the transformed values of the initial conditions to infer

Ψ¯¯(0,p,k)=Gy,t[0]=0,

Ψ¯¯(s,0,k)=𝒢x,t[sinxsinh2t]=sk(s2+1)(k21),

Ψ¯¯(s,p,0)=Gx,y[0]=0,

Ψ¯¯(0,p,k)x=𝒢y,t[cosysinh2t]=2p2k(p2+1)(k21),

Ψ¯¯(s,0,k)y=Gx,t[0]=0 ,

Ψ¯¯(s,p,0)t=𝒢x,y[2sinxcosy]=2sp2(s2+1)(p2+1).

Thus, we have the solution, in the triple ARA space, as

Ψ(s,p,k)=2sp2k(s2+1)(p2+1)(k24).(26)

Now, we reach the solution to Example 3 in the original space, by applying the inverse TARAT 𝒢x,y,t1 to Eq. (26) to have

ψ(x,y,t)=𝒢x,y,t1[2sp2k(p2+1)(s2+1)(k24)]=sinxcosysinh2t.

Our example has been solved.

The following Fig. 3 illustrates the contours 3D plot of the exact solution to Example 3, with different values of the variables as follows.

images

Figure 3: Contour 3D plot of the solution to Example 3, ψ(x,y,t) with the values x[2,2],y[2,2] and three values of t in (a) t[1,2], (b) t=1,2,3, (c) t=1,2,3

Example 4: Consider the following Laplace partial differential equation:

2ψ(x,y,t)x2+2ψ(x,y,t)y2+2ψ(x,y,t)t2=0,(27)

where x,y and t are positive real numbers subject to the conditions:

ψ(0,y,t)=0,ψ(x,0,t)=0,ψ(x,y,0)=0,

ψ(0,y,t)x=sinysinh2t,ψ(x,0,t)y=sinxsinh2t,

ψ(x,y,0)t=2sinxsiny.(28)

Solution: Following steps alike to those used in the previous examples, then by applying the TARAT to the conditions in (28) and operating the triple ARA transform to Eq. (27) we establish that

Ψ¯(0,p,k)=Ψ¯¯(s,0,k)=Ψ¯¯(s,p,0)=0,

Ψ¯¯(0,p,k)x=2pk(p2+1)(k22),

Ψ¯¯(s,0,k)y=2sk(s2+1)(k22),

Ψ¯¯(s,p,0)t=2sp(s2+1)(p2+1).

Thus, we get the solution in the triple ARA space in the form

Ψ(s,p,k)=2spk(s2+1)(p2+1)(k22).(29)

Following that, we get a solution to Example 4 in the original space after applying the inverse triple ARA transform 𝒢x,y,t1 to Eq. (29) as

ψ(x,y,t)=𝒢x,y,t1[2spk(s2+1)(p2+1)(k22)]=sinxsinysinh(2t).

Our example has been solved.

The following Fig. 4 illustrates the contours 3D plot of the exact solution to Example 4, with different values of the variables as follows.

images

Figure 4: Contour 3D plot of the solution to Example 4, ψ(x,y,t) with the values x[2,2],y[2,2] and three values of t in (a) t[1,2], (b) t=1,2,3, (c) t=1,2,3

Example 5: Consider the following partial differential equation

3ψ(x,y,t)xyt+ψ(x,y,t)=cosxcosycostsinxsinysint,(30)

where x,y and t are positive real numbers subject to the initial conditions

ψ(x,y,0)=cosxcosy,ψ(x,0,t)=cosxcost,ψ(0,y,t)=cosycost,

ψ(x,0,0)=cosx,ψ(0,y,0)=cosy,

ψ(0,0,t)=cost,ψ(0,0,0)=1.(31)

Solution: Applying the triple ARA transform to the condition (31) infers

Ψ¯¯(s,p,0)=𝒢x,y[cosxcosy]=s2p2(s2+1)(p2+1).

Ψ¯¯(s,0,k)=𝒢x,t[cosxcost]=s2k2(s2+1)(k2+1).

Ψ¯¯(0,p,k)=𝒢y,t[cosycost]=p2k2(p2+1)(k2+1).

Ψ¯(s,0,0)=𝒢x[cosx]=s2s2+1.

Ψ(0,p,0)=𝒢y[cosy]=p2p2+1.

Ψ¯(0,0,k)=𝒢t[cost]=k2k2+1.

Now, allowing the triple ARA transform to act on both sides of Eq. (31) and using the transformed values of the conditions with simple calculations reveal

𝒢x,y,t[ψ(x,y,t)xyt]+𝒢x,y,t[ψ(x,y,t)]=𝒢x,y,t[cosxcosycostsinxsinysint].

This indeed implies

Ψ(s,p,k)=s2p2k2(s2+1)(p2+1)(k2+1).(32)

Operating the inverse TARAT 𝒢x,y,t1 to Eq. (32) gives

ψ(x,y,t)=𝒢x,y,t1[s2p2k2(s2+1)(p2+1)(k2+1)]=cosxcosycost.

Our example has then been solved.

The following Fig. 5 illustrates the contours 3D plot of the exact solution to Example 5, with different values of the variables as follows.

images images

Figure 5: Contour 3D plot of the solution to Example 4, ψ(x,y,t) with the values x[2,2],y[2,2] and three values of t in (a) t[1,2], (b) t=1,2,3, (c) t=1,2,3

6  Conclusion

In this paper, a new triple transform called the triple ARA transform is presented and several properties and theorems including linearity, existence, partial derivatives, and the multiple convolution theorem are introduced. Tables that summarize the new results are established to illustrate the new approach. The methodology of using the triple ARA transform in solving partial differential equations is given and further examples of several types of partial differential equations involving heat, wave, and Poisson equations are discussed. The outcomes have shown the simplicity of using the proposed transform in different aspects. In future work, we aim to solve integral equations and nonlinear partial differential equations based on the triple ARA transform.

Acknowledgement: Not applicable.

Funding Statement: This research received no external funding.

Author Contributions: All authors have read and agreed to the published version of the manuscript.

Availability of Data and Materials: Not applicable.

Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.

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Cite This Article

APA Style
Saadah, R., Amleh, M., Qazza, A., Al-Omari, S., Akdemir, A.O. (2024). Results involving partial differential equations and their solution by certain integral transform. Computer Modeling in Engineering & Sciences, 138(2), 1593-1616. https://doi.org/10.32604/cmes.2023.029180
Vancouver Style
Saadah R, Amleh M, Qazza A, Al-Omari S, Akdemir AO. Results involving partial differential equations and their solution by certain integral transform. Comput Model Eng Sci. 2024;138(2):1593-1616 https://doi.org/10.32604/cmes.2023.029180
IEEE Style
R. Saadah, M. Amleh, A. Qazza, S. Al-Omari, and A.O. Akdemir, “Results Involving Partial Differential Equations and Their Solution by Certain Integral Transform,” Comput. Model. Eng. Sci., vol. 138, no. 2, pp. 1593-1616, 2024. https://doi.org/10.32604/cmes.2023.029180


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