# Approximate Analytical Solution of Two-Dimensional Nonlinear Time-Fractional Damped Wave Equation in the Caputo Fractional Derivative Operator

Authors: Alemayehu Tamirie Deresse; Yesuf Obsie Mussa; Ademe Kebede Gizaw
Journal: Mathematical Problems in Engineering
 (2022)

---

## Abstract

In this work, we proposed a new method called Laplace–Padé–Caputo fractional reduced differential transform method (LPCFRDTM) for solving a two-dimensional nonlinear time-fractional damped wave equation subject to the appropriate initial conditions arising in various physical models. LPCFRDTM is the amalgamation of the Laplace transform method (LTM), Padé approximant, and the well-known reduced differential transform method (RDTM) in the Caputo fractional derivative senses. First, the solution to the problem is gained in the convergent power series form with the help of the Caputo fractional-reduced differential transform method. Then, the Laplace–Padé approximant is applied to enlarge the domain of convergence. The advantage of this method is that it solves equations simply and directly without requiring enormous amounts of computational work, perturbations, or linearization, and it expands the convergence domain, leading to the exact answer. To confirm the effectiveness, accuracy, and convergence of the proposed method, four test-modeling problems from mathematical physics nonlinear wave equations are considered. The findings and results showed that the proposed approach may be utilized to solve comparable wave equations with nonlinear damping and source components and to forecast and enrich the internal mechanism of nonlinearity in nonlinear dynamic events.


---

## Body

## 1. Introduction

Linear and nonlinear fractional differential equations can successfully simulate fractional derivatives in a range of scientific and technical domains, including electrical networks, chemical physics, control theory of dynamical systems, reaction-diffusion, signal processing, and heat transform [1–7]. Because fractional differential equations (FDEs) often exist in several fields of engineering and science, many researchers focus their efforts on obtaining exact/approximate solutions to these dynamic fractional differential equations utilizing a variety of powerful established approaches, including the finite difference method [8], Caputo fractional-reduced differential transform method [9–11], Padé–Sumudu–Adomian decomposition method [12], triple Laplace transform method [13–15], double Sumudu transform iterative method [16], shifted Chebyshev polynomial-based method [17], Laplace decomposition method [18, 19], homotopy analysis method [20], double Laplace transform method [21], homotopy perturbation method [22, 23], conformable reduced differential transform method [24], conformable fractional-modified homotopy perturbation, Adomian decomposition method [25], differential transform method [26–28], and the new function method based on the Jacobi elliptic functions [29].Among the approaches listed above, Keskin and Oturance were the first ones to present the Caputo FRDTM, which has been successfully utilized to solve linear and nonlinear fractional differential equations [9, 30]. Many intellectuals have implemented this method for solving various sorts of equations in recent years. For example, Kenea [31] used CFRDM to find closed solutions and accurate solutions to one-dimensional time-fractional diffusion equations with beginning conditions in the form of infinite fractional power series. CFRDTM offers the benefit of minimizing the number of computations required and offering analytic approximations, in many cases exact answers, in the form of a fast-converging power series with elegantly computed terms [32–35]. Furthermore, CFRDTM has an alternative plan to solve problems to overcome the drawbacks of well-known numerical and analytical methods such as Adomian decomposition, differential transform, homotopy perturbation, and variational iteration, which suffer from discretization, linearization, or perturbations [36–38].The main purpose of this study is to introduce the LPCFRDTM, which is a new approach for solving the two-dimensional time-fractional nonlinear damped wave equation. The CFRDTM, the Laplace transform method, and the Padé approximant are all jointly used in this procedure. The Padé approximation has been used in a variety of domains to approximate rational series solutions; it was invented by Henri Padé [39] circa 1980. Baker [40] established the existence and convergence of subsequences. The Padé approximant method outperforms other series approximation methods and is used to manage series convergence. The authors of the paper [41] used the multivariate Padé approximation method for solving the European vanilla call option pricing problem. According to the relationships of “smaller than” or “greater than” between stock price and option exercise price, the Padé polynomials have appeared in the fractional Black-Scholes equation using the provided method. Using these polynomials, they applied the multivariate Padé approximation method and calculated numerical solutions of the fractional Black–Scholes equation for both situations. The obtained results reveal that the multivariate Padé approximation is a very quick and accurate method for the fractional Black–Scholes equation. The Padé approximants, in other words, heighten the domain of convergence of the truncated power series solution achieved via CFRDTM or other methods, leading to the exact solutions in many cases [12, 42–45].The proposed LPCFRDTM technique has been utilized to solve the problems as follows: The CFRDTM is used to derive the solution to FDEs in convergent power series form. Second, even though the CFRDTM solution series has a high number of terms, it may converge in a narrow area. As a result, the LPCFRDTM magnifies the truncated power series’ convergence domain, typically resulting in the exact solution. We use the Laplace transform to improve the solution of convergent series generated by the CFRDTM and then form its Padé approximant to turn the transformed series into a meromorphic function. Finally, to achieve the approximate analytical solution, we use the inverse Laplace transform of the Padé approximant function. The capacity to widen the domain of convergence of solutions or include discovering exact answers is a major benefit of using this method. Also, the LPCFRDTM can obtain exact solutions without any perturbation parameters like HPM [27, 46, 47].The generalized two-dimensional dynamical time-fractional nonlinear damped wave equation with a source term in the Caputo fractional derivative operator is taken into account in this article [48]: (1)Dt2α0Cux,y,t+β0CDtαux,y,t=γDx2ux,y,t+Dy2ux,y,t+hx,y,t,u,x,y∈Ω,t≥0,where Ω=x,y:a≤x≤b,c≤y≤d.The initial condition associated with equation (1) is given by(2)ux,y,0=φ0x,y,x,y∈Ω,Dtα0Cux,y,0=φ1x,y,x,y∈Ω,,where ux,y,t is the scalar variable, t is the time, and the parameters γandβ are supposed to be real numbers with γ,β≥0.β is the alleged dissipative term. When β=0, Equation (1) reduces to the undamped wave equation, while β>0 to the damped one. The known functions φ0x,y and φ1x,y represent wave kinks or modes and velocity, respectively.The remainder of this work is divided into the following sections. The fundamental definitions of fractional calculus are provided in Section2. CFRDTM is introduced in Section 3 along with definitions and its convergence analysis in subsection 3.1. The main idea behind the Padé approximant is explained in Section 4. Section 5 explains the underlying premise of the Laplace–Padé resummation method. In Section 6, we demonstrate the proposed method’s reliability, convergence, and efficiency using four illustrative instances. Approximate analytical answers and numerical simulations are presented in tables and graphs in Sections 6.1 and 6.2, respectively. In Section 7, we have a quick discussion. Finally, a conclusion is formed in section 8.


## 2. Preliminaries

In this section, we will go over some essential fractional calculus definitions, which we will use in the present investigation (see [49–53]).Definition 1.
The Riemann–Liouville fractional derivative operator offx is given by(3)Dxα0Rfx=1Γm−α∫0xx−tm−α−1ftdt,m−1<α≤m,m∈ℕ,where the gamma function Γz is simply a generalization of the factorial real arguments and defined by(4)Γz=∫0∞e−ttz−1dt,z>0.Definition 2.
The Riemann–Liouville fractional integral of orderα>0 for a function f∈ℂ−1m with m−1<α≤m, m∈ℕ and is defined as(5)Jxαfx=1Γα∫0xx−tα−1ftdt,​α>0,t>0,Jx0fx=fx.
When trying to describe real-world issues with fractional differential equations, the Riemann–Liouville derivative has several drawbacks because it necessitates the definition of fractional order beginning conditions, which have yet to be physically explained. In their work on the theory of viscoelasticity, Caputo and Mainardi [52] suggested a modified fractional differentiation operator Dxα to address this mismatch. With the Caputo fractional derivative, which has unambiguous physical implications, initial and boundary conditions involving integer-order derivatives can be employed.Definition 3.
From Caputo’s perspective, the fractional derivative offx is defined as(6)Dxα0Cfx=m∈ℕ,m∈ℕ.
In particular, if0<α≤1, then the Caputo fractional derivative of f is(7)Dxα0Cfx=1Γ1−α∫0xx−t−αf′tdt.Lemma 1.
Ifm−1<α≤m,m∈ℕ,t>0 and f∈ℂμm,μ≥−1, then(8)Dtα0CDtβ0Cft=0CDtα+βft=Dtβ0CDtα0Cft,Dtα0Ctγ=Γ1+γΓ1+γ−αtγ−α,t>0,Dtα0CJtαft=ft,t>0,JtαDtα0Cft=ft−∑k=0mfk0+tkk!,t>0.


## 3. Caputo Fractional-Reduced Differential Transform Method (CFRDTM)

In this section, the fundamental necessary concepts and operations of the2+1− dimensional CFRDTM [30, 32, 43] are presented. Additionally, the convergence analysis of the CFRDTM is also presented in subsection 3.1.Definition 4.
If a functionux,y,t is analytic and differentiated continuously with respect to the time variable ″t″ and space variables x,y in the domain of interest, then let(9)Ukx,y=1Γkα+1Dtkα0Cux,y,tt=t0,where Ukx,y is the t-dimensional spectrum function or the transformed function,Dtkα0C=∂kα/∂tkα and the parameter α indicates the order of the time fractional derivative. The original function is represented by lowercase ux,y,t in this article, whereas the transformed function is represented by uppercase Ukx,y.Definition 5.
The inverse CFRDT of a sequenceUkx,yk=0∞ at initial time variable t=t0 is provided by(10)ux,y,t=∑k=0∞Ukx,yt−t0kα,Then, combining Equations (8) and (9), we get(11)ux,y,t=∑k=0∞1Γkα+1Dtkα0Cux,y,tt=t0t−t0kα.Remark 1.
The functionux,y,t is represented in real life by an infinite series of Equation (9) about t0=0 and can be expressed as u˜nx,y,t=∑k=0nUkx,ykα+Rnx,y,t, where the tail function Rnx,y,t=∑k=n+1∞Ukx,ytkα is negligibly small.
Moreover, the inverse CFRDT of the set ofUkx,yk=0n yields an approximation solution as follows:(12)u˜nx,y,t=∑k=0nUkx,ytkα,where ′n′ is the approximate solution’s order. As a result, the actual answer to the problem is obtained as follows:(13)ux,y,t=limn⟶∞u˜nx,y,t=∑k=0∞Ukx,ytkα=U0x,y+U1x,ytα+U2x,yt2α+U3x,yt3α+⋯.From Equation (11), the principle of the CFRDTM can be determined to be derived from the power series expansion.Theorem 1.
(see [40]). Assume that Fkx,y,Gkx,y, and Ukx,y are the RDT of the functions fx,y,t,gx,y,t, and ux,y,t respectively, then we have the following equations:(i)
Iffx,y,t=sinux,y,t, then(14)Fkx,y=sinU0,ifk=0,∑k1=0k−11−k1kGk1x,yUk−k1x,y,ifk≥1.(ii)
Ifgx,y,t=cosux,y,t, then(15)Gkx,y=cosU0,ifk=0,−∑k1=0k−11−k1kFk1x,yUk−k1x,y,ifk≥1.Remark 2.
The Mittag–Leffler function, which is a generalization of the exponential function, is defined as [54](16)Eαt=∑k=0∞tkΓkα+1=∑k=0∞tkkα!,α∈ℂ,Reα>0.
Forα=1,Eαt reduces to et and CFRDTEαλtα=λk/Γkα+1.Theorem 2.
forα=1,see) Caputo fractional-reduced differential transform of the initial condition.
If∂nα/∂tnαux,y,0=φx,y, then Unx,y=φx,y/Γnα+1.
To validate the fundamental concepts of the CFRDTM, we consider the nonlinear damping wave equation (1) with the initial condition (2) by applying the features of CFRDTM listed in Table 1 and Theorem 8 on both sides of problem (1).(17)CFRDTD0Ct2αux,y,t+βD0Ctαux,y,t=CFRDTγDx2ux,y,t+Dy2ux,y,t+hx,y,t,u.
The CFRDT of each term in (17) is given as follows.(18)CFRDTDt2α0Cux,y,t=CFRDT∂2α∂t2αux,y,t=Γαk+2α+1Γαk+1Uk+2x,y,CFRDTβDtα0Cux,y,t=βCFRDT∂α∂tαux,y,t=βΓkα+α+1Γkα+1Uk+1x,y,CFRDTγDx2ux,y,t+Dy2ux,y,t=γ∂2∂x2Ukx,y+∂2∂y2Ukx,y,CFRDThx,y,t,u=Hkx,y,k≥0.
Substituting equation (13) into equation (12), we may construct the following iteration formula:(19)Γαk+2α+1Γαk+1Uk+2x,y+βΓkα+α+1Γkα+1Uk+1x,y=γ∂2∂x2Ukx,y+∂2∂y2Ukx,y+Hkx,y.
Solving forUk+2x,y, we obtain(20)Uk+2x,y=Γαk+1Γαk+2α+1−βΓkα+α+1Γkα+1Uk+1x,y+γ∂2∂x2Ukx,y+∂2∂y2Ukx,y+Hkx,y.When we apply the CFRDT to condition (2) according to Theorem 2, we get (21)U0x,y=φ0x,y,U1x,y=φ1x,y.
Using Equation (15) into Equation (14) and solving the resulting system for k=0,1,2,…, we get the following Uk+2x,y values:(22)U2x,y=1Γ2α+1−βΓα+1φ1x,y+γ∂2∂x2φ0x,y+∂2∂y2φ0x,y+H0x,y,U3x,y=Γα+1Γ3α+1−βΓ2α+1Γα+1U2x,y+γ∂2∂x2φ1x,y+∂2∂y2φ1x,y+H1x,y,and so on.
Then, the inverse CFRDT of the set of valuesUkx,yk=0n gives the approximate solution:(23)u˜nx,y,t=∑k=0nUkx,ytkα.
In the equation,″n″ is the order of approximation solution. Therefore, the exact solution to the considered problem can be obtained as follows:(24)ux,y,t=limn⟶∞u˜nx,y,t=φ0x,y+φ1x,ytα+1Γ2α+1−βΓα+1φ1x,y+γ∂2∂x2φ0x,y+∂2∂y2φ0x,y+H0x,yt2α+Γα+1Γ3α+1−βΓ2α+1Γα+1U2x,y+γ∂2∂x2φ1x,y+∂2∂y2φ1x,y+H1x,yt3α+⋯.Table 1
The fundamental mathematical operations of CFRDTM [32, 33, 35, 36, 43].
OriginalfunctionTransformedfunctionux,y,tUkx,y=1/Γkα+1D0Ctkαux,y,tt=t0wx,y,t=αux,y,t±βvx,y,tWkx,y=αUkx,y±βVkx,y,where αand are constantswx,y,t=xmyntpWkx,y=xmynδk−p, where δk−p=1,ifk=p,0,ifk≠p,wx,y,t=xmyntpux,y,tWkx,y=xmynUk−px,y,k≥n,0,else,wx,y,t=ux,y,tvx,y,tWkx,y=∑r=0kUrx,yVk−rx,y=∑r=0kVrx,yUk−rx,ywx,y,t=∂Nα/∂tNαux,y,tWkx,y=Γkα+Nα+1/Γkα+1Uk+Nx,ywx,y,t=∂nr/∂xnrux,y,tWkx,y=∂nr/∂xnrUkx,ywx,y,t=eλtWkx,y=λk/k!, where λ is a constantwx,y,t=sinαx+βy+ωtWkx,y=ωk/k!sinkπ/2!+αx+βy, where α, β, and ω are constantswx,y,t=cosαx+βy+ωtWkx,y=ωk/k!coskπ/2!+αx+βy, where α, β, and ω are constants


### 3.1. Convergence of the Method

Theorem 3.
(see [40, 55]). If φkx,y,t=Ukx,ytk, then the series solution ∑k=0∞φkx,y,t stated in Equation (9) around t0=0, ∀k∈ℕ∪0.(i)
It is convergent if∃λ,0<λ<1, such that φk+1≤λφk+1.(ii)
It is divergent if∃λ>1, such that φk+1≥λφk+1.Definition 6.
(see [40, 55]). For k∈ℕ∪0, we define(25)γk=φk+1φk=Uk+1x,ytk+1Ukx,ytk,ifφk≠0,0,ifφk=0,and then, the series solution ∑k=0∞φkx,y,t converges to the exact solution ux,y,t when 0≤γk<1 for k=0,1,2,….


## 3.1. Convergence of the Method

Theorem 3.
(see [40, 55]). If φkx,y,t=Ukx,ytk, then the series solution ∑k=0∞φkx,y,t stated in Equation (9) around t0=0, ∀k∈ℕ∪0.(i)
It is convergent if∃λ,0<λ<1, such that φk+1≤λφk+1.(ii)
It is divergent if∃λ>1, such that φk+1≥λφk+1.Definition 6.
(see [40, 55]). For k∈ℕ∪0, we define(25)γk=φk+1φk=Uk+1x,ytk+1Ukx,ytk,ifφk≠0,0,ifφk=0,and then, the series solution ∑k=0∞φkx,y,t converges to the exact solution ux,y,t when 0≤γk<1 for k=0,1,2,….


## 4. Pade Approximate

In numerical mathematics, Padé approximation [56] is believed to be the best approximation of a function by rational functions of a given order. Under this technique, the approximate power series agrees with the power series of the function it is approximating. The Padé approximant often gives a better approximation of the function than truncating its Taylor series, and it may still work where the Taylor series does not converge and also enlarges the domain of convergence of the truncated power series solution. For such reasons, Padé approximants are often used in many fields of computations.Letux,y,t be an analytical function with Maclaurin’s expansion.(26)ux,y,t=∑k=0∞ukx,y,ttk,0≤t≤T,Then, the Padé approximant to ux,y,t of order K/L, which is denoted as K/Lux,y,t, is defined as follows [40, 42]:(27)KLux,y,t=p0+p1t+p2t2+⋯+pKtKq0+q1t+q2t2+⋯+qLtL,where we considered q0=1, and the numerator and denominator have no common factors. The numerator and the denominator in equation (19) are built in such a way that ux,y,t, K/Lux,y,t, and their derivatives agree at t=0 up to K+L. That is,(28)∑k=0∞ukx,y,ttk=p0+p1t+p2t2+⋯+pKtK1+q1t+q2t2+⋯+qLtL+OtK+L+1.By cross multiplying, we find that(29)u0+u1t+u2t2+⋯.1+q1t+q2t2+⋯+qLtL=p0+p1t+p2t2+⋯+pKtK+OtK+L+1.Equating the coefficients oftK+1,tK+2,…,tK+L from (29), we find(30)uKq1+uK+1q2+uK+2q3+⋯+uK−L+1qL=−uK+1,uK+1q1+uK+2q2+uK+3q3+⋯+uK−L+2qL=−uK+2,uK+2q1+uK+3q2+uK+4q3+⋯+uK−L+3qL=−uK+3,...uK+L−1q1+uK+L−2q2+uK+L−3q3+⋯+uKqL=−uK+L.Comparing the equal power of″t″ on both sides of (29), we get(31)p0=u0,p1=u1+u0q1,p2=u2+u1q1+u0q2,p3=u3+u2q1+u1q2+u0q3,...pK=uK+uK−1q1+uK−2q2+uK−3q3+⋯+u0qK.Moreover, from Equation (22), first, we calculate all the coefficients qn,1≤n≤L, and then, we determine the coefficients pn,0≤n≤K from Equation (23).Remark 3.
For a fixed value ofK+L+1, error equation (20) is the smallest when the numerator and denominator of Equation (14) have the same degree or when the numerator has one degree higher than the denominator.


## 5. Laplace–Pade Resummation Method

The LPCFRDTM, which is a combination of the CFRDTM and the Laplace–Padé resummation method, is described as follows:(i)
By means of CFRDTM, we first find the series solution of the given equation that is similar to series (16)(ii)
Second, power series (16) is transformed using the Laplace transformation(iii)
Next, we replaces by 1/t in the resulting equation(iv)
The series produced from (3) is then transformed into a meromorphic function by constructing its Padé approximant of order K/L, where K and L are chosen at random but should be smaller than the series’ order(v)
After that, in the resulting equation, we substitutet by 1/s(vi)
As a final point, we obtain the precise or approximate answer by applying the inverse Laplaces transformation


## 6. Illustrative Examples

We offer four examples to establish the validity and efficiency of the suggested strategy. These cases’ solutions are also compared with exact solutions.


### 6.1. Analytical Solution to Illustrative Examples

Example 1.
Consider the following2+1−dimensional hyperbolic time-fractional telegraph equation in the region Ω=0,22 as follows:(32)Dt2α0Cux,y,t+2Dtα0Cux,y,t=Dx2ux,y,t+Dy2ux,y,t−u2+e2x+yEα−2tα2−2ex+yEα−2tα,
Under the initial condition, we get(33)ux,y,0=ex+y,Dtα0Cux,y,0=−2Γα+1ex+y.
Applying the properties of CFRDT listed in Table1 on both sides of Equation (24), we obtain(34)Uk+2x,y=Γαk+1Γαk+2α+1−2Γkα+α+1Γkα+1Uk+1x,y+∂2∂x2Ukx,y+∂2∂y2Ukx,y−∑r=0kUrx,yUk−rx,y+e2x+y∑r=0kVrx,yVk−rx,y−2ex+y−2kΓkα+1,where Vkx,y is the Caputo fractional-reduced transform of Eα−2tα, and it is obtained by using Remark 2.
When the CFRDT is applied to initial condition (25) in view of Theorem 2, it produces(35)U0x,y=ex+y,U1x,y=−2Γα+1ex+y.
We get the followingUkx,y values for k=0,1,2,… using Equation (27) into Equation (26), recursively.
Fork=0, using the values of U0x,y and U1x,y from (35) into Equation (26), we get(36)U2x,y=Γ1Γ2α+1−2Γα+1Γ1U1x,y+∂2∂x2U0x,y+∂2∂y2U0x,y−U0x,yU0x,y+e2x+yV0x,yV0x,y−2ex+y−20Γ1=1Γ2α+1−2Γα+1−2Γα+1ex+y+∂2∂x2ex+y+∂2∂y2ex+y−ex+yex+y+e2x+y−20Γ1−20Γ1−2ex+y−20Γ1=1Γ2α+14ex+y+2ex+y−e2x+y+e2x+y−2ex+y=2Γ2α+1ex+y,for k=1, using the values of U1x,y and U2x,y in Equation (26), we obtain(37)U3x,y=Γα+1Γ3α+1−2Γ2α+1Γα+1U2x,y+∂2∂x2U1x,y+∂2∂y2U1x,y−∑r=01Urx,yU1−rx,y+e2x+y∑r=01Vrx,yV1−rx,y−2ex+y−21Γα+1=Γα+1Γα+2α+1−2Γ2α+1Γα+1−22Γ2α+1ex+y+−2Γα+1∂2∂x2ex+y+−2Γα+1∂2∂y2ex+y−2U0x,yU1x,y+2e2x+yV0x,yV1x,y−2ex+y−21Γα+1=Γα+1Γα+2α+1−23Γα+1ex+y−4ex+yΓα+1+4ex+yΓα+1−4e2x+yΓα+1+4ex+yΓα+1=−23ex+yΓα+2α+1.
Continuing in the same manner fork≥2, we obtain(38)U4x,y=−24Γ4α+1ex+y,U5x,y=−25Γ5α+1ex+y,…,Ukx,y=−2kΓkα+1ex+y.
Then, using the inverse CFRDT (9) as a definition, we get.(39)ux,y,t=∑k=0∞Ukx,ytkα=ex+y1+−2tαΓα+1+−22t2αΓ2α+1+−23t3αΓ3α+1+−24t4αΓ4α+1+−25t5αΓ5α+1+⋯=ex+y∑k=0∞−2ktkαΓkα+1.
In this situation, the truncated series is approximated using the Laplace–Padé approximant up tonth order of perturbation. That is,(40)u˜nx,y,t=ex+y∑k=0n−2ktkαΓkα+1.
After that, we apply the Laplace transform to both sides of (40), and we get(41)Lu˜nx,y,t=ex+y∑k=0n−2ksαk+1.
For straightforwardness, we write1/t in the place of s in Equation (31), and we get(42)Lu˜nx,y,t=ex+y∑k=0n−2ktkα+1,=ex+yt−2tα+1+22t2α+1−23t3α+1+24t4α+1−25t5α+1+…..
Now, we convert transformed series (32) into a meromorphic function by forming its Padé approximant of K/L with k≥1,L≥1 and K+L=7k=3,L=4.(43)KLt−2tα+1+22t2α+1−23t3α+1+24t4α+1−25t5α+1+26t6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1.
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
Using (23) and (22), we get(44)p0=u0=0,p1=1,p2=−2+q1,p3=4−2q1+q2,and4q1−2q2+q3+0=8,−8q1+4q2−2q3+q4=−16,16q1−8q2+4q3−2q4=32,−32q1+16q2−8q3+4q4=−64..
Solving forp and q, we obtain p0=0,p1=1,p2=0,p3=1,q1=2,q2=1,q3=2,andq4=0.
Then, we obtain(45)KLt−2tα+1+22t2α+1−23t3α+1+24t4α+1−25t5α+1+26t6α+1=t+t2α+11+2t+tα+1+2t2α+1=t1+2tα.
Therefore, allK/LPadé approximants of the eqaution with L≥1,K≥1 and K+L≤n give(46)KLu˜nx,y,t=ex+yt1+2tα.
Now, by changing1/s into t in the equation, we obtain K/Lu˜n in terms of s as follows:(47)KLu˜nx,y,t=ex+ysα−1sα+2.
Finally, on both sides of Equation (34), using the inverse Laplace transform, we have the exact solution as follows:(48)ux,y,t=ex+yL−1sα−1sα+2=ex+yEα−2tα.
Puttingα=1 in Equation (35), we obtain(49)ux,y,t=ex+yL−11s+2=ex+y−2t.Example 2.
Consider the following two-dimensional hyperbolic nonlinear time-fractional sine-Gordon equation on the domainΩ=0,22,t≥0(50)Dt2α0Cux,y,t+Dtα0Cux,y,t=Dx2ux,y,t+Dy2ux,y,t−2sinu+2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπy.
Under the initial condition, we get(51)ux,y,0=1−cosπx1−cosπy,Dtα0Cux,y,0=−1Γα+11−cosπx1−cosπy.
Applying the properties of CFRDT listed in Table1, Theorem 1, and Definition 4 on both sides of Equation (37), we obtain(52)Uk+2x,y=Γαk+1Γαk+2α+1−Γkα+α+1Γkα+1Uk+1x,y+∂2∂x2Ukx,y+∂2∂y2Ukx,y−2Fkx,y+Hkx,y,where Fkx,y and Hkx,y are transformed forms of nonlinear term sinux,y,t and 2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπy, respectively. Applying the CFRDT to initial condition (38) in view of Theorem 2, we get(53)U0x,y=1−cosπx1−cosπy,U1x,y=−1Γα+11−cosπx1−cosπy.
Substituting Equation (40) into Equation (39) and applying Theorem 1, Definition 6, and properties of CFRDT listed in Table 1, we obtain the following successive iterated values for kk=0,1,2,….
Fork=0, using the values of U0x,y and U1x,y from (53) into (52), we get(54)U2x,y=Γ1Γ2α+1−Γα+1Γ1U1x,y+∂2∂x2U0x,y+∂2∂y2U0x,y−2F0x,y+H0x,y=1Γ2α+1−Γα+1Γ1−1Γα+11−cosπx1−cosπy+∂2∂x21−cosπx1−cosπy+∂2∂y21−cosπx1−cosπy−2F0x,y+H0x,y.
According to Theorem (1), F0x,y=sinU0=sin1−cosπx1−cosπy.
Using Definition (5), we get(55)H0x,y=1Γ1Dt00C2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπyt=0=2sin1−cosπx1−cosπy−π2cosπx+cosπy−2cosπxcosπy.
Consequently, we obtain(56)U2x,y=Γ1Γ2α+11−cosπx1−cosπy+π2cosπx+cosπy−2cosπxcoπy−2sin1−cosπx1−cosπy+2sin1−cosπx1−cosπy−π2cosπx+cosπy−2cosπxcosπy=1−cosπx1−cosπyΓ2α+1,for k=1, using the values of U1x,y and U2x,y in (52), we get(57)U3x,y=Γα+1Γ3α+1−Γ2α+1Γα+1U2x,y+∂2∂x2U1x,y+∂2∂y2U1x,y−2F1x,y+H1x,y=Γα+1Γ3α+1−Γ2α+1Γα+11Γ2α+11−cosπx1−cosπy−1Γα+1π2cosπx+cosπy−2cosπxcosπy−2F1x,y+H1x,y.
According to Theorem (1), we get(58)F1x,y=G0x,yU1x,y=cosU0x,yU1x,y=−1Γα+1cos1−cosπx1−cosπy1−cosπx1−cosπy.
Using Definition (5), we get(59)H1x,y=1Γα+1D0Ctα2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπyt=0=1Γα+1π2cosπx+cosπy−2cosπxcosπy−2cos1−cosπx1−cosπy1−cosπx1−cosπy.
Consequently, we obtain(60)U3x,y=Γα+1Γ3α+1−1−cosπx1−cosπyΓα+1−π2Γα+1cosπx+cosπy−2cosπxcosπy+2Γα+1cos1−cosπx1−cosπy1−cosπx1−cosπy+π2Γα+1cosπx+cosπy−2cosπxcosπy−2Γα+1cos1−cosπx1−cosπy1−cosπx1−cosπy=−1Γ3α+11−cosπx1−cosπy.
Continuing in the same manner fork≥2, we obtain(61)U4x,y=1Γ4α+11−cosπx1−cosπx,…,Ukx,y=−1kΓkα+11−cosπx1−cosπy.
Then, by using inverse CRDT (9), we get(62)ux,y,t=∑k=0∞Ukx,yt−t0kα=1−cosπx1−cosπy1−tαΓα+1+t2αΓ2α+1−t3αΓ3α+1+t4αΓ4α+1−t5αΓ5α+1+⋯=1−cosπx1−cosπy∑k=0∞−1ktkαΓkα+1.
The Laplace–Padé approximant is used to approximate the truncated series up tonth order of perturbation in this case. That is,(63)u˜nx,y,t=1−cosπx1−cosπy∑k=0n−1ktkαΓkα+1.
Then, on both sides of (63), we use the Laplace transform to get(64)Lu˜nx,y,t=1−cosπx1−cosπy∑k=0n−1ksαk+1.
For effortlessness, we write1/t instead of s in Eq., and we get(65)Lu˜nx,y,t=1−cosπx1−cosπy∑k=0n−1ktkα+1=1−cosπx1−cosπyt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1+⋯.
Now, we convert transformed series (45) into a meromorphic function by forming its Padé approximant of K/L with and K+L=7k=3,L=4.(66)KLt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1.
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
By means of (23) and (22), we get(67)p0=u0=0,p1=1,p2=−1+q1,p3=1−q1+q2,andq1−q2+q3=1,−q1+q2−q3+q4=−1,q1−q2+q3−q4=1,−q1+q2−q3+q4=−1..
Solving forp and q, we get p0=0,p1=1,p2=0,p3=1,q1=1,q2=1,q3=1,andq4=0.
Then,K/Lt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=t+t2α+1/1+t+tα+1+t2α+1=t/1+tα.
Therefore, allK/L Padé approximants of the equation with L≥1,K≥1 and K+L≤n give(68)KLu˜nx,y,t=1−cosπx1−cosπyt1+tα.
Now, by changing1/s into t in the equation, we obtain K/Lu˜n in terms of s as follows:(69)KLu˜nx,y,t=1−cosπx1−cosπysα−1sα+1.
Finally, on both sides of Equation (47), using the inverse Laplace transform, we have the exact solution as follows:(70)ux,y,t=1−cosπx1−cosπyL−1sα−1sα+1=1−cosπx1−cosπyEα−tα.
Puttingα=1 in Equation (48), we have(71)ux,y,t=1−cosπx1−cosπyL−11s+1=e−t1−cosπx1−cosπy.
This finding is in perfect accord with the one obtained in the previous study as shown in [35, 48].Example 3.
Consider the following2+1−dimensional hyperbolic nonlinear time-fractional telegraph equation with variable coefficients in the area Ω=0,π2 as follows:(72)Dt2α0Cux,y,t−uDtα0Cux,y,t−uDx2ux,y,t+Dy2ux,y,t−1+x2∂2u∂x2−1+y2∂2u∂y2=Eα−tαcoshxcoshy−sinhx+y−1−x2−y2×Eα−tαcoshxcoshy,with the initial condition(73)ux,y,0=coshxcoshy,Dtα0Cux,y,0=−1Γα+1coshxcoshy.
Applying the properties of CFRDT and the related rules in Table1 on both sides of Equation (50), we obtain(74)Uk+2x,y=Γαk+1Γαk+2α+1∑r=0kΓkα+α+1Γkα+1Uk+1x,yUk−rx,y+∑r=0k∂∂xUrx,yUk−rx,y+∑r=0k∂∂yUrx,yUk−rx,y+1+x2∂2∂x2Ukx,y+1+y2∂2∂y2Ukx,y+∑r=0kVrx,yVk−rx,ycosh2xcosh2y−sinhx+ycoshxcoshy−−1kΓkα+11+x2+y2coshxcoshy,where Vkx,y is the fractional reduced transform of Eα−tα.
Applying the CFRDT to initial condition (51) according to Theorem 10, we get(75)U0x,y=coshxcoshy,U1x,y=−1Γα+1coshxcoshy.
Now, taking the values ofkk=0,1,2,⋯ and using Equation (53) into equation (52), we obtain the following successive iterative values.
Fork=0, using the values of U0x,y and U1x,y from (75) into (74), we get(76)U2x,y=Γ1Γ2α+1Γα+1Γ1U1x,yU0x,y+∂∂xU0x,yU0x,y+∂∂yU0x,yU0x,y+1+x2∂2∂x2U0x,y+1+y2∂2∂y2U0x,y+V0x,yV0x,ycosh2xcosh2y−sinhx+ycoshxcoshy−−10Γ11+x2+y2coshxcoshy=1Γ2α+1−Γα+11Γα+1coshxcoshycoshxcoshy+sinhxcoshxcosh2y+sinhycoshycosh2x+1+x2coshxcoshy+1+y2coshxcoshy+cosh2xcosh2y−sinhx+ycoshxcoshy−1+x2+y2coshxcoshy=1Γ2α+1−cosh2xcosh2y+sinhxcoshxcosh2y+sinhycoshycosh2x+cosh2xcosh2y−sinhxcoshxcosh2y−sinhycoshycosh2x+1+x2+y2coshxcoshy+coshxcoshy−1+x2+y2coshxcoshy=1Γ2α+1coshxcoshy.
Fork=1, using the values of U1x,y and U2x,y in (74), we get(77)U3x,y=Γα+1Γ3α+1∑r=01Γ2α+1Γα+1U2x,yU1−rx,y+∑r=01∂∂xUrx,yU1−rx,y+∑r=01∂∂yUrx,yU1−rx,y+1+x2∂2∂x2U1x,y+1+y2∂2∂y2U1x,y+∑r=01Vrx,yV1−rx,ycosh2xcosh2y−sinhx+ycoshxcoshy−−11Γα+11+x2+y2coshxcoshy=Γα+1Γ3α+12Γα+1cosh2xcosh2y−2Γα+1sinhxcoshxcosh2y−2Γα+1coshxcoshysinh2x−1Γα+1coshxcoshy+1Γα+11+x2+y2coshxcoshy−2Γα+1cosh2xcosh2y+2Γα+1sinhxcoshxcosh2y+2Γα+1coshxcoshysinh2x−1Γα+11+x2+y2coshxcoshy=−1Γ3α+1coshxcoshy.
Continuing in the same manner fork≥2, we obtain(78)U4x,y=1Γ4α+1coshxcoshy,…,Ukx,y=−1kΓkα+1coshxcoshy.Then, by (9), we get(79)ux,y,t=∑k=0∞Ukx,yt−t0kα=coshxcoshy1−tαΓα+1+t2αΓ2α+1−t3αΓ3α+1+t4αΓ4α+1−t5αΓ5α+1+⋯=coshxcoshy∑k=0∞−1ktkαΓkα+1.Here, the Laplace–Padé approximant is applied to the truncated series up to nth order of perturbation, that is,(80)u˜nx,y,t=coshxcoshy∑k=0n−1ktkαΓkα+1.
Applying the Laplace transform on both sides of (80) yields(81)Lu˜nx,y,t=coshxcoshy∑k=0n−1ksαk+1.
For simplicity, we write1/t instead of s i equation (57), we get(82)Lu˜nx,y,t=coshxcoshy∑k=0n−1ktkα+1=coshxcoshyt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1+⋯.
Now, we convert transformed series (58) into a meromorphic function by forming its Padé approximant of K/L with k≥1,L≥1, and K+L=7k=3,L=4.(83)KLt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1,
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
Using (23) and (22), we get(84)p0=u0=0,p1=1,p2=−1+q1,p3=1−q1+q2,andq1−q2+q3=1,−q1+q2−q3+q4=−1,q1−q2+q3−q4=1,−q1+q2−q3+q4=−1..
Solving forp and q, we get p0=0,p1=1,p2=0,p3=1,q1=1,q2=1,q3=1,andq4=0.
Then,K/Lt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=t+t2α+1/1+t+tα+1+t2α+1=t/1+tα.Therefore, all K/L Padé approximants of equation (58) with L≥1,K≥1 and K+L≤n give(85)KLu˜nx,y,t=coshxcoshyt1+tα.
Now, by changing1/s into t in equation (59, we obtain K/Lu˜n in terms of s as follows:(86)KLu˜nx,y,t=coshxcoshysα−1sα+1.
Finally, applying the inverse Laplace transform on both sides of (86), we have the exact solution as follows:(87)ux,y,t=coshxcoshyL−1sα−1sα+1=coshxcoshyEα−tα.
Puttingα=1 in the (87), we have(88)ux,y,t=coshxcoshyL−11s+1=e−tcoshxcoshy.
The obtained result is the same as obtained by Hafez [57].Example 4.
Consider the following2+1−dimensional hyperbolic time-fractional differential equation in the area Ω=0,12 as follows:(89)Dt2α0Cux,y,t+Dtα0Cux,y,t=Dx2ux,y,t+Dy2ux,y,t+2π2u+2πsinπx+ytEα−x+ytα+x+y2−x+ysinπxsinπyEα−x+ytα−2sinπxsinπyt2Eα−x+ytα.
With the initial condition, we obtain(90)ux,y,0=sinπxsinπy,D0Ctαux,y,0=−x+yΓα+1sinπxsinπy.
Applying the CFRDT to initial condition (63) according to Theorem 2, we obtain(91)Uk+2x,y=Γαk+1Γαk+2α+1−Γkα+α+1Γkα+1Uk+1x,y+∂2∂x2Ukx,y+∂2∂y2Ukx,y+2π2Ukx,y+2πsinπx+yVkx,y+x+y2−x+ysinπxsinπy−x+ykΓαk+1−2sinπxsinπyWkx,y,where Vkx,y and Wkx,y are the transformed forms of tEα−x+ytα and t2Eα−x+ytα.
Applying the CRDT to initial condition (64), we get(92)U0x,y=sinπxsinπy,U1x,y=−x+yΓα+1sinπxsinπy.
Substituting equation (66) into equation (65) and applying Definition 9, we obtain the following successive Ukx,y values for k=0,1,2,….(93)U2x,y=Γ1Γ2α+1−Γα+1Γ1U1x,y+∂2∂x2U0x,y+∂2∂y2U0x,y+2π2U0x,y+2πsinπx+yV0x,y+x+y2−x+ysinπxsinπy−x+y0Γ1−2sinπxsinπyW0x,y=Γ1Γ2α+1x+ysinπxsinπy−2π2sinπxsinπy+2π2sinπxsinπy+x+y2sinπxsinπyx+ysinπxsinπy=x+y2Γ2α+1sinπxsinπy.U3x,y=Γα+1Γ3α+1−Γ2α+1Γα+1U2x,y+∂2∂x2U1x,y+∂2∂y2U1x,y+2π2U1x,y+2πsinπx+yV1x,y+x+y2−x+ysinπxsinπy−x+y1Γα+1−2sinπxsinπyW1x,y=Γα+1Γ3α+1−Γ2α+1Γα+1x+y2Γ2α+1sinπxsinπy−∂2∂x2x+yΓα+1sinπxsinπy−∂2∂y2x+yΓα+1sinπxsinπy−2π2x+yΓα+1sinπxsinπy+2πsinπx+yV1x,y+x+y2−x+ysinπxsinπy−x+y1Γα+1−2sinπxsinπyW1x,y=Γα+1Γ3α+1−x+y2Γα+1sinπxsinπy−2πcosπxsinπyΓα+1−2πsinπxcosπyΓα+1+2π2x+yΓα+1sinπxsinπy−2π2x+yΓα+1sinπxsinπy+2πcosπxsinπyΓα+1+2πsinπxcosπyΓα+1−x+y3Γα+1sinπxsinπy+x+y2Γα+1sinπxsinπy=−x+y3Γ3α+1sinπxsinπy.
Continuing in the same manner fork≥2, we obtain(94)U4x,y=x+y4Γ4α+1sinπxsinπy,…,Ukx,y,=−1kx+ykΓkα+1sinπxsinπy.
Then, by using inverse CFRDT (9), we get(95)ux,y,t=∑k=0∞Ukx,yt−t0kα=sinπxsinπy1−x+yΓα+1tα+x+y2t2αΓ2α+1−x+y3t3αΓ3α+1+x+y4t4αΓ4α+1−x+y5t5αΓ5α+1+...=sinπxsinπy∑k=0∞−1kx+yktkαΓkα+1.
Here, the Laplace–Padé approximant is applied to the truncated series up tonth order of perturbation, that is,(96)u˜nx,y,t=sinπxsinπy∑k=0n−1kx+yktkαΓkα+1.
Applying the Laplace transform on both sides of (96) yields(97)Lu˜nx,y,t=sinπxsinπy∑k=0n−1kx+yksαk+1.
For simplicity, we write1/t instead of s in equation (70), we get(98)Lu˜nx,y,t=sinπxsinπy∑k=0n−1kx+yktkα+1=sinπxsinπyt−x+ytα+1+x+y2t2α+1−x+y3t3α+1+....
Now, we convert transformed series (71) into a meromorphic function by forming its Padé approximant of K/L with k≥1,L≥1 and K+L=7k=3,L=4.(99)KLt−x+ytα+1+x+y2t2α+1−x+yt3α+1+x+y4t4α+1−x+y5t5α+1+x+yt6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1,
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
By using (30) and (31), we get(100)p0=u0=0,p1=1,p2=−x+y+q1,p3=x+y2−x+yq1+q2,x+y2q1−x+yq2+q3=x+y3,−x+y3q1+x+y2q2−x+yq3+q4=−x+y4,x+y4q1−x+y3q2+x+y2q3−x+yq4=x+y5,−x+y5q1+x+y4q2−x+y3q3+x+y2q4=−x+y6..
Solving forp and q, we get p0=0,p1=1,p2=0,p3=1,q1=x+y,q2=1,q3=x+yandq4=0.
Then, we obtain(101)KLt−x+ytα+1+x+y2t2α+1−x+y3t3α+1+x+y4t4α+1−x+y5t5α+1+x+y6t6α+1=t+t2α+11+x+yt+tα+1+x+yt2α+1=t1+x+ytα.
Therefore, allK/L Padé approximants of equation (84) with L≥1,K≥1 and K+L≤n give(102)KLu˜nx,y,t=sinπxsinπyt1+x+ytα.
Now, by changing1/s into t in equation (72), we obtain K/Lu˜n in terms of s as follows:(103)KLu˜nx,y,t=sinπxsinπysα−1sα+x+y.
To conclude, applying the inverse Laplace transform on both sides of (103), we have the exact solution as follows:(104)ux,y,t=sinπxinπyL−1sα−1sα+x+y=sinπxsinπyEα−x+ytα.
Puttingα=1 in Equation (74), we have(105)ux,y,t=sinπxsinπyL−11s+x+y=e−x+ytsinπxsinπy.
The obtained result is the same as obtained by Li et al. [48].


### 6.2. Numerical Simulation to Illustrative Examples

In this subsection, the numerical simulation of the considered problems was depicted using tables and figures. Mathematica software or program has been used to create all 3D figures in this manuscript.Numerical results corresponding to Example 1 are depicted in Table2 and Figures 1 and 2.Table 2
Six-term approximate solution by CFRDTM of Example 1 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.13.9290683804.4956555075.3234236446.0496474836.0496474641.83E-080.24.5205771713.6602304884.2433341814.9530347114.9530324242.29E-060.36.7870350813.2500986893.5030354784.0552381234.0551999673.82E-050.410.989007473.1829331902.9669044083.3203961663.3201169232.79E-040.517.417310273.5299297732.5801835542.7195831482.7182818281.30E-030.626.346588664.4332471132.3271642272.2300998882.2255409284.56E-030.738.031634096.0838518472.2201211501.8352372691.8221188001.31 E-020.852.709286768.7127228342.2965986141.5245126821.4918246983.27 E-020.970.6009305412.586394612.6198486801.2943793801.2214027587.30 E-02191.914709018.004269173.2806929481.14940872711.49 E-01Figure 1
3D view of the solution behaviour of example 1 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 2
3D view of the solution behaviour of example 1 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)Numerical results corresponding to Example 2 are depicted in Table3 and Figures 3 and 4.Table 3
Six-term approximate solutions by CFRDTM of Example 2 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.12.7124571373.0715029693.3845871833.6193496723.6193496727.84E-110.22.4518912332.7139248893.0025024043.2749230223.2749230129.91E-090.32.3001368722.4695243272.7035464672.9632730502.9632728831.67E-070.42.2041669412.2842392592.4574552042.6812814222.6812801841.24E-060.52.1469975922.1370598742.2494907752.4261284722.4261226395.83E-060.62.1215684552.0177020722.0708506242.1952672002.1952465442.07E-050.72.1246840771.9206787611.9157424491.9864012721.9863412156.01E-050.82.1549543241.8430761341.7801662331.7974670221.7973158561.51E-040.92.2119327361.7835439341.6613149101.6266194501.6262786393.41E-0412.2957044411.7417757121.5572479541.4722222221.4715177657.04E-03Figure 3
3D view of the solution behaviour of Example 2 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 4
3D view of the solution behaviour of Example 1 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)Numerical results corresponding to Example 3 are described in Table4 and Figures 5 and 6.Table 4
Six-term approximate solutions by CFRDTM of Example 3 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.11.6146564611.8283872762.0147583122.1545064272.1545064274.67E-110.21.4595482331.6155301761.7873130011.9494780381.9494780325.90E-090.31.3692127371.4700447641.6093521671.7639607691.7639606699.96E-080.41.3120842891.3597492951.4628603231.5960983541.5960976177.37E-070.51.2780528351.2721371661.3390644101.4442123201.4442088473.47E-060.61.2629155191.2010865141.2327244901.3067865001.3067742041.23E-050.71.2647701691.1433310151.1403925551.1824539471.1824181973.58E-050.81.2827892741.0971361531.0596874961.0699862141.0698962289.00E-050.91.3167070681.0616981540.9889383380.9682850170.9680821412.03E-0411.3665742241.0368345990.9269899370.8763762900.8759569454.19E-04Figure 5
3D view of the solution behaviour of Example 3 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 6
3D view of the solution behaviour of example 3 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)Numerical results corresponding to Example4 are depicted in Table 5 and Figures 7 and 8.Table 5
Six-term approximate solutions by CFRDTM of Example 4 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.17.97E-339.12E-331.08E-321.23E-321.23E-323.72E-410.29.18E-337.43E-338.61E-331.01E-321.01E-324.64E-390.31.38E-326.60E-337.11E-338.23E-338.23E-337.74E-380.42.23E-326.46E-336.02E-336.74E-336.74E-335.67E-370.53.54E-327.16E-335.24E-335.52E-335.52E-332.64E-360.65.35E-329.00E-334.72E-334.53E-334.54E-339.25E-360.77.72E-321.23E-324.51E-333.72E-333.70E-332.66E-350.81.07E-311.77E-324.66E-333.09E-333.03E-336.63E-350.91.43E-312.55E-325.34E-332.63E-332.48E-331.48E-3411.87E-313.65E-326.66E-332.33E-332.03E-333.03E-34Figure 7
3D view of the solution behaviour of Example 4 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 8
3D view of the solution behaviour of example 4 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)


## 6.1. Analytical Solution to Illustrative Examples

Example 1.
Consider the following2+1−dimensional hyperbolic time-fractional telegraph equation in the region Ω=0,22 as follows:(32)Dt2α0Cux,y,t+2Dtα0Cux,y,t=Dx2ux,y,t+Dy2ux,y,t−u2+e2x+yEα−2tα2−2ex+yEα−2tα,
Under the initial condition, we get(33)ux,y,0=ex+y,Dtα0Cux,y,0=−2Γα+1ex+y.
Applying the properties of CFRDT listed in Table1 on both sides of Equation (24), we obtain(34)Uk+2x,y=Γαk+1Γαk+2α+1−2Γkα+α+1Γkα+1Uk+1x,y+∂2∂x2Ukx,y+∂2∂y2Ukx,y−∑r=0kUrx,yUk−rx,y+e2x+y∑r=0kVrx,yVk−rx,y−2ex+y−2kΓkα+1,where Vkx,y is the Caputo fractional-reduced transform of Eα−2tα, and it is obtained by using Remark 2.
When the CFRDT is applied to initial condition (25) in view of Theorem 2, it produces(35)U0x,y=ex+y,U1x,y=−2Γα+1ex+y.
We get the followingUkx,y values for k=0,1,2,… using Equation (27) into Equation (26), recursively.
Fork=0, using the values of U0x,y and U1x,y from (35) into Equation (26), we get(36)U2x,y=Γ1Γ2α+1−2Γα+1Γ1U1x,y+∂2∂x2U0x,y+∂2∂y2U0x,y−U0x,yU0x,y+e2x+yV0x,yV0x,y−2ex+y−20Γ1=1Γ2α+1−2Γα+1−2Γα+1ex+y+∂2∂x2ex+y+∂2∂y2ex+y−ex+yex+y+e2x+y−20Γ1−20Γ1−2ex+y−20Γ1=1Γ2α+14ex+y+2ex+y−e2x+y+e2x+y−2ex+y=2Γ2α+1ex+y,for k=1, using the values of U1x,y and U2x,y in Equation (26), we obtain(37)U3x,y=Γα+1Γ3α+1−2Γ2α+1Γα+1U2x,y+∂2∂x2U1x,y+∂2∂y2U1x,y−∑r=01Urx,yU1−rx,y+e2x+y∑r=01Vrx,yV1−rx,y−2ex+y−21Γα+1=Γα+1Γα+2α+1−2Γ2α+1Γα+1−22Γ2α+1ex+y+−2Γα+1∂2∂x2ex+y+−2Γα+1∂2∂y2ex+y−2U0x,yU1x,y+2e2x+yV0x,yV1x,y−2ex+y−21Γα+1=Γα+1Γα+2α+1−23Γα+1ex+y−4ex+yΓα+1+4ex+yΓα+1−4e2x+yΓα+1+4ex+yΓα+1=−23ex+yΓα+2α+1.
Continuing in the same manner fork≥2, we obtain(38)U4x,y=−24Γ4α+1ex+y,U5x,y=−25Γ5α+1ex+y,…,Ukx,y=−2kΓkα+1ex+y.
Then, using the inverse CFRDT (9) as a definition, we get.(39)ux,y,t=∑k=0∞Ukx,ytkα=ex+y1+−2tαΓα+1+−22t2αΓ2α+1+−23t3αΓ3α+1+−24t4αΓ4α+1+−25t5αΓ5α+1+⋯=ex+y∑k=0∞−2ktkαΓkα+1.
In this situation, the truncated series is approximated using the Laplace–Padé approximant up tonth order of perturbation. That is,(40)u˜nx,y,t=ex+y∑k=0n−2ktkαΓkα+1.
After that, we apply the Laplace transform to both sides of (40), and we get(41)Lu˜nx,y,t=ex+y∑k=0n−2ksαk+1.
For straightforwardness, we write1/t in the place of s in Equation (31), and we get(42)Lu˜nx,y,t=ex+y∑k=0n−2ktkα+1,=ex+yt−2tα+1+22t2α+1−23t3α+1+24t4α+1−25t5α+1+…..
Now, we convert transformed series (32) into a meromorphic function by forming its Padé approximant of K/L with k≥1,L≥1 and K+L=7k=3,L=4.(43)KLt−2tα+1+22t2α+1−23t3α+1+24t4α+1−25t5α+1+26t6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1.
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
Using (23) and (22), we get(44)p0=u0=0,p1=1,p2=−2+q1,p3=4−2q1+q2,and4q1−2q2+q3+0=8,−8q1+4q2−2q3+q4=−16,16q1−8q2+4q3−2q4=32,−32q1+16q2−8q3+4q4=−64..
Solving forp and q, we obtain p0=0,p1=1,p2=0,p3=1,q1=2,q2=1,q3=2,andq4=0.
Then, we obtain(45)KLt−2tα+1+22t2α+1−23t3α+1+24t4α+1−25t5α+1+26t6α+1=t+t2α+11+2t+tα+1+2t2α+1=t1+2tα.
Therefore, allK/LPadé approximants of the eqaution with L≥1,K≥1 and K+L≤n give(46)KLu˜nx,y,t=ex+yt1+2tα.
Now, by changing1/s into t in the equation, we obtain K/Lu˜n in terms of s as follows:(47)KLu˜nx,y,t=ex+ysα−1sα+2.
Finally, on both sides of Equation (34), using the inverse Laplace transform, we have the exact solution as follows:(48)ux,y,t=ex+yL−1sα−1sα+2=ex+yEα−2tα.
Puttingα=1 in Equation (35), we obtain(49)ux,y,t=ex+yL−11s+2=ex+y−2t.Example 2.
Consider the following two-dimensional hyperbolic nonlinear time-fractional sine-Gordon equation on the domainΩ=0,22,t≥0(50)Dt2α0Cux,y,t+Dtα0Cux,y,t=Dx2ux,y,t+Dy2ux,y,t−2sinu+2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπy.
Under the initial condition, we get(51)ux,y,0=1−cosπx1−cosπy,Dtα0Cux,y,0=−1Γα+11−cosπx1−cosπy.
Applying the properties of CFRDT listed in Table1, Theorem 1, and Definition 4 on both sides of Equation (37), we obtain(52)Uk+2x,y=Γαk+1Γαk+2α+1−Γkα+α+1Γkα+1Uk+1x,y+∂2∂x2Ukx,y+∂2∂y2Ukx,y−2Fkx,y+Hkx,y,where Fkx,y and Hkx,y are transformed forms of nonlinear term sinux,y,t and 2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπy, respectively. Applying the CFRDT to initial condition (38) in view of Theorem 2, we get(53)U0x,y=1−cosπx1−cosπy,U1x,y=−1Γα+11−cosπx1−cosπy.
Substituting Equation (40) into Equation (39) and applying Theorem 1, Definition 6, and properties of CFRDT listed in Table 1, we obtain the following successive iterated values for kk=0,1,2,….
Fork=0, using the values of U0x,y and U1x,y from (53) into (52), we get(54)U2x,y=Γ1Γ2α+1−Γα+1Γ1U1x,y+∂2∂x2U0x,y+∂2∂y2U0x,y−2F0x,y+H0x,y=1Γ2α+1−Γα+1Γ1−1Γα+11−cosπx1−cosπy+∂2∂x21−cosπx1−cosπy+∂2∂y21−cosπx1−cosπy−2F0x,y+H0x,y.
According to Theorem (1), F0x,y=sinU0=sin1−cosπx1−cosπy.
Using Definition (5), we get(55)H0x,y=1Γ1Dt00C2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπyt=0=2sin1−cosπx1−cosπy−π2cosπx+cosπy−2cosπxcosπy.
Consequently, we obtain(56)U2x,y=Γ1Γ2α+11−cosπx1−cosπy+π2cosπx+cosπy−2cosπxcoπy−2sin1−cosπx1−cosπy+2sin1−cosπx1−cosπy−π2cosπx+cosπy−2cosπxcosπy=1−cosπx1−cosπyΓ2α+1,for k=1, using the values of U1x,y and U2x,y in (52), we get(57)U3x,y=Γα+1Γ3α+1−Γ2α+1Γα+1U2x,y+∂2∂x2U1x,y+∂2∂y2U1x,y−2F1x,y+H1x,y=Γα+1Γ3α+1−Γ2α+1Γα+11Γ2α+11−cosπx1−cosπy−1Γα+1π2cosπx+cosπy−2cosπxcosπy−2F1x,y+H1x,y.
According to Theorem (1), we get(58)F1x,y=G0x,yU1x,y=cosU0x,yU1x,y=−1Γα+1cos1−cosπx1−cosπy1−cosπx1−cosπy.
Using Definition (5), we get(59)H1x,y=1Γα+1D0Ctα2sinEα−tα1−cosπx1−cosπy−π2Eα−tαcosπx+cosπy−2cosπxcosπyt=0=1Γα+1π2cosπx+cosπy−2cosπxcosπy−2cos1−cosπx1−cosπy1−cosπx1−cosπy.
Consequently, we obtain(60)U3x,y=Γα+1Γ3α+1−1−cosπx1−cosπyΓα+1−π2Γα+1cosπx+cosπy−2cosπxcosπy+2Γα+1cos1−cosπx1−cosπy1−cosπx1−cosπy+π2Γα+1cosπx+cosπy−2cosπxcosπy−2Γα+1cos1−cosπx1−cosπy1−cosπx1−cosπy=−1Γ3α+11−cosπx1−cosπy.
Continuing in the same manner fork≥2, we obtain(61)U4x,y=1Γ4α+11−cosπx1−cosπx,…,Ukx,y=−1kΓkα+11−cosπx1−cosπy.
Then, by using inverse CRDT (9), we get(62)ux,y,t=∑k=0∞Ukx,yt−t0kα=1−cosπx1−cosπy1−tαΓα+1+t2αΓ2α+1−t3αΓ3α+1+t4αΓ4α+1−t5αΓ5α+1+⋯=1−cosπx1−cosπy∑k=0∞−1ktkαΓkα+1.
The Laplace–Padé approximant is used to approximate the truncated series up tonth order of perturbation in this case. That is,(63)u˜nx,y,t=1−cosπx1−cosπy∑k=0n−1ktkαΓkα+1.
Then, on both sides of (63), we use the Laplace transform to get(64)Lu˜nx,y,t=1−cosπx1−cosπy∑k=0n−1ksαk+1.
For effortlessness, we write1/t instead of s in Eq., and we get(65)Lu˜nx,y,t=1−cosπx1−cosπy∑k=0n−1ktkα+1=1−cosπx1−cosπyt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1+⋯.
Now, we convert transformed series (45) into a meromorphic function by forming its Padé approximant of K/L with and K+L=7k=3,L=4.(66)KLt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1.
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
By means of (23) and (22), we get(67)p0=u0=0,p1=1,p2=−1+q1,p3=1−q1+q2,andq1−q2+q3=1,−q1+q2−q3+q4=−1,q1−q2+q3−q4=1,−q1+q2−q3+q4=−1..
Solving forp and q, we get p0=0,p1=1,p2=0,p3=1,q1=1,q2=1,q3=1,andq4=0.
Then,K/Lt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=t+t2α+1/1+t+tα+1+t2α+1=t/1+tα.
Therefore, allK/L Padé approximants of the equation with L≥1,K≥1 and K+L≤n give(68)KLu˜nx,y,t=1−cosπx1−cosπyt1+tα.
Now, by changing1/s into t in the equation, we obtain K/Lu˜n in terms of s as follows:(69)KLu˜nx,y,t=1−cosπx1−cosπysα−1sα+1.
Finally, on both sides of Equation (47), using the inverse Laplace transform, we have the exact solution as follows:(70)ux,y,t=1−cosπx1−cosπyL−1sα−1sα+1=1−cosπx1−cosπyEα−tα.
Puttingα=1 in Equation (48), we have(71)ux,y,t=1−cosπx1−cosπyL−11s+1=e−t1−cosπx1−cosπy.
This finding is in perfect accord with the one obtained in the previous study as shown in [35, 48].Example 3.
Consider the following2+1−dimensional hyperbolic nonlinear time-fractional telegraph equation with variable coefficients in the area Ω=0,π2 as follows:(72)Dt2α0Cux,y,t−uDtα0Cux,y,t−uDx2ux,y,t+Dy2ux,y,t−1+x2∂2u∂x2−1+y2∂2u∂y2=Eα−tαcoshxcoshy−sinhx+y−1−x2−y2×Eα−tαcoshxcoshy,with the initial condition(73)ux,y,0=coshxcoshy,Dtα0Cux,y,0=−1Γα+1coshxcoshy.
Applying the properties of CFRDT and the related rules in Table1 on both sides of Equation (50), we obtain(74)Uk+2x,y=Γαk+1Γαk+2α+1∑r=0kΓkα+α+1Γkα+1Uk+1x,yUk−rx,y+∑r=0k∂∂xUrx,yUk−rx,y+∑r=0k∂∂yUrx,yUk−rx,y+1+x2∂2∂x2Ukx,y+1+y2∂2∂y2Ukx,y+∑r=0kVrx,yVk−rx,ycosh2xcosh2y−sinhx+ycoshxcoshy−−1kΓkα+11+x2+y2coshxcoshy,where Vkx,y is the fractional reduced transform of Eα−tα.
Applying the CFRDT to initial condition (51) according to Theorem 10, we get(75)U0x,y=coshxcoshy,U1x,y=−1Γα+1coshxcoshy.
Now, taking the values ofkk=0,1,2,⋯ and using Equation (53) into equation (52), we obtain the following successive iterative values.
Fork=0, using the values of U0x,y and U1x,y from (75) into (74), we get(76)U2x,y=Γ1Γ2α+1Γα+1Γ1U1x,yU0x,y+∂∂xU0x,yU0x,y+∂∂yU0x,yU0x,y+1+x2∂2∂x2U0x,y+1+y2∂2∂y2U0x,y+V0x,yV0x,ycosh2xcosh2y−sinhx+ycoshxcoshy−−10Γ11+x2+y2coshxcoshy=1Γ2α+1−Γα+11Γα+1coshxcoshycoshxcoshy+sinhxcoshxcosh2y+sinhycoshycosh2x+1+x2coshxcoshy+1+y2coshxcoshy+cosh2xcosh2y−sinhx+ycoshxcoshy−1+x2+y2coshxcoshy=1Γ2α+1−cosh2xcosh2y+sinhxcoshxcosh2y+sinhycoshycosh2x+cosh2xcosh2y−sinhxcoshxcosh2y−sinhycoshycosh2x+1+x2+y2coshxcoshy+coshxcoshy−1+x2+y2coshxcoshy=1Γ2α+1coshxcoshy.
Fork=1, using the values of U1x,y and U2x,y in (74), we get(77)U3x,y=Γα+1Γ3α+1∑r=01Γ2α+1Γα+1U2x,yU1−rx,y+∑r=01∂∂xUrx,yU1−rx,y+∑r=01∂∂yUrx,yU1−rx,y+1+x2∂2∂x2U1x,y+1+y2∂2∂y2U1x,y+∑r=01Vrx,yV1−rx,ycosh2xcosh2y−sinhx+ycoshxcoshy−−11Γα+11+x2+y2coshxcoshy=Γα+1Γ3α+12Γα+1cosh2xcosh2y−2Γα+1sinhxcoshxcosh2y−2Γα+1coshxcoshysinh2x−1Γα+1coshxcoshy+1Γα+11+x2+y2coshxcoshy−2Γα+1cosh2xcosh2y+2Γα+1sinhxcoshxcosh2y+2Γα+1coshxcoshysinh2x−1Γα+11+x2+y2coshxcoshy=−1Γ3α+1coshxcoshy.
Continuing in the same manner fork≥2, we obtain(78)U4x,y=1Γ4α+1coshxcoshy,…,Ukx,y=−1kΓkα+1coshxcoshy.Then, by (9), we get(79)ux,y,t=∑k=0∞Ukx,yt−t0kα=coshxcoshy1−tαΓα+1+t2αΓ2α+1−t3αΓ3α+1+t4αΓ4α+1−t5αΓ5α+1+⋯=coshxcoshy∑k=0∞−1ktkαΓkα+1.Here, the Laplace–Padé approximant is applied to the truncated series up to nth order of perturbation, that is,(80)u˜nx,y,t=coshxcoshy∑k=0n−1ktkαΓkα+1.
Applying the Laplace transform on both sides of (80) yields(81)Lu˜nx,y,t=coshxcoshy∑k=0n−1ksαk+1.
For simplicity, we write1/t instead of s i equation (57), we get(82)Lu˜nx,y,t=coshxcoshy∑k=0n−1ktkα+1=coshxcoshyt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1+⋯.
Now, we convert transformed series (58) into a meromorphic function by forming its Padé approximant of K/L with k≥1,L≥1, and K+L=7k=3,L=4.(83)KLt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1,
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
Using (23) and (22), we get(84)p0=u0=0,p1=1,p2=−1+q1,p3=1−q1+q2,andq1−q2+q3=1,−q1+q2−q3+q4=−1,q1−q2+q3−q4=1,−q1+q2−q3+q4=−1..
Solving forp and q, we get p0=0,p1=1,p2=0,p3=1,q1=1,q2=1,q3=1,andq4=0.
Then,K/Lt−tα+1+t2α+1−t3α+1+t4α+1−t5α+1+t6α+1=t+t2α+1/1+t+tα+1+t2α+1=t/1+tα.Therefore, all K/L Padé approximants of equation (58) with L≥1,K≥1 and K+L≤n give(85)KLu˜nx,y,t=coshxcoshyt1+tα.
Now, by changing1/s into t in equation (59, we obtain K/Lu˜n in terms of s as follows:(86)KLu˜nx,y,t=coshxcoshysα−1sα+1.
Finally, applying the inverse Laplace transform on both sides of (86), we have the exact solution as follows:(87)ux,y,t=coshxcoshyL−1sα−1sα+1=coshxcoshyEα−tα.
Puttingα=1 in the (87), we have(88)ux,y,t=coshxcoshyL−11s+1=e−tcoshxcoshy.
The obtained result is the same as obtained by Hafez [57].Example 4.
Consider the following2+1−dimensional hyperbolic time-fractional differential equation in the area Ω=0,12 as follows:(89)Dt2α0Cux,y,t+Dtα0Cux,y,t=Dx2ux,y,t+Dy2ux,y,t+2π2u+2πsinπx+ytEα−x+ytα+x+y2−x+ysinπxsinπyEα−x+ytα−2sinπxsinπyt2Eα−x+ytα.
With the initial condition, we obtain(90)ux,y,0=sinπxsinπy,D0Ctαux,y,0=−x+yΓα+1sinπxsinπy.
Applying the CFRDT to initial condition (63) according to Theorem 2, we obtain(91)Uk+2x,y=Γαk+1Γαk+2α+1−Γkα+α+1Γkα+1Uk+1x,y+∂2∂x2Ukx,y+∂2∂y2Ukx,y+2π2Ukx,y+2πsinπx+yVkx,y+x+y2−x+ysinπxsinπy−x+ykΓαk+1−2sinπxsinπyWkx,y,where Vkx,y and Wkx,y are the transformed forms of tEα−x+ytα and t2Eα−x+ytα.
Applying the CRDT to initial condition (64), we get(92)U0x,y=sinπxsinπy,U1x,y=−x+yΓα+1sinπxsinπy.
Substituting equation (66) into equation (65) and applying Definition 9, we obtain the following successive Ukx,y values for k=0,1,2,….(93)U2x,y=Γ1Γ2α+1−Γα+1Γ1U1x,y+∂2∂x2U0x,y+∂2∂y2U0x,y+2π2U0x,y+2πsinπx+yV0x,y+x+y2−x+ysinπxsinπy−x+y0Γ1−2sinπxsinπyW0x,y=Γ1Γ2α+1x+ysinπxsinπy−2π2sinπxsinπy+2π2sinπxsinπy+x+y2sinπxsinπyx+ysinπxsinπy=x+y2Γ2α+1sinπxsinπy.U3x,y=Γα+1Γ3α+1−Γ2α+1Γα+1U2x,y+∂2∂x2U1x,y+∂2∂y2U1x,y+2π2U1x,y+2πsinπx+yV1x,y+x+y2−x+ysinπxsinπy−x+y1Γα+1−2sinπxsinπyW1x,y=Γα+1Γ3α+1−Γ2α+1Γα+1x+y2Γ2α+1sinπxsinπy−∂2∂x2x+yΓα+1sinπxsinπy−∂2∂y2x+yΓα+1sinπxsinπy−2π2x+yΓα+1sinπxsinπy+2πsinπx+yV1x,y+x+y2−x+ysinπxsinπy−x+y1Γα+1−2sinπxsinπyW1x,y=Γα+1Γ3α+1−x+y2Γα+1sinπxsinπy−2πcosπxsinπyΓα+1−2πsinπxcosπyΓα+1+2π2x+yΓα+1sinπxsinπy−2π2x+yΓα+1sinπxsinπy+2πcosπxsinπyΓα+1+2πsinπxcosπyΓα+1−x+y3Γα+1sinπxsinπy+x+y2Γα+1sinπxsinπy=−x+y3Γ3α+1sinπxsinπy.
Continuing in the same manner fork≥2, we obtain(94)U4x,y=x+y4Γ4α+1sinπxsinπy,…,Ukx,y,=−1kx+ykΓkα+1sinπxsinπy.
Then, by using inverse CFRDT (9), we get(95)ux,y,t=∑k=0∞Ukx,yt−t0kα=sinπxsinπy1−x+yΓα+1tα+x+y2t2αΓ2α+1−x+y3t3αΓ3α+1+x+y4t4αΓ4α+1−x+y5t5αΓ5α+1+...=sinπxsinπy∑k=0∞−1kx+yktkαΓkα+1.
Here, the Laplace–Padé approximant is applied to the truncated series up tonth order of perturbation, that is,(96)u˜nx,y,t=sinπxsinπy∑k=0n−1kx+yktkαΓkα+1.
Applying the Laplace transform on both sides of (96) yields(97)Lu˜nx,y,t=sinπxsinπy∑k=0n−1kx+yksαk+1.
For simplicity, we write1/t instead of s in equation (70), we get(98)Lu˜nx,y,t=sinπxsinπy∑k=0n−1kx+yktkα+1=sinπxsinπyt−x+ytα+1+x+y2t2α+1−x+y3t3α+1+....
Now, we convert transformed series (71) into a meromorphic function by forming its Padé approximant of K/L with k≥1,L≥1 and K+L=7k=3,L=4.(99)KLt−x+ytα+1+x+y2t2α+1−x+yt3α+1+x+y4t4α+1−x+y5t5α+1+x+yt6α+1=p0+p1t+p2tα+1+p3t2α+11+q1t+q2tα+1+q3t2α+1+q4t3α+1,
We obtainu0=0,u1=1,u2=−2,u3=4,u4=−8,u5=16,u6=−32,u7=64.
By using (30) and (31), we get(100)p0=u0=0,p1=1,p2=−x+y+q1,p3=x+y2−x+yq1+q2,x+y2q1−x+yq2+q3=x+y3,−x+y3q1+x+y2q2−x+yq3+q4=−x+y4,x+y4q1−x+y3q2+x+y2q3−x+yq4=x+y5,−x+y5q1+x+y4q2−x+y3q3+x+y2q4=−x+y6..
Solving forp and q, we get p0=0,p1=1,p2=0,p3=1,q1=x+y,q2=1,q3=x+yandq4=0.
Then, we obtain(101)KLt−x+ytα+1+x+y2t2α+1−x+y3t3α+1+x+y4t4α+1−x+y5t5α+1+x+y6t6α+1=t+t2α+11+x+yt+tα+1+x+yt2α+1=t1+x+ytα.
Therefore, allK/L Padé approximants of equation (84) with L≥1,K≥1 and K+L≤n give(102)KLu˜nx,y,t=sinπxsinπyt1+x+ytα.
Now, by changing1/s into t in equation (72), we obtain K/Lu˜n in terms of s as follows:(103)KLu˜nx,y,t=sinπxsinπysα−1sα+x+y.
To conclude, applying the inverse Laplace transform on both sides of (103), we have the exact solution as follows:(104)ux,y,t=sinπxinπyL−1sα−1sα+x+y=sinπxsinπyEα−x+ytα.
Puttingα=1 in Equation (74), we have(105)ux,y,t=sinπxsinπyL−11s+x+y=e−x+ytsinπxsinπy.
The obtained result is the same as obtained by Li et al. [48].


## 6.2. Numerical Simulation to Illustrative Examples

In this subsection, the numerical simulation of the considered problems was depicted using tables and figures. Mathematica software or program has been used to create all 3D figures in this manuscript.Numerical results corresponding to Example 1 are depicted in Table2 and Figures 1 and 2.Table 2
Six-term approximate solution by CFRDTM of Example 1 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.13.9290683804.4956555075.3234236446.0496474836.0496474641.83E-080.24.5205771713.6602304884.2433341814.9530347114.9530324242.29E-060.36.7870350813.2500986893.5030354784.0552381234.0551999673.82E-050.410.989007473.1829331902.9669044083.3203961663.3201169232.79E-040.517.417310273.5299297732.5801835542.7195831482.7182818281.30E-030.626.346588664.4332471132.3271642272.2300998882.2255409284.56E-030.738.031634096.0838518472.2201211501.8352372691.8221188001.31 E-020.852.709286768.7127228342.2965986141.5245126821.4918246983.27 E-020.970.6009305412.586394612.6198486801.2943793801.2214027587.30 E-02191.914709018.004269173.2806929481.14940872711.49 E-01Figure 1
3D view of the solution behaviour of example 1 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 2
3D view of the solution behaviour of example 1 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)Numerical results corresponding to Example 2 are depicted in Table3 and Figures 3 and 4.Table 3
Six-term approximate solutions by CFRDTM of Example 2 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.12.7124571373.0715029693.3845871833.6193496723.6193496727.84E-110.22.4518912332.7139248893.0025024043.2749230223.2749230129.91E-090.32.3001368722.4695243272.7035464672.9632730502.9632728831.67E-070.42.2041669412.2842392592.4574552042.6812814222.6812801841.24E-060.52.1469975922.1370598742.2494907752.4261284722.4261226395.83E-060.62.1215684552.0177020722.0708506242.1952672002.1952465442.07E-050.72.1246840771.9206787611.9157424491.9864012721.9863412156.01E-050.82.1549543241.8430761341.7801662331.7974670221.7973158561.51E-040.92.2119327361.7835439341.6613149101.6266194501.6262786393.41E-0412.2957044411.7417757121.5572479541.4722222221.4715177657.04E-03Figure 3
3D view of the solution behaviour of Example 2 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 4
3D view of the solution behaviour of Example 1 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)Numerical results corresponding to Example 3 are described in Table4 and Figures 5 and 6.Table 4
Six-term approximate solutions by CFRDTM of Example 3 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.11.6146564611.8283872762.0147583122.1545064272.1545064274.67E-110.21.4595482331.6155301761.7873130011.9494780381.9494780325.90E-090.31.3692127371.4700447641.6093521671.7639607691.7639606699.96E-080.41.3120842891.3597492951.4628603231.5960983541.5960976177.37E-070.51.2780528351.2721371661.3390644101.4442123201.4442088473.47E-060.61.2629155191.2010865141.2327244901.3067865001.3067742041.23E-050.71.2647701691.1433310151.1403925551.1824539471.1824181973.58E-050.81.2827892741.0971361531.0596874961.0699862141.0698962289.00E-050.91.3167070681.0616981540.9889383380.9682850170.9680821412.03E-0411.3665742241.0368345990.9269899370.8763762900.8759569454.19E-04Figure 5
3D view of the solution behaviour of Example 3 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 6
3D view of the solution behaviour of example 3 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)Numerical results corresponding to Example4 are depicted in Table 5 and Figures 7 and 8.Table 5
Six-term approximate solutions by CFRDTM of Example 4 for different values of fractional orderα and the absolute error E=uexact−uα=1.
tα=0.4α=0.6α=0.8α=1ExactAbsolute erroruExact−uα=10.17.97E-339.12E-331.08E-321.23E-321.23E-323.72E-410.29.18E-337.43E-338.61E-331.01E-321.01E-324.64E-390.31.38E-326.60E-337.11E-338.23E-338.23E-337.74E-380.42.23E-326.46E-336.02E-336.74E-336.74E-335.67E-370.53.54E-327.16E-335.24E-335.52E-335.52E-332.64E-360.65.35E-329.00E-334.72E-334.53E-334.54E-339.25E-360.77.72E-321.23E-324.51E-333.72E-333.70E-332.66E-350.81.07E-311.77E-324.66E-333.09E-333.03E-336.63E-350.91.43E-312.55E-325.34E-332.63E-332.48E-331.48E-3411.87E-313.65E-326.66E-332.33E-332.03E-333.03E-34Figure 7
3D view of the solution behaviour of Example 4 att = 0.3 when (a) α = 0.4, (b) α = 0.6, (c) α = 0.8, and (d) α = 1.
(a)(b)(c)(d)Figure 8
3D view of the solution behaviour of example 4 att = 0.3: (a) exact solution and (b) absolute errors.
(a)(b)


## 7. Discussion

In this article, we showed how to solve the two-dimensional fractional wave equation using the LPCFRDTM. The CFRDTM changed the FPDEs into an easily solvable linear algebraic recursion system for the coefficient functions of the power series solution for each of the fundamental problems examined here. The CFRDTM does not require any lengthy computation, which is a major drawback of perturbation methods such as HPM. The main benefit of using this CFRDTM is that it minimizes the computation’s size. As previously stated, a Laplace–Padé resummation was applied to the truncated series gained by CFRDTM to increase the domain of convergence of the CFRDTM power series solution, resulting in the precise solution. It is noted that even though the Laplace–Padé resummation technique fails to reach the exact solution of the fractional PDEs under investigation on occasion, it can provide a fair approximation in the larger domain of convergence. Figures1–8 provide the comparison 3D plots of the current solution with the exact solution as well as their related absolute errors, for two test examples at t=0.3 and different values of α. A comparative study between the approximate solution of Examples 1–4 for different values of αandt, at x=y=1, with the exact solutions and their corresponding absolute errors are given in Tables 2–5. As seen in the tables, the proposed approach gives small error near t=0, but the error increases as t grows. This means that a superior approximation can be achieved for small values of time t whatever the values of x and y are within the domain of interest.


## 8. Conclusion

LPCFRDTM, a hybridization of CFRDTM and a resummation method based on Laplace transform and the Padé approximant, was introduced in this paper. First, CFRDTM is used to obtain the solution of two-dimensional NLFPDEs in convergent series form. To extend the domain of convergence of the truncated power series, a post-treatment combining Laplace transforms and the Padé approximant is employed. This approach significantly enhances the convergence rate and yields an exact solution. Furthermore, unlike HPM, CFRDTM is a powerful method because it calculates solutions without taking any perturbation parameters into account. It is evident from the outcomes of instances1−4 that the current procedure produces exact results. As a result, LPCFRDTM is quite valuable, as it allows us to improve accuracy and efficiency and provides a mathematical tool for NLFDEs. Finally, in [54, 58], given the continued use of nonlinear damped differential equations as models in a variety of domains, it is critical to examine this method of solving some intriguing problems in every branch of science and engineering, and we anticipate that this work is a step in that direction.



---

*Source: 7004412-2022-11-04.xml*