# Homotopy Analysis Method for Three Types of Fractional Partial Differential Equations

Authors: Haidong Qu; Zihang She; Xuan Liu
Journal: Complexity
 (2020)

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## Abstract

In this paper, three types of fractional order partial differential equations, including the fractional Cauchy–Riemann equation, fractional acoustic wave equation, and two-dimensional space partial differential equation with time-fractional-order, are considered, and these models are obtained from the standard equations by replacing an integer-order derivative with a fractional-order derivative in Caputo sense. Firstly, we discuss the fractional integral and differential properties of several functions which are derived from the Mittag-Leffler function. Secondly, by using the homotopy analysis method, the exact solutions for fractional order models mentioned above with suitable initial boundary conditions are obtained. Finally, we draw the computer graphics of the exact solutions, the approximate solutions (truncation of finite terms), and absolute errors in the limited area, which show that the effectiveness of the homotopy analysis method for solving fractional order partial differential equations.


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## Body

## 1. Introduction

The concept of fractional derivatives can be traced back to a question raised by Marquis de L’Hopital to Gottfried Wilhelm Leibniz three hundred years ([1–3]) ago. Due to the lack of geometric or physical background support, fractional calculus has not entered the field of vision of most researchers. Until recent decades, some researchers have found that the fractional models are better than integer models in describing the chemical process, diffusion reaction, financial vibration, and other fields. Therefore, the fractional order problems gradually attracted the interest of many researchers, and the application expanded to many scientific fields including fluid flow, rheology, dynamical processes in self-similar and porous structures, diffusive transport akin to diffusion, electrical networks, probability and statistics, control theory of dynamical systems, viscoelasticity, electrochemistry of corrosion, optics and signal processing [4–8], and so on. These applications in interdisciplinary sciences motivate us to try to find out numerical or analytic solutions for the fractional differential equations.Now, many effective methods for fractional differential equations have been presented, such as the finite difference method [9], spectral method [5], matrix approach [10], homotopy analysis method (HAM) [11], and homotopy perturbation method [12]. Especially, the HAM was first proposed by Liao in [13]. This method has been successfully applied to solve various linear or nonlinear problems [8, 14–18]. The following is a brief survey.In 2007, firstly, the HAM that was developed for an integer-order differential equation was directly extended to derive explicit and numerical solutions of nonlinear fractional differential equations by by Song and Zhang [19]. In 2008, Xu and Cang [20] employed the HAM to derive the solutions of the time fractional wave-like differential equations with a variable coefficient. In [21], HAM is applied to solve linear and nonlinear fractional initial-value problems by Hashim ea al\enleadertwodots. In 2010, Dehghan et al. [22] applied HAM to solve several fractional KDV equations and showed the high accuracy and efficiency of the proposed technique. In [23], Abbasbandy et al. used HAM to obtain approximate solutions of fractional integrodifferential equations and gave some examples to illustrate the high efficiency and precision of this method. Very recently, Morales-Delgado et al. [24] have presented an analysis based on a combination of the Laplace transform and homotopy methods in order to provide new analytical approximated solutions of the fractional partial differential equations in the Liouville–Caputo and Caputo–Fabrizio senses. In 2019, an optimal homotopy analysis approach [25] was proposed to deal with nonlinear fractional differential equations, and the corresponding optimal initial approximation was discussed. For further information about the HAM, refer to [26–28].In this paper, we use the HAM to solve fractional Cauchy–Riemann equations [29] and fractional acoustic wave equations [30], which are given by(1)D0,tαcvx,t+D0,xβcρx,t=0,D0,tαcρx,t−D0,xβcvx,t=0,(2)D0,tαcvx,t+c02b0D0,xβcρx,t=0,D0,tαcρx,t+b0D0,xβcvx,t=0,where 0<α,β≤1 and b0 and c0 are positive numbers and to solve the partial differential equation with a time-fractional-order of the following form:(3)D0,tαcux,y,t=λΔux,y,t,with a time-fractional order 1<α≤2, and D0,tαc⋅ is the Caputo fractional derivative.By settingα=1 and β=1, the models (1) and (2) were transformed into Cauchy–Riemann equations and acoustic wave equations, respectively. Suppose that v and ρ satisfy the Cauchy–Riemann equations in an open subset of R2, and consider the vector field v,ρT. In fluid dynamics, such a vector field is a potential flow. In the second model, b0 and c0 denote the medium density and the propagation velocity without an acoustic disturbance, respectively. Model (3) is a two-dimensional space and time-fractional-order model, which has important applications in many fields.The rest of this paper is organized as follows. In Section2, the definitions and properties of fractional derivatives of some functions are introduced. Section 3 gives an introduction of the HAM which is used to solve fractional order differential equations. In Section 4, the analytic solutions for three types of fractional order differential equations were obtained by using the HAM, and the computer graphics of the exact solutions, the approximate solutions, and absolute errors were drawn in the limited area to clarify the effectiveness of the HAM. Finally, Section 5 offers some concluding remarks.


## 2. Definitions and Lemmas

Definition 1.
(see [3]). A real function fx, x>0, is said to be in the space Cμ, μ∈R, if there exists a real number p>μ, such that fx=xpf1x, where f1x∈C0,∞, and it is said to be in the space Cμn, if and only if fn∈Cμ, n∈N.Definition 2.
(see [3]). The Riemann–Liouville fractional integral of order α∈R, α>0 of a function fx∈Cμ,μ≥−1 is defined as(4)I0+αftx=1Γα∫0xftdtx−t1−α,x>0.Definition 3.
(see [3]). The Riemann–Liouville fractional derivative of order α∈R,α>0, on the usual Lebesgue space L1a,b, is given by(5)D0+αftx=ddxnI0+n−αftx=1Γn−αddxn∫0xftdtx−tα−n+1,where n=α+1,x>0.Definition 4.
(see [3]). The Caputo fractional derivative of fx∈Cμ−1,n∈N, is defined as(6)D0,xαcfx=I0+n−αddxnfx,n−1<α<n,ddxnfx,α=n.Definition 5.
(see [3]). The classical Mittag-Leffler function is defined by(7)EαZ≔∑k=0∞ZkΓαk+1,Z∈C,α>0.
The generalized Mittag-Leffler function is defined by(8)Eα,βZ≔∑k=0∞ZkΓαk+β,Z,β∈C,α>0.Definition 6.
(see [3]). The functions sinαZ, cosαZ, sinα,βZ, and cosα,βZ (Z,β∈C,α>0) are defined by(9)sinαZ=∑k=1∞−1k+1Z2k−1Γα2k−1+1,cosαZ=∑k=0∞−1kZ2kΓα2k+1,sinα,βZ=∑k=1∞−1k+1Z2k−1Γα2k−1+β,cosα,βZ=∑k=0∞−1kZ2kΓα2k+β.
Obviously, Euler’s equations have the following forms:(10)EαiZ=cosαZ+isinαZ,Eα,βiZ=cosα,βZ+isinα,βZ.
In particular, whenα=1, we have(11)sin1Z=sinZ,cos1Z=cosZ,and when β=1, we have(12)sinα,1Z=sinαZ,cosα,1Z=cosαZ.Definition 7.
(see [3]). The functions sinhαZ, coshαZ, sinhα,βZ, and coshα,βZ (Z,β∈C,α>0) are defined by(13)sinhαZ=∑k=1∞Z2k−1Γα2k−1+1,coshαZ=∑k=0∞Z2kΓα2k+1,sinhα,βZ=∑k=1∞Z2k−1Γα2k−1+β,coshα,βZ=∑k=0∞Z2kΓα2k+β.
Obviously, the hyperbolic sine’s and cosine’s equations have the following forms:(14)sinhαZ=EαZ−Eα−Z2,coshαZ=EαZ+Eα−Z2,sinhα,βZ=Eα,βZ−Eα,β−Z2,coshα,βZ=Eα,βZ+Eα,β−Z2.
In particular, whenα=1, we have(15)sinh1Z=sinhZ,cosh1Z=coshZ.
Whenβ=1, we have(16)sinhα,1Z=sinhαZ,coshα,1Z=coshαZ.Lemma 1.
IfsinαZ, cosαZ, sinα,βZ, and cosα,βZ are defined as in Definition 6, then(17)Ia+αt−aβ−1sinμ,βλt−aμx=x−aα+β−1sinμ,α+βλt−aμ,(18)Ia+αt−aβ−1cosμ,βλt−aμx=x−aα+β−1cosμ,α+βλt−aμ,(19)Da+αt−aβ−1sinμ,βλt−aμx=x−aβ−α−1sinμ,β−αλt−aμ,(20)Da+αt−aβ−1cosμ,βλt−aμx=x−aβ−α−1cosμ,β−αλt−aμ.Proof.
We denote the beta function byβ˜, and then, according to the definition of the Riemann–Liouville fractional integral, we have(21)Ia+αt−aβ−1sinμ,βλt−aμx=1Γα∫a+xt−aβ−1sinμ,βλt−aμx−t1−αdt,=1Γα∫0+x−aξβ−1sinμ,βλξμx−ξ−a1−αdξ,ξ=t−a,=1Γα∫0+1tβ−1x−aβ−1sinμ,βλtμx−aμx−a1−α1−t1−α⋅x−adt,ξ=tx−a,=x−aα+β−11Γα∫0+1tβ−11−t1−α⋅∑k=1∞−1k+1λtμx−aμ2k−1Γμ2k−1+βdt,=x−aα+β−11Γα∑k=1∞−1k+1λx−aμ2k−1⋅∫0+1tβ−1tμ2k−11−t1−αΓμ2k−1+βdt,=x−aα+β−11Γα∑k=1∞−1k+1λx−aμ2k−1Γμ2k−1+β⋅β˜α,μ2k−1+β,=x−aα+β−11Γα∑k=1∞−1k+1λx−aμ2k−1Γμ2k−1+β⋅ΓαΓμ2k−1+βΓμ2k−1+β+α,=x−aα+β−1∑k=1∞−1k+1λx−aμ2k−1Γμ2k−1+β+α,=x−aα+β−1sinμ,α+βλt−aμ.
Similarly, we obtain equation (18). Next, we will prove equation (19). From the definition of the Riemann–Liouville fractional derivative, we get(22)Da+αt−aβ−1sinμ,βλt−aμx=1Γn−αddxn∫a+xt−aβ−1sinμ,βλt−aμx−tα−n+1dt=1Γn−αddxn∫0+x−aξβ−1sinμ,βλξμx−ξ−aα−n+1dξ=ddxn∫0+1tβ−1x−aβ−1sinμ,βλtμx−aμx−aα−n+11−tα−n+1⋅1Γn−αx−adtξ=tx−a=ddxnx−aβ−α+n−1Γn−α⋅∫0+1tβ−11−tα−n+1∑k=1∞−1k+1λtμx−aμ2k−1Γμ2k−1+βdt=ddxnx−aβ−α+n−1Γn−α∑k=1∞−1k+1λx−aμ2k−1Γμ2k−1+β⋅Γn−αΓμ2k−1+βΓμ2k−1+β−α+n=x−aβ−α−1∑k=1∞−1k+1λx−aμ2k−1Γμ2k−1+β−α=x−aβ−α−1sinμ,β−αλt−aμ.
Similarly, we obtain equation (20). In particular, when β=1, μ=α, we have(23)Da+αsinαλt−aαx=λcosαλx−aα,Da+αcosαλt−aα−1x=−λsinαλx−aα.
While replacing the Riemann–Liouville fractional derivative with the Caputo derivative, we get the following forms:(24)Da+αcsinαλt−aαx=λcosαλx−aα,Da+αccosαλt−aαx=−λsinαλx−aα.Lemma 2.
IfsinhαZ, coshαZ, sinhα,βZ, and coshα,βZ are defined as in Definition 7, then(25)Ia+αt−aβ−1sinhμ,βλt−aμx=x−aα+β−1sinhμ,α+βλt−aμ,(26)Ia+αt−aβ−1coshμ,βλt−aμx=x−aα+β−1coshμ,α+βλt−aμ,(27)Da+αt−aβ−1sinhμ,βλt−aμx=x−aβ−α−1sinhμ,β−αλt−aμ,(28)Da+αt−aβ−1coshμ,βλt−aμx=x−aβ−α−1coshμ,β−αλt−aμ.Proof.
Similar to the proof of Lemma1, we obtain equations (25)–(28). In particular, when β=1 and μ=α,(29)Da+αsinhαλt−aαx=λcoshαλx−aα,Da+αcoshαλt−aα−1x=λsinhαλx−aα.
While, replacing the Riemann–Liouville fractional derivative with the Caputo derivative, we get the following forms:(30)Da+αcsinhαλt−aαx=λcoshαλx−aα,Da+αccoshαλt−aαx=λsinhαλx−aα.


## 3. HAM

In this section, we consider a linear or nonlinear equation in a general form:(31)Nux,t=0,where ux,t is an unknown function and x and t are independent variables. Let u0x,t denote an initial approximation of the solution of equation (31), h a nonzero auxiliary parameter, Hx,t a nonzero auxiliary function, and L is an auxiliary linear operator. Then, we construct the HAM deformation equation in the following form:(32)1−qLΦx,t;q−u0x,t=qhHx,tNΦx,t;q,where q∈0,1 is an embedding parameter. Obviously, when q=0 and q=1, the abovementioned HAM deformation equation (32) has the solutions(33)Φx,t;0=u0x,t,Φx,t;1=ux,t,respectively. Thus, as q increases from 0 to 1, Φx,t;q varies from the initial guesses Φx,t;0 to the solution Φx,t;1 of equation (31). Expanding Φx,t;q in Taylor’s series with respect to q, we have(34)Φx,t;q=u0x,t+∑m=1∞umx,tqm,where(35)umx,t=1m!∂mΦx,t;q∂qmq=0.For brevity, we define a vector(36)um⟶=u0,u1,…,um.Differentiating the HAM deformation equation (32) m times with respect to q, then setting q=0, and finally dividing it by m!, we obtain the mth-order deformation equation:(37)Lum−χmum−1=hHx,tRmum−1⟶,where Rmum−1⟶=1/m−1!∂m−1NΦx,t;q/∂qm−1q=0, and(38)χm=0,m=1,1,m>1.Operating the inverse operator ofL on both sides of equation (37), we have(39)umx,t=χmum−1x,t+hHx,tL−1Rmum−1x,t⟶.In this way, it is easy to obtainu1x,t,u2x,t,… one after another, and then, we get an exact solution of the original equation (31) of the series form:(40)ux,t=∑m=0∞umx,t.


## 4. Applying HAM

In this section, we present three examples to illustrate the applicability of HAM for solving FCPDEs introduced in Section3. For more details about the HAM, the reader can refer to [8], [13–16].


### 4.1. The Fractional Cauchy–Riemann Equations

In this section, we consider the fractional Cauchy–Riemann equation (1) with the following initial conditions:(41)vx,0=sinβxβ=v0x,t,ρx,0=cosβxβ=ρ0x,t.First, we choose two linear fractional order operators:(42)L1Θx,t;q=D0,tαcΘx,t;q,L2Λx,t;q=D0,tαcΛx,t;q.Secondly, we define two linear operators as(43)N1Θx,t;q,Λx,t;q=D0,tαcΘx,t;q+D0,xβcΛx,t;q,N2Λx,t;q,Θx,t;q=D0,tαcΛx,t;q−cD0,xβΘx,t;q.Using the abovementioned definitions and with the assumptionHx,t=1, we construct the zeroth-order deformation equations (ZDE):(44)1−qL1Θx,t;q−v0x,t=qhN1Θx,t;q,Λx,t;q,(45)1−qL2Λx,t;q−ρ0x,t=qhN2Λx,t;q,Θx,t;q.Thirdly, differentiating the ZDEsm times with respect to q, then setting q=0, and dividing it by m!, we get the mth-order deformation equations:(46)L1vmx,t−χmvm−1x,t=hN1Rm⟶vm−1x,t,ρm−1x,t,L2ρmx,t−χmρm−1x,t=hN2Rm⟶ρm−1x,t,vm−1x,t,where(47)χm=0,m=1,1,m>1,N1Rm⟶vm−1x,t,ρm−1x,t=D0,tαcvm−1x,t;q+D0,xβcρm−1x,t;q,N2Rm⟶ρm−1x,t,vm−1x,t=D0,tαcρm−1x,t;q−D0,xβcvm−1x,t;q.Finally, operating the operatorsL1−1 and L2−1 on both the sides of equations (44) and (45), respectively, we have(48)vmx,t=χmvm−1x,t+hL1−1N1Rm⟶vm−1x,t,ρm−1x,t,ρmx,t=χmρm−1x,t+hL2−1N2Rm⟶ρm−1x,t,vm−1x,t.From Lemma1 and calculating one by one, we get(49)v0x,t=sinβxβ,ρ0x,t=cosβxβ,v1x,t=0+hL1−1N1R1⟶v0x,t,ρ0x,t,=hL1−1D0,tαcv0x,t;q+D0,xβcρ0x,t;q,=hI0,tα0−sinβxβ,=−hsinβxβtαΓ1+α,ρ1x,t=0+hL2−1N1R1⟶ρ0x,t,v0x,t,=hL2−1D0,tαcρ0x,t−D0,xβcv0x,t,=hI0,tα0−cosβxβ,=−hcosβxβtαΓ1+α,v2x,t=v1x,t+hL1−1N1R2⟶v1x,t,ρ1x,t,=v1x,t+hL1−1D0,tαcv1x,t+D0,xβcρ1x,t,=v1x,t+hL1−1L1v1x,t+hI0,tα−hsinβxβtαΓ1+α,=−hsinβxβtαΓ1+α−h2sinβxβtαΓ1+α+h2sinβxβt2αΓ1+2α,ρ2x,t=ρ1x,t+hL2−1N2R2⟶ρ1x,t,v1x,t,=ρ1x,t+hL2−1D0,tαcρ1x,t−D0,xβcv1x,t,=ρ1x,t+hL2−1L2ρ1x,t+hI0,tαhcosβxβtαΓ1+α,=−hcosβxβtαΓ1+α−h2cosβxβtαΓ1+α+h2cosβxβt2αΓ1+2α,⋯.By repeating this procedure forh=−1 and noting Definition 5, we obtain the exact solution:(50)vx,t=∑k=0∞vkx,th=−1,=sinβxβ1+tαΓ1+α+t2αΓ1+2α+⋯,=sinβxβEαtα,ρx,t=∑k=0∞ρkx,th=−1,=cosβxβ1+tαΓ1+α+t2αΓ1+2α+⋯,=cosβxβEαtα.More interestingly, whenα=1 and β=1,(51)vx,t=sinxet,ρx,t=cosxetare the exact solutions of Cauchy–Riemann equations(52)vtx,t+ρxx,t=0,ρtx,t−vxx,t=0,with initial conditions vx,0=sinx and ρx,0=cosx. We draw the computer graphics of the exact solutions vx,t and ρx,t by using MATLAB software; see Figures 1–4. Figures 5–8 show the approximate solution obtained by the HAM with intercepting 15 items, and the absolute errors of exact solutions and approximate solutions are shown in Figures 9–12, with different values α and β. In the process of compiling the program, we calculate cosα,βx and sinα,βx by the formulas cosα,βx=1/2Eα,βix+Eα,β−ix and sinα,βx=1/2iEα,βix−Eα,β−ix, respectively. The MATLAB code for the Mittag-Leffler function was provided by Podlubny [31].Figure 1
The exact solutionvx,t to model (1) with α=β=1.Figure 2
The exact solutionvx,t to model (1) with α=β=0.9.Figure 3
The exact solutionρx,t to model (1) with α=β=1.Figure 4
The exact solutionρx,t to model (1) with α=β=0.9.Figure 5
The approximate solution ofvx,t to model (1) with α=β=1 (truncate 15 items that were obtained by the HAM).Figure 6
The approximate solutionvx,t to model (1) with α=β=0.9 (truncate 15 items that were obtained by the HAM).Figure 7
The approximate solution ofρx,t to model (1) with α=β=1 (truncate 15 items that were obtained by the HAM).Figure 8
The approximate solution ofρx,t to model (1) with α=β=0.9 (truncate 15 items that were obtained by the HAM).Figure 9
Absolute error of the exact solution and approximate solutionvx,t to model (1) with α=β=1.Figure 10
Absolute error of the exact solution and approximate solutionvx,t to model (1) with α=β=0.9.Figure 11
Absolute error of the exact solution and approximate solutionρx,t to model (1) with α=β=1.Figure 12
Absolute error of the exact solution and approximate solutionρx,t to model (1) with α=β=0.9.


### 4.2. Fractional Acoustic Wave Equations

In this section, we consider fractional acoustic wave equation (2) with initial conditions:(53)vx,0=Eβxβ=v0x,t,ρx,0=sinβxβ=ρ0x,t.First, the choose two linear fractional order operators:(54)L1Θx,t;q=D0,tαcΘx,t;q,L2Λx,t;q=D0,tαcΛx,t;q.Secondly, we define two linear operators as(55)N1Θx,t;q,Λx,t;q=D0,tαcΘx,t;q+c02b0D0,xβcΛx,t;q,N2Λx,t;q,Θx,t;q=D0,tαcΛx,t;q+b0D0,xβcΘx,t;q.Using the abovementioned definitions and Lemma2, similar to the method of the fractional Cauchy–Riemann equations, we get(56)v0x,t=Eβxβ,ρ0x,t=sinβxβ,v1x,t=hc02b0cosβxβtαΓ1+α,ρ1x,t=hb0EβxβtαΓ1+α,v2x,t=1+hv1+h2c02Eβxβt2αΓ1+2α,ρ2x,t=1+hρ1−h2c02sinβxβt2αΓ1+2α,v3x,t=1+hv2+c02b0I0αD0,tαc1+hρ1−h3c04b0cosβxβt3αΓ1+3α,ρ3x,t=1+hρ2+b0I0αD0,tαc1+hv1+h3c02b0Eβxβt3αΓ1+3α,v4x,t=1+hv3+1+h⋯+h4c04Eβxβt4αΓ1+4α,ρ4x,t=1+hρ3+1+h⋯+h4c04sinβxβt4αΓ1+4α,⋯.By repeating this procedure forh=−1, noting Definitions 6 and 7, we obtain the exact solutions:(57)vx,t=∑k=0∞v2kx,t+∑k=1∞v2k−1x,t,=Eβxβ1+c0tα2Γ1+2α+c0tα4Γ1+4α+⋯−c0b0cosβxβc0tαΓ1+α−c0tα3Γ1+3α+⋯,=Eβxβcoshαc0tα−c0b0cosβxβsinαc0tαρx,t=∑k=0∞ρ2kx,t+∑k=1∞ρ2k−1x,t,=sinβxβ1−c0tα2Γ1+2α+c0tα4Γ1+4α+⋯−b0c0Eβxβc0tαΓ1+α+c0tα3Γ1+3α+⋯,=sinβxβcosαc0tα−b0c0Eβxβsinhαc0tα.More interestingly, whenα=1 and β=1,(58)vx,t=excoshc0t−c0b0cosxsinc0t,ρx,t=sinxcosc0t−b0c0exsinhc0t,are the exact solutions of the following equations:(59)vtx,t+c02b0ρxx,t=0,ρtx,t+b0vxx,t=0,with initial conditions vx,0=ex, ρx,0=sinx. In Figures 13–16, the absolute errors of exact solutions and approximate solutions for vx,t and ρx,t with differential fractional order α and β are shown, which indicates the high efficiency and precision of this method. In the process of compiling the program, we calculate sinhα,βx and coshα,βx by the formulas sinhα,βZ=Eα,βZ−Eα,β−Z/2 and coshα,βZ=Eα,βZ+Eα,β−Z/2, respectively.Figure 13
Absolute error of the exact solution and approximate solutionvx,t to fractional acoustic wave equation (2) with α=β=1.Figure 14
Absolute error of the exact solution and approximate solutionvx,t to fractional acoustic wave equation (2) with α=β=0.9.Figure 15
Absolute error of the exact solution and approximate solutionρx,t to fractional acoustic wave equation (2) with α=β=1.Figure 16
Absolute error of the exact solution and approximate solutionρx,t to fractional acoustic wave equations with (2) α=β=0.9.


### 4.3. Two-Dimensional Space Partial Differential Equation with a Time-Fractional-Order

In this section, we use the HAM to solve the two-dimensional space partial differential equation with a time-fractional-order of the following form:(60)D0,tαcux,y,t=1φ2+ψ2Δux,y,t,with initial conditions(61)ux,y,0=cos2πφxcos2πψy,utx,y,0=0,and boundary conditions(62)u0,y,t=uφ,y,t=ux,0,t=ux,ψ,t=1,where 1<α≤2. First, we choose the linear fractional order operator(63)LΘx,y,t;q=D0,tαcΘx,y,t;q.Secondly, we define the linear operator as(64)NΘx,y,t;q=D0,tαcΘx,y,t;q−Θxx+Θyyφ2+ψ2.Using the abovementioned definitions and with assumptionHx,t=1, we construct the zeroth-order deformation equation(ZDE):(65)1−qLΘx,y,t;q−u0x,y,t=qhNΘx,y,t;q.Obviously, whenq=0 and q=1, it holds(66)Θx,y,t;0=u0x,y,t=ux,y,0=cos2πφxcos2πψy,φx,y,t;1=ux,y,t.Thirdly, differentiating the ZDEm times with respect to q, then setting q=0, and dividing it by m!, we get the mth-order deformation equation:(67)Lum−χmum−1=hNRmum−1⟶,where(68)NRmum−1⟶=D0,tαcum−1−um−1,xx+um−1,yyφ2+ψ2,um=1m!∂mφx,y,t;q∂qmq=0,χm=0,m=1,1,m>1.Finally, operating the operatorL−1 on both the sides of equation (67), we have(69)um=χmum−1+hL−1NRmum−1⟶.Calculating one by one, we get(70)u0=cos2πφxcos2πψy,u1=h2πφψ2cos2πφxcos2πψytαΓ1+α,u2=1+hu1+h22πφψ4cos2πφxcos2πψyt2αΓ1+2α,u3=1+hu2−hφ2+ψ2I0α1+hu1,xx+1+hu1,yy+h32πab6cos2πφxcos2πψyt3αΓ1+3α,u4=1+hu3−hI0α1+hu2,xx+1+hu2,yyφ2+ψ2+h2φ2+ψ22I0αI0α1+hu1,xx+1+hu1,yyxx+I0α1+hu1,xx+1+hu1,yyyy+h42πφψ8cos2πφxcos2πψyt4αΓ1+4α,⋯.By repeating this procedure forh=−1, we get the exact solution,(71)ux,y,t=∑k=0∞uk=∑k=0∞u2k+∑k=1∞u2k−1,=cos2πφxcos2πψycoshα2πφψ2tα−cos2πφxcos2πψysinhα2πφψ2tα.for example 3. We omit the images of example 3.


## 4.1. The Fractional Cauchy–Riemann Equations

In this section, we consider the fractional Cauchy–Riemann equation (1) with the following initial conditions:(41)vx,0=sinβxβ=v0x,t,ρx,0=cosβxβ=ρ0x,t.First, we choose two linear fractional order operators:(42)L1Θx,t;q=D0,tαcΘx,t;q,L2Λx,t;q=D0,tαcΛx,t;q.Secondly, we define two linear operators as(43)N1Θx,t;q,Λx,t;q=D0,tαcΘx,t;q+D0,xβcΛx,t;q,N2Λx,t;q,Θx,t;q=D0,tαcΛx,t;q−cD0,xβΘx,t;q.Using the abovementioned definitions and with the assumptionHx,t=1, we construct the zeroth-order deformation equations (ZDE):(44)1−qL1Θx,t;q−v0x,t=qhN1Θx,t;q,Λx,t;q,(45)1−qL2Λx,t;q−ρ0x,t=qhN2Λx,t;q,Θx,t;q.Thirdly, differentiating the ZDEsm times with respect to q, then setting q=0, and dividing it by m!, we get the mth-order deformation equations:(46)L1vmx,t−χmvm−1x,t=hN1Rm⟶vm−1x,t,ρm−1x,t,L2ρmx,t−χmρm−1x,t=hN2Rm⟶ρm−1x,t,vm−1x,t,where(47)χm=0,m=1,1,m>1,N1Rm⟶vm−1x,t,ρm−1x,t=D0,tαcvm−1x,t;q+D0,xβcρm−1x,t;q,N2Rm⟶ρm−1x,t,vm−1x,t=D0,tαcρm−1x,t;q−D0,xβcvm−1x,t;q.Finally, operating the operatorsL1−1 and L2−1 on both the sides of equations (44) and (45), respectively, we have(48)vmx,t=χmvm−1x,t+hL1−1N1Rm⟶vm−1x,t,ρm−1x,t,ρmx,t=χmρm−1x,t+hL2−1N2Rm⟶ρm−1x,t,vm−1x,t.From Lemma1 and calculating one by one, we get(49)v0x,t=sinβxβ,ρ0x,t=cosβxβ,v1x,t=0+hL1−1N1R1⟶v0x,t,ρ0x,t,=hL1−1D0,tαcv0x,t;q+D0,xβcρ0x,t;q,=hI0,tα0−sinβxβ,=−hsinβxβtαΓ1+α,ρ1x,t=0+hL2−1N1R1⟶ρ0x,t,v0x,t,=hL2−1D0,tαcρ0x,t−D0,xβcv0x,t,=hI0,tα0−cosβxβ,=−hcosβxβtαΓ1+α,v2x,t=v1x,t+hL1−1N1R2⟶v1x,t,ρ1x,t,=v1x,t+hL1−1D0,tαcv1x,t+D0,xβcρ1x,t,=v1x,t+hL1−1L1v1x,t+hI0,tα−hsinβxβtαΓ1+α,=−hsinβxβtαΓ1+α−h2sinβxβtαΓ1+α+h2sinβxβt2αΓ1+2α,ρ2x,t=ρ1x,t+hL2−1N2R2⟶ρ1x,t,v1x,t,=ρ1x,t+hL2−1D0,tαcρ1x,t−D0,xβcv1x,t,=ρ1x,t+hL2−1L2ρ1x,t+hI0,tαhcosβxβtαΓ1+α,=−hcosβxβtαΓ1+α−h2cosβxβtαΓ1+α+h2cosβxβt2αΓ1+2α,⋯.By repeating this procedure forh=−1 and noting Definition 5, we obtain the exact solution:(50)vx,t=∑k=0∞vkx,th=−1,=sinβxβ1+tαΓ1+α+t2αΓ1+2α+⋯,=sinβxβEαtα,ρx,t=∑k=0∞ρkx,th=−1,=cosβxβ1+tαΓ1+α+t2αΓ1+2α+⋯,=cosβxβEαtα.More interestingly, whenα=1 and β=1,(51)vx,t=sinxet,ρx,t=cosxetare the exact solutions of Cauchy–Riemann equations(52)vtx,t+ρxx,t=0,ρtx,t−vxx,t=0,with initial conditions vx,0=sinx and ρx,0=cosx. We draw the computer graphics of the exact solutions vx,t and ρx,t by using MATLAB software; see Figures 1–4. Figures 5–8 show the approximate solution obtained by the HAM with intercepting 15 items, and the absolute errors of exact solutions and approximate solutions are shown in Figures 9–12, with different values α and β. In the process of compiling the program, we calculate cosα,βx and sinα,βx by the formulas cosα,βx=1/2Eα,βix+Eα,β−ix and sinα,βx=1/2iEα,βix−Eα,β−ix, respectively. The MATLAB code for the Mittag-Leffler function was provided by Podlubny [31].Figure 1
The exact solutionvx,t to model (1) with α=β=1.Figure 2
The exact solutionvx,t to model (1) with α=β=0.9.Figure 3
The exact solutionρx,t to model (1) with α=β=1.Figure 4
The exact solutionρx,t to model (1) with α=β=0.9.Figure 5
The approximate solution ofvx,t to model (1) with α=β=1 (truncate 15 items that were obtained by the HAM).Figure 6
The approximate solutionvx,t to model (1) with α=β=0.9 (truncate 15 items that were obtained by the HAM).Figure 7
The approximate solution ofρx,t to model (1) with α=β=1 (truncate 15 items that were obtained by the HAM).Figure 8
The approximate solution ofρx,t to model (1) with α=β=0.9 (truncate 15 items that were obtained by the HAM).Figure 9
Absolute error of the exact solution and approximate solutionvx,t to model (1) with α=β=1.Figure 10
Absolute error of the exact solution and approximate solutionvx,t to model (1) with α=β=0.9.Figure 11
Absolute error of the exact solution and approximate solutionρx,t to model (1) with α=β=1.Figure 12
Absolute error of the exact solution and approximate solutionρx,t to model (1) with α=β=0.9.


## 4.2. Fractional Acoustic Wave Equations

In this section, we consider fractional acoustic wave equation (2) with initial conditions:(53)vx,0=Eβxβ=v0x,t,ρx,0=sinβxβ=ρ0x,t.First, the choose two linear fractional order operators:(54)L1Θx,t;q=D0,tαcΘx,t;q,L2Λx,t;q=D0,tαcΛx,t;q.Secondly, we define two linear operators as(55)N1Θx,t;q,Λx,t;q=D0,tαcΘx,t;q+c02b0D0,xβcΛx,t;q,N2Λx,t;q,Θx,t;q=D0,tαcΛx,t;q+b0D0,xβcΘx,t;q.Using the abovementioned definitions and Lemma2, similar to the method of the fractional Cauchy–Riemann equations, we get(56)v0x,t=Eβxβ,ρ0x,t=sinβxβ,v1x,t=hc02b0cosβxβtαΓ1+α,ρ1x,t=hb0EβxβtαΓ1+α,v2x,t=1+hv1+h2c02Eβxβt2αΓ1+2α,ρ2x,t=1+hρ1−h2c02sinβxβt2αΓ1+2α,v3x,t=1+hv2+c02b0I0αD0,tαc1+hρ1−h3c04b0cosβxβt3αΓ1+3α,ρ3x,t=1+hρ2+b0I0αD0,tαc1+hv1+h3c02b0Eβxβt3αΓ1+3α,v4x,t=1+hv3+1+h⋯+h4c04Eβxβt4αΓ1+4α,ρ4x,t=1+hρ3+1+h⋯+h4c04sinβxβt4αΓ1+4α,⋯.By repeating this procedure forh=−1, noting Definitions 6 and 7, we obtain the exact solutions:(57)vx,t=∑k=0∞v2kx,t+∑k=1∞v2k−1x,t,=Eβxβ1+c0tα2Γ1+2α+c0tα4Γ1+4α+⋯−c0b0cosβxβc0tαΓ1+α−c0tα3Γ1+3α+⋯,=Eβxβcoshαc0tα−c0b0cosβxβsinαc0tαρx,t=∑k=0∞ρ2kx,t+∑k=1∞ρ2k−1x,t,=sinβxβ1−c0tα2Γ1+2α+c0tα4Γ1+4α+⋯−b0c0Eβxβc0tαΓ1+α+c0tα3Γ1+3α+⋯,=sinβxβcosαc0tα−b0c0Eβxβsinhαc0tα.More interestingly, whenα=1 and β=1,(58)vx,t=excoshc0t−c0b0cosxsinc0t,ρx,t=sinxcosc0t−b0c0exsinhc0t,are the exact solutions of the following equations:(59)vtx,t+c02b0ρxx,t=0,ρtx,t+b0vxx,t=0,with initial conditions vx,0=ex, ρx,0=sinx. In Figures 13–16, the absolute errors of exact solutions and approximate solutions for vx,t and ρx,t with differential fractional order α and β are shown, which indicates the high efficiency and precision of this method. In the process of compiling the program, we calculate sinhα,βx and coshα,βx by the formulas sinhα,βZ=Eα,βZ−Eα,β−Z/2 and coshα,βZ=Eα,βZ+Eα,β−Z/2, respectively.Figure 13
Absolute error of the exact solution and approximate solutionvx,t to fractional acoustic wave equation (2) with α=β=1.Figure 14
Absolute error of the exact solution and approximate solutionvx,t to fractional acoustic wave equation (2) with α=β=0.9.Figure 15
Absolute error of the exact solution and approximate solutionρx,t to fractional acoustic wave equation (2) with α=β=1.Figure 16
Absolute error of the exact solution and approximate solutionρx,t to fractional acoustic wave equations with (2) α=β=0.9.


## 4.3. Two-Dimensional Space Partial Differential Equation with a Time-Fractional-Order

In this section, we use the HAM to solve the two-dimensional space partial differential equation with a time-fractional-order of the following form:(60)D0,tαcux,y,t=1φ2+ψ2Δux,y,t,with initial conditions(61)ux,y,0=cos2πφxcos2πψy,utx,y,0=0,and boundary conditions(62)u0,y,t=uφ,y,t=ux,0,t=ux,ψ,t=1,where 1<α≤2. First, we choose the linear fractional order operator(63)LΘx,y,t;q=D0,tαcΘx,y,t;q.Secondly, we define the linear operator as(64)NΘx,y,t;q=D0,tαcΘx,y,t;q−Θxx+Θyyφ2+ψ2.Using the abovementioned definitions and with assumptionHx,t=1, we construct the zeroth-order deformation equation(ZDE):(65)1−qLΘx,y,t;q−u0x,y,t=qhNΘx,y,t;q.Obviously, whenq=0 and q=1, it holds(66)Θx,y,t;0=u0x,y,t=ux,y,0=cos2πφxcos2πψy,φx,y,t;1=ux,y,t.Thirdly, differentiating the ZDEm times with respect to q, then setting q=0, and dividing it by m!, we get the mth-order deformation equation:(67)Lum−χmum−1=hNRmum−1⟶,where(68)NRmum−1⟶=D0,tαcum−1−um−1,xx+um−1,yyφ2+ψ2,um=1m!∂mφx,y,t;q∂qmq=0,χm=0,m=1,1,m>1.Finally, operating the operatorL−1 on both the sides of equation (67), we have(69)um=χmum−1+hL−1NRmum−1⟶.Calculating one by one, we get(70)u0=cos2πφxcos2πψy,u1=h2πφψ2cos2πφxcos2πψytαΓ1+α,u2=1+hu1+h22πφψ4cos2πφxcos2πψyt2αΓ1+2α,u3=1+hu2−hφ2+ψ2I0α1+hu1,xx+1+hu1,yy+h32πab6cos2πφxcos2πψyt3αΓ1+3α,u4=1+hu3−hI0α1+hu2,xx+1+hu2,yyφ2+ψ2+h2φ2+ψ22I0αI0α1+hu1,xx+1+hu1,yyxx+I0α1+hu1,xx+1+hu1,yyyy+h42πφψ8cos2πφxcos2πψyt4αΓ1+4α,⋯.By repeating this procedure forh=−1, we get the exact solution,(71)ux,y,t=∑k=0∞uk=∑k=0∞u2k+∑k=1∞u2k−1,=cos2πφxcos2πψycoshα2πφψ2tα−cos2πφxcos2πψysinhα2πφψ2tα.for example 3. We omit the images of example 3.


## 5. Conclusions

In this paper, using the HAM, we obtained exact solutions for the fractional Cauchy–Riemann equations, fractional acoustic wave equations, and partial differential equation with a time-fractional-order. The absolute errors of the approximate solutions obtained by the HAM show that the approximate solutions are in good agreement with the exact solutions. Fractional differential equations are the generalization of integral ones. In fact, fractional models provide us with an adjustable parameter, which is the order of derivatives.



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