# Applications of OHAM and MOHAM for Fractional Seventh-Order SKI Equations

Authors: Jafar Biazar; Saghi Safaei
Journal: Journal of Applied Mathematics
 (2021)

---

## Abstract

In this article, a comparative study between optimal homotopy asymptotic method and multistage optimal homotopy asymptotic method is presented. These methods will be applied to obtain an approximate solution to the seventh-order Sawada-Kotera Ito equation. The results of optimal homotopy asymptotic method are compared with those of multistage optimal homotopy asymptotic method as well as with the exact solutions. The multistage optimal homotopy asymptotic method relies on optimal homotopy asymptotic method to obtain an analytic approximate solution. It actually applies optimal homotopy asymptotic method in each subinterval, and we show that it achieves better results than optimal homotopy asymptotic method over a large interval; this is one of the advantages of this method that can be used for long intervals and leads to more accurate results. As far as the authors are aware that multistage optimal homotopy asymptotic method has not been yet used to solve fractional partial differential equations of high order, we have shown that this method can be used to solve these problems. The convergence of the method is also addressed. The fractional derivatives are described in the Caputo sense.


---

## Body

## 1. Introduction

Some phenomena in many disciplines are usually modeled by fractional differential equations or fractional integrodifferential equations. Fractional differential equations have been solved by some series methods, homotopy analysis transform [1], Adomian decomposition [2–4], fractional Jacobi collocation [5], variational iteration [6, 7], differential transform [8, 9], homotopy perturbation (HP) [10–12], homotopy analysis (HA) [13–15], least squares [16], and others [17–21].Perturbation methods are used for solving problems in sciences and engineering [10–15]; however, most perturbation techniques require a small parameter in the equation. An improper choice of the parameter may leads to very bad results. Solutions obtained through perturbation methods can only be valid when a small amount of the parameter is used. Therefore, it is necessary to check the validity of the approximations through numerical processes. In 2008, Marinca and Herisanu introduced a new analytic method known as the optimal homotopy asymptotic method or for short (OHAM) [22]. An advantage of OHAM, in comparison with HAM, is that there is no need of h-curves study. This method provides us with a convenient way to control the convergence of the solution series and allows the adjustment of the convergence region, wherever it is needed. Several authors have demonstrated the effectiveness, generalizability, and reliability of this method. Another advantage of OHAM is built in convergence a criterion that is controllable. OHAM results in to satisfactory solutions on short domains, but when the interval becomes longer, the accuracy of the method decreases, so a new approach was proposed by Anakira et al., which is called multistage optimal homotopy asymptotic method (MOHAM) that is suitable for analytic approximate solutions for any long interval [23, 24]. Finally, the approximate solutions obtained from both methods are compared with the exact solution.


## 2. Basic Definitions

In this section, some basic definitions of fractional calculus are explained briefly [25].Definition 1.
A real-valued functionft with t>0 can be defined on the space Cμ, μ∈ℝ, if there is a real number ρ>μ such that ft=tρf1t, where f1t∈0,+∞ and it is defined on the space Cμn, if and only if fnt∈Cμ, for n ∈ ℕ.Definition 2.
The Riemann–Liouville fractional integral operator of orderα>0, for a continuous function f∈Cμ,μ>−1, is defined as
(1)Jαft=1Γα∫0tt−sα−1fsds,α>0,J0ft=ft.112121Some main properties of the operatorJα are listed below.Forf∈Cμ,μ>−1,α,β≥0, and γ≥−1,(2)JαJβft=Jα+βft,JαJβft=JβJαft,Jαtγ=Γγ+1Γα+γ+1tα+γ.Definition 3.
The fractional derivative offt in Caputo sense is defined as the following:
(3)Dαft=1Γm−α∫0tt−sm−α−1fmsds,for m−1<α≤m,m∈ℕ,t>0,f∈C−1m.Lemma 4.
Ifm−1<α≤m,m∈ℕ,t>0,f∈C−1m,μ≥−1, then the following two properties will be set:
(4)DαJαft=ft,JαDαft=ft−∑k=0m−1fk0+tkk!.


## 3. The Proposed Techniques

In this section, first the basic idea of OHAM is discussed; then, MOHAM is introduced with its convergence analysis will be stated.


### 3.1. OHAM

Let us consider the following fractional equation with the boundary condition:(5)Luξ+Nuξ+fξ=0.Buξ,dudξ=0,whereξ is an independent variable L,N,f, and uξ represent a linear operator, a nonlinear operator, a known function, and an unknown function, respectively. Also, Buξ stands for a boundary operator.An optimal convex homotopyHφξ;p,ci:Ω×0,1⟶ℝ can be constructed as follows:
(6)Hφξ;p,ci=1−pLφξ;p,ci+fξ=Hξ,p,ciLφξ;p,ci+fξ+Nφξ;p,ci,Bφξ;p.,ci=0,Let us consider the following zeroth-order deformation equation(7)1−pLφξ,p+fξ=HpLφξ,p+fξ+Nφξ,p,Bφξ,p,dφξ.pdξ=0,wherep is an embedding parameter in the interval 0,1, Hξ.p is an auxiliary function with nonzero and zero outputs for p≠0 and p=0, respectively, u0ξ represents the initial condition of uξ, and φξ,p is an unknown function. By inserting p=0 and 1 into Equation (7), one can obtain the following boundary conditions:
(8)φξ,0=u0ξ,(9)φξ,1=uξ.Therefore,φξ,p will change continuously from the initial guess to the solution, u0ξ to uξ when p increases from 0 to1.Puttingp=0 into Equation (7), the initial solution u0ξ is determined as a solution for the problem
(10)Lu0ξ+fξ=0,Bu0,du0dξ=0.Next, choose an auxiliary functionHp as the following form:
(11)Hp=pc1+p2c2+p3c3+⋯,wherec1,c2,c3,⋯ are called the convergence control parameter that will be determined shortly. The auxiliary function Hp can be expressed in other forms, see, for exampl,e Herisanu and Marinca [7].Expansion ofφξ;p,ci, in Taylor series aboutp, reads
(12)φξ;p,ci=u0ξ+∑m=1∞umξ,c1,c2,⋯,cmpm.Substituting following Equations (14)–(18) into (7), and equating the coefficients of the terms with identical powers of p, one will obtain the governing equation of the initial approximation u0ξ, given by Equation (8); then, the governing equation of the first order problem is defined as follows:
(13)Lu1ξ+fξ=c1N0u0ξ,Bu1,du1dξ=0.The governing equation of themth-order is defined as
(14)Lumξ−Lum−1ξ=cmN0u0ξ+∑j=1m−1cjLum−jξ+Nm−ju0ξ,u1ξ,⋯,um−1ξBum,dumdξ=0,m=2,3,⋯,where Nmu0ξ,u1ξ,⋯,umξ is the coefficient of pm in the expansion of Nφξ,p about the embedding parameter p(15)Nφξ,p=N0u0ξ+∑m=1∞Nmu0ξ,u1ξ,⋯,umξpm.Solving Equation (14) gives various approximate solutions umξ,c1,c2,c3,⋯.,cm, but there exist still m unknowns, auxiliary parameters, c1,c2,c3,⋯.,cm in the obtained solutions. It is assumed that the auxiliary parameters c1,c2,c3,⋯.,cm, the linear operator L, and the deformation equation of the zeroth order, (10), are appropriately determined in order to ensure the convergence of series (12), at p=1. Hence, substituting Equations (8) and (9) into Equation (12) for p=1 gives the solution uξ as follows:
(16)uξ,c1,c2,c3,⋯=u0ξ+∑m=1∞umξ,c1,c2,c3,⋯,cm.The approximate solution of Equation (5) will be as follows:
(17)u^ξ,ci=u0ξ+∑k=1muiξ,c1,c2,c3,⋯,ck.Substituting Equation (17) in Equation (5), results in the following residual:
(18)Rξ,ci=Lu^ξ,ci+Nu^ξ.,ci+fξ,i=1,2,3,⋯.,m.By consideringRξ,ci=0, the exact solution is u^ξ,ci. However, such a case could not be true for a non-linear equation. By least square technique, the functional Jci should be minimized
(19)Jci=∫abR2ξ,cidξ,where a and b are the bounds of the interval in hand. The optimal values of the unknown coefficients cii=1,2,⋯,m can be determined based on the following conditions:
(20)∂J∂c1=∂J∂c2=∂J∂c3=⋯=∂J∂cm=0.In order to get an analytic approximate solution at the levelm, the obtained optimal coefficients will be substituted in Equation (17).


### 3.2. MOHAM

Although the OHAM is used to obtain approximate solutions of problems, it has some disadvantage in nonlinear problems with long domains. To control this drawback, let us introduce the multistage OHAM to obtain the nonlinear problem in some small intervals. A simple way to confirm the validity of the approximate solutions of a large domain,T is by dividing the interval 0,T in to subinterval as t0,t1,…, tj−1,tj, where tj=T and utilizing OHAM on each subinterval. The initial approximation in each interval is given from the solution in the last interval. First, by considering the following initial condition:
(21)yitj=αi.Then, we can obtain the initial approximationyi,0t=α and the following zero-order equation:
(22)1−pLφit,p−yi,0t=ΗipLφit,p+fit+Niφit,p.Next, we pick out the auxiliary functionΗip in the form
(23)Ηip=C1,jp+C2,jp2+C3,jp3+⋯,or
(24)Ηip=C1,j+C2,jt+C3,jt2+⋯p.Then, the first, second andmth-order-approximate solutions can be generated subject to initial conditions yi,1tj=yi,2tj=⋯=yi,mtj=0 and the approximate solutions as follows:
(25)y~it,C1,j,C2,j,⋯,Cm,j=yi,0t+∑k=1myi,kt,C1,j,C2,j,⋯,Ck,j.Substituting Equation (25) into Equation (5) yields the following residual:
(26)Rit,C1,j,C2,j,⋯,Cm,j=Ly~it,C1,j,C2,j,⋯,Cm,j+fit+Niy~it,C1,j,C2,j,⋯,Cm,j.IfRi=0, then y~i will be the exact solution. Generally, such a case will not arise for nonlinear problems, but we can minimize the functional
(27)JiC1,j,C2,j,⋯,Cm,j=∫tjtj+hRi2t,C1,j,C2,j,⋯,Cm,jdt,where h is the length of the subinterval tj,tj+1 and N=T/h is the number of subinterval. Now, we can solve Equtaion (27) for j=0,1,⋯.,N with changing the initial approximation αi in each subinterval from the solution in the last point of the prior interval. The unknown convergence control parameters Ci,ji=1,2,3,⋯,m,j=1,2,3,⋯,N can be obtained by solving the following system of equations:
(28)∂Ji∂C1.j=∂Ji∂C2.j=⋯=∂Ji∂Cm.j=0.Thus, the analytic approximate solution will be as follows:(29)y~t=y~1t,t0≤t<t1,y~2t,t1≤t<t2,⋮y~Nt,tN−1≤t≤T.In such a way, we successfully gain the solution of the initial value problem for a large intervalT. It should be noted that if j=0, then the MOHAM reduces to OHAM. One of the benefits of MOHAM is that it provides a simple way to control the convergence, regulate convergence region, and adjust the convergence region though the auxiliary function Hip, involving several convergent control parameters Ci,j′. In general, this method eliminates the difficulty of finding approximate solutions in large ranges.Theorem 5.
If the series (17) converges to uξ, where unξ∈LR+ (linear and continuous functions on real numbers whose absolute value integral is finite), is produced by Equations (10)–(13) and satisfies the nth-order deformation equation (14), then uξ is the exact solution of (5).Proof.
This theorem is proved by Liao, in his valuable book [26]; here, we prove it with more details. Since the series ∑n=1∞unξ,c1,c2,⋯,cn is convergent, it can be written as follows:
(30)Sξ=∑n=1∞unξ,c1,c2,⋯,cn.
Therefore, we have(31)limn⟶∞unξ,c1,c2,⋯,cn=0.
The left hand-side of (14) satisfies
(32)u1ξ,c1+∑j=2nujξ,c⟶j−∑j=2nuj−1ξ,c⟶j−1=unξ,c⟶n.
According to (32) and the limit, we have
(33)u1ξ,c1+∑j=2∞ujξ,c⟶j−∑j=2∞uj−1ξ,c⟶j−1=limn⟶∞unξ,c⟶n=0.
Applying the linear operator,(34)Lu1ξ,c1+L∑j=2∞ujξ,c⟶j−L∑j=2∞uj−1ξ,c⟶j−1=0.
Equation (34) can be written as the following:
(35)Lu1ξ,c1+∑j=2∞Lujξ,c⟶j−Luj−1ξ,c⟶j−1=c1N0u0+fξ+∑j=2∞cjN0u0+∑k=1j−1ckLuj−k+Nj−ku⟶j−1=c1N0u0+fξ+∑j=2∞cjN0u0+∑j=2∞∑k=1j−1ckLuj−k+Nj−ku⟶j−1=∑j=2∞∑k=1j−1ckLuj−k+∑j=2∞∑k=1j−1cjN0u0+ckNj−ku⟶j−1+fξ=0.
So by choosing the optimalck,k=1,2,⋯, Equation (35) is converted to the following:
(36)Luξ+Nuξ+fξ=0,
which is the exact solution to the problem.


## 3.1. OHAM

Let us consider the following fractional equation with the boundary condition:(5)Luξ+Nuξ+fξ=0.Buξ,dudξ=0,whereξ is an independent variable L,N,f, and uξ represent a linear operator, a nonlinear operator, a known function, and an unknown function, respectively. Also, Buξ stands for a boundary operator.An optimal convex homotopyHφξ;p,ci:Ω×0,1⟶ℝ can be constructed as follows:
(6)Hφξ;p,ci=1−pLφξ;p,ci+fξ=Hξ,p,ciLφξ;p,ci+fξ+Nφξ;p,ci,Bφξ;p.,ci=0,Let us consider the following zeroth-order deformation equation(7)1−pLφξ,p+fξ=HpLφξ,p+fξ+Nφξ,p,Bφξ,p,dφξ.pdξ=0,wherep is an embedding parameter in the interval 0,1, Hξ.p is an auxiliary function with nonzero and zero outputs for p≠0 and p=0, respectively, u0ξ represents the initial condition of uξ, and φξ,p is an unknown function. By inserting p=0 and 1 into Equation (7), one can obtain the following boundary conditions:
(8)φξ,0=u0ξ,(9)φξ,1=uξ.Therefore,φξ,p will change continuously from the initial guess to the solution, u0ξ to uξ when p increases from 0 to1.Puttingp=0 into Equation (7), the initial solution u0ξ is determined as a solution for the problem
(10)Lu0ξ+fξ=0,Bu0,du0dξ=0.Next, choose an auxiliary functionHp as the following form:
(11)Hp=pc1+p2c2+p3c3+⋯,wherec1,c2,c3,⋯ are called the convergence control parameter that will be determined shortly. The auxiliary function Hp can be expressed in other forms, see, for exampl,e Herisanu and Marinca [7].Expansion ofφξ;p,ci, in Taylor series aboutp, reads
(12)φξ;p,ci=u0ξ+∑m=1∞umξ,c1,c2,⋯,cmpm.Substituting following Equations (14)–(18) into (7), and equating the coefficients of the terms with identical powers of p, one will obtain the governing equation of the initial approximation u0ξ, given by Equation (8); then, the governing equation of the first order problem is defined as follows:
(13)Lu1ξ+fξ=c1N0u0ξ,Bu1,du1dξ=0.The governing equation of themth-order is defined as
(14)Lumξ−Lum−1ξ=cmN0u0ξ+∑j=1m−1cjLum−jξ+Nm−ju0ξ,u1ξ,⋯,um−1ξBum,dumdξ=0,m=2,3,⋯,where Nmu0ξ,u1ξ,⋯,umξ is the coefficient of pm in the expansion of Nφξ,p about the embedding parameter p(15)Nφξ,p=N0u0ξ+∑m=1∞Nmu0ξ,u1ξ,⋯,umξpm.Solving Equation (14) gives various approximate solutions umξ,c1,c2,c3,⋯.,cm, but there exist still m unknowns, auxiliary parameters, c1,c2,c3,⋯.,cm in the obtained solutions. It is assumed that the auxiliary parameters c1,c2,c3,⋯.,cm, the linear operator L, and the deformation equation of the zeroth order, (10), are appropriately determined in order to ensure the convergence of series (12), at p=1. Hence, substituting Equations (8) and (9) into Equation (12) for p=1 gives the solution uξ as follows:
(16)uξ,c1,c2,c3,⋯=u0ξ+∑m=1∞umξ,c1,c2,c3,⋯,cm.The approximate solution of Equation (5) will be as follows:
(17)u^ξ,ci=u0ξ+∑k=1muiξ,c1,c2,c3,⋯,ck.Substituting Equation (17) in Equation (5), results in the following residual:
(18)Rξ,ci=Lu^ξ,ci+Nu^ξ.,ci+fξ,i=1,2,3,⋯.,m.By consideringRξ,ci=0, the exact solution is u^ξ,ci. However, such a case could not be true for a non-linear equation. By least square technique, the functional Jci should be minimized
(19)Jci=∫abR2ξ,cidξ,where a and b are the bounds of the interval in hand. The optimal values of the unknown coefficients cii=1,2,⋯,m can be determined based on the following conditions:
(20)∂J∂c1=∂J∂c2=∂J∂c3=⋯=∂J∂cm=0.In order to get an analytic approximate solution at the levelm, the obtained optimal coefficients will be substituted in Equation (17).


## 3.2. MOHAM

Although the OHAM is used to obtain approximate solutions of problems, it has some disadvantage in nonlinear problems with long domains. To control this drawback, let us introduce the multistage OHAM to obtain the nonlinear problem in some small intervals. A simple way to confirm the validity of the approximate solutions of a large domain,T is by dividing the interval 0,T in to subinterval as t0,t1,…, tj−1,tj, where tj=T and utilizing OHAM on each subinterval. The initial approximation in each interval is given from the solution in the last interval. First, by considering the following initial condition:
(21)yitj=αi.Then, we can obtain the initial approximationyi,0t=α and the following zero-order equation:
(22)1−pLφit,p−yi,0t=ΗipLφit,p+fit+Niφit,p.Next, we pick out the auxiliary functionΗip in the form
(23)Ηip=C1,jp+C2,jp2+C3,jp3+⋯,or
(24)Ηip=C1,j+C2,jt+C3,jt2+⋯p.Then, the first, second andmth-order-approximate solutions can be generated subject to initial conditions yi,1tj=yi,2tj=⋯=yi,mtj=0 and the approximate solutions as follows:
(25)y~it,C1,j,C2,j,⋯,Cm,j=yi,0t+∑k=1myi,kt,C1,j,C2,j,⋯,Ck,j.Substituting Equation (25) into Equation (5) yields the following residual:
(26)Rit,C1,j,C2,j,⋯,Cm,j=Ly~it,C1,j,C2,j,⋯,Cm,j+fit+Niy~it,C1,j,C2,j,⋯,Cm,j.IfRi=0, then y~i will be the exact solution. Generally, such a case will not arise for nonlinear problems, but we can minimize the functional
(27)JiC1,j,C2,j,⋯,Cm,j=∫tjtj+hRi2t,C1,j,C2,j,⋯,Cm,jdt,where h is the length of the subinterval tj,tj+1 and N=T/h is the number of subinterval. Now, we can solve Equtaion (27) for j=0,1,⋯.,N with changing the initial approximation αi in each subinterval from the solution in the last point of the prior interval. The unknown convergence control parameters Ci,ji=1,2,3,⋯,m,j=1,2,3,⋯,N can be obtained by solving the following system of equations:
(28)∂Ji∂C1.j=∂Ji∂C2.j=⋯=∂Ji∂Cm.j=0.Thus, the analytic approximate solution will be as follows:(29)y~t=y~1t,t0≤t<t1,y~2t,t1≤t<t2,⋮y~Nt,tN−1≤t≤T.In such a way, we successfully gain the solution of the initial value problem for a large intervalT. It should be noted that if j=0, then the MOHAM reduces to OHAM. One of the benefits of MOHAM is that it provides a simple way to control the convergence, regulate convergence region, and adjust the convergence region though the auxiliary function Hip, involving several convergent control parameters Ci,j′. In general, this method eliminates the difficulty of finding approximate solutions in large ranges.Theorem 5.
If the series (17) converges to uξ, where unξ∈LR+ (linear and continuous functions on real numbers whose absolute value integral is finite), is produced by Equations (10)–(13) and satisfies the nth-order deformation equation (14), then uξ is the exact solution of (5).Proof.
This theorem is proved by Liao, in his valuable book [26]; here, we prove it with more details. Since the series ∑n=1∞unξ,c1,c2,⋯,cn is convergent, it can be written as follows:
(30)Sξ=∑n=1∞unξ,c1,c2,⋯,cn.
Therefore, we have(31)limn⟶∞unξ,c1,c2,⋯,cn=0.
The left hand-side of (14) satisfies
(32)u1ξ,c1+∑j=2nujξ,c⟶j−∑j=2nuj−1ξ,c⟶j−1=unξ,c⟶n.
According to (32) and the limit, we have
(33)u1ξ,c1+∑j=2∞ujξ,c⟶j−∑j=2∞uj−1ξ,c⟶j−1=limn⟶∞unξ,c⟶n=0.
Applying the linear operator,(34)Lu1ξ,c1+L∑j=2∞ujξ,c⟶j−L∑j=2∞uj−1ξ,c⟶j−1=0.
Equation (34) can be written as the following:
(35)Lu1ξ,c1+∑j=2∞Lujξ,c⟶j−Luj−1ξ,c⟶j−1=c1N0u0+fξ+∑j=2∞cjN0u0+∑k=1j−1ckLuj−k+Nj−ku⟶j−1=c1N0u0+fξ+∑j=2∞cjN0u0+∑j=2∞∑k=1j−1ckLuj−k+Nj−ku⟶j−1=∑j=2∞∑k=1j−1ckLuj−k+∑j=2∞∑k=1j−1cjN0u0+ckNj−ku⟶j−1+fξ=0.
So by choosing the optimalck,k=1,2,⋯, Equation (35) is converted to the following:
(36)Luξ+Nuξ+fξ=0,
which is the exact solution to the problem.


## 4. Numerical Result

Consider time-fractional SKI equation as [27]
(37)Dtαu+252u3ux+63ux3+378uuxuxx+126u2uxxx+63uxxuxxx+42uxuxxxx+21uxxxxx+uxxxxxxx=0,t>0,0<α≤1.with initial condition
(38)ux,0=1752−3tanh20.1x.The exact solution forα=1 is
(39)ux,t=1752−3tanh20.1x−6446875t.


### 4.1. Approximate Solution by OHAM

Choosing the linear operatorL=Dtα, and applying the algorithm based on OHAM, as presented in Section 3, the initial guess u0x,t will be obtained as Dtαu0x,t=0, withux,o=1/752−3tanh20.1x.We obtain(40)u0x,t=1752−3tanh20.1x.Substitutingm=1,2,3,⋯ successively, the problems of various orders are as follows:
(41)u1x,t=c143690666665tanhx/10t63999999997558594−6990506665tanh3x/10t10239999997558594+tanh5x/10t50000000000000000000000−tanh7x/10t33333333333333333333333−tanh9x/10t100000000000000000000000.In this paper, three terms approximation ofux,t is considered. By substituting zeroth-, first-, and second-order solutions in (7), and by using technique in Section 3, we find the following values for c1 and c2:
(42)c1=−7.989551225007650511610−15,c2=−7.7704946584917721471.


### 4.2. Approximate Solution by MOHAM

In this section, we utilize MOHAM for Equation (37). We will consider the auxiliary function Hp,t as the following:
(43)Hp,t=c1,j+c2,jtp.As in OHAM, we obtain two approximate solutions to Equation (37):
(44)u~jx,t=u0jx,t+u1jx,t.This approach leads to the following sequence of equations:(45)p0:Dtαu0jx,t=αj,p1:Dtαu1jx,t=−c1j252u0j3x,tDxu0jx,t+63Dx3u0jx,t+378u0jx,tDxu0jx,tDxxu0jx,t+126u0j2x,tDxxxu0jx,t+63Dxxu0jx,tDxxxu0jx,t+42Dxu0jx,tDxxxxu0jx,t+21Dxxxxxu0jx,t+Dxxxxxxxu0jx,t,By applying the least squares technique, the parametersc1,j and c2,j are determined for α=1,h=0.5, and t from t0=0 up to t2=T=1. The results are presented in Table 1.Table 1
The values of the control parameterscij.
αjc1,jc2,jα=1j=1-4.656060764540397481 e -16-20.370432288203407083j=2-3.6335619254749103832 e-14-5.9545639368933960096Approximate solution forα=1 is in the following form:
(46)1.8×10−54tanh160.1x−8.9×10−54tanh140.1x+1.9×10−53tanh120.1x−2.1×10−53tanh100.1x+1.4×10−54tanh80.1x−5.2×10−54tanh60.1x−1.8×10−42tanh40.1x+2.5×10−42tanh20.1x−6.3×10−43tanh160.1xt2+2.04×10−22tanh90.1x−2.03×10−22tanh70.1x+1.8×10−21tanh50.1x+0.0000tanh30.1x−0.04tanh20.1x−0.00001tanh0.1xt+0.03,0<x<0.5,1.1×10−50tanh160.1x−5.5×10−50tanh140.1x+1.1×10−49tanh120.1x−1.3×10−49tanh100.1x+8.3×10−50tanh80.1x−3.2×10−50tanh60.1x−1.2×10−38tanh40.1x+1.5×10−38tanh20.1x−3.8×10−39t26×10−23tanh90.1x−6×10−23tanh70.1x+5.4×10−22tanh50.1x+0.000004tanh30.1x−0.000004tanh0.1xt−0.04tanh20.1x,0.5≤x≤1.From Figures1–3, one can see that the solutions obtained by OHAM and MOHAM are nearly identical with the exact solution.Figure 1
The results of OHAM, MOHAM, and exact solution forα=1 att=0.2.Figure 2
The results of OHAM, MOHAM, and exact solution forα=0.2 att=0.4.Figure 3
The results of OHAM, MOHAM, and exact solution forα=1.


## 4.1. Approximate Solution by OHAM

Choosing the linear operatorL=Dtα, and applying the algorithm based on OHAM, as presented in Section 3, the initial guess u0x,t will be obtained as Dtαu0x,t=0, withux,o=1/752−3tanh20.1x.We obtain(40)u0x,t=1752−3tanh20.1x.Substitutingm=1,2,3,⋯ successively, the problems of various orders are as follows:
(41)u1x,t=c143690666665tanhx/10t63999999997558594−6990506665tanh3x/10t10239999997558594+tanh5x/10t50000000000000000000000−tanh7x/10t33333333333333333333333−tanh9x/10t100000000000000000000000.In this paper, three terms approximation ofux,t is considered. By substituting zeroth-, first-, and second-order solutions in (7), and by using technique in Section 3, we find the following values for c1 and c2:
(42)c1=−7.989551225007650511610−15,c2=−7.7704946584917721471.


## 4.2. Approximate Solution by MOHAM

In this section, we utilize MOHAM for Equation (37). We will consider the auxiliary function Hp,t as the following:
(43)Hp,t=c1,j+c2,jtp.As in OHAM, we obtain two approximate solutions to Equation (37):
(44)u~jx,t=u0jx,t+u1jx,t.This approach leads to the following sequence of equations:(45)p0:Dtαu0jx,t=αj,p1:Dtαu1jx,t=−c1j252u0j3x,tDxu0jx,t+63Dx3u0jx,t+378u0jx,tDxu0jx,tDxxu0jx,t+126u0j2x,tDxxxu0jx,t+63Dxxu0jx,tDxxxu0jx,t+42Dxu0jx,tDxxxxu0jx,t+21Dxxxxxu0jx,t+Dxxxxxxxu0jx,t,By applying the least squares technique, the parametersc1,j and c2,j are determined for α=1,h=0.5, and t from t0=0 up to t2=T=1. The results are presented in Table 1.Table 1
The values of the control parameterscij.
αjc1,jc2,jα=1j=1-4.656060764540397481 e -16-20.370432288203407083j=2-3.6335619254749103832 e-14-5.9545639368933960096Approximate solution forα=1 is in the following form:
(46)1.8×10−54tanh160.1x−8.9×10−54tanh140.1x+1.9×10−53tanh120.1x−2.1×10−53tanh100.1x+1.4×10−54tanh80.1x−5.2×10−54tanh60.1x−1.8×10−42tanh40.1x+2.5×10−42tanh20.1x−6.3×10−43tanh160.1xt2+2.04×10−22tanh90.1x−2.03×10−22tanh70.1x+1.8×10−21tanh50.1x+0.0000tanh30.1x−0.04tanh20.1x−0.00001tanh0.1xt+0.03,0<x<0.5,1.1×10−50tanh160.1x−5.5×10−50tanh140.1x+1.1×10−49tanh120.1x−1.3×10−49tanh100.1x+8.3×10−50tanh80.1x−3.2×10−50tanh60.1x−1.2×10−38tanh40.1x+1.5×10−38tanh20.1x−3.8×10−39t26×10−23tanh90.1x−6×10−23tanh70.1x+5.4×10−22tanh50.1x+0.000004tanh30.1x−0.000004tanh0.1xt−0.04tanh20.1x,0.5≤x≤1.From Figures1–3, one can see that the solutions obtained by OHAM and MOHAM are nearly identical with the exact solution.Figure 1
The results of OHAM, MOHAM, and exact solution forα=1 att=0.2.Figure 2
The results of OHAM, MOHAM, and exact solution forα=0.2 att=0.4.Figure 3
The results of OHAM, MOHAM, and exact solution forα=1.


## 5. Conclusion

In this study, the optimal homotopy asymptotic method and the multistage optimal homotopy asymptotic method are used to derive an analytic approximate solution to the time-fractional seventh order Sawada-Kotera-Ito equation. The results obtained from these methods show that multistage optimal homotopy asymptotic method converges better than the optimal homotopy asymptotic method. One observes that the results agree very well with the exact solution. The multistage optimal homotopy asymptotic method by dividing the interval0,T into some subintervals can obtain more exact solution than the optimal homotopy asymptotic method. The disadvantage of the multistage optimal homotopy asymptotic method is that it is time consuming and requires more time to solve problems than the optimal homotopy asymptotic method. As far as the authors are aware, the multistage optimal homotopy asymptotic method has not been used to solve fractional partial differential equations yet. In this study, the method has been tested on fractional PDE and yields to satisfactory results. Figure 4 and Table 2 expose that the multistage optimal homotopy asymptotic method results to more accurate solution as compared to the optimal homotopy asymptotic method. The convergence of the method is addressed.Figure 4
The absolute error of OHAM and MOHAM, forα=1 andt=0.2.Table 2
Absolute error of OHAM and MOHAM forα=1 and t=0.2.
x00.20.40.60.81Abs(exact-OHAM)1.17e−132.4e−84.78e−87.15e−89.5e−81.18e−7Abs(exact-MOHAM)1.17e−135.8e−81.16e−85.7e−87.5e−89.4e−8



---

*Source: 6898282-2021-12-31.xml*