solving nonlinear fractional partial differential ... · solving nonlinear fractional partial...

Solving Nonlinear Fractional Partial DifferentialEquations Using the Homotopy Analysis MethodMehdi Dehghan,1 Jalil Manafian,1 Abbas Saadatmandi2

1Department of Applied Mathematics, Faculty of Mathematics and Computer Science,Amirkabir University of Technology, Tehran 15914, Iran

2Department of Mathematics, Faculty of Science, University of Kashan, Kashan, Iran

Received 30 September 2008; accepted 20 January 2009Published online 2 July 2009 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/num.20460

In this article, the homotopy analysis method is applied to solve nonlinear fractional partial differentialequations. On the basis of the homotopy analysis method, a scheme is developed to obtain the approx-imate solution of the fractional KdV, K(2, 2), Burgers, BBM-Burgers, cubic Boussinesq, coupled KdV,and Boussinesq-like B(m, n) equations with initial conditions, which are introduced by replacing someinteger-order time derivatives by fractional derivatives. The homotopy analysis method for partial differen-tial equations of integer-order is directly extended to derive explicit and numerical solutions of the fractionalpartial differential equations. The solutions of the studied models are calculated in the form of convergentseries with easily computable components. The results of applying this procedure to the studied cases showthe high accuracy and efficiency of the new technique. © 2009 Wiley Periodicals, Inc. Numer Methods PartialDifferential Eq 26: 448–479, 2010

Keywords: analytical solution; coupled KdV and Boussinesq-like B(m, n) equations; fractional KdV,K(2, 2), Burgers, BBM-Burgers, cubic Boussinesq; fractional partial differential equations (FPDEs);homotopy analysis method (HAM)

I. INTRODUCTION

As mentioned by several researchers in fractional calculus, derivatives of noninteger order arevery effective for the description of many physical phenomena such as rheology, damping laws,and diffusion process [1–5]. Some fundamental works on various aspects of the fractional cal-culus are given by Abbasbandy [6], Caputo [7], Debanth [8], Diethelm et al. [9], Hayat et al.[10], Jafari and Seifi [11, 12], Kemple and Beyer [13], Kilbas and Trujillo [14], Kiryakova [15],Miller and Ross [2], Momani and Shawagfeh [16], Oldham and Spanier [17], Podlubny [3], etc.Several methods have been used to solve fractional partial differential equations, such as Laplacetransform method [3], Fourier transform method [13], Adomian’s decomposition method (ADM)[16, 18] and so on. We refer the interested reader to [19–22] to study the main idea behind

Correspondence to: Mehdi Dehghan, Department of Applied Mathematics, Faculty of Mathematics and Computer Science,Amirkabir University of Technology, Tehran 15914, Iran (e-mail: [email protected])

© 2009 Wiley Periodicals, Inc.

NONLINEAR FRACTIONAL PARTIAL DIFFERENTIAL EQUATIONS 449

Adomian’s decomposition approach. A substantial amount of research work has been directed forthe study of the nonlinear fractional KdV, K(2, 2), Burgers, cubic Boussinesq and BBM-Burgers,coupled KdV and Boussinesq-like B(m, n) equations given by

Dαt u − 3(u2)x + uxxx = 0, (1.1)

Dαt u + (u2)x + (u2)xxx = 0, (1.2)

Dαt u + 1

2(u2)x − uxx = 0, (1.3)

D2αt u − uxx + 2(u3)xx − uxxxx = 0, (1.4)

Dαt u + ux + uux − auxx − buxxt = 0, (1.5)

and

Dαt u = a[uxxx + 6u(u)x] + 2bv(v)x ,

Dαt v = −vxxx − 3uvx , (1.6)

and

D2αt u − (un)xx − (um)xxxx = 0, m, n > 1, (1.7)

respectively. In the case of α = 1, Equation (1.1) is the pioneering equation that gives rise tosolitary wave solutions. Solitons waves with infinite support are generated as a result of the bal-ance between the nonlinear convection (un)x and the linear dispersion uxxx in these equations[23]. Solitons are localized waves [24] that propagate without change of their shape and velocityproperties and are stable against mutual collisions [23].

The fractional K(n, n) equation

Dαt u + (un)x + (un)xxx = 0, (1.8)

in the case of α = 1, as pointed by [23], developed in [25], is the pioneering equation for com-pactons. In solitary waves theory, compactons are defined as solitons with finite wavelengths orsolitons free of exponential tails [25].

The Burgers’ equation appears in fluid mechanics. This equation incorporates both convectionand diffusion in fluid dynamics, and is used to describe the structure of shock waves [23].

The cubic Boussinesq equation, gives rise to solitons and appeared in the works of Priestlyand Clarkson [26]. Kaya [27] examined this equation by using Adomian decomposition methoddeveloped in [18], and used thoroughly in [19–22, 28–32]. This equation has been investigatedfor solitary [30] waves and for rational [31] solutions as well.

The fractional Benjamin-Bona-Mahony-Burgers (BBM-Burgers) equation studied by Songand Zhang [33] by using the homotopy analysis method (HAM). Zhu and Xuan [34] investigatedBBM-Burgers equation with dissipative term.

The fractional KdV type of equations have been an important class of nonlinear evolution equa-tions with numerous applications in physical sciences and engineering fields. In plasma physicsthese equations give rise to the ion acoustic solutions [35]. In geophysical fluid dynamics, theydescribe a long wave in shallow seas and deep oceans [36].

Boussinesq-like B(m, n) equation appears in nonlinear dispersion in the formation of patternsin liquid drops, and gives many similarity reductions and a compaction solution. Author of [37]extended the decomposition method to seek more compacton solutions of Eq. (1.7) when m = n

Numerical Methods for Partial Differential Equations DOI 10.1002/num

450 DEHGHAN, MANAFIAN, AND SAADATMANDI

and α = 1. When m = 1, n = 2 and α = 1, Eq. (1.7) becomes the well-known Boussinesqequation. When m = 1, n = 3 and α = 1, Eq. (1.7) reduces to the modified Boussinesq equationwhich arises from the famous Fermi-Pasta-Ulam problem [38]. Although m = 1 and n ≥ 4, Eq.(1.7) is the higher-order modified Boussinesq equation. Authors of [39] investigated exact solitarysolutions with compact support for the nonlinear dispersive Boussinesq-like B(m, n) equations.Wazwaz [23], examined the KdV, the K(2, 2), the Burgers, and the cubic Boussinesq equationsby using variational iteration method (VIM) [40–51]. In the case of α = 1, the above equationsreduce to the classical nonlinear partial differential equations. Although there are a lot of stud-ies for the classical form of the above equations, it seems that detailed studies of the nonlinearfractional differential equation are only beginning.

In this work, the homotopy analysis method (HAM) developed by Liao in [52–61] will be usedto conduct an analytic study on the fractional KdV, K(2, 2), Burgers, BBM-Burgers, cubic Boussi-nesq, coupled KdV and Boussinesq-like B(m, n) equations. It is worth to point out that HAMsuccessfully applied to partial differential equations and extended by authors [11, 12, 33, 62–64]to solve different types of nonlinear partial differential equations [65]. The method gives rapidlyconvergent successive approximations of the exact solution if such a solution exists, otherwiseapproximations can be used for numerical purposes. The homotopy analysis method, a new ana-lytic technique is proposed to solve nonlinear partial differential equations with fractional order.The homotopy analysis method is useful to obtain exact and approximate solutions of linearand nonlinear partial differential equations. In this article, we illustrate the validity of the HAM[57–61] for the nonlinear fractional partial differential equations.

The current paper is organized as follows: In Section II, we describe basic definitions. InSection III, the homotopy analysis method will be introduced briefly and this technique will beapplied to solve fractional partial differential equations. Section IV contains some test problemsto show the efficiency and accuracy of the new method. Also a conclusion is given in Section V.Finally some references are given at the end of this paper.

II. BASIC DEFINITIONS

In this section, we give some definitions and properties of the fractional calculus [3].

Definition 1. A real function f (t), t > 0, is said to be in the space Cµ, µ ∈ R, if there exists areal number p > µ, such that f (t) = tpf1(t), where f1(t) ∈ C(0, ∞), and it is said to be in thespace Cn

µ, if and only if f (n) ∈ Cµ, n ∈ N [3].

Definition 2. The Riemann-Liouville fractional integral operator (J α) of order α ≥ 0, of afunction f ∈ Cλ, λ ≥ −1, is defined as [3]

J αf (t) = D−αf (t) = 1

�(α)

∫ t

0(t − τ)α−1f (τ)dτ , (α > 0), (2.1)

J 0f (t) = f (t), (2.2)

where �(z) is the well-known Gamma function. Some of the properties of the operator (J α), whichwe will need here, are given in the following:

For f ∈ Cλ, λ ≥ −1, α, β ≥ 0 and γ ≥ −1:

(1) J αJ βf (t) = J α+βf (t),



(2) J αJ βf (t) = J βJ αf (t),(3) J αtγ = �(γ+1)

�(α+γ+1)tα+γ .

Definition 3. The fractional derivative (Dα) of f (t) in the Caputo’s sense is defined as [3]

Dαf (t) = 1

�(n − α)

∫ t

0(t − τ)n−α−1f (n)(τ )dτ , (α > 0), (2.3)

for n − 1 < α ≤ n, n ∈ N , t > 0, f ∈ Cn−1. The following are two basic properties of the

Caputo’s fractional derivative [7]:

(1) Let f ∈ Cn−1, n ∈ N . Then Dαf , 0 ≤ α ≤ n is well defined and Dαf ∈ C−1.

(2) Let n − 1 < α ≤ n, n ∈ N and f ∈ Cnλ , λ ≥ −1. Then

(J αDα)f (t) = f (t) −n−1∑k=0

f k(0+)tk

k! , t > 0. (2.4)

In this article only real and positive α will be considered. Similar to integer-order differentiation,Caputo’s fractional differentiation is a linear operation [33, 64]

Dα(λf (t) + µg(t)) = λDαf (t) + µDαg(t), (2.5)

where λ, µ are constants, and satisfy the so-called Leibnitz rule

Dα(f (t)g(t)) =∞∑

k=0

(α

k

)g(k)(t)Dα−kf (t), (2.6)

if f (τ) is continuous in [0, t] and g(τ) has (n + 1) continuous derivatives in [0, t].Definition 4. For n to be the smallest integer that exceeds α, the Caputo time-fractionalderivative operator of order α > 0, is defined as [3]

Dαt u(x, t) = ∂αu(x, t)

∂tα=

1

�(n − α)

∫ t

0(t − τ)n−α−1 ∂nu(x, τ)

∂τ ndτ , if n − 1 < α < n,

∂nu(x, t)

∂tn, if α = n ∈ N .

(2.7)

For more information on the mathematical properties of fractional derivatives and integrals onecan consult [3, 7].

For more details of this section, the interested reader can also see [2–5, 8, 9, 11, 14, 17].

III. THE HOMOTOPY ANALYSIS METHOD

In this article, we use the homotopy analysis method to solve the problems described in Section I.This method proposed by a Chinese mathematician Liao [52]. We apply Liao’s basic ideas to thenonlinear fractional partial differential equations. Let us consider the nonlinear fractional partialdifferential equation

NFD(u(x, t)) = 0, (3.1)



where NFD is a nonlinear fractional partial differential operator, x and t denote independentvariables and u(x, t) is an unknown function. For simplicity, we ignore all boundary or initialconditions, which can be treated in the same way. On the basis of the constructed zero-orderdeformation equation by Liao [57], we give the following zero-order deformation equation in thesimilar way

(1 − q)L[v(x, t ; q) − u0(x, t)] = qhNFD[v(x, t ; q)], (3.2)

where q ∈ [0, 1] is the embedding parameter, h is a nonzero auxiliary parameter, L is an auxiliarylinear noninteger order operator and it possesses the property L(C) = 0, u0(x, t) is an initialguess of u(x, t), v(x, t ; q) is an unknown function on independent variables x, t , q. It is importantto that one has great freedom to choose auxiliary parameter h in HAM. The q = 0 and q = 1,give respectively [53–56]

v(x, t ; 0) = u0(x, t), v(x, t ; 1) = u(x, t). (3.3)

Thus as q increases from 0 to 1, the solution v(x, t ; q) varies from the initial guess u0(x, t) to thesolution u(x, t). Expanding v(x, t ; q) in Taylor series with respect to q, one has

v(x, t ; q) = u0(x, t) +∞∑

m=1

um(x, t)qm, (3.4)

where

um(x, t) = 1

m!∂mv(x, t ; q)

∂qm

∣∣∣∣q=0

. (3.5)

If the auxiliary linear noninteger order operator, the initial guess, and the auxiliary parameter h

are so properly chosen, the series Eq. (3.4), converges at q = 1. Hence we have [58–61]

u(x, t) = u0(x, t) +∞∑

m=1

um(x, t), (3.6)

which must be one of the solution of the original nonlinear equation, as proved by [57]. Ash = −1, Eq. (3.2) becomes

(1 − q)L[v(x, t ; q) − u0(x, t)] + qNFDv(x, t ; q) = 0, (3.7)

which is used mostly in the homotopy perturbation method (HPM) [66–70]. Thus, HPM is a spe-cial case of HAM. The comparison between HAM and HPM can be found in [71,72]. Accordingto Eq. (3.4), the governing equation can be deduced from the zero-order deformation Eq. (3.2).Define the vector [57]

�un(x, t) = {u0(x, t), u1(x, t), . . . , un(x, t)}. (3.8)

Differentiating Eq. (3.2), m times with respect to the embedding parameter q and then settingq = 0 and finally dividing them by m!, we have the so-called mth-order deformation equation[57]

L[um(x, t) − χmum−1(x, t)] = hNFR(�um−1(x, t)), (3.9)



where

NFR(�um−1(x, t)) = 1

(m − 1)!∂m−1NFD(v(x, t ; q))

∂qm−1

∣∣∣∣q=0

, (3.10)

and

χm ={

0, m ≤ 1,1, m > 1.

(3.11)

The mth-order deformation Eq. (3.9), is linear and thus can be easily solved, especially by meansof a symbolic computation software such as Mathematica, Maple, Matlab, Maxima and so on.We would like to refer the interested reader to [73–76] for more details.

IV. TEST PROBLEMS

In this section, we present several examples [33,64] to illustrate the applicability of HAM to solvenonlinear fractional partial differential equations introduced in Section I.

A. The Fractional KdV Equation

We first consider the fractional KdV equation [23, 76]

Dαt u(x, t) + a(u2)x(x, t) + buxxx(x, t) = g(x, t), (4.1)

where a, b are constant, 0 < α ≤ 1 and g(x, t) is a function of x and t . We solve the generalnonhomogeneous nonlinear equation with using the HAM method. In the following, we considerEq. (4.1) with the initial condition

u(x, 0) = f (x). (4.2)

We choose the linear noninteger order operator

L[v(x, t ; q)] = Dαt v(x, t ; q). (4.3)

Furthermore, Eq. (4.1), suggests to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = Dαt v(x, t ; q) + a(v2)x(x, t ; q) + bvxxx(x, t ; q) − g(x, t). (4.4)

Using the above definition, we construct the zeroth-order deformation equation

(1 − q)L[v(x, t ; q) − u0(x, t)] = qhNFDv(x, t ; q). (4.5)

Obviously, when q = 0 and q = 1 respectively, we have

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.6)

According to Eqs. (3.9)–(3.11), we gain the mth-order deformation equation




where

NFR(�um−1(x, t)) = Dαt um−1(x, t)

+ a

m−1∑i=0

(uium−1−i )x(x, t) + b(um−1)xxx(x, t) − (1 − χm)g(x, t). (4.8)

Now the solution of Eq. (4.7), for m ≥ 1 becomes

um(x, t) = χmum−1(x, t) + hL−1NFR(�um−1(x, t)). (4.9)

From Eqs. (4.1), (4.2) and (4.9) we now successively obtain

u0(x, t) = u(x, 0) = f (x), (4.10)

u1(x, t) = hD−αt

[Dα

t u0 + a(u20)x + b(u0)xxx − g(x, t)

]= hD−α

t

[Dα

t f (x) + a(f 2(x))x + b(f (x))xxx − g(x, t)]. (4.11)

Define

A1 = a(f 2)x(x) + b(f )xxx(x) − g(x, t),

then, we have

u1(x, t) = hD−αt

[a(f 2)x(x) + b(f )xxx(x) − g(x, t)

] = hD−αt A1,

u2(x, t) = h(h + 1)D−αt A1 + hD−α

t

[hDα

t (2a(f A1)x + b(A1)xxx)]

= h(h + 1)D−αt A1 + h2D−2α

t A2, (4.12)

where

A2 = 2a(f A1)x + b(A1)xxx .

Thus, we have

u3(x, t) = (h + 1)u2 + hD−αt

[a(u2

1 + 2u0u2

)x+ b(u2)xxx

], (4.13)

u3(x, t) = h(h + 1)2D−αt A1 + h2(h + 1)D−2α

t [A2 + (2af A1)x + (bA1)xxx] + h3D−αt a

× ((D−α

t A1)2)x+ h3D−3α

t [(2af A2)x + (bA2)xxx]. (4.14)

Also define,

A3 = A2 + (2af A1)x + (bA1)xxx , A4 = (2af A2)x + (bA2)xxx ,

u3(x, t) = h(h + 1)2D−αt A1 + h2(h + 1)D−2α

t A3 + h3D−αt a

((D−α

t A1

)2)x+ h3D−3α

t A4,(4.15)

u4(x, t) = (h + 1)u3 + hD−αt [2a(u0u3 + u1u2)x + b(u3)xxx], (4.16)

...



In the above terms we substitute h = −1, the dominant terms will be remained and the rest termsvanish, because they include factor of hm(h + 1)n, m, n ∈ N ,

u0(x, t) = f (x),

u1(x, t) = −D−αt (A1),

u2(x, t) = (D−2αt )(A2),

u3(x, t) = −[D−α

t a((

D−αt A1

)2)x+ D−3α

t A4

], (4.17)

and with using Eq. (3.6), we have

u(x, t) = f (x) − D−αt (A1) + (D−2α

t )(A2) − (D−α

t a((

D−αt A1

)2)x+ D−3α

t A4

) + · · · (4.18)

Now consider the following example:

Dαt u − 3(u2)x + (u)xxx = 0, u(x, 0) = 6x, 0 < α < 1. (4.19)

Starting with initial condition, the source term and the auxiliary operator u(x, 0) = f (x) = 6x,g(x, t) = 0 and Lu(x, t) = Dα

t u(x, t), respectively and put h = −1, we have

A1 = −63x, A2 = 2 × 65x, A3 = 4 × 65x, A4 = −4 × 66x.

Now we have

u1(x, t) = −D−αt (−63x) = 63x

�(α + 1)tα , (4.20)

u2(x, t) = D−2αt (2 × 65x) = 2 × 65x

�(2α + 1)t2α , (4.21)

u3(x, t) = −[D−3α

t (−4 × 66x) + D−αt 3

((D−α

t (−63x))2)

x

]= 4 × 66x

�(3α + 1)t3α + 67x

�(α + 1)2

2α + 1

3α + 1t3α , (4.22)

...

For α = 1, u(x, t) reduces to:

u(x, t) = u0 + u1 + u2 + · · · = 6x + 63xt + 65xt2 + 67xt3 + · · ·

= 6x(1 + 36t + 362t2 + 363t3 + · · · ) = 6x

1 − 36t, |36t | < 1, (4.23)

where u(x, t) = 6x

1−36t, is the exact solution [23].

B. The Fractional K (2,2) Equation

We next consider the following K(2, 2) equation [23]

Dαt u(x, t) + (u2)x(x, t) + (u2)xxx(x, t) = 0, u(x, 0) = x, 0 < α ≤ 1, (4.24)



that was examined for compactions. To solve the general homogeneous nonlinear equation withthe HAM method we consider the linear noninteger order operator

L[v(x, t ; q)] = Dαt v(x, t ; q). (4.25)

Furthermore, Eq. (4.24) suggests to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = Dαt v(x, t ; q) + (v2)x(x, t ; q) + (v2)xxx(x, t ; q). (4.26)

Applying the above definition, we construct the zeroth-order deformation equation


It is worth to note that, when q = 0 and q = 1 respectively, we can write

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.28)



where

NFR(�um−1(x, t)) = Dαt um−1(x, t) +

m−1∑i=0

(uium−1−i )x(x, t) +m−1∑i=0

(uium−1−i )xxx(x, t). (4.30)

Now the solution of Eq. (4.29), for m ≥ 1 becomes


From Eqs. (4.24), (4.28), and (4.31), we now successively obtain

u0(x, t) = u(x, 0) = x, (4.32)

u1(x, t) = hD−αt

[Dα

t u0 + (u2

0

)x+ (

u20

)xxx

] = hD−αt (0 + 2x + 0) = 2xh

�(α + 1)tα , (4.33)

u2(x, t) = (h + 1)u1(x, t) + hD−αt [(2u0u1)x + (2u0u1)xxx]

= 2xh(h + 1)

�(α + 1)tα + 8xh2

�(2α + 1)t2α , (4.34)

u3(x, t) = (h + 1)u2 + hD−αt

[(u2

1 + 2u0u2

)x+ (

u21 + 2u0u2

)xxx

], (4.35)



u3(x, t) = 2xh(1 + h)2

�(α + 1)tα + 16xh2(1 + h)

�(2α + 1)t2α +

(8xh3

�(α + 1)2+ 32xh3

�(2α + 1)

)�(2α + 1)

�(3α + 1)t3α ,

(4.36)

u4(x, t) = (h + 1)u3 + hD−αt [(2u1u2 + 2u0u3)x + (2u1u2 + 2u0u3)xxx], (4.37)

u4(x, t) = 2xh(1 + h)3

�(α + 1)tα + 16xh2(1 + h)2

�(2α + 1)t2α +

(8xh3

�(α + 1)2+ 32xh3

�(2α + 1)

)(h + 1)

× �(2α + 1)

�(3α + 1)t3α +

(16x

�(α + 1)2 + 32

�(2α + 1)

)h3(h + 1)

�(2α + 1)

�(3α + 1)t3α

+ 8xh2(1 + h)

�(2α + 1)t2α + 64xh4�(3α + 1)

�(α + 1)�(2α + 1)�(3α + 1)t4α

+(

16h4

�(α + 1)2 + 64h4

�(2α + 1)

)�(2α + 1)

�(4α + 1)t4α , (4.38)

...

In the above terms we substitute h = −1, the dominant terms will be remained and the rest termsvanish, because they include factor of hm(h + 1)n, m, n ∈ N and for each α = 1, we have

u0(x, t) = x,

u1(x, t) = −2xt ,

u2(x, t) = 4xt2,

u3(x, t) = −8xt3,

u4(x, t) = 16xt4,

... (4.39)

Thus, the exact solution of this test problem is as follows [23]:

u(x, t) = x(1 − 2t + 4t2 − 8t3 + · · · ) = x

1 + 2t. (4.40)

C. The Fractional Burgers’ Equation

We consider the modified KdV (mKdV) equation [23]

Dαt u(x, t) + 1

2(u2)x(x, t) − (u)xx(x, t) = 0, u(x, 0) = x, 0 < α ≤ 1. (4.41)

To solve the above problem with the HAM method we choose the linear noninteger order operator

L[v(x, t ; q)] = Dαt v(x, t ; q). (4.42)

Eq. (4.41), approaches us to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = Dαt v(x, t ; q) + 1

2(v2)x(x, t ; q) − (v)xx(x, t ; q). (4.43)



Using (4.43) we have the zeroth-order deformation equation


Also note that, when q = 0 and q = 1 respectively, we get

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.45)

Eqs. (3.9)–(3.11) yield the mth-order deformation equation


where

NFR(�um−1(x, t)) = Dαt u(m−1)(x, t) + 1

2

m−1∑i=0

(uium−1−i )x(x, t) − (um−1)xx(x, t). (4.47)

Thus the solution of Eq. (4.46), for m ≥ 1 becomes [33]


Employing Eqs. (4.41), (4.45), and (4.48), we now successively obtain

u0(x, t) = u(x, 0) = x, (4.49)

u1(x, t) = hD−αt

[Dα

t u0 + 1

2

(u2

0

)x− (u0)xx

]= hD−α

t (0 + x − 0) = xh

�(α + 1)tα , (4.50)

u2(x, t) = (h + 1)u1(x, t) + hD−αt

[1

2(2u0u1)x − (u1)xx

]= xh(h + 1)

�(α + 1)tα + 2xh2

�(2α + 1)t2α ,

(4.51)

u3(x, t) = (h + 1)u2 + hD−αt

[1

2

(u2

1 + 2u0u2

)x− (u2)xx

], (4.52)

u3(x, t) = (h + 1)u2 + 2xh2(1 + h)

�(2α + 1)t2α +

(xh3

�(α + 1)2+ 4xh3

�(2α + 1)

)�(2α + 1)

�(3α + 1)t3α , (4.53)

...

In the above terms we substitute h = −1, the dominant terms will be remained and the rest termsvanish, because they include factor of hm(h + 1)n, m, n ∈ N and for α = 1 we have

u0(x, t) = x,

u1(x, t) = −xt ,

u2(x, t) = xt2,

u3(x, t) = −xt3,

u4(x, t) = xt4,

... (4.54)



Thus, we get the exact solution as [23]

u(x, t) = x(1 − t + t2 − t3 + t4 − · · · ) = x

1 + t(4.55)

D. The Fractional BBM-Burgers’ Equation

We consider the fractional BBM-Burgers’ equation [33],

Dαt u(x, t) + (u)x(x, t) + u(u)x(x, t) − a(u)xx(x, t) − b(u)xxt (x, t) = 0,

u(x, 0) = x2, t > 0, 0 < α ≤ 1. (4.56)

To solve the general homogeneous nonlinear equation with the HAM method we choose the linearnoninteger order operator

L[v(x, t ; q)] = Dαt v(x, t ; q). (4.57)

Eq. (4.56) directs us to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = Dαt v(x, t ; q) + (v)x(x, t ; q) + v(v)x(x, t ; q)

− a(v)xx(x, t ; q) − b(v)xxt (x, t ; q). (4.58)

The zeroth-order deformation equation can be constructed [33, 64] as


Obviously, q = 0 and q = 1 respectively, yield

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.60)

Considering Eqs. (3.9)–(3.11), we gain the mth-order deformation equation


where

NFR(�um−1(x, t)) = Dαt um−1(x, t) + u(m−1)x(x, t) +

m−1∑i=0

ui(um−1−i )x(x, t)

− a(um−1)xx(x, t) − b(um−1)xxt (x, t). (4.62)

Therefore


Using Eqs. (4.56), (4.60), and (4.63), we now successively obtain

u0(x, t) = u(x, 0) = x2, (4.64)

u1(x, t) = hD−αt

[Dα

t u0 + (u0)x + u0(u0)x − a(u0)xx − b(u0)xxt

]= hD−α

t (2x + 2x3 − 2a) = (2x + 2x3 − 2a)h

�(α + 1)tα . (4.65)



Define, A1 = 2x + 2x3 − 2a,

u2(x, t) = (h + 1)u1(x, t) + hD−αt [(u1)x + u0(u1)x + u1(u0)x − a(u1)xx − b(u1)xxt ] (4.66)

u2(x, t) = (h + 1)u1(x, t) + h2(10x4 + 12x2 − 16ax + 2)

�(2α + 1)t2α − 12xh2b

�(2α)t2α−1. (4.67)

Again define, A2 = 10x4 + 12x2 − 16ax + 2,

u3(x, t) = (h + 1)u2 + hD−αt [(u2)x + u0(u2)x + u1(u1)x + u2(u0)x − a(u2)xx − b(u2)xxt ],

(4.68)

u3(x, t) = (h + 1)u2 + h(h + 1)D−αt (u1)x + h3D−3α

t (A2)x − h3D−3αt

12(2x2 + x + 1)b

�(2α)

× t2α−1 − h(h + 1)D−αt (u1)xx − h3D−3α

t (A2)xx + 2h2(h + 1)D−2αt x(A1)

+ h3D−αt

[(D−α

t A1

)(D−α

t (A1)x

] + h2(h + 1)D−2αt x2(A1)x

+ h3D−3αt [x2(A2)x + 2xA2], (4.69)

u4(x, t) = (1 + h)u3 + hD−αt

[(u3)x +

3∑i=0

ui(u3−i )x − a(u3)xx − b(u3)xxt

], (4.70)

...

In the above terms, we substitute h = −1, a = b = 1, only will be remained dominant terms andthe rest terms vanish, because they include factor of hm(h + 1)n, m, n ∈ N and for each α = 1,we have

u0(x, t) = x2,

u1(x, t) = −(2x3 + 2x − 2)t ,

u2(x, t) = (5x4 + 6x2 − 8x + 1)t2 − 12xt ,

u3(x, t) = 1

6(−18x5 − 36x3 + 192x2 − 28x + 104)t3,

... (4.71)

Thus, the solution of the given problem is as follows [33]

u(x, t) = x2 − (2x3 + 14x − 2)t + (5x4 + 6x2 − 8x + 1)t2

+ 1

6(−18x5 − 36x3 + 192x2 − 28x + 104)t3 + · · · (4.72)



E. The Fractional Cubic Boussinesq Equation

We consider the fractional cubic Boussinesq equation [23]

D2αt u(x, t) − (u)xx(x, t) + 2(u3)xx(x, t) − (u)xxxx(x, t) = 0, u(x, 0) = 1

x, ut(x, 0) = −1

x2.

(4.73)

The given initial values admit the use of

u0(x, t) = 1

x− t

x2. (4.74)

As before, we choose the linear noninteger order operator

L[v(x, t ; q)] = D2αt v(x, t ; q). (4.75)

Furthermore Eq. (4.73), suggests to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = D2αt v(x, t ; q) − (v)xx(x, t ; q) + 2(v3)xx(x, t ; q) − (v)xxxx(x, t ; q). (4.76)



Obviously, when q = 0 and q = 1 respectively, we can write

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.78)



where

NFR(�um−1(x, t)) = D2αt um−1(x, t) − u(m−1)xx(x, t)

+ 2

(m−1∑j=0

um−1−j

j∑i=0

uiuj−i

)xx

(x, t) − (um−1)xxxx(x, t). (4.80)

Hence we have




From Eqs. (4.73), (4.78) and (4.81), we now successively obtain

u0(x, t) = u(x, 0) = 1

x− t

x2, (4.82)

u1(x, t) = hD−2αt

[D2α

t u0 − (u0)xx + 2(u3

0

)xx

− (u0)xxxx

]= hD−2α

t

[−2!x3

+ 3!tx4

+ +180t2

x7− 84t3

x8

]

= h

[−−2!

x3

t2α

�(2α + 1)+ 3!

x4

t2α+1

�(2α + 2)+ 180

x7

t2α+2

�(2α + 3)− 84

x8

t2α+3

�(2α + 4)

].

(4.83)

u2(x, t) = (h + 1)u1(x, t) + hD−2αt

[−(u1)xx + 6(u2

0u1

)xx

− (u1)xxxx

], (4.84)

u3(x, t) = (h + 1)u2 + hD−2αt

[−(u2)xx + 6(u2u

20 + u0u

21

)xx

− (u2)xxxx

], (4.85)

...

In the above terms we substitute h = −1, a = b = 1, the dominant terms will be remained andthe rest terms vanish, because they include factor of hm(h + 1)n, m, n ∈ N and for α = 1, wehave

u0(x, t) = 1

x− t

x2, (4.86)

u1(x, t) = t2

x3− t3

x4− 180t4

4!x7+ 84t5

5!x8,

u2(x, t) = t4

x5− t5

x6+ small terms,

u3(x, t) = t6

x7− t7

x8+ small terms,

...

Thus we get

u(x, t) = 1

x− t

x2+ t2

x3− t3

x4+ t4

x5− t5

x6+ t6

x7− t7

x8+ · · · = 1

x

∞∑n=0

(−1)n

(t

x

)n

= 1

x + t,

(4.87)

which is the exact solution [23].

F. The Fractional Coupled KdV Equations

We consider the fractional coupled KdV equations [74, 76]

Dαt w = a[wxxx + 6w(w)x] + 2bv(v)x ,

Dαt v = −vxxx − 3wvx , (4.88)



with initial conditions

w(x, 0) = − 1 + a

3 + 6ak2 + 4k2 exp(kx)

(1 + exp(kx))2, v(x, 0) = M

exp(kx)

(1 + exp(kx))2, (4.89)

where, a �= −12 , ab < 0, M = (−24 a

b)

12 k2, and k is constant. To solve the general homogeneous

system of nonlinear equations with the HAM method we choose the linear noninteger orderoperators

L1[(x, t ; q)] = Dαt (x, t ; q), L2[(x, t ; q)] = Dα

t (x, t ; q). (4.90)

Furthermore, Eq. (4.88), suggests to define the nonlinear fractional partial differential operators

N1FD[(x, t ; q)] = Dαt (x, t ; q) − a[xxx(x, t ; q) + 6x(x, t ; q)] − 2bx(x, t ; q),

N2FD[(x, t ; q)] = Dαt (x, t ; q) + xxx(x, t ; q) + 3x(x, t ; q). (4.91)


(1 − q)L1[(x, t ; q) − w0(x, t)] = qhN1FD(x, t ; q),

(1 − q)L2[(x, t ; q) − v0(x, t)] = qhN2FD(x, t ; q). (4.92)

Obviously, when q = 0 and q = 1 we have

(x, t ; 0) = w0(x, t) = w(x, 0), (x, t ; 1) = w(x, t),

(x, t ; 0) = v0(x, t) = v(x, 0), (x, t ; 1) = v(x, t), (4.93)

respectively. According to Eqs. (3.9)–(3.11), we gain the mth-order deformation equations

L1[wm(x, t) − χmwm−1(x, t)] = hN1FR1( �wm−1(x, t)),

L2[vm(x, t) − χmvm−1(x, t)] = hN2FR2(�vm−1(x, t)), (4.94)

where

N1FR1( �wm−1(x, t)) = Dαt wm−1(x, t) − a

[w(m−1)xxx(x, t) + 6

m−1∑j=0

wjw(m−1−j)x(x, t)

]

− 2b

m−1∑j=0

vjv(m−1−j)x(x, t), (4.95)

N2FR2(�vm−1(x, t)) = Dαt vm−1(x, t) + v(m−1)xxx(x, t) + 3

m−1∑j=0

wjv(m−1−j)x(x, t).

Now the solutions of Eqs. (4.94) and (4.95) for m ≥ 1, become

wm(x, t) = χmwm−1(x, t) + hL−1N1FR1( �wm−1(x, t)),

vm(x, t) = χmvm−1(x, t) + hL−1N2FR2(�vm−1(x, t)). (4.96)



From Eqs. (4.88), (4.93), and (4.96), we now successively obtain

w0(x, t) = w(x, 0) = f (x) = − 1 + a

3 + 6ak2 + 4k2 exp(kx)

(1 + exp(kx))2, (4.97)

v0(x, t) = v(x, 0) = g(x) = Mexp(kx)

(1 + exp(kx))2,

w1(x, t) = hD−αt

[Dα

t w0 − aw0xxx − 6aNw0 − 2bMv0

], (4.98)

where we define

Nwm−1 =m−1∑i=0

wi(wm−1−i )x , Mvm−1 =m−1∑i=0

vi(vm−1−i )x ,

thus

Nw0 = w0(w0)x = ffx , Mv0 = v0(v0)x = ggx ,

w1(x, t) = hD−αt (−a(fxxx) − 6affx − 2bggx) = h

�(α + 1)tαA1,

A1 = −a(fxxx) − 6affx − 2bggx ,

v1(x, t) = hD−αt (gxxx + 3fgx) = h

�(α + 1)tαB1, (4.99)

B1 = gxxx + 3fgx ,

w2(x, t) = (h + 1)w1(x, t) + hD−αt [−aw1xxx − 6(w0w1x + w1w0x) − 2b(v0v1x + v1v0x)].

(4.100)

w2(x, t) = (h + 1)w1(x, t) + hD−αt (A2), (4.101)

A2 = −aw1xxx − 6(f w1x + w1fx) − 2b(gv1x + gxv1),

v2(x, t) = (h + 1)v1(x, t) + hD−αt [v1xxx + 3f v1x + 3gxw1] = (h + 1)v1(x, t) + hD−α

t (B2)

(4.102)

and

B2 = v1xxx + 3f v1x + 3gxw1,

A1 = −4ak5 exp(kx)(−1 + exp(kx))

(1 + 2a)(1 + exp(kx))3,

B1 = −Mak5 exp(kx)(1 − exp(kx))

(1 + 2a)(1 + exp(kx))3,

A2 = 4a2k8 exp(kx)(1 − 4 exp(kx) + exp(2kx))

(1 + 2a)2(1 + exp(kx))4

htα

�(α + 1),

B2 = Ma2k8 exp(kx)(1 − 4 exp(kx) + exp(2kx))

(1 + 2a)2(1 + exp(kx))4

htα

�(α + 1),

...



Thus we have

w1(x, t) = −h4ak5 exp(kx)(−1 + exp(kx))

(1 + 2a)(1 + exp(kx))3

tα

�(α + 1), (4.103)

v1(x, t) = −hMak5 exp(kx)(−1 + exp(kx))

(1 + 2a)(1 + exp(kx))3

tα

�(α + 1), (4.104)

w2(x, t) = (1 + h)w1(x, t) + h2 4a2k8 exp(kx)(−1 + exp(kx))

(1 + 2a)2(1 + exp(kx))4

t2α

�(α + 1)�(2α + 1), (4.105)

v2(x, t) = (1 + h)v1(x, t) + h2 Ma2k8 exp(kx)(−1 + exp(kx))

(1 + 2a)(1 + exp(kx))3

t2α

�(α + 1)�(2α + 1), (4.106)

and so on, the other components can be determined in a similar way. By repeating this procedurefor h = −1 and α = 1 we get to the solution as follows

w(x, t) = − 1 + a

3 + 6ak2 + 4k2 exp(kx)

(1 + exp(kx))2+ 4ak5 exp(kx)(−1 + exp(kx))

(1 + 2a)(1 + exp(kx))3t

+ 2a2k8 exp(kx)(1 − 4 exp(kx) + exp(2kx))

(1 + 2a)(1 + exp(kx))3t2

+ 2a3k11 exp(kx)(−1 + 11 exp(kx) − 11 exp(2kx) + exp(3kx))

3(1 + 2a)3(1 + exp(kx))5t3 + · · · , (4.107)

v(x, t) = − 1 + a

3 + 6ak2 + 4k2 exp(kx)

(1 + exp(kx))2+ Mak5 exp(kx)(−1 + exp(kx))

(1 + 2a)(1 + exp(kx))3t

+ Ma2k8 exp(kx)(−1 + exp(kx))

2(1 + 2a)2(1 + exp(kx))4t2

+ Ma3k11 exp(kx)(−1 + 11 exp(kx) − 11 exp(2kx) + exp(3kx))

6(1 + 2a)3(1 + exp(kx))5t3 + · · · , (4.108)

therefore using the Taylor series we obtain the closed form solutions [73, 76]

w(x, t) = − 1 + a

3 + 6ak2 + 4k2 exp[k(x + ct)]

(1 + exp[k(x + ct)])2 , (4.109)

v(x, t) = Mexp[k(x + ct)]

(1 + exp[k(x + ct)])2,

where M , k, a, and c are arbitrary constants.

G. The Fractional Boussinesq-Like B(m,n) Equation

The Boussinesq-like B(m, n) equation with fully nonlinear dispersion reads [39, 74]

D2αt u(x, t) − (un)xx(x, t) − (um)xxxx(x, t) = 0, m, n > 1. (4.110)

In this section, we would like to choose two special equations, namely, B(2, 2), and B(3, 3), withspecific initial conditions to illustrate efficiency of the homotopy analysis method.



Case 1. m = n = 2.

We consider the B(2, 2) equation with initial conditions [39, 74]

D2αt u(x, t) − (u2)xx(x, t) − (u2)xxxx(x, t) = 0,

u(x, 0) = 4

3a2 sin2

(x

4

),

∂u

∂t(x, 0) = a3

3sin

(x

2

), (4.111)

where a is an arbitrary constant. To solve the above problem with the HAM method we choosethe linear non-integer order operator

L[v(x, t ; q)] = D2αt v(x, t ; q). (4.112)

Furthermore Eq. (4.111), suggests to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = D2αt v(x, t ; q) − (v2)xx(x, t ; q) − (v2)xxxx(x, t ; q). (4.113)



Obviously, when q = 0 and q = 1 respectively, we have

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.115)



where

NFR(�um−1(x, t)) = D2αt um−1(x, t) −

m−1∑j=0

(ujum−1−j )xx(x, t) −m−1∑j=0

(ujum−1−j )xxxx(x, t).

(4.117)

Solution of Eq. (4.116), for m ≥ 1 becomes


From Eqs. (4.111) and (4.118), we now successively obtain

u0(x, t) = u(x, 0) + tut (x, 0) = 4

3a2 sin2

(x

4

)+ a3

3t sin

(x

2

), (4.119)


(D2α

t u0 − (u2

0

)xx

− (u2

0

)xxxx

) = htu0t + hD−2αt (−Nu0 − Mu0)

= htu0t + hD−2αt (A1), (4.120)



define Ak , as follows

Ak = −(Nuk−1 + Muk−1), ∀k = 1, 2, 3, . . . , (4.121)

u2(x, t) = (h + 1)u1(x, t) + hD−2αt (−Nu1 − Mu1), (4.122)

u2(x, t) = (h + 1)u1(x, t) + hD−2αt (−(2f A1)xx − (2f A1)xxxx)

= h(h + 1)D−2αt (A1) + hD−2α

t (A2), (4.123)

u3(x, t) = (h + 1)u2(x, t) + hD−2αt (−Nu2 − Mu2) = h(h + 1)2D−2α

t (A1)

+ h(h + 1)D−2αt (A2) + hD−2α

t (A3), (4.124)

...

The nonlinear operators N(un) and M(un) are the nonlinear terms and can be expressed in termsof Adomian’s polynomials in the following form

Num−1 =m−1∑i=0

(uium−1−i )xx , Mum−1 =m−1∑i=0

(uium−1−i )xxxx , (4.125)

Nu0 = (u20)xx , Mu0 = (u2

0)xxxx ,

Nu1 = (2u0u1)xx , Mu1 = (2u0u1)xxxx ,

Nu2 = (2u0u2 + u2

1

)xx

, Mu2 = (2u0u2 + u2

1

)xxxx

,

A1 = 1

6a4

(at

2sin

(x

2

)− cos

(x

2

)),

A2 = a5

24h

[2t sin

(x

2

)− a cos

(x

2

) t2α

�(2α + 1)+ a2t2α

2�(2α + 2)sin

(x

2

)],

A3 = a5h

96

[(−4a(h + 1)

t2α

�(2α + 1)− a3h

t4α

�(2α + 1)�(4α + 1)

)cos

(x

2

)

+(

2a2(2h + 1)t2α+1

�(2α + 2)+ 8(h + 1)t + a4h

2

t4α+1

�(2α + 2)�(4α + 2)

)sin

(x

2

)],

...

Thus we have

u0(x, t) = 4

3a2 sin2

(x

4

)+ a3

3t sin

(x

2

), (4.126)

u1(x, t) = a3

3ht sin

(x

2

)+ a4h

6

[a

2sin

(x

2

) t2α+1

�(2α + 2)− cos

(x

2

) t2α

�(2α + 1)

], (4.127)



u2(x, t) =[(

h(h + 1)a3t

3+ h(h + 2)

a5t2α+1

12�(2α + 2)+ h2a7t4α+1

48�(2α + 2)�(4α + 2)

)sin

(x

2

)

−(

h(h + 1)a4t2α

6�(2α + 1)+ h2a6

24

t4α

�(2α + 1)�(4α + 1)

)cos

(x

2

)], (4.128)

u3(x, t) =[a3h(2h + 1)t

3+ a5h(2h + 1)

12

t2α+1

�(2α + 2)+ a7h2(3h + 2)

48

t4α+1

�(2α + 2)�(4α + 2)

+ a5h2(3h + 2)

12

t2α+1

�(2α + 2)+ a9h3

96

t6α+1

�(2α + 2)�(4α + 2)�(6α + 2)

]sin

(x

2

)

+[a4h(h + 1)2

6

t2α

�(2α + 1)+ a6h2(h + 1)2

24

t4α

�(2α + 1)�(4α + 1)

+ a8h3

96

t6α

�(2α + 1)�(4α + 1)�(6α + 1)

]cos

(x

2

). (4.129)

By repeating this procedure for h = −1 and α = 1 we get to solution as follows

u(x, t) =∞∑i=0

ui(x, t) = 4

3a2sin2

(x

4

)+ a2

3

(at − 1

2

(at)3

3! − 1

192 × 5!(at)7

7!)

sin(x

2

)

+ a2

6

((at)2

2! − 1

8

(at)4

4! − 1

16 × 48

(at)6

6!)

cos(x

2

)+ · · · (4.130)

Using the Taylor series gives the exact solutions [39]

u(x, t) = 4

3a2 sin2

(x + at

4

), −2π < x + at < 2π , (4.131)

where a is an arbitrary constant.

Case 2. m = n = 3.

We consider the B(3, 3) equation

D2αt u(x, t) − (u3)xx(x, t) − (u3)xxxx(x, t) = 0, (4.132)

with initial conditions [39]

u(x, 0) =√

6

2ab sin

(x

3

)+

√6

2ab cos

(x

3

), ut(x, 0) =

√6

2ab2 cos

(x

3

)−

√6

2ab2 sin

(x

3

),

where a and b are arbitrary constants. To solve the general homogeneous nonlinear equation withthe HAM method we choose the linear noninteger order operator

L[v(x, t ; q)] = D2αt v(x, t ; q). (4.133)

Furthermore Eq. (4.132) suggests to define the nonlinear fractional partial differential operator

NFD[v(x, t ; q)] = D2αt v(x, t ; q) − (v3)xx(x, t ; q) − (v3)xxxx(x, t ; q). (4.134)





Obviously, when q = 0 and q = 1 respectively, we get

v(x, t ; 0) = u0(x, t) = u(x, 0), v(x, t ; 1) = u(x, t). (4.136)



where

NFR(�um−1(x, t)) = D2αt um−1(x, t) −

(m−1∑i=0

um−1−i

i∑j=0

ujui−j

)xx

(x, t)

−(

m−1∑i=0

um−1−i

i∑j=0

ujui−j

)xxxx

(x, t). (4.138)

Therefore we have


From Eqs. (4.132) and (4.139), we now successively obtain

u0(x, t) = u(x, 0) + tut (x, 0) =√

6

2ab

[(1 − bt) sin

(x

3

)+ (1 + bt) cos

(x

3

)], (4.140)


(D2α

t u0 − (u3

0

)xx

− (u3

0

)xxxx

) = htu0t + hD−2αt (−Nu0 − Mu0)

= htu0t + hD−2αt (A1). (4.141)

Define Ak , as given in the following

Ak = −(Nuk−1 + Muk−1), ∀k = 1, 2, 3, . . . , (4.142)

u2(x, t) = (h + 1)u1(x, t) + hD−2αt (−Nu1 − Mu1), (4.143)

u2(x, t) = (h + 1)u1(x, t) + hD−2αt (−(2f A1)xx − (2f A1)xxxx)

= h(h + 1)D−2αt (A1) + hD−2α

t (A2), (4.144)

u3(x, t) = (h + 1)u2(x, t) + hD−2αt (−Nu2 − Mu2) = h(h + 1)2D−2α

t (A1)

+ h(h + 1)D−2αt (A2) + hD−2α

t (A3), (4.145)

...



The nonlinear operators N(un) and M(un) are the nonlinear terms and can be expressed in termsof Adomian’s polynomials as:

Num−1 =(

m−1∑i=0

um−1−i

i∑j=0

ujui−j

)xx

, Mum−1 =(

m−1∑i=0

um−1−i

i∑j=0

ujui−j

)xxxx

, (4.146)

Nu0 = (u3

0

)xx

, Mu0 = (u30)xxxx ,

Nu1 = (3u2

0u1

)xx

, Mu1 = (3u2

0u1

)xxxx

,

Nu2 = (3u2

0u2 + 3u0u21

)xx

, Mu2 = (3u2

0u2 + 3u0u21

)xxxx

,

A1 =√

6

9a3b3

[(1 + bt + b2t2 + b3t3) cos

(x

3

)+ (1 − bt + b2t2 − b3t3) sin

(x

3

)],

A2 =√

6

81a3b4h

{[2a2b(3 + 2bt + b2t2)

(t2α

�(2α + 1)+ b2 t2α+2

�(2α + 3)

)

+2a2b2(1 + 2bt + 3b2t2)

(t2α+1

�(2α+2)+b2 t2α+3

�(2α+4)

)+9t(1 + 2bt+3b2t2)

]cos

(x

3

)

+[

2a2b(3 − 2bt + b2t2)

(t2α

�(2α + 1)+ b2 t2α+2

�(2α + 3)

)− 2a2b2(1 − 2bt + 3b2t2)

×(

t2α+1

�(2α + 2)+ b2 t2α+3

�(2α + 4)

)− 9t(1 − 2bt + 3b2t2)

]sin

(x

3

)}.

Thus we have

u0(x, t) =√

6

2ab

[(1 − bt) sin

(x

3

)+ (1 + bt) cos

(x

3

)], (4.147)

u1(x, t)

= √6abh

{[b

2t+

(ab

3

)2 (t2α

�(2α + 1)+b

t2α+1

�(2α+2)+b2 t2α+2

�(2α+3)+b3 t2α+3

�(2α+4)

)]cos

(x

3

)

+[−b

2t+

(ab

3

)2 (t2α

�(2α + 1)−b

t2α+1

�(2α + 2)+b2 t2α+2

�(2α + 3)−b3 t2α+3

�(2α + 4)

)]sin

(x

3

)},

(4.148)

u2(x, t) = √6abh(h + 1)

{[b

2t+

(ab

3

)2 (t2α

�(2α + 1)+ b

t2α+1

�(2α + 2)+ b2 t2α+2

�(2α + 3)

+ b3 t2α+3

�(2α + 4)

)]cos

(x

3

)+

[−b

2t +

(ab

3

)2 (t2α

�(2α + 1)− b

t2α+1

�(2α + 2)

+ b2 t2α+2

�(2α + 3)− b3 t2α+3

�(2α + 4)

)]sin

(x

3

)}



+√

6

81a3b4h2

{[2a2b

�(2α + 1)

(3

t4α

�(4α + 1)+ 2b

t4α+1

�(4α + 2)+ b2 t4α+2

�(4α + 3)

)

+ 2a2b2

�(2α + 2)

(t4α+1

�(4α + 2)+ 2b

t4α+2

�(4α + 3)+ 3b2 t4α+3

�(4α + 4)

)

+ 2a2b3

�(2α + 3)

(3

t4α+2

�(4α + 3)+ 2b

t4α+3

�(4α + 4)+ b2 t4α+4

�(4α + 5)

)

+ 2a2b4

�(2α + 4)

(t4α+3

�(4α + 4)+ 2b

t4α+4

�(4α + 5)+ 3b2 t4α+5

�(4α + 6)

)

+ 9

(t2α+1

�(2α + 2)+ 2b

t2α+2

�(2α + 3)+ 3b2 t2α+3

�(2α + 4)

)]cos

(x

3

)

+[

2a2b

�(2α + 1)

(3

t4α

�(4α + 1)− 2b

t4α+1

�(4α + 2)+ b2 t4α+2

�(4α + 3)

)

− 2a2b2

�(2α + 2)

(t4α+1

�(4α + 2)− 2b

t4α+2

�(4α + 3)+ 3b2 t4α+3

�(4α + 4)

)

+ 2a2b3

�(2α + 3)

(3

t4α+2

�(4α + 3)− 2b

t4α+3

�(4α + 4)+ b2 t4α+4

�(4α + 5)

)

− 2a2b4

�(2α + 4)

(t4α+3

�(4α + 4)− 2b

t4α+4

�(4α + 5)+ 3b2 t4α+5

�(4α + 6)

)

− 9

(t2α+1

�(2α + 2)− 2b

t2α+2

�(2α + 3)+ 3b2 t2α+3

�(2α + 4)

)]sin

(x

3

)}. (4.149)

By repeating above procedure for h = −1 and α = 1 we get to solution as follows

u(x, t) =√

6

2ab

[sin

(x

3

)+ cos

(x

3

)]−

√6

9a3b

{[5∑

k=2

(bt)k

k!(

cos(x

3

)+ (−1)k sin

(x

3

))]

+[

5∑k=3

(k − 2)(bt)k

k!(

cos(x

3

)+ (−1)k sin

(x

3

))]}

× 2√

6

81a5b

[(3

2!(bt)4

4! + 7

3!(bt)5

5! + 23

4!(bt)6

6! + 71

5!(bt)7

7! + 7

5!(bt)8

8! + 3

5!(bt)9

9!)

cos(x

3

)

+(

3

2!(bt)4

4! − 7

3!(bt)5

5! + 23

4!(bt)6

6! − 71

5!(bt)7

7! + 7

5!(bt)8

8! − 3

5!(bt)9

9!)

sin(x

3

)]+ · · ·

(4.150)

According to the above procedure and by using the Taylor series we have [39]

u(x, t) =√

6

2ab sin

(x + bt

3

)+

√6

2ab cos

(x + bt

3

), (4.151)

where a and b are arbitrary constants.



FIG. 1. The figures show the 3rd-order approximation solution u to Eq. (4.56) when a = b = 1,h = −1. (a) α = 1, (b) α = 0.5. [Color figure can be viewed in the online issue, which is available atwww.interscience.wiley.com.]

We illustrate the accuracy and efficiency of HAM by applying the method to some fractionalpartial differential equations and comparing the approximate solutions with the exact solutions.For this purpose, we calculate the numerical results of the exact solutions (for the cases where exactsolutions are available) and the multi-terms approximate solutions of HAM. At the same time, thesurface graphics of the exact and multi-terms approximate solutions are plotted in Figs. 1–7. One

FIG. 2. The figures show the 3rd-order approximation solution u to Eq. (4.56) when a = b = 1, α = 0.2.(a) h = −0.5, (b) h = −0.2. [Color figure can be viewed in the online issue, which is available atwww.interscience.wiley.com.]



FIG. 3. The surface of the exact and the 3rd-order approximation solutions u to Eq. (4.73) obtained in thiswork. [Color figure can be viewed in the online issue, which is available at www.interscience.wiley.com.]

FIG. 4. The surface of the exact and the 3rd-order approximation solutions w to Eqs. (4.88) fora = −1.5, c = 0.1, b = 0.1 and k = 0.1, obtained in this work. [Color figure can be viewed in theonline issue, which is available at www.interscience.wiley.com.]



FIG. 5. The surface of the exact and the 3rd-order approximation solutions v to Eqs. (4.88) for a =−1.5, c = 0.1, b = 0.1, and k = 0.1, obtained in this work. [Color figure can be viewed in the online issue,which is available at www.interscience.wiley.com.]

FIG. 6. The surface of the exact and the 3rd-order approximation solutions u to Eq. (4.111) fora = 2 obtained in this work. [Color figure can be viewed in the online issue, which is available atwww.interscience.wiley.com.]



FIG. 7. The surface of the exact and the 3rd-order approximation solutions u to Eq. (4.132) fora = b = 1 obtained in this work. [Color figure can be viewed in the online issue, which is available atwww.interscience.wiley.com.]

can see that the approximate solutions obtained by HAM are quite close to their exact solutions.Also Tables I-IV present some computational results.

V. CONCLUSION

In this paper, based on the HAM, a new analytic technique is proposed to solve the nonlinearfractional partial differential equations. Different from all other analytic methods, it provides uswith a simple way to adjust and control the convergence region of solution series by introducing

TABLE I. Numerical results of w to Eq. (4.88) for a = −1.5, c = 0.1, b = 0.1, and k = 0.1.

(x, t) wapprox = ∑3i=0 wi wexa |werror|

(−30, 2) 0.9712810448E −3 0.7256710761E −2 0.6285429716E −2(−20, 2) 0.3361614082E −2 0.9680689767E −2 0.6319075685E −2(−10, 2) 0.7025690948E −2 0.1335354099E −1 0.6327850042E −2

(0, 2) 0.9166661042E −2 0.1541566673E −1 0.6249005688E −2(10, 2) 0.7036593870E −2 0.1320818177E −1 0.6171587900E −2

TABLE II. Numerical results of v to Eq. (4.88) for a = −1.5, c = 0.1, b = 0.1, and k = 0.1.

(x, t) vapprox = ∑3i=0 vi vexa |verror|

(−30, 2) 0.8571552209E −2 0.8728095501E −2 0.156543292E −3(−20, 2) 0.1992090467E −1 0.2022603750E −1 0.30513283E −3(−10, 2) 0.3730423281E −1 0.3764790055E −1 0.34366774E −3

(0, 2) 0.4743416463E −1 0.4742942180E −1 0.474283E −5(10, 2) 0.3730474996E −1 0.3695840120E −1 0.34634876E −3



TABLE III. Numerical results of Eq. (4.111) for a = 1 and m = n = 2.

(x, t) uapprox = ∑3i=0 ui uexa |uerror|

(−20, 0.5) 5.325792416 5.292329374 0.33463042E −1(−10, 0.5) 3.228788798 3.175821501 0.52967297E −1

(0, 0.5) 0.3264465017 0.3298599055 0.0034134038(10, 0.5) 0.7768806021 0.8317844067 0.549038046E −1(20, 0.5) 3.934765141 3.962500004 0.027734863

TABLE IV. Numerical results of Eq. (4.132) for a = 1 and m = n = 3.

(x, t) uapprox = ∑3i=0 ui uexa |uerror|

(−30, 0.2) −0.3639924506 −0.4734041674 0.1094117168(−20, 0.2) 0.6815406637 0.7822338065 0.1006931428(−10, 0.2) −0.9741090582 −1.062393021 0.882839628E −1

(0, 0.2) 1.230974414 1.303613415 0.72639001E −1(10, 0.2) −1.442722108 −1.497053783 0.54331675E −1

an auxiliary parameter h. This is an obvious advantage of the homotopy analysis method. Thevalidity of the method has been successful shown by applying it for different kinds of nonlinearfractional partial differential equations in mathematical physics, namely KdV, K(2, 2), Burgers,BBM-Burgers, cubic Boussinesq, coupled KdV and Boussinesq-like B(m, n) equations. We cansimply choose the fractional operator Dα as the auxiliary linear operator. In this way, we obtainsolutions in power series. Also, we obtained the exact solutions in special case α = 1, h = −1,for some equations. However, it is well-known that a power series often has a small convergenceradius. It should be emphasized that, in the frame of the homotopy analysis method, we have greatfreedom to choose the initial guess and the auxiliary linear operator L = Dα . This work showsthat the homotopy analysis method is a very efficient and powerful tool for solving the nonlinearfractional partial differential equations.

The authors are very grateful to the three reviewers for carefully reading the paper and for theirconstructive comments and suggestions which have improved the paper.

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