# An Iterative Method for Time-Fractional Swift-Hohenberg Equation

Authors: Wenjin Li; Yanni Pang
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

We study a type of iterative method and apply it to time-fractional Swift-Hohenberg equation with initial value. Using this iterative method, we obtain the approximate analytic solutions with numerical figures to initial value problems, which indicates that such iterative method is effective and simple in constructing approximate solutions to Cauchy problems of time-fractional differential equations.


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## Body

## 1. Introduction

In 1695, L’Hopital wrote a letter to Leibniz, and he proposed a problem: “What is the result ofdny/dxn if n=1/2?” Lcebniz answered to L’Hopital “d1/2x will be equal to xdx:x. This is an apparent paradox, from which, one day useful consequences will be drawn.” [1, 2] Later, as the development of mathematics, especially, the theory of operator, researchers started to have a new recognition of the fractional derivative. They found the fractional derivative has a wide applications in many fields, such as physics, chemistry, and many other sciences [3, 4]. It should be emphasised that the fractional derivative is defined by integral and it is a nonlocal operator with a singular kernel; hence it can provide an excellent instrument for description of memory and hereditary properties of various physical processes. For example, half-order derivatives and integrals proved to be more useful for the formulation of certain electrochemical problems than the classical models [5]. However, the definition of fractional derivative has not been unified; there are many kinds of fractional integral and fractional derivative, such as in the sense of Riemann-CLiouville, Caputo, Riesz, and Weyl [6–9]. The Riemann-Liouville fractional integral and the Caputo fractional derivative are the most commonly used.For a fractional system with respect to some initial or boundary conditions, one of the fundamental problems is naturally that what is the exact solution or approximate solution to such system. Solving fractional system is usually more difficult than the classical system, for its operator is defined by integral. Luckily, there are some different effective methods which have been developed to construct approximate solutions of fractional systems and even obtain the exact solutions such as the homotopy analysis method [10–12], the residual power series method [13–16], the differential transform method [17, 18], the Laplace transform method [19], the perturbation method [20, 21]. In addition, using polynomials to approximate the fractional system is an effective method as well, such as Jacobi polynomials [22], Bernstein polynomials [23], and Chebyshev and Legendre polynomials [24]. In this paper, we introduce a type of iterative method, based on decomposing the nonlinearity term, for solving a class of functional equations [25–30].Outline of Paper. In Section 2, we introduce some necessary concepts and lemmas on fractional differential equations. In Section 3, a type of iterative method for solving a class of functional equation is presented. Also, we obtain the convergence analysis of this iterative method. In Sections 4 and 5, we take the linear time-fractional S-H equations, including the term of dispersion, with respect to differential initial conditions as examples to illustrate the strong power of such iterative method, respectively.


## 2. Notations of Fractional Calculus

In subsection, we introduce some concepts and lemmas we need in this paper, such as the Gamma function, the Mittag-Leffler function, the Riemann-Liouville fractional integral, and the Caputo fractional derivative. It should be emphasised that there are many kinds of fractional integral and fractional derivative, such as in the sense of Riemann-Liouville, Caputo, Riesz, and Weyl [6–9]. The Riemann-Liouville fractional integral and Caputo fractional derivative are the most commonly used version.Definition 1.
The Gamma function is defined by [6, 29, 30](1)Γx≔∫0∞tx-1e-tdtRe⁡x>0.Definition 2.
The Mittag-Leffler function is defined by [31, 32](2)Eαx≔∑k=0∞xkΓαk+1α>0.Definition 3.
The (left sided) Riemann-Liouville fractional integral of orderβ(β>0) of a function u(x,t)∈Cp(p≥-1) is denoted by Iβu(x,t) (with respect to t) and defined as [29, 30](3)Iβux,t≔1Γβ∫0tt-τβ-1ux,τdτ.Definition 4.
The (left sided) Caputo fractional derivative of orderβ(β>0) of a function u(x,t)∈Clm is denoted by Dβu(x,t) (with respect to t) and defined as [29, 30] (4)Dβux,t≔∂tmux,tβ=m∈N∗Im-β∂tmux,tm-1<β<m.Lemma 5.
For anym-1<β≤m∈N∗, one has [6–9, 29, 30](5)IβDβux,t=ux,t-∑k=0m-1∂kux,0tkk!.Lemma 6.
For anyβ,γ, one has [6–9, 29](6)Iβtγ=Γγ+1Γβ+γ+1tβ+γ.


## 3. A Type of Iterative Method

In this section, we introduce a generalized iterative method for solving a class of functional equation (7) (see below). Some more specific details about this type of iterative method could be found in [25–30] and the references therein.Now we state this iterative method as the following lemma together convergence analysis.Lemma 7.
Consider the nonlinear functional equation(7)ut,x=ft,x+Lut,x+Nut,x,where u(t,x) is an unknown function, i.e., solution of functional equation (7), f(t,x) is a known function, (t,x)∈D=t,x:t∈R,x∈Rn,n∈N, and L and N are linear and nonlinear operators from a Banach space B to itself. We are looking for a solution of functional equation (7) having a series form (8)ut,x=∑n=0∞unt,x=u0+u1+u2+⋯, where un are defined by letting(9)u0=f,u1=Lu0+Nu0,(10)um+1=Lum+N∑n=0mun-N∑n=0m-1un,m=1,2⋯.Furthermore, if the operator L and N are contractive, then the series ∑n=0∞un(t,x) converges absolutely and uniformly.Proof.
Obviously, the nonlinear operatorN can be decomposed (11)Nut,x=N∑n=0∞unt,x=Nu0+∑n=1∞N∑j=0nuj-N∑j=0n-1uj. Similarly, linear operator L can also be decomposed (12)Lut,x=L∑n=0∞unt,x=Lu0+∑n=1∞L∑j=0nuj-L∑j=0n-1uj. Set (13)Mut,x≔Lut,x+Nut,x. then (14)Mut,x=Mu0+∑n=1∞M∑j=0nuj-M∑j=0n-1uj. Define the following recurrence equations: (15)u0=f,u1=Mu0,um+1=M∑n=0mun-M∑n=0m-1un,m=1,2⋯. Since operators L and N are contractive, then M is also contractive; i.e., there exists a constant 0<K<1, such that (16)Mvi-Mvj≤Kvi-vj,∀vi,vj∈B, where · denotes the usual norm on Banach space B. What is more, for um+1, one can obtain (17)um+1=M∑n=0mun-M∑n=0m-1un≤Kum≤Km+1u0. Since 0<K<1, the series ∑m=1∞Km+1u0 converges absolutely as well as uniformly.According to the Weierstrass M-test, one can obtain that the series∑i=0∞ui converges absolutely as well as uniformly.


## 4. Linear Swift-Hohenberg Equation

The Swift-Hohenberg (for short S-H) equation(18)∂tu=ru-1+∇22u+Nu is a model pattern-forming equation which was derived from the equations for thermal convection by Jack Swift and Pierre Hohenberg [33]. Here u=u(x,t) is a scalar function defined on the line or the plane, r is a real bifurcation parameter, and N(u) is some smooth nonlinearity. The S-H equation plays an important role in pattern formation theory. In [34], Braaksma et al. proved the existence of quasipatterns for the S-H equation. Also, wave process described by the S-H equation is important as well. For example, it describes the patterns inside thin vibrated granular layers, the mechanism of the amplitude of optical electric field inside the cavity, and so on [35].Fractional S-H equation(19)Dαu=-2uxx-uxxxx-1-ru-u30<α≤1 was firstly introduced in [36], and Khan et al. obtained analytical approximation of this equation. Later, Vishal et al. in [37] constructed the approximate analytic solution with respect initial value u(x,0)=1/10sin⁡(πx/l) using the homotopy analysis method. Furthermore, Vishal et al. considered the time-fractional S-H equation with dispersion [38](20)Dα=ru-1+∂xx2u+σ∂xxxu+2u2-u30<α≤1 and obtained the approximate analytic solution. Here σ is the dispersive parameter. Lately, in [39], Merdan applied the fractional variational iteration method to obtain the approximate solution to time-fractional S-H equation with respect to initial condition u(x,0)=1/10sin⁡(πx/l). Also, homotopy analysis method is valid for S-H equation as well [40].In this subsection, we apply the iterative method introduced in Section3 to linear time-fractional S-H equation with different initial values, such as ex, sin⁡x, and cos⁡x.


### 4.1. Linear Time-Fractional S-H Equation with Initial Valueux,0=ex

Consider the following linear time-fractional S-H equation:(21)Dαu+1-ru+2uxx+uxxx=00<α≤1with initial condition(22)ux,0=ex.Clearly, applying Iα to both sides of (21), then initial value problem (21)-(22) is equivalent to the following integral equation:(23)ux,t=fx+Lux,t+Nux,t, where(24)fx=ex,Lu=Iαr-1u-2uxx-uxxxx,Nu=0. The solution to system (21)-(22) what we are looking for has the form(25)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (26)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1=r-4ex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα=r-42ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α=r-4nex1Γnα+1tnαn=2,3,⋯ Hence the N-th approximate solution to (21)-(22) is(27)uNx,t=ex∑n=0Nr-4n1Γnα+1tnα and the exact solution to (21)-(22) is(28)ux,t=limN→∞uNx,t=exEαr-4tα.Numerical Simulation. See Figure 1.Figure 1 (a) (b) (c)


### 4.2. Linear Time-Fractional S-H Equation with Initial Valueu(x,0)=sin⁡x

Consider the following linear time-fractional S-H equation:(29)Dαu+1-ru+2uxx+uxxxx=00<α≤1with initial condition(30)ux,0=sin⁡x.Applying Iα to both sides of (29), then initial value problem (29)-(30) is equivalent to the following integral equation:(31)ux,t=fx+Lux,t+Nux,t, where(32)fx=sin⁡x,Lu=Iαr-1u-2uxx-uxxxx,Nu=0. The solution of system (29)-(30) what we are looking for has the form(33)ux,t=∑n=0∞unx,t. We shall distinguish the following two cases.Case 1 (r=0).
According to iterative scheme (9)-(10), we have (34)u0=f=sin⁡x;u1=Lu0+Nu0=-u0-2u0xx-u0xxxxIα1=0;un=0,n=2,3,⋯. Hence the solution to (29)-(30) is(35)ux,t=sin⁡x.Case 2 (r≠0).
By iterative scheme (9)-(10), we have (36)u0=f=sin⁡x;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1=r-1+1sin⁡x1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα=r-12+r-1+1sin⁡x1Γ2α+1t2α;u3=Lu2+Nu0+u1+u2-Nu0+u1=r-1u2-2u2xx-u2xxxxIα1Γ2α+1t2α=r-13+r-12+r-1+1sin⁡x1Γ3α+1t3α;u4=Lu3+Nu0+u1+u2+u3-Nu0+u1+u2=r-1u3-2u3xx-u3xxxxIα1Γ3α+1t3α=r-14+r-13+r-12+r-1+1sin⁡x1Γ4α+1t4α. Generally, for n=1,2,⋯, by induction, one can derive that (37)un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α=r-1n+r-1n-1+⋯+r-12+r-1+1sin⁡x1Γn-1α+1tn-1α=-1r1-r-1n+1sin⁡x1Γnα+1tnαHence the N-th approximate solution to (29)-(30) is(38)uNx,t=-1rsin⁡x∑n=0N1-r-1n+11Γnα+1tnα and the exact solution to (29)-(30) is(39)ux,t=limN→∞uNx,t=-1rsin⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnα. As a consequence, the solution of (29)-(30) is(40)ux,t=sin⁡xr=0;-1rsin⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnαr≠0.Numerical Simulation. See Figure 2.Figure 2 (a) (b) (c)


### 4.3. Linear Time-Fractional S-H Equation with Initial Valueu(x,0)=cos⁡x

Consider the following linear time-fractional S-H equation:(41)Dαu+1-ru+2uxx+uxxxx=00<α≤1with initial condition(42)ux,0=cos⁡x.Applying Iα to both sides of (41), then initial value problem (41)-(42) is equivalent to the following integral equation:(43)ux,t=fx+Lux,t+Nux,t, where(44)fx=sin⁡x,Lu=Iαr-1u-2uxx-uxxxx,Nu=0. The solution to system (41)-(42) what we are looking for has the form(45)ux,t=∑n=0∞unx,t. We shall distinguish the following two cases.Case 1 (r=0).
According to iterative scheme (9)-(10), we have (46)u0=f=cos⁡x;u1=Lu0+Nu0=-u0-2u0xx-u0xxxxIα1=0;un=0,n=2,3,⋯.Hence the solution to (41)-(42) is(47)ux,t=cos⁡x.Case 2 (r≠0).
According to iterative scheme (9)-(10), we have (48)u0=f=cos⁡x;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1=r-1+1cos⁡x1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα=r-12+r-1+1cos⁡x1Γ2α+1t2α;u3=Lu2+Nu0+u1+u2-Nu0+u1=r-1u2-2u2xx-u2xxxxIα1Γ2α+1t2α=r-13+r-12+r-1+1cos⁡x1Γ3α+1t3α;u4=Lu3+Nu0+u1+u2+u3-Nu0+u1+u2=r-1u3-2u3xx-u3xxxxIα1Γ3α+1t3α=r-14+r-13+r-12+r-1+1cos⁡x1Γ4α+1t4α. Generally, for n=1,2,⋯, one can derive that(49)un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α=r-1n+r-1n-1+⋯+r-12+r-1+1cos⁡x1Γn-1α+1tn-1α=-1r1-r-1n+1cos⁡x1Γnα+1tnαHence the N-th approximate solution to (41)-(42) is(50)uNx,t=-1rcos⁡x∑n=0N1-r-1n+11Γnα+1tnα and the exact solution to (41)-(42) is(51)ux,t=limN→∞uNx,t=-1rcos⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnα.As a consequence, the solution of (41)-(42) is(52)ux,t=cos⁡xr=0;-1rcos⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnαr≠0.Numerical Simulation. See Figure 3.Figure 3 (a) (b) (c)


### 4.4. Linear S-H Equation with Dispersion and Initial Valueu(x,0)=ex

Consider the linear S-H equation with dispersion [41] as follows:(53)Dαu+1-ru+2uxx-σuxxx+uxxxx=00<α≤1with initial value(54)ux,0=ex.Here σ∈R is a parameter. Clearly, applying Iα on both sides of (53), then initial value problem (53)-(54) is equivalent to the following integral equation:(55)ux,t=fx+Lux,t+Nux,t, where(56)fx=ex,Lu=Iαr-1u-2uxx+σuxxx-uxxxx,Nu=0. The solution to system (53)-(54) what we are looking for has the form(57)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (58)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx+σu1xxx-u0xxxxIα1=r-4+σex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx+σu1xx-u1xxxxIα1Γα+1tα=r-4+σ2ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx+σun-1xxx-un-1xxxxIα1Γn-1α+1tn-1α=r-4+σnex1Γnα+1tnαn=2,3,⋯ Hence the N-th approximate solution to (53)-(54) is(59)uNx,t=ex∑n=0Nr-4+σn1Γnα+1tnα and the exact solution to (53)-(54) is(60)ux,t=limN→∞uNx,t=exEαr-4+σtα.Numerical Simulation. See Figure 4.Figure 4 (a) (b) (c)


## 4.1. Linear Time-Fractional S-H Equation with Initial Valueux,0=ex

Consider the following linear time-fractional S-H equation:(21)Dαu+1-ru+2uxx+uxxx=00<α≤1with initial condition(22)ux,0=ex.Clearly, applying Iα to both sides of (21), then initial value problem (21)-(22) is equivalent to the following integral equation:(23)ux,t=fx+Lux,t+Nux,t, where(24)fx=ex,Lu=Iαr-1u-2uxx-uxxxx,Nu=0. The solution to system (21)-(22) what we are looking for has the form(25)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (26)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1=r-4ex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα=r-42ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α=r-4nex1Γnα+1tnαn=2,3,⋯ Hence the N-th approximate solution to (21)-(22) is(27)uNx,t=ex∑n=0Nr-4n1Γnα+1tnα and the exact solution to (21)-(22) is(28)ux,t=limN→∞uNx,t=exEαr-4tα.Numerical Simulation. See Figure 1.Figure 1 (a) (b) (c)


## 4.2. Linear Time-Fractional S-H Equation with Initial Valueu(x,0)=sin⁡x

Consider the following linear time-fractional S-H equation:(29)Dαu+1-ru+2uxx+uxxxx=00<α≤1with initial condition(30)ux,0=sin⁡x.Applying Iα to both sides of (29), then initial value problem (29)-(30) is equivalent to the following integral equation:(31)ux,t=fx+Lux,t+Nux,t, where(32)fx=sin⁡x,Lu=Iαr-1u-2uxx-uxxxx,Nu=0. The solution of system (29)-(30) what we are looking for has the form(33)ux,t=∑n=0∞unx,t. We shall distinguish the following two cases.Case 1 (r=0).
According to iterative scheme (9)-(10), we have (34)u0=f=sin⁡x;u1=Lu0+Nu0=-u0-2u0xx-u0xxxxIα1=0;un=0,n=2,3,⋯. Hence the solution to (29)-(30) is(35)ux,t=sin⁡x.Case 2 (r≠0).
By iterative scheme (9)-(10), we have (36)u0=f=sin⁡x;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1=r-1+1sin⁡x1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα=r-12+r-1+1sin⁡x1Γ2α+1t2α;u3=Lu2+Nu0+u1+u2-Nu0+u1=r-1u2-2u2xx-u2xxxxIα1Γ2α+1t2α=r-13+r-12+r-1+1sin⁡x1Γ3α+1t3α;u4=Lu3+Nu0+u1+u2+u3-Nu0+u1+u2=r-1u3-2u3xx-u3xxxxIα1Γ3α+1t3α=r-14+r-13+r-12+r-1+1sin⁡x1Γ4α+1t4α. Generally, for n=1,2,⋯, by induction, one can derive that (37)un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α=r-1n+r-1n-1+⋯+r-12+r-1+1sin⁡x1Γn-1α+1tn-1α=-1r1-r-1n+1sin⁡x1Γnα+1tnαHence the N-th approximate solution to (29)-(30) is(38)uNx,t=-1rsin⁡x∑n=0N1-r-1n+11Γnα+1tnα and the exact solution to (29)-(30) is(39)ux,t=limN→∞uNx,t=-1rsin⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnα. As a consequence, the solution of (29)-(30) is(40)ux,t=sin⁡xr=0;-1rsin⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnαr≠0.Numerical Simulation. See Figure 2.Figure 2 (a) (b) (c)


## 4.3. Linear Time-Fractional S-H Equation with Initial Valueu(x,0)=cos⁡x

Consider the following linear time-fractional S-H equation:(41)Dαu+1-ru+2uxx+uxxxx=00<α≤1with initial condition(42)ux,0=cos⁡x.Applying Iα to both sides of (41), then initial value problem (41)-(42) is equivalent to the following integral equation:(43)ux,t=fx+Lux,t+Nux,t, where(44)fx=sin⁡x,Lu=Iαr-1u-2uxx-uxxxx,Nu=0. The solution to system (41)-(42) what we are looking for has the form(45)ux,t=∑n=0∞unx,t. We shall distinguish the following two cases.Case 1 (r=0).
According to iterative scheme (9)-(10), we have (46)u0=f=cos⁡x;u1=Lu0+Nu0=-u0-2u0xx-u0xxxxIα1=0;un=0,n=2,3,⋯.Hence the solution to (41)-(42) is(47)ux,t=cos⁡x.Case 2 (r≠0).
According to iterative scheme (9)-(10), we have (48)u0=f=cos⁡x;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1=r-1+1cos⁡x1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα=r-12+r-1+1cos⁡x1Γ2α+1t2α;u3=Lu2+Nu0+u1+u2-Nu0+u1=r-1u2-2u2xx-u2xxxxIα1Γ2α+1t2α=r-13+r-12+r-1+1cos⁡x1Γ3α+1t3α;u4=Lu3+Nu0+u1+u2+u3-Nu0+u1+u2=r-1u3-2u3xx-u3xxxxIα1Γ3α+1t3α=r-14+r-13+r-12+r-1+1cos⁡x1Γ4α+1t4α. Generally, for n=1,2,⋯, one can derive that(49)un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α=r-1n+r-1n-1+⋯+r-12+r-1+1cos⁡x1Γn-1α+1tn-1α=-1r1-r-1n+1cos⁡x1Γnα+1tnαHence the N-th approximate solution to (41)-(42) is(50)uNx,t=-1rcos⁡x∑n=0N1-r-1n+11Γnα+1tnα and the exact solution to (41)-(42) is(51)ux,t=limN→∞uNx,t=-1rcos⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnα.As a consequence, the solution of (41)-(42) is(52)ux,t=cos⁡xr=0;-1rcos⁡xlimN→∞∑n=0N1-r-1n+11Γnα+1tnαr≠0.Numerical Simulation. See Figure 3.Figure 3 (a) (b) (c)


## 4.4. Linear S-H Equation with Dispersion and Initial Valueu(x,0)=ex

Consider the linear S-H equation with dispersion [41] as follows:(53)Dαu+1-ru+2uxx-σuxxx+uxxxx=00<α≤1with initial value(54)ux,0=ex.Here σ∈R is a parameter. Clearly, applying Iα on both sides of (53), then initial value problem (53)-(54) is equivalent to the following integral equation:(55)ux,t=fx+Lux,t+Nux,t, where(56)fx=ex,Lu=Iαr-1u-2uxx+σuxxx-uxxxx,Nu=0. The solution to system (53)-(54) what we are looking for has the form(57)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (58)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx+σu1xxx-u0xxxxIα1=r-4+σex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx+σu1xx-u1xxxxIα1Γα+1tα=r-4+σ2ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx+σun-1xxx-un-1xxxxIα1Γn-1α+1tn-1α=r-4+σnex1Γnα+1tnαn=2,3,⋯ Hence the N-th approximate solution to (53)-(54) is(59)uNx,t=ex∑n=0Nr-4+σn1Γnα+1tnα and the exact solution to (53)-(54) is(60)ux,t=limN→∞uNx,t=exEαr-4+σtα.Numerical Simulation. See Figure 4.Figure 4 (a) (b) (c)


## 5. Nonlinear Time-Fractional Swift-Hohenberg Equation

In this subsection, we apply the iterative method introduced in Section3 to nonlinear time-fractional S-H equation with such as ex.


### 5.1. Nonlinear S-H Equation with Initial Valueu(x,0)=ex

Consider the following nonlinear S-H equation with dispersion:(61)Dαu+1-ru+2uxx+uxxxx=ul-uxl0<α≤1with initial value(62)ux,0=ex.Clearly, applying Iα on both sides of (61), then initial value problem (61)-(62) is equivalent to the following integral equation:(63)ux,t=fx+Lux,t+Nux,t, where(64)fx=ex,Lu=Iαr-1u-2uxx-uxxxx,Nu=Iαul-uxl. The solution to system (61)-(62) what we are looking for has the form(65)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (66)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1+u0l-u0xlIα1=r-4ex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα+Iαu0+u12-u0+u1x2-u02+u0x2=r-42ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α++Iαu0+⋯+un-1l-u0+⋯+un-1xl-u0+⋯+un-2l+u0+⋯+un-2xl=r-4nex1Γnα+1tnαn=2,3,⋯. Hence the N-th approximate solution to (61)-(62) is(67)uNx,t=ex∑n=0Nr-4n1Γnα+1tnα and the exact solution to (61)-(62) is(68)ux,t=limN→∞uNx,t=exEαr-4tα.Numerical Simulation. See Figure 5.Figure 5 (a) (b) (c)


### 5.2. Nonlinear Time-Fractional S-H Equation with Dispersion and Initial Valueu(x,0)=ex

Consider the following nonlinear S-H equation with dispersion:(69)Dαu+1-ru+2uxx-σuxxx+uxxxx=ul-uxl0<α≤1with initial value(70)ux,0=ex.Clearly, applying Iα to both sides of (69), then initial value problem (69)-(70) is equivalent to the following integral equation:(71)ux,t=fx+Lux,t+Nux,t, where(72)fx=ex,Lu=Iαr-1u-2uxx+σuxxx-uxxxx,Nu=Iαul-uxl. The solution to system (69)-(70) what we are looking for has the form(73)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (74)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx+σu0xxx-u0xxxxIα1+Iαu0l-u0xl=r-4+σex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα+Iαu0+u1l-u0+u1xl-u0l+u0xl=r-4+σ2ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α++Iαu0+⋯+un-1l-u0+⋯+un-1xl-u0+⋯+un-2l+u0+⋯+un-2xl=r-4+σnex1Γnα+1tnαn=2,3,⋯ Hence the N-th approximate solution to (69)-(70) is(75)uNx,t=ex∑n=0Nr-4+σn1Γnα+1tnα and the exact solution to (61)-(70) is(76)ux,t=limN→∞uNx,t=exEαr-4+σtα.Numerical Simulation. See Figure 6.Figure 6 (a) (b) (c)


## 5.1. Nonlinear S-H Equation with Initial Valueu(x,0)=ex

Consider the following nonlinear S-H equation with dispersion:(61)Dαu+1-ru+2uxx+uxxxx=ul-uxl0<α≤1with initial value(62)ux,0=ex.Clearly, applying Iα on both sides of (61), then initial value problem (61)-(62) is equivalent to the following integral equation:(63)ux,t=fx+Lux,t+Nux,t, where(64)fx=ex,Lu=Iαr-1u-2uxx-uxxxx,Nu=Iαul-uxl. The solution to system (61)-(62) what we are looking for has the form(65)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (66)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx-u0xxxxIα1+u0l-u0xlIα1=r-4ex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα+Iαu0+u12-u0+u1x2-u02+u0x2=r-42ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α++Iαu0+⋯+un-1l-u0+⋯+un-1xl-u0+⋯+un-2l+u0+⋯+un-2xl=r-4nex1Γnα+1tnαn=2,3,⋯. Hence the N-th approximate solution to (61)-(62) is(67)uNx,t=ex∑n=0Nr-4n1Γnα+1tnα and the exact solution to (61)-(62) is(68)ux,t=limN→∞uNx,t=exEαr-4tα.Numerical Simulation. See Figure 5.Figure 5 (a) (b) (c)


## 5.2. Nonlinear Time-Fractional S-H Equation with Dispersion and Initial Valueu(x,0)=ex

Consider the following nonlinear S-H equation with dispersion:(69)Dαu+1-ru+2uxx-σuxxx+uxxxx=ul-uxl0<α≤1with initial value(70)ux,0=ex.Clearly, applying Iα to both sides of (69), then initial value problem (69)-(70) is equivalent to the following integral equation:(71)ux,t=fx+Lux,t+Nux,t, where(72)fx=ex,Lu=Iαr-1u-2uxx+σuxxx-uxxxx,Nu=Iαul-uxl. The solution to system (69)-(70) what we are looking for has the form(73)ux,t=∑n=0∞unx,t. According to iterative scheme (9)-(10), one can obtain (74)u0=f=ex;u1=Lu0+Nu0=r-1u0-2u0xx+σu0xxx-u0xxxxIα1+Iαu0l-u0xl=r-4+σex1Γα+1tα;u2=Lu1+Nu0+u1-Nu0=r-1u1-2u1xx-u1xxxxIα1Γα+1tα+Iαu0+u1l-u0+u1xl-u0l+u0xl=r-4+σ2ex1Γ2α+1t2α⋮un=Lun-1+Nu0+u1+⋯+un-1-Nu0+u1+⋯+un-2=r-1un-1-2un-1xx-un-1xxxxIα1Γn-1α+1tn-1α++Iαu0+⋯+un-1l-u0+⋯+un-1xl-u0+⋯+un-2l+u0+⋯+un-2xl=r-4+σnex1Γnα+1tnαn=2,3,⋯ Hence the N-th approximate solution to (69)-(70) is(75)uNx,t=ex∑n=0Nr-4+σn1Γnα+1tnα and the exact solution to (61)-(70) is(76)ux,t=limN→∞uNx,t=exEαr-4+σtα.Numerical Simulation. See Figure 6.Figure 6 (a) (b) (c)


## 6. Concluding Remark

This paper introduce an iterative method which has considerable power in constructing approximated solutions and even exact solutions to time-fractional differential equation. We take the linear and nonlinear Swift-Hohenberg equations with different initial conditions to illustrate the effectiveness of such method.



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