# Application of Variational Iteration Method to Fractional Hyperbolic Partial Differential Equations

Authors: Fadime Dal
Journal: Mathematical Problems in Engineering
 (2009)

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## Abstract

The solution of the fractional hyperbolic partial differential equation is obtained by means of the variational iteration method. Our numerical results are compared with those obtained by the modified Gauss elimination method. Our results reveal that the technique introduced here is very effective, convenient, and quite accurate to one-dimensional fractional hyperbolic partial differential equations. Application of variational iteration technique to this problem has shown the rapid convergence of the sequence constructed by this method to the exact solution.


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## Body

## 1. Introduction

It is known that various problems in fluid mechanics (dynamics, elasticity) and other areas of physics lead to fractional partial differential equations. Methods of solutions of problems for fractional differential equations have been studied extensively by many researchers (see [1–11]).The variational iteration method (VIM), which was proposed by He (see, e.g., [12–21]), was successfully applied to autonomous ordinary and partial differential equations and other fields. He [15] was the first research who applied the VIM to fractional differential equations. Odibat and Momani [22] extended the application of this method to provide approximate solutions for initial value problems of nonlinear partial differential equations of fractional order. VIM [23–25] is relatively a new approach to provide an analytical approximation to linear and nonlinear problems which is particularly valuable tools for scientists and applied mathematicians. Yulita et al. [26] used the VIM to obtain analytical solutions of fractional heat- and wave-like equations with variable coefficients. In the Ashyralyev et al. [27], the mixed boundary value problem for the multidimensional fractional hyperbolic equation is considered. The first order of accuracy in t and the second order of accuracy in space variables for the approximate solution of problem were presented. The stability estimates for the solution of this difference scheme and its first- and second-order difference derivatives were established. A procedure of modified Gauss elimination method [28] was used for solving this difference scheme in the case of one-dimensional fractional hyperbolic partial differential equations.In this paper, we apply variational iteration method to fractional hyperbolic partial differential equations and then we compare the results with those obtained using modified Gauss elimination method [27].


## 2. Definitions

Definition 2.1.
A reel functionf(x), x>0, is said to be in the space CM, M∈R, if there exists a real number p(>M), such that f(x)=xpf1(x), where f1(x)∈C[0,∞), and it is said to be in the space CMm if fm∈CM, m∈N.Definition 2.2.
Iff(x)∈C[a,b] and a<x<b, then
(2.1)Ia+αf(x)=1Γ(α)∫axf(t)(x-t)1-αdt,
where -∞<α<∞ is called the Riemann-Liouville fractional integral operator of order α.Definition 2.3.
For0<α<1, we let
(2.2)Da+αf(x)=1Γ(1-α)ddx∫axf(t)(x-t)αdt,
which is called the Riemann-Liouville fractional derivative operator of order α.


## 3. Variational teration Method

In the present paper, the mixed boundary value problem for the multidimensional fractional hyperbolic equation(3.1)∂2u(x,t)∂t2-∑r=1m(ar(x)uxr)xr+Dt1/2u(x,t)=f(x,t),x=(x1,…,xm)∈Ω,0<t<1,u(x,0)=0,ut(x,0)=0,x∈Ω¯,u(x,t)=0,x∈S
is considered. Here Ω is the unit open cube in the m-dimensional Euclidean space ℝm:Ω={x=(x1,…,xm):0<xj<1,1≤j≤m} with boundary S, Ω¯=Ω∪S, ar(x),  (x∈Ω), and f(x,t)(t∈(0,1),x∈Ω) are given smooth functions and ar(x)≥a>0.The correction functional for (3.1) can be approximately expressed as follows:
(3.2)un+1(x,t)=un(x,t)+∫0tλ[∂2u(x,s)∂s2-∑r=1m(ar(x)ũxr)xr+Ds1/2ũ(x,s)-f(x,s)]ds,
where λ is a general Lagrangian multiplier [29] and ũ is considered as a restricted variation as a restricted variation [21], that is, δũ=0, and u0(x,t) is its initial approximation. Using the above correction functional stationary and noticing that δũ=0, we obtain
(3.3)δun+1(x,t)=δun(x,t)+∫0tδλ[∂un2(x,s)∂s2]ds,δun+1(x,t)=δun(x,t)-∂λ∂sδun(x,s)|s=t+λ∂∂sδun(x,s)|s=t+∫0t∂2λ(t,s)∂s2δun(x,s)ds=0.
From the above relation for any δun, we get the Euler-Lagrange equation:
(3.4)∂λ2(t,s)∂s2=0
with the following natural boundary conditions:
(3.5)1-∂λ(t,s)∂s|s=t=0,λ(t,s)|s=t=0.
Therefore, the Lagrange multiplier can be identified as follows:
(3.6)λ(t,s)=s-t.
Substituting the identified Lagrange multiplier into (3.2), following variational iteration formula can be obtained:(3.7)un+1(x,t)=un(x,t)+∫0t(s-t)[∂2u(x,s)∂s2-∑r=1m(ar(x)uxr)xr+Ds1/2u(x,s)-f(x,s)]ds.
In this case, let an initial approximation beu0(x,t)=u(x,0)+tut(x,0). Then approximate solution takes the form u(x,t)=limn→∞un(x,t).


### 3.1. The Difference Scheme

The discretization of problem (3.1) is carried out in two steps. In the first step, let us define the grid space
(3.8)Ω̃h={x=xr=(h1r1,…,hmrm),r=(r1,…,rm),0≤rj≤Nj,hjNj=1,j=1,…,m},Ωh=Ω̃h∩Ω,Sh=Ω̃h∩S.
We introduce the Banach space L2h=L2(Ω̃h) of the grid functions φh(x)={φ(h1r1,…,hmrm)} defined on Ω̃h, equipped with the norm
(3.9)∥φh∥L2(Ω̃h)=(∑x∈Ωh¯|φh(x)|2h1⋯hm)1/2.
To the differential operator Ax generated by problem (3.1), we assign the difference operator Ahx by the formula
(3.10)Ahxuxh=-∑r=1m(ar(x)uxr̅h)xr,jr
acting in the space of grid functions uh(x), satisfying the conditions uh(x)=0 for all x∈Sh. It is known that Ahx is a self-adjoint positive definite operator in L2(Ω̃h). With the help of Ahx we arrive at the initial boundary value problem
(3.11)d2vh(x,t)dt2+Ahxvh(x,t)+Dt1/2vh(x,t)=fh(x,t),0≤t≤1,x∈Ωh,vh(x,0)=0,dvh(x,0)dt=0,x∈Ω̃
for an finite system of ordinary fractional differential equations.In the second step, applying the first order of approximation formula(1/π)∑m=1k(Γ(k-m+1/2)/(k-m)!)((u(tk)-u(tk-1))/τ1/2) for Dt1/2u(t) and using the first order of accuracy stable difference scheme for hyperbolic equations (see [30]), one can present the first order of accuracy difference scheme:
(3.12)uk+1h(x)-2ukh(x)+uk-1h(x)τ2+Ahxuk+1h+1π∑m=1kΓ(k-m+1/2)(k-m)!(umh-um-1h)τ1/2=fkh(x),x∈Ω̃h,fkh(x)=f(xn,tk),tk=kτ,1≤k≤N-1,Nτ=1,u1h(x)-u0h(x)τ=0,u0h(x)=0,x∈Ω̃h
for the approximate solution of problem (3.1). Here Γ(k-m+1/2)=∫0∞tk-m-1/2e-tdt.


### 3.2. Example  1

For the numerical result, the mixed problem(3.13)Dt2u(x,t)-Dt1/2u(x,t)-uxx(x,t)=f(x,t),f(x,t)=(2-8t3/23π+(πt)2)sinπx,0<t,x<1,u(x,0)=0,ut(x,0)=0,0≤x≤1,u(t,0)=u(t,1)=0,0≤t≤1
for solving the one-dimensional fractional hyperbolic partial differential equation is considered.According to the formula (3.7), the iteration formula for (3.13) is given by
(3.14)un+1(x,t)=un(x,t)+∫0t(s-t)[∂un2(x,s)∂s2-Ds1/2un(x,s)-∂un2(x,s)∂x2-(2-8s3/23π+(πs)2)sin(πx)]ds.
Now we start with an initial approximation:(3.15)u0(x,t)=u(x,0)+tut(x,0).
Using the above iteration formula (3.14), we can obtain the other components as
(3.16)u0(x,t)=0,u1(x,t)=(-128420πt7/2+π2t412+t2)sin(πx),u2(x,t)=141580πsin(πx)t5/2[512π2t3+12672t-693πt5/2]+12810395sin(πx)t11/2π3/2-1360π4sin(πx)t6-112π2sin(πx)t4+1420π[-128t7/2+35π5/2t4+420t2π]sin(πx),⋮
and so on; in the same manner the rest of the components of the iteration formula (3.14) can be obtained using the Maple package.


### 3.3. Example  2

We consider one-dimensional fractional hyperbolic partial differential equation as follows:(3.17)Dt2u(x,t)+Dt1/2u(x,t)-uxx(x,t)+u(x,t)=f(x,t),f(x,t)=(2(1-x-exp(-x)))+8t3/23π(1-x-exp(-x))+(1-x)t2x,0<t,x<1,u(x,0)=0,ut(x,0)=0,0≤x≤1,u(t,0)=u(t,1)=0,0≤t≤1.
The iteration formula for (3.17) is given by(3.18)un+1(x,t)=un(x,t)+∫0t(s-t)[∂un2(x,s)∂s2+Ds1/2un(x,s)-∂un2(x,s)∂x2+u(x,s)-[(2(1-x-exp(-x)))8t3/23π+8t3/23π(1-x-exp(-x))+(1-x)t2]]ds.
When an initial approximation is u0(x,t)=u(x,0)+tut(x,0), we have the other components as
(3.19)uo(x,t)=0,u1(x,t)=(1-x-exp(-x))(32105t7/2+t2)+t412(1-x),u2(x,t)=(1-x-exp(-x))(32105t7/2+t2)+t412(1-x)+1420[128t7/2(-1+x+exp(-x))+35t4(x-1)+420t2(-1+x+exp(-x))]+141580π[12672t7/2(-1+x+exp(-x))+512t11/2(-1+x)+693t5(-1+x+exp(-x))]-141580(512exp(-x)t11/2+3465t4exp(-x))+183160[1024t11/2(-1+x+exp(-x))+231t6(-1+x)+6930t4(-1+x+exp(-x))]-1420π[128t7/2(-1+x+exp(-x))+35t4π(-1+x)+420t2π(-1+x+exp(-x))]⋮
and so on. For (3.18), the rest of the components of the iteration formula can be obtained using the Maple 10 package.


## 3.1. The Difference Scheme

The discretization of problem (3.1) is carried out in two steps. In the first step, let us define the grid space
(3.8)Ω̃h={x=xr=(h1r1,…,hmrm),r=(r1,…,rm),0≤rj≤Nj,hjNj=1,j=1,…,m},Ωh=Ω̃h∩Ω,Sh=Ω̃h∩S.
We introduce the Banach space L2h=L2(Ω̃h) of the grid functions φh(x)={φ(h1r1,…,hmrm)} defined on Ω̃h, equipped with the norm
(3.9)∥φh∥L2(Ω̃h)=(∑x∈Ωh¯|φh(x)|2h1⋯hm)1/2.
To the differential operator Ax generated by problem (3.1), we assign the difference operator Ahx by the formula
(3.10)Ahxuxh=-∑r=1m(ar(x)uxr̅h)xr,jr
acting in the space of grid functions uh(x), satisfying the conditions uh(x)=0 for all x∈Sh. It is known that Ahx is a self-adjoint positive definite operator in L2(Ω̃h). With the help of Ahx we arrive at the initial boundary value problem
(3.11)d2vh(x,t)dt2+Ahxvh(x,t)+Dt1/2vh(x,t)=fh(x,t),0≤t≤1,x∈Ωh,vh(x,0)=0,dvh(x,0)dt=0,x∈Ω̃
for an finite system of ordinary fractional differential equations.In the second step, applying the first order of approximation formula(1/π)∑m=1k(Γ(k-m+1/2)/(k-m)!)((u(tk)-u(tk-1))/τ1/2) for Dt1/2u(t) and using the first order of accuracy stable difference scheme for hyperbolic equations (see [30]), one can present the first order of accuracy difference scheme:
(3.12)uk+1h(x)-2ukh(x)+uk-1h(x)τ2+Ahxuk+1h+1π∑m=1kΓ(k-m+1/2)(k-m)!(umh-um-1h)τ1/2=fkh(x),x∈Ω̃h,fkh(x)=f(xn,tk),tk=kτ,1≤k≤N-1,Nτ=1,u1h(x)-u0h(x)τ=0,u0h(x)=0,x∈Ω̃h
for the approximate solution of problem (3.1). Here Γ(k-m+1/2)=∫0∞tk-m-1/2e-tdt.


## 3.2. Example  1

For the numerical result, the mixed problem(3.13)Dt2u(x,t)-Dt1/2u(x,t)-uxx(x,t)=f(x,t),f(x,t)=(2-8t3/23π+(πt)2)sinπx,0<t,x<1,u(x,0)=0,ut(x,0)=0,0≤x≤1,u(t,0)=u(t,1)=0,0≤t≤1
for solving the one-dimensional fractional hyperbolic partial differential equation is considered.According to the formula (3.7), the iteration formula for (3.13) is given by
(3.14)un+1(x,t)=un(x,t)+∫0t(s-t)[∂un2(x,s)∂s2-Ds1/2un(x,s)-∂un2(x,s)∂x2-(2-8s3/23π+(πs)2)sin(πx)]ds.
Now we start with an initial approximation:(3.15)u0(x,t)=u(x,0)+tut(x,0).
Using the above iteration formula (3.14), we can obtain the other components as
(3.16)u0(x,t)=0,u1(x,t)=(-128420πt7/2+π2t412+t2)sin(πx),u2(x,t)=141580πsin(πx)t5/2[512π2t3+12672t-693πt5/2]+12810395sin(πx)t11/2π3/2-1360π4sin(πx)t6-112π2sin(πx)t4+1420π[-128t7/2+35π5/2t4+420t2π]sin(πx),⋮
and so on; in the same manner the rest of the components of the iteration formula (3.14) can be obtained using the Maple package.


## 3.3. Example  2

We consider one-dimensional fractional hyperbolic partial differential equation as follows:(3.17)Dt2u(x,t)+Dt1/2u(x,t)-uxx(x,t)+u(x,t)=f(x,t),f(x,t)=(2(1-x-exp(-x)))+8t3/23π(1-x-exp(-x))+(1-x)t2x,0<t,x<1,u(x,0)=0,ut(x,0)=0,0≤x≤1,u(t,0)=u(t,1)=0,0≤t≤1.
The iteration formula for (3.17) is given by(3.18)un+1(x,t)=un(x,t)+∫0t(s-t)[∂un2(x,s)∂s2+Ds1/2un(x,s)-∂un2(x,s)∂x2+u(x,s)-[(2(1-x-exp(-x)))8t3/23π+8t3/23π(1-x-exp(-x))+(1-x)t2]]ds.
When an initial approximation is u0(x,t)=u(x,0)+tut(x,0), we have the other components as
(3.19)uo(x,t)=0,u1(x,t)=(1-x-exp(-x))(32105t7/2+t2)+t412(1-x),u2(x,t)=(1-x-exp(-x))(32105t7/2+t2)+t412(1-x)+1420[128t7/2(-1+x+exp(-x))+35t4(x-1)+420t2(-1+x+exp(-x))]+141580π[12672t7/2(-1+x+exp(-x))+512t11/2(-1+x)+693t5(-1+x+exp(-x))]-141580(512exp(-x)t11/2+3465t4exp(-x))+183160[1024t11/2(-1+x+exp(-x))+231t6(-1+x)+6930t4(-1+x+exp(-x))]-1420π[128t7/2(-1+x+exp(-x))+35t4π(-1+x)+420t2π(-1+x+exp(-x))]⋮
and so on. For (3.18), the rest of the components of the iteration formula can be obtained using the Maple 10 package.


## 4. Conclusions

Variational iteration method is a powerful and efficient technique in finding exact and approximate solutions for one-dimensional fractional hyperbolic partial differential equations. The solution procedure is very simple by means of variational theory, and only a few steps lead to highly accurate solutions which are valid for the whole solution domain. The results of applying variational iteration method are exactly the same as those obtained by modified Gauss elimination method [27]. All Computations are performed by Maple 10 package program.Figure1 shows the exact solution of (3.13). Figure 2 shows difference scheme solution of (3.13). Figure 3 shows approximate solution by VIM for (3.13). Figure 4 shows the exact solution of (3.17). Figure 5 shows approximate solution by VIM for (3.17).Figure 1
The surface shows the exact solutionu(x,t) for (3.13).Figure 2
Difference scheme solution [27] for (3.13).Figure 3
Variational iteration method for (3.13).Figure 4
The surface shows the exact solutionu(x,t) for (3.17).Figure 5
Variational iteration method for (3.17).



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