# On the Solution of Local Fractional Differential Equations Using Local Fractional Laplace Variational Iteration Method

Authors: Pranay Goswami; Rubayyi T. Alqahtani
Journal: Mathematical Problems in Engineering
 (2016)

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## Abstract

We present iteration formulae of a fractional space-time telegraph equation using the combination of fractional variational iteration method and local fractional Laplace transform.


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## Body

## 1. Introduction

Fractional differential equations have number of properties and play master role in different fields of study. These equations and their solutions gained remarkable interest due to its striking role in every science and technology [1–4]. Various fields like diffusion, transport theory, scattering theory, rheology, quantitative biology, and so forth are successfully applying fractional differential equations to define and explain number of phenomena but fractional calculus is not perfectly applicable in the case of fractal functions. In order to deal with fractal problems in various fields, the concept of local fractional derivative was developed. The local fractional calculus was introduced by Yang and further applications of this derivative can be found in the references [5–8]. There are variety of analytical methods for solving them like Adomian decomposition method, generalized differential transform method, homotopy perturbation method, separating variables, local fractional Laplace method, local fractional Sumudu method, variational iteration method, and fractional variational iteration method [5, 6, 9–14].In this paper, we develop an iteration formula to solve generalized fractional space-time telegraph equation by using the combination of fractional variational iteration method and local fractional Laplace transform.


## 2. Definitions and Mathematical Preliminaries

Definition 1 (see [7, 8]).
The partition of interval [a,b] is denoted as tj,tj+1, j=0,…,N-1, t0=a, and tN=b with Δtj=tj+1-tj and Δt=max⁡{Δt0,Δt1,…}. Local fractional integral of f(x) in the interval [a,b] is given by [7, 8]:(1)Iabαfx=1Γ1+α∫abftdtα=1Γ1+αlimΔt→0⁡∑j=0N-1ftjΔtjα.Definition 2.
The local fractional derivative off(x)∈Cα(a,b) of order α at x=x0 is defined [see [7, 8]] as(2)fαx0=dαdxαfxx=x0=limx→x0⁡Δαfx-fx0x-x0α,where Δαfx-fx0≅Γα+1fx-fx0.Definition 3.
The local fractional Laplace transform off(x) of order α is defined [15] as(3)L-αfx=1Γ1+α∫0∞Eα-sαxαfxdxα,0<α≤1,where x∈C and Eαx is Mittage Leffler function ∑k=0∞xk/ak!.And inverse local fractional Laplace transform is defined as(4)L-α-1fs=fx,0<α≤1.On using the above definition of local fractional Laplace transform, the following result can easily be obtained [15]:(5)L-αfnαx=snαL-αfx-∑k=1nsk-1αfn-kα0,L-αxkα=Γ1+kαsk+1α.


## 3. The Local Fractional Laplace Variational Iteration Method

This section introduces the idea of local fractional Laplace variational method for the following fractional space-time telegraph equation:(6)D0+nαvx,t=Lαvx,t+Nαvx,t+fx,t,where 0<α≤1, x,t≥0, Lα is a linear operator, Nα is nonlinear operator, and f is a source term.Taking local fractional Laplace transform of (6), both sides with respect to “x” are as follows:(7)snαL-αfx-∑k=1nsk-1αfn-kα0=L-αLαvx,t+Nαvx,t+fx,t.For an algebraic equation, the iteration formula can be constructed as(8)xn+1=xn+λαfxn.The optimality condition for the extreme δxn+1/δxn=0, leads to(9)λα=-1f′xn,where δ is the classical variational operator.By the formula of (8), we get the iteration formula for (7) as follows: (10)L-αvn+1s,t=L-αvns,t+λαsnαL-αvns,t-∑k=1nsk-1αvnn-kα0,t-L-αLαvnx,t+Nαvnx,t+fx,t.Put λα=-s-nα, the Lagrange multiplier [16] in (10):(11)L-αvn+1s,t=∑k=1nsk-1-nαvnn-kα0,t+s-nαL-αLαvnx,t+Nαvnx,t+fx,t.Taking inverse local fractional Laplace transform into account, we arrived at(12)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+s-nαL-αLαvnx,t+Nαvnx,t+fx,t.This is the iteration formula for (6).Example 4.
To illustrate the above method, we can consider the following linear equation:(13)dαvx,tdxα=avx,t,0<x≤T,0<α≤1,a∈R.We construct the following iteration formula with the help of (12):(14)vn+1x,t=vn0,t-L-α-1s-αL-αavnx,t.Now, applying local fractional Laplace transform to the above equation, find(15)vn+1x,t=vn0,t-L-α-1s-2αavns,t.This is the iteration formula for (13).
Let us start fromv0x,t=v0,t=v0.
Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (15), and solving inverse local fractional Laplace transform, we have(16)v1x,t=v0-axαv0Γ1+α.For n=1, v2x is given by the following iteration:(17)v2x,t=v10,t-L-α-1s-2αav1s,t.Using (16) and after easy calculations, we get (18)v2x,t=v0-axαv0Γ1+α+axα2v0Γ1+2α.Similarly for n=2, v2x is given by the following iteration:(19)v3x=v0-axαv0Γ1+α+axα2v0Γ1+2α-axα3v0Γ1+3α⋮vx=∑n=0∞vnx=v0∑k=0∞-akxkαΓ1+kα.This is the exact solution of (13).Example 5.
The following space-time fractional homogeneous telegraph equation can also be solved by the above introduced local fractional Laplace variational iteration method:(20)∂2αv∂x2αvx,t=∂2αv∂t2αx,t+∂αv∂tαx,t-vx,t,x,t≥0,0<α≤1with initial conditions(21)v0,t=Eα-tα,vα0,t=Eα-tα.We can find the iteration formula for the above with the help of (12) as(22)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α∂2αvn∂t2αx,t+∂αvn∂tαx,t-vnx,t.Consider initial iteration as follows:(23)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=Eα-tα+xαΓ1+αEα-tα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (22); we have (24)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α∂2αv0∂t2αx,t+∂αv0∂tαx,t-v0x,t.Using (23) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (25)v1x,t=Eα-tα+xαΓ1+αEα-tα-x2αΓ1+2αEα-tα-x3αΓ1+3αEα-tα.Similarly, we can find(26)v2x,t=Eα-tα+xαΓ1+αEα-tα-x2αΓ1+2αEα-tα-x3αΓ1+3αEα-tα+x4αΓ1+4αEα-tα+x5αΓ1+5αEα-tα,v3x,t=Eα-tα+xαΓ1+αEα-tα-x2αΓ1+2αEα-tα-x3αΓ1+3αEα-tα+x4αΓ1+4αEα-tα+x5αΓ1+5αEα-tα-x6αΓ1+6αEα-tα-x7αΓ1+7αEα-tα.Consequently, we obtain(27)vx,t=Eα-tα1-x2αΓ1+2α+x4αΓ1+4α-x6αΓ1+6α+⋯+Eα-tαxαΓ1+α-x3αΓ1+3α+x5αΓ1+5α-x7αΓ1+7α+⋯,vx,t=Eα-tαcosα⁡xα+Eα-tαsinα⁡xα.Remark 6.
The result in (27) is the same as the result obtained by Jafari and Jassim [17].Example 7.
The following space-time fractional homogeneous telegraph equation can also be solved by the above introduced local fractional Laplace variational iteration method:(28)∂2αv∂x2αvx,t=∂2αv∂t2αx,t+2∂αv∂tαx,t+vx,t,x,t≥0,0<α≤1with initial conditions(29)v0,t=1-Eα-tα,∂α∂xαv0,t=0.Then, we can find the iteration formula for the above with the help of (12) as(30)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α∂2αvn∂t2αx,t+2∂αvn∂tαx,t+vnx,t.Consider initial iteration as follows:(31)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=1-Eα-tα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (30); we have (32)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α∂2αv0∂t2αx,t+2∂αv0∂tαx,t+v0x,t.Using (31) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (33)v1x,t=1-Eα-tα+x2αΓ1+2α.Similarly, we can find(34)v2x,t=1-Eα-tα+x2αΓ1+2α-x4αΓ1+4α,v3x,t=1-Eα-tα+x2αΓ1+2α+x4αΓ1+4α+x6αΓ1+6α⋮Consequently, we obtain(35)vx,t=1-Eα-tα1+x2αΓ1+2α+x4αΓ1+4α+x6αΓ1+6α+⋯,vx,t=coshα⁡xα-Eα-tα.Example 8.
We consider the following local fractional Laplace equation:(36)∂2αv∂x2αvx,t+∂2αv∂t2αx,t=0,x,t≥0,0<α≤1with the initial condition(37)v0,t=-Eαtα,∂α∂xαv0,t=0.Then, we can find the iteration formula for the above with the help of (12) as(38)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α-∂2αvn∂t2αx,t.Consider initial iteration as follows:(39)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=-Eαtα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (38); we have (40)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α-∂2αv0∂t2αx,t.Using (39) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (41)v1x,t=-Eαtα+x2αΓ1+2αEαtα.Similarly, we can find(42)v2x,t=-Eαtα+x2αΓ1+2αEαtα-x4αΓ1+4αEαtα,v3x,t=-Eαtα+x2αΓ1+2αEαtα-x4αΓ1+4αEαtα+x6αΓ1+6αEαtα⋮Consequently, we obtain(43)vx,t=-Eαtα1-x2αΓ1+2α+x4αΓ1+4α-x6αΓ1+6α+⋯,vx,t=-Eαtαcosα⁡xα.Example 9.
We consider the following local fractional Laplace equation:(44)∂2αv∂x2αvx,t+∂2αv∂t2αx,t=0with the initial condition(45)v0,t=0,∂α∂xαv0,t=-Eαtα.Then, we can find the iteration formula for the above with the help of (12) as(46)vn+1x,t=L-α-1∑k=1nsk-1-nαvnn-kα0,t+L-α-1s-2αL-α-∂2αvn∂t2αx,t.Consider initial iteration as follows:(47)v0x,t=L-α-1∑k=1nsk-1-nαvn-kα0,t,L-α-1s-αv0,t+s-2αvα0,t=-xαΓ1+αEαtα.Now, by putting the values of “n,” we get the iterations; for v1x,t, put n=0 in (46); we have (48)v1x,t=L-α-1∑k=1nsk-1-nαv0n-kα0,t+L-α-1s-2αL-α-∂2αv0∂t2αx,t.Using (47) and applying local fractional Laplace and inverse local fractional Laplace transform, we get (49)v1x,t=-xαΓ1+αEαtα+x3αΓ1+3αEαtα.Similarly, we can find(50)v2x,t=-xαΓ1+αEαtα+x3αΓ1+3αEαtα-x5αΓ1+5αEαtα,v3x,t=-xαΓ1+αEαtα+x3αΓ1+3αEαtα-x5αΓ1+5αEαtα+x7αΓ1+7αEαtα⋮Consequently, we obtain(51)vx,t=-EαtαxαΓ1+2α-x3αΓ1+4α+x5αΓ1+6α-⋯,vx,t=-Eαtαsinα⁡xα.


## 4. Conclusion

In this work, we have presented successful demonstration of the local fractional Laplace variational iteration method for solutions of a wide class of problems. Analytical solutions of the telegraph and Laplace equations on Cantor sets involving local fractional derivatives are efficiently developed.



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