Solution of Some Fractional Order Telegraph Equations · Solution of Some Fractional Order Telegraph Equations Soluci on de algunas ecuaciones telegr a cas de orden fraccional Leda

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Revista Colombiana de MatematicasVolumen 48(2014)2, paginas 247-267

Solution of Some Fractional Order

Telegraph Equations

Solucion de algunas ecuaciones telegraficas de orden fraccional

Leda Galue

Universidad del Zulia, Maracaibo, Venezuela

Abstract. In recent years, there has been a great interest in fractional differ-ential equations due to their frequent appearance in various fields, and theirmore accurate models of systems under consideration provided by fractionalderivatives. In particular, fractional order telegraph equations have been con-sidered and solved for many researchers, using different methods. In this paperwe derived the solution of two homogeneous space-time fractional telegraphequations using the generalized differential transform method. The derivativesare considered in Caputo sense and the solutions are given in terms of gen-eralized Mittag-Leffler function and the generalized Wright function. Further,various graphics are included which show the behavior of the solution ob-tained, and results given earlier by Momani, Odibat and Momani, Yildrim,Garg and Sharma, and Garg et al. are obtained as particular cases of onesour.

Key words and phrases. Fractional order telegraph equation, Generalized dif-ferential transform method, Caputo fractional derivative, Generalized Mittag-Leffler function, Generalized Wright function.

2010 Mathematics Subject Classification. 35C05, 35C10.

Resumen. En los ultimos anos, ha habido un gran interes en las ecuacionesdiferenciales fraccionales debido a su frecuente aparicion en diversos cam-pos, y a sus modelos mas precisos de los sistemas en estudio proporciona-dos por las derivadas fraccionales. En particular, las ecuaciones telegraficasfraccionales han sido consideradas y resueltas por muchos investigadores, uti-lizando diferentes metodos. En este trabajo se derivo la solucion de dos ecua-ciones telegraficas homogeneas, con espacio-tiempo fraccionales, utilizando elmetodo de la transformada diferencial generalizada. Las derivadas se consid-eran en el sentido Caputo y las soluciones se dan en terminos de la funcion

247

248 LEDA GALUE

generalizada de Mittag-Leffler y la funcion de Wright generalizada. Ademas,se incluyen varias graficas que muestran el comportamiento de la solucionobtenida, y los resultados dados anteriormente por Momani, Odibat y Mo-mani, Yildrim, Garg y Sharma, y Garg et al. se obtienen como casos particu-lares de los nuestros.

Palabras y frases clave. Ecuacion telegrafica de orden fraccional, metodo de latransformada diferencial generalizada, derivada fraccional de Caputo, funciongeneralizada de Mittag-Leffler, funcion de Wright generalizada.

1. Introduction

The telegraph equation is a partial differential equation with constant coeffi-cients given by

utt − c2uxx + aut + bu = 0, (1)

where a, b and c are constants [5].

This equation is used in: signal analysis for transmission and propagation ofelectrical signals, modeling reaction diffusion in various branches of engineeringsciences [5, 22], in the propagation of pressure waves in the study of pulsatileblood flow in arteries and in one-dimensional random motion of bugs alonga hedge [5]. Also, the experimental results described in [7, 8, 20] seem to bebetter modelled by the telegraph equation than by the heat equation [3].

In recent years, there has been a great interest in fractional differential equa-tions due to their frequent appearance in various fields and their more accuratemodels of systems under consideration provided by fractional derivatives. Forexample, fractional derivatives have been used successfully to model frequencydependent damping behavior of many viscoelastic materials [17] and the socalled anomalous phenomena [13]. They are also used in modeling of manychemical processed, mathematical biology and many other problems in engi-neering [16, 20, 21]. In particular, fractional order telegraph equation modelsthe transport of thermal neutrons inside a nuclear reactor with slab geometry[33].

Many authors have solved the classical and fractional order telegraph equa-tions, among them we have: Biazar et al. [1], Cascaval et al. [3], Kaya [19],Momani [24], Odibat and Momani [26], Orsingher and Zhao [29], Orsingherand Beghin [28], Sevimlican [31], Yildirim [37], Huang [17], Garg et al. [13],Garg and Sharma [14], Hayat and Mohyud-Din [16], Hariharan et al. [15], El-Azab and El-Gamel [9], Gao and Chi [12], Das et al. [4], Dehghan et al. [6],Ford et al. [11], Karimi et al. [18], Figueiredo et al. [10], Xindong et al. [35],Yakubovich and Rodrigues [36], etc.

Volumen 48, Numero 2, Ano 2014

SOLUTION OF SOME FRACTIONAL ORDER TELEGRAPH EQUATIONS 249

On the other hand, different methods such as, variational iteration method,transform method, Adomian decomposition method, juxtaposition of trans-forms, generalized differential transform method, homotopy perturbation meth-od (HPM), wavelet method, matrix method, homotopy analysis method (HAM),are been used in order to solve fractional telegraph equations.

In this paper we derived the solution of two homogeneous space-time frac-tional telegraph equations using the generalized differential transform method.The derivatives are considered in Caputo sense and the solutions are given interms of generalized Mittag-Leffler function and the generalized Wright func-tion. Further, various graphics are included which show the behavior of thesolutions obtained, and results given earlier by Momani [24], Odibat and Mo-mani [26], Yildrim [37], Garg and Sharma [14], and Garg et al. [13] are obtainedas particular cases of ones our.

Following we present some necessary definitions for the development for thenext section.

Caputo Fractional Derivative Caputo fractional derivative of order α isdefined as [2, 30]

Dαa f(x) =

1

Γ(m− α)

∫ x

a

f (m)(ξ)

(x− ξ)α−m+1dξ, m− 1 < α ≤ m, m ∈ N. (2)

The generalized Wright function An interesting generalization of the se-ries pFq is given by [32]

pΨq

[(α1, A1), . . . , (αp, Ap)

(β1, B1), . . . , (βq, Bq); z

]=

∞∑n=0

Γ(α1 +A1n) . . .Γ(αp +Apn)

Γ(β1 +B1n) . . .Γ(βq +Bqn)

zn

n!, (3)

where Aj > 0 (j = 1, . . . , p); Bj > 0 (j = 1, . . . , q); 1 +q∑j=1

Bj −p∑j=1

Aj ≥ 0, for

suitably bounded values of |z|.

Mittag-Leffler function [23]

Eα(z) =

∞∑k=0

zk

Γ(αk + 1)= 1Ψ1

[(1,1)(1,α)

; z], z ∈ C, <(α) > 0. (4)

For α = 1, Eα(z) reduces to ez.

Generalized Mittag-Leffler function Given by Wiman [34]

Eα,β(z) =

∞∑k=0

zk

Γ(αk + β)= 1Ψ1

[(1,1)(β,α)

; z], z ∈ C, <(α) > 0, <(β) > 0. (5)

Revista Colombiana de Matematicas

250 LEDA GALUE

2. Generalized Two-Dimensional Differential Transform

Consider a function of two variables u(x, y) and suppose that it can be repre-sented as a product of two single variable functions, i.e. u(x, y) = f(x)g(y). Ifu(x, y) is analytic and differentiated continuously with respect to x and y in thedomain of interest, then the generalized two-dimensional differential transformof the function u(x, y) is given by ([25, 26, 27])

Uα,β(k, h) =1

Γ(αk + 1)Γ(βh+ 1)

[(Dαx0

)k(Dβy0

)hu(x, y)

](x0,y0)

, (6)

where 0 < α, β ≤ 1,(Dαx0

)k= Dα

x0Dαx0. . . Dα

x0(k times), Dα

x0is defined by (2)

and Uα,β(k, h) is the transformed function.

The generalized differential transform inverse of Uα,β(k, h) is given by

u(x, y) =

∞∑k=0

∞∑h=0

Uα,β(k, h)(x− x0)αk(y − y0)βh. (7)

Basic properties of the generalized two-dimensional differential transform:

Let Uα,β(k, h), Vα,β(k, h), Wα,β(k, h) be generalized two-dimensional dif-ferential transform of the functions u(x, y), v(x, y) and w(x, y), respectively.Then

i) If u(x, y) = v(x, y)± w(x, y) then Uα,β(k, h) = Vα,β(k, h)±Wα,β(k, h).

ii) If u(x, y) = av(x, y), a is constant, then Uα,β(k, h) = aVα,β(k, h).

iii) If u(x, y) = Dγx0v(x, y), where m − 1 < γ ≤ m, m ∈ N, then Uα,β(k, h) =

Γ(αk+γ+1)Γ(αk+1) Vα,β

(k + γ

α , h).

iv) If u(x, y) = Dµy0v(x, y), where n − 1 < µ ≤ n, n ∈ N, then Uα,β(k, h) =

Γ(βh+µ+1)Γ(βh+1) Vα,β

(k, h+ µ

β

).

3. Main Results

We consider the homogeneous space-time fractional telegraph equation

D(l+j)/lx u(x, t) = Dpβ

t u(x, t) +Drβt u(x, t) + u(x, t), 0 < x < 1, t > 0 (8)

where l, j ∈ N, j ≤ l, β = 1/q, p, q, r ∈ N, 1 < pβ ≤ 2, 0 < rβ ≤ 1,

D(l+j)/lx ≡ D

1/lx D

1/lx · · ·D1/l

x

((l + j) times

), Dpβ

t ≡ Dβt D

βt · · ·D

βt (p times),

Drβt ≡ D

βt D

βt · · ·D

βt (r times), D

1/lx and Dβ

t are Caputo fractional derivatives,p+ r is odd, with the initial conditions:

u(0, t) = Eβ(−tβ), (9)

Dxu(0, t) = Eβ(−tβ). (10)



Applying generalized two dimensional differential transform with α = 1/l,x0 = t0 = 0, to both side of (8), we obtain

U1/l,β(k + l + j, h) =Γ(kl + 1

)Γ(kl + l+j

l + 1)[Γ(βh+ βp+ 1)

Γ(βh+ 1)U1/l,β(k, h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ 1)U1/l,β(k, h+ r) + U1/l,β(k, h)

], (11)

where we have used properties iii) and iv).

The application of (6) (with α = 1/l, x0 = t0 = 0) to (9), and the use ofthe Caputo fractional derivative given in (2) joint with (4), lead us to

U1/l,β(0, h) =(−1)h

Γ(βh+ 1). (12)

From (6), (10), the property iii), (2) and (4), we get

U1/l,β(l, h) =(−1)h

Γ(βh+ 1). (13)

Now, making in (11) k = 0 we have

U1/l,β(l + j, h) =1

Γ(l+jl + 1

)[Γ(βh+ βp+ 1)

Γ(βh+ 1)U1/l,β(0, h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ 1)U1/l,β(0, h+ r) + U1/l,β(0, h)

], (14)

and from (12)

U1/l,β(l + j, h) =(−1)h

Γ(l+jl + 1

)Γ(βh+ 1)

, (15)

where we have used the condition p+ r odd.

From (11) with k = l and using (13)

U1/l,β(2l + j, h) =(−1)h

Γ(l+jl ×1 + 2

)Γ(βh+ 1)

, (16)

and for k = l + j in (11), by means of (15), we get

U1/l,β(2l + 2j, h) =(−1)h

Γ(l+jl ×2 + 1

)Γ(βh+ 1)

. (17)


252 LEDA GALUE

Continuing with this procedure

U1/l,β(3l + 2j, h) =(−1)h

Γ(l+jl ×2 + 2

)Γ(βh+ 1)

. (18)

U1/l,β(3l + 3j, h) =(−1)h

Γ(l+jl ×3 + 1

)Γ(βh+ 1)

. (19)

U1/l,β(4l + 3j, h) =(−1)h

Γ(l+jl ×3 + 2

)Γ(βh+ 1)

. (20)

U1/l,β(4l + 4j, h) =(−1)h

Γ(l+jl ×4 + 1

)Γ(βh+ 1)

, (21)

and so successively.

Now, from the result (7) with α = 1/l, x0 = t0 = 0

u(x, t) =

∞∑k=0

∞∑h=0

U1/l,β(k, h) xk/l tβh (22)

and, making use of the equations (12), (13), (15)-(21) we can write

u(x, t) =

∞∑h=0

(−1)h tβh

Γ(βh+ 1)

[1 +

x

Γ(l+jl ×0 + 2

) +x(l+j)/l

Γ(l+jl ×1 + 1

) +

x(2l+j)/l

Γ(l+jl ×1 + 2

) +x(2l+2j)/l

Γ(l+jl ×2 + 1

) +x(3l+2j)/l

Γ(l+jl ×2 + 2

) +

x(3l+3j)/l

Γ(l+jl ×3 + 1

) +x(4l+3j)/l

Γ(l+jl ×3 + 2

) +x(4l+4j)/l

Γ(l+jl ×4 + 1

) + · · ·

], (23)

that is,

u(x, t) =

∞∑h=0

(−1)h tβh

Γ(βh+ 1)

{[1 +

x(l+j)/l

Γ(l+jl ×1 + 1

) +

(x2)(l+j)/l

Γ(l+jl ×2 + 1

) +

(x3)(l+j)/l

Γ(l+jl ×3 + 1

) +

(x4)(l+j)/l

Γ(l+jl ×4 + 1

) + · · ·

]+ x

[1

Γ( l+1l ×0 + 2)

+

x(l+j)/l

Γ(l+jl ×1 + 2

) +

(x2)(l+j)/l

Γ(l+jl ×2 + 2

) +

(x3)(l+j)/l

Γ(l+jl ×3 + 2

) + · · ·

]}(24)

which from the definitions given in (4) and (5) can be written as

u(x, t) ={E l+j

l

(x(l+j)/l) + xE l+j

l ,2

(x(l+j)/l

)}Eβ(− tβ

). (25)



Particular cases From (8)-(10) and (25):

i) If we put l = j = 1, the problem

D2xu(x, t) = Dpβ


u(0, t) = Eβ(−tβ),

Dxu(0, t) = Eβ(−tβ)

has the solution

u(x, t) ={E2

(x2)

+ xE2,2

(x2)}Eβ(− tβ

)= E1(x)Eβ

(− tβ

)= exEβ

(− tβ

). (27)

This is same as obtained by Garg et al. [13].

ii) If l = 2, j = 1, the problem

D3/2x u(x, t) = Dpβ


u(0, t) = Eβ(−tβ),

Dxu(0, t) = Eβ(−tβ)

has as solution

u(x, t) ={E3/2

(x3/2

)+ xE3/2,2

(x3/2

)}Eβ(− tβ

). (29)

So, (29) is the same solution obtained by Garg et al. [13] using the gener-alized differential transform method, and by Garg and Sharma [14] employingthe Adomian decomposition method.

Now, if we make p = 2, q = r = 1 in (28) the space-time fractional telegraphequation reduces to

D3/2x u(x, t) = D2

t u(x, t) +Dtu(x, t) + u(x, t), 0 < x < 1, t > 0 (30)

and the solution is the same obtained by Momani [24], Odibat and Momani [26],and Yildrim [37].

Next we consider the homogeneous space-time fractional telegraph equation

D2αx u(x, t) = Dpβ


where 1 < 2α ≤ 2, β = 1/q, p, q, r ∈ N, 1 < pβ ≤ 2, 0 < rβ ≤ 1, D2αx ≡ Dα

xDαx ,

Dpβt ≡ D

βt D

βt · · ·D

βt (p times), Drβ

t ≡ Dβt D

βt · · ·D

βt (r times), Dα

x and Dβt are

Caputo fractional derivatives, with the initial conditions


254 LEDA GALUE

u(0, t) = Eβ,γ(− tβ

). (32)

Dαxu(0, t) = Eβ,γ

(− tβ

), (33)

being <(γ) > 0.

Applying generalized two dimensional differential transform to both side of(31) and using properties iii) and iv), we obtain

Uα,β(k + 2, h) =Γ(αk + 1)

Γ(αk + 2α+ 1)

[Γ(βh+ βp+ 1)

Γ(βh+ 1)Uα,β(k, h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ 1)Uα,β(k, h+ r) + Uα,β(k, h)

]. (34)

Following an analogue procedure to that used in order to establish (12) wehave,

Uα,β(0, h) =1

Γ(βh+ 1)

[(Dβt0

)hEβ,γ

(− tβ

)](x0,t0)=(0,0)

=(−1)h

Γ(βh+ γ). (35)

On the other hand, the application of (6) to the initial condition (33), usingthe property iii), yields

Uα,β(1, h) =(−1)h

Γ(α+ 1)Γ(βh+ γ). (36)

From (34) with k = 0

Uα,β(2, h) =1

Γ(2α+ 1)

[Γ(βh+ βp+ 1)

Γ(βh+ 1)Uα,β(0, h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ 1)Uα,β(0, h+ r) + Uα,β(0, h)

],

now using (35)

Uα,β(2, h) =(−1)h

Γ(2α+ 1)

[Γ(βh+ βp+ 1)

Γ(βh+ βp+ γ)

(−1)p

Γ(βh+ 1)+

Γ(βh+ βr + 1)

Γ(βh+ βr + γ)

(−1)r

Γ(βh+ 1)+

1

Γ(βh+ γ)

]. (37)

For convenience we will use the following notation:

Ti,j(h) =Γ(βh+ β(ip+ jr) + 1)

Γ(βh+ β(ip+ jr) + γ)

(−1)h+ip+jr

Γ(βh+ 1), (38)



so,

Uα,β(0, h) =(−1)h

Γ(βh+ γ)= T0,0(h), (39)

Uα,β(1, h) =1

Γ(α+ 1)T0,0(h). (40)

It is easy to prove that

Γ(βh+ βp+ 1)

Γ(βh+ 1)Ti,j(h+ p) = Ti+1,j(h). (41)

Γ(βh+ βr + 1)

Γ(βh+ 1)Ti,j(h+ r) = Ti,j+1(h). (42)

From (37) and (38) we can write

Uα,β(2, h) =1

Γ(2α+ 1)

[T1,0(h) + T0,1(h) + T0,0(h)

]. (43)

Making in (34) k = 1, and taking into account (40), we have

Uα,β(3, h) =1

Γ(3α+ 1)

[Γ(βh+ βp+ 1)

Γ(βh+ βp+ γ)T0,0(h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ βr + γ)T0,0(h+ r) + T0,0(h)

],

this, according to (41)-(42), can be written as

Uα,β(3, h) =1

Γ(3α+ 1)

[T1,0(h) + T0,1(h) + T0,0(h)

]. (44)

If k = 2 in (34)

Uα,β(4, h) =Γ(2α+ 1)

Γ(4α+ 1)

[Γ(βh+ βp+ 1)

Γ(βh+ 1)Uα,β(2, h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ 1)Uα,β(2, h+ r) + Uα,β(2, h)

],

then from (43), (41) and (42) we get

Uα,β(4, h) =1

Γ(4α+ 1)

[T2,0(h) + 2T1,1(h) + 2T1,0(h) +

2T0,1(h) + T0,2(h) + T0,0(h)]. (45)

For k = 3 in (34) we have


256 LEDA GALUE

Uα,β(5, h) =Γ(3α+ 1)

Γ(5α+ 1)

[Γ(βh+ βp+ 1)

Γ(βh+ 1)Uα,β(3, h+ p) +

Γ(βh+ βr + 1)

Γ(βh+ 1)Uα,β(3, h+ r) + Uα,β(3, h)

]

and, from (44), (41) and (42), we obtain

Uα,β(5, h) =1

Γ(5α+ 1)

[T2,0(h) + 2T1,1(h) + 2T1,0(h) +

2T0,1(h) + T0,2(h) + T0,0(h)]. (46)

Analogously we obtain the following expressions:

Uα,β(6, h) =1

Γ(6α+ 1)

[T3,0(h) + 3T2,1(h) + 3T2,0(h) + 6T1,1(h) +

3T1,2(h) + 3T1,0(h) + 3T0,2(h) + T0,3(h) + 3T0,1(h) + T0,0(h)]. (47)

Uα,β(7, h) =1

Γ(7α+ 1)

[T3,0(h) + 3T2,1(h) + 3T2,0(h) + 6T1,1(h) +

3T1,2(h) + 3T1,0(h) + 3T0,2(h) + T0,3(h) + 3T0,1(h) + T0,0(h)]. (48)

Uα,β(8, h) =1

Γ(8α+ 1)

[T4,0(h) + 4T3,1(h) + 4T3,0(h) + 12T2,1(h) +

6T2,2(h) + 6T2,0(h) + 12T1,2(h) + 4T1,3(h) + 12T1,1(h) + 4T1,0(h) +

4T0,3(h) + T0,4(h) + 6T0,2(h) + 4T0,1(h) + T0,0(h)]. (49)

Uα,β(9, h) =1

Γ(9α+ 1)

[T4,0(h) + 4T3,1(h) + 4T3,0(h) + 12T2,1(h) +

6T2,2(h) + 6T2,0(h) + 12T1,2(h) + 4T1,3(h) + 12T1,1(h) + 4T1,0(h) +

4T0,3(h) + T0,4(h) + 6T0,2(h) + 4T0,1(h) + T0,0(h)]. (50)

Substituting the earlier results (39), (40), (43)-(50)) in (7), we have



u(x, t) =

∞∑h=0

tβh

[(1 +

xα

Γ(α+ 1)

)T0,0(h) +

(x2α

Γ(2α+ 1)+

x3α

Γ(3α+ 1)

)×

[T1,0(h) + T0,1(h) + T0,0(h)

]+

(x4α

Γ(4α+ 1)+

x5α

Γ(5α+ 1)

)×[

T2,0(h) + 2T1,1(h) + 2T1,0(h) + 2T0,1(h) + T0,2(h) + T0,0(h)]

+(x6α

Γ(6α+ 1)+

x7α

Γ(7α+ 1)

)[T3,0(h) + 3T2,1(h) + 3T2,0(h) + 6T1,1(h) +

3T1,2(h) + 3T1,0(h) + 3T0,2(h) + T0,3(h) + 3T0,1(h) + T0,0(h)]

+(x8α

Γ(8α+ 1)+

x9α

Γ(9α+ 1)

)[T4,0(h) + 4T3,1(h) + 4T3,0(h) + 12T2,1(h) +

6T2,2(h) + 6T2,0(h) + 12T1,2(h) + 4T1,3(h) + 12T1,1(h) + 4T1,0(h) +

4T0,3(h) + T0,4(h) + 6T0,2(h) + 4T0,1(h) + T0,0(h)]

+ · · ·

], (51)

that is,

u(x, t) =

∞∑h=0

tβh

[(1 +

xα

Γ(α+ 1)+ · · ·+ x8α

Γ(8α+ 1)+

x9α

Γ(9α+ 1)+ · · ·

)×

T0,0(h) +

(x2α

Γ(2α+ 1)+

x3α

Γ(3α+ 1)+ · · ·+ x8α

Γ(8α+ 1)+

x9α

Γ(9α+ 1)+ · · ·

)×[

T1,0(h) + T0,1(h)]

+(x4α

Γ(4α+ 1)+

x5α

Γ(5α+ 1)+ · · ·+ x8α

Γ(8α+ 1)+

x9α

Γ(9α+ 1)+ · · ·

)×[

T2,0(h) + 2T1,1(h) + T1,0(h) + T0,1(h) + T0,2(h)]

+(x6α

Γ(6α+ 1)+

x7α

Γ(7α+ 1)+

x8α

Γ(8α+ 1)+

x9α

Γ(9α+ 1)+ · · ·

)×[

T3,0(h) + 3T2,1(h) + 2T2,0(h) + 4T1,1(h) +

3T1,2(h) + T1,0(h) + 2T0,2(h) + T0,3(h) + T0,1(h)]

+(x8α

Γ(8α+ 1)+

x9α

Γ(9α+ 1)+ · · ·

)[T4,0(h) + 4T3,1(h) + 3T3,0(h) +

9T2,1(h) + 6T2,2(h) + 3T2,0(h) + 9T1,2(h) + 4T1,3(h) + 6T1,1(h) +

T1,0(h) + 3T0,3(h) + T0,4(h) + 3T0,2(h) + T0,1(h)]

+ · · ·

]. (52)

This can be written as


258 LEDA GALUE

u(x, t) =

∞∑h=0

T0,0(h) tβh∞∑k=0

xkα

Γ(kα+ 1)+

∞∑h=0

[T1,0(h) + T0,1(h)

]tβh

∞∑k=2

xkα

Γ(kα+ 1)+

∞∑h=0

[T2,0(h) + 2T1,1(h) + T1,0(h) + T0,1(h) + T0,2(h)

]tβh

∞∑k=4

xkα

Γ(kα+ 1)+

∞∑h=0

[T3,0(h) + 3T2,1(h) + 2T2,0(h) + 4T1,1(h) + 3T1,2(h) +

T1,0(h) + 2T0,2(h) + T0,3(h) + T0,1(h)]tβh

∞∑k=6

xkα

Γ(kα+ 1)+ · · · . (53)

Making changes of indexes we have

u(x, t) =

∞∑h=0

T0,0(h) tβh∞∑k=0

xkα

Γ(kα+ 1)+ x2α

∞∑h=0

[T1,0(h) + T0,1(h)

]tβh ×

∞∑k=0

xkα

Γ(kα+ 2α+ 1)+ x4α

∞∑h=0

[T2,0(h) + 2T1,1(h) + T1,0(h) +

T0,1(h) + T0,2(h)]tβh

∞∑k=0

xkα

Γ(kα+ 4α+ 1)+

x6α∞∑h=0

[T3,0(h) + 3T2,1(h) + 2T2,0(h) + 4T1,1(h) + 3T1,2(h) + T1,0(h) +

2T0,2(h) + T0,3(h) + T0,1(h)]tβh

∞∑k=0

xkα

Γ(kα+ 6α+ 1)+ · · · . (54)

From (54), (4) and (5) we have

u(x, t) =

∞∑h=0

T0,0(h) tβh Eα(xα) +

x2α∞∑h=0

[T1,0(h) + T0,1(h)

]tβhEα,2α+1(xα) +

x4α∞∑h=0

[T2,0(h) + 2T1,1(h) + T1,0(h) + T0,1(h) + T0,2(h)

]tβhEα,4α+1(xα) +



x6α∞∑h=0

[T3,0(h) + 3T2,1(h) + 2T2,0(h) + 4T1,1(h) + 3T1,2(h) +

T1,0(h) + 2T0,2(h) + T0,3(h) + T0,1(h)]tβhEα,6α+1(xα) + · · · . (55)

Observe that from (38) and (3) we have

∞∑h=0

Ti,j(h) tβh = (−1)ip(−1)jr 2Ψ2

[(βip+ βjr + 1, β), (1, 1)

(βip+ βjr + γ, β), (1, β);−tβ

].

Therefore,

u(x, t) = Eα(xα)Eβ,γ(− tβ

)+ x2αEα,2α+1(xα) ×{

(−1)p 2Ψ2

[(βp+1,β),(1,1)(βp+γ,β),(1,β)

;−tβ]

+ (−1)r 2Ψ2

[(βr+1,β),(1,1)(βr+γ,β),(1,β)

;−tβ]}

+

x4αEα,4α+1(xα)

{2Ψ2

[(2βp+1,β),(1,1)(2βp+γ,β),(1,β)

;−tβ]

+

2(−1)p+r 2Ψ2

[(βp+βr+1,β),(1,1)(βp+βr+γ,β),(1,β)

;−tβ]

+

(−1)p 2Ψ2

[(βp+1,β),(1,1)(βp+γ,β),(1,β)

;−tβ]

+ (−1)r 2Ψ2

[(βr+1,β),(1,1)(βr+γ,β),(1,β)

;−tβ]

+

2Ψ2

[(2βr+1,β),(1,1)(2βr+γ,β),(1,β)

;−tβ]}

+ x6αEα,6α+1(xα) ×{(−1)p 2Ψ2

[(3βp+1,β),(1,1)(3βp+γ,β),(1,β)

;−tβ]

+

3(−1)r 2Ψ2

[(2βp+βr+1,β),(1,1)(2βp+βr+γ,β),(1,β)

;−tβ]

+ 2 2Ψ2

[(2βp+1,β),(1,1)(2βp+γ,β),(1,β)

;−tβ]

+

4(−1)p+r 2Ψ2

[(βp+βr+1,β),(1,1)(βp+βr+γ,β),(1,β)

;−tβ]

+

3(−1)p 2Ψ2

[(βp+2βr+1,β),(1,1)(βp+2βr+γ,β),(1,β)

;−tβ]

+

(−1)p 2Ψ2

[(βp+1,β),(1,1)(βp+γ,β),(1,β)

;−tβ]

+ 2 2Ψ2

[(2βr+1,β),(1,1)(2βr+γ,β),(1,β)

;−tβ]

+

(−1)r 2Ψ2

[(3βr+1,β),(1,1)(3βr+γ,β),(1,β)

;−tβ]

+ (−1)r 2Ψ2

[(βr+1,β),(1,1)(βr+γ,β),(1,β)

;−tβ]}

+ · · · . (56)

As a particular case, if γ = 1 and we use (4)-(5), then

u(x, t) = Eβ(− tβ

){Eα(xα) +

[(−1)p + (−1)r

]x2αEα,2α+1(xα) +[

2 + 2(−1)p+r + (−1)p + (−1)r]x4αEα,4α+1(xα) +[

5(−1)p + 5(−1)r + 4(−1)p+r + 4]x6αEα,6α+1(xα) + · · ·

}. (57)

Now, from (57) we get


260 LEDA GALUE

i) If p + r is odd: u(x, t) = Eβ(− tβ

)Eα(xα), which was given by Garg et

al. [13].

ii) If p+ r is even, being p and r even:


) {Eα(xα) + 2x2αEα,2α+1(xα) +

6x4αEα,4α+1(xα) + 18x6αEα,6α+1(xα) + · · ·}. (58)

iii) If p+ r is even, being p and r odd, we have


) {Eα(xα)− 2x2αEα,2α+1(xα) +

2x4αEα,4α+1(xα)− 2x6αEα,6α+1(xα) + · · ·}

(59)

4. Graphic Representation

In this section some figures are presented which show the behavior of the solu-tion u(x, t) for different values of the parameters, with 0 < x < 1, 0 < t < 1.

i) Figures of u(x, t) as defined by (25) (Figure 1 and Figure 2):

β = 1.0

β = 0.5

β = 0.25

β = 0.1

Figure 1. u(x, t) for l = 10, j = 5 and diferent values of β.



j = 8

j = 6

j = 4

j = 2

Figure 2. u(x, t) for l = 10, β = 0.25 and diferent values of j.

ii) Figures of u(x, t) as defined by (56) (Figure 3 and Figure 4):

α = 0.9

α = 0.75

α = 0.6

Figure 3. u(x, t) for β = 0.5, γ = 2, p = 3, r = 1 and different values of α.


262 LEDA GALUE

α = 0.9

α = 0.75

α = 0.6

Figure 4. u(x, t) for β = 0.5, γ = 2, p = 3, r = 2 and different values of α.

iii) Figures of u(x, t) as defined by (58) (Figure 5 and Figure 6):

β = 0.5

β = 0.25

β = 0.1

Figure 5. u(x, t) for α = 0.75 and different values of β.



α = 0.6

α = 0.8

α = 1.0

Figure 6. u(x, t) for β = 0.5 and different values of α.

iv) Figures of u(x, t) = Eβ(−tβ)Eα(xα) (Figure 7 and Figure 8):

β = 0.25

β = 0.5β = 1.0

Figure 7. u(x, t) for α = 0.75 and different values of β.


264 LEDA GALUE

α = 0.6

α = 0.8

α = 1.0

Figure 8. u(x, t) for β = 0.5 and different values of α.

Acknowledgement. The author would like to thanks to CONDES-Universidaddel Zulia for financial support.

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(Recibido en mayo de 2014. Aceptado en septiembre de 2014)

Centro de Investigacion de Matematica Aplicada

Universidad del Zulia

Facultad de Ingenierıa

Maracaibo, Venezuela

e-mail: [email protected]


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