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arXiv:1011.3692v1 [math-ph] 16 Nov 2010

Lie group classifications and exact solutions for time-fractional

Burgers equation

Guo-cheng Wu∗

College of Textile, Donghua University, Shanghai 201620, P.R. China;

Modern Textile Institute, Donghua University, Shanghai 200051, P.R. China.

——————————————————————————————————————–

Abstract

Lie group method provides an efficient tool to solve nonlinear partial differential equations.

This paper suggests a fractional Lie group method for fractional partial differential equations.

A time-fractional Burgers equation is used as an example to illustrate the effectiveness of the

Lie group method and some classes of exact solutions are obtained.

PACS

02.20.Tw; 45.10.Hj

Key words

Lie group method; Fractional Burgers equation; Fractional characteristic method

——————————————————————————————————————–

1Introduction

Many methods of mathematical physics have been developed to solve differential equations,

among which Lie group method is an efficient approach to derive the exact solution of nonlinear

partial differential equations.

Since Sophus Lie’s group analysis work more than 100 years ago, Lie group theory has

become more and more pervasive in its influence on other mathematical disciplines [1, 2]. There

are, however, there few applications of Lie method in fractional calculus. Then a question may

naturally arise: is there a fractional Lie group method for fractional differential equations?

Some researchers investigated Lie group method for fractional differential equations in sense

of the Caputo derivative and derived scaling transformation and similarity solutions [3–5]. Con-

sidering the classical Lie group method, method of characteristic is used to solve symmetry

equations. Recently, with the modified Riemann-Liouville derivative [6–8], we first propose a

∗Corresponding author, E-mail: wuguocheng2002@yahoo.com.cn. (G.C. Wu)

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more generalized fractional characteristic method [9] than Jumarie’s Lagrange method [10]. Us-

ing our fractional characteristic method, the generalized symmetry equations generating by the

prolongation technique can be solved, and a fractional Lie group method was presented for an

anomalous diffusion equation [9].

In this study, we investigate a simplified version of the fractional Burgers equations [5]

u(α)

t

= uxx+ u2

x, x ∈ (0, ∞), 0<t, 0<α < 1,

(1)

and derive its group classifications. In order to investigate the local behaviors of the above

equation, the fractional derivative is in the sense of the modified Riemann-Liouville [6–8].

2Fractional Calculus and Some Properties

From the viewpoint of Brown motion, Jumarie proposed the modified Riemann-Liouville deriva-

tive [6–8],

Dα

xf(x) =

1

Γ(n − α)

dn

dxn

?x

0

(x − ξ)n−α−1(f(ξ) − f(0)) dξ, n − 1<α < n,

(2)

where the derivative on the right-hand side is the Riemann-Liouville fractional derivative and

n ∈ Z+.

(a) Fractional Taylor series

Recently, Jumarie-Taylor series [11] was proposed

df(x) =

∞

?

i=1

hkα

(kα)!f(kα)(x), 0<α < 1.

(3)

Here f(x) is a kα-differentiable function and k is an arbitrary positive integer.

Taking k = 1, f(x) is a α-differentiable function. We can derive that

df(x) =Dα

xf(x)(dx)α

Γ(1 + α)

.

(4)

(b) Fractional Leibniz product law

If we set Dα

xu(x) and Dα

xv(x) exist, we can readily find that

Dα

x(uv) = u(α)v + uv(α).

(5)

The properties of Jumarie’s derivative were summarized in [11]. The extension of Jumarie’s

fractional derivative and integral to variations approach by Almeida et al. [12, 13]. Fractional

variational interactional method and Adomian decomposition method are proposed for fractional

differential equations [14, 15].

(c) Integration with respect to (dx)α

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0Iα

xf(x) =

1

Γ(α)

?x

0

(x − ξ)α−1f(ξ)dξ =

1

Γ(α + 1)

?x

0

f(ξ)(dξ)α,0 < α ≤ 1.

(6)

(d) Generalized Newton-Leibniz Law

Assume Dα

xf(x) is an integrable function in the interval [0,a]. Obviously,

1

Γ(1 + α)

?a

0

Dα

xf(x)(dx)α= f(a) − f(0),0 < α < 1,

(7)

1

Γ(1 + α)

?x

0

Dα

ξf(ξ)(dξ)α= f(x) − f(a),

(8)

and

Dα

x

Γ(1 + α)

?x

0

f(ξ)(dξ)α= f(x), 0 < α < 1.

(9)

(e) Some other useful properties

f(α)([x(t)]) =df

dxx(α)(t), 0 < α < 1,

Γ(1 + β)

Γ(1 + β − α)xβ−α, 0 < β < 1,

(10)

Dα

xxβ=

(11)

?

(dx)

β= x

β.

(12)

The above properties (a)–(d) can be found in Ref. [11]. We must point out that f(x) should

be differentiable w.r.t x in Eq. (10), and x

βis an α order function in Eq. (11).

3A Characteristic Method for Fractional Differential Equations

It is well known that the method of characteristics has played a very important role in math-

ematical physics. Preciously, the method of characteristics is used to solve the initial value

problem for general first order. With the modified Riemann-Liouville derivative, Jumaire ever

gave a Lagrange characteristic method [10], in which the time-fractional order equals to the

space-fractional order. We present a more generalized fractional method of characteristics and

use it to solve linear fractional partial equations.

Consider the following first order equation,

a(x,t)∂u(x,t)

∂x

+ b(x,t)∂u(x,t)

∂t

= c(x,t).

(13)

The goal of the method of characteristics is to change coordinates from (x, t) to a new coor-

dinate system (x0, s) in which the partial differential equation becomes an ordinary differential

equation along certain curves in the x−t plane. The curves are called the characteristic curves.

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More generally, we consider to extend this method to linear space-time fractional differential

equations

a(x,t)∂

∂xβ

∂tα

With the fractional Taylor’s series in two variables [11]

βu(x,t)

+ b(x,t)∂αu(x,t)

= c(x,t), 0 < α,β < 1.

(14)

du =

∂

βu(x,t)

Γ(1 + β)∂xβ(dx)

β+

∂αu(x,t)

Γ(1 + α)∂tα(dt)α, 0 < α, β < 1,

(15)

similarly, we derive the generalized characteristic curves

du

ds= c(x,t),

(16)

(dx)

β

Γ(1 + β)ds= a(x,t),

(17)

(dt)α

Γ(1 + α)ds= b(x,t).

(18)

Eqs. (16)–(18) can be reduced as Jumaire’s Lagrange method of characteristic if α = β in [10].

4A Fractional Lie Group Method

In the classical Lie method for partial differential equations, the one-parameter Lie group of

transformations in (x, t, u) is given by

˜ x = x + εξ(x,t,u) + O(ε2),

˜t = t + ετ(x,t,u) + O(ε2),

˜ u = u + εφ(x,t,u) + O(ε2),

where ε is the group parameter.

Use the set of fractional vector fields instead of the one of integer order

V = ξ(x,t,u)Dβ

x+ τ(x,t,u)Dα

t+ φ(x,t,u)Du, 0 < α < 1, 0 < β < 1.

(19)

For the fractional second order prolongation Pr(2β)V of the infinitesimal generators, we

proposed [9]

Pr(2β)V = V + φ[t] ∂φ

∂Dα

tu+ φ[x] ∂φ

∂Dβ

xu

+ φ[tt]

∂φ

tu+ φ[xx]

∂D2α

∂φ

∂D2β

x u

+ φ[xt]

∂φ

∂Dβ

xDα

tu.

(20)

As a result, we can have

Pr(2β)V (∆[u]) = 0,

(21)

on∆[u] = 0.

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In the time-fractional Burgers equation, Eq. (1), we only need to consider the case of the

fractional order of space β = 1. Thus, the corresponding Lie algebra of infinitesimal symmetries

is the set of fractional vector fields in the form

V = ξ(x,t,u)Dx+ τ(x,t,u)Dα

t+ φ(x,t,u)Du.

(22)

We assume the one-parameter Lie group of transformations in (x, t, u) given by

˜ x = x + εξ(x,t,u) + O(ε),

˜tα

Γ(1+α)=

˜ u = u + εφ(x,t,u) + O(ε),

tα

Γ(1+α)+ ετ(x,t,u) + O(ε),

(23)

where ε is the group parameter.

The generalized second prolongation satisfies

Pr(2)V = V + φt ∂φ

∂Dα

tu+ φx∂φ

∂Dxu+ φtt

∂φ

tu+ φxx∂φ

∂D2α

∂uxx

+ φxt

∂φ

tux.

∂Dα

(24)

Using the following condition

Pr(2)V (∆[u]) = 0, ∆[u] = 0,

(25)

we can have

(φt− φxx− 2uxφx)??∆[u]=0= 0.

(26)

The generalized prolongation vector fields are reduced as

φt= Dα

φx= Dxφ − (Dxξ)Dxu − (Dxτ)Dα

φxx= D2

tφ − (Dα

tξ)Dxu − (Dα

tτ)Dα

tu,

tu,

xφ − 2(Dxξ)D2

xu − (D2

xξ)Dxu − 2(Dxτ)DxDα

tu − (D2

xτ)Dα

tu.

(27)

Substituting Eq. (27) into Eq. (26) and setting the coefficients of uxu(α)

and 1 to zero. Solve the equations with maple software, we can have

xt, u(α)

xt, uxxux, ux

ξ(x,t,u) = c1+ c4x + 2c5

τ(x,t,u) = c2+ 2c4

tα

Γ(1+α)+ 4c6

Γ(1+α)+ 4c6

2c6tα

Γ(1+α)− c6x2+ a(x,t)eu,

xtα

Γ(1+α),

tα

t2α

Γ2(1+α),

φ(x,t,u) = c3− c5x +

(28)

where k(α)

The Lie algebra of infinitesimal symmetries of Eq. (1) is spanned by the vector field

t

= kxx.

V1=

V5= −x∂

V6=

∂

∂x, V2=∂α

∂u+

4xtα

Γ(1+α)

∂tα, V3=

2tα

Γ(1+α)

∂x+

Γ2(1+α)

∂

∂u, V4= x∂

∂u+

2tα

Γ(1+α)

∂α

∂tα,

∂

∂x,

4t2α

∂

∂α

∂tα− (x2+

2tα

Γ(1+α))∂

∂u,

(29)

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and the infinite-dimensional subalgebra

Vk= k(x,t)e−u∂

∂u.

It is easy to check the two vector fields {V1,V2,V3,V4,V5,V6} are closed under the Lie bracket

[a,b] = ab − ba. In fact, we have

[Vi,Vi] = 0 (i = 0,...,6), [V1,V2] = [V1,V3] = 0, [V1,V4] = −V1, [V1,V5] = V3,

[V1,V6] = −2V5, [V2,V3] = 0, [V2,V4] = −2V2, [V2,V5] = −2V1, [V2,V6] = 2V3− 4V4,

[V3,V4] = [V3,V5] = [V3,V6] = 0, [V4,V5] = −V5, [V4,V6] = −2V6, [V5,V6] = 0.

[V1,Vk] = −Vkx, [V2,Vk] = −Vkt, [V3,Vk] = −Vk, [V4,Vk] = −Vk′, [V5,Vk] = −Vk′′,

[V6,Vk] = −Vk′′′,

where k

t

, k

Take the characteristic equation V5as an example. The characteristic curve of V5can be

given

′= xkx+

2tα

Γ(1+α)k(α)

′′=

2tα

Γ(1+α)kx+xk and k

′′′=

4xtα

Γ(1+α)kx+

4t2α

Γ2(1+α)k(α)

t

+(x2+

2tα

Γ(1+α))k.

du

dε= −x,

(30)

dx

dε=

2tα

Γ(1 + α),

(31)

(dt)α

Γ(1 + α)dε= 0.

(32)

Solve the above ordinary equations with the initial value u = u(x,t,ε)|ε=0, x = x(ε)|ε=0and

t = t(ε)|ε=0. The one-parameter group Gigenerated by the Vi(i = 1,...,6,α) are given as

g1: (x,

g2: (x,

g3: (x,

g4: (x,

g5: (x,

tα

Γ(1+α), u) → (x + ε,

tα

Γ(1+α), u) → (x,

tα

Γ(1+α), u) → (x,

tα

Γ(1+α), u) → (xeε,

tα

Γ(1+α), u) → (x +

tα

Γ(1+α), u) → (

tα

Γ(1+α), u),

Γ(1+α)+ ε, u),

tα

Γ(1+α), u + ε),

tαe2ε

Γ(1+α), u),

2εtα

Γ(1+α),

x

1−4ε

tα

Γ(1+α), log(eu+ εk)).

tα

tα

Γ(1+α), u −

tα

1−4ε

ε2tα

Γ(1+α)− xε),

g6: (x,

tα

Γ(1+α),

tα

Γ(1+α), u −

x2ε

Γ(1+α)+ log

1−4ε

tα

?

1 − 4ε

tα

Γ(1+α)),

gα: (x,

tα

Γ(1+α), u) → (x,

(33)

Take α = 1 in the above classifications. We can derive the results for the case of integer

order. Since giis a symmetry, if u = f(x,

Γ(1+α)) is a solution of Eq. (1), then the following ui

are also the solutions of Eq. (1)

tα

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u1= f(x − ε,

u2= f(x,

u3= f(x,

u4= f(xe−ε,

u5= f(x −

tα

Γ(1+α)),

tα

Γ(1+α)− ε),

tα

Γ(1+α)) + ε,

tαe−ε

Γ(1+α)),

2εtα

Γ(1+α),

x

1+4ε

tα

Γ(1+α)) +

tα

1+4ε

Γ(1+α)

ε2tα

Γ(1+α)− xε,

) −

1−4ε

u6= f(

tα

Γ(1+α),

Γ(1+α))+ εk).

tα

x2ε

tα

Γ(1+α)

− log

?

1 − 4ε

tα

Γ(1+α),

uα= log(ef(x,

tα

(34)

Now we consider the applications of the above transformations. From u1to u4, we can only

obtain trivial solutions. Therefore, we start from the use of u5

g5: (x,

tα

Γ(1 + α), u) → (x +

2εtα

Γ(1 + α),

tα

Γ(1 + α), u −

ε2tα

Γ(1 + α)− xε),

(35)

Assume u5,0= u5,0(x,

where c is a arbitrary constant and also a trivial solution. We can get a new nontrivial exact

solution as

tα

Γ(1+α)) = f(x,

tα

Γ(1+α)) is one exact solution of Eq. (1). Take u5,0= c,

u5,1= c +

ε2tα

Γ(1 + α)− xε.

(36)

Further more, continue this iteration process, we can derive a new exact solution of Eq. (1)

u5,2= u5,1(x −

2εtα

Γ(1 + α),

tα

Γ(1 + α)) +

ε2tα

Γ(1 + α)− xε = c − 2xε +

4ε2tα

Γ(1 + α).

(37)

Similarly, take u6,0= c, then we can have

u6,1= c −

x2ε

1 + 4ε

tα

Γ(1+α)

− log

?

1 + 4ε

tα

Γ(1 + α),

(38)

and

u6,2= u6,1(

x

1 + 4ε

tα

Γ(1+α)

,

tα

1 + 4ε

tα

Γ(1+α)

) −

x2ε

1 + 4ε

tα

Γ(1+α)

− log

?

1 + 4ε

tα

Γ(1 + α)

(39)

= c −

2εx2

1 +

8εtα

Γ(1+α)

− log

?

1 +

8εtα

Γ(1 + α).

We can readily verify that u5,1, u5,2, u6,1 and u6,2 are four exact solutions of Eq. (1).

Assuming the fractional order α = 1, the exact solutions we give here can be reduced as the

exact iteration solution in Ref. [16] if we set the coefficients a = b = 1.

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5 Conclusions

Fractional differential equations have caught considerable attention due to their various appli-

cations in real physical problems. However, there is no a systematic method to derive the exact

solutions. Now, the problem is partly solved in this paper. The presented paper can be applied

to other fractional partial differential equations and investigate their non-smooth properties.

Acknowledgments

The author would like to give his thanks to Prof. En-gui Fan (Math. Dep., Fudan University,

China) and the referee’s sincere suggestions. The author also feels grateful to Dr. Hong-li An’s

helpful discussions when visiting Hongkong Polytechnic University this summer.

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