Exact solutions to the various nonlinear evolution equations

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Exact solutions to the various nonlinear evolution equations

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Exact solutions to the various nonlinear evolution equations

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IOP PUBLISHING PHYSICASCRIPTA Phys. Scr. 79 (2009) 045005 (7pp) doi:10.1088/0031-8949/79/04/045005

Exact solutions to the various nonlinear

evolution equations

Dogan Kaya

1

_{and Ibrahim E Inan}

2

1_{Engineering Faculty, Ardahan University, 75100 Ardahan, Turkey} 2_{Firat University, Faculty of Education, 23119 Elazig, Turkey} E-mail:dkaya@firat.edu.trandieinan@yahoo.com

Received 23 November 2008

Accepted for publication 10 February 2009 Published 31 March 2009

Online atstacks.iop.org/PhysScr/79/045005

Abstract

In this paper, we implemented the exp-function method for the exact solutions of the generalized Fisher’s equations, the Burgers-like equation and the Sharma–Tasso–Olver equation. By using this scheme, we found some exact solutions of the above-mentioned equations.

PACS numbers: 02.30.Jr, 02.70.Wz.

1. Introduction

In this work, we applied the exp-method for some nonlinear partial differential equations. The basic idea of the

exp-method was suggested by He [1] and was implemented

by He and Wu in 2006. The method was further developed some other scientists [2–4]. The method is generally used for differential equations, but it is Zhou [5] who first applied the method to the differential-difference equation with great success. Following Zhu, Dai obtained some excellent results for the discrete nonlinear Schrödinger equation and the hybrid lattice equation [6]. Xu and Zhang [7,8] contributed much to the development of the method.

We aim here to solve the generalized Fisher’s equations, the Burgers-like equation and the Sharma–Tasso–Olver equation by using the exp-function method. The original Fisher equation

ut− vux x= ku

1 −u_κ

is well known in population dynamics, where v > 0 is the diffusion constant, k> 0 is the linear growth rate, and k > 0 is the carrying capacity of the environment. The right-hand side function ku(1 −u_κ) represents a nonlinear growth rate [9,10]. This well-known equation was first proposed by Fisher [11] to a model for the advance of a mutant gene in an infinite one-dimensional habitat. In recent years, the equation has been used as a basis for a wide variety of models for the spatial spread of genes in a population and for chemical wave propagation [11]. The Burgers equation is a model of flow through a shock wave in a viscous fluid [12] and in the Burgers model of turbulence [13]. The Sharma–Tasso–Olver equation

with its fission and fusion [14] is

ut+ 3αu2x+ 3αu

2_u

x+ 3αuux x+αux x x= 0.

Attention has been focused on the Sharma–Tasso–Olver equation in [15, 16] and the references therein due to its appearance in scientific applications [17].

Nonlinear phenomena play a crucial role in applied mathematics and physics. Explicit solutions to nonlinear equations are of fundamental importance. Various methods for obtaining explicit solutions to nonlinear evolution

equations have been proposed [18–30]. Among them are

the homogeneous balance (HB) method [18, 24–30], the

auto Bäcklund transformations [18–22], and the similarity

reductions [21, 22] of the Bäcklund transformation in

mathematical physics. The nonlinear iterative principle from Bäcklund transformations converts the problem of solving a nonlinear partial differential equation to an ordinary differential equation [30, 31]. In a recent paper [32], the authors have further extended the HB method so that it can deal with other cases whose balance constant is a fraction or negative integer. Other methods are the tanh function method, the sine–cosine method, the Painlevé method, Cole–Hopf transformation, Darboux transformation and so on. A feature common to all these methods is that they use the transformations to reduce the equation into a more simple equation and then solve it.

2. An analysis of the method and applications

Before starting to give the exp-function method, we will give a simple description of the exp-function method. For doing this,

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Phys. Scr. 79 (2009) 045005 D Kaya and I E Inan

one can consider in a two-variable general form of nonlinear PDE

Q(u, ut, ux, ux x, . . .) = 0, (1)

and transform equation (1) with

u(x, t) = u(ξ),

ξ = kx + wt,

where k andw are constants. After the transformation, we get a nonlinear ordinary differential equation for u(ξ):

Q0 u0, u00, u000, . . . = 0. (2)

Then, the solution of equation (2) we are looking for is

expressed as u(ξ) = Pd n=−canexp(nξ) Pq m=− pbmexp(mξ) , (3)

where c, d, p, and q are positive integers, which are unknown

and to be further determined, and an and bm are unknown

constants. We suppose that the solution of equation (2) can be expressed as

u(ξ) = acexp(cξ) + · · · + a−dexp(−dξ) apexp(pξ) + · · · + a−qexp(−qξ)

, (4)

where p and c are positive integers that can be determined by balancing the highest order derivative and with the highest nonlinear terms into equation (2). Substituting solution (4) into equation (2) yields a set of algebraic equations for exp(ξ); then all coefficients of exp(ξ) have to vanish. After

this separated algebraic equation, we can find an and bm

constants.

Example 1. In this example we consider the generalized Fisher’s equation [31]

ut− ux x− u + u3= 0. (5)

Let us consider solutions u(x, t) = u(ξ), ξ = kx + wt; then equation (5) becomes

wu0_{− k}2_u00_{+ u + u}3

= 0. (6)

When balancing u3with u00,

c1exp [(3p + c) ξ] + · · · c2exp [4 pξ] + · · ·

= c3exp [cξ] + · · ·

c4exp [ pξ] + · · ·

then gives p = c. Similarly, to determine values of d and q when balancing u2with u0_,

· · · + d1exp [−(3q + d) ξ] · · · + d2exp [−4qξ ]

=· · · + d3exp [−dξ ] · · · + d4exp [−qξ ] then gives q = d. For simplicity, we set p = c = 1 and

q = d = 1, so equation (4) reduces to

u(ξ) =a1exp(ξ) + a0+ a−1exp(−ξ)

exp(ξ) + b0+ b−1exp(−ξ)

. (7)

Substituting equation (7) into (6) yields a set of algebraic equations for a0, a1, a−1, b0, b−1. These systems are

a3 −1− a−1b−12 (b0+ b−1exp(−ξ) + exp(ξ))3 = 0, " 3a₋₁2 a0+ −1 − k2+w a0b2₋₁ + −2 + k2− w a−1b−1b0 # (b0+ b−1exp(−ξ) + exp(ξ))3 = 0, " 3a2 −1a1+ b−1 1 + 4k2− 2w a1b−1 + −2 + k2+w a0b0 + a−1|3a02+ −2 + 4k 2 − 2w b−1 # (b0+ b−1exp(−ξ) + exp(ξ))3 − 1 + k 2₊_{w b}2 0 (b0+ b−1exp(−ξ) + exp (ξ))3 = 0. " a3₀− 2 + 3k2+ 3w a₋₁+ 2 + 3k2− 3w a1b−1 b0 +a06a−1a1+ −2 + 6k2 b−1− b02 # (b0+ b−1exp(−ξ) + exp(ξ))3 = 0, " 3a2₀a1− a−1 1 + 4k2+ 2w − 3a12 + −2 + k2− w a0b0 +2a1b−1 −1 + 2k2+w # (b0+ b−1exp(−ξ) + exp(ξ))3 + 2a1b 2 0 −1 + 2k 2₊_w (b0+ b−1exp(−ξ) + exp(ξ))3 = 0, −a0 1 + k2+w − 3a12 + −2 + k 2₊_{w a} 1b0 (b0+ b−1exp(−ξ) + exp (ξ))3 = 0, a1 −1 + a12 (b0+ b−1exp(−ξ) + exp (ξ))3 = 0. (8)

From the solutions of the system, we obtain the following with the aid of Mathematica.

Case 1. a1= − 1, b−16= 0, k = − 1 √ 2, (9) w = 3 2, a−1= 0, a0= 0. Case 2. a1= 0, b0= 0, b−16= 0, a−1= −b−1, (10) a0= − i p b₋₁, k = −√1 2, w = − 3 2. Case 3. a1= 0, b0= 0, b−16= 0, a₋₁= −b−1, a0=1₂ −b0− q −4b−1+ b20 , (11) k = −√1 2, w = − 3 2. 2

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Phys. Scr. 79 (2009) 045005 D Kaya and I E Inan Case 4. a1= −1, b0= 0, b−16= 0, a−1= 0, a0= −i √ b₋₁, k = −√1 2, w = 3 2. (12) Case 5. a1= 0, b0= 0, b−16= 0, a−1= −b−1, a0= 0, k = − 1 2√2, w = − 3 4. (13) Case 6. a1= 0, b06= 0, b−1= 0, a−1= 0, a0= b0, k = 1 √ 2, w = − 3 2. (14) Case 7. a1= 1, b0= 0, b−16= 0, a−1= 0, a0= −i √ b₋₁, k =√1 2, w = 3 2. (15)

Substituting (9)–(15) into (7) we obtain the following solutions of equation (5). These solutions are

(i) When we choose case 1, the solution is

u(x, t) = − exp(−(1/

√

2)x + (3/2)t) exp(−(1/√2)x + (3/2)t) + b0

. (16)

(ii) In this case, if we choose case 2, the solution will be

u(x, t) = −i √ b₋₁− b−1exp((1/ √ 2)x + (3/2)t) exp(−(1/√2)x − (3/2)t) + b₋₁ . (17) (iii) Again, when we choose case 3, the solution is

u(x, t) = (1/2)(−√−4b−1) − b−1exp((1/ √ 2)x + (3/2)t) exp(−(1/√2)x − (3/2)t) + b₋₁exp((1/√2)x + (3/2)t). (18) (iv) When we choose case 4, the solution is

u(x, t) =

− exp(−(1/√2)x + (3/2)t) − i√b₋₁

exp(−(1/√2)x + (3/2)t) + b₋₁exp((1/√2)x − (3/2)t). (19) (v) When we choose case 5, the solution is

u(x, t) = −b−1exp((1/2 √ 2)x + (3/4)t) exp(−(1/2√2)x −(3/4)t)+b₋₁exp((1/2√2)x +(3/4)t). (20)

(vi) When we choose case 6, the solution is

u(x, t) = b0

exph 1/√2x −(3/2) ti+ b0

. (21)

(vii) When we choose case 7, the solution is

u(x, t) =

exp((1/√2)x + (3/2)t) − i√b₋₁

exp((1/√2)x + (3/2)t) + b₋₁exp(−(1/√2)x − (3/2)t). (22)

Example 2. In this example we next consider the nonlinear equation of Fisher type [31],

ut− ux x− u(1 − u) (u − α) = 0, 0 < α < 1. (23)

wu0_{− k}2_u00₋_{(1 + α) u}2_{+ u} _{α + u}2

. (24)

Balancing the highest order derivative and with the highest nonlinear terms into equation (24),

c1exp [(3p + c) ξ] + · · · c2exp [4 pξ] + · · ·

= c3exp [3cξ] + · · ·

c4exp [3 pξ] + · · · , then gives p = c; where ci are determined coefficients only

for simplicity. Similarly, to determine values of d and q · · · + d1exp [−(3q + d) ξ]

· · · + d2exp [−4qξ ]

=· · · + d3exp [−3dξ ] · · · + d4exp [−3qξ ] ,

then gives q = d, where di are determined coefficients

only for simplicity. We set p = c = 1 and q = d = 1, so equation (4) reduces to

u(ξ) = a1exp(ξ) + a0+ a−1exp(−ξ)

. (25)

a₋₁(a₋₁− b−1) (a−1− αb−1) (exp(ξ) + b−1exp(ξ) + b0)3 = 0, (−k2₊_{w + α)a} 0b2₋₁+ a₋₁2 (3a0− (1 + α)b0) +a₋₁b₋₁(−2(1 + α)a0+(k2− w + 2α)b0) (exp(ξ) + b−1exp(ξ) + b0)3 = 0, 1 (exp(ξ) + b−1exp(ξ) + b0)3 −a2 −1(1 + α − 3a1) + b₋₁(−(1 + α)a2₀+(−4k2+ 2w + α)a1b−1 +(k2+w + 2α)a0b0) − a−1(−3a20+ 2(−2k 2₊_w − α + (1 + α)a1)b−1+ 2(1 + α)a0b0 +(k2+w − α)b2₀) = 0, 3

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Phys. Scr. 79 (2009) 045005 D Kaya and I E Inan 1 (exp(ξ) + b−1exp(ξ) + b0)3 a3 0− (1 + α)a 2 0b0 +(−3k2+ 3w + 2α)a1b−1b0− a−1(2a0(1 + α − 3a1) +(3k2+ 3w − 2α + 2(1 + α)a1)b0) + a0(2(3k2+α − (1 + α)a1)b−1+αb20) = 0, 1 (exp(ξ) + b−1exp(ξ) + b0)3 _(a2 0(1 + α − 3a1) +a₋₁(4k2+ 2w − α + 2(1 + α)a1− 3a21) + a0(−k2+w −2α + 2(1 + α)a1)b0+ a1 −2(2k2+w + α) +(1 + α)a1)b−1+(k2− w − α)b20 = 0, 1 (exp(ξ) + b−1exp(ξ) + b0)3 − a0 k2+w − α +2(1 + α) a1− 3a12 + a1 −k2− w − 2α +(1 + α) a1) b0) = 0, −(α − a1) (−1 + a1) a1 (exp(ξ) + b−1exp(ξ) + b0)3 = 0.

From solutions of the system, we obtain the following with the aid of Mathematica

Case 1. a0= 1 2 b0∓ q −4b−1+ b20 , a−1= b−1, a1= 0, k = − √ 2 2 , w = −1 + 2α 2 , α 6= 0, b−16= 0. (26) Case 2. a0= 1 2 αb0∓ q −4α2_b −1+α2b02 , a₋₁= αb−1, a1= 0, k = − α √ 2, (27) w = 2α − α2 2 , α 6= 0, b−16= 0.

Substituting (26)–(28) into (25) we obtain the following solutions of equation (23).

(i) For case 1,

u(x, t) =    1 2 b0∓ q −4b−1+ b20 + b₋₁ × exp(√2/2)x + ((1 − 2α)/2)t      exp(−(√2/2)x + ((−1 + 2α)/2)t) + b0 +b₋₁exp(√2/2)x + ((1 − 2α)/2)t   . (28)

(ii) For case 2,

u(x, t) =    1 2 αb0∓ q −4α2_b −1+α2b02 +αb₋₁exp(√2α/2)x + ((−2α + α2)/2)t      exp−(√2α/2)x + ((2α − α2_)/2)t_{+ b} 0 +b₋₁exp(√2α/2)x + ((−2α + α2_)/2)t   . (29) Example 3. Let us consider the Burgers-like equation [10,11],

ut+ ux+ uux+1₂ux x= 0. (30)

wu0_{+ ku}0_{+ kuu}0₊k2u00

2 = 0, (31)

and integrating (31) yields (w + k) u +ku2

2 +

k2_u0

2 = 0. (32)

When balancing u2_{with u}0_, c1exp [(p + c) ξ] + · · ·

c2exp [2 pξ] + · · · =

c3exp [2cξ] + · · · c4exp [2 pξ] + · · ·

then gives p = c. Similarly, to determine values of d and q when balancing u2with u0

· · · + d1exp [−(q + d) ξ] · · · + d2exp [−2qξ ]

=· · · + d3exp [−2dξ ] · · · + d4exp [−2qξ ] then gives q = d. For simplicity, we set p = c = 1 and q =

d = 1, so equation (4) reduces to

. (33)

a₁2k/2 + a1k+ a1w (b0+ b−1exp(−ξ) + exp(ξ))2 = 0, a0a1k − (a0k2/2) + a0k+ a1b0k +(a1b0k2/2) + a0w + a1b0w (b0+ b−1exp(−ξ) + exp(ξ))2 = 0, (a2 0k/2) + a1a−1k − a−1k2+ a1b−1k2+ a−1k+ a0b0k +a1b−1k+ a−1w + a0b0w + a1b−1w (b0+ b₋₁exp(−ξ) + exp(ξ))2 = 0, a0a−1k+(a0b−1k2/2) + a−1b0k+ a0b−1k −(a−1b0k2/2) + a−1b0w + a0b−1w (b0+ b−1exp(−ξ) + exp(ξ))2 = 0, 4

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a2

−1k/2 + a−1b−1k+ a−1b−1w (b0+ b−1exp(−ξ) + exp(ξ))2

= 0.

From the solutions of the system, we obtain the following with the aid of Mathematica.

Case 1. a1= − 2(k + w) k , b0= − a0k 2(k + w), b−1= − a₋₁k (k + w), k 6= 0, k+w 6= 0, 2k + k2+ 2w 6= 0. (34) Case 2. a1= − 2(k + w) k , b0= − a0k 2(k + w), b−1= 0, a₋₁= 0, 2k + k2+ 2w 6= 0, k + k2+w 6= 0. (35) Case 3. a1= 2k, b0= 0, a0= 0, a₋₁= 0, w = (−1 − k) k, 1 + k 6= 0. (36) Case 4. a1= 2k, b−1= 0, a06= 0, a−1= 0, b0= a0 2k, w = (−1 − k) k, k 6= 0, 1 + k 6= 0. (37) Case 5. a1= 0, a−1= a0(a0+ b0k) k , b−1= − a0(a0+ b0k) k2 , k 6= 0, w =(−2 + k) k 2 . (38) Case 6. a1= k, a−1= 0, b−1= − a0(a0− b0k) k2 , k 6= 0, w = (−2 − k)k 2 . (39)

Substituting (34)–(39) into (33), we obtain the following solutions of equation (30). These solutions are as follows.

(i) When we choose case 1, the solution is

u(x, t) = −(2(k + w)/k) exp(kx + wt) + a0+ a−1exp(−kx − wt) exp(kx + wt) − (a0k/2(k + w)) −(a−1k/2(k + w)) exp(−kx − wt) . (40)

(ii) In this case, if we choose case 2, the solution will be

u(x, t) = −(2(k + w)/k) exp (kx + wt) + a0

exp(kx + wt) − (a0k/2 (k + w))

. (41)

(iii) Again, when we choose case 3, the solution is

u(x, t) = 2k exp[kx +(−1 − k)kt]

exp[kx +(−1 − k)kt]

+b₋₁[−kx − (−1 − k)kt]

. (42)

(iv) When we choose case 4, the solution is

u(x, t) = 2k exp [kx +(−1 − k)kt] + a0

exp [kx +(−1 − k)kt] + (a0/2k)

. (43)

(v) When we choose case 5, the solution is

u(x, t) = a0+(a0(a0+ b0k)/k) exp[−kx − ((−2 + k)kt/2)] exp[−kx − ((−2 + k)kt/2)] + b0− (a0(a0+ b0k)/k2) × exp[−kx − ((−2 + k)kt/2)] . (44) (vi) When we choose case 6, the solution is

u(x, t) = a0+ k exp [kx +((−2 − k)kt/2)] exp [kx +((−2 − k)kt/2)] + b0− (a0(a0+ b0k)/k2) × exp[−kx − ((−2 − k)kt/2)] . (45)

Example 4. In this example we consider the Sharma– Tasso–Olver equation [15,16],

ut+ 3αu2x+ 3αu

2

ux+ 3αuux x+αux x x= 0. (46)

wu0_{+ 3k}2 _u02

+ 3ku2u0+ 3k2uu00+ k3u000= 0. (47) Balancing the highest order derivative and with the highest nonlinear terms into equation (47),

c1exp [(7p + c) ξ] + · · · c2exp [8 pξ] + · · · =

c3exp [(3p + 2c) ξ] + · · · c4exp [5 pξ] + · · ·

then gives p = c, where ci are determined coefficients only

for simplicity. Similarly, to determine the values of d and q, · · · + d1exp [−(7q + d) ξ]

· · · + d2exp [−8qξ ] =

· · · + d3exp [−(3q + 2d) ξ] · · · + d4exp [−5qξ ]

then gives q = d, where di are determined coefficients

only for simplicity. We set p = c = 1 and q = d = 1, so equation (4) reduces to

. (48)

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−(a−1b0− a0b−1)[3a−12 kα + 3a−1b−1k2α +b2 −1(w + k 3_α)] (b0+ b−1exp(−ξ) + exp(ξ))4 = 0, 2[−3a3 −1kα + 3a 2 −1k(−a0b0+ a1b−1+(b20− 2b−1)k)α +b2₋₁(3a2₀k2_{α + a} 0b0(w − 2k3α)) (b0+ b−1exp(−ξ) + exp (ξ))4 + a1b₋₁3 (w + 4k3α) − a₋₁b₋₁(−3a2₀kα + 6a0b0k2α +b2 0(w − 2k3α) + b−1(w − 6a1k2α + 4k3α))] (b0+ b−1exp(−ξ) + exp (ξ))4 = 0, −3a2 −1(5a0+ b0(a1− 7k))kα − a−1[−3a0b20k 2_α +6a0b−1k(−3a1+ 7k)α + b30(w + k 3_α) (b0+ b−1exp(−ξ) + exp (ξ))4 + 3b0(a02kα + 2b−1(w + (a1− 3k)k2α))] +b₋₁[3a₀2kα − 3a₀2b0k2α + 5a1b0b−1(w + k3α) (b0+ b−1exp(−ξ) + exp (ξ))4 +a0 b−1 w + (27a1− 23k) k 2_{α + b}2 0 w + k 3_α (b0+ b−1exp(−ξ) + exp (ξ))4 = 0, 4[3a2 −1k(−a1+ 2k)α − a−1(3a02kα − 3a0b0k2α +b2 0(w + k 3_{α) + b} −1(w + k(−3a12+ 12a1k − 8k2)α)) (b0+ b−1exp(−ξ) + exp (ξ))4 + b₋₁_((−6a2 0k 2_{α + 6a}2 1b−1k2al pha+ a1(b−1w + 3a02kα +3a0b0k2α − 8b−1k3α + b20(w + k 3_α))))] (b0+ b−1exp(−ξ) + exp (ξ))4 = 0, −3a3 0kα + 3a 2 0b0(a1− k)kα + a1b0(3b−1(2w +(7a1− 6k)k2α + b2₀(w + k3α))) (b0+ b−1exp(−ξ) + exp (ξ))4 − a0[b20(w + k 2_(−3a 1+ k)α) + b−1(w + (−15a12 +42a1k − 23k2)α)] − a−1(9a0(2a1− 3k)kα

(b0+ b−1exp(−ξ) + exp(ξ))4 +b0 5w + k −3a 2 1+ 6a1k+ 5k2 _α (b0+ b−1exp(−ξ) + exp (ξ))4 = 0, −2[3a2 0(a1− k)kα + a0b0(w + k(−3a21+ 6a1k − 2k2)α) +a₋₁(w + k(3a2 1− 6a1k+ 4k2)α) (b0+ b₋₁exp(−ξ) + exp(ξ))4 − a1[b20(w + (−3a1− 2k)k2α) +b₋₁(w + k(3a₁2− 6a1k+ 4k2)α)]] (b0+ b−1exp(−ξ) + exp(ξ))4 = 0, −(a0− a1b0) w + k 3a12− 3a1k+ k2 α (b0+ b−1exp(−ξ) + exp (ξ))4 = 0. From solutions of the system, we obtain the following with the aid of Mathematica

Case 1. a0= 2a1b0− b0k − q b2 0− 4b−1k 2 , a−1= b−1(a1− k) , w = − k 3a2 1− 3a1k+ k2 α, a16= 0, b06= 0, (49) c 6= 0, a1− k 6= 0, k 6= 0, α 6= 0. Case 2. a0= − p −a1b−1k2+ b−1k3 √ a1− k , w =−k 3a2 1− 3a1k+ k2α, b0= 0, k 6= 0, α 6= 0, a₋₁=1 3 −3a1b−1+ 6a2 1b−1 a1− k − a 2 1b−1k (a1− k)2 −8a1b−1k a1− k +2a1b−1k (a1− k)2 +2b−1k 2 a1− k − b−1k 3 (a1− k)2 . (50) Case 3. a0= q b2 0k2− 4b−1k2, a1= k, w = −k3α, b06= 0, k 6= 0, α 6= 0, b−16= 0, a₋₁= 1 −2b20+ 9b−1 2b2₀b₋₁k − 9b2₋₁k − 2b30 q b₀2k2_{− 4b} −1k2 + 8b0b₋₁ q b2 0k2− 4b−1k2+ 2b0 b20k 2 − 4b−1k2 3/2 k2 ! . (51) Substituting (49)–(51) into (48) we obtain the following solutions of equation (46). These solutions are as follows.

(i) For case 1,

u(x, t) =   a1expkx − k 3a12− 3a1k+ k2 αt + (2a1b0− b0k − q b₀2− 4b−1k)/2   exp [kx − k(3a2 1− 3a1k+ k2)αt] + b0 +b₋₁exp [−kx + k(3a2 1− 3a1k+ k2)αt] +b−1(a1− k) exp [−kx + k(3a 2 1− 3a1k+ k2)αt] exp [kx − k(3a2 1− 3a1k+ k2)αt] + b0 +b₋₁exp [−kx + k(3a2 1− 3a1k+ k2)αt] . (52) (ii) For case 2,

u(x, t) =

"a1exp [kx − k(3a21− 3a1k+ k2)αt] −p−a1b−1k2+ b−1k3/ √ a1− k # exp[kx − k(3a2 1− 3a1k+ k2)αt] +b₋₁exp [−kx + k(3a2 1− 3a1k+ k2)αt] +   1 3(−3a1b−1+(6a 2 1b−1/a1− k)

−(a12b−1k/(a1− k)2) − (8a1b−1k/a1− k)) × exp [−kx + k(3a21− 3a1k+ k2)αt]   exp [kx − k(3a2 1− 3a1k+ k2)αt] +b₋₁exp [−kx + k(3a2 1− 3a1k+ k2)αt] 6

(9)

Phys. Scr. 79 (2009) 045005 D Kaya and I E Inan +   1 3  2a1b−1k (a1− k)2 +2b−1k 2 a1− k − b−1k 3 (a1− k)2 × exp[−kx + k(3a12− 3a1k+ k2)αt]   exp[kx − k(3a2 1− 3a1k+ k2)αt] +b₋₁exp [−kx + k(3a2 1− 3a1k+ k2)αt] = 0. (53) (iii) For case 3,

u(x, t) =    kexp [kx − k3_{αt] +}q_b2 0k2− 4b−1k2+_−2b21 0+9b−1 × 2b2 0b−1k − 9b2−1k − 2b 3 0 q b2 0k2− 4b−1k2    exp[kx − k3αt] + b 0+ b−1exp [−kx + k3αt] +   8b0b−1 q b2₀k2_{− 4b} −1k2+ 2b0(b20k 2 − 4b−1k2)3/2 k2 × exp [−kx + k3_αt]   exp [kx − k3αt] + b 0+ b−1exp [−kx + k3αt] . (54)

3. Conclusions

Nonlinear phenomena play a crucial role in applied mathematics and physics. Furthermore, when an original nonlinear equation is directly calculated, the solution will preserve the actual physical characters of solutions. Explicit solutions to nonlinear equations are of fundamental importance. Various effective methods have been developed to understand the mechanisms of these physical models, to help physicians and engineers and to ensure knowledge for physical problems and its applications.

Many explicit exact methods have been introduced

in the literature [18–30]. Some of them are Bäcklund

transformation, Hopf–Cole transformation, generalized

Miura transformation, inverse scattering method, Darboux transformation, Painlevé method, HB method, similarity reduction method, tanh method, exp-function method, sine–cosine method and so on. There are also many numerical methods implemented in these equations [31–42]. Some of them are finite elements method, finite difference methods and some approximate methods such as Adomian decomposition method, homotopy perturbation method, variational perturbation method, sinc–Galerkin method and so on.

In this study, we considered two Fisher’s equations: a Burgers-like equation and a Sharma–Tasso–Olver equation. For doing this we implemented an exp-function method for the exact solution of these nonlinear equations. Of course this method can be implemented in more complicated nonlinear equations by using symbolic computations. In the future, scientists will be busy with applications of exact and numerical solutions of nonlinear evolution equations.

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