DOI 10.1007/s12043-016-1292-9
Comparison between the generalized tanh–coth and the ( G
/G )-expansion methods for solving NPDEs and NODEs
JALIL MANAFIAN1, MEHRDAD LAKESTANI1and AHMET BEKIR2,∗
1Department of Applied Mathematics, Faculty of Mathematical Science, University of Tabriz, Tabriz, Iran
2Department of Mathematics-Computer, Eski¸sehir Osmangazi University, 26480, Eski¸sehir, Turkey
∗Corresponding author. E-mail: abekir@ogu.edu.tr
MS received 15 July 2015; revised 2 February 2016; accepted 7 March 2016; published online 4 November 2016
Abstract. In this paper, we find exact solutions of some nonlinear evolution equations by using generalized tanh–coth method. Three nonlinear models of physical significance, i.e. the Cahn–Hilliard equation, the Allen–
Cahn equation and the steady-state equation with a cubic nonlinearity are considered and their exact solutions are obtained. From the general solutions, other well-known results are also derived. Also in this paper, we shall compare the generalized tanh–coth method and generalized (G/G)-expansion method to solve partial differential equations (PDEs) and ordinary differential equations (ODEs). Abundant exact travelling wave solutions including solitons, kink, periodic and rational solutions have been found. These solutions might play important roles in engineering fields. The generalized tanh–coth method was used to construct periodic wave and solitary wave solutions of nonlinear evolution equations. This method is developed for searching exact travelling wave solutions of nonlinear partial differential equations. It is shown that the generalized tanh–coth method, with the help of symbolic computation, provides a straightforward and powerful mathematical tool for solving nonlinear problems.
Keywords. Generalized tanh–coth method; generalized (G/G)-expansion method; Cahn–Hilliard equation;
Allen–Cahn equation; steady-state equation.
PACS Nos 02.30.Jr; 02.70.−c; 02.70.Wz 1. Introduction
Considerable attention has been directed towards the study of nonlinear problems in all areas of physics and engineering. Nonlinear partial differential equations are widely used to describe complex phenomena in many fields of applied sciences, such as chemistry, physics and the engineering. Over the last few decades, seek- ing new travelling wave solutions of nonlinear partial differential equations became a mandatory task, to sig- nificantly comprehend and describe complex pheno- mena. Well-designed mathematical models accurately describing the studied phenomena, can only enhance the chances of achieving analytical solutions, thereby yielding a better physical understanding of the phe- nomena [1]. The construction of the explicit and exact solutions to nonlinear evolution equations (NEEs), by using different methods, is the goal for many research- ers. Due to the complexity of nonlinear system, find- ing explicit and exact solutions for a real nonlinear
physical model equation is often difficult. Fortunately, an enormous number of powerful methods for seeking explicit and exact solutions of NEEs have been pro- posed and developed. Among them are the inverse scat- tering transform [2], Hirota’s bilinear method [3], homotopy analysis method [4,5], modified homotopy analysis method with Fourier transform [6], variational iteration method [7,8], homotopy perturbation method [9], homotopy perturbation transform method [10], sine–cosine method [11], tanh method [12], tanh–coth method [13], tanh–sech method [14], fractional sub- equation method [15,16], Bäcklund transformation [17], (G/G)-expansion method [18], exp-function method [19–21], modified simple equation method [22] and so on. Here, we use two effective methods for construc- ting a range of exact solutions for the following nonlin- ear partial differential equations that in this article we developed solutions as well. The standard tanh method is a well-known analytical method which was first pre- sented by Malfliet [23] and developed in [23,24]. In 1
[13], we applied the generalized tanh–coth method for solving some nonlinear partial differential equations.
Also in [25], the new approach of generalized (G/G)- expansion method to obtain exact travelling wave solutions of NLEEs is presented. In this paper, the gene- ralized tanh–coth and generalized (G/G)-expansion methods are presented to look for travelling wave solutions of nonlinear evolution equations. Heris and Lakestani [26] obtained exact solutions for the inte- grable sixth-order Drinfeld–Sokolov–Satsuma–Hirota system by the generalized tanh–coth and generalized (G/G)-expansion methods. Chand and Malik [27]
have applied the (G/G)-expansion method for finding exact solutions of some nonlinear evolution equations.
For further information about these methods, refer to [28–32]. To illustrate the basic idea of the generalized tanh–coth method, we consider the Cahn–Hilliard, Allen–Cahn and steady-state equations in the form ut−γux−6u(ux)2−(3u2−1)uxx+uxxxx=0, (1.1)
ut −uxx +u3−u=0, (1.2)
αu(x)−βu(x)(u(x)−m)(u(x)+m)=0, (1.3) whereα, β, γ andmare constants. The Cahn–Hilliard equation was introduced by Cahn and Hilliard in [33]
to describe the complicated phase separation and coars- ening phenomena in a solid. On the other hand, the Allen–Cahn equation was originally introduced by Allen and Cahn in [34] to describe the motion of antiphase boundaries in crystalline solids. The Allen–
Cahn and Cahn–Hilliard equations have been widely used in many complicated moving interface problems in materials science and fluid dynamics through a phase-field approach [35–41].
This article is organized as follows: In §2 and 3, first we briefly give the steps of these methods and apply these methods to solve the nonlinear partial differential equations. In §4, comparison of these methods is pre- sented. In §5, 6 and 7 we examine the Cahn–Hilliard equation, the Allen–Cahn equation and the steady-state equation respectively. Conclusion is given in §8.
2. Basic idea of the generalized tanh–coth method Step1. Suppose the given nonlinear partial differential equation foru(x, t)is in the form
P(u, ut, ux, utt, uxx, . . .)=0, (2.1) which can be converted to an ODE
Q(u,−cu, u, u, u, u, . . .)=0, (2.2) where ξ =x−ct is the wave variable. Also, c is a constant to be determined later.
Step2. We introduce the Riccati equation as follows:
=r+p+q2, =(ξ ), ξ=x−ct, (2.3) and this leads to the change of derivatives
d
dξ =(r+p+q2) d
d, (2.4)
d2
dξ2 =(r+p+q2)
×
(p+2q) d
d+(r+p+q2) d2 d2
, (2.5) d3
dξ3 =(r+p+q2)
×
(6q22+6pq+2rq+p2) d d +(6q23+9pq2
+3(p2+2rq)+3rp) d2 d2 +(r+p+q2)2 d3
d3
, (2.6)
which admits the use of a finite series of functions of the form:
u(ξ )=S()= m k=0
akk+ m k=1
bk−k, (2.7) whereak(k=0,2, . . . , m),bk (k=1,2, . . . , m),p,r and q are constants to be determined later. But, the positive integer m can be determined by considering a homogeneous balance between the highest-order derivatives and nonlinear terms appearing in eq. (2.2).
If m is not an integer, then a transformation formula should be used to overcome this difficulty. For the aforementioned method, expansion (2.7) reduces to the standard tanh method [23] forbk=0,1≤k≤m. Step 3. Substitute eqs (2.3)–(2.6) into eq. (2.2) with the value ofmobtained in Step 2. Collecting the coef- ficients of k (k = 0, 1, 2, . . .), then setting each coefficient to zero, we can get a set of overdetermined equations fora0, ai (i=1, 2,. . .,m),bi(i=1, 2,. . ., m),p, qandrwith the aid of the symbolic computation Maple13.
Step 4. Solve the algebraic equations in Step 3, then substitutea0, a1, b1, . . . , am, bm, cin eq. (2.7).
We shall consider the following special solutions of the Riccati equation (2.3):
Family1: For each p, r and q =0, eq. (2.3) has the following solutions:
(ξ )= −p 2q +
√− 2q tan
√
−ξ
2 +C
,
(ξ )= −p 2q −
√− 2q cot
√
−ξ
2 +C
, (2.8) (ξ )= −p
2q −
√ 2q tanh
√ ξ 2 +C
,
(ξ )= −p 2q −
√ 2q coth
√ ξ 2 +C
, (2.9)
where=p2−4qr, ξ =x−ct andCis a constant.
Family2: For p=0, r=0 and q =0, eq. (2.3) has the following solution:
(ξ )=Cepξ − r
p, ξ =x−ct. (2.10)
Family3: For r =0, p=0 and q =0, eq. (2.3) has the following solution:
(ξ )= r
qtan(√
rqξ+C). (2.11)
But, we know tanh
ξ
2 =coth(ξ )−csch(ξ ), coth
ξ
2 =coth(ξ )+ csch(ξ ). (2.12) Family4: For r =0, p=0 and q =0, eq. (2.3) has the following solution:
(ξ )= p
−q+Ce−pξ. (2.13)
3. Description of the new generalized (G/G)-expansion method
Step1. Consider the general nonlinear partial differen- tial equation
P(u, ut, ux, utt, uxx, . . .)=0, (3.1) which can be converted to an ODE
Q(u,−cu, u, u, u, u, . . .)=0, (3.2) where ξ =x−ct is the wave variable. Also, c is a constant to be determined later.
Step 2. Suppose the travelling wave solution of (3.2) can be expressed as follows:
u(ξ )= N
i=0
di(p+M(ξ ))i+ N i=1
ei(p+M(ξ ))−i, (3.3) where eitherdN oreN may be zero, but both of them could not be zero at the same time. di(i = 0, 1, 2, . . ., N), ei (i = 1, 2, . . ., N) and p are constants to be determined later and M(ξ )=(G/G), where G = G(ξ ) satisfies the following auxiliary nonlinear ordinary differential equation:
k1GG−k2GG−k3(G)2−k4G2 =0, (3.4) where the prime stands for derivative with respect to ξ;k1, k2, k3andk4are real parameters.
Step 3. Determine the positive integer N, taking the homogeneous balance between the highest-order non- linear terms and the derivatives of the highest-order come out in (3.2).
Step 4. Substitute eq. (3.4) including eq. (3.3) into eq. (3.2) and the value of N obtained in Step 3, we obtain polynomials in (p+M(ξ))N (N=1, 2,. . .) and (p+M(ξ ))−N(N =1,2, . . .). Then, we collect each coefficient of the resulted polynomials to zero to yield a set of algebraic equations fordi(i =0,1,2, . . . , N) andei (i=1,2, . . . , N),pandc.
Step5. Suppose that the value of the constantsdi (i= 0, 1, 2,. . .,N) and ei (i=1, 2, . . ., N),p andc can be found by solving the algebraic equations obtained in Step 4. As the general solution of eq. (3.4) is well known, by inserting the values ofdi,ei, p andc into eq. (3.2), we obtain more general type and new exact travelling wave solutions of the nonlinear partial differ- ential eq. (3.1). Using the general solution of eq. (3.4), we have the following solutions of eq. (3.3):
Family 1: When k2 =0, α=k1−k3, β =k22+4k4
andα >0. M(ξ ) = G(ξ )
G(ξ ) = k2
2α +
√β 2α
⎛
⎝Asinh √
βξ 2k1
+Bcosh √
βξ 2k1
Acosh
√ βξ 2k1
+Bsinh √
βξ 2k1
⎞
⎠.(3.5)
Family 2: When k2=0, α=k1−k3, β =k22+4k4
andα <0. M(ξ )=G(ξ )
G(ξ ) =k2
2α +
√−β 2α
⎛
⎝−Asin √
−βξ 2k1
+Bcos √
−βξ 2k1
Acos
√
−βξ 2k1
+Bsin √
−βξ 2k1
⎞
⎠. (3.6) Family 3: When k2=0, α=k1−k3, β =k22+4k4
andα =0. M(ξ )= G(ξ )
G(ξ ) = k2
2α + C2
C1+C2ξ. (3.7) Family 4: When k2 = 0, α = k1 − k3 and γ = αk4 >0.
M(ξ )=G(ξ ) G(ξ )
=
√γ α
⎛
⎝Asinh √
γ ξ k1
+Bcosh √
γ ξ k1
Acosh
√ γ ξ k1
+Bsinh √
γ ξ k1
⎞
⎠. (3.8)
Family 5: When k2 = 0, α = k1 − k3 and γ = αk4 <0.
M(ξ ) = G(ξ ) G(ξ ) =
√−γ α
×
⎛
⎝−Asin √
−γ ξ k1
+Bcos √
−γ ξ k1
Acos
√−γ ξ k1
+Bsin √−γ ξ
k1
⎞
⎠. (3.9)
4. Comparison of the generalized tanh–coth and the (G/G)-expansion methods
We consider (3.4) as follows:
k1GG−k2GG−k3(G)2−k4G2=0
⇒ k1GG−k2GG−k3(G)2−k4G2
G2 =0
(4.1) or
k1
G G −k2
G G −k3
G G
2
−k4 =0. (4.2)
SetF =G/Gand then F=GG−G2
G2 = G
G −
G G
2
= G
G −F2. (4.3) Thus, we conclude that relation (4.2) is transformed to k1F+(k1−k3)F2−k2F −k4 =0 (4.4) or
F = k4
k1 +k2
k1F +k3−k1
k1 F2
= r+pF+qF2, (4.5)
where r =k4/k1, p =k2/k1 and q =(k3−k1)/k1. We found that relation (3.3) is the same as relation (2.3). Thus, we obtain that the exact solutions derived by the generalized (G/G)-expansion are the same as the ones by the generalized tanh–coth methods. Hence we use only the generalized tanh–coth method.
5. The Cahn–Hilliard equation
We consider the Cahn–Hilliard equation in the form [38,41]
ut−γ ux−6u(ux)2−(3u2−1)uxx+uxxxx =0. (5.1) Using the wave variableξ =x−ct, we get,
−(c+γ )u−3u2u+u+u =0 (5.2) which is obtained by integrating and neglecting the constant of integration. In order to determine the value of m, we balance the linear term of the highest- order u with the highest-order nonlinear term u2u in eq. (5.2) and by using eq. (2.7) we obtain m=1.
We can suppose that the solutions of eq. (5.1) is of the form
u(ξ )=a0+a1+ b1
. (5.3)
Using Family 1–4 in §2, substituting (5.3) into eq. (5.2) and collecting all terms with the same order of (ξ ) together, we can obtain a set of algebraic equations for a0, a1, b1, p, q, randcas follows:
Coefficients ofk:
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
0:6b1r3−3b31r =0,
1:12b1r2p−6a0b21r −3b31p=0,
2: −3a1b21r+8r2qb1+7rp2b1+rb1−3b13q
−3a02rb1−6a0b21p=0,
3:8rpqb1+p3b1+Bb1−6a0b12q−3a1b21p
−3a02pb1+cb1+γ b1 =0,
4:γ a0+ca0−2r2qa1+3a02ra1−rp2a1
+2q2rb1−ra1+qp2b1−3a1b21q +qb1
+3a12rb1−3a20qb1 =0,
5:3a20pa1−8rpqa1+ca1+6a0a21r −p3a1
−pa1+γ a1+3a21pb1 =0,
6: −7qp2a1+3a12qb1+3a13r+3a02qa1
+6a0a21p−8q2ra1−qa1 =0, 7: −12pq2a1+3a13p+6a0a12q =0, 8:3a31q−6q3a1=0.
(5.4) Solving eqs (5.4), we have the following sets of coeffi- cients for the solutions of (5.3)
SetI a0= ± p
√2, a1 =0, b1 =b1, c= −γ , =p2∓(p2−2), u(ξ )= ±√p
2 +b1−1. (5.5) By using Family 1 we have
u1(x, t)= ± p
√2 +
∓ p (p2−2)√
2
±
−2p2±2(p2−2) p2−2
×tan
−p2±(p2−2)(x+γt)
2 +C
−1
,(5.6) u2(x, t)= ±√p
2 +
∓ p (p2−2)√
2
∓
−2p2±2(p2−2) p2−2
×cot
−p2±(p2−2)(x+γt)
2 +C
−1
,(5.7) u3(x, t)= ± p
√2+
∓ p (p2−2)√
2
∓
2p2∓2(p2−2) p2−2
×tanh
p2∓(p2−2)(x+γt)
2 +C
−1
,(5.8)
u4(x, t)= ±√p 2+
∓ p (p2−2)√
2
∓
2p2∓2(p2−2) p2−2
×coth
p2∓(p2−2)(x+γt)
2 +C
−1 .(5.9) Set II
a0 = ± p
√2, a1 =a1, b1 =0, c= −γ , = p2∓(p2−2), u(ξ )= ± p
√2 +a1. (5.10) By using Family 1 we have
u5(x, t)= ± p
√2+
∓ p
√2±
−p2±(p2−2)
√2
×tan
−p2±(p2−2)(x+γ t)
2 +C
, (5.11) u6(x, t)= ±√p
2+
∓√p 2∓
−p2±√(p2−2) 2
×cot
−p2±(p2−2)(x+γ t)
2 +C
, (5.12) u7(x, t)= ± p
√2+
∓ p
√2∓
p2∓(p2−2)
√2
×tanh
p2∓(p2−2)(x+γ t)
2 +C
, (5.13) u8(x, t)= ± p
√2+
∓ p
√2∓
p2∓(p2−2)
√2
×coth
p2∓(p2−2)(x+γ t)
2 +C
. (5.14) SetIII
a0 = ± p
√2, a1=a1, b1= −p2−2 4a1 , c= −γ , =p2∓(p2−2),
u(ξ )= ±√p
2+a1−p2−2
4a1 −1, (5.15)
By using Family 1 we have u9(x, t)= ± p
√2 +
⎡
⎢⎣∓ p
√2 ±
−p2∓(p22−2)
√2
×tan
⎛
⎜⎝
−p2∓(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−p2−2 4
⎡
⎢⎣∓ p
√2±
−p2∓ (p22−2)
√2
×tan
⎛
⎜⎝
−p2∓(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
, (5.16) u10(x, t)= ± p
√2 +
⎡
⎢⎣∓ p
√2 ∓
−p2∓(p22−2)
√2
×cot
⎛
⎜⎝
−p2∓(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−p2−2 4
⎡
⎢⎣∓ p
√2 ∓
−p2∓(p22−2)
√2
×cot
⎛
⎜⎝
−p2∓(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
, (5.17) u11(x, t)= ±√p
2 +
⎡
⎢⎣∓√p 2 ∓
p2±(p22−2)
√2
×tanh
⎛
⎜⎝
p2±(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−p2−2 4
⎡
⎢⎣∓√p 2 ∓
p2± (p22−2)
√2
×tanh
⎛
⎜⎝
p2± (p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
, (5.18)
u12(x, t)= ±√p 2 +
⎡
⎢⎣∓√p 2∓
p2± (p22−2)
√2
×coth
⎛
⎜⎝
p2±(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−p2−2 4
⎡
⎢⎣∓√p 2 ∓
p2±(p22−2)
√2
×coth
⎛
⎜⎝
p2±(p22−2)(x+γ t)
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
. (5.19) SetIV
a0 = ±1, a1 =0, b1 =b1, c= −γ ,
u(ξ )= ±1+b1−1. (5.20) By using Family 2 we have
u13(x, t)=±1+b1
Cexp[±√
2(x+γ t)] −b1
2
−1
. (5.21) SetV
a0 =0, a1 =a1, b1 = 1
4a1, c= −γ , u(ξ )=a1+ 1
4a1−1. (5.22)
By using Family 3 we have u14(x, t)=1+16a14 itan
i 2√
2(x+γ t)+C
!2
8a12 itan i
2√
2(x+γ t)+C
!
=1+16a14tanh2 1
2√
2(x+γ t)+C
−8a12tanh 1
2√
2(x+γ t)+C
.(5.23) SetVI
a0 = ±1, a1 =a1, b1 =0, c= −γ ,
u(ξ )= ±1+a1. (5.24)
By using Family 4 we have u15(x, t) = ±1+a1
×
⎛
⎝ ±√ 2
∓√12a1+Cexp[∓√
2(x+γ t)]
⎞
⎠. (5.25)
6. The Allen–Cahn equation
Now we consider the Allen–Cahn equation in the form [39,41]
ut −uxx +u3−u=0. (6.1)
Using the wave variableξ =x−ct, we get
cu+u−u3+u=0. (6.2)
In order to determinem, we balance the linear term of the highest-orderu with the highest-order nonlinear termu3 in eq. (6.2) and by using eq. (2.7) we obtain m=1. We can suppose that the solutions of eq. (6.1) is of the form
u(ξ )=a0+a1+b1
. (6.3)
Using Family 1–4 in §2 and substituting (6.3) into eq. (6.2) and collecting all terms with the same order of (ξ ) together, we can obtain a set of algebraic equations fora0, a1, b1, p, q, randcas follows:
Coefficients ofk:
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
0:2b1r2−b31 =0,
1: −crb1−3a0b12+3rpb1 =0, 2: −3a1b21+p2b1−3a02b1−cpb1
+2qb1r+b1=0,
3:cra1−a03+rpa1−6a0a1b1+a0
−cqb1+pqb1=0,
4:2rqa1+p2a1+a1+cpa1−3a21b1
−3a02a1 =0,
5:3pqa1−3a0a12+cqa1 =0, 6: −a13+2q2a1=0.
(6.4)
Solving eqs (6.4), we have the following sets of coeffi- cients for the solutions of (6.3):
SetI
a0= ± 1
√2
p∓ 1
√2 , a1 =0, b1 =b1, c= ± 3
√2, =p2∓(2p2−1)
2 ,
u(ξ )= ± 1
√2
p∓ 1
√2 +b1−1. (6.5)
By using Family 1 we have u1(x, t) = ± 1
√2
p∓ 1
√2
+
⎡
⎢⎣∓ 2√ 2p 2p2−1 ±
−8p2±4(2p2−1) 2p2−1
×tan
⎛
⎜⎝
−p2±(2p22−1) x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
, (6.6) u2(x, t) = ± 1
√2
p∓ 1
√2
+
⎡
⎢⎣∓ 2√ 2p 2p2−1 ∓
−8p2±4(2p2−1) 2p2−1
×cot
⎛
⎜⎝
−p2±(2p22−1) x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
, (6.7) u3(x, t) = ± 1
√2
p∓ 1
√2
+
⎡
⎢⎣∓ 2√ 2p 2p2−1 ∓
8p2∓4(2p2−1) 2p2−1
×tanh
⎛
⎜⎝
p2∓(2p22−1) x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
, (6.8) u4(x, t)= ± 1
√2
p∓ 1
√2
+
⎡
⎢⎣∓ 2√ 2p 2p2−1∓
8p2∓4(2p2−1) 2p2−1
×coth
⎛
⎜⎝
p2∓(2p22−1)
x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦
−1
. (6.9) SetII
a0 = ± 1
√2
p± 1
√2 , a1=a1, b1=0, c= ± 3
√2, =p2∓ (2p2−1)
2 ,
u(ξ )= ± 1
√2
p± 1
√2 +a1. (6.10)
By using Family 1 we have
u5(x, t)= ± 1
√2
p± 1
√2
+
⎡
⎢⎣∓√p 2±
−p2± (2p22−1)
√2
×tan
⎛
⎜⎝
−p2±(2p22−1)
x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦,
(6.11)
u6(x, t)= ± 1
√2
p± 1
√2
+
⎡
⎢⎣∓√p 2∓
−p2± (2p22−1)
√2
×cot
⎛
⎜⎝
−p2±(2p22−1)
x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦,
(6.12)
u7(x, t)= ± 1
√2
p± 1
√2
+
⎡
⎢⎣∓ p
√2∓
p2∓(2p22−1)
√2
×tanh
⎛
⎜⎝
p2∓(2p22−1)
x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦,
(6.13) u8(x, t)= ± 1
√2
p± 1
√2
+
⎡
⎢⎣∓ p
√2∓
p2∓(2p22−1)
√2
×coth
⎛
⎜⎝
p2∓(2p22−1)
x∓√32t
2 +C
⎞
⎟⎠
⎤
⎥⎦.
(6.14)
SetIII a0 = ±1
2, a1=a1, b1=0, c = ± 3
√2, r = ∓ 1
4√ 2a1
, p =0, q = ± 1
√2a1, u(ξ )= ±1
2+a1. (6.15)
By using Family 3 we have u9(x, t) = ±1
2 + i 2tan
1 2√
2i
x∓ 3
√2t +C
= ±1 2 −1
2tanh 1
2√ 2
x∓ 3
√2t +C
. (6.16) SetIV
a0 = −1
2, a1=a1, b1= 1
16a1, c= ∓ 3
√2, r = ∓ 1
16√ 2a1
, p=0, q = ±a1
√2, u(ξ )= −1
2+a1+ 1
16a1−1. (6.17) By using Family 3 we have
u10(x, t) = −1 2+ i
4tan 1
4√ 2i
x± 3
√2t +C
−i 4cot
1 4√
2i
x± 3
√2t +C
= −1 2−1
4tanh 1
4√ 2
x± 3
√2t +C
−1 4coth
1 4√
2
x± 3
√2t +C
.(6.18) SetV
a0 = 1
2, a1=a1, b1= 1
16a1, c= ± 3
√2, r = ∓ 1
16√ 2a1
, p=0, q = ±a1
√2, u(ξ )= −1
2+a1+ 1
16a1−1. (6.19)
By using Family 3 we have u11(x, t) = 1
2 + i 4tan
1 4√
2i
x∓ 3
√2t +C
−i 4cot
1 4√
2i
x∓ 3
√2t +C
= 1 2 −1
4tanh 1
4√ 2
x∓ 3
√2t +C
−1 4coth
1 4√
2
x∓ 3
√2t +C
. (6.20)
7. The steady-state equation
At last, we consider the steady-state equation as fol- lows [40,41]:
αu(x)−βu(x)(u(x)−m)(u(x)+m)=0, (7.1) where α, β and m are constants. By making the transformation
v(x)=m−1(u(εx)+m), (7.2) whereε=
α/βm2, eq. (7.1) becomes
v(x)+v(x)+v3(x)=0. (7.3) By balancing term v with term v3 in eq. (7.3) we obtainm = 1. We can suppose that the solutions of eq. (7.1) is of the form
v(x)=a0+a1+b1
. (7.4)
Using Family 1–4 in §2 and substituting (7.4) into eq. (7.3) and collecting all terms with the same order of (ξ ) together, we can obtain a set of algebraic equations fora0, a1, b1, p, q, randcas follows:
Coefficients ofk:
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
0: −b13+2b1r2=0,
1:3b21+3rpb1−3a0b21 =0, 2: −3a02b1+6a0b1+p2b1+2qb1r
−2b1−3a1b12=0,
3:rpa1−6a0a1b1−2a0+pqb1−a03 +3a02+6a1b1 =0,
4: −3a02a1−2a1+p2a1+2rqa1
−3a12b1+6a0a1 =0, 5:3pqa1−3a0a12+3a21 =0, 6: −a13+2q2a1=0.
(7.5)
Solving eqs (7.5), we have the following sets of coeffi- cients for the solutions of (7.4) as given below:
SetI
a0 =1± 1
√2p, a1=0, b1=b1, =p2∓(p2−2), v(x)=1± 1
√2p+b1−1. (7.6) By using Family 1 we have
v1(x) =1± 1
√2p+
∓
√2p p2−2
±
−2p2±2(p2−2) p2−2
×tan
−p2±(p2−2)
2 x+C
−1
,
u1(x) = ±mp√ 2 +m
∓
√2p p2−2
±
−2p2±2(p2−2) p2−2
× tan
−p2±(p2−2) 2
×m β
αx+C −1
, (7.7)
v2(x)=1± 1
√2p+
∓
√2p p2−2
∓
−2p2±2(p2−2) p2−2
×cot
−p2±(p2−2)
2 x+C
−1
,
u2(x) = ±mp√ 2 +m
∓
√2p p2−2
∓
−2p2±2(p2−2) p2−2
×cot
−p2±(p2−2)
2 m
β αx+C
−1
,(7.8)
v3(x) = 1± 1
√2p+
∓
√2p p2−2
∓
2p2∓2(p2−2) p2−2
×tanh
p2∓(p2−2)
2 x+C
−1
,
u3(x) = ±√mp 2 +m
∓
√2p p2−2
∓
2p2∓2(p2−2) p2−2
×tanh
p2∓(p2−2)
2 m
× β
αx+C −1
, (7.9)
v4(x) = 1± 1
√2p+
∓
√2p p2−2
∓
2p2∓2(p2−2) p2−2
×coth
p2∓(p2−2)
2 x+C
−1
,
u4(x) = ±mp
√2 +m
∓
√2p p2−2
∓
2p2∓2(p2−2) p2−2
×coth
p2∓(p2−2)
2 m
× β
αx+C −1
. (7.10)
SetII
a0=1± 1
√2p, a1 =a1, b1 =0, =p2∓(p2−2), v(x)=1± 1
√2p+a1. (7.11)
By using Family 1 we have v5(x, t) = 1± 1
√2p+
∓√p 2
±
−p2±√(p2−2) 2
×tan
−p2±(p2−2)x
2 +C
,
u5(x, t) = ±mp
√2 +m
∓ p
√2±
−p2±(p2−2)
√2
×tan
−p2±(p2−2)
2 m
β αx+C
, (7.12) v6(x, t) = 1± 1
√2p+
∓√p 2 ∓
−p2±√(p2−2) 2
×cot
−p2±(p2−2)x
2 +C
,
u6(x, t) = ±mp
√2 +m
∓ p
√2∓
−p2±(p2−2)
√2
×cot
−p2±(p2−2)
2 m
× β
αx+C
, (7.13)
v7(x, t) = 1± 1
√2p+
∓√p 2 ∓
p2∓√(p2−2) 2
×tanh
p2∓(p2−2)
2 x+C
,
u7(x, t) = ±mp
√2 +
∓ p
√2 ∓
p2∓(p2−2)
√2
×tanh
p2∓(p2−2)
2 m
β αx+C
, (7.14)
v8(x) = 1± 1
√2p+
∓ p
√2 ∓
p2∓(p2−2)
√2
×coth
p2∓(p2−2)
2 x+C
,
u8(x) = ±√mp 2 +
∓√p 2 ∓
p2∓√(p2−2) 2
×coth
p2∓(p2−2)
2 m
β αx+C
. (7.15) SetIII
a0=1, a1= ±√
2q, b1= ±
√2
8q, q =q, r = − 1
8q, p=0, u(x)=1±√ 2q±
√2 8q−1.
(7.16) By using Family 3 we have
v9(x) = 1±qi 2 tan
1 2√
2ix+C
∓ i 2q cot
1 2√
2ix+C
= 1∓q 2tanh
1 2√
2x+C
∓ 1 2q coth
1 2√
2x+C
, u9(x) = ±mqi
2 tan
1 2√
2im β
αx+C
∓ i 2q cot
1 2√
2im β
αx+C
= ∓mq 2 tanh
1 2√
2m β
αx+C
∓m 2q coth
1 2√
2m β
αx+C
. (7.17)
SetIV
a0=1, a1= ±√
2q, b1= ±
√2
8q, q =q, r = 1
4q, p=0, u(x)=1±√ 2q±
√2 8q −1.
(7.18)
−10 0
10
−10 0 10
−40
−20 0 20 40 60 80
(a) t x u1(x,t)
−20
−10 0 10 20 30 40 50 60 70
−10 0
10
−10 0 10
−80
−60
−40
−20 0 20
(b) t x u2(x,t)
−60
−50
−40
−30
−20
−10 0 10
Figure 1. Graphs of Cahn–Hilliard equation (a) u1− and (b) u2− real values of (5.6) and (5.7) are demonstrated at p=1/2 andγ =2, when−10< x <10,−10< t <10.
By using Family 3 we have v10(x) = 1±√q
2tan 1
2x+C
∓
√2 4q cot
1 2x+C
, u10(x) = ±mq
√2tan
1 2m
β αx+C
∓
√2m 4q cot
1 2m
β αx+C
. (7.19)
Please note that all the obtained results have been checked withMaple13 by putting them back into the original equation and found to be correct.
−10 0
10
−10 0 10
−3
−2
−1 0 1
(a) t x u3(x,t)
−10 0
10
−10 0 10
−2.5
−2
−1.5
−1
(b) t x u4(x,t)
−2.5
−2
−1.5
−1
−0.5 0
−2.2
−2
−1.8
−1.6
−1.4
−1.2
Figure 2. Graphs of Cahn–Hilliard equation (a) u3− and (b) u4− real values of (5.8) and (5.9) are demonstrated at p=2 andγ =2, when−10< x <10,−10< t <10.
