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Symbolic computation and a uniform direct ansätze method for constructing travelling wave solutions of nonlinear evolution equations
Nonlinear Analysis 69 (2008) 2748–2760 www.elsevier.com/locate/na
Short communication
Symbolic computation and a uniform direct ans¨atze method for constructing travelling wave solutions of nonlinear evolution equations Zhi Hongyan a,∗ , Zhang Hongqing b a College of Mathematics and Computational Science, China University of Petroleum, Dongying 257061, Shandong Province, China b Department of Applied Mathematics, Dalian University of Technology, Dalian 116024, China
Received 7 December 2006; accepted 26 April 2007
Abstract On the basis of the computer symbolic system Maple and the tanh method, the Riccati equation method as well as all kinds of improved versions of these methods, we present a further uniform direct ans¨atze method for constructing travelling wave solutions of nonlinear evolution equations. Compared with the existing methods, the presented method can be used to construct more new general solutions. And we give some examples to illustrate the key step of our method. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Travelling wave solution; Direct ans¨atze method; Nonlinear evolution equation
1. Introduction The investigation of exact travelling wave solutions to nonlinear evolution equations plays an important role in the study of nonlinear physical phenomena. Recently, much effort has been spent on the construction of exact solutions of nonlinear evolution equations and many powerful methods have been developed such as the inverse scattering method [1], B¨acklund and Darboux transformation methods [2–4], Hirota bilinear method [5], Lie group reduction method [6–9], homogeneous balance method [10,11], etc. With the development of symbolic computation, the tanh function method, Riccati method [12,13], projective Riccati equation method [14,15], sine–cosine function method [16], Jacobi elliptic function method [17,18] and all kinds of auxiliary equation methods attract more and more researchers. Many authors have improved these methods [19–23]. In this paper, we give a further uniform direct ans¨atze method to construct the exact solutions of nonlinear evolution equations. 2. Introduction of the new uniform direct ans¨atze method Consider a system of N nonlinear evolution equations (N ≥ 1) with n independent variables x = (x1 , x2 , . . . , xn ) and m dependent variables u = (u 1 , u 2 , . . . , u m ), given by ∗ Corresponding author.
E-mail address: [email protected] (H. Zhi). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.04.041
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
Aµ (u, ∂u, ∂ 2 u, . . . , ∂ r u) = 0,
µ = 1, 2, . . . , N
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(2.1)
where ∂ j u denotes the coordinates with components ∂ j u/∂ xi1 ∂ xi2 · · · ∂ xi j = u i1 i2 ···i j , i j = 1, 2, . . . , n, for j = 1, 2, . . . , k, corresponding to all jth-order partial derivatives of u with respect to x. Because of the importance of travelling wave solutions, throughout this paper, we seek the nonlinear evolution equations’ travelling wave solutions. Then the travelling wave transformation ! n X ξ = λ1 x 1 + λ2 x 2 + · · · + λn x n , λi2 6= 0 (2.2) i=1
transforms (2.1) into nonlinear ODEs: A¯ µ (u, u 0 , u 00 , . . . , u (k) ) = 0,
µ = 1, 2, . . . , N .
(2.3)
Here we give a uniform direct ans¨atze, u i = Fi (θ1 , θ2 , . . . , θl ) +
Pi (θ1 , θ2 , . . . , θl ) , Q i (θ1 , θ2 , . . . , θl )
(i = 1, 2, . . . , m)
(2.4)
where θi (i = 1, 2, . . . , l) are functionally independent functions of ξ , their derivative is a rational polynomial function of θ1 , θ2 , . . . , θl . Fi , Pi , Q i (i = 1, 2, . . . , m) are polynomials of the corresponding variables. deg(Pi ) < deg(Q i ) and gcd(Pi , Q i ) = 1 (i = 1, 2, . . . , m). The leading coefficient of the Q i is equal to 1. If θiJi = Ri (θ1 , θ1 , . . . , θi−1 ) (i = 1, 2, . . . , m), then the highest power of θi in Fi , Pi , Q i is Ji − 1. Substituting (2.4) into (2.3) and simplifying, we obtain X µ ai1 ,i2 ,...,il θ1i1 θ2i2 · · · θlil = 0, µ = 1, 2, . . . , m. (2.5) 0≤i 1 ,i 2 ,...,il
During the simplifying process, we stick to a principle: if θiJi = Ri (θ1 , θ2 , . . . , θi−1 ), then the highest power of θi in (2.5) is Ji − 1. Because of the functional independence of θ1i1 θ2i2 · · · θlil , we obtain the algebraic equations µ ai1 ,i2 ,...,in = 0 with respect to λ j and the coefficients of Fi , Pi , Q i (i = 1, 2, . . . , m, j = 1, 2, . . . , n). Solving the equations, we can determine these explicit values. Using also (2.2) and (2.4), we can obtain the solutions of Eq. (2.1). Remark: (i) Let deg(Fi ) = ki1 , deg(Pi ) = ki2 , deg(Q i ) = ki3 . n i (i = 1, 2, . . . , m) are obtained by the homogeneous balance principle [10,11]. Generally speaking, ki1 ≤ n i , ki2 ≤ n i . (ii) We give some examples to illustrate how to choose appropriate θ1 , θ2 , . . . , θl . When l = 1, we can set θ1 = tanh(ξ ). Similarly, we can set θ1 = sech(ξ ), coth(ξ ), csch(ξ ), tan(ξ ), sec(ξ ), cot(ξ ), csc(ξ ) . . .. Of course, we can set θ1 as a solution of φ 0 = δ + φ 2 [12,13] or φ 0 2 = c0 + c1 φ + c2 φ 2 + c3 φ 3 + c4 φ 4 [24]. When l = 2, we can set θ1 = tanh(ξ ), θ2 = sech(ξ ). Because θ22 = 1 − θ12 , the highest power of θ2 in Fi , Pi , Q i is equal to 1. Similarly we can set θ1 = tan(ξ ), θ2 = sec(ξ ) or θ1 = coth(ξ ), θ2 = csch(ξ ), . . . . And we can set θ1 , θ2 as the solutions of the projective Riccati equations σ 0 (ξ ) = σ (ξ )τ (ξ ), where “0 ” =
d dξ ,
τ 0 (ξ ) = R + τ 2 (ξ ) − µσ (ξ ),
(2.6)
= ±1, µ and R are undetermined constants [14,15]. That is to say θ1 = σ, θ2 = τ . Since
µ2 − 1 2 σ (ξ ) , τ 2 (ξ ) = − R − 2µσ (ξ ) + R
(2.7)
the highest power of θ2 in Fi , Pi , Q i is equal to 1. When l = 3, we can set θ1 = sn(ξ, k), θ2 = cn(ξ, k), θ3 = dn(ξ, k), where the k is the module of the Jacobi elliptic function. Since θ22 = 1 − θ12 , θ32 = 1 − k 2 θ12 , the highest power of θ2 , θ3 in Fi , Pi , Q i is equal to 1. We can also set θ1 = ns(ξ, k), θ2 = nc(ξ, k), θ3 = dn(ξ, k), . . .. In fact, θ1 , θ2 , θ3 can also be solutions of ordinary differential
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equations, for example, θ10 = δ[kθ2 (1 − 2Bθ2 − 2Cθ3 − 2 Aθ3 ) + Bθ12 ], θ20 = δθ1 (1 − Bθ2 − 2Cθ3 ), θ30 = δθ1 (2Aθ2 + Bθ3 ), Aθ12 Aθ22
(2.8)
= kθ3 (1 − Bθ2 − Aθ3 − Cθ3 ), = θ3 (1 − Bθ2 − Cθ3 ),
d where A 6= 0, B, C are arbitrary constants, k = ±1, δ = ±1, = ±1, “0 ” = dξ [25]. Up to now, we have not considered the situation l > 3. This is a problem worth considering. (iii) On the basis of the theory that any proper fraction can be presented as the sum of simplest terms, we can always express u as a polynomial and some simplest fractions. For example, when l = 1, θ1 = tanh(ξ ), one can set ) ( ni X bi j (di j + ei j tanh(ξ )) j ci j j u i = ai0 + ai j tanh (ξ ) + + ··· + + tanh j (ξ ) (µi0 + tanh(ξ )) j (µi1 + µi2 tanh(ξ ) + tanh2 (ξ )) j j=1
(i = 1, 2, . . . , m), where the n i are no larger than the balance constants, ai j , bi j , ci j , di j , ei j , µi0 , µi1 , µi2 are constants to be determined, and 1 ≤ j ≤ n i is a positive integer. Generally, we can only choose parts of these terms in order to obtain more new solutions which have not been presented otherwise. In the following, we present some examples to illustrate the above mentioned method. 3. Examples 3.1. Exact solutions of the (2 + 1)-dimensional Burgers equation The (2 + 1)-dimensional Burgers equation reads −u t + uu y + αvu x + βu yy + αβu x x = 0, u x − v y = 0.
(3.1)
An equivalent form of the Burgers equation (3.1) is derived from the generalized Painl´eve integrability classification in Ref. [26]. On introducing the travelling wave transformation u(x, y, t) = u(ξ ),
v(x, y, t) = v(ξ ),
ξ = k(x + ly + λt),
(3.2)
Eq. (3.1) takes the form −λu 0 + luu 0 + αvu 0 + βkl 2 u 00 + αβku 00 = 0, 0
0
u − lv = 0.
(3.3a) (3.3b)
From (3.3b), we have (3.4)
u = lv. Here we set the integral constant to be zero. Substituting (3.4) into (3.3a) we get −λv 0 + l 2 vv 0 + αvv 0 + βkl 2 v 00 + αβkv 00 = 0.
(3.5)
If we have solved Eq. (3.5), together with (3.4), we can obtain the solutions of Eq. (3.1). Now we turn to solving Eq. (3.5). Balancing the term v 00 with the term vv 0 in Eq. (3.5) gives n = 1. Then we can assume v = a0 + a1 tanh (ξ ) +
b1 c1 + d1 tanh (ξ ) + . µ1 + tanh (ξ ) µ2 + µ3 tanh (ξ ) + tanh2 (ξ )
(3.6)
With the aid of Maple, substituting (3.6) into (3.5) yields a rational polynomial equation in tanh(ξ ). Setting the numerator of the rational polynomial equation to be zero, we get a polynomial equation in tanh(ξ ).
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Setting the coefficients of tanh(ξ ) to be zero yields a set of overdetermined algebraic equations with respect to a0 , a1 , b0 , b1 , c1 , µ1 , µ2 , µ3 , λ, l, k. Solving the overdetermined algebraic equations, we get the following results. Case 1. λ a1 = b1 , α = l 2, a0 = 2 , c1 = b1 µ1 + d1 µ3 , µ2 = µ2 , µ3 = µ3 , d1 = 0, 2l b1 µ1 = 0, k= , l = l, c1 = 0. 2β Case 2. a1 = b1 ,
µ1 = 0,
k=
b1 , 2β
µ3 = 0,
l = l,
µ2 = 1,
d1 = 4b1 ,
c1 = 0.
Case 3. µ3 = 0,
l = l, a1 =
3 (b1 + d1 ), 8
d1 = d1 ,
1 , 3
µ2 =
k=
3(b1 + d1 ) , 16β
µ1 = RootOf 3 Z 2 + 1 ,
c1 = b1 RootOf 3 Z 2 + 1 .
Case 4. d1 = d1 ,
l = l,
c1 = µ1 (b1 + d1 ),
b1 + d1 , 2β a1 = b1 + d1 .
µ3 = µ1 ,
k=
µ1 = µ1 ,
µ2 = 0,
Case 5. µ2 = 0,
l = l,
µ1 = −RootOf
d1 =
Z2 + 1 ,
3b1 , 2
b1 , µ3 = RootOf Z 2 + 1 , 4β b1 b1 c1 = RootOf Z 2 + 1 , a1 = . 2 2 k=
Case 6. µ3 = µ3 ,
l = l,
µ1 = µ1 ,
µ2 = −µ1 (−µ3 + µ1 ) ,
d1 = −b1 ,
k = k,
a1 = 2kβ.
Case 7. µ1 = RootOf 3 Z 2 + 1 ,
l = l,
µ3 = RootOf 3 Z 2 + 1 ,
a1 =
µ2 =
2 , 3
d1 = b1 ,
k=
3b1 , 8β
3 b1 . 4
Case 8. µ1 = RootOf 3 Z 2 + 1 ,
l = l, µ3 =
d1 = b1 ,
3b1 , 8β
a1 =
3 b1 , 4
µ2 =
d1 = b1 ,
a1 =
3 b1 , 4
k=
k=
10 , 21
3 RootOf 3 Z 2 + 1 . 7
Case 9. µ3 = µ3 , µ2 =
l = l,
µ1 = RootOf 3 Z 2 + 1 ,
3b1 , 8β
1 − RootOf 3 Z 2 + 1 µ3 . 3
Case 10. b1 = b1 ,
c1 = µ1 b1 − b1 µ3 ,
µ2 = −µ1 2 + µ1 µ3 ,
µ1 = µ1 ,
k = k,
l = l,
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d1 = −b1 , a1 = 2βk,
a2 = 2βk,
λ = l 2 a0 + α a0 .
According to (3.6) and (3.4) and Cases 1–10, we find solutions of Eq. (3.1) as follows: λ lb1 + lb1 tanh (ξ ) + , 2l tanh (ξ ) λ b1 v1 = 2 + b1 tanh (ξ ) + , tanh (ξ ) 2l u1 =
where ξ =
b1 2β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
lb1 4lb1 tanh (ξ ) λ + lb1 tanh (ξ ) + + , 2l tanh (ξ ) 1 + tanh2 (ξ ) λ b1 4b1 tanh (ξ ) v2 = 2 + b1 tanh (ξ ) + , + tanh (ξ ) 1 + tanh2 (ξ ) 2l u2 =
where ξ =
b1 2β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
3lb1 RootOf 3 Z 2 + 1 + d1 tanh (ξ ) λ 3 lb1 + l (b1 + d1 ) tanh (ξ ) + + , 2l 8 RootOf 3 Z 2 +1 + tanh (ξ ) 1 + 3 tanh2 (ξ ) 3b1 RootOf 3 Z 2 + 1 + d1 tanh (ξ ) 3 b1 λ + v3 = 2 + (b1 + d1 ) tanh (ξ ) + , 8 2l RootOf 3 Z 2 +1 + tanh (ξ ) 1 + 3 tanh2 (ξ ) u3 =
where ξ =
b1 +d1 16β
(x + ly + λt), and l, λ, b1 , d1 are arbitrary constants;
lb1 lb1 µ1 + lµ1 d1 + ld1 tanh (ξ ) λ + l (b1 + d1 ) tanh (ξ ) + + , 2l µ1 + tanh (ξ ) µ1 tanh (ξ ) + tanh2 (ξ ) b1 b1 µ1 + µ1 d1 + d1 tanh (ξ ) λ + , v4 = 2 + (b1 + d1 ) tanh (ξ ) + µ1 + tanh (ξ ) 2l µ1 tanh (ξ ) + tanh2 (ξ ) u4 =
where ξ =
b1 +d1 2β
(x + ly + λt), and l, λ, b1 , d1 , µ1 are arbitrary constants;
lb1 RootOf Z 2 + 1 + 3lb1 tanh (ξ ) b1 + , Z 2 + 1 + tanh (ξ ) 2(RootOf ( Z 2 + 1) tanh (ξ ) + tanh2 (ξ )) b1 RootOf Z 2 + 1 + 3b1 tanh (ξ ) b1 b1 λ + v5 = 2 + tanh (ξ ) + , 2 2l −RootOf Z 2 + 1 + tanh (ξ ) 2(RootOf ( Z 2 + 1) tanh (ξ ) + tanh2 (ξ )) λ lb1 u5 = + tanh (ξ ) + 2l 2 −RootOf
where ξ =
b1 4β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
λ lb1 µ1 − lµ3 b1 − lb1 tanh (ξ ) b1l + 2klβ tanh (ξ ) + + , 2l µ1 + tanh (ξ ) µ1 (µ3 − µ1 ) + µ3 tanh (ξ ) + tanh2 (ξ ) λ b1 b1 µ1 − µ3 b1 − b1 tanh (ξ ) v6 = 2 + 2kβ tanh (ξ ) + + , µ1 + tanh (ξ ) µ1 (µ3 − µ1 ) + µ3 tanh (ξ ) + tanh2 (ξ ) 2l u6 =
where ξ = k (x + ly + λt), and k, l, λ, b1 , µ1 , µ3 are arbitrary constants; u7 =
3lb1 lb1 3lb1 (2RootOf (3 Z 2 + 1) + l tanh (ξ )) λ + tanh (ξ ) + + , 2l 4 RootOf 3 Z 2 + 1 + tanh (ξ ) 2 + 3RootOf 3 Z 2 + 1 tanh (ξ ) + 3 tanh2 (ξ )
v7 =
λ 3b1 b1 3b1 (2RootOf (3 Z 2 + 1) + tanh (ξ )) , + tanh (ξ ) + + 2 2 4 2l RootOf 3 Z + 1 + tanh (ξ ) 2 + 3RootOf 3 Z 2 + 1 tanh (ξ ) + 3 tanh2 (ξ )
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
where ξ =
3b1 8β
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(x + ly + λt), and l, λ, b1 are arbitrary constants;
λ 3lb1 lb1 + tanh (ξ ) + 2 2l 4 RootOf 3 Z + 1 + tanh (ξ ) 3l 10 b1 RootOf 3 Z 2 + 1 + b1 tanh (ξ ) + , 10 + 9 RootOf 3 Z 2 + 1 tanh (ξ ) + 21 tanh2 (ξ ) b1 λ 3b1 tanh (ξ ) + v8 = 2 + 2 4 2l RootOf 3 Z + 1 + tanh (ξ ) 3 10 b1 RootOf 3 Z 2 + 1 + b1 tanh (ξ ) , + 10 + 9 RootOf 3 Z 2 + 1 tanh (ξ ) + 21 tanh2 (ξ ) u8 =
where ξ =
3b1 8β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
3lb1 b1 l λ + tanh (ξ ) + 2 2l 4 RootOf 3 Z + 1 + tanh (ξ ) 3b1l(RootOf (3 Z 2 +1) + µ3 + tanh (ξ )) + , 1 − 3RootOf 3 Z 2 + 1 µ3 + 3µ3 tanh (ξ ) + 3 tanh2 (ξ ) 3b1 b1 λ tanh (ξ ) + v9 = 2 + 2 4 2l RootOf 3 Z + 1 + tanh (ξ ) 3b1 (RootOf (3 Z 2 +1) + µ3 + tanh (ξ )) + , 1 − 3RootOf 3 Z 2 + 1 µ3 + 3µ3 tanh (ξ ) + 3 tanh2 (ξ ) u9 =
where ξ =
3b1 8β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
2lβk lb1 l(µ1 b1 − b1 µ3 − b1 tanh (ξ )) + + 2 tanh (ξ ) µ1 + tanh (ξ ) −µ1 + µ1 µ3 + µ3 tanh (ξ ) + tanh2 (ξ ) βk b1 µ1 b1 − b1 µ3 − b1 tanh (ξ ) v10 = a0 + 2βk tanh (ξ ) + 2 + + 2 tanh (ξ ) µ1 + tanh (ξ ) −µ1 + µ1 µ3 + µ3 tanh (ξ ) + tanh2 (ξ ) where ξ = k x + ly + l 2 a0 + α a0 t , k, l, a0 , µ1 , µ3 , b1 are arbitrary constants. u 10 = la0 + 2l βk tanh (ξ ) +
3.2. New solutions of the (2 + 1)-dimensional nonlinear Davey–Stewartson system On the basis of the idea of the uniform direct ans¨atze method, we use the Riccati equation as the auxiliary equation to find new formal solutions of the coupled (2 + 1)-dimensional nonlinear system of Davey–Stewartson equations. The Riccati equation reads φ0 = h + φ2,
(3.7)
φ 0 = a + bφ + cφ 2 .
(3.8)
or
Under the transformation φ +
b 2c
= 1c Φ(c 6= 0), (3.8) becomes
4ac − b2 + Φ2, 4 which is formally the same as (3.7). It is apparent that (3.7) is simpler than (3.8). So it is more convenient to use Eq. (3.7) as an auxiliary equation. Then we only need to construct solutions of (3.7). As we know, Eq. (3.7) has the following solutions: √ √ −√−h tanh(√−hξ ), (I) φ = when h < 0, (3.9) − −h coth( −hξ ), Φ0 =
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(√
√ h tan( hξ ), √ √ (II) φ = − h cot( hξ ), (III) φ = −
1 , ξ +c
when h > 0,
(3.10)
when h = 0.
(3.11)
Now we use Eq. (3.7) as an auxiliary equation to construct new solutions of the coupled (2+1)-dimensional nonlinear system of Davey–Stewartson equations (DSE) [1,27,28], iu t + pu x x + u yy + r |u|2 u − 2uv = 0,
(3.12a)
vx x − pv yy − r (|u|2 )x x = 0,
(3.12b)
where the u(x, y, t) and v(x, y, t) are complex valued functions and p = ±1, r is constant. Eq. (3.12) where p = 1 and p = −1 is called the DS I equation and the DS II equation, respectively. We find it more convenient to rewrite the corresponding system in a real form by separating v = e + f i,
u = a + bi,
(3.13)
into real and imaginary parts. Then we obtain a system of four real partial differential equations: at + pbx x + b yy + r (a 2 b + b3 ) − 2(a f + be) = 0,
(3.14a)
−bt + pax x + a yy + r (a 3 + ab2 ) + 2(b f − ae) = 0,
(3.14b)
ex x − pe yy − 2r (a + b )x x = 0,
(3.14c)
f x x − p f yy = 0.
(3.14d)
2
2
The travelling wave transformation a(x, y, t) = a(ξ ),
b(x, y, t) = b(ξ ),
e(x, y, t) = e(ξ ),
f (x, y, t) = f (ξ ),
ξ = kx + ly + λt,
(3.15)
transforms Eq. (3.12) into λa 0 + ( pk 2 + l 2 )b00 + r (a 2 b + b3 ) − 2(a f + be) = 0,
(3.16a)
−λb + ( pk + l )a + r (a + ab ) + 2(b f − ae) = 0,
(3.16b)
(k 2 − pl 2 )e00 − 2r k 2 (a 2 + b2 )00 = 0,
(3.16c)
(k − pl ) f = 0.
(3.16d)
2
0
2
2
2
00
3
2
00
From (3.16c) and (3.16d) we may assume e=
2r k 2 (a 2 + b2 ) + α1 = M(a 2 + b2 ) + α1 , k 2 − pl 2
Throughout this paper we write
2r k 2 k 2 − pl 2
f = α2 .
(3.17)
= M. Substituting (3.17) into (3.16a) and (3.16b), we obtain
λa 0 + ( pk 2 + l 2 )b00 + r (a 2 b + b3 ) − 2aα2 − 2b(M(a 2 + b2 ) + α1 ) = 0,
(3.18a)
−λb + ( pk + l )a + 2bα2 + r (ab + a ) − 2a(M(a + b ) + α1 ) = 0.
(3.18b)
0
2
2
00
2
3
2
2
Then we assume Eq. (3.18) to have the following formal solutions: a = β0 + β1 φ +
β2 , µ+φ
b = δ0 + δ1 φ +
δ2 , µ+φ
(3.19)
with φ solving (3.7). With the aid of Maple, substituting (3.19) into Eqs. (3.18) yields two rational polynomial equations in φ. Setting the numerators of the two rational polynomial equations to be zero, we get two polynomial
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
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equations in φ. Setting the coefficients of φ to be zero yields a set of overdetermined algebraic equations with respect to β0 , β1 , β2 , δ0 , δ1 , δ2 , µ, k, l, λ, α1 , α2 . Solving the overdetermined algebraic equations, we get the following results (there are other results; we only list some of them for simplicity). Case 1. β1 = β2 = δ1 = 0, λ=
β0 δ2 (2M − r ) , µ2 + h
β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) , α1 = − 2(µ2 + h) β0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 α2 = , µ2 + h
where β0 , δ2 , δ0 , k, µ, l are arbitrary constants. Case 2. β0 2 h + β2 2 (β2 β0 + δ0 δ2 ) (2M − r ) , α2 = h (β0 δ2 − β2 δ0 )
µ = β1 = δ1 = 0, √ s 2 2Mδ2 2 + 2Mβ2 2 − 2h 2l 2 − r δ2 2 − r β2 2 , k= 2h p
β0 δ2 2 + 2β2 δ0 δ2 + 3β0 β2 2 (2M − r ) λ= , hδ2 3hδ0 β0 2 β2 + hβ0 δ0 2 δ2 − hβ0 3 δ2 + β0 δ2 3 + β2 δ0 δ2 2 + 3β0 β2 2 δ2 + hδ0 3 β2 − δ0 β2 3 (2M − r ) α1 = , 2h (β0 δ2 − β2 δ0 ) where β0 , δ2 , δ0 , β2 , l are arbitrary constants. Case 3. β0 2 h (2M − r ) , α1 = 2µ2
β1 = δ0 = δ1 = δ2 = 0, √ s 2 2β0 2 M − 2µ2l 2 − r β0 2 k=− , 2µ p
β0 µ2 + h β2 = − , µ
where β0 , µ, l, λ, α2 are arbitrary constants. Case 4. 2δ0 2 µ2 h + δ0 2 µ4 − hβ2 2 + δ0 2 h 2 (2M − r ) β2 µ , β0 = − 2 , α1 = − 2 µ +h 2 µ2 + h s √ 2 −2µ4l 2 − 4 µ2 hl 2 + 2Mβ2 2 − 2h 2l 2 − r β2 2 δ2 = α2 = β1 = δ1 = 0, k=− , p 2 µ2 + h where β2 , µ, l, λ, δ0 are arbitrary constants. Case 5. s δ2 = β0 = α2 = β2 = δ1 = 0,
β1 = − −
2(l 2 + k 2 p) , r − 2M
α1 =
δ0 2r + l 2 h + pk 2 h − δ0 2 M, 2
where µ, l, k, λ, δ0 are arbitrary constants. Case 6. µδ2 δ0 = − 2 , µ +h β0 = 0,
s α2 = −δ2 l 2 + k 2 p
2M − r , 2(l 2 + k 2 p)
s δ2 l 2 + k 2 p µ 2(2M − r ) λ= , µ2 + h l2 + k2 p
s β1 = −
2(l 2 + k 2 p) , 2M − r
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α1 =
r µ2 δ2 2 − 2µ2 δ2 2 M + 2µ4 pk 2 h + 2l 2 h 3 + 4 µ2 pk 2 h 2 + 2 pk 2 h 3 + 4 µ2l 2 h 2 + 2µ4l 2 h , 2 2 µ2 + h
where µ, l, k, δ2 are arbitrary constants. Case 7. β1 = µ = β2 = δ2 = 0, α2 = −δ0 β0 (r − 2M) , 1 2 α1 = − λ = δ1 β0 (r − 2M) , δ1 h + δ0 2 − β0 2 (r − 2M) , 2 where l, k, β0 , δ0 , δ1 are arbitrary constants. Case 8. β1 = µ = β2 = 0, λ=−
δ2 δ1 = − , h −4 δ2 2 + hβ0 2 − hδ0 2 (r − 2M) α1 = , 2h
α2 = −δ0 β0 (r − 2M) ,
β0 δ2 (r − 2M) , h
where l, k, β0 , δ0 , δ2 are arbitrary constants. Case 9. β1 = δ1 = µ = β2 = 0,
α2 = β0 δ0 (2M − r ) ,
λ=
β0 δ2 (2M − r ) , h
hβ0 2 − hδ0 2 − δ2 2 (2M − r ) , α1 = − 2h where l, k, β0 , δ0 , δ2 are arbitrary constants. Case 10. r r r √ 6 pk 2 + l 2 hN hN N λ= , β0 = − , β1 = − , δ0 = −δ1 −h, δ1 r − 2M r − 2M r − 2M r 8h p 2 k 4 + 2l 2 pk 2 + l 4 4h l 2r − 2 pk 2 M + pk 2r − 2l 2 M δ1 2 α2 (r − 2M) δ1 N − + , α1 = N N 2M − r N where N = δ1 2r − 2δ1 2 M + 2l 2 + 2 pk 2 , and l, k, α2 , δ1 are arbitrary constants. According to (3.13), (3.17) and (3.19), Case 1 and the solutions (3.9)–(3.11) of (3.7), we get the solutions of Eq. (3.12) : when h < 0, the solitary solutions δ2 u 11 = β0 + i δ0 + , √ √ µ − −h tanh( −hξ ) " 2 # δ2 2 v11 = M β0 + δ0 + √ √ µ − −h tanh( −hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 + ; µ2 + h δ2 , u 12 = β0 + i δ0 + √ √ µ − −h coth( −hξ )
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
" v12 = M
β02
+ δ0 +
√
δ2
2757
2 #
√ µ − −h coth( −hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 + ; µ2 + h
when h > 0, the periodic solutions δ2 u 13 = β0 + i δ0 + , √ √ µ + h tan( hξ ) " 2 # δ2 2 v13 = M β0 + δ0 + √ √ µ + h tan( hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 , + µ2 + h δ2 u 14 = β0 + i δ0 + , √ √ µ − h cot( −hξ ) " 2 # δ2 2 v14 = M β0 + δ0 + √ √ µ − h cot( hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 + ; µ2 + h when h = 0, the rational solution δ2 (ξ + c) = β0 + i δ0 + , µ(ξ + c) + 1 " 2 # β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 (2M − r ) δ2 (ξ + c) 2 − = M β0 + δ0 + µ(ξ + c) + 1 2µ2 iβ0 (2µMδ0 − µr δ0 − r δ2 + 2 Mδ2 ) + , µ
u 15 v15
)t where ξ = β0 δ2µ(2M−r + kx +ly, and β0 , δ2 , δ0 , k, µ, l, c are arbitrary constants. In a similar way, from (3.13), (3.17) 2 +h and (3.19) and Cases 2–10, we have other solutions of Eqs. (3.12) as follows:
δ2 β2 u 2 = β0 + + i δ0 + , φ (ξ ) φ (ξ ) " # β2 2 δ2 2 v2 = M β0 + + δ0 + φ (ξ ) φ (ξ ) 3hδ0 β0 2 β2 + hβ0 δ0 2 δ2 − hβ0 3 δ2 + β0 δ2 3 + β2 δ0 δ2 2 + 3β0 β2 2 δ2 + hδ0 3 β2 − δ0 β2 3 (2M − r ) 2h (β0 δ2 − β2 δ0 ) i β0 2 h + β2 2 (β2 β0 + δ0 δ2 ) (2M − r ) + , h (β0 δ2 − β2 δ0 )
+
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H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
where √ s β0 δ2 2 + 2β2 δ0 δ2 + 3β0 β2 2 (2M − r ) t 2 2Mδ2 2 + 2Mβ2 2 − 2h 2l 2 − r δ2 2 − r β2 2 ξ= + ly, x+ 2h p hδ2 and β0 , δ2 , δ0 , β2 , l are arbitrary constants and φ solves (3.7); β0 µ2 + h u 3 = β0 − , µ (µ + φ (ξ )) !2 β0 µ2 + h β0 2 h (2M − r ) v3 = M β0 − + α2 i, + µ (µ + φ (ξ )) 2µ2 where √ s 2 2β0 2 M − 2µ2l 2 − r β0 2 ξ = λt − x + ly, 2µ p and β0 , µ, l, λ, α2 are arbitrary constants and φ solves (3.7); β2 β2 µ + + iδ0 , 2 µ + φ (ξ ) µ +h # " 2 2δ0 2 µ2 h + δ0 2 µ4 − hβ2 2 + δ0 2 h 2 (2M − r ) µ 1 2 2 + δ0 − + v4 = M β2 , 2 µ + φ (ξ ) µ2 + h 2 µ2 + h u4 = −
where √ 2 ξ = λt − 2 2 µ +h
s
−2µ4l 2 − 4 µ2 hl 2 + 2Mβ2 2 − 2h 2l 2 − r β2 2 x + ly, p
and β2 , µ, l, λ, δ0 are arbitrary constants and φ solves (3.7); s 2l 2 + 2k 2 p φ (ξ ) + iδ0 , u5 = − − r − 2M 2l 2 + 2k 2 p 2 δ0 2r 2 v5 = M − φ (ξ ) + δ0 + + l 2 h + pk 2 h − δ0 2 M, r − 2M 2 where ξ = λ t + kx + ly and µ, l, k, λ, δ0 are arbitrary constants and φ solves (3.7); s 2(l 2 + k 2 p) µδ2 δ2 u6 = − φ (ξ ) + i − 2 + , 2M − r µ + φ (ξ ) µ +h s " 2 # 2(l 2 + k 2 p) 2 µδ2 2M − r δ2 2 2 v6 = M φ (ξ ) + − 2 − iδ2 l + k p + 2M − r µ + φ (ξ ) µ +h 2(l 2 + k 2 p) +
r µ2 δ2 2 − 2µ2 δ2 2 M + 2µ4 pk 2 h + 2l 2 h 3 + 4 µ2 pk 2 h 2 + 2 pk 2 h 3 + 4 µ2l 2 h 2 + 2µ4l 2 h , 2 2 µ2 + h
where s δ2 l 2 + k 2 p µ 2(2M − r ) ξ= t + kx + ly, µ2 + h l2 + k2 p and µ, l, k, δ2 are arbitrary constants and φ solves (3.7); u 7 = β0 + i (δ0 + δ1 φ (ξ )) ,
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
2759
h i 1 v7 = M β02 + (δ0 + δ1 φ (ξ ))2 − δ1 2 h + δ0 2 − β0 2 (r − 2M) − iδ0 β0 (r − 2M) , 2 where ξ = (δ1rβ0 − 2δ1 β0 M) t + kx + ly, and l, k, β0 , δ0 , δ1 are arbitrary constants and φ solves (3.7); δ2 φ (ξ ) δ2 u 8 = β0 + i δ0 − + , h φ (ξ ) " # −4 δ2 2 + hβ0 2 − hδ0 2 (r − 2M) δ2 2 δ2 φ (ξ ) 2 + − iδ0 β0 (r − 2M) , + v8 = M β0 + δ0 − h φ (ξ ) 2h where β0 δ2 (r − 2M) t + kx + ly, h and l, k, β0 , δ0 , δ2 are arbitrary constants and φ solves (3.7); δ2 , u 9 = β0 + i δ0 + φ (ξ ) " # hβ0 2 − hδ0 2 − δ2 2 (2M − r ) δ2 2 2 − + iβ0 δ0 (2M − r ) , v9 = M β0 + δ0 + φ (ξ ) 2h ξ =−
where β0 δ2 (2M − r ) t + kx + ly, h and l, k, β0 , δ0 , δ2 are arbitrary constants and φ solves (3.7); ! r √ hN + iδ1 φ (ξ ) − −h , u 10 = 2M − 2r √ 2 N v10 = M φ (ξ ) − −h + δ12 + iα2 2M − r r 8h p 2 k 4 + 2l 2 pk 2 + l 4 4h l 2r − 2 pk 2 M + pk 2r − 2l 2 M δ1 2 α2 (r − 2M) δ1 N − + , + N N 2M − r N ξ=
where 6 pk 2 + l 2 ξ= δ1
r
hN t + kx + ly, r − 2M
N = δ1 2r − 2δ1 2 M + 2l 2 + 2 pk 2 , and l, k, α2 , δ1 are arbitrary constants and φ solves (3.7). Remark: In [27,28], the authors introduced u = u(ξ )eiη , v = v(ξ ), ξ = k(x + py + λt), η = αx + βy + γ t, to seek the explicit solutions of (3.12), and assumed u(ξ ), v(ξ ) were real functions. So they could not obtain solutions like v1 ∼ v3 , v6 ∼ v7 , where the vi are complex functions. 4. Conclusion In this paper, we present a further uniform direct ans¨atze method for finding exact solutions of nonlinear evolution equations and give two examples to illustrate the method. As a result, we find abundant solutions of the two equations. And we construct new formal solutions of the Riccati equation. On the basis of the idea of the direct ans¨atze method, we may construct other solutions. In addition, this method is also computerizable, which allows us to perform complicated and tedious algebraic calculation on a computer.
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Acknowledgements We wish to express our gratitude to the reviewer for his/her valuable comments. The work was partially supported by the National Key Basic Research Project of China (Grant No. 2004CB318000). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28]
M.J. Ablowitz, P.A. Clarkson, Solitons, Nonlinear Evolution Equations and Inverse Scattering, Cambridge University Press, Cambridge, 1991. M. Wadati, H. Sanuki, K. Konno, Progr. Theoret. Phys. 53 (1975) 419. K. Konno, M. Wadati, Progr. Theoret. Phys. 53 (1975) 1652. C.H. Gu, et al., Soliton Theory and its Application, Springer, Berlin, 1995. R. Hirota, Phys. Rev. Lett. 27 (1971) 1192. P.J. Olver, Applications of Lie Groups to Differential Equations, Springer, Berlin, 1986. G. Bluman, S. Kumei, Symmetries and Differential Equations, Springer, New York, 1989. P.A. Clarkson, M.D. Kruskal, J. Math. Phys. 30 (1989) 2201. S.Y. Lou, X.Y. Tang, J. Lin, J. Math. Phys. 41 (2000) 8286. Z.B. Li, M.L. Wang, J. Phys. A 26 (1993) 6027. M.L. Wang, Phys. Lett. A 199 (1995) 169. W. Hereman, Comput. Phys. Comm. 65 (1991) 143. E.J. Parkes, B.R. Duffy, Comput. Phys. Comm. 98 (1996) 288. R. Conte, M. Musette, J. Phys. A 25 (1992) 5609. T.C. Bountis, V. Papageorgiou, P. Winternitz, J. Math. Phys. 27 (1986) 1215. C. Yan, Phys. Lett. A 224 (1996) 77. S.K. Liu, Z.T. Fu, S.D. Liu, Q. Zhao, Phys. Lett. A 289 (2001) 69. Z.T. Fu, S.K. Liu, S.D. Liu, Q. Zhao, Phys. Lett. A 290 (2001) 72. E. Fan, Phys. Lett. A 277 (2000) 212; E. Fan, Phys. Lett. A 294 (2002) 26. Z.Y. Yan, Phys. Lett. A 292 (2001) 100. B. Li, Y. Chen, H.Q. Zhang, Z. Natur. A 57 (11) (2002) 874. Y. Chen, B. Li, H.Q. Zhang, Int. J. Mod. Phys. C 13 (2003) 99. Q. Wang, Y. Chen, B. Li, H.Q. Zhang, Appl. Math. Comput. 160 (2005) 77. E.G. Fan, J. Phys. A 36 (2003) 7009. H.Y. Zhi, X.Q. Zhao, H.Q. Zhang, Chinese Physics 14 (2005) 1296–1302. K.Z. Hong, B. Wu, X.F. Chen, Commun. Theor. Phys. (Beijing) 39 (2003) 393. E.G. Fan, J. Phys. A 35 (2002) 6853. Z.Y. Yan, Commun. Theor. Phys. (Beijing) 42 (2004) 645.
Short communication
Symbolic computation and a uniform direct ans¨atze method for constructing travelling wave solutions of nonlinear evolution equations Zhi Hongyan a,∗ , Zhang Hongqing b a College of Mathematics and Computational Science, China University of Petroleum, Dongying 257061, Shandong Province, China b Department of Applied Mathematics, Dalian University of Technology, Dalian 116024, China
Received 7 December 2006; accepted 26 April 2007
Abstract On the basis of the computer symbolic system Maple and the tanh method, the Riccati equation method as well as all kinds of improved versions of these methods, we present a further uniform direct ans¨atze method for constructing travelling wave solutions of nonlinear evolution equations. Compared with the existing methods, the presented method can be used to construct more new general solutions. And we give some examples to illustrate the key step of our method. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Travelling wave solution; Direct ans¨atze method; Nonlinear evolution equation
1. Introduction The investigation of exact travelling wave solutions to nonlinear evolution equations plays an important role in the study of nonlinear physical phenomena. Recently, much effort has been spent on the construction of exact solutions of nonlinear evolution equations and many powerful methods have been developed such as the inverse scattering method [1], B¨acklund and Darboux transformation methods [2–4], Hirota bilinear method [5], Lie group reduction method [6–9], homogeneous balance method [10,11], etc. With the development of symbolic computation, the tanh function method, Riccati method [12,13], projective Riccati equation method [14,15], sine–cosine function method [16], Jacobi elliptic function method [17,18] and all kinds of auxiliary equation methods attract more and more researchers. Many authors have improved these methods [19–23]. In this paper, we give a further uniform direct ans¨atze method to construct the exact solutions of nonlinear evolution equations. 2. Introduction of the new uniform direct ans¨atze method Consider a system of N nonlinear evolution equations (N ≥ 1) with n independent variables x = (x1 , x2 , . . . , xn ) and m dependent variables u = (u 1 , u 2 , . . . , u m ), given by ∗ Corresponding author.
E-mail address: [email protected] (H. Zhi). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.04.041
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
Aµ (u, ∂u, ∂ 2 u, . . . , ∂ r u) = 0,
µ = 1, 2, . . . , N
2749
(2.1)
where ∂ j u denotes the coordinates with components ∂ j u/∂ xi1 ∂ xi2 · · · ∂ xi j = u i1 i2 ···i j , i j = 1, 2, . . . , n, for j = 1, 2, . . . , k, corresponding to all jth-order partial derivatives of u with respect to x. Because of the importance of travelling wave solutions, throughout this paper, we seek the nonlinear evolution equations’ travelling wave solutions. Then the travelling wave transformation ! n X ξ = λ1 x 1 + λ2 x 2 + · · · + λn x n , λi2 6= 0 (2.2) i=1
transforms (2.1) into nonlinear ODEs: A¯ µ (u, u 0 , u 00 , . . . , u (k) ) = 0,
µ = 1, 2, . . . , N .
(2.3)
Here we give a uniform direct ans¨atze, u i = Fi (θ1 , θ2 , . . . , θl ) +
Pi (θ1 , θ2 , . . . , θl ) , Q i (θ1 , θ2 , . . . , θl )
(i = 1, 2, . . . , m)
(2.4)
where θi (i = 1, 2, . . . , l) are functionally independent functions of ξ , their derivative is a rational polynomial function of θ1 , θ2 , . . . , θl . Fi , Pi , Q i (i = 1, 2, . . . , m) are polynomials of the corresponding variables. deg(Pi ) < deg(Q i ) and gcd(Pi , Q i ) = 1 (i = 1, 2, . . . , m). The leading coefficient of the Q i is equal to 1. If θiJi = Ri (θ1 , θ1 , . . . , θi−1 ) (i = 1, 2, . . . , m), then the highest power of θi in Fi , Pi , Q i is Ji − 1. Substituting (2.4) into (2.3) and simplifying, we obtain X µ ai1 ,i2 ,...,il θ1i1 θ2i2 · · · θlil = 0, µ = 1, 2, . . . , m. (2.5) 0≤i 1 ,i 2 ,...,il
During the simplifying process, we stick to a principle: if θiJi = Ri (θ1 , θ2 , . . . , θi−1 ), then the highest power of θi in (2.5) is Ji − 1. Because of the functional independence of θ1i1 θ2i2 · · · θlil , we obtain the algebraic equations µ ai1 ,i2 ,...,in = 0 with respect to λ j and the coefficients of Fi , Pi , Q i (i = 1, 2, . . . , m, j = 1, 2, . . . , n). Solving the equations, we can determine these explicit values. Using also (2.2) and (2.4), we can obtain the solutions of Eq. (2.1). Remark: (i) Let deg(Fi ) = ki1 , deg(Pi ) = ki2 , deg(Q i ) = ki3 . n i (i = 1, 2, . . . , m) are obtained by the homogeneous balance principle [10,11]. Generally speaking, ki1 ≤ n i , ki2 ≤ n i . (ii) We give some examples to illustrate how to choose appropriate θ1 , θ2 , . . . , θl . When l = 1, we can set θ1 = tanh(ξ ). Similarly, we can set θ1 = sech(ξ ), coth(ξ ), csch(ξ ), tan(ξ ), sec(ξ ), cot(ξ ), csc(ξ ) . . .. Of course, we can set θ1 as a solution of φ 0 = δ + φ 2 [12,13] or φ 0 2 = c0 + c1 φ + c2 φ 2 + c3 φ 3 + c4 φ 4 [24]. When l = 2, we can set θ1 = tanh(ξ ), θ2 = sech(ξ ). Because θ22 = 1 − θ12 , the highest power of θ2 in Fi , Pi , Q i is equal to 1. Similarly we can set θ1 = tan(ξ ), θ2 = sec(ξ ) or θ1 = coth(ξ ), θ2 = csch(ξ ), . . . . And we can set θ1 , θ2 as the solutions of the projective Riccati equations σ 0 (ξ ) = σ (ξ )τ (ξ ), where “0 ” =
d dξ ,
τ 0 (ξ ) = R + τ 2 (ξ ) − µσ (ξ ),
(2.6)
= ±1, µ and R are undetermined constants [14,15]. That is to say θ1 = σ, θ2 = τ . Since
µ2 − 1 2 σ (ξ ) , τ 2 (ξ ) = − R − 2µσ (ξ ) + R
(2.7)
the highest power of θ2 in Fi , Pi , Q i is equal to 1. When l = 3, we can set θ1 = sn(ξ, k), θ2 = cn(ξ, k), θ3 = dn(ξ, k), where the k is the module of the Jacobi elliptic function. Since θ22 = 1 − θ12 , θ32 = 1 − k 2 θ12 , the highest power of θ2 , θ3 in Fi , Pi , Q i is equal to 1. We can also set θ1 = ns(ξ, k), θ2 = nc(ξ, k), θ3 = dn(ξ, k), . . .. In fact, θ1 , θ2 , θ3 can also be solutions of ordinary differential
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equations, for example, θ10 = δ[kθ2 (1 − 2Bθ2 − 2Cθ3 − 2 Aθ3 ) + Bθ12 ], θ20 = δθ1 (1 − Bθ2 − 2Cθ3 ), θ30 = δθ1 (2Aθ2 + Bθ3 ), Aθ12 Aθ22
(2.8)
= kθ3 (1 − Bθ2 − Aθ3 − Cθ3 ), = θ3 (1 − Bθ2 − Cθ3 ),
d where A 6= 0, B, C are arbitrary constants, k = ±1, δ = ±1, = ±1, “0 ” = dξ [25]. Up to now, we have not considered the situation l > 3. This is a problem worth considering. (iii) On the basis of the theory that any proper fraction can be presented as the sum of simplest terms, we can always express u as a polynomial and some simplest fractions. For example, when l = 1, θ1 = tanh(ξ ), one can set ) ( ni X bi j (di j + ei j tanh(ξ )) j ci j j u i = ai0 + ai j tanh (ξ ) + + ··· + + tanh j (ξ ) (µi0 + tanh(ξ )) j (µi1 + µi2 tanh(ξ ) + tanh2 (ξ )) j j=1
(i = 1, 2, . . . , m), where the n i are no larger than the balance constants, ai j , bi j , ci j , di j , ei j , µi0 , µi1 , µi2 are constants to be determined, and 1 ≤ j ≤ n i is a positive integer. Generally, we can only choose parts of these terms in order to obtain more new solutions which have not been presented otherwise. In the following, we present some examples to illustrate the above mentioned method. 3. Examples 3.1. Exact solutions of the (2 + 1)-dimensional Burgers equation The (2 + 1)-dimensional Burgers equation reads −u t + uu y + αvu x + βu yy + αβu x x = 0, u x − v y = 0.
(3.1)
An equivalent form of the Burgers equation (3.1) is derived from the generalized Painl´eve integrability classification in Ref. [26]. On introducing the travelling wave transformation u(x, y, t) = u(ξ ),
v(x, y, t) = v(ξ ),
ξ = k(x + ly + λt),
(3.2)
Eq. (3.1) takes the form −λu 0 + luu 0 + αvu 0 + βkl 2 u 00 + αβku 00 = 0, 0
0
u − lv = 0.
(3.3a) (3.3b)
From (3.3b), we have (3.4)
u = lv. Here we set the integral constant to be zero. Substituting (3.4) into (3.3a) we get −λv 0 + l 2 vv 0 + αvv 0 + βkl 2 v 00 + αβkv 00 = 0.
(3.5)
If we have solved Eq. (3.5), together with (3.4), we can obtain the solutions of Eq. (3.1). Now we turn to solving Eq. (3.5). Balancing the term v 00 with the term vv 0 in Eq. (3.5) gives n = 1. Then we can assume v = a0 + a1 tanh (ξ ) +
b1 c1 + d1 tanh (ξ ) + . µ1 + tanh (ξ ) µ2 + µ3 tanh (ξ ) + tanh2 (ξ )
(3.6)
With the aid of Maple, substituting (3.6) into (3.5) yields a rational polynomial equation in tanh(ξ ). Setting the numerator of the rational polynomial equation to be zero, we get a polynomial equation in tanh(ξ ).
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Setting the coefficients of tanh(ξ ) to be zero yields a set of overdetermined algebraic equations with respect to a0 , a1 , b0 , b1 , c1 , µ1 , µ2 , µ3 , λ, l, k. Solving the overdetermined algebraic equations, we get the following results. Case 1. λ a1 = b1 , α = l 2, a0 = 2 , c1 = b1 µ1 + d1 µ3 , µ2 = µ2 , µ3 = µ3 , d1 = 0, 2l b1 µ1 = 0, k= , l = l, c1 = 0. 2β Case 2. a1 = b1 ,
µ1 = 0,
k=
b1 , 2β
µ3 = 0,
l = l,
µ2 = 1,
d1 = 4b1 ,
c1 = 0.
Case 3. µ3 = 0,
l = l, a1 =
3 (b1 + d1 ), 8
d1 = d1 ,
1 , 3
µ2 =
k=
3(b1 + d1 ) , 16β
µ1 = RootOf 3 Z 2 + 1 ,
c1 = b1 RootOf 3 Z 2 + 1 .
Case 4. d1 = d1 ,
l = l,
c1 = µ1 (b1 + d1 ),
b1 + d1 , 2β a1 = b1 + d1 .
µ3 = µ1 ,
k=
µ1 = µ1 ,
µ2 = 0,
Case 5. µ2 = 0,
l = l,
µ1 = −RootOf
d1 =
Z2 + 1 ,
3b1 , 2
b1 , µ3 = RootOf Z 2 + 1 , 4β b1 b1 c1 = RootOf Z 2 + 1 , a1 = . 2 2 k=
Case 6. µ3 = µ3 ,
l = l,
µ1 = µ1 ,
µ2 = −µ1 (−µ3 + µ1 ) ,
d1 = −b1 ,
k = k,
a1 = 2kβ.
Case 7. µ1 = RootOf 3 Z 2 + 1 ,
l = l,
µ3 = RootOf 3 Z 2 + 1 ,
a1 =
µ2 =
2 , 3
d1 = b1 ,
k=
3b1 , 8β
3 b1 . 4
Case 8. µ1 = RootOf 3 Z 2 + 1 ,
l = l, µ3 =
d1 = b1 ,
3b1 , 8β
a1 =
3 b1 , 4
µ2 =
d1 = b1 ,
a1 =
3 b1 , 4
k=
k=
10 , 21
3 RootOf 3 Z 2 + 1 . 7
Case 9. µ3 = µ3 , µ2 =
l = l,
µ1 = RootOf 3 Z 2 + 1 ,
3b1 , 8β
1 − RootOf 3 Z 2 + 1 µ3 . 3
Case 10. b1 = b1 ,
c1 = µ1 b1 − b1 µ3 ,
µ2 = −µ1 2 + µ1 µ3 ,
µ1 = µ1 ,
k = k,
l = l,
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d1 = −b1 , a1 = 2βk,
a2 = 2βk,
λ = l 2 a0 + α a0 .
According to (3.6) and (3.4) and Cases 1–10, we find solutions of Eq. (3.1) as follows: λ lb1 + lb1 tanh (ξ ) + , 2l tanh (ξ ) λ b1 v1 = 2 + b1 tanh (ξ ) + , tanh (ξ ) 2l u1 =
where ξ =
b1 2β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
lb1 4lb1 tanh (ξ ) λ + lb1 tanh (ξ ) + + , 2l tanh (ξ ) 1 + tanh2 (ξ ) λ b1 4b1 tanh (ξ ) v2 = 2 + b1 tanh (ξ ) + , + tanh (ξ ) 1 + tanh2 (ξ ) 2l u2 =
where ξ =
b1 2β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
3lb1 RootOf 3 Z 2 + 1 + d1 tanh (ξ ) λ 3 lb1 + l (b1 + d1 ) tanh (ξ ) + + , 2l 8 RootOf 3 Z 2 +1 + tanh (ξ ) 1 + 3 tanh2 (ξ ) 3b1 RootOf 3 Z 2 + 1 + d1 tanh (ξ ) 3 b1 λ + v3 = 2 + (b1 + d1 ) tanh (ξ ) + , 8 2l RootOf 3 Z 2 +1 + tanh (ξ ) 1 + 3 tanh2 (ξ ) u3 =
where ξ =
b1 +d1 16β
(x + ly + λt), and l, λ, b1 , d1 are arbitrary constants;
lb1 lb1 µ1 + lµ1 d1 + ld1 tanh (ξ ) λ + l (b1 + d1 ) tanh (ξ ) + + , 2l µ1 + tanh (ξ ) µ1 tanh (ξ ) + tanh2 (ξ ) b1 b1 µ1 + µ1 d1 + d1 tanh (ξ ) λ + , v4 = 2 + (b1 + d1 ) tanh (ξ ) + µ1 + tanh (ξ ) 2l µ1 tanh (ξ ) + tanh2 (ξ ) u4 =
where ξ =
b1 +d1 2β
(x + ly + λt), and l, λ, b1 , d1 , µ1 are arbitrary constants;
lb1 RootOf Z 2 + 1 + 3lb1 tanh (ξ ) b1 + , Z 2 + 1 + tanh (ξ ) 2(RootOf ( Z 2 + 1) tanh (ξ ) + tanh2 (ξ )) b1 RootOf Z 2 + 1 + 3b1 tanh (ξ ) b1 b1 λ + v5 = 2 + tanh (ξ ) + , 2 2l −RootOf Z 2 + 1 + tanh (ξ ) 2(RootOf ( Z 2 + 1) tanh (ξ ) + tanh2 (ξ )) λ lb1 u5 = + tanh (ξ ) + 2l 2 −RootOf
where ξ =
b1 4β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
λ lb1 µ1 − lµ3 b1 − lb1 tanh (ξ ) b1l + 2klβ tanh (ξ ) + + , 2l µ1 + tanh (ξ ) µ1 (µ3 − µ1 ) + µ3 tanh (ξ ) + tanh2 (ξ ) λ b1 b1 µ1 − µ3 b1 − b1 tanh (ξ ) v6 = 2 + 2kβ tanh (ξ ) + + , µ1 + tanh (ξ ) µ1 (µ3 − µ1 ) + µ3 tanh (ξ ) + tanh2 (ξ ) 2l u6 =
where ξ = k (x + ly + λt), and k, l, λ, b1 , µ1 , µ3 are arbitrary constants; u7 =
3lb1 lb1 3lb1 (2RootOf (3 Z 2 + 1) + l tanh (ξ )) λ + tanh (ξ ) + + , 2l 4 RootOf 3 Z 2 + 1 + tanh (ξ ) 2 + 3RootOf 3 Z 2 + 1 tanh (ξ ) + 3 tanh2 (ξ )
v7 =
λ 3b1 b1 3b1 (2RootOf (3 Z 2 + 1) + tanh (ξ )) , + tanh (ξ ) + + 2 2 4 2l RootOf 3 Z + 1 + tanh (ξ ) 2 + 3RootOf 3 Z 2 + 1 tanh (ξ ) + 3 tanh2 (ξ )
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
where ξ =
3b1 8β
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(x + ly + λt), and l, λ, b1 are arbitrary constants;
λ 3lb1 lb1 + tanh (ξ ) + 2 2l 4 RootOf 3 Z + 1 + tanh (ξ ) 3l 10 b1 RootOf 3 Z 2 + 1 + b1 tanh (ξ ) + , 10 + 9 RootOf 3 Z 2 + 1 tanh (ξ ) + 21 tanh2 (ξ ) b1 λ 3b1 tanh (ξ ) + v8 = 2 + 2 4 2l RootOf 3 Z + 1 + tanh (ξ ) 3 10 b1 RootOf 3 Z 2 + 1 + b1 tanh (ξ ) , + 10 + 9 RootOf 3 Z 2 + 1 tanh (ξ ) + 21 tanh2 (ξ ) u8 =
where ξ =
3b1 8β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
3lb1 b1 l λ + tanh (ξ ) + 2 2l 4 RootOf 3 Z + 1 + tanh (ξ ) 3b1l(RootOf (3 Z 2 +1) + µ3 + tanh (ξ )) + , 1 − 3RootOf 3 Z 2 + 1 µ3 + 3µ3 tanh (ξ ) + 3 tanh2 (ξ ) 3b1 b1 λ tanh (ξ ) + v9 = 2 + 2 4 2l RootOf 3 Z + 1 + tanh (ξ ) 3b1 (RootOf (3 Z 2 +1) + µ3 + tanh (ξ )) + , 1 − 3RootOf 3 Z 2 + 1 µ3 + 3µ3 tanh (ξ ) + 3 tanh2 (ξ ) u9 =
where ξ =
3b1 8β
(x + ly + λt), and l, λ, b1 are arbitrary constants;
2lβk lb1 l(µ1 b1 − b1 µ3 − b1 tanh (ξ )) + + 2 tanh (ξ ) µ1 + tanh (ξ ) −µ1 + µ1 µ3 + µ3 tanh (ξ ) + tanh2 (ξ ) βk b1 µ1 b1 − b1 µ3 − b1 tanh (ξ ) v10 = a0 + 2βk tanh (ξ ) + 2 + + 2 tanh (ξ ) µ1 + tanh (ξ ) −µ1 + µ1 µ3 + µ3 tanh (ξ ) + tanh2 (ξ ) where ξ = k x + ly + l 2 a0 + α a0 t , k, l, a0 , µ1 , µ3 , b1 are arbitrary constants. u 10 = la0 + 2l βk tanh (ξ ) +
3.2. New solutions of the (2 + 1)-dimensional nonlinear Davey–Stewartson system On the basis of the idea of the uniform direct ans¨atze method, we use the Riccati equation as the auxiliary equation to find new formal solutions of the coupled (2 + 1)-dimensional nonlinear system of Davey–Stewartson equations. The Riccati equation reads φ0 = h + φ2,
(3.7)
φ 0 = a + bφ + cφ 2 .
(3.8)
or
Under the transformation φ +
b 2c
= 1c Φ(c 6= 0), (3.8) becomes
4ac − b2 + Φ2, 4 which is formally the same as (3.7). It is apparent that (3.7) is simpler than (3.8). So it is more convenient to use Eq. (3.7) as an auxiliary equation. Then we only need to construct solutions of (3.7). As we know, Eq. (3.7) has the following solutions: √ √ −√−h tanh(√−hξ ), (I) φ = when h < 0, (3.9) − −h coth( −hξ ), Φ0 =
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(√
√ h tan( hξ ), √ √ (II) φ = − h cot( hξ ), (III) φ = −
1 , ξ +c
when h > 0,
(3.10)
when h = 0.
(3.11)
Now we use Eq. (3.7) as an auxiliary equation to construct new solutions of the coupled (2+1)-dimensional nonlinear system of Davey–Stewartson equations (DSE) [1,27,28], iu t + pu x x + u yy + r |u|2 u − 2uv = 0,
(3.12a)
vx x − pv yy − r (|u|2 )x x = 0,
(3.12b)
where the u(x, y, t) and v(x, y, t) are complex valued functions and p = ±1, r is constant. Eq. (3.12) where p = 1 and p = −1 is called the DS I equation and the DS II equation, respectively. We find it more convenient to rewrite the corresponding system in a real form by separating v = e + f i,
u = a + bi,
(3.13)
into real and imaginary parts. Then we obtain a system of four real partial differential equations: at + pbx x + b yy + r (a 2 b + b3 ) − 2(a f + be) = 0,
(3.14a)
−bt + pax x + a yy + r (a 3 + ab2 ) + 2(b f − ae) = 0,
(3.14b)
ex x − pe yy − 2r (a + b )x x = 0,
(3.14c)
f x x − p f yy = 0.
(3.14d)
2
2
The travelling wave transformation a(x, y, t) = a(ξ ),
b(x, y, t) = b(ξ ),
e(x, y, t) = e(ξ ),
f (x, y, t) = f (ξ ),
ξ = kx + ly + λt,
(3.15)
transforms Eq. (3.12) into λa 0 + ( pk 2 + l 2 )b00 + r (a 2 b + b3 ) − 2(a f + be) = 0,
(3.16a)
−λb + ( pk + l )a + r (a + ab ) + 2(b f − ae) = 0,
(3.16b)
(k 2 − pl 2 )e00 − 2r k 2 (a 2 + b2 )00 = 0,
(3.16c)
(k − pl ) f = 0.
(3.16d)
2
0
2
2
2
00
3
2
00
From (3.16c) and (3.16d) we may assume e=
2r k 2 (a 2 + b2 ) + α1 = M(a 2 + b2 ) + α1 , k 2 − pl 2
Throughout this paper we write
2r k 2 k 2 − pl 2
f = α2 .
(3.17)
= M. Substituting (3.17) into (3.16a) and (3.16b), we obtain
λa 0 + ( pk 2 + l 2 )b00 + r (a 2 b + b3 ) − 2aα2 − 2b(M(a 2 + b2 ) + α1 ) = 0,
(3.18a)
−λb + ( pk + l )a + 2bα2 + r (ab + a ) − 2a(M(a + b ) + α1 ) = 0.
(3.18b)
0
2
2
00
2
3
2
2
Then we assume Eq. (3.18) to have the following formal solutions: a = β0 + β1 φ +
β2 , µ+φ
b = δ0 + δ1 φ +
δ2 , µ+φ
(3.19)
with φ solving (3.7). With the aid of Maple, substituting (3.19) into Eqs. (3.18) yields two rational polynomial equations in φ. Setting the numerators of the two rational polynomial equations to be zero, we get two polynomial
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
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equations in φ. Setting the coefficients of φ to be zero yields a set of overdetermined algebraic equations with respect to β0 , β1 , β2 , δ0 , δ1 , δ2 , µ, k, l, λ, α1 , α2 . Solving the overdetermined algebraic equations, we get the following results (there are other results; we only list some of them for simplicity). Case 1. β1 = β2 = δ1 = 0, λ=
β0 δ2 (2M − r ) , µ2 + h
β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) , α1 = − 2(µ2 + h) β0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 α2 = , µ2 + h
where β0 , δ2 , δ0 , k, µ, l are arbitrary constants. Case 2. β0 2 h + β2 2 (β2 β0 + δ0 δ2 ) (2M − r ) , α2 = h (β0 δ2 − β2 δ0 )
µ = β1 = δ1 = 0, √ s 2 2Mδ2 2 + 2Mβ2 2 − 2h 2l 2 − r δ2 2 − r β2 2 , k= 2h p
β0 δ2 2 + 2β2 δ0 δ2 + 3β0 β2 2 (2M − r ) λ= , hδ2 3hδ0 β0 2 β2 + hβ0 δ0 2 δ2 − hβ0 3 δ2 + β0 δ2 3 + β2 δ0 δ2 2 + 3β0 β2 2 δ2 + hδ0 3 β2 − δ0 β2 3 (2M − r ) α1 = , 2h (β0 δ2 − β2 δ0 ) where β0 , δ2 , δ0 , β2 , l are arbitrary constants. Case 3. β0 2 h (2M − r ) , α1 = 2µ2
β1 = δ0 = δ1 = δ2 = 0, √ s 2 2β0 2 M − 2µ2l 2 − r β0 2 k=− , 2µ p
β0 µ2 + h β2 = − , µ
where β0 , µ, l, λ, α2 are arbitrary constants. Case 4. 2δ0 2 µ2 h + δ0 2 µ4 − hβ2 2 + δ0 2 h 2 (2M − r ) β2 µ , β0 = − 2 , α1 = − 2 µ +h 2 µ2 + h s √ 2 −2µ4l 2 − 4 µ2 hl 2 + 2Mβ2 2 − 2h 2l 2 − r β2 2 δ2 = α2 = β1 = δ1 = 0, k=− , p 2 µ2 + h where β2 , µ, l, λ, δ0 are arbitrary constants. Case 5. s δ2 = β0 = α2 = β2 = δ1 = 0,
β1 = − −
2(l 2 + k 2 p) , r − 2M
α1 =
δ0 2r + l 2 h + pk 2 h − δ0 2 M, 2
where µ, l, k, λ, δ0 are arbitrary constants. Case 6. µδ2 δ0 = − 2 , µ +h β0 = 0,
s α2 = −δ2 l 2 + k 2 p
2M − r , 2(l 2 + k 2 p)
s δ2 l 2 + k 2 p µ 2(2M − r ) λ= , µ2 + h l2 + k2 p
s β1 = −
2(l 2 + k 2 p) , 2M − r
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α1 =
r µ2 δ2 2 − 2µ2 δ2 2 M + 2µ4 pk 2 h + 2l 2 h 3 + 4 µ2 pk 2 h 2 + 2 pk 2 h 3 + 4 µ2l 2 h 2 + 2µ4l 2 h , 2 2 µ2 + h
where µ, l, k, δ2 are arbitrary constants. Case 7. β1 = µ = β2 = δ2 = 0, α2 = −δ0 β0 (r − 2M) , 1 2 α1 = − λ = δ1 β0 (r − 2M) , δ1 h + δ0 2 − β0 2 (r − 2M) , 2 where l, k, β0 , δ0 , δ1 are arbitrary constants. Case 8. β1 = µ = β2 = 0, λ=−
δ2 δ1 = − , h −4 δ2 2 + hβ0 2 − hδ0 2 (r − 2M) α1 = , 2h
α2 = −δ0 β0 (r − 2M) ,
β0 δ2 (r − 2M) , h
where l, k, β0 , δ0 , δ2 are arbitrary constants. Case 9. β1 = δ1 = µ = β2 = 0,
α2 = β0 δ0 (2M − r ) ,
λ=
β0 δ2 (2M − r ) , h
hβ0 2 − hδ0 2 − δ2 2 (2M − r ) , α1 = − 2h where l, k, β0 , δ0 , δ2 are arbitrary constants. Case 10. r r r √ 6 pk 2 + l 2 hN hN N λ= , β0 = − , β1 = − , δ0 = −δ1 −h, δ1 r − 2M r − 2M r − 2M r 8h p 2 k 4 + 2l 2 pk 2 + l 4 4h l 2r − 2 pk 2 M + pk 2r − 2l 2 M δ1 2 α2 (r − 2M) δ1 N − + , α1 = N N 2M − r N where N = δ1 2r − 2δ1 2 M + 2l 2 + 2 pk 2 , and l, k, α2 , δ1 are arbitrary constants. According to (3.13), (3.17) and (3.19), Case 1 and the solutions (3.9)–(3.11) of (3.7), we get the solutions of Eq. (3.12) : when h < 0, the solitary solutions δ2 u 11 = β0 + i δ0 + , √ √ µ − −h tanh( −hξ ) " 2 # δ2 2 v11 = M β0 + δ0 + √ √ µ − −h tanh( −hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 + ; µ2 + h δ2 , u 12 = β0 + i δ0 + √ √ µ − −h coth( −hξ )
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
" v12 = M
β02
+ δ0 +
√
δ2
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2 #
√ µ − −h coth( −hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 + ; µ2 + h
when h > 0, the periodic solutions δ2 u 13 = β0 + i δ0 + , √ √ µ + h tan( hξ ) " 2 # δ2 2 v13 = M β0 + δ0 + √ √ µ + h tan( hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 , + µ2 + h δ2 u 14 = β0 + i δ0 + , √ √ µ − h cot( −hξ ) " 2 # δ2 2 v14 = M β0 + δ0 + √ √ µ − h cot( hξ ) β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 − δ0 2 h (2M − r ) − 2(µ2 + h) iβ0 −δ0 hr + 2δ0 h M + 2µ2 Mδ0 − µ2r δ0 − µr δ2 + 2µMδ2 + ; µ2 + h when h = 0, the rational solution δ2 (ξ + c) = β0 + i δ0 + , µ(ξ + c) + 1 " 2 # β0 2 µ2 − 2µδ0 δ2 − δ0 2 µ2 + β0 2 h − δ2 2 (2M − r ) δ2 (ξ + c) 2 − = M β0 + δ0 + µ(ξ + c) + 1 2µ2 iβ0 (2µMδ0 − µr δ0 − r δ2 + 2 Mδ2 ) + , µ
u 15 v15
)t where ξ = β0 δ2µ(2M−r + kx +ly, and β0 , δ2 , δ0 , k, µ, l, c are arbitrary constants. In a similar way, from (3.13), (3.17) 2 +h and (3.19) and Cases 2–10, we have other solutions of Eqs. (3.12) as follows:
δ2 β2 u 2 = β0 + + i δ0 + , φ (ξ ) φ (ξ ) " # β2 2 δ2 2 v2 = M β0 + + δ0 + φ (ξ ) φ (ξ ) 3hδ0 β0 2 β2 + hβ0 δ0 2 δ2 − hβ0 3 δ2 + β0 δ2 3 + β2 δ0 δ2 2 + 3β0 β2 2 δ2 + hδ0 3 β2 − δ0 β2 3 (2M − r ) 2h (β0 δ2 − β2 δ0 ) i β0 2 h + β2 2 (β2 β0 + δ0 δ2 ) (2M − r ) + , h (β0 δ2 − β2 δ0 )
+
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where √ s β0 δ2 2 + 2β2 δ0 δ2 + 3β0 β2 2 (2M − r ) t 2 2Mδ2 2 + 2Mβ2 2 − 2h 2l 2 − r δ2 2 − r β2 2 ξ= + ly, x+ 2h p hδ2 and β0 , δ2 , δ0 , β2 , l are arbitrary constants and φ solves (3.7); β0 µ2 + h u 3 = β0 − , µ (µ + φ (ξ )) !2 β0 µ2 + h β0 2 h (2M − r ) v3 = M β0 − + α2 i, + µ (µ + φ (ξ )) 2µ2 where √ s 2 2β0 2 M − 2µ2l 2 − r β0 2 ξ = λt − x + ly, 2µ p and β0 , µ, l, λ, α2 are arbitrary constants and φ solves (3.7); β2 β2 µ + + iδ0 , 2 µ + φ (ξ ) µ +h # " 2 2δ0 2 µ2 h + δ0 2 µ4 − hβ2 2 + δ0 2 h 2 (2M − r ) µ 1 2 2 + δ0 − + v4 = M β2 , 2 µ + φ (ξ ) µ2 + h 2 µ2 + h u4 = −
where √ 2 ξ = λt − 2 2 µ +h
s
−2µ4l 2 − 4 µ2 hl 2 + 2Mβ2 2 − 2h 2l 2 − r β2 2 x + ly, p
and β2 , µ, l, λ, δ0 are arbitrary constants and φ solves (3.7); s 2l 2 + 2k 2 p φ (ξ ) + iδ0 , u5 = − − r − 2M 2l 2 + 2k 2 p 2 δ0 2r 2 v5 = M − φ (ξ ) + δ0 + + l 2 h + pk 2 h − δ0 2 M, r − 2M 2 where ξ = λ t + kx + ly and µ, l, k, λ, δ0 are arbitrary constants and φ solves (3.7); s 2(l 2 + k 2 p) µδ2 δ2 u6 = − φ (ξ ) + i − 2 + , 2M − r µ + φ (ξ ) µ +h s " 2 # 2(l 2 + k 2 p) 2 µδ2 2M − r δ2 2 2 v6 = M φ (ξ ) + − 2 − iδ2 l + k p + 2M − r µ + φ (ξ ) µ +h 2(l 2 + k 2 p) +
r µ2 δ2 2 − 2µ2 δ2 2 M + 2µ4 pk 2 h + 2l 2 h 3 + 4 µ2 pk 2 h 2 + 2 pk 2 h 3 + 4 µ2l 2 h 2 + 2µ4l 2 h , 2 2 µ2 + h
where s δ2 l 2 + k 2 p µ 2(2M − r ) ξ= t + kx + ly, µ2 + h l2 + k2 p and µ, l, k, δ2 are arbitrary constants and φ solves (3.7); u 7 = β0 + i (δ0 + δ1 φ (ξ )) ,
H. Zhi, H. Zhang / Nonlinear Analysis 69 (2008) 2748–2760
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h i 1 v7 = M β02 + (δ0 + δ1 φ (ξ ))2 − δ1 2 h + δ0 2 − β0 2 (r − 2M) − iδ0 β0 (r − 2M) , 2 where ξ = (δ1rβ0 − 2δ1 β0 M) t + kx + ly, and l, k, β0 , δ0 , δ1 are arbitrary constants and φ solves (3.7); δ2 φ (ξ ) δ2 u 8 = β0 + i δ0 − + , h φ (ξ ) " # −4 δ2 2 + hβ0 2 − hδ0 2 (r − 2M) δ2 2 δ2 φ (ξ ) 2 + − iδ0 β0 (r − 2M) , + v8 = M β0 + δ0 − h φ (ξ ) 2h where β0 δ2 (r − 2M) t + kx + ly, h and l, k, β0 , δ0 , δ2 are arbitrary constants and φ solves (3.7); δ2 , u 9 = β0 + i δ0 + φ (ξ ) " # hβ0 2 − hδ0 2 − δ2 2 (2M − r ) δ2 2 2 − + iβ0 δ0 (2M − r ) , v9 = M β0 + δ0 + φ (ξ ) 2h ξ =−
where β0 δ2 (2M − r ) t + kx + ly, h and l, k, β0 , δ0 , δ2 are arbitrary constants and φ solves (3.7); ! r √ hN + iδ1 φ (ξ ) − −h , u 10 = 2M − 2r √ 2 N v10 = M φ (ξ ) − −h + δ12 + iα2 2M − r r 8h p 2 k 4 + 2l 2 pk 2 + l 4 4h l 2r − 2 pk 2 M + pk 2r − 2l 2 M δ1 2 α2 (r − 2M) δ1 N − + , + N N 2M − r N ξ=
where 6 pk 2 + l 2 ξ= δ1
r
hN t + kx + ly, r − 2M
N = δ1 2r − 2δ1 2 M + 2l 2 + 2 pk 2 , and l, k, α2 , δ1 are arbitrary constants and φ solves (3.7). Remark: In [27,28], the authors introduced u = u(ξ )eiη , v = v(ξ ), ξ = k(x + py + λt), η = αx + βy + γ t, to seek the explicit solutions of (3.12), and assumed u(ξ ), v(ξ ) were real functions. So they could not obtain solutions like v1 ∼ v3 , v6 ∼ v7 , where the vi are complex functions. 4. Conclusion In this paper, we present a further uniform direct ans¨atze method for finding exact solutions of nonlinear evolution equations and give two examples to illustrate the method. As a result, we find abundant solutions of the two equations. And we construct new formal solutions of the Riccati equation. On the basis of the idea of the direct ans¨atze method, we may construct other solutions. In addition, this method is also computerizable, which allows us to perform complicated and tedious algebraic calculation on a computer.
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Acknowledgements We wish to express our gratitude to the reviewer for his/her valuable comments. The work was partially supported by the National Key Basic Research Project of China (Grant No. 2004CB318000). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28]
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