Generalized method and new exact
wave solutions for (2+1)-dimensional
Broer–Kaup–Kupershmidt system q
Ying Wan
a,b,*,Lina Song a ,Li Yin a,b ,Hongqing Zhang a a Department of Applied Mathematics,Dalian University of Technology,Dalian 116024,PR China
b College of Science,Dalian Fisheries University,Dalian 116024,PR China
Abstract
In this paper,a new algebra method,i.e.,generalized elliptic equation rational expansion method is devised to uni-formly construct a series of exact solutions for nonlinear partial di?erential https://www.doczj.com/doc/9c11670481.html,pared with most existing Tanh methods,the proposed method not only recover some known solutions,but also ?nd some new and general solutions.The e?ciency of the method can be demonstrated on the (2+1)-dimensional Broer–Kaup–Kupershmidt system.As a result,we obtain many new types of solutions include rational formal solitary wave solutions,rational formal triangular periodic wave solutions,rational formal Jacobi and Weierstrass double period solutions and rational wave solutions.ó2006Published by Elsevier Inc.
Keywords:The generalized elliptic equation rational expansion method;Rational formal solitary wave solutions;Rational formal triangular periodic wave solutions;Symbolic computation
1.Introduction
It is well known that the investigation of the exact solutions for nonlinear partial di?erential equations (NPDEs),which describe many important phenomena and dynamical processes in physics,chemistry,biology,etc.,plays an important role in the study of soliton theory.Many e?ective methods have been presented [1–30].Among them,the Tanh method is a very e?ective algorithm to construct exact solutions for a large number of NPDEs.Recently,in [8],Fan presented an algebraic method,comparing with other Tanh method,this method easily provides us with new and more general travelling wave solutions in terms of special functions such as solitary wave,triangular function wave,Weierstrass and Jacobi elliptic double periodic functions wave.More recently,based on the Fan method and the more general ansa ¨tz,on the one hand,Chen presented a more general method [9]to uniformly construct more general travelling wave solutions;On the other hand,0096-3003/$-see front matter ó2006Published by Elsevier Inc.doi:10.1016/j.amc.2006.08.082
q
The work is partially supported by the National Key Basic Research Project of China under the Grant No.2004CB318000.*Corresponding author.Address:Department of Applied Mathematics,Dalian University of Technology,Dalian 116024,PR China.E-mail address:wan_cherry2006@https://www.doczj.com/doc/9c11670481.html, (Y.
Wan).
Applied Mathematics and Computation xxx (2006)
xxx–xxx
Wang presented a new rational expansion methods to uniformly construct exact rational formal solutions [15].In this paper,based on the idea of rational expansion method,we present a new algebraic method:general ellip-tic equation rational expansion (EERE)method.We use it to solve the (2+1)-dimension Broer–Kaup–Kupershmidt equation and successfully construct new and more general solutions including rational formal solitary wave,rational formal triangular periodic,rational formal Jacobi and Weierstrass doubly periodic solutions.
2.Summary of the Riccati equation rational expansion method
In the following we would like to outline the main steps of our method:
Step 1.For a given NPDE system with some physical ?elds u i (x ,y ,t )in three variables x ,y ,t ,
F i eu i ;u it ;u ix ;u iy ;u itt ;u ixt ;u iyt ;u ixx ;u iyy ;u ixy ;...T?0;e2:1Tby using the wave transformation
u i ex ;y ;t T?U i en T;n ?k ex tly tk t T;e2:2Twhere k ,l and k are constants to be determined later.Then the nonlinear partial di?erential Eq.(2.1)is reduced to a nonlinear ordinary di?erential equation (ODE)
G i eU i ;U 0i ;U 00i ;...T?0:
e2:3TStep 2.We introduce a new ansa ¨tz in terms of ?nite rational formal expansion in the following forms:U i en T?a i 0tX m i j ?1a ij /j en Ttb ij /j à1en T/0en Tel/en Tt1Tj e2:4T
and the new variable /=/(n )satisfying [1]/02?d /d n 2?h 0th 1/th 2/2th 3/3th 4/4;e2:5Twhere h q ,a i 0,a ij and b ij (q =0,1,...,4;i =1,2,...;j =1,2,...,m i )are constants to be determined later.Step 3.
Determine the m i of the rational formal polynomial solutions (2.4)by,respectively,balancing the highest nonlinear terms and the highest-order partial derivative terms in the given system equations,and then give the formal solutions.Step 4.
Substitute Eq.(2.4)into Eq.(2.3)along with Eq.(2.5)and then set all coe?cients of /i en T?????????????????????P 4q ?0h q /q q ei ?1;2;...Tof the resulting system’s numerator to be zero to get an over-deter-mined system of nonlinear algebraic equations with respect to k ,l ,a i 0,a ij and b ij (i =1,2,...;j =1,2,...,m i ).Step 5.
Solving the over-determined system of nonlinear algebraic equations by use of Maple ,we would end up with the explicit expressions for k ,l ,a i 0,a ij and b ij (i =1,2,...;j =1,2,...,m i ).
Step 6.By using the results obtained in the above step,we can deriver a series of fundamental solutions in rational form such as polynomial,exponential,solitary wave,rational,triangular periodic,Jacobi and Weierstrass doubly periodic solutions.Because we interested in solitary wave,Jacobi and Weierstrass doubly periodic solutions.On the other hand,tan and cot type solutions appear in pairs with tanh and coth type solutions respectively,polynomial,rational triangular periodic solu-tions are omitted in this paper.By considering the di?erent values of h 0,h 1,h 2,h 3and h 4,Eq.(2.5)has many kinds of solitary wave,Jacobi and Weierstrass doubly periodic solutions which are listed as follows.
Case A .If h 3=h 4=0,Eq.(2.5)possesses following solutions:/??????h 0p n ;h 1?h 2?0;h 0>0;e2:6T2Y.Wan et al./Applied Mathematics and Computation xxx (2006)xxx–xxx
/?àh 0h 1t14
h 1n 2;h 2?0;h 1?0;e2:7T/?àh 12h 2texp ?????h 2p n ;h 0?h 2
14h 2;h 2>0;e2:8T/?àh 12h 2th 12h 2sin ????????àh 2p n ;h 0?0;h 2<0;e2:9T/?àh 12h 2th 12h 2sinh ?????h 2p n ;h 0?0;h 2>0:e2:10T
Case B .If h 1=h 3=0,Eq.(2.5)possesses following solutions:/??????????àh 2h 4r sech ?????h 2p n ;h 0?0;h 2>0;h 4<0;e2:11T/?????????????àh 22h 4r tanh ?????????àh 22r n !;h 0?h 2
24h 4
;h 2<0;h 4>0;e2:12T/??????????àh 2h 4r sec ????????àh 2p n ;h 0?0;h 2<0;h 4>0;e2:13T/????????h 22h 4r tan ?????h 22r n !;h 0?h 2
24h 4
;h 2>0;h 4>0;e2:14T/?à1?????h 4p n
;h 0?h 2?0;h 4>0;e2:15T/?sn en T;
h 0?1;h 2?àem 2t1T;h 4?m 2;e2:16T/?cd en T;
h 0?1;h 2?àem 2t1T;h 4?m 2;e2:17T/?cn en T;
h 0?1àm 2;h 2?2m 2à1;h 4?àm 2;e2:18T/?dn en T
h 0?m 2à1;h 2?2àm 2;h 4?à1;e2:19T/?ns en T;
h 0?m 2;h 2?àem 2t1T;h 4?1;e2:20T/?dc en T;
h 0?m 2;h 2?àem 2t1T;h 4?1;e2:21T/?nc en T;
h 0?àm 2;h 2?2m 2à1;h 4?1àm 2;e2:22T/?nd en T;
h 0?à1;h 2?2àm 2;h 4?1àm 2;e2:23T/?cs en T;
h 0?1àm 2;h 2?2àm 2;h 4?1;e2:24T/?sc en T;
h 0?1;h 2?2àm 2;h 4?1àm 2;e2:25T/?sd en T;
h 0?1;h 2?2m 2à1;h 4?m 2em 2à1T;e2:26T/?ds en T;h 0?m 2em 2à1T;h 2?2m 2à1;h 4?1;e2:27T
/?ns en T?cs en T;h 0?14;h 2?1à2m 22;h 4?14;e2:28T
/?nc en T?sc en T;h 0?1àm 2;h 2?1tm 2;h 4?1àm 2;e2:29T
/?ns en T?ds en T;h 0?m 24;h 2?m 2à22;h 4?14;e2:30T
/?sn en T?icn en T;h 0?m 24;h 2?m 2à22;h 4?m 2
4
;e2:31TY.Wan et al./Applied Mathematics and Computation xxx (2006)xxx–xxx 3
where m is a modulus.The Jacobi elliptic functions are doubly periodic and possess properties of triangular functions
sn2enTtcn2enT?1;dn2enT?1àm2sn2n;
sn0enT?cnenTdnenT;cn0enT?àsnenTdnenT;dn0enT?àm2snenTcnenT:
When m!1,the Jacobi functions degenerate to the hyperbolic functions,i.e.
snenT!tanhenT;cnenT!sechenT:
When m!0,the Jacobi functions degenerate to the triangular functions,i.e.
snenT!sinenT;cnenT!cosenT:
The more detailed notations for the Weierstrass and Jacobi elliptic functions can be found in Refs.
[23–25].
Case C.If h4=0,Eq.(2.5)possesses following solutions:
/?àh2
h3
sech2
?????
h2
p
2
n
;h0?h1?0;h2>0;e2:32T
/?àh2
h3
sec2
????????
àh2
p
2
n
;h0?h1?0;h2<0;e2:33T
/?
4
h3n
;h0?h1?h2?0;e2:34T
/?}
?????
h3
p
2
n;g2;g3
;h2?0;h3>0;e2:35T
where g2?à4h1
h3and g3?à4h0
h3
are called invariants of Weierstrass elliptic function.
Case D.If h0=h1=0,Eq.(2.5)possesses following solutions:
/?à
h2sec2
??????
àh2
p
2
n
2
?????????????
àh2h4
p
tan
??????
àh2
p
2
n
th3
;h2<0;e2:36T
/?
h2sech2
???
h2
p
n
2
?????????
h2h4
p
tanh
???
h2
p
n
àh3
;h2>0;e2:37T
Thus according to Eqs.(2.2),(2.4)–(2.37)and the conclusions in Step5,we can obtain some rational formal travelling-wave solutions of Eq.(2.1).
Remark.The more general the ansa¨tz is,the more general and more formal the solutions of the NPDEs will be.If we set the parameters in Eq.(2.5)to di?erent values,the above methods can be recovered by the EERE method.The concrete case is as follows:Setting l=b1=0,we just recover the solutions obtained by the generalized method[8,9].
3.New families of rational form solitary wave solutions to the(2+1)-dimensional Broer–Kaup–Kupershmidt system
In[26],the(2+1)-dimensional Broer–Kaup–Kupershmidt(BKK)system
H tyàH xxyt2eHH xT
y
t2G xx?0;e3:1:1T
G ttG xxt2eHGT
x ?0;e3:1:2T
4Y.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx
may be derived from the inner parameter dependent symmetry constraint of the Kadomtsev–Petviashvili (KP)equation.Though the integrability of the BKK system can be guaranteed by the integrability of the KP equa-tion (because it is a symmetry constraint of the KP equation),some authors have exactly proven its integra-bility in some di?erent sense.For more details about the results about this system,the reader is advised to see the achievements in Refs.[26–30].
In order to get some families of rational form solitary wave solutions to the (2+1)-dimensional BKK sys-tem,by considering the wave transformations H (x ,y ,t )=U (n ),G (x ,y ,t )=V (n )and n =k (x +ly +k t ),we change Eq.(3.1)to the form
k lU 00àlkU 000t2l eUU 0T0t2V 00?0;
e3:2:1Tk V 0tkV 00t2eVU T0?0:
e3:2:2TWe suppose that BKK system have the following formal travelling wave solution:
U en T?a i 0tX m u j ?1
a j /j en Tt
b j /j à1en T/0en Tel/en Tt1Tj ;e3:3:1TV en T?A 0tX m v j ?1A j /j en TtB j /j à1en T/0
el/en Tt1Tj ;e3:3:2T
and the new vriable /=/(n )satisfying (2.5)where a 0,a i ,b i ,A 0,A i and B i are constants to be determined later.For the BKK system,by balancing the highest nonlinear terms and the highest order partial derivative terms in (2.1),gives m u =1and m v =2.So we have
U en T?a 0ta 1/en Ttb 1/0
l/en Tt1
;e3:4:1TV en T?A 0tA 1/en TtB 1/0l 1m /en Ttl 2/0t1tA 2/2en TtB 2/en T/0
el/en Tt1T;e3:4:2Twhere /(n )satis?es Eq.(2.3).With the aid of Maple ,substituting (2.4)along with (2.3)into (2.1),yields a set of algebraic equations for /i en Te?????????????????????P 4q ?0h q /q q Tj ,(i =0,1,...;j =0,1).Setting the coe?cients of these terms /i (n )to zero yields a set of over-determined algebraic equations with respect to a 0,a 1,b 1,A 0,A 1,B 1,A 2,B 2,l and k .
By use of the Maple soft package ‘‘Charsets’’by Dongming Wang,which is based on the Wu-elimination method [21],solving the over-determined algebraic equations,we get the following results:
k ??a 1?????????????????????????????????????????????????????????????h 0l 4th 2l 2àh 3l th 4àh 1l 3
p ;a 1?a 1;l ?l ;b 1?0;a 0?2h 1l 3k à2h 0l 4k à4h 0l 3a 1t3h 1l 2a 1t2h 3lk th 3a 1à2h 2l 2k à2h 2l a 1à2k h 44eh 0l 4th 2l 2àh 3l th 4àh 1l 3T
;A 0?à
e8h 20l 6àh 23t12h 2l 4h 0à16h 3h 0l 3à4h 1l 3h 2t6h 1h 3l 2Ta 21l 16eh 0l 4th 2;l 2àh 3l th 4àh 1l 3Tàe3h 2
1l 4à12h 1l 5h 0t4h 4h 2t24h 4h 0l 2à12h 1h 4l Ta 21l 16eh 0l 4th 2;l 2àh 3l th 4àh 1l 3T;A 1?e4l 3h 0à3l 2h 1àh 3t2l h 2Ta 21l 04223413;B 1??a 21l 2?????????????????????????????????????????????????????????????h 0l 4th 2l 2àh 3l th 4àh 1l 3
p ;A 2?à12a 21l ;B 2??a 21l l 2?????????????????????????????????????????????????????????????h 0l 4th 2l 2àh 3l th 4àh 1l 3
p :e3:5T
Then according to Eq.(3.5),we obtain following solutions of the BKK system.Y.Wan et al./Applied Mathematics and Computation xxx (2006)xxx–xxx 5
Family1.When h3=h4=0and h0?àh21
4h2,we obtain the following solutions for the BKK system,as
follows:
H11?a0t
a1eàh1t2h2expe
?????
h2
p
nTT
àl h1t2l h2expe
?????
h2
p
nTt2h2
;e3:6:1T
G11?A0t
el3h2
1
à3l2h1h2t2l h2
2
Ta2
1
leàh1t2h2expe
?????
h2
p
nTT
eh2
1
l4t4h2
2
l2à4h1h2l3Teàl h1t2h2l expe
?????
h2
p
nTt2h2T
à
a2
1
leàh1th2expe
?????
h2
p
nTT2
2eàl h1t2l h2expe
?????
h2
p
nTt2h2T2
?
a2
1
h2l
?????
h2
p
expe
?????
h2
p
nT
????????????????????????????????????????????????????????????????????????????????????????????????????
h2
1
l4
4h2
th2l2àh1l3eàl h1t2l h2expe
?????
h2
p
nTt2h2T
q
?
a2
1
l làh1h2t2h2
2
exp
?????
h2
p
n
àá
àá?????
h2
p
exp
?????
h2
p
n
àá
????????????????????????????????????
h2
1
l4
4h2
th2l2àh1l3
q
àl h1t2l h2exp
?????
h2
p
n
àá
t2h2
àá2;e3:6:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,h2>0,h1,a1,l and k are arbi-trary constants.
Family2.When h0=h3=h4=0,we obtain the following solutions for the BKK system,as follows:
H21?2h1l3kt3h1l2a1à2h2l2kà2h2l a1
4h2l2àh1l3
eT
t
a1àh1th1sinh
?????
h2
p
n
àá
àá
àl h1tl h1sinh
?????
h2
p
n
àá
t2h2
;e3:7:1T
G21?A0t
à3l2h1t2l h2
eTa2
1
làh1th1sinh
?????
h2
p
n
àá
àá4h2l2àh1l3
eTàl h1tl h1sin h
?????
h2
p
n
àá
t2h2
àá
à
a2
1
làh1th1sinh
?????
h2
p
n
àá
àá2
2àl h1tl h1sinh
?????
h2
p
n
àá
t2h2
àá2?
a2
1
lh1h2cosh
?????
h2
p
n
àá
2
?????
h2
p????????????????????????
h2l2àh1l3
p
àl h1tl h1sinh
?????
h2
p
n
àá
t2h2
àá
?
a2
1
l làh1h2th1h2sinh
?????
h2
p
n
àá
àá
h1cosh
?????
h2
p
n
àá
2
????????????????????????
h2l2àh1l3
p?????
h2
p
àl h1tl h1sinh
?????
h2
p
n
àá
t2h2
àá2;e3:7:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),h2>0,h1,l,a1,l and k are arbi-trary constants.
Family3.When h0=h1=h3=0,we obtain the following solutions for the BKK system,as follows:
H31?à2h2l2kt2h2l a1t2k h4
2h2l2th4
eT
t
a1
????????
àh2
4
q
sech
?????
h2
p
n
àá
l
????????
àh2
h4
q
sech
?????
h2
p
n
àá
t1
;e3:8:1T
G31?à
h4h2a2
1
l
4h2l2th4
eT
t
l h2a2
1
l
????????
àh2
4
q
sech
?????
h2
p
n
àá
2h2l2th4
eTl
????????
àh2
h4
q
sech
?????
h2
p
n
àá
t1
à
a2
1
lh2sech2
?????
h2
p
n
àá
2h4l
????????
àh2
h4
q
sech
?????
h2
p
n
àá
t1
2?
a2
1
l l h32
2
sech2
?????
h2
p
n
àá
tanh
?????
h2
p
n
àá
2h4
???????????????????
h2l2th4
p
l
????????
àh2
h4
q
sech
?????
h2
p
n
àá
t1
2
?
a2
1
l
????????
àh2
4
q
sech
?????
h2
p
n
àá
tanh
?????
h2
p
n
àá?????
h2
p
2
???????????????????
h2l2th4
p
l
????????
àh2
h4
q
sech
?????
h2
p
n
àá
t1
;e3:8:2T
where n=k(x+ly+k t),k is determined by(3.5),h2>0,h4<0,l,a1,l and k are arbitrary
constants.
6Y.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx
Family4.When h1=h3=0and h0?àh22
4h4,we obtain the following solutions for the BKK system,as
follows:
H41?a0t
a1
??????????
à2h2
h4
q
tanh
???????
à2h2
p
2
n
l
??????????
à2h2
4
q
tanh
???????
à2h2
p
n
t2
;e3:9:1T
G41?A0à
àh2
2
l3à2l h2h4
àá
a2
1
l
??????????
à2h2
4
q
tanh
???????
à2h2
p
n
h2
2
l4t4h2h4l2t4h2
4
àá
l
??????????
à2h2
h4
q
tanh
???????
à2h2
p
2
n
t2
t
a2
1
lh2tanh2
???????
à2h2
p
2
n
h4l
??????????
à2h2
h4
q
tanh
???????
à2h2
p
2
n
t2
2?
a2
1
l
????????
àh2
h4
q????????
àh2
p
sech2
???????
à2h2
p
2
n
4
???????????????????????????????
h2
2
l4
2h4
th2l2th4
q
l
??????????
à2h2
h4
q
tanh
???????
à2h2
p
2
n
t2
?
a2
1
l l h2
???????????
à2h2
p
tanh
???????
à2h2
p
2
n
sech2
???????
à2h2
p
2
n
2h4
???????????????????????????????
h2
2
l4
4
th2l2th4
q
l
???????????????????????
à2frach2h4
p
tanh
???????
à2h2
p
n
t2
;e3:9:2T
where n=k(x+ly+k t),k,a0,A0are determined by(3.5),h2<0,h4>0,l,a1,l and k are arbi-trary constants.
Family5.When h0=h1=h4=0,we obtain the following solutions for the BKK system,as follows:
H51?àà2h3lkàh3a1t2h2l2kt2h2l a1
4h2l2àh3l
eT
à
a1h2sech2
???
h2
p
2
n
àl h2sech2
???
h2
p
2
n
th3
;e3:10:1T
G51?
h2
3
a2
1
l
16h2l2àh3l
eT
à
àh3t2l h2
eTa2
1
lh2sech2
???
h2
p
n
4eh2l2àh3lTàl h2sech2
???
h2
p
2
n
th3
à
a2
1
lh2
2
sech4
???
h2
p
2
n
2àl h2sech2
???
h2
p
n
th3
?
a2
1
lh32
2
sech2
???
h2
p
2
n
tanh
???
h2
p
2
n
2
??????????????????????
h2l2àh3l
p
àl h2sech2
???
h2
p
n
th3
?
a2
1
l l h52
2
sech4
???
h2
p
2
n
tanh
???
h2
p
2
n
2
??????????????????????
h2l2àh3l
p
àl h2sech2
???
h2
p
2
n
th3
2;e3:10:2Twhere n=k(x+ly+k t),k is determined by(3.5),h2>0,h3,l,a1,l and k are arbitrary constants.
Family6.When h0=h1=0,we obtain the following solutions for the BKK system,as follows:
H61?a0t
h2a1sech2
???
h2
p
n
l h2sech2
???
h2
p
n
t2
?????????
h4h2
p
tanh
???
h2
p
n
àh3
;e3:11:1TY.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx7
G61?A0t
h2la2
1
àh3t2l h2
eTsech2
???
h2
p
2
n
4l h2sech2
???
h2
p
2
n
t2
?????????
h4h2
p
tanh
???
h2
p
2
n
àh3
h2l2àh3lth4
eT
?
1
2
2
?????????
h4h2
p
à
?????????
h4h2
p
sech2
???
h2
p
2
n
àh3tanh
???
h2
p
2
n
sech2
???
h2
p
2
n
h32
2
a2
1
l
l h2sech2
???
h2
p
2
n
t2
?????????
h4h2
p
tanh
???
h2
p
2
n
àh3
2
?????????
h4h2
p
tanh
???
h2
p
2
n
àh3
??????????????????????????????
h2l2àh3lth4
p
à
h2
2
a2
1
l sech2
???
h2
p
n
2l h2sech2
???
h2
p
n
t2
?????????
h4h2
p
tanh
???
h2
p
n
àh3
?
1à2
?????????
h4h2
p
t
?????????
h4h2
p
sech2
???
h2
p
n
th3tanh
???
h2
p
n
sech4
???
h2
p
n
h52
2
l a2
1
l
l h2sech2
???
h2
p
n
t2
?????????
h4h2
p
tanh
???
h2
p
n
àh3
2
?????????
h4h2
p
tanh
???
h2
p
n
àh3
h2l2àh3lth4
;
e3:11:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),h2>0,h3,h4,l,a1,l and k are arbi-trary constants.
Family7.When h2=h4=0,we obtain the following solutions for the BKK system,as follows:
H71?à2h0l4kt4h0l3a1à2h1l3kà3h1l2a1à2h3lkàh3a1
4h0l4àh3làh1l3
eT
t
a1}
???
h3
p
n;g2;g3
l}
???
h3
p
n;g2;g3
t1
;e3:12:1T
G71?A0t
4l3h0à3l2h1àh3
eTa2
1
l}
???
h3
p
2
n;g2;g3
4h0l4àh3làh1l3
eTl}
???
h3
p
2
n;g
2
;g
3
t1
à
a2
1
l}2
???
h3
p
2
n;g2;g3
2l}
???
h3
p
2
n;g2;g3
t1
2
?a2
1
l
?????????????????????????????????????????????????????????????????????????????????????????????
h0th1}
???
h3
p
n;g
2
;g
3
th3}3
???
h3
p
n;g
2
;g
3
s
2
????????????????????????????????????
h0l4àh3làh1l3
p
l}
???
h3
p
n;g
2
;g
3
t1
?a2
1
l l}
???
h3
p
2
n;g2;g3
?????????????????????????????????????????????????????????????????????????????????????????????
h0th1}
???
h3
p
2
n;g2;g3
th3}3
???
h3
p
2
n;g2;g3
s
2
????????????????????????????????????
h0l4àh3làh1l3
p
l}
???
h3
p
n;g2;g3
t1
;e3:12:2T
where n=k(x+ly+k t),g2?à4h1
h3,g3?à4h0
h3
,k and A0are determined by(3.5),h3>0,h0,h1,l,
a1,l and k are arbitrary constants.
Family8.When h1=h3=0,h0=1,h2=à(m2+1)and h4=m2,we obtain the following solutions for the BKK system,as follows:
H81?2l4kt4l3a1à2m2t1
eTl2kà2em2t1Tl a1t2k m2
4eàl4àeàm2à1Tl2àm2T
t
a1snenT
l snenTt1
;e3:13:1T
8Y.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx
G81?A0t
eà4l3t2lem2t1TTa2
1
l snenT
4eàl4tem2t1Tl2àm2Tel snenTt1T
à
a2
1
l sn2enT
2l snenTt1
eT2
?
a2
1
l cnenTdnenT
2
???????????????????????????????????????????
l4àem2t1Tl2tm2
p
el snenTt1T
?
a2
1
l l snenTcnenTdnenT
2
???????????????????????????????????????????
l4àem2t1Tl2tm2
p
el snenTt1T2
;e3:13:2T
where n=k(x+ly+k t),k and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family9.When h1=h3=0,h0=1,h2=à(m2+1)and h4=m2,we obtain the following solutions for the BKK system,as follows:
H91?2l4kt4l3a1à2em2t1Tl2kà2em2t1Tl a1t2k m2
4222
t
a1cdenT
;e3:14:1T
G91?A0t
e4l3à2lem2t1TTa2
1
l cdenT
4el4àem2t1Tl2tm2Tel cdenTt1T
?
a2
1
l snenTe1àm2T
2
???????????????????????????????????????????
l4àem2t1Tl2tm2
p
dn2enTel cdenTt1T
à
a2
1
l cd2enT
2el cdenTt1T2
?
a2
1
l l cdenTsnenTe1àm2T
2
???????????????????????????????????????????
l4àem2t1Tl2tm2
p
dn2enTel cdenTt1T2
;e3:14:2T
where n=k(x+ly+k t),k and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family10.When h1=h3=0,h0=1àm2,h2=2m2à1and h4=àm2,we obtain the following solutions for the BKK system,as follows:
H101?a0t
a1cnenT
l cnenTt1
;e3:15:1T
G101?A0t
e4l3e1àm2Tt2le2m2à1TTa2
1
l cnenT
4ee1àm2Tl4te2m2à1Tl2àm2Tel cnenTt1T
à
a2
1
l cn2enT
2el cnenTt1T2
?
a2
1
l dnenTcnenT
2
???????????????????????????????????????????????????????????????
e1àm2Tl4te2m2à1Tl2àm2
p
el cnenTt1T
?
a2
1
l l cnenTdnenTsnenT
2
???????????????????????????????????????????????????????????????
e1àm2Tl4te2m2à1Tl2àm2
p
el cnen;mTt1T2
;e3:15:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary
constants.
Family11.When h1=h3=0,h0=m2à1,h2=2àm2and h4=à1,we obtain the following solutions for the BKK system,as follows:
H111?a0t
a1dnenT
l dnenTt1
;e3:16:1T
G111?A0t
e4l3em2à1Tt2le2àm2TTa2
1
l dnenT
4eem2à1Tl4te2àm2Tl2à1Tel dnenTt1T
à
a2
1
l dn2enT
2el dnenTt1T
?
a2
1
lm2cnenTsnenT
2
??????????????????????????????????????????????????????????
em2à1Tl4te2àm2Tl2à1
p
el dnenTt1T
?
a2
1
l l dnenTm2cnenTsnenT
2
??????????????????????????????????????????????????????????
em2à1Tl4te2àm2Tl2à1
p
el dnenTt1T
;e3:16:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary
constants.
Family12.When h1=h3=0,h0=m2,h2=à(1+m2)and h4=1,we obtain the following solutions for the BKK system,as follows:
H121?à2m2l4kt4m2l3a1à2em2t1Tl2kà2em2t1Tl a1t2k
4em2l4teàm2à1Tl2t1T
t
a1nsenT
l nsenTt1
;e3:17:1TY.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx9
G121?A0t
e4l3m2à2lem2t1TTa2
1
l nsenT
4em2l4àem2t1Tl2t1Tel nsenTt1T
à
a2
1
l ns2enT
2el nsenTt1T2
?
a2
1
l cnenTdnenT
2
??????????????????????????????????????????????
m2l4àem2t1Tl2t1
p
sn2enTel nsenTt1T
?
a2
1
l l nsenTcnenTdnenT
2
??????????????????????????????????????????????
m2l4àem2t1Tl2t1
p
sn2enTel nsenTt1T
;e3:17:2T
where n=k(x+ly+k t),k and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family13.When h1=h3=0,h0=m2,h2=à(1+m2)and h4=1,we obtain the following solutions for the BKK system,as follows:
H131?a0t
a1dcenT
l dcenTt1
;e3:18:1T
G131?A0t
e4l3m2à2lem2t1TTa2
1
l dcenT
4em2l4em2t1Tl2t1Tel dcenTt1T
à
a2
1
l dc2enT
el dcenTt1T2
?
a2
1
l snenTe1àm2T
2
??????????????????????????????????????????????
m2l4àem2t1Tl2t1
p
cn2enTel dcenTt1T
?
a2
1
l l dcenTsnenTe1àm2T
2
??????????????????????????????????????????????
m2l4àem2t1Tl2t1
p
cn2enTel dcenTt1T2
;e3:18:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family14.When h1=h3=0,h0=àm2,h2=2m2à1and h4=1àm2,we obtain the following solutions for the BKK system,as follows:
H141?a0ta1ncenT
;e3:19:1T
G141?A0t
eà4l3m2t2le2m2à1TTa2
1
l ncenT
4eàm2l4te2m2à1Tl2t1àm2Tel ncenTt1T
à
a2
1
l nc2enT
2el ncenTt1T
?
a2
1
l dnenTsnenT
2
???????????????????????????????????????????????????????????????
àm2l4te2m2à1Tl2t1àm2
p
cn2enTel ncenTt1T
?
a2
1
l l ncenTdnenTsnenT
???????????????????????????????????????????????????????????????
àm2l4te2m2à1Tl2t1àm2
p
cn2enTel ncenTt1T
;e3:19:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary
constants.
Family15.When h1=h3=0,h0=à1,h2=2àm2and h4=m2à1,we obtain the following solutions for the BKK system,as follows:
H151?2l4kt4l3a1à2e2àm2Tl2kà2e2àm2Tl a1à2kem2à1T
4eàl4te2àm2Tl2à1tm2T
t
a1nd n
l ndenTt1
;e3:20:1T
G151?A0t
eà4l3t2le2àm2TTa2
1
l ndenT
4eàl4te2àm2Tl2à1tm2Tel ndenTt1T
à
a2
1
l nd2enT
2el ndenTt1T2
?
a2
1
lm2cnenTsnenT
2
???????????????????????????????????????????????????????
àl4te2àm2Tl2à1tm2
p
dn2enTel nden;mTt1T
?
a2
1
l l ndenTm2cnenTsnenT
2
???????????????????????????????????????????????????????
àl4te2àm2Tl2à1tm2
p
dn2enTel ndenTt1T2
;e3:20:2T
where n=k(x+ly+k t),k and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family16.When h1=h3=0,h0=1àm2,h2=2àm2and h4=1,we obtain the following solutions for the BKK system,as follows:
10Y.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx
H161?a0t
a1csenT
l csenTt1
;e3:21:1T
G161?A0t
e4l3e1àm2Tt2le2àm2Ta2
1
l csenT
4ee1àm2Tl4te2àm2Tl2t1Tel csenTt1T
à
a2
1
l cs2enT
2el csenTt1T2
?
a2
1
l dnenT
2
??????????????????????????????????????????????????????????
e1àm2Tl4te2àm2Tl2t1
p
sn2enTel csenTt1T
?
a2
1
l l csenTdnenT
2
??????????????????????????????????????????????????????????
e1àm2Tl4te2àm2Tl2t1
p
sn2enTel csenTt1T
;e3:21:2T
where n=k(x+ly+k t),k,a0and A0re determined by(3.5),l,a1,l and k are rbitrary constants. Family17.When h1=h3=0,h0=1,h2=2àm2and h4=1àm2,we obtain the following solutions for the BKK system,as follows:
H171?à2l4kt4l3a1t2e2àm2Tl2kt2e2àm2Tl a1t2ke1àm2T
4el4te2àm2Tl2t1àm2T
t
a1scenT
l scenTt1
;e3:22:1T
G171?A0t
e4l3t2le2àm2TTa2
1
l scenT
4el4te2àm2Tl2t1àm2Tel scenTt1T
à
a2
1
l sc2enT
2el scenTt1T2
?
a2
1
l dnenT
2
????????????????????????????????????????????????????
l4te2àm2Tl2t1àm2
p
cn2enTel scenTt1T
?
a2
1
l l scenTdnenT
2
????????????????????????????????????????????????????
l4te2àm2Tl2t1àm2
p
cn2enTel scenTt1T
;e3:22:2T
where n=k(x+ly+k t),k and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family18.When h1=h3=0,h0=1,h2=2m2à1and h4=m2(m2à1),we obtain the following solutions for the BKK system,as follows:
H181?a0t
a1sdenT
l sdenTt1
;e3:23:1T
G181?A0t
eà4l3à2le2m2à1TTa2
1
l sdenT
4eàl4àe2m2à1Tl2àm2em2à1TTel sdenTt1T
à
a2
1
l sd2enT
2el sdenTt1T
?
a2
1
l cnenT
2
???????????????????????????????????????????????????????????????
l4te2m2à1Tl2tm2em2à1T
p
dnenTel sdenTt1T
?
a2
1
l l sdenTcnenT
2
???????????????????????????????????????????????????????????????
l4te2m2à1Tl2tm2em2à1T
p
dn2enTel sdenTt1T
;e3:23:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary constants. Family19.When h1=h3=0,h0=m2(m2à1),h2=2m2à1and h4=1,we obtain the following solutions for the BKK system,as follows:
H191?a0t
a1dsenT
l dsenTt1
;e3:24:1T
G191?A0t
e4l3m2em2à1Tt2le2m2à1TTa2
1
l dsenT
4em2em2à1Tl4te2m2à1Tl2t1Tel dsenTt1T
à
a2
1
l ds2enT
2el dsenTt1T
?
a2
1
l cnenT
2
?????????????????????????????????????????????????????????????????
m2em2à1Tl4te2m2à1Tl2t1
p
sn2enTel dsenTt1T
?
a2
1
l l dsenTcnenT
2
?????????????????????????????????????????????????????????????????
m2em2à1Tl4te2m2à1Tl2t1
p
sn2enTel dsenTt1T
;e3:24:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary constants.
Y.Wan et al./Applied Mathematics and Computation xxx(2006)xxx–xxx11
Family 20.When h 1=h 3=0,h 0?14,h 2?1à
2m 22and h 4?14,we obtain the following solutions for the BKK system,as follows:
H 201?a 0t
a 1ens en T?cs en TTl ens en T?cs n Tt1;e3:25:1TG 201?A 0tel 3tl e1àm 2TTa 21l ens en T?cs en TTel 4t2e1àm 2Tl 2tTel ens en T?cs en TTt1Tàa 21l ens en T?cs en TT22el ens en T?cs en TTt1T?a 21l ecn en Tdn en T?dn en TT???????????????????????????????????????????l 4t2e1àm 2Tl 2t1p sn 2en Tel ens en T?cs en TTt1T?a 21l l ens en T?cs en TTecn en Tdn en T?dn en TT???????????????????????????????????????????l 4t2e1àm 2Tl 2t1p sn 2en Tel ens en T?cs en TTt1T;e3:25:2T
where n =k (x +ly +k t ),k ,a 0and A 0are determined by (3.5),l ,a 1,l and k are arbitrary constants.Family 21.When h 1=h 3=0,h 0?1à
m 2,h 2?1t
m 2and h 4?1à
m 2,we obtain the following solutions for the
BKK system,as follows:
H 211?a 0t
a 1enc en T?sc en TTl enc en T?sc en TTt1;e3:26:1TG 211?A 0tel 3e1àm 2Ttl e1tm 2TTa 21l enc en T?sc en TT
ee1àm 2Tl 4t2e1tm 2Tl 2t1àm 2Tel enc en T?sc en TTt1Tàa 21l enc en T?sc en TT22el enc en T?sc en TTt1T
2?a 21l edn en Tsn en ;m T?dn en TT???????????????????????????????????????????????????????????????????????e1àm 2Tl 4t2e1tm 2Tl 2t1àm 2p cn 2en ;m Tel enc en T?sc en TTt1T?a 21l l enc en T?sc en TTedn en Tsn en T?dn en TT???????????????????????????????????????????????????????????????????????e1àm 2Tl 4t2e1tm 2Tl 2t1àm 2p cn 2en Tel enc en T?sc en TTt1T2;e3:26:2T
where n =k (x +ly +k t ),k ,a 0and A 0are determined by (3.5),l ,a 1,l and k are arbitrary constants.Family 22.When h 1=h 3=0,h 0?m 44,h 2?m 2à22and h 4?14,we obtain the following solutions for the BKK system,as follows:
H 221?a 0t
a 1ens en T?ds en TTl ens en T?ds en TTt1;e3:27:1TG 221?A 0tel 3m 4tl em 2à2TTa 21l ens en T?ds en TTem 4l 4t2em 2à2Tl 2t1Tel ens en T?ds en TTt1Tàa 21l ens en T?ds en TT22el eds en T?ds en TTt1T?a 21l ecn en Tdn en T?cn en TT????????????????????????????????????????????????m 4l 4t2em 2à2Tl 2t1p sn 2en Tel ens en T?ds en TTt1T?a 21l l ens en T?ds en TTecn en Tdn en T?cn en TT????????????????????????????????????????????????m 4l 4t2em 2à2Tl 2t1p sn 2en Tel ens en T?ds en TTt1T;e3:27:2T
where n =k (x +ly +k t ),k ,a 0and A 0are determined by (3.5),l ,a 1,l and k are arbitrary constants.Family 23.When h 1=h 3=0,h 0?m 2,h 2?m
2à2and h 4?m 2,we obtain the following solutions for the BKK system,as follows:
H 231?a 0ta 1esn en T?icn en TTl esn en T?icn en TTt1;e3:28:1T12Y.Wan et al./Applied Mathematics and Computation xxx (2006)xxx–xxx
G231?A0t
el3m2tlem2à2TTa2
1
lesnenT?icnenTT
em2l4t2em2à2Tl2tm2TelesnenT?icnenTTt1T
à
a2
1
lesnenT?icnenTT2
2elesnenT?icnenTTt1T2
?
a2
1
lecnenTdnenT?idnenTsnenTT
???????????????????????????????????????????????????
m2l4t2em2à2Tl2tm2
p
elesnenT?icnenTTt1T
?
a2
1
l lesnenT?icnenTTecnenTdnenT?idnenTsnenTT
???????????????????????????????????????????????????
m2l4t2em2à2Tl2tm2
p
elesnenT?icnenTTt1T
;e3:28:2T
where n=k(x+ly+k t),k,a0and A0are determined by(3.5),l,a1,l and k are arbitrary constants.
4.Summary and conclusions
Improvements and extension of Tanh method can be classed into two ways,one is to present a general ansatz,another is to use a more general subequation,which has more rich and general solutions,to reduce the equation solved to obtained more styles of solutions.Taking the(2+1)-dimensional Broer–Kaup–Kupershmidt(BKK)system as a simple example,some families of rational formal solitary wave solutions,tri-angular periodic wave solutions and rational wave solutions are constructed by using of the EERE method. The method can also be applied to solve more nonlinear partial di?erential equation or equations.In fact,we naturally present a more general ansa¨tz,in the rational formal polynomial solutions(2.2)taking a0,a i,b i,A0, A i,B i(i=1,2,...,m i)and n as di?erentiable function to be determined later.Therefore,for some nonlinear equations,more types of non-travelling solutions,such as soliton-like solutions,would be expected.We would like to study the more general ansa¨tz for?nding new formal soliton-like solutions of some nonlinear partial di?erential equation or equations further.
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