Solitary Wave Solutions of High Order Scalar Fields and Coupled Scalar Fields
Jian-song Yang and Sen-yue Loua
Physics Department of Hangzhou Normal College, Hangzhou 310000, P. R. China
aInstitute of Mathematical Physics, Ningbo University, Ningbo 315211, P. R. China Z. Naturforsch. 54 a, 195–203 (1999); received December 13, 1998
An arbitrary Klein-Gordon field with a quite general constrained condition (which contains an arbitrary function) can be used as an auxilialy field such that some special types of solutions of high order scalar fields can be obtain by solving an ordinary differential equation (ODE). For a special type of constraint, the general solution of the ODE can be obtained by twice integrating.
The solitary wave solutions of the5model are treated in an alternative simple way. The obtained solutions of the5model can be changed to those of the8field and coupled scalar fields.
1. Introduction
The Lagrangian density of a generalized nonlinear Klein-Gordon (NKG) field
(x
1;x
2;:::x
D;t
) in(
D
+ 1)-dimensions has the formL[
;
∂] =∂∂;V
();
(1)and the corresponding equation of wave motion reads
2
XDi=1
xixi;tt=F
();
F
() dV
d
:
(2)For not very large
, the potentialV
V
() in (1)can be replaced by a polynomial function of
:V
=XNn=0
1
nv
nn;
v
n= (n
;11)!dnv
d
n
=0
;
(3)and then the related wave motion equation becomes
2
=XNn=1
v
nn;1:
(4)Some well known NKG models are just spe- cial cases of (3) (or (4)), say, the
4 model cor-responding to
N
= 4; v
1 =v
3 = 0 [1], the 6model to
N
= 6; v
1 =v
3 =v
5 = 0 [2], and the Reprint requests to Dr. Sen-yue Lou;E-mail: sylou@public.nbptt.zj.cn.
0932–0784 / 99 / 0300–0195 $ 06.00c Verlag der Zeitschrift f¨ur Naturforschung, T¨ubingenwww.znaturforsch.com
3+4 (or Friedberg-Lee (FL)) model [3] is related toN
= 4; v
1= 0. In [4], the authors discuss the inter- esting properties of the self-exited soliton motion by using a10(N
= 10; v
2i+1 = 0;
(i
= 0:::
4)) model.In [5] the generalized
2mmodel (N
= 2m; v
2i+1= 0;
(i
= 0;
1;:::;m
;1)) is used to study the effective kink-kink interaction mediated by phonon exchange.However, to our knowledge there are no known exact solutions of (4) for
N
= 5 andN >
6. In Sects. 2 and 3 of this paper we study the exact solitary wave solutions of (4) forN
= 5. In Sect. 2, using an arbi- trary scalar field with a quite general constraint as a basic equation system, we solve the (D
+ 1)-dimen- sional5 model by a second order ordinary differ- ential equation (ODE). For a special constraint, the general solution of the ODE can be expressed sim- ply by an integration. The solitary wave solutions of the 5 model are discussd in an alternative way in Section 3. In Sect. 4, we give special solutions of the 8 model (v
1 =v
3 =v
5 =v
7 = 0 in (4)) by means of the5 model. In Sect. 5, we discuss some special solutions of a coupled scalar field model by means of the5model. The last section is a short summary.2. Base Equation Approach for the
5ModelFor the
5model, the motion eq. (4) becomes2
=v
2+v
32+v
43+v
54;
(5)where we have set
v
1 = 0 without loss of generality because we can make transformation! +c
byselecting the constant
c
appropriately. In order to get some interesting special solutions of (5), we introduce a basic equation system [4, 5]2 =
A
( );
(6)∂ ∂
D
X
i=1( xi)2;( t)2 =
B
( );
(7)where
A
A
( ) andB
B
( ) are arbitrary func- tions of . Now we suppose thatis only a function of . In other words, the space-time dependence of =( ) results from the auxiliary field , which is given by (6) and (7) with two arbitrary functionsA
and
B
. Using the basic eqs. (6) and (7) and the above ansatz, (5) becomes an ODE:A
+B
=v
2+v
32+v
43+v
54:
(8)Furthermore, after introducing
C
( ) = expn2ZB
(10)h
A
( 0);12dB
( 0)d 0
i
d 0
o
;
(9) the ODE is changed to
= (v
2+v
32+v
43+v
54)C
( ());
(10)where
and are related by =Zd
0p
B
( 0)C
( 0):
(11)Generally, to solve the ODE (10) is still quite difficult.
However, if we select
A
= 12dB
d
;
(12)then
C
( ) = 1, and (10) is reduced to =v
2+v
32+v
43+v
54;
=Zd
0p
B
( 0)
(13) with the general solution
=Zd
0q
C
1+v
202+23v
303+12v
404+25v
505+
C
2;
(14)where
C
1andC
2are two arbitrary integral constants.Now the remaining key problems are how to solve the basic eqs. (6) and (7) and to finish the integra- tion (14). For the first problem, the concrete solutions of (6) are dependent on the selection of the arbitrary function
B
. If we selected the functionB
as the poten- tial of the known NKG fields, say, sine-Gordon (sG), 4 or6 models, various types of special solutions have been given[6]. The simplest nontrivial selection of (6) and (7) reads2 =
2;
(15)∂ ∂ =
2 2:
(16)In this simple case we have
= ln:
(17)The simplest solution of (15) and (16) has the form
=
hXN =1
c
exp1
i1
;
(18)where
;
1andc ;
= 1;
2;:::;N
are arbitrary con- stants and =XDi=1
P
ix
i;! t;
(19)while
P
iand!
satisfy the conditions DX
i=1
P
iP
i0;! !
0 = 1; ;
0 = 1;
2;:::;N:
(20)3. Solitary Wave Solutions of the
5ModelGenerally, the integration of (14) can not be ex- pressed by simple functions. For the solitary wave solutions with some special parameters (
v
i) and inte- gration constant (C
1) we can take an alternative way proposed, in [7] to express (14) by some known func- tions.Equation (14) is equivalent to
2 =C
1+v
22+23v
33+12v
44+25v
55:
(21)To find the solitary wave solutions of (21), we take a
truncated series expansion of
around an extended singular manifold= 0, whileis determined by [7] =XKk=0
a
kk;
(22)where
a
k; k
= 0;
1;:::K
are arbitrary functions of.If the integer
K
is fixed, the possible expansions ofread
=2 3(XK;1)
p=0
b
pp (23)with
b
p; p
= 0;
1;:::
23(K
;1) being functions of.For simplicity, we take
K
= 4:
(24)Substituting (23) with (22) and (24) into (21) and putting to zero the coefficients of
k;k
= 0;:::
, wehave eleven complicated equations of the functions
a
0;a
1;a
2;a
3;a
4;b
0;b
1;b
2and the constantC
1. If theparameters
v
2;v
3;v
4andv
5satisfy the conditionsW
11232v
25v
24v
23;5400v
5v
44v
3+ 9504v
52v
2v
43; 35712
v
35v
2v
4v
3+ 31104v
45v
22 (25) + 675v
64;2048v
35v
33 6= 0and
(640
v
5v
33;150v
24v
23+ 675v
34v
2;3240v
3v
5v
4v
2+ 5832
v
52v
22) (;128v
52v
33;45v
44v
3+ 222v
32v
5v
24+ 27
v
5v
43v
2;504v
3v
52v
4v
2+ 216v
53v
22) = 0;
(26)we have an unique solution for
a
i;b
i andC
1 beingconstants:
a
4= 101p10v
5b
2b
2; a
3= 15b
1p10v
5b
2;
(27)a
2= 8p101v
5b
2(7v
5b
21+ 20v
5b
0b
2+ 5v
4b
2);
(28)a
1=b
18
q
10
v
5b
32(;
v
5b
21+ 20v
5b
0b
2+ 5v
4b
2);
(29)a
0= 1384
q
10
v
35b
52
9
v
52b
41;120v
52b
0b
21b
2; 30
v
4b
21b
2v
5+ 720v
52b
20b
22 (30) + 360v
4b
0b
22v
5;75v
42b
22+ 320v
3b
22v
5; C
1=;23v
3b
30;2
5
v
5b
50;v
2b
20+b
21a
20;1
2
v
4b
40;
(31)b
0 =;4b
2v
15W
51840
v
52b
2v
2v
44;9504v
53v
2v
34b
21+ 35712
v
54v
2v
4b
21v
3+ 98304v
45b
2v
2v
32; 31104
v
55v
22b
21;233856v
35b
2v
2v
24v
3+ 196992
v
45b
2v
22v
4+ 2048v
33v
45b
21 (32) + 5400v
52v
44v
3b
21;11232v
53v
42v
23b
21; 51200
v
35b
2v
4v
33;30600v
5b
2v
54v
3+ 81120
v
52b
2v
34v
23+ 3375b
2v
47;675v
46b
21v
5while
b
1 andb
2remain free.Now the remaining problem is to solve the ODE (22) with (27) - (30) and (32). According to the dif- ferent relations among the model parameters
v
2,v
3,v
4,v
5and the constantsb
1andb
2, there may be nine types of possiblesolutions:1.) If the quartic equation
4
X
k=0
a
kk = 0;
(33)a
i being given by (27) - (30), possesses four non- degenerate real roots, sayfc
1;c
2;c
3;c
4g, the general solution of (22) reads4
X
k=0
ln(
;c
k)Qj6=k(
c
j;c
k);a
4(;0) = 0;
(34)where
a
iare related toc
ibya
0=a
4=c
1c
2c
3c
4;
(35)a
1=a
4=;c
1c
2c
4;c
1c
2c
3;c
2c
3c
4;c
1c
3c
4;
(36)a
2=a
4=c
1c
2+c
2c
4+c
2c
3+c
1c
4+c
1c
3+c
3c
4;
(37)a
3=a
4=;c
1;c
2;c
3;c
4:
(38)ln(
;c
2)(
c
2;c
1)2(c
3;c
2);(c
2;2c
1+c
3) ln(;c
1)(;
c
1+c
2)2(;c
1+c
3)2 + ln(;c
3)(
c
2;c
3)(c
3;c
1)2+(c
2;c
1)(c
3;1c
1)(;c
1)+a
4(;0)= 0
:
(39)3.) If three of the four roots of (33) are equal,
c
3 =c
4 =c
1, the general solution of (22) becomes ln(;c
1)(;
c
1+c
2)3 +(c
2;c
1)12(c
1;) ;2(c
2;c
11)(;c
1)2 ;ln(;c
2)(
c
2;c
1)3 +a
4(;0) = 0:
(40)4.) If all four real roots of (33) are degenerate, the function
becomes =;(3a
4(;10))1=3 +c
1:
(41)5.) If there are two sets of degenerate roots of (33) say,
c
4 =c
1; c
3=c
2, the solution of (22) has the formc
1;2c
2ln;c
1 ;c
2 +;1c
1 +;1c
2 +a
4(c
2;c
1)2(;0) = 0:
(42)6.) If two of the roots are conjugate complex and the other two are nondegenerate real, we have
d
1c
3+ 2c
3c
4+ 2d
2+d
21+d
1c
4(
d
1c
3+c
23;d
2)(d
2;d
1c
4;c
24)q
;4
d
2;d
21arctan
2
+d
1q
;4
d
2;d
21
;
ln(
;c
4)(
c
3;c
4)(d
2;d
1c
4;c
24) + ln(;c
3)(
c
3;c
4)(d
2;d
1c
3;c
23);(
c
3+d
1+c
4) ln(2+d
1;d
2)(
d
2;d
1c
4;c
24)(;d
1c
3;c
23+d
2) +a
4(;0) = 0:
(43)where
d
21+ 4d
2<
0;
(44)and the
a
iare linked withfd
1;d
2gandfc
3;c
4gbya
0=a
4=;d
2c
3c
4;
(45)a
1=a
4=d
2c
4+d
2c
3+d
1c
3c
4;
(46)a
2=a
4=d
2;d
1(c
4+c
3) +c
3c
4;
(47)a
3=a
4=d
1;c
3;c
4:
(48)7.) If two of the roots are conjugate complex and the other two are degenerate real (
c
4=c
3in (45) - (48)), the related solution of (22) reads;
2
d
1c
3+d
21+ 2c
23+ 2d
2(
d
2;d
1c
3;c
23)2q
;4
d
2;d
21arctan 2
+d
1q
;4
d
2;d
21+(2
c
3+d
1) ln(;c
3)(
d
2;d
1c
3;c
23)2;
(2
c
3+d
1) ln(2+d
1;d
2)2(
d
2;d
1c
3;c
23)2;
1
(
d
2;d
1c
3;c
23)(;c
3)+a
4(;0) = 0:
(49)2.) If (33) possesses four real roots and two roots are degenerate, say,
c
4=c
1, then the solution of (22) has the form8.) If (33) possesses two sets of nondegenerate conjugate complex roots, the general solution of (22) becomes (2
d
4+d
1d
3;d
21;2d
2)q
;4
d
2;d
21arctan 2
+d
1q
;4
d
2;d
21+1
2(
d
3;d
1) ln(2+d
1;d
2)+ 2
d
2;2d
4;d
23+d
1d
3q
;4
d
4;d
23arctan 2
+d
3q
;4
d
4;d
23;
1
2(
d
3;d
1) ln(2+d
3;d
4) (50)+ (
d
24;2d
2d
4;d
23d
2+d
1d
3d
4+d
22;d
21d
4+d
1d
2d
3)a
4(;0) = 0;
where
d
21+ 4d
2<
0; d
23+ 4d
4<
0;
(51)and
a
iandd
iare related bya
0=a
4=d
2d
4; a
1=a
4=;d
2d
3;d
1d
4;
(52)a
2=a
4=;d
2;d
4+d
3d
1; a
3=a
4 =d
1+d
3:
(53)9.) Finally, if
d
1=d
3andd
2=d
4in (52) and (53), thefunction should satisfy the relation 2+d
1(
2+d
1;d
2)+q;4d
42;d
21arctan 2
+d
1q
;4
d
2;d
21+
a
4(;4d
2;d
21)(;0) = 0:
(54)If the condition (25) is not satisfied, we have five further possible cases:
(i) If the parameters
v
2;v
3andC
1are taken asC
1= 0; v
3 = 15v
2464
v
5; v
2 = 0;
(55)then the
function is given by 1p
10(2
b
2+b
1);p
v
55
p
;
b
2v
4 arctanhp
2
v
5(2b
2+b
1)p
;5
b
2v
4
+
v
432
p
v
5b
2(;0) = 0;
(56)and the corresponding solution of the
5model with (55) reads =b
214
b
2 +b
1+b
22 (57)with three arbitrary constants
b
1; b
2and0.(ii) If the parameters
v
2;v
3andC
1are given byv
2 =;9v
4332
v
52; v
3 =;3v
4216
v
5; C
1 = 27v
45640
v
54;
(58)the related
obeys the form;
1 2
b
2+b
1 ;p
v
5p
5
b
2v
4arctanh
;
p
v
5(2b
2+b
1)p
5
b
2v
4
+
p
10
v
416
p
v
5b
2(;0) = 0;
(59)and
has the form = ;3v
4b
2+v
5b
214
v
5b
2 +b
1+b
22:
(60)(iii) The constraints on the parameters for the third special solution read
C
1= 0; v
2 = 25v
43864
v
52; v
3 = 5v
4216
v
5;
(61)while the corresponding
andare given by; p
10
100(2
b
2+b
1); 150v
5500
p
b
2v
4arctanh
; p
15
v
5(2b
2+b
1)p
b
2v
4
=;
v
4480
p
v
5b
2(;0) (62)and
= 3v
5b
21;5v
4b
212
v
5b
2 +b
1+b
22;
(63)respectively.
(iv) For the fourth special solution we have constraints on the parameters
v
2; v
3 and the first integration constantC
1:v
3 =3(p
5 + 1)
v
4232
v
5; v
2 =;v
34(29p
5;60) 16
v
25(7p
5;27)
; C
1= 0:
(64)The related
function is given bys
1
p
5+ 3 arctanh
q
v
5(5 + 3p5)(2b
2+b
1)p
10
v
4b
2
; p
2 arctanh
5p
v
5(2b
2+b
1)q
p
5
v
4b
2
;
v
43=2(p
5;1)(
;0)51=4
v
5 = 0;
(65)and the solution of the
equation is =v
5b
21;p
5
v
4b
24
v
5b
2 +b
1+b
22:
(66)(v) The final special solution has the form
=v
5b
21;2v
4b
2+p5v
4b
24
v
5b
2 +b
1+b
22;
(67)where the function
is related toimplicitly bys
5 +
p
5 2 arctanh
q
v
5(5 + 3p5)(2b
2+b
1)p
;
b
2v
4
;51=4arctanh
v
5(2b
2+b
1)q
; p
5
b
2v
4
+
p
10(
p
5;1)
v
4p;v
4(;0)32
v
5 = 0;
(68)while the constrained conditions for the model parameters
v
2; v
3and the integral constantC
1arev
3= 3(p
5 + 1)
v
4232
v
5; v
2 =;v
34(29p
5;60) 16
v
52(7p
5;27)
; C
1=v
45(4525655302009p
5;10119671799191) 1280
v
54(204037668464p
5;456248189781)
:
(69)The parameters
b
1;b
2, and0in the above special cases are all arbitrary constants.4. Solutions of the
8ModelIf we take
N
= 8; v
1=v
3 =v
5 =v
7 = 0 in (4), we get the8model2
=V
2+V
43+V
65+V
87;
(70)where we have written
v
i;i
= 2;
4;
6;
8 and asV
i and for convenience later. By means of the basic equation approach proposed in Sect. 2, some types of special solutions of the8model (70) can be obtained by solving a similar ODE:A
+B
=V
2+V
43+V
65+V
87;
(71)where the auxilialy field and the arbitrary functions
A
=A
( ) andB
=B
( ) are related by (6) and (7).Especially, if
A
andB
satisfy (12), we have 2 =V
22+12V
44+13V
66+14V
88 (72);
c
12
24
V
2+ 18V
4c
+ 16V
6c
2+ 15V
8c
3;
where
is given in (13) andc
is an arbitrary integration constant.It is interesting and straightforword to see that the solutions of (72) can be obtained simply by using the transformation
=pc
+;
(73)where
is a solution of the5model (21) withv
5 =52V
8;
(74)v
4 =83V
6+ 10cV
8;
(75)v
3 = 3V
4+ 16cV
6+ 15c
2V
8;
(76)v
2 = 4V
2+ 6cV
4+ 8c
2V
6+ 10c
3V
8;
(77)C
1= 4c
2(V
2+cV
4+c
2V
6+c
3V
8):
(78)So, all the special solutions obtained in Sects. 2 and 3 can be transformed to those of the
8model.5. Some Special Solutions of Coupled Scalar Fields Actually, the special solutions of the
5model ob-tained in sections 2 and 3 may be used to get exact
solutions of other physically significant models, say the coupled nonlinear scalar field model
2
f
=A
1f
+A
2f
3+A
3fg
2;
(79)2
g
=B
1g
+B
2g
3+B
3gf
2 (80)which appears in some physical fields such as parti- cle physics and field theory [8] and condense matter physics [9].
After some complicated but direct calculations, we can change the solutions of the
5model obtained in Sections 2 and 3 to those of the coupled scalar fieldsf
andg
for some special parametersA
iandB
i. Here are five possible examples.Case 1
If the parametes
B
1 andB
3 are related to other parameters byB
1=;A
1B
211
B
2;6A
3; B
3= 59A
2;
(81)a special type of the coupled scalar fields reads
g
=;
(82)f
2= 950
A
2pA
1
3(6
A
3;11B
2)(A
3;5B
2)(15
B
2;8A
3)1=23+259A
2(15B
2;8A
3)2; 9 25
A
2s
3
A
1(5B
2;A
3)(15B
2;8A
3)(11
B
2;6A
3);
27 25
(15
B
2;8A
3)A
1(11
B
2;6A
3)A
2;
(83)where
is an arbitrary solution of the5model given in Sect. 3, or more generally by (14) withC
1 =24(15B
2;8A
3)A
2125(11
B
2;6A
3)2;
(84)v
3=p
3
A
1(5B
2;A
3)(15B
2;8A
3)5
p
(11