the riccati equation method with variable expansion coefficients. i. solving the burgers equation

7/22/2019 The Riccati equation method with variable expansion coefficients. I. Solving the Burgers equation

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The Riccati equation methodwith variable expansion coefficients.

I. Solving the Burgers equation

Solomon M. Antoniou

SKEMSYSScientific Knowledge Engineering

and Management Systems

37 oliatsou Street, Corinthos 20100, [email protected]


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Abstract

We introduce the Riccati equation method with variable expansion coefficients.

This method is used to find travelling wave solutions to the Burgers equation

0uauuu xxxt =+ . The important new feature of the method lies in the fact that

the ( dependent) coefficients A and B of the Riccati equation 2BYAY +=

satisfy their own nonlinear ODEs, which can be further solved by one of the

known methods, like Jacobi's elliptic equation method, the )G/G( expansion

method, the projective Riccati equation method, and numerous other methods, like

the reduction to Ermakov's equation. We also introduce the )G/G( expansion

method with variable expansion coefficients. The second method presupposes

Riccati's equation method with variable expansion coefficients. The solutions

obtained by the )G/G( expansion method are expressed in terms of

hypergeometric function even in the simplest case. More complicated cases might

require hypercomputing facilities with symbolic capabilities. Both methods are

new, computationally demanding, the use of which reveals a rich number of

solutions not known previously.

Keywords: Riccati method, nonlinear evolution equations, traveling wave

solutions, Burgers equation, exact solutions.


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1. Introduction.

Nonlinear partial differential equations arise in a number of areas of Mathematics

and Physics in an attempt to model physical processes, like Chemical Kinetics

(Gray and Scott [98]), Fluid Mechanics (Whitham [204]), or biological processes

like Population Dynamics (Murray [154]). In the recent past, there is a number of

new methods which have been invented in solving these equations. Among the

new methods are

(1)the inverse scattering transform method(Ablowitz, Kaup, Newell andSegur [17], Ablowitz and Clarkson [18], Ablowitz and Segur [19], Drazin

and Johnson [50], Gardner et. al. [90]-[92], Gelfand and Levitan [93], Kay

and Moses [121], Lax [132], Marchenko [147], Miura [149]-[152],

Novikov, Manakov, Pitaevskii and Zakharov [158], Ramm [170], Wadati

[187], Wadati, Sanuki and Konno [188] )

(2)Hirotas bilinear method(Hirota [111] and [112], Hientarinen [110],Hereman and Zhuang [109] )

(3)theAlgebro-Geometricapproach (Belokolos, Bobenko, Enolskii, Its andMatveev [32] and references)

(4) theBcklund transformation method(Miura [149], Rogers and Shadwick

[172])

(5)thetanh-coth method(Malfliet [142], Malfliet and Hereman [144] and[145], Bekir and Cevikel [31], Wazwaz [199], Fan [72], Abdel-All, Razek

and Seddeek [3])(6)thesn-cn method(Baldwin et al [27])(7)theF-expansion method(Wang and Li [192], Abdou [6] and [7])(8)theJacobi elliptic function method(Abbott, Parkes and Duffy [2], Chen

and Wang [44], Fan and Zhang [82], Liu and Li [135] and [136], Liu, Fu,

Liu and Zhao [137] and [138], Lu and Shi [141], Xiang [206], Yan [209]

and [210],Yang [214])


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(9)theRiccati equation method(Yan and Zhang [213])(10) the Weierstrass elliptic function method(Kudryashov [124])(11) theExp-function method(He and Wu [104], Naher, Abdullah and Akbar

[155] and [156], Mohyuddin, Noor and Noor [153], Yildirim and Pinar [218],

Aslan [26], Bekir and Boz [30])

(11)the G/G expansion method(Abazari and Abazari [1], Borhanibar andMoghanlu [37], Borhanibar and Abazari [38], Elagan, Sayed and Hamed

[52], Feng, Li and Wan [84], Jabbari, Kheiri and Bekir [118], Naher,

Abdullah and Akbar [157], Ozis and Aslan [163], Wang, Li and Zhang

[194], Zayed [220], Zayed and Gepreel [222] )

(12)thehomogeneous balance method (Fan [71], Wang, Zhou and Li [191],Yan and Zhang [213])

(13)thedirect algebraic method(Soliman and Abdou [180])(14)thebasic equation method(Kudryashov [125] ) and its variants, like the

simplest equation method (Vitanov [184]-[186], Yefimova [217])

(15)the Cole-Hopftransformation method (Cole [48], Hopf [115], Salas andGomez [174])

(16)theAdomian decomposition method(Adomian [21] and [22], Abdou [4],Cherruault [46], Cherruault and Adomian [47], El-Wakil, Abdou and

Elhanbaly [58], El-Wakil and Abdou [59], Rach, Baghdasarian and Adomian

[169], Wazwaz [196] and [197])

(17)thePainleve truncated method(Weiss, Tabor and Carnevale [202] and[203], Zhang, Wu and Lou [223], Estevez and Gordoa [69] and [70])

(18)thehomotopy perturbation method(Taghizadeh, Akbari andGhelichzadeh [183], Yahya et al [207], Liao [134])

(19)thereduced differential transformation method(Keskin and Oturanc[122], Arora, Siddiqui and Singh [25])

(20)theLie symmetry method(Lie point symmetries, potential symmetries,nonclassical symmetries, the direct method) (Bluman and Kumei [33],


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Bluman and Anco [34], Bluman and Cole [35], Bluman, Cheviakov and

Anco [36], Cantwell [40], Hydon [116], Olver [160], Ovsiannikov [162],

Schwarz [175], Steeb [181], Stephani [182])

(21)the variational iteration method(He [102], Abdou [8], Wazwaz [198])(22)thefirst integral method(Raslan [171], Feng [85], Feng and Wang [86])(23)the integral bifurcation method(Rui, Xie, Long and He [173] and

references)

The implementation of most of these methods was made possible only using

Symbolic Languages (Grabmeier, Kaltofen and Weispfenning [97]) like

Mathematica (Baldwin et al [28] and [29], Gktaand Hereman [96]), Macsyma

(Hereman [106], Hereman and Takaoka [108], Hereman and Zhuang [109]),

Maple, etc. In many cases some special elimination methods (Wang [189]) and

computational algorithms were also used (Wu [205]).

In this paper we introduce the Riccati equation method with variable expansion

coefficients and we find traveling wave solutions of the Burgers equation.The paper is organized as follows: In Section 2 we introduce the basic ingredients

of the method used. In Section 3 we consider Burgers equation and Riccatis

equation method of solution where the expansion coefficients are not constants,

i.e. they depend on the variable . In this section we consider first the Riccati

expansion method and then the extended Riccati method. In the first case we find

closed-form solutions for )(u expressed in terms of hyperbolic tangent functions.

In the second case we establish that A and B are proportional each other. We

also find a closed form expression for the function )(u expressed in terms of

)(A . The function )(A is proved to satisfy a second order nonlinear ordinary

differential equation solved in Appendices A, B and C. In Section 4 we consider

the various forms the function )(u takes using the various solutions found in the


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Appendices. In Section 5 we consider the )G/G( expansion method with

variable coefficients as a method of solution of the Burgers equation.

2. The Method.

We consider an evolution equation of the general form

),u,u,u(Gu xxxt L= or ),u,u,u(Gu xxxtt L= (2.1)

where u is a sufficiently smooth function. We introduce a new variable given

by

)tx(k = (2.2)

where k and are constants. Changing variables, since

)(u)k(u t = , )(uku x = , )(uku2

xx = , (2.3)

equations (2.1) become ordinary nonlinear differential equations

=

L,

d

udk,

d

duk,uG

d

du)k(

2

22 (2.4)

or

=

L,

d

udk,

d

duk,uG

d

udk

2

22

2

22 (2.5)

Solutions of evolution equations depending on )tx(k are calledtraveling

wavesolutions.

Equations (2.4) or (2.5) will be solved considering expansions of the form

=

=n

0k

kkYa)(u (2.6)

or

==

+=n

0kkk

n

0k

kk

Y

bYa)(u (2.7)

where all the expansion coefficients depend on the variable ,

)(aa kk , )(bb kk for every n,,2,1,0k L=


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contrary to the previously considered research where the expansion coefficients

were considered as constants. The function )(YY satisfy Riccatis

equations

2BYA)(Y += (2.8)

or

2YRYQP)(Y ++= (2.9)

where again all coefficients A, B and P, Q, R depend on the variable .

In solving equations (2.4) or (2.5), we consider the expansions (2.6) or (2.7) and

then we balance the nonlinear term with the highest derivative term of the function

)(u , which determines n (the number of the expansion terms). Equating similar

powers of the function )(Y , we can determine the various coefficients and thus

find the solution of the equation considered.

Our motivation in using Riccati's method with variable expansion coefficients,

was to obtain traveling wave solutions expressed in terms of special functions

(Hypergeometric, Bessel, etc). Introducing variable expansion coefficients, weobtain variable coefficients A and B of the Riccati equation (2.8). Converting

Riccati equation into a second order linear ordinary differential equation with

variable coefficients, its solutions might be expressed in terms of special

functions.

3. The Burgers equation and its solutions.

Burgers equation (Burgers [39]) was introduced in an attempt to model a theory of

turbulence. There is a number of explicit solutions based on Lie symmetries, either

classical (Antoniou [24], Liu, Li and Zhang [139], Ouhadan, Mekkaoui and El

Kinani [161]), or nonclassical (Gandarias [88], Mansfield [146], Qin, Mei and Xu

[168]). Some solutions have also been found using the tanh-method (Hereman and

Malfliet [107]) or some of its descendants, like the Exp-Function Method (Ebaid

[51]) and the Homotopy perturbation method (Taghizadeh, Akbari and

Ghelichzadeh [183]). On the other hand Burgers equation can be reduced to the


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heat equation through the famous Cole-Hopf transformation (Cole [48], Hopf

[115], Olver [160] and Mansfield [146]).

We consider next Burgers equation in the form

0uauuu xxxt =+ (3.1)

and try to find traveling wave solutions of this equation. We introduce a new

variable given by

)tx(k = (3.2)

where k and are constants. Changing variables, since

)(u)k(u t = , )(uku x = and )(uku2

xx =

equation (3.1) becomes an ordinary nonlinear differential equation

0)(uka)(u)(uk)(u)k( 2 =+ (3.3)

Integrating once the above equation, we obtain

02 c)(uka)(u

2

1)(u)( =+ (3.4)

where 0c is a constant.

We consider the following cases in solving equation (3.4).

3.1. First Case. The Riccati Method.

We consider the solution of equation (3.4) to be of the form

=

=n

0k

kkYa)(u (3.5)

where )(aa kk for every n,,2,1,0k L= .

3.1.1. Method I. We first consider that )(YY satisfies Riccatis equation

2BYA)(Y += (3.6)

where A and B depend on the variable : )(AA and )(BB .

We substitute (3.5) into (3.4) and take into account Riccatis equation (3.6). We

then balance the first order derivative term with that of the nonlinear term. The


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order of the nonlinear term )(u2 is n2 and that of the first order derivative term

is 1n2)1n( +=+ . We thus get the equation 1nn2 += from which we obtain1n= . Therefore

Yaa)(u 10+= (3.7)

We first calculate )(u from the above equation, taking into account Riccatis

equation (3.6) and that the various coefficients 0a , 1a , A and B depend on .

We find

2

1110

YBaYa)Aaa()(u +++= (3.8)

Because of (3.7) and (3.8), equation (3.4) becomes, arranged in powers of Y,

Y}a)a(aka{cakaaa2

1Aaka 101000

201 +

+

0Y)Bka2a(a2

1 211 = (3.9)

We now have to determine the coefficients 0a , 1a , A and B from the above

equation, equating to zero the coefficients of Y.

Equating to zero the coefficient of 2Y , we determine the coefficient 1a :

Bka2a1= (3.10)

Equate to zero the coefficient of Y,

0a)a(aka 101 =

we can determine the coefficient 0a :

BBka

aakaa

1

10 +=+= (3.11)

Equate now to zero the constant term:

0cakaaa2

1Aaka 000

201 =+

(3.12)

This equation takes on the form


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Dka2

)2a(aaAa 0001 +

+= (3.13)

where D is a constant (ka

cD 0 ).

Equation (3.13) - because of (3.10) and (3.11) - can be written as

Dka2

B

Bka

B

Bka

B

BkaA)Bka2( +

+

+

+

=

or

MB

B

4

1

B

B

2

1AB

2

+

+

= (3.14)

where M is another constant (22

20

ka4

c2M

).

What we need, in order to determine )(u , is 0a , 1a and Y (see equation (3.7)).

On the other hand, Y is determined by Riccati's equation 2BYA)(Y += once

A and B are known quantities. We thus need four quantities: 0a , 1a , A and B.

We do have however three equations only, (3.10), (3.11) and (3.14). There is one

equation missing. However there is no need for an extra equation, as we shall see

below. First of all, Riccatis equation 2BYA)(Y += under the substitution

)(w

)(w

)(B

1)(Y

= (3.15)

takes on the form of a linear second order ordinary differential equation

0)(wBA)(wB

B)(w =+

(3.16)

with unknown function )(w .

We now transform equation (3.16) under the substitution

)(y)(B)(w = (3.17)


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The derivatives of the function )(u transform as

+= yBBy2

1

B

1

)(w (3.18)

++

= yByBy

B

)B(

4

1B

2

1

B

1)(w

2

(3.19)

Equation (3.16), because of (3.14) and (3.17)-(3.19), takes on the form

+

+

++

yBBy

2

1

B

B

B

1yByBy

B

)B(

4

1B

2

1

B

12

0yBMB

B

4

1

B

B

2

12

=

+

+

+

which gives us upon multiplying by B , the simple equation (some miraculous

cancellations take place)

0)(yM)(y =+ (3.20)

Considering various forms of the constant M, we can determine )(y and then

)(w from (3.16). For 2nM= , equation (3.20) admits the general solution

)nsin(C)ncos(C)(y 21 += , whereas for2nM = admits the general solution

)nsinh(C)ncosh(C)(y 21 += . If 0M= then 21 CC)(y += .

The function )(Y can be determined from (3.15) and then )(u from (3.7),

calculating first 1a and 0a from (3.10) and (3.11) respectively.

We thus find that

+

+=+=

)(w

)(w

)(B

1Bka2

B

BkaYaa)(u 10

or

)(w

)(wka2

B

Bka)(u

+= (3.21)

Since )(y)(B)(w = , we obtain


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)(y

)(y

)(B

)(B

2

1

)(w

)(w

+

=

Using the above expression, we obtain from (3.21) that

)(y

)(yka2)(u

= (3.22)

This is quite aremarkable result :no matter what the coefficients of Riccatis

equation are, we arrive at the equation (3.22) where )(y satisfies equation (3.20).

We thus obtain the following three solutions, depending on the values of the

constant M (which appears in equation (3.20))

)ntan(C

)ntan(C1nka2)(u

+

= for 2nM= (3.23)

)ntanh(C

)ntanh(C1nka2)(u

+

+= for 2nM = (3.24)

+=

C1

Cka2)(u for 0M= (3.25)

where 21 C/CC = , with 1C and 2C are the two arbitrary constants of the general

solution of the differential equation (3.20).

3.1.2. Method II. We consider instead of (3.6), the equation

2YRYQPY ++= (3.26)

where P, Q and R are functions depending on .

In this case considering again ( )(aa 00 = , )(aa 11 = )

Yaa)(u 10+= (3.27)

we find

211110 RYaY)Qaa()Paa()(u ++++= (3.28)

taking into account equation (3.26). Equation (3.4) then becomes

+++ Y)}Qaa(kaa)a{(Y)Rka2a(a2

11110

211


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01000 cPakaaka)2a(a2

1=+ (3.29)

Equating to zero the coefficients of the powers of Y, we obtain the following

system of equations

Rka2a1= (3.30)

0)Qaa(kaa)a( 1110 =+ (3.31)

01000 cPakaaka)2a(a2

1= (3.32)

Equation (3.31), because of (3.30), takes on the form

++=

R

RQkaa0 (3.33)

From (3.32) we obtain

Dka2

)2a(aaPa 0001 +

+= (D is a constant)

Using (3.30) and (3.33), we obtain from the above equation

MRRQ

41

RRQ

21PR

2

+

++

+= (3.34)

Using the usual substitution

w

w

R

1Y

= (3.35)

equation (3.26) transforms into

0wPRw

R

RQw =+

+ (3.36)

Substituting the product PR in the above equation given by (3.34), we arrive at

0wMR

RQ

4

1

R

RQ

2

1w

R

RQw

2

=

+

++

++

+ (3.37)

We introduce another function f by


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+=

R

RQ

f

f (3.38)

Equation (3.37) then becomes

0wMf

f

4

1

f

f

2

1w

f

fw

2

=

+

+

+

(3.39)

Under the substitution

yfw= (3.40)

equation (3.39) takes on the form

0yMy =+ (3.41)

We thus have obtained

++=

R

RQkaa0 , Rka2a1= and

w

w

R

1Y

=

where

yfw= with

+=

R

RQ

f

f and 0yMy =+ .

Therefore

+

+=

w

w

R

1Rka2

f

fka)(u

or

w

wka2

f

fka)(u

+= (3.42)

Since, using (3.40),

y

y

f

f

2

1

w

w +

=

(3.43)

we obtain from (3.39) that

y

yka2)(u

= (3.44)


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We thus arrive again at the same result: no matter what the coefficients of

Riccatis equation (3.26) are, we arrive at the equation (3.41), where )(y satisfies

equation (3.41), which are equations (3.22) and (3.20) respectively of Method I.

Conclusion. Burgers equation 0uauuu xxxt =+ under the substitution

)tx(k = transforms into 02 c)(uka)(u

2

1)(u)( =+ . Considering

the expansion Yaa)(u 10+= where Y satisfies either one of the Riccati

equations 2BYA)(Y += or 2YRYQPY ++= , where all the coefficients

depend on , we obtain the three solutions (3.23)-(3.25), depending on the values

of the constant M.

3.1.3. Method III. Equation (3.4) is a Riccati equation by itself and under the

substitutionw

w)ka2(u

= transforms into the linear second order equation

0wka2

cw

kaw 0 =

+ (3.45)

with constant coefficients. Let22

02

ka

cka2+ be the discriminant of the

auxiliary to (3.45) equation. (I) If 202 ncka2 =+ , n real, then

+

=

ka2exp

ka2

nsinC

ka2

ncosCw 21 and thus

+

=

ka2

ntanC

ka2

ntanC1

nu , 21 C/CC = . (II) If 202 ncka2 =+ , n real,

then

+

=

ka2exp

ka2

nsinhC

ka2

ncoshCw 21


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and thus

+

+

=

ka2

ntanhC

ka2

ntanhC1

nu , 21 C/CC = . (III) If 0cka2 02

=+ ,

then

+=

ka2exp)CC(w 12 and thus

+=

C1

Cka2u ,

21 C/CC = . We thus see that we derive the same set of solutions as in the

previous two sections: it suffices to perform the rescaling n)ka2(n .

The reader might wonder why we have considered all these three cases instead toconsider only the last one, since we get identical solutions. The reason is that the

first two methods will be used as a prelude to the extended Riccati equation

method, considered next, where we obtain quite different solutions.

3.2. Second Case. The Extended Riccati Method.

In this case we consider the expansion

==

+=n

1k k

kn

0k

kk

Y

bYa)(u

and balance the first order derivative term with the second order nonlinear term of

(3.4). We then find 1n= and thus

Y

bYaa)(u 110 ++= (3.46)

where again require all the coefficients 0a , 1a and 1b depend on , and Y

satisfies Riccatis equation2

BYAY += . From equation (3.42) we obtain (taking

into account 2BYAY += )

2112

11110Y

bA

Y

bBYaYa)BbAaa()(u

++++= (3.47)

Therefore equation (3.4), under the substitution (3.46) and (3.47), becomes

++++ )BbAaa(ka)ba2a(2

1a)( 11011

200


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+++ 211110 Y)Bka2a(a2

1Y}akaa)a{(

0211110 c

Y

)Aka2a(b

2

1

Y

bkab)a(=

++

+ (3.48)

Equating the coefficients of Y to zero, obtain a system of differential equations

from which we can determine the various coefficients. Equating to zero the

coefficients of 2Y and 2Y/1 , we find that

Bka2a1= and Aka2b1 = (3.49)

respectively. Equating to zero the coefficients of Y and Y/1 and taking into

account the values of 1a and 1b , we find

B

Bkaa0

+= and

A

Akaa0

+= (3.50)

respectively. The above two equations imply that A and B are proportional each

other:

AsB 2 = (3.51)

with 2s being the proportionality factor with s real. We do not consider the

case AsB 2 = , because this choice leads to imaginary type of solutions (see

equation (3.58) below).

The zeroth order term appearing in (3.48) takes on the form, taking into account

the values of 1a and 1b :

2

)2a(aakaAB)ka8( 000

22 +=

The above equation, after introducing the value of 0a expressed in terms of A,

takes on the form

22

m2A

A

2

1

A

ABA8

+

= (3.52)

where m is defined by


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ka2

1m

= , 0m> (3.53)

From equation (3.52), since AsB 2 = , we obtain the following second order

nonlinear ordinary differential equation

0)(As8)(Am2))(A(2

3)(A)(A 42222 =+ (3.54)

Equation (3.54) can be solved using a number of methods (the auxiliary equation

method, the G/G expansion method, the reduction to Ermakovs equation, the

projective Riccati equation expansion method, etc.). This has been done in

Appendix A, where we have found a considerable number of different solutions.

We now have to determine the function )(w from the equation (3.16)

0)(wBA)(wB

B)(w =+

(3.55)

This equation written in terms of A, using (3.47), takes on the form

0)(wAs)(w

A

A)(w 22 =

(3.56)

The previous equation can be written as (see also the Notebelow, for another

method of solution) 0wAsA

Aw

A

w 22

=

which is equivalent to

0wAsA

w 2 =

(3.57)

Multiplying byA

w, we obtain 0wws

A

w

A

w 2 =

which is equivalent to

0)w(sA

w 222

=

and from this by integration

0wsA

w 222

=

(3.58)

where we have put the constant of integration equal to zero. We thus get


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)(As)(w

)(w=

(3.59)

(Note. Another method of solution of equation (3.57) would be the following:

Under the substitutionA

w= , since

2A

wA

A

w

= , equation (3.57) takes on the

form 0ww

s2 =

, which is equivalent to 0)w(s)( 222 = and by

integration (we have put the integration constant equals to zero) 0ws 222 =

and then )(ws)( = , i.e. )(wsA

w

=

, which is (3.59)).

Therefore

s

1

)(w

)(w

)(As

1

)(w

)(w

)(B

1)(Y

2 =

=

= (3.60)

We then obtain the following expression for the function )(u using (3.46), (3.50),

(3.51) and (3.60):

)(Aska4)(A)(Aka)(u +=

(3.61)

We thus arrive at the following Theorem:

Theorem. Burgers equation 0uauuu xxxt =+ under the substitution

)tx(k = transforms into 02 c)(uka)(u

2

1)(u)( =+ . Considering

the expansionY

bYaa)(u 110 ++= , where Y satisfies the Riccati equation

2YBA)(Y += , with all the coefficients 0a , 1a , 1b and A, B depending on ,

we obtain that AsB 2 = and )(u given by the formula

)(Aska4)(A

)(Aka)(u

+= , where )(A satisfies the differential equation

0)(As8)(Am2))(A(2

3)(A)(A

42222 =+


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4. The final solutions.

Using the various solutions of the function )(A calculated in Appendices A, B

and C, we are able to find the different forms of the function )(u from equation

(3.61).

4.I. First Family of Solutions.

Using (A.10) for )(A , we obtain from (3.61)

)]mtanh()Cmaa()maCa[(])mtanh(C[

)]m(tanh1[maka)C1()(u

1010

221

2

++

+=

+

+

+

)mtanh(C

1)mtanh(Cmaaska4 10

where the coefficients 0a and 1a satisfy the relations (four different

combinations)2

221

s16a

= ,

2

220

s16

ma = .

(4.Ia)For

s4

a1

= and

s4

ma0 = , we obtain

)]mtanh(1[])mtanh(C[

)]m(tanh1[mka)C1()(u

2

++

++=

+

+

)mtanh(C

1)mtanh(C1mka (4.1)

(4.Ib)Fors4

a1

= ands4

ma0 = , we obtain



2

++

=

+

++

)mtanh(C

1)mtanh(C1mka (4.2)

(4.Ic)Fors4

a1

= ands4

ma0 = , we obtain


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21



2

++

=

+

++

)mtanh(C

1)mtanh(C1mkam (4.3)

(4.Id)Fors4

a1

= ands4

ma0 = , we obtain



2

+

+=

+++)mtanh(C1)mtanh(C1mka (4.4)

4.II. Second Family of Solutions.


+

+

+=

)(H

)(Faa)(Ha

)(FDaa

am4exp)}(Fa)(Hma4{m4

ka)(u

102

0

00

120

231

2

+

)(H

)(Faaska4 10

where2

21

s16

1a = and

2

220

s4

ma = (four different combinations).

(4.IIa) For s4

1

a1= and s2

m

a0= , we obtain

+

+

+=m

m2m22

e)(H)(H

)(Fm2

e)(H

)(FDs2e

)(H

)(Fs8mm

s

ka2)(u


22/93

22

+

)(H

)(Fm2ka (4.5)

where = m22

es

mD)(F and ++= m2e

s2

mDE)(H (E, D constants).

Since

)m(tanhDs2

mED

s2

mE

)m(tanhs

mD

s

mD

)(H

)(F

22

++

++

++

=

equation (4.5) can be written as

+=s

ka2)(u

)(Z)(Z

)(Xm2

)mtanh()(Z

)(X)Dm4(s2m

)(Z

)(X)Dm4(s2m 223223

+

++

+

+

)(Z

)(Xm2ka (4.6)

where

)m(tanhs

mD

s

mD)(X

22

++

= (4.7)

)m(tanhD

s2

mED

s2

mE)(Z

++

++= (4.8)

(4.IIb) Fors4

1a1= and

s2

ma0 = , we obtain

+

+=m

m2m22

e)(H)(H

)(Fm2

e)(H

)(FDs2e

)(H

)(Fs8mm

s

ka2)(u


23/93

23

+

)(H

)(Fm2ka (4.9)

where = m22

es

mD)(F and += m2e

s2


Since

)m(tanhDs2

mED

s2

mE

)m(tanhs

mD

s

mD

)(H

)(F

22

++

+

+

=


+=s

ka2)(u

)(Z)(Z

)(Xm2

)mtanh()(Z

)(X)Dm4(s2m

)(Z

)(X)Dm4(s2m 223223

++

+

)(Z

)(Xm2ka (4.10)

where

)m(tanhs

mD

s

mD)(X

22

+

= (4.11)

)m(tanhD

s2

mED

s2

mE)(Z

++

+= (4.12)

(4.IIc) Fors4

1a1 = and

s2

ma0 = , we obtain

+

+=m

m2m22

e)(H)(H

)(Fm2

e)(H

)(FDs2e

)(H

)(Fs8mm

s

ka2)(u


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24

)(H

)(Fm2ka (4.13)

where += m22

es

mD)(F and ++= m2e

s2


Since

)m(tanhDs2

mED

s2

mE

)m(tanhs

mD

s

mD

)(H

)(F

22

+

++

+

=


=s

ka2)(u

)(Z)(Z

)(Xm2

)mtanh()(Z

)(X)Dm4(s2m

)(Z

)(X)Dm4(s2m 223223

++

)(Z

)(Xm2ka (4.14)

where

)m(tanhs

mD

s

mD)(X

22

+= (4.15)

)m(tanhD

s2

mED

s2

mE)(Z

+

++= (4.16)

(4.IId) Fors4

1a1 = and

s2

ma0 = , we obtain

+

+

=m

m2m22

e)(H)(H

)(Fm2

e)(H

)(FDs2e

)(H

)(Fs8mm

s

ka2)(u


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25

+

)(H

)(Fm2ka (4.17)

where += m22

es

mD)(F and += m2e

s2


Since

)m(tanhDs2

mED

s2

mE

)m(tanhs

mD

s

mD

)(H

)(F

22

+++

+

+

+

=


=s

ka2)(u

)(Z)(Z

)(Xm2

)mtanh()(Z

)(X)Dm4(s2m

)(Z

)(X)Dm4(s2m 223223

+

++

+

+

)(Z

)(Xm2ka (4.18)

where

)m(tanhs

mD

s

mD)(X

22

+

+= (4.19)

)m(tanhD

s2

mED

s2

mE)(Z

+++

+= (4.20)

4.III. Third Family of Solutions.


+=

222m212

m421

222

2

sm16)eCCm2(

]eC)s4C(m4[m2ka)(u


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26

222m212

m21

2

sm16)eCCm2(

eCm4ska4

(4.21)

The above expression can also be written as

}Cmska16);s,m(U{);s,m(V

)]m(tanh1[mka2)(u 1

22

++= (4.22)

where

)m2tanh(]sm16C)1m4[(sm16C)1m4();s,m(U 22212222

12 += (4.23)

= 21212 )}mtanh()CCm2()CCm2({);s,m(V m

)]m(tanh1[)]m2(tanh1[sm16 222 + (4.24)

4.IV. Fourth Family of Solutions.

The projective Riccati equation method provides us with eight families of

solutions, each family containing some subfamilies of solutions. In Appendix A,

(Section A.IV) we have found twenty one solutions in total. To save space, the

reader can easily write down the various expressions for )(u , using the different

solutions of )(A found in that Section and (3.61). The complete set is to be

reported in electronic form somewhere else.

4.V. Fifth Family of Solutions.

We now use (A.71)-(A.74) for )(A and the general expression (3.61) and we

find four additional families of solutions.

(4.Va)Using (A.71) and (3.61), we have

+

++=

m2

m1

m

m2

m1

m2

eCeC

eska4

eCeC

eCm2ka)(u

or

)mtanh()CC()CC(

)]mtanh(1[Cmka2)(u

2121

2

++

+=


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27

)mtanh()CC()CC(

)mtanh(1ska4

2121 ++

+ (4.25)

(4.Vb)Using (A.72) and (3.61), we have

=

mm1

m

mm1

m1

2

es2emC

emska4

es2emC

eCm2ka)(u

or

)mtanh()s2mC()s2mC(

)]mtanh(1[Ckam2)(u

11

12

++

+=

)mtanh()s2mC()s2mC()mtanh(1skam4

11 ++ (4.26)

(4.Vc)Using (A.73) and (3.61), we have

+

+=

mm1

m

mm1

m1

2

es2emC

emska4

es2emC

eCm2ka)(u

or

)mtanh()s2mC()s2mC(

)]mtanh(1[Ckam2)(u

11

12

++

+=

)mtanh()s2mC()s2mC(

)mtanh(1skam4

11 ++

(4.27)

(4.Vd)Using (A.74) and (3.61), we have (we find 0)(A = in this case)

++

++=

m2

22m1

m2

m1

2

eCs4s4eCsm2

eCsm2sm2eCmska4)(u

or

kam2)(u = (4.28)

4.VI. Sixth Family of Solutions.

Using the results of Appendix B and (3.61), we have

])(aa[ska4)(aa

)(aka)(u 10

10

1 ++

+= (4.29)


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28

where )( is a solution of the differential equation (B.2), the various coefficients

corresponding to the Solutions 1-11 of Appendix B. The final formulas are too

lengthy to write them down here.

4.VII. Seventh Family of Solutions.

Using the results of Appendix C, we obtain

++

+=

)s,m(F

)s,m(F

eDE

e)s,m(FD

G

G and

++=

)s,m(F

)s,m(F2

eDE

e)s,m(F

G

G

We thus have

++

+=

+=

)s,m(F

)s,m(F010

eDE

e)s,m(FD

s4

1aG

Gaa)(A

and

=

+

+

=

G

Gaa

G

G

G

Gaa

)(A

)(A

10

2

10

++

+

++

+

++

=

)s,m(F

)s,m(F

0

2

)s,m(F

)s,m(F

)s,m(F

)s,m(F2

0

eDE

e)s,m(FD

s4

1a

eDE

e)s,m(FD

eDE

e)s,m(F

s4

1a

The final expression for )(u , follows from (3.61) and the above formulas.

The coefficient 0a satisfies equation (C.7), which is actually equation (A.1) and

therefore admits allthe solutions reported in Appendices A, B and C.

5. Solution of the Burgers equation using the G'/G-expansion.

In this section we solve Burgers equation using the

G

Gexpansion. It will be

clear by the end of this section that we need the solutions found previously using

the Riccati equation methods. Using the expansion


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29

+=

G

Gaa)(u 10 (5.1)

assuming that the coefficients 0a and 1a are dependent, Burgers equation

(3.4) becomes

++

+

2

1010G

Gaa

2

1

G

Gaa)(

0

2

110 cG

G

G

Ga

G

Gaaka =

+

+ (5.2)

Upon expanding the above equation and equating the coefficients of G to zero,

we obtain

Coefficient of 0G :

00200 cakaa

2

1a)( =+ (5.3)

Coefficient of 1G :

0GakaGakaGaaa)(11101

=+ (5.4)

Coefficient of 2G :

0)G(aka)G(a2

1 21

221 =+ (5.5)

From equation (5.3) we conclude that )(aa 00 is a solution of the equation

(3.4). Assuming that 0a1 we obtain from (5.5) that

ka2a1 = (5.6)

showing that 1a is a constant.

Finally from equation (5.4) we obtain

0G)aka(Gka 0 =++ (5.7)

The above equation can be reduced to a linear first order differential equation and

can be solved in principle provided that 0a is a known quantity. However the

coefficient 0a satisfies equation (5.3), which has been solved, using the Riccati


30/93

30

equation method with variable expansion coefficients. The solutions of equation

(5.3) are actually the various expressions of the function )(u of Section 4. The

solutions of (5.3) substituted into (5.7) lead to equations with rather complicated

solutions, despite the fact that equation (5.7) is a linear differential equation.

It is obvious that we could use some other methods in solving equation (5.3),

using, say, the Riccati equation method with constant coefficients. Using the

expansion Ybba 100 += with2

YMLY += ( 0b , 1b , L, and M constants) and

substituting into (5.3), after equating to zero the coefficients of Y, we obtain the

following set of equations

0bMkab2

111 =

, 0)b(b 01 = , 0

2010 cb

2

1bLkab =+

From the previous equations we obtain the following two sets of solutions

Solution 1.

=0b , Mka2b1= ,M

1

ka4

c2L

220

2

+

=

Solution 2.

=0b ,L

1

ka2

c2b 0

2

1 +

= ,L

1

ka4

c2M

220

2

+

=

For Solution 1, Riccati's equation 2YMLY += becomes

2

220

2

YMM

1

ka4

c2Y +

+=

which admits the solution

+

++= )C(

ka2

c2tanh

Mka2

c2Y 1

02

02

Therefore

+

++= )C(

ka2

c2tanhc2a 1

02

02

0


31/93

31

For Solution 2, Riccati's equation 2YMLY += becomes

222

0

2

YL1

ka4c2LY +=

which admits the solution

+

+

+= )C(

ka2

c2tanh

c2

Lka2Y 1

02

02

Therefore

+++= )C(

ka2c2tanhc2a 1

0

2

02

0 (5.8)

We thus see that 0a is given by the same expression in both solutions.

Because of the expression (5.8) for 0a , equation (5.7) becomes

KG2

KtanhKK1G =

++ (5.9)

where we have put 0c0 = , 0C1= and kaK = for simplicity.

Equation (5.9) is a rather complicated equation, which admits a closed-form

solution expressed in terms of the hypergeometric function. Details of the solution

are given in Appendix D. Using the results of that Appendix, we obtain the

following expression for the function )(u :

= ka2

2

Ktanh)(u

22)K1(

12

2)K1(1

22

K)1K3K2(K2);K(ZeK)1K(4);K(VK

2

Khsec]eK)1K3K2(K2);K(UK[

+++

++

(5.10)

The functions );K(U , );K(V and );K(Z are defined in Appendix D. The

function );K(Z is expressed in terms of the hypergeometric function.


32/93

32

We thus see that even in the simplest case, we obtain very complicated equations.

In more complex situations, we might need supercomputing facilities with

symbolic capabilities. We also conclude that in order to find solutions using the

)G/G( expansion method with variable expansion coefficients, we have to

consider the Riccati expansion method with variable coefficients first.

Appendix A.

In this Appendix we shall solve equation (3.54):

0)(As8)(Am2))(A(

2

3)(A)(A 42222 =+ (A.1)

A.I) First Method. We consider an expansion of the form

=

=n

0k

kk )(a)(A (A.2)

where )( satisfies Jacobi's differential equation

42)(

d

d++=

(A.3)

(In Appendix B we consider a full fourth order polynomial).

Upon substitution of (A.2) into (A.1) and balancing 4A with either AA or

2)A( , and taking into account (A.3), we obtain 1n= . We thus substitute

)(aa)(A 10 += (A.4)

Since

4211 a)(a)(A ++== and )2(a)(A

31 +=

equation (A.1) becomes

++++ 3212

1042

122

1 )as16(aa2)2

1as8(a

++++++ )4

1mas8(aa4)

12

1m

3

1as8(a6

220

210

2220

221

0a2

3as8am2

21

40

220

2 =+ (A.5)


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33

Equating to zero the coefficients of the different powers of , we find

= 22

1 s16

1

a ,2

2

2

0 ms16

1

a = ,2

m2= ,4

m= (A.6)

We thus obtain that

4222

2

4

m2m

)( +

= (A.7)

where we have redefined 2 .

The last equation can be written as

=

22 m)( (A.8)

This is a Riccati equation which can be transformed into a second order ODE,

admitting solution

)mtanh(C

1)mtanh(Cm)(

+

+

= (A.9)

Therefore )(A is given by (A.4) with )( given by (A.9):

+

+

+=

)mtanh(C

1)mtanh(Cmaa)(A 10 (A.10)

where the coefficients 0a and 1a satisfy the relations (four different

combinations)

2

221

s16a

= ,

2

220

s16

ma = (A.11)

A.II) Second Method. We consider the )G/G( expansion of the form

kn

0kk

G

Ga)(A

=

= (A.12)

with ka ( n,,1,0k L= ) constants and )(GG = .

(In Appendix C we consider variable expansion coefficients: )(aa kk ).


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Upon substitution of (A.12) into (A.1) and balancing 4A with either AA or

2

)A( , we obtain 1n= . We thus substitute

+=

G

Gaa)(A 10 (A.13)

Notice that 0a and 1a have nothing to do with the similar coefficients appearing

in (3.7), (3.24) or (3.42).

Since

=

2

1 G

G

G

G

a)(A and

+

=

3

21 G

G

2G

GG

3G

G

a)(A


+

+

3

2101 G

G2

G

GG3

G

G

G

Gaaa

+

++

+

22110

20

2422

21

G

Ga

G

Gaa2am2

G

G

G

G

G

G2

G

Ga

2

3

0G

Ga

G

Gaa4

G

Gaa6

G

Gaa4as8

441

3310

221

201

30

40

2 =

+

+

+

+

From the above equation we obtain the following set of coefficients, equating each

one of them to zero:

Coefficient of 4G : 0)G(as8)G(a2

3)G(a2 441

2421

421 = (A.14)

Coefficient of 3G : 0)G(aas32)G(aa2 33102310 = (A.15)

Coefficient of 2G :

221

2221

2110 )G(am2)G(a

2

3)G()G(a)G()G(aa)3( ++

0)G(aas48 22120

2 = (A.16)

Coefficient of 1G : 0Gaas32Gaam4Gaa 130

210

210 =+ (A.17)


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Coefficient of 0G : 0as8am2 4022

02 = (A.18)

The above system (A.14)-(A.18) is equivalent to

2

21

s16

1a = ,

2

220

s4

ma = , Gm4G 2 = ,

0am4G

Ga2

G

Ga 1

20

2

1 =+

+

(A.19)

The last equation is a quadratic equation with respect to the ratio G/G and

because its discriminant is zero, we obtain

1

0

a

a

G

G=

(A.20)

The above equation combined with the equation Gm4G 2 = , gives us

0

12

a

am4

G

G=

(A.21)

and by integration

++=

0

12321aam4expCCCG (A.22)

Therefore

++

+

=

0

12321

0

12

0

1232

a

am4expCCC

a

am4exp

a

am4CC

G

G

or, adjusting the above expression and redefining the constants (i.e. 320 C/CaD =

and 310 C/CaE= ), we obtain

++

=

0

120

0

121

2

a

am4expaDE

a

am4expam4D

G

G

We thus have


36/93

36

++

+=

+=

0

120

0

121

2

1010

a

am4expaDE

a

am4expam4D

aaG

G

aaA (A.23)

where2

21

s16

1a = and

2

220

s4

ma = (four different combinations).

Equation (A.23) can also be written as

)(H

)(FaaA 10

+= (A.24)

where

=

0

121

2

a

am4expam4D)(F (A.25)

++=

0

120

a

am4expaDE)(H (A.26)

A.III) Third Method. Using the transformation

)(w

1)(A2

= (A.27)

equation (A.1) transforms into the Ermakov equation (Ermakov [68])

322 ws4)(wm)(w = (A.28)

Equation 0)(wm)(w 2 = admits two independent solutions = m1 ew and

= m2 ew . We consider one of them,= m2 ew and introduce the

transformation =

2w

d, i.e.

= m2em2

1, )m2(ln

m2

1= and )(wez m = , )(zz = .

Since

})(w)(wm{ed

dz m +=

and })(w)(wm{ed

zd 2m32

2

+=


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37

Ermakovs equation (A.28) transforms into the equation

32

2

2

zs4d

zd = (A.29)

Using one of the standard methods, i.e. multiplying both members byd

dz2 or

introducing the variable p by

=d

dzp , we arrive at the equation

12

22

C

z

s4

d

dz+=

(A.30)

From the above equation we obtain the two equations

12

2

Cz

s4

d

dz+=

and 12

2

Cz

s4

d

dz+=

The first of the above equations admits the solution 2122

1 CCs4zC +=+

while the second 2122

1 CCs4zC +=+ .

We thus have, squaring both the previous two equations and solving with respect

to 2z

12

222m2122

Cm4

sm16)eCCm2(z

=

and then, since )(wez m = ,

= m2

12

222m2122 e

Cm4

sm16)eCCm2()(w

We thus obtain, using the above expression and (A.27), that )(A is given by

222m212

m21

2

sm16)eCCm2(

eCm4)(A

=

(A.31)

The above expression can be written in terms of )mtanh( . In fact since


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38

2m2m1

m2

12

)esm4()eCeCm2(

Cm4)(A

=

we can easily transform the above expression into

22221212

21

2

)]m(tanh1[sm16)]m(tanh)CCm2()CCm2([

)]m(tanh1[Cm4)(A

=

m

A.IV. Fourth Method. We shall now use the projective Riccati equation

method (see for example Abdou [5]) in solving equation (A.1). We consider the

expansion

])(gb)(fa[)(fa)(AN

1iii

1i0

=

++= (A.32)

where the functions )(f and )(g satisfy the system

)(g)(fp)(f = (A.33)

)(fr)(gpq)(g2 += (A.34)

++= )(f

q

r)(fr2q

p

1)(g 2

22 (A.35)

The system of equations (A.33) and (A.34) admitsfive familiesof solutions given

by

(I) If 22 = and 0qp < then

++=

qpcoshqpsinhr

q)(f (A.36)

+++=qpcoshqpsinhr

qpsinhqpcoshp

qp)(g (A.37)

with

++= )(f

q

r)(fr2q

p

1)(g 2

2222 (A.38)

(II) If 22 = and 0qp > then


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39

++=

qpcosqpsinr

q)(f (A.39)

++

+=

qpcosqpsinr

qpsinqpcos

p

qp)(g (A.40)

with

+= )(f

q

r)(fr2q

p

1)(g 2

2222 (A.41)

(III) If 0q= , then

++

=2

2

rp1)(f (A.42)

++

+=

2

2

rp

rp

p

1)(g (A.43)

with

)(fp

r2

p)(fp

r2

)(g2

2

22

+= (A.44)

where and are free parameters.

(IV) If 1p = and 2r= then

)(rp

2

r6

q)(f += (A.45)

)(12q

)(12)(g

+

= (A.46)

where )( satisfies Weirstrassequation

216

qp)(

12

q)(4))((

3232 =

with solution )()( = .

The relation between f and g is given by


40/93


41/93

41

Upon expanding and rearranging, we obtain an equation which contains a constant

term and powers of the functions )(f and )(g . We further substitute )(g2 by

(A.35) wherever powers of the function )(g enter. We thus obtain finally an

equation which will contain powers of g no greater than one. Equating the

coefficients of this equation to zero, we obtain the following.

Constant coefficient:

+ 2122

12

042

02 )b(m

p

q2)b()a(s

p

q48)a(m2

0)a(s8)b(sp

q8 40

441

4

2

2 = (A.53)

Coefficient of f:

+++ 212

042

11044

14

2102 )b()a(s

p

r96)b(aas

p

q96)b(s

p

rq32aam4

0aaqp)b(rqa)a(s32)b(mp

r410

211

30

421

2 =+ (A.54)

Coefficient of g:

0bam4b)a(s32)b(asp

q3210

21

30

4310

4 =+ (A.55)

Coefficient of 2f :

++

++ 2

12

22

12

142

12

04

2

)b(mqp

)r(2)b()a(s

p

q48)b()a(s

qp

)r(48

+

+ 414

2

41

4

2

22

11042

1

2

)b(s

p

16)b(s

p

r48)b(aas

p

r192)b(

2

r5

0aarp3)a(2

qp)a()a(s48)b(2)a(m2 10

21

21

20

421

21

2 =++++ (A.56)

Coefficient of 3f

++

+

10

22

12

144

14

2

2

aaq

)r(p2)b()a(s

p

r96)b(s

qp

)r(r32


42/93

42

++

+ 31042

1

22

1104 )a(as32)b(

q

)r(r2)b(aas

qp

d96

0)b(aasqp

r96 2110

42

=+ (A.57)

Coefficient of 4f

+

+

+ 2

12

22

1

22

12

12

14

2

)b(q2

)a(q2

rp)a(

q2

p)b()a(s

qp

)r(48

+

+ 414

22

24

14

22

44

142

12

42

12

2

)b(sqp

8)b(s

qp

r8)a(s8)b(

q2

r)b(

q

r

0)b(sqp

r16 41

4

22

2

=

(A.58)

Coefficient of gf

++ 3104

11112

04

10112 )b(as

p

r64baqpba)a(s96baprbam4

0)b(as

p

q32 311

4 =+ (A.59)

Coefficient of gf2

++

+

10

23

104

2

baq

)r(p2)b(as

qp

)r(32

0b)a(as96)b(asp

r64bapr 1

210

4311

411 =+ (A.60)

Coefficient of gf3

0b)a(s32baq

)r(p)b(as

qp

)r(321

31

411

23

114

2

=+

+

(A.61)

Solving the system of the nine simultaneous equations (A.53)-(A.61), using any of

the known Computer Algebra Systems (Axiom, Macsyma, Maple, Mathematica),

we obtain a considerable number of solutions, from which we select only the

nontrivial ones:


43/93

43

Solution I.20 s4

ma = ,

m2

ba 11

= , 11 bb = , 1

2bs8p= ,

12

2

bs2

mq =

where is any root of the equation 22 r+= .

Solution II.20

s4

ma = ,

m2

ba 11

= , 11 bb = , 1

2bs8p = ,1

2

2

bs2

mq =


Solution III.20 s4

ma = , 11 aa = , 11 bb = , 2

1

21

221

2

)b(

)b(r)a(m4 = ,

12bs8p= ,

12

2

bs2

mq =

Solution IV.20

s4

ma = , 11 aa = , 11 bb = , 2

1

21

221

2

)b(

)b(r)a(m4 = ,

12bs8p = ,

12

2

bs2

mq=

Solution V.20 s4

ma = , 0a1= , 11 bb = , 1

2bs4p= ,

12

2

bs4

mq =

Solution VI.20

s4

ma = , 0a1= , 11 bb = , 1

2bs8p= ,

12

2

bs2

mq =

and r is any root of 0r2 =+ .

Solution VII. 20 s4

m

a = , 0a1= , 11 bb = , 12

bs8p = ,1

2

2

bs2

m

q=


Solution VIII. 0a0= ,ms8

pa

21

= , 0b1= , pp = ,

p

m4q

2

=

and is any root of 22 r+= .


44/93

44

For each set of coefficients and parameters (Solution I Solution VIII) there

correspond in principle five families of solutions. However not all of them appear

in the final solutions, because there is a conflict between the values of the

coefficients and the values of the parameters of the various solutions.

Family I. This family corresponds to the Solution I

20 s4

ma = ,

m2

ba 11

= , 11 bb = , 1

2bs8p= ,

12

2

bs2

mq =


From (A.51) and Solution I, we obtain

)(gb)(fm2

b

s4

m)(A 1

12

+

+= (A.62)

with 12bs8p= ,

12

2

bs2

mq = and satisfies the equation 22 r+= .

Sub-family Ia. Since 0m4qp 2 >= , we consider first the choice 22 = .

The functions )(f and )(g in (A.62) are given by (A.36) and (A.37)respectively:

)(U

1

bs2

m)(f

12

2

= and

)(U

)(U2

bs2

m)(g

12

2

=

where

)|m|2(cosh)|m|2(sinhr)(U ++=

We thus obtain

)(U

)(Um

s4

1

s4

m)(A

22

=



45/93

45

Sub-family Ib. If 1p = and 2r= , we have 0= and then 0a1= . In this

case we have )(gbs4m)(A 12 += where )(g is given by (A.46).

(Ib.a) For 1p= , we have21

s8

1b = and 2m4q = . Therefore

2m4)(12

)(12)(g

= and thus

222 m4)(12

)(

s2

3

s4

m)(A

+=

where )( is Weirstrassfunction satisfying the equation

27

m8)(3

m4)(4))((6432 +=

(Ib.b) For 1p = , we have21 s8

1b = and 2m4q= . Therefore

2m4)(12

)(12)(g

+

= and thus

222m4)(12

)(

s2

3

s4

m)(A

+

=

Sub-family Ic. If 1p = and25

r2= , is satisfies the equation

25

r24 22 =

while )(f and )(g are given by (A.48) and (A.49) respectively.

(Ic.a) For 1p= , we have21

s8

1

m2a

= ,

21s8

1b = and 2m4q = . Therefore

)(9

m10

r3

m10)(f

42

+= and

)(]m4)(12[

)(m4)(g

2

2

=

We thus have the following expression for )(A :

]m4)(12[)(

)(

s2

m

r

1

)(3

m

s24

m5

s4

m)(A

22

22

22

+

+=


27

m8)(

3

m4)(4))((

6432 +=


46/93

46

(Ic.b) For 1p = , we have21 s8

1

m2a

= ,

21 s8


)(9

m10

r3

m10)(f

42

= and

)(]m4)(12[

)(m4)(g

2

2

=

We thus have the following expression for )(A :

]m4)(12[)(

)(

s2

m

r

1

)(3

m

s24

m5

s4

m)(A

22

22

22

+

+=

Note.If 22 = and 0qp > leads to 0m2 < . The case 0q= cannot be

considered either, since it leads to 0m = and then the coefficient of )(f in (A.62)

becomes infinite.

Family II. This family corresponds to the Solution II

20 s4

ma = ,

m2

ba 11

= , 11 bb = , 1

2bs8p = ,1

2

2

bs2

mq =


From (A.51) and Solution II, we obtain

)(gb)(fm2

b

s4

m)(A 1

12

+

+= (A.63)

with 12bs8p = ,

12

2

bs2

mq = and satisfies the equation 22 r+= .

Sub-family IIa. Since 0m4qp 2 >= , we first consider the case 22 = .

The functions )(f and )(g are given by (A.39) and (A.40) respectively:

)(V

1

bs2

m)(f

12

2

= and

)(V

)(Z

bs4

|m|2)(g

12

=

where

)|m|2(sin)|m|2(cos)(Z +=

)|m|2(cos)|m|2(sinr)(V ++=


47/93

47

We thus obtain

)(V

)(Z|m|m

s4

1

s4

m

)(A 22

+

=

where is any root of the equation 2222 r = .

Sub-family IIb. If 1p = and 2r= then 0= and thus 0a1= .

In this case we have )(gbs4

m)(A 12

+= where )(g is given by (A.46).

(IIb.a) For 1p= , we have2

1

s8


2m4)(12

)(12)(g

+

= and thus

222 m4)(12

)(

s2

3

s4

m)(A

+

=


27

m8)(

3

m4)(4))((

6432 =

(IIb.b) For 1p = , we have21 s8

1b =

and 2m4q = . Therefore

2m4)(12

)(12)(g

= and thus

222m4)(12

)(

s2

3

s4

m)(A

+=

Sub-family IIc. If 1p = and25

r2= then satisfies the equation

25

r24 22 = .

The functions )(f and )(g are given by (A.48) and (A.49) respectively.

(IIc.a) For 1p= , we have 21 s8

1

b = , 21 s8

1

m2a

= and2

m4q= . We then

have the following expressions for )(f and )(g

)(9

m10

r3

m10)(f

42

+= and

)()](12m4[

)(m4)(g

2

2

+

=

Therefore


48/93

48

)](12m4[)(

)(

s2

m

)(3

m

r

1

s24

m5

s4

m)(A

22

22

22 +

+

+

=


27

m8)(

3

m4)(4))((

6432 =

(IIc.b) For 1p = , we have21 s8

1b = ,

21 s8

1

m2a

= and 2m4q = . We then

have the following expressions for )(f and )(g

)(9

m10

r3

m10)(f

42

= and

)()](12m4[

)(m4)(g

2

2

+

=

Therefore

)](12m4[)(

)(

s2

m

)(3

m

r

1

s24

m5

s4

m)(A

22

22

22 +

+

+

=

Note.If 22 = and 0qp < leads to 0m2 < . The case 0q= cannot be

considered either, since it leads to 0m = and then the coefficient of )(f in (A.63)

becomes infinite.

Family III. This family corresponds to the Solution III

20s4

ma = , 11 aa = , 11 bb = , 2

1

21

221

2

)b(

)b(r)a(m4 = ,

12bs8p= ,

12

2

bs2

mq =

From (A.51) and Solution III, we obtain

)(gb)(fas4

m)(A 112

++= (A.64)

where2

1

21

221

2

)b(

)b(r)a(m4 = , 1

2bs8p= ,1

2

2

bs2

mq =


49/93

49

Sub-family IIIa. Since 0m4qp 2


50/93

50

222 m)(3

)(

s8

3

s4

m)(A

=

Sub-family IIIc. If 1p = and25

r2= , i.e.

21

21

221

22

)b(

)b(r)a(m4

25

r =

then )(f and )(g are given by (A.48) and (A.49) respectively.

(IIIc.a) For 1p= , we have21

s8

1b = and 2m4q = and thus

)(9

m10

r3

m10)(f

42

+= and

)(]m4)(12[

)(m4)(g

2

2

=

Therefore

]m)(3[)(

)(

s8

m

)(3

m

r

1

3

ma10

s4

m)(A

22

2221

2

+

++=


27

m8)(

3

m4)(4))((

6432 +=

with21

221

2br24am100 = .

(IIIc.b) For 1p = , we have21 s8

1b = and 2m4q= and thus

)(9

m10

r3

m10)(f

42

= and

)(]m4)(12[

)(m4)(g

2

2

=

Therefore

]m)(3[)(

)(

s8

m

)(3

m

r

1

3

ma10

s4

m)(A

22

2221

2

+

+=

Note. If 22 = and 0qp > leads to 0m2 < . The choice 0q= leads to

0m = .

Family IV. This family corresponds to the Solution IV


51/93

51

20 s4

ma = , 11 aa = , 11 bb = , 2

1

21

221

2

)b(

)b(r)a(m4 = , 1

2bs8p = ,

12

2

bs2

mq=

From (A.51) and Solution IV, we obtain

)(gb)(fas4

m)(A 112

++= (A.65)

where21

21

221

2

b

bram4 = , 1

2bs8p = ,

12

2

bs2

mq=

Sub-family IVa. Since 0m4qp 2


52/93

52

where )( is the Weirstrassfunction, satisfying the equation

27

m8)(3

m4)(4))((

64

32 +=

(IVb.b) For 1p = , we have21 s8

1b = , 2m4q= and thus from (A.46) we get

2m4)(12

)(12)(g

+

= and then

222m)(3

)(

s8

3

s4

m)(A

+

+=

Sub-family IVc. If 1p = and25r

2

= , i.e. 21

2

1

22

1

22

bbram4

25r = then )(f and

)(g are given by (A.48) and (A.49) respectively.

(IVc.a) For 1p= , we have21

s8

1b = , 2m4q = and then

)(9

m10

r3

m10)(f

42

+= and

)(]m4)(12[

m4)(g

2

2

=

Therefore

]m)(3[)(

)(

s8

m

)(3

m

r

1

3

ma10

s4

m)(A

22

2221

2

++=

where )( is the Weirstrassfunction, satisfying the equation

27

m8)(

3

m4)(4))((

6432 +=

with42

221

sm800r3a = . This last equation comes from equating the two different

expressions of and using the value of 1b .

(IVc.b) For 1p = , we have21 s8

1b = , 2m4q= and then

)(9

m10

r3

m10)(f

42

= and

)(]m4)(12[

m4)(g

2

2

=


53/93

53

Therefore

]m)(3[)(

)(

s8

m

)(3

m

r

1

3

ma10

s4

m)(A 22

222

12

+=

Note. The case 22 = and 0qp > leads to 0m2 < . The choice 0q=

leads to 0m = .

Family V. This family corresponds to the Solution V

20s4

ma = , 0a1= , 11 bb = , 1

2bs4p= ,

12

2

bs4

mq =

From (A.51) and Solution V, we obtain

)(gbs4

m)(A 12

+= (A.66)

where 12bs4p= ,

12

2

bs4

mq =

Sub-family Va. Since 0m4qp 2


54/93

54


216

m)(12

m)(4))((

64

32 +=

(Vb.a) For 1p = , we have21 s4

1b = , 2mq= and then

2m)(12

)(12)(g

+

= .

Therefore222

m)(12

)(

s

3

s4

m)(A

+

=

Sub-family Vc. If 1p = and25

r2= then )(g is given by (A.49).

(Vc.a) For 1p= , we get21

s4

1b = , 2mq = and then

]m)(12[)(

)(m)(g

2

2

= .

Therefore

]m)(12[)(

)(

s4

m

s4

m)(A

22

2

2

+=

(Vc.b) For 1p = , we get2

1

s4

1b = , 2mq= and then

]m)(12[)(

)(m)(g

2

2

= . Therefore

]m)(12[)(

)(

s4

m

s4

m)(A

22

2

2

+=

Note. The case 22 = and 0qp > leads to 0m2 < . The case 0q= leads

to 0m = .

Family VI. This family corresponds to the Solution VI

20s4

ma = , 0a1= , 11 bb = , 12bs8p= ,

12

2

bs2mq =


From (A.51) and Solution VI, we obtain

)(gbs4

m)(A 12

+= (A.67)


55/93

55

where 12bs8p= ,

12

2

bs4

mq = and r is any root of 0r2 =+ .

Sub-family VIa. Since 0m4qp 2


56/93

56

(VIc.a) For 1p= , we get21 s8


]m4)(12[)(

)(m4)(g

2

2

= .Therefore

]m)(3[)(

)(

s8

m

s4

m)(A

22

2

2

+=

(VIc.b) For 1p = , we get21

s8


]m4)(12[)(

)(m4)(g

2

2

= .

Therefore]m)(3[)(

)(

s8

m

s4

m)(A22

2

2 += .


to 0m = .

Family VII. This family corresponds to the Solution VII

20s4

ma = , 0a1= , 11 bb = , 1

2bs8p = ,

1

2

2

bs2

mq=


From (A.51) and Solution VII, we obtain

)(gbs4

m)(A 12

+= (A.68)

where 12bs8p = ,

12

2

bs4

mq= and r is any root of 0r2 =+ .

Sub-family VIIa. Since 0m4qp 2


57/93

57

Therefore)(U

)(U

s4

1

s4

m)(A

22

= where r satisfies the equation

0r 222 =+ .

Sub-family VIIb. If 1p = and 2r= (i.e. r any real) then )(g is given by

(A.46).

(VIIb.a) For 1p= , we have21

s8

1b = , 2m4q = and thus

2m4)(12

)(12)(g

= . Therefore

222 m)(3

)(

s8

3

s4

m)(A

=


27

m8)(

3

m4)(4))((

6432 +=

(VIIb.b) For 1p = , we have21 s8

1b = , 2m4q= and thus

2m4)(12

)(12)(g

+

= .

Therefore 222 m)(3

)(

s8

3

s4

m

)(A +

+=

Sub-family VIIc. If 1p = and25

r2= , i.e. 0

25

rr

22 = then )(g is given by

(A.49).

(VIIc.a) For 1p= , we get21 s8


]m4)(12[)()(m4)(g

2

2

= .Therefore

]m)(3[)()(

s8m

s4m)(A

22

2

2 =

(VIIc.b) For 1p = , we get21 s8


]m4)(12[)(

)(m4)(g

2

2

= . Therefore


58/93

58

]m)(3[)(

)(

s8

m

s4

m)(A

22

2

2

= .


to 0m = .

Family VIII. This family corresponds to the Solution VIII

0a0 = ,ms8

pa

21

= , 0b1= , pp = ,p

m4q

2

=

and is any root of 22 r+= .

From (A.51) and Solution VIII, we obtain

)(fms8

p)(A

2

= (A.69)

where pp = ,p

m4q

2

= and is any root of the equation 22 r+= .

Sub-family VIIIa. Since 0m0qp 2 >< , the choice 22 = , i.e.

2222 r+= , then )(f is given by (A.36):)(U

1

p

m4)(f

2

= where

)|m|2(cosh)|m|2(sinhr)(U ++= .Therefore)(U

1

s2

m)(A

2

=

where satisfies the equation 2222 r+= .

Sub-family VIIIb. If 1p = and25

r2= , leads to 22 r

25

24= and then )(f is

given by (A.48).

(VIIIb.a) For 1p= , we have 2m4q = and thus)(9

m10

r3

m10)(f

42

+= .


27

m8)(

3

m4)(4))((

6432 +=


59/93

59

Therefore

+

=

)(3

m

r

1

s12

m5)(A

2

2, and is any root of the equation

25

r24 22 = .

(VIIIb.b) For 1p = , we have 2m4q= and thus)(9

m10

r3

m10)(f

42

= .

Therefore

=

)(3

m

r

1

s12

m5)(A

2

2.


to 0m = . The case 2r= cannot be considered since in this case 0= and then

0a1= (i.e. 0)(A = ).

A.V) Fifth Method. We consider now an expression of the form

++

++=

m

10

m

1

m10

m1

edcec

ebaea)(A (A.70)

and substitute back into (A.1). We then derive an equation which contains powers

of the exponentials me . Equating all the coefficients of these exponentials to

zero, we obtain the following set of solutions

11 aa = , 0a0 = , 0b1= , 11 cc = , 0c0 = , 11 dd =

11 aa = , 00 aa = , 11 bb = ,m

sa2c 11 = ,

m

sa2c 00 = ,

m

sb2d 11 =

0a1= , 0a0 = , 11 bb = , 11 cc = , 0c0 = ,m

sb2d 11 =

0a1= , 0a0 = , 11 bb = , 11 cc = , 0c0 = ,m

sb2d 11=

s2

cma 11= , 00 aa = , 11 bb = , 11 cc = ,

m

sa2c 00 = ,

m

sb2d 11=

We thus have


60/93

60

+=

m1

m1

m1

edec

ea)(A ,

=m1m

1

m1

em

bs2ec

eb)(A

+

=m1m

1

m1

em

bs2ec

eb)(A ,

++

++

=m10m

1

m10

m1

em

bs2

m

as2ec

ebaes2

cm

)(A

or

+

=m

2

m

1

m

eCeC

e)(A ,

1

11

a

cC = ,

1

12

a

dC = (A.71)

=

mm1

m

es2eCm

em)(A ,

1

11

b

cC = (A.72)

+=

mm1

m

es2eCm

em)(A ,

1

11

b

cC = (A.73)

2m2

2m1

m2

m1

2

s4eCs4eCsm2

sm2eCsm2eCm)(A

++

++=

,0

11

a

cC = ,

0

12

a

bC = (A.74)

Appendix B. We consider an expansion of the form

)(aa)(A 10 += (B.1)

where )( satisfies Jacobi's differential equation

44

33

2210 nnnnn)(

d

d++++=

(B.2)

Contrary to the previously considered case (Appendix A, equation (A.3)) we

consider the right hand side of (B.2) to be a complete fourth degree polynomial.

Since

44

33

221011 nnnnna)(a)(A ++++==

and

)n4n3n2n(a2

1)(A 34

23211 +++=


61/93

61


+++

3

4

2

1

2

10

42

1

2

4

2

1 )nas16(aa2as8n2

1

a

+

++ 2310

21

20

22

21

21

2 naa2

3aas48na

2

1am2

+++ )aas32naanaaam4( 130

22101

2110

2

0naa2

1na

2

3as8am2 1100

21

40

220

2 =++ (B.3)

Equating to zero all the coefficients, we find

0as8n2

1a 21

24

21 =

0)nas16(aa2 421

210 =+

0naa2

3aas48na

2

1am2 310

21

20

22

21

21

2 =+

0aas32naanaaam4 130

22101

2110

2 =+

0naa2

1na

2

3as8am2 1100

21

40

220

2 =+

From the above system we consider only the following set of nontrivial solutions,

assuming that 0a1 :

Solution 1.

00 aa = , 11 aa = , 31

03120

21

220

0a

)anaas48am4(an

+= ,

21

03120

21

20

1a

)an3aas128am8(an

+= ,

1

03120

21

2

2a

an3aas96am4n

+=

33 nn = ,21

24 as16n = .

Solution 2.

00 aa = ,2

220

2

031

nm4as96

an3a

+= ,


62/93

62

23

22

220

220

22

2

0n27

)nm4as96()as48nm8(n

++= ,

3

222

022

02

22

1n3

)nm4as96()as32nm4(n

++= ,

22 nn = , 33 nn = , 22

220

2

23

20

2

4)nm4as96(

nas144n

+=

Solution 3.

00aa = ,

1

20

22

20

1 n

)as32nm4(aa

+= ,

22

220

2

21

20

22

2

0 )nm4as32(3

n)as48nm8(n

+=

11 nn = , 22 nn = ,1

20

22

220

22

2

3n3

)as96nm4()as32nm4(n

+++= ,

21

22

220

220

2

4n

)nm4as32(as16n

=

Solution 4.

00 aa = ,s4

a1 = , 24

420242302020

n

)nas12nmnas(as64n +=

4

20

22300

1n

)as32m2nas3(as16n

+=

4

420

24

230

2n

)nas24nmnas3(4n

+=

where is any root of the equation 0n4

2 = .

Solution 5.

00 aa = ,s4

a1

= ,4

20

22

220

2

0n3

)as48nm8(as16n

+=

+=

)as32nm4(as4n

20

22

20

1 ,0

20

22

2

3as12

)as96nm4(n

++=


63/93

63

where is any root of the equation 0n42 = .

Solution 6.

00 aa = ,s4

a1

= , )as64mas16n(n

as

3

4n 30

3201

4

00 +=

0

201

30

3

2as4

mas16nas128n

+= ,

20

2

412

030

3

3as48

nnmas32as512n

+=


Solution 7.

00 aa = ,s4

a1

= ,0

4020

2240

4

1as4

nn3ams64as256n

+=

20

2

4020

2240

4

2as16

nn3ams128as768n

+= ,

30

3

4020

2240

4

3as64

)nnams64as768(n

+=


Solution 8.

00 aa = , 01 aa = ,2

120

22 m4nas32n += ,

)ans16nm4nm24nans384(n9

1n 201

21

20

221

200

2

03 ++=

)as16m4n(n

as

3

16n 20

221

0

20

2

4 +=

where is any root of the equation 0m4as16nn3 2202

12

0 =+

Solution 9.

00 aa = , 01 aa = ,3

220

2

20

2211

0n3m8as128

)as16m4n(n

3

1n

+=

21

20

22 m4nas32n += , )n3m8as128(

n

as16n 3

220

2

1

20

2

4 =

where is any root of the equation 0n3)m8as128(n 322

022

1 =+


64/93

64

Solution 10.

00 aa = , 01 aa = ,2

120

22 m4nas32n +=

)ans16nm4nm24nans384(n9

1n 201

21

20

221

200

2

03 ++=

)as16m4n(n

as

3

16n 20

221

0

20

2

4 +=

where is any root of the equation 0m4as16nn3 2202

12

0 =+ .

Solution 11.

00 aa = , 01 aa = ,

+=

220

20

2

1

m4as16n3n

220

20

22 m8as48n3n += ,

220

24 as16n =

where is any root of the equation 0n)m4as48(n 322

023

0 =+ .

B.1. Solutions of the differential equation (B.2).

For the first three Solutions, we use the following Lemma:

Lemma: If (B.2) is expressed as

)Dqp()Dqp()n(r)(d

d++=

(B.4)

then its solution is given by

Dq4e)pn(4e

Dqn4e)pnpDq(4en2KrKr2

2Kr22Kr2

+++

++=

(B.5)

where

Dq)pn(K22 += (B.6)

The proof of the Lemma is based on the following formula

+

=++

Dq)pn(

1

)Dqp()Dqp()n(

d

22

+++++

+ }Dq)p(Dq)pn()pn()n(Dq)pn({

n

2ln

222222


65/93

65

Integrating equation (B.4) using the above integral, we find (we set the integration

constant equals to zero)

=++++++

}Dq)p(Dq)pn()pn()n(Dq)pn({n

2 222222

)Dq)pn(r(exp 22 +=

Solving the above equation with respect to , we obtain (B.5).

The solution (B.5) can also be expressed in terms of the hyperbolic tangent

function as

)Kr(tanh1)pn(4)Krtanh()Dq41()Dq41(

)Kr(tanh1)pnpDq(4)Krtanh()Dq41(n)Dq41(n

222

22222

+++

++=

(B.7)

For the Solution 1, we have

+=++++

2

1

021

244

33

2210

a

aas16nnnnn

+21

221

2

3210

21

221

2

3210

as32

D

as32

nsaa32

as32

D

as32

nsaa32

where

21

23

210 )asm16()nsaa64(D

Equation (B.2) then gives by integration either (B.5) or (B.7) where

2

1

2

32

10

as32

nsaa32p

= and

2

1

2

as32

1q=



66/93

66

+

=++++2

222

02

23

20

24

43

32

210)nm4as96(

nas144nnnnn

++

2

3

222

02

n3

nm4as96

++

+

320

2

222

02

320

2

22

222

02

nas288

D)nm4as96(

nas288

)nm4()nm4as96(

+

+

3

2

0

22

220

2

3

2

0

22

22

220

2

nas288

D)nm4as96(

nas288

)nm4()nm4as96(

where

20

22

220

2 )asm48()nm4as96(D +


320

2

22

222

02

nas288

)nm4()nm4as96(p

+= ,

320

2

222

02

nas288

nm4as96q

+=


=++++21

22

220

220

24

43

32

210n

)nm4as32(as16nnnnn

++

2

20

22

21

as32nm4

n

+

)nm4as32(as96

Dn

)nm4as32(as96

)m4n(n

222

022

02

1

222

022

02

221

)nm4as32(as96

Dn

)nm4as32(as96

)m4n(n

222

022

02

1

222

022

02

221

where

20

22

220

2 )asm48()nm4as96(D +



67/93

67

)nm4as32(as96

)m4n(np

222

022

02

221

= ,

)nm4as32(as96

nq

222

022

02

1

=

For the rest of the Solutions (Solution 4-Solution 11), we may use Section 11 (pp.

209-211), of Chapter 7 of H. D. Davis: "Introduction to Nonlinear Differential

and Integral Equations", Dover 1962. In all these cases we shall also need the

algorithm of determining the roots of a fourth degree polynomial. The reader may

consult my paper "Solving Third, Fourth and Fifth Degree Polynomial Equations"

available from www.docstoc.com. The solutions are then expressed in terms ofJacobi's or Weirstrass functions or even in terms of Elliptic Integrals of various

kinds.

Appendix C. In this Appendix we shall solve equation (A.1) using the G/G

expansion method with variable coefficients. We substitute

+=

G

Gaa)(A 10 (C.2)

where 0a and 1a are dependent quantities, )(aa 00 = and )(aa 11 = , while

)(GG = . Since

+

the riccati equation method with variable expansion coefficients. i. solving the burgers equation

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