Some General Solutions to the Painlevé – PIV Equation

8/12/2019 Some General Solutions to the Painlev PIV Equation

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Some General Solutions to the

Painlev PIV Equation

Solomon M. Antoniou

SKEMSYSScientific Knowledge Engineering

and Management Systems

Corinthos 20100, [email protected]

Abstract

We initiate a solution procedure to the Painlev equations. The Riccati equation

method with variable expansion coefficients is used to find closed-form solutions

to the Painlev-PIV equation. The solutions are expressed in terms of Whittaker's

W and M functions.

Keywords: Painlev equations, Painlev-PIV equation,extended Riccati equation

method, nonlinear equations, exact solutions, Whittaker's functions.


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1. Introduction.

The Painlev equations (numbered as PI-PVI) is a special class of second order

nonlinear ODEs which have no movable critical points (branch points or essential

singularities). These equations were discovered under a number of assumptions at

the end of nineteenth century/beginning of twentieth century by Painlev and

Gambier. They also appear in many physical applications. Some reviews and

further developments the reader can consult, are the classical book by E. L. Ince

(Ince [10], Chapter XIV) and the articles by A. S. Fokas and M. J. Ablowitz

(Fokas and Ablowitz [9]) and P. A. Clarkson (Clarkson [7]). Closely related toPainlev equations is the so-called ARS conjecture (after Ablowitz, Ramani and

Segur [1], [2] and [11]) which is an integrability test for ordinary differential

equations. The ARS conjecture was extended to PDEs by J. Weiss, M. Tabor and

G. Carnevale (Weiss, Tabor and Carnevale [13]) who introduced the so-called

singular manifold method (for a review and further examples see Ramani,

Grammaticos and Bountis [12]). This method (also named as Painlev truncation

method) serve also as a solution method (Weiss [14], [15] and [16]) in the sense

that it can determine the Lax pairs and the Bcklund transformations.

No explicit solutions of the Painlev equations have been found so far. In some

cases only rational solutions are available (Airlaut [4] and Wenjum and Yezhou

[17]). Some relations have also been established between solutions (Fokas and

Ablowitz [9], Clarkson [7]). We have been able to find explicit solutions for the

Painlev PII equation (Antoniou [6]) expressed in terms of the Airy functions.

In this paper we introduce a solution method and find some closed-form solutions

to the Painlev PIV equation using the extended Riccati equation method with

variable expansion coefficients. The solutions are expressed in terms of

Whittaker's W and M functions. The paper is organized as follows: In Section 2

we describe the extended Riccati equation method with variable expansion

coefficients. This method was introduced in Antoniou ([5]). In Section 3 we apply

our method to the Painleve PIV equation


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3

ww)x(2wx4w

2

3

dx

dw

w2

1

dx

wd 2232

2

2 ++++

=

and we set up a system of nine ordinary differential equations. In solving that

system we consider four cases. This is done in the next four Sections (Section 4-

Section 7). In Sections 4 and 5 we find two families of solutions which are

expressed in terms of Whittaker's functions (modulo some compatibility

conditions). To each family there correspond an infinite number of subfamilies. In

Sections 6 and 7 we find two more families of solutions, expressed again in terms

of Whittaker's functions. However it is not known if these two solutions have to beaccepted or not because it is not clear if they satisfy the compatibility conditions.

In Section 8 we consider another method of solution, the )G/G( expansion

method with variable expansion coefficients. We obtain a system of five

differential equations which, when solved, can in principle determine the unknown

function G and then )x(u .

2. The Method.

We suppose that a nonlinear ordinary differential equation

0),u,u,u,x(F xxx =L (2.1)

with unknown function )x(u admits solution expressed in the form

==

+=n

1kkk

n

0k

kk

Y

bYa)x(u (2.2)

whereallthe expansion coefficientsdepend on the variable x:

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

=

The function )x(YY satisfies Riccatis equation

2YBA)x(Y += (2.3)

where the coefficients A and B depend on the variable x as well.

In solving the nonlinear ODE (2.1), we substitute the expansion (2.2) in (2.1) and

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


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4

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

coefficients of the different powers of the function )x(Y to zero, we find a

system of nonlinear ordinary differential equations from which we can determine

the various expansion coefficients )x(a k , )x(bk and the functions )x(A , )x(B .

We finally solve Riccati's equation and then find the solutions of the equation

considered.

3. The Painleve PIV equation and its solutions.

We consider the Painleve PIV equation

ww)x(2wx4w

2

3

dx

dw

w2

1

dx

wd 2232

2

2 ++++

= (3.1)

where )x(ww is the unknown function and x the independent variable,

considered complex in general. We shall use the extended Riccati equation method

in solving equation (3.1). In this case we consider the expansion

==

+=n

1k

kk

n

0k

kk

Y

bYa)z(w (3.2)

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

of (3.1). We then find 1n= and thus

Y

bYaa)x(w 110 ++= (3.3)

where all the coefficients 0a , 1a and 1b depend on x, and Y satisfies Riccatis

equation

)x(Y)x(B)x(A)x(Y 2+= (3.4)

The prime will always denote derivative with respect to the variable x. From

equation (3.3) we obtain, taking into account 2BYAY +=

2

2112

110Y

)BYA(b

Y

b)BYA(aYaa)x(w

+

++++= (3.5)

)BYA(YB2YBA(a)BYA(a2Yaa)x(w 2212

110 +++++++=


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5

2

21

21

211

Y

)BYA(BYb2)YBA(b)BYA(b2

Y

b +++++

+

3

221

Y

)BYA(b2 ++ (3.6)

Therefore equation (3.1), written as

02w)x(4wx8w3)w(ww2 22342 =

under the substitutions (3.3), (3.5) and (3.6), becomes

3

22112

1101

10

Y

)BYA(b2

Y

b)BYA(a2Yaa

Y

bYaa2

++

++++

++

)BYA(YB2YBA(a 221 ++++

+++++

2

21

21

21

Y

)BYA(BYb2)YBA(b)BYA(b2

41

10

2

2

2112

110Y

bYaa3

Y

)BYA(b

Y

b)BYA(aYaa

++

+

++++

02Y

bYaa)x(4

Y

bYaax8

21

102

31

10 =

++

++ (3.7)

Upon expanding and equating the coefficients of Y to zero, we obtain from the

above equation a system of nine nonlinear ordinary differential equations from

which we can determine the various expansion coefficients. We obtain

coefficient of 4Y :

0a3Ba3 41221 = (3.8)

coefficient of 3Y :

0ax8)Ba2Ba(a2Baa4Baa2aa12 311112

1011310 =+++ (3.9)

coefficient of 2Y :

211101

221 )a(Ba)BbaAa(2)ax(a4 +


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6

211

210

21

20

21

2011 Bba4aax24aa12a)aab2(6 ++

0)ABa2a(a2)Ba2Ba(a2 111110 =+

+

+

+ (3.10)coefficient of Y:

1020111

210

210 aa)aba2(12baa12)x(aa8 +

)AaAa2aBb2Bb(a2)Ba2Ba(b2 110111111 ++++++

)BbaAa(a2Bba2)BAa2a(a2 101111110 +++

0)aaab(x24 120

211 =+ (3.11)

coefficient of 0Y :

)AaAa2aBb2Bb(a2)BAa2a(b2 110110111 +++++

21011111111 )BbaAa(ab2ABba2)ABb2b(a2 ++++

21

21

30110

22011 ba6)abaa6(x8)x()aba2(4 ++

02)aba2(3baa24 220111120 =+ (3.12)

coefficient of 1Y

:

102011

2110

210 ba)aba2(12baa12)x(ba8 +

)AaAa2aBb2Bb(b2)Ab2Ab(a2 110111111 +++++

)BbaAa(b2Aab2)ABb2b(a2 101111110 +++

0)baba(x24 120

211 =+ (3.13)

coefficient of 2Y :

211101221 )b(Ab)BbaAa(2)x(b4 ++

211

210

21

20

21

2011 Aba4bax24ba12b)aab2(6 ++

0)ABb2b(b2)Ab2Ab(a2 111110 =+++ (3.14)

coefficient of 3Y

:

0bx8)Ab2Ab(b2Aba4Abb2ba12 311112

1011310 =+++ (3.15)


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coefficient of 4Y :

0b3Ab3

4

1

22

1 = (3.16)We are to solve the system of equations (3.8)-(3.16), supplemented by Riccati's

equation 2BYAY += . From equations (3.8) and (3.16), ignoring the trivial

solutions, we obtain

Ba1 = and Ab1 = (3.17)

respectively. We thus consider the following four cases.

4. Case I. First Solution of the Painlev PIV Equation.

We first consider the case

Ba1= and Ab1= (4.1)

We then obtain from (3.9) and (3.15)

x2B

Ba2 0

= and x2

A

Aa2 0

= (4.2)

respectively. Equating the two different expressions of 0a we have 0A

A

B

B=

+

and by integration, we find that

pBA = (4.3)

where p is a constant to be determined.

From equation (3.10), using (4.1) we obtain

0)x(4ax24a2a18AB4B

B

B

Ba6

B

B2

200

20

2

0 =

+

From the above equation, using the first of (4.2) and (4.3), we get

0)x(2B

B

2

3

B

B 22

=++

(4.4)

where

1p22 += (4.5)

From equation (3.14), using (4.1) we get


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8

0)ax(4ax24a2a18BA4A

A

A

Aa6

A

A2 200

20

2

0 =+

From the above equation, using the second of (4.2) and (4.3), we obtain

0)x(2A

A

2

3

A

A 22

=++

(4.6)

where

1p22 = (4.7)

From equation (3.11) we get, using (4.1)

300000 a12BA4BA8BAa32a2

BBa2

BBa2 ++

0)aAB(x24)ax(a8 202

0 =+


+

23

B

B

2

3

B

Bx2

B

B

2

3

B

B

B

B

B

B

B

B

B

B

0)p22x(x4B

B)2p44x2(

22

=+

+++ (4.8)

From equation (3.13) we get, using (4.1)

300000 a12BA8BA4BAa32a2

A

Aa2

A

Aa2 ++

0)aAB(x24)ax(a8 202

0 =+

From the above equation, using the second of (4.2) and (4.3), we get

+

+

23

2 AA

23

AAx2

AA

23

AAA3

AA

0)p22x(x4A

A)1p22x(2 22 =++

++ (4.9)

From equation (3.12), using (4.1) and the first of (4.2), we obtain

4

2

22

B

B

16

9

B

BB

2

1

B

B

B

B

B

B

4

1

B

B

+

+


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+

+

+

A

A

B

B)BA(2

B

B

2

3

B

BB3

B

Bx

3

2

+

+

+B

B

A

A)AB(3

B

B)2x(x2)AB(

B

B5

A

Ax2 2

)AB(8x4xB

B1x

2

1

B

B)AB(8 24

22

2

+++

++

012)BA(8)AB(x4 22 =+ (4.10)

The previous equation can be further transformed taking into account B

B

A

A

=

,

pBA = and2

B

B2

A

A

B

B

=

+

. We also express the derived equation in terms of

the function

B

BF

= (4.11)

taking into account the identities 2FF

B

B+=

and 3FFF3F

B

B++=

. We thus

have that (4.10) takes the form

F)p42x(x2F2

1FxF

2

1FF

2

1 233 +

F)F(4

1F

16

1F)p22x(

2

1 2422 ++++

0)12p8p8xp4x4x( 2224 =++++ (4.12)

Before we go on to solving equation (4.4), we first prove that equations (4.4) and

(4.8) are compatible each other: every solution of (4.4) satisfies equation (4.8). To

prove that, we express both equations in terms of F (introduced in (4.11)). We

find that (4.4) and (4.8) are equivalent to

0)1p22x(2F2

1F 22 =+++ (4.13)

and


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0)p22x(x4F)1p22x(2)F2

1F(x2F

2

1F 2223 =++++ (4.14)

respectively. Upon substituting the expression 2F2

1F derived from (4.13) into

(4.14) we obtain the equation

0x4F)1p22x(2F2

1F 23 =++++ (4.15)

Multiplying (4.13) by F and rearranging terms, we obtain

FFF)1p22x(2F2

1 23 =+++

Substituting the above expression into (4.15) we get

0x4FFF =+ (4.16)

Differentiation of (4.13) gives 0x4FFF =+ which is (4.16). We have thus

proved that (4.4) and (4.8) are compatible each other.

We now come into solving equation (4.4), i.e. equation (4.13):

0)x(2F

2

1F 22 =++ (4.17)

The above equation is a Riccati differential equation and thus under the standard

substitution

u

u2F = (4.18)

takes on the form

0u)x(u 2 =+ (4.19)

The above is a Weber differential equation (Abramowitz and Stegun [3], 19.1-

19.3) and its solution can be expressed in terms of Parabolic Cylinder functions,

Kummer's confluent hypergeometric functions or Whittaker's functions. We prefer

to express the solution in terms of Whittaker's W and M functions (Abramowitz

and Stegun [3], 13.1.31-13.1.34) since the notation is more compact:

})x(MC)x(WC{x

1)x(u 2,2

2,1 += (4.20)


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where

4

= , 4

1

= (4.21)

The function )x(BB = can easily be calculated. Sinceu

u2

B

B =

, we obtain by

integration

)x(u

C)x(B

2= (4.22)

where C is a constant. We come next to evaluate the function Y solving

Riccati's equation 2BYAY += . Under the substitution

v

v

B

1Y

= (4.23)

Riccati's equation takes on the form

0v)AB(vB

Bv =+

(4.24)

Sinceu

u2

B

B =

and pAB = , equation (4.24) becomes 0vupvu2vu =++

which can also be written as 0)vu(pvu)vu( =+ . Under the substitution

vuU= (4.25)

we get the differential equation 0Uu

upU =

+ and since +=

2xu

u(see

equation (4.19)) we get the equation

0U)px(U 2 =+ (4.26)

The above differential equation admits the general solution

})x(MC~

)x(WC~

{x

1)x(U

2,2

2,1 += (4.27)

where

4

p = ,

4

1= (4.28)


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We have furtheru

u

U

U

v

v

=

and then we obtain the solution of Riccati's equation

from (4.23), using also (4.22):

=

)x(u

)x(u

)x(U

)x(U)x(u

C

1Y 2 (4.29)

where )x(u and )x(U are given by (4.20) and (4.27) respectively.

We now come to consider the differential equations (4.6) and (4.9) satisfied by the

function A. These equations can be expressed in terms of the function

A

AH

= (4.30)

as

0)1p22x(2H2

1H 22 =++ (4.31)

and

H)1p22x(2H2

1Hx2H

2

1H 223 ++

+

0)p22x(x4 2 =++ (4.32)

The above two equations are compatible each other. The proof is along the lines of

proving that (4.13) and (4.14) are compatible. We thus have to combine (4.4) with

(4.6) and establish a compatibility condition. Adding (4.4) and (4.6) and taking

into account thatB

B

A

A =

and

2

B

B2

A

A

B

B

=

+

, we obtain the equation

0)p22x(4BB 22 =++

, which in view of

uu2

BB = , takes the form

0)p22x(u

u 22

=+

(4.33)

This is thefirst compatibility condition.

We consider next the compatibility issue between (4.4), (4.8) and (4.10). We could

instead examine the compatibility between (4.13), (4.15) and (4.12).


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Solving (4.13) with respect to 2F2

1F squaring both members and rearranging,

we obtain

22422 )1p22x(F16

1FF

4

1)F(

4

1++=+ (4.34)

Adding (4.12), (4.34) and substituting 3F2

1F derived from (4.15) into (4.12)

simultaneously, we get the equation

22222 F)1p22x(

2

1Fxp4F

2

1FF

2

1FF

4

1+++

+

0p44422p16p12x2xp8 2222 =++++ (4.35)

Upon substituting 2F2

1F derived from (4.13) into (4.35) we obtain the equation

02p442p8p6)x2F(xp2 22 =+++ (4.36)

Sinceu

u2F = with u given by (4.20), equation (4.36) takes the form

0)2p442p8p6()x(u

)x(ux)x(uxp4 22 =++++

+ (4.37)

The above equation can be written as

0)x(uk])x(ux)x(u[xp4 =+ (4.38)

where

2p442p8p6k 22 ++++= (4.39)

Equation (4.38) will be called thesecond compatibility condition.

Both compatibility conditions are examined in Appendix A. According to the

results of Appendix A, the compatibility conditions hold for every x if

n43 =+ ( L,3,2,1,0n= ) and 0C

4

1

2C 12 =

+

+ (4.40)


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We thus see that to the first family of solutions there correspond an infinite

number of subfamilies. Since is given by (4.5), we can build a Table of the

various values of the parameters. The values of the parameters affect both

functions )x(u and )x(U given by the pairs (4.20), (4.21) and (4.27), (4.28).

We consider below a Table (Table 1) for the various parameters corresponding to

the first four values of n ( 3,2,1,0n= ).

n p

+

4

1 0C

4

1

2C 12 =

+

+

0 3 2+

4

5+ 2

0CC 12 =+

1 7 4+

4

11+

3

4 0C

2

3C 12 =+

2 11 6+

4

17+

5

8 0C

4

5C 12 =+

3 15 8+

4

23+

10516 0C8

105C 12 =+

Table 1

Therefore we have the following

For 0n= , 2p += and

)}x(M)x(W{

x

C)x(u 2,

2,

1 = where

4

3= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += where

4

5+= ,

4

1=

For 1n= , 4p += and

+= )x(M2

3)x(W

x

C)x(u 2,

2,

1 where4

7= ,

4

1=


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15

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += where

4

11+= ,

4

1=

For 2n= , 6p += and

= )x(M4

5)x(W

x

C)x(u 2,

2,

1 where4

11= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += where

4

17+= ,

4

1=

For 3n= , 8p += and

+= )x(M8

105

)x(Wx

C

)x(u2

,

2

,

1

where 4

15

= , 4

1

=

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += where

4

23+= ,

4

1=

Conclusion. The solution of Painlev PIV equation

ww)x(2wx4w

2

3

dx

dw

w2

1

dx

wd 2232

2

2 ++++

=

is given byY

bYaa)x(w 110 ++= where

+

=

= x

)x(u

)x(ux

B

B

2

1a0 ,

)x(u

C)x(Ba

21 == , )x(u

C

pAb 21 ==

and

=

)x(u

)x(u

)x(U

)x(U)x(u

C

1Y 2 . Therefore we have the solution

1

)x(u

)x(u

)x(U

)x(Upx

)x(U

)x(U)x(w

+

= (4.41)

where )x(U and )x(u are given by (4.27) and (4.20) respectively.

Below we list the following four solutions only.

(I) For 0n= ,

1

)x(u

)x(u

)x(U

)x(U)2(x

)x(U

)x(U)x(w

+

+

=


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16

)}x(M)x(W{x

C)x(u 2,

2,

1 = with

4

3= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += with

4

5+= ,

4

1=

(II) For 1n= ,

1

)x(u

)x(u

)x(U

)x(U)4(x

)x(U

)x(U)x(w

+

+

=

+= )x(M2

3)x(W

x

C)x(u 2,

2,

1 with4

7= ,

4

1=

})x(MC

~

)x(WC

~

{x

1

)x(U2

,2

2

,1 += with 4

11+

= , 4

1

=

(III) For 2n= ,

1

)x(u

)x(u

)x(U

)x(U)6(x

)x(U

)x(U)x(w

+

+

=

= )x(M4

5)x(W

x

C)x(u 2,

2,

1 with4

11= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += with

4

17+= ,

4

1=

(IV) For 3n= ,

1

)x(u

)x(u

)x(U

)x(U)8(x

)x(U

)x(U)x(w

+

+

=

+= )x(M8

105)x(W

x

C)x(u 2,

2,

1 with4

15= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(U 2,2

2,1 += with

4

23+= ,

4

1=

5. Case II. Second Solution of the Painlev PIV Equation.We next consider the case

Ba1 = and Ab1 = (5.1)

We then obtain from (3.9) and (3.15)

x2B

Ba2 0

= and x2

A

Aa2 0

= (5.2)


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17

respectively. Equating the two different expressions of 0a and integrating, we find

that

pBA = (5.3)

where p is a constant. From equation (3.10), using (3.26), we obtain

0)x(4BA4ax24a18a2B

B

B

Ba6

B

B2 20

200

2

0 =+


0)1p22x(2B

B

2

3

B

B 22

=++

(5.4)

From equation (3.14) we get

0)x(4ax24a2a18BA4A

A

A

Aa6

A

A2 200

20

2

0 =

+

From the above equation, using the second of (5.2) and (5.3) we obtain

0)1p22x(2A

A

2

3

A

A 22

=+++

(5.5)


300000 a12BA4BA8BAa32a2

B

Ba2

B

Ba2 ++

0)aBA(x24)x(a8 202

0 =+

From the above equation, using the first of (5.2) and (5.3) we obtain

B

B)1p22x(2

B

B

2

3

B

Bx2

B

B

2

3

B

BB3

B

B 223

2

++

+

+

0)p22x(x4 2 =++ (5.6)

From equation (3.13) we obtain

200

30000 ax12BAx12BAa16a6a

A

Aa

A

Aa +

0)x(a4BA2BA4 20 =+


18/47

18

From the above equation, using the first of (5.2) and (5.3) we obtain

A

Ax2A

Ax3A

A

2

3

A

AA3A

A23

2

+

+

0)p22x(x4A

A)1p22x(2 22 =+

+++ (5.7)

From equation (3.12), using (5.1) and the first of (5.2), we obtain

4

2

22

B

B

16

9

B

BB

2

1

B

B

B

B

B

B

4

1

B

B

+

+

++

++

AA

BB)BA(2

BB

23

BBB3

BBx

3

2

++

+

B

B

A

A)AB(3

B

B)2x(x2)AB(

B

B5

A

Ax2 2

)AB(8x4xB

B1x

2

1

B

B)AB(8 24

22

2

+++

+++

012)BA(8)AB(x4 22 =+ (5.8)

The previous equation can be further transformed taking into accountB

B

A

A =

,

pBA = ,2

B

B2

A

A

B

B

=

+

. We also express the derived equation in terms of the

function F defined by

B

BF

= (5.9)

taking into account the identities 2FFB

B+=

and 3FFF3F

B

B++=

. We thus

have that (5.8) takes the form

F)p42x(x2F2

1FxF

2

1FF

2

1 233 ++

+

F)F(4

1F

16

1F)p22x(

2

1 2422 +++


19/47

19

0)12p8p8xp4x4x( 2224 =++++ (5.10)

The above equation can be simplified using equations (5.4) and (5.6). First of all,equations (5.4) and (5.6) can be expressed in terms of F and they are equivalent

to

0)1p22x(2F2

1F 22 =++ (5.11)

and

0)p22x(x4F)1p22x(2)F2

1F(x2F

2

1F

2223 =+++++ (5.12)

respectively. Substituting 2F2

1F derived from (5.11) into (5.12) we obtain the

equation

0x4F)1p22x(2F2

1F

23 =+++ (5.13)

Equations (5.12) and (5.13) are compatible each other. The proof goes along the

lines of the previous Section. Solving (5.11) with respect to 2F

2

1F , squaring

both members and rearranging, we obtain

22422)1p22x(F

16

1FF

4

1)F(

4

1+=+ (5.14)

Adding (5.14) to (5.10) and substituting 3F2

1F derived from (5.13) into (5.10)

simultaneously, we obtain the equation

Fxp4F)1p22x(21F

21FF

21FF

41 22222 +

022p444p16p12x2xp8 2222 =+++ (5.15)

We substitute 2F2

1F derived from (5.11) into the above equation and we obtain

0)2p442p8p6()x4F2(xp 22 =+++++ (5.16)


20/47

20

The above equation isthesecond compatibility conditionbetween (5.4), (5.6) and

(5.8).

We come next to the solution of (5.4), which is equivalent to (5.11). Under the

substitution

f

f2F = (5.17)

equation (5.11) becomes

0f)x(f 2 =+ (5.18)

where

1p22 = (5.19)

The general solution of (5.18) is given by

)}x(MC)x(WC{x

1)x(f 2,2

2,1 += (5.20)

where

4

= ,

4

1= (5.21)

Sincef

f2

B

B =

, we obtain by integration

)x(f

C)x(B

2= (5.22)

The above relation expresses the function B in terms of )x(f .

Equations (5.5) and (5.7) are compatible each other. Every solution of (5.5)

satisfies equation (5.7). These two equations can be expressed in terms of the

function H defined by

A

AH

= (5.23)

as

0)1p22x(2H2

1H 22 =+++ (5.24)

and


21/47

21

H)1p22x(2H2

1Hx2H

2

1H 223 +++

0)p22x(x4 2 =+ (5.25)

In fact, multiplying (5.24) by F and rearranging, we obtain

H)1p22x(2HHH2

1 23 ++= (5.26)

Upon substitution of 3H2

1 using (5.26) and 2H

2

1H derived from (5.24) into

(5.25) we obtain the equation 0x4HHH =+ . This last equation however

comes by differentiation of (5.24).

We next examine the compatibility between equations (5.4) and (5.5). Adding

these two equations and taking into accountB

B

A

A =

and

2

B

B2

B

B

A

A

=

+

, we

obtain the equation 0)p22x(4B

B 22

=++

, which in view of

f

f2

B

B =

takes on the form

0)p22x(f

f 22

=+

(5.27)

The above equation will be calledthefirst compatibility condition.

In Appendix B we consider both equations (5.27) and (5.16) and we end up with a

pair of compatibility conditions.

We are now ready to solve Riccati's equation 2BYAY += . This equation under

the substitution

v

v

B

1Y

= (5.28)

takes the form 0v)AB(vB

Bv =+

. This equation, since

f

f2

B

B =

and

pAB= , becomes 0vfpvf2vf =++ , which is equivalent to


22/47

22

0)vf(pvf)vf( =+ (5.29)

The substitution

vfX= (5.30)

transforms (5.29) into 0Xf

fpX =

+ and since +=

2xf

f, we get the

equation

0X)px(X 2 =+ (5.31)

The general solution of (5.31) is given by

)}x(MC~

)x(WC~

{x

1)x(X 2,2

2,1 += (5.32)

where

4

p = ,

4

1= (5.33)

Sincef

f

X

X

v

v

=

, we have that

=

ff

XX

C)x(fY

2

(5.34)

Both compatibility conditions (5.27) and (5.16) are examined in Appendix B.

According to the results of Appendix B, the compatibility conditions hold for

every x if

n43 =+ ( L,3,2,1,0n= ) and 0C

4

1

2C 12 =

+

(5.35)

We thus see that to the first family of solutions there correspond an infinite

number of subfamilies. Since is given by (5.19), we can build a Table of the

various values of the parameters. The values of the parameters affect both

functions )x(w and )x(X given by the pairs (5.20), (5.21) and (5.32), (5.33).

We consider below a Table (Table 2) for the various parameters corresponding to

the first four values of n ( 3,2,1,0n= ).


23/47

23

n p

+

4

1 0C

4

1

2C 12 =

+

0 3 1+

4

4+ 2

0CC 12 =+

1 7 3+

4

10+

3

4 0C

2

3C 12 =

2 11 5+

4

16+

5

8 0C

4

5C 12 =+

3 15 7+

4

22+ 105

16 0C

8

105C 12 =

Table 2

Therefore we have the following

For 0n= , we have 1p += and

)}x(M)x(W{x

C

)x(f

2

,

2

,

1

= where 4

3

= , 4

1

=

})x(MC~

)x(WC~

{x

1)x(X 2,2

2,1 += where

4

4+= ,

4

1=


+= )x(M2

3)x(W

x

C)x(f 2,

2,

1 where4

7= ,

4

1=

})x(MC

~

)x(WC

~

{x

1

)x(X

2

,2

2

,1 += where 4

10+

= , 4

1

=


= )x(M4

5)x(W

x

C)x(f 2,

2,

1 where4

11= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(X 2,2

2,1 += where

4

16+= ,

4

1=



24/47

24

+= )x(M8

105)x(W

x

C)x(f 2,

2,

1 where4

15= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(X 2,2

2,1 += where

4

22+= ,

4

1=

Conclusion. The solution of the Painleve PIV equation

ww)x(2wx4w

2

3

dx

dw

w2

1

dx

wd 2232

2

2 ++++

=

is given byY


x)x(f

)x(fx

B

B

2

1a0

=

= ,

)x(f

C)x(Ba

21 == , )x(f

C

p)x(Ab 21 ==

and

=

)x(f

)x(f

)x(X

)x(X

C

)x(fY

2

. Therefore we have the solution

1

)x(f

)x(f

)x(X

)x(Xpx

)x(X

)x(X)x(w

+

=

Below we list the following four solutions only.

For 0n= we have the solution

1

)x(f

)x(f

)x(X

)x(X)1(x

)x(X

)x(X)x(w

++

=

where

)}x(M)x(W{x

C)x(f 2,

2,

1 = with

4

3= ,

4

1=

})x(MC

~

)x(WC

~

{x

1

)x(X

2

,2

2

,1 += with 4

4+

= , 4

1

=


1

)x(f

)x(f

)x(X

)x(X)3(x

)x(X

)x(X)x(w

++

=

where

+= )x(M2

3)x(W

x

C)x(f 2,

2,

1 with4

7= ,

4

1=


25/47

25

})x(MC~

)x(WC~

{x

1)x(X 2,2

2,1 += with

4

10+= ,

4

1=


1

)x(f

)x(f

)x(X

)x(X)5(x

)x(X

)x(X)x(w

++

=

where

= )x(M4

5)x(W

x

C)x(f 2,

2,

1 with4

11= ,

4

1=

})x(MC~

)x(WC~

{x

1)x(X 2,2

2,1 += with

4

16+= ,

4

1=


1

)x(f

)x(f

)x(X

)x(X)7(x

)x(X

)x(X)x(w

++

=

where

+= )x(M8

105)x(W

x

C)x(f 2,

2,

1 with4

15= ,

4

1=

})x(MC~

)x(WC~

{

x

1)x(X 2,2

2,1 += with

4

22+= ,

4

1=

6. Case III. Third Solution of the Painlev PIV Equation.

We continue considering the case

Ba1= and Ab1 = (6.1)

From (3.9) and (3.15) we obtain

x2B

Ba2 0

= and x2

A

Aa2 0

= (6.2)


that

BrA 2= (6.3)

where r is real. From equation (3.10), using (6.1), we obtain

0)x(4ax24a2a18BA8B

B

B

Ba6

B

B2 200

20

2

0 =+

+


26/47

26

The above equation, using the first of (6.2) and the relation (6.3), becomes

0)12x(2Br8B

B

2

3

B

B 2222

=+++

(6.4)

From equation (3.14), using (6.1), we obtain

0)x(4ax24a2a18BA8A

A

A

Aa6

A

A2

200

20

2

0 =+

+

The above equation, using the second of (6.2) and (6.3), takes on the form

0)12x(2Br8

A

A

2

3

A

A 2222

=+++

(6.5)

Equations (6.4) and (6.5) are equivalent in view of (6.3).


300000 a12BA8BA4BAa40a2

B

Ba2

B

Ba2 +++

0)aBA(x24)x(a8 202

0 =+

The previous equation, taking into account the first of (6.2) and (6.3) becomes

+

23

2 B

B

2

3

B

Bx2

B

B

2

3

B

B

B

B

B

BB

B

B

0)2x(x4Bxr16BBr24B

B)12x(2 22222 =++

+++ (6.6)

From equation (3.13) we obtain

300000 a12BA4BA8BAa40a2

A

Aa2

A

Aa2 +++

0)aBA(x24)ax(a8 202

0 =+

The above equation using the second of (6.2) and (6.3) takes on the form

A

A)12x(2

A

A

2

3

A

Ax2

A

A

2

3

A

AA3

A

A 223

2

+++

+

0)2x(x4Bxr16BBr24 2222 =++ (6.7)


27/47

27

We now prove the compatibility of (6.4) and (6.6). These two equations expressed

in terms of

B

BG

= (6.8)

take the form

0)12x(2Br8G2

1G 2222 =+++ (6.9)

and

BBr24G)12x(2G

2

1Gx2G

2

1G 2223 +++

0)2x(x4Bxr16 222 =++ (6.10)

respectively. Substituting 2G2

1G obtained from (6.9) into (6.10) we get the

equation

0x4G)12x(2BBr24G2

1G 223 =++++ (6.11)

Multiplying (6.9) by G and rearranging we obtain the equation

G)12x(2GBr8GGG2

1 2223 +++=

We substitute in (6.11) the above expression and get the equation

0x4BBr16GGG 2 =+ (6.12)

Upon differentiation of (6.9) we obtain

x4BBr16GGG 2 =

which when combined with (6.12) gives an identity. Therefore (6.9) and (6.10),

i.e. (6.4) and (6.6) are compatible each other.

Equations (6.5) and (6.7) expressed in terms ofA

AH

= as

0)12x(2Br8H2

1H 2222 =+++

and


28/47

28

H)12x(2H2

1Hx2H

2

1H 223 +++

0)2x(x4BBr24Bxr16 2222 =++

respectively, are proved similarly to be compatible each other.

We now come into solving equation (6.4). Under the substitution

)x(g

1B

2= (6.13)

since

g

g2

B

B =

and

2

g

g6

g

g2

B

B

+

=

, equation (6.4) becomes

3

22

g

r4g)12x(g =++ (6.14)

which is Ermakov's equation. Two linearly independent solutions of the equation

0g)12x(g 2 =++

are given by Whittaker's W and M functions

)x(Wx

1g 2,1 = and )x(Mx

1g 2,2 = (6.15)

where

4

12 += ,

4

1= (6.16)

According to the standard procedure (Ermakov [8]) the solutions of Ermakov's

equation (6.14) are given by

++=

2

22,

12222

,2

1 dx))x(W(

xCCr4))x(W(x

1)x(gC (6.17)

or

++=

2

22,

12222

,2

1 dx))x(M(

xCCr4))x(M(

x

1)x(gC (6.18)


29/47

29

Equation (3.12) takes the form, because of (6.1) and the first of (6.2)

4

2

22

B

B

16

9

B

BB

2

1

B

B

B

B

B

B

4

1

B

B

+

+

+

+

A

A

B

B2)BA(2

B

B

2

3

B

BB3

B

Bx

3

2

+

+

+

B

B

A

A)AB(5

B

B)2x(x2)AB(

B

B3

A

Ax6 2

)AB(8x4xB

B1x

2

1

B

B)AB(14 24

22

2

++

++

+

012)BA(32)AB(4)AB(x4 22 =+

The above equation, taking into account BrA 2= and introducing the function

G byB

BG

= , takes the form

4233 G16

1G)G(

4

1G

2

1GxG

2

1GG

2

1++

2222222 G)2x(2

1G)2x(x2BBr6BBxr24)B(r19 +++++

012x4xBr32B)12x(r4 2444222 =++++ (6.19)

Let us now determine the function Y which is the solution of Riccati's equation

2YBAY += . Under the standard substitutionu

u

B

1Y

= Riccati's equation

takes the form 0uABuBBu =+ which because of BrA 2= becomes


30/47

30

0uBruB

Bu 22 =

or dividing by B, 0uBr

B

Bu

B

u 22

=

. The last

equation can be written as 0uBrB

u 2 =

and then multiplying by

B

uwe get

the equation 0uurB

u

B

u 2 =

which is equivalent to 0)u(r

B

u 222

=

from which we get 0urB

u 222

=

(the integration constant has been set equal

to zero). We thus have urB

u=

and then Br

u

u=

. Because of the last equality,

we obtain r)Br(B

1

u

u

B

1Y ==

= .

We come next to consider the compatibility between the equations (6.6), (6.4) (i.e.

(6.9), (6.10)) and (6.19). Solving (6.9) with respect to 2G2

1G , squaring both

members and rearranging, we have

=+ 422 G16

1GG

4

1)G(

4

1

2222244 )12x(B)12x(r8Br16 +++++= (6.20)

We add (6.19), (6.20) and simultaneously substitute 3G2

1G derived from

(6.11) into (6.19). We then obtain the equation

BBr6)B(r7G2

1GG2

1GG4

1 222222 +

+

4422222 Br48B)12x(r12G)12x(2

1+++++

012)12x(x4x4x 22224 =+++++ (6.21)


31/47

31

We substitute 2G2

1G derived from (6.9) into the previous equation and we get

the equation

22222442

22 Br8B)12x(r12Br48B

B

2

3

B

BBr6 ++++

0)242(2 2 =+++ (6.22)

In the above expression we substitute

2

B

B

2

3

B

B

derived from (6.4) and we get

the equation

0)242(Br4 222 =+++ (6.23)

which is the compatibility condition.

It is rather hard to find relations between the various parameters and the constants

due to the complexity of the solutions. It is not even known if there are any such

relations. One way out of this problem is to substitute the found expressions to the

original equation and check if they satisfy the equation, a task which may require

supercomputing facilities with symbolic capabilities.


ww)x(2wx4w

2

3

dx

dw

w2

1

dx

wd 2232

2

2 ++++

=

is given byY


+== x

)x(g

)x(gxB

B

2

1a0 ,)x(g

1)x(Ba21

== ,)x(g

rBr)x(Ab2

221 ===

and rY = . Therefore we have the solutions

)x(g

r2x

)x(g

)x(g)x(w

2+

+

= if rY=


32/47

32

)x(g

r2x

)x(g

)x(g)x(w

2

+

= if rY =

In both cases the function )x(g is given by (6.17) or (6.18) provided that the

compatibility condition (6.23) is satisfied.

7. Case IV. Fourth Solution of the Painlev PIV Equation.

We finally consider the case

Ba1 = and Ab1= (7.1)

From (3.9) and (3.15) we obtain

x2BBa2 0 = and x2

AAa2 0 = (7.2)


that

BrA 2= (7.3)

where r is real. From equation (3.10), using (7.1), we obtain

0)x(4ax24a2a18BA8B

B

B

Ba6B

B2 20020

2

0 =++

From the above equation, taking into account the first of (7.2) and (7.3), we obtain

0)12x(2Br8B

B

2

3

B

B 2222

=++

(7.4)

The previous equation can also be written in terms ofB

BG

= as

0)12x(2Br8G2

1G

2222

=++ (7.5)


0)x(4ax24a2a18BA8A

A

A

Aa6

A

A2 200

20

2

0 =++

From the above equation, taking into account the second of (7.2) and (7.3), we

obtain


33/47

33

0)12x(2Br8A

A

2

3

A

A 2222

=++

(7.6)

Equations (7.4) and (7.6) are equivalent in view of (7.3).


300000 a12BA8BA4BAa40a2

B

Ba2

B

Ba2 +++

0)aBA(x24)x(a8 202

0 =+

From the above equation, taking into account the first of (7.2) and (7.3), we obtain

BB)12x(2

BB

23

BBx2

BB

23

B

BB3B

B 2232

++

+

+

0)2x(x4Bxr16BBr24 2222 =++ (7.7)

The previous equation can be expressed in terms ofB

BG

= as

G)12x(2G2

1Gx2G

2

1G 223 ++

+

0)2x(x4Bxr16BBr24 2222 =++ (7.8)

Substituting

2G

2

1G derived from (7.5) into (7.8) we obtain the equation

0x4BBr24G)12x(2G2

1G 223 =+++ (7.9)

We can prove - using also (7.9) - that equations (7.5) and (7.8) are compatible

each other, i.e. every solution of (7.5) satisfies (7.8). The proof goes along the

lines of the previous Section.

From equation (3.13) we obtain using (7.1)

200

30000 ax12BAa20a6a

A

Aa

A

Aa ++

0)x(4BA2BA4BAx12 2 =++

The above equation, because of the second of (7.2) and (7.3) becomes


34/47

34

A

A)12x(2

A

A

2

3

A

Ax2

A

A

2

3

A

AA3

A

A 223

2

++

+

+

0)2x(x4Bxr16BBr24 2222 =++ (7.10)

Equations (7.6) and (7.10) expressed in terms ofA

AH

= as

0)12x(2Br8H2

1H 2222 =++

and

H)12x(2H21Hx2H

21H 223 ++

+

0)2x(x4BBr24Bxr16 2222 =++

respectively, are proved similarly to be compatible each other.

We now come into solving equation (7.4). Under the substitution

)x(g

1B

2= (7.11)

sinceg

g2

B

B =

and

2

g

g6

g

g2

B

B

+

=

, equation (7.4) becomes

3

22

g

r4g)12x(g =+ (7.12)

which is Ermakov's equation. Two linearly independent solutions of the equation

0g)12x(g 2 =+

are given by Whittaker's W and M functions

)x(Wx

1g 2,1 = and )x(M

x

1g 2,2 = (7.13)

where

4

12 = ,

4

1= (7.14)


35/47

35

According to the standard procedure (Ermakov [8]) the solutions of Ermakov's

equation (7.12) are given by

++=

2

22,

12222

,2

1 dx))x(W(

xCCr4))x(W(

x

1)x(gC (7.15)

or

++=

2

22,

12222

,2

1 dx))x(M(

xCCr4))x(M(

x

1)x(gC (7.16)

We find, using the method of the previous Section that the solution of Riccati's

equation is given by

rY = (7.17)

Equation (3.12) takes the form, because of (7.1) and the first of (7.2)

4

2

22

B

B

16

9

B

BB

2

1

B

B

B

B

B

B

4

1

B

B

+

+

+

++

AA

BB2)BA(2

BB

23

B

BB3BBx

3

2

+

++

+

+B

B

A

A)AB(5

B

B)2x(x2)AB(

B

B3

A

Ax6 2

)AB(8x4xB

B1x

2

1

B

B)AB(14 24

22

2

++

+++

+

012)BA(32)AB(4)AB(x4 22 =

The above equation, taking into account BrA 2= and introducing the function G

byB

BG

= , takes the form

4233 G16

1G)G(

4

1G

2

1GxG

2

1GG

2

1+

+


36/47

36

2222222 G)2x(2

1G)2x(x2BBr6BBxr24)B(r19 +++++

012x4xBr32B)12x(r4 2444222 =++++ (7.18)

Solving (7.5) with respect to 2G2

1G , squaring both members and rearranging,

we have

=+ 422 G16

1GG

4

1)G(

4

1

2222244 )12x(B)12x(r8Br16 +++= (7.19)

We add (7.18), (7.19) and simultaneously substitute 3G2

1G derived from (7.9)

into (7.18). We then obtain the equation

BBr6)B(r7G2

1GG

2

1GG

4

1 222222 +

4422222 Br48B)12x(r12G)12x(2

1+++

012)12x(x4x4x 22224 =+++

In the above equation we substitute 2G2

1G derived from (7.5) and we obtain

the equation

44222222

22 Br48Br8B)12x(r12B

B

2

3

B

BBr6 ++

0)242(2 2 =++

Upon substituting

2

B

B

2

3

B

B

derived from (7.4) into the above equation we

obtain

0)242(Br4 222 =+++ (7.20)

which is the compatibility condition.


37/47

37

It is rather hard to find relations between the various parameters and the constants

due to the complexity of the solutions. It is not even known if there are any such

relations. One way out of this problem is to substitute the found expressions to the

original equation and check if they satisfy the equation, a task which may require

supercomputing facilities with symbolic capabilities.


ww)x(2wx4w

2

3

dx

dw

w2

1

dx

wd 2232

2

2 ++++

=

is given byYbYaa)x(w 110 ++= where

x)x(g

)x(gx

B

B

2

1a0

=

= ,

)x(g

1)x(Ba

21 == ,

)x(g

rBr)x(Ab

2

22

1 ===

and rY = . Therefore we have the solutions

)x(g

r2x

)x(g

)x(g)x(w

2+

= if rY=

)x(g

r2x

)x(g

)x(g)x(w

2= if rY =

In both cases the function )x(g is given by (7.15) or (7.16) provided that the

compatibility condition (7.20) is satisfied.

8. The

G

Gexpansion method with variable expansion

coefficients.

We consider that equation (3.1) admits a solution of the form

+=

)x(G

)x(G)x(b)x(b)x(w 10 (8.1)

Upon substituting (8.1) into (3.1) we obtain an equation, which when arranged in

powers of G, takes the form

2b)x(4bx8b3)b(bb2 2023

040

2000


38/47

38

+++

++ )x(bb4bb6bbx12

G

G)bb4bb(

)x(G

2 201

301

2011001

)x(Gbb2G

Gbb2bb2b 1010011

+

+++

++

2

1122

12

1010012 G

Gbb)x(b4)bb(18bb4bb2

])x(G[

1

221011

21011 ))x(G(

G

Gb2

G

G)b3b2(b2bbx24bb2

+

++

32111

310

31013

))x(G(GGb4bb2bb12bx8bb4

])x(G[1

+

421

214

))x(G()b1(b])x(G[

3+ (8.2)

Equating to zero all the coefficients of )x(G in the above equation, we obtain the

following system of ordinary differential equations

02b)x(4bx8b3)b(bb2 2023

040

2000 = (8.3)

)x(bb4bb6bbx12G

G)bb4bb( 201

301

2011001 +++

++

0bb2G

Gbb2bb2b 1010011 =+

+++ (8.4)

2

1122

12

101001G

Gbb)x(b4)bb(18bb4bb2

+

0G

Gb2

G

G)b3b2(b2bbx24bb2 2

1011

2

1011 =

+

++ (8.5)

0G

Gb4bb2bb12bx8bb4 2111

310

3101 =

(8.6)

0b1 21 = (8.7)

Equation (8.3) is essentially equation (3.1) and thus admitsallthe solutions found

in previous Sections. From equation (8.7), we obtain that


39/47

39

1b1 = (8.8)

From equation (8.6), taking into account the above values of 1b , we derive the

equations

)bx(b2G

G01 +=

(8.9)

From equation (8.4), we obtain

0b3bx6)x(2b

b

G

G

b2

b

G

G 200

2

0

0

0

0 =+++

+

+

(8.10)

From equation (8.5), we obtain

0b9bx12)x(2b

b

G

G

2

1

G

G

b

b3

G

G 200

2

1

02

1

0 =

+

(8.11)

From equations (8.10) and (8.11), we obtain because of (8.9), the following two

expressions for the ratioG

G

++

+

=

2x2b3bx6

b

b)bx(b

b

b

G

G 2200

0

001

0

0 (8.12)

and

+++

=

2x4bx10b5

b

b

G

G 20

20

1

0 (8.13)

respectively.

Upon equating the two expressions (8.12) and (8.13), we obtain the following

differential equation for the coefficient 0b :

0)4x6bx16b8(bbbxbb2bb 20200100001 =+++ (8.14)

We thus have to solve the following two systems depending on the value of the

coefficient 1b :

(I) For 1b1= , we have

)bx(2G

G0+=

(8.15)


40/47

40

+++=

2x4bx10b5b

G

G 20

200 (8.16)

0b8bx16b)2x3(2bxbb2b 30200

20000 =+++ (8.17)

(II) For 1b1 = , we have

)bx(2G

G0+=

(8.18)

+++=

2x4bx10b5b

G

G 20

200 (8.19)

0b8bx16b)2x3(2bxbb2b3

0

2

00

2

0000 =+++++ (8.20)

Each one of the above systems should be supplemented by equation (8.3).

Upon elimination of 0b between (8.3) and (8.17), (8.20) we obtain the equations

02b)3x4(4bx40b19b)xb2(b2)b(20

230

40000

20 =+++++ (8.21)

and

02b)3x4(4bx40b19b)xb2(b2)b(20

230

40000

20 =++++++ (8.22)

respectively. These two equations are the compatibility conditions. Every solutionwhich satisfies (8.3) should satisfy each one of (8.21) and (8.22).

We thus have the two solutions:

Solution 1. This solution corresponds to 1b1= . Dividing (8.16) by (8.15) we

obtain

)bx(2

2x4bx10b5b

G

G

0

20

200

+

+++=

(8.23)

where 0b satisfies the compatibility condition (8.21). Solving (8.23) we determine

)x(G .

Solution 2. This solution corresponds to 1b1 = . Dividing (8.19) by (8.18) we

obtain

)bx(2

2x4bx10b5b

G

G

0

20

200

+

+++=

(8.24)


41/47

41

where 0b satisfies the compatibility condition (8.22). Solving (8.24) we determine

)x(G .

Because of the complexity of the expressions for 0b , the integration of the

differential equations (8.23) and (8.24) might require supercomputing facilities

with symbolic capabilities. A remarkable feature of the )G/G( expansion with

variable expansion coefficients is that a repeated use of the method in determining

0b leads to a proliferation of solutions, a rather unique feature of the Riccati

)G/G( expansion.

Appendix A.

We examine the first compatibility condition given by equation (4.33). This

equation splits into a pair of equations

0)x(up22x)x(u 2 =+ (A.1)

and

0)x(up22x)x(u 2 =++ (A.2)

where )x(u is given by (4.20).

Expanding (A.1) into power series we obtain

+

+

+

121 C

4

1

2CC

4

3)p22(

xC

4

1

2Cp22C

4

3121

+

+

+


42/47

42

2

121xC

4

1

2Cp22C

4

3]1)p(2[

p222

1

+

+

+

)x(O 3+ (A.3)

Expanding (A.2) into power series we obtain

+

+

+

121 C

4

1

2CC

4

3)p22(

xC

4

1

2Cp22C

4

3121

+

+

+

2121 xC

4

1

2Cp22C

4

3]1)p(2[

p222

1

+

+

+

+

+

)x(O 3+ (A.4)

We next consider the second compatibility condition (4.38):

0)x(uk])x(ux)x(u[xp4 =+ (A.5)

Taking into account the solution )x(u given by (4.20) and expanding, we get from

(A.5)

21121 xC

4

32

]p8)p8k[(xC

4

1

2C)p4k(C

4

3k

+

+

+

+++

+

312 xC

4

1

2C]p24)p12k[(

6

1

+

+++


43/47

43

41

2

xC

4

3

24

]k2p32p48)p16k[(

+

+++

)x(OxC

4

1

2C]k6p120p80)p20k[(

120

1 6512

2 +

+

+++++ (A.6)

A remarkable feature of the above compatibility conditions, expressed by the

expansions (A.3), (A.4) and (A.6) is that there are two patterns appearing

repeatedly:

+

4

3

1 and

+

+ 12 C

4

1

2C

Therefore the compatibility conditions hold for every x if

n43 =+ ( L,3,2,1,0n= ) and 0C

41

2C 12 =

+

+ .

It is 0

4

3

1=

+

for n43 =+ ( L,3,2,1,0n= ) in view of the fact that the

function )z( has a pole at 0z= and to the set of negative integers (Abramowitz

and Stegun [3], 6.1.7):

0)z(

1

limnz = , L,3,2,1,0n=

Appendix B. We examine the first consistency condition (5.27) written as a

couple of equations

0)x(fp22x)x(f 2 =+ (B.1)

and


44/47

44

0)x(fp22x)x(f 2 =++ (B.2)

where )x(f is given by (5.20).

Expanding (B.1) into power series we obtain

+

+

+

121 C

4

1

2CC

4

3)p22(

xC

4

1

2Cp22C

4

3121

+

+

+

2121 xC

4

1

2Cp22C

4

3]1)p(2[

p222

1

+

+

+

)x(O 3+ (B.3)

Expanding (B.2) into power series we obtain

+

+

+

121 C

4

1

2CC

4

3)p22(

xC

41

2Cp22C

43

121

+

+

+

2121 xC

4

1

2Cp22C

4

3]1)p(2[

p222

1

+

+

+

+

+

)x(O 3+ (B.4)


45/47

45

We next consider the consistency condition (5.16). This condition in view of

f

f

2F

= , takes the form

0)x(fk)]x(fx)x(f[xp4 =+ (B.5)

where

242p)84(p6k 22 ++++= (B.6)

and )x(f is given by (5.20).

Upon expanding )x(f we obtain from (B.5)

2112

1 x

4

3

C]p8)p8k([

2

1xC

4

1

2C)kp4(

4

3

Ck

+

+

+

+

312 xC

4

1

2C]p24)kp12[(

6

1

+

++

412 x

4

3

C]p32k2p48)p16k[(

24

1

+

+

)x(OxC

4

1

2C]k6p120p80)kp20[(

120

1 6512

2 +

+

+++ (B.6)

We see that there are two patterns appearing repeatedly:

+

4

3

1 and

+

12 C

4

1

2C

Therefore expression (B.5) holds for every x if


46/47

46

n43 =+ ( L,3,2,1,0n= ) and 0C

4

1

2C 12 =

+

References

[1] M. J. Ablowitz, A. Ramani and H. Segur: "A connection between

nonlinear evolution equations and ordinary differential equations of

P-type. I"J. Math. Phys. 21(1980) 715-721

[2] M. J. Ablowitz, A. Ramani and H. Segur: "A connection between

nonlinear evolution equations and ordinary differential equations of

P-type. II"J. Math. Phys. 21(1980) 1006-1015

[3] M. Abramowitz and I. A. Stegun: "Handbook of Mathematical

Functions".Dover 1972

[4] H. Airault: "Rational solutions of Painlev equations"

Stud. Appl. Math. 61(1979) 31-53

[5] S. Antoniou: The Riccati equation method with variable expansioncoefficients. I. Solving the Burgers equation. submitted for publication

[6] S. Antoniou: Some General Solutions to the Painlev PII Equation.

submitted for publication

[7] P. A. Clarkson: "Painlev equations-nonlinear special functions"

J. Comp. Appl. Math. 153(2003) 127-140

[8] V. P. Ermakov: Second-Order Differential Equations: Conditions of

Complete Integrability.Appl. Anal. Discr. Math.2(2008) 123-145

Translation from the original Russian article:

Universitetskiye IzvestiyaKiev No. 9(1880) 1-25

[9] A. S. Fokas and M. J. Ablowitz: "On a unified approach to

transformations and elementary solutions of Painlev equations"

J. Math. Phys. 23(1982) 2033-2042


47/47

[10] E. L. Ince: "Ordinary Differential Equations". Dover 1956

[11] A. Ramani, B. Dorizzi and B. Grammaticos: "Painlev conjecture

revisited". J. Math. Phys. 21 (1980) 715-721

[12] A. Ramani, B. Grammaticos and T. Bountis: "The Painleve Property and

Singularity Analysis of Integrable and Non-Integrable Systems".

Phys. Rep. 180(1989) 159-245

[13] J. Weiss, M. Tabor and G. Carnevale: "The Painlev property for partial

differential equations".J. Math. Phys. 24(1983) 522-526

[14] J. Weiss: "The Painlev property for partial differential equations. II:

Bcklund transformation, Lax pairs, and the Schwarzian derivative"

J. Math. Phys. 24(1983) 1405-1413

[15] J. Weiss: "On classes of integrable systems and the Painlev property "

J. Math. Phys. 25(1984) 13-24

[16] J. Weiss: "The Painlev property and Bcklund transformations for the

sequence of Boussinesq equations".

J. Math. Phys. 26(1985) 258-269[17] Y. Wenjum and Li Yezhou: "Rational Solutions of Painlev Equations"

Canadian J. Math. 54(2002) 648-670

Documents

Some General Solutions to the Painlevé – PIV Equation