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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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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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)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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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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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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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 =+
From the above equation, using the first of (4.2) and (4.3), we get
+
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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})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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)}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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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 =+
From the above equation, using the first of (3.27) and (3.28), we get
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)
From equation (3.11) we get
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 =+
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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 +++
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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)
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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
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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
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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= ).
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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=
For 1n= , we have 3p += and
+= )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
=
For 2n= , we have 5p += and
= )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=
For 3n= , we have 7p += and
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+= )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
bYaa)x(w 110 ++= where
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
=
For 1n= we have the solution
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=
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})x(MC~
)x(WC~
{x
1)x(X 2,2
2,1 += with
4
10+= ,
4
1=
For 2n= we have the solution
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=
For 3n= we have the solution
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)
respectively. Equating the two different expressions of 0a and integrating, we find
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 =+
+
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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).
From equation (3.11), using (6.1), we obtain
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)
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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
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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)
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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
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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)
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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.
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
bYaa)x(w 110 ++= where
+== 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=
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)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)
respectively. Equating the two different expressions of 0a and integrating, we find
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)
From equation (3.14), using (7.1), we obtain
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
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0)12x(2Br8A
A
2
3
A
A 2222
=++
(7.6)
Equations (7.4) and (7.6) are equivalent in view of (7.3).
From equation (3.11) we get
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
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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)
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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+
+
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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.
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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.
Conclusion. The solution of the Painleve PIV equation
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
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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
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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)
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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)
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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
+
+
+
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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
+
+++
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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
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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)
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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
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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
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[3] M. Abramowitz and I. A. Stegun: "Handbook of Mathematical
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Universitetskiye IzvestiyaKiev No. 9(1880) 1-25
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[10] E. L. Ince: "Ordinary Differential Equations". Dover 1956
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[14] J. Weiss: "The Painlev property for partial differential equations. II:
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[15] J. Weiss: "On classes of integrable systems and the Painlev property "
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