Schlesinger transformations for the second and third...
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Schlesinger transformations for the second
and third members of a third Painleve hier-
archy
A. H. Sakka
Department of Mathematics,
Islamic University of Gaza,
Gaza, Palestine
e-mail: [email protected]
Fax Number: (+972)(7)2863552
JNMP Conference 4-14 June 2013
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Abstract
In this article, we derive the Schlesinger trans-
formations for second and third members of a
third Painleve hierarchy.
Using the Schlesinger transformations, we ob-
tained the corresponding Backlund transforma-
tions for each of the considered equations.
Furthermore we discussed some special solutions
of the second and third members of the third
Painleve hierarchy.
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Introduction
The six Painleve equations, PI-PVI, are foundby P. Painleve and B. Gambier as the only irre-ducible second-order ordinary differential equa-tions (ODEs) whose general solutions are freefrom movable critical points.
One of the important properties of the Painleveequations is the existence of Schlesinger trans-formations, that is transformations that trans-form the solutions of the associated linear prob-lem but preserve the monodromy data.
The Schlesinger transformations of PII-PVI havebeen studied by M. Jimbo, T. Miwa, and K.Ueno A., Fokas, U. Mugan , A. Sakka.
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Recently there are much interest in higher or-
der analogues of the Pianleve equations. Hier-
archies of Painleve equations are important ex-
amples of higher order analogues of the Pianleve
equations because the connection between these
hierarchies of Painleve equations and integrable
partial differential equations.
A first Painleve hierarchy was given by Kudryashov,
and Airault was the first to derive a second
Painleve hierarchy. After that, Gordoa, Joshi
and Pickering have used non-isospectral scat-
tering problems to derive new second Painleve
hierarchies and new fourth Painleve hierarchies.
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In this article, we present a method to obtainthe Schlesinger transformations for the secondand third members of a third Painleve hierar-chies given in
”Linear problems and hierarchies of Painleve equa-tions”,
J. Phys. A: Math. Theor., 42 (2009) 025210.
These transformations lead to new Backlundtransformations for these equations.
The Schlesinger transformations for some mem-bers of second and fourth Pianleve hierarchiesare derived by A. Sakka and M. EL-KAHLOUT.
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PIII hierarchy
We will consider the following PIII hierarchy
RnIII
(uv
)+ 2
n−2∑i=1
KiRn−iIII
(uv
)
+x
(ux − 2u2v
−vx − 2u(v2 − γ3)
)+ γ1
(uv
)
=
(1− uγ2
), n ≥ 2,
(1)
where RIII is the operator
RIII =
(Dx − 2uv + 2uD−1
x vx−2(v2 − γ3) + 2vD−1
x vx− 2u2 + 2uD−1
x ux−Dx − 2uv + 2vD−1
x ux
).
(2)
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This hierarchy can be obtain as the compatibility
condition of the following linear system
∂Φ
∂λ= A(λ)Φ(λ),
∂Φ
∂x= B(λ)Φ(λ),
(3)
where
B = B0λ+B1, A =n+1∑j=0
Ajλn−j−1. (4)
The first member of the PIII hierarchy (1), that
is n = 1, is the PIII equation.
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The second member of the PIII hierarchy
The second member of PIII hierarchy (1) reads
uxx = (6uv − x)ux − 6u3v2 + 2xu2v
+ 2γ3u3 − (γ1 + 1)u+ 1,
vxx = −(6uv − x)vx − 2u(3uv − x)(v2 − γ3)− γ1v + γ2.
(5)
We set γ3 = 1, without loss of generality (by
scaling u, v and x if necessary).
Equation (5) can be obtained as the compati-
bility condition of the following linear system of
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equations
∂Φ
∂λ= A(λ)Φ(λ),
∂Φ
∂x= B(λ)Φ(λ),
(6)
where B = B0λ+B1 and A =3∑
j=0
Ajλ1−j, with
B0 = 12σ3, B1 =
(0 uw
−1w [vx + u(v2 − 1)] 0
),
A0 = 12σ3, A3 =
(v w
−1w (v2 − 1) −v
),
Aj =
(aj bjcj −aj
), j = 1,2,
(7)
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σ3 is the Pauli matrix
σ3 =
(1 00 −1
)(8)
and aj, bj, cj are given as by
a1 = 12x, b1 = uw,
c1 =−1
w[vx + u(v2 − 1)],
a2 = u[vx + u(v2 − 1)] + 12γ1,
b2 = w[ux − 2u2v + xu],
c2 =1
w[(ux − 4u2v + xu)(v2 − 1)− 2uvvx − γ1v + γ2].
(9)
The auxiliary function w satisfies
wx = −2uvw. (10)
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Direct Problem
The aim of the direct problem is to establish the
analytic structure of Φ with respect to λ in the
entire complex λ-plane. Since (6.a) is a linear
ODE in λ, the analytic structure is completely
determined by its singular points.
The equation (6.a) has irregular singularities at
λ =∞ and λ = 0.
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Solution about λ =∞
The solution Φ(λ) of (6) in the neighborhood
of the irregular singular point λ = ∞ has the
formal expansion
Φ∞ = Φ∞λD∞eQ∞(λ), (11)
where
Φ∞ = (I + φ∞1λ−1 + . . . ),
D∞ = 12γ1σ3,
Q∞(λ) = 14(λ2 + 2xλ)σ3,
[A0, φ∞1] +A1 = 12xσ3.
(12)
The actual asymptotic behavior of Φ changes in
certain sectors of the complex λ-plane.
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These sectors are determined by
Re(1
4λ2 +
1
2xλ) = 0;
thus for large λ the sectors are asymptotic tothe rays argλ = π
4(2j − 3), j = 1,2, . . . ,4.
Let Φ∞,j(λ), j = 1,2,3,4 be solutions of (6)such that det Φ∞,j(λ) = 1 and Φ∞,j(λ) ∼ Φ∞as |λ| → ∞ in the sectorS∞,j : π
4(2j − 3) ≤ argλ < π4(2j − 1).
Then the solutions Φ∞,j(λ) are related by theStokes matrices, G∞,j, as follows
Φ∞,j+1(λ) = Φ∞,j(λ)G∞,j, j = 1,2,3,
Φ∞,1(λ) = Φ∞,4(λe2πi)G∞,4e−2πiD∞,
(13)
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where
G∞,2j−1 =
(1 a2j−10 1
),
G∞,2j =
(1 0a2j 1
), j = 1,2.
(14)
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Solution about λ = 0
The solution Φ(λ) of (6) in the neighborhood of
the irregular singular point λ = 0 has the formal
expansion
Φ0 = P Φ0λD0eQ0(λ), (15)
where
Φ0 = I + φ01λ−1 + . . . ,
P =
(wρ1 wρ2
−(v − 1)ρ1 −(v + 1)ρ2
),
D0 = 12γ2σ3, Q0(λ) = −λσ3.
(16)
The actual asymptotic behavior of Φ changes in
certain sectors of the complex λ-plane.
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These sectors are determined by Re(−θ0λ) = 0;
thus the sectors are asymptotic to the rays
argλ = π2(2j − 3), j = 1,2.
Let Φ0,j(λ), j = 1,2, be solutions of (6) such
that det Φ0,j(λ) = 1 and Φ0,j(λ) ∼ Φ0 as |λ| → 0
in the sector S0j : π
2(2j − 3) ≤ argλ < π2(2j − 1).
Then the solutions Φ0,j(λ) are related by the
Stokes matrices, G0,j, as follows
Φ0,2(λ) = Φ0,1(λ)G0,1,
Φ0,1(λ) = Φ0,2(λe2πi)G0,2e−2πiD0,
(17)
where
G0,1 =
(1 b10 1
), G0,2 =
(1 0b2 1
). (18)
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Monodromy Data
The relation between Φ0(λ) and Φ∞(λ) is givenby
Φ∞(λ) = Φ0(λ)E, (19)
where
E =
(c1 c2c3 c4
), det(E) = 1. (20)
The monodromy data
a1, a2, a3, a4, b1, b2, c1, c2, c3, c4
satisfies the consistency condition
E−1G0,1G0,2e−2πiD0E =
4∏j=1
G∞,je−2πiD∞. (21)
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Schlesinger Transformations
Let Φ(λ) be solution of (6) with parameters
γ1, γ2 and let Φ(λ) be solution of (6) with pa-
rameters γ1, γ2.
We consider transformation
Φ(λ) = R(λ)Φ(λ) (22)
such that Φ(λ) and Φ(λ) have the same mon-
odromy data.
Let γ1 = γ1 + n,
γ2 = γ2+m. Then (21) is invariant if n±m ∈ 2Z.
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All possible Schlesinger transformations admit-
ted by equation (6) may be generated by the
following transformationsγ1 = γ1 + 1γ2 = γ2 + 1
,
R(1)(λ) =
(1 00 0
)λ1/2 +
(u(v − 1) uw
v−1w 1
)λ−1/2,
γ1 = γ1 − 1γ2 = γ2 + 1
,
R(2)(λ) =
(0 00 1
)λ1/2 +
(1 w
v−1r1w
r1v−1
)λ−1/2,
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γ1 = γ1 + 1γ2 = γ2 − 1
,
R(3)(λ) =
(1 00 0
)λ1/2 +
(u(v + 1) uw
v+1w 1
)λ−1/2,
γ1 = γ1 − 1γ2 = γ2 − 1
,
R(4)(λ) =
(0 00 1
)λ1/2 +
(1 w
v+1r1w
r1v+1
)λ−1/2,
where r1 = vx + u(v2 − 1).
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We have
R(i)(λ, x; u, v, γ1, γ2)R(j)(λ, x;u, v, α, β) = I,
(23)
for (i, j) = (3,2) and (i, j) = (1,4). Moreover
R(1)(λ)R(2)(λ) =
(0 00 1
)λ+
(−1
2u −12w
2w 0
).
(24)
The Schlesinger transformation (24) shifts the
parameters as γ1 = γ1 + 2, γ2 = γ2
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Backlund Transformations
The linear equation (6.a) is transformed under
the Schlesinger transformations defined by the
transformation matrices R(j), j = 1,2,3,4 as
follows:
∂Φ
∂λ= A(λ)Φ(λ),
A(λ) = [R(j)(λ)A(λ) +∂
∂λR(j)(λ)]R−1
(j)(λ).
(25)
Using (60.b) we can derive the Backlund trans-
formations between solutions u(x) and v(x) of
(5), with parameters γ1 and γ2, and solutions
u(x) and v(x) of (5), with parameters γ1 and
γ2.
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The Backlund transformations corresponding tothe Schlesinger transformations R(j), j = 1,2,3,4may be listed as follows:
R(1) : v = 1− uΩ2,
u =ux − u2(v + 1)
u[uΩ2 − 2],
γ1 = γ1 + 1, γ2 = γ2 + 1,
(26)
R(2) : v = −1−[vx + u(v2 − 1)]Ω2
(v − 1)2,
u =1− vΩ2
,
γ1 = γ1 − 1, γ2 = γ2 + 1,
(27)
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R(3) : v = −1 + uΓ2,
u =−ux + u2(v − 1)
u[uΓ2 − 2],
γ1 = γ1 + 1, γ2 = γ2 − 1,
(28)
R(4) : v = 1 +[vx + u(v2 − 1)]Γ2
(v + 1)2,
u =(v + 1)
Γ2,
γ1 = γ1 − 1, γ2 = γ2 − 1,
(29)
where
Ω2 = 2uvx − 2(v − 1)(ux + xu)+ 2u2(v − 1)(3v + 1) + γ1 − γ2
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Γ2 = 2uvx − 2(v + 1)(ux + xu)+ 2u2(v + 1)(3v − 1) + γ1 + γ2
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Special Solutions
In this section, we will derive special solutions for
(5). The Backlund transformation (26) breaks
down when
ux − u2(v + 1) = 0 (30)
and
uΩ2 − 2 = 0. (31)
Solving (30) for v and substituting into (31) ,
we obtain
uxx = (6u+x)ux−4u3−2xu2 +1
2(γ2− γ1)u+ 1.
(32)
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However u and v satisfy (5). This implies that
γ1 and γ2 must satisfy
γ2 + γ1 + 2 = 0.
Therefore we have shown that if γ2 = −(γ1+2),
then (5) admits special solution v = u−2ux − 1
and u is a solution of equation
uxx = (6u+x)ux−4u3−2xu2−(γ1+1)u+1. (33)
The Backlund transformation (27) breaks down
when v = 1 and γ2 = γ1. Substituting these
values into (5) we obtain
uxx = (6u−x)ux−4u3+2xu2−(γ1+1)u+1. (34)
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Therefore we have shown that if γ2 = γ1, then(5) admits special solution v = 1 and u is asolution of equation (34).
The Backlund transformation (28) breaks downwhen
ux − u2(v − 1) = 0 (35)
and
u2[2uvx−2(v+1)(ux+xu)+2u2(v+1)(3v−1)+γ1+γ2]−2u = 0.(36)
Solving (35) for v and substituting into (36) ,we obtain
uxx = −(6u−x)ux−4u3 +2xu2−1
2(γ2−γ1)u+1.
(37)
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However u and v satisfy (5). This implies thatγ1 and γ2 must satisfyγ2 − γ1 − 2 = 0. Therefore we have shown thatif γ2 = γ1 + 2, then (5) admits special solutionv = u−2ux + 1 and u is a solution of equation
uxx = −(6u− x)ux− 4u3 + 2xu2− (γ1 + 1)u+ 1.(38)
The Backlund transformation (29) breaks downwhen v = −1 and γ2 = −γ1. Substituting thesevalues into (5) we obtain
uxx = −(6u+ x)ux− 4u3− 2xu2− (γ1 + 1)u+ 1.(39)
Therefore we have shown that if γ2 = −γ1, then(5) admits special solution v = −1 and u is asolution of equation (39).
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Application of the Backlund transformations
One can use the transformations (26)-(29) toobtain infinite hierarchies of elementary solu-tions of (5). For example, let us apply the trans-formation (26) to the following solution of (5):
γ2 = γ1 = α, v = 1 and u is a solution of (34).
Then we obtain a new solution
γ1 = α+ 1, γ2 = γ1, v = 1
and u is a solution of the equation
uxx = (6u−x)ux−4u3+2xu2−(γ1+1)u+1. (40)
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The application of the transformation (29) to
the solution γ2 = γ1 = α, v = 1 and u is a
solution of (34) yields the new solution
γ1 = α− 1, γ2 = γ1, v = 1 and u is a solution of
the equation (40).
Thus we can obtain a hierarchy of special solu-
tions γ1 = α+ n,
γ2 = γ1, n ∈ Z, v = 1 and u is a solution of (34).
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The third member of the PIII hierarchy
The third member of PIII hierarchy (1) reads
uxxx = 2(4uv −K1)uxx + 6vu2x + (4uvx − 30u2v2 + 6u2 + 12K1uv − x)ux
+ 2u2vxx + 20u4v3 − 12K1u3v2
− 12u4v + 4K1u3 + 2xu2v − (γ1 + 1)u+ 1,
vxxx = −2(4uv −K1)vxx − 6uv2x + (4vux + 30u2v2 − 6u2 − 8K1uv + x)vx
+ 2(v2 − 1)(uxx + 10u3v2 − 6K1u2v − 2u3 + xu) + γ1v − γ2.
(41)Equation (41) can be obtained as the compati-bility condition of the following linear system ofequations
∂Φ
∂λ= A(λ)Φ(λ),
∂Φ
∂x= B(λ)Φ(λ), (42)
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where B = B0λ+B1 and A =4∑
j=0
Ajλ2−j, with
B0 = 12σ3, B1 =
(0 uw
−1w [vx + u(v2 − 1)] 0
),
A0 = 12σ3, A4 =
(v w
−1w (v2 − 1) −v
),
Aj =
(aj bjcj −aj
), j = 1,2,3.
(43)
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Direct Problem
Solution about λ =∞
The solution Φ(λ) of (42) in the neighborhood
of the irregular singular point λ = ∞ has the
formal expansion
Φ∞ = Φ∞λD∞eQ∞(λ) = (I+φ∞1λ−1+. . . )λD∞eQ∞(λ),
(44)
where
D∞ = 12γ1σ3,
Q∞(λ) = 16(λ3 + 3K1λ
2 + 3xλ)σ3,
[A0, φ∞1] +A1 = K1σ3.
(45)
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The actual asymptotic behavior of Φ changes incertain sectors of the complex λ-plane. Thesesectors are determined by
Re1
6(λ3 + 3K1λ
2 + 3xλ) = 0;
thus for large λ the sectors are asymptotic tothe raysargλ = π
6(2j−3), j = 1,2, . . . ,6. Let Φ∞,j(λ), j =1, · · · ,6 be solutions of (42) such that det Φ∞,j(λ) =1 and Φ∞,j(λ) ∼ Φ∞ as |λ| → ∞ in the sectorS∞,j : π
6(2j − 3) ≤ argλ < π6(2j − 1) (see Figure
1). Then the solutions Φ∞,j(λ) are related bythe Stokes matrices, G∞,j, as follows
Φ∞,j+1(λ) = Φ∞,j(λ)G∞,j, j = 1,2, · · · ,5Φ∞,1(λ) = Φ∞,6(λe2πi)G∞,6e−2πiD∞,
(46)
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where
G∞,2j−1 =
(1 a2j−10 1
),
G∞,2j =
(1 0a2j 1
), j = 1,2,3.
(47)
Solution about λ = 0
The solution Φ(λ) of (42) in the neighborhood
of the irregular singular point λ = 0 has the
formal expansion
Φ0 = P Φ0λD0eQ0(λ) = P (I+φ01λ
−1+. . . )λD0eQ0(λ),
(48)
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where
P =
(wρ1 wρ2
−(v − 1)ρ1 −(v + 1)ρ2
),
D0 = 12γ2σ3, Q0(λ) = −λσ3.
(49)
The actual asymptotic behavior of Φ changes in
certain sectors of the complex λ-plane. These
sectors are determined by Re(−λ) = 0; thus the
sectors are asymptotic to the rays argλ = π2(2j−
3), j = 1,2. Let Φ0,j(λ), j = 1,2, be solutions of
(42) such that det Φ0,j(λ) = 1 and Φ0,j(λ) ∼ Φ0
as |λ| → 0 in the sector S0j : π
2(2j − 3) ≤ argλ <π2(2j−1). Then the solutions Φ0,j(λ) are related
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by the Stokes matrices, G0,j, as follows
Φ0,2(λ) = Φ0,1(λ)G0,1,
Φ0,1(λ) = Φ0,2(λe2πi)G0,2e−2πiD0,
(50)
where
G0,1 =
(1 b10 1
), G0,2 =
(1 0b2 1
). (51)
Monodromy Data
The relation between Φ0(λ) and Φ∞(λ) is given
by
Φ∞(λ) = Φ0(λ)E, (52)
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where
E =
(c1 c2c3 c4
), det(E) = 1. (53)
The monodromy data a1, a2, a3, a4, a5, a6, b1, b2, c1, c2, c3, c4satisfies the consistency condition
E−1G0,1G0,2e−2πiD0E =
6∏j=1
G∞,je−2πiD∞. (54)
Schlesinger Transformations
Let Φ(λ) be solution of (42) with parameters
γ1, γ2 and let Φ(λ) be solution of (42) with
parameters γ1, γ2. We consider transformation
Φ(λ) = R(λ)Φ(λ) (55)
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such that Φ(λ) and Φ(λ) have the same mon-
odromy data. Let γ1 = γ1 + n, γ2 = γ2 + m.
Then (21) is invariant if n±m ∈ 2Z.
All possible Schlesinger transformations admit-
ted by equation (6) may be generated by the
following transformationsγ1 = γ1 + 1γ2 = γ2 + 1
,
R(1)(λ) =
(1 00 0
)λ1/2 +
(u(v − 1) uw
v−1w 1
)λ−1/2,
(56)
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γ1 = γ1 − 1γ2 = γ2 + 1
,
R(2)(λ) =
(0 00 1
)λ1/2 +
(1 w
v−1r1w
r1v−1
)λ−1/2,
(57)γ1 = γ1 + 1γ2 = γ2 − 1
,
R(3)(λ) =
(1 00 0
)λ1/2 +
(u(v + 1) uw
v+1w 1
)λ−1/2,
(58)
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γ1 = γ1 − 1γ2 = γ2 − 1
,
R(4)(λ) =
(0 00 1
)λ1/2 +
(1 w
v+1r1w
r1v+1
)λ−1/2,
(59)
where r1 = vx + u(v2 − 1).
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Backlund Transformations
The linear equation (42.a) is transformed under
the Schlesinger transformations defined by the
transformation matrices R(j), j = 1,2,3,4 as
follows:
∂Φ
∂λ= A(λ)Φ(λ),
A(λ) = [R(j)(λ)A(λ) +∂
∂λR(j)(λ)]R−1
(j)(λ).
(60)
Using (60.b) we can derive the Backlund trans-
formations between solutions u(x) and v(x) of
(41), with parameters γ1 and γ2, and solutions
u(x) and v(x) of (41), with parameters γ1 and
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γ2. The Backlund transformations correspond-ing to the Schlesinger transformations R(j), j =1,2,3,4 may be listed as follows:
R(1) : v = 1 + uΩ3,
u =−[ux − u2(v + 1)]
u[uΩ3 + 2],
γ1 = γ1 + 1, γ2 = γ2 + 1,
(61)
R(2) : v = −1 +[vx + u(v2 − 1)]Ω3
(v − 1)2,
u =v − 1
Ω3,
γ1 = γ1 − 1, γ2 = γ2 + 1,
(62)
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R(3) : v = −1 + uΓ3,
u =ux − u(v + 1)
u[2− uΓ3],
γ1 = γ1 + 1, γ2 = γ2 − 1,
(63)
R(4) : v = 1 +[vx + u(v2 − 1)]Γ3
(v + 1)2,
u =(v + 1)
Γ3,
γ1 = γ1 − 1, γ2 = γ2 − 1,
(64)
where
Ω3 = 2(v − 1)a3 − (v − 1)2b3w
+ wc3
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Γ3 = 2(v + 1)a3 − (v + 1)2b3w
+ wc3
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Special Solutions
The Backlund transformation (27) breaks down
when v = 1 and γ2 = γ1. Substituting these
values into (5) we obtain
uxxx = 2(4u− k1)uxx + 6u3 − (24u2 − 12u+ x)ux+ 8u4 − 8K1u
3 + 2xu2 − (γ1 + 1)u+ 1.(65)
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Thank you for your attention