Schlesinger transformations for the second and third...

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Schlesinger transformations for the second

and third members of a third Painleve hier-

A. H. Sakka

Department of Mathematics,

Islamic University of Gaza,

Gaza, Palestine

e-mail: asakka@mail.iugaza.edu

Fax Number: (+972)(7)2863552

JNMP Conference 4-14 June 2013

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.

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.

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.

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.

PIII hierarchy

We will consider the following PIII hierarchy

n−2∑i=1

KiRn−iIII

(ux − 2u2v

−vx − 2u(v2 − γ3)

)+ γ1

(1− uγ2

), n ≥ 2,

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

This hierarchy can be obtain as the compatibility

condition of the following linear system

∂λ= A(λ)Φ(λ),

∂x= B(λ)Φ(λ),

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.

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.

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

equations

∂λ= A(λ)Φ(λ),

∂x= B(λ)Φ(λ),

where B = B0λ+B1 and A =3∑

Ajλ1−j, with

B0 = 12σ3, B1 =

−1w [vx + u(v2 − 1)] 0

A0 = 12σ3, A3 =

−1w (v2 − 1) −v

(aj bjcj −aj

), j = 1,2,

σ3 is the Pauli matrix

(1 00 −1

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],

w[(ux − 4u2v + xu)(v2 − 1)− 2uvvx − γ1v + γ2].

The auxiliary function w satisfies

wx = −2uvw. (10)

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.

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)

Φ∞ = (I + φ∞1λ−1 + . . . ),

D∞ = 12γ1σ3,

Q∞(λ) = 14(λ2 + 2xλ)σ3,

[A0, φ∞1] +A1 = 12xσ3.

The actual asymptotic behavior of Φ changes in

certain sectors of the complex λ-plane.

These sectors are determined by

4λ2 +

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∞,

G∞,2j−1 =

(1 a2j−10 1

G∞,2j =

(1 0a2j 1

), j = 1,2.

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)

Φ0 = I + φ01λ−1 + . . . ,

(wρ1 wρ2

−(v − 1)ρ1 −(v + 1)ρ2

D0 = 12γ2σ3, Q0(λ) = −λσ3.

certain sectors of the complex λ-plane.

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,

G0,1 =

(1 b10 1

), G0,2 =

(1 0b2 1

). (18)

Monodromy Data

The relation between Φ0(λ) and Φ∞(λ) is givenby

Φ∞(λ) = Φ0(λ)E, (19)

(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)

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.

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 +

v−1r1w

r1v−1

)λ−1/2,

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 +

v+1r1w

)λ−1/2,

where r1 = vx + u(v2 − 1).

We have

R(i)(λ, x; u, v, γ1, γ2)R(j)(λ, x;u, v, α, β) = I,

for (i, j) = (3,2) and (i, j) = (1,4). Moreover

R(1)(λ)R(2)(λ) =

(0 00 1

2u −12w

The Schlesinger transformation (24) shifts the

parameters as γ1 = γ1 + 2, γ2 = γ2

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:

∂λ= A(λ)Φ(λ),

A(λ) = [R(j)(λ)A(λ) +∂

∂λR(j)(λ)]R−1

(j)(λ).

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

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,

R(2) : v = −1−[vx + u(v2 − 1)]Ω2

(v − 1)2,

u =1− vΩ2

γ1 = γ1 − 1, γ2 = γ2 + 1,

R(3) : v = −1 + uΓ2,

u =−ux + u2(v − 1)

u[uΓ2 − 2],

γ1 = γ1 + 1, γ2 = γ2 − 1,

R(4) : v = 1 +[vx + u(v2 − 1)]Γ2

(v + 1)2,

u =(v + 1)

γ1 = γ1 − 1, γ2 = γ2 − 1,

Ω2 = 2uvx − 2(v − 1)(ux + xu)+ 2u2(v − 1)(3v + 1) + γ1 − γ2

Γ2 = 2uvx − 2(v + 1)(ux + xu)+ 2u2(v + 1)(3v − 1) + γ1 + γ2

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)

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.

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)

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)

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.

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).

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)

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).

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

∂λ= A(λ)Φ(λ),

∂x= B(λ)Φ(λ), (42)

where B = B0λ+B1 and A =4∑

Ajλ2−j, with

B0 = 12σ3, B1 =

−1w [vx + u(v2 − 1)] 0

A0 = 12σ3, A4 =

−1w (v2 − 1) −v

(aj bjcj −aj

), j = 1,2,3.

Direct Problem

Solution about λ =∞

of the irregular singular point λ = ∞ has the

formal expansion

Φ∞ = Φ∞λD∞eQ∞(λ) = (I+φ∞1λ−1+. . . )λD∞eQ∞(λ),

D∞ = 12γ1σ3,

Q∞(λ) = 16(λ3 + 3K1λ

2 + 3xλ)σ3,

[A0, φ∞1] +A1 = K1σ3.

The actual asymptotic behavior of Φ changes incertain sectors of the complex λ-plane. Thesesectors are determined by

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∞,

G∞,2j−1 =

(1 a2j−10 1

G∞,2j =

(1 0a2j 1

), j = 1,2,3.

Solution about λ = 0

of the irregular singular point λ = 0 has the

formal expansion

Φ0 = P Φ0λD0eQ0(λ) = P (I+φ01λ

−1+. . . )λD0eQ0(λ),

(wρ1 wρ2

−(v − 1)ρ1 −(v + 1)ρ2

D0 = 12γ2σ3, Q0(λ) = −λσ3.

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

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,

G0,1 =

(1 b10 1

), G0,2 =

(1 0b2 1

). (51)

Monodromy Data

The relation between Φ0(λ) and Φ∞(λ) is given

Φ∞(λ) = Φ0(λ)E, (52)

(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)

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,

γ1 = γ1 − 1γ2 = γ2 + 1

R(2)(λ) =

(0 00 1

)λ1/2 +

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,

γ1 = γ1 − 1γ2 = γ2 − 1

R(4)(λ) =

(0 00 1

)λ1/2 +

v+1r1w

)λ−1/2,

where r1 = vx + u(v2 − 1).

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:

∂λ= A(λ)Φ(λ),

A(λ) = [R(j)(λ)A(λ) +∂

∂λR(j)(λ)]R−1

(j)(λ).

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

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,

R(2) : v = −1 +[vx + u(v2 − 1)]Ω3

(v − 1)2,

u =v − 1

γ1 = γ1 − 1, γ2 = γ2 + 1,

R(3) : v = −1 + uΓ3,

u =ux − u(v + 1)

u[2− uΓ3],

γ1 = γ1 + 1, γ2 = γ2 − 1,

R(4) : v = 1 +[vx + u(v2 − 1)]Γ3

(v + 1)2,

u =(v + 1)

γ1 = γ1 − 1, γ2 = γ2 − 1,

Ω3 = 2(v − 1)a3 − (v − 1)2b3w

Γ3 = 2(v + 1)a3 − (v + 1)2b3w

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)

Thank you for your attention

Schlesinger transformations for the second and third...

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