Symmetry reductions of a 2+1 Lax pair

Physics Letters A 343 (2005) 40–47 www.elsevier.com/locate/pla

Symmetry reductions of a 2 + 1 Lax pair P.G. Estévez a,∗ , M.L. Gandarias b , J. Prada c a Area de Fisica Teorica, Universidad de Salamanca, Spain b Departamento de Matemáticas, Universidad de Cádiz, Spain c Departamento de Matemáticas, Universidad de Salamanca, Spain

Received 15 March 2005; received in revised form 20 May 2005; accepted 20 May 2005 Available online 13 June 2005 Communicated by C.R. Doering

Abstract In this Letter we present the reductions arising from the classical Lie symmetries of a Lax pair in 2 + 1 dimensions. We obtain several interesting reductions and prove that, by analyzing not only a PDE but also its associated linear problem, it is possible to obtain the reduction of the PDE together with the reduced Lax pair. Specially relevant is the fact that the spectral parameter in 1 + 1 dimensions appears as a natural consequence of the reduction itself and is related to the symmetry of the 2 + 1 eigenfunction.  2005 Elsevier B.V. All rights reserved. Keywords: Exactly solvable and integrable systems

1. Introduction The identification of the Lie symmetries of a given partial differential equation (PDE) is an instrument of primary importance in order to solve such an equation [1]. A standard method for finding solutions of PDEs is that of reduction using Lie symmetries: each Lie symmetry allows a reduction of the PDE to a new equation with the number of independent variables reduced by one [2,3]. In a certain way this procedure gives rise to the ARS conjecture [4] which establishes that a PDE is integrable in the Painlevé sense [5] if all its reductions pass the Painlevé test [6]. This means that solutions of a PDE can be achieved by solving its reductions to ordinary differential equations (ODE). Classical [1] and nonclassical [2,3] Lie symmetries are the usual way for identifying the reductions.

* Corresponding author.

E-mail address: [email protected] (P.G. Estévez). 0375-9601/$ – see front matter  2005 Elsevier B.V. All rights reserved. doi:10.1016/j.physleta.2005.05.089

P.G. Estévez et al. / Physics Letters A 343 (2005) 40–47

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Nevertheless, let us recall that there are some methods for solving PDEs that are more effective in 2 + 1 than in 1 + 1 dimensions [7]. A good example of this is the following equation in 2 + 1 dimensions   
3 h2xz + 3hx hz = hyz , hxxz − (1) 4 hz x which some of us have studied in a recent paper [7] proving that the singular manifold method [6] is a very effective method for solving the equation. It is quite straightforward to determine the associated linear problem through this method. In fact, for (1) we obtained the following Lax pair: 3 2hz ψxz − hxz ψz + 2h2z ψ = 0. −ψy + ψxxx + 3hx ψx + hxx ψ = 0, (2) 2 Notice that there are a lot of papers related with the reduction of 2 + 1 equations through the classical Lie method but the application of this method to the Lax pair is much less frequent [8]. Nevertheless we consider that for integrable equations, it is of primary importance to determine the reduction, not only of the equation, but of the Lax pair. Actually the reduction process should introduce a spectral parameter that is absolutely essential in 1 + 1 dimensions. Therefore, our plan in this Letter is to a certain extent exactly the opposite of the usual approach: we try to obtain 1 + 1 spectral problems arising from a 2 + 1 Lax pair. Once we have solved the problem in 2 + 1 dimensions in [7], in the sense that we have determined its Lax pair, we shall identify the classical symmetries of the Lax pair [8]. This is done in Section 2. In Section 3 we use these symmetries to obtain a reduced Lax pair in 1 + 1 dimensions whose compatibility condition should be a reduction of (1). Actually, there are five possible reductions. Two of them yield linear equations that can be easily integrated. The other three reductions yield 1 + 1 spectral problems that include, as particular cases, well-known equations such as the modified Korteveg–de Vries, Drinfel’d–Sokolov or Ermakov–Pinney equations. It is interesting to note that each of these reductions yields respectively two, three and fourth order spectral problems which exhibit a spectral parameter as a natural output of the Lie method. We close with a section of conclusions.

2. Classical symmetries In order to apply the classical Lie method to the system of PDEs (2) with three independent variables and two fields, we consider the one-parameter Lie group of infinitesimal transformations in x, y, z, h, ψ, given by:     y  = y + εξ2 (x, y, z, h, ψ) + O ε 2 , x  = x + εξ1 (x, y, z, h, ψ) + O ε 2 ,     h = h + εφ1 (x, y, z, h, ψ) + O ε 2 , z = z + εξ3 (x, y, z, h, ψ) + O ε 2 ,   ψ  = ψ + εφ2 (x, y, z, h, ψ) + O ε 2 , (3) where ε is the group parameter. It is therefore necessary that this one transformation leaves the set of solutions of (2) invariant. This yields an overdetermined linear system of equations for the infinitesimals ξ1 (x, y, z, h, ψ), ξ2 (x, y, z, h, ψ), ξ3 (x, y, z, h, ψ), φ1 (x, y, z, h, ψ) and φ2 (x, y, z, h, ψ). The associated Lie algebra of infinitesimal symmetries is the set of vector fields of the form: X = ξ1

∂ ∂ ∂ ∂ ∂ + ξ2 + ξ3 + φ1 + φ2 . ∂x ∂y ∂z ∂h ∂ψ

(4)

By applying the classical method [1] to the system of PDEs (2), we obtain the following system of determining equations (we have used MACSYMA and MAPLE independently to handle the calculations): 0=

∂ξ1 ∂ξ1 ∂ξ1 ∂ξ2 ∂ξ2 ∂ξ2 ∂ξ2 ∂ξ3 ∂ξ3 ∂ξ3 ∂ξ3 = = = = = = = = = = , ∂z ∂h ∂ψ ∂x ∂z ∂h ∂ψ ∂x ∂y ∂h ∂ψ

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P.G. Estévez et al. / Physics Letters A 343 (2005) 40–47

0=

∂φ1 ∂ 2 φ1 ∂φ1 ∂φ2 ∂φ2 ∂ 2 φ2 = = = = = , ∂z ∂ψ ∂z ∂h ∂h2 ∂ψ 2

∂ 2 ξ1 ∂ 2 φ2 ∂ξ1 ∂φ1 ∂ξ1 ∂ξ2 ∂ 2 φ2 ∂ 2 φ1 − −2 = + =3 − , = 2 ∂x ∂ψ ∂x ∂h ∂x ∂ψ ∂x ∂h ∂x ∂y ∂x   ∂φ1 ∂φ2 ∂ξ1 ∂φ1 ∂ 3 φ2 ∂ξ1 ∂ 3 ξ1 0 = φ2 + ψ +3 + 3 − + = − , ∂h ∂ψ ∂x ∂y ∂x ∂x 3 ∂x 2 ∂ψ   2 ∂φ2 ∂ φ1 ∂ 2 ξ1 ∂ 3 φ2 ∂ 2 φ1 ∂φ2 = −2 +ψ 2 − + 2 + 3ψ , 0=2 ∂x ∂x ∂h ∂y ∂x 2 ∂x 3 ∂x 2 0=

whose solution is: ξ1 = −2

3 dA1 (y) x − A2 (y), dy 2

ξ2 = −6A1 (y),

ξ3 = β(z),

1 d 2 A1 (y) 2 1 dA2 (y) dA1 (y) x + A3 (y) + 2 h, φ1 = x + 2 3 dy 2 dy dy

  dA1 (y) φ2 = λ + ψ. dy

(5)

Note that the corresponding Lie symmetry algebra depends on three A1 , A2 and A3 arbitrary functions of y and an arbitrary function β(z) of z. The only constant that appears in (5) is λ that, as we shall show in the next section, plays the role of the spectral parameter in the 1 + 1 reductions of (2). Having determined the infinitesimals of (3) in (5), the symmetry variables are found by solving the invariant surface conditions Φ1 ≡ ξ 1

∂h ∂h ∂h + ξ2 + ξ3 − φ1 = 0, ∂x ∂y ∂z

∂ψ ∂ψ ∂ψ + ξ2 + ξ3 − φ2 = 0 ∂x ∂y ∂z

(6)

dψ = . dA1  dA1 2 λ + ψ x 2 + 12 dA x + A + 2 h 3 dy dy dy

(7)

Φ2 ≡ ξ 1

or the corresponding characteristic equations −dx 1 2 dA dy x

+

3 2 A2

−dy dz = = 6A1 β

=

dh

1 3

d 2 A1 dy 2

In the next section we solve (7) for the different possibilities.

3. Reductions There are five independent reductions that we determine in the following way: Case 1. A1 = 0, β = 0. Integration of (7) provides the reduced variables:    1 x 1 1 A2 dz + z1 = 1/3 − dy, z = 6 dy 2 4/3 4 A β A1 A 1

(8)

1

and the following reduction for the fields − λ6



dy A1

ψ(x, y, z) =

e

h(x, y, z) =

1 U (z1 , z2 ) − 1/3 1/3 A1 6A1

1/6

A1 1

G(z1 , z2 ),

(9) 

1/3

A1 M(z1 , y) dy,

(10)

P.G. Estévez et al. / Physics Letters A 343 (2005) 40–47

where M(z1 , y) is the function   2    d A1 1 dA2 1 d 2 A1 1 A2 2 z + + dy z1 M(z1 , y) = 4/3 dy 2A2/3 dy 2 3A1/3 1 dy 2 6A1/3 A 1 1 1 1  2  d 2 A1 1 dA2 1 A3 A2 A2 + dy + dy + . 2 1/3 4/3 2/3 4/3 dy 8A A1 dy 48A A1 A1 1 1

43

(11)

Substitution of the reduction ansatz (8)–(10) in (2) gives us: 3 λ 0 = Gz1 z1 z1 − Gz2 + 3Uz1 Gz1 + Uz1 z1 G + G, 2 6 0 = 2Uz2 Gz2 z1 − Uz2 z1 Gz2 + 2Uz22 G.

(12) (13)

Solving (12) for Gz2 and substituting in (13), we obtain the fourth order spectral problem: 3 λ Gz2 = Gz1 z1 z1 + 3Uz1 Gz1 + Uz1 z1 G + G, 2 6   3 1 λ 9 Uz2 Gz1 z1 z1 z1 = Uz2 z1 Gz1 z1 z1 − 3Uz1 Uz2 Gz1 z1 + Uz1 Uz2 z1 − Uz2 − Uz1 z1 Uz2 Gz1 2 2 6 2   λ 3 3 + Uz z − Uz22 − Uz1 z1 z1 Uz2 + Uz1 z1 Uz2 z1 G, 12 2 1 2 4 whose compatibility condition gives us the 1 + 1 equation   3 Uz22 z1 Uz2 z1 z1 − + 3Uz2 Uz1 = Uz2 z2 4 Uz2 z1

(14)

(15)

(16)

that can be written as the following nonlocal KdV equation [9], proposed by Drinfel’d and Sokolov [10]. −Pz2 + Pz1 z1 z1 + 2P Vz1 + 4V Pz1 = 0,

Vz2 = −

3  2 P z , 1 4

(17)

where we have done the change 4 Uz1 = V , 3

Uz2 = −P 2 .

Case 2. A1 = 0, β = 0. Integration of (7) provides the reduced variables:  x 1 A2 dy, z2 = z z1 = 1/3 − 4/3 4 A1 A1

(18)

(19)

and the reduction for the fields is exactly the same as in Case 1. By substituting the reduction ansatz in (2) we obtain the third order spectral problem: 3 λ 0 = Gz1 z1 z1 + 3Uz1 Gz1 + Uz1 z1 G + G, 2 6 0 = 2Uz2 Gz2 z1 − Uz2 z1 Gz2 + 2Uz22 G, whose compatibility gives us the 1 + 1 equation   3 Uz22 z1 Uz2 z1 z1 − + 3Uz2 Uz1 = 0. 4 Uz2 z1

(20) (21)

(22)

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P.G. Estévez et al. / Physics Letters A 343 (2005) 40–47

An alternative form of the above equation arises from the following definitions 4 Uz1 = V , Uz2 = −P 2 , 3 which allow us to write (22) as the system

(23)

3  + 2P 2 V + F (z2 ) = 0, (24) Vz2 = − P 2 z , 1 2 4 where F (z2 ) is an arbitrary function. Eq. (24) is the Ermakov–Pinney equation [11]. It has been proved in [12] that this equation is related through a reciprocal transformation to the Degasperis–Procesi equation [13]. P Pz1 z1 −

Pz21

Case 3. A1 = 0, A2 = 0, β = 0. Integration of (7) yields the following reduction:  3 1 z1 = x + A2 dz, z2 = y, 2 β ψ(x, y, z) = e

2λ (− 3A )x 2

G(z1 , z2 ),

h(x, y, z) = U (z1 , z2 ) −

d 2 A2 dy 2

1 2A3 x2 − x. 6 A2 3A2

Substitution of the reduction ansatz (25)–(27) in (2) gives us:   2λ 4λ2 2A3 dA2 z1 Gz1 0 = Gz1 z1 z1 − Gz2 − Gz1 z1 + 3Uz1 + − − A2 A2 dz2 A2 3A22   3 2λ 4λA3 8λ3 1 dA2 G, + Uz1 + − − Uz1 z1 − 2 A2 3A22 27A32 2A2 dz2   1 2λ 0 = Uz1 Gz1 z1 − Uz Gz1 + Uz21 G. Uz z + 2 1 1 3A2 1 Solving (29) for Gz1 z1 and substituting in (28), we obtain the second order linear system:   2λ Uz z Gz1 z1 = + 1 1 Gz1 − Uz1 G, 3A2 2Uz1    2  Uz21 z1 4λ λ Uz1 z1 z1 dA2 Uz1 z1 z1 A3 Gz1 + − − − + 2Uz1 − 2 2 Gz1 Gz2 = 3A2 Uz1 A2 dz2 2Uz1 4Uz21 9A22 A2   3 2λ 8λ 1 dA2 4λA3 G, + − Uz1 − − 2 3 3A 2A 3A2 27A2 2 2 dz2 whose compatibility condition yields the 1 + 1 equation   3 Uz21 z1 2A3 (z1 Uz1 + U ) dA2 Uz2 − Uz1 z1 z1 + − 3Uz21 + Uz1 + = 0. 4 Uz1 A2 A2 dz2 z1 Setting Uz1 = −P 2 we have the equation: Pz2 − Pz1 z1 z1 + 6P 2 Pz1 +

2A3 1 dA2 Pz + (P + z1 Pz1 ) = 0. A2 1 A2 dz2

(25) (26) (27)

(28) (29)

(30)

(31)

(32)

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The explicit dependence on A2 and A3 can be removed by doing the following change of variables:   1 A3 Q(T , X) dz2 , X = z − 2 dz2 , P (z1 , z2 ) = , T= 1 A2 A2 A22 A32 that transforms the above equation in the modified Korteweg–de Vries equation QT − QXXX + 6Q2 QX = 0.

(33)

Case 4. A1 = 0, A2 = 0, β = 0. The reduction is z1 = y,

z2 = z,

ψ(x, y, z) = e

(34)

2λ − 3A x

G(z1 , z2 ), 1 dA2 2 2 A3 x − h(x, y, z) = U (z1 , z2 ) − x, 6A2 dy 3 A2

(35)

2

whose substitution in (2) gives us   4λA3 8λ3 1 dA2 Gz1 = − − G, 3A22 27A32 A2 dz1

(36)

Gz2 =

3A2 Uz G. 2λ 2

(37)

The compatibility of (37) provides the linear equation A2 Uz1 z2 +

dA2 K1 (z2 ) Uz2 = 0 ⇒ U (z1 , z2 ) = + F1 (z1 ), dz1 A2

(38)

where F1 (z1 ) and K1 (z2 ) are arbitrary functions. With the aid of (38) we can integrate (37) as:     1 3K1 4λ 8λ3 A3 1 G = α0 √ exp dz − dz + 1 1 , 2λ 3 27 A2 A22 A32

(39)

where α0 is an arbitrary constant. The resulting solution for (2) is K1 1 dA2 2 2 A3 x − + F1 − x, A2 6A2 dy 3 A2     α0 3K1 4λ 8λ3 2λ A3 1 + ψ = √ exp dy − dy − x , 2λ 3 27 3A2 A2 A22 A32

h=

(40) (41)

depending on two arbitrary constants α0 and λ and five arbitrary functions A1 (y), A2 (y), A3 (y), F1 (y), and K1 (z). Case 5. A1 = 0, A2 = 0, β = 0. Integration of (7) provides the following reduction: z1 = x,

z2 = y,

ψ(x, y, z) = e

λ



dz β

G(z1 , z2 ),

 h(x, y, z) = U (z1 , z2 ) + A3

(42) dz . β

(43)

Substitution of the reduction ansatz (42), (43) in (2) gives us: 3 0 = Gz1 z1 z1 − Gz2 + 3Uz1 Gz1 + Uz1 z1 G, 2

0 = λGz1 + A3 G.

(44)

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P.G. Estévez et al. / Physics Letters A 343 (2005) 40–47

The compatibility condition yields the linear equation λUz1 z1 z1 − 2A3 Uz1 z1 +

2 dA3 =0 3 dz2

(45)

that can be easily integrated as: U (z1 , z2 ) = F3 (z2 )e +

2A3 λ z1

+

  1 dA3 2 λ dA3 1 λ2 dA3 λ z1 + + F (z ) + F1 (z2 ) 1 2 z1 + 2 3 6A3 dz2 2A3 6A3 dz2 4A23 12A3 dz2

1 F2 (z2 ), 2A3

where F1 (z2 ), F2 (z2 ) and F3 (z2 ) are arbitrary functions. Expression (46) allows us to integrate (44) as:      3 1 A3 G = α0 exp − F1 dz2 + 3 A33 dz2 + z1 , 2λ λ λ

(46)

(47)

where α0 is an arbitrary constant. We obtain the following solution for (2):    2A3 λ2 dA3 1 λ 1 dA3 2 λ dA3 1 dz x λ x + + + , + F1 x + F1 + F2 + A3 h = F3 e 2 3 2 6A3 dy dy 2A dy 2A β 6A3 4A3 12A3 3 3       dz 1 A3 3 3 ψ = α0 exp λ − F1 dy + 3 A3 dy + x , β 2λ λ λ

(48) (49)

depending on two arbitrary constants α0 and λ and six arbitrary functions A1 (y), A2 (y), A3 (y), F1 (y), F2 (y), F3 (y) and β(z).

4. Conclusions • A spectral problem in 2 + 1 dimensions is presented. It should be noted that this Lax pair was obtained using the Singular Manifold Method that, surprisingly, does not work properly for some 1 + 1 reductions of the system. It is easier to solve the problem in 2 + 1 than in 1 + 1 dimensions. • For the above reason we attempt to go from 2 + 1 to 1 + 1 dimensions by using the reductions arising from the classical Lie symmetries of the Lax pair. This means that we obtain symmetries that are symmetries of both the field h and the eigenfunction ψ . The symmetries that we have obtained include several arbitrary functions as well as an arbitrary constant that plays the role of the spectral parameter of the reduced spectral problems. • Five possible reductions arise from the classical symmetries. Two of them yield linear equations that can be easily integrated, providing us with nontrivial solutions of (2). • The other three reductions yield interesting Lax pairs of second, third and fourth order, respectively. An important point is that the introduction of the spectral parameters in the 1 + 1 reductions arise in an absolutely natural way. The reduced Eqs. (16), (22) and (33) do not depend of Ai (y), β(z) which means that every different functional form of these functions gives raise to a different solution of the 2 + 1 solution h(x, y, z) by means of the corresponding reduction ansatze.

Acknowledgements This research has been supported in part by the DGICYT under projects BFM2002-02609, BFM2003-04174 and BFM2003-00078.

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