Darboux transformations on a space scale

J. Math. Anal. Appl. 434 (2016) 1690–1718

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Darboux transformations on a space scale Gro Hovhannisyan ∗ , Oliver Ruff

a r t i c l e

i n f o

Article history: Received 1 September 2015 Available online 9 October 2015 Submitted by Goong Chen Keywords: Darboux transformation Lax equation Korteweg de Vries-like equation Time scale

a b s t r a c t Using the time/space-scale calculus introduced by Hilger, we describe a Darboux transformation for partial differential equations on a space scale, and extend the classical theorems of Darboux and Crum to this setting. As an application, we use the Lax equation to deduce a system of Korteweg–de Vries equations on a space scale, and obtain a solution to this system by using the Darboux transformation. © 2015 Elsevier Inc. All rights reserved.

1. Introduction In 1834 John Scott Russell observed the wave of translation (solitary wave, soliton) in the Union Canal near Edinburgh. In response to Russell’s observation, in 1872 Boussinesq introduced a new equation that modeled it. Then in 1895 Korteweg and his student de Vries deduced another equation that describes the solitary water waves. In 1967 Gardner, Green, Kruskal, and Miura [9] discovered the connection between the Korteweg–de Vries equation (referred to as the KdV equation from here on) and the spectral problem for the Schrödinger equation, and they developed the inverse scattering method of solving KdV-like equations. In 1968 P. Lax [14] discovered that the nonlinear KdV equation is equivalent to a compatibility condition on two linear equations. The importance of Lax’s observation is that it may be used to generate other nonlinear dynamic equations having some of the nice properties of KdV-like equations. In 1882 Darboux [5] found a covariant transformation for eigenfunctions and potentials of Sturm–Liouville (Schrödinger) equations with the application to the theory of surfaces. Numerous applications of Darboux transformations have subsequently been found in the modern theory of solitons (see for example [16]). In 1955 Crum [4] obtained a generalization of Darboux’s theorem by solving a recurrence relation generated by an N -fold application of the Darboux transformation. The remarkable formula for the N -soliton solution of the KdV equation was discovered in [9] by other means, and then in [20] this formula was deduced from Crum’s theorem. A discrete version of this formula was obtained in [12,15]. * Corresponding author. E-mail address: [email protected] (G. Hovhannisyan). http://dx.doi.org/10.1016/j.jmaa.2015.10.004 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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In this paper we obtain some extensions of Darboux’s and Crum’s results in the context of Stefan Hilger’s time-scale calculus, introduced in 1988 [11], which is an elegant theory that aims to unify continuous and discrete analysis. A great deal of work has been done to extend Hilger’s theory, for which our main reference is the book of Bohner and Peterson [3] – however, Darboux transformations in this setting have not been considered before. We tend to use the equivalent term “space scale” rather than “time scale” throughout (both just mean an arbitrary nonempty subset of the real numbers), although the only reason to prefer one term over the other is the applications that one has in mind. Our extensions of the Darboux and Crum theorems may be of use in modeling of fluid and plasma waves and networks (see [6–8,19,21]). The rest of the paper is structured as follows. In Section 2 we review the definitions we will need from Hilger’s time-scale calculus, for which our main reference is the book [3]. In Section 3 we define the Darboux transformation on a space scale in the case of the Sturm–Liouville equation, and generalize the theorems of Darboux and Crum. In Section 4, we consider the case of the non-stationary Schrödinger equation, in Section 5 we consider a general higher-order equation, and in Section 6 we consider first-order matrix equations. Finally, in Section 7 we use our results to derive a non-linear KdV system (that was found earlier in [10,2,1,18]) and solve it. The authors would like to thank the anonymous reviewer for the comments and suggestions that helped to improve the quality of the paper. 2. Time, space, and time–space scales We recall the basic definitions of Hilger’s time-scale calculus [11,3]. As already noted, the distinction between a time scale and a space scale is purely notational: each is a term for an arbitrary nonempty closed subset of R. We typically write T for a time scale and X for a space scale, and call the set T × X a time–space scale. For t ∈ T and x ∈ X we define backward jump operators σ(t) : T → T, ρ(x) : X → X σ(t) := sup{s ∈ T : s < t},

ρ(x) := sup{y ∈ X : y < x}.

(2.1)

We emphasize that this usage of σ differs from that which is found in [3]. For t ∈ T we define the forward jump operator β(t) : T → T by β(t) := inf{s ∈ T : s > t}.

(2.2)

The graininess functions μ(t), α(t) : T → [0, ∞), ν(x) : X → [0, ∞) are defined as μ(t) = t − σ(t),

ν(x) = x − ρ(x),

α(t) = β(t) − t.

(2.3)

We are considering nabla time and space derivatives [3] (instead of delta derivatives [11]), since in physics applications nabla derivatives with respect to the time variable are causal. If T contains a minimal element m with the property that β(m) > m, define Tκ := T − {m}; otherwise, set Tκ = T. For w : T → R and t ∈ Tκ , define the nabla derivative [3] of w with respect to t, denoted w(t) (t), to be the number (if one exists) with the property that given any ε > 0, there is a neighborhood U of t such that |w(σ(t)) − w(s) − w(t) (t)[σ(t) − s]| ≤ ε|ρ(t) − s| for all s ∈ U . In the case where T = R, we have w(t) (t) = wt (t), the usual derivative, and for T = Z it instead becomes the backward difference operator, w(t) (t) := w(t) − w(t − 1). The nabla derivative w(x) (x) with respect to x is defined in the same way, using ρ instead of σ.

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For the remainder of the paper we will, where convenient, consider functions from X to R implicitly as functions from T × X to R, constant in the first variable. We will often suppress one or more input variables to simplify our formulas. We will use the following notation for partial nabla derivatives: w(t, y) − wρ (t, x) , y→x y − ρ(x)

(2.4)

w(s, x) − wσ (t, x) , s→t s − σ(t)

(2.5)

w(x) (t, x) = ∂ x w(t, x) := lim w(t) (t, x) := ∂ t w(t, x) = lim where we define wρ and wσ by wρ (t, x) = Ew(t, x) := w(t, ρ(x)),

wσ (t, x) := w(σ(t), x).

(2.6)

For X = hZ, ρ(x) = x − h, h ∈ N we have ∂ x w(x) =

w(x) − w(x − h) w(x) − w(ρ(x)) = . x − ρ(x) h

For X = R, where ν(x) = 0, we have w(s) − w(x) = ∂x w(x) = wx (x). s→x s−x

∂ x w(x) = lim

Frequently we are going to use the product rule (see [3]) (f (x)g(x))(x) = f (x) (x)g(x) + f ρ (x)g (x) (x) = f (x) (x)g ρ (x) + f (x)g (x) (x).

(2.7)

We say that a function θ : X → R is x-regressive provided θ is ld-continuous and 1 − ν(x)θ(x) = 0 for all x ∈ X holds. If θ(x) is x-regressive one can define the nabla x-exponential function on a time scale (see [11,3]): x eˆθ (x, x0 ) = exp

log(1 − qθ(y)) ∇y. −q qν(y) lim

x0

(2.8)

Note that eˆθ (x, x0 ) is the unique solution of the initial value problem (x)

eˆθ (x, x0 ) = θ(t, x)ˆ eθ (x, x0 ),

eˆθ (x0 , x0 ) = 1.

(2.9)

3. Darboux transformations and factorization in the Sturm–Liouville equation Let f1 , . . . , fn be functions having nth-order left-dense continuous nabla derivatives with respect to x (see [3]). Define their Wronskian as follows:

W [f1 , . . . , fn ] =: det (∂ x )k−1 fj nk,j=1

  f1   f1(x) =  ...   (∂ x )n−1 f

1

  ... ϕn  (x)  ... fn .  ... ...  x n−1 . . . (∂ ) fn 

(3.1)

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Lemma 3.1. The following Jacobi formulas are true: W [a(t, x)f (t, x), a(t, x)g(t, x)] = aρ (t, x)a(t, x)W [f (t, x), g(t, x)],    W [f1 , f2 , . . . , fn , g] W [f1 , f2 , . . . , fn , h]  ρ   W x [f1 , f2 , . . . , fn , g] W x [f1 , f2 , . . . , fn , h]  = W [f1 , . . . , fn ]W [f1 , . . . , fn , g, h],

(3.2) (3.3)

or   W [f1 , f2 , . . . , fn , g]   W ρ [f1 , f2 , . . . , fn , g]

 W [f1 , f2 , . . . , fn , h]  = −νW ρ [f1 , . . . , fn ]W [f1 , . . . , fn , g, h]. W ρ [f1 , f2 , . . . , fn , h] 

Define the Casoratian

Cas[f1 , . . . , fn ] := det E k−1 ϕj nk,j=1

  f1   f1ρ =   ...  E n−1 f

1

 ... fn   ... fnρ  . ... . . .   . . . E n−1 fn

(3.4)

There is a connection between the Wronskian and the Casoratian, for example:   f1  Cas[f1 , f2 , f3 ] =  f1ρ  f ρρ 1

f2 f2ρ f2ρρ

 f3  f3ρ  = (−1)3 ν 2 ν ρ W [f1 , f2 , f3 ]. f3ρρ 

Remark 3.1. Using the connection

Cas[f1 , . . . , fn ] = W [f1 , . . . , fn ](−1)

(n−1)n 2

n−2 

(E k ν(x))n−k−1 ,

k=0

one can rewrite (3.3) in terms of the Casoratian: Cas[Cas[f1 , . . . , fn , g], Cas[f1 , . . . , fn , h]] = Casρ [f1 , . . . , fn ] Cas[f1 , . . . , fn , g, h]

n−1  k=1

E k ν(x) . ν(x)

(n−1)n

(−1) 2 ] Cas[f1 , . . . , fn ] into (3.3). n−2 k n−k−1 k=0 (E ν) Note that formula (3.5) in the case in which ν is constant is well-known (see for example [17]).

Indeed, one can get (3.5) by substituting W [f1 , . . . , fn ] =

In the case n = 1 (3.3) becomes W [W [f1 (t, x), g(t, x)], W [f1 (t, x), h(t, x)]] = f1ρ (t, x)W [f1 (t, x), g(t, x), h(t, x)]. Proof of Lemma 3.1. Formula (3.2) follows from the product rule: W [a(t, x)f (t, x), a(t, x)g(t, x)] = af (ag)x − (af )x ag = af (aρ g x + ax g) − (aρ f x + ax f )ag = aρ a(f g x − f g x ).

(3.5)

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The product rule extends to the following rule for differentiation of determinants:   f1  x  x  f1 ∂   ...  f (n) 1

 ρ   f1 . . . fn   (x)ρ   f . . . fnx   1 = ... . . . . . .   (n−1)ρ f (n) . . . fn   1(n+1) f1

 ... fnρ  (x)ρ   ... fn  , ...  (n−1)ρ  . . . f1  (n+1)  . . . f1

where we use the shortcut f (n) = (∂ x )n f . To prove (3.3) note that the left hand side of (3.3) may be rewritten in the form W [W [f1 , f2 , . . . , fn , g], W [f1 , f2 , . . . , fn , h]]  (x)  f1 f2 . . . fn h   (x)  (x) (x)  f1 f2 . . . fn h(x)   = W [f1 , f2 , . . . , fn , g]  ... ... ... . . .   ...  f (n) f (n) . . . f (n) h(n)  n 1 2  (x)  f1 f2 . . . fn g   (x)  (x) (x)  f1 f2 . . . fn g (x)   − W [f1 , f2 , . . . , fn , h]  ... ... ... . . .   ...  f (n) f (n) . . . g (n) h(n)  1 2 and (3.3) turns to    f1ρ ... fnρ hρ   (x)ρ (x)ρ  f . . . fn h(x)ρ   1  ... ... ... . . .  W [f1 , f2 , . . . , fn , g]   (n−1)ρ  (n)ρ  f1 . . . fn−1 h(n−1)ρ   (n+1)  (n+1) f . . . fn+1 h(n+1)  1    f1ρ ... fnρ g ρ    f xρ ... fnxρ g xρ  1   ... ... ... . . .  W [f1 , f2 , . . . , fn , h] = W ρ [f1 , f2 , . . . , fn ]W [f1 , f2 , . . . , fn , g, h]. −   (n−1)ρ (n)ρ  f1 . . . fn−1 g (n−1)ρ    (n+1) (n+1) f . . . fn+1 g (n+1)  1 In this expression the terms next to h(n+1) of the both sides are the same. Since (3.3) is invariant with (n+1) respect to the transpositions fj ↔ fk , g ↔ h it is enough to prove that the terms next to the f1 are the same, which in view of the property f (n−1) = f (n−1)ρ + νf (n) turns into the requirement that:   f2ρ  (x)ρ  n+2  f2 (−1)  ...   f (n−1)ρ 2

... fnρ (x)ρ . . . fn ... ... (n)ρ . . . fn

 ρ   f1   (x)ρ   f1   ...   (n−1)ρ  (n−1)ρ   f1 h  (n) f1 hρ hxρ ...

... fnρ (x)ρ . . . fn ... ... (n)ρ . . . fn (n) . . . fn

  (x)ρ  g  . . .  g (n−1)ρ   g (n) gρ

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

 ρ  f2  (x)ρ  f2  n+2  − (−1) ...  (n−1)ρ f  2  (n) f2

 ρ   f1   (x)ρ   f1   ...  (n−1)ρ   (n−1)ρ g   f1 (n) (n)   g f1

... fnρ (x)ρ . . . fn ... ... (n)ρ . . . fn (n) . . . fn

gρ g xρ ...

... fnρ (x)ρ . . . fn ... ... (n)ρ . . . fn (n) . . . fn

= (−1)n+3 W ρ [f1 , f2 , . . . , fn ]W [f2 , . . . , fn , g, h]  ρ ... fnρ    f2  f1ρ ... fnρ   (x)ρ (x)ρ  (x)ρ . . . fn (x)ρ   f2  f . . . f n  ... ... ... = (−1)n+3  1 ... . . .   (n−1)ρ (n)ρ  ... f . . . f n  f (n−1)ρ . . . f (n−1)ρ   2 n  (n) (n) 1 f2 . . . fn

gρ g (x)ρ ... g (n−1)ρ g (n)

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 hρ   h(x)ρ  . . .  (n−1)ρ  h   h(n)      .  (n−1)ρ  h   h(n) hρ h(x)ρ

To prove this we use the decomposition of W x [f1 , . . . fn , g] and W x [f1 , . . . fn , h] by the last row. (n) In this decomposition the terms next to f1 in the left hand side are equal to 0 and are missing in the right hand side. (n) The terms next to the h(n) are the same and those next to f2 are equal: (−1)n+3 W ρ [f2 , . . . , fn , h]W ρ [f1 , f3 , . . . , fn , g] − (−1)n+3 W ρ [f2 , . . . , fn , g]W ρ [f1 , f3 , . . . , fn , h] = −(−1)n+2 W ρ [f1 , f3 , . . . , fn ]W ρ [f3 , . . . , fn , g, h], since W [f2 , . . . , fn , h]W [f1 , f3 , . . . , fn , g] − W [f2 , . . . , fn , g]W [f1 , f3 , . . . , fn , h] = W [f1 , f2 , . . . , fn ]W [f3 , . . . , fn , g, h]. (n)

The equality of terms next to the g (n) , fj , j = 3, . . . n can also be demonstrated in this way by applying (n)

the transpositions g ↔ h, fj

(n)

↔ f2 .

2

Consider the operators P1 = −∂ x − q(t, x),

Q1 = ∂ x − q(t, x),

L = L[0] = P1 Q1 + λ1 ,

λ1 ∈ C.

(3.6)

In view of P1 Q1 y = −y (xx) + (q ρ − q)y (x) + (q 2 + q (x) )y,

Q1 P1 y = −y (xx) + (q − q ρ )y (x) + (q 2 − q (x) )y,

we have [P1 , Q1 ] = P1 Q1 − Q1 P1 = 2q (x) (t, x)(1 − ν(x)∂ x ),

(3.7)

and L = L[0] = −(∂ x )2 + u[0](t, x),

u[0](t, x) = −νq (x) ∂ x + q 2 + q (x) + λ1 .

(3.8)

Assume λ, λj ∈ C are the eigenvalues of L = L[0] with the corresponding eigenvectors ψ, ψj : L[0]ψ(t, x) = λψ(t, x),

L[0]ψj (t, x) = λj ψj (t, x),

j = 1, 2, . . . , n.

(3.9)

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If the eigenvector ψ1 of L (with corresponding eigenvector λ1 ) is known one can factor L − λ1 by choosing (x)

q=

ψ1 ψ1

: (x)

L[0] − λ1 = P1 Q1 ,

ψ1 , ψ1

P1 = −∂ x −

(x)

Q1 = ∂ x −

ψ1 , ψ1

and construct the new operator L[1] = Q1 P1 + λ1 = P1 Q1 + λ1 + [Q1 , P1 ] = L[0] − 2q (x) (1 − ν∂ x ), L[1] = −(∂ x )2 + u[1],

u[1] = Q1 P1 + λ1 + (∂ x )2 = [Q1 , P1 ] + u[0] = νq (x) ∂ x + q 2 − q (x) + λ1 ,   (x) (x) ψ1 u[0] − u[1] = [P1 , Q1 ] = 2 . ψ1

Since L[1] = Q1 P1 + λ1 , L[0] = P1 Q1 + λ1 we get L[1]Q1 = Q1 L[0]. If ψj are eigenvectors of L with eigenvalues λj for j = 2, 3, then  ψj [1] = Q1 ψj =

ψx ∂ − 1 ψ1



x

ψj =

W [ψ1 , ψj ] ψ1

(3.10)

is an eigenvector of L[1] = Q1 P1 + λ1 with eigenvalue λj : (L[1] − λj )ψj [1] = (L[1] − λj )Q1 ψj = (Q1 L[0] − λj Q1 )ψj = Q1 (L[0] − λj )ψj = 0. The following theorem is an extension of Darboux’s theorem [5] to the space scale X. Theorem 3.2. If Lψ(t, x) = λψ(t, x) then the function ψ[1](t, x) = Q1 ψ(t, x) =

W [ψ1 , ψ] , ψ1 (t, x)

Q1 = ∂ x −

ψ1x (t, x) ψ1 (t, x)

(3.11)

is a solution of L[1]ψ[1](t, x) = λψ[1](t, x), where  L[1] = −(∂ ) + u[1](t, x), x 2

u[1](t, x) = u[0](t, x) − 2

(x)

ψ1 ψ1

(x) (1 − ν∂ x ).

(3.12)

In other words, the Darboux transformation takes the eigenfunction ψ of L[0] to the eigenfunction ψ[1] of L[1]. Define  ψ3 [2](t, x) = Q2 Q1 ψ3 =

∂x −

ψ2x [1] ψ2 [1]

 ∂x −

ψ1x ψ1

ψ3 .

Note that in view of ψ3 [2](t, x) =

W [ψ2 [1], ψ3 [1]] W [W [ψ1 , ψ2 ]/ψ1 , W [ψ1 , ψ3 ]/ψ1 W [ψ2 [1], Q1 ψ3 ] = = , ψ2 [1] ψ2 [1] W [ψ1 , ψ2 ]/ψ1

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we have ψ3 [2](t, x) =

W [ψ1 , ψ2 , ψ3 ] . W [ψ1 , ψ2 ]

ψ2x [1](t, x) , ψ2 [1](t, x)

P2 = −∂ x −

Further using the factors Q2 = ∂ x −

ψ2x [1](t, x) , ψ2 [1](t, x)

we rewrite L[1] = −(∂ x )2 + u[1] = P2 Q2 + λ2 ,

L[2] = −(∂ x )2 + u[2] = Q2 P2 + λ2 , (x)  (x) ψ2 [1](t, x) u[1](t, x) − u[2](t, x) = [P2 , Q2 ] = 2 . ψ2 [1](t, x)

Note that if the eigenvector ψk [k − 1] of L[k − 1] = (∂ x )2 + u[k − 1] with the eigenvalue λk is known, then one can factor L[k − 1] − λk : (x)

L[k − 1] − λk = Pk Qk ,

Pk = −∂x −

ψk [k − 1] , ψk [k − 1]

(x)

Qk = ∂x −

ψk [k − 1] , ψk [k − 1]

(3.13)

and construct the new operator L[k] = Qk Pk + λk = −(∂ x )2 + u[k],

k = 1, 2, . . . , n.

(3.14)

We have L[k]Qk − Qk L[k − 1] = (Qk Pk + λk )Qk − Qk (Pk Qk + λk ) = 0, that is L[k]Qk = Qk L[k − 1],

k = 1, 2, . . . , n.

(3.15)

Define ψj [k](t, x) = Qk . . . Q1 ψj (t, x),

k = 1, 2, . . . , n.

(3.16)

One can check that ψj [k] is an eigenvector of L[k] with eigenvalue λj : (L[k] − λj ) ψj [k](t, x) = (L[k] − λj )Qk . . . Q1 ψj = Qk (L[k − 1] − λj )Qk−1 . . . Q1 ψj = Qk . . . Q1 (L[0] − λj )ψj = 0. The following theorem is an extension of Crum’s theorem [4] to the space scale X. Theorem 3.3. If Lψ(t, x) = λψ(t, x), Lψk (t, x) = λk ψk (t, x), k = 1, . . . , n then the function ψ[n](t, x) = Qn . . . Q2 Q1 ψ(t, x) =

W [ψ1 , ψ2 , . . . , ψn , ψ] W [ψ1 , . . . , ψn ]

(3.17)

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is a solution of L[n]ψ[n](t, x) = λψ[n](t, x), where

L[n] = −(∂ ) + u[n](t, x),

u[n](t, x) = u[0] − 2

x 2

n



k=1

(x)

ψk [k − 1] ψk [k − 1]

(x) (1 − ν(x)∂ x ).

(3.18)

Proof. From (3.7) we get  [Pk , Qk ] = 2

(x)

ψk [k − 1] ψk [k − 1]

(x) (1 − ν(x)∂ x ).

Further we have  uk−1 − uk = L[k − 1] − L[k] = [Pk , Qk ] = 2

(x)

ψk [k − 1] ψk [k − 1]

(x) (1 − ν∂ x ),

k = 1, 2, . . . , n.

Adding these expressions for k = 1, . . . n we get (3.18):

u[0](t, x) − u[n](t, x) = 2

n



k=1

(x)

ψk [k − 1] ψk [k − 1]

(x) (1 − ν∂ x ).

Let prove that Qk−1 Qk−2 . . . Q1 ψj =

W [ψ1 , . . . , ψk−1 , ψj ] , W [ψ1 , . . . , ψk−1 ]

j ≥ k, k = 1, . . . , n + 1.

For k = 1 it becomes the identity ψj = ψj , and for k = 2 it becomes (3.11). Now assume (3.19) is true for some k − 1 > 2, and consider Qk Qk−1 Qk−2 . . . Q1 ψj = Qk ψj [k − 1] =

W [ψk [k − 1], ψj [k − 1]] ψk [k − 1]

=

W [W [ψ1 , . . . , ψk ]/W [ψ1 , . . . , ψk−1 ], W [ψ1 , . . . , ψk−1 , ψj ]]/W [ψ1 , . . . , ψk−1 ] W [ψ1 , . . . , ψk ]/W [ψ1 , . . . ψk−1 ]

=

W [ψ1 , . . . , ψk , ψj ] W [W [ψ1 , . . . , ψk ], W [ψ1 , . . . , ψk−1 , ψj ]] = , W ρ [ψ1 , . . . , ψk−1 ]W [ψ1 , . . . , ψk−1 ] W [ψ1 , . . . , ψk ]

which establishes (3.19) by induction. (3.17) now follows from (3.19). Also from (3.19) ψ1 [0] = ψ1 ,

ψ2 [1] =

W [ψ1 , ψ2 ] , ψ1

ψn [n − 1] =

W [ψ1 , ψ2 , . . . , ψn ] , W [ψ1 , ψ2 , . . . , ψn−1 ]

so ψ1 ψ2 [1] . . . ψn [n − 1] = W [ψ1 , ψ2 , . . . , ψn ]. Considering the case ν(x) ≡ 0 we have un (t, x) = u0 − 2

n

k=1



(x)

ψk [k − 1] ψk [k − 1]

 = u0 − 2∂x2 ln (ψ1 ψ2 [1] . . . ψn [n − 1]) , x

(3.19)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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that is un (t, x) = u0 (t, x) − 2∂x2 ln W [ψ1 , ψ2 , . . . , ψn ].

2

Example 3.1. Consider the Schrödinger operator L = L[0] = −(∂ x )2 + u[0](x),

u[0](x) = xρ(x)

with the starting potential u[0](x), the eigenvalues λ1 = −1, λ2 = 1, and the corresponding eigenfunctions ψ1 (x) = eˆx (x, x0 ),

ψ2 (x) = eˆ−x (x, x0 ).

Indeed Lψ1 (x) = (xρ(x) − (∂ x )2 )ˆ ex (x, x0 ) = xρ(x)ˆ ex (x, x0 ) − (xˆ ex (x, x0 ))(x) = −ˆ ex (x, x0 ), Lϕ2 (x) = (xρ(x) − (∂ x )2 )ˆ e−x (x, x0 ) = xρ(x)ˆ e−x (x, x0 ) + (xˆ e−x (x, x0 ))(x) = eˆ−x (x, x0 ). The operator  L[1] = −(∂ ) + u[1](x), x 2

u[1](x) = u[0](x) − 2

(x)

ψ1 ψ1

(x) (1 − ν∂ x ) = xρ(x) − 2(1 − ν(x)∂ x )

has the eigenfunction (x)

ψ2 [1](x) = (∂ x −

ψ1 )ψ2 = (∂ x − x)ψ2 = −2xψ2 , ψ1

(x)

ψ2 [1] 1 (xψ2 )(x) (x) = = − ρ(x), ψ2 [1] xψ2 x

that satisfies L[1]ψ2 [1](x) = λ2 ψ2 [1](x). Further the operator L[2] = −(∂ x )2 + u[2](x), has the potential  u[2](x) = u[1](x) − 2

(x)

ψ2 [1] ψ2 [1]

(x)

 (1 − ν∂ x ) = u[1] − 2

(x) 1 − ρ(x) (1 − ν∂ x ) x



(x) 1 (1 − ν∂ x ) − ρ(x) x  (x) 1 = xρ(x) − 2 x + − ρ(x) (1 − ν∂ x ), x

= xρ(x) − 2(1 − ν∂ x ) − 2

that is  (x) 1 u[2](x) = xρ(x) − 2 ν(x) + (1 − ν∂ x ). x

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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Furthermore the operator L[3] = −(∂ x )2 + u[3](x) has the eigenfunction (x)

ψ3 [2](x) = (∂ x −

(x)

ψ2 [1] x ψ1 )(∂ − )ψ3 = ψ2 [1] ψ1

 ∂x −

1 + ρ(x) (∂ x − x)ψ3 , x

that is L[2]ψ3 [2](x) = λ3 ψ3 [2](x). Example 3.2. Introducing Bohner–Peterson’s trigonometric functions on a space scale (see [3]) sinhp (x, x0 ) =

eˆp (x, x0 ) − eˆ−p (x, x0 ) , 2

coshp (x, x0 ) =

eˆp (x, x0 ) + eˆ−p (x, x0 ) , 2

we have sinhp(x) (x, x0 ) = p · coshp (x, x0 ),

coshp(x) (x, x0 ) = p · sinhp (x, x0 ),

and the function ψ1 (x) = coshp (x, x0 ) satisfies L[0]ψ(x) = −ψ xx (x) = −p2 ψ(x),

λ1 = −p2 .

Further (x)



u[0] = 0,

(x)

ψ1 (x) ψ1 (x)

(x)

ψ1 (x) = p · tanhp (x, x0 ), ψ1 (x) = p · tanhp(x) (x, x0 ) =

p2 (cosh2p (x, x0 ) − sinh2p (x, x0 )) , coshp (ρ(x), x0 ) coshp (x, x0 )

u[1] = 2p tanhp(x) (x, x0 )(1 − ν∂ x ) =

2p2 eˆ−νp2 (x, x0 ) (1 − ν∂ x ) coshp (ρ(x), x0 ) coshp (x, x0 )

The function ψ2 (x) = eˆip (x, x0 ) satisfies Lψ2 (x) = −∂ xx ψ2 (x) = p2 ψ2 (x),

λ2 = p 2

and (x)

ψ2 [1](x) = (∂ x −

ψ1 )ψ2 = (∂ x − p tanhp (x, x0 ))ˆ eip (x, x0 ) = (i − tanhp (x, x0 ))pˆ eip (x, x0 ) ψ1

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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satisfies L[1]ψ2 (x) = (−∂ xx + u[1])ψ2 (x) = p2 ψ2 (x), that is  −∂ xx +

L[1]ψ2 (x) =

2p2 eˆ−νp2 (x, x0 ) (1 − ν∂ x ) ψ2 (x) = p2 ψ2 (x). coshp (ρ(x), x0 ) coshp (x, x0 )

Furthermore  (x) ψ2 [1] = ip2 1 − tanhp (x, x0 ) + (x)



i − pν(x) ρ coshp (x, x0 ) coshp (x, x0 )

eˆip (x, x0 ),

i − pν(x) 1 − tanhp (x, x0 ) + coshρp (x, x0 ) coshp (x, x0 ) (x)  (x) (x) ψ2 [1] ψ1 + . u[2](x) = −2 ψ1 ψ2 [1]

ip ψ2 [1] = ψ2 [1] i − tanhp (x, x0 )

,

If ν(x) ≡ 0 the function u[1](x) =

2p2 eˆ−νp2 (x, x0 ) 2p2 = = 2p2 sech2 (x, x0 ) 2 coshp (ρ(x), x0 ) coshp (x, x0 ) cosh (x, x0 )

is the continuous solitary wave. Likewise, the expression for u[2](x) gives two solitary waves. 4. Darboux transformations for partial differential equations Consider the nonstationary Schrödinger equation iΦ(t) (t, x) = LΦ(t, x),

L = L0 = −(∂ x )2 + u[0](t, x)

(4.1)

Since L = P1 Q1 + λ1 ,

L[1] = Q1 P1 + λ1 = P2 Q2 + λ2 ,

Qj = ∂ x −

− 1](x) , ψj [j − 1](x)

ψjx [j

L[2] = Q2 P2 + λ2

j = 1, 2, . . . ,

(4.2) (4.3)

we get L[j]Qj = Qj L[j − 1],

j = 1, 2, . . . .

(4.4)

We are looking for solutions ϕj (t, x), j = 1, 2, 3 of (4.1) in the form ϕj (t, x) = ψj (x)ˆ e−iλj (t, t0 ): (t)

iϕ1 (t, x) = Lϕ1 (t, x) = λ1 ϕ1 (t, x), (t)

iϕ2 (t, x) = L0 ϕ2 (t, x) = λ2 ϕ2 (t, x),

ϕ1 (t, x) = ψ1 (x)ˆ e−iλ1 (t, t0 ), ϕ2 (t, x) = ψ2 (x)ˆ e−iλ2 (t, t0 ),

and obviously (x)

(x)

ψ (x) ϕ1 (t, x) = 1 . ϕ1 (t, x) ψ1 (x)

(4.5)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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The function generated by the Darboux transformation: 

(x)

ψ (x) ∂ − 1 ψ1 (x)



x

ϕ2 [1](t, x) = Q1 ϕ2 =

(x)

Q1 = ∂ x −

ϕ2 (t, x),

ψ1 (x) ψ1 (x)

(4.6)

satisfies iΦ(t) (t, x) = L[1]Φ(t, x),

(4.7)

where  L[1] = −(∂ ) + u[1](t, x), x 2

(t)

Indeed since Q1 = 0,

u[1](t, x) = u[0] − 2

(x)

ψ1 (x) ψ1 (x)

(x) (1 − ν(x)∂ x ).

(4.8)

(t)

iϕ2 (t, x) = L0 ϕ2 (t, x) we have (t)

(t)

iϕ2 [1] = iQ1 ϕ2 = Q1 Lϕ2 (t, x) = L[1]Q1 ϕ2 (t, x) = L[1]ϕ2 (t, x). Introduce (x)

Q2 = ∂ x −

ϕ2 [1] (Q1 ϕ2 (t, x))(x) (Q1 ψ2 (x))(x) = ∂x − = ∂x − , ϕ2 [1] Q1 ϕ2 (t, x) Q1 ψ2 (x)

(t)

Q2 = 0.

(4.9)

Applying the Darboux transformation again we construct the function ϕ3 [2](t, x) = Q2 (x)Q1 (x)ϕ3 (t, x)

(4.10)

that satisfies the equation (t)

iϕ3 [2](t, x) = L[2]ϕ3 [2](t, x).

(4.11)

Indeed (t)

(t)

iϕ3 [2](t, x) = iQ2 Q1 ϕ3 (t, x) = Q2 Q1 Lϕ3 (t, x) = Q2 L[1]Q1 ϕ3 (t, x) = L[2]ϕ3 [2](t, x). Note that if T = R then the Hilger time-scale exponential function eˆ−iλ1 (t, t0 ) in (4.5) reduces to the usual exponential function e−iλ1 (t−t0 ) . 5. Darboux transformations for higher order equations Consider the nth-order operator Ln =

n

uk (x)(∂ x )k ,

x ∈ X.

(5.1)

k=0

In this section we are going to use the Leibniz formula (see [3]):

(dy)k =

k

j=0

⎛ ⎝

M ∈Sjk

⎞ dM ⎠ y (j) ,

(5.2)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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where Sjk is defined to be the set containing all possible strings of length k that contain exactly j ρs and exactly k − j ∂ x s. For example (dy)(xx) = d(xx) y + (d(x)ρ + dρ(x) )y (x) + dρρ y (xx) , (dy)(xxx) = d(xxx) y + (d(xx)ρ + d(x)ρx + dρ(xx) )y (x) + (dρρ(x) + dρ(x)ρ + d(x)ρρ )y (xx) + dρρρ y (xxx) . The following theorem is an extension of the similar theorem from [16] to the space scale X. Theorem 5.1. If the functions ψ1 , ψ satisfy the equations Ln ψ1 (x) = λ1 ψ1 (x),

Ln ψ(x) = λψ(x),

(5.3)

then the function

ψ[1](x) = Q1 ψ =

W [ψ1 , ψ] , ψ1

(x)

Q1 := ∂ x − d(x),

d(x) :=

ψ1 (x) , ψ1 (x)

(5.4)

is a solution of Ln [1]ψ[1](x) = λψ[1](x),

(5.5)

where Ln [1] =

n

uk [1](x)(∂ x )k ,

(5.6)

k=0

up [1](x) = uρp (x) + Q1 up+1 (x) +

n



uk [1](x)

dM (x),

p = 0, . . . , n.

(5.7)

k M ∈Sp+1

k=p+1

For example un [1](x) = uρn (x),

un−1 [1](x) = uρn−1 (x) + Q1 un (x) + uρn (x)(E n d(x)).

Proof. Assume that ψ1 (x) = eˆd (x, x0 ) is a solution of (Ln − λ1 )ψ1 (x) = 0 that is n

uk (x)(∂ x )k ψ1 (x) = λ1 ψ1 (x)

k=0 (x)

Then the function d(x) =

ψ1 (x) ψ1 (x)

satisfies the characteristic equation

Ln eˆd (x, x0 ) = u0 (x) − λ1 + uk (x)dk−1 (x) = 0, eˆd (x, x0 ) n

Charn =

k=1

(5.8)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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where d2 (x) ≡ d(x) + dρ d,

d1 (x) = d(x),

d3 (x) = d(xx) + d(x)ρ d + dρ(x) d + dρρ (d(x) + dρ d),

since ψ1 (x) = eˆd (x, x0 ) and (x)

ψ1

(xx)

= dψ1 ,

ψ1

(x)

= dρ ψ1 + d(x) ψ1 ,

(xx)

ψ1 ψ1

d2 =

(xxx)

ψ1

(xx)

(x)

= dρρ ψ1

+ (dxρ + dρ(x) )ψ1 + d(xx) ψ1

(xxx)

= (d(x) + dρ d),

ψ1 ψ1

d3 =

= d(xx) + (d(x)ρ + dρ(x) )d + (d(x) + dρ d)dρρ .

Note that obviously Q1 (Ln − λ1 )ψ1 = (Ln [1] − λ1 )Q1 ψ1 or Q1 Ln ψ1 = Ln [1]Q1 ψ1 . If Ln ψ2 (x) = λψ2 (x) and Q1 Ln = Ln [1]Q1 ,

(5.9)

then ψ2 [1](x) = Q1 ψ2 (x) is a solution of Ln [1]ψ2 [1](x) = Ln [1]Qψ2 (x) = QLψ2 (x) = λ2 Qψ2 (x) = λ2 ψ2 (x). From the assumption (5.9) one can find the coefficients of the operator Ln [1]: (∂ x − d(x))

n

k=0

n

n

uk (∂ x )k y(x) =

(uk (x)y (k) (x))(x) − d(x)

k=0

uk [1](∂ x )k (∂ x − d(x))y(x)

k=0 n

uk (x)y (k) (x) =

k=0

n

n

uk [1](y (k+1) −

k=0

uk [1](dy)(k) ).

k=0

Applying the Leibniz formula we get n

(x)

(uk y (k) + uρk y (k+1) ) − d

k=0 n

n

uk y (k) =

k=0

(uk [1] − uρk )y (k+1) =

k=0

n

k=0

uk [1]y (k+1) −

k=0

(x)

uk y (k) − d

k=0 n

n

(uk [1] − uρk )y (k+1) =

n

k=0

uk [1]

uk y (k) +

Q1 uk y (k) +

n

uk [1]

uk [1]

k=0

k

⎛ ⎝

j=0

k=0 n

k

j=0

k=0

k=0 n

n

k

⎛ ⎝

j=0





M ∈Sjk

dM (x).

⎞ dM ⎠ y (j) ⎞

dM ⎠ y (j)

dM ⎠ y (j)

By equating the terms next to y (n) we get

M ∈Sjn

M ∈Sjk

⎞

un [1](x) − uρn (x) = 0.



⎝



M ∈Sjk

By equating the terms next to y (n+1) we get

un−1 [1](x) − uρn−1 (x) = Q1 un (x) + uρn (x)

⎛

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1705

By equating the terms next to y (n−1) we get n

un−2 [1](x) − uρn−2 (x) = Q1 un−1 (x) +



uk [1](x)

dM (x).

k M ∈Sn−1

k=n−1

By equating the terms next to y (p+1) we get up [1](x) − uρp (x) = Q1 up+1 (x) +

n

uk [1](x)

k=p+1



dM (x),

p = 0, . . . , n.

2

k M ∈Sp+1

Remark 5.1. For the second order operator L2 = u0 (x) + u1 (x)∂ x + u2 (x)(∂ x )2 we have from (5.7) u2 [1](x) = uρ2 (x),

u1 [1](x) = uρ1 (x) + Q1 u2 (x) + u2 [1](x)dρρ (x)

u0 [1](x) = uρ0 (x) + Q1 u1 (x) + u1 [1]dρ + u2 [1](x)dρρ (x).

(5.10)

Assume eˆd (x, x0 ) is an eigenfunction of L, that is Lˆ ed (x, x0 ) = λ1 eˆd (x, x0 ), then (u2 (∂ x )2 + u1 ∂ x + u0 )ˆ ed (x, x0 ) = λ1 eˆd (x, x0 ), and so u2 (d(x) + dρ d) + u1 d + u0 = λ1 . In the case u2 (x) = −1,

u1 (x) = 0,

u0 (x) = u[0](x)

we have d(x) (x) + dρ (x)d(x) = u0 (x) − λ1

(x)

dxx (x) + dρx (x)dρ (x) + dρ (x)d(x) = u0 (x).

Further u2 [1](x) = −1,

u1 [1](x) = d − dρρ = d − dρ + dρ − dρρ = ν(d + dρ )(x) ,

u0 [1](x) = uρ0 + ν(d + dρ )(x) dρ − d(x)ρ − dρ(x) = uρ0 + ν(d(x) + dρ(x) )dρ − (d(x) − νd(xx) ) − (d − νd(x) )(x) (x)

= u0 − 2d(x) + ν(−u0 + d(x) dρ + dρ(x) dρ + d(xx) ) + (νd(x) )(x) , that is u2 [1](x) = −1,

u1 [1](x) = ν(d + dρ )(x) (x),

u0 [1](x) = u0 (x) − (dρ (x) + d(x))(x) .

(5.11)

1706

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Remark 5.2. For the third order operator L3 = u0 (x) + u1 (x)∂ x + u2 (x)(∂ x )2 + u3 (x)(∂ x )3 we have L3 [1] = u0 [1](x) + u1 [1](x)∂ x + u2 [1](x)(∂ x )2 + u3 [1](x)(∂ x )3 , where from (5.7) we get (x)

u3 [1](x) = uρ3 (x),

u2 [1](x) = uρ2 (x) + Q1 u3 (x) + uρ3 (x)dρρρ (x),

d(x) :=

ψ1 (x) , ψ1 (x)

u1 [1](x) = uρ1 + Q1 u2 + u2 [1]dρρ + u3 [1](d(x)ρρ + dρ(x)ρ + dρρ(x) ) u0 [1](x) = uρ0 + Q1 u1 + u1 [1]dρ + u2 [1](d(x)ρ + dρ(x) ) + u3 [1](d(xx)ρ + d(x)ρ(x) + dρ(xx) ).

(5.12)

Applying the Darboux transformation again we obtain  ψ3 [2](x) = Q2 Q1 ψ3 (x) =

(x)

ψ [1] ∂x − 2 ψ2 [1]



(x)

ψ ∂x − 1 ψ1



(x)

ψ3 ,

d2 (x) =

ψ2 [1] , ψ2 [1]

(5.13)

and so (x)

u3 [2](x) = uρ3 [1],

u2 [2](x) − uρ2 [1](x) = u3 [1] − d2 u3 [1] + uρ3 [1]dρρρ 2 (x)

(x)ρρ

u1 [2](x) − uρ1 [1](x) = u2 [1] − du2 [1] + u2 [2]dρρ 2 + u3 [2](d2 (x)

(x)ρ

u0 [2](x) − uρ0 [1](x) = u1 [1] − d2 u1 [1] + u1 [2]dρ2 + u2 [2](d2

ρ(x)

+ d2

ρ(x)ρ

+ d2

(xx)ρ

) + u3 [2](d2

ρρ(x)

+ d2

)

(x)ρ(x)

+ d2

ρ(xx)

+ d2

).

6. Darboux transformations for first order matrix equations Assume that Λ = diag{λ1 , . . . λr }, λj ∈ C is a diagonal constant matrix. Let Mat(r) be the set of ld-continuous r × r-matrix-valued functions on X. Consider the system (x)

Ψ

n

(x) =

Vk (x)Ψ(x)Λk ,

m, n ∈ N,

Vk (x) ∈ Mat(r).

(6.1)

k=−m

The following theorem is an extension of the similar theorem from [16] to the space scale X. Theorem 6.1. If the matrix-valued functions Ψ1 (x), Ψ(x) ∈ Mat(r) satisfy (5.1) and (x)

Ψ1 (x) =

n

Vk (x)Ψ1 (x)Λk1 ,

Λ1 = diag{λ11 , . . . , λr1 },

(6.2)

k=−m

then the matrix-valued function Ψ[1](x) = Ψ(x)Λ − DΨ(x),

D(x) := Ψ1 (x)Λ1 Ψ−1 1 (x),

(6.3)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

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satisfies n

Ψ(x) [1](x) =

m, n ∈ N,

Vk [1](x)Ψ[1](x)Λk ,

(6.4)

k=−m

where n

Vs [1](x) = Vs +

(Vk D − Dρ Vk )Dk−s−1 ,

s = 1, . . . , n,

(6.5)

k=s+1

Vp [1](x) = Dρ Vp D−1 +

p−1

(Dρ Vk − Vk D)Dk−p−1 ,

p = −1, −2, . . . − m,

m ∈ N,

(6.6)

k=−m 0

V0 [1](x) =

(D Vk − Vk D)D ρ

k−1

n

+

k=−m

 [Vk , D]D

k

n

D−ν

k=−m

−1 DVk D

k

.

(6.7)

k=−m

Proof. Note that from the definition of D = D(x) k = −m, . . . , n,

Ψ1 Λk1 = Dk Ψ1 , (x)

Ψ1

n

=

Vk Ψ1 Λk1 =

k=−m

n

Vk Dk Ψ1 ,

k=−m (x)

(DΨ1 )(x) = Ψ1 Λ1 . From the product rule (x)

(x)

(x)

(x)

(x)

(x)

DΨ1 + D(x) (Ψ1 − νΨ1 ) = Ψ1 Λ1 , D(x) (Ψ1 − νΨ1 ) = Ψ1 Λ1 − DΨ1   n n



(x) k D I −ν Vk D Ψ1 = [Vk , D]Dk Ψ1 , k=−m

D

(x)

=

n

k=−m

 [Vk , D]D

n

I −ν

k

k=−m

−1 Vk D

k

.

k=−m

By substitution of (6.3) into (6.4) we get n

Vk [1](ΨΛ − DΨ)Λk = (ΨΛ − DΨ)(x) = Ψ(x) Λ − D(x) Ψ − Dρ Ψ(x)

k=−m n

Vk [1](ΨΛk+1 − DΨΛk ) =

k=−m n

k=−m

n

(Vk ΨΛk+1 − Dρ Vk ΨΛk ) − D(x) Ψ

k=−m

(Vk [1] − Vk )ΨΛk+1 =

n

(Vk [1]D − Dρ Vk )ΨΛk − D(x) Ψ.

k=−m

By equating the terms next to Λn+1 in (6.8) we get Vn [1] − Vn = 0.

(6.8)

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To prove (6.5) note first that if s = n it becomes Vn [1] = Vn , which is true. We now proceed by downward induction on s. Assume that the inductive hypothesis

Vs+1 [1] = Vs+1 +

n

(Vk D − Dρ Vk )Dk−s−2

k=s+2

is true. By equating the terms in (6.8) next to Λs+1 , s + 1 ∈ N we get Vs [1] = Vs − Dρ Vs+1 + Vs+1 [1]D = Vs − Dρ Vs+1 + Vs+1 D +

n

(Vk D − Dρ Vk )Dk−s−1

k=s+2

= Vs +

n

(Vk D − Dρ Vk )Dk−s−1

k=s+1

that is, (6.5) is true. Note that for the terms next to Λ−m : we get 0 = (V−m [1]D − Dρ V−m )ΨΛ−m that is V−m [1] = Dρ V−m D−1 Similarly, to prove (6.6), first note that if p = −m it becomes V−m [1] = Dρ V−m D−1 , which is true. Now we use induction on p. Assume that the inductive hypothesis Vp−1 [1] = Dρ Vp−1 D−1 +

p−2

(Dρ Vk − Vk D)Dk−p

k=−m

is true. By equating the terms next to Λp , −1 − p ∈ N we get Vp [1]D − Dρ Vp = Vp−1 [1] − Vp−1 Vp [1] = Dρ Vp D−1 − Vp−1 D−1 + Vp−1 [1]D−1 and by using the inductive hypothesis we get Vp [1] = Dρ Vp D−1 − Vp−1 D−1 + Dρ Vp−1 D−2 +

p−2

k=−m

= Dρ Vp D−1 +

p−1

k=−m

that is (6.6).

(Dρ Vk − Vk D)Dk−p−1 ,

(Dρ Vk − Vk D)Dk−p−1

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1709

By equating the terms next to Λ0 in (6.8) we get V−1 [1] − V−1 = V0 [1]D − Dρ V0 − D(x) V0 [1] = Dρ V0 D−1 + D(x) D−1 − V−1 D−1 + V−1 [1]D−1 = Dρ V0 D−1 + D(x) D−1 − V−1 D−1 + Dρ V−1 D−2 +

−2

(Dρ Vk − Vk D)Dk−1

k=−m

V0 [1] = D(x) D−1 +

0

(Dρ Vk − Vk D)Dk−1 .

k=−m 0

V0 [1] =

(Dρ Vk − Vk D)Dk−1 +

k=−m

that is (6.7).

n



[Vk , D]Dk

k=−m

n

I −ν

−1 Vk D k

D−1

k=−m

2

7. Deduction and solution of a KdV system Lemma 7.1. (See [14].) If the nontrivial function ψ(t, x) satisfies the equations Lψ(t, x) = λψ(t, x),

ψ (t) (t, x) = A(t, x)ψ(t, x),

then the Lax equation L(t) = AL − LAσ = [A, L] + (L − Lσ )A,

(7.1)

is satisfied as well, where [A, L] = AL − LA is the usual commutator of the operators A and L. Note that one can derive integrable equations on a time–space scale otherwise by using the matrix version of the Lax equation (see [13]). Proof. By differentiation of Lψ = λψ we have L(t) ψ + Lσ ψ (t) = λψ (t) or L(t) ψ = λψ (t) − Lσ ψ (t) = λAψ − Lσ Aψ = (AL − LAσ )ψ.

2

To find a nonlinear KdV system we are looking for the operators L, A in the Lax equation (7.1) in the form L = u2 (∂ x )2 + u1 (t, x)∂ x + u0 (t, x),

u1 (t, x) ≡ w(t, x)

A = v3 (∂ x )3 + v2 (t, x)(∂ x )2 + v1 (t, x)∂ x + v0 (t, x),

(7.2)

where it is assumed that u2 , v3 are complex constants. To simplify the calculations we consider the continuous time scale where μ(t) ≡ 0 and a space scale with ν(x) = x − ρ(x) > 0. In this case (7.1) reverts to the usual Lax equation Lt = AL − LA.

(7.3)

1710

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

Lemma 7.2. The following formulas for commutators are true: [∂ x , u(t, x)] = u(x) (t, x) + (uρ (t, x) − u(t, x))∂ x , x 2

[(∂ ) , u(t, x)] = (u

ρρ

− u)(∂ ) + (u x 2

ρ(x)

+u

(x)ρ

x

)∂ + u

(7.4) (xx)

,

(7.5)

[(∂ x )3 , u(t, x)] = (uρρρ − u)(∂ x )3 + (uρρ(x) + uρ(x)ρ + u(x)ρρ )(∂ x )2 + (uρ(xx) + u(x)ρ(x) + u(xx)ρ )∂ x + u(xxx) ,

(7.6)

[p(t, x)(∂ x )2 , w(t, x)∂ x ] = (pwρρ − wpρ )(∂ x )3 + (pw(x)ρ + pwρ(x) − wp(x) )(∂ x )2 + pw(xx) ∂ x ,

(7.7)

(x)

[v1 (t, x)∂ x , w(t, x)∂ x ] = (v1 wρ − wv1ρ )(∂ x )2 + (v1 w(x) − wv1 )∂ x .

(7.8)

Proof. By direct calculation, [∂ x , u]y(t, x) = (uy)(x) − uy (x) = u(x) y + (uρ − u)y (x) , [(∂ x )2 , u]y(t, x) = (uy)(xx) − uy (xx) = u(xx) y + (uρ(x) + u(x)ρ )y (x) + uρρ y (xx) − uy (xx) = ((uρρ − u)(∂ x )2 + (uρ(x) + u(x)ρ )∂ x + u(xx) )y, [(∂ x )3 , u]y(t, x) = (uy)(xxx) − uy (xxx) = u(xxx) y + (uρ(xx) + u(x)ρ(x) + u(xx)ρ )y x + (uρρ(x) + uρ(x)ρ(x) + u(x)ρρ )y (xx) + (uρρρ − u)y (xxx) = ((uρρρ − u)(∂ x )3 + (uρρ(x) + uρ(x)ρ + u(x)ρρ )(∂ x )2 + (uρ(xx) + uρ(x)ρ + uxρρ )∂ x + uxxx ))y p(t, x)(∂ x )2 w(t, x)∂ x y(t, x) − w(t, x)∂ x p(t, x)(∂ x )2 y(t, x) = p(wy (x) )(xx) − w(py (xx) )(x) = p(w(x) y (x) + wρ y (xx) )(x) − w(p(x) y (xx) + pρ y (xxx) ) = p(w(xx) y (x) + w(x)ρ y (xx) + wρ(x) y (xx) + wρρ y (xxx) ) − w(p(x) y (xx) + pρ y (xxx) ) = [(pwρρ − wpρ )(∂ x )3 + (pw(x)ρ + pwρ(x) − wp(x) )(∂ x )2 + pw(xx) ∂ x ]y [v1 (t, x)∂ x , w(t, x)∂ x ]y(t, x) = v1 (wy (x) )(x) − w(v1 y (x) )(x) (x)

= v1 (w(x) y (x) + wρ y (xx) ) − w(v1 y (x) + v1ρ y (xx) ) (x)

= [(v1 wρ − wv1ρ )(∂ x )2 + (v1 w(x) − wv1 )∂ x ]y.

2

From the Lax equation (7.3) using Lemma 7.2 we get wt ∂ x + ut = [v3 (∂ x )3 + v2 (∂ x )2 + v1 ∂ x + v0 , u2 (∂ x )2 + w∂ x + u] = (v3 wρρρ − v3 w + u2 v2 − u2 v2ρρ )(∂ x )4 + (v3 wρρ(x) + v3 wρ(x)ρ (x)ρ

+ v3 w(x)ρρ + v3 uρρρ − v3 u − u2 v2

ρ(x)

− u2 v2

+ v2 wρρ − wv2ρ + u2 v1 − u2 v1ρρ )(∂ x )3 (xx)

+ (v3 wρ(xx) + v3 w(x)ρ(x) + v3 w(xx)ρ + v3 uρρ(x) + v3 uρ(x)ρ + v3 u(x)ρρ − u2 v2 (x)

ρ(x)

+ v2 wρ(x) − wv2 )(∂ x )2 + (v2 uρρ − v2 u − u2 v1

(x)ρ

− u2 v1

+ v2 w(x)ρ

+ v1 wρ − wv1ρ + u2 v0 − u2 v0ρρ )(∂ x )2 (xx)

+ (v3 w(xxx) + v3 uρ(xx) + v3 u(x)ρ(x) + v3 u(xx)ρ + v2 w(xx) + v2 uρ(x) + v2 u(x)ρ − u2 v1 (x)

(x)ρ

+ v1 w(x) − wv1 )∂ x + (v1 uρ − v1 u − u2 v0 (xx)

+ v3 u(xxx) + v2 u(xx) + v1 u(x) − u2 v0

ρ(x)

− u2 v0 (x)

− wv0 .

+ wv0 − wv0ρ )∂ x

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1711

Now, by equating the coefficients of each power of ∂ x in this expression, we get v3 (wρρρ − w) = u2 (v2ρρ − v2 ), v3 (wρρ(x) + wρ(x)ρ + w(x)ρρ + uρρρ − u) − v3 (w

ρ(xx)

+w

(x)ρ(x)

+w

(xx)ρ

+u

(x)

ρρ(x)

ρ(x)

− wv2 + v2 (uρρ − u) − u2 (v1

(x)ρ u2 (v2

+u

ρ(x)ρ

(x)ρ

+ v1

+

ρ(x) v2 )

+u

(7.9)

+ v2 wρρ −

(x)ρρ

)−

wv2ρ

(xx) u2 v2

+ u2 (v1 −

+ v2 (w

(x)ρ

v1ρρ )

+w

= 0,

ρ(x)

)

) + v1 wρ − wv1ρ + u2 (v0 − v0ρρ ) = 0,

wt = v3 (w(xxx) + uρ(xx) + u(x)ρ(x) + u(xx)ρ ) + v2 (w(xx) + uρ(x) + u(x)ρ ) − (x)

(x)ρ

+ v1 w(x) − wv1 + v1 (uρ − u) − u2 (v0

ρ(x)

+ v0

(xx)

(7.11)

(xx) u2 v1

) + w(v0 − v0ρ ),

ut (t, x) = v3 u(xxx) + v2 u(xx) + v1 u(x) − u2 v0

(7.10)

(x)

− wv0 .

(7.12) (7.13)

Solving equations (7.9)–(7.11) gives us a2 (x) u1 (t, x) = w(t, x) = a0 (t, x) + aρ0 (t, x), u(t, x) = a1 + aρ1 + 0 + a0 , u2  v3 a0 (t, x)w(t, x) (x) ρ w (t, x) + + a1 (t, x) + u (t, x) , v1 (t, x) = u2 u2 v3 a2 + v3 u(x) (a0 (t, x) + wρ (t, x)), v0 (t, x) = , u2 u2 v3 (x) (a0 (u − uρ ) + w(aρ1 + uρ )) − v3 a1 , a2 (t, x) + aρ2 (t, x) = u2 v2 (t, x) =

a2 − aρ2 v3 w ρ(x) (x)ρ (xx) (x) w − a2 − a2 − v3 a1 + 2 (a0 (uρ − u) − a0 wρ ) u2 u2 v3 (x) + [(a1 + uρ )(uρ − u) + (wρ − w)(uρ(x) + u(x)ρ ) + w(u(x) − a1 )] u2 v3 (x) (x)ρ (xx) + [w(x) (a0 − a0 + a1 + 2uρ − u) + a0 (u(x)ρ + uρ(x) ) − a0 w], u2 v3 (xx) ρρ v3 (x) ρ(x) a0 w w (x) (x) (xx) u a0 + u (a0 + a1 + a0 + + uρ ) − a2 − a2 , ut (t, x) = u2 u2 u2 u2

(7.14) (7.15) (7.16) (7.17)

wt (t, x) =

where a0 (t, x), a1 (t, x), a2 (t, x) are arbitrary functions that may be defined by (7.14) and (7.17). Indeed from (7.9) we have v3 (wρρ + wρ + w)ρ − v3 (wρρ + wρ + w) = u2 (v2ρ + v2 )ρ − u2 (v2ρ + v2 ) v3 (wρρ + wρ + w) = u2 (v2ρ + v2 ) or by substitution v3 ρ (w + a0 ) u2

v2 = we get w = aρ0 + a0 ,

v2 =

v3 ρρ (a + aρ0 + a0 ). u2 0

By substituting these formulas into (7.10) we get (x)ρ

v3 (w(x)ρρ − a0

ρ(x)

− a0

) + v2 wρρ − wv2ρ = u2 (v1ρρ − v1 ) + v3 (u − uρρρ ),

(7.18) (7.19)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1712 ρ(x)ρρ

v3 (a0

(x)ρρ

+ a0

(x)ρ

− a0

ρ(x)

− a0

)+

v3 ρρ v3 ρ ρ (a0 + aρ0 + a0 )(aρ0 + a0 )ρρ − (aρ0 + a0 )(aρρ 0 + a0 + a0 ) u2 u2

= u2 (v1ρρ − v1 ) + v3 (u − uρρρ ), that is v3 ρρ ρρρ v3 ρ ρ a0 (a0 + aρρ a (a + a0 ) = u2 (v1ρρ − v1 ) + v3 (u − uρρρ ), 0 )− u2 u2 0 0 1 1 ρ ρ u2 ρρ (x)ρ ρ(x) ρρρ + aρρ a0 (a0 + a0 ) = (v − v1 ) + (u − uρρρ ), (aρ0 + a0 )(x)ρρ − a0 − a0 + aρρ 0 (a0 0 )− u2 u2 v3 1  ρρ u2 a0 w 1 u2 ρ(x) (x) (x)ρ ρ(x) ρ a0 + a0 − v1 + u + − aρ0 w = a0 + a0 − v1 + u. v3 u2 u2 v3 (x)ρ

v3 (aρ0 + a0 )(x)ρρ − v3 a0

ρ(x)

− v3 a0

+

Denoting  −a1 :=

ρ(x) a0

+

(x) a0

u2 a0 w − v1 + uρ + v3 u2



u2 a0 w v1 + uρ + , v3 u2

= w(x) −

that is v3 v1 = u2

 w

(x)

a0 w ρ + + a1 + u , u2

we have −aρρ 1 =

1 ρ u2 1 ρ a0 w (x)ρ ρ(x) (x)ρ ρ(x) a w + a0 + a0 − v1 + u = a w + a0 + a0 + u − w(x) − uρ − − a1 u2 0 v3 u2 0 u2 a1 − aρρ 1 =

(aρ0 − a0 )w (x)ρ ρ(x) + a0 + a0 + u − uρ − w(x) u2

a1 − aρρ 1 =

(aρ0 − a0 )w (x)ρ (x) + a0 − a0 + u − uρ u2 (x)

(a1 + aρ1 ) − (a1 + aρ1 )ρ = −ν

a0 (xx) w − νa0 + νu(x) u2

(x)

ν(a1 + aρ1 )(x) = −ν

a0 (xx) w − νa0 + νu(x) , u2

(a1 + aρ1 )(x) = −

(a20 )(x) (xx) − a0 + u(x) , u2

so under the assumption ν(x) = 0 we get the second formula in (7.14): u = a1 + aρ1 +

a20 (x) + a0 + C, u2

(x)

(x)

ρ(x)

u(x) = a1 + a1

+

a0 w (xx) + a0 . u2

ρ ρ Now, by using the formulas for v2 , v1 , u(x) , aρρ 0 = a0 + w − w in (7.11) we get (xx)

u2 (v0ρρ − v0 ) = v3 (wρ(xx) + w(x)ρ(x) + w(xx)ρ + uρρ(x) + uρ(x)ρ + u(x)ρρ ) − u2 v2 (x)

ρ(x)

(x)ρ

+ v2 (w(x)ρ + wρ(x) ) − wv2 + v2 (uρρ − u) − u2 (v1 + v1 ) + v1 wρ − wv1ρ v3 (x) (x)ρ = v3 (a1 − a1 − u(x) + u(x)ρρ ) + (a0 + wρ )(uρρ − u) u2   v3 a0 − aρ0 (x)ρ ρ ρ ρ ρ ρρ x a1 w − a1 w + u w − wu + wρ . w + a0 − a0 + u2 u2

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1713

Further (u2 v0 − v3 u(x) )ρρ − (u2 v0 − v3 u(x) ) v3 (x) (x)ρ = v3 (a1 − a1 ) + (a0 + wρ )(uρρ − u) + u2   v3 a0 − aρ0 (x) (x)ρ + w + a0 − a0 a1 wρ − aρ1 w + uρ wρ − wuρρ + wρ u2 u2 By the substitution a2 = u2 v0 − v3 ux (since u − uρρ = ν(u + uρ )(x) ), we get v3 (x) (x)ρ aρρ ) + (a0 + wρ )(uρρ − u) + 2 − a2 = v3 (a1 − a1 u2   v3 a0 − aρ0 (x) (x)ρ ρ ρ ρ ρ ρρ + a1 w − a1 w + u w − wu + w + a0 − a0 wρ u2 u2 v3 (xx) − ν(a2 + aρ2 )(x) = νv3 a1 − ν (a0 + wρ )(u + uρ )(x) u2     (x) v3 a (xx) ρ ρ 0 + (a1 − a1 )wρ + a1 (wρ − w) + (uρ − uρρ )wρ + uρρ (wρ − w) + ν w + a0 wρ u2 u2 v3 (xx) − (a2 + aρ2 )(x) = v3 a1 − (a0 + wρ )(u + uρ )(x) u2     (x) v3 a0 (x) ρ (xx) ρ (x) ρ(x) ρ ρρ (x) + w −u w + w + a0 a1 w − a 1 w + u wρ . u2 u2 Since  (x) a1 w ρ

−

aρ1 w(x)

+u

ρ(x)

w −u w ρ

ρρ

(x)

+

(x)

a0 (xx) w + a0 u2

 wρ − (a0 + wρ )(u + uρ )(x)

= (a0 uρ − a0 u − waρ1 − wuρ )(x) we get (x)

−(a2 + aρ2 ) = v3 a1 +

v3 (a0 uρ − a0 u − waρ1 − wuρ ), u2

that is a2 + aρ2 =

v3 (x) (a0 (u − uρ ) + w(aρ1 + uρ )) − v3 a1 . u2

Indeed  (x)

a1 wρ − aρ1 w(x) + uρ(x) wρ − uρρ w(x) +  =

(x)

(x) a1

a (xx) + 0 w + a0 − u x u2

ρ(x)

= −a1



ax0 (xx) wρ − (a0 + wρ )(u + uρ )(x) w + a0 u2

wρ − w(x) (a1 + uρ )ρ − a0 (u + uρ )(x)

wρ − w(x) (a1 + uρ )ρ − a0 (u + uρ )(x)

= −(waρ1 )(x) − w(x) uρρ − a0 (u + uρ )(x) = −(waρ1 )(x) − (wuρ )(x) + wuρ(x) − a0 (u + uρ )(x)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1714

= −(waρ1 + wuρ )(x) + (w − a0 )uρ(x) − a0 u(x) = (−waρ1 − wuρ )(x) + aρ0 uρ(x) − a0 u(x) (x)

= (−waρ1 − wuρ )(x) + (a0 uρ )(x) − a0 uρ − a0 u(x) = (−waρ1 − wuρ + a0 uρ − a0 u)(x) . ρ Further from (7.12) by using the formulas for v2 , v1 , aρρ 0 = a0 + w − w we get

wt =

a2 − aρ2 v3 w ρ(x) (x)ρ (xx) (x) w − a2 − a2 − v3 a1 + 2 (a0 (uρ − u) − a0 wρ ) u2 u2 v3 (x) + [(a1 + uρ )(uρ − u) + (wρ − w)(uρ(x) + u(x)ρ ) + w(u(x) − a1 ) u2 v3 (x) (x)ρ (xx) + [w(x) (a0 − a0 + a1 + 2uρ − u) + a0 (u(x)ρ + uρ(x) ) − a0 w]. u2

(7.20)

From (7.13) by using the formulas for v2 , v1 , v0 we get ut =

v3 (xx) ρρ v3 (x) ρ(x) w (x) (xx) u a0 + u (a0 − aρ1 + u + uρ + aρ0 a0 /u2 ) − a2 − a2 , u2 u2 u2

that is ut =

v3 (xx) ρρ v3 (x) ρ(x) w (x) (x) (xx) u a0 + u (a0 + a1 + a0 + a20 /u2 + uρ + aρ0 a0 /u2 ) − a2 − a2 , u2 u2 u2  v3  (xx) ρρ w (x) (x) (xx) a0 + u(x) (∂ x aρ0 + a1 + a0 + a0 w/u2 + uρ ) − a2 − a2 . ut = u u2 u2

(7.21)

Example 7.1. Consider the equations L[k]ψ[k](t, x) = λψ[k](t, x),

ψt [k](t, x) = A3 [k]ψ[k](t, x),

k ∈ N,

(7.22)

where L[k] = u2 [k](∂ x )2 + u1 [k]∂ x + u[k],

u2 [k] = u2 = const,

A3 [k] = v3 [k](∂ x )3 + v2 [k](∂ x )2 + v1 [k]∂ x + v0 [k],

v3 [k] = v3 = const.

(7.23)

For the starting equations (7.22) and k = 0, say we choose u2 [0] = −1,

u1 [0] = u0 [0] = 0,

v3 [0] = −4,

v2 [0] = v1 [0] = v0 [0] = 0.

(7.24)

If the functions ψ1 (t, x), ψ1 [0](t, x) satisfy both equations (7.22) for k = 0, then applying the Darboux transformation we construct the function (x)

ψ1 [1](t, x) = Q1 ψ1 [0](t, x),

Q1 = ∂ x − d(t, x),

d(t, x) =

ψ1 (t, x) , ψ1 (t, x)

(7.25)

that satisfies L[1]ψ1 [1](t, x) = λψ1 [1](t, x),

ψ1t [1](t, x) = A3 [1]ψ1 [1](t, x),

where L[1] = u2 [1](∂ x )2 + u1 [1]∂ x + u[1],

A3 [1] = v3 [1](∂ x )3 + v2 [1](∂ x )2 + v1 [1]∂ x + v0 [1].

(7.26)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1715

Using the Darboux transformation formulas for second order equations (5.10), and in view of (7.24), we have u2 [1] = uρ2 [0] = −1,

u1 [1] = uρ1 [0] + Q1 u2 [0] + u2 [1]dρρ = d − dρρ

u0 [1] = uρ0 [0] + Q1 u1 [0] + u1 [1]dρ + u2 [1](dxρ + dρx ) = (d − dρρ )dρ − dxρ − dρx , that is u1 [1] = d − dρρ ,

u0 [1] = (d − dρρ )dρ − dxρ − dρx , ρ

or since d(x) = −dρ d, d(xx) = −dρ (d + dρ )(x) = − d u1 [1] = −

ν(x)d(xx) (x) , dρ (x)

(d−dρρ ) , ν

we get

u0 [1] = νd(xx) − dρ(x) − d(x)ρ .

(7.27)

Since ψ1 [1](t, x) satisfies both equations (7.22) for k = 1 by Lemma 7.1 the Lax equation Lt [1] = A[1]L[1] − L[1]A[1]

(7.28)

is also satisfied and from the formulas (7.14)–(7.17) we get u1 [1] = w[1] = a0 [1] + aρ0 [1], a2 [1] + aρ2 [1] =

u0 [1] = a1 [1] + aρ1 [1] +

a20 [1] (x) + a0 [1], u2

(7.29)

v3 (a0 [1](u0 [1] − uρ0 [1]) + w[1](aρ1 [1] + uρ0 [1])) − v3 ax1 [1]. u2

(7.30)

(x)

v3 a2 [1] + v3 u0 [1] (a0 [1] + aρ0 [1] + aρρ v0 [1] = , 0 [1]), u2 u2  v3 a0 [1]w[1] w(x) [1] + + a1 [1] + uρ0 [1] . v1 [1] = u2 u2

v2 [1] =

(7.31) (7.32)

Assuming ν(x) > 0, by integration (see [3]) of (7.29), (7.30) we get x a0 [1](t, x) = −

eˆ2/ν (ρ(y), x0 )

u1 [1]((y) ∇y, ν(y)

x0

x a1 [1](t, x) = −

eˆ2/ν (ρ(y), x0 ) x0

x a2 [1](t, x) =

eˆ2/ν (ρ(y), x0 ) x0



ay0 [1] − u[1](y) + a20 [1]/u2 ∇y, ν(y)

(x)

v3 v3 a1 [1] v3 w[1](a1 [1] + u[1])ρ − a0 [1]uy [1] − ν u2 u2 ν

 ∇y.

(7.33)

These formulas mean that the functions aj [1](t, x), vj [1](t, x), j = 0, 1, 2 in formulas (7.31), (7.32) depend on u1 [1](t, x), u0 [1](t, x). Assuming that ψ1 (t, x) = eˆd (x, x0 ) satisfies both equations (7.22) for k = 0 from characteristic equations we get Char2 = Char3 =

Lˆ ed (x, x0 ) = u2 (d(x) + dρ d) + u1 d + u0 = 0, eˆd (x, x0 )

Aˆ ed (x, x0 ) = v0 + v1 d + v2 (d(x) + dρ d) + v3 (d(xx) + d(x)ρ d + dρ(x) d + dρρ (d(x) + dρ d)) = 0, eˆd (x, x0 )

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1716

and in view of (7.27) d(x) + dρ d = 0,

d(xx) + (dρ(x) + d(x)ρ )d + dρρ (d(x) + dρ d) = 0.

(7.34)

Remark 7.1. From the Lax equation (7.28) it follows that the functions u0 [1], u1 [1] given by (7.27) are the solutions of the nonlinear (KdV-like) system (see (7.12)–(7.13)): (xxx)

u1t [1](t, x) = v3 (u1

ρ(xx)

[1] + u0

(xx)

+ v2 [1](u1

(x)ρ(x)

[1] + u0

ρ(x)

[1] + u0

(x)ρ

[1] + u0

(x)ρ

+ v1 [1](uρ0 [1] − u0 [1]) − u2 (v0 u0t [1](t, x) =

(xxx) v3 u0 [1]

+

(xx) v2 [1]u0 [1]

(xx)ρ

[1] + u0

(xx)

[1]) − u2 v1 ρ(x)

[1] + v0

+

[1]) (x)

(x)

[1] + v1 [1]u1 [1] − u1 [1]v1 [1]

[1]) + u1 [1](v0 [1] − v0ρ [1]),

(x) v1 [1]u0 [1]

−

(xx) u2 v0 [1]

−

(x) u1 [1]v0 [1],

(7.35) (7.36)

with the coefficients vj [1](t, x), j = 0, 1, 2 given by (7.31), (7.32). In the special case a0 [1] ≡ 0 from (7.29)–(7.31) we get u1 [1] = w[1] = v2 [1] = 0, v1 [1] = (xx)

and in view of −v3 a1

ρ(x)

[1] − a2

(x)

(x)ρ

= a2 [1] − a2

(x)

a2 [1] + aρ2 [1] = −v3 a1 [1] (x)

v3 (a1 [1] + uρ0 [1]), u2

v0 [1] =

a2 [1] + v3 u0 [1] u2

(x)

[1] = a2 [1] from (7.20) we get

(xx)

0 = u1t [1] = −v3 a1

u0 [1] = a1 [1] + aρ1 [1],

ρ(x)

[1] − a2

(x)ρ

[1] − a2

[1] +

v3 (a1 [1] + uρ [1])(uρ [1] − u[1]) u2

v3 (a1 [1] + uρ [1])(uρ [1] − u[1]) u2 v3 (xx) 0 = νa2 [1] + (a1 [1] + uρ [1])νu(x) [1]. u2

[1] +

In the case ν(x) = 0 we get (xx)

a2

[1] = −

v3 (a1 [1] + uρ [1])u(x) [1], u2

and from (7.21) we get the non-linear equation: 2v3 (x) u0t [1] = u [1](a1 [1] + uρ0 [1]), u2 0 ⎛ ⎞ x 2v3 (x) ⎝ ρ u0 (y)[1] eˆ2/ν (ρ(y), x0 )∇y ⎠ u [1] u0 [1] − u0t [1] = u2 0 ν(y)

(7.37)

x0

with the solution (see (7.27)): u0 [1](t, x) = ν(x)d(xx) (t, x) − dρ(x) (t, x) − d(x)ρ (t, x).

(7.38)

If ν(x) = μ(t) ≡ 0 then from 2a1 = u0 [1], 2a2 [1] = −v3 a1x [1], (7.35) is satisfied with w = u1 [1] ≡ 0 and (7.36) turns to u0t [1] =

v3 3v3 u0x [1]u0 [1] v3 u0xxx [1] . u0x [1](a1 [1] + uρ [1]) − a2xx [1] = + u2 2u2 4

(7.39)

G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

1717

Choosing v3 = −4,

u2 = −1,

(7.39) turns to the famous KdV equation u0t [1](t, x) = 6u0x [1](t, x)u0 [1](t, x) − u0xxx [1](t, x).

(7.40)

In this special case, and on the continuous time–space scale where ν(x) = μ(t) ≡ 0, the function ψ1 (t, x) = cosh(ax − 4a3 t) satisfies both equations (7.22) with k = 0: L[0]ψ1 = −(∂x )2 ψ1 = λ1 ψ1 ,

ψ1t = A[0]ψ1 = −4∂x3 ψ1 ,

λ1 = −a2 ,

and the second function (7.27):  u0 [1](t, x) = −2dx (t, x) = −2

ψ1x (t, x) ψ1 (t, x)

= −2a2 sech2 (ax − 4a3 t) x

yields the well-known 1-soliton solution of the KdV equation (7.40). 8. Conclusion In the paper the standard Darboux transformation is extended to Hilger’s time-scale calculus. This extension to arbitrary time or space scales provides access to a wider range of integrable non-linear dynamic equations. This is meaningful in quantum physics in the cases X = hZ or X = Kq , and we hope it will also be helpful in numerical experiments involving variable graininess. References [1] M. Blaszak, M. Gurses, B. Silindir, B.M. Szablikowski, Integrable discrete systems on R and related dispersionless systems, J. Math. Phys. 49 (2008) 072702. [2] M. Blaszak, B. Silindir, B.M. Szablikowski, The R-matrix approach to integrable systems on time scales, J. Phys. A 41 (2008) 385203. [3] M. Bohner, A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhäuser, Boston, 2001. [4] M.M. Crum, Associated Sturm–Liouville systems, Quart. J. Math. 6 (1) (1955) 121–127. [5] G. Darboux, Sur une proposition relative aux equation lineaires, C. R. Acad. Sci. Paris 94 (1882) 1456–1459. [6] X.Y. Gao, Variety of the cosmic plasmas: general variable-coefficient Korteweg–de Vries–Burgers equation with experimental/observational support, Europhys. Lett. 110 (2015) 15002. [7] X.Y. Gao, Bäcklund transformation and shock-wave-type solutions for a generalized (3 +1)-dimensional variable-coefficient B-type Kadomtsev–Petviashvili equation in fluid mechanics, Ocean Eng. 96 (2015) 245–247. [8] X.Y. Gao, Incompressible-fluid symbolic computation and Bäcklund transformation: (3 +1)-dimensional variable-coefficient Boiti–Leon–Manna–Pempinelli model, Z. Naturforsch. A 70 (1) (2015) 59–61. [9] C.S. Gardner, J.M. Green, M.D. Kruskal, R.M. Muira, Method of solving the Korteweg–de Vries equation, Phys. Rev. Lett. 19 (1967) 1995–1997. [10] M. Gurses, G.Sh. Guseinov, B. Silindir, Integrable equations on time scales, J. Math. Phys. 46 (2005) 113510. [11] S. Hilger, Analysis on measure chains—a unified approach to continuous and discrete calculus, Results Math. 18 (1990) 18–56. [12] R. Hirota, Nonlinear partial difference equations. I. A difference analogue of the Korteweg–de Vries equation, J. Phys. Soc. Jpn. 43 (1977) 1424–1433. [13] G. Hovhannisyan, Ablowitz–Ladik hierarchy of integrable equations on a time–space scale, J. Math. Phys. 55 (10) (2014) 102701. [14] P.D. Lax, Integrals of nonlinear equations of evolution and solitary waves, Comm. Pure Appl. Math. 21 (5) (1968) 467–490.

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G. Hovhannisyan, O. Ruff / J. Math. Anal. Appl. 434 (2016) 1690–1718

[15] V.B. Matveev, Darboux transformation and the explicit solutions of differential–difference and difference–difference evolution equations I, Lett. Math. Phys. 3 (3) (1979) 217–222. [16] V.B. Matveev, M.A. Salle, Darboux Transformations and Solitons, Springer Ser. Nonlinear Dynam., Springer Verlag, Berlin, 1991. [17] S. Odake, R. Sasaki, Casoratian identities for the Wilson and Askey–Wilson polynomials, J. Approx. Theory 193 (2015) 184–209. [18] B.M. Szablikowski, M. Blaszak, B. Silindir, Bi-Hamiltonian structures for integrable systems on regular time scales, J. Math. Phys. 50 (2009) 073502. [19] F. Tan, Y. Xia, Z. Wei, Robust-yet-fragile nature of interdependent networks, Phys. Rev. E 91 (2015) 023205. [20] M. Wadati, H. Samuki, K. Konno, Relationships among inverse method, Bäcklund transformation and an infinite number of conservation laws, Progr. Theoret. Phys. 53 (2) (1975) 419–436. [21] H.L. Zhen, B. Tian, Y.F. Wang, D.Y. Liu, Soliton solutions and chaotic motions of the Zakharov equations for the Langmuir wave in the plasma, Phys. Plasmas 22 (2015) 032307.