Classification of Multidimensional Darboux Transformations: First Order and Continued Type

We analyze Darboux transformations in very general settings for multidimensional linear partial differential operators. We consider all known types of Darboux transformations, and present a new type. We obtain a full classification of all operators that admit Wronskian type Darboux transformations o...

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Date:	2017
Main Authors:	Hobby, D., Shemyakova, E.
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Language:	English
Published:	Інститут математики НАН України 2017
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Cite this:	Classification of Multidimensional Darboux Transformations: First Order and Continued Type / D. Hobby, E. Shemyakova // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 40 назв. — англ.

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spelling	irk-123456789-1486052019-02-19T01:25:53Z Classification of Multidimensional Darboux Transformations: First Order and Continued Type Hobby, D. Shemyakova, E. We analyze Darboux transformations in very general settings for multidimensional linear partial differential operators. We consider all known types of Darboux transformations, and present a new type. We obtain a full classification of all operators that admit Wronskian type Darboux transformations of first order and a complete description of all possible first-order Darboux transformations. We introduce a large class of invertible Darboux transformations of higher order, which we call Darboux transformations of continued Type I. This generalizes the class of Darboux transformations of Type I, which was previously introduced. There is also a modification of this type of Darboux transformations, continued Wronskian type, which generalize Wronskian type Darboux transformations. 2017 Article Classification of Multidimensional Darboux Transformations: First Order and Continued Type / D. Hobby, E. Shemyakova // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 40 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 16S32; 37K35; 37K25 DOI:10.3842/SIGMA.2017.010 http://dspace.nbuv.gov.ua/handle/123456789/148605 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	We analyze Darboux transformations in very general settings for multidimensional linear partial differential operators. We consider all known types of Darboux transformations, and present a new type. We obtain a full classification of all operators that admit Wronskian type Darboux transformations of first order and a complete description of all possible first-order Darboux transformations. We introduce a large class of invertible Darboux transformations of higher order, which we call Darboux transformations of continued Type I. This generalizes the class of Darboux transformations of Type I, which was previously introduced. There is also a modification of this type of Darboux transformations, continued Wronskian type, which generalize Wronskian type Darboux transformations.
format	Article
author	Hobby, D. Shemyakova, E.
spellingShingle	Hobby, D. Shemyakova, E. Classification of Multidimensional Darboux Transformations: First Order and Continued Type Symmetry, Integrability and Geometry: Methods and Applications
author_facet	Hobby, D. Shemyakova, E.
author_sort	Hobby, D.
title	Classification of Multidimensional Darboux Transformations: First Order and Continued Type
title_short	Classification of Multidimensional Darboux Transformations: First Order and Continued Type
title_full	Classification of Multidimensional Darboux Transformations: First Order and Continued Type
title_fullStr	Classification of Multidimensional Darboux Transformations: First Order and Continued Type
title_full_unstemmed	Classification of Multidimensional Darboux Transformations: First Order and Continued Type
title_sort	classification of multidimensional darboux transformations: first order and continued type
publisher	Інститут математики НАН України
publishDate	2017
url	http://dspace.nbuv.gov.ua/handle/123456789/148605
citation_txt	Classification of Multidimensional Darboux Transformations: First Order and Continued Type / D. Hobby, E. Shemyakova // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 40 назв. — англ.
series	Symmetry, Integrability and Geometry: Methods and Applications
work_keys_str_mv	AT hobbyd classificationofmultidimensionaldarbouxtransformationsfirstorderandcontinuedtype AT shemyakovae classificationofmultidimensionaldarbouxtransformationsfirstorderandcontinuedtype
first_indexed	2025-07-12T18:57:36Z
last_indexed	2025-07-12T18:57:36Z
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fulltext	Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (2017), 010, 20 pages Classification of Multidimensional Darboux Transformations: First Order and Continued Type David HOBBY and Ekaterina SHEMYAKOVA 1 Hawk dr., Department of Mathematics, State University of New York at New Paltz, USA E-mail: hobbyd@newpaltz.edu, shemyake@newpaltz.edu Received October 10, 2016, in final form February 14, 2017; Published online February 24, 2017 https://doi.org/10.3842/SIGMA.2017.010 Abstract. We analyze Darboux transformations in very general settings for multidimen- sional linear partial differential operators. We consider all known types of Darboux trans- formations, and present a new type. We obtain a full classification of all operators that admit Wronskian type Darboux transformations of first order and a complete description of all possible first-order Darboux transformations. We introduce a large class of invertible Darboux transformations of higher order, which we call Darboux transformations of con- tinued Type I. This generalizes the class of Darboux transformations of Type I, which was previously introduced. There is also a modification of this type of Darboux transformations, continued Wronskian type, which generalize Wronskian type Darboux transformations. Key words: Darboux transformations; Laplace transformations; linear partial differential operators; continued Darboux transformations 2010 Mathematics Subject Classification: 16S32; 37K35; 37K25 1 Introduction A Darboux transformation (DT), in the general sense of the word, is a transformation be- tween differential operators that simultaneously transforms their kernels (solution spaces) or eigenspaces. DTs originated in the work of Darboux and others on the theory of surfaces, as in [10], while particular examples were known to Euler and Laplace. They were rediscov- ered in integrable systems theory in the 1970s, where they were used for obtaining solutions of soliton-type equations with remarkable properties. They have also been studied by physicists for quantum-mechanical applications [17, 28]. The very name ‘Darboux transformations’ was introduced by V.B. Matveev in [24], where he played a central role in creating the Darboux transformation method in soliton theory. This theory was elaborated in the fundamental mono- graph [25] of V.B. Matveev and M.A. Salle (see also [11]). As has been pointed out by experts in the field such as Novikov, Darboux transformations play a more important role than merely as technical tools for constructing solutions. While a large amount of work has been devoted to Darboux transformations, it has concen- trated mainly on particular examples and specific operators, while a general theory was largely missing. Although starting from Darboux’s own work, there have been ideas, observations, and conjectures that can be interpreted as elements of this sought-after general theory of Darboux transformations. In view of the numerous applications of Darboux transformations, the impor- tance of developing such a theory cannot be overestimated. In recent years, important steps in this direction have been made. They are based on the algebraic approach in the center of which are so-called intertwining relations. The model example of Darboux transformations is the following transformation of single variable Sturm–Liouville operators: L→ L1, where L = ∂2x + u(x), L1 = ∂2x + u1(x), and u1(x) is obtained from u(x) by the formula u1(x) = u(x)+2(lnϕ0(x))xx. Here ϕ0(x) is a ‘seed’ solution mailto:hobbyd@newpaltz.edu mailto:shemyake@newpaltz.edu https://doi.org/10.3842/SIGMA.2017.010 2 D. Hobby and E. Shemyakova of the Sturm–Liouville equation Lϕ0 = λ0ϕ0 (with some fixed λ0). Then the transformation ϕ 7→ (∂x − (lnϕ0)x)ϕ sends solutions of Lϕ = λϕ to solutions of L1ψ = λψ (with the same λ). The seed solution is mapped to zero. (According to [26], this transformation was already known to Euler.) This example is a model in two ways. First, the formula for the transformation of solutions can be written in terms of Wronskian determinants: ϕ1 = W (ϕ0, ϕ)/ϕ0. This generalizes to a construction based on several linearly independent seed solutions and higher-order Wronskian determinants (Crum [9] for Sturm– Liouville operators and Matveev [24] for general operators on the line). Second, if one lets M = ∂x − (lnϕ0)x, then the following identity is satisfied: ML = L1M. (1.1) This identity is equivalent to the relation between the old potential u(x) and the new poten- tial u1(x). In an abstract framework, if two operators with the same principal symbol, L and L1, sat- isfy (1.1) for some M , then (1.1) is called the intertwining relation and M is (often) called the transformation operator. One can see that if (1.1) is satisfied, then the operator M defines a linear transformation of the eigenspaces of L to the eigenspaces of L1 (for all eigenvalues). The relation (1.1) can be taken as a definition of the DT. The intertwining relation (1.1) appeared, for Sturm–Liouville operators, in work of Shabat [29, equation (19)], Veselov–Shabat [39], and Bagrov–Samsonov [2]. Intertwining relation (1.1) is also related with supersymmetric quan- tum mechanics initiated by E. Witten [40], see in particular [7, 8]. In 2D case intertwining relation (1.1) was studied in the series of papers by A. Andrianov, F. Cannata, M. Ioffe, see, e.g., [18] and references therein. It also appeared for higher dimensions, in Berest–Veselov [4, 5] for the Laplace type operators L = −∆ + u. The same relation was used in [16] for differential operators on the superline. The natural task that arises for such an algebraic definition of DTs, is the classification of the DTs satisfying it. In the case of differential operators on the line, it was established (in steps) that all DTs defined this way arise from seed solutions and are given by Wronskian formulas. For the Sturm–Liouville operators, this was proved in [39, Theorem 5] when the new potential u1(x) differs from the initial u(x) by a constant, u1(x) = u(x) + c. It was proved in [30] for transformations of order two; and, finally in the general case, in [2] and the follow-up paper [3], see also [27, Section 3]. For general operators on the line, it was proved in [1]. For the superline, the classification was obtained in [16]. The intertwining relation (1.1), with a single transformation operator M , turns out to already be too restrictive for differential operators in higher dimensions. We use the more flexible intertwining relation NL = L1M. (1.2) This can be extracted from the work of Darboux himself [10] (see for example [38, equation (2)]). There are several important differences between the two kinds of intertwining relations. If (1.1) is satisfied, then, as already mentioned, the operator M transforms each eigenspace of L to the eigenspace of L1 with the same eigenvalue. In contrast with that, if (1.2) is satisfied, then we only have that M maps the kernel of L to the kernel of L1. (It is in general false that M maps eigenspaces of L with nonzero eigenvalues to eigenspaces of L1.) In [31, 33], a general framework for Darboux transformations defined by intertwining relation (1.2) was put forward, which in particular allowed the proof of a long-standing conjecture of Darboux on factorization of the DTs for 2D second-order hyperbolic operators (the ‘2D Schrödinger operator’)1. As we explain in greater detail in the paper, it is natural to introduce an equivalence relation between 1Calling it Schrödinger is some abuse of language justified by the fact that on the formal algebraic level it is equivalent to the actual elliptic Schrödinger operator [26]. Classification of Multidimensional Darboux Transformations 3 pairs (M,N) in conjunction with the intertwining relation (1.2). This gives extra flexibility and (as explained below, see Section 2) makes it possible to have invertible DTs. Such an equivalence relation does not exist in the case of the intertwining relation (1.1) with M = N . It remains an open problem to analyze fully the connections between the intertwining relations (1.1) and (1.2). In [34], a completely new class of transformations, Darboux transformations of Type I, was described. These are for operators of very general form, and are analogous to Laplace transformations. In [16], a complete classification of Darboux transformations of arbitrary nondegenerate operators on the superline was obtained. In [20] it was proved that every Darboux transformation can be obtained in terms of super-Wronskians, which are based on Berezinians, analogues of determinants in the super case. Among DTs defined by our intertwining relation (1.2) there are non-Wronskian type Darboux transformations such as the classical Laplace transformations for the hyperbolic second-order operator in the plane, sometimes called the 2D Schrödinger operator. In [31, 33] it was proved that every Darboux transformation for the class of operators is a composition of atomic Darboux transformations of two types: Wronskian type and Laplace transformations (a conjecture that can be traced back to Darboux). Since (1.2) includes (1.1) as a special case, we will henceforth use intertwining relation to refer to (1.2). In this paper we recall the general algebraic framework for Darboux transformations based on intertwining relations and then analyze the various types of Darboux transformations that may arise. We partially answer the following questions in the general theory of Darboux trans- formations: • Which classes of operators admit Darboux transformations of Wronskian type? • Are there any new types of invertible Darboux transformations? • What is the qualitative difference between these two types? For more than a century, only two types of transformations of linear partial differential op- erators satisfying our intertwining relation (Darboux transformations) were known: the trans- formations defined by Wronskian formulas, and Laplace transformations. Note that Wronskian formulas define DTs for several different kinds of operators, see [24, 25], while Laplace transfor- mations are defined only for 2D Schrödinger operators. Laplace transformations have a number of good properties including invertibility. Recently, two different generalizations of Laplace transformations were proposed: “intertwining Laplace transformations” in [15] and “Type I transformations”. (The latter are always invertible.) In the present paper we clarify their rela- tion and introduce a further generalization, which we call transformations of “continued Type I”. (This term comes from the construction, which has some resemblance with that of continued fractions.) A similar construction can also be applied starting from first-order DTs of Wronskian type, we say the resulting DTs are of “continued Wronskian type”. We also obtain a full description of first-order DTs and the differential operators that admit them2. Moreover, we show that first-order DTs always fall into two types: Wronskian type and Type I. These two different cases possess a unified algebraic presentation (see Theorem 4.2). Our test makes it possible to tell which type of DTs are admitted by a given operator. The structure of the paper is as follows. In Section 2, we recall key results on the category of Darboux transformations as introduced in [31, 33] and list known types of Darboux transforma- tions. In Section 3, we give a criterion for a general operator to admit a first-order Wronskian Darboux transformation (Theorem 3.4). In Section 4 we classify first-order DTs. In Section 5, we introduce continued Type I Darboux transformations and prove they are always invertible. In Section 6, we give a similar construction of DTs of continued Wronskian type. 2Actually, we consider transformations with M = ∂t + m, where m ∈ K and t is a distinguished variable. It should be possible to reduce a general first-order operator M to this form by a change of variables. 4 D. Hobby and E. Shemyakova 2 Types of Darboux transformations In this section we recall some general facts about Darboux transformations. Consider a diffe- rential field K of characteristic zero with commuting derivations ∂t, ∂x1 , . . . , ∂xn . We use one distinguished letter t to denote the variable which will play a special role. The letter t need not represent time, and was chosen for convenience. By K[∂t, ∂x1 , . . . , ∂xn ], we denote the corresponding ring of linear partial differential operators overK. Operators inK[∂t, ∂x1 , . . . , ∂xn ] of order 0 will often be considered as elements of K, and called “functions”. One can either assume the field K to be differentially closed, or simply assume that K contains the solutions of those partial differential equations that we encounter on the way. Darboux transformations viewed as mappings of linear partial differential operators can be defined algebraically as follows [33], where we write σ(L) for the principal symbol of L. Definition 2.1. Consider a category, with objects all operators in K[∂t, ∂x1 , . . . , ∂xn ] and mor- phisms the Darboux transformations defined as follows. A morphism from an object L to an object L1 with σ(L) = σ(L1) is a pair (M,N) ∈ K[∂t, ∂x1 , . . . , ∂xn ]×K[∂t, ∂x1 , . . . , ∂xn ] satisfying the intertwining relation NL = L1M. (2.1) Composition of Darboux transformations is defined by (M,N) ◦ (M1, N1) = (M1M,N1N), where the left DT is applied first. For each object L, its identity morphism 1L is (1, 1). Observe that given NL = L1M , we have σ(L) = σ(L1) iff σ(N) = σ(M). Note also that since there are only morphisms between objects with the same principal symbol, this category is partitioned into subcategories for each principal symbol. Definition 2.2. Two morphisms (M,N) and (M ′, N ′) from L to L1 are equivalent, written (M,N) ∼ (M ′, N ′), iff there exists A ∈ K[∂t, ∂x1 , . . . , ∂xn ] with M ′ = M + AL and N ′ = N + L1A. Lemma 2.3. The relation ∼ in Definition 2.2 is an equivalence relation. When (M,N) is a morphism from L to L1, so is every pair of the form (M +AL,N + L1A). The relation ∼ is compatible with the composition. Proof. By construction, ∼ is clearly an equivalence relation. To see the next statement, suppose a pair (M,N) satisfies (2.1), and consider an equivalent pair (M + AL,N + L1A). We have (N +L1A)L = NL+L1AL = L1M +L1AL = L1(M +AL), so it also satisfies the intertwining relation. Finally, we show that ∼ is compatible with composition. Consider the Darboux transforma- tions (M,N) : L → L1, and (M1, N1) : L1 → L2, where all operators are in K[∂t, ∂x1 , . . . , ∂xn ]. Taking different representatives in the same equivalence classes, we have (M +AL,N +L1A) : L→ L1, and (M1 +BL1, N1 + L2B) : L1 → L2 for some A,B ∈ K[∂t, ∂x1 , . . . , ∂xn ]. That is we have: (N + L1A)L = L1(M + AL), and (N1 + L2B)L1 = L2(M1 + BL1). For the composition we have (M̂, N̂) : L→ L2 with N̂ = (N1 +L2B)(N +L1A), M̂ = (M1 +BL1)(M +AL). Using NL = L1M and N1L1 = L2M1 we have that N̂ = N1N + L2C, and M̂ = M1M + CL, where C = BN +BL1A+M1A. This concludes the proof. � Every Darboux transformation (M,N) in a given equivalence class defines the same linear mapping from kerL to kerL1, where each function φ goes to M [φ]. For if φ ∈ kerL then (M +AL)[φ] = M [φ] +A[L[φ]] = M [φ] +A[0] = M [φ], so M and M +AL give the same linear map. This motivates our use of the equivalence relation ∼, as in Definition 2.2. Classification of Multidimensional Darboux Transformations 5 Note that one may also view the ∼ classes as being the Darboux transformations, rather than the individual pairs of operators. This approach was taken in [31, 32, 33] and [34]. We define the order of a Darboux transformation (M,N) as the minimum possible order of a transformation in its equivalence class, that is the least possible order of an operator of the form M + AL. We also define invertibility in terms of equivalence classes, and say that the Darboux transformation (M,N) : L → L1 is invertible iff there exists (M ′, N ′) : L1 → L such that the compositions (M,N) ◦ (M ′, N ′) and (M ′, N ′) ◦ (M,N) are equivalent to identity morphisms. Since equivalent morphisms give the same kernel maps, this is justified. Continuing this abuse of terminology, we say that the morphisms (M,N) and (M ′, N ′) are inverses, and so on. Note that with this definition, a morphism may have multiple inverses, which will be equivalent but not equal. This definition gives us that (M,N) : L → L1 and (M ′, N ′) : L1 → L are inverses iff there are some A,B ∈ K[∂t, ∂x1 , . . . , ∂xn ] with M ′M = 1 +AL, (2.2) N ′N = 1 + LA, (2.3) MM ′ = 1 +BL1, (2.4) NN ′ = 1 + L1B. (2.5) Lemma 2.4. Invertible Darboux transformations induce isomorphisms on the kernels of the operators L and L1. Proof. Consider ψ ∈ kerL, then from (2.2) we obtain that M ′M [ψ] = ψ. Similarly for ψ1 ∈ kerL1, (2.4) gives us MM ′[ψ1] = ψ1. Therefore, M and M ′ induce mutually invertible maps between kerL and kerL1. � In particular, (2.2) implies that kerL∩kerM = {0} is necessary for a Darboux transformation to be invertible. Also note that the order of a Darboux transformation is not related in an obvious way to the order of its inverse. The first steps in the general study of invertible Darboux transformations along with the analysis of one particular invertible class of Darboux transformations – Darboux transformations of Type I – can be found in [32, 34]. Lemma 2.5. 1. Given the intertwining relations NL = L1M and N ′L1 = LM ′, the equalities for N , N ′ follow from the equalities for M , M ′. Specifically, (2.3) follows from (2.2), and (2.5) follows from (2.4), 2. If (M,N) and (M ′, N ′) are mutually inverse morphisms satisfying (2.2) through (2.5), then BN = MA. Proof. 1. Assume that (2.2) holds, so M ′M = 1 +AL. Then N ′NL = LM ′M = L(1 +AL) = (1 +LA)L. Since K[∂t, ∂x1 , . . . , ∂xn ] has no zero divisors, we get N ′N = 1 +LA by cancellation, so (2.3) holds. That (2.4) implies (2.5) is proved analogously. 2. We know that NN ′ = 1 + L1B. Then NN ′N = N + L1BN , which implies N(1 + LA) = N+L1BN . Thus, NLA = L1BN , which implies L1MA = L1BN using the intertwining relation NL = L1M . Therefore, BN = MA by cancellation. � Recall that a gauge transformation of an operator L is defined as Lg = g−1Lg, where g ∈ K. Gauge transformations commute with Darboux transformations in the sense that if there is a Darboux transformation (M,N) : L→ L1, then there are also Darboux transformations( Mg, Ng ) : Lg → Lg1, (Mg,Ng) : Lg → L1, ( g−1M, g−1N ) : L→ Lg1. 6 D. Hobby and E. Shemyakova Definition 2.6. If a pair (M,N) defines a Darboux transformation from L to L1, then the same pair defines a different Darboux transformation from L+AM to L1 +NA for every A ∈ K[∂t, ∂x1 , . . . , ∂xn ]. We shall say that Darboux transformations L→ L1 and L+AM → L1+NA are related by a shift. Note that this is a symmetric relation. The shift of the DT (M,N) : L→ L1 is a DT. We have that N(L+AM) = (L1 +NA)M for the intertwining relation, and σ(L) = σ(L1) implies σ(M) = σ(N) which implies σ(L+AM) = σ(L1 +NA). In general, shifts of Darboux transformations do not commute with compositions of Darboux transformations. However, shifts are useful when dealing with invertible Darboux transforma- tions. Lemma 2.7. Shifts of Darboux transformations preserve invertibility. Proof. Suppose we have a Darboux transformation (M,N) : L→ L1 with NL = L1M , which is invertible, and there is a Darboux transformation (M ′, N ′) : L′ → L with N ′L1 = LM ′, and (2.2) and (2.4) hold, giving M ′M = 1+AL and MM ′ = 1+BL1. Then for any C ∈ K[∂t, ∂x1 , . . . , ∂xn ] we have the Darboux transformation (M,N) : L̃→ L̃1, where L̃ = L+CM , and L̃1 = L1 +NC. The inverse Darboux transformation for the shifted Darboux transformation is then given by M̃ ′ = M ′ + AC and Ñ ′ = N ′ + CB. We first show Ñ ′L̃1 = L̃M̃ ′, or (N ′ + CB)(L1 + NC) = (L+CM)(M ′+AC). This is equivalent to N ′L1 +N ′NC+CBL1 +CBNC = LM ′+CMM ′+ CMAC + LAC. Taking into account that N ′L1 = LM ′ and BN = MA, the last equality is equivalent to N ′NC + CBL1 = CMM ′ + LAC, which is equivalent to (1 + LA)C + CBL1 = C(1 +BL1) + LAC, which is true. Let us now prove that M̃ ′M = 1 + AL̃. This equality is equivalent to (M ′ + AC)M = 1 +A(L+ CM), which is true since M ′M = 1 +AL. It remains to show MM̃ ′ = 1 +BL̃1, or M(M ′ +AC) = 1 +B(L1 +NC). Since MA = BN by Lemma 2.5, this is equivalent to MM ′ = 1+BL1, which is true. The equalities for N and N ′ follow from Lemma 2.5 as well. � Another useful idea is what we call the dual of a Darboux transformation. Definition 2.8. If (M,N) : L → L1 is a Darboux transformation, its dual is the Darboux transformation (L,L1) : M → N . This defines a Darboux transformation. If (M,N) : L → L1 is a DT, then NL = L1M and σ(L) = σ(L1). Thus L1M = NL, and σ(M) = σ(N), and (L,L1) : M → N is also a DT. Note that this is not a duality in the sense of category theory; our dual is quite unusual in the sense that it switches objects and morphisms. This duality gives a useful perspective on the category of DTs. For example, we have that DTs are related by a shift if and only if their duals are equivalent. To see this, consider the morphism (M,N) from L to L1, and its shift where (M,N) is a morphism from L + AM to L1 + NA. Taking duals gives us two morphisms from M to N , one given by the pair (L,L1) and the other by the pair (L+AM,L1 +NA), which the reader may verify are equivalent. The proof in the other direction is similar. More importantly, we have the following lemma. Lemma 2.9. A Darboux transformation is invertible if and only if its dual is. Proof. Since the dual of the dual of a DT is the original transformation, it suffices to show that the dual of an invertible DT is invertible. So let (M,N) : L → L1 be an invertible DT. This gives us a DT (M ′, N ′) : L1 → L with M ′M = 1 + AL and MM ′ = 1 + BL1 for some operators A and B. Lemma 2.5 gives us N ′N = 1 + LA and BN = MA. Now we must show that (L,L1) : M → N is invertible. That is, we must show that (M̃, Ñ) : L̃ → L̃1 is invertible, Classification of Multidimensional Darboux Transformations 7 where we have M̃ = L, Ñ = L1, L̃ = M and L̃1 = N . So we need a DT (M̃ ′, Ñ ′) : L̃1 → L̃, with M̃ ′M̃ = 1 + ÃL̃ and M̃M̃ ′ = 1 + B̃L̃1 for some operators Ã and B̃. Now we take M̃ ′ = −A, Ñ ′ = −B, Ã = −M ′ and B̃ = −N ′. We have Ñ ′L̃1 = −BN = M(−A) = L̃M̃ ′, so (M̃ ′, Ñ ′) : L̃1 → L̃ is a DT. Also M̃ ′M̃ = −AL = 1 + (−M ′)M = 1 + ÃL̃, and M̃M̃ ′ = L(−A) = 1 + (−N ′)N = 1 + B̃L̃1, as required. � There are several known types of Darboux transformations. 1. Darboux transformations of Wronskian type. These are the classical DTs. The operator M is given by M(f) = W (f1, f2, . . . , fm, f) W (f1, f2, . . . , fm) , where W (f1, f2, . . . , fm) denotes a Wronskian determinant with respect to one of the variables t, x1, . . . , xn of m linearly independent fj ∈ K, which are elements of kerL. Definition 2.10. We say that a DT (M,N) : L→ L1 is of Wronskian type when M can be put in the above form by a change of variables, and we say that the DT is a multiple of Wronskian type if M is the result of multiplying an operator of the above form on the left by a function in K. Wronskian type Darboux transformations are proved to be admitted by several different types of operators [16, 19, 24, 25], and direct applications of these transformations to solve famous nonlinear equations are well known. Wronskian type transformations are never invertible, since kerL ∩ kerM 6= {0}. (Wronskian type DT formulas in abstract frameworks were introduced in [12] and [19]. An analog of Wronskian DTs for super Sturm–Liouvlle operators was discovered in [21, 22, 23].) Note that in the 1D case with intertwining relation having M = N , as in (1.1), a similar Wronskian construction is used (it has the same formula), but the fi are arbitrary eigenfunctions with not necessarily zero eigenvalues. For the 1D case, every DT is of Wronskian type in this extended sense, see [1]. 2. Darboux transformations obtained from a factorization. Suppose L = CM for some C,M ∈ K[∂t, ∂x1 , . . . , ∂xn ] \ K. Then for any operator N with σ(N) = σ(M), there is a DT (M,N) : CM → NC, since N(CM) = (NC)M . Since kerM ⊆ kerL, these transformations are never invertible. A common trick in proofs is to establish using some reasoning that the transformation opera- tor M can be considered in a form where it is effectively an ordinary differential operator while everything else is multidimensional. In this case, if the Darboux transformation is obtained from a factorization with this particular M , then it is also of Wronskian type. This uses the fact that every linear ordinary differential operator can be expressed by a Wronskian formula. 3. Laplace transformations. These are another type of Darboux transformations, in- troduced in [10]. They are distinct from the Wronskian type. They are only defined for 2D Schrödinger type operators, which have the form L = ∂x∂y + a∂x + b∂y + c, (2.6) where a, b, c ∈ K. If the Laplace invariant k = by + ab− c is nonzero, then L admits a Darboux transformation with M = ∂x + b (in “x-direction”). (Explicit formulas for L1 and N are given below.) If the other Laplace invariant, h = ax + ab − c is nonzero, then L admits a Darboux transformation with M = ∂y + a (in “y-direction”). 8 D. Hobby and E. Shemyakova Laplace transformations are invertible, and are (almost) inverses of each other. This has been mentioned in the literature, e.g., [36]. Classically, the invertibility of Laplace transformations was understood to mean that they induced isomorphisms of the kernels of the operators in question. Interestingly, it is exactly the equivalence relation on Darboux transformations that makes it possible to understand the invertibility of Laplace transformations in the precise algebraic sense. The following statement appeared in a brief form in [32]. Theorem 2.11. 1. The composition of two consecutive Laplace transformations applied to L, first in x di- rection, and then in y direction is equal to the gauge transformation L → L1/k. If the transformation is first in x direction, and then in y direction then the composition is equal to the gauge transformation L→ L1/h. 2. The inverse for the Laplace transformation L → L1 given by the operator M = ∂x + b is (M ′, N ′) : L1 → L, where M ′ = −k−1(∂y+a), N ′ = −k−1(∂y+a−kyk−1). The inverse for the Laplace transformation L→ L1 given by the operator M = ∂y+a is (M ′, N ′) : L1 → L, where M ′ = −h−1(∂x + b), N ′ = −h−1(∂x + b− hyh−1). Proof. Consider the Laplace transformation of L of the form (2.6) in “x-direction”, that is given by the operator M = ∂x+ b. The operators M and L in this case completely define opera- tors L1 and N . (This is an instance of our Theorem 4.7.) Note that expressions for L1 and N are much shorter, when expressed in terms of k (substitute c = by+ab−k): N = ∂x+b−kxk−1, L1 = ∂x∂y + a1∂x + b1∂y + c1 = ∂x∂y + a∂x + (b− kxk−1)∂y + ab− akxk−1 − k − ax. Now consider the Laplace transformation for L1 but in the y direction (which “returns back”), i.e., given by the operator M1 = ∂y + a1 (in this case a1 for L1 equals a for L). Then the intertwining relation N1L1 = L2M1 implies N1 = ∂y + a− kyk−1, and L2 = L1/k. The composition of these transformations is (M̂, N̂) : L → L1/k with M̂ = M1M = (∂y + a)(∂x + b) = ∂x∂y + a∂x + b∂y + ab + by = k + L, and N̂ = N1N = k + L1/k = k + L2. From here we can readily see how M1 and N1 can be changed to make the composition an identity morphism from L to L: the inverse Darboux transformation for the Laplace transformation with M = ∂x + b is ( 1 k (∂y + a1), 1 k (∂y − h1xh−11 ) ) : L1 → L. � (Note that the formulas for the inverse for Laplace transformations can be generalized, as we do below, for Darboux transformations of Type I, which is a generalization of Laplace transfor- mations to operators of a more general form.) 4. Intertwining Laplace transformations. These were introduced in [15], and gener- alize Laplace transformations to linear partial differential operators L ∈ K[∂t, ∂x1 , . . . , ∂xn ] of very general form. One starts with any representation L = X1X2 − H, where L,X1, X2, H ∈ K[∂t, ∂x1 , . . . , ∂xn ]. Then it was proved that there is a Darboux transformation for the opera- tor L, (X2, X2 + ω) : L→ L1, where L1 = X2X1+ωX1−H, and ω = −[X2, H]H−1. The latter is a pseudodifferential operator in the general case, an element of the skew Ore field over K that extends K[∂t, ∂x1 , . . . , ∂xn ]. In [15], E. Ganzha then adds the requirement that ω ∈ K[∂t, ∂x1 , . . . , ∂xn ]. This is a very general class of transformations and can be used for theoretical investigations. The class contains both invertible and non-invertible Darboux transformations. 5. Darboux transformations of Type I. These were introduced by the second author in [34], and are admitted by operators in K[∂t, ∂x1 , . . . , ∂xn ] that can be written in the form L = CM + f , where C,M ∈ K[∂t, ∂x1 , . . . , ∂xn ], and f ∈ K. We have( M,M1/f ) : L→ L1, Classification of Multidimensional Darboux Transformations 9 where L1 = M1/fC+f , writing M1/f for fM(1/f). The inverse Darboux transformation always exists and is as follows:( − 1 f C,−C 1 f ) : L1 → L. The original theory of Laplace transformations is very naturally formulated in terms of differen- tial invariants. In [34], analogous ideas were developed for Darboux transformations of Type I for operators of third order in two independent variables. This can also be done for operators of a general form using regularized moving frames [13, 14] and ideas from [35]. The classical Laplace transformations are special case of transformations of Type I. We will see that Dar- boux transformations of Type I can be identified with a subclass of the Intertwining Laplace transformations defined above. Note that the composition of two Darboux transformations of Type I is (in general) not of Type I. Note also that in the Physics literature there are examples of Darboux transformations for concrete differential operators such as the non-stationary Schrödinger operator or the Fokker– Planck operator, see [7]. 3 Operators of general form admitting Wronskian type Darboux transformations Several different types of operators have been proved to admit Wronskian type Darboux trans- formations. From classical results, it is known that arbitrary 1D operators and 2D Schrödinger operator admit Wronskian type Darboux transformations. Also operators in two independent variables of the form L = ∂t − ai∂ i x, for i = 1, . . . , n (the Einstein summation convention is used) admit Wronskian type Darboux transformations with M of the form M = ∂x + m [24]. Analogues of Wronskian type Darboux transformations are admitted by super Sturm–Liouville operators [21, 22, 23], and, as was recently proved, non-degenerate operators on the super- line [16, 20]. However, there is an abundance of examples where a Wronskian type operator based on a element of kerL does not define a Darboux transformation. Example 3.1. For the following operator L = ∂t + ∂2x + t∂x − 1 t , and the element of its kernel ψ = t, M = ∂t−ψtψ−1 does not generate a Darboux transformation. This mean that it is not possible to find L1 = ∂t + a1∂ 2 x + a2∂x + a3 and N = ∂t + n such that NL = L1M . Thus we ask the following: Given an arbitrary operator in K[∂t, ∂x1 , . . . , ∂xn ] with ψ in his kernel, when does it admit a Wronskian type Darboux transformation with M = ∂t − ψtψ−1? Let us take t a fixed but arbitrary variable. We do not assume a priori any special depen- dence on ∂t. The idea is that we shall start by treating operators L in K[∂t, ∂x1 , . . . , ∂xn ] as ordinary differential operators with respect to t with coefficients in the ringK[∂x1 , . . . , ∂xn ]. Now K[∂x1 , . . . , ∂xn ] is embedded inK[∂t, ∂x1 , . . . , ∂xn ], so we may say that elements ofK[∂x1 , . . . , ∂xn ] are the t-free operators in K[∂t, ∂x1 , . . . , ∂xn ]. Definition 3.2. For a given variable t, call a differential operator t-free if it does not contain ∂t and none of its coefficients depend on t. 10 D. Hobby and E. Shemyakova Theorem 3.3. Given an operator L ∈ K[∂t, ∂x1 , . . . , ∂xn ], let M = ∂t. Then we have the DT (M,N) : L → L1 for some operator L1 and a first-order operator N if and only if there exist some A,B ∈ K[∂t, ∂x1 , . . . , ∂xn ] and c ∈ K, c 6= 0, such that L = A∂t + cB, where B is t-free. If the Darboux transformation exists, then L1 and N are given by the formulas L1 = L−A∂t +NA = NA+ cB, N = M1/c, where M1/c is cM(1/c). Here either B = 0, or we may take any non-zero coefficient of our choice in B to be 1 by a suitable choice of c. Proof. Let L be given, and assume NL = L1M for some N,L1 ∈ K[∂t, ∂x1 , . . . , ∂xn ], and M = ∂t. Since σ(L) = σ(L1), we have σ(M) = σ(N) so N = ∂t + n, for some n ∈ K. We shall now construct a certain shift of this Darboux transformation. By division with remainder in K[∂x1 , . . . , ∂xn ][∂t], we can choose A ∈ K[∂t, ∂x1 , . . . , ∂xn ] such that L′ = L− A∂t does not contain ∂t. If L′ = 0, then L = A∂t = AM and the conclusion easily follows. Henceforth assume L′ 6= 0. We write L′ = cp1p2...pn∂ p1 x1∂ p2 x2 · · · ∂ pn xn , p1, p2, . . . , pn ≥ 0, cp1p2...pn ∈ K, where we assume summation in upper and lower indices. We also write L′1 = L1−NA 6= 0. We then get the Darboux transformation (M,N) : L′ → L′1, since NL′ = NL − NA∂t = L1M − NAM = (L1 − NA)M . Now, in NL′ = L′1∂t every term of NL′ must be a left multiple of ∂t, and all the terms that do not contain ∂t must have their coefficients zero. So, (∂t + n)[L′] = 0, where the square brackets mean application of ∂t+n to all the coefficients of L′. So, for each term cp1p2...pn∂ p1 x1∂ p2 x2 · · · ∂ pn xn in L′ we must have (∂t +n)[cp1p2...pn ] = 0. Writing c for cp1p2...pn , we have ct+nc = 0 or −n = ct/c. Thus ct/c = ∂t(ln(c)) is the same for all of the cp1p2...pn , so all the ln(c) differ by functions that do not depend on t, and all the c are equal to within multiplication by functions which do not depend on t. We may also take some non-zero coefficient in B of our choice to be 1, and write all the cp1p2...pn in the form cb with b t-free, using a common function c that satisfies ct/c = −n, hence N = ∂t − ct/c, and L′ is of the form L′ = cB, where B is t-free. We have NL′ = L′1∂t, and calculate NL′ = (∂t− ct/c)cB = ∂tcB− ctB = c∂tB+ ctB− ctB = c∂tB = cB∂t, since B is t-free. Thus, L′1∂t = cB∂t, and cancelling ∂t gives us L′1 = cB. So L′1 = cB = L′. Thus, L = L′ +A∂t = A∂t + cB. Also L1 = L′1 +NA = cB +NA. The other direction is an easy calculation. � Classification of Multidimensional Darboux Transformations 11 This theorem also gives us a similar result for general transformations of Wronskian type. Theorem 3.4 (criterion). Given an operator L ∈ K[∂t, ∂x1 , . . . , ∂xn ] and a ψ ∈ kerL, the operator M = ∂t−ψtψ−1 generates a DT if and only if there exists some A ∈ K[∂t, ∂x1 , . . . , ∂xn ], and a t-free B ∈ K[∂t, ∂x1 , . . . , ∂xn ] such that Lψ = A∂t + cB. Here either B = 0, or any one non-zero coefficient in B may be taken to be 1. If the DT exists, then L1 and N are given by Lψ1 = Lψ −A∂t +NψA, Nψ = ∂ 1/c t , where ∂ 1/c t is c∂t(1/c). Proof. Let L be given, together with ψ ∈ ker(L), and let M be ∂t − ψtψ −1. Note that Mψ = ψ−1(∂t − ψtψ−1)ψ = ψ−1(ψ∂t + ψt − ψt) = ∂t, so we conjugate NL = L1M by ψ. This gives NψLψ = Lψ1M ψ = Lψ1 ∂t. Now apply the previous theorem. � Corollary 3.5 (necessary condition). Let an operator L ∈ K[∂t, ∂x1 , . . . , ∂xn ], and ψ ∈ kerL be given, where Mψ = ∂t − ψtψ−1 generates a DT. Then there exists some A ∈ K[∂t, ∂x1 , . . . , ∂xn ], and a t-free B ∈ K[∂t, ∂x1 , . . . , ∂xn ] such that L = AMψ + cB, [Mψ, B] = 0. Here B is not necessarily t-free. Proof. From the previous theorem, Lψ = A∂t + cB and [∂t, B]. Conjugating by ψ−1, L = Aψ −1 ∂ψ −1 t +cψ −1 Bψ−1 , with [∂ψ −1 t , Bψ−1 ] = 0. Note that ∂ψ −1 t = ψ∂tψ −1 = ψ(ψ−1∂t−ψtψ−2) = ∂t − ψtψ−1 = Mψ, so L = Aψ −1 Mψ + cBψ−1 , and [Mψ, B ψ−1 ] = 0. Now rename Aψ −1 as A, and Bψ−1 as B. � 4 Classif ication of Darboux transformations of first order We know that Type I Darboux transformations are never of Wronskian type, because they have kerL ∩ kerM = {0}. For first-order transformations, we can say more. Theorem 4.1. Suppose there is a first-order Darboux transformation (M,N) : L → L1 with N,M,L1 ∈ K[∂t, ∂x1 , . . . , ∂xn ], and M = ∂t + m. Then this transformation is either of Type I or of Wronskian type. Proof. While M = ∂t + m for some m ∈ K, it is more convenient to write M as ∂v −1 t = v∂tv −1 = ∂t − vtv−1, which is easily done by solving −mu = vt. Then we have NL = L1∂ v−1 t , and conjugate by v to get NvLv = Lv1∂t. By Theorem 3.3, we have Lv = A∂t + cB, where B is t-free. We have three cases. If cB = 0 then L = Av −1 M and we have a Darboux transformation obtained from a factor- ization. Then v ∈ kerL ∩ kerM , and M = ∂t − vtv−1, showing the DT is of Wronskian type. Henceforth, assume cB 6= 0. Suppose first that cB is a non-zero function. Then we may take cB = c, since we are free to pick one coefficient of B to be 1. Then Lv = A∂t+c, and conjugation gives L = vAv−1M+c. We let C be vAv−1 and f be c, giving L = CM+f . Theorem 3.3 also gives Nv = ∂t−ctc−1 = c∂tc −1, so N = cv∂tv −1c−1 = fMf−1, as expected. Similarly, Lv1 = Lv − A∂t + (∂t − ctc −1)A = Lv − A∂t + NvA, yielding L1 = L − vAv−1M + NvAv−1 = CM + f − CM + NC = NC + f . This shows the transformation is of Type I. We now suppose that B is a t-free differential operator that is not a function. Then we have a non-zero t-free function φ in ker(cB). We let ψ be vφ, and claim ψ is in kerL. We have L = vA∂tv −1 + vcBv−1, so L[ψ] is (vA∂tv −1 + vcBv−1)[vφ] = vA∂t[φ] + vcB[φ] = 0, because φ is t-free. The Wronskian operator for ψ is ∂t − ψtψ−1 = ∂t − (vφ)t(vφ)−1 = ∂t − vtφv−1φ−1 = ∂t − vtv−1 = M . This shows the transformation is of Wronskian type. � 12 D. Hobby and E. Shemyakova We can put first-order Darboux transformations of all types on a common footing, and show that they depend on the existence of a representation of L in a special form. Theorem 4.2. 1. A Darboux transformation (M,N) : L→ L1 where M , N are first order exists if and only if L = CM + cB for some operators C and B and some c ∈ K, where [M,B] = 0. 2. If L, M , C, B and c are as above, we have the following: • If cB = 0 then (M,N) is a Darboux transformation obtained from a factorization, and is also of Wronskian type. • If cB ∈ K and cB 6= 0 then (M,N) is a transformation of Type I. • If cB /∈ K (is an operator of order larger than zero) then (M,N) is a transformation of Wronskian type. The corresponding operators N and L1 for the Darboux transformation are N = cMc−1, L1 = NC + cB, if cB 6= 0, (4.1) and in the case cB = 0, N is any operator with σ(N) = σ(M), and L1 = NC. Proof. For (1), we first assume that L = CM + cB, where MB = BM . If cB = 0, then L = CM , and taking any N with σ(N) = σ(M), and L1 = NC, we have NL = NCM = L1M . Now assuming cB 6= 0, we let N = cMc−1, and L1 = NC + cB. This gives us NL = cMc−1(CM + cB) = cMc−1CM + cMB = NCM + cBM = (NC + cB)M = L1M . Since σ(M) = σ(N), we get σ(L) = σ(L1) showing that this is a Darboux transformation. Now let us prove the statement in the opposite direction and assume (M,N) : L→ L1 is a DT with M and N first order. As before we may change variables and apply gauge transformations to make g−1Mg = ∂u. Conjugating NL = L1M , we get NgLg = Lg1∂u. By Theorem 3.3, this gives Lg = C ′∂u + cB′, where B′ is u-free. Thus [∂u, B ′] = 0, which implies [M, gB′g−1] = 0. Then Lg = C ′∂u + cB′ becomes L = gC ′g−1M + cgB′g−1, which we write as L = CM + cB by taking C = gC ′g−1 and B = gB′g−1. For (2), note that when we changed variables and applied a gauge transformation in the proof of (1) to make g−1Mg = ∂u, that the operator ∂u we obtained was unique. Then the C ′ and cB′ we obtained were unique, and thus the C and cB in L = CM + cB are uniquely determined. In our first case, we assume cB /∈ K. As before, we change variables and apply a gauge transformation to get Lg = C ′∂u+cB′, where cB′ /∈ K. As in the proof of Theorem 4.1, we have a non-zero u-free function φ in ker(cB′), since B′ is u-free. We now let ψ be gφ, and have that ψ is in kerL. The Wronskian operator for ψ and ∂u is then ∂u − ψuψ−1 = ∂u − (guφ)(gφ)−1 = ∂u−gug−1 = g∂ug −1 = M , so M is obtained from a Wronskian operator by a change of variables, and the transformation is of Wronskian type by Definition 2.10. Next assume cB = 0, so L = CM . Taking any N with σ(N) = σ(M), we let L1 be NC and have NL = NCM = L1M , making the DT one obtained from a factorization. The same DT is also of Wronskian type. We change variables and transform, getting Lg = C ′∂u, where M = g∂ug −1. Then 1 is in the kernels of Lg and ∂u, and g is in the kernels of L and M . As in the previous paragraph, ∂u−gug−1 = g∂ug −1 = M , showing the transformation is of Wronskian type. Finally, assume cB = f is a non-zero function in K. Then L = CM + f . If N = fMf−1, then NL = NCM +Nf = NCM + fM = (NC + f)M and the DT is of Type I. We must also show that when cB 6= 0, that N must be cMc−1. To see this, we have Lg = C ′∂u + cB′, as before. Considering (∂u, N ′) : Lg → L′1, we get from Theorem 3.3 that N ′ must be c∂uc −1. Transforming back to the DT (M,N) : L→ L1, N = gN ′g−1 = gc∂uc −1g−1 = cg∂ug −1c−1 = cMc−1, as desired. � Classification of Multidimensional Darboux Transformations 13 Remark 4.3. As Type I Darboux transformations do, the transformations of Wronskian type of order one fit the framework developed by E.I. Ganzha in [15]. Given L = CM + cB with [M,B] = 0, we initially assume cB 6= 0, and write L = X1X2 −H by taking X1 = C, X2 = M and H = −cB. We need ω = −[X2, H]H−1 to be a strictly differential operator, and calculate ω = −[M,−cB](−cB)−1 = (cBM −McB)B−1c−1 = (cMB−McB)B−1c−1 = (cM −Mc)c−1 = cMc−1 −M = N −M . This also gives us that N is M + (N −M) = X2 + ω, as required. In case cB = 0, then L = CM = (C − 1)M + M , and we take X1 = C − 1, X2 = H = M , which works. Remark 4.4. Let us call an expression of L as L = CM + cB, with [M,B] = 0 and c ∈ K a quasi-factorization of L. A natural question is which operators L have quasi-factorizations, and if they do, how many do they have? We do not have uniqueness of factorization for partial differential operators, as, e.g., shown by Landau’s example (mentioned in [6] as well as in [37] and our Example 4.8). Thus the general question of how many quasi-factorizations an operator L has may be difficult. We may however note the following for Darboux transformations of Type I, where cB = c. For the quasi-factorization L = CM + c where C and M are not functions, we have that the principal symbols of C and M are factors of the principal symbol of L. If M is first order, it is almost determined by its principal symbol. Given a first-order M , we may change variables if need be, and write M as a∂t + b. Grouping a with C, we may assume that the principal symbol of M is pt = σ(∂t). A quasi-factorization of L is determined by M , so it is natural to ask if there can be more than one first-order quasi-factor M of L with the same principal symbol pt. For Darboux transformations of Type I, the answer is no, unless the principal symbol of L has a repeated factor of pt. For suppose we have two first-order quasi-factors M1 and M2 of L, that both have principal symbol pt. Applying an appropriate gauge transformation, we may assume M1 = ∂t and M2 = ∂t + n, where n 6= 0. Then L = C∂t + f = E(∂t + n) + g, where f and g are in K. Rearranging, En = (C − E)∂t + f − g. Thus pt divides the principal symbol of En, and hence of E. Thus p2t divides the principal symbol of L. This gives us that when the principal symbol of L has no repeated factors, that there is at most one first-order Type I transformation per factor, and thus at most deg(L) many first-order Type I Darboux transformations of L. In the case of Darboux transformations which are not of Type I, the situation is not as simple, since the principal symbol of M does not necessarily divide that of L. See the following example. Example 4.5. We let L be ∂xxy + ∂xyy + (1− x/2)∂xx + (3− x)/2∂xy + ( −1/x+ 1/2x2 ) ∂yy + 1/2∂x + ( −1/x+ 1/x2 ) ∂y. Letting M = x∂x + ∂y we have L = CM + cB, where BM = MB. Here c = x(x− 1)/(8e3y), C = (1/8)((1− x)∂xx + (4 + 4/x)∂xy + (1/x − 1/x2)∂yy + (1 + 3/x)∂x − (2/x + 2/x2)∂y), and B = e3yx−3(x3∂xxx − 3x2∂xxy + 3x∂xyy − ∂yyy − 3x2∂xx + 9x∂xy − 6∂yy + 3x∂x − 8∂y). Another natural question about Darboux transformations is to what extent L and M deter- mine L1 and N . As a partial answer, we have the following. Definition 4.6. We say that operators L and M uniquely determine their Darboux transfor- mation iff there is at most one pair of operators N and L1 so that (M,N) is a DT from L to L1. 14 D. Hobby and E. Shemyakova Theorem 4.7. If M /∈ K is a first-order operator, and L is any operator, then L and M uniquely determine their Darboux transformation iff L can not be written as L = AM for any operator A. Proof. Let M /∈ K be a first-order operator. If L = AM , we may take any N with σ(N) = σ(M), let L1 = NA, and have a Darboux transformation. For the other direction, assume that L 6= AM for every operator A, and that we have N1, N2, L1 and L2 that give two distinct DTs, so we have N1L = L1M , N2L = L2M , and σ(N1) = σ(M) = σ(N2). If N1 = N2, then L1M = N1L = N2L = L2M , and L1 = L2 by cancellation. So we assume that N1 6= N2. We have σ(N1) = σ(N2) where N1 and N2 are first order, so N1−N2 is a nonzero function f . Now we subtract intertwining relations, and get (N1−N2)L = N1L−N2L = L1M−L2M = (L1−L2)M . Then letting A = f−1(L1 − L2), we have AM = f−1(L1 − L2)M = f−1(N1 − N2)L = L, a contradiction. � The following example is derived from Landau’s famous example [6, 37] of non-unique factor- ization of linear partial differential operators. The resulting Darboux transformation of higher order exhibits a different pattern than Darboux transformations of order one. Example 4.8. Landau’s example of non-unique factorization of operators is RQ = QQP, where P = ∂x +x∂y, Q = ∂x + 1 and the operator R = ∂xx +x∂xy + ∂x + (2 +x)∂y is irreducible over any extension of the differential field of rational functions in x and y. This gives us a Darboux transformation, (M,N) : Q→ Q, where M = QP and N = R. Now formula (4.1) in not satisfied for this M and N . Indeed, assuming N = cMc−1 or R = cQPc−1, we obtain R = ∂xx+x∂xy+(1−2cxc −1−xcyc−1)∂x+(x−xcxc−1 +1)∂y, which implies 2cxc −1 +xcyc −1 = 0 and xcxc −1 = −1. The latter gives c = f/x where f does not depend on x, but plugging this into the former gives c = 0. 5 Continued Type I Darboux transformations The following is an example of a second-order Darboux transformation that is not of Type I, nor obtained from a factorization, nor a multiple of a DT of Wronskian type. Example 5.1. Let M = ∂xx + 1, F = ∂x, C = ∂y + x, L = CM + F and L1 = MC + F , and consider the Darboux transformation (M,M) : L→ L1. Indeed, NL = M(CM + F ) = MCM +MF = MCM + FM = (MC + F )M = L1M . Observe that for any operators A, B and C, that kerA∩ ker(B+CA) = kerA∩ kerB. Thus kerL∩kerM = ker(CM +F )∩kerM = kerF ∩kerM = kerF ∩ker(M −∂xF ) = kerF ∩ker 1 = {0}, so this transformation is not obtained from a factorization, nor a multiple of a DT of Wronskian type. F is not in K, so the transformation is also not of Type I. Our long term goal is to classify Darboux transformations for an operator of general form. The example above is of neither type known to us. It can be understood in terms of the following general phenomenon. Suppose that we have operators L and M so that L = CM + F for some operators C,F /∈ K. So the corresponding Darboux transformation cannot be of Type I. We can Classification of Multidimensional Darboux Transformations 15 rewrite this as F = L−CM , where we have written F as a particular kind of linear combination of L and M . If F were a function f , we could get a Type I transformation where NL = L1M . Now suppose F is not a function, but that we have M = AF + f for some operator A and nonzero function f . We can write this as f = M −AF , and view this as reaching f in two linear combination steps. The first gives F , and the second gives f . As in Example 5.1, we have kerL ∩ kerM = ker(CM + F ) ∩ kerM = kerF ∩ kerM = kerF ∩ ker(AF − f) = kerF ∩ ker f = {0}. Thus a Darboux transformation with this L and M cannot be obtained from a factorization, nor is it a multiple of a DT of Wronskian type. These new Darboux transformations constitute a class which is appreciably “larger” than that of Type I transformations. Example 5.2. Consider operators on the two variables x and y, and let L have the form a001∂xxy + a00∂xx + a01∂xy + a0∂x + a1∂y + a. Not every such L has a Type I transformation (M,N) : L → L1, for if we for instance write L = CM + f = (b00∂xx + b0∂x + b)(c1∂y + c) + f we get a001 = b00c1 and five other equations for the five unknown functions b00, b0, b, c1 and c. Such a system usually has no solution, and trying the other possible forms of C and M is little help. On the other hand, almost every such L does have a DT (M,N) : L → L1. We will write L = CM+F , where M = AF+f and A is a first-order operator. Then letting N = fFf−1A+f , we have σ(N) = σ(M) andNL = N(CM+F ) = NCM+(fFf−1A+f)F = NCM+fFf−1AF+ fF = NCM + fFf−1AF + fFf−1f = NCM + (fFf−1)(AF + f) = (NC+ fFf−1)M . Setting L1 = NC + fFf−1, this is the desired intertwining relation. We now let C = g∂y +h, F = p∂x+q, and A = b∂x+ c, where g, h, p, q, b and c are unknown functions. We want L = CM+F = C(AF+f)+F = (g∂y+h)((b∂x+c)(p∂x+q)+f)+p∂x+q = (g∂y + h)((bp∂xx + r∂x + s) + p∂x + q, where we write r = cp+ bq + bpx and s = cq + bqx + f . Multiplying this out and equating coefficients, we obtain a001 = gbp, a00 = hbp + g(bp)y, a01 = rg, a0 = hr+ gry + p, a1 = gs, and a = hs+ gsy + q. Nothing is lost by setting bp = 1, so we do so. This gives g = a001 and h = a00. Then r = a01/g = a01/a001 and s = a1/g = a1/a001. Thus p and q are determined, as p = a0 − hr − gry and q = a − hs − gsy, and bp = 1 gives b = 1/p. Now we get c and f from r and s, giving c = (r− bq− bpx)/p and then f = s− cq− bqx. While the above examples had two linear combination steps, this can be extended to any number of linear combination steps, giving the following theorem. Theorem 5.3. Let L ∈ K[∂t, ∂x1 , . . . , ∂xn ], and suppose there are nonzero operators A1, A2, . . . , Ak, M = M1,M2, . . . ,Mk ∈ K[∂t, ∂x1 , . . . , ∂xn ] for some k ≥ 1 and a nonzero f = Mk+1∈K so that Mi−1 = AiMi +Mi+1, 1 ≤ i ≤ k, where M0 = L, i.e., L = A1M1 +M2, M1 = A2M2 +M3, · · · · · · · · · · · · · · · · · · Mi−1 = AiMi +Mi+1, for 1 ≤ i ≤ k, and finally · · · · · · · · · · · · · · · · · · Mk−1 = AkMk + f. Then there exists a Darboux transformation for operator L (M1, N) : L→ L1 defined as follows. Define Nk+1 = Mk+1 = f andNk = fMkf −1, and define Ni for 0 ≤ i ≤ k−1 by downward recursion using Ni = Ni+1Ai+1 +Ni+2. 16 D. Hobby and E. Shemyakova Finally let N = N1, and let L1 = N0. Then M and N have the same principal symbol, and the intertwining relation NL = L1M holds. The corresponding Darboux transformation is not obtained from a factorization, nor a multiple of a DT of Wronskian type, and if k > 1 it is not of Type I. Proof. Fix k. If k = 1, we have M2 = f , so L = M0 = A1M1 + M2 = A1M1 + f . Then N = N1 = fM1f −1, and L1 = N0 = N1A1 +N2 = fM1f −1A1 + f . We have NL = L1M , since this is a Darboux transformation of Type I. Henceforth assume k > 1. We have that ker(L) ∩ ker(M) = ker(M0) ∩ ker(M1) ⊆ ker(M1) ∩ ker(M2), the last step because M2 is a linear combination of M0 and M1. Now Mi−1 = AiMi + Mi+1 for all i, so each Mi+1 is a linear combination of Mi−1 and Mi, giving us similarly that ker(Mi−1)∩ker(Mi) ⊆ ker(Mi)∩ker(Mi+1) for all 1 ≤ i ≤ k. Thus ker(M0)∩ker(M1) ⊆ ker(M1)∩ ker(M2) ⊆ ker(M2) ∩ ker(M3) ⊆ · · · ⊆ ker(Mk) ∩ ker(Mk+1). But ker(Mk+1) = ker(f) = {0}, so ker(L) ∩ ker(M) = {0}, and our transformation will not be obtained from a factorization, nor a multiple of a DT of Wronskian type. We will now do two inductive proofs of statements Si, starting with i = k as the basis and showing Si ⇒ Si−1 for all i ≥ 1. We first show NiMi−1 = Ni−1Mi for 1 ≤ i ≤ k + 1. When i = 1, we have N = N1, L = M0, L1 = N0 and M = M1, so this will be NL = L1M . Our basis is when i = k + 1. Then we have NiMi−1 = Nk+1Mk = fMk = fMkf −1f = NkMk+1 = Ni−1Mi, as desired. Now let 1 < i ≤ k + 1, and assume NiMi−1 = Ni−1Mi. Adding Ni−1Ai−1Mi−1 to both sides, we get NiMi−1 + Ni−1Ai−1Mi−1 = Ni−1Mi + Ni−1Ai−1Mi−1. But Ni + Ni−1Ai−1 = Ni−2 and Mi + Ai−1Mi−1 = Mi−2, so we have Ni−2Mi−1 = NiMi−1 + Ni−1Ai−1Mi−1 = Ni−1Mi + Ni−1Ai−1Mi−1 = Ni−1Mi−2. Reversing the order of equality, Ni−1Mi−2 = Ni−2Mi−1, which is the required statement with i replaced by i − 1 throughout. Thus NiMi−1 = Ni−1Mi for 1 ≤ i ≤ k + 1 by induction. We can now show that Mi and Ni have the same principal symbols for 0 ≤ i ≤ k + 1, which will give σ(L) = σ(L1) since M0 = L and N0 = L1. To start with, we have Nk+1 = f = Mk+1, so σ(Nk+1) = σ(Mk+1). Now we let 0 ≤ i ≤ k + 1, and assume σ(Ni) = σ(Mi). Since NiMi−1 = Ni−1Mi from the previous paragraph, we get σ(Mi−1) = σ(Ni−1). Thus the desired conclusion follows by induction. � This theorem then gives us a new type of Darboux transformation, which generalize Type I transformations. Definition 5.4. Darboux transformations defined as in Theorem 5.3 we shall call continued Type I Darboux transformations. A natural question is how Darboux transformations of continued Type I relate to the theory developed by E.I. Ganzha in [15]. Let us consider transformations of continued Type I for k = 2, so we have L = A1M1 + M2 and M1 = A2M2 + M3 where M3 = f . Intertwining Laplace transformations depend on writing L in a particular form, where L = X1X2−H. In our notation, the definition from [15] then gives M = X2, ω = −[X2, H]H−1 and N = X2+ω, where we would like ω to be a differential operator and not merely pseudo-differential. Substituting M = M1 for X2, we have L = A1M1 + M2 = X1X2 − H. So it would be natural to try setting X1 = A1, giving H = −M2. In this case, we get ω = −[M1, H]H−1 = −(M1(−M2) − (−M2)M1)(−M2) −1 = (M2M1−M1M2)M −1 2 . Expanding this using M1 = A2M2+f , it becomes (M2(A2M2 + f) − (A2M2 + f)M2)M −1 2 = ((M2A2 − A2M2 − f)M2 + M2f)M−12 , which is not a differential operator unless M2 and f commute, which is seldom true. We have that transformations of continued Type I are not obtained from a factorization, nor a multiple of a DT of Wronskian type. We also note that when M is first order, that the Classification of Multidimensional Darboux Transformations 17 continued Type I Darboux transformation NL = L1M must be of Type I. It turns out that all continued Type I DTs are invertible. Some insight into why this is can be obtained by looking at the construction of the continued Type I DT (M1, N1) : L → L1 by alternately forming shifts and duals of a simple starting DT. With notation as in Theorem 5.3, we have Mk+1 = Nk+1 = f , and Nk = fMkf −1. Then (Mk, Nk) : Mk+1 → Nk+1 is an invertible DT, since it is Type I. Now (Mk, Nk) : Mk+1+AkMk → Nk+1 + NkAk is a shift of (Mk, Nk) : Mk+1 → Nk+1, and also an invertible DT. But Mk+1 + AkMk = Mk−1 and Nk+1 + NkAk = Nk−1, so this DT is actually (Mk, Nk) : Mk−1 → Nk−1. Then its dual (Mk−1, Nk−1) : Mk → Nk is also an invertible DT. Now we shift again, getting that (Mk−1, Nk−1) : Ak−1Mk−1 +Mk → Nk−1Ak−1 +Nk or (Mk−1, Nk−1) : Mk−2 → Nk−2 is an invertible DT. Then we take the dual of this, and continue the process, eventually obtaining that (M1, N1) : M0 → N0 or (M,N) : L→ L1 is an invertible DT. The above process is constructive, in the sense that the inverse of (M,N) : L → L1 can be recovered using the descriptions of the inverses of a shift and a dual in Lemmas 2.7 and 2.9. Doing this gives recursive definitions, as in the inductive proof of the following result. Theorem 5.5. All continued Type I Darboux transformations are invertible. Proof. Let NL = L1M be of continued Type I, and let k, L = M0, A1, A2, . . . , Ak, M = M1,M2, . . . ,Mk and a nonzero function f = Mk+1 be as in Theorem 5.3. As there, we have Mi−1 = AiMi + Mi+1 for 1 ≤ i ≤ k, and also define Ni by Nk+1 = Mk+1 = f , Nk = fMkf −1, and Ni = Ni+1Ai+1 +Ni+2. for 0 ≤ i ≤ k − 1. Using Mi−1 = AiMi + Mi+1 repeatedly, we can write L = M0 = PkMk + QkMk+1. The operator Pk is of interest to us, so we define it inductively as follows. We let P0 = 1, P1 = A1, and Pi+1 = PiAi+1 + Pi−1 for all i ≥ 1. For example, P2 = P1A2 + P0 = A1A2 + 1. We also need related operators P ′i , created by reversing the order of all the products in Pk. For example, P ′2 is A2A1 + 1. We define these inductively by letting P ′0 = 1, P ′1 = A1, and P ′i+1 = Ai+1P ′ i + P ′i−1 for all i ≥ 1. We need the fact that PiP ′ i+1 = Pi+1P ′ i for all i ≥ 0, and prove this by induction. For the basis, we have i = 0 and get P0P ′ 1 = 1A1 = A11 = P1P ′ 0. Now assume that i ≥ 1 and that Pi−1P ′ i = PiP ′ i−1. We get PiP ′ i+1 = Pi(Ai+1P ′ i + P ′i−1) = PiAi+1P ′ i + PiP ′ i−1 = PiAi+1P ′ i + Pi−1P ′ i = (PiAi+1 + Pi−1)P ′ i = Pi+1P ′ i , which proves the claim. We have L = M0 = PkMk+Pk−1Mk+1, and show this by proving that M0 = PiMi+Pi−1Mi+1 for 1 ≤ i ≤ k by induction on i. Our basis is where i = 1. Here we have M0 = A1M1 + 1M2 = P1M1+P0M2. Now assume that M0 = PiMi+Pi−1Mi+1, we will show M0 = Pi+1Mi+1+PiMi+2. We have M0 = PiMi+Pi−1Mi+1 = Pi(Ai+1Mi+1 +Mi+2) +Pi−1Mi+1 = (PiAi+1 +Pi−1)Mi+1 + PiMi+2 = Pi+1Mi+1 + PiMi+2, as required. A similar inductive proof shows that L1 = N0 = NkP ′ k+Nk+1P ′ k−1. We leave it to the reader. To construct the inverse transformation N ′L1 = LM ′, we let N ′ = (−1)kPkf −1 and M ′ = (−1)kf−1P ′k. Then N ′L1 = (−1)kPkf −1(NkP ′ k + Nk+1P ′ k−1) = (−1)kPkf −1(fMkf −1P ′k + fP ′k−1) = (−1)kPk(Mkf −1P ′k+P ′k−1) = (−1)k(PkMkf −1P ′k+PkP ′ k−1). Similarly, we have LM ′ = (PkMk+Pk−1Mk+1)((−1)kf−1P ′k) = (−1)k(PkMkf −1P ′k+Pk−1ff −1P ′k) = (−1)k(PkMkf −1P ′k+ Pk−1P ′ k) = (−1)k(PkMkf −1P ′k + PkP ′ k−1), where the last step follows from Pk−1P ′ k = PkP ′ k−1. Thus N ′L1 = LM ′ as required. To establish invertibility, we also need to show that there are operators A and B with M ′M = 1 +AL and MM ′ = 1 +BL1 as in equations (2.2) and (2.4). Similarly to the definitions of Pi and P ′i , we define the sequences Ri and R′i recursively. We take R1 = R′1 = 1, R2 = R′2 = A2 and set Ri+1 = RiAi+1 +Ri−1 and R′i+1 = Ai+1R ′ i +R′i−1 for i ≥ 2. We can now prove M1 = RiMi + Ri−1Mi+1 for 2 ≤ i ≤ k, by induction. We have M1 = A2M2 + M3 = R2M2 + R1M3 for our basis. And assuming M1 = RiMi + Ri−1Mi+1, we get 18 D. Hobby and E. Shemyakova M1 = RiMi+Ri−1Mi+1 = Ri(Ai+1Mi+1+Mi+2)+Ri−1Mi+1 = (RiAi+1+Ri−1)Mi+1+RiMi+2 = Ri+1Mi+1 +RiMi+2. This gives us M = M1 = RkMk +Rk−1Mk+1. Next we need several identities, and will prove by induction that P ′iRi = R′iPi, P ′ iRi−1 = R′iPi−1 + (−1)i and R′i−1Pi = P ′i−1Ri + (−1)i for 2 ≤ i ≤ k. When i = 2, these become (A2A1 + 1)A2 = A2(A1A2+1), (A2A1+1)1 = A2A1+(−1)2 and 1(A1A2+1) = A1A2+(−1)2, all of which are true. Now we assume P ′iRi = R′iPi, P ′ iRi−1 = R′iPi−1 + (−1)i and R′i−1Pi = P ′i−1Ri + (−1)i, and calculate as follows. First, P ′i+1Ri+1 = (Ai+1P ′ i +P ′i−1)(RiAi+1 +Ri−1) = Ai+1P ′ iRiAi+1 + Ai+1P ′ iRi−1 + P ′i−1RiAi+1 + P ′i−1Ri−1) = Ai+1R ′ iPiAi+1 + Ai+1(R ′ iPi−1 + (−1)i) + (R′i−1Pi − (−1)i)Ai+1 +R′i−1Pi−1) = Ai+1R ′ iPiAi+1 +Ai+1R ′ iPi−1 +R′i−1PiAi+1 +R′i−1Pi−1) = (Ai+1R ′ i + R′i−1)(PiAi+1 +Pi−1) = R′i+1Pi+1, as desired. Next, P ′i+1Ri = (Ai+1P ′ i +P ′i−1)Ri = Ai+1P ′ iRi + P ′i−1Ri = Ai+1R ′ iPi + (R′i−1Pi − (−1)i) = (Ai+1R ′ i + R′i−1)Pi + (−1)i+1 = R′i+1Pi + (−1)i+1, showing the second equality. Finally, R′iPi+1 = R′i(PiAi+1 + Pi−1) = R′iPiAi+1 + R′iPi−1 = P ′iRiAi+1+(P ′iRi−1−(−i)i) = P ′i (RiAi+1+Ri−1)+(−i)i+1) = P ′iRi+1+(−i)i+1, giving the third equality. Thus we have P ′kRk = R′kPk, P ′ kRk−1 = R′kPk−1+(−1)k andR′k−1Pk = P ′k−1Rk+(−1)k. We also need that Rk−1P ′ k = RkP ′ k−1 + (−1)k. We leave the easier inductive proof of this to the reader. Now we have M ′M = (−1)kf−1P ′k(RkMk + Rk−1Mk+1) = (−1)kf−1P ′kRkMk + (−1)kf−1P ′k ×Rk−1Mk+1 = (−1)kf−1R′kPkMk + (−1)kf−1(R′kPk−1 + (−1)k)Mk+1 = (−1)kf−1R′k(PkMk + Pk−1) + (−1)kf−1(−1)kMk+1 = f−1Mk+1 + (−1)kf−1R′k(PkMk + Pk−1) = 1 + ((−1)kf−1R′k)L. This is 1 +AL, where A is (−1)kf−1R′k. Similarly, we get MM ′ = (RkMk +Rk−1Mk+1)(−1)kf−1P ′k = (−1)k(Rkf −1fMk +Rk−1f −1f ×Mk+1)f −1P ′k = (−1)k(Rkf −1Nk + Rk−1f −1Nk+1)P ′ k = (−1)k(Rkf −1Nk + Rk−1)P ′ k, since Nk = fMkf −1 and Nk+1 = f . Continuing, (−1)k(Rkf −1Nk + Rk−1)P ′ k = (−1)k(Rkf −1NkP ′ k + Rk−1P ′ k) = (−1)k(Rkf −1NkP ′ k + RkP ′ k−1 + (−1)k) = (−1)k(Rkf −1NkP ′ k + Rkf −1Nk+1P ′ k−1) + (−1)2k = (−1)kRkf −1(NkP ′ k + Nk+1P ′ k−1) + 1 = (−1)kRkf −1(NkP ′ k + Nk+1P ′ k−1) + 1 = 1 + (−1)kRkf −1L1. This is 1 +BL1, where B = (−1)kRkf −1. � 6 Continued Wronskian type Darboux transformations We may generalize the previous construction slightly, by removing the requirement that Mk+1 be a function. Much of the theory still works, provided Mk+1 = cB where B and Mk commute. We say that the resulting transformations are of continued Wronskian type. They generalize first-order Wronskian type transformations. Their relation with high-order Wronskians remains to be clarified. Here is the analogue of Theorem 5.3. Theorem 6.1. Let k ≥ 1 be given, and suppose there are nonzero operators L = M0, A1, A2, . . . , Ak, M = M1,M2, . . . ,Mk+1 so that Mi−1 = AiMi +Mi+1 for 1 ≤ i ≤ k, and there is a function c with Mk+1 = cB /∈ K and MkB = BMk. Then we can define a Darboux transformation NL = L1M as follows. Define Nk+1 = Mk+1 and Nk = cMkc −1, and define Ni for 0 ≤ i ≤ k − 1 by downward recursion using Ni = Ni+1Ai+1 + Ni+2. Finally let N = N1, and let L1 = N0. Then M and N have the same principal symbol, and NL = L1M is a Darboux transformation. Classification of Multidimensional Darboux Transformations 19 Proof. The proof is essentially the same as that of Theorem 5.3, so we will only indicate the parts that change. When k = 1 and Mk+1 /∈ K, we will not have a Type I Darboux transformation. The proof of NL = L1M is by induction. We need the hypothesis that Mk and B commute for the basis, and have Nk+1Mk = Mk+1Mk = cBMk = cMkB = cMkc −1cB = NkMk+1. 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Classification of Multidimensional Darboux Transformations: First Order and Continued Type

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