RECURSION OPERATORS AND CLASSIFICATION OF INTEGRABLE NONLINEAR EQUATIONS

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1 RECURSION OPERATORS AND CLASSIFICATION OF INTEGRABLE NONLINEAR EQUATIONS a thesis submitted to the department of mathematics and the graduate school of engineering and science of bilkent university in partial fulfillment of the requirements for the degree of master of science By Ergün Bilen September, 2012

2 I certify that I have read this thesis and that in my opinion it is fully adequate, in scope and in quality, as a thesis for the degree of Master of Science. Prof. Dr. Metin Gürses(Advisor) I certify that I have read this thesis and that in my opinion it is fully adequate, in scope and in quality, as a thesis for the degree of Master of Science. Asst. Prof. Bülent Ünal I certify that I have read this thesis and that in my opinion it is fully adequate, in scope and in quality, as a thesis for the degree of Master of Science. Asst. Prof. Kostyantyn Zheltukhin Approved for the Graduate School of Engineering and Science: Prof. Dr. Levent Onural Director of the Graduate School ii

3 ABSTRACT RECURSION OPERATORS AND CLASSIFICATION OF INTEGRABLE NONLINEAR EQUATIONS Ergün Bilen M.S. in Mathematics Supervisor: Prof. Dr. Metin Gürses September, 2012 Recursion operators, if they exist, of nonlinear partial differential equations map symmetries to symmetries of these equations. It is this property that the integrable nonlinear partial differential equations possess infinitely many symmetries. In this work we studied two other properties of recursion operators. We shall use the recursion operators as Lax operators in the Gelfand-Dikii formalism and to classify certain integrable equations. Keywords: symmetries, Gelfand-Dikii formalism, recursion operator, classification. iii

4 ÖZET ADIM SİMETRİ OPERATÖRLERİ VE İNTEGRE EDİLEBİLİR LİNEER OLMAYAN DENKLEMLERİN SINIFLANDIRILMASI Ergün Bilen Matematik, Yüksek Lisans Tez Yöneticisi: Prof. Dr. Metin Gürses Eylül, 2012 Lineer olmayan kısmi diferansiyel denklemlerin, eğer varsa, simetri adım operatörlerı denklemin simetrilerini simetrilerine gönderirler. Bu yüzden integre edilebilir lineer olmayan kısmi diferansiyel denklemler sonsuz simetriye sahiptirler. Bu tezde simetri adım operatörlerinin diğer iki özelliğini çalıştık. Simetri adım operatörlerini Gelfand-Dikii formülasyonunda Lax operatörü olarak ve integre edilebilir denklemleri sınıflandırmak için kullandık. Anahtar sözcükler: simetriler, Gelfand-Dikii formülasyonları, simetri adım operatörü, sınıflandırma. iv

5 Acknowledgement My supervisor Metin Gürses helped me to a great extent, encouraged me to the very end and made this thesis possible. I take great pleasure in expressing my sincere gratitude to Prof. M. Gürses. I would like to thank Asst. Prof. Burcu Silindir and Asst. Prof. Aslı Pekcan for their helps about Latex and corrections in this thesis. My deepest gratitude further goes to my family for being with me in any situation, their encouragement, endless love and trust. I have been receiving a scholarship about for two years from Yurt İçi Yüksek Lisans Burs Programı of TÜBİTAK. I would like to thank to TÜBİTAK. Finally, with my best feelings I would like to thank all my close friends whose names only a special chapter could encompass. v

6 Contents 1 Introduction 1 2 Symmetries and Recursion Operators Symmetry Groups and Infinitesimal Generators Generalized Symmetries Pseudo-Differential Algebra Recursion Operators Gel fand-dikii Formalism Lax Representations Consistent Lax Hierarchies A Method to Find Recursion Operator Symmetric and Skew-Symmetric Reduction of a Differential Operator 31 5 Recursion Operators in Gelfand-Dikii Formalism Recursion Operators as Lax Operators in Gelfand-Dikii Formalism 39 vi

7 CONTENTS vii 5.2 Recursion Operators Produce Themselves A General Type Lax Operator of Order Two Classification Problem Classification of Nonlinear PDEs by the Use of Recursion Operators Classification for Burgers Type Equations Case 1: β uxxuxx Case 2: β uxxuxx = 0, β uxx Case i. c u = Case ii. c ux = Case 3: β uxx = Conclusion 69 A Well Known PDEs and Recursion Operators 70

8 Chapter 1 Introduction A symmetry group of differential equation produces solutions from a known solution. That is, if G is a symmetry group of some system of differential equations and if f is a solution to the system then for every g G, g.f is also a solution to this system. To deduce a given group of transformations is a symmetry group of the system we use the infinitesimal generators of the group actions and they are vector fields over the space X U, where X is the space of independent variables and U is the space of dependent variables. Using symmetry conditions for group of transformations we define recursion operator which is first introduced by Olver in 1977 [10]. In chapter 2, we study the symmetries of differential equations and recursion operators. We consider the evolutionary type of equations u t = F (x, u, u x, u xx,..., u n ). (1.1) An operator R is the recursion operator of (1.1) if the following is satisfied R t = [F, R], (1.2) where F is the Frechet derivative of (1.1). For example, the operator R = D 2 x + 4u + 2u x D 1 x (1.3) 1

9 CHAPTER 1. INTRODUCTION 2 is the recursion operator of the KdV equation u t = u xxx + 6uu x. where F = D 3 x + 6uD x + 6u x. Here in this chapter, we introduce the concept of pseudo-differential operator. In chapter 3, we study Gel fand-dikii Formalisms which make use of pseudo differential algebra. Operators of pseudo-differential algebra are defined as follows. [1] Definition A pseudo-differential operator of order n is a infinite series n L = P i [u]dx, i (1.4) i= where P i [u] is a differentiable function and the operator D 1 x of D x (i.e, D x.d 1 x = D 1 x.d x = 1). is the formal inverse We will consider equations of the form which were introduced by Peter D. Lax in [11] L t = [A, L], (1.5) where commutator [A, L] is defined as [A, L] = A.L L.A. Equation (1.5) is called Lax equation and the operators L and A are called Lax pair. From Lax equations we can get evolution equations for suitable A and L. Pseudo-differential algebra uses the Lax operators of type L = D m x + u m 2 (x, t)d m 2 x u 1 (x, t)d x + u 0, where u m 2, u m 3,..., u 1, u 0 are functions of x and t. Next we consider the

10 CHAPTER 1. INTRODUCTION 3 fractional powers of L and the hierarchy L tn = [A n, L] where A n = (L n N ) k, (1.6) where n is a positive integer n = 0, 1, 2... and N is the order of L, (L n N ) k is the differential operator with powers k. The case k = 0 was first introduced by Gelfand in 1976 [7] and the cases k = 1, 2 were introduced by Kuperschmidt in [8] The operator L = Dx 2 + u with A = (L 3 2 ) 0 gives KdV equation in (1.6) and the Lax pair L = Dx 2 + 2uD x, A = (L 3 2 ) 1 gives the MKdV equation [4] u t = 1 4 u xxx 3 2 u2 u x. Moreover, in section 3.2, we analyze the form of L in Lax hierarchy so that (1.6) is consistent for all n 1. Blaszak [4] found that (1.6) is consistent if and only if k is either 0, 1 or 2 and he provided the forms of L for each case L = c m D m x L = c m D m x L = u m D m x + c m 1 D m 1 x + u m 1 D m 1 x + u m 1 D m 1 x u 0 + u 1 D 1 x +..., k = u 0 + u 1 D 1 x +..., k = u 0 + u 1 D 1 x +..., k = 2 where c m and c m 1 are functions which do not depend on variable t. In chapter 4, we present a method to find Recursion operator of (1.1) if its Lax representation (1.6) is given. This method was introduced by Gürses, Karasu and Sokolov in 1999 citeon construction.we use the equation L tn+m = L.L tn + [R n, L], (1.7) which is called the recursion relation and the term R n is called the remainder and it is defined as R n = (L.(L n m ) ) + and ord(r n ) m 1. The cases where L is a differential operator and L is a pseudo-differential operator are considered and

11 CHAPTER 1. INTRODUCTION 4 in each case the formula to find recursion operator is given in [2]. For example, using (1.7) and the Lax pair L = D 2 x + u with A = (L 3 2 ) 0 we find that KdV equation admits the recursion operator (1.3). Recursion operators are also Lax operators because they satisfy the equation R t = [F, R]. (1.8) In chapter 5, we discuss the case when the recursion operator is a Lax operator in Gelfand-Dikii formalism. We can use it in Gelfand-Dikii formalism if F = (R n m ) k, (1.9) is satisfied for some k = 0, 1, 2. Furthermore, the adjoint operators should satisfy the condition (F ) + = ((R + ) n m ) k, (1.10) for some k = 0, 1, 2. We check (1.9) and (1.10) for the recursion operators of the KdV, MKdV, Harry Dym and Burgers equations. We show that recursion operators of KdV and MKdV satisfy these conditions but recursion operators of Harry Dym and Burgers fail to satisfy conditions (1.9) or (1.10). Recursion operators should also produce themselves as recursion operators since they satisfy the Lax equation (1.8). This property will be used for the purpose of classification of certain PDEs. In section 5.3, we consider the 2nd order recursion operators of the type L = D 2 x + α + βd 1 x (1.11) where α and β are some functions of (x, t), using the Lax equation the Lax equation L tn = [L, A] where A = (L 3 2 )+,

12 CHAPTER 1. INTRODUCTION 5 we find α and β must satisfy the following conditions β = 1 2 α x and α t = α xxx αα x. Hence we deduce that α satisfies the KdV eqation. Secondly, since β = 1 2 α x the pseudo differential operator (1.11) is itself a recursion operator and also the Lax operator. There are certain ways of classifying the hierarchies of evolution equations u tn = R n (0), u tn = R n (u x ) or u tn = R n (c) where c is a nonzero constant and R is the recursion operator of the evolution equation, u t1 = R(0), u t1 = R(u x ) or u t1 = R(c) respectively. In chapter 6, we classifying 2nd order partial differential equations of the type u t = R(0) with recursion operator of the first degree with coefficients depend at most 2nd order derivatives. Similar classification was done before.[5] We find all equations of the type u t = β(u, u x, u xx ) with the recursion operators of the form R = γd x + α + βdx 1 (ρ), where γ, α and ρ are all functions of u, u x and u xx. We use (1.2) to find the formulas for γ, α, β and ρ and we assume ρ 0 to get evolution equation directly from recursion operator. By solving necessary equations we obtain 3 conditions, i. β uxxu xx 0, ii. β uxx 0 but β uxxu xx = 0, iii. β uxx = 0. Neither case i nor case iii produces any class of equations because in each case we find ρ = 0. However, case ii produces evolution equations of the type u t = c(u)u xx 1 2 c(u) uu 2 x + c(u)c(u) uu c(u) u u 2 x + c 4 c 2 c(u)u x, (1.12)

13 CHAPTER 1. INTRODUCTION 6 which admits recursion operators of the type R = 2c 2 c(u) 1 2 Dx + 2c 4 c(u) (2c2 c(u) 1 2 c(u) uu c(u) u c 2 c(u) u c(u) 1 2 )u x + u t D 1 x ( c 2c(u) u ). c(u) 3 2 Diffusion equation and nonlinear diffusion equations u t = u 2 u xx with R = ud x + u 2 u xx D 1 x ( 1 u 2 ), u t = D x ( u x u ) with R = 1 2 u D x 2 u x u + 2 (2u2 x u u xx 3 u 2 )D 1 x are in this class with c(u) = u 2, c 2 = 1 2, c 4 = 0 and c(u) = 1 u 2, c 2 = 1 2, c 4 = 0, respectively. In the Appendix, a list of integrable nonlinear partial differential equations with their recursion operators are given.

14 Chapter 2 Symmetries and Recursion Operators In this chapter we review the symmetries of partial differential equations and their recursion operators. The main reference on this subject is the book of P. Olver [1]. 2.1 Symmetry Groups and Infinitesimal Generators Consider a system of differential equations, k (x, u (n) ) = 0, k = 1,..., l, (2.1) which depend on the indenpendent variables x = (x 1,..., x p ) X and the derivatives of dependent variables u = (u 1,..., u q ) U with respect to x up to order n. The space X U n, which has independent variables and the derivatives of the dependent variables up to order n as coordinates is called the n-th order jet space. The functions (x, u (n) ) = ( 1 (x, u (n) ),..., l (x, u (n) )) are smooth maps 7

15 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 8 from X U n to R l. Hence we have, : X U n R l. (2.2) Definition A symmetry group of the system (2.1) is defined to be a local group of transformations G acting on X U which transforms solutions of the system to other solutions of the same system. That is, if f is a solution to (2.1) then for every g G g.f is also a solution to (2.1). Rather than studying group actions, it is more easier to study with infinitesimal generators of group actions. Infinitesimal generators are vector fields that act on the space X U. By using them we can easily check whether a group of transformation is a symmetry group of a system of differential equations or not. Example Consider the rotation group SO(2) acting on X U θ(x, u) = (x cos θ u sin θ, x sin θ + u cos θ), (2.3) infinitesimal generator of this group is v = u x + x u. (2.4) Definition Let v be infinitesimal generator of a one-parameter group G. Then n-th prolongation of v is defined as the infinitesimal generator of the corresponding n-th prolongation of the one parameter group G and denoted by pr (n) v. Theorem [1] Let v = p i=1 ξ i (x, u) x i + q α=1 φ α (x, u) u α (2.5) be a vector field. Then the n-th prolongation of v is pr (n) v = v + q φ J α(x, u n ), (2.6) u α J α=1 that acts on the jet space X U n and the second sum is over all indices J = J

16 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 9 (j 1,..., j k ) with 1 j k p, 1 k n and we have the formula for each φ J α, φ J α(x, u (n) ) = D j (φ α p ξ i u α i ) + i=1 p ξ i u α J,i, (2.7) i=1 where u α i = uα x i and u α J,i = uα J x i. Example If we compute the 2nd prolongation of the infinitesimal generator of the SO(2) in Example (2.1.2) we get pr (2) v = u x + x u + (1 + u2 x) + (3u x u xx ). u x u xx Now we state a theorem that gives a condition for a group to be a symmetry group of a system. Theorem [1] Suppose k (x, u (n) ) = 0, k = 1,..., l (2.8) is a system of differential equations. Then G is a symmetry group of the system if G is a local group of transformations acting over X U and pr (n) v[ k (x, u (n) )] = 0, k = 1,..., l whenever (x, u (n) ) = 0 (2.9) for every infinitesimal generator v of G. We can use theorem (2.1.5) to find symmetry groups of a given system. Firstly, using (2.9), we can find infinitesimal generators of group actions and then we find the corresponding symmetry groups of the system. The set of all infinitesimal symmetries of the system also forms a Lie algebra of vector fields over X U. Now we give examples of symmetry groups of KdV and heat equations, see [1]. Here we use the notation x instead of x. Example KdV equation u t + u xxx + uu x = 0 has 4 symmetry groups with

17 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 10 infinitesimal generators, v 1 = x v 2 = t v 3 = t x + u v 4 = x x + 3t t 2u u space translation, time translation, Galilean boost, scaling, and the corresponding group actions are G 1 : (x, t, u) (x + ɛ, t, u), G 2 : (x, t, u) (x, t + ɛ, u), G 3 : (x, t, u) (x + ɛt, t, u + ɛ), G 4 : (x, t, u) (e ɛ x, e 3ɛ t, e 2ɛ u). Example Heat equation u t = u xx has 7 symmetry groups with infinitesimal generators, v 1 = x, v 2 = t, v 3 = u u, v 4 = x x + 2t t, v 5 = 2t x xu u, v 6 = 4tx x + 4t 2 t (x 2 + 2t)u u, v 7 = α(x, t) u, where α is any solution to the heat equation and the corresponding group actions

18 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 11 are G 1 : (x, t, u) (x + ɛ, t, u), G 2 : (x, t, u) (x, t + ɛ, u), G 3 : (x, t, u) (x, t, e ɛ u), G 4 : (x, t, u) (e ɛ x, e 2ɛ t, u), ( G 5 : (x, t, u) x + 2ɛt, t, u exp { ɛx ɛ 2 t }), ( { } ) x G 6 : (x, t, u) 1 4ɛt, t ɛx 2 1 4ɛt, u 1 4ɛt exp, 1 4ɛt G 7 : (x, t, u) (x, t, u + ɛα(x, t)). 2.2 Generalized Symmetries In the previous section, symmetries(infinitesimal generators of group actions) is defined in the space X U, that is v = p i=1 ξ i (x, u) x i + q α=1 φ α (x, u) u α, (2.10) and ξ and φ are functions of x, u. In this section, we will consider more general case of this symmetries. ξ and φ will also depend on derivatives of u. We denote the space of smooth functions P (x, u n ) by A and the functions in A are called differential functions. We will use the notation P [u] = P (x, u n ). Also, the space of l-tuples of differential functions P [u] = (P 1 [u],..., P l [u]) where each P j A will be denoted by A l. Now we define what a generalized vector field is. [1] Definition An expression of the form v = p i=1 ξ i [u] x i + q φ α [u] (2.11) u α α=1 is called a generalized vector field.

19 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 12 Definition A generalized vector field v is a generalized infinitesimal symmetry of a system of differential equations, k [u] = k (x, u (n) ) = 0, k = 1,..., l, (2.12) if and only if prv[ k ] = 0, k = 1,..., l (2.13) for every smooth solution u of (2.12). Definition Let Q[u] = (Q 1 [u],..., Q q [u]) A q be a q-tuple of differential functions. An evolutionary vector field is a generalized vector field of the form v = and Q is called its characteristic. q Q α [u] u, (2.14) α α=1 For every generalized vector field of the form (2.11) we have an evolutionary representative v Q, where its characteristic Q is defined by (2.14) and it is given by the formula where u α i = uα x i. Q α = φ α p ξ i u α i, α = 1,..., q, (2.15) i=1 Corollary A generalized vector field v is a symmetry of a system of differential equations if and only if its evolutionary representative v Q is a symmetry. Computation of evolutionary symmetries of a given system of differential equations is more than the method in section (2.1). However, it is almost impossible to find all evolutionary symmetries because it is not easy to find all higher order evolutionary symmetries. For example, for the heat equation the symmetry u x u is evolutionary representative of the space translational symmetry generator x. Similarly, Galilean generator 2t x + xu u has evolutionary representative (2tu x + xu) u.

20 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 13 Example Burgers equation u t = u xx + u 2 x has 6 second order generalized symmetries and characteristic of their evolutionary representatives are given by Q 0 = 1, Q 1 = u x, Q 2 = tu x x, Q 3 = u xx + u 2 x, Q 4 = t(u xx + u 2 x) xu x, Q 5 = t 2 (u xx + u 2 x) + xtu x + ( 1 2 t x2 ) From now on we will consider differential equations of type u t = P [u], (2.16) and they are called evolution equations. The differential functions P [u] in (2.16) are only functions of x and derivatives of u with respect to x. Definition Let v Q and v R be evolutionary vector fields. Then their Lie bracket is defined as [v Q, v R ] = v S, (2.17) where v S is also an evolutionary vector field with characteristic S = prv Q (R) prv R (Q). (2.18) Proposition An evolutionary vector field v Q is a symmetry of the system of evolution equations u t = P [u] if and only if v Q t = [v P, v Q ], (2.19) where v Q t is the evolutionary vector field with characteristic Q t.

21 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS Pseudo-Differential Algebra Pseudo differential algebra is very useful tool to study the symmetries of partial differential equations, to use the Gelfand-Dikii formalism and to construct the recursion operators. The main reference we used is [3] to review this subject. Definition A differential operator of order n is a finite sum D = n P i [u]dx, i (2.20) i=0 where the coefficients P i [u] are differentiable functions. To multiply two differential operators we have the following equation, where i, j 0 D i x.d j x = D i+j x, (2.21) For the operator D x, derivational property is given by the Leibniz rule D x.q = Q x + QD x, (2.22) where Q is a differentiable function. Definition A pseudo-differential operator of order n is a infinite series n D = P i [u]dx, i (2.23) i= where P i [u] is a differentiable function and the operator D 1 x of D x (D x.d 1 x = D 1 x.d x = 1). is the formal inverse Corollary We have the following formula for the operator D 1 x, D 1 x.q = k=0 ( 1) k Q k D k 1 x. (2.24)

22 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 15 Proof. If we multiply (2.22) on both left and the right by Dx 1, it reduces to where Q of (2.25), we get Dx 1.Q = QDx 1 Dx 1.Q Dx 1, (2.25) = D x Q. If we apply (2.25) to the second term on the right hand side Dx 1.Q = QDx 1 Q Dx 2 + Dx 1.Q Dx 2. Applying (2.25) to the third term on the right hand side of (2.3) and continuing this process we find Dx 1.Q = QDx 1 Q Dx 2 + Q Dx 3... = k=0 ( 1) k Q k D k 1 x. (2.26) The reason why we define pseudo-differential operator is that we want to take roots of the differential operators. It is not possible to take roots of differential operators in the set of differential operators but as we will prove we can take roots of any pseudo-differential operator. Lemma Every nonzero pseudo-differential operator of order n > 0 has an n-th root. Proof. Let D be a pseudo differential operator such that, n D = P i [u]dx i, P n 0. i= Then the n-th root of D will be a first order pseudo-differential operator. Let D = D 1 n. Then D has the form, D = (P n ) 1 n Dx + Q 0 + Q 1 D 1 x +... if we take the n-th power of D, we can compare the coefficients of (D ) n and D and we can find Q k for each k =..., 2, 1, 0 in terms of P i s, i =..., 1, 0,..., n.

23 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 16 Example Consider the operator D = D 2 x + u which corresponds to the KdV equation. Then the square root of D is D 1 2 = Dx ud 1 x 1 4 u xd 2 x (u xx u 2 )D 3 x Recursion Operators It is impossible to find all generalized symmetries of system of evolution equations by using the method in Section (2.2). Hence, in this section we present a new method to find new symmetries of evolution equations using recursion operators. Recursion operators provide a method to find infinite hierarchies of generalized symmetries. On the other hand, it may not be possible to find all symmetries using recursion operator. Definition Let be a system of differential equations. A recursion operator for is a linear operator R : A q A q in the space of q-tuples of differential functions that satisfies the following condition, if v Q is an evolutionary symmetry of then v Q is also a symmetry where Q = RQ. Hence if we know the recursion operator of a system then we can generate infinitely many symmetries by applying R to a known symmetry. In other words, if Q 0 is a symmetry of the system then Q n = R n Q 0 for n = 1, 2,.., are also a symmetries of this system and they constitute an infinite family of symmetries of the system. Now the question arises : how can we check whether a given differential operator is a recursion operator of the system? To solve this problem, we define Frechet derivatives. Definition The Frechet derivative of P [u] A r is the differential operator D P : A q A r which is defined as D P (Q) = d P [u + ɛq[u]] (2.27) dɛ ɛ=0

24 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 17 for any Q A q. Also we can say that D P is a q r matrix differential operator with entries (D P ) µν = J where the sum is taken over all multi-indices J. ( ) P µ D u ν J, µ = 1,..., r, ν = 1,..., q, (2.28) J Proposition If P A r and Q A q, then we have D P (Q) = prv Q (P ). (2.29) One can obtain from Proposition (2.4.3) that if D (P ) = 0 then P is a symmetry of the system v [u] = v (x, u (n) ) = 0, v = 1,..., l. Theorem Suppose [u] = 0 is a system of differential equations. If for all solutions u of, R : A q A q is a linear operator such that D.R = R.D, (2.30) where R : A q A q is a linear differential operator, then R is a recursion operator for the system. As a special case of Theorem 2.4.4, we consider evolution equations [u] = u t K[u] = 0. Then we have D = D t D K. Also, in this case we have R = R and the condition (2.30) is equivalent to R t = [D K, R]. (2.31) The equality (2.31) has the form of a Lax equation but in most cases recursion operators are not usual operators appearing in Lax pairs. In Chapter 5, we will concentrate on the usage of recursion operators as Lax operators.

25 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 18 Example The operator is a recursion operator for the KdV equation R = D 2 x + 4u + 2u x D 1 x, (2.32) u t = u xxx + 6uu x. (2.33) Proof. We will check the condition (2.31). First we find the Frechet derivative of K[u] = u xxx + uu x D K = D 3 x + ud x + u x, (2.34) then we find the required commutator as [D K, R] = 2 3 (u xxx + uu x ) (u xxxx + uu xx + u 2 x)d 1 x, (2.35) which is consistent with the condition (2.31). Hence we can use this recursion operator to find more symmetries of KdV equation. We begin with the space translational symmetry which corresponds to characteristic Q 0 = u x, we find Q 1 = RQ 0 = u xxx + uu x, (2.36) which corresponds to the characteristic of time translational symmetry. Next we obtain Q 2 = RQ 1 = u xxxxx uu xxx u xu xx u2 u x, (2.37) by continuing this process one can find infinite number of symmetries of the KdV equation. In the following chapters, instead of D F we will use the notation F for Frechet derivative of differential function F. That is, R t = [F, R] (2.38)

26 CHAPTER 2. SYMMETRIES AND RECURSION OPERATORS 19 will be the condition for R to be a recursion operator of evolution equation u t = F [u].

27 Chapter 3 Gel fand-dikii Formalism Using the pseudo-differential algebra in Gel fand-dikii formalism, given the Lax representation, we are able to obtain classes of nonlinear partial differential equations. It was recently shown that recursion operators can also be constructed by the use of this formalism. 3.1 Lax Representations Let L be a differential operator of order m and A be a differential operator having coefficients, which are functions of x and t. Consider the Lax representation of the form. L t = [A, L], (3.1) where commutator [A, L] is defined as [A, L] = A.L L.A. Let L = D m x + u m 2 (x, t)d m 2 x u 1 (x, t)d x + u 0, (3.2) 20

28 CHAPTER 3. GEL FAND-DIKII FORMALISM 21 consider L 1 m and its fraction L n m, where n Z and n am. Let L n m = n i= v i D i x = (L n m )+ + (L n m ), where (L n m )+ = n v i Dx i and (L n m ) = i=0 1 i= v i D i x. Since [L, L n m ] = 0, we have [L, (L n m )+ ] = [L, (L n m ) ]. (3.3) The left hand side of (3.3) is a differential operator of order n + m 1, but the right hand side is series of order m 1. Therefore, there are n number of terms canceling each other and that gives us system of evolution equations for u i (x, t) s for i = 0, 1,..., m 2. To have a hierarchy of nonlinear system of differential equations at (3.3), define [2] A n := (L n m )+, (3.4) and consider L tn = [A n, L]. (3.5) As an example for Lax pairs of form, we will give the following (3.5) Example KdV equation has the following two Lax representations. [2] i. L = D 2 + u, A = (L 3 2 ) +, ii. L = (D 2 + u)d 1 x, A = (L 3 ) +. Firstly, we will show each of these lax pairs gives the KdV equation i) L = D 2 + u, A = (L 3 2 ) + We have L t = [A, L], (3.6) and L 1 2 = Dx ud 1 x 1 4 u xd 2 x +...

29 CHAPTER 3. GEL FAND-DIKII FORMALISM 22 using (3.6) and (3.7) we find A = D 3 x ud x u x, (3.7) [A, L] = 1 4 u xxx uu x, and so we have u t = 1 4 u xxx uu x, which is the KdV equation. ii) L = (D 2 + u)d 1 x, A = (L 3 ) + We have L t = [A, L], (3.8) and A = D 3 x + 3uD x + 3u x, (3.9) using (3.8) and (3.9) we find [A, L] = (u xxx + 6uu x )D 1 x, and so we have, u t = u xxx + 6uu x, which is the KdV equation. 3.2 Consistent Lax Hierarchies Consider the Lax equations for n = 1, 2,... L tn = [A n, L] where A n = (L n N ) k, (3.10)

30 CHAPTER 3. GEL FAND-DIKII FORMALISM 23 what type of Lax operator can be used in (3.10) to obtain a consistent evolution equation? Let us consider the Lax operators in the following general form L = u N D N x + u N 1 D N 1 x u 0 + u 1 D 1 x +... (3.11) where u k s are functions of x and t. Firstly to obtain a consistent Lax equation, order of the right hand side of the Lax equation (3.10) should not exceed the order of L, which is N. Since [L, L n N ] = 0 we have where (L n N ) <k = a k 1 Dx k 1 + a k 2 Dx k Now let s look at the equation, [L, (L n N ) k ] = [L, (L n N )<k ], (3.12) L tn = [L, (L n N )<k ] = [u N D N x +..., a k 1 D k 1 x ], (3.13) the right hand side of (3.13) has order at most N +k 2 but the left hand side has order N. Therefore, the cases k = 0, 1, 2 can produce consistent Lax hierarchies. Hence there are 3 consistent Lax hierarchies that satisfy the Lax equation, L tn = [A n, L] where A n = (L n N ) k k = 0, 1, 2. (3.14) For each k = 0, 1, 2 we have some restrictions on the form of the Lax operator L so that (3.14) is consistent. To have a consistent Lax pair in each case L should be in the following form, [4] L = c m D m x L = c m D m x L = u m D m x + c m 1 D m 1 x + u m 1 D m 1 x + u m 1 D m 1 x u 0 + u 1 D 1 x +..., k = 0 (3.15) u 0 + u 1 D 1 x +..., k = 1 (3.16) u 0 + u 1 D 1 x +..., k = 2 (3.17)

31 CHAPTER 3. GEL FAND-DIKII FORMALISM 24 where c m and c m 1 are functions which do not depend on t. If number of u i s are finite, then L should be in the following form, [4] L = c m D m x L = c m D m x L = u m D m x + c m 1 D m 1 x u 0, k = 0 + u m 1 D m 1 x + u m 1 D m 1 x u 0 + D 1 x (u 1 ), k = u 0 + Dx 1 (u 1 ) + Dx 2 (u 2 )k = 2 where c m and c m 1 are functions which do not depend on t. For the case k = 1 there can be reductions on formula (3.16), and they are given by [4] L = c m D m x + u m 1 D m 1 x u 1 D x + u 0, L = c m D m x + u m 1 D m 1 x u 1 D x + λ, where λ is a constant parameter. For the case k = 2 there can be reductions on formula (3.17), and they are given by, [4] L = u m Dx m L = u m Dx m L = u m Dx m L = u m Dx m + u m 1 D m 1 x u 1 D x + u 0 + D 1 x (u 1 ), + u m 1 D m 1 x u 1 D x + u 0, + u m 1 D m 1 x u 1 D x + (λ 1 + λ 2 x), + u m 1 D m 1 x (λ 1 + λ 2 x)d x + λ 3, where λ 1, λ 2 and λ 3 are constant parameters. As an example for each case, [4] i. k = 0, L = Dx 2 + u, A = (L 3 2 ) 0 gives KdV equation ii. k = 1, L = Dx 2 + 2uD x, A = (L 3 2 ) 1 gives MKdV equation iii. k = 2, L = u 2 Dx, 2 A = (L 3 2 ) 2 gives Harry Dym equation In the previous section we derived KdV equation which corresponds to the case k = 0. Now we will derive MKdV and Harry-Dym equations which correspond to the cases k = 1 and k = 2, respectively. MKdV equation Let L = Dx 2 + 2uD x i (3.18) L t = [A, L] where A = (L 3 2 ) 1. (3.19)

32 CHAPTER 3. GEL FAND-DIKII FORMALISM 25 We find L 1 2 = Dx + u 1 2 (u x + u 2 )D 1 x +... and Then we find A = D 3 x + 3uD 2 x (u x + u 2 )D x. [A, L] = ( 1 2 u xxx 3u 2 u x )D x = 2u t D x. Hence we get the MKdv equation, u t = 1 4 u xxx 3 2 u2 u x. Harry-Dym Equation We have the Lax operator L = w 2 D 2 x, and we find and L t = [A, L] where A = (L 3 2 ) 2, (3.20) L 1 2 = wdx 1 2 w x +... A = w 3 D 3 x w2 w x D 2 x, (3.21) by using (3.21) in (3.20), we find [A, L] = 1 2 w4 w xxx D 2 x = 2ww t, hence we get w t = 1 2 w3 w xxx. which is Harry-Dym eqaution. Now we will derive some system of differential equations, [4] 1) Boussinesq Equation Consider L = Dx 3 + ud x + v,

33 CHAPTER 3. GEL FAND-DIKII FORMALISM 26 and we find L t = [A, L] with A = (L 2 3 ) 0, A = D 2 x u, and we have ( ) u v t ( = ) u xx + 2v x 2u 3 xxx + v xx 2uu 3 x using (3.22) we find that u satisfies the Boussinesq equation (3.22) u tt (u xxx + 4uu x ) x = 0. 2) Consider and we find, and we have u v w t L = D 4 x + ud 2 x + vd x + w, L t = [A, L] with A = (L 2 4 ) 0, = A = D 2 x u, 2u xx + 2v x 2u xxx + v xx + 2w x uu x 1 2 u xxxx + w xx 1 2 uu xx 1 2 u xv. 3) Consider and we find, L = D x + u + D 1 x v, (3.23) L t = [A, L] with A = (L 2 ) 1, A = D 2 x + 2uD x,

34 CHAPTER 3. GEL FAND-DIKII FORMALISM 27 and we have ( u ) ( uxx + 2v x + 2uu x ) v t = v xx + 2(uv) x In (3.23) if we choose v = 0, then we get Burgers equation. u t = u xx + 2uu x, with the following Lax pair L = D x + u with A = (L 2 ) 1. 4) Consider and we find, and we have u v w t = L = D 2 x + ud x + v + D 1 x w, (3.24) L t = [A, L] with A = (L 3 2 ) 1, A = D 3 x ud2 x (2u x + u 2 + 4v)D x, 1 4 u xxx v xx + 3w x 3 8 u2 u x (uv) x v xxx u(v) xx + 6u x v x u xw + 3uw x vv x u2 v x w xxx 3 2 u(w) xx 9 4 u xw x 3 4 u xxw (vw) x (u2 w) x. (3.25) If we consider the cases w = v = 0 in (3.24),the system (3.25) reduces to MKdV equation u t = 1 4 u xxx 3 8 u2 u x. 5) Consider and L = ud x + v + Dx 1 w + Dx 1 z, (3.26) L t = [A, L] with A = (L 2 ) 2,

35 CHAPTER 3. GEL FAND-DIKII FORMALISM 28 we find, A = u 2 D 2 x, and we have u v w z u 2 u xx + 2u 2 v x = u 2 v xx 2u(uw) x (u 2 w) xx + 2(u 2 z) x. (u 2 z) xx t If we choose z = 0 in (3.26), then we have the following reduction, if z = w = 0 u v w ( u v t ) = t = u 2 u xx + 2u 2 v x u 2 v xx + 2u(uw) x (u 2 w) xx ( u 2 u xx + 2u 2 v x u 2 v xx ), if z = w = 0 and v = λx, we get the partial differential equations,, u t = u 2 u xx + 2λu 2, where λ is a constant.

36 Chapter 4 A Method to Find Recursion Operator In chapter 3, we considered the hierarchy L tn = [A n, L], where A n := (L n m )+. By using this hierarchy it is possible to derive recursion operators of evolution equations. This method was introduced by Gürses, Karasu and Sokolov [2] Proposition For any n A n+m = L.A n + R n, (4.1) where R n is a differentiable operator and ord(r n ) m 1. Proof. then A n+m = (L.L n m )+ = (L.[(L n m )+ + (L n m ) ]) +, A n+m = L.(L n m )+ + (L.(L n m ) ) + = L.A n + R n, 29

37 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 30 since (L.(L n m ) + ) + = L.(L n m ) + and R n = (L.(L n m ) ) +. Furthermore, ordr n m 1 because (L n m ) does not have any differential part. Corollary For any n L tn+m = L.L tn + [R n, L], (4.2) where R n is as defined in Proposition (4.0.1). Proof. From Proposition (4.0.1), L tn+m = [A n+m, L] = [L.A n + R n, L] = L.[A n, L] + [R n, L] = L.L tn + [R n, L]. The equation (4.2) is called the recursion relation and R n is called the remainder. If we write A n+m = (L n m.l) +, then we get a new formula L tn+m = L.L tn + [R n, L], (4.3) where R n = ((L n m ) L) + is a differential operator and ord(r n ) m 1. If we equate the coefficients of powers of D x in (4.2), we can find R n in terms of L and L tn from Dx i for i = 2m 2,..., m 1. For the case i = m 2,..., 0 the comparison of coefficients gives us the relation u 0 u 1.. = R u 0 u 1.. (4.4) u m 2 t n+m u m 2 t n where R is the recursion operator. Also the recursion operators for (4.2) and (4.3) are the same. [2]

38 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR Symmetric and Skew-Symmetric Reduction of a Differential Operator Definition Let L be a differential operator given by L = m a i Dx. i (4.5) i=0 Then its adjoint L + is defined as L + = m ( D x ) i.a i. (4.6) i=0 If L + = L, in which case ord(l) = m must be even, then (3.4) and Proposition (4.0.1) remains valid, but if L + = L then m should be odd and we take A n+2m = (L n+2m m )+ = (L 2 L n m )+. (4.7) In this case we have the following proposition Proposition If L + = L then A n+2m = L 2 A n + R n, (4.8) and we have the recursion relation L tn+2m = L 2.L tn + [R n, L] (4.9) where R n = (L 2 (L n m ) ) + and ord(r n ) 2ord(L) 1. [2] Also we can use the following recursion relations instead of (4.9) L tn+2m = L.L tn.l + [ R n, L], (4.10) L tn+2m = L tn.l 2 + [ R n, L]. (4.11)

39 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 32 If L is a differential operator, recursion operators obtained from (4.9), (4.10) and (4.11) are same. However, in pseudo-differential case, they are not equivalent. Let us consider the case L = MD 1, where M is a differential operator and define L = D x L + Dx 1. Now we have a lemma that shows how to find recursion operators in some special cases. Lemma Let L = ɛl where ɛ = ±1. Then R n = a m 1 D m 1 x a 0, for ɛ = 1, (4.12) where R n is defined by Proposition (4.0.1) R n = a 2m 1 D 2m 1 x where R n is defined by (4.10). [2] a 1 D 1 x, for ɛ = 1, (4.13) Now we will derive the recursion operators of some evolution equations from their Lax representations. Firstly, we will find recursion operator of the KdV equation by using two different Lax representations. [2] 1. L = D 2 + u, 2. L = (D 2 + u)d 1 x. 1) L = D 2 x + u To find the recursion operator, we use L tn+2 = L.L tn + [R n, L], we choose R n as in Proposition (4.0.1) and we take R n = a n D x + b n. Since L tn = u tn, we have u tn+2 = (D 2 x + u).u tn + [a n D x + b n, D 2 x + u]. (4.14)

40 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 33 We compare the coefficients in (4.14) a n = 1 2 D 1 x (u tn ), (4.15) then by using (4.15) and (4.16) in (4.14) we find b n = 3 4 u t n, (4.16) which gives us the recursion operator u tn+2 = ( 1 4 D2 x + u u xd 1 x ).u tn, R = D 2 x + 4u + 2u x D 1 x. 2) L = D x + ud 1 x We find and L + = D x Dx 1 (u), L = L, Then we have ɛ = 1 and we will use the recursion relation (4.10) and we will take R n as (4.13) with R n = a n D x + b n + c n Dx 1. Then we find L tn+2 = L.L tn.l + [ R n, L], (4.17) a n = D 1 x (u tn ), b n = u tn, if we use a n, b n and c n in (4.17), we get c n = u tnx ud 1 x (u tn ), u tn+2 = u tnxx + 4uu tn + 2u x D 1 x (u tn ),

41 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 34 Hence we find, R = D 2 x + 4u + 2u x D 1 x. 3) We will now find the recursion operator of MKdV equation from its lax representation. Recall that Lax operator of MKdV is L = D 2 x + 2uD x. To find the recursion operator, we use L tn+2 = L.L tn + [R n, L], (4.18) with R n = c n D 2 x + a n D x + b n. By replacing R n into (4.18) we find c n = D 1 x (u tn ), a n = 3 2 u t n Dx 1 (uu tn ) + 2uDx 1 (u tn ), b n = 0. Hence we get and we find 2u tn+2 = 1 2 u t nxx 2u 2 u tn 2u x D 1 x (uu tn ), R = Dx 2 4u 2 4u x Dx 1 (u). 4) Recursion operator of the Harry-Dym equation from its Lax representation : L = w 2 D 2 x, (4.19) we use L tn+2 = L.L tn + [R n, L], (4.20) with R n = d n D 3 x + c n D 2 x + a n D x + b n. Using R n and (4.19) in (4.20) we find d n = w 3 Dx 1 ( w t n w ), 2 c n = 3 2 w2 w x D 1 x ( w t n w 2 ) ww t n, (4.21)

42 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 35 a n = 2b nx, b nxx = 0, Hence we can take a n = b n = 0. Then, substitutingc n and d n into (4.20) w tn+2 = 1 ( 2 w3 w xxx Dx 1 wtn ) ( + 1 w 2 2 ww xw tnx ww xxw tn + 1 ) 2 w2 w tnxx, hence we find ( 1 ) R = w 2 Dx 2 ww x D x + ww xx + w 3 w xxx Dx 1, w 2 5) Recursion operator of Sawada-Kotera equation [2] : u t = u 5x + 5uu xxx + 5u x u xx + 5u 2 u x, from its Lax representation L = D 3 x + ud x, A = (L 5 3 )+. We find and L + = D 3 x ud x u x, L = L, Then we have ɛ = 1 and we will use the recursion relation (4.10) and we will take R n as (4.13) L tn+6 = L.L tn.l + [R n, L], (4.22) with R n = a n Dx 5 + b n Dx 4 + c n Dx 3 + d n Dx 2 + e n D x + f n. Then we find a n = 1 3 D 1 x (u tn ), b n = 5 3 u t n, c n = 1 9 (5uD 1 x (u tn ) + 29u tnx ),

43 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 36 d n = 1 9 (5u xd 1 x (u tn ) + 14uu tn + 26u tnxx ), e n = 1 27 (10u xxd 1 x (u tn ) 2D 1 x +28u tnxxx + 32uu tnx 32u x u tn ), (u xx u tn ) D 1 x (u 2 u tn ) + 5u 2 D 1 x (u tn ) Hence we find recursion operator f n = 0. R = D 6 x + 6uD 4 x + 9u x D 3 x + 9u 2 D 2 x + 11u xx D 2 x + 10u xxx D x + 12uu x D x +4u uu xx + 6u 2 x + 5u 4x +u x Dx 1 (2u xx + u 2 ) + u t Dx 1. 6) Recursion operator of the DSIII (Drinfeld-Sokolov III) system [2] u t = u xxx + 6uu x + 6v x, v t = 2v xxx 6uv x.. from its Lax representation L = (D 5 x 2uD 3 x 2D 3 xu 2D x w 2wD x )D 1 x, A = (L 3 4 )+, where w = v u 2x. We find L + = D 1 x (D 5 x 2uD 3 x 2D 3 xu 2D x w 2wD x ), and L = L. Then applying lemma (4.1.3) with ɛ = 1 we use the recursion relation (4.2) and we will take R n as in (4.12) L tn+4 = L.L tn + [R n, L], (4.23)

44 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 37 with R n = a n D 3 x + b n D 2 x + c n D x + d n. By replacing R n and L into (4.23) we find c n = 1 2 (6uD 1 x a n = D 1 x (u tn ), b n = 4u tn, (u tn ) 11u tn,x 2D 1 x (uu tn ) 2D 1 x (v tn )), d nx = 1 2 (6u xxd 1 x (u tn ) + 10u x u tn 5u tn,xxx + 4uu tn,x 6v tn,x ). Hence we get the equation ( utn+4 ) ( utn ) = R v tn+4 v tn with the recursion operator where R = ( R11 R 12 R 21 R 22 ), (4.24) R 11 = D 4 x 8uD 2 x 8u 2x + 16v 12u x D x + 16u 2 + (12uu x 2u 3x + 12v x )D 1 x + 4u x D 1 x u, R 12 = 10D 2 x + 8u + 4u x D 1 x, R 21 = 12v 2x + 10v x D x + (4v 3x 12uv x )D 1 x R 22 = 4D 4 x + 16uD 2 x + 8u x D x + 16v + 4v x D 1 x. + 4v x D 1 x u, 7) Recursion operator of Boussinesq System. [2], Boussinesq equation (4.25) u tt = 1 3 (u 4x + 2(u 2 ) xx ), can be represented by the system u t = v x, v t = 1 3 (u xxx + 8uu x )..

45 CHAPTER 4. A METHOD TO FIND RECURSION OPERATOR 38 Its Lax representation is given by L = D 3 x + 2uD x + u x + v, A = (L 2 3 )+, We use recursion relation (4.2) and we will take R n as in (4.0.1) L tn+3 = L.L tn + [R n, L], (4.26) with R n = a n D 2 x + b n D x + c n. By replacing R n and L into (4.26) we find a n = 2 3 D 1 x (u tn ), b n = 1 3 (5u t n + D 1 x (v tn )), c n = 1 9 (6v t n + 8uD 1 x (u tn ) + 10u tn,x ), d nx = 1 2 (6u xxd 1 x (u tn ) + 10u x u tn 5u tn,xxx + 4uu tn,x 6v tn,x ). Hence we obtain the recursion operator R = ( R11 R 12 R 21 R 22 ), (4.27) where R 11 = 3v + 2v x D 1 x, R 12 = D 2 x + 2u + u x D 1 x, R 21 = ( 1 3 D4 x ud2 x + 5u x D x + 3u xx u2 + ( 2 3 u xxx uu x)d 1 x ), R 22 = 3v + v x D 1 x.

46 Chapter 5 Recursion Operators in Gelfand-Dikii Formalism 5.1 Recursion Operators as Lax Operators in Gelfand-Dikii Formalism Let u t = F (u, u x, u xx,...), (5.1) be an integrable nonlinear partial differential equation where u is a function of x and t. Then we know that if R t = [F, R] (5.2) is satisfied then R is the recursion operator of (5.1). Hence, recursion operators are also Lax operators of the evolutionary type partial differential equations. In this chapter we discuss if we can use recursion operators in Gelfand-Dikii formalism to derive the evolution equation itself. Remark [4] If the Lax equation R t = [F, R] (5.3) 39

47 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM40 is satisfied then the adjoint equation is given by (R + ) t = [ (F ) +, R + ], (5.4) where (F ) + and R + are both adjoint operators. The consistent Lax operators has the form L t = [A n, L], A n = (L n m ) k, (5.5) where k is either 0 or 1 or 2. (L n m ) k. Hence we have For recursion operators we have F instead of F = (R n m ) k, (5.6) for some k = 0, 1, 2 we can use R in Gelfand-Dikii formalism. Also the adjoint equation should be consistent, that is we should have (F ) + = ((R + ) n m ) k, (5.7) for some k = 0, 1, Recursion Operators Produce Themselves Since recursion operators are also Lax operators, we can use them in Proposition or in Lemma Hence they should reproduce themselves as recursion operators. That is, using recursion relation (4.2), L tn+m = L.L tn + [R n, L] (5.8) we should find R = L. Now we will check (5.6) and (5.7) for the recursion operators of KdV, MKdV, Burgers and Harry Dym equations. Firstly, we will list all required operators and their adjoints and then we will check if (5.6) and (5.7) are satisfied for any

48 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM41 k = 0, 1, 2 or not. If both of them are satisfied we will conclude that, that recursion operator can be used as Lax operator in Gelfand-Dikii formalism. But if any of them is not satisfied we will conclude that, that recursion operator cannot be used as Lax operator in Gelfand-Dikii formalism. Also we will show that recursion operators of these PDEs reproduce themselves. 1) KdV Equation : u t = u xxx + 6uu x Here we list all required operators and their adjoints, i) R R = D 2 x + 4u + 2u x D 1 x, F = D 3 x + 6uD x + 6u x, R 3 2 = D 3 x + 6uD x + 6u x + (2u xx + 6u 2 )D 1 x +... ii) R + R + = D 2 x + 4u D 1 x (2u x ), (F ) + = D 3 x 6uD x, (R + ) 3 2 = D 3 x + 6uD x +.. By using the above list, we see that for the Lax equation, F = (R 3 2 ) 0 is satisfied and for the adjoint Lax equation, (F ) + = ((R + ) 3 2 ) 1 is satisfied and both of these Lax equations are consistent. Hence we can use the recursion operator of KdV in Gelfand-Dikii formalism. a ) Recursion Operator is the Lax operator, L = R L = D 2 x + 4u + 2u x D 1 x, L t = [A, L] and A = (L 3 2 )+. (5.9) We find A = D 3 x + 6uD x + 6u x,

49 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM42 If we compare coefficients in (5.9), we find u t = u xxx + 6uu x, which is KdV equation. b) Adjoint Lax equation, L = R + L = D 2 x + 4u D 1 x (2u x ) L t = [A, L] and A = (L 3 2 ) 1 i (5.10) we find A = D 3 x + 6uD x. If we compare coefficients in (5.10), we find u t = u xxx + 6uu x, which is KdV equation. Now if we take the recursion operator of KDV as a Lax operator, we must find its recursion operator as itself. c) Recursion Operator Produces Itself, L = R Recursion operator of KdV is L = D 2 x + 4u + 2u x D 1 x we find and L + = D 2 x + 4u 2D 1 x (u x ), L = L. Then applying Lemma (4.1.3) with ɛ = 1 and we use the recursion relation (4.2). Also we will take R n as in (4.12) L tn+2 = L.L tn + [R n, L], R n = a n D x + b n. (5.11)

50 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM43 Then we find a n = 2D 1 x (u tn ), If we put a n and b n into (5.11), we get b n = 4u tn. u tn+2 = u tnxx + 4uu tn + 2u x D 1 x (u tn ). Then we find R = D 2 x + 4u + 2u x D 1 x, which is the initial lax operator. 2) MKdV Equation : u t = u xxx 6u 2 u x Here we list all required operators and their adjoints, i) R R = Dx 2 4u 2 4u x Dx 1 (u), F = Dx 3 6u 2 D x 12uu x, R 3 2 = D 3 x 6u 2 D x 12uu x + (6u 4 4uu xx + 2u 2 x)d 1 x +... ii) R + R + = D 2 x + 4u 2 + 4uD 1 x (u x ), (F ) + = D 3 x + 6u 2 D x, (R + ) 3 2 = D 3 x 6u 2 D x 8uu x +.. By using the above list, we see that for the Lax equation F = (R 3 2 ) 0 is satisfied and for the adjoint Lax equation (F ) + = ((R + ) 3 2 ) 1 is satisfied. Both of these Lax equations are consistent. Hence we can use the recursion operator of MKdV in Gelfand-Dikii formalism. a ) Recursion Operator is the Lax operator, L = R L = Dx 2 4u 2 4u x Dx 1 (u),

51 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM44 L t = [A, L] where A = (L 3 2 )+. (5.12) We find A = D 3 x 6u 2 D x 12uu x. (5.13) By using (5.12) and (5.13) we find u t = u xxx 6u 2 u x, which is MKdV equation. b) Adjoint Lax equation, L = R + L = D 2 x + 4u 2 + 4uD 1 x (u x ), L t = [A, L] and A = (L 3 2 ) 1. (5.14) We find A = D 3 x 6u 2 D x. If we compare coefficients in (5.14), we find u t = u xxx 6u 2 u x, which is MKdV equation c) Recursion Operator Produces Itself, L = R Recursion operator of MKdV equation is L = Dx 2 4u 2 4u x Dx 1 (u), we find and L + = D 2 x 4u 2 + 4uD 1 x (u x ), L = L.

52 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM45 Then applying Lemma (4.1.3) with ɛ = 1 we use the recursion relation (4.2) and we will take R n as in (4.12) L tn+2 = L.L tn + [R n, L], (5.15) with R n = a n D x + b n. By replacing R n and L into (5.15) we find a n = 4D 1 x (uu tn ), b n = 8uu tn. Hence we get this yields to u tn+2 = 2u tnxx 8u 2 u tn 8u x D 1 x (uu tn ), R = Dx 2 4u 2 4u x Dx 1 (u). 3) Burgers Equation : u t = u xx + 2uu x Here we list all required operators and their adjoints, i) R R = D x + u + u x D 1 x, F = D 2 x + 2uD x + 2u x, R 2 = D 2 x + 2uD x + (3u x + u 2 ) + (2uu x + u xx )D 1 x ii) R + R + = D x + u D 1 x (u x ), (F ) + = D 2 x 2uD x, (R + ) 2 = D 2 x 2uD x + (u x + u 2 ) +... By using the above list, we see that for the Lax equation F = (R 2 ) k is not satisfied for any k = 0, 1, 2 and for the adjoint Lax equation (F ) + = ((R + ) 2 ) k

53 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM46 is not satisfied for any k = 0, 1, 2. Hence we cannot use recursion operator of Burgers equation in Gelfand-Dikii formalism as the Lax operator. There is an interesting situation here. We have (F ) + = ((R + ) 2 ) 1 but for a consistent Lax representation we should have (F ) + = ((R + ) 2 ) k for some k. We will check if this interesting case in adjoint case is consistent or not. Firstly we show the inconsistency in Lax equation. a ) Recursion Operator is the Lax operator, L = R L = D x + u + u x D 1 x, L t = [A, L] and A = (L 3 2 ) 1. (5.16) We find A = D 2 x + 2uD x.. If we compare coefficients in (5.16), we find u t = 3u xx + 2uu x, (5.17) but coefficients of D 1 x gives u tx = u xxx + 2uu xx + 2u 2 x, which is not compatible with (5.17). Also the left hand side of (5.16) include terms with powers Dx 2, Dx 3,... but the right hand side does not have terms with these powers. b) Adjoint Lax equation, L = R + L = D x + u D 1 x (u x ), we do not have (F ) + = ((L + ) 2 ) k for any k but if we take A = ((L + ) 2 ) 1 L t = [A, L] and A = (L 3 2 ) 1, we find u t = u xx 2uu x,

54 CHAPTER 5. RECURSION OPERATORS IN GELFAND-DIKII FORMALISM47 which is not Burgers equation. Thus, there is a minus difference between adjoint Lax equation and corresponding Gelfand-Dikii formalism. c) Recursion Operator Produces Itself, L = R Recursion operator of Burgers s equation is L = D x + u + u x D 1 x, L tn+1 = L.L tn + [R n, L], (5.18) with R n = a n D x + b n. By replacing R n and L into (5.18) we find a n = D 1 x (u tn ), b n = u tn. Hence we get then we find u tn+1 = uu tn + u tnx + u x D 1 x (u tn ), R = D x + u + u x D 1 x. 4) Harry-Dym Equation : w t = w 3 w xxx Here we list all required operators and their adjoints, i) R R = w 2 D 2 x ww x D x + ww xx + w 3 w xxx D 1 x ( 1 w 2 ), F = w 3 D 3 x + 3w 2 w xxx, R 3 2 = w 3 D 3 x + (w 2 w xx 1 2 ww2 x)d x + (2w 2 w xxx ww x w xx w3 x) +... ii) R + R + = w 2 D 2 x + 3ww x D x + (3w 2 x + 4ww xx ) 1 w 2 D 1 x (w 3 w xxx ), (F ) + = w 3 D 3 x 9w 2 w x D 2 x (19ww 2 x + 9w 2 w xx )D x (18ww x w xx + 6w 3 x), (R + ) 3 2 = w 3 D 3 x + 6w 2 w x D 2 x + ( 17 2 w2 w xx + 10ww 2 x)d x + ( 7 4 w2 w xxx 49 4 ww xw xx + 4w 3 x) +... By using the above list, we see that for the Lax equation F = (R 2 ) k