Continuous and Discrete Homotopy Operators with Applications in Integrability Testing. Willy Hereman

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1 Continuous and Discrete Homotopy Operators with Applications in Integrability Testing Willy Hereman Department of Mathematical and Computer Sciences Colorado School of Mines Golden, Colorado, USA home/whereman/ Colloquium Talk Centre de Recherches Mathématiques Montréal, Canada Thursday, December 4, 2003, 15:30 Collaborator: Michael Colagrosso Students: Adam Ringler, Ingo Kabirschke Francis Martin & Kara Namanny Talk dedicated to Ryan Sayers Research supported in part by NSF under Grant CCR

2 OUTLINE Part I: Continuous Case Integration by Parts on the Jet Space (by hand) + Mathematica Experiment Exactness or Integrability Criterion: Continuous Euler Operator Continuous Homotopy Operator Application of Continuous Homotopy Operator Demo of Mathematica software Part II: Discrete Case Inverting the Total Difference Operator (by hand) Exactness or Total Difference Criterion: Discrete Euler Operator Discrete Homotopy Operator Application of Discrete Homotopy Operator Demo of Mathematica Software Future Research 2

3 Problem Statement For continuous case: Given, for example, f = 3 u v 2 sin(u) u 3 sin(u) 6 v v cos(u) + 2 u u cos(u) + 8 v v Find F so that f = D x F or F = f dx. Result: F = 4 v 2 + u 2 cos(u) 3 v 2 cos(u) Can this be done without integration by parts? Can the problem be reduced to a single integral in one variable? For discrete case: Given, for example, f n = +1 v n vn v n+1 + vn v n+2 +1 v n Find F n so that f n = F n = F n+1 F n or F n = 1 f n. Result: F n = vn v n + +1 v n + +2 v n+1. How can this be done algorithmically? Can this be done in the same way as the continuous case? 3

4 Part I: Continuous Case Integration by Parts on the Jet Space Given f involving u(x) and v(x) and their derivatives f = 3 u v 2 sin(u) u 3 sin(u) 6 v v cos(u)+2 u u cos(u)+8 v v Find F so that f = D x F or F = f dx. Integrate by parts (compute F by hand) f F 8 v v 4 v 2 2 u u cos(u) u 2 cos(u) u 3 sin(u) 6 v v cos(u) 3 v 2 cos(u) 3 u v 2 sin(u) Integral: F = 4 v 2 + u 2 cos(u) 3 v 2 cos(u) Remark: For simplicity we denote f(u, u, u,, u (m) ) as f(u). 4

5 Exactness Criterion: Continuous Euler Operator (variational derivative) Definition (exactness): A function f(u) is exact, i.e. can be integrated fully, if there exists a function F (u), such that f(u) = D x F (u) or equivalently F (u) = D 1 x f(u) = f(u) dx. x D x is the (total) derivative with respect to x. Theorem (exactness or integrability test): A necessary and sufficient condition for a function f to be exact, i.e. the derivative of another function, is that L (0) u (f) 0 where L (0) u is the continuous Euler operator (variational derivative) defined by L (0) u = m 0 ( D x ) k k=0 u (k) = u D x u + D2 x u + + ( 1)m 0 D m 0 x where m 0 is the order (of f). u (m 0) Proof: See calculus of variations (derivation of Euler-Lagrange equations the forgotten case!). 5

6 Example: Apply the continuous Euler operator to f(u) = 3u v 2 sin(u) u 3 sin(u) 6v v cos(u)+2u u cos(u)+8v v Here u = (u, v). For component u (order 2): L (0) u (f) = u (f) D x u (f) + D2 x u (f) = 3u v 2 cos(u) u 3 cos(u) + 6v v sin(u) 2u u sin(u) D x [3v 2 sin(u) 3u 2 sin(u) + 2u cos(u)] + D 2 x[2u cos(u)] = 3u v 2 cos(u) u 3 cos(u) + 6v v sin(u) 2u u sin(u) 0 [3u v 2 cos(u) + 6v v sin(u) 3u 3 cos(u) 6u u sin(u) 2u u sin(u) + 2u cos(u)] +[ 2u cos(u) 6u u sin(u) + 2u cos(u)] For component v (order 2): L (0) v (f) = v (f) D x v (f) + D2 x v (f) = 6u v sin(u) 6v cos(u) D x [ 6v cos(u) + 8v ] + D 2 x[8v ] = 6u v sin(u) 6v cos(u) [6u v sin(u) 6v cos(u)+8v ]+8v 0 6

7 Computation of the integral F Definition (higher Euler operators): The continuous higher Euler operators are defined by L (i) u = m i k ( D x ) k i i u (k) k=i These Euler operators all terminate at some maximal order m i. Examples (for component u) : L (0) u = u D x u + D2 x u D3 x u + + ( 1)m 0 D m 0 x u (m 0) L (1) u = u 2D x u +3D2 x u 4D3 x u + (4) ( 1)m 1 m 1 D m 1 1 x L (2) u = u 3D x u +6D2 x u (4) 10D3 x u (5)+ +( 1)m 2 L (3) u = u 4D x u (4)+10D2 x u (5) 20D3 x u + (6) ( 1)m 3 Similar formulae for component L (i) v u (m 1) m 2 D m 2 2 x 2 u (m 2) m 3 D m 3 3 x 3 u (m 3) 7

8 Definition (homotopy operator): The continuous homotopy operator is defined by where H u = 1 0 N f r (u) = p r r=1 f r (u)[λu] dλ λ i=0 D i x[u r L (i+1) u r ] p r is the maximum order of u r in f N is the number of dependent variables f r (u)[λu] means that in f r (u) one replaces u λu, u λu, etc. Example: For a two-component system (N = 2) where u = (u, v): with and H u = 1 0 {f 1(u)[λu] + f 2 (u)[λu]} dλ λ f 1 (u) = p 1 i=0 f 2 (u) = p 2 i=0 D i x[ul (i+1) u ] D i x[vl (i+1) v ] Theorem (integration via homotopy operator): Given an integrable function f F = D 1 x f = f dx = H u (f) Proof: Olver s book Applications of Lie Groups to Differential Equations, p Proof is given in terms of differential forms. Work of Henri Poincaré ( ), George de Rham (1950), and Ian Anderson & Tom Duchamp (1980). Proofs based on calculus: Deconinck and Hereman. 8

9 Example: Apply the continuous homotopy operator to integrate f(u) = 3u v 2 sin(u) u 3 sin(u) 6vv cos(u) + 2u u cos(u) + 8v v For component u (order 2): i L (i+1) u (f) D i x ( ul (i+1) u (f) ) 0 u f 2D x ( u f) =3v 2 sin u 3u 2 sin u + 2u cos u 2D x (2u cos u) =3v 2 sin u + u 2 sin u 2u cos u 3uv 2 sin u + uu 2 sin u 2uu cos u 1 u f =2u cos u D x [2uu cos u] =2u 2 cos u + 2uu cos u 2uu 2 sin u Hence, f 1 (u)(f) = 3uv 2 sin(u) uu 2 sin(u) + 2u 2 cos(u) For component v (order 2): i L (i+1) v (f) D i x[vl (i+1) v (f)] 0 6v cos(u) + 8v 2D x [8v ] = 6v cos(u) 8v 6v 2 cos(u) 8vv 1 8v D x [8vv ] = 8v 2 + 8vv Hence, f 2 (u)(f) = 6v 2 cos(u) + 8v 2 The homotopy operator leads to an integral for (auxiliary) variable λ. (Use standard integration by parts to work the integral). F (u) = H u (f) = 1 0 {f 1(u)(f)[λu] + f 2 (u)(f)[λu]} dλ λ = 1 0 [3λ2 uv 2 sin(λu) λ 2 uu 2 sin(λu) + 2λu 2 cos(λu) 6λv 2 cos(λu) + 8λv 2 ] dλ = 4v 2 + u 2 cos(u) 3v 2 cos(u) 9

10 Application: Conserved densities and fluxes for PDEs with transcendental nonlinearities Definition (conservation law): D t ρ + D x J = 0 conserved density ρ and flux J. (on PDE) Example: Sine-Gordon system (type u t = F) has scaling symmetry In terms of weights: u t v t = v = u xx + α sin(u) (t, x, u, v, α) (λ 1 t, λ 1 x, λ 0 u, λv, λ 2 α) w(d x ) = 1, w(d t ) = 1, w(u) = 0, w(v) = 1, w(α) = 2 Conserved densities and fluxes ρ (1) = 2α cos(u) + v 2 + u x 2 J (1) = 2u x v ρ (2) = u x v J (2) = [ 1 2 v u x 2 α cos(u)] ρ (3) = 12 cos(u)vu x + 2v 3 u x + 2vu x 3 16v x u 2x ρ (4) = 2 cos 2 (u) 2 sin 2 (u) + v 4 + 6v 2 u 2 x + u 4 x + 4 cos(u)v cos(u)u 2 x 16v 2 x 16u 2 2x. are all scaling invariant! Remark: J (3) and J (4) are not shown (too long). 10

11 Algorithm for Conserved Densities and Fluxes Example: Density and flux of rank 2 for sine-gordon system Step 1: Construct the form of the density ρ = αh 1 (u) + h 2 (u)v 2 + h 3 (u)u x 2 + h 4 (u)u x v where h i (u) are unknown functions. Step 2: Determine the functions h i Compute E = D t ρ = ρ t + ρ (u)[f] = ρ t + m1 k=0 (on PDE) ρ D k u xu t + m2 kx k=0 ρ v kx D k xv t Since E = D t ρ = D x J, the expression E must be integrable. Require that L (0) u (E) 0 and L (0) v (E) 0. Solve the system of linear mixed system (algebraic eqs. and ODEs): h 2 (u) h 3 (u) = 0 h 2 (u) = 0 h 3 (u) = 0 h 4 (u) = 0 h 2 (u) = 0 h 4 (u) = 0 2h 2 (u) h3 (u) = 0 2h 2 (u) h3 (u) = 0 h 1 (u) + 2 sin(u)h2 (u) = 0 h 1 (u) + 2 sin(u)h2 (u) + 2 cos(u)h2 (u) = 0 11

12 Solution: h 1 (u) = 2c 1 cos(u) + c 3 h 2 (u) = h 3 (u) = c 1 h 4 (u) = c 2 (with arbitrary constants c i ). Substitute in ρ ρ = c 1 (2α cos(u) + v 2 + u x 2 ) + c 2 (u x v) + c 3 α Step 3: Compute the flux J First, compute E =D t ρ = c 1 ( 2αu t sin u + 2vv t + 2u x u xt ) + c 2 (u xt v + u x v t ) = c 1 ( 2αv sin u + 2v (u 2x + α sin u) + 2u x v x ) +c 2 (v x v + u x (u 2x + α sin u)) = c 1 (2u 2x v + 2u x v x ) + c 2 (vv x + u x u 2x + αu x sin u) Since E = D t ρ = D x J, one must integrate f = E. Apply the homotopy operator for each component of u = (u, v). For component u (order 2): i L (i+1) u (f) D i x ( ul (i+1) u (f) ) 0 2c 1 v x + c 2 (u 2x α sin u) 2c 1 uv x + c 2 (uu 2x αu sin u) 1 2c 1 v c 2 u x 2c 1 (u x v + uv x ) c 2 (u 2 x + uu 2x ) Hence, f 1 (u)(f) = 2c 1 u x v c 2 (u 2 x + αu sin u) 12

13 For component v (order 1): i L (i+1) v (f) D i ( x vl (i+1) v (f) ) 0 2c 1 u x c 2 v 2c 1 u x v c 2 v 2 Hence, f 2 (u)(f) = 2c 1 u x v c 2 v 2 The homotopy operator leads to an integral for (one) variable λ : J(u) = H u (f) = 1 (f 0 1(u)(f)[λu] + f 2 (u)(f)[λu]) dλ λ = ( 1 4c1 λu 0 x v + c 2 (λu 2 x + αu sin(λu) + λv 2 ) ) dλ = 2c 1 u x v c v u2 x α cos u Split the density and flux in independent pieces (for c 1 and c 2 ): ρ (1) = 2α cos u + v 2 + u x 2 J (1) = 2u x v ρ (2) = u x v J (2) = 1 2 v2 1 2 u2 x + α cos u Remark: Computation of J (3) and J (4) requires integration with the homotopy operator! 13

14 Computer Demos (1) Use continuous homotopy operator to integrate f = 3 u v 2 sin(u) u 3 sin(u) 6 v v cos(u)+2 u u cos(u)+8 v v (2) Compute densities of rank 8 and fluxes for 5th-order Korteweg-de Vries equation with three parameters: u t + αu 2 u x + βu x u 2x + γuu 3x + u 5x = 0 (α, β, γ are nonzero constant parameters). (3) Compute density of rank 4 and flux for sine-gordon system: u t v t = v = u xx + α sin(u) 14

15 Analogy PDEs and DDEs Continuous Case (PDEs) Semi-discrete Case (DDEs) System u t = F(u, u x, u 2x,...) =F(..., 1,, +1,...) Conservation Law D t ρ + D x J = 0 ρ n + J n+1 J n = 0 Symmetry D t G = F (u)[g] D t G = F ( )[G] = F(u + ɛg) ɛ ɛ=0 = F(u ɛ n + ɛg) ɛ=0 Recursion Operator D t R + [R, F (u)] = 0 D t R + [R, F ( )] = 0 Table 1: Conservation Laws and Symmetries KdV Equation Volterra Lattice Equation u t = 6uu x + u 3x = (+1 1 ) Densities ρ = u, ρ = u 2 ρ n =, ρ n = ( 1u 2 n+ +1 ) ρ = u u2 x ρ n = 1 3 u3 n+ +1 ( ) Symmetries G=u x, G=6uu x + u 3x G = +1 ( ) G=30u 2 u x + 20u x u 2x 1 ( ) +10uu 3x + u 5x Recursion Operator R = D 2 x + 4u + 2u x D 1 x R = (I + D)( D D 1 ) (D I) 1 1 Table 2: Prototypical Examples 15

16 Part II: Discrete Case Definitions (shift and total difference operators): D is the up-shift (forward or right-shift) operator if for F n DF n = F n+1 = F n n n+1 D 1 the down-shift (backward or left-shift) operator if D 1 F n = F n 1 = F n n n 1 = D I is the total difference operator F n = (D I)F n = F n+1 F n D (up-shift operator) corresponds the differential operator D x D x F (x) F n+1 F n x = F n x (set x = 1) For k > 0, D k = D D D (k times). Similarly, D k = D 1 D 1 D 1. Problem to be solved: Continuous case: Given f. Find F so that f = D x F or F = D 1 x f = f dx. Discrete case: Given f n. Find F n so that f n = F n = F n+1 F n or F n = 1 f n. 16

17 Inverting the Operator Given f n involving and v n and their shifts: f n = +1 v n vn+u 2 n+1 +2 v n+1 +vn+1+u 2 n+3 v n+2 +1 v n Find F n so that f n = F n = F n+1 F n or F n = 1 f n. Invert the operator (compute F n by hand) f n F n v 2 n v 2 n v 2 n+1 +1 v n +1 v n v n+1 +1 v n +1 v n ++2 v n+1 +2 v n+1 +2 v n+1 +3 v n+2 Result: F n = v 2 n + +1 v n + +1 v n + +2 v n+1. Remarks: We denote f(, +1, +2,, +p ) as f( ). Assume that all negative shifts have been removed via up-shifting Replace f n = 2 v n v n+3 by f n = D 2 f n = v n+2 v n+5. 17

18 Total Difference Criterion: Discrete Euler Operator (variational derivative) Definition (exactness): A function f n ( ) is exact, i.e. a total difference, if there exists a function F n ( ), such that f n = F n or equivalently F n = 1 f n. D is the up-shift operator. Theorem (exactness or total difference test): A necessary and sufficient condition for a function f n to be exact, i.e. a total difference, is that L (0) (f n ) 0, where L (0) is the discrete Euler operator (variational derivative) defined by L (0) = m 0 D k k=0 +k = + D 1 ( ) + D 2 ( ) + + D m 0 ( ) m0 = ( m0 D k ) k=0 L (0) = (I + D 1 + D D m 0 ) where m 0 is the highest forward shift (in f n ). 18

19 Example: Apply the discrete Euler operator to f n ( ) = +1 v n v 2 n v n+1 +v 2 n+1++3 v n+2 +1 v n Here = (, v n ). For component (highest shift 3): L (0) (f n ) = [I + D 1 + D 2 + D 3 ](f n ) =[ +1 v n ] + [ 1 v n v n v n 1 ]+[ 1 v n 1 ]+[v n 1 ] 0 For component v n (highest shift 2): L (0) v n (f n ) = [I + D 1 + D 2 ](f n ) v n = [ +1 2v n +1 ] + [ v n ] + [+1 ] 0 19

20 Computation of F n Definition (higher Euler operators): The discrete higher Euler operators are defined by L (i) = ( mi k D k ) i These Euler operators all terminate at some maximal shifts m i. Examples (for component ): L (0) = L (1) = L (2) = L (3) = k=i (I + D 1 + D 2 + D D m 0 ) (D 1 + 2D 2 + 3D 3 + 4D m 1 D m 1 ) (D 2 + 3D 3 + 6D D m 2(m 2 1) D m 2 ) (D 3 + 4D D D 6 m D m 3 ) 3 Similar formulae for L (i) v n. 20

21 Definition (homotopy operator): The discrete homotopy operator is defined by where H un = 1 0 f r,n ( ) = p r N r=1 f r,n ( )[λ ] dλ λ i=0 (D I) i [u r,n L (i+1) u r,n ] p r is the maximum shift of u r,n in f n N is the number of dependent variables f r,n ( )[λ ] means that in f r,n ( ) one replaces λ, +1 λ+1, etc. Example: For a two-component system (N = 2) where = (, v n ): with and H un = 1 0 {f 1,n( )[λ ] + f 2,n ( )[λ ]} dλ λ f 1,n ( ) = p 1 (D I) i [ L (i+1) i=0 ] f 2,n ( ) = p 2 (D I) i [v n L (i+1) i=0 v n ] Theorem (total difference via homotopy operator): Given a function f n which is a total difference, then F n = 1 f n = H un (f n ) Proof: Recent work by Mansfield and Hydon on discrete variational bi-complexes. Proof is given in terms of differential forms. Proof based on calculus: Deconinck and Hereman. 21

22 Higher Euler Operators Side by Side Continuous Case (for component u) L (0) u = u D x + D 2 x D 3 x + u x u 2x u 3x 2D x + 3D 2 x 4D 3 x + u x u 2x u 3x u 4x 3D x + 6D 2 x 10D 3 x + u 2x u 3x u 4x u 5x 4D x + 10D 2 x 20D 3 x + u 3x u 4x u 5x u 6x L (1) u = L (2) u = L (3) u = Discrete Case (for component ) L (0) = L (1) = L (2) = L (3) = (I + D 1 + D 2 + D 3 + ) (D 1 + 2D 2 + 3D 3 + 4D 4 + ) (D 2 + 3D 3 + 6D D 5 + ) (D 3 + 4D D D 6 + ) 22

23 Homotopy Operators Side by Side Continuous Case (for components u and v) with and H u = 1 0 {f 1(u)[λu] + f 2 (u)[λu]} dλ λ f 1 (u) = p 1 i=0 f 2 (u) = p 2 i=0 D i x[ul (i+1) u ] D i x[vl (i+1) v ] Discrete Case (for components and v n ) with and H un = 1 0 {f 1,n( )[λ ] + f 2,n ( )[λ ]} dλ λ f 1,n ( ) = p 1 (D I) i [ L (i+1) i=0 ] f 2,n ( ) = p 2 (D I) i [v n L (i+1) i=0 v n ] 23

24 Example: Apply the discrete homotopy operator to f n ( ) = +1 v n v 2 n v n+1 +v 2 n+1++3 v n+2 +1 v n For component (highest shift 3): i L (i+1) (f n ) (D I) i [ L (i+1) (f n )] 0 1 v n v n +2v n 1 1 v n v n +2 v n v n 1 +3v n 1 +1 v n +3+1 v n 1 v n 1 3 v n 1 2 v n 1 +2 v n+1 +1 v n +1 v n + v n 1 Hence, f 1,n ( )(f n ) = 2 +1 v n + +1 v n + +2 v n+1 For component v n (highest shift 2): i L (i+1) v n (f n ) (D I) i [v n L (i+1) v n (f n )] v n v n +2v 2 n+2+1 v n v n+1 +1 v n Hence, f 2,n ( )(f n ) = +1 v n + 2v 2 n + +1 v n + +2 v n+1 The homotopy operator leads to an integral for variable λ. (Use standard integration by parts to work the integral). F n ( ) = H un (f n ) = 1 {f 0 1,n( )(f n )[λ ] + f 2,n ( )(f n )[λ ]} dλ λ = 1 0 [2λv2 n + 3λ 2 +1 v n + 2λ+1 v n + 2λ+2 v n+1 ] dλ = vn v n + +1 v n + +2 v n+1 24

25 Application: Conserved densities and fluxes for DDEs Definition (conservation law): D t ρ n + J n = D t ρ n + J n+1 J n = 0 conserved density ρ n and flux J n. Example The Toda lattice (type = F) : has scaling symmetry In terms of weights: = v n 1 v n v n = v n ( +1 ) (t,, v n ) (λ 1 t, λ, λ 2 v n ). (on DDE) w( d dt ) = 1, w() = w(+1 ) = 1, w(v n ) = w(v n 1 ) = 2. Conserved densities and fluxes ρ (0) n = ln(v n ) J n (0) ρ (1) n = J n (1) ρ (2) n = 1 2 u2 n + v n J n (2) ρ (3) n = 1 3 u3 n + (v n 1 + v n ) J n (3) are all scaling invariant! = = v n 1 = v n 1 = 1 v n 1 + v 2 n 1 25

26 Algorithm for Conserved Densities and Fluxes Example: Density of rank 3 for Toda system Step 1: Construct the form of the density. ρ n = c 1 u 3 n + c 2 v n 1 + c 3 v n where c i are unknown constants. Step 2: Determine the constants c i. Compute E n = D t ρ n = ρ n t + ρ n( )[F] (on DDE) = (3c 1 c 2 )u 2 nv n 1 + (c 3 3c 1 )u 2 nv n + (c 3 c 2 )v n 1 v n +c 2 1 v n 1 + c 2 v 2 n 1 c 3 +1 v n c 3 v 2 n Compute Ẽn = DE n to remove negative shift n 1. Since Ẽn = J n, the expression Ẽn must be a total difference. Require L (0) (Ẽn) = (I + D 1 + D 2 )(Ẽn) = = 2(3c 1 c 2 ) v n 1 + 2(c 3 3c 1 ) v n +(c 2 c 3 ) 1 v n 1 + (c 2 c 3 )+1 v n 0 (D + I + D 1 )(E n ) and L (0) v n (Ẽn) = (I + D 1 )(Ẽn) = (D + I)(E n ) v n v n = (3c 1 c 2 )u 2 n+1 + (c 3 c 2 )v n+1 + (c 2 c 3 ) +1 +2(c 2 c 3 )v n + (c 3 3c 1 )u 2 n + (c 3 c 2 )v n

27 Solve the linear system S = {3c 1 c 2 = 0, c 3 3c 1 = 0, c 2 c 3 = 0}. Solution: 3c 1 = c 2 = c 3 Choose c 1 = 1 3, and c 2 = c 3 = 1. Substitute in ρ n ρn = 1 3 u3 n + (v n 1 + v n ) Step 3: Compute the flux J n. Start from Ẽn = +1 v n v 2 n v n+1 + v 2 n+1 Apply the discrete homotopy operator to f n = Ẽn. For component (highest shift 2): i L (i+1) (f n ) (D I) i ( L (i+1) ( Ẽn)) 0 1 v n v n 1 v n v n 1 1 v n 1 +1 v n 1 v n 1 Hence, f 1,n ( )(f n ) = 2 +1 v n For component v n (highest shift 1): i L (i+1) v n (f n ) (D I) i (v n L (i+1) v n ( Ẽn)) v n v n +1 +2vn 2 Hence, f 2,n ( )(f n ) = +1 v n + 2v 2 n J n = H un (f n ) = 1 (f 0 1,n( )(f n )[λ ] + f 2,n (f n )( )[λ ]) dλ λ = 1 0 (3λ2 +1 v n + 2λvn) 2 dλ = +1 v n + vn. 2 Final Result: J n = D 1 Jn = 1 v n 1 + v 2 n 1 27

28 Computer Demos (1) Use discrete homotopy operator to compute F n = 1 f n for f n = +1 v n vn+u 2 n+1 +2 v n+1 +vn+1+u 2 n+3 v n+2 +1 v n (2) Compute density of rank 4 and flux for Toda system: = v n 1 v n v n = v n ( +1 ) (3) Compute density of rank 2 for Ablowitz-Ladik system: i = κu n ( ) (u n is the complex conjugate of ). This is an integrable discretization of the NLS equation: iu t + u xx + κu 2 u = 0 Take equation and its complex conjugate. Treat and v n = u n as dependent variables. Absorb i in t: = v n ( ) v n = (v n+1 2v n + v n 1 ) v n (v n+1 + v n 1 ). 28

29 Future Research Generalize continuous homotopy operator in multi-dimensions (x, y, z,...). Problem (in three dimensions): Given E = J = J (1) x + J (2) y + J (3) z. Find J = (J (1), J (2), J (3) ). Application: Compute densities and fluxes of multi-dimensional systems of PDEs (in t, x, y, z). Generalize discrete homotopy operator in multi-dimensions (n, m,...). 29

30 Higher Euler Operators Side by Side Continuous Case (for component u) L (0) u = u D x + D 2 x D 3 x + u x u 2x u 3x 2D x + 3D 2 x 4D 3 x + u x u 2x u 3x u 4x 3D x + 6D 2 x 10D 3 x + u 2x u 3x u 4x u 5x 4D x + 10D 2 x 20D 3 x + u 3x u 4x u 5x u 6x L (1) u = L (2) u = L (3) u = Discrete Case (for component ) L (0) = L (1) = L (2) = L (3) = (I + D 1 + D 2 + D 3 + ) (D 1 + 2D 2 + 3D 3 + 4D 4 + ) (D 2 + 3D 3 + 6D D 5 + ) (D 3 + 4D D D 6 + ) 30

31 Homotopy Operators Side by Side Continuous Case (for components u and v) with and H u = 1 0 {f 1(u)[λu] + f 2 (u)[λu]} dλ λ f 1 (u) = p 1 i=0 f 2 (u) = p 2 i=0 D i x[ul (i+1) u ] D i x[vl (i+1) v ] Discrete Case (for components and v n ) with and H un = 1 0 {f 1,n( )[λ ] + f 2,n ( )[λ ]} dλ λ f 1,n ( ) = p 1 (D I) i [ L (i+1) i=0 ] f 2,n ( ) = p 2 (D I) i [v n L (i+1) i=0 v n ] 31