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

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 http://www.mines.edu/fs home/whereman/ whereman@mines.edu 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-9901929 1

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

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 2 + +1 +2 v n+1 + vn+1 2 + +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. Result: F n = vn 2 + +1 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

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

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

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

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

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. 372. Proof is given in terms of differential forms. Work of Henri Poincaré (1854-1912), George de Rham (1950), and Ian Anderson & Tom Duchamp (1980). Proofs based on calculus: Deconinck and Hereman. 8

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

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 v2 + 1 2 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 2 +20 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

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

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

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 1 2 2 v2 + 1 2 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

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

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 3 1 2 u2 x ρ n = 1 3 u3 n+ +1 ( ++1 ++2 ) Symmetries G=u x, G=6uu x + u 3x G = +1 ( + +1 + +2 ) G=30u 2 u x + 20u x u 2x 1 ( 2 + 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

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

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 +1 +2 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

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 ( ) +1 +2 +m0 = ( m0 D k ) k=0 L (0) = (I + D 1 + D 2 + + D m 0 ) where m 0 is the highest forward shift (in f n ). 18

Example: Apply the discrete Euler operator to f n ( ) = +1 v n v 2 n++1 +2 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 1 ++1 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 ] + [ +1 + 2v n ] + [+1 ] 0 19

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 3 + + D m 0 ) (D 1 + 2D 2 + 3D 3 + 4D 4 + + m 1 D m 1 ) (D 2 + 3D 3 + 6D 4 + 10D 5 + + 1 2 m 2(m 2 1) D m 2 ) (D 3 + 4D 4 + 10D 5 + 20D 6 m 3 + + D m 3 ) 3 Similar formulae for L (i) v n. 20

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

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 4 + 10D 5 + ) (D 3 + 4D 4 + 10D 5 + 20D 6 + ) 22

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

Example: Apply the discrete homotopy operator to f n ( ) = +1 v n v 2 n++1 +2 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 1 ++1 v n +2v n 1 1 v n 1 + +1 v n +2 v n 1 1 1 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 )] 0 +1 +2v n +2+1 +1 v n +2v 2 n+2+1 v n 1 +1 +2 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 2 + +1 v n + +1 v n + +2 v n+1 24

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

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 1 0. 26

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 + +1 +2 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 1 ++1 v n 1 v n 1 + +1 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)) 0 +1 +2v 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

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 = +1 2 + 1 + κu n (+1 + 1 ) (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: = +1 2 + 1 + v n (+1 + 1 ) v n = (v n+1 2v n + v n 1 ) v n (v n+1 + v n 1 ). 28

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