Symbolic Computation of Conserved Densities and Symmetries of Nonlinear Evolution and Differential-Difference Equations WILLY HEREMAN

. Symbolic Computation of Conserved Densities and Symmetries of Nonlinear Evolution and Differential-Difference Equations WILLY HEREMAN Joint work with ÜNAL GÖKTAŞ and Grant Erdmann Graduate Seminar Department of Mathematical & Computer Sciences Colorado School of Mines November 5, 1997, 16:00 1

Purpose Design and implement algorithms to compute polynomial conservation laws and symmetries for nonlinear systems of evolution equations and differential-difference equations Motivation Conservation laws describe the conservation of fundamental physical quantities such as linear momentum and energy. Compare with constants of motion (first integrals) in mechanics For nonlinear PDEs and DDEs, the existence of a sufficiently large (in principal infinite) number of conservation laws and symmetries assures complete integrability Conservation laws provide a simple and efficient method to study both quantitative and qualitative properties of equations and their solutions, e.g. Hamiltonian structures Conservation laws can be used to test numerical integrators Symmetries can be used to build new integrable equations

PART I: Evolution Equations Conservation Laws for PDEs Consider a single nonlinear evolution equation u t = F(u, u x, u x,..., u nx ) or a system of N nonlinear evolution equations where u = [u 1,..., u N ] T and u t def = u t, u t = F(u, u x,..., u nx ) All components of u depend on x and t Conservation law: ρ is the density, J is the flux u(n) def = u nx = n u x n D t ρ + D x J = 0 Both are polynomial in u, u x, u x, u 3x,... Consequently if J vanishes at infinity P = + ρ dx = constant 3

The Euler Operator (calculus of variations) Useful tool to verify if an expression is a total derivative Theorem: If then f = f(x, y 1,..., y (n) 1,..., y N,..., y (n) N ) L y (f) 0 if and only if where Notations: f = D x g g = g(x, y 1,..., y (n 1) 1,..., y N,..., y (n 1) N ) y = [y 1,..., y N ] T (T for transpose) and Euler Operator: L y (f) = [L y1 (f),..., L yn (f)] T 0 = [0,..., 0] T L yi = d y i dx ( y ) + d dn i dx ( ) + + ( 1)n y i dx n( y ) (n) i 4

Example: Korteweg-de Vries (KdV) equation Conserved densities: u t + uu x + u 3x = 0 ρ 1 = u, (u) t + ( u + u x) x = 0 ρ = u, (u ) t + ( u3 3 + uu x u x ) x = 0 ρ 3 = u 3 3u x, ( u 3 3u ) x. t + 3 4 u4 6uu x +3u u x +3u x 6u x u 3x x = 0 ρ 6 = u 6 60u 3 u x 30u x 4 + 108u u x. + 70 7 u x 3 648 7 uu 3x + 16 7 u 4x,... long... Note: KdV equation and conservation laws are invariant under dilation (scaling) symmetry (x, t, u) (λx, λ 3 t, λ u) u and t carry the weights of and 3 derivatives with respect to x u x, t 3 x 3 5

Key Steps of the Algorithm 1. Determine weights (scaling properties) of variables & parameters. Construct the form of the density (building blocks) 3. Determine the unknown constant coefficients Example: KdV equation Compute the density of rank 6 u t + uu x + u 3x = 0 (i) Compute the weights by solving a linear system w(u) + w( ) = w(u) + w(x) = w(u) + 3w(x) t With w(x) = 1, w( t ) = 3, w(u) =. Thus, (x, t, u) (λx, λ 3 t, λ u) (ii) Take all the variables, except ( t ), with positive weight and list all possible powers of u, up to rank 6 : [u, u, u 3 ] Introduce x derivatives to complete the rank u has weight, introduce 4 x 4 u has weight 4, introduce x u 3 has weight 6, no derivatives needed 6

Apply the derivatives and remove terms that are total derivatives with respect to x or total derivative up to terms kept earlier in the list [u 4x ] [ ] empty list [u x, uu x ] [u x ] since uu x = (uu x ) x u x [u 3 ] [u 3 ] Combine the building blocks: ρ = c 1 u 3 + c u x (iii) Determine the coefficients c 1 and c 1. Compute D t ρ = 3c 1 u u t + c u x u xt. Replace u t by (uu x + u 3x ) and u xt by (uu x + u 3x ) x 3. Apply the Euler operator or integrate by parts D t ρ = [ 3 4 c 1u 4 (3c 1 c )uu x + 3c 1 u u x c u x + c u x u 3x ] x (3c 1 + c )u x 3 4. The non-integrable term must vanish. Thus, c 1 = 1 3 c. Set c = 3, hence, c 1 = 1 Result: Expression [...] yields ρ = u 3 3u x J = 3 4 u4 6uu x + 3u u x + 3u x 6u x u 3x 7

Example: Boussinesq equation u tt u x + 3uu x + 3u x + αu 4x = 0 with nonzero parameter α. Can be written as u t + v x = 0 v t + u x 3uu x αu 3x = 0 The terms u x and α u 3x are not uniform in rank Introduce auxiliary parameter β with weight. Replace the system by u t + v x = 0 v t + βu x 3uu x αu 3x = 0 The system is invariant under the scaling symmetry (x, t, u, v, β) (λx, λ t, λ u, λ 3 v, λ β) Hence w(u) =, w(β) =, w(v) = 3 and w( t ) = or u β x, Form ρ of rank 6 v 3 x 3, t x ρ = c 1 β u + c βu + c 3 u 3 + c 4 v + c 5 u x v + c 6 u x Compute the c i. At the end set β = 1 ρ = u u 3 + v + αu x which is no longer uniform in rank! 8

Application: A Class of Fifth-Order Evolution Equations u t + αu u x + βu x u x + γuu 3x + u 5x = 0 where α, β, γ are nonzero parameters, and u x Special cases: α = 30 β = 0 γ = 10 Lax α = 5 β = 5 γ = 5 Sawada Kotera α = 0 β = 5 γ = 10 Kaup Kupershmidt α = β = 6 γ = 3 Ito Under what conditions for the parameters α, β and γ does this equation admit a density of fixed rank? Rank : No condition Rank 4: Condition: β = γ ρ = u (Lax and Ito cases) ρ = u 9

Rank 6: Condition: 10α = β + 7βγ 3γ (Lax, SK, and KK cases) Rank 8: ρ = u 3 + 15 ( β + γ) u x 1. β = γ (Lax and Ito cases) ρ = u 4 6γ α uu x + 6 α u x. α = β 7βγ 4γ 45 (SK, KK and Ito cases) Rank 10: ρ = u 4 135 β + γ uu x + 675 (β + γ) u x Condition: and β = γ 10α = 3γ (Lax case) ρ = u 5 50 γ u u x + 100 γ uu x 500 7γ 3u 3x 10

What are the necessary conditions for the parameters α, β and γ for this equation to admit infinitely many polynomial conservation laws? If α = 3 10 γ and β = γ then there is a sequence (without gaps!) of conserved densities (Lax case) If α = 1 5 γ and β = γ then there is a sequence (with gaps!) of conserved densities (SK case) If α = 1 5 γ and β = 5 γ then there is a sequence (with gaps!) of conserved densities (KK case) If or α = β 7βγ + 4γ 45 β = γ then there is a conserved density of rank 8 Combine both conditions: α = γ 9 and β = γ (Ito case) 11

PART II: Differential-difference Equations Conservation Laws for DDEs Consider a system of DDEs, continuous in time, discretized in space u n = F(..., u n 1, u n, u n+1,...) u n and F are vector dynamical variables Conservation law: ρ n is the density, J n is the flux ρ n = J n J n+1 Both are polynomials in u n and its shifts d dt ( n ρ n) = n ρ n = n (J n J n+1 ) If J n is bounded for all n, with suitable boundary or periodicity conditions Definitions n ρ n = constant Define: D shift-down operator, U shift-up operator Dm = m n n 1 Um = m n n+1 For example, Du n+ v n = u n+1 v n 1 Uu n v n 1 = u n 1 v n 1

Compositions of D and U define an equivalence relation All shifted monomials are equivalent, e.g. Use equivalence criterion: u n 1 v n+1 u n+ v n+4 u n 3 v n 1 If two monomials, m 1 and m, are equivalent, m 1 m, then for some polynomial M n m 1 = m + [M n M n+1 ] For example, u n u n u n 1 u n+1 since u n u n = u n 1 u n+1 +[u n u n u n 1 u n+1 ] = u n 1 u n+1 +[M n M n+1 ] with M n = u n u n Main representative of an equivalence class; the monomial with label n on u (or v) For example, u n u n+ is the main representative of the class with elements u n 1 u n+1, u n+1 u n+3, etc. Use lexicographical ordering to resolve conflicts For example, u n v n+ (not u n v n ) is the main representative of the class with elements u n 3 v n 1, u n+ v n+4, etc. 13

Algorithm: Toda Lattice mÿ n = a[e (y n 1 y n ) e (y n y n+1 ) ] Take m = a = 1 (scale on t), and set u n = ẏ n, v n = e (y n y n+1 ) u n = v n 1 v n, v n = v n (u n u n+1 ) Simplest conservation law (by hand): u n = ρ n = v n 1 v n = J n J n+1 with J n = v n 1 First pair: ρ (1) n = u n, J n (1) = v n 1 Second pair: ρ () n = 1 u n + v n, J () n = u n v n 1 Key observation: The DDE and the two conservation laws, ρ n = J n J n+1, with ρ (1) n = u n, J n (1) ρ () n = 1 u n + v n, J n () are invariant under the scaling symmetry = v n 1 = u n v n 1 (t, u n, v n ) (λt, λ 1 u n, λ v n ) Dimensional analysis: u n corresponds to one derivative with respect to t For short, u n d dt, and similarly, v n d dt 14

Our algorithm exploits this symmetry to find conserved densities: 1. Determining the weights. Constructing the form of density 3. Determining the unknown coefficients Step 1: Determine the weights The weight, w, of a variable is equal to the number of derivatives with respect to t the variable carries. Weights are positive, rational, and independent of n. Requiring uniformity in rank for each equation u n = v n 1 v n, v n = v n (u n u n+1 ) allows one to compute the weights of the dependent variables. Solve the linear system w(u n ) + w( d dt ) = w(v n) w(v n ) + w( d dt ) = w(v n) + w(u n ) Set w( d dt ) = 1, then w(u n) = 1, and w(v n ) = which is consistent with the scaling symmetry (t, u n, v n ) (λt, λ 1 u n, λ v n ) 15

Step : Construct the form of the density The rank of a monomial is the total weight of the monomial. For example, compute the form of the density of rank 3 List all monomials in u n and v n of rank 3 or less: G ={u n 3, u n, u n v n, u n, v n } Next, for each monomial in G, introduce enough t-derivatives, so that each term exactly has weight 3. Use the DDE to remove u n and v n d 0 d 0 dt 0(u n 3 ) = u 3 n, dt 0(u nv n ) = u n v n, d dt (u n d ) = u n v n 1 u n v n, dt (v n) = u n v n u n+1 v n, d dt (u n) = u n 1 v n 1 u n v n 1 u n v n + u n+1 v n Gather the resulting terms in a set H = {u n 3, u n v n 1, u n v n, u n 1 v n 1, u n+1 v n } Identify members of the same equivalence classes and replace them by the main representatives. For example, since u n v n 1 u n+1 v n both are replaced by u n v n 1. H is replaced by I = {u n 3, u n v n 1, u n v n } containing the building blocks of the density. Form a linear combination of the monomials in I with constant coefficients c i ρ n = c 1 u n 3 + c u n v n 1 + c 3 u n v n 16

Step 3: Determine the unknown coefficients Require that the conservation law, ρ n = J n J n+1, holds Compute ρ n and use the equations to remove u n and v n. Group the terms ρ n = (3c 1 c )u n v n 1 + (c 3 3c 1 )u n v n + (c 3 c )v n 1 v n +c u n 1 u n v n 1 + c v n 1 c 3 u n u n+1 v n c 3 v n Use the equivalence criterion to modify ρ n Replace u n 1 u n v n 1 by u n u n+1 v n + [u n 1 u n v n 1 u n u n+1 v n ]. The goal is to introduce the main representatives. Therefore, ρ n = (3c 1 c )u n v n 1 + (c 3 3c 1 )u n v n +(c 3 c )v n v n+1 + [(c 3 c )v n 1 v n (c 3 c )v n v n+1 ] +c u n u n+1 v n + [c u n 1 u n v n 1 c u n u n+1 v n ] +c v n + [c v n 1 c v n ] c 3 u n u n+1 v n c 3 v n Group the terms outside of the square brackets and move the pairs inside the square brackets to the bottom. Rearrange the latter terms so that they match the pattern [J n J n+1 ]. Hence, ρ n = (3c 1 c )u n v n 1 + (c 3 3c 1 )u n v n +(c 3 c )v n v n+1 + (c c 3 )u n u n+1 v n + (c c 3 )v n +[{(c 3 c )v n 1 v n + c u n 1 u n v n 1 + c v n 1 } {(c 3 c )v n v n+1 + c u n u n+1 v n + c v n }] 17

The terms inside the square brackets determine: J n = (c 3 c )v n 1 v n + c u n 1 u n v n 1 + c v n 1 The terms outside the square brackets must vanish, thus S = {3c 1 c = 0, c 3 3c 1 = 0, c c 3 = 0} The solution is 3c 1 = c = c 3. Choose c 1 = 1 3, thus c = c 3 = 1 ρ n = 1 3 u n 3 + u n (v n 1 + v n ), J n = u n 1 u n v n 1 + v n 1 Analogously, conserved densities of rank 5: ρ (1) n = u n, ρ () n = 1 u n + v n ρ (3) n = 1 3 u n 3 + u n (v n 1 + v n ) ρ (4) n = 1 4 u n 4 + u n (v n 1 + v n ) + u n u n+1 v n + 1 v n + v n v n+1 ρ (5) n = 1 5 u n 5 + u n 3 (v n 1 + v n ) + u n u n+1 v n (u n + u n+1 ) +u n v n 1 (v n + v n 1 + v n ) + u n v n (v n 1 + v n + v n+1 ) 18

Application: A parameterized Toda lattice u n = α v n 1 v n, v n = v n (β u n u n+1 ) α and β are nonzero parameters. The system is integrable if α = β = 1 Compute the compatibility conditions for α and β, so that there is a conserved densities of, say, rank 3. In this case, we have S: {3αc 1 c = 0, βc 3 3c 1 = 0, αc 3 c = 0, βc c 3 = 0, αc c 3 = 0} A non-trivial solution 3c 1 = c = c 3 will exist iff α = β = 1 Analogously, the parameterized Toda lattice has density ρ (1) n = u n of rank 1 if α = 1 and density ρ () n = β u n + v n of rank if α β = 1 Only when α = β = 1 will the parameterized system have conserved densities of rank 3 19

Example: Nonlinear Schrödinger (NLS) equation Ablowitz and Ladik discretization of the NLS equation: i u n = u n+1 u n + u n 1 + u nu n (u n+1 + u n 1 ) where u n is the complex conjugate of u n. Treat u n and v n = u n as independent variables, add the complex conjugate equation, and absorb i in the scale on t u n = u n+1 u n + u n 1 + u n v n (u n+1 + u n 1 ) v n = (v n+1 v n + v n 1 ) u n v n (v n+1 + v n 1 ) Since v n = u n, w(v n ) = w(u n ). No uniformity in rank! Circumvent this problem by introducing an auxiliary parameter α with weight, u n = α(u n+1 u n + u n 1 ) + u n v n (u n+1 + u n 1 ) v n = α(v n+1 v n + v n 1 ) u n v n (v n+1 + v n 1 ). Uniformity in rank requires that w(u n ) + 1 = w(α) + w(u n ) = w(u n ) + w(v n ) = 3w(u n ) w(v n ) + 1 = w(α) + w(v n ) = w(v n ) + w(u n ) = 3w(v n ) which yields w(u n ) = w(v n ) = 1, w(α) = 1 0

Uniformity in rank is essential for the first two steps of the algorithm. After Step, you can already set α = 1. The computations now proceed as in the previous examples Conserved densities: ρ (1) n = c 1 u n v n 1 + c u n v n+1 ρ () n = c 1 ( 1 u n v n 1 + u n u n+1 v n 1 v n + u n v n ) + c ( 1 u n v n+1 + u n u n+1 v n+1 v n+ + u n v n+ ) ρ (3) n = c 1 [ 1 3 u n 3 v n 1 3 +u n u n+1 v n 1 v n (u n v n 1 + u n+1 v n + u n+ v n+1 ) +u n v n 1 (u n v n + u n+1 v n 1 ) +u n v n (u n+1 v n + u n+ v n 1 ) + u n v n 3 ] + c [ 1 3 u n 3 v n+1 3 +u n u n+1 v n+1 v n+ (u n v n+1 + u n+1 v n+ + u n+ v n+3 ) +u n v n+ (u n v n+1 + u n+1 v n+ ) +u n v n+3 (u n+1 v n+1 + u n+ v n+ ) + u n v n+3 ] 1

Scope and Limitations of Algorithm & Software Systems of PDEs or DDEs must be polynomial in dependent variables Only one space variable (continuous x for PDEs, discrete n for DDEs) is allowed No terms should explicitly depend on x and t for PDEs, or n for DDEs Program only computes polynomial conserved densities; only polynomials in the dependent variables and their derivatives; no explicit dependencies on x and t for PDEs (or n for DDEs) No limit on the number of PDEs or DDEs. In practice: time and memory constraints Input systems may have (nonzero) parameters. Program computes the compatibility conditions for parameters such that densities (of a given rank) exist Systems can also have parameters with (unknown) weight. Allows one to test PDEs or DDEs of non-uniform rank For systems where one or more of the weights are free, the program prompts the user to enter values for the free weights Negative weights are not allowed Fractional weights and ranks are permitted Form of ρ can be given in the data file (testing purposes)

Conserved Densities Software Conserved densities programs CONSD and SYMCD by Ito and Kako (Reduce, 1985, 1994 & 1996). Conserved densities in DELiA by Bocharov (Pascal, 1990) Conserved densities and formal symmetries FS by Gerdt and Zharkov (Reduce, 1993) Formal symmetry approach by Mikhailov and Yamilov (MuMath, 1990) Recursion operators and symmetries by Roelofs, Sanders and Wang (Reduce 1994, Maple 1995, Form 1995-present) Conserved densities condens.m by Hereman and Göktaş (Mathematica, 1996) Conservation laws, based on CRACK by Wolf (Reduce, 1995) Conservation laws by Hickman (Maple, 1996) Conserved densities by Ahner et al. (Mathematica, 1995). Project halted. Conserved densities diffdens.m by Göktaş and Hereman (Mathematica, 1997) 3

PART III: Symmetries of PDEs and DDEs Symmetries of PDEs Consider the system of PDEs space variable x, time variable t u t = F(x, t, u, u x, u x,..., u mx ) dynamical variables u = (u 1, u,..., u n ) and F = (F 1, F,..., F n ) Definition of Symmetry Vector function G(x, t, u, u x, u x,...) is a symmetry if and only if the PDE is invariant for the replacement u u + ɛg within order ɛ. Hence (u + ɛg) = F(u + ɛg) t must hold up to order ɛ, or G t = F (u)[g] where F is the Gateaux derivative of F F (u)[g] = ɛ F(u + ɛg) ɛ=0 Equivalently, G is a symmetry if the compatibility condition τ F(x, t, u, u x, u x,..., u nx ) = t G(x, t, u, u x, u x,...) 4

is satisfied, where τ is the new time variable such that u τ = G(x, t, u, u x, u x,...) Example: The KdV Equation has infinitely many symmetries: u t = 6uu x + u 3x G (1) = u x G () = 6uu x + u 3x G (3) = 30u u x + 0u x u x + 10uu 3x + u 5x G (4) = 140u 3 u x + 70u 3 x + 80uu x u x + 70u u 3x + 70u x u 3x + 4u x u 4x G (5) +14uu 5x + u 7x = 630u 4 u x + 160uu x 3 + 50u u x u x + 130u x u x + 40u 3 u 3x +966u x u 3x + 160uu x u 3x + 756uu x u 4x + 5u 3x u 4x +16u u 5x + 168u x u 5x + 7u x u 6x + 18uu 7x + u 9x The recursion operator connecting them is: Algorithm (KdV equation) R = D + 4u + u x D 1 Use the dilation symmetry (t, x, u) (λ 3 t, λ 1 x, λ u) λ is arbitrary parameter. Hence, u and x t 3 x 3 Step 1: Determine the weights of variables We choose w(x) = 1, then w(u) = and w(t) = 3 Step : Construct the form of the symmetry Compute the form of the symmetry with rank 7 5

List all monomials in u of rank 7 or less L = {1, u, u, u 3 } Introduce x-derivatives so that each term has weight 7 x (u3 ) = 3u 3 u x, ) = 6u x 3(u x u x +uu 3x, Gather the non-zero resulting terms in a set 5 x 5(u) = u 5x, 7 x 7(1) = 0 R = {u u x, u x u x, uu 3x, u 5x } which contains the building blocks of the symmetry Linear combination of the monomials in R determines the symmetry G = c 1 u u x + c u x u x + c 3 uu 3x + c 4 u 5x Step 3: Determine the unknown coefficients in the symmetry Requiring that τ F(x, t, u, u x, u x,..., u nx ) = t G(x, t, u, u x, u x,...) holds. Compute G t and F τ Use the PDE, to replace u t, u tx.u txx,... Use to replace u τ, u τx, u τxx,... u t = F u τ = G(x, t, u, u x, u x,...) 6

After grouping the terms F τ G t = (1c 1 18c )u xu x + (6c 1 18c 3 )uu x + (6c 1 18c 3 )uu x u 3x + This yields (3c 60c 4 )u 3x + (3c + 3c 3 90c 4 )u x u 4x + (3c 3 30c 4 )u x u 5x 0 S = {1c 1 18c = 0, 6c 1 18c 3 = 0, 3c 60c 4 = 0, 3c + 3c 3 90c 4 = 0, 3c 3 30c 4 = 0} Choosing c 4 = 1, the solution is c 1 = 30, c = 0, c 3 = 10 Hence G = 30u u x + 0u x u x + 10uu 3x + u 5x which leads to Lax equation (in the KdV hierarchy) u t + 30u u x + 0u x u x + 10uu 3x + u 5x x t Dependent Symmetries Algorithm can be used provided the degree in x or t is given Compute the symmetry of the KdV equation with rank (linear in x or t) Build list of monomials in u, x and t of rank or less L = {1, u, x, xu, t, tu, tu } Introduce the correct number of x-derivatives to make each term weight x (xu) = u + xu 3 x, x (tu ) = tuu x, x 3(tu) = tu 3x, 3 5 x(1) = x3(x) = x 5(t) = 0 7

Gather the non-zero resulting terms R = {u, xu x, tuu x, tu 3x } Linearly combine the monomials to obtain G = c 1 u + c xu x + c 3 tuu x + c 4 tu 3x Determine the coefficients c 1 through c 4 Compute G t and F τ and remove all t and τ derivatives (as before) Group the terms F τ G t = (6c 1 + 6c c 3 )uu x + (3c 3 18c 4 )tu x + (3c c 4 )u 3x + (3c 3 18c 4 )tu x u 3x 0 This yields S = {6c 1 + 6c c 3 = 0, 3c 3 18c 4 = 0, 3c c 4 = 0} The solution is c 1 = 3, c = 1 3, c 3 = 6, c 4 = 1 Hence G = 3 u + 1 3 xu x + 6tuu x + tu 3x These are two x t dependent symmetries (of rank 0 and ) G = 1 + 6tu x and G = u + xu x + t(6uu x + u 3x ) 8

Symmetries of DDEs Consider a system of DDEs (continuous in time, discretized in space) u n = F(..., u n 1, u n, u n+1,...) u n and F have any number of components bf Definition of Symmetry A vector function G(..., u n 1, u n, u n+1,...) is called a symmetry of the DDE if the infinitesimal transformation u u + ɛg(..., u n 1, u n, u n+1,...) leaves the DDE invariant within order ɛ Equivalently d dτ F(..., u n 1, u n, u n+1,...) = d dt G(..., u n 1, u n, u n+1,...) where τ is the new time variable such that d dτ u = G(..., u n 1, u n, u n+1,...) Algorithm Consider the one-dimensional Toda lattice Change the variables ÿ n = exp (y n 1 y n ) exp (y n y n+1 ) u n = ẏ n, v n = exp (y n y n+1 ) to write the lattice in algebraic form u n = v n 1 v n, v n = v n (u n u n+1 ) 9

This system is invariant under the scaling symmetry (t, u n, v n ) (λ 1 t, λu n, λ v n ) λ is an arbitrary parameter. Hence, u n d dt and v n d dt Step 1: Determine the weights of variables Set w(t) = 1. Then w(u n ) = 1, and w(v n ) = Step : Construct the form of the symmetry Compute the form of the symmetry of ranks {3, 4} List all monomials in u n and v n of rank 3 or less and of rank 4 or less L 1 = {u n 3, u n, u n v n, u n, v n } L = {u n 4, u n 3, u n v n, u n, u n v n, u n, v n, v n } For each monomial in both lists, introduce the adjusting number of t-derivatives so that each term exactly has weight 3 and 4, resp. For the monomials in L 1 d 0 dt 0(u n 3 ) = u n 3, d 0 dt 0(u nv n ) = u n v n, d dt (u n ) = u n u n = u n v n 1 u n v n, d dt (v n) = v n = u n v n u n+1 v n, d dt (u n) = d dt ( u n) = d dt (v n 1 v n ) = u n 1 v n 1 u n v n 1 u n v n + u n+1 v n Gather the resulting terms in a set R 1 = {u n 3, u n 1 v n 1, u n v n 1, u n v n, u n+1 v n } 30

Similarly R = {u 4 n, u n 1 v n 1, u n 1 u n v n 1, u n v n 1, v n v n 1, v n 1, u n v n, u n u n+1 v n, u n+1 v n, v n 1 v n, v n, v n v n+1 } Linear combination of the monomials in R 1 and R determines G 1 = c 1 u n 3 + c u n 1 v n 1 + c 3 u n v n 1 + c 4 u n v n + c 5 u n+1 v n G = c 6 u n 4 + c 7 u n 1 v n 1 + c 8 u n 1 u n v n 1 + c 9 u n v n 1 + c 10 v n v n 1 + c 11 v n 1 + c 1 u n v n + c 13 u n u n+1 v n + c 14 u n+1 v n + c 15 v n 1 v n + c 16 v n + c 17 v n v n+1 Step 3: Determine the unknown coefficients in the symmetry Requiring that F τ = G t holds Compute d dt G 1, d dt G d, dτ F 1 and d dτ F and remove all u n, v n, d dτ u n, d dτ v n Require that which gives d dτ F 1 d dt G 1 0, d dτ F d dt G 0 c 1 = c 6 = c 7 = c 8 = c 9 = c 10 = c 11 = c 13 = c 16 = 0, c = c 3 = c 4 = c 5 = c 1 = c 14 = c 15 = c 17 With c 17 = 1 the symmetry is G 1 = u n v n u n 1 v n 1 + u n+1 v n u n v n 1 G = u n+1v n u nv n + v n v n+1 v n 1 v n 31

Conclusions and Further Research Two Mathematica programs are available: condens.m for evolution equations (PDEs) diffdens.m for differential-difference equations (DDEs) Usefulness Testing models for integrability Study of classes of nonlinear PDEs or DDEs Comparison with other programs Parameter analysis is possible Not restricted to uniform rank equations Not restricted to evolution equations provided that one can write the equation(s) as a system of evolution equations Future work Generalization towards broader classes of equations (e.g. u xt ) Generalization towards more space variables (e.g. KP equation) Conservation laws with time and space dependent coefficients Conservation laws with n dependent coefficients 3

Exploit other symmetries in the hope to find conserved densities of non-polynomial form Constants of motion for dynamical systems (e.g. Lorenz and Hénon-Heiles systems) Research supported in part by NSF under Grant CCR-96541 In collaboration with Ünal Göktaş and Grant Erdmann Papers submitted to: J. Symb. Comp., Phys. Lett. A and Physica D Software: available via FTP, ftp site mines.edu in subdirectories pub/papers/math cs dept/software/condens pub/papers/math cs dept/software/diffdens or via the Internet URL: http://www.mines.edu/fs home/whereman/ 33

Additional Examples Nonlinear Schrödinger Equation iq t q x + q q = 0 Program can not handle complex equations Replace by where q = u + iv u t v x + v(u + v ) = 0 v t + u x u(u + v ) = 0 Scaling properties u v x, First seven conserved densities: t x ρ 1 = u + v ρ = vu x ρ 3 = u 4 + u v + v 4 + u x + v x ρ 4 = u vu x + 1 3 v3 u x 1 6 vu 3x 34

ρ 5 = 1 u6 3 u4 v 3 u v 4 1 v6 5 u u x 1 v u x 3 u v x 5 v v x + uv u x 1 4 u x 1 4 v x ρ 6 = 3 4 u4 vu x 1 u v 3 u x 3 0 v5 u x + 1 4 vu x 3 1 4 vu xv x +uvu x u x + 1 4 u vu 3x + 1 1 v3 u 3x 1 40 vu 5x ρ 7 = 5 4 u8 + 5u 6 v + 15 u4 v 4 + 5u v 6 + 5 4 v8 + 35 u4 u x 5u v u x + 5 v4 u x 7 4 u x 4 + 15 u4 v x + 5u v v x + 35 v4 v x 5 u x v x 7 4 v x 4 10u 3 v u x 5uv 4 u x 5uv x u x + 7 u u x + 1 v u x + 5 u v x + 7 v v x v u x u 3x + 1 4 u 3x + 1 4 v 3x + uv u 4x 35

The Ito system u t u 3x 6uu x vv x = 0 v t u x v uv x = 0 u x, v x ρ 1 = c 1 u + c v ρ = u + v ρ 3 = u 3 + uv u x ρ 4 = 5u 4 + 6u v + v 4 10uu x + v u x + u x ρ 5 = 14u 5 + 0u 3 v + 6uv 4 70u u x + 10v u x 4v v x + 0uv u x + 14uu x u 3x + v u 4x and more conservation laws 36

The dispersiveless long-wave system u t + vu x + uv x = 0 v t + u x + vv x = 0 u v w(v) is free choose u x and v x ρ 1 = v ρ = u ρ 3 = uv ρ 4 = u + uv ρ 5 = 3u v + uv 3 ρ 6 = 1 3 u3 + u v + 1 6 uv4 ρ 7 = u 3 v + u v 3 + 1 10 uv5 ρ 8 = 1 3 u4 + u 3 v + u v 4 + 1 15 uv6 and more Always the same set irrespective the choice of weights 37

A generalized Schamel equation n u t + (n + 1)(n + )unu x + u 3x = 0 where n is a positive integer ρ 1 = u, ρ = u ρ 3 = 1 u x n u+ n For n 1, no further conservation laws 38

Three-Component Korteweg-de Vries Equation u t 6uu x + vv x + ww x u 3x = 0 v t vu x uv x = 0 w t wu x uw x = 0 Scaling properties First five densities: u v w x, ρ 1 = c 1 u + c v + c 3 w ρ = u v w ρ 3 = u 3 + uv + uw + u x t 3 x 3 ρ 4 = 5 u4 + 3u v 1 v4 + 3u w v w 1 w4 +5uu x + v u x + w u x 1 u x ρ 5 = 7 10 u5 + u 3 v 3 10 uv4 + u 3 w 3 5 uv w 3 10 uw4 + 7 u u x + 1 v u x + 1 w u x + 1 5 v v x 1 5 w v x + 1 5 w w x + uv u x + uw u x 7 10 uu x 1 5 vw v x + 1 0 u 3x + 1 10 v u 4x + 1 10 w u 4x 39

The Deconinck-Meuris-Verheest equation Consider the modified vector derivative NLS equation: B t + x (B B ) + αb 0 B 0 B x + e x B x = 0 Replace the vector equation by u t + ( u(u + v ) + βu v x )x = 0 v t + ( v(u + v ) + u x )x = 0 u and v denote the components of B parallel and perpendicular to B 0 and β = αb 0 u x, First 6 conserved densities v x, β x ρ 1 = c 1 u + c v ρ = u + v ρ 3 = 1 (u + v ) uv x + u x v + βu ρ 4 = 1 4 (u + v ) 3 + 1 (u x + v x ) u 3 v x + v 3 u x + β 4 (u4 v 4 ) 40

ρ 5 = 1 4 (u + v ) 4 5 (u xv x u x v x ) + 4 5 (uu x + vv x ) + 6 5 (u + v )(u x + v x ) (u + v ) (uv x u x v) + β 5 (u x 4u 3 v x + u 6 + 3u 4 v v 6 ) + β 5 u4 ρ 6 = 7 16 (u + v ) 5 + 1 (u x + v x) 5 (u + v )(u x v x u x v x ) + 5(u + v )(uu x + vv x ) + 15 4 (u + v ) (u x + v x ) 35 16 (u + v ) 3 (uv x u x v) + β 8 (5u8 + 10u 6 v 10u v 6 5v 8 + 0u u x 1u 5 v x + 60uv 4 v x 0v v x ) + β 4 (u6 + v 6 ) 41