SYMBOLIC SOFTWARE FOR SOLITON THEORY: INTEGRABILITY, SYMMETRIES CONSERVATION LAWS AND EXACT SOLUTIONS. Willy Hereman

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1 . SYMBOLIC SOFTWARE FOR SOLITON THEORY: INTEGRABILITY, SYMMETRIES CONSERVATION LAWS AND EXACT SOLUTIONS Willy Hereman Dept. of Mathematical and Computer Sciences Colorado School of Mines Golden, Colorado Kruskal Fest Symposium in Applied Mathematics: Nonlinear Waves, Dynamics, Asymtotic Analysis and Physical Applications Boulder, Colorado August 3-6, 1995

2 I. INTRODUCTION Symbolic Software Painlevé test for systems of ODEs and PDEs (Macsyma & Mathematica) Conservation laws of systems of evolution equations (Mathematica) Solitons via Hirota s method (Macsyma & Mathematica) Lie symmetries for systems of ODEs and PDEs (Macsyma) Purpose of the programs Study of integrability of nonlinear PDEs Exact solutions as bench mark for numerical algorithms Classification of nonlinear PDEs Lie symmetries solutions via reductions

3 Collaborators Ünal Göktaş, Chris Elmer, Wuning Zhuang (MS students) Ameina Nuseir (Ph.D student) Mark Coffey (CU-Boulder) Tony Miller & Tracy Otto (BS students)

4 II. SYMBOLIC SOFTWARE Program 1 Macsyma Painlevé Integrability Test for Systems of ODEs and PDEs Integrability of (a systems of) ODEs or PDEs requires that the only movable singularities in its solution are poles Definition: A single equation or system has the Painlevé Property if its solution in the complex plane has no worse singularities than movable poles Aim: Verify whether or not the system of equations satisfies the necessary criteria to have the Painlevé Property For simplicity, consider the case of a single PDE The solution f expressed as a Laurent series f = g α should only have movable poles k=0 u k g k

5 Steps of the Painlevé Test Step 1: 1. Substitute the leading order term into the given equation f u 0 g α 2. Determine the integer α < 0 by balancing the most singular terms in g 3. Calculate u 0 Step 2: 1. Substitute the generic terms f u 0 g α + u r g α+r into the equation, retaining its most singular terms 2. Require that u r is arbitrary 3. Determine the corresponding values of r > 0 called resonances

6 Step 3: 1. Substitute the truncated expansion f = g α R k=0 u k g k into the complete equation (R represents the largest resonance) 2. Determine u k unambiguously at the non-resonance levels 3. Check whether or not the compatibility condition is satisfied at resonance levels An equation or system has the Painlevé Property and is conjectured to be integrable if: 1. Step 1 thru 3 can be carried out consistently with α < 0 and with positive resonances 2. The compatibility conditions are identically satisfied for all resonances The above algorithm does not detect essential singularities

7 Painlevé Integrability Test Painlevé test for 3rd order equations by Hajee (Reduce, 1982) Painlevé program (parts) by Hlavatý (Reduce, 1986) ODE Painlevé by Winternitz & Rand (Macsyma, 1986) PDE Painlevé by Hereman & Van den Bulck (Macsyma, 1987) Painlevé test by Conte & Musette (AMP, 1988) Painlevé analysis by Renner (Reduce, 1992) Painlevé test for systems of ODEs and PDEs by Hereman, Elmer and Göktaş (Macsyma, , under development) Painlevé test for single ODEs and PDEs by Hereman, Miller and Otto (Mathematica, 1995, under development) Painlevé test for systems of ODEs and PDEs by Hereman and Miller (Mathematica, , planned)

8 Painlevé Test of Systems System of ODEs due to Akhiezer u xx + 2u 3 2au + 2bv = 0 v xx + 2v 3 2av + 2bu = 0 u(x) = g α 1 (x) k=0 v(x) = g α 2 (x) Leading orders: α 1 = α 2 = 1 Resonances: r = 1, r = 4 Coefficients: k=0 u k g k (x) v k g k (x) u 0 = v 0 = i u 1 = v 1 = 0 u 2 = v 2 = i(b a)/3 u 3 = v 3 = 0 At level 4: compatibility condition is satisfied Coefficients u 4 and v 4 are arbitrary and independent The system passes the Painlevé test

9 Carleman system (simplified version) u t + u x c uv = 0 v t v x + c uv = 0 u(x, t) = g α 1 (x, t) k=0 v(x, t) = g α 2 (x, t) k=0 Leading orders: α 1 = α 2 = 1 Resonances: r = 1, r = 1 Coefficients: u k (x, t) g k (x, t) v k (x, t) g k (x, t) u 0 = 1 c (g t g x ), v 0 = 1 c (g t + g x ) At level 1: compatibility condition is satisfied However u 1 and v 1 dependent on each other u 1 = g xx g tt + c v 1 (g t g x ) c (g t + g x ) The system passes the Painlevé test

10 Coupled KdV Equations (Hirota-Satsuma system) u t 3uu x + 6vv x 1 2 u xxx = 0 v t + v xxx + 3uv x = 0 Use Kruskal simplification, g(x, t) = x h(t), in u(x, t) = g α 1 (x, t) k=0 v(x, t) = g α 2 (x, t) k=0 u k (x, t) g k (x, t) v k (x, t) g k (x, t) Leading orders: First Branch α 1 = α 2 = 2 Resonances: r = 2, 1, 3, 4, 6, 8 Coefficients: u 0 = 4, v 0 = 2 u 1 = v 1 = 0 u 2 = h t /3, v 2 = 2h t /3 u 5 = h tt + 30v 3 h t, v 5 = h tt 12v 3 h t u 7 = v 4t + 12v 3 v 4, v 7 = 4h th tt + v 3 (8h 2 t 84v 4 ) + 21v 4t At level 3: compatibility condition is satisfied Here u 3 = v 3, one function is arbitary

11 At level 4: compatibility condition is satisfied Here u 4 = 2v 4, one function is arbitrary At level 6: compatibility condition is satisfied Here u 6 = v 2 3t + 24v 6 3v 3, 12 one function is arbitrary At level 8: compatibility condition is satisfied Here u 8 = h ttt + 6v 3 h tt + (12v 3t + 198v3)h 2 2 t 3024v 8 756v 4, 756 one function is arbitrary

12 Leading orders: Second Branch α 1 = 2, α 2 = 1 Resonances: r = 1, 0, 1, 4, 5, 6 Coefficients: u 0 = 2, v 0 free u 1 = 0, v 1 free u 2 = (2h t + 3v 2 0)/6, v 2 = (4v 0 h t + 3v 3 0)/12 u 3 = v 0 v 1, v 3 = v 0t + 3v 2 0v 1 12 At level 0: compatibility condition is satisfied Coefficient v 0 is arbitary At level 1: compatibility condition is satisfied Coefficient v 1 is arbitary At level 4: compatibility condition is satisfied Here u 4 = 16v 0h 2 t + 24v0h 3 t 24v 1t + 9v u 4 v 0, 288 one function is arbitrary

13 At level 5: compatibility condition is satisfied Here u 5 = 2h tt 12v 0 v 1 h t + 3v 0 v 0t 9v 0 3 v 1 18 one function is arbitrary At level 6: compatibility condition is satisfied Here u 6 = ( 64v 0h 3 t 144v 3 0h 2 t +(96v 1t 108v u 4 v 0 )h t +24v 0 2 v 1t 16v 0tt + 288v 3 0v v u 4 v u 6 v 0 ) one function is arbitrary The system passes the Painlevé test

14 Program 2 Mathematica Conserved Densities Purpose Compute polynomial-type conservation laws of single evolution equations and systems of evolution equations For simplicity, consider a single evolution equation u t = F(u, u x, u xx,..., u nx ) Conservation law is of the form ρ t + J x = 0 both ρ(u, u x, u 2x,..., u nx ) and J(u, u x, u 2x,..., u nx ) are polynomials in their arguments Consequently P = + provided J vanishes at infinity ρ dx = constant

15 Conservation laws describe the conservation of fundamental physical quantities such as mass, linear momentum, total energy (compare with constants of motion in mechanics) For nonlinear PDEs, the existence of a sufficiently large (in principal infinite) number of conservation laws assures complete integrability Tool to test numerical integrators for PDEs Example Consider the KdV equation, u t + uu x + u 3x = 0 Conserved densities: ρ 1 = u ρ 2 = u 2 ρ 3 = u 3 3u 2 x. ρ 6 = u 6 60u 3 u 2 x 30u 4 x + 108u 2 u 2 2x u3 2x uu2 3x u2 4x. Integrable equations have many conservation laws

16 Algorithm and Implementation Consider the scaling (weights) of the KdV u 2 x 2, Compute building blocks of ρ 3 (i) Start with building block u 3 t 3 x 3 Divide by u and differentiate twice (u 2 ) 2x Produces the list of terms [u 2 x, uu 2x ] [u 2 x] Second list: remove terms that are total derivative with respect to x or total derivative up to terms earlier in the list Divide by u 2 and differentiate twice (u) 4x Produces the list: [u 4x ] [ ] [ ] is the empty list

17 Gather the terms: ρ 3 = u 3 + c[1]u 2 x where the constant c 1 must be determined (ii) Compute ρ 3 t = 3u2 u t + 2c 1 u x u xt Replace u t by (uu x + u xxx ) and u xt by (uu x + u xxx ) x (iii) Integrate the result with respect to x Carry out all integrations by parts ρ 3 t = [ 3 4 u4 + (c 1 3)uu 2 x + 3u 2 u xx c 1 u 2 xx+ 2c 1 u x u xxx ] x (c 1 + 3)u 3 x The last non-integrable term must vanish Thus, c 1 = 3 Result: (iv) Expression [...] yields ρ 3 = u 3 3u 2 x J 3 = 3 4 u4 6uu 2 x + 3u 2 u xx + 3u 2 xx 6u x u xxx

18 Computer building blocks of ρ 6 (i) Start with u 6 Divide by u and differentiate twice (u 5 ) 2x produces the list of terms [u 3 u 2 x, u 4 u 2x ] [u 3 u 2 x] Next, divide u 6 by u 2, and compute (u 4 ) 4x Corresponding list: [u 4 x, uu 2 xu 2x, u 2 u 2 2x, u 2 u x u 3x, u 3 u 4x ] [u 4 x, u 2 u 2 2x] Proceed with ( u6 u 3 ) 6x = (u 3 ) 6x, ( u6 u 4 ) 8x = (u 2 ) 8x and ( u6 u 5 ) 10x = (u) 10x Obtain the lists: [u 3 2x, u x u 2x u 3x, uu 2 3x, u 2 xu 4x, uu 2x u 4x, uu x u 5x, u 2 u 6x ] [u 3 2x, uu 2 3x] [u 2 4x, u 3x u 5x, u 2x u 6x, u x u 7x, uu 8x ] [u 2 4x] and [u 10x ] [ ] Gather the terms: ρ 6 = u 6 + c 1 u 3 u 2 x + c 2 u 4 x + c 3 u 2 u 2 2x + c 4 u 3 2x + c 5 uu 2 3x + c 6 u 2 4x

19 where the constants c i must be determined (ii) Compute t ρ 6 Replace u t, u xt,..., u nx,t by (uu x + u xxx ),... (iii) Integrate the result with respect to x Carry out all integrations by parts Require that non-integrabe part vanishes Set to zero all the coefficients of the independent combinations involving powers of u and its derivatives with respect to x Solve the linear system for unknowns c 1, c 2,..., c 6 Result: ρ 6 = u 6 60u 3 u 2 x 30u 4 x + 108u 2 u 2 2x u3 2x uu2 3x u2 4x (iv) Flux J 6 can be computed by substituting the constants into the integrable part of ρ 6

20 Conserved Densities Conserved densities by Ito & Kako (Reduce, 1985, 1994) Conserved densities in DELiA by Bocharov (Pascal, 1990) Conserved densities by Gerdt (Reduce, 1993) Conserved densities by Roelofs, Sanders and Wang (Reduce 1994, Maple 1995) Conserved densities by Hereman and Göktaş (Mathematica, ) Conservation laws by Wolf (Reduce, 1995)

21 A Class of Fifth-order Evolution Equations Special cases: u t + αu 2 u x + βu x u 2x + γuu 3x + u 5x = 0 α = 30 β = 20 γ = 10 Lax α = 5 β = 5 γ = 5 Sawada Kotera or Caudry Dodd Gibbon α = 20 β = 25 γ = 10 Kaup Kuperschmidt α = 2 β = 6 γ = 3 Ito Under what conditions for the parameters α, β, and γ does this equation admit many conservation laws? If α = 3 10 γ2 and β = 2γ then there is a sequence (without gaps!) of conserved densities (Lax case) If α = 1 5 γ2 and β = γ then there is a sequence (with gaps!) of conserved densities (SK case) If α = 1 5 γ2 and β = 5 2γ then there is a sequence (with gaps!) of conserved densities (KK case) If α = 2β2 +7βγ 4γ 2 45 or β = 2γ then there is a conserved density of degree 4 Combining both conditions gives α = 2γ2 9 and β = 2γ (Ito case)

22 Conserved Densities of Systems of Evolution Eqs. Coupled KdV Equations (Hirota-Satsuma system) u t a(u xxx + 6uu x ) 2bvv x = 0 v t + v xxx + 3uv x = 0 u 2 x 2, v 2 x 2 ρ 1 = u ρ 2 = u bv2 ρ 3 = (1 + a)(u u2 x) + b(uv 2 v 2 x) and e.g. ρ 4 = u 4 2uu 2 x u2 xx b(u2 v uvv xx uv2 x 2 3 v2 xx) b2 v 4 provided a = 1 2 There are infinitely many more conservation laws

23 The Ito system u t u xxx 6uu x 2vv x = 0 v t 2u x v 2uv x = 0 u 2 x 2, v 2 x 2 ρ 1 = c 1 u + c 2 v ρ 2 = u 2 + v 2 ρ 3 = 2u 3 + 2uv 2 u 2 x ρ 4 = 5u 4 + 6u 2 v 2 + v 4 10uu 2 x + 2v 2 u 2x + u 2 2x and there are infinitely many more conservation laws

24 The Drinfel d-sokolov system u t + 3vv x = 0 v t + vu x + 2uv x + 2v 3x = 0 u 2 x 2, 2 v free, choose v x 2 ρ 1 = u ρ 2 = v 2 ρ 3 = 2 9 u3 + uv u2 x 3 2 v2 x ρ 4 = 1 2 u4 1 6 u2 v v4 1 6 uu2 x + uv 2 x 5 12 v2 u 2x u2 2x 3 4 v2 2x and there are presumably infinitely many more conservation laws

25 The dispersiveless long-wave system u t + vu x + uv x = 0 v t + u x + vv x = 0 u free, v free, but u 2v choose u x and 2v x ρ 1 = v ρ 2 = u ρ 3 = uv ρ 4 = u 2 + uv 2 ρ 5 = 3u 2 v + uv 3 ρ 6 = 1 3 u3 + u 2 v uv4 ρ 7 = u 3 v + u 2 v uv5 ρ 8 = 1 3 u4 + 2u 3 v 2 + u 2 v uv6 always the same set irrespective the choice of weights

26 Further Examples A generalized Schamel equation n 2 u t + (n + 1)(n + 2)unu 2 x + u xxx = 0 n positive integer ρ 1 = u, ρ 2 = u 2 no further conservation laws The Boussinesq equation ρ 3 = 1 2 u2 x n2 2 u2+ 2 n u tt bu xx + 3uu xx + 3u 2 x + au 4x = 0 a and b are real constants Write as system of evolution equations u t + v x = 0 v t + bu x 3uu x au 3x = 0 u 2 x 2, b 2 x 2, v free, take v 3 x 3

27 ρ 1 = u ρ 2 = u ρ 3 = bu ρ 4 = bc 1 u + c 2 uv ρ 5 = b 2 c 1 u + c 2 (bu 2 u 3 + v 2 + au 2 x) and there are infinitely many more conservation laws

28 A modified vector derivative NLS equation B t + x (B2 B ) + αb 0 B 0 B x + e x 2 B x 2 = 0 Replace vector equation by u t + ( u(u 2 + v 2 ) + βu v x )x = 0 v t + ( v(u 2 + v 2 ) + u x )x = 0 u and v denote the components of B parallel and perpendicular to B 0 and β = αb 2 0 2u x, 2v x, The first 6 conserved densities are: ρ 1 = c 1 u + c 2 v ρ 2 = u 2 + v 2 β x ρ 3 = 1 2 (u2 + v 2 ) 2 uv x + u x v + βu 2 ρ 4 = 1 4 (u2 + v 2 ) (u2 x + v 2 x) u 3 v x + v 3 u x + β 4 (u4 v 4 )

29 ρ 5 = 1 4 (u2 + v 2 ) (u xv xx u xx v x ) (uu x + vv x ) (u2 + v 2 )(u 2 x + v 2 x) (u 2 + v 2 ) 2 (uv x u x v) + β 5 (2u2 x 4u 3 v x + 2u 6 + 3u 4 v 2 v 6 ) + β2 5 u4 ρ 6 = 7 16 (u2 + v 2 ) (u2 xx + v 2 xx) 5 2 (u2 + v 2 )(u x v xx u xx v x ) + 5(u 2 + v 2 )(uu x + vv x ) (u2 + v 2 ) 2 (u 2 x + v 2 x) (u2 + v 2 ) 3 (uv x u x v) + β 8 (5u8 + 10u 6 v 2 10u 2 v 6 5v u 2 u 2 x 12u 5 v x + 60uv 4 v x 20v 2 v 2 x) + β2 4 (u6 + v 6 )

30 Table 1 Conserved Densities for Sawada-Kotera and Lax equations Density Sawada-Kotera equation Lax equation ρ 1 u u ρ u2 ρ u3 u 2 x 1 3 u3 1 6 u2 x ρ u4 9 4 uu2 x u2 2x 1 4 u4 1 2 uu2 x u2 2x ρ u5 u 2 u 2 x uu2 2x 1 70 u2 3x ρ u u3 u 2 x 17 8 u4 x + 6u 2 u 2 2x 1 6 u6 5 3 u3 u 2 x 5 36 u4 x u2 u 2 2x +2u 3 2x 21 8 uu2 3x u2 4x u3 2x 1 14 uu2 3x u2 4x ρ u7 9u 4 u 2 x 54 5 uu4 x u3 u 2 2x 1 7 u7 5 2 u4 u 2 x 5 6 uu4 x + u 3 u 2 2x u2 xu 2 2x uu3 2x u2 u 2 3x u2 xu 2 2x uu3 2x 3 14 u2 u 2 3x u 2xu 2 3x uu2 4x 9 35 u2 5x 5 42 u 2xu 2 3x uu2 4x u2 5x ρ u8 7 2 u5 u 2 x u2 u 4 x u4 u 2 2x uu2 xu 2 2x u2 u 3 2x u4 2x u3 u 2 3x 1 4 u2 xu 2 3x 5 6 uu 2xu 2 3x u2 u 2 4x u 2xu 2 4x uu2 5x u2 6x

31 Table 2 Conserved Densities for Kaup-Kuperschmidt and Ito equations Density Kaup-Kuperschmidt equation Ito equation ρ 1 u u ρ u 2 2 ρ 3 u u2 x ---- ρ 4 u uu2 x u2 2x u uu2 x u2 2x ρ ρ 6 u u3 u 2 x u4 x u2 u 2 2x u3 2x uu2 3x u2 4x ρ 7 u u4 u 2 x uu4 x u3 u 2 2x u2 xu 2 2x uu3 2x u2 u 2 3x u 2xu 2 3x uu2 4x u2 5x ρ

32 Example 3 Macsyma/Mathematica Solitons Hirota s Method Hirota s Direct Method allows to construct soliton solutions of nonlinear evolution equations wave equations coupled systems Test conditions for existence of soliton solutions Examples: Korteweg-de Vries equation (KdV) u t + 6uu x + u 3x = 0 Kadomtsev-Petviashvili equation (KP) (u t + 6uu x + u 3x ) x + 3u 2y = 0 Sawada-Kotera equation (SK) u t + 45u 2 u x + 15u x u 2x + 15uu 3x + u 5x = 0

33 Hirota s Method Korteweg-de Vries equation u t + 6uu x + u 3x = 0 Substitute u(x, t) = 2 2 ln f(x, t) x 2 Integrate with respect to x Bilinear form ff xt f x f t + ff 4x 4f x f 3x + 3f 2 2x = 0 B(f f) def = ( D x D t + D 4 x) (f f) = 0 Introduce the bilinear operator D m x D n t (f g) = ( x x ) m ( t t ) n f(x, t) g(x, t ) x =x,t =t Use the expansion f = 1 + n=1 ɛn f n Substitute f into the bilinear equation

34 Collect powers in ɛ (book keeping parameter) Start with O(ɛ 0 ) : B(1 1) = 0 O(ɛ 1 ) : B(1 f 1 + f 1 1) = 0 O(ɛ 2 ) : B(1 f 2 + f 1 f 1 + f 2 1) = 0 O(ɛ 3 ) : B(1 f 3 + f 1 f 2 + f 2 f 1 + f 3 1) = 0 O(ɛ 4 ) : B(1 f 4 + f 1 f 3 + f 2 f 2 + f 3 f 1 + f 4 1) = 0 O(ɛ n ) : B( n j=0 f j f n j ) = 0 with f 0 = 1 f 1 = N exp(θ i) = N exp (k i x ω i t + δ i ) i=1 i=1 k i, ω i and δ i are constants Dispersion law ω i = k 3 i (i = 1, 2,..., N) If the original PDE admits a N-soliton solution then the expansion will truncate at level n = N

35 Consider the case N=3 Terms generated by B(f 1, f 1 ) determine f 2 = a 12 exp(θ 1 + θ 2 ) + a 13 exp(θ 1 + θ 3 ) + a 23 exp(θ 2 + θ 3 ) = a 12 exp [(k 1 + k 2 ) x (ω 1 + ω 2 ) t + (δ 1 + δ 2 )] + a 13 exp [(k 1 + k 3 ) x (ω 1 + ω 3 ) t + (δ 1 + δ 3 )] + a 23 exp [(k 2 + k 3 ) x (ω 2 + ω 3 ) t + (δ 2 + δ 3 )] Calculate the constants a 12, a 13 and a 23 a ij = (k i k j ) 2 (k i + k j ) 2 i, j = 1, 2, 3 Terms from B(f 1 f 2 + f 2 f 1 ) determine with f 3 = b 123 exp(θ 1 + θ 2 + θ 3 ) = b 123 exp [(k 1 +k 2 +k 3 )x (ω 1 +ω 2 +ω 3 )t+(δ 1 +δ 2 +δ 3 )] b 123 = a 12 a 13 a 23 = (k 1 k 2 ) 2 (k 1 k 3 ) 2 (k 2 k 3 ) 2 (k 1 + k 2 ) 2 (k 1 + k 3 ) 2 (k 2 + k 3 ) 2 Subsequently, f i = 0 for i > 3 Set ɛ = 1 f = 1 + exp θ 1 + exp θ 2 + exp θ 3 + a 12 exp(θ 1 + θ 2 ) + a 13 exp(θ 1 + θ 3 ) + a 23 exp(θ 2 + θ 3 ) + b 123 exp(θ 1 + θ 2 + θ 3 )

36 Return to the original u(x, t) Single soliton solution u(x, t) = 2 2 ln f(x, t) x 2 f = 1 + e θ, k, ω and δ are constants and ω = k 3 Substituting f into θ = kx ωt + δ Take k = 2K u(x, t) = 2 2 ln f(x, t) x 2 = 2( f xxf f 2 x f 2 ) u = 2K 2 sech 2 K(x 4K 2 t + δ)

37 Two-soliton solution f = 1 + e θ 1 + e θ 2 + a 12 e θ 1+θ 2 θ i = k i x ω i t + δ i with ω i = k 3 i, (i = 1, 2) and a 12 = (k 1 k 2 ) 2 (k 1 +k 2 ) 2 Select e δ i = c2 i k i e k ix ω i t+ i f = 1 4 fe 1 2 ( θ 1 + θ 2 ) θ i = k i x ω i t + i c 2 i = k 2 + k 1 k 2 k 1 k i Return to u(x, t) u(x, t) = ũ(x, t) = 2 2 ln f(x, t) x 2 = k 2 2 k k2cosech 2 2 θ2 2 + k1sech 2 2 θ1 2 (k 2 coth θ 2 2 k 1 tanh θ 1 2 ) 2

38 Program 4 Macsyma Lie-point Symmetries System of m differential equations of order k i (x, u (k) ) = 0, i = 1, 2,..., m with p independent and q dependent variables x = (x 1, x 2,..., x p ) IR p u = (u 1, u 2,..., u q ) IR q The group transformations have the form x = Λ group (x, u), ũ = Ω group (x, u) where the functions Λ group and Ω group are to be determined Look for the Lie algebra L realized by the vector field α = p i=1 ηi (x, u) x i + q l=1 ϕ l (x, u) u l

39 Procedure for finding the coefficients Construct the k th prolongation pr (k) α of the vector field α Apply it to the system of equations Request that the resulting expression vanishes on the solution set of the given system pr (k) α i j =0 i, j = 1,..., m This results in a system of linear homogeneous PDEs for η i and ϕ l, with independent variables x and u ( determining equations) Procedure thus consists of two major steps: deriving the determining equations solving the determining equations

40 Procedure for Computing the Determining Equations Use multi-index notation J = (j 1, j 2,..., j p ) IN p, to denote partial derivatives of u l u l J where J = j 1 + j j p J u l x 1 j 1 x2 j 2... xp j p, u (k) denotes a vector whose components are all the partial derivatives of order 0 up to k of all the u l Steps: (1) Construct the k th prolongation of the vector field pr (k) α = α + q l=1 u l J J ψj l (x, u (k) ), 1 J k The coefficients ψ J l of the first prolongation are: ψ J i l = D i ϕ l (x, u) p j=1 ul J j D i η j (x, u), where J i is a p tuple with 1 on the i th position and zeros elsewhere D i is the total derivative operator D i = + q x i l=1 J ul J+J i u l J, 0 J k

41 Higher order prolongations are defined recursively: ψ J+J i l = D i ψ J l p j=1 ul J+J j D i η j (x, u), J 1 (2) Apply the prolonged operator pr (k) α to each equation i (x, u (k) ) = 0 Require that pr (k) α vanishes on the solution set of the system pr (k) α i j =0 = 0 i, j = 1,..., m (3) Choose m components of the vector u (k), say v 1,..., v m, such that: (a) Each v i is equal to a derivative of a u l (l = 1,..., q) with respect to at least one variable x i (i = 1,..., p). (b) None of the v i is the derivative of another one in the set. (c) The system can be solved algebraically for the v i in terms of the remaining components of u (k), which we denoted by w: (d) The derivatives of v i, v i = S i (x, w), i = 1,..., m. v i J = D J S i (x, w),

42 where D J D j 1 1 D j D j p p, can all be expressed in terms of the components of w and their derivatives, without ever reintroducing the v i or their derivatives. For instance, for a system of evolution equations u i t(x 1,..., x p 1, t) = F i (x 1,..., x p 1, t, u (k) ), i = 1,..., m, where u (k) involves derivatives with respect to the variables x i but not t, choose v i = u i t. (4) Eliminate all v i and their derivatives from the expression prolonged vector field, so that all the remaining variables are independent (5) Obtain the determining equations for η i (x, u) and ϕ l (x, u) by equating to zero the coefficients of the remaining independent derivatives u l J.

43 Example: The Korteweg-de Vries Equation one equation (parameter a) u t + auu x + u xxx = 0 two independent variables t and x one dependent variable u vector field α = η x x + ηt t + ϕu u Format for SYMMGRP.MAX variables x[1] = x, x[2] = t, u[1] = u equation e1 : u[1, [0, 1]] + a u[1] u[1, [1, 0]] + u[1, [3, 0]] variable to be eliminated v1 : u[1, [0, 1]] coefficients of vectorfield in SYMMGRP.MAX: eta[1] = η x, eta[2] = η t and phi[1] = ϕ u

44 There are only eight determining equations phi[1] x[2] eta[2] u[1] eta[2] x[1] eta[1] u[1] = 0 = 0 = 0 2 phi[1] u[1] 2 = 0 2 phi[1] u[1] x[1] 2 eta[1] = 0 x[1] phi[1] x[1] 3 phi[1] + u[1] x[1] = 0

45 3 3 phi[1] u[1] x[1] 2 eta[1] x[2] 3 eta[1] x[1] u[1] eta[1] x[1] + phi[1] = 0 u[1] eta[2] x[2] phi[1] u[1] x[1] 2 The solution in the original variables eta[1] x[2] η x = k 1 + k 3 t k 4 x η t = k 2 3k 4 t ϕ u = k k 4 u The four infinitesimal generators are 3 eta[1] x[1] 3 u[1] eta[1] x[1] + phi[1] = 0 G 1 = x G 2 = t G 3 = t x + u G 4 = x x + 3 t t 2 u u

46 Equation is invariant under: translations G 1 and G 2 Galilean boost G 3 scaling G 4 Computation of the flows corresponding to G 1 thru G 4 shows that for any solution u = f(x, t) of the KdV equation the transformed solutions ũ = f(x ɛ, t) ũ = f(x, t ɛ) ũ = f(x ɛ, t) + ɛ will solve the KdV equation ũ = e 2ɛ f(e ɛ x, e 3ɛ t) Note that ɛ is the parameter of the transformation group

47 III. PLANS FOR THE FUTURE Extension of Symbolic Software Packages (Macsyma/Mathematica) Lie symmetries of differential-difference equations Solver for systems of linear, homogeneous PDEs (Hereman) Painlevé test for systems of PDEs (Elmer, Göktaş & Coffey) Solitons via Hirota s method for bilinear equations (Zhuang) Simplification of Hirota s method (Hereman & Nuseir) Conservation laws of PDEs with variable coefficients (Göktaş) Lax pairs, special solutions,... New Software Wavelets (prototype/educational tool) Other methods for Differential Equations