Solving Nonlinear Wave Equations and Lattices with Mathematica 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 Departement Toegepaste Wiskunde Universiteit Stellenbosch Matieland, Zuid-Afrika Woensdag, 21 Februarie 2001, 16:00 Collaborators: Ünal Göktaş (WRI), Doug Baldwin (CSM) Ryan Martino, Joel Miller, Linda Hong (REU NSF 99) Steve Formenac, Andrew Menz (REU NSF 00) Research supported in part by NSF under Grant CCR-9901929 1
OUTLINE Purpose & Motivation Typical Examples Algorithm for Tanh Solutions Algorithm for Sech Solutions Extension: Tanh Solutions for Differential-difference Equations (DDEs) Solving/Analyzing Systems of Algebraic Equations with Parameters Implementation Issues Demo Mathematica Software Package Future Work 2
Purpose & Motivation Develop various symbolic algorithms to compute exact solutions of nonlinear (systems) of partial differential equations (PDEs) and differential-difference equations (DDEs, lattices). Fully automate the tanh and sech methods to compute closed form solitary wave solutions. Solutions of tanh or sech type model solitary waves in fluid dynamics, plasmas, electrical circuits, optical fibers, bio-genetics, etc. Class of nonlinear PDEs and DDEs solvable with the tanh/sech method includes famous evolution and wave equations. Typical examples: Korteweg-de Vries, Fisher and Boussinesq PDEs, Toda and Volterra lattices (DDEs). Research aspect: Design a high-quality application package for the computation of exact solitary wave solutions of large classes of nonlinear evolution and wave equations. Educational aspect: Software as course ware for courses in nonlinear PDEs, theory of nonlinear waves, integrability, dynamical systems, and modeling with symbolic software. Users: scientists working on nonlinear wave phenomena in fluid dynamics, nonlinear networks, elastic media, chemical kinetics, material science, bio-sciences, plasma physics, and nonlinear optics. 3
Typical Examples of Single PDEs and Systems of PDEs The Korteweg-de Vries (KdV) equation: Solitary wave solution: or, equivalently, u t + αuu x + u 3x = 0. u(x, t) = 8c3 1 c 2 6αc 1 2c2 1 α tanh2 [c 1 x + c 2 t + δ], u(x, t) = 4c3 1 + c 2 6αc 1 + 2c2 1 α sech2 [c 1 x + c 2 t + δ]. The modified Korteweg-de Vries (mkdv) equation: Solitary wave solution: u t + αu 2 u x + u 3x = 0. u(x, t) = ± 6 α c 1 sech [ c 1 x c 3 1t + δ ]. The Fisher equation: u t u xx u (1 u) = 0. Solitary wave solution: u(x, t) = 1 4 ± 1 2 tanhξ + 1 4 tanh2 ξ, with ξ = ± 1 2 6 x ± 5 12 t + δ. 4
The generalized Kuramoto-Sivashinski equation: Solitary wave solutions u t + uu x + u xx + σu 3x + u 4x = 0. (ignoring symmetry u u, x x, σ σ) : For σ = 4 : u(x, t) = 9 2c 2 15 tanhξ (1 + tanhξ tanh 2 ξ), with ξ = x 2 + c 2t + δ. For σ = 0 : u(x, t) = 2 19 11 c 2 135 11 19 19 with ξ = 1 11 2 19 x + c 2t + δ. For σ = 12 47 : u(x, t) = 45 4418c 2 47 47 with ξ = ± 1 2 47 x + c 2t + δ. For σ = 16/ 73 : tanhξ + 165 19 ± 45 47 47 tanhξ 11 19 tanh3 ξ, 45 47 47 tanh2 ξ ± 15 47 47 tanh3 ξ, u(x, t) = 2 (30 5329c 2) 73 ± 75 73 73 73 tanhξ 60 73 73 tanh2 ξ ± 15 73 73 tanh3 ξ, with ξ = ± 1 2 73 x + c 2t + δ. 5
Three-dimensional modified Korteweg-de Vries equation: Solitary wave solution: u t + 6u 2 u x + u xyz = 0. u(x, y, z, t) = ± c 2 c 3 sech [c 1 x + c 2 y + c 3 z c 1 c 2 c 3 t + δ]. The Boussinesq (wave) equation: u tt βu 2x + 3uu 2x + 3u x 2 + αu 4x = 0, or written as a first-order system (v auxiliary variable): Solitary wave solution: u t + v x = 0, v t + βu x 3uu x αu 3x = 0. u(x, t) = βc2 1 c 2 2 + 8αc 4 1 3c 2 1 4αc 2 1 tanh 2 [c 1 x + c 2 t + δ] v(x, t) = b 0 + 4αc 1 c 2 tanh 2 [c 1 x + c 2 t + δ]. The Broer-Kaup system: Solitary wave solution: u ty + 2(uu x ) y + 2v xx u xxy = 0, v t + 2(uv) x + v xx = 0. u(x, t) = c 3 2c 1 + c 1 tanh [c 1 x + c 2 y + c 3 t + δ] v(x, t) = c 1 c 2 c 1 c 2 tanh 2 [c 1 x + c 2 y + c 3 t + δ] 6
Typical Examples of DDEs (lattices) The Toda lattice: ü n = (1 + u n ) (u n 1 2u n + u n+1 ). Solitary wave solution: u n (t) = a 0 ± sinh(c 1 ) tanh [c 1 n ± sinh(c 1 ) t + δ]. The Volterra lattice: u n = u n (v n v n 1 ) v n = v n (u n+1 u n ). Solitary wave solution: u n (t) = c 2 coth(c 1 ) + c 2 tanh [c 1 n + c 2 t + δ] v n (t) = c 2 coth(c 1 ) c 2 tanh [c 1 n + c 2 t + δ]. The Relativistic Toda lattice: u n = (1 + αu n )(v n v n 1 ) v n = v n (u n+1 u n + αv n+1 αv n 1 ). Solitary wave solution: u n (t) = c 2 coth(c 1 ) 1 α + c 2 tanh [c 1 n + c 2 t + δ] v n (t) = c 2 coth(c 1 ) α c 2 α tanh [c 1n + c 2 t + δ]. 7
Algorithm for Tanh Solutions for PDE system Given is a system of PDEs of order n (u(x), u (x), u (x), u (n) (x)) = 0. Dependent variable u has components u i (or u, v, w,...) Independent variable x has components x i (or x, y, z, t) Step 1: Seek solution of the form u i (x) = U i (T ), with T = tanh [c 1 x + c 2 y + c 3 z + c 4 t] = tanh ξ. Observe cosh 2 ξ sinh 2 ξ = 1, ( tanhξ) = 1 tanh 2 ξ or T = 1 T 2. Repeatedly apply the operator rule c i (1 T 2 ) d x i dt This produces a coupled system of Legendre equations of type P(T, U i, U i,..., U (n) i ) = 0 for U i (T ). Example: For Boussinesq system u t + v x = 0 v t + βu x 3uu x αu 3x = 0, we obtain after cancelling common factors 1 T 2 c 2 U + c 1 V = 0 c 2 V + βc 1 U 3c 1 UU +αc 3 [ 1 2(1 3T 2 )U + 6T (1 T 2 )U (1 T 2 ) 2 U ] = 0 8
Step 2: Seek polynomial solutions U i (T ) = M i j=0 a ij T j Balance the highest power terms in T to determine M i. Example: Powers for Boussinesq system gives M 1 = M 2 = 2. M 1 1 = M 2 1, 2M 1 1 = M 1 + 1 Hence, U 1 (T ) = a 10 + a 11 T + a 12 T 2, U 2 (T ) = a 20 + a 21 T + a 22 T 2. Step 3: Determine the algebraic system for the unknown coefficients a ij by balancing the coefficients of the various powers of T. Example: Boussinesq system a 11 c 1 (3a 12 + 2α c 2 1) = 0 a 12 c 1 (a 12 + 4α c 2 1) = 0 a 21 c 1 + a 11 c 2 = 0 a 22 c 1 + a 12 c 2 = 0 βa 11 c 1 3a 10 a 11 c 1 + 2αa 11 c 3 1 + a 21 c 2 = 0 3a 2 11 c 1 + 2β a 12 c 1 6a 10 a 12 c 1 + 16α a 12 c 3 1 + 2a 22 c 2 = 0. 9
Step 4: Solve the nonlinear algebraic system with parameters. Reject complex solutions? Test the solutions. Example: Solution for Boussinesq case Step 5: a 10 = βc2 1 c 2 2 + 8αc 4 1 3c 2 1 a 11 = 0 a 12 = 4αc 2 1 = free a 20 a 21 = 0 a 22 = 4αc 1 c 2. Return to the original variables. Test the final solution in the original equations Example: Solitary wave solution for Boussinesq system: u(x, t) = βc2 1 c 2 2 + 8αc 4 1 3c 2 1 4αc 2 1 tanh 2 [c 1 x + c 2 t + δ] v(x, t) = a 20 + 4αc 1 c 2 tanh 2 [c 1 x + c 2 t + δ]. 10
Algorithm for Sech Solutions for PDE system Given is a system of PDEs of order n (u(x), u (x), u (x), u (n) (x)) = 0. Dependent variable u has components u i (or u, v, w,...) Independent variable x has components x i (or x, y, z, t) Step 1: Seek solution of the form u i (x) = U i (S), with S = sech [c 1 x + c 2 y + c 3 z + c 4 t] = tanh ξ. Observe ( sech ξ) = tanh ξ sech ξ or S = T S = 1 S 2 S. Also, cosh 2 ξ sinh 2 ξ = 1, hence, T 2 + S 2 = 1 and dt ds = S T. Repeatedly apply the operator rule c i 1 S2 S d x i ds This produces a coupled system of Legendre type equations of type P(S, U i, U i,..., U (m) i ) + 1 S 2 Q(S, U i, U i,..., U (n) i ) = 0 for U i (S). For every equation one must have P i = 0 or Q i = 0. Only odd derivatives produce the extra factor 1 S 2. Conclusion: The total number of derivatives in each term in the given system should be either even or odd. No mismatch is allowed. Example: For the 3D mkdv equation u t + 6u 2 u x + u xyz = 0. 11
we obtain after cancelling a common factor 1 S 2 S c 4 U +6c 1 U 2 U +c 1 c 2 c 3 [(1 6S 2 )U +3S(1 2S 2 )U +S 2 (1 S 2 )U ] = 0 Step 2: Seek polynomial solutions U i (S) = M i j=0 a ij S j Balance the highest power terms in S to determine M i. Example: Powers for the 3D mkdv case 3M 1 1 = M 1 + 1 gives M 1 = 1. Hence, U(S) = a 10 + a 11 S. Step 3: Determine the algebraic system for the unknown coefficients a ij by balancing the coefficients of the various powers of S. Example: System for 3D mkdv case a 11 c 1 (a 2 11 c 2 c 3 ) = 0 a 11 (6a 2 10 c 1 + c 1 c 2 c 3 + c 4 ) = 0 a 10 a 2 11 c 1 == 0 12
Step 4: Solve the nonlinear algebraic system with parameters. Reject complex solutions? Test the solutions. Example: Solution for 3D mkdv case a 10 = 0 a 11 = ± c 1 c 3 c 4 = c 1 c 2 c 3 Step 5: Return to the original variables. Test the final solution in the original equations Example: Solitary wave solution for the 3D mkdv equation u(x, y, z, t) = ± c 2 c 3 sech(c 1 x + c 2 y + c 3 z c 1 c 2 c 3 t). 13
Extension: Tanh Solutions for DDE system Given is a system of differential-difference equations (DDEs) of order n (..., u n 1, u n, u n+1,..., u n,..., u (m) n,...) = 0. Dependent variable u n has components u i,n (or u n, v n, w n,...) Independent variable x has components x i (or n, t). No derivatives on shifted variables are allowed! Step 1: Seek solution of the form u i,n (x) = U i,n (T (n)), with T (n) = tanh [c 1 n + c 2 t + δ] = tanh ξ. Note that the argument T depends on n. Complicates matters. Repeatedly apply the operator rule on u i,n t c 2(1 T 2 ) d dt This produces a coupled system of Legendre equations of type for U i,n (T ). Example: Toda lattice transforms into P(T, U i,n, U i,n,...) = 0 ü n = (1 + u n ) (u n 1 2u n + u n+1 ). c 2 2(1 T 2 ) [ 2T U n (1 T 2 )U n ] + [ 1 + c 2 (1 T 2 )U n] [Un 1 2U n + U n+1 ] = 0 14
Step 2: Seek polynomial solutions U i,n (T (n)) = M i j=0 For U n+p, p 0, there is a phase shift: U i,n±p (T (n ± p) = M i j=0 a ij T (n) j a i,j [T (n + p)] j = M i a i,j j=0 Balance the highest power terms in T (n) to determine M i. Example: Powers for Toda lattice gives M 1 = 1. Hence, 2M 1 1 = M 1 + 1 U n (T (n)) = a 10 + a 11 T (n) T (n) ± tanh(pc 1 ) 1 ± T (n) tanh(pc 1 ) U n 1 (T (n 1)) = a 10 + a 11 T (n 1) = a 10 + a 11 T (n) tanh(c 1 ) 1 T (n) tanh(c 1 ) U n+1 (T (n + 1)) = a 10 + a 11 T (n + 1) = a 10 + a 11 T (n) + tanh(c 1 ) 1 + T (n) tanh(c 1 ). Step 3: Determine the algebraic system for the unknown coefficients a ij by balancing the coefficients of the various powers of T (n). Example: Algebraic system for Toda lattice c 2 2 tanh 2 (c 1 ) a 11 c 2 tanh 2 (c 1 ) = 0, c 2 a 11 = 0 j 15
Step 4: Solve the nonlinear algebraic system with parameters. Reject complex solutions? Test the solutions. Example: Solution of algebraic system for Toda lattice a 10 = free, a 11 = c 2 = ± sinh(c 1 ) Step 5: Return to the original variables. Test the final solution in the original equations Example: Solitary wave solution for Toda lattice: u n (t) = a 0 ± sinh(c 1 ) tanh [c 1 n ± sinh(c 1 ) t + δ]. 16
Example: System of DDEs: Relativistic Toda lattice u n = (1 + αu n )(v n v n 1 ) v n = v n (u n+1 u n + αv n+1 αv n 1 ). Step 1: Change of variables with gives u n (x, t) = U n (T (n)), v n (x, t) = V n (T (n)), T (n) = tanh [c 1 n + c 2 t + δ] = tanh ξ. c 2 (1 T 2 )U n (1 + αu n )(V n V n 1 ) = 0 c 2 (1 T 2 )V n V n (U n+1 U n + αv n+1 αv n 1 ) = 0. Step 2: Seek polynomial solutions U n (T (n)) = M 1 j=0 V n (T (n)) = M 2 j=0 a 1j T (n) j a 2j T (n) j. Balance the highest power terms in T (n) to determine M 1, and M 2 : gives M 1 = M 2 = 1. Hence, M 1 + 1 = M 1 + M 2, M 2 + 1 = M 1 + M 2 U n = a 10 + a 11 T (n), V n = a 20 + a 21 T (n). 17
Step 3: Algebraic system for a ij : a 11 c 2 + a 21 tanh(c 1 ) + α a 10 a 21 tanh(c 1 ) = 0 a 11 tanh(c 1 ) (α a 21 + c 2 ) = 0 a 21 c 2 + a 11 a 20 tanh(c 1 ) + 2α a 20 a 21 tanh(c 1 ) = 0 tanh(c 1 ) (a 11 a 21 + 2α a 2 21 a 11 a 20 tanh(c 1 )) = 0 a 21 tanh 2 (c 1 ) (c 2 a 11 ) = 0 Step 4: Solution of the algebraic system a 10 = c 2 coth(c 1 ) 1 α, a 11 = c 2, a 20 = c 2 coth(c 1 ), a 21 = c 2 α α. Step 5: Solitary wave solution in original variables: u n (t) = c 2 coth(c 1 ) 1 α + c 2 tanh [c 1 n + c 2 t + δ] v n (t) = c 2 coth(c 1 ) α c 2 α tanh [c 1n + c 2 t + δ]. 18
Solving/Analyzing Systems of Algebraic Equations with Parameters Class of fifth-order evolution equations with parameters: u t + αγ 2 u 2 u x + βγu x u 2x + γuu 3x + u 5x = 0. Well-Known Special cases Lax case: α = 3 10, β = 2, γ = 10. Two solutions: u(x, t) = 4c 2 1 6c 2 1 tanh 2 [ c 1 x 56c 5 1t + δ ] and u(x, t) = a 0 2c 2 1 tanh 2 [ c 1 x 2(15a 2 0c 1 40a 0 c 3 1 + 28c 5 1)t + δ ] where a 0 is arbitrary. Sawada-Kotera case: α = 1 5, β = 1, γ = 5. Two solutions: u(x, t) = 8c 2 1 12c 2 1 tanh 2 [ c 1 x 16c 5 1t + δ ] and u(x, t) = a 0 6c 2 1 tanh 2 [ c 1 x (5a 2 0c 1 40a 0 c 3 1 + 76c 5 1)t + δ ] where a 0 is arbitrary. Kaup-Kupershmidt case: α = 1 5, β = 5 2, γ = 10. Two solutions: u(x, t) = c 2 1 3 2 c2 1 tanh 2 [ c 1 x c 5 1t + δ ] and no free constants! u(x, t) = 8c 2 1 12c 2 tanh 2 [ c 1 x 176c 5 1t + δ ], Ito case: α = 2 9, β = 2, γ = 3. One solution: u(x, t) = 20c 2 1 30c 2 1 tanh 2 [ c 1 x 96c 5 1t + δ ]. 19
What about the General case? Q1: Can we retrieve the special solutions? Q2: What are the condition(s) on the parameters α, β, γ for solutions of tanh-type to exist? Tanh solutions: u(x, t) = a 0 + a 1 tanh [c 1 x + c 2 t + δ] + a 2 tanh 2 [c 1 x + c 2 t + δ]. Nonlinear algebraic system must be analyzed, solved (or reduced!): a 1 (αγ 2 a 2 2 + 6γa 2 c 2 1 + 2βγa 2 c 2 1 + 24c 4 1) = 0 a 1 (αγ 2 a 2 1 + 6αγ 2 a 0 a 2 + 6γa 0 c 2 1 18γa 2 c 2 1 12βγa 2 c 2 1 120c 4 1) = 0 αγ 2 a 2 2 + 12γa 2 c 2 1 + 6βγa 2 c 2 1 + 360c 4 1 = 0 2αγ 2 a 2 1a 2 + 2αγ 2 a 0 a 2 2 + 3γa 2 1c 2 1 + βγa 2 1c 2 1 + 12γa 0 a 2 c 2 1 8γa 2 2c 2 1 8βγa 2 2c 2 1 480a 2 c 4 1 = 0 a 1 (αγ 2 a 2 0c 1 2γa 0 c 3 1 + 2βγa 2 c 3 1 + 16c 5 1 + c 2 ) = 0 αγ 2 a 0 a 2 1c 1 + αγ 2 a 2 0a 2 c 1 γa 2 1c 3 1 βγa 2 1c 3 1 8γa 0 a 2 c 3 1 + 2βγa 2 2c 3 1 +136a 2 c 5 1 + a 2 c 2 = 0 Unknowns: a 0, a 1, a 2. Parameters: c 1, c 2, α, β, γ. Solve and Reduce cannot be used on the whole system! 20
Assumptions: All c i 0 Strategy to Solve/Reduce Nonlinear Systems Parameters (α, β, γ,...) are nonzero. Otherwise the highest powers M i may change. All a j Mi 0. Coefficients of highest power in U i are present. Solve for a ij, then c i, then find conditions on parameters. Strategy followed by hand: Solve all linear equations in a ij first (cost: branching). Start with the ones without parameters. Capture constraints in the process. Solve linear equations in c i if they are free of a ij. Solve linear equations in parameters if they free of a ij, c i. Solve quasi-linear equations for a ij, c i, parameters. Solve quadratic equations for a ij, c i, parameters. Eliminate cubic terms for a ij, c i, parameters, without solving. Show remaining equations, if any. Alternatives: Use (adapted) Gröbner Basis Techniques. Use combinatorics on coefficients a ij = 0 or a ij 0. 21
Actual Solution: Two major cases: CASE 1: a 1 = 0, two subcases Subcase 1-a: a 2 = 3 2 a 0 c 2 = c 3 1(24c 2 1 βγa 0 ) where a 0 is one of the two roots of the quadratic equation: αγ 2 a 2 0 8γa 0 c 2 1 4βγa 0 c 2 1 + 160c 4 1 = 0. Subcase 1-b: If β = 10α 1, then a 2 = 6 αγ c2 1 where a 0 is arbitrary. CASE 2: a 1 0, then c 2 = 1 α (α2 γ 2 a 2 0c 1 8αγa 0 c 3 1 + 12c 5 1 + 16αc 5 1) α = 1 392 (39 + 38β + 8β2 ) and a 2 = 168 γ(3 + 2β) c2 1 provided β is one of the roots of (104β 2 + 886β + 1487)(520β 3 + 2158β 2 1103β 8871) = 0 22
Subcase 2-a: If β 2 = 1 104 (886β + 1487), then α = 2β + 5 26 a 0 = 49c2 1(9983 + 4378β) 26γ(8 + 3β)(3 + 2β) 2 a 1 = ± 336c2 1 γ(3 + 2β) a 2 = 168c2 1 γ(3 + 2β) c 2 = 364 c5 1 (3851 + 1634β). 6715 + 2946β Subcase 2-b: If β 3 = 1 520 (8871 + 1103β 2158β2 ), then α = 39 + 38β + 8β2 392 a 0 = 28 c2 1 (6483 + 5529β + 1066β 2 ) (3 + 2β)(23 + 6β)(81 + 26β)γ a 2 1 = 28224 c 4 1 (4β 1)(26β 17) (3 + 2β) 2 (23 + 6β)(81 + 26β)γ 2 a 2 = 168c2 1 γ(3 + 2β) c 2 = 8 c5 1 (1792261977 + 1161063881β + 188900114β 2 ) 959833473 + 632954969β + 105176786β 2. 23
Implementation Issues Software Demo Future Work Demonstration of Mathematica package for tanh/sech methods. Long term goal: Develop PDESolve for closed form solutions of nonlinear PDEs and DDEs. Implement various methods: Lie symmetry (similarity) methods. Look at other types of explicit solutions involving hyperbolic functions sinh, cosh, tanh,... other special functions. complex exponentials combined with sech or tanh. Example: Set of ODEs from quantum field theory u xx = u + u 3 + auv 2 Try solutions (c 2 = 0 for ODEs) u i (x, t) = M i or j=0 u i (x, t) = M i j=0 Solitary wave solutions: provided b = v xx = bv + cv 3 + av(u 2 1). a ij tanh j [c 1 x + c 2 t + δ] + N i j=0 b ij sech 2j+1 [c 1 x + c 2 t + δ]. (ã ij + b ij sech[c 1 x + c 2 t + δ]) tanh j [c 1 x + c 2 t + δ]. u = ± tanh[ v = ± 1 a a c sech[ a 2 c 2(a c). a 2 c 2(a c) x + δ] a 2 c 2(a c) x + δ], 24
Example: Nonlinear Schrödinger equation (focusing/defocusing): Bright soliton solution (+ sign): i u t + u xx ± u 2 u = 0. u(x, t) = k 2 exp[i ( c 2 x + (k2 c2 4 )t)] sech[k(x ct x 0)] Dark soliton solution ( sign): u(x, t) = 1 exp[i(kx (2k 2 + 3K 2 2Kc + c2 2 2 )t)] {k tanh[k(x ct x 0 )] i(k c } 2 ). Example: Nonlinear sine-gordon equation (light cone coordinates): u xt = sin u. Setting Φ = u x, Ψ = cos(u) 1, gives Solitary wave solution (kink): Φ xt Φ Φ Ψ = 0 2Ψ + Ψ 2 + Φ 2 t = 0. Φ = u x = ± 1 1 sech[ (x ct) + δ], c c Ψ = cos(u) 1 = 1 2 sech 2 1 [ (x ct) + δ], c in final form: 1 u(x, t) = ±4 arctan exp( (x ct) + δ) c. 25
Example: Coupled nonlinear Schrödinger equations: Seek particular solutions i u t = u xx + u( u 2 + h v 2 ) i v t = v xx + v( v 2 + h u 2 ) u(x, t) = a tanh(µx) exp(iat) v(x, t) = b sech(µx) exp(ibt). Seek solutions u(x, t) = U(F (ξ)), where derivatives of F (ξ) are polynomial in F. Now, F (ξ) = 1 F 2 (ξ) F = tanh(ξ). Other choices are possible. Add the constraining differential equations to the system of PDEs directly. Why are tanh and sech solutions so prevalent? Other applications: Computation of conservation laws, symmetries, first integrals, etc. leading to linear parameterized systems for unknowns coefficients (see InvariantsSymmetries by Göktaş and Hereman). 26