Approximate symmetry analysis

Transcription

1 Approximate symmetry analysis Saturday June 7, 10:00 10:35 am Greg Reid 1 Ian Lisle and Tracy Huang 2 CMS Summer Meeting Session on Gröbner Bases and Computer Algebra Winnipeg 1 Ontario Research Centre for Computer Algebra, Western University, Canada. 2 University of Canberra

2 Table of contents Introduction & Historical Comments Symmetries and Introductory Examples Direct use of floating point numbers in exact differential elimination methods Difficulties with symbolic approaches Prolongation and Projection y xx yy x + 5 y 3 = 0 Symbolic-numeric algorithm for structure constants References

3 Considerable progress in the theory and computer implementation of symbolic computation algorithms to automatically determine and exploit exact symmetries of exact DE. Underlying such algorithms are differential elimination algorithms (e.g. RifSimp and diffalg).

4 Considerable progress in the theory and computer implementation of symbolic computation algorithms to automatically determine and exploit exact symmetries of exact DE. Underlying such algorithms are differential elimination algorithms (e.g. RifSimp and diffalg). Often DE describing a model are only approximately known they may contain parameters that are only known approximately. Symbolic methods are unstable.

5 Considerable progress in the theory and computer implementation of symbolic computation algorithms to automatically determine and exploit exact symmetries of exact DE. Underlying such algorithms are differential elimination algorithms (e.g. RifSimp and diffalg). Often DE describing a model are only approximately known they may contain parameters that are only known approximately. Symbolic methods are unstable. In earlier work described a stable method for determining the size of symmetry groups of approximate DE.

6 Considerable progress in the theory and computer implementation of symbolic computation algorithms to automatically determine and exploit exact symmetries of exact DE. Underlying such algorithms are differential elimination algorithms (e.g. RifSimp and diffalg). Often DE describing a model are only approximately known they may contain parameters that are only known approximately. Symbolic methods are unstable. In earlier work described a stable method for determining the size of symmetry groups of approximate DE. Today we extend this numerical method to determining structure of the Lie algebra.

7 Considerable progress in the theory and computer implementation of symbolic computation algorithms to automatically determine and exploit exact symmetries of exact DE. Underlying such algorithms are differential elimination algorithms (e.g. RifSimp and diffalg). Often DE describing a model are only approximately known they may contain parameters that are only known approximately. Symbolic methods are unstable. In earlier work described a stable method for determining the size of symmetry groups of approximate DE. Today we extend this numerical method to determining structure of the Lie algebra.

8 Introduction & Historical Comments Symmetries and Introductory Examples Prolongation and Projection References

9 Symmetry defining system The (exact) symmetries of a differential equation are usually not known a priori and have to be determined. The Infinitesimal Lie symmetries of a DE = 0 are the linearized form of such symmetries about the identity transformation: ˆx = x + ɛξ(x, y) + O(ɛ 2 ) ŷ = y + ɛη(x, y) + O(ɛ 2 ) (1) that leave the invariant: pr(l)( ) =0 = 0 (2) and modulo certain regularity conditions result in a linear homogeneous system of over-determined PDE for the functions ξ, η and η [5, 19], which are called the symmetry defining system.

10 Consider the much studied class of ODE y xx = g(x, y, y x ) (3) The Symmetry Defining System for (3) is: ξ yx = 0, η yx = 0, g y η + g ( 3 y x ξ y 2 ξ x ) g x ξ (4) +g yx ( yx η y + y x 2 ξ y + y x ξ x η x ) +η x,x + y x 2 η y,y y x 3 ξ y,y 2y x 2 ξ x,y + 2y x η x,y y x ξ x,x = 0

11 In particular for our illustrative example we consider: y xx yy x + 5 y 3 = 0 (5) The symmetry defining system of (5) is: ξ yy = 0, η yy 2 ξ xy yξ y = 0, (6) 2 η xy ξ xx + 15 y 3 ξ y yξ x η = 0, η xx 5 y 3 η y + 10 y 3 ξ x yη x + 15 y 2 η = 0

12 Apply the differential-elimination packages RifSimp and DifferentialAlg for y xx yy x + 5 y 3 = 0. The defining system: ξ x = 0.0, ξ y = 0.0, η = 0.0 (7) which represents a 1 parameter translation symmetry in x for (5).

13 Apply the differential-elimination packages RifSimp and DifferentialAlg for y xx yy x + 5 y 3 = 0. The defining system: ξ x = 0.0, ξ y = 0.0, η = 0.0 (7) which represents a 1 parameter translation symmetry in x for (5). Convert the defining system to rationals: ξ x = 0, ξ y = 0, η = 0, (8) Convert the ODE to rationals, then create the defining system: ξ xx = 0, ξ y = 0, η = yξ x. (9)

14 Apply the differential-elimination packages RifSimp and DifferentialAlg for y xx yy x + 5 y 3 = 0. The defining system: ξ x = 0.0, ξ y = 0.0, η = 0.0 (7) which represents a 1 parameter translation symmetry in x for (5). Convert the defining system to rationals: ξ x = 0, ξ y = 0, η = 0, (8) Convert the ODE to rationals, then create the defining system: ξ xx = 0, ξ y = 0, η = yξ x. (9)

15 Apply the differential-elimination packages RifSimp and DifferentialAlg for y xx yy x + 5 y 3 = 0. The defining system: ξ x = 0.0, ξ y = 0.0, η = 0.0 (7) which represents a 1 parameter translation symmetry in x for (5). Convert the defining system to rationals: ξ x = 0, ξ y = 0, η = 0, (8) Convert the ODE to rationals, then create the defining system: ξ xx = 0, ξ y = 0, η = yξ x. (9) Then ξ = ax + b, η = ay correspoding to obvious symmetries of translation in x and a mutual scaling in (x, y) (x/c, cy) possessed by the input ode. Thus the symbolic simplification methods are not continuous with respect to small changes in the coefficients. Ideas, anyone?

16 Apply the differential-elimination packages RifSimp and DifferentialAlg for y xx yy x + 5 y 3 = 0. The defining system: ξ x = 0.0, ξ y = 0.0, η = 0.0 (7) which represents a 1 parameter translation symmetry in x for (5). Convert the defining system to rationals: ξ x = 0, ξ y = 0, η = 0, (8) Convert the ODE to rationals, then create the defining system: ξ xx = 0, ξ y = 0, η = yξ x. (9) Then ξ = ax + b, η = ay correspoding to obvious symmetries of translation in x and a mutual scaling in (x, y) (x/c, cy) possessed by the input ode. Thus the symbolic simplification methods are not continuous with respect to small changes in the coefficients. Ideas, anyone?

17 Symbolic substitution strategy For example embed the given ode in the one parameter class of ode: y xx + αyy x + 5y 3 = 0 (10) Application of RifSimp and diffalg yields: Case 1 (α 2 45, dim G = 2): Case 2 (α 2 = 45, dim G = 8): ξ xx = 0, ξ y = 0, η = yξ x (11) η = 2 15 ξ xxxy + 2 y 2 ξ xy yξ x 2 15 y αξ xxy αξ xx 1 3 αξ y y 3, ξ xxxx = 30 y 2 α ξ xxxy 30 y α ξ xxx, ξ yy = 0. (12) Indeed second order ode are linearizable [20] if and only if they have an eight dimensional group. Hence the ode (5) is very close to a linearizable ode.

18 Difficulties with symbolic approaches Thus care has to be taken with symbolic approaches Produce unstable results when used with numeric coefficients.

19 Difficulties with symbolic approaches Thus care has to be taken with symbolic approaches Produce unstable results when used with numeric coefficients. Symbolic replacement, although powerful in certain situations, can be impractical due to their greater complexity.

20 Difficulties with symbolic approaches Thus care has to be taken with symbolic approaches Produce unstable results when used with numeric coefficients. Symbolic replacement, although powerful in certain situations, can be impractical due to their greater complexity. It s useful to integrate symbolic and numeric methods, and such methods need to consider close-by systems.

21 Difficulties with symbolic approaches Thus care has to be taken with symbolic approaches Produce unstable results when used with numeric coefficients. Symbolic replacement, although powerful in certain situations, can be impractical due to their greater complexity. It s useful to integrate symbolic and numeric methods, and such methods need to consider close-by systems. Want a symbolic-numeric approach to determine that the case α = is close to a desirably large (8) dimensional symmetry group of a linearizable ode.

22 Difficulties with symbolic approaches Thus care has to be taken with symbolic approaches Produce unstable results when used with numeric coefficients. Symbolic replacement, although powerful in certain situations, can be impractical due to their greater complexity. It s useful to integrate symbolic and numeric methods, and such methods need to consider close-by systems. Want a symbolic-numeric approach to determine that the case α = is close to a desirably large (8) dimensional symmetry group of a linearizable ode.

23 Prolongation A q-th order system R = {R 1 = 0,..., R s = 0} where its j-th equation has order d j q has prolongation to order q + k : D k (R) = { K R j : K N n with K + d j (q + k)} (13) So the k-th prolongation consists of the set of all possible partial derivatives of the equations of R of order (q + k). Symmetry defining systems of form of (differential) order q are considered as a system R in matrix form: A (q) (z)v (q) = 0 (14) where v (q) is a column vector of all partial derivatives of infinitesimals of order q.

24 Prolongations DR, D 2 R,... yield the sequence of matrix systems: A (q) (z)v (q) = 0, A (q+1) (z)v (q+1) = 0,... (15) Then substitute a random point z = z 0 in the space of independent variables. This leads to a sequence of constant matrix systems: A (q) (z 0 )v (q) = 0, A (q+1) (z 0 )v (q+1) = 0,... (16) Note that v (q) has N(n, q, m) = m ( q+n q ) coordinates and ker A (q) (z 0 ) is a subspace of J q R N(n,q,m).

25 y xx yy x + 5 y 3 = 0 The symmetry defining system of our illustrative ODE above in matrix form with z = (x, y) is: y y y y y 2 v (2) = y y where 0 = [0, 0, 0, 0] T and v (2) = [ξ yy, η yy, ξ xy, η xy, ξ xx, η xx, ξ y, η y, ξ x, η x, ξ, η] T

26 Projection Define projection π by on v (q) J q by πv (q) J q 1 where πv (q) is obtained by deleting coordinates in v (q) of order q. Successive projections are defined by iteration: π 2 v (q) = ππv (q) by deleting coordinates of order q and q 1, etc. Define π l R := {π l v (q) J q k : A (q) (z 0 )v (q) = 0} (17) The main step of our geometric involutive form algorithm computes π l D k R (18) Symbolically: just compute symbolic prolongations and after subsituting z = z 0 use symbolic Gauss elimination to obtain the projections. But this is not numerically stable.

27 Numeric computation of π l D k R Apply the SVD to the matrix of each prolongation. Then an approximate basis is obtained for the kernel of the matrix by setting singular values to zero below a user input tolerance. To numerically compute projections coordinates corresponding to higher order derivatives are deleted in the basis, and this yields a spanning set for the projection as described in [9]. These spanning sets can be converted to a basis by another SVD calculation. These systems are tested for the property of projective involutivity which as shown in [9] is equivalent to Cartan involutivity. In summary in the finite d dimensional case we get d approximate basis vectors for a projective involutive system. These are used in the numerical computation of the structure constants outlined in Section 2.

28 y xx yy x + 5 y 3 = 0 Applying the symbolic-numeric method to case α = 7.1 yields Table 2: dim π l D k R for y xx yy x + 5 y 3 = 0 k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 l = l = l = l = l = l = l = 6 2 Guess the dimension of the symmetry group.

29 y xx yy x + 5 y 3 = 0 Applying the symbolic-numeric method to case α = 7.1 yields Table 2: dim π l D k R for y xx yy x + 5 y 3 = 0 k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 l = l = l = l = l = l = l = 6 2 Guess the dimension of the symmetry group. System π l D k R is approximately involutive for k = 4 and l = 4.

30 y xx yy x + 5 y 3 = 0 Applying the symbolic-numeric method to case α = 7.1 yields Table 2: dim π l D k R for y xx yy x + 5 y 3 = 0 k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 l = l = l = l = l = l = l = 6 2 Guess the dimension of the symmetry group. System π l D k R is approximately involutive for k = 4 and l = 4.

31 y xx yy x + 5 y 3 = 0 Applying the symbolic-numeric method to case α = 7.1 yields Table 2: dim π l D k R for y xx yy x + 5 y 3 = 0 k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 l = l = l = l = l = l = l = 6 2 Guess the dimension of the symmetry group. System π l D k R is approximately involutive for k = 4 and l = 4. Get a 2 dimensional solution space and symmetry group.

32 y xx yy x + 5 y 3 = 0 Applying the symbolic-numeric method to case α = 7.1 yields Table 2: dim π l D k R for y xx yy x + 5 y 3 = 0 k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 l = l = l = l = l = l = l = 6 2 Guess the dimension of the symmetry group. System π l D k R is approximately involutive for k = 4 and l = 4. Get a 2 dimensional solution space and symmetry group.

33 y xx yy x + 5 y 3 = 0 The symmetry defining system of our illustrative ODE above in matrix form with z = (x, y) is: y y y y y 2 v (2) = y y where 0 = [0, 0, 0, 0] T and v (2) = [ξ yy, η yy, ξ xy, η xy, ξ xx, η xx, ξ y, η y, ξ x, η x, ξ, η] T

34 y xx yy x + 5 y 3 = 0 The symmetry defining system of our illustrative ODE above in matrix form with z = (x, y) is: y y y y y 2 v (2) = y y where 0 = [0, 0, 0, 0] T and v (2) = [ξ yy, η yy, ξ xy, η xy, ξ xx, η xx, ξ y, η y, ξ x, η x, ξ, η] T Table 1: dim π l D k R for (5) k = 0 k = 1 k = 2 l = l = l = l = 3 6

35 y xx yy x + 5 y 3 = 0 The symmetry defining system of our illustrative ODE above in matrix form with z = (x, y) is: y y y y y 2 v (2) = y y where 0 = [0, 0, 0, 0] T and v (2) = [ξ yy, η yy, ξ xy, η xy, ξ xx, η xx, ξ y, η y, ξ x, η x, ξ, η] T Table 1: dim π l D k R for (5) k = 0 k = 1 k = 2 l = l = l = l = 3 6

36 We get the table of dimensions: Table 1: dim π l D k R for (5) k = 0 k = 1 k = 2 l = l = l = l = 3 6

37 We get the table of dimensions: Table 1: dim π l D k R for (5) k = 0 k = 1 k = 2 l = l = l = l = 3 6 For finite type systems the symbol is involutive if dim(s π l D k R) = 0 dim π l (D k R) = dim D l 1 (D k R) So πdr, πd 2 R and π 2 D 2 R have involutive symbols.

38 We get the table of dimensions: Table 1: dim π l D k R for (5) k = 0 k = 1 k = 2 l = l = l = l = 3 6 The system involutive if its symbol is involutive and dim π l+1 (D k+1 R) = dim π l (D k R) So πdr is an involutive system and we expect approximately an 8 dimensional solution space and symmetry group. This coincides with the dimension for the case α =

39 ode k l dim(g) 1 y y 2 = y 6 y 2 x 4 = y 6 y 2 x = y 6 y y = y + y x + 3 = y 2 y 3 yx + 1 = y y 3 = y 2 y yx 2 = y yx + 3 y + y 3 = y y y + y 3 = y + x 4 y 9 = y y 1 = y y = y y x 2 = (1 + x 2 )y xy 2 (x + 4y ) + 2(x + y )y 2y = ( 16 y 1 x 3 y ) 3 = y y 2 y+4 y x+2 = y y = y + 9 y = y 3 y y 2 2 y = y 7 y y y = y + 5 y 6 y y = y 3 y 2 y 3 + ( 2 y = 0) y 7 2 y y y 9 1 = y + y + 2 y 8 = y + y + 2 x 3 y 8 = x 4 y + y 8 ( = 0 ) x 4 y x x y y + 4 y 2 = x 4 y x 2 ( x + y ) y + 4 y 2 =

40 Symbolic-numeric algorithm for structure constants Input a DE. Compute its symmetry defining system. Make the defining system involutive. Use SVD for numerical stability. Read-off the cij k from the commutation relations.

41 Make the defining system involutive

42 Make the defining system involutive We suppose we have identified an involutive finite-type system π l D k R is where D k R has matrix form A (q) (z 0 ).

43 Make the defining system involutive We suppose we have identified an involutive finite-type system π l D k R is where D k R has matrix form A (q) (z 0 ).

44 Make the defining system involutive We suppose we have identified an involutive finite-type system π l D k R is where D k R has matrix form A (q) (z 0 ). The q = q l order involutive system π l D k R means that local values of derivatives of solutions are uniquely determined by π l D k R up to order q.

45 Make the defining system involutive We suppose we have identified an involutive finite-type system π l D k R is where D k R has matrix form A (q) (z 0 ). The q = q l order involutive system π l D k R means that local values of derivatives of solutions are uniquely determined by π l D k R up to order q.

46 Make the defining system involutive We suppose we have identified an involutive finite-type system π l D k R is where D k R has matrix form A (q) (z 0 ). The q = q l order involutive system π l D k R means that local values of derivatives of solutions are uniquely determined by π l D k R up to order q. To determine the structure constants in the comutation relations we need derivatives up to order q + 1 to be uniquely determined. This means that we need a pair of neighboring involutive systems π l D k R and π l 1 D k R.

47 Make the defining system involutive We suppose we have identified an involutive finite-type system π l D k R is where D k R has matrix form A (q) (z 0 ). The q = q l order involutive system π l D k R means that local values of derivatives of solutions are uniquely determined by π l D k R up to order q. To determine the structure constants in the comutation relations we need derivatives up to order q + 1 to be uniquely determined. This means that we need a pair of neighboring involutive systems π l D k R and π l 1 D k R.

48 SVD for stability

49 SVD for stability Input a tolerance ɛ.

50 SVD for stability Input a tolerance ɛ.

51 SVD for stability Input a tolerance ɛ. At random point z 0, compute the SVD A (q) (z 0 ) = UΣV T. Singular values below an input tolerance are replaced with zeroes, giving a nearby matrix A = UΣ V T = [ ] [ ] [ ] Σ U 1 U 1 0 V T where Σ 1 contains the numerically nonzero singular values. See Trefethen & Bau 06 and Akritas 06. V T 0

52 SVD for stability Input a tolerance ɛ. At random point z 0, compute the SVD A (q) (z 0 ) = UΣV T. Singular values below an input tolerance are replaced with zeroes, giving a nearby matrix A = UΣ V T = [ ] [ ] [ ] Σ U 1 U 1 0 V T where Σ 1 contains the numerically nonzero singular values. See Trefethen & Bau 06 and Akritas 06. V T 0

53 Project to subspace for involutive system Project the (column) vectors of V 0 to get Ṽ0 = π l 1 V 0 = {π l 1 (v) : v V 0 } by simply deleting the relevant coordinates in these column vectors.

54 Project to subspace for involutive system Project the (column) vectors of V 0 to get Ṽ0 = π l 1 V 0 = {π l 1 (v) : v V 0 } by simply deleting the relevant coordinates in these column vectors.

55 Project to subspace for involutive system Project the (column) vectors of V 0 to get Ṽ0 = π l 1 V 0 = {π l 1 (v) : v V 0 } by simply deleting the relevant coordinates in these column vectors. So Ṽ 0 is a spanning set for the involutive system π l 1 D k R at z = z 0. Convert it to a basis ˆV (q +1) 0 of π l 1 D k R. Then dim Ṽ 0 = dim L. Application of one more projection to this basis gives a basis ˆV (q ) 0 for π l D k R.

56 Project to subspace for involutive system Project the (column) vectors of V 0 to get Ṽ0 = π l 1 V 0 = {π l 1 (v) : v V 0 } by simply deleting the relevant coordinates in these column vectors. So Ṽ 0 is a spanning set for the involutive system π l 1 D k R at z = z 0. Convert it to a basis ˆV (q +1) 0 of π l 1 D k R. Then dim Ṽ 0 = dim L. Application of one more projection to this basis gives a basis ˆV (q ) 0 for π l D k R.

57 Project to subspace for involutive system Project the (column) vectors of V 0 to get Ṽ0 = π l 1 V 0 = {π l 1 (v) : v V 0 } by simply deleting the relevant coordinates in these column vectors. So Ṽ 0 is a spanning set for the involutive system π l 1 D k R at z = z 0. Convert it to a basis ˆV (q +1) 0 of π l 1 D k R. Then dim Ṽ 0 = dim L. Application of one more projection to this basis gives a basis ˆV (q ) 0 for π l D k R. Each basis vector v (q +1) i Ṽ0 uniquely determines local symmetry vector field of form X = ξ j and these vector fields x j close under the commutator bracket: [U, U ] = (ξ j ξ i x j ξ j ξi x j ) x i (19)

58 Project to subspace for involutive system Project the (column) vectors of V 0 to get Ṽ0 = π l 1 V 0 = {π l 1 (v) : v V 0 } by simply deleting the relevant coordinates in these column vectors. So Ṽ 0 is a spanning set for the involutive system π l 1 D k R at z = z 0. Convert it to a basis ˆV (q +1) 0 of π l 1 D k R. Then dim Ṽ 0 = dim L. Application of one more projection to this basis gives a basis ˆV (q ) 0 for π l D k R. Each basis vector v (q +1) i Ṽ0 uniquely determines local symmetry vector field of form X = ξ j and these vector fields x j close under the commutator bracket: [U, U ] = (ξ j ξ i x j ξ j ξi x j ) x i (19)

59 The structure constants c k ij are given by [X i, X j ] = r cij k X k, 1 i, j d (20) k=1 If Y = [U, U ] has components η i (z) then by (19) we have η i = ξ j ξ i ξ j ξi. x j x j By repeated symbolic differentiation one obtains any derivative I η i as a sum of bilinear skew terms of the form b ( J ξ i J ξ j J ξ i J ξ j) with J + J = I + 1 and coefficients b being integers. Compute these derivatives for order 0 I q.

60 Substitution of v (q +1) i, v (q +1) w (q ) ij j Ṽ 0 for I η i gives a vector of jet variable values uniquely determining Lie bracket. Since the solution vector fields are to form a Lie algebra, these vectors w (q) ij should lie in the numerical nullspace of π l D k R: w (q ) ij = k c k ij v (q ) k Projecting the vectors v (q +1) i Ṽ 0 yields {π 2 v (q +1) } which is a basis of d vector corresponding to the Lie algebra vector. Convert this to an orthonormal basis V (q 1) 0. As a consequence: cij k = w (q 1) ) T ij (v (q 1) k

61 Concluding Remarks We extended an algorithm that computes structure in the exact case to the approximate case. Exact methods lack properties of continuity, so we needed to consider nearby problems. Recast as matrix problem, using geometric techniques from Linear algebra (SVD) and geometry of differential equations. Many new unexplored approximate problems (approximate equivalence,... ) Increasingly mixture of algebra, analysis, geometry,... is needed Other methods for approximate symmetries Gaziov & Ibragimov,... Looking into the future will everything be done numerically?

62 Thank you

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