Characterization of the Newtonian free particle system in m 2 dependent variables

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1 Characterization of the Newtonian free particle system in m dependent variables Joël MERKER I Introduction II Proof of Lie s theorem III Systems of second order ordinary differential equations equivalent to free particles IV Compatibility conditions for the second auxiliary system Linearizability of a system of second order ordinary differential equations Applications to nonlinear Newtonian mechanics, especially in threedimensional space Let K = R or C, let x K, let m, let y := (y,, y m ) K m and let y xx = F (x, y, y x ),, y m xx = F m (x, y, y x ), be m analytic second order ordinary differential equations (nonlinear) Problem A Characterize linearizability Problem B Characterize equivalence, under a local diffeomorphism (x, y) (X, Y ), viz (x, y,, y m ) ( X(x, y), Y (x, y),, Y m (x, y) ), to the Newtonian free particle system YXX = = Y XX m = 0 Striing: m =, Lie, 883, System of order Two m, system of order One

2 Formal calculus Prolongation formulas Combinatorics Differential algebra Expression swelling I Introduction I Scalar equation Let K = R of C Let x K and y K Consider a local second order ordinary differential equation y xx = F(x, y, y x ), possibly nonlinear Theorem I (Lie, 883) The following four conditions are equivalent: () y xx = F(x, y, y x ) is equivalent under a local point transformation (x, y) (X, Y ) to the free particle equation Y XX = 0; () y xx = F(x, y, y x ) is equivalent to some linear equation Y XX = G 0 (X) G (X) Y H(X) Y X ; (3) the local Lie symmetry group of y xx = F(x, y, y x ) is eightdimensional, locally isomorphic to the group PGL(, K) of all projective transformations of P (K) (4) F yx y x y x y x = 0, or equivalently F = G y x H (y x ) L (y x ) 3 M, where G, H, L, M are functions of (x, y) that satisfy: 0 = G yy 4 3 H xy 3 L xx (GL)y G x M 4 GM x 3 H L x 4 3 H H y, 0 = 3 H yy 4 3 L xy M xx GM y 4 G y M (H M) x 3 H y L 4 3 L L x Lie s Grail would comprise:

3 a complete classification of all Lie algebras of local vector fields; Lie achieved this tas in dimension over C (real case: [?]); however, as soon as the dimension is 3, the complete classification is unnown, due to the intrinsic richness of imprimitive Lie algebras of vector fields; a list of all the possible Lie algebras that can be realized as infinitesimal Lie symmetry algebras of partial differential equations, together with their Levi-Malčev decomposition; an explicit Gröbner basis of the (noncommutative) algebra of all differential invariants of each equation in the list 3 I3 Systems Let x K, let m, let y := (y,, y m ) K m and consider yxx (x) =F (x, y(x), y x (x)), yxx(x) m =F m (x, y(x), y x (x)) Leach s conjecture (980): every linear system y j xx = Gj 0 (x) m l= y l G j,l (x) m j m is equivalent to free particles False: González-López (988) l = y l x H j l (x), Chern (939): É Cartan equivalence algorithm for fiber-preserving (x, y) (X(x), Y (x, y)) transformations (easy) Fels (995): É Cartan equivalence algorithm for general (x, y) (X(x, y), Y (x, y)) Not explicit Compute tensors partially Get at least: equivalence between () and (3) below Neut-Petitot (003): m =, complete, explicit, true (after corrections) Hachtroudi (937): Projective connexion Explicit for m = Duality General m Theorem I4 Suppose m The following three conditions are equivalent:

4 4 () the system y j xx = F j (x, y, y x ), j =,, m, is equivalent, under a local point transformation (x, y j ) (X, Y j ) to the free particle equation Y j XX = 0; () the local Lie symmetry group of y j xx = F j (x, y, y x ) is (m 4 m 3)-dimensional and locally isomorphic to PGL(m, K); (3) the right hand sides F j (x, y, y x ) are of a special form, described as follows (i) There exist local K-analytic functions G j, H j l, L j l,l and M l,l, where j, l, l =,, m, enjoying the symmetries L j l,l = L j l,l and M l,l = M l,l and depending only on (x, y) such that F j (x, y, y x ) may be written as the following specific cubic polynomial with respect to y x : y j xx =G j m l = y j x y l x H j l m l = m l = m l = m l = y l x y l x M l,l y l x y l x L j l,l (ii) The functions G j, H j l, L j l,l and M l,l satisfy the following explicit system of four families of first order partial differential equations: (I) 0 = G j y l δj l G l y l Hj l,x δj l H l m G L j l, m δj l G L l l, = δj l m = H l H l = m Hl H j, = l,x

5 5 where the indices j, l vary in {,,, m}; 0 = Hj l,y l 6 δj l H l l,y l 3 δj l H l l,y l L j l,l,x 3 δj l L l l,l,x 3 δj l,x (II) G j M l,l 3 δj l G l M l,l 3 δj l G l M 3 δj l m m = ( δ j l 6 ( δ j l 3 = G M l, 3 δj l H j,l m = m = m = H l L l,l 6 H l,l 3 m = H l L j l, m = m = G M l, H l L l l, H l l, ) ), where the indices j, l, l vary in {,,, m}; 0 = L j l,l,y l Lj 3 l,l 3,y l δj l 3 M l,l,x δ j l M l,l 3,x (III) Hj l 3 M l,l Hj l M l,l 3 m δj l Hl 3 M l, δj l = m δj l 3 Hl M l, δj l = = = m = m = m m L l,l 3 L j l, L l,l L j l 3,, Hl M l3, Hl M l3, (IV) where the indices j, l, l, l 3 vary in {, m}; and { m m 0 = M l,l,y l 3 M l,l 3,y l L l,l M l3, L l,l 3 M l,, = = where the indices l, l, l 3 vary in {,, m}

6 6 Corollary I5 For m, a linear (nonhomogeneous) system y j xx = G j 0 (x) m l= y l G j,l (x) m l = y l x H j l (x) is equivalent to Y j XX = 0 if and only if there exists a function B(x) such that the m m matrix G may be written under the specific form: G j,l = Hj l m H j 4 H l δ j l B = Detail each intermediate computation step Manual computations Combinatorial synthesis Failure of Maple when m 3 Power of mental induction II PROOF OF S LIE S THEOREM II Jacobians Let (x, y) (X(x, y), Y (x, y)) be a local K-analytic invertible transformation, defined near 0 K, transforming y xx = F(x, y, y x ) to Y XX = 0 (to be better understood soon) (x y) := X x X y 0 0 Y x Curve (x, y(x)) ( X(x, y(x)), Y (x, y(x)) ) : Y X := dy dx Total differentiation operator: D := x y x Y y =dx Y (x, y(x))/ x dx X(x, y(x))/ x = Y x y x Y y X x y x X y y y xx y x

7 7 First derivative Y X = DY DX Second derivative: Y XX := d Y dx DY X DX = D [ (Y x y x Y y )(X x y x X x ) ] X x y x X y = [X x y x X y ] {y 3 xx [X x Y y Y x X y ] X x Y xx Y x X xx y x [(X x Y xy Y x X xy ) (X xx Y y Y xx X y )] y x y x [X x Y yy Y x X yy (X xy Y y Y xy X y )] y x y x y x [(X yy Y y Y yy X y )]} What is it? is it complicated? ( 0 = Y XX ) = remove the numerator! Hidden determinant: X x Y xx Y x X xx = X x Y x X xx Y xx Appropriate notation: (xx y) := X xx Y xx X y Y y (x xy) := X x Y x X xy Y xy Combinatorics: 6 modifications of (x y): (xx y) (xy y) (yy y) (x xx) (x xy) (x yy) Reformulation: 0 = y xx X x X y Y x Y y X x X xx Y x Y xx { y x X x X xy Y x Y xy X xx Y xx { X y x y x x X yy Y x Y yy X xy Y xy { y x y x y x X } yy X y Y yy Y y X y Y y } } X y Y y

8 8 Equivalently: (divide by Jacobian, solve y xx ): { (x xy) (x y) (xx y) (x y) y xx = (x xx) (x y) y x { (y x ) (x yy) (x y) (xy y) (x y) { } (yy y) (y x ) 3 (x y) } } Notational convention: y 0 x, y y Define square functions: 0 xx := (xx y) (x y), xx := (x xx) (x y), 0 xy := (xy y) (x y), xy := (x xy) (x y), 0 yy := (yy y) (x y), y := (x yy) (x y) Summary: The equation y xx = F(x, y, y x ) is equivalent to the free equation Y XX = 0 if and only if there exist two local K-analytic functions X(x, y) and Y (x, y) such that it may be written under the form y xx = xx y x ( xy 0 xx) (y x ) ( y 0 xy) (yx ) 3 0 yy Observation: system under the form: y xx = G y x H (y x ) L (y x ) 3 M Where G, H, L, M functions of (x, y): G = xx, H = xy 0 xx, L = y 0 xy, M = 0 yy Squares = functions of (J X, J Y ) = Hidden nonlinear system of PDE s

9 = Not all systems 9 y xx = G y x H (y x ) L (y x ) 3 M are equivalent to Y XX = 0 First auxiliary system: 0 xx = Π0 0,0, 0 xy = Π0 0,, 0 yy = Π0,, xx = Π 0,0, xy = Π 0,, y = Π, where Π j,j given functions of (x, y): It is complete, approximatively equal to X xx = Λ 0 0,0, X xy = Λ 0 0,, X yy = Λ 0,, Y xx = Λ 0,0, Y xy = Λ 0,, Y yy = Λ, Cross differentiation: ( ) 0 xx y ( ) 0 xy = ( ) (xx y) ( ) (xy y) = x y (x y) x (x y) { = (xxy y) (x y) [ (x y)] a (xx yy) (x y) = (xy y) (xx y) b (x yy) (xx y) (xxy y) (x y) a (xy xy) (x y) c } (xy y) (xx y) b (xy y) (x xy) = { (xx yy) (x y) (x yy) (xx y) [ (x y)] (xy y) (x xy)} Plücer identity: A B A B C D C D = A D A D C B C B B D B D C A C A

10 0 Remove doubly differentiated determinants: (xx xy) (x y) = (xx y) (x xy) (xy y) (x xx), (xx yy) (x y) = (xx y) (x yy) (yy y) (x xx), (xy yy) (x y) = (xy y) (x yy) (yy y) (x xy) Insert: ( 0 xx ) y ( 0 xy )x = { (xx y) (x yy) (yy y) (x xx) [ (x y)] (x yy) (xx y) (xy y) (x xy)} = { (yy y) (x xx) (xy y) (x xy)} [ (x y)] = 0 yy xx 0 xy xy Lemma Identifying y 0 with x and y with y, for every 0 j, j, j 3,, we have : ( ) ( ) y j y j y j y 3 j y j 3 y j = =0 y j y j y j 3y =0 y j y j 3 y j y Equivalently: ( ) 0 xx y ( ) 0 xy = x xx 0 yy xy 0 xy ( ), 0 ( ) xy 0 y yy = x 0 xy 0 yx xy 0 yy 0 yy 0 xx y 0 xy, ( ) xx y ( ) xy = x 0 xx x xx y 0 xy xx xy xy ( ), ( ) xy yy = x 0 xy x 0 yy xx Proposition The first auxiliary system 0 xx = Π0 0,0, 0 xy = Π0 0,, 0 yy = Π0,, xx = Π 0,0, xy = Π 0,, y = Π,

11 has admits solutions (X(x, y), Y (x, y)) iff ( ) Π 0 ( ) 0,0 Π 0 y 0, = x Π 0,0 Π 0, Π 0, Π 0 0,, ( ) Π 0 ( ) 0, Π 0 y, = x Π0 0, Π0 0, Π 0, Π0, Π0, Π0 0,0 Π, Π0 0, ( ), Π ( ) 0,0 Π 0, = x Π0 0,0 Π 0, Π 0,0 Π, Π 0 0, Π, Π 0, Π 0,, ( ) Π ( ) 0, Π, = x Π0 0, Π 0, Π0, Π 0,0 Remind that true system is: G = xx, H = xy 0 xx, L = y 0 xy, M = 0 yy 4 functions G, H, L, M but 6 functions Π j,j : Quasi inversion: Π 0,0 = xx = G Π 0, = xy = H Θ0, Π 0 0, = 0 xy = L Θ, Π 0, = 0 yy = M 6 4 = : principal unnowns Θ 0 Π 0 0,0 and Θ Π, Insert in the compatibility conditions of the first auxiliary system: Solve in first derivatives of Θ 0, Θ : Get: second auxiliary system:

12 Θ = L y M x HM L M Θ 0 Θ 0 x = G y H x GL H G Θ ( Θ ), ( Θ 0 ), Θ x = 3 H y 3 L x GM H L H Θ L Θ0 Θ0 Θ, Θ 0 y = 3 H y 3 L x GM H L H Θ L Θ0 Θ0 Θ It is again complete: compatibilities: 0 = ( ) Θ 0 x y ( ) Θ 0 y 0 = ( ) Θ x y ( ) Θ Insert plainly without simplifying: x, x 0 = ( ) Θ 0 x y ( ) Θ 0 y x = G yy H xy G y L GL y H H y G y Θ G Θ Θ0 Θ 0 y 3 H xy 3 L xx G x M GM x H x L H L x H x Θ H Θ x L x Θ 0 L Θ0 x Θ0 x Θ Θ0 Θ x Caramba! must replace Θ 0 x, Θ 0 y, Θ x and Θ : Write: respect order: do not simplify yet: Brute expression:

13 3 0 = G yy 4 3 H xy 3 L xx G y L GL y H H y G y Θ GL y GM x GH M G (L) GM Θ 0 G (Θ ) 3 H y Θ 0 3 L x Θ 0 GM Θ 0 H L Θ0 H Θ0 Θ L (Θ0 ) (Θ0 ) Θ G x M GM x H x L H L x H x Θ 6 H L x 3 H H y GH M 4 (H) L 4 (H) Θ 4 H L Θ0 4 H Θ0 Θ L x Θ 0 G y L H x L G (L) 4 H L GLΘ 4 L ( Θ 0) Gy Θ H x Θ GLΘ 4 H Θ G (Θ ) 4 ( Θ 0 ) Θ 6 L x Θ 0 3 H y Θ 0 GM Θ 0 4 H L Θ0 4 H Θ0 Θ 4 L (Θ0 ) 4 (Θ0 ) Θ Simplify: 0 = G yy 4 3 H xy 3 L xx (GL) y G x M 4 GM x 3 H L x 4 3 H H y,

14 4 INTERLUDE ABOUT HAND COMPUTATIONS: m : manipulate terms m : each term 0 0 characters How to control a sea of signs? Color tric Keep a written trac of every computation Living coral tree of computations Keep a trac of every simplification Avoid copying too much Reorganize fastly after mistaes Blan corrector Several other virtual conventions PARADOXUS OF MECHANISMUS: TGV much better used than Maple!! III Passage to several dependent variables Assume yxx j = F j (x, y, y x ), j =,, m, is equivalent under (x, y) (X(x, y), Y (x, y)) to a free particle Y j XX = 0, j =,, m Jacobian determinant: (x y y m ) := X x X y X y m Yx Yy Y m Yy m m Y m x Y m y

15 Total differentiation: 5 D := x m l= y l x y m l l= y l xx y l x Transform of first order derivatives: Y j X = DY j DX = Y j x m l=l x Y j y l X x m l=l x X y l, Transform of second order derivatives: Y j XX = D ( ) Y j X DX = DDY j DX DDX DY j [DX] 3 Compact expression not understood ( 0 = Y j XX) = remove the numerator! Be human! Start with m = : Develope plainly the numerator:

16 6 0 = X xx Yx j Y xx j X x [ ] yx X xx Y j y Y j xx X y X xy Yx j Y j xy X x [ ] yx X xx Y j y Y j xx X y X xy Yx j Y j xy X x [ ] yx x X xy Y j y Y j xy X X y Y j x Y j y X x [ yx x X xy Y j y Y j xy X X xy Y j y Y j xy X y ] X y Y x j Y j y X x [ ] yx yx X xy Y j y Y j xy X y X y y Y x j Y j y y X x [ ] yx x y x X y Y j Y j X [ ] yx x y x X y Y j Y j y X X y Y j y Y j X y [ ] yx x y x X y y Y j y Y j y X y X y Y j y Y j y X y [ ] yx yx yx X y y Y j y Y j y y X y [ { }] yxx X y Yx j Y j y X x yx X y Y j y Y j y X [ { }] yxx X y Yx j Y j y X x yx X y Y j y Y j X y Bad luc! the system is not directly solved with respect to y xx and y xx! 0 = A y xx B y xx B, 0 = A y xx B y xx B, Cramer rule: No Maple: Discover many hidden determinants:

17 7 0 = yxx X x X y X y Yx Yy Y Yx Yy Y X x X xx X y Yx Yxx Y y Y x Yxx Yy yx X x X xy X y Yx Yxy Y Yx Yxy Y X xx X y X y Yxx Yy Y Y xx Yy Y yx X x X xy X y Yx Yxy Y y Yx Yxy Y y yx x X x X y X y Yx Yy Y Yx Yy Y X xy X y X y Yxy Y Y Y xy Y Y yx x X x X y X y Yx Yy Y y Yx Yy Y X xy X y X y Yxy Y y Y y Y xy Y y Y yx y x X x X y y X y Yx Yy y Y y Yx Yy y Y y yx x x X y X y X y Yy Y Y Yy Y Y y x yx x X y X y X y Yy Y y Y Y y Y y Y yx x yx X y y X y X y Yy y Y y Y Yy y Y y Y

18 8 Similarly: 0 = yxx X x X y X y Yx Yy Y Yx Yy Y X x X y X xx Yx Yy Y xx Y x Yy Y xx yx X x X y X xy Yx Yy Y xy Yx Yy Y xy yx X x X y X xy Yx Yy Y xy Yx Yy Y X xx X y X y Yxx Yy Y xy Y xx Yy Y yx x X x X y X y Yx Yy Y Yx Yy Y yx x X x X y X y Yx Yy Y Yx Yy Y X xy X y X y Yxy Y Y Y xy Y Y yx y x X x X y X y y Yx Yy Y y Yx Yy Y X xy X y X y Yxy Y y Y y Y xy Y y Y yx x y x X y X X Yy Y Y Yy Y Y y x y x y x X y X y X Yy Y y Y Y y Y y Y yx y x y x X y y X y X y Yy y Y y Y Yy y Y y Y Modifications of Jacobian determinant: examples: X y (y y y X y X y ) := Yy Y y Y Yy Y y Y X x X y X xy (x y xy ) := Yx Yy Y xy Yx Yy Y xy and

19 6 3 = 8 possible modifications: (xx y y ) (x xx y ) (x y xx) (xy y y ) (x xy y ) (x y xy ) (xy y y ) (x xy y ) (x y xy ) (y y y ) (x y y ) (x y y ) (y y y ) (x y y ) (x y y ) (y y y y ) (x y y y ) (x y y y ), 9 Square functions: 0 j, j, : (3) 0 xx := (xx y y ) (x y y ) 0 y := (y y y ) (x y y ) xx := (x xx y ) (x y y ) := (x y y ) (x y y ) xx := (x y xx) (x y y ) y := (x y y ) (x y y ) 0 xy := (xy y y ) (x y y ) 0 y := (y y y ) (x y y ) xy := (x xy y ) (x y y ) := (x y y ) (x y y ) xy := (x y xy ) (x y y ) y := (x y y ) (x y y ) 0 xy := (xy y y ) (x y y ) 0 y y := (y y y y ) (x y y ) xy := (x xy y ) (x y y ) y := (x y y y ) (x y y ) xy := (x y xy ) (x y y ) y y := (x y y y ) (x y y ) Lemma The system of two second order ordinary differential equations yxx = F (x, y, y x ) and yxx = F (x, y, y x ) is equivalent, under a point transformation, to the free particle system YXX = 0 and Y XX = 0 if and only if there exist three local K-analytic functions X(x, y), Y (x, y) and Y (x, y) such that it may be written under the form: 0 = y xx xx y x ( xy 0 xx) y x ( xy ) y x y x ( 0 xy ) y x y x ( 0 xy ) y x y x ( y ) y x y x y x ( 0 y ) y x y x y x ( 0 y ) y x y x y x ( 0 y y ), 0 = y xx xx y x ( xy ) y x ( xy 0 xx) y x y x ( y ) y x y x ( y 0 xy ) y x y x ( y y 0 xy ) y x y x y x ( 0 y ) y x y x y x ( 0 y ) y x y x y x ( 0 y y ) General m : (x y y m ) (x y l y l y m )

20 0 Convention: y 0 x Square functions: y l y l := (x y l y l y m ) (x ym ) Lemma The system yxx j = F j (x, y, y x ), j =,, m, is equivalent to the free particle system Y j XX = 0, j =,, m, if and only if there exist local K-analytic functions X(x, y) and Y j (x, y), j =,, m, such that it may be written under the specific form m 0 = yxx j j xx [ ] y l x j xy l δj l 0 xx l = m l = y j x m l = m l = [ ] y l x y l x j y l y l δj l 0 xy l δ j l 0 xy l m l = [ ] y l x y l x 0 y l y l Set: G j := j xx, H j l := j xy l δj l 0 xx, L j l,l := j y l y l δj l 0 xy l δ j l 0 xy, l M l,l := 0 y l y, l General form: y j xx =Gj First auxiliary system: m l = y j x y l x H j l m l = m l = m l = m l = y l x y l x M l,l y l x y l x L j l,l 0 xx = Π 0 0,0, 0 xy l = Π 0 0,l, 0 y l y l = Π 0 l,l, j xx = Π j 0,0, j xy l = Πj 0,l, j y l y l = Πj l,l

21 Lemma For all j, l, l, l 3 = 0,,, m, we have the cross differentiation relations : ( ) ( ) m m j j = y l y l y l y l 3 y l y l j y l 3y y l y l 3 j y l y y l 3 y l =0 =0 Proof sipped: Plücer identities: Matrices ( Proposition Existence and uniqueness of a solution (X, Y j ) = X(x, y j ), Y j (x, y j )) to the first auxiliary system if and only if the following 6 families of compatibility conditions hold true : ( ) Π j 0,0 (Π j y l 0,l )x m = Π0 0,0 Πj l,0 Π 0,0 Πj l, Π0 0,l Π j 0,0 m Π 0,l Π j 0,, = = ( ) (Π j l,l )x m Π j l,0 = Π 0 y l l,l Π j 0,0 m Π l,l Π j 0, Π0 l,0 Πj l,0 Π l,0 Πj l,, (Π j l,l ) y l 3 ) (Π j l,l3 y l = = m = Π 0 l,l Π j l 3,0 m Π l,l Π j l 3, Π0 l,l 3 Π j l,0 Π l,l 3 Π j l,, = ( Π 0 0,0 )y ( Π 0 l 0,l )x = Π0 0,0 Π0 l,0 m Π 0,0 a Π0 l, Π0 0,l Π 0 0,0 m Π 0,l a Π 0 0,, = ( ) Π 0 l,l x ( Π 0 l,0 )y m = l Π0 l,l Π 0 0,0 m Π l,l Π 0 0, Π0 l,0 Π0 l,0 Π l,0 Π0 l,, = ( Π 0 l,l )y ( Π 0 l3 l,l 3 )y m = l Π0 l,l Π 0 l 3,0 m Π l,l Π 0 l 3, Π0 l,l 3 Π 0 l,0 Π l,l 3 Π 0 l, Principal unnowns: = Θ 0 Π 0 0,0 and Θ j Π j j,j = = = = Quasi-inversion: Π j 0,0 = j xx = Gj, Π j 0,l = j xy l = Hj l δj l Θ 0, Π j l,l = j y l y l = Lj l,l δj l L l l,l δj l δj l Θ l δj l, Π 0 0,l = 0 xy l = Ll Θl, Π 0 l,l = 0 y l y l = M l,l Insert in the 6 families of compatibility conditions:

22 For instance, insert in the third family: (Π jl,l ) y l 3 ) (Π jl,l3 y l = = L j l,l,y l 3 δj l L l l,l,y l 3 δj l l,l,y l 3 δj l Θ l y l 3 δj l y l 3 L j l,l 3,y l δj l l 3,l 3,y l δj l 3 l,l,y l δj l y l δj l 3 y l = = Π 0 l,l Π j l 3,0 Π l,l Π j l 3, Π0 l,l 3 Π j l,0 = M l,l ( ) Θ 0 Hjl3 δjl3 δ l δ l Θ l ) δ l δj Ll 3 l3,l 3 δj l 3 Θ δj Θl 3 ) Π l,l 3 Π j l, = ( L l,l δ l L l l,l ( L j l 3, δj l 3 L, M l,l 3 ( ) Θ 0 Hjl δjl δ l 3 δ l ) δ l 3 δj Ll l,l δj l Θ ) δj Θl ( L l,l 3 δ l l3,l 3 ( L j l, δj l L,

23 Develope the products and order lexicographically: 3 = Hj l 3 M l,l δj l 3 M l,l Θ 0 6 Lj l,l l3,l 3 a δj l 3 L l,l Θ 3 L l,l L j l 3, 6 Lj l,l 4 δj l 3 L l l,l 4 4 δj l L l l,l l3,l 3 d 4 δj l 3 L l l,l δj l 3 b L l,l L, Ll l,l L j l 3,l c e 4 δj l L l l,l Ll L j l 3,l g 4 δj l 3 L l l,l h 4 δj l l3,l 3 i 4 δj l 3 Θ l 7 f j 4 δj l Lj l 3,l Θ l l 4 δj l 3 Θ l 0 4 δj l l3,l 3 Θ l m 4 δj l 3 Θ l n 4 δj l Θ l o Lj l 3,l p 4 δj l 3 L l l,l 4 δj l l3,l 3 Hj l M l,l 3 q 4 δj l 3 Θ l 8 δj l M l,l 3 Θ 0 5 Lj l,l 3 L l l,l c δj l L l,l 3 Θ 4 δj l 4 L l,l 3 L j l, r Lj l,l 3 Θ l 5 l δj l L l,l 3 L, Ll 3 l3,l 3 L j l,l a 8 4 δj l l3,l δj l l3,l 3 L l l,l d 4 δj l l3,l 3 q 4 δj l l3,l 3 Θ l Ll L j l,l 3 g 4 δj l l3,l 3 i 4 δj l 3 L l l,l h 4 δj l 4 δj l 3 Θ l j Lj l,l b 4 δj l 4 δj l L l l,l f 9 m 4 δj l 7 4 δj l Θ l o Lj l,l 3 p 4 δj l l3,l 3 4 δj l 3 L l l,l e 4 δj l r 4 δj l 3 Θ l n

24 4 Simplify: Solve first derivatives of Θ y : δj l Θ l y l 3 δj l y l δj l y l 3 δj l 3 y l = = L j l,l,y l 3 Lj l,l 3,y l δj l l 3,l 3,y l δj l L l l,l,y l 3 δj l 3 l,l,y l δj l l,l,y l 3 Hj l 3 M l,l Hj l M l,l 3 4 δj l l3,l 3 4 δj l 3 L l l,l L l,l 3 L j l, L l,l L j l 3, δj l 3 L l,l L, δj l L l,l 3 L, 4 δj l 4 δj l 3 Θ l 4 δj l l3,l 3 4 δj l 3 L l l,l δj l 3 L l,l Θ δj l L l,l 3 Θ δj l M l,l 3 Θ 0 δj l 3 M l,l Θ 0 4 δj l 4 δj l 3 Θ l Do the same for the 6 families: Get linear system involiving the first derivatives Θ j y l, 0 j, l m: Solve these partial derivatives: Firstly: Θ 0 x = Gl y l Hl l,x G l, G L, H l H l G Θ Θ0 Θ 0

25 Secondly: 5 Thirdly: Θ 0 y l = 3 Ll,x 3 Hl l,y l 3 Gl M H l l, G M l, 3 Ll Θ 0 Θ0 x = 3 Hl l,y l 3 Ll,x Fourthly: 4 3 Gl M 3 3 H l l, H l L, G M l, 3 Ll Θ 0 Θ0 y l = Ll l,l,y l M l,l,x H l L, H l,l H l Θ H l,l H l Θ H l M l, Ll L l l,l L l,l L, Ll Θ l Ll l,l L l,l Θ M l,l Θ 0 Θl Θ l Observation: Overdetermined system: Re-insert these solutions in the six families: Get (I), (II), (III), (IV) of the theorem: Conclusion: these are necessary conditions:

26 6 COMPATIBILITY CONDITIONS FOR THE SECOND AUXILIARY SYSTEM Complete system of four families of PDE s satisfied by Θ j y l : Attac first: 0 = ( ( ) Θ 0 x )y Θ 0 l l x = G l y l y l Hl l,xy l Θ 0 Θ 0 y l G y l l, H l,y l H l G l,,y l H l H l,y l G y l L, G y l Θ G L,,y l G Θ y l 3 3 Ll,xx 3 Hl l,y l x 3 Gl x M 3 Gl M,x 4 3 H l,x,l 3 H l,x L, H l,l,x 3 Hl L,,x G x M l, 4 3 Hl,x l, 3 H l,x Θ G M l,,x H l l,,x H l Θ x,l,x Θ 0,l Θ 0 x Θ0 x Θl Θ0 x Must replace the first derivatives of Θ: 0 = G l y l y l 4 3 Hl l,xy l 3 Ll,xx G y l Ll l, 4 H l,y l Hl 0 G l,,y l 6 H l H l,y l G y l L, 5 G y l Θ α G L,,y l a

27 G L,,y l a G L, Ll b G Θ d G M,l,x 8 G L p,l L p p,p p G H p M l,p 3 G L p,l Θ p p ε 3 Ll,x Θ0 g 3 Hl l,y l Θ0 h 3 Gl M Θ 0 4 G M l, Θ 0 i 3 e Hl L, Θ0 l 3 Gl x M 3 Gl M,x H l,x L l,l 9 H l,x L, Hl H,y Hl G p M,p p Hl H p Lp p,p p 3 p 9 G L, Θl G M,l Θ 0 e H l Θ 0 j 3 c 7 G Θ Hl Θ Θ 0 Ll Θ 0 Θ 0 n Θ0 Θ 0 o m 4 3 H l L l,l,x Hl L,,x Hl L,,x 6 3 H l Θ 0 Θ m Ll,x Θ0 g G x M l, Hl Hp L p, p 4 3 G M l,,x H l,x Ll l, Hl,x Θ γ Hl G M, f H l l, Θ0 8 H l l,,x Hl H p L,p Hl 4 H p Θp Hl 4 L, Θ0 p η l p 5 3 G l y l Ll 3 Hl l,x Ll 6 G l, G L, G Θ 4 Ll Θ 0 Θ 0 n d b 4 H l H l 7 G l y l Θl β Hl l,x Θl G l, Θl 4 Θ0 Θ 0 o ζ δ G L, Θl c 4 H l H l θ G Θ f

28 8 3 Hl l,y l Θ0 h 6 Ll,x Θ0 g 3 Gl M Θ 0 G M l, Θ 0 i 3 e Hl 4 L, Θ0 4 l We are far from the end! 3 H l L l,l Θ 0 j 3 Hl Θ 0 Θ 4 Ll Θ 0 Θ 0 n 4 Θ0 Θ 0 o m Main Proposition: follows from (I), (II), (III), (IV) Simplify: Underline: Mae unavoidable mistaes: Continue: Find disharmonies and incoherences: Coo-up a guide of computations: Focus on highest order terms: Chec correctness: Suffer: Return: Correct: Suffer: Return: Correct: H l l, Θ0