Travaux sur les symétries de Lie des équations aux ... - DMA - Ens

246It then suffices to replace Y XX above by F(X, Y, Y X ) and to solve y xx :(3.5) ( (1 [Xx ]y xx =∣ X ∣ 3x X y ∣∣∣+ y x X y F X, Y, Y )x + y x Y y−X x + y x X y∣ X ∣x X xx ∣∣∣+Y x Y xxY x Y y{+y x · −2∣ X ∣ x X xy ∣∣∣ +Y x Y xy∣ X ∣}xx X y ∣∣∣+Y xx Y y{+y x y x · −∣ X ∣ x X yy ∣∣∣ + 2Y x Y yy∣ X ∣}xy X y ∣∣∣+Y xy Y y{∣ ∣})∣∣∣ X+y x y x y x · yy X y ∣∣∣Y yy Y y=: f(x, y, y x ).Open problem 3.6. Find general formulas expressing the F j α in terms of F ′ jα,x ′i , y ′j .Conversely, given two such systems (E ) and (E ′ ), when do they transformto each other ? Let π κ,p ′ denote the projection from J ′ κ+1n,m to ∆ E ′ definedby((3.7) π κ,p′ x ′i , y ′j , y ′ ji 1, . . .,y ′ j) ( )i 1 ,...,i κ+1 := x ′i , y ′j , y ′ j(q).β(q)Let ϕ (κ+1) be the (κ + 1)-th prolongation of ϕ (Section 1(II)).Lemma 3.8. ([Ol1986, BK1989, Ol1995]) The following three conditionsare equivalent:(1) ϕ transforms (E ) to (E ′ );(2) its (κ+1)-th prolongation ϕ (κ+1) : Jn,m κ+1 → J ′ κ+1n,m maps ∆ E to ∆ E ′;(3) ϕ (κ+1) : Jn,m κ+1 → J ′ κ+1n,m maps ∆ E to ∆ E ′ and the associated map(3.9) Φ E ,E ′ := π κ,p ′ ◦ ( ϕ (κ+1)∣ )∣∆Esends every leaf of F ∆E to some leaf of F ∆E ′.Equivalence problem 3.10. Find an algorithm to decide whether two given(E ) and (E ′ ) are equivalent.Élie Cartan’s widely applicable method (not reviewed here; [Ca1937,Ste1983, G1989, HK1989, Fe1995, Ol1995]) provides an answer “in principle”to this question by reducing to an {e}-structure an initial G-structureassociated to (E ). Due to the incredible size-length-complexity of the underlyingcomputations, this approach almost never abutes: it is forced toincompleteness. But in fact, the main question is to classify.