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

362By inspecting (2.5) above, we see that the equivalence between (i), (ii)and (iii) just below is obvious:Lemma 2.6. The following conditions are equivalent:(i) the differential equations Y X j Xk = 0 hold for 1 j, k n;(ii) the matrix equations D k (Y X ) = 0 hold for 1 k n;(iii) the matrix equations DX · D k (Y X ) = 0 hold for 1 k n;(iv) the matrix equations 0 = D k (DX)·Y X −D k (DY ) hold for 1 k n.Formally, in the sequel, it will be more convenient to achieve the explicitcomputations starting from condition (iv), since no matrix inversion at all isinvolved in it.Proof. Indeed, applying the total differentiation operator D k to the matrixequation (2.4) written under the equivalent form 0 = DX · Y X − DY , weget:(2.7) 0 = D k (DX) · Y X + DX · D k (Y X ) − D k (DY ),so that the equivalence between (iii) and (iv) is now clear.2.8. An explicit formula in the case n = 2. Thus, we can start to developeexplicitely the matrix equations(2.9) 0 = D k (DX) · Y X − D k (DY ).In it, some huge formal expressions are hidden behind the symbol D k . Proceedinginductively, we start by examinating the case n = 2 thoroughly. Bydirect computations which require to be clever, we reconstitute some 3 × 3determinants in the four (in fact three) developed equations (2.9). Aftersome work, the first equation is:(2.10)∣∣ ∣ ∣∣∣∣∣ X 10 = y x 1 x 1 · xX 1 1 xX 1 2 y ∣∣∣∣∣ X 2 xX 2 1 xX 2 2 y +Y x 1 Y x 2 Y y∣⎧ ∣⎨ ∣∣∣∣∣ X 1+ y x 1 ·⎩ 2 xX 1 1 xX 1 2 x 1 yX 2 xX 2 1 xX 2 2 x 1 yY x 1 Y x 2 Y x 1 y∣X 1 xX 1 1 xX 1 ∣∣∣∣∣ 2 x 1 x 1X 2 xX 2 1 xX 2 2 x 1 x+ 1Y x 1 Y x 2 Y x 1 x 1∣ ⎫∣∣∣∣∣ X 1 ∣ − x 1 xX 1 1 xX 1 2 y⎬X 2 x 1 xX 2 1 xX 2 2 yY x 1 x 1 Y x 2 Y ∣⎭ +y⎧ ∣ ⎫⎨ ∣∣∣∣∣ X 1+ y x 2 ·⎩ − xX 1 1 x 1 xX 1 1 y⎬X 2 xX 2 1 x 1 xX 2 1 yY x 1 Y x 1 x 1 Y ∣⎭ +y