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

70they consists of the seven partial differential relations (4.32), (4.33), (4.35),(4.37), (4.39), (4.43) and (4.45) below. At the end, we shall make the addition(4.47) below, producing the desired subgoal (4.29) := (4.32) + (4.33)+ (4.35) + (4.37) + (4.39) + (4.43) + (4.45), with the numerotation of termscorresponding to the order of appearance of the terms of (4.29), as usual.Firstly, put j := k, l 1 := l 1 and l 2 := l 1 in (3.106):(4.31)0 = −2G k y + l 1 2δk l 1G l 1y + l 1 Hk l 1 ,x − δk l 1H l 1l1 ,x ++ 2 ∑ pG p L k l 1 ,p − 2δk l 1∑pG p L l 1l1 ,p − 1 2∑H p l 1Hp k + 1 ∑2 δk l 1ppH p l 1H l 1 p .∑Apply the operator 1 2 k Lk k,k (·) to the preceding equality, namely compute∑12 k Lk k,k · (4.31). This yields:(4.32)0 = − ∑ G k y l 1 Lk k,k + Gl 1y + 1 ∑H l 1 l k 2 1 ,x Lk k,k − 1 2 Hl 1l1 ,x Ll 1l1 ,l 1+kk+ ∑ ∑G p L k k,k Lk l 1 ,p − ∑ G p L l 1l1 ,l 1L k l 1 ,p − 1 ∑ ∑H p4l 1Hp k L k k,k +k ppk p+ 1 ∑H p4l 1H l 1 p L l 1l1 ,l 1.pSecondly, apply the operator − ∑ k Ll 2l2 ,k(·) to (4.32), namely compute− ∑ k Ll 2 · (4.32). This yields:(4.33)0 = 2 ∑ kl2 ,k− 2 ∑ k− 1 ∑2G k y l 1 Ll 2l2 ,k − 2Gl 1y l 1 Ll 2l2 ,l 1− ∑ kp∑pG p L l 2l2 ,k Lk l 1 ,p + 2 ∑ pH p l 1H l 1 p L l 2l2 ,l 1.H k l 1 ,x Ll 2l2 ,k + Hl 1l1 ,x Ll 2l2 ,l 1−G p L l 2l2 ,l 1L l 1l1 ,p + 1 2∑ ∑kpH p l 1H k p L l 2l2 ,k −Thirdly, put j := k, l 1 := l 2 and l 2 := l 1 with l 2 ≠ l 1 in (3.106):(4.34)0 = −2G k y + l 2 2δk l 2G l 1y + l 1 Hk l 2 ,x − δk l 2H l 1l1 ,x ++ 2 ∑ pG p L k l 2 ,p − 2δk l 2∑pG p L l 1l1 ,p − 1 2∑H p l 2Hp k + 1 ∑2 δk l 2ppH p l 1H l 1 p .