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

68Finally, compute the linear combination (4.23)+(4.24)+2·(4.25)−2·(4.26):(4.27)0 = −2G l 1y l 1y l 1 + 4 3 Hl 1l 1 ,xy l 1 − 2 3 Ll 1l1 ,l 1 ,xx −− 2 3 Gl 1 x M l1 ,l 1− 4 3+ 2 ∑ k+ ∑ k+ 1 ∑2k− 1 ∑2+ ∑ kG k y l 1 Ll 1l1 ,k − 2 ∑ kHl k 1 ,x Ll 2l2 ,k + 1 ∑3k∑G k x M l 1 ,k+kkH k l 2 ,y l 1 Hl 2k+ 1 2H k l 1H l 1k,y l 1 − 1 2L l 2l2 ,k,x Hk l 1+ 1 3∑G k y l 1 Ll 2l2 ,k + 2 ∑ kH l 1k,xL k l 1 ,l 1− 1 3∑k∑+ 2 ∑ L l 1l 1 ,k,y l 1 Gk + 2 ∑kk− 2 3 M l 1 ,l 1 ,x G l 1− 4 ∑3kkk∑kH l 2k,y l 1 Hk l 2− 1 2H k l 1 ,y l 2 Hl 1k− 1 2L k l 1 ,l 1 ,x Hl 1k− 1 3∑kG k y l 2 Ll 2l1 ,k +H k l 1 ,x Ll 1l1 ,k − ∑ k∑k∑kL l 2l 1 ,k,y l 2 Gk − 2 ∑ kM l1 ,k,x G k .H k l 1 ,y l 1 Hl 1k−H k l 1H l 2k,y l 2 +L l 1l1 ,k,x Hk l 1− ∑ kL l 2l 2 ,k,y l 1 Gk −H k l 2 ,x Ll 2l1 ,k +L l 2l1 ,k,x Hk l 2+In this partial differential relation, importantly, the second order terms areexactly the same as in our goal (4.5). Unfortunately, the first order and thezero order terms are not the same.4.28. Formulation of a new goal. Thus, in order to get rid of the second orderexpression 2 G l 1y l 1y + 4 l 1 3 Hl 1− 2l 1 ,xy l 1 3 Ll 1l1 ,l 1 ,xx, we substract: (4.5)−(4.27).In the result, we write the first order terms in a certain way, adapted in advanceto our subsequent computations. For this substraction yielding (4.29)just below, we have not underlined the terms in (4.5) and in (4.27). However,they may be underlined with a pencil to check that the result (4.29) is