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

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

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In<strong>de</strong>ed, the collection of equations (3.2) is a system of partial differentialequations with unknowns X l , Y , by virtue of the <strong>de</strong>finition of the squarefunctions.3.4. First <strong>aux</strong>iliary system. To proceed further, we observe that there are(m + 1) more square functions than functions G j1 ,j 2, H k 1j 1 ,j 2, L k 1j 1and M k 1.In<strong>de</strong>ed, a simple counting yields:⎧#{□⎪⎨k 1x j 1x } = n2 (n + 1), #{□ k j 122x j 1y } = n2 ,(3.5)#{□ k 1yy } = n,n(n + 1)#{□n+1 } = ,x⎪⎩j 1x j 22#{□ n+1x j 1y } = n,#{□n+1 yy } = 1,whereas⎧⎨#{G j 1,j 2n(n + 1)} = , #{H k 1j(3.6)21 ,j 2} = n2 (n + 1),2⎩#{L k 1j 1} = n 2 , #{M k 1} = n.Here, the indices j 1 , j 2 , k 1 satisfy 1 j 1 , j 2 , k 1 n. Similarly asin [Me2004], to transform the system (3.2) in a true complete system, letus introduce functions Π k 1j 1 ,j 2of (x l 1, y), where 1 j 1 , j 2 , k 1 n+1, whichsatisfy the symmetry Π k 1j 1 ,j 2= Π k 1j 1 ,j 1, and let us introduce the following first<strong>aux</strong>iliary system:{□k 1x(3.7)j 1x = j 2 Πk 1j 1 ,j 2, □ k 1x j 1y = Πk 1j 1 ,n+1 , □k 1yy = Πk 1n+1,n+1 ,□ n+1x j 1x = j 2 Πn+1 j 1 ,j 2, □ n+1x j 1y = Πn+1 j 1 ,n+1 , □n+1 yy = Π n+1n+1,n+1 .It is complete. The necessary and sufficient conditions for the existence ofsolutions X l , Y follow by cross differentiations.Lemma 3.8. For all j 1 , j 2 , j 3 , k 1 = 1, 2, . . ., n+1, we have the cross differentiationrelations(3.9)(□k 1)x j 1x −( )□ k j 12 x j 3 x j 1x j 3= − ∑n+1x j 2k 2 =1□ k 2x j 1x j 2 □k 1x j 3x + ∑n+1k 2k 2 =1371□ k 2x j 1x j 3 □k 1x j 2x k 2 .The proof of this lemma is exactly the same as the proof of Lemma 3.40in [Me2004].As a direct consequence, we <strong>de</strong>duce that a necessary and sufficient conditionfor the existence of solutions Π k 1j 1 ,j 2to the first <strong>aux</strong>iliary system is thatthey satisfy the following compatibility partial differential relations:(3.10)∂Π k 1j 1 ,j 2∂x j 3− ∂Πk 1j 1 ,j 3∂x j 2= −∑n=1k 2 =1∑n=1Π k 2j 1 ,j 2 · Π k 1j 3 ,k 2+k 2 =1Π k 2j 1 ,j 3 · Π k 1j 2 ,k 2,

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