11.07.2015 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

175R j containing at least one partial <strong>de</strong>rivative with respect to u i 1:⎧⎨ 0 = Q l x k1 x k2 x k3= Q l x k1 x k2 u(7.28)i 1= Q l x k1 u i 1u i 2= Q l u i 1u i 2u 3, i⎩ 0 = R j = x k1 x k2 u i 1 Rj x k1 u i 1u = i 2 Rj u i 1u i 2u . i 3It follows from the equations (7.28) and (7.28) that all the functions Q l arepolynomials of <strong>de</strong>gree ≤ 2 with respect to the variab<strong>les</strong> x k1 and all the functionsR j are a sum of a polynomial of <strong>de</strong>gree ≤ (κ − 1) in the variab<strong>les</strong> x k1and of monomials of the form u i 1and x k1 u i 1. Let us <strong>de</strong>velop now the relations(7.28) separately for j = i 1 and j ≠ i 1 . We obtain the five equations:(7.28)⎧0 = Cκ ⎪⎨1 Rj , j ≠ i x k1 u i 1 1,0 = R i 1− x k1 u i 1 C1 κ−1 Q k 2x k1 x k2, k 2 ≠ k 1 ,⎪⎩0 = C 1 κ R i 1x k1 u i 1 − C2 κ Q k 1x k1 x k1,0 = R j x k1 u i 1 , j ≠ i 1,0 = − Q k 3x k1 x k2, k 3 ≠ k 1 , k 3 ≠ k 2 .According to the equations (7.28), (7.28), (7.28), we have the followingform of the general solution:(7.28) ⎧n∑n∑Q l (x, u) = A l + Bk l 1x k1 + C k1 x k1 x l ,⎪⎨R j (x, u) =⎪⎩k 1 =1k 1 =1n∑(κ − 1) C k1 x k1 u j +k 1 =1+ · · · +∑1≤k 1 ≤···≤k κ−1 ≤nm∑D j i 1u i 1+ E j,0 +i 1 =1E j,κ−1k 1 ,...,k κ−1x k1 · · ·x kκ−1 .n∑k 1 =1E j,1k 1x k1 +Here the n+n 2 +n+m 2 +m Cn+κ−1 constants A l , Bk l 1, C k1 , D j i 1, E j,0 , E j,1k 1,. . . , E j,κ−1k 1 ,...,k κ−1∈ K are arbitrary. Moreover one can check that the vectorspace spanned by the vector fields(7.28)⎧⎪⎨∂∂x k1, x k1∂,∂x k2∂+ · · · + x n∂x 1((∂x k1 x 1+ (κ − 1) u 1 ∂∂x n ∂u + · · · + ∂ ))1 um ,∂u m⎪⎩ u i ∂1∂u , ∂i 2 ∂u , x ∂i k 11 ∂u , . . ...., x ∂i k 1· · ·x kκ−11 ∂u , i 1is stable un<strong>de</strong>r the <strong>Lie</strong> bracket action and that the flow of each of thesegenerators is in<strong>de</strong>ed a <strong>Lie</strong> symmetry of the system (E 0 ). Finally the <strong>Lie</strong>

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!