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.

326the fourth lines of (3.18), we may permute freely certain indices in some ofthe terms insi<strong>de</strong> the brackets. This yields the passage from lines 2, 3 and 4of (3.18) to lines 2, 3 and 4 of (3.20).It remains to explain how we pass from the fifth (last) line of (3.18) to thefifth (last) line of (3.20). The bracket in the fifth line of (3.18) contains threeterms: [−T 1 − T 2 − T 3 ]. The term T 3 involves the product δ k 1,k 3i 2 , i 1, which werewrite as δ k 3,k 1i 1 , i 2, in or<strong>de</strong>r that i 1 appears before i 2 . Then, we rewrite the threeterms in the new or<strong>de</strong>r [−T 2 − T 3 − T 1 ], which yields:(3.22)∑ [−δ k 1,k 3i 1 , i 2X k 2y − δ k 3,k 1i 1 , i 2X k 2y − δ k 2,k 3i 1 , i 2X k 1yk 1 ,k 2 ,k 3]y k1 y k2 ,k 3.It remains to observe that we can permute k 2 and k 3 in the first term −T 2 ,which yields the last line of (3.20). The <strong>de</strong>tailed proof is complete.3.23. Final perfect expression of Y i1 ,i 2 ,i 3. Thanks to similar (longer) computations,we have obtained an expression of Y i1 ,i 2 ,i 3which we consi<strong>de</strong>r tobe in final harmonious shape. Without copying the intermediate steps, let uswrite down the result. The comments which are necessary to read it and tointerpret it start just below.Y i1 ,i 2 ,i 3= Y x i 1x i 2x i 3 + ∑ k 1[δ k 1i 1Y x i 2x i 3y + δ k 1i 2Y x i 1x i 3y + δ k 1i 3Y x i 1x i 2y − X k 1x i 1x i 2x i 3]y k1 +(3.24)+ ∑ k 1 ,k 2[δ k 1,k 2i 1 , i 2Y x i 3y 2 + δ k 1,k 2i 3 , i 1Y x i 2y 2 + δ k 1,k 2i 2 , i 3Y x i 1y 2−]−δ k 1i 1X k 2x i 2x i 3y − δk 1i 2X k 2x i 1x i 3y − δk 1i 3X k 2yx i 1x i 2y k1 y k2 ++ ∑ []δ k 1,k 2 ,k 3i 1 , i 2 , i 3Y y 3 − δ k 1,k 2i 1 , i 2X k 3− δ k 1,k 2x i 3y 2 i 1 , i 3X k 3− δ k 1,k 2x i 2y 2 i 2 , i 3X k 3yx i 1y 2 k1 y k2 y k3 +k 1 ,k 2 ,k 3+ ∑ [−δ k 1,k 2 ,k 3i 1 , i 2 , i 3X k 4]yy 3 k1 y k2 y k3 y k4 +k 1 ,k 2 ,k 3 ,k 4+ ∑ k 1 ,k 2[δ k 1,k 2i 1 , i 2Y x i 3y + δ k 1,k 2i 3 , i 1Y x i 2y + δ k 1,k 2i 2 , i 3Y x i 1y −]−δ k 1i 1X k 2x i 2x − i 3 δk 1i 2X k 2x i 1x − i 3 δk 1i 3X k 2yx i 1x i 2 k1 ,k 2++ ∑ [δ k 1,k 2 ,k 3i 1 , i 2 , i 3Y y 2 + δ k 3,k 1 ,k 2i 1 , i 2 , i 3Y y 2 + δ k 2,k 3 ,k 1i 1 , i 2 , i 3Y y 2−k 1 ,k 2 ,k 3−δ k 1,k 2i 1 , i 2X k 3x i 3y − δk 3,k 1i 1 , i 2X k 2x i 3y − δk 2,k 3i 1 , i 2X k 1x i 3y −−δ k 1,k 2i 1 , i 3X k 3x i 2y − δk 3,k 1i 1 , i 3X k 2x i 2y − δk 2,k 3i 1 , i 3X k 1x i 2y −−δ k 1,k 2i 2 , i 3X k 3x i 1y − δk 3,k 1i 2 , i 3X k 2x i 1y − δk 2,k 3i 2 , i 3X k 1x i 1y]y k1 y k2 ,k 3+

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

Saved successfully!

Ooh no, something went wrong!