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

We have to compute:( ) ⎛ ⎞∑ ∂(3.11)y i2,k 1⎝ ∑ [ ]−δ k1i∂y 1Xyk2y k1 y k2⎠ .k1k 1 k 1,k 2This term is of the form( ) ⎛ ⎞∑ ∂(3.12)A k1⎝ ∑[B k1,k∂y2] y k1 y k2⎠ ,k1k 1 k 1,k 2where the terms B k1 ,k 2are independent of the pure first jet variables y x k. Bythe rule of Leibniz for the differentiation of a product, we may write(3.13)( ) ⎛ ⎞∑ ∂A k1⎝ ∑[B k1,k∂y2]y k1 y k2⎠ =k1k 1 k 1,k 2⎛⎞⎛⎞= ∑ [B k1,k 2] y k2⎝ ∑ ∂A k ′1(y k1 ) ⎠ + ∑ [B k1,k∂y2] y k1⎝ ∑ ∂A k ′k 1,k 2 k 1′ k ′ 2(y k2 ) ⎠1∂yk 1,k 2 k 2′ k ′2= ∑ k 1,k 2[B k1,k 2] y k2 A k1 + ∑ k 1,k 2[B k1,k 2] y k1 A k2 .This is how we have written line 11 of (3.10).Next, the first term Y x i 1y y i2 in line 10 of (3.10) is not in a suitable shape.For reasons of harmony and coherence, we must insert it inside a sum of theform ∑ k 1[·] y k1 . Hence, using the Kronecker symbol, we transform:(3.14) Y x i 1y y i2 ≡ ∑ k 1[δk 1i 2Y x i 1y]yk1 .Also, we must “summify” the seven other terms, remaining in lines 10, 11and 12 of (3.10). Sometimes, we use the symmetry y i2 ,k 1≡ y k1 ,i 2withoutmention. Similarly, we get:∑(3.15)[δk 1]i 1Y yy − X k 1x i 1 y yk1 y i2 ≡ ∑ [k 1 k 1 ,k 2∑ ]yk1 y k2 y i2 ≡ ∑[−δk 1i 1X k 2yyk 1 ,k 2∑[ δk 1i 1Y y − X k 1x i 1k 1δ k 1i 1δ k 2i 2Y yy − δ k 2[−δk 1i 1δ k 3i 2X k 2yyk 1 ,k 2 ,k 3i 2X k 1x i 1y]yk1 y k2 y k3 ,]y k1 y k2 ,]yk1 ,i 2≡ ∑ k 1 ,k 2[δk 1i 1δ k 2i 2Y y − δ k 2i 2X k 1x i 1]yk1 ,k 2,∑ [−δk 1]i 1X k 2y yk2 y k1 ,i 2= ∑ [−δk 2]i 1X k 1y yk1 y k2 ,i 2k 1 ,k 2 k 1 ,k 2≡∑ [−δk 2]i 1δ k 3i 2X k 1y yk1 y k2 ,k 3,k 1 ,k 2 ,k 3323