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

the values of the partial derivatives R j′x j ′1u i′ 1179and Q l′x k ′1x k ′2with k ′ 1 ≠ k′ 2 orl ′ ≠ k 1 ′ , l′ ≠ k 2 ′ appearing in the expressions Π of the second memberof (7.28) [1] by their values just obtained from the fifth, the sixth and theseventh equations of (7.28). This gives the first equation of (7.28).Then we specify the indices in (7.28) [2] as follows: l = k κ = · · · = k 3 =k 2 = k 1 ; then l = k κ = · · · = k 3 = k 2 ≠ k 1 ; then l = k κ = · · · = k 3 ,k 3 ≠ k 2 , k 3 ≠ k 1 ; and finally l = k κ = · · · = k 4 , l ≠ k 1 , l ≠ k 2 , l ≠ k 3 .This gives four equations, whose members on the right hand side are thesame as those in (7.28) and the members on the left hand side are the sameas those in (7.28) [2] :(7.28) ⎧⎪⎨⎪⎩ΠΠΠΠ((((x,u,Q l′ ,Q l′x k ′1,Q l′x k ′1x k ′2,R j′ ,R j′x k ′1,... ,R j′x k ′1 ···x k ′κ−1,R j′= C 2 κ R j x k1 x k1 u i 1 − C3 κ δ j i 1Q k 1x k1 x k1 x k1,x,u,Q l′ ,Q l′x k ′1,Q l′x k ′1x k ′2,R j′ ,R j′x k ′1,... ,R j′x k ′1 ···x k ′κ−1,R j′u i′ 1u i′ 1,R j′x k ′1u i′ 1,R j′x k ′1u i′ 1= C 1 κ−1 Rj x k1 x k2 u i 1 − C2 κ−1 δj i 1Q k 2x k1 x k2 x k2, k 2 ≠ k 1 ,x,u,Q l′ ,Q l′x k ′1,Q l′x k ′1x k ′2,R j′ ,R j′x k ′1,... ,R j′x k ′1 ···x k ′κ−1,R j′u i′ 1,R j′x k ′1u i′ 1= R j x k1 x k2 u − i 1 C1 κ−2 δ j i 1Q k 3x k1 x k2 x k3, k 3 ≠ k 1 , k 3 ≠ k 2 ,)x,u,Q l′ ,Q l′x k ′1,Q l′x k ′1x k ′2,R j′ ,R j′x k ′1,... ,R j′x k ′1 ···x k ′κ−1,R j′= −δ j i 1Q l x k1 x k2 x k3, l ≠ k 1 , l ≠ k 2 , l ≠ k 3 .u i′ 1,R j′x k ′1u i′ 1Using the fifth, the sixth and the seventh equations of (7.28) just obtained,we may replace the partial derivatives R j′and Q l′x k ′ x kwith k′ 1 ′ ≠ k′ 2 or1 2x j ′1u i′ 1l ′ ≠ k 1, ′ l ′ ≠ k 2 ′ appearing in the expressions Π of (7.28), providing four newequations in which the arguments of Π are the desired ones: (x, u, J), whereJ is defined in (7.28):(7.28) ⎧Π(x, u, J) = Cκ 2 R j − x k1 x k1 u i 1 C3 κ δ j i 1Q k 1⎪⎨x k1 x k1 x k1,Π(x, u, J) = C 1 κ−1 Rj x k1 x k2 u i 1 − C2 κ−1 δj i 1Q k 2x k1 x k2 x k2, k 2 ≠ k 1 ,Π(x, u, J) = R j − x k1 x k2 u i 1 C1 κ−2 δj i 1Q k 3x k1 x k2 x k3, k 3 ≠ k 1 , k 3 ≠ k 2 ,⎪⎩ Π(x, u, J) = − δ j i 1Q l x k1 x k2 x k3, l ≠ k 1 , l ≠ k 2 , l ≠ k 3 .)))====