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

linear equations:(7.28) ⎧0 = Rx j k1 x k2···x kκ,0 = ∑δ l,.........,lk σ(1) ,...,k σ(κ−2)R j −x kσ(κ−1) x kσ(κ) u i 1⎪⎨⎪⎩σ∈S κ−2κ0 = ∑σ∈S κ−1κ− δ j i 1⎛⎝ ∑σ∈S κ−3κδ l,.........,lk σ(1) ,...,k σ(κ−1)R j x kσ(κ) u i 1 −− δ j i 1⎛⎝ ∑σ∈S κ−2κ0 = κ δ l,......,lk 1 ,...,k κR j u i 1u i 1 − κ δj i 1⎛⎝ ∑0 = 2κ δ l,......,lk 1 ,...,k κR j u i 1u i 2 − κ δj i 1⎛σ∈S κ−1κ⎝ ∑− κ δ j i 2⎛⎝ ∑0 = − C 2 κ+1 δ j i 1δ l,......,lk 1 ,...,k κQ l u i 1.173δ l,.........,lk σ(1) ,...,k σ(κ−3)Q l x kσ(κ−2) x kσ(κ−1) x kσ(k)⎞⎠,δ l,.........,lk σ(1) ,...,k σ(κ−2)Q l x kσ(κ−1) x kσ(κ)⎞⎠ ,σ∈S κ−1κσ∈S κ−1κδ l,.........,lk σ(1) ,...,k σ(κ−1)Q l x kσ(κ) u i 1⎞δ l,.........,lk σ(1) ,...,k σ(κ−1)Q l x kσ(κ) u i 2δ l,.........,lk σ(1) ,...,k σ(κ−1)Q l x kσ(κ) u i 1⎞⎠ ,⎞⎠ −⎠ , i 1 ≠ i 2 ,To solve the system (7.28) we fix the indices k 1 = · · · = k κ = l and j = i 1 inthe sixth equation, implying Q l = 0. Hence the terms following u i 1 δj i 1and δ j i 2in the fourth and in the fifth equations vanish identically. Let us choose theindices k 1 = · · · = k κ in the fourth and the fifth equations (this last equationis satisfied only for i 1 ≠ i 2 ). We obtain first three simple equations, withoutany restriction on the indices:⎧0 = Rx ⎪⎨j k1 x k2 ···x kκ,(7.28)0 = Q l u 1,⎪⎩i 0 = R j u i 1u . i 2Finally we specify the indices in the third equation of (7.28) as follows:l = k κ = · · · = k 3 = k 2 = k 1 ; then l = k κ = · · · = k 3 = k 2 ≠ k 1 ;finally l = k κ = · · · = k 3 , k 3 ≠ k 2 , k 3 ≠ k 1 . This gives the three following