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

of the cancellation properties (11.29), we get:⎧Jz λ ϕ( Γ 1 ([xa] 1 ) ) (≡ Φ λ 0 Γ1 ([xa] 1 ), J κ∗ +λc h ( Γ 1 ([xa] 1 ) ))(≡ Φ λ 0 Γ1 ([xa] 1 ), J κ∗ +λc h(0)) )⎪⎨(=: Φ λ 1 [xa]1 , J κ∗ +λc h(0) ) ,(11.34)Jc λ h( Γ ∗ 1 ([ax] 1) ) (≡ H0λ Γ∗1 ([ax] 1 ), Jz κ+λ ϕ ( Γ ∗ 1 ([ax] 1) ))(≡ H0λ Γ∗1 ([ax] 1 ), Jz κ+λ ϕ(0) )⎪⎩(=: H1λ [ax]1 , Jz κ+λ ϕ(0) ) .Here, the third line defines Φ λ 1 and the sixth line defines Hλ 1 . Thus, theproposition holds for 2k + 1 = 1.The rest of the proof proceeds by induction. We treat only the inductionstep from an odd chain-length 2k + 1 to an even chain-length 2k + 2, theother induction step being similar.To this aim, we replace the variables (z, c) in the first line of (11.27)by Γ ∗ 2k+2 ([ax] 2k+2). Taking account of the cancellation property and of theinduction assumption:(11.35)Jz λ ϕ ( Γ ∗ ( ))2k+2 [ax]2k+2 ≡ Φλ0(Γ ∗ 2k+2 ([ax] 2k+2), J κ∗ +λc h ( Γ ∗ 2k+2 ([ax] 2k+2) ))(≡ Φ λ 0 Γ ∗ 2k+2 ([ax] 2k+2), J κ∗ +λc h ( Γ ∗ 2k+1 ([ax] 2k+1) ))(())≡ Φ λ 0 Γ ∗ 2k+2 ([ax] 2k+2), H κ∗ +λ[ax] 2k+1 , J (k+1)(κ+κ∗ )+λc ϕ(0)=: Φ λ 2k+2(2k+1[ax] 2k+2 , J (k+1)(κ+κ∗ )+λc2k+1)ϕ(0) ,The last line defines Φ λ 2k+2 . Similarly, we replace (z, c) in the second lineof (11.27) by Γ 2k+2 ([xa] 2k+2 ). Taking account of the cancellation propertyand of the induction assumption:(11.36)Jc λ h ( Γ 2k+2 ([xa] 2k+2 ) ) (≡ H0λ Γ 2k+2 ([xa] 2k+2 ), Jc κ+λ ϕ ( Γ 2k+2 ([xa] 2k+2 ) ))(≡ H0λ Γ 2k+2 ([xa] 2k+2 ), Jc κ+λ ϕ ( Γ 2k+1 ([xa] 2k+1 ) ))(())≡ H0λ Γ 2k+2 ([xa] 2k+2 ), Φ κ+λ [xa] 2k+1 , J (k+1)(κ+κ∗ )+λc h(0)This completes the proof.=: H λ 2k+2([xa] 2k+2 , J (k+1)(κ+κ∗ )+λc)h(0) .End of the proof of Theorem 11.6. With (µ, µ ∗ ) being the type of (F v , F p )and with [ax] 0 2µ ∗ given by Corollary 10.31, the rank property (10.32) insuresthe existence of an affine (n+m)-dimensional space H ⊂ K µ∗ (p+n) passing303