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

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368possible modified Jacobian <strong>de</strong>terminants:(2.27)⎧⎪⎨⎪⎩∆(x 1 x 1 |x 2 |y) ∆(x 1 |x 1 x 1 |y) ∆(x 1 |x 2 |x 1 x 1 )∆(x 1 x 2 |x 2 |y) ∆(x 1 |x 1 x 2 |y) ∆(x 1 |x 2 |x 1 x 2 )∆(x 1 y|x 2 |y) ∆(x 1 |x 1 y|y) ∆(x 1 |x 2 |x 1 y)∆(x 2 x 2 |x 2 |y) ∆(x 1 |x 2 x 2 |y) ∆(x 1 |x 2 |x 2 x 2 )∆(x 2 y|x 2 |y) ∆(x 1 |x 2 y|y) ∆(x 1 |x 2 |x 2 y)∆(yy|x 2 |y) ∆(x 1 |yy|y) ∆(x 1 |x 2 |yy).Next, we observe that if we want to solve with respect to y x 1 x1 in (2.10),with respect to y x 1 x 2 in (2.22) and with respect to y x 2 x2 in (2.23), we have todivi<strong>de</strong> by the Jacobian <strong>de</strong>terminant ∆(x 1 |x 2 |y). Consequently, we introduce18 new square functions as follows:(2.28)⎧⎪⎨⎪⎩□ 1 x 1 x 1 := ∆(x1 x 1 |x 2 |y)∆(x 1 |x 2 |y)□ 1 x 2 x 2 := ∆(x2 x 2 |x 2 |y)∆(x 1 |x 2 |y)□ 2 x 1 x 1 := ∆(x1 |x 1 x 1 |y)∆(x 1 |x 2 |y)□ 2 x 2 x 2 := ∆(x1 |x 2 x 2 |y)∆(x 1 |x 2 |y)□ 3 x 1 x 1 := ∆(x1 |x 2 |x 1 x 1 )∆(x 1 |x 2 |y)□ 3 x 2 x 2 := ∆(x1 |x 2 |x 2 x 2 )∆(x 1 |x 2 |y)□ 1 x 1 x 2 := ∆(x1 x 2 |x 2 |y)∆(x 1 |x 2 |y)□ 1 x 2 y := ∆(x2 y|x 2 |y)∆(x 1 |x 2 |y)□ 2 x 1 x 2 := ∆(x1 |x 1 x 2 |y)∆(x 1 |x 2 |y)□ 2 x 2 y := ∆(x1 |x 2 y|y)∆(x 1 |x 2 |y)□ 3 x 1 x 2 := ∆(x1 |x 2 |x 1 x 2 )∆(x 1 |x 2 |y)□ 3 x 2 y := ∆(x1 |x 2 |x 2 y)∆(x 1 |x 2 |y)□ 1 x 1 y := ∆(x1 y|x 2 |y)∆(x 1 |x 2 |y)□ 1 yy := ∆(yy|x2 |y)∆(x 1 |x 2 |y)□ 2 x 1 y := ∆(x1 |x 1 y|y)∆(x 1 |x 2 |y)□ 2 yy := ∆(x1 |yy|y)∆(x 1 |x 2 |y)□ 3 x 1 y := ∆(x1 |x 2 |x 1 y)∆(x 1 |x 2 |y)□ 3 yy := ∆(x1 |x 2 |yy)∆(x 1 |x 2 |y) .Thanks to these notations, we can rewrite the three equations (2.10),(2.22) and (2.23) in a more compact style.Lemma 2.29. A completely integrable system of three second or<strong>de</strong>r partialdifferential equations(2.30)⎧⎪⎨y x 1 x 1(x) = F ( 1,1 x 1 , x 2 , y(x), y x 1(x), y x 2(x) ) ,y x⎪⎩1 x 2(x) = F ( 1,2 x 1 , x 2 , y(x), y x 1(x), y x 2(x) ) ,y x 2 x 2(x) = F ( 2,2 x 1 , x 2 , y(x), y x 1(x), y x 2(x) ) ,is equivalent to the simp<strong>les</strong>t system Y X 1 X 1 = 0, Y X 1 X 2 = 0, Y X 2 ,X 2 = 0, ifand only if there exist local K-analytic functions X 1 , X 2 , Y such that it may

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