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

explicit computation, so let us rewrite the formula for G yxy xusing this newlyintroduced formalism:{1 [G yxy x=∆(a|b) 3 T aa Qb Q b ∆(a|b) ] [+ T ab − 2Qa Q b ∆(a|b) ] [+ T bb Qa Q a ∆(a|b) ] +211+ T a[Qa Q a ∆(b|bb) − 2Q a Q b ∆(b|ab) + Q b Q b ∆(b|aa) ] ++ T b[− Qa Q a ∆(a|bb) + 2Q a Q b ∆(a|ab) − Q b Q b ∆(a|aa) ]} .Then the twelve partial derivatives with respect to a and with respect to b ofall the six determinants ∆ ( ∗|∗ ) appearing in the the second line are easy towrite down:[ ( ) ( )∂∂b ∆ b|bb ] = ∆ bb|bb + ∆( b|bbb ) [ ( ) ( ) ( )∂0 ∂a ∆ b|bb ] = ∆ ab|bb + ∆ b|abb∂∂b∂∂b∂∂b∂∂b∂∂b[∆(b|ab)] = ∆(bb|ab)+ ∆(b|abb)[∆(b|aa)] = ∆(bb|aa)+ ∆(b|aab)[∆(a|bb)] = ∆(ab|bb)+ ∆(a|bbb)∂∂a∂∂a∂∂a[∆(b|ab)] = ∆(ab|ab)0 + ∆( b|aab )[∆(b|aa)] = ∆(ab|aa)+ ∆(b|aaa)[∆(a|bb)] = ∆(aa|bb)+ ∆(a|abb)[ ( ) ( )∆ a|ab ] = ∆ ab|ab + ∆( a|abb ) [ ( ) ( ) ( )∂0 ∂a ∆ a|ab ] = ∆ aa|ab + ∆ a|aab[ ( ) ( ) ( ) [ ( ) ( )∆ a|aa ] = ∆ ab|aa + ∆ a|aab∂∂a ∆ a|aa ] = ∆ aa|aa + ∆( a|aaa ) ,0and the underlined terms vanish for the trivial reason that any 2 ×2 determinant,two columns of which coincide, vanishes. Consequently, we may nowendeavour the computation of the third order derivative:(G yxy xy x=− Q b∆∂∂a + Q a∆∂∂bWhen applying the two derivations in parentheses to:G yxy x= 1 ∆ 3 {expression}) [Gyxyx].we start out by differentiating1 multiplied by expression, and then we∆ 3differentiate expression. Before any contraction, the full expansion of:∆ 5 G yxy xy x=(we indeed clear out the denominator ∆ 5 ) is then:= T aaˆ3Qb Q b Q b ∆(a|b)∆(aa|b) + 3Q b Q b Q b ∆(a|b)∆(a|ab) − 3Q aQ b Q b ∆(a|b)∆(ab|b) − 3Q aQ b Q b ∆(a|b)∆(a|bb)˜++ T abˆ− 6QaQ b Q b ∆(a|b)∆(aa|b) − 6Q aQ b Q b ∆(a|b)∆(a|ab) + 6Q aQ aQ b ∆(a|b)∆(ab|a) + 6Q aQ aQ b ∆(a|b)∆(a|bb)˜++ T bbˆ3QaQ aQ b ∆(a|b)∆(aa|b) + 3Q aQ aQ b ∆(a|b)∆(a|ab) − 3Q aQ aQ a∆(a|b)∆(ab|b) − 3Q aQ aQ a∆(a|b)∆(a|bb)˜++T ah3Q aQ aQ b ∆(b|bb)∆(aa|b) + 3Q aQ aQ b ∆(b|bb)∆(a|ab) − 3Q aQ aQ a∆(b|bb)∆(ab|b) − 3Q aQ aQ a∆(b|bb)∆(a|bb)−− 6Q aQ b Q b ∆(b|ab)∆(aa|b) − 6Q aQ b Q b ∆(b|ab)∆(a|ab) + 6Q aQ aQ b ∆(b|ab)∆(ab|b) + 6Q aQ aQ b ∆(b|ab)∆(a|bb)+i+ 3Q b Q b Q b ∆(b|aa)∆(aa|b) + 3Q b Q b Q b ∆(b|aa)∆(a|ab) − 3Q aQ b Q b ∆(b|aa)∆(ab|b) − 3Q aQ b Q b ∆(b|aa)∆(a|bb) +