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

523.97. Solving Θ 0 x , Θ0 , y l 1 Θl 1 x and Θ l 1. It is now easy to solve all first orderpartial derivatives of the functions Θ 0 and Θ l . Equation (3.93) alreadyy l 2provides the solution for Θ l 1. We state the result as an independent proposition.y l 2Proposition 3.98. As a consequence of the six families of equations (3.69),(3.86), (3.89), (3.91), (3.93) and (3.96) the first order derivatives Θ 0 x , Θ0 y l 1 ,Θ l 1 x and Θ l 1y l 2of the principal unknowns are given by:(3.99)⎧⎪⎨Θ 0 x = −2 G l 1y l 1 + Hl 1l1 ,x ++ 2 ∑ kG k L l 1l1 ,k − ∑ kG k L k k,k − 1 2∑kH k l 1H l 1k−⎪⎩− ∑ kG k Θ k + 1 2 Θ0 Θ 0 .(3.100)⎧Θ 0 y = 2 l 13 Ll 1l1 ,l 1 ,x − 1 3 Hl 1+ l 1 ,y l 1⎪⎨⎪⎩+ 2 3 Gl 1M l1 ,l 1+ 4 ∑G k M l1 ,k − 1 ∑H l 133 kL k l 1 ,l 1+kk+ 1 ∑Hl k 3 1L l 1l1 ,k − 1 ∑Hl k 2 1L k k,k − 1 ∑Hl k 2 1Θ k +k+ 1 2 Ll 1l1 ,l 1Θ 0 + 1 2 Θ0 Θ l 1.kk(3.101)⎧Θ l 1x = − 2 3 Hl 1+ 1l 1 ,y l 13 Ll 1l1 ,l 1 ,x +⎪⎨+ 4 3 Gl 1M l1 ,l 1+ 2 ∑G k M l1 ,k − 2 ∑H l 133 kL k l 1 ,l 1+kk+ 2 ∑Hl k 3 1L l 1l1 ,k − 1 ∑Hl k 2 1L k k,k − 1 ∑Hl k 2 1Θ k +⎪⎩k+ 1 2 Ll 1l1 ,l 1Θ 0 + 1 2 Θ0 Θ l 1.kk