By Hendriks P.A.
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Extra resources for Algebraic Aspects of Linear Differential and Difference Equations
A eld isomorphism : L ! M is a dierential K -isomorphism, if (a) = a for all a 2 K and (a) = (a) for all a 2 L. If M = L then is a dierential K -automorphism. Consider the system of linear dierential equations (A) : y = Ay, where A is a n n-matrix with coecients in K . 3 Dierential eld (L; L) is called a Picard-Vessiot extension of K associated with (A) if 1. L K and L jK = . 2. CL = CK = C . 30 3. (A) has n linear independent solutions over C in Ln . 4. L is minimal with respect to the conditions 1,2 and 3 or equivalently if U = (uij )i;j=1;:::;n 2 Lnn is a fundamental matrix of the system (A) then L = K (u11; : : : ; unn).
Let 1 1 l = 12 v(b) We write a = P an tn (al = 0 is allowed), b = P bntn and n=l n=2l 1 P n u = unt and try to solve the equation (y) : u(u) + au + b = 0. n=l The (2l)th-coecient of (y) gives us the equation u2l + al ul + b2l p= 0. Let D = a2l 4b2l . If D 6= 0 then we nd two solutions ul 2 k( D) of the quadratic equation. If D = 0 then we nd exactly one solution ul 2 k of the quadratic equation. The (2l + 1)th-coecient of (y) gives us the equation (2ul + al )ul+1 lu2l + al+1 ul + b2l+1 = 0. 55 If D 6= 0 then p 2ul + al 6= 0.
That means, if we did not need a quadratic extension of our eld of constants in step 1, it is possible that we really need a quadratic extension of our eld of constants in step 3 for solving the Riccati equation. 2. 2 G is reducible If the Riccati equation has a solution in K^ , then we can compute an equivalent system (A) : y = Ay, where A has the form ( a0 db ). Moreover we can assume that b = 0 if the Riccati equation has two or innitely many solutions. We denote by U the subgroup of Gl(2; Q) consisting of upper triangular matrices and by D the subgroup of diagonal matrices.