Differential Resultants and Subresultants

Transcription

1 1 Differential Resultants and Subresultants Marc Chardin Équipe de Calcul Formel SDI du CNRS n 6176 Centre de Mathématiques et LIX École Polytechnique F Palaiseau (France) chardin@polytechniquefr Abstract Consider two differential operators L 1 = a i d i and L 2 = b j d j with coefficients in a differential field, say C(t) with d = t for example If the a i and b j are constants, the condition for the existence of a solution y of L 1 (y) = L 2 (y) = 0 is that the resultant in X of the polynomials (in C[X]) a i X i and b j X j is zero A natural question is: how one could extend this for the case of non constant coefficients? One of the main motivation in the resultant techniques is the universality of the computation result When you calculate the resultant, or the subresultants, you always stay in the ring where the coefficients lies and the result can be specialized to every particular case (if the equations depends on parameters for example) to get the information in theses cases In this paper, we define all the differential equivalents of these classical objects In theorem 4 we describe the universal properties of the differential subresultants They are due to the fact that, as in the polynomial case, we can wor out the problem by means of linear algebra As it was nown by Ritt ([R]), the differential resultant is a polynomial in the a i, b j and their derivatives that tells us in the general algebraic context of linear differential equations over a differential field if they have a common solution In particular, if you now in a given context that there exists a base of solutions (eg you have C coefficients and you wor locally (Cauchy s theorem)), this universal calculation answers this existence problem in this particular context We here give a Sylvester style expression for the resultant and the subresultants So, as theses objects are determinants, you can get size information on them if you have any reasonable notion of size in the differential ring and, as a corollary, complexity estimates for the calculation At least three techniques may be used: Gauss elimination type techniques (eg the Bareiss algorithm) if the ring does not contain zero divisors, Berowitz fast parallel algorithm ([Be]) or the natural extension of the subresultant algorithm As in the classical case, the differential resultant can be expressed in terms of the values of one of the operators on a base of the solutions of the other, we exhibit this formula and most of the differential equivalents of the classical ones

2 I The differential resultant 2 Let K be a differential field and L 1, L 2 two differential operators with coefficients in K A natural problem is to discover if there exists a differential extension of K, and an element y in this extension, such that L 1 (y) = L 2 (y) = 0 In order to give an algebraic answer to this question, it is very convenient to assume that any linear differential equation defined by an element of K [ ] admits a solution in a suitable extension of K ; that is the case in a very natural fashion if the characteristic of K is zero (see eg [Ka]) As we will see the sentence have a common zero may be replaced by the one have a non trivial left gcd if you assume the preceeding fact, and that is the convenient algebraic condition As in the polynomial case, there exists a universal polynomial that gives the answer, this result is probably due to J F Ritt [R] We shall call this polynomial the differential resultant The basic result is the following easy theorem: Theorem 1 The ring K [ ] is left euclidean (for the stathme defined by the order) Let L 1 and L 2 be two differential operators of respective orders m and n, and suppose, for example, that m n The operator L 1 (p/q) m n L 2, where p and q are the leading coefficients of L 1 and L 2, is of order less than m This shows the existence by induction; the uniqueness is also easy Corollary Every left ideal of K [ ] is left principal By successive euclidean divisions (on the left) we get successive differential operators F 1 = L 1, F 2 = L 2, F 3, of decreasing orders, every pair (F, F 1 ) having the same common solutions as the pair (L 1, L 2 ) The non zero operator of minimal order F in this sequence is the left gcd of L 1 and L 2 ; the common solutions of L 1 = 0 and L 2 = 0 are exactly the ones of F F can be expressed in a unique way as F = A L 1 + B L 2, where the orders of A and B satisfy the natural restrictions on their orders As in the polynomial case, the search of A and B is equivalent to the resolution of a linear system, and a necessary and sufficient condition for the existence of common solutions is given by a determinant, the differential resultant Let L = m i=0 a i i, and let S () (L, n) be the following (m+n ) (n ) matrix: ( ) a 0 ( +1 ) a 0 ( ) ( a m n 1 ) a 0 ( +1 ) a m ) a m ( n 1

3 3 ( ) 0 S and let S () (L, n) be the (m + n) n bloc matrix () (L, n) 0 0 The following easy lemma, is just what we need to determine the coefficients of the linear system corresponding to the Bézout relation mentioned above Lemma 1 Let a ij be defined by j L 1 = i a ij i (notice that a ij = 0 if i > m + j) Consider the matrix A of the a ij, i from 0 to m + n 1 and j from 0 to n 1 We then have: n 1 A = S(L 1, n) = S () (L 1, n) Let S(L 1, L 2 ) be the square matrix obtained by concatenation of S(L 1, n) and S(L 2, m) =0 The next result is just a rephrasing of what we explained: Theorem 2 Let L 1 and L 2 be two differential operators of orders m et n respectively Then L 1 and L 2 have a non trivial left gcd if and only if det(s(l 1, L 2 )) = 0 Let us mae two easy remars: (1) In the case of operators with constant coefficients, we recognize the classical resultant (2) This determinant is an irreducible polynomial of Z[a (j) i, b (l) ], because it is ho- to mogeneous and irreducible when you specialize all the variables but a (0) i and b (0) j zero Exemple Let L 1 = a a 1 + a 0 and L 2 = b b 1 + b 0, then: a 0 a 0 b 0 b 0 a det S(L 1, L 2 ) = 1 a 1 + a 0 b 1 b 1 + b 0 a 2 a 2 + a 1 b 2 b 2 + b 1 0 a 2 0 b 2 If you develop, you will find: a 2 0b 2 2 +a 0 a 1b 2 2 a 0 a 1 b 1 b 2 a 0 a 2b 1 b 2 +a 0 a 2 b 2 1 +a 0 a 2 b 1 b 2 2a 0 a 2 b 0 b 2 a 0 a 2 b 1b 2 a 1 a 0b a 2 1b 0 b 2 +a 1 a 2b 0 b 2 a 1 a 2 b 0 b 1 a 1 a 2 b 0 b 2+a 1 a 2 b 0b 2 +a 2 a 0b 1 b 2 a 2 a 1b 0 b 2 +a 2 2b 2 0+a 2 2b 0 b 1 a 2 2b 0b 1 II Differential subresultants As in the polynomial case, we will now define the subresultants Let us define E[l] as the vector space of differential operators of K [ ] of order strictly less than l, equipped with the base (1,,, l 1 ) Remember that the resultant is the determinant of the map φ 0 defined by: E[n ] E[m ] (A, B) φ E[m + n ] φ AL1 + BL 2

4 The matrix S of φ loos lie this: 4 a 0 a 0 a (n 1) 0 b 0 b 0 b (m 1) 0 a m bn a m b n It is the concatenation of S(L 1, n ) and S(L 2, m ) Let s call S i the square minor of S obtained by erasing lines 1 to + 1 except line i + 1 and put S ( ) = det(s) i i i=0 We will now derive the ey result for our purpose: S is the left gcd of L 1 and L 2 if theses operators have a left gcd of degree, and in fact finest results But first let s give a definition and a lemma from linear algebra: Definition 1 If A is a matrix we will denote by M j 1,,j p i 1,,i p (A) the determinant of the submatrix of A corresponding to lines i 1,, i p and columns j 1,, j p Lemma 2 Let A M p q, B M q r be two matrices and suppose that p q r, then : M 1,,r i 1,,i r (AB) = M j 1,,j r i 1,,i r (A)M 1,,r j 1,,j r (B) j 1,,j r [1,q] We are sure that this is not a new result! But we will setch the proof for lac of reference If we denote by A I (resp (AB) I ) the submatrix of A (resp AB) constituted by the lines i 1,, i r, we have (AB) I = A I B, so we are reduced to the following statement: if u 1,, u r, v 1,, v r are 2r vectors of a q dimensional vector space E q, then det(< u i, v j >) 1 i r = 1 j r i 1,,i r [1,q] < u 1 u r, e i1 e ir >< e i1 e ir, v 1 v r > where <, > denotes the standard scalar product This is easy to show when you remar these two expressions defines a map that factors through r E q r E q, and from that observes these two elements are both equal to < u 1 u r, v 1 v r > Let L 1, L 2 and L be three differential operators of respective orders m, n and We remar that S +l (L 1 L, L 2 L) is the product of two matrices, corresponding to the following decomposition of φ +l : φ l (L 1,L 2 ) E[n l] E[m l] E[m + n l] L E[m + m l + ] (A, B) AL 1 + BL 2 AL 1 L + BL 2 L so S +l (L 1 L, L 2 L) = S(L, m + n) S l (L 1, L 2 )

5 Remember that if L = c + + c 0, S(L, t) = 5 c 0 c 0 c (t 1) 0 c c so that one immediatly sees that for l = 0 we have: S (L 1 L, L 2 L)( ) = c m+n 1 S 0 (L 1, L 2 )L( ) In the same vein, we will show that in the general case you have: Theorem 3 Let L 1, L 2 and L be three differential operators of respective orders m, n and For 0 l min{m, n}, we have: S +l (L 1 L, L 2 L) = c m+n 2l 1 S l (L 1, L 2 )L Let s put r = m + n 2l To prove the theorem, we need to calculate M 1,,r i,+l+2,,+l+r (S +l(l 1 L, L 2 L)) that is the determinant of the product: c(i, 0) c(i, l) c(i, l + 1) c(i, s) 0 0 c 0 S l (L 1, L 2 ) c where c(i, j) is defined by j L = +j i=0 c(i, j) i, (c(i, j) = 0 if i > + j) and s = + l + r 1 As all the maximal square submatrices of this matrix must contain one of the first l + 1 columns and has determinant zero if it contains two of them we get from lemma 2: det S i +l(l 1 L, L 2 L) = l j=0 c(i, j)c r 1 det S j l (L 1, L 2 )

6 6 summing over i we have: and the theorem follows S +l (L 1 L, L 2 L) = c r 1 = c r 1 = c r 1 +l i=0 j=0 l det S j l (L 1, L 2 )c(i, j) i l det S j l (L +l 1, L 2 ) c(i, j) i j=0 i=0 l det S j l (L 1, L 2 ) j L, j=0 To emphases the universality of the subresultant and the analogy with the polynomial case, let us state most what we have seen in a theorem: Theorem 4 Let m, n and min{m, n} be three positive integers Then for any differential ring A and all pairs of linear differential operators (L 1, L 2 ) in A [ ] of respective orders m m and n n ; there exists a (unique) element S m,n (L 1, L 2 ) of A [ ] such that: (1) For every morphism of differential rings φ : A B, where φ is the natural extension of φ φ(s m,n (L 1, L 2 )) = S m,n ( φ(l 1 ), φ(l 2 )) (2) If A is a field and m = m or n = n, L 1 and L 2 have a left gcd of degree if and only if S m,n i (L 1, L 2 ) = 0 for i < and S m,n (L 1, L 2 ) 0 (3) If A is a field, m = m or n = n and the left gcd G of L 1 and L 2 is of degree, then S m,n (L 1, L 2 ) = G (4) In the generic case, S m,n homogeneous polynomial of degree n in the a (j) i is a linear differential operator of order and an and m in the b (j) i in the differential ring Z[a (j) i, b (l) ] [ ] (5) S m,n ( m, n + ) = (6) S m,n (L 1, L 2 ) = 0 if m < m and n < n Moreover (2) may be replaced by: That is L1 = m A = Z[a (j) i, b (l) ] i=0 a(0) i i, L 2 = n i=0 b(0) i i with a (j) i = a (j+1) i, b (j) i = b (j+1) i and

7 7 (2 ) If A is a field such that L 1 admits m linearly independent solutions in A, L 2 admits exactly linearly independant common solutions with L 1 in A if and only if S m,n i (L 1, L 2 ) = 0 for i < and S m,n (L 1, L 2 ) 0 We have put the word unique in bracets because we will not prove the uniqueness here Except that point, everything is immediate from what we have seen In analogy to the polynomial case, let us give as an exercise one more property If a is an elements of K, one can consider the operators L a i defined by La i (y) = L i(ay) Then one has: S 0 (L a 1, L a 2) = a mn S 0 (L 1, L 2 ) We will not write down the differential subresultant algorithm: it is exactly the same as in the classical case, except that you must do the reductions (pseudo-divisions) using left multiplication only The proofs are, word for word, the same, as you can easily chec in the exposition of [GLRR] or the original paper of Loos [L] the signs of the subresultants may not be the same, but this is of no importance in our context and easy to rectify (multiply by ( 1) (m )(n ) ) So one can use the subresultant algorithm as well for linear differential equations, the only change is to replace multiplication in K[X] by left multiplication in K [ ] To conclude this paper, we will now give the differential equivalent to the expression of the resultant in terms of the values of one of the polynomials on the roots of the other Let us first recall a definition: Definition 2 Let (y 1,, y m ) be a m-uple of elements of a differential field L The wronsian of (y 1,, y m ) is defined by the formula: y 1 y m y W (y 1,, y m ) = 1 y m m 1 y 1 m 1 y m It is a classical exercise to show that this determinant is zero if and only if the elements y i are linearly dependent over the constants of L With this definition, we have: Theorem 5 Let L 1 and L 2 be two differential operators of respective orders m and n, and let y 1,, y m (resp z 1,, z n ) be linearly independent solutions of L 1 (y) = 0 (resp L 2 (z) = 0) in an extension of K, then: S 0 (L 1, L 2 ) = a n mb m W (y 1,, y m, z 1,, z n ) n W (y 1,, y m )W (z 1,, z n ) = b m W (L 1 (z 1 ),, L 1 (z n )) n W (z 1,, z n ) = ( 1) mn a n W (L 2 (y 1 ),, L 2 (y m )) m, W (y 1,, y m )

8 where a m (resp b n ) is the leading coefficient of L 1 (resp L 2 ) We will do an inductive proof, based on the following two lemmas: 8 Lemma 3 Under the hypotheses of Theorem 5, if L 1 is a differential operator of order m m such that L 1 = LL 2 + L 1, then: S 0 (L 1, L 2 ) = b m m n S 0 (L 1, L 2 ) Lemma 4 Under the hypotheses of Theorem 5, a n mb m W (y 1,, y m, z 1,, z n ) n W (y 1,, y m )W (z 1,, z n ) = W (L 1 (z 1 ),, L 1 (z n )) bm n W (z 1,, z n ) = ( 1) mn a n W (L 2 (y 1 ),, L 2 (y m )) m W (y 1,, y m ) Proof of Lemma 3 Let L 1 = α m m + + α 0 By the column manipulations corresponding to the relations i L 1 = ( i L)L 2 + i L 1, you have: α 0 α (n 1) 0 b 0 b (m 1) 0 α m S 0 (L 1, L 2 ) = bn, α m b n the lemma follows Proof of Lemma 4 Remar that the wronsian determinant is multiplied by c if one changes the line [ i x 1 i x N ] by [L(x 1 ) L(x N )] where L is a differential operator of order i and leading coefficient c So: a n mw (y, z) = y 1 y m z 1 z n m 1 y 1 m 1 y m m 1 z 1 m 1 z n L 1 (z 1 ) L 1 (z n ) n 1 L 1 (z 1 ) n 1 L 1 (z n ) and from that one has a n mw (y, z) = W (y)w (L 1 (z)) that gives the first relation The second is obtained in the same fashion

9 9 Let us now prove the theorem by induction When the order of one of the operators is zero the assertion is clear Else we can reduce the order of one of them by euclidean division, say L 1 = LL 2 + L 1 with ord(l 1 ) = m < m So we have: S 0 (L 1, L 2 ) = b m m n S 0 (L 1, L 2 ) (by Lemma 3) = b m m n b m W (L 1(z)) n (by induction hypothesis) W (z) = b m W (L 1 (z)) n (because L 2 (z i ) = 0), W (z) so we are done (in the case where n > m we remar that exchanging the roles of L 1 and L 2 multiplies the expressions on both sides by ( 1) mn ) I am very grateful to F Ollivier who suggested me to wor on this problem and contributed to the early important results

10 References 10 [Be] Berowitz, On computing the determinant in small parallel time using a small number of processors, Information processing letter 18 pp , 1984 [C] Chardin, Thèse de l Université Paris VI, prépublication du Centre de Mathématiques de l Ecole Polytechnique, 1990 [GLRR] González, Lombardi, Recio & Roy, Spécialisation de la suite de Sturm et sous-résultants, Publicaciones Universidad de Cantabria 8/90, Agosto 1990 [Ka] Kaplansy, An Introduction to Differential Algebra, Hermann 1957 [Ko] Kolchin, Differential Algebra and Algebraic Groups, Academic Press, New Yor, 1973 [L] Loos, Generalized polynomial remainder sequences, Symbolic and Algebraic computation pp Buchberger, Collins & Loos ed, Springer Verlag 1982 [P] Poole, Introduction to the theory of linear differential equations, Dover Public, New Yor, 1960 [R] Ritt, Differential equations from the algebraic standpoint, AMS Colloquium Publications Vol XIV, New Yor, 1932