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1 Introduction 1 A dierential intermediate value theorem by Joris van der Hoeven D pt. de Math matiques (B t. 425) Universit Paris-Sud Orsay Cedex France June 2000 Abstract Let T be the eld of grid-based transseries or the eld of transseries with nite logarithmic depths. In our PhD. we announced that given a dierential polynomial P with coecients in T and transseries ' < with P(') <0 and P( ) >0, there exists an f 2 ('; ), such that P(f) = 0. In this note, we will prove this theorem. 1 Introduction 1.1 Statement of the results Let C be a totally ordered exp-log eld. In chapter 2 of [vdh97], we introduced the eld T = C[[[x]]] of transseries in x of nite logarithmic and exponential depths. In chapter 5, we then gave an (at least theoretical) algorithm to solve algebraic dierential equations with coecients in T. By that time, the following theorem was already known to us (and stated in the conclusion), but due to lack of time, we had not been able to include the proof. Theorem 1.1. Let P be a dierential polynomial with coecients in T. Given ' < in T, such that P (') P ( ) < 0, there exists an f 2 ('; ) with P (f ) = 0. In the theorem, ('; ) stands for the open interval between ' and. The proof that we will present in this note will be based on the dierential Newton polygon method as described in chapter 5 of [vdh97]. We will freely use any results from there. We recall (and renew) some notations in section 2. In chapter 1 of [vdh97], we also introduced the eld of grid-based C[[[x]]] C[[[x]]] transseries in x. In chapter 12, we have shown that our algorithm for solving algebraic dierential equations preserves the grid-based property. Therefore, it is easily checked that theorem 1.1 also holds for T = C[[[x]]]. Similarly, it may be checked that the theorem holds if we take for T the eld of transseries of nite logarithmic depths (and possibly countable exponential depths). 1.2 Proof strategy Assume that P is a dierential polynomial with coecients in T, which admits a sign change on a non empty interval ('; ) of transseries. The idea behind the proof of theorem 1.1 is very simple: using the dierential Newton polygon method, we shrink the interval ('; ) further and further while preserving the sign change property. Ultimately, we end up with an interval which is reduced to a point, which will then be seen to be a zero of P.

2 2 Section 2 However, in order to apply the above idea, we will need to allow non standard intervals ('; ) in the proof. More precisely, ' and may generally be taken in the compactication of T, as constructed in section 2.6 of [vdh97]. In this paper we will consider non standard ' (resp. ) of the following forms: ' =, with 2 T; ' = f, with 2 T; ' = m, with 2 T and where m is a transmonomial. ' = m, with 2 T and where m is a transmonomial. ' =, with 2 T and = (x log x log log x )?1. Here and f respectively designate the innitely small and large constants 1 T?1 and 1 T in the compactication of T. Similarly, and designate the innitely small and large constants 1 C?1 and 1 C in the compactication of C. We may then interpret ' as a cut of the transline T into two pieces T = ff 2 Tjf < 'g q ff 2 Tjf > 'g. Notice that ff 2 T + jf < g = ff 2 T + j9g 2 T + : g 1 ^ f = g 0 g; ff 2 T + jf > g = ff 2 T + j9g 2 T + : g 1 ^ f = g 0 g: Remark 1.2. Actually, the notations f, m, and so on are redundant. Indeed, f does not depend on, we have + m= + m whenever? m, etc. Now consider a generalized interval I = ('; ), where ' and may be as above. We have to give a precise meaning to the statement that P admits a sign change on I. This will be the main object of sections 3 and 4. We will show there that, given a cut ' of the above type, the function P (f ) = sign P (f ) may be prolongated by continuity into ' from at least one direction: If ' = +, then P is constant on ('; ) = (; ) for some > '. If ' = + f, then P is constant on (; ') for some < '. If ' = + m, then P is constant on (; ') for some < '. If ' = + m, then P is constant on ('; ) for some > '. If ' = +, then P is constant on ('; ) for some > '. (In the cases ' =?, ' =? f and so on, one has to interchange left and right continuity in the above list.) Now we understand that P admits a sign change on a generalized interval ('; ) if P (') P ( ) < 0. 2 List of notations Asymptotic relations. Logarithmic derivatives. f g, f = o(g); f 4 g, f = O(g); f g, log jf j log jgj; f g, log jf j 4 log jgj: f y = f 0 /f ; f hii = f y y (i times):

3 List of notations 3 Natural decomposition of P. P (f ) = i P i f (i) (2:1) Here we use vector notation for tuples i = (i 0 ; ; i r ) and j = (j 0 ; ; j r ) of integers: Decomposition of P along orders. P (f ) = jij = r; i 6 j, i 0 6 j 0 ^ ^ i r 6 i r ; f i = f i 0 (f 0 ) i 1 (f (r) ) i r ; j j1 jr = : i i 1 i r! P [!] f [!] (2:2) In this notation,! runs through tuples! = (! 1 ; ;! l ) of integers in f0; ; rg of length l at most d, and P [!] = P [!(1) ; ;! (l) ] for all permutations of integers. We again use vector notation for such tuples We call k! the weight of! and the weight of P. j! j = l; k!k =! 1 + +! j! j ;! 6, j!j = j j ^! ^ ^! j! j 6 j j ; f [!] = f (! 1) (! f j!j ) ; 1 j j =!:!! 1! j! j kp k = max!jp [!] 0 Additive, multiplicative and compositional conjugations or upward shifting. Additive conjugation: Multiplicative conjugation: k!k P +h (f ) = P (h + f ); P h (f ) = P (h f ); P "(f ") = P (f )": P +h;i = j>i P h;[!] = >! j i! h j?i P j : (2:3) h [?!] P [ ] : (2:4) Upward shifting (compositional conjugation): (P ") [!] = s ;! e?k kx (P [ ] "); (2:5) >! where the s ;! are generalized Stirling numbers of the rst kind: s ;! = s 1 ;! 1 s j j ;! j!j ; j (f (log x)) (j) = s j;i x?j f (i) (log x): i=0

4 4 Section 3 3 Behaviour of P near zero and innity 3.1 Behaviour of P near innity Lemma 3.1. Let P be a dierential polynomial with coecients in T. Then P (f ) has constant sign for all suciently large f 2 T. Proof. If P = 0, then the lemma is clear, so assume that P 0. Using the rules f = f ; f 0 = f y f ; f 00 = (f y ) 2 f + f yy f y f ; f 000 = (f y ) 3 f + 3 f yy (f y ) 2 f + (f yy ) 2 f y f + f yyy f yy f y f ; we may rewrite P (f ) as an expression of the form P (f ) = i=(i 0 ; ;i r ) P hii f hii ; (3:1) where P hii 2 T and f hii = f i 0 (f y ) i 1 (f hri ) i r for each i. Now consider the lexicographical ordering 6 lex on N r+1, dened by i < lex j (i 0 < j 0 ) _ (i 0 = j 0 ^ i 1 < j 0 ) _ (i 0 = j 0 ^ ^ i r?1 = j r?1 ^ i r < j r ): This ordering is total, so there exists a maximal i for 6 lex, such that P hii 0. Now let k > 1 be suciently large such that P hj i exp k x for all j. Then P (f ) = (1) i 0 sign P hii (3:2) for all postive, innitely large f exp k+r x, since exp k x f hri f y f for all such f. 3.2 Behaviour of P near zero Lemma 3.2. Let P be a dierential polynomial with coecients in T. Then P (") has constant sign for all suciently small " 2 T +. Proof. If P = 0, then the lemma is clear. Assume that P 0 and rewrite P (f ) as in (3.1). Now consider the twisted lexicographical ordering 6 tl on N r+1, dened by i < tl j (i 0 > j 0 ) _ (i 0 = j 0 ^ i 1 < j 0 ) _ (i 0 = j 0 ^ ^ i r?1 = j r?1 ^ i r < j r ): This ordering is total, so there exists a maximal i for 6 tl, such that P hii 0. If k > 1 is suciently large such that P hj i exp k x for all j, then for all postive innitesimal " exp k+r x. P (") = (1) i 0 sign P i (3:3)

5 Behaviour of P near constants Canonical form of dierential Newton polynomials Assume that P has purely exponential coecients. In what follows, we will denote by N P ;m the purely exponential dierential Newton polynomial associated to a monomial m, i.e. where N P ;m (c) = i P m;i;d(pm) c i ; (3:4) d P = max i;4 d P i : (3:5) The following theorem shows how N P = N P ;1 looks like after suciently many upward shiftings: Theorem 3.3. Let P be a dierential polynomial with purely exponential coecients. Then there exists a polynomial Q 2 C[c] and an integer, such that for all i > kp k, we have N P "i = Q (c 0 ). Proof. Let be minimal, such that there exists an! with k!k = and (N P ") [!] 0. Then we have d(n P ") = e?x and N P " (c) = k!k= >! s ;! N P ;[ ]!c [!] ; (3:6) by formula (2.5). Since N P " 0, we must have 6 kn P k. Consequently, kn P k > = kn P " k>kn P "" k>. Hence, for some i6kp k, we have kn P "i+1 k = kn P "i k. But then (3.6) applied on P " i instead of P yields N P "i+1 = N P "i. This shows that N P "i is independent of i, for i > kp k. In order to prove the theorem, it now suces to show that N P " = N P implies N P " = Q (c 0 ) for some polynomial Q 2 C[c]. For all dierential polynomials R of homogeneous weight, let R = j ([c j (c 0 ) ] R) c j (c 0 ) : (3:7) Since N P " = NP, it suces to show that P = 0 whenever N P = 0. Now NP = 0 implies that N P (x) = 0. Furthermore, (2.5) yields N P " = e?x N P : (3:8) Consequently, we also have N P (e x ) = e x (N P ")(e x ) = e x (N P (x))" = 0. By induction, it follows that N P (exp i x) = 0 for any iterated exponential of x. We conclude that N P = P = 0, by the lemma 3.1. Remark 3.4. Given any dierential polynomial P with coecients in T, this polynomial becomes purely exponential after suciently many upward shiftings. After at most kp k more upward shiftings, the purely exponential Newton polynomial stabilizes. The resulting purely exponential dierential Newton polynomial, which is in C[c] (c 0 ) N, is called the dierential Newton polynomial of P. 4 Behaviour of P near constants In the previous section, we have seen how to compute P ( ) and P ( f) for all 2 T. In this section, we show how to compute P ( m) and P ( m) for all 2 T and all transmonomials m. Modulo an additive and a multiplicative conjugation with resp. m, we may assume without loss of generality that = 0 and m= 1. Hence it will suce to study the behaviour of P (c ") for c 2 C and positive innitesimal (but suciently large) ", as well as the behaviour of P (f ) for positive innitely large (but suciently small) f.

6 6 Section 4 Modulo suently upward shiftings (we have P (c + ") = P " (c + "") and P (f ) = P " (f ")), we may assume that P has purely exponential coecients. By theorem 3.3 and modulo at most kp k more upward shiftings, we may also assume that N P (c) = Q(c) (c 0 ) ; (4:1) for some polynomial Q 2 C[c] and k 2 N. We will denote by the multiplicity of c as a root of Q. Finally, modulo division of P by its dominant monomial (this does not alter P ), we may assume without loss of generality that d P = Behaviour of P in between constants Lemma 4.1. For all 0 < " 1 with " e x, the signs of P (c? ") and P (c + ") are independent of " and given by (?1) P (c? ) = (?1) P (c + ) = Q () (c): (4:2) Proof. Since P is purely exponential and d P = 1, there exists an > 0 such that P (c + ")? N P (c + ") e?x (4:3) for all " 1. Let " > 0 be such that e?x Q(c ") 1! Q() (c) ("), whence " 1, where = /( + ). Then Furthermore,? e?x " 0?, whence e?x 4 Q(c + ") 4 1: (4:4) e?x (" 0 ) : (4:5) Put together, (4.4) and (4.5) imply that N P (c) e?x. Hence P (c + ") = NP (c + "), by (4.3). Now P (c ") = Q (c ") sign ((c ") 0 ) = (1) Q () (c) (1) ; (4:6) since " 0 < 0 for all positive innitesimal ". Corollary 4.2. If P is homogeneous of degree i, then for all 0 < " 1 with " e x. P () = P (") = RP ;i (" y ) = Rp;i (?); (4:7) Corollary 4.3. Let c 1 < c 2 be constants such that P (c 1 + ) P (c 2? ) < 0. Then there exists a constant c 2 (c 1 ; c 2 ) with P (c? ) P (c + ) < 0. Proof. In the case when is odd, then P (c? ) P (c + ) < 0 holds for any c > c 1 with Q(c) 0, by (4.2). Assume therefore that is even and let 1 ; 2 denote the multiplicities of c 1 ; c 2 as roots of Q. From (4.2) we deduce that (? 1) 2 Q ( 1 ) (c 1) Q ( 2 ) (c 2) < 0: (4:8) In other words, the signs of Q(c) for c#c 1 and c"c 2 are dierent. Hence, there exists a root c of Q between c 1 and c 2 which has odd multiplicity. For this root c, (4.2) again implies that P (c? ) P (c + ) < 0.

7 Proof of the intermediate value theorem Behaviour of P before and after the constants Lemma 4.4. For all 0 < f 1 with f e x, the signs of P (?f ) and P (f ) are independent of f and given by (?1) deg Q+ P (?) = P () = sign Q deg Q: (4:9) Proof. Since P is purely exponential and d P = 1, there exists an > 0 such that P (f )? N P (f ) e?x ; (4:10) since f ; f 0 ; f 00 ; e x. Furthermore Q(f ) Q deg Q (f ) deg Q e x and (f 0 ) e x, whence N P (f ) e x. In particular, N P (f ) e?x, so that P (f ) = NP (f ), by (4.10). Now P (f ) = Q (") sign (f 0 ) = sign Q deg Q (1) deg Q+ ; (4:11) since f 0 > 0 for positive innitely large f. Corollary 4.5. If P is homogeneous of degree i, then for all 0 < f 1 with f e x. P () = P (f ) = RP ;i (f y ) = RP ;i (); (4:12) Corollary 4.6. Let c 1 be a constant such that P (c 1 + ) P () < 0. Then there exists a constant c > c 1 with P (c? ) P (c + ) < 0. Proof. In the case when is odd, then P (c? ) P (c + ) < 0 holds for any c > c 1 with Q(c) 0, by (4.2). Assume therefore that is even and let 1 be the multiplicity of c 1 as a root of Q. From (4.2) and (4.9) we deduce that Q ( 1 ) (c 1) sign Q deg Q < 0: (4:13) In other words, the signs of Q(c) for c#c 1 and c" are dierent. Hence, there exists a root c > c 1 of Q which has odd multiplicity. For this root c, (4.2) implies that P (c? ) P (c + ) < 0. 5 Proof of the intermediate value theorem It is convenient to prove the following generalizations of theorem 1.1. Theorem 5.1. Let and v be a transseries resp. a transmonomial in T. Assume that P changes sign on an open interval I of one of the following forms: a) I = (; ), for some > with d(? ) = v. b) I = (? v; ). c) I = (; + v). d) I = (? v; + v). Then P changes sign at some f 2 I. Theorem 5.2. Let and v be a transseries resp. a transmonomial in T. Assume that P changes sign on an open interval I of one of the following forms: a) I = ( + ;? ), for some > with d(? ) = v. b) I = (? v;? ).

8 8 Section 5 c) I = ( + ; + v). d) I = (? v; + v). Then P changes sign on (f? ; f + ) for some f 2 I with (f? ; f + ) I. Proof. Let us rst show that cases a, b and d may all be reduced to case c. We will show this in the case of theorem 5.1; the proof is similar in the case of theorem 5.2. Let us rst show that case a may be reduced to cases b, c and d. Indeed, if P changes sign on (; ), then P changes sign on (; + v), ( + v;? v) or (? v; ). In the second case, modulo a multiplicative conjugation and upward shifting, corollary 4.3 implies that there exists a 0 < < (? ) v such that P admits a sign change on (( + v)? v; ( + v) + v). Similarly, case d may be reduced to cases b and c by splitting the interval in two parts. Finally, cases b and c are symmetric when replacing P (f ) by P (?f ). Without loss of generality we may assume that = 0, modulo an additive conjugation of P by. We prove the theorem by a triple induction over the order r of P, the Newton degree d of the asymptotic algebraic dierential equation P (f ) = 0 (f v) (5:1) and the maximal length l of a sequence of privileged renements of Newton degree d (we have l 6 (r + 1) d, by proposition 5.12 in [vdh97]). Let us show that, modulo upward shiftings, we may assume without loss of generality that P and v are purely exponential and that N P 2 C[c] (c 0 ) N. In the case of theorem 5.1, we indeed have P " (0) = P (0) and P " ( v") = P ( v). In the case of theorem 5.2, we also have P "e?x () = P " (") = P (). Furthermore, if f 2 (; v" e x ) = I" e x is such that P " e x changes sign on (f? ; f + ) I " e x, then f #/x 2 (; v) = I is such that P changes sign on (f #/x? ; f #/x + ) I. Case 1: (5.1) is quasi-linear. Let m be the potential dominant monomial relative to (5.1). We may assume without loss of generality that m = 1, modulo a multiplicative conjugation with m. Since By N P 2 C[c] (c 0 ) N, we have N P = c + or N P = c 0 for certain constants ; 2 C. In the case when N P = c +, there exists a solution to (5.1) with f? / 0. Now P (0) = sign and P () = sign. We claim that P () = RP ;1 () and RP ;1 (v y? ) = P ( v) must be equal. Otherwise R P ;1 would admit a solution between and v y?, by the induction hypothesis. But then the potential dominant monomial relative to (5.1) should have been er, if is the largest such solution. Our claim implies that (sign ) (sign ) = P (0) P ( v) < 0, so that f > 0. Finally, lemma 3.2 implies that P admits a signchange at f. Lemma 4.1 also shows that P (f? ) P (f + ) = P (f? ) P (f + ) < 0. In the case when N P = c 0, then any constant 2 C is a root of N P. Hence, for each > 0, there exists a solution f to (5.1) with f. Again by lemmas 3.2 and 4.1, it follows that P admits a sign change at f and on (f? ; f + ). Case 2: d > 1. Let m be the largest classical potential dominant monomial relative to (5.1). Since P (0) P ( v) < 0 (resp. P () P ( v) < 0), one of the following always holds: Case 2a. We have P (0) P ( m) < 0 (resp. P () P ( m) < 0). Case 2b. We have P ( m) P ( m) < 0. Case 2c. We have P ( m) P ( v) < 0. For the proof of theorem 5.2, we also assume that m in the above three cases and distinguish a last case 2d in which m.

9 Proof of the intermediate value theorem 9 Case 2a. We are directly done by the induction hypothesis, since the equation has a strictly smaller Newton degree than (5.1). P (f ) = 0 (f m): (5:2) Case 2b. Modulo multiplicative conjugation with m, we may assume without loss of generality that m= 1. By corollary 4.6, there exists a c > 0 such that P (c? ) P (c + ) < 0. Actually, for any transseries ' c we then have P ('? ) P (' + ) < 0. Take ' such that P +' (f ~ ) = 0 (f ~ 1) (5:3) is a privileged renement of (5.1). Then either the Newton degree of (5.3) is strictly less than d, or the longest chain of renements of (5.3) of Newton degree d is strictly less than l. We conclude by the induction hypothesis. Case 2c. Since m is the largest classical dominant monomial relative to (5.1), the degree of the Newton polynomial associated to any monomial between m and v must be d. Consequently, P ( m) P ( v) = Pd ( m) Pd ( v) = RP ;d (m y + ) RP ;d (v y? ) < 0: (5:4) By the induction hypothesis, there exists a monomial n with m y + < n y < v y? and RP ;d (n y? ) RP ;d (n y + ) < 0: (5:5) In other words, n is a dominant monomial, such that m n v and Pd ( n) Pd ( n) < 0: (5:6) We conclude by the same argument as in case 2b, where we let n play the role of m. Case 2d. Since m is the largest classical dominant monomial relative to (5.1), the degree of the Newton polynomial associated to any monomial between and v must be d. Consequently, P () P ( v) = Pd () Pd ( v) = RP ;d (x y + ) RP ;d (v y? ) < 0: (5:7) By the induction hypothesis, there exists a monomial n with x y + < n y < v y? and RP ;d (n y? ) RP ;d (n y + ) < 0: (5:8) In other words, n is a dominant monomial, such that x n v and We again conclude by the same argument as in case 2b. Pd ( n) Pd ( n) < 0: (5:9) Corollary 5.3. Any dierential polynomial of odd degree and with coecients in T admits a root in T. Proof. Let P be a polynomial of odd degree with coecients in T. Then formula (3.2) shows that for suciently large f 2 T + we have P (? f ) P (f ) < 0, since i 0 is odd in this formula. We now apply the intermediate value theorem between? f and f. Bibliography [vdh97] J. van der Hoeven. Automatic asymptotics. PhD thesis, cole polytechnique, France, 1997.

A differential intermediate value theorem

A differential intermediate value theorem by Joris van der Hoeven Dépt. de Mathématiques (Bât. 425) Université Paris-Sud 91405 Orsay Cedex France June 2000 Abstract Let T be the field of grid-based transseries