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Riccati Differential Equation for Hypergeometric Differential Equation

TAKAHIRONAKAGAWA

ABSTRACT- In this paper, we study the solutions of the Riccati differential equation collesponding to ap-adic differential equation which is solvable on the generic disc. As an application, we consider the Grothendieck conjecture for Riccati differential equations. We see that the Riccati differential equation for some globally nilpotent differential equation with coefficients inQ(t) have, for almost all prime number, a solution inFp(t), butdo nothave any algebraic solutions.

1. Introduction

Letpbe a prime andK a complete non-archimedean valued field with mixed characteristics. In this paper, we consider the solutions of the Ric- cati differential equation for ap-adic differential equationLwhich is sol- vable on the generic discD(t;1 ) (see the Definition 2.5).

We attach the Riccati differential equationC(L)1for each linear ordinary differential operatorLin section 3. LetEbe the completion ofK(t) under the Gauss norm. We see that a solution of the Riccati differential equationC(L)1 forLonEis connected with a bounded solution ofLat the generic point.

LetbddKert(L) be the vector space consisting of the bounded solutions of L. If its dimension is one, then C(L)1 has a unique solution on E. In particular if we can reduce Lmodulop, thenC(L)1 has a solution on the rational function fieldK(t) over the residue field K ofK.

Let L be an n-th order linear ordinary differential operator with coefficients in Q(t). The Grothendieck conjecture (in its simplest form) asserts the following statements are equivalent (see [Put]):

(i) the equation L(y)ˆ0 has n linear independent(over Qalg) solutions which are algebraic overQ(t).

(ii) For almostallp, the equationL(y)ˆ0 hasnlinear independent solutions over the fieldFp(tp) in the fieldFp(t).

(*) Indirizzo dell'A.: Mathematical Institute CHIBA University, 263-8522 Japan.

E-mail: [email protected]

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Van der Put studied the Grothendieck conjecture for non-linear differ- ential equations. He proved that the differential equationy0ˆ(1=t2)y‡1 has no algebraic solutions but there exists a solution inFp(t) for all primesp. He also proved that the conjecture is true for some Risch equations [Put], [Put2].

We will prove the following theorem.

THEOREM 1.1. Let L be a monic differential operator with coeffi- cients in Q(t). We assume that when we regard L as a p-adic differ- ential equation, L is solvable on the generic disc D(t;1) for almost all prime p, and the dimension of the vector space which consists of the bounded solutions of L is one. If L has no algebraic solutions, then C1(L)modulo p has a solution onFp(t)for almost all prime p and no algebraic solutions.

In particular some hypergeometric differential equations satisfy the condition of Theorem 1.1.

The author is grateful to Professor S. Matsuda for his helpful advice and thank referee for his suggestion.

2. Notation

2.1 ±Analytic elements

Let K be a complete non-archimedean valued field with mixed char- acteristics and K the residue field ofK. Let K(t) be the field of rational functions. For a polynomialf ˆP

aitiwith coefficients inK, we set jfj0ˆmaxjaij:

(2:1)

The field of rational functionsK(t) has the Gauss norm j j0 defined by jfj0ˆ jgj0=jhj0 if f ˆg=h and g;h2K[t]. Let E be a completion ofK(t) under the Gauss norm. An element ofEis called an analytic element. The derivationd=dtofK(t) is extended toE. We regard (E;d=dt) as a differ- ential field.

Fora2Eandr2R>0, let

D(a;r )ˆ fz2Ej jz aj05rg (2:2)

and

D(a;r)ˆ fz2Ej jz aj0rg:

(2:3)

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We define the ring of analytic functions on the discD(a;r ) t o be A(a;r)ˆ X1

nˆ0

an(x a)n2E[[x a]]an2E;n!1lim janjsnˆ0 for all 0s5r

( )

:

We define the subringE[[x a]]0of the ring of analytic functionsA(a;1) by E[[x a]]0ˆ X1

nˆ0

an(x a)n 2E[[x a]] sup

0n51janj51

( )

:

PROPOSITION2.1 ([Christol, 2.5.1]). We define a ring homomorphism t:E!E[[x t]]0 so that, for f 2E,

t:E!E[[x t]]0 f 7!X1

nˆ0

(d=dt)n(f)

n! (x t)n:

Thentis an isometric isomorphism from E to the subfield of E[[x t]]0 consisting of elements f which satisfy:

t((d=dt)(f))ˆ(d=dx)(t(f)):

(2:4)

2.2 ±Differential modules

Let(H;D) be a differential field of characteristic zero and CH the constant field of H. We say differential field (L;D0) is a differential ex- tensionof (H;D) ifLHandD0jHˆD. AnH-differential module (M;r) is a free H-module M of finite type endowed with a CH-linear homo- morphismr(D) which satisfiesr(D)(am)ˆDa:m‡ar(D)(m) fora2H andm2M.

Let(M;r) be anH-differential module andLa differential extension of H. AnH-linear maps:M!Lis saidhorizontal ifs(rm)ˆD(s(m)) for m2M. We denote the CH-vector space by Homr(M;L):ˆ

fs2HomCH(M;K)js is horizontalg.

PROPOSITION 2.2. Let (M;r) be an H-differential module. For any differential extension L=H,

dimCH(Homr(M;L))dimH(M) (2:5)

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REMARK2.3. By the existence of Picard-Vessiot field of HCHCalgH , there exists a differential extension L=H such that equality (2.5) holds (see [PS]).

PROPOSITION2.4. Let G be a matrix in Mn(E). There exists a unique matrix U of GLn(E[[x t]])such that

U(t)ˆIn; (d=dx)(U)ˆt(G)U:

(2:6)

LetfGigi2Nbe the sequence of matrix in Mn(E)defined by G0ˆIn; Gi‡1ˆ(d=dt)(Gi)‡GiG:

Then we have

UˆX1

iˆ0

Gi(x t)i i! :

DEFINITION2.5. Let(M;r) be anE-differential module of rankn. Let e1; ;en be a basis of M and Gˆ(gij) the matrix such that r(D)eiˆPn

jˆ0gijej. We say (M;r) is solvableover D(t;r) if the matrix U which is defined by (2.6) is an elementof GLn(A(t;r)).

For a differential field (H;D) which satisfiesK(t)HE, we call an H-differential module (M;r) is solvable overD(t;r) if t heE-differential module (MHE;r) is solvable overD(t;r).

3. Riccati differential equation

Let(H;D) be a differential field of characteristic zero andCHa constant field ofH. LetLˆDn gnDn 1 g1be a differential operator with coefficients inH. Then (M;r)ˆH[D]=H[D]Lis anH-differential module of rankn. LetGbe the matrix of representation ofr(D) with respect to the basis 1; ;Dn 1of (M;r), i.e,

0 1 0 0

0 0 1 0

0... 1

g1 g2 gn

0 BB BB

@

1 CC CC A: (3:1)

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We suppose thatLhas a decomposition

LˆL1R; L1;R2H[D] andR is monic.

We write RˆDk ckDk 1 c1. Since the canonical morphism MˆH[D]=H[D]L!H[D]=H[D]R is surjective, the induced morphism (H[D]=H[D]R)_!M_is injective.

Let f1; ;fn (resp. l1; ;lk) be the dual basis of the basis 1;D; ;Dn 1ofM(resp. (H[D]=H[D]R)). If we regard (H[D]=H[D]R)_as a submodule ofM_then eachliis a linear combination offi. In fact, by the definition of dual basis, we have

l1

l2

l...k 0 BB B@

1 CC CAˆC

f1 f2 f...n 0 BB B@

1 CC CA (3:2)

such thatCˆ fci;jgis theknmatrix defined by

Dm 1ˆck;mDk 1‡ck 1;mDk 2‡ ‡c1;m (modR):

(3:3)

In particularcj;k‡1ˆcjand ifjkthenci;jˆdi;j.

Let F=Hbe a differential extension such that dimCH(Homr(M;F))ˆ dimH(M) (see Remark 2.3). Lets1;. . .;snbe a basis of Homr(M;F). By the equality (3.2), we have

s1(l1) s2(l1) sn(l1) s1(l2) s2(l2) sn(l2)

... ...

s1(lk) s2(lk) sn(lk) 0

BB B@

1 CC CAˆC

s1(f1) s2(f1) sn(f1) s1(f2) s2(f2) sn(f2)

... ...

s1(fn) s2(fn) sn(fn) 0

BB B@

1 CC CA: (3:4)

OperatingDon both sides the equation (3.4), we have 0 c1

1 ...

1 ck 0

BB BB

@

1 CC CC A

s1(l1) s2(l1) sn(l1) s1(l2) s2(l2) sn(l2)

... ...

s1(lk) s2(lk) sn(lk) 0

BB BB

@

1 CC CC A

ˆ D(C) C

0 g1

1 g2

... ...

0 1 gn

0 BB BB

@

1 CC CC A 0

BB BB

@

1 CC CC A

s1(f1) s2(f1) sn(f1) s1(f2) s2(f2) sn(f2)

... ...

s1(fn) s2(fn) sn(fn) 0

BB BB

@

1 CC CC A

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Since the matrix (si(fj))i;jis invertible, we have 0 c1

1 ...

1 ck 0

BB B@

1 CC

CACˆD(C) C

0 g1

1 g2

... ...

0 1 gn

0 BB B@

1 CC CA:

Therefore we have the equations

c1ck;n‡D(c1;n) g1 c1;k‡1gk‡1 c1;ngnˆ0;

c1;n‡c2ck;n‡D(c2;n) g2 c2;k‡1gk‡1 c2;ngnˆ0;

ck 1;n‡ckck;n‡D(ck;n) gk ...ck;k‡1gk‡1 ck;ngn ˆ0:

(3:5)

We define the differential systemC(L)kink-variabley1;. . .;ykby y1yk;n‡D(y1;n) g1 y1;k‡1gk‡1 y1;ngnˆ0;

y1;n‡y2yk;n‡D(y2;n) g2 y2;k‡1gk‡1 y2;ngn ˆ0;

yk 1;n‡ykyk;n‡D(yk;n) gk ... yk;k‡1gk‡1 yk;ngnˆ0 (3:6)

whereyi;j (1ik; j1) is defined inductively by y1;1ˆ1;yj;1ˆ0 (2jk);

y1;m‡1ˆD(y1;m)‡yk;m:y1; yj;m‡1ˆD(yj;m)‡yk;m:yj‡yj 1;m (2jk):

By definition, the vector (c1; ;ck) is solutions of differential systemC(L)k. In particular,C(L)1is a differential equation

y1y1;n‡D(y1;n) g1 y1;2g2 y1;ngnˆ0 (3:7)

wherey1;jis defined inductively by

y1;1 ˆ1; y1;m‡1 ˆD(y1;m)‡y1;m:y1:

Conversely we assume differential system C(L)k has a solution (c1; ;ck)2Hk. Let F=H be a differential extension which satisfies dimCH(fy2Fj(Dk ckDk 1 c1)yˆ0g)ˆk (see Remark 2.3). If y2Fsatisfies the equality (Dk ckDk 1 c1)yˆ0 then we have

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G y Dy

...

Dn 1y 0 BB BB

@

1 CC CC AˆG:tC

y Dy

...

Dk 1y 0 BB BB

@

1 CC CC A

ˆD(tC) y Dy

...

Dk 1y 0 BB BB

@

1 CC CC A‡tC:

0 1 0

0 1 0

...

c1 c2 ck

0 BB BB

@

1 CC CC A

y Dy

...

Dk 1y 0 BB BB

@

1 CC CC A

ˆD(tC) y Dy

...

Dk 1y 0 BB BB

@

1 CC CC A‡tC:

Dy D2y ...

Dky 0 BB BB

@ 1 CC CC Aˆ

Dy D2y

...

Dny 0 BB BB

@ 1 CC CC A:

Thereforeysatisfies a solution of differential equation DnyˆgnDn 1y‡gn 1Dn 2y‡ ‡g1y:

(3:8)

By proposition 2.2, we have a decomposition

Dn gnDn 1 g12H[D](Dk ckDk 1 c1):

(3:9)

As a result, we have the following proposition.

PROPOSITION3.1. Let H be a differential field such that the character- istic of H is zero. Let LˆDn gnDn 1 . . . g1 be a differential poly- nomial with coefficients in H.

There is a one to one correspondence between a k-th order differential operator R2H[D] which satisfies L2H[D]R and a solution of differ- ential system C(L)k in Hk.

If there exists a differential polynomial RˆDk ckDk 1 c1

such that L2H[D]R then (c1;c2;. . .;ck)2Hk is a solution of C(L)k. Conversely, if (c1;c2;. . .;ck)2Hk is a solution of C(L)k then the differ- ential polynomial RˆDk ckDk 1 c1satisfies L2H[D]R.

By induction, we have the following lemma.

LEMMA 3.2. We assume HˆE. Let LˆDn gnDn 1 . . . g1 be a differential polynomial with coefficients in E. Ifjgjj01for jˆ1; ;n, then, for any solution y2E of C1(L), we havejyj01.

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EXAMPLE3.3. For a differential operatorLˆD2 g2D g1of rank 2, C(L)1 is a Riccati differential equation c2‡c0 g1 g2cˆ0. Young con- sidered in this case [Young].

4. Analytic element solution of Riccati differential equation

LetL be a differential operator with coefficients inEandDˆd=dt. LetC(L)k be the differential system defined byL. We suppose thatLis solvable onD(t;1) (i.e.,E[D]=E[D]Lis solvable onD(t;1)). We denote the vector space which is generated by the solutions of (resp. the bounded solutions) ofLby

Kert(L)ˆ fy2 A(t;1)jt(L)yˆ0g;

(resp. Kerbddt(L)ˆ fy2E[[x t]]0jt(L)yˆ0g):

In general, sinceLis solvable onD(t;1), we have dimK(Kerbddt(L))1 by [Robba, Theorem 3.5]. If dimK(Kerbddt(L))n 1, then there exists a monic differential operatorR2E[D] such thatL2E[D]Rand Kert(R)ˆ Kerbddt(L) by [Robba, Theorem 2.6].

PROPOSITION 4.1. Let L2E[D] be a monic n-th order differential operator which is solvable on D(t;1).

(i) I f kˆdimK(Kerbddt(L))n 1, then C(L)khas a solution on E.

(ii) If1ˆdimK(Kerbddt(L)), C(L)1has a unique solution on E.

(iii) If there exists fundamental system of solutions y1;. . .ynof L at t such that yiis exactly of log-growth i 1(see the definition of log-growth in [C-T]) then, for each j, C(L)j has a unique solution over E.

PROOF. There exists a monic differential operatorR2E[D] such that L2E[D]Rand Kert(R)ˆ fy2E[[x t]]jt…L†yˆ0 andy is log-growthig by [Robba, Theorem 2.6]. By Proposition 3.1, there exists a solution of C(L)konEwherekˆdeg(R). The uniqueness of case (ii),(iii) is confirmed

by [Dwork1, Theorem 1]. p

By [D-R, Theorem 2.4.3 (iii)], we have the following proposition.

PROPOSITION4.2. Let L2E[D]be a monic n-th differential operator which is solvable on D(t;1). Assume kˆdimK(Kerbddt(L))n 1. Let S be the setfa2K jL has no pole on D(a;1 )g. Then there exists a solution

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(y1;. . .;yk)2Ek of C(L)k such that

fa2Sjdim(Kerbddt(L))>dim(Kerbdda(L))g fa2Sjyiis notanalytic on D(a;1 ) for someig

5. Hypergeometric differential equation and Riccati differential equation modp

Letpbe a prime number andDˆd=dx. Fora;b;c2Q\Zp, we setthe hypergeometric differential operator

L(a;b;c)ˆD2‡ c (1‡a‡b)x

x(1 x) D ab

x(1 x): (5:1)

We assume the following condition:

c a;c b;b;a all lie outsideZ:

(5:2)

c2Q f0; 1; 2;. . .g:

(5:3)

By [Dwork2, p. 11, Remark], we see the Riccati differential equation y2‡y0‡c (1‡a‡b)x

x(1 x) y ab

x(1 x)ˆ0 (5:4)

has no rational solutions.

For a rational number a2Q\Zp, we set ma2[0;p 1] such that maˆ a (modp). According t o Dwork, we say (a;b;c)2Qissplit(resp.

unsplit) if mc5minfma;mbg or mc >maxfma;mbg (resp. ma5mc5mb or mb5mc5ma). There exists an integer msuch that

F:ˆpm Xp 1

nˆ0

(a;n)(b;n)

(c;n)n! xn2Zp[x] pZp[x];

(5:5)

where (a;n) is defined

(a;n)ˆ 1 (nˆ0)

a(a‡1) (a‡n 1) (n1):

If (a;b;c) is split , t henFis uniquely determined. SinceFis a solution of hypergeometric differential equation

D2‡c (1‡a‡b)x

x(1 x) D ab

x(1 x)

yˆ0 (modp);

(5:6)

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by Lemma 5.2,F0=Fis a solution of the Riccati differential equation y2‡y0‡c (1‡a‡b)x

x(1 x) y ab

x(1 x)ˆ0 (modp):

(5:7)

Under the special condition, Dwork proved that if (a;b;c) is split, then L(a;b;c) has a unbounded solution and if (a;b;c) is unsplit, thenL(a;b;c) has two independent bounded solutions (see [Dwork2, Theorem 6.6, 9.4]).

We consider solutions of the Riccati differential equation (5.7).

LEMMA 5.1. We assume that(a;b;c)2(Q\Zp)3 is unsplit. Then the Riccati differential equation(5.7)has infinite solutions.

PROOF. In this case, the hypergeometric differential equation (5.6) has solution

uminfmXa;mbg

nˆ0

(a;n)(b;n)

(c;n)n! xn‡vpmmaxfmXa;mbg

nˆmc‡1

(a;n)(b;n)

(c;n)n! xn (u;v2Fp(xp)) (5:8)

with m2Z which satisfies pmmaxfmPa;mbg

nˆmc‡1

(a;n)(b;n)

(c;n)n! xn2Zp[x] pZp[x]. By Lemma 5.2, the Riccati differential equation (5.7) has infinite solutions. p LEMMA 5.2. Let RˆDn gnDn 1 gn 1Dn 2 . . . g1 be a differ- ential operator with coefficients Fp(x). Then R has a decomposition RˆQ(D c)where Q2Fp(x)[D]and c2Fp(x)if and only if c2Fp(x) is a solution of the Riccati differential equation C(R)1.

PROOF. Assume that Rˆ(Dn 1‡bn 1Dn 2‡ ‡b1)(D c) where b1; ;bn 1;c2Fp(x). Comparing the both sides, we have the equalities

gnˆc bn 1

gn 1ˆ n 1 1

c(1)‡cbn 1 bn 2

...

giˆ n 1 n i

c(n i)‡ n 2 n i 1

c(n i 1)bn 1‡ ‡c0 i

1 bi‡1‡cbi bi 1

...

g2ˆ n 1 n 2

c(n 2)‡ n 2 n 3

c(n 3)bn 1‡ ‡c0 2

1 b3‡cb2 b1

g1ˆc(n 1)‡c(n 2)bn 1‡ ‡c(1)b2‡cb1:

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We setcm;Cmby

cm‡1ˆc0m‡ccm; c1ˆ1 CmˆXm

iˆ1

m 1

i 1

cm i‡1c(i 1) cm‡1:

We have the equality

g1 c2g2 cngnˆcn‡1‡Cn‡Xn 1

iˆ1

Cibi:

To prove thatcis a solution ofC(R)1, we have only to proveCm ˆ0 for allm.

We use induction onm. Formˆ1,C1ˆc1c(0) c2ˆ0. Therefore the case mˆ1 is true.

We assume that the assertion is true formˆk 1.

c0kˆ Xk 1

iˆ1

k 2 i 1

ck ic(i 1)

!0

ˆXk 1

iˆ1

k 2 i 1

(c0k ic(i 1)‡ck ic(i))

ˆXk 1

iˆ1

k 2 i 1

((ck i‡1 cck i)c(i 1)‡ck ic(i))

ˆXk 1

iˆ1

k 2 i 1

(ck i‡1c(i 1)‡ck ic(i)) cck

ˆXk 1

iˆ2

k 2 i 1

‡ k 2 i 2

ck i‡1c(i 1)‡c(k 1)

ˆXk 1

iˆ2

k 1 i 1

ck i‡1c(i 1)‡c(k 1)ˆCk cck‡ck‡1

This completes the induction.

Conversely let c2Fp(x) be a solution of C(R)1. We define

b1;. . .;bn 12Fp(x) as above. Then we have the decomposition

Rˆ(Dn 1‡bn 1Dn 2‡ ‡b1)(D c). p

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LEMMA5.3. We assume that(a;b;c)2 Q\Zp3

is split. The Riccati differential equation(5.7)has a unique solution if and only if

xmc(1 x) 1‡ma‡mb mc F2

!(p 1)

6ˆ0 (modp) (5:9)

PROOF. Substituting yˆz‡F0=F for y in the Riccati differential equation (5.7), we have

z2‡z0‡ 2 F0

F‡c (a‡b‡1)x x(1 x)

zˆ0 (modp):

(5:10)

Substitutingzˆhxmc(1 x) 1‡ma‡mb mc=F2forzin the Riccati differential equation (5.7), we have

xmc(1 x) 1‡ma‡mb mch2

F2‡h0ˆ0 (modp):

(5:11)

We supposeh6ˆ0. By this equality, 1

h

0ˆxmc(1 x) 1‡ma‡mb mc

F2 (modp):

(5:12)

The differential equation (5.12) has a solution if and only if xmc(1 x) 1‡ma‡mb mc

F2 2Fp(xp)xFp(xp). . .xp 2Fp(xp)

(5:13) i.e.

xmc(1 x) 1‡ma‡mb mc F2

!(p 1)

ˆ0 (modp):

(5:14)

p PROPOSITION5.4. We assume thatmcˆp 1and(a;b;c)is split. Then the Riccati differential equation(5.7)has a unique solution F0=F.

PROOF. Let vx be a valuation ofFp(x) with respect to a prime ideal xF[x] of F[x]. Since p 1ˆmc>maxfma;mbg by assumption, vx(F)ˆ0.

Therefore we have vx 1 h 0

ˆvx xmc(1 x) 1‡ma‡mb mc F2

!

ˆp 1:

(5:15)

Since there exists noh2F(x) which satisfies (5.15), we havehˆ0. p

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6. Algebraic solution of Riccati differential equation

Let L be a monic differential operator with coefficients in Q(t) and Dˆd=dt. For almostall primep, we can reduce L modulo p. If we can reduceLmodulop, then we denoteLmodulopbyLp.

THEOREM 6.1. (i) If L has an algebraic solution, then C(L)1 has an algebraic solution.

(ii) We assume that for almost all prime p, L is solvable on the generic disc D(t;1) as a p-adic differential equation. If C(L)1 has a ra- tional solution then L has an algebraic solution.

(iii) We assume that for almost all prime p, L is solvable on the generic disc D(t;1) and dim(bddKert(L))ˆ1 as a p-adic differential equation. Then for almost all prime p, C(L)1has a solution inFp(t). And if C(L)1 has an algebraic solution, then L has an algebraic solution.

PROOF. IfLhas an algebraic solution then there exists an algebraic function h such that L2Q(t)alg[D](D h). Hence the Proposition 3.1 implies (i).

Suppose that C(L)1 has a rational solution y. By Proposition 3.1, we haveL2Q(t)[D](D y). Letpbe a prime such that when we regardLas a p-adic differential equation,Lis solvable on the generic disc D(t;1) and that we can reduceLandD ymodulop. By [DGS, 3. Proposition 5.1],Lp is nilpotent. And by [DGS, 3. Theorem 2.1], (D y)pis also nilpotent and therefore has a solution ofFp(t). Since for all almostprimep, (D y)phas a solution ofFp(t), by Honda's theorem([Put2, Proposition 2.2]),D yhas an algebraic solution. This implies (ii).

We will prove (iii). By Lemma 3.2 and Proposition 4.1(ii), for all almost prime p, C(L)1 has a solution on Fp(t). We assume that C(L)1 has an al- gebraic solution y which is not a rational function. Let G be the Galois group which corresponds to a finite Galois extension ofQ(t) containsy. For alls2G,s(y) is a solution ofC(L)1. Sinceyis nota rational function,C(L)1 has at least two algebraic solutions. By Proposition 3.1, this implies dim(bddKert(L))>1. This is a contradiction. p COROLLARY 6.2. Let L be a monic differential operator with coeffi- cients inQ(t). We assume that for almost all primep, when we regardLas ap-adic differential equation,Lis solvable on the generic discD(t;1)and dim(bddKert(L))ˆ1. I f L has no algebraic solutions, then C(L)1 has a solution onFp(t)for almost all primepand no algebraic solutions.

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REFERENCES

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Manoscritto pervenuto in redazione il 20 maggio 2011.

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