Annales Mathématiques Blaise Pascal

My Driss Aouragh

Naji Yebari

Stabilisation exponentielle d’une équation des poutres d’Euler-Bernoulli à coefficients variables

Abdelmalek Azizi

Mohamed Ayadi

Moulay Chrif Ismaili

Mohamed Talbi

Sur les unités des extensions cubiques cycliques non ramifiées sur certains sous-corps de $\mathbf{Q}(\sqrt{d},\sqrt{-3})$

Katsuya Miyake

Twists of Hessian Elliptic Curves and Cubic Fields

Toru Nakahara

ANNALES MATHÉMATIQUES
BLAISE PASCAL
Toru Nakahara
Hasse’s problem for monogenic fields
Volume 16, no 1 (2009), p. 47-56.
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Annales mathématiques Blaise Pascal 16, 47-56 (2009)
Hasse’s problem for monogenic fields
Toru Nakahara
Abstract
In this article we shall give a survey of Hasse’s problem for integral power bases
of algebraic number fields during the last half of century. Specifically, we developed
this problem for the abelian number fields and we shall show several substantial
examples for our main theorem [7] [9], which will indicate the actual method to
generalize for the forthcoming theme on Hasse’s problem [15].
1. Introduction
In 1960’s Hasse proposed to characterize number fields whose rings of
integers have power integral bases. First, we define a power integral basis.
Definition 1.1. Let ZK be the ring of integers in an algebraic number
field K of extension degree n. When the ring ZK is generated by a primitive
element α in K, namely ZK = Z[α] = Z[1, α, · · · , αn−1], we call that ZK
has a power basis or K is monogenic.
Let ζn be a primitive n-th root of unity. When K is any cyclotomic
number field kn = Q(ζn), its maximal real subfield Q(ζn + ζ−1n ) or any
quadratic number field Q(
√
m) for a square-free m 6= 0, 1, the ring ZK of
integers has a power basis;
Z[ζn], Z[ζn + ζ−1n ], Z[ω]
respectively. Here
ω =
d +
√
d
2
, d =
{
m if m ≡ 1 (mod 4),
4m if m ≡ 2, 3 (mod 4).
If K is a certain cubic cyclic quartic abelian or a maximal imaginary
abelian subfield of a cyclotomic field kn, such families of infinitely many
Partially supported by grant (]16540029) from the Japan Society for the Promotion of
Science.
Keywords: Power integral basis, monogenic fields, Hasse’s problem.
Math. classification: 11R27, 11R29, 11R37.
47
T. Nakahara
fields have power integral bases [1, 2, 4, 10, 11, 12, 13, 14, 17]. On the
other hand, Dedekind showed that a non-Galois cubic field K = Q(θ) is
non-monogenic, where θ is a root of f(θ) = θ3− θ2− 2θ− 8 = 0 with the
discriminant d(θ) = −NK/Qf ′(θ) = −22 · 503. If K is non-monogenic, by
Stickelberger’s theorem the field discriminant d(K) would equal to −503.
In fact, we can seek for an integer η as a third generator of ZK . Let η =
(θ+θ2)/2 6∈ Z[θ] Then θη = θ2 +θ+4 ∈ Z[θ]. Hence η2 +η = 2θ2 +4θ+6.
Thus η ∈ Z̄K ∩K = ZK , where Z̄K is the integral closure of ZK . Then
 1θ
η
 = M
 1θ
θ2
 with M =
 1 1
1
2
1
2
 .
Hence d(1, θ, η) = (detM)2d(1, θ, θ2) namely, d(1, θ, η) = 14 · (−4 · 503) =
−503. Thus we have ZK = Z[1, θ, η].
Let K be a cyclic quartic extension field over Q with prime conductor.
Then, before a quarter of a century, K has no power integral basis except
for the 5-th cyclotomic field Q(ζ5) (see [10]).
Next we quote a criterion for non-monogenic phenomena.
Lemma 1.2 ([14, 17]). Let ` be a prime number and let F/Q be a Galois
extension of degree n = efg with ramification index e and the relative
degree f with respect to `. If one of the following conditions is satisfied,
then ZF has no power integral basis, i.e. F is non-monogenic;
(1) e`f < n if f = 1;
or
(2) e`f 5 n + e− 1 if f = 2.
Any cyclic extension F over Q with prime degree ` = [F : Q] = 5 is
non-monogenic except for the maximal real subfield F of the (2` + 1)-th
cyclotomic field with prime conductor 2` + 1, by M.-N. Gras [3] and it is
proved by us that some type of imaginary extension has no power integral
basis [17, 8, 16].
Finally we will propose a few open problems concerning Hasse’s prob-
lem.
48
Hasse’s problem for monogenic fields
2. The rank r ≤ 2 or r ≥ 4
In the case of any quadratic field K = Q(
√
a) (r = 1), where a 6= 0, 1
square-free, is monogenic. Namely, the ring ZK of integers in K, ZK =
Z[1, ω], where ω is defined in the Section 1.
If K is a biquadratic extension field (r = 2), the author showed that
there exist infinitely many monogenic fields and non-monogenic ones within
the estimation of the field indices:
m̃(K) = min
primitive α∈K
{Ind(α)}, where Ind(α) =
√∣∣∣∣dK(α)dK
∣∣∣∣
for the discriminant dK(α) of a number α and the field discriminant dK
[11]. M. N. Gras and T. Tanoé obtained a necessary and sufficient con-
dition such that F = Q(
√
a1,
√
a2) is a monogenic biquadratic field [4].
Specifying their result, Y.Motoda proved that there exist infinitely many
such fields [5].
Applying Lemma 1.2 for the prime number ` = 2, by the ideal decom-
position of a principal ideal (2) in K, we obtain for any such a field K of
higher rank r = 4.
Proposition 2.1 ([7]). Let a1, a2, · · · , ar be square-free rational integers
and F be the field Q(
√
a1,
√
a2, · · · ,
√
ar) of degree 2r, r = 4. Then F is
non-monogenic.
Next we obtained the followings for the octic field over Q whose Galois
group is 2-elementary abelian.
Theorem 2.2 ([15]). Let F = Q(
√
a1,
√
a2,
√
a3) be any octic field over Q.
Then F is non-monogenic except for the field Q(
√
−1,
√
2,
√
−3), namely
the cyclotomic field Q(ζ24) of conductor 24.
In this article we explain the basic idea and show prospective examples
for the theorem.
3. The rank r=3
If an octic field Q(
√
a1,
√
a2,
√
a3) with aj ≡ 1 (mod 4), that is, K has an
odd conductor, then we have e · `f 5 1 · 22 < 8 by Lemma 1.2 since prime
number (2) is not ramified in K and the inert group T with respect to 2
in the Galois group G(K/Q) is cyclic, hence the order f of T is less or
49
T. Nakahara
equal to 2. Thus K is non-monogenic. On the other hand, we have in the
case a1 ≡ 3 (mod 4), a2 ≡ a3 ≡ 1 (mod 4), e = 2 and f 5 2 with rspect
to (2). Then
e · `f = 2 · 21 < 8 if f = 1,
e · `f = 2 · 22 5 8 + 2− 1 if f = 2.
Then without loss of generality, for any octic field K, it is enough for us
to investigate
K = Q
(√
4a1,
√
4a2,
√
a3
)
where a1 = mn ≡ 3 (mod 4), a2 = dn ≡ 2 (mod 4), a3 = d1m1n1` ≡
1 (mod 4), d = d1d2, m = m1m2, n = n1n2, d2 ≡ 2 (mod 4), d1,m1, n1 = 1
and dmn` is square-free.
Let k = Q
(√
d1m1n1`
)
and L = Q
(√
4mn,
√
4dn.
)
. When k and L are
linearly disjoint, namely d1m1n1 = 1, it is known that any such an octic
field K = kL is non-monogenic except for K = Q
(√
−4,
√
8,
√
−3
)
=
Q(ζ24) (see [6]).
In general, for d1m1n1 = 1 we obtain
Theorem 3.1 ([15]). Let
K = Q
(√
4mn,
√
4dn,
√
d1m1n1l
)
where a1 = mn ≡ 3 (mod 4), a2 = dn ≡ 2 (mod 4), a3 = d1m1n1 ≡ 1 ≡
3 (mod 4), d = d1d2, m = m1m2 n = n1n2 d2 ≡ 2 (mod 4), d1,m1, n1 = 1
and dmnl is square-free. Then K is non-monogenic except for the 24-th
cyclotomic number field Q(ζ24).
In this section, we show a prospective example for Theorem 3.1. We
consider the octic field
K = Q
(√
4 · 3,
√
4 · 2,
√
21
)
,
where m = m1m2 = 3, n = n1n2 = 1, d = d1d2 = 2 and d1m1n1` =
1 · 3 · 1 · 7. Then K contains seven quadratic subfields;
50
Hasse’s problem for monogenic fields
k1 = Q(
√
4 ·mn) = Q(
√
4 · 3),
k2 = Q(
√
4 · dn) = Q(
√
4 · 2),
k3 = Q(
√
4 ·mn · dn) = Q(
√
4 · 6),
k4 = Q(
√
d1m1n1`) = Q(
√
21),
k5 = Q(
√
4mn · d1m1n1`) = Q(
√
4 · 7),
k6 = Q(
√
4dn · d1m1n1`) = Q(
√
4 · 42),
k7 = Q(
√
4mn · 4dn · d1m1n1`) = Q(
√
4 · 14),
and seven biquadratic subfields;
L1 = k1k2 = Q(
√
4 · 3,
√
4 · 2), L2 = k3k5 = Q(
√
4 · 6,
√
4 · 7),
L3 = k2k4 = Q(
√
4 · 2,
√
21), L4 = k4k5 = Q(
√
21,
√
4 · 7),
L5 = k3k7 = Q(
√
4 · 6,
√
4 · 14), L6 = k6k7 = Q(
√
4 · 42,
√
4 · 14),
L7 = k4k7 = Q(
√
21,
√
4 · 14).
Let G =< τ, σ, ρ > be the Galois group of the octic extension K over Q,
where three automorphisms are defined by
τ :
√
3 7→ −
√
3,
√
2 7→
√
2,
√
21 7→
√
21,
σ :
√
3 7→
√
3,
√
2 7→ −
√
2,
√
21 7→
√
21,
ρ :
√
3 7→
√
3,
√
2 7→
√
2,
√
21 7→ −
√
21.
Let G(L/M) be the Galois group of an extension field L over an alge-
braic number field M . Denote G(L/Q) by G(L) and G(K) by G. Then
we have
G(k1) ∼= G/ < σ, ρ >∼=< τ̃ >, G(k2) ∼= G/ < τ, ρ >∼=< σ̃ >,
G(k3) ∼= G/ < τσ, ρ >∼=< τ̃ >, G(k4) ∼= G/ < τ, σ >∼=< ρ̃ >,
G(k5) ∼= G/ < τρ, σ >∼=< τ̃ >, G(k6) ∼= G/ < τ, σρ >∼=< σ̃ >,
G(k7) ∼= G/ < τσ, τρ >∼=< τ̃ >,
and
G(L1) ∼= G/{< σ, ρ > ∩ < τ, ρ >} ∼= G/ < ρ > ∼= < τ̃, σ̃ >,
G(L2) ∼= G/{< τσ, ρ > ∩ < τρ, σ >} ∼= G/ < τσρ > ∼= < τ̃, ρ̃ >,
G(L3) ∼= G/{< τ, ρ > ∩ < τ, σ >} ∼= G/ < τ > ∼= < σ̃, ρ̃ >,
G(L4) ∼= G/{< τ, σ > ∩ < τρ, σ >} ∼= G/ < σ > ∼= < τ̃, ρ̃ >,
G(L5) ∼= G/{< τσ, ρ > ∩ < τρ, τσ >} ∼= G/ < τσ > ∼= < τ̃, ρ̃ >,
G(L6) ∼= G/{< τ, σρ > ∩ < τσ, τρ >} ∼= G/ < σρ > ∼= < τ̃, σ̃ >,
G(L7) ∼= G/{< τ, σ > ∩ < τσ, τρ >} ∼= G/ < τσ > ∼= < τ̃, ρ̃ >,
where for a subgroup H of G and α ∈ G, α̃ means a coset αH in the
residue class group G/H.
51
T. Nakahara
Assume the ring ZK has a power basis; ZK = Z[ξ] for a suitable prim-
itive integer in K. Then the different
dK(ξ) = (ξ − ξτ )(ξ − ξσ)(ξ − ξρ)(ξ − ξτσ)(ξ − ξτρ)(ξ − ξσρ)(ξ − ξτσρ)
of a number ξ is equal to the field different dK of the field K. Then it
holds that
(dK(ξ)) = (NK(dK(ξ))) = NK(dK) = (dK),
where for a number α and an ideal a in an algebraic number field F/Q,
NF (α) and NF (a) mean the norm map of α and of a with respect to
F/Q, respectively and for a number β in K, (β) means the principal ideal
generated by β. We denote the discriminant of a number α and of a field
F by dF (α) and dF , respectively.
By Hasse’s discriminant-conductor formula, we have
dK =
7∏
j=1
dkj = (4 · 3)(4 · 2)(4 · 6)(21)(4 · 7)(4 · 42)(4 · 14) = 2
16 · 34 · 74,
We consider the following identity whose terms are fixed by the subgroup
< σ, ρ >= Hk1 in G;
(ξ − ξσ)(ξ − ξσ)ρ − (ξ − ξρ)(ξ − ξρ)σ + (ξ − ξσρ)(ξ − ξσρ)σ = 0. (3.1)
Since the difference ξ − ξσ is divisible by the relative different dK/L4 ,
the product (ξ−ξσ)(ξ−ξσ)ρ is divisible by dK/L4 ·d
ρ
K/L4
By the transitive
law of different,
dK = dL4 · dK/L4 ,
we have NKdK = NK/L4(NL4(dL4))NL4(NK/L4(dK/L4)), where for a field
tower Q ⊂ F ⊂ L of algebraic number fields, NL/F means the rel-
ative norm map with respect to L/F and we denote the relative dis-
criminant NL/F dL/F by dL/F . Then we obtain (dK) = (dL4)2(dK/L4)
4,
namely 216 · 34 · 74 = (24 · 32 · 72)2(dK/L4)4. Then (22) = dK/L4 , hence
(2)(2) = dK/L4(dK/L4)
ρ. In the same way since the differences ξ − ξρ,
ξ − ξσρ are divisible by dK/L1 , dK/L6 , respectively, we have (7) = dK/L1
and (1) = dK/L6 , hence (7) = dK/L1d
σ
K/L1
and (1) = dK/L6d
σ
K/L6
.
Since the number ξ generates a power basis, then using the identity (3.1)
we obtain
7E1 + 22E2 + 1E3 = 0
for suitable units Ej(1 5 j 5 3) in the fixed field F<σ,ρ> = k1 = Q(
√
4 · 3)
with notations 7 = l, 22 = 2d2, 1 = d1, for three partial products in the
52
Hasse’s problem for monogenic fields
equation (3.1), because it should follow that (ξ − ξρ)(ξ − ξρ)σ = dK/L1 ,
(ξ − ξσ)(ξ − ξσ)ρ = dK/L4 , (ξ − ξσρ)(ξ − ξσρ)σ = dK/L6 as ideals.
With general notations 7 = `, 22 = 2d2, 1 = d1, we have
`E1 + 2d2E2 + d1E3 = 0 in Q
(√
4mn
)
, (3.2)
`Ē1 + 2d2Ē2 + d1Ē3 = 0 in Q
(√
4mn
)
. (3.3)
If the rank r4·3 of the equations (3.2), (3.3) in Q(
√
4 · 3) is one, we have
` ± 2d2 ± d1 = 0 i.e. ` = ±2d2 ± d1 < 2d2 + d1 = 5, which is impossible.
Then the rank r4·3 is two. By Ej = εej , let e1 = min15j53{ej}. Thus
` + 2d2εe + d1εf = 0 and e, f = 0 holds.
Put
εg = ug + vg
√
3.
Then, for unknown valuables `, 2d2 and d1, it holds that
2d2 : d1 =
∣∣∣∣∣εf 1ε̄f 1
∣∣∣∣∣ :
∣∣∣∣1 εe1 ε̄e
∣∣∣∣ = 2vf√3 : −2ve√3.
Hence, 2d2/d1 = 4/1 = vf/−ve, namely |vf | = 4|ve|.
` : d1 =
∣∣∣∣∣εe εfε̄e ε̄f
∣∣∣∣∣ :
∣∣∣∣1 εe1 ε̄e
∣∣∣∣ = εe · ε̄e
∣∣∣∣∣1 εf−e1 ε̄f−e
∣∣∣∣∣ :
∣∣∣∣1 εe1 ε̄e
∣∣∣∣ = −2vf−e : −2ve.
Hence, `/d1 = 7/1 = vf−e/ve, namely |vf−e| = 7|ve| and f − e > e, which
is a contradiction to 0 < |vf−e| < |vf | = 4|ve|.
Then the equations (3.2), (3.3) are impossible in Q(
√
3). Therefore the
octic field Q(
√
3,
√
2,
√
21) is non-monogenic.
However we could not always determine the monogenesis of the quartic
subfields in K from the following necessary condition (3.4). In fact we can
find an integral power basis for the field L1 = k1k2 = Q(
√
4 · 3,
√
4 · 2) as
follows.
We can confirm that the ring ZL1 of integers in L1 has an integral power
basis Z[1, α, β, (α + 1)β/2] for α =
√
3 and β =
√
2. We select an integer
ξ = β − (α + 1)β/2. Then it holds that
dL1(ξ) = 3
2 · 28 = 1 · (4 · 3)(4 · 2)(4 · 6) = dL1 .
Then the field L1 is monogenic. In fact, the identity (−6)−(−4)−(−2) = 0
holds for (ξ− ξτ )(ξ− ξτ )σ = −6, (ξ− ξσ)(ξ− ξσ)τ = −4 and (ξ− ξτσ)(ξ−
ξτσ)σ = −2.
53
T. Nakahara
Next we consider the second quartic subfield L2 := Q(
√
4 · 6,
√
4 · 7) in
K. Since the Galois group G(L2/Q) coincides with < τ, % >, we have for
an integer ξ ∈ ZL2 ,
(ξ − ξτ )(ξ − ξτ )% − (ξ − ξ%)(ξ − ξ%)τ − (ξ − ξτ%)(ξ − ξτ%)% = 0 (3.4)
For three quadratic subfields
k3 = Q(
√
4 · 6), k5 = Q(
√
4 · 7), k6 = Q(
√
4 · 42),
in L2, we calculate each of the relative discriminants
±dL2/k3 =
√
dL2/d
2
k3
=
√
28 · 32 · 72/(23 · 3)2 = 2 · 7,
±dL2/k5 =
√
dL2/d
2
k5
=
√
28 · 32 · 72/(22 · 7)2 = 22 · 3,
±dL2/k6 =
√
dL2/d
2
k6
=
√
28 · 32 · 72/(23 · 3 · 7)2 = 2.
If the ring ZL2 has an integral power basis, which is generated by ξ, then
the equation (3.4) should hold. However since
2± 2 · 7± 22 · 3 = 2− 14 + 12 = 0!
would happen, we can not deduce a contradiction. But, on the first partial
different (ξ− ξτ )(ξ− ξτ )ρ for any integer ξ = aα+ bβ + cγ in ZL2 , we have
the value (2a + c)2 · 6− (2b)2 · 7 = 2[3(2a + c)2 − 14b2]. Then we consider
the Diophantine equation 3X2 − 14Y 2 = ±1. Assume that this equation
has an integral solution. Then in the case of −1, Y 2 ≡ −1 (mod 3); which
is impossible. In the case of +1, it holds that X21 − 3 · 2 · 7Y 2 = 3 with
X1 = 3X. Since 3 is a quadratic non-residue modulo 7, this case is also
impossible. Then we obtain that |(ξ−ξτ )(ξ−ξτ )ρ| > 2. Namely the integral
closure ZL2 in the second quartic subfield has no integral power basis.
Problems.
• Characterize Hasse’s Problem for the cyclic quartic fields over the ratio-
nals Q.
• Let the fields K run through all the real octic fields whose Galois groups
are 2-elementary abelian. Then evaluate the values of
inf
K
m̃(K) and inf
K
m(K),
respectively. Here, m̃(K) denotes the field index of K and m(K) the com-
mon index gcd(Ind(α);α ∈ ZK) for the integral closure ZK of the field
K.
54
Hasse’s problem for monogenic fields
Acknowledgement. I wish to express my gratitude Prof. Mohammed
Ayadi for his courteous invitation to CIANTA 2006 at Oujda.
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[17] S. Shah & T. Nakahara – Monogenesis of the rings of integers in
certain imaginary abelian fields, Nagoya Math. J. 168 (2002), p. 85–
92.
Toru Nakahara
Department of Mathematics, Faculty of
Science and Engineering, Saga University,
Saga 840-8502, Japan.
Current address:
NUCES, Peshawar Campus,
160-Industrial Estate, Hayatabad,
Peshawar, N.W.F.P.
The Islamic Republic of Pakistan
nakahara@ms.saga-u.ac.jp,
toru.nakahara@nu.edu.pk
56


Hasse’s problem for monogenic fields

Francesco Russo

Nadir Trabelsi

On minimal non-<i>PC</i>-groups

Noureddine Ghiloufi

Khalifa Dabbek

Prolongement d’un courant positif quasi-plurisurharmonique

Yaroslav Kopylov

$L_{p,q}$-cohomology of warped cylinders

Athanasios Katsaras

P-adic Spaces of Continuous Functions I

Mladen Božičević

Constant term in Harish-Chandra’s  limit formula

Salah Hajji

ANNALES MATHÉMATIQUES
BLAISE PASCAL
Salah Hajji
Stochastic fractional partial differential equations driven
by Poisson white noise
Volume 15, no 1 (2008), p. 43-55.
<http://ambp.cedram.org/item?id=AMBP_2008__15_1_43_0>
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Annales mathématiques Blaise Pascal 15, 43-55 (2008)
Stochastic fractional partial differential equations
driven by Poisson white noise
Salah Hajji
Abstract
We study a stochastic fractional partial differential equations of order α > 1
driven by a compensated Poisson measure. We prove existence and uniqueness of
the solution and we study the regularity of its trajectories.
Équations aux dérivées partielles fractionnaires stochastiques
dirigées par un bruit poissonnien
Résumé
On étudie une équation aux dérivées partielles stochastiques fractionnaires
d’ordre α > 1 dirigée par une mesure de Poisson compensée. On montre l’existence
et l’unicité de la solution et on étudie la régularité de ses trajectoires.
1. Introduction
In recent years fractional calculus knew a great progress considering their
applications in different fields of science, including numerical analysis,
physics, engineering, biology, economics and finance.
Partial differential equations is one of the mathematical domains where the
fractional calculus is strongly used (see Podlubny [8]). However, few pub-
lications treat stochastic partial differential equations involving fractional
derivatives. Most of them investigate evolution type equations, driven by a
fractional power of the Laplacian (see [3], [10]). These operators generate
symmetric stable semigroups when the order of derivation is less than 2.
Our idea is inspired by the paper of Debbi and Dozzi [5] in which the au-
thors studied a nonlinear stochastic fractional partial differential equations
of high order containing also derivatives of entire order and perturbed by
Keywords: Stochastic partial differential equations; fractional derivative operator; Pois-
son measure.
Math. classification: 26A33, 60H15.
43
S. Hajji
Gaussian white noise.
The aim of this paper is to generalize the result of [1] to nonlinear sto-
chastic fractional partial differential equations. We study the existence,
uniqueness and regularity of the solution of the following one-dimensional
stochastic fractional partial differential equation, formally given by
∂u
∂t
(t, x, ω) = Dδαu(t, x, ω) + f(t, u(t, x)) +
∫
U
g(t, u(t, x), z)Ñ(dt, dx, dz)
u(0, x) = u0(x) (1.1)
for t ∈ (0,+∞), x ∈ R, where (U ,B(U), q) is a σ finite measure space, Dαδ
is the fractional differential operator with respect to the spacial variable,
f : (0,+∞)× R −→ R and g : (0,+∞)× R× U −→ R are measurable.
One rigorous formulations of the equation (1.1) can be given by the fol-
lowing integral equation
u(t, x) =
∫
R
Gα(t, x− y)u0(y)dy
+
∫ t
0
∫
R
Gα(t− s, x− y)f(s, u(s, y))dyds (1.2)
+
∫ t+
0
∫
R
∫
U
Gα(t− s, x− y)g(s, u(s−, y), z)Ñ(ds, dy, dz)
for t ∈ [0,+∞) and x ∈ R, where {Gα(t, x), x ∈ R, t ∈ [0,+∞)} stands
for the fundamental solution of the operator Dαδ .
Some equations similar to equation (1.1) are studied also in [6] and [2],
but with the Laplacian operator in place of the operator Dαδ .
We assume that the functions f and g satisfy the following growth and
Lipschitz conditions:
(H)
For all T > 0, there exists a constant K = KT , such that for all 0 ≤ t ≤
T, x, y ∈ R, z ∈ U
• |f(t, x)|2 +
∫
U |g(t, x, z)|2q(dz) ≤ KT (1 + |x|2)
• |f(t, x)− f(t, y)|2 +
∫
U |g(t, x, z)− g(t, y, z)|2q(dz) ≤ KT |x− y|2.
The remaining of the paper is structured as follows. In Section 2, we
give the definitions of stochastic integral with respect to Ñ(ds, dy, dz)
and the derivative operator Dαδ , then we recall some inequalities satisfied
by the fundamental solution Gα and we precise the notion of solutions
44
Fractional SPDEs driven by Poisson white noise
of the equation (1.1). Section 3 is dedicated to the proof of existence
and uniqueness of the solution. In Section 4 we prove the spatial Hölder
continuity of the solution.
2. Preliminaries and definitions
Let (Ω,F , P ) be a probability space equipped with a filtration {Ft} sat-
isfying the usual conditions. Let (U ,B(U), q) be a σ finite measure space
and let N be a Poisson measure on R+ × R × U with intensity measure
ν(ds, dx, dz) = dsdxq(dz). The compensated Poisson measure is denoted
by Ñ = N − ν.
Now, we define the stochastic integrals we will use.
2.1. Stochastic integral with respect to Ñ(ds, dy, dz)
Let us introduce the following class:
Fpred := {h(t, y, z, ω) : h is Ft predictable and ∀t > 0,
E
∫ t
0
∫
R
∫
U
|h(s, y, z)|2ν(ds, dy, dz) <∞
}
it is known (see Ikeda and Watanabe [7]) that for any t > 0, the stochastic
integral
∫ t+
0
∫
R
∫
U h(s, y, z)Ñ(ds, dy, dz), for h ∈ Fpred, can be well defined.
The stochastic integral has the following isometry property
E
{∫ t+
0
∫
R
∫
U
h(s, y, z)Ñ(ds, dy, dz)
}2
=
E
∫ t
0
∫
R
∫
U
|h(s, y, z)|2ν(ds, dy, dz) .
Thus for any t > 0,
∫ t+
0
∫
R
∫
U h(s, y, z)Ñ(ds, dy, dz) ∈ L2(Ω).
This integral can be extended to a more general class F of integrands
without the predictable property, in the following manner :
A function h is said to be of class F if is Ft adapted, and there exists
a sequence hn ∈ Fpred such that for any t > 0
E
∫ t
0
∫
R
∫
U
|hn(s, y, z)− h(s, y, z)|2ν(ds, dy, dz) → 0 as n→∞
45
S. Hajji
For h ∈ F , for any t > 0,
∫ t+
0
∫
R
∫
U h(s, y, z)Ñ(ds, dy, dz) is defined as the
L2(Ω)-limit of the following Cauchy sequence
{
∫ t+
0
∫
R
∫
U
hn(s, y, z)Ñ(ds, dy, dz)}n∈N
(see Albeverio et al. [1] for more details).
2.2. Definition of the operator Dαδ
The fractional differential operator Dαδ is an extension of the inverse of
the generalized Riesz-Feller Potential when α > 2. It is given for α > 0 by
Definition 2.1. The fractional differential Dαδ is given by
Dαδ ϕ = F−1{ψα.Fϕ}
where
ψα(λ) = −|λ|αe−iδ
π
2
sgn(λ).
|δ| ≤ min{α − [α]2, 2 + [α]2 − α}, [α]2 is the largest even integer less or
equal to α, and δ = 0 when α ∈ 2N+1. F (respectively F−1) is the Fourier
(respectively Fourier inverse) transform.
The operator Dαδ is a closed, densely defined operator on L
2(R) and it
is the infinitesimal generator of a semigroup which is in general not sym-
metric and not a contraction. It is selfadjoint only when δ = 0 and in this
case, it coincides with the fractional power of the Laplacian. When α = 2
it is the Laplacian itself (see [4, 5]).
The Green function Gα(t, x) associated to the equation (1.1) is the
fundamental solution of the Cauchy problem
∂
∂t
G(t, x) = DδαG(t, x), t > 0, x ∈ R,
G(0, x) = δ0(x).
Using Fourier’s calculus we obtain
Gα(t, x) = F−1{eδψα(λ)t;x} =
1
2π
∫ +∞
−∞
exp(−iλx− t|λ|αe−iδ
π
2
sgn(λ))dλ.
Let us recall some well-known properties, (see [5]), of the Green kernel Gα
which will be used later.
Lemma 2.2. For α ∈ (0,+∞)
46
Fractional SPDEs driven by Poisson white noise
(1)
∫
RGα(t, x)dx = 1.
(2) Gα(t, x) satisfy the semigroup property, i.e. for 0 < s < t
Gα(t+ s, x) =
∫
R
Gα(t, y)Gα(s, x− y)dy.
(3) For 0 < α ≤ 2, the function Gα(t, .) is the density of a Lévy stable
process in time t.
(4) For fixed t, Gα(t, .) ∈ S∞={f ∈ C∞ and ∂
k
∂xk
f is bounded and
tends to zero when |x| tend to ∞, ∀k ∈ N}.
(5) ∂
nGα
∂xn (t, x) = t
−n+1
α
∂nGα
∂ξn (1, ξ)|ξ=t− 1α x, for all n ≥ 0.
(6) Let α ∈ (1,+∞). Then there exists a constant Kα such that
• |Gα(1, x)| ≤ Kα(1 + |x|1+α)−1,
• |Gnα(1, x)| ≤ Kα
1+|x|α+n−1
(1+|x|α+n)2 .
(7) For α ∈ (1,+∞), for any fixed n ∈ N, and T ≥ 0 , for γ such that
1
α+n+1 < γ <
α+1
n+1 ,∫ T
0
∫ +∞
−∞
|∂
nGα
∂yn
(s, y)|γdyds <∞.
In this paper, we restrict ourselves to the case α > 1 in order that∫ T
0
∫
RG
2
α(t, x)dxds <∞ (see Lemma 2.2, property (7)).
Let us now give a precise formulation of solutions for Equation (1.1).
Definition 2.3. Let 0 < T <∞. A measurable stochastic field
{u(t, x), t ∈ [0, T ], x ∈ R}
is said to be a mild solution of the equation (1.1) on [0, T ] if
• u(t, x) is adapted for every x.
• {u(t, x)}, as a family of L2(Ω,F , P )-valued random variables, is
right continuous and has left limits in the variable t ∈ [0, T ],
namely,
u(t−, x, .) = L2(Ω)− lim
s↑t
u(s, x, .), t ∈ [0, T ].
47
S. Hajji
In this paper we simply call such a u modified càdlàg in t (after
the French acronym).
• Equation (1.2) holds a.s. for all t ∈ [0, T ] and x ∈ R.
A solution u on [0, T ] is said to be Lp(Ω)- bounded if
sup
t≤T
sup
x∈R
E|u(t, x)|p <∞.
All positive constants appearing in this paper are called C. They may
change from one line to the next one.
3. Existence and uniqueness
The main result of this section is the following
Theorem 3.1. Let p ≥ 2 and T > 0 be fixed. Under condition (H) and
the assumption that the initial condition u0 is Lp(Ω) bounded, equation
(1.1) has a unique mild solution u on [0, T ] which is Lp(Ω) bounded.
For the proof we need the following lemma.
Fix T > 0 and let
E := {{v(t, x), t ∈ [0, T ], x ∈ R} : v is measurable ,Ft adapted,
modified càdlàg in t such that, sup
t≤T
sup
x∈R
E|v(t, x)|p <∞
}
.
Lemma 3.2. Let v ∈ E and set
X(t, x) :=
∫ t+
0
∫
R
∫
U
Gα(t− s, x− y)g(s, v(s−, y), z)Ñ(ds, dy, dz),
Y (t, x) :=
∫ t
0
∫
R
Gα(t− s, x− y)f(s, v(s, y))dyds.
Then X and Y belong to E.
Proof. Let us first show that X is well defined.
For any fixed t ∈ [0, T ], let, for (s, y, z) ∈ [0, t]× R× U ,
ht,x(s, y, z) := Gα(t− s, x− y)g(s, v(s−, y), z),
vn(s, y) := v(0, y)1{0}(s) +
2n−1∑
k=0
v(
kt
2n
, y)1
( kt
2n
,
(k+1)t
2n
]
(s), n ∈ N,
48
Fractional SPDEs driven by Poisson white noise
and
hnt,x(s, y, z) := Gα(t− s, x− y)g(s, vn(s, y), z).
Clearly vn is Fs predictable, so hn is also Fs predictable. Let us show that
hn ∈ Fpred.
By condition(H) we have :
E
∫ t
0
∫
R
∫
U
|hnt,x(s, y, z)|2ν(ds, dy, dz)
≤ KT
∫ t
0
∫
R
Gα(t− s, x− y)2(E|vn(s, y)|2 + 1)dyds <∞.
Thus hnt,x ∈ Fpred. On the other hand, by the condition (H) we have
E
∫ t
0
∫
R
∫
U
|hnt,x(s, y, z)− ht,x(s, y, z)|2ν(ds, dy, dz)
≤ KT
∫ t
0
∫
R
Gα(t− s, x− y)2E|vn(s, y)− v(s−, y)|2dsdy
By the definition of vn and the fact that v is modified càdlàg we have
E|vn(s, y)− v(s−, y)|2 −→ 0
n→∞
, for all (s, y) ∈ [0, t]× R
and
E|vn(s, y)− v(s−, y)|2 ≤ 4 sup
r≤T
sup
x∈R
E|v(r, x)|2, ∀(s, y, n) ∈ [0, t]× R× N.
Furthermore by Lemma 2.2, property (7), we have∫ t
0
∫
R
Gα(t− s, x− y)2dsdy <∞.
Then by the Lebesgue’s dominated convergence theorem we obtain that∫ t
0
∫
R
Gα(t− s, x− y)2E|vn(s, y)− v(s−, y)|2dsdy −→ 0
n→∞
.
Hence ht,x ∈ F for any fixed t ∈ [0, T ], so the integral∫ t+
0
∫
R
∫
U Gα(t− s, x− y)g(s, v(s−, y), z)Ñ(ds, dy, dz) is well defined.
Using the same techniques, as in [1], we show that X is Ft adapted and
modified càdlàg.
49
S. Hajji
Further more, by Burkholder’s inequality and condition (H), we have
E|X(t, x)|p ≤ CE(
∫ t
0
∫
R
∫
U
G2α(t− s, x− y)g2(s, v(s−, y), z)dsdyq(dz))
p
2
≤ CE(
∫ t
0
∫
R
G2α(t− s, x− y)(1 + v2(s−, y)dsdy)
p
2
≤ C(1 + sup
s≤T
sup
x∈R
E|v(s, x)|p)
∫ T
0
∫
R
G2α(t− s, x− y)dsdy)
p
2
By Lemma 2.2, property (7), the last integral is finite. Then
sup
t≤T
sup
x∈R
E|X(t, x)|p) <∞
Finally X ∈ E .
It is well known that Y is measurable Ft adapted and modified càdlàg.
By Hölder inequality with respect to the measure |Gα(t − s, x − y)|dsdy
and the growth assumption on f , we have
E|Y (t, x)|p ≤ E(
∫ t
0
∫
R
|Gα(t− s, x− y)|fp(s, v(s, y))dsdy)
×
{∫ t
0
∫
R
|Gα(t− s, x− y)|dsdy
}p−1
≤ CE
∫ t
0
∫
R
|Gα(t− s, x− y)|(1 + vp(s, y))dsdy
≤ C(1 + sup
s≤T
sup
x∈R
E|v(s, x)|p)
∫ T
0
∫
R
|Gα(t− s, x− y)|dsdy.
So
sup
t≤T
sup
x∈R
E|Y (t, x)|p) <∞.
and finally Y ∈ E . 
Proof of Theorem 3.1.
• Proof of the existence: Let the sequence (un, n ≥ 1) be given by:
u1(t, x) =
∫
R
Gα(t, x− y)u0(y)dy
50
Fractional SPDEs driven by Poisson white noise
un+1(t, x) = u1(t, x) +
∫ t
0
∫
R
Gα(t− s, x− y)f(s, un(s, y))dyds
+
∫ t+
0
∫
R
∫
U
Gα(t− s, x− y)g(s, un(s−, y), z)Ñ(ds, dy, dz)
It is easy to see that u1 ∈ E , so by Lemma 3.2, we have for all n ≥ 1, un ∈ E .
Let
Fn(t, x) := E|un+1(t, x)− un(t, x)|p,
and
Hn(t) := sup
s≤t
sup
x∈R
Fn(s, x).
we have ∀t ∈ [0, T ],∀x ∈ R
Fn(t, x) ≤ 2p−1[An(t, x) +Bn(t, x)] (3.1)
with
An(t, x) :=
E
(∫ t
0
∫
R
Gα(t− s, x− y)(f(s, un(s, y))− f(s, un−1(s, y)))dyds
)p
and
Bn(t, x) := E
{∫ t+
0
∫
R
∫
U
Gα(t− s, x− y)
(g(s, un(s−, y), z)− g(s, un−1(s−, y), z))Ñ(ds, dy, dz)
}p
.
By condition (H) and Hölder inequality with respect to the measure
|Gα(t− s, x− y)|dsdy, we obtain
An(t, x) ≤ CE(
∫ t
0
∫
R
|Gα(t− s, x− y)||(un − un−1)(s, y)|pdyds)
×
(∫ t
0
∫
R
|Gα(t− s, x− y)|dyds
)p−1
.
≤ C
∫ t
0
sup
y∈R
E|(un − un−1)(s, y)|p(
∫
R
|Gα(t− s, x− y)|dy)ds
≤ C
∫ t
0
Hn−1(s)ds. (3.2)
51
S. Hajji
By condition (H), Burkholder inequality and Hölder inequality with re-
spect to the measure G2α(t− s, x− y)dsdy we obtain
Bn(t, x) ≤ CE(
∫ t
0
∫
R
G2α(t− s, x− y)|un − un−1(s−, y)|2dsdy)
p
2
≤ C
∫ t
0
∫
R
G2α(t− s, x− y)E|un − un−1(s−, y)|pdsdy
×
(∫ t
0
∫
R
G2α(t− s, x− y)dsdy
) p
2
−1
≤ C
∫ t
0
sup
y∈R
|(un − un−1)(s, y)|p(
∫
R
G2α(t− s, x− y)dy)ds
≤ C
∫ t
0
(t− s)−
1
αHn−1(s)ds. (3.3)
Hence, by (3.1), (3.2), (3.3), we obtain that
Hn(t) ≤ C
∫ t
0
(t− s)−
1
αHn−1(s)ds
for some positive constant C.
By Lemma 3.3 in [9], the series
∑
n≥1(Hn(t))
1
p converges uniformly on
[0, T ]. Hence the sequence (un) converges in Lp(Ω) uniformly on [0, T ]×R.
Let
u(t, x) = Lp(Ω)− lim
n
un(t, x)
It is easy to see that u ∈ E and moreover u satisfies equation (1.2). Thus
u is a solution to equation (1.1) on [0, T ].
• Proof of the uniqueness: Let u and v be two mild solutions of the
equation (1.1) on [0, T ] and set
H(t) = sup
s≤t
sup
x∈R
E|u(s, x)− v(s, x)|2.
By the same argument as in the proof of existence, we get :
H(t) ≤ C
∫ t
0
(t− s)−
1
αH(s)ds.
Applying Gronwall Lemma, we obtain that H(T ) = 0. This clearly implies
that u(t, x, .) = v(t, x, .) a.s. 
52
Fractional SPDEs driven by Poisson white noise
4. Regularity of the solution
In this section we give the spatial regularity of the solution. This regularity
is based on the following lemma proved in [5].
Lemma 4.1. For α+12 < γ < α+ 1, we have∫ +∞
0
∫ +∞
−∞
|Gα(v, 1 + z)−Gα(v, z)|γdzdv <∞.
Theorem 4.2. Suppose that u0 is Lp(Ω) bounded for all p ≥ 2 and let u
denote the mild solution of the equation (1.1) on [0, T ], then we have:
For α < 3 and for fixed t > 0 the process {u(t, x), x ∈ R} has Hölder
continuous trajectories with exponent α−12 − ε, for any ε > 0, P-a.s.
Proof. The proof is similar to one given by Debbi and Dozzi in [5]. We
have
u(t, x) =∫
R
Gα(t, x− y)u0(y)dy +
∫ t
0
∫
R
Gα(t− s, x− y)f(s, u(s, y))dyds
+
∫ t+
0
∫
R
∫
U
Gα(t− s, x− y)g(s, u(s−, y), z)Ñ(ds, dy, dz)
= u(t, x) + v(t, x) + w(t, x) .
It is easy to see that u1 is smooth function with respect to t and to x.
By using Lemma 2.2, property (4), we can show that x → v(t, x) is a
smooth function, so it is sufficient to estimate the spatial regularity of w.
By Burkholder-Davis-Gundy inequality, condition (H) and the fact the
solution is Lp(Ω) bounded, we have:
E|w(t+ h, x)− w(t, x)|p
≤ C(1 + sup
[0,T ]×R
E|u(t, x)|p)× {
∫ t
0
∫
R
|Gα(v, y + h)−Gα(v, y)|2dydv}
p
2
≤ Chp
α−1
2α {
∫ +∞
0
∫ +∞
−∞
|Gα(v, 1 + z)−Gα(v, z)|2}
p
2 dzdv .
By Lemma 4.1, the last integral converge ( we have taken α < 3 in
order that α+12 < 2 < α+ 1 ).
Hence,
E|w(t+ h, x)− w(t, x)|p ≤ Chp
α−1
2α
53
S. Hajji
we can choose p as large as we please, so that the result follows from
Kolmogorov’s Theorem. 
Acknowledgment. I would like to thank the anonymous referee for
the careful reading of the first version of the paper and for the valuable
suggestions for improving the quality of this article.
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Salah Hajji
Department of Mathematics
Faculty of Sciences Semlalia
Cadi Ayyad University
BP. 2390
Marrakesh, MOROCCO.
s.hajji@ucam.ac.ma
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Stochastic fractional partial differential equations driven by Poisson white noise

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