arXiv:1507.06784v1 [math.AP] 24 Jul 2015
On the existence of solutions for a nonlinear stochastic partial differential equation arising as a model of phytoplankton aggregation Nadjia El Saadi (1)∗& Zakia Benbaziz (2) (1) LAMOPS, ENSSEA, Algiers, Algeria (2) TPFA, ENS Kouba, Algiers, Algeria
Abstract In this paper, we are interested in the analytical study of a nonlinear Stochastic Partial Differential Equation (SPDE) arising as a model of phytoplankton aggregation. This SPDE consists in a diffusion equation with a chemotaxis term responsible of self-attraction of phytoplankton cells and a multiplicative branching noise. Existence of mild solutions is established through weak and tightness arguments.
Keywords: Phytoplankton aggregation, Nonlinear stochastic partial differential equation, Semigroups, Chemotaxis, Gaussian space-time white noise, Weak convergence, Tightness, Skorohod representation theorem.
Introduction In this paper, we are interested in the following nonlinear stochastic partial differential equation: . p ∂ ∂ ∂2 u(t, x) G ∗ u0 (t, .) (x) + λu+ (t, x)W (t, x), u(t, x) = D 2 u(t, x) − ∂t ∂x ∂x (1) in [0, T ] × Ω, where Ω = ]0, L[ is a bounded domain with boundary ∂Ω in R, x is a one dimensional coordinate, t is time. The motivation in studying this SPDE is that equation (1) arises as a model of phytoplankton aggregation ([9],[10],[11],[12]). In fact, the authors in ([9],[10],[11],[12]) have investigated an individual-based model (IBM) of a population of phytoplankton that takes into account small scales biological mechanisms for phytoplankton cells. These processes are: random dispersal of phytoplankton cells due to turbulence, spatial interactions between phytoplankton cells caused by chemical signals and random division and death of phytoplankton cells. The aim of such modelling was ∗ Corresponding
author. E-mail address:
[email protected].
1
to capture the main features of the individual dynamics at small scales that are responsible at a larger scale for the formation of aggregating patterns. An Eulerian version of the IBM has been derived in ([9],[11]) and the SPDE (1) is obtained as a continuum limit of this IBM. u(t, x) represents the spatio-temporal distribution density of phytoplankton on the vertical water column. We denote by u+ (t, x) the positive part of u(t, x), that is u+ (t, x) = max(u(t, x), 0). The diffusion term in (1) takes into account the random spread of the phytoplankton cells with the coefficient of diffusion D; while the transport term, which is a chemotaxis term, describes the interaction mechanisms between the phytoplankton cells via the velocity G ∗ u0 . The latter has the form of convolution as in [23], i.e., Z G (x − x′ ) u0 (t, x′ )dx′ , G ∗ u0 (t, .) (x) = R
where G is the attractive force given by:
G (x) = [(|x| − r1 ) (|x| − r0 )] 1[−r1 ,−r0 ]∪[r0 ,r1 ] (x) . and 0
u (x) =
u(x) 0
0 0. The restrictions T (t)/D(B) send D(B) into itself and are uniformly bounded in D(B) (that is, there exists C1 ≥ 0, such that, T (t)/D(B) D(B) ≤ C1 , for t ≥ 0).
Proof. The proof is similar to that one in [2]. We propose to solve, on time interval [0, T ], equation (4) in integrated form by using the stochastic generalization of the classical variation of constants formula Z t Z t u(t) = T (t)u0 − T (t − s)B [u(s)gG (u(s))] ds + T (t − s)C (u(s)) dW (s). 0
0
(13) We remind that a predictable X-valued stochastic process (u(t))t∈[0,T ] defined on the probability space (Λ, F , (Ft )t∈[0,T ] , P ) is called a mild solution of the differential equation (4) if u(t) is a solution of (13). On another hand, the stochastic process (u(t))t∈[0,T ] is called a weakened solution of (4) if u(t) is a solution of Zt Zt Zt u(t) = u0 + A( u(s)ds) − B [u(s)gG (u(s))] ds + C(u(s))dW (s). 0
(14)
0
0
It is well known that solutions of (13) and (14) are equivalent ([6],[7]). Note Z t that the stochastic integral T (t − s)C (u(s)) dW (s) is well defined since 0 Z t 2 E( kT (t − s)C (u(s))k2 ds) < ∞, ∀t ∈ [0, T ] . This is due to the fact that 0
C (u(s)) is Hilbert-Schmidt operator and T (t − s) is linear bounded operator then, basing on the theory of Hilbert-Schmidt operators (e.g. [16], Chapter I), T (t − s)C (u(s)) is Hilbert-Schmidt operator too.
5
2
Existence of solutions
This section is concerned with the existence of mild solutions for (4). For this purpose, we need first to establish several propositions and lemmas. We start by giving some useful estimates. Lemma 2 1) There exists a constant M, such that, for all u, v ∈ D (B) , we have kB [ugG (u)] − B [vgG (v)]k ≤ M max(|u|D(B) , |v|D(B) ) |u − v|D(B) . 2) There exists a positive constant Q, such that, for all u ∈ D (B) , it holds that kB [ugG (u)]k ≤ Q |u|D(B) kuk . 3) There exists a positive constant C, such that, for all u ∈ X, it holds that C kBT (t)uk ≤ √ kuk , ∀t > 0. t
(15)
Proof. For the claim, the proof is the same as the one given in (Lemma 2.1, [1]). By using Lemma 2, we can prove Proposition 3 Consider the following abstract problem du(t) = (Au(t) − B [u(t)gG (u(t))]) dt + a(u(t))dW (t) u(0) = u0 .
(16)
Suppose that the operator a satisfies a : X → LHS (D (B) , X) and there exists a constant K such that for all x, y ∈ X, ka(x) − a(y)k2 6 K kx − yk . Then, for each u0 ∈ D(B), problem (16) has a unique mild solution u(t), t ∈ 2 [0, T ] such that sup E(|u(t)|D(B) ) < ∞. t∈[0,T ]
Proof. We use the method of successive approximations. Let u0 ∈ D(B) and define the sequence (un )n≥1 by un+1 (t) = T (t)u0 (t) −
Zt
0
T (t − s)B[un (s)gG (un (s))]ds +
Zt
0
T (t − s)a(un (s))dW (s).
Let us denote by Y the Banach space of all nonanticipating D(B)-valued continuous stochastic processes {U (t)}t∈[0,T ] endowed with the norm |||U ||| =
2
sup E |U (t)|D(B)
0≤t≤T
6
1/2
0 such that |(un (t)|D(B) ≤ R, ∀t ∈ [0, T ], ∀n ≥ 1. Hence (un (t), t ∈ [0, T ])n≥1 is also bounded in Y , since 1/2 2 |||un ||| = sup E |un (t)|D(B) ≤ R. 0≤t≤T
Let
2
hn+1 (t) = E(|un+1 (t) − un (t)|D(B) ),
n≥0
We have 2 t Z hn+1 (t) ≤ 2E T (t − s) [a(un (s)) − a(un−1 (s))] dW (s) 0 D(B) 2 t Z +2E T (t − s) (B[un (s)gG (un (s))] − B[un−1 (s)gG (un−1 (s))]) ds 0
D(B)
Zt 2 ≤ 2E( kT (t − s) [a(un (s)) − a(un−1 (s))]k2 ds) |
0
{z } I 2 t Z +2E T (t − s) (B[un (s)gG (un (s))] − B[un−1 (s)gG (un−1 (s))]) ds 0 D(B) {z } | II
Since ∃ C1 ≥ 0 such that T (t)/D(B) D(B) ≤ C1 , for t ≥ 0, hence I
≤
Zt
2C12 E(
≤
0
ka(un (s)) − a(un−1 (s))k22 ds) Zt
2C12 K 2 E(
0
2
kun (s) − un−1 (s)k ds)
≤ 2C12 K 2 T E( sup |un (s) − un−1 (s)|2D(B) ). 0≤s≤T
7
To calculate II, we have t Z T (t − s) (B[un (s)gG (un (s))] − B[un−1 (s)gG (un−1 (s))]) ds 0 D(B)
t
Z
= T (t − s) (B[un (s)gG (un (s))] − B[un−1 (s)gG (un−1 (s))]) ds
0 t
Z
T (t − s) (B[u (s)g (u (s))] − B[u (s)g (u (s))]) ds + B n G n n−1 G n−1
0 Zt ≤ M max(|un (s)|D(B) , |un−1 (s)|D(B) ) |un (s) − un−1 (s)|D(B) ds 0
Zt
C √ kB[un (s)gG (un (s))] − B[un−1 (s)gG (un−1 (s))]k ds t−s 0 √ ≤ M R(T + 2 T C) sup |un (t) − un−1 (t)|D(B) . +
0≤t≤T
Therefore √ 2 II ≤ 2M 2 R2 (T + 2 T C)2 E( sup |un (t) − un−1 (t)|D(B) ). 0≤t≤T
Hence, we obtain √ 2 ≤ 2[M 2 R2 ( T + 2 T C)2 + C12 K 2 T ]E( sup |un (t) − un−1 (t)|D(B) )
hn+1 (t)
0≤t≤T
By choosing T > 0 small enough so that √ 1 2[M 2 R2 ( T + 2 T C)2 + C12 K 2 T ] < , 2 we obtain hn+1 (t) ≤
(17)
1 2 E( sup |u1 (t) − u0 (t)|D(B) ). 2n 0≤t≤T
un+1 (t) − un (t) is the general term of an absolutely convergent series in the Banach space Y. Hence, {un (.)} is a convergent sequence in Y. The limit in Y, u∞ (t) = lim un (t) is a solution of (16) for fixed t. To complete the proof of n−→∞
the proposition, we have to show that the sequence un (t) remains in Y, i.e. 2 E( sup |un+1 (t)|D(B) ) < ∞. 0≤t≤T
√ 2 2 E(|un+1 (t)|D(B) ) ≤ 2C12 |u0 (t)|D(B) + 2M 2 R2 (T + 2C T )2 E( sup |un (t)|2D(B) ) + 2C12 K12 T E( sup |un (s)|2D(B) ) 0≤t≤T 0≤s≤T h i √ ≤ 2C12 R2 + 2R2 M 2 R2 (T + 2C T )2 + C12 K12 T , ∀t ∈ [0, T ]. 8
Using T > 0 such that (17) holds, we have 1 2 sup E(|un+1 (t)|D(B) ) ≤ 2C12 R2 + R2 < ∞ . 2 0≤t≤T To prove the uniqueness of the solution (up to equivalence), consider u and v two mild solutions of (16) on [0, T ]. We might show that 2
sup E |u(t) − v(t)|D(B) = 0.
0≤t≤T
We have 2
E |u(t) − v(t)|D(B)
2 t Z ≤ 2E T (t − s)(B[u(s)gG (u(s))] − B[v(s)gG (v(s))])ds 0 D(B) | {z } (1)
2 t Z +2E T (t − s)[a(u(s)) − a(v(s))]dW (s) 0 D(B) | {z } (2)
We obtain
2
(2) ≤ 2C12 K 2 T E( sup |u(s) − v(s)|D(B) ) 0≤s≤T
To calculate (1), we have t Z T (t − s)(B[u(s)gG (u(s))] − B[v(s)gG (v(s))])ds 0 D(B) √ ≤ M R(T + 2C T ) sup |u(t) − v(t)|D(B) 0≤t≤T
Then
√ 2 (1) ≤ 2M 2 γ 2 R2 (T + 2C T )2 E( sup |u(t) − v(t)|D(B) ) 0≤t≤T
Finally, we obtain 2
E |u(t) − v(t)|D(B)
i h √ ≤ 2M 2 R2 (T + 2C T )2 + 2C12 K 2 T 2
E( sup |u(t) − v(t)|D(B) ) 0≤t≤T
∀t ∈ [0, T ]
h i √ Since T is chosen such that 2M 2 R2 (T + 2C T )2 + 2C12 K 2 T < 21 , hence we have 2 sup E(|u(t) − v(t)|D(B) ) = 0 0≤t≤T
9
which leads to u(t) = v(t), ∀t ∈ [0, T ],
P − a.s.
We return to problem (4). We propose as in [5] to construct a sequence of Lipschitz√continuous mappings (an (u))n∈N to approximate the non-Lipschitzian function λu+ on R. For n ∈ N, define an (.) by 0 √ nu an (u) = √ λu
u 0 such that 2
kan (u) − an (v)k2
≤ K2
k=1 0 +∞ P
k=1
Using the hypothesis (12), we have
k=1
|ek |2D(B) =
+∞ X
k=1
2
k(u − v)ek k
≤ K 2 ku − vk2
+∞ X
2
+∞ P
k=1
|ek |2D(B)
kJek k2 = kJk22 < ∞,
10
2
|u(x)) − v(x)| |ek (x)| dx
which leads to the fact that there exists M1 > 0 such that 2
kan (u) − an (v)k2 ≤ M1 K 2 ku − vk
2
Let us now consider the following approximating SPDE’s:
un (t) = I0 (t) − In (t) + Jn (t), with
n∈N
(22)
I0 (t) = T (t)u0 (t) Zt In (t) = T (t − s)B[un (s)gG (un (s))]ds 0
Jn (t) =
Zt
0
T (t − s)an (un (s))dW (s)
where the sequence (an (.))n is defined by (18) and satisfies (19). We aim to prove weak convergence of the sequence {un (t), t ∈ [0, T ]}n∈N by proving the tightness of {In (t), t ∈ [0, T ]}n∈N and {Jn (t), t ∈ [0, T ]}n∈N since I0 (t) is deterministic. Proposition 5 Let the initial condition u0 ∈ D(B) be a continuous deterministic mapping on Ω.The sequences {In (t), t ∈ [0, T ]}n∈N and {Jn (t), t ∈ [0, T ]}n∈N are tight in C([0, T ], D(B)). Proof. By Lemma 4, it follows that for each n ∈ N, the sequence (an (.))n satisfies the conditions of Proposition 3. Hence, there exists a unique mild solution un (t), t ∈ [0, T ] for problem (22) such that sup E(|un (t)|2D(B) ) < ∞. t∈[0,T ]
For given 0 ≤ t′ ≤ t ≤ T and η ∈ [0, 12 ], it follows from Lemma 7 (in Appendix A) that 2
E |Jn (t) − Jn (t′ )|D(B)
Zt 2 ≤ 2E( k(T (t − s) − T (t′ − s))an (un (s))k2 ds) 0
′
Zt 2 +2E( k(T (t′ − s)an (un (s))k2 ds) t
≤ 2C ′ K 2 E( sup |un (s)|2D(B) ) |t − t′ |η 0≤s≤T
Zt′ +2E( |B(T (t′ − s)|2D(B) |un (s)gG (un (s))|2D(B) ds t
η
≤ C1 |t − t′ | . 11
On another hand, by using Lemma 8 (in Appendix A), we have for given 0 ≤ t′ ≤ t ≤ T and η ∈ [0, 12 ] E |In (t) − In (t
′
2 )|D(B)
Zt 2 ≤ 2E( |(T (t − s) − T (t′ − s))B[un (s)gG (un (s))]|D(B) ds 0
Zt′ 2 +2E( |B(T (t′ − s)[un (s)gG (un (s))]|D(B) t
Zt 2 2 ≤ 2E( |B(T (t − s) − T (t′ − s))|D(B) |un (s)gG (un (s))|D(B) ds 0
′
Zt 2 2 +2E( |B(T (t′ − s)|D(B) |un (s)gG (un (s))|D(B) ds t
using (9), we obtain 2
E |In (t) − In (t′ )|D(B)
≤ 2C ′′ δ 2 E( sup |un (s)|4D(B) ) |t − t′ |
η
0≤s≤T
+2C1 δ 2 E( sup |un (s)|4D(B) ) |t − t′ |η ′′
0≤s≤T η
≤ C2 |t − t′ | . It follows that E
ZT ZT 0
0
|Jn (t) − Jn (t′ )|D(B) |t − t′ |
γ
!2
dtdt
′
≤
ZT ZT
0
0
C1 |t − t′ |
0, we define AR ⊂ Λ ZT ZT AR = ω ∈ Λ / 0
0
|Jn (t, ω) − Jn (t′ , ω)|D(B) |t − t′ |
γ
η−2γ
dtdt′
for all 0 < γ
0 !2 ZT ZT |J (t) − J (t′ )| n n D(B) ′ P (AR ) = P ω ∈ Λ / dtdt ≤ R γ |t − t′ | 0 0 o n ≤ P kJn (., ω)kC δ ([0,T ],D(B)) ≤ C ′′ R1/2 Let us now show that {Jn (t), t ∈ [0, T ]}n∈N is tight on C([0, T ], D(B)). For R > 0, we define o n BR = f ∈ C([0, T ], D(B))/ kf kC δ ([0,T ],D(B)) ≤ R . BR is a compact set of C([0, T ], D(B)) (see the Proof B1 in Appendix B) and we have the following: supP [Jn ∈ B R ] = supP [kJn (., ωkC δ ([0,T ],D(B)) > R] n n # " ≤ supP A R2 n ′′ TC T2 Z Z ≤ supP n 0
J |n (t) − Jn (t′ )|D(B) |t − t′ |γ
0
13
!2
2 R dtdt′ > ′′ 2 C
Using Markov inequality, we obtain T T ′′ Z Z C 2 supP [Jn ∈ B R ] ≤ 2 supE R n n Since E
T T Z Z
0
|Jn (t) − Jn (t′ )|D(B)
0
|t − t′ |
0
0
|t − t′ |
|Jn (t) − Jn (t′ )|D(B)
γ
!2
dtdt′
γ
!2
dtdt′
.
(23)
< +∞, (23) leads to
′′
supP [Jn ∈ B R ] ≤ n
M R2
0, such that Zs
0
2
|T (t − u) − T (s − u)|D(B) du ≤ C ′ |t − s|
and
Zt
s
2
|T (t − u)|D(B) du ≤ C1′ |t − s|
η
η
with 0 ≤ s < t ≤ T. Proof. We represent T (t) in terms of the eigenvalues of A = D −wj2 to which correspond the eigenvectors ϕj : T (t) =
∞ X j=1
exp{−wj2 t} ., ϕj ϕj
17
(26)
(27)
d2 denoted dx2
Then, we have Zs
0
|T (t − u) − T (s − u)|2D(B) du
Zs X ∞ exp{−w2 (t − u)} − exp{−w2 (s − u)} 2 ≤ j
≤ ≤
0 j=1 Zs X ∞
0 j=1 ∞ X j=1
j
D(B)
., ϕj 2
D(B)
exp{−w2 (s − u)} 2 1 − exp{−w2 (t − s)} 2 ., ϕj 2 j
2 2 ϕj w (t − s) 2λ j D(B) D(B)
., ϕj 2
j
D(B)
Zs
exp{−2wj2 (s − u)}du
0
2 ϕj
D(B)
du
2 ϕj
D(B)
du
∞ X 2
2 −(2−4λ) 2λ wj ≤ sup ., ϕj D(B) sup ϕj D(B) |t − s)| j
j
≤ C0 |t − s)|
∞ X 2λ
j=1
−(2−4λ) wj
j=1
where we have used the fact that for all u ∈ [0, 1], 1 − exp{−x} ≤ xu For λ ∈ [0,
1 4 ],
we have
∞ X
−(2−4λ)
wj
j=1
Zs
0
(x > 0).
< ∞. If we set η = 2λ (η ∈ [0, 12 ]), hence 2
|T (t − u) − T (s − u)|D(B) du ≤ C ′ |t − s)|
η
Similarly, for the same η ∈ [0, 21 ] as above, we have Zt
s
|T (t −
u)|2D(B)
∞ X
., ϕj 2 du ≤ j=1
2 ϕj D(B) D(B)
Zt
s
exp{−2wj2 (t − u)}du
∞ X
2 2 1 ≤ sup ., ϕj D(B) sup ϕj D(B) wj−2 (1 − exp{−2wj2 (t − s)}) 2 j j j=1 ∞ X −(2−2η) η wj ≤ C0′ |t − s|
≤ C1′ |t − s)|
j=1 η
′′
Lemma 8 For η ∈ [0, 21 ], there exists constants C ′′ , C1 > 0,such that 18
Zs
0
2
η
|B [T (t − u) − T (s − u)]|D(B) du ≤ C ′′ |t − s| Zt
s
2
(28)
η
′′
|BT (t − u)|D(B) du ≤ C1 |t − s|
(29)
with 0 ≤ s < t ≤ T. Proof. From the representation T (t) =
∞ X
exp{−wj2 t} ., ϕj ϕj .
∞ X
exp{−wj2 t} ., ϕj Bϕj .
j=1
we obtain BT (t) =
j=1
Hence Zs
0
2
|B [T (t − u) − T (s − u)]|D(B) du
Zs X ∞ exp{−w2 (t − u)} − exp{−w2 (s − u)} 2 ≤ j
≤ ≤
0 j=1 Zs X ∞
0 j=1 ∞ X j=1
j
D(B)
., ϕ 2 j
D(B)
exp{−w2 (s − u)} 2 1 − exp{−w2 (t − s)} 2 ., ϕ 2 j
2 Bϕj 2 w (t − s) 2λ j D(B) D(B)
., ϕj 2
2 ≤ sup ., ϕj j
≤ C2 |t − s)|
D(B)
∞ X 2λ
j
j
2 sup Bϕj
D(B)
j
|t − s)|
2λ
Zs
0
D(B)
Bϕ 2
j D(B)
du
Bϕ 2
j D(B)
exp{−2wj2 (s − u)}du
∞ X
−(2−4λ)
wj
j=1
−(2−4λ) wj
j=1
where we have used the fact that for all u ∈ [0, 1], 1 − exp{−x} ≤ xu For λ ∈ [0, 14 ], we have Zs
0
∞ X j=1
−(2−4λ)
wj
(x > 0).
< ∞. If we set η = 2λ (η ∈ [0, 21 ]), hence 2
η
|B [T (t − u) − T (s − u)]|D(B) du ≤ C1 |t − s)| . 19
du
Similarly, for the same η ∈ [0, 12 ] as above, we have Zt
s
|BT (t −
u)|2D(B)
Zt ∞ X 2
2 exp{−2wj2 (t − u)}du ., ϕj D(B) Bϕj D(B) du ≤ j=1
s
∞ X
2 2 wj−2 (1 − exp{−2wj2 (t − s)}) ≤ 12 sup ., ϕj D(B) sup Bϕj D(B) j
j
≤ C ′ |t − s| ′
∞ X η
≤ C1 |t − s)|
j=1
−(2−2η) wj
j=1 η
Lemma 9 (Garsia-Rodemich-Rumsey) Let the function ψ : [0, ∞[→ [0, ∞[ non-decreasing with lim ψ(u) = +∞. Let the function p : [0, 1] → [0, 1] be u→+∞
continuous and non-decreasing with p(0) = 0. (
Set
ψ −1 (u) = sup v if ψ(0) ≤ u < ∞ . Let f be a continuous function on [0, 1] ψ(v)≤u
and suppose that Z1 Z1 0
0
ψ(
|f (x) − f (y)| )dxdy ≤ B < ∞ p(x − y)
Then, for all x, y ∈ [0, 1], we have |f (x) − f (y)| ≤ 8
|y−x| Z
ψ −1 (
4B )dp(u). u2
0
Appendix B Proof B1 Since kf kC δ ([0,T ],D(B)) = Sup |f (t)|D(B) + t∈[0,T ]
|f (t) − f (t′ )|D(B)
Sup t,t′ ∈[0,T ],t6=t′
|t − t′ |
δ
,
this means that ∀ t, t′ ∈ [0, T ], t 6= t′ |f (t) − f (t′ )|D(B) ≤ kf kC δ ([0,T ],D(B)) |t − t′ |
δ
As f ∈ BR , we have |f (t) − f (t′ )|D(B) ≤ R |t − t′ |
δ
This means that BR is uniformly equicontinuous. By the theorem of AscoliArzela, BR is a compact of C([0, T ], D(B)). 20
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