Abstract
In this paper, we consider the existence of a pullback attractor for the random dynamical system generated by stochastic two-compartment Gray-Scott equation for a multiplicative noise with the homogeneous Neumann boundary condition on a bounded domain of space dimension n ≤ 3. We first show that the stochastic Gray-Scott equation generates a random dynamical system by transforming this stochastic equation into a random one. We also show that the existence of a random attractor for the stochastic equation follows from the conjugation relation between systems. Then, we prove pullback asymptotical compactness of solutions through the uniform estimate on the solutions. Finally, we obtain the existence of a pullback attractor.
1 Introduction
In this paper, we consider the following coupled stochastic two-compartment Gray-Scott equation, which is a reaction-diffusion system with multiplicative noise:
for t > 0, on a bounded domain D. Here D is an open bounded set of ℝn (n ≤ 3), and it has a locally Lipschitz coutinuous boundary ∂D. Suppose that the equations have the following homogeneous Neumann boundary condition:
where
Here d1, d2, F, k, D1 and D2 are positive constants; σ is a positive parameter; Δ is the Laplacian operator with respect to
the Borel sigma-algebra
The Gray-Scott equation is a kind of very important reaction-diffusion system, which arises from many chemical or biological systems [2-5]. This equation has been researched by many authors (see [2-10]). One of the most important problems in mathematical physics is the asymptotic behavior of dynamical system, which has been developed greatly in recent years. For the deterministic system, the global attractor is a very important tool to study the asymptotic behavior of dynamical system (see [9-16]). If σ = 0, system (1)-(4) reduces to the two-compartment Gray-Scott equation without random terms, which has been investigated by You [10], where we proved the existence of the global attractor for the coupled two-compartment Gray-Scott equations with homogeneous Neumann boundary condition on a bounded domain.
Stochastic differential equations of this type arise from many chemical or biological systems when random spatiotemporal force is taken into consideration. These random perturbations play important roles in macroscopic phenomena. To study the properties of stochastic dynamical systems, the concept of pullback random attractor is introduced [1, 17, 18]. The existence of random attractors for stochastic dynamical systems has been studied [6, 7, 19-21]. In this paper, we study the existence of random attractor for stochastic two-compartment Gray-Scott equation on bounded domain D of space dimension n ≤ 3.
The paper is organized as follows. In Section 2, we recall a theorem about the existence of random pullback attractor for random dynamical system, and transform the stochastic system (1)-(6) into a continuous random dynamical system (18)-(19) Ornstein-Uhlenbeck process. Moreover, we show that, for each ω, the random dynamical system has a unique global solution. In Section 3, we obtain some uniform estimates of solutions for system (18)-(19) as t → ∞. These estimates are used to prove the existence of bounded absorbing sets and the asymptotic compactness of the solutions. In the last section, we obtain the existence of a pullback random attractor.
The following notations will be used throughout this paper. ‖ · ‖ and (·, ·) denote the norm and the inner product in L2(D) or [L2(D)]4 respectively. ‖ · ‖Lp and ‖ · ‖H1 are used to denote the norm in Lp(D) and H1(D).
By the Poincaré’s inequality, there is a constant γ > 0 such that
Note that
2 RDS Generated by Stochastic Gray-Scott equation
In this section, we first recall a theorem for the existence of random attractors. Please note that here we omit the basic knowledge about random dynamical systems (RDS) and random attractor. Reader can refer to [1, 11, 17-19] for these knowledge.
Suppose that (X, ‖ · ‖X) is a separable Banach space with Borel σ—algebra Β(X), and (Ω,
Suppose D is the collection of random subsets of X, and {K(ω)}ω∈Ω, ∈ D is a random absorbing set for RDS ϕ in D and ϕ is D-pullback asymptotically compact in X. Then ϕ has a unique D-random attractor {A(ω)}ω∈Ωwhich has the following form
Next, we shall show that system (1)-(6) generates a random dynamical system. For our purpose, we first transform this stochastic system into a deterministic dynamical with random attractor. Assume that (Ω, ℱ, ℙ) is the probability space defined in Section 1. Define (θt) t∈ℝ on Ω by
then (Ω, ℱ, ℙ, (θt) t∈ℝ) is a metric dynamical system.
Set
where
To transform the stochastic system into a deterministic system with random parameter, we introduce the following one-dimensional Ornstein-Uhlenbeck process:
From [13], we know that the stationary solution of Ornstein-Uhlenbeck process has the following form:
.
Moreover, the random variable z(θtω) is tempered, and ℙ—a.e. ω ∈ Ω, t ↦ z(θtω) is continuous in t, and satisfies the properties (see [1, 11, 13]):
.
Set
that is, g = (u, v, w, y)T satisfies
with
Notice that for ℙ—a.e. ω ∈ Ω, F(g, ω) is locally Lipschtiz continuous with respect to g. In [10], You proved that the deterministic system has a unique solution by the Galekin method. Similar to deterministic system, by the Galekin method, for ℙ—a.e. ω ∈ Ω, we can prove that for g0 ∈ [L2(D)]4, (18)-(19) has a unique solution g(·, ω, g0) ∈ C([0, ∞), [L2(D)]4) ∩ L2((0, ∞), [H1(D)]4) with g(0, ω, g0) = g0. Moreover, similarly to Lemma 3 of [10], we can prove that g(t, ω, g0) is a unique, global, weak solution with respect to g0 ∈ [L2(D)]4, for t ∈ [0, ∞). This shows that (18) and (19) generate a continuous random dynamical system (φ(t))t ≥ 0 over (Ω, ℱ, ℙ, (θt) t∈ℝ) with
Now assume that ϕ : ℝ+ × Ω × [L2(D)]4 → [L2(D)]4 is given by
Then ϕ is a continuous dynamical system associated to (10)-(11). Notice that two dynamical systems are conjugate to each other. Thus, in the following sections, we consider only the existence of a random attractor of φ.
3 Uniform estimates of solutions
To find the existence of the random attractor, we first need to obtain some uniform estimates of the solutions. Therefore in this section we first prove the uniform estimates about the solution of the two-compartment stochastic Gray-Scott equation on D, as t → + ∞. We assume that D is a collection of all tempered random subsets of [L2(D)]4. First, we define some functions which will be used in this section. Set
The next lemma shows that φ has a random absorbing set in D.
Random dynamical system φ has a random absorbing set {K(ω)}ω∈Ω, in D, that is, for any {K(ω)}ω∈Ω ∈ D, and for ℙ — a.e. ω ∈ Ω, there is TB (ω) > 0 such that φ(t, θ—tω, B(θ—tω)) ⊂ K(ω), for any t > TB (ω).
Taking the inner products of (15) and (17) with v and y respectively, and adding them up, we get
Gronwall’s inequality yields that
Replacing ω by θ—tω in the above inequality, we obtain that, for all t ≥ 0,
Using the properties of the Ornstein-Uhlenbeck process (13), for any g0 (θ—tω) ∈ B(θ—tω), we obtain that
and
Thus, there exists a TB (ω) > 0, such that, for all t > TB (ω),
It is easy to check that
Taking the inner product of (32) and Y1, then apply the Holder’s inequality and Poincaré’s inequality (7), we get:
Therefore,
Applying Gronwall’s inequality, we get that
By replacing ω by θ—tω in the above inequality, it follows that
Similar to (28), we have, for any g0 (θ—tω) ∈ B(θ—tω),
By (27)-(30), we obtain that, for all t ≥ TB (ω)
Let
Therefore (29), (36) and (39) imply that
Hence, we obtain from (30) and (40) that
It is easy to check that Y3 (t, x) satisfies the following equation
Taking the inner product of the last equation with Y3, we obtain that
In the last step we used Poincaré’s inequality (7). Thus,
with
Replacing ω by θ—tω, we find that, for all t ≥ 0,
Similar to (28) and (39), we obtain that
and
Set
It follows from (30) and (48) that, for all t ≥ TB (ω),
with c4 = 2c3 + 4. Consequently, we obtain that, for all t ≥ TB (ω),
with c5 = 2c1 + 8 + c4. We finally obtain that, for all t ≥ TB (ω),
It is easy to check that
□
There exists a random variableρ1 (ω), such that, for anyB(ω) ∈ D, and g0 (ω) ∈ B(ω), for ℙ—a.e. ω∈ Ω, there is aTB (ω) > 0, such that, for any t ≥ TB (ω), the following estimate holds
Because the unique global weak solution of system (18)-(19) satisfies g(t, ω, g0 (ω)) ∈ C((0, ∞), [L2(D)]4) ∩ L2((0, ∞), [H1(D)]4), then, for any initial value g0 ∈ [L2(D)]4, there exists a small time t0 ∈ (0, 1) such that,
This means that the weak solution g(t, ω, g0 (ω)) becomes a strong solution on [t0, +∞) and satisfies g(t, ω, g0 (ω)) ∈ C([t0, ∞), [H1(D)]4) ∩ L2([t0, ∞), [H2(D)]4) ⊂ C([t0, ∞), [L6(D)]4). Thus, without loss of generality, we can assume that g0 ∈ [L6(D)]4. Taking the inner products of (15) and (17) with v5 and y5 respectively, and adding them up, we obtain
We now estimate all terms on the right hand side of (54). For the fourth term, by Young’s inequality, we get that
For the first term, by Young’s inequality,
By (54)-(56), we arrive at the following estimate, for all t ≥ 0
By Gronwall’s inequality, we obtain that, for all t ≥ 0,
Replacing ω by θ—tω in the above inequality, we have that, for all t ≥ 0,
For
and
Set
Then there exists a TB(ω) > 0, independent of σ, such that for all t > TB(ω),
□
There exists a random variable ρ2(ω) > 0 such that, for any
Using (27), (28) and (29), we obtain:
Setting t = t + 1, we have:
Let TB(ω) be the constant defined in Lemma 3.1. It follows that, for all t ≥ TB(ω),
with
where
and
Therefore, by (67) and (68), we get
Finally, by (66) and (69), we obtain that, for all t ≥ TB(ω),
□
There exists a random variable ρ3(ω), such that, for any
Taking the inner products of (14) and (16) with –Δu and –Δw respectively, and then summing them up, we obtain that,
Now, we estimate each term on the right hand side of (72). For the second term,
For the first term, by Hölder’s inequality and (8), we have that
It follows from (72)-(74) that
Hence, for all t ≥ 0
Replacing t by s, and replacing ω by θ-t-1 ω in the last inequality, for every s < 0, we obtain that
which can be rewritten as
with
We use uniform Gronwall inequality to estimate
We need to compute ∫t+1tβ(s)ds and ∫t+1tα(s)ds first. By (50) and (52), we have that, for t < TB(ω)
and
It follows from (80) and (81) that
Next, we use Lemma 4.3 to estimate
Therefore, applying uniform Gronwall inequality, we can get that
□
There exists a random variableρ4(ω), such that, for any
Taking the inner products of (15) and (17) with –Δu and –Δw respectively, and then summing them up, we obtain that,
Now we estimate each term on the right hand side of (86). For the first and fourth term, by Green’s formula,
As for the third term,
It follows from Young’s inequality that
Replacing t by s and replacing ω by θ-t-1ω in the last inequality, we obtain:
It can be rewritten as
,
with
Similarly to Lemma 4.4, we need to estimate
By (80), (81) and Lemma 4.4, we have that, for
.
It follows that
.
By using uniform Gronwall inequality and (66), (93) and (91), we get that
.
□
4 Existence of random attractors
In this section we use Proposition 2.1 to prove the existence of a pullback attractor.
The random dynamical system ϕ has a unique D–pullback random attractor in [L2(D)]4.
In Lemma 3.1, we find that the system has a bounded absorbing set. By Lemma 3.4 and Lemma 3.5, we know that for any
Since ϕ and φ are conjugated by the random homeomorphism
.
□
Competing interests
The authors declare that there is no conflict of interest regarding the publication of this paper.
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