# Some Properties of Solutions for the Sixth-Order Cahn-Hilliard-Type Equation

Authors: Zhao Wang; Changchun Liu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study the initial boundary value problem for a sixth-order Cahn-Hilliard-type equation which describes the separation properties of oil-water mixtures, when a substance enforcing the mixing of the phases is added. We show that the solutions might not be classical globally. In other words, in some cases, the classical solutions exist globally, while in some other cases, such solutions blow up at a finite time. We also discuss the existence of global attractor.


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## Body

## 1. Introduction

We consider the following equation:(1.1)ut=γΔ3u+Δ[-a(u)Δu-a′(u)2|∇u|2+f(u)],inΩ×(0,T),
where Ω is a bounded domain in ℝn(n≤3) with smooth boundary and γ>0. Equation (1.1) is supplemented by the boundary value conditions:
(1.2)∂u∂n|∂Ω=∂Δu∂n|∂Ω=∂Δ2u∂n|∂Ω=0,t>0,
and the initial value condition:
(1.3)u(x,0)=u0(x).Equation (1.1) describes dynamics of phase transitions in ternary oil-water-surfactant systems [1–3]. The surfactant has a character that one part of it is hydrophilic and the other lipophilic is called amphiphile. In the system, almost pure oil, almost pure water, and microemulsion which consist of a homogeneous, isotropic mixture of oil and water can coexist in equilibrium. Here u(x,t) is the scalar order parameter which is proportional to the local difference between oil and water concentrations. The amphiphile concentration a(u) is approximated by the quadratic function [1]
(1.4)a(u)=a2u2+a0.
From the physical consideration, we prefer to consider a typical case of the volumetric free energy F(u), that is, F'(u)=f(u), in the following form:
(1.5)F(u)=∫0u‍f(s)ds=γ1(u+1)2(u2+h0)(u-1)2.During the past years, many authors have paid much attention to the sixth-order-parabolic equation, such as the existence, uniqueness, and regularity of the solutions [4–8]. However, as far as we know, there are few investigations concerned with the sixth order Cahn-Hilliard equation. Pawłow and Zajączkowski [9] proved that the initial-boundary value problem (1.1)–(1.5) with γ1=1 admits a unique global smooth solution which depends continuously on the initial datum. Schimperna and Pawłow [10] studied (1.1) with viscous term Δut and logarithmic potential:
(1.6)F(r)=(1-r)log(1-r)+(1+r)log(1+r)-λ2r2,λ>0.
They investigated the behavior of the solutions to the sixth-order system as the parameter γ tends to 0. The uniqueness and regularization properties of the solutions have been discussed. Liu studied the following equation:
(1.7)∂u∂t-div[m(u)(k∇Δ2u+∇(-a(u)Δu-a′(u)2|∇u|2+f(u)))]=0,
and he proved the existence of classical solutions for two dimensions [11]. Korzec et al. [12] established the stationary solutions of the sixth-order convective Cahn-Hilliard type equation, which arose from epitaxial growth of nanostructures on crystal surfaces.The dynamic properties of (1.1), such as the global asymptotical behaviors of solutions and existence of global attractors, are important for the study of higher order parabolic system. During the past years, many authors have paid much attention to the attractors. Nicolaenko et al. [13] proved the existence of global compact and finite-dimensional attractors for Cahn-Hilliard equation (see also [14–16]).In this paper, we consider the problem (1.1)–(1.3). The purpose of the present paper is devoted to the investigation of properties of solutions with γ1 not restricted to be positive. We first discuss the regularity. We show that the solutions might not be classical globally. In other words, in some cases, the classical solutions exist globally, while in some other cases, such solutions blow up at a finite time. We also discuss the existence of global attractor. We will use the regularity estimates for the linear semigroups, combining with the iteration technique and the classical existence theorem of global attractors, to prove that the problem (1.1)–(1.3) possesses a global attractor in Hk(k≥0) space.The plan of the paper is as follows. In Section2, we investigate the global existence of the solution when γ1>0. The blowup of the solution is obtained in Section 3 when γ1<0. In Section 4, we obtain the existence of the global attractor in Hk(k≥0) space.Throughout the paper, we useQT to denote Ω×(0,T), and
(1.8)H6,1(QT)={u;∂u∂t∈L2(QT),Diu∈L2(QT),0≤i≤6}.
The norms of L∞(Ω), L2(Ω), and Hs(Ω) are denoted by ∥·∥∞, ∥·∥, and ∥·∥s.


## 2. Global Existence

Now, we deal with problem (1.1)–(1.3) for n=2,3. The one-dimensional case is similar. From the classical approach, it is not difficult to conclude that the problem admits a unique classical solution local in time. So it is sufficient to make a priori estimates.Theorem 2.1.
For the initial datau0∈H3(Ω), (∂u0/∂n)|∂Ω=0, and T>0,(i)
ifγ1>0 and a2>0, then the problem (1.1)–(1.3) exists a unique global solution  u∈H6,1(QT);(ii)
ifγ1>0, a2<0, and γ are sufficiently large, then the problem (1.1)–(1.3) also  admits a unique global solution u∈H6,1(QT).Proof.
(i) First, we set(2.1)E(u)=∫Ω‍(γ2|Δu|2+a(u)2|∇u|2+F(u))dx.
Integrating by parts and using (1.1) itself and the boundary value condition (1.2), we see that
(2.2)dE(t)dt=∫Ω‍[γΔuΔut+a(u)∇u∇ut+a′(u)2|∇u|2ut+f(u)ut]dx=∫Ω‍[γΔ2u-a(u)Δu-a′(u)2|∇u|2+f(u)]utdx=-∫Ω‍[γ∇Δ2u+∇(-a(u)Δu-a′(u)2|∇u|2+f(u))]2dx≤0.
This implies
(2.3)E(t)≤E(0),
where
(2.4)E(0)=∫Ω‍(γ2|Δu0|2+a(u0)2|∇u0|2+F(u0))dx.
On the other hand, we have
(2.5)∫Ω‍|∇u|2dx≤ε∫Ω‍|Δu|2dx+C(ε)∫Ω‍u2dx.
By Young’s inequality, we derive
(2.6)u2≤εu6+C1ε,u4≤εu6+C2ε.
Combining the above inequalities, we get
(2.7)sup0<t<T∫Ω‍|Δu|2dx≤C,(2.8)sup0<t<T∫Ω‍|∇u|2dx≤C,(2.9)sup0<t<T∫Ω‍u6dx≤C.
From (2.9), we know
(2.10)sup0<t<T∫Ω‍u2dx≤C.
By (2.7), (2.8), and (2.10), we obtain
(2.11)∥u∥H2≤C.
By Sobolev embedding theorem, it follows from (2.11) that
(2.12)∥u∥∞≤C,(n=2,3),∥∇u∥Lq≤C,for  anyq<∞,(n=2),∥∇u∥L6≤C,(n=3).
The second step: multiplying (1.1) by Δu and integrating with respect to x, we obtain
(2.13)12ddt∫Ω‍|∇u|2dx+γ∫Ω‍|Δ2u|2dx=-∫Ω‍Δf(u)Δudx+∫Ω‍a(u)ΔuΔ2udx+∫Ω‍a′(u)2|∇u|2Δ2udx.
Thus, it follows from (2.11) and (2.12) that
(2.14)12ddt∫Ω‍|∇u|2dx+γ∫Ω‍|Δ2u|2dx=-∫Ω‍(f′(u)Δu+f′′(u)|∇u|2)Δudx+∫Ω‍(a2u2+a0)ΔuΔ2udx=+∫Ω‍a′(u)2|∇u|2Δ2udx≤C∫Ω‍|Δu|2dx+C∫Ω‍|∇u|4dx+C∫Ω‍|Δu|2dx≤+C∫Ω‍|Δu|2dx+γ4∫Ω‍|Δ2u|2dx+C∫Ω‍|∇u|4dx+γ4∫Ω‍|Δ2u|2dx≤γ2∫Ω‍|Δ2u|2dx+C.
By the Gronwall’s inequality, (2.14) implies
(2.15)∬QT‍|Δ2u|2dxdt≤C.
The third step: multiplying (1.1) by Δ2u and integrating with respect to x, we obtain
(2.16)12ddt∫Ω‍|Δu|2dx+γ∫Ω‍|∇Δ2u|2dx=-∫Ω‍f′(u)∇u∇Δ2udx+∫Ω‍∇(a(u)Δu)∇Δ2udx=+∫Ω‍∇(a′(u)2|∇u|2)∇Δ2udx=-∫Ω‍f′(u)∇u∇Δ2udx+∫Ω‍a(u)∇Δu∇Δ2udx=+2∫Ω‍a′(u)∇uΔu∇Δ2udx+12∫Ω‍a′′(u)|∇u|2∇u∇Δ2udx.
On the other hand, by the Nirenberg’s inequality and (2.12), we have
(2.17)∥∇u∥∞≤C∥∇Δ2u∥a∥∇u∥Lq1-a,wherea=11+3q/2,(n=2),∥∇u∥∞≤C∥∇Δ2u∥1/6∥∇u∥L65/6,(n=3),∥∇Δu∥≤∥∇Δ2u∥1/2∥∇u∥1/2,(n=2),∥∇Δu∥≤∥∇Δ2u∥1/3∥∇u∥L62/3,(n=3).
Using (2.7), (2.12), and the above inequality, we derive
(2.18)|-∫Ω‍f′(u)∇u∇Δ2udx|≤C∫Ω‍|∇u|2dx+γ8∫Ω‍|∇Δ2u|2dx,|∫Ω‍a(u)∇Δu∇Δ2udx|≤C∫Ω‍|∇Δu|2dx+γ16∫Ω‍|∇Δ2u|2dx≤γ8∫Ω‍|∇Δ2u|2dx+C,|2∫Ω‍a′(u)∇uΔu∇Δ2udx|≤C∥u∥∞∥Δu∥∥∇Δ2u∥≤γ8∫Ω‍|∇Δ2u|2dx+C,|12∫Ω‍a′′(u)|∇u|2∇u∇Δ2udx|≤C∫Ω‍|∇u|6dx+γ8∫Ω‍|∇Δ2u|2dx≤γ8∫Ω‍|∇Δ2u|2dx+C.
From these inequalities we finally arrive at
(2.19)12ddt∫Ω‍|Δu|2dx+γ∫Ω‍|∇Δ2u|2dx≤γ2∫Ω‍|∇Δ2u|2dx+C.
A Gronwall’s argument now gives
(2.20)∬QT‍|∇Δ2u|2dxdt≤C.
Similarly, multiplying (1.1) by Δ3u and using
(2.21)∥∇u∥∞≤∥Δ3u∥b∥∇u∥Lq1-b,b=11+2q,(n=2),∥∇u∥∞≤∥Δ3u∥1/8∥∇u∥L67/8,(n=3),∥Δu∥∞≤∥Δ3u∥2/5∥∇u∥3/5,(n=2),∥Δu∥∞≤∥Δ3u∥3/8∥∇u∥L65/8,(n=3),∥Δ2u∥∞≤∥Δ3u∥4/5∥∇u∥1/5,(n=2),∥Δ2u∥∞≤∥Δ3u∥7/8∥∇u∥L61/8,(n=3),
we obtain
(2.22)12ddt∫Ω‍|∇Δu|2dx+γ∫Ω‍|Δ3u|2dx≤γ2∫Ω‍|Δ3u|2dx+C.
So that
(2.23)sup0<t<T∫Ω‍|∇Δu|2dx≤C,(2.24)∬QT‍|Δ3u|2dxdt≤C.
Define the linear spaces
(2.25)X={u∈H6,1(QT);∂u∂n|∂Ω=∂Δu∂n|∂Ω=∂Δ2u∂n|∂Ω=0,u(x,0)=u0(x)}
and the associated operator W:X→X,u→w, where w is determined by the following linear problem:
(2.26)∂w∂t-γΔ3w=-Δ(a(u)Δu+a′(u)2|∇u|2)+Δf(u),x∈Ω,∂w∂n|∂Ω=∂Δw∂n|∂Ω=∂Δ2w∂n|∂Ω=0,t>0,w(x,0)=u0(x).
From the discussions above and by the contraction mapping principle, W has a unique fixed point u, which is the desired solution of the problem (1.1)–(1.3).
Now, we show the uniqueness. Assumeu and v are two solutions of the problem (1.1)–(1.3). Let w=u-v. Then w satisfies
(2.27)wt=γΔ3w+Δφ(u)-Δφ(v),inΩ×(0,T),∂w∂n|∂Ω=∂Δw∂n|∂Ω=∂Δ2w∂n|∂Ω=0,w(x,0)=0,
where φ(u)=-a(u)Δu-(a'(u)/2)|∇u|2+f(u).
Multiplying the above equation byw and integrating with respect to x, integrating by parts, and using the boundary value condition, we have
(2.28)ddt∥w∥2+2γ∥∇Δw∥2≤|∫Ω‍φ(u)Δwdx|+|∫Ω‍φ(v)Δwdx|≤C∥Δw∥2≤γ2∥∇Δw∥2+C∥w∥2.
The Gronwall inequality yields
(2.29)∥w∥2≤C∥w0∥2=0.
Therefore, ∥w∥2=0,thatis,w=0.
(ii) From the proof of (i), we know(2.30)E(t)≤E(0),
which together with the Young inequality gives
(2.31)∫Ω‍(γ2|Δu|2+γ12u6)dx≤∫Ω‍|a0|2|∇u|2dx-∫Ω‍a22u2|∇u|2dx+E(u0)+C0.
Note that(2.32)∫Ω‍u2|∇u|2dx=-∫Ω‍(2u2|∇u|2+u3Δu)dx,
that is,
(2.33)∫Ω‍u2|Du|2dx=-13∫Ω‍u3Δudx.
Hence,
(2.34)|a22∫Ω‍u2|∇u|2dx|=|a26∫Ω‍u3Δudx|≤|a2|6(ε∫Ω‍u6dx+C(ε)∫Ω‍|Δu|2dx).
Taking into account (2.31) and (2.34), we see
(2.35)∫Ω‍(γ2|Δu|2+γ12u6)dx≤∫Ω‍|a0|2|∇u|2dx-∫Ω‍|a2|2u2|∇u|2dx+E(u0)+C0≤ε1∫Ω‍|a0|2|Δu|2dx≤+C(ε1)∫Ω‍u2dx+|a2|6(ε∫Ω‍u6dx+C(ε)∫Ω‍|Δu|2dx)+E(u0)+C≤C1∫Ω‍|Δu|2dx+γ14∫Ω‍u6dx+E(u0)+C.
Hence, when γ is sufficiently large such that γ/2-C1>0, we obtain the estimates (2.7)–(2.9). The other steps are similar to the proof of (i), so the details are omitted here.


## 3. Blow Up

In the previous sections, we have seen that the solution of the problem is globally existent, provided thatγ1>0. The following theorem shows that the solution of the problem blows up at a finite time for γ1<0.Theorem 3.1.
Ifγ1<0, -∫Ω‍((γ/2)|Δu0|2+a((u0)/2)|∇u0|2+F(u0)dx) is sufficiently large.(i)
Ifa2<0, then the solution u of (1.1)–(1.3) blows up in finite time, that is, for T>0,
(3.1)limt→T∥u(t)∥=+∞.(ii)
Ifa2>0 and a0>0, the solution must blow up in a finite time.(iii)
Ifa2>0, a0<0, and γ are large enough, the solution blows up in finite time.Proof.
(i) From the proof of Theorem2.1, we know
(3.2)E(t)=∫Ω‍(γ2|Δu|2+a(u)2|∇u|2+F(u))dx≤E(0).
Hence,
(3.3)-∫Ω‍γ|Δu|2dx≥∫Ω‍a(u)|∇u|2dx+2∫Ω‍F(u)dx-2E(0).
Let w be the unique solution of the following problem:
(3.4)Δw=u,x∈Ω,∂w∂n=0,on∂Ω,∫Ω‍wdx=0.
It is easily seen that
(3.5)∥∇w∥2≤C∥u∥2.
Multiplying (1.1) by w and integrating with respect to x, we obtain
(3.6)ddt∫Ω‍|∇w|2dx=-2γ∫Ω‍(Δu)2dx+2∫Ω‍a(u)uΔudx+∫Ω‍a′(u)|∇u|2udx-2∫Ω‍f(u)udx≥2∫Ω‍a(u)|∇u|2dx+2∫Ω‍a(u)uΔudx+∫Ω‍a′(u)|∇u|2udx=+4∫Ω‍F(u)dx-2∫Ω‍f(u)udx-4E(0)=∫Ω‍4a2u2|∇u|2dx+∫Ω‍2a0|∇u|2dx+∫Ω‍2a2u3Δudx+2a0∫Ω‍uΔudx=+4∫Ω‍F(u)dx-2∫Ω‍f(u)udx-4E(0)=-2a2∫Ω‍u2|∇u|2dx+4∫Ω‍F(u)dx-2∫Ω‍f(u)udx-4E(0)=-2a2∫Ω‍u2|∇u|2dx+γ1∫Ω‍(-8u6-4(h0-2)u4+4h0)dx-4E(0).
Owing to a2<0,γ1<0, and (3.5), it follows from the above inequality that
(3.7)ddt∫Ω‍|∇w|2dx≥-γ1∫Ω‍u6dx-4E(0)-C(h0,γ1,|Ω|)≥-γ1(∫Ω‍u2dx)3-4E(0)-C(h0,γ1,|Ω|)≥-γ1(∫Ω‍|∇w|2dx)3-4E(0)-C(h0,γ1,|Ω|),
where C1>0. Hence, when -E(0) is sufficiently large, such that -4E(0)-C(h0,γ1,|Ω|)≥0, then by γ1<0, we know that ∥u∥ has to blow up.
From the proof of (i), we obtain(3.8)ddt∫Ω‍|∇w|2dx≥-2a2∫Ω‍u2|∇u|2dx+4∫Ω‍F(u)dx-2∫Ω‍f(u)udx-4E(0).
On the other hand, we have
(3.9)∫Ω‍u2|∇u|2dx=-13∫Ω‍u3Δudx≤13(ε∫Ω‍u6dx+C(ε)∫Ω‍(Δu)2dx).
From the above inequality, we know
(3.10)C(ε)∫Ω‍(Δu)2dx≥3∫Ω‍u2|∇u|2dx-ε∫Ω‍u6dx.
Using (3.3) and (3.10), we see that
(3.11)-γC(ε)(3∫Ω‍u2|∇u|2dx-ε∫Ω‍u6dx)≥-γ∫Ω‍(Δu)2dx≥∫Ω‍a(u)|∇u|2dx+2∫Ω‍F(u)dx-2E(0)=∫Ω‍(a2u2|∇u|2+a0|∇u|2)dx+2∫Ω‍F(u)dx-2E(0).
It follows that(3.12)-(3γC(ε)+a2)∫Ω‍u2|∇u|2dx≥∫Ω‍a0|∇u|2dx+εγC(ε)∫Ω‍u6dx+2∫Ω‍F(u)dx-2E(0).
Again by (3.12) and (3.3), we get
(3.13)ddt∫Ω‍|∇w|2dx≥-2a2∫Ω‍u2|∇u|2dx+4∫Ω‍F(u)dx-2∫Ω‍f(u)udx-4E(0)≥2a2C(ε)3γ+a2C(ε)(∫Ω‍a0|∇u|2dx+εγC(ε)∫Ω‍u6dx+2∫Ω‍F(u)dx-2E(0))≥+4∫Ω‍F(u)dx-2∫Ω‍f(u)udx-4E(0).
By a0>0,a2>0,C(ε)>0,γ>0, choosing C(ε) large enough, we obtian
(3.14)ddt∫Ω‍|∇w|2dx≥-γ1∫Ω‍u6dx-CE(0)-C(h0,γ1,|Ω|).
Similar to the proof of (i), we easily see that (ii) holds.(iii)
The crucial term is(3.15)∫Ω‍a0|∇u|2dx.By the Young inequality, we have
(3.16)∫Ω‍|∇u|2dx≤12∫Ω‍(Δu)2dx+12∫Ω‍u2dx.
By (3.3) and (3.16), we have
(3.17)-γ(2∫Ω‍|∇u|2dx-∫Ω‍u2dx)≥-∫Ω‍γ|Δu|2dx≥∫Ω‍a(u)|∇u|2dx+2∫Ω‍F(u)dx-2E(0)=∫Ω‍a2u2|∇u|2dx+∫Ω‍a0|∇u|2dx+2∫Ω‍F(u)dx-2E(0)≥∫Ω‍a0|∇u|2dx+2∫Ω‍F(u)dx-2E(0).
It follows that
(3.18)(-2γ-a0)∫Ω‍|∇u|2dx≥-γ∫Ω‍u2dx+2∫Ω‍F(u)dx-2E(0).
If 2γ+a0>0, substituting the (3.18) into (3.13), similarly, we know that the solution must blow up in finite time.


## 4. Global Attractor inHk Space

In this section, we will give the existence of the global attractors of the problem (1.1)–(1.3) in any kth order space Hk(Ω).First of all, we will prove the existence of attractor forγ1>0. We define the operator semigroup {S(t)}t≥0 in H3(Ω) space by
(4.1)S(t)u0=u(t),t≥0,
where u(t) is the solution of the problem (1.1)–(1.3).We let(4.2)Xm={u∣u∈H3(Ω),|∫Ω‍udx|≤m},
where m>0 is a constant, and the {S(t)} on Xm is a well-defined semigroup.Theorem 4.1.
For everym chosen as above, the semiflow associated with the solution u of the problem (1.1)–(1.3) possesses in Xm a global attractor Am, which attracts all the bounded set in Xm.In order to prove Theorem4.1, we need to establish some a priori estimates for the solution u of problem (1.1)–(1.3). In what follows, we always assume that {S(t)}t≥0 is the semigroup generated by the weak solutions of (1.1) with initial data u0∈H3(Ω).Lemma 4.2.
There exists a bounded setBm whose size depends only on m and Ω in Xm such that for all the orbits staring from any bounded set B in Xm, ∃t1=t1(X)≥0, such that for all t≥t1 all the orbits will stay in Bm.Proof.
It suffices to prove that there is a positive constantC such that for large t, then the following holds:
(4.3)∥u∥3≤C.
From (2.22), we have
(4.4)ddt∫Ω‍|∇Δu|2dx+γ∫Ω‍|Δ3u|2dx≤C.
On the other hand, we know that
(4.5)∫Ω‍|∇Δu|2dx=-∫Ω‍uΔ3udx≤C∫Ω‍|u|2dx+γ2∫Ω‍|Δ3u|2dx,∫Ω‍|u|2dx≤ε∫Ω‍|∇Δu|2dx.
Hence, we see that
(4.6)ddt∫Ω‍|∇Δu|2dx+C1∫Ω‍|∇Δu|2dx≤C2,
which gives
(4.7)∫Ω‍|∇Δu|2dx≤e-C1t∫Ω‍|∇Δu0|2dx+C2C1.
The proof is completed.Lemma 4.3.
For any initial datau0 in any bounded set B⊂Xm, there exists t2=t2(B)>0 such that
(4.8)∥u(t)∥H4≤C,∀t≥t2>0,
which turns out that ⋃t≥t2‍u(t) is relatively compact in Xm.Proof.
From Theorem2.1, we know when γ1>0,a2>0 or γ1>0,a2<0 that (2.14) holds. Integrating (2.14) over (t,t+1), we obtain
(4.9)∫tt+1‍∫Ω‍(Δ2u)2dxdt≤C.
On the other hand, Differentiating (1.1) gives
(4.10)Δut=γΔ4u+Δ2[-a(u)Δu-a′(u)2|∇u|2+f(u)].
Multiplying (4.10) by Δ3u and integrating on Ω, using the boundary conditions, we obtain
(4.11)12ddt∫Ω‍|Δ2u|2dx+γ∫Ω‍|∇Δ3u|2dx≤γ2∫Ω‍|∇Δ3u|2dx+C.
From (4.11), (4.9), and the uniform Gronwall inequality, we have
(4.12)∥Δ2u∥≤C,t2≥t2(B).Then by [17, Theorem I.1.1], we immediately conclude that Am=ω(Bm); the ω-limit set of absorbing set Bm is a global attractor in Xm. By Lemma 4.3, this global attractor is a bounded set in H3(Ω). Thus, we complete the proof of Theorem 4.1.Secondly, we consider the existence of the global attractors of the problem (1.1)–(1.3) in any kth order space Hk(Ω). Because both γ1>0,a2>0 and γ1>0,a2<0 lead to Theorem 4.1, the following proofs are based on Theorem 4.1, hence for simplicity, we let γ=1.In order to consider the global attractor for (1.1) in Hk space, we introduce the definition as follows:
(4.13)H={u∈L2(Ω);|∫Ω‍udx|≤m},H1/2={u∈H3(Ω)∩H;∂u∂n|∂Ω=∂Δu∂n|∂Ω=0},H1={u∈H6(Ω)∩H;∂u∂n|∂Ω=∂Δu∂n|∂Ω=∂Δ2u∂n|∂Ω=0},
where m>0 is a constant.We rewrite (1.1) as
(4.14)ut-Δ3u=Δg(u),
where
(4.15)g(u)=(-a(u)Δu-a′(u)2|∇u|2+f(u)).Let(4.16)L=Δ3:H1→H,G=Δg:H1→H.
Then, (4.14) and (4.15) can be rewritten as
(4.17)dudt=Lu+Gu.The linear operatorL is a sectorial operator which generates an analytic semigroup etL. Without loss of generality, we assume that L generates the fractional power operators ℒα and the fractional order spaces Hα as follows:
(4.18)ℒα=(-L)α:Hα→H,α∈R,
where Hα=D(ℒα) is the domain of ℒα and Hβ⊂Hα is a compact inclusion for any β>α [see Pazy [18]].The spaceH1/6 is given by H1/6= the closure of H1/2 in H1(Ω) and Hk=H6k∩H1 for k≥1.The following lemmas which can be found in [19, 20] are crucial to our proof.Lemma 4.4.
Letu(t,u0)=S(t)u0  (u0∈H,t≥0) be the solution of (4.17) and S(t) the semigroup generated by (4.17). Assume Hα is the fractional order space generated by L and(1)
for someα≥0 there is a bounded set B~⊂Hα, for any u0∈Hα there exists tα>0 such that
(4.19)u(t,u0)∈B~,∀t>tu0;(2)
there is aβ>α, for any bounded set U⊂Hβ, ∃T>0 and C>0 such that
(4.20)∥u(t,u0)∥Hβ≤C,∀t>T,u0∈U.Then (4.17) has a global attractor A which attracts any bounded set Hα in the Hα-norm.Lemma 4.5.
LetL:H1→H be a sectorial operator which generates an analytic semigroup T(t)=etL. If all eigenvalues λ of L satisfyReλ<-λ0 for some real number λ0>0,   then for ℒα(ℒ=L) we have(1)
T(t):H1→H is bounded for all α∈ℝ and t>0;(2)
T(t)ℒαx=ℒαT(t)x, for all x∈Hα;(3)
for eacht>0, T(t)ℒα:H→H is bounded, and
(4.21)∥ℒαT(t)∥≤Cαt-αe-σt,
where some σ>0, Cα>0 is a constant only depending on α;(4)
theHα-norm can be defined by
(4.22)∥x∥Hα=∥ℒαx∥H.Now, we give the main theorem.Theorem 4.6.
For anyα≥0 the problem (1.1)–(1.3) has a global attractor A in Hα, and A attracts any bounded set of Hα in the Hα-norm.Proof.
Owing to (4.17), the solution of the system (1.1)–(1.3) can be written as
(4.23)u(t,u0)=etL+∫0t‍e(t-τ)LG(u)dτ.
First of all, we are going to prove that for anyα≥0, the solution u(t,u0) of the problem is uniformly bounded in Hα, that is, for any bounded set U⊂Hα, there exists C such that
(4.24)∥u(t,u0)∥Hα≤C,∀t≥0,u0∈U⊂Hα,α≥0.
From Theorem4.1, we have known that, for any bounded set U⊂H1/2 there is a constant C>0 such that
(4.25)∥u∥H1/2≤C,∀t≥0,u0∈U⊂H1/2.
Next, according to Lemma 4.4, we prove (4.24) for any α>1/2 in the following steps.
Step1. We are going to show that for any bounded set U⊂Hα(0<α<1), there exists C>0 such that
(4.26)∥u(t,u0)∥Hα≤C,∀t≥0,u0∈U⊂Hα,α<1.
In fact, by the embedding theorems of fractional order spaces [18], we have
(4.27)H3(Ω)↪W2,2(Ω),H3(Ω)↪W1,4(Ω),H1/2↪C0(Ω)∩H3(Ω).
From (4.23), we have
(4.28)∥u(t,u0)∥Hα=∥etL+∫0t‍e(t-τ)LG(u)dτ∥Hα≤∥u0∥Hα+∫0t‍∥Lαe(t-τ)LG(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥Lα+(1/3)e(t-τ)L∥∥g(u)∥Hdτ.
We deduce that
(4.29)∥g(u)∥=∫Ω‍|Δu-a(u)Δu-a′(u)2|∇u|2+f(u)|2dx≤∫Ω‍|a2u2Δu+(a0+1)Δu+a2|∇u|2+Cu6+C|2dx≤C∫Ω‍|u4|Δu|2+|Δu|2+|∇u|4+u12|dx+C≤C(∥u∥H1/24∥u∥H1/22+∥u∥H1/22+∥u∥H1/24+∥u∥H1/212)+C,
which means that g:H1/2→H is bounded.
Hence, from (4.25), (4.28), (4.29), and Lemma 4.5, we have
(4.30)∥u∥Hα≤∥u0∥Hα+C∫0t‍τ-α-(1/3)e-στdτ≤C,
where 0<α<2/3.
Step2. We prove that for any bounded set U⊂Hα(2/3≤α<5/6), there is a constant C>0 such that
(4.31)∥u(t,u0)∥Hα≤C,∀t≥0,u0∈U⊂Hα,α<56.
In fact, by the embedding theorems, we have(4.32)H3(Ω)↪W1,4(Ω),H3(Ω)↪W2,4(Ω),H3(Ω)↪W1,6(Ω),H3(Ω)↪W1,2(Ω),Hα(Ω)↪C0(Ω)∩H3(Ω),α>12.
We deduce that
(4.33)∥g(u)∥H1/6=∫Ω‍|∇(Δu-a(u)Δu-a′(u)2|∇u|2+f(u))|2dx≤C∫Ω‍(u2|∇u|2|Δu|2+u4|∇Δu|2+|∇Δu|2+|∇u|6+u10|∇u|2)dx+C≤C∫Ω‍((supx∈Ω|u|2)|∇u|2|Δu|2+(supx∈Ω|u|4)|∇Δu|2=+|∇Δu|2+|∇u|6+(supx∈Ω|u|10)|∇u|2)dx+C≤C∫Ω‍((supx∈Ω|u|2)|∇u|4+(supx∈Ω|u|2)|Δu|4+(supx∈Ω|u|4)|∇Δu|2=+|∇Δu|2+|∇u|6+(supx∈Ω|u|10)|∇u|2)dx+C≤∥u∥Hα2∥u∥Hα4+∥u∥Hα2∥u∥Hα2+∥u∥Hα4∥u∥Hα2+∥u∥Hα2+∥u∥Hα6+∥u∥Hα10∥u∥Hα4+C.
It implies that
(4.34)g:Hα→H1/6is  bounded  for12≤α<23.
Therefore, it follows from (4.26) and (4.33) that
(4.35)∥g(u(t,u0))∥H1/6≤C,∀t≥0,u0∈U,12≤α<23.
Thus, by the same method as that in Step 1, we get from (4.35) that
(4.36)∥u(t,u0)∥Hα=∥etL+∫0t‍e(t-τ)LG(u)dτ∥Hα≤∥u0∥Hα+∫0t‍∥Lαe(t-τ)LG(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥L(α+(1/3)-(1/6))e(t-τ)L∥∥∇g(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥L(α+(1/3)-(1/6))e(t-τ)L∥∥g(u)∥H(1/6)dτ≤∥u0∥Hα+C∫0t‍τ-βe-στdτ≤C,∀t≥0,u0∈U⊂Hα,
where β=α+1/3-1/6(0<β<1).
Step3. We prove that for any bounded set U⊂Hα(5/6≤α<1), there is a constant C>0 such that
(4.37)∥u(t,u0)∥Hα≤C,∀t≥0,u0∈U⊂Hα,α<1.
In fact, by the embedding theorems, we have(4.38)H4(Ω)↪W1,8(Ω),H4(Ω)↪W2,4(Ω),H4(Ω)↪W1,4(Ω),H4(Ω)↪W3,4(Ω),H4(Ω)↪W2,2(Ω),Hα(Ω)↪C0(Ω)∩H4(Ω),α≥23.
We deduce
(4.39)∥g(u)∥H1/3=∫Ω‍|Δ(Δu-a(u)Δu-a′(u)2|∇u|2+f(u))|2dx≤C∫Ω‍||∇u|2Δu+u|Δu|2+u∇u∇Δu+u2Δ2u+Δ2u+u4|∇u|2+u5Δu|2dx+C≤C∫Ω‍(|∇u|4|Δu|2+(supx∈Ω|u|2)|Δu|4≤+(supx∈Ω|u|2)(|∇u|2)|∇Δu|2dx+(supx∈Ω|u|4)|Δ2u|2+|Δ2u|2≤+(supx∈Ω|u|8)|∇u|4+(supx∈Ω|u|10)|Δu|2)dx+C≤∥u∥W1,88+∥u∥W2,44+∥u∥Hα2∥u∥W2,44+∥u∥Hα2∥u∥W1,44+∥u∥Hα2∥u∥W3,42≤+∥u∥Hα4∥u∥W4,22+∥u∥W4,22+∥u∥Hα8∥u∥W1,44+∥u∥Hα10∥u∥W2,22+C≤∥u∥Hα8+∥u∥Hα4+∥u∥Hα2∥u∥Hα4+∥u∥Hα2∥u∥Hα4+∥u∥Hα2∥u∥Hα2≤+∥u∥Hα4∥u∥Hα2+∥u∥Hα2+∥u∥Hα8∥u∥Hα4+∥u∥Hα10∥u∥Hα2+C.
It implies that(4.40)g:Hα→H1/3is  bounded  for23≤α<56.
Therefore, it follows from (4.31) and (4.40) that
(4.41)∥g(u(t,u0))∥H1/3≤C,∀t≥0,u0∈U,23≤α<56.
Thus, by the same method as those in Steps 1 and 2, we get from (4.41) that
(4.42)∥u(t,u0)∥Hα=∥etL+∫0t‍e(t-τ)LG(u)dτ∥Hα≤∥u0∥Hα+∫0t‍∥Lαe(t-τ)LG(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥Lαe(t-τ)L∥∥Δg(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥Lαe(t-τ)L∥∥g(u)∥H1/3dτ≤∥u0∥Hα+C∫0t‍τ-βe-στdτ≤C,∀t≥0,u0∈U⊂Hα,
where β=α(0<β<1).
Step4. We prove that for any bounded set U⊂Hα(1≤α<(7/6)), there is a constant C>0 such that
(4.43)∥u(t,u0)∥Hα≤C,∀t≥0,u0∈U⊂Hα,1≤α<76.
In fact, by the embedding theorems, we have(4.44)H5(Ω)↪W1,4(Ω),H5(Ω)↪W2,8(Ω),H5(Ω)↪W1,8(Ω),H5(Ω)↪W3,4(Ω),H5(Ω)↪W2,4(Ω),H5(Ω)↪W4,4(Ω),H5(Ω)↪W1,6(Ω),H5(Ω)↪W3,2(Ω),Hα(Ω)↪C0(Ω)∩H5(Ω),α≥56.
Hence, similar to the above, we have
(4.45)∥g(u)∥H1/2=∫Ω‍|∇Δ(Δu-a(u)Δu-a′(u)2|∇u|2+f(u))|2dx≤C∫Ω‍(∇u|Δu|2+|∇u|2∇Δu+uΔu∇Δu+u∇uΔ2u+u2∇Δ2u+∇Δ2u+u3|∇u|2∇u+u4∇uΔu+u5∇Δu)2dx+C≤∥u∥Hα4+∥u∥Hα8+∥u∥Hα8+∥u∥Hα4+∥u∥Hα2∥u∥Hα4+∥u∥Hα2∥u∥Hα4=+∥u∥Hα2∥u∥Hα4+∥u∥Hα2∥u∥Hα4+∥u∥Hα4∥u∥Hα2+∥u∥Hα2+∥u∥Hα6∥u∥Hα6=+∥u∥Hα8∥u∥Hα4+∥u∥Hα8∥u∥Hα4+∥u∥Hα10∥u∥Hα4+C.
It implies that
(4.46)g:Hα→H1/2is  bounded  for56≤α<1.
Therefore, it follows from (4.37) and (4.46) that
(4.47)∥g(u(t,u0))∥H1/2≤C,∀t≥0,u0∈U,56≤α<1.
Thus, by the same method as that in Steps 1, 2, and 3, we get from (4.47) that
(4.48)∥u(t,u0)∥Hα=∥etL+∫0t‍e(t-τ)LG(u)dτ∥Hα≤∥u0∥Hα+∫0t‍∥Lαe(t-τ)LG(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥L(α+1/3-1/2)e(t-τ)L∥∥∇Δg(u)∥Hdτ≤∥u0∥Hα+∫0t‍∥L(α+1/3-1/2)e(t-τ)L∥∥g(u)∥H1/2dτ≤∥u0∥Hα+C∫0t‍τ-βe-στdτ≤C,∀t≥0,u0∈U⊂Hα,
where β=α+1/3-1/2(0<β<1).
In the same fashion as in the proof of (4.43), by iteration we can prove that for any bounded set U⊂Hα(α>0), there is a constant C>0 such that
(4.49)∥u∥Hα≤C,∀t≥0,u0∈U⊂Hα,α≥0.
That is, for all α≥0 the semigroup S(t) generated by the problem (1.1)–(1.3) is uniformly compact in Hα.
Secondly, we are going to show that for anyα≥0, the problem (1.1)–(1.3) has a bounded absorbing set in Hα; that is, for any bounded set U⊂Hα(α≥0) there are T>0 and a constant C>0 independent of u0, such that
(4.50)∥u(t,u0)∥Hα≤C,∀t≥T,u0∈U⊂Hα.
Forα=1/2, this follows from Theorem 4.1. Now, we will prove (4.13) for any α≥1/2 in the following steps.
Step1. We will prove that for any 1/2≤α<4/3, the problem (1.1)–(1.3) has a bounded absorbing set in Hα.
By (4.23), we have
(4.51)u(t,u0)=e(t-T)Lu(T,u0)+∫Tt‍e(t-τ)Lg(u)dτ.
Let D⊂H1/2 be the bounded absorbing set of the problem (1.1)–(1.3) in H1/2, and T0>0 is the time such that
(4.52)u(t,u0)∈D,∀t>T0,u0∈U⊂Hα,α≥12.
On the other hand, it is known that(4.53)∥etL∥≤Ce-λ13t,
where λ1>0 is the first eigenvalue of
(4.54)-Δu=λu,∂u∂n|∂Ω=0,∫Ωudx=0.
For any given T>0 and u0∈U⊂Hα(α≥1/2), we have
(4.55)∥e(t-T)Lu(T,u0)∥Hα→0,ast→∞.
By assertion (3) of Lemma 4.5, it follows from (4.51) and (4.29) that for any 1/2≤α<2/3, we have
(4.56)∥u(t,u0)∥Hα=∥e(t-T0)Lu(T0,u0)∥Hα+∫T0t‍∥Lαe(t-τ)L∥·∥Δg(x,u)∥Hdτ=∥e(t-T0)Lu(T0,u0)∥Hα+∫T0t‍∥Lα+1/3e(t-τ)L∥·∥g(x,u)∥Hdτ≤∥e(t-T0)Lu(T0,u0)∥Hα+∫T0t‍∥Lα+1/3e(t-τ)L∥·∥g(x,u)∥Hdτ≤∥e(t-T0)Lu(T0,u0)∥Hα+C∫T0t‍∥Lα+1/3e(t-τ)L∥dτ≤∥e(t-T0)Lu(T0,u0)∥Hα+C∫0t-T0‍τ-(α+1/3)e-στdτ≤∥e(t-T0)Lu(T0,u0)∥Hα+C,
where C>0 is a constant independent of u0. Then, we infer from (4.55) and (4.56) that (4.50) holds for all 1/2≤α<2/3.
Step2. We will show that for any 2/3≤α<5/6, the problem (1.1)–(1.3) has a bounded absorbing set in Hα.
By (4.33) and (4.51), we deduce that
(4.57)∥u(t,u0)∥Hα=∥e(t-T0)Lu(T0,u0)∥Hα+∫T0t‍∥Lαe(t-τ)L∥·∥Δg(x,u)∥Hdτ≤∥e(t-T0)Lu(T0,u0)∥Hα+∫T0t‍∥Lα+1/3-1/6e(t-τ)L∥·∥g(x,u)∥H1/6dτ≤∥e(t-T0)Lu(T0,u0)∥Hα+C∫T0t‍∥Lα+1/3-1/6e(t-τ)L∥dτ≤∥e(t-T0)Lu(T0,u0)∥Hα+C∫0t-T0‍τ-(α+1/3-1/6)e-στdτ≤∥e(t-T0)Lu(T0,u0)∥Hα+C,
where C>0 is a constant independent of u0. Thus, we verify from (4.55) and (4.57) that (4.50) is true for all 2/3≤α<5/6.
By iteration, we can obtain (4.50) for all α≥0. Hence, (1.1)–(1.3) has a bounded absorbing set in Hα for all α≥0.
Finally, this theorem follows from (4.24), (4.50), and Lemma 4.4. The proof is completed.Remark 4.7.
The attractorsAα⊂Hα in Theorem 4.6 are the same for all α≥0, that is, Aα=A, for all  α≥0. Hence, A⊂C∞(Ω). Theorem 4.6 implies that for any u0∈H, the solution u(t,u0) of the problem (1.1)–(1.3) satisfies that
(4.58)limt→∞infv∈A∥u(t,u0)-v∥Ck=0,∀k≥1.



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*Source: 414590-2012-10-21.xml*