On the Cauchy problem of a sixth-order Cahn–Hilliard equation arising in oil-water-surfactant mixtures

Abstract

We study the well-posedness and asymptotic behavior of solutions to the Cauchy problem of a three-dimensional sixth-order Cahn–Hilliard equation arising in oil-water-surfactant mixtures. First, by using the pure energy method and a standard continuity argument, we prove that there exists a unique global strong solution provided that the $H^{2}$ -norm of the initial data is sufficiently small. Moreover, we establish suitable negative Sobolev norm estimates and obtain the optimal decay rates of the higher-order spatial derivatives of the strong solution.

Keywords

Sixth-order Cahn–Hilliard equation global well-posedness asymptotic behavior decay estimates

1. Introduction

In 1990s, Gompper et. al. [7,8] introduced the following free energy functional: $\begin{matrix} (1.1) & F {u} = \int G (u, \nabla u, Δ u) d x, \end{matrix}$ with density given by $\begin{matrix} G (u, \nabla u, Δ u) = f (u) + \frac{1}{2} a (u) | \nabla u |^{2} + \frac{δ}{2} | Δ u |^{2}, \end{matrix}$ to describe the dynamics of phase transitions in ternary oil-water-surfactant systems, where $u (x, t)$ describes a scalar parameter that is proportional to the local difference between oil and water concentrations; δ denotes the mobility and the second gradient energy coefficient; $a (u)$ is the first gradient energy coefficient, which may be of arbitrary sign; and $f (u)$ denotes the multiwell volumetric free energy density [18,19,24].

Let M be the mobility, and μ the chemical potential difference between the oil and water phases. Applying mass conservation, we obtain $\begin{matrix} \frac{\partial u}{\partial t} = - div j, \end{matrix}$ with the mass flux j given by $\begin{matrix} j = - M \nabla μ . \end{matrix}$ Moreover, the chemical potential can be defined by the constitutive equation $\begin{matrix} μ = \frac{δ F {u}}{δ u}, \end{matrix}$ where $\frac{δ F {u}}{δ u}$ is the first variation of the function $F {u}$ . Let the mobility $M \equiv const$ , and we end up with the following sixth-order Cahn–Hilliard type equation: $\begin{matrix} (1.2) & \{\begin{matrix} u_{t} = M Δ μ, \\ μ = δ Δ^{2} u - a (u) Δ u - \frac{1}{2} a^{'} (u) | \nabla u |^{2} + f (u) . \end{matrix} \end{matrix}$

Many papers have studied the initial boundary value problem of equation (1.2) from the point of view of global well-posedness. Pawlow and Zajaczkowski [18] assumed that the considered space $Ω \subset R^{3}$ is a bounded domain with a boundary of class $C^{6}$ , and proved the existence of a unique global smooth solution which depends continuously on the initial data. Moreover, by using the Bäcklund transformation and the Leray–Schauder fixed point theorem, Pawlow and Zajaczkowski [20] proved the global unique solvability of equation (1.2) in the Sobolev space $H^{6, 1} (Ω \times (0, T))$ under the assumption that the initial datum is in $H^{3} (Ω)$ . Schimperna and Pawlow [24] discussed the existence, uniqueness, and parabolic regularization of a weak solution to the initial boundary value problem of equation (1.2) with a singular potential. The authors also supposed that the parameter $δ = 0$ , investigated a fourth-order system, and considered the existence of weak solutions under very general conditions by means of a fixed-point argument. In 2013, Schimperna and Pawlow [23] continued to study the fourth-order system with singular diffusion. The authors proved that for any final time T, the system admits a unique energy type weak solution, and for any $τ > 0$ , such a solution is classical. Later, based on Leray–Schauder’s fixed-point theorem and Campanato spaces, Liu and Wang [14] proved the existence of time-periodic solutions to the initial-boundary value problem of equation (1.2) in two spatial dimensions.

There are also some results on equation (1.2) with a viscous term. Pawlow and Zajaczkowski [19] applied Leray–Schauder’s fixed-point theorem and suitable estimates to establish the existence and uniqueness of a global in-time regular solution. Very recently, based on the estimates of weighted Sobolev spaces, Duan and Zhao [6] considered the existence of a global attractor for equation (1.2) with a viscous term in a 2D unbounded domain.

A Cauchy problem in mathematics asks for the solution of a partial differential equation that satisfies certain conditions that are given on a hypersurface in the domain. It is worth pointing out that the study of the Cauchy problem on higher-order nonlinear diffusion equations is also interesting. Many classical results are related to this topic (e.g., Caffarelli and Muler [1], Dlotko, Kania, and Sun [4], Cholewa and Rodriguez-Bernal [2] for the global existence of higher-order diffusion equations; Dlotko and Sun [5] and Savostianov and Zelik [22] for the global dynamics of higher-order diffusion equations). As far as we know, there is no reference concerning this aspect of the sixth-order Cahn–Hilliard equation arising in oil-water-surfactant mixtures. Can we establish some well-posedness results for the Cauchy problem of the sixth-order Cahn–Hilliard equation (1.2)? We will answer that question in this paper.

Remark 1.1.
Due to its important applications in everyday life, the model of oil-water-surfactant mixtures is increasingly used in many fields. The Cauchy problem of equation (1.2) may be used to describe the handling process of offshore oil spills, where the oil pollution region may be large and irregular. To handle a spill, we must draw support from the surfactant, and consider the relations between oil, water, and surfactant.

Consider the Cauchy problem of the sixth-order viscous Cahn–Hilliard equation (1.2) in $R^{3}$ . The problem is stated as follows: $\begin{matrix} (1.3) & \{\begin{matrix} u_{t} - δ Δ^{3} u = - Δ [a (u) Δ u + \frac{1}{2} a^{'} (u) | \nabla u |^{2} - f^{'} (u)], x \in R^{3}, \\ u (x, 0) = u_{0} (x) . \end{matrix} \end{matrix}$
Remark 1.2.
There are some different choices of the functions $f (s)$ and $a (s)$ . For example, on the basis of Landau–Ginzburg free energy, Gompper et al. [7,8] chose $f (s)$ and $a (s)$ as $\begin{matrix} (1.4) & f (s) = {(s + 1)}^{2} (s^{2} + h_{0}) {(s - 1)}^{2}, a (s) = g_{0} + g_{2} s^{2}, \end{matrix}$ where $h_{0}$ , $g_{0}$ , and $g_{2} > 0$ are constants. It is easy to see that the free energy is nonconvex, and it has at least two ground states, 1 and $- 1$ . Based on the fourth-order gradient free energy, Pawlow and Zajaczkowski [20] supposed that $\begin{matrix} (1.5) & f (s) = (1 - α) \frac{s^{2}}{2} + \frac{s^{4}}{4}, a (s) = - 2 . \end{matrix}$ A specific free energy with composition-dependent gradient energy coefficient $a (u)$ also will arise in the modeling of phase separation in polymers [3,19]. Known as Flory–Huggins–de Gennes energy, it has the form (1.1) with $δ = 0$ , $\begin{matrix} F (s) = \int_{0}^{s} f (τ) d τ = (1 - s) log (1 - s) + (1 + s) log (1 + s) - \frac{λ}{2} s^{2}, λ ⩾ 0, \end{matrix}$ and the singular coefficient $\begin{matrix} a (s) = \frac{1}{(1 - s) (1 + s)} . \end{matrix}$
Remark 1.3.
The sixth-order Cahn–Hilliard equation model is only a phenomenological model. Hence, various modifications, e.g., [12,13,15,16] and references therein, have been proposed to better capture the dynamical picture of the phase transition phenomena.

In this paper, we suppose that $f (s)$ and $a (s)$ satisfy (1.4). There are two cases to be considered: First, if the parameter $g_{0} > 0$ , then equation (1.3)₁ can be rewritten as $\begin{matrix} (1.6) & u_{t} - δ Δ^{3} u + g_{0} Δ^{2} u = - Δ [g_{2} u^{2} Δ u + g_{2} u | \nabla u |^{2} - f^{'} (u)] . \end{matrix}$ Second, if $g_{0} ⩽ 0$ , then we would like to rewrite problem (1.3) as $\begin{array}{l} u_{t} - δ Δ^{3} u + (1 - g_{0}) Δ^{2} u \\ (1.7) & = - Δ [g_{2} (u + \sqrt{\frac{1 - 2 g_{0}}{g_{2}}}) (u - \sqrt{\frac{1 - 2 g_{0}}{g_{2}}}) Δ u + g_{2} u | \nabla u |^{2} - f^{'} (u)] . \end{array}$ For simplicity, we consider the following Cauchy problem in $R^{3}$ : $\begin{matrix} (1.8) & \{\begin{matrix} u_{t} - δ Δ^{3} u + κ_{0} Δ^{2} u = - Δ [κ_{1} (u + κ_{2}) (u - κ_{2}) Δ u + κ_{1} u | \nabla u |^{2} - f^{'} (u)], \\ u (x, 0) = u_{0} (x), \end{matrix} \end{matrix}$ where $δ > 0$ , $κ_{0} > 0$ , $κ_{1} > 0$ , $κ_{2} ⩾ 0$ , and $h_{0} \in R$ are constants. If $κ_{0} = g_{0} > 0$ , $κ_{1} = g_{2}$ , and $κ_{2} = 0$ , then equation (1.8)₁ is equivalent to (1.6). Moreover, if $κ_{0} = 1 - g_{0} > 0$ , $κ_{1} = g_{2}$ , and $κ_{2} = \sqrt{\frac{1 - 2 g_{0}}{g_{2}}}$ , then we obtain equation (1.7).
Remark 1.4.
Since we consider problem (1.8) in $R^{3}$ , the Laplacian ${(- Δ)}^{δ}$ ( $δ \in R$ ) can be defined through the Fourier transform, i.e., $\begin{matrix} (1.9) & {(- Δ)}^{δ} f (x) = Λ^{2 δ} f (x) = \int_{R^{3}} | x |^{2 δ} \hat{f} (ξ) e^{2 π i x \cdot ξ} d ξ, \end{matrix}$ where $\hat{f}$ is the Fourier transform of f. Moreover, $\nabla^{l}$ with the integer $l ⩾ 0$ stands for the usual spatial derivatives of order l. If $l < 0$ or l is not a positive integer, then $\nabla^{l}$ stands for $Λ^{l}$ , as defined by (1.9). We use the notation $A ≲ B$ to mean that $A ⩽ c B$ for a universal constant $c > 0$ that only depends on the parameters coming from the problem, and the indices N and s come from the regularity of the data. We also employ C as a positive constant, depending again on the initial data.

The first purpose of this paper is to consider the global well-posedness of solutions to problem (1.8). Let $ω_{0} = \pm 1$ . Then, we have the following theorem.
Theorem 1.5.
Let $u_{0} - ω_{0} \in H^{2} (R^{3})$ . Suppose that $\begin{matrix} (1.10) & ‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} < ε_{0} \equiv \frac{1}{2} \sqrt{1 + \frac{4 min {δ, κ_{0}}}{C}} - \frac{1}{2}, \end{matrix}$ where $C$ is a positive constant to be determined in ( 3.28 ). Then there exists a unique global solution $u (x, t)$ such that for all $t ⩾ 0$ , $\begin{matrix} (1.11) & {‖ u (t) - ω_{0} ‖}_{H^{2}}^{2} + \int_{0}^{t} ({‖ \nabla Δ u (s) ‖}_{H^{2}}^{2} + {‖ Δ u (s) ‖}_{H^{2}}^{2}) d s ⩽ C ‖ u_{0} ‖_{H^{2}}^{2} . \end{matrix}$ Moreover, if $u_{0} \in H^{N} (R^{3})$ for any integer $N ⩾ 3$ , then $\begin{matrix} (1.12) & {‖ u (t) - ω_{0} ‖}_{H^{N}}^{2} + \int_{0}^{t} ({‖ \nabla Δ u (s) ‖}_{H^{N}}^{2} + {‖ Δ u (s) ‖}_{H^{N}}^{2}) d s ⩽ C ‖ u_{0} ‖_{H^{N}}^{2} . \end{matrix}$

The asymptotic behavior of solutions is also an interesting topic in the study of the Cauchy problem of dissipative equations. In this paper, we show that the solutions of problem (1.8) satisfy some negative exponent decay rate.
Theorem 1.6.
Under the assumptions of Theorem 1.5 , and assuming that $u_{0} - ω_{0} \in {\dot{H}}^{- s} (R^{3})$ for some $s \in [0, \frac{1}{2}]$ , then for all $t ⩾ 0$ , $\begin{matrix} (1.13) & {‖ Λ^{- s} [u (t) - ω_{0}] ‖}_{L^{2}} ⩽ C, \end{matrix}$ and $\begin{matrix} (1.14) & {‖ \nabla^{l} [u (t) - ω_{0}] ‖}_{H^{N - l}} ⩽ C {(1 + t)}^{- \frac{l + s}{4}}, for l = 0, 1, \dots, N - 1 . \end{matrix}$

Note that the Hardy–Littlewood–Sobolev theorem implies that for $p \in [\frac{3}{2}, 2]$ , $L^{p} (R^{3}) \subset {\dot{H}}^{- s} (R^{3})$ with $s = 3 (\frac{1}{p} - \frac{1}{2}) \in [0, \frac{1}{2}]$ . Therefore, based on Theorem 1.5, we obtain the following corollary on the optimal decay estimates.
Corollary 1.7.
Under the assumptions of Theorem 1.5 , if we replace the ${\dot{H}}^{- s} (R^{3})$ assumption by $u_{0} \in L^{p} (R^{3}) (\frac{3}{2} ⩽ p ⩽ 2)$ , then the following decay estimate holds: $\begin{matrix} (1.15) & {‖ \nabla^{l} [u (t) - ω_{0}] ‖}_{H^{N - l}} ⩽ C {(1 + t)}^{- σ_{l}}, for l = 0, 1, \dots, N - 1, \end{matrix}$ where $\begin{matrix} σ_{l} = \frac{3}{4} (\frac{1}{p} - \frac{1}{2}) + \frac{l}{4} . \end{matrix}$
Remark 1.8.
The temporary decay rate of problem (1.8) is optimal because it is equivalent to the decay rate of the fourth-order generalized heat equation, $\begin{matrix} (1.16) & \{\begin{matrix} u_{t} + Δ^{2} u = 0, x \in R^{3}, t ⩾ 0, \\ u (x, 0) = u_{0} (x) . \end{matrix} \end{matrix}$

The main difficulties regarding the Cauchy problem of the sixth-order Cahn–Hilliard equation arising in oil-water-surfactant mixtures in $R^{3}$ are dealing with the second order nonlinear term $Δ f (u) = Δ {(u - 1)}^{2} {(u + 1)}^{2} (u^{2} + h_{0})$ and obtaining a negative Sobolev estimate to study the decay rate of solutions, since the principle part of problem (1.8) is a sixth-order linear term, and the nonlinear term $Δ f (u)$ is only second-order. Due to Sobolev’s embedding $L^{\frac{6}{3 - 2 s}} (R^{3}) \subset {\dot{H}}^{s} (R^{3})$ , we cannot control $‖ \nabla^{k} f (u) ‖_{L^{p}}$ through $‖ \nabla^{k + 3} u ‖_{L^{2}}$ . To overcome this difficulty, we borrow a fourth-order term from $Δ (a (u) Δ u)$ , and rewrite (1.3)₁ as (1.8). Moreover, to consider the temporal decay rate of solutions of dissipative equations, one of the main tools is the standard Fourier splitting method. In this paper, since there exists the lower-order linear term on the right side of problem (1.8), it is difficult to use the Fourier splitting method to study the decay rate of solutions. By using the pure energy method [10,26,27], i.e., using a family of scaled energy estimates with minimum derivative counts and interpolations among them, we overcome the difficulty caused by the lower-order linear term, obtain suitable a priori estimates in the Sobolev space $H^{N}$ and negative Sobolev space ${\dot{H}}^{- s}$ ( $0 ⩽ s ⩽ \frac{1}{2}$ ), and establish the optimal decay rate of problem (1.8) in $R^{3}$ .

The structure of this paper is organized as follows. In Section 2, we introduce some preliminary results, which are useful to prove our main results. Section 3 is devoted to prove the small data global well-posedness of problem (1.8). In the last section, we establish the time decay rate of solutions.
2. Preliminaries

First of all, we introduce the Kato–Ponce inequality which is of great importance in this paper.

Lemma 2.1 ([11]).

Let $1 < p < \infty$ , $s > 0$ . There exists a positive constant C such that $\begin{matrix} (2.1) & {‖ Λ^{s} (f g) - f Λ^{s} g ‖}_{L^{p}} ⩽ C (‖ \nabla f ‖_{L^{p_{1}}} {‖ Λ^{s - 1} g ‖}_{L^{p_{2}}} + {‖ Λ^{s} f ‖}_{L^{q_{1}}} ‖ g ‖_{L^{q_{2}}}), \end{matrix}$ and $\begin{matrix} (2.2) & {‖ Λ^{s} f g ‖}_{L^{p}} ⩽ C (‖ f ‖_{L^{p_{1}}} {‖ Λ^{s} g ‖}_{L^{p_{2}}} + {‖ Λ^{s} f ‖}_{L^{q_{1}}} ‖ g ‖_{L^{q_{2}}}, \end{matrix}$ where $p_{2}, q_{2} \in (1, \infty)$ satisfying $\frac{1}{p} = \frac{1}{p_{1}} + \frac{1}{p_{2}} = \frac{1}{q_{1}} + \frac{1}{q_{2}}$ .

The following Gagliardo–Nirenberg inequality was proved in [17, p. 125, Theorem].

Lemma 2.2 ([17]).

Let $0 ⩽ m, α ⩽ l$ , then we have $\begin{matrix} (2.3) & {‖ \nabla^{α} f ‖}_{L^{p} (R^{3})} ≲ {‖ \nabla^{m} f ‖}_{L^{q} (R^{3})}^{1 - θ} {‖ \nabla^{l} f ‖}_{L^{r} (R^{3})}^{θ}, \end{matrix}$ where $θ \in [0, 1]$ and α satisfies $\begin{matrix} (2.4) & \frac{α}{3} - \frac{1}{p} = (\frac{m}{3} - \frac{1}{q}) (1 - θ) + (\frac{l}{3} - \frac{1}{r}) θ . \end{matrix}$ Here, when $p = \infty$ , we require that $0 < θ < 1$ .

There’s a Sobolev embedding for the homogeneous space ${\dot{H}}^{s}$ .

Lemma 2.3 ([21]).

There exists a constant c such that for $0 ⩽ s < \frac{3}{2}$ , $\begin{matrix} (2.5) & ‖ u ‖_{L^{\frac{6}{3 - 2 s}}} ⩽ c ‖ u ‖_{{\dot{H}}^{s}} for all u \in {\dot{H}}^{s} (R^{3}) . \end{matrix}$

We also introduce the Hardy–Littlewood–Sobolev theorem, which implies the following $L^{p}$ type inequality.

Lemma 2.4 ([9,25]).

Let $0 ⩽ s < \frac{3}{2}$ , $1 < p ⩽ 2$ and $\frac{1}{2} + \frac{s}{3} = \frac{1}{p}$ , then $\begin{matrix} (2.6) & ‖ f ‖_{{\dot{H}}^{- s} (R^{3})} ≲ ‖ f ‖_{L^{p} (R^{3})} . \end{matrix}$

The following special Sobolev interpolation lemma will be used in the proof of Theorem 1.5.

Lemma 2.5 ([10,27]).

Let $s, k ⩾ 0$ and $l ⩾ 0$ , then $\begin{matrix} (2.7) & {‖ \nabla^{l} f ‖}_{L^{2} (R^{3})} ⩽ {‖ \nabla^{l + k} f ‖}_{L^{2} (R^{3})}^{1 - θ} ‖ f ‖_{{\dot{H}}^{- s} (R^{3})}^{θ}, with θ = \frac{2}{l + k + s} . \end{matrix}$

We also give the following inequality on the $L^{\infty}$ -norm of $u (x, t)$ , which can be seen as a corollary of Lemma 2.5. In fact, if we set $α = 0$ , $p = \infty$ , $m = 1$ , $l = 2$ and $θ = \frac{1}{2}$ in Lemma 2.5, the following inequality is easily to proved.

Corollary 2.6.
If $u \in H^{2} (R^{3})$ , then u is (almost everywhere equal to) a continuous function and $\begin{matrix} (2.8) & ‖ u ‖_{L^{\infty} (R^{3})} ≲ ‖ \nabla u ‖_{L^{2} (R^{3})}^{\frac{1}{2}} ‖ Δ u ‖_{L^{2} (R^{3})}^{\frac{1}{2}} . \end{matrix}$

3. Proof of Theorem 1.5

3.1. Energy estimates

The purpose of this subsection is to establish the a priori nonlinear energy estimates for problem (1.8).

We begin with the energy estimates including $u - ω_{0}$ itself.

Lemma 3.1.
Suppose that all assumptions in Theorem 1.5 hold. Then for $k = 0, 1, \dots, N$ , we have $\begin{array}{l} \frac{d}{d t} \int_{R^{3}} {| \nabla^{k} (u - ω_{0}) |}^{2} d x + ({‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2}) \\ (3.1) & ⩽ C {‖ (u - ω_{0}) ‖}_{H^{2}}^{2} ({‖ (u - ω_{0}) ‖}_{H^{2}}^{2} + 1) ({‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2}) . \end{array}$
Proof.
Applying $\nabla^{k}$ to (1.8)₁, multiplying the resulting identity by $\nabla^{k} (u - ω_{0})$ , and then integrating over $R^{3}$ by parts, we arrive at $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ \nabla^{k} (u - ω_{0}) ‖}_{L^{2}}^{2} + δ {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + κ_{0} {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2} \\ = - κ_{1} \int_{R^{3}} \nabla^{k} {Δ [(u + κ_{2}) (u - κ_{2}) Δ u]} \cdot \nabla^{k} (u - ω_{0}) d x \\ - κ_{1} \int_{R^{3}} \nabla^{k} [Δ (u | \nabla u |^{2})] \cdot \nabla^{k} (u - ω_{0}) d x \\ + \int_{R^{3}} \nabla^{k} Δ f^{'} (u) \cdot \nabla^{k} (u - ω_{0}) d x \\ (3.2) & = I_{1} + I_{2} + I_{3} . \end{array}$ We can estimate the first term of the right hand side of (3.2) as $\begin{array}{l} I_{1} = & - κ_{1} \int_{R^{3}} \nabla^{k} {Δ [(u + κ_{2}) (u - κ_{2}) Δ u]} \cdot \nabla^{k} (u - ω_{0}) d x \\ = & - κ_{1} \int_{R^{3}} \nabla^{k} [(u + κ_{2}) (u - κ_{2}) Δ u] \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ = & - κ_{1} \sum_{0 ⩽ l ⩽ k} \sum_{0 ⩽ m ⩽ l} C_{k}^{l} C_{l}^{m} \int_{R^{3}} \nabla^{m} (u + κ_{2}) \cdot \nabla^{l - m} (u - κ_{2}) \cdot \nabla^{k - l} Δ u \cdot \nabla^{k} Δ u d x \\ ≲ & {‖ \nabla^{m} (u + κ_{2}) ‖}_{L^{6}} {‖ \nabla^{l - m} (u - κ_{2}) ‖}_{L^{6}} {‖ \nabla^{k - l + 2} u ‖}_{L^{6}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ ≲ & {‖ \nabla^{m + 1} u ‖}_{L^{2}} {‖ \nabla^{l - m + 1} u ‖}_{L^{2}} {‖ \nabla^{k - l + 3} u ‖}_{L^{2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ ≲ & {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{\frac{m}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{1 - \frac{m}{k + 2}} {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{\frac{l - m}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{1 - \frac{l - m}{k + 2}} {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{1 - \frac{l}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{\frac{l}{k + 2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ (3.3) & ≲ & ‖ \nabla u ‖_{L^{2}}^{2} ({‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2}) . \end{array}$ For the second term of the right hand side of (3.2), we have $\begin{array}{l} I_{2} = & - κ_{1} \int_{R^{3}} \nabla^{k} [Δ (u | \nabla u |^{2})] \cdot \nabla^{k} (u - ω_{0}) d x \\ = & - κ_{1} \int_{R^{3}} \nabla^{k} (u | \nabla u |^{2}) \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ = & - κ_{1} \sum_{0 ⩽ l ⩽ k} \sum_{0 ⩽ m ⩽ l} C_{k}^{l} C_{l}^{m} \int_{R^{3}} \nabla^{m} u \cdot \nabla^{l - m} \nabla u \cdot \nabla^{k - l} \nabla u \cdot \nabla^{k} Δ u d x \\ ≲ & {‖ \nabla^{m} u ‖}_{L^{6}} {‖ \nabla^{l + 1 - m} u ‖}_{L^{6}} {‖ \nabla^{k + 1 - l} u ‖}_{L^{6}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ ≲ & {‖ \nabla^{m + 1} u ‖}_{L^{2}} {‖ \nabla^{l + 2 - m} u ‖}_{L^{2}} {‖ \nabla^{k + 2 - l} u ‖}_{L^{2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ ≲ & {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{\frac{m}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{1 - \frac{m}{k + 2}} {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{\frac{l + 1 - m}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{1 - \frac{l + 1 - m}{k + 2}} \\ \times {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{\frac{k + 1 - l}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{1 - \frac{k + 1 - l}{k + 2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ (3.4) & ≲ & ‖ \nabla u ‖_{L^{2}}^{2} ({‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2}) . \end{array}$ Moreover, the last term of the right hand side of (3.2) satisfies $\begin{array}{l} I_{3} = & \int_{R^{3}} \nabla^{k} f^{'} (u) \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ = & \int_{R^{3}} \nabla^{k} [2 (u + 1) (u^{2} + h_{0}) {(u - 1)}^{2}] \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ + \int_{R^{3}} \nabla^{k} [2 (u - 1) (u^{2} + h_{0}) {(u + 1)}^{2}] \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ + \int_{R^{3}} \nabla^{k} [2 u {(u + 1)}^{2} {(u - 1)}^{2}] \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ (3.5) & = & I_{31} + I_{32} + I_{33} . \end{array}$ Note that $\begin{array}{l} I_{31} = & \int_{R^{3}} \nabla^{k} [2 (u + 1) (u^{2} + h_{0}) {(u - 1)}^{2}] \cdot \nabla^{k} Δ (u - ω_{0}) d x \\ = & 2 \sum_{0 ⩽ l ⩽ k} \sum_{0 ⩽ m ⩽ l} C_{k}^{l} C_{l}^{m} \int_{R^{3}} \nabla^{m} (u + 1) \cdot \nabla^{l - m} u^{2} \cdot \nabla^{k - l} {(u - 1)}^{2} \cdot \nabla^{k} Δ u d x \\ + 2 h_{0} \sum_{0 ⩽ l ⩽ k} C_{k}^{l} \int_{R^{3}} \nabla^{l} (u + 1) \cdot \nabla^{k - l} {(u - 1)}^{2} \cdot \nabla^{k} Δ u d x \\ ≲ & {‖ \nabla^{m} (u + 1) ‖}_{L^{6}} {‖ \nabla^{l - m} u^{2} ‖}_{L^{6}} {‖ \nabla^{k - l} {(u - 1)}^{2} ‖}_{L^{6}} {‖ \nabla^{k} Δ u ‖}_{L^{2}} \\ + {‖ \nabla^{l} (u + 1) ‖}_{L^{6}} {‖ \nabla^{k - l} {(u - 1)}^{2} ‖}_{L^{3}} {‖ \nabla^{k} Δ u ‖}_{L^{2}} \\ ≲ & {‖ \nabla^{m} (u + 1) ‖}_{L^{6}} ‖ u ‖_{L^{\infty}} {‖ \nabla^{l - m} u ‖}_{L^{6}} ‖ u - 1 ‖_{L^{\infty}} {‖ \nabla^{k - l} (u - 1) ‖}_{L^{6}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ + {‖ \nabla^{l} (u + 1) ‖}_{L^{6}} ‖ u - 1 ‖_{L^{6}} {‖ \nabla^{k - l} u - 1 ‖}_{L^{6}} {‖ \nabla^{k} Δ u ‖}_{L^{2}} \\ ≲ & {(‖ \nabla u ‖_{L^{2}}^{\frac{1}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2}})}^{2} {‖ \nabla^{m + 1} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{l - m + 1} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{k - l + 1} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ + ‖ \nabla u ‖_{L^{2}} {‖ \nabla^{l + 1} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{k - l + 1} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{k} Δ u ‖}_{L^{2}} \\ ≲ & (‖ \nabla u ‖_{L^{2}}^{2} + {‖ \nabla^{2} u ‖}_{L^{2}}^{2}) \\ \times ({‖ \nabla^{k + 2} u ‖}_{L^{2}}^{\frac{m + 1}{k + 2}} {‖ (u - ω_{0}) ‖}_{L^{2}}^{1 - \frac{m + 1}{k + 2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{\frac{l - m + 1}{k + 2}} {‖ (u - ω_{0}) ‖}_{L^{2}}^{1 - \frac{l - m + 1}{k + 2}} \\ \times {‖ \nabla^{k + 3} u ‖}_{L^{2}}^{\frac{k - l}{k + 2}} ‖ \nabla u ‖_{L^{2}}^{1 - \frac{k - l}{k + 2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ + ‖ \nabla u ‖_{L^{2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{\frac{l + 1}{k + 2}} {‖ (u - ω_{0}) ‖}_{L^{2}}^{1 - \frac{l + 1}{k + 2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{\frac{k - l + 1}{k + 2}} {‖ (u - ω_{0}) ‖}_{L^{2}}^{1 - \frac{k - l + 1}{k + 2}} {‖ \nabla^{k + 2} u ‖}_{L^{2}} \\ (3.6) & ≲ & (‖ u - ω_{0} ‖_{H^{2}}^{4} + ‖ u - ω_{0} ‖_{H^{2}}^{2}) ({‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2}) . \end{array}$ Similarly, we also have $\begin{matrix} (3.7) & I_{32} + I_{33} ≲ (‖ u - ω_{0} ‖_{H^{2}}^{4} + ‖ u - ω_{0} ‖_{H^{2}}^{2}) ({‖ \nabla^{k + 3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{k + 2} u ‖}_{L^{2}}^{2}) . \end{matrix}$ Plugging the estimates (3.3)–(3.7) into (3.2), we then obtain (3.1) and complete the proof. □

3.2. Local well-posedness

In this subsection, we prove the local well-posedness of solution $u (t)$ in $H^{2}$ -norm.

We first construct the solution ${(u^{j})}_{j ⩾ 0}$ by solving iteratively the Cauchy problem: $\begin{matrix} (3.8) & \{\begin{matrix} \partial_{t} u^{j + 1} - δ Δ^{3} u^{j + 1} + κ_{0} Δ^{2} u^{j + 1} \\ = - Δ [κ_{1} (u^{j} + κ_{2}) (u^{j} - κ_{2}) Δ u^{j + 1} \\ + κ_{1} u^{j} \nabla u^{j} \cdot \nabla u^{j + 1}] + \nabla \cdot [f^{″} (u^{j}) \nabla u^{j + 1}], \\ u^{j + 1} |_{t = 0} = u_{0} (x), x \in R^{3}, \end{matrix} \end{matrix}$ for $j ⩾ 0$ , where $u^{0} - ω_{0} \equiv 0$ holds. One denote ${(u^{j} - ω_{0})}_{j ⩾ 0}$ in short hand by ${(A_{j})}_{j ⩾ 0}$ and denote $u_{0}$ as $A_{0}$ . In the following, we shall show that ${(A^{j})}_{j ⩾ 0}$ is a Cauchy sequence in Banach space $C ([0, T_{1}]; H^{2})$ with $T_{1} > 0$ suitable small. Then, by take the limit and a continuous argument, one propose to prove that $u (t)$ is a global solution to Cauchy problem (1.8).

Lemma 3.2.
Suppose that all assumptions in Theorem 1.5 hold. There are constants $ε_{1} > 0$ and $T_{1} > 0$ such that if $‖ A_{0} ‖_{H^{2}} ⩽ ε_{1}$ , then for each $j ⩾ 0$ , $A^{j} \in C ([0, T_{1}]; H^{2})$ is well defined and $\begin{matrix} (3.9) & sup_{0 ⩽ t ⩽ T_{1}} {‖ A^{j} (t) ‖}_{H^{2}} ⩽ ε_{1}, j ⩾ 0 . \end{matrix}$ Moreover, ${(A^{j})}_{j ⩾ 0}$ is a Cauchy sequence in Banach space $C ([0, T_{1}]; H^{2})$ , the corresponding limit function denoted by $A (t)$ belongs to $C ([0, T_{1}]; H^{2})$ with $\begin{matrix} (3.10) & sup_{0 ⩽ t ⩽ T_{1}} {‖ A (t) ‖}_{H^{2}} ⩽ ε_{1}, \end{matrix}$ and $A = u (t) - ω_{0}$ is a solution over $[0, T_{1}]$ to problem ( 1.8 ). Finally, for the Cauchy problem ( 1.8 ), there exists at most one solution $u (t) - ω_{0}$ in $C ([0, T_{1}]; H^{2})$ satisfying ( 3.10 ).
Proof.
The inequality(3.9) will be proved by induction. By using the assumption at the initial step, we get $A_{0} = 0$ , which means $j = 0$ holds. Next, suppose that (3.9) holds for $j ⩾ 0$ with $ε_{1} > 0$ small enough to be determined later, we are going to prove it also holds for $j + 1$ . Hence, we need some energy estimates on $A^{j + 1}$ . On the basis of (3.8)₁, we obtain for $k = 2$ and $0 ⩽ l ⩽ k$ , $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ \nabla^{l} (u^{j + 1} - ω_{0}) ‖}_{L^{2}}^{2} + δ {‖ \nabla^{l + 3} u^{j + 1} ‖}_{L^{2}}^{2} + κ {‖ \nabla^{l + 2} u^{j + 1} ‖}_{L^{2}}^{2} \\ = - κ_{1} \int_{R^{3}} \nabla^{l} [(u^{j} + κ_{2}) (u^{j} - κ_{2}) Δ u^{j + 1}] \cdot \nabla^{l} Δ (u^{j + 1} - ω_{0}) d x \\ - κ_{1} \int_{R^{3}} \nabla^{l} [u^{j} \nabla u^{j} \cdot \nabla u^{j + 1}] \cdot \nabla^{l} Δ (u^{j + 1} - ω_{0}) d x \\ + \int_{R^{3}} \nabla^{l} \nabla \cdot [f^{″} (u^{j}) \nabla u^{j + 1}] \cdot \nabla^{l} (u^{j + 1} - ω_{0}) d x \\ (3.11) & = J_{1} + J_{2} + J_{3} . \end{array}$ Since it is trivial for the case $l = 0$ , thus, we only need to consider $l = 1, 2$ . For the term $J_{1}$ , if $l = 1$ , we estimate as follows $\begin{array}{l} J_{1} = & - κ_{1} \int_{R^{3}} \nabla [(u^{j} + κ_{2}) (u^{j} - κ_{2}) Δ u^{j + 1}] \cdot \nabla Δ (u^{j + 1} - ω_{0}) d x \\ ≲ & {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}} {‖ \nabla [(u^{j} + κ_{2}) (u^{j} - κ_{2}) Δ u^{j + 1}] ‖}_{L^{2}} \\ ≲ & \sum_{s = 0}^{1} \sum_{m = 0}^{s} C_{1}^{s} C_{s}^{m} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}} {‖ \nabla^{m} (u^{j} + κ_{2}) ‖}_{L^{6}} {‖ \nabla^{s - m} (u^{j} - κ_{2}) ‖}_{L^{6}} {‖ \nabla^{1 - s + 2} u^{j + 1} ‖}_{L^{6}} \\ ≲ & \sum_{s = 0}^{1} \sum_{m = 0}^{s} C_{1}^{s} C_{s}^{m} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}} {‖ \nabla^{m + 1} u^{j} ‖}_{L^{2}} {‖ \nabla^{s - m + 1} u^{j} ‖}_{L^{2}} {‖ \nabla^{1 - s + 3} u^{j + 1} ‖}_{L^{2}} \\ (3.12) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ For the case $l = 2$ , we can estimate $J_{1}$ as $\begin{array}{l} J_{1} ≲ & {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{2} [(u^{j} + κ_{2}) (u^{j} - κ_{2}) Δ u^{j + 1}] ‖}_{L^{\frac{6}{5}}} \\ ≲ & {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} {‖ \sum_{s = 0}^{2} \sum_{m = 0}^{s} \nabla^{m} (u^{j} + κ_{2}) \cdot \nabla^{s - m} (u^{j} - κ_{2}) \cdot \nabla^{4 - s} u^{j + 1} ‖}_{L^{\frac{6}{5}}} \\ ≲ & {‖ \nabla^{5} u^{j + 1} ‖}_{L^{2}} (\underset{s = m = 0}{\underset{︸}{‖ u^{j} + κ_{2}) ‖_{L^{6}} {‖ u^{j} - κ_{2} ‖}_{L^{6}} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{2}}}} \\ + \underset{s = 1, m = 0}{\underset{︸}{{‖ u^{j} + κ_{2} ‖}_{L^{6}} {‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}}}} + \underset{s = m = 1}{\underset{︸}{{‖ \nabla u^{j} ‖}_{L^{6}} {‖ u^{j} - κ_{2} ‖}_{L^{6}} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}}}} \\ + \underset{s = 2, m = 0}{\underset{︸}{{‖ u^{j} + κ_{2} ‖}_{L^{6}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{6}}}} + \underset{s = 2, m = 1}{\underset{︸}{{‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{6}}}} \\ + \underset{s = 2, m = 2}{\underset{︸}{{‖ \nabla^{2} (u^{j} + κ_{2}) ‖}_{L^{2}} {‖ u^{j} - κ_{2} ‖}_{L^{6}} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{6}}}}) \\ (3.13) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ For the term $J_{2}$ , if $l = 1$ , we have $\begin{array}{l} J_{2} = & - κ_{1} \int_{R^{3}} \nabla [u^{j} \nabla u^{j} \cdot \nabla u^{j + 1}] \cdot \nabla Δ u^{j} d x \\ ≲ & {‖ \nabla^{3} u^{j + 1} ‖}_{L^{6}} {‖ \nabla [u^{j} \nabla u^{j} \cdot \nabla u^{j + 1}] ‖}_{L^{\frac{6}{5}}} \\ ≲ & \sum_{0 ⩽ s ⩽ 1} \sum_{0 ⩽ m ⩽ s} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{m} u^{j} \cdot \nabla^{s - m} \nabla u^{j} \cdot \nabla^{1 - s} \nabla u^{j + 1} ‖}_{L^{\frac{6}{5}}} \\ ≲ & \sum_{0 ⩽ s ⩽ 1} \sum_{0 ⩽ m ⩽ s} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{m} u^{j} ‖}_{L^{6}} {‖ \nabla^{s + 1 - m} u^{j} ‖}_{L^{2}} {‖ \nabla^{1 - s + 1} u^{j + 1} ‖}_{L^{6}} \\ ≲ & \sum_{0 ⩽ s ⩽ 1} \sum_{0 ⩽ m ⩽ s} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{2}} {‖ \nabla^{m + 1} u^{j} ‖}_{L^{2}} {‖ \nabla^{s + 1 - m} u^{j} ‖}_{L^{2}} {‖ \nabla^{1 - s + 2} u^{j + 1} ‖}_{L^{2}} \\ (3.14) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ For the case $l = 2$ of $J_{2}$ , we estimate as $\begin{array}{l} J_{2} ≲ & \sum_{s = 0}^{1} \sum_{m = 0}^{s} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{m} u^{j} \cdot \nabla^{s - m} \nabla u^{j} \cdot \nabla^{2 - s} \nabla u^{j + 1}] ‖}_{L^{\frac{6}{5}}} \\ - \underset{s = 2}{\underset{︸}{κ_{1} \sum_{0 ⩽ m ⩽ 2} C_{2}^{m} \int_{R^{3}} \nabla^{m} u^{j} \cdot \nabla^{2 - m} \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x}} \\ (3.15) & = & J_{21} + J_{22} . \end{array}$ Note that $\begin{array}{l} J_{21} = & \sum_{s = 0}^{1} \sum_{m = 0}^{s} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{m} u^{j} \cdot \nabla^{s - m} \nabla u^{j} \cdot \nabla^{2 - s} \nabla u^{j + 1}] ‖}_{L^{\frac{6}{5}}} \\ ≲ & {‖ \nabla^{5} u^{j + 1} ‖}_{L^{2}} (\underset{s = m = 0}{\underset{︸}{{‖ u^{j} ‖}_{L^{6}} {‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}}}} + \underset{s = 1, m = 0}{\underset{︸}{{‖ u^{j} ‖}_{L^{6}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{6}}}} \\ + \underset{s = m = 1}{\underset{︸}{{‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{6}}}}) \\ (3.16) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}), \end{array}$ and $\begin{array}{l} J_{22} = & - κ_{1} \sum_{0 ⩽ m ⩽ 2} C_{2}^{m} \int_{R^{3}} \nabla^{m} u^{j} \cdot \nabla^{2 - m} \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x \\ = & - κ_{1} \int_{R^{3}} u^{j} \cdot \nabla^{2} \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x - 2 κ_{1} \int_{R^{3}} \nabla u^{j} \cdot \nabla \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x \\ - κ_{1} \int_{R^{3}} \nabla^{2} u^{j} \cdot \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x \\ = & κ_{1} \int_{R^{3}} \nabla u^{j} \cdot \nabla \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x + κ_{1} \int_{R^{3}} u^{j} \cdot \nabla \nabla u^{j} \cdot \nabla^{2} u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x \\ + κ_{1} \int_{R^{3}} u^{j} \cdot \nabla \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{3} Δ u^{j + 1} d x - 2 κ_{1} \int_{R^{3}} \nabla u^{j} \cdot \nabla \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x \\ - κ_{1} \int_{R^{3}} \nabla^{2} u^{j} \cdot \nabla u^{j} \cdot \nabla u^{j + 1} \cdot \nabla^{2} Δ u^{j + 1} d x \\ ≲ & {‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} + {‖ u^{j} ‖}_{L^{6}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} \\ + {‖ u^{j} ‖}_{L^{\infty}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla u^{j + 1} ‖}_{L^{\infty}} {‖ \nabla^{5} u^{j + 1} ‖}_{L^{2}} + {‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} \\ + {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla u^{j} ‖}_{L^{6}} {‖ \nabla u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{4} u^{j + 1} ‖}_{L^{6}} \\ (3.17) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ Combining (3.15)–(3.17) together, if $l = 2$ , we obtain $\begin{matrix} (3.18) & J_{2} ≲ {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{matrix}$ Next, for $J_{3}$ , if $l = 1$ , we have $\begin{array}{l} J_{3} = & \int_{R^{3}} \nabla \nabla \cdot [f^{″} (u^{j}) \nabla u^{j + 1}] \cdot \nabla (u^{j + 1} - ω_{0}) d x \\ = & \int_{R^{3}} \nabla^{2} \cdot {[2 ({(u^{j})}^{2} + h_{0}) {(u^{j} - 1)}^{2} + 4 u^{j} (u^{j} + 1) {(u^{j} - 1)}^{2} \\ + 4 (u^{j} - 1) (u^{j} + 1) ({(u^{j})}^{2} + h_{0}) \\ + 2 ({(u^{j})}^{2} + h_{0}) {(u^{j} + 1)}^{2} + 4 (u^{j} - 1) u^{j} {(u^{j} + 1)}^{2} + 4 (u^{j} - 1) (u^{j} + 1) ({(u^{j})}^{2} + h_{0}) \\ + 2 {(u^{j} + 1)}^{2} {(u^{j} - 1)}^{2} + 4 u^{j} (u^{j} + 1) {(u^{j} - 1)}^{2} \\ + 4 u^{j} (u^{j} - 1) {(u^{j} + 1)}^{2}] \nabla u^{j + 1}} \cdot \nabla u^{j + 1} d x \\ ≲ & ({‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} - 1 ‖}_{L^{6}}^{2} + {‖ u^{j} ‖}_{L^{6}} {‖ u^{j + 1} ‖}_{L^{6}} {‖ u^{j} - 1 ‖}_{L^{\infty}}^{2} \\ + {‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} - 1 ‖}_{L^{6}} {‖ u^{j} + 1 ‖}_{L^{6}} + {‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} + 1 ‖}_{L^{6}}^{2} \\ + {‖ u^{j} + 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} ‖}_{L^{6}} {‖ u^{j} - 1 ‖}_{L^{6}} + {‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} - 1 ‖}_{L^{6}} {‖ u^{j} + 1 ‖}_{L^{6}} \\ + {‖ u^{j} + 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} - 1 ‖}_{L^{6}}^{2} + {‖ u^{j} - 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} ‖}_{L^{6}} {‖ u^{j} + 1 ‖}_{L^{6}} \\ + {‖ u^{j} + 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} ‖}_{L^{6}} {‖ u^{j} - 1 ‖}_{L^{6}}) {‖ \nabla u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}} \\ ≲ & {‖ \nabla u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla u^{j} ‖}_{L^{2}}^{2} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{2}} {‖ \nabla^{3} u^{j + 1} ‖}_{L^{2}} \\ (3.19) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{4} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ By Lemma 2.1, we consider the case $l = 2$ of $J_{3}$ : $\begin{array}{l} J_{3} = & \int_{R^{3}} \nabla^{2} \nabla \cdot [f^{″} (u^{j}) \nabla u^{j + 1}] \cdot \nabla^{2} u^{j + 1} d x \\ = & \int_{R^{3}} \nabla^{3} \cdot {[2 ({(u^{j})}^{2} + h_{0}) {(u^{j} - 1)}^{2} + 4 u^{j} (u^{j} + 1) {(u^{j} - 1)}^{2} \\ + 4 (u^{j} - 1) (u^{j} + 1) ({(u^{j})}^{2} + h_{0}) \\ + 2 ({(u^{j})}^{2} + h_{0}) {(u^{j} + 1)}^{2} + 4 (u^{j} - 1) u^{j} {(u^{j} + 1)}^{2} + 4 (u^{j} - 1) (u^{j} + 1) ({(u^{j})}^{2} + h_{0}) \\ + 2 {(u^{j} + 1)}^{2} {(u^{j} - 1)}^{2} + 4 u^{j} (u^{j} + 1) {(u^{j} - 1)}^{2} \\ + 4 u^{j} (u^{j} - 1) {(u^{j} + 1)}^{2}] \nabla u^{j + 1}} \cdot \nabla^{2} u^{j + 1} d x \\ ≲ & ({‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} - 1 ‖}_{L^{6}}^{2} + {‖ u^{j} ‖}_{L^{6}} {‖ u^{j + 1} ‖}_{L^{6}} {‖ u^{j} - 1 ‖}_{L^{\infty}}^{2} \\ + {‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} - 1 ‖}_{L^{6}} {‖ u^{j} + 1 ‖}_{L^{6}} + {‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} + 1 ‖}_{L^{6}}^{2} \\ + {‖ u^{j} + 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} ‖}_{L^{6}} {‖ u^{j} - 1 ‖}_{L^{6}} + {‖ {(u^{j})}^{2} + h_{0} ‖}_{L^{\infty}} {‖ u^{j} - 1 ‖}_{L^{6}} {‖ u^{j} + 1 ‖}_{L^{6}} \\ + {‖ u^{j} + 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} - 1 ‖}_{L^{6}}^{2} + {‖ u^{j} - 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} ‖}_{L^{6}} {‖ u^{j} + 1 ‖}_{L^{6}} \\ + {‖ u^{j} + 1 ‖}_{L^{\infty}}^{2} {‖ u^{j} ‖}_{L^{6}} {‖ u^{j} - 1 ‖}_{L^{6}}) {‖ \nabla u^{j + 1} ‖}_{L^{6}} {‖ \nabla^{5} u^{j + 1} ‖}_{L^{2}} \\ ≲ & {‖ \nabla u^{j} ‖}_{L^{2}} {‖ \nabla^{2} u^{j} ‖}_{L^{2}} {‖ \nabla u^{j} ‖}_{L^{2}}^{2} {‖ \nabla^{2} u^{j + 1} ‖}_{L^{2}} {‖ \nabla^{5} u^{j + 1} ‖}_{L^{2}} \\ (3.20) & ≲ & {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{4} ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ Now, summing up the estimates (3.11), (3.12), (3.13), (3.14), (3.18), (3.19) and (3.20), we arrive at $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ u^{j + 1} - ω_{0} ‖}_{H^{2}}^{2} + δ {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2} + κ {‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} \\ (3.21) & ⩽ C ({‖ u^{j} - ω_{0} ‖}_{H^{2}}^{2} + {‖ u^{j} - ω_{0} ‖}_{H^{2}}^{4}) ({‖ \nabla^{2} u^{j + 1} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} u^{j + 1} ‖}_{H^{2}}^{2}) . \end{array}$ By taking time integration, we have $\begin{array}{l} {‖ u^{j + 1} - ω_{0} ‖}_{H^{2}}^{2} + 2 \int_{0}^{t} (δ {‖ \nabla^{3} u^{j + 1} (s) ‖}_{H^{2}}^{2} + κ {‖ \nabla^{2} u^{j + 1} (s) ‖}_{H^{2}}^{2}) d s \\ ⩽ ‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} + C \int_{0}^{t} ({‖ u^{j} (s) - ω_{0} ‖}_{H^{2}}^{2} + {‖ u^{j} (s) - ω_{0} ‖}_{H^{2}}^{4}) \\ \times ({‖ \nabla^{3} u^{j + 1} (s) ‖}_{H^{2}}^{2} + {‖ \nabla^{2} u^{j + 1} (s) ‖}_{H^{2}}^{2}) d s, \end{array}$ which from the inductive assumption implies $\begin{array}{l} {‖ u^{j + 1} - ω_{0} ‖}_{H^{2}}^{2} + 2 \int_{0}^{t} (δ {‖ \nabla^{3} u^{j + 1} (s) ‖}_{H^{2}}^{2} + κ {‖ \nabla^{2} u^{j + 1} (s) ‖}_{H^{2}}^{2}) d s \\ (3.22) & ⩽ ε_{1}^{2} + C (ε_{1}^{2} + ε_{1}^{4}) \int_{0}^{t} ({‖ \nabla^{3} u^{j + 1} (s) ‖}_{H^{2}}^{2} + {‖ \nabla^{2} u^{j + 1} (s) ‖}_{H^{2}}^{2}) d s, \end{array}$ for any $0 ⩽ t ⩽ T_{1}$ . Take $T_{1} > 0$ and $ε_{1} > 0$ sufficiently small, such that $\begin{matrix} C (ε_{1}^{2} + ε_{1}^{4}) < 2 min {δ, κ} . \end{matrix}$ Then, by (3.22), we find that $\begin{matrix} (3.23) & {‖ u^{j + 1} - ω_{0} ‖}_{H^{2}}^{2} + C \int_{0}^{t} {‖ \nabla^{3} u^{j + 1} (s) ‖}_{H^{2}}^{2} + {‖ \nabla^{2} u^{j + 1} (s) ‖}_{H^{2}}^{2}) d s ⩽ ε_{1}^{2}, \end{matrix}$ for any $t \in [0, T_{1}]$ . Therefore, (3.9) is true for $j + 1$ if so for j, which implies (3.9) is proved for all $j ⩾ 0$ .

Next, by using (3.21), we deduce that $\begin{array}{l} | {‖ A^{j + 1} (t) ‖}_{H^{2}}^{2} - {‖ A^{j + 1} (s) ‖}_{H^{2}}^{2} | \\ = | \int_{s}^{t} \frac{d}{d τ} {‖ A^{j + 1} (τ) ‖}_{H^{2}}^{2} d τ | \\ ⩽ C \int_{s}^{t} ({‖ A^{j} (s) ‖}_{H^{2}}^{2} + {‖ A^{j} (s) ‖}_{H^{2}}^{4}) ({‖ \nabla^{3} A^{j + 1} (s) ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j + 1} (s) ‖}_{H^{2}}^{2}) d s \\ ⩽ C (ε_{1}^{2} + ε_{1}^{4}) \int_{s}^{t} ({‖ \nabla^{3} A^{j + 1} (s) ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j + 1} (s) ‖}_{H^{2}}^{2}) d s, \end{array}$ for any $t \in [0, T_{1}]$ . Therefore, due to (3.23), the time integral in the last inequality is finite, and hence $‖ A^{j + 1} (t) ‖_{H^{2}}^{2}$ is continuous in t for each $j ⩾ 1$ . On the other hand, we also need to consider the convergence of the sequence ${(A^{j})}_{j ⩾ 0}$ . Taking the difference of (3.8)₁ for j and $j - 1$ , it yields that $\begin{array}{l} \partial_{t} (u^{j + 1} - u^{j}) + δ \nabla Δ (u^{j + 1} - u^{j}) + κ_{0} Δ (u^{j + 1} - u^{j}) \\ = - κ_{1} Δ {(u^{j} + κ_{2}) (u^{j} - κ_{2}) (Δ u^{j + 1} - Δ u^{j}) \\ + [(u^{j} + κ_{2}) (u^{j} - κ_{2}) - (u^{j - 1} + κ_{2}) (u^{j - 1} - κ_{2})] Δ u^{j}} \\ - κ_{1} Δ [u^{j} \nabla u^{j} \cdot (\nabla u^{j + 1} - \nabla u^{j}) + (u^{j} \nabla u^{j} - u^{j - 1} \nabla u^{j - 1}) \cdot \nabla u^{j}] \\ (3.24) & + \nabla \cdot [f^{″} (u^{j}) \nabla u^{j + 1}] - \nabla \cdot [f^{″} (u^{- 1} j) \nabla u^{j}], \end{array}$ Appealing to the same energy estimate as before, we get $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ u^{j + 1} - u^{j} ‖}_{H^{2}}^{2} + δ {‖ \nabla Δ u^{j + 1} - \nabla Δ u^{j} ‖}_{H^{2}}^{2} + κ {‖ Δ u^{j + 1} - Δ u^{j} ‖}_{H^{2}}^{2} \\ ⩽ C ({‖ u^{j} - u^{j - 1} ‖}_{H^{2}}^{2} + {‖ u^{j} - u^{j - 1} ‖}_{H^{2}}^{4}) (\nabla Δ u^{j} ‖_{H^{2}}^{2} + ‖ Δ u^{j} ‖_{H^{2}}^{2}) \\ (3.25) & + C ({‖ u^{j} ‖}_{H^{2}}^{2} + {‖ u^{j} ‖}_{H^{2}}^{4}) ({‖ \nabla Δ u^{j + 1} - \nabla Δ u^{j} ‖}_{H^{2}}^{2} + {‖ Δ u^{j + 1} - Δ u^{j} ‖}_{H^{2}}^{2}), \end{array}$ which is equivalent to $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ A^{j + 1} - A^{j} ‖}_{H^{2}}^{2} + {‖ \nabla^{3} A^{j + 1} - \nabla^{3} A^{j} ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j + 1} - \nabla^{2} A^{j} ‖}_{H^{2}}^{2} \\ ⩽ C {‖ A^{j} - A^{j - 1} ‖}_{H^{2}}^{2} ({‖ A^{j} ‖}_{H^{2}}^{2} + {‖ A^{j - 1} ‖}_{H^{2}}^{2} + 1) ({‖ \nabla^{3} A^{j} ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j} ‖}_{H^{2}}^{2}) \\ (3.26) & + C ({‖ A^{j} ‖}_{H^{2}}^{2} + {‖ A^{j} ‖}_{H^{2}}^{4}) ({‖ \nabla^{3} A^{j + 1} - \nabla^{3} A^{j} ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j + 1} - \nabla^{2} A^{j} ‖}_{H^{2}}^{2}), \end{array}$ Based on (3.23), by taking time integration, it holds that $\begin{array}{l} {‖ (A^{j + 1} - A^{j}) (t) ‖}_{H^{2}}^{2} + \int_{0}^{t} ({‖ \nabla^{3} A^{j + 1} - \nabla^{3} A^{j} ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j + 1} - \nabla^{2} A^{j} ‖}_{H^{2}}^{2}) d s \\ ⩽ C (ε_{1}^{2} + ε_{1}^{4}) sup_{0 ⩽ s ⩽ T_{1}} {‖ A^{j} - A^{j - 1} ‖}_{H^{2}}^{2} \\ + C (ε_{1}^{2} + ε_{1}^{4}) \int_{0}^{t} ({‖ \nabla^{3} A^{j + 1} - \nabla^{3} A^{j} ‖}_{H^{2}}^{2} + {‖ \nabla^{2} A^{j + 1} - \nabla^{2} A^{j} ‖}_{H^{2}}^{2}) d s . \end{array}$ Let $ε_{1}$ be sufficiently small, then there exists a constant $λ \in (0, 1)$ such that $\begin{matrix} (3.27) & sup_{0 ⩽ t ⩽ T_{1}} {‖ A^{j + 1} (t) - A^{j} (t) ‖}_{H^{2}}^{2} ⩽ λ sup_{0 ⩽ t ⩽ T_{1}} {‖ A^{j} (t) - A^{j - 1} (t) ‖}_{H^{2}}^{2}, \end{matrix}$ for any $j ⩾ 1$ . Hence, ${(A^{j})}_{j ⩾ 0}$ is a Cauchy sequence in the Banach space $C ([0, T_{1}]; H^{2})$ , the limit function $\begin{matrix} A (t) = A_{0} + lim_{n \to \infty} \sum_{j = 0}^{n} (A^{j + 1} - A^{j}) \end{matrix}$ exists in $C ([0, T_{1}]; H^{2})$ , satisfies $\begin{matrix} sup_{0 ⩽ t ⩽ T_{1}} {‖ A (t) ‖}_{H^{2}}^{2} ⩽ sup_{0 ⩽ t ⩽ T_{1}} \underset{j \to \infty}{lim inf} {‖ A (t) ‖}_{H^{2}} ⩽ ε_{1} . \end{matrix}$ Hence, (3.10) is proved. Finally, suppose that $A (t)$ and $\tilde{A} (t)$ are two solutions in $C ([0, T_{1}]; H^{2})$ satisfying (3.10). By using the same process as in (3.27) to prove the convergence of ${(A^{j})}_{j ⩾ 0}$ , we find that $\begin{matrix} sup_{0 ⩽ t ⩽ T_{1}} {‖ A (t) - \tilde{A} (t) ‖}_{H^{2}}^{2} ⩽ λ sup_{0 ⩽ t ⩽ T_{1}} {‖ A (t) - \tilde{A} (t) ‖}_{H^{2}}^{2}, \end{matrix}$ for $λ \in (0, 1)$ , which implies that $A (t) = \tilde{A} (t)$ . The proof of uniqueness is complete and thus the proof of Lemma 3.2 is complete too. □

3.3. Global well-posedness

In this subsection, we shall combine all the energy estimates that we have derived in the previous sections and the Sobolev interpolation to prove Theorem 1.5.

Proof of Theorem 1.5.
Summing up the estimates (3.2) of Lemma 3.1 from $k = 0$ to 2, we easily obtain $\begin{array}{l} \frac{d}{d t} ‖ u - ω_{0} ‖_{H^{2}}^{2} + δ {‖ \nabla^{3} u ‖}_{H^{2}}^{2} + κ_{0} {‖ \nabla^{2} u ‖}_{H^{2}}^{2} \\ (3.28) & ⩽ C ‖ u - ω_{0} ‖_{H^{2}}^{2} (‖ u - ω_{0} ‖_{H^{2}}^{2} + 1) ({‖ \nabla^{3} u ‖}_{H^{2}}^{2} + {‖ \nabla^{2} u ‖}_{H^{2}}^{2}), \end{array}$ where $C$ is a positive constant. Keeping in mind the condition (1.10), that is $\begin{matrix} ‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} < ε_{0} = \frac{1}{2} \sqrt{1 + \frac{4 min {δ, κ_{0}}}{C}} - \frac{1}{2}, \end{matrix}$ one may deduce that there exists a time $T > 0$ such that for all $t \in [0, T)$ , $\begin{matrix} (3.29) & {‖ u (t) - ω_{0} ‖}_{H^{2}}^{2} ⩽ ‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} . \end{matrix}$ Indeed, we argue (3.29) by contradiction. Suppose that (3.29) is not true. Then there exists a time $T * < T$ (the first time) such that $\begin{matrix} (3.30) & {‖ u (T_{}) - ω_{0} ‖}_{H^{2}}^{2} = ε_{0} = \frac{1}{2} \sqrt{1 + \frac{4 min {δ, κ_{0}}}{C}} - \frac{1}{2} . \end{matrix}$ Hence, $\begin{matrix} {‖ u (t) - ω_{0} ‖}_{H^{2}}^{2} ⩽ ε_{0} = \frac{1}{2} \sqrt{1 + \frac{4 min {δ, κ_{0}}}{C}} - \frac{1}{2}, 0 ⩽ t ⩽ T_{} . \end{matrix}$ Since $x = \frac{1}{2} \sqrt{1 + \frac{4 min {δ, κ_{0}}}{C}} - \frac{1}{2}$ is one of the solutions for equation $x^{2} + x - \frac{min {δ, κ_{0}}}{C} = 0$ . It follows from (3.28) that $\begin{array}{l} \frac{d}{d t} ‖ u - ω_{0} ‖_{H^{2}}^{2} + δ {‖ \nabla^{3} u ‖}_{H^{2}}^{2} + κ_{0} {‖ \nabla^{2} u ‖}_{H^{2}}^{2} \\ (3.31) & ⩽ min {δ, κ_{0}} ({‖ \nabla^{3} u ‖}_{H^{2}}^{2} + {‖ \nabla^{2} u ‖}_{H^{2}}^{2}), 0 ⩽ t ⩽ T_{}, \end{array}$ which implies $\begin{matrix} {‖ u (T_{}) - ω_{0} ‖}_{H^{2}}^{2} ⩽ ‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} < ε_{0} = \frac{1}{2} \sqrt{1 + \frac{4 min {δ, κ_{0}}}{C}} - \frac{1}{2} . \end{matrix}$ This is a contradiction with (3.30). In other words, (3.29) is true. It follows from (3.28) that for all $t \in [0, T)$ , if $‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} < ε_{0}$ , $\begin{matrix} (3.32) & {‖ u (t) - ω_{0} ‖}_{H^{2}}^{2} + \int_{0}^{t} ({‖ \nabla Δ u (s) ‖}_{H^{2}}^{2} + {‖ Δ u (s) ‖}_{H^{2}}^{2}) d s ⩽ C ‖ u_{0} - ω_{0} ‖_{H^{2}}^{2} . \end{matrix}$ Therefore, we obtain the estimate (1.11). Summing up the estimates (3.2) of Lemma 3.1 from $k = 0$ to N, we easily obtain $\begin{array}{l} \frac{d}{d t} ‖ u - ω_{0} ‖_{H^{N}}^{2} + δ {‖ \nabla^{3} u ‖}_{H^{N}}^{2} + κ_{0} {‖ \nabla^{2} u ‖}_{H^{N}}^{2} \\ (3.33) & ⩽ C ‖ u - ω_{0} ‖_{H^{2}}^{2} (‖ u - ω_{0} ‖_{H^{2}}^{2} + 1) ({‖ \nabla^{3} u ‖}_{H^{N}}^{2} + {‖ \nabla^{2} u ‖}_{H^{N}}^{2}) . \end{array}$ On the basis of the condition (1.10), we easily obtain $\begin{matrix} (3.34) & \frac{d}{d t} ‖ u - ω_{0} ‖_{H^{N}}^{2} + C_{0} ({‖ \nabla^{3} u ‖}_{H^{N}}^{2} + {‖ \nabla^{2} u ‖}_{H^{N}}^{2}) ⩽ 0 . \end{matrix}$ Then integrating it directly in time, we obtain (1.11). The proof of Theorem 1.5 is complete. □

4. Proof of Theorem 1.6

4.1. Negative Sobolev estimates

In this subsection, we derive the evolution of the negative Sobolev norms of the solution.

Lemma 4.1.
Suppose that all assumptions in Theorem 1.5 hold. Then, for $s \in [0, \frac{1}{2}]$ , we have $\begin{matrix} (4.1) & \frac{d}{d t} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}}^{2} + {‖ Λ^{- s} \nabla Δ u ‖}_{L^{2}}^{2} + {‖ Λ^{- s} Δ u ‖}_{L^{2}}^{2} ≲ {‖ \nabla^{2} u ‖}_{H^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} . \end{matrix}$
Proof.
Applying $Λ^{- s}$ to (1.8), multiplying the resulting identity by $Λ^{- s} u$ , integrating over $R^{3}$ by parts, we deduce that $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}}^{2} + δ {‖ Λ^{- s} \nabla Δ u ‖}_{L^{2}}^{2} + κ_{0} {‖ Λ^{- s} Δ u ‖}_{L^{2}}^{2} \\ = - κ_{1} \int_{R^{3}} Λ^{- s} {Δ [(u + κ_{2}) (u - κ_{2}) Δ u]} \cdot Λ^{- s} (u - ω_{0}) d x \\ - κ_{1} \int_{R^{3}} Λ^{- s} [Δ (u | \nabla u |^{2})] \cdot Λ^{- s} (u - ω_{0}) d x \\ + \int_{R^{3}} Λ^{- s} Δ [f^{'} (u)] \cdot Λ^{- s} (u - ω_{0}) d x \\ (4.2) & = K_{1} + K_{2} + K_{3} . \end{array}$ We will estimate the three terms of the right hand side of (4.2) one by one. For the first term, we have $\begin{array}{l} K_{1} = & - κ_{1} \int_{R^{3}} Λ^{- s} {Δ [(u + κ_{2}) (u - κ_{2}) Δ u]} \cdot Λ^{- s} (u - ω_{0}) d x \\ ≲ & \int_{R^{3}} | Λ^{- s} (| \nabla u |^{2} Δ u) | | Λ^{- s} u | d x + \int_{R^{3}} | Λ^{- s} (u | Δ u |^{2}) | | Λ^{- s} (u - ω_{0}) | d x \\ + \int_{R^{3}} | Λ^{- s} (u \nabla u \nabla Δ u) | | Λ^{- s} u | d x + \int_{R^{3}} | Λ^{- s} ((u + κ_{2}) (u - κ_{2}) Δ^{2} u) | | Λ^{- s} (u - ω_{0}) | d x \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} (| \nabla u |^{2} Δ u) ‖}_{L^{2}} + {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} (u | Δ u |^{2}) ‖}_{L^{2}} \\ + {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} (u \nabla u \nabla Δ u) ‖}_{L^{2}} \\ + {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} ((u + κ_{2}) (u - κ_{2}) Δ^{2} u) ‖}_{L^{2}} \\ (4.3) & = & K_{11} + K_{12} + K_{13} + K_{14} . \end{array}$ We estimate $K_{11}$ as $\begin{array}{l} K_{11} = & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} (| \nabla u |^{2} Δ u) ‖}_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ | \nabla u |^{2} Δ u ‖}_{L^{\frac{1}{\frac{1}{2} + \frac{s}{3}}}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ \nabla u ‖_{L^{\infty}} ‖ \nabla u ‖_{L^{\frac{3}{s}}} ‖ Δ u ‖_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2}} {‖ \nabla^{3} u ‖}_{L^{2}}^{\frac{1}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2} + s} {‖ \nabla^{3} u ‖}_{L^{2}}^{\frac{1}{2} - s} {‖ \nabla^{2} u ‖}_{L^{2}} \\ ≲ & {‖ \nabla^{2} u ‖}_{L^{2}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{1 + s} {‖ \nabla^{3} u ‖}_{L^{2}}^{1 - s} \\ (4.4) & ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ({‖ \nabla^{2} u ‖}_{L^{2}}^{2} + {‖ \nabla^{3} u ‖}_{L^{2}}^{2}), \end{array}$ where we have used the fact $‖ \nabla^{2} u ‖_{L^{2}} ⩽ C$ , which was proved in Theorem 1.5. $K_{12}$ can be estimated as $\begin{array}{l} K_{12} = & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} (u | Δ u |^{2}) ‖}_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ u | Δ u |^{2} ‖}_{L^{\frac{1}{\frac{1}{2} + \frac{s}{3}}}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ u ‖_{L^{\infty}} ‖ Δ u ‖_{L^{\frac{3}{s}}} ‖ Δ u ‖_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} (‖ \nabla u ‖_{L^{2}}^{\frac{1}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2}}) {‖ \nabla^{3} u ‖}_{L^{2}}^{\frac{1}{2} + s} {‖ \nabla^{4} u ‖}_{L^{2}}^{\frac{1}{2} - s} {‖ \nabla^{2} u ‖}_{L^{2}} \\ ≲ & ‖ \nabla u ‖_{H^{1}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{3} u ‖}_{L^{2}}^{\frac{1}{2} + s} {‖ \nabla^{4} u ‖}_{L^{2}}^{\frac{1}{2} - s} {‖ \nabla^{2} u ‖}_{L^{2}} \\ (4.5) & ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ({‖ \nabla^{2} u ‖}_{L^{2}}^{2} + {‖ \nabla^{3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{4} u ‖}_{L^{2}}^{2}), \end{array}$ where we have used the fact $‖ \nabla u ‖_{H^{1}} ⩽ C$ , which was proved in Theorem 1.5. We also have $\begin{array}{l} K_{13} = & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} (u \nabla u \nabla Δ u) ‖}_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ u \nabla u \nabla Δ u ‖_{L^{\frac{1}{\frac{1}{2} + \frac{s}{3}}}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ u ‖_{L^{\infty}} ‖ \nabla u ‖_{L^{\frac{3}{s}}} ‖ \nabla Δ u ‖_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} (‖ \nabla u ‖_{L^{2}}^{\frac{1}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2}}) {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2} + s} {‖ \nabla^{3} u ‖}_{L^{2}}^{\frac{1}{2} - s} {‖ \nabla^{3} u ‖}_{L^{2}} \\ ≲ & ‖ \nabla u ‖_{H^{1}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2} + s} {‖ \nabla^{3} u ‖}_{L^{2}}^{\frac{3}{2} - s} \\ (4.6) & ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ({‖ \nabla^{2} u ‖}_{L^{2}}^{2} + {‖ \nabla^{3} u ‖}_{L^{2}}^{2}), \end{array}$ and $\begin{array}{l} K_{14} = & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ Λ^{- s} ((u + κ_{2}) (u - κ_{2}) Δ^{2} u) ‖}_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ (u + κ_{2}) (u - κ_{2}) Δ^{2} u ‖}_{L^{\frac{1}{\frac{1}{2} + \frac{s}{3}}}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ u + κ_{2} ‖_{L^{\infty}} ‖ u - κ_{2} ‖_{L^{\frac{3}{s}}} {‖ Δ^{2} u ‖}_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ \nabla u ‖_{L^{2}}^{\frac{1}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2}} ‖ \nabla u ‖_{L^{2}}^{\frac{1}{2} + s} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2} - s} {‖ \nabla^{4} u ‖}_{L^{2}} \\ = & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ \nabla u ‖_{L^{2}}^{1 + s} {‖ \nabla^{2} u ‖}_{L^{2}}^{1 - s} {‖ \nabla^{4} u ‖}_{L^{2}} \\ ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ‖ u ‖_{L^{2}}^{\frac{1}{2} + \frac{s}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2} + \frac{s}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{1 - s} {‖ \nabla^{4} u ‖}_{L^{2}} \\ ≲ & ‖ u ‖_{L^{2}}^{\frac{1}{2} + \frac{s}{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{\frac{1}{2} - \frac{s}{2}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{2} u ‖}_{L^{2}} {‖ \nabla^{4} u ‖}_{L^{2}} \\ (4.7) & ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ({‖ \nabla^{2} u ‖}_{L^{2}}^{2} + {‖ \nabla^{4} u ‖}_{L^{2}}^{2}), \end{array}$ where we have used the fact $‖ u ‖_{H^{2}} ⩽ C$ , which was proved in Theorem 1.5. Combining (4.3)–(4.7) together gives $\begin{matrix} (4.8) & K_{1} ≲ {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ({‖ \nabla^{2} u ‖}_{L^{2}}^{2} + {‖ \nabla^{3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{4} u ‖}_{L^{2}}^{2}) . \end{matrix}$ Next, on the basis of the estimates (4.4)–(4.6), we estimate $K_{2}$ as $\begin{array}{l} K_{2} = & - κ_{1} \int_{R^{3}} Λ^{- s} [Δ (u | \nabla u |^{2})] \cdot Λ^{- s} (u - ω_{0}) d x \\ ≲ & \int_{R^{3}} Λ^{- s} (| \nabla u |^{2} Δ u) \cdot Λ^{- s} (u - ω_{0}) d x + \int_{R^{3}} Λ^{- s} (u | Δ u |^{2}) \cdot Λ^{- s} (u - ω_{0}) d x \\ + \int_{R^{3}} Λ^{- s} (u \nabla u \nabla Δ u) \cdot Λ^{- s} (u - ω_{0}) d x \\ (4.9) & ≲ & {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} ({‖ \nabla^{2} u ‖}_{L^{2}}^{2} + {‖ \nabla^{3} u ‖}_{L^{2}}^{2} + {‖ \nabla^{4} u ‖}_{L^{2}}^{2}) . \end{array}$ For $K_{3}$ , we have $\begin{array}{l} K_{3} = & \int_{R^{3}} Λ^{- s} Δ [f^{'} (u)] \cdot Λ^{- s} (u - ω_{0}) d x \\ = & 2 \int_{R^{3}} Λ^{- s} Δ [(u + 1) (u^{2} + h_{0}) {(u - 1)}^{2}] \cdot Λ^{- s} (u - ω_{0}) d x \\ + 2 \int_{R^{3}} Λ^{- s} Δ [(u - 1) (u^{2} + h_{0}) {(u + 1)}^{2}] \cdot Λ^{- s} (u - ω_{0}) d x \\ + 2 \int_{R^{3}} Λ^{- s} Δ [u {(u + 1)}^{2} {(u - 1)}^{2}] \cdot Λ^{- s} (u - ω_{0}) d x \\ (4.10) & = & K_{31} + K_{32} + K_{33} . \end{array}$ Note that $\begin{array}{l} K_{31} = & \int_{R^{3}} Λ^{- s} [2 (u^{2} + h_{0}) {(u - 1)}^{2} Δ u + 4 u {(u - 1)}^{2} | \nabla u |^{2} + 4 (u^{2} + h_{0}) (u - 1) | \nabla u |^{2} \\ + 4 (u + 1) u {(u - 1)}^{2} Δ u + 4 u {(u - 1)}^{2} | \nabla u |^{2} + 4 (u + 1) {(u - 1)}^{2} | \nabla u |^{2} \\ + 8 (u + 1) u (u - 1) | \nabla u |^{2} + 4 (u + 1) (u - 2) (u^{2} + h_{0}) Δ u + 4 (u - 1) (u^{2} + h_{0}) | \nabla u |^{2} \\ + 4 (u + 1) (u^{2} + h_{0}) | \nabla u |^{2} + 8 (u + 1) (u - 2) u | \nabla u |^{2}] \cdot Λ^{- s} (u - ω_{0}) d x \\ (4.11) & = & K_{31, 1} + K_{31, 2} + \dots + K_{31, 11} . \end{array}$ For $K_{31, 1}$ and $K_{31, 2}$ , we have $\begin{array}{l} K_{31, 1} ≲ & {‖ Λ^{- s} [(u^{2} + h_{0}) {(u - 1)}^{2} Δ u] ‖}_{L^{2}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & {‖ (u^{2} + h_{0}) {(u - 1)}^{2} Δ u ‖}_{L^{\frac{1}{\frac{1}{2} + \frac{s}{3}}}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & ‖ Δ u ‖_{L^{2}} {‖ u^{2} + h_{0} ‖}_{L^{\infty}} ‖ u - 1 ‖_{L^{\infty}} ‖ u - 1 ‖_{L^{\frac{3}{s}}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & ‖ Δ u ‖_{L^{2}} {(‖ \nabla u ‖_{L^{2}} ‖ Δ u ‖_{L^{2}})}^{\frac{3}{2}} ‖ \nabla u ‖_{L^{2}}^{\frac{1}{2} + s} ‖ Δ u ‖_{L^{2}}^{\frac{1}{2} - s} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & (‖ \nabla u ‖_{L^{2}}^{2 + s} ‖ Δ u ‖_{L^{2}}^{1 - s}) ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ (4.12) & ≲ & ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}}, \end{array}$ and $\begin{array}{l} K_{31, 2} ≲ & {‖ Λ^{- s} [u {(u - 1)}^{2} | \nabla u |^{2}] ‖}_{L^{2}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & {‖ u {(u - 1)}^{2} | \nabla u |^{2} ‖}_{L^{\frac{1}{\frac{1}{2} + \frac{s}{3}}}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & ‖ \nabla u ‖_{L^{6}}^{2} ‖ u ‖_{L^{6}} ‖ u - 1 ‖_{L^{\infty}} ‖ u - 1 ‖_{L^{\frac{3}{s}}} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & ‖ Δ u ‖_{L^{2}}^{2} ‖ \nabla u ‖_{L^{2}} {(‖ \nabla u ‖_{L^{2}} ‖ Δ u ‖_{L^{2}})}^{\frac{1}{2}} ‖ \nabla u ‖_{L^{2}}^{\frac{1}{2} + s} ‖ Δ u ‖_{L^{2}}^{\frac{1}{2} - s} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ ≲ & (‖ \nabla u ‖_{L^{2}}^{2 + s} ‖ Δ u ‖_{L^{2}}^{1 - s}) ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} \\ (4.13) & ≲ & ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}}, \end{array}$ where we have used the facts that $‖ u ‖_{H^{N}} ⩽ ‖ u_{0} ‖_{H^{N}}$ and Sobolev’s embedding $‖ u ‖_{L^{\infty}} ≲ ‖ \nabla u ‖_{L^{2}}^{\frac{1}{2}} ‖ Δ u ‖_{L^{2}}^{\frac{1}{2}}$ in (4.12) and (4.13). Similarly, we can easily obtain $\begin{matrix} (4.14) & K_{31, 3} + K_{31, 4} + \dots + K_{31, 11} ≲ ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} . \end{matrix}$ Combining (4.11)–(4.14) together gives $\begin{matrix} (4.15) & K_{31} ≲ ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} . \end{matrix}$ Similarly, we obtain $\begin{matrix} (4.16) & K_{32} + K_{33} ≲ ‖ Δ u ‖_{L^{2}}^{2} {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} . \end{matrix}$ It then follows from (4.15)–(4.16) that $\begin{matrix} (4.17) & K_{3} ≲ {‖ Λ^{- s} (u - ω_{0}) ‖}_{L^{2}} {‖ \nabla^{2} u ‖}_{L^{2}}^{2} . \end{matrix}$ Plugging (4.8), (4.9) and (4.17) into (4.2), we deduce (4.1). □

4.2. Decay estimates

In the following, we prove Theorem 1.6 for $s \in [0, \frac{1}{2}]$ .

Proof of Theorem 1.6.

Define $\begin{matrix} E_{- s} (t) : = {‖ Λ^{- s} [u (t) - ω_{0}] ‖}_{L^{2}}^{2} . \end{matrix}$ For inequality (4.1), integrating in time, by the bound (1.11), we have $\begin{array}{l} E_{- s} (t) ⩽ & E_{- s} (0) + C \int_{0}^{t} {‖ \nabla^{2} u ‖}_{H^{2}}^{2} \sqrt{E_{- s} (τ)} d τ \\ (4.18) & ⩽ & C_{0} (1 + sup_{0 ⩽ τ ⩽ t} \sqrt{E_{- s} (τ)} d τ), \end{array}$ which implies (1.13) for $s \in [0, \frac{1}{2}]$ , that is $\begin{matrix} (4.19) & {‖ Λ^{- s} [u (t) - ω_{0}] ‖}_{L^{2}}^{2} ⩽ C_{0} . \end{matrix}$ Moreover, if $l = 1, 2, \dots, N - 1$ , we may use Lemma 2.5 to have $\begin{matrix} {‖ \nabla^{l + 2} f ‖}_{L^{2}} ⩾ C {‖ Λ^{- s} f ‖}_{L^{2}}^{- \frac{2}{l + s}} {‖ \nabla^{l} f ‖}_{L^{2}}^{1 + \frac{2}{l + s}} . \end{matrix}$ Then, by this facts and (4.19), we get $\begin{matrix} (4.20) & {‖ \nabla^{l + 2} u ‖}_{L^{2}}^{2} ⩾ C_{0} {({‖ \nabla^{l} u ‖}_{L^{2}}^{2})}^{1 + \frac{2}{k + s}} . \end{matrix}$ Hence, for $1 = 1, 2, \dots, N - 1$ , $\begin{matrix} {‖ \nabla^{l + 2} u ‖}_{H^{N - l}}^{2} ⩾ C_{0} {({‖ \nabla^{l} u ‖}_{H^{N - l}}^{2})}^{1 + \frac{2}{l + s}} . \end{matrix}$ Thus, we deduce from (3.34) the following inequality $\begin{matrix} (4.21) & \frac{d}{d t} {‖ u (t) - ω_{0} ‖}_{H^{N}}^{2} + C_{0} {({‖ u (t) - ω_{0} ‖}_{H^{N}}^{2})}^{1 + \frac{2}{l + s}} ⩽ 0, for l = 1, 2, \dots, N - 1 . \end{matrix}$ Solving this inequality directly gives $\begin{matrix} (4.22) & {‖ u (t) - ω_{0} ‖}_{H^{N}}^{2} ⩽ C_{0} {(1 + t)}^{- \frac{1}{2} (l + s)}, for l = 1, 2, \dots, N - 1, \end{matrix}$ which means (1.14) holds. Hence, we complete the proof of Theorem 1.6. □

Footnotes

Acknowledgements

This paper was supported by the Fundamental Research Funds for the Central Universities (grant No. N2005031). The author would like to thank the anonymous referee for his/her useful comments and suggestions, which led to the final version of this paper. The author also thank Prof. Hao Wu from Fudan University for his encourage and useful suggestions.

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On the Cauchy problem of a sixth-order Cahn–Hilliard equation arising in oil-water-surfactant mixtures

Abstract

Keywords

1. Introduction

Lemma 2.1 ([11]).

Lemma 2.2 ([17]).

Lemma 2.3 ([21]).

Lemma 2.4 ([9,25]).

Lemma 2.5 ([10,27]).

Corollary 2.6. If u ∈ H 2 ( R 3 ) , then u is (almost everywhere equal to) a continuous function and (2.8) ‖ u ‖ L ∞ ( R 3 ) ≲ ‖ ∇ u ‖ L 2 ( R 3 ) 1 2 ‖ Δ u ‖ L 2 ( R 3 ) 1 2 . 3. Proof of Theorem 1.5

3.1. Energy estimates

4.1. Negative Sobolev estimates

Footnotes

Acknowledgements

References

Corollary 2.6.
If $u \in H^{2} (R^{3})$ , then u is (almost everywhere equal to) a continuous function and $\begin{matrix} (2.8) & ‖ u ‖_{L^{\infty} (R^{3})} ≲ ‖ \nabla u ‖_{L^{2} (R^{3})}^{\frac{1}{2}} ‖ Δ u ‖_{L^{2} (R^{3})}^{\frac{1}{2}} . \end{matrix}$

3. Proof of Theorem 1.5