Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 46, pp. 1–22. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu EXISTENCE AND NONEXISTENCE OF GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION WITH VARIABLE EXPONENT SOURCES QUACH V. CHUONG, LE C. NHAN, LE X. TRUONG Abstract. In this article, we study the existence and nonexistence of global solutions to the Cahn-Hilliard equation with variable exponent sources and arbitrary initial energy. We also study the asymptotic behavior of weak solu- tions. Our results extend some recent results of Han [10] to PDEs with variable exponent sources. 1. Introduction Let Ω ⊂ RN be a bounded domain with smooth boundary ∂Ω. In this article, we study the initial-boundary value problem ut + ∆2u−∆p(x)u = |u|q(x)−2u, (x, t) ∈ QT , u(x, t) = ∂u ∂ν (x, t) = 0, (x, t) ∈ ΓT , u(x, 0) = u0(x), x ∈ Ω, (1.1) where QT := Ω × (0, T ), ΓT := ∂Ω × (0, T ), ν is the unit outward normal vector on ∂Ω and initial data u0 ∈ H2 0 (Ω). It will also be assumed throughout this paper that p(·) and q(·) are measurable functions satisfying 1 < p− := ess infx∈Ω p(x) ≤ p+ := ess supx∈Ω p(x) < { ∞, if N ≤ 2, 2N N−2 , if N > 2, (1.2) and max{2, p+} < q− := ess infx∈Ω q(x) ≤ q+ := ess supx∈Ω q(x) < { ∞, if N ≤ 4, 2N N−4 , if N > 4. (1.3) It is well known that the fourth-order parabolic equations ut + ∆2u−∇ · f(∇u) = h(x, t, u) (1.4) can be used to model a variety of important physical processes. For example, it can be used to describe the evolution of the epitaxial growth of nanoscale thin films, see [19, 25, 30] and references therein. Equation (1.4) is also known as classical 2020 Mathematics Subject Classification. 35B35, 35B40, 35K57, 35Q92, 92C17. Key words and phrases. Fourth-order parabolic equation; potential well method; global solution; variable exponents. ©2022. This work is licensed under a CC BY 4.0 license. Submitted September 4, 2021. Published July 5, 2022. 1 2 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 Cahn-Hilliard equation arising from modeling of phase transitions in binary systems such as alloys, glasses, thin film epitaxy, and polymer-mixtures; see for example [2, 24, 26, 29]. When the nonlinearities f and h satisfy some constant growth conditions, there have been many results on the existence, uniqueness, and some other properties of the solutions of (1.4). We refer the interested readers to the bibliography given in [3, 4, 14, 12, 16, 15]. Concerning to the blow-up property, Han [10] used the potential well method proposed by Sattinger [23] (see [20, 17]) to study the problem ut + ∆2u−∆pu = |u|q−2u, (x, t) ∈ QT , u(x, t) = ∂u ∂ν (x, t) = 0, (x, t) ∈ ΓT , u(x, 0) = u0(x), x ∈ Ω, In that paper, the author showed a threshold result on global solutions, and on the blow-up in finite time when the initial energy is subcritical critical and supercritical, respectively. In addition, the author also studied the decay rate of the L2-norm of global solutions. This result was then improved by Zhou [31] in studying the exponential decay property of the energy functional. To the best of our knowledge there are very few results on parabolic equations involving variable exponent sources. In [21] the author used the eigenfunction argu- ment of Kaplan [11] to study the blow-up property of solutions of the homogeneous Dirichlet problem for the semilinear parabolic equation ut −∆u = f(x, u), where the source term is either f(x, u) = a(x)up(x) or f(x, u) = a(x) ∫ Ω uq(x)(y, t)dy. Then in [32, 27] the authors established the blow-up results for solutions of the evolution m-Laplace equation involving the variable exponent sources, ut −∆mu = |u|p(x)−1u. Recently, in [22] by using the concavity method, the authors established a blow up in finite time result, in the case of non-positive initial energy J(u0), for fourth-order parabolic equation ut + ∆2u = uq(x). Motivated by the above papers, we consider a more general problem with variable exponent nonlinearities of the form (1.1). Our results are twofold: Firstly, different from the previous results [21, 32, 27, 22] in which the authors only concerned about the blow-up property (global nonexistence). We establish in this paper some sharp results on the existence and nonexistence of global weak solutions for arbitrary initial energy J(u0). As far as we know, such results in the case of PDEs with variable exponent sources are new. Secondly, the decay rate of H2 0 -norm of global solutions which start from potential wells is also concerned. It is noticed that even in the case of constant exponent, Han [10] only proved decay rate of L2-norm of global solutions in case J(u0) < d where d is the potential well depth (see (3.5)). Then Zhou [31] proved the decay of energy functional when J(u0) < d0 ≤ d with d0 = q − 2 2q S− 2q q−2 , EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 3 where S > 0 is the optimal constant of the embedding H2 0 (Ω) ↪→ Lq(Ω). In this paper, we improve this result by showing the exponential decay of energy functional under the condition J(u0) < d. In addition, our techniques are different from [31] where the author’s arguments depend either on p ≤ 2 or p > 2. Because (1.3), we shall prove in Theorem 4.5 that the decay property of energy functional depends strongly on q(x) instead of p(x). It is also worth noticing that this is not a trivial generalization of similar problems with the constant exponent sources. The substantial difficulties in treating the above problem are caused by the complicated nonlinearities −∆p(x)u and |u|q(x)−2u (it is non-homogeneous) and the lack of a maximum principle and comparison principle for fourth-order equations. The key point is to treat the gap between the norm and the integral in variable exponent spaces. Our method presented here can be used to treat the problem in [21, 22, 27, 32]. This article is organized as follows: In Section 2 we recall some facts about H2 0 (Ω) space and Ozlicz-Sobolev type spaces. In Section 3 we study the stationary state of (1.1) and construct the stable sets and unstable sets; In Section 4 we present our main results on the evolution problem. The proof of main results are given in the rest of the paper. 2. Preliminaries Let Ω be as in Section 1. We denote by ‖ · ‖r the usual norm of the space Lr(Ω) for 1 ≤ r ≤ ∞ and 〈·, ·〉 the usual inner product of the Hilbert space L2(Ω). We also denote by ‖ · ‖H2 0 the norm of H2 0 (Ω). That is ‖u‖H2 0 = √ ‖u‖22 + ‖∇u‖22 + ‖∆u‖22. As in [10], H2 0 (Ω) is a Hilbert space with inner product 〈u, v〉H2 0 = 〈∆u,∆v〉 . Then H2 0 (Ω) is uniformly convex and the norm ‖ · ‖H2 0 is equivalent to the norm ‖∆(·)‖2 due to Poincare’s inequality. We next introduce some preliminary results on Lebesgue and Sobolev spaces with variable exponents (see [5, 6, 7, 8, 13]). Denote by P(Ω) the set of all measurable functions p : Ω → [1,∞]. Define the Lebesgue space with a variable exponent p(·) which is the so-called Nakano space and a special case of Musielak-Orlicz spaces (see [18]), as follows: Lp(·)(Ω) := { u : Ω→ R measurable, ρ(u) := ∫ Ω |u(x)|p(x) dx <∞ } , where p ∈ P(Ω). The space Lp(·)(Ω) is equipped with the Luxemburg-type norm ‖u‖p(·) := inf { λ > 0 : ∫ Ω |u(x) λ |p(x) dx ≤ 1 } . The following proposition shows the relation between the norm ‖u‖p(·) and the modular ρ(u). Proposition 2.1 ([5]). Let p ∈ P(Ω). It holds that min { ‖u‖p − p(·), ‖u‖ p+ p(·) } ≤ ρ(u) ≤ max { ‖u‖p − p(·), ‖u‖ p+ p(·) } , for all u ∈ Lp(·)(Ω). 4 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 For p+ <∞, the dual space of Lp(·)(Ω) is identified with Lp ′(·)(Ω) with the dual variable exponent p′ ∈ P(Ω) given by 1 p(x) + 1 p′(x) = 1 for a.e. x ∈ Ω, where we write 1/∞ = 0. The Hölder inequality also holds for variable Lebesgue spaces. Proposition 2.2 (Hölder inequality,[5]). Let p, q, s ∈ P(Ω), it holds that ‖uv‖s(·) ≤ 2‖u‖p(·)‖v‖q(·) for all u ∈ Lp(·)(Ω), v ∈ Lq(·)(Ω), provided that 1 s(x) = 1 p(x) + 1 q(x) for a.e. x ∈ Ω. Proposition 2.3 ([5]). Let p, q ∈ P(Ω). If p(x) ≤ q(x) for a.e. x ∈ Ω, then the embedding Lq(·)(Ω) ↪→ Lp(·)(Ω) is continuous. We next define variable exponent Sobolev spaces W 1,p(·)(Ω) = {u ∈ Lp(·)(Ω) : |∇u| ∈ Lp(·)(Ω)}, endowed with the norm ‖u‖W 1,p(·)(Ω) = ( ‖u‖2p(·) + ‖∇u‖2p(·) )1/2 . Furthermore, let W 1,p(·) 0 (Ω) be the closure of C∞0 (Ω) in W 1,p(·)(Ω). It is noticed that if 1 < p− ≤ p+ < ∞, then Lp(·)(Ω) and W 1,p(·)(Ω) are uniformly convex Banach spaces and therefore they are reflexive. 3. Stationary problem and potential wells In this section, we consider the stationary solutions of (1.1) which solve the problem ∆2u−∆p(x)u = |u|q(x)−2u in Ω, u(x) = ∂u ∂ν (x) = 0 on ∂Ω, (3.1) where p(x) and q(x) satisfy (1.2)-(1.3). Consider the energy functional J and the Nehari functional I given by J(u) = 1 2 ‖∆u‖22 + ∫ Ω 1 p(x) |∇u|p(x) dx− ∫ Ω 1 q(x) |u|q(x) dx, I(u) = ‖∆u‖22 + ∫ Ω |∇u|p(x) dx− ∫ Ω |u|q(x) dx. Then J and I are of class C1 over H2 0 (Ω) and critical points of J are weak solutions of (3.1). Moreover, we can estimate J and I as follows: J(u) ≥ 1 2 ‖∆u‖22 + 1 p+ ∫ Ω |∇u|p(x) dx− 1 q− ∫ Ω |u|q(x) dx = (1 2 − 1 q− ) ‖∆u‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u|p(x) dx+ 1 q− I(u), (3.2) EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 5 J(u) ≤ 1 2 ‖∆u‖22 + 1 p− ∫ Ω |∇u|p(x) dx− 1 q+ ∫ Ω |u|q(x) dx = (1 2 − 1 q+ ) ‖∆u‖22 + ( 1 p− − 1 q+ ) ∫ Ω |∇u|p(x) dx+ 1 q+ I(u), (3.3) J(u) = (1 2 − 1 q− ) ‖∆u‖22 + ∫ Ω ( 1 p(x) − 1 q− ) |∇u|p(x) dx + ∫ Ω ( 1 q− − 1 q(x) ) |u|q(x) dx+ 1 q− I(u). (3.4) Let u ∈ H2 0 (Ω)\{0} and consider the fibering map λ 7→ j(λ) := J(λu) for λ > 0 given by j(λ) = λ2 2 ‖∆u‖22 + ∫ Ω λp(x) p(x) |∇u|p(x) dx− ∫ Ω λq(x) q(x) |u|q(x) dx. Lemma 3.1. Let (1.2)-(1.3) hold and u ∈ H2 0 (Ω)\{0}. Then the following results hold: (i) limλ→0+ j(λ) = 0 and limλ→∞ j(λ) = −∞. (ii) There exists a λ∗ = λ∗(u) > 0 such that j(λ) attains the maximum at λ = λ∗, then I(λ∗u) = 0. In addition, we have 0 < λ∗ < 1, λ∗ = 1 and λ∗ > 1 provided that I(u) < 0, I(u) = 0 and I(u) > 0, respectively. Proof. It is easily seen that j(λ) ≥ 1 2 λ2‖∆u‖22 + min{λp − , λp + } ∫ Ω 1 p(x) |∇u|p(x) dx −max{λq − , λq + } ∫ Ω 1 q(x) |u|q(x) dx, and j(λ) ≤ 1 2 λ2‖∆u‖22 + max{λp − , λp + } ∫ Ω 1 p(x) |∇u|p(x) dx −min{λq − , λq + } ∫ Ω 1 q(x) |u|q(x) dx. This, together with q− > max{2, p+} and ∫ Ω 1 q(x) |u| q(x) dx > 0, implies (i). Fur- thermore, we also have j(λ) > 0 for sufficiently small λ > 0. Hence, there exists a λ∗ > 0 such that j(λ∗) = supλ>0 j(λ). Then by Fermat’s Theorem, we obtain j′(λ∗) = 0. This gives I(λ∗u) = 0 by the relation I(λu) = λj′(λ). Finally, we prove the last statement of (ii). By the definition of I, we obtain 0 = I(λ∗u) = λ2 ∗‖∆u‖22 + ∫ Ω λ p(x) ∗ |∇u|p(x) dx− ∫ Ω λ q(x) ∗ |u|q(x) dx = ( λ2 ∗ − λq − ∗ ) ‖∆u‖22 + ∫ Ω ( λ p(x) ∗ − λq − ∗ ) |∇u|p(x) dx + ∫ Ω (λq − ∗ − λ q(x) ∗ )|u|q(x) dx+ λq − ∗ I(u), which can be rewritten in the form λq − ∗ I(u) = ( λq − ∗ − λ2 ∗ ) ‖∆u‖22 + ∫ Ω ( λq − ∗ − λ p(x) ∗ ) |∇u|p(x) dx 6 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 + ∫ Ω ( λ q(x) ∗ − λq − ∗ ) |u|q(x) dx. Since q− > max{2, p+}, the above equality shows that 0 < λ∗ < 1, λ∗ = 1 and λ∗ > 1 provided that I(u) < 0, I(u) = 0 and I(u) > 0, respectively. This completes the proof. � We now define the so-called Nehari manifold associated to the energy functional J by N = {u ∈ H2 0 (Ω)\{0} : I(u) = 0}. It follows from Lemma 3.1 that N is not empty set. Thus, we can define d = inf u∈N J(u). (3.5) The following lemma plays an important role in the proofs of our main results for the low initial energy case. Lemma 3.2. Let (1.2)-(1.3) hold and u ∈ H2 0 (Ω)\{0}. Then J(u)− 1 q− I(u) ≥ d max{λ2 ∗, λ p− ∗ , λq + ∗ } , where λ∗ is as in Lemma 3.1. Proof. For any u ∈ H2 0 (Ω)\{0}, by Lemma 3.1, there exists λ∗ ∈ (0,∞) such that I(λ∗u) = 0. By the definition of d and replacing u by λ∗u in (3.4), one has d ≤ J(λ∗u) = (1 2 − 1 q− ) λ2 ∗‖∆u‖22 + ∫ Ω λ p(x) ∗ ( 1 p(x) − 1 q− ) |∇u|p(x) dx + ∫ Ω λ q(x) ∗ ( 1 q− − 1 q(x) ) |u|q(x) dx ≤ max{λ2 ∗, λ p− ∗ , λq + ∗ } [(1 2 − 1 q− ) ‖∆u‖22 + ∫ Ω ( 1 p(x) − 1 q− ) |∇u|p(x) dx + ∫ Ω ( 1 q− − 1 q(x) ) |u|q(x) dx ] = max { λ2 ∗, λ p− ∗ , λq + ∗ } [J(u)− 1 q− I(u)]. This implies the required result. The proof is complete. � Based on the above two lemmas, we can prove the following lemma, which shows that d is positive and is actually attained at some u ∈ N . Lemma 3.3. Let (1.2)–(1.3) hold. Then we have (i) d = infu∈H2 0 (Ω)\{0} supλ>0 J(λu). (ii) d is a positive number. (iii) There exists u∗ ∈ N , u∗(x) ≥ 0 a.e. in Ω such that J(u∗) = d. Proof. For any u ∈ H2 0 (Ω)\{0}. By Lemma 3.1, we have sup λ>0 J(λu) = J(λ∗u). (3.6) By the definition of N , it follows from Lemma 3.1 that λ∗u ∈ N . Hence J(λ∗u) ≥ inf u∈N J(u) = d. (3.7) EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 7 Combining (3.6) and (3.7), one has inf u∈H2 0 (Ω)\{0} sup λ>0 J(λu) ≥ d. (3.8) On the other hand, for any u ∈ N , by Lemma 3.1, one has λ∗ = 1, that is, sup λ>0 J(λu) = J(u). Therefore, inf u∈H2 0 (Ω)\{0} sup λ>0 J(λu) ≤ inf u∈N sup λ>0 J(λu) = inf u∈N J(u) = d. (3.9) We deduce from (3.8) and (3.9) that (i) holds. We next prove (ii). Since q(x) satisfies (1.3), H2 0 (Ω) can be embedded into Lq(·)(Ω) continuously. Denote by Sq(·) the optimal embedding constant, i.e., Sq(·) = sup u∈H2 0 (Ω)\{0} ‖u‖q(·) ‖∆u‖2 . Let any u ∈ H2 0 (Ω)\{0} such that I(u) ≤ 0. Then it follows that ‖∆u‖22 ≤ ∫ Ω |u|q(x) dx ≤ max { ‖u‖q − q(·), ‖u‖ q+ q(·) } ≤ max { Sq − q(·)‖∆u‖ q− 2 , Sq + q(·)‖∆u‖ q+ 2 } . Taking this fact into account and notice that ‖∆u‖2 > 0 and q− > 2, we obtain ‖∆u‖2 ≥ δ1, (3.10) where δ1 = min { S q− 2−q− q(·) , S q+ 2−q+ q(·) } . Fixing u ∈ N , we have u ∈ H2 0 (Ω)\{0} and I(u) = 0. By using (3.2) and (3.10), we obtain J(u) ≥ (1 2 − 1 q− ) ‖∆u‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u|p(x) dx+ 1 q− I(u) ≥ (1 2 − 1 q− ) ‖∆u‖22 ≥ (1 2 − 1 q− ) δ2 1 . (3.11) Then by the definition of d, we obtain d ≥ (1 2 − 1 q− ) δ2 1 > 0. Finally, we prove (iii). By (3.5), there exists {un}∞n=1 ⊂ N is a minimizing sequence of J such that limn→∞ J(un) = d. Clearly, |un| ∈ N and J(|un|) = J(un). For this reason, we may assume that un(x) ≥ 0 a.e. in Ω for all n ∈ N∗. Since limn→∞ J(un) = d and using (3.11), we infer that {un} is bounded in H2 0 (Ω). Then, since H2 0 (Ω) is reflexive, H2 0 (Ω) ↪→W 1,p(·) 0 (Ω) and H2 0 (Ω) ↪→ Lq(·)(Ω) 8 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 are compact embeddings (by (1.2) and (1.3)), there exists a sub-sequence of {un}, still denoted by {un} and a u∗ ∈ H2 0 (Ω) such that un ⇀ u∗ weakly in H2 0 (Ω), un → u∗ strongly in W 1,p(·) 0 (Ω), un → u∗ strongly in Lq(·)(Ω), un(x)→ u∗(x) a.e. in Ω. Then we have u∗(x) ≥ 0 a.e. in Ω and ‖∆u∗‖2 ≤ lim inf n→∞ ‖∆un‖2,∫ Ω |∇u∗|p(x) dx = lim n→∞ ∫ Ω |∇un|p(x) dx,∫ Ω ( 1 p(x) − 1 q− ) |∇u∗|p(x) dx = lim n→∞ ∫ Ω ( 1 p(x) − 1 q− ) |∇un|p(x) dx,∫ Ω |u∗|q(x) dx = lim n→∞ ∫ Ω |un|q(x) dx,∫ Ω ( 1 q− − 1 q(x) ) |u∗|q(x) dx = lim n→∞ ∫ Ω ( 1 q− − 1 q(x) ) |un|q(x) dx. Replacing u by un in (3.4) and notice that un ∈ N , one has d = lim inf n→∞ J(un) = lim inf n→∞ [(1 2 − 1 q− ) ‖∆un‖22 + ∫ Ω ( 1 p(x) − 1 q− ) |∇un|p(x) dx + ∫ Ω ( 1 q− − 1 q(x) ) |un|q(x) dx] ≥ (1 2 − 1 q− ) ‖∆u∗‖22 + ∫ Ω ( 1 p(x) − 1 q− ) |∇u∗|p(x) dx + ∫ Ω ( 1 q− − 1 q(x) ) |u∗|q(x) dx = J(u∗)− 1 q− I(u∗). (3.12) Suppose that I(u∗) < 0. Then by Lemmas 3.1 and 3.2, there exists λ∗ ∈ (0, 1) such that J(u∗)− 1 q− I(u∗) ≥ d max{λ2 ∗, λ p− ∗ , λq + ∗ } > d. This contradicts (3.12), and so I(u∗) ≥ 0. (3.13) Since un ∈ N , we have I(un) = 0, it follows that 0 = lim inf n→∞ I(un) = lim inf n→∞ ( ‖∆un‖22 + ∫ Ω |∇un|p(x) dx− ∫ Ω |un|q(x) dx ) ≥ ‖∆u∗‖22 + ∫ Ω |∇u∗|p(x) dx− ∫ Ω |u∗|q(x) dx = I(u∗), EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 9 which, together with (3.13), implies that I(u∗) = 0. We now prove u∗ ∈ N . It remains to show that u∗ 6= 0. Indeed, since un ∈ N and (3.10), we obtain∫ Ω |un|q(x) dx = ‖∆un‖22 + ∫ Ω |∇un|p(x) dx ≥ δ2 1 . Passing to the limit, we have ∫ Ω |u∗|q(x) dx ≥ δ2 1 > 0, which gives u∗ 6= 0. Hence, u∗ ∈ N and therefore J(u∗) ≥ d. By (3.12) and I(u∗) = 0, we have J(u∗) ≤ d. So, J(u∗) = d. The proof is complete. � We now define the so-called stable set W and unstable set U which is similar to Sattinger [23], Payne and Sattinger [20]. W = {u ∈ H2 0 (Ω) : J(u) < d, I(u) > 0} ∪ {0}, U = {u ∈ H2 0 (Ω) : J(u) < d, I(u) < 0}, We also introduce N− = {u ∈ H2 0 (Ω) : I(u) < 0}, N+ = {u ∈ H2 0 (Ω) : I(u) > 0}, and the open sub levels of J , Jk = {u ∈ H2 0 (Ω) : J(u) < k}. The variational characterization of d also shows that Nk := N ∩ Jk 6= ∅ for all k > d. For k > d, we now define λk = inf{‖u‖2 : u ∈ Nk} and Λk = sup{‖u‖2 : u ∈ Nk}. (3.14) It is obvious that k 7→ λk is non-increasing, and k 7→ Λk is non-decreasing. The next lemma shows that λk and Λk are finite positive numbers, and therefore the result of Theorem 4.9 is nontrivial. Lemma 3.4. Let (1.2)-(1.3) hold. Then for any k > d, λk and Λk defined in (3.14) satisfy 0 < λk ≤ Λk <∞. Proof. Firstly, we prove Λk < ∞. For any k > d and u ∈ Nk, we have J(u) < k and I(u) = 0. Then by (3.2) and using the embedding H2 0 (Ω) ↪→ L2(Ω), we obtain k > J(u) ≥ (1 2 − 1 q− ) ‖∆u‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u|p(x) dx+ 1 q− I(u) ≥ (1 2 − 1 q− ) ‖∆u‖22 ≥ (1 2 − 1 q− ) S−2 2 ‖u‖22, (3.15) which yields ‖u‖2 ≤ S2 √ 2kq− q− − 2 , where S2 > 0 is the optimal embedding constant, i.e., S2 = sup u∈H2 0 (Ω)\{0} ‖u‖2 ‖∆u‖2 . (3.16) 10 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 This shows that Λk ≤ S2 √ 2kq− q− − 2 <∞. Secondly, we prove that λk > 0. By the Gagliardo-Nirenberg inequality, there exists a positive constant A0 depending only on Ω, N, q− and q+ such that ‖u‖q − q− ≤ A0‖∆u‖θ −q− 2 ‖u‖(1−θ −)q− 2 , ‖u‖q + q+ ≤ A0‖∆u‖θ +q+ 2 ‖u‖(1−θ +)q+ 2 , where θ± = N(q±−2) 4q± ∈ (0; 1) by (1.3). Then, since u ∈ Nk and Nk ⊂ N , it follows that ‖∆u‖22 ≤ ∫ Ω |u|q(x) dx ≤ ∫ Ω ( |u|q − + |u|q +) dx ≤ 2 max { ‖u‖q − q− , ‖u‖ q+ q+ } ≤ 2A0 max { ‖∆u‖θ −q− 2 ‖u‖(1−θ −)q− 2 , ‖∆u‖θ +q+ 2 ‖u‖(1−θ +)q+ 2 } . Taking this into account and noticing that ‖∆u‖2 > 0 and θ± < 1, we obtain ‖u‖2 ≥ min { (2A0) 1 (θ−−1)q− ‖∆u‖ 2−θ−q− (1−θ−)q− 2 , (2A0) 1 (θ+−1)q+ ‖∆u‖ 2−θ+q+ (1−θ+)q+ 2 } . (3.17) On the other hand, it follows from (3.10) and (3.15) that δ1 ≤ ‖∆u‖2 ≤ √ 2kq− q− − 2 := δ2, for all u ∈ Nk. This, together with (3.17), implies ‖u‖2 ≥ min { (2A0) 1 (θ−−1)q− min { δ 2−θ−q− (1−θ−)q− 1 , δ 2−θ−q− (1−θ−)q− 2 } , (2A0) 1 (θ+−1)q+ min { δ 2−θ+q+ (1−θ+)q+ 1 , δ 2−θ+q+ (1−θ+)q+ 2 }} > 0. Hence, λk > 0 by the definition of λk. This completes the proof. � Finally, we give the following lemma, which is necessary for our proofs of the main results in case of the high initial energy. Lemma 3.5. Let (1.2)-(1.3) hold. Then we have (i) 0 is away from both N and N−, that is, dist(0,N ) > 0 and dist(0,N−) > 0. (ii) The set N+ ∩ Jk is bounded in H2 0 (Ω) for any k > 0. Proof. By (3.10), it is easy to see that dist(0,N ) = inf u∈N ‖∆u‖2 ≥ δ1 > 0, dist(0,N−) = inf u∈N− ‖∆u‖2 ≥ δ1 > 0. EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 11 We now prove (ii). For any u ∈ N+ ∩ Jk, we have J(u) < k and I(u) > 0. Then by using (3.2), it follows that k > J(u) ≥ (1 2 − 1 q− ) ‖∆u‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u|p(x) dx+ 1 q− I(u) ≥ (1 2 − 1 q− ) ‖∆u‖22, which implies that ‖∆u‖2 < √ 2kq− q− − 2 and completes the proof. � 4. Evolution problem We first give the precise meaning of solution to problem (1.1). Definition 4.1. Let T > 0, a function u = u(t) ∈ L∞(0, T ;H2 0 (Ω)) with ut ∈ L2(0, T ;L2(Ω)) is said to be a weak solution to (1.1) in Ω × [0, T ), if u(0) = u0 ∈ H2 0 (Ω) and satisfies 〈ut, v〉+ 〈∆u,∆v〉+ 〈|∇u|p(x)−2∇u,∇v〉 = 〈|u|q(x)−2u, v〉, a.e. t ∈ (0, T ), (4.1) for any v ∈ H2 0 (Ω). Moreover,∫ t 0 ‖u′(s)‖22 ds+ J(u(t)) = J(u0), 0 ≤ t < T. (4.2) Definition 4.2. Let u(t) be a weak solution to the problem (1.1). We define the maximal existence time Tmax of u(t) as follows (i) If u(t) exists for 0 ≤ t <∞, then Tmax =∞. (ii) If there exists t0 > 0 such that u(t) exists for 0 ≤ t < t0, but does not exist at t0, then Tmax = t0. Remark 4.3. Under the assumption u0 ∈ H2 0 (Ω), by using the standard Galerkin’s method as in [10, 31], we can prove immediately the existence of local weak solutions u(t). For the uniqueness of solution to (1.1), it can be obtained under some suitable assumptions on the initial data u0, p(·) and q(·). For example, in [10] Han proved the uniqueness of bounded weak solution (L∞-bounded solution), such kind of solution can be obtained by either in one dimensional space or p(·) > N . The next lemma shows the invariant of stable and unstable sets. Lemma 4.4. Let (1.2)-(1.3) hold and J(u0) < d. Then we possess the following statements: (i) If I(u0) < 0, then I(u(t)) < 0 for all t ∈ [0, Tmax). (ii) If I(u0) ≥ 0, then I(u(t)) ≥ 0 for all t ∈ [0, Tmax). Proof. Note that u(t) /∈ N , for all t ∈ [0, Tmax) since J(u(t)) ≤ J(u0)) < d. For (i). Assume to the contrary that there exists t0 ∈ (0, Tmax) such that I(u(t)) < 0 for all t ∈ [0, t0) and I(u(t0)) = 0. Then by (3.10), we obtain ‖∆u(t)‖2 ≥ δ1, for all t ∈ [0, t0). Letting t → t0, we have ‖∆u(t0)‖2 ≥ δ1, which gives u(t0) 6= 0. Hence, u(t0) ∈ N . We thus arrive at a contradiction. For (ii). By contradiction, we assume that there exists t1 ∈ (0, Tmax) such that I(u(t1)) < 0. This and I(u0) ≥ 0 imply that there exists t2 ∈ [0, t1) such that 12 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 I(u(t2)) = 0. It gives u(t2) = 0 due to u(t2) /∈ N . Then we obtain u(t) = 0 for t ∈ [t2, Tmax). Thus u(t1) = 0. This contradicts I(u(t1)) < 0. The proof is complete. � We introduce the set S = {φ ∈ H2 0 (Ω) : φ is a stationary solution of (1.1)}, and define the ω-limit set ω(u0) of the initial data u0 ∈W 1,p(·) 0 (Ω) by ω(u0) = {w ∈ H2 0 (Ω) : ∃{tn} with tn →∞ such that u(tn)→ w}. Let u(t) be a solution to (1.1) associated with u0 ∈ H2 0 (Ω) on the maximal existence time interval [0, Tmax). We then introduce the sets G = {u0 ∈ H2 0 (Ω) : u(t) exists globally, i.e. Tmax =∞}, G0 = {u0 ∈ G : u(t)→ 0 in H2 0 (Ω) as t→∞}, B = {u0 ∈ H2 0 (Ω) : u(t) blows up in finite time, i.e. Tmax <∞}. Our main results read as follows. Theorem 4.5. Let (1.2)-(1.3) hold. If J(u0) < d and I(u0) ≥ 0, then the maximal existence time Tmax =∞. Moreover, u(t) holds following decay estimates: ‖u(t)‖2 ≤ ‖u0‖2e−αt, ‖∆u(t)‖2 ≤ √ 2q− q− − 2 (J(u0) + ‖u0‖22)e−βt,√ J(u(t)) + ‖u(t)‖22 ≤ √ J(u0) + ‖u0‖22e−βt, where α and β are some positive constants. Theorem 4.6. Let (1.2)-(1.3) hold. Then (i) If u0 ∈ H2 0 (Ω)\{0} holds J(u0) ≤ 0, then Tmax <∞. Furthermore, we can get an upper bound for the maximal existence time Tmax ≤ C max { ‖u0‖2−q − 2 , ‖u0‖2−q + 2 } , where C = q−max{Sq − q(·),2, S q+ q(·),2} (q− − 2)(q− −max{2, p+}) > 0, (4.3) and Sq(·),2 is the optimal embedding constant of Lq(·)(Ω) ↪→ L2(Ω) when q− > 2, i.e., Sq(·),2 = sup u∈Lq(·)(Ω)\{0} ‖u‖2 ‖u‖q(·) . (4.4) (ii) If 0 < J(u0) < d and I(u0) < 0, then Tmax <∞. Remark 4.7. As a consequence of Theorem 4.5 and 4.6, we have a sharp result in the case J(u0) < d, that is, the weak solution to (1.1) exists globally and blows up in finite time provided that I(u0) ≥ 0 and I(u0) < 0, respectively. Theorem 4.5 shows that any global weak solution which starts from the potential wells W tends to zero. And the next theorem shows the asymptotic behavior of any global weak solution of (1.1). EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 13 Theorem 4.8. Let (1.2)-(1.3) hold and u(t) be a global solution of (1.1). Then there exists a sequence {tn} with tn →∞ as n→∞ and φ ∈ S such that lim n→∞ ‖∆u(tn)−∆φ‖2 = 0. Our next result gives an abstract criterion for vanishing and global nonexistence of solutions to (1.1) in terms of the variational values λk and Λk. Theorem 4.9. Let (1.2)-(1.3) hold and J(u0) > d. If u0 ∈ N+ and ‖u0‖2 ≤ λJ(u0), then u0 ∈ G0. If u0 ∈ N− and ‖u0‖2 ≥ ΛJ(u0), then u0 ∈ B. As a consequence, one has a characterization on the data u0 with arbitrary high energy J(u0) that leads to blow-up in finite time phenomena. Theorem 4.10. Let (1.2)-(1.3) hold and assume that u0 ∈ H2 0 (Ω) holds J(u0) > d and ‖u0‖22 ≥ 2q−S2 2 q− − 2 J(u0), (4.5) then u0 ∈ N− ∩ B. Here S2 is the constant given in (3.16). 5. Proof of Theorem 4.5 Let u(t) := u(x, t) be a solution of (1.1) on the interval [0, Tmax) associated with to the initial data u0. We first prove the uniform boundedness in time of u(t) in H2 0 (Ω), which implies Tmax = ∞ by the continuation principle. Indeed, since J(u0) < d and I(u0) ≥ 0, by Lemma 4.4 we have that I(u(t)) ≥ 0. Then by using the non-increasing property of J(u(t)) and (3.2), we obtain J(u0) ≥ J(u(t)) ≥ (1 2 − 1 q− ) ‖∆u(t)‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u(t)|p(x) dx+ 1 q− I(u(t)) ≥ (1 2 − 1 q− ) ‖∆u(t)‖22, (5.1) which implies ‖∆u(t)‖2 ≤ √ 2q−J(u0) q− − 2 . We next show the decay estimates of u(t). If u(t0) = 0 for some t0 ≥ 0, then we have u(t) = 0 for all t ≥ t0, and the proof is complete. So we may assume u(t) 6= 0 for all t ≥ 0. Then due to I(u(t)) ≥ 0, by Lemma 3.1, there exists λ∗ ≥ 1 such that I(λ∗u(t)) = 0. And therefore λq − ∗ I(u(t)) = λq − ∗ I(u(t))− I(λ∗u(t)) = ( λq − ∗ − λ2 ∗ ) ‖∆u(t)‖22 + ∫ Ω ( λq − ∗ − λ p(x) ∗ ) |∇u(t)|p(x) dx + ∫ Ω ( λ q(x) ∗ − λq − ∗ ) |u(t)|q(x) dx ≥ ( λq − ∗ − λ2 ∗ ) ‖∆u(t)‖22 + ( λq − ∗ − λp + ∗ ) ∫ Ω |∇u(t)|p(x) dx. 14 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 Dividing the above inequality by λq − ∗ , we obtain I(u(t)) ≥ (1− λ2−q− ∗ )‖∆u(t)‖22 + ( 1− λp +−q− ∗ ) ∫ Ω |∇u(t)|p(x) dx. (5.2) We next estimate for λ∗. By applying Lemma 3.2 and notice that λ∗ ≥ 1, one has J(u(t))− 1 q− I(u(t)) ≥ d max{λ2 ∗, λ p− ∗ , λq + ∗ } = d λq + ∗ . (5.3) On the other hand, by using the non-increasing property of J(u(t)) and notice that I(u(t)) ≥ 0, we have J(u(t))− 1 q− I(u(t)) ≤ J(u0). This together with (5.3), implies that λ∗ ≥ ( d J(u0) )1/q+ > 1. (5.4) It follows from (5.2) and (5.4) that I(u(t)) ≥ ( 1− ( d J(u0) ) 2−q− q+ ) ‖∆u(t)‖22 + ( 1− ( d J(u0) ) p+−q− q+ )∫ Ω |∇u(t)|p(x) dx, which yields I(u(t)) ≥ C1‖∆u(t)‖22 and I(u(t)) ≥ C2 ∫ Ω |∇u(t)|p(x) dx, (5.5) where C1 = 1− ( d J(u0) ) 2−q− q+ and C2 = 1− ( d J(u0) ) p+−q− q+ . We now consider the exponential decay of ‖u(t)‖2. Taking v = u in (4.1), we have d dt ‖u(t)‖22 = −2 ( ‖∆u(t)‖22 + ∫ Ω |∇u(t)|p(x) dx− ∫ Ω |u(t)|q(x) dx ) = −2I(u(t)). From this and (5.5), it follows that d dt ‖u(t)‖22 ≤ −2C1‖∆u(t)‖22 ≤ −2C1S −2 2 ‖u(t)‖22, where S2 is the constant given in (3.16). This implies that ‖u(t)‖2 ≤ ‖u0‖2e−αt, where α = C1S −2 2 > 0. We next consider the exponential decay of J(u(t)) and ‖∆u(t)‖2. Using (3.3), we obtain J(u(t)) ≤ (1 2 − 1 q+ ) ‖∆u(t)‖22 + ( 1 p− − 1 q+ ) ∫ Ω |∇u(t)|p(x) dx+ 1 q+ I(u(t)). This together with (5.5) immediately yields J(u(t)) ≤ C3I(u(t)), (5.6) EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 15 where C3 = 1 C1 (1 2 − 1 q+ ) + 1 C2 ( 1 p− − 1 q+ ) + 1 q+ > 0. Let us define an auxiliary function L(t) = J(u(t)) + ‖u(t)‖22, for t ≥ 0. (5.7) Then by (5.1) and (5.7), we obtain L(t) ≤ J(u(t)) + S2 2‖∆u(t)‖22 ≤ C4J(u(t)). (5.8) Here C4 = 1 + 2q− q−−2S 2 2 > 0 and S2 is the constant given in (3.16). It follows from (5.6), (5.7) and (5.8) that d dt L(t) = −‖u′(t)‖22 − 2I(u(t)) ≤ − 2 C3 J(u(t)) ≤ − 2 C3C4 L(t), which implies that L(t) ≤ L(0)e−2βt, where β = 1 C3C4 > 0. The above inequality can be rewritten as J(u(t)) + ‖u(t)‖22 ≤ (J(u0) + ‖u0‖22)e−2βt. (5.9) By (5.1) and (5.9), we obtain ‖∆u(t)‖22 ≤ 2q− q− − 2 J(u(t)) ≤ 2q− q− − 2 ( J(u0) + ‖u0‖22 ) e−2βt. The proof is complete. 6. Proof of Theorem 4.6 We consider following two cases by using different methods: Case 1: u0 ∈ H2 0 (Ω)\{0} with J(u0) ≤ 0. We define the function f(t) = ‖u(t)‖22, for all t ∈ [0, Tmax). By the definition of J and I, we have J(u(t)) ≥ 1 2 ‖∆u(t)‖22 + 1 p+ ∫ Ω |∇u(t)|p(x) dx− 1 q− ∫ Ω |u(t)|q(x) dx ≥ 1 max{2, p+} ( ‖∆u(t)‖22 + ∫ Ω |∇u(t)|p(x) dx ) − 1 q− ∫ Ω |u(t)|q(x) dx = ( 1 max{2, p+} − 1 q− )∫ Ω |u(t)|q(x) dx+ 1 max{2, p+} I(u(t)). Taking this fact into account and notice that J(u(t)) ≤ J(u0) ≤ 0, one has f ′(t) = −2I(u(t)) ≥ −2 max{2, p+}J(u(t)) + 2 ( 1− max{2, p+} q− )∫ Ω |u(t)|q(x) dx ≥ 2 ( 1− max{2, p+} q− )∫ Ω |u(t)|q(x) dx. (6.1) From this and q− > max{2, p+}, we obtain that f ′(t) ≥ 0 for all t ∈ [0, Tmax), which implies that f(t) ≥ f(0) = ‖u0‖22 > 0, for all t ∈ [0, Tmax). (6.2) 16 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 Then by (6.2), we can estimate ∫ Ω |u(t)|q(x) dx as follows:∫ Ω |u(t)|q(x) dx ≥ min { ‖u(t)‖q − q(·), ‖u(t)‖q + q(·) } ≥ min { S−q − q(·),2‖u(t)‖q − 2 , S−q + q(·),2‖u(t)‖q + 2 } ≥ min { S−q − q(·),2, S −q+ q(·),2 } min { ‖u(t)‖q − 2 , ‖u(t)‖q + 2 } = min { S−q − q(·),2, S −q+ q(·),2 } min { 1, f q+−q− 2 (t) } f q− 2 (t) ≥ min { S−q − q(·),2, S −q+ q(·),2 } min { 1, ‖u0‖q +−q− 2 } f q− 2 (t), (6.3) where Sq(·),2 is defined in (4.4). It follows from (6.1) and (6.3) that f ′(t) ≥ C0f q− 2 (t), (6.4) where C0 = 2 ( 1− max{2, p+} q− ) min { S−q − q(·),2, S −q+ q(·),2 } min { 1, ‖u0‖q +−q− 2 } > 0. Since f(t) > 0, dividing the inequality (6.4) by f q− 2 (t), we obtain f ′(t)f−q −/2(t) ≥ C0. Integrating the above inequality over [0, t], one has f1− q − 2 (t) ≤ f1− q − 2 (0)− (q− 2 − 1 ) C0t, for all t ∈ [0, Tmax). This and f1− q − 2 (t) > 0 imply t < 2 (q− − 2)C0 ‖u0‖2−q − 2 , for all t ∈ [0, Tmax). Thus, we obtain Tmax ≤ 2 (q− − 2)C0 ‖u0‖2−q − 2 = C max{‖u0‖2−q − 2 , ‖u0‖2−q + 2 }, where C is the constant given in (4.3). Case 2: 0 < J(u0) < d and I(u0) < 0. By contradiction, we assume that Tmax = ∞. Thanks to I(u0) < 0, by Lemma 4.4 we have I(u(t)) < 0. Then by Lemma 3.1 and 3.2, there exists λ∗ ∈ (0, 1) such that J(u(t))− 1 q− I(u(t)) ≥ d max { λ2 ∗, λ p− ∗ , λq + ∗ } > d, which implies that d dt ‖u(t)‖22 = −2I(u(t)) > 2q−(d− J(u(t))) ≥ 2q−(d− J(u0)). Then we have ‖u(t)‖22 = ‖u0‖22 + ∫ t 0 d ds ‖u(s)‖22 ds ≥ ‖u0‖22 + 2q−(d− J(u0))t. EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 17 From this and J(u0) < d, we obtain limt→∞ ‖u(t)‖22 = ∞. Hence, we can choose sufficiently large t0 > 0 such that ‖u(t0)‖22 > q− q− − 2 ‖u0‖22. Let T = ∫ t0 0 ‖u(s)‖22 ds( q− 2 − 1 ) (‖u(t0)‖22 − q− q−−2‖u0‖22) + t0 ≥ t0 > 0. (6.5) We now define the auxiliary function F : [0, T ]→ (0,∞) by F (t) = ∫ t 0 ‖u(s)‖22 ds+ (T − t)‖u0‖22. (6.6) Then F ′(t) = ‖u(t)‖22 − ‖u0‖22 = 2 ∫ t 0 〈u′(s), u(s)〉ds, and F ′′(t) = 2〈u′(t), u(t)〉 = −2I(u(t)) > 2q−(d− J(u(t))) = 2q−(d− J(u0)) + 2q− ∫ t 0 ‖u′(s)‖22 ds ≥ 2q− ∫ t 0 ‖u′(s)‖22 ds. (6.7) We deduce from (6.6) and (6.7) that F (t)F ′′(t) ≥ 2q− ∫ t 0 ‖u′(s)‖22 ds ∫ t 0 ‖u(s)‖22 ds. (6.8) On the other hand, by Cauchy-Schwarz inequality, we have∫ t 0 ‖u′(s)‖22 ds ∫ t 0 ‖u(s)‖22 ds ≥ (∫ t 0 〈u′(s), u(s)〉ds )2 = 1 4 (F ′(t))2. (6.9) Combining (6.8)–(6.9), we obtain F (t)F ′′(t) ≥ q− 2 (F ′(t))2, for all t ∈ [0, T ]. (6.10) Setting G(t) = F 1− q − 2 (t), we obtain G′(t) = ( 1− q− 2 ) F ′(t) F q− 2 (t) , G′′(t) = ( 1− q− 2 )F (t)F ′′(t)− q− 2 (F ′(t))2 F 1+ q− 2 (t) . Then we have G′′(t) ≤ 0, for all t ∈ [0, T ] due to (6.10). Thus, G(t) is concave on [0, T ]. This implies that G(t) ≤ G(t0) +G′(t0)(t− t0), for all t ∈ [0, T ]. Replacing t by T in the above inequality and notice that (6.5), we obtain G(T ) ≤ G(t0) +G′(t0)(T − t0) = F− q− 2 (t0)[F (t0)− (q− 2 − 1 ) (T − t0)F ′(t0)] = 0. This contradicts G(T ) > 0, and the proof is complete. 18 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 7. Proof of Theorem 4.8 Assume that u = u(t) is a global weak solution to (1.1). Then by (i) in Theorem 4.6, we obtain J(u(t)) ≥ 0 for all t ≥ 0. Therefore,∫ t 0 ‖u′(s)‖22 ds = J(u0)− J(u(t)) ≤ J(u0). Letting t→∞, one has ∫ ∞ 0 ‖u′(s)‖22 ds ≤ J(u0) <∞. Therefore, there exists a sequence {tn} with tn →∞ as n→∞ such that lim n→∞ ‖u′(tn)‖2 = 0, (7.1) which implies that ‖u′(tn)‖2 ≤ A for all n ∈ N, for some a constant A. Then |I(u(tn))| = |〈u′(tn), u(tn)〉| (7.2) ≤ ‖u′(tn)‖2‖u(tn)‖2 (7.3) ≤ ‖u′(tn)‖2S2‖∆u(tn)‖2 (7.4) ≤ AS2‖∆u(tn)‖2, (7.5) where S2 is the constant given in (3.16). Using the non-increasing property of J(u(t)), (7.5) and replacing u by u(tn) in (3.2), we obtain J(u0) ≥ J(u(tn)) ≥ (1 2 − 1 q− ) ‖∆u(tn)‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u(tn)|p(x) dx+ 1 q− I(u(tn)) ≥ (1 2 − 1 q− ) ‖∆u(tn)‖22 − AS2 q− ‖∆u(tn)‖2, which implies that ‖∆u(tn)‖2 ≤ AS2 + √ A2 2S 2 2 + 2q−(q− − 2)J(u0) q− − 2 . (7.6) The above inequality ensures that {u(tn)} is bounded in H2 0 (Ω). Then, since H2 0 (Ω) is reflexive, H2 0 (Ω) ↪→ W 1,p(·) 0 (Ω) and H2 0 (Ω) ↪→ Lq(·)(Ω) are compact embeddings (by (1.2) and (1.3)), there exists a sub-sequence of {u(tn)}, still denoted by {u(tn)} and a φ ∈ H2 0 (Ω) such that u(tn) ⇀ φ weakly in H2 0 (Ω), (7.7) u(tn)→ φ strongly in W 1,p(·) 0 (Ω), (7.8) u(tn)→ φ strongly in Lq(·)(Ω). (7.9) For any v ∈ H2 0 (Ω). Replacing u by u(tn) in the equation (1.1), by multiplying (1.1) by v and integrating by parts, we have |〈∆u(tn),∆v〉+ 〈|∇u(tn)|p(x)−2∇u(tn),∇v〉 − 〈|u(tn)|q(x)−2u(tn), v〉| = |〈u′(tn), v〉| ≤ ‖u′(tn)‖2‖v‖2. From this and (7.1), it follows that lim n→∞ ( 〈∆u(tn),∆v〉+ 〈|∇u(tn)|p(x)−2∇u(tn),∇v〉 − 〈|u(tn)|q(x)−2u(tn), v〉 ) = 0, EJDE-2022/46 GLOBAL SOLUTIONS TO THE CAHN-HILLIARD EQUATION 19 which, together with (7.7), (7.8) and (7.9) yields φ ∈ S. (7.10) By (7.1), (7.4) and (7.6), we obtain lim n→∞ I(u(tn)) = 0, which, together with (7.8), (7.9) and (7.10), implies lim n→∞ ‖∆u(tn)‖2 = − lim n→∞ ∫ Ω |∇u(tn)|p(x) dx+ lim n→∞ ∫ Ω |u(tn)|q(x) dx = − ∫ Ω |∇φ|p(x) dx+ ∫ Ω |φ|q(x) dx = ‖∆φ‖2. (7.11) Note that H2 0 (Ω) is uniformly convex. Then by (7.7) and (7.11), we imply (see [1, Proposition 3.32]) u(tn)→ φ strongly in H2 0 (Ω). The proof is complete. 8. Proof of Theorems 4.9 and 4.10 By borrowing the ideas from [9, 28] we can prove the Theorem 4.9 and 4.10 as follows. Proof of Theorem 4.9. Assume that u0 ∈ N+ and ‖u0‖2 ≤ λJ(u0). We first prove that u(t) ∈ N+ for all t ∈ [0, Tmax). Indeed, assume on the contrary that there is t0 > 0 such that u(t) ∈ N+ for all t ∈ [0, t0) and u(t0) ∈ N . Then for all t ∈ [0, t0), we have 0 < |I(u(t))| = |〈u′(t), u(t)〉| ≤ ‖u′(t)‖2‖u(t)‖2, which gives ‖u′(t)‖2 > 0. From this and (4.2), we obtain J(u(t0)) < J(u0), i.e., u(t0) ∈ JJ(u0). So ‖u(t0)‖2 ≥ λJ(u0). On the other hand, for all t ∈ [0, t0), we have d dt ‖u(t)‖22 = −2I(u(t)) < 0, which implies that ‖u(t0)‖2 < ‖u0‖2 ≤ λJ(u0). We thus arrive at a contradiction and therefore it proves the claim u(t) ∈ N+ for all t ∈ [0, Tmax). This gives u(t) ∈ N+ ∩ JJ(u0) for all t ∈ [0, Tmax) by using the strictly decreasing property of J(u(t)). Then by (ii) in Lemma 3.5, u(t) remains bounded in H2 0 (Ω) for all t ∈ [0, Tmax) so that Tmax = ∞, i.e., u0 ∈ G. We next prove u0 ∈ G0. Let any w ∈ ω(u0), we obtain ‖w‖2 < λJ(u0) and J(w) < J(u0), which implies that ω(u0) ∩ N = ∅ by definition of λJ(u0). And therefore, ω(u0) = {0}, i.e., u0 ∈ G0. Now we assume that u0 ∈ N− and ‖u0‖2 ≥ ΛJ(u0). By analogous arguments as above, we also have u(t) ∈ N− for all t ∈ [0, Tmax). Assume on the contrary that Tmax =∞, then for every w ∈ ω(u0), one has ‖w‖2 > ΛJ(u0) and J(w) < J(u0), which gives ω(u0)∩N = ∅ by the definition of ΛJ(u0). However, since dist(0,N−) > 0, we also have 0 /∈ ω(u0). And hence ω(u0) = ∅, which contradicts Tmax = ∞. Thus u0 ∈ B. The proof is complete. � 20 Q. V. CHUONG, L. C. NHAN, L. X. TRUONG EJDE-2022/46 Proof of Theorem 4.10. Let any u ∈ H2 0 (Ω)\{0}. By using (3.2), we obtain J(u) ≥ (1 2 − 1 q− ) ‖∆u‖22 + ( 1 p+ − 1 q− ) ∫ Ω |∇u|p(x) dx+ 1 q− I(u) > (1 2 − 1 q− ) ‖∆u‖22 + 1 q− I(u) ≥ (1 2 − 1 q− ) S−2 2 ‖u‖22 + 1 q− I(u). (8.1) where S2 is defined in (3.16). Replacing u by u0 in (8.1) and using (4.5), we obtain J(u0) > (1 2 − 1 q− ) S−2 2 ‖u0‖22 + 1 q− I(u0) ≥ J(u0) + 1 q− I(u0), which gives I(u0) < 0, i.e., u0 ∈ N−. (8.2) For any u ∈ NJ(u0), we have I(u) = 0 and J(u) < J(u0). Then by using (8.1), we obtain ‖u‖22 ≤ 2q−S2 2 q− − 2 J(u0), which, together with (4.5), implies ‖u‖2 ≤ ‖u0‖2. 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Yang; Upper bound estimate for the blow-up time of an evolution m-Laplace equa- tion involving variable source and positive initial energy, Comput. Math. Appl., 69 (2015), 1463–1469. Quach V. Chuong Faculty of Mathematics and Computer Science, University of Science, Ho Chi Minh City, Vietnam. Vietnam National University, Ho Chi Minh City, Vietnam. Department of Mathematics, Dong Nai University, Bien Hoa City, Dong Nai Province, Vietnam Email address: quachuong1812@dnpu.edu.vn Le C. Nhan (corresponding author) Faculty of Applied Sciences, Ho Chi Minh City University of Technology and Education (UTE), Ho Chi Minh City, Vietnam Email address: nhanlc@hcmute.edu.vn Le X. Truong Department of Mathematics, Univerisity of Economics Ho Chi Minh city (UEH), Ho Chi Minh city, Vietnam Email address: lxuantruong@ueh.edu.vn 1. Introduction 2. Preliminaries 3. Stationary problem and potential wells 4. Evolution problem 5. Proof of Theorem 4.5 6. Proof of Theorem 4.6 7. Proof of Theorem 4.8 8. Proof of Theorems 4.9 and 4.10 Acknowledgments References