A FOURTH-ORDER DEGENERATE PARABOLIC ...

A FOURTH-ORDER DEGENERATE PARABOLIC EQUATION WITH VARIABLE EXPONENT

Chao Zhang and Shulin Zhou LMAM, School of Mathematical Sciences, Peking University Beijing 100871, P. R. CHINA

Abstract. In this paper we establish the existence, uniqueness and long-time behavior of weak solutions for the initial-boundary value problem of a fourthorder degenerate parabolic equation with variable exponent of nonlinearity.

1. Introduction Suppose that Ω is a bounded open domain of RN with a C 1,1 boundary ∂Ω, T is a given positive number. Denote the cylinder Q ≡ Ω × (0, T ], the lateral surface Γ ≡ ∂Ω × (0, T ]. In this paper, we consider the following fourth-order parabolic equation ∂u + div |∇∆u|p(x)−2 ∇∆u = 0, (x, t) ∈ Q, (1) ∂t where p : Ω → (1, +∞) is a continuous function (called the variable exponent). Based on the physical consideration, as usual Eq. (1) is supplemented with the natural boundary conditions u Γ = ∆u Γ = 0 (2) and the initial value condition u(x, 0) = u0 (x),

x ∈ Ω.

(3)

When p > 1 is a constant, J. R. King in [19] firstly derived the equation (1) which is relevant to capillary driven flows of thin films of power-law fluids, where u(x, t) denotes the height from the surface of the oil to the surface of the solid. The exponent p is related to the rheological properties of the liquid: p = 2 corresponds to a Newtonian liquid, whereas p 6= 2 emerges when considering “power-law” liquids. When p > 2 the liquid is said to be shear-thinning. The existence, uniqueness and qualitative properties of solutions which are related to Eq. (1) with constant exponent have been studied in many papers. We refer to [4, 5, 19, 22, 27] for details. The study of differential equations and variational problems with nonstandard growth conditions arouses much interest with the development of elastic mechanics, electro-rheological fluid dynamics and image processing, etc. We refer the readers to [8, 25, 26, 30] and references therein. p(x)-growth conditions can be regarded as a very important class of nonstandard (p, q)-growth conditions. There are already numerous results for such kind of problems (see [1, 2, 3, 6, 14, 15, 16, 23]). The Key words and phrases. Variable exponents, Fourth-order, Existence, Uniqueness, Long-time behavior. This work was supported in part by the NBRPC under Grant 2006CB705700, the NSFC under Grant 60532080, and the KPCME under Grant 306017. 1

2

CHAO ZHANG AND SHULIN ZHOU

functional spaces to deal with these problems are the generalized Lebesgue space Lp(x) (Ω) and the generalized Lebesgue-Sobolev space W k,p(x) (Ω). In [27], Xu and Zhou have established the existence, uniqueness and some results on the regularity of weak solutions when p(x) is a constant. This paper is a further step to extend the constant p to the variable exponent p(x). As far as we know, there are few papers concerned with the fourth-order nonlinear parabolic equation involving multiple anisotropic exponents. It is not a trivial generalization of similar problems studied in the constant case. So, the study of Eq. (1) is a new and interesting topic. The main difficulties for treating the problem are caused by the complicated nonlinearities (it is nonhomogeneous) of Eq. (1) and the lack of a maximum principle for fourth-order equations. In this paper, a key observation is an energy type estimate (9), which is the starting point of our arguments. Under the appropriate definition of weak solutions, we first combine the difference and variation techniques to construct an approximation solution sequence for problem (1)–(3) and establish some a priori estimates. Next, we draw a subsequence to obtain a limit function, and prove this function is a weak solution. For the convenience of the readers, we first recall some definitions and basic properties of the generalized Lebesgue space Lp(x) (Ω) and the generalized LebesgueSobolev space W k,p(x) (Ω). Roughly speaking, anisotropic Lebesgue and Sobolev spaces are functional spaces of Lebesgue’s and Sobolev’s type in which different space directions play different roles. Set C+ (Ω) = {h ∈ C(Ω) : minx∈Ω h(x) > 1}. For any h ∈ C+ (Ω) we define h+ = sup h(x)

and h− = inf h(x). x∈Ω

x∈Ω

For any p ∈ C+ (Ω), we introduce the variable exponent Lebesgue space Lp(·) (Ω) to consist of all measurable functions such that Z |u(x)|p(x) dx < ∞, Ω

endowed with the Luxemburg norm Z u(x) p(x) dx ≤ 1 , |u|p(·) = inf λ > 0 : λ Ω which is a separable and reflexive Banach space. The dual space of Lp(x) (Ω) is 0 Lp (x) (Ω), where 1/p(x) + 1/p0 (x) = 1. If p is a constant function, then the variable exponent Lebesgue space coincides with the classical Lebesgue space and so the notation will give rise to no confusion. The variable exponent Lebesgue space is a special case of Orlicz-Musielak spaces treated by Musielak in [24]. For any positive integer k, set W k,p(x) (Ω) = {u ∈ Lp(x) (Ω) : Dα u ∈ Lp(x) (Ω), |α| ≤ k}. We can define the norm on W k,p(x) (Ω) by X kukW k,p(x) = |Dα u|p(x) |α|≤k k,p(x)

and W (Ω) also becomes a Banach space. We call it generalized LebesgueSobolev space, it is a special generalized Orlicz-Sobolev space. An interesting feature of generalized Lebesgue-Sobolev space is that smooth functions are not dense in it without additional assumptions on the exponent p(·). This was observed by

DEGENERATE PARABOLIC EQUATION WITH VARIABLE EXPONENT

3

Zhikov [29] in connection with Lavrentiev phenomenon. However, when the exponent satisfies the log-H¨ older continuity condition, i.e., there is a positive constant C such that −|p(x) − p(y)| log |x − y| ≤ C

(4)

for all x, y ∈ Ω, then smooth functions C ∞ (Ω) are dense in variable exponent Sobolev space W k,p(x) (Ω) and ˙ k,p(x) (Ω) := W k,p(x) (Ω) ∩ W k,1 (Ω) = W k,p(x) (Ω) W 0 0 (see [15]). Moreover (4) implies that |p(x) − p(y)| ≤

C − log |x − y|

(5)

for every x, y ∈ Ω with |x − y| ≤ 1/2, then there is no confusion in defining the 1,p(·) Sobolev space with zero boundary values, W0 (Ω), as the completion of C0∞ (Ω) with respect to the norm kukW 1,p(·) (see [17]). Under the condition (5), if Ω is bounded, the Hardy-Littlewood maximal operator is bounded from Lp(·) (Ω) to itself (see [11]). In this paper we assume that u0 ∈ H01 (Ω).

(6)

Because of the degeneracy, problem (1)–(3) does not admit classical solutions in general. So, we introduce weak solutions in the following sense. Definition 1.1. A function u is said to be a weak solution of problem (1)–(3) if the following conditions are satisfied: − (i) u ∈ C([0, T ]; L2 (Ω)) ∩ L∞ (0, T ; H01 (Ω)) ∩ Lp 0, T ; W 2,p(x) (Ω) with ∆u ∈ N − 1,p(x) Lp 0, T ; W0 (Ω) and ∇∆u ∈ Lp(x) (Q) ; (ii) For any ϕ ∈ C 1 (Q) with ϕ(·, T ) = 0, we have Z Z TZ − u0 (x)ϕ(x, 0) dx − [uϕt + |∇∆u|p(x)−2 ∇∆u · ∇ϕ] dxdτ = 0. Ω

0

(7)

Ω

Remark 1. Let u be a weak solution of problem (1)–(3). If p(x) satisfies the log-H¨ older continuity condition (4), then u ∈ W 1,p(x) (Ω) ∩ H01 (Ω) ⊂ W 1,p(x) (Ω) ∩ ˙ 1,p(x) (Ω) = W 1,p(x) (Ω) (see [15]). By using the approxiW01,1 (Ω) and thus u ∈ W 0 mation technique (see [9], Chapter 3 or [10], Chapter 2) we have, for each t ∈ [0, T ] and every ϕ ∈ C 1 (Q), Z t Z tZ uϕ dx − [uϕt + |∇∆u|p(x)−2 ∇∆u · ∇ϕ] dxdτ = 0. (8) Ω

0

0

C0∞ (Ω)

Ω

1,p(x)

Remark 2. Since is dense in W0 (Ω) due to the log-Hölder continuity condition (4), we can choose ∆u as a test function in (7) and (8). Indeed we may use the Steklov averages Z 1 t+h [v]h (x, t) = v(x, τ ) dτ h t

4


of the function v(x, t) to replace the corresponding function, and then pass to the limits. Therefore, we obtain from (8) an energy type estimate Z tZ 1 1 |∇∆u|p(x) dxdτ = k∇u0 k2L2 (Ω) . (9) k∇u(t)k2L2 (Ω) + 2 2 0 Ω Next, we state our main results as follows. Theorem 1.2. Under assumption (6), let p(x) ∈ C+ (Ω), p(x) satisfies the logH¨ older continuity condition (4). Then the initial-boundary value problem (1)–(3) admits a unique weak solution. Theorem 1.3. Let u be a weak solution of problem (1)–(3). Assume that p(x) − satisfies the log-H¨ older continuity condition (4). Then, when max{1, N2N +4 } < p ≤ N p+ < 2 , we have lim ku(t)kH01 (Ω) = 0. t→+∞

The rest of this paper is organized as follows. In Section 2, we state some basic results for the generalized Lebesgue space Lp(x) (Ω) and the generalized LebesgueSobolev space W k,p(x) (Ω). We will prove the main results in Section 3 and Section 4. 2. Preliminaries In this section, we present some elementary results for the generalized Lebesgue space Lp(x) (Ω) and the generalized Lebesgue-Sobolev space W k,p(x) (Ω). The basic properties of these spaces can be found from the paper of O. Kováˇcik and J. R´ akosn´ık [20], and many of these properties were independently established by Fan and Zhao [15]. In the following sections C will represent a generic constant that may change from line to line even if in the same inequality. Lemma 2.1 ([15, 20]). (1) The space Lp(·) (Ω), | · |p(·) is a separable, uniform convex Banach space, and its conjugate space is Lq(·) (Ω), where

1 p(·)

1 + q(·) = 1. For

any u ∈ Lp(·) (Ω) and v ∈ Lq(·) (Ω), we have Z 1 1 uv dx ≤ − + − |u|p(x) |v|q(x) ≤ 2|u|p(x) |v|q(x) ; p q Ω (2) If p1 , p2 ∈ C+ (Ω), p1 (x) ≤ p2 (x) for any x ∈ Ω, then there exists the continuous embedding Lp2 (x) (Ω) ,→ Lp1 (x) (Ω), whose norm does not exceed |Ω| + 1. Lemma 2.2 ([15]). If we denote Z ρ(u) = |u|p(x) dx

∀ u ∈ Lp(x) (Ω),

Ω

then the following relations hold: 1) |u|p(x) < 1(= 1; > 1) ⇔ ρ(u) < 1(= 1; > 1); −

+

−

+

2) |u|p(x) > 1 ⇒ |u|pp(x) ≤ ρ(u) ≤ |u|pp(x) ; |u|p(x) < 1 ⇒ |u|pp(x) ≥ ρ(u) ≥ |u|pp(x) ; 3) |u|p(x) → 0 ⇔ ρ(u) → 0; |u|p(x) → ∞ ⇔ ρ(u) → ∞. Lemma 2.3 ([15]). W k,p(x) (Ω) is a separable and reflexive Banach space.


5

1,p(·)

Lemma 2.4 ([18, 20]). For u ∈ W0 (Ω) with p ∈ C+ (Ω), the p(·)-Poincaré inequality |u|p(x) ≤ C|∇u|p(x) holds, where the positive constant C depends on p and Ω. Remark 3. Note that the following inequality Z Z p(x) |u| dx ≤ C |∇u|p(x) dx Ω

Ω

in general does not hold. Lemma 2.5 ([16]). Let Ω be an open domain (that may be unbounded) in RN with cone property. If p : Ω → R is Lipschitz continuous function satisfying 1 < p− ≤ p+ < N k and q(x) : Ω → R is measurable and satisfies p(x) ≤ q(x) ≤ pe(x) :=

N p(x) , N − kp(x)

a.e. x ∈ Ω,

then there is a continuous embedding W k,p(x) (Ω) ,→ Lq(x) (Ω). Remark 4. Actually, the Lipschitz continuity of p(x) in Lemma 2.5 has been weekend to the log-H¨ older continuity condition (4) (see [12], Corollary 5.3). Let Ω ⊂ RN be an open, bounded set with Lipschitz boundary and p is a bounded exponent on Ω with 1 < p− ≤ p+ < N which satisfies the log-Hölder continuity condition 1 1 1 ∗ (4). Define ∗ = − . Then W 1,p(x) (Ω) ,→ Lp (x) (Ω) continuously. p (x) p(x) N Next, we prove a lemma which claims the W 2,p(x) -norm of a function in W 2,p(x) ∩ can be controlled by the Lp(x) -norm of its Laplacian.

1,p(x) W0

Lemma 2.6. Let Ω be a bounded open domain of RN with a C 1,1 boundary ∂Ω. Suppose that p(x) ∈ C+ (Ω) and satisfies the log-H¨ older continuity condition (4). Then there exists a positive constant C depending only on p, N, Ω such that, for 1,p(x) every v ∈ W 2,p(x) (Ω) ∩ W0 (Ω), kvkW 2,p(x) ≤ C|∆v|p(x) .

(10)

Proof. Recalling the definition of norm of W 2,p(x) (Ω), we have X |Dα v|p(x) . kvkW 2,p(x) = |v|p(x) + |∇v|p(x) + |α|=2 1,p(x)

From v ∈ W0

(Ω) and the p(·)-Poincaré inequality, we get |v|p(x) ≤ C|∇v|p(x)

and X kvkW 2,p(x) ≤ C |∇v|p(x) + |Dα v|p(x) .

(11)

|α|=2

Since p(x) satisfies the log-Hölder continuity condition (4) and ∂Ω is C 1,1 , from 1,p(x) the proof of Theorem 4.4 in [28], we know that for any v ∈ W 2,p(x) (Ω)∩W0 (Ω), X |Dα v|p(x) ≤ C|∆v|p(x) . (12) |α|=2

6


Combining (12) with (11), there exists a positive constant C such that kvkW 2,p(x) ≤ C |∇v|p(x) + |∆v|p(x) .

(13)

In order to prove (10), we only need to show that kvkW 1,p(x) ≤ C|∆v|p(x) .

(14) 1,p(x)

2,p(x) If (14) is violated, then there exists a sequence {vn }∞ (Ω) ∩ W0 n=1 ⊂ W such that

kvn kW 1,p(x) > n|∆vn |p(x) .

(Ω)

(15)

Without loss of generality, we assume that kvn kW 1,p(x) = 1.

(16)

Then it follows from (13) and (15) that 1 . n 2,p(x) We draw a subsequence (we still denote it by {vn }∞ (Ω)∩ n=1 ) and a function v ∈ W 1,p(x) W0 (Ω) such that vn * v weakly in W 2,p(x) (Ω), which implies that 1,p(x) vn → v strongly in W0 (Ω). Therefore, we deduce from (16) that kvn kW 2,p(x) ≤ C,

|∆vn |p(x) ≤

kvkW 1,p(x) = 1.

(17)

On the other hand, by the weak convergence of D2 vn we know that |∆v|p(x) ≤ lim inf |∆vn |p(x) = 0, n→∞

−

1,p(x)

which implies that ∆v = 0. As we know that v ∈ W0 (Ω) ⊂ W01,p (Ω), we conclude that v = 0 a.e. in Ω. This is a contradiction to (17). Thus (14) holds and then (10) is true. 3. Existence and uniqueness In this section, we are going to prove the existence and uniqueness of weak solutions. Let n be a positive integer. Now we let h = T /n be a sequence of time steps and consider the following elliptic problems   uk − uk−1 + div |∇∆u |p(x)−2 ∇∆u = 0, in Ω, k k (18) h u = ∆u = 0, k = 1, 2, . . . , n, on ∂Ω. k k We first consider the existence of weak solutions of the following elliptic problem   u − u0 + div |∇∆u|p(x)−2 ∇∆u = 0, in Ω, (19) h u = ∆u = 0, k = 1, 2, . . . , n, on ∂Ω, with h > 0 and u0 ∈ H01 (Ω), which is the case of (18) when k = 1. We introduce the space 1,p(x)

W = {u ∈ H01 (Ω) ∩ W 2,p(x) (Ω)|∆u ∈ W0

(Ω)}


7

with the norm kukW = kukH01 (Ω) + kukW 2,p(x) (Ω) + k∆ukW 1,p(x) (Ω) . 0

It is easy to verify that W is a Banach space. Definition 3.1. A function u ∈ W is called a weak solution of problem (19), if for 1,p(x) (Ω), we have any φ ∈ C 1 (Ω) ∩ W0 Z Z u − u0 φ dx − (|∇∆u|p(x)−2 ∇∆u) · ∇φ dx = 0. (20) h Ω Ω Theorem 3.2. There exists at least one weak solution for problem (19). Proof. Consider the variation problem min{J(u)|u ∈ W } and functional J defined by Z Z Z 1 1 2 p(x) |∇u| dx + |∇∆u| dx − f ∆u dx, J(u) = 2h Ω Ω p(x) Ω where f ∈ H01 (Ω) is a known function. We will establish that J(u) has a minimizer u1 (x) in W . Using the Young’s inequality, there exists a constant C = C(h) > 0, such that Z Z Z 1 1 p(x) 2 |∇∆u| dx + |∇u| dx − C |∇f |2 dx. J(u) ≥ + p 4h Ω Ω Ω Furthermore, we need to check that J(u) satisfies the coercive condition. Recall1,p(x) ing u ∈ W0 (Ω) ∩ W 2,p(x) (Ω) and Lemma 2.6, we have kukW 2,p(x) ≤ C|∆u|p(x) . Again by ∆u ∈

1,p(x) W0 (Ω)

and the p(·)-poincaré inequality, we get |∆u|p(x) ≤ C|∇∆u|p(x) .

Therefore kukW ≤ C kukH01 (Ω) + |∇∆u|p(x) .

(21)

According to Lemma 2.2, we have J(u) → +∞, if kukW → +∞. On the other hand, J(u) is clearly weakly lower semicontinuous on W . So it follows from the critical point theory in [7] that there exists u1 ∈ W , such that J(u1 ) = inf J(w). w∈W

1 h u0 ,

Therefore, by choosing f = the function u1 is a weak solution of the corresponding Euler-Lagrange equation of J(u), which is problem (19). The proof is complete. Proof of Theorem 1.2. First we construct an approximation solution sequence {uh } for problem (1)–(3). When k = 1, it implies from Theorem 3.2 that there is a weak solution u1 ∈ W . Following the same procedures, we find weak solutions uk ∈ W of (18), k = 2, . . . , n. It follows that, for every η ∈ W , Z Z 1 ∇(uk − uk−1 ) · ∇η dx + |∇∆uk |p(x)−2 ∇∆uk · ∇∆η dx = 0. (22) h Ω Ω

8


Next, we take η = uk as a test function in (22) to obtain a priori estimate for the function uk that Z 1 1 k∇uk k2L2 (Ω) + h (23) |∇∆uk |p(x) dx ≤ k∇uk−1 k2L2 (Ω) . 2 2 Ω Thus, we obtain k∇uk k2L2 (Ω) ≤k∇u0 k2L2 (Ω) . Now for every h = T /n, we define  u0 (x),     u1 (x),    · · · · · · , uh (x, t) =  uj (x),      ······ ,    un (x),

(24)

t = 0, 0 < t ≤ h, ······ , (j − 1)h < t ≤ jh, ······ , (n − 1)h < t ≤ nh = T.

(25)

For each t ∈ (0, T ], there exists some k = {1, . . . , n} such that t ∈ ((k − 1)h, kh]. Thus, recalling (24) we have k∇uh k2L2 (Ω) ≤ k∇u0 k2L2 (Ω) . Therefore, we conclude that k∇uh kL∞ (0,T ;L2 (Ω)) ≤ C,

(26)

where C is a positive constant. Summing up inequalities in (23), we have Z TZ 0

|∇∆uh |p(x) dxdt = h

Ω

n Z X k=1

|∇∆uk |p(x) dx ≤

Ω

1 k∇u0 k2L2 (Ω) ≤ C. 2

p(x)−2

Define w = |∇∆uh | ∇∆uh . Recalling Lemma 2.2, we have Z T Z TZ + − min{|∇∆uh |pp(x) , |∇∆uh |pp(x) } dt ≤ |∇∆uh |p(x) dxdt ≤ C 0

0

Ω

and Z

T

(p0 )−

(p0 )+

min{|w|p0 (x) , |w|p0 (x) } dt

0

Z TZ ≤ 0

0

|w|p (x) dxdt =

Ω

Z TZ 0

|∇∆uh |p(x) dxdt ≤ C.

Ω

Thus, we conclude that N ∇∆uh ∈ Lp(x) (Q) ,

N 0 w ∈ Lp (x) (Q) ,

k∇∆uh kLp− (0,T ;Lp(x) (Ω)) ≤ C, kwkL(p0 )− (0,T ;Lp0 (x) (Ω)) ≤ C.

(27)

Employing the same technique as in the proof of (21), we obtain kuh kL∞ (0,T ;H01 (Ω)) + kuh kLp− (0,T ;W 2,p(x) (Ω)) + k∆uh kLp− (0,T ;W 1,p(x) (Ω)) ≤ C. (28) 0

Therefore, from (27) and (28), we may choose a subsequence (we also denote it by the original sequence for simplicity) such that uh * u,

weakly-* in L∞ (0, T ; H01 (Ω)),


9

−

weakly in Lp (0, T ; W 2,p(x) (Ω)), N ∇∆uh * ∇∆u, weakly in Lp(x) (Q) , uh * u,

|∇∆uh |p(x)−2 ∇∆uh * ξ,

0 −

weakly in L(p )

0 0, T ; Lp (x) (Ω) ,

(29)

which follows that (see [13]) kukL∞ (0,T ;H01 (Ω)) + kukLp− (0,T ;W 2,p(x) (Ω)) + k∆ukLp− (0,T ;W 1,p(x) (Ω)) ≤ C.

(30)

0

Next, we prove that the function u is a weak solution of problem (1)–(3). For each ϕ ∈ C 1 (Q) with ϕ(·, T ) = 0 and ϕ(x, t)|Γ = 0 and for every k ∈ {1, 2, . . . , n}, we solve the equation −∆ηk (x) = ϕ(x, kh) to find a function ηk ∈ W and let it be a test function in (22) to have Z Z 1 1 uk (x)ϕ(x, kh) dx − uk−1 (x)ϕ(x, kh) dx h Ω h Ω Z − (|∇∆uk |p(x)−2 ∇∆uk )(x) · ∇ϕ(x, kh) dx = 0. Ω

Summing up all the equations and recalling the definition of uh (x, t) and ϕ(·, T ) = ϕ(·, nh) = 0, we have Z n−1 XZ ϕ(x, kh) − ϕ(x, (k + 1)h) h uh (x, kh) dx − u0 (x)ϕ(x, h) dx h Ω k=1 Ω n Z X (|∇∆uh |p(x)−2 ∇∆uh )(x, kh) · ∇ϕ(x, kh) dx = 0. (31) −h k=1

Ω

Recalling (29) and ϕ ∈ C 1 (Q), we have n Z X (|∇∆uh |p(x)−2 ∇∆uh )(x, kh) · ∇ϕ(x, kh) dx h Ω

k=1

Z TZ

(|∇∆uh |p(x)−2 ∇∆uh )(x, τ ) · ∇ϕ(x, τ ) dxdτ

= 0

+

Ω

n Z X

kh

Z

(|∇∆uh |p(x)−2 ∇∆uh )(x, τ )

(k−1)h Ω

k=1

· (∇ϕ(x, kh) − ∇ϕ(x, τ )) dxdτ Z TZ → ξ · ∇ϕ(x, τ ) dxdτ, as h → 0. 0

Ω

It follows from (31) that Z Z TZ Z TZ ∂ϕ dxdτ − u0 (x)ϕ(x, 0) dx − ξ · ∇ϕ dxdτ = 0. − u ∂t Ω 0 Ω 0 Ω Now, we choose ϕ ∈ C0∞ (Q) to have Z TZ Z TZ ∂ϕ − u dxdτ = ξ · ∇ϕ dxdτ, ∂t 0 Ω 0 Ω which implies that 0 − 0 ∂u ∈ L(p ) 0, T ; W −1,p (x) (Ω) . ∂t

(32)

(33)

10

CHAO ZHANG AND SHULIN ZHOU 0

Thus we can find a large integer s such that W −1,p (x) (Ω) ⊂ H −s (Ω), and then obtain 0 − ∂u ∈ L(p ) (0, T ; H −s (Ω)), ∂t which follows (see [31]) that u ∈ C([0, T ]; H −s (Ω)). For each ε > 0 and all t, t0 ∈ [0, T ], by (30) there exists a positive number δ > 0 such that ε δk∇u(t) − ∇u(t0 )kL2 (Ω) ≤ . 2 From the compact imbedding relation H01 (Ω) ,→ L2 (Ω) ,→ H −s (Ω), we have, for all t, t0 ∈ [0, T ], ku(t) − u(t0 )kL2 (Ω) ≤ δku(t) − u(t0 )kH01 (Ω) + C(δ)ku(t) − u(t0 )kH −s (Ω) ≤ δk∇u(t) − ∇u(t0 )kL2 (Ω) + C(δ)ku(t) − u(t0 )kH −s (Ω) ε ≤ + C(δ)ku(t) − u(t0 )kH −s (Ω) , 2 where the first inequality is guaranteed by Lemma 5.1 in Chapter 1 of [21]. It follows from the above inequalities that u ∈ C([0, T ]; L2 (Ω)). Therefore, the function u satisfies condition (i) and (ii) in Definition 1.1. We only need to show that ξ = |∇∆u|p(x)−2 ∇∆u, a.e. in Q. Now taking ∆u as a test function in (32), we have k∇u0 k2L2 (Ω) − k∇u(T )k2L2 (Ω) Z TZ − ξ · ∇∆u dxdτ = 0. (34) 2 0 Ω Denote Au = |∇∆u|p(x)−2 ∇∆u. Recalling an elementary inequality (|ξ|p−2 ξ − |η|p−2 η)(ξ − η) ≥ 0, for all ξ, η ∈ RN , we have that Z (Auk − Av(τ )) · (∇∆uk − ∇∆v(τ )) dx ≥ 0,

(35)

Ω

− for each k = 1, 2, . . . , n and every v ∈ Lp 0, T ; W 2,p(x) (Ω) with ∆v ∈ − 1,p(x) (Ω) . Choosing uk as a test function in (22), we have Lp 0, T ; W0 Z Z 1 − ∇(uk − uk−1 ) · ∇uk dx − Auk · ∇∆uk dx = 0. h Ω Ω In view of (35) we obtain Z Z 1 − ∇(uk − uk−1 ) · ∇uk dx − Auk · ∇∆v(τ ) dx h Ω Ω Z − Av · (∇∆uk − ∇∆v(τ )) dx ≥ 0. Ω

(36)


11

Since Z −

Z

Z

∇(uk − uk−1 ) · ∇uk dx = Ω

∇uk−1 · ∇uk dx − Ω

≤

k∇uk−1 k2L2 (Ω) −

∇uk Ω k∇uk k2L2 (Ω)

· ∇uk dx

, 2 by integrating inequality (36) over ((k − 1)h, kh) we have k ∇uk−1 k2L2 (Ω) − k ∇uk k2L2 (Ω) Z kh Z − Auk · ∇∆v dxdτ 2 (k−1)h Ω Z kh Z − Av · (∇∆uk − ∇∆v) dxdτ ≥ 0. (k−1)h Ω

Summing up the above inequality for k = 1, 2, . . . , n, we obtain k ∇u0 k2L2 (Ω) − k ∇uh (T ) k2L2 (Ω) Z TZ − Auh · ∇∆v dxdτ 2 0 Ω Z TZ − Av · (∇∆uh − ∇∆v) dxdτ ≥ 0. 0

Ω

Passing to limits as h → 0 and noting that k ∇u(T ) kL2 (Ω) ≤ lim inf k ∇u(T ) kL2 (Ω) , h→0

we obtain k ∇u0 k2L2 (Ω) − k ∇u(T ) k2L2 (Ω) 2

Z TZ −

ξ · ∇∆v dxdτ 0

Ω

Z TZ −

Av · (∇∆u − ∇∆v) dxdτ ≥ 0. 0

(37)

Ω

Combining (37) with (34), we have Z TZ (ξ − Av) · (∇∆u − ∇∆v) dxdτ ≥ 0. 0

Ω

N Next we choose v = u − λw for any λ > 0, ∇∆w ∈ Lp(x) (Q) in the above inequality to have Z TZ (ξ − A(u − λw)) · ∇∆w dxdτ ≥ 0. 0

Ω +

Passing to limits as λ → 0 and using Lebesgue’s dominated convergence theorem, we obtain Z TZ N (ξ − Au) · ψ dxdτ ≥ 0, ∀ ψ ∈ Lp(x) (Q) . 0

Ω

We conclude that ξ = Au, a.e. in Q. Therefore, we finish the proof of the existence of weak solutions. Now we prove the uniqueness of weak solutions. Suppose there exist two weak solutions u and v of problem (1)–(3). Using Remark 1, we have Z t (u − v)ϕ dx Ω

0

12


Z tZ − 0

(u − v)ϕt + |∇∆u|p(x)−2 ∇∆u − |∇∆v|p(x)−2 ∇∆v · ∇ϕ dxds = 0,

Ω

for each t ∈ [0, T ] and every ϕ ∈ C 1 (Q). Choosing ϕ = ∆(u − v) as a test function in the above integral equality and using Remark 2, we have Z Z tZ |∇u − ∇v|2 (t) dx + |∇∆u|p(x)−2 ∇∆u − |∇∆v|p(x)−2 ∇∆v 2 Ω 0 Ω · (∇∆u − ∇∆v) dxds = 0. (38) Since the two terms on the left-hand side are nonnegative, we have ∇u = ∇v a.e. in Q. Since u − v = 0 on Γ, we conclude u − v = 0 a.e. in Q, which implies u = v a.e. in Q. Thus, we obtain the uniqueness of weak solutions. Remark 5. Since T is arbitrary, we can obtain that a unique global weak solu− tion u ∈ C([0, ∞]; L2 (Ω)) ∩ L∞ (0, ∞; H01 (Ω)) ∩ Lp 0, ∞; W 2,p(x) (Ω) with ∆u ∈ − 1,p(x) Lp 0, ∞; W0 (Ω) . Remark 6. Indeed, we can obtain the existence and uniqueness theorem for the following general problem  ∂u p(x)−2   ∇∆u = − div |F |p(x)−2 F in Q,  ∂t + div |∇∆u| (39) u = 0, ∆u = 0 on Γ,    u(x, 0) = u0 (x) on Ω, where the data (u0 , F ) satisfy u0 ∈ H01 (Ω),

F ∈ L∞ 0, T ; Lp(x) (Ω) .

(40)

The proof of the existence and uniqueness theorem for problem (39) under Definition 1.1 of weak solutions is almost the same as that of Theorem 1.2. Since the procedure is long and tedious, we omit the standard proof. p(x) Remark 7. Furthermore, if we assume that we can get ∇∆u0 ∈ L ∞(Ω). Then ∞ 2,p(x) the weak solution u ∈ L 0, T ; W (Ω) and ∇∆u ∈ L 0, T ; Lp(x) (Ω) . Choosing η = ∆uk − ∆uk−1 in the integral equality (22) and integrating by parts, we have Z Z 1 |∇uk − ∇uk−1 |2 dx + |∇∆uk |p(x)−2 ∇∆uk · ∇∆(uk − uk−1 ) dx = 0. h Ω Ω

Since the first term is nonnegative, it follows that Z Z |∇∆uk |p(x) dx ≤ |∇∆uk |p(x)−2 ∇∆uk · ∇∆uk−1 dx Ω Ω Z Z 1 p(x) − 1 p(x) |∇∆uk | dx + |∇∆uk−1 |p(x) dx, ≤ p(x) Ω p(x) Ω which implies that Z Ω

1 |∇∆uk |p(x) dx ≤ p(x)

Z Ω

1 |∇∆uk−1 |p(x) dx. p(x)

For any m, 1 ≤ m ≤ n, summing the above inequality for k from 1 to m, we have Z Z 1 1 p(x) |∇∆um | dx ≤ |∇∆u0 |p(x) dx, p(x) p(x) Ω Ω


13

that is

Z Z 1 1 p(x) |∇∆um | dx ≤ − |∇∆u0 |p(x) dx ≤ C. p+ Ω p Ω It follows from Lemma 2.6, the p(·)-Poincaré inequality and Lemma 2.2 that kum kW 2,p(x) ≤ C|∆um |p(x) ≤ C|∇∆um |p(x) n o − + ≤ C max (ρ(∇∆um ))1/p , (ρ(∇∆um ))1/p ≤ C. After taking the supremum over [0, T ], we get kuh kL∞ (0,T ;W 2,p(x) (Ω)) ≤ C,

k∇∆uh kL∞ (0,T ;Lp(x) (Ω)) dx ≤ C,

(41)

where C is a constant independent of h, k. Therefore, we can choose a subsequence (we also denote it by the original sequence for simplicity) such that uh * u, weakly-* in L∞ 0, T ; W 2,p(x) (Ω) , ∇∆uh * ∇∆u, weakly-* in L∞ 0, T ; Lp(x) (Ω) , (42) which follows that (see [13]) kukL∞ (0,T ;W 2,p(x) (Ω)) + k∇∆ukL∞ (0,T ;Lp(x) (Ω)) ≤ C. 4. Long-time behavior of solutions This section is devoted to the long-time behavior of weak solutions. Proof of Theorem 1.3. It follows from (9) that k∇u(t)k2L2 (Ω) is a non-increasing function with respect to t. Therefore, its derivative with respect to t exists almost everywhere on (0, ∞). After differentiating (9) with respect to t, we have Z Z 1 d 2 |∇u(x, t)| dx + |∇∆u(x, t)|p(x) dx = 0, a.e. t ∈ (0, ∞). 2 dt Ω Ω Denote v = ∇u. It follows that Z Z 1 d 2 |v(x, t)| dx + |∆v(x, t)|p(x) dx = 0, a.e. t ∈ (0, ∞). (43) 2 dt Ω Ω − 1,p(x) By Remark 4 and Lemma 2.6, recalling ∆u ∈ Lp 0, ∞; W0 (Ω) , we ob− ∗ p(x) tain ∆u ∈ Lp (0, ∞; Lp (x) (Ω)), where p∗ (x) = NN−p(x) is the Sobolev embedding exponent with respect to p(x). Thus we have kukW 2,p∗ (x) ≤ C|∆u|p∗ (x) ≤ C|∇∆u|p(x) . Using Sobolev embedding theorem in Remark 4 again and Lemma 2.1, we have kvkL2 (Ω) = |v|2 ≤ C|v|p∗∗ (x) ≤ CkvkW 1,p∗ (x) ≤ C|∆v|p(x) N − + ∗∗ for max{1, N2N +4 } < p ≤ p < 2 , where p (x) = ∗ ding exponent with respect to p (x). We distinguish two cases to prove the theorem.

N p(x) N −2p(x)

is the Sobolev embed-

−

(1) If |∇∆u|p(x) > 1, then by Lemma 2.2, we have |∇∆u|pp(x) ≤ ρ(∇∆u). Therefore we obtain from (43) that Z Z p2− 1 d 2 2 ≤ 0, |v(x, t)| dx + C |v(x, t)| dx 2 dt Ω Ω

a.e. t ∈ (0, ∞).

(44)

14


Denote

Z

|v(x, t)|2 dx.

G(t) = Ω

Then we get from (44) that G0 (t) + CG

p− 2

≤ 0,

a.e. t ∈ (0, ∞).

(45)

− Case I: max{1, N2N +4 } < p < 2. Denote T0 = sup{t ∈ (0, ∞)|G(t) > 0}. Then it follows from (45) that 1− p− 0 G 2 + C ≤ 0, a.e. t ∈ (0, T0 ).

(46)

Integrating (46) over (0, t) with t ∈ (0, T0 ), we have 1− p2− 1− p2− G(t) ≤ G(0) − Ct, as long as the right-hand side is nonnegative. Therefore, we conclude that −

2−p T0 ≤ Cku0 kH 1 (Ω) 0

and G(t) = 0,

t ≥ T0 ,

which implies that Z

u2 dx ≤ C

Ω

Z

|∇u|2 dx = 0,

t ≥ T0 .

Ω

That is, u(x, t) = 0,

for x ∈ Ω, t ≥ T0 .

Case II: p− = 2. It follows from (45) that G0 (t) + CG(t) ≤ 0,

a.e. t ∈ (0, ∞).

This implies that 0 G(t)eCt ≤ 0, a.e. t ∈ (0, ∞). Integrating the above inequality over (0, t) with t ∈ (0, ∞), we have G(t) ≤ e−Ct G(0), which implies that ku(t)kH01 (Ω) ≤ e−Ct ku0 kH01 (Ω) ,

for t > 0.

Case III: p− > 2. It follows from (45) that 1− p− 0 2 − p− G 2 + C ≥ 0, a.e. t ∈ (0, ∞). 2 Integrating (47) over (0, t) with t ∈ [0, ∞), we have 1− p2− 1− p2− G(t) ≥ Ct + G(0) ,

(47)


15

which implies that 1 2−p− 2−p− k∇u(t)kL2 (Ω) ≤ Ct + k∇u0 kL , 2 (Ω)

for t > 0.

Therefore, we obtain 1 2−p− 2−p− ku(t)kH01 (Ω) ≤ Ct + ku0 kH , 1 (Ω)

for t > 0.

0

+

(2) If |∇∆u|p(x) < 1, then by Lemma 2.2, we have |∇∆u|pp(x) ≤ ρ(∇∆u). Therefore we obtain from (43) that Z Z p2+ 1 d 2 2 ≤ 0, |v(x, t)| dx + C |v(x, t)| dx 2 dt Ω Ω

a.e. t ∈ (0, ∞).

In this case we can also obtain the similar results for p+ by the same method, i.e., − + 1) If max{1, N2N +4 } < p ≤ p < 2, then there exist two positive number C and T0 with + T0 ≤ Cku0 k2−p H 1 (Ω) 0

such that u(x, t) = 0, for x ∈ Ω, t ≥ T0 . 2) If p+ = 2, then there exists a positive constant C such that ku(t)kH01 (Ω) ≤ e−Ct ku0 kH01 (Ω) ,

for t > 0.

+

3) If p > 2, then there exists a positive constant C such that h i 1+ + 2−p , for t > 0. ku(t)kH01 (Ω) ≤ Ct + ku0 k2−p 1 H (Ω) 0

Therefore, when max{1,

2N N +4 }

−

+