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Well-posedness of damped Kirchhoff-type wave equation with fractional Laplacian

  • Shaohua Chen , Jiangbo Han , Runzhang Xu , Chao Yang EMAIL logo and Meina Zhang
Published/Copyright: March 13, 2025

Abstract

In the present paper, we study the well-posedness of the solution to the initial boundary value problem for the damped Kirchhoff-type wave equation with fractional Laplacian. First, the existence and uniqueness of the local solution are established by the Banach fixed point theorem. Then, the global existence and finite time blowup of the solution are derived at the subcritical and critical initial energy levels. Finally, the finite time blowup of the solution and upper bound and lower bound estimate of blowup time are given at the arbitrarily positive initial energy level.

2020 MSC: 35R11; 35L05; 35A01; 35B40; 35B44

1 Introduction

We consider the initial boundary value problem (IBVP) of the damped Kirchhoff-type wave equation with fractional Laplacian

(1.1) u t t + M [ u ] s 2 ( Δ ) s u + ( Δ ) s u t + | u t | q 2 u t = | u | p 2 u ,  in  Ω × R + , u ( x , 0 ) = u 0 , u t ( x , 0 ) = u 1 ,  in  Ω , u = 0 ,  in  ( R N \ Ω ) × R 0 + ,

where Ω R N is a bounded domain with smooth boundary Ω, s ∈ (0, 1),

(1.2) 3 < 2 θ < p < 2 s * 2 N N 2 s ,

and the power q of the weak damping and the dimension N satisfy

(1.3) 2 q < 2 + 4 s N , 2 s < N < 6 s .

Combining (1.2) and (1.3) ensures

(1.4) 3 < 2 θ , 2 q < 2 θ < p < 2 s * .

The fractional Laplacian operator is defined by

( Δ ) s u 2 lim ε 0 + R N \ B ε ( x ) u ( x ) u ( y ) | x y | N + 2 s d y

for any u C 0 ( R N ) , where s ∈ (0, 1), B ɛ (x) is a ball with x R N as the center and ɛ > 0 as the radius. The Kirchhoff function M ( m ) : R + R + is given by

(1.5) M ( m ) a + m θ 1 , a > 0 ,

which satisfies

  1. M(m) is an increasing and continuous function;

  2. M(m) is continuously differentiable for all m R + and its derivative M′(m) satisfies the condition |M′(m)m| ≤ CM(m) for all m R + and some C > 0. By (1.5), we can see this inequality must hold for Cθ − 1.

In problem (1.1), the Gagliardo seminorm of u is defined by

[ u ] s Q | u ( x ) u ( y ) | 2 | x y | N + 2 s d x d y 1 2 ,

where Q = R 2 N \ ( C ( Ω ) × C ( Ω ) ) , C ( Ω ) = R N \ Ω .

The Kirchhoff problems can be used to describe the transverse oscillations of a stretched string with nonlocal flexural rigidity, which continuously depend on the Sobolev deflection norm of the displacement u via the Kirchhoff function M [1], [2]. The original version of the equation in (1.1) is

(1.6) ρ h u t t P 0 + E h 2 L 0 L u x ( x , t ) 2 d x 2 u x 2 + δ u t = f ,

where t ≥ 0, 0 < x < L, u(x, t) is the lateral displacement at the space x and time t, E is the Young modulus, h is the cross-section area, ρ is the mass density, L is the length, P 0 is the initial axial tension, δ is the resistance modulus, and f is the external force. Equation (1.6) with f = 0 was first presented by Kirchhoff in [3], which is an extension of the classical D’Alembert wave equation considering the changing effects in the length of the string during the vibrations, then Kirchhoff in [4] introduced equation (1.6) with δ = f = 0. Ebihara et al. in [5] studied (1.6) with P 0 = δ = f = 0 and obtained the existence and uniqueness of the local solution to the corresponding IBVP.

To systematically recall the related work of problem (1.1), we give the following general form of the Kirchhoff-type wave equation

(1.7) u t t M ( u 2 ) Δ u + g ( x , u , u t ) = f ( u ) .

For g = f = 0 and M is non-degenerate, D’Ancona and Spagnolo in [1] proved the global existence of the solution to the Cauchy problem of (1.7) in the analytic periodic case. Lasiecka and Ong in [6] obtained the global existence, uniqueness and uniform decay rate of the solution to the IBVP of (1.7). Miranda and Jutuca in [7] established the global existence and the boundary stability to the initial-inhomogeneous boundary value problem of (1.7). Baldi and Haus in [8] gave a lower bound of the existence time of the solution to the IBVP of (1.7) with periodic boundary condition. For g = u t , f = 0 and M is non-degenerate, Vicente in [9] proved the well-posedness and exponential stability of the energy to the IBVP of (1.7) with non-porous acoustic boundary condition. For g = u t , f = |u| α u, α > 0 and M is mildly degenerate, Ono in [10] studied the constraint mechanism between damping and nonlinear source term and obtained the global existence and finite time blowup of the solution to the IBVP of (1.7) in the framework of the potential well theory. For g = 1 ( 1 + t ) p u t , p 0 , f = 0 and M is degenerate, Ghisi and Gobbino in [11] considered the perturbation Cauchy problem of (1.7) and proved the global existence and uniqueness of the solution for suitable values of the parameters, then they continued to study the corresponding non-degenerate case in [12] and proved that the results in [11] cannot be true for p > 1 but hold for p = 1. For g = u t + u + a(x, u), a C ( R , R n ) and M is non-degenerate, Yang in [13] investigated the Cauchy problem of (1.7) and obtained the global existence of the solution for f L 2 ( R n ) and global attractor for f H 1 ( R n ) . For g = λu with λ R , f = |u| p−1 u and M is non-degenerate, Chen et al. in [14] considered the IBVP of (1.7) in the framework of the potential well theory and proved the finite time blowup of the solution at the subcritical initial energy level. For g = −σ(‖∇u2u t + h(x), f(u) is a C 1 function such that f(0) = 0 and M is degenerate, Chueshov in [15] proved on the long-time dynamics of the IBVP to (1.7). Later, Li and Yang in [16] and Ma et al. in [17] extended the results in [15] to the fractional version and exponential source case, respectively. For more related work on Kirchhoff-type wave equations, we can refer to the papers on the fourth-order case [18], [19], memory term [20], p-Laplacian [21] and the references therein.

It is worthy mentioning that many mathematical models involving fractional operators have attracted lots of attention. The fractional partial differential equations arise in a quite natural way in many applications, such as in the fields of physics, chemistry, biology and finance [22]. In Applebaum [23], the fractional Laplacian is viewed as the infinitesimal generator of stable radially symmetric Lévy processes. We point out that (−Δ) s can be reduced to the classical Laplace operator −Δ as s → 1, and we can see Proposition 4.4 in Nezza et al. [24] for more details. In particular, thanks to the pioneering work of Caffarelli and Silvestre [25], many interesting results in some classic elliptic problems have been extended in the fractional Laplacian setting [26], [27]. Following Caffarelli and Roquejoffre [28], Fiscella and Valdinoci in [29] proposed a stationary Kirchhoff variational model in bounded regular domains of R n , which takes into account the nonlocal aspect of the tension arising from nonlocal measurements of the fractional length of the string. Pan et al. in [30] considerd the following Kirchhoff-type wave equation with the fractional Laplacian and nonlinear weak damping

u t t + M u s 2 ( Δ ) s u + | u t | q 2 u t + u = | u | p 2 u ,

where M(m) = m γ−1, 2 < q < 2 γ < p < 2 s * . They obtained the global existence, asymptotic behavior and finite time blowup of solutions at the subcritical initial energy level in the framework of the potential well theory, which proposed by Payne and Sattinger in [31], [32] to study the evolution equation without positive definite energy.

For the investigation of the Kirchhoff problem with the classical Laplacian, there have been plentiful related results, and only a few works are concerned with the fractional Laplacian case, especially both the Kirchhoff term and strong damping with fractional Laplacian. The significant feature of the fractional Laplacian is the nonlocal property, which leads to more complicated nonlinearity for the Kirchhoff term. From the perspective of the energy of the system, the dampings and nonlinear source term are the factors causing decay and nonlinear growth of the energy, respectively, which lead to global existence and blowup of the solution, respectively also. The strong damping with fractional Laplacian, nonlinear weak damping and nonlinear source term in (1.1) not only bring substantial difficulties to constructing the variational structure but also to analyzing the constraint relationship between dampings and nonlinear source term. Inspired by these challenges and the works we recalled, in the present paper, we shall systematically study the well-posedness of solutions to problem (1.1) at three initial energy levels, i.e., subcritical, critical and arbitrarily positive initial energy levels, and further prove the asymptotic behavior of the global solution and estimate the blowup time of the finite time blowup solution. Since we consider two types of dampings, i.e., nonlinear weak damping and strong damping, especially the strong damping is nonlocal, the standard concavity method is ineffective, which means we need propose new strategies to construct appropriate auxiliary functional to establish corresponding concave inequalities for problem (1.1).

The rest of the paper is organized as follows. In Section 2, we give some notations and recall necessary definitions and properties of the fractional Sobolev spaces and potential well theory. In Section 3, we establish the local existence of the solution by Banach fixed point theorem. In Section 4 and Section 5, we prove the global existence, asymptotic behavior and finite time blowup of the solution to problem (1.1) at the subcritical initial energy level, i.e., E(0) < d and the critical initial energy level, i.e., E(0) = d, respectively. In Section 6, we consider the case of a > 1 and q = 2, give the finite time blowup of the solution to problem (1.1) at the arbitrarily positive initial energy level, i.e., E(0) > 0 and estimate the upper bound and lower bound of blowup time.

2 Preliminary results

2.1 Fractional Sobolev spaces

In this subsection, we first recall some necessary definitions and properties of the fractional Sobolev spaces, see also in [24], [33] for further detail.

In the following, for the sake of convenience, we denote by ‖ ⋅‖ p the usual L p (Ω) norm for 2 s * > p > 2 , when p = 2, we denote by ‖ ⋅‖ the L 2(Ω) norm especially. Let (⋅, ⋅) denote by the inner product of L 2(Ω). In the sequel, we denote by C > 0 a generic constant that may vary even from line to line within the same formula.

Let W be the linear space of Lebesgue measurable functions u from R N to R such that the restriction to Ω of any function u in W belongs to L 2(Ω) and

Q | u ( x ) u ( y ) | 2 | x y | N + 2 s d x d y < + .

The space W is equipped with the norm

(2.1) u W u 2 + Q | u ( x ) u ( y ) | 2 | x y | N + 2 s d x d y 1 2 .

We shall work in the closed linear subspace

W 0 { u W : u = 0 a.e. in R N \ Ω } .

As shown in [34], an equivalent norm on W 0 is

(2.2) u W 0 2 [ u ] s 2 = Q | u ( x ) u ( y ) | 2 | x y | N + 2 s d x d y , u W 0 .

and

( u , v ) W 0 = Q u ( x ) u ( y ) v ( x ) v ( y ) | x y | N + 2 s d x d y .

Taking derivative of (2.2) with respect to t, we have

(2.3) d u W 0 2 d t = d [ u ] s 2 d t = 2 ( u , u t ) W 0 .

Specially, ⟨⋅, ⋅⟩ denotes the duality pairing between W 0 and its dual space W s . Next, we state a functional property as following Proposition, proved in detail in [35].

Proposition 2.1.

For any r 1 , 2 s * , there exists a positive constant C ̃ = C ̃ ( N , r , s ) such that for any uW 0

(2.4) u r 2 C ̃ Ω × Ω | u ( x ) u ( y ) | 2 | x y | N + 2 s d x d y C 0 Q | u ( x ) u ( y ) | 2 | x y | N + 2 s d x d y .

2.2 Potential well theory

In this subsection, we give some necessary definitions and lemmas related to potential well theory. We define the total, potential and Nehari energy functionals as follows

(2.5) E ( t ) 1 2 u t 2 + a 2 u W 0 2 + 1 2 θ u W 0 2 θ 1 p u p p ,

(2.6) J ( u ) a 2 u W 0 2 + 1 2 θ u W 0 2 θ 1 p u p p ,

(2.7) I ( u ) a u W 0 2 + u W 0 2 θ u p p .

The stable set U and unstable set V are defined by

U = { u W 0 I ( u ) > 0 } { 0 }

and

V = { u W 0 I ( u ) < 0 } .

Finally, we define the depth of potential well

d inf { J ( u ) u W 0 \ { 0 }  and  I ( u ) = 0 } .

From (2.6) and (2.7), we have

(2.8) J ( u ) = 1 p I ( u ) + a ( p 2 ) 2 p u W 0 2 + p 2 θ 2 θ p u W 0 2 θ .

For the depth of the potential well d, we have the following knowledge.

Lemma 2.2.

(Depth of the potential well) The potential well depth is

(2.9) d > C 0 p θ 2 θ p p 2 θ 2 θ p ,

where

(2.10) C 0 = sup u W 0 , u 0 u p 2 u W 0 2 .

Proof.

From the definition of d, for I(u) = 0, we have

a u W 0 2 + u W 0 2 θ = u p p ,

which together with (2.10) gives

u W 0 2 θ u p p C 0 p 2 u W 0 p .

Then we have

(2.11) u W 0 2 θ C 0 p θ 2 θ p .

Combining (2.8), (2.11) and I(u) = 0, there holds

J ( u ) = a ( p 2 ) 2 p u W 0 2 + p 2 θ 2 θ p u W 0 2 θ > C 0 p θ 2 θ p p 2 θ 2 θ p ,

which implies (2.9). □

Lemma 2.3.

(Non-increasing energy) Let u(x, t) be the weak solution to problem (1.1) with (u 0, u 1) ∈ W 0 × L 2(Ω). Then

E ( t ) E ( 0 ) , t 0 .

Proof.

Testing the equation in (1.1) with u t , we obtain

(2.12) E ( t ) + 0 t u t W 0 2 d τ + 0 t u t q q d τ = E ( 0 ) .

Obviously, for any t ∈ [0, ), the energy is non-increasing. □

Lemma 2.4.

Let a > 0, (1.2), (1.3) and uW 0 \{0} hold, then:

  1. lim λ 0 J ( λ u ) = 0 , lim λ + J ( λ u ) = .

  2. For 0 < λ < , there exists a unique λ* = λ*(u) such that

d d λ J ( λ u ) λ = λ * = 0 .

  1. J(λu) is increasing on 0 < λλ*, decreasing on λ* < λ < and takes the maximum at λ = λ*.

  2. I(λu) > 0 for 0 < λ < λ*, I(λu) < 0 for λ* < λ < , and I(λ*u) = 0.

Proof.

(i) From the definition of J(u), we know

(2.13) J ( λ u ) = a λ 2 2 u W 0 2 + λ 2 θ 2 θ u W 0 2 θ λ p p u p p .

Due to u W 0 0 , a > 0 and (1.2), we have

lim λ 0 J ( λ u ) = 0

and

lim λ + J ( λ u ) = .

(ii) Taking the derivative of (2.13) with respect to λ yields

(2.14) d d λ J ( λ u ) = a λ u W 0 2 + λ 2 θ 1 u W 0 2 θ λ p 1 u p p = λ a u W 0 2 + λ 2 θ 2 u W 0 2 θ λ p 2 u p p .

We set

h ( λ ) λ 2 θ 2 u W 0 2 θ λ p 2 u p p = λ 2 θ 2 u W 0 2 θ λ p 2 θ u p p

with (1.2), we know

lim λ 0 h ( λ ) > 0 , lim λ + h ( λ ) =

and there exists

λ ̃ = u W 0 2 θ u p p 1 p 2 θ

such that h ( λ ̃ ) = 0 . Obviously, h(λ) is monotonically decreasing function for λ > λ ̃ > 0 . Therefore, there exists a unique λ * > λ ̃ > 0 such that

a u W 0 2 + λ 2 θ 2 u W 0 2 θ λ p 2 u p p = 0 ,

i.e.,

d d λ J ( λ u ) λ = λ * = 0 .

(iii) By a direct calculation, (2.14) gives

d d λ J ( λ u ) > 0  for  0 < λ < λ * , d d λ J ( λ u ) < 0  for  λ * < λ < .

(iv) The conclusion follows from

I ( λ u ) = a λ 2 u W 0 2 + λ 2 θ u W 0 2 θ λ p u p p = λ 2 a u W 0 2 + λ 2 θ 2 u W 0 2 θ λ p 2 u p p = λ d d λ J ( λ u ) .

Hence, when 0 < λ < λ*, I(λu) > 0; when λ* < λ < , I(λu) < 0; when λ = λ*, I(λ*u) = 0. □

Proposition 2.5.

(Gronwall inequality) Suppose y(t) ∈ L 1[0, T], if there exist constants b and c such that

y ( t ) b + c 0 t y ( z ) d z , 0 t T .

Then

y ( t ) b e c t .

3 Local solution

We first give the definition of weak solution to problem (1.1).

Definition 3.1.

(Weak solution) A function u = u(x, t) is said to be a weak solution to problem (1.1), if uC([0, T], W 0) ∩ C 1([0, T], L 2(Ω)) ∩ C 2([0, T], W s ) with u t L 2([0, T], W 0), and the initial data satisfy

(3.1) u ( , 0 ) = u 0  in  W 0 , u t ( , 0 ) = u 1  in  L 2 ( Ω ) ,

such that

(3.2) u t t ( , t ) , ϕ ( , t ) + M ( [ u ( , t ) ] s 2 ) ( u ( , t ) , ϕ ( , t ) ) W 0 + ( | u t ( , t ) | q 2 u t ( , t ) , ϕ ( , t ) ) + ( u t ( , t ) , ϕ ( , t ) ) W 0 = ( | u ( , t ) | p 2 u ( , t ) , ϕ ( , t ) )

for any ϕW 0 and a.e. t ∈ [0, T], where

( u ( , t ) , ϕ ( , t ) ) W 0 = Q u ( x , t ) u ( y , t ) ϕ ( x , t ) ϕ ( y , t ) | x y | N + 2 s d x d y .

For a given T > 0, consider the space

(3.3) H = C ( [ 0 , T ] , W 0 ) C 1 ( [ 0 , T ] , L 2 ( Ω ) )

endowed with the norm

u H 2 = max t [ 0 , T ] u t 2 + a u W 0 2 ,

where the positive constant a was defined in (1.5). Next, we prove the existence and uniqueness of the local solution to problem (1.1). Firstly, we give an estimate of nonlinear term |u| p−2 u by Gâteaux derivative.

Lemma 3.2.

If (1.2) holds, for any u 1(x, t), u 2(x, t) with (x, t) ∈ Ω × [0, T], |u 1| + |u 2| > 0 and u 1u 2, there holds

| u 1 | p 2 u 1 | u 2 | p 2 u 2 ( p 1 ) ( | u 1 | + | u 2 | ) p 2 | u 1 u 2 | .

Proof.

Let S(u)≔|u| p−2 u. Define ς(x, t)≔u 1u 2 for (x, t) ∈ Ω × [0, T]. By the property of Gâteaux derivative, we know

S ( u 1 ) S ( u 2 ) = S ( u 2 + ς ) S ( u 2 ) = 0 1 d S ( u 2 + η 0 ς ; ς ) d η 0

for η 0 ∈ (0, 1), which together with the definition of Gâteaux derivative

d S ( u 2 + η 0 ς ; ς ) = lim τ 0 S ( u 2 + η 0 ς + τ ς ) S ( u 2 + η 0 ς ) τ = d d τ S ( u 2 + η 0 ς + τ ς ) τ = 0

gives

S ( u 1 ) S ( u 2 ) = 0 1 d d τ S ( u 2 + η 0 ς + τ ς ) τ = 0 d η 0 = 0 1 d d τ | u 2 + η 0 ς + τ ς | p 2 ( u 2 + η 0 ς + τ ς ) τ = 0 d η 0 0 1 ( p 1 ) | u 2 + η 0 ς + τ ς | p 2 | ς | τ = 0 d η 0 = 0 1 ( p 1 ) | η 0 u 1 + ( 1 η 0 ) u 2 | p 2 | u 1 u 2 | d η 0 ( p 1 ) ( | u 1 | + | u 2 | ) p 2 | u 1 u 2 | .

The proof is completed. □

In order to prove the local existence of the solution to the nonlinear problem (1.1) by the Banach fixed point theorem, we first convert problem (1.1) into the linearized problem. For the wave equation with nonlinear weak damping |u t | q−2 u t , Georgiev and Todorova [36] inspired by Lions [37] gave a standard method for linearizing the nonlinear problem and proved the existence and uniqueness of the solution to the linearized problem. However, for the model equation in (1.1) considered in the present paper, the method proposed by [36], [37] does not work due to the appearance of the fractional Laplacian strong damping (−Δ) s u t and the fractional Kirchhoff term M [ u ] s 2 ( Δ ) s u . Hence, we shall propose a new strategy to linearize problem (1.1), especially, we propose a new version of the linearized problem, i.e., (3.4), and give the following corresponding lemma about the existence and uniqueness of solutions to this linearized problem. It is worth noting that our method for dealing with linearized problems can not be applied to improve the conclusions in [36], [37], because this method depends on the terms (−Δ) s u t and M [ u ] s 2 ( Δ ) s u , which do not appear considered in the models of [36], [37].

Next, we give the following version of the linearized problem to (1.1) and consider the existence and uniqueness of its solution.

Lemma 3.3.

(Linearized problem to (1.1) with 2 q < 2 + 4 s N and 2s < N < 6s) Let (1.2) hold. For every T > 0, u H with u t L 2([0, T], W 0) and initial data satisfying (u 0, u 1) ∈ W 0 × L 2(Ω), there exists a unique v H C 2 ( [ 0 , T ] , W s ) such that v t L 2([0, T], W 0), which solves the following linear problem of (1.1)

(3.4) v t t + M [ u ] s 2 ( Δ ) s v + ( Δ ) s v t = | u | p 2 u | u t | q 2 u t ,  in  Ω × R + , v ( x , 0 ) = u 0 , v t ( x , 0 ) = u 1 ,  in  Ω , v ( x , t ) = 0 ,  in  ( R N \ Ω ) × R 0 + .

Proof.

By [33], there exists a sequence { ω j } C 0 ( Ω ) of eigenfunctions of the fractional Laplace operator (−Δ) s , which is an orthogonal basis of L 2(Ω) and W 0 such that ‖ω j ‖ = 1. We construct the approximate solutions to problem (3.4)

(3.5) v h ( t ) = j = 1 h γ j h ( t ) ω j ,

which solves the problem

(3.6) Ω v ̈ h + M [ u ] s 2 ( Δ ) s v h + ( Δ ) s v ̇ h + | u t | q 2 u t | u | p 2 u η d x = 0 , v h ( 0 ) = u 0 h = j = 1 h Ω u 0 ω j d x ω j , v ̇ h ( 0 ) = u 1 h = j = 1 h Ω u 1 ω j d x ω j

for every η ∈ {ω j } and t ≥ 0. For j = 1, …, h, taking η = ω j in problem (3.6), we estimate the terms of the equation in (3.6). The first term becomes

(3.7) Ω v ̈ h ω j d x = Ω j = 1 h γ ̈ j h ( t ) ω j ω j d x = γ ̈ j h ( t ) Ω | ω j | 2 d x = γ ̈ j h ( t ) .

For the second term, we get

(3.8) Ω M [ u ] s 2 ( Δ ) s v h ω j d x = M [ u ] s 2 Q ( v h ( x , t ) v h ( y , t ) ) ( ω j ( x ) ω j ( y ) ) | x y | N + 2 s d x d y = M [ u ] s 2 Q j = 1 h γ j h ( t ) ( ω j ( x ) ω j ( y ) ) ( ω j ( x ) ω j ( y ) ) | x y | N + 2 s d x d y = M [ u ] s 2 γ j h ( t ) ω j W 0 2 .

For the third term, we obtain

(3.9) Ω ( Δ ) s v ̇ h ω j d x = Q j = 1 h γ ̇ j h ( t ) ( ω j ( x ) ω j ( y ) ) ( ω j ( x ) ω j ( y ) ) | x y | N + 2 s d x d y = γ ̇ j h ( t ) ω j W 0 2 .

Then by (3.7), (3.8) and (3.9), we obtain the following linear ordinary differential problem with unknown γ j h

(3.10) γ ̈ j h + M [ u ] s 2 ω j W 0 2 γ j h ( t ) + ω j W 0 2 γ ̇ j h ( t ) = ψ j ( t ) , γ j h ( 0 ) = Ω u 0 h w j d x , γ ̇ j h ( 0 ) = Ω u 1 h w j d x ,

where ψ j (t) = Ω|u| p−2 j − |u t | q−2 u t ω j dxC[0, T] due to u H . According to ([38], page 199), we can obtain that (3.10) admits the unique solution γ j h C 2 [ 0 , T ] , where T is finite. Then we can obtain the existence of the approximate solution v h defined in (3.5) solving problem (3.6).

Taking η = v ̇ h ( t ) into (3.6), combining (2.3), and integrating both sides over [0, t] ⊂ [0, T], we get

0 t d d τ v ̇ h ( τ ) 2 + d d τ M [ u ] s 2 v h ( τ ) W 0 2 d τ + 2 0 t v ̇ h ( τ ) W 0 2 d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v ̇ h ( τ ) d x d τ + 0 t 2 v h ( τ ) W 0 2 M ( [ u ( τ ) ] s 2 ) ( u , u t ) W 0 d τ 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ̇ h ( τ ) d x d τ ,

i.e.,

(3.11) v ̇ h ( t ) 2 + M ( [ u ( t ) ] s 2 ) v h ( t ) W 0 2 + 2 0 t v ̇ h ( τ ) W 0 2 d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v ̇ h ( τ ) d x d τ + 0 t 2 v h ( τ ) W 0 2 M ( [ u ( τ ) ] s 2 ) ( u , u t ) W 0 d τ + u 1 h 2 + M [ u 0 ] s 2 u 0 h W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ̇ h ( τ ) d x d τ ,

for every h ≥ 1. By Cauchy-Schwarz inequality and Hölder inequality, we estimate the first term on the right hand side in (3.11)

(3.12) 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v ̇ h ( τ ) d x d τ 2 0 t Ω | u ( τ ) | p 1 v ̇ h ( τ ) d x d τ 2 0 t Ω | u ( τ ) | p 1 2 N N + 2 s d x N + 2 s 2 N Ω | v ̇ h ( τ ) | 2 s * d x N 2 s 2 N d τ = 2 0 t u ( τ ) 2 N ( p 1 ) N + 2 s p 1 v ̇ h ( τ ) 2 s * d τ .

In particular, (3.5) implies that v ̇ h ( t ) W 0 for every t ∈ [0, T]. From Proposition 2.1, we know

(3.13) v ̇ h ( t ) 2 s * C 0 1 2 v ̇ h ( t ) W 0

for every t ∈ [0, T]. Due to (1.2), then

2 N ( p 1 ) N + 2 s < 2 N 2 N N 2 s 1 N + 2 s = 2 N N 2 s = 2 s * .

Therefore

W 0 L 2 s * ( Ω ) L 2 N ( p 1 ) N + 2 s ( Ω ) ,

further we get

(3.14) u ( t ) 2 N ( p 1 ) N + 2 s p 1 C 0 p 1 2 u ( t ) W 0 p 1 .

Since u H , then u ( t ) W 0 is bounded. By (3.13) and (3.14), (3.12) becomes

(3.15) 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v ̇ h ( τ ) d x d τ 0 t 2 C u ( τ ) W 0 p 1 v ̇ h ( τ ) W 0 d τ 2 0 t C u ( τ ) W 0 2 ( p 1 ) d τ + 1 2 0 t v ̇ h ( τ ) W 0 2 d τ C T + 1 2 0 t v ̇ h ( τ ) W 0 2 d τ .

By Cauchy-Schwarz inequality and Hölder inequality, we estimate the last term on the right hand side in (3.11)

(3.16) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ̇ h ( τ ) d x d τ 2 0 t Ω | u t ( τ ) | q 1 v ̇ h ( τ ) d x d τ 2 0 t Ω | u t ( τ ) | q 1 2 N N + 2 s d x N + 2 s 2 N Ω | v ̇ h ( τ ) | 2 s * d x N 2 s 2 N d τ = 2 0 t u t ( τ ) 2 N ( q 1 ) N + 2 s q 1 v ̇ h ( τ ) 2 s * d τ .

In order to deal with (3.16), we divide the range of the power q, i.e., 2 q < 2 + 4 s N , into two cases, i.e., Case I ( 2 q 2 + 2 s N ) and Case II ( 2 + 2 s N < q < 2 + 4 s N ) , to use the two different strategies to treat it.

Case I:

2 q 2 + 2 s N .

For the case 2 q 2 + 2 s N , since it gives 2 N ( q 1 ) N + 2 s 2 , we know that (3.16) becomes

(3.17) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ̇ h ( τ ) d x d τ C 0 t u t ( τ ) 2 q 1 v ̇ h ( τ ) 2 s * d τ C 0 t u t ( τ ) 2 q 1 v ̇ h ( τ ) W 0 d τ .

Meanwhile, u H also ensures that ‖u t (t)‖2 is bounded. Then, by Young inequality, (3.17) gives

(3.18) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ̇ h ( τ ) d x d τ 0 t C 2 u t ( τ ) 2 2 ( q 1 ) + 1 2 v ̇ h ( τ ) W 0 2 d τ C T + 1 2 0 t v ̇ h ( τ ) W 0 2 d τ .

Case II:

2 + 2 s N < q < 2 + 4 s N .

For the case 2 + 2 s N < q < 2 + 4 s N , we have 2 < 2 N ( q 1 ) N + 2 s < 2 s * . Then, by interpolation inequality, we know

(3.19) u t ( τ ) 2 N ( q 1 ) N + 2 s q 1 u t ( τ ) 2 α ( q 1 ) u t ( τ ) 2 s * ( 1 α ) ( q 1 ) ,

where

(3.20) α N + 2 s ( q 1 ) ( N 2 s ) ( q 1 ) 2 s

and 0 < α < 1 is guaranteed by the condition 2 + 2 s N < q < 2 + 4 s N . By substituting (3.19) into (3.16), we have

(3.21) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ̇ h ( τ ) d x d τ 2 0 t u t ( τ ) 2 α ( q 1 ) u t ( τ ) 2 s * ( 1 α ) ( q 1 ) v ̇ h ( τ ) 2 s * d τ 2 0 t max 0 τ T u t ( τ ) 2 α ( q 1 ) u t ( τ ) 2 s * ( 1 α ) ( q 1 ) v ̇ h ( τ ) 2 s * d τ 2 C max 0 τ T u t ( τ ) 2 α ( q 1 ) 0 t u t ( τ ) W 0 ( 1 α ) ( q 1 ) v ̇ h ( τ ) W 0 d τ .

By Hölder inequality and Young inequality, we know (3.21) becomes

(3.22) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ˙ h ( τ ) d x d τ 2 C max 0 τ T u t ( τ ) 2 α ( q 1 ) 0 t u t ( τ ) W 0 ( 1 α ) ( q 1 ) v ˙ h ( τ ) W 0 d τ C max 0 τ T u t ( τ ) 2 2 α ( q 1 ) 0 t u t ( τ ) W 0 2 ( 1 α ) ( q 1 ) d τ + 0 t v ˙ h ( τ ) W 0 2 d τ C max 0 τ T u t ( τ ) 2 2 α ( q 1 ) 0 t u t ( τ ) W 0 2 ( 1 α ) ( q 1 ) r 1 d τ 1 r 1 0 t 1 r 2 d τ 1 r 2 + 1 2 0 t v ˙ h ( τ ) W 0 2 d τ ,

where

(3.23) r 1 1 ( 1 α ) ( q 1 ) ,

(3.24) r 2 1 1 ( 1 α ) ( q 1 ) ,

α is defined by (3.20), and the condition 2 + 2 s N < q < 2 + 4 s N ensures that r 1 > 1 and r 2 > 1. Here, from the known function u H with u t L 2([0, T], W 0), we know (3.22) becomes

(3.25) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v ˙ h ( τ ) d x d τ C max 0 τ T u t ( τ ) 2 2 α ( q 1 ) 0 t u t ( τ ) W 0 2 d τ ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) + 1 2 0 t v ˙ h ( τ ) W 0 2 d τ C T 1 ( 1 α ) ( q 1 ) + 1 2 0 t v ˙ h ( τ ) W 0 2 d τ .

Then for 2 q < 2 + 4 s N , by the convergence of u 0 h and u 1 h , u H with u t L 2([0, T], W 0), combining (3.11), (3.15), (3.18) and (3.25), we get

(3.26) v ̇ h ( t ) 2 + M ( [ u ( t ) ] s 2 ) v h ( t ) W 0 2 + 0 t v ̇ h ( τ ) W 0 2 d τ C T + 0 t 2 M ( [ u ( τ ) ] s 2 ) v h ( τ ) W 0 2 ( u , u t ) W 0 d τ + u 1 h 2 + M [ u 0 ] s 2 u 0 h W 0 2 b + 0 t 2 ( θ 1 ) u ( τ ) W 0 2 ( θ 2 ) v h ( τ ) W 0 2 ( u , u t ) W 0 d τ b + C 0 t v h ( τ ) W 0 2 d τ ,

where b = C T + C T 1 ( 1 α ) ( q 1 ) + C u 1 2 + C M [ u 0 ] s 2 u 0 W 0 2 > 0 is independent of h for every h ≥ 1. Next, we estimate the second term on the right hand side of (3.26). By M(m) ≥ a > 0, we have

a v h ( t ) W 0 2 M ( [ u ( t ) ] s 2 ) v h ( t ) W 0 2 b + C 0 t v h ( τ ) W 0 2 d τ ,

then from Proposition 2.5, we obtain

v h ( t ) W 0 2 b a e C t a .

Therefore, we know

0 t v h ( τ ) W 0 2 d τ b C e C t a 1 .

Further, (3.26) becomes

v ̇ h ( t ) 2 + a v h ( t ) W 0 2 + 0 t v ̇ h ( τ ) W 0 2 d τ b + b C e C T a 1 .

By this uniform estimate and (3.6), we have

{ v h } is bounded in L ( [ 0 , T ] , W 0 ) ; { v ̇ h } is bounded in L ( [ 0 , T ] , L 2 ( Ω ) ) L 2 ( [ 0 , T ] , W 0 ) ; { v ̈ h } is bounded in L 2 ( [ 0 , T ] , W s ) ) .

Therefore, up to a subsequence, we could pass to the limit in (3.6) satisfying the above regularity. Then the existence of weak solution to problem (3.4) can be obtained. Since vL ([0, T], W 0) and v t L 2([0, T], W 0), we get vC([0, T], W 0). Moreover since v t L 2([0, T], W 0)) and v tt L 2([0, T], W s )), we have v t C([0, T], L 2(Ω)). And (3.4) implies that v tt C([0, T], W s ). The following is the proof of the uniqueness. Arguing by contradiction, if v and w are two solutions of (3.4) sharing the same initial data. By subtracting the equations, we have

(3.27) ( v w ) t t + M [ u ] s 2 ( Δ ) s ( v w ) + ( Δ ) s ( v w ) t = 0 .

Testing (3.27) with v t w t and integrating over [0, t] ⊂ [0, T], we have

(3.28) 0 t 1 2 d d τ v t w t 2 2 + M [ u ] s 2 ( Δ ) s ( v w ) , v t w t d τ + 0 t ( Δ ) s ( v w ) t , v t w t d τ = 0 .

By direct calculation of (3.28), there holds

(3.29) 0 t 1 2 d d τ v t w t 2 2 + 1 2 M [ u ] s 2 d d τ v w W 0 2 d τ + 0 t v t w t W 0 2 d τ = 0 .

By using (2.3), we know that (3.29) gives

(3.30) 1 2 v t w t 2 2 1 2 v 1 w 1 2 2 + 1 2 M [ u ] s 2 v w W 0 2 1 2 M [ u 0 ] s 2 v 0 w 0 W 0 2 0 t M [ u ] s 2 ( u , u t ) W 0 v w W 0 2 d τ 0 .

From (3.30), u H , v 0 = w 0 = u 0 and v 1 = w 1 = u 1, there holds

1 2 v t w t 2 2 + 1 2 M [ u ] s 2 v w W 0 2 0 t M [ u ] s 2 ( u , u t ) W 0 v w W 0 2 d τ C 0 t 1 2 v t w t 2 2 + a 2 v w W 0 2 d τ ,

which together with (1.5) gives

(3.31) 1 2 v t w t 2 2 + a 2 v w W 0 2 C 0 t 1 2 v t w t 2 2 + a 2 v w W 0 2 d τ .

Combining (3.31) and Gronwall inequality, we have

1 2 v t w t 2 2 + a 2 v w W 0 2 0 ,

which yields vw. The proof is completed. □

Next, we give the theorem about the local existence of the solution to problem (1.1) based on Lemma 3.3.

Theorem 3.4.

(Local existence for 2 q < 2 + 4 s N and 2s < N < 6s) Let the conditions in Lemma 3.3 hold.

Then problem (1.1) admits a weak local solution over [0, T] satisfying u H , i.e., (3.3), with u t L 2([0, T], W 0).

Proof.

We define the map Φ1, such that v = Φ1(u), where v is the unique solution to problem (3.4). Let (u 0, u 1) satisfy (3.1) and

(3.32) R 2 = 2 u 1 2 + a u 0 W 0 2 ,

then for any T > 0 we define

(3.33) M T u H , u t L 2 ( [ 0 , T ] , W 0 ) : u ( , 0 ) = u 0 , u t ( , 0 ) = u 1 , u M T 2 R 2 ,

where

u M T 2 u H 2 + 0 T u t ( t ) W 0 2 d t .

Next, we prove that Φ 1 ( M T ) M T is a contractive mapping. We first prove that M T is a complete metric space. By [35], we know that W 0 is a uniformly convex Banach space. Let {u n } be a Cauchy sequence in M T . Thus, for any ɛ > 0, there exists μ ɛ such that if n, mμ ɛ , then

u n u m M T 2 = max t [ 0 , T ] u n t u m t 2 + a u n u m W 0 2 + 0 t u n t u m t W 0 2 d τ ε .

By the completeness of L 2(Ω) and W 0, there exists uL 2(Ω) such that u n u strongly in L 2(Ω) as n, and uW 0 such that u n u strongly in W 0 as n, i.e.,

u n u M T 2 = max t [ 0 , T ] u n t u t 2 + a u n u W 0 2 + 0 t u n t u t W 0 2 d τ ε .

That is M T is complete metric space. Given u M T , the corresponding solution v = Φ1(u) satisfies the energy identity (3.11)

(3.34) v t ( t ) 2 + M ( [ u ( t ) ] s 2 ) v ( t ) W 0 2 + 2 0 t v t ( τ ) W 0 2 d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 0 t 2 M ( [ u ( τ ) ] s 2 ) v ( τ ) W 0 2 ( u , u t ) W 0 d τ + u 1 2 + M [ u 0 ] s 2 u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ .

We denote M ̃ ( m ) m θ 1 . Then, we notice that

M ̃ ( m ) = M ( m ) = ( θ 1 ) m θ 2 ,

which together with (3.34) gives

(3.35) v t ( t ) 2 + M ( [ u ( t ) ] s 2 ) v ( t ) W 0 2 + 2 0 t v t ( τ ) W 0 2 d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 0 t 2 M ̃ ( [ u ( τ ) ] s 2 ) v ( τ ) W 0 2 ( u , u t ) W 0 d τ + u 1 2 + M [ u 0 ] s 2 u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 0 t v ( τ ) W 0 2 d d τ M ̃ ( [ u ( τ ) ] s 2 ) d τ + u 1 2 + M [ u 0 ] s 2 u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ 0 t M ̃ ( [ u ( τ ) ] s 2 ) d d τ v ( τ ) W 0 2 d τ + M ̃ ( [ u ( t ) ] s 2 ) v W 0 2 M ̃ [ u 0 ] s 2 u 0 W 0 2 + u 1 2 + M [ u 0 ] s 2 u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ = 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ 0 t 2 M ̃ ( [ u ( τ ) ] s 2 ) ( v , v t ) W 0 d τ + M ̃ ( [ u ( t ) ] s 2 ) v W 0 2 M ̃ [ u 0 ] s 2 u 0 W 0 2 + u 1 2 + M [ u 0 ] s 2 u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ .

By Hölder inequality and Young inequality, we know that (3.35) becomes

(3.36) v t ( t ) 2 + a v ( t ) W 0 2 + 2 0 t v t ( τ ) W 0 2 d τ 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 0 t 2 M ̃ ( [ u ( τ ) ] s 2 ) ( v , v t ) W 0 d τ + u 1 2 + a u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 0 t 2 M ̃ ( [ u ( τ ) ] s 2 ) v W 0 v t W 0 d τ + u 1 2 + a u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 2 0 t M ̃ 2 ( [ u ( τ ) ] s 2 ) v W 0 2 d τ + 1 2 0 t v t W 0 2 d τ + u 1 2 + a u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ + 2 0 t a 2 ( θ 1 ) R 4 ( θ 1 ) v W 0 2 d τ + 1 2 0 t v t W 0 2 d τ + u 1 2 + a u 0 W 0 2 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ .

For the first term and the last term on the right hand side of (3.36), we argue it in the same spirit (although slightly differently) as for (3.15), (3.18) and (3.25) and get

(3.37) 2 0 t Ω | u ( τ ) | p 2 u ( τ ) v t ( τ ) d x d τ 2 0 t C u ( τ ) W 0 p 1 v t ( τ ) W 0 d τ 2 C T R 2 ( p 1 ) a ( p 1 ) + 1 4 0 t v t ( τ ) W 0 2 d τ ,

(3.38) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ 0 t C 2 u t ( τ ) 2 2 ( q 1 ) + 1 4 v t ( τ ) W 0 2 d τ C T R 2 ( q 1 ) 2 + 1 4 0 t v t ( τ ) W 0 2 d τ

for the case 2 q 2 + 2 s N , and

(3.39) 2 0 t Ω | u t ( τ ) | q 2 u t ( τ ) v t ( τ ) d x d τ C R 2 ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) + 1 4 0 t v t ( τ ) 2 s * 2 d τ

for the case 2 + 2 s N < q < 2 + 4 s N , respectively, for all t ∈ (0, T], where α is defined by (3.20) and the condition 2 + 2 s N < q < 2 + 4 s N gives 1 − (1 − α)(q − 1) > 0. According to (3.36), (3.37), (3.38) and (3.39), we have

(3.40) v t ( t ) 2 + a v ( t ) W 0 2 + 0 t v t ( τ ) W 0 2 d τ C T R 2 ( p 1 ) a ( p 1 ) + max C T R 2 ( q 1 ) , C R 2 ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) + 0 t 2 a 2 ( θ 1 ) R 4 ( θ 1 ) v W 0 2 d τ + u 1 2 + a u 0 W 0 2 C T R 2 ( p 1 ) a ( p 1 ) + max C T R 2 ( q 1 ) , C R 2 ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) + 2 a 2 ( θ 1 ) R 4 ( θ 1 ) T v H 2 + 1 2 R 2

for any time t ∈ (0, T], which means

v H 2 + 0 T v t ( t ) W 0 2 d t C T R 2 ( p 1 ) a ( p 1 ) + max C T R 2 ( q 1 ) , C R 2 ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) + 2 a 2 ( θ 1 ) R 4 ( θ 1 ) T v H 2 + 1 2 R 2 ,

i.e.,

(3.41) 1 2 a 2 ( θ 1 ) R 4 ( θ 1 ) T v H 2 + 0 T v t ( t ) W 0 2 d t C T R 2 ( p 1 ) a ( p 1 ) + max C T R 2 ( q 1 ) , C R 2 ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) + 1 2 R 2 .

Choosing T sufficiently small such that 1 − 2a −2(θ−1) R 4(θ−1) T > 0 and

C T R 2 ( p 1 ) a ( p 1 ) + max C T R 2 ( q 1 ) , C R 2 ( 1 α ) ( q 1 ) T 1 ( 1 α ) ( q 1 ) 1 2 a 2 ( θ 1 ) R 4 ( θ 1 ) T + R 2 2 1 2 a 2 ( θ 1 ) R 4 ( θ 1 ) T R 2 ,

then (3.41) gives

v H 2 + 1 1 2 a 2 ( θ 1 ) R 4 ( θ 1 ) T 0 T v t ( t ) W 0 2 d t R 2 ,

which shows that

(3.42) v M T 2 = v H 2 + 0 T v t ( t ) W 0 2 d t R 2 ,

i.e., Φ 1 ( M T ) M T .

Next, we prove that Φ1 is contractive. Now taking w 1 and w 2 in M T , subtracting the two equations (3.4) for v 1 = Φ1(w 1) and v 2 = Φ1(w 2) and setting v = v 1v 2, we obtain that for all ηW 0 and a.e. t ∈ [0, T], there holds

v t t ( , t ) , η + M [ w 1 ] s 2 ( ( Δ ) s v 1 ( , t ) , η ) M [ w 2 ] s 2 ( ( Δ ) s v 2 ( , t ) , η ) + ( v t ( , t ) , η ) W 0 = | w 1 ( , t ) | p 2 w 1 ( , t ) | w 2 ( , t ) | p 2 w 2 ( , t ) , η | w 1 t ( , t ) | q 2 w 1 t ( , t ) | w 2 t ( , t ) | q 2 w 2 t ( , t ) , η .

Then, we have

(3.43) v t t ( , t ) , η + a ( ( Δ ) s ( v 1 ( , t ) v 2 ( , t ) ) , η ) + [ w 1 ] s 2 ( θ 1 ) ( ( Δ ) s v 1 ( , t ) , η ) [ w 2 ] s 2 ( θ 1 ) ( ( Δ ) s v 2 ( , t ) , η ) + ( v t ( , t ) , η ) W 0 = | w 1 ( , t ) | p 2 w 1 ( , t ) | w 2 ( , t ) | p 2 w 2 ( , t ) , η | w 1 t ( , t ) | q 2 w 1 t ( , t ) | w 2 t ( , t ) | q 2 w 2 t ( , t ) , η .

Taking η = v t and integrating both sides of (3.43) on [0, t] ⊂ [0, T], we get

0 t 1 2 d d τ v t 2 2 + a 2 d d τ v W 0 2 + v t W 0 2 d τ = 0 t | w 1 ( , τ ) | p 2 w 1 ( , τ ) | w 2 ( , τ ) | p 2 w 2 ( , τ ) , v t d τ 0 t [ w 1 ] s 2 ( θ 1 ) ( v 1 , v t ) W 0 [ w 1 ] s 2 ( θ 1 ) ( v 2 , v t ) W 0 d τ 0 t [ w 1 ] s 2 ( θ 1 ) ( v 2 , v t ) W 0 [ w 2 ] s 2 ( θ 1 ) ( v 2 , v t ) W 0 d τ 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ 0 t | w 1 ( , τ ) | p 2 w 1 ( , τ ) | w 2 ( , τ ) | p 2 w 2 ( , τ ) , v t d τ + 0 t [ w 1 ] s 2 ( θ 1 ) ( v , v t ) W 0 + [ w 1 ] s 2 ( θ 1 ) [ w 2 ] s 2 ( θ 1 ) ( v 2 , v t ) W 0 d τ 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ .

which gives

(3.44) 1 2 v t 2 2 + a 2 v W 0 2 + 0 t v t W 0 2 d τ 0 t | w 1 ( , τ ) | p 2 w 1 ( , τ ) | w 2 ( , τ ) | p 2 w 2 ( , τ ) , v t d τ + 0 t [ w 1 ] s 2 ( θ 1 ) v W 0 v t W 0 d τ + 0 t [ w 1 ] s 2 ( θ 1 ) [ w 2 ] s 2 ( θ 1 ) v 2 W 0 v t W 0 d τ 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ

by Hölder inequality. Here, if w 1 = w 2 at some time t, we know

[ w 1 ] s 2 ( θ 1 ) [ w 2 ] s 2 ( θ 1 ) v 2 W 0 v t W 0 = 0

in (3.44). For the case w 1w 2, by the property of Gâteaux derivative, we have

(3.45) [ w 1 ] s 2 ( θ 1 ) [ w 2 ] s 2 ( θ 1 ) = w 1 W 0 2 ( θ 1 ) w 2 W 0 2 ( θ 1 ) = 0 1 d d ι w 2 W 0 + ι w 1 W 0 w 2 W 0 2 ( θ 1 ) d ι = 0 1 2 ( θ 1 ) w 2 W 0 + ι w 1 W 0 w 2 W 0 2 θ 3 w 1 W 0 w 2 W 0 d ι 0 1 2 ( θ 1 ) ι w 1 W 0 + ( 1 ι ) w 2 W 0 2 θ 3 w 1 W 0 w 2 W 0 d ι 0 1 2 ( θ 1 ) ι w 1 W 0 + ( 1 ι ) w 2 W 0 2 θ 3 d ι w 1 w 2 W 0 0 1 2 ( θ 1 ) w 1 W 0 + w 2 W 0 2 θ 3 d ι w 1 w 2 W 0 = 2 ( θ 1 ) w 1 W 0 + w 2 W 0 2 θ 3 w 1 w 2 W 0 .

Then, by (3.45) and Young inequality, we know that (3.44) gives

(3.46) 1 2 v t 2 2 + a 2 v W 0 2 + 0 t v t W 0 2 d τ 0 t | w 1 ( , τ ) | p 2 w 1 ( , τ ) | w 2 ( , τ ) | p 2 w 2 ( , τ ) , v t d τ + 0 t [ w 1 ] s 2 ( θ 1 ) v W 0 v t W 0 d τ + 2 0 t ( θ 1 ) w 1 W 0 + w 2 W 0 2 θ 3 w 1 w 2 W 0 v 2 W 0 v t W 0 d τ 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ 0 t | w 1 ( , τ ) | p 2 w 1 ( , τ ) | w 2 ( , τ ) | p 2 w 2 ( , τ ) , v t d τ + 0 t [ w 1 ] s 4 ( θ 1 ) v W 0 2 d τ + 1 4 0 t v t W 0 2 d τ + 0 t 4 ( θ 1 ) 2 w 1 W 0 + w 2 W 0 2 ( 2 θ 3 ) w 1 w 2 W 0 2 v 2 W 0 2 d τ + 1 4 0 t v t W 0 2 d τ 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ .

Since w 1 and w 2 are taken from the space M T , we know w 1 H R and w 2 H R , which mean w 1 W 0 a 1 2 R , w 2 W 0 a 1 2 R and w 1 w 2 W 0 w 1 W 0 + w 2 W 0 2 a 1 2 R . Due to w 1 , w 2 M T , we notice that w 1 W 0 and w 2 W 0 are continuous with respect to t on any interval [0, T] with the maximum and minimum. And due to Φ ( M T ) M T , we notice that v 1 H R and v 2 H R , which mean v 1 W 0 a 1 2 R , v 2 W 0 a 1 2 R and v 1 v 2 W 0 v 1 W 0 + v 2 W 0 2 a 1 2 R . Based on these facts, we know that (3.46) becomes

(3.47) 1 2 v t 2 2 + a 2 v W 0 2 + 0 t v t W 0 2 d τ 0 t | w 1 ( , τ ) | p 2 w 1 ( , τ ) | w 2 ( , τ ) | p 2 w 2 ( , τ ) , v t + C v W 0 2 d τ + 0 t 1 4 v t W 0 2 d τ 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ + C 0 t w 1 w 2 W 0 2 d τ + 0 t 1 4 v t W 0 2 d τ .

In order to estimate the right hand side of (3.47), by the similar process of proving Lemma 3.2, we have

| w 1 t | q 2 w 1 t | w 2 t | q 2 w 2 t ( q 1 ) ( | w 1 t | + | w 2 t | ) q 2 | w 1 t w 2 t | ,

which gives

(3.48) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ 0 t Ω w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) v t d x d τ 0 t Ω ( q 1 ) ( | w 1 t | + | w 2 t | ) q 2 | w 1 t w 2 t | v t d x d τ .

From the condition 2 q < 2 + 4 s N , we know that exists a 1 2 < β < 1 such that

(3.49) 2 q 2 + 4 s β N .

By using Hölder inequality, we know that (3.48) becomes

(3.50) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ ( q 1 ) 0 t Ω ( | w 1 t | + | w 2 t | ) k 1 ( q 2 ) d x 1 k 1 Ω | w 1 t w 2 t | k 2 d x 1 k 2 Ω | v t | k 3 d x 1 k 3 = ( q 1 ) 0 t | w 1 t | + | w 2 t | k 1 ( q 2 ) q 2 w 1 t w 2 t k 2 v t k 3 d τ ,

where k 1 N 2 s β , k 2 2 N N 2 ( 2 β 1 ) s < 2 N N 2 s and k 3 2 s * = 2 N N 2 s , with 1 k 1 + 1 k 2 + 1 k 3 = 1 . We can find 0 < σ < 1 such that

(3.51) 2 N N 2 ( 2 β 1 ) s = 2 N σ N 2 s = σ 2 s * < 2 s * .

Noticing the condition 2 q 2 + 4 s β N , we know k 1(q − 2) ≤ 2, which means that (3.50) becomes

(3.52) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ ( q 1 ) C 0 t | w 1 t | + | w 2 t | 2 q 2 w 1 t w 2 t σ 2 s * v t 2 s * d τ ( q 1 ) C 0 t w 1 t 2 + w 2 t 2 q 2 w 1 t w 2 t σ 2 s * v t 2 s * d τ .

Due to w 1 and w 2 being taken from the space M T , we have

w 1 t 2 2 R 2 , w 2 t 2 2 R 2 ,

which means (3.52) gives

(3.53) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ ( q 1 ) C R q 2 0 t w 1 t w 2 t σ 2 s * v t 2 s * d τ .

Here, from (3.51) and 1 2 < β < 1 , we have

(3.54) 2 < 2 N N 2 ( 2 β 1 ) s = σ 2 s * < 2 s * .

Then, by interpolation inequality, we know that (3.53) gives

(3.55) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ ( q 1 ) C R q 2 0 t w 1 t w 2 t 2 ρ w 1 t w 2 t 2 s * 1 ρ v t 2 s * d τ ,

where

(3.56) ρ 2 2 σ 2 s * σ 2 σ

and we know 0 < ρ < 1 according to (3.54). Using Hölder inequality and Young inequality, we know that (3.55) becomes

(3.57) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ ( q 1 ) C R q 2 0 t w 1 t w 2 t 2 ρ w 1 t w 2 t W 0 1 ρ v t W 0 d τ ( q 1 ) 2 C R 2 ( q 2 ) 0 t w 1 t w 2 t 2 2 ρ w 1 t w 2 t W 0 2 ( 1 ρ ) d τ + 1 8 0 t v t W 0 2 d τ ( q 1 ) 2 C R 2 ( q 2 ) 0 t max 0 τ T w 1 t w 2 t 2 2 ρ w 1 t w 2 t W 0 2 ( 1 ρ ) d τ + 1 8 0 t v t W 0 2 d τ ( q 1 ) 2 C R 2 ( q 2 ) 0 t w 1 w 2 H 2 ρ w 1 t w 2 t W 0 2 ( 1 ρ ) d τ + 1 8 0 t v t W 0 2 d τ ( q 1 ) 2 C R 2 ( q 2 ) w 1 w 2 H 2 ρ 0 t w 1 t w 2 t W 0 2 ( 1 ρ ) k 4 d τ 1 k 4 0 t 1 k 5 d τ 1 k 5 + 1 8 0 t v t W 0 2 d τ ,

where k 4 1 1 ρ , k 5 1 ρ , and ρ is defined by (3.56). Then, we know that (3.57) becomes

(3.58) 0 t | w 1 t ( , τ ) | q 2 w 1 t ( , τ ) | w 2 t ( , τ ) | q 2 w 2 t ( , τ ) , v t d τ ( q 1 ) 2 C R 2 ( q 2 ) w 1 w 2 H 2 ρ 0 t w 1 t w 2 t W 0 2 d τ 1 ρ T ρ + 1 8 0 t v t W 0 2 d τ ( q 1 ) 2 C R 2 ( q 2 ) w 1 w 2 H 2 ρ w 1 w 2 M T 2 ( 1 ρ ) T ρ + 1 8 0 t v t W 0 2 d τ ( q 1 ) 2 C R 2 ( q 2 ) w 1 w 2 M T 2 ρ w 1 w 2 M T 2 ( 1 ρ ) T ρ + 1 8 0 t v t W 0 2 d τ ( q 1 ) 2 C R 2 ( q 2 ) w 1 w 2 M T 2 T ρ + 1 8 0 t v t W 0 2 d τ .

And using Lemma 3.2, we obtain

| w 1 | p 2 w 1 | w 2 | p 2 w 2 ( p 1 ) ( | w 1 | + | w 2 | ) p 2 | w 1 w 2 |

for 2 s * > p > 2 . Arguing as above, we obtain

(3.59) | w 1 ( , t ) | p 2 w 1 ( , t ) | w 2 ( , t ) | p 2 w 2 ( , t ) , v t C w 1 w 2 W 0 2 + 1 8 v t W 0 2 .

By (3.58) and (3.59), we know (3.47) becomes

(3.60) 1 2 v t 2 2 + a 2 v W 0 2 + 1 4 0 t v t W 0 2 d τ . C 0 t v W 0 2 + C w 1 w 2 W 0 2 d τ + C T ρ w 1 w 2 M T 2 C T v H 2 + w 1 w 2 H 2 + C T ρ w 1 w 2 M T 2 C T v M T 2 + w 1 w 2 M T 2 + C T ρ w 1 w 2 M T 2 ,

where ρ is defined by (3.56). Then, from (3.60), we have

v M T 2 C T v M T 2 + C T w 1 w 2 M T 2 + C T ρ w 1 w 2 M T 2 ,

i.e.,

(3.61) ( 1 C T ) v M T 2 C T w 1 w 2 M T 2 + C T ρ w 1 w 2 M T 2 .

For sufficiently small T, we know 1 − CT > 0 and δ 1 C T + C T ρ 1 C T < 1 , which means (3.61) gives

v 1 v 2 M T 2 δ 1 w 1 w 2 M T 2 .

By the Banach fixed point theorem, there exists a unique local solution for problem (1.1) defined on [0, T]. Here, we can deduce that u t L q ([0, T], L q (Ω)). For the case 2 q 2 + 2 s N , replacing v ̇ h ( τ ) by u t (τ) and taking t = T in the proof of (3.18), we have

(3.62) 0 T Ω | u t ( τ ) | q d x d τ = 0 T Ω | u t ( τ ) | q 1 | u t ( τ ) | d x d τ C T + 1 2 0 T u t ( τ ) W 0 2 d τ < + .

For the case 2 + 2 s N < q < 2 + 4 s N , replacing v ̇ h ( τ ) by u t (τ) and taking t = T in the proof of (3.25), we have

(3.63) 0 T Ω | u t ( τ ) | q d x d τ = 0 T Ω | u t ( τ ) | q 1 | u t ( τ ) | d x d τ C T 1 ( 1 α ) ( q 1 ) + 0 T u t ( τ ) W 0 2 d τ < + .

Hence, the proof of Theorem 3.4 is completed. □

4 Subcritical initial energy level E(0) < d

In this section, we prove the global existence, asymptotic behavior and blowup of the solution to problem (1.1) for subcritical initial energy level E(0) < d.

4.1 Global existence for subcritical initial energy E(0) < d

In this subsection, as an application of Galerkin method, we prove the global existence of the solution to problem (1.1) for subcritical initial energy level E(0) < d. First, we prove the invariance of set U for E(0) < d as follows.

Lemma 4.1.

(Invariant set U for E(0) < d). Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Then all the solutions to problem (1.1) with E(0) < d belong to U, provided u 0U.

Proof.

Let u be a solution to problem (1.1) with E(0) < d, u 0U and T be the maximum existence time of u(t). Arguing by contradiction, due to the continuity of I(u(t)) in t, we suppose that t* is the first time such that I(u(t*)) = 0 and I(u(t)) > 0 for t ∈ [0, t*). By Lemma 2.3, we have E(t) ≤ E(0) < d, which together with the definition of the depth of potential well d gives the following contradiction

d > E ( 0 ) E ( t * ) J ( u ( t * ) ) d .

Hence, the proof is completed. □

Theorem 4.2.

(Global existence for E(0) < d). Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Suppose that 0 < E(0) < d and u 0U. Then problem (1.1) admits a global solution.

Proof.

Let ω j (x) be a sequence of eigenfunctions of the fractional Laplacian operator (−Δ) s , which is an orthogonal basis of L 2(Ω) and W 0. By Galerkin method, we construct approximate solution for problem (1.1)

u m ( x , t ) = j = 1 m g j m ( t ) ω j ( x ) , m = 1,2 , ,

satisfying

(4.1) u mtt , ω h + M [ u m ] s 2 ( u m , ω h ) W 0 + ( u m t , ω h ) W 0 + | u m t | q 2 u m t , ω h = | u m | p 2 u m , ω h

and

(4.2) u m ( x , 0 ) = j = 1 m g j m ( 0 ) ω j ( x ) u 0 ( x )  in  W 0  as  m , u m t ( x , 0 ) = j = 1 m g j m ( 0 ) ω j ( x ) u 1 ( x )  in  L 2 ( Ω )  as  m ,

where h = 1, 2, …, m. Multiplying (4.1) by g h m ( t ) and summing for h, we can obtain

(4.3) u mtt , u m t + M [ u m ] s 2 ( u m , u m t ) W 0 + ( u m t , u m t ) W 0 + | u m t | q 2 u m t , u m t = | u m | p 2 u m , u m t .

Integrating (4.3) with respect to t, we have

1 2 u m t 2 + a 2 u m W 0 2 + 1 2 θ u m W 0 2 θ + 0 t u m τ W 0 2 d τ + 0 t u m τ q q d τ = 1 p u m p p + 1 2 u m t ( 0 ) 2 + a 2 u m ( 0 ) W 0 2 + 1 2 θ u m ( 0 ) W 0 2 θ 1 p u m ( 0 ) p p .

Combining (2.5) with (2.6), we can obtain

1 2 u m t 2 + J ( u m ) + 0 t u m τ W 0 2 d τ + 0 t u m τ q q d τ = E m ( 0 ) .

From (2.8), we see that

1 2 u m t 2 + 1 p I ( u m ) + a ( p 2 ) 2 p u m W 0 2 + p 2 θ 2 θ p u m W 0 2 θ + 0 t u m τ W 0 2 d τ + 0 t u m τ q q d τ = E m ( 0 ) .

By (4.2), we get E m (0) → E(0) and u m (0) → u 0 as m. From 0 < E(0) < d and u 0U, we have u m (0) ∈ U and E m (0) < d, i.e.,

(4.4) 1 2 u m t 2 + 1 p I ( u m ) + a ( p 2 ) 2 p u m W 0 2 + p 2 θ 2 θ p u m W 0 2 θ + 0 t u m τ W 0 2 d τ + 0 t u m τ q q d τ < d

for sufficiently large m and t ≥ 0. Therefore, we get u m (t) ∈ U by the argument in the proof of Lemma 4.1. Since u m (t) ∈ U, i.e., I(u m (t)) > 0, then (4.4) gives

u m t 2 < 2 d , u m W 0 2 θ < 2 θ p p 2 θ d , 0 t u m τ W 0 2 d τ < d , 0 t u m τ q q d τ < d

for sufficiently large m. Due to u m W 0, Proposition 2.1 gives

u m p p C 0 p 2 u m W 0 p C 0 p 2 2 θ p p 2 θ d p 2 θ .

Hence, there exists a subsequence of {u m }, such that

(4.5) u m u  in  L ( 0 , ; W 0 )  weakly  , u m t u t  in  L ( 0 , ; L 2 ( Ω ) ) L 2 ( 0 , ; W 0 )  weakly , | u m t | q 2 u m t | u | p 2 u  in  L q ( 0 , ; L q )  weakly , | u m | p 2 u m | u t | q 2 u t  in  L ( 0 , ; L p )  weakly 

for sufficiently large m, where 1 q + 1 q = 1 and 1 p + 1 p = 1 . Integrating (4.1) from 0 to t, we have

(4.6) ( u m t , ω h ) + 0 t M [ u m ] s 2 ( u m , ω h ) W 0 d τ + ( u m , ω h ) W 0 + 0 t ( | u m τ | q 2 u m τ , ω h ) d τ = 0 t ( | u m | p 2 u m , ω h ) d τ + ( u 0 m , w h ) W 0 + ( u 1 m , w h ) .

Hence, by (4.5), we can pass to the limit of (4.6) to obtain a weak solution to problem (1.1) satisfying (3.3). By (4.2), we obtain u(x, 0) = u 0W 0 and u t (x, 0) = u 1L 2(Ω). □

4.2 Asymptotic behavior for subcritical initial energy E(0) < d

In this subsection, we are concerned with the asymptotic behavior of the global solution to problem (1.1) with E(0) < d. First, we explore the relationship between the norm  u W 0 and the energy E(t), which will be used to prove the asymptotic behavior of the global solution.

Lemma 4.3.

Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Assume that u 0U, E(0) < d, and u is the solution to problem (1.1). Then

(4.7) u W 0 2 θ 2 θ p p 2 θ E ( t )

for t ∈ [0, + ).

Proof.

By Lemma 2.3, we know that the energy E(t) is non-increasing. Since E(0) < d, combining (2.5), (2.6) and (2.8), we get

(4.8) d > E ( 0 ) E ( t ) J ( u ) = 1 p I ( u ) + a ( p 2 ) 2 p u W 0 2 + p 2 θ 2 θ p u W 0 2 θ .

By Lemma 4.1 and u 0U, we deduce uU, i.e., I(u) > 0 for t ∈ [0, ). Thus (4.8) shows that

E ( t ) > a ( p 2 ) 2 p u W 0 2 + p 2 θ 2 θ p u W 0 2 θ

for 2 s * > p > 2 θ 3 and a > 0. Obviously, (4.7) holds. □

Theorem 4.4.

(Asymptotic behavior for E(0) < d) Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. If E(0) < d and u 0U, then any global solution u to problem (1.1) has the following decay estimate

E ( t ) κ e γ t ,

where κ, γ > 0 will be discussed later.

Proof.

By Theorem 4.2, we know that the solution u(t) of problem (1.1) exists globally.

Let

H ( t ) E ( t ) + ε ( u , u t ) + 1 2 u W 0 2

for ɛ > 0 and t ∈ [0, ). Next, we claim that for a suitable ɛ > 0, there exist two positive constants δ 1, δ 2 > 0 such that

(4.9) δ 1 E ( t ) H ( t ) δ 2 E ( t ) .

Testing equation (1.1) with u, we obtain

(4.10) u , u t t = u p p a u W 0 2 u W 0 2 θ ( u , u t ) W 0 Ω | u t | q 2 u t u d x .

Taking derivative of H(t) with respect to t, combining Lemma 2.3 with (4.10), we get

(4.11) H ( t ) = E ( t ) + ε u t 2 + u , u t t + ( u , u t ) W 0 = u t W 0 2 u t q q + ε u t 2 + u , u t t + ( u , u t ) W 0 = u t W 0 2 u t q q + ε u t 2 + u p p a u W 0 2 u W 0 2 θ Ω | u t | q 2 u t u d x .

Next, we estimate each term to the right side of (4.11). Firstly, we deal with u p p . By Proposition 2.1 and (4.7), we have

(4.12) u p p C 0 p 2 u W 0 p = C 0 p 2 u W 0 p 2 θ u W 0 2 θ C 0 p 2 2 θ p p 2 θ E ( 0 ) p 2 θ 2 θ u W 0 2 θ = ξ 0 u W 0 2 θ ,

where C 0 is defined in (2.10) and

ξ 0 C 0 p 2 2 θ p p 2 θ E ( 0 ) p 2 θ 2 θ .

By E(0) < d and Lemma 2.2, we have

0 < ξ 0 < C 0 p 2 2 θ p p 2 θ d p 2 θ 2 θ < 1 .

Combining (2.5) and (4.12), for 0 < β 0 < 1, we can get

(4.13) u p p = ( 1 β 0 ) u p p + β 0 u p p ξ 0 ( 1 β 0 ) u W 0 2 θ + β 0 u p p ξ 0 ( 1 β 0 ) u W 0 2 θ + β 0 p E ( t ) + p 2 u t 2 + a p 2 u W 0 2 + p 2 θ u W 0 2 θ

Substituting (4.13) into (4.11), we have

(4.14) H ( t ) u t W 0 2 u t q q + ε u t 2 a u W 0 2 u W 0 2 θ Ω | u t | q 2 u t u d x + ε ξ 0 ( 1 β 0 ) u W 0 2 θ + ε β 0 p E ( t ) + p 2 u t 2 + a p 2 u W 0 2 + p 2 θ u W 0 2 θ = u t W 0 2 u t q q + ε 1 + p 2 β 0 u t 2 + a ε β 0 p 2 1 u W 0 2 ε β 0 p E ( t ) + ε p β 0 2 θ + ξ 0 ( 1 β 0 ) 1 u W 0 2 θ ε Ω | u t | q 2 u t u d x .

Next, we estimate the term Ω|u t | q−2 u t udx in (4.14). By Hölder inequality

(4.15) 0 t | f ( s ) g ( s ) | d s 0 t | f ( s ) | ρ d s 1 ρ 0 t | g ( s ) | ϱ d s 1 ϱ ,

where ρ, ϱ > 0 and 1 ρ + 1 ϱ = 1 and ɛ-Young’s inequality

(4.16) X Y δ ̃ k 1 k 1 X k 1 + δ ̃ k 2 k 2 Y k 2 , X , Y 0 , δ ̃ > 0 , 1 k 1 + 1 k 2 = 1 .

Let f ( s ) = | u t | q 1 , g ( s ) = u , ρ = q q 1 , ϱ = q in (4.15) and pick k 1 = q q 1 and k 2 = q in (4.16), we have

(4.17) Ω | u t | q 2 u t u d x u t q q 1 u q q 1 q δ ̃ q q 1 u t q q + 1 q δ ̃ q u q q .

Substituting (4.17) into (4.14), we have

(4.18) H ( t ) u t W 0 2 u t q q + ε 1 + p 2 β 0 u t 2 + a ε β 0 p 2 1 u W 0 2 ε β 0 p E ( t ) + ε p β 0 2 θ + ξ 0 ( 1 β 0 ) 1 u W 0 2 θ + ε q 1 q δ ̃ q q 1 u t q q + ε q δ ̃ q u q q = u t W 0 2 + ε q 1 q δ ̃ q q 1 1 u t q q + ε 1 + p 2 β 0 u t 2 ε β 0 p E ( t ) + a ε β 0 p 2 1 u W 0 2 + ε p β 0 2 θ + ξ 0 ( 1 β 0 ) 1 u W 0 2 θ + ε q δ ̃ q u q q .

Using Proposition 2.1, we have u t 2 C 0 u t W 0 2 . Thus (4.18) becomes

(4.19) H ( t ) ε C 0 ( 2 + p β 0 ) 2 1 u t W 0 2 + ε q 1 q δ ̃ q q 1 1 u t q q + a ε β 0 p 2 1 u W 0 2 ε β 0 p E ( t ) + ε p β 0 2 θ + ξ 0 ( 1 β 0 ) 1 u W 0 2 θ + ε q δ ̃ q u q q .

Now for 0 < ξ 0 < 1 and (1.2), we have

0 < θ ( 1 ξ 0 ) p 2 θ ξ 0 = θ ( 1 ξ 0 ) p 2 θ + 2 θ ( 1 ξ 0 ) < θ p < 1 2

such that

β 0 2 θ ( 1 ξ 0 ) p 2 θ + 2 θ ( 1 ξ 0 ) , 2 θ p ( 0,1 ) ,

which means

(4.20) p β 0 2 θ + ξ 0 ( 1 β 0 ) 1 0

and

(4.21) β 0 p 2 1 < 0 .

Then combining (4.20) with (4.7)(4.19) becomes

(4.22) H ( t ) ε C 0 ( 2 + p β 0 ) 2 1 u t W 0 2 + ε q 1 q δ ̃ q q 1 1 u t q q + a ε β 0 p 2 1 u W 0 2 ε β 0 p E ( t ) + ε 2 θ p p 2 θ p β 0 2 θ + ξ 0 ( 1 β 0 ) 1 E ( t ) + ε q δ ̃ q u q q = ε C 0 1 + p 2 β 0 1 u t W 0 2 + ε q 1 q δ ̃ q q 1 1 u t q q + a ε β 0 p 2 1 u W 0 2 2 θ p ε p 2 θ ( 1 β 0 ) ( 1 ξ 0 ) E ( t ) + ε q δ ̃ q u q q .

For the term u q q in (4.22), by Proposition 2.1 and (4.7), we have

u q q C 0 q 2 u W 0 q C 0 p 2 2 θ p p 2 θ E ( 0 ) p 2 θ 2 θ 2 θ p p 2 θ E ( t ) ,

which combining (4.22) indicates

(4.23) H ( t ) ε C 0 ( 2 + p β 0 ) 2 1 u t W 0 2 + ε q 1 q δ ̃ q q 1 1 u t q q + a ε β 0 p 2 1 u W 0 2 2 θ p ε p 2 θ ( 1 β 0 ) ( 1 ξ 0 ) E ( t ) + ε q δ ̃ q C 0 p 2 2 θ p p 2 θ E ( 0 ) p 2 θ 2 θ 2 θ p p 2 θ E ( t ) = ε C 0 ( 2 + p β 0 ) 2 1 u t W 0 2 + ε q 1 q δ ̃ q q 1 1 u t q q + a ε β 0 p 2 1 u W 0 2 2 θ p ε p 2 θ ( 1 β 0 ) ( 1 ξ 0 ) C 0 p 2 q δ ̃ q 2 θ p p 2 θ E ( 0 ) p 2 θ 2 θ E ( t ) = λ 1 ̃ u t W 0 2 + λ 2 ̃ u t q q + λ 3 ̃ u W 0 2 2 θ p ε p 2 θ λ 4 ̃ E ( t ) ,

where

λ 1 ̃ = ε C 0 ( 2 + p β 0 ) 2 1 ,

λ 2 ̃ = ε q 1 q δ ̃ q q 1 1 ,

λ 3 ̃ = a ε β 0 p 2 1

and

λ 4 ̃ = ( 1 β 0 ) ( 1 ξ 0 ) C 0 p 2 q δ ̃ q 2 θ p p 2 θ E ( 0 ) p 2 θ 2 θ .

Since 0 < ξ < 1, we choose large enough δ ̃ such that λ 4 ̃ > 0 . After fixing δ ̃ , we choose small enough ɛ such that

(4.24) λ 1 ̃ < 0 , λ 2 ̃ < 0 .

Combining (4.21), (4.24) and (4.9), (4.23) becomes

H ( t ) 2 θ p ε p 2 θ λ 4 ̃ E ( t ) 2 θ p ε δ 2 ( p 2 θ ) λ 4 ̃ H ( t ) .

Therefore, by Gronwall inequality, we can obtain

H ( t ) H ( 0 ) e γ t , γ = 2 θ p ε δ 2 ( p 2 θ ) λ 4 ̃ > 0 .

Further by (4.9), there holds

E ( t ) κ e γ t , κ = δ 2 δ 1 E ( 0 ) .

The proof is completed. □

4.3 Finite time blowup for subcritical energy E(0) < d

In this subsection, we prove the invariance of set V with E(0) < d, and then based on this property, we prove the finite time blowup of the solution.

Lemma 4.5.

(Invariant set V for E(0) < d). Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. If E(0) < d and u 0V, then all the weak solutions to problem (1.1) belong to V.

Proof.

The proof of is similar to Lemma 4.1, we omit it here. □

Lemma 4.6.

For I(u) < 0 and uW 0, there holds

I ( u ) < p ( J ( u ) d )

and

d < p 2 θ 2 θ p u W 0 2 θ + a ( p 2 ) 2 p u W 0 2

satisfying (1.4).

Proof.

From uW 0 and I(u) < 0, (iv) of Lemma 2.4 can be applied to ensure that there exists a λ* such that 0 < λ* < 1 and I(λ*u) = 0. From the definition of d, we have

d J ( λ * u ) = 1 p I ( λ * u ) + p 2 θ 2 θ p λ * u W 0 2 θ + a ( p 2 ) 2 p λ * u W 0 2 = p 2 θ 2 θ p λ * u W 0 2 θ + a ( p 2 ) 2 p λ * u W 0 2 < p 2 θ 2 θ p u W 0 2 θ + a ( p 2 ) 2 p u W 0 2 ,

which means

d < p 2 θ 2 θ p u W 0 2 θ + a ( p 2 ) 2 p u W 0 2 = J ( u ) 1 p I ( u ) .

The proof is completed. □

Next, we prove the finite time blowup of the solution to problem (1.1) for E(0) < d.

Theorem 4.7.

(Finite time blowup for E(0) < d with q = 2). Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Suppose that E(0) < d and u 0V, then the solution to problem (1.1) blows up in finite time.

Proof.

Let u(x, t) be a weak solution to problem (1.1) with I(u 0) < 0 and E(0) < d. Arguing by contradiction, we suppose that u(x, t) is a global solution, i.e., T = +. For any T 0 > 0, we introduce the following auxiliary function

(4.25) F ( t ) : = u ( t ) 2 + 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ + ( T 0 t ) ( u 0 W 0 2 + u 0 2 ) .

It is obviously that F(t) > 0 for all t ∈ [0, T 0]. From the continuity of F(t) with respect to t, it is easy to see that there exists ρ > 0 (independent of the choice of T 0) such that

F ( t ) ρ ,  for all  t [ 0 , T 0 ] .

Moreover, for t ∈ [0, T 0], direct calculation gives

(4.26) F ( t ) = 2 ( u ( t ) , u t ( t ) ) + u ( t ) W 0 2 + u ( t ) 2 u 0 W 0 2 + u 0 2 = 2 ( u ( t ) , u t ( t ) ) + 2 0 t ( ( u ( τ ) , u t ( τ ) ) W 0 + ( u ( τ ) , u t ( τ ) ) ) d τ

and

(4.27) ( F ( t ) ) 2 = 4 ( u ( t ) , u t ( t ) ) 2 + 4 0 t ( ( u ( τ ) , u t ( τ ) ) W 0 + ( u ( τ ) , u t ( τ ) ) ) d τ 2 + 8 ( u ( t ) , u t ( t ) ) 0 t ( ( u ( τ ) , u t ( τ ) ) W 0 + ( u ( τ ) , u t ( τ ) ) ) d τ .

Next, we estimate the right hand side of (4.27). By Cauchy-Schwarz inequality, we obtain

( u ( t ) , u t ( t ) ) 2 u ( t ) 2 u t ( t ) 2

and

(4.28) 0 t ( ( u ( τ ) , u t ( τ ) ) W 0 + ( u ( τ ) , u t ( τ ) ) ) d τ 2 0 t ( u ( τ ) W 0 u t ( τ ) W 0 + u ( τ ) u t ( τ ) ) d τ 2 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ .

Combining (4.28) and Young inequality, we derive

2 ( u ( t ) , u t ( t ) ) 0 t ( ( u ( τ ) , u t ( τ ) ) W 0 + ( u ( τ ) , u t ( τ ) ) ) d τ 2 u ( t ) 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ 1 2 × u t ( t ) 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ 1 2 u ( t ) 2 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ + u t ( t ) 2 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ .

Then (4.27) becomes

(4.29) ( F ( t ) ) 2 4 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ + 4 u ( t ) 2 u t ( t ) 2 + 4 u ( t ) 2 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ + 4 u t ( t ) 2 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ = 4 u ( t ) 2 + 0 t ( u ( τ ) W 0 2 + u ( τ ) 2 ) d τ × u t ( t ) 2 + 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ 4 F ( t ) u t ( t ) 2 + 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ .

Set ϕ = u(⋅, t) in (3.2), using (2.7), we obtain

(4.30) u ( t ) , u t t ( t ) = u ( t ) p p a u ( t ) W 0 2 u ( t ) W 0 2 θ ( u ( t ) , u t ( t ) ) W 0 ( u ( t ) , u t ( t ) ) = I ( u ( t ) ) ( u ( t ) , u t ( t ) ) W 0 ( u ( t ) , u t ( t ) ) .

Then combining (4.26) and (4.30), we have

(4.31) F ( t ) = 2 u t ( t ) 2 + 2 u ( t ) , u t t ( t ) + 2 ( u ( t ) , u t ( t ) ) W 0 + 2 ( u ( t ) , u t ( t ) ) = 2 u t ( t ) 2 2 I ( u ( t ) ) 2 ( u ( t ) , u t ( t ) ) W 0 2 ( u ( t ) , u t ( t ) ) + 2 ( u ( t ) , u t ( t ) ) W 0 + 2 ( u ( t ) , u t ( t ) ) = 2 u t ( t ) 2 2 I ( u ( t ) ) .

By (4.25), (4.29) and (4.31), we get

(4.32) F ( t ) F ( t ) p + 2 4 ( F ( t ) ) 2 ( p + 2 ) F ( t ) u t ( t ) 2 + 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ + F ( t ) ( 2 u t ( t ) 2 2 I ( u ( t ) ) ) = F ( t ) p u t ( t ) 2 2 I ( u ( t ) ) ( p + 2 ) 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ = F ( t ) ξ ( t ) ,

where

(4.33) ξ ( t ) p u t ( t ) 2 2 I ( u ( t ) ) ( p + 2 ) 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ .

From (2.12) and q = 2, we get

E ( 0 ) = E ( t ) + 0 t ( u t ( τ ) 2 + u t ( τ ) W 0 2 ) d τ = 0 t ( u t ( τ ) 2 + u t ( τ ) W 0 2 ) d τ + 1 2 u t ( t ) 2 + J ( u ( t ) ) = 0 t ( u t ( τ ) 2 + u t ( τ ) W 0 2 ) d τ + 1 2 u t ( t ) 2 + 1 p I ( u ( t ) ) + p 2 θ 2 θ p u ( t ) W 0 2 θ + a ( p 2 ) 2 p u ( t ) W 0 2 ,

hence

(4.34) 2 p E ( 0 ) = 2 p 0 t u t ( τ ) 2 + u t ( τ ) W 0 2 d τ + p u t ( t ) 2 + 2 I ( u ( t ) ) + p 2 θ θ u ( t ) W 0 2 θ + a ( p 2 ) u ( t ) W 0 2 .

From (4.33) and (4.34), we obtain

(4.35) ξ ( t ) = p 2 θ θ u ( t ) W 0 2 θ + a ( p 2 ) u ( t ) W 0 2 2 p E ( 0 ) + ( p 2 ) 0 t ( u t ( τ ) W 0 2 + u t ( τ ) 2 ) d τ

for 2 s * > p > 2 θ . From Lemma 4.6 and E(0) < d, we obtain

2 p E ( 0 ) < 2 p d < p 2 θ θ u ( t ) W 0 2 θ + a ( p 2 ) u ( t ) W 0 2 ,

then (4.35) gives

ξ ( t ) > σ 1 > 0 .

So (4.32) becomes

F ( t ) F ( t ) p + 2 4 ( F ( t ) ) 2 > F ( t ) σ 1 > 0 , t [ 0 , T 0 ] .

Let y ( t ) F ( t ) p 2 4 , then we have

y ( t ) = p 2 4 p 2 4 1 F ( t ) p 2 4 2 ( F ( t ) ) 2 p 2 4 F ( t ) p 2 4 1 F ( t ) = p 2 4 F ( t ) p 2 4 2 F ( t ) F ( t ) p + 2 4 ( F ( t ) ) 2 < 0

for every t ∈ [0, T 0], which means that

lim t T y ( t ) = 0 ,

for some finite time T > 0, i.e.,

lim t T F ( t ) = +

The proof is completed. □

5 Critical initial energy level E(0) = d

In this section, we shall study the global existence, asymptotic behavior of global solution and blowup of solution to problem (1.1) at critical initial energy level E(0) = d.

5.1 Global existence for critical initial energy E(0) = d

Theorem 5.1.

(Global existence for E(0) = d). Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Suppose that E(0) = d and u 0U, then problem (1.1) admits a global solution u with uL ([0, ), W 0) and u t L 2([0, ), W 0).

Proof.

We prove the theorem in two cases.

Case 1:

u 0 W 0 2 0

Let λ m = 1 1 m and u 0m = λ m u 0, m = 2, 3, …. Consider new initial conditions

u ( x , 0 ) = u 0 m ( x )  and  u t ( x , 0 ) = u 1 ( x )

for problem (1.1). From u 0U, i.e., I(u 0) > 0 and (iv) in Lemma 2.4, we know λ* = λ*(u 0) > 1. It implies 1 1 m < 1 < λ * and I(u 0m ) > 0. Combining (iii) in Lemma 2.4, we get J(u 0m ) < J(u 0). Therefore,

E m ( 0 ) = 1 2 u 1 2 + J ( u 0 m ) < 1 2 u 1 2 + J ( u 0 ) = E ( 0 ) = d .

Together with sufficiently large m, it is easy to see u m L ([0, ), W 0) and u mt L 2([0, ), W 0) from Theorem 4.2. The rest of the proof is similar to Theorem 4.2.

Case 2:

u 0 W 0 2 = 0

Let λ m = 1 1 m and u 1m = λ m u 1, m = 2, 3, …. Obviously, λ m → 1 as m. Consider the following initial conditions

u ( x , 0 ) = u 0 ( x )  and  u t ( x , 0 ) = u 1 m ( x )

for problem (1.1). By u 0 W 0 2 = 0 , we have J(u 0) = 0 and

1 2 u 1 2 + J ( u 0 ) = 1 2 u 1 2 = E ( 0 ) = d ,

which yields

E m ( 0 ) = 1 2 u 1 m 2 + J ( u 0 ) = 1 2 u 1 m 2 = 1 2 λ m u 1 2 = λ m 2 2 u 1 2 < 1 2 u 1 2 = E ( 0 ) = d .

Combining Theorem 4.2, we can deduce u m L ([0, ), W 0) and u mt L 2([0, ), W 0) for sufficiently large m. The rest proof is similar to Theorem 4.2. □

5.2 Asymptotic behavior for critical initial energy E(0) = d

Next, we give a lemma as follows to prove the asymptotic behavior of global solution to problem (1.1) with E(0) = d.

Lemma 5.2.

Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. If E(0) = d, assume that u(x, t) is the solution to problem (1.1). Then we have

0 t u t W 0 2 + u t q q d τ > 0

for t ∈ (0, ).

Proof.

Let u(t) be any nontrivial solution for problem (1.1) with E(0) = d. Arguing by contradiction, we suppose that

0 t u t W 0 2 + u t q q d τ 0

for t ∈ (0, ), which means u t W 0 2 + u t q q = 0 for t ∈ (0, ), i.e., u(x, t) is the steady-state solution, which is a contradiction. □

Theorem 5.3.

(Asymptotic behavior for E(0) = d) Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. If E(0) = d and u 0U, then any global (weak) solution u for problem (1.1) has the following decay estimate

E ( t ) κ e γ t ,

where κ, γ > 0.

Proof.

We have proved that the solution u to problem (1.1) is global for E(0) = d in Theorem 5.1, then we claim that uU for t ∈ (0, ). Arguing by contradiction, we suppose that t 0 > 0 is the first time such that I(u(t 0)) = 0. By the definition of d, we get J(u(t 0)) ≥ d. By (2.12), we have

d J ( u ( t 0 ) ) 1 2 u t ( t 0 ) 2 + J ( u ( t 0 ) ) = d 0 t 0 u t W 0 2 d τ 0 t 0 u t q q d τ d

for any t 0 > 0. Hence we deduce J(u(t 0)) = d and 1 2 u t ( t 0 ) 2 = 0 , i,e., u t ≡ 0 for t ∈ [0, t 0]. Set ϕ = u(⋅, t) in (3.2), using (2.7), we obtain (4.30). Combining u t ≡ 0 and (4.30), we know I(u) = 0 for t ∈ [0, t 0]. It contradicts I(u 0) > 0. Hence we have uU for t ∈ (0, ).

By the continuity of J(u) and I(u) with respect to t, we take a t 1 > 0 as the initial time, then uU for t > t 1. Therefore, by the same method with Theorem 4.4, we have E(t) ≤ κeγt , t 0t. The proof is completed. □

5.3 Finite time blowup for critical initial energy E(0) = d

Theorem 5.4.

(Finite time blowup for E(0) = d with q = 2). Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Suppose that E(0) = d and u 0V, then the solution to problem (1.1) blows up in finite time.

Proof.

Let u(t) be any local weak solution to problem (1.1) with E(0) = d > 0 and u 0V. Next, we prove u(t) ∈ V for t ∈ (0, ). Arguing by contradiction, from the continuity of J(u) and I(u) with respect to t, there exists a sufficiently small t 0 > 0 such that I(u(t 0)) < 0 and E(t 0) > 0. Combining with (4.30), we have u t ≠ 0 for 0 < tt 0. Hence, by Lemma 2.3, there holds

E ( t 0 ) = 1 2 u t ( t 0 ) 2 + J ( u ( t 0 ) ) = d 0 t 0 u t W 0 2 d τ 0 t 0 u t q q d τ d .

By Lemma 4.5 and taking t = t 0 as the initial time, we get uV for t > t 0. The rest of the proof is similar to Theorem 4.7. □

6 Arbitrarily positive initial energy level E(0) > 0

6.1 Finite time blowup for arbitrarily positive initial energy E(0) > 0

In this part, we consider the case of q = 2. In order to prove the blowup of solution to problem (1.1) with E(0) > 0, we first give the following lemmas.

Lemma 6.1.

Let (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) and E(0) > 0 hold, and the initial data satisfy

(6.1) u 0 2 + u 0 W 0 2 + 2 ( u 0 , u 1 ) > 2 p ( 1 + 2 C 0 ) p 2 E ( 0 )

hold, where C 0 > 0 is given in Proposition 2.1, then

F ̃ ( t ) u 2 + u W 0 2 + 2 ( u , u t )

is positive and strictly increasing provided u(t) ∈ V.

Proof.

Taking derivative of F ̃ ( t ) with respect to t, there holds

F ̃ ( t ) = 2 ( u , u t ) + 2 ( u , u t ) W 0 + 2 u t 2 + 2 u , u t t = 2 u t 2 2 I ( u ) .

By u(t) ∈ V, we know F ̃ ( t ) > 0 for any t ∈ [0, + ), i.e., F ̃ ( t ) is increasing with respect to t. From (6.1), obviously we can get F ̃ ( 0 ) > 0 , which combines with F ̃ ( t ) > 0 giving that F ̃ ( t ) is positive and strictly increasing. □

Lemma 6.2.

(Invariant set V with E(0) > 0.) Let a > 1, (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold, and T be the maximum existence time of solution. If E(0) > 0, u 0V and (6.1) hold, then all the weak solutions to problem (1.1) belong to V.

Proof.

We prove u(t) ∈ V by contradiction. Due to the continuity of I(u(t)) with respect to t, we suppose that there exists the first time t 0 ∈ (0, T) such that I(u(t 0)) = 0 and I(u(t)) < 0 for t ∈ [0, t 0). Hence, from Lemma 6.1, we know

(6.2) u 2 + u W 0 2 + 2 ( u , u t ) > u 0 2 + u 0 W 0 2 + 2 ( u 0 , u 1 ) > 2 p ( 1 + 2 C 0 ) p 2 E ( 0 )

for t ∈ [0, t 0) and C 0 > 0. Combining (2.5), (2.6), (2.8) and (2.12), we obtain

(6.3) E ( 0 ) = E ( t ) + 0 t u t W 0 2 d τ + 0 t u t 2 d τ = 1 2 u t 2 + J ( u ) + 0 t u t W 0 2 d τ + 0 t u t 2 d τ = 1 2 u t 2 + 1 p I ( u ) + a ( p 2 ) 2 p u W 0 2 + p 2 θ 2 p θ u W 0 2 θ + 0 t u t W 0 2 d τ + 0 t u t 2 d τ 1 2 u t 2 + 1 p I ( u ) + a ( p 2 ) 2 p u W 0 2

by a > 1 and (1.2). By Lemma 2.3 and the fact I(u(t 0)) = 0, (6.3) becomes

(6.4) E ( 0 ) E ( t 0 ) 1 2 u t ( t 0 ) 2 + 1 p I ( u ( t 0 ) ) + a ( p 2 ) 2 p u ( t 0 ) W 0 2 = 1 2 u t ( t 0 ) 2 + a ( p 2 ) 2 p u ( t 0 ) W 0 2 p 2 2 p ( 1 + 2 C 0 ) u t ( t 0 ) 2 + u ( t 0 ) W 0 2 + a ( p 2 ) 2 p p 2 2 p ( 1 + 2 C 0 ) u ( t 0 ) W 0 2 p 2 2 p ( 1 + 2 C 0 ) u t ( t 0 ) 2 + u ( t 0 ) W 0 2 + ( p 2 ) ( a ( 1 + 2 C 0 ) 1 ) 2 p ( 1 + 2 C 0 ) C 0 u ( t 0 ) 2 p 2 2 p ( 1 + 2 C 0 ) u t ( t 0 ) 2 + u ( t 0 ) W 0 2 + 2 u ( t 0 ) 2 p 2 2 p ( 1 + 2 C 0 ) u ( t 0 ) 2 + u ( t 0 ) W 0 2 + 2 ( u ( t 0 ) , u t ( t 0 ) )

by a > 1 and (1.2). It is easy to see that (6.4) contradicts (6.1). The proof is completed. □

Theorem 6.3.

(Finite time blowup for E(0) > 0 with q = 2). Let a > 1, (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Suppose that E(0) > 0, u 0V (6.1) hold, then the solution to problem (1.1) blows up in finite time.

Proof.

Arguing by contradition, we suppose that solution exists globally, i.e., T = +. Recalling auxiliary function F(t) as (4.25) and

F ( t ) F ( t ) η 1 4 ( F ( t ) ) 2 η 1 F ( t ) u t ( t ) 2 + 0 t u t ( τ ) W 0 2 + u t ( τ ) 2 d τ + F ( t ) 2 u t ( t ) 2 2 I ( u ( t ) ) = F ( t ) ξ ̃ ( t ) ,

where η 1 > 4 will be chosen later, and

(6.5) ξ ̃ ( t ) ( 2 η 1 ) u t ( t ) 2 2 I ( u ( t ) ) η 1 0 t u t ( τ ) W 0 2 + u t ( τ ) 2 d τ .

By (6.3), we get

(6.6) 2 p E ( 0 ) = 2 p 0 t u t ( τ ) W 0 2 + u t ( τ ) 2 d τ + p u t ( t ) 2 + 2 I ( u ) + a ( p 2 ) u W 0 2 + p 2 θ θ u W 0 2 θ ,

where (1.2) and a > 1 hold. Combining (6.5) and (6.6) and setting

η 1 = 4 + ( p + 2 ) 2 C 0 1 + 2 C 0 > 4 ,

we further derive

(6.7) ξ ̃ ( t ) = ( p + 2 η 1 ) u t ( t ) 2 2 p E ( 0 ) + ( 2 p η 1 ) 0 t u t ( τ ) W 0 2 + u t ( τ ) 2 d τ + a ( p 2 ) u W 0 2 + p 2 θ θ u ( t ) W 0 2 θ ( p + 2 η 1 ) u t ( t ) 2 2 p E ( 0 ) + a ( p 2 ) u ( t ) W 0 2 p 2 1 + 2 C 0 u t ( t ) 2 2 p E ( 0 ) + a ( p 2 ) u ( t ) W 0 2 = 2 p p 2 2 p ( 1 + 2 C 0 ) u t ( t ) 2 + a ( p 2 ) 2 p u ( t ) W 0 2 E ( 0 ) = 2 p p 2 2 p ( 1 + 2 C 0 ) u t ( t ) 2 + p 2 2 p ( 1 + 2 C 0 ) u ( t ) W 0 2 + a ( p 2 ) 2 p p 2 2 p ( 1 + 2 C 0 ) u ( t ) W 0 2 E ( 0 ) 2 p p 2 2 p ( 1 + 2 C 0 ) u t ( t ) 2 + p 2 2 p ( 1 + 2 C 0 ) u ( t ) W 0 2 + a ( p 2 ) 2 p C 0 p 2 2 p ( 1 + 2 C 0 ) C 0 u ( t ) 2 E ( 0 ) 2 p κ 1 u t ( t ) 2 + u ( t ) W 0 2 + 2 κ 2 u ( t ) 2 E ( 0 ) ,

where κ 1 = p 2 2 p ( 1 + 2 C 0 ) and κ 2 = ( p 2 ) ( a ( 1 + 2 C 0 ) 1 ) 4 p ( 1 + 2 C 0 ) C 0 . By a > 1, we have κ 1 < κ 2. By (6.2), Young, Hölder inequalities and Lemma 6.1, (6.7) becomes

ξ ̃ ( t ) 2 p ( κ 1 F ̃ ( t ) E ( 0 ) ) 2 p ( κ 1 F ̃ ( 0 ) E ( 0 ) ) > 0 .

Hence

F ( t ) F ( t ) η 1 4 F ( t ) 2 > F ( t ) ξ ̃ ( t ) > 0 .

Let y ( t ) = F ( t ) η 1 4 4 , then a direct calculation on t yields

y ( t ) = η 1 4 4 F ( t ) η 1 4 4 1 F ( t )

and

y ( t ) = η 1 4 4 η 1 4 4 1 F ( t ) η 1 4 4 2 F ( t ) 2 η 1 4 4 F ( t ) η 1 4 4 1 F ( t ) = η 1 4 4 F ( t ) η 1 4 4 2 F ( t ) F ( t ) η 1 4 F ( t ) 2 < 0 ,

which means that the function y(t) is concave. Therefore, there exists a time T 0 > 0 such that

lim t T 0 y ( t ) = 0 ,

i.e.,

lim t T 0 F ( t ) = + ,

which contradicts T max = +. Hence, the solution to problem (1.1) blows up in finite time. The proof is completed. □

6.2 Upper and lower bounds of blowup time for arbitrarily positive initial energy E(0) > 0

Finally, we estimate the upper bound and lower bound of blowup time of the blowup solution to problem (1.1) with E(0) > 0. In order to obtain the upper bound of blowup time, we first introduce the following lemma.

Lemma 6.4.

([39]) Support that a positive and twice-differentiable function Φ(t) satisfies the inequality

Φ ( t ) Φ ( t ) ( 1 + α 1 ) Φ ( t ) 2 0 ,

where t > 0 and α 1 > 0. If Φ(0) > 0 and Φ′(0) > 0, then there exists 0 < T Φ ( 0 ) α 1 Φ ( 0 ) such that Φ(t) → + as tT.

Theorem 6.5.

(Upper bound of blowup time for E(0) > 0) Let a > 1, (1.2), (1.3) and (u 0, u 1) ∈ W 0 × L 2(Ω) hold. Assume that u 0V, E(0) > 0, (u 0, u 1) > 0 and (6.1) hold. Let u be the solution to problem (1.1), which blows up at a finite time with

(6.8) 0 < T 2 ( 1 + 2 C 0 ) F ( 0 ) C 0 ( p 2 ) F ( 0 ) ,

where F is defined in (4.25).

Proof.

Let u be the solution to problem (1.1). Assume the initial data satisfy (u 0, u 1) ∈ W 0 × L 2(Ω), u 0V, E(0) > 0 and (6.1) hold. By Theorem 6.3, we know that the solution u blows up in finite time T. Recalling the auxiliary function F(t) given in (4.25), F(0) > 0 and F′(0) > 0 and the inequality as follow

F ( t ) F ( t ) η 1 4 F ( t ) 2 > 0 , t [ 0 , T ) ,

where

η 1 4 = 4 + ( p + 2 ) 2 C 0 4 ( 1 + 2 C 0 ) > 1 .

By Lemma 6.4, we get the upper bound of blowup time T of solution to problem (1.1) with E(0) > 0, i.e.,

0 < T F ( 0 ) η 1 4 4 F ( 0 ) = 4 F ( 0 ) ( η 1 4 ) F ( 0 ) ,

which yields (6.8). □

Theorem 6.6.

(Lower bound of blowup time for E(0) > 0) Assume that a > 1, (1.2), (1.3), p < 2 s * + 2 2 , 2s < N < 4s, (u 0, u 1) ∈ W 0 × L 2(Ω) and E(0) > 0 hold. Let u be the solution for problem (1.1), which blows up at a finite time T. Let

G ( t ) : = u p p .

Then, one has

(6.9) T max G ( 0 ) 1 p E ( 0 ) + y + 2 2 p 4 p C 0 p 1 a p 1 E ( 0 ) p 1 + 1 p y p 1 d y , G ( 0 ) 1 p E ( 0 ) + y + 2 p 1 θ 2 p C 0 p 1 ( 2 θ ) p 1 θ E ( 0 ) p 1 θ + 1 p y p 1 θ d y ,

where C 0 is the constant given in (2.4), and

G ( 0 ) u 0 p p .

Proof.

By Lemma 2.3, we know E(t) ≤ E(0) and

(6.10) 1 2 u t 2 + a 2 u W 0 2 + 1 2 θ u W 0 2 θ = E ( t ) + 1 p G ( t ) E ( 0 ) + 1 p G ( t ) .

Then (6.10) gives

(6.11) u t 2 2 E ( 0 ) + 1 p G ( t ) ,

(6.12) u W 0 2 2 a E ( 0 ) + 1 p G ( t ) ,

and

(6.13) u W 0 2 2 θ E ( 0 ) + 1 p G ( t ) 1 θ ,

By Hölder and Young inequalities, we know

(6.14) G ( t ) = p Ω | u | p 2 u u t d x p Ω | u t | 2 d x 1 2 Ω | u | 2 ( p 1 ) d x 1 2 p 2 Ω | u t | 2 d x + Ω | u | 2 ( p 1 ) d x .

As 2 s * + 2 2 > p > 2 θ > 2 , it is easy to know W 0L 2(p−1)(Ω) by Proposition 2.1, i.e.,

(6.15) Ω | u | 2 ( p 1 ) d x C 0 p 1 u W 0 2 ( p 1 ) .

Hence by (6.15)(6.14) becomes

(6.16) G ( t ) p 2 ( u t 2 + C 0 p 1 u W 0 2 ( p 1 ) ) .

Next, we substitute (6.12) and (6.13) into (6.16), separately.

  1. By (6.12) and (6.11), (6.16) becomes

G ( t ) p E ( 0 ) + G ( t ) + C 0 p 1 p a p 1 2 p 2 E ( 0 ) + 1 p G ( t ) p 1 p E ( 0 ) + G ( t ) + C 0 p 1 p a p 1 2 2 p 4 E ( 0 ) p 1 + 1 p G ( t ) p 1 ,

which implies

(6.17) G ( t ) p E ( 0 ) + G ( t ) + C 0 p 1 p a p 1 2 2 p 4 E ( 0 ) p 1 + 1 p G ( t ) p 1 1 .

Integrating (6.17) over [0, T], we have

(6.18) T 0 T G ( t ) p E ( 0 ) + G ( t ) + C 0 p 1 p a p 1 2 2 p 4 E ( 0 ) p 1 + 1 p G ( t ) p 1 d t .

By Theorem 6.3, there holds

(6.19) lim t T G ( t ) = ,

which together with (6.18) gives

(6.20) T G ( 0 ) 1 p E ( 0 ) + y + C 0 p 1 p a p 1 2 2 p 4 E ( 0 ) p 1 + 1 p y p 1 d y .

  1. By (6.13) and (6.11), (6.16) becomes

G ( t ) p E ( 0 ) + G ( t ) + C 0 p 1 p 2 ( 2 θ ) p 1 θ E ( 0 ) + 1 p G ( t ) p 1 θ p E ( 0 ) + G ( t ) + C 0 p 1 p 2 ( 2 θ ) p 1 θ 2 p 1 θ 1 E ( 0 ) p 1 θ + 1 p G ( t ) p 1 θ ,

which implies

G ( t ) p E ( 0 ) + G ( t ) + C 0 p 1 p 2 ( 2 θ ) p 1 θ 2 p 1 θ 1 E ( 0 ) p 1 θ + 1 p G ( t ) p 1 θ 1 .

From (6.19), we obtain

T G ( 0 ) 1 p E ( 0 ) + y + 2 p 1 θ 2 C 0 p 1 p ( 2 θ ) p 1 θ E ( 0 ) p 1 θ + 1 p y p 1 θ d y ,

which together with (6.20) yields (6.9). The proof is completed. □

Remark 6.7

(Open problems). In this section, we only consider the special case of non-degenerate Kirchhoff-type, i,e, a > 1 in (1.5), however the degenerate Kirchhoff-type case have still not been resolved. For the subcritical, critical and arbitrarily positive initial energy levels, we only established the finite time blowup of the solution for the case of linear weak damping, i.e., q = 2, and the case of nonlinear weak damping, i.e., q > 2 is still an open problem.


Corresponding author: Chao Yang, College of Mathematical Sciences, Harbin Engineering University, 150001 Harbin, People’s Republic of China, E-mail: 

Funding source: NSERC (Canada) Grant

Award Identifier / Grant number: RGPIN-2019-05940

Funding source: Fundamental Research Funds for the Central Universities

Award Identifier / Grant number: 3072023GIP2401

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: 12271122

Funding source: China Scholarship Council

Award Identifier / Grant number: 202306680038

Acknowledgements

The authors would like to appreciate the referees for taking their valuable time to provide constructive and valuable comments, which led to the improvement of our paper. Part of this work was done while Chao Yang was studying in the Faculty of Applied Mathematics, AGH University of Krakow, Poland.

  1. Research ethics: Not applicable.

  2. Informed consent: Not applicable.

  3. Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.

  4. Use of Large Language Models, AI and Machine Learning Tools: None declared.

  5. Conflict of interest: The authors state no conflict of interest.

  6. Research funding: Shaohua Chen was supported by the NSERC (Canada) Grant RGPIN-2019-05940. Jiangbo Han was supported by the Fundamental Research Funds for the Central Universities (3072023GIP2401). Runzhang Xu was supported by the National Natural Science Foundation of China (12271122), and the Fundamental Research Funds for the Central Universities. Chao Yang was supported by the China Scholarship Council (No. 202306680038).

  7. Data availability: Not applicable.

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Received: 2024-06-22
Accepted: 2025-01-30
Published Online: 2025-03-13

© 2025 the author(s), published by De Gruyter, Berlin/Boston

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