Global Attractor for Nonlinear Wave Equations with Critical Exponent on Unbounded Domain

In this paper, we study the asymptotic behavior of solutions of a damped semilinear wave equation with nonlinearity of a critical growth exponent over the Euclidean space ℝⁿ of arbitrary dimension n ≥ 3,

$u_{t t} - Δ u + β u_{t} + f (x, u) + α u = g (x)$ $$\begin{array}{} \displaystyle u_{tt} -\Delta u + \beta u_t +f(x, u)+\alpha u = g(x) \end{array}$$(1)

for t ≥ 0, with the initial condition

$u (x, 0) = u_{0} (x), u_{t} (x, 0) = u_{1} (x),$ $$\begin{array}{} \displaystyle u(x, 0)=u_{0}(x), \ \ \ \ u_t(x, 0)=u_{1}(x), \end{array}$$(2)

where α and β are arbitrary positive constants, g is a given functions defined on ℝⁿ and f (x, u) is a nonlinear function satisfying some typical dissipative conditions to be specified.

The asymptotic dynamics of global weak solutions for deterministic nonlinear wave equations and for more general nonlinear hyperbolic evolutionary equations with linear or nonlinear damping have been studied in last three decades by many authors, e.g. [1]- [4], [6]- [8], [12]- [14], [16], [19]- [22], [26]. The obtained results focused on the existence of global attractors under certain assumptions.

In the arena of stochastic wave equations driven by additive or multiplicative noise, the solution mapping defines a random dynamical system or called a cocycle on a state space with a parametric base space. The existence of random attractors for stochastic damped wave equations has been studied in [9], [11], [15], [17],[18], [23]- [25].

However, the existence problem of global attractors remains open for damped nonlinear wave equations with nonlinearity of a critical growth exponent and on the unbounded domain Rⁿ with arbitrary dimension. This is the topic of this work.

In case of nonlinearity with higher or critical growth exponents and on the unbounded domain, the issue of asymptotic compactness for the weak or mild solutions of nonlinear damped wave equations becomes difficult to handle due to not only the lack of compactness of the Sobolev embeddings but also the necessarily involved high-order integrable function spaces, in addition to the local existence and regularity of solutions in such spaces. In this work we shall tackle this challenging problem and prove the existence of a global attractor by means of

1) the uniform estimates for absorbing property and norm-smallness of solutions outside a large ball,

2) the esimates of the extended energy functional for the compactness in the space H¹(ℝⁿ) × L²(ℝⁿ) and

3) the Vitali-type convergence criterion (Theorem 8) for the function space L^p(ℝⁿ) shown in the paper. This new approach has potential applications to many other nonlinear and stochastic PDEs and to longtime and asymptotic dynamics of various problems with complex and nonlinear interactions.

In Section 2, we briefly recall basic concepts and results related to semiflow and global attractors. In Section 3, we shall conduct uniform estimates of the weak solutions for absorbing sets and for tail parts. In Section 4, we shall establish the intricate asymptotic compactness of the solution semiflow with respect to the Hilbert energy space H¹(ℝⁿ) × L²(ℝⁿ). In Section 5, we prove the crucial asymptotic compactness of the first component of solutions in L^p(ℝⁿ). Then the existence of a global attractor for this nonlinear damped wave equation is finally proved.

In this paper, we shall use || · || and 〈·,·〉 to denote the norm and inner product of L²(ℝⁿ), respectively. The norm of L^r(ℝⁿ) with r ≠ 2 or a Banach space X will be denoted by || · ||_r or || · ||_X . We use c, C or C_i to denote generic or specific positive constants.

Preliminaries and Assumptions

Let (X,|| · ||_X ) be a real Banach space. The following are the basic concepts and result on the topic of global attractor for infinite dimensional dynamical systems, cf. [2], [8], [20] and [22].

Definition 1

A mapping Φ : ℝ⁺ × X → X is called a semiflow on X, if the following conditions are satisfied:

(i) Φ(0, ·) is the identity on X.

(ii) Φ(t + s, ·) = Φ(t, Φ(s, ·)), for any t, s ≥ 0.

(iii) Φ : ℝ⁺ × X → X is a continuous mapping.

Definition 2

Let Φ be a semiflow on X. A bounded set K ⊂ X is called an absorbing set for Φ if for any bounded subset B ⊂ X there exists a finite time T_B > 0 such that

$Φ (t, B) = {Φ (t, x) : x \in B} \subset K, for all t > T_{B} .$ $$\begin{array}{} \displaystyle \Phi (t,B) = \left\{ {\Phi (t,x):x \in B} \right\} \subset K,{\text{for all }}t \gt {T_B}. \end{array}$$

Φ is called asymptotically compact in X if for any given bounded set B ⊂ X it holds that

${Φ (t_{m}, x_{m})}_{m = 1}^{\infty} has a convergent subsequence in X,$ $$\begin{array}{} \displaystyle \left\{ {\Phi ({t_m},{x_m})} \right\}_{m = 1}^\infty \,{\text{has a convergent subsequence in}}\,X, \end{array}$$

whenever t_m → ∞ and ${x_{m}}_{m = 1}^{\infty} \subset B$ $\begin{array}{} \displaystyle \{x_m\}_{m=1}^\infty \subset B \end{array}$.

Definition 3

Let F be a semiflow on X. A set 𝒜 ⊂ X is called a global attractor for Φ, if the following conditions are satisfied:

(i) 𝒜 is a compact and invariant set in the sense that Φ(t, 𝒜 ) = 𝒜, for all t ≥ 0.

(iii) 𝒜 attracts every bounded set B in X,

$\lim_{t \to \infty} d i s t_{X} (Φ (t, B), A) = 0,$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{t \to \infty } dis{t_X}(\Phi (t,B),{\mathscr A}) = 0, \end{array}$$

where dist_X (·,·) is the Hausdorff semi-distance with respect to the X-norm.

Theorem 1

Let Φ be a semiflow on a Banach space X. If the following two conditions are satisfied:

(1) there is a bounded absorbing set K for the semiflow Φ in X, and

(2) Φ is asymptotically compact in X,

then there exists a global attractor 𝒜 in X for the semiflow Φ, which is given by

$A = ω (K) = \underset{τ \geq 0}{\cap} \bar{\underset{t \geq τ}{\cup} Φ (t, K)} .$ $$\begin{array}{} \displaystyle \mathscr{A} = \omega (K) =\bigcap_{\tau \geq 0}\, \overline{\bigcup_{t\geq \tau}\Phi(t, K)}. \end{array}$$(3)

Now we formulate the original initial value problem of the nonlinear damped wave equation (1)-(2). Let ξ = u_t + δ u, where δ is a positive number to be specified later. Then (1)-(2) can be rewritten as

$\begin{matrix} u_{t} + δ u = v, \\ v_{t} - δ v + (δ^{2} + α + A) u + β (v - δ u) + f (x, u) = g (x) \\ u (x, τ) = u_{0} (x), v (x, τ) = v_{0} (x) = u_{1} (x) + δ u_{0} (x), \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} u_t \,+ \, &\delta u =v, \\ v_{t}-\delta v+ (\delta^2+\alpha+A) u &+\beta (v-\delta u)+f(x, u)=g(x) \\ u(x, \tau) =u_{0} (x), \;\, v (x, \tau &) =v_{0}(x) =u_{1}(x)+\delta u_{0}(x), \end{split} \end{array}$$(4)

where the linear operator A = −Δ : H²(ℝⁿ) → L²(ℝⁿ).

Standing Assumption. Throughout the paper, assume that the nonlinear term f ∈ C¹(ℝⁿ × ℝ, ℝ), n ≥ 3, and its antiderivative $F (x, u) = \int_{0}^{u} f (x, s) d s$ $\begin{array}{} \displaystyle F(x, u)=\int^u_0 f(x, s)ds \end{array}$ satisfy the following conditions:

$| f (x, u) | \leq C_{1} | u |^{p - 1} + ϕ_{1} (x), ϕ_{1} (x) \in H^{1} (ℝ^{n}),$ $$\begin{array}{} \displaystyle |f(x, u)|\leq C_1 |u|^{p -1}+\phi_1(x), \quad \phi_1(x)\in H^1(\mathbb{R}^n), \label{33} \end{array}$$(5)

$f (x, u) u - C_{2} F (x, u) \geq ϕ_{2} (x), ϕ_{2} (x) \in L^{1} (ℝ^{n}),$ $$\begin{array}{} \displaystyle f(x, u)u - C_2F(x, u)\geq \phi_2(x), \quad \phi_2(x)\in L^1(\mathbb{R}^n), \end{array}$$(6)

$F (x, u) \geq C_{3} | u |^{p} - ϕ_{3} (x), ϕ_{3} (x) \in L^{1} (ℝ^{n}),$ $$\begin{array}{} \displaystyle F(x, u)\geq C_3 |u|^{p}-\phi_3(x), \quad \phi_3(x)\in L^1(\mathbb{R}^n), \end{array}$$(7)

where C₁, C₂ and C₃ are positive constants and $1 \leq p \leq \frac{n + 2}{n - 2}$ $\begin{array}{} \displaystyle 1 \leq p \leq \frac{n+2}{n-2} \end{array}$ is arbitrarily given. Assume that g ∈ H¹(ℝⁿ).

Define the phase space

$E = (H^{1} (ℝ^{n}) \cap L^{p} (ℝ^{n})) \times L^{2} (ℝ^{n})$ $$\begin{array}{} \displaystyle E= \left(H^1(\mathbb{R}^n) \cap L^p (\mathbb{R}^n)\right) \times L^2(\mathbb{R}^n) \end{array}$$

endowed with the norm

$‖ (u, v) ‖_{(H^{1} \cap L^{p}) \times L^{2}} = {(‖ \nabla u ‖^{2} + ‖ u ‖^{2} + ‖ v ‖^{2})}^{\frac{1}{2}} + ‖ u ‖_{L^{p}}, for (u, v) \in E .$ $$\begin{array}{} \displaystyle {\left\| {(u,v)} \right\|_{({H^1} \cap {L^p}) \times {L^2}}} = {\left( {{{\left\| {\nabla u} \right\|}^2} + {{\left\| u \right\|}^2} + {{\left\| v \right\|}^2}} \right)^{\frac{1}{2}}} + {\left\| u \right\|_{{L^p}}},\quad {\text{for}}\,(u,v) \in E. \end{array}$$(8)

Lemma 2

For any given g₀ = (u₀, v₀) ∈ E, the initial value problem (4) has a unique global weak solution

$(u (\cdot, u_{0}), v (\cdot, v_{0})) \in C ([0, \infty), E) .$ $$\begin{array}{} \displaystyle (u(\cdot, u_{0}), v(\cdot, v_{0}))\in C([0, \infty), E). \end{array}$$

Moreover, for any t ≥ 0, the solution (u(t, u₀), v(t, v₀)) is weakly continuous with respect to g₀ = (u₀, v₀) ∈ E in the sense that

$(u (t, u_{0, m}), v (t, v_{0, m})) ⇀ (u (t, u_{0}), v (t, v_{0}))$ $$\begin{array}{} \displaystyle (u(t, u_{0,m}), v(t, v_{0,m})) \rightharpoonup (u(t, u_{0}), v(t, v_{0})) \end{array}$$

weakly in E, provided that g_{0, m} = (u_0,m, v_{0, m}) ⇀ g₀ = (u₀, v₀) weakly in E.

Proof. The local existence and uniqueness of a weak solution for this problem (4) in the phase space E = (H¹(ℝⁿ) ⋂ L^p(ℝⁿ)) × L²(ℝⁿ) and its weakly continuous dependence on the initial data can be established by the Galerkin approximation method as in [8, Chapter XV] and [3]. Also see [20], [22] and [25]. Here the detail is omitted. The proof of the global existence of weak solutions will be included in the proof of Lemma 3 below.

Uniform Estimates of Solution Trajectories

In this section, we shall derive uniform estimates on the solutions of the nonlinear damped wave equation(4) defined on ℝⁿ in a long run. These a priori estimates pave the way to proving the existence of absorbing set and the asymptotic compactness of the semiflow Φ. In particular, we will show that tails of the solutions for large spatial variables are uniformly small when time is sufficiently large.

Define a new norm of E by

$‖ (u, v) ‖_{E} = {(‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2})}^{\frac{1}{2}} + ‖ u ‖_{L^{p}},$ $$\begin{array}{} \displaystyle \|(u, v)\|_E= \left(\|v\|^2+(\alpha+\delta^2- \beta \delta)\|u\|^2+\|\nabla u\|^2\right)^{\frac 12} + \|u\|_{L^p}, \end{array}$$(9)

in which and hereafter let δ be a fixed positive constant satisfying

$α + δ^{2} - β δ > 0 and β - 3 δ > 0.$ $$\begin{array}{} \displaystyle \alpha + {\delta ^2} - \beta \delta \gt 0\quad \quad {\text{and}}\quad \quad \beta - 3\delta \gt 0. \end{array}$$(10)

Obviously the norm || · ||_E in (9) and the Sobolev norm || · ||_{(H¹⋂L^p)×L²} in (8) are equivalent.

3.1

Absorbing Set

The next lemma shows that there exists an absorbing set in the Banach space E for the semiflow Φ generated by the weak solutions (u(t, u₀), (v(t, v₀)) to the problem (4),

$Φ (t, g_{0}) = (u (t, u_{0}), v (t, v_{0})), t \geq 0, g_{0} = (u_{0}, v_{0}) .$ $$\begin{array}{} \displaystyle \Phi (t, g_0) = (u(t, u_0), v(t, v_0)), \quad t \geq 0, \;\; g_0= (u_0, v_0). \end{array}$$

Lemma 3

There exists an absorbing set K ⊂ E for the solution semiflow Φ of the problem (4). For any bounded set B ⊂ E, there exists a finite T_B > 0, such that

$Φ (t, B) \subset K, for all t > T_{B} .$ $$\begin{array}{} \displaystyle \Phi (t,B) \subset K,\quad {\text{for all}}\,\,t \gt {T_B}. \end{array}$$

Proof. Take the inner product of the second equation of (4) with v in L²(ℝⁿ) to get

$\frac{1}{2} \frac{d}{d t} ‖ v ‖^{2} - δ ‖ v ‖^{2} + (α + δ^{2}) 〈 u, v 〉 + 〈 A u, v 〉 + 〈 f (x, u), v 〉 = - 〈 β (v - δ u, v 〉 + 〈 g (x), v 〉 .$ $$\begin{array}{} \displaystyle \frac{1}{2}\frac{d}{{dt}}{\left\| v \right\|^2} - \delta {\left\| v \right\|^2} + (\alpha + {\delta ^2}){\rm{\;}}\langle u,v\rangle + \langle Au,v\rangle + \langle f(x,u),v\rangle = - \langle \beta (v - \delta u,v\rangle + \langle g(x),v\rangle . \end{array}$$(11)

Then we find that

$〈 u, v 〉 = 〈 u, u_{t} + δ u 〉 = \frac{1}{2} \frac{d}{d t} ‖ u ‖^{2} + δ ‖ u ‖^{2} and 〈 A u, v 〉 = \frac{1}{2} \frac{d}{d t} ‖ \nabla u ‖^{2} + δ ‖ \nabla u ‖^{2} .$ $$\begin{array}{} \displaystyle \langle u,v\rangle = {\rm{\;}}\langle u,{u_t} + \delta u\rangle = \frac{1}{2}\frac{d}{{dt}}{\left\| u \right\|^2} + \delta {\left\| u \right\|^2}\,\,{\rm{\;and}}\,\,{\rm{\;}}\langle Au,v\rangle = \frac{1}{2}\frac{d}{{dt}}{\left\| {\nabla u} \right\|^2} + \delta {\left\| {\nabla u} \right\|^2}. \end{array}$$

For the last term on the left-hand side of (11), we have

$〈 f (x, u), v 〉 = \frac{d}{d t} \int_{ℝ^{n}} F (x, u) d x + δ 〈 f (x, u), u 〉 .$ $$\begin{array}{} \displaystyle \langle f(x, u), v\rangle = \frac{d}{dt}\int_{\mathbb{R}^n}F(x, u)dx + \delta\langle f(x, u), u\rangle. \end{array}$$

By (6), we get

$δ 〈 f (x, u), u 〉 \geq δ C_{2} \int_{ℝ^{n}} F (x, u) d x + δ \int_{ℝ^{n}} ϕ_{2} d x .$ $$\begin{array}{} \displaystyle \delta \langle f(x,u),u\rangle \ge \delta {C_2}\int_{{\mathbb{R}^n}} F (x,u)dx + \delta \int_{{\mathbb{R}^n}} {{\phi _2}} {\mkern 1mu} dx. \end{array}$$

For the last term on the right-hand side of (11),

$〈 g, v 〉 ‖ g ‖ ‖ v ‖ \leq \frac{‖ g ‖^{2}}{2 (β - δ)} + \frac{β - δ}{2} ‖ v ‖^{2} .$ $$\begin{array}{} \displaystyle \langle g, v \rangle \|g\| \|v\|\leq \frac{\|g\|^2}{2(\beta-\delta)} + \frac{\beta-\delta}{2}\|v\|^2. \end{array}$$

Substitute the above inequalities into (11) to obtain

$\begin{matrix} \frac{1}{2} \frac{d}{d t} [‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2} + 2 \int_{ℝ^{n}} F (x, u) d x] \\ + \frac{δ}{2} [‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2}] + δ C_{2} \int_{ℝ^{n}} F (x, u) d x \\ \leq \frac{3 δ - β}{2} ‖ v ‖^{2} + \frac{‖ g ‖^{2}}{2 (β - δ)} + δ ‖ ϕ_{2} ‖_{L^{1}} \leq \frac{‖ g ‖^{2}}{2 (β - δ)} + δ ‖ ϕ_{2} ‖_{L^{1}}, t \geq 0. \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\frac 12\frac{d}{dt}\left[\|v\|^2+ \left(\alpha+\delta^2-\beta\delta \right)\|u\|^2+\|\nabla u\|^2+2\int_{\mathbb{R}^n}F(x, u)\, dx\right] \\ & +\, \frac{\delta}{2} \left[\|v\|^2+ \left(\alpha+\delta^2-\beta \delta \right)\|u\|^2+\|\nabla u\|^2 \right] + \delta C_2 \int_{\mathbb{R}^n} F(x, u)\, dx \\ \leq &\, \frac{3\delta-\beta}{2}\|v\|^2+\frac{\|g\|^2}{2(\beta-\delta)} + \delta \|\phi_2\|_{L^1} \leq \frac{\|g\|^2}{2(\beta -\delta)} + \delta \|\phi_2\|_{L^1}, \quad t \geq 0. \end{split} \end{array}$$(12)

where the term (3δ − β)||v||²/2 ≤ 0 due to (10). Let σ be a fixed positive constant:

$σ = \min {δ, δ C_{2}} > 0.$ $$\begin{array}{} \displaystyle \sigma=\min \left\{\delta, \, \delta C_2\right\} \gt 0. \end{array}$$(13)

Note that ∫_ℝⁿ (F(x, u) + ϕ₃(x))dx ≥ 0 due to (7). It follows from (12) and (13) that

$\begin{matrix} \frac{d}{d t} [‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2} + 2 \int_{ℝ^{n}} (F (x, u) + ϕ_{3} (x)) d x] \\ + σ [‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2} + 2 \int_{ℝ^{n}} (F (x, u) + ϕ_{3} (x)) d x] \\ \leq \frac{‖ g ‖^{2}}{β - δ} + 2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}), t \geq 0. \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \frac{d}{dt}\left[\|v\|^2+ \left(\alpha+\delta^2-\beta\delta\right)\|u\|^2+\|\nabla u\|^2+2\int_{\mathbb{R}^n} (F(x, u) + \phi_3 (x))\, dx\right] \\ &\, + \sigma \left[\|v\|^2+ \left(\alpha+\delta^2-\beta\delta\right)\|u\|^2+\|\nabla u\|^2 +2 \int_{\mathbb{R}^n} (F(x, u) + \phi_3 (x))\, dx\right] \\ \leq &\, \frac{\|g\|^2}{\beta-\delta} +2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}), \quad t \geq 0. \end{split} \end{array}$$(14)

Apply Gronwall inequality to (14) and see that every weak solution of (4) satisfies

$\begin{matrix} ‖ v (t) ‖^{2} + (α + δ^{2} - β δ) ‖ u (t) ‖^{2} + ‖ \nabla u (t) ‖^{2} + 2 \int_{ℝ^{n}} (F (x, u (t)) + ϕ_{3} (x)) d x \\ \leq e^{- σ (t - τ)} [‖ v_{0} ‖^{2} + (α + δ^{2} - β δ) ‖ u_{0} ‖^{2} + ‖ \nabla u_{0} ‖^{2} + 2 \int_{ℝ^{n}} F (x, u_{0}) d x] \\ + 2 e^{- σ (t - τ)} ‖ ϕ_{3} ‖_{L^{1}} + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \|v(t)\|^2+ \left(\alpha+\delta^2-\beta \delta\right)\|u(t)\|^2 + \|\nabla u(t)\|^2+2\int_{\mathbb{R}^n} (F(x, u( t)) + \phi_3 (x))\, dx \\ \leq &\, e^{-\sigma (t - \tau)} \left[\|v_0\|^2+ \left(\alpha+\delta^2-\beta \delta\right)\|u_0\|^2 +\|\nabla u_0\|^2+2\int_{\mathbb{R}^n}F(x, u_0)\, dx\right] \\ &\, + 2e^{-\sigma (t - \tau)} \|\phi_3\|_{L^1} +\frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) +\frac{\|g\|^2}{\beta -\delta}\right). \end{split} \end{array}$$(15)

Thus for any given bounded set B ⊂ E and any (u₀, v₀) ∈ B, we have

$\begin{matrix} ‖ v (t) ‖^{2} + (α + δ^{2} - β δ) ‖ u (t ‖^{2} + ‖ \nabla u (t) ‖^{2} + 2 \int_{ℝ^{n}} (F (x, u (t)) + ϕ_{3} (x)) d x \\ \leq e^{- σ t} [‖ v_{0} ‖^{2} + (α + δ^{2} - β δ) ‖ u_{0} ‖^{2} + ‖ \nabla u_{0} ‖^{2} + 2 \int_{ℝ^{n}} F (x, u_{0}) d x + 2 ‖ ϕ_{3} ‖_{L^{1}}] \\ + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}) t \geq 0. \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\|v(t)\|^2+ \left(\alpha+\delta^2-\beta\delta\right)\|u(t \|^2 +\|\nabla u(t)\|^2 +2\int_{\mathbb{R}^n} (F(x, u(t)) + \phi_3 (x))\, dx \\ \leq &\, e^{-\sigma t} \left[\|v_0\|^2+ \left(\alpha+\delta^2-\beta\delta\right) \|u_0\|^2+\|\nabla u_0\|^2 + 2\int_{\mathbb{R}^n} F(x, u_0)\, dx + 2\|\phi_3\|_{L^1}\right] \\ &\, + \frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) +\frac{\|g\|^2}{\beta -\delta}\right) \quad t \geq 0. \end{split} \end{array}$$(16)

According to the assumption (5) and (6), there exists a constant c = c(C₁, C₂, ϕ₁, ϕ₂) > 0 such that

$\int_{ℝ^{n}} F (x, u_{0}) d x \leq c (1 + ‖ u_{0} ‖^{2} + ‖ u_{0} ‖_{L^{p}}^{p}) .$ $$\begin{array}{} \displaystyle \int_{\mathbb{R}^n}F(x, u_0)\, dx\leq c \left(1 +\|u_0\|^2 +\|u_0\|^{p}_{L^p}\right). \end{array}$$

It follows that, for any given bounded set B ⊂ E and (u₀, v₀) ∈ B, there exist a constant C > 0 and a finite T_B > 0 such that

$\begin{matrix} e^{- σ t} [‖ v_{0} ‖^{2} + (α + δ^{2} - β δ) ‖ u_{0} ‖^{2} + ‖ \nabla u_{0} ‖^{2} + 2 \int_{ℝ^{n}} F (x, u_{0}) d x + 2 ‖ ϕ_{3} ‖_{L^{1}}] \\ \leq C e^{- σ t} (1 + ‖ v_{0} ‖^{2} + ‖ u_{0} ‖_{H^{1}}^{2} + ‖ u_{0} ‖_{L^{p}}^{p}) \leq 1, for all t > T_{B} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{l}} {{e^{ - \sigma t}}{\rm{\;}}\left[ {{{\left\| {{v_0}} \right\|}^2} + \left( {\alpha + {\delta ^2} - \beta \delta } \right){{\left\| {{u_0}} \right\|}^2} + {{\left\| {\nabla {u_0}} \right\|}^2} + 2{\int _{{\mathbb{R}^n}}}F(x,{u_0})dx + 2{{\left\| {{\phi _3}} \right\|}_{{L^1}}}} \right]}\\ {\qquad \le C{e^{ - \sigma t}}\left( {1 + {{\left\| {{v_0}} \right\|}^2} + \left\| {{u_0}} \right\|_{{H^1}}^2 + \left\| {{u_0}} \right\|_{{L^p}}^p} \right) \le 1,\quad \,{\text{for all}}\,\,t \gt {T_B}.} \end{array} \end{array}$$(17)

Substitute (17) into the right-hand side of the last equality in (16) and note that (7) implies

$2 \int_{ℝ^{n}} (F (x, u (t, u_{0}) + ϕ_{3} (x)) d x \geq 2 C_{3} ‖ u (t, u_{0}) ‖_{L^{p}}^{p} .$ $$\begin{array}{} \displaystyle 2\int_{\mathbb{R}^n} (F(x, u(t, u_0) + \phi_3 (x))\, dx \geq 2C_3 \|u(t, u_0)\|_{L^p}^p. \end{array}$$

Then it results in

$\begin{matrix} ‖ v (t, v_{0}) ‖^{2} + (α + δ^{2} - β δ) ‖ u (t, u_{0}) ‖^{2} + ‖ \nabla u (t, u_{0}) ‖^{2} + 2 C_{3} ‖ u (t, u_{0}) ‖_{L^{p}}^{p} \\ \leq 1 + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}), for t > T_{B} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {{{\left\| {v(t,{v_0})} \right\|}^2} + \left( {\alpha + {\delta ^2} - \beta \delta } \right){{\left\| {u(t,{u_0})} \right\|}^2} + {{\left\| {\nabla u(t,{u_0})} \right\|}^2} + 2{C_3}\left\| {u(t,{u_0})} \right\|_{{L^p}}^p}\\ {\;\; \le 1 + \frac{1}{\sigma }\left( {2\delta ({C_2}{{\left\| {{\phi _3}} \right\|}_{{L^1}}} + {{\left\| {{\phi _2}} \right\|}_{{L^1}}}) + \frac{{{{\left\| g \right\|}^2}}}{{\beta - \delta }}} \right),\quad {\rm{\;for }}\,t \gt {T_B}.} \end{array} \end{array}$$(18)

The inequality (18) show that Φ(t, B) ⊂ K = B_E(0, R) for t > T_B, where the radius of the ball B_E(0, R) in E is

$\begin{matrix} R = {(\frac{1}{\min {1, (α + δ^{2} - β δ)}} [1 + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ})])}^{\frac{1}{2}} \\ + {(\frac{1}{2 C_{3}} [1 + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ})])}^{\frac{1}{p}} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} R= &\,\left(\frac{1}{\min \{1, (\alpha + \delta^2 -\beta \delta)\}} \left[1 + \frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) +\frac{\|g\|^2}{\beta -\delta}\right)\right]\right)^{\frac{1}{2}} \\ &+ \left(\frac{1}{2C_3}\left[1 + \frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) +\frac{\|g\|^2}{\beta -\delta}\right)\right]\right)^{\frac{1}{p}}. \end{split} \end{array}$$(19)

Therefore, this set K = B_E(0, R) is an absorbing set in the phase space E for the solution semiflow Φ. The proof is completed.

3.2

Tail Estimate

Next we conduct uniform estimates on the tail parts of the weak solutions for large spatial and time variables. These estimates play key roles in proving the asymptotic compactness in the space E of the dynamical systems F generated by the nonlinear wave equation (4) on the unbounded domain ℝⁿ.

Lemma 4

For every bounded set B ⊂ E and 0 < η ≤ 1, there exists T = T (B, η) > 0 and V = V (η) ≥ 1 such that the semiflow Φ generated by the nonlinear damped wave equation (4) satisfies

$\begin{matrix} ∥ Φ (t, B) ∥_{E (R^{n} ∖ B_{r})} = max_{g_{0} \in B} ∥ Φ (t, g_{0}) ζ_{B_{r}^{c}} ∥_{E} < η, \end{matrix}$ $$\begin{array}{l} \| \Phi (t, B)\|_{E(\mathbb{R}^n \backslash B_r)} = \max\limits_{g_0 \in B} \|\Phi (t, g_0)\zeta_{B_r^c}\|_E \lt \eta, \end{array}$$(20)

for all t > T and every r > V , where $ζ_{B_{r}^{c}} (x)$ $\begin{array}{} \displaystyle \zeta_{B_r^c } (x) \end{array}$is the characteristic function of the set {x ∈ ℝⁿ : |x| > r}.

Proof. Choose a smooth and nondecreasing function ρ such that 0 ≤ ρ(s) ≤ 1 for all s ∈ [0, ∞) and

$ρ (s) = {\begin{matrix} 0, i f 0 \leq s < 1, \\ 1, i f s > 2, \end{matrix}$ $$\begin{array}{} \displaystyle \rho(s)=\left\{ \begin{array}{l} 0, \qquad if \;\; 0 \leq s <1,\\ \\ 1, \qquad if \;\; s >2, \end{array} \right. \end{array}$$(21)

with 0 ≤ ρ^′(s) ≤ 2 for s ≥ 0. Taking the inner product of the second equation of (4) with ρ(|x|²/r²)v in L²(ℝⁿ), we get

$\begin{matrix} \frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x - δ \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x \\ + (α + δ^{2}) \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) u v d x + \int_{ℝ^{n}} (A u) ρ (\frac{| x |^{2}}{r^{2}}) v d x + \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) f (x, u) v d x \\ = \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) g v d x - \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) β (v - δ u) v d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|v|^2\, dx -\delta \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|v|^2 \, dx \\ &+(\alpha+\delta^2)\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) u\, v \, dx + \int_{\mathbb{R}^n}(Au)\rho\left(\frac{|x|^2}{r^2}\right)v \, dx +\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)f(x, u)\, v \, dx \\ =& \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) g\, v\, dx -\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) \beta (v -\delta u)\, v\, dx. \end{split} \end{array}$$(22)

Hence we have

$\begin{matrix} \frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x + (α + δ^{2} - β δ) \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) u v d x \\ + (β - δ) \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x + \int_{ℝ^{n}} (A u) ρ (\frac{| x |^{2}}{r^{2}}) v d x + \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) f (x, u) v d x \\ \leq \frac{δ}{2} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x + \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) g v d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|v|^2\, dx + (\alpha+\delta^2-\beta\delta)\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2} \right)u\,v\, dx \\ + &\, (\beta -\delta) \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|v|^2\, dx + \int_{\mathbb{R}^n}(Au)\rho\left(\frac{|x|^2}{r^2}\right)v\, dx+\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)f(x, u)\,v\, dx \\ \leq &\; \frac{\delta}{2} \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) |v|^2\, dx +\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) g\, v \, dx. \end{split} \end{array}$$(23)

For the second term on the left-hand side of (23), by (4) we have

$\begin{matrix} (α + δ^{2} - β δ) \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) u v d x = (α + δ^{2} - β δ) \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) u (u_{t} + δ u) d x \\ \geq (α + δ^{2} - β δ) (\frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | u |^{2} d x + δ \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | u |^{2} d x) \\ - \frac{δ}{2} (α + δ^{2} - β δ) \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | u |^{2} d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &(\alpha+\delta^2-\beta\delta)\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2} {r^2}\right)u\,v\, dx = (\alpha+\delta^2-\beta\delta)\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2} \right)u(u_t+\delta u)\, dx \\ \geq &\, (\alpha+\delta^2-\beta\delta) \left(\frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|u|^2\, dx +\delta\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|u|^2\, dx\right) \\ &\, -\frac{\delta}{2} (\alpha+\delta^2-\beta\delta) \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|u|^2\, dx. \end{split} \end{array}$$(24)

For the fourth term on the left-hand side of (23),

$\begin{matrix} \int_{ℝ^{n}} (A u) ρ (\frac{| x |^{2}}{r^{2}}) v d x = \int_{ℝ^{n}} (A u) ρ (\frac{| x |^{2}}{r^{2}}) (u_{t} + δ u) d x = \int_{ℝ^{n}} (\nabla u) \nabla (ρ (\frac{| x |^{2}}{r^{2}}) (u_{t} + δ u)) d x \\ = \int_{ℝ^{n}} (\nabla u) \frac{2 x}{r^{2}} ρ^{'} (\frac{| x |^{2}}{r^{2}}) v d x + \int_{ℝ^{n}} (\nabla u) ρ (\frac{| x |^{2}}{r^{2}}) \nabla (u_{t} + δ u) d x \\ = \int_{ℝ^{n}} (\nabla u) \frac{2 x}{r^{2}} ρ^{'} (\frac{| x |^{2}}{r^{2}}) v d x + \frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | \nabla u |^{2} d x + δ \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | \nabla u |^{2} d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{align*} &\int_{\mathbb{R}^n}(Au)\, \rho\left(\frac{|x|^2}{r^2}\right)v \, dx = \int_{\mathbb{R}^n}(Au)\, \rho\left(\frac{|x|^2}{r^2}\right)(u_t+\delta u)\, dx = \int_{\mathbb{R}^n}(\nabla u)\nabla\left(\rho\left(\frac{|x|^2}{r^2}\right)(u_t+\delta u)\right) dx \\ =&\, \int_{\mathbb{R}^n}(\nabla u)\frac{2x}{r^2}\rho'\left(\frac{|x|^2}{r^2}\right)v\, dx+\int_{\mathbb{R}^n}(\nabla u)\rho\left(\frac{|x|^2}{r^2}\right)\nabla(u_t+\delta u)\, dx \\ =&\, \int_{\mathbb{R}^n}(\nabla u)\frac{2x}{r^2}\rho'\left(\frac{|x|^2}{r^2}\right)v\, dx+\frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|\nabla u|^2\, dx +\delta\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|\nabla u|^2 \, dx. \end{align*} \end{array}$$

Since 0 ≤ ρ^′(s) ≤ 2, it follows that

$\begin{matrix} \int_{ℝ^{n}} (A u) ρ (\frac{| x |^{2}}{r^{2}}) v d x \geq - \int_{r \leq | x | \leq \sqrt{2} r} \frac{4 | x |}{r^{2}} | (\nabla u) v | d x \\ + \frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | \nabla u |^{2} d x + δ \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | \nabla u |^{2} d x \\ \geq - \frac{2 \sqrt{2}}{r} \int_{r \leq | x | \leq \sqrt{2} r} (| \nabla u |^{2} + | v |^{2}) d x + \frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | \nabla u |^{2} d x + \frac{δ}{2} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | \nabla u |^{2} d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\int_{\mathbb{R}^n}(Au)\, \rho\left(\frac{|x|^2}{r^2}\right)v\, dx \geq -\, \int_{r\leq |x|\leq \sqrt{2}r}\frac{4|x|}{r^2} \, |(\nabla u)\, v|\, dx \\ &+ \frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|\nabla u|^2\, dx +\delta\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|\nabla u|^2 \, dx \\ &\geq -\, \frac{2\sqrt{2}}{r} \int_{r \leq |x| \leq \sqrt{2}r} (|\nabla u|^2+ |v |^2)\, dx +\frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|\nabla u|^2\, dx +\frac{\delta}{2}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|\nabla u|^2\, dx. \end{split} \end{array}$$(25)

For the fifth term on the left-hand side of (23), by (5)-(7), we have

$\begin{matrix} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) f (x, u) v d x = \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) f (x, u) (u_{t} + δ u) d x \\ \geq \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) F (x, u) d x + δ \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (C_{2} F (x, u) + ϕ_{2} (x)) d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)f(x, u)v \, dx =\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)f(x, u)(u_t+\delta u ) \, dx \\ \geq &\, \frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)F(x, u)\, dx +\delta \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) (C_2 F(x, u) + \phi_2(x))\, dx. \end{split} \end{array}$$(26)

For the last term on the right-hand side of (23), we see

$\int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) g v d x \leq \frac{1}{2 (β - δ)} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | g |^{2} d x + \frac{β - δ}{2} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x .$ $$\begin{array}{} \displaystyle \begin{split} \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) g\, v\, dx \leq \frac{1}{2(\beta -\delta)}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) |g|^2\, dx + \frac{\beta -\delta}{2}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)|v|^2\, dx. \end{split} \end{array}$$(27)

Now substitute (24)-(27) into (23), we obtain

$\begin{matrix} \frac{1}{2} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (| v |^{2} + (α + δ^{2} - β δ) | u |^{2} + | \nabla u |^{2} + 2 F (x, u)) d x \\ + \frac{δ}{2} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) | v |^{2} d x + \frac{δ}{2} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) ((α + δ^{2} - β δ) | u |^{2} + | \nabla u |^{2}) d x \\ + δ C_{2} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) F (x, u) d x \\ \leq \frac{2 \sqrt{2}}{r} \int_{r \leq | x | \leq \sqrt{2} r} (| \nabla u |^{2} + | v |^{2}) d x + \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (\frac{| g |^{2}}{β - δ} + δ | ϕ_{2} |) d x . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \frac 12\frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) \left(|v|^2+(\alpha+\delta^2-\beta \delta)|u|^2+|\nabla u|^2+2F(x, u)\right)dx \\ &\, +\frac{\delta}{2}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) |v|^2\,dx +\frac{\delta}{2}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) ((\alpha+\delta^2-\beta \delta)|u|^2+|\nabla u|^2)\, dx \\ &\, + \delta C_2 \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) F(x, u)\, dx\\ \leq &\,\frac{2\sqrt{2}}{r} \int_{r \leq |x| \leq \sqrt{2}r} (|\nabla u|^2+ |v |^2)\, dx +\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) \left(\frac{|g|^2}{\beta -\delta} + \delta|\phi_2|\right)\, dx. \end{split} \end{array}$$(28)

Since g ∈ L²(ℝⁿ) and ϕ₂, ϕ₃ ∈ L¹(ℝⁿ), for any η > 0, there exists K₀ = K₀(η) 1 such that for all r ≥ K₀,

$\int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (\frac{| g |^{2}}{β - δ} + δ | ϕ_{2} | + 2 σ | ϕ_{3} |) d x \leq \int_{| x | \geq r} (\frac{| g |^{2}}{β - δ} + δ | ϕ_{2} | + 2 σ | ϕ_{3} |) d x < η .$ $$\begin{array}{} \displaystyle \int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) \left(\frac{|g|^2}{\beta -\delta} + \delta|\phi_2| + 2\sigma |\phi_3|\right)\, dx \leq \int_{|x| \geq r} \left(\frac{|g|^2}{\beta -\delta} + \delta|\phi_2| + 2\sigma |\phi_3|\right)\, dx \lt \eta. \end{array}$$(29)

By (13) and (28)-(29), there exists K₁ = K₁(η) ≥ 1 such that for all r > K₁,

$\begin{matrix} \frac{d}{d t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (| v |^{2} + (α + δ^{2} - β δ) | u |^{2} + | \nabla u |^{2} + 2 (F (x, u) + ϕ_{3})) d x \\ + σ \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (| v |^{2} + (α + δ^{2} - β δ) | u |^{2} + | \nabla u |^{2} + 2 (F (x, u) + ϕ_{3})) d x \\ \leq η [1 + \int_{r \leq | x | \leq \sqrt{2} r} (| \nabla u |^{2} + | v |^{2}) d x] . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} & \frac{d}{dt}\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)\left(|v|^2 +(\alpha+\delta^2-\beta\delta)|u|^2+|\nabla u|^2+2(F(x, u) + \phi_3)\right) dx \\ & + \sigma\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right) \left(|v|^2+(\alpha+\delta^2-\beta\delta)|u|^2+|\nabla u|^2+2(F(x, u) + \phi_3)\right) dx \\ \leq &\,\eta \left[1 + \int_{r \leq |x| \leq \sqrt{2}r} (|\nabla u|^2+ |v |^2)\, dx\right]. \end{split} \end{array}$$

Therefore, for any t > 0 and r > K₁, it holds that

$\begin{matrix} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) [| v (t) |^{2} + (α + δ^{2} - β δ) | u (t) |^{2} + | \nabla u (t) |^{2} + 2 (F (x, u (t)) + ϕ_{3} (x))] d x \\ \leq e^{- σ t} \int_{ℝ^{n}} ρ (| v_{0} |^{2} + (α + δ^{2} - β δ) | u_{0} |^{2} + | \nabla u_{0} |^{2} + 2 (F (x, u_{0}) + ϕ_{3} (x))) d x \\ + \frac{η}{σ} + η \int_{0}^{t} e^{- σ (t - s)} \int_{r \leq | x | \leq \sqrt{2} r} (| \nabla u (s) |^{2} + | v (s) |^{2}) d x d s . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{l}} {\int_{{^n}} \rho \left( {\frac{{|x{|^2}}}{{{r^2}}}} \right)\left[ {|v(t){|^2} + \left( {\alpha + {\delta ^2} - \beta \delta } \right)|u(t){|^2} + |\nabla u(t){|^2} + 2(F(x,u(t)) + {\phi _3}(x))} \right]dx}\\ { \le {\rm{ }}{\mkern 1mu} {e^{ - \sigma t}}\int_{{^n}} \rho (|{v_0}{|^2} + (\alpha + {\delta ^2} - \beta \delta )|{u_0}{|^2} + |\nabla {u_0}{|^2} + 2(F(x,{u_0}) + {\phi _3}(x))){\mkern 1mu} dx}\\ {{\mkern 1mu} + \frac{\eta }{\sigma } + \eta \int_0^t {{e^{ - \sigma (t - s)}}} \int_{r \le |x| \le \sqrt 2 r} {(|\nabla u(s){|^2} + |v(s){|^2})} {\mkern 1mu} dx{\mkern 1mu} ds.} \end{array} \end{array}$$(30)

Next we conduct estimates of the terms on the right-hand side of in (30). For the first term, there exists T₁ = T₁(B, η) > 0 and a constant C₄ > 0 such that

$\begin{matrix} e^{- σ t} \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (| v_{0} |^{2} + (α + δ^{2} - β δ) | u_{0}) |^{2} + | \nabla u_{0} |^{2} + 2 (F (x, u_{0}) + ϕ_{3})) d x \\ \leq e^{- σ t} \int_{ℝ^{n}} (| v_{0} |^{2} + (α + δ^{2} - β δ) | u_{0} |^{2} + | \nabla u_{0} |^{2}) d x \\ + 2 e^{- σ t} \int_{ℝ^{n}} [\frac{1}{C_{2}} (C_{1} | u_{0} |^{p} + | u_{0} | | ϕ_{1} (x) | + | ϕ_{2} (x) |) + | ϕ_{3} (x) |] d x \\ \leq C_{4} e^{- σ t} (‖ (v_{0}, u_{0}) ‖^{2} + ‖ \nabla u_{0} ‖^{2} + ‖ u_{0} ‖_{L^{p}}^{p} + ‖ ϕ_{1} ‖^{2} + ‖ ϕ_{2} ‖_{L^{1}} + ‖ ϕ_{3} ‖_{L^{1}}) < η \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} e^{-\sigma t} &\int_{\mathbb{R}^n}\rho\left(\frac{|x|^2}{r^2}\right)(|v_0|^2 + (\alpha+\delta^2-\beta\delta) |u_0)|^2 + |\nabla u_0|^2 + 2(F(x, u_0) + \phi_3))\, dx \\ \leq &\, e^{-\sigma t} \int_{\mathbb{R}^n} \left(|v_0|^2 + (\alpha+\delta^2-\beta\delta) |u_0|^2 +|\nabla u_0|^2\right) dx \\ & + 2e^{-\sigma t} \int_{\mathbb{R}^n} \left[\frac{1}{C_2}\left(C_1 |u_0|^p + |u_0| |\phi_1 (x)| + |\phi_2 (x)| \right) + |\phi_3 (x)| \right] dx \\ \leq &\, C_{4} \, e^{-\sigma t} (\|(v_0, u_0)\|^2+ \|\nabla u_0\|^2 +\|u_0\|_{L^p}^p + \|\phi_1 \|^2 + \|\phi_2 \|_{L^1} + \|\phi_3 \|_{L^1}) \lt \eta \end{split} \end{array}$$(31)

for all t > T₁. For the second integral term on the right-hand side of (30), applying the Gronwall inequality to(14) while taking the spatial integral over the region $r \leq | x | \leq \sqrt{2} r$ $\begin{array}{} \displaystyle r \leq |x| \leq \sqrt{2} r \end{array}$, with (13) in mind, we get

$\begin{matrix} \int_{- t}^{0} e^{σ s} \int_{r \leq | x | \leq \sqrt{2} r} (| \nabla u (s) |^{2} + | v (s) |^{2}) d x d s \\ \leq e^{- σ (s + t)} (‖ v_{0} ‖^{2} + (α + δ^{2} - β δ) ‖ u_{0} ‖^{2} + ‖ \nabla u_{0} ‖^{2}) \\ + 2 e^{- σ (s + t)} \int_{ℝ^{n}} (F (x, u_{0}) + ϕ_{3} (x)) d x + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{1}{β - δ} ‖ g ‖^{2}) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\int_{-t}^0 e^{\sigma s} \int_{r \leq |x| \leq \sqrt{2}r} (|\nabla u(s)|^2 + |v(s)|^2)\, dx\, ds \\ \leq &\, e^{-\sigma (s + t)} \left(\| v_0 \|^2 + (\alpha + \delta^2 - \beta \delta) \|u_0\|^2 + \|\nabla u_0\|^2\right) \\ &\, + 2e^{-\sigma (s + t)} \int_{\mathbb{R}^n} \left(F(x, u_0) + \phi_3 (x)\right)\, dx + \frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{1}{\beta - \delta} \|g\|^2\right). \end{split} \end{array}$$(32)

Based on (31) and (32), there exists T₂ = T₂(B, η) > 0 such that

$\begin{matrix} \int_{- t}^{0} e^{σ s} \int_{r \leq | x | \leq \sqrt{2} r} (| \nabla u (s) |^{2} + | v (s) |^{2}) d x d s \\ \leq C t e^{- σ t} [‖ (u_{0}, v_{0}) ‖^{2} + ‖ \nabla u_{0} ‖^{2} + \int_{ℝ^{n}} (F (x, u_{0}) + ϕ_{3} (x)) d x] + \frac{1}{σ} (C_{6} + \frac{1}{β - δ} ‖ g ‖^{2}) \\ \leq C t e^{- σ t} [‖ (u_{0}, v_{0}) ‖^{2} + ‖ \nabla u_{0} ‖^{2} + ‖ ϕ_{3} ‖_{L^{1}} + \frac{1}{C_{2}} (C_{1} ‖ u_{0} ‖_{L^{p}}^{p} + ‖ u_{0} ‖^{2} + ‖ ϕ_{1} ‖^{2} + ‖ ϕ_{2} ‖_{L^{1}})] \\ + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{1}{β - δ} ‖ g ‖^{2}) \leq M, for all t \geq T_{2}, \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} & \int^0_{-t} e^{\sigma s} \int_{r \leq |x| \leq \sqrt{2}r} \left(|\nabla u(s)|^2+|v(s)|^2 \right) dx\, ds \\ \leq &\, C t \, e^{-\sigma t} \left[\| (u_0, v_0) \|^2 + \|\nabla u_0 \|^2 + \int_{\mathbb{R}^n} (F(x, u_0)+ \phi_3 (x))\, dx\right] + \frac{1}{\sigma} \left(C_6 + \frac{1}{\beta - \delta}\|g \|^2\right) \\ \leq &\,C t \, e^{-\sigma t} \left[\| (u_0, v_0)\|^2 + \|\nabla u_0\|^2 + \|\phi_3 \|_{L^1}+ \frac{1}{C_2} (C_1\|u_0 \|_{L^p}^p + \|u_0 \|^2 + \|\phi_1\|^2 + \|\phi_2 \|_{L^1})\right] \\ &\, + \frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{1}{\beta - \delta} \|g \|^2\right) \leq M, \quad {\text{for all}} \;\; t \geq T_2, \end{split} \end{array}$$(33)

where the constant

$M = 1 + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{1}{β - δ} ‖ g ‖^{2}) .$ $$\begin{array}{} \displaystyle M = 1 + \frac{1}{\sigma} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{1}{\beta - \delta} \|g\|^2\right). \end{array}$$

Now assemble all these estimates and substitute (31) and (33) into (30). It shows that for any bounded set B and any 0 < η ≤ 1, as long as r > V = max {K₀, K₁} and t > max {T₁, T₂}, one has

$\begin{matrix} \int_{| x | \geq \sqrt{2} r} (| v (t) |^{2} + (α + δ^{2} - β δ) | u (t) |^{2} + | \nabla u (t) |^{2} + | u (t) |^{p}) d x \\ \leq \int_{ℝ^{n}} ρ (\frac{| x |^{2}}{r^{2}}) (| v (t) |^{2} + (α + δ^{2} - β δ) | u (t) |^{2} + | \nabla u (t) |^{2}) d x \\ + \frac{2}{C_{3}} \int_{ℝ^{n}} ρ (F (x, u (t) + ϕ_{3} (x)) d x \leq (1 + \frac{1}{C_{3}}) (2 + M) η . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \int_{|x|\geq \sqrt{2}r} (|v(t)|^2 + (\alpha + \delta^2 - \beta \delta)|u(t)|^2 +|\nabla u(t)|^2 + |u(t)|^p)\, dx \\ \leq &\, \int_{\mathbb{R}^n} \rho\left(\frac{|x|^2}{r^2}\right) \left(|v(t)|^2 +(\alpha+\delta^2-\beta\delta) |u(t)|^2 + |\nabla u(t)|^2 \right) dx \\ &\, + \frac{2}{C_3} \int_{\mathbb{R}^n} (F(x, u(t) + \phi_3 (x))\, dx \leq \left(1 + \frac{1}{C_3}\right) (2 + M)\eta. \end{split} \end{array}$$(34)

By (9), the above inequality (34) demonstrates that for any bounded B ⊂ E it holds that

$\begin{matrix} {((Φ (t, B)))}_{E (R^{n} ∖ B_{R})} = max_{g_{0} \in B} Φ (t, g_{0})_{E (R^{n} ∖ B_{R})} \\ \leq {((1 + \frac{1}{C_{3}}) (2 + M) η)}^{1 / 2} + {((1 + \frac{1}{C_{3}}) (2 + M) η)}^{1 / p}, \end{matrix}$ $$\begin{array}{*{20}{c}} {{{\left\| {\left| {\Phi (t,B)} \right|} \right\|}_{E({\mathbb{R}^n}\backslash {B_R})}} = {\max\limits_{{g_0} \in B}}\Phi (t,{g_0}){_{E({\mathbb{R}^n}\backslash {B_R})}}}\\ { \le {\mkern 1mu} {{\left[ {\left( {1 + \frac{1}{{{C_3}}}} \right)(2 + M)\eta } \right]}^{1/2}} + {{\left[ {\left( {1 + \frac{1}{{{C_3}}}} \right)(2 + M)\eta } \right]}^{1/p}},} \end{array}$$(35)

where $R = \sqrt{2} r$ $\begin{array}{} \displaystyle R = \sqrt{2}r \end{array}$. (35) implies that (20) is satisfied as stated in this theorem just by renaming r to be R and η to be ((1 + 1/C₃)(2 + M)η)^1/2 + ((1 + 1/C₃)(2 + M)η)¹/p. The proof is completed.

Asymptotic Compactness in H¹(ℝⁿ) × L²(ℝⁿ)

In this section, we shall prove the asymptotic compactness in the space H¹(ℝⁿ) × L²(ℝⁿ) of the solution semiflow Φ associated with the nonlinear damped wave equation (4).

Lemma 5

The following statements hold for L^p(ℝⁿ).

1) For 1 ≤ p < ∞, let {ψ_m} be a sequence and ψ be a function in L^p(ℝⁿ) such that ||ψ_m − ψ||_L^p → 0 as m → ∞. Then there exists a subsequence {ψ_mk} such that

$\lim_{k \to \infty} ψ_{m_{k}} (x) = ψ (x), a.e. on ℝ^{n} .$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{k \to \infty } {\psi _{{m_k}}}(x) = \psi (x),\quad {\text{a}}{\rm{.e}}{\rm{. on}}\,\,{\mathbb{R}^n}. \end{array}$$

2) For 1 < p < ∞, if a sequence {ψ_m} and a function ψ in L^p(ℝⁿ) satisfy the following two conditions:

$\lim_{m \to \infty} ψ_{m} (x) = ψ (x), a.e. on ℝ^{n} and {ψ_{m}} is bounded in L^{p} (ℝ^{n}),$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{m \to \infty } {\psi _m}(x) = \psi (x),\,{\text{a}}{\rm{.e}}{\rm{. on}}\,{\rm{ }}{\mathbb{R}^n}{\rm{ }}\quad {\text{and }}\left\{ {{\psi _m}} \right\}\,{\text{is bounded in }}\,{L^p}({\mathbb{R}^n}), \end{array}$$(36)

then ψ_m → ψ weakly in L^p(ℝⁿ), as m → ∞.

3) For 1 < p < ∞, if a sequence {ψ_m} and a function ψ in L^p(ℝⁿ) satisfy the following two conditions:

$\lim_{m \to \infty} ψ_{m} (x) = ψ (x), a.e. on ℝ^{n} and \lim_{m \to \infty} ‖ ψ_{m} ‖_{L^{p}} = ‖ ψ ‖_{L^{p}},$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{m \to \infty } {\psi _m}(x) = \psi (x),\,{\text{a}}{\rm{.e}}{\rm{. on}}\,\,{\mathbb{R}^n}\quad {\text{and}}\quad \mathop {\lim }\limits_{m \to \infty } {\psi _m}{_{{L^p}}} = \psi {_{{L^p}}}, \end{array}$$(37)

then lim_m→∞ ||ψ_m − ψ||_L^p = 0.

Proof. Since ℝⁿ with the Lebesgue measure is a σ-finite measure space, the first item is a standard result in Real and Functional Analysis.

For the second item, since L^p(ℝⁿ) is a reflexive Banach space for 1 < p < ∞, the boundedness of {ψ_m} in L^p(ℝⁿ) implies that there is φ ∈ L^p(ℝⁿ) such that ψ_m → φ weakly as m → ∞. By Mazur’s lemma, this weak convergence implies there exists a sequence {ζ_m} ⊂ L^p(ℝⁿ) such that

$ζ_{m} \in covn {ψ_{m}, ψ_{m + 1}, \dots} and ζ_{m} \to φ strongly in L^{p} (ℝ^{n}) .$ $$\begin{array}{} \displaystyle {\zeta _m} \in {\text{covn}}\left\{ {{\psi _m},{\psi _{m + 1}}, \cdots } \right\}\,{\text{and}}\,{\zeta _m} \to \varphi \,{\text{strongly in}}\,{L^p}({\mathbb{R}^n}). \end{array}$$(38)

From the condition ψ_m → ψ a.e. and $ζ_{m} \in c o n v (\cup_{i = m}^{\infty} ψ_{i})$ ${\zeta _m} \in {\rm{\;conv }}(\cup_{i = m}^\infty {\psi _i})$, it follows that

$ζ_{m} \to ψ a.e. in ℝ^{n} .$ $$\begin{array}{} \displaystyle {\zeta _m} \to \psi \,\,{\text{a}}{\rm{.e}}{\rm{. in }}{\mathbb{R}^n}. \end{array}$$(39)

On the other hand, by the first statement in this lemma, the strong convergence in (38) implies that there exists a subsequence {ζ_mk} such that ζ_mk → σ a.e. as k → ∞. Therefore, (39) leads to ψ = φ a.e. on ℝⁿ so that ψ_m → ψ weakly as m → ∞. The third item is a known result in Functional Analysis, cf. [5, Chapter 4]. Thus the proof is completed.

Let us define the following energy functional on E: for (u, v) ∈ E,

$Γ (u, v) = ‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2} + 2 \int_{ℝ^{n}} (F (x, u) + ϕ_{3} (x)) d x .$ $$\begin{array}{} \displaystyle \Gamma (u, v)=\|v\|^2+ \left(\alpha+\delta^2-\beta \delta\right)\|u\|^2 + \|\nabla u\|^2+2\int_{\mathbb{R}^n} (F(x, u) + \phi_3 (x))\, dx. \end{array}$$(40)

Compare (9) and (40), we see that

$Γ (u, v) \leq ‖ (u, v) ‖_{E}^{2} + 2 \int_{ℝ^{n}} (F (x, u) + ϕ_{3} (x)) d x .$ $$\begin{array}{} \displaystyle \Gamma (u, v) \leq \|(u, v)\|_E^2 + 2\int_{\mathbb{R}^n} (F(x, u) + \phi_3 (x))\, dx. \end{array}$$(41)

Lemma 6

For every bounded set B ⊂ E and any integer k > 0, there exists a constant M₁ = M₁(B, k) > 0 such that for all m > M₁ one has t_m > k with the property that

$Γ (u (t_{m} - t, u_{0, m}), v (t_{m} - t, v_{0, m})) \leq R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]$ $$\begin{array}{} \displaystyle \Gamma (u(t_m - t, u_{0, m}), v(t_m - t, v_{0, m})) \leq R + 1 + \frac{1}{\sigma} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right] \end{array}$$(42)

for all t ∈ [0, k] and (u_{0, m}, v_0,m) ∈ B, where the constant R is the same as in (19).

Proof. Integrate the inequality (14) over the time interval [0, t] ⊂ [0, k], where δ ≥ σ by (13). Similar to (18), there exists M₁ = M₁(B, k) > 0 such that for all m > M₁ one has t_m > k and

$\begin{matrix} Γ (u (t_{m} - t, u_{0, m}), v (t_{m} - t, v_{0, m})) \leq e^{- σ (k - t)} Γ (u (t_{m} - k, u_{0, m}), v (t_{m} - k, v_{0, m})) \\ + \int_{t}^{k} e^{- σ (s - t)} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}) d s \\ \leq R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}], t \in [0, k] . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \Gamma (u(t_m - t, u_{0, m}), v(t_m - t, v_{0, m})) \leq e^{-\sigma (k - t)} \Gamma (u(t_m -k, u_{0,m}), v(t_m -k, v_{0, m})) \\ &\, + \int_{t}^k e^{-\sigma (s-t)} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right) ds \\ \leq &\, R + 1 + \frac{1}{\sigma} \left[ 2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta} \right], \quad t \in [0, k]. \end{split} \end{array}$$(43)

Therefore, (42) is proved.

Theorem 7

For every bounded set B and for any sequences t_m → ∞ and g_0,m = (u_0,m, v_0,m) ∈ B, the sequence ${Φ (t_{m}, g_{0, m})}_{m = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\Phi(t_m, g_{0,m})\}^\infty_{m=1} \end{array}$has a strongly convergent subsequence in H¹(ℝⁿ) × L²(ℝⁿ), where Φ is the solution semiflow generated by the nonlinear damped wave equation (4).

Proof. The proof goes through the following steps.

Step 1

By Lemma 3, there is a constant M₂ = M₂(B) > 0 such that for all m ≥ M₂ and g_0,m ∈ B, we have

$‖ Φ (t_{m}, g_{0, m}) ‖_{E} \leq R + 1$ $$\begin{array}{} \displaystyle \|\Phi(t_m, \, g_{0,m})\|_E\leq R+1 \end{array}$$(44)

where R > 0 is given by (19). Then there is $(\tilde{u}, \tilde{v}) \in E$ $\begin{array}{} \displaystyle (\tilde{u}, \tilde{v}) \in E \end{array}$ such that, up to a subsequence and relabeled as the same,

$\begin{matrix} Φ (t_{m}, g_{0, m}) \to (\tilde{u}, \tilde{v}) weakly in E and \\ Φ (t_{m}, g_{0, m}) \to (\tilde{u}, \tilde{v}) weakly in H^{1} (ℝ^{n}) \times L^{2} (ℝ^{n}) . \end{matrix}$ $$\begin{array}{*{20}{l}} {\Phi ({t_m},{g_{0,m}}) \to (\tilde{u} ,\tilde{v} )\quad {\text{weakly in }}\,\,E\quad {\text{and}}}\\ {\Phi ({t_m},{g_{0,m}}) \to (\tilde{u} ,\tilde{v} )\quad {\text{weakly in }}\,\,\,{H^1}({\mathbb{R}^n}) \times {L^2}({\mathbb{R}^n}).} \end{array}$$(45)

Since E is a reflexive and separable Banach space, the weak lower-semicontinuity of the E-norm and of the norm of H¹(ℝⁿ) × L²(ℝⁿ) as well implies that

$\begin{matrix} \underset{m \to \infty}{\lim \inf} ‖ Φ (t_{m}, g_{0, m}) ‖_{E} \geq ‖ (\tilde{u}, \tilde{v}) ‖_{E}, \\ \underset{m \to \infty}{\lim \inf} ‖ Φ (t_{m}, g_{0, m}) ‖_{H^{1} \times L^{2}} \geq ‖ (\tilde{u}, \tilde{v}) ‖_{H^{1} \times L^{2}} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\liminf_{m\rightarrow \infty} \, \|\Phi(t_m, \, g_{0,m})\|_E\geq \|(\tilde{u}, \tilde{v})\|_E, \\ &\liminf_{m\rightarrow \infty} \, \|\Phi(t_m, \, g_{0,m})\|_{H^1 \times L^2} \geq \|(\tilde{u}, \tilde{v})\|_{H^1 \times L^2}. \end{split} \end{array}$$(46)

Next we want to prove that in the Hilbert space H¹(ℝⁿ) × L²(ℝⁿ),

$Φ (t_{m}, g_{0, m}) \to (\tilde{u}, \tilde{v}) strongly.$ $$\begin{array}{} \displaystyle \Phi ({t_m},{g_{0,m}}) \to (\tilde{u} ,\tilde{v})\,{\text{strongly}}{\rm{.}} \end{array}$$(47)

It suffices to show that

$\underset{m \to \infty}{\lim \sup} ‖ Φ (t_{m}, g_{0, m}) ‖_{H^{1} \times L^{2}} \leq ‖ (\tilde{u}, \tilde{v}) ‖_{H^{1} \times L^{2}} .$ $$\begin{array}{} \displaystyle \limsup_{m\rightarrow \infty}\, \|\Phi(t_m, \, g_{0,m})\|_{H^1 \times L^2}\leq \|(\tilde{u}, \tilde{v})\|_{H^1 \times L^2}. \end{array}$$(48)

If so, then (46) and (48) will lead to ${lim}_{m \to \infty} ∥ Φ (t_{m}, g_{0, m}) ∥_{H^{1} \times L^{2}} = ∥ (\tilde{u}, \tilde{v}) ∥_{H^{1} \times L^{2}}$ $\lim\nolimits_{m\rightarrow \infty}\|\Phi(t_m, g_{0,m})\|_{H^1 \times L^2} = \|(\tilde{u}, \tilde{v})\|_{H^1 \times L^2}$. By the item 3 of Lemma 5, we shall obtain (47).

Step 2

By Lemma 6 and (7), there exists a constant C > 0 such that, for any given integer k > 0 and all m ≥ M₁(B, k), one has t_m > k and

$\begin{matrix} ‖ (u (t_{m} - t, u_{0, m}), v (t_{m} - t, v_{0, m})) ‖_{E} \\ \leq C {[R + 1 + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ})]}^{1 / 2} \\ + C {[R + 1 + \frac{1}{σ} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ})]}^{1 / p}, t \in [0, k], \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \|(u(t_m - t, u_{0, m}), v(t_m - t, v_{0, m}))\|_E \\ \leq &\, C\left[ R+1 + \frac{1}{\sigma} \, \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right)\right]^{1/2} \\ + &\, C\left[ R+1 + \frac{1}{\sigma} \, \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right)\right]^{1/p}, \;\; t \in [0, k], \end{split} \end{array}$$(49)

for any (u_0,m, v_0,m) ∈ B. In particular, (49) is satisfied for t = k.

According to Banach-Alaoglu theorem, there exists a sequence ${{\tilde{u}}_{k}, {\tilde{v}}_{k}}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\tilde{u}_k, \tilde{v}_k\}^\infty_{k=1} \end{array}$ in the space E and subsequences of ${t_{m}}_{m = 1}^{\infty}$ $\begin{array}{} \displaystyle \{t_m\}_{m=1}^\infty \end{array}$ and ${(u_{0, m}, v_{0, m})}_{m = 1}^{\infty}$ $\begin{array}{} \displaystyle \{(u_{0, m}, v_{0, m})\}^\infty_{m=1} \end{array}$ again relabeled as the same, such that for every integer k ≥ 1,

$(u (t_{m} - k, u_{0, m}), v (t_{m} - k, v_{0, m})) \to ({\tilde{u}}_{k}, {\tilde{v}}_{k}) Weakly in E,$ $$\begin{array}{} \displaystyle (u({t_m} - k,{u_{0,m}}),v({t_m} - k,{v_{0,m}})) \to \left( {{{\tilde u}_k},{{\tilde v}_k}} \right){\text{ Weakly in }}E, \end{array}$$(50)

as m → ∞, which can be extracted through a diagonal selection procedure as in Real Analysis.

By the weakly continuous dependence on the initial data of the weak solutions stated in Lemma 2, here the weak convergence (50) together with the concatenation

$(u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) = (u (k, u (t_{m} - k, u_{0, m})), v (k, v (t_{m} - k, v_{0, m})))$ $$\begin{array}{} \displaystyle (u(t_m, u_{0, m}), \;\, v(t_m, v_{0, m})) = (u(k, u(t_m -k, u_{0, m})), \;\, v(k, v(t_m -k, v_{0, m}))) \end{array}$$(51)

implies that for all integers k ≥ 1, when m → ∞,

$(u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \to (u (k, {\tilde{u}}_{k}), v (k, {\tilde{v}}_{k})) weakly in E .$ $$\begin{array}{} \displaystyle (u({t_m},{u_{0,m}}),v({t_m},{v_{0,m}})) \to (u(k,{\tilde u_k}),v(k,{\tilde v_k}))\,\,{\text{weakly in }}E. \end{array}$$(52)

Thus (45) and (52) validate the following equality that for all positive integers k,

$(\tilde{u}, \tilde{v}) = (u (k, {\tilde{u}}_{k}), v (k, {\tilde{v}}_{k})) .$ $$\begin{array}{} \displaystyle (\tilde{u}, \tilde{v})=(u(k, \tilde{u}_k), v(k, \tilde{v}_k)). \end{array}$$(53)

By the similar argument from (11) to(12), the weak solutions (u, v) of (4) satisfy

$\frac{d}{d t} Γ (u (t, u_{0}), v (t, v_{0})) + 2 σ Γ (u (t, u_{0}), v (t, v_{0})) \leq G (u (t, u_{0}), v (t, v_{0})),$ $$\begin{array}{} \displaystyle \frac{d}{dt} \,\Gamma (u(t, u_0), v(t, v_0)) +2\sigma \,\Gamma (u(t, u_0), v(t, v_0)) \leq G (u(t, u_0), v(t, v_0)), \end{array}$$(54)

where

$\begin{matrix} G (u, v) = - 2 (β - δ - σ) ‖ v ‖^{2} - 2 (δ - σ) (α + δ^{2} - β δ) ‖ u ‖^{2} \\ - 2 (δ & - σ) ‖ \nabla u ‖^{2} + 4 σ \int_{ℝ^{n}} (F (x, u) + ϕ_{3} (x)) d x - 2 δ 〈 f (x, u), u 〉 + 2 〈 g, v 〉 . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{array}{*{20}{c}} {G(u,v) = - 2(\beta - \delta - \sigma )v{^2} - 2(\delta - \sigma )\left( {\alpha + {\delta ^2} - \beta \delta } \right)u{^2}}\\ { - 2(\delta {\rm{\;\& }} - \sigma )\nabla u{^2} + 4\sigma {\int _{{\mathbb{R}^n}}}(F(x,u) + {\phi _3}(x))dx - 2\delta \langle f(x,u),u\rangle + 2\langle g,v\rangle .} \end{array} \end{array}$$(55)

From (53) and (54), for any integer k ≥ 1 we have

$Γ (\tilde{u}, \tilde{v}) \leq e^{- 2 σ k} Γ ({\tilde{u}}_{k}, {\tilde{v}}_{k}) + \int_{0}^{k} e^{- 2 σ ξ} G (u (ξ, {\tilde{u}}_{k}), v (ξ, {\tilde{v}}_{k})) d ξ .$ $$\begin{array}{} \displaystyle \Gamma (\tilde{u}, \,\tilde{v}) \leq e^{-2\sigma k} \,\Gamma (\tilde{u}_k, \tilde{v}_k) +\int_0^{k} e^{-2\sigma \xi} \, G (u(\xi, \,\tilde{u}_k), v(\xi, \,\tilde{v}_k ))\, d\xi. \end{array}$$(56)

Step 3

On the other hand, from the concatenation (51) and the inequality (54), we obtain

$\begin{matrix} Γ (u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \leq e^{- 2 σ k} Γ (u (t_{m} - k, u_{0, m}), v (t_{m} - k, v_{0, m})) \\ - 2 (β - δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ v (ξ, v (t_{m} - k, v_{0, m})) ‖^{2} d ξ \\ - 2 (δ - σ) (α + δ^{2} - β δ) \int_{0}^{k} e^{- 2 σ ξ} ‖ u (ξ, u (t_{m} - k, u_{0, m})) ‖^{2} d ξ \\ - 2 (δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ \nabla u (ξ, u (t_{m} - k, u_{0, m})) ‖^{2} d ξ \\ + 4 σ \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} (F (x, u (ξ, u (t_{m} - k, u_{0, m}))) + ϕ_{3} (x)) d x d ξ \\ - 2 δ \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} f (x, u (ξ, u (t_{m} - k, u_{0, m}))) u (ξ, u (t_{m} - k, u_{0, m})) d x d ξ \\ + 2 \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} g (x) v (ξ, v (t_{m} - k, v_{0, m})) d x d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} \Gamma &\, (u(t_m, \, u_{0, m}), v(t_m, \, v_{0, m})) \leq e^{-2\sigma k}\, \Gamma (u(t_m - k, \, u_{0, m}), v(t_m -k, \, v_{0, m})) \\ & -2(\beta-\delta-\sigma)\int_0^{k}e^{-2\sigma \xi}\|v(\xi, v(t_m -k, \, v_{0, m}))\|^2d\,\xi \\ &-2(\delta-\sigma)\left(\alpha+\delta^2-\beta\delta\right)\int_0^{k} e^{-2\sigma \xi}\|u(\xi, u(t_m -k, \, u_{0, m}))\|^2d\xi \\ & -2(\delta-\sigma)\int_0^{k} e^{-2\sigma \xi} \|\nabla u(\xi, u(t_m -k, \, u_{0, m}))\|^2 \, d\xi \\ &+4\sigma\int_0^{k} e^{-2\sigma \xi} \int_{\mathbb{R}^n} (F(x, u(\xi, u(t_m -k, \, u_{0, m}))) + \phi_3 (x))\, dx\,d\xi \\ &-2\delta\int_0^{k} e^{-2\sigma \xi} \int_{\mathbb{R}^n}f(x, u(\xi, u(t_m -k, \, u_{0, m}))) u(\xi, u(t_m -k, \, u_{0, m}))\, dx\, d\xi \\ & +2\int_0^{k}e^{-2\sigma \xi}\int_{\mathbb{R}^n} g(x)\, v(\xi, v(t_m -k, \, v_{0, m}))\, dx\, d\xi. \end{split} \end{array}$$(57)

Below we treat all these terms on the right-hand side of the inequality (57).

1) For the first term on the right-hand side of (57), by (42) in Lemma 6, for all m ≥ M₁(B, k) we have

$\begin{matrix} e^{- 2 σ k} Γ (u (t_{m} - k, u_{0, m}), v (t_{m} - k, v_{0, m})) \\ \leq e^{- 2 σ k} (R + 1 + \frac{1}{σ} e^{\frac{1}{2} σ k} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]) \\ \leq e^{- σ k} (R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, e^{-2\sigma k} \, \Gamma (u(t_m -k, u_{0, m}), v(t_m -k, v_{0, m})) \\ \leq &\, e^{-2\sigma k} \left(R + 1 + \frac{1}{\sigma} \, e^{\frac{1}{2} \sigma k} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right]\right) \\ \leq &\, e^{-\sigma k} \left(R + 1 + \frac{1}{\sigma} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right]\right). \end{split} \end{array}$$(58)

2) For the second term on the right-hand side of (57), by (50) and the weakly continuous dependence of solutions on the initial data stated in Lemma 2, we find that for any ξ ∈ [0, k], when m → ∞,

$v (ξ, v (t_{m} - k, v_{0, m})) \to v (ξ, {\tilde{v}}_{k}) weakly in L^{2} (ℝ^{n}),$ $$\begin{array}{} v(\xi ,v({t_m} - k,{v_{0,m}})) \to v(\xi ,{\tilde v_k})\quad \text{weakly in}\,{L^2}({\mathbb{R}^n}), \end{array}$$

which implies that for all ξ ∈ [0, k],

$\underset{m \to \infty}{\lim \inf} ‖ v (ξ, v (t_{m} - k, v_{0, m})) ‖^{2} \geq ‖ v (ξ, {\tilde{v}}_{k}) ‖^{2} .$ $$\begin{array}{} \displaystyle \liminf_{m\rightarrow \infty} \, \|v(\xi, \,v(t_m -k, \, v_{0, m}))\|^2 \geq \|v(\xi, \,\tilde{v}_{k})\|^2. \end{array}$$(59)

By (59) and Fatou’s lemma we obtain

$\begin{matrix} \underset{m \to \infty}{\lim \inf} \int_{0}^{k} e^{- 2 σ ξ} ‖ v (ξ, v (t_{m} - k, v_{0, m})) ‖^{2} d ξ \\ \geq \int_{0}^{k} e^{- 2 σ ξ} \underset{m \to \infty}{\lim \inf} ‖ v (ξ, v (t_{m} - k, v_{0, m})) ‖^{2} d ξ \geq \int_{0}^{k} e^{- 2 σ ξ} ‖ v (ξ, {\tilde{v}}_{k}) ‖^{2} d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\liminf_{m\rightarrow \infty}\, \int_0^{k} e^{-2\sigma \xi}\|v(\xi, \, v(t_m -k, \, v_{0, m}))\|^2\, d\xi \\ \geq &\, \int_0^{k} e^{-2\sigma \xi}\liminf_{m\rightarrow \infty}\, \|v(\xi, \, v(t_m -k, \,v_{0, m}))\|^2\, d\xi \geq \int_0^{k} e^{-2\sigma \xi}\|v(\xi, \,\tilde{v}_{k})\|^2\, d\xi. \end{split} \end{array}$$(60)

Since (10) and (13) implies β − δ − σ ≥ β − 2δ > 0, (60) leads to

$\begin{matrix} \underset{m \to \infty}{\lim \sup} - 2 (β - δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ v (ξ, v (t_{m} - k, v_{0, m})) ‖^{2} d ξ \\ = - 2 (β - δ - σ) \underset{m \to \infty}{\lim \inf} \int_{0}^{k} e^{- 2 σ ξ} ‖ v (ξ, v (t_{m} - k, v_{0, m})) ‖^{2} d ξ \\ \leq - 2 (β - δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ v (ξ, {\tilde{v}}_{k}) ‖^{2} d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\limsup_{m\rightarrow \infty}\, -2(\beta -\delta-\sigma)\int_0^k e^{-2\sigma \xi} \|v(\xi, \, v(t_m -k, \,v_{0, m}))\|^2\, d\xi \\ =&\, -2(\beta -\delta-\sigma)\liminf_{m\rightarrow \infty}\, \int_0^k e^{-2\sigma \xi} \|v(\xi, \, v(t_m -k, \, v_{0, m}))\|^2\, d\xi \\ \leq &\, -2(\beta -\delta-\sigma)\int_0^k e^{-2\sigma \xi} \|v(\xi, \, \tilde{v}_{k})\|^2\, d\xi. \end{split} \end{array}$$(61)

Similarly for the third and fourth terms, by (50) and Fatou’s lemma we obtain

$\begin{matrix} \underset{m \to \infty}{\lim \sup} - 2 (δ - σ) (α + δ^{2} - β δ) \int_{0}^{k} e^{- 2 σ ξ} ‖ u (ξ, u (t_{m} - k, u_{0, m})) ‖^{2} d ξ \\ \leq - 2 (δ - σ) (α + δ^{2} - β δ) \int_{0}^{k} e^{- 2 σ ξ} ‖ u (ξ, {\tilde{u}}_{k}) ‖^{2} d ξ, \\ \underset{m \to \infty}{\lim \sup} - 2 (δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ \nabla u (ξ, u (t_{m} - k, u_{0, m})) ‖^{2} d ξ \\ \leq - 2 (δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ \nabla u (ξ, {\tilde{u}}_{k}) ‖^{2} d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} \limsup_{m\rightarrow \infty} &\, -2(\delta-\sigma)\left(\alpha+\delta^2-\beta\delta\right)\int_0^k e^{-2\sigma \xi}\|u(\xi, \, u(t_m -k, \, u_{0, m}))\|^2\, d\xi \\ &\leq -2(\delta-\sigma)\left(\alpha+\delta^2-\beta\delta\right)\int_0^k e^{-2\sigma \xi}\|u(\xi, \,\tilde{u}_{k})\|^2\, d\xi , \\ \limsup_{m\rightarrow \infty} &\, -2(\delta-\sigma)\int_0^k e^{-2\sigma \xi}\|\nabla u(\xi, \, u(t_m -k, \, u_{0, m}))\|^2\, d\xi \\ &\leq -2(\delta-\sigma)\int_0^k e^{-2\sigma \xi}\|\nabla u(\xi, \, \tilde{u}_{k})\|^2\, d\xi. \end{split} \end{array}$$(62)

3) For the fifth term on the right-hand side of (57), we have

$\begin{matrix} | \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} (F (x, u (ξ, u (t_{m} - k, u_{0, m}))) - F (x, u (ξ, {\tilde{u}}_{k}))) d x d ξ | \\ \leq \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | > r} | F (x, u (ξ, u (t_{m} - k, u_{0, m}))) - F (x, u (ξ, {\tilde{u}}_{k})) | d x d ξ \\ + \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | \leq r} | F (x, u (ξ, u (t_{m} - k, u_{0, m}))) - F (x, u (ξ, {\tilde{u}}_{k})) | d x d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\left|\int_0^k e^{-2\sigma \xi}\int_{\mathbb{R}^n} \left(F(x, u(\xi, u(t_m -k, u_{0, m})))-F(x, u(\xi, \tilde{u}_k))\right) dx\,d\xi \right| \\ \leq& \int_0^k e^{-2\sigma \xi}\int_{|x|> r} \left|F(x, u(\xi, u(t_m -k, u_{0, m})))-F(x, u(\xi, \tilde{u}_k))\right|dx\,d\xi \\ +& \int_0^k e^{-2\sigma \xi}\int_{|x|\leq r} \left|F(x, u(\xi, u(t_m -k, u_{0, m})))-F(x, u(\xi,\tilde{u}_k))\right|dx\,d\xi. \end{split} \end{array}$$(63)

A) For any given η > 0, by the proof of Lemma 4 adapted to the time interval [k, ∞), there exist M₃ = M₃(B, η) > M₂ and K = K(B, η) ≥ 1 such that for ξ ∈ [0, k], whenever r > K and m > M₃, one has

$\int_{| x | > r} (| u (t_{m} - ξ, u_{0, m}) |^{2} + | u (t_{m} - ξ, u_{0, m}) |^{p} + | ϕ_{1} |^{2} + | ϕ_{2} | + | ϕ_{3} |) d x < η .$ $$\begin{array}{} \displaystyle \int_{|x| \gt r} \left(|u(t_m - \xi, u_{0, m})|^2 + |u(t_m - \xi, u_{0, m})|^p + |\phi_1|^2 + |\phi_2| + |\phi_3|\right) dx \lt \eta. \end{array}$$(64)

In view of (5) and (6), there is a constant L₁ > 0 such that for any ξ ∈ [0, k] one has

$\begin{matrix} \int_{| x | > r} | F (x, u (t_{m} - ξ, u_{0, m})) | d x \\ \leq \int_{| x | > r} L_{1} (| u (t_{m} - ξ, & u_{0, m}) |^{2} + | u (t_{m} - ξ, u_{0, m}) |^{p} + | ϕ_{1} |^{2} + | ϕ_{2} | + | ϕ_{3} |) d x < L_{1} η, \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\int_{|x|> r} |F(x, u(t_m - \xi, u_{0, m}))|\, dx \\ \leq \int_{|x| \gt r} L_1 (|u(t_m - \xi, &\, u_{0, m}) |^2 + |u(t_m - \xi, u_{0, m})|^p + |\phi_1|^2 + |\phi_2| + |\phi_3|) dx \lt L_1 \eta, \end{split} \end{array}$$

for all r > K and m > M₃.

B) Since (50) shows that

${\tilde{u}}_{k} = (weak) \lim_{m \to \infty} u (t_{m} - k, u_{0, m}) in L^{2} (ℝ^{n}) \cap L^{p} (ℝ^{n}),$ $$\begin{array}{} \displaystyle {\tilde u_k} = ({\text{weak}})\mathop {\lim }\limits_{m \to \infty } u({t_m} - k,{u_{0,m}})\,\,{\text{in}}\,\,{L^2}({\mathbb{R}^n})\mathop \cap \nolimits^ {L^p}({\mathbb{R}^n}), \end{array}$$

by the weakly continuous dependence of solutions on intial data stated in Lemma 2, by the weak lower-semicontinuity of the L² and L_P norms, it yields from (64) that

$\begin{matrix} \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | > r} | F (x, u (k - ξ, {\tilde{u}}_{k})) | d x d ξ \\ \leq \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | > r} L_{1} (| u (k - ξ, {\tilde{u}}_{k}) |^{2} + | u (k - ξ, {\tilde{u}}_{k}) |^{p} + | ϕ_{1} |^{2} + | ϕ_{2} | + | ϕ_{3} |) d x d ξ \\ = \int_{0}^{k} e^{- 2 σ ξ} L_{1} (‖ u (k - ξ, {\tilde{u}}_{k}) ‖_{L^{2} (ℝ^{n} \ B_{r})}^{2} + ‖ u (k - ξ, {\tilde{u}}_{k}) ‖_{L^{p} (ℝ^{n} \ B_{r})}^{p}) d ξ \\ + \int_{0}^{k} e^{- 2 σ ξ} L_{1} \int_{| x | > r} (| ϕ_{1} |^{2} + | ϕ_{2} | + | ϕ_{3} |) d x d ξ \\ \leq \int_{0}^{k} e^{- 2 σ ξ} L_{1} [\underset{m \to \infty}{\lim \inf} ‖ u (k - ξ, {\tilde{u}}_{k}) ‖_{L^{2} (ℝ^{n} \ B_{r})}^{2} + \underset{m \to \infty}{\lim \inf} ‖ u (k - ξ, {\tilde{u}}_{k}) ‖_{L^{p} (ℝ^{n} \ B_{r})}^{p}] d ξ \\ + \int_{0}^{k} e^{- 2 σ ξ} L_{1} \int_{| x | > r} (| ϕ_{1} |^{2} + | ϕ_{2} | + | ϕ_{3} |) d x d ξ \leq \frac{L_{1}}{2 σ} η, f o r r > K, m > M_{3} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\int_0^k e^{-2\sigma \xi} \int_{|x|> r} |F(x,\, u(k - \xi, \,\tilde{u}_k))|\, dx\, d\xi \\ \leq &\, \int_0^k e^{-2\sigma \xi} \int_{|x|> r} L_1 (|u(k-\xi, \,\tilde{u}_k)|^2 + |u(k-\xi, \,\tilde{u}_k)|^p + |\phi_1|^2 + |\phi_2| + |\phi_3|)\, dx\, d\xi \\ = &\, \int_0^k e^{-2\sigma \xi} L_1 \left(\|u(k-\xi, \,\tilde{u}_k)\|_{L^2 ({\mathbb{R}^n} \backslash B_r)}^2 + \|u(k-\xi, \,\tilde{u}_k)\|_{L^p ({\mathbb{R}^n} \backslash B_r)}^p \right) d\xi \\ &\, + \int_0^k e^{-2\sigma \xi} L_1 \int_{|x|>r} (|\phi_1|^2 + |\phi_2| + |\phi_3|)\, dx\, d\xi \\ \leq &\, \int_0^k e^{-2\sigma \xi} L_1 \left[\liminf_{m\to\infty}\, \|u(k-\xi, \,\tilde{u}_k)\|_{L^2 ({\mathbb{R}^n} \backslash B_r)}^2 + \liminf_{m\to\infty}\, \|u(k-\xi, \,\tilde{u}_k)\|_{L^p ({\mathbb{R}^n} \backslash B_r)}^p \right] d\xi \\ &\, + \int_0^k e^{-2\sigma \xi} L_1 \int_{|x|>r} (|\phi_1|^2 + |\phi_2| + |\phi_3|)\, dx\, d\xi \leq \frac{L_1}{2\sigma}\, \eta, \quad {\text{for}} \;\; r \gt K,\, m \gt M_3. \end{split} \end{array}$$

The above two inequalities show that there exists a constant L₂ = L₁ (1 +1/(2σ)) > 0 such that the first term on the right-hand side of (63) satisfies

$\begin{matrix} \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | > r} | F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) - F (x, u (k - ξ, {\tilde{u}}_{k})) | d x d ξ \\ \leq \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | > r} (| F (x, u (t_{m} - ξ, u_{0, m})) | + | F (x, u (k - ξ, {\tilde{u}}_{k})) |) d x d ξ \leq L_{2} η, \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\int_0^k e^{-2\sigma \xi}\int_{|x|> r} |F(x, u(k-\xi, u(t_m -k, u_{0, m})))-F(x, u(k-\xi, \tilde{u}_k))|\, dx\,d\xi \\ \leq &\int_0^k e^{-2\sigma \xi}\int_{|x|> r} \left(|F(x, u(t_m - \xi, u_{0, m}))| + |F(x, u(k- \xi, \tilde{u}_{k}))|\right) dx\,d\xi \leq L_2 \eta, \end{split} \end{array}$$(65)

for all r > K and m > M₃.

C) For the second term on the right-hand side of (63), by (50) we have

$u (k - ξ, u (t_{m} - k, u_{0, m})) \to u (k - ξ, {\tilde{u}}_{k}) weakly in H^{1} (B_{r}) \cap L^{p} (B_{r}) .$ $$\begin{array}{} \displaystyle u(k-\xi, \,u(t_m -k, u_{0, m})) \longrightarrow u(k-\xi, \,\tilde{u}_k) \;\; \mathrm{weakly } \;\, \mathrm{in} \;\, H^1(\mathbb{B}_r) \cap L^p(\mathbb{B}_r). \end{array}$$

Since $H^{1} (B_{r})$ $\begin{array}{} \displaystyle H^1(\mathbb{B}_r) \end{array}$ is compactly embedded in $L^{2} (B_{r})$ $\begin{array}{} \displaystyle L^2(\mathbb{B}_r) \end{array}$, it follows that for any ξ ∈ [0, k],

$u (k - ξ, u (t_{m} - k, u_{0, m})) \to u (k - ξ, {\tilde{u}}_{k}) strongly in L^{2} (B_{r}) .$ $$\begin{array}{} \displaystyle u(k -\xi, \, u(t_m -k, u_{0, m})) \longrightarrow u(k- \xi, \,\tilde{u}_k) \;\; \mathrm{strongly } \;\, \mathrm{in} \;\, L^2(\mathbb{B}_r). \end{array}$$(66)

Then by the first item of Lemma 5 and the continuity of F (x, u),

$F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) \to F (x, u (k - ξ, {\tilde{u}}_{k})) in B_{r}, as m \to \infty .$ $$\begin{array}{} \displaystyle F(x,u(k - \xi ,u({t_m} - k,{u_{0,m}}))) \to F(x,u(k - \xi ,{\tilde u_k}))\,\,{\text{in }}\,{\mathbb{B}_r},\,\,{\text{as}}\,\,m \to \infty . \end{array}$$(67)

On the other hand, by the Standing Assumption and Lemma 6, we have the following uniform bound that there is a constant L₃> 0 such that

$\begin{matrix} \int_{| x | < r} | F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) | d x \leq L_{1} (‖ u (k - ξ, u (t_{m} - k, u_{0, m})) ‖_{L^{2} (B_{r})}^{2} \\ + ‖ u (k - ξ, u (t_{m} - k, u_{0, m})) ‖_{L^{p} (B_{r})}^{p} + ‖ ϕ_{1} ‖^{2} + ‖ ϕ_{2} ‖_{L^{1} (ℝ^{n})} + ‖ ϕ_{3} ‖_{L^{1} (ℝ^{n})}) \\ \leq L_{3} [R + 1 + \frac{1}{σ} e^{\frac{1}{2} σ k} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}) + ‖ ϕ_{1} ‖^{2} + ‖ ϕ_{2} ‖_{L^{1}} + ‖ ϕ_{3} ‖_{L^{1}}] \end{matrix}$ $$\begin{equation} \begin{split} &\int_{|x|< r} \left|F(x, u(k -\xi, \, u(t_m -k, u_{0, m})))\right| \,dx \leq L_1 \left(\|u (k- \xi, \, u(t_m -k, u_{0, m}))\|_{L^2 (B_r)}^2 \right. \\[3pt] &\, \left. + \|u (k- \xi, \, u(t_m -k, u_{0, m}))\|_{L^p (B_r)}^p + \|\phi_1\|^2 + \|\phi_2 \|_{L^1(\mathbb {R}^n)} + \|\phi_3\|_{L^1 (\mathbb {R}^n)}\right) \\ \leq &\, L_3 \left[R + 1 + \frac{1}{\sigma} \, e^{\frac{1}{2} \sigma k} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right)+ \|\phi_1\|^2 + \|\phi_2 \|_{L^1} + \|\phi_3\|_{L^1}\right] \end{split} \end{equation}$$(68)

for any ξ ∈ [0, k] and m > M₁. By the second item of Lemma 6, it follows from (67) and (68) that

$F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) \to F (x, u (k - ξ, {\tilde{u}}_{k})) weakly in L^{1} (B_{r}),$ $$\begin{array}{} \displaystyle F(x, \, u(k-\xi, \, u(t_m -k, u_{0, m}))) \longrightarrow F(x, \, u(k-\xi, \tilde{u}_k)) \;\; \text{weakly in} \;\, L^1 (\mathbb{B}_r), \end{array}$$

as m → ∞. Consequently, when m → ∞,

$\int_{| x | < r} F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) d x \to \int_{| x | < r} F (x, u (k - ξ, {\tilde{u}}_{k})) d x .$ $$\begin{array}{} \displaystyle \int_{|x|< r} F(x, \, u(k-\xi, \, u(t_m -k, u_{0, m})))dx \longrightarrow \int_{|x|< r} F(x, \, u(k-\xi, \tilde{u}_k))\, dx. \end{array}$$(69)

Furthermore, (68) shows that

$\begin{matrix} | \int_{| x | < r} [F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) - F (x, u (k - ξ, {\tilde{u}}_{k}))] d x | \\ \leq L_{3} [R + 1 + \frac{1}{σ} e^{\frac{1}{2} σ k} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}) + ‖ ϕ_{1} ‖^{2} + ‖ ϕ_{2} ‖_{L^{1}} + ‖ ϕ_{3} ‖_{L^{1}}] \\ + ‖ F (\cdot, u (k - ξ, {\tilde{u}}_{k})) ‖_{L^{1} (ℝ^{n})} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\left|\int_{|x|< r} \left[F(x, \, u(k-\xi, \, u(t_m -k, u_{0, m})))-F(x, \, u(k-\xi, \tilde{u}_k)) \right] dx\,\right| \\ \leq &\, L_3 \left[R + 1 + \frac{1}{\sigma} \, e^{\frac{1}{2} \sigma k} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right)+ \|\phi_1\|^2 + \|\phi_2 \|_{L^1} + \|\phi_3\|_{L^1}\right] \\ &\, +\|F(\cdot, \, u(k- \xi, \tilde{u}_k))\|_{L^1 (\mathbb{R}^n)}. \end{split} \end{array}$$(70)

According to Lebesgue dominated convergence theorem, (69) and (70) imply that for every integer k ≥ 1 and any r ≥ K,

$\begin{matrix} \lim_{m \to \infty} \int_{0}^{k} & e^{- 2 σ ξ} \int_{| x | < r} F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) d x d ξ \\ = \int_{0}^{k} e^{- 2 σ ξ} \int_{| x | < r} F (x, u (k - ξ, {\tilde{u}}_{k})) d x d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} \lim_{m\to \infty} \int_0^k &\, e^{-2\sigma \xi} \int_{|x|<r} F(x, \, u(k-\xi, \, u(t_m -k, \, u_{0, m})))\, dx\, d\xi \\ & = \int_0^k e^{-2\sigma \xi}\int_{|x|< r} F(x, \, u(k-\xi, \, \tilde{u}_k))\, dx\, d\xi. \end{split} \end{array}$$(71)

Put together (63), (65) and (71). Then we obtain

$\begin{matrix} \lim_{m \to \infty} \int_{0}^{k} & e^{- 2 σ ξ} \int_{ℝ^{n}} (F (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) + ϕ_{3} (x)) d x d ξ \\ = \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} (F (x, u (k - ξ, {\tilde{u}}_{k})) + ϕ_{3} (x)) d x d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} \lim_{m\rightarrow \infty}\int_0^k &\, e^{-2\sigma \xi}\int_{\mathbb{R}^n} \left(F(x,\, u(k-\xi, \, u(t_m -k, \, u_{0, m}))) + \phi_3 (x)\right)\, dx\, d\xi \\ & = \int_0^k \, e^{-2\sigma \xi}\int_{\mathbb{R}^n} \left(F(x, \, u(k-\xi, \,\tilde{u}_k)) + \phi_3 (x)\right)\, dx\, d\xi. \end{split} \end{array}$$(72)

4) By an argument similar to the proof of (72), we can also prove the convergence of the sixth term on the right-hand side of (57),

$\begin{matrix} \lim_{m \to \infty} \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} f (x, u (k - ξ, u (t_{m} - k, u_{0, m}))) u (k - ξ, u (t_{m} - k, u_{0, m})) d x d ξ \\ = \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} f (x, u (k - ξ, {\tilde{u}}_{k})) u (k - ξ, {\tilde{u}}_{k}) d x d ξ, \\ \lim_{m \to \infty} \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} g (x) v (k - ξ, v (t_{m} - k, v_{0, m})) d x d ξ = \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} g (x) v (k - ξ, {\tilde{v}}_{k}) d x d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{gather*} \lim_{m\to \infty} \int_0^k e^{-2\sigma \xi} \int_{\mathbb{R}^n}f(x, \, u(k-\xi, \, u(t_m -k, u_{0, m})))\, u(k-\xi, \, u(t_m -k, u_{0, m}))\, dx\, d\xi \\a = \int_0^k e^{-2\sigma \xi}\int_{\mathbb{R}^n}f(x, \, u(k-\xi, \tilde{u}_{k})) \, u(k-\xi, \tilde{u}_{k})\, dx\, d\xi, \\ \lim_{m\to \infty} \int_0^k e^{-2\sigma \xi}\int_{\mathbb{R}^n} g(x) \, v(k-\xi, \, v(t_m -k, \, v_{0, m}))\, dx\, d\xi = \int_0^k e^{-2\sigma \xi}\int_{\mathbb{R}^n} g(x)\, v(k-\xi, \,\tilde{v}_{k})\, dx\, d\xi. \end{gather*} \end{array}$$

Step 4

Take the limit of (57) as m → ∞ and assemble together the results shown above in the items 1) through 5) of Step 3. Then we get

$\begin{matrix} \underset{m \to \infty}{\lim \sup} Γ (u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \leq e^{- σ k} (R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]) \\ - 2 (β - δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ v (k - ξ, {\tilde{v}}_{k}) ‖^{2} d ξ - 2 (δ - σ) (α + δ^{2} - β δ) \int_{0}^{k} e^{- 2 σ ξ} ‖ u (k - ξ, {\tilde{u}}_{k}) ‖^{2} d ξ \\ - 2 (δ - σ) \int_{0}^{k} e^{- 2 σ ξ} ‖ \nabla u (k - ξ, {\tilde{u}}_{k}) ‖^{2} d ξ + 4 σ \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} (F (x, u (k - ξ, {\tilde{u}}_{k})) + ϕ_{3} (x)) d x d ξ \\ + 2 \int_{0}^{k} e^{- 2 σ ξ} \int_{ℝ^{n}} [g (x) v (k - ξ, {\tilde{v}}_{k}) - δ f (x, u (k - ξ, {\tilde{u}}_{k})) u (k - ξ, {\tilde{u}}_{k})] d x d ξ . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\limsup_{m\rightarrow \infty} \, \Gamma (u(t_m, u_{0, m}), \, v(t_m, v_{0, m})) \leq e^{-\sigma k} \left(R+ 1 + \frac{1}{\sigma} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right]\right) \\ &\, -2(\beta -\delta-\sigma)\int_0^k e^{-2\sigma \xi}\|v(k-\xi, \tilde{v}_{k})\|^2\, d\xi -2(\delta-\sigma) \left(\alpha+\delta^2-\beta\delta\right)\int_0^k e^{-2\sigma \xi}\|u(k-\xi, \tilde{u}_{k})\|^2\, d\xi \\ &- 2(\delta-\sigma) \int_0^k e^{-2\sigma \xi}\|\nabla u(k-\xi, \tilde{u}_{k})\|^2\, d\xi + 4\sigma \int_0^k e^{-2\sigma \xi}\int_{\mathbb{R}^n} (F(x, \, u(k-\xi, \tilde{u}_k)) + \phi_3 (x))\, dx\,d\xi \\ &+2\int_0^k e^{-2\sigma \xi}\int_{\mathbb{R}^n} \left[g(x)\, v(k-\xi, \tilde{v}_{k}) - \delta f(x, \, u(k-\xi, \tilde{u}_{k})) u(k-\xi, \tilde{u}_{k})\right] dx\,d\xi. \end{split} \end{array}$$(73)

From (56) and (73) it follows that

$\begin{matrix} \underset{m \to \infty}{\lim \sup} Γ (u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \\ \leq e^{- σ k} (R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]) + \int_{0}^{k} e^{- 2 σ ξ} G (u (k - ξ, {\tilde{u}}_{k}), v (k - ξ, {\tilde{v}}_{k})) d ξ \\ = e^{- σ k} (R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]) + Γ (\tilde{u}, \tilde{v}) - e^{- 2 σ k} Γ ({\tilde{u}}_{k}, {\tilde{v}}_{k}) \\ \leq e^{- σ k} (R + 1 + \frac{1}{σ} [2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖^{2}}{β - δ}]) + Γ (\tilde{u}, \tilde{v}) . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\limsup_{m\to \infty}\, \Gamma (u(t_m, u_{0, m}), \, v(t_m, v_{0, m})) \\ \leq&\, e^{-\sigma k} \left(R + 1 + \frac{1}{\sigma} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right]\right) +\int_0^k e^{-2\sigma \xi}\, G(u(k-\xi, \tilde{u}_k), \, v(k-\xi, \tilde{v}_k))\, d\xi \\ =&\, e^{-\sigma k} \left(R + 1 + \frac{1}{\sigma} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right]\right) + \Gamma (\tilde{u}, \tilde{v}) - e^{-2\sigma k} \Gamma (\tilde{u}_k, \tilde{v}_k) \\ \leq&\, e^{-\sigma k} \left(R+ 1 + \frac{1}{\sigma} \left[2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|^2}{\beta - \delta}\right]\right) +\Gamma (\tilde{u}, \tilde{v}). \end{split} \end{array}$$

Take limit k → ∞ of the above inequality to obtain limsup

$\underset{m \to \infty}{\lim \sup} Γ (u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \leq Γ (\tilde{u}, \tilde{v}) .$ $$\begin{array}{} \displaystyle \limsup_{m\rightarrow \infty} \, \Gamma\,(u(t_m, u_{0, m}), v(t_m, v_{0, m})) \leq \Gamma\,(\tilde{u}, \tilde{v}). \end{array}$$(75)

On the other hand, from (53), (67) and (68) we have

$\lim_{m \to \infty} \int_{ℝ^{n}} F (x, u (t_{m}, u_{0, m})) d x = \int_{ℝ^{n}} F (x, \tilde{u}) d x,$ $$\begin{array}{} \displaystyle \lim_{m\to \infty} \int_{\mathbb{R}^n} F(x, \, u(t_m, u_{0, m}))\, dx = \int_{\mathbb{R}^n}F(x, \,\tilde{u})\, dx, \end{array}$$(76)

which along with (74) shows that

$\begin{matrix} \underset{m \to \infty}{\lim \sup} & (‖ v (t_{m}, v_{0, m}) ‖^{2} + (α + δ^{2} - β δ) ‖ u (t_{m}, u_{0, m}) ‖^{2} + ‖ \nabla u (t_{m}, u_{0, m}) ‖^{2}) \\ \leq ‖ \tilde{v} ‖^{2} + (α + δ^{2} - β δ) ‖ \tilde{u} ‖^{2} + ‖ \nabla \tilde{u} ‖^{2} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} \limsup_{m\to \infty} &\, \left(\|v(t_m, v_{0, m})\|^2 + (\alpha+\delta^2-\beta \delta) \|u(t_m, u_{0, m})\|^2 + \|\nabla u(t_m, u_{0, m})\|^2 \right) \\ &\, \leq \|\tilde{v}\|^2 +(\alpha+\delta^2 - \beta \delta) \|\tilde{u}\|^2 +\|\nabla \tilde{u}\|^2. \end{split} \end{array}$$(76)

Step 5

Note that the norm of H¹(ℝⁿ) × L²(ℝⁿ) is equivalent to

$‖ (u, v) ‖_{Π} \overset{def}{=} Γ (u, v) - 2 \int_{R^{n}} (F (x, u) + ϕ_{3} (x)) d x = ‖ v ‖^{2} + (α + δ^{2} - β δ) ‖ u ‖^{2} + ‖ \nabla u ‖^{2} .$ $$\begin{array}{} \displaystyle \|(u,v)\|{_\Pi }\mathop = \limits^{{\text{def}}} \Gamma (u,v) - 2{\smallint _{{^n}}}(F(x,u) + {\phi _3}(x))dx = \|v\|{^2} + (\alpha + {\delta ^2} - \beta \delta )\|u\|{^2} + \|\nabla u\|{^2}. \end{array}$$

Same as the second inequality in (46), from the weak convergence shown by (45), for any sequence ${g_{0, m} = (u_{0, m}, v_{0, m})}_{m = 1}^{\infty} \subset B$ $\begin{array}{} \displaystyle \{g_{0,m} = (u_{0,m}, v_{0,m})\}_{m=1}^\infty \subset B \end{array}$, we have

$\underset{m \to \infty}{\lim \inf} ‖ Φ (t_{m}, g_{0, m}) ‖_{Π} \geq ‖ (\tilde{u}, \tilde{v}) ‖_{Π} .$ $$\begin{array}{} \displaystyle \liminf_{m\to \infty} \, \| \Phi (t_m, \,g_{0,m} )\|_\Pi \geq \|(\tilde{u}, \tilde{v})\|_\Pi. \end{array}$$

Meanwhile, (76) implies that limsup

$\underset{m \to \infty}{\lim \sup} ‖ Φ (t_{m}, g_{0, m}) ‖_{Π} \leq ‖ (\tilde{u}, \tilde{v}) ‖_{Π} .$ $$\begin{array}{} \displaystyle \limsup_{m\to \infty} \, \| \Phi (t_m, \,g_{0,m})\|_\Pi \leq \|(\tilde{u}, \tilde{v})\|_\Pi. \end{array}$$

Thus we have proved lim

$\lim_{m \to \infty} ‖ Φ (t_{m}, g_{0, m}) ‖_{Π} = ‖ (\tilde{u}, \tilde{v}) ‖_{Π} .$ $$\begin{array}{} \displaystyle \lim_{m\to \infty} \, \| \Phi (t_m, \, g_{0,m} )\|_\Pi = \|(\tilde{u}, \tilde{v})\|_\Pi. \end{array}$$(77)

Finally, for the Hilbert space H¹(ℝⁿ) × L²(ℝⁿ), the weak convergence (45) and the norm convergence (77) imply the strong convergence. Therefore, up to finite steps of subsequence selections (always relabeled as the same), we reach the conclusion that

$\begin{matrix} lim_{m \to \infty} Φ (t_{m}, g_{0, m}) = (\tilde{u}, \tilde{v}) strongly H^{1} (R^{n}) \times L^{2} (R^{n}) . \end{matrix}$ $$\begin{array}{l} \mathop {\lim }\limits_{m \to \infty } \Phi ({t_m},{g_{0,m}}) = (\tilde{u}, \tilde{v} )\quad {\text{strongly }}\,{H^1}(\mathbb{R}^n) \times {L^2}(\mathbb{R}^n). \end{array}$$

Thus the proof is completed.

The Existence of Random Attractor

In this section we shall first prove an instrumental convergence theorem in the space L^p(X,ℳ, μ) of Vitalitype. It will pave the way to prove asymptotic compactness of the first component of the semiflow Φ in the space L^p(ℝⁿ) for any exponent $1 \leq p \leq \frac{n + 2}{n - 2}$ $\begin{array}{} \displaystyle 1 \le p \le \frac{{n + 2}}{{n - 2}} \end{array}$. This is the crucial and final step to accomplish the proof of the existence of a global attractor for this dynamical system F for the nonlinear damped wave equation (1).

Theorem 8

Let (X, ℳ, μ) be a σ-finite measure space and assume that a sequence ${f_{m}}_{m = 1}^{\infty} \subset L^{p} (X, M, μ)$ $\begin{array}{} \displaystyle \left\{ {{f_m}} \right\}_{m = 1}^\infty \subset {L^p}(X,{\mathscr M},\mu ) \end{array}$with 1 ≤ p < ∞ satisfies

$\lim_{m \to \infty} f_{m} (x) = f (x), a.e.$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{m \to \infty } {f_m}(x) = f(x),\,\,{\rm{\;a}}{\rm{.e}}{\rm{.}} \end{array}$$(78)

Then f ∈ L^p(X,ℳ, μ) and

$\lim_{m \to \infty} ‖ f_{m} - f ‖_{L^{p} (X, M, μ)} = 0$ $$\begin{array}{} \displaystyle \lim_{m \to \infty} \|f_m - f \|_{L^p (X, \mathscr{M},\, \mu)} = 0 \end{array}$$(79)

if and only if the following two conditions are satisfied:

(a) For any given ε > 0, there exists a set A_ε ∈ ℳ such that μ(A_ε) < ∞ and

$\int_{X \ A_{ε}} | f_{m} (x) |^{p} d μ < ε, for all m \geq 1.$ $$\begin{array}{} \displaystyle {\int _{X\backslash {A_\varepsilon }}}|{f_m}(x){|^p}d\mu \lt \varepsilon ,\quad {\text{for all}}\,\,m \ge 1. \end{array}$$(80)

(b) The absolutely continuous property of the L^p integrals is satisfied uniformly, i.e.

$\lim_{μ (Y) \to 0} \int_{Y} | f_{m} (x) |^{p} d μ = 0, uniformly for all m \geq 1.$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{\mu (Y) \to 0} {\int _Y}|{f_m}(x){|^p}d\mu = 0,\quad \,\,{\text{uniformly for all }}m \ge 1. \end{array}$$(81)

Proof. First we prove the necessity.

Statement (a): Under the condition (79), for an arbitrarily given ε > 0 there exists an integer N = N(ε) ≥ 1 such that

$‖ f_{m} - f ‖_{L^{p} (X, M, μ)}^{p} < \frac{ε}{2^{p}}, for all m > N .$ $$\begin{array}{} \|f_m - f\|_{L^p (X, \mathcal{M},\, \mu)}^p \lt \frac{\varepsilon}{2^p}, \quad \text{for all} \;\, m \gt N. \end{array}$$(82)

Since f ∈ L^p(X,ℳ, μ), there exist measurable sets B_ε and S_ε both of finite measure, such that

$\int_{X \ B_{ε}} | f (x) |^{p} d μ < \frac{ε}{2^{p}} and \int_{X \ S_{ε}} | f_{m} (x) |^{p} d μ < ε, for m = 1, \dots, N .$ $$\begin{array}{} \displaystyle {\int _{X\backslash {B_\varepsilon }}}|f(x){|^p}d\mu \lt \frac{\varepsilon }{{{2^p}}}\quad {\rm{\;and}}\quad {\int _{X\backslash {S_\varepsilon }}}|{f_m}(x){|^p}d\mu \lt \varepsilon ,\,\,{\rm{\;for }}\,m = 1, \cdots ,N. \end{array}$$(83)

Put A_ε = B_ε ∪ S_e. Then μ(A_ε) < ∞ and we have

$\begin{matrix} \begin{matrix} \int_{X ∖ A_{ε}} | f_{m} (x) |^{p} d μ = \int_{X ∖ A_{ε}} (| f_{m} (x) - f (x) | + | f (x) |)^{p} d μ \\ \leq & 2^{p - 1} (\int_{X} | f_{m} (x) - f (x) |^{p} d μ + \int_{X ∖ B_{ε}} | f (x) |^{p} d μ) < \frac{ε}{2} + \frac{ε}{2} = ε, for m > N . \end{matrix} \end{matrix}$ $$\begin{array}{c} \begin{split} &\int_{X \backslash A_\varepsilon} |f_m (x)|^p\, d\mu = \int_{X \backslash A_\varepsilon} (|f_m (x) - f(x)| + |f(x)|)^p\, d\mu \\ \leq &\, 2^{p-1} \left(\int_{X} |f_m (x) - f(x)|^p\, d\mu + \int_{X \backslash B_\varepsilon} |f(x)|^p\, d\mu\right) \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \;\; \text{for}\;\, m \gt N. \end{split} \end{array}$$

Besides, from the second inequality in (83) it follows that

$\int_{X \ A_{ε}} | f_{m} (x) |^{p} d μ \leq \int_{X \ S_{ε}} | f_{m} (x) |^{p} d μ < ε, for m = 1, \dots, N .$ $$\begin{array}{} \displaystyle {\int _{X\backslash {A_\varepsilon }}}\left| {{f_m}(x){|^p}d\mu \le {\int _{X\backslash {S_\varepsilon }}}} \right|{f_m}(x){|^p}d\mu \lt \varepsilon ,\quad {\rm{\;for }}\,m = 1, \cdots ,N. \end{array}$$

for m = 1,··· , N.

Therefore, the statement (a) is valid.

Statement (b): By the absolutely continuous property of Lebesgue integral on a σ-finite measure space, for any given ε > 0, there exists δ₀ = δ₀(ε) > 0 such that whenever μ(Y ) < δ₀ one has

$\int_{Y} | f (x) |^{p} d μ < \frac{ε}{2^{p}} and \int_{Y} | f_{m} (x) |^{p} d μ < ε, for m = 1, \dots, N,$ $$\begin{array}{} \displaystyle {\int _Y}|f(x){|^p}d\mu \lt \frac{\varepsilon }{{{2^p}}}\quad {\rm{ and}}\quad {\int _Y}|{f_m}(x){|^p}d\mu \lt \varepsilon ,\quad {\text{for }}\,m = 1, \cdots ,N, \end{array}$$(84)

where N = N(ε) is the same integer in (82). Then for any measurable set Y ⊂ X with μ(Y) < δ₀ one also has

$\int_{Y} | f_{m} (x) |^{p} d μ \leq 2^{p - 1} (\int_{X} | f_{m} (x) - f (x) |^{p} d μ + \int_{Y} | f (x) |^{p} d μ) < ε, for m > N .$ $$\begin{array}{} \displaystyle {\int _Y}|{f_m}(x){|^p}d\mu \le {2^{p - 1}}\left( {{\int _X}|{f_m}(x) - f(x){|^p}d\mu + {\int _Y}|f(x){|^p}d\mu } \right)\lt {\varepsilon ,\quad {\text{for }}\,m} \gt N. \end{array}$$

Thus the statement (b) is valid.

Next we prove the sufficiency. Suppose the two conditions (a) and (b) are satisfied. First of all, by the condition (a) and Fatou’s Lemma, for an arbitrarily given ε > 0 there exists a set A_ε of finite measure with

$\sup_{m \geq 1} \int_{X \ A_{ε}} | f_{m} (x) |^{p} d μ < ε,$ $$\begin{array}{} \displaystyle \sup_{m\geq 1} \, \int_{X \backslash A_\varepsilon} |f_m (x)|^p\, d\mu \lt \varepsilon, \end{array}$$

which implies that the limit function f in the assumption (78) satisfies

$\int_{X \ A_{ε}} | f (x) |^{p} d μ \leq \underset{m \to \infty}{\lim \inf} \int_{X \ A_{ε}} | f_{m} (x) |^{p} d μ < ε .$ $$\begin{array}{} \displaystyle \int_{X \backslash A_\varepsilon} |f (x)|^p\, d\mu \leq \liminf_{m\to \infty} \, \int_{X \backslash A_\varepsilon} |f_m (x)|^p\, d\mu \lt \varepsilon. \end{array}$$(85)

Hence it follows that

$f \in L^{p} (X \ A_{ε}) and ‖ f_{m} - f ‖_{L^{p} (X \ A_{ε})} < 2 ε^{1 / p}, for all m \geq 1.$ $$\begin{array}{} \displaystyle f \in {L^p}(X\backslash {A_\varepsilon })\quad {\text{and}}\quad {f_m} - f{_{{L^p}(X\backslash {A_\varepsilon })}} \lt 2{\varepsilon ^{1/p}},\,\,{\rm{ for all }}\,m \ge 1. \end{array}$$(86)

Therefore, the proof of f ∈ L^P(X, ℳ, μ) and (79) is reduced to proving that

$f \in L^{p} (Y) and \lim_{m \to \infty} ‖ f_{m} - f ‖_{L^{p} (Y)} = 0,$ $$\begin{array}{} \displaystyle f \in {L^p}(Y)\quad {\text{and}}\quad \mathop {\lim }\limits_{m \to \infty } {f_m} - f{_{{L^p}(Y)}} = 0, \end{array}$$(87)

for any given measurable set Y ⊂ X with μ(Y) < ∞.

By the condition (b), for any given ε > 0, there exists δ₁ = δ₁(ε) > 0 such that for any S ⊂ X with μ(S) < δ₁ one has

$\int_{S} | f_{m} (x) |^{p} d μ < ε^{p}, uniformly in m \geq 1.$ $$\begin{array}{} \displaystyle {\int _S}|{f_m}(x){|^p}d\mu \lt {\varepsilon ^p},\,\,{\text{uniformly in }}\,m \ge 1. \end{array}$$(88)

Consequently, by Fatou’s lemma,

$\int_{S} | f (x) |^{p} d μ \leq \underset{m \to \infty}{\lim \inf} \int_{S} | f_{m} (x) |^{p} d μ < ε^{p} .$ $$\begin{array}{} \displaystyle \int_S \, |f (x)|^p\, d\mu \leq \liminf_{m\to\infty}\, \int_S \, |f_m (x)|^p\, d\mu \lt \varepsilon^p. \end{array}$$(89)

By Egorov’s theorem on Lebesgue integral over such a set Y of finite measure in the space (X,ℳ, μ), there exists a measurable subset B ⊂ Y with μ(Y\B) < δ₁ such that

$\lim_{m \to \infty} f_{m} (x) = f (x), uniformly a.e. on B,$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{m \to \infty } {f_m}(x) = f(x),\,\,{\rm{\;uniformly a}}\;{\rm{.e}}{\rm{. on }}\,B, \end{array}$$

so that there exists an integer m₀ = m₀(ε) ≥ 1 such that

$‖ f_{m} - f ‖_{L^{p} (B)} < ε, for all m > m_{0} .$ $$\begin{array}{} \displaystyle {f_m} - f{_{{L^p}(B)}}\lt {\varepsilon ,\,\,{\text{for all }}\,\,m} \gt {m_0}. \end{array}$$(90)

Combining (88), (89) and (90), we get

$‖ f_{m} - f ‖_{L^{p} (Y)} \leq ‖ f_{m} ‖_{L^{p} (Y \ B)} + ‖ f ‖_{L^{p} (Y \ B)} + ‖ f_{m} - f ‖_{L^{p} (B)} < 3 ε, for m > m_{0} .$ $$\begin{array}{} \displaystyle \|{f_m} - f\|{_{{L^p}(Y)}} \le \|{f_m}\|{_{{L^p}(Y\backslash B)}} + \|f\|{_{{L^p}(Y\backslash B)}} + \|{f_m} - f\|{_{{L^p}(B)}}\lt {3\varepsilon ,\,\,{\text{for}}\,\,m} \gt {m_0}. \end{array}$$

Therefore, (87) is proved. The proof is completed.

Finally we present and prove the main result of this work on the existence of a global attractor for this semiflow Φ generated by the nonlinear damped wave equation (1) on the product Banach space with critical exponent and arbitrary space dimension.

Theorem 9

Under the Standing Assumption, the semiflow Φ generated by the nonlinear damped wave equation (1) in the converted problem (4) on the space E = (H¹(ℝⁿ) ⋂ L^p(ℝⁿ)) × L²(ℝⁿ) has a global attractor A in E.

Proof.Lemma 3 shows that there exists an absorbing set, the K = B_E(0, R) in the space E for the semiflow Φ. It suffices to prove that the semiflow F is asymptotically compact in E.

(1) Theorem 7 shows that for any given bounded set B ⊂ E and any sequences t_m → ∞ and {g_0,m = (u_0,m, v_0,m)} ⊂ B, the sequence ${Φ (t_{m}, g_{0, m})}_{m = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\Phi (t_m, g_{0,m})\}_{m=1}^\infty \end{array}$ has a convergent subsequence, which is relabeled by the same, such that

$Φ (t_{m}, g_{0, m}) \to (\tilde{u}, \tilde{v}) strongly in H^{1} (ℝ^{n}) \times L^{2} (ℝ^{n}),$ $$\begin{equation} \Phi(t_m, \, g_{0,m}) \longrightarrow (\tilde{u}, \,\tilde{v}) \;\; \text{strongly in} \;\, H^1 (\mathbb{R}^n) \times L^2 (\mathbb{R}^n), \end{equation}$$(91)

and consequently

$ℙ_{u} Φ (t_{m}, g_{0, m}) \to \tilde{u} strongly in L^{2} (ℝ^{n}) .$ $$\begin{array}{} \displaystyle {\mathbb{P}_u}\Phi ({t_m},{g_{0,m}}) \to \tilde{u} \,\,{\text{ strongly in }}\,{L^2}({\mathbb{R}^n}). \end{array}$$(92)

Here ℙ_u : (u, v) ↦ u is the component projection.

(2) Applying the first item in Lemma 5 to the space L²(ℝⁿ), it follows from (91) that there exists a subsequence ${Φ (t_{m_{k}}, g_{0, m_{k}})}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\Phi(t_{m_k}, \, g_{0, m_k})\}_{k=1}^\infty \end{array}$ of ${Φ (t_{m}, g_{0, m})}_{m = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\Phi(t_m, \, g_{0, m})\}_{m=1}^\infty \end{array}$ such that

$\lim_{k \to \infty} Φ (t_{m_{k}}, g_{0, m_{k}}) (x) = (\tilde{u} (x), \tilde{v} (x)), a.e. in ℝ^{n} .$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{k \to \infty } \Phi ({t_{{m_k}}},{g_{0,{m_k}}})(x) = (\tilde{u} (x),\tilde{v} (x)),\,\,{\text{a}}{\rm{.e}}{\rm{. in}}\,{\rm{ }}{\mathbb{R}^n}. \end{array}$$(93)

Hence we have

$\lim_{k \to \infty} ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x) = \tilde{u} (x), a.e. in ℝ^{n} .$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{k \to \infty } {\mathbb{P}_u}\Phi ({t_{{m_k}}},{g_{0,{m_k}}})(x) = \tilde{u} (x),\,\,{\text{a}}{\rm{.e}}{\rm{. in}}\,{\rm{\;}}{\mathbb{R}^n}. \end{array}$$(94)

Therefore, the assumption (78) in Theorem 8 is satisfied by the sequence of functions ${ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x)}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\mathbb{P}_u \Phi (t_{m_k}, \, g_{0, m_k}) (x)\}_{k=1}^\infty \end{array}$ in L^p(ℝⁿ).

(3) By Lemma 4, for any ε > 0, there exists an integer k₀ = k₀(B, ε) > 0 and V = V (ε) ≥ 1 such that for all k > k₀ one has

$\int_{ℝ^{n} \ B_{V}} | ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x) |^{p} d x \leq ‖ Φ (t_{m_{k}}, g_{0, m_{k}}) ‖_{E (ℝ^{n} \ B_{V})}^{p} < ε,$ $$\begin{equation} \int_{\mathbb{R} \backslash B_V} |\mathbb{P}_u \Phi (t_{m_k}, \, g_{0, m_k}) (x)|^p \, dx \leq \| \Phi(t_{m_k}, \, g_{0, m_k}) \|^p_{E(\mathbb{R} \backslash B_V)} \lt \varepsilon, \end{equation}$$(95)

for any g_0,mk ∈ B, where B_V is the ball centered at the origin with radius V in ℝⁿ. Then there exists V₀ = V₀(ε) > 0 such that

$\int_{ℝ^{n} \ B_{V_{0}}} | ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x) |^{p} d x < ε, for k = 1, \dots, k_{0} .$ $$\begin{equation} \int_{\mathbb{R} \backslash B_{V_0}} |\mathbb{P}_u \Phi (t_{m_k},\, g_{0, m_k}) (x)|^p \, dx \lt \varepsilon, \quad \text{for} \;\, k = 1, \cdots, k_0. \end{equation}$$(96)

Thus (95) and (96) confirm that with A_ε = B_{max{V, V0}} the condition (a) in Theorem 8 is satisfied by the sequence of functions ${ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x)}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\mathbb{P}_u \Phi (t_{m_k}, g_{0, m_k}) (x)\}_{k=1}^\infty \end{array}$ in L^p(ℝⁿ).

(4) Finally we prove that the uniform absolutely continuous condition (b) of Theorem 8 is also satisfied by the sequence of functions ${ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x)}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\mathbb{P}_u \Phi (t_{m_k}, g_{0, m_k}) (x)\}_{k=1}^\infty \end{array}$ in L^p(ℝⁿ).

According to the Standing Assumption, for any measurable set Y ⊂ ℝⁿ, we have

$C_{3} \int_{Y} | u |^{p} d x \leq \int_{Y} (F (x, u) + ϕ_{3} (x)) d x \leq Γ_{Y} (u, v), for (u, v) \in E,$ $$\begin{array}{} \displaystyle {C_3}{\int _Y}|u{|^p}dx \le {\int _Y}(F(x,u) + {\phi _3}(x))dx \le {\Gamma _Y}(u,v),\quad {\text{for}}\,(u,v) \in E, \end{array}$$

where Γ_Y (u, v) is analogous to (40) and defined by

$Γ_{Y} (u, v) = ‖ v ‖_{L^{2} (Y)}^{2} + (α + δ^{2} - β δ) ‖ u ‖_{L^{2} (Y)}^{2} + ‖ \nabla u ‖_{L^{2} (Y)}^{2} + 2 \int_{Y} (F (x, u) + ϕ_{3} (x)) d x .$ $$\begin{array}{} \displaystyle \Gamma_Y (u, v)=\|v\|_{L^2(Y)}^2+ \left(\alpha+\delta^2-\beta \delta\right)\|u\|_{L^2(Y)}^2 + \|\nabla u\|_{L^2(Y)}^2+2\int_{Y} (F(x, u) + \phi_3 (x))\, dx. \end{array}$$(97)

We can integrate the inequality (14) over the time interval [0,t_m] to get

$\begin{matrix} Γ_{Y} (u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \\ \leq e^{- σ t_{m}} Γ_{Y} (u_{0, m}, v_{0, m}) + \int_{0}^{t_{m}} e^{- σ t} (2 δ (C_{2} ‖ ϕ_{3} ‖_{L^{1}} + ‖ ϕ_{2} ‖_{L^{1}}) + \frac{‖ g ‖_{L^{2} (Y)}^{2}}{β - δ}) d t . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, \Gamma_Y (u(t_m, u_{0,m}), \, v(t_m, v_{0,m})) \\ \leq &\, e^{-\sigma t_m} \, \Gamma_Y (u_{0,m}, v_{0,m}) + \int_0^{t_m} e^{-\sigma t} \left(2\delta (C_2 \|\phi_3 \|_{L^1} + \|\phi_2 \|_{L^1}) + \frac{\|g\|_{L^2 (Y)}^2}{\beta - \delta}\right) dt. \end{split} \end{array}$$(98)

Substitute the expression of Γ_Y (u_0,m, v_0,m) with (u_0,m, v_0,m) ∈ B into the inequality (98). Since (5)-(6) yield

$\int_{Y} (F (x, u) + ϕ_{3} (x)) d x \leq \frac{1}{C_{2}} [C_{1} ‖ u ‖_{L^{p} (Y)}^{p} + ‖ u ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{2} ‖_{L^{1} (Y)}],$ $$\begin{array}{} \displaystyle \int_Y (F(x, u) + \phi_3 (x))\, dx \leq \frac{1}{C_2} \left[C_1 \|u\|_{L^p (Y)}^p + \|u\|_{L^2 (Y)}^2 + \|\phi_1 \|_{L^2 (Y)}^2 + \|\phi_2 \|_{L^1 (Y)} \right], \end{array}$$

for any g_0,m = (u_0,m, v_0,m) ∈ B, we get

$\begin{matrix} C_{3} \int_{Y} | u (t_{m}, u_{0, m}) |^{p} d x \leq Γ_{Y} (u (t_{m}, u_{0, m}), v (t_{m}, v_{0, m})) \\ \leq e^{- σ t_{m}} [‖ v_{0, m} ‖_{L^{2} (Y)}^{2} + (α + δ^{2} - β δ) ‖ u_{0, m} ‖_{L^{2} (Y)}^{2} + ‖ \nabla u_{0, m} ‖_{L^{2} (Y)}^{2}] \\ + e^{- σ t_{m}} \frac{1}{C_{2}} [C_{1} ‖ u_{0, m} ‖_{L^{p} (Y)}^{p} + ‖ u_{0, m} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{2} ‖_{L^{1} (Y)}] \\ + 2 δ C_{2} \int_{0}^{t_{m}} e^{- σ t} (‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{3} ‖_{L^{1} (Y)}) d t + \int_{0}^{t_{m}} \frac{e^{- σ t}}{β - δ} ‖ g ‖_{L^{2} (Y)}^{2} d t . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} &\, C_3 \int_Y |u(t_m, u_{0,m})|^p\, dx \leq \Gamma_Y (u(t_m, u_{0,m}), v(t_m, v_{0,m})) \\ \leq &\, e^{-\sigma t_m} \left[\|v_{0,m}\|_{L^2 (Y)}^2 + (\alpha + \delta^2 - \beta \delta)\|u_{0,m}\|_{L^2 (Y)}^2 + \|\nabla u_{0,m}\|_{L^2 (Y)}^2\right] \\ &\, + e^{-\sigma t_m} \frac{1}{C_2} \left[C_1\|u_{0,m}\|_{L^p (Y)}^p + \|u_{0,m}\|_{L^2 (Y)}^2 + \|\phi_1 \|_{L^2(Y)}^2 + \|\phi_2 \|_{L^1 (Y)} \right] \\a &\, + 2\delta C_2 \int_0^{t_m} e^{-\sigma t} \left(\|\phi_1 \|_{L^2 (Y)}^2 + \|\phi_3 \|_{L^1 (Y)}\right) dt + \int_0^{t_m} \frac{e^{-\sigma t}}{\beta - \delta} \|g\|_{L^2 (Y)}^2\, dt. \end{split} \end{array}$$(99)

Due to the absolute continuity of the respective Lebesgue integrals of the functions ϕ₁(x), ϕ₂(x), ϕ₃(x) and g involved in the above inequality (99), for an arbitrarily given η > 0, there exists μ₀ = μ₀(η) > 0 such that for any measurable set Y ⊂ ℝⁿ with μ(Y ) < μ₀ one has

$\begin{matrix} e^{- σ t_{m}} \frac{1}{C_{2}} (‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{2} ‖_{L^{1} (Y)}) + 2 δ C_{2} \int_{0}^{t_{m}} e^{- σ t} (‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{3} ‖_{L^{1} (Y)}) d t + \int_{0}^{t_{m}} \frac{e^{- σ t}}{β - δ} ‖ g ‖_{L^{2} (Y)}^{2} d t \\ \leq \frac{1}{C_{2}} (‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{2} ‖_{L^{1} (Y)}) + \frac{2 δ}{σ} C_{2} (‖ ϕ_{1} ‖_{L^{2} (Y)}^{2} + ‖ ϕ_{3} ‖_{L^{1} (Y)}) + \frac{1}{σ (β - δ)} ‖ g ‖_{L^{2} (Y)}^{2} < \frac{η}{2} . \end{matrix}$ $$\begin{array}{} \displaystyle \begin{split} e^{-\sigma t_m} & \frac{1}{C_2} \left(\|\phi_1 \|_{L^2(Y)}^2 + \|\phi_2 \|_{L^1 (Y)} \right) + 2\delta C_2 \int_0^{t_m} e^{-\sigma t} \left(\|\phi_1 \|_{L^2 (Y)}^2 + \|\phi_3 \|_{L^1 (Y)}\right) dt + \int_0^{t_m} \frac{e^{-\sigma t}}{\beta - \delta} \|g\|_{L^2 (Y)}^2\, dt \\ &\leq \frac{1}{C_2} \left(\|\phi_1 \|_{L^2(Y)}^2 + \|\phi_2 \|_{L^1 (Y)} \right) + \frac{2\delta}{\sigma} C_2 \left(\|\phi_1 \|_{L^2 (Y)}^2 + \|\phi_3 \|_{L^1 (Y)}\right) + \frac{1}{\sigma (\beta - \delta)} \|g\|_{L^2 (Y)}^2 \lt \frac{\eta}{2}. \end{split} \end{array}$$(100)

Moreover, since B ⊂ E is a bounded set, there exists a constant Ĉ > 0 such that

$\begin{matrix} e^{- σ t_{m}} [‖ v_{0, m} ‖_{L^{2} (Y)}^{2} + (α + δ^{2} - β δ) ‖ u_{0, m} ‖_{L^{2} (Y)}^{2} + ‖ \nabla u_{0, m} ‖_{L^{2} (Y)}^{2}] \\ + \frac{1}{C_{2}} e^{- σ t_{m}} [C_{1} ‖ u_{0, m} ‖_{L^{p} (Y)}^{p} + ‖ u_{0, m} ‖_{L^{2} (Y)}^{2}] \leq e^{- σ t_{m}} \hat{C} (‖ B ‖_{E (Y)}^{2} + ‖ B ‖_{E (Y)}^{p}), \end{matrix}$ $$\begin{align*} &\, e^{-\sigma t_m} \left[\|v_{0,m}\|_{L^2 (Y)}^2 + (\alpha + \delta^2 - \beta \delta)\|u_{0,m}\|_{L^2 (Y)}^2 + \|\nabla u_{0,m}\|_{L^2 (Y)}^2\right] \\ + \frac{1}{C_2} &\, e^{-\sigma t_m} \left[C_1\|u_{0,m}\|_{L^p (Y)}^p + \|u_{0,m}\|_{L^2 (Y)}^2\right] \leq e^{-\sigma t_m} \hat{C} \left(\|B\|_{E(Y)}^2 + \|B\|_{E(Y)}^p\right), \end{align*}$$

where ||B||_E(Y) = max_{g0∈B(θ_{−t_m}ω}) ||g₀ζ_Y||_E with ζ_Y being the characteristic function for the set Y . Clearly,

$\lim_{t \to \infty} e^{- σ t} ‖ B ‖_{E} = 0.$ $$\begin{array}{} \displaystyle \lim_{t \to \infty} e^{-\sigma t} \|B\|_E = 0. \end{array}$$

For the aforementioned arbitrary η > 0, there exists an integer m₀ = m₀(B, η) ≥ 1 such that

$e^{- σ t_{m}} \hat{C} (‖ B ‖_{E (Y)}^{2} + ‖ B ‖_{E (Y)}^{p}) \leq e^{- σ t_{m}} \hat{C} (‖ B ‖_{E}^{2} + ‖ B ‖_{E}^{p}) < \frac{η}{2}$ $$\begin{equation} e^{-\sigma t_m} \hat{C} \left(\|B\|_{E(Y)}^2 + \|B\|_{E(Y)}^p\right) \leq e^{-\sigma t_m} \hat{C} \left(\|B\|_E^2 + \|B\|_E^p\right) \lt \frac{\eta}{2} \end{equation}$$(101)

for all m > m₀. Then there esists μ₁ = μ₁(B, m₀, η) > 0 such that for any set Y with μ(Y ) < μ₁ one has

$e^{- σ t_{j}} \hat{C} (‖ B ‖_{E (Y)}^{2} + ‖ B ‖_{E (Y)}^{p}) < \frac{η}{2}, j = 1, \dots, m_{0} .$ $$\begin{array}{} \displaystyle {e^{ - \sigma {t_j}}}\hat C\left( {\|B\|_{E(Y)}^2 + \|B\|_{E(Y)}^p} \right) \lt \frac{\eta }{2},\quad j = 1, \cdots ,{m_0}. \end{array}$$(102)

Put together (100), (101) and (102) with (99). It shows that

$C_{3} \int_{Y} | u (t_{m}, u_{0, m}) |^{p} d x \leq \frac{η}{2} + \frac{η}{2} = η, for all m \geq 1,$ $$\begin{array}{} \displaystyle {C_3}{\int _Y}|u({t_m},{u_{0,m}}){|^p}dx \le \frac{\eta }{2} + \frac{\eta }{2} = \eta ,\,\,{\text{for all }}\,m \ge 1, \end{array}$$(103)

whenever a measurable set Y ⊂ ℝⁿ satisfies μ(Y) < min{μ₀, μ₁}. Therefore,

$\lim_{μ (Y) \to 0} \int_{Y} | ℙ_{u} Φ (t_{m_{k}}, g_{0, m}) (x) |^{p} d x = 0 uniformly for all k \geq 1,$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{\mu (Y) \to 0} {\int _Y}|{\mathbb{P}_u}\Phi ({t_{{m_k}}},{g_{0,m}})(x){|^p}dx = 0\,\,\,{\text{uniformly for all }}\,k \ge 1, \end{array}$$(104)

so that the condition (b) of Theorem 8 is also satisfied by the sequence of functions ${ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x)}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\mathbb{P}_u \Phi (t_{m_k}, \, g_{0,m_k})(x)\}_{k=1}^\infty \end{array}$ in L^p(ℝⁿ).

As checked by the above steps (2), (3) and (4) in this proof, all the conditions in Theorem 8 are satisfied by the sequence of functions ${ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) (x)}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\mathbb{P}_u \Phi (t_{m_k}, \, g_{0, m_k}) (x)\}_{k=1}^\infty \end{array}$ in L^p(ℝⁿ). Then we apply Theorem 8 to obtain

$\lim_{k \to \infty} ℙ_{u} Φ (t_{m_{k}}, g_{0, m_{k}}) = \tilde{u} strongly in L^{p} (ℝ^{n}) .$ $$\begin{array}{} \displaystyle \mathop {\lim }\limits_{k \to \infty } {\mathbb{P}_u}\Phi ({t_{{m_k}}},{g_{0,{m_k}}}) = \tilde u \,\,\,{\text{strongly in }}\,{L^p}({\mathbb{R}^n}). \end{array}$$(105)

Finally, combination of (91) and (105) shows that there exists a convergent subsequence ${Φ (t_{m_{k}}, g_{0, m_{k}})}_{k = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\Phi(t_{m_k}, g_{0, m_k})\}_{k=1}^\infty \end{array}$ of the sequence ${Φ (t_{m}, g_{0, m})}_{m = 1}^{\infty}$ $\begin{array}{} \displaystyle \{\Phi (t_m, g_{0,m})\}_{m=1}^\infty \end{array}$ in the space E = (H¹(ℝⁿ) ⋂ L^p(ℝⁿ)) × L²(ℝⁿ). Therefore, the semiflow Φ on the Banach space E is asymptotically compact.

According to Theorem 1, we conclude that there exists a global attractor 𝒜 in E for this semiflow Φ generated by the original nonlinear damped wave equation (1). The proof is completed.

eISSN:: 2444-8656
Language:: English

Publication timeframe:: Volume Open
Journal Subjects:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

Journal RSS Feed

Global Attractor for Nonlinear Wave Equations with Critical Exponent on Unbounded Domain

Published Online: Dec 02, 2016

Page range: 581 - 602

Received: Sep 17, 2016

Accepted: Nov 29, 2016

DOI: https://doi.org/10.21042/AMNS.2016.2.00045

KeywordsNonlinear wave equation, global attractor, asymptotic compactness, critical exponent, Vitali-type convergence criterion

© 2016 Yuncheng You, published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

Keywords
Nonlinear wave equation, global attractor, asymptotic compactness, critical exponent, Vitali-type convergence criterion