Consider the abstract nonlinear evolution equation defined on X, given by
$$ \left \{ \begin{array}{l} \frac{d^{2}u}{dt^{2}}+k\frac{du}{dt}=G(u), \quad k>0, \\ u(x,0)=\varphi(x), \\ u_{t}(x,0)=\psi(x), \end{array} \right . $$
(2.1)
where \(G:X_{2}\times\mathbf{R}^{+}\rightarrow{X_{1}}^{*}\) is a mapping, \(X_{2}\subset X_{1}\), \(X_{1}\), \(X_{2}\) are Banach spaces and \(X_{1}^{*}\) is the dual space of \(X_{1}\), \(\mathbf{R}^{+}=[0,\infty)\), \(u=u(x,t)\) is an unknown function.
First we introduce a sequence of function spaces:
$$ \left \{ \begin{array}{l} X\subset H_{2}\subset X_{2}\subset X_{1}\subset H, \\ X_{2}\subset H_{1}\subset H, \end{array} \right . $$
(2.2)
where H, \(H_{1}\), \(H_{2}\) are Hilbert spaces, X is a linear space, \(X_{1}\), \(X_{2}\) are Banach spaces and all inclusions are dense embeddings.
Suppose that
$$ \left \{ \begin{array}{l} L:X\rightarrow X_{1} \mbox{ is a one to one dense linear operator}, \\ \langle Lu,v\rangle_{H} =\langle u,v\rangle_{H_{1}},\quad \forall u,v\in X. \end{array} \right . $$
(2.3)
In addition, the operator L has an eigenvalue sequence
$$ Le_{k}=\lambda_{k} e_{k} \quad (k=1,2, \ldots) $$
(2.4)
such that \(\{e_{k}\}\subset X\) is the common orthogonal basis of H and \(H_{2}\).
Definition 2.1
[17]
Set \((\varphi,\psi)\in X_{2}\times H_{1}\), \(u \in W_{\mathrm{loc}}^{1,\infty }((0,\infty),H_{1})\cap L_{\mathrm{loc}}^{\infty}((0,\infty),X_{2})\) is called a globally weak solution of (2.1), if \(\forall v \in X_{1}\), we have
$$ \langle u_{t},v\rangle_{H}+k\langle u,v \rangle_{H}=\int_{0}^{t}\langle Gu,v \rangle \, dt+k\langle \varphi ,v\rangle_{H}+\langle \psi,v \rangle_{H}. $$
(2.5)
Definition 2.2
[17]
Let \(Y_{1}\), \(Y_{2}\) be Banach spaces, the solution \(u(t,\varphi,\psi)\) of (2.1) is called uniformly bounded in \(Y_{1}\times Y_{2}\), if for any bounded domain \(\Omega_{1}\times\Omega_{2}\subset Y_{1}\times Y_{2}\), there exists a constant C which only depends on the domain \(\Omega_{1}\times\Omega_{2}\), such that
$$\|u\|_{Y_{1}}+\|u_{t}\|_{Y_{2}}\leq C,\quad \forall( \varphi,\psi)\in\Omega _{1}\times\Omega_{2} \mbox{ and } t \geq0. $$
Suppose that \(G=A+B:X_{2}\times\mathbf{R}^{+}\rightarrow{X_{1}}^{*}\). Throughout this paper, we assume that:
-
(i)
There exists a functional \(F \in C^{1}:X_{2}\rightarrow\mathbf {R}^{1} \) such that
$$ \langle Au,Lv\rangle=\bigl\langle -DF(u),v\bigr\rangle ,\quad \forall u,v \in X. $$
(2.6)
-
(ii)
The functional F is coercive, i.e.
$$ F(u)\rightarrow\infty\quad \Leftrightarrow\quad {\|u\|}_{X_{2}} \rightarrow\infty . $$
(2.7)
-
(iii)
There exist constants \(C_{1}>0\) and \(C_{2}>0\) such that
$$ \bigl\vert \langle Bu,Lv\rangle\bigr\vert \leq C_{1}F(u)+C_{2}{ \|v\|}_{H_{1}}^{2},\quad \forall u,v \in X. $$
(2.8)
Lemma 2.1
[17]
Set
\(G:X_{2}\times\mathbf{R}^{+}\rightarrow{X_{1}}^{*}\)
to be weakly continuous, \((\varphi,\psi)\in X_{2}\times H_{1}\), then we obtain the following results:
-
(1)
If
\(G=A\)
satisfies the assumptions (i) and (ii), then there exists a globally weak solution of (2.1),
$$u\in W_{\mathrm{loc}}^{1,\infty}\bigl((0,\infty),H_{1}\bigr)\cap L_{\mathrm{loc}}^{\infty }\bigl((0,\infty),X_{2}\bigr), $$
and
u
is uniformly bounded in
\(X_{2}\times H_{1}\).
-
(2)
If
\(G=A+B\)
satisfies the assumptions (i), (ii) and (iii), then there exists a globally weak solution of (2.1),
$$u\in W_{\mathrm{loc}}^{1,\infty}\bigl((0,\infty),H_{1}\bigr)\cap L_{\mathrm{loc}}^{\infty }\bigl((0,\infty),X_{2}\bigr). $$
-
(3)
Furthermore, if
\(G=A+B\)
satisfies
$$ \bigl\vert \langle Gu,v\rangle\bigr\vert \leq\frac{1}{2}\|v \|_{H}^{2}+CF(u)+g(t) $$
(2.9)
for some
\(g\in L_{\mathrm{loc}}^{1}(0,\infty)\), then
\(u\in W_{\mathrm{loc}}^{2,2}((0,\infty),H)\).
A family of operators \(S(t): X\rightarrow X\) (\(t\geq0\)) is called a semigroup generated by (2.1) if it satisfies the following properties:
-
(1)
\(S(t): X\rightarrow X\) is a continuous map for any \(t\geq0\),
-
(2)
\(S(0)=\mathrm{id}: X\rightarrow X\) is the identity,
-
(3)
\(S(t+s)=S(t)\cdot S(s)\), \(\forall t, s\geq0\). Then the solution of (2.1) can be expressed as
$$u(t,u_{0})=S(t)u_{0}. $$
Introducing the expression of the abstract semilinear wave equation:
$$ \left \{ \begin{array}{l} \frac{d^{2}u}{dt^{2}}+2k\frac{du}{dt}=Lu+T(u), \quad k\geq0, \\ u(x,0)=\varphi(x), \\ u_{t}(x,0)=\psi(x), \end{array} \right . $$
(2.10)
where \(X_{1}\), X are Banach spaces, \(X_{1}\subset X\) is a dense inclusion, \(L:X_{1}\rightarrow X\) is a sectorial linear operator, and \(T:X_{1}\rightarrow X\) is a nonlinear bounded operator.
Lemma 2.2
[19]
Set
\(L:X_{1}\rightarrow X\), a sectorial linear operator and
\(T:X_{1}\rightarrow X\), a nonlinear bounded operator, \({\mathcal {L}}=L+k^{2}I\), then the solution of (2.9) can be expressed as follows:
$$\begin{aligned}& u = e^{-kt}\biggl[\cos t{(-\mathcal{L})}^{\frac{1}{2}}\varphi+k{(- \mathcal {L})}^{-\frac{1}{2}} \sin{(-\mathcal{L})}^{\frac{1}{2}}\varphi+{(- \mathcal{L})}^{-\frac {1}{2}}\sin t{(-\mathcal{L})}^{\frac{1}{2}}\psi \\& \hphantom{u =}{} + \int_{0}^{t}e^{-k(t-\tau)}{(- \mathcal{L})}^{-\frac{1}{2}}\sin(t-\tau ){(-\mathcal{L})}^{\frac{1}{2}}T(u)\, d \tau\biggr], \\& u_{t} = -ku+e^{-kt}\biggl[-{(-\mathcal{L})}^{\frac{1}{2}} \sin t{(-\mathcal {L})}^{\frac{1}{2}}\varphi+ k\cos t{(-\mathcal{L})}^{\frac{1}{2}} \varphi+\cos t{(-\mathcal {L})}^{\frac{1}{2}}\psi \\& \hphantom{u_{t} =}{} + \int_{0}^{t}e^{-k(t-\tau)} \cos(t-\tau){(-\mathcal{L})}^{\frac {1}{2}}T(u)\, d\tau\biggr]. \end{aligned}$$
Next, we introduce the concepts and definitions of invariant sets, global attractors, and ω-limit compactness sets for the semigroup \(S(t)\).
Definition 2.3
Let \(S(t)\) be a semigroup defined on X. A set \(\Sigma\subset X\) is called an invariant set of \(S(t)\) if \(S(t)\Sigma= \Sigma\), \(\forall t\geq 0\). An invariant set Σ is an attractor of \(S(t)\) if Σ is compact, and there exists a neighborhood \(U\subset X\) of Σ such that, for any \(u_{0}\in U\),
$$\inf_{v\in\Sigma}\bigl\Vert S(t)u_{0}-v\bigr\Vert _{X}\rightarrow0, \quad \mbox{as } t\rightarrow0. $$
In this case, we say that Σ attracts U. Especially, if Σ attracts any bounded set of X, Σ is called a global attractor of \(S(t)\) in X.
Definition 2.4
Let X be an infinite dimensional Banach space and A be a bounded subset of X. The measure of noncompactness \(\gamma(A)\) of A is defined by
$$\gamma(A)=\inf\{\delta>0\mid \mbox{for }A \mbox{ there exists a finite cover by sets whose diameter}\leq\delta\}. $$
Lemma 2.3
[11]
If
\({A_{n}}\subset X\)
is a sequence bounded and closed sets, \(A_{n}\neq \emptyset\), \(A_{n+1}\subset A_{n}\), and
\(\gamma(A_{n})\rightarrow0\) (\(n\rightarrow\infty\)), then the set
\(A=\bigcap_{n=1}^{\infty} A_{n}\)
is a nonempty compact set.
Definition 2.5
[16]
A semigroup \(S(t): X\rightarrow X\) (\(t\geq0\)) in X is called ω-limit compact, if for any bounded set \(B\subset X\) and \(\forall \varepsilon>0\), there exists \(t_{0}\) such that
$$\gamma\biggl(\bigcup_{t\geq t_{0}}S(t)B\biggr)\leq \varepsilon, $$
where γ is a noncompact measure in X.
For a set \(D\subset X\), we define the ω-limit set of D as follows:
$$\omega(D)=\bigcap_{s\geq0}\overline{\bigcup _{t\geq s}S(t)D}, $$
where the closure is taken in the X-norm.
Lemma 2.4
[19]
Let
\(S(t)\)
be a semigroup in
X, then
\(S(t)\)
has a global attractor
\(\mathcal{A}\)
in
X
if and only if
-
(1)
\(S(t)\)
has
ω-limit compactness, and
-
(2)
there is a bounded absorbing set
\(B\subset X\).
In addition, the
ω-limit set of
B
is the attractor
\(\mathcal {A}=\omega(B)\).
Remark 2.1
Although the lemma has been proved partly in [19], we still give a proof here. Our proof is different from that in [20] but is similar to that in [16]. We adopt and present the proof also because we will use the same method to obtain the existence of the global attractor.
Proof
Step 1. To prove the sufficiency of Lemma 2.4.
(a) \(S(t)\) has ω-limit compactness, i.e., for any bounded set \(B\subset X\) and \(\forall\varepsilon>0\), there exists a \(t_{0}\), such that
$$\gamma\biggl(\bigcup_{t\geq t_{0}}S(t)B\biggr)\leq \varepsilon. $$
So, we know that \(\omega(B)=\bigcap_{t_{0}= 0}^{\infty}\overline{\bigcup_{t\geq t_{0}}S(t)B}\) is a compact set from Lemma 2.3.
(b) \(\omega(B)\) is nonempty.
For \(B\neq\emptyset\), so \(\overline{\bigcup_{t\geq s}S(t)B}\neq \emptyset\), \(\forall s\geq0\), and
$$\overline{\bigcup_{t\geq s_{1}}S(t)B}\subset\overline{ \bigcup_{t\geq s_{2}}S(t)B},\quad \forall s_{1}\geq s_{2}, $$
we can obtain
$$\omega(B)=\bigcap_{s\geq0}^{\infty}\overline{ \bigcup_{t\geq s}S(t)B}\neq\emptyset. $$
(c) \(\omega(B)\) is invariant.
For \(x\in\omega(B)\) ⇔ there exist \(\{x_{n}\}\in B\) and \(t_{n}\rightarrow\infty\), such that \(S(t_{n})x_{n}\rightarrow x\).
If \(y\in S(t)\omega(B)\), then for some \(x\in\omega(B)\), \(y=S(t)x\).
Hence, there exist \(\{x_{n}\}\subset B\), \(t_{n}\rightarrow\infty\), such that
$$S(t)S(t_{n})x_{n}=S(t+t_{n})x_{n} \rightarrow S(t)x=y. $$
In conclusion, \(y\in\omega(B)\), \(S(t)\omega(B)\in\omega(B)\), \(\forall t\geq0\).
If \(x\in\omega(B)\), fix \(\{x_{n}\}\subset B\) and \({t_{n}}\), such that
$$S(t)x_{n}\rightarrow x, \quad \mbox{as } t_{n}\rightarrow \infty, n\rightarrow \infty. $$
\(S(t)\) is ω-limit compact, i.e., there exists a \(y\in H\), such that
$$S(t)\bigcap_{t_{n}\geq0}\overline{\bigcup _{t\geq t_{n}}S(t_{n})x_{n}}\rightarrow y, \quad n \rightarrow\infty. $$
Therefore \(y\in\omega(B)\).
For
$$\bigcap_{t_{n}\geq0}\overline{\bigcup _{t\geq t_{n}}S(t_{n})x_{n}} =\bigcap _{t_{n}\geq0}\overline{\bigcup_{t\geq t_{n}}S(t)S(t_{n}-t)x_{n}} \rightarrow\bigcap_{t_{n}\geq0}\overline {\bigcup _{t\geq t_{n}}S(t)y} $$
and
$$S(t_{n})x_{n}\rightarrow x\in\omega(B), $$
which implies that
$$S(t)y\rightarrow x, \qquad \omega(B)\subset S(t)\omega(B). $$
In conclusion, combining (a)-(c) and condition (2), Step 1 has been proved.
Step 2. To prove the necessity of Lemma 2.4.
If \({\mathcal{A}}\) is a global attractor, then the ε-neighborhood \(U_{\varepsilon}({\mathcal{A}})\subset X\) is an absorbing set. So we need only to prove \(S(t)\) has ω-limit compactness.
Since \(U_{\varepsilon}({\mathcal{A}})\) is an absorbing set, for any bounded set \(B\subset X\) and \(\varepsilon> 0\), there exists a time \(t_{\varepsilon}(B)> 0\) such that
$$\bigcup_{t\geq t_{\varepsilon}(B)}S(t)B\subset U_{\frac{\varepsilon }{4}}( \mathcal{A})=\biggl\{ x\in X\Bigm| \operatorname{dist}(x,\mathcal{A})< \frac{\varepsilon }{4}\biggr\} . $$
On the other hand, \(\mathcal{A}\) is a compact set, and there exist finite elements \(x_{1}, x_{2}, \ldots, x_{n} \in X\) such that
$$\mathcal{A}\subset\bigcup_{k=1}^{n}U \biggl(x_{k}, \frac{\varepsilon}{4}\biggr). $$
Then
$$U_{\frac{\varepsilon}{2}}(\mathcal{A})\subset\bigcup_{k=1}^{n}U \biggl(x_{k}, \frac{\varepsilon}{2}\biggr), $$
which implies that
$$\gamma\biggl(\bigcup_{t\geq t_{\varepsilon}(B)}S(t)B\biggr)\leq\gamma \bigl(U_{\frac {\varepsilon}{4}}(\mathcal{A})\bigr)\leq\varepsilon. $$
Hence, Lemma 2.4 has been proved. □