Exponential Attractors for the Sup-Cubic Wave Equation with Nonlocal Damping

Abstract

We study the long-time dynamics of a wave equation with nonlocal weak damping, nonlocal weak anti-damping and sup-cubic nonlinearity. Based on the Strichartz estimates in a bounded domain, we obtain the global well-posedness of the Shatah–Struwe solutions. To overcome the difficulties brought by the nonlinear weak damping term, we present a new-type Gronwall’s lemma to obtain the dissipative for the Shatah–Struwe solutions semigroup of this equation. Finally, we establish the existence of a time-dependent exponential attractor with the help of a more general criteria constructed by the quasi-stable technique.

Appendix

1.1 Proof of Lemma 2.5

Suppose $\tau<s<t<T$, let $\theta _{m}$ be a piecewise continuous linear function and satisfying

$$\begin{aligned} \theta _{m}(r)= {\left\{ \begin{array}{ll} 1,\quad r\in (s+\frac{2}{m},t-\frac{2}{m});\\ 0,\quad r\in [t-\frac{1}{m},T]\text { or }r\in [\tau ,s+\frac{1}{m}]. \end{array}\right. } \end{aligned}$$

Let $\rho _{n}\in \mathscr {D}(\mathbb {R})$ be a regularizing sequence of even functions,

$$\begin{aligned} \rho _{n}(r)=\rho _{n}(-r),\quad \int \limits _{-\infty }^{+\infty }\rho _{n}(r)dr=1,\quad \text {Supp}\rho _{n}\subset \left[ -\frac{1}{n},\frac{1}{n}\right] . \end{aligned}$$

Let

$$\begin{aligned} v=\left( (\theta _{m}u^{\prime })*\rho _{n}*\rho _{n}\right) \theta _{m}. \end{aligned}$$

Obviously, we have

$$\begin{aligned} v=\left( (\theta _{m}u)^{\prime }*\rho _{n}*\rho _{n}\right) \theta _{m} -\left( (\theta _{m}^{\prime }u)*\rho _{n}*\rho _{n}\right) \theta _{m}. \end{aligned}$$

Taking the $L^{2}$-inner product between (2.7)$^{1}$ and v, we have

$$\begin{aligned} P_{n,m}+Q_{n,m}=\int \limits _{\tau }^{T}\left\langle f,v\right\rangle dr, \end{aligned}$$

(5.1)

where

$$\begin{aligned} P_{n,m}&=\int \limits _{\tau }^{T}\theta _{m}\left\langle u^{\prime \prime },(\theta _{m}u^{\prime })*\rho _{n}*\rho _{n}\right\rangle dr,\\ Q_{n,m}&=\int \limits _{\tau }^{T}\theta _{m}\left\langle \nabla u,\nabla (\theta _{m}u^{\prime })\right\rangle *\rho _{n}*\rho _{n})dr. \end{aligned}$$

Firstly, we have

$$\begin{aligned} P_{n,m}&=\int \limits _{\tau }^{T}\left\langle (\theta _{m}u^{\prime })^{\prime }*\rho _{n},(\theta _{m}u^{\prime })*\rho _{n}\right\rangle dr-\int \limits _{\tau }^{T}\left\langle (\theta _{m}^{\prime }u^{\prime })*\rho _{n},(\theta _{m}u^{\prime })*\rho _{n}\right\rangle dr \nonumber \\&=-\int \limits _{\tau }^{T}\left\langle (\theta _{m}^{\prime }u^{\prime })*\rho _{n},(\theta _{m}u^{\prime })*\rho _{n}\right\rangle dr \nonumber \\&\rightarrow - \int \limits _{\tau }^{T}\theta _{m}\theta _{m}^{\prime }\Vert u^{\prime }(r)\Vert ^{2}dr, \quad \text {as }n\rightarrow \infty . \end{aligned}$$

(5.2)

Secondly, we can get that

$$\begin{aligned} Q_{n,m}&=\int \limits _{\tau }^{T}\left\langle \left\langle (\theta _{m}u)*\rho _{n},(\theta _{m}u)^{\prime }*\rho _{n}\right\rangle \right\rangle dr-\int \limits _{\tau }^{T}\left\langle \left\langle (\theta _{m}u)*\rho _{n},\theta _{m}^{\prime }u*\rho _{n}\right\rangle \right\rangle dr\nonumber \\&=-\int \limits _{\tau }^{T} \left\langle \left\langle (\theta _{m}u)*\rho _{n},(\theta _{m}^{\prime }u)*\rho _{n}\right\rangle \right\rangle dr\nonumber \\&\rightarrow -\int \limits _{\tau }^{T}\theta _{m}\theta _{m}^{\prime }\Vert \nabla u(r)\Vert ^{2}dr\quad \text {as }n\rightarrow \infty , \end{aligned}$$

(5.3)

here $\left\langle \left\langle \cdot ,\cdot \right\rangle \right\rangle$ stands for the $\mathcal {H}^{1}$-inner product. Combining now (5.1)-(5.3) and letting $m\rightarrow \infty$, we find

$$\begin{aligned} -\int \limits _{\tau }^{T}\theta _{m}\theta _{m}^{\prime }\Vert u^{\prime }(r)\Vert ^{2}dr- \int \limits _{\tau }^{T}\theta _{m}\theta _{m}^{\prime }\Vert \nabla u(r)\Vert ^{2}dr= \int \limits _{\tau }^{T}\left\langle f,\theta _{m}^{2}u^{\prime }(r)\right\rangle dr. \end{aligned}$$

(5.4)

On the other hand, if $k\in L^{1}(\tau ,T)$, then Lebesgue dominated Theorem implies that

$$\begin{aligned} -\int \limits _{\tau }^{T}\theta _{m}\theta _{m}^{\prime }k(r)dr&=\,m\int \limits _{t-\frac{2}{m}}^{t-\frac{1}{m}}(-mr+mt-1)k(r)dr-m\int \limits _{s+\frac{1}{m}}^{s+\frac{2}{m}}(mr-ms-1)k(r)dr \nonumber \\&\rightarrow \frac{1}{2}k(t)-\frac{1}{2}k(s)\quad \text {as }m\rightarrow \infty . \end{aligned}$$

(5.5)

Letting $m\rightarrow \infty$ in (5.4) and applying (5.5), we deduce

$$\begin{aligned} \frac{1}{2}\Vert \xi _{u}(t)\Vert _{\mathscr {E}}^{2}= \frac{1}{2}\Vert \xi _{u}(s)\Vert _{\mathscr {E}}^{2}+\int \limits _{s}^{t}\left\langle f,\partial _{t}u\right\rangle dr\quad \text {for a.e. }s,t\in [\tau ,T]. \end{aligned}$$

(5.6)

Recalling $(u,\partial _{t}u)\in L^{\infty }(\tau ,T;\mathscr {E})$ and satisfying (2.7), we discover

$$\begin{aligned} {\left\{ \begin{array}{ll} u(s)\rightharpoonup z_{1}\quad \text {weakly in }\mathcal {H},\\ \partial _{t}u(s)\rightharpoonup z_{2}\quad \text {weakly in }\mathcal {H}^{0}, \end{array}\right. } \text {as }s\rightarrow \tau . \end{aligned}$$

Since $(u(s),\partial _{t}u(s))$ is continuous in $\mathcal {H}^{0}\times \mathcal {H}^{-1}$, then we can get $z_{1}=u_{\tau }^{0}$, $z_{2}=u_{\tau }^{1}$. Then we can get (2.8) by letting $s\rightarrow \tau$ in (5.6) and applying $\Vert \cdot \Vert _{\mathscr {E}}$ is weak lower semi-continuous.

1.2 Proof of New-Type Gronwall’s Inequality

Proof

Using condition (3.4), we can choose $\varepsilon$ so small such that

$$\begin{aligned} \Psi (u(\tau ))<\left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }} -\frac{3Q}{e^{\alpha (\varepsilon )}-1}. \end{aligned}$$

(5.7)

We assert now that (5.7) implies

$$\begin{aligned} \Psi (u(t))<\left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }},\quad \forall t\ge \tau . \end{aligned}$$

(5.8)

If (5.8) is not true, let $t^{*}=\inf \{t\ge \tau ,\Psi (u(t)) \ge \left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }}\}$, obviously we have $\Psi (u(t^{*}))=[\frac{\kappa }{\beta (\varepsilon )}]^{\frac{1}{\gamma }}$ and

$$\begin{aligned} \Psi (u(t))\le \left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }}. \end{aligned}$$

(5.9)

Combining (3.3) and (5.9), we have

$$\begin{aligned} \Psi (u(t^{*}))&\le e^{-\alpha (\varepsilon )(t^{*}-\tau )}\Psi (u(\tau )) +\int \limits _{\tau }^{t^{*}}e^{-\alpha (\varepsilon )(t^{*}-r)}|q(r)|dr \\&\le e^{-\alpha (\varepsilon )(t^{*}-\tau )}\Psi (u(\tau )) +\frac{3Q}{e^{\alpha (\varepsilon )}-1} \\&<\left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }}. \end{aligned}$$

So we have a contradiction and thereby (5.8) can exist.

Now applying (3.3) and (5.8), we can obtain

$$\begin{aligned} \Psi (u(t)) \le e^{-\alpha (\varepsilon )(t-\tau )}\Psi (u(\tau )) +\frac{3Q}{e^{\alpha (\varepsilon )}-1}. \end{aligned}$$

(5.10)

Using Taylor’s formula, we can easily find

$$\begin{aligned} \left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }} -\frac{3Q}{e^{\alpha (\varepsilon )}-1} \ge \left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }} -\frac{3Q}{\alpha (\varepsilon )}. \end{aligned}$$

(5.11)

Choosing $\frac{1}{\alpha (\varepsilon )}=\frac{2(\lambda +1)}{\kappa }R^{\gamma }$ and applying (3.4) we can get

$$\begin{aligned} \left[ \frac{\kappa }{\beta (\varepsilon )}\right] ^{\frac{1}{\gamma }} -\frac{3Q}{\alpha (\varepsilon )}\ge R \end{aligned}$$

for any R large enough. Then applying (5.10), we observe

$$\begin{aligned} \Psi (u(t)) \le N(1+R^{\gamma }),\quad \forall t-\tau \ge T'(R), \end{aligned}$$

(5.12)

where $N:=N_{Q,\lambda ,\kappa }$ may depend on Q, $\lambda$ and $\kappa$. We can assume without loss of generality that $N>1$ and $R>1$. It follows from (5.12) and $0<\gamma <1$ that there exists $T=T(R)$ such that

$$\begin{aligned} \Psi (u(t)) \le (2N)^{\frac{1}{1-\gamma }}+1,\quad \forall t-\tau \ge T\text { and }\Psi (u(\tau ))\le R. \end{aligned}$$

(5.13)

Let $R_{0}=(2N)^{\frac{1}{1-\gamma }}+1$ and thus the lemma is proved. $\square$