Micro-inertia effects on existence of attractors for Form II Mindlin’s strain gradient viscoelastic plate

Abstract

In this paper, we study the long-time dynamics of solutions to strain gradient viscoelastic plates equations taking into account micro-inertia effects and subjected to three different types of external nonlinear terms. First we derive briefly the equations of strain gradient viscoelastic plate corresponding to anti-plane shear deformations which is assumed to be consistent with the Mindlin Form II. Based on semigroup theory, we prove the existence and uniqueness of global solution. Then, we show that the existence of finite dimensional global attractors depends on the value of the micro-inertia parameter (whether or not zero) and on the assumptions on the external non-linearities. Sufficient conditions on existence of exponential and global minimal attractors can be deduced. Finally, we show the upper-semicontinuity of global attractors with respect to the micro-inertia parameter.

Appendix

In this appendix we show in the case $i=1$ when $h=0$, two interesting properties to $C^2$-vector field. Then we provide two examples of vector fields illustrating both properties. To this end, we first quote a lemma due to Zhou and Fu [37].

Lemma 6.1

Assume that $G(z_1,\ldots ,z_n)$ is a k-times continuously differentiable function with respect to variables $z_1,\ldots ,z_n$ and $z_j\in L^\infty ([0,T]; H^k(\Omega ))$ $(j=1,\ldots ,n)$. Then

$$\begin{aligned} \Vert \frac{\partial ^k}{\partial x^k}G(z_1(\cdot ,t),\ldots ,z_n(\cdot ,t))\Vert ^2\le C({\overline{M}},k,n)\sum _{j=1}^{n}\Vert z_j(t)\Vert ^2_{H^k} \end{aligned}$$

(6.1)

where ${\overline{M}}=\max _{1\le j\le n}\max _{(x,t)\in \Omega \times [0,T]} |z_j(x,t)|,$ $C({\overline{M}},k,n)$ is a positive constant depending only on ${\overline{M}},\ k$ and n.

Now we state our main result of this section.

Lemma 6.2

Let $F\ :\ {\mathbb {R}}^2 \rightarrow {\mathbb {R}}^2$ be a $C^2$-vector field given by $F = (F_1, F_2)$. Then, there exist two constants $K_i = K_i({\overline{M}},k,n)> 0,\ j = 1, \ldots ,n,\ i=1,2$ ($n=1,2$) such that

$$\begin{aligned} \Vert \text {div}( F (\nabla \ u))\Vert \le K_1 \Vert \Delta u\Vert , \end{aligned}$$

(6.2)

and

$$\begin{aligned} \Vert \text {div}( F (\nabla \ u_1))-\text {div}( F (\nabla \ u_2))| \le K_2 \Vert \Delta u_1- \Delta u_2\Vert , \end{aligned}$$

(6.3)

where ${\overline{M}}=\max _{1\le j\le 2}\max _{(x,t)\in \Omega \times [0,T]} |\frac{\partial u_j}{\partial x_j}(x,t)|.$

Proof

Since $n=1,2$, for $u \in L^\infty (0, T; V^2)$ we have

$$\begin{aligned} \text {div}\ F(\nabla u) =\sum _{j=1}^{2}\frac{\partial }{\partial x_j}F_j \left( \frac{\partial u_j}{\partial x_j}\right) . \end{aligned}$$

For simplicity, the same constant $C_{{\overline{M}}}> 0$ will be used to denote several different constants depending on ${{\overline{M}}}$ in the next estimates. From (6.1), by taking $z=\nabla u$, one can conclude that

$$\begin{aligned} \Vert \text {div}\ F(\nabla u)\Vert ^2\le C_{{\overline{M}}}\Vert \nabla u\Vert ^2_{H^1}\le C_{{\overline{M}}}\Vert \Delta u\Vert ^2 \end{aligned}$$

from which (6.2) follows.

Since $F\in C^2({\mathbb {R}}^2)$, by applying the mean value inequality we obtain

$$\begin{aligned} |\text {div}\ F(\nabla u^1)-\text {div}\ F(\nabla u^2)|\le \Big |\Big (\text {div}\ F\left( (1-\chi )\nabla u^1+\chi \nabla u^2\right) \Big )'\Big ||\nabla u^1-\nabla u^2|_{H^1}. \end{aligned}$$

(6.4)

We infer from (6.1) that

$$\begin{aligned} \Big \Vert \Big (\text {div}\ F\left( (1-\chi )\nabla u^1+\chi \nabla u^2\right) \Big )'\Big \Vert ^2\le C_{{\overline{M}}}\Vert (1-\chi )\nabla u^1+\chi \nabla u^2)\Vert ^2_{H^2}. \end{aligned}$$

Since $|\nabla u^1|,\ |\nabla u^1|\le {\overline{M}}$, we get

$$\begin{aligned} \Big \Vert \Big (\text {div}\ F\left( (1-\chi )\nabla u^1+\chi \nabla u^2\right) \Big )'\Big \Vert \le C_{{\overline{M}}}. \end{aligned}$$

Consequently, (6.4) becomes

$$\begin{aligned} \Vert \text {div}\ F(\nabla u^1)-\text {div}\ F(\nabla u^2)\Vert \le C_{{\overline{M}}}|\Delta u^1-\Delta u^2| \end{aligned}$$

which gives (6.3) immediately. $\square$

Remark 6.1

One can prove the above lemma from (2.29) (see [24, Lemma 4.1] for more details).

In the following we give two examples of vector fields (inspired from [19, 24]) satisfying (2.29), (2.31), (6.2) and (6.3).

• 1. Let us consider $F(z) =\lambda z,\ \lambda \in (0,1]$. It follows that condition (2.29) holds true for any $k_j,\ p_j\ge 1$, $j=1,2$ (see [19, 24]). Moreover, we note that $F = \lambda \nabla \wp$ is a conservative vector field, such that $\wp (z)=\frac{\lambda }{2}|z|^2$ and then condition (2.31) is readily verified for any $\eta _3,\ \beta _3 \ge 0$ [24]. Now, we show that (6.2) and (6.3) are satisfied. Since the rotational of a gradient is zero, the vector field $F(\nabla u) =\lambda \nabla u$ generates the following operator $\text {div}\ F(\nabla u) = \lambda \Delta u$, which satisfies (6.2). Also we have

  $$\begin{aligned} \Vert \text {div}(F(\nabla u^1))-\text {div}(F(\nabla u^2))\Vert =\lambda \left\| \Delta u^1-\Delta u^2\right\| \le \left\| \Delta u^1-\Delta u^2\right\| \end{aligned}$$

  which satisfies (6.3).

• 2. We give another example of vector field satisfying (2.29), (2.31), (6.2) and (6.3). Jorge Silva et al. [24] presented an example (Example 4.14) of vector field

  $$\begin{aligned} F(z)=\frac{z}{|z|^2+1},\quad \forall z\in {\mathbb {R}}^n\end{aligned}$$

  (6.5)

  that satisfies (2.29). Moreover, they showed that $F=\nabla f$ where the potential function given by $f =(z)=\ln (|z|^2+1)$ satisfies (2.31) for any $z \in {\mathbb {R}}^n$ as well. First, we will show that the vector field given by (6.5) satisfies (6.2). It is easy to check that $F\in C^2({\mathbb {R}}^2)$. Taking account of $u \in L^\infty (0, T; V^2)$ and $z=\nabla u$, we have

  $$\begin{aligned} \Vert \text {div}(F(\nabla u))\Vert= & {} \left\| \text {div}(\frac{\nabla u}{|\nabla u|^2+1})\right\| \\= & {} \left\| \frac{\Delta u}{|\nabla u|^2+1}-2\frac{\nabla (|\nabla u|^2)\cdot \nabla u}{( |\nabla u|^2+1)^2}\right\| \\\le & {} \Vert {\Delta u}\Vert +2\left\| \frac{\nabla (|\nabla u|^2)\cdot \nabla u}{( |\nabla u|^2+1)^2}\right\| \\= & {} \Vert \Delta u\Vert +2\left\| \frac{u_x^2u_{xx}+2u_xu_yu_{xy}+u_y^2u_{yy}}{( u_x^2+u_y^2+1)^2}\right\| \\\le & {} \Vert \Delta u\Vert +C_1\Vert \Delta u\Vert \le C\Vert \Delta u\Vert , \end{aligned}$$

  where $\nabla (|\nabla u|^2)=2(u_xu_{xx}+u_yu_{yx},u_xu_{xy}+u_yu_{yy})$. Now, we will show that (6.5) satisfies (6.3). We have

  $$\begin{aligned}&\Vert \text {div}(F(\nabla u^1))-\text {div}(F(\nabla u^2))\Vert \\&\quad =\left\| \text {div}(\frac{\nabla u^1}{|\nabla u^1|^2+1})-\text {div}(\frac{\nabla u^2}{|\nabla u^2|^2+1})\right\| \\&\quad \le \left\| \frac{\Delta u^1}{|\nabla u^1|^2+1}-2\frac{\nabla (|\nabla u^1|^2)\cdot \nabla u^1}{(|\nabla u^2|^2+1)^2}-\frac{\Delta u^2}{|\nabla u^2|^2+1}+2\frac{\nabla (|\nabla u^2|^2)\cdot \nabla u^2}{( |\nabla u^2|^2+1)^2}\right\| \\&\quad \le \left\| \frac{\Delta u^1-\Delta u^2}{|\nabla u^1|^2+1}-\frac{\Delta u^2(|\nabla u^2|^2-|\nabla u^1|^2)}{(|\nabla u^1|^2+1)(|\nabla u^2|^2+1)}\right\| \\&\qquad +\, 2\left\| \frac{\nabla (|\nabla u^2|^2)\cdot \nabla u^2}{( |\nabla u^2|^2+1)^2}-\frac{\nabla (|\nabla u^1|^2)\cdot \nabla u^1}{( |\nabla u^2|^2+1)^2}\right\| \end{aligned}$$

  Since $\Vert \nabla u^1\Vert$, $\Vert \nabla u^2\Vert \le {\overline{M}}$, we obtain

  $$\begin{aligned}&\left\| \frac{\Delta u^1-\Delta u^2}{|\nabla u^1|^2+1}-\frac{\Delta u^2(|\nabla u^2|^2-|\nabla u^1|^2)}{(|\nabla u^1|^2+1)(|\nabla u^2|^2+1)}\right\| \\&\quad \le \Vert \Delta u^1-\Delta u^2\Vert +\Vert \Delta u^2(|\nabla u^2|^2-|\nabla u^1|^2)\Vert \\&\quad \le \Vert \Delta u^1-\Delta u^2\Vert +\Vert \Delta u^2\Vert \Vert \nabla u^2+\nabla u^1\Vert \Vert \nabla u^2-\nabla u^1\Vert \\&\quad \le \Vert \Delta u^1-\Delta u^2\Vert +C_{{\overline{M}}}\Vert \Delta u^1-\Delta u^2\Vert \\&\quad \le C_{{\overline{M}}}\Vert \Delta u^1-\Delta u^2\Vert . \end{aligned}$$

  By applying the above procedure, one can obtain

  $$\begin{aligned} 2\left\| \frac{\nabla (|\nabla u^2|^2)\cdot \nabla u^2}{( |\nabla u^2|^2+1)^2}-\frac{\nabla (|\nabla u^1|^2)\cdot \nabla u^1}{( |\nabla u^2|^2+1)^2}\right\| \le C_{{\overline{M}}}\Vert \Delta u^1-\Delta u^2\Vert . \end{aligned}$$

  By collecting the above estimates, we get

  $$\begin{aligned} \Vert \text {div}( F (\nabla u^1))-\text {div}( F (\nabla u^2))\Vert \le C_{{\overline{M}}}\Vert \Delta u^1-\Delta u^2\Vert \end{aligned}$$

  from which we conclude that (6.3) is verified.