Well-Posedness of a Fourth Order Parabolic Equation Modeling MEMS

Lai Baishun,, Luo Qing,

School of Mathematics and Statistics, Hunan Normal University, Changsha 410081

 基金资助: 国家自然科学基金.  11971148

 Fund supported: the NSFC.  11971148

Abstract

In this paper, we consider a fourth order evolution equation involving a singular nonlinear term $\frac{\lambda}{(1-u)^{2}}$ in a bounded domain $\Omega \subset \mathbb{R}^{n}$. This equation arises in the modeling of microelectromechanical systems. We first investigate the well-posedness of a fourth order parabolic equation which has been studied in [1], where the authors, by the semigroup argument, obtained the well-posedness of this equation for $n\leq2$. Instead of semigroup method, we use the Faedo-Galerkin technique to construct a unique solution of the fourth order parabolic equation for $n\leq7$, which completes the result of [1].

Keywords： Electrostatic MEMS ; Fourth order evolution equation ; Well-posedness

Lai Baishun, Luo Qing. Well-Posedness of a Fourth Order Parabolic Equation Modeling MEMS. Acta Mathematica Scientia[J], 2021, 41(6): 1718-1733 doi:

1 研究背景与主要结果

$$$\left \{ \begin{array}{ll} { } \gamma u_{tt}+a u_{t}+\beta \Delta^{2}u-\tau \Delta u = \frac{\lambda}{(1-u)^2}, \\ u(x, 0) = u^{0}(x), \ \ u_{t}(x, 0) = u^{1}(x), \\ \mbox{适当边界条件}, \\ \end{array} \right.$$$

Dirichlet边界条件

Navier边界条件

Dirichlet边界条件在物理上又称为固支(clamped) 边界条件, 它表示器件沿着边界固定, 边界部分不产生任何垂直方向的位移以及不产生任何倾斜. Navier边界条件又称为pinned边界条件, 在物理上它表示器件沿着边界在垂直方向不产生位移, 但是器件沿着边界可以自由旋转.

$\gamma\ll 1$, 即粘性力占优的情形, 那么方程(1.1) 可以简化为下列四节抛物型初边值问题(为简单起见我们假设$a = 1$):

$$$\left \{ \begin{array}{ll} { } u_{t}+\beta \Delta^{2}u-\tau \Delta u = \frac{\lambda}{(1-u)^{2}}, & x\in\Omega\subset{{\Bbb R}}^{n}, t>0, \\ u(x, 0) = u^{0}(x), & x\in\Omega\subset{{\Bbb R}}^{n}, \\ \mbox{适当边界条件}, & x\in \partial\Omega, \, t>0. \end{array} \right.$$$

$\rm ($1$\rm ) $$\lambda \in {\mathbb{R}^ + }$$ T > 0$足够小;

$\rm ($2$\rm ) $$T = \infty$$ \lambda \in {\mathbb{R}^ + }$足够小.

(i) 方程(1.2) 在Dirichlet边界条件下存在唯一弱解$u\in\chi_T$. 对于Navier边界条件, 也有类似的结论, 但此时我们是在以下框架下

(ii) 如果$\lambda$充分大, 对于Dirichlet边界条件且区域为球形区域(或者是Navier边界条件下任意一般光滑区域) 那么我们构造的解一定在有限时间爆破.

2 预备知识

$$$(u, v)\mapsto (u, v)_{W_{0}^{2, 2}}\triangleq\beta(\Delta u, \Delta v)_{2}+\tau(\nabla u, \nabla v)_{2}, \ \ \ \mbox{对所有的}\ u, v\in W_{0}^{2, 2}(\Omega).$$$

(2) 所有特征值构成一个可数集${\lambda _k}$, 其中

$k \to \infty $${\lambda _k} \to \infty , 即特征值依次递增趋于 \infty . (3) 存在一组函数列 \{w_k\} _{k = 1}^\infty , 它是 L^2 函数空间的的一组正交基地, 且 w_k\in W_0^{2, 2}(\Omega ) (或 W^{2, 2}(\Omega ) \cap W_0^{1, 2}(\Omega )) 并满足 由Lax-Milgram定理, 椭圆算子的 {L^2} 理论, 以及紧嵌入定理, 我们知道 S\buildrel \Delta \over = {L^{ - 1}}$$ L^2 \left( \Omega \right)$到自身的一个有界线性紧算子, 设$Sf = u, Sg = v,$通过分部积分我们有

3 方程(1.2) 解的存在唯一性

$$$\left \{ \begin{array}{ll} u_{t}+\beta \Delta^{2}u-\tau \Delta u = f(x, t), \ \ &(x, t)\in \Omega\times (0, T), \\ u(x, 0) = u^{0}(x), \ \ &x\in \Omega, \\ { } u = \frac{\partial u}{\partial n} = 0\ (\mbox{or}\ u = \Delta u = 0), \ \ &(x, t)\in \Omega\times (0, T). \end{array}\right.$$$

$$$\left \{ \begin{array}{ll} (u_{k}'(t), \omega_{j})_{2}+\beta (\Delta u_{k}, \Delta\omega_{j} )_{2}+\tau (\nabla u_{k}, \nabla \omega_{j})_{2} = (f(t), \omega_{j})_{2}, \\ j = 1, \cdots, k, \ \mbox{ a.e.}\ t\in (0, T), \\ u_{k}(x, 0) = u^{0k}(x), x\in\Omega. \end{array} \right.$$$

$$$\left \{ \begin{array}{ll} { } (g_{i}^{k}(t))'+\sum\limits_{i = 1}^{k}\lambda_{i}g_{i}^{k}(t) = (f(t), \omega_{i})_{2}, \\ [3mm] { } g_{i}^{k}(0) = (u^{0k}, \omega_{i})_{2}. \end{array} \right.$$$

$$$\frac{1}{2}\frac{\rm{d}}{\rm{d}s}\|u_{k}(s)\|_{2}^{2}+\beta \|\Delta u_{k}\|_{2}^{2}+\tau \|\nabla u_{k}\|_{2}^{2} = (f(s), u_{k})_{2}.$$$

$\begin{eqnarray} \|u_{k}\|_{L^{\infty}(0, T;L^{2}(\Omega))}^{2}+\|u_{k}\|_{L^{2}(0, T;W^{2, 2}(\Omega))}^{2}&\leq& \|u_{0}^{k}\|_{2}^{2} +4\|f\|^{2}_{L^{2}(0, T;L^{2}(\Omega))}{}\\& \leq &4\left(\|u^{0}\|_{2}^{2}+\|f\|^{2}_{L^{2}(0, T;L^{2}(\Omega))}\right). \end{eqnarray}$

$\begin{eqnarray} &&\int_{0}^{T}\int_{\Omega}|u_{k}'(t)|^{2}\, {\rm d}x{\rm d}t+\frac{\beta}{2}\|\Delta u_{k}(\cdot, T)\|_{L^{2}(\Omega)}^{2}+\frac{\tau}{2}\|\nabla u_{k}(\cdot, T)\|_{L^{2}(\Omega)}^{2}{}\\ &\leq &\frac{\beta}{2}\|\Delta u_{k}(\cdot, 0)\|_{L^{2}(\Omega)}^{2}+\frac{\tau}{2}\|\nabla u_{k}(\cdot, 0)\|_{L^{2}(\Omega)}^{2}+\int_{0}^{T}\int_{\Omega}|f|^{2}\, {\rm d}x{\rm d}t{}\\ &\leq &C(\beta, \tau) \|u^{0}\|_{W^{2, 2}(\Omega)}^{2}+\int_{0}^{T}\int_{\Omega}|f|^{2}\, {\rm d}x{\rm d}t. \end{eqnarray}$

$\begin{eqnarray} &&{\rm ess}\sup\limits_{0\leq t\leq T}(\|u'(t)\|_{2}^{2}+\|u\|_{W^{4, 2}(\Omega)}^{2})+\int_{0}^{T}\|u'(t)\|_{W^{2, 2}(\Omega)}^{2}\, {\rm d}t\\ &\leq &C(\|f\|_{W^{1, 2}(0, T;L^{2}(\Omega))}^{2}+\|u^{0}\|_{W^{4, 2}(\Omega)}), \end{eqnarray}$

固定$k\geq 1$, 对方程(3.4) 关于时间$t$微分, 可得

$$$(\tilde{u}_{k}', \omega_{j})_{2}+\beta(\Delta\tilde{u}_{k}, \Delta \omega_{j})_{2}+ \tau(\nabla\tilde{u}_{k}, \nabla\omega_{j})_{2} = (f', \omega_{j})_{2}, \ \ j = 1, \cdots, k,$$$

$\begin{eqnarray} &&\sup\limits_{0\leq t\leq T}\|u_{k}'(t)\|_{2}^{2}+2\beta\int_{0}^{T}\|\Delta u_{k}'(t)\|_{2}^{2}\, {\rm d}t +2\tau\int_{0}^{T}\|\nabla u_{k}'(t)\|_{2}^{2}\, {\rm d}t{}\\ &\leq & 4(\|u_{k}'(0)\|_{2}^{2}+\|u_{k}'\|^{2}_{L^{2}(0, T;L^{2}(\Omega))}+\|f'\|^{2}_{L^{2}(0, T;L^{2}(\Omega))}){}\\ &\leq &C(\tau, \beta, \Omega)(\|f\|^{2}_{W^{1, 2}(0, T;L^{2}(\Omega))}+\|u_{k}(0)\|_{W^{4, 2}(\Omega)}^{2}). \end{eqnarray}$

$\begin{eqnarray} \|v_{i}-w\|_{{\cal X}_{T}}\leq C\lambda T^{\frac{1}{2}}(k(r)+r)\|u_{i}\|_{{\cal X}_{T}} \leq \lambda T^{\frac{1}{2}} C(k(r)+r)r, \end{eqnarray}$

$u(x, t)$是方程(1.2) 在区间$[0, T_{m})$的解, 对任意$t\in [0, T_{m})$, 我们定义

$\begin{eqnarray} \frac{\rm{d}M}{\rm{d}t}& = &-\int_{ {\mathbb B}}(\beta\Delta^{2}\phi_{1}-\tau\Delta\phi_{1})u\, {\rm d}x+\lambda\int_{ {\mathbb B}}\frac{\phi_{1}}{(1-u)^{2}}\, {\rm d}x{}\\& \geq&-\lambda_{1}\int_{ {\mathbb B}}\phi_{1}u\, {\rm d}x+\frac{\lambda}{(1-\int_{ {\mathbb B}}\phi_{1}u\, {\rm d}x)^{2}}{}\\& = &-\lambda_{1}M+\frac{\lambda}{(1-M)^{2}}: = g(M) \end{eqnarray}$

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