## Blow-Up Phenomena of Solutions for a Class of Viscous Cahn-Hilliard Equation with Gradient Dependent Potentials and Sources

Long Qunfei,1, Chen Jianqing2

 基金资助: 国家自然科学基金（青年项目）.  11401100国家自然科学基金.  11371091国家自然科学基金.  11161057非线性分析及其应用创新团队.  IRTL1206

 Fund supported: the NSFC (Youth Fund).  11401100the NSFC.  11371091the NSFC.  11161057the Innovation Group of 'Nonlinear Analysis and Its Applications'.  IRTL1206

Abstract

In this manuscript, we study the blow-up phenomena of solutions for a class of viscous Cahn-Hilliard equation with gradient dependent potentials and source. We establish a criterion for blow-up and determine the upper bound for blow-up time by the energy method, the differential inequality and the derivative formula of the product; We determine the lower bounds for blow-up time by the differential inequality and the derivative formula of the product.

Keywords： Viscous Cahn-Hilliard equation ; Bounds for blow-up time ; Criteria for the low-up ; Differential inequality ; Derivative formula of the product

Long Qunfei, Chen Jianqing. Blow-Up Phenomena of Solutions for a Class of Viscous Cahn-Hilliard Equation with Gradient Dependent Potentials and Sources. Acta Mathematica Scientia[J], 2019, 39(3): 510-517 doi:

## 1 引言

$\begin{eqnarray} && u_t + k_{1}u_{xxxx}- k_{2} u_{xxt} - \big(\phi(u_{x})\big)_{x} + \psi(u) = 0, \quad (x, t)\in\Omega\times(0, T_{0}), \end{eqnarray}$

$\begin{eqnarray} && u(0, t) = u(1, t) = u_{xx}(0, t) = u_{xx}(1, t) = 0, \quad t\in(0, T_{0}), \end{eqnarray}$

$\begin{eqnarray} && u(x, 0) = u_0(x), \quad x \in \Omega \end{eqnarray}$

$$$\hbox{对于}\ s \leq 0 \ \hbox{有} \ \phi(s) \leq B_{1}|s|^{q - 1}s \quad \hbox{和}\quad \hbox{对于}\ s \geq 0\ \hbox{有}\ \phi(s) \geq B_{1}|s|^{q - 1}s, \quad q > 1, \quad \quad$$$

$$$\hbox{对于}\ s \leq 0\ \hbox{有}\ \psi(s) \leq - B_{2}|s|^{p - 1}s \quad \hbox{和}\quad \hbox{对于}\ s \geq 0\ \hbox{有}\ \psi(s) \geq - B_{2}|s|^{p - 1}s, \quad p > 1,$$$

$$$(q + 1)\Phi(s) \geq s\phi(s), \quad \ s \in {\Bbb R} ,$$$

$$$(p + 1)\Psi(s) \geq s\psi(s), \quad\ s \in {\Bbb R} ,$$$

(1)当$g(s) = h(s) $$h(s) = \hbox{arc}\cot s + \frac{1}{2} ,有 m = \frac{1}{2} \leq h(s) \leq \frac{3\pi}{2} = M 和对任意的 s\in {\Bbb R}$$ \Gamma^{\prime}(s) = - (- \frac{1}{s^2 + 1}) |s|^{p + 1} \geq 0$.$B_{1} = m $$\phi(s) = g(s) ,获得 (1.5) 式,以及使用 \Gamma (s)$$ \Gamma^{\prime}(s) \geq 0$获得对任意的$s\in {\Bbb R} $$(1.7) 式成立. (2)当 g(s) = - h(s)$$ h(s) = \arctan s + \pi$时,对任意的$s\in {\Bbb R} $$- M = - \frac{3\pi}{2} \leq h(s) \leq - \frac{\pi}{2} = - m$$ \Gamma^{\prime}(s) = \frac{1}{s^2 + 1} |s|^{p + 1} \geq 0$.$B_{2} = M $$\psi(s) = g(s) 获得 (1.6) 式,以及使用 \Gamma (s)$$ \Gamma^{\prime}(s) \geq 0$获得对于任意的$s\in {\Bbb R} $$(1.8) 式成立. 这篇文章的主要结果如下: ● 在 J(u_{0}) \leq 0 的条件下,我们给出了初边值问题 (1.1)$$ (1.3)$解的爆破准则和确定爆破时间的上界;

●  在初边值问题$(1.1) $$(1.3) 的解在有限爆破的条件下,我们确定了爆破时间的下界. 这篇文章的主要结构被组织如下:在第二部分建立爆破准则和确定爆破时间的上界;在第三部分确定爆破时间的下界. ## 2 爆破准则和爆破时间的上界 贯穿全文,对空间 L^{p}(\Omega) : = L^{p} , W^{l, p}(\Omega): = W^{l, p}$$ H_{0}^{l}(\Omega): = H_{0}^{l}$分别赋予范数$\|\cdot\|_{p}$, $\|\cdot\|_{W^{l, p}} $$\|\cdot\|_{H_{0}^{l}} ,其中 1 \leq l, p < +\infty . 定义2.1 如果具有 u_{t}(x, t) \in L^{2}\big([0, T); H_{0}^{1}\big) 的函数 且满足: (ⅰ)对任意的 \varphi \in L^{\infty}\big([0, T);W^{1, q}_{0}\cap W^{2, 2}_{0}\cap L^{p}\big) 都有 (ⅱ)在 W^{1, q}_{0}\cap W^{2, 2}_{0}\cap L^{p}$$ u(x, 0) = u_{0}(x)$.

$$$\frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} \geq (p - 1) \|u_{xx}\|_{2}^{2} + \frac{2(p - q)}{q + 1} \int_{\Omega}\phi(u_{x})u_{x}{\rm d}x.$$$

$\begin{eqnarray} \frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} &\geq& (p - 1) \|u_{xx}\|_{2}^{2} + \frac{2(p - q)}{q + 1} B_{1}\int_{\Omega}|u_{x}|^{q + 1}{\rm d}x\\ & \geq& (p - 1)\pi^{2} \|u_{x}\|_{2}^{2} + \frac{2(p - q)}{q + 1} B_{1}\|u_{x}\|^{q + 1}_{2}. \end{eqnarray}$

$\begin{eqnarray} \|u_{x}\|_{2}^{2} & = & \frac{1 + \pi^{2}}{1 + \pi^{2}}\|u_{x}\|_{2}^{2} = \frac{1}{1 + \pi^{2}}\|u_{x}\|_{2}^{2} + \frac{\pi^{2}}{1 + \pi^{2}}\|u_{x}\|_{2}^{2}\\ & \geq& \frac{\pi^{2}}{1 + \pi^{2}}\|u\|_{2}^{2} + \frac{\pi^{2}}{1 + \pi^{2}}\|u_{x}\|_{2}^{2} = \frac{\pi^{2}}{1 + \pi^{2}}\|u\|_{H_{0}^{1}}^{2}. \end{eqnarray}$

$(2.9)$式插入$(2.8)$式且应用引理2.1和$(2.3)$式推出

$\begin{eqnarray} \frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} & \geq& (p - 1)\frac{\pi^{4}}{1 + \pi^{2}} \|u\|_{H_{0}^{1}}^{2} + \frac{2(p - q)}{q + 1} B_{1}\Big(\frac{\pi^{2}}{1 + \pi^{2}}\Big) ^{\frac{q + 1}{2}}\|u\|_{H_{0}^{1}}^{q + 1}\\ & = & C_{1}\|u\|_{H_{0}^{1}}^{2} + C_{2}\|u\|_{H_{0}^{1}}^{q + 1}, \end{eqnarray}$

$(2.10)$式推出

$$$\frac{\rm d}{{\rm d}t}\frac{1}{\big(\|u\|^{2}_{H_{0}^{1}}{\rm e}^{-C_{1} t}\big)^{\frac{q - 1}{2}}} \leq - \frac{q - 1}{2}C_{2} {\rm e}^{\frac{C_{1}(q - 1)}{2} t}.$$$

$(0, t)$上积分$(2.11)$式推出

$$$\frac{1}{\|u\|^{q - 1}_{H_{0}^{1}}} \leq \bigg(\frac{1}{\|u_{0}\|^{q - 1}_{H_{0}^{1}}} + \frac{C_{2}}{C_{1}}\bigg) {\rm e}^{- \frac{(q - 1)C_{1}}{2}t} - \frac{C_{2}}{C_{1}}.$$$

## 3 爆破时间的下界

(ⅰ)如果$0 < \|u_{0}\|_{H_{0}^{1}}< 1$,那么

$$$T_{0} \geq \frac{2}{(p - 1)C_{3}} \ln{\frac{C_{4}\|u_{0}\|^{p - 1}_{H_{0}^{1}}} {C_{4}\|u_{0}\|^{p - 1}_{H_{0}^{1}} - C_{3}}};$$$

(ⅱ)如果$\|u_{0}\|_{H_{0}^{1}} \geq 1$,那么

$$$T_{0} \geq \frac{2}{(p - 1)C_{5}} \ln{\frac{C_{4}\|u_{0}\|^{p - 1}_{H_{0}^{1}}} {C_{4}\|u_{0}\|^{p - 1}_{H_{0}^{1}} - C_{5}}},$$$

$$$C_{3} = 2 \frac{\pi^{4}}{1 + \pi^{2}}, \quad C_{4} = 2 B_{2}S_{p + 1}^{p + 1}, \quad C_{5} = \bigg( 2 \frac{\pi^{4}} {1 + \pi^{2}} + 2 B_{1}\bigg(\frac{\pi^{2}} {1 + \pi^{2}}\bigg)^{\frac{q + 1}{2}}\bigg),$$$

$S_{p + 1} $$H_{0}^{1} \hookrightarrow L^{p + 1} 的最佳嵌入常数,以及下列的 (3.10) 式和 (3.13) 式分别隐含了 C_{4}\|u_{0}\|^{p - 1}_{H_{0}^{1}} - C_{3} > 0$$ C_{4}\|u_{0}\|^{p - 1}_{H_{0}^{1}} - C_{5} > 0$.

结合$(2.4)$式, $(1.5)$式和$(1.6)$式推出

$$$\frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} \leq - 2 \|u_{xx}\|_{2}^{2} - 2 B_{1}\int_{\Omega}|u_{x}|^{q + 1}{\rm d}x + 2 B_{2}\int_{\Omega}|u|^{p + 1}{\rm d}x.$$$

$$$\frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} \leq - 2 \pi^{2}\|u_{x}\|_{2}^{2} - 2 B_{1}\|u_{x}\|_{2}^{q + 1} + 2 B_{2}S^{p + 1}_{p + 1}\|u\|_{H_{0}^{1}}^{p + 1}.$$$

$(2.9)$式代入$(3.5)$式推出

$$$\frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} \leq - 2 \frac{\pi^{4}}{1 + \pi^{2}}\|u\|_{H_{0}^{1}}^{2} - 2 B_{1}\Big(\frac{\pi^{2}}{1 + \pi^{2}}\Big) ^{\frac{q + 1}{2}}\|u\|_{H_{0}^{1}}^{q + 1} + 2 B_{2}S_{p + 1}^{p + 1}\|u\|_{H_{0}^{1}}^{p + 1}.$$$

$$$\frac{\rm d}{{\rm d}t}\|u\|^{2}_{H_{0}^{1}} \geq 0, \quad t \in[0, T_{0}).$$$

(ⅰ)当$0 < \|u_{0}\|_{H_{0}^{1}}< 1$时,从$(3.7)$式推出$\|u\|_{H_{0}^{1}}^{2} \geq \|u_{0}\|_{H_{0}^{1}}^{2} > 0, t \in [0, T_{0}).$因此, $(3.6)$式和$(3.3)$式隐含了

$$$\frac{\rm d}{{\rm d}t}\frac{1}{\big(\|u\|^{2}_{H_{0}^{1}}{\rm e}^{C_{3} t}\big)^{\frac{p - 1}{2}}} \geq - \frac{p - 1}{2} C_{4} {\rm e}^{ - \frac{(p - 1)C_{3}}{2} t}.$$$

$(0, t)$上积分$(3.8)$式推出

$$$\frac{1}{\|u\|^{p - 1}_{H_{0}^{1}}} \geq \bigg(\frac{1}{\|u_{0}\|^{p - 1}_{H_{0}^{1}}} - \frac{C_{4}}{C_{3}}\bigg){\rm e}^{ \frac{(p - 1)C_{3}}{2} t} + \frac{C_{4}}{C_{3}}.$$$

$t \rightarrow T_{0}^{-}$,从$(3.9)$式推出

$$$0 \geq \bigg(\frac{1}{\|u_{0}\|^{p - 1}_{H_{0}^{1}}} - \frac{C_{4}}{C_{3}}\bigg){\rm e}^{ \frac{(p - 1)C_{3}}{2} T_{0}} + \frac{C_{4}}{C_{3}} \quad \Leftrightarrow \quad \bigg(\frac{\|u_{0}\|^{p - 1}_{H_{0}^{1}} C_{4} - C_{3}}{\|u_{0}\|^{p - 1}_{H_{0}^{1}} C_{3}}\bigg) {\rm e}^{ \frac{(p - 1)C_{3}}{2} T_{0}} \geq \frac{C_{4}}{C_{3}}.$$$

(ⅱ)当$\|u_{0}\|_{H_{0}^{1}} \geq 1$时,使用$(3.7)$式推出$\|u\|_{H_{0}^{1}}^{2} \geq \|u_{0}\|_{H_{0}^{1}}^{2}, \ t \in [0, T_{0}).$因此,从$(3.6)$式和$(3.3)$式推出

$$$\frac{\rm d}{{\rm d}t}\frac{1}{\big(\|u\|^{2}_{H_{0}^{1}}{\rm e}^{C_{5} t}\big)^{\frac{p - 1}{2}}} \geq - \frac{p - 1}{2} C_{4} {\rm e}^{ - \frac{(p - 1)C_{5}}{2} t}.$$$

$(0, t)$上积分$(3.11)$式推出

$$$\frac{1}{\|u\|^{p - 1}_{H_{0}^{1}}} \geq \bigg(\frac{1}{\|u_{0}\|^{p - 1}_{H_{0}^{1}}} - \frac{C_{4}}{C_{5}}\bigg){\rm e}^{ \frac{(p - 1)C_{5}}{2} t} + \frac{C_{4}}{C_{5}}.$$$

$t = T_{0}$,从$(3.12)$式推出

$$$0 \geq \bigg(\frac{1}{\|u_{0}\|^{p - 1}_{H_{0}^{1}}} - \frac{C_{4}}{C_{5}}\bigg){\rm e}^{ \frac{(p - 1)C_{5}}{2} T_{0}} + \frac{C_{4}}{C_{5}} \quad \Leftrightarrow \quad \bigg(\frac{\|u_{0}\|^{p - 1}_{H_{0}^{1}} C_{4} - C_{5}}{\|u_{0}\|^{p - 1}_{H_{0}^{1}} C_{5}}\bigg) {\rm e}^{ \frac{(p - 1)C_{5}}{2} T_{0}} \geq \frac{C_{4}}{C_{5}}.$$$

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