ABSTRACT

In this paper, we study the initial boundary value of a thermally coupled beam equation with a memory term under the action of an axial force.

Keywords:Axial Force, Memory Terms, Thermoelastic, Beam

轴向力作用下具有记忆项的热弹耦合梁方程组的初边值问题

刘帅博，张建文^*

太原理工大学，数学学院，江西 太原

收稿日期：2020年3月25日；录用日期：2020年4月10日；发布日期：2020年4月17日

摘 要

本文研究了轴向力作用下具有记忆项的热弹耦合梁方程的初边值的问题，通过先验估计证明了系统弱解和正则解的存在唯一性。

关键词 :轴向力，记忆项，热弹耦合，梁

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

1968年，Gurtin Monton E等 [1] 提出了具有记忆项的热传导理论。之后关于含记忆性的各类非线性梁方程的初边值问题的研究有若干的进展 [2] - [8]，比如2009年，Wang Junmin等 [2] 研究了如下具有记忆项的热传导方程

${\theta }_t+{\displaystyle {\int }_0^{t}g\left(t-\tau \right){\theta }_{xx}\left(x,\tau \right)\text{d}\tau }=0$

解的存在性。2011年，Lazo P P D [3] 运用Galerkin方法证明了如下具有记忆项的波动方程

$u_{tt}-M\left({\left|u_x\right|}^2\right)u_{xx}+{\displaystyle {\int }_0^{t}g\left(t-\tau \right)u_{xx}\left(x,\tau \right)\text{d}\tau }=0$

整体解存在唯一性和稳定性。2011年，Santos M L [4] 考虑热记忆项，研究了如下热弹耦合杆方程组

$\{\begin{array}{l}{u}_{tt}-u_{xx}+\alpha {\theta }_x=0\\ {\theta }_t-{\theta }_{xx}-{\displaystyle {\int }_0^{t}g\left(t-\tau \right){\theta }_{xx}\left(x,\tau \right)\text{d}\tau }+\beta u_{tx}=0\end{array}$

解的存在唯一性及指数衰减率。

本文在前人的基础上考虑如下轴向力作用下具有记忆项热弹耦合梁方程组

$\{\begin{array}{l}{u}_{tt}+u_{xxxx}-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)u_{xxxx}\left(x,\tau \right)\text{d}\tau }-M\left({\displaystyle {\int }_0^L{\left|u_x\right|}^2\text{d}x}\right)u_{xx}+\alpha {\theta }_{xx}=q\left(x\right)\\ {\theta }_t-{\theta }_{xx}-{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right){\theta }_{xx}\left(x,\tau \right)\text{d}\tau }-\alpha u_{xxt}=0\end{array}$ (1)

在边界条件

$u\left(0,t\right)=u\left(L,t\right)=u_{xx}\left(0,t\right)=u_{xx}\left(L,t\right)=0$,$\theta \left(0,t\right)=\theta \left(L,t\right)=0$ (2)

和初始条件

$u\left(x,0\right)=u_0\left(x\right)$,$u_t\left(x,0\right)=u_1\left(x\right)$,$\theta \left(x,0\right)={\theta }_0\left(x\right)$ (3)

下解的存在唯一性。

2. 预备知识

本文的讨论是基于以下的Hilbert空间：

$S_1=\left\{u\in H^4\left(0,L\right)|u,u_{xx}\in H_0^1\left(0,L\right)\right\}$

$S_2=H_0^1\left(0,L\right)\cap H^2\left(0,L\right)$

且 $S_1\subset S_2\subset L^2$。

方程中的函数满足如下基本假设

(h1) 对函数M，假设 $M\in C^1\left(\left[0,\infty \right)\right)$，且M为非负函数，有

$M\left(s\right)\ge a+bs\left(a,b>0\right)$,$\left|M^{\prime }\left(s\right)\right|<\sigma$ (4)

其中 $\hat{M}\left(s\right)={\displaystyle {\int }_0^{s}M\left(z\right)\text{d}z}$，其中 $\sigma$ 为大于0的正常数。

(h2) 假设 $k_1\left(t\right)\in C^1\left[\left[0,\infty \right),\left[0,\infty \right)\right]$,$k_1\left(t\right)\ge 0$,${k^{\prime }}_1\left(t\right)\le 0$,${\displaystyle {\int }_0^{\infty }{k}_1\left(t\right)\text{d}t}<a$。

(h3) 假设 $k_2\left(t\right)\in C^1\left[\left[0,\infty \right),\left[0,\infty \right)\right]$,$k_2\left(0\right)=0$，且 $1-{\displaystyle {\int }_0^{\infty }{k}_2\left(t\right)\text{d}t=l}>0$,$\exists \xi >0$，使得 ${k^{\prime }}_2\left(t\right)\le -\xi k_2\left(t\right)$。

(h4) 假设 $q\left(x\right)\in L^2\left(0,L\right)$。

3. 主要结论

3.1. 弱解的存在性

定理1若 $u_0\in S_2,u_1,{\theta }_0\in H_0^1\left(0,L\right)$，且假设(h1)~(h4)成立，则存在 $u\equiv u\left(x,t\right),\theta \equiv \theta \left(x,t\right)$ 使得 $u\in L^{\infty }\left(0,T;S_2\right)$,$u_t\in L^{\infty }\left(0,T;L^2\right)$,$\theta \in L^{\infty }\left(0,T;L^2\right)$,$T>0$。且在下述定义下满足方程(1)，即对几乎处处的 $t\in \left(0,T\right)$ 有

$\{\begin{array}{l}\frac{\text{d}}{\text{d}t}\left(u_t,\omega \right)+\left(u_{xx},{\omega }_{xx}\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx},{\omega }_{xx}\right)\text{d}\tau }+M\left({\Vert u_x\left(t\right)\Vert }_2^2\right)\left(u_x,{\omega }_x\right)+\alpha \left({\theta }_{xx},\omega \right)=\left(q\left(x\right),\omega \right)\\ \frac{\text{d}}{\text{d}t}\left(\theta ,\tilde{\omega }\right)+\left({\theta }_x,{\tilde{\omega }}_x\right)+{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\theta }_x,{\tilde{\omega }}_x\right)\text{d}\tau }-\alpha \left(u_{xxt},\tilde{\omega }\right)=0\end{array}$,$\forall \omega \in S_2$,$\tilde{\omega }\in H_0^1$ (5)

证明：1) 构造近似解：设 ${\left\{{\omega }_j\right\}}_{j=1}^{\infty }$ 和 ${\left\{{\tilde{\omega }}_j\right\}}_{j=1}^{\infty }$ 分别是 $S_2$ 和 $H_0^1$ 的标准正交基，使得 $u_0,u_1\in span\left\{{\omega }_1,{\omega }_2\right\}$，${\theta }_0={\tilde{\omega }}_1$，记

$u^m\left(x,t\right)={\displaystyle \underset{j=1}{\overset{m}{\sum }}{r}_j\left(t\right){\omega }_j\left(x\right)}$,${\theta }^m\left(t\right)={\displaystyle \underset{j=1}{\overset{m}{\sum }}{h}_j\left(t\right){\tilde{\omega }}_j(x)}$

其中 $r_j\left(t\right)$,$h_j\left(t\right)$ 为未知函数，仿(5)作如下关于 $u^m\left(x,t\right)$ 的方程组

$\{\begin{array}{l}\left(u_{tt}^m,{\omega }_j\right)+\left(u_{xx}^m,{\omega }_{jxx}\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx}^m,{\omega }_{jxx}\right)\text{d}\tau }+M\left({\Vert u_x^m\left(t\right)\Vert }_2^2\right)\left(u_x^m,{\omega }_{jx}\right)+\alpha \left({\theta }_{xx}^m,{\omega }_j\right)=\left(q\left(x\right),{\omega }_j\right)\\ \left({\theta }_t^m,{\tilde{\omega }}_j\right)+\left({\theta }_x^m,{\tilde{\omega }}_{jx}\right)+{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\theta }_x^m,{\tilde{\omega }}_{jx}\right)\text{d}\tau }-\alpha \left(u_{xxt}^m,{\tilde{\omega }}_j\right)=0\end{array}$ (6)

另外有

$\{\begin{array}{l}{u}^m\left(x,0\right)=u_0^m\to u_0\in S_2\\ u_t^m\left(x,0\right)=u_1^m\to u_1\in L^2\\ {\theta }^m\left(x,0\right)={\theta }_0^m\to {\theta }_0\in L^2\end{array}$ (7)

根据常微分的解的存在唯一性理论可知，方程(6) (7)存在唯一解，下面进行先验估计。

2) 先验估计

以下讨论中C表示与 $m,T$ 无关的常数，它在不同的地方代表不同的表达式中可能有不同的值。

记

$\left(k\circ \phi \right)\left(t\right)={\displaystyle {\int }_0^{t}k\left(t-\tau \right){\displaystyle {\int }_0^L\left|\phi \left(t\right)-\phi \left(\tau \right)\right|\text{d}x\text{d}\tau }}$ (8)

令方程组(6)中两式分别与 $r_{jt}$ 和 $h_j$ 做积，并对 $j=1,2,3,\cdots ,m$ 求和，得

$\begin{array}{l}\left(u_{tt}^m,u_t^m\right)+\left(u_{xx}^m,u_{txx}^m\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx}^m\left(\tau \right),u_{txx}^m\left(t\right)\right)\text{d}\tau }\\ +M\left({\Vert u_x^m\left(t\right)\Vert }_2^2\right)\left(u_x^m,u_{tx}^m\right)+\alpha \left({\theta }_{xx}^m,u_t^m\right)=\left(q\left(x\right),u_t^m\right)\end{array}$ (9)

$\left({\theta }_t^m,{\theta }^m\right)+\left({\theta }_x^m,{\theta }_x^m\right)+{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\theta }_x^m\left(\tau \right),{\theta }_x^m\left(t\right)\right)\text{d}\tau }-\alpha \left(u_{xxt}^m,{\theta }_x^m\right)=0$ (10)

两式相加，并从0到t积分所得两式相加，由假设(h1)及(8)式，得

$\begin{array}{l}{F}_m\left(t\right)+\left(k_1\circ u_{xx}^m\right)\left(t\right)+{\displaystyle {\int }_0^t\left[k_1\left(s\right){\Vert u_{xx}^m\Vert }_2^2-\left({k^{\prime }}_1\circ u_{xx}^m\right)\left(s\right)\right]\text{d}s}\\ +2{\displaystyle {\int }_0^t\left[{\displaystyle {\int }_0^s{k}_2\left(s-\tau \right)\left({\theta }_x^m\left(\tau \right),{\theta }_x^m\left(s\right)\right)\text{d}\tau }\right]\text{d}s}+2{\displaystyle {\int }_0^t{\Vert {\theta }_x^m\left(s\right)\Vert }_2^2\text{d}s}\\ =F_m\left(0\right)+{\displaystyle {\int }_0^t\left(q\left(x\right),u_t^m\right)\text{d}s}\end{array}$ (11)

其中

$F_m\left(t\right)={\Vert u_t^m\Vert }_2^2+{\Vert u_{xx}^m\Vert }_2^2+\hat{M}\left({\Vert u_x^m\Vert }_2^2\right)+{\Vert {\theta }^m\Vert }_2^2-{\displaystyle {\int }_0^t{k}_1\left(s\right)\text{d}s}{\Vert u_{xx}^m\Vert }_2^2$

$F_m\left(0\right)={\Vert u_1^m\Vert }_2^2+{\Vert u_{0xx}^m\Vert }_2^2+\hat{M}\left({\Vert u_{0x}^m\Vert }_2^2\right)+{\Vert {\theta }_0^m\Vert }_2^2-{\displaystyle {\int }_0^t{k}_1\left(s\right)\text{d}s}{\Vert u_{0xx}^m\Vert }_2^2$

结合假设(h3)，由Schwarz不等式和Young不等，得到

$\begin{array}{l}\left|{\displaystyle {\int }_0^s{k}_2\left(s-\tau \right)\left({\theta }_x^m\left(\tau \right),{\theta }_x^m\left(s\right)\right)\text{d}\tau }\right|\\ =\left|{\displaystyle {\int }_0^L{\displaystyle {\int }_0^s{k}_2\left(s-\tau \right){\theta }_x^m\left(\tau \right){\theta }_x^m\left(s\right)\text{d}\tau }\text{d}x}\right|\\ \le \eta {\Vert {\theta }_x^m\left(s\right)\Vert }_{\text{2}}^2+\frac{1}{4\eta }{\displaystyle {\int }_0^L{\left[{\displaystyle {\int }_0^s{k}_2\left(s-\tau \right){\theta }_x^m\left(\tau \right)\text{d}\tau }\right]}^2\text{d}x}\\ \le \eta {\Vert {\theta }_x^m\left(s\right)\Vert }_{\text{2}}^2+\frac{1}{4\eta }{\displaystyle {\int }_0^s{k}_2\left(\tau \right)\text{d}\tau {\displaystyle {\int }_0^s{k}_2\left(s-\tau \right){\displaystyle {\int }_0^L{\left|{\theta }_x^m\left(\tau \right)\right|}^2\text{d}x}\text{d}\tau }}\\ \le \eta {\Vert {\theta }_x^m\left(s\right)\Vert }_{\text{2}}^2+\frac{\left(1-l\right)k_2\left(0\right)}{4\eta }{\displaystyle {\int }_0^s{\Vert {\theta }_x^m\left(\tau \right)\Vert }_2^2\text{d}\tau }\end{array}$ (12)

其中 $0<\eta <1$。则

$\left|2{\displaystyle {\int }_0^t\left[{\displaystyle {\int }_0^s{k}_2\left(s-\tau \right)\left({\theta }_x^m\left(\tau \right),{\theta }_x^m\left(s\right)\right)\text{d}\tau }\right]\text{d}s}\right|\le \text{2}\eta {\displaystyle {\int }_0^t{\Vert {\theta }_x^m\left(s\right)\Vert }_2^{2}ds}$ (13)

由假设(h4)，得

$2{\displaystyle {\int }_0^t\left(q\left(x\right),u_t^m\left(s\right)\right)\text{d}s}\le 2{\displaystyle {\int }_0^t\left[\Vert q\left(x\right)\Vert \cdot \Vert u_t^m\left(s\right)\Vert \right]\text{d}s}\le {\Vert q\left(x\right)\Vert }_2^2+{\displaystyle {\int }_0^t{F}_m\left(s\right)\text{d}s}$ (14)

由(12)~(14)，得

$F_m\left(t\right)+2\left(1-\eta \right){\displaystyle {\int }_0^t{\Vert {\theta }_x^m\left(s\right)\Vert }_2^{\text{2}}\text{d}s}\le F_m\left(0\right)+{\Vert q\left(x\right)\Vert }_2^2+C{\displaystyle {\int }_0^t{F}_m\left(s\right)\text{d}s}$ (15)

故由Gronwall引理可得

$F_m\left(t\right)+2\left(1-\eta \right){\displaystyle {\int }_0^t{\Vert {\theta }_x^m\left(s\right)\Vert }^2\text{d}s}\le \left(F_m\left(0\right)+{\Vert q\left(x\right)\Vert }_2^2\right){\text{e}}^{Ct}$

因此得到

$\Vert u_t^m\Vert \le C$,$\Vert u_{xx}^m\Vert \le C$,$\Vert {\theta }^m\Vert \le C$,${\displaystyle {\int }_0^t{\Vert {\theta }_x^m\left(s\right)\Vert }_2^{\text{2}}\text{d}s}<C$.

由于上述关于t一致有界，故对 $\forall T,t\in \left(0,T\right)$ 内均成立。由Poincare不等式可得 $\Vert u^m\Vert \le C$。

3) 收敛：

$u^m\to u$ 在 $L^{\infty }\left(0,T;S_2\right)$ 弱*收敛

$u_t^m\to u_t$ 在 $L^{\infty }\left(0,T;L^2\right)$ 弱*收敛

${\theta }^m\to \theta$ 在 $L^{\infty }\left(0,T;L^2\right)$ 弱*收敛

${\displaystyle {\int }_0^t{\Vert {\theta }_x^m\left(s\right)\Vert }_2^{\text{2}}\text{d}s}$ 在 $L^{\infty }\left(0,T;L^2\right)$ 弱*收敛

$u^m\to u$ 在 $L^2\left(0,T\right)$ 中强收敛且几乎处处收敛

对 $\forall \phi \in L^1\left(0,T\right)$，有 $\omega \phi \in L^1\left(0,T;L^2\right)$，那么有

${\displaystyle {\int }_0^T\left(u_{xx}^m,{\omega }_{xx}\right)\phi \text{d}t}={\displaystyle {\int }_0^T{\displaystyle {\int }_0^L{u}_{xx}^m{\omega }_{xx}\phi \text{d}x\text{d}t}}={\displaystyle {\int }_0^T\left(u^m,{\omega }_{xxxx}\phi \right)\text{d}t}\to {\displaystyle {\int }_0^T\left(u,{\omega }_{xxxx}\phi \right)\text{d}t}={\displaystyle {\int }_0^T\left(u_{xx},{\omega }_{xx}\right)\phi \text{d}t}$

即 $\left(u_{xx}^m,{\omega }_{xx}\right)\to \left(u_{xx},{\omega }_{xx}\right)$ 在 $L^{\infty }\left(0,T\right)$ 弱*收敛。

${\displaystyle {\int }_0^T\left({\theta }_x^m,{\tilde{\omega }}_x\right)\phi \text{d}t}={\displaystyle {\int }_0^T{\displaystyle {\int }_0^L{\theta }_x^m{\tilde{\omega }}_x\phi \text{d}x\text{d}t}}={\displaystyle {\int }_0^T\left({\theta }^m,{\tilde{\omega }}_{xx}\phi \right)\text{d}t}\to {\displaystyle {\int }_0^T\left(\theta ,{\tilde{\omega }}_{xx}\phi \right)\text{d}t}={\displaystyle {\int }_0^T\left({\theta }_x,{\tilde{\omega }}_x\right)\phi \text{d}t}$

即 $\left({\theta }_x^m,{\omega }_x\right)\to \left({\theta }_x,{\omega }_x\right)$ 在 $L^{\infty }\left(0,T\right)$ 弱*收敛。

同理可得

${\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx}^m,{\omega }_{xx}\right)\text{d}\tau }\to {\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx},{\omega }_{xx}\right)\text{d}\tau }$ 在 $L^{\infty }\left(0,T\right)$ 弱*收敛。

$M\left({\Vert u_t^m\left(t\right)\Vert }_2^2\right)\left(u_x^m,{\omega }_x\right)\to M\left({\Vert u_x\left(t\right)\Vert }_2^2\right)\left(u_x,{\omega }_x\right)$ 在 $L^{\infty }\left(0,T\right)$ 弱*收敛。

由以上可知

$\{\begin{array}{l}\left(u_{tt},\omega \right)+\left(u_{xx},{\omega }_{xx}\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx},{\omega }_{xx}\right)\text{d}\tau }+M\left({\Vert u_x\left(t\right)\Vert }_2^2\right)\left(u_x,{\omega }_x\right)=q\left(L\right)\\ \left({\theta }_t,\tilde{\omega }\right)+\left({\theta }_x,{\tilde{\omega }}_x\right)-{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\theta }_x,{\tilde{\omega }}_x\right)\text{d}\tau }=0\end{array}$ (16)

成立。定理1证毕。

定理2设 $\left(u,\theta \right)$,$\left(v,\eta \right)$ 是方程(5)的两个解，且 $u,v\in L^{\infty }\left(0,T;S_2\right)$,$\theta ,\eta \in L^{\infty }\left(0,T;L^2\right)$,$u_t,v_t\in L^{\infty }\left(0,T;L^2\right)$，以及假设 $\left(u,\theta \right)$,$\left(v,\eta \right)$ 满足初始条件： $u\left(x,0\right)=u_0$,$u_t\left(x,0\right)=u_1$,$\theta \left(x,0\right)={\theta }_0$,$v\left(x,0\right)=v_0$,$v_t\left(x,0\right)=v_1$,$\eta \left(x,0\right)={\eta }_0$ 及 $u_0,v_0\in S_2$,${\theta }_0,{\eta }_0\in L^2$ 且 $u_1,v_1\in L^2$,$u_0,v_0\in S_2$，令 $\left(z,\lambda \right)=\left(u-v,\theta -\eta \right)$，则有

$\{\begin{array}{l}\left(z_{tt},{\omega }_j\right)+\left(z_{xx},{\omega }_{jxx}\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(z_{xx},{\omega }_{jxx}\right)\text{d}\tau }+M\left({\Vert z_x\left(t\right)\Vert }_2^2\right)\left(z_x,{\omega }_{jx}\right)+\alpha \left({\gamma }_{xx},{\omega }_j\right)=0\\ \left({\gamma }_t,{\tilde{\omega }}_j\right)+\left({\gamma }_x,{\tilde{\omega }}_{jx}\right)+{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\gamma }_x,{\tilde{\omega }}_{jx}\right)\text{d}\tau }-\alpha \left(z_{txx},{\tilde{\omega }}_j\right)=0\end{array}$

证明:与定理1先验估计类似，可得到

${\Vert z_t\Vert }_2^2+{\Vert z_{xx}\Vert }_2^2+{\Vert \lambda \Vert }_2^2\le \left\{{\Vert u_1-v_1\Vert }_2^2+{\Vert u_{0xx}-v_{0xx}\Vert }_2^2+{\Vert {\theta }_0-{\eta }_0\Vert }_2^2\right\}{\text{e}}^{Ct}$ (17)

其中C是关于 $\Vert u_1\Vert ,\Vert v_1\Vert ,\Vert u_{0xx}\Vert ,\Vert v_{0xx}\Vert ,\Vert {\theta }_0\Vert ,\Vert {\eta }_0\Vert$ 的连续函数。

推论方程(1)-(3)的解是唯一的。

证明:由(17)得，当 $u_0=v_0,u_1=v_1,{\theta }_0={\eta }_0$ 时，得到 ${\Vert z_t\Vert }_2^2+{\Vert z_{xx}\Vert }_2^2+{\Vert \lambda \Vert }_2^2\le \text{0}$，则有 $\Vert z_t\Vert =0,\Vert z_{xx}\Vert =0,\Vert \lambda \Vert =0$，从而有 $u=v,\theta =\eta$。解的唯一性得证。

3.2. 正则性

本节我们假设

(h5) $k_1\left(0\right)=0$,${k^{\prime }}_1\left(0\right)=0$,${k^{\prime }}_2\left(t\right)\le 0$,${k^{″}}_2\left(t\right)\ge 0$。

定理3若 $u_0\in S_1,u_1,{\theta }_0\in S_2$，且假设(h1)—(h5)成立，则存在唯一的函数 $u\equiv u\left(x,t\right),\theta \equiv \theta \left(x,t\right)$ 使得 $u\in L^{\infty }\left(0,T;S_1\right)$,$u_t\in L^{\infty }\left(0,T;S_2\right)$,$u_{tt}\in L^{\infty }\left(0,T;L^2\left(0,L\right)\right)$,$\theta \in L^{\infty }\left(0,T;S_2\right)$,${\theta }_t\in L^{\infty }\left(0,T;L^2\left(0,L\right)\right)$,$T>0$。且在下述定义下满足方程(1)，即对几乎处处的 $t\in \left(0,t\right)$ 有

$\{\begin{array}{l}\left(u_{tt},\omega \right)+\left(u_{xx},{\omega }_{xx}\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xx},{\omega }_{xx}\right)\text{d}\tau }+M\left({\Vert u_x\left(t\right)\Vert }_2^2\right)\left(u_x,{\omega }_x\right)+\alpha \left({\theta }_{xx},\omega \right)=\left(q\left(x\right),\omega \right)\\ \left({\theta }_t,\tilde{\omega }\right)+\left({\theta }_x,{\tilde{\omega }}_x\right)+{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\theta }_x,{\tilde{\omega }}_x\right)\text{d}\tau }-\alpha \left(u_{xxt},\tilde{\omega }\right)=0\end{array}$ $\forall \omega \in S_1,\tilde{\omega }\in S_2$ (18)

证明：1) 构建近似解:设 ${\left\{{\omega }_j\right\}}_{j=1}^{\infty }$ 和 ${\left\{{\tilde{\omega }}_j\right\}}_{j=1}^{\infty }$ 分别是 $S_1$ 和 $S_2$ 的标准正交基，使得 $u_0,u_1\in span\left\{{\omega }_1,{\omega }_2\right\}$,${\theta }_0={\tilde{\omega }}_1$，记

$u^m\left(x,t\right)={\displaystyle \underset{j=1}{\overset{m}{\sum }}{r}_j\left(t\right){\omega }_j\left(x\right)}$,${\theta }^m\left(t\right)={\displaystyle \underset{j=1}{\overset{m}{\sum }}{h}_j\left(t\right){\tilde{\omega }}_j(x)}$

其中 $r_j\left(t\right)$,$h_j\left(t\right)$ 为未知函数，仿(5)作如下关于 $u^m\left(x,t\right)$ 的方程组

$\{\begin{array}{l}\left(u_{tt}^m,{\omega }_j\right)+\left(u_{xxxx}^m,{\omega }_j\right)-{\displaystyle {\int }_0^t{k}_1\left(t-\tau \right)\left(u_{xxxx}^m,{\omega }_j\right)\text{d}\tau }-M\left({\Vert u_x^m\left(t\right)\Vert }_2^2\right)\left(u_{xx}^m,{\omega }_j\right)+\alpha \left({\theta }_{xx}^m,{\omega }_j\right)=\left(q\left(x\right),{\omega }_j\right)\\ \left({\theta }_t^m,{\tilde{\omega }}_j\right)-\left({\theta }_{xx}^m,{\tilde{\omega }}_j\right)-{\displaystyle {\int }_0^t{k}_2\left(t-\tau \right)\left({\theta }_{xx}^m,{\tilde{\omega }}_j\right)\text{d}\tau }-\alpha \left(u_{xxt}^m,{\tilde{\omega }}_j\right)=0\end{array}$ (19)

另外有

$\{\begin{array}{l}{u}^m\left(x,0\right)=u_0^m\to u_0\in S_1\\ u_t^m\left(x,0\right)=u_1^m\to u_1\in S_2\\ {\theta }^m\left(x,0\right)={\theta }_0^m\to {\theta }_0\in S_2\end{array}$ (20)

根据常微分的解的存在唯一性理论可知，方程(19) (20)存在唯一解，下面进行先验估计。

2) 先验估计

与之前一样可得 $\Vert u_t^m\Vert \le C$,$\Vert u_{xx}^m\Vert \le C$,$\Vert {\theta }^m\Vert \le C$。

将 $r_j=r_{jtt}\left(0\right),h_j=h_{jt}\left(0\right)$ 两式分别与方程组(19)中两式做积，对 $j=1,2,3,\cdots ,m$ 求和后，得

${\Vert u_{tt}^m\left(0\right)\Vert }_2^2+\left(u_{xxxx}^m\left(0\right),u_{tt}^m\left(0\right)\right)-M\left({\Vert u_x^m\left(0\right)\Vert }_2^2\right)\left(u_{xx}^m\left(0\right),u_{tt}^m\left(0\right)\right)+\alpha \left({\theta }_{xx}^m\left(0\right),u_{tt}^m\left(0\right)\right)=\left(q\left(x\right),u_{tt}^m\left(0\right)\right)$ (21)

${\Vert {\theta }_t^m\left(0\right)\Vert }_2^2+\left({\theta }_{xx}^m\left(0\right),{\theta }_t^m\left(0\right)\right)-\alpha \left(u_{xxt}^m\left(0\right),{\theta }_t^m\left(0\right)\right)=0$ (22)

结合假设(h4)应用Schwarz不等式，得

${\Vert u_{tt}^m\left(0\right)\Vert }_2^2\le \left[\Vert u_{0xxxx}^m\Vert +M\left({\Vert u_{0x}^m\Vert }_2^2\right)\Vert u_{0xx}^m\Vert +\alpha \Vert {\theta }_{0xx}\Vert +\Vert q\left(x\right)\Vert \right]\Vert u_{tt}^m\left(0\right)\Vert$ (23)

${\Vert {\theta }_t^m\left(0\right)\Vert }_2^2\le \left[\Vert {\theta }_{0xx}^m\Vert +\alpha \Vert u_{1xx}^m\Vert \right]\Vert {\theta }_t^m\left(0\right)\Vert$ (24)

因此得到

$\begin{array}{l}{u}_{ttt}^m+u_{xxxxt}^m-\left[M^{\prime }\left({\Vert u_x^m\left(t\right)\Vert }_2^2\right)+2M\left(u_x^m,u_{xt}^m\right)\right]u_{xx}^m-M\left({\Vert u_x\left(t\right)\Vert }_2^2\right)u_{xxt}^m\\ -{\displaystyle {\int }_0^t{k^{\prime }}_1\left(t-\tau \right)u_{xxxx}^m\left(\tau \right)\text{d}\tau }-k_1\left(0\right)u_{xxxxt}^m+\alpha {\theta }_{xxt}^m=0\end{array}$ (25)

${\theta }_{tt}^m-{\theta }_{xxt}^m-{\displaystyle {\int }_0^t{k^{\prime }}_2\left(t-\tau \right){\theta }_{xx}^m\left(\tau \right)\text{d}\tau }-k_2\left(0\right){\theta }_{xx}^m-\alpha u_{xxt}^m=0$ (26)

(25) (26)两式分别与 $u_{tt}^m$ 与 ${\theta }_t^m$ 做积，并从0到t积分，所得两式相加，结合假设(h1)及(8)式，得

$\begin{array}{l}{F}_m\left(t\right)+2{\displaystyle {\int }_0^t{\displaystyle {\int }_0^s{k^{\prime }}_1\left(s-\tau \right)\left(u_{xxx}^m\left(\tau \right),u_{ttx}^m\left(s\right)\right)\text{d}\tau \text{d}s}}-\left(k_2\circ {\theta }_x^m\right)\left(t\right)+{\displaystyle {\int }_0^t\left[-{k^{\prime }}_2\left(s\right){\Vert {\theta }_x^m\Vert }_2^2+\left({k^{″}}_2\circ {\theta }_x^m\right)\left(s\right)\right]\text{d}s}+2{\displaystyle {\int }_0^t{\Vert {\theta }_{xt}^m\Vert }_2^2\text{d}s}\\ =F_m(0)+{\displaystyle {\int }_0^{t}M(\text{||}{u}_x(t){\text{||}}_2^2)(u_{xxt}^m,u_{tt}^m)+\left[M^{\prime }\left({\Vert u_x\left(t\right)\Vert }_2^2\right)+2M\left(u_x^m,u_{xt}^m\right)\right]\left(u_{xx}^m,u_{tt}^m\right)\text{d}s}\end{array}$ (27)

其中 $F_m\left(t\right)={\Vert u_{tt}^m\Vert }_2^2+{\Vert u_{xxt}^m\Vert }_2^2+{\Vert {\theta }_t^m\Vert }_2^2+{\Vert {\theta }_x^m\Vert }_2^2{\displaystyle {\int }_0^t{k^{\prime }}_2\left(\tau \right)\text{d}\tau }$。

由假设(h1)及定理1结论，并由Schwarz不等式和Young不等式，得到

$\begin{array}{l}M\left({\Vert u_x^m\Vert }_2^2\right)\left(u_{xxt}^m,u_{tt}^m\right)+\left[M^{\prime }\left({\Vert u_x^m\Vert }_2^2\right)+2M\left(u_x^m,u_{xt}^m\right)\right]\left(u_{xx}^m,u_{tt}^m\right)\\ \le \frac{1}{2}\sigma \left({\Vert u_{xx}^m\Vert }_2^2+{\Vert u_{tt}^m\Vert }_2^2\right)+2M{\Vert u_x^m\Vert }_2^2\cdot {\Vert u_{xt}^m\Vert }_2^2\cdot \frac{1}{2}\left({\Vert u_{xx}^m\Vert }_2^2+{\Vert u_{tt}^m\Vert }_2^2\right)\\ +M\left({\Vert u_x^m\Vert }_2^2\right)\cdot \frac{1}{2}\left({\Vert u_{xxt}^m\Vert }_2^2+{\Vert u_{tt}^m\Vert }_2^2\right)\\ \le C\left({\Vert u_{xx}^m\Vert }_2^2+{\Vert u_{tt}^m\Vert }_2^2+{\Vert u_x^m\Vert }_2^2+{\Vert u_{xt}^m\Vert }_2^2+{\Vert u_{xxt}^m\Vert }_2^2\right)\\ \le C{F}_m\left(t\right)\end{array}$ (28)

由假设(h5)，及并结合Schwarz不等式和Young不等式，得到

$\begin{array}{l}\left|{\displaystyle {\int }_0^s{k^{\prime }}_1\left(s-\tau \right)\left(u_{xxx}^m\left(\tau \right),u_{xtt}^m\left(s\right)\right)\text{d}\tau }\right|\\ \le \left|{\displaystyle {\int }_0^L{\displaystyle {\int }_0^s{k^{\prime }}_1\left(s-\tau \right)u_{xxx}^m\left(\tau \right)u_{xtt}^m\left(s\right)\text{d}\tau }\text{d}x}\right|\\ \le \eta {\Vert u_{xtt}^m\left(s\right)\Vert }_2^2+\frac{1}{4\eta }{\displaystyle {\int }_0^L{\left({\displaystyle {\int }_0^s{k^{\prime }}_1\left(s-\tau \right)u_{xxx}^m\left(\tau \right)\text{d}\tau }\right)}^2\text{d}x}\\ \le \eta {\Vert u_{xtt}^m\left(s\right)\Vert }_2^2+\frac{1}{4\eta }{\displaystyle {\int }_0^s{k^{\prime }}_1\left(\tau \right)\text{d}\tau {\displaystyle {\int }_0^s{k^{\prime }}_1\left(s-\tau \right){\displaystyle {\int }_0^L{\left|u_{xxx}^m\left(\tau \right)\right|}^2}\text{d}x}}\\ \le \eta {\Vert u_{xtt}^m\left(s\right)\Vert }_2^2+\frac{\left(1-l\right){k^{\prime }}_1\left(0\right)}{4\eta }{\displaystyle {\int }_0^s{\Vert u_{xxx}^m\left(\tau \right)\Vert }_2^2\text{d}\tau }\end{array}$ (29)

其中 $0<\eta <1$。故

$\left|{\displaystyle {\int }_0^t{\displaystyle {\int }_0^s{k^{\prime }}_1\left(s-\tau \right)\left(u_{xxxx}^m\left(\tau \right),u_{xtt}^m\left(s\right)\right)\text{d}s\text{d}\tau }}\right|\le \text{2}\eta {\displaystyle {\int }_0^t{\Vert u_{xtt}^m\left(s\right)\Vert }_{\text{2}}^{\text{2}}\text{d}s}\le \text{2}\eta {\displaystyle {\int }_0^t{\Vert F_m\left(s\right)\Vert }_{\text{2}}^{\text{2}}\text{d}s}$ (30)

由(28)~(31)，可得

$F_m\left(t\right)+2{\displaystyle {\int }_0^t{\Vert {\theta }_{xt}^m\Vert }_2^2\text{d}s}\le F_m\left(0\right)+C{\displaystyle {\int }_0^t{F}_m\left(s\right)\text{d}s}$ (31)

由Gronwall引理，得

$F_m\left(t\right)+2{\displaystyle {\int }_0^t{\Vert {\theta }_{xt}^m\Vert }_2^2\text{d}s}\le F_m\left(0\right){\text{e}}^{Ct}$

因此得到

$\Vert u_{tt}^m\Vert \le C$,$\Vert u_{txx}^m\Vert \le C$,$\Vert {\theta }_t^m\Vert \le C$,${\displaystyle {\int }_0^t{\Vert {\theta }_{xt}^m\Vert }_2^2\text{d}s}\le C$