给定$ T>0 $.设$ Q = (0, 1)\times(0, T) $.考虑如下一类具有非局部项的抛物型偏微分方程

$\left\{ \begin{array}{l} {y_t}(x,t) - {y_{xx}}(x,t) = a\int_0^t {{y_{xx}}} (x,s){\rm{d}}s + b(x)u(t),\quad (x,t) \in {\rm{Q,}}\\ y(0,t) = y(1,t) = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad t \in (0,T),\\ y(x,0) = {y_0}(x),{\rm{ }}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad x \in (0,1), \end{array} \right.$

(1.1)

这里$ a\in\mathbb R (a\neq0) $, $ y $是状态函数, 给定函数$ b\in L^2(0, 1), $控制函数$ u\in L^2(0, T), $并且$ y_0\in L^2(0, 1) $是初始条件.从而, 方程(1.1)存在唯一的解$ y\in L^2(Q) $.系统(1.1)主要用来描述一类含有关于时间的非局部反应项的扩散现象^{[1, 2]}, 在物理、生物、天文、信息等领域有着广泛的应用.

当$ a = 0 $时, 系统(1.1)是经典抛物型偏微分方程, 它是近似能控且零能控, 这是一个熟知的结果.因此, 本文主要讨论的问题是系统(1.1)的近似能控性和零能控性.首先, 给出相关定义.

定义 1.1   对于任意$ y_0, y_1\in L^2(0, 1) $和$ \varepsilon>0 $, 存在控制$ u\in L^2(0, T) $使得(1.1)相应的解$ y $满足$ \lVert y(\cdot, T)-y_1\rVert_{L^2(0, T)}\leq\varepsilon, $则称系统(1.1)在$ T $时刻近似能控.

定义 1.2   对于任意$ y_0\in L^2(0, 1) $, 存在控制$ u\in L^2(0, T) $使得系统(1.1)相应的解$ y $满足$ y(x, T) = 0\quad{\rm a.e.\; }x\in(0, 1), $则称系统(1.1)在$ T $时刻零能控.

关于这类系统的能控性问题已有一些相关的讨论^[3-6]. 2000年, Barbu和Innelli ^[7]得到了如下具有二阶导数记忆项的系统的内部近似能控性

$ \begin{equation} y_t-\gamma \Delta y = \int_0^t a(t-s)\Delta y(x, s)ds+\chi_{\omega}u. \end{equation} $

(1.2)

为了应用Laplace变换给出对偶系统解的表达式, 需要假设记忆核$ a $满足一定的条件.在文献[8]中, 作者给出了记忆核$ a\equiv 1 $时, 受控系统(1.2)在$ L^2(0, 1) $空间中不零能控.随后, Halanay和Pandolfi^[9]将近似能控性和不零能控两个结果推广到一般记忆核的情况.以上讨论的结果都是建立在内部控制或者边界控制的基础上.2018年, 文献[10]讨论了在施加双线性控制下系统的零能控性, 利用对偶理论、泰勒展式和反证法得出当系统施加双线性控制时, 系统(1.1)在$ L^2(0, 1) $空间中不零能控.即存在初使条件$ y_0\in L^2(0, 1) $使得对于任意控制$ u\in L^2(0, T) $, 系统(1.1)相应的解$ y $在时刻$ T $都不能达到目标零.那么一个自然的问题是, 这样的结论在相对较小的$ H^m $空间中是否成立呢?本文给出了明确的答案.另一方面, 利用对偶原理给出在施加双线性控制下的近似能控性的充分必要条件, 它恰好是不零能控性的充分条件.

全文分为四部分, 第二部分给出控制系统关于近似能控性和零能控性的两个主要结果；第三部分主要应用对偶原理将近似能控性转化为相应对偶系统的唯一延拓性, 并得到相关结果；第四部分利用泰勒展式和反证法证明控制系统不零能控.

为了给出主要结果, 先引入一些记号.设

$ \begin{equation} \omega_j(x) = \sin(j\pi x), \quad\lambda_j = (j\pi)^2, \; j\in\mathbb{Z}^+, \end{equation} $

(2.1)

则$ \{\omega_j\}_{j\geq1} $构成空间$ L^2(0, 1) $中的一组正交基.系统(1.1)的对偶系统如下

$ \left\{ \begin{array}{l} {\varphi _t}(x,t) + {\varphi _{xx}}(x,t) + a\int_t^T {{\varphi _{xx}}} (x,s){\rm{d}}s = 0,\;\;\;\quad (x,t) \in Q,\\ \varphi (0,t) = \varphi (1,t) = 0,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad t \in (0,T),\\ \varphi (x,T) = {\varphi _T}(x),{\rm{ }}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\quad x \in (0,1).{\rm{ }} \end{array} \right. $

(2.2)

引理 2.1   设$ \varphi_T(x) = \sum\limits_{j\geq1}\beta_j\omega_j(x), \; \forall x\in(0, 1). $则对偶系统(2.2)的解可以表示为

$ \varphi(x, t) = \sum\limits_{j\geq1} \left(\tilde{C}_{1, j}e^{\tilde{r}_{1, j}(T-t)}+ \tilde{C}_{2, j}e^{\tilde{r}_{2, j}(T-t)}\right)\beta_j\omega_j(x), \quad\forall (x, t)\in\; Q, $

其中

$ \tilde{r}_{1, j} = \frac{-\lambda_j(1+\sqrt{1-4a/\lambda_j})}{2}, \quad\tilde{r}_{2, j} = \frac{-\lambda_j(1-\sqrt{1-4a/\lambda_j})}{2}, $

$ \tilde{C}_{i, j} = \frac{(-1)^{i-1}+ \sqrt{1-4a/\lambda_j}}{2\sqrt{1-4a/\lambda_j}}, \quad i = 1, 2. $

证  由于$ \varphi_T\in L^2(0, 1) $, 因此存在$ {\{\beta_j\}_{j\geq1}}\in l^2 $使得$ \varphi_T(x) = \sum\limits_{j\geq1}\beta_j\omega_j(x), \forall x\in(0, 1), $又由于系统(2.2)的解$ \varphi\in L^2(0, T;L^2(0, 1)) $, 因此可以表示为

$ \varphi(x, t) = \sum\limits_{j\geq1}\alpha_j(t)\omega_j(x), \forall x\in(0, 1)\times(0, T). $

代入系统(2.2)第一个方程有$ \sum\limits_{j\geq1}\omega_j(x) (\alpha^\prime_j(t)-\lambda_j\alpha_j(t)-a\lambda_j \int_t^T\alpha_j(s){\rm d}s) = 0. $因此关于$ t $的函数$ \alpha_j $满足

$\left\{ \begin{array}{l} \alpha _j^\prime (t) - {\lambda _j}{\alpha _j}(t) - a{\lambda _j}\int_t^T {{\alpha _j}} (s){\rm{d}}s\\ {\alpha _j}(T) = {\beta _j}.{\rm{ }} \end{array} \right.$

(2.3)

令$ \int_t^T \alpha_j(s){\rm d}s = \eta_j(t) $, 则$ \eta^\prime_j(t) = -\alpha_j(t) $, $ \alpha_j(t) = -\eta^\prime_j(t) $, 代入(2.3)式得

$ -\eta^{\prime\prime}_j(t)+\lambda_j\eta^\prime_j(t)-a\lambda_j\eta_j(t) = 0. $

因此特征方程为$ -r^2+\lambda_jr-a\lambda_j = 0 $, 解得

$ r_{1, j} = \frac{-\lambda_j-\sqrt{\lambda_j^2-4a\lambda_j}}{-2}, \quad r_{2, j} = \frac{-\lambda_j+\sqrt{\lambda_j^2-4a\lambda_j}}{-2}. $

设$ \eta_j(t) = {C}_{1, j}e^{{r}_{1, j}t}+{C}_{2, j}e^{{r}_{2, j}t}. $由于$ \eta_j(T) = 0, \eta^\prime_j(T) = -\alpha_j(T) = -\beta_j, $所以有

$\left\{ \begin{array}{l} {C_{1,j}}{e^{{r_{1,j}}T}} + {C_{2,j}}{e^{{r_{2,j}}T}} = 0,\\ {r_{1,j}}{C_{1,j}}{e^{{r_{1,j}}T}} + {r_{2,j}}{C_{2,j}}{e^{{r_{2,j}}T}} = - {\beta _j}. \end{array} \right. $

解线性方程组得

$ C_{1, j} = -\frac{\beta_j}{r_{1, j}-r_{2, j}}e^{-r_{1, j}T}, \quad C_{2, j} = \frac{\beta_j}{r_{1, j}-r_{2, j}}e^{-r_{2, j}T}. $

因此

$ \begin{eqnarray*} \alpha_j(t)& = &-\eta'_j(t) = -r_{1, j}C_{1, j}e^{r_{1, j}t}- r_{2, j}C_{2, j}e^{r_{2, j}t}\\ &{ = }&\beta_j\frac{\lambda_j+\sqrt{\lambda_j^2-4a\lambda_j}} {2\sqrt{\lambda_j^2-4a\lambda_j}} e^{\frac{-\lambda_j-\sqrt{\lambda_j^2-4a\lambda_j}} {2}(T-t)}+\beta_j\frac{-\lambda_j+\sqrt{\lambda_j^2-4a\lambda_j}} {2\sqrt{\lambda_j^2-4a\lambda_j}} e^{\frac{-\lambda_j+\sqrt{\lambda_j^2-4a\lambda_j}} {2}(T-t)}. \end{eqnarray*} $

令

$ \tilde{r}_{1, j} = \frac{-\lambda_j(1+\sqrt{1-4a/\lambda_j})}{2}, \quad\tilde{r}_{2, j} = \frac{-\lambda_j(1-\sqrt{1-4a/\lambda_j})}{2}, $

$ \tilde{C}_{i, j} = \frac{(-1)^{i-1}+\sqrt{1-4a/\lambda_j}} {2\sqrt{1-4a/\lambda_j}}, \quad i = 1, 2. $

于是方程(2.2)的解可以表示为

$ \begin{equation} \varphi(x, t) = \sum\limits_{j\geq1}\left(\tilde{C}_{1, j}e^{\tilde{r}_{1, j}(T-t)}+\tilde{C}_{2, j}e^{\tilde{r}_{2, j}(T-t)}\right)\beta_j\omega_j(x). \end{equation} $

(2.4)

引理2.1得证.

对于任意给定$ m\in\mathbb{N} $, 定义Sobolev空间

$ H^m(0, 1) = \left\{f\in L^2(0, 1)\mid f(x) = \sum\limits_{j = 1}^\infty f_j\omega_j(x), \sum\limits_{j = 1}^\infty\lambda_j^m|f_j|^2<\infty\right\}. $

显然$ H^0(0, 1) = L^2(0, 1), $并且$ H^{m_2}(0, 1)\subset H^{m_1}(0, 1), \forall m_1\leq m_2. $全文设$ C $为只与$ T $有关的常数.主要结果是下面的两个定理.

定理 2.1  下面三种陈述等价

(i)  系统(1.1)在$ T $时刻近似能控;

(ii)  设$ \varphi $是对偶系统对应于$ \varphi_T $的解, 则

$ \int_0^T\int_0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t = 0, \quad\forall u\in L^2(0, T) \Rightarrow\varphi_T = 0; $

(iii)  $ b_j = \int_0^1b(x)\omega_j(x){\rm d}x\neq0, \forall j\geq1. $

定理 2.2  任意给定$ m\in\mathbb{N} $, 设$ b_j = \int_0^1b(x)\omega_j(x){\rm d}x\neq0, \forall j\geq1, $则存在初值$ y_0\in H^m(0, 1) $使得对于任意$ u\in L^2(0, T) $, 系统(1.1)相应的解都不满足

$ \begin{equation} y(x, T) = 0\quad{\rm a.e.\; }x\in(0, 1). \end{equation} $

(2.5)

注  定理2.2中的条件不能去掉, 因为它与近似能控性等价.

证  第一步  证明(i)与(ii)等价.

充分性  要证明的结论是系统(1.1)的近似能控性.不失一般性, 假设$ y_0(x)\equiv0. $由文献[10]可得

$ \begin{equation} \int_0^1\varphi_T(x)y(x, T){\rm d}x = \int_0^T\int_0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t. \end{equation} $

(3.1)

设能达集$ R(T) = \{y(\cdot, T)\mid y\text{是系统}(1.1)\text{中对应于}u\in L^2(0, T)\text{的解}\}. $由定义可知, 系统(1.1)在$ T $时刻近似能控当且仅当$ R(T) $在$ L^2(0, 1) $中稠密.采用反证法, 假设系统(1.1)在$ T $时刻不近似能控, 则$ R(T) $不在空间$ L^2(0, 1) $中稠密, 由Hahn-Banach定理得, 存在$ \varphi_T\in L^2(0, 1), \varphi_T\neq0 $使下式成立

$ \int_0^1y(x, T)\varphi_T(x){\rm d}x = 0, \quad\forall y\in(\cdot, T)\in R(T). $

由(3.1)式可得

$ \int_0^T\int_0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t = 0, \quad\forall u\in L^2(0, T). $

因此$ \varphi_T = 0, $这与假设$ \varphi_T $的选取产生矛盾, 故假设不成立, 充分性得证.

必要性  设

$ \int_0^T\int_0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t = 0, \quad\forall u\in L^2(0, T). $

由(3.1)式可得

$ \int_0^1y(x, T)\varphi_T(x){\rm d}x = 0, \quad\forall y\in(\cdot, T)\in R(T). $

由于$ \overline{R(T)} = L^2(0, 1), $所以$ \varphi_T = 0. $综上(i)与(ii)等价得证.

第二步  证明(ii)与(iii)等价.

充分性  假设

$ \begin{equation} \int_0^T\int_0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t = 0, \quad\forall u\in L^2(0, T). \end{equation} $

(3.2)

由(ii)可知, 只需要证明$ \beta_j = 0, \forall j\geq1 $即可.将(2.4)式代入(3.2)式得

$ 0 = \int_0^T\int_0^1b(x)\sum\limits_{j\geq1} \left(\tilde{C}_{1, j}e^{\tilde{r}_{1, j}(T-t)}+\tilde{C}_{2, j} e^{\tilde{r}_{2, j}(T-t)}\right)\beta_j\omega_j(x)u(t){\rm d}x{\rm d}t $

$ \; \; \; \; = \sum\limits_{j\geq1}\beta_j\int_0^T\int_0^1b(x)\omega_j(x){\rm d}x \left(\tilde{C}_{1, j}e^{\tilde{r}_{1, j}(T-t)}+\tilde{C}_{2, j}e^{\tilde{r}_{2, j}(T-t)}\right) u(t){\rm d}t $

$ = \int_0^T\sum\limits_{j\geq1}\beta_j b_j \left(\tilde{C}_{1, j}e^{\tilde{r}_{1, j}(T-t)}+\tilde{C}_{2, j} e^{\tilde{r}_{2, j}(T-t)}\right)u(t){\rm d}t. $

从而

$ \sum\limits_{j\geq1}\beta_j b_j\left(\frac{1+\sqrt{1-4a/\lambda_j}}{2\sqrt{1-4a/\lambda_j}} e^{\frac{-\lambda_j(1+\sqrt{1-4a/\lambda_j})}{2}(T-t)} +\frac{-1+\sqrt{1-4a/\lambda_j}}{2\sqrt{1-4a/\lambda_j}} e^{\frac{-\lambda_j(1-\sqrt{1-4a/\lambda_j})}{2}(T-t)}\right) = 0. $

因此由文献[11]可得$ \beta_jb_j = 0, \; \forall j\geq 1. $又由$ b_j\neq 0, $可得$ \beta_j = 0, \; \forall j\geq 1. $即$ \varphi_T = 0. $故当$ b_j\neq 0, \; \forall j\geq 1 $, 系统(1.1)在$ T $时刻近似能控.

必要性  假设存在$ j_0\in \mathbb {N}, $使得$ b_{j_0} = 0 $, 则利用充分性的证明过程得不到$ \beta_{j_0} $一定为0, 故$ \varphi_T\neq 0 $.这与条件矛盾, 所以必要性得证.故(ii)与(iii)等价, 综上定理2.1得证.

注  这一结论与经典抛物方程$ (a = 0) $相同.

证  第一步  构造对偶系统(2.2)终端时刻的是一个充分大的正整数值.任意给定正整数$ m $, 设$ p>m+\frac{5}{2} $是一个正整数.令正整数$ N>\frac{3p-1}{2} $, $ M $是一个充分大的正整数, 且

$ \varphi_T(x) = \sum\limits_{k = 1}^N\beta_{NM+k}\omega_{NM+k}(x), \quad\forall x\in(0, 1), $

其中$ \{{\beta_{NM+k}}\}_{k = 1}^N $满足

$ \begin{equation} \sum\limits_{k = 1}^N b_{NM+k} \beta_{NM+k} (\frac{1+\sqrt{1-4a/\lambda_{NM+k}}}{2\sqrt{1-4a/\lambda_{NM+k}}} (\tilde{r}_{2, {NM+k}}+k^2\pi^2+2NMk\pi^2)^l = 0, l = {0, 1, \cdots{p-1}, } \end{equation} $

(4.1)

并且

$ \begin{equation} \sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\lambda_{NM+k}^{-i} = 0, \quad i = {1, 2\cdots, [\frac{p+1}{2}]-1}. \end{equation} $

(4.2)

联立(4.1)和(4.2)式是关于$ N $个未知数$ \{\beta_{NM+k}\}_{k = 1}^N $的$ p+[\frac{p-1}{2}] $个方程的代数方程组, 所以存在界与$ M $无关的解$ \{\beta_{NM+k}\}_{k = 1}^N $.进一步, 由引理2.1得, 方程(2.2)的解为

$ \begin{equation} \varphi(x, t) = \sum\limits_{k = 1}^N\beta_{NM+k} (\tilde{C}_{1, {NM+k}}e^{\tilde{r}_{1, {NM+k}}(T-t)} +\tilde{C}_{2, {NM+k}}e^{\tilde{r}_{2, {NM+k}}(T-t)})\omega_{NM+k}(x), \forall(x, t)\in Q, \end{equation} $

(4.3)

其中

$ \tilde{r}_{i, {NM+k}} = \frac{-\lambda_{NM+k}}{2} (1\pm\sqrt{1-4a/\lambda_{NM+k}}), \quad i = 1, 2, $

$ \tilde{C}_{i, {NM+k}} = \frac{(-1)^{i-1}+\sqrt{1-4a/\lambda_{NM+k}}} {2\sqrt{1-4a/\lambda_{NM+k}}}, \quad i = 1, 2. $

第二步   给出系统(1.1)零能控的充要条件.由文献[10]可得, 系统(1.1)在时刻$ T $零能控的充要条件是, 对于任意$ y_0\in L^2(0, 1) $存在控制$ u\in L^2(0, T), $使得下述的等式成立

$ \begin{equation} \int_0^T\int_0^1b(x)\varphi(x, t){\rm d}x{\rm d}t = -\int_0^1y_0(x)\varphi(x, 0){\rm d}x, \; \forall\varphi_T\in L^2(0, 1), \end{equation} $

(4.4)

其中$ \varphi $是对偶系统(2.2)对应于$ \varphi_T $的解.

第三步  估计(4.4)式的左端.由(4.3)式可得

$ \begin{eqnarray*} && \int_0^T| \int_0^1b(x)\varphi(x, t){\rm d}x|^2{\rm d}t \\ & = &\int_0^T|\int_0^1\sum \limits_{k = 1}^N\beta_{NM+k}(\tilde{C}_{1, NM+k}e^{\tilde{r}_{1, NM+k}(T-t)}+ \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}(T-t)})\omega_{NM+k}(x) b(x){\rm d}x|^2{\rm d}t\\ & = & \int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}(\tilde{C}_{1, NM+k}e^{\tilde{r}_{1, NM+k}(T-t)}+ \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}(T-t)})|^2{\rm d}t\\ &\leq&2\int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}e^{\tilde{r}_{1, NM+k}t}|^2{\rm d}t\\ &&+2\int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k} \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}t}|^2{\rm d}t. \end{eqnarray*} $

记上式最后两项分别为$ 2E_1, 2E_2. $注意到$ \tilde{r}_{1, NM+k}+ \tilde{r}_{2, NM+k} = -\lambda_{NM+k}, $从而

$ \begin{eqnarray} E_1& = &\int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}e^{\tilde{r}_{1, NM+k}t}|^2{\rm d}t\\ & = &\int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}e^{-(\tilde{r}_{2, NM+k}+ \lambda_{NM+k})t}|^2{\rm d}t\\ & = &\int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}e^{(-\tilde{r}_{2, NM+k}- 2NMk\pi^2-k^2\pi^2)t}e^{-N^2M^2\pi^2t}|^2{\rm d}t\\ & = &\int_0^T e^{-2N^2M^2\pi^2t}|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}e^{(-\tilde{r}_{2, NM+k}- 2NMk\pi^2-k^2\pi^2)t}|^2{\rm d}t. \end{eqnarray} $

(4.5)

令$ h_M(t) = \sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}e^{(-\tilde{r}_{2, NM+k}- 2NMk\pi^2-k^2\pi^2)t}, $且$ g_M(t) = h_M^2(t) $从而有,

$ h_M^{(r)}(t) = \sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{1, NM+k}\left(-\tilde{r}_{2, NM+k}- 2NMk\pi^2-k^2\pi^2\right)^r e^{(-\tilde{r}_{2, NM+k}- 2NMk\pi^2-k^2\pi^2)t}. $

因此$ |h_M^{(r)}(t)|\leq CM^r. $由(4.1)式可得, 当$ 0\leq r\leq p-1 $时, 有$ h_M^{(r)}(0) = 0. $进一步, 存在$ r $个常数$ \{c_i\}_{i = 1}^r $使得$ g_M^{(r)}(t) = c_1 h_M(t)h_M^{(r)}(t)+c_2 h^\prime_M(t)h_M^{(r-1)}(t)+\cdots +c_r h_M^{(r)} (t)h_M(t). $于是有

$ \begin{equation} |g_M^{(r)}(t)|\leq CM^r\text{且}g_M^{(r)}(0) = 0, \quad\forall0\leq r\leq 2p-2. \end{equation} $

(4.6)

从而结合(4.5)式和(4.6)式并分部积分可得

$ \begin{eqnarray} E_1& = &-\sum\limits_{l = 0}^{2p-3}\frac{e^{-2N^2M^2\pi^2T}g_M^{(l)}(T)} {(2N^2M^2\pi^2)^{l+1}}+\int_0^T\frac{e^{-2N^2M^2\pi^2t}} {(-2N^2M^2\pi^2)^{2p-2}}g_M^{(2p-2)}(t){\rm d}t\\ &\leq&C e^{-2N^2M^2\pi^2T}\sum\limits_{l = 0}^{2p-3}\frac{M^l}{M^{2l+2}}+ \frac{CM^{2p-2}}{{(2N^2M^2\pi^2)}^{2p-2}}\int_0^T e^{-2N^2M^2\pi^2t}{\rm d}t \leq\frac{C}{M^{2p}}. \end{eqnarray} $

(4.7)

将$ \frac{1}{\sqrt{1-4a/\lambda_{NM+k}}} $泰勒展开代入得

$ \begin{eqnarray*} \tilde{C}_{2, NM+k}& = &-\frac{1}{2}\times (1+\sum\limits_{n = 2}^{\infty} \frac{(2n-3)!!}{(n-1)!}\cdot(\frac{2a}{\lambda_{NM+k}})^{n-1}) +\frac{1}{2}\\ & = &-\sum\limits_{n = 2}^{\infty} \frac{(2n-3)!!\cdot2^{n-2}}{(n-1)!}\cdot(\frac{a}{\lambda_{NM+k}})^{n-1}. \end{eqnarray*} $

同理将$ \sqrt{1-4a/\lambda_{NM+k}} $泰勒展开代入得

$ \begin{eqnarray*} \tilde{r}_{2, NM+k}& = &-\frac{\lambda_{NM+k}}{2}[1-(1-\frac{2a} {\lambda_{NM+k}} -\cdots-\frac{(2a)^{[\frac{p+1}{2}]}(2[\frac{p+1}{2}]-3)!!} {[\frac{p+1}{2}]!\lambda_{NM+k}^{[\frac{p+1}{2}]}}+ O(\lambda_{NM+k}^{-[\frac{p+1}{2}]-1}))]\\ & = &-a-\sum\limits_{n = 1}^{[\frac{p+1}{2}]-1}\frac{(2a)^{n+1}(2n-1)!!} {2\times(n+1)!\lambda_{NM+k}^n}+O(\lambda_{NM+k}^{-[\frac{p+1}{2}]}), \end{eqnarray*} $

于是存在一列关于$ t $的函数$ \{C_i(\cdot)\}_{i = 1}^{[p+1]/2} $使得

$ \tilde{C}_{2, {NM+k}}e^{\tilde{r}_{2, NM+k}t} = e^{-at}(\frac{C_1(t)} {\lambda_{NM+k}}+\cdots+\frac{C_{[(p-1)/2]}(t)} {\lambda_{NM+k}^{[(p-1)/2]}}+\frac{C_{[(p+1)/2]}(t)} {\lambda_{NM+k}^{[(p+1)/2]}}+O(\lambda_{NM+k}^{-[\frac{p+1}{2}]-1})). $

由上式及(4.2)式可得

$ \begin{eqnarray} E_2& = &\int_0^T|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}\tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}t}|^2{\rm d}t\\ & = &\int_0^T e^{-2at}|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}(\frac{C_1(t)} {\lambda_{NM+k}}+\cdots+\frac{C_{[(p+1)/2]}(t)} {\lambda_{NM+k}^{[(p+1)/2]}}+O(\lambda_{NM+k}^ {-[\frac{p+1}{2}]-1}))|^2{\rm d}t\\ & = &\int_0^T e^{-2at}|\sum\limits_{k = 1}^N b_{NM+k}\beta_{NM+k}(\frac{C_{[(p+1)/2]}(t)} {\lambda_{NM+k}^{[(p+1)/2]}}+O(\lambda_{NM+k}^ {-[\frac{p+1}{2}]-1}))|^2{\rm d}t\\ &\leq&CM^{-[(p+1)/2]\times 4}\int_0^T e^{-2at}{\rm d}t\leq\frac{C}{M^{[(p+1)/2]\times 4}}. \end{eqnarray} $

(4.8)

由(4.7)和(4.8)式可得

$ \int _0^T|\int _0^1b(x)\varphi(x, t){\rm d}x|^2{\rm d}t\leq 2E_1+2E_2\leq \frac{C}{M^{2p}}. $

第四步  估计(4.4)式的右端.注意到

$ \lim\limits_{M\rightarrow\infty}\tilde{C}_{1, NM+k} = \lim\limits_ {M\rightarrow\infty}\frac{1+\sqrt{1-4a/\lambda_{NM+k}}} {2\sqrt{1-4a/\lambda_{NM+k}}} = 1, $

$ \lim\limits_{M\rightarrow\infty}\frac{\tilde{r}_{1, NM+k}} {-\lambda_{NM+k}} = \lim\limits_ {M\rightarrow\infty}\frac{1+\sqrt{1-4a/\lambda_{NM+k}}}{2} = 1. $

进而由(4.4)式可得

$ \begin{eqnarray*} \int_0^1|\varphi(x, 0)|^2{\rm d}x & = &\int_0^1|\sum \limits_{k = 1}^N\beta_{NM+k}(\tilde{C}_{1, NM+k}e^{\tilde{r}_ {1, NM+k}T}+ \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}T})\omega_{NM+k}(x) |^2{\rm d}x\\ & = &\frac{1}{2}\sum \limits_{k = 1}^N\beta_{NM+k}^2(\tilde{C}_{1, NM+k}e^{\tilde{r}_ {1, NM+k}T}+ \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}T})^2\\ &\geq&\frac{1}{2}\sum \limits_{k = 1}^N\beta_{NM+k}^2(\frac{3}{4}\tilde{C}_{2, NM+k}^2 e^{2\tilde{r}_{2, NM+k}T}-3 \tilde{C}_{1, NM+k}^2e^{2\tilde{r}_{1, NM+k}T})\\ &\geq&\frac{1}{2}\sum \limits_{k = 1}^N[\frac{3}{4}\beta_{NM+k}^2e^{-2aT} (\frac{a^2}{\lambda_{NM+k}^2}+O({\lambda_{NM+k}^{-3}})) -3\beta_{NM+k}^2e^{-2\lambda_{NM+k}T}]\\ &\geq&C(\frac{1}{M^4}-\frac{1}{M^6}-e^{-CM^2})\geq\frac{C}{M^4}. \end{eqnarray*} $

故存在$ 1\leq {k_0}\leq N $使得

$ \begin{equation} \frac{1}{2}{\beta^2_{NM+k_0}}\left(\tilde{C}_{1, NM+k_0}e^{\tilde{r}_ {1, NM+k_0}T}+ \tilde{C}_{2, NM+k_0}e^{\tilde{r}_{2, NM+k_0}T}\right)^2\geq\frac{C}{NM^4}. \end{equation} $

(4.9)

第五步  构造适当初值$ y_0\in H^m(0, 1) $, 得到结论, 即定理2.2.取$ y_0 = \sum\limits_{l\geq1}\frac{1}{l^\delta}\omega_{Nl+k_0}, $ 则$ 2\delta-2m>1. $于是

$ \begin{eqnarray*} &&| \int_0^1y_0(x)\varphi(x, 0){\rm d}x|\\ & = &(\sum \limits_{l\geq1}\frac{1}{l^\delta}\omega_{Nl+k_0}, \sum \limits_{k = 1}^N\beta_{NM+k}(\tilde{C}_{1, NM+k}e^{\tilde{r}_ {1, NM+k}T}+ \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}T})\omega_{NM+k})\\ & = &\sum \limits_{k = 1}^N(\sum \limits_{l\geq1}\frac{1}{l^\delta}\omega_{Nl+k_0}, \beta_{NM+k}(\tilde{C}_{1, NM+k}e^{\tilde{r}_ {1, NM+k}T}+ \tilde{C}_{2, NM+k}e^{\tilde{r}_{2, NM+k}T})\omega_{NM+k} ). \end{eqnarray*} $

设$ Nl+k_0 = NM+k, $即$ N(l-M) = k-k_0. $由$ |k-k_0|<N, N|l-M|\in\{0, N, 2N, \cdots\}, $ 故$ l = M, k = k_0. $所以由(4.9)式得

$ |\int_0^1y_0(x)\varphi(x, 0){\rm d}x| = \frac{1}{2}\frac{1}{M^\delta}{\beta_{NM+k_0}}|\tilde{C}_{1, NM+k_0}e^{\tilde{r}_ {1, NM+k_0}T}+ \tilde{C}_{2, NM+k_0}e^{\tilde{r}_{2, NM+k_0}T}|\geq\frac{C}{NM^{2+\delta}}. $

假设(1.1)在时刻$ T $零能控, 则由证明的第二步可得, 存在$ u\in L^2(0, T) $使得

$ \int_0^1y_0(x)\varphi(x, 0){\rm d}x = -\int _0^T\int _0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t. $

由H$ {\rm\ddot{o}} $lder不等式可知

$ \begin{eqnarray*} |\int _0^T\int _0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t| &\leq&\parallel u\parallel(\int _0^T|\int _0^1b(x)\varphi(x, t){\rm d}x|^2{\rm d}t)^\frac{1}{2}\\ &\leq&\parallel u\parallel(2E_1+2E_2)^\frac{1}{2}\leq \frac{C}{M^p}. \end{eqnarray*} $

也就是说, 存在于$ M $无关的两个常数$ C_1, C_2 $使得

$ \frac{C_1}{M^{2+\delta}}\leq|\int_0^1y_0(x)\varphi(x, 0){\rm d}x| = |\int _0^T\int _0^1b(x)\varphi(x, t)u(t){\rm d}x{\rm d}t|\leq\frac{C_2}{M^p}. $

只要$ 2+\delta<p, $即$ m+\frac{1}{2}<\delta<p-2 $时, 上式对于充分大的$ M $是矛盾的, 即给定一个$ m $ 的值, 可找到属于空间$ H^m $的一类初值使得系统(1.1)不零能控, 证毕.