具有非局部项的p-Laplace方程的边界最优控制

Boundary Optimal Control of the p-Laplace Equation with Non-Local Terms

摘要: 本文研究了一类具有非局部项的p-Laplace方程的边界最优控制问题，通过对成本泛函的极小化序列取极限给出p-Laplace方程初边值问题最优控制函数的存在性。首先利用能量估计方法研究该问题解的存在唯一性，其次利用紧性估计和紧嵌入定理分析成本泛函极小化序列的收敛性，最后由成本泛函的弱下半连续性证明最优控制函数的存在性。

Abstract: In this paper, we study the boundary optimal control problem of a class of p-Laplace equations with non-local terms, and the existence of the optimal control function of the initial boundary value problem of the p-Laplace equation is given by taking the limit of the minimization sequence of the cost function. Firstly, the energy estimation method is used to study the existence uniqueness of the solution of the problem, then the tightness estimation and the tight embedding theorem are used to analyze the convergence of the cost functional minimization sequence, and finally the existence of the optimal control function is proved by the weak lower semi-continuity of the cost function.

1. 引言

最优控制理论是基于经典的变分学而发展起来的，其来源可以追溯到三百多年前。关于最优控制早期的发展情况可参阅文献[1]-[3]。最优控制问题是人们在实际生活中常见的一类问题，事实上就是一种求最优解的问题。关于最优控制问题在生活中的应用研究不断涌现[4] [5]。最优控制可以提高资源的利用率，减少不必要的浪费，在机器人[6]、车辆[7]、航空[8] [9]以及医疗化工[10]等专业领域也有着广泛的应用。与此同时，许多研究者针对不同类型系统的最优控制问题开展了大量研究。

带约束条件的偏微分方程最优控制问题可以广泛应用于流体控制、动物种群演化等领域的数学模型中。因此，解决这些问题对于航空航天、生物保护等领域都具有重要意义和价值。作为一个数学分支，这些问题与偏微分方程、控制理论、优化、数值计算和泛函分析密切相关，是一个活跃且具有重要理论意义和应用价值的研究领域。

文献[11]研究了发展型p-Laplace方程

$\{\begin{array}{l}\frac{\partial u}{\partial t}-div\left({\left|\nabla u\right|}^{p-2}\nabla u\right)+c\left(x,t\right)u=0,\left(x,t\right)\in Q_T=\Omega \times \left(0,T\right),\\ {\left|\nabla u\right|}^{p-2}\nabla u\cdot n=-hu,\left(x,t\right)\in \partial \Omega \times \left(0,T\right),\\ u\left(x,0\right)=u_0\left(x\right),x\in \Omega \end{array}$

的边界最优控制问题，利用能量估计方法研究了该问题解的存在唯一性，并通过紧性估计和紧嵌入定理分析了成本泛函极小化序列的收敛性，最终证明了最优控制的存在性。

近年来，非局部问题受到研究者们的重视，在种群动力学，生物学[12] [13]和物理学[14] [15]等领域广泛的应用。以Laplace算子作为扩散项的经典反应扩散方程仅能反映空间上的局部行为，即相邻空间上位置的移动。例如物质从浓度高的部分向浓度低的部分进行扩散。实际上，在自然界中，受多种因素影响，空间中的非局部行为是普遍存在的。例如，生物种群的位置移动涉及范围较大，因此需用非局部扩散进行描述。文献[16]研究了一类是由含非局部算子的变分不等式约束的最优控制问题。

据笔者了解，对于具有非局部项的p-Laplace方程的边界最优控制问题目前还没有任何结果，本文受文献[11]和[16]的启发，考虑具有非局部项的p-Laplace方程

$\{\begin{array}{l}\frac{\partial y}{\partial t}-div\left({\left|\nabla y\right|}^{p-2}\nabla y\right)+a\left(l\left(y\left(t\right)\right)\right)y=0,\left(x,t\right)\in \Omega \times \left(0,T\right),\\ {\left|\nabla y\right|}^{p-2}\nabla y\cdot v=-hy,\left(x,t\right)\in \partial \Omega \times \left(0,T\right),\\ y\left(x,0\right)=y_0\left(x\right),x\in \Omega \end{array}$

的边界最优控制问题，其中$\Omega \in R^n\left(n\ge 1\right)$ 是具有光滑边界$\partial \Omega$ 的有界域，$p\ge 2$ ，$a(\cdot )$ 是非负非减函数且$a(\cdot )\le M$ ，$a\left(l\left(y\left(t\right)\right)\right)={\displaystyle {\int }_{\Omega }{y}^q\left(x,t\right)\text{d}x}$ ，$q\ge 1$ ，h是控制函数，$v$ 是$\partial \Omega$ 上的单位外法向量，初值$y^0\in L^2\left(\Omega \right)$ 。

设

$U_M:=\left\{h|0\le h\le M,h\in L^{\infty }\left(\partial \Omega \times \left(0,T\right)\right)\right\}$

是允许控制集，定义成本泛函为

$J\left(h\right):=\frac{1}{2}{\displaystyle {\int }_Q{\left|y-y_d\right|}^2\text{d}x\text{d}t}+\frac{\mu }{2}{\displaystyle {\int }_{{\omega }_T}{\left|h\right|}^2\text{d}x\text{d}t},$

其中$\omega$ 是$\Omega$ 的一个非空子集，$W_T=\omega \times \left(0,T\right)$ ，$Q:=\Omega \times \left(0,T\right)$ ，$y_d\in L^{\infty }\left(Q\right)$ ，$\mu >0$ ，本文主要结果是证明最优控制的存在性，

即证存在$h^{\ast }\in U_M$ ，使得

$J\left(h^{\ast }\right)=\underset{h\in U_M}{\inf }J\left(h\right).$

为此，首先利用能量估计方法研究该问题解的存在唯一性，其次利用紧性估计和紧嵌入定理分析成本泛函极小化序列的收敛性，最后由成本泛函的弱下半连续性证明最优控制函数的存在性。

2. 准备工作

本节给出一些定义及引理。

定义1.1 令$V=W^{1,p}\left(\Omega \right)$ ，$V^{\prime }$ 表示V的对偶空间，$\langle u,v\rangle \left(u\in V^{\prime },v\in V\right)$ 表示$V^{\prime }-V$ 的对偶积。

定义1.2 如果对$\forall \varphi \in H\left(Q\right):=\left\{\varphi \in W\left(0,T\right):\varphi \left(\cdot ,T\right)=0\right\}$ ，若

$-{\displaystyle {\int }_0^T\langle {\varphi }_t,y\rangle \text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y\right|}^{p-2}\nabla y\nabla \varphi \text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }a\left(l\left(y\left(t\right)\right)\right)\varphi y\text{d}x\text{d}t}+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }hy\phi }}={\displaystyle {\int }_{\Omega }{y}^0\varphi \left(\cdot ,0\right)\text{d}x}$

成立，则称函数$y\in L^2\left(0,T;H^1\left(\Omega \right)\right)$ 为问题(1)的弱解，其中$y^0\in L^2\left(\Omega \right)$ ，$\tau \in \left(0,T\right)$ 。

引理1 [1]对任意$u,v\in L^p\left(0,T;W^{1,p}\left(\Omega \right)\right)$ ，有

${\displaystyle \underset{Q_{\tau }}{\iint }\left({\left|\nabla u\right|}^{p-2}\nabla u-{\left|\nabla v\right|}^{p-2}\nabla v\right)\cdot \nabla \left(u-v\right)\text{d}x\text{d}t}\ge 0.$

3. 状态方程的适定性

研究(1)的初边值问题，定义成本泛函为：

$J\left(h\right)=\frac{1}{2}\left(\beta {\displaystyle \underset{Q_T}{\iint }{\left(y-z_d\right)}^2}\text{d}x\text{d}t+\gamma {\displaystyle {\int }_0^T{\displaystyle {\int }_{\partial \Omega }{h}^2}\text{d}S\text{d}t}\right),h\in U_M.$

主要结果是证明最优控制的存在性，即存在$h^{\ast }\in U_M$ ，使得$J\left(h^{\ast }\right)=\underset{h\in U_M}{\inf }J\left(h\right)$ 。

定义2.1 称非负函数$y\in C\left(0,T;L^2\left(\Omega \right)\right)\cap L^p\left(0,T;V\right)$ 为问题(1)的弱解，

若$\forall \phi \in C\left(0,T;L^2\left(\Omega \right)\right)\cap L^p\left(0,T;V\right)\cap L^2\left(0,T;V^{\prime }\right)$ ，有如下积分等式成立：

$\begin{array}{l}{\displaystyle {\int }_{\Omega }y\left(x,\tau \right)\phi }\left(x,\tau \right)\text{d}x+{\displaystyle {\iint }_{Q_{\tau }}{\left|\nabla y\right|}^{p-2}}\nabla y\nabla \phi \text{d}x\text{d}t+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }hy\phi \text{d}S\text{d}t}}\\ +{\displaystyle {\iint }_{Q_{\tau }}{y}^q\left(x,\tau \right)y\phi \text{d}x\text{d}t}-{\displaystyle {\int }_{\Omega }{y}_0\left(x\right)\phi }\left(x,0\right)\text{d}x-{\displaystyle {\int }_0^{\tau }\langle {\phi }_t,y\rangle \text{d}t}=0,\end{array}$

其中$\tau \in \left(0,T\right)$ 。

命题1. 设${\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\to {\omega }_1$ 弱收敛于$L^{\frac{p}{p-1}}\left(Q_T,R^n\right)$ ，${\left|\nabla y_k\right|}^{p-2}\nabla y_k\to {\omega }_2$ 弱收敛于$L^{\frac{p}{p-1}}\left(Q_T,R^n\right)$ ，则${\omega }_1={\omega }_2=\omega$ 。

证明 由牛顿莱布尼茨公式可知

${\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k={\left|\nabla y_k\right|}^{p-2}\nabla y_k+\frac{p-2}{2k}{\displaystyle {\int }_0^1{\left({\left|\nabla y_k\right|}^2+\frac{s}{k}\right)}^{\frac{p-4}{2}}}\text{d}s\nabla y_k,$

当$2<p<4$ 时，

$\begin{array}{c}{\displaystyle {\int }_0^1{\left({\left|\nabla y_k\right|}^2+\frac{s}{k}\right)}^{\frac{p-4}{2}}}\text{d}s\le {\displaystyle {\int }_0^1{\left(\frac{s}{k}\right)}^{\frac{p-2}{2}}\text{d}s}=k\cdot {\displaystyle {\int }_0^1{\left(\frac{s}{k}\right)}^{\frac{p-4}{2}}\text{d}}\left(\frac{s}{k}\right)={k\cdot \frac{1}{\frac{p-4}{2}+1}{\left(\frac{s}{k}\right)}^{\frac{p-4}{2}+1}|}_0^1\\ ={k\cdot \frac{2}{p-2}{\left(\frac{s}{k}\right)}^{\frac{p-2}{2}}|}_0^1=\frac{2}{p-2}k\cdot {\left(\frac{1}{k}\right)}^{\frac{p-2}{2}}.\end{array}$

利用Hölder不等式有

$\begin{array}{l}\left|{\displaystyle \underset{Q_{\tau }}{\iint }\frac{p-2}{2k}}{\displaystyle {\int }_0^1{\left({\left|\nabla y_k\right|}^2+\frac{s}{k}\right)}^{\frac{p-4}{2}}}\text{d}s\nabla y_k\cdot \phi \text{d}x\text{d}t\right|\\ \le \frac{p-2}{2k}\cdot \frac{2}{p-2}k\cdot {\left(\frac{1}{k}\right)}^{\frac{p-2}{2}}{\displaystyle \underset{Q_{\tau }}{\iint }\nabla y_k\cdot \phi }\text{d}x\text{d}t\le {\left(\frac{1}{k}\right)}^{\frac{p-2}{2}}{\Vert \nabla y_k\Vert }_{L^p\left(Q_T,R^n\right)}\to 0,\left(k\to \infty \right),\\ \forall \nabla y_k\in L^p\left(Q_T\right),\phi \in L^{\frac{p}{p-1}}\left(Q_T\right),\end{array}$

当$p>4$ 时，同理，由Hölder不等式有

$\begin{array}{l}\left|{\displaystyle \underset{Q_{\tau }}{\iint }\frac{p-2}{2k}}{\displaystyle {\int }_0^1{\left({\left|\nabla y_k\right|}^2+\frac{s}{k}\right)}^{\frac{p-4}{2}}}\text{d}s\nabla y_k\cdot \phi \text{d}x\text{d}t\right|\\ \le \left|{\displaystyle \underset{Q_{\tau }}{\iint }\frac{p-2}{2k}}{\displaystyle {\int }_0^1\left[{\left|\nabla y_k\right|}^{p-4}+{\left(\frac{1}{k}\right)}^{\frac{p-4}{2}}\right]\text{d}s}\nabla y_k\cdot \phi \text{d}x\text{d}t\right|\\ \le {\displaystyle \underset{Q_{\tau }}{\iint }\frac{p-2}{2k}}\left[{\left|\nabla y_k\right|}^{p-3}\left|\phi \right|+{\left(\frac{1}{k}\right)}^{\frac{p-4}{2}}\nabla y_k\left|\phi \right|\right]\text{d}x\text{d}t\\ \le \frac{p-2}{2k}{\Vert \nabla y_k\Vert }_{L^p}^{p-3}{\Vert \phi \Vert }_{L^{\frac{p}{3}}}+{\left(\frac{1}{k}\right)}^{\frac{p-4}{2}}{\Vert \nabla y_k\Vert }_{L^p}{\Vert \phi \Vert }_{L^{\frac{p}{p-1}}}\to 0,\left(k\to \infty \right),\end{array}$

所以$\forall \phi \in L^p\left(Q_T,R^n\right)$ ，

$\underset{n\to \infty }{\lim }{\displaystyle \underset{Q_T}{\iint }{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\cdot \phi \text{d}x\text{d}t}-{\displaystyle \underset{Q_T}{\iint }{\left|\nabla y_k\right|}^{p-2}}\nabla y_k\cdot \phi \text{d}x\text{d}t=0,$

所以有

${\displaystyle \underset{Q\tau }{\iint }{\omega }_1\cdot \phi }\text{d}x\text{d}t={\displaystyle \underset{Q\tau }{\iint }{\omega }_2\cdot \phi }\text{d}x\text{d}t,$

所以${\omega }_1={\omega }_2$ 。证毕。

定理1 对任意非负函数$y\in L^2\left(\Omega \right)$ ，$h\in U_M$ 。问题(1)存在唯一弱解y，满足

${\Vert y\Vert }_{L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)}+{\Vert \nabla y\Vert }_{L^p\left(Q_T;R^n\right)}\le C_1{\Vert y_0\Vert }_{L^2\left(\Omega \right)},$

其中$C_1$ 是一个与$y_0$ 和M无关的常数。

证明 取$h_k\in C^{\infty }\left(\partial \Omega \times \left(0,T\right)\right)$ ，$y_{0,k}\in C_0^{\infty }\left(\Omega \right)$ 满足$h_k$ 在$L^{\infty }\left(\partial \Omega \times \left(0,T\right)\right)$ 中弱收敛于h，$y_{0,k}$ 在$L^2\left(\Omega \right)$ 中强收敛于$y_0$ ，而且

${\Vert h_k\Vert }_{L^{\infty }\left(\partial \Omega \times \left(0,T\right)\right)}\le {\Vert h\Vert }_{L^{\infty }\left(\partial \Omega \times \left(0,T\right)\right)},{\Vert y_{0,k}\Vert }_{L^2\left(\Omega \right)}\le 2{\Vert y_0\Vert }_{L^2\left(\Omega \right)}.$

对于问题

$\{\begin{array}{l}\frac{\partial y_k}{\partial t}-div\left({\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\right)+a\left(l\left(y\left(t\right)\right)\right)y=0,\left(x,t\right)\in \Omega \times \left(0,T\right),\\ {\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\cdot v=-h_k{y}_k,\left(x,t\right)\in \partial \Omega \times \left(0,T\right),\\ y_k\left(x,0\right)=y_{0,k}\left(x\right),x\in \Omega \end{array}$ (2)

由经典理论知，问题(2)存在唯一解$y_k\in C^{\infty }\left(Q_T\right)$ ，在(2)的第一个式子两边同乘$y_k$ ，在$Q_{\tau }$ 上积分得

${\displaystyle \underset{Q_{\tau }}{\iint }\left(\frac{\partial y_k}{{\partial }_t}\cdot y_k+{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}{\left|\nabla y_k\right|}^2+y_k^{q+2}\right)\text{d}x\text{d}t+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}\text{d}S\text{d}t}=0,}$

由分部积分公式得：

$\frac{1}{2}{\displaystyle \underset{\Omega }{\int }{y}_k^2\left(x,\tau \right)\text{d}x}-\frac{1}{2}{\displaystyle \underset{\Omega }{\int }{y}_{0,k}^2\left(x\right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }\left({\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}{\left|\nabla y_k\right|}^2+y_k^{q+2}\right)\text{d}x\text{d}t}+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}\text{d}S\text{d}t}=0.$

则有

$\frac{1}{2}{\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}+{\displaystyle {\iint }_{Q_{\tau }}{\left|\nabla y_k\right|}^p}\text{d}x\text{d}t\le \frac{1}{2}{\displaystyle {\int }_{\Omega }{y}_{0,k}^2\left(x\right)\text{d}x}+C{\displaystyle {\iint }_{Q_{\tau }}{y}_k^2}\text{d}x\text{d}t,$ (3)

由Young不等式可知

${\displaystyle {\iint }_{Q_{\tau }}{\left[{\left({\left|\nabla y_n\right|}^2+\frac{1}{n}\right)}^{\frac{p-2}{2}}\left|\nabla y_n\right|\right]}^{\frac{p}{p-1}}}\le C{\displaystyle \iint {\left|\nabla y_n\right|}^p+{\left(\frac{1}{n}\right)}^{\frac{p-2}{2}\cdot \frac{p}{p-1}}\cdot {\left|\nabla y_n\right|}^{\frac{p}{p-1}}}\le C{\displaystyle \iint {\left|\nabla y_n\right|}^p}+{\left(\frac{1}{n}\right)}^{\frac{p}{2}}\le C.$

由Gronwall不等式，

${\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}\le C{\displaystyle {\int }_{\Omega }{y}_0^2\left(x\right)\text{d}x},$ (4)

由(3)和(4)，可得

${\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}+{\displaystyle {\iint }_{Q_{\tau }}{\left|\nabla y_k\right|}^p}\text{d}x\text{d}t\le C_1{\displaystyle {\int }_{\Omega }{y}_0^2\left(x\right)\text{d}x},$ (5)

其中$C_1$ 是一个与$y_0$ 和M无关的常数。

下面证明$y_k$ 在$W^{1,p}\left(\Omega \right)$ 中有界，即${\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }\left({\left|\nabla y_k\right|}^p+{\left|y_k\right|}^p\right)\text{d}x\text{d}t}}\le c$ ，

由Poincare不等式，有

${\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }{\left|y_k-y_{\Omega }\right|}^p}}\text{d}x\text{d}t\le c{\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }{\left|\nabla y_k\right|}^p}}\text{d}x\text{d}t,$

$\left|y_{\Omega }\right|\le \frac{1}{\left|\Omega \right|}\left|{\displaystyle {\int }_{\Omega }y\left(x\right)\text{d}x}\right|\le \frac{1}{\left|\Omega \right|}{\displaystyle {\int }_{\Omega }\left|y\left(x\right)\right|\text{d}x}\le \frac{1}{\left|\Omega \right|}{\left({\displaystyle {\int }_{\Omega }y{\left(x\right)}^2}\text{d}x\right)}^{\frac{1}{2}}\le c,$

${\displaystyle {\int }_{\Omega }{\left|y_k\right|}^p}\text{d}x\le {\displaystyle {\int }_{\Omega }{\left|y_k-y_{\Omega }\right|}^p}\text{d}x+{\displaystyle {\int }_{\Omega }{\left|y_{\Omega }\right|}^p}\text{d}x\le c{\displaystyle {\int }_{\Omega }{\left|D{y}_k\right|}^p}\text{d}x+{\left|y_{\Omega }\right|}^p\cdot \left|\Omega \right|\le c,$

则${\displaystyle {\iint }_{Q_T}{\left|y_k\right|}^p}\text{d}x\text{d}t\le c$ ，

对于$\forall \phi \in L^2\left(0,T;V\right)$ ，

$\begin{array}{c}{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\Omega }{\left(y_k\right)}_t\phi }}\text{d}x\text{d}t={\displaystyle \underset{Q_{\tau }}{\iint }div\left({\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\left|\nabla y_k\right|\right)\phi }\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\phi }\text{d}x\text{d}t\\ =-{\displaystyle \underset{Q_{\tau }}{\iint }{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\left|\nabla y_k\right|\nabla \phi \text{d}x\text{d}t}-{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\phi }\text{d}x\text{d}t+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k\phi }}\text{d}S\text{d}t\\ \le C{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^{p-1}\left|\nabla \phi \right|\text{d}x\text{d}t}+C{\left(\frac{1}{k}\right)}^{\frac{p-2}{2}}{\displaystyle \underset{Q_{\tau }}{\iint }\left|\nabla y_k\right|\left|\nabla \phi \right|\text{d}x\text{d}t}+M{\displaystyle {\int }_{\text{0}}^{\tau }{\Vert y_k\Vert }_{L^2\left(\Omega \right)}}{\Vert \phi \Vert }_{L^2\left(\partial \Omega \right)}\text{d}t\\ \le C{\Vert \nabla y_k\Vert }_{L^p\left(Q_{\tau }\right)}^{p-1}{\Vert \nabla \phi \Vert }_{L^p\left(Q_{\tau }\right)}+C{\left(\frac{1}{k}\right)}^{\frac{p}{2}}{\Vert \nabla y_k\Vert }_{L^p\left(Q_{\tau }\right)}{\Vert \nabla \phi \Vert }_{L^p\left(Q_{\tau }\right)}+M{\displaystyle {\int }_{\text{0}}^{\tau }{\Vert y_k\Vert }_{H^1\left(\Omega \right)}}{\Vert \phi \Vert }_{H^1\left(\Omega \right)}\text{d}t\\ \le C_2{\Vert \phi \Vert }_{L^2\left(0,T;W^{1,p}\left(\Omega \right)\right)},\end{array}$

其中$C_2$ 为一个与$y_0$ 和M有关的常数，因此

${\Vert {\left(y_k\right)}_t\Vert }_{L^2\left(0,T;V^{\prime }\right)}\le C_2$ ，

故在$C\left(\left[0,T\right];L^2\left(\Omega \right)\right)$ 中，存在弱收敛的子列$\left\{y_k\right\}$ ，以及函数$y\in L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)\cap L^p\left(0,T;W^{1,p}\left(\Omega \right)\right)$ ，使得$y_k\to y$ 强收敛于$L^p\left(Q_T\right)$ ，则对$\forall t\in \left(0,T\right)$ 有$y_k\to y$ 弱收敛于$L^2\left(\Omega \right)$ ，故有$y_k\to y$ 强收敛于$L^2\left(\partial \Omega \times \left(0,T\right)\right)$ 。

由于$\nabla y_k\to \nabla y$ 弱收敛于$L^p\left(Q_T\right)$ ，则有${\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\left|\nabla y_k\right|\to {\left|\nabla y\right|}^{p-2}\nabla y$ 弱收敛于$L^{p/\left(p-1\right)}\left(Q_T\right)$ ，因此${\left(y_k\right)}_t\to y_t$ 弱收敛于$L^2\left(0,T;V^{\prime }\right)$ ，其中$y_t\in L^2\left(0,T;V^{\prime }\right)$ 。

接着证明y为问题(1)的弱解，由于$y_k$ 是(2)~(4)的弱解，对于$\forall \phi \in C_0^{\infty }\left(Q_T\right)$ ，

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{y}_k\left(x,\tau \right)\phi \left(x,\tau \right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\nabla \phi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\phi }\text{d}x\text{d}t\\ -{\displaystyle {\int }_{\Omega }{y}_k^0\left(x\right)\phi \left(x,0\right)\text{d}x}-{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k\left(x,t\right){\phi }_t\left(x,t\right)\text{d}x\text{d}t}=0.\end{array}$

在上式中令$k\to \infty$ ，利用$y_k$ 的收敛性，得

$\begin{array}{l}{\displaystyle {\int }_{\Omega }y\left(x,\tau \right)\phi \left(x,\tau \right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y\right|}^{p-2}\nabla y\nabla \phi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{q+1}\phi }\text{d}x\text{d}t+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }hy\phi }}\text{d}S\text{d}t\\ -{\displaystyle {\int }_{\Omega }{y}^0\left(x\right)\phi \left(x,0\right)\text{d}x}-{\displaystyle \underset{Q_{\tau }}{\iint }y\left(x,t\right){\phi }_t\left(x,t\right)\text{d}x\text{d}t}=0.\end{array}$

由此可知，y是问题(1)的弱解。证毕。

最后证明唯一性，若$y_1,y_2$ 为问题(1)的两个弱解，由(1)得

$\begin{array}{l}{\displaystyle \underset{Q_{\tau }}{\iint }\left(y_1^q-y_2^q\right)\phi \text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }\left({\left|\nabla y_1\right|}^{p-2}\nabla y_1-{\left|\nabla y_2\right|}^{p-2}\nabla y_2\right)}\nabla \phi \text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }\left(y_1-y_2\right){\phi }_t\text{d}x\text{d}t}\\ +{\displaystyle {\int }_{\partial \Omega \times \left(0,\tau \right)}h{\left(y_1-y_2\right)}^2}\text{d}S\text{d}t={\displaystyle {\int }_{\Omega }\left(y_2\left(x,\tau \right)-y_1\left(x,\tau \right)\right)\phi \left(x,\tau \right)\text{d}x},\end{array}$

令$\phi =y_1-y_2$ ，得

$\begin{array}{l}{\displaystyle \underset{Q_{\tau }}{\iint }{\left(y_1-y_2\right)}^2\left(y_1^{q-1}+y_1^{q-2}{y}_2+\cdots +y_1{y}_2^{q-2}+y_2^{q-1}\right)\text{d}x\text{d}t}\\ +{\displaystyle \underset{Q_{\tau }}{\iint }\left({\left|\nabla y_1\right|}^{p-2}\nabla y_1-{\left|\nabla y_2\right|}^{p-2}\nabla y_2\right)\nabla \left(y_1-y_2\right)\text{d}x\text{d}t}+{\displaystyle {\int }_{\partial \Omega \times \left(0,\tau \right)}h{\left(y_1-y_2\right)}^2}\text{d}S\text{d}t\\ ={\displaystyle \underset{Q_{\tau }}{\iint }\left(y_1-y_2\right){\left(y_1-y_2\right)}_t}\text{d}x\text{d}t-{\displaystyle {\int }_{\Omega }{\left(y_1\left(x,\tau \right)-y_2\left(x,\tau \right)\right)}^2}\text{d}x,\end{array}$

由引理1有

${\displaystyle \underset{Q_{\tau }}{\iint }\left({\left|\nabla y_1\right|}^{p-2}\nabla y_1-{\left|\nabla y_2\right|}^{p-2}\nabla y_2\right)\cdot \nabla \left(y_1-y_2\right)\text{d}x\text{d}t}\ge 0,$

则

$\begin{array}{l}{\displaystyle {\int }_{\partial \Omega \times \left(0,\tau \right)}h{\left(y_1-y_2\right)}^2}\text{d}S\text{d}t+\frac{1}{2}{\displaystyle {\int }_{\Omega }{\left(y_1-y_2\right)}^2\left(x,\tau \right)\text{d}x}\\ \le -{\displaystyle \underset{Q_{\tau }}{\iint }{\left(y_1-y_2\right)}^2\left(y_1^{q-1}+y_1^{q-2}{y}_2+\cdots +y_1{y}_2^{q-2}+y_2^{q-1}\right)\text{d}x\text{d}t},\end{array}$

即

${\displaystyle {\int }_{\Omega }{\left(y_1-y_2\right)}^2\left(x,\tau \right)\text{d}x}\le C{\displaystyle \underset{Q_{\tau }}{\iint }{\left(y_1-y_2\right)}^2}\text{d}x\text{d}t,$

由Gronwall不等式，有

${\displaystyle {\int }_{\Omega }{y}_n^2}\left(x,\tau \right)\text{d}x\le 0,$

即$y_1\left(x,t\right)=y_2\left(x,t\right)$ ，a.e. $\left(x,t\right)\in Q_T$ 。证毕。

4. 最优控制的存在性

定理2 若$Z_d\in L^2\left(Q_T\right)$ ，$y_0\in L^2\left(\Omega \right)$ ，$a(\cdot )$ 是非负非减函数且$a(\cdot )\le M$ ，$a\left(l\left(y\left(t\right)\right)\right)={\displaystyle {\int }_{\Omega }{y}^q\left(x,t\right)\text{d}x}$ ，$q\ge 1$ ，且满足兼容性条件

${\left|\nabla y_0\right|}^{p-2}\nabla y_0\cdot v=-h\left(x,0\right)y_0,x\in \partial \Omega ,$

则存在最优控制$h^{\ast }\in U_M$ ，使得成本泛函$J\left(h\right)$ 最小。

证明 由于$J\left(h\right)\ge 0$ ，所以$J\left(h\right)$ 必有下确界。设$\left\{h_k\right\}$ 为$U_M$ 中的极小化序列，即

$\underset{k\to \infty }{\lim }J\left(h_k\right)=\underset{h\in U_M}{\inf }J\left(h\right).$

由于$h_k\in U_M$ ，故在$L^{\infty }\left(\partial \Omega \times \left(0,T\right)\right)$ 中，存在$\left\{h_k\right\}$ 的子列$h_k$ 弱收敛于h。

根据定理1，当h为$h_k$ 时问题(1)存在唯一弱解$y_k$ ，并且满足

${\Vert y_k\Vert }_{L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)}+{\Vert \nabla y{}_k\Vert }_{L^p\left(Q_T;R^n\right)}+{\Vert {\left(y_k\right)}_t\Vert }_{L^2\left(0,T;V^{\prime }\right)}\le C,$

其中C为常数。类似定理1的证明，若存在$y^{\ast }\in L^{\infty }\left(0,T;L^2\left(\Omega \right)\right)\cap L^p\left(0,T;W^{1,p}\left(\Omega \right)\right)$ 和$y_t^{\ast }\in L^2\left(0,T;V^{\prime }\right)$ ，使得

$y_k\to y^{\ast }$ 强收敛于$L^p\left(Q_T\right)$ ，

则对$\forall t\in \left(0,T\right)$ 有$y_k\to y^{\ast }$ 弱收敛于$L^2\left(\Omega \right)$ ，故有$y_k\to y^{\ast }$ 强收敛于$L^p\left(Q_T\right)$ ，因此${\left|\nabla y_k\right|}^{p-2}\left|\nabla y_k\right|\to {\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }$ 弱收敛于$L^{\frac{p}{p-1}}\left(Q_T\right)$ 。

下证$y_k$ 是问题(1)的弱解，对于$\forall \phi \in C_0^{\infty }\left(Q_T\right)$ ，

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{y}_k\left(x,\tau \right)\phi }\left(x,\tau \right)\text{d}x+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^{p-2}\nabla y_k\cdot \nabla \phi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^q\left(x,t\right)\phi }\text{d}x\text{d}t\\ +{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k\phi }}\text{d}S\text{d}t-{\displaystyle {\int }_{\Omega }{y}^0\left(x\right)\phi }\left(x,0\right)\text{d}x-{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k\left(x,t\right){\phi }_t\left(x,t\right)\text{d}x\text{d}t}=0.\end{array}$ (6)

在上式中令$k\to \infty$ ，利用$y_k$ 的收敛性得

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{y}^{\ast }\left(x,\tau \right)\phi \left(x,\tau \right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }\cdot \nabla \phi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{\left(y^{\ast }\right)}^q\left(x,t\right)\cdot \phi }\text{d}x\text{d}t\\ +{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}^{\ast }{y}^{\ast }\phi }}\text{d}x\text{d}t-{\displaystyle {\int }_{\Omega }{y}^0\left(x\right)\phi \left(x,0\right)\text{d}x}-{\displaystyle \underset{Q_{\tau }}{\iint }y\left(x,t\right){\phi }_t\left(x,t\right)\text{d}x\text{d}t}=0.\end{array}$ (7)

由此可知，$y^{\ast }$ 是h为$h^{\ast }$ 时问题(1)的弱解。

在(6)式中，取$y_k$ 为测试函数，则有

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^{p-2}\nabla y_k\cdot \nabla y_k}\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }\left(y_k^q\left(x,t\right)\right)y_k}\text{d}x\text{d}t+{\displaystyle {\int }_{\text{0}}^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}\text{d}S}\text{d}t\\ -{\displaystyle {\int }_{\Omega }{y}^0\left(x\right)y_k\left(x_0\right)\text{d}x}-{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k{\left(y_k\right)}_t}\text{d}x\text{d}t=0,\end{array}$

其中$y_k^0\left(x\right)=y\left(x,0\right)=y^0\left(x\right)$ ，

整理得

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^{p-2}\nabla y_k\cdot \nabla y_k}\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\left(x,t\right)\text{d}x\text{d}t}+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}}\text{d}S\text{d}t\\ -{\displaystyle {\int }_{\Omega }{y}^0\left(x\right)y_k\left(x_0\right)\text{d}x}-\frac{1}{2}{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\Omega }{\left(y_k\right)}_t^2}\text{d}x\text{d}t}=0,\end{array}$

于是有

$\begin{array}{l}{\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}-{\displaystyle {\int }_{\Omega }{y}^0\left(x\right)y_k\left(x_0\right)\text{d}x}-\frac{1}{2}{\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}+\frac{1}{2}{\displaystyle {\int }_{\Omega }{y}^0{}^{{}^2}\left(x\right)\text{d}x}\\ +{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\left(x,t\right)\text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^{p-2}\nabla y_k\cdot \nabla y_k}\text{d}x\text{d}t+{\displaystyle {\int }_0^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}}\text{d}S\text{d}t=0,\end{array}$

整理得

$\frac{1}{2}{\displaystyle {\int }_{\Omega }{y}_k^2\left(x,\tau \right)\text{d}x}-\frac{1}{2}{\displaystyle {\int }_{\Omega }{y}^0{}^{{}^2}\left(x\right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\left(x,t\right)\text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^{p-2}\nabla y_k\cdot \nabla y_k}\text{d}x\text{d}t+{\displaystyle {\int }_{\text{0}}^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}}\text{d}S\text{d}t=0,$

所以有

$\frac{1}{2}{\displaystyle {\int }_{\Omega }\left(y_k^2\left(x,\tau \right)-y^0{}^{{}^2}\left(x\right)\right)\text{d}x}+{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\left(x,t\right)\text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left|\nabla y_k\right|}^p}\text{d}x\text{d}t+{\displaystyle {\int }_{\text{0}}^{\tau }{\displaystyle {\int }_{\partial \Omega }{h}_k{y}_k^2}}\text{d}S\text{d}t=0.$

由于(7)式中的后两项是非负的，所以有

${\displaystyle {\int }_{\Omega }{y}_k^2}\text{d}x\le {\displaystyle {\int }_{\Omega }{y}^0{}^{{}^2}}\text{d}x$ 和${{\displaystyle {\iint }_{Q_{\tau }}\left|\nabla y_k\right|}}^p\text{d}x\text{d}t\le \frac{1}{2}{\displaystyle {\int }_{\Omega }{y}^0{}^{{}^2}}\text{d}x$ ，$\tau \in \left(0,T\right)$ 。

故有

${\displaystyle {\iint }_{Q_{\tau }}{y}^{\ast }{\phi }_t}\text{d}x\text{d}t-{\displaystyle {\iint }_{Q_{\tau }}w\nabla \phi }\text{d}x\text{d}t-{\displaystyle {\iint }_{Q_{\tau }}{y}^{q+1}\left(x\right)\phi }\text{d}x\text{d}t=0$ ，

其中$w={\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }$ 。

对于$\forall \phi \in C_0^{\infty }\left(Q_T\right)$ ，下证

${{\displaystyle {\iint }_{Q_{\tau }}\left|\nabla y^{\ast }\right|}}^{p-2}\nabla y^{\ast }\cdot \nabla \phi \text{d}x\text{d}t={\displaystyle {\iint }_{Q_{\tau }}w\nabla \phi }\text{d}x\text{d}t.$

由引理1，可得对$\forall v\in L^p\left(0,T;W_0^{1,p}\left(\Omega \right)\right)$ ，$\psi \in C_0^{\infty }\left(Q_T\right)$ ，$0\le \psi \le 1$ ，$\text{supp}\psi \subset \Omega$

${\displaystyle {\iint }_{Q_T}\psi \left({\left|\nabla y_k\right|}^{p-2}\nabla y_k-{\left|\nabla v\right|}^{p-2}\nabla v\right)\cdot \nabla \left(y_k-v\right)\text{d}x\text{d}t}\ge 0.$

在问题(1)的第一个式子两边乘$\psi y_k$ ，$\psi \in C_0^{\infty }\left(Q_T\right)$ ，得

$\begin{array}{l}{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}{\left|\nabla y_k\right|}^2}\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\nabla \psi y_k}\text{d}x\text{d}t\\ +{\displaystyle {\int }_{\text{0}}^{\tau }{\displaystyle {\int }_{\partial \Omega }{y}_k^{q+1}\left(x,\tau \right)\psi }}\text{d}S\text{d}t=\frac{\text{1}}{\text{2}}{\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }{y}_k^2{\psi }_t}}\text{d}x\text{d}t.\end{array}$

整理得

${\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}{\left|\nabla y_k\right|}^2\text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\nabla \psi y_k\text{d}x\text{d}t}+{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\left(x\right)\psi }\text{d}x\text{d}t=\frac{\text{1}}{\text{2}}{\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }{y}_k^2{\psi }_t}\text{d}x\text{d}t}\text{,}$

于是有

$\begin{array}{l}\frac{1}{2}{\displaystyle {\int }_0^T{\displaystyle {\int }_{\Omega }{y}_k^2{\psi }_t}}\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }{\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}\nabla y_k\nabla \psi y_k}\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }{y}_k^{q+1}\left(x\right)\psi }\text{d}x\text{d}t\\ -{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left|\nabla y_k\right|}^{p-2}\nabla y_k\nabla v}\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left|\nabla v\right|}^{p-2}\nabla v\nabla \left(y_k-v\right)\text{d}x\text{d}t}\\ ={\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left({\left|\nabla y_k\right|}^2+\frac{1}{k}\right)}^{\frac{p-2}{2}}{\left|\nabla y_k\right|}^2}\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left|\nabla y_k\right|}^{p-2}\nabla y_k\nabla v}\text{d}x\text{d}t\\ -{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left|\nabla v\right|}^{p-2}\nabla v\nabla \left(y_k-v\right)\text{d}x\text{d}t}.\end{array}$

在上式中，令$k\to \infty$ 可得

$\frac{1}{2}{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }{}^2{\psi }_t}\text{d}x\text{dt}-{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }w\nabla \psi }\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }{}^{q+1}\left(x\right)\psi }\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }\psi w\nabla v}\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left|\nabla v\right|}^{p-2}\nabla v\nabla \left(y^{\ast }-v\right)\text{d}x\text{d}t}\ge 0.$ (8)

在(8)式中取$\phi =\psi y^{\ast }$ ，有

${\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }w\cdot \nabla \psi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }w\cdot \nabla y^{\ast }\psi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }{}^{q+1}\psi }\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }{\psi }_t{y}^{\ast }}\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }\psi {\left(y^{\ast }\right)}_t}\text{d}x\text{d}t=0,$

其中$\psi \in C_0^{\infty }\left(Q_T\right)$ 。

所以有

${\displaystyle \underset{Q_{\tau }}{\iint }w\cdot \nabla \psi y^{\ast }}\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }w\cdot \nabla y^{\ast }\psi }\text{d}x\text{d}t+{\displaystyle \underset{Q_{\tau }}{\iint }{y}^{\ast }{}^{q+1}\psi }\text{d}x\text{d}t-\frac{1}{2}{\displaystyle \underset{Q_{\tau }}{\iint }{\left(y^{\ast }\right)}^2{\psi }_t}\text{d}x\text{d}t=0,$

结合(8)式，得

${\displaystyle \underset{Q_{\tau }}{\iint }w\cdot \nabla y^{\ast }\psi }\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }w\cdot \nabla v\psi }\text{d}x\text{d}t-{\displaystyle \underset{Q_{\tau }}{\iint }\psi {\left|\nabla v\right|}^{p-2}\nabla v\cdot \nabla \left(y^{\ast }-v\right)\text{d}x\text{d}t}\ge 0,$

整理得

${\displaystyle \underset{Q_{\tau }}{\iint }\psi \left(w-{\left|\nabla v\right|}^{p-2}\nabla v\right)\cdot \nabla \left(y^{\ast }-v\right)\text{d}x\text{d}t}\ge 0,$

在上式中取$v=y^{\ast }-\mu \phi$ ，有

$\mu {\displaystyle \underset{Q_{\tau }}{\iint }\psi \left(w-{\left|\nabla \left(y^{\ast }-\mu \phi \right)\right|}^{p-2}\nabla \left(y^{\ast }-\mu \phi \right)\right)\nabla \phi }\text{d}x\text{d}t\ge 0,\phi \in C_0^{\infty }\left(Q_T\right),$

令$\mu \to 0^{+}$ ，得

${\displaystyle \underset{Q_{\tau }}{\iint }\psi \left(w-{\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }\right)\cdot \nabla \phi }\text{d}x\text{d}t\ge 0,$

令$\mu \to 0^{-}$ ，得

${\displaystyle \underset{Q_{\tau }}{\iint }\psi \left(w-{\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }\right)\cdot \nabla \phi }\text{d}x\text{d}t\le 0,$

所以有

${\displaystyle \underset{Q_{\tau }}{\iint }\psi \left(w-{\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }\right)\cdot \nabla \phi }\text{d}x\text{d}t=0,$

在上式中取$\psi$ 使得$\text{supp}\phi \subset \text{supp}\psi$ 成立，则有

${\displaystyle \underset{Q_{\tau }}{\iint }\left(w-{\left|\nabla y^{\ast }\right|}^{p-2}\nabla y^{\ast }\right)\cdot \nabla \phi }\text{d}x\text{d}t=0,$

因此，可得$w={\left|\nabla y\right|}^{p-2}\nabla y$ 。

又由$J\left(h\right)$ 的弱下半连续性，有

$\begin{array}{c}J\left(h^{\ast }\right)=\frac{1}{2}{\displaystyle {\int }_Q{\left|y-y_d\right|}^2}\text{d}x\text{d}t+\frac{\mu }{2}{\displaystyle {\int }_{{\omega }_T}{\left|h^{\ast }\right|}^2}\text{d}x\text{d}t\\ \le \underset{k\to \infty }{\lim }\inf \left(\frac{1}{2}{\displaystyle {\int }_Q{\left(y_k-y_d\right)}^2}\text{d}x\text{d}t+\frac{\mu }{2}{\displaystyle {\int }_{{\omega }_T}{\left|h_k\right|}^2}\text{d}x\text{d}t\right)\\ =\underset{k\to \infty }{\lim }J\left(h_k\right)\\ =\underset{h\in U_M}{\inf }J\left(h\right).\end{array}$

因此，$h^{\ast }$ 是$J\left(h\right)$ 在$U_M$ 上的最优控制。证毕。

基金项目

国家自然科学基金项目(11961039)；甘肃省自然科学基金项目(1310RJZA070)。