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  北京化工大学学报(自然科学版)  2018, Vol. 45 Issue (6): 106-110   DOI: 10.13543/j.bhxbzr.2018.06.017
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王世杰, 常延贞. 一类抛物最优控制问题的有限元误差估计[J]. 北京化工大学学报(自然科学版), 2018, 45(6): 106-110. DOI: 10.13543/j.bhxbzr.2018.06.017.
WANG ShiJie, CHANG YanZhen. Error estimates of the finite element method for a class of parabolic optimal control problems[J]. Journal of Beijing University of Chemical Technology (Natural Science), 2018, 45(6): 106-110. DOI: 10.13543/j.bhxbzr.2018.06.017.

第一作者

王世杰, 男, 1994年生, 硕士生.

通信联系人

常延贞, E-mail:changyz@mail.buct.edu.cn

文章历史

收稿日期:2017-11-24
一类抛物最优控制问题的有限元误差估计
王世杰 , 常延贞     
北京化工大学 理学院, 北京 100029
摘要:对一类抛物最优控制问题给出了有限元逼近格式,其中控制约束集为积分受限的形式K={u(t)∈L2(Ω):a$ \int_\mathit{\Omega } u $(t)≤b}。对问题的状态变量和伴随状态变量用线性连续函数离散,而控制变量使用分片常数近似;最后得到控制和状态变量逼近的先验误差估计Oh2+k)。
关键词有限元逼近    积分受限最优控制    误差估计    抛物型方程    
Error estimates of the finite element method for a class of parabolic optimal control problems
WANG ShiJie , CHANG YanZhen     
Faculty of Science, Beijing University of Chemical Technology, Beijing 100029, China
Abstract: In this paper, we study a finite element approximation scheme for a class of parabolic optimal control problems. The control constraint is given in an integral sense:K={u(t)∈L2(Ω):a$ \int_\mathit{\Omega } u $(t)≤b}, where the state and co-state variables are discretized by piecewise linear continuous functions and the control variable is approximated by piecewise constant functions. Some error estimates are derived for both control and state approximations. It is proven that these approximations have convergence order O(h2+k).
Key words: finite element approximation    integral constrained optimal control    error estimates    parabolic equation    
引言

作为数学尤其是应用数学的一个分支, 偏微分方程最优控制涵盖了晶体增长、化学反应、材料设计等领域的多种类型的问题,如回馈控制、时间控制、最优形状设计、流体方程控制等[1-3]。实际最优控制问题中的状态方程往往涉及线性或非线性、常定或时变的情况。在数值求解最优控制问题时,普遍存在很多计算瓶颈,而有限元方法是解决这类问题最有效的方法。近年来,出现了使用有限元方法解决各种不同类型最优控制问题的研究成果,如付红斐等对一类非线性抛物控制问题进行了先验误差估计[4],文献[5-6]提出了控制积分受限问题的自适应有限元方法,文献[7-8]对参数全计控制问题进行了分析。然而关于积分受限的抛物最优控制问题的有限元先验误差研究还未见文献报道。该问题由于控制约束集的光滑性,可以在理论上得到比其他类型问题更高阶的误差估计。因此本文研究控制约束集为{u(t)∈L2(Ω):a$ \int_\mathit{\Omega } u $(t)≤b}的抛物最优控制问题的有限元逼近格式,并求出控制变量和状态变量在L2模范数下的先验误差估计。

1 积分受限的抛物最优控制问题

考虑如下抛物最优控制问题

$ \left\{ \begin{array}{l} \mathop {\min }\limits_{u \in X,u\left( t \right) \in K} \frac{1}{2}\int_0^T {\left\{ {\left\| {y - {y_{\rm{d}}}} \right\|_{{L^2}\left( \mathit{\Omega } \right)}^2 + \left\| u \right\|_{{L^2}\left( \mathit{\Omega } \right)}^2} \right\}} {\rm{d}}t\\ \frac{{\partial y}}{{\partial t}} - \Delta y = f + u,\left( {x,t} \right) \in \mathit{\Omega } \times \left( {0,T} \right]\\ y|\partial \mathit{\Omega } = 0, \ \ \ t \in \left[ {0,T} \right]\\ y\left( {x,0} \right) = {y_0}\left( x \right),\ \ \ x \in \mathit{\Omega } \end{array} \right. $ (1)

式(1)中,Ω是在Rn(n≥2)上的凸多边形区域,y0H01(Ω),fL2(0, T; L2(Ω)),U=L2(Ω),X=L2(0, T; U)。

令控制集K={u(t)∈L2(Ω):a$ \int_\mathit{\Omega } u $(t)≤b}是一个封闭凸集,其中ab是已知系数。则利用式(1)可以模拟模型锻造等实际问题。

本文采用一般记号Wm, q(Ω)来表示Sobolev空间,其中范数为‖·‖Wm, q(Ω),半范数为|·|Wm, q(Ω);并规定W0m, q(Ω)≡{wWm, q(Ω):w|Ω=0}, Hm(Ω)表示Wm, 2(Ω)。此外,文中的cC表示一般的正常数,与离散参数无关。

2 有限元离散

定义状态空间W=L2(0, T; V),其中V=H01(Ω);定义控制空间X=L2(0, T; U),其中U=L2(Ω)。

再定义

$ \left\{ {\begin{array}{*{20}{l}} {a\left( {v,w} \right) = \int_\mathit{\Omega } {\left( {\nabla v} \right) \cdot \nabla w} ,}&{\forall v,w \in V}\\ {\left( {v,w} \right) = \int_\mathit{\Omega } {vw,} }&{\forall v,w \in U} \end{array}} \right. $ (2)

并且满足

$ \left\{ \begin{array}{l} c\left\| v \right\|_1^2 \le a\left( {v,v} \right)\\ \left| {a\left( {v,w} \right)} \right| \le C{\left\| v \right\|_1}{\left\| w \right\|_1},\forall v,w \in V \end{array} \right. $

那么求解最优控制问题式(1)等价于求解yH1(0, T; L2(Ω))∩W使得(OCP)成立,如式(3)

$ \left\{ \begin{array}{l} \mathop {\min }\limits_{u \in X,u\left( t \right) \in K} \frac{1}{2}\int_0^T {\left\{ {\left\| {y - {y_{\rm{d}}}} \right\|_{{L^2}\left( \mathit{\Omega } \right)}^2 + \left\| u \right\|_{{L^2}\left( \mathit{\Omega } \right)}^2} \right\}} {\rm{d}}t\\ \left( {\frac{{\partial y}}{{\partial t}},w} \right) + a\left( {y,w} \right) = \left( {f + u,w} \right),\forall w \in V,\\ \;\;\;\;\;\;t \in \left( {0,T} \right]\\ y\left( {x,0} \right) = {y_0}\left( x \right),x \in \mathit{\Omega } \end{array} \right. $ (3)

由文献[1]可知,(OCP)存在唯一弱解(y, u), 当且仅当存在伴随状态pW时,(y, p, u)满足式(4)所示的最优性条件(OCP-OPT)

$ \left\{ \begin{array}{l} \left( {\frac{{\partial y}}{{\partial t}},w} \right) + a\left( {y,w} \right) = \left( {f + u,w} \right),\;\;\forall w \in V,\\ \;\;\;\;\;\;t \in \left( {0,T} \right]\\ y\left( {x,0} \right) = {y_0}\left( x \right),\;\;x \in \mathit{\Omega }\\ - \left( {\frac{{\partial p}}{{\partial t}},q} \right) + a\left( {q,p} \right) = \left( {y - {y_{\rm{d}}},q} \right),\forall q \in V\\ p\left( {x,T} \right) = 0,\;\;x \in \mathit{\Omega }\\ \int_0^T {\left( {u + p,v - u} \right){\rm{d}}t \ge 0,\forall v\left( t \right) \in K,v \in X} \end{array} \right. $ (4)

下面考虑(OCP)的全离散有限元逼近。令Th表示Ω的正则三角形剖分,剖分单元记为τ,剖分单元的最大直径为h,并且满足Ω= ∪τThτ

定义状态有限元空间Vh={vhSh:vh|Ω=0},其中Sh={χC(Ω):χτP1, ∀τTh};定义控制有限元空间Uh={χL2(Ω):χ|τP0, ∀τTh},此处对控制有限元空间没有连续性要求。

Wh=L2(0, T; Vh),Xh=L2(0, T; Uh),Kh={uh(t)∈Uh:a$ \int_\mathit{\Omega } u_h $(t)≤b},显然VhVWhWXhX。再将时间T离散化,使0=t0t1<…<tN=Tki=ti-ti-1i=1, 2, …, N。将问题的全离散格式记为(OCP)hk,那么(OCP)等价于求解(yhi, uhi)∈Vh×Khi=1, 2, …, N,使得式(5)成立

$ \left\{ \begin{array}{l} \mathop {\min }\limits_{u_h^i \in {K^h}} \frac{1}{2}\sum\limits_{i = 1}^N {{k_i}\left\{ {\left\| {y_h^i - {y_{\rm{d}}}} \right\|_{{L^2}\left( \mathit{\Omega } \right)}^2 + \left\| {u_h^i} \right\|_{{L^2}\left( \mathit{\Omega } \right)}^2} \right\}} \\ \left( {\frac{{y_h^i - y_h^{i - 1}}}{{{k^i}}},{w_h}} \right) + a\left( {y_h^i,{w_h}} \right) = \left( {{f^i} + u_h^i,{w_h}} \right),\\ \;\;\;\;\;\;\forall {w_h} \in {V_h},i = 1,2, \cdots ,N\\ y_0^h = y_0^h\left( x \right),x \in \mathit{\Omega } \end{array} \right. $ (5)

为了方便表达,将式(5)及后文中的dtyhi用向后Euler差商$ \frac{{y_h^i - y_h^{i - 1}}}{{{k^i}}}$来表示。

同样地,(OCP)hk存在唯一弱解(Yhi, Uhi)(i=1, 2, …, N),当且仅当存在伴随离散状态Phi-1Vh(i=1, 2, …, N)时,可使(Yhi, Pi-1, Uhi)∈Vh×Vh×Kh(i=1, 2, …, N)满足式(6)所示的离散最优性条件(OCP-OPT)hk

$ \left\{ \begin{array}{l} \left( {\frac{{Y_h^i - Y_h^{i - 1}}}{{{k^i}}},{w_h}} \right) + a\left( {Y_h^i,{w_h}} \right) = \left( {{f^i} + U_h^i,{w_h}} \right),\\ \;\;\;\;\;\;\forall {w_h} \in {V_h},i = 1,2, \cdots ,N\\ {Y_h}\left( {x,0} \right) = y_0^h\left( x \right),x \in \mathit{\Omega }\\ - \left( {\frac{{P_h^i - P_h^{i - 1}}}{{{k^i}}},{q_h}} \right) + a\left( {{q_h},P_h^{i - 1}} \right) = \\ \;\;\;\left( {Y_h^i - {y_d},{q_h}} \right),\forall {q_h} \in {V_h},i = N,N - 1, \cdots ,1\\ P_h^N = 0,x \in \mathit{\Omega }\\ \left( {U_h^i + P_h^{i - 1},{v_h} - U_h^i} \right) \ge 0,\forall {v_h} \in {K^h},i = 1,2, \cdots ,N \end{array} \right. $ (6)

定义Πh为从空间UUhL2投影算子,即

$ \int_\mathit{\Omega } {\left( {{\Pi _h}v - v} \right)\phi = 0} ,\forall \phi \in {U^h} $

由于Uh为分片常数元空间,所以有

$ {\Pi _h}v{|_\tau } = \frac{1}{{\left| \tau \right|}}\int_\tau {v,\forall \tau \in {T^h}} $ (7)

式(7)中|τ|为τ的测度。当vK时,显然有a$ \int_\mathit{\Omega } $Πhvb,因此ΠhvKhK

3 收敛性分析与误差估计

下面将讨论最优控制问题(OCP)的有限元逼近的先验误差估计,并最终求出并证明控制和状态逼近在l2(0, T; L2(Ω))模意义下的最优收敛阶。

yyh分别是(OPC)、(OPC)h的解,那么可以定义目标泛函

$ \left\{ \begin{array}{l} J\left( u \right) = \frac{1}{2}\int_\mathit{\Omega } {{{\left( {y - {y_d}} \right)}^2} + \frac{1}{2}} \int_\mathit{\Omega } {{u^2}} \\ {J_h}\left( {{u_h}} \right) = \frac{1}{2}\int_{{\mathit{\Omega }^h}} {{{\left( {{y_h} - {y_d}} \right)}^2} + \frac{1}{2}} \int_{{\mathit{\Omega }^h}} {u_h^2} \end{array} \right. $ (8)

由于(OCP)是一个线性控制问题,所以其目标泛函是一致凸的。假设存在与h无关的常数c>0, 对所有的u, vX,满足

$ J'\left( u \right) - J'\left( v \right) \ge c{\left\| {u - v} \right\|^2} $ (9)

进而对于充分小的h,可得出

$ {{J'}_h}\left( u \right) - {{J'}_h}\left( v \right) \ge c{\left\| {u - v} \right\|^2} $ (10)

那么在式(9)、(10)的基础上可计算得到

$ \left\{ \begin{array}{l} \left( {J'\left( u \right),v} \right) = \left( {u + p,v} \right)\\ \left( {{{J'}_h}\left( {{U_h}} \right),v} \right) = \left( {{U_h} + {P_h},v} \right)\\ \left( {{{J'}_h}\left( u \right),v} \right) = \left( {u + {P_h}\left( u \right),v} \right) \end{array} \right. $ (11)

式(11)中Ph(u)是式(12)辅助问题的解

$ \left\{ \begin{array}{l} \left( {\frac{{\partial Y_h^i\left( u \right)}}{{\partial t}},{w_h}} \right) + a\left( {Y_h^i\left( u \right),{w_h}} \right) = \left( {{f^i} + {u^i},{w_h}} \right)\\ \;\;\;\;\;\;\forall {w_h} \in {V^h}\\ Y_h^0\left( u \right) = y_0^h\left( x \right),x \in \mathit{\Omega }\\ a\left( {q,P_h^{i - 1}\left( u \right)} \right) - \left( {\frac{{\partial P_h^{i - 1}\left( u \right)}}{{\partial t}},{q_h}} \right) = \left( {Y_h^i\left( u \right) - } \right.\\ \;\;\;\;\;\;\left. {{y_d},{q_h}} \right),\forall {q_h} \in {V^h}\\ P_h^N\left( u \right) = 0,x \in \mathit{\Omega } \end{array} \right. $ (12)

引理1  令Πh为式(7)定义的投影算子,则存在常数C,使得式(13)对于所有的vH2(Ω),k=0, 1成立。

$ {\left| {v - {\Pi _h}v} \right|_{{H^k}\left( \tau \right)}} \le C{h^{2 - k}}{\left| v \right|_{{H^2}\left( \tau \right)}} $ (13)

引理2  令(Yh, Ph)和(Yh(u), Ph(u))分别为式(6)和式(12)的解,则有

$ \begin{array}{l} \;\;\;\;\;\;\;{\left\| {{Y_h} - {Y_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{H^1}\left( \mathit{\Omega } \right)} \right)}} \le C\left\| {u - } \right.\\ {\left. {{U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \end{array} $ (14)
$ \begin{array}{l} \;\;\;\;\;\;\;{\left\| {{P_h} - {P_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{H^1}\left( \mathit{\Omega } \right)} \right)}} \le C\left\| {u - } \right.\\ {\left. {{U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \end{array} $ (15)

证明  为方便表达,令θi=Yhi-Yhi(u),ηi=yhi-Yhi(u),i=1, 2, …, Nζi=Phi-Phi(u),ξi=phi-Phi(u),i=N, N-1, …, 1。显然有θ0=ζN=ξN=0。在不失一般性的前提下,假设M是一个正整数, 使得‖ρM‖≡‖ρl2(0, T; L2(Ω))=$ \mathop {\max }\limits_{0 \le i \le N} $ρi‖。

首先证明式(14)。由式(6)、(12)可以得出

$ \left( {\frac{{{\theta ^i} - {\theta ^{i - 1}}}}{{{k^i}}},{w_h}} \right) + a\left( {{\theta ^i},{w_h}} \right) = \left( {U_h^i - {u^i},{w_h}} \right) $ (16)

取检验函数wh=θi,由a(a-b)≥$ \frac{1}{2}$(a2-b2)和Cauchy-不等式得到

$ \begin{array}{l} \;\;\;\;\;\frac{1}{{2{k^i}}}\left\| {{\theta ^i}} \right\|_{0,\mathit{\Omega }}^2 - \frac{1}{{2{k^i}}}\left\| {{\theta ^{i - 1}}} \right\|_{0,\mathit{\Omega }}^2 + \left\| {{\theta ^i}} \right\|_a^2 \le C\left( \varepsilon \right)\\ \left\| {{u^i} - U_h^i} \right\|_{0,\mathit{\Omega }}^2 + \varepsilon \left\| {{\theta ^i}} \right\|_{0,\mathit{\Omega }}^2 \end{array} $ (17)

式(17)中‖θa2=a(θ, θ)。

ε=0.5,由Poincaré-不等式和式(17)可得出

$ \begin{array}{l} \;\;\;\;\;{\left\| {{d_t}{\theta ^i}} \right\|^2} + 2\left\| {{\theta ^i}} \right\|_{1,\mathit{\Omega }}^2 \le C\left\| {{u^i} - U_h^i} \right\|_{0,\mathit{\Omega }}^2 + \\ \left\| {{\theta ^i}} \right\|_{0,\mathit{\Omega }}^2 \end{array} $ (18)

对式(18)左右两端同乘2ki,并对k从1到N求和,则由离散的Gronwall-引理可得到

$ \begin{array}{l} \;\;\;\;\;{\left\| {{\theta ^M}} \right\|^2} + \sum\limits_{i = 1}^N {{k^i}\left\| {{\theta ^i}} \right\|_{1,\mathit{\Omega }}^2} \le C\sum\limits_{i = 1}^N {{k^i}} \left\| {{u^i} - } \right.\\ \left. {U_h^i} \right\|_{0,\mathit{\Omega }}^2 + \left\| {{\theta ^0}} \right\|_{0,\mathit{\Omega }}^2 \end{array} $ (19)

由于θ0=0,所以由式(19)可以推导出式(14)。式(14)得证。

然后证明式(15)。由式(6)、(12)可以得出

$ \left( {\frac{{{\zeta ^{i - 1}} - {\zeta ^i}}}{{{k^i}}},{q_h}} \right) + a\left( {\zeta _h^{i - 1},{q_h}} \right) = \left( {{\theta ^i},{q_h}} \right) $

取检验函数qh=ζi-1,可以得到

$ \begin{array}{l} \;\;\;\;\;\frac{1}{{2{k^i}}}\left\| {{\zeta ^{i - 1}}} \right\|_{0,\mathit{\Omega }}^2 - \frac{1}{{2{k^i}}}\left\| {{\zeta ^i}} \right\|_{0,\mathit{\Omega }}^2 + \left\| {{\zeta ^{i - 1}}} \right\|_a^2 \le \\ C\left( \varepsilon \right)\left\| {{\theta ^i}} \right\|_{0,\mathit{\Omega }}^2 + \varepsilon \left\| {{\zeta ^{i - 1}}} \right\|_{0,\mathit{\Omega }}^2 \end{array} $ (20)

ε=0.5,运用Poincaré-不等式和离散的Gronwall-引理,对式(20)两端同乘ki,并对iNM+1求和,可得

$ \begin{array}{l} \;\;\;\;\;{\left\| {{\zeta ^M}} \right\|^2} + \sum\limits_{i = M + 1}^N {{k^i}\left\| {{\zeta ^{i - 1}}} \right\|_{1,\mathit{\Omega }}^2} \le C\sum\limits_{i = M + 1}^N {{k^i}} \left\| {{\theta ^i}} \right\|_{0,\mathit{\Omega }}^2 + \\ \left\| {{\zeta ^N}} \right\|_{0,\mathit{\Omega }}^2 \end{array} $ (21)

由于ζN=0,则结合(21)、(14)可以推导出式(15),式(15)得证。引理2证毕。

引理3  令(y, p, u)和(Yh, Ph, Uh)分别表示式(4)和式(6)的解,并假设ul2(0, T; H2(Ω)),pl2(0, T; H2(Ω))∩H1(0, T; L2(Ω)),则有

$ \begin{array}{l} \;\;\;\;\;{\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le C\left( {{h^2} + k + \left\| {p - } \right.} \right.\\ \left. {{{\left. {{P_h}\left( u \right)} \right\|}_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}}} \right) \end{array} $ (22)

证明  由Jh(·)的定义式(9)~(11)可得

$ \begin{array}{l} \ \ \ \ \ c{\left\| {u - {U_h}} \right\|_{{l^2}(0,T;{L^2}(\mathit{\Omega }))}} \le {\rm{ }}\sum\limits_{i = 1}^N {{k^i}({{J'}_h}({u^i}) - } \\ {{J'}_h}(U_h^i),{u^i} - U_h^i) = \sum\limits_{i = 1}^N {{k^i}({u^i} + P_h^{i - 1}\left( u \right),{u^i} - } \\ U_h^i) + \sum\limits_{i = 1}^N {{k^i}({U_h^i} + P_h^{i - 1},U_h^i - {u^i})} = \sum\limits_{i = 1}^N {{k^i}({u^i} + {p^i},} \\ {u^i} - U_h^i) + \sum\limits_{i = 1}^N {{k^i}(P_h^{i - 1}\left( u \right) - {p^i},{u^i} - U_h^i) + } \sum\limits_{i = 1}^N {{k^i}} \\ (U_h^i + P_h^{i - 1},U_h^i - {\Pi _h}{u^i}) \le {\rm{ }}0 + \sum\limits_{i = 1}^N {{k^i}(P_h^{i - 1}\left( u \right) - {p^i},} \\ {u^i} - U_h^i) + 0 + \sum\limits_{i = 1}^N {{k^i}(U_h^i + P_h^{i - 1},{\Pi _h}{u^i} - {u^i}) = } \sum\limits_{i = 1}^N {{k^i}} \\ (U_h^i,{\Pi _h}{u^i} - {u^i}) + \sum\limits_{i = 1}^N {{k^i}(P_h^{i - 1} + {p^{i - 1}} - {p^{i - 1}},{\Pi _h}{u^i} - } \\ {u^i}) + \sum\limits_{i = 1}^N {{k^i}(P_h^{i - 1}\left( u \right) - P_h^{i - 1}\left( u \right),{\Pi _h}{u^i} - {u^i}) + } \\ \sum\limits_{i = 1}^N {{k^i}(P_h^{i - 1}\left( u \right) - {p^i},{u^i} - U_h^i).} \end{array} $

由算子Πh的定义可知

$ \left\{ \begin{array}{l} \sum\limits_{i = 1}^N {{k^i}\left( {U_h^i,{\Pi _h}{u^i} - {u^i}} \right) = 0} \\ \sum\limits_{i = 1}^N {{k^i}\left( {{p^{i - 1}},{\Pi _h}{u^i} - {u^i}} \right) = } \\ \;\;\;\;\;\;\;\sum\limits_{i = 1}^N {{k^i}\left( {{p^{i - 1}} - {\Pi _h}{p^{i - 1}},{\Pi _h}{u^i} - {u^i}} \right)} \end{array} \right. $

那么

$ \begin{array}{l} \;\;\;\;\;c{\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le 0 + \sum\limits_{i = 1}^N {{k^i}} \left( {{p^{i - 1}} - } \right.\\ \left. {{\Pi _h}{p^{i - 1}},{\Pi _h}{u^i} - {u^i}} \right) + \sum\limits_{i = 1}^N {{k^i}\left( {{p^{i - 1}} - P_h^{i - 1}\left( u \right),{u^i} - {\Pi _h}{u^i}} \right)} + \\ \sum\limits_{i = 1}^N {{k^i}} \left( {P_h^{i - 1}\left( u \right) - P_h^{i - 1},{u^i} - {\Pi _h}{u^i}} \right) + \sum\limits_{i = 1}^N {{k^i}\left( {P_h^{i - 1}\left( u \right) - } \right.} \\ \left. {p_h^{i - 1},{u^i} - U_h^i} \right) + \sum\limits_{i = 1}^N {{k^i}\left( {{p^{i - 1}} - p_h^i,{u^i} - U_h^i} \right) = {I^1} + {I^2} + } \\ {I^3} + {I^4} + {I^5} \end{array} $

由式(13) ~(15)以及Cauchy-不等式可得

$ \begin{array}{l} \ \ \ \ \ \ {I^1} \le {C_1}\sum\limits_{i = 1}^N {{k^i}\left\| {{p^{i - 1}} - {\Pi _h}{p^{i - 1}}} \right\|_{0,\mathit{\Omega }}^2} + {C_1}\sum\limits_{i = 1}^N {{k^i}} \\ \left\| {{\Pi _h}{u^i} - {u^i}} \right\|_{0,\mathit{\Omega }}^2 \le {C_1}{h^4}\left\| p \right\|_{{l^2}\left( {0,T;{H^2}\left( \mathit{\Omega } \right)} \right)}^2 + {C_1}{h^4}\\ \left\| u \right\|_{{l^2}\left( {0,T;{H^2}\left( \mathit{\Omega } \right)} \right)}^2 \end{array} $
$ \begin{array}{l} \ \ \ \ \ \ {I^2} \le {C_2}\sum\limits_{i = 1}^N {{k^i}\left\| {{p^{i - 1}} - p_h^{i - 1}\left( u \right)} \right\|_{0,\mathit{\Omega }}^2} + {C_2}\sum\limits_{i = 1}^N {{k^i}} \\ \left\| {{u^i} - {\Pi _h}{u^i}} \right\|_{0,\mathit{\Omega }}^2 \le {C_2}\left\| {p - P\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}^2 + {C_2}{h^4}\\ \left\| u \right\|_{{l^2}\left( {0,T;{H^2}\left( \mathit{\Omega } \right)} \right)}^2 \end{array} $
$ \begin{array}{l} \ \ \ \ \ \ {I^3} \le {\varepsilon _3}\sum\limits_{i = 1}^N {{k^i}\left\| {p_h^{i - 1}\left( u \right) - p_h^{i - 1}} \right\|_{0,\mathit{\Omega }}^2} + \\ {C_3}\left( \varepsilon \right)\sum\limits_{i = 1}^N {{k^i}} \left\| {{u^i} - {\Pi _h}{u^i}} \right\|_{0,\mathit{\Omega }}^2 \le {C_3}{\left( \varepsilon \right)}{h^4}\\ \left\| u \right\|_{{l^2}\left( {0,T;{H^2}\left( \mathit{\Omega } \right)} \right)}^2 + {\varepsilon _3}{\left\| {{P_h} - {P_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{H^1}\left( \mathit{\Omega } \right)} \right)}} \le \\ {C_3}\left( \varepsilon \right){h^4}\left\| u \right\|_{{l^2}\left( {0,T;{H^2}\left( \mathit{\Omega } \right)} \right)}^2 + {\varepsilon _3}{\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \end{array} $
$ \begin{array}{l} \ \ \ \ \ \ {I^4} \le {C_4}\left( \varepsilon \right)\sum\limits_{i = 1}^N {{k^i}\left\| {p_h^{i - 1}\left( u \right) - p_h^{i - 1}} \right\|_{0,\mathit{\Omega }}^2} + \\ {\varepsilon _4}\sum\limits_{i = 1}^N {{k^i}} \left\| {{u^i} - U_h^i} \right\|_{0,\mathit{\Omega }}^2 \le {C_4}\left( \varepsilon \right)\left\| {p - P\left( u \right)} \right.\\ \left\| {_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}^2} \right. + {\varepsilon _4}{\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \end{array} $
$ \begin{array}{l} \ \ \ \ \ \ {I^5} \le {C_5}\left( \varepsilon \right)\sum\limits_{i = 1}^N {{k^i}\left\| {{p^{i - 1}} - {p^i}} \right\|_{0,\mathit{\Omega }}^2} + {\varepsilon _5}\sum\limits_{i = 1}^N {{k^i}} \\ \left\| {{u^i} - U_h^i} \right\|_{0,\mathit{\Omega }}^2 \le {C_5}\left( \varepsilon \right){k^2}\left\| {\frac{{\partial p}}{{\partial t}}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}^2 + {\varepsilon _5}\left\| {u - } \right.\\ {\left. {{U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \end{array} $

ε3+ε4+ε5=0.5c时,有C=max{C1, C2, C3, C4, C5},由此可以得到式(22),引理3得证。

引理4  令(y, p)与(Yh(u), Ph(u))分别为式(4)与式(12)的解,则以下估计成立

$ {\left\| {y - {Y_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le C\left( {{h^2} + k} \right) $ (23)
$ {\left\| {p - {P_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le C\left( {{h^2} + k} \right) $ (24)

证明  由式(4)减去式(12)可以得到误差方程$ \left( {\frac{{\partial (y - {Y_h}\left( u \right))}}{{\partial t}}, {w_h}} \right)$+a(y-Yh(u), wh)=0,∀whVh,注意到Yh(u)是y的标准有限元解,运用一般的有限元误差估计可以得到式(23)[9-10];类似地,可以得出式(24)。引理4得证。

定理1  令(y, p, u)和(Yh, Ph, Uh)是式(4)和式(6)的解,假设引理2~4的条件成立,则有

$ \begin{array}{l} {\left\| {y - {Y_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + {\left\| {p - {P_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + \\ {\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le C\left( {{h^2} + k} \right) \end{array} $ (25)

证明  由式(22)、(24)可知

$ \begin{array}{l} \;\;\;\;\;{\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le C\left( {{h^2} + k + \left\| {p - } \right.} \right.\\ \left. {{{\left. {{P_h}\left( u \right)} \right\|}_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}}} \right) \le C\left( {{h^2} + k} \right) \end{array} $ (26)

利用引理2、4以及式(26)可以推出

$ \begin{array}{l} \;\;\;\;\;{\left\| {p - {P_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + {\left\| {y - {Y_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le \\ {\left\| {p - {P_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + {\left\| {{P_h} - {P_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + \\ {\left\| {y - {Y_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + {\left\| {{Y_h} - {Y_h}\left( u \right)} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} \le \\ C{\left\| {u - {U_h}} \right\|_{{l^2}\left( {0,T;{L^2}\left( \mathit{\Omega } \right)} \right)}} + C\left( {{h^2} + k} \right) \le C\left( {{h^2} + k} \right) \end{array} $ (27)

结合式(26)、(27)可以得到式(25),定理1得证。

4 结束语

本文通过有限元离散,研究了一类控制约束集为积分受限形式的最优控制问题,并得到该问题先验误差的最优收敛阶。由于文中问题控制约束集的光滑性较好,所以得出的先验误差为O(h2+k),而不仅仅为O(h+k)。

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