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≤
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,y0∈H01(Ω),f∈L2(0, T; L2(Ω)),U=L2(Ω),X=L2(0, T; U)。
令控制集K={u(t)∈L2(Ω):a≤
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)≡{w∈Wm, q(Ω):w|∂Ω=0}, Hm(Ω)表示Wm, 2(Ω)。此外,文中的c或C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y∈H1(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p∈W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= ∪τ∈Thτ。
定义状态有限元空间Vh={vh∈Sh:vh|∂Ω=0},其中Sh={χ∈C(Ω):χτ∈P1, ∀τ∈Th};定义控制有限元空间Uh={χ∈L2(Ω):χ|τ∈P0, ∀τ∈Th},此处对控制有限元空间没有连续性要求。
令Wh=L2(0, T; Vh),Xh=L2(0, T; Uh),Kh={uh(t)∈Uh:a≤
$ \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
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-1∈Vh(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U到Uh的L2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v∈K时,显然有a≤
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。
若y、yh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, v∈X,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v∈H2(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, …, N;ζi=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‖≡‖ρ‖l2(0, T; L2(Ω))=
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)≥
$ \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‖θ‖a2=a(θ, θ)。
取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) |
取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i从N到M+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u∈l2(0, T; H2(Ω)),p∈l2(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} $ |
取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
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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