Time Optimal Controls of the Lengyel–Epstein Model with Internal Control

Abstract

This paper is concerned with the time optimal control problem governed by the internal controlled Lengyel–Epstein model. We prove the existence of optimal controls. Moreover, we give necessary optimality conditions for an optimal control of our original problem by using one of the approximate problems.

Appendix

Let $Y=H^{2,1}(Q)\cap L^\infty (0,T; V)$ and $(y^{w,v}, z^{w,v})\in Y\times Y$ be the solution to the following linear system

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {y_t-\triangle y+cy+4\xi z-a+\phi =\chi _\omega v(x, t)~~\hbox {in}~~ Q,}\\ \displaystyle {z_t-\delta \triangle z-b\theta y+\theta \xi z-\theta \phi =\chi _\omega w(x, t)~~~~~\hbox {in}~~ Q,}\\ y(x, 0)=y_0(x),~~ z(x,0)=z_0(x) ~~\hbox {in}~~ \Omega ,\\ y(x, t)=z(x, t)=0~~~~\hbox {on}~~\Sigma \equiv \partial \Omega \times (0,T). \end{array}\right. \end{aligned}$$

(4.1)

For $\varepsilon > 0$, let $(y_\varepsilon ,z_\varepsilon ,v_\varepsilon ,w_\varepsilon )$ be the optimal for problem

$$\begin{aligned} \min L_\varepsilon (y_\varepsilon ,z_\varepsilon , v_\varepsilon ,w_\varepsilon ) \end{aligned}$$

(4.2)

with

$$\begin{aligned}&L_\varepsilon (y,z, v,w):=\left\{ \frac{1}{2}\int \limits _{Q} e^{-\frac{3}{2}s\alpha }(v^2 + w^2)dxdt \right. \\&\qquad \left. +\frac{1}{2\varepsilon }\int \limits _{\Omega }(y^2(x,T) + z^2(x, T))dx\right\} \hbox {subject to}~(4.1). \end{aligned}$$

Then, we have

$$\begin{aligned} v_\varepsilon =\chi _\omega p_\varepsilon e^{\frac{3}{2}\alpha s}, w_\varepsilon =\chi _\omega q_\varepsilon e^{\frac{3}{2}\alpha s} ~~~\hbox {all most}~~~(x,t)\in Q, \end{aligned}$$

(4.3)

where $(p_\varepsilon , q_\varepsilon )$ is the solution to the dual backward system

$$\begin{aligned} \left\{ \begin{array}{ll} (p_\varepsilon )_t+(\triangle -c)p_\varepsilon +b\theta q_\varepsilon =0~~\hbox {in}~~ Q,\\ (q_\varepsilon )_t+(\delta \triangle -\xi \theta )q_\varepsilon -4\xi p_\varepsilon =0~~~~~\hbox {in}~~ Q,\\ p_\varepsilon (T)=-\dfrac{1}{\varepsilon }y_\varepsilon (T), q_\varepsilon (T)=-\dfrac{1}{\varepsilon }z_\varepsilon (T) ~~\hbox {in}~~ \Omega ,\\ p_\varepsilon (x,t)=q_\varepsilon (x,t)=0~~~~\hbox {on}~~\Sigma . \end{array}\right. \end{aligned}$$

(4.4)

Proof

Let $\bar{Y}=\{y\in Y,y(0)=0\}$, $(y,\bar{y})\in \bar{Y}\times \bar{Y}$ and $\chi _\omega v,\chi _\omega w\in L^2(0,T;H)$, setting $y_\varepsilon ^\lambda =y_\varepsilon +\lambda y,z_\varepsilon ^\lambda =z_\varepsilon +\lambda \bar{z}, v_\varepsilon ^\lambda =v_\varepsilon +\lambda v,w_\varepsilon ^\lambda =w_\varepsilon +\lambda w$, where $(y_\varepsilon ,z_\varepsilon ,v_\varepsilon ,w_\varepsilon )$ is optimal for problem $L_\varepsilon (y_\varepsilon ,z_\varepsilon , v_\varepsilon ,w_\varepsilon )$. It is clear that $(y_\varepsilon ^\lambda ,z_\varepsilon ^\lambda ,\chi _\omega v_\varepsilon ^\lambda ,\chi _\omega w_\varepsilon ^\lambda )\in \bar{Y}\times \bar{Y}\times L^2(0,T;H)\times L^2(0,T;H)$ and

$$\begin{aligned} (y_\varepsilon ^\lambda ,z_\varepsilon ^\lambda ,\chi _\omega v_\varepsilon ^\lambda ,\chi _\omega w_\varepsilon ^\lambda )\rightarrow (y_\varepsilon ,z_\varepsilon ,\chi _\omega v_\varepsilon , \chi _\omega w_\varepsilon ),~ \hbox {in}~ Y^2\times (L^2(0,T; H))^2~\hbox {as}~\lambda \rightarrow 0. \end{aligned}$$

(4.5)

It follows from (4.5) that

$$\begin{aligned} \begin{array}{rl} &{}\displaystyle {\frac{1}{2\varepsilon \lambda }\int \limits _{0}^T\left| \frac{dy_\varepsilon ^\lambda }{dt}-\triangle y_\varepsilon ^\lambda +cy_\varepsilon ^\lambda +4\xi z_\varepsilon ^\lambda -a+\phi -\chi _\omega v_\varepsilon ^\lambda \right| _{2}^2dt}\\ &{}\qquad -\displaystyle {\frac{1}{2\varepsilon \lambda }\int \limits _{0}^T\left| \frac{dy_\varepsilon }{dt}-\triangle y_\varepsilon +cy_\varepsilon +4\xi z_\varepsilon -a+\phi -\chi _\omega v_\varepsilon \right| _{2}^2dt}\\ &{}\quad \rightarrow \displaystyle {\int \limits _{0}^T\left\langle \tilde{p}_\varepsilon ,\frac{dy}{dt}-\triangle y+cy+4\xi z-\chi _\omega v\right\rangle dt~~~\hbox {as}~~\lambda \rightarrow 0} \end{array} \end{aligned}$$

(4.6)

and

$$\begin{aligned} \begin{array}{ll} &{}\displaystyle {\frac{1}{2\varepsilon \lambda }\int \limits _{0}^T\left| \frac{dz_{\varepsilon }^\lambda }{dt}-\delta \Delta z_{\varepsilon }^\lambda -b\theta y_{\varepsilon }^\lambda +\theta \xi z_{\varepsilon }^\lambda -\theta \phi -\chi _\omega w_{\varepsilon }^\lambda \right| _{2}^2dt}\\ &{}\qquad -\displaystyle {\frac{1}{2\varepsilon \lambda }\int \limits _{0}^T\left| \frac{dz_{\varepsilon }}{dt}-\delta \triangle z_{\varepsilon }-b\theta y_{\varepsilon }+\theta \xi z_{\varepsilon }-\theta \phi -\chi _\omega w_{\varepsilon }\right| _{2}^2dt}\\ &{}\quad \rightarrow \displaystyle {\int \limits _{0}^T\left\langle \tilde{q}_\varepsilon ,\frac{dz}{dt}-\delta \triangle z-b\theta y+\theta \xi z-\chi _\omega w\right\rangle dt~~~\hbox {as}~~\lambda \rightarrow 0}, \end{array} \end{aligned}$$

(4.7)

where $\tilde{p}_\varepsilon$ is given by

$$\begin{aligned} \tilde{p}_\varepsilon =\frac{1}{\varepsilon }\left[ \frac{dy_\varepsilon }{dt}-\triangle y_\varepsilon +cy_\varepsilon +4\xi z_\varepsilon -a+\phi -\chi _\omega v_\varepsilon \right] \end{aligned}$$

(4.8)

and $\tilde{q}_\varepsilon$ is given by

$$\begin{aligned} \tilde{q}_\varepsilon =\frac{1}{\varepsilon }\left[ \frac{dz_{\varepsilon }}{dt}-\delta \triangle z_{\varepsilon }-b\theta y_{\varepsilon }+\theta \xi z_{\varepsilon }-\theta \phi -\chi _\omega w_{\varepsilon }\right] . \end{aligned}$$

(4.9)

Now, employing the same arguments as in the proof of [21], we conclude that

$$\begin{aligned} \lim _{\lambda \rightarrow 0}\frac{1}{2\lambda }\int \limits _Qe^{-\frac{s\alpha }{2}}\Big [(v_\varepsilon ^\lambda )^2+(w_\varepsilon ^\lambda )^2- v_\varepsilon ^2-w_\varepsilon ^2 dx\Big ]dt=\int \limits _Qe^{-\frac{s\alpha }{2}}(v_\varepsilon v+w_\varepsilon w)dxdt \end{aligned}$$

(4.10)

$$\begin{aligned} \begin{array}{rl} &{}\displaystyle \lim _{\lambda \rightarrow 0}\displaystyle {\frac{1}{2\varepsilon \lambda }\int \limits _\Omega (y_\varepsilon ^\lambda )^2(x,T)+ (z_\varepsilon ^\lambda )^2(x,T)-y_\varepsilon ^2(x,T)-z_\varepsilon ^2(x,T)dx}\\ &{}\qquad =\displaystyle {\frac{1}{\varepsilon }\int \limits _\Omega y_\varepsilon (x,T)y(x,T)+z_\varepsilon (x,T)z(x,T)dx}. \end{array} \end{aligned}$$

(4.11)

Let

$$\begin{aligned} ~~~~p_\varepsilon =\tilde{p}_\varepsilon \in L^2(0,T;H),~~q_\varepsilon =\tilde{q}_\varepsilon \in L^2(0,T;H). \end{aligned}$$

(4.12)

By the Pontryagin’s maximum principle, we get

$$\begin{aligned} \displaystyle {\frac{L_\varepsilon (y_\varepsilon ^\lambda ,z_\varepsilon ^\lambda , v_\varepsilon ^\lambda ,w_\varepsilon ^\lambda )-L_\varepsilon (y_\varepsilon ,z_\varepsilon , v_\varepsilon ,w_\varepsilon )}{\lambda }\ge 0}~~~~\hbox {for all}~~~~\lambda >0. \end{aligned}$$

It then follows from (4.8)–(4.12) that for any $(y,z,v,w)\in \bar{Y}\times \bar{Y}\times L^2(0,T; H)\times L^2(0,T; H)$,

$$\begin{aligned} 0&\le \displaystyle {\int \limits _0^T\langle e^{-\frac{s\alpha }{2}}v_\varepsilon ,v\rangle +\langle e^{-\frac{s\alpha }{2}}w_\varepsilon ,w\rangle dt+\frac{1}{\varepsilon }\int \limits _\Omega y_\varepsilon (x,T)y(x,T)+z_\varepsilon (x,T)z(x,T)dx}\nonumber \\&+\displaystyle {\int \limits _{0}^T\left\langle p_\varepsilon ,\frac{dy}{dt}- \triangle y+cy+4\xi z-\chi _\omega v\right\rangle dt}\nonumber \\&+\displaystyle {\int \limits _{0}^T\left\langle q_\varepsilon ,\frac{dz}{dt}-\delta \triangle z-b\theta y+\theta \xi z-\chi _\omega w\right\rangle dt.} \end{aligned}$$

(4.13)

Let $w=v=y=0$ and $w=v=z=0$ in (4.13), we have

$$\begin{aligned} e^{-\frac{s\alpha }{2}}w_\varepsilon =q_\varepsilon \chi _\omega w \end{aligned}$$

and

$$\begin{aligned} e^{-\frac{s\alpha }{2}}v_\varepsilon =p_\varepsilon \chi _\omega v, \end{aligned}$$

which imply that

$$\begin{aligned} v_\varepsilon =\chi _\omega p_\varepsilon e^{\frac{3}{2}\alpha s}, w_\varepsilon =\chi _\omega q_\varepsilon e^{\frac{3}{2}\alpha s} ~~~\hbox {all most}~~~(x,t)\in Q. \end{aligned}$$

(4.14)

Now, taking $w=v=0$ in (4.13), then for any $(y,z)\in \bar{Y}\times \bar{Y}$, we have

$$\begin{aligned} \begin{array}{rl} 0=&{}\displaystyle {\frac{1}{\varepsilon }\int \limits _\Omega y_\varepsilon (x,T)y(x,T)+z_\varepsilon (x,T)z(x,T)dx}\\ &{}+\displaystyle {\int \limits _{0}^T\left\langle p_\varepsilon ,\frac{dy}{dt}-\triangle y+cy+4\xi z\right\rangle dt}\\ &{}+\displaystyle {\int \limits _{0}^T\left\langle q_\varepsilon ,\frac{dz}{dt}-\delta \triangle z-b\theta y+\theta \xi z\right\rangle dt.}\\ \end{array} \end{aligned}$$

(4.15)

Let $(\bar{p}_\varepsilon ,\bar{q}_\varepsilon )\in Y\times Y$ be the unique solution to

$$\begin{aligned} \left\{ \begin{array}{ll} -\displaystyle {\bar{p}_{\varepsilon ,t} -\triangle \bar{p}_\varepsilon +c\bar{p}_\varepsilon -b\theta \bar{q}_\varepsilon = 0},\\ -\displaystyle {\bar{q}_{\varepsilon ,t} -\delta \triangle \bar{q}_\varepsilon + 4\xi \bar{p}_\varepsilon +\xi \theta \bar{q}_\varepsilon = 0},\\ \bar{p}_\varepsilon (T)=-\frac{1}{\varepsilon }y_\varepsilon (T),~~\bar{q}_\varepsilon (T)=-\frac{1}{\varepsilon }z_\varepsilon (T),\\ \bar{p}_\varepsilon (x,t)=\bar{q}_\varepsilon (x,t)=0~~~~\hbox {on}~~\Sigma . \end{array}\right. \end{aligned}$$

(4.16)

Multiplying (4.16)$_1$ and (4.16)$_2$ by $y$ and $z$, respectively, integrating over $\Omega \times (0, T)$, we have

$$\begin{aligned} \frac{1}{\varepsilon }\int \limits _\Omega y_\varepsilon (x,T)y(x,T)dx+\int \limits _{0}^T\left\langle \bar{p}_{\varepsilon },\frac{dy}{dt}- \triangle y+cy\right\rangle dt+\int \limits _{0}^T\left\langle \bar{q}_{\varepsilon }, -b\theta y\right\rangle dt=0 \end{aligned}$$

(4.17)

and

$$\begin{aligned} \frac{1}{\varepsilon }\int \limits _\Omega z_\varepsilon (x,T)z(x,T)dx+\int \limits _{0}^T\left\langle \bar{q}_{\varepsilon },\frac{dz}{dt}-\delta \triangle z+\xi \theta z\right\rangle dt+\int \limits _{0}^T\left\langle \bar{p}_{\varepsilon }, 4\xi z\right\rangle dt=0, \end{aligned}$$

(4.18)

which, together with (4.15) imply that

$$\begin{aligned} \displaystyle {\int \limits _{0}^T\left\langle p_{\varepsilon }-\bar{p}_{\varepsilon },\frac{dy}{dt}- \triangle y+cy\right\rangle dt+\int \limits _{0}^T\langle q_{\varepsilon }-\bar{q}_{\varepsilon },-b\theta y\rangle dt=0} \end{aligned}$$

(4.19)

and

$$\begin{aligned} \displaystyle {\int \limits _{0}^T\left\langle q_{\varepsilon }-\bar{q}_{\varepsilon },\frac{dz}{dt}-\delta \triangle z+\xi \theta z\right\rangle dt+\int \limits _{0}^T\langle p_{\varepsilon }-\bar{p}_{\varepsilon },4\xi z\rangle dt=0}, \end{aligned}$$

(4.20)

which imply that

$$\begin{aligned} \begin{array}{ll} &{}\displaystyle {\int \limits _{0}^T\left\langle p_{\varepsilon }-\bar{p}_{\varepsilon },\frac{dy}{dt}- \triangle y+cy+4\xi z\right\rangle dt}\\ &{}\quad +\displaystyle {\int \limits _{0}^T\left\langle q_{\varepsilon }-\bar{q}_{\varepsilon },\frac{dz}{dt}-\delta \triangle z+\xi \theta z-b\theta y\right\rangle dt=0.}\\ \end{array} \end{aligned}$$

(4.21)

It is well known that for each $f,g\in L^2(0, T; H)$, there exists $(y,z)\in \bar{Y}\times \bar{Y}$ such that

$$\begin{aligned} \left\{ \begin{array}{ll} \ \displaystyle {\frac{dy}{dt}- \triangle y+cy+4\xi z=f~~t\in (0, T),}\\ \displaystyle {\frac{dz}{dt}-\delta \triangle z+\xi \theta z -b\theta y=g~~t\in (0, T)},\\ \end{array}\right. \end{aligned}$$

which, combined with (4.21) implies that $p_{\varepsilon }=\bar{p}_{\varepsilon },q_{\varepsilon }=\bar{q}_{\varepsilon }$ for almost $t\in (0, T)$. That is (4.4). This completes the proof. $\square$