Bilinear Control of Convection-Cooling: From Open-Loop to Closed-Loop

Abstract

This paper is concerned with a bilinear control problem for enhancing convection-cooling via an incompressible velocity field. Both optimal open-loop control and closed-loop feedback control designs are addressed. First and second order optimality conditions for characterizing the optimal solution are discussed. In particular, the method of instantaneous control is applied to establish the feedback laws. Moreover, the construction of feedback laws is also investigated by directly utilizing the optimality system with appropriate numerical discretization schemes. Computationally, it is much easier to implement the closed-loop feedback control than the optimal open-loop control, as the latter requires to solve the state equations forward in time, coupled with the adjoint equations backward in time together with a nonlinear optimality condition. Rigorous analysis and numerical experiments are presented to demonstrate our ideas and validate the efficacy of the control designs.

Appendix

Proof of Corollary 2.5

Without loss of generality, we assume that $\langle T_0\rangle =0$. First, using the optimality condition (2.25) together with (2.8) and (2.27), we get

$$\begin{aligned}&\int ^{t_f}_0\Vert {\mathbf {v}}\Vert ^2_{H^2} \,dt\le C\int ^{t_f}_0\Vert q\nabla T\Vert ^2_{L^2}\,dt \le C\sup _{t\in [0, t_f]}\Vert q\Vert ^2_{L^\infty }\int ^{t_f}_0\Vert \nabla T\Vert ^2_{L^2}\,dt\le C(T_0, t_f). \end{aligned}$$

(5.1)

Moreover, by (2.14) and (2.27) we have

$$\begin{aligned}&\sup _{t\in [0, t_f]}\Vert {\mathbf {v}}\Vert _{H^2} \le C\sup _{t\in [0, t_f]}\Vert q\nabla T\Vert _{L^2} \le C\sup _{t\in [0, t_f]}\Vert q\Vert _{L^\infty }\sup _{t\in [0, t_f]} \Vert \nabla T\Vert _{L^2}\le C(T_0, t_f). \end{aligned}$$

(5.2)

To obtain a higher regularity of T, we take the inner product of (2.20) with $(-\Delta )^2T$ and get

$$\begin{aligned}&\frac{1}{2}\frac{d\Vert \Delta T\Vert ^2_{L^2}}{dt}+\kappa \Vert \nabla (-\Delta T)\Vert ^2_{L^2} =-({\mathbf {v}}\cdot \nabla T, (-\Delta )^2T) =(\nabla ({\mathbf {v}}\cdot \nabla T), \nabla ((-\Delta )T))\\&\quad \le C \Vert \nabla ({\mathbf {v}}\cdot \nabla T)\Vert _{L^2}\Vert \nabla ((-\Delta )T\Vert _{L^2} \le C (\Vert \nabla {\mathbf {v}}\cdot \nabla T\Vert _{L^2}\\&\quad \,\, +\Vert {\mathbf {v}}\cdot \nabla (\nabla T)\Vert _{L^2})\Vert \nabla ((-\Delta )T\Vert _{L^2}\\&\quad \le C (\Vert \nabla {\mathbf {v}}\Vert ^2_{H^1}\Vert \Delta T\Vert ^2_{L^2}+\Vert {\mathbf {v}}\Vert ^2_{L^\infty }\Vert \Delta T\Vert ^2_{L^2}) +\frac{\kappa }{2}\Vert \nabla ((-\Delta )T\Vert ^2_{L^2}. \end{aligned}$$

This follows

$$\begin{aligned}&\frac{d\Vert \Delta T\Vert ^2_{L^2}}{dt}+\kappa \Vert \nabla ((-\Delta )T\Vert ^2_{L^2} \le C (\Vert \nabla {\mathbf {v}}\Vert ^2_{H^1}+\Vert {\mathbf {v}}\Vert ^2_{L^\infty })\Vert \Delta T\Vert ^2_{L^2} \le C\Vert {\mathbf {v}}\Vert ^2_{H^2}\Vert \Delta T\Vert ^2_{L^2}, \end{aligned}$$

(5.3)

where we used Agmon’s inequality (3.32) in the last inequality. Therefore, applying (5.1) to (5.3) yields

$$\begin{aligned}&\sup _{t\in [0, t_f]} \Vert \Delta T\Vert _{L^2} \le e^{C\int ^{t_f}_0\Vert {\mathbf {v}}\Vert ^2_{H^2}\,dt}\Vert \Delta T_0\Vert _{L^2}<\infty \end{aligned}$$

(5.4)

and

$$\begin{aligned}&\kappa \int ^{t_f}_0 \Vert \nabla ((-\Delta )T\Vert ^2_{L^2} \,dt \le C\int ^{t_f}_0\Vert {\mathbf {v}}\Vert ^2_{H^2}\Vert \Delta T\Vert ^2_{L^2}\,dt+\Vert \Delta T_0\Vert ^2_{L^2}<\infty . \end{aligned}$$

(5.5)

This completes the proof. $\square$

Proof of Theorem 2.6

Let $h_i\in U_{\text {ad}}$ and $z_i=T'({\mathbf {v}})\cdot h_i, i=1,2$. Then we have

$$\begin{aligned} \begin{aligned}&\frac{\partial z_i}{\partial t}=\kappa \Delta z_i - {\mathbf {v}}\cdot \nabla z_i - h_i \cdot \nabla T, \quad \frac{\partial z_i}{\partial n}|_{\Gamma }=0,\\&z_i(x, 0)=0. \end{aligned} \end{aligned}$$

(5.6)

In light of Corollary 2.5, we can also obtain a higher regularity of $z_i, i=1,2,$ than (2.19). To see this, taking the inner produce of (5.6) with $-\Delta z_i$ and noting that $\langle z_i\rangle =0$ by (2.18), we obtain

$$\begin{aligned}&\frac{1}{2}\frac{d \Vert \nabla z_i\Vert ^2_{L^2}}{d t}+\kappa \Vert \Delta z_i\Vert ^2_{L^2}\le \Vert {\mathbf {v}}\Vert _{L^\infty } \Vert \nabla z_i\Vert _{L^2}\Vert \Delta z_i\Vert _{L^2} +\Vert h_i\Vert _{L^4}\Vert \nabla T\Vert _{L^4} \Vert \Delta z_i\Vert _{L^2}\nonumber \\&\qquad \le C \Vert {\mathbf {v}}\Vert ^2_{L^\infty } \Vert \nabla z_i\Vert ^2_{L^2} +C\Vert \nabla h_i\Vert ^2_{L^2}\Vert \Delta T\Vert ^2_{L^2} +\frac{\kappa }{2}\Vert \Delta z_i\Vert _{L^2}. \end{aligned}$$

(5.7)

Thus

$$\begin{aligned}&\frac{d \Vert \nabla z_i\Vert ^2_{L^2}}{d t}+\kappa \Vert \Delta z_i\Vert ^2_{L^2} \le C \Vert {\mathbf {v}}\Vert ^2_{L^\infty } \Vert \nabla z_i\Vert ^2_{L^2} +C\Vert \nabla h_i\Vert ^2_{L^2}\Vert \Delta T\Vert ^2_{L^2}, \end{aligned}$$

where by (5.4),

$$\begin{aligned}\int ^{t_f}_0\Vert \nabla h_i\Vert ^2_{L^2}\Vert \Delta T\Vert ^2_{L^2}\,dt \le \sup _{t\in [0, t_f]}\Vert \Delta T\Vert ^2_{L^2} \int ^{t_f}_0\Vert \nabla h_i\Vert ^2_{L^2}\,dt \le C(T_0, t_f) \Vert h_i\Vert ^2_{U_{\text {ad}}}. \end{aligned}$$

Consequently,

$$\begin{aligned}&\sup _{t\in [0, t_f]} \Vert \nabla z_i\Vert ^2_{L^2}\le \int ^{t_f}_0e^{C\int ^{t_f}_{\tau } \Vert {\mathbf {v}}\Vert ^2_{L^\infty }\,ds}\Vert \nabla h_i\Vert ^2_{L^2}\Vert \Delta T\Vert ^2_{L^2} \, d\tau \le C(T_0, t_f) \Vert h_i\Vert ^2_{U_{\text {ad}}} \end{aligned}$$

(5.8)

and

$$\begin{aligned} \kappa \int ^{t_f}_0\Vert \Delta z_i\Vert ^2_{L^2}\le C \int ^{t_f}_0 (\Vert {\mathbf {v}}\Vert ^2_{L^\infty } \Vert \nabla z_i\Vert ^2_{L^2} +\Vert \nabla h_i\Vert ^2_{L^2}\Vert \Delta T\Vert ^2_{L^2})\, dt\le C(T_0, t_f) \Vert h\Vert ^2_{U_{\text {ad}}}. \end{aligned}$$

Next, let $Z=z'_1({\mathbf {v}})\cdot h_2$. Then Z satisfies

$$\begin{aligned}&\frac{\partial Z}{\partial t}= \kappa \Delta Z - h_2 \cdot \nabla z_1 -{\mathbf {v}}\cdot \nabla Z-h_1 \cdot \nabla z_2, \quad \frac{\partial Z}{\partial n}|_{\Gamma }=0,\nonumber \\&Z(x, 0)=0, \end{aligned}$$

(5.9)

and $\langle Z\rangle =0$. Applying an $L^2$-estimate for Z gives

$$\begin{aligned}&\frac{1}{2}\frac{d\Vert Z\Vert ^2_{L^2}}{dt}+\kappa \Vert \nabla Z\Vert ^2_{L^2} \le \Vert \nabla h_2\Vert _{L^2} \Vert \nabla z_1\Vert _{L^2}\Vert \nabla Z\Vert _{L^2} +\Vert \nabla h_1\Vert _{L^2}\Vert \nabla z_2\Vert _{L^2}\Vert \nabla Z\Vert _{L^2}\\&\quad \le \Vert \nabla h_2\Vert ^2_{L^2} \Vert \nabla z_1\Vert ^2_{L^2}+\frac{\kappa }{4}\Vert \nabla Z\Vert ^2_{L^2} +\Vert \nabla h_1\Vert ^2_{L^2} \Vert \nabla z_2\Vert ^2_{L^2}+\frac{\kappa }{4}\Vert \nabla Z\Vert ^2_{L^2}, \end{aligned}$$

which, together with (5.8), follows

$$\begin{aligned}&\frac{d\Vert Z\Vert ^2_{L^2}}{dt}+\kappa \Vert \nabla Z\Vert ^2_{L^2} \le C(\Vert \nabla h_2\Vert ^2_{L^2} \Vert \nabla z_1\Vert ^2_{L^2}+\Vert \nabla h_1\Vert ^2_{L^2} \Vert \nabla z_2\Vert ^2_{L^2} )\\&\quad \le C(T_0, t_f)( \Vert \nabla h_2\Vert ^2_{L^2} \Vert h_1\Vert ^2_{U_{\text {ad}}} + \Vert \nabla h_1\Vert ^2_{L^2} \Vert h_2\Vert ^2_{U_{\text {ad}}}). \end{aligned}$$

Therefore,

$$\begin{aligned}&\Vert Z\Vert ^2_{L^2}+\kappa \int ^{t}_0\Vert \nabla Z\Vert ^2_{L^2}\,dt \le C(T_0, t_f) \Vert h_1\Vert ^2_{U_{\text {ad}}} \Vert h_2\Vert ^2_{U_{\text {ad}}}, \quad t\in [0,t_f]. \end{aligned}$$

(5.10)

By Lemma 2.2, (2.19) and (5.10), it can be easily verified that the terms on the right hand side of (5.9) are all in $L^1(0, t_f; (H^{1}(\Omega ))')$, and hence $\frac{\partial Z}{\partial t} \in L^1(0, t_f; (H^{1}(\Omega ))')$. Thus there exists a unique solution to (5.9), which implies that $T({\mathbf {v}})$ is twice G${\hat{a}}$teaux differentiable at ${\mathbf {v}}\in U_{\text {ad}}$ satisfying the optimality condition (2.22), with respect to $h_1$ and $h_2$, so is $J({\mathbf {v}})$.

Now differentiating $J'({\mathbf {v}})\cdot h_1$ once again in the direction $h_2\in U_{\text {ad}}$ gives

$$\begin{aligned} J''({\mathbf {v}}) \cdot (h_1,h_2)&=\alpha (D^*Dz_2(t_f), z_1(t_f)) + \alpha (D^*DT(t_f), Z(t_f)) \nonumber \\&\quad +\beta \int ^{t_f}_0(D^*Dz_2, z_1)\,dt\nonumber \\&\quad + \beta \int ^{t_f}_{0} (D^*DT, Z)\,dt + \gamma \int ^{t_f}_0(Ah_2, h_1)\,dt. \end{aligned}$$

(5.11)

Next taking the inner product of (5.9) with q and applying (2.3), we get

$$\begin{aligned} \alpha (D^*DT(t_f), Z(t_f))&-\int ^{t_f}_0\left( Z, \frac{\partial q}{\partial t}\right) \,dt=\kappa \int ^{t_f}_0(Z, \Delta q)\,dt +\int ^{t_f}_0( z_1, h_2 \cdot \nabla q) \,dt\\&+ \int ^{t_f}_0 ( Z, {\mathbf {v}}\cdot \nabla q)\,dt +\int ^{t_f}_0( z_2, h_1\cdot \nabla q)\,dt. \end{aligned}$$

With the help of the adjoint equations (2.21), we obtain

$$\begin{aligned}&\alpha (D^*D T(t_f), Z(t_f))+\beta \int ^{t_f}_0(Z, D^*DT)\,dt\\&\quad =\int ^{t_f}_0( z_1, h_2 \cdot \nabla q)\,dt+\int ^{t_f}_0( z_2, h_1\cdot \nabla q)\,dt. \end{aligned}$$

Therefore, (5.11) becomes

$$\begin{aligned} J''({\mathbf {v}}) \cdot (h_1,h_2)&= \alpha (D^*D z_2(t_f), z_1(t_f)) +\beta \int ^{t_f}_0(D^*D z_2, z_1)\,dt\\&\quad +\int ^{t_f}_0( z_1, h_2 \cdot \nabla q)\,dt\\&\quad +\int ^{t_f}_0( z_2, h_1\cdot \nabla q))\,dt + \gamma \int ^{t_f}_0(Ah_2, h_1)\,dt. \end{aligned}$$

Setting $h_1=h_2=h$ and $z_1=z_2=z=T'({\mathbf {v}})\cdot h$ follows

$$\begin{aligned} J''({\mathbf {v}}) \cdot (h, h)&=\alpha \Vert Dz(t_f)\Vert ^2_{L^2}+\beta \int ^{t_f}_0\Vert Dz\Vert ^2_{L^2}\,dt\nonumber \\&\quad +2\int ^{t_f}_0( z, h\cdot \nabla q)\,dt +\gamma \int ^{t_f}_0\Vert A^{1/2}h\Vert ^2_{L^2}\,dt. \end{aligned}$$

(5.12)

Furthermore, by (2.1), (2.19), (2.26) and (5.8), we get

$$\begin{aligned}&\Vert Dz(t_f)\Vert ^2_{L^2} \le \frac{C}{\kappa } \Vert T_0\Vert ^2_{L^\infty } \Vert h\Vert ^2_{U_{\text {ad}}}, \\&\int ^{t_f}_0\Vert Dz\Vert ^2_{L^2}\,dt \le C \int ^{t_f}_0\Vert \nabla z\Vert ^2_{L^2}\,dt \le \frac{C}{\kappa ^2} \Vert T_0\Vert ^2_{L^\infty } \Vert h\Vert ^2_{U_{\text {ad}}}, \end{aligned}$$

and

$$\begin{aligned}&\left| \int ^{t_f}_0( z, h\cdot \nabla q)\,dt\right| \le C\int ^{t_f}_0 \Vert \nabla z\Vert _{L^2}\Vert \nabla h\Vert _{L^2}\Vert \nabla q\Vert _{L^2}\,dt\\&\quad \le C\sup _{t\in [0, t_f]}\Vert \nabla z\Vert _{L^2}\left( \int ^{t_f}_0 \Vert \nabla h\Vert ^2_{L^2}\right) ^{1/2} \left( \int ^{t_f}_0 \Vert \nabla q\Vert ^2_{L^2}\,dt\right) ^{1/2} \le C(T_0, t_f) \Vert h\Vert ^2_{U_{\text {ad}}}. \end{aligned}$$

As a result,

$$\begin{aligned}&|J''({\mathbf {v}}) \cdot (h, h)| \le C(T_0, t_f)\left( \frac{\alpha }{\kappa }+ \frac{\beta }{\kappa ^2}\right) \Vert T_0\Vert ^2_{L^\infty } \Vert h\Vert ^2_{U_{\text {ad}}} +\gamma \Vert h\Vert ^2_{U_{\text {ad}}}\\&\quad =( C(T_0, t_f, \kappa , \alpha , \beta )+\gamma )\Vert h\Vert ^2_{U_{\text {ad}}} \end{aligned}$$

and

$$\begin{aligned} J''({\mathbf {v}}) \cdot (h, h)&\ge -2\int ^{t_f}_0( z, h\cdot \nabla q)\,dt +\gamma \int ^{t_f}_0\Vert A^{1/2}h\Vert ^2_{L^2}\,dt \\&= (\gamma -C(T_0, t_f, \kappa , \alpha , \beta ))\Vert h\Vert ^2_{U_{\text {ad}}}. \end{aligned}$$

Therefore, letting $\gamma$ large enough such that

$$\begin{aligned} \gamma -C(T_0, t_f, \kappa , \alpha , \beta ) \ge c_0>0, \end{aligned}$$

(5.13)

we obtain (2.29). $\square$

Remark 5.1

For $T_0\in L^2(\Omega )$ and ${\mathbf {v}}\in L^2(0, \infty ; H)$, $\Vert DT\Vert _{L^2}$ obeys an exponential decay rate in time.

Proof

Taking the inner product of (1.1) with $D^*DT$ and applying Greens’ formula and (2.3), we have

$$\begin{aligned} \frac{1}{2} \frac{d \Vert DT\Vert ^2_{L^2}}{dt}&=\kappa ( \Delta T, D^*DT)-({\mathbf {v}}\cdot \nabla T, D^*DT)\nonumber \\&=\kappa \langle \frac{\partial T}{\partial n}, D^*DT\rangle \rangle _{\Gamma } -\kappa (\nabla T, \nabla (D^*DT) )+( T, {\mathbf {v}}\cdot \nabla (D^*DT)). \end{aligned}$$

(5.14)

Since $\langle T\rangle =\langle T_0\rangle$ and $D^*D=D$, we have $\nabla (D^*DT) =\nabla (DT) =\nabla (T-\langle T\rangle )=\nabla T,$ and hence using Stokes formula follows

$$\begin{aligned} ( T, {\mathbf {v}}\cdot \nabla (D^*DT))=\frac{1}{2}\int _{\Omega }{\mathbf {v}}\cdot \nabla (T^2)\,dx=0. \end{aligned}$$

Therefore, (5.14) becomes

$$\begin{aligned}&\frac{1}{2} \frac{d \Vert DT\Vert ^2_{L^2}}{dt}+\kappa \Vert \nabla (DT)\Vert ^2_{L^2}=0. \end{aligned}$$

(5.15)

Further applying Grönwall’s inequality and Poincaré inequality we derive that

$$\begin{aligned} \Vert DT\Vert ^2_{L^2}\le e^{-C\kappa t}\Vert DT_0\Vert ^2_{L^2}, \end{aligned}$$

(5.16)

which establishes the claim. $\square$