Learning an Optimal Feedback Operator Semiglobally Stabilizing Semilinear Parabolic Equations

Abstract

Stabilizing feedback operators are presented which depend only on the orthogonal projection of the state onto the finite-dimensional control space. A class of monotone feedback operators mapping the finite-dimensional control space into itself is considered. The special case of the scaled identity operator is included. Conditions are given on the set of actuators and on the magnitude of the monotonicity, which guarantee the semiglobal stabilizing property of the feedback for a class of semilinear parabolic-like equations. Subsequently an optimal feedback control minimizing the quadratic energy cost is computed by a deep neural network, exploiting the fact that the feedback depends only on a finite dimensional component of the state. Numerical simulations demonstrate the stabilizing performance of explicitly scaled orthogonal projection feedbacks, and of deep neural network feedbacks.

Appendix

1.1 A.1 Proof of Lemma 3.3

Using Lemma 3.2 with $(y_1,y_2)=(y,0)$, we find

$$\begin{aligned} 2({{\mathcal {N}}}(y),Ay)_H&\le \gamma _2 \left| y\right| _{\text {D}(A)}^{2} +D_1 \sum \limits _{j=1}^n \left| y\right| _{\text {D}(A)}^\frac{2\zeta _{2j}}{1-\delta _{2j}} \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}}},\quad \text{ for } \text{ all }\quad \gamma _2>0, \end{aligned}$$

(A.1)

with $D_1{:}{=}\left( 1+\gamma _2^{-\frac{1+\Vert \delta _2\Vert }{1-\Vert \delta _2\Vert } }\right) {\overline{C}}_{{{\mathcal {N}}}1}$. Next, we recall the young inequality in the form

$$\begin{aligned} ab\le \tfrac{1}{s}\eta ^sa^s+\tfrac{s-1}{s}\eta ^{-\frac{s}{s-1}}a^\frac{s}{s-1}, \quad \text{ for } \text{ all }\quad a\ge 0,\;b\ge 0,\;\eta>0,\;s>1, \end{aligned}$$

(cf. [31, Appendix A.1]), which allow us to obtain, with

$$\begin{aligned} s=s_j=\tfrac{1-\delta _{2j}}{\zeta _{2j}}>1, \qquad \tfrac{s_j-1}{s_j}=\tfrac{1-\delta _{2j}-\zeta _{2j}}{1-\delta _{2j}},\qquad \text{ in } \text{ case }\quad \zeta _{2j}>0, \end{aligned}$$

the inequality

$$\begin{aligned}&\left| y\right| _{\text {D}(A)}^\frac{2\zeta _{2j}}{1-\delta _{2j}} \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}}}\\&\quad \le \tfrac{\zeta _{2j}}{1-\delta _{2j}}\eta _j^\frac{1-\delta _{2j}}{\zeta _{2j}}\left| y\right| _{\text {D}(A)}^2+\tfrac{1-\delta _{2j}-\zeta _{2j}}{1-\delta _{2j}} \eta _j^{-\frac{1-\delta _{2j}}{1-\delta _{2j}-\zeta _{2j}}} \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}}, \quad \text{ for } \text{ all }\quad \eta _j>0. \end{aligned}$$

Now, for any given $\gamma _3>0$, we choose

$$\begin{aligned} \eta _j= (\tfrac{1-\delta _{2j}}{\zeta _{2j}}\gamma _3)^\frac{\zeta _{2j}}{1-\delta _{2j}},\quad \text{ in } \text{ case }\quad \zeta _{2j}>0. \end{aligned}$$

which gives us

$$\begin{aligned}&\left| y\right| _{\text {D}(A)}^\frac{2\zeta _{2j}}{1-\delta _{2j}} \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}}}\\&\quad \le \gamma _3\left| y\right| _{\text {D}(A)}^2+\tfrac{1-\delta _{2j}-\zeta _{2j}}{1-\delta _{2j}} (\tfrac{1-\delta _{2j}}{\zeta _{2j}}\gamma _3)^{-\frac{\zeta _{2j}}{1-\delta _{2j}-\zeta _{2j}}} \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}}, \quad \text{ in } \text{ case }\quad \zeta _{2j}>0. \end{aligned}$$

Recalling (A.1), it follows that

$$\begin{aligned} 2({{\mathcal {N}}}(y),Ay)_H&\le \gamma _2 \left| y\right| _{\text {D}(A)}^{2} +D_1 \sum \limits _{\begin{array}{c} 1\le j\le n\\ \zeta _{2j}=0 \end{array}} \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}}}\nonumber \\&\quad +D_1 \sum \limits _{ \begin{array}{c} 1\le j\le n\\ \zeta _{2j}>0 \end{array}}\left( \gamma _3\left| y\right| _{\text {D}(A)}^2+ D_{2}\left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}}\right) \nonumber \\&\le (\gamma _2 +D_1\gamma _3 n)\left| y\right| _{\text {D}(A)}^{2} +D_1(1+D_2) \sum \limits _{j=1}^n \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}}, \end{aligned}$$

(A.2a)

for all $\gamma _2>0$, $\gamma _3>0$, with

$$\begin{aligned} D_2&=\max _{\begin{array}{c} 1\le j\le n\\ \zeta _{2j}\ne 0 \end{array}} \tfrac{1-\delta _{2j}-\zeta _{2j}}{1-\delta _{2j}} (\tfrac{1-\delta _{2j}}{\zeta _{2j}}\gamma _3)^{-\frac{\zeta _{2j}}{1-\delta _{2j}-\zeta _{2j}}}. \end{aligned}$$

(A.2b)

For an arbitrary given $\gamma _4>0$, we may set $\gamma _2=\frac{\gamma _4}{2}$ and $\gamma _3=\frac{\gamma _4}{2nD_1}$, leading us to

$$\begin{aligned} 2({{\mathcal {N}}}(y),Ay)_H \le \gamma _4 \left| y\right| _{\text {D}(A)}^{2} +D_1(1+D_2) \sum \limits _{j=1}^n \left| y\right| _{V}^{\frac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}}, \quad \text{ for } \text{ all }\quad \gamma _4>0. \end{aligned}$$

Observe that

$$\begin{aligned} \tfrac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}\ge 2 \quad&\Longleftrightarrow \quad \zeta _{1j}+\delta _{1j}\ge 1-\delta _{2j}-\zeta _{2j}\\&\Longleftrightarrow \quad \delta _{2j}+\delta _{1j}-1\ge -\zeta _{1j}-\zeta _{2j}, \end{aligned}$$

hence, from Assumption 2.4, we have that

$$\begin{aligned} p_j{:}{=}2-\tfrac{2\zeta _{1j}+2\delta _{1j}}{1-\delta _{2j}-\zeta _{2j}}\ge 0 \end{aligned}$$

and can write

$$\begin{aligned} 2({{\mathcal {N}}}(y),Ay)_H&\le \gamma _4 \left| y\right| _{\text {D}(A)}^{2} +D_1(1+D_2) \left| y\right| _{V}^2\sum \limits _{j=1}^n \left| y\right| _{V}^{p_j}\\&\le \gamma _4 \left| y\right| _{\text {D}(A)}^{2} +nD_1(1+D_2) \left| y\right| _{V}^2(1+\left| y\right| _{V}^{\Vert p\Vert }) \quad \text{ for } \text{ all }\quad \gamma _4>0, \end{aligned}$$

with

$$\begin{aligned} \Vert p\Vert&=\max _{1\le j\le n}p_j,\qquad D_1=\left( 1+(\tfrac{\gamma _4}{2})^{-\frac{1+\Vert \delta _2\Vert }{1-\Vert \delta _2\Vert } }\right) {\overline{C}}_{{{\mathcal {N}}}1},\\ \quad \text{ and }\quad D_2&=\max _{\begin{array}{c} 1\le j\le n\\ \zeta _{2j}\ne 0 \end{array}} \tfrac{1-\delta _{2j}-\zeta _{2j}}{1-\delta _{2j}} (\tfrac{1-\delta _{2j}}{\zeta _{2j}}\tfrac{\gamma _4}{2nD_1})^{-\frac{\zeta _{2j}}{1-\delta _{2j}-\zeta _{2j}}}. \end{aligned}$$

This ends the proof. $\square$