Exhaustion approximation for the control problem of the heat or Schrödinger semigroup on unbounded domains

Abstract

We consider the control problem of the heat equation on bounded and unbounded domains, and more generally the corresponding inhomogeneous equation for the Schrödinger semigroup. We show that if the sequence of null-controls associated to an exhaustion of an unbounded domain converges, then the solutions do in the same way, and that the control cost estimate carries over to the limiting problem on the unbounded domain. This allows to infer the controllability on unbounded domains by studying the control problem on a sequence of bounded domains.

Appendix A. The optimal feedback operator

As in the main part of the paper, let $\mathcal {H}$ and $\mathcal {U}$ be Hilbert spaces, H a lower semibounded self-adjoint operator on $\mathcal {H}$, $B:\mathcal {U}\rightarrow \mathcal {H}$ a bounded linear operator, and $T>0$. Denote by $\mathcal {B}^T:L^2((0,T),\mathcal {U})\rightarrow \mathcal {H}$ the controllability map associated to the system

$$\begin{aligned} \partial _t u(t) + H u(t) = Bf(t) \quad \text { for }\quad 0<t<T,\quad u(0)=u_0, \end{aligned}$$

(A.1)

with $u_0\in \mathcal {H}$ and $f\in L^2((0,T),\mathcal {U})$.

Clearly, the controllability map $\mathcal {B}^T$ is linear and bounded. In particular, given two null-controls f and $\tilde{f}$ for the initial datum $u_0$, one has

$$\begin{aligned} 0=(\mathrm {e}^{-TH}u_0 + \mathcal {B}^T f) - (\mathrm {e}^{-TH}u_0+\mathcal {B}^T \tilde{f}) = \mathcal {B}^T(f-\tilde{f}), \end{aligned}$$

(A.2)

so that the set of all null-controls for the initial datum $u_0$ is given by

$$\begin{aligned} f + {{\,\mathrm{Ker}\,}}\mathcal {B}^T \end{aligned}$$

(A.3)

for any such null-control f.

Recall that a mapping $F:\mathcal {H}\rightarrow L^2((0,T),\mathcal {U})$ is called a feedback operator associated to the system (A.1) if

$$\begin{aligned} \mathrm {e}^{-TH}+\mathcal {B}^T F=0, \end{aligned}$$

that is, if F maps every initial datum $u_0$ to a corresponding null-control.

The following well known abstract result from [21] guarantees the existence of bounded linear feedback operators for null-controllable systems. For convenience of the reader, we give the whole statement, but reproduce only the part of the proof that we need.

Proposition A.1

([21, Proposition 12.1.2]). Let $\mathcal {H}_1, \mathcal {H}_2,\mathcal {H}_3$ be Hilbert spaces, and let $X:\mathcal {H}_1 \rightarrow \mathcal {H}_3$, $Y:\mathcal {H}_2 \rightarrow \mathcal {H}_3$ be bounded linear operators. Then the following are equivalent:

• (a) ${{\,\mathrm{Ran}\,}}X \subset {{\,\mathrm{Ran}\,}}Y$;

• (b) There is $c>0$ such that $\Vert X^* z\Vert \le c \Vert Y^* z\Vert$ for all $z\in \mathcal {H}_3$;

• (c) There is a bounded linear operator $Z:\mathcal {H}_1 \rightarrow \mathcal {H}_2$ satisfying $X=YZ$.

Proof of$(a)\Rightarrow (c)$. By hypothesis, for every $x \in \mathcal {H}_1$, there is a unique $y \in ( {{\,\mathrm{Ker}\,}}Y )^\perp$ with $X x = Y y$, and we define $Z :\mathcal {H}_1 \rightarrow \mathcal {H}_2$ by $Z x = y$. By construction, this operator Z satisfies $X = YZ$, and it is easy to see that it is linear. It remains to show that Z is bounded. Since Z is everywhere defined, by the closed graph theorem, it suffices to show that Z is closed. To this end, let $(x_n)_n$ be a sequence in $\mathcal {H}_1$ such that $x_n \rightarrow x$ in $\mathcal {H}_1$ and $Z x_n \rightarrow z$ in $\mathcal {H}_2$. Since X and Y are bounded, this yields, on the one hand, that $X x_n \rightarrow X x$ in $\mathcal {H}_3$ and, on the other hand, that $X x_n = YZ x_n \rightarrow Yz$ in $\mathcal {H}_3$, so that $X x = Y z$. Taking into account that $( {{\,\mathrm{Ker}\,}}Y )^\perp$ is closed and $Z x_n \in ( {{\,\mathrm{Ker}\,}}Y )^\perp$, one has $z \in ( {{\,\mathrm{Ker}\,}}Y )^\perp$ and, hence, $Z x = z$, which proves the claim.$\square$

If the system (A.1) is null-controllable in time $T>0$, then the implication (a)$\Rightarrow$(c) in Proposition A.1 with $X=\mathrm {e}^{-TH}:\mathcal {H}\rightarrow \mathcal {H}$ and $Y=\mathcal {B}^T:L^2((0,T),\mathcal {U})\rightarrow \mathcal {H}$ yields that $F=-Z$ is a bounded linear feedback operator for this system.

Remark 4.2

As established in the proof of Lemma 4.2, we have $(\mathcal {B}^T)^*=B^*\mathrm {e}^{-(T-\cdot )H}$, so that

$$\begin{aligned} \Vert (\mathcal {B}^T)^*v_0 \Vert _{L^2((0,T),\mathcal {U})}^2 = \int \limits _0^T \Vert B^*\mathrm {e}^{-(T-s)H}v_0 \Vert _\mathcal {U}^2 \,\mathrm {d}s = \int \limits _0^T \Vert B^*\mathrm {e}^{-tH}v_0 \Vert _\mathcal {U}^2 \,\mathrm {d}s \end{aligned}$$

for all $v_0\in \mathcal {H}$. The equivalence (a)$\Leftrightarrow$(b) in Proposition A.1 with the choice $X=\mathrm {e}^{-TH}$ and $Y=\mathcal {B}^T$ as above therefore yields the equivalence between the null-controllability of (A.1) and the so-called final-state-observability of the corresponding homogeneous system as mentioned in Remark 4.7.

Given any two feedback operators F and $\tilde{F}$ for the system (A.1), one sees analogously to (A.2) that $0 = (\mathrm {e}^{-TH} + \mathcal {B}^T F) - (\mathrm {e}^{-TH} + \mathcal {B}^T \tilde{F}) = \mathcal {B}^T(F-\tilde{F})$, that is,

$$\begin{aligned} {{\,\mathrm{Ran}\,}}(F-\tilde{F}) \subset {{\,\mathrm{Ker}\,}}\mathcal {B}^T. \end{aligned}$$

Hence, denoting by P the orthogonal projection in $L^2((0,T),\mathcal {U})$ onto ${{\,\mathrm{Ker}\,}}\mathcal {B}^T$, the operator $\mathcal {F}^T:=(\mathrm {Id}-P)F$ is again a feedback operator for the system (A.1) and does not depend on the choice of F. In particular, one has $\Vert \mathcal {F}^T\Vert \le \Vert F\Vert$ for every bounded linear feedback operator F. Thus, $\mathcal {F}^T$ is a bounded linear feedback operator with minimal operator norm. Moreover, by definition of the orthogonal projection P, for every $u_0\in \mathcal {H}$, one has

$$\begin{aligned} \Vert \mathcal {F}^T u_0\Vert _{L^2((0,T),\mathcal {U})}&= \Vert Fu_0 - PFu_0\Vert _{L^2((0,T),\mathcal {U})}\\&= \min \{ \Vert Fu_0 - g\Vert _{L^2((0,T),\mathcal {U})} \mid g\in {{\,\mathrm{Ker}\,}}\mathcal {B}^T \}\\&= \min \{ \Vert f\Vert _{L^2((0,T),\mathcal {U})} \mid \mathrm {e}^{-TH}u_0 + \mathcal {B}^T f = 0 \}, \end{aligned}$$

where for the last equality, we have taken into account that by (A.3) the set of all null-controls for the initial datum $u_0$ is given by $Fu_0 + {{\,\mathrm{Ker}\,}}\mathcal {B}^T$. Thus, $\mathcal {F}^T u_0\in ({{\,\mathrm{Ker}\,}}\mathcal {B}^T)^\perp$ is the uniquely determined null-control associated to $u_0$ with minimal norm.