2
$\begingroup$

Let $A: D(A) \subset H \rightarrow H$ generate a strongly continuous semigroup $T(t)$ on a Hilbert space $H$ and $B\in \mathcal{B}(H)$. Consider the two control systems: $$(1)\; x'(t)=Ax(t)+ Bu(t) \qquad \text{ and } \qquad (2)\; x'(t)=R(\lambda_0,A)x(t)+Bu(t),$$ where $\lambda_0 \in \rho(A): \Re \lambda_0 \ge\omega>\omega_0(T)$ (the type of $T(t)$).

In

H. O. Fattorini, Some remarks on complete controllability, Siam J. Control, 4 (1966), pp. 686-694.

it is proved that the approximate controllability of $(1)$ and of $(2)$ are equivalent. See for instance Proposition 2.3.

My question: is there any relation between exact null controllability of $(1)$ and of $(2)$? If so, any reference that consider this topic would be helpful.

$\endgroup$

1 Answer 1

3
$\begingroup$

No. Many PDE systems, e.g. the heat and wave equation, allow exact null controllability for controls restricted to a proper subset of the physical domain. On the other hand, resolvent operators are smoothing, so if the initial condition for (2) has a singularity outside the controlled region, there is no way for the control to remove it.

$\endgroup$
2
  • $\begingroup$ Thank you. Could you please clarify the last claim, say for heat equation on $L^2(\Omega)$ e.g., why (2) is not exactly null controllable ? $\endgroup$
    – S. Euler
    Commented Feb 28, 2021 at 17:29
  • $\begingroup$ If the initial condition has a singularity in the uncontrolled region, it will persist in time, because $R(\lambda_0,A)x$ is smoother than $x$ and $Bu$ is zero at the location of the singularity. $\endgroup$ Commented Feb 28, 2021 at 19:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .