From 6f4def5414f9519d1d565076a609b73ec38b5c0d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Thu, 25 Apr 2024 23:11:36 +0200 Subject: [PATCH] final updates ICT --- .../Introduction_to_control_theory.tex | 153 ++++++++++++------ 1 file changed, 100 insertions(+), 53 deletions(-) diff --git a/Mathematics/5th/Introduction_to_control_theory/Introduction_to_control_theory.tex b/Mathematics/5th/Introduction_to_control_theory/Introduction_to_control_theory.tex index ff5ab32..f05f5db 100644 --- a/Mathematics/5th/Introduction_to_control_theory/Introduction_to_control_theory.tex +++ b/Mathematics/5th/Introduction_to_control_theory/Introduction_to_control_theory.tex @@ -9,7 +9,7 @@ A function $\alpha: \RR_{\geq 0} \to \RR_{\geq 0}$ is said to be of \emph{class $\mathcal{K}$} if it is continuous, strictly increasing and $\alpha(0) = 0$. If, moreover, $\displaystyle \lim_{s \to \infty} \alpha(s) = \infty$, then $\alpha$ is said to be of \emph{class $\mathcal{K}^\infty$}. \end{definition} \begin{definition} - A function $\beta: \RR_{\geq 0} \times \RR_{\geq 0} \to \RR_{\geq 0}$ is said to be of \emph{class $\mathcal{KL}$} if, for each fixed $t \geq 0$, the function $\beta(\cdot, t)$ is of class $\mathcal{K}$ and, for each fixed $s \geq 0$, the function $\beta(s, \cdot)$ is decreasing and $\displaystyle \lim_{t \to \infty} \beta(s, t) = 0$. + A function $\beta: \RR_{\geq 0} \times \RR_{\geq 0} \to \RR_{\geq 0}$ is said to be of \emph{class $\mathcal{KL}$} if it is continuous, for each fixed $t \geq 0$, the function $\beta(\cdot, t)$ is of class $\mathcal{K}$ and, for each fixed $s \geq 0$, the function $\beta(s, \cdot)$ is decreasing and $\displaystyle \lim_{t \to \infty} \beta(s, t) = 0$. \end{definition} \begin{remark} An example of a function class $\mathcal{K}$ not in $\mathcal{K}^\infty$ is for example $\alpha(s)=\arctan(s)$. Examples of functions of class $\mathcal{KL}$ are for instance $\beta(s, t) = s\exp{-t}$ or $\beta(s, t) = \arctan(s/(t+1))$. @@ -61,7 +61,7 @@ \norm{\vf{X}(\vf{x}_0, t)}\leq k\norm{\vf{x}_0} \exp{-\lambda t}\quad\forall t\geq 0 $$ \end{itemize} - Moreover, in the last two cases, if $\mu$ can be picked as large as we want, then the equilibrium is said to be \emph{globally stable}. + Moreover, in the last two cases, if $\mu$ can be picked as large as we want, then the equilibrium is said to satisfy that property \emph{globally}. \end{definition} \begin{remark} Note that exponential stability implies asymptotic stability, which implies stability and attractivity. Moreover, it can be seen that asymptotically stability is equivalent to stability and attractivity. @@ -109,7 +109,7 @@ $$ Thus: $$ - \norm{\vf{X}(\vf{x}_0,t)}\leq k \exp{-\lambda t}\norm{\vf{x}_0}+\frac{\lambda}{2}\int_0^t\exp{-\lambda(t-s)}\norm{\vf{X}(\vf{x}_0,s)}\dd s + \norm{\vf{X}(\vf{x}_0,t)}\leq k \exp{-\lambda t}\!\norm{\vf{x}_0}+\frac{\lambda}{2}\int_0^t\!\exp{-\lambda(t-s)}\!\norm{\vf{X}(\vf{x}_0,s)}\dd s $$ And so: $$ @@ -146,7 +146,7 @@ with $w:\mathcal{O}\to \RR_{\geq 0}$ continuous and positive definite, then the origin is globally asymptotically stable. \end{theorem} \begin{proof} - As in the previous proof, we define $\mu:=\alpha_2^{-1}(\alpha_1(R/2))$ and we get $\dot{v}(t)\leq -w(\vf{X}(\vf{x}_0, t))$. Since $w$ is continuous and positive definite, $\exists \alpha_3\in\mathcal{K}^\infty$ such that $\alpha_3(\norm{\vf{x}})\leq w(\vf{x})$ for all $\vf{x}\in \overline{B(0,R)}$. Thus, $\dot{v}(t) \leq -\alpha_3(\norm{\vf{X}(\vf{x}_0, t)})\leq -\alpha_3(\alpha_2^{-1}(V(\vf{X}(x_0, t))))$. Now, in this case, one can proove that $\exists \beta \in \mathcal{KL}$ such that $v(t)\leq \beta(v(0), t)$ for all $t\geq 0$. But: + As in the previous proof, we define $\mu:=\alpha_2^{-1}(\alpha_1(R/2))$ and we get $\dot{v}(t)\leq -w(\vf{X}(\vf{x}_0, t))$. Since $w$ is continuous and positive definite, $\exists \alpha_3\in\mathcal{K}^\infty$ such that $\alpha_3(\norm{\vf{x}})\leq w(\vf{x})$ for all $\vf{x}\in \overline{B(0,R)}$. Thus, $\dot{v}(t) \leq -\alpha_3(\norm{\vf{X}(\vf{x}_0, t)})\leq -\alpha_3(\alpha_2^{-1}(V(\vf{X}(x_0, t))))$. Now, in this case, one can prove that $\exists \beta \in \mathcal{KL}$ such that $v(t)\leq \beta(v(0), t)$ for all $t\geq 0$. But: \begin{multline*} \alpha_1(\norm{\vf{X}(\vf{x}_0, t)})\leq V(\vf{X}(\vf{x}_0, t))=v(t) \leq \beta(v(0), t)=\\ = \beta(V(\vf{x}_0), t)\leq \beta(\alpha_2(\norm{\vf{x}_0}), t) \end{multline*} @@ -168,7 +168,7 @@ Let $V:\mathcal{O}\to \RR_{\geq 0}$ be a locally Lipschitz function such that: \begin{itemize} \item $0\in \Fr{G}$, with $G:=\{\vf{x}\in \mathcal{O}: V(\vf{x})=0\}$. - \item There exists a neighbourhood $U$ (called Chetaev surface) of the origin such that $D_f^+V(\vf{x})>0$ for all $\vf{x}\in U\cap G$ + \item There exists a neighbourhood $U$ (called Chetaev surface) of the origin such that $D_f^+V(\vf{x})>0$ for all $\vf{x}\in U\cap G$. \end{itemize} Then, the origin is unstable. \end{theorem} @@ -176,9 +176,9 @@ If the origin is asymptotically stable, then $\forall\varepsilon>0$ $\{f(\vf{x}): \norm{\vf{x}}\leq \varepsilon\}$ is a neighbourhood of the origin. \end{theorem} \begin{theorem} - If the origin is locally asymptotically stable with basin of attraction $\mathcal{A}$, then $\exists \lambda>0$ and $v:\mathcal{A}\to \RR_{\geq 0}$ $\mathcal{C}^\infty$, positive definite and proper (that is, $\displaystyle \lim_{d(\vf{x}, \Fr{\mathcal{A}})\to 0}v(\vf{x})=\infty$) such that: + If the origin is locally asymptotically stable with basin of attraction $\mathcal{A}$, then $\exists \lambda>0$ and $V\in \mathcal{C}^\infty(\mathcal{A},\RR_{\geq 0})$ positive definite and proper (that is, $\displaystyle \lim_{d(\vf{x}, \Fr{\mathcal{A}})\to 0}V(\vf{x})=\infty$) such that: $$ - D_f^+v(\vf{x})\leq -\lambda v(\vf{x})\quad\forall \vf{x}\in \mathcal{A} + D_f^+V(\vf{x})\leq -\lambda V(\vf{x})\quad\forall \vf{x}\in \mathcal{A} $$ \end{theorem} \subsubsection{Control design and stabilization of equilibrium points} @@ -194,7 +194,7 @@ \dot{\vf{\vf\chi}} = \vf\psi(t, \vf{x}, \vf\chi) \end{cases} \end{equation} - If $q=0$, then the feedback control law is called \emph{static}, whereas if $q>0$ it is called \emph{dynamic}. Moreover if both $\vf\varphi$ and $\vf\psi$ are independent of $t$, then the control law is called \emph{stationary} and if $\vf\psi$ and $\vf\chi$ are independent of $\vf{x}$, it is called \emph{open-loop control}. + If $q=0$, then the feedback control law is called \emph{static}, whereas if $q>0$ it is called \emph{dynamic}. Moreover if both $\vf\varphi$ and $\vf\psi$ are independent of $t$, then the control law is called \emph{stationary} and if $\vf\psi$ and $\vf\chi$ are independent of $\vf{x}$, it is called \emph{open-loop control}. The last two equations are called the \emph{feedback control laws}. \end{definition} \begin{theorem}[Kalmann's theorem] Consider the linear system $\dot{\vf{x}} = \vf{Ax} + \vf{Bu}$ with $\vf{A}\in \RR^{n\times n}$ and $\vf{B}\in \RR^{n\times p}$. Then, the system is controllable (or the pair $(\vf{A},\vf{B})$ is controllable) if and only if @@ -227,18 +227,21 @@ \begin{equation}\label{ICT:inputaffine} \dot{\vf{x}} = \vf{a}(\vf{x}) + \vf{b}(\vf{x})\vf{u} \end{equation} - there exists a continuous static feedback control law asymptotically stabilizing the origin. + there exists a continuous static stationary feedback control law asymptotically stabilizing the origin. \end{theorem} - \begin{theorem} + \begin{theorem}[Sonntag's theorem] Let $V$ be a SCLF for an input-affine system (\mcref{ICT:inputaffine}) Then, a stabilizing control law is given by: $$ \vf\psi=\begin{cases} - \vf{0} & \text{if } L_bV(\vf{x})=0 \\ - -\frac{L_aV(\vf{x})+\sqrt{(L_aV(\vf{x}))^2+\abs{L_bV(\vf{x})}^4}}{\abs{L_bV(\vf{x})}^2}\transpose{L_bV(\vf{x})} & \text{otherwise} + \vf{0} & \!\!\!\!\!\!\!\!\!\!\!\!\!\text{if } L_bV(\vf{x})=0 \\ + -\frac{L_aV(\vf{x})+\sqrt{(L_aV(\vf{x}))^2+\abs{L_bV(\vf{x})}^4}}{\abs{L_bV(\vf{x})}^2}\transpose{L_bV(\vf{x})} & \!\text{otherwise} \end{cases} $$ where $L_aV(\vf{x}):=\pdv{V}{\vf{x}}\vf{a}(\vf{x})$ and $L_bV(\vf{x}):=\pdv{V}{\vf{x}}\vf{b}(\vf{x})$. \end{theorem} + \begin{remark} + This $\psi$ is as smooth as $L_aV$ and $L_bV$ on $\RR^n\setminus\varnothing$. And if $V$ is a SCLF continuously at the origin, then $\vf\psi$ is continuous at the origin. + \end{remark} \subsubsection{Backstepping} Consider a system of the form: \begin{equation}\label{ICT:backstepping} @@ -253,12 +256,12 @@ We define $\vf\eta$ such that $$ \begin{cases} - \pdv{V}{\vf{y}}(\vf{x}, \vf{\eta}(x)) = \vf{0} \\ + \pdv{V}{\vf{y}}(\vf{x}, \vf{\eta}(\vf{x})) = \vf{0} \\ \vf\eta(\vf{0}) = \vf{0} \end{cases} $$ \begin{lemma} - If $V$ is a $\mathcal{C}^2$ function and $\vf\eta$ is a $1/2$-Hölder continuous function, then $W(\vf{x}):=V(\vf{x}, \vf{\eta}(\vf{x}))$ is a SCLF for the system $\dot{\vf{x}} = \vf{f}(\vf{x}, \vf{v})$. + If $V$ is a $\mathcal{C}^2$ function and $\vf\eta$ is a locally $1/2$-Hölder continuous function, then $W(\vf{x}):=V(\vf{x}, \vf{\eta}(\vf{x}))$ is a SCLF for the system $\dot{\vf{x}} = \vf{f}(\vf{x}, \vf{v})$. \end{lemma} Finally we consider $$ @@ -267,8 +270,21 @@ with $\vf\varphi$ such that $\vf\varphi(\vf{x}, \vf{y}) = \vf{0}\iff \vf{y} = \vf{\eta}(\vf{x})$. Then, this $V$ is a SCLF for the system of \mcref{ICT:backstepping}. \begin{remark} Usually we take $\vf\varphi(\vf{x}, \vf{y}) = \vf{y}-\vf{\eta}(\vf{x})$ and consider + \begin{equation}\label{ICT:backsteppingV} + V(\vf{x}, \vf{y}) = V(\vf{x}, \vf{\eta}(\vf{x})) + \frac{1}{2}\norm{\vf{y}-\vf{\eta}(\vf{x})}^2 + \end{equation} + \end{remark} + In practice, we first look for a SCLF $W$ for the system $\dot{\vf{x}} = \vf{f}(\vf{x}, \vf{v})$, and then we find $v=\vf\eta(\vf{x})$ such that $\dot{W} < 0$. Finally, we construct $V$ as in \mcref{ICT:backsteppingV}. And we could iterate this process. + \begin{remark} + This method is only valid for systems in \emph{strict-feedback form}, that is, systems of the form: $$ - V(\vf{x}, \vf{y}) = W(\vf{x}) + \frac{1}{2}\left(\vf{y}-\vf{\eta}(\vf{x})\right)^2 + \begin{cases} + \dot{x}_1 = f_1(x_1, x_2) \\ + \dot{x}_2 = f_2(x_1, x_2, x_3) \\ + \quad\;\;\vdots \\ + \dot{x}_{n-1} = f_{n-1}(x_1, \ldots, x_n) \\ + \dot{x}_n = f_n(x_1, \ldots, x_n, u) + \end{cases} $$ \end{remark} \subsection{Control theory in PDEs} @@ -311,7 +327,7 @@ $$ \end{proof} \begin{definition}[Feedback stabilization] - Given $\dot{\vf{x}}=\vf{Ax}+\vf{Bu}$, the \emph{feedback stabilization process} consists in finding an operator $K:\mathcal{X}\to\mathcal{U}$ such that $\dot{\vf{x}}=\vf{Ax}+\vf{B}\vf{K}\vf{x}$ has a stable (or asymptotically stable) equilibrium at the origin? + Given $\dot{\vf{x}}=\vf{Ax}+\vf{Bu}$, the \emph{feedback stabilization process} consists in finding an operator $K:\mathcal{X}\to\mathcal{U}$ such that $\dot{\vf{x}}=\vf{Ax}+\vf{B}\vf{K}\vf{x}$ has a stable (or asymptotically stable) equilibrium at the origin. \end{definition} \begin{definition}[Optimal control] Let $J$ be a cost function, $J=J(\vf{x}, \vf{u},\vf{x}(T))$. The \emph{optimal control problem} consists in finding $\vf{u}: [0,T]\to \mathcal{U}$ such that $J$ is minimized, where $\vf{x}$ satisfies $\dot{\vf{x}} = \vf{Ax} + \vf{Bu}$ with $\vf{x}(0) = \vf{x}_0$. @@ -362,7 +378,7 @@ $$ So we have uniqueness and a formula: \begin{align}\label{ICT:heat_equation_solution} - v(t,x) & =\sum_{i\in\NN}\exp{-\lambda_i t}v_i(0)e_i(x)+\sum_{i\in\NN}\int_0^t\exp{-\lambda_i(t-s)} f_i(s)e_i(x)\dd s \\ + v(t,x) & =\sum_{i\in\NN}\exp{-\lambda_i t}v_i(0)e_i(x)+\sum_{i\in\NN}\int_0^t\!\exp{-\lambda_i(t-s)} f_i(s)e_i(x)\dd s \\ & =:v_a(t,x)+v_b(t,x)\nonumber \end{align} For the existence, it suffices to check that the solution in \mcref{ICT:heat_equation_solution} belongs to the desired space. We first check that $v\in \mathcal{C}^0([0,T]; H_0^1(\Omega))$. We have: @@ -401,12 +417,12 @@ $$ \end{proposition} \begin{proof} - We can sppose that all functions are smooth (otherwise we replace them by a linear combination of $e_i$ and pass to the limit using the fact that $(v_0,f)\mapsto v$ is continuous from $L^2(\Omega)\times L^2((0,T)\times \Omega)\to \mathcal{C}^0([0,T]; L^2(\Omega))\cap L^2((0,T); L^2(\Omega))$). Now, multiplying \mcref{ICT:heat_equation_control} by $\theta$ and integrating we get: + We can suppose that all functions are smooth (otherwise we replace them by a linear combination of $e_i$ and pass to the limit using the fact that $(v_0,f)\mapsto v$ is continuous from $L^2(\Omega)\times L^2((0,T)\times \Omega)\to \mathcal{C}^0([0,T]; L^2(\Omega))\cap L^2((0,T); L^2(\Omega))$). Now, multiplying \mcref{ICT:heat_equation_control} by $\theta$ and integrating we get: \begin{multline*} \int_0^T\int_\Omega \indi{\omega}u\theta = \int_0^T\int_\Omega \partial_t v\theta + \int_0^T\int_\Omega \nabla v \nabla \theta=\\=\int_\Omega v\theta\bigg|_0^T-\int_0^T\int_\Omega \partial_t \theta v + \int_0^T\int_\Omega \nabla v \nabla \theta = \int_\Omega v\theta\bigg|_0^T \end{multline*} \end{proof} - \begin{definition} + \begin{definition}[Observability inequality]\label{ICT:observability_inequality} We will say that the dual problem \mcref{ICT:heat_equation_control_dual} satisfies the \emph{finite-time observability inequality} if $\exists C>0$ such that $\forall \theta_T\in L^2(\Omega)$ the solution $\theta$ satisfies: $$ \norm{\theta(0)}_{L^2(\Omega)}^2\leq C\int_0^T\int_\omega \theta^2 @@ -416,19 +432,20 @@ If the dual problem \mcref{ICT:heat_equation_control_dual} is finite-time observable, then the control problem \mcref{ICT:heat_equation_control} is null controllable. \end{proposition} \begin{proof} - Note that the null controllability condition is equivalent to $\forall \theta_T\in L^2(\Omega)$ we have (by \mnameref{ICT:duality}): + Note that the null controllability condition is equivalent to $\forall \theta_T\in L^2(\Omega)$ we have (by \mcref{ICT:duality}): $$ -\langle \theta(0),v_0\rangle_{L^2(\Omega)} = \int_0^T\int_\omega u \theta $$ Now let's define: - $$ - A:=\overline{\{\indi{\omega}\theta:\theta \text{ solution of \mcref{ICT:heat_equation_control_dual} for some }\theta_T\in L^2(\Omega)\}}^{{}_{L^2((0,T)\times \omega)}} - $$ + \begin{align*} + B & :=\{\indi{\omega}\theta:\theta \text{ solution of \mcref{ICT:heat_equation_control_dual} for some }\theta_T\in L^2(\Omega)\} \\ + A & :=\overline{B}^{{}_{L^2((0,T)\times \omega)}} + \end{align*} We equip $A$ with the norm $\norm{\cdot}_{L^2((0,T)\times \omega)}$. Now consider: $$ \function{\Phi}{L^2(\Omega)}{L^2((0,T)\times \omega)}{\theta_T}{\indi{\omega}\theta} $$ - Note that $\overline{\Phi}=A$. Now, to any $\phi\in\im(\Phi)$ we could a priori associate several $\theta_T$, but all of them would generate the same $\theta_0$ due to the observability condition. So we may consider the map: + Note that $\overline{\im\Phi}=A$. Now, to any $\phi\in\im(\Phi)$ we could a priori associate several $\theta_T$, but all of them would generate the same $\theta(0)$ due to the observability condition. So we may consider the map: $$ \function{}{\im(\Phi)}{L^2(\Omega)}{\indi{\omega}\theta}{\theta(0)} $$ @@ -477,11 +494,11 @@ $$ \end{lemma} \begin{lemma}[Interior regulariy]\label{ICT:interiorregularity} - Let $w$ be a solution of the heat equation ($w\in \mathcal{C}^0([0,T]; H_0^1(\Omega))\cap L^2((0,T); H^2(\Omega)\cap H_0^1(\Omega))$). Then: + Let $w$ be a solution of the heat equation ($w\in \mathcal{C}^0([0,T]; H_0^1(0,1))\cap L^2((0,T); H^2(0,1)\cap H_0^1(0,1))$). Then: $$ \norm{w}_{L^\infty([T/2,T]\times [a,b])}\leq C\norm{w}_{L^2([T/4,T] \times [c,d])} $$ - with $c0$ such that for any solution $\theta$ of \mcref{ICT:wave_equation_control_dual2} we have: $$ - \norm{\theta(0)}_{H_0^1}+\norm{\partial_t\theta(0)}_{L^2}\leq C\norm{\partial_{\vf{n}}\theta}_{L^2(\Sigma_T)} + \norm{\theta(T)}_{H_0^1}+\norm{\partial_t\theta(T)}_{L^2}\leq C\norm{\partial_{\vf{n}}\theta}_{L^2(\Sigma_T)} $$ \end{definition} \begin{remark} @@ -655,10 +673,10 @@ \begin{proof} Suppose \mcref{ICT:wave_equation_control_dual2} is exactly observable. We make the choice to find $u$ of the form $u=\partial_{\vf{n}}\tilde{\theta}$ for some $\tilde{\theta}$ solution of \mcref{ICT:wave_equation_control_dual2} (in order to put the problem in the standard Riesz's form). Consider now $E:= H_0^1(\Omega)\times L^2(\Omega)$ equipped with the norm $\norm{(\theta_0,\theta_1)}_E:=\norm{\partial_{\vf{n}}\theta_0}_{L^2(\Sigma_T)}$, where $\theta$ is the solution of \mcref{ICT:wave_equation_control_dual2}. This is an equivalent norm to the standard one: \begin{align*} - \norm{(\theta_0,\theta_1)}_E & \gtrsim \norm{\theta_0}_{H_0^1}+\norm{\theta_1}_{L^2} \text{(by observability inequality)} \\ - \norm{(\theta_0,\theta_1)}_E & \lesssim \norm{\theta_0}_{H_0^1}+\norm{\theta_1}_{L^2} \text{(by \mnameref{ICT:hiddenregularity})} + \norm{(\theta_0,\theta_1)}_E\! & \gtrsim \!\norm{\theta_0}_{H_0^1}\!\!+\!\!\norm{\theta_1}_{L^2} \text{(\mnameref{ICT:observability_inequality})} \\ + \norm{(\theta_0,\theta_1)}_E\! & \lesssim \!\norm{\theta_0}_{H_0^1}\!\!+\!\!\norm{\theta_1}_{L^2} \text{(\mnameref{ICT:hiddenregularity})} \end{align*} - $E$ is Hilber with this norm. Now, given $(\hat{v}_0,\hat{v}_1)\in E$, the left hand side is a continuous linear form on $(\theta(T),\partial_t\theta(T))\in E$. So $\exists (\overline{\theta}_0,\overline{\theta}_1)\in E$ such that with $\overline{\theta}$ the corresponding solution of \mcref{ICT:wave_equation_control_dual2} one has: $\forall (\theta(T),\partial_t\theta(T))\in E$ with the corresponding solution $\theta$ we have: + $E$ is Hilbert with this norm. Now, given $(\hat{v}_0,\hat{v}_1)\in E$, the left hand side is a continuous linear form on $(\theta(T),\partial_t\theta(T))\in E$. So $\exists (\overline{\theta}_0,\overline{\theta}_1)\in E$ such that with $\overline{\theta}$ the corresponding solution of \mcref{ICT:wave_equation_control_dual2} one has: $\forall (\theta(T),\partial_t\theta(T))\in E$ with the corresponding solution $\theta$ we have: $$ \langle \hat{v}_1,\theta(T)\rangle_{H^{-1}\times H_0^1}-\langle \hat{v}_0,\partial_t\theta(T)\rangle_{L^2\times L^2}=\int_{\Sigma_T}\partial_{\vf{n}}\overline{\theta}\partial_{\vf{n}}\theta $$ @@ -698,7 +716,7 @@ \end{definition} \subsubsection{Semigroups} From now on we will assume $X=Y$. - \begin{definition} + \begin{definition}\label{ICT:semigroup} A one-parameter family of unbounded operators $(T(t))_{t\geq 0}$ is a \emph{semigroup of operators} if: \begin{itemize} \item $T(0)=\id$ @@ -713,9 +731,9 @@ \end{itemize} \end{definition} \begin{definition} - Let $(T(t))_{t\geq 0}$ be a semigroup of operators. We call \emph{infinitessimal generator} of $(T(t))_{t\geq 0}$ the unbounded operator $(D(A),A)$ where + Let $(T(t))_{t\geq 0}$ be a semigroup of operators. We call \emph{infinitesimal generator} of $(T(t))_{t\geq 0}$ the unbounded operator $(D(A),A)$ where $$ - D(A)=\{ x\in X:\exists \lim_{t\to 0^+}\frac{T(t)x-x}{t}\} + D(A)=\left\{ x\in X:\exists \lim_{t\to 0^+}\frac{T(t)x-x}{t}\right\} $$ and $\forall x\in D(A)$ we define: $$ @@ -726,18 +744,47 @@ Let $(T(t))_{t\geq 0}$ be a strongly continuous semigroup of operators. Then: \begin{enumerate} \item $\forall x\in X$, $t\mapsto T(t)x$ is continuous. - \item $\forall x\in X$ and all $t\geq 0$: + \item\label{ICT:item2} $\forall x\in X$ and all $t\geq 0$: $$ - \int_0^t T(s)x\dd s\in D(A)\text{ and } T(t)x-x=A\left( \int_0^t T(s)x\dd s\right) + \int_0^t T(s)x\dd s\in\! D(A)\text{ and } T(t)x-x=A\!\left( \int_0^t\! T(s)x\dd s\!\right) $$ \item $\forall x\in D(A)$, $\dv{}{t}T(t)x=AT(t)x=T(t)Ax$. \item $\forall x\in D(A)$ and all $t,s\geq 0$: $$ T(t)x-T(s)x=\int_s^tA T(r)x\dd r=\int_s^t T(r)Ax\dd r $$ - \item $\exists c,C>0$ such that $\norm{T(t)}\leq C\exp{ct}$. + \item $\exists \alpha,C>0$ such that $\norm{T(t)}\leq C\exp{\alpha t}$. \end{enumerate} \end{proposition} + \begin{proof}\hfill + \begin{enumerate} + \item Take $x\in X$, $s,t\geq 0$ and $y:=T(s)x$. Then: + \begin{multline*} + \norm{T(t)x-T(s)x}=\norm{T(t-s)T(s)x-T(s)x}=\\ + =\norm{T(t-s)y-y}\underset{t\to s^+}{\longrightarrow} 0 + \end{multline*} + because of the strong continuity of the semigroup and \mcref{ICT:semigroup}. + \item We have: + \begin{multline*} + \frac{T(h)-\id}{h}\!\! \int_0^t\!\! T(s)x\dd s = \frac{1}{h}\!\int_0^t\!\!T(s+h) x\dd s - \frac{1}{h}\!\int_0^t\!\! T(s)x\dd s \\ + = \frac{1}{h}\int_0^{t+h}T(s)\dd{s}-\frac{1}{h}\int_0^t T(s)\dd s=\\=\frac{1}{h}\int_t^{t+h}T(s)\dd s-\frac{1}{h}\int_0^h T(s)\dd s\underset{h\to 0^+}{\longrightarrow} T(t)x-x + \end{multline*} + \item Let $x\in D(A)$ then the following limit + \begin{multline*} + \lim_{h\to 0^+}\frac{T(t+h)x-T(t)x}{h}=\lim_{h\to 0^+}T(t) \frac{T(h)x-x}{h}=\\=T(t)Ax + \end{multline*} + exists because of $x\in D(A)$. Moreover using the properties of the semigroup we have it is also equal to $AT(t)x$. Now assume $h\to 0^-$ (so $h<0$). Then: + \begin{equation*} + \lim_{h\to 0^-}\frac{T(t+h)x-T(t)x}{h}=T(t+h)\frac{T(-h)x-x}{-h} + \end{equation*} + which exists and is equal to $AT(t)x$ because $x\in D(A)$. + \item We prove it for $s=0$, and then the general case follows by the linearity of the integral. But then by \mcref{ICT:item2}: + \begin{equation*} + T(t)x-x= A\int_0^t T(s)x\dd s=\int_0^t AT(s)x\dd s + \end{equation*} + and the exchange of the limit and the integral is justified by the existence of both limits. + \end{enumerate} + \end{proof} \begin{theorem}\hfill \begin{enumerate} \item If $A$ is the infinitesimal generator of a strongly continuous semigroup, then it is closed and densely defined. @@ -750,10 +797,10 @@ \end{enumerate} \end{theorem} \begin{remark} - Duhamel's formula is still valid if $x_0\in X$ and $f\in L^1((0,T),X)$. In this case, the resulting solution $x(t)$ is called \emph{mild solution}. Any mild solution is a limit of classical solutions. + Duhamel's formula is still valid even if $x_0\in X$ and $f\in L^1((0,T),X)$. In this case, the resulting solution $x(t)$ is called \emph{mild solution}. Any mild solution is a limit of classical solutions. \end{remark} \begin{theorem} - Suppose $X$ is reflexive. Then, if $(T(t))_{t\geq 0}$ is a strongly continuous semigroup with infinitesimal generator $A$, then $A^*$ is the infinitesimal generator of the dual semigroup $(T(t)^*)_{t\geq 0}$. + Suppose $X$ is reflexive. Then, if $(T(t))_{t\geq 0}$ is a strongly continuous semigroup with infinitesimal generator $A$, then $A^*$ is the infinitesimal generator of the adjoint semigroup $(T(t)^*)_{t\geq 0}$. \end{theorem} \begin{remark} We will denote by $T(t)^*$ also by $T^*(t)$. @@ -762,7 +809,7 @@ A semigroup $(T(t))_{t\geq 0}$ is called \emph{contraction} if $\norm{T(t)}\leq 1$ $\forall t\geq 0$. \end{definition} \begin{definition} - Let $(D(A),A)$ be an unbounded operator. The \emph{resolvency set} of $A$ is the set $\rho(A):=\{ \lambda\in \CC:\lambda I-A\text{ is bijective}\}$. Given $\lambda\in\rho(A)$, the \emph{resolvent operator} is the operator $R_\lambda(A):=(\lambda I-A)^{-1}$. + Let $(D(A),A)$ be an unbounded operator. The \emph{resolvent set} of $A$ is the set $\rho(A):=\{ \lambda\in \CC:\lambda I-A\text{ is bijective}\}$. Given $\lambda\in\rho(A)$, the \emph{resolvent operator} is the operator $R_\lambda(A):=(\lambda I-A)^{-1}$. \end{definition} \begin{theorem}[Hille-Yosida] Let $(D(A),A)$ be an operator closed and densely defined. Then, it is the infinitesimal generator of a contraction semigroup if and only if: @@ -771,7 +818,7 @@ $$ \end{theorem} \begin{corollary} - $(D(A),A)$ is the infinitesimal generator of a semigroup $(T(t))_{t\geq 0}$ such that $\norm{T(t)}\leq \exp{ct}$ $\forall t\geq 0$ if and only if: + Let $(D(A),A)$ be an operator closed and densely defined. $(D(A),A)$ is the infinitesimal generator of a semigroup $(T(t))_{t\geq 0}$ such that $\norm{T(t)}\leq \exp{ct}$ $\forall t\geq 0$ if and only if: $$ (c,\infty)\subseteq \rho(A)\text{ and }\forall \lambda>c, \norm{R_\lambda(A)}\leq \frac{1}{\lambda-c} $$ @@ -806,7 +853,7 @@ \function{F_T}{L^2(0,T;L^2(\omega))}{L^2(\Omega)}{u}{\int_0^T S(T-s)Bu(s)\dd s} $$ \begin{remark} - Note that exact controllability at time $T$ is equivalent to the surjectivity of $F_T$; approximate controllability at time $T$ is equivalent to $\im F_T$ being dense in $L^2(\Omega)$, and null controllability at time $T$ is equivalent to $\im F_T\supseteq \im S(T)$. + Note that exact controllability at time $T$ is equivalent to controllability starting from 0 which in turn is equivalent to the surjectivity of $F_T$; approximate controllability at time $T$ is equivalent to $\im F_T$ being dense in $L^2(\Omega)$, and null controllability at time $T$ is equivalent to $\im F_T\supseteq \im S(T)$. \end{remark} \begin{theorem}\label{ICT:controltheorem} Let $S:H_1\to H$ and $T:H_2\to H$ be bounded linear operators between Hilbert spaces. Then, $\im(S)\subseteq \im(T)$ if and only if $\exists c>0$ such that $\forall x\in H$, $\norm{S^*x}_{H_1}\leq c\norm{T^*x}_{H_2}$. @@ -877,7 +924,7 @@ $$ \norm{y_T}_{H}\leq c\norm{B^*x}_{L^2(0,T;L^2(\omega))} $$ - \item The system \mcref{ICT:control_system} is approximately controllable at time $T$ if and only if the system \mcref{ICT:dual_control_system} satifies the unique continuation property, that is, if $x$ solution of \mcref{ICT:ICT:dual_control_system} satisfies $B^*x(t)=0$ $\forall t\in [0,T]$ then $y_T=0$, i.e.\ $x=0$. + \item The system \mcref{ICT:control_system} is approximately controllable at time $T$ if and only if the system \mcref{ICT:dual_control_system} satisfies the unique continuation property, that is, if $x$ solution of \mcref{ICT:dual_control_system} satisfies $B^*x(t)=0$ $\forall t\in [0,T]$ then $y_T=0$, i.e.\ $x=0$. \item The system \mcref{ICT:control_system} is null controllable at time $T$ if and only if the system \mcref{ICT:dual_control_system} is initial time observable: $\exists c>0$ such that for all solution $x$ of \mcref{ICT:dual_control_system} we have: $$ \norm{x(0)}_{H}\leq c\norm{B^*x}_{L^2(0,T;L^2(\omega))} @@ -933,7 +980,7 @@ Finally, we need to define our control law. Imposing $w_z(1,t)=0$ in \mcref{ICT:backstepping_transformation} we get: \begin{gather*} - 0=w_z(1,t)=x_z(1,t)-K(1,1)x(1,t)-\int_0^1 K_z(1,y)x(y,t)\dd y\\ + 0\!=\!w_z(1,t)\!=\!x_z(1,t)\!-\!K(1,1)x(1,t)\!-\!\int_0^1\! K_z(1,y)x(y,t)\dd y\\ u(t)=K(1,1)x(1,t)+\int_0^1 K_z(1,y)x(y,t)\dd y \end{gather*} So, we are left to find if a suitable $K$ exists. Recall that the condition $w(1,t)=0$ is automatically satisfied. We need to make use of the PDE of $w$. Using \mcref{ICT:backstepping_transformation} to compute $w_t$ and $w_{zz}$, and equating both equations it suffices to find $K$ such that: