diff --git a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex index 44f04a5..e580343 100644 --- a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex +++ b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex @@ -28,7 +28,7 @@ Let $\Sigma$ be a $\sigma$-algebra over a set $\Omega$. A \emph{measure} over $\Omega$ is any function $$\mu:\Sigma\longrightarrow[0,\infty]$$ satisfying the following properties: \begin{enumerate}[ref = $\sigma$-additivity] \item $\mu(\varnothing)=0$. - \item\label{RFA:sigmaadditivity} \emph{$\sigma$-additivity}: If $(A_n)\in\Sigma$ are pairwise disjoint, then: $$\mu\left(\bigsqcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n)$$ + \item\label{RFA:sigmaadditivity} \emph{$\sigma$-additivity}: If $(A_n)\in\Sigma$ are pairwise disjoint, then: $$\mu\left(\bigsqcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n)$$ \end{enumerate} \end{definition} \begin{definition} @@ -39,8 +39,8 @@ \begin{enumerate} \item If $A\subseteq B$, then $\mu(B\setminus A)=\mu(B)-\mu(A)$. \item If $A\subseteq B$, then $\mu(A)\leq\mu(B)$. - \item\label{RFA:incresingseq} If $A_n\nearrow A$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. - \item\label{RFA:decresingseq} If $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. + \item\label{RFA:incresingseq} If $A_n\nearrow A$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. + \item\label{RFA:decresingseq} If $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. \end{enumerate} \end{proposition} \begin{sproof} @@ -113,12 +113,12 @@ The outer measure has the following properties: \begin{enumerate} \item $\om{\varnothing}=0$. - \item\label{RFA:measureB} If $A\subseteq B\subseteq\RR^n$, then $\om{A}\leq \om{B}$. + \item\label{RFA:measureB} If $A\subseteq B\subseteq\RR^n$, then $\om{A}\leq \om{B}$. \item \label{RFA:measureC} If $(A_k)\subseteq \RR^n$, then: $$\om{\bigcup_{k=1}^\infty A_k}\leq \sum_{k=1}^\infty \om{A_k}$$ \item \label{RFA:measureD}If $I\subseteq \RR^n$ is an open interval and $I\subseteq A\subseteq \cl{I}$, then $\om{A}=\vol(I)$. - \item\label{RFA:measureE} If $I_1,\ldots,I_N\subseteq \RR^n$ are disjoint intervals, then: $$\displaystyle \om{\bigsqcup_{k=1}^N I_k}= \sum_{k=1}^N \vol(I_k)$$ + \item\label{RFA:measureE} If $I_1,\ldots,I_N\subseteq \RR^n$ are disjoint intervals, then: $$\displaystyle \om{\bigsqcup_{k=1}^N I_k}= \sum_{k=1}^N \vol(I_k)$$ \item If $A,B\subseteq \RR^n$ and $d(A,B):=\inf\{d(a,b):a\in A,b\in B\}>0$, then $\om{A\sqcup B}=\om{A}+\om{B}$. - \item\label{RFA:measureG} If $A\subseteq\RR^n$ and $x\in\RR^n$, then $\om{A+x}=\om{-A}=\om{A}$\footnote{Here $A+x:=\{a+x:a\in A\}$ and $-A:=\{-a:a\in A\}$}. + \item\label{RFA:measureG} If $A\subseteq\RR^n$ and $x\in\RR^n$, then $\om{A+x}=\om{-A}=\om{A}$\footnote{Here $A+x:=\{a+x:a\in A\}$ and $-A:=\{-a:a\in A\}$}. \end{enumerate} \end{theorem} \begin{sproof} @@ -658,7 +658,7 @@ \begin{enumerate}[ref = Triangular inequality] \item $\|\vf{u}\|=0\iff \vf{u}=0$ \item $\|\lambda \vf{u}\|=|\lambda|\|\vf{u}\|$ - \item\label{RFA:triangularineq} $\|\vf{u}+\vf{v}\|\leq \|\vf{u}\|+\|\vf{v}\|\quad$(\emph{triangular inequality}) + \item\label{RFA:triangularineq} $\|\vf{u}+\vf{v}\|\leq \|\vf{u}\|+\|\vf{v}\|\quad$(\emph{triangular inequality}) \end{enumerate} We define a \emph{normed vector space} as a pair $(E,\|\cdot\|)$ that satisfy the previous properties. \end{definition} @@ -857,11 +857,11 @@ The case of $L^\infty(E)$ is easy and the first two properties for $L^p(E)$, $p\geq 1$, too (remember \mcref{RFA:postmonotoneE}). It's missing to prove the \mref{RFA:triangularineq} (also called \emph{Minkowski inequality} in this case): $$\norm{f+g}_p\leq \norm{f}_p+\norm{g}_p$$ We have that: \begin{align*} - {\norm{f+g}_p}^p & =\int_E\abs{f+g}\abs{f+g}^{p-1} \\ - & \leq \int_E\abs{f}\abs{f+g}^{p-1}+\int_E\abs{g}\abs{f+g}^{p-1} \\ + {\norm{f+g}_p}^p & =\int_E\abs{f+g}\abs{f+g}^{p-1} \\ + & \leq \int_E\abs{f}\abs{f+g}^{p-1}+\int_E\abs{g}\abs{f+g}^{p-1} \\ \begin{split} - & \leq \left[{\left(\int_E\abs{f}^p\right)}^{1/p}+{\left(\int_E\abs{g}^p\right)}^{1/p}\right]\cdot \\ - & \hspace{2.5cm}\cdot{\left(\int_E\abs{f+g}^{(p-1)\frac{p}{p-1}}\right)}^{1-1/p} + & \leq \left[{\left(\int_E\abs{f}^p\right)}^{1/p}+{\left(\int_E\abs{g}^p\right)}^{1/p}\right]\cdot \\ + & \hspace{2.5cm}\cdot{\left(\int_E\abs{f+g}^{(p-1)\frac{p}{p-1}}\right)}^{1-1/p} \end{split} \\ & =(\norm{f}_p+\norm{g}_p)\frac{{\norm{f+g}_p}^p }{{\norm{f+g}_p}} \end{align*} @@ -1056,14 +1056,14 @@ \item\label{RFA:TcontinuousC} $T(B_E)$ is bounded on $F$, where $B_E:=\{x\in E:\norm{x}_E\leq 1\}$. \item\label{RFA:TcontinuousD} $\norm{T}<\infty$. \item\label{RFA:TcontinuousE} $\exists C\geq 0$ such that $\forall x\in E$ we have: $$\norm{Tx}_F\leq C\norm{x}_E$$ - If, moreover, $T$ is continuous, $\norm{T}$ is the least of such constants $C$. + If, moreover, $T$ is continuous, $\norm{T}$ is the least of such constants $C$. \end{enumerate} \end{theorem} \begin{sproof} \begin{enumerate}[leftmargin=1.5cm] \item[\mref{RFA:TcontinuousA}$\implies$\mref{RFA:TcontinuousB}:] Let $x\in E$ and $(x_n)\in E$ such that $\displaystyle \lim_{n\to\infty}x_n=x$. Then $\displaystyle \lim_{n\to\infty}(x_n - x)=0$ and the continuity and linearity imply $\displaystyle \lim _{n\to\infty}(Tx_n-Tx)=0$. \item[\mref{RFA:TcontinuousB}$\implies$\mref{RFA:TcontinuousC}:] The continuity at the origin of $T$ implies that given $\varepsilon =1$, $\exists \delta>0$ such that: $$T(B_E(0,\delta))\subseteq B_F(0,1)$$ - The linearity of $T$ implies that $T(B_E(0,1))\subseteq B_F(0,1/\delta)$. + The linearity of $T$ implies that $T(B_E(0,1))\subseteq B_F(0,1/\delta)$. \item[\mref{RFA:TcontinuousC}$\implies$\mref{RFA:TcontinuousD}:] Consequence of \mcref{RFA:normT}. \item[\mref{RFA:TcontinuousD}$\implies$\mref{RFA:TcontinuousE}:] By the definition of supremum we have: $$\norm{T\left(\frac{x}{\norm{x}_E}\right)}_F\leq \norm{T}$$ And so $\norm{Tx}_F\leq \norm{T}\norm{x}_E$. \item[\mref{RFA:TcontinuousE}$\implies$\mref{RFA:TcontinuousA}:] Evident. @@ -1354,7 +1354,7 @@ \begin{theorem}[Open mapping theorem] Let $E$, $F$ be a Banach spaces and $T:E\rightarrow F$ be a surjective bounded operator. Then, $T$ is open. \end{theorem} - \begin{theorem}[Closed graph theorem] + \begin{theorem}[Closed graph theorem]\label{RFA:closedgraph} Let $E$, $F$ be a Banach spaces and $T:E\rightarrow F$ be an operator. Consider the graph of $T$: $$\graph(T)=\{(x,y)\in E\times F:y=Tx\}$$ Then, $T$ is bounded if and only if $\graph(T)$ is a closed set on $E\times F$. \end{theorem} 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 da1621d..f438458 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 @@ -670,6 +670,220 @@ \begin{remark} If $\Sigma=\Fr{\Omega}$, then the system is controllable of $T>\diam(\Omega)$. \end{remark} - \subsubsection{Abstract systems} + \subsection{Abstract systems} + \subsubsection{Basic definitions} + \begin{definition} + Let $X$, $Y$ be Banach. A \emph{bounded operator} is a couple $(D(A), A)$ where $D(A)\subseteq X$ is a subspace and $A:D(A)\to Y$ is a continuous linear operator. An \emph{unbounded operator} is a couple $(D(A), A)$ where $D(A)\subseteq X$ is a subspace and $A:D(A)\to Y$ is a linear operator. + \end{definition} + \begin{remark} + Usually we will omit specifying the domain $D(A)$. + \end{remark} + \begin{remark} + Note that bounded operators are also unbounded operators. + \end{remark} + \begin{definition} + Let $A$ be an unbounded operator between Banach spaces $X$ and $Y$. $A$ is \emph{densely defined} if $\overline{D(A)}=X$. $A$ is \emph{closed} if $\graph(A)=\{(x,y)\in D(A)\times Y:y=Ax\}$ is closed. + \end{definition} + Form now on we will assume $X$, $Y$ are Hilbert. + \begin{definition} + Let $(D(A),A)$ be a densely unbounded operator. We define the \emph{adjoint operator} $(D(A^*),A^*)$ by: + \begin{multline*} + D(A^*)=\{y\in Y^*:\exists c>0\text{ with}\\ + \abs{\langle y,Ax\rangle_{Y^*\times Y}}\leq c \norm{x}_X\ \forall x\in D(A)\} + \end{multline*} + and $\forall y\in D(A^*)$ $\forall x\in D(A)$, $A^*y$ is given in order to satisfy: + $$ + \langle A^*y,x\rangle_{X^*\times X}=\langle y,Ax\rangle_{Y^*\times Y} + $$ + \end{definition} + \subsubsection{Semigroups} + From now on we will assume $X=Y$. + \begin{definition} + A one-parameter family of unbounded operators $(T(t))_{t\geq 0}$ is a \emph{semigroup of operators} if: + \begin{itemize} + \item $T(0)=\id$ + \item $T(t+s)=T(t)\circ T(s)$ $\forall t,s\geq 0$ + \end{itemize} + \end{definition} + \begin{definition} + A semigroup of operators $(T(t))_{t\geq 0}$ is called + \begin{itemize} + \item \emph{uniformly continuous} if $\norm{T(t)-\id} \underset{t\to 0^+}{\longrightarrow} 0$. + \item \emph{strongly continuous} if $\forall x\in X$ $\norm{T(t)x-x}\underset{t\to 0^+}{\longrightarrow} 0$. + \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 + $$ + D(A)=\{ x\in X:\exists \lim_{t\to 0^+}\frac{T(t)x-x}{t}\} + $$ + and $\forall x\in D(A)$ we define: + $$ + Ax:=\lim_{t\to 0^+}\frac{T(t)x-x}{t} + $$ + \end{definition} + \begin{proposition} + 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 $t\geq 0$: + $$ + \int_0^t T(s)x\dd s\in D(A)\text{ and } T(t)x-x=A\left( \int_0^t T(s)x\dd s\right) + $$ + \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 $\norm{T(t)}\leq C\exp{ct}$. + \end{enumerate} + \end{proposition} + \begin{theorem}\hfill + \begin{enumerate} + \item If $A$ is the infinitesimal generator of a strongly continuous semigroup, then it is closed and densely defined. + \item If $(T(t))_{t\geq 0}$, $(S(t))_{t\geq 0}$ are two strongly continuous semigroups with the same infinitesimal generator, then $T(t)=S(t)$ $\forall t\geq 0$. + \item $\forall x_0\in D(A)$, there exists a unique solution $x\in \mathcal{C}^0([0,+\infty),D(A))\cap \mathcal{C}^1([0,+\infty),X)$ of $\dv{}{t}x(t)=Ax(t)$ with $x(0)=x_0$, and it is given by: $$x(t)=T(t)x_0$$ + Moreover, $\forall f\in \mathcal{C}^1([0,T],X)$, there exists a unique solution $x\in \mathcal{C}^0([0,T],D(A))\cap \mathcal{C}^1([0,T],X)$ of $\dv{}{t}x(t)=Ax(t)+f(t)$ with $x(0)=x_0$, and it is given by: $$ + x(t)=T(t)x_0+\int_0^t T(t-s)f(s)\dd s + $$ + This last formula is called the \emph{variation of constants formula} or \emph{Duhamel's formula}. + \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. + \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 $(T(t)^*)_{t\geq 0}$. + \end{theorem} + \begin{remark} + We will denote by $T(t)^*$ also by $T^*(t)$. + \end{remark} + \begin{definition} + 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 $R_\lambda(A):=(\lambda I-A)^{-1}$. + \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: + $$ + (0,\infty)\subseteq \rho(A)\text{ and }\forall \lambda>0, \norm{R_\lambda(A)}\leq \frac{1}{\lambda} + $$ + \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: + $$ + (c,\infty)\subseteq \rho(A)\text{ and }\forall \lambda>c, \norm{R_\lambda(A)}\leq \frac{1}{\lambda-c} + $$ + \end{corollary} + \begin{definition} + An operator $(D(A),A)$ is called \emph{dissipative} if $\forall x\in D(A)$, $\langle Ax,x\rangle\leq 0$. + \end{definition} + \begin{theorem}[Lümmer-Phillips] + Let $(D(A),A)$ be an operator closed and densely defined. Then: + \begin{enumerate} + \item If $A$ is dissipative and $\exists \lambda_0>0$ such that $\im(\lambda_0 I-A)=R$, then $\forall \lambda>0$, $\im(\lambda I-A)=R$ and $A$ generates a semigroup of contractions. + \item If $A$ generates a semigroup of contractions, then $\im(\lambda I-A)=X$ $\forall \lambda>0$. + \end{enumerate} + \end{theorem} + \begin{corollary} + Let $(D(A),A)$ be an operator closed and densely defined. If $A$ and $A^*$ are both dissipative, then $A$ generates a semigroup of contractions. + \end{corollary} + \subsubsection{Applications to control theory} + In this section we consider the control system: + \begin{equation}\label{ICT:control_system} + \begin{cases} + \dot{x}(t)=Ax(t)+Bu(t) & \text{in } [0,T] \\ + x(0)=x_0 + \end{cases} + \end{equation} + where $A$ is the infinitesimal generator of a strongly continuous semigroup $(S(t))_{t\geq 0}$ and $B$ is a bounded operator, with: + $$ + S(t):L^2(\Omega)\to L^2(\Omega)\qquad B:L^2(\omega)\to L^2(\Omega) + $$ + We will first consider the interior control in a region $\omega\subseteq \Omega$. We will denote by $F_T$ the operator: + $$ + \function{F_T}{L^2(0,T;L^2(\omega))}{L^2(\Omega)}{u}{\int_0^T S(T-s)Bu(s)\dd s} + $$ + \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 $\im F_T\supseteq \im S(T)$. + \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 $\forall x\in H$, $\norm{S^*x}_{H_1}\leq c\norm{T^*x}_{H_2}$. + \end{theorem} + \begin{proof} + \begin{itemizeiff} + If $\im(S)\subseteq \im(T)$, then $\forall x\in H_1$ $\exists! y\in \ker(T)^\perp$ such that $Sx=Ty$. + \begin{itemize} + \item \textit{Existence}: for $x\in H_1$, we find $y\in H_2$ such that $Sx=Ty$ and we define $z:=\pi_{\ker(T)^\perp}y$, where $\pi_{\ker(T)^\perp}$ is the orthogonal projection on $\ker(T)^\perp$. Then, $z-y\in (\ker(T)^\perp)^\perp=\ker(T)$, and so $T(z-y)=0$, and thus $Tz=Ty=Sx$. + \item \textit{Uniqueness}: if $Sx=Ty_1=Ty_2$, then $T(y_1-y_2)=0$, and so $y_1-y_2\in \ker(T)$, but also $y_1,y_2\in \ker(T)^\perp$ (by hypothesis). Thus, $y_1=y_2$. + \end{itemize} + We define $G:H_1\to \ker(T)^\perp\subset H_2$ such that to $x\in H_1$ we associate the unique $y\in \ker(T)^\perp$ such that $Sx=Ty$. We have that $G$ is linear. To see that it is continuous we used the \mnameref{RFA:closedgraph}. We need to prove that if $x_n\overset{H_1}{\longrightarrow} x$ and $Gx_n\overset{H_2}{\longrightarrow} y$, then $Gx=y$. We have that $G(x_n)\in \ker(T)^\perp$ $\forall n$ and $\ker(T)^\perp$ is closed, so $y\in \ker(T)^\perp$. We have that $T(G(x_n))=Sx_n$ $\forall n$. Taking the limit and using the continuity of $S$ and $T$ we get $Ty=Sx$, which by uniqueness implies $y=Gx$. So $S=TG$, and thus $S^*=G^*T^*$, with $G^*$ continuous. Thus, $\forall x\in H$: + $$ + \norm{S^*x}_{H_1}=\norm{G^*T^*x}_{H_1}\leq \norm{G^*}_{\mathcal{L}(H_2,H_1)}\norm{T^*x}_{H_2} + $$ + \item We will prove that there exists an operator $D:H_2\to H_1$ such that $S^*=DT^*$. If $y\in\im(T^*)$, let $x\in H$ be such that $y=T^*x$, and then we define $Dy:=S^*x$. This definition is independent of the choice of $x$ because if $x_1,x_2\in H$ are such that $y=T^*x_1=T^*x_2$, then $T^*(x_1-x_2)=0$, and so (by hypotheses) $S^*(x_1-x_2)=0$, and thus $S^*x_1=S^*x_2$. So $D:\im(T^*)\to H_1$ is well-defined, it is linear and continuous: + $$ + \norm{Dy}_{H_1}=\norm{S^*x}_{H_1}\leq c\norm{T^*x}_{H_2}=c\norm{y}_{H_2} + $$ + So $D$ can be uniquely extended as a continuous linear map on $\overline{\im(T^*)}$. We decide to set $D|_{\im(T^*)^\perp}= 0$ and we get a continuous linear map $D:H_2\to H_1$ such that $S^*=DT^*$. Taking adjoints we get $S=TD^*$, so $\im(S)\subseteq \im(T)$. + \end{itemizeiff} + \end{proof} + \begin{theorem} + Let $A:H_1\to H_2$ be a bounded linear operator between Hilbert spaces. Then: + \begin{enumerate} + \item $\im(A)$ is dense $\iff$ $\ker(A^*)=\{0\}$. + \item $\im(A)=H_2\iff \exists c>0$ such that $\forall x\in H_2$, $\norm{x}_{H_2}\leq c\norm{A^*x}_{H_1}$. + \end{enumerate} + \end{theorem} + \begin{proof}\hfill + \begin{enumerate} + \item Recall that $\im(A)^{\perp}=\ker(A^*)$: + \begin{align*} + \overline{\im(A)}=H_2 & \iff (\im(A)^{\perp})^\perp=H_2 \\ + & \iff\im(A)^{\perp}=\{0\} \\ + & \iff \ker(A^*)=\{0\} + \end{align*} + \item Use \cref{ICT:controltheorem} with $H_1=H$ and $S=\id_H$. + \end{enumerate} + \end{proof} + \begin{proposition} + The adjoint of $F_T$ is given by: + $$ + \function{F_T^*}{L^2(\Omega)}{L^2(0,T;L^2(\omega))}{y_T}{s\mapsto B^*S^*(T-s)y_T} + $$ + \end{proposition} + \begin{proof} + Let $y_T\in L^2(\Omega)$ and $u\in L^2(0,T;L^2(\omega))$. Then: + \begin{align*} + \langle F_T^*y_T,u\rangle_{L^2(0,T;L^2(\omega))} & = \int_0^T\langle B^*S^*(T-s)y_T,u(s)\rangle_{L^2(\omega)} \dd s \\ + & = \int_0^T\langle S^*(T-s)y_T,Bu(s)\rangle_{L^2(\Omega)} \dd s \\ + & = \int_0^T\langle y_T,S(T-s)Bu(s)\rangle_{L^2(\Omega)} \dd s \\ + & = \langle y_T, \int_0^T S(T-s)Bu(s)\dd s\rangle_{L^2(\Omega)} \\ + & = \langle y_T, F_Tu\rangle_{L^2(\Omega)} + \end{align*} + \end{proof} + \begin{theorem} + Consider the control system \mcref{ICT:control_system} and its dual system: + \begin{equation}\label{ICT:dual_control_system} + \begin{cases} + -\dot{x}(t)=A^*x(t) & \text{in } [0,T] \\ + x(T)=y_T + \end{cases} + \end{equation} + Then: + \begin{enumerate} + \item The system \mcref{ICT:control_system} is exactly controllable at time $T$ if and only if the system \mcref{ICT:dual_control_system} is final time observable by means of $B^*$, that is, if $\exists c>0$ such that for all solution $y$ of \mcref{ICT:dual_control_system} we have: + $$ + \norm{y_T}_{H}\leq c\norm{B^*x}_{L^2(0,T;L^2(\omega))} + $$ + \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 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: + $$ + \norm{x(0)}_{H}\leq c\norm{B^*x}_{L^2(0,T;L^2(\omega))} + $$ + \end{enumerate} + \end{theorem} + \subsection{Backstepping for boundary control in PDEs} \end{multicols} \end{document} \ No newline at end of file diff --git a/preamble_general.sty b/preamble_general.sty index 40f01f6..01c9fb9 100644 --- a/preamble_general.sty +++ b/preamble_general.sty @@ -74,7 +74,7 @@ {NMPDE}{\apl} % Numerical methods for pdes {SCO}{\pro} % Stochastic control {JP}{\pro} % Jump processes - {ICT}{\apl} % Introduction to control theory + {ICT}{\ana} % Introduction to control theory }{\col}% } \ExplSyntaxOff