Examples of control systems modeled by linear PDE's - davidar/scholarpedia GitHub Wiki

We return to the transport equation presented in the Section Examples of control systems modeled by PDE's. Let <math>L>0\ .</math> The linear control system we study is

<math>\label{eq1}

y_t+y_x=0,\, t\in (0,T),\, x\in (0,L), </math>

<math>\label{eq2}

y(t,0)=u(t),\, t\in(0,T), </math>

where, at time <math>t\ ,</math> the control is <math>u(t)\in \mathbb{R}</math> and the state is <math>y(t,\cdot):(0,L)\rightarrow \mathbb{R}\ .</math>

One can put this linear control system in the general framework detailed at this link in the following way. For the Hilbert space <math>H\ ,</math> we take <math>H:=L^2(0,L)\ .</math> For the operator <math>A: D(A)\rightarrow H</math> we take

<math>

D(A):=\{f\in H^1(0,L);\, f(0)=0\}, </math>

<math>

Af:=-f_x,\, \forall f\in D(A). </math> Then <math>D(A)</math> is dense in <math>L^2(0,L)\ ,</math> <math>A</math> is closed. Moreover

<math>

(Af,f)_{L^2(0,L)}=-\frac{1}{2}f(L)^2, \, \forall f \in D(A), </math> showing that <math>A</math> is dissipative. The adjoint <math>A^*</math> of <math>A</math> is defined by

<math>

D(A^*):=\{f\in H^1(0,L);\, f(L)=0\}, </math>

<math>

A^*f:=f_x,\, \forall f\in D(A^*). </math> As the operator <math>A\ ,</math> the operator <math>A^*</math> is also dissipative. Hence, by the Lumer-Phillips theorem, the operator <math>A</math> is the infinitesimal generator of a strongly continuous semigroup <math>S(t),\, t\in[0,+\infty)\],</math> of continuous linear operators on <math>H\ .</math>

For the Hilbert space <math>U\ ,</math> we take <math>U:=\mathbb{R}\ .</math> The operator <math>B:\mathbb{R}\rightarrow D(A^*)'</math> is defined by

<math>\label{eq3}

(Bu)z=u z(0), \, \forall u \in \mathbb{R}, \forall z\in D(A^*). </math>

Note that <math>B^*:D(A^*)\rightarrow \mathbb{R}</math> is defined by

<math>

B^*z=z(0), \, \forall z \in D(A^*). </math> Let us deal with the regularity property:

<math>\label{eqReg}

\forall T>0, \exists C_T>0 \text{ such that } \int_0^T\|B^*S(t)^* z\|_{U}^2dt \leqslant C_T \|z\|^2_H, \, \forall z\in D(A^*). </math>

Let <math>z^0\in D(A^*)\ .</math> Let

<math>

z\in C^0([0,T]; D(A^*))\cap C^1([0,T];L^2(0,L)) </math> be defined by <math>z(t,\cdot)=S(t)^*z^0\ .</math> Inequality \eqref{eqReg} is equivalent to

<math>\label{eq4}

\int_0^Tz(t,0)^2dt \leqslant C_T \int_0^L z^0(x)^2 dx. </math>

Let us prove this inequality for<math>C_T:=1\ .</math> We have

<math>\label{eq5}

z_t=z_x, \, t\in(0,T), \, x\in (0,L), </math>

<math>\label{eq6}

z(t,L)=0, \, t\in (0,T), </math>

<math>\label{eq7}

z(0,x)=z^0(x), \, x\in (0,L). </math>

We multiply \eqref{eq5} by <math>z</math> and integrate on <math>[0,T]\times[0,L]\ .</math> Using \eqref{eq6}, \eqref{eq7} and integrations by parts, we get

<math>\label{eq8}

\int_0^Tz(t,0)^2dt= \int_0^Lz^0(x)^2dx - \int_0^Lz(T,x)^2dx\leqslant \int_0^Lz^0(x)^2dx, </math>

which shows that \eqref{eq4} holds for<math>C_T:=1\ .</math>

In fact, as one can easily check, the solution to the following Cauchy problem

<math>\label{eq9}

y_t+y_x=0, \, t\in(0,T),\, x\in (0,L), </math>

<math>\label{eq10}

y(t,0)=u(t),\,t\in(0,T), </math>

<math>\label{eq11}

y(0,x)=y^0(x), \, x\in (0,L), </math>

where <math>T>0\ ,</math> <math>y^0\in L^2(0,L)</math> and <math>u\in L^2(0,T)</math> are given data, is

<math>\label{eq12}

y(t,x)=y^0(x-t), \, \forall (t,x) \in [0,T]\times(0,L) \text{ such that } t\leqslant x, </math>

<math>\label{eq13}

y(t,x)=u(t-x), \, \forall (t,x) \in [0,T]\times(0,L)\text{ such that } t > x. </math>

For the controllability of the linear control system \eqref{eq1}-\eqref{eq2}, see at this link.

We return to the linear Korteweg-de Vries equation already mentioned at this link in the Section Examples of control systems modeled by PDE's. Let <math>L>0\ .</math> The linear control system we study is

<math>\label{eq14}

y_t+y_x+y_{xxx}=0,\, t\in (0,T), \, x\in (0,L), </math>

<math>\label{eq15}

y(t,0)=y(t,L)=0,\, y_x(t,L)=u(t), \, t\in (0,T), </math>

where, at time <math>t\ ,</math> the control is <math>u(t)\in \mathbb{R}</math> and the state is <math>y(t,\cdot):(0,L)\rightarrow \mathbb{R}\ .</math>

One can put this linear control system in the general framework detailed at this link in the following way. For the Hilbert space <math>H\ ,</math> we take <math>H=L^2(0,L)\ .</math> For the operator <math>A: D(A)\rightarrow H\ ,</math> we take

<math>

D(A):=\{f\in H^3(0,L);\, f(0)=f(L)=f_x(L)=0\}, </math>

<math>

Af:=-f_x-f_{xxx},\, \forall f\in D(A). </math> Then <math>D(A)</math> is dense in <math>L^2(0,L)\ ,</math> <math>A</math> is closed. Simple integrations by parts give

<math>

(Af,f)_{L^2(0,L)}= -\frac{1}{2}f_x(0)^2, \, \forall f\in L^2(0,L), </math> which shows that <math>A</math> is dissipative. The adjoint <math>A^*</math> of <math>A</math> is defined by

<math>

D(A^*):=\{f\in H^3(0,L);\, f(0)=f(L)=f_x(0)=0\}, </math>

<math>

A^*f:=f_x+f_{xxx},\, \forall f\in D(A^*). </math> As <math>A\ ,</math> the operator <math>A^*</math> is also dissipative. Hence, by the Lumer-Phillips theorem, the operator <math>A</math> is the infinitesimal generator of a strongly continuous semigroup <math>S(t),\, t\in[0,+\infty)\],</math> of continuous linear operators on<math>L^2(0,L)\ .</math>

For the Hilbert space <math>U\ ,</math> we take <math>U:=\mathbb{R}\ .</math> The operator <math>B:\mathbb{R}\rightarrow D(A^*)'</math> is defined by

<math>\label{eq16}

(Bu)z=u z_x(L), \, \forall u \in \mathbb{R}, \forall z\in D(A^*). </math>

Note that <math>B^*:D(A^*)\rightarrow \mathbb{R}</math> is defined by

<math>

B^*z=z_x(L), \, \forall z \in D(A^*). </math> Let us check the regularity property \eqref{eqReg}. Let <math>z^0\in D(A^*)\ .</math> Let

<math></math>

z\in C^0([0,+\infty);]D(A^*))\cap C^1([0,+\infty);L^2(0,L)) be]defined by

<math>\label{eq17}

z(t,\cdot)=S(t)^*z^0. </math>

The regularity property \eqref{eqReg} is equivalent to

<math>\label{eq18}

\int_0^T|z_x(t,L)|^2dt \leqslant C_T \int_0^L |z^0(x)|^2 dx. </math>

From \eqref{eq17}, one has

<math>\label{eq19}

z_t-z_x-z_{xxx}=0 \text{ in } C^0([0,+\infty);]L^2(0,L)), </math>

<math></math>\label{eq20}

z(t,0)=z_x(t,0)=z(t,L)=0, \, t \in [0,+\infty),

<math>\label{eq21}

z(0,x)=z^0(x),]multiply \eqref{eq19} by <math>z</math> and integrate on <math>(0,T)\times(0,L)\ .</math> Using \eqref{eq20},

\eqref{eq21} and simple integrations by parts one gets

<math>\label{eq22}

\int_0^T |z_x(t,L)|^2 dt=\int_0^L|z^0(x)|^2dx- \int_0^L|z(T,x)|^2dx\leqslant \int_0^L|z^0(x)|^2dx, </math>

which shows that \eqref{eq18} holds with <math>C_T:=1\ .</math> For the controllability of the linear control system \eqref{eq14}-\eqref{eq15}, see at this link.

We return to the linear heat equation already considered at this link in the Section Examples of control systems modeled by PDE's. Let <math>\Omega</math> be a non empty open subset of <math>\mathbb{R}^l</math> and let <math>\omega</math> be a non empty open subset of <math>\Omega\ .</math> The linear heat equation considered in this section is

<math>\label{eq23}

y_t-\Delta y = u(t,x),\, t\in (0,T), \, x\in \Omega, </math>

<math>\label{eq24}

y=0 \text{ on } (0,T)\times \partial \Omega, </math>

where, at time <math>t \in [0,T]\ ,</math> the state is <math>y(t,\cdot) \in L^2(\Omega)</math> and the control is <math>u(t,\cdot) \in L^2 (\Omega)\ .</math> We require that

<math>\label{eq25}

u(\cdot,x)=0, \, x\in \Omega\setminus \omega. </math>

One can put this linear control system in the general framework detailed at this link in the following way. One chooses

<math>

H:=L^2(\Omega), </math> equipped with the usual scalar product. Let <math>A:D(A)\subset H\rightarrow H</math> be the linear operator defined by

<math>

D(A):=\left\{y\in H^1_0(\Omega);\, \Delta y \in L^2(\Omega) \right\}, </math>

<math>

Ay:=\Delta y \in H. </math> Note that, if <math>\Omega</math> is smooth enough (for example of class <math>C^2</math>), then

<math>\label{eq26}

D(A)=H^1_0(\Omega)\cap H^2(\Omega). </math>

However, without any regularity assumption on <math>\Omega\ ,</math> \eqref{eq26} is wrong in general (see in particular Theorem 2.4.3, page 57, in (Pierre Grisvard,1992)). One easily checks that

<math>\label{eq27}

D(A) \text{ is dense in }L^2(\Omega), </math>

<math>\label{eq28}

A \text{ is closed.} </math>

Moreover,

<math>\label{eq29}

(Ay,y)_{H}=-\int_\Omega |\nabla y |^2 dx,\, \forall y \in D(A). </math>

Let <math>A^*</math> be the adjoint of<math>A\ .</math> One easily checks that

<math>\label{eq30}

A^*=A. </math>

From the Lumer-Phillips theorem, \eqref{eq27}, \eqref{eq28}, \eqref{eq29} and \eqref{eq30}, <math>A</math> is the infinitesimal generator of a strongly continuous semigroup of linear contractions <math>S(t)\ ,</math> <math>t\in [0,+\infty)\],</math> on <math>H\ .</math> For the Hilbert space <math>U</math> we take <math>L^2(\omega)\ .</math> The linear map <math>B\in \mathcal{L}(U;D(A^*)')</math> is the map which is defined by

<math>

(Bu)\varphi=\int_\omega u\varphi dx. </math> Note that <math>B\in \mathcal{L}(U;H)\ .</math> Hence the regularity property \eqref{eqReg} is automatically satisfied. For the controllability of the linear control system \eqref{eq23}-\eqref{eq24}, see at this link.