Control of partial differential equations/Examples for controllability of linear control systems

From Scholarpedia
Jump to: navigation, search

    Contents

    A transport equation

    We return to the transport control system, already considered at this link,

    <math eqS1>

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

    <math eqS2>

    y(t,0)=u(t),\, t\in(0,T). </math> This example is pedagogically interesting since one can give explicitly the solution to the Cauchy problem where \(T>0\), \(y^0\in L^2(0,L)\) and \(u\in L^2(0,T)\) are given. This solution is given by

    <math eqC1>

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

    <math eqC2>

    y(t,x)=u(t-x), \, \forall (t,x) \in [0,T]\times(0,L)\text{ such that } t > x. </math> From this explicit solution one readily gets the following proposition.

    Proposition 1. The control system (<ref>eqS1</ref>)-(<ref>eqS2</ref>) is

    Let us show how to use the multiplier method in order to prove that if

    <math eq1>

    T>L </math> then the control system (<ref>eqS1</ref>)-(<ref>eqS2</ref>) is exactly controllable. By Theorem 3, the exact controllability in time \(T\) is equivalent to the existence of \(c>0\) such that

    <math eq2>

    \int_0^T z(t,0)^2 dt\geqslant c \int_0^Lz^0(x)^2dx, </math> where \(z:[0,T]\times[0,L]\rightarrow \mathbb{R}\) is the solution of the Cauchy problem

    <math eq3>

    z_t-z_x =0,\, </math>

    <math eq4>

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

    <math eq5>

    z(0,\cdot)=z^0. </math> Let us prove prove (<ref>eq2</ref>). With simple density arguments, we may assume that \(z\) is of class \(C^1\). Let us multiply (<ref>eq3</ref>) by the multiplier \(z\) and integrate the obtained equality on \([0,L]\). Using (<ref>eq4</ref>), one gets

    <math eq6>

    \frac{\text{d}}{\text{d}t}\left(\int_0^L|z(t,x)|^2dx\right) =-|z(t,0)|^2. </math> Let us now multiply (<ref>eq3</ref>) by the multiplier \(xz\) and integrate the obtained equality on \([0,L]\). Using (<ref>eq4</ref>), one gets

    <math eq7>

    \frac{\text{d}}{\text{d}t}\left(\int_0^Lx |z(t,x)|^2dx\right) =-\int_0^L|z(t,x)|^2dx. </math> For \(t\in [0,T]\), let \(e(t):=\int_0^L|z(t,x)|^2dx\). From (<ref>eq7</ref>), we have

    <math eq8>

    \int_0^Te(t)dt=-\int_0^Lx|z(T,x)|^2dx+\int_0^Lx|z(0,x)|^2dx\leqslant L \int_0^L|z(0,x)|^2dx=Le(0). </math>

    From (<ref>eq6</ref>), we get <math eq9> e(t)=e(0)-\int_0^t|z(\tau,0)|^2d\tau \geqslant e(0)-\int_0^T|z(\tau,0)|^2d\tau. </math> From (<ref>eq5</ref>), (<ref>eq8</ref>) and (<ref>eq9</ref>), we get

    <math eq10>

    (T-L)\|z^0\|_{L^2(0,L)}^2\leqslant T\int_0^T|z(\tau,0)|^2d\tau, </math> which proves the observability inequality (<ref>eq2</ref>) with \(c\) given by

    <math eq11>

    c:=\sqrt{\frac{T-L}{T}}. </math>

    A linear Schrödinger equation

    We return to the nonlinear Schrödinger control system considered at this link in the Section [[../Examples_of_control_systems_modeled_by_PDE%27s|Examples of control systems modeled by PDE's]]. For simplicity, we forget the variables \(S\) and \(D\): the control system is

    <math jmk1>

    \psi_{t}=i\psi_{xx} +i u(t) x \psi ,\, (t ,x) \in (0,T)\times I, </math>

    <math jmk2>

    \psi(t,-1)=\psi(t,1)=0,\, t \in (0,T), </math> In this section we explain how the moment theory can be used in order to prove the controllability of linearized control system around important trajectories of the control system (<ref>jmk1</ref>)-(<ref>jmk2</ref>).

    Let \(I\) be the open interval\((-1,1)\). For \(\gamma \in \mathbb{R}\), let \(A_{\gamma}:D(A_{\gamma})\subset L^2(I;\mathbb{C}) \rightarrow L^2(I;\mathbb{C})\)be the operator defined on

    <math eq12>

    D(A_{\gamma}):=H^{2}(I;\mathbb{C}) \cap H^{1}_{0}(I;\mathbb{C}) </math> by

    <math eq13>

    A_{\gamma} \varphi := -\varphi_{xx} - \gamma x \varphi. </math> In (<ref>eq12</ref>), as usual, \[ H^{1}_{0}(I;\mathbb{C}):=\{\varphi \in H^1((0,L);\mathbb{C});\, \varphi(0)=\varphi(L)=0\}. \] We denote by \(\langle \cdot, \cdot\rangle\) the usual Hermitian scalar product in the Hilbert space \(L^{2}(I;\mathbb{C})\): <math eq14> \langle\varphi,\psi\rangle:= \int_I\varphi (x)\overline {\psi(x)}dx, </math> where \(\overline z\) denotes the complex conjugate of the complex number\(z\). Note that

    <math eq15>

    D(A_{\gamma}) \text{ is dense in }L^2(I;\mathbb{C}), </math>

    <math eq16>

    A_{\gamma} \text{ is closed}, </math>

    <math eq17>

    A_{\gamma}^*=A_{\gamma}, \text{ (i.e., } A_{\gamma} \text{ is self-adjoint)}, </math>

    <math eq18>

    A_{\gamma} \text{ has compact resolvent}. </math> Let us recall that (<ref>eq18</ref>) means that there exists a real \(\alpha\) in the resolvent set of \(A_{\gamma}\) such that the operator \((\alpha\text{Id}-A_{\gamma})^{-1}\) is compact from \(L^2(I;\mathbb{C})\) into \(L^2(I;\mathbb{C})\), where \(\text{Id}\) denotes the identity map on \(H\) (see, for example, pages 36 and 187 in (Tosio Kato, 1995)). Then (see, for example, page 277 in (Tosio Kato, 1995)), the Hilbert space \(L^{2}(I;\mathbb{C})\) has a complete orthonormal system \((\varphi_{k,\gamma})_{k \in \mathbb{N}\setminus\{0\}}\) of eigenfunctions for the operator \(A_{\gamma}\): \[ A_{\gamma} \varphi_{k,\gamma} = \lambda_{k,\gamma} \varphi_{k,\gamma}, \] where \((\lambda_{k,\gamma})_{k \in \mathbb{N} \setminus\{0\}}\) is an increasing sequence of positive real numbers. Let \({\mathbb{S}}\) be the unit sphere of \(L^2(I;\mathbb{C})\):

    <math eq19>

    \mathbb{S}:=\{\phi\in L^2(I;\mathbb{C}); \, \int_I|\phi(x)|^2dx=1 \} </math> and, for \(\phi \in \mathbb{S}\), let \(T_{\mathbb{S}}\phi\) be the tangent space to \({\mathbb{S}}\) at \(\phi\):

    <math eq20>

    T_{\mathbb{S}}\phi :=\{\Phi\in L^2(I;\mathbb{C}); \, \Re\langle\Phi,\phi\rangle=0\}, </math> where, as usual, \(\Re z\) denotes the real part of the complex number \(z\). Let

    <math eq21>

    \psi_{1,\gamma}(t,x):=e^{-i\lambda_{1,\gamma}t}\varphi_{1,\gamma}(x), \, (t,x)\in (0,T)\times I. </math> Note that \[ \psi_{1,\gamma t}=i\psi_{1,\gamma xx}+i\gamma x \psi_{1,\gamma}, \, t\in (0,T),\, x\in I, \] \[ \psi_{1,\gamma}(t,-1)=\psi_{1,\gamma}(t,1)=0, \, t\in(0,T), \] \[ \int_I|\psi_{1,\gamma}(t,x)|^2dx =1,\, t\in(0,T). \] Hence \((\psi,u)=(\psi_{1,\gamma},\gamma)\) is a trajectory of the control system (<ref>jmk1</ref>)-(<ref>jmk2</ref>). The linearized control system around this trajectory is the following linear control system:

    <math eq22>

    \Psi_ t = i \Psi_{xx}+ i \gamma x \Psi +i u x \psi_{1,\gamma}, \, (t,x)\in (0,T)\times I, </math>

    <math eq23>

    \Psi(t,-1)=\Psi(t,1)=0,\, t\in (0,T). </math> This is a control system where, at time \(t\in [0,T]\),

    • The state is \(\Psi(t,\cdot)\in L^2(I;\mathbb{C})\) with \(\Psi(t,\cdot) \in T_{\mathbb{S}}(\psi_{1,\gamma}(t,\cdot))\).
    • The control is \(u(t)\in\mathbb{R}\).

    Let us first deal with the Cauchy problem

    <math eq24>

    \Psi_ t = i \Psi_{xx}+ i \gamma x \Psi +i u x \psi_{1,\gamma}, \, (t,x)\in (0,T)\times I, </math>

    <math eq25>

    \Psi(t,-1)=\Psi(t,1)=0,\, t\in (0,T), </math>

    <math eq26>

    \Psi(0,x)=\Psi^0(x), </math> where \(T>0\), \(u\in L^1(0,T)\) and \(\Psi^0\in L^2(I;\mathbb{C})\) are given. By (<ref>eq17</ref>), \[ (-iA_{\gamma})^*=-(-iA_{\gamma}). \] Therefore, by the Lumer-Phillips theorem, \(-iA_{\gamma}\) is the infinitesimal generator of a strongly continuous group of linear isometries on \(L^2(I;\mathbb{C})\). We denote by \(S_\gamma(t)\), \(t\in \mathbb{R}\), this group.

    Note that, since \(\psi_{1,\gamma}\) depends on time, Theorem 3 cannot be applied. Our notion of solution to the Cauchy problem (<ref>eq24</ref>)-(<ref>eq25</ref>)-(<ref>eq26</ref>) is given in the following definition.

    Definition. Let \(T>0\), \(u\in L^1(0,T)\) and \(\Psi^0\in L^2(I;\mathbb{C})\). A solution \(\Psi:[0,T]\times I\rightarrow \mathbb{C}\) to the Cauchy problem (<ref>eq24</ref>)-(<ref>eq25</ref>)-(<ref>eq26</ref>) is the function \(\Psi\in C^0([0,T]; L^2(I;\mathbb{C}))\) defined by

    <math eq27>

    \Psi(t)=S_\gamma(t)\Psi^0+\int_0^tS_\gamma(t-\tau)iu(\tau)x\psi_{1,\gamma}(\tau,\cdot) d\tau . </math>

    With this definition and standard arguments, one can show that the Cauchy problem (<ref>eq24</ref>)-(<ref>eq25</ref>)-(<ref>eq26</ref>) is well posed (see e.g., Theorem A.7, page 375 in (Jean-Michel Coron, 2007)).

    Let us now study the controllability of the linear Schrödinger control system (<ref>eq22</ref>)-(<ref>eq23</ref>). Let

    <math eq28>

    H^3_{(0)}(I;\mathbb{C}):=\{\psi \in H^3(I;\mathbb{C});\, \psi(-1)=\psi_{xx}(-1)=\psi(1)=\psi_{xx}(1)=0\}. </math> The goal of this section is to prove the following controllability result (Theorem 5 on page 862 in (Karine Beauchard, 2005)).

    Theorem 7. There exists \(\gamma_0>0\) such that, for every \(T>0\), for every \(\gamma \in (0,\gamma_0]\), for every \(\Psi^0\in T_{\mathbb{S}}\psi_{1,\gamma}(0,\cdot)\cap H^3_{(0)}(I;\mathbb{C})\) and for every \(\Psi^1\in T_{\mathbb{S}} \psi_{1,\gamma}(T,\cdot)\cap H^3_{(0)}(I;\mathbb{C})\), there exists \(u\in L^2(0,T)\) such that the solution of the Cauchy problem

    <math eq29>

    \Psi_t=i\Psi_{xx}+i \gamma x\Psi +i u(t)x\psi_{1,\gamma}, \, t\in(0,T),\, x\in I, </math>

    <math eq30>

    \Psi(t,-1)=\Psi(t,1)=0,\, t\in (0,T), </math>

    <math eq31>

    \Psi(0,x)=\Psi^0(x), \, x\in I, </math> satisfies

    <math eq32>

    \Psi(T,x)=\Psi^1(x),\, x\in I. </math>

    We are also going to see that the conclusion of Theorem 7 does not hold for \(\gamma=0\) (as already noticed in (Pierre Rouchon, 2003).

    Proof of Theorem 7. Let \(T>0\), \(\Psi^0 \in T_{\mathbb{S}}(\psi_{1,\gamma}(0,\cdot))\) and \(\Psi^1 \in T_{\mathbb{S}}(\psi_{1,\gamma}(T,\cdot))\). Let \(u \in L^{2}(0,T)\). Let \(\Psi\) be the solution of the Cauchy problem (<ref>eq29</ref>)-(<ref>eq30</ref>)-(<ref>eq31</ref>). Let us decompose \(\Psi(t,\cdot)\) in the complete orthonormal system \((\varphi_{k,\gamma})_{k \in \mathbb{N}\setminus\{0\}}\) of eigenfunctions for the operator \(A_{\gamma}\): \[ \Psi(t,\cdot)=\sum_{k=1}^{\infty}y_k(t)\varphi_{k,\gamma}. \] Taking the Hermitian product of (<ref>eq29</ref>) with \(\varphi_{k,\gamma}\), one readily gets, using (<ref>eq30</ref>) and integrations by parts,

    <math eq33>

    \dot y_k= -i\lambda_{k,\gamma} y_k +i b_{k,\gamma}u(t)e^{-i\lambda_{1,\gamma}t}, </math> with

    <math eq34>

    b_{k,\gamma}:=\langle \varphi_{k,\gamma},x \varphi_{1,\gamma}\rangle \in \mathbb{R}. </math> Note that (<ref>eq31</ref>) is equivalent to

    <math eq35>

    y_k(0)=\langle \Psi^{0},\varphi_{k,\gamma}\rangle , \, \forall k\in \mathbb{N}\setminus\{0\}. </math> From (<ref>eq34</ref>) and (<ref>eq35</ref>) one gets

    <math eq36>

    y_k(T)=e^{-i\lambda_{k,\gamma}T}(\langle \Psi^{0},\varphi_{k,\gamma}\rangle + i b_{k,\gamma}\int_0^Tu(t) e^{i(\lambda_{k,\gamma}-\lambda_{1,\gamma})t}dt). </math> By (<ref>eq36</ref>), (<ref>eq32</ref>) is equivalent to the following so-called moment problem on \(u\):

    <math eq37>

    b_{k,\gamma}\int_{0}^{T} u(t)e^{i (\lambda_{k,\gamma}-\lambda_{1,\gamma})t}dt = i \left( \langle \Psi^{0},\varphi_{k,\gamma}\rangle -\langle \Psi^{1},\varphi_{k,\gamma}\rangle e^{i \lambda_{k,\gamma}T} \right), \forall k \in \mathbb{N}\setminus\{0\}. </math>

    Let us now explain why for \(\gamma=0\) the conclusion of Theorem 7 does not hold. Indeed, one has

    <math eq38>

    \varphi_{n,0}(x):= \sin( n \pi x/2), \, n\in \mathbb{N}\setminus\{0\}, \, \text{if }n\text{ is even,} </math>

    <math eq39>

    \varphi_{n,0}(x):= \cos( n \pi x/2), \, n\in \mathbb{N}\setminus\{0\}, \, \text{if }n\text{ is odd}. </math> In particular, \(x\varphi_{1,0}\varphi_{k,0}\) is an odd function if \(k\) is odd. Therefore \[ b_{k,0}=0 \text{ if }k\text{ is odd}. \] Hence, by (<ref>eq37</ref>), if there exists \(k\) odd such that \[\langle \Psi^{0},\varphi_{k,0}\rangle -\langle \Psi^{1},\varphi_{k,0}\rangle e^{i \lambda_{k,0}T}\not = 0, \] there is no control \(u\in L^2(0,T)\) such that the solution of the Cauchy problem (<ref>eq29</ref>)-(<ref>eq30</ref>)-(<ref>eq31</ref>) (with \(\gamma =0\)) satisfies (<ref>eq32</ref>).

    Let us now turn to the case where \(\gamma\) is small but not \(0\). Since \(\Psi^{0}\) is in \(T_{\mathbb{S}}(\psi_{1,\gamma}(0,\cdot))\),

    <math eq40>

    \Re \langle \Psi^{0},\varphi_{1,\gamma}\rangle =0. </math> Similarly, the fact that \(\Psi^{1}\) is in \(T_{\mathbb{S}}(\psi_{1,\gamma}(T,\cdot))\) tells us that

    <math eq41>

    \Re(\langle \Psi^{1},\varphi_{1,\gamma}\rangle e^{i\lambda_{1,\gamma} T})=0. </math> The key ingredient to prove Theorem 7 is the following theorem.

    Theorem 8. Let \((\mu_i)_{i\in \mathbb{N}\setminus\{0\}}\) be a sequence of real numbers such that

    <math eq42>

    \mu_1=0, </math>

    <math eq43>

    \text{there exists }\rho>0\text{ such that } \mu_{i+1}-\mu_i\geqslant \rho, \, \forall i \in \mathbb{N}\setminus\{0\}. </math> Let \(T>0\) be such that

    <math eq44>

    \lim_{x\rightarrow+\infty} \frac{N(x)}{x}<\frac{T}{2\pi}, </math> where, for every \(x>0\), \(N(x)\) is the largest number of \(\mu_j\)'s contained in an interval of length \(x\). Then there exists \(C>0\) such that, for every sequence \((c_k)_{k\in \mathbb{N}\setminus\{0\}}\) of complex numbers such that

    <math eq45>

    c_1\in \mathbb{R}, </math>

    <math eq46>

    \sum_{k=1}^{\infty}|c_k|^2<\infty, </math> there exists a (real-valued) function \(u\in L^2(0,T)\) such that

    <math eq47>

    \int_0^Tu(t) e^{i\mu_k t}dt=c_k, \, \forall k \in \mathbb{N}\setminus\{0\}, </math>

    <math eq48>

    \int_0^T u(t)^2dt \leqslant C \sum_{k=1}^{\infty}|c_k|^2. </math>

    Theorem 8 is a special case of Theorem III.6.1, page 114 in (Jean-Pierre Kahane, 1962); see also pages 341--365 in (Arne Beurling, 1989). See also, in the context of control theory, Section 3 in (David Russell, 1967) which uses the prior works (Albert Ingham, 1936), (Ray Redheffer, 1951) and (Laurent Schwartz, 1959). For the proof of this theorem, see, for example, Section 1.2.2 in (Werner Krabs, 1992), Chapter 9 of (Vilmos Komornik and Paola Loreti, 2005) or Chapter II, Section 4 in the book (Sergei Avdonin and Sergei Ivanov, 1995). Improvements of this theorem have been obtained in (Stéphane Jaffard, Marius Tucsnak and Enrique Zuazua, 1997; 1998), (Stéphane Jaffard and Sorin Micu, 2001), in (Claudio Baiocchi, Vilmos Komornik and Paola Loreti, 2002),and in (Vilmos Komornik and Paola Loreti, 2004; 2005).

    Note that, by (<ref>eq40</ref>) and (<ref>eq41</ref>),

    <math eq49>

    i(\langle \Psi^{0},\varphi_{1,\gamma}\rangle -\langle \Psi^{1},\varphi_{1,\gamma}\rangle e^{i \lambda_{1,\gamma}T})\in \mathbb{R}. </math> Hence, in order to apply Theorem 8 to our moment problem, it remains to estimate \(\lambda_{k,\gamma}\) and \(b_{k,\gamma}\). This is done in the following propositions, due to Karine Beauchard (Proposition 41 pages 937-938 and page 860 in (Karine Beauchard, 2005)).

    Proposition 2. There exist \(\gamma_0>0\) and \(C_0>0\) such that, for every \(\gamma \in[-\gamma_0,\gamma_0]\) and for every \(k\in \mathbb{N}\setminus\{0\}\), \[ \left|\lambda_{k,\gamma}-\frac{\pi^2k^2}{4}\right|\leqslant C_0\frac{\gamma^2}{k}. \] Proposition 3. There exist \(\gamma_{1}>0\) and \(C>0\) such that, for every \(\gamma \in (0,\gamma_{1}]\) and for every even integer \(k \geqslant 2\), \[ \left| b_{k,\gamma}- \frac{ (-1)^{\frac{k}{2}+1} 8 k }{ \pi^{2}(k^{2}-1)^{2} } \right| < \frac{ C \gamma}{k^{3}}, \] and for every odd integer \(k \geqslant 3\), \[\left| b_{k,\gamma} - \gamma \frac{ 2 (-1)^{\frac{k-1}{2}} ( k^{2}+1) }{\pi^{4} k(k^{2}-1)^2} \right| < \frac{ C \gamma ^{2}}{k^{3}}. \]

    It is a classical result that \[ \varphi:=\sum_{k=1}^{+\infty} d_k\varphi_{k,\gamma} \in H^3_{(0)}(I;\mathbb{C})\] if and only if \[ \sum_{k=1}^{+\infty}k^6|d_k|^2<+\infty. \] Hence, Theorem 7 readily follows from Theorem 8 applied to the moment problem (<ref>eq37</ref>) with the help of Proposition 2 and Proposition 3.

    Remark 4. The moment method does not work well for the Schrödinger in dimension larger than 1 (see, however, (Stéphane Jaffard, 1988) in dimension 2). For these dimensions controllability results have been obtained by means of other methods. Let us mention, in particular,

    For a survey on these results, see, in particular, (Enrique Zuazua, 2003).

    A linear Korteweg-de Vries equation

    We go back to the linear Korteweg-de Vries equation control system

    <math yt+yxkdvabs>

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

    <math bcykdvabs>

    y(t,0)=y(t,L)=0,\, y_x(t,L)=u(t), \, t\in (0,T), </math> already considered at this link in the Section [[../Examples_of_control_systems_modeled_by_PDE%27s|Examples of control systems modeled by PDE's]]. Let

    <math eq50>

    \mathcal{N}:=\left\{2\pi \sqrt{\frac{j^2+l^2+jl}{3}};\, j,l\in \mathbb{N}\setminus\{0\}\right\}. </math> Then one has the following theorem (Lionel Rosier, 1997).

    Theorem 9: Let \(T>0\). The control system (<ref>yt+yxkdvabs</ref>)-(<ref>bcykdvabs</ref>) is exactly controllable in time \(T\) if and only if \(L\not \in \mathcal{N}\). Moreover, if \(L\in \mathcal{N}\), then the control system (<ref>yt+yxkdvabs</ref>)-(<ref>bcykdvabs</ref>) is neither approximately controllable nor null controllable in time \(T\).

    For a proof of Theorem 9, see the original paper (Lionel Rosier, 1997) or Section 2.2.2, pages 42-48 in (Jean-Michel Coron, 2007). Let us only point out that \(2\pi \in \mathcal{N}\) (take \(j=l=1\) in (<ref>eq50</ref>)) and that for \(L=2\pi\) and for every \(T>0\) the control system (<ref>yt+yxkdvabs</ref>)-(<ref>bcykdvabs</ref>) is neither approximately controllable nor null controllable in time \(T\). This readily follows from the following observation. Let \(T>0\), let \(y^0\in L^2(0,L)\), and let \(u\in L^2(0,L)\). Let \(y\in C^0([0,T];L^2(0,L))\) be the solution to the Cauchy problem (see the definition given at this link)

    <math eq51>

    y_t+y_x+y_{xxx}=0,\, t\in (0,T), \, x\in (0,2\pi), </math>

    <math eq52>

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

    <math eq53>

    y(0,x)=y^0(x),\, x\in (0,2\pi). </math> We multiply (<ref>eq51</ref>)) by \((1-\cos(x))\) and integrate on \((0,2\pi)\). Using (<ref>eq52</ref>) together with integrations by parts, one gets (first when \(y\) is smooth enough and by density for the general case) \[ \frac{d}{dt}\int_0^{2\pi}(1-\cos(x))ydx=0, \] which shows that the control system (<ref>yt+yxkdvabs</ref>)-(<ref>bcykdvabs</ref>) is neither approximately controllable nor null controllable in time \(T\).

    A heat equation

    Let \(\Omega\) be a non empty open subset of \(\mathbb{R}^l\) and let \(\omega\) be a non empty open subset of \(\Omega\). We go back to the control system

    <math h1>

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

    <math h2>

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

    <math h3>

    u(\cdot,x)=0, \, x\in \Omega\setminus \omega. </math> already considered at this link in the Section Examples of control systems modeled by linear PDE's and show how Carleman estimates allow to prove the following theorem ((Hector Fattorini and David Russell, 1971) if \(n=1\), (Oleg Imanuvilov, 1993; 1995) and (Gilles Lebeau and Luc Robbiano, 1995) for \(n>1\); see also the book (Andrei Fursikov and Oleg Imanuvilov, 1996)).

    Theorem 10. Let us assume that \(\Omega\) is of class \(C^2\) and connected. Then, for every \(T>0\) the control system (<ref>h1</ref>)-(<ref>h2</ref>)-(<ref>h3</ref>) is null controllable and approximately controllable in time \(T\).

    Sketch of the proof of Theorem 10. Let \(T>0\). With the above [[../Examples_of_control_systems_modeled_by_linear_PDE%27s#heatlineargeneralframework|notations]], for\(y^0\in L^2(\Omega)\),

    <math eq54>

    (B^*S^*(t)y^0)(x)=y(t,x), \, t\in (0,T), \, x\in \omega, </math> where \(y:(0,+\infty)\times \Omega \rightarrow \mathbb{R}\) is defined by

    <math eq55>

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

    <math eq56>

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

    <math eq57>

    y(0,x)=y^0(x), \, x\in \Omega, </math> The first step is the following lemma (Lemma 1.2 in (Oleg Imanuvilov, 1995)) (see also Lemma 1.1 on page 4 in (Andrei Fursikov and Oleg Imanuvilov, 1996) and Lemma 2.68 on page 80 in (Jean-Michel Coron, 2007)) and whose proof is omitted.

    Lemma. There exists \(\psi \in C^2(\overline \Omega)\) such that

    <math eq58>

    \psi >0 \text{ in }\Omega, \,\psi=0 \text{ on } \partial \Omega, </math>

    <math eq59>

    |\nabla \psi (x)|>0,\, \forall x\in \overline \Omega \setminus \omega_0. </math>

    Remark: In the case \(n=1\), \(\Omega =(a,b)\) for some real numbers \(a<b\). Let us take \(c\in \omega\). Then \(\psi :\bar \Omega \rightarrow \mathbb{R}\) defined by \[ \psi(x):= (b-c)^3- (x-c)^3, \text{ if } x\in [c,b], \, \psi(x):= (c-a)^3- (c-x)^3, \text{ if } x\in [a,c], \] satisfies (<ref>eq58</ref>)-(<ref>eq59</ref>).

    Let us fix \(\psi\) as in the above lemma. Let \(\alpha: (0,T)\times \overline\Omega\rightarrow (0,+\infty)\) and \(\phi : (0,T)\times \overline\Omega\rightarrow (0,+\infty)\) be defined by

    <math eq60>

    \alpha(t,x)=\frac{e^{2\lambda \|\psi\|_{C^0(\overline \Omega)}}-e^{\lambda \psi(x)}}{t(T-t)} , \,\forall (t, x)\in (0,T)\times \overline \Omega, </math>

    <math eq61>

    \phi(t,x)=\frac{e^{\lambda \psi(x)}}{t(T-t)}, \,\forall (t, x)\in (0,T)\times \overline \Omega, </math> where \(\lambda \in [1,+\infty)\) will be chosen later on. Let \(z:[0,T]\times\overline \Omega\rightarrow \mathbb{R}\) be defined by

    <math eq62>

    z(t,x):=e^{-s\alpha(t,x)}y(t,x),\,(t, x)\in (0,T)\times \overline \Omega, </math>

    <math eq63>

    z(0,x)=z(T,x)=0 ,\, x\in \overline \Omega, </math> where \(s\in [1, +\infty)\) will be chosen later on. From (<ref>eq55</ref>), (<ref>eq60</ref>), (<ref>eq61</ref>) and (<ref>eq62</ref>), we have

    <math eq64>

    P_1+P_2=P_3 </math> with

    <math eq65>

    P_1:=-\Delta z -s^2\lambda^2\phi^2|\nabla\psi |^2z+s \alpha_t z, </math>

    <math eq66>

    P_2:=z_t+2s\lambda \phi \nabla \psi\nabla z +2s\lambda^2\phi|\nabla\psi|^2z, </math>

    <math eq67>

    P_3:=-s\lambda \phi(\Delta\psi)z + s\lambda^2\phi |\nabla \psi|^2z. </math> Let \(Q:=(0,T)\times \Omega\). From (<ref>eq64</ref>), we have

    <math eq68>

    2\int\!\!\int_Q P_1P_2dxdt\leqslant \int\!\!\int_Q P_3^2dxdt. </math> Let \(n\) denote the outward unit normal vector field on \(\partial \Omega\). Note that \(z\) vanishes on \([0,T]\times \partial \Omega\) (see (<ref>eq56</ref>) and (<ref>eq62</ref>)) and on \(\{0,T\}\times\overline\Omega\) (see (<ref>eq63</ref>)). Then straightforward computations using integrations by parts lead to

    <math eq69>

    2\int\!\!\int_Q P_1P_2 dxdt=I_1+I_2 </math> with

    <math eq70>

    I_1:=\int\!\!\int_Q(2 s^3\lambda^4\phi^3|\nabla\psi|^4|z|^2+4 s\lambda^2\phi |\nabla\psi|^2 |\nabla z|^2) dxdt-\int_0^T\int_{\partial\Omega}2s\lambda\phi\frac{\partial \psi}{\partial n} \left(\frac{\partial z}{\partial n}\right)^2d\sigma dt, </math> \[ I_2:=\int\!\!\int_Q \Big(4 s\lambda(\phi \psi_{i})_jz_iz_j-2s\lambda (\phi\psi_i)_i|\nabla z|^2 \] \[ +2s^3\lambda^3\phi^3(|\nabla\psi|^2\psi_i)_iz^2 -2s\lambda^2(\phi |\nabla\psi|^2)_{ii}z^2 \]

    <math eq71>

    -s\alpha_{tt}z^2-2s^2\lambda (\phi\psi_i\alpha_t)_i z^2 +4s^2\lambda^2\phi \alpha_t|\nabla\psi|^2z^2+2s^2\lambda^2\phi\phi_t|\nabla\psi|^2z^2 \Big)dxdt. </math> In (<ref>eq71</ref>) and until the end of the proof, we use the usual repeated-index sum convention. By (<ref>eq59</ref>) and (<ref>eq61</ref>), there exists \(\Lambda\) such that, for every \(\lambda\geqslant \Lambda\), we have, on \((0,T)\times (\Omega\setminus\omega_0)\),

    <math eq72>

    -4\lambda ^2\phi |\nabla\psi|^2|a|^2\leqslant 4 \lambda(\phi \psi_{i})_ja_ia_j-2\lambda (\phi\psi_i)_i|a|^2,\, \forall a=(a_1,\ldots,a_l)^\text{tr}\in \mathbb{R}^l, </math>

    <math eq73>

    -\lambda^4\phi^3|\nabla\psi|^4\leqslant 2\lambda^3\phi^3(|\nabla\psi|^2\psi_i)_i. </math> We take \(\lambda:=\Lambda\). Note that, by (<ref>eq58</ref>),

    <math eq74>

    \frac{\partial \psi}{\partial n}\leqslant 0 \text{ on } \partial \Omega. </math> Moreover, using (<ref>eq60</ref>) and (<ref>eq61</ref>), one gets the existence of \(C>0\) such that, for every \((t,x)\in (0,T)\times \Omega\),

    <math eq75>

    |\alpha_{tt}|+|(\phi\psi_i\alpha_t)_i |+|\phi \alpha_t|\nabla\psi|^2|+|\phi\phi_t|\nabla\psi|^2|+|\phi^3(|\nabla\psi|^2\psi_i)_i|+|\phi^3|\nabla\psi|^4|\leqslant \frac{C}{t^3(T-t)^3}, </math>

    <math eq76>

    |\phi(\Delta\psi)| + |\phi |\nabla \psi|^2|+ |(\phi\psi_i)_i|+|(\phi\psi_i)_i|+|(\phi |\nabla\psi|^2)_{ii}|\leqslant \frac{C}{t(T-t)}, </math>

    <math eq77>

    |(\phi \psi_{i})_j|\leqslant \frac{C}{t(T-t)}, \, \forall (i,j)\in \{1,\ldots,l\}^2. </math> From (<ref>eq59</ref>) and (<ref>eq61</ref>), one gets the existence of \(C>0\) such that

    <math eq78>

    \frac{1}{t^3(T-t)^3}\leqslant C \phi^3|\nabla\psi|^4(t,x),\, \forall (t,x)\in (0,T)\times (\Omega \setminus \omega_0). </math> Using (<ref>eq68</ref>) to (<ref>eq78</ref>), we get the existence of \(C>0\) such that, for every \(s\geqslant 1\) and for every \(y^0\),

    <math eq79>

    s^3 \int_{(0,T)}\int_{\Omega\setminus \omega_0}\frac{ |z|^2}{t^3(T-t)^3}dxdt\leqslant Cs^2\int\!\!\int_Q\frac{|z|^2}{t^3(T-t)^3}dxdt + Cs^3\int_{(0,T)}\int_{\omega_0}\frac{|\nabla z |^2+|z|^2}{t^3(T-t)^3}dxdt. </math> Taking for \(s\geqslant 1\) large enough in (<ref>eq79</ref>), one gets

    <math eq80>

    s^3 \int_{(0,T)}\int_{\Omega\setminus \omega_0}\frac{ |z|^2}{t^3(T-t)^3}dxdt\leqslant Cs^3\int_{(0,T)}\int_{\omega_0}\frac{|\nabla z |^2+|z|^2}{t^3(T-t)^3}dxdt. </math>

    From Theorem 4, (<ref>eq62</ref>), and (<ref>eq80</ref>), one gets the approximate controllability in time \(T\) (with \(H:=L^2(\Omega)\) and \(U:=L^2(\omega)\)) of the control system (<ref>h1</ref>)-(<ref>h2</ref>)(<ref>h2</ref>).

    Let us now deal with the null controllability. Taking \(s\geqslant 1\) large enough in (<ref>eq79</ref>), one gets the existence of \(c_0>0\) independent of \(y^0\) such that

    <math eq81>

    \int_{T/3}^{2T/3}\int_\Omega |z|^2dxdt\leqslant c_0\int_0^T\int_{\omega_0}\frac{|\nabla z |^2+|z|^2}{t^3(T-t)^3}dxdt. </math> We choose such an \(s\) and such a \(c_0\). Coming back to \(y\) using (<ref>eq60</ref>) and (<ref>eq62</ref>), we deduce from (<ref>eq81</ref>) the existence of \(c_1>0\) independent of \(y^0\) such that

    <math eq82>

    \int_{T/3}^{2T/3}\int_\Omega |y|^2dxdt\leqslant c_1\int_0^T\int_{\omega_0}t(T-t)(|\nabla y |^2+|y|^2)dxdt. </math> Let \(\rho\in C^\infty(\overline \Omega)\) be such that \[ \rho =1 \text{ in } \omega_0, \] \[ \rho=0 \text{ in } \overline \Omega \setminus \omega. \] We multiply (<ref>eq55</ref>) by \(t(T-t)\rho y\) and integrate on \(Q\). Using (<ref>eq56</ref>) and integrations by parts, we get the existence of \(c_2>0\) independent of \(y^0\) such that

    <math eq83>

    \int_0^T\int_{\omega_0}t(T-t)(|\nabla y |^2+|y|^2)dxdt\leqslant c_2\int_0^T\int_{\omega}|y|^2dxdt. </math> From (<ref>eq82</ref>) and (<ref>eq83</ref>), we get

    <math eq84>

    \int_{T/3}^{2T/3}\int_\Omega |y|^2dxdt\leqslant c_1c_2\int_0^T\int_{\omega}|y|^2dxdt. </math> Let us now multiply (<ref>eq55</ref>) by \(y\) and integrate on \(\Omega\). Using integrations by parts together with (<ref>eq56</ref>), we get

    <math eq85>

    \frac{\text{d}}{\text{d} t}\int_\Omega |y(t,x)|^2 dx \leqslant 0. </math> From (<ref>eq84</ref>) and (<ref>eq85</ref>), one gets that (15) of Theorem 5 holds with \[ M:=\frac{T}{3c_1c_2}. \] With Theorem 5 and Theorem 6, this concludes the proof of Theorem 10.

    Personal tools
    Namespaces

    Variants
    Actions
    Navigation
    Focal areas
    Activity
    Tools