Feedback stabilization for unbounded bilinear systems using bounded control

Feedback stabilization for unbounded bilinear systems using bounded control Abstract In this paper, we deal with the distributed bilinear system $$ \frac{d z(t)}{d t}= A z(t) + v(t)Bz(t), $$ where A is the infinitesimal generator of a semigroup of contractions on a real Hilbert space H. The linear operator B is supposed bounded with respect to the graph norm of A. Then we give sufficient conditions for weak and strong stabilizations. Illustrating examples are provided. 1. Introduction In this paper, we deal with the infinite dimensional bilinear system \begin{equation} \displaystyle\frac{d z(t)}{d t}= A z(t) + v(t)Bz(t),\quad\;z(0)=z_0\in H, \end{equation} (1.1) where A is an unbounded operator of H with domain D(A) and generates a semigroup of contractions $$(S(t))_{t\geqslant 0}$$ on a real Hilbert space H, whose norm and scalar products are denoted by ∥⋅∥ and ⟨⋅, ⋅⟩, respectively; the linear operator B, with domain D(A) ⊂ D(B), is A-bounded in the sense that there exists $$\alpha ,\beta>0$$ such that $$\|Bz\|\leqslant \alpha \|Az\|+\beta \|z\|,\; \forall z\in D(A), $$ or equivalently (see Desch & Schappacher 1984) \begin{equation} \|Bz\|\leqslant M\left(\|Az\|+\|z\|\right),\quad\; \forall z\in D(A)\; (\mbox{for some }\; M>0). \end{equation} (1.2) The real valued function v(⋅) denotes the control and z(t) is the corresponding mild solution of (1.1). In the case where B ∈ L(H) (L(H) is the set of bounded linear operators on H), the problem of stabilization by non-linear feedback controls has been studied by many authors (see e.g. Ball & Slemrod, 1979; Berrahmoune, 1999; Bounit & Hammouri 1999; Ouzahra, 2008). In Ball & Slemrod, 1979, a result of weak stabilization was obtained under the following condition: \begin{equation} \langle BS(t)y,S(t)y \rangle=0,\quad\;\; \forall t\geqslant 0 \Longrightarrow y=0, \end{equation} (1.3) by using the quadratic feedback \begin{equation} v(t) = -\langle z(t),Bz(t)\rangle\cdot \end{equation} (1.4) On the other hand, if (1.3) is replaced by the following inequality: \begin{equation} \int_0^T|\langle BS(s)y, S(s)y \rangle|\, \mathrm{d}s \geqslant \boldsymbol{\mu} \|y\|^2,\quad\;\; \forall y\in H\;(\mbox{for some}\; \boldsymbol{\mu},T>0), \end{equation} (1.5) then (see Berrahmoune, 1999; Ouzahra, 2008) the feedback (1.4) is a strongly stabilizing control and guarantees the following decay estimate: \begin{equation} \|z(t)\|=O\left(t^{\frac{-1}{2}}\right),\quad\; \mbox{as}\; t\to +\infty\cdot \end{equation} (1.6) In the study by Bounit & Hammouri (1999), the authors considered the control \begin{equation} v(t)=-\frac{\langle Bz(t),z(t) \rangle}{1+|\langle Bz(t),z(t) \rangle|} \end{equation} (1.7) and showed that if the resolvent of A is compact and B is bounded, self-adjoint and monotone, then the feedback (1.7) strongly stabilizes (1.1) provided that (1.3) holds. In the study by Ouzahra (2010), it has been shown that the control \begin{equation} v(t)=-\displaystyle\frac{\left \langle z(t),Bz(t)\right \rangle}{\|z(t)\|^2}\textbf{1}_{\{t\geqslant0; z(t)\ne0\}}, \end{equation} (1.8) weakly stabilizes (1.1) provided that B is a compact operator such that (1.3) holds. On the other hand, it has been shown that if B is a bounded operator satisfying (1.5), then (see the studies by Ouzahra, 2010, 2011) the control (1.8) exponentially stabilizes (1.1). The question of well-posedness of linear and bilinear systems with unbounded control operator has been treated in the studies by Weiss (1989, 1994), Idrissi (2003), Bounit & Idrissi (2005), Idrissi & Bounit (2008), Berrahmoune (2010) and El Ayadi et al. (2012). In the study by Berrahmoune (2010), the author considered the case where A is self-adjoint, and B is positive self-adjoint and bounded from the subspace $$V=D((I-A)^{\frac{1}{2}})$$ of H to its dual space $$V^{^{\prime}}.$$ Then he established the weak and strong stabilities of the closed-loop system \begin{equation} \displaystyle\frac{d z(t)}{d t}=Az(t)+f(\langle Bz(t),z(t)\rangle)Bz(t),\quad\;z(0)=z_0, \end{equation} (1.9) for an appropriate function $$f : \mathbb{R}\longrightarrow \mathbb{R}$$. Moreover, in the study by El Ayadi et al. (2012), it has been supposed that B is an unbounded linear operator from H to a Banach extension X of H with a continuous embedding H ↪ X. Then, it has been shown that (1.1) is strongly stabilizable, and a polynomial decay estimate of the stabilized state has been provided. It is noted that the above-mentioned results of El Ayadi et al. (2012) use spectral decomposition of the system at hand which reduces the class of considered systems to parabolic ones. Here, we deal with control operators which are relatively bounded. Such class of operators is very interesting either in theoretical or practical point of view. Indeed, various properties of semigroup under bounded perturbations are preserved when dealing with relatively bounded perturbations of the generator (see e.g. Pazy, 1978; Engel & Nagel, 2000). On the other hand, relatively bounded control operators arise as models for many dynamical processes (see e.g. Hislop & Sigal, 1996; Yarotsky, 2006). The article is organized as follows: the next section provides background material on non-linear semigroups and unbounded operators. In the third section we present an existence and uniqueness result. In the fourth section we present our main results and we study the weak and strong stabilizations. In the last section, we give illustrating examples. 2. Review on non-linear semigroups and unbounded operators In this section, we recall some existing results related to non-linear semigroups and unbounded operators. Let us begin with the following definitions and results concerning certain classes of unbounded operators (see Hille & Phillips, 1957). Definition 2.1 (Hille & Phillips, 1957, p. 391) Let A be the infinitesimal generator of a $$C_0$$-semigroup. A linear operator C is said to belong to the class F(A) if D(A) = D(C) and $$CR(\lambda _0,A)\in L(H)$$ for some $$\lambda _0\in \boldsymbol{\rho} (A),$$ where $$\boldsymbol{\rho}(A)$$ is the resolvent set of A and $$R(\lambda _0,A)$$ is its resolvent operator. Remark 2.1 (Hille & Phillips, 1957, p. 391) If C ∈ F(A), then $$CR(\lambda ,A)=CR(\lambda _0,A)+(\lambda _0-\lambda )CR(\lambda _0,A)R(\lambda ,A)$$. Then, the operator $$CR(\lambda ,A)$$ is bounded for all $$\lambda \in \boldsymbol{\rho} (A)$$. If C is A-bounded (i.e. C verifies (1.2) for some M > 0), then for all y ∈ D(A) and $$\lambda \in \boldsymbol{\rho} (A) $$ we have $$\|Cy\|\leqslant M\|(-A+\lambda I)y\|+M|\lambda | \|y\|+M\|y\|.$$ Thus, for all x ∈ H, we have $$\|CR(\lambda ,A)x\|\leqslant M\|x\|+(M|\lambda |+M)\|R(\lambda ,A)x\| \leqslant (2M+|\lambda | M)\|x\|.$$ Then the operator $$C|_{D(A)}, $$ restriction of C to D(A), is an element of F(A). Definition 2.2 (Hille & Phillips, 1957, p. 392) Let A be the infinitesimal generator of a $$C_0$$-semigroup. A linear operator C is said to belong to the class $$\widetilde{F}(A)$$ if D(A) ⊂ D(C), $$CR(\lambda _0,A)\in L(H),$$ for some $$\lambda _0\in \boldsymbol{\rho}(A),$$ for all x ∈ H, we have x ∈ D(C) if and only if the limit $$\lim _{\lambda \to \infty }\lambda CR(\lambda ,A)x=y$$ exists, in which case Cx = y. In the sequel, for linear operators $$\Lambda $$ and C with domains $$D(\Lambda )$$ and D(C) (respectively) such that $$D(C)\supset D(\Lambda )$$, we set $$\|C\|_{\Lambda }=\sup \left \{\|Cx\| \;/\; x\in D(\Lambda ), \|x\|\leqslant 1\right \}.$$ Theorem 2.1 (Hille & Phillips, 1957, p. 392) Let C ∈ F(A). Then C has a unique extension $$\widetilde{C}\in \widetilde{F}(A)$$, called A-extension of C and is defined by $$\widetilde{C}x=\lim _{\lambda \rightarrow +\infty }\lambda CR(\lambda ,A)x,\;\forall x\in D(\widetilde{C}):=\{x\in H \; /\;\lim _{\lambda \rightarrow +\infty }\lambda CR(\lambda ,A)x\; \mbox{exists} \}$$. If C is the restriction of a closed operator $$C_1$$, then $$C\subset \widetilde{C} \subset C_1$$. Remark 2.2 If C is bounded, then $$\widetilde{C}$$ is bounded and $$\|\widetilde{C}\|=\|C\|_A$$. If C is A-bounded then $$C|_{D(A)}$$ has a unique extension $$\widetilde{C}\in \widetilde{F}(A)$$. Proposition 2.1 (Hille & Phillips, 1957, p. 394) Let $$C\in \widetilde{F}(A)$$ and suppose that $$CS(t_0)$$ is bounded on D(A) for some $$t_0>0 $$. Then for all $$t\geqslant t_0$$, we have S(t)H ⊂ D(C) and CS(t) is bounded with $$\|CS(t)\|=\|CS(t)\|_A$$. Moreover, the function t ↦ CS(t) is strongly continuous for $$t>t_0, $$ and we have \begin{equation} \limsup_{t\to +\infty} t^{-1}\ln(\|CS(t)\|)\leqslant \omega_0, \end{equation} (2.1) where $$\boldsymbol{\omega}_0$$ is the growth bound of the semigroup S(t). Remark 2.3 If C is an A-bounded operator such that $$\|CS(t_0)\|$$ is bounded on D(A) for some $$t_0>0$$, then $$C|_{D(A)}$$ has a unique extension $$\widetilde{C}$$, which satisfies $$S(t)H\subset D(\widetilde{C})$$ for all $$t\geqslant t_0$$. This property will be useful to establish our weak stabilization result. Let us now recall the notion of non-linear semigroups (Pazy, 1978). Definition 2.3 (Pazy, 1978) Let H be a Hilbert space. A (generally non-linear) strongly continuous semigroup $$(T(t))_{t\geqslant 0}$$ on H is a family of continuous maps T(t) : H→H satisfying T(0) = identity, $$T(t + s) = T(t) T(s)$$, for all $$t, s\in \mathbb{R}^+,$$ for every y ∈ H, T(t)y → y, as $$t\to 0^+.$$ In this case, the mapping defined by $${\mathcal A}y = \lim _{h\to 0^+}\frac{T(h)y - y}{h}$$ for all $$y\in D({\mathcal A}):=\{y\in H/\; \lim _{h\to 0^+}\frac{T(h)y - y}{h} $$ exists in H} is called the infinitesimal generator of the semigroup T(t). If in addition $$\|T(t)y_1-T(t)y_2\|\leqslant \|y_1-y_2\|$$, for every $$t\geqslant 0$$ and $$y_1,y_2\in H$$, then T(t) is said to be a contraction semigroup (or a semigroup of contractions) on H. In this case, $${\mathcal A}$$ is dissipative, i.e. $$\langle{\mathcal A}y_1-{\mathcal A}y_2,y_1-y_2\rangle \leqslant 0,\;$$ for all $$y_1,y_2\in D({\mathcal A}).$$ For $$\phi \in H$$, define the positive orbit through $$\phi $$ by $$ O^+(\phi )=\cup _{t>0}T(t)\phi $$. The $$\omega $$-limit set of $$\phi $$ is the (possibly empty) set given by $$\omega (\phi )= \{\psi \in H;\; $$ there exists a sequence $$ t_n\to +\infty ,\; $$ such that $$ T(t_n)\phi \to \psi , \; $$ as $$ n\to +\infty \}.$$ The weak $$\omega $$-limit set of $$\phi $$ is the (possibly empty) set given by $$\omega _w(\phi )= \{\psi \in H;\; $$ there exists a sequence $$ t_n\to +\infty ,\; $$ such that $$ T(t_n)\phi \rightharpoonup \psi , \; \mbox{as}\; n\to +\infty \}.$$ The sets $$\omega (\phi )$$ and $$\omega _w(\phi )$$ are invariant under the action of any contraction semigroup $$(T(t))_{t\geqslant 0}$$ (see Pazy, 1978). Moreover, given a semigroup of contraction $$S(t)=e^{{\mathcal A}t},$$ we denote by $$E_{\mathcal A}$$ the set of equilibrium states given by $$E_{\mathcal A}={\mathcal A}^{-1}(0)=\{y\in D({\mathcal A}),\;{\mathcal A}y=0\}.$$ We have $$E_{\mathcal A}=\{y\in H;\; e^{t{\mathcal A}}y=y,\, \forall t\geqslant 0\}.$$ Indeed, for y ∈ H such that $$e^{t{\mathcal A}}y=y, \; \forall t\geqslant 0$$, we have $$\lim\nolimits _{t\to 0^+}\dfrac{e^{t{\mathcal A}}y-y}{t}=0$$, then $$y\in D({\mathcal A})$$ and $${\mathcal A}y=0$$. Now, let $$y\in D({\mathcal A})$$ be such that $${\mathcal A}y=0$$. We know that $$\frac{d^+}{dt} e^{t{\mathcal A}}y={\mathcal A}e^{t{\mathcal A}}y, \, \forall t\geqslant 0$$ (see the study by Komura, 1967). It follows that \begin{align*} \left\|y-e^{t{\mathcal A}}y\right\|^2& = \int_{0}^{t}\frac{\mathrm{d}^+}{\mathrm{d}s}\left\|y-e^{s{\mathcal A}}y\right\|^2 \mathrm{d}s\\ & = \int_{0}^{t}2\left\langle{\mathcal A}y-{\mathcal A}e^{s{\mathcal A}}y\;,\;y-e^{s{\mathcal A}}y \right\rangle \mathrm{d}s, \end{align*} from which, we deduce that $$\|y-e^{t{\mathcal A}}y\|^2= 0,\; \forall t\geqslant 0$$. Thus, $$e^{t{\mathcal A}}y=y, \; \forall t\geqslant 0$$. The following result concerns the asymptotic behaviour of the system (1.1) in connection with the structure of the weak $$\omega $$-limit set. Theorem 2.2 (see Pazy, 1978) Let $${\mathcal A}$$ be an infinitesimal generator of a non-linear semigroup of contractions $$(T(t))_{t\geqslant 0}$$ on H and let $$\phi \in H$$. The following conditions are necessary and sufficient for the existence of the weak limit of $$T(t)\phi $$, as $$t\to +\infty $$: $$E_{{\mathcal A}}={\mathcal A}^{-1}(0)\neq \emptyset $$, $$\omega _{w}(\phi )\subset E_{{\mathcal A}}.$$ The next result discusses the case of strong $$\omega $$-limit set. Theorem 2.3 (see Pazy, 1978) Let $${\mathcal A}$$ be an infinitesimal generator of a non-linear semigroup of contractions $$(T(t))_{t\geqslant 0}$$ on H and let $$\phi \in H.$$ The following conditions are necessary and sufficient for the existence of the strong limit of $$T(t)\phi $$, as $$t\to +\infty $$: $$E_{{\mathcal A}} \neq \emptyset $$ and $$\omega (\phi )\neq \emptyset $$, $$\omega (\phi )\subset E_{{\mathcal A}}.$$ 3. Considered systems and well-posedness In this section, we reconsider the system (1.1) with the same hypotheses on A and B. The purpose of this section is to study the feedback stabilization of the system (1.1) using the bounded control \begin{equation} v(t)=-\boldsymbol{\rho}\;\frac{\langle By(t),y(t)\rangle}{1+\langle By(t),y(t)\rangle}, \end{equation} (3.1) where $$\boldsymbol{\rho}>0$$ is the gain control and y is the solution of the corresponding closed-loop system, i.e. \begin{equation} \displaystyle\frac{d z(t)}{d t}={\mathcal A} z(t), \end{equation} (3.2) where $${{\mathcal A}} y = A y+V(y)By,\, \forall y\in D({\mathcal A})=D(A)$$ and $$V(y)=-\boldsymbol{\rho} \;\frac{\langle By,y\rangle }{1+\langle By,y\rangle }, \, \forall y\in D(A)$$. Remark 3.1 For $$B=B^{\ast }\geqslant 0$$ on D(A), we have the following Cauchy–Schwartz-like inequality: $$ |\langle By_1,y_2 \rangle|\leqslant \sqrt{\boldsymbol{\varphi}(y_1)}\sqrt{\boldsymbol{\varphi}(y_2)},\quad \, \forall (y_1,y_2)\in D(A)^2, $$ with $$\boldsymbol{\varphi}(y_1)=\langle By_1,y_1 \rangle $$ and $$\boldsymbol{\varphi}(y_2)=\langle By_2,y_2 \rangle $$. Indeed, since $$B=B^{\ast }\geqslant 0$$ on D(A), we have \begin{equation} \boldsymbol{\mu}^2\varphi(y_2)+2\boldsymbol{\mu}<By_1,y_2>+\varphi(y_1)=\langle B(y_1+\boldsymbol{\mu} y_2),y_1+\boldsymbol{\mu} y_2\rangle\geqslant 0,\quad \, \forall \boldsymbol{\mu}\in\mathbb{R}, \end{equation} (3.3) then $$|\langle By_1,y_2 \rangle |\leqslant \sqrt{\boldsymbol{\varphi} (y_1)}\sqrt{\boldsymbol{\varphi} (y_2)}.$$ For all $$y_1\in D(A),$$ we have $$\langle By_1,y_1\rangle =0 \Rightarrow By_1=0.$$ Indeed, by simplifying with $$\mu>0 $$ and $$\mu <0$$ in (3.3), and letting $$\boldsymbol{\mu} \to 0^{\pm }$$, we obtain $$\langle By_1, y_2\rangle = 0, \, \forall y_2\in D(A)$$. Thus, we conclude by using the density of D(A) in H. In the sequel, we will analyse the well-posedness of the system (3.2). Theorem 3.1 Let A generate a semigroup S(t) of contractions on H, and let B : D(B)→H be a linear A-bounded operator such that (i) ⟨By, z⟩ = ⟨y, Bz⟩, ∀ y, z ∈ D(A), (ii) ⟨By, y⟩ ⩾ 0, ∀ y ∈ D(A). Then for any $$0<\boldsymbol{\rho} <\min (1,\frac{1}{M} )$$, (where M is the constant given in (1.2)) and for all $$z_{0}\in H$$, the system (3.2) admits a unique solution $$z\in{\mathcal C}([0,+\infty [;H)$$. Furthermore, $${{\mathcal A}}$$ generates a contraction semigroup $$e^{t{{\mathcal A}}}$$ on H, and for all $$z_{0}\in H$$ the solution of (3.2) is given by $$z(t) = e^{t {{\mathcal A}}}z_{0}\cdot $$ Proof. Let us set $$\boldsymbol{\varphi} (y)= \langle By,y \rangle ,\, \forall y\in D(A)$$ and let us consider the map $$ \boldsymbol{\phi} =g(\boldsymbol{\varphi}),\;\mbox{with}\;g(z)=\dfrac{1}{2}(z-\ln(1+z)).$$ Since B is self-adjoint, we have $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)=t^2\boldsymbol{\varphi}(y_1)+2t(1-t) \langle By_1,y_2 \rangle+(1-t)^2\boldsymbol{\varphi}(y_2), \forall t\in [0,1], \; \forall (y_1,y_2)\in D({A})^2.$$ Taking into account Remark 3.1, 1., we deduce that $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)=t^2\boldsymbol{\varphi}(y_1)+2t(1-t) \sqrt{\boldsymbol{\varphi}(y_1)} \sqrt{\boldsymbol{\varphi}(y_2)}+(1-t)^2\boldsymbol{\varphi}(y_2), \forall t\in [0,1], \; \forall (y_1,y_2)\in D({A})^2.$$ Hence, $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)\leqslant \big(t\sqrt{\boldsymbol{\varphi}(y_1)}+(1-t)\sqrt{\boldsymbol{\varphi}(y_2)}\big)^2.$$ This together with the convexity of $$t\mapsto t^2$$ gives $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)\leqslant t\boldsymbol{\varphi}(y_1)+(1-t)\boldsymbol{\varphi}(y_2).$$ Thus, the map $$\boldsymbol{\varphi} $$ (and so is $$\boldsymbol{\phi} $$) is convex. It follows that the Gâteaux derivative $$\boldsymbol{\phi} ^{\prime} : D({A} )\longrightarrow H$$ of $$\boldsymbol{\phi} $$ is monotone. On the other hand, for all h ∈ H and y ∈ D(A) we have \begin{align*} \boldsymbol{\phi}^{\prime}(y)\cdot h &=\lim\limits_{t\to 0}\dfrac{\boldsymbol{\phi}(y+th)-\boldsymbol{\phi}(y)}{t}\\ & =\dfrac{1}{2}\;\dfrac{\langle By,y \rangle}{1+\langle By,y\rangle }\langle (B+B^{\ast})y,h\rangle.\end{align*} Then using the fact that B is symetric, we obtain $$ \boldsymbol{\phi}^{\prime}(y)=\frac{\langle By,y\rangle}{1+\langle By,y\rangle}By.$$ On the other hand, we have the following expression regarding the Gâteaux derivative of $$\boldsymbol{\phi} $$: $$ \langle \boldsymbol{\phi}^{\prime}(tx+(1-t)y),x-y\rangle= \dfrac{\big(t\langle Bx,y\rangle+(1-t)\langle By,y\rangle \big)\big(t\langle Bx,x-y\rangle+(1-t)\langle By,x-y\rangle \big)}{1+t^2\langle Bx,x\rangle+2t(1-t)\langle Bx,y\rangle+(1-t)^2\langle By,y\rangle}, \, \forall t\in[0,1],$$ from which, we deduce that $$\boldsymbol{\phi} ^{\prime}$$ is hemicontinous, i.e. for any x, y ∈ D(A), the mapping $$ : t \mapsto \langle \boldsymbol{\phi} ^{\prime}(tx+(1-t)y),x-y) \rangle $$ is continuous on [0, 1]. Now, it comes from (1.2) that for all y ∈ D(A), we have $$ \|V(y)By\|\leqslant \boldsymbol{\rho}\;M(\|Ay\|+\|y\|).$$ Since A generates a semigroup of contractions then − A is maximal monotone and $$\rho \phi ^{\prime}(\cdot )=-V(\cdot ) B(\cdot )$$ is monotone hemicontinuous and $$\boldsymbol{\rho} M<1$$, we have that (see the book by Brezis, 1973) the operator $$-{\mathcal A} : y\longrightarrow -Ay-V(y) By $$ is maximal monotone, and hence $${\mathcal A}$$ generates a semigroup of contractions $$e^{t{{\mathcal A}}}z_0$$, and the function $$ z(t)=e^{t{{\mathcal A}}}z_0$$ is a solution of (3.2) (see Brezis, 1973). Remark 3.2 For all $$z_{0}\in D(A )$$, we have \begin{equation} \|z(t)\|\leqslant \|z_0\|,\quad\; \forall t\geqslant 0 \end{equation} (3.4) and z(t) ∈ D(A) admits a right derivative at t (see the study by Komura, 1967) as well as \begin{equation} \displaystyle\frac{d^+z(t)}{d t}={{\mathcal A}} z(t)\cdot \end{equation} (3.5) \begin{equation} \|{{\mathcal A}} z(t)\|\leqslant \|{{\mathcal A}} z_0\|\cdot \end{equation} (3.6) Remark 3.3 Under the assumptions of Theorem 3.1, we have For all y ∈ D(A) \begin{align*} \|By\|\ & \leqslant M\|{\mathcal A}y-V(y) By\|+M \|y\|\\ &\leqslant M\|{\mathcal A}y\|+M\boldsymbol{\rho}\|By\|+M \|y\| \end{align*} and hence \begin{equation} \|By\|\leqslant \frac{M}{(1-M \boldsymbol{\rho})}\|{{\mathcal A}}y\|+\frac{M}{(1-M \boldsymbol{\rho})}\|y\|, \end{equation} (3.7) For all y ∈ D(A) \begin{align*} \|Ay\|\ & \leqslant \|{\mathcal A}y\|+\boldsymbol{\rho} \|By\|\\ &\leqslant \|{\mathcal A}y\|+M\boldsymbol{\rho}\|Ay\|+ M \boldsymbol{\rho}\|y\| \end{align*} then \begin{equation} \|Ay\|\leqslant \frac{1}{1-\boldsymbol{\rho} M}\|{\mathcal A}y\|+ \frac{\boldsymbol{\rho} M}{1-\boldsymbol{\rho} M}\|y\|, \end{equation} (3.8) The stabilizing control (3.1) is uniformly bounded with respect to initial state $$ |v(t)| \leqslant \boldsymbol{\rho} $$ and if the system (3.2) is subject to the control constraint $$|v(t)| \leqslant \boldsymbol{\alpha} , \; \boldsymbol{\alpha}>0,$$ then one may take $$\boldsymbol{\rho} \leqslant \alpha $$. 4. Stabilization problem In this section, we give sufficient conditions to obtain weak and strong stabilizations of (1.1) using the control (3.1). 4.1 Weak stability In the sequel, we assume that B is A-bounded so $$B|_{D(A)}$$ admits an A-extension denoted by $$\tilde{B}$$. Let us now introduce the following sets: $${\mathcal M} = \{\boldsymbol{\varphi} \in D( A) \;/\; \langle Be^{tA}\boldsymbol{\varphi} ,e^{tA}\boldsymbol{\varphi} \rangle =0, \; \forall t\geqslant 0 \} $$ and $$\widetilde{{\!\mathcal M}} = \{\boldsymbol{\varphi} \in H \;/\; \langle Be^{tA}\boldsymbol{\varphi} ,e^{tA}\boldsymbol{\varphi} \rangle =0, \; \forall t\geqslant 0 \} $$ and let us consider the following hypotheses: $$(\boldsymbol{C}_1)$$: For all sequence $$(x_n)\subset D(A),$$ we have $$\left .\begin{array}{l} 1.\; x_n\rightharpoonup x\in H, \\ 2.\; \|Ax_n\|\; \mbox{is bounded,} \\ 3.\; \langle Bx_n,x_n\rangle \to 0. \end{array} \right \}\;\Longrightarrow\ x\in E_{{{\mathcal A}}}.$$ $$(\boldsymbol{C}_2)$$: If there exists $$t_0>0$$ such that $$\|BS(t_0)\| $$ is bounded in D(A) and for all sequence $$(y_n)\subset D( A)$$ and y ∈ H such that $$y_n\rightharpoonup y$$ in H, there exists a subsequence $$(y_{\gamma (n)})$$ such that for all $$t\geqslant t_0$$, we have $$BS(t)y_{\gamma (n)}\rightarrow \widetilde{B}S(t)y$$ in H, as $$n\to +\infty $$. Remark 4.1 As a class of operators that satisfy the assumption $$(\boldsymbol{C}_1)$$, the set of operators B satisfying $$ \langle By,y\rangle \geqslant \mu \|y\|^s$$, for all y ∈ D(B) (for some $$\mu ,s>0$$). As an other example of operators that satisfies $$(\boldsymbol{C}_1)$$, let $$A=\Delta $$ with Newmann boundary conditions and $$By=-\Delta y+ Ny, \, \forall y\in D(A)$$ with $$Ny=\sum _{j=1}^{+\infty }\frac{1}{j^2}\langle y,\boldsymbol{\varphi} _j\rangle \boldsymbol{\varphi} _j$$ where $$(\boldsymbol{\varphi} _j)_{j\geqslant 0}$$ are the orthonormal basis of $$L^2(\boldsymbol{\Omega})$$ formed with the eigenfunctions of A. Here, B is A-bounded and for any sequence $$(x_n)\subset D(A)$$ such that $$x_n\rightharpoonup x\in H$$ and $$\langle Bx_n,x_n\rangle \to 0,$$ we have $$ \langle -\Delta x_n,x_n\rangle +\langle Nx_n,x_n\rangle \to 0$$. Then, it follows from the positivity of the operators $$-\Delta $$ and N that$$ \nabla x_n \to 0 \;\mbox{and}\; \langle Nx_n,x_n\rangle \to 0.$$ Now using that N is a compact operator, we derive that ⟨Nx, x⟩ = 0. Finally, it follows from the definition of the operator N that $$x=\lambda \textbf{1}\in \ker A\cap \ker B.$$ As an example where the assumption $$(\boldsymbol{C}_2)$$ holds, one can take $$B=A=\Delta $$ (see the study by Haraux, 2001). We note that $$E_{\mathcal A}=\ker (A)\cap \ker (B)$$. Indeed, we have $$y\in E_{\mathcal A} \Longleftrightarrow Ay+V(y)By=0,$$ so the inclusion ($$\supset $$) is clear. Moreover, $$y\in E_{\mathcal A} \Rightarrow \langle Ay,y \rangle +V(y)\langle By,y\rangle =0$$. Thus, since A and − B are dissipative, we have $$ y\in E_{\mathcal A} \Rightarrow \langle Ay,y\rangle = \langle By,y\rangle =0. $$ Then, taking into account Remark 3.1, 2., it comes By = 0 and then Ay = 0. The following theorem concerns the weak stability of (3.2). Theorem 4.1 Suppose that the hypotheses of Theorem 3.1 are verified. 1. If $$(\boldsymbol{C}_1)$$ holds then for all $$z_0 \in D(A)$$ the solution of (3.2) is weakly convergent to $$z^{\ast }\in E_{\mathcal A}$$, as $$t\to +\infty $$. 2. If $$(\boldsymbol{C}_2)$$ holds, then for all $$z_0 \in D(A)$$ the solution of (3.2) satisfies (a) $$z(t)\rightharpoonup z^{\ast }\in E_{A}$$, if $$\widetilde{{\mathcal M}} \subset E_A,$$ (b) $$z(t)\rightharpoonup 0$$, if $$\widetilde{{\mathcal M}} =\{0\}$$. Proof. 1. Let $$z_0 \in D({{A}} )$$. According to Remark 3.2, the function t → z(t) admits a right derivative at all time, then we have $$ \dfrac{d^+\|z(\tau)\|^2}{d\tau}=2\langle{{\mathcal A}} z(\tau),z(\tau)\rangle,\;\; \forall \tau \geqslant 0.$$ By integrating this equality between s and t, we obtain \begin{equation} \frac{1}{2}\left[\|z(t)\|^2-\|z(s)\|^2\right]=\int_s^t \langle{{\mathcal A}} z(\tau),z(\tau)\rangle\;\mathrm{d}\tau,\quad \; \forall t\geqslant s \geqslant 0. \end{equation} (4.1) Since A is dissipative, we have $$ \int_s^t \langle{{\mathcal A}} z(\tau),z(\tau)\rangle\;\mathrm{d}\tau \leqslant -\boldsymbol{\rho} \int_s^t \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle}\, \mathrm{d}\tau,\quad \; \forall t\geqslant s \geqslant 0. $$ It comes from (4.1) that $$ \int_0^t \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle} \, \mathrm{d}\tau \leqslant \frac{1}{2\boldsymbol{\rho}}\|z_0\|^2. $$ Thus, \begin{equation} \int_{0}^{+\infty}\frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle}\;\mathrm{d}\tau <+\infty. \end{equation} (4.2) Moreover, from Remark 3.2 and inequality (3.7) we get \begin{equation} \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\left[\frac{M}{(1-M \boldsymbol{\rho})}\|{{\mathcal A}}z_0\|+\frac{M}{(1-M \boldsymbol{\rho})}\|z_0\|\right]\|z_0\|} \leqslant \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle},\quad\;\forall \tau>0. \end{equation} (4.3) Hence, \begin{equation} \int_{0}^{+\infty}\langle Bz(\tau),z(\tau)\rangle^2\;\mathrm{d}\tau <+\infty. \end{equation} (4.4) From (3.4), it follows that $$\boldsymbol{\omega} _w(z_0)\neq \emptyset $$. Let $$\boldsymbol{\varphi} _0 \in \boldsymbol{\omega} _w(z_0)$$ and let $$t_n\to +\infty $$ such that $$z(t_n)=e^{t_n{{\mathcal A}}}z_0\rightharpoonup \boldsymbol{\varphi} _0$$, as $$n\to +\infty $$. We shall prove that there exists a sequence $$(s_{n})$$ such that $$s_n\to +\infty $$, $$z(s_n)\rightharpoonup \boldsymbol{\varphi} _0$$ and $$~{\lim _{n\to +\infty }\langle Bz(s_{n}),z(s_{n})\rangle =0.}$$ For this end, let us consider $$\boldsymbol{\varepsilon}>0$$ and $$H_{\boldsymbol{\varepsilon}} =\{t>0 \; /\; \langle Bz(t),z(t)\rangle \geqslant \boldsymbol{\varepsilon} \}$$ and let us define the map G : t →< Bz(t), z(t) >. From (4.4), we have $$G \in L^2(0,+\infty )$$, thus the Lebesgue measure of $$H_{\boldsymbol{\varepsilon}} $$ is finite. Then for any A > 0, the set $$E=\{k\in \mathbb{N}\; /\; \; t_k>A\}$$ is infinite, hence $$\bigcup _{k\in E}[t_k,t_k+\boldsymbol{\varepsilon} ]\not \subset H_{\boldsymbol{\varepsilon}} .$$ It follows that for $$\boldsymbol{\varepsilon}>0$$ and A > 0, there exist $$t_k>A$$ and $$ s\in [t_k,t_k+\boldsymbol{\varepsilon} ]$$ such that $$G(s)<\boldsymbol{\varepsilon} $$. Let us now proceed to the construction of sequence $$(s_j)_{j\geqslant 1}$$. For $$\boldsymbol{\varepsilon} =\dfrac{1}{1}$$ and A = 1, then there exist $$t_{k_1}>1$$ and $$s_1\in [t_{k_1},t_{k_1}+\dfrac{1}{1} ]$$ such that $$G(s_1) < \dfrac{1}{1}$$. Now, let $$j\in \mathbb{N}^{\ast }$$ and suppose that the terms $$t_{k_j}$$ and $$s_j$$ are constructed. For $$\boldsymbol{\varepsilon} =\dfrac{1}{j+1}$$ and $$A=t_{k_j}+j>0$$, there exist $$t_{k_{j+1}}>A$$ and $$s_{j+1}\in [t_{k_{j+1}},t_{k_{j+1}}+\frac{1}{j+1} ]$$ such that $$G(s_{j+1}) < \dfrac{1}{j+1}$$. Then from the construction of $$(s_j)_{j\geqslant 1}$$ and $$(t_{k_j})_{j\geqslant 1}, $$ we have $$G(s_j)\leqslant \dfrac{1}{j}$$. On the other hand, we have $$t_{k_{j+1}}-t_{k_j}>j$$, then $$t_{k_{q+1}}-t_{k_1}>\sum _{j=1}^{q}j=\frac{q(q+1)}{2}$$ for all $$q\geqslant 1$$, and so $$\lim _{j\to +\infty }t_{k_j}=+\infty $$. Thus, $$\lim_{j\to +\infty }s_j=+\infty $$ (recall that $$s_j\geqslant t_{k_j}$$). We can therefore choose a subsequence $$(t_{k_j})$$ of $$(t_{n})$$ and a sequence $$(s_j)$$ such that $$t_{k_j} \leqslant s_j \leqslant t_{k_j}+\frac{1}{j}$$, $$G(s_j) < \frac{1}{j}$$ and $$s_j\to +\infty $$. From (3.5) for all $$t\geqslant 0$$, we have $$\|{\mathcal A}e^{t{{\mathcal A}}}z_0\|\leqslant \|{{\mathcal A}}z_0\|$$. Then \begin{equation} \left\|e^{s_j{{\mathcal A}}}z_0-e^{t_{k_j}{{\mathcal A}}}z_0\right\|\leqslant \|{{\mathcal A}}z_0\|\left|s_j-t_{k_j}\right|\leqslant \frac{\|{{\mathcal A}}z_0\|}{j}. \end{equation} (4.5) We deduce that $$z(s_j)\rightharpoonup \boldsymbol{\varphi} _0$$ and $$\langle Bz(s_j),z(s_j)\rangle \to 0$$, as $$j\to +\infty .$$ Moreover, according to the Remark 3.3, 2., $$(\|Az(s_j)\|)_{j\geqslant 1}$$ is bounded. It follows from $$(\boldsymbol{C}_1)$$ that $$\varphi _0\in E_{{\mathcal A}}$$, which gives $$\boldsymbol{\omega} _w(z_0)\subset E_{{\mathcal A}}$$. Finally, Theorem 2.2 implies that z(t) is weakly convergent as $$t\to +\infty $$. 2. (a) Let $$\boldsymbol{\varphi} _0$$ and $$(s_j)_{j\geqslant 1}$$ be as in 1. and suppose that the condition $$(\boldsymbol{C}_2)$$ holds. From (4.2) we have \begin{equation} \lim_{j\to+\infty}\int_{s_{j}}^{t+s_{j}} \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle}\;\mathrm{d}\tau=0,\quad\;\forall t>0. \end{equation} (4.6) It follows from the invariance of $$\boldsymbol{\omega} _w(z_0)$$ under the semigroup $$e^{t{\mathcal A}}$$ that for all t > 0, we have \begin{equation} \langle Bz(t+s_j),z(t+s_j)\rangle\to 0,\quad\; \mbox{as}\; j\to+\infty. \end{equation} (4.7) Moreover, from the variation of constants formula, we have $$ z(t+s_j)=S(t)z(s_j)+\int_{s_j}^{t+s_j} v(z(\tau))S(t+s_j-\tau)Bz(\tau)\, \mathrm{d}\tau,\quad\;\forall t>0.$$ It follows that $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \int_{s_j}^{t+s_j} |v(z(\tau))|\;\|Bz(\tau)\|\, \mathrm{d}\tau,\quad\;\forall t>0\cdot$$ Using (3.4)–(3.7), we deduce that \begin{equation} \|Bz(\tau)\|\leqslant \frac{M}{(1-M \boldsymbol{\rho})}\|{{\mathcal A}}z_0\|+\frac{M}{(1-M \boldsymbol{\rho})}\|z_0\|=:M_{z_0,\boldsymbol{ \rho}}\cdot \end{equation} (4.8) Then $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \; M_{z_0, \boldsymbol{\rho}} \int_{s_j}^{t+s_j} |v(z(\tau))|d\tau,\quad\;\forall t>0\cdot$$ The Schwartz’s inequality gives $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \boldsymbol{\rho}\; \sqrt{t}\; M_{z_0, \boldsymbol{\rho}}\; \sqrt{\int_{s_j}^{t+s_j} \frac{<Bz(\tau),z(\tau)>^2}{(1+<Bz(\tau),z(\tau)>)^2}\, \mathrm{d}\tau},\quad\;\forall t>0,$$ and hence $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \boldsymbol{\rho}\; \sqrt{t}\; M_{z_0, \rho}\; \sqrt{\int_{s_j}^{t+s_j} \frac{<Bz(\tau),z(\tau)>^2}{1+<Bz(\tau),z(\tau)>}\ \textrm{d}\tau},\quad\;\forall t>0.$$ Thus, using (4.2) we deduce that \begin{equation} \lim_{j\to +\infty}[z(t+s_j)-S(t)z(s_j)]=0,\quad\;\forall t>0. \end{equation} (4.9) Using the fact that B is self-adjoint, we can write $$\langle BS(t)z(s_j), S(t)z(s_j)\rangle =\langle S(t)z(s_j)-z(t+s_j), BS(t)z(s_j)\rangle +\langle Bz(t+s_j), S(t)z(s_j)-z(t+s_j)\rangle + \langle Bz(t+s_j), z(t+s_j)\rangle$$. Furthermore, from (1.2), (3.4), (3.8) and the fact that S(t) is of contractions, we get \begin{align*} \|BS(t)z(s_j)\| & \leqslant M\left(\|AS(t)z(s_j)\|+\|S(t)z(s_j)\|\right)\\ &\leqslant \frac{M}{1-\boldsymbol{\rho} M}\| ({\mathcal A}z_0\|+ \|z_0\|)\cdot \end{align*} This, together with (4.7), (4.8) and (4.9) gives \begin{equation} \langle BS(t)z(s_j),S(t)z(s_j)\rangle\to 0,\quad\; \mbox{as}\; j\to +\infty. \end{equation} (4.10) Now, from the condition $$(\boldsymbol{C}_2)$$, there exists a subsequence of $$(s_j), $$ still denoted by $$(s_j), $$ such that for all $$t\geqslant t_0$$ we have \begin{equation} \langle BS(t)z(s_j),S(t)z(s_j)\rangle\to \langle \widetilde{B}S(t){\boldsymbol{\varphi}_0},S(t){\boldsymbol{\varphi}_0}\rangle,\quad\; \mbox{as}\; j\to +\infty. \end{equation} (4.11) We conclude that \begin{equation} \langle \widetilde{B}S(t){\boldsymbol{\varphi}_0},S(t){\boldsymbol{\varphi}_0}\rangle=0. \end{equation} (4.12) In other words, $$\boldsymbol{\varphi} _0\!\in\! \widetilde{{\mathcal M}}$$ and hence $$\boldsymbol{\omega} _w(z_0)\!\subset \!\widetilde{{\mathcal M}}$$. But $$\widetilde{{\mathcal M}}\subset E_A,$$ then $$S(t)\boldsymbol{\varphi} _0=\boldsymbol{\varphi} _0$$ and so $$\langle \widetilde{B}{\boldsymbol{\varphi} _0},{\boldsymbol{\varphi} _0}\rangle =0$$. Moreover, since $$\boldsymbol{\omega} _w(z_0)$$ is invariant by $$e^{t{{\mathcal A}}}$$, we have $$\boldsymbol{\varphi} (t)=e^{t{{\mathcal A}}}\boldsymbol{\varphi} _0\in \boldsymbol{\omega} _w(z_0)$$, thus $$\langle \widetilde{B}\boldsymbol{\varphi} (t),\varphi (t)\boldsymbol{\varphi} =0.$$ Consequently, $$\boldsymbol{\varphi} (t)=e^{t{{\mathcal A}}}\boldsymbol{\varphi} _0=S(t)\boldsymbol{\varphi} _0=\boldsymbol{\varphi} _0$$, and hence $$\boldsymbol{\varphi} _0\in E_{{\mathcal A}}$$. We deduce that $$ \boldsymbol{\omega} _w(z_0)\subset E_{{\mathcal A}}$$. Finally, from Theorem 2.2, there exists $$z^{\ast }\in E_A$$ such that $$z(t)\rightharpoonup z^{\ast },$$ as $$t\to +\infty$$. (b) In the case $$\widetilde{{\mathcal M}}=\{0\}$$, it follows from $$\boldsymbol{\omega} _w(z_0)\subset \widetilde{{\mathcal M}}$$ that $$z(t)\rightharpoonup 0,$$ as $$t\to +\infty .$$ 4.2 Strong stability In this part, we study the strong stability of the closed-loop system (3.2). Let us start with the two following lemmas. Lemma 4.1 Suppose that the assumptions of Theorem 3.1 hold and that the embedding D(A) ↪ H is compact. Then the operator $$(I-{{\mathcal A}})^{-1}$$ is compact from H to itself. Proof. Since $${\mathcal A}$$ generates a non-linear semigroup of contractions, the operator $$(I-{{\mathcal A}})^{-1}$$ is defined on H (see Theorem 2, in the study by Komura, 1967). In order to prove the compactness of the operator $$(I-{{\mathcal A}})^{-1}$$, let us consider a bounded sequence $$(v_n)$$ in H. We shall prove that the sequence $$u_n=(I-{{\mathcal A}})^{-1}v_n$$, defined in D(A), has a converging subsequence (see Definition 2.5, p. 13 in the study by Toledano et al., 1997). We have $$ \langle v_n,u_n\rangle=\|u_n\|^2-\langle{{\mathcal A}}u_n,u_n\rangle.$$ Then, using the fact that the operator $${{\mathcal A}}$$ is dissipative, we get $$\|u_n\|^2 \leqslant \langle v_n,u_n\rangle \leqslant \|v_n\| \|u_n\|$$. It follows that the sequence $$(u_n)$$ is bounded. Since $${\mathcal A}u_n=u_n-v_n,$$ then the sequence $$({\mathcal A}u_n)$$ is bounded. From (3.7) we can see that the sequence $$(Bu_n)$$ is bounded. It follows from the equality $$ Au_n={\mathcal A}u_n+\boldsymbol{\rho} \frac{\langle B{u_n},u_n\rangle}{1+\langle B{u_n},u_n\rangle}B{u_n}$$ that $$(Au_n)$$ is bounded. Furthermore, $$(u_n)\subset D(A)$$, then we can conclude by the fact that D(A) is compactly embedded in H. Remark 4.2 Under the assumptions of Lemma 4.1, we have $$\boldsymbol{\omega} (z_0)\neq \emptyset $$, for all $$z_0\in D(\mathcal A)$$ (see the study by Pazy, 1978). Now, we are ready to establish the strong stabilization result. This is the subject of the two next theorems. Theorem 4.2 Let assumptions of Theorem 3.1 be verified, let D(A) be compactly embedded in H and let us set $$U(A)=\{y\in H\;/\; \|e^{tA}y\|=\|e^{tA^{\ast }}y\|=\|y\|\}$$. Then for all $$z_0 \in D(A)$$, we have $$\boldsymbol{\omega} (z_0)\subset{\mathcal M} \cap U(A)$$. If $$U(A)\cap{\mathcal M}\subset E_A$$, then for all $$z_0 \in D(A)$$ we have $$ z(t)\to z^{\ast } \in E_A$$, as $$t\to +\infty $$. Proof. Let $$z_0\in D(A)$$, $$\boldsymbol{\varphi} _0\in \boldsymbol{\omega} (z_0)$$ and let $$(t_n)_{n\geqslant 0 }$$ be such that $$t_n\to +\infty $$ and $$e^{t_n{\mathcal A}}z_0\to \boldsymbol{\varphi} _0$$ as $$n\to +\infty .$$ For t > 0, there exists $$N\in \mathbb{N}$$ such that for all $$n\geqslant N$$, we have $$t_n\geqslant t.$$ Using the fact that $$e^{t{\mathcal A}}$$ is a semigroup of contractions, we obtain $$ \|e^{t_n{\mathcal A}}z_0\|\leqslant \|e^{t{\mathcal A}}z_0\|$$; and letting $$n\to +\infty ,$$ we get \begin{equation} \|\boldsymbol{\varphi}_0\|\leqslant \left\|e^{t{\mathcal A}}z_0\right\|=\|z(t)\|. \end{equation} (4.13) It follows from the inequality (4.13) that $$\|\boldsymbol{\varphi} _0\|\leqslant \|e^{(t+t_n){\mathcal A}}z_0\|$$. Then, letting $$n\to +\infty $$ we obtain $$\|\boldsymbol{\varphi}_0\|\leqslant \|e^{t{\mathcal A}}\boldsymbol{\varphi} _0\|$$. This together with the fact that $$(e^{t{\mathcal A}})_{t\geqslant 0}$$ is a contraction semigroup leads to $$\|e^{t{\mathcal A}}\boldsymbol{\varphi} _0\|=\|\boldsymbol{\varphi} _0\|$$ for all $$t\geqslant 0$$. Furthermore, we have $$\boldsymbol{\omega} (z_0)\subset D({\mathcal A})$$ (see Theorem 5, in the study by Dafermos & Slemrod, 1973) and so $$\boldsymbol{\varphi} _0 \in D(A)$$. From (4.5) we have $$e^{s_j{\mathcal A}}z_0\to \boldsymbol{\varphi} _0$$, as $$j\to +\infty $$. On the other hand, \begin{align*} \langle BS(t)z(s_j),S(t)z(s_j)\rangle & =\langle BS(t)z(s_j),S(t)(z(s_j)-\boldsymbol{\varphi}_0)\rangle+\langle BS(t)\boldsymbol{\varphi}_0,S(t)z(s_j)\rangle \\ &=\langle BS(t)z(s_j),S(t)(z(s_j)-\boldsymbol{\varphi}_0)\rangle+\langle S(t)^{\ast}BS(t)\boldsymbol{\varphi}_0,z(s_j)\rangle.\end{align*} Since $$(\|BS(t)z(s_j)\|)_{j\geqslant 1}$$ is bounded, the semigroup $$(S(t))_{t\geqslant 0}$$ is of contractions and $$z(s_j)\to \varphi _0$$, as $$j\to +\infty $$, then $$\langle BS(t)z(s_j),S(t)z(s_j)\rangle \to \langle BS(t)\boldsymbol{\varphi} _0,S(t)\boldsymbol{\varphi} _0\rangle $$, as $$j\to +\infty $$, this combined with (4.10), gives $$\boldsymbol{\varphi} _0\in{\mathcal M}$$ and so $$e^{t{\mathcal A}}\boldsymbol{\varphi} _0=e^{tA}\boldsymbol{\varphi} _0$$. We conclude that $$\boldsymbol{\varphi} _0\in U(A)\cap{\mathcal M}.$$ Let $$\boldsymbol{\varphi} _0\in \boldsymbol{\omega} (z_0)$$. From 1. we have $$\boldsymbol{\varphi} _0\in U(A)\cap{\mathcal M} \subset E_A$$. Since $$\boldsymbol{\varphi} (t)=e^{t{\mathcal A}}\boldsymbol{\varphi} _0\in \boldsymbol{\omega} (z_0)$$, then $$\boldsymbol{\varphi} (t)\in U(A)\cap{\mathcal M} \subset E_A.$$ It follows that $${\mathcal A}\boldsymbol{\varphi} (t)=A\boldsymbol{\varphi} (t)=0$$. In particular, $${\mathcal A}\boldsymbol{\varphi}_0=0$$. In other words, $$\boldsymbol{\varphi} _0\in E_{\mathcal A}$$. From Theorem 2.3, we deduce the existence of the strong limit of z(t), as $$t\to +\infty $$. Thus, $$z(t)\to \boldsymbol{\varphi} _0\in E_A$$, as $$t\to +\infty $$. Theorem 4.3 Suppose that the hypotheses of Theorem 3.1 are verified. Let D(A) be compactly embedded in H, $${\mathcal M}\cap U(A) =\{0\}$$. Then for all $$z_0 \in H$$, we have z(t) → 0, as $$t\to +\infty $$. Proof. Let $$z_0\in D(A)$$. It follows from Theorem 4.2 that $${\mathcal M}\cap U(A)=\{0\}\subset E_A$$. Moreover, we have $$\boldsymbol{\omega} (z_0)\subset{\mathcal M}\cap U(A)$$. Then $$\boldsymbol{\omega} (z_0)=\{0\}$$ and hence z(t) → 0, as $$t\to +\infty $$. Now let us consider $$z_0\in \overline{D({{\mathcal A}} )}$$ and let $$\boldsymbol{\varepsilon}>0$$. Then there exists $$z_{\boldsymbol{\varepsilon}} \in D({{\mathcal A}} )$$ such that $$\|z_0-z_\varepsilon \|\leqslant \frac{\boldsymbol{\varepsilon} }{2}$$. It follows from the fact that $$e^{t{\mathcal A}}$$ is a contraction semigroup that \begin{equation} \left\|e^{t{\mathcal A}}z_0-e^{t{\mathcal A}}z_\varepsilon\right\|\leqslant \frac{\boldsymbol{\varepsilon}}{2}, \;\;\;\forall t\geqslant 0. \end{equation} (4.14) Since $$z_\varepsilon \in D({{\mathcal A}} )$$ and $${\mathcal M}\cap U(A)=\{0\}$$, we deduce from Theorem 4.2, 1. that $$\boldsymbol{\omega} (z_{\boldsymbol{\varepsilon}} )=\{0\} $$. Thus, there exists $$T_{\boldsymbol{\varepsilon}}>0$$ such that \begin{equation} \left\|e^{t{\mathcal A}}z_{\boldsymbol{\varepsilon}}\right\|\leqslant \frac{\boldsymbol{\varepsilon}}{2},\;\;\;\forall t\geqslant T_{\boldsymbol{\varepsilon}} . \end{equation} (4.15) It follows from (4.14) and (4.15) that for all $$t\geqslant T_{\boldsymbol{\varepsilon}} ,$$ we have $$\|e^{t{\mathcal A}}z_0\|\leqslant \boldsymbol{\varepsilon} $$. We conclude that for all $$z_0\in \overline{D({\mathcal A})}=H$$, we have z(t) → 0, as $$t\to +\infty .$$ 5. Applications 5.1 Heat equation Let $$\boldsymbol{\Omega} =]0,1[$$ and let us consider the bilinear system given by the following heat equation: \begin{equation} \left\{ \begin{array}{lll} \frac{\partial z}{\partial t}(t,x)=\Delta z (t,x)+v(t)a(x)z(t,x), \; \mbox{on} \; ]0,\quad+\infty[\times\boldsymbol{\Omega}&& \\ z^{\prime}(t,0)=z^{\prime}(t,1)=0,\quad\; \forall \;t>0 && \\ z(0,x)=z_0(x),\quad\; \mbox{on} \;\boldsymbol{\Omega} && \end{array} \right. \end{equation} (5.1) where $$a\in L^{\gamma }(\boldsymbol{\Omega} )\;(\gamma> 2)$$ and $$\;a \notin L^{\gamma +2}(\boldsymbol{\Omega} )$$ (as an example, one can take $$a(x)=x^{-\frac{1}{\gamma +1}}, \; \forall x\in ]0,1[$$). Let $$ H= L^2(\boldsymbol{\Omega} )$$ and $$Az=\Delta z$$, for all $$z\in D(A) =\{z\in L^2(\boldsymbol{\Omega} )\;/\;\Delta z \in L^2(\boldsymbol{\Omega} ),\;z^{\prime}(0)=z^{\prime}(1)=0\}$$. Observing that $$Ba^{\frac{\gamma }{2}}=a^{\frac{\gamma +2}{2}}\not \in L^2(\boldsymbol{\Omega} )$$, we deduce that the operator defined by Bz = az, for all $$z\in D(B):=H^1(\boldsymbol{\Omega} )$$ is not bounded from $$L^2(\boldsymbol{\Omega} ) $$ to $$L^2(\boldsymbol{\Omega} )$$. Now, we have $$\|az\|_{L^{2}(\boldsymbol{\Omega} )}\leqslant \|a\|_{L^{\gamma }(\boldsymbol{\Omega} )}\|z\|_{L^{\gamma ^{\ast }}(\boldsymbol{\Omega} ), }\; \mbox{for all}\; z\in H^1(\boldsymbol{\Omega} )$$ with $$\boldsymbol{\gamma} ^{\ast }=\frac{2\boldsymbol{\gamma} }{\boldsymbol{\gamma} -2}$$. By the Sobolev embedding $$H^1(\boldsymbol{\Omega} )\hookrightarrow L^q(\boldsymbol{\Omega} ),\; 1 \leqslant q < +\infty $$ (see Corollary 9.14 in the book by Brezis, 2010), and the fact that for all z ∈ D(A) we have $$\|\nabla z\|^2=|\langle \Delta z,z \rangle | \leqslant \frac{1}{2}(\|\Delta z\|^2+\|z\|^2).$$ We conclude that B is A-bounded. We can state the following stabilization result: Proposition 5.1 Suppose that $$a(x)\geqslant 0,\;\mbox{a.e. on}\; \boldsymbol{\Omega} $$ and $$\int _{\Omega }a(x)\, \mathrm{d}x \neq 0$$. Then there exists $$\boldsymbol{\rho} _0>0$$ such that, for any $$0<\boldsymbol{\rho} < \boldsymbol{\rho} _0$$ and for all $$z_0\in L^2(0,1)$$, the control $$v(t)=- \boldsymbol{\rho} \; \frac{\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\;\mathrm{d}x}{1+\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\;\mathrm{d}x} $$ strongly stabilizes (5.1). Proof. Here, A and B satisfy the assumptions of Theorem 3.1 and we have $$U(A)=E_A\!=\!\{c 1_{\boldsymbol{\Omega}} \; / \; c\!\in\! \mathbb{R}\}$$. Since $$\int _{\Omega }a(x)\, \mathrm{d}x \neq 0,$$ we have $${\mathcal M}\cap U(A)=\{0\}$$, then we conclude by Theorem 4.2. 5.2 Transport equation Let $$\boldsymbol{\Omega} =]0,+\infty [$$ and let us consider the following bilinear transport equation: \begin{equation} \left\{ \begin{array}{lll} \frac{\partial z}{\partial t}(t,x)=-\frac{\partial z}{\partial x} (t,x)+v(t)a(x)z(t,x),\; \mbox{on} \; ]0,+\infty[\times\boldsymbol{\Omega}&& \\ z(t,0)=0,\; \mbox{on} \; ]0,+\infty[&& \\ z(0,x)=z_0(x)\; \mbox{on} \;\boldsymbol{\Omega}&& \end{array} \right. \end{equation} (5.2) where a is such that a(x) > 0, a.e. $$x\in \boldsymbol{\Omega} $$$$\int _0^{+\infty }x\;a^2(x)\;\mathrm{d}x<+\infty $$ (for example, $$a(x)=\frac{1}{\sqrt{x(x^2+1)}}$$). Let $$ H= L^2(\boldsymbol{\Omega} )$$ and $$Az=-\frac{\partial z}{\partial x}$$, for all $$z\in D(A) =\{y\in H^1(\boldsymbol{\Omega} )\;/\; y(0)=0 \; \}$$ and Bz = az, for all z ∈ D(B) := D(A). Thus, B is unbounded from $$L^2(\boldsymbol{\Omega} ) $$ to $$L^2(\boldsymbol{\Omega} )$$. Moreover, by Morrey’s inequality (see Corollary 9.14 in the book by Brezis, 2010), there exists C > 0 such that for all z ∈ D(A) we have $$|z(x)|\leqslant C\sqrt{x}\;\|\nabla z\|_{L^2(\boldsymbol{\Omega} )}$$, a.e. $$x\in \boldsymbol{\Omega} $$, then $$\int _{0}^{+\infty }|a(x)z(x)|^2\, \mathrm{d}x\leqslant C^2\; \|\nabla z\|_{L^2(\boldsymbol{\Omega} )}^2\; \int _0^{+\infty }xa^2(x)\;\mathrm{d}x$$, then B is A-bounded i.e. (1.2) holds for $$M=C\sqrt{\int _0^{+\infty }x\;a^2(x)\;\mathrm{d}x}$$. Furthermore, for all sequence $$(y_n)_{n\in \mathbb{N}} \subset H_0^1(\boldsymbol{\Omega} )$$ such that $$y_n\rightharpoonup y$$ in H and $$<By_n,y_n>\rightarrow 0$$, as $$n\rightarrow +\infty $$, there exists a subsequence of $$(y_n)_{n\in \mathbb{N}}$$ still denoted by $$(y_n)_{n\in \mathbb{N}}$$ such that $$a(x)y_n^2(x)\rightarrow 0$$, a.e. $$x\in \boldsymbol{\Omega} $$ as $$n\to +\infty $$. If in addition $$(Ay_n)_{n\in \mathbb{N}}$$ is bounded in H, then for all test function $$\varphi $$ we have $$|y_n(x)\;\boldsymbol{\varphi} (x)|\leqslant C\|\nabla y_n\|_{L^2(\boldsymbol{\Omega} )}\sqrt{x}\;|\boldsymbol{\varphi} (x)|$$. This combined with the dominated convergence theorem gives $$\langle y_n,\varphi \rangle \rightarrow 0$$ as $$n\rightarrow +\infty $$, we conclude that $$y_n\rightharpoonup 0$$ in H, as $$n\to +\infty $$. Then y = 0, thus the condition $$(\boldsymbol{C}_1)$$ is verified. As a consequence of Theorem 4.1, we have the following result: Proposition 5.2 Suppose that a(x) > 0, a.e. $$x\in \boldsymbol{\Omega} $$ and $$\int _0^{+\infty }x\;a^2(x)\;\mathrm{d}x<+\infty $$ . Then there exists $$\boldsymbol{\rho} _0>0$$ such that for any $$0<\boldsymbol{\rho} < \rho _0$$ and for all $$z_0\in L^2(\boldsymbol{\Omega} )$$, the system (5.2) controlled by the feedback $$v(t)=- \boldsymbol{\rho} \; \frac{\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\;\mathrm{d}x}{1+\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\, \mathrm{d}x}$$ admits a unique solution $$z\in{\mathcal C}(0,+\infty ;H)$$, and we have $$z(t)\rightharpoonup 0$$ in $$L^2(\boldsymbol{\Omega} )$$, as $$t\to +\infty $$. 6. Conclusion In this paper, a set of stabilization results for unbounded bilinear systems has been given. Sufficient conditions for weak and strong stabilizations have been given. The stabilization control is uniformly bounded with respect to time and initial states. The established results can be applied to different type of bilinear system including parabolic and hyperbolic case. Though the established results enable us to discuss various types of stabilization problems as illustrated above, there are other interesting situations which are not covered by the present study. This is the case when the control is exercised through the boundary or a point for systems governed by partial differential equations. Acknowledgements The authors would like to thank the anonymous referees for their valuable comments. References Ball , J. & Slemrod , M. ( 1979 ) Feedback stabilization of distributed semilinear control systems . Appl. Math. Optim. , 5 , 169 – 179 . Google Scholar CrossRef Search ADS Berrahmoune , L. ( 1999 ) Stabilization and decay estimate for distributed bilinear systems . Systems Control Lett. , 36 , 167 – 171 . Google Scholar CrossRef Search ADS Berrahmoune , L. ( 2010 ) Stabilization of unbounded bilinear control systems in Hilbert space . J. Math. Anal. Appl. , 372 , 645 – 655 . Google Scholar CrossRef Search ADS Bounit , H. & Hammouri , H. ( 1999 ) Feedback stabilization for a class of distributed semilinear control systems . Nonlinear Anal. , 37 , 953 – 969 . Google Scholar CrossRef Search ADS Bounit , H. & Idrissi , A. ( 2005 ) Regular bilinear systems . IMA J. Math. Control Inform., 22 , 26 – 57 . Google Scholar CrossRef Search ADS Brezis , H. ( 1973 ) Opérateurs Maximaux Monotones et Semi-groupes de Contractions dans les Espaces de Hilbert . Mathematics Studies, vol. 5. North-Holland . Brezis , H. ( 2010 ) Functional Analysis, Sobolev Spaces and Partial Differential Equations . New York : Springer . Google Scholar CrossRef Search ADS Dafermos , C. M. & Slemrod , M. ( 1973 ) Asymptotic behavior of nonlinear contraction semigroups . New York : J. Funct. Anal ., 13 , 97 – 106 . Google Scholar CrossRef Search ADS Desch , W. & Schappacher , W. ( 1984 ) On relatively bounded perturbations of linear $$c_0$$-semigroups . Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) , 11 , 327 – 341 . El Ayadi , R. , Ouzahra , M. & Boutoulout , A. ( 2012 ) Strong stabilisation and decay estimate for unbounded bilinear systems . Int. J. Control , 85 , 1497 – 1505 . Google Scholar CrossRef Search ADS Engel , K.-J. & Nagel , R. ( 2000 ) One-Parameter Semigroups for Linear Evolution Equations . Graduate Texts in Mathematics , vol. 194. New York : Springer . Haraux , A . ( 2001 ) Some sharp estimates for parabolic equations . J. Funct. Anal. , 187 , 110 – 128 . Google Scholar CrossRef Search ADS Hille , E. & Phillips , R. S. ( 1957 ) Functional Analysis and Semi-groups . Unite States of America : American Mathematical Society Colloquium Publications , vol. 31. Providence, R.I. Revised ed . MR 0089373 . Hislop , P. D. & Sigal , I. M. ( 1996 ) Perturbation theory: relatively bounded perturbations . Introduction to Spectral Theory . New York, NY : Springer , pp. 149 – 159 . ISO 690 . Idrissi , A . ( 2003 ) On the unboundedness of control operators for bilinear systems . Quaestiones Math. , 26 , 105 – 123 . Google Scholar CrossRef Search ADS Idrissi , A. & Bounit , H. ( 2008 ) Time-varying bilinear systems . SIAM J. Control Optim. , 47 , 1097 – 1126 . Google Scholar CrossRef Search ADS Komura , Y. ( 1967 ) Nonlinear semigroups in Hilbert space . J. Math. Soc. Japan , 19 , 493 – 507 . Google Scholar CrossRef Search ADS Ouzahra , M. ( 2008 ) Strong stabilization with decay estimate of semilinear systems . Systems Control Lett. , 57 , 813 – 815 . Google Scholar CrossRef Search ADS Ouzahra , M. ( 2010 ) Exponential and weak stabilization of constrained bilinear systems . SIAM J. Control Optim. , 48 , 3962 – 3974 . Google Scholar CrossRef Search ADS Ouzahra , M. ( 2011 ) Exponential stabilization of distributed semilinear systems by optimal control . J. Math. Anal. Appl. , 380 , 117 – 123 . Google Scholar CrossRef Search ADS Pazy , A. ( 1978 ) On the asymptotic behavior of semigroups of nonlinear contrations in Hilbert space . J. Funct. Anal. , 27 , 292 – 307 . Google Scholar CrossRef Search ADS Toledano , J. M. A. , Benavides , T. D. & Acedo , G. L. ( 1997 ) Measures of Noncompactness in Metric Fixed Point Theory , vol. 99. Birkhäuser Basel : Springer Science & Business Media . Google Scholar CrossRef Search ADS Weiss , G. ( 1989 ) Admissibility of unbounded control operators . SIAM J. Control Optim. , 27 , 527 – 545 . Google Scholar CrossRef Search ADS Weiss , G . ( 1994 ) Regular linear systems with feedback . Math. Control Signals Systems , 7 , 23 – 57 . Google Scholar CrossRef Search ADS Yarotsky , D. A. ( 2006 ) Ground states in relatively bounded quantum perturbations of classical lattice systems . Comm. Math. Phys. , 261 , 799 – 819 . Google Scholar CrossRef Search ADS © The Author(s) 2018. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved. 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Feedback stabilization for unbounded bilinear systems using bounded control

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Abstract

Abstract In this paper, we deal with the distributed bilinear system $$ \frac{d z(t)}{d t}= A z(t) + v(t)Bz(t), $$ where A is the infinitesimal generator of a semigroup of contractions on a real Hilbert space H. The linear operator B is supposed bounded with respect to the graph norm of A. Then we give sufficient conditions for weak and strong stabilizations. Illustrating examples are provided. 1. Introduction In this paper, we deal with the infinite dimensional bilinear system \begin{equation} \displaystyle\frac{d z(t)}{d t}= A z(t) + v(t)Bz(t),\quad\;z(0)=z_0\in H, \end{equation} (1.1) where A is an unbounded operator of H with domain D(A) and generates a semigroup of contractions $$(S(t))_{t\geqslant 0}$$ on a real Hilbert space H, whose norm and scalar products are denoted by ∥⋅∥ and ⟨⋅, ⋅⟩, respectively; the linear operator B, with domain D(A) ⊂ D(B), is A-bounded in the sense that there exists $$\alpha ,\beta>0$$ such that $$\|Bz\|\leqslant \alpha \|Az\|+\beta \|z\|,\; \forall z\in D(A), $$ or equivalently (see Desch & Schappacher 1984) \begin{equation} \|Bz\|\leqslant M\left(\|Az\|+\|z\|\right),\quad\; \forall z\in D(A)\; (\mbox{for some }\; M>0). \end{equation} (1.2) The real valued function v(⋅) denotes the control and z(t) is the corresponding mild solution of (1.1). In the case where B ∈ L(H) (L(H) is the set of bounded linear operators on H), the problem of stabilization by non-linear feedback controls has been studied by many authors (see e.g. Ball & Slemrod, 1979; Berrahmoune, 1999; Bounit & Hammouri 1999; Ouzahra, 2008). In Ball & Slemrod, 1979, a result of weak stabilization was obtained under the following condition: \begin{equation} \langle BS(t)y,S(t)y \rangle=0,\quad\;\; \forall t\geqslant 0 \Longrightarrow y=0, \end{equation} (1.3) by using the quadratic feedback \begin{equation} v(t) = -\langle z(t),Bz(t)\rangle\cdot \end{equation} (1.4) On the other hand, if (1.3) is replaced by the following inequality: \begin{equation} \int_0^T|\langle BS(s)y, S(s)y \rangle|\, \mathrm{d}s \geqslant \boldsymbol{\mu} \|y\|^2,\quad\;\; \forall y\in H\;(\mbox{for some}\; \boldsymbol{\mu},T>0), \end{equation} (1.5) then (see Berrahmoune, 1999; Ouzahra, 2008) the feedback (1.4) is a strongly stabilizing control and guarantees the following decay estimate: \begin{equation} \|z(t)\|=O\left(t^{\frac{-1}{2}}\right),\quad\; \mbox{as}\; t\to +\infty\cdot \end{equation} (1.6) In the study by Bounit & Hammouri (1999), the authors considered the control \begin{equation} v(t)=-\frac{\langle Bz(t),z(t) \rangle}{1+|\langle Bz(t),z(t) \rangle|} \end{equation} (1.7) and showed that if the resolvent of A is compact and B is bounded, self-adjoint and monotone, then the feedback (1.7) strongly stabilizes (1.1) provided that (1.3) holds. In the study by Ouzahra (2010), it has been shown that the control \begin{equation} v(t)=-\displaystyle\frac{\left \langle z(t),Bz(t)\right \rangle}{\|z(t)\|^2}\textbf{1}_{\{t\geqslant0; z(t)\ne0\}}, \end{equation} (1.8) weakly stabilizes (1.1) provided that B is a compact operator such that (1.3) holds. On the other hand, it has been shown that if B is a bounded operator satisfying (1.5), then (see the studies by Ouzahra, 2010, 2011) the control (1.8) exponentially stabilizes (1.1). The question of well-posedness of linear and bilinear systems with unbounded control operator has been treated in the studies by Weiss (1989, 1994), Idrissi (2003), Bounit & Idrissi (2005), Idrissi & Bounit (2008), Berrahmoune (2010) and El Ayadi et al. (2012). In the study by Berrahmoune (2010), the author considered the case where A is self-adjoint, and B is positive self-adjoint and bounded from the subspace $$V=D((I-A)^{\frac{1}{2}})$$ of H to its dual space $$V^{^{\prime}}.$$ Then he established the weak and strong stabilities of the closed-loop system \begin{equation} \displaystyle\frac{d z(t)}{d t}=Az(t)+f(\langle Bz(t),z(t)\rangle)Bz(t),\quad\;z(0)=z_0, \end{equation} (1.9) for an appropriate function $$f : \mathbb{R}\longrightarrow \mathbb{R}$$. Moreover, in the study by El Ayadi et al. (2012), it has been supposed that B is an unbounded linear operator from H to a Banach extension X of H with a continuous embedding H ↪ X. Then, it has been shown that (1.1) is strongly stabilizable, and a polynomial decay estimate of the stabilized state has been provided. It is noted that the above-mentioned results of El Ayadi et al. (2012) use spectral decomposition of the system at hand which reduces the class of considered systems to parabolic ones. Here, we deal with control operators which are relatively bounded. Such class of operators is very interesting either in theoretical or practical point of view. Indeed, various properties of semigroup under bounded perturbations are preserved when dealing with relatively bounded perturbations of the generator (see e.g. Pazy, 1978; Engel & Nagel, 2000). On the other hand, relatively bounded control operators arise as models for many dynamical processes (see e.g. Hislop & Sigal, 1996; Yarotsky, 2006). The article is organized as follows: the next section provides background material on non-linear semigroups and unbounded operators. In the third section we present an existence and uniqueness result. In the fourth section we present our main results and we study the weak and strong stabilizations. In the last section, we give illustrating examples. 2. Review on non-linear semigroups and unbounded operators In this section, we recall some existing results related to non-linear semigroups and unbounded operators. Let us begin with the following definitions and results concerning certain classes of unbounded operators (see Hille & Phillips, 1957). Definition 2.1 (Hille & Phillips, 1957, p. 391) Let A be the infinitesimal generator of a $$C_0$$-semigroup. A linear operator C is said to belong to the class F(A) if D(A) = D(C) and $$CR(\lambda _0,A)\in L(H)$$ for some $$\lambda _0\in \boldsymbol{\rho} (A),$$ where $$\boldsymbol{\rho}(A)$$ is the resolvent set of A and $$R(\lambda _0,A)$$ is its resolvent operator. Remark 2.1 (Hille & Phillips, 1957, p. 391) If C ∈ F(A), then $$CR(\lambda ,A)=CR(\lambda _0,A)+(\lambda _0-\lambda )CR(\lambda _0,A)R(\lambda ,A)$$. Then, the operator $$CR(\lambda ,A)$$ is bounded for all $$\lambda \in \boldsymbol{\rho} (A)$$. If C is A-bounded (i.e. C verifies (1.2) for some M > 0), then for all y ∈ D(A) and $$\lambda \in \boldsymbol{\rho} (A) $$ we have $$\|Cy\|\leqslant M\|(-A+\lambda I)y\|+M|\lambda | \|y\|+M\|y\|.$$ Thus, for all x ∈ H, we have $$\|CR(\lambda ,A)x\|\leqslant M\|x\|+(M|\lambda |+M)\|R(\lambda ,A)x\| \leqslant (2M+|\lambda | M)\|x\|.$$ Then the operator $$C|_{D(A)}, $$ restriction of C to D(A), is an element of F(A). Definition 2.2 (Hille & Phillips, 1957, p. 392) Let A be the infinitesimal generator of a $$C_0$$-semigroup. A linear operator C is said to belong to the class $$\widetilde{F}(A)$$ if D(A) ⊂ D(C), $$CR(\lambda _0,A)\in L(H),$$ for some $$\lambda _0\in \boldsymbol{\rho}(A),$$ for all x ∈ H, we have x ∈ D(C) if and only if the limit $$\lim _{\lambda \to \infty }\lambda CR(\lambda ,A)x=y$$ exists, in which case Cx = y. In the sequel, for linear operators $$\Lambda $$ and C with domains $$D(\Lambda )$$ and D(C) (respectively) such that $$D(C)\supset D(\Lambda )$$, we set $$\|C\|_{\Lambda }=\sup \left \{\|Cx\| \;/\; x\in D(\Lambda ), \|x\|\leqslant 1\right \}.$$ Theorem 2.1 (Hille & Phillips, 1957, p. 392) Let C ∈ F(A). Then C has a unique extension $$\widetilde{C}\in \widetilde{F}(A)$$, called A-extension of C and is defined by $$\widetilde{C}x=\lim _{\lambda \rightarrow +\infty }\lambda CR(\lambda ,A)x,\;\forall x\in D(\widetilde{C}):=\{x\in H \; /\;\lim _{\lambda \rightarrow +\infty }\lambda CR(\lambda ,A)x\; \mbox{exists} \}$$. If C is the restriction of a closed operator $$C_1$$, then $$C\subset \widetilde{C} \subset C_1$$. Remark 2.2 If C is bounded, then $$\widetilde{C}$$ is bounded and $$\|\widetilde{C}\|=\|C\|_A$$. If C is A-bounded then $$C|_{D(A)}$$ has a unique extension $$\widetilde{C}\in \widetilde{F}(A)$$. Proposition 2.1 (Hille & Phillips, 1957, p. 394) Let $$C\in \widetilde{F}(A)$$ and suppose that $$CS(t_0)$$ is bounded on D(A) for some $$t_0>0 $$. Then for all $$t\geqslant t_0$$, we have S(t)H ⊂ D(C) and CS(t) is bounded with $$\|CS(t)\|=\|CS(t)\|_A$$. Moreover, the function t ↦ CS(t) is strongly continuous for $$t>t_0, $$ and we have \begin{equation} \limsup_{t\to +\infty} t^{-1}\ln(\|CS(t)\|)\leqslant \omega_0, \end{equation} (2.1) where $$\boldsymbol{\omega}_0$$ is the growth bound of the semigroup S(t). Remark 2.3 If C is an A-bounded operator such that $$\|CS(t_0)\|$$ is bounded on D(A) for some $$t_0>0$$, then $$C|_{D(A)}$$ has a unique extension $$\widetilde{C}$$, which satisfies $$S(t)H\subset D(\widetilde{C})$$ for all $$t\geqslant t_0$$. This property will be useful to establish our weak stabilization result. Let us now recall the notion of non-linear semigroups (Pazy, 1978). Definition 2.3 (Pazy, 1978) Let H be a Hilbert space. A (generally non-linear) strongly continuous semigroup $$(T(t))_{t\geqslant 0}$$ on H is a family of continuous maps T(t) : H→H satisfying T(0) = identity, $$T(t + s) = T(t) T(s)$$, for all $$t, s\in \mathbb{R}^+,$$ for every y ∈ H, T(t)y → y, as $$t\to 0^+.$$ In this case, the mapping defined by $${\mathcal A}y = \lim _{h\to 0^+}\frac{T(h)y - y}{h}$$ for all $$y\in D({\mathcal A}):=\{y\in H/\; \lim _{h\to 0^+}\frac{T(h)y - y}{h} $$ exists in H} is called the infinitesimal generator of the semigroup T(t). If in addition $$\|T(t)y_1-T(t)y_2\|\leqslant \|y_1-y_2\|$$, for every $$t\geqslant 0$$ and $$y_1,y_2\in H$$, then T(t) is said to be a contraction semigroup (or a semigroup of contractions) on H. In this case, $${\mathcal A}$$ is dissipative, i.e. $$\langle{\mathcal A}y_1-{\mathcal A}y_2,y_1-y_2\rangle \leqslant 0,\;$$ for all $$y_1,y_2\in D({\mathcal A}).$$ For $$\phi \in H$$, define the positive orbit through $$\phi $$ by $$ O^+(\phi )=\cup _{t>0}T(t)\phi $$. The $$\omega $$-limit set of $$\phi $$ is the (possibly empty) set given by $$\omega (\phi )= \{\psi \in H;\; $$ there exists a sequence $$ t_n\to +\infty ,\; $$ such that $$ T(t_n)\phi \to \psi , \; $$ as $$ n\to +\infty \}.$$ The weak $$\omega $$-limit set of $$\phi $$ is the (possibly empty) set given by $$\omega _w(\phi )= \{\psi \in H;\; $$ there exists a sequence $$ t_n\to +\infty ,\; $$ such that $$ T(t_n)\phi \rightharpoonup \psi , \; \mbox{as}\; n\to +\infty \}.$$ The sets $$\omega (\phi )$$ and $$\omega _w(\phi )$$ are invariant under the action of any contraction semigroup $$(T(t))_{t\geqslant 0}$$ (see Pazy, 1978). Moreover, given a semigroup of contraction $$S(t)=e^{{\mathcal A}t},$$ we denote by $$E_{\mathcal A}$$ the set of equilibrium states given by $$E_{\mathcal A}={\mathcal A}^{-1}(0)=\{y\in D({\mathcal A}),\;{\mathcal A}y=0\}.$$ We have $$E_{\mathcal A}=\{y\in H;\; e^{t{\mathcal A}}y=y,\, \forall t\geqslant 0\}.$$ Indeed, for y ∈ H such that $$e^{t{\mathcal A}}y=y, \; \forall t\geqslant 0$$, we have $$\lim\nolimits _{t\to 0^+}\dfrac{e^{t{\mathcal A}}y-y}{t}=0$$, then $$y\in D({\mathcal A})$$ and $${\mathcal A}y=0$$. Now, let $$y\in D({\mathcal A})$$ be such that $${\mathcal A}y=0$$. We know that $$\frac{d^+}{dt} e^{t{\mathcal A}}y={\mathcal A}e^{t{\mathcal A}}y, \, \forall t\geqslant 0$$ (see the study by Komura, 1967). It follows that \begin{align*} \left\|y-e^{t{\mathcal A}}y\right\|^2& = \int_{0}^{t}\frac{\mathrm{d}^+}{\mathrm{d}s}\left\|y-e^{s{\mathcal A}}y\right\|^2 \mathrm{d}s\\ & = \int_{0}^{t}2\left\langle{\mathcal A}y-{\mathcal A}e^{s{\mathcal A}}y\;,\;y-e^{s{\mathcal A}}y \right\rangle \mathrm{d}s, \end{align*} from which, we deduce that $$\|y-e^{t{\mathcal A}}y\|^2= 0,\; \forall t\geqslant 0$$. Thus, $$e^{t{\mathcal A}}y=y, \; \forall t\geqslant 0$$. The following result concerns the asymptotic behaviour of the system (1.1) in connection with the structure of the weak $$\omega $$-limit set. Theorem 2.2 (see Pazy, 1978) Let $${\mathcal A}$$ be an infinitesimal generator of a non-linear semigroup of contractions $$(T(t))_{t\geqslant 0}$$ on H and let $$\phi \in H$$. The following conditions are necessary and sufficient for the existence of the weak limit of $$T(t)\phi $$, as $$t\to +\infty $$: $$E_{{\mathcal A}}={\mathcal A}^{-1}(0)\neq \emptyset $$, $$\omega _{w}(\phi )\subset E_{{\mathcal A}}.$$ The next result discusses the case of strong $$\omega $$-limit set. Theorem 2.3 (see Pazy, 1978) Let $${\mathcal A}$$ be an infinitesimal generator of a non-linear semigroup of contractions $$(T(t))_{t\geqslant 0}$$ on H and let $$\phi \in H.$$ The following conditions are necessary and sufficient for the existence of the strong limit of $$T(t)\phi $$, as $$t\to +\infty $$: $$E_{{\mathcal A}} \neq \emptyset $$ and $$\omega (\phi )\neq \emptyset $$, $$\omega (\phi )\subset E_{{\mathcal A}}.$$ 3. Considered systems and well-posedness In this section, we reconsider the system (1.1) with the same hypotheses on A and B. The purpose of this section is to study the feedback stabilization of the system (1.1) using the bounded control \begin{equation} v(t)=-\boldsymbol{\rho}\;\frac{\langle By(t),y(t)\rangle}{1+\langle By(t),y(t)\rangle}, \end{equation} (3.1) where $$\boldsymbol{\rho}>0$$ is the gain control and y is the solution of the corresponding closed-loop system, i.e. \begin{equation} \displaystyle\frac{d z(t)}{d t}={\mathcal A} z(t), \end{equation} (3.2) where $${{\mathcal A}} y = A y+V(y)By,\, \forall y\in D({\mathcal A})=D(A)$$ and $$V(y)=-\boldsymbol{\rho} \;\frac{\langle By,y\rangle }{1+\langle By,y\rangle }, \, \forall y\in D(A)$$. Remark 3.1 For $$B=B^{\ast }\geqslant 0$$ on D(A), we have the following Cauchy–Schwartz-like inequality: $$ |\langle By_1,y_2 \rangle|\leqslant \sqrt{\boldsymbol{\varphi}(y_1)}\sqrt{\boldsymbol{\varphi}(y_2)},\quad \, \forall (y_1,y_2)\in D(A)^2, $$ with $$\boldsymbol{\varphi}(y_1)=\langle By_1,y_1 \rangle $$ and $$\boldsymbol{\varphi}(y_2)=\langle By_2,y_2 \rangle $$. Indeed, since $$B=B^{\ast }\geqslant 0$$ on D(A), we have \begin{equation} \boldsymbol{\mu}^2\varphi(y_2)+2\boldsymbol{\mu}<By_1,y_2>+\varphi(y_1)=\langle B(y_1+\boldsymbol{\mu} y_2),y_1+\boldsymbol{\mu} y_2\rangle\geqslant 0,\quad \, \forall \boldsymbol{\mu}\in\mathbb{R}, \end{equation} (3.3) then $$|\langle By_1,y_2 \rangle |\leqslant \sqrt{\boldsymbol{\varphi} (y_1)}\sqrt{\boldsymbol{\varphi} (y_2)}.$$ For all $$y_1\in D(A),$$ we have $$\langle By_1,y_1\rangle =0 \Rightarrow By_1=0.$$ Indeed, by simplifying with $$\mu>0 $$ and $$\mu <0$$ in (3.3), and letting $$\boldsymbol{\mu} \to 0^{\pm }$$, we obtain $$\langle By_1, y_2\rangle = 0, \, \forall y_2\in D(A)$$. Thus, we conclude by using the density of D(A) in H. In the sequel, we will analyse the well-posedness of the system (3.2). Theorem 3.1 Let A generate a semigroup S(t) of contractions on H, and let B : D(B)→H be a linear A-bounded operator such that (i) ⟨By, z⟩ = ⟨y, Bz⟩, ∀ y, z ∈ D(A), (ii) ⟨By, y⟩ ⩾ 0, ∀ y ∈ D(A). Then for any $$0<\boldsymbol{\rho} <\min (1,\frac{1}{M} )$$, (where M is the constant given in (1.2)) and for all $$z_{0}\in H$$, the system (3.2) admits a unique solution $$z\in{\mathcal C}([0,+\infty [;H)$$. Furthermore, $${{\mathcal A}}$$ generates a contraction semigroup $$e^{t{{\mathcal A}}}$$ on H, and for all $$z_{0}\in H$$ the solution of (3.2) is given by $$z(t) = e^{t {{\mathcal A}}}z_{0}\cdot $$ Proof. Let us set $$\boldsymbol{\varphi} (y)= \langle By,y \rangle ,\, \forall y\in D(A)$$ and let us consider the map $$ \boldsymbol{\phi} =g(\boldsymbol{\varphi}),\;\mbox{with}\;g(z)=\dfrac{1}{2}(z-\ln(1+z)).$$ Since B is self-adjoint, we have $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)=t^2\boldsymbol{\varphi}(y_1)+2t(1-t) \langle By_1,y_2 \rangle+(1-t)^2\boldsymbol{\varphi}(y_2), \forall t\in [0,1], \; \forall (y_1,y_2)\in D({A})^2.$$ Taking into account Remark 3.1, 1., we deduce that $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)=t^2\boldsymbol{\varphi}(y_1)+2t(1-t) \sqrt{\boldsymbol{\varphi}(y_1)} \sqrt{\boldsymbol{\varphi}(y_2)}+(1-t)^2\boldsymbol{\varphi}(y_2), \forall t\in [0,1], \; \forall (y_1,y_2)\in D({A})^2.$$ Hence, $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)\leqslant \big(t\sqrt{\boldsymbol{\varphi}(y_1)}+(1-t)\sqrt{\boldsymbol{\varphi}(y_2)}\big)^2.$$ This together with the convexity of $$t\mapsto t^2$$ gives $$ \boldsymbol{\varphi}(ty_1+(1-t)y_2)\leqslant t\boldsymbol{\varphi}(y_1)+(1-t)\boldsymbol{\varphi}(y_2).$$ Thus, the map $$\boldsymbol{\varphi} $$ (and so is $$\boldsymbol{\phi} $$) is convex. It follows that the Gâteaux derivative $$\boldsymbol{\phi} ^{\prime} : D({A} )\longrightarrow H$$ of $$\boldsymbol{\phi} $$ is monotone. On the other hand, for all h ∈ H and y ∈ D(A) we have \begin{align*} \boldsymbol{\phi}^{\prime}(y)\cdot h &=\lim\limits_{t\to 0}\dfrac{\boldsymbol{\phi}(y+th)-\boldsymbol{\phi}(y)}{t}\\ & =\dfrac{1}{2}\;\dfrac{\langle By,y \rangle}{1+\langle By,y\rangle }\langle (B+B^{\ast})y,h\rangle.\end{align*} Then using the fact that B is symetric, we obtain $$ \boldsymbol{\phi}^{\prime}(y)=\frac{\langle By,y\rangle}{1+\langle By,y\rangle}By.$$ On the other hand, we have the following expression regarding the Gâteaux derivative of $$\boldsymbol{\phi} $$: $$ \langle \boldsymbol{\phi}^{\prime}(tx+(1-t)y),x-y\rangle= \dfrac{\big(t\langle Bx,y\rangle+(1-t)\langle By,y\rangle \big)\big(t\langle Bx,x-y\rangle+(1-t)\langle By,x-y\rangle \big)}{1+t^2\langle Bx,x\rangle+2t(1-t)\langle Bx,y\rangle+(1-t)^2\langle By,y\rangle}, \, \forall t\in[0,1],$$ from which, we deduce that $$\boldsymbol{\phi} ^{\prime}$$ is hemicontinous, i.e. for any x, y ∈ D(A), the mapping $$ : t \mapsto \langle \boldsymbol{\phi} ^{\prime}(tx+(1-t)y),x-y) \rangle $$ is continuous on [0, 1]. Now, it comes from (1.2) that for all y ∈ D(A), we have $$ \|V(y)By\|\leqslant \boldsymbol{\rho}\;M(\|Ay\|+\|y\|).$$ Since A generates a semigroup of contractions then − A is maximal monotone and $$\rho \phi ^{\prime}(\cdot )=-V(\cdot ) B(\cdot )$$ is monotone hemicontinuous and $$\boldsymbol{\rho} M<1$$, we have that (see the book by Brezis, 1973) the operator $$-{\mathcal A} : y\longrightarrow -Ay-V(y) By $$ is maximal monotone, and hence $${\mathcal A}$$ generates a semigroup of contractions $$e^{t{{\mathcal A}}}z_0$$, and the function $$ z(t)=e^{t{{\mathcal A}}}z_0$$ is a solution of (3.2) (see Brezis, 1973). Remark 3.2 For all $$z_{0}\in D(A )$$, we have \begin{equation} \|z(t)\|\leqslant \|z_0\|,\quad\; \forall t\geqslant 0 \end{equation} (3.4) and z(t) ∈ D(A) admits a right derivative at t (see the study by Komura, 1967) as well as \begin{equation} \displaystyle\frac{d^+z(t)}{d t}={{\mathcal A}} z(t)\cdot \end{equation} (3.5) \begin{equation} \|{{\mathcal A}} z(t)\|\leqslant \|{{\mathcal A}} z_0\|\cdot \end{equation} (3.6) Remark 3.3 Under the assumptions of Theorem 3.1, we have For all y ∈ D(A) \begin{align*} \|By\|\ & \leqslant M\|{\mathcal A}y-V(y) By\|+M \|y\|\\ &\leqslant M\|{\mathcal A}y\|+M\boldsymbol{\rho}\|By\|+M \|y\| \end{align*} and hence \begin{equation} \|By\|\leqslant \frac{M}{(1-M \boldsymbol{\rho})}\|{{\mathcal A}}y\|+\frac{M}{(1-M \boldsymbol{\rho})}\|y\|, \end{equation} (3.7) For all y ∈ D(A) \begin{align*} \|Ay\|\ & \leqslant \|{\mathcal A}y\|+\boldsymbol{\rho} \|By\|\\ &\leqslant \|{\mathcal A}y\|+M\boldsymbol{\rho}\|Ay\|+ M \boldsymbol{\rho}\|y\| \end{align*} then \begin{equation} \|Ay\|\leqslant \frac{1}{1-\boldsymbol{\rho} M}\|{\mathcal A}y\|+ \frac{\boldsymbol{\rho} M}{1-\boldsymbol{\rho} M}\|y\|, \end{equation} (3.8) The stabilizing control (3.1) is uniformly bounded with respect to initial state $$ |v(t)| \leqslant \boldsymbol{\rho} $$ and if the system (3.2) is subject to the control constraint $$|v(t)| \leqslant \boldsymbol{\alpha} , \; \boldsymbol{\alpha}>0,$$ then one may take $$\boldsymbol{\rho} \leqslant \alpha $$. 4. Stabilization problem In this section, we give sufficient conditions to obtain weak and strong stabilizations of (1.1) using the control (3.1). 4.1 Weak stability In the sequel, we assume that B is A-bounded so $$B|_{D(A)}$$ admits an A-extension denoted by $$\tilde{B}$$. Let us now introduce the following sets: $${\mathcal M} = \{\boldsymbol{\varphi} \in D( A) \;/\; \langle Be^{tA}\boldsymbol{\varphi} ,e^{tA}\boldsymbol{\varphi} \rangle =0, \; \forall t\geqslant 0 \} $$ and $$\widetilde{{\!\mathcal M}} = \{\boldsymbol{\varphi} \in H \;/\; \langle Be^{tA}\boldsymbol{\varphi} ,e^{tA}\boldsymbol{\varphi} \rangle =0, \; \forall t\geqslant 0 \} $$ and let us consider the following hypotheses: $$(\boldsymbol{C}_1)$$: For all sequence $$(x_n)\subset D(A),$$ we have $$\left .\begin{array}{l} 1.\; x_n\rightharpoonup x\in H, \\ 2.\; \|Ax_n\|\; \mbox{is bounded,} \\ 3.\; \langle Bx_n,x_n\rangle \to 0. \end{array} \right \}\;\Longrightarrow\ x\in E_{{{\mathcal A}}}.$$ $$(\boldsymbol{C}_2)$$: If there exists $$t_0>0$$ such that $$\|BS(t_0)\| $$ is bounded in D(A) and for all sequence $$(y_n)\subset D( A)$$ and y ∈ H such that $$y_n\rightharpoonup y$$ in H, there exists a subsequence $$(y_{\gamma (n)})$$ such that for all $$t\geqslant t_0$$, we have $$BS(t)y_{\gamma (n)}\rightarrow \widetilde{B}S(t)y$$ in H, as $$n\to +\infty $$. Remark 4.1 As a class of operators that satisfy the assumption $$(\boldsymbol{C}_1)$$, the set of operators B satisfying $$ \langle By,y\rangle \geqslant \mu \|y\|^s$$, for all y ∈ D(B) (for some $$\mu ,s>0$$). As an other example of operators that satisfies $$(\boldsymbol{C}_1)$$, let $$A=\Delta $$ with Newmann boundary conditions and $$By=-\Delta y+ Ny, \, \forall y\in D(A)$$ with $$Ny=\sum _{j=1}^{+\infty }\frac{1}{j^2}\langle y,\boldsymbol{\varphi} _j\rangle \boldsymbol{\varphi} _j$$ where $$(\boldsymbol{\varphi} _j)_{j\geqslant 0}$$ are the orthonormal basis of $$L^2(\boldsymbol{\Omega})$$ formed with the eigenfunctions of A. Here, B is A-bounded and for any sequence $$(x_n)\subset D(A)$$ such that $$x_n\rightharpoonup x\in H$$ and $$\langle Bx_n,x_n\rangle \to 0,$$ we have $$ \langle -\Delta x_n,x_n\rangle +\langle Nx_n,x_n\rangle \to 0$$. Then, it follows from the positivity of the operators $$-\Delta $$ and N that$$ \nabla x_n \to 0 \;\mbox{and}\; \langle Nx_n,x_n\rangle \to 0.$$ Now using that N is a compact operator, we derive that ⟨Nx, x⟩ = 0. Finally, it follows from the definition of the operator N that $$x=\lambda \textbf{1}\in \ker A\cap \ker B.$$ As an example where the assumption $$(\boldsymbol{C}_2)$$ holds, one can take $$B=A=\Delta $$ (see the study by Haraux, 2001). We note that $$E_{\mathcal A}=\ker (A)\cap \ker (B)$$. Indeed, we have $$y\in E_{\mathcal A} \Longleftrightarrow Ay+V(y)By=0,$$ so the inclusion ($$\supset $$) is clear. Moreover, $$y\in E_{\mathcal A} \Rightarrow \langle Ay,y \rangle +V(y)\langle By,y\rangle =0$$. Thus, since A and − B are dissipative, we have $$ y\in E_{\mathcal A} \Rightarrow \langle Ay,y\rangle = \langle By,y\rangle =0. $$ Then, taking into account Remark 3.1, 2., it comes By = 0 and then Ay = 0. The following theorem concerns the weak stability of (3.2). Theorem 4.1 Suppose that the hypotheses of Theorem 3.1 are verified. 1. If $$(\boldsymbol{C}_1)$$ holds then for all $$z_0 \in D(A)$$ the solution of (3.2) is weakly convergent to $$z^{\ast }\in E_{\mathcal A}$$, as $$t\to +\infty $$. 2. If $$(\boldsymbol{C}_2)$$ holds, then for all $$z_0 \in D(A)$$ the solution of (3.2) satisfies (a) $$z(t)\rightharpoonup z^{\ast }\in E_{A}$$, if $$\widetilde{{\mathcal M}} \subset E_A,$$ (b) $$z(t)\rightharpoonup 0$$, if $$\widetilde{{\mathcal M}} =\{0\}$$. Proof. 1. Let $$z_0 \in D({{A}} )$$. According to Remark 3.2, the function t → z(t) admits a right derivative at all time, then we have $$ \dfrac{d^+\|z(\tau)\|^2}{d\tau}=2\langle{{\mathcal A}} z(\tau),z(\tau)\rangle,\;\; \forall \tau \geqslant 0.$$ By integrating this equality between s and t, we obtain \begin{equation} \frac{1}{2}\left[\|z(t)\|^2-\|z(s)\|^2\right]=\int_s^t \langle{{\mathcal A}} z(\tau),z(\tau)\rangle\;\mathrm{d}\tau,\quad \; \forall t\geqslant s \geqslant 0. \end{equation} (4.1) Since A is dissipative, we have $$ \int_s^t \langle{{\mathcal A}} z(\tau),z(\tau)\rangle\;\mathrm{d}\tau \leqslant -\boldsymbol{\rho} \int_s^t \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle}\, \mathrm{d}\tau,\quad \; \forall t\geqslant s \geqslant 0. $$ It comes from (4.1) that $$ \int_0^t \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle} \, \mathrm{d}\tau \leqslant \frac{1}{2\boldsymbol{\rho}}\|z_0\|^2. $$ Thus, \begin{equation} \int_{0}^{+\infty}\frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle}\;\mathrm{d}\tau <+\infty. \end{equation} (4.2) Moreover, from Remark 3.2 and inequality (3.7) we get \begin{equation} \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\left[\frac{M}{(1-M \boldsymbol{\rho})}\|{{\mathcal A}}z_0\|+\frac{M}{(1-M \boldsymbol{\rho})}\|z_0\|\right]\|z_0\|} \leqslant \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle},\quad\;\forall \tau>0. \end{equation} (4.3) Hence, \begin{equation} \int_{0}^{+\infty}\langle Bz(\tau),z(\tau)\rangle^2\;\mathrm{d}\tau <+\infty. \end{equation} (4.4) From (3.4), it follows that $$\boldsymbol{\omega} _w(z_0)\neq \emptyset $$. Let $$\boldsymbol{\varphi} _0 \in \boldsymbol{\omega} _w(z_0)$$ and let $$t_n\to +\infty $$ such that $$z(t_n)=e^{t_n{{\mathcal A}}}z_0\rightharpoonup \boldsymbol{\varphi} _0$$, as $$n\to +\infty $$. We shall prove that there exists a sequence $$(s_{n})$$ such that $$s_n\to +\infty $$, $$z(s_n)\rightharpoonup \boldsymbol{\varphi} _0$$ and $$~{\lim _{n\to +\infty }\langle Bz(s_{n}),z(s_{n})\rangle =0.}$$ For this end, let us consider $$\boldsymbol{\varepsilon}>0$$ and $$H_{\boldsymbol{\varepsilon}} =\{t>0 \; /\; \langle Bz(t),z(t)\rangle \geqslant \boldsymbol{\varepsilon} \}$$ and let us define the map G : t →< Bz(t), z(t) >. From (4.4), we have $$G \in L^2(0,+\infty )$$, thus the Lebesgue measure of $$H_{\boldsymbol{\varepsilon}} $$ is finite. Then for any A > 0, the set $$E=\{k\in \mathbb{N}\; /\; \; t_k>A\}$$ is infinite, hence $$\bigcup _{k\in E}[t_k,t_k+\boldsymbol{\varepsilon} ]\not \subset H_{\boldsymbol{\varepsilon}} .$$ It follows that for $$\boldsymbol{\varepsilon}>0$$ and A > 0, there exist $$t_k>A$$ and $$ s\in [t_k,t_k+\boldsymbol{\varepsilon} ]$$ such that $$G(s)<\boldsymbol{\varepsilon} $$. Let us now proceed to the construction of sequence $$(s_j)_{j\geqslant 1}$$. For $$\boldsymbol{\varepsilon} =\dfrac{1}{1}$$ and A = 1, then there exist $$t_{k_1}>1$$ and $$s_1\in [t_{k_1},t_{k_1}+\dfrac{1}{1} ]$$ such that $$G(s_1) < \dfrac{1}{1}$$. Now, let $$j\in \mathbb{N}^{\ast }$$ and suppose that the terms $$t_{k_j}$$ and $$s_j$$ are constructed. For $$\boldsymbol{\varepsilon} =\dfrac{1}{j+1}$$ and $$A=t_{k_j}+j>0$$, there exist $$t_{k_{j+1}}>A$$ and $$s_{j+1}\in [t_{k_{j+1}},t_{k_{j+1}}+\frac{1}{j+1} ]$$ such that $$G(s_{j+1}) < \dfrac{1}{j+1}$$. Then from the construction of $$(s_j)_{j\geqslant 1}$$ and $$(t_{k_j})_{j\geqslant 1}, $$ we have $$G(s_j)\leqslant \dfrac{1}{j}$$. On the other hand, we have $$t_{k_{j+1}}-t_{k_j}>j$$, then $$t_{k_{q+1}}-t_{k_1}>\sum _{j=1}^{q}j=\frac{q(q+1)}{2}$$ for all $$q\geqslant 1$$, and so $$\lim _{j\to +\infty }t_{k_j}=+\infty $$. Thus, $$\lim_{j\to +\infty }s_j=+\infty $$ (recall that $$s_j\geqslant t_{k_j}$$). We can therefore choose a subsequence $$(t_{k_j})$$ of $$(t_{n})$$ and a sequence $$(s_j)$$ such that $$t_{k_j} \leqslant s_j \leqslant t_{k_j}+\frac{1}{j}$$, $$G(s_j) < \frac{1}{j}$$ and $$s_j\to +\infty $$. From (3.5) for all $$t\geqslant 0$$, we have $$\|{\mathcal A}e^{t{{\mathcal A}}}z_0\|\leqslant \|{{\mathcal A}}z_0\|$$. Then \begin{equation} \left\|e^{s_j{{\mathcal A}}}z_0-e^{t_{k_j}{{\mathcal A}}}z_0\right\|\leqslant \|{{\mathcal A}}z_0\|\left|s_j-t_{k_j}\right|\leqslant \frac{\|{{\mathcal A}}z_0\|}{j}. \end{equation} (4.5) We deduce that $$z(s_j)\rightharpoonup \boldsymbol{\varphi} _0$$ and $$\langle Bz(s_j),z(s_j)\rangle \to 0$$, as $$j\to +\infty .$$ Moreover, according to the Remark 3.3, 2., $$(\|Az(s_j)\|)_{j\geqslant 1}$$ is bounded. It follows from $$(\boldsymbol{C}_1)$$ that $$\varphi _0\in E_{{\mathcal A}}$$, which gives $$\boldsymbol{\omega} _w(z_0)\subset E_{{\mathcal A}}$$. Finally, Theorem 2.2 implies that z(t) is weakly convergent as $$t\to +\infty $$. 2. (a) Let $$\boldsymbol{\varphi} _0$$ and $$(s_j)_{j\geqslant 1}$$ be as in 1. and suppose that the condition $$(\boldsymbol{C}_2)$$ holds. From (4.2) we have \begin{equation} \lim_{j\to+\infty}\int_{s_{j}}^{t+s_{j}} \frac{\langle Bz(\tau),z(\tau)\rangle^2}{1+\langle Bz(\tau),z(\tau)\rangle}\;\mathrm{d}\tau=0,\quad\;\forall t>0. \end{equation} (4.6) It follows from the invariance of $$\boldsymbol{\omega} _w(z_0)$$ under the semigroup $$e^{t{\mathcal A}}$$ that for all t > 0, we have \begin{equation} \langle Bz(t+s_j),z(t+s_j)\rangle\to 0,\quad\; \mbox{as}\; j\to+\infty. \end{equation} (4.7) Moreover, from the variation of constants formula, we have $$ z(t+s_j)=S(t)z(s_j)+\int_{s_j}^{t+s_j} v(z(\tau))S(t+s_j-\tau)Bz(\tau)\, \mathrm{d}\tau,\quad\;\forall t>0.$$ It follows that $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \int_{s_j}^{t+s_j} |v(z(\tau))|\;\|Bz(\tau)\|\, \mathrm{d}\tau,\quad\;\forall t>0\cdot$$ Using (3.4)–(3.7), we deduce that \begin{equation} \|Bz(\tau)\|\leqslant \frac{M}{(1-M \boldsymbol{\rho})}\|{{\mathcal A}}z_0\|+\frac{M}{(1-M \boldsymbol{\rho})}\|z_0\|=:M_{z_0,\boldsymbol{ \rho}}\cdot \end{equation} (4.8) Then $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \; M_{z_0, \boldsymbol{\rho}} \int_{s_j}^{t+s_j} |v(z(\tau))|d\tau,\quad\;\forall t>0\cdot$$ The Schwartz’s inequality gives $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \boldsymbol{\rho}\; \sqrt{t}\; M_{z_0, \boldsymbol{\rho}}\; \sqrt{\int_{s_j}^{t+s_j} \frac{<Bz(\tau),z(\tau)>^2}{(1+<Bz(\tau),z(\tau)>)^2}\, \mathrm{d}\tau},\quad\;\forall t>0,$$ and hence $$ \|z(t+s_j)-S(t)z(s_j)\|\leqslant \boldsymbol{\rho}\; \sqrt{t}\; M_{z_0, \rho}\; \sqrt{\int_{s_j}^{t+s_j} \frac{<Bz(\tau),z(\tau)>^2}{1+<Bz(\tau),z(\tau)>}\ \textrm{d}\tau},\quad\;\forall t>0.$$ Thus, using (4.2) we deduce that \begin{equation} \lim_{j\to +\infty}[z(t+s_j)-S(t)z(s_j)]=0,\quad\;\forall t>0. \end{equation} (4.9) Using the fact that B is self-adjoint, we can write $$\langle BS(t)z(s_j), S(t)z(s_j)\rangle =\langle S(t)z(s_j)-z(t+s_j), BS(t)z(s_j)\rangle +\langle Bz(t+s_j), S(t)z(s_j)-z(t+s_j)\rangle + \langle Bz(t+s_j), z(t+s_j)\rangle$$. Furthermore, from (1.2), (3.4), (3.8) and the fact that S(t) is of contractions, we get \begin{align*} \|BS(t)z(s_j)\| & \leqslant M\left(\|AS(t)z(s_j)\|+\|S(t)z(s_j)\|\right)\\ &\leqslant \frac{M}{1-\boldsymbol{\rho} M}\| ({\mathcal A}z_0\|+ \|z_0\|)\cdot \end{align*} This, together with (4.7), (4.8) and (4.9) gives \begin{equation} \langle BS(t)z(s_j),S(t)z(s_j)\rangle\to 0,\quad\; \mbox{as}\; j\to +\infty. \end{equation} (4.10) Now, from the condition $$(\boldsymbol{C}_2)$$, there exists a subsequence of $$(s_j), $$ still denoted by $$(s_j), $$ such that for all $$t\geqslant t_0$$ we have \begin{equation} \langle BS(t)z(s_j),S(t)z(s_j)\rangle\to \langle \widetilde{B}S(t){\boldsymbol{\varphi}_0},S(t){\boldsymbol{\varphi}_0}\rangle,\quad\; \mbox{as}\; j\to +\infty. \end{equation} (4.11) We conclude that \begin{equation} \langle \widetilde{B}S(t){\boldsymbol{\varphi}_0},S(t){\boldsymbol{\varphi}_0}\rangle=0. \end{equation} (4.12) In other words, $$\boldsymbol{\varphi} _0\!\in\! \widetilde{{\mathcal M}}$$ and hence $$\boldsymbol{\omega} _w(z_0)\!\subset \!\widetilde{{\mathcal M}}$$. But $$\widetilde{{\mathcal M}}\subset E_A,$$ then $$S(t)\boldsymbol{\varphi} _0=\boldsymbol{\varphi} _0$$ and so $$\langle \widetilde{B}{\boldsymbol{\varphi} _0},{\boldsymbol{\varphi} _0}\rangle =0$$. Moreover, since $$\boldsymbol{\omega} _w(z_0)$$ is invariant by $$e^{t{{\mathcal A}}}$$, we have $$\boldsymbol{\varphi} (t)=e^{t{{\mathcal A}}}\boldsymbol{\varphi} _0\in \boldsymbol{\omega} _w(z_0)$$, thus $$\langle \widetilde{B}\boldsymbol{\varphi} (t),\varphi (t)\boldsymbol{\varphi} =0.$$ Consequently, $$\boldsymbol{\varphi} (t)=e^{t{{\mathcal A}}}\boldsymbol{\varphi} _0=S(t)\boldsymbol{\varphi} _0=\boldsymbol{\varphi} _0$$, and hence $$\boldsymbol{\varphi} _0\in E_{{\mathcal A}}$$. We deduce that $$ \boldsymbol{\omega} _w(z_0)\subset E_{{\mathcal A}}$$. Finally, from Theorem 2.2, there exists $$z^{\ast }\in E_A$$ such that $$z(t)\rightharpoonup z^{\ast },$$ as $$t\to +\infty$$. (b) In the case $$\widetilde{{\mathcal M}}=\{0\}$$, it follows from $$\boldsymbol{\omega} _w(z_0)\subset \widetilde{{\mathcal M}}$$ that $$z(t)\rightharpoonup 0,$$ as $$t\to +\infty .$$ 4.2 Strong stability In this part, we study the strong stability of the closed-loop system (3.2). Let us start with the two following lemmas. Lemma 4.1 Suppose that the assumptions of Theorem 3.1 hold and that the embedding D(A) ↪ H is compact. Then the operator $$(I-{{\mathcal A}})^{-1}$$ is compact from H to itself. Proof. Since $${\mathcal A}$$ generates a non-linear semigroup of contractions, the operator $$(I-{{\mathcal A}})^{-1}$$ is defined on H (see Theorem 2, in the study by Komura, 1967). In order to prove the compactness of the operator $$(I-{{\mathcal A}})^{-1}$$, let us consider a bounded sequence $$(v_n)$$ in H. We shall prove that the sequence $$u_n=(I-{{\mathcal A}})^{-1}v_n$$, defined in D(A), has a converging subsequence (see Definition 2.5, p. 13 in the study by Toledano et al., 1997). We have $$ \langle v_n,u_n\rangle=\|u_n\|^2-\langle{{\mathcal A}}u_n,u_n\rangle.$$ Then, using the fact that the operator $${{\mathcal A}}$$ is dissipative, we get $$\|u_n\|^2 \leqslant \langle v_n,u_n\rangle \leqslant \|v_n\| \|u_n\|$$. It follows that the sequence $$(u_n)$$ is bounded. Since $${\mathcal A}u_n=u_n-v_n,$$ then the sequence $$({\mathcal A}u_n)$$ is bounded. From (3.7) we can see that the sequence $$(Bu_n)$$ is bounded. It follows from the equality $$ Au_n={\mathcal A}u_n+\boldsymbol{\rho} \frac{\langle B{u_n},u_n\rangle}{1+\langle B{u_n},u_n\rangle}B{u_n}$$ that $$(Au_n)$$ is bounded. Furthermore, $$(u_n)\subset D(A)$$, then we can conclude by the fact that D(A) is compactly embedded in H. Remark 4.2 Under the assumptions of Lemma 4.1, we have $$\boldsymbol{\omega} (z_0)\neq \emptyset $$, for all $$z_0\in D(\mathcal A)$$ (see the study by Pazy, 1978). Now, we are ready to establish the strong stabilization result. This is the subject of the two next theorems. Theorem 4.2 Let assumptions of Theorem 3.1 be verified, let D(A) be compactly embedded in H and let us set $$U(A)=\{y\in H\;/\; \|e^{tA}y\|=\|e^{tA^{\ast }}y\|=\|y\|\}$$. Then for all $$z_0 \in D(A)$$, we have $$\boldsymbol{\omega} (z_0)\subset{\mathcal M} \cap U(A)$$. If $$U(A)\cap{\mathcal M}\subset E_A$$, then for all $$z_0 \in D(A)$$ we have $$ z(t)\to z^{\ast } \in E_A$$, as $$t\to +\infty $$. Proof. Let $$z_0\in D(A)$$, $$\boldsymbol{\varphi} _0\in \boldsymbol{\omega} (z_0)$$ and let $$(t_n)_{n\geqslant 0 }$$ be such that $$t_n\to +\infty $$ and $$e^{t_n{\mathcal A}}z_0\to \boldsymbol{\varphi} _0$$ as $$n\to +\infty .$$ For t > 0, there exists $$N\in \mathbb{N}$$ such that for all $$n\geqslant N$$, we have $$t_n\geqslant t.$$ Using the fact that $$e^{t{\mathcal A}}$$ is a semigroup of contractions, we obtain $$ \|e^{t_n{\mathcal A}}z_0\|\leqslant \|e^{t{\mathcal A}}z_0\|$$; and letting $$n\to +\infty ,$$ we get \begin{equation} \|\boldsymbol{\varphi}_0\|\leqslant \left\|e^{t{\mathcal A}}z_0\right\|=\|z(t)\|. \end{equation} (4.13) It follows from the inequality (4.13) that $$\|\boldsymbol{\varphi} _0\|\leqslant \|e^{(t+t_n){\mathcal A}}z_0\|$$. Then, letting $$n\to +\infty $$ we obtain $$\|\boldsymbol{\varphi}_0\|\leqslant \|e^{t{\mathcal A}}\boldsymbol{\varphi} _0\|$$. This together with the fact that $$(e^{t{\mathcal A}})_{t\geqslant 0}$$ is a contraction semigroup leads to $$\|e^{t{\mathcal A}}\boldsymbol{\varphi} _0\|=\|\boldsymbol{\varphi} _0\|$$ for all $$t\geqslant 0$$. Furthermore, we have $$\boldsymbol{\omega} (z_0)\subset D({\mathcal A})$$ (see Theorem 5, in the study by Dafermos & Slemrod, 1973) and so $$\boldsymbol{\varphi} _0 \in D(A)$$. From (4.5) we have $$e^{s_j{\mathcal A}}z_0\to \boldsymbol{\varphi} _0$$, as $$j\to +\infty $$. On the other hand, \begin{align*} \langle BS(t)z(s_j),S(t)z(s_j)\rangle & =\langle BS(t)z(s_j),S(t)(z(s_j)-\boldsymbol{\varphi}_0)\rangle+\langle BS(t)\boldsymbol{\varphi}_0,S(t)z(s_j)\rangle \\ &=\langle BS(t)z(s_j),S(t)(z(s_j)-\boldsymbol{\varphi}_0)\rangle+\langle S(t)^{\ast}BS(t)\boldsymbol{\varphi}_0,z(s_j)\rangle.\end{align*} Since $$(\|BS(t)z(s_j)\|)_{j\geqslant 1}$$ is bounded, the semigroup $$(S(t))_{t\geqslant 0}$$ is of contractions and $$z(s_j)\to \varphi _0$$, as $$j\to +\infty $$, then $$\langle BS(t)z(s_j),S(t)z(s_j)\rangle \to \langle BS(t)\boldsymbol{\varphi} _0,S(t)\boldsymbol{\varphi} _0\rangle $$, as $$j\to +\infty $$, this combined with (4.10), gives $$\boldsymbol{\varphi} _0\in{\mathcal M}$$ and so $$e^{t{\mathcal A}}\boldsymbol{\varphi} _0=e^{tA}\boldsymbol{\varphi} _0$$. We conclude that $$\boldsymbol{\varphi} _0\in U(A)\cap{\mathcal M}.$$ Let $$\boldsymbol{\varphi} _0\in \boldsymbol{\omega} (z_0)$$. From 1. we have $$\boldsymbol{\varphi} _0\in U(A)\cap{\mathcal M} \subset E_A$$. Since $$\boldsymbol{\varphi} (t)=e^{t{\mathcal A}}\boldsymbol{\varphi} _0\in \boldsymbol{\omega} (z_0)$$, then $$\boldsymbol{\varphi} (t)\in U(A)\cap{\mathcal M} \subset E_A.$$ It follows that $${\mathcal A}\boldsymbol{\varphi} (t)=A\boldsymbol{\varphi} (t)=0$$. In particular, $${\mathcal A}\boldsymbol{\varphi}_0=0$$. In other words, $$\boldsymbol{\varphi} _0\in E_{\mathcal A}$$. From Theorem 2.3, we deduce the existence of the strong limit of z(t), as $$t\to +\infty $$. Thus, $$z(t)\to \boldsymbol{\varphi} _0\in E_A$$, as $$t\to +\infty $$. Theorem 4.3 Suppose that the hypotheses of Theorem 3.1 are verified. Let D(A) be compactly embedded in H, $${\mathcal M}\cap U(A) =\{0\}$$. Then for all $$z_0 \in H$$, we have z(t) → 0, as $$t\to +\infty $$. Proof. Let $$z_0\in D(A)$$. It follows from Theorem 4.2 that $${\mathcal M}\cap U(A)=\{0\}\subset E_A$$. Moreover, we have $$\boldsymbol{\omega} (z_0)\subset{\mathcal M}\cap U(A)$$. Then $$\boldsymbol{\omega} (z_0)=\{0\}$$ and hence z(t) → 0, as $$t\to +\infty $$. Now let us consider $$z_0\in \overline{D({{\mathcal A}} )}$$ and let $$\boldsymbol{\varepsilon}>0$$. Then there exists $$z_{\boldsymbol{\varepsilon}} \in D({{\mathcal A}} )$$ such that $$\|z_0-z_\varepsilon \|\leqslant \frac{\boldsymbol{\varepsilon} }{2}$$. It follows from the fact that $$e^{t{\mathcal A}}$$ is a contraction semigroup that \begin{equation} \left\|e^{t{\mathcal A}}z_0-e^{t{\mathcal A}}z_\varepsilon\right\|\leqslant \frac{\boldsymbol{\varepsilon}}{2}, \;\;\;\forall t\geqslant 0. \end{equation} (4.14) Since $$z_\varepsilon \in D({{\mathcal A}} )$$ and $${\mathcal M}\cap U(A)=\{0\}$$, we deduce from Theorem 4.2, 1. that $$\boldsymbol{\omega} (z_{\boldsymbol{\varepsilon}} )=\{0\} $$. Thus, there exists $$T_{\boldsymbol{\varepsilon}}>0$$ such that \begin{equation} \left\|e^{t{\mathcal A}}z_{\boldsymbol{\varepsilon}}\right\|\leqslant \frac{\boldsymbol{\varepsilon}}{2},\;\;\;\forall t\geqslant T_{\boldsymbol{\varepsilon}} . \end{equation} (4.15) It follows from (4.14) and (4.15) that for all $$t\geqslant T_{\boldsymbol{\varepsilon}} ,$$ we have $$\|e^{t{\mathcal A}}z_0\|\leqslant \boldsymbol{\varepsilon} $$. We conclude that for all $$z_0\in \overline{D({\mathcal A})}=H$$, we have z(t) → 0, as $$t\to +\infty .$$ 5. Applications 5.1 Heat equation Let $$\boldsymbol{\Omega} =]0,1[$$ and let us consider the bilinear system given by the following heat equation: \begin{equation} \left\{ \begin{array}{lll} \frac{\partial z}{\partial t}(t,x)=\Delta z (t,x)+v(t)a(x)z(t,x), \; \mbox{on} \; ]0,\quad+\infty[\times\boldsymbol{\Omega}&& \\ z^{\prime}(t,0)=z^{\prime}(t,1)=0,\quad\; \forall \;t>0 && \\ z(0,x)=z_0(x),\quad\; \mbox{on} \;\boldsymbol{\Omega} && \end{array} \right. \end{equation} (5.1) where $$a\in L^{\gamma }(\boldsymbol{\Omega} )\;(\gamma> 2)$$ and $$\;a \notin L^{\gamma +2}(\boldsymbol{\Omega} )$$ (as an example, one can take $$a(x)=x^{-\frac{1}{\gamma +1}}, \; \forall x\in ]0,1[$$). Let $$ H= L^2(\boldsymbol{\Omega} )$$ and $$Az=\Delta z$$, for all $$z\in D(A) =\{z\in L^2(\boldsymbol{\Omega} )\;/\;\Delta z \in L^2(\boldsymbol{\Omega} ),\;z^{\prime}(0)=z^{\prime}(1)=0\}$$. Observing that $$Ba^{\frac{\gamma }{2}}=a^{\frac{\gamma +2}{2}}\not \in L^2(\boldsymbol{\Omega} )$$, we deduce that the operator defined by Bz = az, for all $$z\in D(B):=H^1(\boldsymbol{\Omega} )$$ is not bounded from $$L^2(\boldsymbol{\Omega} ) $$ to $$L^2(\boldsymbol{\Omega} )$$. Now, we have $$\|az\|_{L^{2}(\boldsymbol{\Omega} )}\leqslant \|a\|_{L^{\gamma }(\boldsymbol{\Omega} )}\|z\|_{L^{\gamma ^{\ast }}(\boldsymbol{\Omega} ), }\; \mbox{for all}\; z\in H^1(\boldsymbol{\Omega} )$$ with $$\boldsymbol{\gamma} ^{\ast }=\frac{2\boldsymbol{\gamma} }{\boldsymbol{\gamma} -2}$$. By the Sobolev embedding $$H^1(\boldsymbol{\Omega} )\hookrightarrow L^q(\boldsymbol{\Omega} ),\; 1 \leqslant q < +\infty $$ (see Corollary 9.14 in the book by Brezis, 2010), and the fact that for all z ∈ D(A) we have $$\|\nabla z\|^2=|\langle \Delta z,z \rangle | \leqslant \frac{1}{2}(\|\Delta z\|^2+\|z\|^2).$$ We conclude that B is A-bounded. We can state the following stabilization result: Proposition 5.1 Suppose that $$a(x)\geqslant 0,\;\mbox{a.e. on}\; \boldsymbol{\Omega} $$ and $$\int _{\Omega }a(x)\, \mathrm{d}x \neq 0$$. Then there exists $$\boldsymbol{\rho} _0>0$$ such that, for any $$0<\boldsymbol{\rho} < \boldsymbol{\rho} _0$$ and for all $$z_0\in L^2(0,1)$$, the control $$v(t)=- \boldsymbol{\rho} \; \frac{\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\;\mathrm{d}x}{1+\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\;\mathrm{d}x} $$ strongly stabilizes (5.1). Proof. Here, A and B satisfy the assumptions of Theorem 3.1 and we have $$U(A)=E_A\!=\!\{c 1_{\boldsymbol{\Omega}} \; / \; c\!\in\! \mathbb{R}\}$$. Since $$\int _{\Omega }a(x)\, \mathrm{d}x \neq 0,$$ we have $${\mathcal M}\cap U(A)=\{0\}$$, then we conclude by Theorem 4.2. 5.2 Transport equation Let $$\boldsymbol{\Omega} =]0,+\infty [$$ and let us consider the following bilinear transport equation: \begin{equation} \left\{ \begin{array}{lll} \frac{\partial z}{\partial t}(t,x)=-\frac{\partial z}{\partial x} (t,x)+v(t)a(x)z(t,x),\; \mbox{on} \; ]0,+\infty[\times\boldsymbol{\Omega}&& \\ z(t,0)=0,\; \mbox{on} \; ]0,+\infty[&& \\ z(0,x)=z_0(x)\; \mbox{on} \;\boldsymbol{\Omega}&& \end{array} \right. \end{equation} (5.2) where a is such that a(x) > 0, a.e. $$x\in \boldsymbol{\Omega} $$$$\int _0^{+\infty }x\;a^2(x)\;\mathrm{d}x<+\infty $$ (for example, $$a(x)=\frac{1}{\sqrt{x(x^2+1)}}$$). Let $$ H= L^2(\boldsymbol{\Omega} )$$ and $$Az=-\frac{\partial z}{\partial x}$$, for all $$z\in D(A) =\{y\in H^1(\boldsymbol{\Omega} )\;/\; y(0)=0 \; \}$$ and Bz = az, for all z ∈ D(B) := D(A). Thus, B is unbounded from $$L^2(\boldsymbol{\Omega} ) $$ to $$L^2(\boldsymbol{\Omega} )$$. Moreover, by Morrey’s inequality (see Corollary 9.14 in the book by Brezis, 2010), there exists C > 0 such that for all z ∈ D(A) we have $$|z(x)|\leqslant C\sqrt{x}\;\|\nabla z\|_{L^2(\boldsymbol{\Omega} )}$$, a.e. $$x\in \boldsymbol{\Omega} $$, then $$\int _{0}^{+\infty }|a(x)z(x)|^2\, \mathrm{d}x\leqslant C^2\; \|\nabla z\|_{L^2(\boldsymbol{\Omega} )}^2\; \int _0^{+\infty }xa^2(x)\;\mathrm{d}x$$, then B is A-bounded i.e. (1.2) holds for $$M=C\sqrt{\int _0^{+\infty }x\;a^2(x)\;\mathrm{d}x}$$. Furthermore, for all sequence $$(y_n)_{n\in \mathbb{N}} \subset H_0^1(\boldsymbol{\Omega} )$$ such that $$y_n\rightharpoonup y$$ in H and $$<By_n,y_n>\rightarrow 0$$, as $$n\rightarrow +\infty $$, there exists a subsequence of $$(y_n)_{n\in \mathbb{N}}$$ still denoted by $$(y_n)_{n\in \mathbb{N}}$$ such that $$a(x)y_n^2(x)\rightarrow 0$$, a.e. $$x\in \boldsymbol{\Omega} $$ as $$n\to +\infty $$. If in addition $$(Ay_n)_{n\in \mathbb{N}}$$ is bounded in H, then for all test function $$\varphi $$ we have $$|y_n(x)\;\boldsymbol{\varphi} (x)|\leqslant C\|\nabla y_n\|_{L^2(\boldsymbol{\Omega} )}\sqrt{x}\;|\boldsymbol{\varphi} (x)|$$. This combined with the dominated convergence theorem gives $$\langle y_n,\varphi \rangle \rightarrow 0$$ as $$n\rightarrow +\infty $$, we conclude that $$y_n\rightharpoonup 0$$ in H, as $$n\to +\infty $$. Then y = 0, thus the condition $$(\boldsymbol{C}_1)$$ is verified. As a consequence of Theorem 4.1, we have the following result: Proposition 5.2 Suppose that a(x) > 0, a.e. $$x\in \boldsymbol{\Omega} $$ and $$\int _0^{+\infty }x\;a^2(x)\;\mathrm{d}x<+\infty $$ . Then there exists $$\boldsymbol{\rho} _0>0$$ such that for any $$0<\boldsymbol{\rho} < \rho _0$$ and for all $$z_0\in L^2(\boldsymbol{\Omega} )$$, the system (5.2) controlled by the feedback $$v(t)=- \boldsymbol{\rho} \; \frac{\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\;\mathrm{d}x}{1+\int _{\boldsymbol{\Omega}} a(x)\;|z(x,t)|^{2}\, \mathrm{d}x}$$ admits a unique solution $$z\in{\mathcal C}(0,+\infty ;H)$$, and we have $$z(t)\rightharpoonup 0$$ in $$L^2(\boldsymbol{\Omega} )$$, as $$t\to +\infty $$. 6. Conclusion In this paper, a set of stabilization results for unbounded bilinear systems has been given. Sufficient conditions for weak and strong stabilizations have been given. The stabilization control is uniformly bounded with respect to time and initial states. The established results can be applied to different type of bilinear system including parabolic and hyperbolic case. 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This article is published and distributed under the terms of the Oxford University Press, Standard Journals Publication Model (https://academic.oup.com/journals/pages/about_us/legal/notices) For permissions, please e-mail: journals. permissions@oup.com

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