\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2006(2006), No. 142, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2006 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2006/142\hfil Exact controllability] {Exact controllability of generalized Hammerstein type integral equation and applications} \author[D. N. Chalishajar, R. K. George\hfil EJDE-2006/142\hfilneg] {Dimplekumar N. Chalishajar, Raju K. George} % in alphabetical order \address{Dimplekumar N. Chalishajar \newline Department of Applied Mathematics, Sardar Vallabhbhai Patel Institute of Technology (SVIT), Gujarat University, Vasad-388306. Gujarat State, India} \email{dipu17370@yahoo.com, dimple.chalishajar@gmail.com} \address{Raju K. George \newline Department of Mathematical Sciences, University of Delaware, Newark, DE 19716, USA} \email{rkgeorgemsu@yahoo.com} \date{} \thanks{Submitted April 23, 2006. Published November 9, 2006.} \subjclass[2000]{93B05, 93C10} \keywords{Exact controllability; Hammerstein type integral equation; \hfill\break\indent monotone operator; solution operator} \begin{abstract} In this article, we study the exact controllability of an abstract model described by the controlled generalized Hammerstein type integral equation $$x(t) = \int_0^t h(t,s)u(s)ds+ \int_0^t k(t,s,x)f(s,x(s))ds, \quad 0 \leq t \leq T <\infty,$$ where, the state $x(t)$ lies in a Hilbert space $H$ and control $u(t)$ lies another Hilbert space $V$ for each time $t \in I=[0,T]$, $T>0$. We establish the controllability result under suitable assumptions on $h, k$ and $f$ using the monotone operator theory. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \section{Introduction} Let $X$ and $V$ be Hilbert spaces and $I=[0,T]$, where $0 0$ is a constant and $b_0(t)\geq 0$ and $b_0 \in L^2(I)$. \end{itemize} \begin{lemma} \label{2l1} The operators $K,H$ and $N$ satisfy the following estimates. \begin{gather} \label{2e5} \| {K(x)y} \|_Y \le k\| y \|_{Y}\quad \forall x,y \in Y. \\ \label{2e6} \| {Hu} \|_Y \le h\| u \|_U \quad u \in U. \\ \label{2e7} \| Nx \|_Y \le \sqrt 2 (a_0\| x \|_Y +b_0)\quad \forall x \in Y, \end{gather} where $b_0 = \| {b_0} \|_{L2(I)}$. \end{lemma} \begin{proof} The estimate (\ref{2e5}) follows from Cauchy-Schwartz inequality as: \begin{align*} \| K(x)y \|_Y^2 &= \int_0^T \| ((Kx)(y)(t)) \|_X^2 dt\\ & \le \;\int_0^T (\int_0^t \| {k(t,s,x)} \|\;\;\| y(s) \|_X ds)^2dt \\ & \le \int_.^T (\int_0^t \| k(t,s,x) \|^2 ds)(\int_0^t \| y(s) \|^2ds)dt \\ & \le k_0^2\| y \|_Y^2 . \end{align*} The inequality (\ref{2e6}) follows in a similar fashion. Now \begin{align*} \| {Nx} \|_Y^2 &= \int_0^T \| {Nx(t)} \|_X^2 dt = \int_0^T \| {f(t,x(t))} \|_X^2 dt \\ & \le 2\int_0^T [a_0^2 \| {x(t)} \|^2 + b_0(t)^2]dt \\ & \le 2 [a_0^2\| x \|_Y^2 + b_0^2]\\ & \le 2[a_0\| x \|_Y + b_0]^2. \end{align*} Hence (\ref{2e7}). \end{proof} \subsection*{Operator form of the equation} With the notation as earlier, we may write the equation \eqref{1e1} as $$\label{2e8} x(t) = (Hu)(t) + (K(x)(Nx))(t)$$ or, equivalently $$\label{2e9} x = Hu + K(x)(Nx).$$ The following theorem gives the existence of solution $x$ of (\ref{2e9}) for a given $u$ which can be proved along the lines as in \cite{JG}. \begin{theorem}[Existence and Uniqueness] \label{2t1} Assume the following: \begin{itemize} \item[(AK1)] There exists a constant $\mu >0$ such that $$\label{2e10} \int_0^T \big\langle \int_0^t k(t,s,x)x(s)ds,x(t) \big\rangle dt \geq \mu \int_0^T \big\| \int_0^t k(t,s,x)x(s)ds \big\|^2dt\quad \forall x \in Y.$$ \item[(AF1)] The function $-f$ is monotone in the sense that $$\label{2e11} \langle f(t,x) - f(t,y),x - y \rangle \le 0 \quad \forall x,y \in X,t \in I.$$ \end{itemize} Then, given $u\in U$, there exists a unique solution $x\in Y$ of (\ref{2e9}) and $x$ satisfies a growth condition $$\label{2e12} \| x \|_Y \leq \frac{b_0}{\mu } + \big( \frac{b_0}{\mu } + \quad 1 \big)h_0 \| u \|_U.$$ \end{theorem} \begin{lemma} \label{2l2} Under the assumptions (AK1), (AF1) and the assumptions (A1)--(A4), the Nemytskii operator $W$ is well-defined and continuous. Moreover it satisfies the following growth condition: $$\label{2e13} \| {Wu} \|_Y \le \sqrt 2 \big( \frac{b_0}{\mu } + 1 \big) a_0h_0 \| u \|_U + \sqrt 2 (\frac{1}{\mu } + 1 ) b_0.$$ \end{lemma} The proof of the above lemma follows from the assumptions and estimate (\ref{2e12}). \section{Compactness of the operator W} We make the following further assumptions in this section to guarantee the compactness of $W$ . \noindent\textbf{Assumptions} \begin{itemize} \item[(B1)] There exists $\tilde{k}>0$ such that $$\big\| \int_s^t k(t,\tau ,x)x(\tau )d\tau \big\|_X \leq \tilde{k} (t-s) \| x \|_Y,\quad 0 \leq s < t \leq T.$$ \item[(B2)] There exists $\tilde{h}>0$ such that $$\big\| \int_s^t h(t,\tau )u(\tau )d\tau\big\|_X \leq \tilde {h} (t-s) \| u \|_U,\quad 0 \le s < t \le T.$$ \item[(B3)] The operators $k$ and $h$ satisfy the uniform continuity in the following sense: Given $\varepsilon>0$ there exists $h>0$ small such that $$\| k(r+h, s, x) - k(r,s,x)\|_{BL(X)} \leq \varepsilon$$ and $$\| h(r+h,s) - h(r,s) \|_{BL(X)} \leq \varepsilon, \quad 0\leq r < r+h\leq T.$$ \item[(B4)] There exists a space $\hat {X}$ such that $X \mapsto \hat{X}$ is a compact imbedding. \item[(B5)] Assume that $f$ can be extended to $I \times \hat{ X} \mapsto X$ such that $f$ is Caratheodory and $x\mapsto f(.,x(.))$ is continuous from $L^2(I;\hat{X})\mapsto L^2(I;X)$. \end{itemize} \begin{theorem} \label{thm3.1} Under the Assumptions (B1)--(B5), the operator $W$ is compact. \end{theorem} \begin{proof} Let $\{u_{n}\}$ be a bounded sequence in $U$. We have to show that $\{Wu_n \} = \{f(.,x_n (.))\}$ has a convergent subsequence. First of all $\{f(.,x_n (.))\}$ is bounded in $Y$ by Lemma \ref{2l2}. Therefore there exists a constant $M>0$ such that $\int_0^T \| {f(t,x_n (t))} \|_X^2 dt \le M^2,$ where, $x_n$ is the solution of \eqref{1e1} corresponding to $u_n$. We show that the family $\{x_{n}(.)\}$ is equicontinuous in $C(I;X)$. $$x_n(t)=\int_0^t k(t,\tau,x_n) f(\tau,x_n(\tau))d(\tau) +\int_0^t h(t,\tau)u_n(\tau)d(\tau)$$ Let $t = r+h_{0}$. We have \begin{align*} \| {x_n (t) - x_n (r)} \| &\le \| {\int_0^r {\{k(t,\tau ,x_n ) - k(r,\tau ,x_n )\}} f(\tau ,x_n (\tau ))d\tau } \|\\ &\quad + \|{\int_r^t {k(t,\tau ,x_n ) f((\tau ,x_n (\tau ))d\tau } } \|\\ &\quad + \| {\int_0^r {\{h(t,\tau ) - h(r,\tau )\}u_n (\tau )d\tau } } \| + \| {\int_r^t {h(t,\tau )u_n (\tau )d\tau } } \|\\ &=I_{1} + I_{2} +I_{3} + I_{4}. \end{align*} Now by (B3) and (B1) respectively, we get $$I_{1} \leq \varepsilon \int_0^r {\| {f(\tau ,x_n (\tau ))} \|} _X d\tau \leq \varepsilon r^{1/2}M \le \varepsilon MT^{1/2}$$ and $$I_{2} \leq \tilde{k}\;h_0 \| {f(.,x_n (.))} \|_{Y}.$$ Similarly $I_3$ and $I_4$ can be estimated as $$I_{3} \leq \varepsilon T^{1/2}\| {u_n } \|_U \quad \mbox{and}\quad I_{4} \leq \tilde {h} h_{0} \| {u_n } \|_U.$$ The above estimates shows that $\{x_n(.)\}$ is equicontinuous in $C(I;X)$ as $\Vert u_n \Vert$ is bounded. Further, $\{x_n(.)\}$ is also uniformly bounded in $C(I;X)$. Now, using the compact inclusion $X\hookrightarrow \hat{X}$ and applying general form of Arzela-Ascoli theorem \cite{AMR}, we deduce that $\{x_n(.)\}$ is relatively compact in $C(I;\hat{X})$. Thus along a subsequence $\{x_{n_k}\}$ converges in $C(I;\hat{X})$ and so converges in $L^2(0,T;\hat{X})$. Then from the assumption (B5), it follows that $f(.,x_{n_k}(.))$ converges in $Y=L^2(0,T;X)$. Thus the operator is compact and the proof is complete. \end{proof} \begin{remark} \label{3r2} \rm If $h(t,s)$ is a compact operator, then it is easy to show that $W$ is compact. In such situations, the exact controllability in the whole space may be impossible \cite{TR1,ST} for different conditions to ensure the compactness of $W$ with non-compact $h(t,s)$. \end{remark} Also, it is possible to give various more specific conditions under which the operator $W$ is compact. When $W$ is assumed to be compact, the assumption [AK1] can be weakened by imposing strong monotonicity on $f$. i.e. by making [AK1] stronger which is shown in the following Lemma. \begin{itemize} \item[(AK2)] $\int_0^T\langle\int_0^t k(t,s,x)x(s)ds,x(t)\rangle_X dt \geq 0$ for all $x \in Y$. \item[(AF2)] There exists a constant $\beta > 0$, such that $$\langle f(t,x)-f(t,y),x-y\rangle \geq \beta \Vert x-y\Vert^2$$ \item[(AF3)] Assumptions of are satisfied. \end{itemize} \begin{lemma} \label{3l3} Assume (AK2), (AF2), (B1)--(B5). Then the operator $W$ is well defined and continuous. Further it satisfies the growth condition $$\Vert Wu\Vert \leq C_0 +C \Vert u\Vert_U,$$ where $C_0=b_0+a_0mbT e^{ma_0T}$ and $C= a_0h_0 T e^{ma_0T}$, with $m$ is a positive constant satisfying $$\Vert k(t,s,x)\Vert \leq m(x) 0  such that$$ \Vert f(t,x)-f(t,y)\Vert \leq \alpha \Vert x-y \Vert \quad \forall x,y, \in X, t\in I $$\item[(AF5)] There exists \beta > 0 such that$$ \langle f(t,x)-f(t,y),x-y\rangle\;\; \leq - \beta\Vert x-y\Vert^2 \quad \forall x,y \in X, t\in I $$\end{itemize} \begin{lemma} \label{3l4} In each of the following cases, the solution operator W is well defined and Lipschitz continuous. \begin{itemize} \item[(i)] Assumption (AF4) holds with  k_0 \alpha < 1 \item[(ii)] Assumption (AF4) and (AF5) hold with \beta > k_0 \alpha^2 \item[(iii)] Assumption (AF4) hold with \Vert k(t,s,x)\Vert \leq m(x) < m \quad \forall t,s \in I, m>0 \item[(iv)] Assumption (AF4) holds. \end{itemize} Further the Lipschitz constants for W in the above cases are respectively, $\frac{\alpha k_0 h_0}{1-k_0 \alpha} ,\quad \frac {k_0 \alpha^3 h_0}{\beta(\beta-k_0 \alpha^2)}, \quad \alpha T h_0 e^{ma_0T},\quad \frac {k_0 h_0 \alpha}{1-\alpha},$ where \varepsilon > 0 is an arbitrary small constant. \end{lemma} \begin{remark} \label{3r5} \rm Here (AF4) is sufficient to prove the existence of W and Lipschitz continuity of W. The additional assumptions only give better estimation on the Lipschitz constant of the solution operator W. \end{remark} When f is locally Lipschitz continuous, then also we can show that W is well-defined, shown in the following Lemma. The proof follows along the same line as in the proof of \cite[Lemma 2.4]{RKG}. \begin{itemize} \item[(i)] There exists a constant \alpha (r)  such that$$ \Vert f(t,x)-f(t,y)\Vert\leq\alpha(r)\Vert x-y \Vert \quad \forall x,y \in X $$such that \Vert x \Vert \leq r, \Vert y \Vert \leq r. \item[(ii)] There exists m>0 such that \Vert k(t,s,x) \Vert\leq m for all t,s \in I \item[(iii)] f satisfies the growth condition (A4). \end{itemize} \begin{lemma} \label{3l6} Under assumptions (i)--(iii) above, the operator W is well-defined and continuous. Moreover, W satisfies a growth condition (A4),$$ \Vert Wu \Vert_Y \leq (b_0 + a_0mbT e^{ma_0T}) + a_0h_0T e^{ma_0T}\Vert u \Vert_U $$\end{lemma} \begin{proof} Since, by the local Lipschitz condition, there exists a unique solution to \eqref{1e1} in a maximal interval [0,t_{\rm max}], t_{\rm max} \leq t. If  t_{\rm max} < t then \lim_{t \mapsto t_{\rm max}} \Vert x(t,s)\Vert_X = \infty(refer \cite{TH}). In other words, if  \lim_{t \mapsto t_{\rm max}}\Vert x(t,s)\Vert_X=\infty , then \exists a unique solution in the interval [0,t]. We have already shown in the proof of Lemma \ref{3l3} that \Vert x(t,s)\Vert_X < \infty for each u. Thus W is well-defined and the growth condition follows from the proof of Lemma \ref{3l3}. \end{proof} We now move on to the exact controllability under the assumption that the operator W is compact. \section{Exact controllability} We, first reduce the controllability problem to a solvability problem which in turn imply the conditions for controllability of system \eqref{1e1}. Define an operator C: U\mapsto X by $$\label{4e1} Cu = \int_0^T {h(T,s) u(s)\,ds}.$$ The operator C is bounded linear and in fact, is known as the control operator for the linear system $$\label{4e2} x(t) = \int_0^t h(t,s)\;u(s)\,ds,\quad x(0) = 0.$$ Let N(C) = \{u \in U:Cu = 0\} be the null space and $Z = [N(C)]^ \bot = \{u \in U: \langle u,v \rangle = 0 \text{ for all }v \in N(C)\}.$ \begin{definition} \label{def4.1}\rm We call a bounded linear operator S: X\mapsto Z, a {\bf Steering Operator} if S steers the linear system \eqref{4e2} from 0 to x_1. In other words, if u=Sx_{1}, (x_{1} \in X), then $x(T) = \int_0^T {h(T,s)(Sx_1 )(s)ds = x_1 }.$ \end{definition} Clearly CS = I, the identity operator on X. Thus, if there exists a steering operator S, then u=Sx_{1} acts as a control and the linear system \eqref{4e2} is controllable. Conversely, if the linear system is controllable, then for any x_{1} \in X there exists u \in U such that Cu=x_1, i.e., C is onto. Thus, we can define a generalized inverse C^{\#}= (C|_Z)^{-1}: X\mapsto Z and S= C^{\#} will be a steering operator. Thus, one gets the following result. \begin{theorem} \label{4t2} The linear system \eqref{4e2} is exactly controllable if and only if there exists a steering operator. \end{theorem} Here we note that C^{\#} Cu=u  for \forall u \in z  and C^{\#} Cu = v for u \in U , where v is the projection of  u on z. We now assume the controllability of the linear system and proceed to prove the exact controllability of the nonlinear system. Define an operator F:Z\mapsto X by$$ Fu = \int_0^T \quad k(T,s,x)(Wu)(s)ds, $$where, x is the solution of the system \eqref{1e1} corresponding to the control u. Let S be the steering operator of the linear system. Let x_1\in X and u_0=Sx_1 be the control which steers the linear system from 0 to x_1. The exact controllability of \eqref{1e1} is equivalent to the existence of u\in Z (let x be the solution of \eqref{1e1} corresponding to u) such that$$ x_{1} = x (T) = \int_0^T k(T,s,x)(Wu)(s)ds+\int_0^T h(T,s)u(s)ds. $$That is,$$ x_1=Fu+Cu. $$Applying S on both sides, we get$$ u_0=SFu+u. $$in z, where u_0 is the control, steering the linear system from 0 to x_1. Thus, the problem of controllability reduces to solvability problem of the operator equation: Solve for u\in Z, $$\label{4e3} (I+SF)u =u_0.$$ We now state our controllability result. For the sake of generality, we state the theorem by imposing indirect conditions on W and F. The explicit conditions on k,h,f can be given to verify the conditions on W and F. The corollaries follow are direct verfication of the conditions of the main theorem. \begin{theorem} \label{4t4} Assume the linear system \eqref{4e2} is exactly controllable with the steering operator S. Further assume that the operator W is well defined and compact and satisfies$$ \| {SFu} \| \le a_0 \| u \| + b_0, \quad \mbox{with}\quad a_0 < 1,\;b_0 \geq 0 $$Then the system \eqref{4e3} is solvable in Z. \end{theorem} \begin{proof} We look for the solvability of the operator R: Z\mapsto U, where$$ Ru = [ I + SF]u. $$Then $\langle Ru,u \rangle = \langle u,u \rangle + \langle SFu,u \rangle \geq \| u \|^2 - a_0\| u \| - b_0 \| u \|,$ which implies$$ \lim_ {\| u \| \to \infty } \frac{\langle Ru,u \rangle}{\| u\|} = \infty . $$Thus, R is coercive operator. Again compactness of W implies that SF is compact. Now, R is compact perturbation of the identity operator and hence R is of {\bf type (M)}. See \cite{JB} for a definition of type(M). Since any coercive operator of type (M) is onto \cite{JB}, the proof of the theorem is complete. \end{proof} \begin{corollary} \label{4c5} Assume the linear system is exactly controllable with a steering operator S. Assume the conditions (AK1) and (AF1) and the assumptions (B1)--(B5). Then the nonlinear system \eqref{1e1} is controllable if$$ \| S \| k_{0} (b_{0} + \mu ) a_{0}h_{0} < \mu . $$\end{corollary} \begin{theorem} \label{4t5} Suppose that the system \eqref{1e1} satisfies: \begin{enumerate} \item The linear part is exactly controllable. \item W is well defined and compact. \item SF is uniformly bounded i.e. \vert SFu \vert\leq C, for some C>0. \end{enumerate} Then the system \eqref{1e1} is exactly controllable. \end{theorem} \begin{proof} Let R be the operator defined in the proof of the Theorem \ref{4t4}. We have$$ \langle Ru,u \rangle > \Vert u \Vert^2 -C\Vert u \Vert\; \Rightarrow \; \lim_{\Vert u\Vert \mapsto\infty} \langle Ru,u \rangle=\infty $$By following the same argument as in the proof of Theorem \ref{4t4}, we have that R is a coercive operator of type (M) and hence it is onto. This completes the proof. \end{proof} In the above result we do not require the Lipschitz continuity of W but we need F to be uniformly bounded. If f is uniformly bounded then it is not hard to show that SF is also uniformly bounded. When f is uniformly bounded, we have the following result which follows as particular case of Theorem \ref{4t5}. \begin{corollary} \label{4c6} Suppose that the linear system \eqref{4e2} is exactly controllable( i.e linear part of \eqref{1e1} is exactly controllable) and the nonlinear term f is uniformly bounded. Further suppose that the assumptions in Theorem \ref{2t1}, Lemma \ref{2l2} and assumptions (B1)--(B5) hold. Then the system \eqref{1e1} is exactly controllable. \end{corollary} When f is Lipschitz continuous, we have the following result. \begin{theorem} \label{4t7} Suppose that the system \eqref{1e1} satisfies the following two conditions: \begin{enumerate} \item The linear part is exactly controllable. \item There exists \alpha \in (0,1) such that \Vert SFu-SFv \Vert \leq \alpha \Vert u-v \Vert for all u,v \in Z. \end{enumerate} Then the system \eqref{1e1} is exactly controllable. Further, if u_0 is the steering control for the linear system \eqref{4e2}, to steer the system from 0 to x_1; then the control u, approximated from the following iterative scheme, steers the state of the nonlinear system \eqref{1e1} from 0 to x_1 in the same time interval [0,T], \begin{gather*} u^{(n+1)}=u_0-SFu^{(n)} \\ u^{(0)}=u_0. \end{gather*} \end{theorem} \begin{proof} Since SF is a contraction, the solvability of \eqref{4e3} and the approximating scheme follow from Banach Contraction Principle \cite{JB}. \end{proof} The next corollary follows from Theorem \ref{4t7} using Lemma \ref{3l4}. \begin{corollary} \label{4c8} Suppose that the linear system \eqref{4e2} is exactly controllable with steering operator S. Then under each of the following cases the nonlinear system \eqref{1e1} is exactly controllable. \begin{itemize} \item[(i)] Assumption (AF4) holds with k(x)\alpha < 1 and  \alpha k_0 h_0k_0\Vert S \Vert < (1-k_0 \alpha) \item[(ii)] Assumption (AF4) and (AF5) hold with \beta > k_0 \alpha^2  and  \Vert S\Vert. k_0 k_0 h_0 \alpha^3 < \beta(\beta-k_0 \alpha^2) \item[(iii)] Assumption (AF4) hold with \Vert k(t,s,x)\Vert \leq m  for all t, s \in I, m>0  and$$ \Vert S \Vert k_0k_0h_0 \alpha e^{ma_0T} <1 $$\item[(iv)] Assumption (AF4) folds with \Vert S \Vert. k_0 k_0 h_0 \alpha <(1-\epsilon) where  \epsilon > 0 being an arbitrary small constant. \end{itemize} \end{corollary} \begin{proof} The proof of all the cases follow by using proof of respective cases of the Lemma \ref{3l3} and by using \cite{RKG}. \end{proof} \section{Applications} One can put nonlinear evolution systems with internal control in above frame work to study the exact controllability. It is also possible to use the above results to study the exact controllability problems associated with the partial differential equations with boundary controls. \subsection*{(a) Nonlinear evolution system with internal control} $$\label{5e2} \begin{gathered} \frac{dx}{dt} = A(t)x + B(t)u + f(t,x), \quad 0< t \leq T < \infty \\ x(0)=0 \end{gathered}$$ where, A(t) is a linear operator for each t\in [0,T], but not necessarily bounded, B(t) is a bounded linear operator and f is a nonlinear operator in a suitable Hilbert space. Let X and U be the state space and space of control functions, respectively. Assume that, for each  t \in [0,T], A(t) generates a strongly continuous evolution system  \Phi (t,s) on X. By using the variation of constant formula, a mild solution of (\ref{5e2}) can be written as as follows \cite[pp.106]{PA}: $$\label{5e3} x(t)=\int_0^t \Phi(t,s)f(s,x(s))ds +\int_0^t \Phi(t,s)B(s) u(s) ds\,.$$ This equation is in the form of \eqref{1e1} and can be written in the form $$\label{5e4} u+K(x) Nx = 0,$$ with k(t,s,x) =\Phi(t,s)  and  h(t,s) = \Phi(t,s)B(s). We apply our main result to deduce controllability. In this case it is easy to show that the linear part of (\ref{5e2}) is exactly controllable if and only if there exists \lambda > 0 such that$$ \big\langle \int_0^T \Phi (T,s)B(s)B^*(s)\Phi^*(T,s)vds, v\big \rangle \geq \lambda \Vert v \Vert^2 \quad \forall v \in X where \Phi^* (t,s),B^*(s) are the adjoint operators of  \Phi(t,s) and B(s), respectively. \begin{lemma} \label{5l1} Under the condition  \langle -A(t)x,x \rangle_X \geq \mu\Vert x \Vert^2 for all x \in D(A(t)), the reduced form of the assumption [AK1], that is $\textrm{(AK3)}\quad \int_0^T \langle \int_0^t \Phi(t,s)x(s)ds,x(t)\rangle_X dt \geq \mu \int_0^T \Vert \int_0^t \Phi(t,s)x(s)ds \Vert^2 dt, \quad \forall x \in Y$ holds good for (\ref{5e6}) \end{lemma} \begin{proof} Let $$\label{5e5} f(t)= \int_0^t\Phi(t,s)x(s),\quad x \in Y$$ Then f'(t) = x(t) + A(t) \int_0^t \Phi(t,s)x(s)ds. Therefore, \label{5e6} \begin{aligned} \int_0^T \langle \int_0^t \Phi(t,s)x(s)ds,x(t) \rangle_Xdt &=\int_0^T \langle f(t),f'(t)-A(t)\int_0^t\Phi(t,s)x(s) ds\rangle dt \\ &=\int_0^T \langle f(t),f'(t) \rangle dt+\int_0^T \langle f(t),-A(t)f(t)\rangle dt \end{aligned} However, $\int_0^T \langle f(t),f'(t)\rangle dt= \langle f(t),f(t)\rangle |_0^T-\int_0^T \langle f'(t), f(t)\rangle dt$ implies $\int_0^T \langle f(t),f'(t)\rangle dt=\frac{1}{2} \Vert f(t) \Vert^2 \geq 0$ Therefore, the right-hand side of \eqref{5e6} is greater than or equal to \begin{align*} \int_0^T \langle f(t),-A(t)f(t) \rangle dt &\geq \mu \int_0^T \Vert f(t)\Vert^2 \quad \text{(by hypothesis)}\\ &\geq \mu \int_0^T \langle \int_0^t \Phi(t,s)x(s)ds,\int_0^t \Phi(t,s)x(s)ds \end{align*} Hence, \int_0^T\langle \int_0^t \Phi(t,s)x(s)ds, x(t)\rangle dt\geq \mu\int_0^T\Vert\int_0^t\Phi(t,s)x(s)ds\Vert^2 dt;\quad\forall x\in Y $$This completes the proof. \end{proof} Similarly one can impose other conditions on A(t),B(t), f(t,x) to verify that the assumptions made on system \eqref{5e2} are not redundant. Thus by using the main theorem, one can obtain different sets of verifiable conditions for exact controllability of the nonlinear system (\ref{5e2}). \subsection*{(b) The autonomous parabolic system with boundary control} $$\label{5e7} \begin{gathered} \frac{dx}{dt}=Ax+f(t,x)\quad \text{on }[0,T]\times\Omega\\ \beta x = u\\ x(0)=0 \end{gathered}$$ where, A is an elliptic differential operator (eg. second order or fourth order), f is a nonlinear operator and \beta is a boundary operator(eg. Dirichlet or Neumann) in some appropriate space. Here u is the boundary control. \Omega is a bounded open domain in {R}^n with boundary \partial\Omega. Assume that D(A) includes homogeneous boundary conditions \beta x=0. Let L^2(\Omega) be the state space X and  L^2(\Gamma) be the control space V for some choice of \Gamma \subset\partial\Omega. Assume that 0 is not an eigenvalue of A. Define a Green's operator D : V \mapsto X with Ax=0, \beta x=u. Now the standard trace and regularity theory for these elliptic operators implies that A^\theta D:V\mapsto X is bounded for \theta < 3/4 . Using the variation of parameter formula, solution of (\ref{5e7}) can be written as$$ x(t)=\int_0^t \Phi(t-s)f(s,x(s))ds + \int_0^t\Phi(t-s)A Du(s)ds $$where \Phi(t-s) is the strongly continuous semigroup generated by the elliptic operator A. Thus the system (\ref{5e7}) can be represented in the form (\ref{5e2}) with  k(t,s,x)=\Phi(t-s) and h(t,s)=\Phi(t-s)AD. Hence we can make the use of the main results of Section 4 to obtain controllability criterion for (\ref{5e7}). \subsection*{(c) Nonlinear Euler-Bernoulli equation with boundary control} $$\label{5e8} \begin{gathered} \frac{\partial^2}{\partial t^2} w(t,y)=\Delta^2 w(t,y)+g(t,w(t,y),w_t (t,y))\quad \textrm{in } [0,T]\times\Omega \\ w(0,y)=w_t (0,y)=0 \quad \textrm{in }\Omega \\ w|_\Sigma =u_1\quad \textrm{in } \sum\equiv[0,T]\times\Gamma \\ \Delta w|_\Sigma =u_2 \quad \textrm{in } \Sigma \end{gathered}$$ where \Omega is an open and bounded domain of {R^{n}} with sufficiently smooth boundary \Gamma. Here u_1 and u_2 are the boundary controls. Let A_1:L^2(\Omega)\mapsto L^2(\Omega) be the positive self-adjoint operator defined by$$ A_1h=\Delta^2 h, \textrm{with}~D(A_1)=\{h \in H^4(\Omega):h|_\Gamma=\Delta h|_\Gamma=0\} $$So that A_{1}^{1/2}h=-\Delta h and A_{1}h=\Delta^2 h. Let X=D(A_{1}^{1/2})\times L^2(\Omega), \textrm{where} D(A_{1}^{1/2})=H^2(\Omega)\cap H_0(\Omega). Define Green's operators G_1 and G_2 as follows: \\ G_1:H^s(\Gamma)\mapsto H^{s+1/2}(\Omega) is continuous such that \begin{gather*} G_1 u_1=h\\ \Delta^2 h=0 \quad \text{in } \Omega\\ h=u_1 \quad \text{on } \Gamma\\ \Delta h=0 \quad \textrm{on } \Gamma. \end{gather*} \noindent G_2:H^s(\Gamma) \mapsto H^{s+5/2}(\Omega) is continuous such that \begin{gather*} G_2u_2=y\\ \Delta^2 y=0 \quad \text{in }\Omega\\ y=0\quad \text{in } \Gamma\\ \Delta y=u_2\quad \text{in }\Gamma. \end{gather*} Define on operator B as $B \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} =\begin{bmatrix} 0 \\ A_{1}(G_1u_1+G_2u_2) \end{bmatrix}$ The operator -A_{1} generates a strongly continuous cosine operator C(t) on L^2(\Omega) with S(t)=\int_0^t C(\tau)d\tau. Define the operator A as follows: ${A}=\begin{bmatrix} 0 & I\\ -A_{1} & 0 \end{bmatrix}$ where  D(A)=D(A_{1})\times D(A_{1}^{1/2}). A generates a unitary strongly continuous semigroup e^{At} given by ${e^{At}}=\begin{bmatrix} C(t) & S(t)\\ -A_{1}S(t) & C(t) \end{bmatrix}$ Using variation of constant formula, the solution of (\ref{5e8}), can be written in the form (\ref{5e3}), where \begin{gather*} x(t)=\begin{bmatrix} w(t)\\ w_t(t) \end{bmatrix},\quad u(t)=\begin{bmatrix} u_1(t)\\ u_2(t)\end{bmatrix}, \quad f(t,x(t))=\begin{bmatrix} 0\\ g(t,w,w_t) \end{bmatrix} \\ h(t,s)u=e^{A(t-s)}Bu=\begin{bmatrix} S(t-s)A_{1}(G_1u_1+G_2u_2)\\ C(t-s)A_{1}(G_1u_1+G_2u_2) \end{bmatrix}, \end{gather*} It is well-known that the linear part is exactly controllable \cite{LT1}. Thus by using the main results of Section 4, one can obtain verifiable assumptions on g to achieve exact controllability for (\ref{5e8}). \begin{remark} \label{5r2} \rm As a particular case of the above example, one can consider the following nonlinear Euler-Bernoulli equations with boundary control only in \Delta w|_\Sigma, $$\label{5e9} \begin{gathered} \frac{\partial^2}{\partial t^2} w(t,y)= \Delta^2 w(t,y)+g(t,w(t,y),w_t(t,y))\quad \text{in }(0,T)\times \Omega\\ w(0,y)= w_t (0,y)=0 \quad\text{in} \Omega\\ w|_\Sigma = 0 \quad\text{in } (0,T)\times\Gamma=\Sigma\\ \Delta w|_\Sigma = u \quad\text{in } \Sigma, \end{gathered}$$ where  \Omega is an open bounded domain in \mathbb{R}^{n} with sufficiently smooth boundary \partial \Omega=\Gamma. Here u is the only boundary control. As in the case of above example, controllability of the linear part is established in Lasiecka and Triggiani \cite{LT2}. Using the main result in Section 4, we can get the verifiable assumptions on g to achieve exact controllability for the system (\ref{5e9}). \end{remark} \begin{remark} \label{5r3} \rm We consider the system governed by parabolic initial boundary-value problem $$\label{5e10} \begin{gathered} \frac{\partial}{\partial t} y(t,x)+Ay(t,x) = u(t,x)+g(t,y(t,x),y_t(t,x))\quad\text{in }Q=(0,t)\times\Omega\\ y(\cdot,x) =0\quad\text{on } \sum=(0,T)\times\partial \Omega \\ y(0) = y_0\quad \text{on }\Omega, \end{gathered}$$ where \Omega is a bounded domain in {R^{n}} with smooth boundary  \partial\Omega. y_0\in H_o^1(\Omega) and u\in L^2(Q). Let A be the second order elliptic differential operator given by$$ Ay=-\sum_{i,j =1}^d \frac{\partial}{\partial x_i} \big( a_{i,j}(x) \frac{\partial y}{\partial x_j}\big)+c(x)y $$with the assumptions that c\geq 0 on \overline\Omega and the matrix (a_{ij}(x)) is symmetric and positive definite. As an exact controllability problem of linear part of system (\ref{5e10}), Cao and Gunzburger \cite{YM} proved that for given function y_0,\hat{y} \in L^2 (\Omega), a function y=y(t,x)  and a control u(t,x) both defined for  (t,x) \in Q  such that  y,u satisfy (\ref{5e10}) together with  y(T,x)=\dot{y}(x) for x \in \Omega . \end{remark} For the nonlinear portion, we can follow the method given in example (c). \subsection*{(d)} Consider the partial functional integro-differential system of the form \label{5e11} \begin{gathered} x_t(y,t)=x_{yy}(y,t)+ e^{(t-s)} u(y,t)+\int_0^t (t-s)\{e^{-\int_{0}^{1}\Vert x(u) \Vert}du\}p(s,x(y,s))ds\\ 00, \end{gathered} where u \in L^2(I,V) and X=L^1[(0,1);{R}]. Let f(t,w(t))(y)=p(t,w(t,y)), 00,b(\cdot)=\Vert b(\cdot)\Vert_{L^2 (I)}.$$ Thus all the conditions of our main theorem are satisfied. 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