\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
\begin{equation} \label{2e8}
x(t) = (Hu)(t) + (K(x)(Nx))(t)
\end{equation}
or, equivalently
\begin{equation} \label{2e9}
x = Hu + K(x)(Nx).
\end{equation}
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
\begin{equation} \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.
\end{equation}
\item[(AF1)] The function $-f$ is monotone in the sense that
\begin{equation}
\label{2e11} \langle f(t,x) - f(t,y),x - y \rangle
\le 0 \quad \forall x,y \in X,t \in I.
\end{equation}
\end{itemize}
Then, given $u\in U$, there exists a unique solution $x\in Y$ of
(\ref{2e9}) and $x$ satisfies a growth condition
\begin{equation} \label{2e12}
\| x \|_Y \leq \frac{b_0}{\mu } +
\big( \frac{b_0}{\mu } + \quad 1 \big)h_0 \| u \|_U.
\end{equation}
\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:
\begin{equation}\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{equation}
\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
\begin{equation} \label{4e1}
Cu = \int_0^T {h(T,s) u(s)\,ds}.
\end{equation}
The operator $C$ is bounded linear and in fact, is known as the
control operator for the linear system
\begin{equation} \label{4e2}
x(t) = \int_0^t h(t,s)\;u(s)\,ds,\quad
x(0) = 0.
\end{equation}
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 $u\in Z$,
\begin{equation} \label{4e3}
(I+SF)u =u_0.
\end{equation}
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}
\begin{equation} \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}
\end{equation}
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}:
\begin{equation} \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\,.
\end{equation}
This equation is in the form of \eqref{1e1} and can be written in
the form
\begin{equation}\label{5e4}
u+K(x) Nx = 0,
\end{equation}
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
\begin{equation} \label{5e5}
f(t)= \int_0^t\Phi(t,s)x(s),\quad x \in Y
\end{equation}
Then $f'(t) = x(t) + A(t) \int_0^t \Phi(t,s)x(s)ds$.
Therefore,
\begin{equation}\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}
\end{equation}
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}
\begin{equation}\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}
\end{equation}
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}
\begin{equation} \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}
\end{equation}
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 $\Delta w|_\Sigma$,
\begin{equation} \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}
\end{equation}
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
\begin{equation} \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}
\end{equation}
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
\begin{equation} \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}
\end{equation}
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. Hence system
(\ref{5e9}) is exactly controllable on I.
\subsection*{Acknowledgements}
The authors would like to express their
sincere gratitude to referee for his/her suggestions and comments which
resulted in the improvement of this paper.
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\end{document}