\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 72, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/72\hfil Almost periodic solutions] {Almost periodic solutions of higher order differential equations on Hilbert spaces} \author[L. T. Nguyen\hfil EJDE-2010/72\hfilneg] {Lan Thanh Nguyen} \address{Lan Thanh Nguyen \newline Department of Mathematics, Western Kentucky University, Bowling Green, KY 42101, USA} \email{Lan.Nguyen@wku.edu} \thanks{Submitted March 26, 2010. Published May 17, 2010.} \subjclass[2000]{34G10, 34K06, 47D06} \keywords{Almost periodic; higher order differential equations} \begin{abstract} We find necessary and sufficient conditions for the differential equation $$ u^{(n)}(t) = Au(t)+f(t), \quad t\in \mathbb{R} $$ to have a unique almost periodic solution. Some applications are also given. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} In this article, we study the almost periodicity of solutions to the differential equation \begin{eqnarray}\label{eq} u^{(n)}(t)=Au(t)+f(t), \quad t\in \mathbb{R}, \end{eqnarray} where $A$ is a linear, closed operator on a Hilbert space $H$ and $f$ is a function from $\mathbb{R}$ to $H$. The asymptotic behavior and, in particular, the almost periodicity of solutions of \eqref{eq} has been a subject of intensive study for recent decades, see e.g. \cite{arba,lezi,minh,lan2,phong,vusc,ruvu} and references therein. A particular condition for almost periodicity is the countability of the spectrum of the solution. In this paper we investigate the almost periodicity of mild solutions of Equation \eqref{eq}, when $A$ is a linear, unbounded operator on a Hilbert space $H$. We use the Hilbert space $AP(\mathbb{R}, H)$ introduced in \cite{ka}, defined by follows: Let $(, )$ be the inner product of $H$ and let $AP_b(\mathbb{R},E)$ be the space of all almost periodic functions from $\mathbb{R}$ to $H$. The completion of $AP_b(\mathbb{R},E)$ is then a Hilbert space with the inner product defined by: $$ \langle f,g\rangle := \lim_{T\to \infty}\frac{1}{2T} \int_{-T}^T(f(s), g(s)) ds. $$ First, we establish the relationship between the Bohr transforms of the almost periodic solutions of \eqref{eq} and those of the inhomogeneity $f$. We then give a necessary and sufficient condition so that \eqref{eq} admits a unique almost periodic solution for each almost periodic inhomogeneity $f$. As applications, in Section 4 we show a short proof of the Gearhart's Theorem: If $A$ is generator of a strongly continuous semigroup $T(t)$, then $1\in \varrho(T(1))$ if and only if $2k\pi i\in \varrho(A)$ and $\sup_{k\in \mathbb{Z}}\|(2k\pi i- A)^{-1}\|< \infty $. \section{Hilbert space of almost periodic functions} Let us fix some notation. Define $S(t)f$ as $(S(t)f)(s) =f(s+t)$. Recall that a bounded, uniformly continuous function $f$ from $\mathbb{R}$ to a Banach space $H$ is almost periodic, if the set $\{S(t)f:t\in \mathbb{R}\}$ is relatively compact in $BUC(\mathbb{R}, H)$, the space of bounded uniformly continuous functions with sup norm topology. Let $H$ be now a complex Hilbert space with $(, )$ and $\|\cdot \|$ be the inner product and the norm in $H$, respectively. Let $AP_b(\mathbb{R},H)$ be the space of all almost periodic functions from $\mathbb{R}$ to $H$. In $AP_b(\mathbb{R},H)$ the following expression $$ \langle f,g\rangle := \lim_{T\to \infty} \frac{1}{2T}\int_{-T}^T(f(s), g(s)) ds $$ exists and defines an inner product. Hence, $AP_b(\mathbb{R},H)$ is a pre-Hilbert space and its completion, denoted by $AP(\mathbb{R},H)$, is a Hilbert space. The inner product and the norm in $AP(\mathbb{R},H)$ are denoted by $\langle, \rangle$ and $\| \cdot \|_{AP}$, respectively. For each function $f\in AP(\mathbb{R},H)$, the Bohr transform is defined by $$ a(\lambda,f):=\lim_{T\to \infty}\frac{1}{2T}\int_{-T}^Tf(s) e^{-i\lambda s}ds. $$ The set $$ \sigma(f):=\{\lambda \in \mathbb{R}:a(\lambda, f)\not= 0\} $$ is called the Bohr spectrum of $f$. It is well known that $\sigma(f)$ is countable for each function $f\in AP(\mathbb{R},H)$. The Fourier-Bohr series of $f$ is $$ \sum_{\lambda \in \sigma(f)}a(\lambda, f)e^{i\lambda t} $$ and it converges to $f$ in the norm topology of $AP(\mathbb{R},H)$. The following Parseval's equality also holds: $$ \|f\|_{AP(\mathbb{R},H)}^2=\sum_{\lambda \in \sigma(f)}\|a(\lambda, f)\|^2. $$ For more information about the almost periodic functions and properties of the Hilbert space $AP(\mathbb{R},H)$, we refer readers to \cite{ka,lezi,phong}. Let $W^k(AP)$ be the space consisting of all almost periodic functions $f$, such that $f'$, $f''$, \dots , $f^{(k)}$ are in $AP(\mathbb{R},H)$. $W^k(AP)$ is then a Hilbert space with the norm $$ \|f\|^2_{W^k(AP)}:=\sum_{i=0}^k\|f^{(i)}\|^2_{AP(\mathbb{R}, H)}. $$ Note that, for $k\ge 0$, the $W^{k+1}(AP)$-topology is stronger than the sup-norm topology in $C_b^{k}(\mathbb{R}, H)$, the space of $k$-times continuously differentiable functions with all derivatives until order $k$ inclusively bounded (see \cite{triebel}). We will use the following lemma in the sequel. (See also \cite[Lemma 2.1]{lan2}). \begin{lemma}\label{lem2} If $F$ is a function in $W^1(AP)$ and $f= F'$, then we have \begin{equation}\label{der} a(\lambda,f)=\lambda i \cdot a(\lambda,F). \end{equation} \end{lemma} \begin{proof} If $\lambda \neq 0$, using integration by part we have \begin{align*} \frac{1}{2T}\int_{-T}^Te^{-i\lambda s}f(s) ds &= \frac{1}{2T}F(t)e^{-i\lambda t}|_{-T}^T +\frac{i\lambda }{2T}\int_{-T}^TF(s)e^{-i\lambda s}ds\\ &= \frac{F(T)e^{-i\lambda T}-F(-T)e^{i\lambda T}}{2T} +i\lambda \frac{1}{2T}\int_{-T}^TF(s)e^{-i\lambda s}ds. \end{align*} Let $T\to \infty$, and note that $F(t)$ is bounded, we have \eqref{der}. If $\lambda =0$, then $$ a(0,f)=\lim_{T\to \infty}\frac{1}{2T}\int_{-T}^Tf(s) ds = \lim_{T\to \infty}\frac{F(T)-F(-T)}{2T}=0, $$ which also satisfies \eqref{der}. \end{proof} Finally, for a linear and closed operator $A$ in a Hilbert space $H$, we denote the domain, the range, the spectrum and the resolvent set of $A$ by $D(A)$, $Range(A)$, $\sigma (A)$ and $\varrho(A)$, respectively. \section{Almost periodic mild solutions of differential equations} We now turn to the differential equation \begin{equation} \label{high} u^{(n)}(t)=Au(t)+f(t), \quad t\in \mathbb{R}, \end{equation} where $n\in \mathbb{N}^+$ and $A$ is a linear and closed operator on $H$. First we define two types of solutions to Equation \eqref{high}. Let $I: C(\mathbb{R}, H) \to C(\mathbb{R}, H)$ be the operator defined by $If(t): = \int_0^tf(s)ds$ and $I^{n} f:= I(I^{n-1}f)$. \begin{definition}\label{def3.1} \rm (a) We say that $u: \mathbb{R} \to H$ is a classical solution of \eqref{high}, if $u $ is $n$-times continuously differentiable , $u(t)\in D(A)$ and \eqref{high} is satisfied for all $t\in \mathbb{R}$. (b) For $f\in C(\mathbb{R}, H)$, a continuous function $u$ is called a mild solution of \eqref{high}, if $I^{n}u(t)\in D(A)$ and there exist $n$ points $ v_0$, $v_1$, \dots , $v_{n-1}$ in $H$ such that \begin{equation}\label{mild} u(t) = \sum_{j=0}^{n-1}\frac{t^j}{j!}v_j + AI^{n}u(t) + I^{n}f(t) \end{equation} for all $t\in \mathbb{R}$. \end{definition} \noindent{\bf Remark.} Using the standard argument, we can prove the following: \begin{itemize} \item[(i)] If a mild solution $u$ is $m$ times differentiable, $0\leq m < n$ , then $v_i$, ($i= 0,1,\dots , m$), are the initial values, i.e. $u(0)= v_0$, $u'(0)=v_1$, \dots , and $u^{(m)}(0)=v_{m}$. \item[(ii)] If $n=1$ and $A$ is the generator of a $C_0$ semigroup $T(t)$, then a continuous function $u:\mathbb{R}\to E$ is a mild solution of \eqref{high} if and only if it has the form $$u(t) =T(t-s)u(s)+\int_s^tT(t-r)f(r)dr$$ for $t\ge s$. \item[(iii)] If $u$ is a bounded mild solution of \eqref{high} corresponding to a bounded inhomogeneity $f$ and $\phi \in L^1(\mathbb{R}, E)$ then $u*\phi$ is a mild solution of \eqref{high} corresponding to $f*\phi$. \end{itemize} The mild solution to \eqref{high} defined by \eqref{mild} is really an extension of classical solution in the sense that every classical solution is a mild solution and conversely, if a mild solution is $n$-times continuously differentiable, then it is a classical solution. That statement is actually contained in the following lemma (see also \cite{lan1}). \begin{lemma}\label{ntime} Suppose $0\le m\le n$ and $u$ is a mild solution of {\rm \eqref{high}}, which is $m$-times continuously differentiable. Then for all $t\in \mathbb{R}$ we have $I^{n-m}u(t)\in D(A)$ and \begin{equation} \label{deri} u^{(m)}(t) = \sum_{j=m}^{n-1}\frac{t^{j-m}}{(j-m)!}v_j +AI^{n-m}u(t) +I^{n-m}f(t), \end{equation} where $v_m, \dots , v_{n-1}$ are given in Definition \ref{def3.1}(b). \end{lemma} \begin{proof} If $m=0$, then \eqref{deri} coincides with \eqref{mild}. We prove for $m=1$: Let $v(t):= AI^nu(t)$. Then, by \eqref{mild}, $v$ is continuously differentiable and $$ v'(t)= u'(t)- \sum_{j=1}^{n-1}\frac{t^{j-1}}{(j-1)!}v_j-I^{n-1}f(t). $$ Let $h>0$ and put $$ v_h :=\frac{1}{h}\int_t^{t+h}I^{n-1}u(s)ds. $$ Then $v_h\to (I^{n-1}u)(t)$ for $h\to 0$ and \begin{align*} \lim_{h\to 0}Av_h &=\lim_{h\to 0}\frac{1}{h} \Big(A\int_0^{t+h}I^{n-1}u(s)ds -A\int_0^t I^{n-1}u(s)ds\Big)\\ &= \frac{1}{h}(v(t+h)-v(t))\\ &= v'(t). \end{align*} Since $A$ is a closed operator, we obtain that $I^{n-1}u(t)\in D(A)$ and $$ AI^{n-1}u(t) =u'(t)- \sum_{j=1}^{n-1} \frac{t^{j-1}}{(j-1)!}v_j-I^{n-1}f(t), $$ from which \eqref{deri} with $m=1$ follows. If $m>1$, we obtain \eqref{deri} by repeating the above process $(m-1)$ times. \end{proof} In particular, if the mild solution $u$ is $n$-times continuously differentiable, then \eqref{deri} becomes $u^{(n)}(t)=Au(t)+f(t)$; i.e. $u$ is a classical solution of \eqref{high}. We now consider the mild solutions of \eqref{high}, which are $(n-1)$ times continuously differentiable. The following proposition describes the connection between the Bohr transforms of such solutions and those of $f(t)$. \begin{proposition}\label{prop1} Suppose $A$ is a linear and closed operator on $H$, $f\in AP(\mathbb{R},H)$ and $u$ is an almost periodic mild solution of \eqref{high}, which belongs to $C_b^{n-1}(\mathbb{R}, H)$. Then \begin{equation}\label{fourier} [(\lambda i)^n -A]a(\lambda,u) = a(\lambda,f) \end{equation} for every $\lambda\in \mathbb{R}$. \end{proposition} \begin{proof} Suppose $u$ is an almost periodic mild solution of \eqref{high}, which belongs to $C_b^{n-1}(\mathbb{R}, H)$ and $\lambda $ is a real number. Using \eqref{deri} with $m=n-1$ we have \begin{equation} \label{n-1} u^{(n-1)}(t) = u^{(n-1)}(0)+AIu(t) +If(t). \end{equation} For $\lambda \neq 0$, multiplying each side of \eqref{n-1} with $e^{-i\lambda t}$ and taking definite integral from $-T$ to $T$ on both sides, we have \begin{equation} \label{f1} \begin{aligned} \int_{-T}^Te^{-i\lambda t}u^{(n-1)}(t)dt &= \int_{-T}^Te^{-i\lambda t}u^{(n-1)}(0)dt +A\int_{-T}^Te^{-i\lambda t}\int_0^tu(s)\,ds\,dt\\ &\quad +\int_{-T}^Te^{-i\lambda t}\int_0^tf(s)\,ds\,dt. \end{aligned} \end{equation} Here we used the fact that $\int_{a}^b Au(t)dt=A\int_a^bu(t)dt$ for a closed operator $A$. It is easy to see that $$ \int_{-T}^Te^{-i\lambda t}u^{(n-1)}(0)dt =-\frac{e^{-i\lambda T}u^{(n-1)}(0) -e^{i\lambda T}u^{(n-1)}(0)}{i\lambda } $$ and, applying integration by part for any integrable function $g(t)$, we have \begin{equation} \label{f2} \begin{aligned} \int_{-T}^Te^{-i\lambda t}\int_0^tg(s)\,ds\,dt &=-\frac{1}{i\lambda}e^{-i\lambda t}\int_0^tg(s)ds|_{-T}^T +\frac{1}{i\lambda }\int_{-T}^Te^{-i\lambda t}g(t)dt\\ &= -\frac{1}{i\lambda}e^{-i\lambda T}\int_0^T g(t)dt +\frac{1}{i\lambda} e^{i\lambda T}\int_0^{-T}g(t)dt\\ &\quad +\frac{1}{i\lambda }\int_{-T}^Te^{-i\lambda t}g(t)dt. \end{aligned} \end{equation} Using \eqref{f2} for $g(t)=u(t)$ and $g(t)=f(t)$ in \eqref{f1}, respectively, we have \begin{equation} \label{d2} \begin{aligned} & \frac{1}{2T}\int_{-T}^Te^{-i\lambda t}u^{(n-1)}(t)dt\\ &= -\frac{e^{-i\lambda T}u^{(n-1)}(0) -e^{i\lambda T}u^{(n-1)}(0)}{2 i\lambda T} -\frac{e^{-i\lambda T}}{2 i\lambda T}\Big(A\int_0^Tu(t)dt +\int_0^Tf(t)dt\Big)\\ &\quad +\frac{e^{i\lambda T}}{2 i\lambda T}\Big(A\int_0^{-T}u(t)dt +\int_0^{-T}f(t)dt\Big)\\ &\quad + \frac{1}{i\lambda 2T}\Big(A\int_{-T}^Te^{-i\lambda t}u(t)dt +\int_{-T}^Te^{-i\lambda t}f(t)dt\Big) \\ &= I_1 +I_2+I_3, \end{aligned} \end{equation} where $$ I_1= -\frac{e^{-i\lambda T}u^{(n-1)}(0) -e^{i\lambda T}u^{(n-1)}(0)}{2 i\lambda T}\to 0 $$ as $T\to \infty$; \begin{align*} I_2&= - \frac{e^{-i\lambda T}}{2i\lambda T}\Big(A\int_0^Tu(t)dt +\int_0^Tf(t)dt\Big) +\frac{e^{i\lambda Ti\lambda}}{2 i\lambda T} \Big(A\int_0^{-T}u(t)dt +\int_0^{-T}f(t)dt\Big)\\ &= -\frac{e^{-i\lambda T}}{2i\lambda T}\Big(u^{(n-1)}(T)-u^{(n-1)}(0)\Big) +\frac{e^{i\lambda T}}{2i\lambda T}\Big(u^{(n-1)}(-T)-u^{(n-1)}(0)\Big)\\ &\quad \to 0 \quad\text{as }T\to \infty, \end{align*} and \begin{equation}\label{f3} I_3= \frac{1}{i\lambda}\Big(\frac{1}{2T}A\int_{-T}^Te^{-i\lambda t}u(t)dt +\frac{1}{2T}\int_{-T}^Te^{-i\lambda t}f(t)dt\Big). \end{equation} Let $u_T:= \frac{1}{2T}\int_{-T}^Te^{-i\lambda t}u(t)dt $. It is clear that \begin{equation}\label{close1} \lim_{T\to \infty}u_T= a(\lambda, u) \end{equation} and from \eqref{f3} we have \begin{equation} \label{close2} \begin{aligned} Au_T&= \frac{1}{2T}A\int_{-T}^Te^{-i\lambda t}u(t)dt\\ &= i\lambda I_3-\frac{1}{2T}\int_{-T}^Te^{-i\lambda t}f(t)dt\\ &= i\lambda\Big(\frac{1}{2T}\int_{-T}^{T}u^{(n-1)}(t)dt-I_1-I_2\Big) -\frac{1}{2T}\int_{-T}^Te^{-i\lambda t}f(t)dt\\ &\quad \to i\lambda a(\lambda , u^{(n-1)})-a(\lambda, f) \quad\text{as } T\to \infty . \end{aligned} \end{equation} Since $A$ is a closed operator, from \eqref{close1} and \eqref{close2}, we obtain $a(\lambda, u) \in D(A)$ and \[ Aa(\lambda, u) = i\lambda a(\lambda, u^{(n-1)})-a(\lambda, f) = (i\lambda)^na(\lambda, u)-a(\lambda, f), \] from which \eqref{fourier} follows. Next, if $\lambda =0$, using Formula \eqref{n-1}, we have \begin{gather*} u^{(n-1)}(T)=v_{n-1}+A\int_0^Tu(t)dt +\int_0^T f(t)dt,\\ u^{(n-1)}(-T)=v_{n-1}+A\int_0^{-T}u(t)dt +\int_0^{-T} f(t)dt. \end{gather*} Hence, \begin{equation} \label{sub} \frac{u^{(n-1)}(T)-u^{(n-1)}(-T)}{2T} =A\frac{1}{2T}\int_{-T}^Tu(t)dt +\frac{1}{2T}\int_{-T}^T f(t)dt. \end{equation} Let $u_T= \frac{1}{2T}\int_{-T}^Tu(s)ds$. Then $\lim_{t\to \infty}u_T=a(0, u)$, and by \eqref{sub}, \begin{align*} Au_T&= \frac{1}{2T}A\int_{-T}^Tu(s)ds\\ &= \frac{u^{(n-1)}(T) -u^{(n-1)}(-T)}{2T} - \frac{1}{2T}\int_{-T}^Tf(s)ds \to -a(0, f) \quad\text{as } T\to \infty . \end{align*} Again, since $A$ is a closed operator, it implies $a(0, u) \in D(A)$ and $Aa(0, u) =-a(0, f)$, from which \eqref{fourier} follows, and this completes the proof. \end{proof} Note that Proposition \ref{prop1} also holds in a Banach space. We are now going to look for conditions that Equation \eqref{high} has an almost periodic mild solution. \begin{theorem}\label{theorem1} Suppose $A$ is a linear and closed operator and $f$ is a function in $AP(\mathbb{R}, H)$. Then the following statements are equivalent \begin{itemize} \item[(i)] Equation \eqref{high} has an almost periodic mild solution, which is in $W^n(AP)$; \item[(ii)] For every $\lambda\in \sigma(f)$, $a(\lambda, f)\in \mathop{\rm Range}((i\lambda)^n-A)$ and there exists a set $\{x_{\lambda}\}_{\lambda\in \sigma(f)}$ in $H$ satisfying $((i\lambda)^n-A)x_{\lambda}=a(\lambda,f)$, for which the following inequalities \begin{equation}\label{ab} \sum_{\lambda\in \sigma(f)} |\lambda |^{2k}\|x_{\lambda}\|^2 < \infty \end{equation} hold for $k=0, 1, 2, \dots , n$. \end{itemize} \end{theorem} \begin{proof} (i) $\Rightarrow$ (ii): Let $u(t)$ be an almost periodic solution to \eqref{high}, which is in $W^n(AP)$. By Proposition \ref{prop1}, $((i\lambda)^n -A)a(\lambda, u)= a(\lambda, f)$. Hence $a(\lambda, f)\in Range((i\lambda)^n-A)$ for all $\lambda \in \sigma(f)$. Put now $x_{\lambda}:= a(\lambda, u)$ for $\lambda \in \sigma(f)$. Then it satisfies $((i\lambda)^n -A)x_{\lambda}= a(\lambda, f)$. Moreover, by Lemma \ref{der}, $(i\lambda)^k x_{\lambda}=a(\lambda, u^{(k)})$. Hence, for $0\le k\le n$ we have \begin{align*} \sum_{\lambda\in \sigma(f)} |\lambda | ^{2k}\|x_{\lambda}\|^2 &= \sum_{\lambda\in \sigma(f)} |a(\lambda, u^{(k)})|^2\\ &\le \sum_{\lambda\in \sigma(f)\cup \sigma(u^{(k)})} |a(\lambda, u^{(k)})|^2\\ & =\sum_{\lambda\in \sigma(u^{(k)})} |a(\lambda, u^{(k)})|^2\\ &= \|u^{(k)}\|_{AP}^2 , \end{align*} from which \eqref{ab} follows. (ii) $\Rightarrow$ (i): Let $\{x_{\lambda}\}_{\lambda\in \sigma(f)}$ be a set in $H$ satisfying $((i\lambda)^n-A)x_{\lambda}=a(\lambda,f)$, for which \eqref{ab} holds. Put \[ f_N(t):= \sum_{\lambda\in \sigma(f), |\lambda|