\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 85, 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 (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/85\hfil Energy estimate for wave equations] {Energy estimate for wave equations with coefficients in some Besov type class} \author[S. Tarama\hfil EJDE-2007/85\hfilneg] {Shigeo Tarama} \address{Lab. of Applied Mathematics, Graduate School of Engineering, Osaka City University, Osaka 558-8585, Japan} \email{starama@mech.eng.osaka-cu.ac.jp} \thanks{Submitted February 14, 2007. Published June 6, 2007.} \subjclass[2000]{35L05, 16D10} \keywords{Wave equation; energy estimate; non regular coefficients} \begin{abstract} In this paper, we obtain an energy estimate for wave equations with coefficients satisfying Besov type conditions. We give an example of a wave equation with continuous and nowhere differentiable coefficients for which the $L^2$ estimate holds. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{example}[theorem]{Example} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Consider a wave equation on $[0,T]\times\mathbb{R}$: \begin{equation} Lu=\partial_t^2u-a(t)\partial_x^2u \end{equation} with a positive coefficient $a(t)\ge \delta_0$ with $\delta_0>0$. It is well known that, if $a(t)$ is Lipschitz continuous, then we have the energy estimate \begin{equation}\label{est-1} \sum_{0\le j+k\le 1}\|\partial_t^j\partial_x^{k}u(t,\cdot )\| \le C\Bigl(\sum_{0\le j+k\le 1}\|\partial_t^j\partial_x^{k}u(0,\cdot )\| +\int_0^t\|Lu(s,\cdot)\|\,ds \Bigr) \end{equation} (see for example \cite[Ch. IX]{H}). Here $\|\cdot\|$ denotes $L^2$ norm. Colombini, De Giorgi and Spagnolo \cite{CDS} (see also \cite{DST}) have shown that the estimate \eqref{est-1} is still valid if the coefficient $a(t)$ has a bounded variation, that is, in the integral form, there exists a constant $C\ge 0$ such that we have \begin{equation}\label{bv} \int_0^{T-\varepsilon}|a(t+\varepsilon)-a(t)|\,dt\le C\varepsilon\quad (0< \varepsilon\le T/2). \end{equation} Furthermore, in the same paper, they have shown that if $a(t)$ satisfies \begin{equation}\label{loglip} \int_0^{T-\varepsilon}|a(t+\varepsilon)-a(t)|\,dt \le C\varepsilon(|\log \varepsilon|+1)\quad (0< \varepsilon\le T/2) \end{equation} with a constant $C\ge 0$, then the Cauchy problem for $L$ is $C^{\infty}$ well posed. According to Yamazaki \cite{YZ}, we have the estimate \eqref{est-1} when $a(t)\in C^2((0,T])$ satisfies $|a(t)|+|ta'(t)|+|t^2a''(t)|\le C$ on $(0,T]$ (see also \cite{HRR}). Then we see that the estimate \eqref{est-1} is valid for $L$ with some coefficient $a(t)$ whose total variation is not finite, for example $a(t)=2+\sin(\log t)$. In this paper we introduce an integral version of the condition $|a(t)|+|ta'(t)|+|t^2a''(t)|\le C$ so that the estimate \eqref{est-1} holds still for $L$ with the coefficient $a(t)$ satisfying such a condition. Namely we show the following. When the coefficient $a(t)$ is a bounded measurable function on $[0,T]$ and satisfies: there exists a constant $C\ge 0$ such that we have \begin{equation}\label{cond1} \int_{\varepsilon}^{T-\varepsilon}|a(t+\varepsilon) +a(t-\varepsilon)-2a(t)|\,dt\le C\varepsilon\quad (0< \varepsilon\le T/2), \end{equation} then the estimate \eqref{est-1} holds. Using the same method, we show also the following. The Cauchy problem for $L$ is $C^{\infty}$ well posed if the coefficient $a(t)$ is a bounded measurable function on $[0,T]$ and satisfies the following: There exists a constant $C\ge 0$ such that \begin{equation}\label{cond2} \int_{\varepsilon}^{T-\varepsilon}|a(t+\varepsilon) +a(t-\varepsilon)-2a(t)|\,dt\le C\varepsilon(|\log \varepsilon|+1)\quad (0< \varepsilon\le T/2). \end{equation} Note that the boundedness of $a(t)$ and the estimate \eqref{cond1} imply \begin{equation}\notag%\label{cond11} \int_{0}^{T-\varepsilon}|a(t+\varepsilon)-a(t)|^2\,dt\le C\varepsilon\quad (0<\varepsilon\le T/2) \end{equation} with some constant $C$. While from the boundedness of $a(t)$ and \eqref{cond2} we obtain \begin{equation}\notag%\label{cond21} \int_{0}^{T-\varepsilon}|a(t+\varepsilon)-a(t)|^2\,dt \le C\varepsilon(|\log \varepsilon|+1)\quad (0<\varepsilon\le T/2) \end{equation} with some constant $C$ (see the next section). We remark that Colombini, Del Santo and Reissig \cite{CDR} (see also \cite{HR1} and \cite{HRR}) have shown that the Cauchy problem for $L$ is $C^{\infty}$ well posed when $a(t)$ satisfies $|a(t)|+|(t\log t) a'(t)|+|(t\log t)^2a''(t)|\le C$ on $(0,T]$. For example the Cauchy problem for $L$ with $a(t)=2+\sin(|\log t|^2)$ is $C^{\infty}$ well posed but this function $a(t)$ does not satisfy the condition \eqref{cond2}. Nonetheless we can find some positive function $a(t)$ which satisfies the estimate \eqref{cond2} with the right hand side replaced with $ C\varepsilon(|\log \varepsilon|+1)^{1+\delta}$ ($\delta>0$), so that the Cauchy problem for $L$ is not $C^{\infty}$ well posed. Indeed Colombini and Lerner \cite{CL} have given an example of a positive function $a(t)$ such that $a(t)$ satisfies $\sup_{\varepsilon\in(0,1],t\in[0,1]}| a(t+\varepsilon)-a(t)|/(\varepsilon(|\log \varepsilon|+1)^{1+\delta})<\infty$ (for any $\delta>0$) but the Cauchy problem on $[0,1]\times\mathbb{R}$ for $\partial_t^2-a(t)\partial_x^2$ is not $C^{\infty}$ well posed. In the next section, in order to study properties of bounded functions that satisfying \eqref{cond1} or \eqref{cond2}, we define the function spaces $Z_{\gamma}(I)$ and show some properties of functions in such spaces. Some properties of examples are discussed in the appendix. In the third section, we state and prove the main theorems. We use the following notation. Let $L^2(\mathbb{R}^d)$ or $L^2$ denote the space of all square integrable functions on $\mathbb{R}^d$ with the norm $\|\cdot\|$ given by $\|f(\cdot)\|^2=\int |f(x)|^2\,dx$. For $s\in\mathbb{R}$ let $H^s$ denote the space that consists of functions $f(x)$ on $\mathbb{R}^d$ satisfying $\int (1+|\xi|^2)^s|\hat{f}(\xi)|^2\,d\xi<\infty$ where $\hat{f}(\xi)$ is the Fourier transform of $f(x)$ and $\|\cdot\|_s$ be its norm, that is, $\|f(\cdot)\|_s^2=\int (1+|\xi|^2)^{s}|\hat{f}(\xi)|^2\,d\xi$. We set $H^{\infty}=\bigcap_{s\in \mathbb{R}}H^s$. For $X=H^s$, $H^{\infty}$ or $C^{\infty}(\mathbb{R}^d)$, the space of indefinitely differentiable functions on $\mathbb{R}^d$, and $T>0$, we denote by $L^1([0,T],X)$ the space of $X$-valued integrable functions on $[0,T]$ and by $C^{j}([0,T],X)$ with an integer $j$ the space of $X$-valued $j$-times continuously differentiable functions on $[0,T]$. We use also the standard notation of multi-index. We use $C$ or $C$ with some suffix in order to denote a non-negative constant that may be different line by line. \section{Space $Z_{\gamma}(I)$} Let $I=(t_0, t_1)\subset \mathbb{R}$ with $t_00$, we have \begin{gather} \int_{t_0}^{t_1}|f(t+\varepsilon)-2f(t)+f(t-\varepsilon)|\,dt \le C\varepsilon (\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}\label{bs-2}\\ \int_{t_0}^{t_1}(|f(t+\varepsilon)-f(t)|^2+|f(t)-f(t-\varepsilon)|^2)\,dt \le C\varepsilon(\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}.\label{bs-3} \end{gather} where the constant $C$ depends only on $\|f\|_{Z_{\gamma}(I)}$. Indeed if $\varepsilon\ge (t_1-t_0)/2$, we see that the right hand side of \eqref{bs-2} is not larger than $8\varepsilon\|f\|_{L^{\infty}(I)}$. While, in the case of $\varepsilon< (t_1-t_0)/2$, we see that, on the right hand side of \eqref{bs-2}, the integral on the interval $[t_0+\varepsilon,t_1-\varepsilon]$ is not larger than $\varepsilon(\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}\|f\|_{Z_{\gamma}(I)}$ and the integral on the remainder part is not larger than $8\varepsilon\|f\|_{L^{\infty}(I)}$. Hence we have \eqref{bs-2}. Similarly we obtain \eqref{bs-3}. Now we consider the regularization of a function $f(t)$ in $Z_{\gamma}(I)$. We take the above mentioned extension $f(t)$. Let $\phi(s)$ be a smooth function on $\mathbb{R}$ satisfying $\phi(-s)=\phi(s)$, $\phi(s)\ge 0$, $\phi(s)=0$ for $|s|\ge1$ and $\int_{\mathbb{R}}\phi(s)\,ds=1$. We denote by $f_{\varepsilon}(t)$ with $\varepsilon>0$ the regularization of $f(t)$ given by \[ f_{\varepsilon}(t)=\frac{1}{\varepsilon}\int_{\mathbb{R}}\phi(\frac{t-s}{\varepsilon})f(s)\,ds. \] Then we have the following result. \begin{lemma}\label{lemma2-1} \begin{gather} \int_I|f_{\varepsilon}(t)-f(t)|\,dt \le C_1\varepsilon(\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}\label{bs-4}\\ \int_I(|f''_{\varepsilon}(t)|+|f'_{\varepsilon}(t)|^2)\,dt \le C_2(\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}/\varepsilon\label{bs-5} \end{gather} where the constants $C_1$ and $C_2$ depend on $\|f\|_{Z_{\gamma}(I)}$ and $\phi(s)$ but not on the length of the interval $I$. Furthermore, for any function $F\in C^2(\mathbb{R})$, setting $h(t)=F(f_{\varepsilon}(t))$, we have \begin{equation}\label{bs-6} \int_I(|h''(t)|+|h'(t)|^2)\,dt \le C(\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}/\varepsilon. \end{equation} Here the constant $C$ is also independent of the length of the interval $I$. \end{lemma} \begin{proof} Since $f_{\varepsilon}(t)-f(t)=\int_{\mathbb{R}} \phi(s)(f(t-\varepsilon s)-f(t))\,ds$ and $\phi(-s)=\phi(s)$, we have \begin{align*} |f_{\varepsilon}(t)-f(t)| &=|\int_{\mathbb{R}} \frac{\phi(s)+\phi(-s)}{2}(f(t-\varepsilon s)-f(t))\,ds|\\ &=\frac{1}{2}|\int_{\mathbb{R}} \phi(s)(f(t+\varepsilon s)+f(t-\varepsilon s)-2f(t))\,ds|, \end{align*} from which and from \eqref{bs-2}, we obtain \[ \int_I|f_{\varepsilon}(t)-f(t)|\,dt\le C\int_{\mathbb{R}}\phi(s)|s|\varepsilon(\log((|s|\varepsilon)^{-1}+1)+1+\gamma)^{\gamma}\,ds. \] Since $s(\log (s^{-1}+1)+1+\gamma)^{\gamma}$ is increasing, the right hand side of the estimate above is not larger than $C\varepsilon(\log (\varepsilon^{-1}+1)+1+\gamma)^{\gamma}$. Similarly, $f''_{\varepsilon}(t)=\varepsilon^{-2}\int_{\mathbb{R}} \phi''(s) (f(t-\varepsilon s)-f(t))\,ds$ and $\phi''(-s)=\phi''(s)$ imply \[ \int_I|f''_{\varepsilon}(t)|\,dt\le C(\log(\varepsilon^{-1}+1) +1+\gamma)^{\gamma}/\varepsilon. \] While it follows from $f'_{\varepsilon}(t)=\varepsilon^{-1}\int_{\mathbb{R}} \phi'(s)(f(t-\varepsilon s) -f(t))\,ds$, \eqref{bs-2} and Schwarz's inequality that \[ |f'_{\varepsilon}(t)|^2\le \varepsilon^{-2}\|\phi'(\cdot)\|_{L^1} \int_I|\phi'(s)||f(t-\varepsilon s)-f(t)|^2\,ds, \] from which and from \eqref{bs-3} we obtain the desired estimate of $\int_I|f'_{\varepsilon}(t)|^2\,dt$. Hence we have \eqref{bs-5}. We obtain \eqref{bs-6} from \eqref{bs-4} and \eqref{bs-5}. \end{proof} \begin{example}\label{exa1} \rm If $f(t)\in C^2((0,1/2])$ satisfies $|f(t)|+|f''(t)|t^2/|\log t|^{\gamma}\le C$ on $I=(0,1/2)$, then $f(t)$ belongs to $Z_{\gamma}(I)$. Indeed, if $\varepsilon0 \end{equation} (see the appendix for detail). Thus we see that $h_{1/2}(t)$ belongs to $ Z_{1}((0,1/2))$ but does not satisfy \eqref{loglip} with $T=1/2$. \end{example} \begin{example}\label{exa2} \rm Here we show that the Weierstrass function \[ w_{\gamma}(t)=\sum_{n=1}^{\infty}2^{-n}n^{\gamma}\cos{2^nt} \] with $\gamma\ge0$, that is continuous and nowhere differentiable (see for example \cite{K} ), belongs to $Z_{\gamma}((0,2\pi))$. Indeed, for any $\varepsilon \in (0,1/2)$ we have $w_{\gamma}(t)=w_{\gamma,1,\varepsilon}(t)+w_{\gamma,2,\varepsilon}(t) $ where $$ w_{\gamma,1,\varepsilon}(t)=\sum_{1\le n \le \frac{|\log \varepsilon|}{\log2}}2^{-n}n^{\gamma}\cos{2^nt} \quad\text{and}\quad w_{\gamma,2,\varepsilon}(t)=\sum_{ n> \frac{|\log \varepsilon|} {\log2}}2^{-n}n^{\gamma}\cos{2^nt}. $$ Since $|w''_{\gamma,1,\varepsilon}(t)|\le C\varepsilon^{-1}| \log \varepsilon|^{\gamma}$ and $|w_{\gamma,2, \varepsilon}(t)| \le C \varepsilon|\log \varepsilon|^{\gamma}$, then we see $|w_{\gamma}(t+\varepsilon)+w_{\gamma}(t-\varepsilon)-2w_{\gamma}(t)| \le C|\log \varepsilon|^{\gamma}\varepsilon$. Hence $w_{\gamma}(t)\in Z_{\gamma}((0,2\pi))$. We remark that $w_0(t)$ satisfies \eqref{loglip}. Indeed, in the expression above $w_0(t)=w_{0,1,\varepsilon}(t)+w_{0,2,\varepsilon}(t) $ we have $|w'_{0,1,\varepsilon}(t)|\le C|\log \varepsilon|$ and $|w_{0,2, \varepsilon}(t)|\le C \varepsilon$. Then we see $|w_0(t+\varepsilon)-w_0(t)|\le C|\log \varepsilon|\varepsilon$. \end{example} \section{Main results} Let $a_{jk}(t)$ ($j,k=1,\dots,d$) be a real-valued bounded measurable function on $(0,T)$ with $T>0$ satisfying $a_{kj}(t)=a_{jk}(t)$ and \begin{equation}\label{hyp} \sum_{j,k=1}^d a_{jk}(t)\xi_j\xi_k \ge C_0|\xi|^2 \text{ for }\xi\in \mathbb{R}^d\text{ and }t\in(0,T)\end{equation} with some positive constant $C_0>0$. Set \begin{equation} P_2(t,\partial_t,\xi)=\partial_t^2+\sum_{j,k=1}^d a_{jk}(t)\xi_j\xi_k \end{equation} where $\xi\in \mathbb{R}^d$. Then we have the following result. \begin{theorem}\label{thm1} Assume that $a_{jk}(t)\in Z_{\gamma}((0,T))$ $(j,k=1,\dots,d)$ with $\gamma\ge0$. Let $\xi\in \mathbb{R}^d$. If $u(t)\in C^1([0,T])$ satisfies $P_2(t,\partial_t,\xi)u=f(t)$ on $(0,T)$ with $f(t)\in L^1([0,T])$, then we have \begin{equation}\label{thm3-1} \begin{aligned} &(|\partial_tu(t_2)|^2+|\xi|^2|u(t_2)|^2)^{1/2}\\ &\le C_1e^{C_2(\log(|\xi|+1)+1+\gamma)^{\gamma}} \bigl((|\partial_tu(t_1)|^2+|\xi|^2|u(t_1)|^2)^{1/2}+\int_{t_1}^{t_2} |f(t)|\,dt\bigr) \end{aligned} \end{equation} for any $0\le t_1\le t_2\le T$. Here constants $C_1$ and $C_2$ depend on $C_0$ of \eqref{hyp} and $Z_{\gamma}$-norm of coefficients $a_{jk}(t)$ but not on the length of the interval $[0,T]$. \end{theorem} Before presenting the proof of Theorem above, we remark the following well known result. Let $L=\partial^2_t+a^2(t)\rho^2$ where $a(t)$ is smooth and positive and $\rho>0$. Noting that $(\partial_t-ia(t)\rho-\frac{a'(t)}{2a(t)}) (\partial_t+i a(t)\rho+\frac{a'(t)}{2a(t)})$ and $(\partial_t+i a(t)\rho-\frac{a'(t)}{2a(t)}) (\partial_t- ia(t)\rho+\frac{a'(t)}{2a(t)})$ are equal to $$ L-(\frac{a'(t)}{2a(t)})^2+(\frac{a'(t)}{2a(t)})', $$ we consider the energy $$ E(u)=\frac{1}{a(t)}|\partial_t u+\frac{a'(t)}{2a(t)}u|^2+a(t)\rho^2|u|^2. $$ Then we have \begin{equation}\label{energ} \frac{d}{dt}E(u)=\frac{2}{a(t)}{\rm Re}\bigl(\overline{\ (\partial_t u +\frac{a'(t)}{2a(t)}u)\ }(Lu-Ru)\bigr) \end{equation} where $R=(\frac{a'(t)}{2a(t)})^2-(\frac{a'(t)}{2a(t)})'$. \begin{proof}[Proof of Theorem \ref{thm1}] If $\xi=0$, $P_2u=f(t)$ is equal to $\partial_t^2u=f(t)$. Then we have immediately \eqref{thm3-1}. In the following, we assume $\xi\ne0$. First we extend the coefficients $a_{jk}(t)$ on $\mathbb{R}$ so that $\|a_{jk}(t)\|_{L^{\infty}(\mathbb{R})}= \|a_{jk}(t)\|_{L^{\infty}((0,T))}$ and the estimate \eqref{hyp} still holds for $t\in\mathbb{R}$. Then we consider the regularization $a_{jk,\varepsilon}(t)$ of $a_{jk}(t)$ given by $\int_{\mathbb{R}}\phi((t-s)/\varepsilon)a_{jk}(s)\,ds/\varepsilon$ with $\varepsilon>0$ using a non-negative, even and smooth function $\phi(s)$ as described in the section 2. Then we see that \eqref{hyp} with $a_{jk,\varepsilon}(t)$ in the place of $a_{jk}(t)$ holds. Then we define $a(t,\xi,\varepsilon)$ by \[ a(t,\xi,\varepsilon)=|\xi|^{-1}\bigl( \sum_{j,k=1}^d a_{jk,\varepsilon}(t)\xi_j\xi_k\bigr)^{1/2} \quad\text{for } \xi\in\mathbb{R}^d\setminus\{0\}. \] We have \begin{equation}\label{est3-11} C_1\ge a(t,\xi,\varepsilon)\ge \sqrt{C_0} \end{equation} with constants $C_0$ appearing in \eqref{hyp} and $C_1$ depending only on $\|a_{jk}(\cdot)\|_{L^{\infty}((0,T))}$. We see from \eqref{bs-6}, that \begin{equation}\label{est3-2} \int_0^T(|\partial_ta(t,\xi,\varepsilon)|^2 +|\partial^2 _ta(t,\xi,\varepsilon)|)\,dt \le C_1\varepsilon^{-1}(|\log(\varepsilon^{-1}+1)|+1+\gamma)^{\gamma} \end{equation} for any $\varepsilon>0$. Furthermore Lemma \ref{lemma2-1} implies that \begin{equation}\label{est3-3} \int_0^T |a(t,\xi,\varepsilon)^2|\xi|^2-\sum_{j,k=1}^d a_{jk}(t)\xi_j\xi_k|\,dt \le C_2\varepsilon(|\log(\varepsilon^{-1}+1)|+1+\gamma)^{\gamma}|\xi|^2. \end{equation} Here the constants above $C_1$ and $C_2$ may depend on $Z_{\gamma}$-norm of $a_{jk} (t)$ and the constant $C_0$ of \eqref{hyp} but not on the length of interval $[0,T]$. Assume that $u(t)\in C^1([0,T])$ satisfies $\partial_t^2u+\sum_{j,k=1}^d a_{jk}(t)\xi_j\xi_ku=f(t)$ on $(0,T)$ with $\xi\in \mathbb{R}^d\setminus\{0\}$ and $f(t)\in L^1([0,T])$. Let \begin{equation} E_{\varepsilon}(t)=\frac{1}{a(t,\xi,\varepsilon)} |\partial_t u+\frac{\partial_ta(t,\xi,\varepsilon)}{2a(t,\xi,\varepsilon)}u|^2 +a(t,\xi,\varepsilon)|\xi|^2|u|^2. \end{equation} Then it follows from \eqref{energ} that \begin{equation}\label{est3-4} \frac{d}{dt}E_{\varepsilon}(t)=\frac{2}{a(t,\xi,\varepsilon)}{\rm Re} \bigl(\overline{\ (\partial_t u+\frac{\partial_ta(t,\xi,\varepsilon)} {2a(t,\xi,\varepsilon)}u)\ }\ (L_{\varepsilon}u-R_{\varepsilon}u)\bigr) \end{equation} where $L_{\varepsilon}u=\partial_t^2-a(t,\xi,\varepsilon)^2|\xi|^2u$ and $R_{\varepsilon}=(\frac{\partial_ta(t,\xi, \varepsilon)}{2a(t,\xi,\varepsilon)})^2-\partial_t(\frac{\partial_t a(t,\xi,\varepsilon)}{2a(t,\xi,\varepsilon)})$. Note that \[ |L_{\varepsilon}u|\le |a(t,\xi,\varepsilon)^2|\xi|^2-\sum_{j,k=1}^d a_{jk}(t)\xi_j\xi_k||u|+|f(t)| \] and \[ |R_{\varepsilon}u|\le C(|\partial_ta(t,\xi,\varepsilon)|^2 +|\partial^2_ta(t,\xi,\varepsilon)|). \] Since $$ |(\partial_t u+\frac{\partial_ta(t,\xi,\varepsilon)}{2 a(t,\xi,\varepsilon)}u)||u|\le \frac{1}{2|\xi|}E_{\varepsilon}(t), $$ we see that \begin{equation} \left|\frac{d}{dt}E_{\varepsilon}(t) \right| \le 2 C(t,\xi,\varepsilon)E_{\varepsilon}(t)+E_{\varepsilon}(t)^{1/2}2C_0^{-1/4}|f(t)| \end{equation} where \begin{align*} C(t,\xi,\varepsilon) &=\frac{1}{2} C_0^{-1/2}\Bigl( |\bigl(a(t,\xi,\varepsilon)^2| \xi|^2-\sum_{j,k=1}^d a_{jk}(t)\xi_j\xi_k\bigr)|\\ &\quad + C|\partial_ta(t,\xi,\varepsilon)|^2+|\partial^2_ta(t,\xi,\varepsilon)| \Bigr)|\xi|^{-1}. \end{align*} Hence for any positive constant $\delta>0$, we have \[ \big|\frac{d}{dt}\bigl(E_{\varepsilon}(t)+\delta\bigr) \big| \le 2 C(t,\xi,\varepsilon)\bigl(E_{\varepsilon}(t)+\delta\bigr) +\bigl(E_{\varepsilon}(t)+\delta\bigr)^{1/2}2C_0^{-1/4}|f(t)|, \] from which we obtain \[ \big|\frac{d}{dt}\bigl(E_{\varepsilon}(t)+\delta\bigr)^{1/2} \big| \le C(t,\xi,\varepsilon)\bigl(E_{\varepsilon}(t)+\delta\bigr)^{1/2} +C_0^{-1/4}|f(t)|. \] Then we see that, for $0\le t_1\le t_2\le T$, \begin{equation*} (E_{\varepsilon}(t_2)+\delta\bigr)^{1/2}\le e^{\int_{t_1}^{t_2}C(t,\xi,\varepsilon) \,dt}(E_{\varepsilon}(t_1)+\delta\bigr)^{1/2}+\int_{t_1}^{t_2}e^{ \int_t^{t_2}C(s,\xi,\varepsilon)\,ds} C_0^{-1/4}|f(t)|\,dt. \end{equation*} It follows from \eqref{est3-2} and \eqref{est3-3} that \[ \int_0^TC(t,\xi,\varepsilon) \,dt\le C(\varepsilon |\xi|+\frac{1}{\varepsilon|\xi|})(\log(\varepsilon^{-1}+1)+1+\gamma)^{\gamma}. \] Now picking $\varepsilon=1/|\xi|$, we obtain \begin{equation} \bigl(E_{1/|\xi|}(t_2)+\delta\bigr)^{1/2}\le e^{C(\log(|\xi|+1)+1+\gamma)^{\gamma} }\Bigl((\bigl(E_{1/|\xi|}(t_1)+\delta\bigr)^{1/2}+\int_{t_1}^{t_2}C_0^{-1/4}|f(t)|\,dt\Bigr). \end{equation} By taking $\delta\rightarrow 0$, we obtain \[ \bigl(E_{1/|\xi|}(t_2)\bigr)^{1/2}\le e^{C(\log(|\xi|+1)+1+\gamma)^{\gamma} }\Bigl((\bigl(E_{1/|\xi|}(t_2))^{1/2}+\int_{t_1}^{t_2}C_0^{-1/4}|f(t)|\,dt\Bigr). \] Since $|\partial_ta_{jk,\varepsilon}(t)|\le C\varepsilon^{-1}\|a_{jk}(\cdot)\|_{L^{\infty}((0,T))}$ and $\varepsilon=1/|\xi|$, we see from \eqref{est3-11} that there exists a constant $C>0$ such that \[ C(|\partial_tu(t)|^2+|\xi|^2|u(t)|^2)\le E_{1/|\xi|}(t) \le C^{-1}(|\partial_tu(t)|^2+|\xi|^2|u(t)|^2) \] for any $t\in[0,T]$ and any $\xi\in \mathbb{R}^d\setminus\{0\}$. Then we obtain the desired estimate \eqref{thm3-1}. \end{proof} Since $u(t_2)=u(t_1)+i\int_{t_1}^{t_2}\partial_tu(t)\,dt$, from \eqref{thm3-1} we obtain \begin{equation}\label{thm3-2} \begin{aligned} &(|\partial_tu(t_2)|^2+(|\xi|^2+1)|u(t_2)|^2)^{1/2}\\ &\le C_Te^{C_2(\log(|\xi|+1)+1+\gamma)^{\gamma}} \bigl((|\partial_tu(t_1)|^2+(|\xi|^2+1)|u(t_1)|^2)^{1/2} +\int_{t_1}^{t_2}|f(t)|\,dt\bigr) \end{aligned} \end{equation} where the constant $C_T$ may depend on the length of the interval $[0,T]$. Now consider $u(t,x)\in C^2([0,T],H^{\infty})$. Let $$ f(t,x)=\partial_t^2u(t,x)-\sum_{j,k=1}^d a_{jk}(t)\partial_{x_j} \partial_{x_k}u(t,x). $$ Then we have $P_2\hat{u}(t,\xi)=\hat{f}(t,\xi)$ where $\hat{u}(t,\xi)$ and $\hat{f}(t,\xi)$ are the Fourier transform of $u(t,x)$ and $f(t,x)$ in variables $x$ respectively. Then from \eqref{thm3-2}, we obtain \begin{equation}\label{thm3-2-1} \begin{aligned} &(|\partial_t\hat{u}(t_2,\xi)|^2+(|\xi|^2+1)|\hat{u}(t_2,\xi)|^2)^{1/2}\\ &\le C_Te^{C_2(\log(|\xi|+1)+1+\gamma)^{\gamma}} \bigl((|\partial_t\hat{u}(t_1,\xi)|^2+(|\xi|^2+1)|\hat{u}(t_1,\xi)|^2)^{1/2} \\ &\quad +\int_{t_1}^{t_2}|\hat{f}(t,\xi)|\,dt\bigr) \quad \text{for $0\le t_10$ satisfying $a_{kj}(t)=a_{jk}(t)$ and \eqref{hyp}. Let $L$ be a second order hyperbolic operator given by \[ L=\partial_t^2-\sum_{j,k=1}^da_{jk}(t)\partial_{x_{j}}\partial_{x_{k}}. \] If $a_{jk}(t)\in Z_0((0,T))$ $(j,k=1,\dots,d)$, then we have the estimate \begin{equation}\label{thm33-l2} \sum_{l+|\alpha|\le 1}\|\partial^l_t\partial_x^{\alpha}u(t_2,\cdot)\| \le C(\sum_{l+|\alpha|\le 1}\|\partial^l_t\partial_x^{\alpha}u(t_1\cdot)\| +\int_{t_1}^{t_2}\|Lu(s,\cdot)\|\,ds) \end{equation} for any $0\le t_1\le t_2\le T$. Here $u(t,x)\in \bigcap_{j=0}^1C^j([0,T],H^{1-j})$ satisfying $Lu\in L^1([0,T],L^2)$. If $a_{jk}(t)\in Z_1((0,T))$ $(1\le j,k\le d)$, then the Cauchy problem for $L$ is $C^{\infty}$ well posed. Namely, for any $u_0(x),u_1(x)\in C^{\infty}(\mathbb{R}^d)$ and $f(t,x)\in L^1([0,T], C^{\infty}(\mathbb{R}^d))$, we have a unique solution $u(t,x)\in C^{1}([0,T],C^{\infty}(\mathbb{R}^d))$ to the equation $Lu=f(t,x)$ on $(0,T)\times \mathbb{R}^d$ with the initial conditions $u(0,x)=u_0(x)$ and $\partial_tu(0,x)=u_1(x)$. \end{theorem} \begin{proof} Assume that $a_{jk}(t)\in Z_0((0,T))$ ($j,k=1,\dots,d$). If $u(t,x)$ belongs to $\bigcap_{j=0}^2C([0,T],H^{2-j})$, the estimate \eqref{thm33-l2} follows from \eqref{thm3-3} with $\gamma=0$ and \eqref{a-est0}. In the case where $u(t,x)\in \bigcap_{j=0}^1C^j([0,T],H^{1-j})$ and $f(t,x)=Lu\in L^1([0,T],L^2)$, we regularize $u(t,x)$ with respect to $x$-variables by setting $u_{\delta}(t,x)=\int e^{i(x-y)\xi} (1+\delta |\xi|^2)^{-1}u(t,y)\,d\xi dy/(2\pi)^d$ with $\delta>0$. We denote this by $(1-\delta \Delta)^{-1}u(t,x)$. Then we regularize $u_{\delta}$ with respect to $t$-variable by setting $u^{\varepsilon}_{\delta}(t,x) =\int_{\mathbb{R}} \psi_{\varepsilon}(t-s)u_{\delta}(s,x)\,ds$ with $\varepsilon>0$ where $\psi_{\varepsilon}(s)$ is given by $\psi_{\varepsilon}(s)=\psi(s/\varepsilon)/\varepsilon$ with a smooth function $\psi(s)$ on $\mathbb{R}$ satisfying \[ \int_{\mathbb{R}} \psi(s)\,ds=1 \text{ and }\psi(s)=0 \quad \text{for $s\ge 0$ or $s\le -1$}. \] We denote this convolution by $\psi_{\varepsilon}*u_{\delta}(t,x)$. Then we see that $Lu^{\varepsilon}_{\delta}(t,x)=F^{\varepsilon}_{\delta}(t,x)$ for $t\in[0,T-\varepsilon]$ where $F^{\varepsilon}_{\delta}(t,x)=f^{\varepsilon}_{\delta}(t,x) +R^{\varepsilon}_{\delta}$ with $f^{\varepsilon}_{\delta}(t,x)=\psi_{\varepsilon}* (1-\delta \Delta)^{-1}f(t,x)$ and \[ R^{\varepsilon}_{\delta}=\sum_{j,k=1}^d [\psi_{\varepsilon}*,a_{jk}(t)] \partial_{x_j}\partial_{x_k}u_{\delta}(t,x). \] Here $[\cdot,\cdot]$ denotes the commutator. Since $u^{\varepsilon}_{\delta}(t,x)\in \bigcap_{j=0}^2 C^j ([0,T-\varepsilon],H^{2-j})$, the estimate \eqref{thm33-l2} is valid for $u^{\varepsilon}_{\delta}(t,x)$ when $0\le t_1 \le t_2 \le T- \varepsilon$. Since $f(t,x)\in L^1([0,T],L^2)$, we see that, for $0\le t_1\le t_20$}, \] then under the condition \eqref{hyp} with $T=\infty$ we have the following estimate for the homogeneous energy $E_0(u)(t)=\|\partial_t u(t,\cdot)\|^2+\sum_{j=1}^d \|\partial_{x_j}u(t,\cdot)\|^2$: \[ E_0(u)(t_1)\le C E_0(u)(t_0) \quad \text{$\bigl(t_0,t_1\in [0,\infty)\bigr)$} \] for any $u(t,x)\in \bigcap_{j=0,1}C^j([0,\infty),H^{1-j})$ satisfying $Lu=0$ on $(0,\infty)\times \mathbb{R}^d$. \end{remark} \section{Appendix} In this section we show \eqref{example}. Let $h_{\gamma}(t)=\sin(|\log t|^{1+\gamma})$ with $\gamma>0$. For any positive integer $n$, let $t_n,t_{n-},t_{n+}\in(0,1)$ be given by \[ t_n=e^{-(2\pi n)^{1/(1+\gamma)}},\quad t_{n-}=e^{-(2\pi n-\pi/4)^{1/(1+\gamma)}},\quad t_{n+}=e^{-(2\pi n+\pi/4)^{1/(1+\gamma)}}. \] We note $t_{n+}t_n-t_{n+}$ from $\frac{d}{ds}e^{-s^{1/(1+\gamma)}}<0$ and $\frac{d^2}{ds^2}e^{-s^{1/(1+\gamma)}}>0$. Since $h_{\gamma}'(t)=-(1+\gamma)\cos(|\log t|^{1+\gamma})|\log t|^{\gamma}/t$ on $(0,1)$, we see that \[ |h_{\gamma}'(t)|\ge C|\log t|^{\gamma}/t \quad \text{ for }t_{n+}\le t\le t_{n-}. \] Since $Cn^{-\gamma/(1+\gamma)}\le |(2n\pm 1/4)^{1/(1+\gamma)} -(2n)^{1/(1+\gamma)}|\le C^{-1}n^{-\gamma/(1+\gamma)}$, we see that \[ t_n-t_{n+}\ge C e^{-(2\pi n)^{1/(1+\gamma)}}n^{-\gamma/(1+\gamma)} \] and $1\le t_{n-}/t_{n+}\le C$. Then when $0<\varepsilon\le t_n-t_{n+}$, we have \[ |h_{\gamma}(t+\varepsilon)-h_{\gamma}(t)| \ge C\varepsilon(|\log t|^{\gamma}/t) \quad \text{for }t_{n+}\le t\le t_n, \] from which we have \[ I_n=\int_{t_{n+}}^{t_n}|h_{\gamma}(t+\varepsilon)-h_{\gamma}(t)|\,dt\ge C\varepsilon(|\log t_{n+}|^{1+\gamma}- |\log t_n|^{1+\gamma}. ) \] Then we have $I_n\ge C\varepsilon$ with some constant $C>0$ for any positive integer $n$ and $\varepsilon\in[0,t_n-t_{n+}]$. We pick a large positive integer $n_0$ so that we have $t_{n-}\le 1/2$ for $n\ge n_0$. For any large positive integer $N>n_0$, pick $\varepsilon=t_N-t_{N+}$. Then we see that \[ \int_{0}^{1/2-\varepsilon}|h_{\gamma}(t+\varepsilon)-h_{\gamma}(t)|\,dt \ge \sum_{n=n_0}^N I_n\ge C(N-n_0+1)\varepsilon. \] Since $t_N-t_{N+}\ge Ce^{-(2\pi N)^{1/(1+\gamma)}}N^{-\gamma/(1+\gamma)}$, we see that $N\ge C|\log \varepsilon|^{1+\gamma}$ for large $N$. Then we see that \[ \limsup_{\varepsilon\rightarrow 0}\frac{1}{\varepsilon| \log \varepsilon|^{1+\gamma}}\int_{0}^{1/2-\varepsilon}|h_{\gamma} (t+\varepsilon)-h_{\gamma}(t)|\,dt>0. \] By a similar argument, we see that $h_{\gamma}(t)\notin Z_{\sigma}((0,1/2))$ when $0\le \sigma<2\gamma$. Indeed noting that $h_{\gamma}''(t) =-(\gamma+1)^2\sin(|\log t|^{\gamma+1})|\log t|^{2\gamma}/t^2+r(t)$ where $|r(t)|\le C|\log t|^{\gamma}/t^2$, we choose $s_n$ and $s_{n\pm}$ in $(0,1)$ so that $|\log s_n|^{\gamma+1}=2\pi n+\pi/2$ and $|\log s_{n\pm}|^{\gamma+1}=2\pi n+\pi(1/2\pm1/4)$. 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