\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 244, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/244\hfil Lyapunov functions]
{Lyapunov functions for general nonuniform
trichotomy with different growth rates}

\author[Y. Jiang, F.-F. Liao \hfil EJDE-2014/244\hfilneg]
{Yongxin Jiang, Fang-Fang Liao}  % in alphabetical order

\address{Yongxin Jiang (corresponding author)\newline
Department of Mathematics, College of Science,
Hohai University, Nanjing 210098, China}
\email{yxinjiang@hhu.edu.cn, fax +86 (025) 8378-6626}

\address{Fang-Fang Liao \newline
Department of Mathematics, College of Science,
Hohai University, Nanjing 210098, China}
\email{liaofangfang8178@sina.com}

\thanks{Submitted April 12, 2014. Published November 20, 2014.}
\subjclass[2000]{34A30, 37B25}
\keywords{Lyapunov functions; autonomous systems; nonuniform trichotomies;
\hfill\break\indent  hyperbolicity}

\begin{abstract}
 In this article, we consider non-autonomous linear equations $x' = A(t)x$
 that may exhibit stable, unstable and central behaviors in different directions.
 We give a complete characterization of nonuniform $(\mu,\nu)$ trichotomies
 in terms of strict  Lyapunov functions. In particular, we obtain an inverse
 theorem giving explicitly Lyapunov functions for each given trichotomy.
 The main novelty of our work is that we consider a very general type of
 nonuniform exponential trichotomy, which admits different growth rates in
 the uniform and nonuniform parts.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction} 

We consider the non-autonomous linear equation
\begin{equation} \label{le}
x'=A(t)x
\end{equation}
where $A:\mathbb{R}_0^{+}\to \mathcal{B}(X)$ is a continuous function with values
in the space of bounded linear operators in a Banach space $X$.
Our main aim is to characterize the existence of nonuniform $(\mu,\nu)$
trichotomies behavior for the solutions of equation \eqref{le}
in terms of strict Lyapunov functions.

We consider a very general type of nonuniform trichotomies,
which generalizes the classical notion of exponential trichotomies in various
ways: besides introducing a nonuniform term, we consider arbitrary rates that in
particular may not be exponential, as well as different growth rates in the uniform
and nonuniform parts. This includes for example the classical notion of uniform
exponential trichotomies, as well as the notions of nonuniform exponential 
trichotomies and $\rho$-nonuniform exponential trichotomy.

Exponential trichotomy is the most complex asymptotic property of dynamical 
systems arising from the central manifold theory. The conception of trichotomy
 was first introduced by Sacker and Sell \cite{SS}. They described SS-trichotomy 
for linear differential systems by linear skew-product flows. 
Later, Elaydi and H\'ajek \cite{EH88,EH90} gave the notions of exponential 
trichotomy for differential systems and for nonlinear differential systems, 
respectively. These notions are stronger notions than SS-trichotomy. 
Recently, Barreira and Valls \cite{bv09jmaa} considered a general concept
of nonuniform exponential trichotomy, from which can see exponential 
trichotomy as a special case of the nonuniform exponential trichotomy. 
 Jiang \cite{jyx} consider a general case of nonuniform $(\mu,\nu)$ trichotomy 
for an arbitrary non-autonomous linear dynamics, and establish the robustness 
of the nonuniform $(\mu,\nu)$ trichotomy in Banach spaces,  based on \cite{chu} 
for nonuniform $(\mu,\nu)$ dichotomy. In \cite{bv10j}, they proposed a 
$\rho$-nonuniform exponential trichotomy and considered the Lyapunov 
functions for the trichotomy.

The importance of Lyapunov functions is well-established, particularly in the
study of the stability of trajectories both under linear and nonlinear perturbations.
This study dates back to the seminal work of Lyapunov in his 1892 thesis \cite{ly92}. 
Among the first accounts of the theory we have the
works by LaSalle and Lefschetz \cite{ll61}, Hahn \cite{h67}, and Bhatia 
and Szeg\"o \cite{bs70}.
 According to \cite{coppel}, the connection between Lyapunov functions and 
uniform exponential dichotomies was first considered by Ma\v{z}el' 
in \cite{ma54}. 
For recent works, we refer the reader to \cite{chu}, \cite{bp}-\cite{bv09}, 
\cite{BS}, for  nonuniform dichotomies 
 \cite{chu13}, \cite{bv09j1}-\cite{bv10j}, \cite{lj},
for the corresponding characterization
of nonuniform exponential contractions, dichotomies and trichotomies
using Lyapunov functions. From there, we follow some of the ideas in the 
proofs of this articles.
We give a complete characterization of nonuniform $(\mu,\nu)$ trichotomies in terms
of strict  Lyapunov functions. In particular, we obtain an inverse theorem
 giving explicitly Lyapunov functions for each given trichotomy.

The remaining part of this paper is organized as follows. 
In Section 2, we introduce some basic definitions.
In Section 3, we establish a criterion for the existence of partially hyperbolic 
behavior in terms of pairs of strict Lyapunov functions.
 A characterization of nonuniform $(\mu,\nu)$
trichotomies in terms of quadratic Lyapunov functions is
presented in Section 4.

\section{Preliminaries}

Let $X$ be a Banach space and denote by $\mathcal{B}(X)$ the space
of bounded linear operators acting on $X$. Given a continuous
function $A:\mathbb{R}_0^{+}\to \mathcal{B}(X)$, we assume that each
solution of \eqref{le} is global and denote the evolution operator
associated with \eqref{le} by $T(t,s)$, i.e., the linear operator such
that \[T(t,s)x(s)=x(t),\quad t,s>0,\] where $x(t)$ is any solution
of \eqref{le}. Clearly, $T(t,t)={\rm Id}$ and
\[
T(t,\tau)T(\tau,s)=T(t,s),\quad t,\tau,s>0.
\]


First we introduce the notion of nonuniform $(\mu,\nu)$
trichotomy. We say that an increasing function $\mu:\mathbb{R}_0^{+}\to [1,+\infty)$ is
a growth rate if
\[
\mu(0)=1\quad\text{and}\quad\lim_{t\to+\infty}\mu(t)=+\infty.
\]

\begin{definition} \label{def2.1} \rm
Given growth rates $\mu$ and $\nu$, we say that \eqref{le} admits a 
nonuniform $(\mu,\nu)$ trichotomy in $I$ if there exist projections
 $P(t), Q (t), R(t)$ for each $t \in I$ such that 
\begin{equation} \label{PQR}
T(t,\tau)P(\tau)=P(t)T(t,\tau),T(t,\tau)Q(\tau)
=Q(t)T(t,\tau),T(t,\tau)R(\tau)=R(t)T(t,\tau)
\end{equation}
and
\begin{equation} \label{PQRI}
P(t)+Q(t)+R(t)=Id
\end{equation}
for every $t,\tau\in I$, and there exist
constants
\begin{equation} \label{abcd}
0\leq \eta<\alpha,\quad  0\leq \xi<\beta, \quad
\varepsilon\geq0, \quad D\geq1
\end{equation}
 such that for every $t,\tau\in I$ with
$t\geq \tau$, we have
\begin{equation} \label{tp}
\begin{gathered}
\|T(t,\tau)P(\tau)\|\leq D\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-\alpha}
\nu^{\varepsilon}(\tau), \\
 \|T(t,\tau)R(\tau)\|\leq D\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\xi}
 \nu^{\varepsilon}(\tau),
\end{gathered}
\end{equation}
and
\begin{equation} \label{tq}
\begin{gathered}
\|T(t,\tau)^{-1}Q(t)\|\leq D\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-\beta}
\nu^{\varepsilon}(t), \\
 \|T(t,\tau)^{-1}R(t)\|\leq D\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\eta}
\nu^{\varepsilon}(t).
\end{gathered}
\end{equation}
 When $\varepsilon= 0$, we say that \eqref{le} admits
a uniform $(\mu,\nu)$ trichotomy.
\end{definition}

Setting $t = \tau$ in \eqref{tp} and \eqref{tq} we obtain
\begin{equation} \label{pqr}
P(t)\leq D \nu^{\varepsilon}(t), Q(t)\leq D
\nu^{\varepsilon}(t),R(t)\leq D \nu^{\varepsilon}(t),
\end{equation}
for every $t\in I$.
We refer the reader to \cite{jyx} for an example with a nonuniform $(\mu,\nu)$ trichotomy which can not be uniform.

Now, we introduce the notion of Lyapunov function.
Let $A :\mathbb{R}_0^{+}\to M_{p}$ be a continuous function, 
where $M_{p}$ is the set of $p\times p$ matrices.
Given a function $V :\mathbb{R}^{p}\to \mathbb{R}$, we consider the cones
$$
C^{u}(V)=\{0\}\cup V^{-1}(0,+\infty),\quad 
C^{s}(V)=\{0\}\cup V^{-1}(-\infty,0).
$$

\begin{definition} \label{def2.2} \rm
We say that a continuous function $V :\mathbb{R}_0^{+} \times \mathbb{R}^{p}\to \mathbb{R}$ 
is a Lyapunov function for \eqref{le} if there exist
integers $r_{u}, r_{s} \in \mathbb{N} \cup\{0\}$ with $r_{u}+r_{s}= p$ 
such that the following properties hold:
\begin{itemize}
\item[(1)] $r_{u}$ and $r_{s}$ are respectively the maximal dimensions of 
the linear subspaces inside the cones $C^{u}(V_{t})$
and $C^{s}(V_{t})$, for every $t\in \mathbb{R}_0^{+}$;

\item[(2)] for every $x \in \mathbb{R}^{p}$ and $t\geq\tau$ we have
$$
T(t,\tau)\overline{C^{u}(V_{\tau})}\subset \overline{C^{u}(V_{t})},\quad 
T(\tau,t)\overline{C^{s}(V_{t})}\subset \overline{C^{s}(V_{\tau})}, 
$$
where $V_{t}= V(t,\cdot)$
\end{itemize}
\end{definition}

In view of the compactness of the closed unit ball in $\mathbb{R}^{p}$, 
if $(V_{t})_{t\in\mathbb{R}}$ is a Lyapunov function for  \eqref{le},
then for each $\tau\in\mathbb{R}_0^{+}$ the sets
\begin{gather} \label{hu}
 H^{u}_{\tau}=\cap_{r\in\mathbb{R}_0^{+}}T(\tau,r)\overline{C^{u}(V_{r})}
\subset \overline{C^{u}(V_{\tau})}, \\
\label{hs} 
H^{s}_{\tau}=\cap_{r\in\mathbb{R}_0^{+}}T(\tau,r)\overline{C^{s}(V_{r})}
\subset \overline{C^{s}(V_{\tau})}
\end{gather}
contain subspaces respectively of dimensions $r_{u}$ and $r_{s}$. 
We note that for every $t,\tau\in\mathbb{R}_0^{+}$,
\begin{equation} \label{hus}
T(t,\tau)H^{u}_{\tau}=H^{u}_{t}\quad\text{and}\quad
 T(t,\tau)H^{s}_{\tau}=H^{s}_{t}
\end{equation}
Next, we introduce the notion of strict Lyapunov functions.
Given growth rates $\mu$ and $\nu$, we denote by $V$ the Lyapunov function.


\begin{definition} \label{def2.3} \rm
Given $\lambda>\sigma> 0$ and $\gamma\geq0$, we say that $V$ is a
$(\lambda,\sigma )$-strict Lyapunov function if there exist 
$C > 0 $ and $ \delta\geq0$ such that
\begin{equation} \label{v1}
|V(t,x)|\leq C \nu^{\delta}(t)\|x\|
\end{equation}
for each $t\in \mathbb{R}_0^{+}$ and $x\in \mathbb{R}^{p}$, the following properties hold:
\begin{itemize}
\item[(1)]  if $x\in H^{u}_{\tau}$  then
\begin{equation} \label{v11}
 V(t,T(t,\tau)x)\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda+\sigma)}
V(\tau,x),\quad t\geq\tau;
\end{equation}

\item[(2)] if $x\in H^{s}_{\tau}$  then
\begin{equation} \label{v12}
|V(t,T(t,\tau)x)|\leq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)}
V(\tau,x),\ \ t\geq\tau;
\end{equation}

\item[(3)] if $x\in H^{u}_{\tau}\cup H^{s}_{\tau}$  then
\begin{equation} \label{v13}
|V(\tau,x)|\geq\nu^{-\gamma}(\tau)\|x\|/C.
\end{equation}
\end{itemize}
\end{definition}

\section{Criterion for nonuniform hyperbolic behavior}\noindent

For each $\tau\in\mathbb{R}_0^{+}$ , we set
\begin{equation} \label{u1}
\chi^{+}_{\tau}(x)= \limsup_{t\to+\infty}\frac{\log\|T(t,\tau)x\|}{\log\mu(t)}.
\end{equation}

\begin{lemma} \label{lem1}
 Let $\mu\geq\nu$ be growth rates with $\mu(t)/\mu(\tau)>\nu(t)/\nu(\tau)$ 
for every $t\geq\tau$. If there exist a $(\lambda,\sigma)$-strict Lyapunov 
function $V$ for \eqref{le} with
\begin{equation} \label{i1}
(\lambda+\sigma)/(\lambda-\sigma)>e^{\delta+\gamma},
\end{equation}
then for each $t,\tau\in\mathbb{R}_0^{+}$:
\begin{itemize}
\item[(1)]  the sets $H^{u}_{\tau}$ and $H^{s}_{\tau}$ in  \eqref{hu}
and \eqref{hs}  are linear subspaces respectively of dimensions
$r_{u}$ and $r_{s}$, with
\begin{gather} \label{Rp}
 \mathbb{R}^{p}=H^{u}_{\tau}\oplus H^{s}_{\tau},\\
\label{ths}
T(t,\tau)H^{s}_{\tau}=H^{s}_{t},\quad   T(t,\tau)H^{u}_{\tau}=H^{u}_{t};
\end{gather}

\item[(2)]
\begin{gather} \label{u21}
\chi^{+}_{\tau}(x)\geq \log(\lambda+\sigma)-\delta \quad\text{for }
 x\in H^{u}_{\tau}, \\
\label{u22}
\chi^{+}_{\tau}(x)\leq \log(\lambda-\sigma)+\gamma \quad\text{for }
 x\in H^{s}_{\tau};
\end{gather}

\item[(3)] for each $t\geq\tau$ we have
\begin{gather} \label{thus}
\|T(t,\tau)^{-1}|H^{u}_{t}\|
\leq C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-\log(\lambda+\sigma)}
\nu^{\delta+\gamma}(t), \\
 \|T(t,\tau)|H^{s}_{\tau}\|
\leq C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)
+\gamma}\nu^{\delta+\gamma}(\tau).
\end{gather}
\end{itemize}
\end{lemma}

\begin{proof}
 It follows from \eqref{v13} that \eqref{hu} and \eqref{hs} can be replaced by
\begin{equation} \label{huhs}
 H^{u}_{\tau}\subset C^{u}(V_{\tau}) \quad\text{and}\quad
 H^{s}_{\tau}\subset C^{s}(V_{\tau}).
\end{equation}
Indeed, if $x\in H^{u}_{\tau}\backslash\{0\}$, then by \eqref{v13} we have
 $V(\tau, x)>0$. This establishes the first inclusion in \eqref{huhs}.
A similar argument establishes the second one. By \eqref{huhs},
the function $V(\tau, \cdot)$ is positive in $H^{u}_{\tau}\backslash\{0\}$
and negative in $H^{s}_{\tau}\backslash\{0\}$. For each $x\in H^{s}_{\tau}$,
it follows from \eqref{v12} and \eqref{v13} that for every $t\geq\tau$ we have
\begin{equation} \label{lp1}
\begin{aligned}
\|T(t,\tau)x\|
&\leq C\nu^{\gamma}(t)|V(t,T(t,\tau)x)| \\
&\leq C\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)}
\nu^{\gamma}(t)|V(\tau,x)| \\
&\leq C\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)}
\mu^{\gamma}(t)|V(\tau,x)|.
\end{aligned}
\end{equation}
Thus, \eqref{u22} holds.

For each $x\in H^{u}_{\tau}$, it follows from \eqref{v1} and \eqref{v11}
 that for every $t\geq\tau$ we have
\begin{equation} \label{lp2}
\begin{aligned}
 \|T(t,\tau)x\|
&\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda+\sigma)}
\frac{\nu^{-\delta}(t)}{C}|V(t,T(t,\tau)x)| \\
&\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda+\sigma)}
\frac{\mu^{-\delta}(t)}{C}|V(\tau,x)|.
\end{aligned}
\end{equation}
Thus, \eqref{u21} holds.
For each $\tau\in \mathbb{R}_0^{+}$, let $D^{u}_{\tau}\subset H^{u}_{\tau}$
be any $r_{u}$-dimensional subspace, and let $D^{s}_{\tau}\subset H^{s}_{\tau}$
be any $r_{s}$-dimensional subspace. By \eqref{huhs}, we have
$H^{u}_{\tau}\cap H^{s}_{\tau}=\{0\}$, and hence
$D^{u}_{\tau}\cap D^{s}_{\tau}=\{0\}$. So we have
 $\mathbb{R}^{p}=H^{u}_{\tau}\oplus H^{s}_{\tau}$. We want to show that
$$
H^{s}_{\tau}= D^{s}_{\tau}\quad\text{and}\quad H^{u}_{\tau}=D^{u}_{\tau}.
$$
If there exists $x\in H^{s}_{\tau}\backslash D^{s}_{\tau}$, then we write
$x = y+ z$ with $y\in D^{s}_{\tau}$
and $z\in D^{u}_{\tau}\backslash \{0\}$. By {\rm\eqref{i1}} we have
\[
\log(\lambda+\sigma)-\delta > \log(\lambda-\sigma)+\gamma.
\]
Hence, it follows from \eqref{u21} and \eqref{u22} that
\[
\chi^{+}_{\tau}(x)=\max\{\chi^{+}_{\tau}(y),\chi^{+}_{\tau}(z)\}
=\chi^{+}_{\tau}(z)\geq\log(\lambda+\sigma)-\delta,
\]
which contradicts to \eqref{u22}. This implies that $H^{s}_{\tau}= D^{s}_{\tau}$
for each $\tau\in \mathbb{R}_0^{+}$. We can show in a similar manner that
$H^{u}_{\tau}= D^{u}_{\tau}$ for each $\tau\in \mathbb{R}_0^{+}$.
By \eqref{v1} and \eqref{lp1}, for every $x\in H^{s}_{\tau}$ and
$t\geq\tau$ we have
\begin{equation} \label{lp3}
\begin{aligned}
 \|T(t,\tau)x\|
&\leq C\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)}
 \nu^{\gamma}(t)|V(\tau,x)| \\
&\leq C\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)}
\Big(\frac{\nu(t)}{\nu(\tau)}\Big)^{\gamma}\nu^{\delta+\gamma}(\tau)\|x\| \\
&\leq C\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda-\sigma)
 +\gamma}\nu^{\delta+\gamma}(\tau)\|x\|.
\end{aligned}
\end{equation}
Moreover, by \eqref{v13} and {\rm\eqref{lp2}}, for every $x\in H^{u}_{\tau}$
and $t\geq\tau$ we have
\begin{equation} \label{lp4}
\begin{aligned}
 \|T(t,\tau)x\|
&\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{\log(\lambda+\sigma)}
 \frac{\nu^{-\delta}(t)}{C}|V(\tau,x)| \\
&\geq \frac{\nu^{-\delta}(t)}{C}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)
 ^{\log(\lambda+\sigma)}\frac{\nu^{-\gamma}(\tau)}{C}\|x\| \\
&\geq \frac{1}{C^{2}}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)
 ^{\log(\lambda+\sigma)}\Big(\frac{\nu(t)}{\nu(\tau)}\Big)^{\gamma}
\nu^{-(\delta+\gamma)}(t)\|x\| \\
&\geq \frac{1}{C^{2}}\Big(\frac{\mu(t)}{\mu(\tau)}
\Big)^{\log(\lambda+\sigma)}\nu^{-(\delta+\gamma)}(t)\|x\|.
\end{aligned}
\end{equation}
Hence, it gives
\begin{equation} \label{lp5}
 \|T(t,\tau)^{-1}x\|\leq
C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-\log(\lambda+\sigma)}
\nu^{\delta+\gamma}(t)\|x\|
\end{equation}
for every $x\in H^{u}_{\tau}$ and $t\geq\tau$. This completes the proof.
\end{proof}

Now we establish a criterion for the existence of partially hyperbolic
 behavior in terms of pairs of strict
Lyapunov functions. Without loss of generality we consider the same constants
 $\delta$ and $\gamma$ for the two functions in each pair.

\begin{theorem} \label{main-a}
Let $\lambda_{1}>\lambda_{2}>0$. If there exist a $(\lambda_{1},\sigma_{1})$-strict
 Lyapunov function $V$ and a $(\lambda_{2},\sigma_{2})$-strict Lyapunov function 
$W$ for \eqref{le} with
\begin{gather} \label{con1} 
(\lambda_{i}+\sigma_{i})/(\lambda_{i}-\sigma_{i})>e^{\delta+\gamma},\quad i=1,2;\\
\label{con2} 
\lambda_{1}-\lambda_{2}\geq|\sigma_{1}-\sigma_{2}|,
\end{gather}
then for each $t,\tau\in\mathbb{R}_0^{+}$, the following statements hold.
\begin{itemize}
\item[(1)] The sets
\begin{equation} \label{set1}
F^{s}_{\tau}=H^{s}_{\tau}(V),\quad F^{u}_{\tau}=H^{u}_{\tau}(W),\quad
\ F^{c}_{\tau}=H^{s}_{\tau}(W)\cap H^{u}_{\tau}(V)
\end{equation}
 are linear subspaces, with
\begin{gather} \label{Rp2}
\mathbb{R}^{p}=F^{u}_{\tau}\oplus F^{s}_{\tau}\oplus F^{c}_{\tau}, \\
 \label{set2}
T(t,\tau)F^{u}_{\tau}=F^{u}_{t},\quad   T(t,\tau)F^{s}_{\tau}=F^{s}_{t},\quad
T(t,\tau)F^{c}_{\tau}=F^{c}_{t};
\end{gather}

\item[(2)] for each $t\geq\tau$ we have
\begin{equation} \label{tpus}
\begin{gathered}
\|T(t,\tau)^{-1}|F^{u}_{t}\|\leq
C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-(\log(\lambda_{2}
+\sigma_{2})-\gamma)}\nu^{\delta+\gamma}(t), \\
 \|T(t,\tau)|F^{s}_{\tau}\|\leq
C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{(\log(\lambda_{1}-\sigma_{1})
+\gamma)}\nu^{\delta+\gamma}(\tau),
\end{gathered}
\end{equation}
and
\begin{equation} \label{tpc}
\begin{gathered}
 \|T(t,\tau)|F^{c}_{\tau}\|\leq
C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{(\log(\lambda_{2}-\sigma_{2})
+\gamma)}\nu^{\delta+\gamma}(\tau), \\
\|T(t,\tau)^{-1}|F^{c}_{t}\|\leq
C^{2}\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-(\log(\lambda_{1}+\sigma_{1})
-\gamma)}\nu^{\delta+\gamma}(t).
\end{gathered}
\end{equation}
\end{itemize}
 \end{theorem}

\begin{proof}
 According to Lemma \ref{lem1}, it is easy to prove Part 2.
Using a similar argument as in \cite{bv10j}, it can be obtained for every 
$\tau\in \mathbb{R}_0^{+}$ that
\begin{gather*}
H^{s}_{\tau}(V)\subset H^{s}_{\tau}(W),\quad 
H^{u}_{\tau}(W)\subset H^{s}_{\tau}(V), \\ %\label{zhihe}
H^{s}_{\tau}(W)\cap H^{u}_{\tau}(V)\oplus H^{s}_{\tau}(V)\oplus H^{u}_{\tau}(W)
=\mathbb{R}^{p}.
\end{gather*}
So Part 1 holds.
This completes the proof.
\end{proof}

Now we consider the particular case of differentiable Lyapunov functions. Set
\begin{align*}
\dot{V}(t,x)
&= \frac{d}{dh}V(t+h,T(t+h,\tau)x)_{|h=0} \\
&= \frac{\partial V}{\partial t}(t,x)
+\frac{\partial V}{\partial x}(t,x)A(t)x.
\end{align*}

\begin{proposition} \label{pro1} 
 Let $V$ and  $\mu$ be $C^{1}$ functions.
\begin{itemize}
\item[(1)] For each $x\in H^{u}_{\tau}$, property \eqref{v11} is equivalent to
\begin{equation} \label{p1}
\dot{V}(t,T(t,\tau)x)\geq V(t,T(t,\tau)x)\frac{\mu'(t)}{\mu(t)}\ln(\lambda+\sigma),
\quad t>\tau.
\end{equation}

\item[(2)] For each $x\in H^{s}_{\tau}$, property \eqref{v12} is equivalent to
\begin{equation} \label{p2}
\dot{V}(t,T(t,\tau)x)\geq V(t,T(t,\tau)x)\frac{\mu'(t)}{\mu(t)}\ln(\lambda-\sigma),
 \quad t>\tau.
\end{equation}
\end{itemize}
\end{proposition}

\begin{proof}
 Let $x\in H^{u}_{\tau}$, by {\rm\eqref{set2}}  we have $T(t,\tau)x\in H^{u}_{t}$
for every $t\in \mathbb{R}_0^{+}$. We assume that \eqref{v11} holds. If
$t > \tau$ and $h > 0$ then
\[
V(t+h,T(t+h,\tau)x)\geq \Big(\frac{\mu(t+h)}{\mu(t)}\Big)^{\ln(\lambda+\sigma)}
V(t,T(t,\tau)x), 
\]
and
\begin{align*}
&\lim_{h\to 0^{+}} \frac{V(t+h,T(t+h,\tau)x)-V(t,T(t,\tau)x)}{h}\\
&\geq V(t,T(t,\tau)x)\lim_{h\to 0^{+}}
 \frac{\Big(\frac{\mu(t+h)}{\mu(t)}\Big)^{\ln(\lambda+\sigma)}-1}{h} \\ 
&=  V(t,T(t,\tau)x)\frac{\mu'(t)}{\mu(\tau)}\ln(\lambda+\sigma).
\end{align*}
Similarly, If  $h <0$ is such that $t+h >\tau$, then 
\[
V(t+h,T(t+h,\tau)x)\leq \Big(\frac{\mu(t+h)}{\mu(t)}
\Big)^{\ln(\lambda+\sigma)}V(t,T(t,\tau)x), 
\]
and
\begin{align*}
&\lim_{h\to 0^{-}} \frac{V(t+h,T(t+h,\tau)x)-V(t,T(t,\tau)x)}{h}\\
&\geq V(t,T(t,\tau)x)\lim_{h\to 0^{-}}
 \frac{\Big(\frac{\mu(t+h)}{\mu(t)}\Big)^{\ln(\lambda+\sigma)}-1}{h} \\
& =  V(t,T(t,\tau)x)\frac{\mu'(t)}{\mu(\tau)}\ln(\lambda+\sigma).
\end{align*}
This establishes \eqref{p1}. Now we assume that \eqref{p1} holds. 
Given $x\in H^{u}_{\tau}\backslash \{0\}$, it follows from \eqref{v13} that
$V(\tau, x)> 0$, and thus, by \eqref{hus}, $V(t, T(t,\tau)x)> 0$ 
for every $t\in\mathbb{R}_0^{+}$. Thus, we can rewrite \eqref{p1} in the
form
\[
 \frac{\dot{V}(t,T(t,\tau)x)}{V(t,T(t,\tau)x}
\geq \frac{\mu'(t)}{\mu(t)}\ln(\lambda+\sigma),
\]
which implies that
\[
\ln\frac{V(t,T(t,\tau)x)}{V(\tau,x)}
=\int^{t}_{\tau}\frac{\dot{V}(s,T(s,\tau)x)ds}{V(s,T(s,\tau)x)}
\geq \ln(\lambda+\sigma)\ln\frac{\mu(t)}{\mu(\tau)}.
\]
Hence, \eqref{v11} holds. Part 2 is true.
\end{proof}

\section{Quadratic Lyapunov functions} 

In this section, we give a complete characterization
of nonuniform $(\mu,\nu)$ trichotomies by using  quadratic Lyapunov functions.
For each $t\in\mathbb{R}_0^{+}$, let $S(t)$ and $T(t)$ be symmetric invertible
 $p \times p$ matrices. We consider the functions
\begin{gather} \label{GV}
G(t,x)=\langle S(t)x,x\rangle, \quad 
V(t,x)=-\operatorname{sign} G(t,x)\sqrt{|G(t,x)}|, \\
\label{HW}
H(t,x)=\langle T(t)x,x\rangle, \quad
 W(t,x)=-\operatorname{sign} H(t,x)\sqrt{|H(t,x)}|.
\end{gather}
Any Lyapunov functions $V$ and $W$ as in \eqref{GV} and \eqref{HW} are 
called quadratic Lyapunov functions. Notice
that when $t\mapsto S(t)$ is differentiable we have
\begin{align*}
\dot{V}(t,x)
&=  \frac{\partial V}{\partial t}(t,x)+\frac{\partial V}{\partial x}(t,x)A(t)x \\
&= \langle S'(t)x,x\rangle+2\langle S(t)x,A(t)x\rangle
\end{align*}
with similar identities for $\dot{W}$. We present in two theorems a 
characterization of nonuniform $(\mu,\nu)$
trichotomies in terms of quadratic Lyapunov functions.

\begin{theorem} \label{main-b}
Assume that \eqref{le} admits a  nonuniform $(\mu,\nu)$ trichotomy. 
Then there exist symmetric invertible $p \times p$ matrices $S(t)$ and 
$T(t)$ for $t\in\mathbb{R}_0^{+}$ such that:
\begin{itemize}
\item[(1)] $t\mapsto S(t)$ and $t\mapsto T(t)$ are of class $C^{1}$, and
\begin{gather} \label{b1}
\limsup_{t\to\pm\infty}\frac{\log\|S(t)\|}{\log\nu(t)}<\infty, \\
\label{b2}
\limsup_{t\to\pm\infty}\frac{\log\|T(t)\|}{\log\nu(t)}<\infty;
\end{gather}

\item[(2)] there exist $K_{1}> K_{2}>0$ and $L_{1}> L_{2}>0$ such that for 
every $t\in\mathbb{R}_0^{+}$ and $x\in\mathbb{R}^{p}$ we have
\begin{gather} \label{b3}
\dot{G}(t,x)\leq \begin{cases}
-K_{1}\frac{\mu'(t)}{\mu(t)}G(t,x)-\frac{1}{2}\frac{\mu'(t)}{\mu(t)}\|x\|^{2}
&\text{if } G(t,x)\geq0,  \\[4pt]
-K_{2}\frac{\mu'(t)}{\mu(\tau)}G(t,x)-\frac{1}{2}\frac{\mu'(t)}{\mu(\tau)}\|x\|^{2}
&\text{if } G(t,x)\leq0, 
\end{cases}\\
 \label{b4}
\dot{H}(t,x)\leq \begin{cases}
L_{2}\frac{\mu'(t)}{\mu(t)}H(t,x)-\frac{1}{2}\frac{\mu'(t)}{\mu(t)}\|x\|^{2}
&\text{if } H(t,x)\geq 0,  \\[4pt]
L_{1}\frac{\mu'(t)}{\mu(\tau)}H(t,x)-\frac{1}{2}\frac{\mu'(t)}{\mu(\tau)}\|x\|^{2}
&\text{if } H(t,x)\leq 0. 
\end{cases}
\end{gather}
\end{itemize}
 \end{theorem}

\begin{proof}
Take $\varrho_{1}\in(0, (\alpha-\eta)/2)$. Let 
$\Gamma_{1}(t) = Q (t)\oplus R(t)$. Consider the matrices
\begin{equation} \label{bp1}
\begin{aligned}
S(t) &=  \int_{t}^{\infty}T(\tau,t)^{\ast}P(\tau)^{\ast}P(\tau)T(\tau,t)
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
\frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad -\int_{-\infty}^{t}T(\tau,t)^{\ast}\Gamma_{1}(\tau)^{\ast}
\Gamma_{1}(\tau)T(\tau,t)
\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
\frac{\mu'(\tau)}{\mu(\tau)}d\tau.
\end{aligned}
\end{equation}
Similarly,
Take $\varrho_{2}\in(0, (\beta-\xi)/2)$, and let $\Gamma_{2}(t) =P(t)\oplus R(t)$.
Consider also the matrices
\begin{equation} \label{bp2}
\begin{aligned}
 T(t) &=  \int_{t}^{\infty}T(\tau,t)^{\ast}\Gamma_{2}
(\tau)^{\ast}\Gamma_{2}(\tau)T(\tau,t)
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{-2(\xi+\varrho_{2})}
\frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad -\int_{-\infty}^{t}T(\tau,t)^{\ast}Q(\tau)^{\ast}Q(\tau)T(\tau,t)
\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{2(\beta-\varrho_{2})}
\frac{\mu'(\tau)}{\mu(\tau)}d\tau.
\end{aligned}
\end{equation}
The matrices $S(t)$ and $T (t)$ are symmetric and invertible for each
$t\in\mathbb{R}_0^{+}$. We define the functions $G$
and $H$ by \eqref{GV} and \eqref{HW}. By \eqref{tp} and \eqref{tq},
since $\mu$ is an increasing function we have
\begin{align*}
|G(t,x)|
&\leq \int_{t}^{\infty}\|T(\tau,t)P(t)x\|^{2}
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad +\int_{-\infty}^{t}\|T(\tau,t)\Gamma_{1}(t)x\|^{2}
\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
\frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\leq D^{2} \|x\|^{2}\nu^{2\epsilon}(t)\int_{t}^{\infty}
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{-2\varrho_{1}}\frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad + 4D^{2} \|x\|^{2}\nu^{2\epsilon}(t)\int_{-\infty}^{t}
\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2\varrho_{1}}\frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&=  \frac{D^{2}}{2\varrho_{1}} \|x\|^{2}\nu^{2\epsilon}(t)
 +\frac{4D^{2}}{2\varrho_{1}} \|x\|^{2}\nu^{2\epsilon}(t)
\Big(1-\lim_{\tau\mapsto -\infty}
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2\varrho_{1}}\Big) \\
&\leq \frac{5D^{2}}{2\varrho_{1}} \|x\|^{2}\nu^{2\epsilon}(t).
\end{align*}
Since
\[
\frac{\partial}{\partial t}T(\tau,t)=-T(\tau,t)A(t) \quad\text{and}\quad
\frac{\partial}{\partial t}T(\tau,t)^{\ast}=-A(t)^{\ast}T(\tau,t)^{\ast},
\]
one can easily verify that $S(t)$ and $T (t)$ are of class $C^{1}$ in $t$.
Moreover, since $S(t)$ is symmetric we obtain
\begin{equation} \label{bp3}
\|S(t)\|
=\sup_{x\neq 0}\frac{|G(t,x)|}{\|x\|}
\leq \frac{5D^{2}}{2\varrho_{1}}\nu^{2\epsilon}(t)
\end{equation}
and \eqref{b1} holds. Similar arguments apply to $T (t)$ to obtain \eqref{b2}.
Furthermore, taking derivatives in \eqref{bp1}
we obtain
\begin{align*}
S'(t)
&=  -P(t)^{\ast}P(t)\frac{\mu'(t)}{\mu(t)} \\
&\quad -\int_{t}^{\infty}A(t)^{\ast}T(\tau,t)^{\ast}P(\tau)^{\ast}P(\tau)T(\tau,t)
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad-\int_{t}^{\infty}T(\tau,t)^{\ast}P(\tau)^{\ast}P(\tau)T(\tau,t)A(t)
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad-2(\alpha-\varrho_{1})\frac{\mu'(t)}{\mu(t)}\int_{t}^{\infty}
 T(\tau,t)^{\ast}P(\tau)^{\ast}P(\tau)T(\tau,t)
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad -\Gamma_{1}(t)^{\ast}\Gamma_{1}(t)\frac{\mu'(t)}{\mu(t)} \\
&\quad +\int_{-\infty}^{t}A(t)^{\ast}T(\tau,t)^{\ast}
 \Gamma_{1}(\tau)^{\ast}\Gamma_{1}(\tau)T(\tau,t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau. \\
&\quad +\int_{-\infty}^{t}T(\tau,t)^{\ast}\Gamma_{1}(\tau)^{\ast}
 \Gamma_{1}(\tau)T(\tau,t)A(t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau. \\
&\quad +2(\eta+\varrho_{1})\frac{\mu'(t)}{\mu(t)}\int_{-\infty}^{t}
 T(\tau,t)^{\ast}\Gamma_{1}(\tau)^{\ast}\Gamma_{1}(\tau)T(\tau,t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&= -[P(t)^{\ast}P(t)+\Gamma_{1}(t)^{\ast}\Gamma_{1}(t)]
 \frac{\mu'(t)}{\mu(t)}-A(t)^{\ast}S(t)-S(t)A(t) \\
&\quad -2(\alpha-\varrho_{1})\frac{\mu'(t)}{\mu(t)}\int_{t}^{\infty}
 T(\tau,t)^{\ast}P(\tau)^{\ast}P(\tau)T(\tau,t)
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad +2(\eta+\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{-\infty}^{t}T(\tau,t)^{\ast}\Gamma_{1}(\tau)^{\ast}\Gamma_{1}(\tau)T(\tau,t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau.
\end{align*}
Thus, we have
\begin{equation} \label{bp4}
\begin{aligned}
& S'(t)+A(t)^{\ast}S(t)+S(t)A(t)+[P(t)^{\ast}P(t)+\Gamma_{1}(t)^{\ast}
 \Gamma_{1}(t)]\frac{\mu'(t)}{\mu(t)} \\
&=  -2(\alpha-\varrho_{1})\frac{\mu'(t)}{\mu(t)}\int_{t}^{\infty}
 T(\tau,t)^{\ast}P(\tau)^{\ast}P(\tau)T(\tau,t)
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad +2(\eta+\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{-\infty}^{t}T(\tau,t)^{\ast}\Gamma_{1}(\tau)^{\ast}
 \Gamma_{1}(\tau)T(\tau,t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&=  -2(\alpha-\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{t}^{\infty}(T(\tau,t)P(t))^{\ast}T(\tau,t)P(t)
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad +2(\eta+\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{-\infty}^{t}(T(\tau,t)\Gamma_{1}(t))^{\ast}T(\tau,t)\Gamma_{1}(t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau.
\end{aligned}
\end{equation}
Since
\begin{equation} \label{bp5}
\begin{aligned}
2\langle (P(t)^{\ast}P(t)+\Gamma_{1}(t)^{\ast}\Gamma_{1}(t))x,x\rangle
&\geq (\|P(t)x\|+\|\Gamma_{1} (t)x\|)^{2} \\
&\geq \|(P(t)+\Gamma_{1}(t))x\|^{2}=\|x\|,
\end{aligned}
\end{equation}
we have
\begin{equation} \label{bp6}
P(t)^{\ast}P(t)+\Gamma_{1}(t)^{\ast}\Gamma_{1}(t)\geq\frac{1}{2}{\rm Id}.
\end{equation}
Given two $p\times p$ matrices $A$ and $B$, we say that $A\geq B$ if
$\langle Ax,x\rangle\geq \langle Bx,x \rangle$ for every $x\in \mathbb{R}^{p}$.
Furthermore,
if $x(t)$ is a solution of \eqref{le}, then we obain
\begin{equation} \label{bp7}
\begin{aligned}
\frac{d}{dt}G(t,x(t))
&=  \langle S'(t)x(t),x(t)\rangle+\langle S(t)x'(t),x(t)\rangle
 +\langle S(t)x(t),x'(t)\rangle \\
&=  \langle (S'(t)+A(t)^{\ast}S(t)+S(t)A(t))x(t),x(t)\rangle.
\end{aligned}
\end{equation}
We note that
\begin{align*}
G(t,x(t))&=  \int_{t}^{\infty}\|T(\tau,t)P(t)\|^{2}
\Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad -\int_{-\infty}^{t}\|T(\tau,t)\Gamma_{1}(t)\|^{2}
\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
\frac{\mu'(\tau)}{\mu(\tau)}d\tau\,.
\end{align*}
If $G(t, x(t))\geq 0$, then by \eqref{bp4}, \eqref{bp6} and \eqref{bp7},
 since $\eta +\varrho_{1} <\alpha -\varrho_{1}$ and $\mu$ is an increasing
function, we obtain
% \label{bp8}
\begin{align*}
\frac{d}{dt}G(t,x(t))
&\leq -\frac{1}{2}\|x(t)\|^{2}\frac{\mu'(t)}{\mu(t)} \\
&\quad -2(\alpha-\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{t}^{\infty}\|T(\tau,t)P(t)\|^{2}
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad+2(\alpha-\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{-\infty}^{t}\|T(\tau,t)\Gamma_{1}(t)\|^{2}
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau   \\
&=  -\frac{1}{2} \|x(t)\|^{2}\frac{\mu'(t)}{\mu(t)}-2(\alpha-\varrho_{1})
 \frac{\mu'(t)}{\mu(t)}G(t,x(t)).
\end{align*}
Thus, we can take $K_{1}=2(\alpha-\varrho_{1})> 0$. On the other hand,
if $G(t, x(t))\leq0$, then by \eqref{bp4}, \eqref{bp6} and \eqref{bp7}
again, since $\eta+\varrho_{1}<\alpha-\varrho_{1}$ and $\mu$  is an
increasing function, we obtain
\begin{align*}
&\frac{d}{dt}G(t,x(t))\\
&\leq -\frac{1}{2}\|x(t)\|^{2}\frac{\mu'(t)}{\mu(t)}
 -2(\eta+\varrho_{1})\frac{\mu'(t)}{\mu(t)}\int_{t}^{\infty}\|T(\tau,t)P(t)\|^{2}
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{2(\alpha-\varrho_{1})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad +2(\eta+\varrho_{1})\frac{\mu'(t)}{\mu(t)}
 \int_{-\infty}^{t}\|T(\tau,t)\Gamma_{1}(t)\|^{2}
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-2(\eta+\varrho_{1})}
  \frac{\mu'(\tau)}{\mu(\tau)}d\tau   \\
&= -\frac{1}{2} \|x(t)\|^{2}\frac{\mu'(t)}{\mu(t)}-2(\eta+\varrho_{1})
 \frac{\mu'(t)}{\mu(t)}G(t,x(t)).
\end{align*}
Take $K_{2}=2(\eta+\varrho_{1}) > 0$. Furthermore, we have
$$
K_{1}-K_{2}= 2(\alpha-\eta-2\varrho_{1}) > 0.
$$
Proceeding in a similar manner with $T (t)$ we deduce that
%\label{bp9}
\begin{align*}
& T'(t)+A(t)^{\ast}T(t)+T(t)A(t)+[\Gamma_{2}(t)^{\ast}\Gamma_{2}(t)
 +Q(t)^{\ast}Q(t)]\frac{\mu'(t)}{\mu(t)} \\
&=  2(\xi+\varrho_{2})\frac{\mu'(t)}{\mu(t)}
 \int_{t}^{\infty}(T(\tau,t)\Gamma_{2}(t))^{\ast}T(\tau,t)\Gamma_{2}(t)
 \Big(\frac{\mu(\tau)}{\mu(t)}\Big)^{-2(\xi+\varrho_{2})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau \\
&\quad -2(\beta-\varrho_{2})\frac{\mu'(t)}{\mu(t)}
 \int_{-\infty}^{t}(T(\tau,t)Q(t))^{\ast}T(\tau,t)Q(t)
 \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{2(\beta-\varrho_{2})}
 \frac{\mu'(\tau)}{\mu(\tau)}d\tau
\end{align*}
Note that inequalities \eqref{b4} hold with $L_{1} = 2(\beta-\varrho_{2}) > 0$
and $L_{2} = 2(\xi+\varrho_{2}) > 0$. Moreover,
by the choice of $\varrho_{2}$ we have
$$
L_{1}-L_{2}= 2(\beta-\xi-2\varrho_{2}) > 0.
$$
This completes the proof.
\end{proof}


\begin{theorem} \label{main-c} 
Assume that there exist constants $\gamma,a, \kappa> 0$ such that
\begin{equation} \label{c1}
\|T (t, s)\|\leq \kappa \nu^{a}(t)\quad\text{whenever }|t-s|\leq \gamma.
\end{equation}
Also assume that $\mu,\nu$ is of class $C^{1}$, and that there exist
symmetric invertible $p\times p$ matrices $S(t)$ and $T(t)$
for $t\in R_0^{+}$, satisfying conditions 1 and 2 in Theorem \ref{main-b} with
$K_{1}-K_{2}>2a $ and $L_{1}-L_{2}>2a$. Then \eqref{le}
admits a nonuniform $(\mu,\nu)$ trichotomy with
\begin{equation} \label{c2}
   \alpha=\frac{K_{1}}{2}-a, \quad \beta=\frac{L_{1}}{2}-a,\quad
 \xi=\frac{L_{2}}{2}+a, \quad  \eta=\frac{K_{2}}{2}+a.
\end{equation}
\end{theorem}

\begin{proof} 
We start with some auxiliary results. Set
\begin{gather*} 
I^{s}_{\tau}=\{0\}\cup \{x\in \mathbb{R}^{p}:G(t,T(t,\tau)x)>0 \text{ for  every }
 t\geq \tau\},\\
 I^{u}_{\tau}=\{0\}\cup \{x\in \mathbb{R}^{p}:G(t,T(t,\tau)x)<0 \text{ for  every }
 t\geq \tau\}.
\end{gather*}

\begin{lemma} \label{lem4}  
If $x\in I^{s}_{\tau}$ then
\begin{equation} \label{l41}
G(t,T(t,\tau)x)\leq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-K_{1}}G(\tau,x),\quad
 t\geq \tau,
\end{equation}
and if $x\in I^{u}_{\tau}$ then
\begin{equation} \label{l42}
 |G(t,T(t,\tau)x)|\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-K_{2}}|G(\tau,x)|,\quad
 t\geq \tau.
\end{equation}
\end{lemma}

\begin{proof}  Given $x\in I^{s}_{\tau}\backslash\{0\}$, since 
$G(t, T(t, \tau)x) > 0$ for every $t\geq\tau$, it follows from \eqref{b3} that
\[
\frac{\dot{G}(t,T(t,\tau)x)}{G(t,T(t,\tau)x)}\leq -K_{1}\frac{\mu'(t)}{\mu(t)},
\quad t>\tau.
\]
This implies that
\[
 \ln\frac{G(t,T(t,\tau)x)}{G(\tau,x)}\leq -K_{1}\int^{t}_{\tau}
\frac{\mu'(s)}{\mu(s)}ds=-K_{1}(\ln\mu(t)-\ln\mu(\tau)).
\]
Hence, \eqref{l41} holds. Similarly, given $x\in I^{u}_{\tau}\backslash\{0\}$, 
since $G(t, T(t, \tau)x) < 0$ for every $t\geq\tau$, it follows from 
\eqref{b3} that
\[
\frac{\dot{G}(t,T(t,\tau)x)}{G(t,T(t,\tau)x)}
\geq -K_{2}\frac{\mu'(t)}{\mu(t)},\ t>\tau.
\]
This implies that
\[
 \ln\big|\frac{G(t,T(t,\tau)x)}{G(\tau,x)}\big|
\geq -K_{2}\int^{t}_{\tau}\frac{\mu'(s)}{\mu(s)}ds
=-K_{2}(\ln\mu(t)-\ln\mu(\tau)).
\]
Thus, \eqref{l42} holds.
\end{proof}


\begin{lemma} \label{lem5}  
If $x\in I^{u}_{\tau}\cup I^{s}_{\tau}$ then there holds
\[
 |G(\tau,x)|\geq \frac{1}{2\kappa^{2}}\max\{\gamma,
(1-e^{-K_{2}\gamma})/K_{2}\}\nu^{-2a}(\tau)\|x\|^{2}.
\]
\end{lemma}

\begin{proof}  Set $x(t)=T(t,\tau)x$. It follows from \eqref{b3} that if 
$x\in I^{s}_{\tau}$, then it holds
\[
 \frac{d}{dt}G(t,x(t))\leq-\frac{1}{2}\|x(t)\|^{2}\frac{\mu'(t)}{\mu(t)}.
\]
Since $\mu$ is a growth rate, given $\tau\in \mathbb{R}_0^{+}$, we take $t >\tau$ such that 
$\ln\mu(t) = \ln\mu(\tau )+\gamma$ (with $\gamma$ as in \eqref{c1}).
 Then
\begin{align*}
G(t,x(t))-G(\tau,x)
&= \int^{t}_{\tau}\frac{d}{ds}G(s,x(s))ds
 \leq -\frac{1}{2}\int_{\tau}^{t}\|x(s)\|^{2}\frac{\mu'(s)}{\mu(s)}ds \\
&=  -\frac{1}{2}\int_{\tau}^{t}\|T(s,\tau)x\|^{2} \frac{\mu'(s)}{\mu(s)}ds \\
&\leq -\frac{1}{2}\|x\|^{2}\int_{\tau}^{t}\frac{1}{\|T(\tau,s)\|^{2}}
 \frac{\mu'(s)}{\mu(s)}ds.
\end{align*}
It follows from \eqref{c1}  that
\begin{align*}
 G(t,x(t))-G(\tau,x)
&\leq -\frac{1}{2}\|x\|^{2}\int_{\tau}^{t}\frac{\nu^{-2a}(\tau)}{\kappa^{2}}
 \frac{\mu'(s)}{\mu(s)}ds \\
&=  -\frac{1}{2\kappa^{2}}\nu^{-2a}(\tau)\|x\|^{2}(\ln\mu(t)- \ln\mu(\tau )) \\
&=  -\frac{\gamma}{2\kappa^{2}}\nu^{-2a}(\tau)\|x\|^{2}.
\end{align*}
So we have
\[
  G(\tau,x)\geq G(\tau,x)-G(t,x(t))\geq\frac{\gamma}{2\kappa^{2}}
 \nu^{-2a}(\tau)\|x\|^{2}.
\]
On the other hand, it follows from \eqref{b3}  that if $x\in I^{u}_{\tau}$, 
then it has
\begin{align*}
  \frac{d}{dt}(\mu^{K_{2}}(t)G(t,x(t))
&=  \mu^{K_{2}}(t)\Big(\frac{d}{dt}G(t,x(t))+K_{2}\frac{\mu'(t)}{\mu(t)}G(t,x(t))
\Big) \\ 
&\leq-\frac{1}{2}\frac{\mu'(t)}{\mu(t)}\mu^{K_{2}}(t)\|x(t)\|^{2}.
\end{align*}
Hence, given $\tau\in \mathbb{R}_0^{+}$ and $t <\tau$ such that
 $\ln\mu(\tau)=\ln\mu(t)+\gamma$ , using \eqref{c1} again we obtain
\begin{align*} 
\mu^{K_{2}}(\tau)G(\tau, x)-\mu^{K_{2}}(t)G(t,x(t))
&\leq -\frac{1}{2\kappa^{2}}\nu^{-2a}(\tau)\|x\|^{2}
 \int^{\tau}_{t}\frac{\mu'(s)}{\mu(s)}\mu^{K_{2}}(s)ds   \\
&=  -\frac{1}{2\kappa^{2}K_{2}}\nu^{-2a}(\tau)(\mu^{K_{2}}(\tau)
-\mu^{K_{2}}(t))\|x\|^{2}.
\end{align*}
Since $G(t, x(t)) < 0$ and $G(\tau, x) <0$, we have
\begin{gather*}
\mu^{K_{2}}(\tau)|G(\tau, x)| \geq -\frac{1}{2\kappa^{2}K_{2}}
\nu^{-2a}(\tau)(\mu^{K_{2}}(\tau)-\mu^{K_{2}}(t))\|x\|^{2}, \\
\begin{aligned}
 |G(\tau, x)|
&\geq -\frac{1}{2\kappa^{2}K_{2}}\nu^{-2a}(\tau)
\Big[1-\Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{K_{2}}\Big]\|x\|^{2}  \\
&=  -\frac{1}{2\kappa^{2}K_{2}}\nu^{-2a}(\tau)
 [1-e^{-K_{2}\gamma}]\|x\|^{2}.
\end{aligned}
\end{gather*}
This completes the proof.
\end{proof}

Now we set
\begin{gather*}
 J^{s}_{\tau}=\{0\}\cup \{x\in \mathbb{R}^{p}:H(t,T(t,\tau)x)>0 
\text{ for  every } t\geq \tau\},\\
J^{u}_{\tau}=\{0\}\cup \{x\in \mathbb{R}^{p}:H(t,T(t,\tau)x)<0 
\text{ for every } t\geq \tau\}.
\end{gather*}
Proceeding in a similar manner to that in the proofs of Lemmas \ref{lem4} and 
\ref{lem5} we obtain the following
statements.

\begin{lemma} \label{lem6}  
If $x\in J^{s}_{\tau}$, then
\begin{equation} \label{l61}
H(t,T(t,\tau)x)\leq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{L_{2}}H(\tau,x),\quad
 t\geq \tau,
\end{equation}
and if $x\in J^{u}_{\tau}$, then
\begin{equation} \label{l62}
|H(t,T(t,\tau)x)|\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{L_{1}}|H(\tau,x)|,\quad
 t\geq \tau.
\end{equation}
\end{lemma}

\begin{lemma} \label{lem7}  
If $x\in J^{u}_{\tau}\cup J^{s}_{\tau}$, then
\[
 |H(\tau,x)|\geq \frac{1}{2\kappa^{2}}\max\{\gamma,(1-e^{-L_{2}\gamma})/L_{2}\}
\nu^{-2a}(\tau)\|x\|^{2}.
\]
\end{lemma}

\begin{lemma} \label{lem8}  
The function $V$ in \eqref{GV} is a $(\lambda_{1},\sigma_{1})$ strict Lyapunov 
function for \eqref{le} with
\[
 \lambda_{1}=\frac{e^{-K_{1}/2}+e^{-K_{2}/2}}{2},\quad
\sigma_{1}=\frac{e^{-K_{2}/2}-e^{-K_{1}/2}}{2}.
\]
\end{lemma}

\begin{proof} By \eqref{b1}, for each $\delta>0$ there exists $d > 0$ such that
\[
 \|S(t)\|\leq d \nu^{\delta}(t),
\]
for every $t\in \mathbb{R}_0^{+}$. So we have
\[
 |G(t,x)|\leq d \nu^{\delta}(t)\|x\|^{2}.
\]
That is,
\[
|V(t,x)|\leq \sqrt{d} \nu^{\delta/2}(t)\|x\|^{2},
\]
and  \eqref{v1} holds. Furthermore, by Lemma {\rm\ref{lem5}}, 
for $x\in I^{u}_{\tau}\cup I^{s}_{\tau}=H^{u}_{\tau}\cup H^{s}_{\tau}$ we have
\begin{equation} \label{l81}
 |V(\tau,x)|\geq \frac{1}{\sqrt{2}\kappa}\max\{\gamma,
(1-e^{-L_{2}\gamma})/L_{2}\}^{1/2}\nu^{-a}(\tau)\|x\|^{2}.
\end{equation}
and \eqref{v13} holds. Finally, by Lemma \ref{lem4}, if
 $x\in I^{s}_{\tau}=H^{s}_{\tau}$ then
\[
 |V(t,T(t,\tau)x)|\leq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-K_{1}/2}|V(\tau,x)|,
\quad t\geq\tau.
\]
That is, \eqref{v12} holds with $\lambda_{1}-\sigma_{1}= e^{-K_{1}/2}$. Moreover,
if $x\in I^{u}_{\tau}=H^{u}_{\tau}$  then
\[
V(t,T(t,\tau)x)\geq \Big(\frac{\mu(t)}{\mu(\tau)}\Big)^{-K_{2}/2} V(\tau,x),\quad
t\geq\tau.
\]
That is, \eqref{v11} holds with $\lambda_{1}+\sigma_{1}= e^{-K_{2}/2}$.
This completes the proof.
\end{proof}

In an analogous manner we can prove the following result.

\begin{lemma} \label{lem9}  
The function $W$ in \eqref{HW} is a $(\lambda_{2},\sigma_{2})$ strict Lyapunov 
function for \eqref{le} with
\[
 \lambda_{2}=\frac{e^{L_{1}/2}+e^{L_{2}/2}}{2},\quad 
\sigma_{2}=\frac{e^{L_{1}/2}-e^{L_{2}/2}}{2}.
\]
\end{lemma}

Since $V$ and $W$ is a strict Lyapunov function, by Lemma \ref{lem1}, there 
exist subspaces $H^{u}_{t}(V)$ and $H^{s}_{t}(V)$, $H^{u}_{t}(W)$ and 
$H^{s}_{t}(W)$ such
that $\mathbb{R}^{p} = H^{u}_{t}(V)\oplus H^{s}_{t}(V)$, 
$\mathbb{R}^{p} = H^{u}_{t}(W)\oplus H^{s}_{t}(W)$ for each $t\in \mathbb{R}_0^{+}$. 
We consider the associated projections
\begin{gather*}
P_{V}(t):\mathbb{R}^{p}\to H^{s}_{t}(V),\quad  Q_{V}(t):\mathbb{R}^{p}\to H^{u}_{t}(V),\\
P_{W}(t):\mathbb{R}^{p}\to H^{s}_{t}(W),\quad  Q_{W}(t):\mathbb{R}^{p}\to H^{u}_{t}(W).
\end{gather*}

\begin{lemma} \label{lem10}  
There exists $K > 0$ such that for each $t\in \mathbb{R}_0^{+}$ we have
\begin{gather*}
\|P_{V}(t)\|=\|Q_{V}(t)\|\leq K\nu^{2a}(t)\|S(t)\|,\\
\|P_{W}(t)\|=\|Q_{W}(t)\|\leq K\nu^{2a}(t)\|T(t)\|.
\end{gather*}
\end{lemma}

\begin{proof}
  We  prove only the statement for the Lyapunov function $V$.
The proof for $W$ is completely analogous. We note that
\begin{gather} \label{l101}
 V(t,P_{V}(t)x)^{2}=\langle S(t)P_{V}(t)x,P_{V}(t)x\rangle,\\
 \label{l102}
 V(t,Q_{V}(t)x)^{2}=-\langle S(t)Q_{V}(t)x,Q_{V}(t)x\rangle.
\end{gather}
Given $x\in \mathbb{R}^{p}$ we write $x = y + z$ with
\[
y =P_{V}(t)x \in H^{s}_{t}(V)\quad\text{and}\quad z =Q_{V}(t)x\in H^{u}_{t}(V).
\]
Now take $b(t) > 0$, and set
$$  
V^{s}(t,y)=-V(t,y)^{2}+b(t)\|y\|^{2}=-\langle S(t)y,y\rangle+b(t)\|y\|^{2}.
$$
By {\rm\eqref{l81}}, there exists $K > 0$ such that
$$ 
 V^{s}(t,y)\leq -K\nu^{-2a}(t)\|y\|^{2}+b(t)\|y\|^{2}
=(b(t)-K\nu^{-2a}(t))\|y\|^{2}.
$$
Similarly, for each $t\in \mathbb{R}_0^{+}$ we set
$$  
V^{u}(t,z)=V(t,z)^{2}-b(t)\|z\|^{2}=-\langle S(t)z,z\rangle-b(t)\|z\|^{2}.
$$
By \eqref{l81}, we have
$$  
V^{u}(t,z)\geq (K\nu^{-2a}(t)-b(t))\|z\|^{2}.
$$
We conclude that if $b(t)\leq K\nu^{-2a}(t)$, then
\[  
-V(t,y)^{2}+b(t)\|y\|^{2}\leq0,\quad V(t,z)^{2}-b(t)\|z\|^{2}\geq0.
\]
Thus,  from \eqref{l101} and \eqref{l102} it follows that
\begin{gather*}
-\langle S(t)P_{V}(t)x,P_{V}(t)x\rangle+b(t)\|P_{V}(t)x\|^{2}\leq 0,\\
-\langle S(t)Q_{V}(t)x,Q_{V}(t)x\rangle-b(t)\|Q_{V}(t)x\|^{2}\geq 0.
\end{gather*}
Since $S(t)$ is symmetric, subtracting the two inequalities we obtain
\begin{align*}
0&\geq b(t)\|P_{V}(t)x\|^{2}+b(t)\|Q_{V}(t)x\|^{2} \\
&\quad -\langle S(t)P_{V}(t)x,P_{V}(t)x\rangle
 +\langle S(t)Q_{V}(t)x,Q_{V}(t)x\rangle \\
&=  b(t)\|P_{V}(t)x\|^{2}+b(t)\|Q_{V}(t)x\|^{2}+\langle S(t)x,x\rangle
 -2\langle S(t)P_{V}(t)x,x\rangle.
\end{align*}
Hence, we have
\begin{align*}
& b(t)\|P_{V}(t)x-\frac{1}{2b(t)}S(t)x\|^{2}
 +b(t)\|Q_{V}(t)x+\frac{1}{2b(t)}S(t)x\|^{2} \\
&=  b(t)\|P_{V}(t)x\|^{2}+b(t)\|Q_{V}(t)x\|^{2}
 +\frac{\|S(t)x\|}{2b(t)}+\langle S(t)x,x\rangle
 -2\langle S(t)P_{V}(t)x,x\rangle  \\
&\geq \frac{\|S(t)x\|}{2b(t)},
\end{align*}
and
\[
\|P_{V}(t)x-\frac{1}{2b(t)}S(t)x\|^{2}+\|Q_{V}(t)x+\frac{1}{2b(t)}S(t)x\|^{2}
\leq \frac{\|S(t)x\|}{2b(t)^{2}}.
\]
This implies 
\begin{align*}
\|P_{V}(t)x\|^{2}
&= \|P_{V}(t)x-\frac{1}{2b(t)}S(t)x+\frac{1}{2b(t)}S(t)x\| \\
&\leq \|P_{V}(t)x-\frac{1}{2b(t)}S(t)x\|+\frac{1}{2b(t)}\|S(t)x\| \\
&\leq \frac{1}{\sqrt{2}b(t)}\|S(t)x\|+\frac{1}{\sqrt{2}b(t)}\|S(t)x\|\\
&\leq \frac{\sqrt{2}}{b(t)}\|S(t)x\|.
\end{align*}
Similarly, we obtain
\begin{align*}
 \|Q_{V}(t)x\|^{2}
&\leq \|Q_{V}(t)x-\frac{1}{2b(t)}S(t)x\|+\frac{1}{2b(t)}\|S(t)x\| \\
&\leq \frac{1}{\sqrt{2}b(t)}\|S(t)x\|+\frac{1}{\sqrt{2}b(t)}\|S(t)x\|\\
&\leq \frac{\sqrt{2}}{b(t)}\|S(t)x\|.
\end{align*}
Taking $b(t)= \frac{\sqrt{2}}{K}\nu^{-2a}(t)$ we obtain the desired statement.
 \end{proof}

Note that by taking $\delta$ sufficiently small we have
\begin{gather*}
\frac{\lambda_{1}+\sigma_{1}}{\lambda_{1}-\sigma_{1}}
=e^{(K_{1}-K_{2})/2}>e^{a+\delta/2},\\
\frac{\lambda_{2}+\sigma_{2}}{\lambda_{2}-\sigma_{2}}
=e^{(L_{1}-L_{2})/2}>e^{a+\delta/2}.
\end{gather*}
Moreover, we can easily verify that
 $\lambda_{2}-\lambda_{1}>|\sigma_{2}-\sigma_{1}|$. 
This allows us to apply Theorem \ref{main-a} (with $V$
and $W$ interchanged). If we set
\begin{gather*}
P(\tau)=P_{W}(\tau):\mathbb{R}\to F^{s}_{\tau}=H^{s}_{\tau}(W),\\
Q(\tau)=Q_{V}(\tau):\mathbb{R}\to F^{u}_{\tau}=H^{u}_{\tau}(V),\\
R(\tau)=P_{V}(\tau)\oplus Q_{W}(\tau):\mathbb{R}\to F^{c}_{\tau}
=H^{s}_{\tau}(V)\cap H^{u}_{\tau}(W).
\end{gather*}
The subspaces $F^{s}_{\tau}$, $F^{u}_{\tau}$ and $F^{c}_{\tau}$
satisfy the properties in Theorem \ref{main-a}. Moreover, for every 
$t\geq \tau $ we
have
\begin{gather*}
\|T(t,\tau)P(\tau)\|\leq \|T(t,\tau)|F^{s}_{\tau}\|\, \|P(\tau)\|,\\
\|T(t,\tau)^{-1}Q(\tau)\|\leq \|T(t,\tau)^{-1}|F^{u}_{\tau}\|\,\|Q(\tau)\|,\\
\|T(t,\tau)R(\tau)\|\leq \|T(t,\tau)|F^{c}_{\tau}\|\, \|R(\tau)\|,\\
\|T(t,\tau)^{-1}R(\tau)\|\leq \|T(t,\tau)^{-1}|F^{c}_{\tau}\|\, \|R(\tau)\|.
\end{gather*}

Hence, by property 2 in Theorem \ref{main-a} and Lemma \ref{lem10} there exist 
constants as in \eqref{abcd} satisfying \eqref{tp}
and \eqref{tq}. In other words, \eqref{le} admits a  nonuniform $(\mu,\nu)$ 
trichotomy. By \eqref{tpus}, \eqref{tpc}, and
Lemma \ref{lem10} we can take the constants $\alpha,\beta,\xi,\eta$ in 
\eqref{abcd} as in \eqref{c2}. This completes the proof.
\end{proof}


\subsection*{Acknowledgments} 
This work is partly supported by the National Natural Science Foundation 
of China under Grant No. 11171090, No. 11401165,
and the State Scholarship Fund of China.
 The first author would like to thank the Department of
Mathematics at the University of Texas-Pan American for its
hospitality and generous support during her visiting from January
2014 to January 2015.


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