Proof. Let $X(t)$ be a fundamental matrix of system (2.2). By QR decomposition, we obtain a real orthogonal matrix $U(t)$ (i.e., $U(t)U(t)^T=U(t)^TU(t)=I$ holds for all $t\in \mathbb {T}$) and a real upper triangular matrix $Y(t)$ such that

\[ X(t)=U(t)Y(t). \]

Since $X(t)$ is rd-continuously differentiable, it is easily seen that $U(t)$ and $Y(t)$ are also. The change of variables $x=U(t)y$ replaces the equation (2.2) by

\[ y^\Delta=B(t)y, \]

where $B=(U^{\sigma })^TAU-(U^{\sigma })^TU^{\Delta }$. Then we have

\begin{align*} I+\mu(t)B(t)& =I+\mu(t)[(U^{\sigma})^TAU-(U^{\sigma})^TU^{\Delta}]\\ & =(U^{\sigma})^T[I+\mu(t)A]U+I-(U^{\sigma})^T\left(U+\mu(t)U^\Delta\right)\\ & =(U^{\sigma})^T[I+\mu(t)A]U+I-(U^{\sigma})^TU^{\sigma}=(U^{\sigma})^T[I+\mu(t)A]U, \end{align*}

which implies that $B\in \mathcal {R}$ since $A\in \mathcal {R}$. Since $Y(t)$ is a fundamental matrix of the transformed equation, $B(t)=Y^{\Delta }(t)Y^{-1}(t)$ is real upper triangular. Note that

(4.1)\begin{equation} U^{\sigma}BU^T=A-U^{\Delta}U^T,\quad U^{\sigma}B^T(U^{\sigma})^T=U^{\sigma}U^TA^T-U^{\sigma}(U^{\Delta})^T. \end{equation}

and

(4.2)\begin{equation} I^{\Delta}=(UU^{T})^{\Delta}=U^\sigma (U^T)^\Delta+U^{\Delta}U^T=O. \end{equation}

From (4.1) and (4.2), we have

(4.3)\begin{equation} U^{\sigma}[B+B^T(U^{\sigma})^TU]U^T=U^{\sigma}BU^T+U^{\sigma}B^T(U^{\sigma})^T=A+U^{\sigma}U^TA^T. \end{equation}

We introduce the norm defined by $\|A\|:=\sup \limits _{x\in \mathbb {R}^n}|Ax|/|x|$. It is well known that ${\|U\|=1}$ if $U$ is an orthogonal matrix. Then, from (4.3), we have

(4.4)\begin{equation} \|B+B^T(U^{\sigma})^TU\|\leq\|A+U^{\sigma}U^TA^T\|\leq\|A\|+\|A^T\|. \end{equation}

We claim that $A^T$ is bounded. In fact, let $\|A\|_{\infty }=\sum |a_{ij}|$ and we obtain $\|A\|_{\infty }=\|A^T\|_{\infty }$. Since $M_n(\mathbb {R})$ is a finite dimensional linear space, there are positive constants $c_1,c_2$, such that $c_1\|\cdot \|\leq \|\cdot \|_{\infty }\leq c_2\|\cdot \|.$ Therefore, we get $\|B+B^T(U^{\sigma })^TU\|\leq (1+c_2)\|A\|$

Suppose that $B(t)$ is unbounded. Then there exists a sequence $\{t_m|m\in \mathbb {N}_+\}$, such that $\|B(t_m)\|\geq m$. Note that

\[\|B\|=\|(U^{\sigma})^TAU-(U^{\sigma})^TU^{\Delta}\|\leq\|A\|+\|U^\Delta\|=\|A\|+\|\displaystyle\frac{U^{\sigma}-U}{\mu(t)}\|\leq\|A\|+\frac{2}{\mu(t)}, \]

which implies that $\mu (t_m)\rightarrow 0$ as $m\rightarrow +\infty$. Let

\[ C(t)=(U^{\sigma})^TU-I={-}(U^{\sigma})^T(U^\sigma-U). \]

Thus, $\|C(t_m)\|\leq \|U(t_m+\mu (t_m))-U(t_m)\|$. Since $U(t)$ is continuous, we have $\|C(t_m)\|\rightarrow 0$ as $m\rightarrow +\infty$. For any $n\times n$ matrix $D=(d_{ij})$, we have

\[ n^{{-}1}\displaystyle\sum_{i,j}|d_{ij}|^2\leq\|D\|^2\leq\sum_{i,j}|d_{ij}|^2. \]

This inequality can be found in the page 88 of [Reference Coppel19]. Since $B$ is upper triangular, we have

\begin{align*} (1+c_2)\|A\|& \geq \|B+B^T(U^{\sigma})^TU\|=\|B+B^T+B^TC\|\geq \|B+B^T\|-\|C\|\|B\|\\ & \geq \left(n^{{-}1}\sum_{j,k}|b_{jk}+b_{kj}|^2\right)^{\frac{1}{2}}-\|C\|\|B\|\\ & \geq \left(n^{-\frac{1}{2}}-\|C\|\right)\|B\|. \end{align*}

Thus, $(1+c_2)\|A(t_m)\|\geq \left (n^{-\frac {1}{2}}-\|C(t_m)\|\right )\|B(t_m)\|\geq m\left (2n^{-\frac {1}{2}}-\|C(t_m)\|\right )$. Since $A(t)$ is bounded, let $m\rightarrow +\infty$ and we have the right side of the above inequality unbounded, which leads to a contradiction. Therefore, $B(t)$ is bounded.

Proof. We only prove the case $Q=O$ and the case $Q=I$ can be proved in a similar way. Let $\Phi (t,s)$ and $\Psi (t,s)$ denote the evolution operators of systems (2.2) and (4.6), respectively. It can be seen that

\[ \|\Phi(t,s)\|\leq K\,{\rm e}^{(\gamma+\alpha)(t-s)},\quad t\leq s. \]

For any $y\in \mathbb {R}^n$, we have

\[\Psi(t,s)y=\Phi(t,s)y+\displaystyle\int_s^t\Phi(t,\sigma(\tau))B(\tau)\Psi(\tau,s)y\Delta\tau, \]

hence for $t\leq s$,

\begin{align*} \|\Psi(t,s)y\|& \leq \|\Phi(t,s)y\|+|\int_s^t\|\Phi(t,\sigma(\tau))\|~\|B(\tau)\|~\|\Psi(\tau,s)y\|\Delta\tau|\\ & \leq K\,{\rm e}^{(\gamma+\alpha)(t-s)}\|y\|+K\delta |\int_s^t\|\Psi(\tau,s)y\|{\rm e}^{(\gamma+\alpha)(t-\sigma(\tau))}\Delta\tau|. \end{align*}

Mutiplying both sides by ${\rm e}^{-(\gamma +\alpha )t}$, we get

\begin{align*} {\rm e}^{-(\gamma+\alpha)t}\|\Psi(t,s)y\|& \leq K\,{\rm e}^{-(\gamma+\alpha) s}\|y\|+K\delta |\int_s^t\|\Psi(\tau,s)y\| {\rm e}^{-(\gamma+\alpha)\tau}\,{\rm e}^{-(\gamma+\alpha)\mu(\tau)}\Delta\tau|\\ & \leq K\,{\rm e}^{-(\gamma+\alpha) s}\|y\|+K\delta |\int_s^t\|\Psi(\tau,s)y\|\,{\rm e}^{-(\gamma+\alpha)\tau}\,{\rm e}^{|\gamma+\alpha|\mu^*}\Delta\tau|. \end{align*}

It can be seen that $-K\delta \,{\rm e}^{|\gamma + \alpha |\mu ^*}$ is positively regressive if $\delta <[\mu ^*K\,{\rm e}^{|\gamma + \alpha |\mu ^*}]^{-1}$. By lemma 4.4, for $\delta <[\mu ^*K\,{\rm e}^{|\gamma + \alpha |\mu ^*}]^{-1}$ we have

\[ {\rm e}^{-(\gamma+\alpha)t}\|\Psi(t,s)y\|\leq K\,{\rm e}^{-(\gamma+\alpha) s}\|y\| e_{{-}K\delta {\rm e}^{|\gamma+\alpha|M}}(t,s). \]

It can be seen that the function $\xi (v)=\log (1-vK\delta \,{\rm e}^{|\gamma +\alpha |\mu ^*})/v$ is decreasing with respect to $v$. Thus,

\[ {\rm e}^{-(\gamma+\alpha)t}\|\Psi(t,s)y\|\leq K\,{\rm e}^{-(\gamma+\alpha) s}\|y\|\exp\{{-}K\delta\,{\rm e}^{|\gamma+\alpha|\mu^*}(t-s)\}. \]

Mutiplyng both sides by ${\rm e}^{(\gamma +\alpha )t}$, we get

\begin{align*} \|\Psi(t,s)y\|& \leq K\,{\rm e}^{(\gamma+\alpha)(t-s)}\|y\|\exp\{{-}K\delta\,{\rm e}^{|\gamma+\alpha|\mu^*}(t-s)\}\\ & \leq K\|y\|\exp\{(\gamma+\alpha-K\delta\,{\rm e}^{|\gamma+\alpha|\mu^*})(t-s)\},\quad t\leq s. \end{align*}

Note that $\alpha _1=\alpha -K\delta \,{\rm e}^{|\gamma +\alpha |\mu ^*}>0$ for $\delta <\delta _1=\min \{[\mu ^*K\,{\rm e}^{|\gamma +\alpha |\mu ^*}]^{-1}, \alpha [K\,{\rm e}^{|\gamma +\alpha |\mu ^*}]^{-1}\}$, therefore system (4.6) admits $\gamma$-ED with projection $Q_1=O$ if $\delta <\delta _1.$ The other assertion can be proved in a similar way.

Proof. Let $k$ be the rank of $Q$. By lemma 4.5, system (2.2) is kinematically similar to the block diagonal system (4.5) by a Lyapunov transformation $x=J(t)y$, where $B_1(t)\in \mathbb {R}^{m\times m}$, $B_2(t)\in \mathbb {R}^{(n-m)\times (n-m)}$ for all $t\in \mathbb {T},$ and there exists a constant $K_0\geq 1$ such that the estimates

(4.7)\begin{equation} \begin{aligned} & \|\Phi_{B_1}(t,s)\|\leq K_0\,{\rm e}^{(\gamma-\alpha)(t-s)},\quad t\geq s,\\ & \|\Phi_{B_2}(t,s)\|\leq K_0\,{\rm e}^{(\gamma+\alpha)(t-s)},\quad t\leq s, \end{aligned} \end{equation}

hold for $t,s\in \mathbb {T}$. Let

\[ B_0(t)=\left(\begin{matrix} B_1(t) & \\ & B_2(t) \end{matrix}\right). \]

Note that

\begin{align*} J^{\Delta}(t)& =A(t)J(t)-J(\sigma(t))B_0(t)\\ & =[A(t)+B(t)]J(t)-J(\sigma(t))[B_0(t)+J^{{-}1}(\sigma(t))B(t)J(t)], \end{align*}

which implies that system (4.6) is kinematically similar to the system

(4.8)\begin{equation} z^{\Delta}(t)=[B_0(t)+J^{{-}1}(\sigma(t))B(t)J(t)]z. \end{equation}

Let

\[ D(t)=J^{{-}1}(\sigma(t))B(t)J(t) \]

and

\[ \mathcal{X}=\{H\in \mathcal{C}_{rd}(\mathbb{T},\mathbb{R}^{n\times n}):\|H\|\leq\infty\}, \]

where $\|H\|:=\sup \limits _{t\in \mathbb {T}}\|H(t)\|$. It can be seen that $\mathcal {X}$ is a Banach space with the norm $\|\cdot \|$. Let $E_k=\mathrm {diag}(I_k,O)$, where $I_k$ is the identity matrix of order $k$. Consider a matrix function $H\in \mathcal {X}$, and the mapping $T$ defined by

\begin{align*} TH(t)& =\int_{-\infty}^t\Phi_{B_0}(t,\sigma(s))E_k(I-H(\sigma(s)))D(s)(I+H(s))(I-E_k)\Phi_{B_0}(s,t)\Delta s\\ & \quad-\int_t^{\infty}\Phi_{B_0}(t,\sigma(s))(I-E_k)(I-H(\sigma(s)))D(s)(I+H(s))E_k\Phi_{B_0}(s,t)\Delta s. \end{align*}

Now we show that $TH\in \mathcal {X}$. It follows from (4.7) that

\begin{align*} \|\Phi_{B_0}(t,s)E_k\|& =\|\Phi_{B_1}(t,s)\|\leq K_0\,{\rm e}^{(\gamma-\alpha)(t-s)},\quad t\geq s,\\ \|\Phi_{B_0}(t,s)(I-E_k)\|& =\|\Phi_{B_2}(t,s)\|\leq K_0\,{\rm e}^{(\gamma+\alpha)(t-s)},\quad t\leq s. \end{align*}

Thus we have

(4.9)\begin{align} \|TH(t)\|& \leq \int_{-\infty}^tK_0\,{\rm e}^{(\gamma-\alpha)(t-s)}\,{\rm e}^{-(\gamma-\alpha)\mu(s)}(1+\|H\|)^2\|D\|K_0\,{\rm e}^{(\gamma+\alpha)(s-t)}\Delta s\nonumber\\ & \quad+\int_t^{\infty}K_0\,{\rm e}^{(\gamma+\alpha)(t-s)}\,{\rm e}^{-(\gamma+\alpha)\mu(s)}(1+\|H\|)^2\|D\|K_0\,{\rm e}^{(\gamma-\alpha)(s-t)}\Delta s\nonumber\\ & =K_0^2(1+\|H\|)^2\|D\|\nonumber\\ & \quad\times \left({\rm e}^{|\gamma-\alpha|\mu^*}\int_{-\infty}^t\,{\rm e}^{2\alpha(s-t)}\Delta s+{\rm e}^{|\gamma+\alpha|\mu^*}\int_t^{\infty}\,{\rm e}^{{-}2\alpha(s-t)}\Delta s\right). \end{align}

Let us define the map $\varphi :\mathbb {R}\rightarrow \mathcal {R}^+$:

\[ \varphi(\gamma):=\lim\limits_{s\searrow\mu(t)}\displaystyle\frac{{\rm e}^{\gamma s}-1}{s}. \]

It can be verified that $\gamma \leq \varphi (\gamma )\leq \frac {{\rm e}^{\gamma \mu ^*}-1}{\mu ^*}$. Note that

(4.10)\begin{equation} \int_{-\infty}^t\,{\rm e}^{2\alpha(s-t)}\Delta s\leq\int_{-\infty}^t\,{\rm e}^{{-}2\alpha(t-s)}\,\mathrm{d}s=\frac{1}{2\alpha} \end{equation}

and

(4.11)\begin{align} \int_t^{\infty}\,{\rm e}^{{-}2\alpha(s-t)}\Delta s=\int_t^{\infty}e_{\varphi({-}2\alpha)}(s,t)\Delta s=\left.\frac{{\rm e}^{{-}2\alpha(s-t)}}{\varphi({-}2\alpha)(s)}\right|_{s=t}^{s=\infty}\leq\frac{\mu^*}{1-{\rm e}^{{-}2\alpha\mu^*}}. \end{align}

Therefore, $TH(t)$ is bounded and $TH\in \mathcal {X}$. Let

\[ \mathcal{X}_0=\left\{H:H\in\mathcal{X},\|H\|\leq\displaystyle\frac{1}{2}\right\}, \]

then for any $H_1,H_2\in \mathcal {X}_0$, we have

\begin{align*} & (I-H_1(\sigma(t)))D(t)(I+H_1(t))-(I-H_2(\sigma(t)))D(t)(I+H_2(t))\\ & =(H_2(\sigma(t))-H_1(\sigma(t)))D(t)-D(t)(H_2(t)-H_1(t))\\ & \quad+(H_2(\sigma(t))-H_1(\sigma(t)))D(t)H_2(t)+H_1(\sigma(t))D(t)(H_2(t)-H_1(t)). \end{align*}

Thus,

(4.12)\begin{align} & \|TH_1(t)-TH_2(t)\|\nonumber\\ & \quad=\|\int_{-\infty}^t\Phi_{B_0}(t,\sigma(s))E_k[(I-H_1(\sigma(s)))D(s)(I+H_1(s))\nonumber\\ & \qquad-(I-H_2(\sigma(s)))D(s)(I+H_2(s))]\nonumber\\ & \qquad\cdot(I-E_k)\Phi_{B_0}(s,t)\Delta s -\int_t^{\infty}\Phi_{B_0}(t,\sigma(s))(I-E_k)\nonumber\\ & \qquad\times [(I-H_1(\sigma(s)))D(s)(I+H_1(s))\nonumber\\ & \qquad-(I-H_2(\sigma(s)))D(s)(I+H_2(s))]E_k\Phi_{B_0}(s,t)\Delta s\|\nonumber\\ & \quad=\|\int_{-\infty}^t\Phi_{B_0}(t,\sigma(s))E_k[(H_2(\sigma(t))-H_1(\sigma(t)))D(t)-D(t)(H_2(t)-H_1(t))\nonumber\\ & \qquad+(H_2(\sigma(t))-H_1(\sigma(t)))D(t)H_2(t)\nonumber\\ & \qquad+H_1(\sigma(t))D(t)(H_2(t)-H_1(t))](I-E_k)\Phi_{B_0}(s,t)\Delta s\nonumber\\ & \qquad -\int_t^{\infty}\Phi_{B_0}(t,\sigma(s))(I-E_k)[(H_2(\sigma(t))\nonumber\\ & \qquad-H_1(\sigma(t)))D(t)-D(t)(H_2(t)-H_1(t))\nonumber\\ & \qquad+(H_2(\sigma(t))-H_1(\sigma(t)))D(t)H_2(t)+H_1(\sigma(t))D(t)(H_2(t)\nonumber\\ & \qquad-H_1(t))]E_k\Phi_{B_0}(s,t)\Delta s\|\nonumber\\ & \quad\leq 3K_0^2\|D\|\,\|H_1-H_2\|\left({\rm e}^{|\gamma-\alpha|\mu^*}\int_{-\infty}^t\,{\rm e}^{2\alpha(s-t)}\Delta s+{\rm e}^{|\gamma+\alpha|\mu^*}\int_t^{\infty}\,{\rm e}^{{-}2\alpha(s-t)}\Delta s\right). \end{align}

Note that $\|D\|\leq \|J\|\|J^{-1}\|\|B\|$. It follows from (4.9), (4.10), (4.11) and (4.12) that there exists a constant $\delta >0$, such that $\|B\|<\delta$ implies that

1. (i) $TH\in \mathcal {X}_0$, if $H\in \mathcal {X}_0$;

2. (ii) $\|TH_1-TH_2\|\leq \frac {1}{2}\|H_1-H_2\|$, if $H_1,H_2\in \mathcal {X}_0.$

Therefore, $T$ is a contraction mapping. Note that $\mathcal {X}_0$ is a closed subspace of $\mathcal {X}$, then $\mathcal {X}_0$ is a Banach space with the norm $\|\cdot \|$. By the contraction mapping principle, there exists a unique fixed point $H\in \mathcal {X}_0$ such that

(4.13)\begin{align} H(t)& =\int_{-\infty}^t\Phi_{B_0}(t,\sigma(s))E_k(I-H(\sigma(s)))D(s)(I+H(s))(I-E_k)\Phi_{B_0}(s,t)\Delta s\nonumber\\ & \quad-\int_t^{\infty}\Phi_{B_0}(t,\sigma(s))(I-E_k)(I-H(\sigma(s)))D(s)(I+H(s))E_k\Phi_{B_0}(s,t)\Delta s. \end{align}

It can be seen that $E_k\Phi _{B_0}(t,s)=\Phi _{B_0}(t,s)E_k$ and $(I-E_k)\Phi _{B_0}(t,s)=\Phi _{B_0}(t,s)(I-E_k)$ since

\[ \Phi_{B_0}(t,s)=\left(\begin{matrix} \Phi_{B_1}(t,s) & \\ & \Phi_{B_2}(t,s)\end{matrix}\right). \]

Therefore, $E_kH(t)=H(t)(I-E_k)$. Then we obtain

\begin{align*} & H^{\Delta}(t)\\ & \quad=B_0(t)H(t)+\int_{-\infty}^t\Phi_{B_0}(\sigma(t),\sigma(s))E_k(I-H(\sigma(s)))D(s)(I+H(s))(I-E_k)\\ & \qquad\times(-\Phi_{B_0}(s,\sigma(t)))B_0(t)\Delta s\\ & \qquad-\int_t^{\infty}\Phi_{B_0}(\sigma(t),\sigma(s))(I-E_k)(I-H(\sigma(s)))D(s)(I+H(s))\\ & \qquad\times E_k(-\Phi_{B_0}(s,\sigma(t)))B_0(t)\Delta s\\ & \qquad+\Phi_{B_0}(\sigma(t),\sigma(t))E_k(I-H(\sigma(t)))D(t)(I+H(t))(I-E_k)\Phi_{B_0}(t,\sigma(t))\\ & \qquad+\Phi_{B_0}(\sigma(t),\sigma(t))(I-E_k)(I-H(\sigma(t)))D(t)(I+H(t))E_k\Phi_{B_0}(t,\sigma(t))\\ & \quad =B_0H(t)-H(\sigma(t))B_0(t)\\ & \qquad+\int_t^{\sigma(t)}\Phi_{B_0}(\sigma(t),\sigma(s))E_k(I-H(\sigma(s)))D(s)(I+H(s))\\ & \qquad\times(I-E_k)\Phi_{B_0}(s,\sigma(t))B_0(t)\Delta s\\ & \qquad+\int_t^{\sigma(t)}\Phi_{B_0}(\sigma(t),\sigma(s))(I-E_k)(I-H(\sigma(s)))D(s)(I+H(s))\\ & \qquad\times E_k\Phi_{B_0}(s,\sigma(t))B_0(t)\Delta s\\ & \qquad+E_k(I-H(\sigma(t)))D(t)(I+H(t))(I-E_k)\Phi_{B_0}(t,\sigma(t))\\ & \qquad+(I-E_k)(I-H(\sigma(t)))D(t)(I+H(t))E_k\Phi_{B_0}(t,\sigma(t))\\ & \quad=B_0H(t)-H(\sigma(t))B_0(t)+[E_k(I-H(\sigma(t)))D(t)(I+H(t))(I-E_k)\\ & \qquad+(I-E_k)(I-H(\sigma(t)))D(t)(I+H(t))E_k]\Phi_{B_0}(t,\sigma(t))(\sigma(t)B_0(t)+I)\\ & \quad=B_0H(t)-H(\sigma(t))B_0(t)+E_k(I-H(\sigma(t)))D(t)(I+H(t))(I-E_k)\\ & \qquad+(I-E_k)(I-H(\sigma(t)))D(t)(I+H(t))E_k\\ & \quad=B_0(t)H(t)-H(\sigma(t))B_0(t)+E_kD(t)(I+H(t))(I-E_k)\\ & \quad -E_kH(\sigma(t))D(t)(I+H(t))(I-E_k)\\ & \qquad+(I-E_k)D(t)(I+H(t))E_k-(I-E_k)H(\sigma(t))D(t)(I+H(t))E_k\\ & \quad=B_0(t)H(t)-H(\sigma(t))B_0(t)+E_kD(t)(I+H(t))(I-E_k)\\ & \qquad-H(\sigma(t))(I-E_k)D(t)(I+H(t))(I-E_k)\\ & \qquad+(I-E_k)D(t)(I+H(t))E_k-H(\sigma(t))E_kD(t)(I+H(t))E_k\\ & \quad=B_0(t)(I+H(t))-(I+H(\sigma(t)))B_0(t)+E_kD(t)(I+H(t))(I-E_k)\\ & \qquad+(I-E_k)D(t)(I+H(t))(I-E_k)\\ & \qquad -(I+H(\sigma(t)))(I-E_k)D(t)(I+H(t))(I-E_k)\\ & \qquad+(I-E_k)D(t)(I+H(t))E_k+E_kD(t)(I+H(t))E_k\\ & \qquad -(I+H(\sigma(t)))E_kD(t)(I+H(t))E_k\\ & \quad= B_0(t)(I+H(t))-(I+H(\sigma(t)))B_0(t)+D(t)(I+H(t))\\ & \qquad-(I+H(\sigma(t)))[(I-E_k)D(t)(I+H(t))(I-E_k)+E_kD(t)(I+H(t))E_k]. \end{align*}

Let

(4.14)\begin{equation} R(t)=I+H(t). \end{equation}

Then $\|R\|\leq \frac {3}{2}$ since $\|H\|\leq \frac {1}{2}$. For any $t\in \mathbb {T}$ and $y\in \mathbb {R}^n$, $y\neq 0$,

\[\|R(t)y\|=\|y+H(t)y\|\geq\|y\|-\|H(t)\|\|y\|\geq\|y\|-\displaystyle\frac{1}{2}\|y\|=\frac{1}{2}\|y\|.\]

It follows from $\|R(t)y\|\neq 0$ that $R(t)$ is invertible for any $t\in \mathbb {T}$. Then we obtain

\[ \|y\|=\|R(t)R^{{-}1}(t)y\|\geq\displaystyle\frac{1}{2}\|R^{{-}1}(t)y\|,\]

i.e., $\|R^{-1}(t)y\|\leq 2\|y\|$. Therefore, $\|R^{-1}(t)\|\leq 2$. Note that

\begin{align*} R^{\Delta}(t)& =H^{\Delta}(t)=[B_0(t)+D(t)]R(t)\\& \quad -R(\sigma(t))[B_0(t)+(I-E_k)D(t)R(t)(I-E_k)+E_kD(t)R(t)E_k], \end{align*}

which implies that system (4.8) is kinematically similar to the system

(4.15)\begin{equation} u^{\Delta}(t)=[B_0(t)+(I-E_k)D(t)R(t)(I-E_k)+E_kD(t)R(t)E_k]u. \end{equation}

Hence, system (4.6) is kinematically similar to system (4.15). It follows from lemma 4.6 that system (4.15) admits $\gamma$-ED when $\|B\|$ is sufficiently small and the rank of projection is $k$. Then system (4.6) also admits $\gamma$-ED and the rank of projection is $k$.

Proof. Let $\gamma \in (b_{k-1},a_k)$, then system (2.2) admits $\gamma$-ED. It follows from lemma 4.5 that system (2.2) is kinematically similar to

(4.17)\begin{equation} x^{\Delta}_1=A_1(t)x_1,\quad x^{\Delta}_2=A_2(t)x_2, \end{equation}

where the first equation of (4.17) admits $\gamma$-ED with the invariant projector $I$ and the second equation of (4.17) admits $\gamma$-ED with the invariant projector $O$. By lemma 4.9, we have $\Sigma (A_1)=\bigcup \limits _{i=1}^{k-1}[a_i,b_i],\quad \Sigma (A_2)=[a_k,b_k].$ Let $B_0=A_1$, $B_k=A_2$, then

\[ \Sigma(B_0)=\bigcup\limits_{i=1}^{k-1}[a_i,b_i],\quad \Sigma(B_k)=[a_k,b_k]. \]

The proof is completed by repeating the above steps.

Proof. (a). For any $\lambda \in [c,d]$, system (2.2) admits $\lambda$-ED, i.e., there exist constants $\alpha _\lambda >0,K_\lambda \geq 1$ and a projection $Q_\lambda$ such that

\begin{align*} \|\Phi_A(t,s)P_\lambda (s)\|\leq K_{\lambda}\,{\rm e}^{(\lambda-\alpha_\lambda)(t-s)},& t\geq s,\\ \|\Phi_A(t,s)(I-P_\lambda (s))\|\leq K_{\lambda}\,{\rm e}^{(\lambda+\alpha_\lambda)(t-s)},& t\leq s, \end{align*}

where $\Phi _A(t,s)$ is the evolution operator of system (2.2). Obviously, system (2.2) also admits $\gamma$-ED with the projector $P_\lambda$ when $\gamma \in (\lambda -\alpha _\lambda,\lambda +\alpha _\lambda )$. Since the family of open intervals $\{(\lambda -\alpha _{\lambda },\lambda +\alpha _{\lambda })| \lambda \in [c,d]\}$ covers the interval $[c,d]$. By Heine–Borel theorem, there are finite open intervals $(\lambda _1-\delta _{\lambda _1},\lambda _1+\delta _{\lambda _1}),\cdots,(\lambda _m-\delta _{\lambda _m},\lambda _m+\delta _{\lambda _m})$ that cover the interval $[c,d]$. Suppose $\lambda _1\leq \lambda _2\leq \cdots \leq \lambda _m$. Then $(\lambda _i-\delta _{\lambda _i},\lambda _{i}+\delta _{\lambda _{i}})\cap (\lambda _{i+1}-\delta _{\lambda _{i+1}},\lambda _{i+1}+\delta _{\lambda _{i+1}})\neq \emptyset$ $(i=1,2\cdots,m-1)$. Therefore, $P_{\lambda _i}=P_{\lambda _{i+1}}$ $(i=1,2\cdots,m-1)$, which implies $P_c=P_d$.

(b). Let $L=\sup \limits _{t\in \mathbb {T}}\|A(t)\|$ and let $M=\max \{L,\lambda \}$. According to lemma 4.11, for any $\lambda >b$, we have

\begin{align*} \|\Phi_A(t,s)x\|& =\|\Phi_A(t,s)x-\Phi_A(t,s)\cdot 0\|\leq\|x\| e_L(t,s)\leq {\rm e}^{L(t-s)}\|x\|\\& \leq {\rm e}^{M(t-s)}\|x\|,\quad t\in[s,+\infty)_\mathbb{T}, \end{align*}

which implies that $\|\Phi _A(t,s)\|\leq {\rm e}^{M(t-s)}$ for $t\in [s,+\infty )_{\mathbb {T}}$. Thus, system (2.2) admits $M$-ED with the projector $I$. Note that $[\lambda,M]\subseteq \mathbb {R}-\Sigma (A)$. From the statement $(a)$ in this lemma, we have system (2.2) that admits $\gamma$-ED with the projector $I$. The other assertion can be proved in a similar way.

Proof. Let $D(t)=\mathrm {diag}(c_{11}(t),c_{22}(t),\cdots,c_{nn}(t))$. Obviously, for any $\lambda \notin \bigcup \limits _{i=1}^{n}\Sigma (c_{ii})$, the diagonal system

\[ x^\Delta=D(t)x \]

admits $\lambda$-ED. From theorem 4.8, there exists $\delta >0$, such that the system

\[ y^\Delta=[\mathrm{diag}(c_{11}(t),c_{22}(t),\cdots,c_{nn}(t))+B(t)]y \]

also admits $\lambda$-ED if $\sup \limits _{t\in \mathbb {T}}\|B(t)\|<\delta$, where $B(t)\in \mathcal {C}_{rd}(\mathbb {T},\mathbb {R}^{n\times n})$. Let $\sup \limits _{t\in \mathbb {T}}\|C(t)\|=L$ and $\eta =\frac {\delta }{2n^2L}$. Taking the transformation $x=\mathrm {diag}(1,\eta,\cdots,\eta ^{n-1})z$, we obtain

\begin{align*} z^\Delta& =[\mathrm{diag}(1,\eta,\cdots,\eta^{n-1})]^{{-}1}x^\Delta\\ & =[\mathrm{diag}(1,\eta,\cdots,\eta^{n-1})]^{{-}1}C(t)x\\ & =[\mathrm{diag}(1,\eta,\cdots,\eta^{n-1})]^{{-}1}C(t)\mathrm{diag}(1,\eta,\cdots,\eta^{n-1})z\\ & =\left[\begin{pmatrix}c_{11}(t) & & & \\ & c_{22}(t) & & \\ & & \ddots & \\ & & & c_{nn}(t)\end{pmatrix}\right.\\& \left.\quad +\begin{pmatrix}0 & \eta c_{12}(t) & \eta^2 c_{13}(t) & \cdots & \eta^{n-1}c_{1n}(t)\\ & 0 & \eta c_{23}(t) & \cdots & \eta^{n-2}c_{2n}(t)\\ & & 0 & \cdots & \eta^{n-3} c_{3n}(t)\\ & & & \ddots & \vdots\\ & & & & 0 \end{pmatrix}\right]z\\ & \mathop{=}\limits^{\Delta}[D(t)+B(t)]z \end{align*}

It is clear that $\|B(t)\|<\delta$. Therefore, $z^{\Delta }=[D(t)+B(t)]z$ admits $\lambda$-ED. Then the system $x^{\Delta }=C(t)x$ also admits $\lambda$-ED, which implies $\Sigma (C)\subseteq \bigcup \limits _{i=1}^n\Sigma (c_{ii})$.

On the other hand, let $\lambda \notin \Sigma (C)=[a,b]$. If $\lambda >b$, by lemma 4.12, the system $x^{\Delta }=C(t)x$ admits $\lambda$-ED with the projector $I$. Then there exist constants $K_\lambda \geq 1, \alpha _\lambda >0$ such that

\[ \|\Phi_C(t,s)\|\leq K_\lambda\,{\rm e}^{(\lambda-\alpha_\lambda)(t-s)},\quad t\in[s,+\infty). \]

It can be easily verified that

\[ \Phi_C(t,s)=\begin{pmatrix}e_{c_{11}}(t,s) & * & \cdots & *\\ & e_{c_{22}}(t,s) & \cdots & *\\ & & \ddots & \vdots\\ & & & e_{c_{nn}}(t,s) \end{pmatrix} \]

and $|e_{c_{ii}}(t,s)|\leq \|\Phi _C(t,s)\|,i=1,2,\cdots,n.$ Therefore,

\[|e_{c_{ii}}(t,s)|\leq K_\lambda {\rm e}^{(\lambda-\alpha_\lambda)(t-s)},\quad t\in[s,+\infty),i=1,2,\cdots,n, \]

which implies that the system $x_i^{\Delta }=c_{ii}x_i$ admits $\lambda$-ED for any $i\in \{1,2,\cdots,n\}$. Thus, $\lambda \notin \bigcup \limits _{i=1}^n\Sigma (c_{ii})$. In a similar way, we can prove that $\lambda \notin \bigcup \limits _{i=1}^n\Sigma (c_{ii})$ if $\lambda < a$. Therefore, $\bigcup \limits _{i=1}^n\Sigma (c_{ii})\subseteq \Sigma (C)$. In conclusion, we have $\Sigma (C)=\bigcup \limits _{i=1}^n\Sigma (c_{ii})$. The proof is completed.

Proof. If $\lambda \notin F$, then $\alpha =\inf \limits _{x\in F}|\lambda -x|>0$ since $F=\bigcup \limits _{i=1}^n\varpi (F_i)$, where $F_i$ are closed intervals $(i=1,2,\cdots,n)$. It follows from the definition of generalized contractible set that there exist $c_1(t),c_2(t),\cdots,c_n(t)\in \mathcal {C}_{rd}(\mathbb {T},\mathbb {R})$ and $B(t)\in \mathcal {C}_{rd}(\mathbb {T},\mathbb {R}^{n\times n})$, such that $\sup \limits _{t\in \mathbb {T}}\|B(t)\|\leq \delta,\mathrm {Im}\,\overline {\xi }_\mu (c_i)\subseteq F_i(i=1,2,\cdots,n)$ and system (2.2) is kinematically similar to

(4.18)\begin{equation} y^{\Delta}=[\mathrm{diag}(c_1(t),c_2(t),\cdots,c_n(t))+B(t)]y. \end{equation}

Without loss of generalization, we assume that

\begin{align*} & \varpi(F_i)<\lambda-\alpha,\quad (i=1,2,\cdots,k),\\ & \varpi(F_j)>\lambda+\alpha,\quad (j=k+1,\cdots,n), \end{align*}

which implies

\begin{align*} & \overline{\xi}_\mu(c_i)(t)<\lambda-\alpha,\quad\forall t\in\mathbb{T}\ (i=1,2,\cdots,k),\\ & \overline{\xi}_\mu(c_j)(t)>\lambda+\alpha,\quad\forall t\in\mathbb{T}\ (j=k+1,\cdots,n). \end{align*}

Therefore, we have

\begin{align*} & |e_{c_i}(t,s)|={\rm e}^{\int_s^t\overline{\xi}_\mu(c_i)(s)\Delta s}<{\rm e}^{(\lambda-\alpha)(t-s)},\quad(t\geq s)(i=1,2,\cdots,k),\\ & |e_{c_j}(t,s)|={\rm e}^{\int_s^t\overline{\xi}_\mu(c_j)(s)\Delta s}<{\rm e}^{(\lambda+\alpha)(t-s)},\quad(t\leq s)(j=k+1,\cdots,n). \end{align*}

Since $\sup \limits _{t\in \mathbb {T}}\|B(t)\|\leq \delta$ and $\delta$ can be sufficiently small, by roughness theorem, we get system (4.18) admits $\lambda$-ED. Therefore, system (2.2) also admits $\lambda$-ED, which implies $\lambda \notin \Sigma (A)$. The proof is completed.

Proof. The proof is divided into several parts. Firstly, we prove that $F\subseteq \Sigma (A)$.

Part 1: System (2.2) is kinematically similar to an upper triangular system.

Suppose $\Sigma (A)=\bigcup \limits _{i=1}^k[a_i,b_i] (1\leq k\leq n),a_1\leq b_1< a_2\leq b_2<\cdots < a_k\leq b_k$. From lemma 4.10, system (2.2) is kinematically similar to system (4.16), where $B_i(t) (i=1,2,\cdots,k)$ are bounded rd-continuous functions and $\Sigma (B_i)=[a_i,b_i]$. Using Perron's transformation, the system

(4.19)\begin{equation} x_i^\Delta=B_i(t)x_i \end{equation}

is kinematically similar to the $n_i\times n_i$ upper triangular system

(4.20)\begin{equation} y_i^\Delta=D_i(t)y_i, \end{equation}

where $\Sigma (D_i)=[a_i,b_i]$. Let $D_i(t)=(d_{ij}^{(i)}(t))$. From lemma 4.13, $\bigcup \limits _{r=1}^{n_i}\Sigma (d_{rr}^{(i)})=[a_i,b_i].$ Then, by lemma 4.12, for any $\delta >0$, the system

\[ u^\Delta=d_{rr}^{(i)}(t)u \]

admits $(b_i+\delta )$-ED with projector $I$ and $(a_i-\delta )$-ED with projector $O$. In consequence, there exist constants $K=1,\alpha >0$, such that

(4.21)\begin{equation} \begin{aligned} & |e_{d_{rr}^{(i)}}(t,s)|\leq {\rm e}^{(b_i+\delta-\alpha)(t-s)},\quad \text{for }t\in[s,+\infty)_{\mathbb{T}},\\ & |e_{d_{rr}^{(i)}}(t,s)|\leq {\rm e}^{(a_i-\delta+\alpha)(t-s)},\quad \text{for }t\in(-\infty,s]_{\mathbb{T}}. \end{aligned} \end{equation}

Part 2: We are going to construct a strictly increasing and unbounded sequence $\{t^{(ir)}_p\}_{p=0}^{+\infty }$ such that for any constant $M>\exp \{(b_i+1-a_i+\delta )(1+\mu ^*)\}$, the estimate

(4.22)\begin{equation} M^{{-}1}\leq {\rm e}^{-\int_{t_0}^th_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|\leq M \end{equation}

holds for all $t\in [t_0,+\infty )_\mathbb {T}$, where $h_i,g_i:[t_0,+\infty )_{\mathbb {T}}\rightarrow \mathbb {R}$ is defined by:

\[ h_i(t)=\begin{cases}a_i & \text{if }t\in[t^{(ir)}_j,t^{(ir)}_{j+1})_{\mathbb{T}},\\ b_i & \text{if }t\in[t^{(ir)}_{j+1},t^{(ir)}_{j+2})_{\mathbb{T}}, \end{cases}\ (j=0,2,4,\cdots) \]

and

\[ g_i(t)=\begin{cases} -\delta & \text{if }t\in[t^{(ir)}_j,t^{(ir)}_{j+1})_{\mathbb{T}},\\ \delta & \text{if }t\in[t^{(ir)}_{j+1},t^{(ir)}_{j+2})_{\mathbb{T}}.\end{cases}\ (j=0,2,4,\cdots)\]

In what follows, we denote $t_p^{(ir)}$ by $t_p$ $(p\in \mathbb {N}_0)$ if there is no ambiguity. Interchanging $t$ by $s$ in the second inequality of (4.21), we obtain

(4.23)\begin{equation} \begin{cases} {\rm e}^{-(b_i+\delta)(t-s)}|e_{d^{(i)}_{rr}}(t,s)|\leq {\rm e}^{-\alpha(t-s)}, & t\in[t_0,+\infty)_{\mathbb{T}},\\ {\rm e}^{-(a_i-\delta)(t-s)}|e_{d^{(i)}_{rr}}(t,s)|\geq {\rm e}^{\alpha(t-s)}, & t\in[t_0,+\infty)_{\mathbb{T}}.\\ \end{cases} \end{equation}

Let

\[ U(t,t_0)={\rm e}^{-(a_i-\delta)(t-t_0)}|e_{d^{(i)}_{rr}}(t,t_0)|,\quad V(t,t_0)={\rm e}^{-(b_i+\delta)(t-t_0)}|e_{d^{(i)}_{rr}}(t,t_0)|. \]

Therefore, by (4.23), we have

(4.24)\begin{equation} \begin{cases} V(t,s)\leq {\rm e}^{-\alpha(t-s)}, & t\in[t_0,+\infty)_{\mathbb{T}},\\ U(t,s)\geq {\rm e}^{\alpha(t-s)}, & t\in[t_0,+\infty)_{\mathbb{T}}.\end{cases} \end{equation}

It can be seen that $U(t,s)\geq 1$ and $V(t,s)\leq 1$ if $t\geq s$ and for any fixed $t_0\in \mathbb {T}$, $U(t,t_0)$ is unbounded on $[t_0,+\infty )_\mathbb {T}$ since $\alpha >0$. Moreover, $V(t,t_0)\rightarrow 0$ as $t\rightarrow +\infty$. In consequence, there is $t_1\geq t_0$ such that

(4.25)\begin{equation} M^{{-}1}<1\leq U(t,t_0)\leq M \end{equation}

for any $t\in [t_0,t_1]$ and

(4.26)\begin{equation} U(\sigma(t_1),t_0)\geq M. \end{equation}

Meanwhile, we assert that $t_1-t_0>1$. In fact, one can see that

\[ U(t,s)\leq {\rm e}^{(b_i+1-a_i+\delta)(t-s)},\quad V(t,s)\leq {\rm e}^{(b_i+1-b_i-\delta)(t-s)}\quad \text{for }t\geq s \]

since $\Sigma (d^{(i)}_{rr})\subseteq [a_i,b_i]$ and $|e_{d_{rr}^{(i)}}(t,s)|\leq {\rm e}^{(b_i+1)(t-s)}$ for $t\geq s$. From (4.26), we get

\[ M\leq {\rm e}^{(b_i+1-a_i+\delta)(\sigma(t_1)-t_0)}, \]

which implies that

\[ \sigma(t_1)-t_0\geq\frac{\ln M}{b_i+1-a_i+\delta}>1+\mu^* \]

since $M>\exp \{(b_i+1-a_i+\delta )(1+\mu ^*)\}$, and then $t_1-t_0>1.$

On the other hand, the function $U(t_1,t_0)V(t,t_1)$ is convergent to zero as $t\rightarrow +\infty$. Then, there exists $t_2>t_1$ such that

\[ U(t_1,t_0)V(\sigma(t_2),t_1)\leq M^{{-}1} \]

and

\[U(t_1,t_0)V(t,t_1)\geq M^{{-}1}\quad\text{for any }t\in[t_1,t_2]_{\mathbb{T}}. \]

Combined with the first inequality of (4.24) and (4.25), we have

(4.27)\begin{equation} M^{{-}1}\leq U(t_1,t_0)V(t,t_1)\leq M\quad\text{for any }t\in[t_1,t_2]_{\mathbb{T}}. \end{equation}

If $t\in [t_1,t_2]_{\mathbb {T}}$, by the definition of $h_i(t)$ and $g_i(t)$, we have

\begin{align*} & {\rm e}^{-\int_{t_0}^th_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|\\ & \quad={\rm e}^{-(\int_{t_0}^{t_1}+\int_{t_1}^t)(h_i(\tau)+g_i(\tau))\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_1)||e_{d_{rr}^{(i)}}(t_1,t_0)|\\ & \quad={\rm e}^{-(a_i-\delta)(t_1-t_0)}|e_{d^{(i)}_{rr}}(t_1,t_0)|\\& \qquad \cdot {\rm e}^{-(b_i+\delta)(t-t_1)}|e_{d^{(i)}_{rr}}(t,t_1)|=U(t_1,t_0)V(t,t_1). \end{align*}

Combined with (4.27), equation (4.22) is verified for any $t\in [t_0,t_2]_\mathbb {T}.$ As inductive hypothesis, we assume that there are $2m+1$ numbers $t_0< t_1<\cdots < t_{2m-1}< t_{2m}$ such that

(4.28)\begin{equation} {\rm e}^{-\int_{t_0}^{\sigma(t_{2m})}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(\sigma(t_{2m}),t_0)|\leq M^{{-}1} \end{equation}

and

\[ M^{{-}1}\leq {\rm e}^{-\int_{t_0}^th_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|\leq M\quad\text{for any }t\in[t_{2m-1},t_{2m}]. \]

Using the second inequality of (4.24), we have that

(4.29)\begin{equation} {\rm e}^{-\int_{t_0}^{t_{2m}}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t_{2m},t_0)|U(t,t_{2m}) \end{equation}

is unbounded on $[t_{2m},+\infty )_{\mathbb {T}}$. Then, there exists $t_{2m+1}\geq t_{2m}$ such that this product is less than $M$ for any $t\in [t_{2m},t_{2m+1}]$ and

(4.30)\begin{equation} {\rm e}^{-\int_{t_0}^{t_{2m}}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t_{2m},t_0)|U(\sigma(t_{2m+1}),t_{2m})\geq M. \end{equation}

In addition, it follows from the inductive hypothesis and $U(t,t_{2m})\geq 1\ (t\geq t_{2m})$ that

\[M^{{-}1}\leq {\rm e}^{-\displaystyle\int_{t_0}^{t_{2m}}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t_{2m},t_0)|U(t,t_{2m})\leq M\quad\text{for any }t\in[t_{2m},t_{2m+1}]_{\mathbb{T}}. \]

Moreover, by (4.31) and (4.30), it is not difficult to verify that $t_{2m+1}-t_{2m}>1$.

Finally, we have that

\[ {\rm e}^{-\displaystyle\int_{t_0}^{t_{2m}}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t_{2m},t_0)|U(t_{2m+1},t_{2m})V(t,t_{2m+1}) \]

converges to zero as $t\rightarrow +\infty$. Then there exists $t_{2m+2}$ such that the product above is greater than $M^{-1}$ when $t\in [t_{2m+1},t_{2m+2}]$. Since $V(t,t_{2m+1})\leq 1$ for $t\geq t_{2m+1}$, the product above is less than $M$. It can be easily verified that

\[ {\rm e}^{-\int_{t_0}^{t}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|= {\rm e}^{-\int_{t_0}^{t_{2m}}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t_{2m},t_0)|U(t,t_{2m}) \]

for $t\in [t_{2m},t_{2m+1}]$ and

\begin{align*} & {\rm e}^{-\int_{t_0}^{t}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|\\& \quad={\rm e}^{-\int_{t_0}^{t_{2m}}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t_{2m},t_0)|U(t_{2m+1},t_{2m})V(t,t_{2m+1}) \end{align*}

for $t\in [t_{2m+1},t_{2m+2}]$. This proves (4.22) and $t_p\rightarrow +\infty$ as $p\rightarrow +\infty$.

Part 3: In a similar way, we can define $h_i(t)$ and $g_i(t)$ on $(-\infty,t_0]$ satisfying

\[ M^{{-}1}\leq {\rm e}^{-\int_{t_0}^{t}h_i(\tau)+g_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|\leq M. \]

Since $h_i(t)$ and $g_i(t)$ are sectionally continuous on $\mathbb {T}$. Then there exist continuous functions $\widehat {h}_i(t)$ and $\widehat {g}_i(t)$ satisfying

\[ a_i\leq\widehat{h}_i(t)\leq b_i,\quad -\delta\leq\widehat{g}_i(t)\leq \delta \]

and

\[\displaystyle\int_{-\infty}^{+\infty}|(\widehat{h}_i(\tau)+\widehat{g}_i(\tau))-(h_i(\tau)+g_i(\tau))|\Delta\tau\leq1. \]

Thus, we have

\[ (eM)^{{-}1}\leq {\rm e}^{-\int_{t_0}^{t}\widehat{h}_i(\tau)+\widehat{g}_i(\tau)\Delta\tau}|e_{d_{rr}^{(i)}}(t,t_0)|\leq eM,\quad t\in\mathbb{T}. \]

Let $S_{ir}(t)= {\rm e}^{-\int _{t_0}^{t}\widehat {h}_i(\tau )+\widehat {g}_i(\tau )\Delta \tau }\cdot e_{d_{rr}^{(i)}}(t,t_0)$ and

\[ L_i(t)=\mathrm{diag}\left(S_{i1}(t),S_{i2}(t),\cdots,S_{in_i}(t)\right). \]

It can be seen that $\|L_i(t)\|\leq eM$ and $\|L^{-1}_i(t)\|\leq eM$ for any $t\in \mathbb {T}$ and system (4.20) is kinematically similar to

(4.31)\begin{equation} z_i^{\Delta}=\Lambda_i(t)z_i \end{equation}

with $y_i=L_i(t)z_i$, where $\Lambda _i(t)=L^{-1}(\sigma (t))D_i(t)L(t)-L^{-1}(\sigma (t))L^{\Delta }(t)$ is a $n_i\times n_i$ matrix whose $rj$-coefficient is defined by

\[ \{\Lambda_i(t)\}_{rj}=\begin{cases} d^{(i)}_{rr}S_{ir}^{{-}1}(\sigma(t))S_{ir}(t)- S_{ir}^{{-}1}(\sigma(t))S_{ir}^{\Delta}(t) & \text{if }r=j,\\ d^{(i)}_{rj}S_{ir}^{{-}1}(\sigma(t))S_{ij}(t) & \text{if }1\leq r< j\leq n_i,\\ 0 & \text{others.} \end{cases} \]

A straightforward calculation leads to

\begin{align*} \{\Lambda_i(t)\}_{rj}& =d^{(i)}_{rj}S_{ir}^{{-}1}(\sigma(t))S_{ij}(t)=d^{(i)}_{rj}\\& \quad\cdot\frac{{\rm e}^{\mu(t)(\widehat{h}_i(t)+\widehat{g}_i(t))}e_{d^{(i)}_{jj}\ominus d^{(i)}_{rr}}(t,t_0)}{1+\mu(t)d^{(i)}_{rr}},\quad (r\leq j)\\ S_{ir}^{{-}1}(\sigma(t))S_{ir}^{\Delta}(t)& =S_{ir}^{{-}1}(\sigma(t))\lim_{s\searrow\mu(t)}\frac{S_{ir}(t+s)-S_{ir}(t)}{s}\\& =\lim_{s\searrow\mu(t)}\left(1-\frac{{\rm e}^{s(\widehat{h}_i(t)+\widehat{g}_i(t))}}{1+sd^{(i)}_{rr}(t)}\right)\cdot s^{{-}1}, \end{align*}

and

(4.32)\begin{equation} \{\Lambda_i(t)\}_{rr}=d^{(i)}_{rr}S_{ir}^{{-}1}(\sigma(t))S_{ir}(t)- S_{ir}^{{-}1}(\sigma(t))S_{ir}^{\Delta}(t)=\lim_{s\searrow\mu(t)}\frac{{\rm e}^{s(\widehat{h}_i(t)+\widehat{g}_i(t))}-1}{s}. \end{equation}

Since $(eM)^{-1}\leq S_{ir}(t)\leq eM$ and $A(t)$ is bounded, we obtain $d_{rr}^{(i)}$ is bounded and $d^{(i)}_{rj}S_{ir}^{-1}(\sigma (t))S_{ij}(t)$ is bounded, i.e., $\{\Lambda _i(t)\}_{rj}$ is bounded if $r< j$.

We define the $\eta$-transformation

\[ z_i(t)=\mathrm{diag}(1,\eta,\cdots,\eta^{n_i-1})w_i(t). \]

It can be seen that (4.31) and (4.20) are kinematically similar to

\[ w_i^{\Delta}=\Gamma_i(t)w_i, \]

where the $rj$-coefficient of $\Gamma (t)$ is

\[ \{\Gamma_i(t)\}_{rj}=\begin{cases} \{\Lambda_i(t)\}_{rj} & \text{if }r=j,\\ \eta^{j-r}\{\Lambda_i(t)\}_{rj} & \text{if }1\leq r< j\leq n_i,\\ 0 & \text{others.}\end{cases} \]

Observe that $\Gamma _i(t)$ can be written as

\[ \Gamma_i(t)=\mathrm{diag}(\{\Lambda_i(t)\}_{11},\{\Lambda_i(t)\}_{22},\cdots,\{\Lambda_i(t)\}_{n_in_i})+W_i(t), \]

where the $rj$-coefficient of $W_i(t)$ is defined by

\[ \{W_i(t)\}_{rj}=\begin{cases}\eta^{j-r}\{\Lambda_i(t)\}_{rj} & \text{if }1\leq r< j\leq n_i,\\ 0 & \text{others.}\end{cases} \]

Since $\{\Lambda _i(t)\}_{rj}$ is bounded if $r< j$, we obtain that $\|W_i(t)\|\rightarrow 0$ as $\eta \rightarrow 0$. On the other hand, it follows from (4.32) that

\begin{align*} \overline{\xi}_\mu(\{\Lambda_i(t)\}_{rr})& =\mathrm{Re}\left(\lim_{s\searrow\mu(t)}\displaystyle\frac{\log(1+s\{\Lambda_i(t)\}_{rr})}{s}\right)\\& =\mathrm{Re}\left(\widehat{h}_i(t)+\widehat{g}_i(t)\right)\in(a_i-\delta,b_i+\delta). \end{align*}

Therefore, it is clear that system (2.2) is $\Delta$-contracted to the set ${G_{\delta }=\bigcup \limits _{i=1}^k(a_i-\delta,b_i+\delta)}$ for any $\delta >0$. Since the constant $\delta$ can be sufficiently small and $G_{\delta }\rightarrow \Sigma (A)$ as $\delta \rightarrow 0$, we have $F\subseteq \Sigma (A)$, where $F$ is the generalized contractible set of (2.2).

On the other hand, by the definition of generalized contractible set and lemma 4.14, we have $\Sigma (A)\subseteq F$. In conclusion, we have $F=\Sigma (A)$. The proof is completed.