Theorem 1 If there exist a symmetric and positive definite matrix P\in {R}^{n\times n}, positive scalar constants {g}_{1}>0, {\u03f5}_{1}>0, {\u03f5}_{2}>0, and scalar constant {g}_{2}\in R such that the following hold:

(1)
PA+{A}^{T}P+P{K}_{1}+{K}_{1}^{T}P+{\u03f5}_{1}{P}^{2}+{\u03f5}_{1}^{1}L+{g}_{1}P\le 0,

(2)
PA+{A}^{T}P+P{K}_{2}+{K}_{2}^{T}P+{\u03f5}_{2}{P}^{2}+{\u03f5}_{2}^{1}L{g}_{2}P\le 0,

(3)
{g}_{1}\tau {g}_{2}(T\tau )>0,
then the origin of the system (3) is exponentially stable, and
\parallel x(t)\parallel <\sqrt{\frac{{\lambda}_{M}(P)}{{\lambda}_{m}(P)}}\parallel {x}_{0}\parallel exp[\gamma (tT)],
where \gamma =\frac{{g}_{1}\tau {g}_{2}(T\tau )}{2T}, for any t>0.
Proof Let us construct the following Lyapunov function:
V(x(t))={x}^{T}(x)Px(t),
(5)
from which we obtain
{\lambda}_{m}(P){\parallel x(t)\parallel}^{2}\le V(x(t))\le {\lambda}_{M}(P)\parallel x(t)\parallel .
(6)
If mT\le t<mT+\tau, then by (3), (4), and (5) we have
\begin{array}{rcl}\dot{V}(x)& =& 2{x}^{T}P\dot{x}\\ =& 2{x}^{T}P[Ax+f(x)+{K}_{1}x]\\ =& 2{x}^{T}PAx+2{x}^{T}Pf(x)+2{x}^{T}P{K}_{1}x\\ =& {x}^{T}[PA+{A}^{T}P+P{K}_{1}+{K}_{1}^{T}P]x+2{x}^{T}Pf(x)\\ \le & {x}^{T}[PA+{A}^{T}P+P{K}_{1}+{K}_{1}^{T}P]x\\ +{\u03f5}_{1}{x}^{T}{P}^{2}x+{\u03f5}_{1}^{1}{x}^{T}Lx\\ =& {g}_{1}V(x)+{x}^{T}[PA+{A}^{T}P+P{K}_{1}+{K}_{1}^{T}P\\ +{\u03f5}_{1}{P}^{2}+{\u03f5}_{1}^{1}L+{g}_{1}P]x\\ \le & {g}_{1}V(x),\end{array}
which implies that
V(x(t))\le V\left(x\left({(mT)}^{}\right)\right)exp({g}_{1}(tmT)).
(7)
Similarly, if mT+\tau \le t<(m+1)T, then we have
\begin{array}{rcl}\dot{V}(x)& =& 2{x}^{T}P\dot{x}\\ \le & {g}_{2}V(x)+{x}^{T}[PA+{A}^{T}P+P{K}_{2}+{K}_{2}^{T}P+{\u03f5}_{2}{P}^{2}+{\u03f5}_{2}^{1}L{g}_{2}P]x\\ \le & {g}_{2}V(x),\end{array}
which implies that
V(x(t))\le V\left(x\left({(mT+\tau )}^{}\right)\right)exp({g}_{2}(tmT\tau )).
(8)
It follows from (7) and (8) that:
(1) If 0\le t<\tau, then we have
V(x(t))\le V({x}_{0})exp({g}_{1}t).
So
V\left(x\left({\tau}^{}\right)\right)\le V({x}_{0})exp({g}_{1}\tau ).
(2) If \tau \le t<T, then we have
\begin{array}{rcl}V(x(t))& \le & V\left(x\left({\tau}^{}\right)\right)exp({g}_{2}(t\tau ))\\ \le & V({x}_{0})exp({g}_{1}\tau +{g}_{2}(t\tau )).\end{array}
So
V\left(x\left({T}^{}\right)\right)\le V({x}_{0})exp({g}_{1}\tau +{g}_{2}(T\tau )).
(3) If T\le t<T+\tau, then we have
\begin{array}{rcl}V(x(t))& \le & V\left(x\left({T}^{}\right)\right)exp({g}_{1}(tT))\\ \le & V({x}_{0})exp({g}_{1}\tau {g}_{1}(tT)+{g}_{2}(T\tau )).\end{array}
So
V\left(x\left({(T+\tau )}^{}\right)\right)\le V({x}_{0})exp(2{g}_{1}\tau +{g}_{2}(T\tau )).
(4) If T+\tau \le t<2T, then we have
\begin{array}{rcl}V(x(t))& \le & V\left(x\left({(T+\tau )}^{}\right)\right)exp({g}_{2}(tT\tau ))\\ \le & V({x}_{0})exp(2{g}_{1}\tau +{g}_{2}(T\tau )+{g}_{2}(tT\tau )).\end{array}
So
V\left(x\left({(2T)}^{}\right)\right)\le V({x}_{0})exp(2{g}_{1}\tau +2{g}_{2}(T\tau )).
(5) If 2T\le t<2T+\tau, then we have
\begin{array}{rcl}V(x(t))& \le & V\left(x\left({(2T)}^{}\right)\right)exp({g}_{1}(t2T))\\ \le & V({x}_{0})exp(2{g}_{1}\tau {g}_{1}(t2T)+2{g}_{2}(T\tau )).\end{array}
So
V\left(x\left({(2T+\tau )}^{}\right)\right)\le V({x}_{0})exp(3{g}_{1}\tau +2{g}_{2}(T\tau )).
(6) If 2T+\tau \le t<3T, then we have
\begin{array}{rcl}V(x(t))& \le & V\left(x\left({(2T+\tau )}^{}\right)\right)exp({g}_{2}(t2T\tau ))\\ \le & V({x}_{0})exp(3{g}_{1}\tau +2{g}_{2}(T\tau )+{g}_{2}(t2T\tau )).\end{array}
So
V\left(x\left({(3T)}^{}\right)\right)\le V({x}_{0})exp(3{g}_{1}\tau +3{g}_{2}(T\tau )).
By induction, we have:
(7) If mT\le t<mT+\tau, i.e., \frac{t\tau}{T}<m\le \frac{t}{T}, then we have
V(x(t))\le V({x}_{0})exp(m{g}_{1}\tau {g}_{1}(tmT)+m{g}_{2}(T\tau )).
(9)
So
V\left(x\left({(mT+\tau )}^{}\right)\right)\le V({x}_{0})exp((m+1){g}_{1}\tau +m{g}_{2}(T\tau )).
(8) If mT+\tau \le t<(m+1)T, i.e., \frac{t}{T}<m+1\le \frac{t+T\tau}{T}, then we have that
\begin{array}{rcl}V(x(t))& \le & V\left(x\left({(mT+\tau )}^{}\right)\right)exp({g}_{2}(tmT\tau ))\\ \le & V({x}_{0})exp((m+1){g}_{1}\tau +m{g}_{2}(T\tau )\\ +{g}_{2}(tmT\tau )).\end{array}
(10)
From (9) we know that
\begin{array}{rcl}V(x(t))& \le & V({x}_{0})exp(m{g}_{1}\tau +m{g}_{2}(T\tau ))\\ =& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))m)\\ <& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{t\tau}{T})\\ <& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{tT}{T}),\end{array}
(11)
where mT\le t<mT+\tau.
From (10) we know that
Case 1. If {g}_{2}>0, then
\begin{array}{rcl}V(x(t))& \le & V({x}_{0})exp((m+1){g}_{1}\tau +(m+1){g}_{2}(T\tau ))\\ <& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{t}{T})\\ \le & V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{t\tau}{T})\\ <& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{tT}{T}).\end{array}
Case 2. If {g}_{2}\le 0, then
\begin{array}{rcl}V(x(t))& \le & V({x}_{0})exp((m+1){g}_{1}\tau +m{g}_{2}(T\tau ))\\ \le & V({x}_{0})exp(m{g}_{1}\tau +m{g}_{2}(T\tau ))\\ =& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))m)\\ <& V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{tT}{T}).\end{array}
So, for any {g}_{2}\in R, we have
V(x(t))<V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{tT}{T}),
(12)
where mT+\tau \le t<(m+1)T.
It follows from (11) and (12) that, for any t>0,
V(x(t))<V({x}_{0})exp(({g}_{1}\tau {g}_{2}(T\tau ))\frac{tT}{T}).
(13)
By (5), (6), and (13), we conclude that
\parallel x(t)\parallel <\sqrt{\frac{{\lambda}_{M}(P)}{{\lambda}_{m}(P)}}\parallel {x}_{0}\parallel exp[\gamma (tT)],
where \gamma =\frac{{g}_{1}\tau {g}_{2}(T\tau )}{2T}, for any t>0.
So we finish the proof. □
From Lemma 2, we know that the two conditions of Theorem 1 are equivalent to the following two LMIs, respectively:
\left[\begin{array}{cc}PA+{A}^{T}P+P{K}_{1}+{K}_{1}^{T}P+{\u03f5}_{1}^{1}L+{g}_{1}P& P\\ P& {\u03f5}_{1}^{1}I\end{array}\right]\le 0,
(14)
\left[\begin{array}{cc}PA+{A}^{T}P+P{K}_{2}+{K}_{2}^{T}P+{\u03f5}_{2}^{1}L{g}_{2}P& P\\ P& {\u03f5}_{2}^{1}I\end{array}\right]\le 0.
(15)