Theorem 1
Suppose that a symmetric and positive definite matrix
\(D\in R^{n\times n}\)
and positive scalar constants
\(h_{1}>0\), \(h_{2}>0\), \(\epsilon_{1}>0\), and
\(\epsilon_{2}>0\)
satisfy the following conditions:
-
(1)
\(DH+H^{T}D+DC+C^{T}D+\epsilon_{1}D^{2}+\epsilon_{1}^{-1}L+h_{1}D\leq 0\),
-
(2)
\(DH+H^{T}D+\epsilon_{2}D^{2}+\epsilon_{2}^{-1}L-h_{2}D\leq 0\),
-
(3)
\(h_{1}\frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln\lambda_{2}>0\),
where
\(\lambda_{1}=\lambda_{M}(D^{-1}(I+J1)^{T}D(I+J1))\), \(\lambda_{2}=\lambda _{M}(D^{-1}(I+J2)^{T}D(I+J2))\). Then the origin of system (2) is exponentially stable.
Proof
First, we construct the Lyapunov function
$$ V \bigl(x(t) \bigr)=x^{T}(t)Dx(t), $$
(4)
so that
$$ \lambda_{m}(D) \bigl\Vert x(t) \bigr\Vert ^{2}\leq V \bigl(x(t) \bigr)\leq\lambda _{M}(D) \bigl\Vert x(t) \bigr\Vert ^{2}. $$
(5)
If \(mT< t<mT+\frac{T}{2}\), then by (2), (3), and (4) we get
$$\begin{aligned} \dot{V}(x) =&2x^{T}D\dot{x} \\ =&2x^{T}D \bigl[Hx+f(x)+Cx \bigr] \\ =&2x^{T}DHx+2x^{T}Df(x)+2x^{T}DCx \\ =& x^{T}[2DH+2DC]x+2x^{T}Df(x) \\ =& x^{T} \bigl[DH+H^{T}D+DC+C^{T}D \bigr]x+2x^{T}Df(x) \\ \leq&x^{T} \bigl[DH+H^{T}D+DC+C^{T}D \bigr]x \\ &{}+\epsilon_{1} x^{T}D^{2}x+ \epsilon_{1}^{-1}x^{T}Lx \\ =&x^{T} \bigl[DH+H^{T}D+DC+C^{T}D+ \epsilon_{1}D^{2}+\epsilon_{1}^{-1}L \bigr]x \\ =&-h_{1}V(x)+x^{T} \bigl[DH+H^{T}D+DC+C^{T}D \\ &{}+\epsilon_{1}D^{2}+\epsilon_{1}^{-1}L+h_{1}D \bigr]x \\ \leq&-h_{1}V(x), \end{aligned}$$
where \(DH+H^{T}D+DC+C^{T}D+\epsilon_{1}D^{2}+\epsilon_{1}^{-1}L+h_{1}D\leq 0\). We get
$$ V \bigl(x(t) \bigr)\leq V \bigl(x \bigl((mT)^{+} \bigr) \bigr)\exp \bigl(-h_{1}(t-mT) \bigr), $$
(6)
where \(mT< t<mT+\frac{T}{2}\).
If \(t=mT+\frac{T}{2}\), then we get
$$\begin{aligned} V(x)|_{t=mT+\frac{T}{2}} =& \bigl(x \bigl(t^{-} \bigr)+J1x \bigl(t^{-} \bigr) \bigr)^{T}D \bigl(x \bigl(t^{-} \bigr)+J1x \bigl(t^{-} \bigr) \bigr) \\ =&x \bigl(t^{-} \bigr)^{T}(I+J1)^{T}D (I+J1)x \bigl(t^{-} \bigr) \\ \leq& \lambda_{1} V \bigl(x \bigl(t^{-} \bigr) \bigr). \end{aligned}$$
(7)
If \(mT+\frac{T}{2}< t<(m+1)T\), then we get
$$\begin{aligned} D^{+}V(x) =&2x^{T}D\dot{x} \\ =&2x^{T}D \bigl[Hx+f(x) \bigr] \\ =& 2x^{T}DHx+2x^{T}Df(x) \\ \leq& x^{T} \bigl[DH+H^{T}D \bigr]x+\epsilon_{2} x^{T}D^{2}x+\epsilon_{2}^{-1}x^{T}Lx \\ =&x^{T} \bigl[DH+H^{T}D+\epsilon_{2}D^{2}+ \epsilon_{2}^{-1}L \bigr]x \\ =&h_{2}V(x) \\ &{}+x^{T} \bigl[DH+H^{T}D+\epsilon_{2}D^{2}+ \epsilon_{2}^{-1}L-h_{2}D \bigr]x \\ \leq& h_{2}V(x), \end{aligned}$$
where \(DH+H^{T}D+\epsilon_{2}D^{2}+\epsilon_{2}^{-1}L-h_{2}D\leq 0\). We get
$$ V \bigl(x(t) \bigr)\leq\lambda_{1}V \biggl(x \biggl( \biggl(mT+\frac{T}{2} \biggr)^{-} \biggr) \biggr)\exp \biggl(h_{2} \biggl(t-mT-\frac {T}{2} \biggr) \biggr), $$
(8)
where \(mT+\frac{T}{2}< t<(m+1)T\).
If \(t=(m+1)T\), then we get
$$\begin{aligned} V(x)|_{t=(m+1)T} =& \bigl(x \bigl(t^{-} \bigr)+J2x \bigl(t^{-} \bigr) \bigr)^{T}D \bigl(x \bigl(t^{-} \bigr)+J2x \bigl(t^{-} \bigr) \bigr) \\ =&x \bigl(t^{-} \bigr)^{T}(I+J2)^{T}D (I+J2)x \bigl(t^{-} \bigr) \\ \leq& \lambda_{2} V \bigl(x \bigl(t^{-} \bigr) \bigr). \end{aligned}$$
(9)
We can do the following mathematical induction through (6), (7), (8), and (9).
Case 1: \(m=0\).
Subcase 1. If \(0< t<\frac{T}{2}\), then we get
$$V \bigl(x(t) \bigr)\leq V(x_{0})\exp(-h_{1}t). $$
So
$$V \biggl(x \biggl(\frac{T}{2}^{-} \biggr) \biggr)\leq V(x_{0}) \exp \biggl(-h_{1}\frac{T}{2} \biggr). $$
Subcase 2. If \(\frac{T}{2}\leq t< T\), then we get
$$\begin{aligned} V \bigl(x(t) \bigr) \leq&\lambda_{1}V \biggl(x \biggl( \frac{T}{2}^{-} \biggr) \biggr)\exp \biggl(h_{2} \biggl(t- \frac{T}{2} \biggr) \biggr) \\ \leq& \lambda_{1}V(x_{0})\exp \biggl(-h_{1} \frac{T}{2}+h_{2} \biggl(t-\frac{T}{2} \biggr) \biggr) \end{aligned}$$
and
$$V \bigl(x \bigl(T^{-} \bigr) \bigr)\leq\lambda_{1}V(x_{0}) \exp \biggl(-h_{1}\frac{T}{2}+h_{2}\frac{T}{2} \biggr). $$
Subcase 3. If \(t=T\), then we get
$$\begin{aligned} V \bigl(x(T) \bigr) \leq&\lambda_{2} V \bigl(x \bigl(T^{-} \bigr) \bigr) \\ \leq& \lambda_{1} \lambda_{2} V(x_{0})\exp \biggl(-h_{1}\frac{T}{2}+h_{2}\frac{T}{2} \biggr). \end{aligned}$$
Case 2: \(m=1\).
Subcase 1. If \(T< t< T+\frac{T}{2}\), then we get
$$\begin{aligned} V \bigl(x(t) \bigr) \leq&V \bigl(x \bigl(T^{+} \bigr) \bigr)\exp \bigl(-h_{1}(t-T) \bigr) \\ \leq&V \bigl(x(T) \bigr)\exp \bigl(-h_{1}(t-T) \bigr) \\ \leq&\lambda_{1} \lambda_{2} V(x_{0})\exp \biggl(-h_{1} \biggl(t-\frac{T}{2} \biggr)+h_{2} \frac{T}{2} \biggr) \end{aligned}$$
and
$$V \biggl(x \biggl( \biggl(T+\frac{T}{2} \biggr)^{-} \biggr) \biggr)\leq \lambda_{1} \lambda_{2} V(x_{0})\exp \biggl(-h_{1}T+h_{2}\frac{T}{2} \biggr). $$
Subcase 2. If \(T+\frac{T}{2}\leq t<2T\), then we get
$$V\bigl(x(t)\bigr)\leq\lambda_{1}^{2} \lambda_{2}V(x_{0}) \exp\bigl(-h_{1}T+h_{2}(t-T)\bigr). $$
So
$$V \bigl(x \bigl((2T)^{-} \bigr) \bigr)\leq\lambda_{1}^{2} \lambda_{2}V(x_{0})\exp(-h_{1}T+h_{2}T). $$
Subcase 3. If \(t=2T\), then we get
$$\begin{aligned} V\bigl(x(2T)\bigr) \leq&\lambda_{2} V\bigl(x\bigl((2T)^{-}\bigr)\bigr) \\ \leq& \lambda_{1}^{2} \lambda_{2}^{2} V(x_{0})\exp(-h_{1}T+h_{2}T). \end{aligned}$$
Case 3: \(m=2\).
Subcase 1. If \(2T< t<2T+\frac{T}{2}\), then we get
$$\begin{aligned} V\bigl(x(t)\bigr) \leq&V\bigl(x\bigl((2T)\bigr)\bigr)\exp\bigl(-h_{1}(t-2T) \bigr) \\ \leq& \lambda_{1}^{2} \lambda_{2}^{2} V(x_{0})\exp\bigl(-h_{1}(t-T)+h_{2}T\bigr). \end{aligned}$$
So
$$V\biggl(x\biggl(\biggl(2T+\frac{T}{2}\biggr)^{-}\biggr)\biggr)\leq \lambda_{1}^{2} \lambda_{2}^{2} V(x_{0})\exp \biggl(-h_{1}\biggl(T+\frac{T}{2} \biggr)+h_{2}T\biggr). $$
Subcase 2. If \(2T+\frac{T}{2}\leq t<3T\), then we get
$$V\bigl(x(t)\bigr) \leq \lambda_{1}^{3} \lambda_{2}^{2}V(x_{0})\exp\biggl(-h_{1} \biggl(T+\frac {T}{2}\biggr)+h_{2}\biggl(t-\frac{3T}{2} \biggr)\biggr). $$
So
$$V\bigl(x\bigl((3T)^{-}\bigr)\bigr) \leq \lambda_{1}^{3} \lambda_{2}^{2}V(x_{0})\exp\biggl(-h_{1} \biggl(T+\frac {T}{2}\biggr)+h_{2}\frac{3T}{2}\biggr). $$
Subcase 3. If \(t=3T\), then we get
$$V\bigl(x(3T)\bigr) \leq \lambda_{1}^{3} \lambda_{2}^{3}V(x_{0})\exp\biggl(-h_{1} \biggl(T+\frac {T}{2}\biggr)+h_{2}\frac{3T}{2}\biggr). $$
Case 4: \(m=3\).
Subcase 1. If \(3T< t<3T+\frac{T}{2}\), then we get
$$\begin{aligned} V\bigl(x(t)\bigr) \leq&V\bigl(x\bigl((3T)\bigr)\bigr)\exp\bigl(-h_{1}(t-3T) \bigr) \\ \leq& \lambda_{1}^{3} \lambda_{2}^{3} V(x_{0})\exp\biggl(-h_{1}\biggl(t-\frac{3T}{2} \biggr)+h_{2}\frac {3T}{2}\biggr). \end{aligned}$$
So
$$V\biggl(x\biggl(\biggl(3T+\frac{T}{2}\biggr)^{-}\biggr)\biggr) \leq \lambda_{1}^{3} \lambda_{2}^{3} V(x_{0})\exp \biggl(-h_{1}2T+h_{2} \frac{3T}{2}\biggr). $$
Subcase 2. If \(3T+\frac{T}{2}\leq t<4T\), then we get
$$V\bigl(x(t)\bigr) \leq \lambda_{1}^{4} \lambda_{2}^{3}V(x_{0})\exp\bigl(-h_{1}2T+h_{2}(t-2T) \bigr). $$
So
$$V\bigl(x\bigl((4T)^{-}\bigr)\bigr) \leq \lambda_{1}^{4} \lambda_{2}^{3}V(x_{0})\exp(-h_{1}2T+h_{2}2T). $$
Subcase 3. If \(t=4T\), then we get
$$V\bigl(x(4T)\bigr) \leq \lambda_{1}^{4} \lambda_{2}^{4}V(x_{0})\exp(-h_{1}2T+h_{2}2T). $$
Through the above induction, we get the following.
Case
\(k+1\): \(m=k\).
Subcase 1. If \(kT< t<kT+\frac{T}{2}\), then we get
$$ V\bigl(x(t)\bigr) \leq \lambda_{1}^{k} \lambda_{2}^{k} V(x_{0})\exp\biggl(-h_{1} \biggl(t-\frac{kT}{2}\biggr)+h_{2}\frac {kT}{2}\biggr). $$
(10)
Subcase 2. If \(kT+\frac{T}{2}\leq t<(k+1)T\), then we get
$$ V\bigl(x(t)\bigr) \leq \lambda_{1}^{{k + 1}} \lambda_{2}^{k} V(x_{0})\exp\biggl(-h_{1} \frac {(k+1)T}{2}+h_{2}\biggl(t-\frac{(k+1)T}{2}\biggr)\biggr). $$
(11)
Subcase 3. If \(t=(k+1)T\), then we get
$$ V\bigl(x(t)\bigr)|_{t=(k+1)T} \leq \lambda_{1}^{{k + 1}} \lambda_{2}^{{k + 1}}V(x_{0}) \exp\biggl(-h_{1}\frac{(k+1)T}{2}+h_{2}\frac{(k+1)T}{2} \biggr). $$
(12)
From (10) we get that if \(kT< t<kT+\frac{T}{2}\), then we let \(t=kT\), so that
$$\begin{aligned} V\bigl(x(t)\bigr) \leq& \lambda_{1}^{k} \lambda_{2}^{k}V(x_{0})\exp\biggl( -h_{1}\biggl(t-\frac {kT}{2}\biggr)+h_{2} \frac{kT}{2}\biggr) \\ \leq&\lambda_{1}^{k} \lambda_{2}^{k}V(x_{0})\exp\biggl(-h_{1} \frac{kT}{2}+h_{2}\frac {kT}{2}\biggr) \\ \leq&\exp(k\ln\lambda_{1}+k\ln\lambda_{2})V(x_{0}) \exp\biggl(-h_{1}\frac {kT}{2}+h_{2}\frac{kT}{2} \biggr) \\ \leq&V(x_{0})\exp\biggl(-\biggl(h_{1} \frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln \lambda _{2}\biggr)k\biggr) \end{aligned}$$
(13)
for \(kT< t<kT+\frac{T}{2}\).
From (11) we get that if \(kT+\frac{T}{2}\leq t<(k+1)T\), then we let \(t=(k+1)T\), so that
$$\begin{aligned} V\bigl(x(t)\bigr) \leq&\lambda_{1}^{{k + 1}} \lambda_{2}^{k}V(x_{0})\exp\biggl(-h_{1} \frac{(k+1)T}{2}+h_{2}\biggl(t-\frac{(k+1)T}{2}\biggr)\biggr) \\ \leq&\lambda_{1}^{{k + 1}} \lambda_{2}^{k}V(x_{0}) \exp\biggl(-h_{1}\frac {(k+1)T}{2}+h_{2}\frac{(k+1)T}{2} \biggr) \\ \leq&V(x_{0})\exp\biggl(-h_{1}\frac{(k+1)T}{2}+h_{2} \frac{(k+1)T}{2}+(k+1)\ln \lambda_{1}+k\ln\lambda_{2} \biggr) \\ \leq&V(x_{0})\exp\biggl(-h_{1}\frac{T}{2}-h_{1} \frac{kT}{2}+h_{2}\frac {kT}{2}+h_{2} \frac{T}{2}+\ln\lambda_{1}+k\ln\lambda_{1}+k\ln \lambda_{2}\biggr) \\ \leq&V(x_{0})\exp\biggl(-h_{1}\frac{T}{2}+h_{2} \frac{T}{2}+\ln\lambda_{1}-\biggl(h_{1} \frac {T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln \lambda_{2}\biggr)k\biggr) \end{aligned}$$
(14)
for \(kT+\frac{T}{2}\leq t<(k+1)T\).
From (12) we get that, for \(t=(k+1)T\),
$$\begin{aligned} V\bigl(x(t)\bigr)|_{t=(k+1)T} \leq& \lambda_{1}^{{k + 1}} \lambda_{2}^{{k + 1}}V(x_{0}) \exp\biggl(-h_{1}\frac{(k+1)T}{2}+h_{2}\frac{(k+1)T}{2} \biggr) \\ \leq&\exp\biggl((k+1)\ln\lambda_{1}+(k+1)\ln\lambda_{2}-h_{1} \frac{(k+1)T}{2}+h_{2}\frac{(k+1)T}{2}\biggr)V(x_{0}) \\ \leq&V(x_{0})\exp\biggl(-\biggl(h_{1} \frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln \lambda _{2}\biggr) (k+1)\biggr). \end{aligned}$$
(15)
From (13), (14), (15), and the conditions of Theorem 1 we conclude that \(k\rightarrow\infty\) as \(t\rightarrow \infty\). So
$$\lim_{t\rightarrow\infty}V \bigl(x(t) \bigr)=0, $$
which ends the proof. □
Corollary 1
As a consequence of Lemma
2, the first two conditions of Theorem
1
are equivalent to the following two LIMs:
$$\begin{aligned}& \begin{bmatrix} DH+H^{T}D+DC+C^{T}D +\epsilon_{1}^{-1}L+h_{1}D & -D \\ -D & -\epsilon_{1}^{-1}I \end{bmatrix} \leq0, \end{aligned}$$
(16)
$$\begin{aligned}& \begin{bmatrix} DH+H^{T}D +\epsilon_{2}^{-1}L-h_{2}D & -D \\ -D & -\epsilon_{2}^{-1}I \end{bmatrix} \leq0. \end{aligned}$$
(17)