Lyapunov theorem for LTV systems

Consider continuous-time LTV systems of the form
$$
\dot{x}=A \mid(t) x(t)
$$
Let $P(t)$ be a time-varying matrix that satisfies
$$
\eta I \leq P(t) \leq \rho I \quad \eta, \rho>0
$$
as well as
$$
A(t)^{T} P(t)+P(t) A(t)+\dot{P}(t) \leq 0
$$
Then the LTV system is uniformly stable. If, in addition,
$$
A(t)^{T} P(t)+P(t) A(t)+\dot{P}(t) \leq-\nu I \quad \nu>0
$$
then the LTV system is uniformly exponentially stable.

UAS is equivalent to UES for LTV systems

Theorem 2: UAS is equivalent to UES
The following two statements are equivalent for LTV systems 1. $\dot{x}=A(t) x$ is uniformly asymptotically stable (UAS)
2. $\dot{x}=A(t) x$ is uniformly exponentially stable (UES)
It is obvious that UES $\Longrightarrow$ UAS. What about the converse?

Proof of UAS $\Longrightarrow$ UES

Let’s assume system is UAS. By definition there exist $\gamma, T>0$ such that
$$
\begin{gathered}
\left\|\Phi\left(t, t_{0}\right)\right\| \leq \gamma \text { for all } t, t_{0} t \geq t_{0} \\
\left\|\Phi\left(t, t_{0}\right)\right\| \leq \frac{1}{2} \text { for all } t, t_{0}, t \geq t_{0}+T
\end{gathered}
$$
Now
$$
\begin{aligned}
\left\|\Phi\left(t_{0}+k T, t_{0}\right)\right\| &=\left\|\Phi\left(t_{0}+k T, t_{0}+(k-1) T\right) \cdots \Phi\left(t_{0}+T, t_{0}\right)\right\| \\
& \leq\left\|\Phi\left(t_{0}+k T, t_{0}+(k-1) T\right)\right\| \cdots\left\|\Phi\left(t_{0}+T, t_{0}\right)\right\| \\
& \leq \frac{1}{2} \cdots \frac{1}{2}=\frac{1}{2^{k}}
\end{aligned}
$$

Therefore
$$
\begin{aligned}
\left\|\Phi\left(t, t_{0}\right)\right\| &=\left\|\Phi\left(t, t_{0}+k T\right) \Phi\left(t_{0}+k T, t_{0}\right)\right\| \\
& \leq\left\|\Phi\left(t, t_{0}+k T\right)\right\|\left\|\Phi\left(t_{0}+k T, t_{0}\right)\right\| \\
& \leq \gamma \frac{1}{2^{k}}=2 \gamma\left(\frac{1}{2}\right)^{k+1} \\
& \leq 2 \gamma\left(\frac{1}{2}\right)^{\frac{t-t_{0}}{T}} \\
&=2 \gamma e^{-\ln (2)\left(\frac{t-t_{0}}{T}\right)}
\end{aligned}
$$
Let $M=2 \gamma$ and $\mu=\frac{\ln 2}{T}$ to obtain
$$
\left\|\Phi\left(t, t_{0}\right)\right\| \leq M e^{-\mu\left(t-t_{0}\right)}
$$

Converse theorem of Lyapunov for LTV

Theorem 3: Converse Lyapunov theorem for UES
Assume that the LTV system
$$
\dot{x}=A(t) x
$$
is UES and assume $\|A(t)\| \leq \alpha .$ Then there exists Lyapunov function $V(t, x)=x^{T} P(t) x$ where
$$
P(t)=\int_{t}^{+\infty} \Phi^{T}(\tau, t) \Phi(\tau, t) d \tau
$$
satisfying all conditions of the UES Lyapunov theorem

Stability implies Robustness

Consider the linear, time-invariant system
$$
\dot{x}=A x
$$
which is exponentially stable. Now consider the perturbed LTV system
$$
\dot{x}=(A+F(t)) x \quad\|F(t)\|_{2}<\beta
$$
Is the perturbed LTV system also uniform exponentially stable? For what values of $\beta$ ?

Since $\dot{x}=A x$ is exponentially stable, we can solve the Lyapunov equation
$$
A^{T} P+P A=-I
$$
Therefore $V(x)=x^{T} P x$ is a Lyapunov function for the LTI system. Can we use this as a Lyapunov function for the perturbed LTV system? To
show that the LTV systems is UES, $V(x)$ must satisfy for $\eta, \rho, \nu>0$
$$
\begin{gathered}
\eta\|x\|_{2}^{2} \leq V(x) \leq \rho\|x\|_{2}^{2} \quad \eta, \rho>0 \\
\dot{V} \leq-\nu\|x\|_{2}^{2} \quad \nu>0
\end{gathered}
$$
From the Raleigh-Ritz inequality we already have
$$
\lambda_{\min }(P) x^{T} x \leq x^{T} P x \leq \lambda_{\max }(P) x^{T} x
$$

$$
\begin{aligned}
\dot{V} &=\dot{x}^{T} P x+x^{T} P \dot{x}=x^{T}(A+F(t))^{T} P x+x^{T} P(A+F(t)) x \\
&=x^{T}\left(A^{T} P+P A\right) x+x^{T}(F(t) P+P F(t)) x \\
&=-x^{T} x+x^{T}(F(t) P+P F(t)) x \\
&=-\nu x^{T} x ?
\end{aligned}
$$
Now the perturbation term can be bounded as follows
$$
x^{T}(F(t) P+P F(t)) x \leq 2\|F(t)\|_{2}\|P\|_{2} x^{T} x \leq 2 \beta \lambda_{\max }(P) x^{T} x
$$
Therefore $\dot{V} \leq-x^{T} x+2 \beta \lambda_{\max }(P) x^{T} x$. Thus if
$$
-1+2 \beta \lambda_{\max }(P)=\nu<0 \Longrightarrow \dot{V} \leq-\nu x^{T} x
$$
This proves the following result

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Math 130A，Ordinary Differential Equations

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