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The system dynamics are defined as $\dot{x}=Ax+B$, where $B$ is a constant vector. I defined a Lyapunov function for my system as $V(x)=x^TPx$, where $P$ is positive definite and $A^TP+PA=-I$.
Differentiating the Lyapunov:
$$ \dot{V}_1=\dot{x}^TPx+x^TP\dot{x}\\ =-x^Tx+B^TPx+x^TPB\\ =-\frac{-x^Tx}{2}-\left(\frac{x}{\sqrt{2}}-\frac{PB}{\sqrt{2}}\right)^T\left(\frac{x}{\sqrt{2}}-\frac{PB}{\sqrt{2}}\right)+\frac{(PB)^T(PB)}{2}\\ \leq -\frac{-x^Tx}{2}+\frac{(PB)^T(PB)}{2} $$ Using the condition: $$ V_1\leq \lambda_{max}(P)||x||^2 $$ I reach the condition: $$ \dot{V}_1\leq \frac{V_1}{2\lambda_{max}(P)}+\frac{(PB)^T(PB)}{4}\\ \leq -\alpha V+ \gamma $$ After using Gronwall's Lemma, I reached the condition:
$$ V(x)\leq V_0(x) + \alpha $$ where $\alpha$ is a constant and $V_o(x)$ is the value of Lyapunov at $x=0$. What does this say about my system?

Kashish Pilyal
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  • Can you give more context? Namely, just a Lyapunov function by itself doesn't say anything about system. From your statement all I can conclude it that it isn't radially unbound, so one can't make any conclusions regarding global stability using this candidate Lyapunov function. – Kwin van der Veen Aug 10 '22 at 00:29

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It says that your system cannot travel arbitrarily far from the starting value, i.e., the trajectories are bounded.

However, note that $x=0$ is not an equilibrium. As your matrix $A$ is Hurwitz -- otherwise, no $P$ could exist -- it is invertible. Define $x_e := -A^{-1}B$, the new state $z := x - x_e$ and consider $V = z^\top P z$.

Arastas
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