If an LTI system is dissipative, does this mean that it is stable?

Question

Suppose that we have the LTI system $$\dot{x}=Ax+Bw\\z=C x$$ and we know it is dissipative with respect to the function $s(x,w)=z^\top w=x^\top C^\top w$, i.e. $\exists X=X^\top >0$ and a quadratic storage function $V(x)=x^\top X x$ such that $$2x^\top X(Ax+Bw)\leq x^\top C^\top w$$ Can we assume that the LTI system is stable (in the Lyapunov sense)? I know that dissipativity is connected to Lyapunov functions and stability but I can't see the connection here.

Answer 1

Lyapunov stability (as well as others) of LTI depends only on $A$ matrix, so we can set $w=0$. Dissipativity with $w=0$ gives that $$ A^TX+XA\le 0\quad\text{(i.e. negative semidefinite).} $$ Consider $\Phi(t)=e^{A^Tt}Xe^{At}$. Then $$ \dot\Phi(t)=e^{A^Tt}[A^TX+XA]e^{At}\le 0. $$ Hence, $\Phi(t)\le\Phi(0)$ $\Rightarrow$ the fundamental matrix $e^{At}$ is bounded (due to $X$ being positive definite). It is Lyapunov stability by definition.