Consider the system
$$\dot{x} = \left(A + \frac{1}{2\varepsilon}BB^TP\right)x + Dg(t,y),\quad y=Cx,$$
where $g(t,y)$ is continuously differentiable and satisfies
$$\Vert g(t,y)\Vert_2 \le k\Vert y\Vert_2,\, \forall t\ge0,\, \forall \Vert y\Vert_y\le r,$$
for some $r>0$. Given the equation ($P,Q$ symmetric positive definite, $\varepsilon>0$, $0<\gamma<1/k$.)
$$PA + A^TP+\varepsilon Q -\frac1\varepsilon PBB^TP+\frac1\gamma PDD^TP + \frac1\gamma C^TC=0.$$
Using $V(x)=x^TPx$ it can be shown that for
$$\dot{V} \le -\varepsilon x^TQx.$$
I have worked through the above in an exam question but the last part says using this information determine whether the system is globally exponentially stable.
Now the Lyapunov condition for exponential stability is
- $V$ is positive definite.
- $\dot{V}$ is negative definite.
- $V(x)$ is radially unbounded.
- $\dot{V}\le -\varepsilon V(x)$.
So the first three conditions are satisfied however i'm not so sure the final condition is satisfied since we have $\dot{V} \le -\varepsilon x^TQx \not= -\varepsilon x^TPx = -\varepsilon V(x)$. So this can't be exponentially stable unless we know $x^TQx\ge x^TPx$. Is there any way to deduce this from the information given?
Since $P$ and $Q$ are both positive definite, for $V(x) = x^TPx$ we have
$$\lambda_{min}(Q)\Vert x\Vert^2\le x^TQx\le \lambda_{max}(Q)\Vert x\Vert^2\tag{1}$$
$$\lambda_{min}(P)\Vert x\Vert^2\le V(x)\le \lambda_{max}(P)\Vert x\Vert^2\tag{2}$$
From $(1)$:
$$-x^TQx\le-\lambda_{min}(Q)\Vert x\Vert^2$$
and from $(2)$:
$$\frac1{\lambda_{max}(P)}V(x)\le \Vert x\Vert^2$$
This yields
$$\dot{V}(x) \le -\varepsilon x^TQx\le-\varepsilon \frac{\lambda_{min}(Q)}{\lambda_{max}(P)}V(x)$$
Hence the origin of this system is globally exponentially stable.