I hope you are having a good day. I am working on a homework and I was looking for some help.
Can anyone please help me with the next step to prove whether the L-2 stability of the system and the bound on L2-gain of the system?
The question and my solution are in this link. Thank you very much.
You're on the right track, let $\tilde{w}_i:=\sqrt{J_i}w_i$ and $\tilde{v}_i:=\sqrt{J_i}v_i$. Then,
$\dot{V} = -k\tilde{w}^T\tilde{w}+\tilde{w}^T\tilde{v}=-\frac{1}{2}(\sqrt{k}\tilde{w}-\frac{1}{\sqrt{k}}\tilde{v})^T(\sqrt{k}\tilde{w}-\frac{1}{\sqrt{k}}\tilde{v})+\frac{1}{k}\tilde{v}^T\tilde{v}-\frac{k}{2}\tilde{w}^T\tilde{w}\leq \frac{1}{k}\tilde{v}^T\tilde{v}-\frac{k}{2}\tilde{w}^T\tilde{w}.$
So,
$V(w(t))-V(w(0))\leq\int^t_0\frac{1}{k}\tilde{v}(\tau)^T\tilde{v}(\tau)-\frac{k}{2}\tilde{w}(\tau)^T\tilde{w}(\tau)d\tau$
But, $V(w(t))\geq 0$, $\tilde{w}^T\tilde{w}\geq J_{min}w^Tw$ and $\tilde{v}^T\tilde{v}\leq J_{max}v^Tv$, so
$\frac{k}{2}||w_t||_2^2=\frac{k}{2}\int^t_0w(\tau)^Tw(\tau)d\tau\leq \frac{k}{2J_{min}}\int^t_0\tilde{w}(\tau)^T\tilde{w}(\tau)d\tau\leq \frac{1}{kJ_{min}}\int^t_0\tilde{v}(\tau)^T\tilde{v}(\tau)d\tau+V(w(0))\leq \frac{J_{max}}{kJ_{min}}\int^t_0v(\tau)^Tv(\tau)d\tau+V(w(0))=\frac{J_{max}}{kJ_{min}}||v_t||^2_2+V(w(0)).$
Then re-arranging and applying the triangle rule gives you the desired result with a gain upper bound of $\frac{\sqrt{2}}{k}\sqrt{\frac{J_{max}}{J_{min}}}$. An interesting question is whether this is the least upper bound (I don't know). By the way, most of this type of exercises require you to complete some square, so if you are ever stuck its handy to see if you can complete any squares.