Passivity of a dynamical system Let a dynamical system , in particular a RLC circuit. I want to study the passivity of this.
In general the notion of $\textbf{passivity}$ comes out form the statics physics for which by a mathematical point of view it can be said that a static system $y=h(t,u)$ is said to be passive if $u^Ty\geq 0$ (for instance let a resistor and let $u=V$, $y=I$, it is passive since $VI\geq 0$). 
Now it is possible to extend the notion of passivity to the dynamical system in the sense that follows with an example:
let a RLC circuit for which the energy absorbed by
the network over any period of time $[0, t]$ such that is greater than or equal to the change in
the energy stored in the network over the same period; that is,
$$\int_{0}^{t}u(s)y(s)ds\geq V(x(t)-V(x(0))$$ where $V(x)=\frac{1}{2}Lx_1^2+\frac{1}{2}Cx_2^2=\text{Total energy stored in the network}$, with $x_1=I_L$ and $x_2=V_C$.
My book indicates that the inequality above it is a matter of passitivity in the sense that this is due by the fact that the total energy stored by the network is greater than the variation of this in the period $[0,t]$.
$\textbf{My problems:}$
$1)$ It is true to think about the term $\int_{0}^{t}u(s)y(s)ds$ (that is the integral of the power flows the network over the period $[0,t]$) $\textit{equal to the energy stored in this period}$?
$2)$I can't understand the fact that the total energy stored by the network is greater than the variation of this in the period $[0,t]$ indicates dissipation of energy.
Can you help me please?
 A: I'll answer the questions first and proceed to my explanations after.

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*The integral term tells you how much work you perform on the system. In principle, if there are no losses of any sort, then this is precisely equal to the change in the system's energy. Of course, energy is relative. You can always view $E(0)$ has being no energy stored and then you could view that integral as the amount of energy the system stores over that period.


*Briefly, the integral tells you exactly how much energy you've put into the system. If the constraint $$V(t) - V(0) \leq \int_{0}^{t} u(s) y(s)\,ds$$ is satisfied, it is telling you that the system is not going to magically increase in energy greater than the amount you are applying. That is, you can strictly bound the growth of the system's energy using your input and, in particular, force it to decrease if you like!
It helps to consider a physical example. Consider the position of a particle $x(t) \in \mathbb{R}^3$ with mass $M$ and velocity $v(t) \in \mathbb{R}^3.$ We will take the output of the system to be $y(t):= v(t).$ The energy of the cart (there is no potential energy here) is $$E(t) = \frac{1}{2} M v(t)\cdot v(t) = \frac{1}{2} M y(t)\cdot y(t).$$ Suppose we could apply an input force $u(t) \in \mathbb{R}^3.$
In this case, with no potential energy, we know that the work is the change in kinetic energy. The work induced by the applied force $u(t)$ is the integral of the applied force inner producted with the velocity. That is, $$W(t) = \int_{0}^{t} u(t) \cdot v(t)\,dt = \int_{0}^{t} u(t) \cdot y(t)\,dt.$$ As discussed, we have that the change in kinetic energy is the work (because of a lack of potential energy) so we get the relation $$E(t) - E(0) = W(t) = \int_{0}^{t} u(t) \cdot y(t)\,dt.$$
The kinetic energy is energy the object we wish to control has (the particle) that is "stored" in the sense that the system has energy it can lose. What happens if there is friction or some other force that "steals" energy from the system? We must have $$E(t) - E(0) \leq W(t) = \int_{0}^{t} u(t) \cdot y(t)\,dt.$$
Why? Suppose we did positive work $W(t) > 0$ over this period. This would be equivalent to adding energy into the system, such as adding force in the direction of motion making the particle move faster. If it was the case that $E(t) - E(0) < W(t)$ then somehow, although we added some $W(t)$ amount of energy, we don't see it in $E(t),$ implying some was stolen through losses. You can see a similar argument in the case $W(t) < 0.$
Of course, if you do satisfy the equality (as in the above example), you of course don't have any internal losses but it still is a useful property for a system to have. It implies that you can essentially determine how much the energy the system gains or loses by your input $u(t).$
Back to your example, the question you must ask yourself --- if you seek the intuition --- is what is the work in a circuit? We have that power $P(t)$ is the derivative of the work, $$P(t) = \frac{dW}{dt}(t).$$ We know the instantaneous power of an ideal linear circuit is $$P(t) = I(t) V(t)$$ where $I(t)$ and $V(t)$ are the instantaneous current and voltage respectively measured from the single input to the single output. Solve for $W(t),$ $$W(t) = \int_{0}^{t} I(t) V(t)\,dt = \int_{0}^{t} u(t) y(t)\,dt$$ and we have that the change in energy, the work, is the integral in question!
Hope this helps.
