# Passivity in State Space Systems

Suppose I have the state space system $$a\dot{x} = -x + \frac{1}{k}h(x) + u$$, and output $$y = h(x)$$, where $$a,k > 0$$. The only information I'm given about $$h(x)$$ is that $$h\in[0,k]$$. I want to show that the system is passive using the storage function $$V(x) = a\int_{0}^{x}h(\sigma)\ d\sigma$$. The definition of a passive system is $$u^\intercal y \geq \dot{V}$$ for all $$(x,u)$$. Thus,

$$\dot{V} = ah(x) \dot{x} = h(x) (-x + h(x)/k + u) = \frac{1}{k}h(x)(h(x) - kx) + h(x) u.$$

Rearranging this gives

$$h(x)u = yu = \dot{V} + \frac{1}{k}h(x)[kx-h(x)].$$

In order to prove passivity, I need to show that $$\frac{1}{k}h(x)[kx - h(x)] \geq 0$$, for all $$x$$. However, all I'm given is that $$h\in[0,k]$$ so how can this be true in general? Am I missing something really simple here?

In your problem, you define the supply rate (I will name it $$s$$) $$s(y,u)=u^{\top}y$$. And it is (strict) passive if $$\dot{V} \leq s(y,u)$$. I also consider that you are working taking the case where this consideration is valid $$u^{\top}y \equiv uy$$.
Saying that, I believe you can re-arrange your equation as follows: \begin{align*} \dot{V}(x) &= uy+\frac{1}{k}y^2-xy \\ &=y(u+\frac{1}{k}y) -xy \end{align*} then define as the new input $$\nu := u+\frac{1}{k}y$$, in that sense you will have the following equation: \begin{align*} \dot{V}(x) &= y\nu-xy\leq \nu y \end{align*}
The system is (strict) passive if $$xh(x)> 0\quad\forall x\in\mathbb{R}$$.