Consider the representation $(\mathcal U, \Sigma, \mathcal Y, s, r)$ of a dynamical system (with times on $\mathbb R_+$) where
- $\mathcal U$ is a set of input functions (usually $\mathbb R_+ \to \mathbb R^{n_i}$)
- $\Sigma$ is a set of states (usually $\mathbb R^n$)
- $\mathcal Y$ is a set of outputs (usually $\mathbb R^{n_o}$)
- $s : \mathbb R_+ \times \mathbb R_+ \times \Sigma \times \mathcal U \to \Sigma$ is a state transition function ($x(t) = s(t, t_0, x_0, u)$)
- $r : \mathbb R_+ \times \Sigma \times \mathcal U \to \mathcal Y$ is an output read-out map ($y(t) = r(t, x(t), u(t)$)
My book says the state transition map must satisfy two axioms:
(i) only values of $u(t)$ for $t \in [t_0, t_1]$ influence $x(t_1)$
$$\forall t_1 \geq t_0, \forall x_0 \in \Sigma, \forall u, \hat u \in \mathcal U, \text{ if } \forall t \in [t_0,t_1], u(t) = \hat u(t), \text{ then } s(t_1,t_0,x_0,u) = s(t_1,t_0,x_0, \hat u)$$
(ii) applying $u_{[t_0, t_1]}$ then $u_{[t_1, t_2]}$ on $x_0$ is the same as applying $u_{[t_0, t_2]}$ on $x_0$
$$\forall t_0 \leq t_1 \leq t_2, \forall x_0 \in \Sigma, \forall u \in \mathcal U, s(t_2, t_1, s(t_1, t_0, x_0, u), u) = s(t_2, t_0, x_0, u)$$
I am wondering though, what guarantees that we have $x(t_0) = s(t_0, t_0, x_0, u(t_0)) = x_0$? Is there a missing axiom, or can this somehow be inferred from the others?