We have simple system of linear differential equations $\dot x = f(t, x)$, where $x = (x_1, x_2, ..., x_n)$, $f$ is defined, continious, satisfies Lipschitz conditions for $x$ in domain $G = (c, +\infty) \times D $, and $D$ is domain of phase space $\mathbb{R}^n $.
Let $x = \phi(t)$ be is solution or motion for this system, defined on interval $[t_0, +\infty)$, where $t_0 > c$.
Now we hold fix(or lock) start moment of time, or $t_0$. Through $x = x(t, x^0)$ we will denote a solution or changed system motion with starting data $t_0, x^0$($x^0 \in D$).
That was some denotion part. Now we take $\textbf{one}$ solution $x = \phi(t)$ of all solutions for this system. Starting data is still ($t_0, x^0)$. Variables $t, t_0 \in (a, +\infty)$. Solution has Lyapunov stability.
Now we don't care about $x^0$, because the main question is about $t^0$, so we forget about $x^0$. Now we take another starting data, let it be $t_1$ and one more - $t_2$.
$\textbf{So, the main question is}$, why Lyapunov stability on chosen solution does not depend on starting data on time, or $t_0$. (Means we choose $t_0$ as starting data on $t$ - solution has Lyapunov stability; we choose $t_1$, $t_1 < t_0$ - has stability, $t_2 > t_0$ - has stability.)
I does not understand, how to prove it. Any help will be appreciated.
Let $\phi(t)$, $t\in [a,+\infty)$ be a solution of the system $$ \dot x= f(t,x). $$ Its Lyapunov stability means that for any $\varepsilon>0, t_0\ge a$ there exists $\delta(\varepsilon,t_0)$ such that all of the solutions $y(t)$ with initial conditions $y(t_0)=y_0$, $\|y_0-\phi(t_0)\|<\delta$, exist and for any $t\in [t_0,+\infty)$ $$\tag{1} \|y(t)-\phi(t)\|<\varepsilon. $$ Now let $t_1\ne t_0$, $t_1\in [a,+\infty)$. Due to the continuous dependence on initial conditions of ode solutions, for any $\delta>0$ there exists $\eta(\delta,t_1)$ such that for any solution $z(t)$ satisfying $\|z(t_1)-\phi(t_1)\|<\eta$ we have $$\tag{2} \|z(t_0)-\phi(t_0)\|<\delta. $$ Combining these two facts, one can show that stability of the solution with initial time $t_0$ implies the stability of the same solution with initial time $t_1$.