Showing that equilibrium is unstable

Question

I just solved an ODE $x'=A(t)x$ which has $x(t)=e^{t/2}(-\cos t, \sin t)^T$ as a solution.

Now I want to show that the equilibrium $\bar x=0$ is unstable.

So according to my book I need to show that $\forall \epsilon >0 \ \exists \delta>0:\|x(t_0)-\bar x\|\le\delta \Rightarrow \|x(t)-\bar x\|\le \epsilon \ \forall t\ge t_0$ does not hold.

So we need a contradiciton of $$\forall \epsilon >0 \ \exists \delta>0:\|x_0\|\le\delta \Rightarrow \|e^{t/2}(-\cos t, \sin t)^T\|\le \epsilon \ \forall t\ge t_0$$

As I'm doing this for the first time, I don't know to reason it correctly. Does the instability simply follow from $ \lim_{t \to \infty}\|x(t)\| = +\infty$?

Answer 1

Formally, what you have to show is that $$\exists \epsilon >0 \ \forall \delta>0:\|x_0\|\le\delta \implies \|e^{t/2}(-\cos t, \sin t)^T\|\ge \epsilon \ \forall t\ge t_0$$ However, showing $\| \lim_{t \to \infty}x(t)\| = +\infty$ is sufficient for showing instability.

Answer 2

Instability means that $$\tag{1}\exists \epsilon >0 \; \forall \delta>0\; \exists x_0,t_0: \|x_0\|\le\delta \wedge \left( \exists t\ge t_0:\; \|x(t)\|\ge \epsilon \right), $$ where $x(t)$ is a solution such that $x(t_0)=x_0$.

Let $x(t)= e^{t/2}(-\cos t, \sin t)^T$. Since $\lim_{t\to -\infty}\|x(t)\|= \lim_{t\to -\infty}e^{t/2}=0$, one can (by the limit definition) for any $\delta>0$ take sufficiently negative $t_0$ such that $$ \|x(t_0)\|=\|x_0\|<\delta. $$ On the other hand, $\lim_{t\to +\infty}\|x(t)\|=\lim_{t\to +\infty}e^{t/2}=+\infty$ means that $\|x(t)\|$ is greater than any fixed number $\epsilon>0$ for any $t$ greater than some number. Hence, (1) is satisfied for any positive $\epsilon$.

There is also a theorem that states that the linear system $$\tag{2} \dot x= A(t)x,\quad x\in\mathbb R^n $$ is stable if and only if all of its solutions are bounded on $I=[t_0,+\infty)$.