# Showing a fixed point of a system is not Liapunov stable or attracting using $\epsilon-\delta$

For several systems, in particular the following (note that $\dot{x}=\frac{dx}{dt}$ and $\dot{y}=\frac{dy}{dt}$):

1. $\dot{x}=2y$, $\dot{y}=x$
2. $\dot{x}=0$, $\dot{y}=x$
3. $\dot{x}=x$, $\dot{y}=y$

I have to show that the origin, $\mathbf{x^{*}=0}$ is neither Liapunov stable nor attracting.

For reference,

• Definition of attracting: Consider a fixed point $\mathbf{x^{*}}$ of a system $\mathbf{\dot{x}=f(x)}$. We say that $\mathbf{x^{*}}$ is attracting if there is a $\delta > 0$ such that $\lim_{t \to \infty}\mathbf{x(t)}=\mathbf{x^{*}}$ whenever $\Vert \mathbf{x(0)}-\mathbf{x^{*}}\Vert < \delta$
• Definition of Liapunov stability: We say that $\mathbf{x^{*}}$ is Liapunov stable if for each $\epsilon > 0$, there is a $\delta > 0$ such that $\Vert \mathbf{x(t)}-\mathbf{x^{*}} \Vert < \epsilon$ whenever $t \geq 0$ and $\Vert \mathbf{x(0)}-\mathbf{x^{*}} \Vert < \delta$

So, I guess in the case of attracting, I need to show that $\forall \delta > 0$, whenever $\Vert \mathbf{x(0)} - \mathbf{x^{*}} \Vert < \delta$, the series does not converge? And in the case of Liapunov, I need to show that $\forall \delta > 0$, $\exists \epsilon > 0$ such that $\Vert \mathbf{x(t)}-\mathbf{x^{*}}\Vert \geq \epsilon$? Did I get the negations correct?

Now, in the case of the third system I mention above, the solution is (if $x(0) = x_{0}$ and $y(0) = y_{0}$), $x(t) = x_{0}e^{t}$, $y(t) = y_{0}e^{t}$, where $\mathbf{x(t)}$ is the vector $\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$.

Clearly, the trajectories are moving away from the origin, due to the exponentials. However, specifically, how would I go about choosing an $\epsilon$ for Liapunov or showing that $\forall \delta > 0$, $\lim_{t \to \infty} \mathbf{x(t)} \neq \mathbf{x^{*}}$ in the case of this system? The more specific you are in your details the better, because then I can use that information to help me figure out the rest of the problems.

Thank you.

• You need to solve the differential equations and then show that the solutions are unbounded. – MrYouMath May 16 '18 at 8:44
• @MrYouMath for the third equation, I did solve it. So then, how would I show its unbounded, and then how would showing that also show me that it is neither liapunov nor attracting? – ALannister May 16 '18 at 8:49
• What happens for $t\to \infty$ for $\boldsymbol{x}_0\neq\boldsymbol{0}$? – MrYouMath May 16 '18 at 9:56
• Sorry, I did not write the not equal sign :D. If the solution escapes to infinity for any $\boldsymbol{x}_0\neq \boldsymbol{0}$ then this implies that the origin is not attracting and not Lyapunov stable. – MrYouMath May 16 '18 at 9:56
• By showing that the solution is unbounded you directly show that the $\varepsilon$ criterion cannot be fulfilled. Which tells us that the solution needs to stay bounded for all $t$. – MrYouMath May 16 '18 at 13:18