I need to determine whether the fixed point $x^{*}=0$ of the system $$\dot{x} = 0 \\ \dot{y}=x $$
is attracting, Liapunov stable, asymptotically stable, or none of the above.
For reference, a fixed point $x^{*}$ is said to be attracting if any trajectory that starts within a distance $\delta$ of $x^{*}$ is guaranteed to converge to $x^{*}$ eventually. In contrast, Liapunov stability requires that nearby trajectories remain close for all time. Finally, asymptotically stable means that it is both attracting and Liapunov stable.
Here, if I set up the matrix equation $$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}$$
Notice that the trace of the matrix $A = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$ is $\tau = 0$, and the determinant $\Delta$ $is$0$ as well.
Moreover, both eigenvalues are equal to $0$. My book says that in the case where both eigenvalues are equal to $0$, the entire plane is filled with fixed points, but what does this tell me about what is going on at the fixed point $x^{*} = 0$?
I even solved the system to get $$x(t) = x_{0} \\ y(t) = x_{0}t+y_{0} $$
But I still do not know how to determine the stability of the origin here. Could somebody please help?
Solutions for this problem are available on the Internet, but they all 1) don't contain any explanation as to why the answers are what they are, 2) disagree with each other about what is going on at $x^{*} = 0$ , and finally 3) even disagree about what the vector field looks like. Our professor didn't really explain to us very well how even to draw vector fields, so I would appreciate if whoever answered this question could please explain how to do so (by hand).
Thank you.
Best Answer
A hand drawn graphical representation of a vector field is constructed by choosing a collection of points in the plane an drawing a short line segment centered on the chosen points that has a slope equal to $\frac{dy}{dx} = \frac{\dot y}{\dot x}$. In your case, all of the slopes away from the $y$-axis are undefined(vertical). The slopes along the $y$-axis are also technically undefined but something drastically different is happening there.
However, consider what happens if your initial conditions are along the $y$-axis. The solution stays fixed at $y_0$. A glance at your solution verifies this to be true. How is this fact related to the stability criteria you have proposed? In other words, if you are displaced a small amount away from the fixed point at the origin in the $y$-direction, what happens?
If you are displaced in any other direction then $x_0$ would not be $0$. What are the properties of your solution now? Look at the asymptotic behavior as $t\to\pm\infty$.
You will ultimately need to classify the stability properties in these two different situations.