Showing that a centre of the 2D linear system $\dot{\mathbf{x}} = A \mathbf x$ is Lyapunov stable

Question

Consider the 2D linear system $\dot{\mathbf{x}} = A \mathbf x$ with $$A = \begin{pmatrix} 0&1\\ -4 & 0\end{pmatrix}.$$ The eigenvalues of this matrix are $\lambda = \pm 2i$, meaning that the phase portrait will be a center. How do I show that the origin $\mathbf x = 0$ is Lyapunov stable for this system? i.e. for any $\epsilon >0$, I need to find a $\delta(\epsilon)>0$ such that $$||\mathbf{x}(0)||< \delta \implies ||\mathbf{x}(t)||<\epsilon .$$ I am really confused about how to show this for an arbitrary $\epsilon$, how can one probe the Lyapunov stability of such systems where the fixed point is a centre?

Answer 1

Since this is a linear system, you can find the solution explicitly. The solutions curves will be ellipses, and if the half-axis lengths are $a<b$, then $\delta = \epsilon a/b$ will work.

Answer 2

If you are allowed to use a Lyapunov function to test for stability (rather than provide an $\epsilon-\delta$ proof), consider the following...

Let ${\bf x} = \pmatrix{x\\y}$. If we write the system out using variables: $$\dot{\bf x} = \pmatrix{0 & 1\\ -4 & 0}{\bf x} \qquad\implies\qquad\begin{cases}\dot x &= y\\\dot y &=-4x\end{cases}$$Suppose we think the function $V(x,y) = 2x^2 + {1\over2}y^2$ might be a good candidate for a Lyapunov function. To show it is, we have that $V(0,0) = 0$, and $V > 0$ for any $(x,y)\in\mathbb R^2\setminus(0,0)$. Then, $$\begin{align}\dot V &= 4x\dot x + y\dot y\\ &=4x(y) + y(-4x) \tag{Substitute from above}\\ &= 4xy - 4xy \\ &= 0.\end{align}$$

This implies that the origin ${\bf x} = {\bf 0}$ is stable (and thus also Lyapunov stable) for this system.