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

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?

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 $$a, then $$\delta = \epsilon a/b$$ will work.
If you are allowed to use a Lyapunov function to test for stability (rather than provide an $$\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.
• I am trying to show specifically with $\epsilon-\delta$ proof – gene Oct 9 '18 at 7:04