Liapunov function for linear ODE

Question

I want to find a Liapunov function at $(0, 0)$ for the system $x'=f(x)$: \begin{align} x_1'=& x_2 \\ \\ x_2'=& -4x_1-2x_2 \end{align}

I saw something similar in a textbook and so decided to try:

\begin{equation} V(x) = x_1^2+\beta x_2^2 \end{equation} Where I will now find $\beta$: \begin{align} \nabla\cdot f =& 2x_1x_2+2\beta x_2(-4x_1-2x_2) \\ \\ =& 2x_1x_2-8\beta x_1x_2 -4\beta x_2^2 \\ \\ =& 2x_1x_2(1-4\beta) - 4\beta x_2^2 \end{align} So, $\beta = \frac{1}{4}$ suffices and we have: \begin{equation} V(x) = x_1^2+\frac{1}{4}x_2^2. \end{equation}

Now checking the Liapunov conditions:

1. $V(0) = 0$. This is clear.

2. $V(x) > 0$ for all $x \neq 0$ This is also clear.

3. $\nabla\cdot f \leq 0$. We have that $\nabla\cdot f = -x_2^2 \leq 0$

$$\therefore V(x) = x_1^2+\frac14x_2^2$$

Is a Liapunov function w.r.t $(0, 0)$. Is this correct?

Answer 1

Yes, the function:

$$V(x) = x_1^2+\frac14x_2^2$$

is a correct Liapunov function.

Note: It is easier to just write $V'$ and leave that $f$ stuff off as it is confusing IMO.

Also, this is telling us that there are no closed orbits in the system. We can plot a phase portrait and this shows the following (of course, we could have done an eigenvalue analysis or found the solution to this uncoupled system too to determine this).