Question regarding global lyapunov function of a biological system?

Question

Suppose I have the following Lyapunov function that is often used in biological model: $V(x) = x - \ln x$, how is this bounded below since as $x \rightarrow -\infty$, $V(x) \rightarrow -\infty$? My guess is that because $x$ is biological, we only consider $x \geq 0$, so $V(x)$ is indeed bounded below. If this is true, my question is: does assuming $x \geq 0$ affect the global stability result obtaining from showing $V'(x) < 0$?

Answer 1

The function $V(x) = x - \ln x$ is defined for $x>0$ and we have

• $V(x) = x - \ln x\implies V'(x)=1-\frac1x=0 \implies x=1$

and

• $V''(x)=\frac 1 {x^2}>0$

then $V(1)=1$ is a global minimum.