so my analysis is quite lacking so I wanted to check with everyone here. I need a sanity check and see if my analysis is grounded.
I am looking at stability of 1-D function to the quadratic-type Lyapunov: $ V(x) = \frac{1}{2} x^2 $. We will assume both conditions are met such that V is positive definite and its partials are continuous.
My system is defined as $$ \dot{x} = (5-x)^5 $$
Which has an equilibrium point $x^* = 5$
My reasoning:
We can consider $(5-x)^5$ as $-x^5$ with an offset of 5. With this we can say if stability holds for $-x^5$ it will hold for $(5-x)^5$.
Therefore we have the following:
$$ \dot{V}(x) = \frac{dV}{dx} -x^5 \\ = x(-x^5) \\ = -x^6 $$
We can see that $ -x^6 < 0 $ satisfying condition that $\dot{V}(x^*) = 0 \& \dot{V}(x) < 0 \forall x \neq 0 $. Which makes our function asymptotically stable. We can concur that our original function with the offset will also be asymptotically stable if we define $x^*=5$
Is this sound in reasoning? Any suggestions on reevaluating this if not?
Edit: Missing square
The key to recognize the system is stable around $x\equiv 5$ is simply: If $x$ is smaller than $5$, then $x'(t)>0$ hence $x(t)$ increases and similarly if $x$ is bigger than $5$, then $x'(t)>0$ hence $x(t)$ decreases. Lyapunov's (second) method for this system is just to formalize the analysis.
As commented by @Lutz Lehmann, it's better to use $V(x)=\frac{1}{2}(x-5)^2$ as the Lyapunov function, for which $V'(x)=(x-5)x'=-(5-x)^6<0$ as long as $x\not=5$. Hence informally the system tends to minimize $V(x)$ which leads to $x=5$ in the long run.