I started reading the first chapter of Eduardo D. Sontag book Mathematical Control Theory.
In the first exercise, the differential equation (obtained from mathematical pendulum linearized at the upper equilibrium point with proportional feedback) is given as
\begin{equation} \ddot{\varphi}-\sin \varphi+\alpha\varphi=0, \end{equation}
in which $\alpha \in \mathbb{R}$. The exercise claims that solutions of this equation typically will not approach zero, no matter how the feedback gain $\alpha$ is picked.
By choosing the energy
\begin{equation} V(\varphi,\dot{\varphi})=\cos \varphi -1 +\dfrac{1}{2}\left[\alpha \varphi^2+\dot{\varphi}^2 \right] \end{equation}
as a candidate Lyapunov function it is easy to show that $\dot{V}=0$. This implies that $V=\text{const.}$ along solutions.
Now, the author states
...that $V(x,0)$ is an analytic function and therefore that its zero at $x=0$ is isolated.
Why does analyticity imply an isolated zero? For example $V=0$ is also analytic and still does not have isolated zeros.
EDIT: This is my attempt to show that analyticity of $f(x)\not\equiv 0$ (to exclude the trivial case such that all the derivative vanish) in the neighbourhood of $x=0$ for which $f(x=0) =0$ implies that the zero is isolated.
As $f(x)$ is analytic we can expand it as a Taylor expansion
$$f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
evaluated at a point $x_0$ close to the point $x=0$. Now, as we assumed $f(x=0)=0$ we can write
$$0=\sum_{n=0}^{\infty}(-1)^n \dfrac{f^{(n)}(x_0)}{n!}x_0^n.$$
If $x=0$ is not an isolated zero, we can find a point $x_0\neq 0$ infinitely close to $x=0$ such that $f(x_0)=0$. If that is the case then we can conclude that $x_0=0$ from the Taylor expansion, hence we obtain a contradiction to $x_0\neq 0$.
Question: Is my way of reasoning correct?
If we assume that $x=0$ is isolated, how can I conclude that there are initial conditions $\varphi(0)=\varepsilon$,$\dot{\varphi}(0)=0$, with arbitrarily small, for which the corresponding solutions do not converge to $\varphi\to 0$ and $\dot{\varphi}\to 0$ for $t\to 0$.
The point to recognize is that even with that controller input the system is still conservative. This you confirmed with computing that $\dot V=0$. You would need a mechanism that acts as friction $\beta\dot x$, $β>0$, to bleed off energy and sink towards a minimum of the potential function $V(x,0)$ ($=f(α,x)$ in the graph below).
Note that potential around $x=0$ has the expansion $$ V(x,0)=\frac12(αx^2-(2\sin x/2)^2)=\frac12\left((α-1)x^2+\frac1{12}x^4+\dots\right) $$ This means that for $α\ge 1$ the potential function has a local minimum at $x=0$ while for $α<1$ this is a local maximum with local minima close by at $x\approx \pm\sqrt{12(1-α)}$ for $α\lessapprox1$.
In the phase plane the solutions move along the level curves of $V$. Around a local minimum these are concentric circles, the solutions thus oscillations with the minimum in the middle.
As $V(ε,0)=V(0,\dot x)$ so that $\dot x=\pm\sqrt{αε^2-(2\sin(ε/2))^2}$ the velocity is never zero at $x=0$ for $α\ge 1$, for $ α<1$ the motion never reaches the maximum if starting at rest at a point below that maximum. There are non-zero solutions of $\sqrtα ε=2\sin(ε/2)$ so that a solution converging to $x=0$ exists, however that is unstable, any small deviation leads to oscillating behavior, which is not what a controller should do.