Trajectory in the Lienard system I'm read about a Lienard System in Perko books, but I don't understand how this applies


I have managed to understand the proof, the shape of the trajectory is intuitively clear, doing an analysis of the vector field $F(x,y)=(y-F(x),-g(x))$, b ut how do I formally justify it? Any suggestion? If anyone knows a book or article where this system and its generalizations are treated in detail, I would be very grateful if you could cite it.
Postscript: The result mentioned in Perko is as follows (Ordinary differential equations by Philip Hartman):

 A: I too can not see the value of the cited theorem without further context.
Let's fixate what is obviously true:

*

*As long as $x>0$ we have $\dot y=-g(x)<0$, so $y$ is at least non-increasing, and with $xg(x)\ge cx^2$ or a similar coercivity condition, $y$ will also be strictly decreasing until $x$ crosses again to negative values.


*Due to $\dot x=y-F(x)$, $x$ will be increasing as long as $y>F(x)$. As $y$ is falling and $F(x)$ eventually increasing towards $+\infty$, there will be a cross-over point.


*If $y\le F_\min-\epsilon$, where $F_\min$ is the minimum value taken by $F$ on $[0,\infty)$, then from this moment on $\dot x\le-\epsilon$, so $x$ has an upper bound that falls to zero in finite time.
It remains to cover the segment from the crossover $y(t_2)=F(x(t_2))$ to some time $t_3$ where $y(t_3)<F_\min$. Consider $v=\dot x=y-F(x)$ with
$$
\dot v =-g(x)-f(x)v,
\\
\frac{d}{dt}(e^{F(x)}v)=-g(x)e^{F(x)}.
$$
At the cross-over point $t_2$ we have $v=0$ and thus $\dot v<0$. With the integrating factor in the second equation we see that $e^{F(x)}v$ is strictly decreasing as long as $x>0$. So as $e^{F(x)}v<-\epsilon$ for $t>t_2+\delta$ for some $\epsilon,\delta>0$, we also get $\dot x=v<-\tilde\epsilon$ as $F$ is bounded on $[0,x(t_2)]$. This again gives an upper bound for $x$ that falls to zero in finite time.
A: To be honest, the theorem you quote is a bit out of context for my taste. For that reason, maybe one can just cook up their own argument (or maybe that argument is analogous to the theorem, I am not sure). So here is what I think, by assessing the situation quickly. I am going to assume we are dealing with the following system of ODEs:
\begin{align}
&\frac{dx}{dt} \,=\, y \, -\, F(x)\\
&\\
&\frac{dy}{dt} \,=\, -\, g(x)
\end{align}
where $F(x)$ and $g(x)$ are continuously differentiable functions with enough continuous derivatives with respect to $x$. I am assuming the graph of the function $y = F(x)$ is as depicted on your diagram, and I am going to assume that
\begin{align}
&g(x) < 0 \,\,\text{ for }\,\, x < 0\\
&g(x) = 0 \,\,\text{ for }\,\, x = 0\\
&g(x) > 0 \,\,\text{ for }\,\, x > 0\\
\end{align}
So basically, to analyse this situation, I would consider the following four curves :
\begin{align}
L \, &=\, \big\{\, (x, y) \in \mathbb{R}^2 \,\, :\,\, x = 0\,\,\text{ and }\,\,  y > 0\, \big\}\\ 
FG_{+} \, &= \, \big\{\, (x, y)\in \mathbb{R}^2\, \,:\, \, y = F(x) + g(x) \text{ where } x > 0\, \big\}\\ 
R \, &= \, \big\{\, (x, y)\in \mathbb{R}^2 \, \, :\,\, y = F(x) \text{ where } x > 0\, \big\}\\
FG_{-} \, &= \, \big\{\, (x, y)\in \mathbb{R}^2\, \,:\, \, y = F(x) - g(x) \text{ where } x > 0\, \big\}
\end{align}
With these four curves as boundaries, define the two bigger domains
\begin{align}
&D_{+} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, y \, > \, F(x) \,\text{ and }\, x > 0\,  \big\}\\
&D_{-} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, y \, < \, F(x) \,\text{ and }\, x > 0\,  \big\}
\end{align}
and the  four subdomains
\begin{align}
&D_{x+} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, y \, > \, F(x) + g(x) \text{ and } x > 0\,  \big\}\\
&D_{y+} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \,  F(x) + g(x) 
\, >\,  y > F(x)\,  \big\}\\
&D_{y-} \,=\, \big\{ \, (x, y)\in \mathbb{R}^2\, \,:\, \, F(x) \, >\, y \,>\, F(x) - g(x)  \text{ and } x > 0\,  \big\}\\
&D_{x-} \,=\, \big\{ \, (x, y) \in \mathbb{R}^2\, \,:\, \, F(x) - g(x)\, >\, y \text{ and } x > 0\,  \big\}
\end{align}
where $D_{x+} \cup D_{y+} \subset D_+$ and $D_{x-} \cup D_{y-} \subset D_{-} \,$.
Assume that for some interval $t \in (\alpha, \beta)$ a solution $\big(\,x(t), \, y(t)\,\big)$ of the original system of differential equations lies in the domain $D_{+}$. Then, since $x > 0$
$$\frac{dx}{dt} \, = \, y - F(x) \, > \, F(x) + g(x) - F(x) \, =\, g(x)
\, >\, 0$$ which means that $x=x(t)$ is a monotonously increasing function, and as such it has a well defined inverse $t = t(x)$ and therefore, $y=y(t)$ can be reparametrized as $y = y\big(t(x)\big) = y(x)$ and so the solution trajectory of the original system can be reparametrized as $\big(x, \, y(x)\big)$, where, by the chain rule, $y=y(x)$ should satisfy the differential equation
$$\frac{dy}{dx} \,=\, \frac{dy}{dt} \frac{dt}{dx} \, =\, \frac{\,\,\frac{dy}{dt}\,\,}{\frac{dx}{dt}} \,=\, \frac{-\,g(x)}{y - F(x)} \, =\,  \frac{g(x)}{F(x)-y} $$ or in short
$$\frac{dy}{dx} \, =\,  \frac{g(x)}{F(x)-y} $$
One can check that in $ D_{+}$ one has $\frac{dy}{dx}\,=\,\frac{g(x)}{F(x)-y} \, < \, 0$, because $g(x) > 0$ and $y > F(x)$ implies $F(x) - y < 0$, which means that $y=y(x)$ is a monotonously decreasing function.
Hence, for $x > 0$ the solution trajectory $(x, \,y(x))$ has the property $y(x) < y(0) = y_0$. The initial condition $y_0$ can be taken to be the $y$-coordinate of the point $P_0 = (0, y_0)$ on the picture.
Next, we will apply the following theorem from the theory of systems of ordinary differential equations:
Theorem. Let $f \, : \, U \, \to \, \mathbb{R}^n$ be a Lipschitz continuous function (this includes continuously differentiable functions) defined on an open set $U \, \subset \, \mathbb{R} \times \mathbb{R}^n$. Furthermore, let $D \subset U$ be a bounded connected open subset of $U$ and let $(x_0, \,y_0) \, \in \, {D}$. Then there exists a unique (local) solution $y = y(x)$ of the initial value problem
\begin{align}
&\frac{dy}{dx} \, =\, f(x,\, y)\\
&y(x_0) \, =\, y_0 
\end{align}
Furthermore, the solution $y = y(x)$ can be extended over a closed interval $x \in [a, b]$, so that $\big(x,\, y(x)\big) \in D$ for all $x \in (a, b)$, while $y(a) \in \partial D$ and $y(b) \in \partial D$.
By applying the theorem above to the equation $$\frac{dy}{dx} \, =\,  \frac{g(x)}{F(x)-y} $$ defined in the open set $U = \big\{\,(x,y) \in \mathbb{R}^2 \,\, :\,\, F(x) \neq y\,\big\}$ and restricted to the open bounded domain $D \,=\, D_{x+} \, \cap\, \big\{\,(x,y) \in \mathbb{R}^2 \,\, : \,\, y < y_0\,\big\} $, we can conclude that any solution in $D$, being decreasing, cannot go over the boundary $y = y_0$ and since it is extended forward, the only boundary component of $D$ it can intersect, as per the theorem, is the curve $FG_{+}$.
Absolutely analogously, in the domain $D_{+}$, any solution trajectory of the original system of differential equations can be reparametrized as $\big(x(y), y\big)$ where $x = x(y)$ is the solution to the differential equation
$$\frac{dx}{dy} \, =\,  \frac{F(x)-y}{g(x)} $$
Again, applying the theorem above, but this time for the latter differential equation defined in the open set $U = \big\{\,(x,y) \in \mathbb{R}^2 \,\, :\,\, g(x) \neq 0\,\big\}$ and restricted to the bounded open set $D = D_{y+} \cap\, \big\{\,(x,y) \in \mathbb{R}^2 \,\, : \,\, y < y_0\,\big\} $  we can conclude that any solution trajectory (x(y), y) in $D$ cannot go over the boundary $y = y_0$,  because the same trajectory can be reparametrized as $(x, y(x))$ with $\frac{dy}{dx} = \frac{g(x)}{F(x) - y}$ and we already know that such solution $y=y(x)$ is decreasing and cannot reach $y=y_0$. Therefore, according to the theorem above, the only boundary component of $D$ the solution trajectory can intersect is the curve $R$. This  point of intersection is denoted by $P_2$ on your picture.
The reason for this ping-pong between the two representations
$$\frac{dy}{dx} \,=\, \frac{g(x)}{F(x) - y} \, \text{ and } \, \frac{dx}{dy} \,=\, \frac{F(x) - y}{g(x)}$$ of a solution trajectory of the original system is because on the curve $L$ at the point $P_0$ we have $\frac{dy}{dx} = 0$ and  $\frac{dx}{dy} = \infty$, while on the curve $R$ at the point $P_2$ we have $\frac{dy}{dx} = \infty$ and  $\frac{dx}{dy} = 0$.
Following absolutely the same method for the other domain $D_{-}$, which splits into the subdomains $D_{y-}$ and $D_{x-}$, ping-ponging between the two differential equation representation of a solution trajectory of the original system, you can complete the argument that the solution trajectory reaches a point $P_4$ on the negative half of the vertical axis $x=0$.
