I am analyzing this problem, and some questions has appeared to me.
1- In case 2 and 3, which Hamiltonian should I choose?
2- I did not understand case 4? Could you enlighten me please.
Let $\alpha$ be non-negative constant. Let $g(x)$ be a smooth real-valued function, and assume $g(0)=0$ and $\forall x \not = 0, \,\, x g(x) >0$. Let $G(x) = \int_0^x g(s) ds$ be one of the primitive of $g$. For an ODE $\ddot{x} + \alpha \dot{x} + g(x) =0 $, (x: real-valued). Set $y= \dot{x}$ and consider the ODE system
\begin{align*}
\dot{x} &=y\\
\dot{y} &= – g(x) – \alpha y\\
\end{align*}
Assume the following:
1)- When $\alpha =0 $, this system is a Hamiltonian system. What is the Hamiltonian $H(x,y)$?
2)- When $\alpha \geq 0$, show that the origin $(x,y) =(0,0)$ is stable in the sense of Lyapunov, by using $H(x,y)$.
3)- When $\alpha >0$, find a strict Lyampunov function and show that the origin is asymptotically stable.
4)-When $ \alpha >0$, can we show asymptotic stability of the origin by using $H(x,y)$ instead of a strict Lyapunov function in (3)? (\textbf{HINT}: LaSalle's invariance principle)
\begin{solution}
\textbf{(1):} When $\alpha =0$ the system is:
\begin{align*}
\dot{x}&=y\\
\dot{y}&= – g(x)
\end{align*}
The Hamiltonian must satisfy $\partial_x H(x,y)= g(x)$ and $\partial_y H(x,y) = y$. So we solve these equations and find $H(x,y)$
\begin{align*}
\partial_x H = g(x) \Rightarrow H &= \int_0^x g(s) ds + C(y)\\
H&= G(x) + C(y)
\end{align*}
and
\begin{align*}
\partial_y H &= C'(y) = y \Rightarrow C(y) = \frac{1}{2} y^2.\\
\end{align*}
Thus, the Hamiltonian is
$$H(x,y) = G(x) + \frac{1}{2} y^2.$$
\end{solution}
\textbf{(2):} When $\alpha \ge 0$, the system is:
\begin{align*}
\dot{x} &= y\\
\dot{y} &= -g(x) – \alpha y\\
\end{align*}
First, we find the Hamiltonian to the system;
\begin{align*}
\partial_y H=y \Rightarrow H =& \frac{1}{2}y^2 + C(x)\\
\partial_x H = & C'(x) = g(x) + \alpha y\\
\Rightarrow &C(x) = G(x) + \alpha x y\\
\end{align*}
So, the Hamiltonian is $H(x,y) = G(x) + \frac{1}{2} y^2 + \alpha x y$. Now, we have two Hamiltonians. However, we are going to use the one in case (1). We will show that the $H(x,y) = G(x) + \frac{1}{2}y^2$ satisfies the conditions of Lyapunov function. Define the set
$$\Omega := \{ (x,y) \in \mathbb{R}^2 ; |G(x)| < \frac{1}{2}y^2 \}$$
1- It is clear to see that $H(x,y)$ is continuous on $\Omega$.
2- $H(0,0) =0$, and $\forall (x,y) \in \Omega \setminus \{(0,0)\}$ we have
$$H(0,0) =0 < H(x,y).$$
3- $\forall (x,y) \in \Omega \setminus \{(0,0)\}$ we have
\begin{align*}
\frac{d H(x,y)}{dt} =& \frac{\partial H}{\partial x} \frac{d x}{dt} + \frac{\partial H}{\partial y} \frac{ d y}{dt}\\
=& g(x) y + y (- g(x) – \alpha y)\\
=& – \alpha y^2 \leq 0.\\
\end{align*}
So, $H(x,y)$ is non-increasing, then $(0,0)$ is stable.
Best Answer
UPD: Some comments on the Lyapunov function. So, we have $$V(x,y) = S(x,y)+G(x),$$ where $S(x,y) = \frac{\alpha^2}{4}x^2+\frac{\alpha}{2}xy+\frac{1}{2}y^2$ and $G(x) = \int_0^x g(s)ds$. Note that $G(x)$ is positive definite by design, e.g., for $g(x)=x$ we have $G(x)=x^2$.
Ok, why $S(x,y)$ is positive definite? There are two ways to show that. First, note that $$S(x,y) = \frac{1}{2}\begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}\frac{\alpha^2}{2} & \frac{\alpha}{2} \\ \frac{\alpha}{2} & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix},$$ and this matrix is positive-definite.
The second way is to write $$ \begin{aligned} S(x,y) &= \frac{\alpha^2}{16}x^2 + \frac{3\alpha^2}{16}x^2 + \frac{\alpha}{2}xy + \frac{1}{3}y^2 + \frac{1}{6}y^2\\ & = \frac{\alpha^2}{16}x^2 + \left(\frac{\sqrt{3}\alpha}{4}x+\frac{1}{\sqrt{3}}y\right)^2+ \frac{1}{6}y^2, \end{aligned} $$ which is also positive definite.