Stability analysis of the dynamical system $\ddot{x}+b\dot{x}+K x-\|\dot{x}\| \frac{x-x_i}{\|x-x_i\|^2}=0$ . Consider the dynamical system described as:
$$\ddot{z}+b\dot{z}+ K z-\|\dot{z}\| \frac{z-z_i}{\|z-z_i\|^3}=0$$
where $z=[x \ \ y]^T$, $K$ is a positive definite matrix and $b \in \mathbb{R}$, I made some simulations and based on the numerical results I concluded that:

*

*if $b>0$, the system converges to either $z=0$, $z=z_i$ or a limit cycle i.e. stable in the sense of lyapunov.

*if $b>\frac{1}{\|z_i\|}$,  the system converges to either $z=0$ or $z=z_i$ i.e. no limit cycles.

I was able to only prove that if $b>\frac{1}{\|z_i\|}$, $z=0$ is a stable fixed point by using lyapunov function as:
\begin{align}& V =\frac{1}{2} z^T K z +\frac{1}{2}\dot{z}^T \dot{z}\\
\implies &  \dot{V}=\|\dot{z}\|^2\left(-b+ \frac{cos(\theta)}{\|z-z_i\|^2}\right)
\end{align}
where $\theta$ is the angle between $\dot{z}$ and $z-z_i$, so if $b>\frac{1}{\|z_i\|}\implies \dot{V}|_{z=0} <0$ independent of $cos(\theta)$ in an open neighborhood of the origin so $z=0$ is a stable fixed point.
I tried to study the system near $z_i$ by using perturbation and introduced the parameter $\mu$ to the system as:
$$\ddot{z}+b\dot{z}+ K z-(\|\dot{z}\|+\mu) \frac{z-z_i}{\|z-z_i\|^3}=0$$
to study the system near $z_i$, I chose $\mu \gg \|\dot{z}(0)\|$ so the system becomes:
$$\ddot{z}+b\dot{z}+ K z-\mu \frac{z-z_i}{\|z-z_i\|^3}=0$$
Choose lyapunov function as:
$$\begin{align}&V=\frac{1}{\frac{1}{2} z^T K z +\frac{1}{2}\dot{z}^T \dot{z}+U_i}\\
\implies &\dot{V}=\frac{b\|\dot{z}\|^2}{(\frac{1}{2} z^T K z +\frac{1}{2}\dot{z}^T \dot{z}+U_i)^2}
\end{align}
$$
where $U_i=\frac{\mu}{\|z-z_i\|}$, so at $z=z_i$ , $V=0$ and $\dot{V}>0$ , so $z=z_i$ is unstable. However, if I check the equilibrium points by letting the derivatives vanish the system is reduced to:
$$K z=\mu \frac{z-z_i}{\|z-z_i\|^3}\implies \|z-z_i\|^3 K z=\mu(z-z_i) \text{ and } z \neq z_i$$
The right hand side can be made arbitrary small by choosing $\mu$ arbitrary small,since $K$ is full rank and $z\neq 0$ so it must be that $\|z-z_i\|$ is getting arbitrary small i.e. $z\rightarrow z_i$. So the system has another equilibrium point $q$ that is getting closer and closer to the unstable node $z_i$. I believe $q$ is a saddle point (I don't know how to prove it) and so I concluded that $z_i$ in my original system is a bifurcation between an unstable node and a saddle node.
My questions are : How to confirm the above claims ? and How to give a qualitative analysis of the behavior of the system near $z_i$ ?
 A: I assume that $x_i$ is a constant and not the $i$th component of $x$.
First, it is always helpful to write the system in a state space form. In this case, defining $z_1 = x$ and $z_2 = \dot{x}$, we have
\begin{equation}
\begin{bmatrix}\dot{z}_1 \\ \dot{z}_2\end{bmatrix} = \begin{bmatrix}0 & I_2\\ -K & -bI_2 \end{bmatrix}\begin{bmatrix}z_1 \\ z_2\end{bmatrix} + \begin{bmatrix} 0\\ \|z_2\|\frac{z_1-x_i}{\|z_1-x_i\|^2} \end{bmatrix}.
\end{equation}
Some observations: 1) the origin is the only equilibrium point. 2) The RHS has a singularity at $x_i$, so it would really surprise me if $z_1$ converged to $x_i$. Note that, for $z_1 \approx x_i$, the $z_2$ equation heuristically looks like
$$
\dot{z}_2\approx \frac{\|z_2\|}{\|z_1-x_i\|}\hat{\xi},
$$
where $\hat{\xi}$ is a unit vector. That's certainly not going to be an equilibrium beacuse the RHS is huge and gets bigger and bigger the closer $z_1$ gets to $x_i$.
There's a lot of detail for the case where you want to get an asymptotic solution near a singularity.
For the origin, you probably only need the indirect Lyapunov method (which is essentially just linearized stability) for most values of $b$ and $K$. Here you only need eigenvalues with negative real parts for $\begin{bmatrix}0 & I_2\\ -K & -bI_2 \end{bmatrix}$. For parameters that make the indirect method unable to conclude stability isL, there's much literature on perturbation theory to help you.
