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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:

  1. 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.
  2. 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 $z_i$ ?

abc1455
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1 Answers1

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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.

abc1455
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JMJ
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  • I know that in fact there are trajectories that converge to $z_1=\mathbf{x}_i$ , as a numerical example let $\mathbf{x_i} =[1 \ \ 0]$ i.e. on the x-axis and let $K$ have an eignvector $v=[1 \ \ 0]$ then if start on the x-axis with $z_2=0$ and $z_1=[2 \ \ 0]$ you will converge to $\mathbf{x}_i$ – abc1455 Jul 21 '20 at 11:14
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    and as for the conclusion about $$ \dot{z}_2\approx \frac{|z_2|}{|z_1-x_i|}\hat{\xi} $$ you have to consider that $| z_2 |$ can also approach zero in a way that makes $\dot{z_2}$ doesn't grow unbounded . – abc1455 Jul 21 '20 at 11:31
  • @abc1455 Here is another way to see that the origin is the only equilibrium. For an equilibrium to exist at a point $e$, you must have $f(e)=0$, where $f$ is the RHS of your equation. That is not true for $z_1 = x_i$ and $z_2=0$, even if you ignore the nonlinear term (which is generally not possible--since, again, the equation has a singularity here). Perhaps this is more acceptable if you do not like the heuristic argument--yes, of course you have a good point that, were the linear part missing, $z_2$ might go to zero in exactly the right way to suppress the blowup in the denominator. – JMJ Jul 21 '20 at 11:52
  • Indeed $z_1=0$ is the only fixed point on the domain except at $z_1=x_i$ , so unless we know the behavior near $z_1=x_i$ , we cannot say that the origin will be globally asymptotically stable . consider this example , if you have a system with a stable equilibrium point $g$ and take your domain $D$ as the whole state space expect at neighborhood of some point $p$ , if the energy system is decreasing in $D$ except $g$ where the energy is constant, then you cannot say that if you start in $D$ you will converge to $g$ as some trajectories may indeed escape $D$ and enter the neighborhood of $p$ . – abc1455 Jul 21 '20 at 12:11
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    @abc1455 Since $z_1 = x_i$ isn't a point in the domain of the function, it can't be a fixed point of the function. I wasn't aware you're interested in GAS rather than just AS. There are many systems with only one equilibrium which is AS but not GAS. If there are more than one equilibrium (in this equation there are not) then there is no equilibrium that is GAS. To investigate the behavior near $x_i$, I would look up literature on singular differential equations and singular perturbation theory. There's a lot of work on these topics, but not simple. – JMJ Jul 21 '20 at 12:20
  • I am interested in proving that the system will converge to either $z_1= 0$ or $x_i$ , so I want to prove that if $b$ is greater that some value the system won't have a limit cycle . I also want to determine the set of the trajectories that converges to $x_i$ , I know indeed that such a set exists as in the example I gave in my first comment . – abc1455 Jul 21 '20 at 12:28
  • If the solution could converge to $x_i$, it would also be possible to define a solution that starts at $x_i$. How could that ever be done? – Hans Engler Jul 24 '20 at 13:58
  • @HansEngler : This is a very difficult technical detail . If you look at only in the domain of the function it is not convergent. However, if you look at in $\mathbb{R}^2$ it converges to $x_i$ – abc1455 Jul 26 '20 at 13:19