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I have a Lagrangian made up of a superposition of $k$ central repulsive potentials centered at $(a_k,b_k)$. Each potential is somewhat strange - they take the form $V(r)=-\ln\left(r\right)$. The whole Lagrangian is $$L=\frac{1}{2}(\dot{x}^2+\dot{y}^2)+\sum_{k=1}^n\ln\left(\sqrt{(x-a_k)^2+(y-b_k)^2}\right).$$ I don't expect to find any sort of exact solution, but I'm interested in finding conserved quantities (besides of course energy) and an algorithm for finding equilibria. I tried a simple algorithm using Newton's laws with negative feedback on the velocity to slow it down, but the equilibria seem to be so unstable that that wasn't an effective method.

At the end of the day, I'm most interested in finding at least approximate locations for the equilibria, and the conserved quantities were just an idea of how I can better study the dynamics to try to find equilibria. Does anyone have an idea for either of those?

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    $\begingroup$ Could you confirm the sign of the potential $V$? Is it supposed to be attractive or repulsive? $\endgroup$
    – Qmechanic
    Nov 16 '18 at 19:59
  • $\begingroup$ I chose the sign to be + because I intended for the potential to be repulsive. The equilibria should be somewhere in the convex hull of the force centers. Given that, I would expect the equilibria to be unstable if the force centers were attractive, since a small nudge would send the particle to the nearest force center. I would expect repulsive potentials, on the other hand, to give negative feedback. $\endgroup$
    – BGreen
    Nov 16 '18 at 20:24
  • $\begingroup$ What do you mean by 'equilibria'? Be more precise. Try to use some basic methods from dynamical systems; write Hamilton equations, find which $(x,p)$ are nullifying $(\dot{x},\dot{p})$, then assess their stability. $\endgroup$
    – Alexander
    Nov 16 '18 at 20:44
  • $\begingroup$ @Qmechanic Thank you for the edit and please excuse my carelessness in transcribing the potential. This potential should generate a force field of the same form as in the Gauss-Lucas theorem, meaning the there should be no net force at the critical points of the polynomial with roots at each $z_k = a_k+ib_k$, unless I've made a mistake. $\endgroup$
    – BGreen
    Nov 17 '18 at 1:30
  • $\begingroup$ @Alexander "Equilibria" has always meant a point at which the first derivative vanishes. Perhaps I wasn't clear in the problem statement, but the problem is that I do not know a good way to the locations at which the derivatives vanish in the first place. $\endgroup$
    – BGreen
    Nov 17 '18 at 1:30
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FWIW, OP's potential of the form $$ V(z) ~=~-\sum_{i=1}^n k_i\ln|z-z_i| , \qquad z~=~x+iy~\in ~\mathbb{C}\backslash\{z_1, \ldots, z_n\}, \qquad k_i~\neq~0,$$ is a non-constant harmonic function in 2D, which means that all critical points are saddle points (i.e. non-stable equilibrium points), and the potential $V$ obeys the maximum principle.

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  • $\begingroup$ Thank you for pointing that out! Now I remember that $\ln(z)$ is the 2D Coulomb Green's function, so this is even more a 2D gravitational analogue than I had at first recognized, making it even more interesting. Evidently I'll have to think of another approach to finding the critical points, but this is an interesting point to start from. $\endgroup$
    – BGreen
    Nov 18 '18 at 19:28

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