1

I deal with analyzing the double pendulum system. in the first stage, I found the motion equation. now, I would like to find the stability condition. I want to do that by using Lyapunov stability theorem I know that there are 3 conditions that should exist for the Lyapunov function: $V(x')=0$, $V(x)>0$, $V'(x)\leq0$. My question is, how can I know which Lyapunov function to choose? Guessing? trial and error? thanks in advance

Sebastiano
  • 8,290
אוריה גולדנברג
  • 11
  • 2
    There is an energy functional, that should serve well as Lyapunov function. – Lutz Lehmann Nov 25 '22 at 08:43
  • thanks' do you mean to potential energy? the kinetic energy? – אוריה גולדנברג Dec 01 '22 at 16:41
  • The sum of both, the total energy where we know that it is invariant. You could reduce it to the lowest order terms to get a quadratic functional. – Lutz Lehmann Dec 01 '22 at 22:58
  • ok, and what if I will use just the potential energy? it is not gonna work? – אוריה גולדנברג Dec 02 '22 at 08:06
  • No, it works perfectly, the energy is constant along the trajectory, so you just have to prove that it has a strict minimum at the rest position. Note that general trajectories need not be bounded in terms of the angles, both links can rotate, the lower link more easily than the upper one. – Lutz Lehmann Dec 02 '22 at 08:19
  • Ok, the potential energy is ( lyapunov function ): -(m1+m2)glcos(Q1)-m2glcos(Q2). the derivative is : (m1+m2)gl sin(Q1)Q1'+ (m2glsin(Q2)*Q2'. let's say that I assume the function to be positive by determining the range of the angle with respect to cos. what about the derivative? ( Q1', Q2') – אוריה גולדנברג Dec 02 '22 at 09:07
  • which is should be negative in order to be a Lyapunov function – אוריה גולדנברג Dec 02 '22 at 09:09
  • That is automatic. With $\dot x = H_p^T$ and $\dot p=-H_x^T$, you get for the Lyapunov function $V=H$ the derivative along a solution $\dot V=V_x\dot x+V_p\dot p=0$. As it should be for a first integral. – Lutz Lehmann Dec 02 '22 at 09:19
  • what exactly is HTp ? – אוריה גולדנברג Dec 02 '22 at 09:39
  • in fact, the original system of equations was of second order. for using the Lyapunov method I had to reduce order. in summary, this is what I got ( find attached). do I have to solve it for psi? ( psi=Q') – אוריה גולדנברג Dec 02 '22 at 09:44
  • $H_p$ is the $p$ derivative of $E=H(x,p)$, where $p=\frac{\partial K}{\partial \dot x}$ in a general framework. The right-side function of a Hamiltonian system is the symplectic gradient of the Hamiltonian function $H$, so one needs the transpose of this part of the Jacobian, for the standard mechanical setup $\dot x=\nabla_pH=(\partial_pH)^T=H_p(x,p)^T$, depending on the notation you like. – Lutz Lehmann Dec 02 '22 at 09:45
  • unfortunately photos cant be added :/ – אוריה גולדנברג Dec 02 '22 at 09:47
  • Yes, it is a lengthy calculation, see https://math.stackexchange.com/questions/3930145/simplistic-simulation-of-double-pendulum-doesnt-work-is-this-approach-valid/3931310#3931310. See https://stackoverflow.com/questions/61044226/solving-a-system-of-2nd-order-differential-equations-from-sympy/61078311#61078311 for the symbolic derivation of the system with impulse variables using sympy. Add some pretty-print statements to the script to actually see all the equations. – Lutz Lehmann Dec 02 '22 at 10:06
  • Hi, I success to find the first-order equation of motion. can you think of a good candidate for the Lyapunov function? I tried several functions but it does not confirm the Lyapunov condition: V>0, V'<0. the equations are the following: (dθ_1)/dt=γ_1 (dθ_2)/dt=γ_2

    (dγ_1)/dt=(-g)/((1-m)l)θ_1+(mg)/((1-m)l)θ_2 (dγ_2)/dt=g/((1-m)l)θ_1-g/((1-m)l)*θ_2

     – אוריה גולדנברג Dec 09 '22 at 11:59

0 Answers0