1
$\begingroup$

I want to study the stability of the equilibrium points of the following ODE in $n=2$:

$x'(t) = y(t)$

$y'(t) = −kx(t) + cx^2(t)$

Where $k,c\geq0$

From the system:

$y = 0$

$−kx + cx^2 = 0$

we find that equilibrium points are $P_1 = (0, 0)$ e $P_2 = (k/c,0)$. Evaluating the jacobian of the equations at $P_1$ we get that it has two eigenvalues with real part equal to $0$. My professor claims that $P_1$ is stable but not asymptotically, but the Hartmann-Grossman theorem says the following:

Consider the nonlinear autonomous ODE in dimension $n$ and assume that $f ∈C^1(X;R)$. Let $\bar x ∈ {\rm Int}X$ be an equilibrium point such that $\max\{Reλ, λ ∈ σ(D f(x)\} \neq 0$. Then :

• If $\max\{Reλ, λ ∈ σ(D f(x))\}> 0$ then $\bar x$ is unstable

• If $\max\{Reλ, λ ∈ σ(D f(x))\}< 0$ then $\bar x$ is stable and AS.

So it looks to me that we can not say anything a-priori looking only at the eigenvalues since the real part is zero.

Can you help me?

$\endgroup$
1
$\begingroup$

Your system is conservative, all trajectories lie on level curves of the first integral/energy function $$E(x,y)=\frac12y^2+\frac{k}2x^2-\frac{c}3x^2,$$ which means that you do not get source or sink equilibrium points, but only centers and saddle points. With $f'(x,y)=\pmatrix{0&1\\-k+2cx&0}$ you get a center at $(x,y)=(0,0)$ and a saddle point at $(x,y)=(\frac kc,0)$.

$\endgroup$
  • $\begingroup$ Thank you, but how do I see that we get a center at $(x,y) =(0,0)$? Also is if we have two eigenvalues with real part equal to zero, is it always true that the system is stable but not a.s.? This seems to be in conflict with the H-G theorem. $\endgroup$ – LearningProb Jan 17 at 14:33
  • $\begingroup$ In general no, but in a conservative system there are no other possibilities. You could use the energy function as a Lyapunov function to directly confirm the (non-asymptotic) stability. $\endgroup$ – Lutz Lehmann Jan 17 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.