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I was asked to linearize the first order ODE for an oscillating spring $$(x,v)^{'}=(v,-av^3-b(x^2-1))$$ where $a,b>0$.

Is it correct to simply find the jacobian of $(v, -av^3-b(x^2-1))$?

For the jacobian I got $$J=\begin{bmatrix}0&1\\-2bx&-3av^2\end{bmatrix}$$ After plugging in the critical points $(1,0)$ and $(-1,0)$ my matrix became $\begin{bmatrix}0&1\\-2b&0\end{bmatrix}$ and $\begin{bmatrix}0&1\\2b&0\end{bmatrix}$

After this, I used tried to find the eigenvalues by using the equations:

$\lambda^2-2b=0$ for $(1,0)$ and $\lambda^2+2b=0$ for $(-1,0)$.

By my calculations, $\lambda=\pm{\sqrt{2b}}$ for the point $(1,0)$ and $\lambda=\pm{\sqrt{-2b}}$ for the point $(-1,0)$.

This should mean $(1,0)$ is a saddle point while $(-1,0)$ is a an ellipse. This result does not seem intuitive though. Is there any point where I made a mistake? Should there not be a stable equilibrium for a spring?

E. Hurd
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  • Your work looks good. There may not be a stable equilibrium for your spring, but the "ellipse" or "center" is a reasonable orbit to be expected (no damping term). – Paichu May 04 '17 at 15:20
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    Well, never say that equilibrium of nonlinear system is a center judging only by eigenvalues, it's wrong. But, in this case, you can prove that this is a center equilibrium: it can be done via symmetries or using the first integral (kind of idea that @Paichu meant noticing that there is no damping term). The description for symmetry method can be found for example here. – Evgeny May 05 '17 at 15:39
  • @Evgeny Your answer for the other question is really nice. :) – Paichu May 05 '17 at 16:42
  • @Paichu Thank you! That was quite interesting problem to solve and I tried to write an entertaining solution :) – Evgeny May 05 '17 at 17:22

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