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Consider the second order ODE $u'' +(u^2-2u+1)u'+u^3-u^5=0$. We'd like to show the $0$ solution is asymptotically stable. Writing it as a first order system and of the form $\vec{u}' = A\vec{u} + F(\vec{u})$ yields $$ \begin{pmatrix} u \\v \end{pmatrix}' = \begin{pmatrix}0 && 1\\ 0 && -1 \end{pmatrix}\begin{pmatrix} u \\v \end{pmatrix} + \begin{pmatrix} 0 \\-u^2v+2uv-u^3+u^5 \end{pmatrix}. $$

Unfortunately, the matrix has eigenvalues $0$ and $-1$, and $0$ is not negative, so we can't get asymptotic stability that way. I'm thinking we might want to find a Lyapunov function, but I do not have much experience with this, so any help would be appreciated.

Curious
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    I am quite sure, that you are not allowed to split the equation like you did and then apply the eigenvalue theorem. If I remember correctly, than you must have F constant in u. (Btw... you used u sometimes as a vector and sometimes as the first component of that vector.) Your F is even nonlinear, therefore you cant use the Eigenvalue theorem the form like you have written above. (Think about it,... you even find a F s.th. A was the Identity... and that was possible for all ode's... This indicates that your F can't be nonlinear) – Börge Jul 08 '17 at 22:01
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    Apologies I was being a little sloppy and forgot the arrows for the vector form. I know of a Theorem that says if $F$ is locally lipschtiz and $||F(x)||/||x||\to 0$ as $||x||\to 0$ then I can apply the eigenvalue theorem. I didn't show the last part for my specific $F$ since I don't have asymptotic stability on the linear part anyway, so it doesn't work. Do you have any hints on how I might proceed then? – Curious Jul 08 '17 at 22:15
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    Are you shire, that the solution to this ode is asymptotically stable? Because I tested it numerically and, it doesn't look that way... It looks more like stability without asymptotical stability... which makes kind of sense , because I was not able to find a Liapunov function... (But it's late, and I could be mistaken....) – Börge Jul 09 '17 at 01:48
  • It's certainly possible it's not. In fact, I decided to put it on Wolfram Alpha and it did plot a solution that seems to blow up in finite time. This was a problem on an old exam, so it could have been an error. – Curious Jul 09 '17 at 04:10
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    @Börge There is absolutely nothing wrong with rewriting the equations this way because we do this in the neighbourhood of $(0, 0)$ point: in that case splitting of linear and non-linear part is absolutely legit because all nonlinear terms in Jacobi matrix will become zero when we plug $(0, 0)$ and all that is left is linear part. – Evgeny Jul 09 '17 at 07:26

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A quite common trick in studying systems of form $\ddot{u} = -\nabla F(u)$ is to multiply both sides by $\dot{u}$: $$\dot{u} \cdot \ddot{u} = \dot{u} \cdot (-\nabla F(u))$$ $$\dfrac{d}{dt} \left ( \dfrac{\dot{u}^2}{2} \right ) = \dfrac{d}{dt} \left ( -F(u) \right ) $$ $$\dfrac{d}{dt} \left ( \dfrac{\dot{u}^2}{2} + F(u) \right ) = 0 $$ $$\dfrac{\dot{u}^2}{2} + F(u) = \dfrac{\dot{u}^2(0)}{2} + F(u(0)) $$

And in that case the system is conservative and has first integral. It also can't have asymptotically stable and completely unstable equilibria. But your case is not the same: you have terms with $\dot{u}$. Still we can apply the trick to your system: $$u'' +(u^2-2u+1)u'+u^3-u^5=0$$ $$ u' \cdot u'' +(u^2-2u+1)\cdot (u')^2 + u' \cdot (u^3-u^5)=0 $$ $$ u' \cdot u'' + u' \cdot (u^3-u^5)= -(u^2-2u+1)\cdot (u')^2 $$ $$ \dfrac{d}{dt} \left ( \dfrac{\dot{u}^2}{2} + \dfrac{u^4}{4} - \dfrac{u^6}{6} \right ) = -(u-1)^2 \cdot (u')^2 $$

So what we got here is that derivative of some function w.r.t. time is non-positive. This is almost a setting of Lyapunov theorem. You can check that $\dfrac{v^2}{2} + \dfrac{u^4}{4} - \dfrac{u^6}{6}$ has strict local minimum at $(0, 0)$ so this function could be used in Lyapunov theorem. At least we have that this equilibrium is Lyapunov stable. As far as I understand, the asymptotical stability can be concluded from LaSalle's invariance principle.

Evgeny
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    Thanks for the help. Multiplying by $u'$ has been coming up a lot in ODE problems I've been studying recently. – Curious Jul 09 '17 at 13:09
  • Yeah, that's very common method for studying second order equations :) You can accept the answer if it solves your problem. Or I can answer your additional questions if something needs to be clarified. – Evgeny Jul 09 '17 at 16:06