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$$x'=y$$ $$y'=-x+x^3$$

from above system, one gets hyperbolic equilibria $(1,0)$ and $(-1,0)$. and both equilibria have same eigenpairs $(\lambda,v)$, such as $(\sqrt{2},(1,\sqrt{2})^T)$ and $(-\sqrt{2},(1,-\sqrt{2})^T)$.

and here I tried to find its stable/unstable manifolds start with letting $y=f(x)=ax+bx^2+...$ and differentiate both side with respect to time $t$. then

$$y'=f'(x)x'$$

$$-x+x^3=(a+2bx+...)(ax+bx^2+...)$$

and I got $a^2=-1$ which is a contradiction. So it makes me to think that this system has no invariant manifolds at hyperbolic points, but I learned that every hyperbolic equilibrium has at least local invariant manifolds. It makes me very confused. Where was I wrong?

rekt
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  • What happens if you take the system with the $x^3$ erased? – Will Jagy May 20 '15 at 04:46
  • @WillJagy then it will have complex eigenvalues. btw, I think I solved it by letting $f(x)=a+bx+cx^2+dx^3+ex^4+...$ – rekt May 20 '15 at 04:48
  • it needs more than 5 coefficients, (I don't know exactly,) anyway I could find polynomial that satisfy $y'=f'(x)x'$ for given system – rekt May 20 '15 at 04:53
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    Near the point $(1,0)$, one would expand $y$ as $y=a(x-1)+b(x-1)^2+c(x-1)^3+\ldots$, not as $y=ax+bx^2+cx^3+\ldots$ If I am not mistaken, this yields $$y=\pm\left(\sqrt2(x-1)+\frac1{\sqrt2}(x-1)^2+\ldots\right).$$ – Did May 20 '15 at 05:29
  • @Did yes right, I was trying to find one manifold tangent to eigenspaces of two equilibria, but in fact there was no reason that they should share one manifold. – rekt May 20 '15 at 08:12

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