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I am not sure the name for this is really "degenerate", but consider the following system of non-linear ODEs: \begin{align*} \frac{dx}{dt} & = a(1-x)z-ex, \\ \frac{dy}{dt} & = -axy, \\ \frac{dz}{dt} & = axy - cz. \end{align*} The critical point is $(x,y,z) = (0,y,0)$, for any $y$. But starting from a certain $y$ I end up with another, final one. Is it possible to obtain the final values from the initial values in any way? I would be very thankful if anyone could point me in the right direction.

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Is this what you mean: if a solution curve $(x(t),y(t),z(t))$ tends to $(0,y_1,0)$ as $t \to -\infty$ and to $(0,y_2,0)$ as $t \to +\infty$, then what's the relation between $y_1$ and $y_2$? –  Hans Lundmark Jul 6 '12 at 11:29
    
Yes. Or, similarly, if I start with y=y0 then what is y at $\infty$. –  Gabriel Landi Jul 6 '12 at 13:57
    
If you start at $(0,y_o,0)$ at $t=t_0$, then you stay there forever, since it's an equilibrium point... –  Hans Lundmark Jul 7 '12 at 6:35
    
If you start at (0.0000001,y0,0) then things happen. The main point is how to deal with situations when the critical point does not tell you the equilibrium points exactly. Or perhaps, what is the "category" of this type of problem. I don't even know what to search. Thanks in advance for any help. –  Gabriel Landi Jul 7 '12 at 13:33
    
Yes, but something else might happen if you start at $(0.0001,y_0,0)$, so you need to say exactly what perturbation you consider. That's why I was trying to make the question precise by specifying the value of $y_0$ as $t \to -\infty$ instead of at some finite time. Anyway, it seems like a tricky problem. If one could find some constant of motion, that might help... –  Hans Lundmark Jul 8 '12 at 16:14
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