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Suppose a dynamical system is described by two variables, $x$ and $y$, and they change over time according to the following two coupled nonlinear differential equations: \begin{equation} \begin{split} &\frac{dx}{dt}=-x^\alpha\\ &\frac{dy}{dt}=-x-y^2 \end{split} \end{equation} where $\alpha>0$ is a parameter for this system, and the initial condition for $x$ is positive $x(t=0)>0$.

This system is referred to as stable if $|y(t\rightarrow\infty)|<\infty$, and unstable otherwise. For a given $\alpha>0$ and $x(t=0)>0$, under what initial condition of $y$ is the system stable?

The following is some qualitative understanding I have.

First, the first term in the second equation tends to destabilize the system (by pushing $y$ to $-\infty$).

Second, if $x(t=0)=0$, then the system is stable if $y(t=0)\geqslant 0$ and unstable if $y(t=0)<0$, while $x=0$ for all $t>0$. That is, there is at least some (possibly measure-zero) regime where the system is stable. If $x(t=0)>0$ and $y(t=0)=0$, the system is unstable because $y\rightarrow-\infty$ as $t\rightarrow\infty$. So one expects there may be a separatrix between the stable regime and the unstable regime. The goal is to understand this separatrix.

Third, it seems we can focus on the vicinity of $(x, y)=(0, 0)$ and understand the separatrix there. In this regime, it seems if $\alpha$ is sufficiently large, $x$ approaches zero too slowly, so in the second equation it always destabilizes $y$ unless $x(t=0)=0$. That is, it seems the stable regime is really a measure-zero line in the two dimensional space of $x$ and $y$. On the other hand, if $\alpha$ is small, $x$ may approach zero sufficiently fast, and it does not destabilize $y$ if $y(t=0)$ is also large. So there seems to be a value of $\alpha_0$, such that when $\alpha>\alpha_0$, there is only a measure-zero stable regime, and when $\alpha<\alpha_0$, there is an extended stable regime.

I would like to understand (i) what is $\alpha_0$? (ii) when $\alpha<\alpha_0$, what is the separatrix (expressed in terms of $y(t=0)$ as a function of $x(t=0)$ and $\alpha$)? (iii) what happens exactly at $\alpha=\alpha_0$?

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The first equation is solvable in closed form, and then the second equation becomes a Riccati equation. For that, you have closed form solutions only for special values of $\alpha$. Some general observations: $y$ remains bounded if and only if it is always nonnegative. A necessary condition for that is that $y(0)$ is positive and $x$ is integrable. Whether $x$ is integrable depends on $\alpha$. If $x$ is not integrable, then $y$ cannot remain nonnegative. It may remain nonnegative if $x$ is integrable and $y(0)$ is large enough.

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  • $\begingroup$ Thank you. Can you provide more details on what it means to say $x$ is integrable, and for which $\alpha$ it is? For example, if $\alpha=1$, does the system have a stable regime? $\endgroup$ – Mr. Gentleman May 31 '20 at 3:48
  • $\begingroup$ By integrable I mean that $\int_0^\infty x(t)\,dt<\infty$. This is the case if $\alpha<2$. $\endgroup$ – Michael Renardy May 31 '20 at 16:15

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