# Dynamical system described by coupled nonlinear differential equations

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: $$$$\begin{split} &\frac{dx}{dt}=-x^\alpha\\ &\frac{dy}{dt}=-x-y^2 \end{split}$$$$ where $$\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 $$\alpha=\alpha_0$$?

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.
• 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? – Mr. Gentleman May 31 '20 at 3:48
• By integrable I mean that $\int_0^\infty x(t)\,dt<\infty$. This is the case if $\alpha<2$. – Michael Renardy May 31 '20 at 16:15