Let $g(y) = \frac{dy}{dx} = y^2$. To find the equilibria, we set $\frac{dy}{dx} = 0$, and we get $y = 0$. To determine its stability, we will look at $g'(0)$. We have $g'(y) = 2y$ and $g'(0) = 0$. We know that the equilibium solution is stable when $g'(0) < 0$ and unstable when $g'(0) > 0$. But what can we say if $g'(0) = 0$?

Here are the graphs of $g(y)$ and $g'(y)$:

Graph of g(y) and g'(y)

Some points are worth noting:

  • $g'(y) = 0$ when $y = 0$
  • $g'(y) < 0$ when $y < 0$
  • $g'(y) > 0$ when $y > 0$

Then, we solve the differential equation by separation of variables, which gives $y = -\frac1{x+c}$, where $c$ is some constant.

Here are some of the solutions:

Some solutions of dy/dx = y^2

Here is the direction field:

Direction field of dy/dx = y^2

Here, the equilibrium solution, $y = 0$, seems to repel any other solution above it and attract any other solution below it. This is somehow consistent with the findings we noted earlier: when $y > 0$, $g'(y) > 0$ and the solutions move further and further away from the equilibrium; when $y < 0$, $g'(y) < 0$ and the solutions move closer and closer to the equilibrium. But should we say that this equilibrium solution is stable or unstable?

When we put $y(0) = 0$ into our general solution, we get $0 = -\frac1c$, and there seems to be no definite solution for $c$, and we can only say that it is some very large value near infinity. This is very confusing, too.

  • $\begingroup$ @Moo So you call that semi-stable? The book I am reading does not introduce that term, and in the exercises, I have to find the equilibrium solutions and discuss their stability. $\endgroup$ – W. Zhu Oct 7 '16 at 15:12
  • $\begingroup$ If $y(0)=0$ then $y(x)=0$ for every $x$. What is confusing? $\endgroup$ – Did Oct 7 '16 at 15:28
  • $\begingroup$ @Moo Thank you. Those are great resources. $\endgroup$ – W. Zhu Oct 7 '16 at 15:28
  • $\begingroup$ @Did You are right. I get it now. I should not have been troubled by that constant $c$. $\endgroup$ – W. Zhu Oct 7 '16 at 15:31
  • $\begingroup$ Stable means that all nearby points are attracted to the equilibrium, so I would definitely call this unstable. (Unless you are, for some reason, only interested in solutions with $y \le 0$.) $\endgroup$ – Hans Lundmark Oct 8 '16 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.