---
title: "stability (ODEs)"
source: https://www.jemoka.com/posts/kbhstability/
---

A stationary point of an ODE is considered &ldquo;stable&rdquo; if, at the stationary point \(y=c\), the function with initial condition.
If you start near a stationary point, the function will either diverge \(t\to \infty\) to that stationary point, or converge to a stationary point. Whether the functions done that makes it &ldquo;stable&rdquo;/&ldquo;unstable&rdquo;.
For an autonomous ODEs \(y&rsquo;(t) = f(y(t))\), suppose \(y(t) = c\) is a stationary solutiona:
\(c\) is stable (i.e. \(t\to \infty, y \to c\) for \(y_0 \approx c\)) if the graph of \(f\) near \(c\) crosses from positive to negative; that is, when \(f&rsquo;( c) &lt; 0\) \(c\) is unstable (i.e. \(t\to -\infty, y \to c\) for \(y_0 \approx c\)) if the graph of \(f\) near \(c\) crosses from negative to positive; that is, when \(f&rsquo;(t) &gt; 0\) \(c\) is semi-stable (i.e. stable on one side, unstable on the other) if the graph of \(f\) near \(c\) has the same sign on both sides; meaning \(f&rsquo;( c) = 0\) and \(f&rsquo;&rsquo;( c)\neq 0\) if \(f&rsquo;( c) = 0\) and \(f&rsquo;&rsquo;( c) \neq 0\), we are sad and should investigate more away from zeros, the concavity of \(y(t)\) could be checked for \(f f&rsquo;\). when its positive, \(y(t)\) is concave up; when its negative \(y(t)\) is concave down.
