Determine stability of the equilibrium of $\dot x=f(x)$ in terms of the derivatives of $f$

Question

Let $ f : \mathbb{R} → \mathbb{R} $ be $C^∞$ suppose the following:

1. $f(0) = 0,$
2. There is a smallest $n \in \mathbb{N} $ so that $f^{n}(0) \neq 0$

Determine stability of the equilibrium at $0$ for the differential equation $ \dot x = f(x) $ in terms of $n$ and the sign of $f^{n}(0)$.

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The exercise sheet provides a hint: Use Taylor’s theorem: $$\dot x = \left( \frac {f^{n}(x)}{n!} + xh(x) \right) x^{n}$$

The problem is the $h$ is not definied (there is a reference to a page in the book for $h$, but the page is wrong!)

Any suggestion on how to proceed? even without using the given hint?

Answer 1

This is more transparent if you think of the phase line. What actually determines stability is not the derivatives of $f$ per se, but the sign of $f$ on both sides of the equilibrium point.

• If $f$ changes sign from positive to negative, the equilibrium is stable. (The arrows on phase line are toward the equilibrium)
• If $f$ changes sign from negative to positive, the equilibrium is unstable. (The arrows point away from equilibrium)
• If $f$ does not change the sign, the equilibrium is semi-stable. (Arrows point toward it on one side, and away on the other.)

The Taylor expansion tells us which of three options holds, because in a small neighborhood of a zero $x_0$, the sign of $f$ is determined by the sign of the nonzero term of lowest degree (the other terms are small in comparison to it). So:

• If $n$ is odd and $f^{(n)}(x_0)<0$, the sign goes from positive to negative: stable.
• If $n$ is odd and $f^{(n)}(x_0)>0$, the sign goes from negative to positive: unstable.
• If $n$ is even, the sign is the same on both sides of $x_0$: semi-stable.