Let’s say some p∈R is a fixed point for a smooth map f:R→R.
[1] If ∣f′(p)∣<1, then p is a sink.
[2] If ∣f′(p)∣>1, then p is a source.
Example
As an example of a 1-dimensional map, consider f(x)=x3 which makes it easy to see that the fixed point f(0)=0 is a sink, and f(1)=1 is a source since f′(x)=3x2.
Proof
Proof of Theorem [1]
Let’s assume a∈(∣f′(p)∣,1).
x→plim∣x−p∣∣f(x)−f(p)∣=∣f′(p)∣
Therefore, for all x∈Nϵ(p),
∣x−p∣∣f(x)−f(p)∣<a
There must exist ϵ>0 such that, meaning ∣f(x)−p∣<a∣x−p∣ but since a<1, it gets closer to p every time f is applied to x. In other words, for all k∈N, fk(x)−p≤ak∣x−p∣, thus p must be a sink.
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Proof of Theorem [2]
Let’s assume a∈(1,∣f′(p)∣).
x→plim∣x−p∣∣f(x)−f(p)∣=∣f′(p)∣
Therefore, for all x∈Nϵ(p),
∣x−p∣∣f(x)−f(p)∣>a
There must exist ϵ>0 such that, meaning ∣f(x)−p∣>a∣x−p∣ but since a>1, it gets further from p every time f is applied to x. In other words, for all k∈N, fk(x)−p≥ak∣x−p∣, thus p must be a source.
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Yorke. (1996). CHAOS: An Introduction to Dynamical Systems: p10. ↩︎