---
tag: 1K8Y6BP097
kind: definition
name: Transcritical bifurcation
msc: [34C23, 37G10]
defines: KFY58YKR3N
uses: [0G4X36AE0X, 0V33BNKZ9S]
sources:
  - url: https://en.wikipedia.org/wiki/Transcritical_bifurcation
---

## Definition

Let $f \colon \R \times \R \to \R$ be [[smooth|0V33BNKZ9S]], and consider the family of scalar equations $x' = f(x, \mu)$ with parameter $\mu$.
Assume $f(0, \mu) = 0$ for all $\mu$, so that $0$ is an [[equilibrium point|0G4X36AE0X]] for every $\mu$.
The family has a *transcritical bifurcation* at $(x, \mu) = (0, 0)$ if:

1. $f_x(0, 0) = 0$;
2. $f_{x\mu}(0, 0) \ne 0$;
3. $f_{xx}(0, 0) \ne 0$.

## Remarks

Under these conditions, besides $x = 0$, the equilibria near $(0, 0)$ form a curve $x = c(\mu)$ with $c(0) = 0$ and $c'(0) = -2 f_{x\mu}(0, 0) / f_{xx}(0, 0) \ne 0$.
So this second branch crosses the branch $x = 0$ at $\mu = 0$.
Since $f_x(0, \mu)$ changes sign at $\mu = 0$, the equilibrium $0$ changes stability there, and the other branch changes stability the opposite way.

Example: $f(x, \mu) = \mu x - x^2$.
Then $f(0, \mu) = 0$, $f_x(0, 0) = 0$, $f_{x\mu} = 1$ and $f_{xx} = -2$.
The second branch is $x = \mu$, with $c'(0) = 1 = -2 \cdot 1 / (-2)$.

Non-example: $f(x, \mu) = \mu x - x^3$.
Here $f_{xx}(0, 0) = 0$, so clause 3 fails; this family has a pitchfork bifurcation instead.