**Proof plan.** The persistent trivial branch allows us to factor $f(y, \mu) = y \cdot g(y, \mu)$ for a smooth [function](/page/Function) $g$. The non-trivial equilibria correspond to zeros of $g$, which are located by the [Implicit Function Theorem](/theorems/52). **Step 1 (Factoring out the trivial branch).** Since $f(0, \mu) = 0$ for all $\mu$, define \begin{align*} g: \mathbb{R} \times \mathbb{R} &\to \mathbb{R} \\ (y, \mu) &\mapsto \int_0^1 \partial_y f(ty, \mu) \, dt. \end{align*} By the [fundamental theorem of calculus](/theorems/632), $f(y, \mu) = y \cdot g(y, \mu)$ for all $(y, \mu)$. The function $g$ is smooth since $f$ is, and the non-trivial equilibria of $f$ in a punctured neighbourhood of $y = 0$ are precisely the zeros of $g$. **Step 2 (Applying the [Implicit Function Theorem](/page/Implicit%20Function%20Theorem)).** Evaluate $g$ and its derivatives at $(0, 0)$: \begin{align*} g(0, 0) &= \partial_y f(0, 0) = l_1 = 0, \\ \partial_y g(0, 0) &= \int_0^1 t\,\partial^2_{yy} f(0, 0) \, dt = \frac{1}{2}\partial^2_{yy} f(0, 0) = l_2 \neq 0. \end{align*} Since $g(0, 0) = 0$ and $\partial_y g(0, 0) = l_2 \neq 0$, the [Implicit Function Theorem](/theorems/52) provides $\delta > 0$ and a smooth function $y^*: (-\delta, \delta) \to \mathbb{R}$ with $y^*(0) = 0$ such that $g(y^*(\mu), \mu) = 0$ for all $|\mu| < \delta$, and $y^*$ is the unique zero of $g(\cdot, \mu)$ near $y = 0$. **Step 3 (Leading-order asymptotics).** Implicit [differentiation](/page/Derivative) of $g(y^*(\mu), \mu) = 0$ at $\mu = 0$ gives \begin{align*} \partial_y g(0, 0)\,(y^*)'(0) + \partial_\mu g(0, 0) = 0. \end{align*} Since $\partial_\mu g(0, 0) = \int_0^1 \partial^2_{y\mu} f(0, 0)\, dt = c$ and $\partial_y g(0, 0) = l_2$: \begin{align*} (y^*)'(0) = -\frac{c}{l_2}. \end{align*} **Step 4 (Exchange of stability).** The linearisation at the trivial branch is $\partial_y f(0, \mu) = g(0, \mu) + 0 \cdot \partial_y g(0, \mu) = g(0, \mu)$. Since $g(0, 0) = 0$ and $\partial_\mu g(0, 0) = c \neq 0$, this gives $\partial_y f(0, \mu) = c\mu + O(|\mu|^2)$. At the non-trivial branch, $\partial_y f(y^*(\mu), \mu) = g(y^*(\mu), \mu) + y^*(\mu)\,\partial_y g(y^*(\mu), \mu) = 0 + y^*(\mu)\,\partial_y g(y^*(\mu), \mu)$. Substituting $y^*(\mu) = -(c/l_2)\mu + O(|\mu|^2)$ and $\partial_y g(0, 0) = l_2$: \begin{align*} \partial_y f(y^*(\mu), \mu) = -\frac{c}{l_2}\,\mu \cdot l_2 + O(|\mu|^2) = -c\mu + O(|\mu|^2). \end{align*} The two linearisations $c\mu$ and $-c\mu$ have opposite signs for $\mu \neq 0$, confirming the exchange of stability.