**Proof plan.** The odd symmetry forces $f(y, \mu) = y \cdot g(y, \mu)$ with $g$ even in $y$, so $g(y, \mu) = h(y^2, \mu)$ for a smooth function $h$. The non-trivial equilibria reduce to solving $h(w, \mu) = 0$ where $w = y^2 \geq 0$, and the [Implicit Function Theorem](/theorems/52) applies to $h$. **Step 1 (Factoring and symmetry reduction).** As in the [Transcritical Bifurcation](/theorems/696), $f(0, \mu) = 0$ gives $f(y, \mu) = y \cdot g(y, \mu)$ where $g(y, \mu) = \int_0^1 \partial_y f(ty, \mu)\, dt$. The odd symmetry $f(-y, \mu) = -f(y, \mu)$ implies $g(-y, \mu) = g(y, \mu)$, so $g$ is even in $y$. By a standard result on even smooth [functions](/page/Function), there exists a smooth function $h: [0, \infty) \times \mathbb{R} \to \mathbb{R}$ such that $g(y, \mu) = h(y^2, \mu)$. **Step 2 (Evaluating $h$ at the bifurcation point).** Using the relation $g(y, \mu) = h(y^2, \mu)$: \begin{align*} h(0, 0) &= g(0, 0) = \partial_y f(0, 0) = 0 \quad \text{(by PF1)}, \\ \partial_\mu h(0, 0) &= \partial_\mu g(0, 0) = \partial^2_{y\mu} f(0, 0) = c \neq 0 \quad \text{(by PF3)}. \end{align*} For $\partial_w h(0, 0)$: differentiating $g(y, \mu) = h(y^2, \mu)$ twice with respect to $y$ and evaluating at $y = 0$ gives $\partial^2_{yy} g(0, 0) = 2\,\partial_w h(0, 0)$. Meanwhile, $\partial^2_{yy} g(0, 0) = \int_0^1 t^2 \partial^3_{yyy} f(0, 0)\, dt = \frac{1}{3}\partial^3_{yyy} f(0, 0) = 2l_3$. Hence \begin{align*} \partial_w h(0, 0) = l_3 \neq 0 \quad \text{(by PF2)}. \end{align*} **Step 3 ([Implicit Function Theorem](/page/Implicit%20Function%20Theorem)).** Since $h(0, 0) = 0$ and $\partial_w h(0, 0) = l_3 \neq 0$, the [Implicit Function Theorem](/theorems/52) provides a smooth function $w^*(\mu)$ with $w^*(0) = 0$ and $h(w^*(\mu), \mu) = 0$. Implicit [differentiation](/page/Derivative) gives \begin{align*} (w^*)'(0) = -\frac{\partial_\mu h(0,0)}{\partial_w h(0,0)} = -\frac{c}{l_3}. \end{align*} Non-trivial equilibria exist when $w^*(\mu) = y^2 > 0$, which requires $w^*(\mu) > 0$. Since $w^*(\mu) = -(c/l_3)\mu + O(|\mu|^2)$, this holds precisely when $c\mu l_3 < 0$, and then $y_\pm = \pm\sqrt{w^*(\mu)} = \pm\sqrt{-c\mu/l_3} + O(|\mu|)$. **Step 4 (Stability).** The linearisation at the trivial branch is $\partial_y f(0, \mu) = g(0, \mu) = h(0, \mu)$. Since $h(0, 0) = 0$ and $\partial_\mu h(0, 0) = c$, this gives $\partial_y f(0, \mu) = c\mu + O(|\mu|^2)$, so the trivial branch is stable when $c\mu < 0$ and unstable when $c\mu > 0$. At the non-trivial branches, $\partial_y f(y_\pm, \mu) = g(y_\pm, \mu) + y_\pm \partial_y g(y_\pm, \mu)$. Since $g(y_\pm, \mu) = h(w^*(\mu), \mu) = 0$, this reduces to $y_\pm \partial_y g(y_\pm, \mu) = y_\pm \cdot 2y_\pm \partial_w h(w^*, \mu) = 2w^*(\mu) l_3 + O(|\mu|^2) = -2c\mu + O(|\mu|^2)$. This is negative (stable) when $c\mu > 0$ — which, combined with the existence condition $c\mu l_3 < 0$, requires $l_3 < 0$ (supercritical). When $l_3 > 0$ (subcritical), the branches exist for $c\mu < 0$ and are unstable ($-2c\mu > 0$).