[proofplan]
We analyse the saddle-node bifurcation by reducing the equilibrium equation $f(y,\mu) = 0$ to a perturbation of the quadratic $c\mu + l_2 y^2 = 0$ via Taylor expansion. When $c\mu l_2 > 0$, both leading terms have the same sign and the remainder is too small to compensate, yielding no equilibria. When $c\mu l_2 < 0$, the [Implicit Function Theorem](/theorems/52) applied near the two approximate roots $y_0^\pm = \pm\sqrt{-c\mu/l_2}$ produces exactly two equilibrium branches. Stability is read off from $\partial_y f$, and semi-stability at $\mu = 0$ follows from the sign structure of $l_2 y^2$.
[/proofplan]
[step:Reduce the equilibrium equation to normal form]
Taylor's theorem at the origin gives $f(y, \mu) = c\mu + l_2 y^2 + R(y, \mu)$, where $l_1 = 0$ by (SN1), $c = \partial_\mu f(0,0) \neq 0$ by (SN3), $l_2 = \frac{1}{2}\partial_y^2 f(0,0) \neq 0$ by (SN2), and $R(y, \mu) = O(|\mu|^2 + |\mu||y| + |y|^3)$. There exist $C_0, r_0 > 0$ such that $|R(y, \mu)| \le C_0(|\mu|^2 + |\mu||y| + |y|^3)$ for $|y| + |\mu| < r_0$.
[/step]
[step:Show there are no equilibria when $c\mu l_2 > 0$]
When $c\mu l_2 > 0$, the terms $c\mu$ and $l_2 y^2$ have the same sign: $|c\mu + l_2 y^2| = |c\mu| + |l_2| y^2 \ge |c\mu| > 0$. It suffices to show $|R(y, \mu)| < |c\mu| + |l_2| y^2$ for small $|y|, |\mu|$.
Using $|\mu|^2 \le \eta_1 |\mu|$, $|y|^3 \le \eta_1 y^2$, and Young's inequality $|\mu||y| \le \frac{1}{2}(\eta_1|\mu| + y^2)$ for $|y|, |\mu| < \eta_1$:
\begin{align*}
|R(y, \mu)| &\le C_0\bigl(\tfrac{3}{2}\eta_1 |\mu| + (\eta_1 + \tfrac{1}{2}) y^2\bigr).
\end{align*}
Choosing $\eta_1 > 0$ small enough that $\frac{3}{2}C_0\eta_1 < |c|$ and $C_0(\eta_1 + \frac{1}{2}) < |l_2|$ ensures $|R| < |c\mu| + |l_2| y^2$, so $f(y, \mu) \neq 0$ for $|y|, |\mu| < \eta_1$ with $c\mu l_2 > 0$.
[guided]
When $c\mu l_2 > 0$, the leading terms $c\mu$ and $l_2 y^2$ have the same sign, so
\begin{align*}
|c\mu + l_2 y^2| = |c\mu| + |l_2|y^2 \geq |c\mu| > 0.
\end{align*}
The remainder $|R(y,\mu)| \leq C_0(|\mu|^2 + |\mu||y| + |y|^3)$ is cubic-or-higher order, and Young's inequality $|\mu||y| \leq \frac{1}{2}(\eta_1|\mu| + y^2)$ converts it to
\begin{align*}
|R(y,\mu)| \leq C_0\bigl(\tfrac{3}{2}\eta_1|\mu| + (\eta_1 + \tfrac{1}{2})y^2\bigr).
\end{align*}
We need this to be strictly less than $|c\mu| + |l_2|y^2$. Comparing coefficients: $\frac{3}{2}C_0\eta_1 < |c|$ and $C_0(\eta_1 + \frac{1}{2}) < |l_2|$. The second condition at $\eta_1 = 0$ reads $C_0/2 < |l_2|$, which holds by shrinking $r_0$ (making $C_0$ as small as desired, since $R = o(|y|^2 + |\mu|)$ near the origin while $|l_2| = |\partial_y^2 f(0,0)|/2 > 0$ is fixed). Then $\eta_1 > 0$ can be chosen small enough that both conditions hold.
[/guided]
[/step]
[step:Localise all equilibria to the parabolic region $|y| \le R|\mu|^{1/2}$]
On $R|\mu|^{1/2} \le |y| < \eta_2$, the term $|l_2| y^2 \ge |l_2| R^2 |\mu|$ dominates $|c\mu| + |R(y, \mu)|$. Using $|\mu| \le y^2/R^2$ to bound all error terms by multiples of $y^2$, and choosing $\eta_2$ small and $R$ large, one obtains $|f(y, \mu)| \ge |l_2| y^2/2 > 0$. Hence every solution with $|y| < \eta_2$ satisfies $|y| \le R|\mu|^{1/2}$.
[/step]
[step:Produce exactly two equilibria when $c\mu l_2 < 0$ via the implicit function theorem]
When $c\mu l_2 < 0$, the quantity $-c\mu/l_2 > 0$, so $y_0^\pm := \pm\sqrt{-c\mu/l_2}$ are well-defined approximate roots with $|y_0^\pm| = O(|\mu|^{1/2})$. By construction, $c\mu + l_2(y_0^\pm)^2 = 0$, so
\begin{align*}
f(y_0^\pm, \mu) = R(y_0^\pm, \mu) = O(|\mu|^{3/2}).
\end{align*}
The derivative $\partial_y f(y, \mu) = 2 l_2 y + \partial_y R(y, \mu)$ satisfies, at $y = y_0^\pm$:
\begin{align*}
|\partial_y f(y_0^\pm, \mu)| \ge 2\sqrt{|l_2||c\mu|} - O(|\mu|) > 0
\end{align*}
for $|\mu|$ sufficiently small (the leading term is $O(|\mu|^{1/2})$ while the error is $O(|\mu|)$). In particular, $\partial_y f(y_0^\pm, \mu) \neq 0$.
The [Implicit Function Theorem](/theorems/52) applied to $f(y, \mu) = 0$ near each $(y_0^\pm, \mu)$ provides a unique smooth solution $y_\pm(\mu)$ with $f(y_\pm(\mu), \mu) = 0$ and
\begin{align*}
y_\pm(\mu) = \pm\sqrt{\frac{-c\mu}{l_2}} + O(|\mu|).
\end{align*}
Localisation ensures these are the only solutions near the origin, and $y_+(\mu) > 0 > y_-(\mu)$ for small $|\mu|$ confirms they are distinct.
[/step]
[step:Determine the stability of the two equilibrium branches]
Evaluating $\partial_y f$ at $y_\pm(\mu)$:
\begin{align*}
\partial_y f(y_\pm, \mu) &= \pm 2 l_2 \sqrt{\frac{-c\mu}{l_2}} + O(|\mu|) = \pm\, 2\,\mathrm{sgn}(l_2)\sqrt{-c\mu l_2} + O(|\mu|).
\end{align*}
Since $c\mu l_2 < 0$ implies $-c\mu l_2 > 0$, the square root is strictly positive. One equilibrium has $\partial_y f < 0$ (stable) and the other has $\partial_y f > 0$ (unstable).
[/step]
[step:Verify semi-stability at $\mu = 0$]
At $\mu = 0$, the equation $f(y, 0) = l_2 y^2 + R(y, 0) = y^2(l_2 + R(y, 0)/y^2)$ has $y = 0$ as its only solution near the origin, since $R(y, 0)/y^2 \to 0$ as $y \to 0$ and $l_2 \neq 0$. The expression $f(y, 0) = l_2 y^2(1 + O(|y|))$ has the sign of $l_2$ for all sufficiently small $y \neq 0$, so the equilibrium is semi-stable: trajectories of $\dot{y} = f(y, 0)$ are attracted from one side and repelled from the other.
[/step]
