[proofplan] We prove that a degenerate equilibrium of order $k$ (first $k-1$ Lyapunov coefficients vanish, $k$-th nonzero) can bifurcate into at most $k$ equilibria under parameter variation. The proof reduces the equilibrium equation to a perturbation of $c\mu + l_k y^k = 0$ via Taylor expansion, introduces a power-law rescaling $y = |\mu|^{1/k} u$ to normalise the equation, shows the rescaled remainder vanishes uniformly as $\mu \to 0$, verifies the $k$-th derivative of the rescaled equation is bounded away from zero (so [Rolle's Theorem](/theorems/185) limits the root count to $k$), and finally confirms all small equilibria lie within the rescaling regime. [/proofplan] [step:Write the equilibrium equation in local normal form] Since $f \in C^\infty(\mathbb{R} \times \mathbb{R}, \mathbb{R})$ and $f(0,0) = 0$, Taylor's theorem at the origin gives \begin{align*} f(y, \mu) = c\mu + \sum_{j=1}^{k} l_j \, y^j + h(y, \mu), \end{align*} where $h \in C^\infty$ satisfies $h(y, \mu) = O(|\mu|^2 + |\mu||y| + |y|^{k+1})$ as $(y, \mu) \to (0, 0)$. By hypothesis 1, $l_1 = \cdots = l_{k-1} = 0$ and $l_k \neq 0$. Setting $c := \partial_\mu f(0, 0) \neq 0$ (hypothesis 2), the equilibrium equation becomes \begin{align*} c\mu + l_k y^k + h(y, \mu) = 0. \end{align*} [/step] [step:Introduce the power-law rescaling and the rescaled equilibrium function] For $\mu \neq 0$, define $u \in \mathbb{R}$ by $y = |\mu|^{1/k} u$ and define the rescaled remainder \begin{align*} \tilde{h}(u, \mu) := \frac{h\bigl(|\mu|^{1/k} u,\, \mu\bigr)}{|\mu|}. \end{align*} Substituting $y = |\mu|^{1/k} u$ into the equilibrium equation and dividing by $|\mu| > 0$: \begin{align*} c\,\mathrm{sgn}(\mu) + l_k u^k + \tilde{h}(u, \mu) = 0. \end{align*} Define the rescaled equilibrium function $F(u, \mu) := c\,\mathrm{sgn}(\mu) + l_k u^k + \tilde{h}(u, \mu)$. The map $y \mapsto u = y / |\mu|^{1/k}$ is a bijection $\mathbb{R} \to \mathbb{R}$ for each $\mu \neq 0$, so the zeros of $F(\cdot, \mu)$ are in exact bijection with those of $f(\cdot, \mu)$ near the origin. [guided] The rescaling $y = |\mu|^{1/k} u$ is chosen so that both the parameter term $c\mu$ and the leading nonlinear term $l_k y^k$ contribute at the same order in $|\mu|$. Substituting: \begin{align*} c\mu &= c\,\mathrm{sgn}(\mu) \cdot |\mu|, \\ l_k y^k &= l_k(|\mu|^{1/k} u)^k = l_k |\mu| u^k. \end{align*} Both are $O(|\mu|)$. Dividing through by $|\mu| > 0$: \begin{align*} c\,\mathrm{sgn}(\mu) + l_k u^k + \tilde{h}(u, \mu) = 0, \end{align*} where $\tilde{h}(u, \mu) = h(|\mu|^{1/k}u, \mu)/|\mu|$ vanishes uniformly on compact sets as $\mu \to 0$ (because $h$ is $O(|\mu|^2 + |\mu||y| + |y|^{k+1})$, each of which contributes a factor tending to zero after dividing by $|\mu|$). This normalisation reduces the equilibrium equation to a perturbation of the degree-$k$ polynomial $c\,\mathrm{sgn}(\mu) + l_k u^k$, whose root count is at most $k$ by [Rolle's Theorem](/theorems/185). [/guided] [/step] [step:Show the rescaled remainder vanishes uniformly on compact sets] [claim:Uniform vanishing of remainder] For every $R > 0$, $\sup_{|u| \le R} |\tilde{h}(u, \mu)| \to 0$ as $\mu \to 0$. [/claim] [proof] Since $h(y, \mu) = O(|\mu|^2 + |\mu||y| + |y|^{k+1})$, there exist $C_1 > 0$ and $r > 0$ such that $|h(y, \mu)| \le C_1(|\mu|^2 + |\mu||y| + |y|^{k+1})$ for $|y| + |\mu| < r$. For $|u| \le R$ and $|\mu|$ small enough that $|\mu|^{1/k} R + |\mu| < r$, substituting $y = |\mu|^{1/k} u$ and dividing by $|\mu|$: \begin{align*} |\tilde{h}(u, \mu)| \le C_1\bigl(|\mu| + |\mu|^{1/k} R + |\mu|^{1/k} R^{k+1}\bigr). \end{align*} Since $k \ge 2$, the exponent $1/k > 0$, so every term tends to $0$ as $\mu \to 0$, uniformly over $|u| \le R$. [/proof] [/step] [step:Show the $k$-th derivative of $F$ is bounded away from zero] [claim:Non-vanishing of the $k$-th derivative] For every $R > 0$, there exists $\delta_2 > 0$ such that for all $\mu \in (-\delta_2, \delta_2) \setminus \{0\}$, \begin{align*} \inf_{|u| \le R} |\partial_u^k F(u, \mu)| \ge \frac{|k!\, l_k|}{2} > 0. \end{align*} [/claim] [proof] Computing the $k$-th derivative: $\partial_u^k F(u, \mu) = k!\, l_k + \partial_u^k \tilde{h}(u, \mu)$. By the chain rule applied $k$ times to $\tilde{h}(u, \mu) = h(|\mu|^{1/k} u, \mu)/|\mu|$, each differentiation in $u$ contributes a factor $|\mu|^{1/k}$, giving \begin{align*} \partial_u^k \tilde{h}(u, \mu) = (\partial_y^k h)\bigl(|\mu|^{1/k} u, \mu\bigr). \end{align*} Now $\partial_y^k h(y, \mu) = \partial_y^k f(y, \mu) - k!\, l_k$, and $\partial_y^k h(0, 0) = 0$ since $\partial_y^k f(0, 0) = k!\, l_k$. By continuity, $|\partial_y^k h(|\mu|^{1/k} u, \mu)| \to 0$ uniformly over $|u| \le R$ as $\mu \to 0$. Choose $\delta_2$ so that this quantity is less than $|k!\, l_k|/2$; then the triangle inequality gives $|\partial_u^k F(u, \mu)| \ge |k!\, l_k|/2$. [/proof] [/step] [step:Bound the root count using Rolle's theorem] By the previous claim, $\partial_u^k F(\cdot, \mu)$ has no zeros on $[-R, R]$ for $|\mu| < \delta_2$. Applying [Rolle's Theorem](/theorems/185) inductively: if $F(\cdot, \mu)$ had $k + 1$ distinct zeros in $[-R, R]$, then $\partial_u F(\cdot, \mu)$ would have at least $k$ zeros, $\partial_u^2 F(\cdot, \mu)$ at least $k - 1$, and so on, until $\partial_u^k F(\cdot, \mu)$ would have at least one zero -- contradicting the claim. Therefore $F(\cdot, \mu)$ has at most $k$ zeros in $[-R, R]$. [/step] [step:Confirm all small equilibria lie within the rescaling regime and conclude] [claim:Localisation] There exist $R > 0$ and $\eta > 0$ such that for all $|\mu| < (\eta / R)^k$, the equation $f(y, \mu) = 0$ has no solutions in the annular region $R|\mu|^{1/k} \le |y| \le \eta$. [/claim] [proof] On this region, $|y|^k \ge R^k |\mu|$ so $|l_k||y|^k \ge |l_k|R^k|\mu|$. Meanwhile, $|c\mu| + |h(y, \mu)|$ is bounded by a constant times $|y|^k \cdot (|c|/R^k + C\eta)$. Choosing $\eta$ small and $R$ large makes this factor less than $|l_k|/2$, so $|f(y, \mu)| \ge |l_k||y|^k/2 > 0$. [/proof] Fix $R$ and $\eta$ from the localisation claim, and set $\delta := \min(\delta_2, (\eta / R)^k)$. For $\mu \in (-\delta, \delta) \setminus \{0\}$, every solution $y$ of $f(y, \mu) = 0$ with $|y| \le \eta$ satisfies $|y| < R|\mu|^{1/k}$, so $u = y/|\mu|^{1/k} \in (-R, R)$. By the Rolle argument, $F(u, \mu) = 0$ has at most $k$ roots on $[-R, R]$, giving at most $k$ solutions for $f$. At $\mu = 0$, the equation $f(y, 0) = y^k(l_k + h(y, 0)/y^k)$ has $y = 0$ as its only small solution since $l_k + h(y, 0)/y^k \to l_k \neq 0$. [/step]