Scalar fold, transcritical, and symmetric pitchfork equilibrium branches

In the fold case choose $n=1$, $v=w=1$ in [F2]. Its hypotheses are exactly $F(0,0)=F_x(0,0)=0$ and $F_{\lambda }{F}_{xx}\mathrm{\ne }0$. Thus all nearby equilibria lie on a unique curve $\lambda =\varphi (x)$ with the stated first and second derivatives.

For the transcritical and pitchfork cases define $G(x,\lambda ):={\int }_0^1{F}_x(tx,\lambda )dt$. Because $F(0,\lambda )=0$ in both cases, [F3] and the change of variable $s=tx$ give $F(x,\lambda )=xG(x,\lambda )$. Since $F\in C^5$, the derivatives of $F_x(tx,\lambda )$ through order four are continuous jointly in $(t,x,\lambda )$; on compact parameter neighborhoods their difference quotients converge uniformly in $t\in [0,1]$, so they may be integrated termwise. Thus $G\in C^4$, with $G_x={\int }_0^{1}t{F}_{xx}(tx,\lambda )dt$, $G_{\lambda }={\int }_0^1{F}_{x\lambda }(tx,\lambda )dt$, and $G_{xx}={\int }_0^1{t}^2{F}_{xxx}(tx,\lambda )dt$.

In the fold case, since ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{\ne }0$ and ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}$ is continuous, shrink $\delta >0$ so ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}$ has the constant sign of $c:={\varphi }^{\mathrm{\prime }\mathrm{\prime }}(0)$ on $[-\delta ,\delta ]$. By [F3], ${\varphi }^{\mathrm{\prime }}(x)={\int }_0^x{\varphi }^{\mathrm{\prime }\mathrm{\prime }}(s)ds$, so ${\varphi }^{\mathrm{\prime }}(x)$ has the sign of $cx$; hence $\varphi$ is strictly monotone on each half-interval. Also $\varphi (x)={\int }_0^x{\varphi }^{\mathrm{\prime }}(s)ds$ has the sign of $c$ for every $x\mathrm{\ne }0$. Thus $\varphi (-\delta )$ and $\varphi (\delta )$ are nonzero and have sign $c$. Every sufficiently small nonzero level with $\lambda c>0$ lies strictly between zero and each endpoint value, so [F5] and strict monotonicity give exactly one solution on each half-interval; no level of the opposite sign is attained.

In the transcritical case, the derivative formulas from step 1.2 give $G(0,0)=F_x(0,0)=0$, $G_x(0,0)=F_{xx}(0,0)\mathrm{/}2\mathrm{\ne }0$, and $G_{\lambda }(0,0)=F_{x\lambda }(0,0)\mathrm{\ne }0$. Apply [F1] with $k=4$, solving $G=0$ for $x=\psi (\lambda )$; differentiating gives ${\psi }^{\mathrm{\prime }}(0)=-G_{\lambda }\mathrm{/}{G}_x=-2F_{x\lambda }\mathrm{/}{F}_{xx}$. Since $F=xG$, every nearby zero lies on at least one of the branches $x=0$ or $x=\psi (\lambda )$; at their common point $(0,0)$ they coincide, and elsewhere sufficiently near zero they are distinct. Their distinct tangent slopes make them transverse.

In the pitchfork case, oddness of $F$ makes $G$ even in $x$. The derivative formulas from step 1.2 give $G(0,0)=0$, $G_x(0,0)=0$, $G_{\lambda }(0,0)=F_{x\lambda }(0,0)\mathrm{\ne }0$, and $G_{xx}(0,0)=F_{xxx}(0,0)\mathrm{/}3\mathrm{\ne }0$. By [F1] with $k=4$, $G=0$ is a unique local curve $\lambda =\varphi (x)$, with ${\varphi }^{\mathrm{\prime }}(0)=0$ and ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}(0)=-G_{xx}\mathrm{/}{G}_{\lambda }=-F_{xxx}\mathrm{/}(3F_{x\lambda })$. Evenness of $G$ and uniqueness imply $\varphi (-x)=\varphi (x)$.

Shrink the interval so ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}$ has the sign of ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}(0)$. Then $\varphi$ is strictly monotone for $x>0$, and by evenness it has exactly one positive and one negative solution of $\varphi (x)=\lambda$ on the side $\lambda {\varphi }^{\mathrm{\prime }\mathrm{\prime }}(0)>0$, and no nonzero solution on the other side. Since $F=xG$, these are exactly the two nontrivial equilibria in addition to the persistent branch $x=0$.