Pseudo arclength crosses the fold $x^2-\lambda =0$

At $(0,0)$, $F_x=0$, $F_{\lambda }=-1$, and $F_{xx}=2$. With $v=w=1$, $a=-1$ and $b=2$, so [F1] gives ${\lambda }^{\mathrm{\prime }\mathrm{\prime }}(0)=2$. The full derivative is $DF(x,\lambda )=[2x,-1]$, which has rank $1$ everywhere, and at the fold its unit kernel tangent is $t=(1,0)$.

Choose $0<\alpha <1\mathrm{/}32$ and use the seed center ${\bar{z}}_0=(-\alpha ,{\alpha }^2)$, radius $R_0=\alpha \mathrm{/}4$, previous orientation ${\tau }_{-1}=(1,0)$, and seed section point $z_{p,-1}=(-\alpha ,0)$, so $G_{-1}(x,\lambda )=(x^2-\lambda ,x+\alpha )$. At the seed center the residual is zero and the inverse of the bordered Jacobian is $\left[\begin{array}{cc}0& 1\\ -1& -2\alpha \end{array}\right]$, whose operator norm is less than $2$. The Jacobian is $2$-Lipschitz, so the exact-data certificate with $\widehat{\beta }=2$, $\widehat{g}=0$, zero error bounds, and radius $R_0$ has $q=4R_0=\alpha <1$ and certifies the seed ball. The ball lies in $V=\{(x,\lambda ):\mathrm{\mid }x\mathrm{\mid }<1\mathrm{/}4\}$, where every zero lies on the regular graph $(x,x^2)$ and $DF=[2x,-1]$ has rank $1$. Parametrize that graph from $x=-\alpha$ by arclength $s(x)={\int }_{-\alpha }^x\sqrt{1+4u^2}du$. Its positive unit tangent is $(1,2x)\mathrm{/}\sqrt{1+4x^2}$, so the previous normal has dot product at least $\kappa =2\mathrm{/}\sqrt{5}$ throughout $\mathrm{\mid }x\mathrm{\mid }\le 1\mathrm{/}4$. The forward chart length to $x=1\mathrm{/}4$ is at least $1\mathrm{/}4+\alpha$.

At the seed root, the exact unit tangent oriented toward increasing $x$ is $t=(1,-2\alpha )\mathrm{/}\sqrt{1+4{\alpha }^2}$. The tangent callback may return $\widehat{t}=(1,0)$ with certified error $e=\mathrm{\parallel }\widehat{t}-t{\mathrm{\parallel }}_2=\sqrt{2-\frac{2}{\sqrt{1+4{\alpha }^2}}}$. With the previous normal $\tau =(1,0)$, the angle tests are ${\tau }^{T}t=1\mathrm{/}\sqrt{1+4{\alpha }^2}\ge 2\mathrm{/}\sqrt{5}$ and $a={\tau }^T\widehat{t}=1$. Set the algorithm's step length to $H=2\alpha +{\alpha }^2$. Then the predictor is $z_p={\bar{z}}_0+H\widehat{t}=(\alpha +{\alpha }^2,{\alpha }^2)$. Its section is $G_H(x,\lambda )=(x^2-\lambda ,x-(\alpha +{\alpha }^2))$, with unique zero $z_{\ast }=(x_{\ast },x_{\ast }^2)$, where $x_{\ast }:=\alpha +{\alpha }^2$. The forward-offset tests hold because $H>R_0$ and $H+R_0=\frac{9}{4}\alpha +{\alpha }^2<\frac{1}{8}<\frac{1}{2\sqrt{5}}\le \kappa r_{\mathrm{c}\mathrm{h}}^{+}$. The section is transverse throughout the chart. At its selected root, set ${\sigma }_{\ast }:=\sqrt{1+4x_{\ast }^2}$, $t_{\ast }=(1,2x_{\ast })\mathrm{/}{\sigma }_{\ast }$, and $A:=DF(x_{\ast },x_{\ast }^2)$. The unit vector $e_{\ast }=(-2x_{\ast },1)\mathrm{/}{\sigma }_{\ast }$ spans $t_{\ast }^{\perp }$, and $[2x_{\ast },-1]e_{\ast }=-{\sigma }_{\ast }$, so $A{\mathrm{\mid }}_{t_{\ast }^{\perp }}$ has inverse norm $1\mathrm{/}{\sigma }_{\ast }\le 1$. Since ${\tau }^T{t}_{\ast }=1\mathrm{/}{\sigma }_{\ast }>0$, [F2] gives a local quadratic Newton corrector; the bordered Jacobian has determinant $1$.

At the predictor, the augmented residual is $(2{\alpha }^3+{\alpha }^4,0)$ and its Jacobian is $\left[\begin{array}{cc}2(\alpha +{\alpha }^2)& -1\\ 1& 0\end{array}\right]$. The Newton equation forces $\mathrm{\Delta }x=0$ and $\mathrm{\Delta }\lambda =2{\alpha }^3+{\alpha }^4$, so one exact correction reaches the section root $(\alpha +{\alpha }^2,(\alpha +{\alpha }^2{)}^2)$.

At the corrected center $z_{\ast }=(x_{\ast },x_{\ast }^2)$, where $x_{\ast }:=\alpha +{\alpha }^2<1\mathrm{/}4$, the inverse bordered Jacobian is $\left[\begin{array}{cc}0& 1\\ -1& 2x_{\ast }\end{array}\right]$. Its operator norm is at most its Frobenius norm $\sqrt{2+4x_{\ast }^2}<2$. Choose corrected-ball radius $R_1=\alpha \mathrm{/}4$. The ball lies in $V$ since $x_{\ast }+R_1<1\mathrm{/}4$. With exact data, $\widehat{\beta }=2$, zero residual and derivative errors, and Jacobian Lipschitz constant $2$, the certificate has $q=4R_1=\alpha <1$ and $\beta \rho =0$. It therefore certifies the next section root and, because the whole ball lies in $V$, identifies it with the unique forward branch point. With $N_{\mathrm{r}\mathrm{e}\mathrm{q}}=N_s=N_c=1$, residual tolerance $ϵ=0$, and exact callbacks, steps 2.1 and 3.1 produce this accepted ball in one attempt and one Newton update.

The certified trace therefore starts at $x=-\alpha$ and accepts a next section root at $x=\alpha +{\alpha }^2>0$, with the fold $x=0$ lying between them on the supplied chart. It is a finite local continuation step through the fold even though ${\lambda }^{\mathrm{\prime }}(0)=0$; it asserts nothing about other components.