Pseudo arclength continuation and fold certificates — Examples

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.

Solving for the parameter gives exactly $\lambda =x^2$. At $x=0$ its derivative is $d\lambda \mathrm{/}dx=0$, and its second derivative is $2$. Thus the projection from the local solution curve to the $\lambda$-axis has zero derivative at the fold.

For every $\lambda <0$ there is no real solution, while for every $\lambda >0$ there are two, $x=\sqrt{\lambda }$ and $x=-\sqrt{\lambda }$. Hence no open interval about $0$ is the parameter domain of a single-valued local solution graph covering the whole nearby solution set. On either one-sided branch, $\mathrm{\mid }dx\mathrm{/}d\lambda \mathrm{\mid }=1\mathrm{/}(2\sqrt{\lambda })$ is unbounded as $\lambda \downarrow 0$.

The fold conditions are $F_x(0,0)=0$, $F_{\lambda }(0,0)=-1\mathrm{\ne }0$, and $F_{xx}(0,0)=2\mathrm{\ne }0$, so [F1] gives the same nonzero curvature conclusion independently of the explicit graph calculation.

The derivative is $DF(0,0)=[10]$, which has rank $1$ and kernel $\mathrm{span}\{t\}$ for the unit tangent $t=(0,1)$.

Take the unit hyperplane normal $\tau =(1,0)$. Then ${\tau }^{T}t=0$ and $\left[\begin{array}{c}DF(0,0)\\ {\tau }^T\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right],$ whose rows are dependent and whose kernel contains $t\mathrm{\ne }0$.

By [F1], the bordered matrix is singular exactly at this failed angle; the inverse bound also cannot hold. This is the tangent-angle failure required by the continuation algorithm.

The equation factors as $(x-1)(x+1)=0$, so its complete solution set is $S=(\{1\}\times \mathbb{R})\cup (\{-1\}\times \mathbb{R})\mathrm{.}$ Each set is connected and open-and-closed relative to $S$, and the two sets are disjoint; therefore they are the two connected components of $S$.

Choose $h>0$ and $0<R_0<\min \{h\mathrm{/}2,1\mathrm{/}4\}$. The seed section is $G_{-1}(x,\lambda )=(x^2-1,\lambda )$ with center $(1,0)$ and radius $R_0$. Its residual is zero, its Jacobian inverse there is $\mathrm{diag}(1\mathrm{/}2,1)$ with norm $1$, and its Jacobian is $2$-Lipschitz. Thus the exact-data certificate has $q=2R_0<1$ and zero residual term, and certifies the seed ball by [F2]. In $V=\{(x,\lambda ):\mathrm{\mid }x-1\mathrm{\mid }<1,-h<\lambda <2h\}$ every zero lies on $c(s)=(1,s)$ for $s\in [-h,2h]$. Rank is one there.

At $(1,0)$, $DF=[20]$ has rank $1$ and unit tangent $t=(0,1)$. Choose previous orientation and candidate tangent $\tau =\widehat{t}=t=(0,1)$. The chart has $a=\kappa =1$, $r_{\mathrm{c}\mathrm{h}}^{+}=2h$, and $h>R_0$, $h+R_0<2h$, so the previous-normal offset and chart tests in [F1] pass. The augmented equation is $(x^2-1,\lambda -h)=0$, whose unique root in any ball of radius less than $1$ centered at $(1,h)$ is $(1,h)$. Its bordered derivative is $\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]$, so the local corrector is nonsingular.

At the exact center $(1,h)$ the augmented residual is zero and the Jacobian is invertible. The augmented Jacobian is $2$-Lipschitz; with exact data, choose $0<R<\min \{1\mathrm{/}4,h\mathrm{/}4\}$, so the whole corrected ball lies in $V$, set $\widehat{\beta }=1$, and use zero residual and derivative errors. Then $q=2R<1$ and $\beta \rho =0$, so [F2] certifies the unique local section root. With $N_{\mathrm{r}\mathrm{e}\mathrm{q}}=N_s=1$ and $N_c=0$, the exact predictor already solves this section and both the seed and next ball are certified. The ball of radius $R<1$ also excludes the other root $(-1,h)$.

The other component $\{-1\}\times \mathbb{R}$ contains no point of that local chart around $(1,0)$, and the accepted one-step trace contains no point with $x=-1$. Nevertheless both components satisfy $F=0$. By [F1], the finite trace correctly certifies its local step but cannot certify that the second component is absent or that all branches have been found.