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Pseudo arclength continuation and fold certificates

1 · Prerequisites

2 · Summary

Objective

This page develops finite-budget pseudo-arclength continuation for $F : R^{n + 1} \to R^{n}$ on an explicit $C^{2}$ tube, with $C^{3}$ regularity for the simple-fold curvature expansion. At a regular solution point, a unit kernel tangent gives the predictor direction, and the hyperplane section uses the previous tangent as its normal. The bordered Jacobian is invertible exactly when the normal has nonzero component along the current tangent; the stated angle margin supplies an inverse-norm bound. The curve chart gives a quadratic exact-tangent predictor defect, and the augmented residual receives a local quadratic Newton correction under the same invariant-ball hypotheses as the exact square-system Newton theorem.

For bounded residual and derivative evaluation errors, a proposed corrected point is accepted only when the explicit contraction inequalities certify a unique zero and the ball lies in the local chart. At a simple fold, the page uses rank $F_{x} = n - 1$ , normalized right and nonzero left nullvectors, and the strict conditions $w^{T} F_{λ} \neq 0$ and $w^{T} F_{x x} [v, v] \neq 0$ . The local Lyapunov–Schmidt chart has $λ^{'} (0) = 0$ and $λ^{''} (0) = - w^{T} F_{x x} [v, v] / (w^{T} F_{λ}) \neq 0$ under the stated normalization. A finite trace, residual tolerance, or Newton budget does not certify global branch completeness or an exact zero by itself.

The separate finite-trace theorem proves that every accepted ball identifies the unique selected section point on its certified chart. It also proves termination conditional on each supplied callback returning in finite time, with explicit attempted-step and corrector budgets, including zero requested steps. The previous orientation remains the section normal while the current tangent candidate gives the predictor; certified seed and tangent errors enter the predictor bound explicitly. These local and conditional conclusions make no global branch-completeness or numerical-success claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Invertibility and a norm bound for the pseudo-arclength bordered matrix

Statement

Let $n \geq 1$ , let $A : R^{n + 1} \to R^{n}$ be linear of rank $n$ , and let $t \in \ker A$ have $∥ t ∥_{2} = 1$ . Set $E = t^{⊥}$ and let $Q = (A ∣_{E})^{- 1} : R^{n} \to E$ . For a unit vector $τ \in R^{n + 1}$ , form the square bordered matrix $B = [\begin{array}{c} A \\ τ^{T} \end{array}] .$ Then $B$ is invertible exactly when $τ^{T} t \neq 0$ . If $0 < c \leq 1$ , $∣ τ^{T} t ∣ \geq c$ , and $K_{A} \geq 0$ satisfies $∥ Q u ∥_{2} \leq K_{A} ∥ u ∥_{2}$ for all $u \in R^{n}$ , then $∥ B^{- 1} ∥_{o p} \leq \sqrt{K_{A}^{2} (1 + c^{- 1})^{2} + c^{- 2}} .$

Here $t$ is the current unit tangent and $τ$ is the normal used in the corrector hyperplane. The lower bound $c$ is the explicit transversality margin; as it tends to zero this estimate need not remain bounded.

Facts & Assumptions

Given: The dimensions, rank, unit kernel vector $t$ , orthogonal complement $E = t^{⊥}$ , unit vector $τ$ , and bound $∣ τ^{T} t ∣ \geq c > 0$ in the Statement.

[F1]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map from a finite-dimensional domain, the domain dimension is the sum of the kernel and image dimensions.

[F2]

In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ : for a subspace $W$ of a finite-dimensional inner-product space, $\dim W + \dim W^{⊥} = \dim V$ .

[F3]

If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ : a subspace of a finite-dimensional space has dimension at most the ambient dimension, and equal dimension forces equality. The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial: a linear map is injective exactly when its kernel is zero. The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ : the coordinate space $R^{n}$ has dimension $n$ .

[F4]

Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ : every Euclidean linear map has a finite operator-norm bound.

[F5]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: for Euclidean vectors, $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ .

[F6]

Rank and nullity of a linear map with finite-dimensional domain: for a linear map with finite-dimensional domain, rank is the dimension of its image.

Proof

Given: The hypotheses in the Statement.

1.1F1F3F6given

By [F1], [F3], and [F6], $\dim \ker A = (n + 1) - n = 1$ . Since $t$ is a unit vector in the kernel, $\ker A = span {t}$ .

2.1F2F3step 1.1

Apply [F2] to $W = \ker A$ in $R^{n + 1}$ to obtain $\dim E = (n + 1) - 1 = n$ . The codomain $R^{n}$ also has dimension $n$ by [F3].

3.1F1F3step 2.1givenalgebra

If $x = α t$ also lies in $E = t^{⊥}$ , then $0 = t^{T} x = α ∥ t ∥_{2}^{2} = α$ , so $E \cap \ker A = {0}$ . Thus $A ∣_{E}$ is injective by [F3]. Its domain and codomain both have dimension $n$ , so [F1] and [F3] make it surjective. Thus $Q$ is well-defined.

4.1givenstep 1.1step 3.1construct

For $(u, s) \in R^{n} \times R$ , write a prospective solution as $v = e + α t$ with $e \in E$ . The first row equation $A v = u$ forces $e = Q u$ , and the last row equation forces $α = \frac{s - τ^{T} Q u}{τ^{T} t} .$ If $τ^{T} t \neq 0$ , these formulas give one solution for every $(u, s)$ , so $B$ is invertible. If $τ^{T} t = 0$ , then $B t = 0$ with $t \neq 0$ , so $B$ is singular.

5.1F4F5step 4.1algebra

Under $∣ τ^{T} t ∣ \geq c$ , [F5] gives $∣ α ∣ \leq c^{- 1} (∣ s ∣ + K_{A} ∥ u ∥_{2}) .$ Consequently $∥ v ∥_{2} \leq K_{A} (1 + c^{- 1}) ∥ u ∥_{2} + c^{- 1} ∣ s ∣ .$ Applying [F5] to the two scalar coefficients of $\sqrt{∥ u ∥_{2}^{2} + ∣ s ∣^{2}}$ gives the stated operator-norm bound for $B^{- 1}$ .

6.1step 4.1step 5.1∎

Step 4.1 proves the exact invertibility criterion, and step 5.1 proves the quantitative inverse bound under the stated angular margin.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Complement splitting and range projection for a simple fold

Definition

Let $J : R^{n} \to R^{n}$ be linear of rank $n - 1$ (Rank and nullity of a linear map with finite-dimensional domain). Let $v$ be a unit vector with $\ker J = span {v}$ , let $w \neq 0$ satisfy $w^{T} J = 0$ , and let $b \in R^{n}$ satisfy $a : = w^{T} b \neq 0$ . Define the state-space complement and range complement by $X : = v^{⊥}, Y : = \ker (w^{T}) .$ Set $q : = b / a$ and define the projection candidate $P : = I - q w^{T} : R^{n} \to R^{n} .$ For a differentiable $F : U \subseteq R^{n} \times R \to R^{n}$ , $F_{x}$ and $F_{λ}$ denote the restrictions of the total derivative The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder to the state and scalar-parameter directions. At a simple fold, the intended data are $J = F_{x} (x_{0}, λ_{0})$ and $b = F_{λ} (x_{0}, λ_{0})$ . The next lemma proves that $J ∣_{X} : X \to Y$ is an isomorphism and that $P$ is the projection onto $Y$ along $span {q}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

A bounded residual and derivative error give a local root certificate

Statement

For a linear map $A$ between Euclidean spaces, write $∥ A ∥_{o p} : = \sup_{∥ u ∥_{2} = 1} ∥ A u ∥_{2}$ . Let $m \geq 1$ , let $V \subseteq R^{m}$ be open, and let $G : V \to R^{m}$ be $C^{1}$ . Let $y \in V$ , $R > 0$ , and suppose the closed ball ${\overline{B}}_{2} (y, R)$ lies in $V$ . Assume that for some $L \geq 0$ , $∥ D G (u) - D G (v) ∥_{o p} \leq L ∥ u - v ∥_{2} (u, v \in {\overline{B}}_{2} (y, R)) .$ Let $\hat{g} \in R^{m}$ and $\hat{A} \in R^{m \times m}$ come with certified bounds $∥ \hat{g} - G (y) ∥_{2} \leq ϵ_{g}, ∥ \hat{A} - D G (y) ∥_{o p} \leq ϵ_{J},$ where $ϵ_{g}, ϵ_{J} \geq 0$ . Suppose $\hat{A}$ is invertible and $∥ {\hat{A}}^{- 1} ∥_{o p} \leq \hat{β}$ , with $\hat{β} ϵ_{J} < 1$ . Define $β : = \frac{\hat{β}}{1 - \hat{β} ϵ_{J}}, ρ : = ∥ \hat{g} ∥_{2} + ϵ_{g}, q : = β L R .$ If $q < 1$ and $β ρ \leq (1 - q) R$ , then $G$ has exactly one zero $y^{*}$ in ${\overline{B}}_{2} (y, R)$ and $∥ y^{*} - y ∥_{2} \leq \frac{β ρ}{1 - q} .$

Facts & Assumptions

Given: $G, V, y, R, L, \hat{g}, \hat{A}, ϵ_{g}, ϵ_{J}$ and $\hat{β}$ as in the Statement.

[F1]

A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point: a contraction of a nonempty complete metric space into itself has exactly one fixed point.

[F2]

$R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ : Euclidean $R^{m}$ is complete. A closed Euclidean ball is complete because every Cauchy sequence in it converges in $R^{m}$ and continuity of the norm keeps the limit in the ball.

[F3]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composite is the composite of its derivatives.

[F4]

On a convex open set, a uniform bound $∥ D f (z) v ∥_{2} \leq M ∥ v ∥_{2}$ implies $∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2}$ : on a convex open set, a uniform derivative-norm bound implies the same Lipschitz bound on the map.

[F5]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean inner product satisfies Cauchy--Schwarz.

[F6]

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ and Factorisation of $b^{n} - a^{n}$ , and the resulting Lipschitz estimate: if $0 \leq a < 1$ , then $a^{k} \to 0$ and $(1 - a) \sum_{j = 0}^{N} a^{j} = 1 - a^{N + 1}$ .

[F7]

Continuously differentiable maps, local inverses, and local diffeomorphisms: $C^{1}$ means total differentiability at every point and continuity of the first derivative matrix entries.

[F8]

The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder: the total derivative is the linear first-order approximation to a map at each point.

Proof

Given: The data and hypotheses in the Statement.

1.1givenalgebra

For matrices $X, Y$ , the definition of the operator norm gives $∥ X Y u ∥_{2} \leq ∥ X ∥_{o p} ∥ Y ∥_{o p} ∥ u ∥_{2}$ for every $u$ ; taking the supremum over unit $u$ proves submultiplicativity. Put $A : = D G (y)$ , $E : = A - \hat{A}$ , $P : = {\hat{A}}^{- 1} E$ , and $a : = \hat{β} ϵ_{J} < 1$ . Then $A = \hat{A} (I + P)$ and $∥ P^{k} ∥_{o p} \leq a^{k}$ by induction.

2.1F2F5F6step 1.1givenalgebra

The finite sums $S_{N} : = \sum_{j = 0}^{N} (- P)^{j}$ satisfy $(I + P) S_{N} = S_{N} (I + P) = I + (- 1)^{N} P^{N + 1}$ . By [F6], the tails are bounded by $a^{N + 1} / (1 - a) \to 0$ . Each matrix entry of $S_{N} - S_{M}$ is bounded in absolute value by $∥ S_{N} - S_{M} ∥_{o p}$ , by [F5] applied to standard unit vectors. Thus the entries are Cauchy; completeness [F2] gives a matrix limit $S$ . Taking entrywise limits in the finite matrix products gives $(I + P) S = S (I + P) = I$ . Also the finite-sum bound in [F6] gives $∥ S ∥_{o p} \leq (1 - a)^{- 1}$ by taking limits on each vector. Hence $A^{- 1} = S {\hat{A}}^{- 1}$ and $∥ A^{- 1} ∥_{o p} \leq β$ .

3.1F3F4F7F8givenstep 2.1algebra

Define $T (x) : = x - A^{- 1} G (x)$ on the ball. Its derivative is $D T (x) = A^{- 1} (A - D G (x))$ . For $x$ in the open ball, the Lipschitz hypothesis gives $∥ D T (x) ∥_{o p} \leq β L ∥ x - y ∥_{2} \leq q$ . The open ball is convex, so [F4] gives $∥ T (u) - T (v) ∥_{2} \leq q ∥ u - v ∥_{2}$ there; continuity extends this inequality to the closed ball.

4.1F1F2step 2.1step 3.1givenalgebra

Since $∥ T (y) - y ∥_{2} = ∥ A^{- 1} G (y) ∥_{2} \leq β ρ$ , every $x \in {\overline{B}}_{2} (y, R)$ satisfies $∥ T (x) - y ∥_{2} \leq ∥ T (x) - T (y) ∥_{2} + ∥ T (y) - y ∥_{2} \leq q R + β ρ \leq R .$ Thus $T$ maps the nonempty complete ball into itself. By [F1] it has a unique fixed point $y^{*}$ there, and invertibility of $A$ makes $T (y^{*}) = y^{*}$ equivalent to $G (y^{*}) = 0$ . Any zero of $G$ is a fixed point, so this zero is unique.

5.1F1step 3.1step 4.1givenalgebra∎

The contraction inequality and fixed-point identity give $∥ y^{*} - y ∥_{2} \leq ∥ T (y^{*}) - T (y) ∥_{2} + ∥ T (y) - y ∥_{2} \leq q ∥ y^{*} - y ∥_{2} + β ρ,$ which rearranges to the stated error bound because $1 - q > 0$ .

Remarks

For an augmented pseudo-arclength residual, use this certificate only when the ball lies in the local curve-chart neighborhood and the corrector hyperplane is transverse to the curve. Failure of either strict contraction condition gives no certificate; a finite Newton budget alone certifies neither existence nor nonexistence of a root.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A regular solution curve has a unique local arclength chart and a quadratic predictor defect

Statement

Let $n \geq 1$ , let $U \subseteq R^{n + 1}$ be open, and let $F : U \to R^{n}$ be $C^{2}$ . Suppose $F (z_{0}) = 0$ and $rank D F (z_{0}) = n$ . Let $t \in \ker D F (z_{0})$ be a unit vector. There is an interval about $0$ and a unique local $C^{2}$ curve $γ (r)$ satisfying $F (γ (r)) = 0, γ (0) = z_{0}, t^{T} (γ (r) - z_{0}) = r .$ There is a neighborhood $V$ of $z_{0}$ in which every solution $z$ of $F (z) = 0$ has the form $z = γ (t^{T} (z - z_{0}))$ . It has $γ^{'} (0) = t$ . After reparameterization by Euclidean arclength, the same local curve is represented by a $C^{1}$ unit-speed curve $c (s)$ with $c (0) = z_{0}$ and $c^{'} (0) = t$ .

Let $K \subseteq U$ be convex, let $c (s)$ and the segment from $c (s)$ to $c (s) + h c^{'} (s)$ lie in $K$ for a step $h > 0$ , and suppose $∥ D F (u) - D F (v) ∥_{o p} \leq L ∥ u - v ∥_{2} (u, v \in K) .$ For $z_{p} : = c (s) + h c^{'} (s)$ , the augmented pseudo-arclength residual $G_{s, h} (z) : = (F (z), c^{'} (s)^{T} (z - z_{p}))$ satisfies $∥ G_{s, h} (z_{p}) ∥_{2} = ∥ F (z_{p}) ∥_{2} \leq \frac{\sqrt{n} L}{2} h^{2} .$ Thus the predictor defect is quadratic in the step length on the stated tube.

Facts & Assumptions

Given: $U, F, z_{0}, t, K, L$ and the hypotheses in the Statement.

[F1]

Invertibility and a norm bound for the pseudo-arclength bordered matrix: if $A : R^{n + 1} \to R^{n}$ has rank $n$ , $t$ is a unit kernel vector, and $τ$ is unit, then $[\begin{array}{c} A \\ τ^{T} \end{array}]$ is invertible exactly when $τ^{T} t \neq 0$ .

[F2]

The parametrized implicit function theorem with $C^{k}$ regularity: for a $C^{2}$ equation whose derivative in the solved variables is invertible, there is a unique local $C^{2}$ implicit solution, and every nearby zero of the equation is on its graph.

[F3]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composite is the composite of the derivatives.

[F4]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: a scalar $C^{1}$ function on a compact interval equals its initial value plus the integral of its derivative.

[F5]

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ : pointwise bounds between integrable real functions pass to their integrals.

[F6]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: a coordinate of a vector has absolute value at most its Euclidean norm.

[F7]

The Euclidean inverse function theorem: a $C^{1}$ map with invertible derivative at a point has a local $C^{1}$ inverse.

[F8]

$C^{k}$ Euclidean maps and diffeomorphisms: $C^{2}$ means each scalar component has continuous coordinate partial derivatives through order two.

[F9]

The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder: $D F (z)$ is the linear map in the first-order approximation $F (z + h) = F (z) + D F (z) h + o (∥ h ∥_{2})$ .

[F10]

Rank and nullity of a linear map with finite-dimensional domain: rank is the dimension of the image of a linear map with finite-dimensional domain.

[F11]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : the kernel dimension plus the image dimension equals the domain dimension.

[F12]

The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive: the integral function of a continuous scalar integrand has derivative equal to that integrand.

[F13]

For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ in its oriented form and The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ : $\int_{a}^{r} v - \int_{a}^{0} v = \int_{0}^{r} v$ , including negative $r$ .

Proof

Given: The data and hypotheses in the Statement.

1.1F1F2F3F8F9F10F11given

Define $Φ (z, r) : = (F (z), t^{T} (z - z_{0}) - r)$ . Its derivative in $z$ at $(z_{0}, 0)$ is $[\begin{array}{c} D F (z_{0}) \\ t^{T} \end{array}]$ , which is invertible by [F1] because $D F (z_{0}) t = 0$ and $∥ t ∥_{2} = 1$ . Apply [F2] with parameter $r$ to obtain the unique local $C^{2}$ solution $z = γ (r)$ and the stated normalization. For each nearby zero $z$ of $F$ , setting $r = t^{T} (z - z_{0})$ makes $Φ (z, r) = 0$ , so the local uniqueness clause in [F2] places $z$ on this graph. Differentiating the equations at $r = 0$ using [F3] gives $D F (z_{0}) γ^{'} (0) = 0$ and $t^{T} γ^{'} (0) = 1$ ; rank-nullity [F11] gives $\dim \ker D F (z_{0}) = (n + 1) - n = 1$ , so the kernel and unit length of $t$ force $γ^{'} (0) = t$ .

2.1F3F4F5F7F12F13step 1.1algebra

The speed $v (r) : = ∥ γ^{'} (r) ∥_{2}$ is continuous and positive on a smaller interval because $v (0) = 1$ . Set $ℓ (r) : = \int_{0}^{r} v (u) d u$ . On a compact subinterval $[a, b]$ with $a < 0 < b$ , continuity makes $v$ integrable by [F4]. Set $H (r) = \int_{a}^{r} v$ ; [F12] gives $H' (r) = v (r)$ . By [F13], $ℓ (r) = H (r) - H (0)$ on both sides of zero, so $ℓ' (r) = v (r) > 0$ ; [F7] therefore gives a local inverse $r = r (s)$ . The curve $c (s) : = γ (r (s))$ is $C^{1}$ , satisfies $∥ c^{'} (s) ∥_{2} = 1$ , and has $c (0) = z_{0}$ , $c^{'} (0) = t$ .

3.1F3F4F5F6F8F9givenstep 2.1

Fix an admissible $s, h$ and put $e : = h c^{'} (s)$ . Differentiating the identity $F (c (r)) = 0$ at $r = s$ by the chain rule gives $D F (c (s)) c^{'} (s) = 0$ , hence $D F (c (s)) e = 0$ . For each coordinate $0 \leq i < n$ , apply [F3] and [F4] to $u \mapsto F_{i} (c (s) + u e)$ and subtract the zero linear term $D F_{i} (c (s)) e$ to obtain $F_{i} (c (s) + e) = \int_{0}^{1} (D F_{i} (c (s) + u e) - D F_{i} (c (s))) e d u .$ The Lipschitz bound, [F6], and $∥ e ∥_{2} = h$ bound the integrand in absolute value by $L u h^{2}$ ; [F5] then gives the coordinate bound $L h^{2} / 2$ .

4.1step 3.1algebra∎

Combining the $n$ coordinate bounds gives $∥ F (z_{p}) ∥_{2} \leq \sqrt{n} L h^{2} / 2$ . The last coordinate of $G_{s, h} (z_{p})$ is zero by the definition of $z_{p}$ , proving the augmented residual bound.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

The simple-fold complement restriction is an isomorphism

Statement

For the data and notation of Complement splitting and range projection for a simple fold, $J ∣_{X} : X \to Y$ is a linear isomorphism. The map $P = I - q w^{T}$ satisfies $P^{2} = P, range P = Y, \ker P = span {q}, P J = J, P b = 0.$

Facts & Assumptions

Given: $J, v, w, b, a, X, Y, q, P$ and their hypotheses in Complement splitting and range projection for a simple fold.

[F1]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a map between finite-dimensional spaces, $\dim V = \dim \ker T + \dim im T$ .

[F2]

In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ : $\dim W + \dim W^{⊥} = \dim V$ in finite-dimensional inner-product spaces.

[F3]

If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ : subspaces of equal finite dimension are equal; The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial: a linear map is injective exactly when its kernel is zero.

[F4]

The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ : $\dim R^{n} = n$ .

[F5]

Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, and Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis: independence means a zero linear combination has only zero coefficients, a basis is an independent spanning set, and dimension is the size of a finite basis.

[F6]

Rank and nullity of a linear map with finite-dimensional domain: rank of a map with finite-dimensional domain is the dimension of its image.

Proof

Given: The fold data in Complement splitting and range projection for a simple fold.

1.1F2F5givenalgebra

Since $v$ is unit it is nonzero. If $α v = 0$ , then $α ∥ v ∥_{2}^{2} = 0$ , so $α = 0$ ; thus ${v}$ is independent, and it spans $span {v}$ by definition. It is therefore a basis and that span has dimension one by [F5]. The orthogonal-complement formula [F2] gives $\dim X = n - 1$ .

1.2F1F4givenalgebra

Since $w^{T} q = 1$ , the map $w^{T}$ is onto $R$ . Rank-nullity [F1] and $\dim R^{n} = n$ [F4] give $\dim Y = n - 1$ ; [F4] also gives $\dim R^{n} = n$ .

1.3givenalgebra

Since $w^{T} q = 1$ , direct multiplication gives $P^{2} = (I - q w^{T})^{2} = I - q w^{T} = P$ . Also $w^{T} P = 0$ , so $range P \subseteq Y$ , while $P y = y$ for every $y \in Y$ ; hence $range P = Y$ . The equation $P z = 0$ is equivalent to $z = q (w^{T} z)$ , so $\ker P = span {q}$ .

2.1F1F3F6step 1.1step 1.2givenalgebra

From $w^{T} J = 0$ we have $im J \subseteq Y$ , and $J$ has rank $n - 1$ by [F6]. Thus $\dim im J = \dim Y$ , so [F3] gives $im J = Y$ . Moreover, if $x \in X \cap \ker J$ , then $x = α v$ and $0 = v^{T} x = α$ , so $J ∣_{X}$ is injective by [F3]. Its image has dimension $n - 1$ by [F1] and therefore equals $Y$ by [F3]. This proves that $J ∣_{X}$ is an isomorphism.

3.1givenstep 1.3algebra∎

Finally $P J = J - q (w^{T} J) = J$ , and $P b = b - q (w^{T} b) = b - q a = 0$ by the definitions.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A transverse predictor hyperplane selects one nearby point of a regular branch

Statement

Let $c : [- δ, δ] \to U \subseteq R^{n + 1}$ be a unit-speed reparameterization of the unique locally normalized solution curve supplied by A regular solution curve has a unique local arclength chart and a quadratic predictor defect, with $c (0) = z_{0}$ . Let $τ \in R^{n + 1}$ be a unit vector, let $0 < r_{-}, r_{+} \leq δ$ , and suppose that for some $κ > 0$ , $τ^{T} c^{'} (s) \geq κ (- r_{-} \leq s \leq r_{+}) .$ Let $z_{p} \in R^{n + 1}$ satisfy $d : = τ^{T} (z_{p} - z_{0}), 0 \leq d \leq κ r_{+} .$ Then there is exactly one $s_{*} \in [- r_{-}, r_{+}]$ such that $τ^{T} (c (s_{*}) - z_{p}) = 0.$ Thus the predictor hyperplane has exactly one intersection with this local branch segment containing its basepoint. The statement is local to the supplied chart.

Facts & Assumptions

Given: The chart $c$ , $z_{0}$ , unit normal $τ$ , $r_{-}$ , $r_{+}$ , $κ$ , and $z_{p}$ with the conditions in the Statement.

[F1]

A regular solution curve has a unique local arclength chart and a quadratic predictor defect: $c$ is a $C^{1}$ unit-speed reparameterization of the unique local solution curve on an interval containing $[- r_{-}, r_{+}]$ .

[F2]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a linear functional composed with a differentiable curve is the functional applied to the curve derivative.

[F3]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: a $C^{1}$ scalar function's increment on a compact interval is the integral of its derivative.

[F4]

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ : an integrable function bounded below by $κ$ on an interval of length $ℓ$ has integral at least $κ ℓ$ .

[F5]

Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ : a continuous real function takes every value between its endpoint values.

Proof

Given: The chart and inequalities in the Statement.

1.1F1F2F3F4F5givenalgebra

Define $g (s) : = τ^{T} (c (s) - z_{p})$ on $[- r_{-}, r_{+}]$ . By [F2], $g^{'} (s) = τ^{T} c^{'} (s) \geq κ$ . The fundamental theorem in [F3] therefore gives $g (r_{+}) - g (0) = \int_{0}^{r_{+}} g^{'} (u) d u \geq κ r_{+}$ by [F4]. Since $g (0) = - d$ and $d \leq κ r_{+}$ , we have $g (0) \leq 0$ and $g (r_{+}) \geq 0$ . If $d = 0$ , then $s_{*} = 0$ is a zero; if $d > 0$ , the intermediate value theorem [F5] gives $s_{*} \in (0, r_{+}]$ with $g (s_{*}) = 0$ .

2.1F2F3F4step 1.1algebra∎

If $- r_{-} \leq s_{1} < s_{2} \leq r_{+}$ , [F2]–[F4] give $g (s_{2}) - g (s_{1}) = \int_{s_{1}}^{s_{2}} g^{'} (u) d u \geq κ (s_{2} - s_{1}) > 0.$ Thus $g$ is strictly increasing, so it has at most one zero on $[- r_{-}, r_{+}]$ . The zero found in step 1.1 is therefore unique.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A simple fold has a unique local Lyapunov–Schmidt chart and nonzero parameter curvature

Statement

Let $n \geq 1$ , let $U \subseteq R^{n} \times R$ be open, and let $F : U \to R^{n}$ be $C^{3}$ . Write $D_{x} F (x, λ) u : = D F (x, λ) (u, 0)$ and $F_{λ} (x, λ) : = D F (x, λ) (0, 1)$ . Define $F_{x x} (x_{0}, λ_{0}) [v, v] : = {\frac{d}{d t} (D_{x} F (x_{0} + t v, λ_{0}) v) ∣}_{t = 0} .$ Suppose $F (x_{0}, λ_{0}) = 0, J : = D_{x} F (x_{0}, λ_{0}), rank J = n - 1.$ Assume that $v \in R^{n}$ is a unit vector with $\ker J = span {v}$ , that $w \in R^{n} ∖ {0}$ satisfies $w^{T} J = 0$ , and that $a : = w^{T} F_{λ} (x_{0}, λ_{0}) \neq 0, b : = w^{T} F_{x x} (x_{0}, λ_{0}) [v, v] \neq 0.$ Then there is $δ > 0$ and a unique local $C^{3}$ curve of solutions $(x (s), λ (s))$ , $∣ s ∣ < δ$ , with $x (0) = x_{0}, λ (0) = λ_{0}, v^{T} (x (s) - x_{0}) = s, x^{'} (0) = v, λ^{'} (0) = 0, λ^{''} (0) = - \frac{b}{a} \neq 0.$ In particular, on a smaller interval there is a finite $M_{3}$ such that $∣ λ (s) - λ_{0} + \frac{b}{2 a} s^{2} ∣ \leq \frac{M_{3}}{6} ∣ s ∣^{3} .$ The full derivative $D_{(x, λ)} F$ has rank $n$ at the fold, and the solution curve tangent there is $(v, 0)$ . Thus an augmented pseudo-arclength hyperplane with unit normal $τ$ remains locally regular whenever $∣ τ^{T} (v, 0) ∣ > 0$ . This is a local statement; it asserts no global branch completeness.

Facts & Assumptions

Given: The dimensions, $U, F, (x_{0}, λ_{0}), J, v, w, a, b$ and hypotheses in the Statement.

[F1]

Complement splitting and range projection for a simple fold: with $X = v^{⊥}$ , $Y = \ker (w^{T})$ , $q = F_{λ} / a$ , and $P = I - q w^{T}$ , the map $P$ is the complement projection notation used below.

[F2]

The simple-fold complement restriction is an isomorphism: $J ∣_{X} : X \to Y$ is a linear isomorphism, $P^{2} = P$ , $range P = Y$ , $\ker P = span {q}$ , $P J = J$ , and $P F_{λ} = 0$ .

[F3]

The parametrized implicit function theorem with $C^{k}$ regularity: a $C^{3}$ equation with invertible derivative in the solved variables has a unique local $C^{3}$ implicit solution in the remaining parameters.

[F4]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : derivatives of the parameterized compositions are obtained by composing their derivatives.

[F5]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: for a $C^{1}$ scalar function, its endpoint difference is the integral of its derivative.

[F6]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real-valued function on a compact metric space is bounded and attains its extrema.

[F7]

Invertibility and a norm bound for the pseudo-arclength bordered matrix: the augmented derivative with last row $τ^{T}$ is invertible when its dot product with the unit tangent is nonzero.

[F8]

If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ : each finite-dimensional subspace has a finite basis, so finite-dimensional coordinates can be fixed on $X$ and $Y$ .

[F9]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map with finite-dimensional domain, the domain dimension is the sum of kernel and image dimensions.

[F10]

$C^{k}$ Euclidean maps and diffeomorphisms: $C^{3}$ means each component has continuous coordinate partial derivatives through order three.

[F11]

The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder: $D F$ is the linear map in the first-order approximation; state and parameter derivatives are its restrictions to the corresponding coordinate directions.

[F12]

Rank and nullity of a linear map with finite-dimensional domain: rank of a map with finite-dimensional domain is the dimension of its image.

Proof

Given: The hypotheses in the Statement.

1.1F2F3F9F10F11F12givenalgebra

Let $c : = F_{λ} (x_{0}, λ_{0})$ . By [F2], $im J = Y = \ker (w^{T})$ ; and $w^{T} c = a \neq 0$ . For any $r \in R^{n}$ , the vector $r - (w^{T} r / a) c$ lies in $Y$ , so the columns of $[J c]$ span $R^{n}$ . Thus the full derivative has rank $n$ . Its kernel is one-dimensional by [F9] and contains $(v, 0)$ , so its unit tangent at the fold is $(v, 0)$ . The local full-rank curve chart follows from [F3], as also recorded in the regular-curve lemma.

1.2F3F4F10F11givenalgebra

First take $n = 1$ . Then $J = 0$ , $v \in {- 1, 1}$ , and $a = w F_{λ} \neq 0$ implies $F_{λ} \neq 0$ . Apply [F3] solving for $λ$ as a function of $x = x_{0} + s v$ . Differentiating $F (x_{0} + s v, λ (s)) = 0$ once gives $λ^{'} (0) = 0$ , and differentiating twice gives $F_{x x} [v, v] + F_{λ} λ^{''} (0) = 0$ . Since $w$ is a nonzero scalar, this is $λ^{''} (0) = - b / a$ . Local uniqueness is the uniqueness clause of [F3].

1.3F1F2F3F8F10F11givenconstruct

Now take $n \geq 2$ . Fix finite-dimensional coordinates on $X$ and $Y$ by [F8]. Apply [F3] to the complement equation $P F (x_{0} + s v + u, λ_{0} + μ) = 0, u \in X .$ Its derivative in $u$ at $(0, 0, 0)$ is $P J ∣_{X} = J ∣_{X}$ , invertible by [F1]–[F2]. It therefore has a unique local $C^{3}$ solution $u = ψ (s, μ) \in X$ with $ψ (0, 0) = 0$ .

2.1F1F2F4F10F11step 1.3algebra

Differentiating the complement equation at $(0, 0)$ and using [F1]–[F2] gives $J ψ_{s} (0, 0) = 0$ and $J ψ_{μ} (0, 0) = 0$ . Both derivatives lie in $X$ , on which $J$ is injective, so $ψ_{s} (0, 0) = ψ_{μ} (0, 0) = 0$ . Define $g (s, μ) : = w^{T} F (x_{0} + s v + ψ (s, μ), λ_{0} + μ) .$ The identity $F = P F + q (w^{T} F)$ shows that $F = 0$ is equivalent, locally, to the complement equation and $g = 0$ .

3.1F4F10F11step 2.1givenalgebra

The chain rule and $w^{T} J = 0$ give $g (0, 0) = 0, g_{s} (0, 0) = 0, g_{μ} (0, 0) = a, g_{s s} (0, 0) = b .$ For the last equality, terms involving $J ψ_{s s}$ vanish after multiplication by $w^{T}$ , and $ψ_{s} (0, 0) = 0$ leaves exactly $w^{T} F_{x x} [v, v]$ .

4.1F3F4F10F11step 2.1step 3.1algebra

Since $a \neq 0$ , apply [F3] to $g (s, μ) = 0$ to obtain a unique local $C^{3}$ function $μ = μ (s)$ . Differentiating this scalar equation gives $μ^{'} (0) = - g_{s} / g_{μ} = 0$ and $μ^{''} (0) = - g_{s s} / g_{μ} = - b / a$ . The resulting $x (s) = x_{0} + s v + ψ (s, μ (s))$ , $λ (s) = λ_{0} + μ (s)$ has $x^{'} (0) = v$ and $λ^{'} (0) = 0$ . The equivalence in step 2.1 proves local uniqueness among solutions.

5.1step 1.2step 1.3step 4.1

Step 1.2 gives the stated curve and derivatives when $n = 1$ ; steps 1.3–4.1 give them when $n \geq 2$ . In both cases $x^{'} (0) = v$ , $λ^{'} (0) = 0$ , and $λ^{''} (0) = - b / a \neq 0$ .

6.1F5F6step 1.2step 4.1step 5.1algebra

In both cases $n = 1$ and $n \geq 2$ , the branch is $C^{3}$ . Choose a compact subinterval $[- δ, δ]$ and use [F6] to set $M_{3} : = \sup_{∣ s ∣ \leq δ} ∣ λ^{'''} (s) ∣ < \infty$ . Repeatedly applying [F5] gives the integral remainder $λ (s) - λ_{0} - λ^{'} (0) s - \frac{1}{2} λ^{''} (0) s^{2} = \int_{0}^{s} \int_{0}^{u} \int_{0}^{t} λ^{'''} (r) d r d t d u,$ whose absolute value is at most $M_{3} ∣ s ∣^{3} / 6$ .

7.1F7step 1.1step 1.2step 4.1step 5.1givenalgebra∎

The tangent $(v, 0)$ has unit length because $∥ v ∥_{2} = 1$ . Taking $τ = (v, 0)$ in [F7] gives angular factor $1$ , so the bordered derivative of the local pseudo-arclength corrector is invertible at the fold. More generally it remains invertible whenever $∣ τ^{T} (v, 0) ∣ > 0$ . The local curve has therefore a regular pseudo-arclength chart through the fold, while $λ^{'} (0) = 0$ and $λ^{''} (0) \neq 0$ show that $λ$ is not a local regular coordinate. No assertion about other components or global branch coverage follows.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

A transverse pseudo-arclength section has a locally quadratic Newton corrector

Statement

Let $n \geq 1$ , let $U \subseteq R^{n + 1}$ be open, and let $F : U \to R^{n}$ be $C^{2}$ . Suppose $z_{*} \in U$ satisfies $F (z_{*}) = 0$ and $A : = D F (z_{*})$ has rank $n$ . Let $t \in \ker A$ be a unit vector, let $τ \in R^{n + 1}$ be a unit vector with $∣ τ^{T} t ∣ \geq c$ for some $0 < c \leq 1$ , and let $z_{p}$ satisfy $τ^{T} (z_{*} - z_{p}) = 0$ . Define $G (z) : = (F (z), τ^{T} (z - z_{p})) .$ Let $K_{A} \geq 0$ satisfy $∥ (A ∣_{t^{⊥}})^{- 1} u ∥_{2} \leq K_{A} ∥ u ∥_{2} (u \in R^{n}),$ and set $β_{*} : = \sqrt{K_{A}^{2} (1 + c^{- 1})^{2} + c^{- 2}} .$ The derivative $D G (z_{*})$ is invertible and $∥ D G (z_{*})^{- 1} ∥_{o p} \leq β_{*}$ . There is a local regular unit-speed solution-curve chart through $z_{*}$ , and after shrinking it, the hyperplane $τ^{T} (z - z_{p}) = 0$ has exactly one intersection with that chart.

Suppose also that for some $R > 0$ and $L \geq 0$ the closed ball ${\overline{B}}_{2} (z_{*}, R)$ lies in $U$ and $∥ D F (u) - D F (v) ∥_{o p} \leq L ∥ u - v ∥_{2} (u, v \in {\overline{B}}_{2} (z_{*}, R)) .$ Then there are $ρ \in (0, R]$ and $K > 0$ such that each exact-Jacobian Newton update for $G$ started in ${\overline{B}}_{2} (z_{*}, ρ)$ is defined, remains in that ball, and satisfies $∥ z_{k + 1} - z_{*} ∥_{2} \leq C ∥ z_{k} - z_{*} ∥_{2}^{2}, C : = K L / 2,$ where $C ρ \leq 1 / 2$ (if $L = 0$ , any sufficiently small $ρ$ works). Continuing the exact recurrence indefinitely gives convergence to $z_{*}$ ; a finite update budget may stop first, and its root status means only that the residual tolerance was met. Separately, whenever the hypotheses of A bounded residual and derivative error give a local root certificate hold for $G$ on a proposed closed ball, that theorem certifies exactly one zero there and its stated residual-to-root error bound. These conclusions are local and imply no global branch-completeness claim.

Facts & Assumptions

Given: $n, U, F, z_{*}, A, t, τ, z_{p}, K_{A}, c, R, L$ and the hypotheses in the Statement.

[F1]

A regular solution curve has a unique local arclength chart and a quadratic predictor defect: at a root where $D F$ has rank $n$ , a unit kernel vector determines a unique locally normalized solution curve, which has a $C^{1}$ unit-speed reparameterization $c (s)$ with $c (0) = z_{*}$ and $c^{'} (0) = t$ ; every sufficiently nearby solution lies on this curve.

[F2]

A transverse predictor hyperplane selects one nearby point of a regular branch: if $τ^{T} c^{'} (s) \geq κ > 0$ on a short two-sided chart interval and the signed predictor offset lies between $0$ and $κ$ times its forward length, the hyperplane intersects that entire local branch segment exactly once.

[F3]

Invertibility and a norm bound for the pseudo-arclength bordered matrix: for rank- $n$ $A$ with unit kernel $t$ , the matrix $[\begin{array}{c} A \\ τ^{T} \end{array}]$ is invertible when $∣ τ^{T} t ∣ \geq c > 0$ and its inverse norm is at most $\sqrt{K_{A}^{2} (1 + c^{- 1})^{2} + c^{- 2}}$ .

[F4]

Updated-Jacobian Newton has local quadratic convergence at a simple root: if a $C^{2}$ square residual has an exact zero with invertible Jacobian and a Lipschitz Jacobian on a closed ball, there is a smaller invariant ball on which every exact Newton update is defined and satisfies the quadratic error recurrence; an infinite recurrence converges, while finite-budget root status records only a residual tolerance.

[F5]

Finite-budget Newton for a square vector residual specifies the residual, trial-domain, singularity and budget branches. Its correction-call bound is proved in Finite-budget Newton preserves its domain and returns a status in bounded outer work; budget exhaustion may precede tolerance attainment.

[F6]

A bounded residual and derivative error give a local root certificate: under its certified residual/Jacobian error, inverse, Lipschitz and strict contraction hypotheses, a square residual has exactly one zero in the proposed ball and the distance from its center is at most $β ρ_{g} / (1 - q)$ , where $ρ_{g}$ is its certified residual upper bound.

[F7]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : the derivative of a composite is the composition of derivatives.

[F8]

$C^{k}$ Euclidean maps and diffeomorphisms: $C^{2}$ means each component has continuous coordinate partial derivatives through order two.

[F9]

[F10]

Rank and nullity of a linear map with finite-dimensional domain: rank of a map with finite-dimensional domain is the dimension of its image.

Proof

Given: The data and hypotheses in the Statement.

1.1F1F2F8F9F10givenchoose

Apply [F1] to obtain a $C^{1}$ unit-speed branch chart $c (s)$ with $c (0) = z_{*}$ and $c^{'} (0) = t$ . Set $\tilde{τ} : = τ$ if $τ^{T} t > 0$ and $\tilde{τ} : = - τ$ if $τ^{T} t < 0$ ; it is a unit vector, its hyperplane is unchanged, and ${\tilde{τ}}^{T} c^{'} (0) = ∣ τ^{T} t ∣ \geq c$ . By continuity of $c^{'}$ , shrink the chart to $[- r, r]$ so that ${\tilde{τ}}^{T} c^{'} (s) \geq c / 2$ . For $g (s) : = {\tilde{τ}}^{T} (c (s) - z_{p})$ the offset in [F2] is zero because $z_{*}$ lies on the hyperplane. The $d = 0$ case of [F2] gives exactly one intersection on this chart segment, at $s = 0$ .

1.2F3F7F9F10givenalgebra

By [F7], the derivative of $G$ at $z_{*}$ is the bordered matrix $D G (z_{*}) = [\begin{array}{c} A \\ τ^{T} \end{array}] .$ The rank, kernel, angle and complement-inverse hypotheses let [F3] give its invertibility and the stated bound $β_{*}$ .

2.1F4F8F9F10givenstep 1.2algebra

The last component of $G$ is affine, so for $u, v$ in the ball, $∥ D G (u) - D G (v) ∥_{o p} = {∥ [\begin{array}{c} D F (u) - D F (v) \\ 0 \end{array}] ∥}_{o p} \leq L ∥ u - v ∥_{2} .$ Also $G (z_{*}) = 0$ by $F (z_{*}) = 0$ and the section equation. Apply [F4] to the square map $G$ on this ball. It supplies a smaller radius $ρ$ , a uniform inverse bound $K$ , and the quadratic estimate with $C = K L / 2$ ; its induction also ensures every attempted update stays in the invariant ball.

3.1F5F6step 2.1given∎

The finite procedure in [F5] uses exactly these updates and its own residual, domain, singularity and budget checks. Hence any update it attempts from the invariant ball has the bound of step 2.1; if it stops for the iteration budget before reaching tolerance, no exact-zero conclusion follows. For a separately supplied approximate residual/Jacobian tuple satisfying [F6], that certificate gives the stated unique local zero and error bound for this same augmented map $G$ .

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Finite-budget pseudo-arclength continuation with a certified local corrector

Definition

Let $n \geq 1$ , let $U \subseteq R^{n + 1}$ be open, and let $F : U \to R^{n}$ be $C^{2}$ in the componentwise sense of $C^{k}$ Euclidean maps and diffeomorphisms on a declared tube; its total derivative $D F$ has the linear first-order meaning of The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder. Require $C^{3}$ on any subtube where a fold check is requested. A valid seed consists of a center ${\overset{ˉ}{z}}_{0}$ , a radius $r_{0}$ , and a certified ball containing exactly one zero $z_{0}^{*}$ of an initial augmented section $G_{- 1} (z) : = (F (z), τ_{- 1}^{T} (z - z_{p, - 1})),$ where $τ_{- 1}$ is unit and the section is transverse to the regular local solution branch. Thus the seed certifies one solution point of $F = 0$ by intersecting the branch with a hyperplane; it does not claim that $F = 0$ has an isolated zero in $R^{n + 1}$ . The seed also includes a neighborhood in which every zero of $F$ lies on the certified local curve chart and a verified rank- $n$ condition on that chart.

Supply a unit previous orientation, step length $h > 0$ , requested step count $N_{r e q} \in N_{0}$ , continuation budget $N_{s} \in N_{0}$ , corrector budget $N_{c} \in N_{0}$ , residual tolerance $ϵ \geq 0$ , and angle margin $c \in (0, 1]$ . At each current certified section ball $B_{k} = {\overline{B}}_{2} ({\overset{ˉ}{z}}_{k}, r_{k})$ , its unique branch point is denoted $z_{k}^{*}$ . A tangent routine must provide a unit candidate ${\hat{t}}_{k}$ and a certified error bound to a unit vector $t_{k} \in \ker D F (z_{k}^{*})$ , plus a verified inverse bound for $D F (z_{k}^{*}) ∣_{t_{k}^{⊥}}$ . Missing rank, tangent, or error bounds are failures; they are never treated as zero error.

At a queried point $z$ , an evaluator returns either exact $(F (z), D F (z))$ or certified data $(\hat{F} (z), η_{F}, \hat{J} (z), η_{J})$ with $∥ \hat{F} (z) - F (z) ∥_{2} \leq η_{F}, ∥ \hat{J} (z) - D F (z) ∥_{o p} \leq η_{J} .$ All supplied tangent, evaluator, chart, certificate and requested fold callbacks must return a result or their declared failure in finite time for a termination claim. Their mathematical bounds must be sound; finite outer budgets cannot force a nonreturning callback to finish. For every root ball proposed by a continuation attempt, the certificate routine must also supply an inverse bound for the approximate augmented Jacobian, a Jacobian-Lipschitz constant on the whole ball, and an isolating branch neighborhood/angle enclosure. If any bound is missing, return root-certificate-inconclusive for that candidate. Initialize the fold field as not-requested when no fold check was asked, and as inconclusive while a requested check remains unresolved.

Output: a finite list of certified local section-root balls; a continuation status from finite-trace, rank-or-tangent-unverified, tangent-angle-failure, tube-exit, singular-corrector, corrector-budget-exhausted, root-certificate-inconclusive, or continuation-budget-exhausted; and a fold field with value not-requested, certified, or inconclusive. The initial seed record retains its supplied unique-section-root certificate, center, radius, section and transverse local chart/rank evidence; its center error is at most $r_{0}$ . Any stronger error or contraction bounds proved by the supplied seed certificate are retained, but no contraction bounds are required or inferred for the seed. Every newly appended ball records its center, radius, section, strict contraction bounds and center-error bound from A bounded residual and derivative error give a local root certificate, the local chart and angle enclosure identifying it with the unique next point on the traced branch, and the evaluation-error bounds used.

Procedure:

On entry only, initialize the list with the valid seed ball, accepted count $a = 0$ and attempted continuation count $s = 0$ , and use the supplied unit previous orientation at the seed. On each visit to this step, denote the current previous orientation by $τ_{k}$ ; retain the list, counts and orientation when returning from step6. If $a = N_{r e q}$ , including $N_{r e q} = 0$ , finalize with status finite-trace. Otherwise, if $s = N_{s}$ , finalize with continuation-budget-exhausted. Increment $s$ before making any callback for the next attempt. Thus $N_{s}$ caps attempts, not just successful steps.
At the unique current root $z_{k}^{*}$ , query the tangent routine. Verify rank $D F (z_{k}^{*}) = n$ , a unit candidate ${\hat{t}}_{k}$ , a unit true kernel vector $t_{k}$ , its certified error $∥ {\hat{t}}_{k} - t_{k} ∥_{2} \leq e_{k}$ , and the complement inverse bound. Missing evidence returns rank-or-tangent-unverified. Orient ${\hat{t}}_{k}$ and its corresponding $t_{k}$ together, keeping $τ_{k}$ fixed. Certify $τ_{k}^{T} t_{k} \geq c$ and $a_{k} : = τ_{k}^{T} {\hat{t}}_{k} > 0$ ; an unavailable or failed angular test returns tangent-angle-failure.
Set $z_{p} = {\overset{ˉ}{z}}_{k} + h {\hat{t}}_{k}$ and retain the previous orientation as section normal: $G_{k} (z) = (F (z), τ_{k}^{T} (z - z_{p})) .$ The chart certificate supplies a unit-speed chart $c_{k}$ with $c_{k} (0) = z_{k}^{*}$ , $c_{k}^{'} (0) = t_{k}$ , an interval $[- r_{c h}^{-}, r_{c h}^{+}]$ , a number $κ > 0$ , and a neighborhood $V_{k}$ in which every zero of $F$ belongs to that chart interval. Verify $τ_{k}^{T} c_{k}^{'} (u) \geq κ$ throughout the interval and $h a_{k} > r_{k}, h a_{k} + r_{k} \leq κ r_{c h}^{+} .$ These checks bound $d = τ_{k}^{T} (z_{p} - z_{k}^{*})$ between $h a_{k} - r_{k}$ and $h a_{k} + r_{k}$ , so the selected section has one forward intersection on the supplied chart by A transverse predictor hyperplane selects one nearby point of a regular branch. A failed chart angle gives tangent-angle-failure; failed offset/length evidence gives root-certificate-inconclusive. If the predictor, proposed ball or required chart leaves the declared tube, return tube-exit.
With exact evaluations, call Finite-budget Newton for a square vector residual on $G_{k}$ from $z_{p}$ , tolerance $ϵ$ and update budget $N_{c}$ . Map singular-jacobian to singular-corrector and invalid-trial to tube-exit. For root or iteration-budget-exhausted, retain the last valid iterate as a candidate. If the predictor is certified inside the invariant ball of A transverse pseudo-arclength section has a locally quadratic Newton corrector, every attempted exact update has its quadratic error bound; otherwise this is only a candidate generator. Apply the root certificate of A bounded residual and derivative error give a local root certificate to the last candidate. Accept only if every inequality passes and the whole ball lies in $V_{k}$ . Unavailable inputs give root-certificate-inconclusive; failed tests after a root residual stop give that same status; failed tests after exhausted updates give corrector-budget-exhausted. Neither residual tolerance nor budget alone is a root proof. A passing ball proceeds to step6, without an earlier append.
With bounded evaluation errors, set $y_{0} = z_{p}$ . For $j = 0, \dots, N_{c}$ , query the supplied certified residual/Jacobian and root-ball data at $y_{j}$ . Unavailable certificate inputs give root-certificate-inconclusive. Apply the same strict contraction tests and whole-ball inclusion in $V_{k}$ as in step4. A passing ball proceeds to step6, without an earlier append. If the tests fail at $j = N_{c}$ , return corrector-budget-exhausted. Otherwise call Newton correction for a square vector residual on the supplied approximate augmented residual and matrix. Map singular-jacobian to singular-corrector; otherwise let $s_{j}$ be its returned correction. A trial $y_{j} + s_{j}$ outside the tube gives tube-exit; otherwise set $y_{j + 1} = y_{j} + s_{j}$ . This bounded-error branch has no convergence promise.
Append the passing ball once, with its section, error bounds and chart/angle record; increase $a$ by one. For the next attempt use the already certified unit candidate ${\hat{t}}_{k}$ as the new previous orientation. It is not silently declared the exact tangent at the new root: step2 must certify its angle against the next root's tangent. Return to step1.
At every finalization, including every early return, retain the continuation status already assigned. For a requested fold check require a returned certified record satisfying all hypotheses of A simple fold has a unique local Lyapunov–Schmidt chart and nonzero parameter curvature at its specified point: $F = 0$ , the required $C^{3}$ neighborhood, $rank F_{x} = n - 1$ , normalized right and nonzero left nullvectors, and strict enclosures excluding zero for $w^{T} F_{λ}$ and $w^{T} F_{x x} [v, v]$ , together with the stated complement/parameter normalization. Missing or failed evidence leaves the fold field inconclusive; otherwise set it to certified and record only the local fold chart and curvature formula. Without a fold request keep not-requested. This finalization makes finitely many callback requests and performs no unbounded search.

Remarks

The local predictor-defect estimate in A regular solution curve has a unique local arclength chart and a quadratic predictor defect applies to an exact root and exact unit tangent. Tangent and seed-center errors in an actual step are handled by the supplied residual/error enclosures and the root certificate, not silently set to zero. The exact Newton branch preserves the supplier's root convention: residual tolerance was met, not necessarily an exact zero. No finite trace certifies that a connected component is exhausted or that another component is absent.

TheoremStatement: AI-adaptedProof: AI-generatedOpen item page →

Certified pseudo-arclength traces are locally sound and conditionally terminate

Statement

For admitted inputs of Finite-budget pseudo-arclength continuation with a certified local corrector, assume its supplied certificate bounds are sound. Every returned ball, including the seed, contains exactly one zero of its recorded augmented section. Each appended ball identifies the unique forward section point on the supplied local chart of the previously certified branch point. Its center error obeys the bound of A bounded residual and derivative error give a local root certificate. A finite-trace return has exactly $N_{r e q}$ appended balls; every other return retains a certified prefix and never asserts that the failed candidate is a root. A fold field certified implies precisely the supplied local simple-fold chart and curvature conclusion, with all hypotheses of A simple fold has a unique local Lyapunov–Schmidt chart and nonzero parameter curvature.

If every invoked tangent, evaluator, chart, certificate and requested fold callback returns a result or declared failure in finite time, the procedure terminates. There are at most $N_{s}$ continuation attempts, at most $N_{c}$ Newton updates per attempt, and at most $N_{c} + 1$ certificate attempts in a bounded-error corrector. When $N_{r e q} = 0$ it returns the seed with continuation status finite-trace; when $N_{s} = 0 < N_{r e q}$ it returns the seed with continuation-budget-exhausted, in both cases followed only by the finite finalization. These are partial correctness and conditional termination, not numerical success, a callback time bound, or global branch completeness.

There is also an error-aware predictor bound. In an admitted attempt write $z_{*} = z_{k}^{*}$ , $τ = τ_{k}$ , $\hat{t} = {\hat{t}}_{k}$ , $t = t_{k}$ , and $r = r_{k}$ , with $∥ \overset{ˉ}{z} - z_{*} ∥_{2} \leq r$ and $∥ \hat{t} - t ∥_{2} \leq e$ . Suppose a convex open tube contains $z_{p} = \overset{ˉ}{z} + h \hat{t}$ , $p_{*} = z_{*} + h t$ and their segment, as well as the exact tangent-predictor segment from $z_{*}$ to $p_{*}$ . Suppose on that tube $∥ D F ∥_{o p} \leq D$ and $D F$ is $L$ -Lipschitz. Then $∥ G_{k} (z_{p}) ∥_{2} = ∥ F (z_{p}) ∥_{2} \leq \frac{\sqrt{n} L}{2} h^{2} + D (r + h e) .$ In particular the exact-seed, exact-tangent branch preserves the quadratic predictor-defect bound even though the section uses the previous orientation.

Facts & Assumptions

Given: The algorithm inputs and soundness hypotheses, and, for termination, the stated finite-callback condition; for the last bound, the stated convex tube and derivative bounds.

[F1]

Finite-budget pseudo-arclength continuation with a certified local corrector fixes the previous section normal, candidate predictor, accepted/attempt counters, finite update loops, common single append, chart inclusion and all return/finalization tests.

[F2]

A transverse predictor hyperplane selects one nearby point of a regular branch selects exactly one point on a two-sided unit-speed chart when its normal derivative is bounded below by $κ > 0$ and its offset is in $[0, κ r_{+}]$ .

[F3]

A bounded residual and derivative error give a local root certificate supplies exactly one augmented zero in the proposed ball and center error at most $β ρ_{g} / (1 - q)$ under its explicit derivative/residual errors, inverse bound, Lipschitz bound and contraction/inclusion inequalities.

[F4]

A simple fold has a unique local Lyapunov–Schmidt chart and nonzero parameter curvature gives its local chart and normalized nonzero parameter curvature when all stated rank, regularity, complement and strict nondegeneracy hypotheses hold.

[F5]

Finite-budget Newton for a square vector residual specifies its singular/domain/budget and residual-stop outputs; Finite-budget Newton preserves its domain and returns a status in bounded outer work proves its correction-call bound. Neither result makes a residual stop a root certificate.

[F6]

A regular solution curve has a unique local arclength chart and a quadratic predictor defect bounds $∥ F (z_{*} + h t) ∥_{2}$ by $\sqrt{n} L h^{2} / 2$ for the exact root and unit tangent on its admitted tube.

[F7]

On a convex open set, a uniform bound $∥ D f (z) v ∥_{2} \leq M ∥ v ∥_{2}$ implies $∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2}$ : on a convex open set, a derivative-operator-norm bound $D$ gives $∥ F (u) - F (v) ∥_{2} \leq D ∥ u - v ∥_{2}$ .

[A1]

Every returned bound used in an accepted seed, chart, root-ball or fold record is sound. Every callback in the conditional-termination claim returns in finite time; a callback may return failure.

Proof

1.1A1F1F2givenalgebra

The valid seed is the base of the certified list by [F1]. In an attempted step, $τ$ is unit and the root lies within distance $r$ of its current center. Thus $∣ τ^{T} (\overset{ˉ}{z} - z_{*}) ∣ \leq r$ . Writing $a = τ^{T} \hat{t}$ , the actual offset is $d = τ^{T} (z_{p} - z_{*}) = h a + τ^{T} (\overset{ˉ}{z} - z_{*})$ . Hence $h a - r \leq d \leq h a + r$ . The tests $h a > r$ and $h a + r \leq κ r_{+}$ in [F1] put $0 < d \leq κ r_{+}$ . With the whole chart angle inequality, [F2] gives a unique intersection and puts its chart parameter strictly forward of zero.

2.1A1F1F2F3F5step 1.1algebrainduction

An append is permitted only after [F3] applies to that same augmented section and the whole ball is contained in the neighborhood where every zero of $F$ belongs to the supplied chart interval. Its certified augmented zero therefore lies on that chart and its section plane. By uniqueness from step 1.1 it is the selected point. [F3] supplies the recorded center error. This argument applies to exact and bounded-error candidate generators alike; their residual stops or exhausted budgets cannot replace the ball certificate. Induction over the common append block proves local soundness for the entire returned list and for each prefix on an early return.

3.1A1F1F4step 2.1algebra

The accepted count increases exactly once per append and starts at zero. The procedure returns finite-trace precisely when it equals $N_{r e q}$ , so that status has the stated list length. Every other continuation return preserves the already certified list. At finalization a fold field can change to certified only after its record establishes the actual fold point, its neighborhood, and every hypothesis of [F4]; applying [F4] gives precisely the local chart and curvature conclusion. Unavailable evidence leaves inconclusive, and a fold request cannot turn an unsuccessful continuation candidate into a certified section root.

4.1A1F1F5step 3.1givenalgebra

At each attempt the attempted counter increases before any callback and cannot exceed $N_{s}$ . In the exact branch [F5] allows at most $N_{c}$ Newton updates and a single subsequent root-ball certificate request. In the bounded-error branch [F1] allows at most $N_{c}$ updates and $N_{c} + 1$ certificate requests. All other checks in an attempt are finite computations or finitely many callbacks, and by [A1] they return in finite time. Finalization likewise has finitely many terminating callback requests and no unbounded search. Hence finitely many attempts and finalization terminate, regardless of whether any certificate succeeds. The initial tests in [F1] assign the stated statuses for $N_{r e q} = 0$ and $N_{s} = 0 < N_{r e q}$ before querying continuation callbacks; a zero corrector budget still allows the specified certificate test but no update.

5.1F1F6F7givenalgebra∎

The exact tangent predictor $p_{*} = z_{*} + h t$ satisfies [F6], giving $∥ F (p_{*}) ∥_{2} \leq \sqrt{n} L h^{2} / 2$ . The actual predictor obeys $z_{p} - p_{*} = (\overset{ˉ}{z} - z_{*}) + h (\hat{t} - t)$ , so $∥ z_{p} - p_{*} ∥_{2} \leq r + h e$ . Apply [F7] on the admitted convex open tube and add the exact predictor bound to obtain $∥ F (z_{p}) ∥_{2} \leq D (r + h e) + \sqrt{n} L h^{2} / 2$ . The last coordinate $τ^{T} (z_{p} - z_{p})$ is zero irrespective of which previous unit orientation is used, so the augmented norm equals $∥ F (z_{p}) ∥_{2}$ . This proves the final claim without setting seed or tangent errors to zero.

5 · Examples, counterexamples and false statements

None yet.

Pseudo arclength continuation and fold certificates

1 · Prerequisites

2 · Summary

Objective

3 · Logical flowchart

4 · Definitions, theorems and proofs

Invertibility and a norm bound for the pseudo-arclength bordered matrix

Statement

Facts & Assumptions

Proof

Complement splitting and range projection for a simple fold

Definition

A bounded residual and derivative error give a local root certificate

Statement

Facts & Assumptions

Proof

Remarks

A regular solution curve has a unique local arclength chart and a quadratic predictor defect

Statement

Facts & Assumptions

Proof

The simple-fold complement restriction is an isomorphism

Statement

Facts & Assumptions

Proof

A transverse predictor hyperplane selects one nearby point of a regular branch

Statement

Facts & Assumptions

Proof

A simple fold has a unique local Lyapunov–Schmidt chart and nonzero parameter curvature

Statement

Facts & Assumptions

Proof

A transverse pseudo-arclength section has a locally quadratic Newton corrector

Statement

Facts & Assumptions

Proof

Finite-budget pseudo-arclength continuation with a certified local corrector

Definition

Remarks

Certified pseudo-arclength traces are locally sound and conditionally terminate

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources