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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.

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Sources

Eugene L. Allgower and Kurt Georg, Numerical Continuation Methods: An Introduction, Springer Series in Computational Mathematics 13 (1990) (standard reference, not scraped)
K. A. Cliffe, A. Spence, and S. J. Tavener, The numerical analysis of bifurcation problems with application to fluid mechanics, Acta Numerica 9 (2000), 39–131 (standard reference, not scraped)