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Symbolic derivative recursion for a rational expression DAG is correct

Statement

For a finite rational expression DAG of Finite rational residual-tube certificate for an ODE trajectory, at every input point where each division denominator is nonzero, the DAG evaluates to a continuously differentiable function of time and state. The recursion assigning zero derivatives to constants, coordinate derivatives to inputs, sum/difference rules, the product rule, and $D (u / v) = \frac{v D u - u D v}{v^{2}}$ for a division node computes its actual total derivative. On any open neighborhood where all denominators stay nonzero, the represented field is locally Lipschitz.

Facts & Assumptions

Given: The finite acyclic rational expression DAG and an input point where all division denominators are nonzero.

[F1]

Finite rational residual-tube certificate for an ODE trajectory: the only DAG operations are constants, input coordinates, addition, subtraction, multiplication, and division.

[F2]

Chain sum product and composition rules for Banach derivatives: sums and bounded-bilinear products of Fréchet-differentiable maps obey the corresponding derivative rules; [F6] gives the same first-order remainder convention for the total derivatives used here.

[F3]

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

[F4]

The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : a scalar differentiable function with derivative bounded by $M$ on an interval is $M$ -Lipschitz there.

[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: $\sum_{j = 1}^{n} ∣ h_{j} ∣ \leq \sqrt{n} ∥ h ∥_{2}$ and finite-dimensional Euclidean norms obey the triangle inequality.

[F6]

The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder: total differentiability gives the expansion $u (x + h) = u (x) + D u (x) h + o (∥ h ∥_{2})$ .

Proof

1.1F1constructinduction

Order the finite acyclic graph topologically. A constant node defines a constant smooth function with derivative zero; an input node is a coordinate projection with its coordinate derivative. Assume inductively that the parent nodes define $C^{1}$ functions with their assigned derivatives.

2.1F2F6step 1.1algebra

For an addition or subtraction node, [F2] gives the sum or difference of the parent derivatives. For a product node, apply its bounded-bilinear product rule to real multiplication, which is bounded by $∣ a b ∣ \leq ∣ a ∣ ∣ b ∣$ , to obtain $v D u + u D v$ . These derivatives are continuous because the parent values and derivatives are continuous.

2.2F6step 1.1algebra

At a division node write $u_{0} = u (x)$ , $v_{0} = v (x) \neq 0$ , $A = D u (x)$ , $B = D v (x)$ , and $Δ u = A h + r_{u}$ , $Δ v = B h + r_{v}$ , where $r_{u}, r_{v} = o (∥ h ∥_{2})$ by [F6]. Continuity of $v$ keeps $v_{0} + Δ v \neq 0$ near $x$ . Put $L (h) = v_{0} A h - u_{0} B h$ and $R = v_{0} r_{u} - u_{0} r_{v}$ . Direct algebra gives $\frac{u (x + h)}{v (x + h)} - \frac{u_{0}}{v_{0}} - \frac{L (h)}{v_{0}^{2}} = \frac{v_{0}^{2} R - v_{0} L (h) Δ v}{v_{0}^{3} (v_{0} + Δ v)}$ . Here $R = o (∥ h ∥_{2})$ , $L (h) = O (∥ h ∥_{2})$ , and $Δ v = O (∥ h ∥_{2})$ , while the denominator stays bounded away from zero. Thus the right side is $o (∥ h ∥_{2})$ , so the quotient derivative is $(v_{0} D u (x) - u_{0} D v (x)) / v_{0}^{2}$ . Its formula is continuous wherever $v \neq 0$ .

3.1F1F3F4F5step 2.1step 2.2constructinduction∎

Induction through the topological order, using steps 2.1–2.2, proves that every node is $C^{1}$ with its recursively assigned derivative, so the root derivative is correct. Continuity and nonvanishing of the finitely many denominator functions give a ball around the input on which all denominators stay nonzero. Shrink it so every partial derivative of every root component is bounded in absolute value by a common finite $M$ , using continuity at the center and the finite number of entries. For two points $u, v$ in this convex ball, apply [F4] to each scalar component along $v + s (u - v)$ : its derivative has magnitude at most $M \sum_{j} ∣ u_{j} - v_{j} ∣ \leq M \sqrt{d + 1} ∥ u - v ∥_{2}$ by [F5]. Combining the finitely many component bounds gives a Lipschitz constant for the root map. Thus the field is locally Lipschitz.

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Sources

Walter Rudin, Principles of Mathematical Analysis, 3rd ed., Chapter 9 (standard reference, not scraped)
Ernst Hairer, course notes, Chapter III: Équations Différentielles Ordinaires (standard reference, not scraped)