TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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The selected two-stage Rosenbrock W-method has a second-order global error bound

Statement

Let $f : D \to R^{d}$ be $C^{2}$ on an open time-state domain and let $y (t)$ solve $y^{'} = f (t, y)$ on $[t_{0}, t_{0} + T]$ . Fix $α = 1 - 1 / \sqrt{2}$ , $h_{0} > 0$ , and a compact convex tube $Q \subset D$ containing the exact and computed states, all stage midpoint segments, and the line segments between states compared below. Use a fixed step size satisfying $0 < h \leq h_{0}$ . Assume on $Q$ that

∥ f ∥_{2} \leq F, ∥ D_{y} f ∥_{2} \leq L, ∥ f_{t} ∥_{2} \leq T_{0}, ∥ D^{2} f ∥_{o p} \leq H,

and that $∥ \tilde{J} (t, y) ∥_{2} \leq B_{J}$ , $∥ \tilde{T} (t, y) ∥_{2} \leq B_{T}$ , with state-Lipschitz bounds $∥ \tilde{J} (t, u) - \tilde{J} (t, v) ∥_{2} \leq L_{J} ∥ u - v ∥_{2}$ and $∥ \tilde{T} (t, u) - \tilde{T} (t, v) ∥_{2} \leq L_{T} ∥ u - v ∥_{2}$ . Suppose $∥ α h \tilde{J} (t, y) ∥_{2} \leq q < 1$ for all $0 < h \leq h_{0}$ and relevant $(t, y)$ ; put $κ = (1 - q)^{- 1}$ . Define the constants $C_{1}, K_{1}, C_{F}, C_{K}, M_{2}, C_{r}$ from The two-stage Rosenbrock W-method has a cubic local defect and explicit stage-residual allowance using $h_{0}$ in $K_{1} = F + α h_{0} κ C_{1}$ , and put $C_{l o c} = C_{K} + M_{2} / 6$ .

Use the method of Two-stage second-order Rosenbrock W-method with shared factorization with stage residual upper bounds $ρ_{1, n} \leq R_{1} h$ and $ρ_{2, n} \leq R_{2} h^{2}$ , where $R_{1}, R_{2}$ are fixed, and assume each call returns success, including both field evaluations, residual certifications, and exact-storage checks. Starting from $y_{0} = y (t_{0})$ and assuming $N h \leq T$ and all stated tube conditions hold, define

G_{1} = F + α h_{0} B_{T}, K_{s} = κ G_{1},

L_{1} = κ (L + α h_{0} L_{T}) + α h_{0} κ^{2} L_{J} G_{1},

V = F + α h_{0} B_{J} K_{s},

$L_{V} = L (1 + h_{0} L_{1} / 2) + α h_{0} (L_{J} K_{s} + B_{J} L_{1}), L_{2} = κ L_{V} + α h_{0} κ^{2} L_{J} V .$ Then the global error satisfies

∥ y_{n} - y (t_{n}) ∥_{2} \leq (C_{l o c} + C_{r} R_{1} + κ R_{2}) T e^{L_{2} T} h^{2} (0 \leq n \leq N) .

This is a conditional order-two theorem on the common tube. Its constants and assumptions permit both the exact-J branch and the bounded W-method branch; scalar A/L-stability is asserted separately only for the exact-J branch.

Facts & Assumptions

Given: The exact solution, compact convex tube, bounds and Lipschitz constants, shifted inverse condition, residual budgets, and exact initial value in the Statement.

[F1]

Two-stage second-order Rosenbrock W-method with shared factorization: the two stages use $W = I - α h \tilde{J}$ , the midpoint field value, and one shared factorization.

[F2]

The two-stage Rosenbrock W-method has a cubic local defect and explicit stage-residual allowance: the exact map has local defect at most $C_{l o c} h^{3}$ , while residual perturbations change the output by at most $C_{r} h^{2} ρ_{1} + κ h ρ_{2}$ .

[F3]

Perturbing a shifted Jacobian perturbs its inverse explicitly: the difference between two shifted inverses is at most $α h$ times their inverse bounds and the difference of their Jacobian matrices.

[F4]

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1: induced matrix norms bound vector actions and products.

[F5]

If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz: a derivative bound on a convex segment gives the corresponding Lipschitz bound between its endpoints.

[F6]

$1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ : $1 + x \leq e^{x}$ for $x \geq 0$ .

[F7]

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative: the componentwise $C^{2}$ assumption makes $f$ totally differentiable at each point of $Q$ .

[F8]

A total derivative computes every directional derivative, and its matrix is the Jacobian: in the joint variable $(t, y)$ , the nonzero standard time and state basis directions identify the derivative blocks with $f_{t}$ and $D_{y} f$ .

[F9]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : composing $f$ with a differentiable state segment gives derivative $D_{y} f$ applied to the segment direction.

Proof

Given: The data in the Statement.

1.1givenF1F4algebra

For fixed $t$ , put $W_{y} = I - α h \tilde{J} (t, y)$ and $g_{y} = f (t, y) + α h \tilde{T} (t, y)$ . Then $∥ g_{y} ∥_{2} \leq G_{1}$ and $k_{1} (y) = W_{y}^{- 1} g_{y}$ , so $∥ k_{1} (y) ∥_{2} \leq K_{s}$ .

1.2givenF5F7F8F9

For $y, z$ in a state slice of $Q$ , set $x (s) = y + s (z - y)$ for $0 \leq s \leq 1$ . By [F7], [F9], and [F8], the curve $s \mapsto f (t, x (s))$ has derivative $D_{y} f (t, x (s)) (z - y)$ , whose norm is at most $L ∥ y - z ∥_{2}$ ; [F5] gives $∥ f (t, y) - f (t, z) ∥_{2} \leq L ∥ y - z ∥_{2}$ . The other two differences satisfy their assumed Lipschitz bounds.

2.1givenF3F4step 1.2algebra

By [F3], $∥ W_{y}^{- 1} - W_{z}^{- 1} ∥_{2} \leq α h κ^{2} L_{J} ∥ y - z ∥_{2}$ . Apply this and step 1.2 to $k_{1} (y) = W_{y}^{- 1} g_{y}$ after adding and subtracting $W_{y}^{- 1} g_{z}$ . The result is $∥ k_{1} (y) - k_{1} (z) ∥_{2} \leq L_{1} ∥ y - z ∥_{2}$ .

3.1givenF5step 2.1algebra

The midpoint state $m_{y} = y + (h / 2) k_{1} (y)$ therefore satisfies $∥ m_{y} - m_{z} ∥_{2} \leq (1 + h L_{1} / 2) ∥ y - z ∥_{2}$ . Applying [F5] along the midpoint-state segment gives $∥ F_{1} (y) - F_{1} (z) ∥_{2} \leq L (1 + h_{0} L_{1} / 2) ∥ y - z ∥_{2}$ .

4.1givenF1F4step 1.1step 2.1step 3.1algebra

The algebraic identity from [F1] is $k_{2} (y) = W_{y}^{- 1} v_{y}$ , with $v_{y} = F_{1} (y) - α h \tilde{J} (t, y) k_{1} (y)$ . Its norm is at most $V$ . Adding and subtracting terms in the product and using steps 2.1 and 3.1 gives $∥ v_{y} - v_{z} ∥_{2} \leq L_{V} ∥ y - z ∥_{2}$ .

5.1givenF3F4step 4.1algebra

Apply [F3] once more to $W_{y}^{- 1} v_{y} - W_{z}^{- 1} v_{z}$ . Steps 4.1 and the inverse bound give $∥ k_{2} (y) - k_{2} (z) ∥_{2} \leq L_{2} ∥ y - z ∥_{2}$ . Hence the exact one-step map $Φ_{h} (t, y) = y + h k_{2} (y)$ obeys $∥ Φ_{h} (t, y) - Φ_{h} (t, z) ∥_{2} \leq (1 + L_{2} h) ∥ y - z ∥_{2}$ .

6.1givenF2F7F8step 5.1algebra

At numerical state $y_{n}$ , [F2] and the two residual budgets bound the difference between the computed output and $Φ_{h} (t_{n}, y_{n})$ by $(C_{r} R_{1} + κ R_{2}) h^{3}$ . At exact state $y (t_{n})$ , [F2] with zero residual gives local defect at most $C_{l o c} h^{3}$ ; the componentwise $C^{2}$ premise of [F2] supplies the total differentiability by [F7], and [F8] identifies its time and state derivative blocks. Insert the exact maps at the numerical and exact states and use step 5.1; the errors satisfy $e_{n + 1} \leq (1 + L_{2} h) e_{n} + (C_{l o c} + C_{r} R_{1} + κ R_{2}) h^{3}$ .

7.1step 6.1givenF6algebra∎

Iterate step 6.1 from $e_{0} = 0$ . Each amplification factor product is at most $e^{L_{2} T}$ by [F6], and there are at most $T / h$ local terms. Therefore $e_{n} \leq (C_{l o c} + C_{r} R_{1} + κ R_{2}) T e^{L_{2} T} h^{2}$ , proving the assertion.

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Sources

WIAS Berlin, Numerical Methods for Stiff Ordinary Differential Equations, §2.4 Rosenbrock Methods (standard reference, not scraped)
Joachim Rang and Lutz Angermann, New Rosenbrock Methods of Order 3 for PDAEs of Index 2 (standard reference, not scraped)