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The two-stage Rosenbrock W-method has a cubic local defect and explicit stage-residual allowance

Statement

Let $D \subseteq R \times R^{d}$ be open, $f : D \to R^{d}$ be $C^{2}$ , and $y (s)$ solve $y^{'} = f (s, y)$ on $[t, t + h]$ . Assume a compact convex tube $Q \subset D$ contains the trajectory, the stage midpoint segments, and all vectors used below. On $Q$ , suppose

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

where $D^{2} f$ is the second total derivative on the joint variable $(t, y)$ , and $∥ D^{2} f ∥_{o p} : = \sup_{∥ u ∥ = ∥ v ∥ = 1} ∥ D^{2} f [u, v] ∥_{2}$ for the Euclidean product norm. Fix $α = 1 - 1 / \sqrt{2}$ , and suppose $∥ \tilde{J} ∥_{2} \leq B_{J}$ , $∥ \tilde{T} ∥_{2} \leq B_{T}$ , and $∥ α h \tilde{J} ∥_{2} \leq q < 1$ . Put $κ = (1 - q)^{- 1}$ and $M = T_{0} + L F$ . The shifted inverse has norm at most $κ$ by A small shifted Jacobian has an explicit inverse bound.

For exact-arithmetic stages $k_{1}, q_{2}, k_{2}$ of Two-stage second-order Rosenbrock W-method with shared factorization (with its exact-storage checks successful), define $C_{1} = B_{J} F + B_{T}, K_{1} = F + α h κ C_{1}, C_{F} = \frac{α κ L C_{1}}{2} + \frac{H (1 + K_{1}^{2})}{8} .$ $C_{K} = α^{2} κ^{2} B_{J} C_{1} + \frac{α κ B_{J} M}{2} + κ C_{F}, M_{2} = H (1 + F^{2}) + L T_{0} + L^{2} F .$ Then the exact one-step defect is at most

∥ y (t) + h k_{2} - y (t + h) ∥_{2} \leq (C_{K} + M_{2} / 6) h^{3} .

If the computed first and second shifted solves have certified update-residual bounds $ρ_{1}$ and $ρ_{2}$ , then the computed update satisfies $∥ {\hat{y}}_{+} - y (t + h) ∥_{2} \leq (C_{K} + M_{2} / 6) h^{3} + C_{r} h^{2} ρ_{1} + κ h ρ_{2}, C_{r} : = κ^{2} (L / 2 + α B_{J}) .$ Thus $ρ_{1} = O (h)$ and $ρ_{2} = O (h^{2})$ preserve a cubic local defect. These are equation-residual conditions, not stage forward-error claims without the inverse bound.

At any base point in the tube, the same two-stage residual perturbation argument gives the solver-to-exact-map bound $∥ {\hat{Φ}}_{h} (t, y) - Φ_{h} (t, y) ∥_{2} \leq C_{r} h^{2} ρ_{1} + κ h ρ_{2}$ , whenever both midpoint evaluations lie in $Q$ .

Facts & Assumptions

Given: The exact solution, tube, derivative bounds, stage data, and stated residual bounds.

[F1]

Two-stage second-order Rosenbrock W-method with shared factorization: $W = I - α h \tilde{J}$ , $W k_{1} = f + α h \tilde{T}$ , $F_{1} = f (t + h / 2, y + h k_{1} / 2)$ , $W q_{2} = F_{1} - k_{1}$ , and $k_{2} = q_{2} + k_{1}$ .

[F2]

A small shifted Jacobian has an explicit inverse bound: $∥ W^{- 1} ∥_{2} \leq (1 - q)^{- 1} = κ$ when $∥ α h \tilde{J} ∥_{2} \leq q < 1$ .

[F3]

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 differentiable vector curve with derivative bounded by $C$ varies by at most $C$ times the interval length.

[F4]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : composition with the differentiable exact solution gives $y^{''} = f_{t} + D_{y} f f$ and the corresponding derivative of this expression.

[F5]

A computed shifted-system residual has a certified upper bound: the solver returns a certified upper bound for each exact linear equation residual.

[F6]

Implicit residuals and their update and derivative units: each stage residual is the value of the defining linear residual map; it is distinct from forward error.

[F7]

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1: the induced norm bounds matrix action and is submultiplicative.

[F8]

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative: componentwise $C^{2}$ implies the first partial derivatives exist and are continuous; applying the theorem to $f$ and to each first-partial component gives the total derivatives used below.

[F9]

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 first derivative blocks with $f_{t}$ and $D_{y} f$ , and identify derivatives of first-partial components with their partial derivatives.

[F10]

Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ : the scalar product and sum rules apply coordinatewise to each finite component of a matrix-vector product along a differentiable curve.

[F11]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral: a continuous scalar component on a compact interval is integrable, and vector integrability is componentwise; therefore continuous vector curves used in the FTC remainders are integrable.

[F12]

For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable: the norm of an integrable vector integral is at most the integral of its Euclidean norm.

Proof

Given: The data in the Statement.

1.1givenF1F2F7algebra

Write $f_{0} = f (t, y (t))$ and $J_{0} = D_{y} f (t, y (t))$ . Since $W = I - α h \tilde{J}$ , the first stage equation in [F1] gives $W (k_{1} - f_{0}) = α h (\tilde{J} f_{0} + \tilde{T})$ . Apply [F2], [F7], and the bounds on $f_{0}, \tilde{J}, \tilde{T}$ to obtain $∥ k_{1} - f_{0} ∥_{2} \leq α h κ C_{1}$ and hence $∥ k_{1} ∥_{2} \leq K_{1}$ (using $h$ in the displayed constants).

1.2givenF2F3F5F6F7

Let $δ_{1} = {\hat{k}}_{1} - k_{1}$ . The first residual equation and [F2], [F5], [F7] give $∥ δ_{1} ∥_{2} \leq κ ρ_{1}$ . Let ${\hat{F}}_{1} = f (t + h / 2, y + h {\hat{k}}_{1} / 2)$ . The state segment lies in $Q$ , so [F3] and $∥ D_{y} f ∥ \leq L$ imply $∥ {\hat{F}}_{1} - F_{1} ∥_{2} \leq (L h / 2) ∥ δ_{1} ∥_{2}$ .

2.1givenF3F8F9F11F12step 1.1algebra

Define $g (s) = f (t + s h / 2, y (t) + s h k_{1} / 2)$ for $0 \leq s \leq 1$ . The componentwise $C^{2}$ hypothesis, [F8], and the basis-direction identification [F9] justify the chain rule and give $g^{'} (0) = (h / 2) (f_{t} + J_{0} k_{1})$ . The derivatives $g^{'}$ and $g^{''}$ are continuous, so [F11] makes them integrable; applying the vector FTC [F3] twice and the integral norm inequality [F12] gives $F_{1} = f_{0} + (h / 2) (f_{t} + J_{0} k_{1}) + R_{F}$ with $∥ R_{F} ∥_{2} \leq H h^{2} (1 + K_{1}^{2}) / 8$ . Replacing $J_{0} k_{1}$ by $J_{0} f_{0}$ adds at most $(h / 2) L ∥ k_{1} - f_{0} ∥_{2} \leq (α κ L C_{1} / 2) h^{2}$ . Thus $F_{1} = f_{0} + (h / 2) G_{0} + E$ , where $G_{0} = f_{t} + J_{0} f_{0}$ and $∥ E ∥_{2} \leq C_{F} h^{2}$ .

3.1step 2.1F1algebra

By [F1], $W k_{2} = W q_{2} + W k_{1} = F_{1} - k_{1} + W k_{1} = F_{1} - α h \tilde{J} k_{1}$ , so $k_{2} = W^{- 1} (F_{1} - α h \tilde{J} k_{1})$ . Also $W^{- 1} - I = α h W^{- 1} \tilde{J}$ . Insert the expansion from step 2.1 and regroup to get $k_{2} - f_{0} - (h / 2) G_{0} = α h W^{- 1} \tilde{J} (f_{0} - k_{1}) + (h / 2) (W^{- 1} - I) G_{0} + W^{- 1} E$ .

4.1givenF2F7step 1.1step 3.1algebra

We have $∥ G_{0} ∥_{2} \leq T_{0} + L F = M$ . Bound the three terms in step 3.1 using [F2], [F7], step 1.1, and $∥ E ∥_{2} \leq C_{F} h^{2}$ . This yields $∥ k_{2} - f_{0} - (h / 2) G_{0} ∥_{2} \leq C_{K} h^{2}$ .

5.1givenF3F4F8F9F10F11F12step 4.1algebra

Let $X (s) = (s, y (s))$ , so $X^{'} = (1, f)$ and $X^{''} = (0, y^{''})$ . Applying the chain rule [F4] twice and the componentwise product rule [F10] to $D f (X) X^{'}$ gives $y^{'''} = D^{2} f [(1, f), (1, f)] + D_{y} f y^{''} = D^{2} f [(1, f), (1, f)] + D_{y} f f_{t} + (D_{y} f)^{2} f$ . The componentwise $C^{2}$ assumption yields differentiability of $D f$ by applying [F8] to the first-partial components, and [F9] identifies their derivative blocks; hence these second-total-derivative expressions are valid. Therefore $∥ y^{'''} ∥_{2} \leq H (1 + F^{2}) + L T_{0} + L^{2} F = M_{2}$ . The curve $y^{'''}$ is continuous by these $C^{2}$ derivatives, so [F11] gives its integrability. Three applications of [F3] give the integral remainder $\int_{0}^{h} (h - s)^{2} y^{'''} (t + s) / 2 d s$ , whose norm is at most $M_{2} h^{3} / 6$ by [F12]. Combine with step 4.1 after multiplying its bound by $h$ to obtain the exact-stage defect.

6.1step 1.2F1F2F5F7step 5.1algebra∎

Subtract the two second-stage solve equations, including the residual vector $r_{2}$ with $∥ r_{2} ∥_{2} \leq ρ_{2}$ , and use ${\hat{k}}_{2} = {\hat{q}}_{2} + {\hat{k}}_{1}$ . The result is ${\hat{k}}_{2} - k_{2} = W^{- 1} ({\hat{F}}_{1} - F_{1}) + (I - W^{- 1}) δ_{1} + W^{- 1} r_{2}$ . Since $I - W^{- 1} = - α h W^{- 1} \tilde{J}$ , steps 1.2 and [F2], [F7] yield $∥ {\hat{k}}_{2} - k_{2} ∥_{2} \leq C_{r} h ρ_{1} + κ ρ_{2}$ . Multiplying by $h$ and adding the exact-stage defect proves the computed bound.

Depends on

Used by

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

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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)