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The ARS(2,2,2) pair is second order under distinct stage residual tolerances

Statement

Use the tableau and stage equations in The ARS(2,2,2) additive tableau and its three stages, with $F = a + b$ . Let $a, b \in C^{2} (U, R^{d})$ on an open set, and suppose the exact solution path and, for every starting state in the tube, all exact stages, approximate candidates, and joining line segments remain in a common compact convex tube on which $a, b$ and their first two derivatives are bounded and $a, b$ are Lipschitz with constants $L_{a}, L_{b}$ . For every starting state in the tube and each implicit stage, assume the selected branch hypotheses of The two implicit ARS stages have selected branches and residual bounds hold with uniform bounds and $∣ γ ∣ h L_{b} \leq 1 / 2$ . All one-step conclusions below are for sufficiently small $h > 0$ , and the global conclusion is for $h = T / N$ in that same range.

For exact stage roots, the exact-start one-step defect is at most $C h^{3}$ , and the exact-root one-step map is Lipschitz with constant $1 + L h$ on the tube, for constants independent of sufficiently small $h$ . If computed stages have unscaled update residuals $r_{2}, r_{3}$ satisfying $∥ r_{2} ∥_{2} \leq C_{2} h^{2}, ∥ r_{3} ∥_{2} \leq C_{3} h^{3},$ then their output differs from the exact-root output at the same input by at most $C_{a l g} h^{3}$ . Consequently, for $h = T / N$ and initial error $∥ y_{0} - y (0) ∥_{2} \leq C_{0} h^{2}$ , the computed method satisfies $\max_{0 \leq n \leq N} ∥ y_{n} - y (t_{n}) ∥_{2} \leq C_{T} h^{2} .$ The second-stage residual has the weaker $O (h^{2})$ budget because its stage error is multiplied by $h$ in the last stage; the final-stage residual must be $O (h^{3})$ . All residuals are in update units.

Facts & Assumptions

Given: The split IVP, ARS tableau and stages, tube, branch balls, residuals, and initial error in the Statement.

[F1]

The ARS(2,2,2) additive tableau and its three stages: $Y_{2} = y + γ h a (y) + γ h b (Y_{2})$ and $Y_{3} = y + h [δ a (y) + (1 - δ) a (Y_{2}) + (1 - γ) b (Y_{2}) + γ b (Y_{3})], y^{+} = Y_{3} .$

[F2]

The two implicit ARS stages have selected branches and residual bounds: on each admitted stage ball, the implicit root is unique and an unscaled stage residual is bounded in forward error by the reciprocal contraction margin.

[F3]

Implicit residuals and their update and derivative units: the unscaled stage residual is measured in the units of the unknown stage; a derivative-scaled residual is divided by $h$ .

[F4]

A first-order initial value problem is equivalent to its Volterra integral equation and 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: an IVP solution satisfies its integral equation, and the norm of a vector integral is bounded by the integral of the norm.

[F5]

Multivariable Taylor formula with a Lagrange remainder along a line segment: on an admitted line segment, a $C^{2}$ field has its first-order Taylor polynomial with a remainder bounded by a constant times the squared segment length.

[F6]

The ARS(2,2,2) coefficients satisfy the coupled second-order identities: $γ > 0$ , $γ (1 - δ) = 1 / 2$ , and $γ (2 - γ) = 1 / 2$ .

[F7]

Finite geometric error bound from a one-step recurrence: a recurrence $e_{n + 1} \leq (1 + L h) e_{n} + C h^{p + 1}$ with $h = T / N$ and $e_{0} \leq C_{0} h^{p}$ gives $\max_{n} e_{n} = O (h^{p})$ .

[A1]

The exact path and all exact and computed stages remain in the stated convex tube; for each stage the closed ball is contained in $U$ , $∥ b ∥_{2} \leq M_{b}$ , $b$ is $L_{b}$ -Lipschitz, $∣ γ ∣ h M_{b} \leq R_{j}$ , and $∣ γ ∣ h L_{b} \leq 1 / 2$ .

Proof

1.1A1F4F5algebra

Let $ϕ_{s}$ be the exact solution starting at $y$ . By [F4], $ϕ_{s} - y = \int_{0}^{s} F (ϕ_{r}) d r$ . Boundedness gives $∥ ϕ_{r} - y ∥ = O (r)$ , and Lipschitz continuity then gives $∥ ϕ_{s} - y - s F (y) ∥ \leq \int_{0}^{s} L_{F} ∥ ϕ_{r} - y ∥ d r = O (s^{2})$ , where $L_{F} = L_{a} + L_{b}$ . Taylor's formula [F5] gives $F (ϕ_{s}) = F (y) + D F (y) (ϕ_{s} - y) + O (s^{2})$ , hence $F (ϕ_{s}) = F (y) + s D F (y) F (y) + O (s^{2})$ . Integrating this identity from $0$ to $h$ yields $ϕ_{h} = y + h F (y) + \frac{1}{2} h^{2} D F (y) F (y) + O (h^{3}) .$

1.2A1F1F6algebra

For two inputs $u, v$ with exact selected second stages $Y_{2} (u), Y_{2} (v)$ , subtract the first equation in [F1] and use the Lipschitz bounds. Since $γ > 0$ from [F6] and $γ h L_{b} \leq 1 / 2$ , this gives $∥ Y_{2} (u) - Y_{2} (v) ∥_{2} \leq \frac{1 + γ h L_{a}}{1 - γ h L_{b}} ∥ u - v ∥_{2} \leq (1 + 2 γ (L_{a} + L_{b}) h) ∥ u - v ∥_{2} .$

1.3A1F1F3F6algebra

Let $e_{2} = ∥ {\tilde{Y}}_{2} - Y_{2} ∥_{2}$ . By $γ > 0$ from [F6] and $∣ γ ∣ h L_{b} \leq 1 / 2$ from [A1], subtracting the second-stage equations and using its residual gives $(1 - γ h L_{b}) e_{2} \leq ∥ r_{2} ∥_{2}$ , hence $e_{2} \leq 2 ∥ r_{2} ∥_{2}$ . For the final stage, subtract the equations with the same initial input; the terms depending on $Y_{2}$ are bounded by $h K_{s} e_{2}$ , where $K_{s} = ∣ 1 - δ ∣ L_{a} + ∣ 1 - γ ∣ L_{b}$ . The final-stage residual therefore gives $∥ {\tilde{Y}}_{3} - Y_{3} ∥_{2} \leq 2 ∥ r_{3} ∥_{2} + 4 K_{s} h ∥ r_{2} ∥_{2} .$ This is at most $C_{a l g} h^{3}$ under the two stated residual budgets.

2.1A1F1F2F5step 1.1algebra

By [F2] and [A1], the selected second-stage root exists in its branch ball. Its equation in [F1] and boundedness first give $Y_{2} - y = O (h)$ . Lipschitz continuity gives $b (Y_{2}) = b (y) + O (h)$ , so the same equation gives $Y_{2} - y = γ h F (y) + O (h^{2})$ . Applying [F5] to $b (Y_{2})$ then gives $b (Y_{2}) = b (y) + γ h D b (y) F (y) + O (h^{2})$ , and substitution into [F1] yields $Y_{2} = y + γ h F (y) + γ^{2} h^{2} D b (y) F (y) + O (h^{3}) .$

2.2A1F1step 1.2algebra

Let $K_{2} = 2 γ (L_{a} + L_{b})$ and $K_{3} = ∣ δ ∣ L_{a} + (∣ 1 - δ ∣ L_{a} + ∣ 1 - γ ∣ L_{b}) (1 + K_{2} h_{0})$ for $h \leq h_{0}$ , where $h_{0}$ is any fixed upper bound on the small steps. The third-stage centers in [F1] then differ by at most $(1 + K_{3} h) ∥ u - v ∥_{2}$ . Applying the same subtraction to the third-stage equation and using $(1 - γ h L_{b})^{- 1} \leq 1 + 2 γ h L_{b}$ gives $∥ Ψ_{h} (u) - Ψ_{h} (v) ∥_{2} \leq (1 + L h) ∥ u - v ∥_{2}$ for $L : = K_{3} + 2 γ L_{b} + 2 γ L_{b} K_{3} h_{0}$ , which is independent of $h$ and the step index.

3.1A1F1F2F5step 2.1algebra

By [F2] and [A1], the selected third-stage root exists after $Y_{2}$ is selected. Since $Y_{2} - y = O (h)$ and the stage equations give $Y_{3} - y = O (h)$ , Lipschitz continuity and the weight sums $δ + (1 - δ) = 1$ and $(1 - γ) + γ = 1$ imply $Y_{3} - y = h F (y) + O (h^{2})$ . Taylor's formula applied to the last-stage terms gives $a (Y_{2}) = a (y) + γ h D a (y) F (y) + O (h^{2}), b (Y_{2}) = b (y) + γ h D b (y) F (y) + O (h^{2}), b (Y_{3}) = b (y) + h D b (y) F (y) + O (h^{2}) .$ Substituting these expansions into [F1] and collecting the coefficients gives $Y_{3} = y + h F (y) + h^{2} {γ (1 - δ) D a (y) F (y) + γ (2 - γ) D b (y) F (y)} + O (h^{3}) .$

4.1F6step 1.1step 3.1algebra

The identities [F6] make both second-order coefficients in step 3.1 equal to $1 / 2$ , so its expansion agrees with the exact expansion in step 1.1. Thus $∥ Ψ_{h} (y) - ϕ_{h} ∥_{2} \leq C h^{3}$ uniformly on the tube.

5.1A1F3F7step 1.3step 2.2step 4.1algebra∎

Put $e_{n} = ∥ y_{n} - y (t_{n}) ∥_{2}$ . Compare the computed output first with the exact-root ARS output at $y_{n}$ , then with that map at $y (t_{n})$ , then with the exact next state. Step 1.3 bounds the first difference, step 2.2 bounds the second, and step 4.1 bounds the third, giving $e_{n + 1} \leq (1 + L h) e_{n} + (C + C_{a l g}) h^{3} .$ Apply [F7] with $p = 2$ and the initial error from the Statement to obtain the claimed fixed-time $O (h^{2})$ bound. By [F3], a derivative-scaled residual must be multiplied by $h$ before it is tested against either stage budget.

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Sources

Uri M. Ascher, Steven J. Ruuth, and Raymond J. Spiteri, Implicit-explicit Runge-Kutta methods for time-dependent partial differential equations (standard reference, not scraped)
Lorenzo Pareschi and Giovanni Russo, Implicit-explicit Runge-Kutta schemes and applications to hyperbolic systems with relaxation (standard reference, not scraped)